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Solutions to problems from the Berkeley Math Tournament 2016 Team Round. The problems cover topics such as recursive definitions, polygon angles, triangle areas, and functions. The solutions provide step-by-step explanations and diagrams to aid understanding.
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Beckman Math Club
Problem 1 Define an such that a 1 =
3 and for all integers i, ai+1 = a^2 i − 2. What is a 2016?
We can find the first three terms through the recursive definition: a 1 =
3 , a 2 = 1, a 3 = − 1. Note that since −1 = (−1)^2 − 2 , we have that an = a 3 = −1 for n ≥ 3. Hence a 2016 = − 1.
Problem 2 Jennifer wants to do origami, and she has a square of side length 1. However, she would prefer to use a regular octagon for her origami, so she decides to cut the four corners of the square to get a regular octagon. Once she does so, what will be the side length of the octagon Jennifer obtains?
First, draw a diagram and label the diagram, as shown above. The interior angle of a polygon with n sides is 180( nn −2). Therefore, the interior angle of a polygon with 8 sides is 180(8 8 −2)= 135 degrees. This means that the triangles in the corner have angles of 180 − 135 = 45 degrees, making them 45 − 45 − 90 triangles.
Furthermore, a regular octagon has equal side lengths. Call this side length x. As seen from the diagram, the hypotenuse of one of the corner triangles is a side length of the octagon. From the 45 − 45 − 90 triangle proportionality, we know that the ratio of the triangle’s leg to the triangle’s hypotenuse if 1 to
From the diagram, it is easy to see that the side length of the square is made up of the legs of two of the corner triangles and one side of the octagon.
side length of square = 2 (leg of corner triangle) + (side length of the octagon)
Substituting in the expressions we derived for the leg of a corner triangle and the side length of the octagon, 1 = 2 x √ 2
1 =
2 x + x
x =
Therefore, the side length of the octagon is
Problem 4 How many graphs are there on 6 vertices with degrees 1,1,2,3,4,5?
Suppose we let the 6 vertices of the graphs be ABCDEF , and let A be the vertex with degree of 5, thus connecting A to each of BCDEF , making them each with a degree of 1. Now we assume B to be the vertex with degree of 5, connecting three of the remaining four vertices, making there a maximum of one vertex with a degree of one, making the requirement of two degree ones impossible, rendering there to be zero solutions.
Problem 5 Let ABC be a right triangle with AB = BC = 2. Let ACD be a right triangle with angle DAC = 30 degrees and angle DCA = 60 degrees. Given that ABC and ACD do not overlap, what is the area of triangle BCD?
We can use properties of 30-60-90 triangles and 45-45-90 triangles to immediately find that AC = 2
obtain the area of the triangle as
2 · 2 · sin(105◦) =
Problem 7 Suppose f (x, y) is a function that takes in two integers and outputs a real number, such that it satisfies f (x, y) =
f (x, y + 1) + f (x, y − 1) 2
f (x, y) = f (x + 1, y) + f (x − 1 , y) 2 What is the minimum number of pairs (x, y) we need to evaluate to be able to uniquely determine f?
This one is a little hard to describe, but imagine placing two points vertically on the graph. This allows you to determine the value of any point on that vertical line. Now, pick a horizontal point with a separate y value than either of the aforementioned points, and using the vertical points find the value of the point between them that shares a y-value with the new point. Now you can determine any point on that horizontal line. Finally choose a point that shares no x or y values with either point predetermined. Using similar techniques, you can find the scale for a new vertical and a new horizontal line. Once you have two vertical lines and two horizontal lines, you can determine the value of any vertical or horizontal line on the graph and thus can find any point. This means that the number of pairs we need is 4.
Problem 8 How many ways are there to divide 10 candies between 3 Berkeley students and 4 Stanford students, if each Berkeley student must get at least one candy? All students are distinguishable from each other; all candies are indistinguishable.
There is a restriction that each Berkeley student much receive at least one piece of candy; to take care of this restriction, first distribute each of the 3 Berkeley students 1 piece of candy. There are 10 − 3 = 7 pieces of candy left. Now, the problem becomes: distribute 7 candies amongst 7 distinguishable people, where it is acceptable for a person to receive zero pieces of candy. Notice that the condition in which 3 students attend Berkeley and 4 students attend Stanford is irrelevant now.
To find the number of ways to distribute 7 candies to 7 distinguishable people, where it is acceptable for a person to receive zero pieces of candy, we utilize Starts and Bars. Let the candies be represented by stars. Since we want to partition the candy into 7 spaces, where a space represents a person. We need 6 bars, as 6 bars creates 7 spaces.
Now, our problem becomes: how many ways are there to rearrange 13 objects (7 stars and 6 bars)? There are 13 spaces for the 7 stars and 6 bars; choose 6 spaces for the bars, and place the stars in the empty spaces. Therefore, there are
6
= 1716 ways to distribute the candies.
