1958 AHSME Problems/Problem 48

Problem

Diameter $\overline{AB}$ of a circle with center $O$ is $10$ units. $C$ is a point $4$ units from $A$ , and on $\overline{AB}$ . $D$ is a point $4$ units from $B$ , and on $\overline{AB}$ . $P$ is any point on the circle. Then the broken-line path from $C$ to $P$ to $D$ :

$\textbf{(A)}\ \text{has the same length for all positions of }{P}\qquad\\ \textbf{(B)}\ \text{exceeds }{10}\text{ units for all positions of }{P}\qquad \\ \textbf{(C)}\ \text{cannot exceed }{10}\text{ units}\qquad \\ \textbf{(D)}\ \text{is shortest when }{\triangle CPD}\text{ is a right triangle}\qquad \\ \textbf{(E)}\ \text{is longest when }{P}\text{ is equidistant from }{C}\text{ and }{D}.$

Solution

If somebody wants to draw a diagram or make this solution better, PLEASE do so. I cannot express how bad I am at this.

If P is on A, then the length is 10, eliminating answer choice (B). If P is equidistant from C and D, the length is 2 $\sqrt{1^2+5^2}$ =2 $\sqrt{26}$ >10, eliminating (A) and (C). If CDP is a right triangle, then CDP will be right or DCP will be right. Assume that DCP is right. Then, APB is right, so CP= $\sqrt{4times6}$ = $\sqrt{24}$ . Then, DP= $\sqrt{28}$ , so the length we are looking for is $\sqrt{24}$ + $\sqrt{28}$ >10, eliminating (D). Thus, our answer is (E). $\fbox{}$

1958 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 47	Followed by Problem 49
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1958 AHSME Problems/Problem 48

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