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 "2013 AIME II Problems/Problem 8"
(→‎Solution)
Line 15: Line 15:
 
Multiplying both sides by <math>r^2</math> and reordering gives us the quadratic <cmath>r^2-10r-242=0</cmath>
 
Multiplying both sides by <math>r^2</math> and reordering gives us the quadratic <cmath>r^2-10r-242=0</cmath>
 
Using the quadratic equation to solve, we get that <math>r=10+\sqrt{267}</math>, so the answer is <math>10+267=\boxed{277}</math>
 
Using the quadratic equation to solve, we get that <math>r=10+\sqrt{267}</math>, so the answer is <math>10+267=\boxed{277}</math>
 +
 +
EDIT: This solution is wrong, a fix will be here soon.

Revision as of 19:33, 4 April 2013

A hexagon that is inscribed in a circle has side lengths $22$, $22$, $20$, $22$, $22$, and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$, where $p$ and $q$ are positive integers. Find $p+q$.

Solution

Let us call the hexagon $ABCDEF$, where $AB=CD=DE=AF=22$, and $BC=EF=20$. We can just consider one half of the hexagon, $ABCD$, to make matters simpler. Draw a line from the center of the circle, $O$, to the midpoint of $BC$, $E$. Now, draw a line from $O$ to the midpoint of $AB$, $F$. Clearly, $\angle BEO=90^{\circ}$, because $BO=CO$, and $\angle BFO=90^{\circ}$, for similar reasons. Also notice that $\angle AOE=90^{\circ}$. Let us call $\angle BFO=\theta$. Therefore, $\angle AOB=2\theta$, and so $\angle AOE=90-2\theta$. Let us label the radius of the circle $r$. This means \[\sin{\theta}=\frac{BF}{r}=\frac{11}{r}\] \[\sin{90-2\theta}=\frac{BE}{r}=\frac{10}{r}\] Now we can use simple trigonometry to solve for $r$. Recall that $\sin{90-\alpha}=\cos(\alpha)$: That means $\sin{90-2\theta}=\cos{2\theta}=\frac{10}{r}$ Recall that $\cos{2\alpha}=1-2\sin^2{\alpha}$: That means $\cos{2\theta}=1-2\sin^2{\theta}=\frac{10}{r}$. Let $\sin{\theta}=x$. Substitute to get $x=\frac{11}{r}$ and $1-2x^2=\frac{10}{r}$ Now substitute the first equation into the second equation: $1-2\left(\frac{11}{r}\right)^2=\frac{10}{r}$ Multiplying both sides by $r^2$ and reordering gives us the quadratic \[r^2-10r-242=0\] Using the quadratic equation to solve, we get that $r=10+\sqrt{267}$, so the answer is $10+267=\boxed{277}$

EDIT: This solution is wrong, a fix will be here soon.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_8&oldid=52111"