Download 2015-16 Meet 2 - Fox Valley Math League

Fox Valley Math League
Meet 2 – November 2nd, 2015
Oshkosh North High School
Score
Event 1 (No calculator)
Student Name_________________________________ School Name_______________________Team #___________
Print your name clearly
Every answer must be exact and completely simplified.
1. The first four pentagonal numbers are 1, 5, 12, 22. What is the next Pentagonal number?
1. __________
2. The points (2, a) and (b, 7) lie on the line with the equation x  3 y  13  0 .
2. __________
Find the sum  a  b  .
3. Each side of an isosceles right triangle is the diameter of a semicircle (picture not drawn to scale).
What is the area of the triangle if the sum of the areas of the three semicircles is 200π?
C
3. __________
A
B
4. In the Fibonacci sequence 1, 1, 2, 3, 5, …, each term after the second is the sum of the
previous two terms. How many of the first 100 terms of the Fibonacci sequence are odd?
4. __________
Fox Valley Math League
Meet 2 – November 2nd, 2015
Oshkosh North High School
Score
Event 2 (No calculator)
Student Name_________________________________ School Name_______________________Team #___________
Print your name clearly
Every answer must be exact and completely simplified.
1. Express the following product as a reduced fraction:
1. __________
1 
 1  1  1  
1   1   1     1 

 2  3  4   2015 
2. Next Halloween, Manuel plans to scare twice as many people as Sierra, and Sierra plans
to scare three times as many people as Rafi. In all they plan to scare at most 2016 people.
If no one is scared more than once, at most how many people does Sierra plan to scare?
2. __________
3. Remove eight toothpicks to leave only three squares.
Answer with a sketch a picture of the remaning toothpicks.
3. _________________
Sketch above
4. In a village of 2029 inhabitants, at least x of the residents have the same
two-letter initials (in English). Find the least possible value of x?
4. __________
Fox Valley Math League
Meet 2 – November 2nd, 2015
Oshkosh North High School
Score
Event 3 (No calculator)
Student Name_________________________________ School Name_______________________Team #___________
Print your name clearly
Every answer must be exact and completely simplified.
1. The rows of Pascal’s triangle provides us with the coefficients of ( x  y )n .
What do the levels in Pascal’s Pyramid represent?
1. _______________
In Pascal’s Pyramid (tetrahedron) each face is a Pascal’s Triangle.
Each number in the interior of Pascal’s Pyramid is the sum of the
three numbers immediately above it.
2. Simplify log  log2  log3 9  
2. __________
3. Pat wants to buy four donuts from an ample supply of three types of donuts:
glazed, chocolate, and powdered. How many different selections are possible?
3. __________
4. The y-intercepts of three parallel lines are 2, 3, and 4. The sum of the x-intercepts
of the three lines is 36. What is the slope of these three parallel lines?
4. __________
Fox Valley Math League
Meet 2 – November 2nd, 2015
Oshkosh North High School
Score
Event 4 (Calculator allowed)
Student Name_________________________________ School Name_______________________Team #___________
Print your name clearly
Every answer must be exact and completely simplified unless specified differently.


1. Find the value of the expression  sin

3
 tan

4
 cos
7
4

3 
 sec
 csc  cot 
6
3
6
2 
1. __________
2. Find the area of the region bounded by the graph of x  y  20 .
2. __________
3. What are all three ordered triples of integers (a, b, c) with 0  a  b  c ,
3. __________
for which
1 1 1
  1 ?
a b c
3. __________
3. __________
3x  3 x 1

