1958 AHSME Problems/Problem 48

Revision as of 23:21, 31 December 2023 by Alexanderruan (talk | contribs) (Solution)

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=$\sqrt{4x6}$=$\sqrt{24}$. Then, DP=$\sqrt(28)$, so the length we are looking for is $\sqrt{24}$+$\sqrt{28}$>10, eliminating (D). Thus, our answer is (E). $\fbox{}$

See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 47
Followed by
Problem 49
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