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Problem 1# Let I = 3 2 ∫ 0 x 1 + cos t 17 − 8 cos t d t \displaystyle I = 3 \sqrt{2} \int_0^x \frac{\sqrt{1+\cos t}}{17 - 8\cos t}
dt I = 3 2 ∫ 0 x 17 − 8 cos t 1 + cos t d t . If 0 < x < π 0 < x < \pi 0 < x < π and tan I = 2 3 \tan I = \frac{2}{\sqrt{3}} tan I = 3 2 , what is x x x ?
Problem 2# Let A B C ABC A BC be a right-angled triangle with ∠ A B C = 9 0 ∘ \angle ABC = 90^\circ ∠ A BC = 9 0 ∘ , and let D D D
be on A B AB A B such that A D = 2 D B AD = 2DB A D = 2 D B . What is the maximum possible value of ∠ A C D \angle
ACD ∠ A C D ?
Problem 3# Define a sequence ( a n ) (a_n) ( a n ) for n ≥ 1 n \ge 1 n ≥ 1 by a 1 = 2 a_1 = 2 a 1 = 2 and a n + 1 = a n 1 + n − 3 / 2 a_{n+1} = a_n^{1 +
n^{-3/2}} a n + 1 = a n 1 + n − 3/2 . Is ( a n ) (a_n) ( a n ) convergent (i.e., lim n → ∞ a n < ∞ \displaystyle \lim_{n \to \infty} a_n
< \infty n → ∞ lim a n < ∞ )?
Problem 4# A positive integer n n n is called special if it can be represented in the form
n = x 2 + y 2 u 2 + v 2 n = \dfrac{x^2+y^2}{u^2+v^2} n = u 2 + v 2 x 2 + y 2 , for some positive integers x , y , u , v x, y, u, v x , y , u , v . Prove
that
25 is special; 2013 is not special; 2014 is not special. Problem 5# Prove that x 1 + x 2 + y 1 + y 2 + z 1 + z 2 ≤ 3 3 2 \displaystyle \frac{x}{\sqrt{1+x^2}} + \frac{y}{\sqrt{1+y^2}} +
\frac{z}{\sqrt{1+z^2}} \le \frac{3\sqrt3}{2} 1 + x 2 x + 1 + y 2 y + 1 + z 2 z ≤ 2 3 3 for any positive real numbers x , y , z x,
y, z x , y , z such that x + y + z = x y z x + y + z = xyz x + y + z = x yz .
Problem 6# Let X = ( 7 8 9 8 − 9 − 7 − 7 − 7 9 ) X = \begin{pmatrix}
7 & 8 & 9 \newline
8 & -9 & -7 \newline
-7 & -7 & 9
\end{pmatrix} X = ⎝ ⎛ 7 8 − 7 8 − 9 − 7 9 − 7 9 ⎠ ⎞ ,
Y = ( 9 8 − 9 8 − 7 7 7 9 8 ) Y = \begin{pmatrix}
9 & 8 & -9 \newline
8 & -7 & 7 \newline
7 & 9 & 8
\end{pmatrix} Y = ⎝ ⎛ 9 8 7 8 − 7 9 − 9 7 8 ⎠ ⎞ ,
let A = Y − 1 − X A = Y^{-1} - X A = Y − 1 − X and let B B B be the inverse of X + A − 1 X + A^{-1} X + A − 1 . Find a matrix
M M M such that M 2 = X Y − B Y M^2 = XY - BY M 2 = X Y − B Y (you may assume that A A A and X − 1 + A − 1 X^{-1} + A^{-1} X − 1 + A − 1
are invertible).
Problem 7# Find ∑ n = 1 ∞ n ( 2 n + 2 − n ) 2 + ( − 1 ) n n ( 2 n − 2 − n ) 2 \displaystyle \sum_{n=1}^\infty \dfrac{n}{(2^n + 2^{-n})^2} +
\dfrac{(-1)^n n}{(2^n - 2^{-n})^2} n = 1 ∑ ∞ ( 2 n + 2 − n ) 2 n + ( 2 n − 2 − n ) 2 ( − 1 ) n n .