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2018-2019 school handbook Check out this year 's math problems on pg. 11! 2018-2019 Sc h o o l H a n d b o o k 2019 MATHCOUNTS National Competition Sponsor National Sponsors: Raytheon Company U.S. Department of Defense Northrop Grumman Foundation National Society of Professional Engineers CNA Insurance Texas Instruments Incorporated 3Mgives Phillips 66 Art of Problem Solving NextThought WestEd has recognized MATHCOUNTS as having one of the nation’s most effective STEM learning programs, listing the Math Video Challenge as an Accomplished Program in STEMworks. Executive Sponsor: General Motors Foundation Official Sponsors: Bentley Systems Incorporated National Council of Examiners for Engineering and Surveying Patron Sponsor: BAE Systems Founding Sponsors: National Society of Professional Engineers National Council of Teachers of Mathematics CNA Insurance The National Association of Secondary School Principals has placed all three MATHCOUNTS programs on the NASSP Advisory List of National Contests and Activities for 2018-2019. How To Use This School Handbook If You’re a New Coach Welcome! We’re so glad you’re a coach this year. Check out the Guide for New Coaches starting on the next page. If You’re a Returning Coach Welcome back! Thank you for coaching again. Get the 2018-2019 Handbook Materials starting on page 8. Guide For New Coaches Welcome to the MATHCOUNTS® Competition Series! Thank you so much for serving as a coach this year. Your work truly does make a difference in the lives of the students you mentor. We’ve created this Guide for New Coaches to help you get acquainted with the Competition Series and understand your role as a coach in this program. If you have questions at any point during the program year, please feel free to contact the MATHCOUNTS national office at [email protected]. The MATHCOUNTS Competition Series in a Nutshell The MATHCOUNTS Competition Series is a national program that provides students the opportunity to compete in live, in-person math contests against and alongside their peers. Created in 1983, it is the longest-running MATHCOUNTS program and is open to all sixth-, seventh- and eighth-grade students. HOW DOES IT WORK? The Competition Series has 4 levels of competition—school, chapter, state and national. Here’s what a typical program year looks like. Schools register in the fall and work with students during the year. Coaches administer the School Competition, usually in January. Any number of students from your school can participate in your team meetings and compete in the School Competition. MATHCOUNTS provides the School Competition to coaches in November. Many coaches use this to determine which student(s) will advance to the Chapter Competition. Between 1 and 10 students from each school advance to the local Chapter Competition, which takes place in February. Each school can send a team of 4 students plus up to 6 individual competitors. All chapter competitors—whether they are team members or individuals—participate in the individual rounds of the competition; then just the 4 team members participate in the team round. Schools also can opt to send just a few individual competitors, rather than forming a full team. Over 500 Chapter Competitions take place across the country. 56 Top students from each Chapter Competition advance to their State Competition, which takes place in March. Your school’s registration fees cover your students as far as they get in the Competition Series. If your students make it to one of the 56 State Competitions, no additional fees are required. Top 4 individual competitors from each State Competition receive an all-expenses-paid trip to the National Competition, which takes place in May. These 224 students combine to form 4-person state teams, while also competing individually for the title of National Champion. 2 MATHCOUNTS 2018-2019 WHAT DOES THE TEST LOOK LIKE? Every MATHCOUNTS competition consists of 4 rounds—Sprint, Target, Team and Countdown Round. Altogether the rounds are designed to take about 3 hours to complete. Here’s what each round looks like. VS Sprint Round 40 minutes 30 problems total no calculators used focus on speed and accuracy Target Round Approx. 30 minutes 8 problems total calculators used focus on problemsolving and mathematical reasoning Team Round 20 minutes 10 problems total calculators used focus on problemsolving and collaboration The problems are given Only the 4 students on to students in 4 pairs. a school’s team can take Students have 6 minutes this round officially. to complete each pair. Countdown Round Maximum of 45 seconds per problem no calculators used focus on speed and accuracy Students with highest scores on Sprint and Target Rounds compete head-to-head. This round is optional at the school, chapter and state level. HOW DO I GET MY STUDENTS READY FOR THESE COMPETITIONS? What specifically you do to prepare your students will depend on your schedule as well as your students’ schedules and needs. But in general, working through lots of different MATHCOUNTS problems and completing practice competitions is the best way to prepare to compete. Each year MATHCOUNTS provides the School Handbook to all coaches, plus lots of additional free resources online. The next sections of this Guide for New Coaches will explain the layout of the MATHCOUNTS School Handbook and other resources, plus give you tips on structuring your team meetings and preparation schedule. The Role of the Competition Coach Your role as the coach is such an important one, but that doesn’t mean you need to know everything, be a math expert or treat coaching like a full-time job. Every MATHCOUNTS coach has a different coaching style and you’ll find the style that works best for you and your students. But in general every good MATHCOUNTS coach must do the following. • Schedule and run an adequate number of practices for participating students. • Help motivate and encourage students throughout the program year. • Select the 1-10 student(s) who will represent the school at the Chapter Competition in February. • Take students to the Chapter Competition or make arrangements with parents and volunteers to get them there. MATHCOUNTS 2018-2019 Looking for tools to help you become a top-notch coach? check out our videos at the coach section of the MATHCOUNTS website! 3 You don’t need to know how to solve every MATHCOUNTS problem to be an effective coach. In fact, many coaches have told us that they themselves improved in mathematics through coaching. Chances are, you’ll learn with and alongside your students throughout the program year. You don’t need to spend your own money to be an effective coach. You can prepare your students using solely the free resources and this handbook. We give coaches numerous detailed resources and recognition materials so you can guide your Mathletes® to success even if you’re new to teaching, coaching or competition math, and even if you use only the free resources MATHCOUNTS provides all competition coaches. Making the Most of Your Resources As the coach of a registered competition school, you already have received what we at MATHCOUNTS call the School Competition Kit. Your kit includes the following materials for coaches. 2018-2019 MATHCOUNTS School Handbook The most important resource included in the School Competition Kit. Includes 250 problems. Student Recognition Ribbons and Certificates 10 participation certificates, 1 Champion ribbon and 1 Second Place ribbon. You’ll also get access to electronic resources. The following resources are available to coaches online at www.mathcounts.org/coaches. This section of the MATHCOUNTS website is restricted to coaches and you already should have received an email with login instructions. If you have not received this email, please contact us at [email protected] to make sure we have your correct email address. Official 2019 MATHCOUNTS School Competition Released in November 2018 Includes all 4 test rounds and the answer key 2018 MATHCOUNTS School, Chapter + State Competitions Released by mid-April 2018 Each level includes all 4 test rounds and the answer key MATHCOUNTS Problem of the Week Released each Monday Each multi-step problem relates to a timely event You can use the 2019 MATHCOUNTS School Competition to choose the students who will represent your school at the Chapter Competition. Sometimes coaches already know which students will attend the Chapter Competition. If you do not need the School Competition to determine your chapter competitors, then we recommend using it as an additional practice resource for your students. The 2018-2019 MATHCOUNTS School Handbook will be your primary resource for the Competition Series this year. It is designed to help your students prepare for each of the 4 rounds of the test, plus build critical thinking and problem-solving skills. This section of the Guide for New Coaches will focus on how to use this resource effectively for your team. WHAT’S IN THE HANDBOOK? There is a lot included in the School Handbook, and you can find a full table of contents on pg. 8 of this book, but below are the sections that you’ll use the most when coaching your students. • Handbook Problems: 250 math problems divided into Warm-Ups, Workouts and Stretches. These problems in4 Check out our online coach resource videos: Making the Most of Your Coaching Resources How to Use the Handbook MATHCOUNTS 2018-2019 • • • crease in difficulty as the students progress through the book. (pg. 11) Solutions to Handbook Problems: complete step-by-step explanations for how each problem can be solved. These detailed explanations are only available to registered coaches. (pg. 56) Answers to Handbook Problems: key available to the general public. Your students can access this key, but not the full solutions to the problems. (pg. 49) Problem Index + Common Core State Standards Mapping: catalog of all handbook problems organized by topic, difficulty rating and mapping to Common Core State Standards. (pg. 53) There are 3 types of handbook problems to prepare students for each of the rounds of the competition. You’ll want to have your students practice all of these types of problems. Warm-Ups 14 Warm-Ups in handbook 10 questions per Warm-Up no calculators used Warm-Ups prepare students particularly for the Sprint and Countdown Rounds. Workouts 8 Workouts in handbook 10 questions per Workout calculators used Workouts prepare students particularly for the Target and Team Rounds. Stretches 3 Stretches in handbook Number of questions and use of calculators vary by Stretch Each Stretch covers a particular math topic that could be covered in any round. These help prepare students for all 4 rounds. VS VS IS THERE A SCHEDULE I SHOULD FOLLOW FOR THE YEAR? On average coaches meet with their students for an hour once a week at the beginning of the year, and more often as the competitions approach. Practice sessions may be held before school, during lunch, after school, on weekends or at other times, coordinating with your school’s schedule and avoiding conflicts with other activities. Designing a schedule for your practices will help ensure you’re able to cover more problems and prepare your students for competitions. We’ve designed the School Handbook with this in mind. Below is a suggested schedule for the program year that mixes in Warm-Ups, Workouts and Stretches from the School Handbook, plus free practice competitions from last year. This schedule allows your students to tackle more difficult problems as the School and Chapter Competition approach. Mid-August – September 2018 Warm-Ups 1, 2 + 3 Workouts 1 + 2 October 2018 Warm-Ups 4, 5 + 6 Workout 3 Measurement Stretch January 2019 Warm-Ups 12, 13 + 14 Workouts 7 + 8 2019 MATHCOUNTS School Competition Select chapter competitors (optional at this time) November 2018 Warm-Ups 7 + 8 Workouts 4 + 5 Expected Value Stretch December 2018 Warm-Ups 9, 10 + 11 Workout 6 Transformations Stretch February 2019 Practice Competition: 2018 School Competition Practice Competition: 2018 Chapter Competition Select chapter competitors (required by this time) 2019 MATHCOUNTS Chapter Competition You’ll notice that in January or February you’ll need to select the 1-10 student(s) who will represent your school at the Chapter Competition. This must be done before the start of your local Chapter Competition. You’ll submit the names of your chapter competitors either online at www.mathcounts.org/coaches or directly to your local Chapter Coordinator. MATHCOUNTS 2018-2019 5 It’s possible you and your students will meet more frequently than once a week and need additional resources. If that happens, don’t worry! You and your Mathletes can work together using the Interactive MATHCOUNTS Platform, powered by NextThought. This free online platform contains numerous MATHCOUNTS School Handbooks and past competitions, not to mention lots of features that make it easy for students to collaborate with each other and track their progress. You and your Mathletes can sign up for free at mathcounts.nextthought.com. And remember, just because you and your students will meet once a week doesn’t mean your students can only prepare for MATHCOUNTS one day per week. Many coaches assign “homework” during the week so they can keep their students engaged in problem solving outside of team practices. Here’s one example of what a 2-week span of practices in the middle of the program year could look like. Check out the Interactive MATHCOUNTS Platform to get even more handbook problems + past competitions! Monday Tuesday Wednesday (Weekly Team Practice) Thursday -Students continue to work individually on Workout 4, due Wednesday -Students continue to work on Workout 4 -Coach emails team to assign new Problem of the Week, due Wednesday -Coach reviews solutions to Workout 4 -Coach gives Warm-Up 7 to students as timed practice and then reviews solutions -Students discuss solutions to Problem of the Week in groups -Coach emails math team to assign Workout 5 as individual work, due Wednesday -Students continue to work individually on Workout 5 -Students continue to work individually on Workout 5, due Wednesday -Students continue to work on Workout 5 -Coach emails team to assign new Problem of the Week, due Wednesday -Coach reviews solutions to Workout 5 -Coach gives Warm-Up 8 to students as timed practice and then reviews solutions -Students discuss solutions to Problem of the Week in groups -Coach emails math team to assign Workout 6 as group work, due Wednesday -Students work together on Workout 6 using online Interactive Platform Friday WHAT SHOULD MY TEAM PRACTICES LOOK LIKE? Obviously every school, coach and group of students is different, and after a few practices you’ll likely find out what works and what doesn’t for your students. Here are some suggestions from veteran coaches about what makes for a productive practice. • Encourage discussion of the problems so that students learn from each other • Encourage a variety of methods for solving problems • Have students write math problems for each other to solve • Use the Problem of the Week (posted online every Monday) • Practice working in groups to develop teamwork (and to prepare for the Team Round) • Practice oral presentations to reinforce understanding On the following page is a sample agenda for a 1-hour practice session. There are many ways you can structure math team meetings and you will likely come up with an agenda that works better for you and your group. It also is probably a good idea to vary the structure of your meetings as the program year progresses. 6 MATHCOUNTS 2018-2019 MATHCOUNTS Team Practice Sample Agenda – 1 Hour Review Problem of the Week (20 minutes) • Have 1 student come to the board to show how s/he solved the first part of the problem. • Discuss as a group other strategies to solve the problem (and help if student answers incorrectly). • Have students divide into groups of 4 to discuss the solutions to the remaining parts of the problem. • Have 2 groups share answers and explain their solutions. Timed Practice with Warm-Up (15 minutes) • Have students put away all calculators and have one student pass out Warm-Ups (face-down). • Give students 12 minutes to complete as much of the Warm-Up as they can. • After 12 minutes is up, have students hold up pencils and stop working. Play Game to Review Warm-Up Answers (25 minutes) • Have students divide into 5 groups (size will depend on number of students in meeting). • Choose a group at random to start and then rotate clockwise to give each group a turn to answer a question. When it is a group’s turn, ask the group one question from the Warm-Up. • Have the group members consult their completed Warm-Ups and work with each other for a maximum of 45 seconds to choose the group’s official answer. • Award 2 points for a correct answer on questions 1-3, 3 points for questions 4-7 and 5 points for questions 8-10. The group gets 0 points if they answer incorrectly or do not answer in 45 seconds. • Have all students check their Warm-Up answers as they play. • Go over solutions to select Warm-Up problems that many students on the team got wrong. Get more resources + activities to make team meetings fun at the coach section of the MATHCOUNTS website! OK I’M READY TO START. HOW DO I GET STUDENTS TO JOIN? Here are some tips given to us from successful competition coaches and club leaders for getting students involved in the program at the beginning of the year. • Ask Mathletes who have participated in the past to talk to other students about participating. • Ask teachers, parent volunteers and counselors to help you recruit. • Reach parents through school newsletters, PTA meetings or Back-to-School-Night presentations. • Advertise around your school by: 1. posting intriguing math questions (specific to your school) and referring students to the first meeting for answers. 2. designing a bulletin board or display case with your MATHCOUNTS poster (included in your School Competition Kit) and/or photos and awards from past years. 3. attending meetings of other extracurricular clubs (such as honor society) so you can invite their members to participate. 4. adding information about the MATHCOUNTS team to your school’s website. 5. making a presentation at the first pep rally or student assembly. Good luck in the competition! If you have any questions during the year, please contact the MATHCOUNTS national office at [email protected]. Coach Resources: www.mathcounts.org/coaches MATHCOUNTS 2018-2019 7 2018-2019 Handbook Materials Thank you for being a coach in the MATHCOUNTS Competition Series this year! We hope participating in the program is meaningful and enriching for you and your Mathletes. Don’t forget to log in at www.mathcounts.org/coaches for additional resources! What's in This Year's Handbook Highlighted Resources .................................................................................................. 9 Best Materials + Tools for Coaches + Mathletes! Critical 2018–2019 Dates ........................................................................................ 10 This Year’s Handbook Problems ....................................................................... 11 250 Math Problems to Boost Problem-Solving Skills Official Rules + Procedures ..................................................................................... 37 All the Ins-and-Outs + Dos-and-Don'ts of Competing Registration .............................................................................................................. 37 Eligibility Requirements .......................................................................................... 38 Levels of Competition ............................................................................................. 40 Competition Components ..................................................................................... 41 Scoring ...................................................................................................................... 42 Results Distribution ................................................................................................. 43 Additional Rules ....................................................................................................... 43 Forms of Answers ................................................................................................... 45 Competition Coach Toolkit ................................................................................. 46 Vocabulary, Formulas + Tips Organized by Math Topic Answers to Handbook Problems ..................................................................... 49 Available to the General Public...Including Students Problem Index + Common Core State Standards Mapping ......................... 53 All 250 Problems Categorized + Mapped to the CCSS Solutions to Handbook Problems.................................................................... 56 Step-by-Step Solution Explanations for Coaches Other MATHCOUNTS Programs ............................................................................ 76 The National Math Club + Math Video Challenge School Registration Form for the National Math Club ........................................ 77 Additional Students Registration Form .................................................................... 79 8 MATHCOUNTS 2018-2019 Highlighted Resources Also access resources at www.mathcounts.org/coaches! OPLET Online database of over 13,000 problems and over 5,000 step-by-step solutions. Create personalized quizzes, flash cards, worksheets and more! C Great for Coaches M Great for Mathletes Practice Competitions for MATHCOUNTS, Vol. I & II Practice books written by repeat national-level coach Josh Frost. Each volume includes 4 complete mock-competitions plus solutions. Save $25 when you buy your subscription by Oct. 12, 2018 Renewers: use code RENEW18 First-Time Subscribers: use code NEW18 A Advanced Level Book $ Free Resource Most Challenging MATHCOUNTS Problems Solved The Most Challenging MATHCOUNTS Problems ion petit l Com unds iona Ro 0 Nat , Target lutions -201 ds So 2001 t Roun y-Step Sprin Step-b and Advanced level practice book with 10 years of national-level Sprint Rounds, plus detailed step-bystep solutions to each problem. C C M C M A www.mathcounts.org/myoplet www.mathcounts.org/store www.mathcounts.org/store All Time Greatest MATHCOUNTS Problems Interactive MATHCOUNTS Platform MATHCOUNTS Trainer App A collection of some of the most creative, interesting and challenging MATHCOUNTS competition problems. ™ C M A Online platform of past and current handbook and competition problems. Interactive features make collaboration easy and fun! C M $ Train your Mathletes with this fun app, featuring real-time leaderboards and lots of past MATHCOUNTS problems. M $ www.mathcounts.org/store mathcounts.nextthought.com aops.com/mathcounts_trainer or download at the App Store Past Competitions Problem of the Week MATHCOUNTS Minis A new, multi-step problem every week! Each problem focuses on a particular set of math skills and coincides with a timely event, holiday or season. Get the problem at the beginning of the week and the step-by-step solution the following week. A fun monthly video series featuring Richard Rusczyk from Art of Problem Solving. Each video looks at a particular math skill and walks through how to solve different MATHCOUNTS problems using creative problem-solving strategies. Last year’s School, Chapter and State competitions are free online! Other years’ competitions can be purchased. C M $ www.mathcounts.org/ pastcompetitions www.mathcounts.org/store MATHCOUNTS 2018-2019 C M $ www.mathcounts.org/potw C M $ www.mathcounts.org/minis 9 Critical 2018-2019 Dates 2018 Aug. 15 – Dec. 14 Submit your school’s registration to participate in the Competition Series and receive this year’s School Competition Kit, which includes a hard copy of the 2018-2019 MATHCOUNTS School Handbook. Kits are shipped on an ongoing basis between mid-August and December 31. The fastest way to register is online at www.mathcounts.org/compreg. You also can download the MATHCOUNTS Competition Series Registration form and mail or email it with payment to: MATHCOUNTS Foundation – Competition Series Registrations 1420 King Street, Alexandria, VA 22314 Email: [email protected] To add students to your school’s registration, log in at www.mathcounts.org/coaches to access the Dashboard. Questions? Email the MATHCOUNTS national office at [email protected]. Nov. 2 The 2019 School Competition will be available online. All registered coaches can log in at www.mathcounts.org/coaches to download the competition. Nov. 2 Deadline to register for the Competition Series at reduced registration rates ($30 per student, $300 for full registration of 10 students). After November 2, registration rates will be $35 per student, $350 for full registration. Dec. 14 Competition Series Registration Deadline In some circumstances, late registrations might be accepted at the discretion of MATHCOUNTS and the local coordinator. Late fees will apply. Register on-time to ensure your students’ participation. (postmark) (postmark) 2019 10 Early Jan. If you have not been contacted with details about your upcoming competition, call your local or state coordinator. Coordinator contact information is available at www.mathcounts.org/findmycoordinator. Late Jan. If you have not received your School Competition Kit, contact the MATHCOUNTS national office at [email protected]. Feb. 1-28 Chapter Competitions March 1-31 State Competitions May 12-13 2019 Raytheon MATHCOUNTS National Competition in Orlando, FL MATHCOUNTS 2018-2019 Measurement Stretch units 1.____________ Merri places weights of 6 units and 28 units on the right side of a balance and weights of 3 units and 19 units on the left side. If she adds an object to the left side that makes the balance level, how many units does the object weigh? 2 tacks The weight of a small clip is 2.____________ 3 the weight of a large clip. If 2 tacks weigh the same amount as a large clip, how many tacks weigh the same amount as 12 small clips? Dems On the planet Klem, 1 Bem plus 7 Dems equals 4 Pems, and 2 Bems plus 1 Dem equals 3.____________ 1 Pem. How many Dems equal 7 Bems? meters If a race car is traveling at 99 mi/h, how many meters does it travel in a second, given that 4.____________ 0.305 meter = 1 foot? Express your answer as a decimal to the nearest tenth. % 5.____________ If the results when reading a measuring stick can be off by at most 1 cm, what is the maximum percent error when 24 cm is measured? Express your answer to the nearest tenth. grams Vijay gives Sanjay a set of four weights of 1, 3, 8 and 26 grams. When Sanjay 6.____________ places weights on either side of a balance, what is the smallest positive integer number of grams that he cannot measure with this set? gallons If Clem has 2 cups, 7 pints, 8 quarts and 11 half-gallons of lemonade, how many total 7.____________ gallons of lemonade does she have? Express your answer as a mixed number. Klegs If 2 Blams equal 15 Droms and 5 Droms equal 28 Klegs, how many Klegs are in a Blam? 8.____________ 9.____________ What is the ratio of 1 ounce to 1000 grams, given that 1 pound equals 454 grams? Express your answer as a decimal to the nearest thousandth. times If one order of fries and five burgers cost twice as much as three orders of fries 10.___________ and two burgers, how many times as much does a burger cost compared to one order of fries? MATHCOUNTS 2018-2019 11 Expected Value Stretch If the outcomes of random variable X have values x1, x2, x3, ..., xn and the probabilities of these outcomes occurring are p1, p2, p3, ..., pn, respectively, then the expected value of the outcome is the sum of the products of the probability of each outcome and the value of that outcome. E( X ) = p1 x1 + p2 x2 + p3 x3 + ⋯ + pn xn 11.___________ An unfair six-sided die with faces labeled 1, 2, 3, 5, 8 and 13 is rolled. The table lists the probability of the die landing with each number showing on the top face. The expected value of the roll is the sum of the products of each face value and its corresponding probability of being rolled. What is the expected value when the die is rolled? Express your answer as a mixed number. Top Face Value Probability 1 3 1 15 1 6 1 5 2 15 1 10 1 2 3 5 8 13 $ 12.___________ Terry plays a game with prizes of 5, 10, 15 and 20 dollars. The graph shows each possible prize amount and its corresponding probability. The expected value of her prize is the sum of the products of each prize and the probability of winning that prize. What is the expected value of Terry’s prize? Prize Amounts Probability 0.5 0.45 0.4 0.3 0.2 0.1 0 0.25 0.20 0.10 $5 $10 $15 $20 Prize 13.___________ A fair 10-sided die with one face labeled 1, two faces labeled 2, three faces labeled 3 and four faces labeled 4 is rolled. What is the expected value when this die is rolled? cm2 Ana has a bowl containing two square tiles, one with side length 2 cm and the other with 14.___________ side length 3 cm. She randomly chooses a tile from the bowl. The expected value of the area of the chosen tile is the sum of the products of each tile’s area and its corresponding probability of being chosen. If the probability of choosing a particular tile is proportional to its area, what is the expected value of the area of the tile Ana chooses? Express your answer as a common fraction. 12 MATHCOUNTS 2018-2019 points For the dartboard shown, the number of points scored when a dart lands 15.___________ in each region is indicated. The innermost circle of the board has radius 1 inch, and each subsequent circle has a radius 2 inches greater than the previous circle. Kane throws a dart that lands randomly somewhere on the board. What is the expected value of the number of points he scores? Express your answer as a decimal to the nearest tenth. 100 50 25 10 16.___________ Gwen randomly draws a card from a deck of 40 cards numbered 1 through 40. What is the expected value of the number on the card she draws? Express your answer as a decimal to the nearest tenth. faces Luke paints each face of a 5 × 5 × 5 cube red. He then cuts the cube into 125 unit cubes 17.___________ and randomly chooses a single unit cube. What is the expected value of the number of painted faces on this unit cube? Express your answer as a decimal to the nearest tenth. A property of E is that it is a linear function of the random variable. So, for random variables X and Y, the expected value of the sum of random variables equals the sum of their expected values. E( X + Y ) = E( X ) + E(Y ) points In each round of a particular game, Dinara can win at most one point. If she has a 70% 18.___________ chance of winning a point in each round, what is the expected value of Dinara’s total score after three rounds? Express your answer as a decimal to the nearest tenth. 19.___________ Jo and her four friends each secretly pick a random integer from −5 to 5, inclusive. What is the expected value of the sum of the five chosen numbers? jelly Allen randomly distributes 1000 jelly beans into 10 jars lined up in a row from left to right. 20.___________ beans What is the expected value of the number of jelly beans in the leftmost jar? MATHCOUNTS 2018-2019 13 Transformations Stretch units A point P(−3, 2) is translated right 4 units to its image P′. The point P′ is then translated 21.___________ up 3 units to its image P″. What is the distance from P to P″? units A segment has endpoints A(0, 0) and B(−3, 4). Point C is the image of point B translated 22.___________ down 4 units and left 3 units. What is the perimeter of DABC? ( , ) A point Q(−3, 4) is reflected across the x-axis, and then the image Q′ is reflected across 23.___________ the line x = 2. What are the coordinates of the image Q″? Express your answer as an ordered pair. 24.___________ A point S(1, 6) is reflected across the line x − 2y = −6. What is the sum of the coordinates of the image S′? ( , ) What are the coordinates of the image of point D(−5, −3) when it is rotated 90 degrees 25.___________ clockwise about the origin? Express your answer as an ordered pair. ( , ) What are the coordinates of the image of the point E(3, −1) when it is rotated 90 degrees 26.___________ counterclockwise about the point F(5, 4)? Express your answer as an ordered pair. 2 27.___________ A segment with endpoints G(−2, 3) and H(4, 7) is dilated by a scale factor of 3 with center of dilation (0, 0). What is the sum of all the coordinates of G′ and H′? 28.___________ Point J(4, 8) is dilated by a scale factor of product of the coordinates of J′? 3 with center of dilation K(2, 2). What is the 2 units2 A point L(−2, 4) is rotated 90 degrees clockwise about the point M(3, 2). Point N is the 29.