# Difference between revisions of "2008 AMC 12B Problems/Problem 9"

## Problem 9

Points $A$ and $B$ are on a circle of radius $5$ and $AB = 6$. Point $C$ is the midpoint of the minor arc $AB$. What is the length of the line segment $AC$?

$\textbf{(A)}\ \sqrt {10} \qquad \textbf{(B)}\ \frac {7}{2} \qquad \textbf{(C)}\ \sqrt {14} \qquad \textbf{(D)}\ \sqrt {15} \qquad \textbf{(E)}\ 4$

## Solution

### Trig Solution:

Let $\alpha$ be the angle that subtends the arc AB. By the law of cosines, $6^2=5^2+5^2-2*5*5cos(\alpha)$

$\alpha = cos^{-1}(7/25)$

The half-angle formula says that $cos(\alpha/2) = \frac{\sqrt{1+cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$ $AC = \sqrt{5^2+5^2-2*5*5*\frac{4}{5}}$

$AC = \sqrt{50-50\frac{4}{5}}$

$AC = \sqrt{10}$, which is answer choice A.

### Other Solution

Define D as the midpoint of AB, and R the center of the circle. R, C, and D are collinear, and since D is the midpoint of AB, $m\angle RDA=90\deg$, and so $RD=\sqrt{5^2-3^2}=4$. Since $RD=4$, $CD=5-4=1$, and so $AC=\sqrt{3^2+1^2}=\sqrt{10} \rightarrow A$