2007 AIME I Problems/Problem 15
Contents
Problem
Let be an equilateral triangle, and let and be points on sides and , respectively, with and . Point lies on side such that angle . The area of triangle is . The two possible values of the length of side are , where and are rational, and is an integer not divisible by the square of a prime. Find .
Solution
Denote the length of a side of the triangle , and of as . The area of the entire equilateral triangle is . Add up the areas of the triangles using the formula (notice that for the three outside triangles, ): . This simplifies to . Some terms will cancel out, leaving .
is an exterior angle to , from which we find that , so . Similarly, we find that . Thus, . Setting up a ratio of sides, we get that . Using the previous relationship between and , we can solve for .
Use the quadratic formula, though we only need the root of the discriminant. This is . The answer is .
Solution 2
First of all, assume , then we can find It is not hard to find , we apply LOC on , getting that , leads to Apply LOC on separately, getting Add those terms together and use the equality , we can find:
According to basic angle chasing, , so , the ratio makes , getting that Now we have two equations with , and values for both equations must be the same, so we can solve for in two equations. , then we can just use positive sign to solve, simplifies to , getting , since the triangle is equilateral, , and the desired answer is
~bluesoul
See also
2007 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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