Difference between revisions of "2013 AIME II Problems/Problem 8"
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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 | + | Let us call <math>\angle BOF=\theta</math>. Therefore, <math>\angle AOB=2\theta</math>, and so <math>\angle BOE=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>. |
Revision as of 23:17, 11 December 2013
Problem 8
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
Solution 1
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 (because gives a negative value), so the answer is .
Solution 2
Using the trapezoid mentioned above, draw an altitude of the trapezoid passing through point onto at point . Now, we can use the pythagorean theorem: . Expanding and combining like terms gives us the quadratic and solving for gives . So the solution is .
Solution 3
Join the diameter of the circle and let the length be . By Ptolemy's Theorem on trapezoid , . Since it is an isosceles trapezoid, both diagonals are equal. Let them be equal to each. Then
Since is subtended by the diameter, it is right. Hence by the Pythagorean Theorem with right :
From the above equations, we have:
Since the radius is half the diameter, it is , so the answer is .
See Also
2013 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.