2006 AIME II Problems/Problem 15
Problem
Given that and
are real numbers that satisfy:
![$x = \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}}$](http://latex.artofproblemsolving.com/4/8/2/482ec834e100a5c445119023117da55b0a08dae6.png)
![$y = \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}}$](http://latex.artofproblemsolving.com/0/7/e/07e3a5bfe68da11f9902c4b34d8381e913e31859.png)
![$z = \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}$](http://latex.artofproblemsolving.com/0/0/b/00b6096d5f7c9f1584a2efea64e687039b0c28ff.png)
and that where
and
are positive integers and
is not divisible by the square of any prime, find
Solution
This solution requires you to disregard rigor. Additional solutions, or justification for the nonrigorous steps, would be appreciated.
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 a 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.
Justification that this triangle is an acute triangle:
Still needed.
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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