Difference between revisions of "2013 AIME II Problems/Problem 8"
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Now substitute the first equation into the second equation: <math>1-2\left(\frac{11}{r}\right)^2=\frac{10}{r}</math> | Now substitute the first equation into the second equation: <math>1-2\left(\frac{11}{r}\right)^2=\frac{10}{r}</math> | ||
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> | + | Using the quadratic equation to solve, we get that <math>r=10+\sqrt{267}</math>, so the answer is <math>5+267=\boxed{272}</math> |
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Revision as of 19:38, 4 April 2013
A hexagon that is inscribed in a circle has side lengths , , , , , and in that order. The radius of the circle can be written as , where and are positive integers. Find .
Solution
Let us call the hexagon , where , and . We can just consider one half of the hexagon, , to make matters simpler. Draw a line from the center of the circle, , to the midpoint of , . Now, draw a line from to the midpoint of , . Clearly, , because , and , for similar reasons. Also notice that . Let us call . Therefore, , and so . Let us label the radius of the circle . This means Now we can use simple trigonometry to solve for . Recall that : That means Recall that : That means . Let . Substitute to get and Now substitute the first equation into the second equation: Multiplying both sides by and reordering gives us the quadratic Using the quadratic equation to solve, we get that , so the answer is