Mathprize2016solutions

The Advantage Testing Foundation
2016 Solutions
Problem 1 Let T be a triangle with side lengths 3, 4, and 5. If P is a point
in or on T, what is the greatest possible sum of the distances from P to each
of the three sides of T?
Answer: 4
Solution: Draw segments from point P to each of the vertices of the trian-
gle T, as shown in the ﬁgure below.
P
3
4
5
Let x, y, and z, respectively, be the distances from P to the sides of
length 3, 4, and 5. The areas of the three subtriangles are 3x/2, 4y/2, and
5z/2. The area of the original triangle is 6. So we get the equation
3x + 4y + 5z = 12.
Hence we have
x + y + z ≤
3x + 4y + 5z
3
=
12
3
= 4.
When P is the vertex opposite the side of length 3, we have x = 4, y = 0,
and z = 0, which achieves x + y + z = 4. So the greatest possible value of
x + y + z is 4 .
1

Math Prize for Girls 2016 Solutions
Problem 2 Katrine has a bag containing 4 buttons with distinct letters M,
P, F, G on them (one letter per button). She picks buttons randomly, one at
a time, without replacement, until she picks the button with letter G. What
is the probability that she has at least three picks and her third pick is the
button with letter M? Express your answer as a fraction in simplest form.
Answer:
1
12
Solution: Even though Katrine stops with G, let’s imagine that she picks
all 4 buttons no matter what. We’ll call this scenario the extended game.
The extended game is just a random permutation of the 4 buttons. For
the original game to have at least three picks with the third pick being M
is equivalent to the extended game having the third pick be M and the
fourth pick be G. So in the extended game, the only successful permutations
are PFMG and FPMG. Hence the probability of success is 2
4!
= 2
24
, which
simpliﬁes to
1
12
.
Problem 3 Compute the least possible value of ABCD − AB × CD, where
ABCD is a 4-digit positive integer, and AB and CD are 2-digit positive
integers. (Here A, B, C, and D are digits, possibly equal. Neither A nor C
can be zero.)
Answer: 109
Solution: Let x be AB and y be CD. (Note that x and y are between 10
and 99.) Then ABCD is 100x + y. So ABCD − AB × CD is 100x + y − xy.
By “completing the product” to factor, we see that
100x + y − xy = 100 + (x − 1)(100 − y) ≥ 100 + 9 · 1 = 109.
Equality is achieved when x = 10 and y = 99. So the least possible value
is 109 .
Problem 4 Compute the smallest positive integer n such that 2016n
does
not divide 2016!.
Answer: 335
Solution: The prime factorization of 2016 is 25
·32
·7. So 2016n
is 25n
·32n
·7n
.
The number of factors of 2 in 2016! is
1008 + 504 + 252 + 126 + 63 + 31 + 15 + 7 + 3 + 1 = 2010.
2

672 + 224 + 74 + 24 + 8 + 2 = 1004.
288 + 41 + 5 = 334.
Because 2016n
does not divide 2016!, we have 5n > 2010 or 2n > 1004 or
n > 334. Those three inequalities simplify to n > 402 or n > 502 or n > 334.
Those three inequalities combine to n > 334. So the smallest n is 335 .
Problem 5 A permutation of a finite set S is a one-to-one function from S
to S. A permutation P of the set {1, 2, 3, 4, 5} is called a W-permutation if
P(1) > P(2) < P(3) > P(4) < P(5). A permutation of the set {1, 2, 3, 4, 5}
is selected at random. Compute the probability that it is a W-permutation.
Express your answer as a fraction in simplest form.
Answer:
2
15
Solution: Say that P(2) and P(4) are the low values; say that P(1), P(3),
and P(5) are the high values. It’s easy to see that 1 must be low. It’s easy to
see that 4 and 5 must be high. So the low values are either {1, 2} or {1, 3}.
Let’s look at the first case: the low values are {1, 2}. There are 2! = 2
ways to assign them to P(2) and P(4). There are 3! = 6 ways to assign
{3, 4, 5} to P(1), P(3), and P(5). So there are 2 · 6 = 12 W-permutations in
this case.
Let’s now look at the second case: the low values are {1, 3}. There are
2! = 2 ways to assign them to P(2) and P(4). For concreteness, suppose that
P(2) = 1. Then P(1) must be 2. There are 2! = 2 ways to assign {4, 5} to
P(3) and P(5). So there are 2 · 2 = 4 W-permutations in this case.
Altogether, there are 12 + 4 = 16 W-permutations. So the probability of
being a W-permutation is 16
5!
= 16
120
, which simplifies to
2
15
.
Problem 6 The largest term in the binomial expansion of (1 + 1
2
)31
is of
the form a
b
, where a and b are relatively prime positive integers. What is the
value of b? As an example of a binomial expansion, the binomial expansion
of an expression of the form (x + y)3
is the sum of four terms
x3
+ 3x2
y + 3xy2
+ y3
.
3

