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Difference between revisions of "2000 AMC 12 Problems/Problem 17"

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== Problem ==
 
== Problem ==
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[[Image:2000_12_AMC-17.png|right]]
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A circle centered at <math>O</math> has radius <math>1</math> and contains the point <math>A</math>. The segment <math>AB</math> is tangent to the circle at <math>A</math> and <math>\angle AOB = \theta</math>. If point <math>C</math> lies on <math>\overline{OA}</math> and <math>\overline{BC}</math> bisects <math>\angle ABO</math>, then <math>OC =</math>
  
−
A [[circle]] centered at <math>O</math> has [[radius]] <math>1</math> and contains the point <math>A</math>. The segment <math>AB</math> is [[tangent (geometry)|tangent]] to the circle at <math>A</math> and <math>\angle AOB = \theta</math>. If point <math>C</math> lies on <math>\overline{OA}</math> and <math>\overline{BC}</math> bisects <math>\angle ABO</math>, then <math>OC =</math>
+
<asy>
 +
import olympiad;
 +
unitsize(1cm);
 +
defaultpen(fontsize(8pt)+linewidth(.8pt));
 +
labelmargin=0.2;
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dotfactor=3;
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pair O=(0,0);
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pair A=(1,0);
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pair B=(1,1.5);
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pair D=bisectorpoint(A,B,O);
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pair C=extension(B,D,O,A);
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draw(Circle(O,1));
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draw(O--A--B--cycle);
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draw(B--C);
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label("$O$",O,SW);
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dot(O);
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label("$\theta$",(0.1,0.05),ENE);
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dot(C);
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label("$C$",C,S);
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dot(A);
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label("$A$",A,E);
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dot(B);
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label("$B$",B,E);</asy>
  
 
<math>\text {(A)}\ \sec^2 \theta - \tan \theta \qquad \text {(B)}\ \frac 12 \qquad \text {(C)}\ \frac{\cos^2 \theta}{1 + \sin \theta}\qquad \text {(D)}\ \frac{1}{1+\sin\theta} \qquad \text {(E)}\ \frac{\sin \theta}{\cos^2 \theta}</math>
 
<math>\text {(A)}\ \sec^2 \theta - \tan \theta \qquad \text {(B)}\ \frac 12 \qquad \text {(C)}\ \frac{\cos^2 \theta}{1 + \sin \theta}\qquad \text {(D)}\ \frac{1}{1+\sin\theta} \qquad \text {(E)}\ \frac{\sin \theta}{\cos^2 \theta}</math>
  
 
== Solution ==
 
== Solution ==
−
Since <math>\overline{AB}</math> is tangent to the circle, <math>\triangle OAB</math> is a [[right triangle]]. Thus since <math>OA = 1</math>, <math>BA = \tan \theta</math> and <math>OB = \sec \theta</math>. By the [[Angle Bisector Theorem]], <math>\frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta</math>. Multiply both sides by <math>\cos \theta</math> to simplify the trigonometric functions. Since <math>AC + OC = 1</math>, <math>1 - OC = OC \sin \theta </math>(from <math>AC \sec \theta = OC \tan \theta</math>) <math>\Longrightarrow</math> <math>OC = \frac{1}{1+\sin \theta} \Rightarrow \mathrm{(D)}</math>.
+
Since <math>\overline{AB}</math> is tangent to the circle, <math>\triangle OAB</math> is a right triangle. This means that <math>OA = 1</math>, <math>BA = \tan \theta</math> and <math>OB = \sec \theta</math>. By the [[Angle Bisector Theorem]], <cmath> \frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta </cmath> We multiply both sides by <math>\cos \theta</math> to simplify the trigonometric functions, <cmath> AC=OC \sin \theta </cmath> Since <math>AC + OC = 1</math>, <math>1 - OC = OC \sin \theta \Longrightarrow</math> <math>OC = \dfrac{1}{1+\sin \theta}</math>. Therefore, the answer is <math>\boxed{\textbf{(D)} \dfrac{1}{1+\sin \theta}}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 09:11, 24 March 2013

Problem

A circle centered at $O$ has radius $1$ and contains the point $A$. The segment $AB$ is tangent to the circle at $A$ and $\angle AOB = \theta$. If point $C$ lies on $\overline{OA}$ and $\overline{BC}$ bisects $\angle ABO$, then $OC =$

[asy] import olympiad; unitsize(1cm); defaultpen(fontsize(8pt)+linewidth(.8pt)); labelmargin=0.2; dotfactor=3; pair O=(0,0); pair A=(1,0); pair B=(1,1.5); pair D=bisectorpoint(A,B,O); pair C=extension(B,D,O,A); draw(Circle(O,1)); draw(O--A--B--cycle); draw(B--C); label("$O$",O,SW); dot(O); label("$\theta$",(0.1,0.05),ENE); dot(C); label("$C$",C,S); dot(A); label("$A$",A,E); dot(B); label("$B$",B,E);[/asy]

$\text {(A)}\ \sec^2 \theta - \tan \theta \qquad \text {(B)}\ \frac 12 \qquad \text {(C)}\ \frac{\cos^2 \theta}{1 + \sin \theta}\qquad \text {(D)}\ \frac{1}{1+\sin\theta} \qquad \text {(E)}\ \frac{\sin \theta}{\cos^2 \theta}$

Solution

Since $\overline{AB}$ is tangent to the circle, $\triangle OAB$ is a right triangle. This means that $OA = 1$, $BA = \tan \theta$ and $OB = \sec \theta$. By the Angle Bisector Theorem, \[\frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta\] We multiply both sides by $\cos \theta$ to simplify the trigonometric functions, \[AC=OC \sin \theta\] Since $AC + OC = 1$, $1 - OC = OC \sin \theta \Longrightarrow$ $OC = \dfrac{1}{1+\sin \theta}$. Therefore, the answer is $\boxed{\textbf{(D)} \dfrac{1}{1+\sin \theta}}$.

See also

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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