1957 AHSME Problems/Problem 18

Problem

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$",P,NW);[/asy]$

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

Solution 1

Draw $MB$ . Since $\angle AMB$ is inscribed on a diameter, $\angle AMB$ is $90^\circ$ . By AA Similarity, $\triangle APO \sim \triangle ABM$ . Setting up ratios, we get $\frac{AP}{AO}=\frac{AB}{AM}$ . Cross-multiplying, we get $AP\cdot AM = AO \cdot AB$ , so the answer is $\boxed{\textbf{(B)}}$ .

Solution 2

By Thales' Theorem, $\measuredangle PMB = \measuredangle AMB = 90^{\circ}$ . Because, from the problem, $\measuredangle POB = 90^{\circ}$ as well, $\measuredangle PMB + \measuredangle POB = 180^{\circ}$ , so $PMBO$ is a cyclic quadrilateral. Thus, because $P$ , $M$ , $O$ , and $B$ lie on a circle, we can use Power of a Point. From this theorem, we get that $AP \cdot AM = AO \cdot AB$ , which is answer choice $\boxed{\textbf{(B)}}$ .

1957 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1957 AHSME Problems/Problem 18

Contents

Problem

Solution 1

Solution 2

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