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Difference between revisions of "1976 AHSME Problems/Problem 18"

(Created page with "Extend <math>\overline{BD}</math> until it touches the opposite side of the circle, say at point <math>E.</math> By power of a point, we have <math>AB^2=(BC)(BE),</math> so <m...")
 
 
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== Problem 18 ==
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<asy>
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//size(100);//local
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size(200);
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real r1=2;
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pair
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O=(0,0),
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D=(.5,.5*sqrt(3)),
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C=(D.x+.5*3,D.y),
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B,
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B_prime=endpoint(arc(D, 3, 0,-2));
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B=B_prime;
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path
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c1=circle(O, r1);
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pair C=midpoint(D--B_prime);
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path arc2=arc(B_prime, 6/2, 158.25,250);
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draw(c1);
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draw(O--D);
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draw(D--C);
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draw(C--B_prime);
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pair A=beginpoint(arc2);
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draw(B_prime--A);
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//dot(O^^D^^C^^A);
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//dot(B_prime);
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label("\scriptsize{$O$}",O,.6dir(D--O));
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label("\scriptsize{$C$}",C,.5dir(-55));
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label("\scriptsize{$D$}", D,.2NW);
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//label("\scriptsize{$B$}",B,S);
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label("\scriptsize{$B$}", B_prime, .5*dir(D--B_prime));
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label("\scriptsize{$A$}",A,.5dir(NE));
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label("\tiny{2}", O--D, .45*LeftSide);
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label("\tiny{3}", D--C, .45*LeftSide);
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label("\tiny{6}", B_prime--A, .45*RightSide);
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label("\tiny{3}", waypoint(C--B_prime,.1), .45*N);
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//Credit to Klaus-Anton for the diagram</asy>
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In the adjoining figure, <math>AB</math> is tangent at <math>A</math> to the circle with center <math>O</math>; point <math>D</math> is interior to the circle;
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and <math>DB</math> intersects the circle at <math>C</math>. If <math>BC=DC=3</math>, <math>OD=2</math>, and <math>AB=6</math>, then the radius of the circle is
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<math>\textbf{(A) }3+\sqrt{3}\qquad
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\textbf{(B) }15/\pi\qquad
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\textbf{(C) }9/2\qquad
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\textbf{(D) }2\sqrt{6}\qquad
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\textbf{(E) }\sqrt{22}</math>
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== Solution ==
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Extend <math>\overline{BD}</math> until it touches the opposite side of the circle, say at point <math>E.</math> By power of a point, we have <math>AB^2=(BC)(BE),</math> so <math>BE=\frac{6^2}{3}=12.</math> Therefore, <math>DE=12-3-3=6.</math>  
 
Extend <math>\overline{BD}</math> until it touches the opposite side of the circle, say at point <math>E.</math> By power of a point, we have <math>AB^2=(BC)(BE),</math> so <math>BE=\frac{6^2}{3}=12.</math> Therefore, <math>DE=12-3-3=6.</math>  
  
 
Now extend <math>\overline{OD}</math> in both directions so that it intersects the circle in two points (in other words, draw the diameter of the circle that contains <math>OD</math>). Let the two endpoints of this diameter be <math>P</math> and <math>Q,</math> where <math>Q</math> is closer to <math>C.</math> Again use power of a point. We have <math>(PD)(DQ)=(CD)(DE)=3 \cdot 6=18.</math> But if the radius of the circle is <math>r,</math> we see that <math>PD=r+2</math> and <math>DQ=r-2,</math> so we have the equation <math>(r+2)(r-2)=18.</math> Solving gives <math>r=\sqrt{22}.</math>
 
Now extend <math>\overline{OD}</math> in both directions so that it intersects the circle in two points (in other words, draw the diameter of the circle that contains <math>OD</math>). Let the two endpoints of this diameter be <math>P</math> and <math>Q,</math> where <math>Q</math> is closer to <math>C.</math> Again use power of a point. We have <math>(PD)(DQ)=(CD)(DE)=3 \cdot 6=18.</math> But if the radius of the circle is <math>r,</math> we see that <math>PD=r+2</math> and <math>DQ=r-2,</math> so we have the equation <math>(r+2)(r-2)=18.</math> Solving gives <math>r=\sqrt{22}.</math>

Latest revision as of 08:05, 9 April 2023

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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