Answer: 1024
Solution: The binomial expansion of (1 + 1
2
)31
is
31
k=0
31
k
1
2k
.
Let ak be the kth term 31
k
1
2k . We can express ak+1 in terms of ak:
ak+1 =
31
k + 1
1
2k+1
=
31 − k
k + 1
31
k
1
2k+1
=
31 − k
2(k + 1)
ak .
So the ratio ak+1
ak
is 31−k
2(k+1)
. This ratio is greater than 1 if k ≤ 9 and is less
than 1 if k ≥ 10. In particular, a10 is the largest term. The value of a10 is
31
10
1
210 . Because 31 is 1 less than a power of 2, every binomial coefficient 31
k
is odd. So the denominator of a10 is 210
, which is 1024 .
Problem 7 Let S be the set of all real numbers x such that 0 ≤ x ≤ 2016π
and sin x < 3 sin(x/3). The set S is the union of a finite number of disjoint
intervals. Compute the total length of all these intervals. Express your
answer in terms of π.
Answer: 1008π
Solution: The triple-angle formula says that
sin 3θ = 3 sin θ − 4 sin3
θ.
In particular, we have
3 sin(x/3) − sin x = 4 sin3
(x/3).
So sin x < 3 sin(x/3) is equivalent to sin(x/3) > 0. The inequality sin(x/3) >
0 is satisfied on the open intervals (0, 3π), (6π, 9π), . . . , (2010π, 2013π).
There are 2016/6 = 336 such intervals, each of length 3π. So their total
length is 336 · 3π, which is 1008π .
Problem 8 A strip is the region between two parallel lines. Let A and B be
two strips in a plane. The intersection of strips A and B is a parallelogram P.
Let A be a rotation of A in the plane by 60◦
. The intersection of strips A
and B is a parallelogram with the same area as P. Let x◦
be the measure
(in degrees) of one interior angle of P. What is the greatest possible value
of the number x?
4

Answer: 150
Solution: Let hA be the width of strip A (the distance between its two
parallel lines). Let hB be the width of strip B. The parallelogram P has
sides of lengths hA/ sin x◦
and hB/ sin x◦
. In particular, the area of P is
hAhB/ sin x◦
. When A is rotated, the angle of measure x◦
becomes an angle
of measure say y◦
. So the area of the new parallelogram is hAhB/ sin y◦
.
Because the area stayed the same, we have sin x◦
= sin y◦
. Hence y is either
x or 180 − x.
When the strip A is rotated by 60◦
, what happens to the angle of mea-
sure x◦
? If 0 < x ≤ 60, then the new angle has measure (60 − x)◦
or
(x + 60)◦
, depending on the direction of the rotation; for the new angle to
be x◦
or (180 − x)◦
, either x = 30 or x = 60. If 60 ≤ x ≤ 120, then the
new angle has measure (x − 60)◦
or (x + 60)◦
; for the new angle to be x◦
or
(180 − x)◦
, either x = 60 or x = 120. If 120 ≤ x < 180, then the new angle
has measure (x−60)◦
or (x−120)◦
; for the new angle to be x◦
or (180−x)◦
,
either x = 120 or x = 150.
Among these four possible values of x, the greatest is 150 .
Problem 9 How many distinct lines pass through the point (0, 2016) and
intersect the parabola y = x2
at two lattice points? (A lattice point is a
point whose coordinates are integers.)
Answer: 36
Solution: The vertical line through (0, 2016) intersects the parabola y =
x2
at only one point. A non-vertical line through (0, 2016) intersects the
parabola at two points, one on the left and one on the right. Call the two
intersection points (−a, a2
) and (b, b2
), where a and b are positive. Note that
the equation
y = x2
− (x + a)(x − b)
is a linear equation that passes through (−a, a2
) and (b, b2
). So it must be the
equation of the given line. Plugging in x = 0 and y = 2016 gives ab = 2016.
To get lattice points, we are looking for positive integers a and b such
that ab = 2016. The prime factorization of 2016 is 25
· 32
· 71
. So 2016
has 6 · 3 · 2 = 36 positive divisors. Each divisor is a choice for a, which
determines b. So the number of solutions is 36 .
Problem 10 How many solutions of the equation tan x = tan tan x are on
5