4. Solve the equation for x. x
3  3 x 2
4. __________
Fox Valley Math League
Meet 2 – November 2nd, 2015
Oshkosh North High School
Score
Event 5 TEAM EVENT (calculator allowed)
Student Name_________________________________ School Name_______________________Team #___________
Print your name clearly
Every answer must be exact and completely simplified unless rounding is specified in a particular problem.
o
1. From the stage of a theater, the angle of elevation of the first balcony is 19 .
1. __________
o
The angle of elevation of the second balcony is 29 . The second balcony is 6.3 meters
above the first balcony. To the nearest tenth, how high above stage level is the first balcony?
2. Find the area of the outer square.
2. __________
3. The triple of positive integers (x, y, z) is called an Almost Pythagorean Triple (APT) if
x  1 and y  1 and x 2  y 2  z 2  1 . For example (5, 5, 7) is an APT.
Determine the values of y and z so that (4, y, z) is an APT.
3. __________
4. A bag contains red and blue marbles. When two marbles are drawn, the probability
that they are both red is equal to the probability they are both blue.
4. __________
The probability that one of each color is drawn is 4 .
7
Find the total number of marbles in the bag.
5. How many triangles, quadrilaterals and pentagons are there in this diagram.
Count only those with vertices on the outer perimeter.
Triangles
5. __________
__________
Quadrilaterals __________
Pentagons
__________
6. In the figure, rectangle ABCD is inscribed with a triangle by selecting points E and F
on the segments AB and BC respectively, so that triangles AED, BEF, and CFD
all have equal area. Find the ratios of AE:EB and BF:FC.
Please note that the diagram (triangles) is not drawn to scale.
D
C
F
A
E
B
6. __________
Fox Valley Math League
Meet 2 – November 2nd, 2015
Oshkosh North High School
Score
Event 1 (No calculator)
Student Name_________________________________ School Name_______________________Team #___________
Print your name clearly
Every answer must be exact and completely simplified.
1. The first four pentagonal numbers are 1, 5, 12, 22. What is the next Pentagonal number?
The next Pentagonal number is 35.
add 4, add 7, add 10, add 13
2. The points (2, a) and (b, 7) lie on the line with the equation x  3 y  13  0 . Find the sum  a  b  .
The sum is 13.
2  3a  13  0
b  21  13  0
3a  15
b8
a 5
so a  b  13
3. Each side of an isosceles right triangle is the diameter of a semicircle (picture not drawn to scale). What is the area of
the triangle if the sum of the areas of the three semicircles is 200π?
If the length of each leg of the isosceles triangle is 2r then the length of the hypotenuse is
C
2r 2 . The sum of the areas of the the three semi-circles is
2
1 2 1 2 1
 r   r   r 2  2 r 2 . Hence 2 r 2  200 . The area of the isosceles right
2
2
A 2


triangle is 2r  200 .
2
B
4. In the Fibonacci sequence 1, 1, 2, 3, 5, …, each term after the second is the sum of the previous two terms. How
many of the first 100 terms of the Fibonacci sequence are odd?
There are 67 odd terms. The parities in the sequence repeat the cycle Odd, Odd, Even. This cycle has a length 3. In
particular the 99th term ends the 33rd cycle. Each cycle contains two odd terms. Therefore the first 99 terms in the
sequence include 2 x 33 = 66 odd terms. Finally, the 100th term in the sequence begins a new cycle, so is odd.
Therefore, the first 100 terms include 66 + 1 = 67 odd terms.
Fox Valley Math League
Meet 2 – November 2nd, 2015
Oshkosh North High School
Score
Event 2 (No calculator)
Student Name_________________________________ School Name_______________________Team #___________
Print your name clearly
Every answer must be exact and completely simplified.
1. Express the following product as a reduced fraction:
1 
 1  1  1  
1   1   1     1 

 2  3  4   2015 
1 2 3 4 2013 2014
1
   


2 3 4 8 2014 2015 2015
2. Next Halloween, Manuel plans to scare twice as many people as Sierra, and Sierra plans to scare three times as many
people as Rafi. In all they plan to scare at most 2016 people. If no one is scared more than once, at most how many
people does Sierra plan to scare?
If Rafi plans to scare n people, then Sierra plans to scare 3n people and Manuel plans to scare double that, 6n.
Altogether, n  3n  6n  2016; so n  201.6 . Since n is an integer, n is at most 201. Since Sierra plans to scare 3n
people, that’s at most 603 people.
3. Remove eight toothpicks to leave only three squares. Answer with a sketch a picture of the remaning toothpicks.
See http://math.sfsu.edu/cm2/papers/Toothpickgeometry.pdf
4. In a village of 2029 inhabitants, at least x of the residents have the same two-letter initials (in English). Find the least
possible value of x?
Since 26  676 and since 2029  3  676  1, the pigeon-hole principle assures us that the minimum value of x is 4.
2
See http://www.mit.edu/~pengshi/math149/talk_pigeonhole.pdf
Pigeon hole principle: If m pigeons are in n holes and m > n, then at least 2 pigeons are in the same hole. In fact, at least
m
 n  (ceiling function = round up) pigeons must be in the same hole. Examples: If there are 10 pigeons and 9 holes at
least one hole must have more than one pigeon. If there are 16 people and 5 possible grades, at least 4 people must
have the same grade.
Fox Valley Math League
Meet 2 – November 2nd, 2015
Oshkosh North High School
Score
Event 3 (No calculator)
Student Name_________________________________ School Name_______________________Team #___________
Print your name clearly
Every answer must be exact and completely simplified.
1. The rows of Pascal’s triangle provides us with the coefficients of ( x  y )n . What do the levels in the Pascal’s Pyramid
represent?
Note: In Pascal’s Pyramid (tetrahedron) each face is a Pascal’s
Triangle. Each number in the interior of Pascal’s Pyramid is the
sum of the three numbers immediately above it.
They represent the coefficients of ( x  y  z )n .
See the trinomial theorem
http://staff.spd.dcu.ie/breens/documents/bitri2.pdf
2. Simplify log  log2  log3 9  
The answer is 0