___________ 3 image of L′ dilated by a scale factor of with center of dilation M. What is the area of 2 DLMN? Express your answer as a common fraction. units A point R(−5, 3) is reflected across the line y = x − 2, and then the image R′ is rotated 30.___________ 90 degrees clockwise about the origin. What is the distance from R to R″? Express your answer in simplest radical form. 14 MATHCOUNTS 2018-2019 Warm-Up 1 combi- Bob has 40 cents in his pocket. If Bob has no pennies, how many different combinations 31. ���������� nations of quarters, dimes and/or nickels could he have? 32.___________ On the number line shown, what is the value of x − y ? Express your answer as a mixed number. x y 8 33. ���������� 10 12 14 Ted flips a coin that is equally likely to land heads up or tails up. Ted flips the coin 10 times, and each time it lands heads up. What is the probability that the next flip will also land heads up? Express your answer as a common fraction. 34. ���������� What is the value of 9 + 5 × 3 – 8 ÷ 2? 35. ���������� If two more than three times x is equal to five less than ten times x, what is the value of x ? cm3 36. ���������� 5 cm 7 cm 3 cm What is the volume of a rectangular prism of height 5 cm, width 7 cm and depth 3 cm? 37. ���������� What is the average of the prime numbers between 20 and 30? lines How many lines of symmetry does an isosceles right triangle have? 38. ���������� 39. ���������� What is the quotient when 1,000,000,000 is divided by 28 × 57? people Of 1000 people surveyed, one-third of the 630 people who reported owning 40. ���������� a cat also own a dog. If each person surveyed owns a cat, a dog or both, how many own a dog? MATHCOUNTS 2018-2019 15 Warm-Up 2 41. ���������� If the value of x is 10, what is the value of 3x + 4x + 5x ? I feet In the figure, two regular pentagons have been attached to a regular 42.___________ hexagon to create a 12-gon. If each regular polygon has side length 3 feet, what is the perimeter of the resulting 12-gon? E J F C A K H D B L ft/s A speed of 60 miles per hour is equal to 88 feet per second. If the speed limit in a school 43. ���������� zone is 15 miles per hour, what is the speed limit in this zone in feet per second? divisors How many divisors of 64 are perfect squares? 44. ���������� 45. ���������� If ab = 2c, bd = c, b ≠ 0 and d = 16, what is the value of a? square 46. ���������� blocks All of the streets in Tom’s city are on a regular rectangular grid and run east-west or north-south. Tom starts at the intersection of Poyntz Avenue and Eleventh Street and walks 2 blocks east, then 3 blocks north, then 4 blocks east, then 5 blocks north. Tom then returns to his starting location by walking 6 blocks west, then 8 blocks south. What is the area, in square blocks, enclosed by Tom’s path? 47. ���������� What is the value of 1 2 1 4 + 31 ? Express your answer as a common fraction. − 51 5 2 7 1 48. ���������� What is the arithmetic mean of 3 , 9 , 4 and 16 ? Express your answer as a common fraction. 49. ���������� If a and b are real numbers such that a + b = a − b and a ≠ b, what is the value of a2b + a + b – ab2 ? a–b 3 50. ���������� Given that 75% of a certain number is 88, what is 8 of the number? 16 MATHCOUNTS 2018-2019 G Warm-Up 3 jelly All the jelly beans in Sammi’s cup are red, yellow or orange. Five are red and seven are 51. ���������� beans orange. If two-thirds are yellow, how many jelly beans are there in Sammi’s cup? 52.___________ What is the value of the sum 0.49 + 0.53 + 0.55 + 0.47 + 0.48? Express your answer as a decimal to the nearest hundredth. outfits If an outfit consists of jeans, a shirt, a sweater and a scarf, how many different outfits can 53. ���������� Allie make from one pair of jeans, four shirts, three sweaters and two scarves? minutes Destiny can run one-eighth of a mile in three-quarters of a minute, and she can walk 54. ���������� one-quarter of a mile in 4 minutes. How many minutes total will it take for Destiny to run one mile and then walk one mile? 55. ���������� What is the value of 272 − 232? C units In circle P, two chords intersect at A, as shown, with AB = 3 units, 56. ���������� AC = 8 units and AD = 6 units. What is the value of AE? Express your answer as a mixed number. B E A D P days Six people are invited to attend a five-day conference scheduled for Monday through 57. ���������� Friday. Grace will attend only on Monday and Thursday. Becca will attend every day except Monday. Carmen will attend every day except Thursday. Davis will attend only on Thursday. Ernie will only attend either Thursday or Friday. Frank will attend every day except Monday and Wednesday. On how many days during the conference will at least three of these six people be in attendance? miles 58. ���������� The bus from Kevin’s home to the middle school travels 8 miles west, then turns and travels 8 miles north, then turns and travels 7 miles west to arrive at school. If the bus were able to travel directly from Kevin’s house to the middle school along a straight path, how much shorter would the trip be? 59. ���������� When twenty-one-and-a-half trillion is written in scientific notation, what is the exponent needed on the base of 10? diagonals If a polygon has 17 sides, how many diagonals does it have? 60. ���������� MATHCOUNTS 2018-2019 17 Warm-Up 4 degrees What is the degree measure of the complement to an angle that is a supplement to an 61. ���������� angle of measure 163 degrees? 62.___________ If a b x 4 = ad − bc, what is the value of x when = 38? c d 3 10 : a.m. Every morning, the Sharetrain arrives in Mountain View at 9:19 a.m. It takes 63. ���������� Miranda between 17 and 21 minutes to walk to the train station from home. If she wants to guarantee that she will arrive at the station with at least 5 minutes to spare, what is the latest time she can leave home? 64. ���������� Ryan picks two different numbers from the set {2, 3, 5, 7} and multiplies them. What is the absolute difference between the greatest and the least products that Ryan can get? $ 65. ���������� Zu’s zoo offers a promotional deal: get a free $3 cotton candy and a free $2 soda with the purchase of five $12 admission tickets. A group of 20 students will each purchase an admission ticket. If half of them want cotton candy and one-fourth of them want soda, how many dollars would they save by using the promotional deal? ounces Preston has four potatoes, each of which weighs a whole number of ounces. The median 66. ���������� weight of his potatoes is 11 ounces, and the mean weight of his potatoes is 12 ounces. What is the greatest possible difference between the weight of the heaviest and lightest of his potatoes? mm Photographers often rely on a rule to choose a shutter speed in seconds that is the 67. ���������� reciprocal of the effective focal length of the lens in millimeters. If Jackie is shooting at a focal length of 80 mm, Clarise is shooting at 200 mm, Sage is shooting at 400 mm, and they all apply this reciprocal rule, what focal length corresponds to the sum of their shutter speeds? in2 Hexagons A and B are geometrically similar. The shortest sides of the two hexagons are 68. ���������� 4 inches and 3 inches, respectively. If the area of hexagon A is 48 in2, what is the area of hexagon B? A B 69. ���������� What is the value of 1 × 12 + 2 × 11 + 3 ×10 + 4 × 9 + 5 × 8 + 6 × 7 + 7 × 6 + 8 × 5 + 9 × 4 + 10 × 3 + 11 × 2 + 12 × 1? degrees What is the measure of an interior angle of a regular polygon with 90 sides? 70. ���������� 18 MATHCOUNTS 2018-2019 Warm-Up 5 $ For each lawn Kayla mows, she charges a flat fee of $15, plus an amount proportional to 71. ���������� the area of the lawn. If Kayla charges the owner of a 20-foot by 40-foot lawn $24, how much would she charge the owner of a 60-foot by 80-foot lawn? 435 446 Nash $ Alicia needs to make her way from Nash to Wiles. Available flights and 72.___________ their respective costs, in dollars, are indicated in the figure shown. If Alicia can take any sequence of flights, what is the total cost of her least expensive route from Nash to Wiles? 335 325 113 Wiles 162 168 89 Euclid 147 162 165 Cantor 70 58 55 98 102 Galois 73. ���������� Ancient Egyptians used the symbols shown in Figure 1 to express certain quantities by placing the symbols in any order. What quantity is represented by the symbols in Figure 2? Staff Heel Rope Lotus Finger 1 10 100 1000 10,000 Figure 1 Figure 2 74. ���������� Yixin thinks of a positive number that is a perfect cube, and Aly thinks of a number that is a perfect square. If the sum of their numbers is 31, what is the product of their numbers? integers Jalacia erases some integers from the following list: 5, 9, 2, 3, 7, 8, 6, 5, 6, 4, 1, 7, 9, 8. 75. ���������� When she is done, the range of the remaining list is 2. What is the least possible number of integers that Jalacia could have erased? $ Esther purchased two rock tumblers and one spy pen for $74. Eli purchased two puzzles 76. ���������� and a spy pen for $50. Sabine purchased a rock tumbler and two puzzles for $57. Based on this, how much does one puzzle cost? B cm In triangle ABC, shown here, segment BD bisects angle ABC. If 77. ���������� CD = 4 cm, AD = 5 cm and BC = 6 cm, what is the value of AB? Express your answer as a common fraction. cm 78. ���������� 15 cm 10 cm 6 x A 5 D 4 C In the figure shown, two sides have lengths 10 cm and 15 cm. If all line segments intersect at right angles, what is the perimeter of the figure? a 3 5 5 and b = 6 + 3 , what is the value of b ? Express your answer as a 4+ 6 4+ 2 common fraction. 79. ���������� If a = 2 + sides If the difference between the measures of an interior angle and an exterior angle of a 80. ���������� regular polygon is 100 degrees, how many sides does the polygon have? MATHCOUNTS 2018-2019 19 Warm-Up 6 inches When Jim was 10 years old, he was 4 feet 8 inches tall. When he was 20 years old, Jim 81.___________ was 6 feet 2 inches tall. By how many inches did Jim’s height increase in those 10 years? 82.___________ Circles A and B intersect, and points E and F are on circle B as shown. If quadrilateral AEBF is a square, what is the probability that a randomly chosen line through point A intersects both circles? Express your answer as a common fraction. E A B F 83.___________ Lauren randomly picks two cards, without replacement, from a group of six cards numbered 2, 3, 5, 6, 7 and 10. What is the probability that the product of the two numbers on the selected cards is a multiple of 10? Express your answer as a common fraction. 84.___________ Alana draws four squares, five pentagons and six octagons. Marie draws n hexagons and notices that the combined number of diagonals in her hexagons is equal to the combined number of diagonals in Alana’s polygons. What is the value of n? 85.___________ Consider the set of all four-digit positive integers less than 2000 whose digits have a sum of 24. What is the median of this set of integers? 86.___________ If x 2 − y 2 = 7, x = 12 and y < 0, what is the value of x 4 + y 4? y $ 87.___________ At the county fair, two hot dogs and an ice cream cone cost $2.50. Two slices of pizza and an ice cream cone cost $3.50. What is the total cost of one hot dog, one slice of pizza and one ice cream cone at the county fair? 88.___________ When the positive integer divisors of 385 are arranged from least to greatest, what is the sum of the 4th, 5th and 6th divisors? minutes Gavin goes for a run at a constant pace of 9 minutes per mile. Ten minutes later, Lars 89.___________ goes for a run, along the same route, at a constant pace of 7 minutes per mile. How many minutes does it take for Lars to reach Gavin? degrees Lines are drawn from the center of a 72-gon to two of its vertices. The smaller central 90.___________ angle formed by these lines includes seven sides of the polygon. What is the degree measure of this central angle? 20 MATHCOUNTS 2018-2019 Warm-Up 7 problems Two days ago, Neil was assigned a set of problems to solve. Today he solved the final 91. ���������� three. Yesterday he solved half of those then remaining plus half a problem. The first day he solved half of those assigned plus half a problem. How many problems were in the assigned set? scores A deck of 14 cards numbered 1 through 14 is dealt to Ken and Gunther so that 92.___________ each gets 7 cards. Each player’s score is the sum of his card values, and the player with the lower score wins. How many different winning scores are possible? 93. ���������� What is the units digit of the sum 1! + 2! + 3! + ⋯ + n! when n = 2019? inches 94. ���������� The sum of the heights of Alex and his two brothers, Evan and Joel, is 221 inches. Alex is 8 inches taller than Evan and 5 inches taller than Joel. What is Alex’s height, in inches? 95. ���������� If all test scores are integers from 0 to 100, inclusive, what is the least possible median of five test scores that add to 204? 2 f(x) 96. ���������� If f(x) = 4 − x and g(x) = x – 2, what is the value of g ( x ) , for x ≠ ±2? x +2 jumps 97. ���������� Freddie Frog traveled 29 yards in 261 jumps, at a constant distance per jump. After hurting a toe, her distance per jump decreased by 1 inch. How many jumps more than 261 will it take for her to travel another 29 yards but with a hurt toe? 98. ���������� On the first day of class, Smiley’s teacher gave her a piece of paper 821 9 902 containing Smiley’s seven-digit locker combination. Smiley noticed that the seven-digit number was a multiple of 11. When she went to her locker, she saw that one digit on the paper had been completely smudged, as shown. Smiley had forgotten the smudged digit. What was Smiley’s seven-digit locker combination? ways How many ways are there to put twelve identical donuts into three differently colored 99. ���������� boxes if there must be at least two donuts in each box? y units2 On a coordinate grid, Micah constructs a pentagon with vertices 100.��������� A(0, 0), B(0, 2), C(3, 4), D(4, 2) and E(2, −1). What is the area of pentagon ABCDE? MATHCOUNTS 2018-2019 C B A D x E 21 Warm-Up 8 kg Don’s signature coffee blend is 60% dark roast and 40% light roast. He has 10 kg of 101.��������� blend A, which is 80% dark roast and 20% light roast, and 10 kg of blend B, which is 20% dark roast and 80% light roast. How many kilograms of blend A will Don need to use to DARK make 10 kg of his signature blend? Express your answer as a mixed number. LIGHT Roast minutes Coordinated Universal Time (UTC) is 5 hours later than Eastern Standard Time (EST). 102.__________ If Nico and Jon start playing a tennis match in London at 8:00 a.m. UTC and finish at 12:15 p.m. EST, for how many minutes did they play? times While working on a history project, Erika writes down a list of years starting with 1809 and 103.��������� ending with 2019. How many times did Erika write the digit 1? 104.��������� What is the probability that a sequence of five flips of a fair coin will not land heads up twice in a row? Express your answer as a common fraction. games Liz, Eva and Ace have played trivia as a team 33 times and won 24 times. What is the 105.��������� minimum number of games they must win to have an overall winning percentage of 80%? ways 106.��������� A certain vending machine accepts nickels, dimes, quarters and dollar bills, and it provides change using nickels, dimes and quarters. If Sarah selects a healthful snack priced at $1.30 after inserting $2.00, in how many ways can the vending machine provide Sarah’s change? 107.��������� What is the geometric mean of the median and mode of the set {23, 25, 3, 25, 20, 22, 21, 2, 1, 14, 12}? Express your answer in simplest radical form. 108.��������� What is the mean of all three-digit positive integers whose digits are in the set {2, 0, 1, 9}? 109.��������� If we define the binary operation as a b = ab + b for all numbers a and b, what positive value of x satisfies the equation x (4 x) = 550? degrees A circle with center P is inscribed in isosceles triangle ABC with apex angle A measuring 110.��������� 34 degrees. What is the degree measure of angle APC? Express your answer as a decimal to the nearest tenth. 22 MATHCOUNTS 2018-2019 Roast Warm-Up 9 points Thomas is a basketball player who is x inches tall. One season he averaged y points per 111.��������� game. The sum of x and y is 98, and x is 11 more than twice y. Based on these statistics, how many points did Thomas score in the 60 games he played that season? : p.m. Pierre needs 39 minutes to paint a painting. The painting will be dry 55 minutes after 112.__________ Pierre finishes. If Pierre starts painting at 1:00 p.m., what is the earliest time when he can have three finished, dry paintings? pounds Blackbeard and Redbeard take a total of 60 pounds of gold from Treasure Cove. Because 113.��������� Blackbeard outranks Redbeard, Blackbeard gets 5 of the gold and Redbeard gets the 8 remaining 3 . How many more pounds of gold does Blackbeard get than Redbeard? 8 114.��������� A pentagonal number is a number that can be represented as the number of dots on a regular pentagon as in these figures, which show the first five pentagonal numbers. What is the 20th pentagonal number? 1 5 12 22 35 115.��������� What is the arithmetic mean of all 3-digit numbers whose digits are distinct and nonzero? 4 2 2 2 116.��������� 6 6 6 2 2 2 The figure shown can be folded into a rectangular prism. What is the absolute difference between the numerical values of the surface area and volume of the prism? 6 4 117.��������� If the sum of three numbers is equal to their product, and two of the numbers are 4 and 1, 3 what is the third number? 118.��������� Alvin plays mini-golf with four friends. The median and unique mode of all their scores are both 10, and the range of their scores is 16. If the mean of all their scores is 15, what is the sum of all possible values of the lowest score? 119.��������� Cayley has a fair six-sided die whose faces are numbered −2, −1, 0, 0, 1 and 2. She rolls the die three times. What is the probability that the sum of the three numbers she rolls is 0? Express your answer as a common fraction. F degrees Triangle ABP is drawn inside a regular octagon as shown. What is 120.��������� the degree measure of acute angle P if mHAP = 75 degrees and mCBP = 85 degrees? MATHCOUNTS 2018-2019 E G H D P 75° 85° A B C 23 Warm-Up 10 121.��������� If x = 2x + 1 and x = 2x − 1, what is the value of π − π ? 122.__________ What is the sum of the two values of x for which |x − 3| + |x − 7| = 6? units2 What is the area of triangle ABC, with vertices A(6, 8), B(9, 2) and C(17, 6)? 123.��������� students 124.��������� A local college has 985 students, of whom 460 play a varsity sport and 571 belong to a club. What is the absolute difference between the least and greatest possible numbers of students who play a varsity sport AND belong to a club? in2 What is the smallest possible area of an isosceles triangle with side lengths 5 inches and 125.��������� 6 inches? Express your answer as a decimal to the nearest tenth. 126.��������� An equilateral triangle is inscribed in a regular hexagon, and a smaller regular hexagon is inscribed inside the triangle so that three of its vertices are each the midpoint of a side of the triangle, as shown. What is the ratio of the area of the smaller hexagon to that of the larger hexagon? Express your answer as a common fraction. divisors How many positive integer divisors of 23,328 are perfect cubes? 127.��������� points Saila took nine exams, each scored out of 100 points. She received a passing score of 128.��������� 60 or above on each of them. If the mean of her nine scores was 80 points, what is the greatest possible value of the median of her nine scores? paths If A(0, 0, 0) and B(2, 2, 2) are points in coordinate space, how many paths are there from 129.��������� A to B that move from one lattice point to another in the positive x-, y- or z-direction? seconds Two race cars travel in opposite directions in separate lanes of a circular track. Car F takes 130.��������� 120 seconds to complete a lap. Car S takes 240 seconds to complete a lap. Once the two cars pass each other, how long does it take for them to pass each other again? 24 MATHCOUNTS 2018-2019 Warm-Up 11 131.��������� What is the greatest common divisor of 2563 and 4147? widgets Hongyi is making a triangular display of widgets. The bottom layer will contain n widgets. 132.__________ The layer immediately above it will contain n − 1 widgets. Each subsequent layer will contain one fewer widget than the previous layer, and the top layer will contain one widget. If widgets are packed 25 per case, and Hongyi must use all of the widgets in every case he opens, what is the smallest positive number of widgets Hongyi can use to build the display? degrees Regular pentagon ABCDE has diagonal BE. If P is the midpoint of side ED, what is the 133.��������� degree measure of angle EBP? prime A prime date is a date for which the number of the month and the number of the day are 134.��������� dates both prime numbers. How many prime dates are there in the month of March? miles Bennie and Flossie are traveling on a straight road, going in the same direction and 135.��������� starting at the same place. Bennie is traveling at an average speed of 62 mi/h, and Flossie is traveling at an average speed of 75 mi/h. After 2 hours, Flossie stops for a 20-minute break before continuing. After a total of 5 hours, what is the distance between them? F cm 136.��������� E A D K B Regular hexagon ABCDEF has side length 4 cm. Point K is the midpoint of side CD. What is the length of AK? Express your answer in simplest radical form. C crackers At every lunch, three friends share their cheesy crackers with each other but agree not to 137.��������� eat them until they all have the same number. At the first lunch, Axel gives Ben and Chloe as many crackers as they each already have. At the second lunch, Ben gives Axel and Chloe as many crackers as they each already have. At the third lunch, Chloe gives Axel and Ben as many crackers as they each already have, after which they each have 8 crackers, so they eat them. How many crackers did Axel start with? sequen138.��������� ces Paul and his friends play a version of tag football in which a team earns 5 points for every touchdown and 3 points for every field goal. If Paul’s team has a total of 25 points, how many different sequences of field goals and touchdowns could the team have scored? 139.��������� Five numbers form an arithmetic sequence with a mean of 18. If the mean of the squares of the five numbers is 374, what is the greatest of the five original numbers? dolphins An oceanographer tags every dolphin that she sees over a one-week 140.��������� period. At the end of the week, she has tagged 1000 dolphins. One month later, when she returns and examines 2000 dolphins, she finds that 400 of them are ones that she tagged. Assuming that each time, she saw a random subset of the local dolphin population, what is the expected number of dolphins in the local population? MATHCOUNTS 2018-2019 25 Warm-Up 12 cubes Exactly 512 small cubes perfectly fill a lidless cubical box. All of the small cubes are 141.��������� removed except those touching the bottom of the box and those touching the sides of the box. How many of the small cubes were removed? 142.__________ To earn Gold Status on a social media site, Max needs 800 points. He earns a point for every 3 posts he reads, and he gets 15 points for each post he writes. Max can write at most 3 posts a day. He can read up to 120 posts on a weekday and 225 posts on a weekend day. If he starts with 0 points on a Monday morning, what day of the week is the day on which he can first earn Gold Status? 143.��������� If a 4 3 = 4 a 3 , what is the value of a2? cm Segment AC is a diameter of circle P, which has radius 13 cm. If B is a point on circle P 144.��������� such that AB = 10 cm, what is the length of segment BC? integers How many positive integers are each a divisor of both 630 and 360 but not a divisor of 60? 145.��������� penta- Seven points are equally spaced around a unit circle. How many non-congruent 146.��������� gons convex pentagons can be drawn by using a subset of the points as vertices of the pentagons? 147.��������� If the value of a is 25 and a – b = a – 1, what is the sum of all possible values of b? b 148.��������� If square WXYZ has vertices W(2, 1), X(a, b), Y(4, 7) and Z(c, d ), what is the value of |ab − cd |? 149.��������� The ten guppies in Vida’s tank are 5, 5, 8, 8, 8, 9, 10, 11, 11 and 33 weeks old. A guppy of age n weeks is added to her tank. The sum of the mean, median and unique mode of the eleven guppies’ ages equals the range. What is the sum of all possible integer values of n? cm If a rhombus with area 26 cm2 has one diagonal of length 4 cm, what is the length of its 150.��������� other diagonal? 26 MATHCOUNTS 2018-2019 Warm-Up 13 151.��������� What is the coefficient of abc when the product (a + 2b)(b + 2c)(c + 2a) is expanded and like terms are combined? candies Jenny has a bag that contains equal numbers of red, green, yellow, orange and purple 152.__________ candies. She likes only the red candies. If she randomly selects a candy from her bag and discards any non-red candy she selects, then she is guaranteed to select a red candy within a minimum of 45 random selections. What is the total number of candies in her bag? feet 153.��������� The dimensions of a packing box are 15 inches by 13 inches by 10 inches. Every edge of the box is to be secured with packing tape. If the length of tape along each edge is equal to the edge length, how many feet of tape are needed? Express your answer as a mixed number. units What is the length of the hypotenuse of a right triangle with legs of length 12 3 units and 154.��������� 23 3 units? Express your answer in simplest radical form. people It takes a crew of eight people three days to paint an aircraft. At this 155.��������� rate, how many people are needed to paint five aircraft in six days? 156.��������� In a particular sequence, the ratio between consecutive terms remains constant. If the 2nd term is 2, and the 5th term is 5, what is the value of the 11th term of the sequence? Express your answer as a common fraction. 157.��������� Ilana rolls a fair 8-sided die with faces labeled 1 through 8, and Yolanda rolls a fair 12-sided die with faces labeled 1 through 12. What is the probability that Ilana rolls a smaller number than Yolanda? Express your answer as a common fraction. units Rectangle PQRS is drawn on a coordinate plane. The y-coordinate of Q is 24 units 158.��������� greater than the y-coordinate of S, the x-coordinate of R is 25 units greater than the x-coordinate of P, and the y-coordinate of P is equal to the y-coordinate of R. What is the perimeter of rectangle PQRS? miles Hannah ran five races, each of which was a different positive integer number of 159.��������� miles. The mean of her distances was 4 miles. What is the maximum possible distance of her longest race? 160.��������� If the coefficients c and d of x 3 – 19x 2 + cx + d = 0 are chosen so that the roots of the equation are positive integers that form a geometric sequence, what is the value of c + d ? MATHCOUNTS 2018-2019 27 Warm-Up 14 units Rectangles ABGH, BCFG and CDEF are adjacent, as shown, with AH = n, 161.__________ AB = n + 3, BC = n – 3 and CD = n – 6. If rectangle BDEG has area 110 units2, what is the perimeter of rectangle ACFH? A B C D H G F E units The larger of two right triangles has area numerically equal to 3 times the sum of its integer 162.__________ leg lengths, a and b. The other right triangle has area numerically equal to 2 times the sum of its integer leg lengths, c and d. If the triangles have areas in the ratio 3:2, what is the value of a + b + c + d ? 163.__________ The vertices of a cube are labeled with the integers 1 through 8, in some order, with each integer used exactly once. Each edge of the cube is labeled with the product of the integers at its two endpoints. Each face of the cube is labeled with the sum of the labels of its four edges. What is the greatest possible absolute difference between the labels of two adjacent faces of the cube? 164.__________ Burt has a set of tiles spelling BANANAS. Ernie randomly chooses a set of five tiles from Burt’s set. What is the probability that Ernie can create a palindrome using all five tiles? Express your answer as a common fraction. 165.__________ Screen sizes of rectangular television sets are measured on the diagonal. If Joe owns a television with a 62-inch screen, and Daniel owns a geometrically similar television with a 31-inch screen, what is the ratio of the area of Daniel’s screen to the area of Joe’s screen? Express your answer as a common fraction. rhombi How many rhombi of any size are in this figure, composed of equally spaced 166.__________ horizontal lines and parallel diagonal lines that are also equally spaced? 167.__________ What is the 15th digit to the right of the decimal point in the decimal expansion of 3 ? 7 168.__________ If 6, a, b, c, d, e, 34 is a list of seven different positive integers written in increasing numerical order such that c is both the mean and the median of the list, what is the least possible value for c? 169.__________ Two real numbers from 0 to 10, inclusive, are to be chosen at random. What is the probability that the absolute difference between the two numbers will be less than 3? Express your answer as a common fraction. permu- How many permutations of the digits 1, 2, 3 and 4 can be arranged in a line, so that no 170.__________ tations adjacent digits differ by more than two? 28 MATHCOUNTS 2018-2019 Workout 1 cm A rectangular sheet of paper is cut in half perpendicular to the longer side. One half is 171.��������� discarded, and the other half is cut into thirds as shown. Two of the thirds are discarded, and the remaining third is cut into fourths by vertical lines. Three of the fourths are discarded, and the remaining fourth is cut into fifths by horizontal lines. Four of the fifths are discarded, and the remaining fifth is cut into sixths by vertical lines. Five of the sixths are discarded, leaving a square piece with side length 2 cm. What is the perimeter of the original sheet of paper? days 172.__________ In 1969, the Apollo 10 mission set the record for the fastest crewed space travel, at 39,897 km/h. At that speed, how many days would it take to travel 54.6 million km, the minimum distance from Earth to Mars? Express your answer to the nearest whole number. 173.��������� Yeong multiplies two-digit positive integers AB and CD. If the digits A, B, C and D are all distinct, what is the greatest possible value of the product? $ 174.��������� Karli paid $3.00 for lunch every day she attended school. During a six-week period, Karli attended school every Monday through Friday, with the exception of one school holiday. What is the total amount that Karli spent on school lunches during this six-week period? 175.��������� What is the sum of the mean, median, mode and range of the numbers 15, 33, 24, 10, 20 and 24? gallons A certain race car consumes 1.3 gallons of fuel during each lap of a race. 176.��������� If each lap is 2.5 miles, and the entire race is 500 miles, how much fuel does the race car consume from start to finish? inches One television screen measures 56 inches long and 33 inches wide. A smaller, 177.��������� geometrically similar television screen measures 48 inches long. What is the width of the smaller screen? Express your answer as a decimal to the nearest tenth. mi/h 178.��������� Lindsay starts at the peak of a mountain, and it takes her 50 minutes to hike 15,000 feet. What was her average walking speed, in miles per hour, given 1 mile = 5280 feet? Express your answer as a decimal to the nearest tenth. 179.��������� Let a be the arithmetic mean of 3.27 and 17.95. Let b be the product of 32.7 and 0.4382. Let c be the quotient of 2.637 and 0.316. Let d be the absolute difference between 793.241 and 804.3692. What is the numerical value of the median of a, b, c and d ? Express your answer as a decimal to the nearest hundredth. units2 Jessie makes butterfly wings by shading four sections of a regular hexagon 180.��������� as shown. If the hexagon has side length 4 3 units, what is the area of the shaded region? Express your answer in simplest radical form. MATHCOUNTS 2018-2019 29 Workout 2 units2 What is the area of a triangle with sides of length 16, 30 and 34? 181.��������� gallons How many gallons are in the volume of a cylinder with height 12 inches and diameter 182.__________ 14 inches if 1 gallon = 231 in3? Express your answer to the nearest whole number. 183.