2008 AMC 10B Problem 10

$C$ sits exactly in the middle of the arc, so the picture is symmetric. The line of symmetry goes through the center and the middle of the chord. Drawing it creates right triangles.

1. Call the center $O$ and the midpoint of chord $AB$ point $M$. Then $AM = MB = 3$.
2. By symmetry, the line from $O$ through $M$ is perpendicular to $AB$, and when you keep going past $M$ it hits the circle exactly at $C$, the middle of the arc.
3. Triangle $OMA$ has a right angle at $M$, with $OA = 5$ (a radius) and $AM = 3$. This is a $3$-$4$-$5$ triangle, so $OM = 4$.
4. $OC$ is also a radius, so $OC = 5$. The piece past the chord is
$$MC = OC - OM = 5 - 4 = 1.$$
5. Triangle $AMC$ has a right angle at $M$, with legs $AM = 3$ and $MC = 1$. By the Pythagorean theorem,
$$AC = \sqrt{3^2 + 1^2} = \sqrt{10}.$$

The answer is $\boxed{\textbf{(A)}\ \sqrt{10}}$.

A line from the center of a circle to the middle of a chord always meets the chord at a right angle, and it continues to the middle of the arc. That one line gives two right triangles that share a side. Whenever a problem has a chord and a circle's radius, draw this line first; the distance from the center to the chord is usually the key number.

Let $O$ be the center and $M$ the midpoint of chord $AB$. By symmetry, the midpoint $C$ of the minor arc lies on line $OM$, on the far side of $M$ from $O$, and $OM \perp AB$.

In right triangle $OMA$, $OA = 5$ and $AM = 3$, so $OM = \sqrt{25 - 9} = 4$. Since $OC$ is a radius, $OC = 5$ and
$$MC = OC - OM = 5 - 4 = 1.$$
Now right triangle $AMC$ has legs $AM = 3$ and $MC = 1$:
$$AC = \sqrt{3^2 + 1^2} = \sqrt{10}.$$

The answer is $\boxed{\textbf{(A)}\ \sqrt{10}}$.