2000 AMC 12 Problems/Problem 17

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 $OC =$

$[asy] import olympiad; size(6cm); unitsize(1cm); defaultpen(fontsize(8pt)+linewidth(.8pt)); labelmargin=0.2; dotfactor=3; pair O=(0,0); pair A=(1,0); pair B=(1,1.5); pair D=bisectorpoint(A,B,O); pair C=extension(B,D,O,A); draw(Circle(O,1)); draw(O--A--B--cycle); draw(B--C); label("$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 1

Since $\overline{AB}$ is tangent to the circle, $\triangle OAB$ is a right triangle. This means that $OA = 1$ , $AB = \tan \theta$ 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 $\boxed{\textbf{(D)} \dfrac{1}{1+\sin \theta}}$ .

Solution 2

Alternatively, one could notice that OC approaches the value 1/2 as theta gets close to 90 degrees. The only choice that is consistent with this is (D).

Solution 3 (with minimal trig)

Let's assign a value to $\theta$ so we don't have to use trig functions to solve. $60$ is a good value for $\theta$ , because then we have a $30-60-90 \triangle$ -- $\angle BAC=90$ because $AB$ is tangent to Circle $O$ .

Using our special right triangle, since $AO=1$ , $OB=2$ , and $AB=\sqrt{3}$ .

Let $OC=x$ . Then $CA=1-x$ . since $BC$ bisects $\angle ABO$ , we can use the angle bisector theorem:

$\frac{2}{x}=\frac{\sqrt{3}}{1-x}$

$2-2x=\sqrt{3}x$

$2=(\sqrt{3}+2)x$

$x=\frac{2}{\sqrt{3}+2}$ .

Now, we only have to use a bit of trig to guess and check: the only trig facts we need to know to finish the problem is:

$\sin\theta =\frac{\text{Opposite}}{\text{Hypotenuse}}$

$\cos\theta =\frac{\text{Adjacent}}{\text{Hypotenuse}}$

$\tan\theta =\frac{\text{Opposite}}{\text{Adjacent}}$ .

With a bit of guess and check, we get that the answer is $\boxed{D}$ .

Solution 4

Let $OC$ = x, $OB$ = h, and $AB$ = y. $AC$ = $OA$ - $OC$ . Because $OC$ = x, and $OA$ = 1 (given in the problem), $AC$ = 1-x. Using the Angle Bisector Theorem, $\frac{h}{y}$ = $\frac{x}{1-x}$ \Longrightarrow h(1-x) = xy. Solving for x gives us x = $\frac{h}{h+y}$ . $\sin\theta = \frac{opposite}{hypotenuse} or \frac{y}{h}$ . Solving for y gives us y = h $\sin\theta$ . Substituting this for y in our initial equation yields x = $\dfrac{h}{h+\sin \theta}$ . Using the distributive property, x = $\dfrac{h}{h(1+\sin \theta)}$ and finally $\dfrac{1}{1+\sin \theta}$ , or $\boxed{\textbf{(D)}}$

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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