��������� If x @ y = LCM(x, y) and x # y = GCD(x, y), what is the value of ((2 @ 7)2 # 42) # 105? minutes On a digital clock that shows hours and minutes, for how many minutes, 184.��������� during a single day, between 4:00 a.m. and 6:00 a.m. will the sum of the digits be 10? cm3 The density of an object is the mass of the object per unit volume. The density of water 185.��������� is 1 gram/cm3, and the density of ice is 91.67% that of water. If 1 kilogram of water completely freezes, what is the volume, in cubic centimeters, of the resulting ice block? Express your answer to the nearest whole number. a b c 2 −5 3 186.��������� If d e f = aek + bfg + cdh – ceg – afh – bdk, what is the value of 0 4 −6 ? −1 8 7 g h k degrees A certain trapezoid has these properties: its diagonals are congruent and perpendicular to 187.��������� each other, and its longer base length is equal to the length of a diagonal. What is the sum of the degree measures of the two acute angles of this trapezoid? sub- How many subsets of {1, 2, 3, 4, 6, 8, 10, 15} are there for which the sum of the elements 188.��������� sets is 15? % The figure shows a shaded regular octagon inscribed in a square. What percent 189.��������� of the figure is shaded? Express your answer to the nearest whole number. minutes 190.��������� 30 Spike can dig 8 holes in 3 hours. Butch can dig 7 holes in 4 hours. Lucky can dig 6 holes in 5 hours. How many minutes will it take them to dig 3 holes if all three work together? Express your answer to the nearest whole number. MATHCOUNTS 2018-2019 Workout 3 191.��������� Let q be the sum of the lengths of all 12 edges of a cube, and let A be the total surface area of the cube. If q × A is k times the volume of the cube, what is the value of k? 192.__________ Alta’s favorite sphere has six times as much surface area compared to her favorite cylinder, which is right circular. If the height of her favorite cylinder is equal to its diameter, what is the ratio of the volume of Alta’s favorite cylinder to the volume of her favorite sphere? Express your answer as a common fraction. triangles How many triangles of any size are in the figure shown here? 193.��������� page 194.��������� Rex reads a novel at a constant rate while Wren writes a report at a constant rate. Rex finishes page 313 of his novel at the same time Wren finishes page 5 of her report. Rex later finishes page 409 of his novel when Wren finishes page 9 of her report. What page of his novel will Rex finish when Wren finishes page 12 of her report? 195.��������� The Fibonacci sequence is defined by the function F (n) = F (n − 2) + F (n − 1), for n >2 and F (1) = F (2) = 1. What is the value of F (11)? 196.��������� What is the ratio of the area of the largest square that can fit inside a circle of diameter 2 cm compared to the area of the largest square that can fit inside a semicircle of diameter 2 cm? Express your answer as a common fraction. 197.��������� If 20! = 2a × b for some integer b, what is the greatest possible value of a? % At a certain middle school, the 6th-grade class of 390 increased by 10%, the 7th-grade 198.��������� class of 350 increased by 22% and the 8th-grade class of 420 increased by 20%. Overall, what was the percentage increase of students for this middle school? Express your answer to the nearest tenth. units3 A 6-8-10 right triangle is rotated about the side of length 10 units. What is the volume of 199.��������� the resulting solid? Express your answer as a decimal to the nearest tenth. 200.��������� MATHCOUNTS 2018-2019 Vikram attempted to calculate 67 × 58, but he entered exactly one digit incorrectly, resulting in a product of 3596. What is the sum of the two values he actually multiplied? 31 Workout 4 $ Alice’s car travels 29 miles per gallon of gas. Phil’s car uses 0.02 gallon of gas per mile 201.��������� traveled. If gas costs $2.50 per gallon, what is the absolute difference in the amounts Alice and Phil would each spend on the gas needed to drive 113 miles? 202.__________ A cheese cube is sliced diagonally from one edge to another, as shown. One piece of the cheese is melted and then reformed into a perfect sphere. What is the ratio of the side length of the original cheese cube to the radius of the cheese sphere? Express your answer as a decimal to the nearest hundredth. : a.m. Three consecutive rounds of an X-game last 24 minutes, 48 minutes and 96 minutes. If the 203.��������� game begins at 6:00 a.m., at what time will the third round end? meters 204.��������� 1.2 1.3 1.4 1.5 1.6 1.7 1 2 0 0 0 1 3 4 1 3 5 6 5 4 4 7 6 5 7 9 8 6 8 8 9 Ms. Santa asks the 25 students in her math class to measure their heights in meters. The students’ heights are recorded in the stem-and-leaf plot shown here, where 1.2 | 1 represents 1.21 meters. What is the median height of the students in Ms. Santa’s class? Express your answer as a decimal to the nearest hundredth. 2 − 3 = 1? ordered How many ordered pairs of positive integers (a, b) satisfy the equation 205.��������� a b 5 pairs steps 206.��������� R.J.’s pedometer indicates that he has walked 10,002 steps and equates that to traveling 4.11 miles. Based on this, how many additional steps must he walk to travel the equivalent of 5 miles total? Express your answer to the nearest whole number. $ A liter of fuel cost £1.33. At that rate, what would be the cost, in U.S. currency, of 1 gallon 207.��������� of fuel, given that 1 gallon ≈ 3.78541 liters and US$1.30 ≈ £1.00? $ A 10-meter by 20-meter garden is going to have a concrete walkway constructed 208.��������� diagonally through the middle as shown. The longest sides of the walkway each start 1 meter from the closest corner of the garden. If the walkway will have a uniform depth of 10 cm and the concrete costs $70 per cubic meter, what will be the cost of the concrete used for the walkway? years On his 15th birthday, Bo invests $10,000 in a bank that pays 7.5% interest, compounded 209.��������� old annually. At this rate and assuming Bo makes no deposits or withdrawals, how old will Bo be on the first birthday when his account balance is at least $30,000? permu- How many different two-letter permutations can be made from two distinct letters in the 210.��������� tations word MATHCOUNTS? 32 MATHCOUNTS 2018-2019 Workout 5 cm A regular octagon with side length 4 cm is concentric with a circle, as shown. If 211.��������� the area of the circle is equal to the area of the shaded region between the shapes, what is the radius of the circle? Express your answer as a decimal to the nearest tenth. 1 ounces Jonas has a 1 2 -inch by 4-inch by 3-inch rectangular block of tofu weighing 16 ounces. He 212.__________ cuts the block into smaller pieces, each of which is a triangular prism having a right triangle base with legs of length 1 31 inch and 43 inch. If each tofu prism has height 21 inch, what is the weight of one tofu prism? Express your answer as a common fraction. sequen213.��������� ces Blake has a repertoire of 15 songs, of which 5 are acoustic. In his concert, he plans to perform 10 distinct songs, of which 4 will be acoustic. He wants to perform the 4 acoustic songs consecutively in the second half of the concert but doesn’t want to end the concert with an acoustic song. How many different sequences of songs can Blake perform? % Each time Chris bats, he has a 37% probability of getting a hit, independent of his 214.��������� previous performance at bat. What is his percent probability of getting at least two hits in four times at bat? Express your answer to the nearest whole number. % The Bobcats had a win-loss record of 41-23 before their star player, Melissa, got injured. 215.��������� Melissa didn’t play in any more games, and the Bobcats finished the season with an overall win-loss record of 54-34. What is the absolute difference in the Bobcats’ win percentages with and without Melissa? Express your answer to the nearest whole number. points On a test, Walter scored 86, Bev scored 72, Claire scored 61 and Rachel scored 93. Tyler 216.��������� didn’t remember his score, but he knew that his score was both the mean and the median of all five students’ scores. What was Tyler’s score? 217.��������� The two values of x that satisfy the equation x 2 + bx + c = 0, for integers b and c, are x = 6 and x = −4. What is the value of b? 218.��������� An eight-digit number N has one 2, three 3s, two 4s and two 5s as digits. If N can be expressed as a power of 2, what is the value of N? $ 219.��������� At Randolf’s furniture store, a standard breakfast table has a price of $1000. Karthik wants to purchase the luxury version of the table, which costs 35% more than the standard one, and he has a coupon that gives him a 15% discount on his purchase. After applying 6% sales tax to the discounted price, how much will Karthik pay for the luxury table? 220.��������� Tarin and Sonja construct the trapezoid shown, which contains an equilateral triangle and three isosceles triangles, two of which are right triangles. What is the ratio of the length of the shorter base to the length of the longer base of the trapezoid? Express your answer as a common fraction in simplest radical form. MATHCOUNTS 2018-2019 33 Workout 6 spells 221.��������� Ronald is attempting to cast a levitation spell. On his first attempt, he has a 20% chance of success. Ronald’s likelihood of success improves by 5% after each failed attempt. For example, his second attempt has a 25% chance of success if he fails on the first attempt. What is the minimum number of spells that he must attempt to guarantee a successful casting? units In a sequence of adjacent squares, a portion of which is shown, the side length of 222.__________ each successive square is two-thirds that of the preceding square. In the completed A sequence, the areas of the largest and smallest squares are 6561 units2 and 256 units2, respectively. What is the length of the segment with endpoints A and B? Express your answer to the nearest whole number. B ways A cheerleading squad consists of ten cheerleaders of ten different heights. How many 223.��������� ways are there for the cheerleaders to line up for a photo in two rows of five people each so that each cheerleader in the back row is taller than the one immediately in front? $ 224.��������� The math club purchased a number of gadgets to sell. Each box of gadgets cost the club $75 and contained 30 gadgets. In the first 15 days, the club members sold 22 gadgets each day at the regular price of $6.80 each. In the next 10 days, they sold 17 gadgets per day at a discount of 25% off. They then sold the remaining 220 gadgets at 30% off the discounted price. After all the gadgets were sold, what was the math club’s total profit? 225.��������� If A, B and C are the digits of the base-five number ABC, which is equal to the base-ten number 47, what digit does B represent? feet 226.��������� 20 ft 5 ft 5 ft The figure, which shows half of the cross section of a pool, can be rotated 360 degrees around the dotted line to create a three-dimensional pool with depth 10 feet in the center and 5 feet at the outer edge. What is the average depth of the pool? Express your answer as a decimal to the nearest tenth. 15 ft games In his first 319 career games, goaltender Braden Holtby had 200 wins and 119 losses. 227.��������� If he were to maintain this rate of winning games, what is the least number of games he would need to play in his career to pass Martin Brodeur’s career record of 691 wins? 228.��������� What is the least positive odd integer with exactly nine natural number divisors? seconds 229.��������� Noa completes 5 31 laps around a track in the same amount of time it takes Lev to complete 6 laps. Lev is running at a pace of 7.5 minutes per mile. How many seconds longer will it take Noa to run 1 mile, compared to Lev? Express your answer to the nearest whole number. in2 What is the area of a triangle with side lengths of 4, 6 and 8 inches? Express your answer 230.��������� in simplest radical form. 34 MATHCOUNTS 2018-2019 Workout 7 231.��������� If a number x plus its reciprocal is equal to 4, what is the absolute difference between x and its reciprocal? Express your answer in simplest radical form. $ 232.__________ For making house numbers, 5-inch digits can be purchased online for $8.79 each for the digits 1, 3, 5 and 6 and $5.98 each for the digits 0, 2, 4, 7, 8 and 9. Ashera wishes to create consecutive house numbers beginning with 5100 and ending with 5200. What is the total cost to purchase the necessary digits online, excluding tax and shipping? D units Rectangle ABCD is shown here with AB = 6 and BC = 8. If X and Y lie on 233.��������� diagonal AC, and XY = 1, what is the total area of shaded triangles BCX and DAY? Express your answer as a decimal to the nearest tenth. C 2 Y X A orders 234.��������� B Fran has four dogs whose ages are 3, 4, 5 and 6 years. She feeds them one at a time according to the rule that a younger dog can never eat right before a dog that is only one year older. In how many different orders can Fran feed her four dogs? 235.��������� What is the arithmetic mean of the terms of a five-term geometric sequence of positive numbers whose second term is 15 and whose fourth term is 135? units 236.��������� 24 12 24 x 12 Five right triangles are constructed adjacent to each other as shown. All five triangles share a vertex. The longer leg of each 24 triangle, except the rightmost triangle, is the hypotenuse of an adjacent triangle. What is the length of the longest hypotenuse, labeled x ? Express your answer in simplest radical form. 12 cm An equilateral triangle and a regular hexagon have a combined perimeter of 42 cm and a 237.��������� combined area of 49 3 cm2. What is the side length of the hexagon? Express your answer as a decimal to the nearest tenth. 238.��������� Ty flips a fair coin 20 times. What is the probability that it lands heads up and tails up equal numbers of times? Express your answer as a decimal to the nearest thousandth. troy One hundred years ago, a newly minted Walking Liberty Half Dollar contained 0.36 troy 239.��������� ounces ounce of silver, a Standing Liberty Quarter contained 90% silver (by mass) and a Mercury Dime was made from the same alloy as the other two coins. If the mass of each coin was proportional to its face value, what was the total mass of a Mercury Dime? Express your answer as a decimal to the nearest hundredth. units Two right triangles have areas in the ratio 1:2. If the smaller triangle has legs of length 240.��������� 3 units and 4 units, and the larger triangle has a side of length 8 units, what is the sum of the possible lengths of the hypotenuse of the larger triangle? Express your answer in simplest radical form. MATHCOUNTS 2018-2019 35 Workout 8 inches Scott thinks that Miguel is 6 feet 3 inches tall. If Miguel’s height is actually 96% of Scott’s 241.��������� estimate, how many inches tall is Miguel? in3 242.__________ A hemisphere of radius 6 inches lies flat on its circular base, and a rectangular prism with two square bases lies inside the hemisphere, with one square base on the base of the hemisphere and all four vertices of the other square base touching the upper boundary of the hemisphere. The height of the prism is one-third of the length of the square base. What is the volume of the prism? Express your answer as a decimal to the nearest tenth. 243.��������� If N is a perfect square and a divisor of 13!, what is the greatest possible value of N? lines When 64 unit cubes are glued together seamlessly to form a 4 × 4 × 4 cube, as shown, 244.��������� there are 76 distinct lines that pass through the centers of exactly four unit cubes. If 1000 unit cubes are glued together seamlessly to form a 10 × 10 × 10 cube, how many distinct lines pass through the centers of ten unit cubes? 245.��������� Suppose (a, b, c) is an ordered triple such that a 2 + b 2 = c 2. If b = x + y 2 when a = 12 + 10 2 and c = 15 + 8 2, what is the positive integer value of x + y? yellow 246.��������� blocks Camsie has a large number of identical yellow blocks, identical green blocks and identical white blocks. Using a balance scale, she finds that 3 green blocks will balance 1 yellow plus 3 white blocks and that 5 white blocks will balance 1 green plus 5 yellow blocks. How many yellow blocks will balance 30 green plus 60 white blocks? 247.��������� A box contains only quarters and dimes. If there were 10% more quarters, there would be 7.5% more money in the box. What is the ratio of the original number of quarters to the original number of dimes in the box? Express your answer as a common fraction. 660 ft feet 248.��������� 2x ft Farmer John’s field has area 60 acres. The field is in the shape of an “L”, with right angles at every vertex and dimensions as shown. If one acre is equal in area to a strip of land 40 rods long and 4 rods wide, and one rod is x ft equal in length to 16.5 feet, what is value of x in feet? 2640 ft ways Notice that 2019 can be written as the sum of consecutive integers in increasing order 249.��������� in numerous ways, for example 1009 +1010 = 2019 and (−2018) + (−2017) + ... + 2018 + 2019 = 2019. Including these two examples, in how many ways can 2019 be written as the sum of at least two consecutive integers in increasing order? units Three parallel chords of lengths 12, k and 16 units are in a circle, as shown. 250.��������� If the chord of length k is equally spaced 2 units away from each of the other chords, what is the value of k ? Express your answer in simplest radical form. 36 16 k 12 MATHCOUNTS 2018-2019 OFFICIAL RULES + PROCEDURES The following rules and procedures govern all MATHCOUNTS competitions. The MATHCOUNTS Foundation reserves the right to alter these rules and procedures at any time. Coaches are responsible for being familiar with the rules and procedures outlined in this handbook. Coaches should bring any difficulty in procedures or in student conduct to the immediate attention of the appropriate chapter, state or national official. Students violating any rules may be subject to immediate disqualification. Any questions regarding the MATHCOUNTS Competition Series Official Rules + Procedures articulated in this handbook should be addressed to the MATHCOUNTS national office at (703) 299-9006 or [email protected]. REGISTRATION The fastest and easiest way to register for the MATHCOUNTS Competition Series is online at www.mathcounts.org/compreg. For your school to participate in the MATHCOUNTS Competition Series, a school representative is required to complete a registration form and pay the registration fees. A school representative can be a teacher, administrator or parent volunteer who has received expressed permission from his/her child’s school administration to register. By completing the Competition Series Registration Form, the coach attests to the school administration’s permission to register students for MATHCOUNTS. School representatives can register online at www.mathcounts.org/compreg or download the Competition Series Registration Form and mail or email a scanned copy of it to the MATHCOUNTS national office. Refer to the Critical 2019-2019 Dates on pg. 10 of this handbook for contact information. WHAT REGISTRATION COVERS: Registration in the Competition Series entitles a school to: 1) send 1-10 student(s)—depending on number registered—to the Chapter Competition. Students can advance beyond the chapter level, but this is determined by their performance at the competition. 2) receive the School Competition Kit, which includes the 2018-2019 MATHCOUNTS School Handbook, one recognition ribbon for each registered student, 10 student participation certificates and a catalog of additional coaching materials. Mailings of School Competition Kits will occur on a rolling basis through December 31, 2018. 3) receive online access to the 2019 School Competition, along with electronic versions of other competition materials at www.mathcounts.org/coaches. Coaches will receive an email notification no later than November 2, 2018 when the 2019 School Competition is available online. Your state or chapter coordinator will be notified of your registration, and then you will be informed of the date and location of your Chapter Competition. If you have not been contacted by mid-January with competition details, it is your responsibility to contact your local coordinator to confirm that your registration has been properly routed and that your school’s participation is expected. Coordinator contact information is available at www.mathcounts.org/findmycoordinator. MATHCOUNTS 2018-2019 37 DEADLINES: The sooner your Registration Form is received, the sooner you will receive your preparation materials. To guarantee your school’s participation, submit your registration by one of the following deadlines: Early Bird Discount Deadline: November 2, 2018 Regular Registration Deadline: December 14, 2018* Online registrations: submitted by 11:59 PST Emailed forms: received by 11:59 PST Mailed forms: postmarked by November 2, 2018 Online registrations: submitted by 11:59 PST Emailed forms: received by 11:59 PST Mailed forms: postmarked by December 14, 2018 *Late Registrations may be accepted at the discretion of the MATHCOUNTS national office and your local coordinators, but are not guaranteed. If a school’s late registration is accepted, an additional $20 processing fee will be assessed. REGISTRATION FEES: The cost of your school’s registration depends on when your registration is postmarked/emailed/submitted online. The cost of your school’s registration covers the students for the entire Competition Series; there are no additional registration fees to compete at the state or national level. Title I schools (as affirmed by a school’s administration) receive a 50% discount off the total cost of their registration. Early Bird Registrations (by November 2, 2018) Regular Registrations (by December 14, 2018) Late Registrations (after December 14, 2018) $30 per student $120 for 1 team of 4 | $300 for 1 team of 4 + 6 individuals $35 per student $140 for 1 team of 4 | $350 for 1 team of 4 + 6 individuals $35 per student + $20 late fee on entire order $160 for 1 team of 4 | $370 for 1 team of 4 + 6 individuals CANCELLATION FEES: Registered schools that need to cancel their Competition Series registration must notify the MATHCOUNTS national office in writing via email or mail. Schools may request and receive a full refund minus a $30 non-refundable cancellation fee to cover refund processing and the cost of materials shipped to the school. MATHCOUNTS will verify a school’s non-participation with local coordinators and reserves the right to refuse a refund request. No cancellations or refund requests will be processed after February 1, 2019. This fee does not apply to schools that reduce their number of registered students but remain registered with at least one student. ELIGIBILITY REQUIREMENTS Eligibility requirements for the MATHCOUNTS Competition Series are different from other MATHCOUNTS programs. Eligibility for the National Math Club or the Math Video Challenge does not guarantee eligibility for the Competition Series. WHO IS ELIGIBLE: •• U.S. students enrolled in the 6th, 7th or 8th grade can participate in MATHCOUNTS competitions. •• Schools that are the students’ official school of record can register. •• Any type of school, of any size, can register—public, private, religious, charter, virtual or homeschools— but virtual and homeschools must fill out additional forms to participate (see pgs. 39-40). •• Schools in 50 U.S. states, District of Columbia, Guam, Puerto Rico and Virgin Islands can register. •• Overseas schools that are affiliated with the U.S. Departments of Defense and State can register. 38 MATHCOUNTS 2018-2019 WHO IS NOT ELIGIBLE: •• Students who are not full-time 6th, 7th or 8th graders cannot participate, even if they are taking middle school math classes. •• Academic centers, tutoring centers or enrichment programs that do not function as students’ official school of record cannot register. If it is unclear whether your educational institution is considered a school, please contact your local Department of Education for specific criteria governing your state. •• Schools located outside of the U.S. states and territories listed on the previous page cannot register. •• Overseas schools not affiliated with the U.S. Departments of Defense or State cannot register. NUMBER OF STUDENTS ALLOWED: A school can register a maximum of one team of four students and six individuals; these 1-10 student(s) will represent the school at the Chapter Competition. Any number of students can participate at the school level. Prior to the Chapter Competition, coaches must notify their chapter coordinator of which students will be team members and which students will compete as individuals. NUMBER OF YEARS ALLOWED: Participation in MATHCOUNTS competitions is limited to 3 years for each student, but there is no limit to the number of years a student may participate in school‑based coaching. WHAT TEAM REGISTRATION MEANS: Members of a school team will participate in the Target, Sprint and Team Rounds. Members of a school team also will be eligible to qualify for the Countdown Round (where conducted). Team members will be eligible for team awards, individual awards and progression to the state and national levels based on their individual and/or team performance. It is recommended that your strongest four Mathletes form your school team. Teams of fewer than four will be allowed to compete; however, the team score will be computed by dividing the sum of the team members’ scores by four (see pg. 43), meaning, teams of fewer than four students will be at a disadvantage. Only one team (of up to four students) per school is eligible to compete. WHAT INDIVIDUAL REGISTRATION MEANS: Students registered as individuals will participate in the Target and Sprint Rounds, but not the Team Round. Individuals will be eligible to qualify for the Countdown Round (where conducted). Individuals also will be eligible for individual awards and progression to the state and national levels. A student registered as an “individual” may not help his/her school’s team advance to the next level of competition. Up to six students may be registered in addition to or in lieu of a school team. HOW STUDENTS ENROLLED PART-TIME AT TWO SCHOOLS PARTICIPATE: A student may compete only for his/her official school of record. A student’s school of record is the student’s base or main school. A student taking limited course work at a second school or educational center may not register or compete for that second school or center, even if the student is not competing for his/her school of record. MATHCOUNTS registration is not determined by where a student takes his or her math course. If there is any doubt about a student’s school of record, the chapter or state coordinator must be contacted for a decision before registering. HOW SMALL SCHOOLS PARTICIPATE: MATHCOUNTS does not distinguish between the sizes of schools for Competition Series registration and competition purposes. Every “brick-and-mortar” school will have the same registration allowance of up to one team of four students and/or up to six individuals. A school’s participants may not combine with any other school’s participants to form a team when registering or competing. HOW HOMESCHOOLS PARTICIPATE: Homeschools and/or homeschool groups in compliance with the homeschool laws of the state in which they are located are eligible to participate in MATHCOUNTS competitions in accordance with all other rules. Homeschool coaches must complete the 2018-2019 Homeschool + Virtual School Participation Form, verifying that students from the homeschool or homeschool group are in the 6th, 7th or 8th grade and that each homeschool complies with applicable state laws. Forms can be downloaded at www.mathcounts.org/competition and must be submitted to the MATHCOUNTS national office in order for registrations to be processed. MATHCOUNTS 2018-2019 39 HOW VIRTUAL SCHOOLS PARTICIPATE: Virtual schools that want to register must contact the MATHCOUNTS national office by December 7, 2018 for specific registration details. Any student registering as a virtual school student must compete in the MATHCOUNTS Chapter Competition assigned according to the student’s home address. Additionally, virtual school coaches must complete the 2018-2019 Homeschool + Virtual School Participation Form, verifying that the students from the virtual school are in the 6th, 7th or 8th grade and that the virtual school complies with applicable state laws. Forms must be submitted to the national office in order for registrations to be processed; forms can be downloaded at www.mathcounts.org/competition. WHAT IS DONE FOR SUBSTITUTIONS OF STUDENTS: Coaches determine which students will represent the school at the Chapter Competition. Coaches cannot substitute team members for the State Competition unless a student voluntarily releases his/her position on the school team. Additional requirements and documentation for substitutions (such as requiring parental release or requiring the substitution request be submitted in writing) are at the discretion of the State Coordinator. A student being added to a team need not be a student who was registered for the Chapter Competition as an individual. Coaches cannot make substitutions for students progressing to the State Competition as individuals. At all levels of competition, student substitutions are not permitted after on-site competition registration has been completed. WHAT IS DONE FOR RELIGIOUS OBSERVANCES: A student who is unable to attend a competition due to religious observances may take the written portion of the competition up to one week in advance of the scheduled competition. In addition, all competitors from that student’s school must take the Sprint and Target Rounds at the same earlier time. If the student who is unable to attend the competition due to a religious observance: (1) is a member of the school team, then the team must take the Team Round at the same earlier time; (2) is not part of the school team, then the team has the option of taking the Team Round during this advance testing or on the regularly scheduled day of the competition with the other school teams. The coordinator must be made aware of the team’s decision before the advance testing takes place. Advance testing will be done at the discretion of the chapter and state coordinators. If advance testing is deemed possible, it will be conducted under proctored conditions. Students who qualify for an official Countdown Round but are unable to attend will automatically forfeit one place standing. WHAT IS DONE FOR STUDENTS WITH SPECIAL NEEDS: Reasonable accommodations may be made to allow students with special needs to participate. However, many accommodations that are employed in a classroom or teaching environment cannot be implemented in the competition setting. Accommodations that are not permissible include, but are not limited to: granting a student extra time during any of the competition rounds or allowing a student to use a calculator for the Sprint or Countdown Rounds. A request for accommodation of special needs must be directed to chapter or state coordinators in writing at least three weeks in advance of the Chapter or State Competition. This written request should thoroughly explain a student’s special need, as well as what the desired accommodation would entail. In conjunction with the MATHCOUNTS Foundation, coordinators will review the needs of the student and determine if any accommodations will be made. In making final determinations, the feasibility of accommodating these needs at the National Competition will be taken into consideration. LEVELS OF COMPETITION There are four levels in the MATHCOUNTS Competition Series: school, chapter (local), state and national. Competition questions are written for 6th, 7th and 8th graders. The competitions can be quite challenging, particularly for students who have not been coached using MATHCOUNTS materials. All competition materials are prepared by the national office. SCHOOL COMPETITIONS (TYPICALLY HELD IN JANUARY 2019): After several months of coaching, schools registered for the Competition Series should administer the 2019 School Competition to all interested 40 MATHCOUNTS 2018-2019 students. The School Competition should be an aid to the coach in determining competitors for the Chapter Competition. Selection of team and individual competitors is entirely at the discretion of the coach and does not need to be based solely on School Competition scores. School Competition materials are sent to the coach of a school, and it may be used by the teachers and students only in association with that school’s programs and activities. The current year’s School Competition questions must remain confidential and may not be used in outside activities, such as tutoring sessions or enrichment programs with students from other schools. For updates or edits, please check www.mathcounts.org/coaches before administering the School Competition. It is important that the coach look upon coaching sessions during the academic year as opportunities to develop better math skills in all students, not just in those students who will be competing. Therefore, it is suggested that the coach postpone selection of competitors until just prior to the Chapter Competition. CHAPTER COMPETITIONS (HELD FEB. 1–28, 2019): The Chapter Competition consists of the Sprint, Target and Team Rounds. The Countdown Round (official or just for fun) may or may not be conducted. The chapter and state coordinators determine the date and location of the Chapter Competition in accordance with established national procedures and rules. Winning teams and students will receive recognition. The winning team will advance to the State Competition. Additionally, the two highest-ranking competitors not on the winning team (who may be registered as individuals or as members of a team) will advance to the State Competition. This is a minimum of six advancing Mathletes (assuming the winning team has four members). Additional teams and/or individuals also may progress at the discretion of the state coordinator, but the policy for progression must be consistent for all chapters within a state. STATE COMPETITIONS (HELD MAR. 1–31, 2019): The State Competition consists of the Sprint, Target and Team Rounds. The Countdown Round (official or just for fun) may or may not be included. The state coordinator determines the date and location of the State Competition in accordance with established national procedures and rules. Winning teams and students will receive recognition. The four highest-ranked Mathletes and the coach of the winning team from each State Competition will receive an all-expenses-paid trip to the National Competition. 