# Difference between revisions of "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 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,

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

$y=16$