2006 AIME II Problems/Problem 15
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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.
This solution is incomplete. You can help us out by completing it.
See also
| 2006 AIME II (Problems • Resources) | ||
| Preceded by Problem 14 | Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