2019 RAYTHEON MATHCOUNTS NATIONAL COMPETITION (HELD MAY 12–13 IN ORLANDO, FL): The National Competition consists of the Sprint, Target, Team and Countdown Rounds (conducted officially). Expenses of the state team and coach to travel to the National Competition will be paid by MATHCOUNTS. The national program does not make provisions for the attendance of additional students or coaches. All national competitors will receive a plaque and other items in recognition of their achievements. Winning teams and individuals also will receive medals, trophies and college scholarships. COMPETITION COMPONENTS The four rounds of a MATHCOUNTS competition, each described below, are designed to be completed in approximately three hours: TARGET ROUND (approximately 30 minutes): In this round eight problems are presented to competitors in four pairs (six minutes per pair). The multi-step problems featured in this round engage Mathletes in mathematical reasoning and problem-solving processes. Problems assume the use of calculators. SPRINT ROUND (40 minutes): Consisting of 30 problems, this round tests accuracy, with the time period allowing only the most capable students to complete all of the problems. Calculators are not permitted. TEAM ROUND (20 minutes): In this round, interaction among team members is permitted and encouraged as they work together to solve 10 problems. Problems assume the use of calculators. MATHCOUNTS 2018-2019 41 Note: The order in which the written rounds (Target, Sprint and Team) are administered is at the discretion of the competition coordinator. COUNTDOWN ROUND: A fast-paced oral competition for top-scoring individuals (based on scores on the Target and Sprint Rounds), this round allows pairs of Mathletes to compete against each other and the clock to solve problems. Calculators are not permitted. At Chapter and State Competitions, a Countdown Round (1) may be conducted officially, (2) may be conducted unofficially (for fun) or (3) may be omitted. However, the use of an official Countdown Round must be consistent for all chapters within a state. In other words, all chapters within a state must use the round officially in order for any chapter within a state to use it officially. All students, whether registered as part of a school team or as individual competitors, are eligible to qualify for the Countdown Round. An official Countdown Round determines an individual’s final overall rank in the competition. If a Countdown Round is used officially, the official procedures as established by the MATHCOUNTS Foundation must be followed, as described below.* •• The top 25% of students, up to a maximum of 10, are selected to compete. These students are chosen based on their Individual Scores. •• The two lowest‑ranked students are paired; a question is read and projected, and students are given 45 seconds to solve the problem. A student may buzz in at any time, and if s/he answers correctly, a point is scored. If a student answers incorrectly, the other student has the remainder of the 45 seconds to answer. •• Three total questions are read to the pair of students, one question at a time, and the student who scores the higher number of points (not necessarily 2 out of 3) progresses to the next round and challenges the next-higher‑ranked student. •• If students are tied in their matchup after three questions (at 1‑1 or 0‑0), questions should continue to be read until one is successfully answered. The first student who answers an additional question correctly progresses to the next round. •• This procedure continues until the 4th-ranked Mathlete and his/her opponent compete. For the final four matchups, the first student to correctly answer three questions advances. •• The Countdown Round proceeds until a 1st place individual is identified. More details about Countdown Round procedures are included in the 2019 School Competition. *Rules for the Countdown Round change for the National Competition. An unofficial Countdown Round does not determine an individual’s final overall rank in the competition, but is done for practice or for fun. The official procedures do not have to be followed. Chapters and states choosing not to conduct the round officially must determine individual winners solely on the basis of students’ scores in the Target and Sprint Rounds of the competition. SCORING MATHCOUNTS Competition Series scores do not conform to traditional grading scales. Coaches and students should view an Individual Score of 23 (out of a possible 46) as highly commendable. INDIVIDUAL SCORE: calculated by taking the sum of the number of Sprint Round questions answered correctly and twice the number of Target Round questions answered correctly. There are 30 questions in the Sprint Round and eight questions in the Target Round, so the maximum possible Individual Score is 30 + 2(8) = 46. If used officially, the Countdown Round yields final individual standings. 42 MATHCOUNTS 2018-2019 TEAM SCORE: calculated by dividing the sum of the team members’ Individual Scores by four (even if the team has fewer than four members) and adding twice the number of Team Round questions answered correctly. The highest possible Individual Score is 46. Four students may compete on a team, and there are 10 questions in the Team Round. Therefore, the maximum possible Team Score is ((46 + 46 + 46 + 46) ÷ 4) + 2(10) = 66. TIEBREAKING ALGORITHM: used to determine team and individual ranks and to determine which individuals qualify for the Countdown Round. In general, questions in the Target, Sprint and Team Rounds increase in difficulty so that the most difficult questions occur near the end of each round. In a comparison of questions to break ties, generally those who correctly answer the more difficult questions receive the higher rank. The guidelines provided below are very general; competition officials receive more detailed procedures. •• Ties between individuals: the student with the higher Sprint Round score will receive the higher rank. If a tie remains after this comparison, specific groups of questions from the Target and Sprint Rounds are compared. •• Ties between teams: the team with the higher Team Round score, and then the higher sum of the team members’ Sprint Round scores, receives the higher rank. If a tie remains after these comparisons, specific questions from the Team Round will be compared. RESULTS DISTRIBUTION Coaches should expect to receive the scores of their students, as well as a list of the top 25% of students and top 40% of teams, from their competition coordinators. In addition, single copies of the blank competition materials and answer keys may be distributed to coaches after all competitions at that level nationwide have been completed. Before distributing blank competition materials and answer keys, coordinators must wait for verification from the national office that all such competitions have been completed. Both the problems and answers from Chapter and State Competitions will be posted on the MATHCOUNTS website following the completion of all competitions at that level nationwide, replacing the previous year’s posted tests. Student competition papers and answers will not be viewed by or distributed to coaches, parents, students or other individuals. Students’ competition papers become the confidential property of MATHCOUNTS. ADDITIONAL RULES All answers must be legible. Pencils and paper will be provided for Mathletes by competition organizers. However, students may bring their own pencils, pens and erasers if they wish. They may not use their own scratch paper or graph paper. Use of notes or other reference materials (including dictionaries and translation dictionaries) is prohibited. Specific instructions stated in a given problem take precedence over any general rule or procedure. Communication with coaches is prohibited during rounds but is permitted during breaks. All communication between guests and Mathletes is prohibited during competition rounds. Communication between teammates is permitted only during the Team Round. Calculators are not permitted in the Sprint and Countdown Rounds, but they are permitted in the Target, Team and Tiebreaker (if needed) Rounds. When calculators are permitted, students may use any calculator (including programmable and graphing calculators) that does not contain a QWERTY (typewriter‑like) keypad. Calculators that have the ability to enter letters of the alphabet but do not have a keypad in a standard typewriter arrangement are acceptable. Smart phones, laptops, tablets, iPods®, personal MATHCOUNTS 2018-2019 43 digital assistants (PDAs) and any other “smart” devices are not considered to be calculators and may not be used during competitions. Students may not use calculators to exchange information with another person or device during the competition. Coaches are responsible for ensuring their students use acceptable calculators, and students are responsible for providing their own calculators. Coordinators are not responsible for providing Mathletes with calculators or batteries before or during MATHCOUNTS competitions. Coaches are strongly advised to bring backup calculators and spare batteries to the competition for their team members in case of a malfunctioning calculator or weak or dead batteries. Neither the MATHCOUNTS Foundation nor coordinators shall be responsible for the consequences of a calculator’s malfunctioning. Pagers, cell phones, tablets, iPods® and other MP3 players should not be brought into the competition room. Failure to comply could result in dismissal from the competition. Should there be a rule violation or suspicion of irregularities, the MATHCOUNTS coordinator or competition official has the obligation and authority to exercise his/her judgment regarding the situation and take appropriate action, which might include disqualification of the suspected student(s) from the competition. 44 MATHCOUNTS 2018-2019 FORMS OF ANSWERS The following rules explain acceptable forms for answers. Coaches should ensure that Mathletes are familiar with these rules prior to participating at any level of competition. Competition answers will be scored in compliance with these rules for forms of answers. Units of measurement are not required in answers, but they must be correct if given. When a problem asks for an answer expressed in a specific unit of measure or when a unit of measure is provided in the answer blank, equivalent answers expressed in other units are not acceptable. For example, if a problem asks for the number of ounces and 36 oz is the correct answer, 2 lb 4 oz will not be accepted. If a problem asks for the number of cents and 25 cents is the correct answer, $0.25 will not be accepted. All answers must be expressed in simplest form. A “common fraction” is to be considered a fraction in the form a ± b , where a and b are natural numbers and GCF(a, b) = 1. In some cases the term “common fraction” is to be A considered a fraction in the form B , where A and B are algebraic expressions and A and B do not have a common factor. A simplified “mixed number” (“mixed numeral,” “mixed fraction”) is to be considered a fraction in the form a ± N b , where N, a and b are natural numbers, a < b and GCF(a, b) = 1. Examples: 2 4 Problem: What is 8 ÷ 12 expressed as a common fraction? Answer: 3 Unacceptable: 6 Problem: What is 12 ÷ 8 expressed as a common fraction? Answer: 2 Unacceptable: 8 , 1 2 3 12 1 1 Problem: What is the sum of the lengths of the radius and the circumference of a circle of diameter 4 unit expressed as a common fraction in terms of π? Answer: Problem: What is 20 ÷ 12 expressed as a mixed number? 1 + 2π 8 2 Answer: 1 3 8 5 Unacceptable: 1 12 , 3 Ratios should be expressed as simplified common fractions unless otherwise specified. Examples: 7 3 4–π Acceptable Simplified Forms: 2 , π , 6 1 1 4 Unacceptable: 3 2 , 3 , 3.5, 2:1 Radicals must be simplified. A simplified radical must satisfy: 1) no radicands have a factor which possesses the root indicated by the index; 2) no radicands contain fractions; and 3) no radicals appear in the denominator of a fraction. Numbers with fractional exponents are not in radical form. Examples: Problem: What is 15 × 5 expressed in simplest radical form? Answer: 5 3 Unacceptable: 75 Answers to problems asking for a response in the form of a dollar amount or an unspecified monetary unit (e.g., “How many dollars...,” “How much will it cost...,” “What is the amount of interest...”) should be expressed in the form ($) a.bc, where a is an integer and b and c are digits. The only exceptions to this rule are when a is zero, in which case it may be omitted, or when b and c are both zero, in which case they both may be omitted. Answers in the form ($) a.bc should be rounded to the nearest cent, unless otherwise specified. Examples: Acceptable Forms: 2.35, 0.38, .38, 5.00, 5 Unacceptable: 4.9, 8.0 2 Do not make approximations for numbers (e.g., π, 3 , 5 3) in the data given or in solutions unless the problem says to do so. Do not do any intermediate rounding (other than the “rounding” a calculator performs) when calculating solutions. All rounding should be done at the end of the calculation process. Scientific notation should be expressed in the form a × 10n where a is a decimal, 1 < |a| < 10, and n is an integer. Examples: Problem: What is 6895 expressed in scientific notation? Problem: What is 40,000 expressed in scientific notation? Answer: 6.895 × 103 Answer: 4 × 104 or 4.0 × 104 An answer expressed to a greater or lesser degree of accuracy than called for in the problem will not be accepted. Whole-number answers should be expressed in their whole-number form. Thus, 25.0 will not be accepted for 25, and 25 will not be accepted for 25.0. The plural form of the units will always be provided in the answer blank, even if the answer appears to require the singular form of the units. MATHCOUNTS 2018-2019 45 Competition Coach Toolkit This is a collection of lists, formulas and terms that Mathletes frequently use to solve problems like those found in this handbook. There are many others we could have included, but we hope you find this collection useful. Fraction ½ ¹⁄ 3 ¼ ¹⁄ 5 ¹⁄ 6 ¹⁄ 8 ¹⁄ 9 ¹⁄ 10 ¹⁄ 11 ¹⁄ 12 Decimal 0.5 0.3 0.25 0.2 0.16 0.125 0.1 0.1 0.09 0.083 Percent 50 33.3 25 20 16.6 12.5 11.1 10 9.09 8.3 Common Arithmetic Series 2 1 + 3 + 5 + 7 + + (2n − 1) = n2 2 + 4 + 6 + 8 + + 2n = n2 + n Combinations & Permutations Cr = n2 n3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 1 8 27 64 125 216 343 512 729 1000 1331 1728 2197 2744 3375 n! Pr = n r! (n − r)! n! (n − r)! 2: units digit is 0, 2, 4, 6 or 8 3: sum of digits is divisible by 3 Geometric Mean a x = x b and 4: two-digit number formed by tens and units digits is divisible by 4 x= ab 5: units digit is 0 or 5 6: number is divisible by both 2 and 3 8: three-digit number formed by hundreds, tens and units digits is divisible by 8 9: sum of digits is divisible by 9 10: units digit is 0 Distance = Rate × Time Quadratic Formula For ax2 + bx + c = 0, where a ≠ 0, Standard Form x = −b ± Ax + By = C Slope-Intercept Form y = mx + b m = slope b = y-intercept Point-Slope Form y − y1 = m(x − x1) m = slope 46 (x1, y1) = point on the line 43 47 53 59 61 67 71 73 79 83 89 97 Divisibility Rules Distance Traveled Equation of a Line 2 3 5 7 11 13 17 19 23 29 31 37 41 n(n + 1) 1+2+3+4++n= n n Prime Numbers b2 − 4ac 2a Pythagorean Triples (3, 4, 5) (5, 12, 13) (8, 15, 17) (9, 40, 41) Difference of Squares a2 − b2 = (a + b) (a − b) (7, 24, 25) (12, 35, 37) Sum and Difference of Cubes a3 − b3 = (a − b) (a2 + ab + b2) a3 + b3 = (a + b) (a2 − ab + b2) MATHCOUNTS 2018-2019 Given A(x1, y1) and B(x2, y2) Circles Circumference 2×π×r=π×d Area π × r2 For radius r Arc Length x ×2×π×r 360 (x2 − x1)2 + (y2 − y1)2 Distance from A to B = Midpoint of AB = x1 + x2 2 Slope of AB = y2 − y1 x2 − x1 Sector Area x × π × r2 360 Special Right Triangles For central angle of x degrees 45° 30° 2a b 2 a 3 Pythagorean Theorem c a a a +b =c 2 2 b 30-60-90 Right Triangle b b 45° 60° 2 y1 + y2 2 , 45-45-90 Right Triangle Area of Polygons Square side length s s2 Polygon Angles Rectangle length l, width w l×w Sum of the interior angle measures: 180 × (n − 2) Parallelogram base b, height h b×h Trapezoid bases b1, b2, height h 1 2 (b1 Rhombus diagonals d1, d2 1 2 × d1 × d2 Triangle base b, height h 1 2 ×b×h Triangle semi-perimeter s, side lengths a, b, c Equilateral Triangle side length s Solid (n sides) Central angle measure of a regular polygon: + b2) ×h 360 n Interior angle measure of a regular polygon: 180 × (n − 2) n s(s − a)(s − b)(s − c) or 180 − 360 n s2 3 4 Dimensions Surface Area Volume Cube side length s 6×s s3 Rectangular Prism length l, width w, height h 2 × (l × w + w × h + l × h) l×w×h Circular Cylinder base radius r, height h 2 × π × r × h + 2 × π × r2 π × r2 × h Circular Cone base radius r, height h π × r2 + π × r × 1 3× π × r2 × h Sphere radius r 4 × π × r2 4 3× π × r3 Pyramid base area B, height h 1 3× B×h MATHCOUNTS 2018-2019 2 r2 + h2 47 Vocabulary & Terms The following list is representative of terminology used in the problems but should not be viewed as all‑inclusive. It is recommended that coaches review this list with their Mathletes. absolute difference absolute value acute angle additive inverse (opposite) adjacent angles apex arithmetic mean arithmetic sequence base ten binary bisect box-and-whisker plot center chord circumscribe coefficient collinear common divisor common factor common fraction complementary angles congruent convex coordinate plane/system coplanar counting numbers counting principle diagonal of a polygon diagonal of a polyhedron digit-sum dilation direct variation divisor domain of a function edge equiangular equidistant expected value exponent exterior angle of a polygon factor finite frequency distribution frustum function 48 GCF (GCD) geometric sequence hemisphere image(s) of a point(s) (under a transformation) improper fraction infinite series inscribe integer interior angle of a polygon intersection inverse variation irrational number isosceles lateral edge lateral surface area lattice point(s) LCM median of a set of data median of a triangle mixed number mode(s) of a set of data multiplicative inverse (reciprocal) natural number obtuse angle ordered pair origin palindrome parallel Pascal’s Triangle percent increase/decrease perpendicular planar polyhedron polynomial prime factorization principal square root proper divisor proper factor proper fraction quadrant quadrilateral random range of a data set range of a function rate ratio rational number ray real number reciprocal (multiplicative inverse) reflection regular polygon relatively prime revolution right angle right polyhedron rotation scalene triangle scientific notation sector segment of a circle segment of a line semicircle semiperimeter sequence set significant digits similar figures slope space diagonal square root stem-and-leaf plot supplementary angles system of equations/inequalities tangent figures tangent line term transformation translation triangular numbers trisect twin primes union unit fraction variable whole number y-intercept MATHCOUNTS 2018-2019 ANSWERS In addition to the answer, we have provided a difficulty rating for each problem. Our scale is 1-7, with 7 being the most difficult. These are only approximations, and how difficult a problem is for a particular student will vary. Below is a general guide to the ratings: Difficulty 1/2/3 - One concept; one- to two-step solution; appropriate for students just starting the middle school curriculum. 4/5 - One or two concepts; multistep solution; knowledge of some middle school topics is necessary. 6/7 - Multiple and/or advanced concepts; multistep solution; knowledge of advanced middle school topics and/or problem-solving strategies is necessary. Measurement Stretch Answer Warm-Up 1 Answer Difficulty (4) 31. 7 7.8 2 (3) (4) 8.42 4.44.3 (3) 5.4.2 (3) Difficulty (2) 36. 105 (2) 32. 4 6 1 (2) 37. 26 (3) (3) 33. 1/2 (2) 38. 1* (2) 9.0.028 (3) 34. 20 (2) 39. 50 (3) 10. 5 (4) 35. 1 (3) 40. 580 (3) 1.12 (1) 6.13 2.16 (4) 3.3 1 Expected Value Stretch Answer Warm-Up 2 Answer Difficulty 1 11. 4 3 Difficulty (2) 16. 20.5 (3) 41. 120 (2) 46. 36 (3) 12. 12.75 (2) 17. 1.2 (4) 42. 36 (2) 47. 50/3 (3) 13. 3 (3) 18. 2.1 (3) 43. 22 (2) 48. 289/576 (3) 14. 97/13 (4) 19. 0 (3) 44. 4 (2) 49. 1 (3) 15. 23.3 (4) 20. 100 (3) 45. 32 (3) 50. 44 (3) Transformations Stretch Answer Warm-Up 3 Difficulty Answer 21. 5 (2) 26. (10, 2) (5) 51. 22. 16 (3) 27. 8 (4) 23. (7, −4) (3) 28. 55 24. 5 (5) 25. (−3, 5) (3) 36 Difficulty 1 (3) 56. 2 4 (4) 52. 2.52 (1) 57. 3 (3) (5) 53. 24 (2) 58. 6 (4) 29. 87/4 (6) 54. 22 (3) 59. 13 (3) 30. 2√17 (5) 55. 200 (4) 60. 119 (4) * The plural form of the units is always provided in the answer blank, even if the answer appears to require the singular form of the units. MATHCOUNTS 2018-2019 49 Warm-Up 4 Answer Warm-Up 7 Difficulty Answer Difficulty 61. 73 (3) 66. 24 (4) 91. 15 (4) 96. −1 (4) 62. 5 (3) 67. 50 (4) 92. 25 (5) 97. 87 (3) 63. 8:53 (2) 68. 27 (4) 93. 3 (3) 98. 8217902 (2) 64. 29 (2) 69. 364 (2) 94. 78 (3) 99. 28 (5) 65. 20 or 20.00 (4) 70. 176 (3) 95. 2 (3) 100.12 (4) Warm-Up 8 Warm-Up 5 Answer 71. 69 or 69.00 Difficulty (4) 72. 392 or 392.00 (3) 73. 52,413 (2) 74. 108 (3) 75. 8 (4) Answer 76. 15 or 15.00 (5) 2 101. 6 3 77. (5) 78. 50 (4) 106. 16 (3) 102. 555 (3) 107. 10√5 (4) (3) 103. 242 (3) 108. 433 (4) 79. 11/29 (3) 104. 13/32 (4) 109. 10 (4) 80. 9 (5) 105. 12 (4) 110. 126.5 (5) 15/2 Warm-Up 6 Answer Difficulty Warm-Up 9 Difficulty Answer Difficulty 81. 18 (2) 337 86. (5) 111.1740 (3) 116.40 (4) 82. 1/2 (5) 3 or 3.00 87. (4) 112.3:52 (2) 117.7 (3) 83. 7/15 (4) 101 88. (3) 113.15 (2) 118.27 (5) 84. 17 (4) 35 89. (4) 114.590 (5) 119.19/108 (5) 85. 1878 (4) 35 90. (4) 115.555 (4) 120.70 (4) 50 MATHCOUNTS 2018-2019 Warm-Up 10 Answer Warm-Up 13 Answer Difficulty Difficulty 121.6 (3) 126.1/4 (5) 151.9 (4) 156.125/4 (5) 122.10 (3) 127.6 (4) 152.55 (4) 157.5/8 (5) 2 153.12 3 (3) 158.70 (5) 154.√2019 (3) 159.10 (4) (4) 160.–102 (6) 123.30 (4) 128.96 (3) 124.414 (4) 129.90 (5) 125.12.0 (4) 130.80 (4) 155.20 Warm-Up 14 Warm-Up 11 Answer Difficulty Answer Difficulty 131.11 (3) 136.2√13 (5) 161.60 (4) 166.55 (3) 132.300 (4) 137.13 (4) 162.50 (7) 167.8 (2) 133. 18 (5) 138.22 (4) 163.121 (5) 168.15 (5) 134.11 (2) 139.28 (5) 164.1/3 (5) 169.51/100 (7) 135.40 (3) 140.5000 (3) 165.1/4 (3) 170.12 (4) Warm-Up 12 Answer Difficulty 141.252 (3) 146.3 (5) 142.Tuesday (3) 147.26 (5) 143.144 (3) 148.18 (5) 144.24 (4) 149.48 (6) 145.4 (4) 150.13 (4) MATHCOUNTS 2018-2019 51 Workout 1 Answer Workout 5 Answer Difficulty Difficulty 171.252 (3) 176.260 (3) 211.3.5 (5) 216.78 (4) 172.57 (3) 177.28.3 (4) 212.2/9 (5) 217.−2 (4) 173.8352 (3) 178. 3.4 (3) 213.18,144,000 (4) 218.33,554,432 (3) 174.87 or 87.00 (2) 179.10.87 (3) 214.47 (5) 219.1216.35 (4) 175.90 180.48√3 (4) 215.10 (4) 220.√3/3 (5) (3) Workout 2 Answer Workout 6 Answer Difficulty Difficulty 181.240 (3) 186.134 (3) 221.17 (3) 226.7.8 (6) 182.8 (3) 187.135 (5) 222.145 (5) 227.1103 (4) 183.7 (3) 188.7 (3) 223.113,400 (5) 228. 225 (4) 184.12 (2) 189.83 (5) 224.2096.40 (4) 229.56 (4) 185.1091 (3) 190.32 (4) 225.4 (3) 230.3√15 (5) Workout 3 Answer Workout 7 Answer Difficulty Difficulty 191.72 (4) 196.5/2 (5) 231.2√3 (5) 236.12√15 (4) 192.1/18 (5) 197.18 (4) 232.3205.53 (4) 237.5.6 (7) 193.60 (3) 198.17.2 (3) 233.21.6 (6) 238.0.176 (5) 194.481 (3) 199.241.3 (5) 234.11 (3) 239.0.08 (4) 195.89 (3) 200.120 (3) 235.121 (4) 240.8 + √73 or √73 + 8 (5) Workout 4 Answer Workout 8 Answer Difficulty Difficulty 201.4.09 (4) 206.2166 (4) 241.72 (3) 246.130 (5) 202.2.03 (5) 207.6.54 (4) 242.150.7 (6) 247. 6/5 (5) 203.8:48 (2) (4) 243.2,073,600 (5) 248.660 (4) 204.1.44 (2) 208. 203 or 203.00 244.364 (5) 249.7 (5) 205.5 (5) 209.31 (6) 245.3 (5) 250.6√6 (7) 210.72 (5) 52 MATHCOUNTS 2018-2019 MATHCOUNTS Problems Mapped to Common Core State Standards (CCSS) Forty-two states, the District of Columbia, four territories and the Department of Defense Education Activity (DoDEA) have voluntarily adopted the Common Core State Standards (CCSS). As such, MATHCOUNTS considers it beneficial for teachers to see the connections between the 2018-2019 MATHCOUNTS School Handbook problems and the CCSS. MATHCOUNTS not only has identified a general topic and assigned a difficulty level for each problem but also has provided a CCSS code in the Problem Index (pages 54-55). A complete list of the Common Core State Standards can be found at www.corestandards.org. The CCSS for mathematics cover K-8 and high school courses. MATHCOUNTS problems are written to align with the NCTM Standards for Grades 6-8. As one would expect, there is great overlap between the two sets of standards. MATHCOUNTS also recognizes that in many school districts, algebra and geometry are taught in middle school, so some MATHCOUNTS problems also require skills taught in those courses. In referring to the CCSS, the Problem Index code for each of the Standards for Mathematical Content for grades K-8 begins with the grade level. For the Standards for Mathematical Content for high school courses (such as algebra or geometry), each code begins with a letter to indicate the course name. The second part of each code indicates the domain within the grade level or course. Finally, the number of the individual standard within that domain follows. Here are two examples: • 6.RP.3 → Standard #3 in the Ratios and Proportional Relationships domain of grade 6 • G-SRT.6 → Standard #6 in the Similarity, Right Triangles and Trigonometry domain of Geometry Some math concepts utilized in MATHCOUNTS problems are not specifically mentioned in the CCSS. Two examples are the Fundamental Counting Principle (FCP) and special right triangles. In cases like these, if a related standard could be identified, a code for that standard was used. For example, problems using the FCP were coded 7.SP.8, S-CP.8 or S-CP.9 depending on the context of the problem; SP → Statistics and Probability (the domain), S → Statistics and Probability (the course) and CP → Conditional Probability and the Rules of Probability. Problems based on special right triangles were given the code G-SRT.5 or G-SRT.6, explained above. There are some MATHCOUNTS problems that either are based on math concepts outside the scope of the CCSS or based on concepts in the standards for grades K-5 but are obviously more difficult than a grade K-5 problem. When appropriate, these problems were given the code SMP for Standards for Mathematical Practice. The CCSS include the Standards for Mathematical Practice along with the Standards for Mathematical Content. The SMPs are (1) Make sense of problems and persevere in solving them; (2) Reason abstractly and quantitatively; (3) Construct viable arguments and critique the reasoning of others; (4) Model with mathematics; (5) Use appropriate tools strategically; (6) Attend to precision; (7) Look for and make use of structure and (8) Look for and express regularity in repeated reasoning. MATHCOUNTS 2018-2019 53 PROBLEM INDEX (2) (4) (3) (3) (4) (5) (5) (5) (4) (5) (6) 6.G.2 7.G.6 7.G.6 G-GMD.3 7.G.6 G-GMD.3 G-GMD.4 G-GMD.3 7.G.6 G-GMD.3 8.G.9 1 Transformations Stretch1 46 (3) 100 (4) 6.G.1 123 (4) 7.G.6 129 (5) SMP 148 (5) 8.G.1 158 (5) 6.G.3 CCSS 8.G.3 54 6.G.1 (2) (4) (4) (4) (4) (3) (5) (3) (5) (5) (4) (4) (5) (4) (4) (5) (5) (5) (4) (5) (4) (3) (3) (4) (3) (5) (5) (3) (5) (5) (5) (5) (6) (4) (5) (7) 4.G.3 G-C.2 8.G.8 4.G.2 G-SRT.5 G-CO.10 7.RP.2 4.MD.3 7.G.5 G-C.2 SMP 7.G.5 G-C.2 7.G.5 8.G.8 G-SRT.5 7.G.5 G-SRT.6 8.G.7 S-CP.9 7.G.6 8.G.7 SMP 7.G.6 6.G.1 8.G.5 7.G.6 SMP 8.G.8 7.G.6 G-SRT.6 7.G.6 7.G.1 8.G.7 8.G.7 G-C.2 ALGEBRAIC EXPRESSIONS & EQUATIONS 36 116 153 182 191 192 199 202 208 212 242 38 56 58 60 68 70 77 78 80 82 84 90 110 120 125 126 133 136 144 146 150 154 166 180 181 187 189 193 196 211 220 230 233 236 240 250 G ENERAL M ATH 6.RP.3 4.MD.2 6.RP.3 4.MD.2 SMP 6.RP.3 4.MD.2 4.MD.2 4.MD.2 6.RP.3 6.G.1 7.G.5 SMP 6.RP.3 4.MD.2 4.OA.2 6.G.1 PLANE G EOMETRY M EASUREMENT (1) (4) (4) (3) (3) (4) (3) (3) (3) (4) (2) (3) (2) (3) (3) (2) (4) COORDINATE G EOMETRY 1 2 3 4 5 6 7 8 9 10 42 61 81 97 102 112 248 SOLID G EOMETRY It is difficult to categorize many of the problems in the MATHCOUNTS School Handbook. It is very common for a MATHCOUNTS problem to straddle multiple categories and cover several concepts. This index is intended to be a helpful resource, but since each problem has been placed in exactly one category and mapped to exactly one Common Core State Standard (CCSS), the index is not perfect. In this index, the code 9 (3) 7.SP.3 refers to problem 9 with difficulty rating 3 mapped to CCSS 7.SP.3. For an explanation of the difficulty ratings refer to page 49. For an explanation of the CCSS codes refer to page 53. 34 35 41 45 49 55 62 67 76 86 87 94 96 109 111 121 122 130 142 147 151 161 176 186 201 217 231 237 245 246 (2) (3) (2) (3) (3) (4) (3) (4) (5) (5) (4) (3) (4) (4) (3) (3) (3) (4) (3) (5) (4) (4) (3) (3) (4) (4) (5) (7) (5) (5) 3.NBT.2 8.EE.8 6.EE.3 8.EE.8 6.EE.9 A-SSE.2 8.EE.7 7.EE.4 8.EE.8 8.EE.8 8.EE.8 6.EE.7 A-SSE.2 6.EE.3 6.EE.9 6.EE.3 8.EE.8 6.RP.3 6.RP.3 A-REI.4 A-APR.5 6.EE.7 6.RP.3 6.EE.2 7.RP.3 A-CED.1 A-REI.4 A-SSE.3 A-REI.2 8.EE.8 52 (1) 7.NS.2 63 (2) SMP 69 (2) 3.NBT.2 174 (2) 6.NS.3 203 (2) 4.MD.2 MATHCOUNTS 2018-2019 F-BF.2 139 (5) F-BF.2 156 (5) F-BF.2 160 (6) A-APR.5 195 (3) F-IF.3 222 (5) 8.G.7 235 (4) F-BF.2 249 (5) F-BF.2 57 (3) SMP 91 (4) 6.EE.7 92 (5) SMP 162 (7) 8.EE.8 218 (3) SMP 232 (4) SMP 234 (3) SMP 244 (5) SMP PROBABILITY, COUNTING & COMBINATORICS Expected Value Stretch2 2 33 (2) 7.SP.7 53 (2) S-CP.9 83 (4) 7.SP.8 99 (5) S-CP.9 103 (3) S-CP.9 104 (4) 7.SP.8 106 (3) S-CP.9 119 (5) 7.SP.8 138 (4) S-CP.9 141 (3) 7.G.6 152 (4) SMP 157 (5) 7.SP.8 164 (5) S-CP.9 169 (7) 7.SP.8 170 (4) S-CP.9 210 (5) S-CP.9 213 (4) S-CP.9 214 (5) 7.SP.8 223 (5) S-CP.9 238 (5) 7.SP.8 (3) (4) (4) (4) (3) (4) (4) (5) (3) (3) (6) (4) (5) (3) (3) (2) (4) (6) 6.SP.2 6.SP.5 6.SP.2 6.SP.2 6.SP.5 6.SP.2 SMP 6.SP.2 6.SP.5 7.SP.6 SMP 6.SP.5 6.SP.5 6.SP.2 6.SP.2 6.SP.4 6.SP.5 G-GMD.4 43 54 71 89 135 155 165 172 177 178 185 190 194 206 207 215 224 227 229 239 (2) (3) (4) (4) (3) (4) (3) (3) (4) (3) (3) (4) (3) (4) (4) (4) (4) (4) (4) (4) 6.RP.3 6.RP.3 6.RP.3 6.RP.3 6.RP.3 6.RP.3 G-SRT.5 7.RP.3 6.RP.1 6.RP.3 7.RP.3 7.RP.3 7.RP.3 6.RP.3 6.RP.3 7.RP.3 6.RP.3 6.RP.3 7.NS.3 7.NS.3 31 40 72 124 137 163 171 200 221 (2) (3) (3) (4) (4) (5) (3) (3) (3) S-CP.9 SMP SMP SMP SMP SMP 7.NS.3 SMP 7.SP.5 NUMBER THEORY (4) 48 66 75 85 95 108 115 118 128 140 149 159 168 175 179 204 216 226 PERCENTS & FRACTIONS 132 STATISTICS SMP PROPORTIONAL REASONING (3) LOGIC SEQUENCES, SERIES & PATTERNS PROBLEM SOLVING (M ISC.) 93 32 37 39 44 59 64 73 74 79 88 98 105 107 113 114 117 127 131 134 145 167 173 183 184 188 197 225 228 243 (2) (3) (3) (2) (3) (2) (2) (3) (3) (3) (2) (4) (4) (2) (5) (3) (4) (3) (2) (4) (2) (3) (3) (2) (3) (4) (3) (4) (5) 3.NBT.2 6.SP.2 5.NBT.2 8.EE.2 8.EE.8 4.OA,3 SMP 8.EE.2 6.NS.1 4.OA.4 SMP 6.RP.3 6.SP.2 6.RP.3 F-BF.2 6.EE.7 N-RN.2 6.NS.4 4.OA.4 6.NS.4 7.NS.2 SMP 6.NS.4 SMP S-CP.9 SMP SMP 4.OA.4 SMP 47 (3) 7.NS.3 50 (3) 7.RP.3 51 (3) 7.NS.3 65 (4) 7.NS.3 101 (4) 6.EE.7 143 (3) 8.EE.7 198 (3) 7.RP.3 205 (5) 7.EE.4 209 (6) 7.NS.3 219 (4) 7.RP.3 241 (3) 7.RP.3 247 (5) 6.RP.3 CCSS S-MD.2 MATHCOUNTS 2018-2019 55 SOLUTIONS The solutions provided here are only possible solutions. It is very likely that you or your students will come up with additional—and perhaps more elegant—solutions. Happy solving! Measurement Stretch 1. When the left and right sides are at the same height or “balanced,” the weight on the left side equals the weight on the right side. Therefore, 3 + 19 + w = 6 + 28, where w is the weight of the object. Solving for w, we get 22 + w = 34 w = 34 − 22 = 12 units. 