1983 AHSME Problems/Problem 13

Problem

If $xy = a, xz =b,$ and $yz = c$ , and none of these quantities is zero, 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

Using the equations, you can get $x = \frac{a}{y}, z = \frac{b}{x},$ and $y = \frac{c}{z}$ . Substitute into the expression given. You get $\frac{a^2}{y^2} + \frac{b^2}{x^2} + \frac{c^2}{z^2}$ . Combine into one large fraction to get $\frac{a^2x^2z^2+b^2y^2z^2+c^2y^2x^2}{x^2y^2z^2}$ . Make everything in terms of a,b, c. $\frac{a^2b^2+b^2c^2+c^2a^2}{x^2y^2z^2}$ . So the answer is $\boxed{\textbf{(E)}\ \frac{(ab)^2+(ac)^2+(bc)^2}{abc}}$ .

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1983 AHSME Problems/Problem 13

Problem

Solution