log log2  log3 32   log  log2  2    log 1  0
3. Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How
many different selections are possible?
There are 15 different selections possible.
The number of possible selections is the number of solutions to the equation g  c  p  4 where g, c, and p
represent, respectively, the number of glazed , chocolate and powdered donuts. The 15 possible solutions to this
equation are:
 4,0,0 ,  0,4,0 ,  0,0,4  ,
 3,0,1 ,  3,1,0  , 1,3,0  ,  0,3,1 , 1,0,3 , 0,1,3
 2,2,0 ,  2,0,2  ,  0,2,2 
 2,1,1 , 1,2,1 , 1,1,2 
4. The y-intercepts of three parallel lines are 2, 3, and 4. The sum of the x-intercepts of the three lines is 36. What is
the slope of these three parallel lines?
4. __________
y  mx  2
The answer is  1 .
4
9  36m  m   1
y  mx  3
y  mx  4
4
0  mx  2  x  2
m

2 3 4
 
 36
m m m
Fox Valley Math League
Meet 2 – November 2nd, 2015
Oshkosh North High School
Score
Event 4 (Calculator allowed)
Student Name_________________________________ School Name_______________________Team #___________
Print your name clearly
Every answer must be exact and completely simplified unless specified differently.


1. Find the value of the expression  sin

3
 tan

4
 cos
7
4

3 
 sec
 csc  cot 
6
3
6
2 
The expression equals 1.
2. Find the area of the region bounded by the graph of x  y  20 .
The area is 800. The region is a diamond shape with intercepts at positive and negative 20. So the area is 200 for
quadrant for a total of 800 square units.
3. What are all three ordered triples of integers (a, b, c) with 0  a  b  c , for which
1 1 1
  1 ?
a b c
Clearly, (a, b, c) = (3, 3, 3) is a solution. In any other solution, at least one fraction must exceed 1 , which
3
means one fraction must equal 1 . Since 0  a  b  c , it follows that, in any other solution a  2 . Now, solve
2
1 1 1
  in positive integers. This is a simpler version of the original equation. This time, an obvious solution is
b c 2
(b, c)  (4,4) . In any other solution, one fraction must exceed 1 . That means that one fraction must equal 1 .
4
3
1 1 1 1
Thus,    . Finally the only positive integer solutions are the ordered triples (3,3,3), (2,4,4), (2,3,6) .
c 2 3 6
4. Solve the equation for x.
3x  3 x 1

3x  3 x 2
1
. This could be solved graphically since this is a calculator round.
2
(3x  3 x ) 3x 1
32 x  1 1



  2  32 x  2  32 x  1 32 x  31  2 x  1  x  1
x
x
x
2x
2
(3  3 ) 3
2
3 1 2
The solution is x 
Fox Valley Math League
Meet 2 – November 2nd, 2015
Oshkosh North High School
Score
Event 5 TEAM EVENT (calculator allowed)
Student Name_________________________________ School Name_______________________Team #___________
Print your name clearly
Every answer must be exact and completely simplified unless rounding is specified in a particular problem.
o
1. From the stage of a theater, the angle of elevation of the first balcony is 19 . The angle of elevation of the second
o
balcony is 29 . The second balcony is 6.3 meters above the first balcony. To the nearest tenth, how high above stage
level is the first balcony?
tan 19  
x
x
y
y
tan 19
tan  29  
x  6.3
x  6.3
y
y
tan  29 

