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2017, AMC 12 A Problem 1 At a gathering of people, there are people who all know each other and people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur? Solution Let the group of people who all know each other be , and let the group of people who know no one be . Handshakes occur between each pair such that and , and between each pair of members in . Thus, the answer is Solution - Complementary Counting The number of handshakes will be equivalent to the difference between the number of total interactions and the number of hugs, which are and , respectively. Thus, the total amount of handshakes is Problem 2 Chloé chooses a real number uniformly at random from the interval . Independently, Laurent chooses a real number uniformly at random from the interval . What is the probability that Laurent's number is greater than Chloe's number? Solution Suppose Laurent's number is in the interval . Then, by symmetry, the probability of Laurent's number being greater is . Next, suppose Laurent's number is in the interval . Then Laurent's number will be greater with probability . Since each case is 1 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 equally likely, the probability of Laurent's number being greater is answer is C. , so the Alternate Solution: Geometric Probability Let be the number chosen randomly by Chloé. Because it is given that the number Chloé choose is interval , . Next, let be the number chosen randomly by Laurent. Because it is given that the number Chloé choose is interval , . Since we are looking for when Laurent's number is Chloé's we write the equation . When these three inequalities are graphed the area captured by and represents all the possibilities, forming a rectangle 2017 in width and 4034 in height. Thus making its area . The area captured by , , and represents the possibilities of Laurent winning, forming a trapezoid with a height 2017 in length and bases 4034 and 2017 length, thus making an area . The simplified quotient of these two areas is the probability Laurent's number is larger than Chloé's, which is . Problem 3 In the figure below, semicircles with centers at and and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter . The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at ? Solution 1 Connect the centers of the tangent circles! (call the center of the large circle 2 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 ) Notice that we don't even need the circles anymore; thus, draw triangle : with cevian and use Stewart's Theorem: From what we learned from the tangent circles, we have , , , , , and , where is the radius of the circle centered at that we seek. Thus: Solution 2 Like the solution above, connecting the centers of the circles results in triangle with cevian . The two triangles and share angle , which means we can use Law of Cosines to set up a system of 2 equations that solve for respectively: 3 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 (notice that the diameter of the largest semicircle is 6, so its radius is 3 and is 3 - r) We can eliminate the extra variable of angle by multiplying the first equation by 3 and subtracting the second from it. Then, expand to find : , so = Solution 3 Let be the center of the largest semicircle and be the radius of . We know that , , , , and . Notice that and are bounded by the same two parallel lines, so these triangles have the same heights. Because the bases of these two triangles (that have the same heights) differ by a factor of 2, the area of must be twice that of , since the area of a triangle is . Again, we don't need to look at the circle and the semicircles anymore; just focus on the triangles. 4 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 Let equal to the area of and equal to the area of states that the area of an triangle with sides . Heron's Formula and is where , or the semiperimeter, is The semiperimeter Formula to obtain of is Use Heron's Using Heron's Formula again, find the area of with sides , , and Now, Solution 4 Let , the center of the large semicircle, to be at Therefore is at Let the radius of circle and is at , and to be at . be . Using Distance Formula, we get the following system of three equations: By simplifying, we get 5 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 . . By subtracting the first equation from the second and third equations, we get which simplifies to When we add these two equations, we get So Problem 4 A square with side length is inscribed in a right triangle with sides of length , , and so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length is inscribed in another right triangle with sides of length , , and so that one side of the square lies on the hypotenuse of the triangle. What is ? Solution 1 Analyze the first right triangle. Note that and . Solving, are similar, so . This can be written as . Now we analyze the second triangle. 6 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 Similarly, and are similar, so , and . Solving for Thus, . Thus, , we get . . Solution 2 (Shortcut using answer choices) As stated in Solution 1, the square in the first triangle has a side length of Then we look at the answers. The only answer with a factor of in the denominator is . . Problem 5 How many ordered pairs such that is a positive real number and is an integer between and , inclusive, satisfy the equation Solution By the properties of logarithms, we can rearrange the equation to read . If , we may divide by it and get , which implies . Hence, we have possible values , namely Since is equivalent to , as we can assign to each , each possible value yields exactly . In total, we have solutions. 