# Difference between revisions of "1983 AHSME Problems/Problem 13"

## Problem

If $xy = a, xz =b,$ and $yz = c$, and none of these quantities is $0$, then $x^2+y^2+z^2$ equals

$\textbf{(A)}\ \frac{ab+ac+bc}{abc}\qquad \textbf{(B)}\ \frac{a^2+b^2+c^2}{abc}\qquad \textbf{(C)}\ \frac{(a+b+c)^2}{abc}\qquad \textbf{(D)}\ \frac{(ab+ac+bc)^2}{abc}\qquad \textbf{(E)}\ \frac{(ab)^2+(ac)^2+(bc)^2}{abc}$

## Solution

From the equations, we deduce $x = \frac{a}{y}, z = \frac{b}{x},$ and $y = \frac{c}{z}$. Substituting these into the expression $x^2+y^2+z^2$ thus gives $\frac{a^2}{y^2} + \frac{b^2}{x^2} + \frac{c^2}{z^2} = \frac{a^2x^2z^2+b^2y^2z^2+c^2y^2x^2}{x^2y^2z^2} = \frac{a^2b^2+b^2c^2+c^2a^2}{x^2y^2z^2}$, so the answer is $\boxed{\textbf{(E)}\ \frac{(ab)^2+(ac)^2+(bc)^2}{abc}}$.

## Solution 2

$x^2$ is $\frac{abc}{c^2}$, $y^2$ is $\frac{abc}{b^2}$, and $z^2$ is $\frac{abc}{a^2}$, so $x^2+y^2+z^2$ is $\frac{abc}{c^2}+\frac{abc}{b^2}+\frac{abc}{a^2}= \boxed{\textbf{(E)}\ \frac{(ab)^2+(ac)^2+(bc)^2}{abc}}$

-purplepenguin2

## See Also

 1983 AHSME (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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