x
x  6.3

 x tan  29   tan 19
tan 19  tan  29 

x tan  29   tan 19
  x  6.3
   6.3tan 19   x  tan
6.3tan 19
  10.3 meters
 29   tan 19 
2. Find the area of the outer square.
2. __________
Let A denote the leftmost point (shown) on the small circle and C be the rightmost point shown.
Let B denote the “top” point shown on the small circle. Let O be the center of the square.
Now let x be the length of AO. The then the length of BO is 2 + x since (6 + x = 4 + BO) and the
length of OC is 6 + x. Now an elementary theorem from geometry tells us that abc  90 . It follows that triangles
ABC and AOB are similar. By similar triangles we have
length of the square is 16 so the area is 256.
OR
6  2x
x2  2  x 
2

x2  2  x 
x
2
 x  2 . The side
Inside the small circle are two intersecting chords. Call the lengths of the segments h, h(vertical ), l , L(horizontal ) .
Clearly L  h  4  6  l since all three are the radius of the large circle. By the intersecting chords theorem,
h 2  lL   h  2 h  4  h 2  2h  8  h  4 This implies the large circle has radius 8 , which means the square
has sides of length 16, hence an area of 256.
3. The triple of positive integers (x, y, z) is called an Almost Pythagorean Triple (APT) if
x  1 and y  1 and x 2  y 2  z 2  1 . For example (5, 5, 7) is an APT. Determine the
values of y and z so that (4, y, z) is an APT.
The only APT with x  4 is  4,7,8 or y = 7 and z = 8
If  4, y, z  is an APT, then 42  y 2  z 2  1 or z 2  y 2  15 and so  z  y  z  y   15 .
Since y and z are positive integers, then  y  z  is a positive integer and thus  z  y  is also a positive integer (since
the product of the two factors is 15). That is  z  y  and  y  z  are the possible pairs of factors of 15, of which there
are two: 1 and 15, and 3 and 5.
Since z  y  z  y , we have the following two systems of equations to solve:
z  y 1
z y 3
z  y  15  2 z  16 or z  8 and so y  7
z  y  5  2 z  8 or z  4 and so y  1
Since y  1 , this second solution is not an APT. The only APT with x  4 is  4,7,8 .
4. A bag contains red and blue marbles. When two marbles are drawn, the probability that they are both red is equal to
the probability they are both blue. The probability that one of each color is drawn is 4 . Find the total number of
7
marbles in the bag.
The answer is 8 marbles. From the given information one can conclude that there are an equal number of red and blue
marbles. If we consider the number of ways to succeed divided by the possible outcomes we get:
ways to get red  ways to get blue
Possible ways to choose 2
xx 4
2x
4
 

x!
7
 x 7
2
( x  2)!2!
 
4 x( x  1)
2x  
 14 x  2 x 2  2 x  x  0 or 8 . Excluding 0, the solution is 8
7
2
5. How many triangles, quadrilaterals and pentagons are there in this diagram.
Count only those with vertices on the outer perimeter.
Triangles
__________
Quadrilaterals __________
Pentagons
__________
 5
 5
 5
triangles     10, quadrilaterals     5, pentagons     1
 3
 4
 5
Consider using a chart to build the solution
Graph
Points
Line
Triangles Quadrilaterals Pentagons
segments
2
1
3
3
1
4
6
4
1
5
10
10
5
1
6. In the figure, rectangle ABCD is inscribed with a triangle by selecting points E and F on the segments AB and BC
respectively, so that triangles AED, BEF, and CFD all have equal area. Find the ratios of AE:EB and BF:FC. Please note
that the diagram (triangles) are not drawn to scale. Round your answer to the nearest thousandth.
The answer is the golden ratio so approximately 1.618
Let x = length AE, y = length EB, s = length BF and t = FC. Since the three areas are equal:
x( s  t )  ys  ( x  y )t  xs  xt  xt  yt  xs  yt 
xs yt
s y
   . So the ratios are the same.
xt xt
t x
xs
into the equal areas formula.
t
2
2
xs 2
s2
s
s 1 5
s
s s
x( s  t )  ys  xs  xt 
 s  t   1     0       1 
The Golden Ratio!
t
t
t
t
2
t
t t
This also leads us to a substitution y 