1951 AMC 12 Problem 44

Question

If

\dfrac{xy}{x+y}=a,

\dfrac{xz}{x+z}=b,

and

\dfrac{yz}{y+z}=c,

where

a,

b,

c

are other than zero, then

x

equals:

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

\dfrac{2abc}{ab+bc+ac}

\dfrac{2abc}{ab+ac-bc}

\dfrac{2abc}{ab+bc-ac}

\dfrac{2abc}{ac+bc-ab}

Invert all three given equations to make them linear in

\frac{1}{x},\frac{1}{y},\frac{1}{z}

Compute

\dfrac1b+\dfrac1a-\dfrac1c=\dfrac2x

Inverting gives

\begin{gathered} \frac1x+\frac1y=\frac1a,\\ \frac1x+\frac1z=\frac1b,\\ \frac1y+\frac1z=\frac1c. \end{gathered}

Adding the first two and subtracting the third yields

\begin{aligned} \frac2x &=\frac1a+\frac1b-\frac1c\\ &=\frac{ac+bc-ab}{abc}. \end{aligned}

Hence

x=\dfrac{2abc}{ac+bc-ab}.

Thus, the correct answer is E.

Problem 44 in other years