Let $ABCDEF$ be a convex equilateral hexagon with sides of length $1$. Let $R_1$ be the area of the region contained within both $ACE$ and $BDF$, and let $R_2$ be the area of the region within the hexagon outside both triangles. Prove that: \[ \min \{ [ACE], [BDF] \} + R_2 - R_1 \le \frac{3\sqrt{3}}{4}. \]
Problem
Source: ELMO 2003 Problem 1
Tags: inequalities, geometry, trigonometry, geometry unsolved
hyperbolictangent
29.12.2012 19:10
Ok so after the unnecessary obfuscation, we basically want \[[ABCDEF] - \max \{[ACE], [BDF]\} \le \frac{3\sqrt{3}}{4}\] If $\theta_A, \theta_B, \dots, \theta_F$ are the corresponding vertex angles, this is equivalently \[\min \{\frac{1}{2}(1)^2[\sin{\theta_A} + \sin{\theta_C} + \sin{\theta_E}], \frac{1}{2}(1)^2[\sin{\theta_B} + \sin{\theta_D} + \sin{\theta_F}] \} \le \frac{3\sqrt{3}}{4}\] By the convexity of $\sin(x)$ on the interval $[0, \pi]$ (where all angles lie, by the convexity of the figure), this is at most \[ \min \{\frac{3}{2} \sin \left({\frac{\theta_A + \theta_C + \theta_E}{3}} \right), \frac{3}{2}\sin \left( {\frac{\theta_B + \theta_D + \theta_F}{3}} \right) \} \] Since $\theta_A + \theta_B + \dots + \theta_F = 4\pi$, this is just \[\frac{3}{2} \min \{\sin{\theta}, \sin \left( {\frac{4\pi}{3} - \theta} \right) \}\] where $\pi / 3 \le \theta \le \pi$ If $\theta \in [\pi / 3, 2\pi /3]$ then \[ \sin \left( \frac{4\pi}{3} - \theta \right) \le \sin \left( \frac{4\pi}{3} - \frac{2\pi}{3} \right) = \sin \left( \frac{2\pi}{3} \right) = \frac{\sqrt{3}}{2}\] Similarly, if $\theta \in [2\pi /3, \pi]$, \[ \sin \left(\theta \right) \le \sin \left(\frac{2\pi}{3} \right) = \frac{\sqrt{3}}{2}\] so in all cases the expression is at most \[\frac{3}{2} \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}\] as desired.
bobthesmartypants
22.04.2015 01:46
Like the above solution, realize that we must prove \[\dfrac{1}{2}\min (\sin{\theta_A} + \sin{\theta_C} + \sin{\theta_E}, \sin{\theta_B} + \sin{\theta_D} + \sin{\theta_F} ) \le \dfrac{3\sqrt{3}}{4}\] or \[\min (\sin{\theta_A} + \sin{\theta_C} + \sin{\theta_E}, \sin{\theta_B} + \sin{\theta_D} + \sin{\theta_F} ) \le \dfrac{3\sqrt{3}}{2}\]
Suppose that \[\min (\sin{\theta_A} + \sin{\theta_C} + \sin{\theta_E}, \sin{\theta_B} + \sin{\theta_D} + \sin{\theta_F} ) >\dfrac{3\sqrt{3}}{2}\]
Then \[\sum\sin\theta > 3\sqrt{3}\]
But by Jenson's Inequality, since $\sin x$ is concave for $0\le x\le \pi$, $$\dfrac{1}{6}\sum\sin\theta \le \sin\left(\dfrac{\sum\theta}{6}\right)=\sin\left(\dfrac{720}{6}\right)=\dfrac{\sqrt{3}}{2}$$ so $$\sum\sin\theta\le 3\sqrt{3}$$, contradiction.
Thus, the original inequality is true.