Difference between revisions of "1951 AHSME Problems/Problem 30"
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==Problem== | ==Problem== | ||
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| + | If two poles <math>20''</math> and <math>80''</math> high are <math>100''</math> apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is: | ||
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| + | <math> \textbf{(A)}\ 50''\qquad\textbf{(B)}\ 40''\qquad\textbf{(C)}\ 16''\qquad\textbf{(D)}\ 60''\qquad\textbf{(E)}\ \text{none of these} </math> | ||
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==Solution== | ==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 <math>\frac1{\frac1{20}+\frac1{80}}</math>, or <math>\frac1{\frac1{16}}=\boxed{16 \textbf{ (C)}}</math>. | |
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== See Also == | == See Also == | ||
{{AHSME 50p box|year=1951|num-b=29|num-a=31}} | {{AHSME 50p box|year=1951|num-b=29|num-a=31}} | ||
Revision as of 14:34, 19 April 2014
Problem
If two poles 
and 
high are 
apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:
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 
, or 
.
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 29 |
Followed by Problem 31 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
