2025 AMC 10B Problem 5

The circle through $A$, $B$, $C$ is the circumcircle, so $OA = OB = OC$. Draw the radii $OA$ and $OC$.

The inscribed angle $\angle ABC = 130^\circ$ stands on the arc $AC$ that does not contain $B$, so that arc measures $2 \cdot 130^\circ = 260^\circ$. The remaining arc $AC$, the one containing $B$, therefore measures
$$360^\circ - 260^\circ = 100^\circ,$$
and this is the ordinary (non-reflex) central angle $\angle AOC = 100^\circ$.

Triangle $OAC$ is isosceles with $OA = OC$, so its base angles are equal:
$$\angle CAO = \frac{180^\circ - 100^\circ}{2} = 40^\circ.$$

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

The inscribed angle theorem gives the central angle, and the circumcentre's defining property turns the rest into an isosceles triangle. The obtuse angle at $B$ is the only subtlety: it forces $O$ onto the opposite side of $AC$ from $B$, so the arc that $\angle B$ subtends is the major one and the central angle is $360^\circ - 2\angle B$, not $2\angle B$. Running the same argument for an acute $B$ gives $\angle CAO = 90^\circ - \angle B$, so in general $\angle CAO = |90^\circ - \angle B|$, which is worth remembering. The side lengths $AB = 10$ and $AC = 18$ are decoration: they fix the size of the circle but not any angle.

Use the extended law of sines: $AC / \sin B = 2R$, so $R = 9/\sin 130^\circ$. Let $M$ be the midpoint of $AC$; then $OM \perp AC$ and in right triangle $OMA$,
$$\sin \angle AOM = \frac{AM}{OA} = \frac{9}{R} = \sin 130^\circ = \sin 50^\circ,$$
so $\angle AOM = 50^\circ$ and $\angle OAM = 40^\circ$. This route uses the given $AC = 18$ and shows explicitly that it cancels.

• Answering the supplement 180 - 130 = 50, which is choice (D), or using 2(130) = 260 as the apex angle of triangle OAC.
• Assuming $O$ lies inside the triangle; with $\angle B$ obtuse it lies outside, on the far side of $AC$.
• Trying to use $AB = 10$ and $AC = 18$ in the law of cosines to find the other angles first, which costs several minutes and is not needed.