35th VTRMC, 2013, Solutions

1. Make the substitution t = 2y, so dt = 2dy. Then I = ∫0x/26√2(√((1 + 1 cos 2y)/2))/(17 - 8 cos 2ydy = ∫0x/23√2(2√2cos y)/(9 + 16 sin2ydy. Now make the substitution z = sin y. Then dz = dy cos y and I = 12∫0sin x/2dz/(32 +42z2) = tan-1(4/3)sin x/2. If tan I = 2/√3, then 2√3 = (4/3)sin x/2 and we deduce that x = 2π/3.

2. Without loss of generality we may assume that BC = 1, and then we set xBD, so AD = 2x. Write θ = ∠CAD, y = AC and z = DC. The area of ADC is both x and (yz sinθ)/2. Also y2 = 1 + 9x2 and z2 = 1 + x2. Therefore 4x2 = (1 + 9x2)(1 + x2)sin2θ. We need to maximize θ, equivalently sin2θ, which in turn is equivalent to minimizing (1 + 9x2)(1 + x2)/(4x2). Therefore we need to find x such that x-2 +9x2 is minimal. Differentiating, we find -2x-3 + 18x = 0, so x2 = 1/3. It follows that sin2θ = 1/4 and we deduce that the maximum value of CAD = θ is 30o.

3. We need to show that an is bounded, equivalently ln an is bounded, i.e. ln 2∑n=1ln(1 + n-3/2) is bounded. But ln(1 + n-3/2) < n-3/2 and n=1n-3/2 is convergent. It follows that (an) is convergent.

4. (a)
25 = 50/2 = (52 +52)/(12 +12).

(b)
Assume that 2013 is special. Then we have

 x2 + y2 = 2013(u2 + v2) (1)

for some positive integers x, y, u, v. Also, we assume that x2 + y2 is minimal with this property. The prime factorization of 2013 is 3·11·61. From (1) it follows 3| x2 + y2. It is easy to check by looking to the residues mod 3 that 3| x and 3| y, hence we have x = 3x1 and y = 3y1. Replacing in (1) we get

 3(x12 + y12) = 11·61(u2 + v2), (2)

i.e. 3| u2 + v2. It follows u = 3u1 and v = 3v1, and replacing in (2) we get

x12 + y12 = 2013(u12 + v12).

Clearly, x12 + y12 < x2 + y2, contradicting the minimality of x2 + y2.

(c)
Observe that 2014 = 2·19·53 and 19 is a prime of the form 4k + 3. If 2014 is special, then we have,

x2 + y2 = 2014(u2 + v2),

for some positive integers x, y, u, v. As in part (b), we may assume that x2 + y2 is minimal with this property. Now, we will use the fact that if a prime p of the form 4k + 3 divides x2 + y2, then it divides both x and y. Indeed, if p does not divide x, then it does not divide y too. We have x2≡ -y2mod p implies (x2)(p-1)/2≡(-y2)(p-1)/2)mod p. Because (p - 1)/2 = 2k + 1, the last relation is equivalent to (x2)(p-1)/2≡ - (y2)(p-1)/2mod p, hence xp-1≡ -yp-1mod p. According to the Fermat's little theorem, we obtain 1≡ -1 mod p, that is p divides 2, which is not possible.

Now continue exactly as in part (b) using the prime 19, and contradict the minimality of x2 + y2.

5. Write x = tan A, y = tan B, z = tan C, where 0 < A, B, C < π/2. Using the formula tan(A + B) = (tan A + tan B)/(1 - tan A tan B) twice, we see that

tan(A + B + C) = (x + y + z -xyz)/(1 -yz -zx -xy) = 0

and therefore A + B + C = π. Now sin A = x(1 + x2), so we need to prove that sin A + sin B + sin C≤3√3/2. However sin t is a concave function, so we may apply Jensen's inequality (or consider the tangent at t = (A + B + C)/3) to deduce that

(sin A + sin B + sin C)/3≤(sin(A + B + C))/3 = sin(π/3) = √3/2,

and the result follows.

6. Let C = X-1 + (Y-1 -X)-1. Observe that (Y-1 -X) = (X -XYX)X-1Y-1, consequently (Y-1 -X)-1 = YX(X -XYX)-1. Therefore C(X -XYX)-1 = I -YX + YX = I and we deduce that XY -BY = (X -X + XYXD)Y = XYXY. Therefore we can take

M = XY = (
 190 81 65 -49 64 -191 -56 74 86
)

7. For | q| < 1, we have k=1qk = q/(q - 1). Therefore for | q| > 1,

 - ∑n=1∞(- 1)n/qn - 1 = -∑n=1∞(- 1)nq-n/(1 -q-n) = -∑n=1∞∑k=1∞(q-n)k(- 1)n+1 = ∑n=1∞(- 1)n+1q-n/(1 -q-n), ∑n=1∞1/(qn + 1) = ∑n=1∞q-n/(1 + q-n) = ∑n=1∞∑k=1∞(- 1)k+1(q-n)k = ∑k=1∞(- 1)k+1q-k/(1 -q-k).

It follows that - ∑n=1(- 1)n/(qn -1) = ∑n=11/(qn + 1). Now

 d /dx 1/(xn - 1) = -n/(x(xn/2 -x-n/2)2) d /dx 1/(xn + 1) = -n/(x(xn/2 + x-n/2)).

We deduce that

- ∑n=1(- 1)nn/(q(qn/2 -q-n/2)2) = ∑n=1n/(q(qn/2 + q-n/2)2)

Now set q = 4. We conclude that n=1n/((2n +2-n)2) + (- 1)nn/(2n -2-n)2 = 0.

Peter Linnell 2013-11-11