2. Let P be a point in the interior of triangle ABC . Extend AP, BP, and CP to meet BC, AC, and AB at D, E, and F, respectively. If triangle APF, triangle BPD and triangle CPE have equal areas, prove that P is the centroid of triangle ABC .
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Tags: geometry
Freeman
04.02.2013 15:03
Let $[APF]=[BPD]=[CPE]=1,[APE]=a,[BPF]=b,[CPD]=c$
$\frac{[APE]}{[CPE]}=\frac{[APB]}{[BPC]},
\frac{[BPF]}{[APF]}=\frac{[BPC]}{[CPA]},
\frac{[CPD]}{[BPD]}=\frac{[CPA]}{[APB]}$
$\frac a1=\frac{b+1}{c+1},\frac b1=\frac{c+1}{a+1},\frac c1=\frac{a+1}{b+1}$
Sovling, $a=b=c=1,P$ is the centroid.
sayantanchakraborty
24.01.2014 22:15
$[APF]=[BPD]=[CPE]=x,[PBF]=y,[PDC]=z,[APE]=w$ Then $\frac{x}{z}=\frac{x+y}{x+w} \implies x^2=xz+yz-xw$ $\frac{x}{w}=\frac{x+z}{z+y} \implies x^2=xw+wz-xy$ $\frac{x}{y}=\frac{x+w}{x+z} \implies x^2=xy+wy-xz$ Multiplying we get $x^3=yzw$ Also on adding the right-side equations we get and applying AM-GM we get $3x^2=yz+wz+wy \ge 3(wyz)^{\frac{2}{3}}=3x^2 \implies y=z=w=x$. Now there's no problem.