35th VTRMC, 2013, Solutions
- Make the
substitution t = 2y, so dt = 2dy. Then
I = ∫0x/26√2(√((1 + 1 cos 2y)/2))/(17 - 8 cos 2y) dy = ∫0x/23√2(2√2cos y)/(9 + 16 sin2y) dy.
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
x = 2π/3.
- 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
y2 = 1 + 9x2 and
z2 = 1 + x2. Therefore
4x2 = (1 + 9x2)(1 + x2)sin2θ. We need to maximize
sin2θ, which in turn is equivalent to
(1 + 9x2)(1 + x2)/(4x2). Therefore we need to find x
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
- We need to show that an is bounded, equivalently ln an is
ln 2∑n=1∞ln(1 + n-3/2) is
ln(1 + n-3/2) < n-3/2 and
∑n=1∞n-3/2 is convergent. It follows that (an) is convergent.
25 = 50/2 = (52 +52)/(12 +12).
- Assume that 2013 is special. Then we have
|x2 + y2 = 2013(u2 + v2)
for some positive integers x, y, u, v. Also, we assume that
x2 + y2 is minimal with this property. The prime factorization of
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)
|3(x12 + y12) = 11·61(u2 + v2),
3| u2 + v2. It follows u = 3u1 and v = 3v1, and replacing
in (2) we get
x12 + y12 = 2013(u12 + v12).
x12 + y12 < x2 + y2, contradicting the minimality of
x2 + y2.
- 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
(p - 1)/2 = 2k + 1, the last relation is equivalent
(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.
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
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.
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
XY -BY = (X -X + XYXD)Y = XYXY. Therefore we can take
| ||M = XY =
- For | q| < 1, we have
∑k=1∞qk = 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).
|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=1∞n/(q(qn/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.