1976 AHSME Problems/Problem 18

Problem 18

$[asy] //size(100);//local size(200); real r1=2; pair O=(0,0), D=(.5,.5*sqrt(3)), C=(D.x+.5*3,D.y), B, B_prime=endpoint(arc(D, 3, 0,-2)); B=B_prime; path c1=circle(O, r1); pair C=midpoint(D--B_prime); path arc2=arc(B_prime, 6/2, 158.25,250); draw(c1); draw(O--D); draw(D--C); draw(C--B_prime); pair A=beginpoint(arc2); draw(B_prime--A); //dot(O^^D^^C^^A); //dot(B_prime); label("\scriptsize{$O$}",O,.6dir(D--O)); label("\scriptsize{$C$}",C,.5dir(-55)); label("\scriptsize{$D$}", D,.2NW); //label("\scriptsize{$B$}",B,S); label("\scriptsize{$B$}", B_prime, .5*dir(D--B_prime)); label("\scriptsize{$A$}",A,.5dir(NE)); label("\tiny{2}", O--D, .45*LeftSide); label("\tiny{3}", D--C, .45*LeftSide); label("\tiny{6}", B_prime--A, .45*RightSide); label("\tiny{3}", waypoint(C--B_prime,.1), .45*N); //Credit to Klaus-Anton for the diagram[/asy]$

In the adjoining figure, $AB$ is tangent at $A$ to the circle with center $O$ ; point $D$ is interior to the circle; and $DB$ intersects the circle at $C$ . If $BC=DC=3$ , $OD=2$ , and $AB=6$ , then the radius of the circle is

$\textbf{(A) }3+\sqrt{3}\qquad \textbf{(B) }15/\pi\qquad \textbf{(C) }9/2\qquad \textbf{(D) }2\sqrt{6}\qquad \textbf{(E) }\sqrt{22}$

Solution

Extend $\overline{BD}$ until it touches the opposite side of the circle, say at point $E.$ By power of a point, we have $AB^2=(BC)(BE),$ so $BE=\frac{6^2}{3}=12.$ Therefore, $DE=12-3-3=6.$

Now extend $\overline{OD}$ in both directions so that it intersects the circle in two points (in other words, draw the diameter of the circle that contains $OD$ ). Let the two endpoints of this diameter be $P$ and $Q,$ where $Q$ is closer to $C.$ Again use power of a point. We have $(PD)(DQ)=(CD)(DE)=3 \cdot 6=18.$ But if the radius of the circle is $r,$ we see that $PD=r+2$ and $DQ=r-2,$ so we have the equation $(r+2)(r-2)=18.$ Solving gives $r=\sqrt{22}.$

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1976 AHSME Problems/Problem 18

Problem 18

Solution