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1960 AHSME Problems/Problem 32

Problem

In this figure the center of the circle is $O$. $AB \perp BC$, $ADOE$ is a straight line, $AP = AD$, and $AB$ has a length twice the radius. Then:

[asy] size(150); defaultpen(linewidth(0.8)+fontsize(10)); real e=350,c=55; pair O=origin,E=dir(e),C=dir(c),B=dir(180+c),D=dir(180+e), rot=rotate(90,B)*O,A=extension(E,D,B,rot); path tangent=A--B; pair P=waypoint(tangent,abs(A-D)/abs(A-B)); draw(unitcircle^^C--B--A--E); dot(A^^B^^C^^D^^E^^P,linewidth(2)); label("$O$",O,dir(290)); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,NE); label("$D$",D,dir(120)); label("$E$",E,SE); label("$P$",P,SW);[/asy]

$\textbf{(A)} AP^2 = PB \times AB\qquad \\ \textbf{(B)}\ AP \times DO = PB \times AD\qquad \\ \textbf{(C)}\ AB^2 = AD \times DE\qquad \\ \textbf{(D)}\ AB \times AD = OB \times AO\qquad \\ \textbf{(E)}\ \text{none of these}$

Solution 1

We claim that $\fbox{A}$ is the right answer.

Let the radius of circle $O$ be $r$, and let the length of $AD=x$. Since $AB \perp BC$, $AB$ is a tangent to circle $O$. Thus, by the tangent-secant theorem, we have $AB^2=AD\times AE$, or, $(2r)^2=x(x+2r)$. Through some algebraic manipulation, we find \begin{align}AD^2=x^2=2r(2r-x)\end{align}.

Since $AD^2=AP^2=x^2$, $PB=AB-AP=2r-x$, and $AB=2r$, we see that $(1)$ is identical to $AP^2=PB\times AB$, hence our answer is $\fbox{A}$

Solution 2 (guess and check)

Let $r$ be the radius of the circle, so $AB = 2r$. By the Pythagorean Theorem, $AO = \sqrt{(2r)^2 + r^2} = r \sqrt{5}$. That means, $AD = AP = r \sqrt{5} - r$, so $PB = 2r - r\sqrt{5} + r = 3r - r\sqrt{5}$.

Substitute values for each answer choice to determine which one is correct for all $r$.

For option A, substitution results in \[(r \sqrt{5} - r)^2 = (3r - r\sqrt{5})2r\] \[5r^2 - 2r^2 \sqrt{5} + r^2 = 6r^2 - 2r^2 \sqrt{5}\] \[6r^2 - 2r^2 \sqrt{5} = 6r^2 - 2r^2 \sqrt{5}\]

For option B, substitution results in \[(r \sqrt{5} - r)r = (3r - r\sqrt{5})(r \sqrt{5} - r)\] \[r^2 \sqrt{5} - r^2 = 4r^2 \sqrt{5} - 8r^2\]

For option C, substitution results in \[(2r)^2 = (r\sqrt{5} - r)2r\] \[4r^2 = 2r^2 \sqrt{5} - 2r^2\]

For option D, substitution results in \[(r \sqrt{5} - r) \cdot 2r = r \cdot r\sqrt{5}\] \[2r^2 \sqrt{5} - 2r^2 = r^2 \sqrt{5}\]

From each option, only option A has both sides equaling each other, so the answer is $\boxed{\textbf{(A)}}$.

See Also

1960 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 31 		Followed by
Problem 33
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All AHSME Problems and Solutions
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