Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Prove it China

Let $\triangle ABC$ be a triangle. Points $D$ and $E$ are on sides $AB$ and $AC$ , respectively, and point $F$ is on line segment $DE$ . Let $\frac{AD}{AB} = x$ , $\frac{AE}{AC} = y$ , $\frac{DF}{DE} = z$ . Prove that

(1) $S_{\triangle BDF} = (1-x)y zS_{\triangle ABC}$ and $S_{\triangle CEF} = x(1-y)(1-z)S_{\triangle ABC}$ ;

(2) $\sqrt[3]{S_{\triangle BDF}} + \sqrt[3]{S_{\triangle CEF}} \le \sqrt[3]{S_{\triangle ABC}}$ .

(posed by Li Weigu)

Solution

Connect $BE$ and $CD$ . Then we have

(1) $S_{\triangle BDF} = zS_{\triangle BDE} = z(1-x)S_{\triangle ABE}$
$= z(1-x)yS_{\triangle ABC}$ and
$S_{\triangle CEF} = (1-z)S_{\triangle CDE}$
$= (1-z)(1-y)S_{\triangle ACD}$
$= (1-z)(1-y)xS_{\triangle ABC}$ .

(2) From (1) we get

$\begin{align*} & \sqrt[3]{S_{\triangle BDF}} + \sqrt[3]{S_{\triangle CEF}} \\ &= (\sqrt[3]{(1-x)yz} + \sqrt[3]{x(1-z)(1-y)})\sqrt[3]{S_{\triangle ABC}} \\ &\le \left(\frac{(1-x)+y+z}{3} + \frac{x+(1-y)+(1-z)}{3}\right)\sqrt[3]{S_{\triangle ABC}} \\ &= \sqrt[3]{S_{\triangle ABC}}. \end{align*}$

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.