1951 AMC 12 Problem 30

Attempt Problem 30 of the 1951 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1951 AMC 12 solutions, or check the answer key.

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30.

If two poles 20′′20'' and 80′′80'' high are 100′′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:
50′′50''
40′′40''
16′′16''
60′′60''
None of these
Answer: C
Concepts:coordinate geometrylinear equation
Small hint:
Place the pole bases at x=0x=0 and x=100x=100
Big hint:
The cross-lines can be written y=4x5y=\frac{4x}{5} and y=20−x5y=20-\frac{x}{5}
Solution:
Put the bases at (0,0)(0,0) and (100,0),(100,0), with tops (0,80)(0,80) and (100,20).(100,20). The cross-lines are y=45xy=\frac45x and y=20−15x.y=20-\frac15x. Equating them gives x=20,x=20, and hence y=16.y=16.
Thus, the correct answer is C.
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