1951 AHSME Problems/Problem 44

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Problem

If $\frac{xy}{x+y}= a,\frac{xz}{x+z}= b,\frac{yz}{y+z}= c$, where $a, b, c$ are other than zero, then $x$ equals:

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

Solution

Note that $\frac{1}{a}=\frac{1}{x}+\frac{1}{y}$, $\frac{1}{b}=\frac{1}{x}+\frac{1}{z}$, and $\frac{1}{c}=\frac{1}{y}+\frac{1}{z}$. Therefore

\[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{2}\]

Therefore

\[\frac{1}{x}=\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}-\frac{1}{c}=\frac{1}{2a}+\frac{1}{2b}-\frac{1}{2c}\]

A little algebraic manipulation yields that

\[x=\boxed{\textbf{(E)}\ \frac{2abc}{ac+bc-ab}}\]

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 43
Followed by
Problem 45
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