A circle centered at $O$ has radius 1 and contains the point $A$. 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 express $OC$ in terms of $s$ and $c,$ where $s = \sin \theta$ and $c = \cos \theta.$

[asy]
pair A,B,C,O;
O=(0,0);
A=(1,0);
C=(0.6,0);
B=(1,2);
label("$\theta$",(0.1,0),NE);
label("$O$",O,S);
label("$C$",C,S);
label("$A$",A,E);
label("$B$",B,E);
draw(A--O--B--cycle,linewidth(0.7));
draw(C--B,linewidth(0.7));
draw(Circle(O,1),linewidth(0.7));
[/asy]
Let $\alpha=\angle CBO=\angle ABC$.  By the Law of Sines on triangle $BCO,$
\[\frac{BC}{\sin\theta} = \frac{OC}{\sin\alpha},\]so $OC=\frac{BC\sin\alpha}{\sin\theta}$.

In right triangle $ABC$,
\[\sin\alpha = \frac{AC}{BC} = \frac{1-OC}{BC}.\]Hence, $OC=\frac{1-OC}{\sin\theta}$. Solving this for $OC$ yields $OC= \frac{1}{1+\sin\theta} = \boxed{\frac{1}{1 + s}}.$
\frac{1}{1+s}
Level 3
Precalculus
Problem: A circle centered at $O$ has radius 1 and contains the point $A$. 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 express $OC$ in terms of $s$ and $c,$ where $s = \sin \theta$ and $c = \cos \theta.$

[asy]
pair A,B,C,O;
O=(0,0);
A=(1,0);
C=(0.6,0);
B=(1,2);
label("$\theta$",(0.1,0),NE);
label("$O$",O,S);
label("$C$",C,S);
label("$A$",A,E);
label("$B$",B,E);
draw(A--O--B--cycle,linewidth(0.7));
draw(C--B,linewidth(0.7));
draw(Circle(O,1),linewidth(0.7));
[/asy]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the length of segment \( OC \), we start by analyzing the geometric configuration. The circle has a radius of 1, and point \( A \) lies on the circumference at coordinates \( (1, 0) \). The tangent line \( AB \) at point \( A \) forms an angle \( \theta \) with the line segment \( OA \).

Since \( \overline{BC} \) bisects \( \angle ABO \), we can apply the angle bisector theorem. This theorem states that the ratio of the lengths of the two segments created by the angle bisector is equal to the ratio of the lengths of the other two sides of the triangle. Specifically, we have:

\[
\frac{AC}{OC} = \frac{AB}{OB}
\]

Next, we need to determine the lengths \( AB \) and \( OB \). The length \( OB \) can be calculated using the coordinates of points \( O \) and \( B \). Since \( B \) lies on the tangent line at \( A \), we can find its coordinates based on the angle \( \theta \). The slope of line \( AB \) is given by \( \tan \theta \), leading to the coordinates of \( B \) being \( (1, \tan \theta) \).

Now, we calculate \( AB \):

\[
AB = \sqrt{(1 - 1)^2 + (\tan \theta - 0)^2} = \tan \theta
\]

Next, we find \( OB \):

\[
OB = \sqrt{(1 - 0)^2 + (\tan \theta - 0)^2} = \sqrt{1 + \tan^2 \theta} = \sec \theta
\]

Using the identity \( \tan \theta = \frac{s}{c} \) and \( \sec \theta = \frac{1}{c} \), we can express \( OB \) in terms of \( s \) and \( c \):

\[
OB = \frac{1}{c}
\]

Now substituting back into the angle bisector theorem:

\[
\frac{AC}{OC} = \frac{\tan \theta}{\sec \theta} = \frac{s/c}{1/c} = s
\]

Let \( OC = x \). Then \( AC = 1 - x \) since \( OA = 1 \). Substituting into the ratio gives:

\[
\frac{1 - x}{x} = s
\]

Cross-multiplying leads to:

\[
1 - x = sx
\]

Rearranging this equation results in:

\[
1 = x + sx \implies 1 = x(1 + s)
\]

Solving for \( x \) yields:

\[
x = \frac{1}{1 + s}
\]

Thus, we have expressed \( OC \) in terms of \( s \) and confirmed that the final answer is:

\[
OC = \frac{1}{1+s}
\]<|END_OF_LATENT|> The final answer is $\frac{1}{1+s}$.