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Student: Matt Superdock
Username: matt276eagles
ID#: 6213
USA Mathematical Talent Search
Year
19
Round
4
Problem
1
Let the labels of the vertices, clockwise, be a1 , a2 , . . . , a8 .
Since 2008 = 23 · 251, the factors of 2008 are 1, 2, 22 , 23 , 251, 2 · 251, 22 · 251, and 23 · 251.
Consider the vertices labelled with 23 and 251. Each vertex is connected to four other vertices.
Only 1, 2, 22 , and 23 · 251 can be connected to 23 , and only 1, 2 · 251, 22 · 251, and 23 · 251 can be
connected to 251, so exactly these numbers must be connected to 23 and 251, respectively.
Since 23 does not divide 251 and 251 does not divide 23 , these two numbers cannot be connected by an edge. Therefore, the 23 and 251 must be three or four vertices apart. We proceed
with casework.
Case 1: 23 and 251 are three vertices apart.
Since rotations and reflections are considered the same labelling, we can choose a vertex for 23
without loss of generality, and then we can choose one of the two vertices that are three away for
251, again without loss of generality. Therefore, we can assume a1 = 23 and a4 = 251. The two
numbers connected to both 23 and 251 are 1 and 23 · 251. Therefore, a2 and a3 are 1 and 23 · 251 in
either order. This leaves 2 · 251 and 22 · 251 for a5 and a6 , and 2 and 22 for a7 and a8 . Of these four
numbers, the only pair that cannot be connected is 22 and 2 · 251. Therefore, the only possibility
for these four numbers is a5 = 2 · 251, a6 = 22 · 251, a7 = 2, and a8 = 22 . We can choose either
permutation for a2 and a3 , so there are two possibilities for this case.
Case 2: 23 and 251 are four vertices apart.
Since rotations are considered the same labelling, we can assume a1 = 23 and a5 = 251. The
two numbers connected to both 23 and 251 are 1 and 23 · 251, so a3 and a7 are 1 and 23 · 251. Now
since reflections are considered the same labelling, we can assume a3 = 1 and a7 = 23 · 251. This
leaves 2 and 22 for a2 and a8 , and 2 · 251 and 22 · 251 for a4 of a6 . Again, of these four numbers,
the only pair that cannot be connected is 22 and 2 · 251, so these two numbers must be opposite
each other. One possibility for the remaining four numbers is a2 = 2 · 251, a4 = 2, a6 = 22 , and
a8 = 22 ·251. The only other possibility is a2 = 22 ·251, a4 = 22 , a6 = 2, and a8 = 2·251. Therefore,
there are two possibilities for this case.
In total there are four possibilities.
Page 1 of Problem 1
Student: Matt Superdock
Username: matt276eagles
ID#: 6213
USA Mathematical Talent Search
Year
19
Round
4
Problem
2
We manipulate the given inequality using bxc = x − {x} to get
nno nno n n o n n o
n n
n
n
+ +
+
−
−
−
−
< n
2
3 11 13
2
3
11
13
859n n n o n n o n n o n n o
−
−
−
−
< n
858
2 o n 3 o n 11 o n 13 o
nn
n
n
n
n
+
+
+
>
2
3
11
13
858
x
x
Let f (x) = { x2 } + { x3 } + { 11
} + { 13
}. Suppose that n is the largest integer such that f (n) >
We claim that n ≡ −1 (mod 2), n ≡ −1 (mod 3), n ≡ −1 (mod 11), and n ≡ −1 (mod 13).
n
858 .
To prove the first part of our claim, suppose n 6≡ −1 (mod 2). Consider f (n) and f (n+3·11·13).
The last three terms are the same for both f (n) and f (n + 3 · 11 · 13). Therefore, since 3 · 11 · 13 ≡ 1
n
+ 21 = n+3·11·13
. This contradicts the maximality
(mod 2), we have f (n+3·11·13) = f (n)+ 12 > 858
858
of n, since n + 3 · 11 · 13 satisfies the same inequality.
Note that 2 · 11 · 13 ≡ 1 (mod 3), 2 · 3 · 13 ≡ 1 (mod 11), and 2 · 3 · 11 ≡ 1 (mod 13). Now we
can achieve similar contradictions for n 6≡ −1 (mod 3), n 6≡ −1 (mod 11), and n 6≡ −1 (mod 13):
1
n
1
n + 2 · 11 · 13
>
+ =
.
3
858 3
858
n
1
n + 2 · 3 · 13
1
>
+
=
.
