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Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 14"

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== Solution ==
 
== Solution ==
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 +
EDIT: The following solution is wrong.  See the discussion page.
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By the [[trigonometric identity|cosine double-angle]] formula,  
 
By the [[trigonometric identity|cosine double-angle]] formula,  
 
<cmath>\cos(2\angle OAC) = 2\cos^2(\angle OAC) - 1 = \frac 38\ \Longrightarrow\ \cos \angle OAC = \sqrt{\frac{11}{16}}.</cmath>
 
<cmath>\cos(2\angle OAC) = 2\cos^2(\angle OAC) - 1 = \frac 38\ \Longrightarrow\ \cos \angle OAC = \sqrt{\frac{11}{16}}.</cmath>
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r &= \frac{36 \pm \sqrt{36^2 + 4 \cdot 1782 \cdot 1}}{2 \cdot 1782} = \frac{2 + \sqrt{26}}{198}\end{align*}</cmath>
 
r &= \frac{36 \pm \sqrt{36^2 + 4 \cdot 1782 \cdot 1}}{2 \cdot 1782} = \frac{2 + \sqrt{26}}{198}\end{align*}</cmath>
 
The answer is thus <math>a+b+c+d = 2 + 1 + 26 + 198 = \boxed{227}</math>.
 
The answer is thus <math>a+b+c+d = 2 + 1 + 26 + 198 = \boxed{227}</math>.
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ALTERNATE SOLUTION:
 
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Let AB=x.  Use power of a point:
 
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\frac{(frac{1}{18})}{(frac{1/18}{2r})}=frac{x}{2x}=frac{1}{2}
 
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\frac{2}{18}=frac{1}{18} + 2r
 
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\r=frac{1}{36}
 
  
 
== See also ==
 
== See also ==

Latest revision as of 14:23, 25 November 2009

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

EDIT: The following solution is wrong. See the discussion page.

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

[asy] size(220); defaultpen(fontsize(10)); real r = (2+26^0.5)/198; pair O = (0,0), A=(r,0), B= A*dir(112.02432), C=2*B-A; pair[] D = intersectionpoints(C--O,Circle(O,r)); dot(O); dot(A); dot(B); dot(C); dot(D[0]); draw(O--A--C--cycle); draw(Circle(O,r)); draw(O--B,dashed); draw(anglemark(B,A,O,0.15)); label("\(A\)",A,NE); label("\(B\)",B,N); label("\(C\)",C,N); label("\(D\)",D[0],W); label("\(\frac{1}{18}\)",(C+D[0])/2,W); label("\(O\)",O,SW); [/asy]

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} (Error compiling LaTeX. Unknown error_msg)

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 $a+b+c+d = 2 + 1 + 26 + 198 = \boxed{227}$.

See also

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