Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2006 AIME II Problems/Problem 15

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 $2 + 7 = \boxed{9}$.


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.

Video solution

https://www.youtube.com/watch?v=M6sC26dzb_I

See also

2006 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_15&oldid=128826"