1983 AMC 12 Problem 13

Attempt Problem 13 of the 1983 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1983 AMC 12 solutions, or check the answer key.

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13.

xy=a,

xz=b,

and

yz=c,

and none of these quantities is

0,

then

x^2+y^2+z^2

equals

\frac{ab+ac+bc}{abc}

\frac{a^2+b^2+c^2}{abc}

\frac{(a+b+c)^2}{abc}

\frac{(ab+ac+bc)^2}{abc}

\frac{(ab)^2+(ac)^2+(bc)^2}{abc}

Multiply two of

xy=a,

xz=b,

and

yz=c,

then divide by the third

Use

x^2=\frac{ab}{c},

y^2=\frac{ac}{b},

and

z^2=\frac{bc}{a}

From the three products,

\begin{aligned} x^2&=\frac{ab}{c},\\ y^2&=\frac{ac}{b},\\ z^2&=\frac{bc}{a}. \end{aligned}

Putting their sum over denominator

abc

gives

\begin{aligned} x^2+y^2+z^2 &=\frac{(ab)^2+(ac)^2}{abc}\\ &\quad+\frac{(bc)^2}{abc}. \end{aligned}

Therefore, the correct answer is E.

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