Competition · AMC preparation · step 4 of 4

AMC 10 · 2012B · #16

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid
area-circlesequilateral-trianglecircular-sectortangent-circles complementary-counting ↑ Prerequisites: area-circles
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
Three circles, each with radius 2, are placed so that every circle just touches the other two. The picture shades all three circles together with the curved patch trapped in the middle between them. Find the total shaded area — the three circles plus the middle region.

Pick an answer.

(A)
$10\pi+4\sqrt{3}$
(B)
$13\pi-\sqrt{3}$
(C)
$12\pi+\sqrt{3}$
(D)
$10\pi+9$
(E)
$13\pi$

AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

The curved shape has no single area formula, so Tool #7 (Identify Subproblems) splits it into two things we can measure: the three whole circles, and the curved patch in the middle. Tool #1 (Draw a Diagram) adds the key hidden line — joining the three centers makes an equilateral triangle that pins down the middle patch. Tool #16 (Change Focus / Count the Complement) handles that patch by measuring the triangle and subtracting the circle slices that stick into it, instead of trying to integrate a curved region directly.

1STEP 1

Split into circles plus middle patch

The shaded figure is two things added: the three full circles, plus the curved patch trapped between them.

Total=(area of 3 circles)+(area of middle patch)
2STEP 2

Add up the three circles

Each circle has area π·2²=4π, and tangent circles meet at one point without overlapping, so the three cover 12π.

3·π(2)²=3·4π=12π
3STEP 3

Frame the middle with a triangle

Join the three centers: each pair sits 2+2=4 apart, an equilateral triangle of side 4, height 2√(3), area 4√(3).

h=√(4²-2²)=2√(3), A_△=1/2·4·2√(3)=4√(3)
4STEP 4

Subtract the circle slices

Each corner holds a 60° sector; the three make 180°, a half circle of area 2π, so the patch is 4√(3)-2π.

patch=4√(3)-1/2π(2)²=4√(3)-2π
5STEP 5

Add the two parts

Add the parts: 12π+(4√(3)-2π); the π terms give 12π-2π=10π, so the total is 10π+4√(3), choice (A).

12π+(4√(3)-2π)=10π+4√(3) → (A)
Answer
10π+4√(3)
The answer must be a bit more than the three circles alone (12π) but the middle patch is small — it is 4√(3)-2π≈6.93-6.28≈0.65, a sliver, so the total should be just above 12π≈37.7. Choice (A) gives 10π+4√(3)≈31.4+6.9≈38.3, which fits. Watch the trap of forgetting to subtract the sectors: that would give 12π+4√(3) (choice C uses +√(3), a similar bait), which overcounts. Removing the half-circle of overlap is what turns 12π into 10π.
💡Key takeaway

Cut the odd shape into three whole circles plus a middle patch, and the patch is just the center triangle minus half a circle.

  • Split into circles plus middle patch
  • Add up the three circles
  • Frame the middle with a triangle
  • Subtract the circle slices
  • Add the two parts

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