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Difference between revisions of "2006 AMC 12B Problems/Problem 23"
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Line 72: Line 72:
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\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 \\
+
\sin^2(\alpha)+\cos^2(\alpha) = 1 &\Rightarrow \frac{s^4-26s^2+169}{144s^2} + \frac{s^4-170s^2+7225}{144s^2} = 1 \\
 
&\Rightarrow 2s^4-340s^2+7394 = 0 \\
 
&\Rightarrow 2s^4-340s^2+7394 = 0 \\
 
&\Rightarrow s^4-170s^2+3697 = 0 \\
 
&\Rightarrow s^4-170s^2+3697 = 0 \\

Revision as of 22:52, 19 February 2012

Problem

Isosceles $\triangle ABC$ has a right angle at $C$. Point $P$ is inside $\triangle ABC$, such that $PA=11$, $PB=7$, and $PC=6$. Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt{2}{$ (Error compiling LaTeX. Unknown error_msg), where $a$ and $b$ are positive integers. What is $a+b$?

[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); [/asy]

$\mathrm{(A)}\ 85 \qquad \mathrm{(B)}\ 91 \qquad \mathrm{(C)}\ 108 \qquad \mathrm{(D)}\ 121 \qquad \mathrm{(E)}\ 127$

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 $\triangle PBC$, we have:

\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*}

Using the Law of Cosines on $\triangle PAC$, we have: \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*}

Now we use $\sin^2(\alpha) + \cos^2(\alpha) = 1$. \begin{align*} \sin^2(\alpha)+\cos^2(\alpha) = 1 &\Rightarrow \frac{s^4-26s^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*}

Note that we know that we want the solution with $s^2 > 85$ since we know that $\sin(\alpha) > 0$. Thus, $a+b=85+42=\boxed{127}$.

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
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