the interval 0 ≤ x ≤ tan−1
942? (Here tan−1
means the inverse tangent
function, sometimes written arctan.)
Answer: 300
Solution: Two angles have the same tangent if and only if their difference
is a multiple of π. So tan x = tan tan x if and only if tan x − x is a multiple
of π. Let T be the function defined by T(x) = tan x − x. We are looking
for solutions to T(x) = nπ, where n is an integer. We will show later that T
is strictly increasing on the interval [0, π/2). On that interval, T goes from
0 to infinity. So for every nonnegative integer n, there is a unique x in that
interval such that T(x) = nπ.
We know that 3.14 < π < 3.145. Multiplying by 300, we get 942 <
300π < 943.5. So
T(tan−1
942) = 942 − tan−1
942 < 942 < 300π.
Also
T(tan−1
942) = 942 − tan−1
942 > 942 −
π
2
> (300π − 1.5) −
π
2
> 299π.
Hence T(tan−1
942) is strictly between 299π and 300π. So for every integer n,
the equation T(x) = nπ has a solution on the interval [0, tan−1
942] if and
only if 0 ≤ n < 300. There are 300 such integers n, so the number of solutions
to our original equation is 300 .
We promised to show that T is strictly increasing on the interval [0, π/2).
We will use the well-known inequality (easy to prove with the unit circle)
that tan θ > θ if 0 < θ < π/2. Let x and y be real numbers such that
0 ≤ x < y < π/2. By the tangent subtraction formula, we have
y − x < tan(y − x) =
tan y − tan x
1 + tan x tan y
≤ tan y − tan x.
By rearranging, we see that T(x) < T(y). So T indeed is strictly increasing.
Problem 11 Compute the number of ordered pairs of complex numbers
(u, v) such that uv = 10 and such that the real and imaginary parts of u and
v are integers.
Answer: 48
6

Solution: Because u and v have integer parts, |u|2
and |v|2
are nonnegative
integers. From uv = 10, it follows that |u|2
· |v|2
= 100. So |u|2
and |v|2
are
positive integers whose product is 100. We will divide the count into three
cases: |u| < |v|, |u| = |v|, and |u| > |v|.
Let’s handle the case |u| < |v| first. In that case, |u|2
is a small divisor
of 100: either 1, 2, 4, or 5. If |u|2
= 1, then we have 4 choices for u: either
±1 or ±i. If |u|2
= 2, then we have 4 choices: ±1 ± i. If |u|2
= 4, then we
have 4 choices: ±2 or ±2i. If |u|2
= 5, then we have 8 choices: ±1 ± 2i or
±2 ± i. Altogether, we have 20 choices for u. Each such choice gives a single
valid choice for v, namely v = 10/u = 10u/|u|2
. So we have 20 pairs in the
case |u| < |v|.
Let’s next handle the case |u| = |v|. In that case, |u|2
= |v|2
= 10. So we
have 8 choices for u: either ±1±3i or ±3±i. Each such choice determines v,
namely v = 10/u = u. So we have 8 pairs in the case |u| = |v|.
Finally, we have the case |u| > |v|. By symmetry, it has the same count
as the first case |u| < |v|. So we have 20 pairs in this case.
Altogether, the number of pairs is 20 + 8 + 20, which is 48 .
Problem 12 Let b1, b2, b3, c1, c2, and c3 be real numbers such that for every
real number x, we have
x6
− x5
+ x4
− x3
+ x2
− x + 1 = (x2
+ b1x + c1)(x2
+ b2x + c2)(x2
+ b3x + c3).
Compute b1c1 + b2c2 + b3c3.
Answer: −1
Solution: Let P be the polynomial defined by
P(x) = x6
− x5
+ x4
− x3
+ x2
− x + 1.
Note that (x + 1)P(x) = x7
+ 1. So the roots of P are on the unit circle.
Hence the roots of each quadratic factor x2
+ bkx + ck are also on the unit
circle. Because each quadratic factor has real coefficients, its roots come in
conjugate pairs. Because the roots are on the unit circle, each ck is 1.
When we expand the product of the three quadratic factors, we get a
polynomial of the form
x6
+ (b1 + b2 + b3)x5
+ · · · .
7

