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

m (Solution (with minimal trig))
m (Solution 3 (with minimal trig))
Line 52: Line 52:
 
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:
 
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:
  
<math>sin\theta =\frac{Opposite}{Hypotenuse}</math>
+
<math>\sin\theta =\frac{\text{Opposite}}{\text{Hypotenuse}}</math>
  
<math>cos\theta =\frac{Adjacent}{Hypotenuse}</math>
+
<math>\cos\theta =\frac{\text{Adjacent}}{\text{Hypotenuse}}</math>
  
<math>tan\theta =\frac{Opposite}{Adjacent}</math>.
+
<math>\tan\theta =\frac{\text{Opposite}}{\text{Adjacent}}</math>.
  
 
With a bit of guess and check, we get that the answer is <math>\boxed{D}</math>.
 
With a bit of guess and check, we get that the answer is <math>\boxed{D}</math>.

Revision as of 11:59, 26 December 2019

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 $\boxed{D}$.

See also

2000 AMC 12 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png