2025 AMC 10B Problem 5

1. Draw the circle and the radii. $O$ is the center of the circle through $A$, $B$, $C$, so $OA$, $OB$, $OC$ are all radii and have the same length. Draw $OA$ and $OC$. We want an angle of triangle $OAC$, so we need its angle at $O$.

2. Use the inscribed angle rule. An angle whose vertex is on the circle is half the arc it cuts off. Angle $B$ has its vertex on the circle and cuts off the arc $AC$ that does not pass through $B$. So that arc is
$$2 \times 130^\circ = 260^\circ.$$

3. Find the other arc. The whole circle is $360^\circ$, so the arc $AC$ that does pass through $B$ is
$$360^\circ - 260^\circ = 100^\circ.$$
This smaller arc is the one the ordinary angle $AOC$ at the center cuts off, so $\angle AOC = 100^\circ$.

1. Finish in the isosceles triangle. In triangle $OAC$, the sides $OA$ and $OC$ are equal, so the two base angles are equal. They share $180^\circ - 100^\circ = 80^\circ$, so each is $40^\circ$.

So $\angle CAO = 40^\circ$.

The answer is $\boxed{\textbf{(C)}\ 40}$.

The center angle and the angle on the circle are linked by the "half the arc" rule, and radii always make isosceles triangles. The only thing to watch is that angle $B$ is obtuse, so it cuts off the big arc, more than half the circle. That puts $O$ on the other side of $AC$ from $B$. The side lengths $10$ and $18$ are never needed: they set the size of the circle, not its angles.

Use a point on the far arc. Because angle $B$ is more than $90^\circ$, the center $O$ lies outside the triangle, across $AC$ from $B$. Any point $X$ on the big arc (away from $B$) sees $AC$ at an angle of $180^\circ - 130^\circ = 50^\circ$, since opposite angles of a quadrilateral drawn in a circle add to $180^\circ$. The center angle on the same arc is twice that, $100^\circ$, which gives the same isosceles triangle with base angles $40^\circ$.