Because the coefficient of x5
in P is −1, we see that b1 + b2 + b3 = −1. So
we have
b1c1 + b2c2 + b3c3 = b1 + b2 + b3 = −1 .
Problem 13 Alice, Beth, Carla, Dana, and Eden play a game in which
each of them simultaneously closes her eyes and randomly chooses two of the
others to point at (one with each hand). A participant loses if she points at
someone who points back at her; otherwise, she wins. Find the probability
that all five girls win. Express your answer as a fraction in simplest form.
Answer:
1
324
Solution: Each of the five participants has 4
2
= 6 choices of where to point.
So the total number of choices is 65
.
Consider a directed graph whose vertices are the five participants. Draw
an edge from one participant to another if the first points to the second. The
number of edges is 5 · 2 = 10. The number of unordered pairs of vertices is
5
2
= 10. In an all-winning game, each unordered pair has at most one edge.
Because the number of edges equals the number of unordered pairs, each
unordered pair must have exactly one edge. In human terms, that means for
each pair of participants, exactly one of them points to the other.
Next, let’s count the number of all-winning outcomes. Alice has 4
2
= 6
choices of where to point. For concreteness, suppose that Alice points to
Beth and Carla. That means Dana and Eden will point to Alice. Between
Beth and Carla, one will point to the other (2 choices); for concreteness,
suppose that Beth points to Carla. Between Dana and Eden, one will point
to the other (2 choices); for concreteness, suppose that Dana points to Eden.
Now there is exactly one way to complete the picture. So the total number
of all-winning outcomes is 6 · 2 · 2 = 24.
Therefore, the probability of an all-winning outcome is 24
65 = 23·3
2535 , which
simplifies to
1
324
.
Problem 14 We call a set X of real numbers three-averaging if for every
two distinct elements a and b of X, there exists an element c in X (different
from both a and b) such that the number (a+b+c)/3 also belongs to X. For
8

instance, the set {0, 1008, 2016} is three-averaging. What is the least possible
number of elements in a three-averaging set with more than 3 elements?
Answer: 5
Solution: Every set of 3 real numbers that form an arithmetic sequence is
three-averaging. For example, the sets {0, 1, 2}, {1, 2, 3}, and {0, 3, 6} are
three-averaging.
We claim that the 5-element set S = {0, 1, 2, 3, 6} is three-averaging.
Most pairs (a, b) in the definition of three-averaging follow from the sets
{0, 1, 2}, {1, 2, 3}, and {0, 3, 6} being three-averaging. The only remaining
pairs that we need to check are (1, 6) and (2, 6). If (a, b) is (1, 6), then c = 2
leads to (a + b + c)/3 = 3, which is in the set S. If (a, b) is (2, 6), then c = 1
also leads to (a + b + c)/3 = 3. So S is indeed three-averaging.
We claim that no 4-element set is three-averaging. Let X = {p, q, r, s}
be a set of 4 real numbers. Without loss of generality, assume that p < q <
r < s. For contradiction’s sake, assume that X is three-averaging. By the
definition of three-averaging (with a = q and b = r), either (p + q + r)/3 or
(q + r + s)/3 is in X. Without loss of generality, assume that (p + q + r)/3 is
in X. Note that (p + q + r)/3 is strictly between p and r. So (p + q + r)/3 is
equal to q. Hence p, q, and r forms an arithmetic sequence. By translating,
we may assume that p = 0. By scaling, we may assume that q = 1. So r = 2.
Choose a = s. If s is an integer, then choose b = s mod 3; otherwise,
choose b = p. By the definition of three-averaging, there is a c in X (different
from a and b) such that (a + b + c)/3 is in X. Note that (a + b + c)/3 is
strictly between p and s. So (a+b+c)/3 is q or r. In particular, (a+b+c)/3
is an integer. So a + b + c is a multiple of 3. Because b and c are integers, a
is an integer too. Working mod 3, we have
a + b + c ≡ a + b + b ≡ b + b + b ≡ 0 (mod 3).
(We know that c ≡ b (mod 3) because b and c are different numbers from
the set {0, 1, 2}.) This contradicts a + b + c being a multiple of 3. Hence
our assumption that X is three-averaging is false. So no 4-element set is
three-averaging.
We constructed a 5-element set that is three-averaging and showed that
no 4-element set is three-averaging. So the answer to the problem is 5 .
Problem 15 Let H be a convex, equilateral heptagon whose angles measure
(in degrees) 168◦
, 108◦
, 108◦
, 168◦
, x◦
, y◦
, and z◦
in clockwise order. Compute
9

