Bariz - Problem

Source: JBMO 2001, Problem 3

Tags: geometry, trigonometry, geometry proposed

Valentin Vornicu

30.10.2005 02:27

Let $ABC$ be an equilateral triangle and $D$, $E$ points on the sides $[AB]$ and $[AC]$ respectively. If $DF$, $EF$ (with $F\in AE$, $G\in AD$) are the interior angle bisectors of the angles of the triangle $ADE$, prove that the sum of the areas of the triangles $DEF$ and $DEG$ is at most equal with the area of the triangle $ABC$. When does the equality hold? Greece

We have \[\frac{[DEF]}{[DEA]} = \frac{EF}{EA} = \frac{DE}{DE+AD}\] Similarly \[\frac{[DEG]}{[DEA]} = \frac{DG}{DA} = \frac{DE}{DE+AE}\] Thus $[DEF] + [DEG] = DE[DEA](1/(DE+AD) + 1/(DE+AE))$. We have $[DEA] = 1/2 AD\cdot AE\sin A$, and $[ABC] = 1/2 AB\cdot AC \sin A$, so \[\frac{[DEF] + [DEG]}{[ABC]} = \frac{DE\cdot AD\cdot AE}{AB^2}\cdot\frac{1}{(DE+AD) + 1/(DE+AE)}\] Obviously $DE \le AB$, so it is sufficient to prove that $AD\cdot AE(1/(DE+AD) + 1/(DE+AE)) \le DE (*)$. The cosine rule applied to $ADE$ gives $DE^2 = AD^2 + AE^2 - AD\cdot AE$. Hence also $DE^2 - AD\cdot AE = (AD - AE)^2\ge 0$. Thus we have $DE(AD - AE)^2 + DE^2(AD + AE) \ge AD\cdot AE(AD + AE)$. So \[AD\cdot AE(AD + AE + 2DE) \le DE(AD^2 + AE^2 + DE(AD + AE)) = DE(DE + AD)(DE + AE)\], which is $(*)$.