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

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

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1997 PMWC Problems/Problem T1

Problem

Solution

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