2025 AMC 10B Problem 5

Below is the professionally curated solution for Problem 5 of the 2025 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 10B solutions, or check the answer key.

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Concepts:circumcircle, circumcenter, and circumradius inscribed angle isosceles triangle

Difficulty rating: 1310

In $\triangle ABC,$ $AB = 10,$ $AC = 18,$ and $\angle B = 130^\circ.$ Let $O$ be the center of the circle containing points $A,$ $B,$ and $C.$ What is the degree measure of $\angle CAO?$

$20$

$30$

$40$

$50$

$60$

Solution:

Since $O$ is the circumcenter, $OA = OB = OC.$ The inscribed angle $\angle B = 130^\circ$ subtends arc $AC,$ and because $B$ is obtuse, the central angle is $\angle AOC = 360^\circ - 2 \cdot 130^\circ = 100^\circ.$ Triangle $OAC$ is isosceles, so $\angle CAO = \tfrac{180^\circ - 100^\circ}{2} = 40^\circ.$ (The lengths $AB$ and $AC$ never enter.) Thus, C is the correct answer.

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