the number y.
Answer: 132
Solution: Label the vertices of the heptagon A, B, C, D, E, F, and G.
Draw a regular pentagon (with new vertex P) so that it shares the two 108◦
angles, as shown below.
A
BC
D
E
F
G
P
108◦
108◦
108◦
108◦
108◦
y◦
Note that PDE is 168◦
−108◦
= 60◦
. Because PD = DE, triangle PDE
is equilateral. By symmetry, triangle PAG is also equilateral. So PEFG is
a rhombus. Hence
y◦
= EPG = 360◦
− 60◦
− 108◦
− 60◦
= 132◦
.
Thus y = 132 .
Problem 16 Let A < B < C < D be positive integers such that every
three of them form the side lengths of an obtuse triangle. Compute the least
possible value of D.
Answer: 14
Solution: The triangle inequality says that C < A + B, D < A + B,
D < A + C, and D < B + C. Of these inequalities, the only one that
matters is D < A + B, because the others follow from it. From obtuseness,
we get the inequalities A2
+ B2
< C2
, A2
+ B2
< D2
, A2
+ C2
< D2
,
and B2
+ C2
< D2
. Of these inequalities, the only two that matter are
A2
+ B2
< C2
and B2
+ C2
< D2
, because the others follow from them.
10

So we are looking for positive integers A < B < C < D such that
D < A + B, A2
+ B2
< C2
, and B2
+ C2
< D2
. Trial and error with small
integers and their squares eventually leads to the solution A = 7, B = 8,
C = 11, and D = 14. In what remains, we will show that every solution
satisﬁes D ≥ 14.
First, we claim that B < 2A − 3. From the obtuseness inequalities A2
+
B2
< C2
and B2
+ C2
< D2
, we deduce that A2
+ 2B2
< D2
. The triangle
inequality D < A + B can be written as D ≤ A + B − 1. So we have
2B2
< D2
− A2
= (D − A)(D + A) ≤ (B − 1)(2A + B − 1).
Dividing both sides by B − 1 gives
2A + B − 1 >
2B2
B − 1
> 2(B + 1) = 2B + 2.
Solving for B gives B < 2A − 3, as claimed.
Second, we claim that B > 5. We know that A < B, which can be written
as A ≤ B − 1. By the previous paragraph, we have
B < 2A − 3 ≤ 2(B − 1) − 3 = 2B − 5.
Solving for B gives B > 5, as claimed.
Assume that B = 6. Then A ≤ B − 1 = 5. Also B < 2A − 3 implies that
A > 4.5. So A = 5. Because A2
+ B2
< C2
, it follows that C ≥ 8. Because
B2
+ C2
< D2
, it follows that D ≥ 11. But that contradicts the triangle
inequality D < A + B.
Assume that B = 7. Then A ≤ B − 1 = 6. Also B < 2A − 3 implies that
A > 5. So A = 6. Because A2
+ B2
< C2
, it follows that C ≥ 10. Because
B2
+ C2
< D2
, it follows that D ≥ 13. But that contradicts the triangle
inequality D < A + B.
So we know that B ≥ 8. Then B < 2A − 3 implies that A ≥ 6. Because
A2
+ B2
< C2
, it follows that C ≥ 11. Because B2
+ C2
< D2
, it follows
that D ≥ 14. That’s what we wanted to show. So the least possible value of
D is 14 .
Problem 17 We deﬁne the weight W of a positive integer as follows: W(1) =
0, W(2) = 1, W(p) = 1 + W(p + 1) for every odd prime p, W(c) = 1 + W(d)
for every composite c, where d is the greatest proper factor of c. Compute
the greatest possible weight of a positive integer less than 100.
11

