Difference between revisions of "2005 AMC 10A Problems/Problem 23"
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==Solution== | ==Solution== | ||
− | {{ | + | http://img501.imageshack.us/img501/5625/tempym3.png |
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+ | Let a side of <math>BCDE</math> be <math>x</math>. | ||
+ | |||
+ | |||
+ | <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
is a square. Point is chosen outside of such that angle and . Point is chosen inside such that the triangles and are congruent. If , compute the area of .
Solution
http://img501.imageshack.us/img501/5625/tempym3.png
Let a side of be .
The area of is .
See also
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 |