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)
(→‎Solution 2)
Line 18: Line 18:
  
 
===Solution 2===
 
===Solution 2===
−
Using the trapezoid <math>ABCD</math> mentioned above, draw an altitude of the trapezoid passing through point <math>B</math> onto <math>AD</math> at point <math>E</math>. Now, we can use the pythagorean theorem: <math>(22^2-(r-10)^2)+10^2=r^2</math>. Expanding and combining like terms gives us the quadratic <cmath>r^2-10r-242</cmath> and solving for <math>r</math> gives <math>r=5+\sqrt{267}</math>. So the solution is <math>5+267=\boxed{272}</math>
+
Using the trapezoid <math>ABCD</math> mentioned above, draw an altitude of the trapezoid passing through point <math>B</math> onto <math>AD</math> at point <math>E</math>. Now, we can use the pythagorean theorem: <math>(22^2-(r-10)^2)+10^2=r^2</math>. Expanding and combining like terms gives us the quadratic <cmath>r^2-10r-242=0</cmath> and solving for <math>r</math> gives <math>r=5+\sqrt{267}</math>. So the solution is <math>5+267=\boxed{272}</math>

Revision as of 20:43, 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

Solution 1

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 $r=5+\sqrt{267}$ (because $5-\sqrt{267}$ gives a negative value), so the answer is $5+267=\boxed{272}$

Solution 2

Using the trapezoid $ABCD$ mentioned above, draw an altitude of the trapezoid passing through point $B$ onto $AD$ at point $E$. Now, we can use the pythagorean theorem: $(22^2-(r-10)^2)+10^2=r^2$. Expanding and combining like terms gives us the quadratic \[r^2-10r-242=0\] and solving for $r$ gives $r=5+\sqrt{267}$. So the solution is $5+267=\boxed{272}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_8&oldid=52125"