Mock AIME 1 2007-2008 Problems/Problem 14

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Problem 14

Points $A$ and $B$ lie on $\odot O$ , with radius $r$ , so that $\angle OAB$ is acute. Extend $AB$ to point $C$ so that $AB = BC$ . Let $D$ be the intersection of $\odot O$ and $OC$ such that $CD = \frac {1}{18}$ and $\cos(2\angle OAC) = \frac 38$ . If $r$ can be written as $\frac{a+b\sqrt{c}}{d}$ , where $a,b,$ and $d$ are relatively prime and $c$ is not divisible by the square of any prime, find $a+b+c+d$ .

Solution

By the cosine double-angle formula, $\cos(2\angle OAC) = 2\cos^2(\angle OAC) - 1 = \frac 38\ \Longrightarrow\ \cos \angle OAC = \sqrt{\frac{11}{16}}.$

The Law of Cosines on $\triangle BAO$ with respect to $\angle BAO$ yields $\begin{align*}r^2 &= r^2 + AB^2 - 2 \cdot AB \cdot r \cos \angle BAO \\AB^2 &= 2 \cdot AB \cdot r \cdot \frac{\sqrt{11}}{4}\\AB &= \frac{r\sqrt{11}}{2}$ Now, $AC = 2AB = r\sqrt{11}$ . The Law of Cosines on $\triangle CAO$ with respect to $\angle CAO$ yields $\begin{align*}\left(r+\frac{1}{18}\right)^2 &= r^2 + \left(r\sqrt{11}\right)^2 - 2 \cdot r\sqrt{11} \cdot r \cos \angle BAO \\\frac{r}{9} + \frac{1}{324} &= 11r^2 - r\sqrt{11} \cdot \left(\frac{\sqrt{11}}{2}\right) \\0 &= 1782 r^2 - 36r - 1 \\r &= \frac{36 \pm \sqrt{36^2 + 4 \cdot 1782 \cdot 1}}{2 \cdot 1782} = \frac{2 + \sqrt{26}}{198}\end{align*}$ The answer is thus $a+b+c+d = 2 + 1 + 26 + 198 = \boxed{227}$ .

Mock AIME 1 2007-2008 (Problems, Source)
Preceded by Problem 13	Followed by Problem 15
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Mock AIME 1 2007-2008 Problems/Problem 14

From AoPSWiki

Problem 14

Solution

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