2. We can solve this problem by using a series of conversion factors (ratios that compare two different units and equal 1). Since the weight of a small clip is 2/3 that of a large clip, it follows that, in terms of weight, 2 large clips = 3 small clips. Since 2 tacks weigh the same amount as a large clip, we have the ratios 2 large clips/3 small clips and 2 tacks/1 large clip, both of which equal 1. Multiplying the given number of small clips by these ratios, we determine that 12 small clips × 2 large clips 3 small clips × 2 tacks 1 large clip = 48/3 = 16 tacks. 3. Let B, D and P represent Bems, Dems and Pems, respectively. From the given information we can write the following equations: B + 7D = 4P and 2B + D = P. It follows that B + 7D = 4(2B + D) B + 7D = 8B + 4D 7B = 3D. Therefore, 7 Bems equal 3 Dems. 4. In a second, the car travels 1 second × 99 miles 1 hour × 5280 feet 1 mile × 1 hour 60 minutes × 1 minute 60 seconds × 0.305 meter 1 foot ≈ 44.3 meters. 5. With the error range being ±1 cm, the maximum percent error when measuring 24 cm is 1/24 × 100% ≈ 4.2%. 6. We can determine the weight of an object on the right side of the balance by placing weight(s) on one or both sides until the sides are balanced. For example, if the sides are balanced by placing the 1-gram and the 3-gram weights on the left side, then the object on the right side weighs 4 grams. Similarly, if the sides are balanced by placing the 3-gram weight on the left side and the 1-gram weight on the ride side with the object, then the object weighs 3 − 1 = 2 grams. In this manner, 1 to 12 grams of weight can be measured using the 1-, 3- and 8-gram weights. With the 26gram weight included, we can measure between 26 − 12 = 14 grams and 26 + 12 = 38 grams. Therefore, the smallest positive integer number of grams that cannot be measured using these four weights is 13 grams. 7. Since 2 cups = 1 pint, 2 cups and 7 pints combined are 8 pints, and 8 pints = 1 gallon. Since 4 quarts = 1 gallon, 8 quarts = 2 gallons. Finally, 11 half-gallons = 11/2 = 5½ gallons. So, the total amount of lemonade Clem has is 1 + 2 + 5½ = 8 ½ gallons. 8. From the problem we can write the following conversion factors: 2 Blams/15 Droms and 5 Droms/28 Klegs. Multiplying, we see that 2 Blams 15 Droms × 5 Droms 28 Klegs = 1 Blam 42 Klegs . So, in a Blam there are 42 Klegs. 9. With the given information, we get a ratio of 1 ounce 1000 grams × 454 grams 1 pound × 1 pound 16 ounces ≈ 0.028. 10. From the given information, we can write the equation f + 5b = 2(3f + 2b), where b and f represent the costs of burgers and fries, respectively. Simplifying, we get f + 5b = 6f + 4b b = 5f. So, a burger costs 5 times as much as an order of fries. Expected Value Stretch 11. The expected value of a roll is (1)(1/3) + (2)(1/15) + (3)(1/6) + (5)(1/5) + (8)(2/15) + (13)(1/10) = 1/3 + 2/15 + 3/6 + 5/5 + 16/15 + 13/10 = (10 + 4 + 15 + 30 + 32 + 39)/30 = 130/30 = 4 ¹⁄ ³. 12. The expected prize value is (5)(0.10) + (10)(0.45) + (15)(0.25) + (20)(0.20) = 0.5 + 4.5 + 3.75 + 4 = $12.75. 13. The expected value of a roll is (1)(1/10) + (2)(2/10) + (3)(3/10) + (4)(4/10) = (1 + 4 + 9 + 16)/10 = 30/10 = 3. 14. The table shows the area and probability of being selected for each tile. Based on this, the expected value of the chosen tile’s area is (4)(4/13) + (9)(9/13) = 16/13 + 81/13 = 97/13 cm2.. Side length 2 cm 3 cm Area 4 cm2 9 cm2 Probability 4/13 9/13 TOTAL AREA = 13 cm2 56 MATHCOUNTS 2018-2019 Points 100 50 25 10 Area π in2 8π in2 16π in2 24π in2 Probability 1/49 8/49 16/49 24/49 15. The successive circles that border the 100-, 50-, 25- and 10-point regions have radii 1, 3, 5 and 7 inches, respectively. The area of the 100-point region is π(12) = π in2. The area of the 50-point region is π(32) − π(12) = 9π − π = 8π in2. The area of the 25-point region is π(52) − π(32) = 25π − 9π = 16π in2. 2 TOTAL AREA = 49π in Finally, the area of the 10-point region is π(72) − π(52) = 49π − 25π = 24π in2. The table shows the area for the region having each point value. Based on this, the expected number of points is (100)(1/49) + (50)(8/49) + (25)(16/49) + (10)(24/49) = (100 + 400 + 400 + 240)/49 = 1140/49 ≈ 23.3 points. 16. For the deck of cards described, since the 40 cards are all equally likely to be drawn, the probability of randomly drawing any one of the 40 cards is 1/40. The expected value of the card drawn, then, is (1)(1/40) + (2)(1/40) + (3)(1/40) + ⋯ + (40)(1/40) = (1/40)(1 + 2 + 3 + ⋯ + 40) = (1/40)(41)(40)/2 = 41/2 = 20.5. 17. Each of the 125 unit cubes can have 0, 1, 2 or 3 painted faces. Each of the 8 unit cubes cut from a corner of the original cube has three painted faces. The probability Luke chooses one of these is 8/125. There are 3 unit cubes cut from each of the 12 edges of the original cube, excluding the corner unit cubes, and all 3 × 12 = 36 of these have two painted faces. The probability Luke chooses one of these is 36/125. Of the remaining unit cubes, there are 9 that are cut from each of the 6 faces of the original cube, all 9 × 6 = 54 of which have one painted face. The probability Luke chooses one of these is 54/125. The remaining 125 − (8 + 36 + 54) = 125 − 98 = 27 unit cubes, which are from the interior of the original cube, have no painted faces. The probability Luke chooses one of these is 27/125. The expected value of the number of painted faces on his chosen unit cube is (3)(8/125) + (2)(36/125) + (1)(54/125) + (0)(27/125) = (24 + 72 + 54 + 0)/125 = 150/125 = 1.2 faces. 18. During each round, Dinara will win 0 points or 1 point. Since there is a 70% = 7/10 probability of winning 1 point in each round, it follows that there is a 1 − 7/10 = 3/10 probability of winning 0 points. A property of E is that it is a linear function of the random variable, so the expected value of the sum of random variables equals the sum of their expected values. In other words, for random variables X and Y, E(X + Y) = E(X) + E(Y). That means the expected value of the total score after three rounds is equal to the sum of the expected point values for the three rounds. We will calculate both the expected value of the total score and the sum of the individual expected values to show that the quantities are equal. After three rounds, Dinara’s total score can be 0, 1, 2 or 3 points. A total score of 0 results from winning 0 points in all three rounds, so P(0) = (3/10)(3/10)(3/10) = 27/1000. A total score of 1 results from winning 1 point in only one round. Since this can occur in three ways, P(1) = (3)(7/10)(3/10)(3/10) = 189/1000. A total score of 2 results from winning 0 points in only one round. Since this can occur in three ways, P(2) = (3)(3/10)(7/10)(7/10) = 441/1000. A total score of 3 results from winning 1 point in all three rounds, so P(3) = (7/10)(7/10)(7/10) = 343/1000. The expected value of Dinara’s total score is (0)(27/1000) + (1)(189/1000) + (2)(441/1000) + (3)(343/1000) = 2.1 points. Now notice that the expected point value for each round is 0(3/10) + 1(7/10) = 7/10, so the sum of the expected point values for the three rounds is (3)(7/10) = 21/10 = 2.1 points. 19. Recall that the expected value of the sum of random variables equals the sum of their expected values. Let’s solve this problem by calculating the sum of the expected values for each of the five people. Since each of the integers from −5 to 5, inclusive is equally likely to be picked, the probability of each of the five people randomly picking any of the 11 integers is 1/11. So, the expected value of the integer each person randomly picks is (−5)(1/11) + (−4)(1/11) + ⋯ + 4(1/11) + 5(1/11) = (1/11)(−5 + (−4) + ⋯ + 4 + 5) = (1/11)(0) = 0. By symmetry, there is no reason for the expected value to be positive or negative, which further explains why the sum of the expected values for all five people is (5)(0) = 0. 20. Let’s look at this as the experiment of placing a single jelly bean in one of 10 jars repeated 1000 times. For each iteration, the probability that the jelly bean is placed in the leftmost jar is 1/10 and the probability it is placed in one of the other 9 jars is 9/10. So, the expected number of jelly beans in the jar for each iteration is 0(9/10) + 1(1/10) = 1/10. Since the expected value of the sum of random variables equals the sum of their expected values, it follows that the expected number of jelly beans in the leftmost jar is (1000)(1/10) = 100 jelly beans. Transformations Stretch 21. When a point is translated right 4 units and up 3 units, we increase the x-coordinate by 4 and increase the y-coordinate by 3. So, the final image of P(−3, 2) is P″(−3 + 4, 2 + 3) = (1, 5). The distance from P to P″ is √[(−3 − 1)2 + (2 − 5)2] = √(16 + 9) = √25 = 5 units. 22. When a point is translated left 3 units and down 4 units, we decrease the x-coordinate by 3 and decrease the y-coordinate by 4. So, the image of B(−3, 4) is C(−3 − 3, 4 − 4) = (−6, 0). Since AB = √[(0 + 3)2 + (0 − 4)2] = √(9 + 16) = √25 = 5 units, BC = √[(−3 + 6)2 + (4 − 0)2] = √(9 + 16) = 5 units and CA = √[(0 + 6)2 + (0 − 0)2] = √(36 + 0) = √36 = 6 units, it follows that DABC has perimeter 5 + 5 + 6 = 16 units. 23. When a point is reflected across the x-axis, the x-coordinate of the image is the same as that of the pre-image, but the y-coordinate of the image is the opposite of that of the pre-image. So, the image of Q(−3, 4) reflected across the x-axis is Q′(−3, −4). A point and its reflection must be equidistant from the line of reflection, in this case x = 2. Since Q′(−3, −4) is | −3 − 2 | = 5 units from this line, it follows that its reflection is Q″(2 + 5, −4) = (7, −4). 24. Rewriting x − 2y = −6 in slope-intercept form as y = (1/2)x +3, we see that this line has slope 1/2. The point S(1, 6) and its reflection across this line must both be points on a line that is perpendicular to that line of reflection. The perpendicular line containing S and S′ has slope −2 and is given by the equation y − 6 = −2(x − 1) y − 6 = −2x + 2 y = −2x + 8. The intersection of these two lines is the midpoint of segment SS′. We can determine the coordinates of the midpoint by setting the two expressions for y equal to each other and solving for x. We get (1/2)x + 3 = −2x + 8 x + 6 = −4x + 16 5x = 10 x = 2. Substituting back into the equation y = −2x + 8, we see that y = −2(2) + 8 = 4. So, the midpoint of segment SS′ has coordinates (2, 4). Using the midpoint formula, we can determine the coordinates of S′ as follows: 2 = (x + 1)/2 4 = x +1 x = 3 and 4 = (y + 6)/2 8 = y + 6 y = 2. The sum of the coordinates of S′(3, 2) is 3 + 2 = 5. MATHCOUNTS 2018-2019 57 25. When a point is rotated 90 degrees clockwise about the origin, in general, the x-coordinate of the image is the y-coordinate of the pre-image, and the y-coordinate of the image is the opposite of the x-coordinate of the pre-image. The image of D(−5, −3) rotated 90 degrees clockwise about (0, 0) is D′(−3, 5). 26. When rotating a figure about a point that is not the origin, it helps to think of the figure’s location as if the rotation point were the origin. Once the location of the image (after rotation) is determined, we can rewrite its coordinates in terms of the actual origin. If we think of F(5, 4) as (5 − 5, 4 − 4) = (0, 0), then we need to think of the coordinates of E(3, −1) as (3 − 5, −1 − 4) = (−2, −5). When a point is rotated 90 degrees counterclockwise about the origin, in general, the x-coordinate of the image is the opposite of the y-coordinate of the pre-image, and the y-coordinate of the image is the x-coordinate of the pre-image. So, the image of (−2, −5) rotated 90 degrees counterclockwise about (0, 0) is (5, −2). In terms of the actual origin, the image has coordinates (5 + 5, −2 + 4) = (10, 2). 27. When a point is dilated by a scale factor of k about the origin, we multiply both the x- and y-coordinate by k to determine the coordinates of the image. So, the image of G(−2, 3) dilated by a factor of 2/3 about (0, 0) is G′((−2)(2/3), (3)(2/3)) = (−4/3, 2). The image of H(4, 7) dilated by a factor of 2/3 about (0, 0) is H′((4)(2/3), (7)(2/3)) = (8/3, 14/3). The sum of the coordinates of G′ and H′ is −4/3 + 2 + 8/3 + 14/3 = (−4 + 6 + 8 + 14)/3 = 24/3 = 8. Alternatively, we could multiply the sum of the coordinates of G and H by 2/3 to get (2/3)(−2 + 3 + 4 + 7) = (2/3)(12) = 8. 28. Similar to rotating a figure about a point that is not the origin, when dilating a figure about a point that is not the origin, let’s think of the figure’s location as if the center of dilation were the origin. Once the location of the image (after dilation) is determined, we’ll rewrite its coordinates in terms of the actual origin. If we think of K(2, 2) as ( 2 − 2, 2 − 2) = (0, 0), then we need to think of the coordinates of J(4, 8) as ( 4 − 2, 8 − 2) = (2, 6). The image of J(4, 8) dilated by 3/2 about (0, 0) is ((2)(3/2), (6)(3/2)) = (3, 9). In terms of the actual origin, the image has coordinates J′(3 + 2, 9 + 2) = (5, 11). The product of the coordinates of J′ is 5 × 11 = 55. 29. If we think of M(3, 2) as (3 − 3, 2 − 2) = (0, 0), then we need to think of the coordinates of L(−2, 4) as (−2 − 3, 4 − 2) = (−5, 2). The image of (−5, 2) rotated 90 degrees clockwise about (0, 0) is (2, 5). The image of (2, 5) dilated 3/2 with center of dilation (0, 0) is ((2)(3/2), (5)(3/2)) = (3, 15/2). The area of the triangle doesn’t change if we translate it back to the right 3 and up 2, so let’s use the current vertices L(−5, 2), M(0, 0) and N(3, 15/2) to determine its area. Using the shoelace method (as shown), we have (1/2)[(−5 × 0 + 0 × 15/2 + 3 × 2) − (2 × 0 + 0 × 3 + 15/2 × (−5)] = (1/2)(6 + 75/2) = (1/2)(12 + 75)/2 = (1/2)(87/2) = 87/4. So the area of DLMN is 87/4 units2. Shoelace Method: The area of a closed figure with vertices A(x1, y1), B(x2, y2) and C(x3, y3) is (x1y2 + x2y3 + x3y1) − (x2y1 + x3y2 + x1y3) 2 It’s easy to remember these products if you notice that the arrows connecting factors resemble shoelaces. A x1 B x2 C x3 A x1 y1 y2 y3 y1 30. The line of reflection y = x − 2 has slope 1. The point R(−5, 3) and its reflection across this line must both be points on a line that is perpendicular to the line of reflection. The perpendicular line containing R and R′ has slope −1 and is given by the equation y − 3 = −1(x − (−5)) y − 3 = −x − 5 y = −x − 2. The intersection of these two lines is the midpoint of segment RR′. Setting the two expressions for y equal to each other and solving for x, we get x − 2 = −x − 2 2x = 0 x = 0. Substituting back into the equation y = x − 2, we see that y = 0 − 2 = −2. So, the midpoint of segment RR′ has coordinates (0, −2). Using the midpoint formula, we can determine the coordinates of R′ as follows: 0 = (x − 5)/2 0 = x − 5 x = 5 and −2 = (y + 3)/2 −4 = y + 3 y = −7. The image R′(5, −7) rotated 90 degrees clockwise about (0, 0) is R″(−7, −5). The distance from R to R″ is √[(−7 + 5)2 + (−5 − 3)2] = √(4 + 64) = √68 = 2√17 units. Warm-Up 1 31. The table shown is an organized list of all 7 combinations of quarters, dimes and/or nickels that have a total value of 40 cents. Q 0 0 0 0 0 1 1 D 0 1 2 3 4 0 1 N 8 6 4 2 0 3 1 32. The scale of the number line is 1/6, so x = 13 3 ⁄ 6 and y = 9 ² ⁄ 6. The difference is x − y = 13 3 ⁄ 6 − 9 ² ⁄ 6 = 4 ¹⁄ 6. 33. The coin that Ted is flipping is equally likely to land heads up or tails up. Since each flip is an independent event, which means that the previous flips have no influence over the next flip, the probability that the next flip will land heads up is 1/2. 34. The scientific order of operations requires that we do the multiplication and division in order from left to right before we do the addition and subtraction in order from left to right. The result is 9 + (5 × 3) – (8 ÷ 2) = 9 + 15 – 4 = 24 – 4 = 20. 35. Translating the English to algebra, we get 2 + 3x = 10x − 5. Now, solve for x by first adding 5 to each side of the equation to get 7 + 3x = 10x. Then subtract 3x from each side to get 7 = 7x. Finally, divide each side by 7 to get 1 = x, or x = 1. 36. The formula for the volume of a rectangular prism is given by a formula usually expressed as V = lwh, which is the product of the length, the width and the height of the prism. In this case, we are given the height, the width and the depth, a slightly different way to name the three dimensions of the prism. The volume is thus V = (3)(5)(7) = 105 cm3. 37. There are only two prime numbers between 20 and 30, namely 23 and 29. Their average is (23 + 29)/2 = 52/2 = 26. 38. There is only 1 line of symmetry in an isosceles right triangle. The line of symmetry is shown as a dotted line in the figure. 58 MATHCOUNTS 2018-2019 39. The number 1,000,000,000 can be written as 109 = [(2)(5)]9 = (29)(59). If we divide this by (28)(57), we get [(29)(59)]/[(28)(57)] = (29 − 8)(59 – 7) = (21)(52) = (2)(25) = 50. 40. One-third of the 630 people who own a cat also own a dog, which accounts for 630/3 = 210 people. Since each person surveyed owns a cat, a dog or both, there are 1000 – 630 = 370 people who own a dog but not a cat. So, a total of 210 + 370 = 580 people own a dog. Warm-Up 2 41. Substituting 10 for x, we get (3)(10) + (4)(10) + (5)(10) = (3 + 4 + 5)(10) = (12)(10) = 120. 42. The 12-gon has 12 sides, each of length 3 feet. Thus, the perimeter is (12)(3) = 36 feet. 43. The school zone speed limit of 15 mi/h is one-quarter of 60 mi/h = 88 ft/s. So, 15 mi/h is equivalent to 88/4 = 22 ft/s. 44. The number 64 is a power of the prime number 2, and the divisors of 64 are all powers of 2: 1, 2, 4, 8, 16, 32 and 64. Of these factors, 1, 4, 16 and 64 are the 4 divisors that are perfect squares. 45. Since d = 16, we can rewrite bd = c as c = 16b. Since c = 16b, we can rewrite ab = 2c as ab = (2)(16b) = 32b. The value of a must be 32. 46. We can think of the figure that Tom’s path enclosed as a 6-by-8 rectangle with a 3-by-4 rectangle missing, as shown. The area is (6)(8) − (3)(4) = 48 – 12 = 36 square blocks. You might also calculate the area of this region by decomposing it into a 5-by-6 rectangle with area (5)(6) = 30 and a 3-by-2 rectangle with area (3)(2) = 6 to get a total area of 30 + 6 = 36 square blocks. 5 3 47. (1/2 + 1/3) ÷ (1/4 − 1/5) = (3/6 + 2/6) ÷ (5/20 − 4/20) = (5/6) ÷ (1/20) = (5/6)(20/1) = (5/3)(10/1) = 50/3. 4 2 48. The sum of the four fractions is 2/3 + 7/9 + 1/4 + 5/16 = (6 + 7)/9 + (4 + 5)/16 = 13/9 + 9/16 = (208 + 81)/144 = 289/144. To find the arithmetic mean of the four numbers, we need to divide this sum by 4 or, equivalently, multiply by 1/4. The result is (289/144)(1/4) = 289/576. 49. If we subtract a from both sides of the equation a + b = a − b, we get b = –b. The only number equal to its own opposite is zero. So, in order for the sum and difference of a and b to be equal, the value of b must be zero. We can simplify the expression as follows: (a2b + a + b − ab2)/(a − b) = a2(0) + a + 0 − a(02)/(a − 0) = a/a = 1. 50. Seventy-five percent of a number is the same as 3/4 = 6/8 of the number. Since 88 is 6/8 of the number and since 3/8 is half of 6/8, 3/8 of the number must be half of 88, which is 44. Warm-Up 3 51. If two-thirds of Sammi’s jelly beans are yellow, then the 5 red and 7 orange jelly beans must account for the remaining one-third of the jelly beans. There must be (5 + 7)(3) = (12)(3) = 36 jelly beans in Sammi’s cup. 52. In the sum 0.49 + 0.53 + 0.55 + 0.47 + 0.48, each addend is about 0.50. So, the sum is about (5)(0.50) = 2.50. To find the exact total, let's examine by how much each addend exceeds or falls short of 0.50. We have, in hundredths, −1, +3, +5, −3 and −2, giving us a net of +2 hundredths. Therefore, the actual total is 2.50 + 0.02 = 2.52. 53. Each outfit will include Allie’s pair of jeans. For each of the four shirts, there are three sweaters Allie can wear, resulting in (4)(3) = 12 outfits so far. Each of these 12 combinations of shirt and sweater can be worn with either of the two scarves, so there are (12)(2) = 24 outfits. The Fundamental Counting Principle states that if independent events M and N can occur in m ways and n ways, respectively, then the event M followed by N can occur in m × n ways. So, we can get the same answer directly by multiplying (1)(4)(3)(2) = 24 outfits. 54. Since it takes Destiny 3/4 minute to run 1/8 mile, it will take her (8)(3/4) = 6 minutes to run a mile. Since it takes her 4 minutes to walk 1/4 mile, it will take her (4)(4) = 16 minutes to walk a mile. So, to run a mile and then walk a mile, it will take her 6 + 16 = 22 minutes. 55. The difference of two perfect squares x2 − y2 can be rewritten as the product (x + y)(x − y). That means the expression 272 − 232 can be rewritten as the product (27 + 23)(27 − 23) = (50)(4) = 200. 56. When two chords intersect in a circle, the point of intersection cuts each chord into two segment, and the product of the lengths of the two segments of one chord is equal to the product of the lengths of the two segments of the other chord. In this case chords BD and CE intersect at A, with AB = 3 units, AD = 6 units and AC = 8 units. So, (3)(6) = (8)(AE) AE = 18/8 = 2 ¼ units. MATHCOUNTS 2018-2019 59 M T W Th F Grace Becca Carmen Davis Ernie ? ? Frank Total 2 3 2 4-5 3-4 57. The table shown organizes the given information and shows that, regardless of which day Ernie chooses to attend, there will be at least three of the six people in attendance on 3 days of the conference. 58. The bus travels a total of 8 + 7 = 15 miles west and 8 miles north. If we make a right triangle with legs of lengths 15 miles and 8 miles, the direct route would be along the hypotenuse of this triangle. We can use the Pythagorean Theorem to find this distance: 152 + 82 = c2 225 + 64 = c2 289 = c 2 c = √289 = 17. Therefore, if the bus traveled from Kevin’s house to the middle school along this straight path, the distance would be 17 miles, which is shorter than the current route by (15 + 8) – 17 = 23 – 17 = 6 miles. 59. A trillion is 1012, so twenty-one-and-a-half trillion can be written as 21.5 × 1012. In scientific notation, it is 2.15 × 1013, and the exponent is 13. 60. A diagonal can be drawn from each of the 17 vertices of a 17-sided polygon to 14 other vertices (no diagonal can be drawn to the two adjacent vertices or to the vertex itself). If we multiply 17 by 14, we have counted each diagonal at both ends, so we need to divide this product by 2. Thus, there are (17)(14) ÷ 2 = 119 diagonals. Warm-Up 4 61. Working backward, the supplement of a 163-degree angle is 180 – 163 = 17 degrees. The complement of a 17-degree angle is 90 – 17 = 73 degrees. The measure of an angle’s complement is always 90 degrees less than the measure of its supplement. Indeed, 163 – 90 = 73 degrees. 62. We have x(10) − (4)(3) = 38. Solving for x, we get 10x − 12 = 38 10x = 50 x = 5. 63. Miranda must leave 21 + 5 = 26 minutes before 9:19 a.m. From 9:00 a.m. to 9:19 a.m. is 19 minutes, so she needs to leave 26 − 19 = 7 minutes before 9:00 a.m., which is 8:53 a.m. 64. The greatest product Ryan can get is (5)(7) = 35, and the least product he can get is (2)(3) = 6. The absolute difference is 35 – 6 = 29. 65. The group of 20 students includes 4 sets of five students. So, if the group purchases 4 sets of five $12 tickets, they will get 4 free $3 cotton candies and 4 free $2 sodas. While this is not enough to accommodate the (1/2)(20) = 10 students who want cotton candy and the (1/4)(20) = 5 students who want soda, with the promotion, the total savings is (4)(3) + (4)(2) = 12 + 8 = $20 or 20.00. 66. Since the mean weight of the four potatoes is 12 ounces, their combined weight is (4)(12) = 48 ounces. The median weight of 11 ounces is the average of the weights of the two potatoes that are neither the lightest nor the heaviest. The sum of their weights must be (11)(2) = 22 ounces. That leaves 48 − 22 = 26 ounces for the combined weight of the lightest and heaviest potatoes. The lightest potato must weigh at least 1 ounce, making the heaviest potato 26 − 1 = 25 ounces. The greatest possible difference in these weights is 25 – 1 = 24 ounces. 67. The sum of the reciprocals of the focal lengths is 1/80 + 1/200 + 1/400 = (5 + 2 + 1)/400 = 8/400 = 1/50. The reciprocal of this corresponds to a focal length of 50 mm. 68. Given that the lengths of the shortest sides of the similar figures are in the ratio A/B = 4/3, it follows that each side of hexagon B is 3/4 the length of the corresponding side of hexagon A. Since area is two-dimensional, the ratio of the hexagons’ areas is proportional to the square of the ratio of their side lengths, meaning the area of hexagon B is (3/4)2 = 9/16 the area of hexagon A. The area of hexagon A is 48 in2, so the area of hexagon B is (48)(9/16) = 27 in2. 69. To find the value of the expression, we must calculate the products before we calculate the sums. Noting that the products in the second half of the expression are the same as those in the first half, we can simply compute (2)(12 + 22 + 30 + 36 + 40 + 42) = (2)(182) = 364. 70. A tour around the perimeter of a regular polygon with 90 sides would make one 360-degree rotation in 90 equal turns of 360/90 = 4 degrees. This is the measure of each exterior angle, so each interior angle must be 180 – 4 = 176 degrees. Warm-Up 5 71. After subtracting the $15 flat fee, we see that Kayla charges 24 – 15 = $9 to mow the 20-foot by 40-foot lawn of total area 800 ft2. The 60-foot by 80-foot lawn has total area 4800 ft2, which is 4800/800 = 6 times the area of the 20-foot by 40-foot lawn. For the larger lawn, Kayla would charge (6)(9) + 15 = 54 + 15 = $69 or 69.00. 72. The least expensive route is from Nash to Cantor to Euclid to Galois to Wiles. The total cost is 147 + 98 + 58 + 89 = $392 or 392.00. 73. The symbols, from left to right, represent 3, 2000, 400, 10 and 50,000. The total represented is 3 + 2000 + 400 + 10 + 50,000 = 52,413. 60 MATHCOUNTS 2018-2019 74. In order, the first few positive perfect cubes are 1, 8, 27, 64, 125. In order, the first few perfect squares are 1, 4, 9, 16, 25, 36, 49. Yixin and Aly can only be thinking of 27 and 4, respectively, since 27 + 4 = 31 and no other pair has this sum. The desired product is (27)(4) = 108. 75. The list of numbers, from least to greatest, is 1, 2, 3, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9. The possible min and max pairs that yield a range of 2 are as follows: 1 and 3 (11 integers erased), 2 and 4 (11 integers erased), 3 and 5 (10 integers erased), 4 and 6 (9 integers erased), 5 and 7 (8 integers erased), 6 and 8 (8 integers erased), 7 and 9 (8 integers erased). The least possible number of integers Jalacia could have erased is 8 integers. 76. If we use r for the price of a rock tumbler, s for the price of a spy pen and p for the price of a puzzle, then we can write a system of three equations: 2r + s = 74, 2p + s = 50, r + 2p = 57. Subtracting the second equation from the third, we get r − s = 7 r = s + 7. Substituting this into the first equation, we get 2(s + 7) + s = 74, which we can solve for s as follows: 2s+ 14 + s = 74 3s = 60 s = 20. Now that we know the spy pen costs $20, we can substitute this into the second equation and solve for p as follows: 2p + 20 = 50 2p = 30 p = 15. One puzzle costs $15 or 15.00. 77. The Angle Bisector Theorem says that the ratio of lengths on one side of the bisector equals the ratio of lengths on the other side of the bisector. In this case, we have x/5 = 6/4. Cross multiplying, we get 4x = 30 x = 30/4 = 15/2 cm. 78. The total horizontal length across the lower half of the figure must equal 10 cm, although it is split into two segments. Likewise, the total vertical length along the right sides of the figure must equal 15 cm. The perimeter of the figure is just 2(15 + 10) = (2)(25) = 50 cm. 79. The expression for a can be simplified to a = 2 + 3/(4 + (5/6)) = 2 + 3/(29/6) = 2 + 18/29 = 76/29. Likewise, b = 6 + 5/(4 + (3/2)) = 6 + 5/(11/2) = 6 + 10/11 = 76/11. The quotient of these two fractions is a/b = (76/29) ÷ (76/11) = (76/29)(11/76) = 11/29. 80. Let the interior angle be i and the exterior angle be e. Then i + e = 180 and i − e = 100. If we add these two equations, we get 2i = 280, so i = 140. The interior angle is 140 degrees, and the exterior angle is 40 degrees. If the regular polygon has n sides, then a walk around the perimeter of the n-gon would consist of n turns of 40 degrees for a total of 360 degrees. Therefore, the polygon must have 360/40 = 9 sides. Warm-Up 6 81. Because 6 feet 2 inches = (6)(12) + 2 = 74 inches and 4 feet 8 inches = (4)(12) + 8 = 56 inches, the increase is 74 – 56 = 18 inches. E 82. Since quadrilateral AEBF is a square, we know that angles AEB and AFB are right angles. This means that line segments AE and AF are tangent to circle B at points E and F, respectively. The diameters of circle A that coincide with segments AE and AF create the two shaded 90-degree sectors shown. Every line through point A that intersects the shaded half of circle A also intersects circle B. So, half of the lines through A intersect both circles, and the requested probability is 1/2. A B F 83. When two items are to be chosen from a group of six, we say that the number of ways this can be done is “six choose two.” We reason as follows: The first pick can be any of the six items, and the second pick can be any of the remaining five items, but the same two items can be chosen in either order, so there are (6)(5)/2 = 15 ways to do it. If Lauren picks the 10, then any of the five other numbers will create a product that is a multiple of 10. If she picks the 5, then there are two even numbers (excluding the 10, which is already accounted for) that will create a product that is a multiple of 10. Thus, there are 5 + 2 = 7 ways to get a product of 10 out of the 15 possible pairs of cards, so the probability is 7/15. 84. Each vertex of an n-gon can be connected to n – 3 other vertices by a diagonal, but n(n − 3) counts every diagonal at both ends, so we need to divide by 2 to get the number of diagonals. Based on this, each square has (4)(1)/2 = 2 diagonals, each pentagon has (5)(2)/2 = 5 diagonals and each octagon has (8)(5)/2 = 20 diagonals. Alana’s four squares, five pentagons and six octagons have a total of (4)(2) + (5)(5) + (6)(20) = 8 + 25 + 120 = 153 diagonals. Marie draws n hexagons, each of which has (6)(3)/2 = 9 diagonals. Marie has the same number of diagonals as Alana has, so n = 153/9 = 17. 85. All four-digit positive integers less than 2000 have the digit 1 in the thousands place, and there are four combinations of four digits that have a sum of 24. They are 1, 5, 9 and 9; 1, 6, 8 and 9; 1, 7, 8 and 8; and 1, 7, 7 and 9, and they can be arranged to form the integers 1599, 1689, 1698, 1779, 1788, 1797, 1869, 1878, 1887, 1896, 1959, 1968, 1977, 1986 and 1995. In this ordered list, the middle number, or median, is 1878. 86. Squaring each side of the equation x2 − y2 = 7, we get (x2 − y2 )2 = 72 x4 − 2x2y2 + y4 = 49. The equation x = 12/y can be rewritten as xy = 12. Squaring each side of this equation gives us (xy)2 = 122 x2y2 = 144. Now substituting 144 for x2y2 in the equation x4 − 2x2y2 + y4 = 49, we get x4 − (2)(144) + y4 = 49 x4 – 288 + y4 = 49 x4 + y4 = 49 + 288 = 337. 87. Let d, c and p represent the costs of a hot dog, an ice cream cone and a slice of pizza, respectively. From the first sentence, we have 2d + c = 2.50, and from the second sentence, we have 2p + c = 3.50. Adding these two equations, we get 2d + 2c + 2p = 6 d + c + p = 3. That means the total cost of one hot dog, one ice cream cone and one slice of pizza is $3 or 3.00. 