If n 6≡ −1 (mod 11), then f (n + 2 · 3 · 13) = f (n) +
11
858 11
858
1
n
1
n + 2 · 3 · 11
If n 6≡ −1 (mod 13), then f (n + 2 · 3 · 11) = f (n) +
>
+
=
.
13
858 13
858
If n 6≡ −1 (mod 3), then f (n + 2 · 11 · 13) = f (n) +
These are all contradictions, so our claim is correct. By our claim, we have
f (n) =
1 2 10 12
3 · 858 − 1
+ +
+
=
2 3 11 13
858
Also, by the Chinese Remainder Theorem, the four parts of our claim together imply that
n
n ≡ −1 (mod 858). Since f (n) = 3·858−1
for all such n, the largest value of n such that f (n) > 858
858
is n = 2 · 858 − 1 .
Page 1 of Problem 2
Student: Matt Superdock
Username: matt276eagles
ID#: 6213
USA Mathematical Talent Search
Year
19
Round
4
Problem
3
Let f (n) be the number of 0’s in the binary representation of n, and let g(n) be the number of
1’s in the binary representation of n. We will prove by induction on the number of digits in the
binary representation of n that an = µf (n) (1 − µ)g(n)−1 . For the base case, we only need to consider
n = 1. We have f (1) = 0 and g(1) = 1, so our formula gives a1 = 1, which is correct. Now assume
that for all n with d digits in their binary representations, an = µf (n) (1 − µ)g(n)−1 . Since the binary
representation of 2n can be obtained by appending a 0 to the end of the binary representation of
n, we have
a2n = µan = µf (n)+1 (1 − µ)g(n)−1 = µf (2n) (1 − µ)g(2n)−1
Similarly, since the binary representation of 2n + 1 can be obtained by appending a 1 to the
end of the binary representation of n, we have
a2n+1 = (1 − µ)an = µf (n) (1 − µ)g(n) = µf (2n) (1 − µ)g(2n)−1
Since we can obtain all integers with d + 1 digits in their binary representations by appending
a 0 or a 1 to an integer with d digits in its binary representation, for all n with d + 1 digits, we also
have an = µf (n) (1 − µ)g(n)−1 . Therefore, the induction is complete, so an = µf (n) (1 − µ)g(n)−1 for
all n.
Using this expression for an , we have
∞
X
a2k a2k+1 =
k=1
∞
X
µf (2k) (1 − µ)g(2k)−1 · µf (2k+1) (1 − µ)g(2k+1)−1
k=1
=
∞
X
µf (k)+1 (1 − µ)g(k)−1 · µf (k) (1 − µ)g(k)
k=1
=
∞
X
µ2f (k)+1 (1 − µ)2g(k)−1
k=1
= µ(1 − µ)
∞
X
(µ2 )f (k) ((1 − µ)2 )g(k)−1
k=1
Now consider the sum
∞
X
(µ2 )f (k) ((1 − µ)2 )g(k)−1 . Let d + 1 be the number of digits of k. Then
k=1
g(k) − 1 = (d + 1 − f (k)) − 1 = d − f (k). Using this, we have
∞
X
k=1
a2k a2k+1 = µ(1 − µ)
∞
X
(µ2 )f (k) ((1 − µ)2 )d−f (k)
k=1
Since f (k) is the number of 0’s in the binary representation of k, and any f (k) digits of the
d
d non-leading digits can be 0’s, the number of terms with given values of d and f (k) is f (k)
.
Therefore, we can rewrite the sum as
∞
∞ X
d
X
X
d
a2k a2k+1 = µ(1 − µ)
(µ2 )f (k) ((1 − µ)2 )d−f (k)
f (k)
k=1
d=0 f (k)=0
Page 1 of Problem 3
Student: Matt Superdock
Username: matt276eagles
ID#: 6213
USA Mathematical Talent Search
Year
19
Round
4
Problem
3
By the Binomial Theorem,
∞
X
a2k a2k+1
∞
X
= µ(1 − µ)
(µ2 + (1 − µ)2 )d
k=1
d=0
∞
X
= µ(1 − µ)
(2µ2 − 2µ + 1)d
d=0
∞
X
Note that
(2µ2 −2µ+1)d is a geometric series with first term 1 and common ratio 2µ2 −2µ+1.
d=0
Therefore,
∞
X
k=1
a2k a2k+1 = µ(1 − µ) ·
µ − µ2
1
1
=
=
2
2
1 − (2µ − 2µ + 1)
2µ − 2µ
2
Page 2 of Problem 3
Student: Matt Superdock
Username: matt276eagles
ID#: 6213
USA Mathematical Talent Search
Year
19
Round
4
Problem
4
Assume for the sake of contradiction that w, x, y, z are positive real numbers such that the three
inequalities w + x < y + z, (w + x)yz < wx(y + z), and (w + x)(y + z) < wx + yz are all satisfied.
