# 1997 PMWC Problems/Problem T1

## Problem

Let $PQR$ be an equilateral triangle with sides of length three units. $U$, $V$, $W$, $X$, $Y$, and $Z$ divide the sides into lengths of one unit. Find the ratio of the area of the shaded quadrilateral $UWXY$ to the area of the triangle $PQR$. $[asy] draw((1/2,0)--(-1/2,0)--(0,sqrt(3)/2)--cycle); dot((1/6,sqrt(3)/3)); dot((-1/6,sqrt(3)/3)); dot((1/3,sqrt(3)/6)); dot((-1/3,sqrt(3)/6)); dot((1/6,0)); dot((1/6,0)); dot((-1/6,0)); filldraw((-1/6,sqrt(3)/3)--(1/3,sqrt(3)/6)--(1/6,0)--(-1/6,0)--cycle); label("P",(0,sqrt(3)/2),N); label("Z",(1/6,sqrt(3)/3),NE); label("Y",(1/3,sqrt(3)/6),NE); label("R",(1/2,0),E); label("X",(1/6,0),S); label("W",(-1/6,0),S); label("Q",(-1/2,0),W); label("V",(-1/3,sqrt(3)/6),NW); label("U",(-1/6,sqrt(3)/3),NW); //Credit to chezbgone2 for the diagram[/asy]$

## Solution

Triangles UWQ, PUY, UWX, and UXY are all right triangles with side lengths 1, $\sqrt{3}$, and 2. Thus $[UWXY]=\sqrt{3}$ and $[PQR]=\frac{9}{4}\sqrt{3}$. $\frac{[UWXY]}{[PQR]}=\frac{1}{\frac{9}{4}}=\boxed{\frac{4}{9}}$