2010/11 AHSMC Part 1 Problem 15

Problem 15 of 16

Triangle $ABC$ has area $1$ . $X$ , $Y$ are points on the side $AB$ and $Z$ a point on the side $AC$ such that $XY = 2AX$ , $XZ$ is parallel to $YC$ and $YZ$ is parallel to $BC$ . The area of $XYZ$ is

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Solution

Denote the area of triangle $T$ by $[T]$ . Since triangles $AXZ$ and $AYC$ are similar, $ZC = 2AZ$ . Since triangles $AYZ$ and $ABC$ are similar, $YB = 2AY$ . It follows that $[XYZ] = \frac{2}{3}[AYZ] = \frac{2}{9}[AYC] = \frac{2}{27}[ABC] = \frac{2}{27}.$ The answer is (B).

University of Alberta

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