# Difference between revisions of "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

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 $\boxed{\textbf{(A)}}$.