# 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 $P$ 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 triangle $CDP$ is right, then angle $CDP$ is right or angle $DCP$ is right. Assume that angle $DCP$ is right. Triangle $APB$ is right, so $CP=\sqrt{4*6}=\sqrt{24}$. Then, $DP=\sqrt{28}$, so the length we are looking for is $\sqrt{24}+\sqrt{28}>10$, eliminating $(D)$.

Thus, our answer is $(E)$. $\fbox{}$

Note: Say you are not convinced that $\sqrt{24}+\sqrt{28}>10$. We can prove this as follows.

Start by simplifying the equation: $\sqrt{6}+\sqrt{7}>5$.

Square both sides: $6+2\sqrt{42}+7>25$.

Simplify: $\sqrt{42}>6$

Square both sides again: $42>36$. From here, we can just reverse our steps to get $\sqrt{24}+\sqrt{28}>10$.