88. The prime factorization of 385 is 5 × 7 × 11. Each of these prime factors is either included or not included in a product that is a divisor of 385, so there are 23 = 8 divisors. From least to greatest, they are 1, 5, 7, 11, 35, 55, 77 and 385 itself. The sum of the 4th, 5th and 6th divisors is 11 + 35 + 55 = 101. MATHCOUNTS 2018-2019 61 89. Gavin’s pace is 9 minutes per mile, which is a speed of 1/9 mile per minute. Gavin’s 10-minute head start means he has already run 10/9 miles when Lars starts running. Lars’ pace is 7 minutes per mile, which is a speed of 1/7 mile per minute. The distance between Lars and Gavin decreases at a rate of 1/7 − 1/9 = (9 − 7)/63 = 2/63 mile per minute. Dividing 10/9 miles by 2/63 mile per minute, we see that it takes Lars (10/9) ÷ (2/63) = (10/9)(63/2) = (5)(7) = 35 minutes to reach Gavin. 90. If we drew all 72 lines from the center of a 72-gon to each of its vertices, the central angle between consecutive lines would be 360/72 = 5 degrees. The central angle that includes seven sides is (5)(7) = 35 degrees. Warm-Up 7 91. Let n be the total number of problems in the assigned set. On the first day, Neil solved (1/2)n + 1/2 = (n + 1)/2 problems, leaving (1/2)n − 1/2 = (n − 1)/2 unsolved. On the second day, he solved (1/2)[(n − 1)/2] + 1/2 = (n − 1)/4 + 2/4 = (n + 1)/4 problems, leaving (1/2)[(n − 1)/2] − 1/2 = (n − 1)/4 − 2/4 = (n − 3)/4 unsolved. The three problems Neil solved on the third day were the (n − 3)/4 unsolved problems from the second day. We have (n − 3)/4 = 3 n − 3 = 12 n = 15. So, the number of assigned problems was 15 problems. 92. The lowest winning score is the smallest possible sum of 7 of these cards, which is 1 + 2 + 3 + ⋯ + 7 = 7(1 + 7)/2 = 56/2 = 28. The corresponding losing score is 8 + 9 + 10 + ⋯ + 14 = 7(8 + 14)/2 = 154/2 = 77. The ordered pairs of scores are (28, 77), (29, 76), (30, 75), …, (52, 53), with the possible winning scores being 28, 29, 30, …, 52. That’s a total of 52 – 27 = 25 scores. 93. Only the units digit of each of the 2019 addends contributes to the units digit of their sum. Since the values of the factorials from 5! through 2019! each have a units digit of zero, we need only consider the units digit of the sum 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24. The sum of the units digits is 1 + 2 + 6 + 4 = 13, so the units digit of the sum of all 2019 factorials is 3. 94. We will use A, E and J for Alex’s, Evan’s and Joel’s heights, respectively. From the first sentence, we have A + E + J = 221. Since we’re asked to determine Alex’s height, let’s try to use the given information to express E and J in terms of A. From the second sentence, we have A = 8 + E A − 8 = E, and A = 5 + J A − 5 = J. Substituting A − 8 for E and A − 5 for J, the first equation can be rewritten as A + (A − 8) + (A − 5) = 221. Solving for A, we get 3A − 13 = 221 3A = 234 A = 234/3 = 78. So, Alex’s height is 78 inches. 95. The median of five ordered scores is the middle, or third, score in the list. By minimizing the third score, we’re essentially minimizing the first three scores. To do this, let’s maximize the fourth and fifth scores, making them each 100. Since the sum of the five scores is 204, it follows that the sum of the first three scores must be 204 − 200 = 4. The possible values for the first three scores, then, are 0, 0 and 4; 0, 1 and 3; 0, 2 and 2; and 1, 1 and 2. Of these, the last two options result in the least possible median, which is 2. 96. Note that 4 − x2 is a “difference of two squares” that can be factored into (2 + x)(2 − x). So, we have f(x) = (2 + x)(2 − x)/(x + 2) = 2 − x, and g(x) = x − 2. The value of f(x)/g(x) is (2 − x)/(x − 2) = (–1)(x − 2)/(x − 2) = –1. 97. There are 36 inches in a yard, so the 29 yards that Freddie Frog jumped are equal to (29)(36) = 1044 inches. Freddie traveled this distance in 261 jumps, so each jump was 1044/261 = 4 inches. With her hurt toe, Freddie’s jumps are now only 3 inches, so it will now take her 1044/3 = 348 jumps to go the same distance, which is 348 – 261 = 87 jumps more. 98. One divisibility rule for 11 states that a number is divisible by 11 if and only if the alternating sum of its digits is divisible by 11. (The alternating sum is the first digit minus the second plus the third an so on.) If we let x represent the missing digit, then 8 − 2 + 1 − x + 9 − 0 + 2 = 18 − x must be a multiple of 11. Solving 18 − x = 11 yields x = 7. Therefore, the seven-digit locker combination was 8217902. Alternatively, to determine if a number is divisible by 11, subtract the units digit from the number formed by the remaining digits. If the new number is divisible by 11, then so is the original number. We start by subtracting 821█ 90 − 2 = 821█ 88. Next, we subtract 821█ 8 − 8 = 821█ 0. Subtracting once more yields 821█ − 0 = 821█ . Now we can perform long division to see what value for the missing digit results in a remainder of 0. Since 821█ ÷ 11 = 747 when the missing digit is 7, the complete combination was 8217902. 99. Let’s first imagine that we put two donuts in each of the three boxes, since that is required. That leaves 12 − 6 = 6 donuts to place into three differently colored boxes. We could make an organized list of ordered triples showing all the possible ways to split those six donuts among the three boxes, starting with (0, 0, 6), (0, 1, 5), (0, 2, 4) and ending with (6, 0, 0). That might take a while. Alternatively, we can imagine that the six identical donuts are lined up in a row and that two “partitions” are used to separate the donuts into three groups. Using “stars” for the donuts and “bars” for the partitions, we can represent the previous examples starting with | | * * * * * *, | * | * * * * *, | * * | * * * * and ending with * * * * * * | |. The number of ways to arrange the eight symbols, six of which are stars and two of which are bars, is 8!/(6! × 2!) = (8)(7)/2 = 28 ways. C B A 100. One strategy for finding the area of irregular shapes on a grid is to subdivide the shape into figures for which we can determine the areas. The figure on the left shows one way of doing this. We have triangle BCD with area (1/2)(4)(2) = 4 units2, square ABFG with area (2)(2) = 4 units2, triangle EFD with area (1/2)(3)(2) = 3 units2 and triangle AGE with area (1/2)(2)(1) = 1 unit2. The total G C area is 4 + 4 + 3 + 1 = 12 units2. Alternatively, Pick’s Theorem says that for a simple polygon whose vertices have integer ) coordinates, its area can be calculated in terms of the number of interior lattice points i and the number of boundary lattice ' ' E B) ' ' ' ) D points b using the formula A = i + b/2 − 1. In the figure on the right, ' and ) are used to denote interior and boundary ) ' ' ' lattice points, respectively. There are 10 interior lattice points and 6 boundary lattice points (one of which is not a vertex). So, we A) ' ' have i = 10, b = 6 and A = 10 + 6/2 − 1 = 10 + 3 − 1 = 12 units2. F D ) E 62 MATHCOUNTS 2018-2019 Warm-Up 8 101. To make 10 kg of his 60% dark roast signature blend, Don will need to combine x kg of blend A, which is 80% dark roast, and (10 – x) kg of blend B, which is 20% dark roast. So, we have the equation 0.8x + 0.2(10 − x) = (0.6)(10). Solving for x, we get 0.8x + 2 − 0.2x = 6 0.6x = 4 x = 4/0.6 = 40/6 = 6 ² ⁄ ³. Don will need 6 ² ⁄ ³ kg of blend A. 102. If the match started at 8:00 a.m. UTC, it was 3:00 a.m. EST. From 3:00 a.m. EST to 12:15 p.m. EST is 9 hours 15 minutes. Nico and Jon must have played for (9)(60) + 15 = 540 + 15 = 555 minutes. 103. From 1809 to 2019 is 2019 – 1808 = 211 years. The 20 years from 2000 to 2019 each have thousands digit 2, leaving 211 – 20 = 191 years with thousands digit 1. There is a tens decade in each of the periods1809 to 1999, 1900 to 1999 and 2000 to 2019. So, that’s 30 years with tens digit 1. Finally, from 1809 to 2019, there are 21 years with units digit 1. Erika wrote the digit 1 a total of 191 + 30 + 21 = 242 times. 104. There are 25 = 32 possible outcomes when a fair coin is flipped five times. One of these outcomes has no heads, and 5 of these outcomes consist of four tails and one head. So, for these 6 outcomes (TTTTT, HTTTT, THTTT, TTHTT, TTTHT, TTTTH) the coin lands heads up fewer than two times. Of the “5 choose 2”, or 5!/(3! × 2!) = (5)(4)/2 = 10 outcomes with three tails and two heads, there are 6 outcomes (HTHTT, HTTHT, HTTTH, THTHT, THTTH, TTHTH) for which the coin doesn’t land heads up twice in a row. Finally, of the “5 choose 3”, or 5!/(2! × 3!) = (5)(4)/2 = 10 outcomes with three heads and two tails, there is only 1 outcome (HTHTH) for which the coin doesn’t land heads up twice in a row. In all, there are 6 + 6 + 1 = 13 desired outcomes out of 32 total outcomes, which is a probability of 13/32. 105. With 24 wins in 33 games, they have won 24/33 = 8/11 of their games so far. If they win the next n games, then they will have 24 + n wins in 33 + n games. Since we want this ratio to equal 80%, or 4/5, we have the proportion (24 + n)/(33 + n) = 4/5. Cross multiplying and solving for n yields 5(24 + n) = 4(33 + n) 120 + 5n = 132 + 4n n = 12. To have a winning percentage of 80%, they must win the next 12 games. 106. Making an organized list like this one shows that the vending machine can provide Sarah's change in 16 ways. Q 2 2 2 1 1 1 1 1 0 0 0 0 0 0 0 0 D 2 1 0 4 3 2 1 0 7 6 5 4 3 2 1 0 N 0 2 4 1 3 5 7 9 0 2 4 6 8 10 12 14 107. The 11 numbers, ordered from least to greatest, are 1, 2, 3, 12, 14, 20, 21, 22, 23, 25, 25. We can now see that the median is 20 and the mode is 25. The geometric mean of two numbers is the square root of their product. The geometric mean of 20 and 25 is √(20 × 25) = √500 = √(100 × 5) = 10√5. 108. We cannot have a three-digit integer that starts with a zero, so the hundreds digit must be 1, 2 or 9. The tens digit and the units digit can each be 0, 1, 2 or 9. We don’t need to calculate the sum or the total number of possible combinations to determine the mean of this set of three-digit numbers. We know that there will be equal occurances of each possible digit in each of the hundreds, tens and units places. The mean of the hundreds digits, then, is (2 + 1 + 9)/3 = 4. The mean of the tens digits is the same as the mean of the units digits, namely (2 + 0 + 1 + 9)/4 = 3. So, the mean of all the specified integers must be 4(100) + 3(10) + 3 = 400 + 30 + 3 = 433. 109. Let’s rewrite the expression x (4 x) based on the definition a b = ab + b = b(a + 1). We get x (4 x) = x x(4 + 1) = x 5x = 5x(x + 1). If 5x(x + 1) = 550, then x(x + 1) = 110. Since 110 = (−11)(−10) = (10)(11), the positive value of x that satisfies the equation is x = 10. 110. Since angle A has measure 34 degrees, the base angles B and C must each have measure (180 − 34)/2 = 146/2 = 73 degrees. As the figure shows, the center of a circle inscribed in a triangle is the intersection of the three angle bisectors of the triangle. It follows, then, that the measure of angle PAC is 34/2 = 17 degrees, and the measure of angle PCA is 73/2 = 36.5 degrees. Again, since the sum of the angle measures of a triangle is 180 degrees, we can conclude that the measure of angle APC is 180 – (17 + 36.5) = 180 − 53.5 = 126.5 degrees. B A P C Warm-Up 9 111. From the information given, we can write the following two equations: x + y = 98 and x = 2y + 11. Substituting the second equation into the first, we get (2y + 11) + y = 98, which leads to 3y + 11 = 98, then 3y = 87 and finally y = 29. Based on an average of 29 points per game, Thomas must have scored (60)(29) = 1740 points in the 60 games he played that season. 112. As soon as he finishes one painting, Pierre can start another painting while the previous one dries. It will take him (3)(39) = 117 minutes to paint the three paintings. All three will be dry 55 minutes after he completes the last one, making the total time 117 + 55 = 172 minutes, or 2 hours 52 minutes. If Pierre starts at 1:00 p.m., the earliest he can have three finished, dry paintings is 3:52 p.m. 113. The question is about the difference in the numbers of pounds of gold that the two pirates get. Let’s start by finding the difference in the fractions of the gold that the two pirates get: 5/8 − 3/8 = 2/8 = 1/4. So, Blackbeard gets one-quarter more of the 60 pounds of gold than Redbeard, which is (1/4)(60) = 15 pounds. Blackbeard gets (5/8)(60) = 75/2 = 37 ½ pounds of gold, and Redbeard gets (3/8)(60) = 45/2 = 22 ½ pounds of gold. MATHCOUNTS 2018-2019 63 114. In the series of pentagons, the nth figure has 3n − 2 more dots than the previous figure (3 new sides, each with n dots, less 2 dots that were counted twice at the intersections of those sides. So, the 20th pentagonal number is 1 + 4 + 7 + ⋯ + (3(20) − 2) = 1 + 4 + 7 + ⋯ + 58 = (1/2)(20)(1 + 58) = (10)(59) = 590. 115. We do not have to list all the three-digit numbers whose digits are distinct and nonzero to find their arithmetic mean (average). We don’t even need to know how many such numbers there are! We only need to realize that each of the digits 1 through 9 will occur in each place value an equal number of times. The average of the digits 1 though 9 is 5, so the average of all these numbers must be 555. 116. Based on the figure, the dimensions of this rectangular prism (or box) are 2 by 4 by 6. The volume of a rectangular prism is the product of its length, width and height, which is, in this case, (2)(4)(6) = 48 units3. The surface area of a rectangular prism is the sum of the areas of its six rectangular faces, or twice the sum of the three pairwise products of the three dimensions. This prism has surface area 2[(2)(4) + (2)(6) + (4)(6)] = 2(8 + 12 + 24) = (2)(44) = 88 units2. Although the units of volume and area are different, we see that the absolute difference of these numerical values is 88 – 48 = 40. 117. Letting x represent the third number, we have the equation 4/3 + 1 + x = (4/3)(1)x. Simplifying and solving for x, we see that 7/3 + x = (4/3)x 7/3 = (1/3)x x = 7. 118. To get a median of 10, the middle number when the five scores are arranged from least to greatest must be 10. To get a unique mode of 10, there must be at least one other score of 10. To get a mean of 15, the sum of the five scores must be (5)(15) = 75. Finally, to get a range of 16, the greatest and the least scores must have a difference of 16. The three possible groups of five scores that meet these criteria are 8, 10, 10, 23, 24; 9, 10, 10, 21, 25 and 10, 10, 10, 19, 26. The three possible values of the lowest score have a sum of 8 + 9 + 10 = 27. 119. When the die is rolled three times, there are (6)(6)(6) = 216 possible outcomes. Let’s examine the five outcomes that yield a sum of 0. Cayley can roll 0, 0 and 0 one way, for a probability of (1)(2/6)(2/6)(2/6) = 8/216. She can roll −1, 1 and 0 in 3! = 6 ways, for a probability of (6)(1/6)(1/6)(2/6) = 12/216. She can roll −2, 2 and 0 in 3! = 6 ways, for a probability of (6)(1/6)(1/6)(2/6) = 12/216. She can roll −2, 1 and 1 in 3 ways, for a probability of (3)(1/6)(1/6)(1/6) = 3/216. She can roll −1, −1 and 2 in 3 ways, for a probability of (3)(1/6)(1/6)(1/6) = 3/216. The total probability is (8 + 12 + 12 + 3 + 3)/216 = 38/216 = 19/108. 120. Each interior angle of a regular octagon has measure 180 − (360/8) =180 − 45 = 135 degrees. The measure of angle BAP is 135 – 75 = 60 degrees, and the measure of angle ABP is 135 – 85 = 50 degrees. The sum of the measures of the angles of a triangle is 180 degrees, so acute angle P must have measure 180 – (60 + 50) = 180 − 110 = 70 degrees. Warm-Up 10 121. For convenience we will write the circle function as C(x) and the triangle function as T(x). We need to determine the value of the difference C(T(C(x))) − T(C(T(x))) when x = π. Starting with the innermost function of the minuend (quantity from which we are subtracting) and working outward, we have C(x) = 2x + 1, then T(C(x)) = T(2x + 1) = 2(2x + 1) − 1 = 4x + 2 − 1 = 4x + 1, and C(T(C(x))) = C(4x + 1) = 2(4x + 1) + 1 = 8x + 2 + 1 = 8x + 3. Likewise, starting with the innermost function of the subtrahend (quantity being subtracted) and working outward, we have T(x) = 2x − 1, then C(T(x)) = C(2x − 1) = 2(2x − 1) + 1 = 4x – 2 + 1 = 4x − 1, and T(C(T(x))) = T(4x − 1) = 2(4x − 1) – 1 = 8x – 2 – 1 = 8x − 3. So, C(T(C(x))) − T(C(T(x))) = 8x + 3 − (8x − 3) = 8x + 3 − 8x + 3 = 6. So, C(T(C(π))) − T(C(T(π))) = 6. 122. The absolute value expression | x − 3 | is a representation of the positive distance from x to 3. Likewise, the expression | x − 7 | is a representation of the positive distance from x to 7. The equation | x − 3 | + | x − 7 | = 6 means that the sum of these two distances is 6. Since the distance from 3 to 7 is only 4, we know that x cannot be between 3 and 7 on the number line. Visually, we can work out that the distances from x to 3 and to 7 are 1 and 5, respectively, when x = 2, and are 5 and 1, respectively, when x = 8, as displayed on the number line shown. The sum of these two values is 2 + 8 = 10. Alternatively, we can solve the given equation algebraically using cases. The value of x − 3 is negative when x < 3, and the value of x − 7 is negative when x < 7. So, we will solve the absolute value equation for three cases. 1+5=6 5 1 0 1 F C(17, 6) D B(9, 2) E 5 7 9 1 5 Case 1: If x < 3, then we have −(x − 3) + (−(x − 7)) = 6 3 − x + 7 − x = 6 −2x + 10 = 6 4 = 2x x = 2. Case 2: If 3 ≤ x < 7, then we have x − 3 + (−(x − 7)) = 6 x − 3 + 7 − x = 6 4 = 6, which is a contradiction. Case 3: If x ≥ 7, then we have (x − 3) + (x − 7) = 6 2x − 10 = 6 2x = 16 x = 8. Again, the sum of the two values of x is 2 + 8 = 10. A(6, 8) 3 5+1=6 123. Rectangle ADEF can be drawn with inscribed triangle ABC, as shown. The area of triangle ABC is the difference when the areas of triangles AFC, CEB and BDA are subtracted from the area of rectangle ADEF. The result is (6)(11) − (1/2)(11)(2) − (1/2)(4)(8) − (1/2)(3)(6) = 66 − 11 − 16 − 9 = 30 units2. Alternatively, using the shoelace method with the points A(6, 8), B(9, 2), C(17, 6) and back to A(6, 8), we get (1/2)[ | (6 × 2 + 9 × 6 + 17 × 8) − (8 × 9 + 2 × 17 + 6 × 6) |] = (1/2)(202 − 142) = (1/2)(60) = 30 units2. 124. At one extreme, it is possible that all 460 students who play a varsity sport also belong to a club. At the other extreme, since the combined total of the students in the two activities is greater than the number of students in the college, there must be at least (460 + 571) − 985 = 1031 − 985 = 46 people who play a varsity sport and belong to a club. The difference between these values is 460 – 46 = 414 students. 64 MATHCOUNTS 2018-2019 11 125. If the isosceles triangle has two sides of length 5 inches and one of length 6 inches, we can use the Pythagorean Theorem to determine the length a of the altitude drawn to the 6-inch base as follows: 32 + a2 = 52 9 + a2 = 25 a2 = 16 a = 4 inches. (Or you might have recognized that the altitude forms two 3-4-5 right triangles.) The area of this triangle is (1/2)(6)(4) = 12.0 in2. If the isosceles triangle has two sides of length 6 inches and one of length 5 inches, we can use the Pythagorean Theorem to determine the length b of the altitude drawn to the 5-inch base as follows: 2.52 + b2 = 62 6.25 + b2 = 36 b2 = 36 − 6.25 b2 = 29.75 b = √29.75 inches. Without a calculator, it is difficult to compute this value. However, we know that it will be between √25 = 5 and √36 = 6, resulting in an area between (1/2)(5)(5) = 12.5 in2 and (1/2)(5)(6) = 15 in2. Any area in this range will be greater than the smallest possible area, which is 12.0 in2. 126. The larger hexagon is actually twice the area of the equilateral triangle inside it. This can be seen in Figure 1, where three of the vertices can be folded to the center to fit within the equilateral triangle with no overlapping regions. As Figure 2 shows, the same can be done with the smaller hexagon in the middle to make a smaller equilateral triangle that is 1/4 the area of the larger equilateral triangle. Finally, Figure 3 shows how the larger hexagon can be subdivided into eight half-hexagons, two of which coincide with the smaller hexagon. The ratio of the areas of the smaller and larger hexagons is 2/8 = 1/4. Figure 1 Figure 2 Figure 3 127. The prime factorization of 23,328 is 25 × 36. In order for a divisor to be a perfect cube, it must contain as a factor each prime factor raised to a power that is a multiple of 3. There are two choices for the exponent of 2, namely 0 and 3; and there are three choices for the exponent of 3, namely 0, 3 and 6. Therefore, 23,328 has (2)(3) = 6 divisors that are perfect cubes: 1, 23, 33, 23 × 33, 36 and 23× 36. 128. With a mean of 80, Saila must have earned a total of (9)(80) = 720 points on the nine exams. The median of nine scores is the fifth score when the scores are ordered from least to greatest. To maximize the median, we will assume that Saila got four scores of 60, the minimum passing score. This accounts for (4)(60) = 240 points, leaving 720 − 240 = 480 points for the remaining five scores. If we make the last five scores all the y same, we get the greatest possible median score of 480/5 = 96 points. 6 1 3 129. To count the paths in three-dimensional space, we will use a method similar to one used 3 12 30 B when counting paths in a two-dimensional plane. The right, center and left planes shown 90 30 6 1 2 3 correspond to z = 0, z = 1 and z = 2, respectively. Starting at A(0, 0, 0) in the right plane, (2, 2, 2) 2 12 6 there is one path (indicated by “1” at each lattice point) to each of (0, 1, 0), (0, 2, 0), (1, 0, 0) A 12 z 3 and (2, 0, 0). The two paths (indicated by “2” at this lattice point) leading to (1, 1, 0) come 30 1 1 1 (0, 0, 0) from (1, 0, 0) and (0, 1, 0). Continuing in this manner, we determine the number of paths 2 3 1 to each lattice point in the right plane. Next, examining the center plane, there is one path to 3 6 1 (0, 0, 1), which is from (0, 0, 0) in the right plane. There is one path to each of (1, 0, 1) and (0, 1, 1) x from (0, 0, 1), but there is also one path to each of (1, 0, 1) and (0, 1, 1) from an adjacent point in the right plane. That means there are a total of two paths leading to each of (1, 0, 1) and (0, 1, 1). Continuing in this manner, we determine the number of paths to each point in the center plane, including those from an adjacent point in the right plane. We then use the same process with the left plane to determine the number of paths to each point in that plane, remembering to include paths from an adjacent point in the center plane. As the figure shows, the total number of paths from A to B is 90 paths. Alternatively, a path can be described by a sequence of two Xs, two Ys and two Zs, such as YXXZYZ, which represents a move in the (positive) Y direction, followed in order by two moves in the X direction, a move in the Z direction, a move in the Y direction and finally a move in the Z direction. Any such sequence will take us from (0, 0, 0) to (2, 2, 2). There are 6C2 = 6!/(4! × 2!) = (6)(5)/2 = 30/2 = 15 choices for the positions of the Xs. After the Xs have been placed, there are 4C2 = 4!/(2! × 2!) = (4)(3)/2 = 12/2 = 6 choices for the positions of the Ys, leaving only 1 choice for the positions of the Zs. That results in (15)(6)(1) = 90 paths. 130. Since Car F is traveling twice as fast as Car S, it will travel twice as far as Car S in the same amount of time. This means that Car F will be 2/3 of the way around the circular track in one direction and Car S will be 1/3 of the way around the track in the other direction when they meet again. The amount of time this will take is 2/3 of the 120 seconds Car F takes for a full lap or 1/3 of the 240 seconds Car S takes for a full lap, each of which is 80 seconds. Alternatively, Car F travels at a rate of 1/120 lap per second, and Car S travels at a rate of 1/240 lap per second. They are traveling toward each other at a combined rate of 1/120 + 1/240 = (2 + 1)/240 = 3/240 = 1/80 lap per second. Therefore, they will meet each other every 80 seconds. Euclidean Algorithm: Use this process Warm-Up 11 131. For each of these numbers, we can quickly rule out factors of 2, 3 and 5 by well-known divisibility rules. We can also find by division that neither of them is divisible by 7. The next prime to try is 11, and a number is divisible by 11 if and only if the alternating sum of its digits is divisible by 11. For 4147, we get 4 − 1 + 4 − 7 = 8 − 8 = 0. Since zero is a multiple of 11, the number is divisible by 11. For 2563, we compute 2 − 5 + 6 − 3 = 8 − 8 = 0, so it is also divisible by 11. The prime factorization of 4147 is 11 × 13 × 29, and the prime factorization of 2563 is 11 × 233. The greatest common divisor of the two numbers is 11. Alternatively, we can use the Euclidean Algorithm to determine the greatest common divisor (GCD) of 4147 and 2563. We get GCD(4147, 2563) = GCD(2563, 1584) = GCD(1584, 979) = GCD(979, 605) = GCD(605, 374) = GCD(374, 231) = GCD(231, 143) = GCD(143, 88) = GCD(88, 55) = GCD(55, 33) = GCD(33, 22) = GCD(22, 11) = 11. to find the greatest common divisor (GCD) of a and b, for positive integers a, b, c and r. Start with a>b Outcome of a ÷ b is a = bc + r NO Remainder r=0 YES GCD(a, b) = GCD(b, r) GCD(a, b) = b Let a = b and b = r End with GCD(a, b) 132. Hongyi needs to find a triangular number that is a multiple of 25. The triangular numbers are 1, 3, 6, 10, 15, 21, etc., and the nth triangular number has value n(n + 1)/2. We are looking for a triangular number whose value is 25k, for some integer k. We can solve the equation n(n + 1)/2 = 25k or, equivalently, n(n + 1) = 50k. The left side shows a product of two consecutive natural numbers, and the right side shows that we need a multiple of 50, which has prime factorization 2 × 5 × 5. (We don’t really care about the value of k.) The factor of 2 does not present an issue since any two consecutive numbers will include one even number. The only way we will get two factors of 5 is for either n or n + 1 to provide both factors. Therefore, the least possible number of widgets Hongyi can use to build the display is (24)(25)/2 = (12)(25) = 300 widgets. MATHCOUNTS 2018-2019 65 D P 133. The interior angle measure of a regular pentagon is 180 − (360/5) = 180 − 72 = 108 degrees. In the figure, the measure of angle A of isosceles triangle EAB is 108 degrees. So, the remaining 180 – 108 = 72 degrees of the triangle must be equally divided E between the base angles AEB and ABE, making them each 36 degrees. The measure of angle PEB must be 108 – 36 = 72 degrees. For right triangle PEB, shown here, it follows that the measure of angle EBP is 180 − (90 + 72) = 180 − 162 = 18 degrees. 108° A C B 134. To find the number of prime dates in March, we must determine how many primes there are from 1 to 31, since there are 31 days in March. The 11 primes less than or equal to 31 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, each of which is associated with one of 11 prime dates in March. 135. The difference between Bennie’s and Flossie’s speeds is 75 – 62 = 13 mi/h. This means that the distance between Flossie and Bennie increases by 13 miles every hour. After 5 hours, that would be (5)(13) = 65 miles, except that Flossie stopped for 20 minutes, or 1/3 of an hour. That’s (75)(1/3) = 25 miles that Flossie did not travel, so the distance between Bennie and Flossie would be 65 − 25 = 40 miles. Alternatively, we can use the fact that distance = rate × time. Traveling at a speed of 62 mi/h for 5 hours, Bennie has gone (62)(5) = 310 miles. Traveling at a speed of 75 mi/h for 4 hours 40 minutes, or 280/60 = 14/3 hours, Flossie has gone (75)(14/3) = (25)(14) = 350 miles. The difference in their distances is 350 − 310 = 40 miles. F A E 136. The figure shows hexagon ABCDEF with K at the midpoint of side CD. We need to find the length of segment AK. To do this, we draw segment AC with midpoint G, as shown. Since the interior angle measure of a regular hexagon is 180 − (360/6) = D 180 − 60 = 120 degrees. It follows that the segment BG divides triangle ABC into two 30-60-90 right triangles. We know that AB = BC = 4 cm, so it follows that, by properties of this special right triangle, BG = 2 cm and AG = GC = 2√3 cm. We have G 2√3 2 K right triangle ACK with leg lengths AC = 2√3 + 2√3 = 4√3 cm and CK = 2 cm. Now we can use the Pythagorean Theorem to 2 determine the length of hypotenuse AK as follows: (4√3)2 + 22 = AK2 48 + 4 = AK2 AK = √52 = 2√13 cm. B 4 C 137. We will solve this problem by working backward. At the third lunch, after Chloe doubled the number of crackers Axel and Ben already had from the second lunch, Axel, Ben and Chloe each had 8 crackers. So, Axel and Ben each must have had 8/2 = 4 crackers, and Chloe must have had 8 + 4 + 4 = 16 crackers from the second lunch. Now the 4 crackers that Axel had and the 16 crackers that Chloe had at the second lunch were the result of Ben doubling the number of crackers Axel and Chloe had from the first lunch. So, Axel must have had 4/2 = 2 crackers, Chloe must have had 16/2 = 8 crackers and Ben must have had 4 + 2 + 8 = 14 crackers at the first lunch. Finally, the 8 crackers Chloe had and the 14 crackers Ben had at the first lunch were the result of Axel doubling the number of crackers Chloe and Ben each started with. So, Chloe must have started with 8/2 = 4 crackers, Ben must have started with 14/2 = 7 crackers and Axel must have started with 2 + 4 + 7 = 13 crackers. 138. There are only two scoring combinations that result in a total score of 25 points: five 5-point touchdowns, TTTTTT, and two 5-point touchdowns with five 3-point field goals, TTFFFFF. While there is only 1 scoring sequence for five touchdowns, there are many sequences for scoring two touchdowns with five field goals. The number of scoring sequences if the team scored seven times, with two times being touchdowns and five times being field goals, is 7C2 = 7!/(5! × 2!) = (7)(6)/2 = 42/2 = 21 ways. So, the number of unique 25-point scoring sequences is 1 + 21 = 22 sequences. 139. Since the five numbers form an arithmetic sequence, the mean of 18 is equal to the middle number. If we use d for the common difference between the terms of this sequence, the five numbers can be written as 18 − 2d, 18 − d, 18, 18 + d, 18 + 2d. The squares of these five numbers are 324 − 72d + 4d 2, 324 − 36d + d 2, 324, 324 + 36d + d 2, 324 + 72d + 4d 2. The sum of these five squares is (5)(324) + 10d 2, so the mean of the squares is [(5)(324) + 10d 2]/5 = 324 + 2d 2. We are told that the mean of the squares is 374, so we have 324 + 2d 2 = 374 2d 2 = 50 d 2 = 25 d = ±5. The five original numbers are 8, 13, 18, 23 and 28. The greatest is 28. 140. The 400 tagged dolphins examined one month later represent 400/1000 = 2/5 of the dolphins that were originally tagged. The idea is that the 2000 dolphins she examined must be 2/5 of the local population of dolphins. The expected number of dolphins in the local population is (2000)/(2/5) = (2000)(5/2) = 5000 dolphins. Warm-Up 12 141. The 512 small cubes make an 8 × 8 × 8 cube themselves. The cubes that are not touching the lidless cubical box form a rectangular prism with dimensions 6 × 6 × 7. Thus, (6)(6)(7) = 252 cubes were removed. 142. On a weekday, Max can earn (3)(15) = 45 points for writing posts and 120/3 = 40 points for reading posts, which is a total of 85 points per weekday. By the end of the day on Friday, Max’s point total can be (5)(85) = 425 points. On the weekend days, he can earn (3)(15) = 45 points for writing posts and 225/3 = 75 points for reading posts, which is a total of 120 points per weekend day. From Monday through Sunday, he can earn 425 + 120 + 120 = 665 points, which is still 800 − 665 = 135 points from Gold Status. It will take Max two more weekdays to earn a total of 665 + 85 + 85 = 835 points. The earliest Max can earn Gold Status is on a Tuesday. 143. If we take the cross product of (a/4)/3 = 4/(a/3), we get (a/4)(a/3) = (3)(4) a2/12 = 12 a2 = 144. 144. Since the radius of the circle is 13 cm, the diameter must be 26 cm, as shown. The measure of an inscribed angle in a circle is half B the measure of the arc that it intercepts. Since segment AC is a diameter, arc AC has measure 180 degrees, so angle ABC must have 10 measure 180/2 = 90 degrees. This means that triangle ABC is a right triangle, and we can use the Pythagorean Theorem to find the A 13 length of leg BC, labeled x. We have x 2 + 102 = 262 x 2 + 100 = 676 x 2 = 576 x = 24. The length of segment BC is 24 cm. 66 x P 13 MATHCOUNTS 2018-2019 C 145. The prime factorization of 630 is 2 × 32 × 5 × 7. The prime factorization of 360 is 23 × 32 × 5. The prime factorization of 60 is 22 × 3 × 5. We note that all three numbers have at least one factor of 2, at least one factor of 3 and at least one factor of 5. The common divisors of all three numbers are 1, 2, 3, 5, 6, 10, 15 and 30. The numbers 630 and 360, however, have a second factor of 3 in common. This factor of 3 can be multiplied by the common divisors of all three numbers to get these common divisors of 630 and 360: 3, 6, 9, 15, 18, 30, 45 and 90. Eliminating the divisors of 60 that are on that list (3, 6, 15 and 30) leaves 9, 18, 45 and 90. These are the 4 positive integers that are each a divisor of both 630 and 360 but not 60. 