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

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

Revision as of 22:24, 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 $600$.

See also

2005 AMC 10A (ProblemsAnswer KeyResources)
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
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