AMC 10 · 2004 · #16

Grade 8 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
equilateral-triangletangent-circlesthirty-sixty-ninety-triangle physical-representationidentify-subproblems ↑ Prerequisites: pythagorean-theorem
📏 Medium solution 💡 3 insights
Problem
Three circles, each of radius 1, are placed so that every pair touches on the outside (externally tangent), and all three sit inside one larger circle, each touching it from within (internally tangent). Find the radius of the larger circle.

Pick an answer.

(A)
$\frac{2 + \sqrt{6}}{3}$
(B)
2
(C)
$\frac{2 + 3\sqrt{2}}{2}$
(D)
$\frac{3 + 2\sqrt{3}}{3}$
(E)
$\frac{3 + \sqrt{3}}{2}$

AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

The whole problem is about how circles touch, so the first move is Tool #1 (Draw a Diagram) and mark every center. Once the picture is drawn, tangency turns into clean distance facts: two touching unit circles have centers exactly 2 apart, so the three small centers form an equilateral triangle of side 2. Tool #4 (Introduce a Variable) lets us drop coordinates onto that triangle and name the big center's height, and Tool #7 (Identify Subproblems) splits the job into two bite-sized pieces: first find the distance from the big center to a small center, then add one small radius to reach the big circle's edge.

1STEP 1

Turn tangency into a triangle

Two touching unit circles have centers 1+1=21+1=2 apart, so the three small centers form an equilateral triangle of side 2.

center distance=1+1=2 → equilateral triangle, side 2
2STEP 2

Place coordinates on the triangle

Set B=(−1,0)B=(-1,0) and C=(1,0)C=(1,0); the altitude 22−12=3\sqrt{2^2-1^2}=\sqrt{3} gives A=(0,3)A=(0,\sqrt{3}), and symmetry puts the big center at O=(0,y)O=(0,y).

B=(-1,0), C=(1,0), h=√(2²-1²)=√(3), A=(0,√(3)), O=(0,y)
3STEP 3

Find the big center by equal distances

Equal distances to AA and BB give (3−y)2=1+y2(\sqrt{3}-y)^2=1+y^2, so y=13y=\frac{1}{\sqrt{3}} and the common reach is d=233d=\frac{2\sqrt{3}}{3}.

(√(3)-y)²=1+y²→ y=1/√3, d=√(1+1/3)=2/√3=2√3/3
4STEP 4

Step out one radius to the rim

Internal tangency adds one small radius, so R=d+1R=d+1 gives 3+233\frac{3+2\sqrt{3}}{3} — choice (D).

R=d+1=2√3/3+3/3=(3+2√3)/3 (D)
Answer
(3 + 2√(3))/3
As a decimal, (3+2√(3))/3≈(3+3.46)/3≈ 2.15. That is comfortably bigger than 1 (it must contain three radius-1 circles) yet not huge, which matches the picture. Checking the other choices numerically, (A) ≈1.48 is too small to hold the arrangement, (B) =2 is a touch too small, (C) ≈3.12 and (E) ≈2.37 are too big; only (D) ≈2.15 fits, so the answer is stable.
💡Key takeaway

Touching circles fix the distances between their centers, so the three unit circles make an equilateral triangle; measure from that triangle's center out to a small center and add one radius to reach the big circle's edge.

  • Turn tangency into a triangle
  • Place coordinates on the triangle
  • Find the big center by equal distances
  • Step out one radius to the rim

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