146. There are “7 choose 5,” or 7C5 = 7!/(5! × 2!) = (7)(6)/2 = 21 ways to select five of the seven points to be vertices of a pentagon. There are 7 congruent convex pentagons with five consecutive points as vertices, 7 with four consecutive points as vertices and 7 with three consecutive points as vertices. As shown, there are 3 non-congruent convex pentagons. 147. First we will substitute 25 for a in the original expression, which gives us 25 – b = 25/b − 1. Simplifying gives us the quadratic equation 25b − b2 = 25 − b b2 − 26b + 25 = 0. For quadratic equations of this form, with a leading coefficient of 1, you may recall that the sum of the solutions is the opposite of the linear term’s coefficient. Thus, the sum of the values of b that are solutions of b2 − 26b + 25 = 0 is −(−26) = 26. Alternatively, we can factor the trinomial into the product of two binomials to get (b − 25)(b − 1) = 0 . This product is equal to zero when b − 25 = 0 b = 25 or when b − 1 = 0 b = 1. So, the sum of all possible values of b is 25 + 1 = 26. Y 148. We can sketch square WXYZ inscribed in square ABCD, as shown, creating four congruent right triangles. Based on the coordinates of W and Y, we see that square ABCD has side length a − c = 7 − 1 = 6 units. Given the coordinates of W, X, Y and Z, and since AY = BX = CW = DZ and ZA = YB = XC = WD, we know that AY = DZ = 4 − c = d − 1 c + d = 5, Z XC = WD = b − 1 = 2 − c b + c = 3. Now, since AY + YB = AB, it follows that 4 − c + 2 − c = 6 6 − 2c = 6 2c = 0 c = 0. Substituting 0 for c in a − c = 6, we get a − 0 = 6 a = 6, and into b + c = 3, we get b + 0 = 3 b = 3, and d−1 X b − 1 into c + d = 5, we get 0 + d = 5 d = 5. So, | ab – cd | = | (6)(3) – (0)(5) | = 18. (Note: If the chosen locations of (a, b) and (c, d) were switched, the answer would be the same because of the absolute value calculation.) Alternatively, notice that the D2 − c C W center of the square is the midpoint of diagonal WY, so its coordinates are (3, 4). If we translate the figure so that the center is at the origin, then W′ = (2 – 3, 1 – 4) = (–1, –3), and Y′ = (4 – 3, 7 – 4) = (1, 3). Next, rotating 90 degrees clockwise about the new origin, we see that W″ = (–3, 1) and Y″ = (3, –1). Finally, translating back, we see that Z = W′′′ = (–3 + 3, 1 + 4) = (0, 5) and X = Y′′′ = (3 + 3, –1 + 4) = (6, 3). Again, | ab – cd | = | (6)(3) – (0)(5) | = 18. A 4−c B 149. The sum of the ages of the original ten guppies is 5 + 5 + 8 + 8 + 8 + 9 + 10 + 11 + 11 + 33 = 108 weeks. The range of their ages is 33 – 5 = 28 weeks. Since the range of the eleven integer ages will be an integer and we know that the median and mode of the integer ages are also integers, it follows, then, that the mean must also be an integer. Since the sum is currently 2 less than a multiple of 11, the value of n will have to be 2 more than a multiple of 11, also known as “2 mod 11.” The table shows four possible values of n, along with the mean, median, mode, range and the Range − (Mean + Median + Mode) Mean Median Mode Range n difference between the range and the sum of the mean, median and mode. 2 10 8 8 31 31 − (10 + 8 + 8) = 31 − 26 = 5 The values that makethe difference 0 are n = 13 and n = 35 weeks. Note that 13 11 9 8 28 28 − (11 + 9 + 8) = 28 − 28 = 0 any larger values of n would increase the mean and range at a rate of 1:11, making it impossible for the range to equal the sum of the mean, median and 24 12 9 8 28 28 − (12 + 9 + 8) = 28 − 29 = −1 mode. The sum of the two possible values of n is 13 + 35 = 48. 35 13 9 8 30 30 − (13 + 9 + 8) = 30 − 30 = 0 150. The area of a rhombus is half the product of the lengths of its diagonals. In other words, twice the area of this rhombus equals the product of its diagonal lengths. So, the unknown diagonal length is (2)(26)/4 = 52/4 = 13 cm. Warm-Up 13 151. When the product of three binomials is expanded, there are 2 × 2 × 2 = 8 terms before like terms are combined. We will get an abc from (a + 2b)(b + 2c)(c + 2a) when we multiply the first terms of the three binomials, and we will get 8abc when we multiply the last terms of the three binomials. Combining these, we get 9abc. There are no other abc terms in the expansion, so the coefficient of abc is 9. 152. In the worst-case scenario, Jenny might draw out all the green, yellow, orange and purple candies before getting a red candy on her 45th draw. There must be 44 non-red candies, or 44/4 = 11 of each of the colors green, yellow, orange and purple. Including the 11 red candies, there must be a total of 55 candies in her bag. 153. The box has 12 edges. The total length of tape is 4(15 + 13 + 10) = (4)(38) = 152 inches, which equals 152/12 = 38/3 = 12 ² ⁄ ³ feet. 154. By the Pythagorean Theorem, the hypotenuse has length √[(12√3)2 + (23√3)2] = √[(144)(3) + (529)(3)] = √(432 + 1587) = √2019 units. 155. Since it takes a crew of eight people to paint an aircraft in three days, then with twice as much time, only half as many people are needed. So, a crew of four people can paint an aircraft in six days. To paint five aircraft in six days, then, requires 5 × 4 = 20 people. 156. We will use r to represent the constant ratio between consecutive terms in this geometric sequence. If the 2nd term is 2, then the 3rd term is 2r, the 4th term is 2r 2 and the 5th term is 2r 3, which is equal to 5. Since 2r 3 = 5, we know that r 3 = 5/2. To get from the 5th term to the 11th term, we need to multiply by r 6, which is the same as (r 3)2. So, the 11th term is (5)(5/2)2 = 125/4. MATHCOUNTS 2018-2019 67 157. There are 8 × 12 = 96 combinations of possible rolls. If Ilana rolls a 1, then as long as Yolanda rolls a 2 or greater, Yolanda will have a larger roll. This gives 11 possible rolls. Similarly, if Ilana rolls a 2, then Yolanda can roll a 3 or greater, which is 10 possible rolls. This continues up through Ilana rolling an 8 and Yolanda rolling a 9 or greater, which is 4 possible rolls. This means Yolanda can roll a number greater than Ilana in 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 = 60 ways. The probability of this is 60/96 = 5/8. y Q 25 20 15 12 25 − x x P 10 R T 12 5 5 10 15 S 20 x 25 158. The figure shows one possible orientation of rectangle PQRS. We can solve for the unknown x in the figure by using similar triangles PTQ and QTR. The ratio PT to QT is equal to the ratio QT to RT, so we have x/12 = 12/(25 − x) x(25 − x) = (12)(12) 25x − x 2 = 144 x 2 − 25x + 144 = 0. We can factor this trinomial into the product of two binomials if 144 has a pair of factors with a sum of 25. Since 9 × 16 = 144 and 9 + 16 = 25, we can rewrite the equation as (x − 9)(x − 16) = 0. Solving for x, we get x − 9 = 0 x = 9 or x − 16 = 0 x = 16. For our figure, we choose x = 9 and that makes 25 – x equal to 16. Now we can use the Pythagorean Theorem to find the outside dimensions of the rectangle. Side PQ is √(92 + 122) = √(81 + 144) = √225 = 15 units, and side QR is √(162 + 122) = √(256 + 144) = √400 = 20 units. The perimeter of rectangle PQRS is 2(15 + 20) = (2)(35) = 70 units. 159. Since the mean of Hannah’s five races was 4 miles, she must have run a total of (5)(4) = 20 miles. To get the maximum possible distance for one of her races, we have to minimize the other four races, each with a different positive integer distance. Four of the races would have been 1, 2, 3 and 4 miles, for a total of 10 miles, giving us as the maximum possible distance for the longest race 20 – 10 = 10 miles. 160. The coefficient of the x 2 term in a cubic equation with leading coefficient 1 is the sum of the roots (with the opposite sign). Since the roots of this equation are known to be positive integers that form a geometric sequence, we need to solve a + ar + ar 2 = 19, for positive integer a and rational common ratio r in order for ar and ar 2 to be positive integers. We know 19/3 = 6 1/3, so we can try the nearest integer, 6, for ar and experiment with pairs of integers that sum to 19 – 6 = 13 to find our sequence. With some trial and error, we get a + ar + ar 2 = 4 + 6 + 9 = 19, for r = 3/2. These are the roots of the equation, so we have (x − 4)(x − 6)(x − 9) = 0 (x 2 − 10x + 24)(x − 9) = 0 x 3 − 19x 2 + 114x − 216 = 0. The sum of c and d is, therefore, 114 + (−216) = −102. Warm-Up 14 161. Rectangle ACFH has dimensions n and n + 3 + n − 3 = 2n, so its perimeter is 2(n + 2n) = 2(3n) = 6n. The dimensions of rectangle BDEG are n and (n − 3) + (n − 6) = 2n − 9. We are told that its area is 110 units2, so we can solve the equation n(2n − 9) = 110 for n. Doing so yields 2n 2 − 9n = 110 2n 2 − 9n − 110 = 0 (2n + 11)(n − 10) = 0. So, 2n + 11 = 0 2n = −11 n = −11/2 or n − 10 = 0 n = 10. Since n represents the length of a side of a rectangle, only the positive solution makes sense. So, the perimeter of rectangle ACFH is 6n = 6(10) = 60 units. 162. The larger right triangle has an area of 0.5ab, which is numerically equal to the length 3(a + b). We can set these expressions equal to each other and solve for a as follows: 0.5ab = 3(a + b) ab = 6a + 6b ab − 6a = 6b a(b − 6) = 6b a = 6b/(b − 6). We can take this last equation a little further if we add a form of zero and rearrange it as follows: a = 6b/(b − 6) − 36/(b − 6) + 36/(b − 6) = (6b − 36)/(b − 6) + 36/(b − 6) = 6(b − 6)/(b − 6) + 36/(b − 6) = 6 + 36/(b − 6). Since a and b must both be positive integers, we need to find positive integer values of b that make b – 6 equal to a factor of 36. Those values of b are 7, 8, 9, 10, 12, 15, 18, 24 and 42. This list is nearly halved when we realize that each value of b produces a value of a from the same list. Without loss of generality, we only need to consider these ordered pairs for (a, b): (7, 42), (8, 24), (9, 18), (10, 15) and (12, 12). For each pair, we can verify that 0.5ab does indeed equal 3(a + b). However, we need to consider whether the smaller triangle meets its required conditions. The smaller right triangle has an area of 0.5cd, which is numerically equal to 2(c + d). Since the ratio of the areas of the triangles is 3:2, we can say that 2 times the larger area is equal to 3 times the smaller area. This means that (2)(0.5ab) = (3)(0.5cd) ab = (3/2)cd or cd = (2/3)ab. It also means that (2)[3(a + b)] = (3)[2(c + d)], which simplifies to a + b = c + d. So, we need an ordered pair of integers (c, d) to have the same sum as a and b but have a product that is two-thirds that of a and b. An lengthy examination (not included here) of our five possible ordered pairs (a, b) reveals that only (10, 15) yields integer values for (c, d). The larger right triangle has area (0.5)(10)(15) = 75, with 3(10 + 15) = 75 also. The smaller right triangle needs to have an area of (2/3)(75) = 50, so we need to find positive integers (c, d) such that (0.5)(cd) = 50 and c + d = 25. The solution is (5, 20). Finally, the value of a + b + c + d = 25 + 25 = 50 units. 8 40 5 8 56 7 163. We start by noting that, in this case, the greatest difference results when we maximize the minuend and minimize the subtrahend. There are (8)(7)/2 = 56/2 = 28 pairs of the integers from 1 to 40 165 42 168 42 48 169 35 8 that can be the labels for adjacent vertices of a face of the cube. (These 28 pairings yield only 24 different products because 6, 8, 12 and 24 are duplicated.) In order to maximize the value of one face, 6 5 7 35 30 6 5 30 we need to maximize the four products calculated for the edges. We’ll 56 8 7 56 8 7 Figure 1 use the four greatest possible values for the vertices of this face: 8, 7, 6 48 169 35 and 5. The three different values for the sum of the edge products are shown here in Figure 1. The greatest possible 48 169 35 30 30 sum is 169. Now in order to minimize the value of an adjacent face, we need to minimize the four products calculated 6 5 6 5 for the edges. We start by making the edge with vertices labeled 5 and 6 the one shared by the two faces. For the 12 49 5 6 48 10 remaining two vertices, we need to use the least possible values: 1 and 2. The two different values for the sum of the edge products for this face are shown here in Figure 2. The least possible sum is 48. Therefore, the greatest possible 2 1 2 1 2 2 absolute difference between the two labels of two adjacent faces is 169 – 48 = 121. 56 7 8 48 6 Figure 2 164. There are “7 choose 5” = 7C5 = [(7)(6)(5)(4)(3)]/[(5)(4)(3)(2)(1)] = [(7)(6)]/2 = 21 ways for Ernie to choose five of Burt’s seven tiles. The six possible palindromes that Ben could create are ANANA, NAAAN, ANBNA, NABAN, ANSNA and NASAN. It’s actually easier to consider the letters that are left out in each case. The first two palindromes leave out the letters B and S. There is only 1 way to choose those two letters, so the probability is 1/21. The next two palindromes leave out the letters A and S. There are three As, so the probability is 3/21. The last two palindromes leave out the letters A and B; again the probability is 3/21. The probability, then, that Ernie will be able to make a palindrome from the five tiles he chooses is 1/21 + 3/21 + 3/21 = 7/21 = 1/3. 68 MATHCOUNTS 2018-2019 165. Whatever the ratio of the side lengths, if the diagonal of Joe’s screen is twice as long as Daniel’s, then Joe’s screen has 22 = (2)(2) = 4 times the area, as shown in the figure. This means the ratio of the area of Daniel’s screen to Joe’s is 1/4. 166. There are 25 of the smallest size rhombi in the figure. Four of the smallest size rhombi combined give us the next size, of which there are 16 rhombi in this figure. Nine of the smallest size rhombi combined give us the next size, of which there are 9 rhombi in this figure. Sixteen of the smallest size rhombi combined give us the next size, of which there are 4 rhombi in this figure. Finally, 25 of the smallest size rhombi combined give us the largest, of which there is 1 rhombus in this figure. That’s a total of 25 + 16 + 9 + 4 + 1 = 55 rhombi. 167. The decimal expansion of 3/7, which repeats after 6 digits, is written in “bar notation” as 0.428571. Since the remainder is 3 when 15 is divided by 6, or 15 is “3 mod 6”, it follows that the 15th digit to the right of the decimal point will be the 3rd digit of the repeating pattern, which is 8. 168. Since we want the least possible value of c and since the integers must be different, we will rewrite the list as 6, 7, 8, c, c + 1, c + 2, 34. The sum of these numbers is 6 + 7 + 8 + c + c + 1 + c + 2 + 34 = 3c + 58 = 7c, since the mean of the numbers is c. Solving the equation 3c + 58 = 7c, we get 4c = 58, and c = 14.5. We cannot lower c to the integer 14 while maintaining a mean of 14, but we can raise it to 15 if we raise one or two of the other unknown values to compensate. Some possible sets are {6, 7, 8, 15, 17, 18, 34}, {6, 7, 9, 15, 16, 18, 34}, {6, 7, 10, 15, 16, 17, 34} and {6, 8, 9, 15, 16, 17, 34}. After some trial and error, we see that the smallest possible value of c is 15. y 169. The figure shows an area model for the probability of two continuous variables. For ordered pairs (x, y) with 0 ≤ x ≤ 10 and 0 ≤ y ≤ 10, if the point is located within the shaded region, then the values x and y have an absolute difference less than 3; otherwise their absolute difference is greater than or equal to 3. The 10-by-10 square has area 102 = 100 units2, and the unshaded region has area 2(1/2)(7)(7) = 49 units2, meaning the shaded region has area 100 − 49 = 51 units2. The probability that (x, y) is located within the shaded region, then, is 51/100. 10 8 6 4 2 0 2 4 6 8 10 x 170. Among the digits 1, 2, 3 and 4, the only digits that differ by more than two are 1 and 4. So, the permutations we must exclude are of the form 14_ _, _ _14, _14_, 41_ _, _ _ 41 and _41_. There are two ways to order the digits 2 and 3 for each of these. So, of the 4! = 24 permutations of the digits 1, 2, 3 and 4, we must exclude the (6)(2) = 12 permutations in which 1 and 4 are adjacent. That leaves 24 − 12 = 12 permutations as having no adjacent digits that differ by more than two. Workout 1 171. The cuts to the rectangular sheet of paper are made both vertically and horizontally. The vertical cuts reduce the length to one-half of one-fourth of one-sixth of the original length, so we have 1/2 × 1/4 × 1/6 × original length = 2 cm. So, the length of the original sheet is (2)(4)(6)(2) = 96 cm. Similarly, the horizontal cuts reduce the width to one-third of one-fifth of the original width, so we have 1/3 × 1/5 × original width = 2 cm. So, the width of the original sheet is (3)(5)(2) = 30 cm. The perimeter of the original sheet is 2(96 + 30) = (2)(126) = 252 cm. 172. Based on the information provided, it would take about 54,600,000 km × 1 hour 39, 897 km × 1 day 24 hours ≈ 57 days. 173. To get the greatest possible product AB × CD, we need A, B, C and D to be as large as possible, namely chosen from the set {6, 7, 8, 9}. Since A and C are tens digits, we’ll chose those digits from the subset {8, 9}. So, the greatest product will be 9B × 8C (or 8B × 9C). Since there are only two options for choosing B and C from the subset {6, 7}, we’ll calculate both possible products to see which is greater. We get (96)(87) = 8352 and (97)(86) = 8342. The greatest product, then, is 8352. 174. Six full weeks of school lunches would be 30 lunches at $3.00 each for a total of (30)(3) = $90.00. Excluding one lunch to account for the holiday, we see that Karli spent 90 − 3 = $87 or 87.00. 175. From least to greatest, the numbers are 10, 15, 20, 24, 24 and 33. Their sum is 126, so the mean is 126/6 = 21. The median is the average of 20 and 24, which is 22. The mode is the value that occurs the most, which is 24, and the range is 33 – 10 = 23. The sum of the mean, median, mode and range is 21 + 22 + 24 + 23 = 90. 176. Since each lap is 2.5 miles, the 500-mile race is a total of 500/2.5 = 200 laps. Since the car consumes 1.3 gallons of fuel per lap, during the entire race it consumes (200)(1.3) = 260 gallons. 177. The two television screens are geometrically similar, so their length-to-width ratios must be equal. We can set up the proportion 56/33 = 48/w, where w represents the width of the smaller screen. Solving for w yields 56w = (33)(48) w = (33)(48)/56 ≈ 28.3 inches. 178. Lindsay’s average walking speed was 15, 000 feet 50 minutes × 60 minutes 1 hour × 1 mile 5280 feet ≈ 3.4 mi/h. 179. Evaluating each variable, we get the average a = (3.27 + 17.95)/2 = 10.6100, the product b = (32.7)(0.4382) ≈ 14.3291, the quotient c = 2.637/0.316 ≈ 8.34494 and the absolute difference d = 804.3692 − 793.241 = 11.1282. From least to greatest, the values are 8.34494, 10.6100, 11.1282 and 14.3291. The median of these numbers is the average of the two middle values, which is (10.6100 + 11.1282)/2 ≈ 10.87. MATHCOUNTS 2018-2019 69 180. Recall that the area, in square units, of an equilateral triangle of side length s is s 2 × √3/4. Since the hexagon is divided into six congruent equilateral triangles of side length 4√3 units, four of which are shaded, it follows that the shaded region has area (4)(4√3)2(√3/4) = 48√3 units2. Workout 2 181. According to Heron’s formula, the area of a triangle with side lengths a, b and c equals s ( s − a )( s − b )( s − c ) , where s = (a + b + c)/2, which is the semiperimeter of the triangle. For this triangle, s = (16 + 30 + 34)/2 = 80/2 = 40, so its area is √[(40)(40 − 16)(40 − 30)(40 − 34)] = √[(40)(24)(10)(6)] = √57,600 = 240 units2. Alternatively, you may recognize that the side lengths 16, 30 and 34 are a multiple of the 8-15-17 Pythagorean Triple, meaning the triangle in question is a right triangle with legs of length 16 and 30. Its area is (1/2)(16)(30) = 240 units2. 182. The radius of the cylinder is 14/2 = 7 inches, so its volume is V = πr 2h = π(72)(12) = 588π in3, which equals (588π)/231 ≈ 8 gallons. 183. We will evaluate each greatest common divisor (GCD) and least common multiple (LCM) expression starting with the innermost parentheses and then working outward. First, we evaluate 2 @ 7 = LCM(2, 7). Since 2 and 7 are relatively prime, it follows that the LCM is 2 × 7, so (2 @ 7)2 = 22 × 72. Next, let’s evaluate (22 × 72) # 42 = GCD(22 × 72, 42). Since 42 = 2 × 3 × 7, it follows that the GCD is 2 × 7. Finally, we evaluate (2 × 7) # 105 = GCD(2 × 7, 105). Since 105 = 3 × 5 × 7, it follows that the GCD is 7. 184. During the 4 o’clock hour, the sum of the digits displayed is 10 for these 6 1-minute intervals: 4:06, 4:15, 4:24, 4:33, 4:42 and 4:51. During the 5 o’clock hour, the sum of the digits displayed is 10 for these 6 1-minute intervals: 5:05, 5:14, 5:23, 5:32, 5:41 and 5:50. The total is 12 minutes. 185. The density of water is 1 gram per cubic centimeter, and 1 kilogram = 1000 grams, so 1 kilogram of water would occupy 1000 cm3 of space. When the same mass of water freezes, it will take up more space. Since the density of ice is 91.67% that of water, or 0.9167 gram/cm3, the space occupied by the frozen 1 kilogram = 1000 grams of water is 1000/0.9167 ≈ 1091 cm3. 186. Evaluating aek + bfg + cdh − ceg − afh − bdk for a = 2, b = −5, c = 3, d = 0, e = 4, f = −6, g = −1, h = 8 and k = 7, we get (2)(4)(7) + (−5)(−6)(−1) + (3)(0)(8) − (3)(4)(−1) − (2)(−6)(8) − (−5)(0)(7) = 56 + (−30) + 0 − (−12) − (−96) − 0 = 56 − 30 + 12 + 96 = 134. 187. The figure shows trapezoid ABCD with diagonals AC and BD that are perpendicular to each other and with AC = BD = CB. So, triangle AED is an isosceles right triangle, with m EAD = mEDA = 45 degrees. In addition, triangles ACB and DBC are congruent isosceles triangles, and mCAB = mCBA = mBDC = mBCD = x degrees. We are asked to find mCBA + mBCD = 2x. Since the sum of the interior angles of a quadrilateral is 360 degrees, it follows that 4x + 45 + 45 = 360 4x = 270 2x = 135 degrees. D A E C 188. The 7 subsets with a sum of 15 are {15}, {10, 4, 1}, {10, 3, 2}, {8, 6, 1}, {8, 4, 3}, {8, 4, 2, 1} and {6, 4, 3, 2}. 189. If the shaded octagon has side length s, then the four unshaded isosceles right triangles each have legs of length s/√2 = (s√2)/2 units, and the square has side length (s√2)/2 + s + (s√2)/2 = s(√2 + 1) units. Together, two of the unshaded triangles form a square of side length (s√2)/2 units. So, the four unshaded triangles have a total area of 2 × ((s√2)/2)2 = s 2 units2. Since the square has area (s(√2 + 1))2 = s 2(3 + 2√2) units2, then s 2/(s 2(3 + 2√2)) = 1/(3 + 2√2) ≈ 0.17 = 17% of the figure is unshaded, meaning the shaded octagon accounts for 100 − 17 = 83% of the figure. 190. Spike digs 8 holes in 3 hours, or 8/3 holes per hour. Similarly, Butch digs 7/4 holes per hour, and Lucky digs 6/5 holes per hour. Working together, they would dig 8/3 + 7/4 + 6/5 = 337/60 holes in an hour. At this rate, digging 3 holes would take Spike, Butch and Lucky 3/(337/60) = 3(60/337) = 180/337 hour, which is equivalent to (180/337)(60) ≈ 32 minutes. Workout 3 191. Let s be the side length of the cube. Since q represents the sum of the edge lengths, q = 12s. Since A represents its total surface area, A = 6s 2. The volume of a cube is V = s 3, and we are told that qA = kV, so (12s)(6s 2) = k(s 3) 72s 3 = ks 3. Thus, k = 72. 192. Let R represent the radius of the sphere, and let r represent the radius of the cylinder. The surface area of the sphere is 4πR 2, and the surface area of the cylinder is 2πr 2 + 2πrh. Since the cylinder’s height is equal to its diameter, or twice its radius, we can rewrite the cylinder’s surface area as 2πr 2 + (2πr)(2r) = 2πr 2 + 4πr 2 = 6πr 2. We know that the sphere has six times the surface area of the cylinder, so 4πR 2 = (6)(6πr 2) R 2 = 9r 2 R = 3r. The cylinder has volume πr 2h = πr 2(2r) = 2πr 3, and the sphere has volume (4/3)πR 3 = (4/3)π(3r)3 = 36πr 3. The ratio of the volume of the cylinder to the volume of the sphere is 2πr 3/(36πr 3) = 1/18. 193. All the triangles in the figure are isosceles right triangles. Let n be the leg length of the smallest triangle. We will categorize and count the triangles based on their leg lengths. There are 32 triangles with legs of length n. There are 18 triangles with legs of length 2n. There are 8 triangles with legs of length 3n. There are 2 triangles with legs of length 4n. That’s a total of 32 + 18 + 8 + 2 = 60 triangles. Alternatively, another counting method is to examine the lattice points of the five rows, starting at the leftmost lattice point of the top row and counting the number of triangles for which that point is the right-angle vertex. The totals, in order, for all five rows are: 4 + 3 + 2 + 1 + 0 = 10, 3 + 4 + 3 + 2 + 1 = 13, 2 + 3 + 4 + 3 + 2 = 14, 1 + 2 + 3 + 4 + 3 = 13 and 0 + 1 + 2 + 3 + 4 = 10. That’s a total of 10 + 13 + 14 + 13 + 10 = 60 different triangles. 70 MATHCOUNTS 2018-2019 B 194. Rex reads 409 – 313 = 96 pages in the time that Wren writes 9 – 5 = 4 pages. That’s a ratio of 96/4 = 24/1, meaning Wren writes a page for every 24 pages Rex reads. In the time it takes Wren to write pages 10, 11 and 12 of her report, Rex will read (3)(24) = 72 more pages. Since 409 + 72 = 481, when Wren finishes page 12 of her report, Rex will finish reading page 481. 195. The function given is recursive, since the value of each term after the second term is based on the values of the two previous terms. We have F(1) = 1, F(2) = 1, F(3) = 1 + 1 = 2, F(4) = 1 + 2 = 3, F(5) = 2 + 3 = 5, F(6) = 3 + 5 = 8, F(7) = 5 + 8 = 13, F(8) = 8 + 13 = 21, F(9) = 13 + 21 = 34, F(10) = 21 + 34 = 55 and F(11) = 34 + 55 = 89. 196. As shown in Figure 1, the largest square that can fit inside a circle of diameter 2 cm is an inscribed square with diagonal of length 2 cm. Recall that the area of a square is half the product of its diagonal lengths. So the square has area (1/2)(2)(2) = 2 cm2. Let the largest square that can fit inside a semicircle of diameter 2 cm have side length s and area s 2. It is inscribed as shown in Figure 2. The center s 2 of the semicircle is the midpoint of a side of the square, and the radius drawn to a vertex on the arc of the semicircle forms s a right triangle with side lengths s/2 , s and 1 cm. Using the Pythagorean Theorem to find s 2, we get (s/2)2 + s 2 = 12 s 2/4 + s 2 = 1 5s 2 = 4 s 2 = 4/5 cm2. So, the desired ratio is 2/(4/5) = 2(5/4) = 5/2. Figure 1 Figure 2 197. The greatest possible value of a is the number of factors of 2 there are in 20! = (1)(2)(3)(4)(5) ⋯ (20). This product has 10 even numbers that each provide a factor of 2. It has 5 multiples of 4 that each provide a second factor of 2. It has 2 multiples of 8 that each provide a third factor of 2. Finally, the number 16 provides 1 more factor of 2. That’s 10 + 5 + 2 + 1 = 18 factors of 2, so the greatest possible value of a is 18. 198. The sixth grade increased by 10% to (390)(1.10) = 429 students. The seventh grade increased by 22% to (350)(1.22) = 427 students. The eighth grade increased by 20% to (420)(1.20) = 504 students. The original population of the middle school was 390 + 350 + 420 = 1160 students, and the new population is 429 + 427 + 504 = 1360. That’s an increase of 1360/1160 − 1 ≈ 0.1724 ≈ 17.2%. C 199. The figure shows that the solid that results from rotating a 6-8-10 right triangle about the side of length 10 units is composed of 6 two cones. Altitude BP creates triangles APB and BPC, which are similar to each other and to triangle ABC. We can use properties B P of similar figures to find BP and CP. Since ratios of corresponding sides of similar figures are proportional, we know that BP/6 = 8/10. Cross-multiplying, we get 10 × BP = 48, so BP = 4.8 units. We also know that CP/6 = 6/10. Again, cross-multiplying, we 8 get 10 × CP = 36, so CP = 3.6 units. It follows, then, that AP = 10 – 3.6 = 6.4 units. The cone of height CP = 3.6 units and base 2 3 radius 4.8 units has volume (1/3)π(4.8) (3.6) = 27.648π units . The cone of height AP = 6.4 units and base radius 4.8 units has volume (1/3)π(4.8)2(6.4) = 49.152π units3. The combined volume of the two cones is 27.648π + 49.152π = 76.8π ≈ 241.3 units3. It is interesting to A note that this equals the volume of the cone with base radius 4.8 units and height 10 units that would result from rotating a right triangle with side lengths 24/5 = 4.8 units, 10 units and (2√769)/5 ≈ 11.1 units about the side of length 10 units, which makes sense because that right triangle has the same area as the 6-8-10 right triangle. 200. Because Vikram entered only one digit incorrectly, one of his factors must have been either 67 or 58. Since 3596/67 is not a whole number, but 3596/58 = 62, his factors must have been 58 and 62. The desired sum is 58 + 62 = 120. Workout 4 201. At 29 miles per gallon, the amount Alice would spend on gas is (113/29)(2.50) ≈ $9.74. At 0.02 gallon per mile, the amount Phil would spend on gas is (113)(0.02)(2.50) = $5.65. The absolute difference in these amounts is 9.74 − 5.65 = $4.09. 202. Since we’re asked to find a ratio, let’s assume the cheese cube has edge length 1 unit and volume 1 unit3. The diagonal slice divides the cheese into two equal pieces, so we need to find the radius of a sphere with a volume of 1/2 unit3. The volume of a sphere is given by the formula V = (4/3)πr 3. So, we have 1/2 = (4/3)πr 3 3/8 = πr 3 3/(8π) = r 3 r = ∛[3/(8π)], and the desired ratio is 1/∛[3/(8π)] ≈ 2.03. 203. Three consecutive rounds last a total of 24 + 48 + 96 = 168 minutes = 2 hours 48 minutes. So, the third round will end 2 hours 48 minutes after 6:00 a.m., which is at 8:48 a.m. 204. The 13th value is the median of the 25 ordered values. Starting with either the first or last leaf, we count to find that 4 is the 13th value, with corresponding stem 1.4. The median value 1.4 | 4 represents a height of 1.44 meters. 205. First, let’s algebraically manipulate 2/a − 3/b = 1/5 so that we have an expression for one of the variables in terms of the other. We get (2b −3a)/(ab) = 1/5 10b − 15a = ab 10b − ab = 15a b(10 − a) = 15a b = 15a/(10 − a). We are looking for positive integer values of a and b, and 1 ≤ a ≤ 9 yields b > 0. The table shows the values of b that result when we substitute a 1 2 3 4 5 6 7 8 9 the integers 1 through 9 for a. The ordered pairs of positive integers (a, b) that satisfy the equation b 5/3 15/4 45/7 10 15 45/2 35 60 135 are (4, 10), (5, 15), (7, 35), (8, 60) and (9, 135). There are 5 such ordered pairs. 206. We need to determine the number of steps s that are equivalent to the additional 5 − 4.11 = 0.89 mile R.J. must walk to reach 5 miles total. We can set up a proportion and solve as follows: 10,002/4.11 = s/0.89 4.11s = (10,002)(0.89) s = (10,002)(0.89)/4.11 ≈ 2166 steps. 207. If we multiply pounds per liter by liters per gallon by dollars per pound, the result is dollars per gallon. Doing so yields (1.33)(3.78541)(1.30) ≈ $6.54. MATHCOUNTS 2018-2019 71 208. The entire rectangular garden has area (10)(20) = 200 m2. The two right-triangular shaded regions each have a pair of perpendicular sides measuring 9 meters and 19 meters. So, the shaded regions have total area (2)(1/2)(9)(19) = 171 m2. The remaining 200 – 171 = 29 m2 area belongs to the walkway. The walkway will have concrete of depth 10 cm = 0.1 meter, which requires a total of (29)(0.1) = 2.9 m3 of concrete. At $70 per cubic meter, the cost of the concrete will be (2.9)(70) = $203 or 203.00. 209. After n years, Bo’s account balance will be $10,000 ×1.075n. Since $30,000 is 3 times the original investment amount, we’re looking for the smallest value of n for which 1.075n ≥ 3. With a calculator and a bit of guessing and checking, we find that 1.07515 ≈ 2.96, which is not quite enough, but 1.07516 ≈ 3.180793. After 16 years Bo’s account balance will be 10,000 ×1.07516 ≈ $31,807.93. He will be 15 + 16 = 31 years old. 210. There are 10 letters in the word MATHCOUNTS, 9 of which are distinct. In a two-letter permutation, the order of the letters matters. There are 9 choices for the first letter and 8 choices for the second letter, for a total of (9)(8) = 72 permutations. Workout 5 P 2 2√ 2 O 211. In the figure shown, four diagonals divide the octagon into eight congruent triangles, one of which is triangle NOP. Let’s start r by extending two sides of the octagon to form 45-45-90 right triangle PQN as shown. We know that NP = 4 cm, and based on properties of 45-45-90 right triangles, NQ = QP = 2√2 cm. Notice that the eight congruent triangles each have height 2 + 2√2 cm. Thus, triangle NOP has area (1/2)(4)(2 + 2√2) = 4 + 4√2 cm2. We are told that the shaded region has area equal to the unshaded N 4 circle. Now, notice that triangle NOP, which is 1/8 of the octagon, is also composed of shaded and unshaded regions that are equal in area. So, the unshaded region of triangle NOP has area (1/2)(4 + 4√2) = 2 + 2√2 cm2. Finally, since the unshaded region of triangle NOP is Q 1/8 of the circle with radius r, it follows that (1/8)πr 2 = 2 + 2√2 πr 2 = 16 + 16√2 r 2 = (16 + 16√2)/π r = √[(16 + 16√2)/π] ≈ 3.5 cm. 1 3″ × 3 = 4″ ¾ ″ × 4 = 3″ ½ ″ × 3 = 1½″ 4 × 3 × 3 = 36 blocks 1 3″ × 3 = 4″ ½ ″ × 6 = 3″ ¾ ″ × 2 = 1½″ 212. The figure shows two ways that Jonas can slice the original block of tofu into 36 smaller ½-inch by ¾-inch by 1¹⁄ ³-inch rectangular blocks. Each of the smaller blocks can be sliced in half to create two of the triangular prisms described, for a total of 36 × 2 = 72 prisms, which makes sense because the original block of tofu has a volume of (1.5)(4)(3) = 18 in3, each triangular prism has volume (1/2)(4/3)(3/4)(1/2) = 1/4 in3 and 18 ÷ 1/4 = 72. So, each prism is 1/72 of the 16 ounces, or 16/72 = 2/9 ounce. 