Multiplying both sides of the third inequality by w + x, we get
(w + x)2 (y + z) < wx(w + x) + yz(w + x)
< wx(w + x) + wx(y + z)
< wx(y + z) + wx(y + z)
< 2wx(y + z)
Therefore, (w + x)2 < 2wx, so w2 + x2 < 0, a contradiction. Therefore, both inequalities cannot
be simultaneously satisfied.
Page 1 of Problem 4
Student: Matt Superdock
Username: matt276eagles
ID#: 6213
USA Mathematical Talent Search
Year
19
Round
4
Problem
5
We can scale the 13-gon without changing the constants ci , so scale the 13-gon so that its
circumradius is 1 and place it on the complex plane with its center on the origin and with P1 = 1.
Then the Pi are the 13th roots of unity. By symmetry, the small 13-gon has the same center as
the large 13-gon. Therefore, the inradius of the small 13-gon is the distance from the origin to the
6π
20π
20π
line through P4 = cos 6π
13 + i sin 13 and P11 = cos 13 + i sin 13 . The real parts of these two points
are equal, so the line through P4 and P11 is a vertical line, and the distance from the origin to this
line is cos 6π
13 . Similarly, the inradius of the large 13-gon is the distance from the origin to the line
12π
14π
14π
π
through P7 = cos 12π
13 + i sin 13 and P8 = cos 13 + i sin 13 , which is cos 13 .
Since both 13-gons are regular and therefore similar, the ratio of their side lengths equals the
π
6π
π
ratio of their circumradius. Therefore, s/d1 = cos 6π
13 / cos 13 , so s = d1 cos 13 / cos 13 .
Now consider the triangle formed by P1 , P2 , and P2+n , for 1 ≤ n ≤ 6. By the Law of Sines, we
have
P1 P2
P2 P2+n
=
sin ∠P2 P1 P2+n
sin ∠P1 P2+n P2
d1 sin nπ
13
dn =
π
sin 13
We claim that s = 2d1 − 2d3 + 2d5 − d6 . We verify this using trigonometric identities and the
relations we derived above.
s = 2d1 − 2d3 + 2d5 − d6
2d1 sin 3π
2d1 sin 5π
d1 sin 6π
13
13
13
= 2d1 −
+
−
π
π
π
sin 13
sin 13
sin 13
6π
π
π
3π
π
5π
π
6π
π
π
cos
= 2 sin
cos
− 2 sin
cos
+ 2 sin
cos
− sin
cos
sin
13
13
13
13
13
13
13
13
13
13
7π
π
π
3π
π
5π
π
sin
= 2 sin
cos
− 2 sin
cos
+ 2 sin
cos
13
13
13
13
13
13
13
6π
π
π
3π
π
5π
π
sin
= 2 sin
cos
− 2 sin
cos
+ 2 sin
cos
13
13
13
13
13
13
13
We need to derive some trigonometric identities to continue to simplify this equation. We have
d1 cos 6π
13
π
cos 13
sin 3x = Im{(cis x)3 } = 3 cos2 x sin x − sin3 x
sin 5x = Im{(cis x)5 } = 5 cos4 x sin x − 10 cos2 x sin3 x + sin5 x
sin 6x = Im{(cis x)6 } = 6 cos5 x sin x − 20 cos3 x sin3 x + 6 cos x sin5 x
π
π
Let a = cos 13
and b = sin 13
. Using these substitutions and applying identities, our equation
becomes
6a5 b − 20a3 b3 + 6ab5 = 2ab − 2a(3a2 b − b3 ) + 2a(5a4 b − 10a2 b3 + b5 )
3a4 − 10a2 b2 + 3b4 = 1 − (3a2 − b2 ) + (5a4 − 10a2 b2 + b4 )
3a4 − 10a2 b2 + 3b4 = (a2 + b2 )2 − (3a2 − b2 )(a2 + b2 ) + (5a4 − 10a2 b2 + b4 )
3a4 − 10a2 b2 + 3b4 = a4 + 2a2 b2 + b4 − (3a4 + 2a2 b2 − b4 ) + (5a4 − 10a2 b2 + b4 )
Page 1 of Problem 5
Student: Matt Superdock
Username: matt276eagles
ID#: 6213
USA Mathematical Talent Search
Year
19
Round
4
Problem
5
This last equation is true after combining like terms on the right side, so s = 2d1 −2d3 +2d5 −d6
as we claimed earlier.
Page 2 of Problem 5