2013 AIME II Problems/Problem 8

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: The answer key says the answer is 272, can someone point out where I went wrong?

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2013 AIME II Problems/Problem 8

Solution