AMC 10 · 2025 · #5
Grade 8 geometry-2dIn △ABC, AB=10, AC=18, and ∠B=130∘. Let O be the center of the circle containing points A,B,C. What is the degree measure of ∠CAO?
Pick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: In triangle $ABC$, side $AB = 10$, side $AC = 18$, and $\angle B = 130^\circ$. The point $O$ is the center of the circle that passes through $A$, $B$, and $C$ (the circumcircle). Find the degree measure of $\angle CAO$.
Givens: $AB = 10$; $AC = 18$; $\angle B = 130^\circ$; $O$ is the center of the circle through $A$, $B$, and $C$
Unknowns: The degree measure of $\angle CAO$
Understand
Restated: In triangle $ABC$, side $AB = 10$, side $AC = 18$, and $\angle B = 130^\circ$. The point $O$ is the center of the circle that passes through $A$, $B$, and $C$ (the circumcircle). Find the degree measure of $\angle CAO$.
Givens: $AB = 10$; $AC = 18$; $\angle B = 130^\circ$; $O$ is the center of the circle through $A$, $B$, and $C$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #4 Introduce a Variable
The word 'circumcenter' hides three equal radii $OA = OB = OC$. Drawing the circle and those radii makes the equal lengths visible and reveals two facts to chase: how the $130^\circ$ angle at $B$ controls the angle at the center $O$, and how the isosceles triangle $AOC$ then splits into two equal base angles. So the plan is: draw it, break it into the central-angle subproblem and the isosceles-triangle subproblem, and name the unknown base angle to finish.
Execute — Answer: C
4.G.A.1 Step 1 Draw the circle and its radii
- Draw the circle that passes through $A$, $B$, and $C$, and mark its center $O$.
- Draw the three radii $OA$, $OB$, and $OC$.
- Because $O$ is the center, these three segments have equal length.
- Notice that $\angle B = \angle ABC$ is an inscribed angle: its vertex sits on the circle, and its two sides cut off the arc $AC$ on the far side from $B$.
💡 Putting the circle and its radii on paper turns the hidden equal lengths into lines you can actually see.
7.G.B.5 Step 2 Find the central angle AOC
- An inscribed angle is always half of the central angle that stands on the same arc.
- The inscribed angle $\angle B = 130^\circ$ stands on arc $AC$ (the arc away from $B$), so the central angle covering that same arc is $2 \times 130^\circ = 260^\circ$.
- That $260^\circ$ is the large way around from $A$ to $C$.
- A full turn around the center $O$ is $360^\circ$, so the ordinary angle $\angle AOC$ inside triangle $AOC$ is what is left over.
💡 An arc looks twice as wide from the center as it does from a point on the edge.
8.G.A.5 Step 3 Set up the isosceles triangle
- Look at triangle $AOC$.
- Its sides $OA$ and $OC$ are both radii, so they are equal, which makes the triangle isosceles.
- In an isosceles triangle the two angles across from the equal sides are equal, so $\angle CAO = \angle ACO$.
- Call each of these unknown base angles $x$.
- The three angles of the triangle add to $180^\circ$.
💡 Equal sides always sit opposite equal angles, so the two mystery angles must be the same.
8.G.A.5 Step 4 Solve for the base angle
- Subtract the $100^\circ$ from $180^\circ$ to see that the two equal angles share $80^\circ$ between them.
- Splitting that evenly gives each one $40^\circ$.
- Since $\angle CAO$ is one of these base angles, $\angle CAO = 40^\circ$.
- That is choice (C).
💡 Once the third angle is known, the two equal angles just split what is left of $180^\circ$.
4.G.A.1 Draw the circle that passes through $A$, $B$, and $C$, and mark its center $O$. 7.G.B.5 An inscribed angle is always half of the central angle that stands on the same a 8.G.A.5 Look at triangle $AOC$. Its sides $OA$ and $OC$ are both radii, so they are equa 8.G.A.5 Subtract the $100^\circ$ from $180^\circ$ to see that the two equal angles share Review
Reasonableness: Add the angles of triangle $AOC$ back up: $40^\circ + 40^\circ + 100^\circ = 180^\circ$, which checks out. The result also makes sense of the numbers we were handed: the answer used only $\angle B$, and the side lengths $10$ and $18$ turned out to be a distraction, which is common in this kind of circle problem. $40^\circ$ is answer (C).
Alternative: Extend $AO$ past $O$ until it meets the circle again at $D$, so $AD$ is a diameter. Then $ABCD$ is a cyclic quadrilateral, so opposite angles add to $180^\circ$, giving $\angle ADC = 180^\circ - 130^\circ = 50^\circ$. Because $AD$ is a diameter, the angle $\angle ACD$ inscribed in the semicircle is $90^\circ$. In triangle $ACD$, $\angle CAD = 180^\circ - 90^\circ - 50^\circ = 40^\circ$, and $\angle CAD$ is the same as $\angle CAO$ because $O$ lies on segment $AD$.
CCSS standards used (min grade 8)
4.G.A.1Draw points, lines, line segments, rays, angles, and identify in figures (Drawing the circumcircle, its center, and the three radii, and identifying the inscribed angle at $B$.)7.G.B.5Use facts about supplementary, complementary, vertical, and adjacent angles (Relating the inscribed angle at $B$ to the central angle and using the full $360^\circ$ turn around $O$ to get $\angle AOC = 100^\circ$.)8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Applying the $180^\circ$ triangle angle sum to the isosceles triangle $AOC$ and splitting the remaining $80^\circ$ into two equal base angles.)
⭐ The center of a circle sees an arc twice as wide as the edge does, so a $130^\circ$ angle at $B$ makes a $100^\circ$ angle at $O$, and splitting the leftover in the equal-radius triangle gives $40^\circ$.
⭐ The center of a circle sees an arc twice as wide as the edge does, so a $130^\circ$ angle at $B$ makes a $100^\circ$ angle at $O$, and splitting the leftover in the equal-radius triangle gives $40^\circ$.
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