2010 AMC 12A Problems/Problem 17
Contents
[hide]Problem
Equiangular hexagon has side lengths and . The area of is of the area of the hexagon. What is the sum of all possible values of ?
Solution 1
It is clear that is an equilateral triangle. From the Law of Cosines on , we get that . Therefore, the area of is .
If we extend , and so that and meet at , and meet at , and and meet at , we find that hexagon is formed by taking equilateral triangle of side length and removing three equilateral triangles, , and , of side length . The area of is therefore
.
Based on the initial conditions,
Simplifying this gives us . By Vieta's Formulas we know that the sum of the possible value of is .
Solution 2
Step 1: Use Law of Cosines in the same manner as the previous solution to get .
Step 2: ~~ via SAS congruency. Using the formula and because of the congruency, the area condition, and the fact is equilateral, .
Step 3: and by Vieta's Formulas , we get .
Note: Since has to be positive we must first check that the discriminant is positive before applying Vieta's. And it indeed is.
See also
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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