1951 AHSME Problems/Problem 30

Revision as of 20:24, 19 March 2017 by Xiej (talk | contribs) (Solution)

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 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

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,

$y=\frac{4\cdot 20}{5}$

$y=16$

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 29
Followed by
Problem 31
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