6 × 2 × 3 = 36 blocks 213. Since the 4 acoustic songs must be performed in the second half of the 10-song concert and since Blake doesn’t want to end the concert with an acoustic song, it follows that the acoustic songs must be the 6th, 7th, 8th and 9th songs that Blake performs. There are 5P4 = (5)(4)(3)(2) = 120 ways to choose and order 4 acoustic songs. There are 10P6 = (10)(9)(8)(7)(6)(5) = 151,200 ways to choose and order 6 other songs. Thus, there are (120)(151,200) = 18,144,000 different sequences of songs Blake can perform. 214. If Chris gets four, three or two hits, he will meet the condition of getting at least two hits. We will consider the probability that he does NOT meet this condition; in other words, he gets zero hits or one hit. The probability that Chris gets zero hits is 0.634 = 0.15752961. If he gets exactly one hit, it can be in any of the four times at bat, so the probability that he gets one hit is (4)(0.37)(0.633) = 0.37006956. That means the probability that he gets at least two hits in four times at bat is 1 − (0.15752961 + 0.37006956) = 0.47240083 ≈ 47%. 215. Before Melissa got injured, the Bobcats won 41/(41 + 23) = 41/64 of their games. At the end of the season, the Bobcats had a win-loss record of 54 to 34, which means that, without Melissa, they won 54 – 41 = 13 games and lost 34 – 23 = 11 games. So, they won 13/(13 + 11) = 13/24 of the games they played without Melissa. The absolute difference in the Bobcats’ win percentages with and without Melissa is 41/64 – 13/24 ≈ 0.09896 ≈ 10%. 216. If t represents Tyler’s score, the list of scores from least to greatest is 61, 72, t, 86, 93 since Tyler’s score is the median. The sum of the five scores is 312 + t. Since Tyler’s score is also the mean, we have (312 + t)/5 = t 312 + t = 5t 312 = 4t t = 312/4 = 78 points, which is the middle score. 217. Since x = 6 and x = −4 are solutions of x 2 + bx + c = 0, then (x − 6)(x + 4) = 0 x 2 + 4x − 6x − 24 = 0 x 2 − 2x − 24 = 0. We now see that b = –2. Alternatively, you may recall that for quadratic equations of this form, with a leading coefficient of 1, b is the opposite of the sum of the solutions. So, b = −(6 − 4) = –2. 218. Using a calculator, we can find the three eight-digit powers of 2. They are 224 = 16,777,216, 225 = 33,554,432 and 226 = 67,108,864. The value of N is 33,554,432. 219. The luxury version of the table costs (1000)(1.35) = $1350. Karthik’s 15% discount means he will pay 85% of the cost, which is (1350)(0.85) = $1147.50. After applying the sales tax, Karthik will pay (1147.50)(1.06) = $1216.35. 220. Let s be the shorter base length of the trapezoid, which also is the side length of the equilateral triangle. Using the known angle measures of the equilateral triangle and the two isosceles right triangles, we determine that the isosceles triangle that shares a side with the longer base of the trapezoid has base angles measuring 30 degrees. We can divide this triangle, as shown, to create two congruent right triangles with hypotenuse length s. Based on properties of 30-60-90 right triangles, each of these right triangles has a short leg of length s/2 and a long leg of length s√3/2. It follows, then, that the longer base of the trapezoid has length (2)(s√3/2) = s√3. The ratio of the lengths of the shorter and longer bases of the trapezoid is s/(s√3) = 1/√3 = √3/3. 72 s 60° 60° 45° 45° 60° 45° 30° √3 s 2 s 2 45° 30° √3 s 2 MATHCOUNTS 2018-2019 Workout 6 221. Ronald’s first attempt has a 20% chance of success. To get to a 100% chance of success, his likelihood of success will have to improve 80%, which can occur in 5% increments with the next 80/5 = 16 attempts. Thus, he must attempt to cast a minimum of 17 spells to guarantee success. (It’s unlikely that it would take this long, since the probability that he fails on each of his first 16 attempts is (0.80)(0.75)⋯(0.10)(0.05) ≈ 3 × 10−8.) 222. The first and largest square in the sequence has a side length of √6561 = 81 units. The second square has a side length of (2/3)(81) = 54 units. The third side length is (2/3)(54) = 36 units. The fourth is (2/3)(36) = 24 units. And the fifth is (2/3)(24) = 16 units. This is the square with an area of 256 units2. The horizontal distance from A to B is 54 + 36 + 24 + 16 = 130 units. The vertical distance from A to B is 81 − 16 = 65 units. So, segment AB is the hypotenuse of a right triangle with legs of lengths 65 units and 130 units. Its length is √(652 + 1302) ≈ 145 units. 223. If we pick any two of the ten cheerleaders, they are guaranteed to be of different heights. We will place the taller cheerleader behind the shorter cheerleader in the leftmost position for the photo. We then choose two of the remaining eight cheerleaders and arrange them, taller behind shorter, just to the right of the first two cheerleaders. Continuing in this manner, we can calculate that there are 10C2 × 8C2 × 6C2 × 4C2 × 2C2 = (45)(28)(15)(6)(1) = 113,400 ways for the cheerleaders to line up for the photo. 224. The math club sold a total of (15)(22) + (10)(17) + 220 = 720 gadgets. Since there are 30 gadgets in a box, the club members must have bought 720/30 = 24 boxes. At $75 per box, this cost them (24)(75) = $1800. The regular price was $6.80 per gadget, so the club members collected (15)(22)(6.80) = $2244 in the first 15 days. The discounted price of 25% off $6.80 was (0.75)(6.80) = $5.10 per gadget, so they collected (10)(17)(5.10) = $867 in the next 10 days. The final price was 30% off $5.10, which is (0.70)(5.10) = $3.57 per gadget, so they collected (220)(3.57) = $785.40 for the remaining gadgets. Their total income was 2244 + 867 + 785.40 = $3896.40. The total profit was the difference between the amount they collected and the amount they spent, which was 3896.40 − 1800 = $2096.40. 225. The value of the base-five number ABC is 25A + 5B + C. We need to convert the base-ten number 47 to base five. Since 47 ÷ 25 is 1 with a remainder of 22, it follows that A = 1. Since 22 ÷ 5 is 4 with a remainder of 2, it follows that B = 4 and C = 2. 226. One way to solve this problem is by calculating a “weighted” average. The top of the pool is a circle of radius 20 feet 20 ft with area 202π = 400π ft2. The bottom of the pool is a circle of radius 15 feet with area 152π = 225π ft2. The top “ring” 15 ft has area 202π − 152π = 400π − 225π = 175π ft2. So, the part of the pool with depth 10 feet is 225π/400π = 9/16 of the total area, and the part of the pool with depth 5 feet is 175π/400π = 7/16 of the total area. Thus, the average depth of the water in the pool is (9/16)(10) + (7/16)(5) ≈ 7.8 feet. Alternatively, we can find the total volume of water in the pool and determine what the depth would be if the pool were circular with radius 20 feet and uniform depth throughout. The total volume of the water is π(202)(5) + π(152)(5) = 2000π + 1125π = 3125π ft3. Solving π(202)h = 3125π for h, we get h ≈ 7.8 feet. 10 ft 227. Let g be the total number of games Braden Holtby would need to play in his career to exceed 691 wins. Assuming that he continues to win 200/319 of the games he plays, we want the minimum value of g that satisfies (200/319)g > 691. Solving this inequality yields g > 691(319/200) = 1102.145. That means he would need to play at least 1103 games. 228. Only square numbers have an odd number of divisors. A number with nine divisors can be the eighth power of a prime or the product of the squares of two primes. We want the least positive odd integer, so we will square the two smallest odd primes, which are 3 and 5, and multiply the results. We get (32)(52) = 225, and its nine divisors are 1, 3, 5, 9, 15, 25, 45, 75 and 225. 229. The ratio of the distance Noa runs (a = 5 ¹⁄ ³ = 16/3 laps) to the distance Lev runs (b = 6 laps) in the same amount of time is a/b = (16/3)/6 = 8/9. So, a = (8/9)b, meaning Noa runs 8/9 of the distance that Lev runs in the same amount of time. It follows, then, that it will take Noa 9/8 as much time as it will take Lev to run the same distance. Since Lev is running a mile in 7.5 minutes = (7.5)(60) = 450 seconds, it will take Noa (450)(9/8) = 506.25 seconds to run a mile. Thus, it will take Noa 506.25 − 450 ≈ 56 seconds longer than Lev to run a mile. 230. Since all we know about this particular triangle is that it has side lengths 4, 6 and 8 inches, we should use Heron’s formula to determine its area. According to Heron’s formula, the area of a triangle with side lengths a, b and c equals s ( s − a )( s − b )( s − c ) , where s = (a + b + c)/2, which is the semiperimeter of the triangle. For this triangle, s = (4 + 6 + 8)/2 = 9 inches, so its area is √[9(9 − 4)(9 − 6)(9 − 8)] = 3√15 in2. Workout 7 231. The statement that x plus its reciprocal equals 4 translates to the equation x + 1/x = 4, which can be rewritten as x 2 + 1 = 4x x 2 − 4x + 1 = 0. The quadratic trinomial cannot by factored, so let’s use the quadratic formula x = [−b ± √(b2 − 4ac)]/(2a), where a, b and c are the coefficients of the terms of the quadratic expression, from left to right. Our solution is x = [4 ± √(42 − (4)(1)(1))]/[(2)(1)] = [4 ± √(16 − 4)]/2 = (4 ± √12)/2 = (4 ± 2√3)/2 = 2 ± √3. If x = 2 + √3, then 2 − √3 is the reciprocal and vice versa. Thus, the absolute difference between x and its reciprocal is (2 + √3) − (2 − √3) = 2 + √3 − 2 + √3 = 2√3. Alternatively, we are given x + 1/x = 4, so let c = x − 1/x be the requested difference. Adding these two equations yields 2x = 4 + c, and subtracting them yields 2/x = 4 − c. If we then multiply the equations 2x = 4 + c and 2/x = 4 − c, we get 4 = 16 − c 2 c 2 = 12 c = √12 = ±2√3. We are asked to find the absolute difference, which is 2√3. MATHCOUNTS 2018-2019 73 232. This range of house numbers will require 101 of the 5s for the first digits, which will cost (101)(8.79) = $887.79. Ashera will need 100 of the 1s and 1 of the 2s for the second digits, which will cost (100)(8.79) + (1)(5.98) = $884.98. From 5100 to 5199, each digit will appear 10 times as the third digit and 10 times as the fourth digit. For the digits 1, 3, 5 and 6, this will cost (4)(20)(8.79) = $703.20. For the digits 0, 2, 4, 7, 8 and 9, this will cost (6)(20)(5.98) = $717.60. Finally, we need 2 more 0s for the third and fourth digits of house number 5200, which will cost (2)(5.98) = $11.96. The total cost is 887.79 + 884.98 + 703.20 + 717.60 + 11.96 = $3205.53. D 233. Rectangle ABCD has side lengths 6 units and 8 units, and diagonal AC has length 10 units, since 6-8-10 is a Pythagorean triple. Let Z be the point in the rectangle whose horizontal position is the same as that of X and whose vertical position is the same as that of Y, as shown in the figure. Triangle YZX is similar to triangle ABC, so YX/YZ = 1/YZ = AC/AB = 10/6, which implies that YZ = 6/10 = 0.6 unit. The total area of the shaded regions is the sum of the areas of two triangles, each of base length 8 units, and with total altitude 6 – 0.6 = 5.4 units. Therefore, the total area of the shaded triangles is (1/2)(8)(5.4) = 21.6 units2. Y 1 X Z A 3456 4356 5346 6345 3465 4365 5364 6354 3546 4536 5436 6435 3564 4563 5463 6453 3645 4635 5634 6534 3654 4653 5643 6543 B 234. The table shows all 4! = (4)(3)(2)(1) = 24 ways to order the ages 3, 4, 5 and 6. Once we exclude the arrangements that contain the groupings 34, 45 and 56, we are left with the 11 orders in which Fran can feed her four dogs. 235. If r is the common ratio between consecutive terms of this geometric sequence, then to go from the second term to the fourth term we must multiply by this common ratio twice. In other words, 15r 2 = 135 r 2 = 9 r = 3. The first term is 15/3 = 5. The third term is (15)(3) = 45. The fifth term is (135)(3) = 405. The arithmetic mean of the terms is (5 + 15 + 45 + 135 + 405)/5 = 121. 236. For each of the first four triangles, its longer leg is the hypotenuse of the adjacent triangle to its right. Using the Pythagorean Theorem, we have x 2 = 242 + 122 + 242 + 122 + 122 + 242 = (3)(122) + (3)(242) = 2160, so x = √2160 = 12√15 units. 237. Let x and y represent the side lengths of the hexagon and equilateral triangle, respectively. Then 6x + 3y = 42 2x + y = 14 y = 14 − 2x. A hexagon of side length x cm can be divided into six congruent equilateral triangles, each of side length x cm. Recall that an equilateral triangle of side length s has area (√3/4)s 2. The hexagon in question has area (6)(√3/4)x 2 = (3√3/2)x 2 cm2, and the equilateral triangle has area (√3/4)y 2 cm2. So, (3√3/2)x 2 + (√3/4)y 2 = 49√3 (3/2)x 2 + (1/4)y 2 = 49 6x 2 + y 2 = 196. Substituting 14 − 2x for y, we get 6x 2 + (14 − 2x)2 = 196 6x 2 + 196 − 56x + 4x 2 = 196 10x 2 − 56x = 0 2x(5x − 28) = 0 x = 0 or 5x − 28 = 0 5x = 28 x = 28/5 = 5.6 cm. 238. Each time Ty flips the fair coin, there are two possible outcomes. Since he flips the coin 20 times, there are 220 = 1,048,576 possible outcomes. We want to know the probability that the coin lands heads up 10 times and lands tails up 10 times, regardless of the order. There are “20 choose 10” favorable outcomes, or 20C10 = 20!/(10! × 10!) = 184,756 outcomes. The probability is 184,756/1,048,576 ≈ 0.176. 239. Since all three coins were 90% silver, the 0.36 troy ounce of silver in a Walking Liberty Half Dollar accounted for 90% of its mass. Its full mass must have been 0.36/0.9 = 0.40 troy ounce. A Mercury Dime was worth 1/5 of a Walking Liberty Half Dollar, so the mass of a Mercury Dime must have been 0.40/5 = 0.08 troy ounce. 240. The smaller right triangle has an area of (1/2)(3)(4) = 6 units2. If the 8-unit side of the larger right triangle is a leg of that triangle, then the other leg has length 3 units to yield the required area of (1/2)(3)(8) = 12 units2. Using the Pythagorean Theorem, we see that in this case, the hypotenuse would have length √(32 + 82) = √73 units. On the other hand, the 8-unit side could be the hypotenuse. Therefore, the sum of the possible lengths of the hypotenuse is 8 + √73 units. Workout 8 241. Scott’s estimate of 6 feet 3 inches is equivalent to (6)(12) + 3 = 75 inches. Miguel’s actual height is 96% of Scott’s estimate, or (0.96)(75) = 72 inches. 242. In the figure, the center of the circular base, C, is also the center of the square base of the prism. The top four vertices of the prism touch the upper boundary of the hemisphere, so radius AC has length 6 inches. If we let x represent the side length A of the square base of the prism, then the prism has height AB = (1/3)x and BC = (√2/2)x. Using the Pythagorean Theorem, we have [(1/3)x]2 + [(√2/2)x]2 = 62 (1/9)x 2 + (1/2)x 2 = 36 (11/18)x 2 = 36 x 2 = 648/11 x = √(648/11) inches. B The prism has volume (1/3)x 3 = (1/3)[√(648/11)]3 ≈ 150.7 in3. 6 C 243. Recall that 13! equals 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1, so its prime factorization is 210 × 35 × 52 × 71 × 111 × 131. Since N is a perfect square factor of 13!, it must be a product of prime factors in this prime factorization, each being the greatest even power of that factor. The greatest possible value of N is 210 × 34 × 52 = 2,073,600. 74 C MATHCOUNTS 2018-2019 244. We will use the figure shown to identify the 76 lines that pass through the centers of exactly four unit cubes in the 4 × 4 × 4 cube. In this figure, the cube on the left shows lines, running from the top face of the cube to the bottom face so that each line passes through the centers of exactly four unit cubes. Through each of the cube’s 6 faces there are 16 such lines that exit the opposite face. So, there are (16)(6)/2 = 48 lines associated with the cube’s faces. Next, the cube in the center of this figure shows lines, running from the left edge of the top face to the right edge of the bottom face so that each line passes through the centers of exactly four unit cubes in the 4 × 4 × 4 cube. Through each of that cube’s 12 edges there are 4 such lines that exit an opposite edge. So, there are (4)(12)/2 = 24 lines associated with the cube’s edges. Lastly, the cube on the right shows a line running from the vertex at the intersection of the top, left and front faces to the vertex at the intersection of the bottom, right and back faces and passing through the centers of exactly four unit cubes. Through each of the cube’s 8 vertices there is 1 line that exits from an opposite vertex. So, there are (8)(1)/2 = 4 lines associated with the cube’s vertices. That accounts for all 48 + 24 + 4 = 76 lines. Based on this, we can determine the number of lines that pass through the centers of exactly ten unit cubes in a 10 × 10 × 10 cube. There would be (100)(6)/2 = 300 lines associated with the cube’s faces, (10)(12)/2 = 60 lines associated with its edges and (8)(1)/2 = 4 lines associated with its vertices. That’s a total of 300 + 60 + 4 = 364 lines. 245. Substituting the given values for a, b and c, we get (12 + 10√2)2 + (x + y√2)2 = (15 + 8√2)2 144 + 240√2 + 200 + (x + y√2)2 = 225 + 240√2 + 128 (x + y√2)2 = 9 x + y√2 = ±3. This equation is true only when either x = 0 or y = 0. If x = 0, then y will be an irrational number, meaning x + y will not be a positive integer. If y = 0, then x = ±3 and x + y = ±3. Thus, the positive integer value of the sum is 3. 246. Let y, g and w represent the weights of a yellow, green and white block, respectively. From Camsie’s experiments with the balance scale, we have the following system of equations: 3g = 1y + 3w and 5w = 1g + 5y. Without loss of generality, we can assign an arbitrary weight to the yellow block. So, letting y = 1, our equations become 3g = 1 + 3w 3g − 3w = 1 and 5w = 1g + 5 1g − 5w = −5. Multiplying this last equation by −3 yields −3g + 15w = 15, and adding the equations 3g − 3w = 1 and −3g + 15w = 15, we get 12w = 16 w = 4/3. Substituting this value into the equation 3g − 3w = 1, we see that 3g − 3(4/3) = 1 3g = 5 g = 5/3. The total weight of 30 green blocks and 60 white blocks, which equals 30(5/3) + 60(4/3) = 130, is balanced with130 yellow blocks. 247. The box originally contains q quarters and d dimes with an initial total value, in cents, of 25q + 10d. Increasing the total value by 7.5% changes the initial total value to 1.075(25q + 10d). Increasing the number of quarters by 10% changes the initial total value to 25(1.10q) + 10d. Equating these two new expressions, we get 25(1.10q) + 10d = 1.075(25q + 10d) 27.5q + 10d = 26.875q + 10.75d 0.625q = 0.75d. So, the desired ratio is q/d = 0.75/0.625 = 6/5. 660 2x 1980 x 248. We can find the area of farmer John’s field by finding the area of a rectangle with dimensions 3x by 2640 feet and then subtracting the area of the rectangle with dimensions 2x by 1980 feet that is not part of his field. The result is 2 3x (3x)(2640) − (2x)(1980) = 7920x − 3960x = 3960x ft . We are told that farmer John has 60 acres total. Since an acre is 40 rods by 4 rods, or 160 square rods, and one rod is 16.5 feet, it follows that an acre is (160)(16.5)(16.5) = 43,560 ft2. So, farmer John’s 60 acres is equivalent to (60)(43,560) = 2,613,600 ft2. We now have the land area in two ways, so we can write 3960x = 2,613,600, so x = 660 feet. 2640 249. Let’s write the sequence as (n + 1) + (n + 2) + ⋯ + (n + k) for some integer k ≥ 2. Since the value of an arithmetic series is 1/2 times the sum of the first and last terms times the number of terms, we have (1/2)(n + 1 + n + k)k = 2019 k(2n + k + 1) = 4038. Note that the prime factorization of 4038 is 2 × 3 × 673. It follows that k must be 2, 3, 6, 673, 1346, 2019 or 4038. The corresponding values of n are 1008, 671, 333, –334, –672, –1009, and –2019. This gives the 7 ways that 2019 can be written as the sum of two or more consecutive integers. 250. The figure shows a circle of radius r with center O. The radius drawn perpendicular to the chords of lengths 16, k and 12 units intersects the chords at P, Q and R, respectively. Now, let OP = x, and we know that PQ = QR = 2 units. A radius is drawn to an endpoint of each chord, as shown, to form right triangles LOP, MOQ and NOR. We can use the Pythagorean Theorem to determine the value of k. For triangles LOP and NOR, respectively, we have x 2 + 82 = r 2 and (x + 4)2 + 62 = r 2. Equating r O the two expressions for r 2, we get x 2 + 82 = (x + 4)2 + 62 x 2 + 64 = x 2 + 8x + 16 + 36 64 = 8x + 52 8x = 12 P L 8 x = 3/2 units. Substituting 3/2 for x in the equation x 2 + 82 = r 2 yields (3/2)2 + 82 = r 2 9/4 + 64 = r 2 265/4 = r 2. For Q triangle MOQ, we have (7/2)2 + (k/2)2 = 265/4. Solving for k, we get 49/4 + k 2/4 = 265/4 49 + k 2 = 265 k 2 = 216 R k = √216 = 6√6 units. N 6 MATHCOUNTS 2018-2019 M k2 75 Other MATHCOUNTS Programs MATHCOUNTS was founded in 1983 as a way to provide new avenues of engagement in math for middle school students. MATHCOUNTS began solely as a competition, but has grown to include 3 unique but complementary programs: the MATHCOUNTS Competition Series, the National Math Club and the Math Video Challenge. Your school can participate in all 3 MATHCOUNTS programs! The National Math Club is a free enrichment program that provides teachers and club leaders with resources to run a math club. The materials provided through the National Math Club are designed to engage students of all ability levels—not just the top students—and are a great supplement for classroom teaching. This program emphasizes collaboration and provides students with an enjoyable, pressure-free atmosphere in which they can learn math at their own pace. Active clubs also can earn rewards by having a minimum number of club members participate (based on school/ organization/group size). There is no cost to sign up for the National Math Club, and registration is open to schools and non-school groups that consist of at least 4 U.S. students in grades 6-8 that have regular inperson meetings. More information can be found at www.mathcounts.org/club, and the 2018-2019 School Registration Form is included on the next page. The Math Video Challenge is an innovative program that challenges students to work in teams of 4 to create a video explaining the solution to a MATHCOUNTS handbook problem and demonstrating its real-world application. This project-based activity builds math, communication and collaboration skills. Students post their videos to the contest website, where the general public votes for the best videos. The 100 videos with the most votes advance to judging rounds, in which 20 semifinalists and, later, 4 finalists are selected. This year’s finalists will present their videos to the students competing at the 2019 Raytheon MATHCOUNTS National Competition, and the 224 Mathletes will vote to determine the winner. Members of the winning team receive college scholarships. Registration is completely free and open to all students in grades 6-8. More information can be found at videochallenge.mathcounts.org. 76 MATHCOUNTS 2018-2019 ! The fastest way to register is online at www.mathcounts.org/clubreg! 2018-2019 SCHOOL REGISTRATION FORM This registration form is for U.S. middle schools only. To register a non-school group (such as a Girl Scout troop, Boys and Girls Club Chapter or math circle) for the National Math Club, please go to www.mathcounts.org/club to review eligibility requirements and register. *indicates required information STEP 1 Tell Us About Your School School (Including Homeschools) with Students in Grades 6-8 X U.S. One school can have multiple clubs, as long as each club has a different club leader. School Name* ______________________________________________________________________ School Type (check one)* Public Charter Private Homeschool Virtual Title I School? (check one)* Yes No Overseas U.S. schools must provide additional information below: My school is sponsored by the U.S. Department of: Defense (DoDDS) State Country ___________________________________________________________________ Estimated Number of Students Participating in Club (Minimum 4)* ___________ Typically, what percentage of your club members are female?* ___________% My school participated in the National Math Club last year (2017-2018). How did you hear about the National Math Club?* Prior Participation Word-of-mouth Mailing (postcard or newsletter) Internet Search Workshop/Conference Email STEP 2 Tell Us Club Leader First & Last Name* ____________________________________________________________________ About You Club Leader Email* ________________________________________________________________________________ Club Leader Alternate Email*_______________________________________________________________________________ Your email and alternate email address will not be made public or shared and will only be used by MATHCOUNTS. Club Street Address* _______________________________________________________________________________________ City, State and ZIP Code* ___________________________________________________________________________________ I participated in MATHCOUNTS when I was in middle school. → Please tell us more info. below! When? (for example, 1999-02) ____________________________ Where? (state/territory) ____________________________ Which Program(s)? MATHCOUNTS Competition Series → Highest Level Reached: ___________________________ The National Math Club (formerly MATHCOUNTS Club Program) Math Video Challenge (formerly Reel Math Challenge) STEP 3 Turn in Your Form ! IMPORTANT! By submitting this form you attest your group consists of at least 4 U.S. students in grades 6-8 who meet in person regularly, and is therefore eligible to participate in the National Math Club. The club leader will receive an emailed confirmation once this registration has been processed. Mail or email a scanned copy of this completed form to: MATHCOUNTS 2018-2019 Address: MATHCOUNTS Registration | 1420 King Street | Alexandria, VA 22314 Email: [email protected] 77 ! The fastest way to add students to your school’s registration is by logging in to your Coach Dashboard at www.mathcounts.org/coaches! 2018–2019 ADDITIONAL STUDENTS REGISTRATION FORM $55 $35 $30 1 $90 $70 $60 2 $125 $105 $90 3 $160 $140 $120 4 $195 $175 $150 5 $230 $210 $180 6 $265 $245 $210 7 $300 $280 $240 8 $335 $315 $270 9 1 (1 individual) 2 (2 ind) 3 (3 ind) 4 (1 team) 5 (1 tm, 1 ind) 6 (1 tm, 2 ind) 7 (1 tm, 3 ind) 8 (1 tm, 4 ind) 9 (1 tm, 5 ind) 10 (1 tm, 6 ind) Mail or email a scanned copy of this completed form to: MATHCOUNTS Registration | 1420 King Street | Alexandria, VA 22314 ! Email: [email protected] IMPORTANT! By submitting this form you (1) agree to adhere to the rules of the MATHCOUNTS Competition Series; (2) attest you have the school administration’s permission to register students for this program under this school’s name; and (3) affirm the above named school is a U.S. school eligible for this program and not an academic or enrichment center. The coach will receive an emailed confirmation and receipt once this additional students registration has been processed. Amount Due $_________ Credit Card Check (payable to MATHCOUNTS Foundation) Money Order Purchase Order #________________ (must include copy of P.O.) Do NOT include any credit card information on this form. Within 5 business days you will receive an email invoice enabling you to pay by credit card. Step 4: Almost done... just fill in payment information and turn in your form! Total # of Registered Students Step 3: Tell us what your school’s FINAL registration should be (including all changes/additions). Principal Name ___________________________________________________________ Principal Signature ______________________________________________________________ My school qualifies for the 50% Title I discount, so the Amount Due in Step 4 will be half the amount I circled above. Principal signature required below to verify Title I eligibility. (postmarked after Dec. 14, 2018) Late Registration (postmarked by Dec. 14, 2018) Regular Rate (postmarked by Nov. 2, 2018) Early Bird Rate # of Students You Are Adding Please circle the number of additional students you will enter in the Chapter Competition and the associated cost below (depending on the date your registration is postmarked). Step 2: Tell us how many students you are adding to your school’s registration. Following the instructions below. School Address ___________________________________________________________________________ City, State ZIP _______________________________________________ Coach Email Address ____________________________________________________________________________________________________________________________________ Coach Name ________________________________________________ School Name _____________________________________________________________________________ Step 1: Tell us about your school so we can find your original registration (please print legibly). MATHCOUNTS 2018-2019 79 Acknowledgments The MATHCOUNTS Foundation wishes to acknowledge the hard work and dedication of those volunteers instrumental in the development of this handbook: the question writers who develop the questions for the handbook and competitions, the judges who review the competition materials and serve as arbiters at the National Competition and the proofreaders who edit the questions selected for inclusion in the handbook and/or competitions. 2017-2018 QUESTION WRITING COMMITTEE Chair: Cody Patterson (STE 94, NAT 95), San Antonio, TX Evan Dummit (STE 99, NAT 00 & 01), Tempe, AZ Nina Otterson, Lakeville, CT Leona Penner, Lincoln, NE Apoorva Rajagopal (STE 02), Palo Alto, CA Alex Rice (STE 00), Jackson, MS Rocke Verser, Loveland, CO 2018-2019 NATIONAL JUDGES Sam Baethge, San Marcos, TX Flavia Colonna, George Mason University, Fairfax, VA Barb Currier, Addison, TX Peter Kohn, James Madison University, Harrisonburg, VA Dave Sundin (STE 84), San Mateo, CA 2018-2019 NATIONAL REVIEWERS William Aldridge, Springfield, VA Erica Arrington, North Chelmsford, MA Rachel Chou (NAT 90), Santa Clara, CA Lars Christensen (STE 89), St. Paul, MN Dan Cory (NAT 84, 85), Seattle, WA John Dempsey, Slingerlands, NY Edward Early (STE 92), Austin, TX Joyce Glatzer, Woodland Park, NJ Helga Huntley (STE 91), Newark, DE Chris Jeuell, Kirkland, WA Doug Keegan (STE 91, NAT 92), Bryan, TX Stanley Levinson, P.E., Lynchburg, VA Howard Ludwig, Ocoee, FL Tom Price, Hickman, ME Randy Rogers (NAT 85), Davenport, IA Patrick Vennebush, Falls Church, VA Craig Volden (NAT 84), Earlysville, VA Judy White, Littleton, MA Special Thanks to: Jerrold Grossman, Rochester, MI Jane Lataille, Los Alamos, NM Leon Manelis, Orlando, FL The Solutions to the problems were written by Kent Findell, William Diamond Middle School, Lexington, MA. MathType software for handbook development contributed by WIRIS, www.wiris.com, Barcelona Spain and Long Beach, CA. MATHCOUNTS FOUNDATION Editor and Contributing Author: Content Editor: Author of Introduction & Program Information: Executive Director: Honorary Chairman: Kera Johnson, Senior Manager of Education Cara Norton, Manager of Outreach Amanda Naar, Communications Manager Kristen Chandler Thomas A. Kennedy, Chairman and CEO of Raytheon Company ©2018 MATHCOUNTS Foundation 1420 King Street, Alexandria, VA 22314 www.mathcounts.org ♦ [email protected] Unauthorized reproduction of the contents of this publication is a violation of applicable laws. Materials may be duplicated for use by U.S. schools. MATHCOUNTS®, The National Math Club®, Math Video Challenge® and Mathlete® are registered trademarks of the MATHCOUNTS Foundation. The MATHCOUNTS Foundation makes its products and services available on a nondiscriminatory basis. MATHCOUNTS does not discriminate on the basis of race, religion, color, creed, gender, sexual orientation, physical disability or ethnic origin. 2019 MATH COUN engagin TS pro g math vides pro U .S. midd ograms f l e s c h o ol stude or ability l eve nts of a ll and imp ls to build con fidence rove att itu math an d probl des towards em solv ing. Foundi ng Spo nsors Nationa l So ciety of Professi onal En gineers Nationa l Counc il o of Math ematics f Teachers CNA In surance Nation al Spo nsors Raytheo nC ompany U.S. De partmen t of Def Northro ense p Grum m a Founda n tion Nationa l Societ y of Professi onal En gineers CNA In surance Texas In strumen ts Incorp 3Mgive orated s Phillips 66 Art of P roblem Solving NextTho ught www.mathcounts.org
