2021 WSMO Speed Round Problems/Problem 4

Problem

A square $ABCD$ with side length $10$ is placed inside of a right isosceles triangle $XYZ$ with $\angle XYZ=90^{\circ}$ such that $A$ and $B$ are on $XZ$ , $C$ is on $YZ$ , and $D$ is on $XY$ . Find the area of $XYZ$ .

Solution 1

We are given that $AB = BC = CD = DA = 10$ . Then $YD = \frac{10}{\sqrt2} = 5\sqrt2$ , and $XD = \sqrt2 \cdot AD = 10\sqrt2$ . This means that $XY = 10\sqrt2+5\sqrt2 = 15\sqrt2$ , for an answer of $\frac{(15\sqrt2)^2}{2} = \boxed{225}$ .

~OlympusHero

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2021 WSMO Speed Round Problems/Problem 4

Problem

Solution 1