Problem

1957 AHSME Problem 18

Circle O has diameters AB and CD perpendicular to each other. AM is any chord intersecting CD at P. Then AP\cdot AM is equal to:

[asy] defaultpen(linewidth(.8pt)); unitsize(2cm); pair O = origin; pair A = (-1,0); pair B = (1,0); pair C = (0,1); pair D = (0,-1); pair M = dir(45); pair P = intersectionpoint(O--C,A--M); draw(Circle(O,1)); draw(A--B); draw(C--D); draw(A--M); label("$A$",A,W); label("$B$",B,E); label("$C$",C,N); label("$D$",D,S); label("$M$",M,NE); label("$O$",O,NE); label("$P$",P,NW);[/asy]

\begin{array}{l} \textbf{(A)}\ AO\cdot OB \qquad \textbf{(B)}\ AO\cdot AB\qquad \\ \textbf{(C)}\ CP\cdot CD \qquad \textbf{(D)}\ CP\cdot PD\qquad \textbf{(E)}\ CO\cdot OP \end{array}

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


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