2006 AIME II Problems/Problem 15
Contents
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 1 (Geometric Interpretation)
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.
Solution 2 (Algebraic)
Video solution
https://www.youtube.com/watch?v=M6sC26dzb_I
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
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