2025 AMC 10B Problem 5

Question

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

O

is the circumcenter, so

OA = OB = OC.

Relate the central angle

\angle AOC

to the inscribed angle

\angle B

Because

\angle B

is obtuse,

\angle AOC = 360^\circ - 2\angle B;

then use the isosceles triangle

OAC

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.

Problem 5 in other years