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 1

The two pole formula says this height is half 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 $\frac1{\frac1{16}}=\boxed{16 \textbf{ (C)}}$ .

Solution 2

The two lines can be represented as $y=\frac{-x}{5}+20$ and $y=\frac{4x}{5}$ . Solving the system,

$\frac{-x}{5}+20=\frac{4x}{5}$

$20=x$

So the lines meet at an $x$ -coordinate of 20.

Solving for the height they meet,

$y=\frac{4\cdot 20}{5}$ $y=16$ $\boxed{16 \textbf{ (C)}}.$

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

Contents

Problem

Solution 1

Solution 2

See Also