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Difference between revisions of "1951 AHSME Problems/Problem 30"

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==Solution 1==
 
==Solution 1==
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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 <math>\frac1{\frac1{20}+\frac1{80}}</math>, or <math>\frac1{\frac1{16}}=\boxed{16 \textbf{ (C)}}</math>.
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The [[two poles 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 <math>\frac1{\frac1{20}+\frac1{80}}</math>, or <math>\frac1{\frac1{16}}=\boxed{16 \textbf{ (C)}}</math>.
  
 
==Solution 2==
 
==Solution 2==
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<math>\frac{-x}{5}+20=\frac{4x}{5}</math>
 
<math>\frac{-x}{5}+20=\frac{4x}{5}</math>
  
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<math>20=x</math>
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<math>20=x.</math>
  
 
So the lines meet at an <math>x</math>-coordinate of 20.
 
So the lines meet at an <math>x</math>-coordinate of 20.

Latest revision as of 18:41, 17 August 2021

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 poles 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,

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

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Problem 31
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All AHSME Problems and Solutions

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