Suppose $\theta_{i}\in(-\frac{\pi}{2},\frac{\pi}{2}), i = 1,2,3,4$. Prove that, there exist $x\in \mathbb{R}$, satisfying two inequalities \begin{eqnarray*} \cos^2\theta_1\cos^2\theta_2-(\sin\theta\sin\theta_2-x)^2 &\geq& 0, \\ \cos^2\theta_3\cos^2\theta_4-(\sin\theta_3\sin\theta_4-x)^2 & \geq & 0 \end{eqnarray*} if and only if \[ \sum^4_{i=1}\sin^2\theta_i\leq2(1+\prod^4_{i=1}\sin\theta_i + \prod^4_{i=1}\cos\theta_i). \]
Problem
Source: china mathematical olympiad cmo 2005 final round - Problem 1
Tags: inequalities, trigonometry, inequalities unsolved
billzhao
22.01.2005 18:19
Is it just me, or is Chinese Olympiad getting easier? This one is straight forward manipulation, using sum-to-product and product-to-sum rules repeatedly. I'll sketch the idea, but I'll leave out the details. Let $p = \cos(\theta_1 + \theta_2)$, $q = \cos(\theta_1 - \theta_2)$, $r = \cos(\theta_3 + \theta_4)$, $s = \cos(\theta_3 - \theta_4)$. The first inequality is equivalent to $-p \leq x \leq q$ and the second inequality is equivalent to $-r \leq x \leq s$. So $x$ exist iff $\max(-p, -r) \leq \min(q, s)$, or $\min(p, r) + \min(q,s) \geq 0$. Since $|\theta_i | \leq \pi / 2$, we have $p + q \geq 0$, and $ r + s \geq 0$. So the first condition is equivalent to $p + s \geq 0$ and $q + r \geq 0$. The second condition rearranges to $(p+s)(q+r) \geq 0$. The only if part is clear. The other way is also easy, since if both $p+s$ and $q+r$ are nonpositive (with at least one negative), then $p+q+r+s < 0 $, contradicting the $p + q \geq 0$ and $r + s \geq 0$ requirement.
yptsoi
22.01.2005 18:32
I think q1 of this year CMO is quite easy if comparing it with the first one of last year's.
3X.lich
22.01.2005 18:50
It got easier because you guys improved skills over the year and therefore it seems easier. It's a good sign, you know.
Valentin Vornicu
22.01.2005 18:51
billzhao wrote: Is it just me, or is Chinese Olympiad getting easier? Maybe it's just you getting better over the year rather than the level going down
kennyworm
25.01.2005 16:54
i think it's the standard going down cos i managed to solve it too. was quite shocked initially.
math_sipo
31.05.2005 11:48
nice solution billzhao!
suweijie
01.06.2005 08:18
it is the easiest problem of the 2005 Chinese MO, but at that time I nearly spent an hour to solve it.