1958 AHSME Problem 48

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:

\begin{array}{l} \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}. \end{array}

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.