AMC 10 · 2025 · #24

Grade 11 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
tangent-circleslaw-of-cosinesnested-radical-denestingequal-spacing identify-subproblems ↑ Prerequisites: law-of-cosinestangent-circles
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
A central circle of radius r is ringed by 12 unit circles. Each small circle touches the central circle on the outside and also touches its two neighbors. Given that r can be written as (√(a) + √(b) + c) for integers a, b, c, find a + b + c.

Pick an answer.

(A)
3
(B)
5
(C)
7
(D)
9
(E)
11
How to solve
Strategy Draw a Diagram

The figure has twelve-fold symmetry, so drawing the centers and exploiting that symmetry collapses the whole ring to a single triangle: the center O and two neighboring small-circle centers. Keeping r as the unknown, that triangle has two known sides, a known included angle, and a known opposite side, which is exactly what the Law of Cosines needs. Solving the resulting equation with radical algebra produces r.

1STEP 1

Isolate one center triangle

Mark O and two adjacent small centers A₁, A₂: tangency gives OA₁ = OA₂ = r+1 and A₁A₂ = 2, and equal spacing gives ∠A₁OA₂ = 30°.

OA₁ = OA₂ = r+1, A₁A₂ = 2, ∠ A₁OA₂ = 360°/12 = 30°
2STEP 2

Apply the Law of Cosines

In triangle OA₁A₂ two sides r+1 enclose a 30° angle opposite side 2, so the Law of Cosines gives 4 = 2(r+1)²(1 - 32\frac{\sqrt{3}}{2}).

2² = 2(r+1)² - 2(r+1)²cos 30° = 2(r+1)²(1 - √(3)/2)
3STEP 3

Solve for (r+1) squared

Divide by 2(1 - 32\frac{\sqrt{3}}{2}): (r+1)² = 42−3\frac{4}{2 - \sqrt{3}}; rationalizing with 2 + √(3) gives (r+1)² = 8 + 4√(3).

(r+1)² = 4/(2-√(3)) = (4(2+√(3)))/(2-√(3))(2+√(3)) = 8 + 4√(3)
4STEP 4

Take the square root by denesting

Denest √(8 + 4√(3)) = √(x) + √(y) via x + y = 8, xy = 12: x = 6, y = 2, so r + 1 = √(6) + √(2) and r = √(6) + √(2) - 1.

√(8+4√(3)) = √(6) + √(2) → r = √(6) + √(2) - 1
5STEP 5

Read off a, b, c

Match r = √(6) + √(2) - 1 to √(a) + √(b) + c: a = 6, b = 2, c = -1, so a + b + c = 7, which is (C).

a = 6, b = 2, c = -1 → a + b + c = 7
Answer
7
Numerically ((r+1)² = 8 + 4√(3) ≈ 14.93), so (r + 1 ≈ 3.86) and (r ≈ 2.86). That matches (√(6) + √(2) - 1 ≈ 2.449 + 1.414 - 1 = 2.863), and it sits right next to the drawing's (r ≈ 2.9). A radius near 2.9 also makes sense: 12 unit circles need a fairly large host circle to ring it without overlapping.
💡Key takeaway

When circles touch, their centers sit a fixed distance apart, so a ring of circles becomes one triangle you can crack with the Law of Cosines.

  • Isolate one center triangle
  • Apply the Law of Cosines
  • Solve for (r+1) squared
  • Take the square root by denesting
  • Read off a, b, c

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