2013 AIME II Problems/Problem 8
Contents
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 .
Solution 4
As we can see this image, it is symmetrical hence the diameter divides the hexagon into two congruent quadrilateral. Now we can apply the Ptolemy's theorem. Denote the radius is r, we can get , after simple factorization, we can get , it is easy to see that are two solutions for the equation, so we can factorize that into so we only need to find the solution for and we can get is the desired answer for the problem, and our answer is .～bluesoul
Solution 5
Using solution 1's diagram, extend line segments and upwards until they meet at point . Let point be the center of the hexagon. By the postulate, . This means , so . We then solve for :
Remember that as well, so . Solving for gives . So the solution is .
~SoilMilk
Solution 6 (Trig Bash)
Let . So, we have and . So, . Let be the foot of the perpendicular from to . We have . Using Pythagorean theorem on , to get , or . Multiplying by , we get . Rearranging and simplifying, we get a quadratic in : which gives us . Because is in the form , we know to choose the larger option, meaning , so and . By inspection, we get , so our answer is .
~Puck_0
Solution 7
We know that is a diameter, hence and are right triangles. Let , and Hence, is a right triangle with legs and hypotenuse, and is a right triangle with legs with hypotenuse . By Ptolemy's we have . We square both sides to get
We solve for via the Quadratic Formula and receive , but we must divide by since we want the radius, and hence ~SirAppel
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 | ||
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