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
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

Difference between revisions of "2013 AIME II Problems/Problem 8"

(→‎Solution)
(→‎Solution)
Line 6: Line 6:
 
We can just consider one half of the hexagon, <math>ABCD</math>, to make matters simpler.  
 
We can just consider one half of the hexagon, <math>ABCD</math>, to make matters simpler.  
 
Draw a line from the center of the circle, <math>O</math>, to the midpoint of <math>BC</math>, <math>E</math>. Now, draw a line from <math>O</math> to the midpoint of <math>AB</math>, <math>F</math>. Clearly, <math>\angle BEO=90^{\circ}</math>, because <math>BO=CO</math>, and <math>\angle BFO=90^{\circ}</math>, for similar reasons. Also notice that <math>\angle AOE=90^{\circ}</math>.  
 
Draw a line from the center of the circle, <math>O</math>, to the midpoint of <math>BC</math>, <math>E</math>. Now, draw a line from <math>O</math> to the midpoint of <math>AB</math>, <math>F</math>. Clearly, <math>\angle BEO=90^{\circ}</math>, because <math>BO=CO</math>, and <math>\angle BFO=90^{\circ}</math>, for similar reasons. Also notice that <math>\angle AOE=90^{\circ}</math>.  
−
Let us call <math>\angle BFO=\theta</math>. Therefore, <math>\angle AOB=2\theta</math>, and so <math>\angle AOE=90-2\theta</math>. Let us label the radius of the circle <math>r</math>. This means <cmath>\sin{\theta}=\frac{BF}{r}=\frac{11}{r}</cmath> <cmath>\sin{90-2\theta}=\frac{BE}{r}=\frac{10}{r}</cmath>
+
Let us call <math>\angle BFO=\theta</math>. Therefore, <math>\angle AOB=2\theta</math>, and so <math>\angle AOE=90-2\theta</math>. Let us label the radius of the circle <math>r</math>. This means <cmath>\sin{\theta}=\frac{BF}{r}=\frac{11}{r}</cmath> <cmath>\sin{(90-2\theta)}=\frac{BE}{r}=\frac{10}{r}</cmath>
 
Now we can use simple trigonometry to solve for <math>r</math>.
 
Now we can use simple trigonometry to solve for <math>r</math>.
−
Recall that  <math>\sin{90-\alpha}=\cos(\alpha)</math>: That means <math>\sin{90-2\theta}=\cos{2\theta}=\frac{10}{r}</math>
+
Recall that  <math>\sin{(90-\alpha)}=\cos(\alpha)</math>: That means <math>\sin{(90-2\theta)}=\cos{2\theta}=\frac{10}{r}</math>
 
Recall that <math>\cos{2\alpha}=1-2\sin^2{\alpha}</math>: That means <math>\cos{2\theta}=1-2\sin^2{\theta}=\frac{10}{r}</math>.
 
Recall that <math>\cos{2\alpha}=1-2\sin^2{\alpha}</math>: That means <math>\cos{2\theta}=1-2\sin^2{\theta}=\frac{10}{r}</math>.
 
Let <math>\sin{\theta}=x</math>.
 
Let <math>\sin{\theta}=x</math>.

Revision as of 20:21, 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 $r=5+\sqrt{267}$, so the answer is $5+267=\boxed{272}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_8&oldid=52118"