AMC 10 · 2007 · #3

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
isosceles-triangleangle-sum-triangleangles-around-a-point identify-subproblems ↑ Prerequisites: angle-sum-triangleisosceles-triangle
📏 Medium solution 💡 2 insights 📊 Diagram
Problem
A circle's centre lies inside a triangle whose three vertices are on the circle. Two of the central angles are given. Find the triangle's angle at the shared vertex.

Pick an answer.

(A)
35
(B)
40
(C)
45
(D)
50
(E)
60
How to solve
Strategy Identify Subproblems

The angle you want, ∠ ABC, is not given directly, but the radius OB cuts it into two pieces, and each piece is a base angle of an isosceles triangle whose apex angle IS given. That is Tool #7 (Identify Subproblems): solve triangle AOB, solve triangle BOC, then add. Tool #1 (Draw a Diagram) does the load-bearing work of reading the arrangement — the three radii OA, OB, OC leave O in three different directions and O sits inside the triangle, which is what licenses adding rather than subtracting the two pieces. Tool #3 (Eliminate Possibilities) is the check on that reading: the rival arrangement, with A and C on the same side of OB, produces 10^°, which appears nowhere in the choices.

1STEP 1

Three radii, two isosceles triangles

Three radii make two isosceles triangles.

OA = OB = OC = r
2STEP 2

Base angles of triangle AOB

The first central angle gives a base angle of 20.

∠ ABO = (180^° - 140^°)/2 = 40^°/2 = 20^°
3STEP 3

Base angles of triangle BOC

The second gives a base angle of 30.

∠ OBC = (180^° - 120^°)/2 = 60^°/2 = 30^°
4STEP 4

Add the two pieces at B

The centre is inside, so the pieces add to give 50, choice (D).

∠ ABC = ∠ ABO + ∠ OBC = 20^° + 30^° = 50^° → (D)
Answer
50
First, the configuration really exists — it is not just asserted. Put B, A, C on a circle at directions 0^°, 140^°, and 240^° from the center. Then ∠ AOB = 140^° and ∠ BOC = 360^° - 240^° = 120^° as required, and the third central angle is ∠ AOC = 240^° - 140^° = 100^°, so the three fill the 360^° turn. Second, finish the whole triangle with the same isosceles rule and see whether it closes: triangle AOC has apex 100^°, so its base angles are 40^° each. That gives ∠ BAC = 20^° + 40^° = 60^° and ∠ BCA = 30^° + 40^° = 70^°, and 50^° + 60^° + 70^° = 180^°. All three angles come out acute, and a triangle has its circumcenter inside exactly when it is acute — so the picture is internally consistent rather than assumed. Finally, 50^° is a plausible size for the angle at B: ∠ AOB and ∠ BOC are the two largest central angles, which leaves ∠ AOC the smallest, so B should be the smallest-but-one vertex angle, and it is.
💡Key takeaway

Every radius has the same length, so the center and two vertices always make an isosceles triangle — and if the center sits inside, the radius to a vertex slices that corner into two base angles you can just add.

  • Three radii, two isosceles triangles
  • Base angles of triangle AOB
  • Base angles of triangle BOC
  • Add the two pieces at B

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