2006 AIME II Problems/Problem 15

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Problem

Given that $x, y,$ and $z$ are real numbers that satisfy:

$x = \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}}$ $y = \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}}$ $z = \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}$

and that $x+y+z = \frac{m}{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime, find $m+n.$

Solution

Let $\triangle XYZ$ be a triangle with sides of length $x, y$ and $z$ , and suppose this triangle is acute (so all altitudes are on the interior of the triangle). Let the altitude to the side of length $x$ be of length $h_x$ , and similarly for $y$ and $z$ . Then we have by two applications of the Pythagorean Theorem that $x = \sqrt{y^2 - h_x^2} + \sqrt{z^2 - h_x^2}$ . As a function of $h_x$ , the RHS of this equation is strictly decreasing, so it takes each value in its range exactly once. Thus we must have that $h_x^2 = \frac1{16}$ and so $h_x = \frac{1}4$ and similarly $h_y = \frac15$ and $h_z = \frac16$ .

Since the area of the triangle must be the same no matter how we measure, $x\cdot h_x = y\cdot h_y = z \cdot h_z$ and so $\frac x4 = \frac y5 = \frac z6 = 2A$ and $x = 8A, y = 10A$ and $z = 12A$ . The semiperimeter of the triangle is $s = \frac{8A + 10A + 12A}{2} = 15A$ so by Heron's formula we have $A = \sqrt{15A \cdot 7A \cdot 5A \cdot 3A} = 15A^2\sqrt{7}$ . Thus $A = \frac{1}{15\sqrt{7}}$ and $x + y + z = 30A = \frac2{\sqrt{7}}$ and the answer is $2 + 7 = \boxed{009}$ .

Justification that there is an acute triangle with sides of length $x, y$ and $z$ :

Note that $x, y$ and $z$ are each the sum of two positive square roots of real numbers, so $x, y, z \geq 0$ . (Recall that, by AIME convention, all numbers (including square roots) are taken to be real unless otherwise indicated.) Also, $\sqrt{y^2-\frac{1}{16}} < \sqrt{y^2} = y$ , so we have $x < y + z$ , $y < z + x$ and $z < x + y$ . But these conditions are exactly those of the triangle inequality, so there does exist such a triangle.

2006 AIME II (Problems • Resources)
Preceded by Problem 14	Followed by Last Question
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2006 AIME II Problems/Problem 15

From AoPSWiki

Problem

Solution

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