Problem 10 What is the smallest possible perimeter of a triangle with integer coordinate vertices, area 12 , and no side parallel to an axis?
According to Pick’s Theorem, for a triangle with integer coordinate vertices, the area given is A = I + B 2 − 1, with I being the amount of interior points and B being the amount of points on the border. Given that this is a triangle with area of 12 , we can substitute the formula as 12 = I + 3+ 2 B 1 −1, with B 1 being extra vertices on the border or the triangle. We then get 1 = 2I + 3 + B 1 − 2, 0 = 2I + B 1 , and since I and B 1 are both non-negative integers, both I and B 1 is 0, implying that there will be no interior or border points in the triangle. Since all points in the area [0, 2]×[0, 2] won’t make a triangle that either fits the given requirement of no parallel to an axis or the requirement above, the smallest triangle that fits both conditions is the triangle with vertices of (0, 0), (1, 1), (2, 3) with side length of
5, and
13, giving the perimeter of
Problem 11
Circles C 1 and C 2 intersect at points X and Y. Point A is a point on C 1 such that the tangent line with respect to C 1 passing through A intersects C 2 at B and C, with A closer to B than C, such that 2016 · AB = BC. Line XY intersects line AC at D. If circles C 1 and C 2 have radii of 20 and 16, respectively, find the ratio of
Problem 13 Consider an urn containing 51 white and 50 black balls. Every turn, we randomly pick a ball, record the color of the ball, and then we put the ball back into the urn. We stop picking when we have recorded n black balls, where n is an integer randomly chosen from { 1 , 2 , ..., 100 } What is the expected number of turns?
To do this question, let us first find the expected value to pick just one black ball (n = 1). The expected value can be written as:
t=
(t + 1)
This is kind of similar to a geometric series, where r = 10151 , but it is much more difficult because each term is also multiplied by a consecutive integer. Well, we know:
∑^ ∞
t=
rt^ = 1 + r + r^2 + · · · =
1 − r
If you take the derivative of both sides, you get:
1 + 2r + 3r^2 + · · · =
(1 − r)^2
Using this formula for our series, the sum of the series is:
50 101
Now, if you know the expected value of turns to do something once, the expected value to do it n amount of times is just the expected value of doing it once times n. The average value between 1 and 100 is 1012 , so the expected value for all n from 1 to 100 is 10150 · 1012 = 102. 01.
Problem 14
Consider the set of axis-aligned boxes in Rd, B(a, b) = {x ∈ Rd^ : ∀i, ai ≤ xi ≤ bi} where a, b ∈ Rd. In terms of d, what is the maximum number n, such that there exists a set of n points S = {x 1 , ..., xn} such that no matter how one partition S = P ∪ Q with P, Q disjoint and P, Q can possibly be empty, there exists a box B such that all the points in P are contained in B, and all the points in Q are outside B?
Now, we can go back to our original goal of finding a 1 a 2 + a 1 a 3 + a 2 a 3.
a 1 a 2 + a 1 a 3 + a 2 a 3 = s^41 s^42 + s^41 s^43 + s^42 s^43 = (s^21 s^22 + s^21 s^23 + s^22 s^23 )^2 − 2(s^21 s^22 s^23 )(s^21 + s^22 + s^23 )
=
Finally, we need to look for a 1 + a 2 + a 3.
a 1 + a 2 + a 3 = s^41 + s^42 + s^43 = (s^21 + s^22 + s^23 )^2 − 2(s^21 s^22 + s^21 s^23 + s^22 s^23 )
Previously, we have already found the values of s^21 + s^22 + s^23 and s^21 s^22 + s^21 s^23 + s^22 s^23 , so we can just plug them in.
a 1 + a 2 + a 3 = s^41 + s^42 + s^43 = (s^21 + s^22 + s^23 )^2 − 2(s^21 s^22 + s^21 s^23 + s^22 s^23 )
= (−8)^2 − 2 ·
Now, we can finally go back to our original initiative, which was to find
∏^3
i=
(4ai + 81) = (4a 1 + 81)(4a 2 + 81)(4a 3 + 81)
Plugging in the values we have found for a 1 a 2 a 3 , a 1 a 2 + a 1 a 3 + a 2 a 3 and a 1 + a 2 + a 3 :
∏^3
i=
(4ai + 81) = (4a 1 + 81)(4a 2 + 81)(4a 3 + 81)
= 64(a 1 a 2 a 3 ) + 16 · 81(a 1 a 2 + a 1 a 3 + a 2 a 3 ) + 4 · 812 (a 1 + a 2 + a 3 ) + 81^3
= 64(9^4 ) + 16 · 81
Therefore,
i=1(4s 4 i + 81) = 2 (^7) · 38 · 5, and the sum of the powers is 7 + 8 + 1 = 16.
