1951 AHSME Problems/Problem 30

Problem

If two poles $20''$ and $80''$ high are $100''$ apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:

$\textbf{(A)}\ 50''\qquad\textbf{(B)}\ 40''\qquad\textbf{(C)}\ 16''\qquad\textbf{(D)}\ 60''\qquad\textbf{(E)}\ \text{none of these}$

Solution

The two pole formula says this height is the harmonic mean of the heights of the two poles. (The distance between the poles is irrelevant.) So the answer is $\frac1{\frac1{20}+\frac1{80}}$ , or $\frac1{\frac1{16}}=\boxed{16 \textbf{ (C)}}$ .

1951 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 29	Followed by Problem 31
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1951 AHSME Problems/Problem 30

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