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

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

Revision as of 18:50, 26 January 2019

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 $\boxed{\textbf{(E)}\ \frac{(ab)^2+(ac)^2+(bc)^2}{abc}}$.

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