# 2005 AMC 10A Problems/Problem 23

## Problem $BCDE$ is a square. Point $A$ is chosen outside of $BCDE$ such that angle $BAC= 120^\circ$ and $AB=AC$. Point $F$ is chosen inside $BCDE$ such that the triangles $ABC$ and $FCD$ are congruent. If $AF=20$, compute the area of $BCDE$. $\mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3}$

## Solution

Let a side of $BCDE$ be $x$. $AS=ST+AT=\frac{x}{2}+\frac{x\sqrt{3}}{6}$ $SF=ST-AT=\frac{x}{2}-\frac{x\sqrt{3}}{6}$ $\left(\frac{x}{2}+\frac{x\sqrt{3}}{6}\right)^2+\left(\frac{x}{2}-\frac{x\sqrt{3}}{6}\right)^2=20^2$ $x=10\sqrt{6}$

The area of $BCDE$ is $600$.

## See also

 2005 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 24 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
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