Answer: 12
Solution: Let’s compute the first 13 weights.
n 1 2 3 4 5 6 7 8 9 10 11 12 13
W(n) 0 1 3 2 5 4 4 3 4 6 6 5 6
In particular, if n ≤ 13, then W(n) ≤ 6.
Suppose that n ≤ 26. We claim that W(n) ≤ 8. If n ≤ 13, then
W(n) ≤ 6 < 8. If n is composite, then its greatest proper factor d is at most
n/2 ≤ 13, so
W(n) = 1 + W(d) ≤ 1 + 6 < 8.
If n is an odd prime, then
W(n) = 1 + W(n + 1) = 1 + 1 + W(
n + 1
2
) ≤ 1 + 1 + 6 = 8.
In every case, W(n) ≤ 8.
In the same way, we can show that if n ≤ 52, then W(n) ≤ 10. Similarly,
if n ≤ 104, then W(n) ≤ 12.
Let’s find a positive integer less than 100 whose weight is 12. It helps to
find a nearly-doubling sequence of primes. Note that 19 is prime, 2·19−1 =
37 is prime, and 2 · 37 − 1 = 73 is prime. So
W(73) = 2 + W(37) = 4 + W(19) = 6 + W(10) = 6 + 6 = 12.
Hence the greatest possible weight of a positive integer less than 100 is 12 .
Problem 18 Let T = {1, 2, 3, . . . , 14, 15}. Say that a subset S of T is handy
if the sum of all the elements of S is a multiple of 5. For example, the empty
set is handy (because its sum is 0) and T itself is handy (because its sum
is 120). Compute the number of handy subsets of T.
Answer: 6560
Solution: We will identify a subset of T with a red-blue coloring of T: color
the elements of the subset red and the nonelements blue. Imagine slicing
every coloring into three parts of length 5 (namely 1 to 5, 6 to 10, and 11
to 15). In every coloring, either all three parts are monochromatic or not.
Let’s deal with the first case: all three parts of the coloring are monochro-
matic. There are 23
= 8 such colorings. Every such coloring corresponds to
a handy subset.
12

Next, let’s deal with the second case: at least one of the three parts of the
coloring is not monochromatic. There are 215
− 23
= 32,760 such colorings.
Given such a coloring, consider the first part that is not monochromatic.
We can apply an arbitrary cyclic shift of this part of the coloring without
affecting which parts are monochromatic. So the set of colorings in this case
can be partitioned into classes of 5 colorings each. In every class, we claim
that exactly one of the 5 colorings corresponds to a handy subset. (Suppose
we have a subset R of 1 to 5, or 6 to 10, or 11 to 15. When we shift it cyclically
one position to the right, its sum goes up by |R| mod 5. So the 5 possible
shifts lead to increases mod 5 of 0, |R|, 2|R|, 3|R|, and 4|R|. Because |R|
is not a multiple of 5, these offsets cover every mod 5 residue exactly once.)
Hence the number of colorings in this case that correspond to a handy subset
is (215
− 23
)/5, which is 6552.
Combining the two cases, we see that the number of handy subsets is
6552 + 8, which is 6560 .
Alternative Solution: We will use a “roots of unity” filter. Let F be
the polynomial defined by F(x) = (1+x1
)(1+x2
) · · · (1+x15
). By expanding,
we see that F(x) is the sum of xsum(S)
over all subsets S of T, where sum(S) is
the sum of the elements of S. Let ω be a 5th root of unity (different from 1).
Note that if n is an integer, then 1+ωn
+ω2n
+ω3n
+ω4n
is equal to 5 when n is
a multiple of 5, and is equal to 0 otherwise. So the number of handy subsets is
[F(1) + F(ω) + F(ω2
) + F(ω3
) + F(ω4
)] /5. We see that F(1) = 215
. Let G
be the polynomial defined by G(y) = (y+ω0
)(y+ω1
)(y+ω2
)(y+ω3
)(y+ω4
).
Its roots are the 5th roots of −1. So G(y) = y5
+ 1. We also see that
F(ω) = G(1)3
= 23
= 8. Similarly, F(ω2
) = F(ω3
) = F(ω4
) = 8. Hence the
number of handy subsets is (215
+ 4 · 23
)/5, which is 6560.
Problem 19 In the coordinate plane, consider points A = (0, 0), B =
(11, 0), and C = (18, 0). Line A has slope 1 and passes through A. Line B is
vertical and passes through B. Line C has slope −1 and passes through C.
The three lines A, B, and C begin rotating clockwise about points A, B,
and C, respectively. They rotate at the same angular rate. At any given
time, the three lines form a triangle. Determine the largest possible area of
such a triangle.
Answer: 85
Solution: Let X be the point where lines B and C intersect, let Y be the
point where lines A and C intersect, and let Z be the point where lines A
13

