2008 AMC 10B Problem 10

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

The perpendicular from the center to a chord bisects it and passes through the arc's midpoint, so one auxiliary line creates two right triangles sharing a leg. The distance from center to chord ($4$) is the key intermediate quantity; the arc midpoint is one radius from the center along that same line.

Let the central angle be $2\theta$ with $\sin\theta = 3/5$. Chord $AC$ subtends half that angle, so $AC = 2 \cdot 5 \sin(\theta/2)$, and the half-angle formula $\sin^2(\theta/2) = (1 - \cos\theta)/2 = (1 - 4/5)/2 = 1/10$ gives $AC = 10 \cdot \sqrt{1/10} = \sqrt{10}$.

• Taking AC to be half the chord (3) or forgetting that the distance from the center to the chord is 4, not 5.
• Placing $C$ on the segment $OM$ (between center and chord) and computing $MC = 4 + 5$ or similar.