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 $[PQR]=\frac{9}{4}\sqrt{3}$. $\frac{[UWXY]}{[PQR]}=\frac{1}{\frac{9}{4}}=\boxed{\frac{4}{9}}$

See Also

1997 PMWC (Problems)
Preceded by
Problem I15
Followed by
Problem T2
I: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
T: 1 2 3 4 5 6 7 8 9 10