2006 AIME II Problems/Problem 15
Problem
Given that and are real numbers that satisfy:
and that where and are positive integers and is not divisible by the square of any prime, find
Solution
Let be a triangle with sides of length and , and suppose this triangle is acute (so all altitudes are on the interior of the triangle). Let the altitude to the side of length be of length , and similarly for and . Then we have by two applications of the Pythagorean Theorem that . As a function of , the RHS of this equation is strictly decreasing, so it takes each value in its range exactly once. Thus we must have that and so and similarly and .
Since the area of the triangle must be the same no matter how we measure, and so and and . The semiperimeter of the triangle is so by Heron's formula we have . Thus and and the answer is .
Justification that there is an acute triangle with sides of length and :
Note that and are each the sum of two positive square roots of real numbers, so . (Recall that, by AIME convention, all numbers (including square roots) are taken to be real unless otherwise indicated.) Also, , so we have , and . But these conditions are exactly those of the triangle inequality, so there does exist such a triangle.
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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