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A1. Find all ordered pairs (a, b) of positive integers for which
1 a
b
Answer. The six ordered pairs are (1009, 2018), (2018, 1009), (1009 · 337 , 674) = (350143, 674), (1009 · 1346 , 673) = (1358114, 673), (674, 1009 · 337) = (674, 350143), and (673, 1009 · 1346) = (673, 1358114). Solution. First rewrite the equation as 2 · 1009(a + b) = 3ab, and note that 1009 is prime, so at least one of a and b must be divisible by 1009. If both a and b are divisible by 1009, say with a = 1009q, b = 1009r, then we have 2(q + r) = 3qr. But qr ≥ q + r for integers q, r ≥ 2, so at least one of q, r is 1. This leads to the solutions q = 1, r = 2 and r = 1, q = 2, corresponding to the ordered pairs (a, b) = (1009, 2018) and (a, b) = (2018, 1009). In the remaining case, just one of a and b is divisible by 1009, say a = 1009q. This yields 2 · 1009(1009q + b) = 3 · 1009 qb, which can be rewritten as 2 · 1009 q = (3q − 2)b. Because the prime 1009 does not divide b, it must divide 3q − 2; say 3q − 2 = 1009k. Then 1009k + 2 = 3 · 336 k + k + 2 is divisible by 3, so k ≡ 1 (mod 3). For k = 1, we get q = 337, a = 1009 · 337 , b = 2q = 674. For k = 4, we get q = 1346, a = 1009 · 1346 , b = q/2 = 673. We now show there is no solution with k > 4. Assuming there is one, the corresponding value of q is greater than 1346, and so the corresponding b =
2 q 3 q − 2
is less than 673. Because b is an integer, it follows that b ≤ 672, which implies 1 b
, contradicting
a
b
Finally, along with the two ordered pairs (a, b) for which a is divisible by 1009 and b is not, we get two more ordered pairs by interchanging a and b.
A2. Let S 1 , S 2 ,... , S 2 n− 1 be the nonempty subsets of { 1 , 2 ,... , n} in some order, and let M be the (2n^ − 1) × (2n^ − 1) matrix whose (i, j) entry is
mij =
0 if Si ∩ Sj = ∅; 1 otherwise. Calculate the determinant of M. Answer. The determinant is 1 if n = 1 and −1 if n > 1. Solution. Note that if we take the nonempty subsets of { 1 , 2 ,... , n} in a different order, this will replace the matrix M by P M P −^1 for some permutation matrix P , and the determinant will stay unchanged. Denote the matrix M by Mn to indicate its dependence on n. Then we will show that det(Mn+1) = −( det(Mn))^2 , from which the given answer follows by induction on n because M 1 has the single entry 1. Using the order S 1 , S 2 ,... , S 2 n− 1 for the nonempty subsets of { 1 , 2 ,... , n}, we arrange the nonempty subsets of { 1 , 2 ,... , n + 1} in the order S 1 , S 2 ,... , S 2 n− 1 , {n + 1}, S 1 ∪ {n + 1}, S 2 ∪ {n + 1},... , S 2 n− 1 ∪ {n + 1}. Then any of the first 2n^ − 1 subsets has empty intersection with {n + 1}, any two of the last 2n^ subsets have nonempty intersection because they both contain n + 1, and for any other choice of two nonempty subsets of { 1 , 2 ,... , n + 1}, whether they have
nonempty intersection is completely determined by the relevant entry in Mn. Thus the matrix Mn+1 has the following block form:
Mn+1 =
Mn 0 n Mn (^0) n^ T^1 1 n^ T Mn 1 n En
where (^0) n, (^1) n denote column vectors of length 2n^ − 1, all of whose entries are 0, 1 respectively, and En is the (2n^ − 1) × (2n^ − 1) matrix, all of whose entries are 1. To find det(Mn+1), we subtract the middle row from each of the rows below it. This will not affect the lower left block, it will change the lower part of the middle column from (^1) n to (^0) n, and it will replace the lower right block En by a block of zeros. Then we can expand the determinant using the middle column (whose only nonzero entry is now the “central” 1) to get
det(Mn+1) = det
Mn Mn Mn O
= − det
Mn O Mn Mn
where the last step is carried out by switching the ith row with the (2n^ − 1 + i)th row for all i = 1, 2 ,... , 2 n^ − 1. (Because this is an odd number of row swaps, the sign of the determinant is reversed.) Finally, the determinant of the block triangular matrix( Mn O Mn Mn
is equal to the product of the determinants of the diagonal blocks, so it
equals (det(Mn))^2 , and the result follows.
