35th VTRMC, 2013, Solutions

1. Make the substitution t = 2y, so dt = 2dy. Then I = ∫₀^x/26√2(√((1 + 1 cos 2y)/2))/(17 - 8 cos 2y) dy = ∫₀^x/23√2(2√2cos y)/(9 + 16 sin²y) dy. Now make the substitution z = sin y. Then dz = dy cos y and I = 12∫₀^sin 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 x ≔BD, 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 30^o.

3. We need to show that a_n is bounded, equivalently ln a_n is bounded, i.e. ln 2∑_n=1^∞ln(1 + n^-3/2) is bounded. But ln(1 + n^-3/2) < n^-3/2 and ∑_n=1^∞n^-3/2 is convergent. It follows that (a_n) 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 x^p-1≡ -y^p-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=1^∞q^k = 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=1^∞1/(qⁿ + 1). Now

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

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

   
   
   We deduce that
   - ∑_n=1^∞(- 1)ⁿn/(q(q^n/2 -q^-n/2)²) = ∑_n=1^∞n/(q(q^n/2 + q^-n/2)²)
   Now set q = 4. We conclude that ∑_n=1^∞n/((2ⁿ +2^-n)²) + (- 1)ⁿn/(2ⁿ -2^-n)² = 0.