AMC 10 · 2025 · #25

Grade 11 geometry-2d

rotation-isometrylaw-of-cosinesequilateral-triangle identify-subproblemsextreme-principle ↑ Prerequisites: polygon-inequality

📏 Long solution 💡 4 insights

Problem

Three circles share one center and have radii 1, 2, and 3. An equilateral triangle has side length s and puts exactly one of its three vertices on each of the three circles. Find s².

Pick an answer.

(A)

(B)

$\frac{25}{4}$

(C)

$\frac{13}{2}$

(D)

$\frac{27}{4}$

(E)

How to solve

Strategy Visualize Spatial Relationships

The three unknown central angles make a coordinate attack messy, so Tool #17 (Visualize Spatial Relationships) drives the solution: an equilateral triangle is exactly the shape that a 60° turn about a vertex maps onto itself, so spin the whole plane 60° about one vertex and watch where the center O lands. Tool #1 (Draw a Diagram) sets up and labels the figure so the three radii and three equal sides are visible at once. Tool #15 (Organize Information in More Ways) re-reads the radii not as 1, 2, 3 but as the relation 1+2=3, which is the hint that a length-1 piece and a length-2 piece will lie end to end along the length-3 radius. Tool #7 (Identify Subproblems) reduces the whole configuration to one small triangle whose three parts we can name. Tool #14 (Extreme Principle) supplies the finishing lock: 1+2=3 is the equality case of the triangle inequality, and equality forces collinearity, which hands us the angle we need.

1STEP 1

Draw and label the figure

Draw it and label the three radii.

OA=1, OB=2, OC=3, AB=BC=CA=s, ∠ BAC=60°

2STEP 2

Re-read the radii as 1+2=3

Notice one plus two is three.

OA + OB = 1 + 2 = 3 = OC

3STEP 3

Spin the plane 60° about A

Rotate the plane 60 degrees about a vertex.

ρ_A,60°: A↦ A, B↦ C, O↦ O' ⟹ △ OAB ≅ △ O'AC

4STEP 4

Measure the two new lengths

The new lengths are 1 and 2.

AO'=AO=1, ∠ OAO'=60° → △ OAO' equilateral → OO'=1; O'C=OB=2

5STEP 5

Equality forces O' onto OC

Summing to three puts the new point on the segment.

OO'+O'C = 1+2 = 3 = OC ⟹ O' ∈ OC

6STEP 6

Read the 120° angle

There the angle reads 120 degrees.

∠ AO'C = 180° - ∠ AO'O = 180° - 60° = 120°

7STEP 7

Finish with the Law of Cosines

The law of cosines gives 7.

s² = AO'² + O'C² - 2 · AO' · O'Ccos 120° = 1 + 4 - 2(1)(2)(-1/2) = 7

Answer

First a bound check: s=AB ≤ OA+OB=3 forces s² ≤ 9, and s=CA ≥ OC-OA=2 forces s² ≥ 4, so s²=7 lands inside the allowed band [4,9]. Better, the configuration can be written down exactly. Take A=(1,0), B=(-1,-√(3)), C=(3/2,-3√(3)/2). Then OA²=1, OB²=1+3=4, and OC²=9/4+27/4=9, so the three vertices really do sit on the circles of radii 1, 2, 3. The sides come out to AB²=2²+(√(3))²=7, BC²=(5/2)²+(√(3)/2)²=25/4+3/4=7, and CA²=(1/2)²+(3√(3)/2)²=1/4+27/4=7. All three sides are √(7), so the triangle is genuinely equilateral and s²=7 exactly, confirming (E). Note also that ∠ AOB=120° here, matching the derivation: the three near-miss choices 25/4, 13/2, 27/4 are all just under 7 and would each fail this coordinate test.

💡Key takeaway

Because the radii satisfy 1+2=3, a 60° spin about one vertex drops the center's image exactly onto the longest radius, and the whole problem shrinks to one triangle with sides 1 and 2 meeting at 120°, giving s²=7.

Draw and label the figure
Re-read the radii as 1+2=3
Spin the plane 60° about A
Measure the two new lengths
Equality forces O' onto OC
Read the 120° angle
Finish with the Law of Cosines