A3. Determine the greatest possible value of
∑^10
i=
cos(3xi) for real numbers x 1 , x 2 ,... , x 10
satisfying
∑^10
i=
cos(xi) = 0.
Answer. The maximum value is
Solution. Let zi = cos(xi). Then the real numbers z 1 , z 2 ,... , z 10 must satisfy − 1 ≤ zi ≤ 1, and the given restriction on the xi is equivalent to the additional restriction z 1 + z 2 + · · · + z 10 = 0 on the zi. Also, note that
cos(3xi) = cos(2xi) cos(xi) − sin(2xi) sin(xi) = (2 cos^2 (xi) − 1) cos(xi) − 2 sin^2 (xi) cos(xi) = 2 cos^3 (xi) − cos(xi) − 2(1 − cos^2 (xi)) cos(xi) = 4z i^3 − 3 zi.
Thus we can rephrase the problem as follows: Find the maximum value of the function f given by
f (z 1 ,... , z 10 ) =
∑^10
i=
(4z^3 i − 3 zi) = 4
∑^10
i=
z^3 i
on the set S = {(z 1 ,... , z 10 ) ∈ [− 1 , 1]^10 | z 1 + · · · + z 10 = 0}.
Because f is continuous and S is closed and bounded (compact), f does take on a maximum value on this set; suppose this occurs at (m 1 ,... , m 10 ) ∈ S. Let
so they are
k 1 =
⌈ (^) n
m
, k 2 =
2 n m
,... , km =
⌈mn
m
= n.
Thus the given relation gha^1 gha^2 · · · ghan^ = e can be rewritten, by omitting all factors h^0 , as gk^1 hgk^2 −k^1 h · · · gkm−km−^1 h = e. Taking the inverse of both sides, we get h−^1 (g−^1 )km−km−^1 · · · h−^1 (g−^1 )k^2 −k^1 h−^1 (g−^1 )k^1 = e. We claim that this new relation is one of the original form, but with g and h replaced by h−^1 and g−^1 respectively, and with the roles of m and n interchanged. Because m < n, it will follow by the induction hypothesis that h−^1 and g−^1 commute, and thus g and h commute. To prove the claim, note that the exponents of g−^1 in the new relation are b 1 = km − km− 1 ,... , bm− 1 = k 2 − k 1 , bm = k 1 , and so, with the convention k 0 = 0, we have
bi = km−i+1 − km−i =
(m − i + 1)n m
(m − i)n m
(m − i + 1)n m
(m − i)n m
in m
− n
(i − 1)n m
− n
in m
(i − 1)n m
as desired. Now consider the remaining case, in which m > n. Write m = qn+r with 0 ≤ r < n. We have ak = bmkn c − bm(k n− 1)c = qk + brkn c − q(k − 1) − br(k n− 1)c = q + a′ k
where a′ k = brkn c − br(k n− 1)c. If we set g′^ = ghq^ then we have
g′ha
′ 1 g′ha
′ 2 · · · g′ha
′ n = gha^1 gha^2 · · · ghan^ = e. This is again a relation of the original form, but with m replaced by r (and g replaced by g′). So by the case considered above, g′^ and h commute. As a result, (gh)hq^ = g′h = hg′^ = (hg)hq^ and it follows that gh = hg, completing the proof.
A5. Let f : R → R be an infinitely differentiable function satisfying f (0) = 0, f (1) = 1, and f (x) ≥ 0 for all x ∈ R. Show that there exist a positive integer n and a real number x such that f (n)(x) < 0. Solution. Suppose, to the contrary, that f (n)(x) ≥ 0 for all integers n ≥ 0 and all x ∈ R. To begin, note that f has a minimum at x = 0, so f ′(0) = 0. Now we show by induction on n that for every function f satisfying the given conditions and such that f (n)(x) ≥ 0 for all n ≥ 0 and all x, we have f (2x) ≥ 2 nf (x) for all n ≥ 0 and all x ≥ 0. Since f ′^ ≥ 0, we have f (2x) ≥ f (x) for x ≥ 0, showing the base case n = 0. Suppose
we have shown for some particular n that f (2x) ≥ 2 nf (x) for all relevant functions f and all x ≥ 0. Note that because f (0) = 0 and f (1) = 1, f ′(x) > 0 for some x ∈ [0, 1]; because f ′^ is nondecreasing, it follows that f ′(1) > 0. Therefore, we can define a function g by g(x) = f ′(x)/f ′(1), and for this infinitely differentiable function we have that g(0) = 0, g(1) = 1, and all derivatives of g are nonnegative. By the induction hypothesis, we then have g(2x) ≥ 2 ng(x), and therefore f ′(2x) ≥ 2 nf ′(x), for all x ≥ 0. Integrating with respect to x from 0 to y gives 12 f (2y) ≥ 2 nf (y) for all y ≥ 0, so this shows that f (2x) ≥ 2 n+1f (x) for all x ≥ 0, completing the induction proof. In particular, we now have f (2) ≥ 2 nf (1) = 2n^ for all n ≥ 0, which is obviously false. So it must be the case that f (n)(x) < 0 for some integer n ≥ 0 and some x ∈ R.
