Difference between revisions of "2006 AMC 12B Problems/Problem 23"
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Note that we know that we want the solution with <math>s^2 > 85</math> since we know that <math>\sin(\alpha) > 0</math>. Thus, <math>a+b=85+42=\boxed{127}</math>. | Note that we know that we want the solution with <math>s^2 > 85</math> since we know that <math>\sin(\alpha) > 0</math>. Thus, <math>a+b=85+42=\boxed{127}</math>. | ||
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+ | ==Solution 2== | ||
+ | Rotate triangle <math>PAC</math> 90 degrees counterclockwise about <math>C</math> so that the image of <math>A</math> rests on <math>B</math>. Now let the image of <math>P</math> be <math>P'</math>. Note that <math>P'C=6</math>, meaning triangle <math>PCP'</math> is right isosceles, and <math>\angle PP'C=45^\circ</math>. Then <math>PP'=6\sqrt{2}</math>. Now because <math>PB=7</math> and <math>P'B=11</math>, we observe that <math>\angle P'PB=90^\circ</math>, by the Pythagorean Theorem on <math>P'PB</math>. Now we have that <math>\angle APC=\angle BP'C=\angle BP'P + \angle PP'C</math>. So we take the cosine of the second equality, using that fact that <math>\angle PP'C=45^\circ</math>, to get <math>\cos(BP'C)=\frac{6\sqrt{2}-7}{11\sqrt{2}}</math>. Finally, we use the fact that <math>\cos(BP'C)=\cos(CPA)</math> and use the Law of Cosines on triangle <math>CPA</math> to arrive at the value of <math>s^2</math>. | ||
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== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=B|num-b=22|num-a=24}} | {{AMC12 box|year=2006|ab=B|num-b=22|num-a=24}} |
Revision as of 21:19, 21 January 2013
Contents
Problem
Isosceles has a right angle at . Point is inside , such that , , and . Legs and have length $s=\sqrt{a+b\sqrt{2}{$ (Error compiling LaTeX. ! Missing } inserted.), where and are positive integers. What is ?
Solution
Using the Law of Cosines on , we have:
Using the Law of Cosines on , we have:
Now we use .
Note that we know that we want the solution with since we know that . Thus, .
Solution 2
Rotate triangle 90 degrees counterclockwise about so that the image of rests on . Now let the image of be . Note that , meaning triangle is right isosceles, and . Then . Now because and , we observe that , by the Pythagorean Theorem on . Now we have that . So we take the cosine of the second equality, using that fact that , to get . Finally, we use the fact that and use the Law of Cosines on triangle to arrive at the value of .
See also
2006 AMC 12B (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 12 Problems and Solutions |