Library / /478 of 1394

, 2015

Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:

Let $ABC$ be a triangle with orthocenter $H$ ; suppose that $AB = 13$ , $BC = 14$ , $CA = 15$ . Let $G_{A}$ be the centroid of triangle $HBC$ , and define $G_{B}, G_{C}$ similarly. Determine the area of triangle $G_{A} G_{B} G_{C}$ .

Solution

Solution:

Answer: $\quad 28 / 3$

Let $D, E, F$ be the midpoints of $BC, CA$ , and $AB$ , respectively. Then $G_{A} G_{B} G_{C}$ is the $DEF$ about $H$ with a ratio of $\frac{2}{3}$ , and $DEF$ is the dilation of $ABC$ about $H$ with a ratio of $-\frac{1}{2}$ , so $G_{A} G_{B} G_{C}$ is the dilation of $ABC$ about $H$ with ratio $-\frac{1}{3}$ . Thus $\left[G_{A} G_{B} G_{C}\right] = \frac{[ABC]}{9}$ .

By Heron's formula, the area of $ABC$ is $\sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84$ , so the area of $G_{A} G_{B} G_{C}$ is $[ABC] / 9 = 84 / 9 = 28 / 3$ .

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