7 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 with solutions Problem 6 A set is constructed as follows. To begin, . Repeatedly, as long as possible, if is an integer root of some polynomial for some , all of whose coefficients are elements of , then is put into . When no more elements can be added to , how many elements does have? Solution 1 At first, . At this point, no more elements can be added to . To see this, let with each in . is a factor of , and is in , so has to be a factor of some element in . There are no such integers left, so there can be no more additional elements. has elements Problem 7 For certain real numbers , , and , the polynomial three distinct roots, and each root of is also a root of the polynomial What is ? has Solution Let and be the roots of . Let Vieta's formulas on the quadratic term of following: be the additional root of and the cubic term of 8 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 . Then from , we obtain the Thus . Now applying Vieta's formulas on the constant term of linear term of , we obtain: Substituting for we obtain: , and the in the bottom equation and factoring the remainder of the expression, It follows that . But Now we can factor Then , the linear term of in terms of so as and Hence . Solution 2 Since all of the roots of more than the degree of for some number are distinct and are roots of , we have that . By comparing , and the degree of coefficients, we see that is one . Thus, Expanding and equating coefficients we get that The third equation yields , and the first equation yields 9 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 . So we have that Problem 8 Quadrilateral is inscribed in circle , and and has side lengths . Let and be points on such that and . Let be the intersection of line and the line through parallel to . Let be the intersection of line and the line through parallel to . Let be the point on circle other than that lies on line . What is ? Solution It is easy to see that with a ratio of First we note that Then with a ratio of , so Now we find the length of quadrilateral is cyclic, we can simply use the Law of Cosines. . Because the By Power of a Point, . Thus -solution by FRaelya Problem 9 Call a positive integer if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, , , and are monotonous, but , , and are not. How many monotonous positive integers are there? Solution 1 Case 1: monotonous numbers with digits in ascending order 10 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 There are ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will always be a leading digit and that is not allowed. The sum is equivalent to Case 2: monotonous numbers with digits in descending order There are ways to choose n digits from the digits 0 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in descending order. Note that 0 is included since we are allowed to end numbers with zeros. The sum is equivalent to is not positive. Thus there are We discard the number 0 since it here. Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9. Thus there are monotonous numbers. Solution 2 Like Solution 1, divide the problem into an increasing and decreasing case: Case 1: Monotonous numbers with digits in ascending order. Arrange the digits 1 through 9 in increasing order, and exclude 0 because a positive integer cannot begin with 0. To get a monotonous number, we can either include or exclude each of the remaining 9 digits, and there are ways to do this. However, we cannot exclude every digit at once, so we subtract 1 to get monotonous numbers for this case. Case 2: Monotonous numbers with digits in descending order. This time, we arrange all 10 digits in decreasing order and repeat the process to find ways to include or exclude each digit. We cannot exclude every digit at once, and we cannot include only 0, so we subtract 2 to get monotonous numbers for this case. At this point, we have counted all of the single-digit monotonous numbers twice, so we must subtract 9 from our total. Thus our final answer is . 11 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 Problem 10 The number has over positive integer divisors. One of them is chosen at random. What is the probability that it is odd? Solution If a factor of is odd, that means it contains no factors of . We can find the number of factors of two in by counting the number multiples of , , , and that are less than or equal to .After some quick counting we find that this number is . If the prime factorization of has factors of , there are choices for each divisor for how many factors of should be included ( to inclusive). The probability that a randomly chosen factor is odd is the same as if the number of factors of is which is . Solution by: vedadehhc Solution 2 We can write as its prime factorization: Each exponent of these prime numbers are one less than the number of factors at play here. This makes sense; is going to have factors: , and the other exponents will behave identically. In other words, has factors. We are looking for the probability that a randomly chosen factor of that do not contain multiples of as factors. will be odd--numbers From our earlier observation, the only factors of that are even are ones with at least one multiplier of , so our probability of finding an odd factor becomes the following: Solution submitted by David Kim Problem 11 A coin is biased in such a way that on each toss the probability of heads is probability of tails is and the . The outcomes of the tosses are independent. A player has the choice 12 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B? The probability of winning Game A is less than the probability of winning Game B. The probability of winning Game A is less than the probability of winning Game B. The probabilities are the same. The probability of winning Game A is greater than the probability of winning Game The probability of winning Game A is greater than the probability of winning Game B. B. Solution The probability of winning Game A is the sum of the probabilities of getting three tails and getting three heads which is . The probability of winning Game B is the sum of the probabilities of getting two heads and getting two tails squared. This gives us Game A is and the probability of winning Game B is . The probability of winning , so the answer is Solution by: vedadehhc Problem 12 The diameter of a circle of radius is extended to a point outside the circle so that . Point is chosen so that and line is perpendicular to line . Segment intersects the circle at a point between and . What is the area of ? 