Difference between revisions of "2005 AMC 10A Problems/Problem 23"

Revision as of 21:24, 4 May 2007

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

http://img501.imageshack.us/img501/5625/tempym3.png

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 $600$ .

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 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

@@ Line 5: / Line 5: @@
 ==Solution==
-{{solution}}
+http://img501.imageshack.us/img501/5625/tempym3.png
+Let a side of <math>BCDE</math> be <math>x</math>.
+<math>AS=ST+AT=\frac{x}{2}+\frac{x\sqrt{3}}{6}</math>
+<math>SF=ST-AT=\frac{x}{2}-\frac{x\sqrt{3}}{6}</math>
+<math>\left(\frac{x}{2}+\frac{x\sqrt{3}}{6}\right)^2+\left(\frac{x}{2}-\frac{x\sqrt{3}}{6}\right)^2=20^2</math>
+<math>x=10\sqrt{6}</math>
+The area of <math>BCDE</math> is <math>600</math>.
 ==See also==

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Difference between revisions of "2005 AMC 10A Problems/Problem 23"

Revision as of 21:24, 4 May 2007

Problem

Solution

See also