Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Solution)
 
m (→‎Solution)
 
(6 intermediate revisions by the same user not shown)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
 +
 +
EDIT: The following solution is wrong.  See the discussion page.
 +
 
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>

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_1_2007-2008_Problems/Problem_14&oldid=32978"