13 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 Solution 1 Let be the center of the circle. Note that . However, by Power of a Point, , so . Now . Since . Solution 2: Similar triangles with Pythagorean is the diameter of the circle, so . is a right angle, and therefore by AA similarity, Because of this, Likewise, , so , so Thus the area of 14 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 . . . Solution 3: Similar triangles without Pythagorean Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation: Draw with on . . . .( ratio applied twice) . Problem 13 Let the integers from to divided by ? be the -digit number that is formed by writing in order, one after the other. What is the remainder when is Solution We will consider this number that the number is and . By looking at the last digit, it is obvious . To calculate the number , note that so it is equivalent to Thus it is and , so it is Problem 14 Real numbers and are chosen independently and uniformly at random from the interval . What is the probability that , where denotes the greatest integer less than or equal to the real number ? 15 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 Solution First let us take the case that . In this case, both the interval . Similarly, in the case that , and the probability is . The probability of this is , and lie in the interval and . It is easy to see that the probabilities for lie in for are the infinite geometric series that starts at and with common ratio . Using the formula for the sum of an infinite geometric series, we get that the probability is . Solution by: vedadehhc Problem 15 The graph of , where is a polynomial of degree , contains points , and . Lines , , and intersect the graph again at points and , respectively, and the sum of the -coordinates of , , and is 24. What is , , , ? Solution First, we can define , which contains points , , and . Now we find that lines , , and are defined by the equations , , and respectively. Since we want to find the coordinates of the intersections of these lines and , we set each of them to , and synthetically divide by the solutions we already know exist (eg. if we were looking at line , we would synthetically divide by the solutions and , because we already know intersects the graph at and , which have -coordinates of and ). After completing this process on all three lines, we get that the -coordinates of , , and are , which gives us , and respectively. Adding these together, we get . Substituting this back into the original equation, we get 16 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 , and Solution by: vedadehhc and tdeng Problem 16 Quadrilateral There is a point is has right angles at in the interior of times the area of and , such that . What is , and . and the area of ? Solution 1 Let Theorem, , , and . Since . Note that . By the Pythagorean , the ratios of side lengths must be equal. Since , a point on such that is an altitude of triangle and . Note that . Let F be . Therefore, and . Since and form altitudes of triangles and , respectively, the areas of these triangles can be calculated. Additionally, the area of triangle can be calculated, as it is a right triangle. Solving for each of these yields: Therefore, the answer is Solution 2 Draw line through , , with on and on , . By weighted average . WLOG let . Meanwhile, . . We obtain , namely 17 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 . , The rest is the same as Solution 1. Problem 17 A set of people participate in an online video basketball tournament. Each person may be a member of any number of -player teams, but no two teams may have exactly the same members. The site statistics show a curious fact: The average, over all subsets of size of the set of participants, of the number of complete teams whose members are among those people is equal to the reciprocal of the average, over all subsets of size of the set of participants, of the number of complete teams whose members are among those people. How many values , , can be the number of participants? Solution Solution by Pieater314159 Let there be teams. For each team, there are different subsets of players including that full team, so the total number of team-(group of 9) pairs is Thus, the expected value of the number of full teams in a random set of players is Similarly, the expected value of the number of full teams in a random set of players is The condition is thus equivalent to the existence of a positive integer 18 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 such that Note that this is always less than have that this is equivalent to , so as long as is integral, It is obvious that divides the RHS, and that does iff divides it iff . One can also bash out that possible residues . is a possibility. Thus, we divides it in . Also, out of the Using all numbers from to , inclusive, it is clear that each possible residue is reached an equal number of times, so the total number of working range is . However, we must subtract the number of "working" , which is . Thus, the answer is . 19 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017 in that