Difference between revisions of "2006 AMC 12B Problems/Problem 23"
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== Solution == | == Solution == | ||
+ | <asy> | ||
+ | pathpen = linewidth(0.7); | ||
+ | pen f = fontsize(10); | ||
+ | size(5cm); | ||
+ | pair B = (0,sqrt(85+42*sqrt(2))); | ||
+ | pair A = (B.y,0); | ||
+ | pair C = (0,0); | ||
+ | pair P = IP(arc(B,7,180,360),arc(C,6,0,90)); | ||
+ | D(A--B--C--cycle); | ||
+ | D(P--A); | ||
+ | D(P--B); | ||
+ | D(P--C); | ||
+ | MP("A",D(A),plain.E,f); | ||
+ | MP("B",D(B),plain.N,f); | ||
+ | MP("C",D(C),plain.SW,f); | ||
+ | MP("P",D(P),plain.NE,f); | ||
+ | MP("\alpha",C,5*dir(80),f); | ||
+ | MP("90^\circ-\alpha",C,3*dir(30),f); | ||
+ | MP("s",(A+C)/2,plain.S,f); | ||
+ | MP("s",(B+C)/2,plain.W,f); | ||
+ | </asy> | ||
+ | Using the Law of Cosines on <math>\triangle PBC</math>, we have: | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | PB^2&=BC^2+PC^2-2\cdot BC\cdot PC\cdot \cos(\alpha) \Rightarrow 49 = 36 + s^2 - 12s\cos(\alpha) \Rightarrow \cos(\alpha) = \dfrac{s^2-13}{12s}. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Using the Law of Cosines on <math>\triangle PAC</math>, we have: | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | PA^2&=AC^2+PC^2-2\cdot AC\cdot PC\cdot \cos(90^\circ-\alpha) \Rightarrow 121 = 36 + s^2 - 12s\sin(\alpha) \Rightarrow \sin(\alpha) = \dfrac{s^2-85}{12s}. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Now we use <math>\sin^2(\alpha) + \cos^2(\alpha) = 1</math>. | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \sin^2(\alpha)+\cos^2(\alpha) = 1 &\Rightarrow \frac{s^4-20s^2+169}{144s^2} + \frac{s^4-170s^2+7225}{144s^2} = 1 \\ | ||
+ | &\Rightarrow 2s^4-340s^2+7394 = 0 \\ | ||
+ | &\Rightarrow s^4-170s^2+3697 = 0 \\ | ||
+ | &\Rightarrow s^2 = \dfrac{170 \pm \sqrt{170^2 - 4\cdot3697}}{2}\\ | ||
+ | &\Rightarrow s^2 = \dfrac{170 \pm \sqrt{28900 - 14788}}{2}\\ | ||
+ | &\Rightarrow s^2 = \dfrac{170 \pm \sqrt{14112}}{2}\\ | ||
+ | &\Rightarrow s^2 = \dfrac{170 \pm \sqrt{2^5\cdot3^2\cdot7^2}}{2}\\ | ||
+ | &\Rightarrow s^2 = \dfrac{170 \pm 84\sqrt{2}}{2} = 85 \pm 42\sqrt2 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | 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>. | ||
== 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 21:00, 16 April 2009
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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. Unknown error_msg), 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, .
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 |