and B intersect. Here is a picture of the starting position.
A
B
C
X
Y
Z
11 7
45◦
45◦
Note that XY Z starts oﬀ as a 45-45-90 triangle, with a right angle
at Y . Because the three lines rotate at the same rate, the angles they form
with each other will remain the same. So XY Z will remain a 45-45-90
triangle, with a right angle at Y . Here is a typical picture after rotation.
A
B
C
X
Y
Z
11
7
45◦
45◦
Other cases for the picture are possible too; we will handle them together.
Let α = CAY , an angle between 0◦
and 90◦
. So BAY = α. Hence BAZ
is either α or 180◦
−α, depending on the case. Note that BZA is either 45◦
or 135◦
. Applying the Law of Sines to BAZ, we see that BZ = 11
√
2 sin α.
Note that ACY is 90◦
− α. So BCY is 90◦
− α. Hence BCX is either
90◦
− α or 90◦
+ α. Note that BXC is either 45◦
or 135◦
. Applying the
Law of Sines to BCX, we see that XB = 7
√
2 cos α. Hence XZ is either
7
√
2 cos α+11
√
2 sin α or 7
√
2 cos α − 11
√
2 sin α , depending on whether B
is between X and Z or not. By the Cauchy-Schwarz inequality, we have
XZ ≤ (7
√
2)2
+ (11
√
2)2
1/2
=
√
340 .
14

Equality is achieved when α is the acute angle such that 7 sin α = 11 cos α,
with B between X and Z, as in the diagram above. So the largest possible
value of XZ is
√
340.
Note that XZ is the hypotenuse of XY Z. The leg XY of XY Z is
XZ/
√
2. So the area of XY Z is (XZ)2
/4. Hence the largest possible area
of XY Z is 340/4, which is 85 .
Problem 20 Let a1, a2, a3, a4, and a5 be random integers chosen indepen-
dently and uniformly from the set {0, 1, 2, . . . , 23}. (Note that the integers
are not necessarily distinct.) Find the probability that
5
k=1
cis
akπ
12
= 0.
(Here cis θ means cos θ+i sin θ.) Express your answer as a fraction in simplest
form.
Answer:
35
27,648
Solution: Let ω = cis π
12
, a 24th root of unity. Note that cis akπ
12
is ωak . So
we’re looking for 5 powers of ω (not necessarily distinct) that have a vanishing
sum (add up to zero).
When does a sum of powers of ω vanish? One example is ω0
+ ω12
, or
any rotation of it such as ω5
+ ω17
. We’ll call such an example a diameter,
because the segment with endpoints ω5
and ω17
in the complex plane is a
diameter of the unit circle. Another example is ω0
+ω8
+ω16
, or any rotation
of it such as ω7
+ω15
+ω23
. We’ll call such an example an equilateral triangle,
because the triangle with vertices ω7
, ω15
, and ω23
is equilateral. A sum of
diameters and equilateral triangles, such as (ω5
+ ω17
) + (ω7
+ ω15
+ ω23
) or
(ω0
+ω12
)+(ω0
+ω8
+ω16
), also vanishes. We claim that the converse is true:
every sum of powers of ω that vanishes can be partitioned into diameters and
equilateral triangles. We will prove this claim later. For now, we shall assume
that it is true.
We are looking at tuples (a1, a2, a3, a4, a5) such that each ak is in the
set {0, 1, 2, . . . , 23}. The number of such 5-tuples is 245
. We would like
to count the number of such tuples for which the sum of the ωak vanishes.
By the claim above, the ωak must split into an equilateral triangle and a
diameter. We have 8 choices for the equilateral triangle. For now, ﬁx the
choice of the equilateral triangle.
15

