1953 AHSME Problems/Problem 49

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Problem

The coordinates of $A,B$ and $C$ are $(5,5),(2,1)$ and $(0,k)$ respectively. The value of $k$ that makes $\overline{AC}+\overline{BC}$ as small as possible is:

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4\frac{1}{2} \qquad \textbf{(C)}\ 3\frac{6}{7} \qquad \textbf{(D)}\ 4\frac{5}{6}\qquad \textbf{(E)}\ 2\frac{1}{7}$

Solution

k will be between 1 and 5 for AC+BC to be minimum. If we mirror A across the Y axis as A' (-5,5), the distance A'C+BC will be same as AC+BC. The minimum of A'C+BC will occur when C is on the straight line connecting A' and B, i.e., C lies on the line A'B. So, we can find the Y-intercept of the line connecting A'(-5,5) and B(2,1), which is 15/7 = 2 1/7. so, the answer is $\boxed{(E) 2 1/7}.$

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 48
Followed by
Problem 50
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