A6. Suppose that A, B, C, and D are distinct points in the Euclidean plane no three of which lie on a line. Show that if the squares of the lengths of the line segments AB, AC, AD, BC, BD, and CD are rational numbers, then the quotient area( 4 ABC) area( 4 ABD) is a rational number. NOTE: I don’t believe this is quite the final wording of this problem. Solution. Let v, w, z be the displacement vectors
AB,
AC,
AD from A to the points B, C, D respectively. Then the dot products v · v = |v|^2 , w · w, (v − w) · (v − w) are all rational, so
v · w =
(v · v + w · w − (v − w) · (v − w))
is also rational; similarly, v · z and w · z are rational. Because A, B, and C are not collinear, v and w form a basis for R^2 , so there are constants λ, μ ∈ R such that z = λv + μw. Then we have the system of linear equations v · z = (v · v) λ + (v · w) μ w · z = (w · v) λ + (w · w) μ for λ and μ. The determinant (v · v)(w · w) − (v · w)^2 of the matrix of this system is positive (because v and w are linearly independent) by the Cauchy-Schwarz inequality. Thus, because all coefficients of the system are rational, λ and μ are also rational. Now we can rewrite the desired quotient as area( 4 ABC) area( 4 ABD)
| det(v w)|/ 2 | det(v z)|/ 2
det(v w) λ det(v v) + μ det(v w)
μ
a rational number. (Note that μ 6 = 0 because A, B, D are not collinear.)
(The B section starts on the next page.)
B2. Let n be a positive integer, and let fn(z) = n + (n − 1)z + (n − 2)z^2 + · · · + zn−^1. Prove that fn has no roots in the closed unit disk {z ∈ C : |z| ≤ 1 }. Solution. If z is a root of fn, then 0 = (1 − z)fn(z) = n −
∑n j=1 z j (^) , so ∑n j=1 z j (^) = n.
On the other hand, when |z| ≤ 1, we have
∑^ n
j=
zj
∑^ n
j=
|z|j^ ≤ n, with equality only
for z = 1. Because z = 1 is not a root of fn, we are done.
B3. Find all positive integers n < 10100 for which simultaneously n divides 2n, n − 1 divides 2n^ − 1, and n − 2 divides 2n^ − 2. Answer. n = 2^2 , 24 , 216 , 2256. Solution. We first prove that for positive integers a and b, 2a^ − 1 divides 2b^ − 1 if and only if a divides b. If a divides b then we can write b = aq, and modulo 2a^ − 1 we then have 2b^ − 1 = (2a)q^ − 1 ≡ 1 q^ − 1 = 0. Conversely, suppose that 2a^ − 1 divides 2b^ − 1. Let b = aq + r with 0 ≤ r < a; we then have 2b^ − 1 = 2r(2aq^ − 1) + (2r^ − 1). Because 2 a^ − 1 divides 2b^ − 1 and 2aq^ − 1, it must also divide 2r^ − 1. Because 0 ≤ r < a, it follows that r = 0, so a divides b. A positive integer n divides 2n^ if and only if n is a power of 2. So we may assume that n = 2m^ for some nonnegative integer m. Now n − 1 = 2m^ − 1 divides 2n^ − 1 = 2^2 m − 1 if and only if m divides 2m, so if and only if m is a power of 2. So we may assume that m = 2l, that is, n = 2^2 l , for some nonnegative integer l, and we have to find all n < 10100 of this form for which n − 2 divides 2n^ − 2. Note that n−2 = 2·(2^2 l− 1 −1), so n−2 divides 2n^ −2 = 2·(2n−^1 −1) = 2·(2^2 2 l^ − 1 −1) if and only if 2l^ − 1 divides 2^2 l − 1, which is if and only if l divides 2l, that is, if and only if l is a power of 2. We can write l = 2k, so that n = 2^2 2 k , and to finish we have to find the values of k ≥ 0 for which n < 10100. Suppose that n is a solution with n < 10100. Then we have 2m^ = n < 10100 < (2^4 )^100 = 2^400. It follows that m = 2l^ < 400 < 29 , so l = 2k^ < 9 < 24 and thus 0 ≤ k ≤ 3. Conversely, 2^2 23 = 2^2 8 = 2^256 < 2300 = (2^3 )^100 < 10100. So the solutions are n = 2^2
2 k with k = 0, 1 , 2 , 3; in other words, n is equal to 2^2 , 2^4 , 2^16 or 2^256.