The diameter either shares an endpoint with the equilateral triangle or is
disjoint from the triangle. In the first case (sharing), there are 3 choices for
the diameter. There are 5!/2! = 60 ways to choose the ak from the triangle
and diameter. So the first case has 3 · 60 = 180 choices. In the second case
(disjoint), there are 12 − 3 = 9 choices for the diameter. There are 5! = 120
ways to choose the ak from the triangle and diameter. So the second case
has 9 · 120 = 1080 choices. Combining the two cases gives 180 + 1080 = 1260
choices. Multiplying by 8 (the number of choices for the equilateral triangle)
gives a total of 10,060 tuples with a vanishing sum. Hence the probability of
vanishing is 10,060
245 , which simplifies to
35
27,648
.
We still need to prove the claim: every sum of powers of ω that vanishes
can be partitioned into diameters and equilateral triangles. We will use some
facts about polynomials. Because x12
+1 factors into x4
+1 times x8
−x4
+1,
it follows that the polynomial x8
− x4
+ 1 has ω as a root. Conversely, it is
known that every polynomial with rational coefficients having ω as a root is
divisible by the 24th cyclotomic polynomial Φ, which is Φ(x) = x8
− x4
+ 1.
Note that every exponent of Φ is a multiple of 4.
Suppose that a sum of powers of ω vanishes. Let P be the polynomial
that is this sum of powers. It is of the form b0 + b1x + · · · + b23x23
, where
the bj are nonnegative integers. It has ω as a root. Let Q be the polynomial
c0 +c1x+· · ·+c11x11
, where cj = bj −bj+12. Because ω12
= −1, it follows that
Q also has ω as a root. Hence Q is divisible by Φ; say Q = R·Φ. Split Q into
four parts according to the exponent mod 4: Q0(x) = c0+c4x4
+c8x8
, Q1(x) =
c1x+c5x5
+c9x9
, Q2(x) = c2x2
+c6x6
+c10x10
, and Q3(x) = c3x3
+c7x7
+c11x11
.
Because every exponent of Φ is a multiple of 4 and Q = R · Φ, it follows that
Q0, Q1, Q2, and Q3 are each divisible by Φ. (Namely, split R into four parts
R0, R1, R2, and R3 according to the exponent mod 4. Then Q0 = R0 · Φ and
so on.)
Let’s look at Q0 first. Because Q0 is divisible by Φ, it follows that c0 =
−c4 = c8. In other words, b0 − b12 = b8 − b20 = b16 − b4. Let’s call their
common value t. If t = 0, then b0 = b12, b8 = b20, and b4 = b16; in other
words, these terms split into diameters. (Namely, b0 copies of ω0
+ ω12
, b8
copies of ω8
+ ω20
, and b16 copies of ω4
+ ω16
.) If t > 0, then after removing
diameters (namely b12 copies of ω0
+ω12
, b20 copies of ω8
+ω20
, and b4 copies
of ω4
+ ω16
), we have t copies of the equilateral triangle ω0
+ ω8
+ ω16
. If
t < 0, then after removing diameters, we have −t copies of the equilateral
16

triangle ω4
+ ω12
+ ω20
. A similar analysis applies to Q1, Q2, and Q3. Hence
we have proved the claim that a vanishing sum of powers of ω splits into
diameters and equilateral triangles.
Note: Our claim that a vanishing sum of 24th roots of unity splits
into vanishing diameters and triangles also follows from a theorem of N.
G. de Bruijn (1953) in his paper “On the Factorization of Cyclic Groups”.
His Theorem 2 proves a more general result, with 24 replaced by a product
of two prime powers.
Credits
The authors of the 20 problems are as follows.
Ravi Boppana: 9
Mathew Crawford: 13, 20
Ken Fan: 15, 19
Oleg Kryzhanovsky: 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 14, 16, 17, 18
17

Mathprize2016solutions

More Related Content

What's hot (17)

Similar to Mathprize2016solutions (20)

Recently uploaded (20)

Mathprize2016solutions