MathDB
Find the radius of circle O

Source: 1976 AHSME Problem 18

May 18, 2014
geometrypower of a pointAMC

Problem Statement

[asy] //size(100);//local size(200); real r1=2; pair O=(0,0), D=(.5,.5*sqrt(3)), C=(D.x+.5*3,D.y), B, B_prime=endpoint(arc(D, 3, 0,-2)); B=B_prime; path c1=circle(O, r1); pair C=midpoint(D--B_prime); path arc2=arc(B_prime, 6/2, 158.25,250); draw(c1); draw(O--D); draw(D--C); draw(C--B_prime); pair A=beginpoint(arc2); draw(B_prime--A); //dot(O^^D^^C^^A); //dot(B_prime); label("\scriptsize{OO}",O,.6dir(D--O)); label("\scriptsize{CC}",C,.5dir(-55)); label("\scriptsize{DD}", D,.2NW); //label("\scriptsize{BB}",B,S); label("\scriptsize{BB}", B_prime, .5*dir(D--B_prime)); label("\scriptsize{AA}",A,.5dir(NE)); label("\tiny{2}", O--D, .45*LeftSide); label("\tiny{3}", D--C, .45*LeftSide); label("\tiny{6}", B_prime--A, .45*RightSide); label("\tiny{3}", waypoint(C--B_prime,.1), .45*N); //Credit to Klaus-Anton for the diagram[/asy]
In the adjoining figure, ABAB is tangent at AA to the circle with center OO; point DD is interior to the circle; and DBDB intersects the circle at CC. If BC=DC=3BC=DC=3, OD=2OD=2, and AB=6AB=6, then the radius of the circle is
<spanclass=latexbold>(A)</span>3+3<spanclass=latexbold>(B)</span>15/π<spanclass=latexbold>(C)</span>9/2<spanclass=latexbold>(D)</span>26<spanclass=latexbold>(E)</span>22<span class='latex-bold'>(A) </span>3+\sqrt{3}\qquad<span class='latex-bold'>(B) </span>15/\pi\qquad<span class='latex-bold'>(C) </span>9/2\qquad<span class='latex-bold'>(D) </span>2\sqrt{6}\qquad <span class='latex-bold'>(E) </span>\sqrt{22}