B4. Given a real number a, we define a sequence by x 0 = 1, x 1 = x 2 = a, and xn+1 = 2 xnxn− 1 − xn− 2 for n ≥ 2. Prove that if xn = 0 for some n, then the sequence is periodic. Solution. Let z be a (complex) root of the polynomial z^2 − 2 az + 1. Then z 6 = 0, and z satisfies 2az = z^2 + 1, so a =
z + z−^1 2
. If the Fibonacci numbers F 0 , F 1 , F 2 ,... are
given, as usual, by F 0 = 0, F 1 = 1, and Fn+1 = Fn + Fn− 1 for n ≥ 1, then we see that
xn =
zFn^ + z−Fn 2 for n = 0, 1 , 2. We now show by induction on n that this equation holds for all n. For the induction step, assuming the equation is correct for n, n − 1 , and n − 2, we have xn+1 = 2xnxn− 1 − xn− 2
=
(zFn^ + z−Fn^ )(zFn−^1 + z−Fn−^1 ) − (zFn−^2 + z−Fn−^2 ) 2
=
zFn+Fn−^1 + z−(Fn+Fn−^1 )^ + zFn−Fn−^1 + z−(Fn−Fn−^1 )^ − (zFn−^2 + z−Fn−^2 ) 2
=
zFn+1^ + z−Fn+ 2
because Fn − Fn− 1 = Fn− 2. Suppose that xn = 0 for some n, say xm = 0. Then z^2 Fm^ = −1, so z is a dth root of unity, where d = 4Fm. Now note that the Fibonacci sequence modulo d is periodic: There are only finitely many (specifically, d^2 ) possibilities for a pair (Fi mod d, Fi+ mod d) of successive Fibonacci numbers modulo d, and when a pair reoccurs, say (Fi mod d, Fi+1 mod d) = (Fj mod d, Fj+1 mod d) with i < j, it is straightforward to show by induction that Fi+k = Fj+k mod d for all k, including negative k for which
i + k ≥ 0. But then, because xn =
zFn^ + z−Fn 2
is determined by the value of Fn modulo d, the sequence (xn) is also periodic, and we are done.
B5. Let f = (f 1 , f 2 ) be a function from R^2 to R^2 with continuous partial derivatives (^) ∂x∂fij
that are positive everywhere. Suppose that ∂f 1 ∂x 1
∂f 2 ∂x 2
∂f 1 ∂x 2
∂f 2 ∂x 1
everywhere. Prove that f is one-to-one. Solution. Consider the Jacobian matrix
J = J(x 1 , x 2 ) =
∂f 1 ∂x 1
∂f 1 ∂x 2 ∂f 2 ∂x 1
∂f 2 ∂x 2
and the related symmetric matrix
A =
(J + JT^ ) =
∂f 1 ∂x 1
∂f 1 ∂x 2
∂f 2 ∂x 1
∂f 2 ∂x 1
∂f 1 ∂x 2
∂f 2 ∂x 2
From the givens, the entries of A are positive everywhere, and the determinant of A is also positive everywhere. So the eigenvalues of A are positive (because their sum tr(A) and their product are both positive), and so A is positive definite. That is, for any nonzero vector v, we have vT^ Av = v·Av > 0, so vT^ (J +JT^ )v = vT^ Jv+(vT^ Jv)T^ > 0.
On the other hand, this probability is less than the probability that each Xn takes on a value in A; as we have seen, for any Xn individually that probability is 2018/2048, and so because the Xn are independent, the probability that they will all take on values in A is (2018/2048)^2018. So we have the inequality
|S| · 2 −^3860 = P(X ∈ S) <
and the desired result follows.
