AMC 10 · 2011 · #18

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid

area-circlestangent-circlescoordinate-geometry identify-subproblemscomplementary-counting ↑ Prerequisites: area-circles

📏 Long solution 💡 3 insights

Problem

Three unit circles A, B, and C are placed so that A and B just touch each other, and C touches the exact midpoint of the segment joining the centers of A and B. Find the area of the part of circle C that lies outside both circle A and circle B.

Pick an answer.

(A)

$3 - \frac{\pi}{2}$

(B)

$\frac{\pi}{2}$

(C)

(D)

$\frac{3\pi}{4}$

(E)

$1+\frac{\pi}{2}$

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The wanted region is an awkward curved shape, but it is really the whole disk C with two overlapping bites removed, so Tool #7 (Identify Subproblems) is the anchor: compute Area(C), then the overlap C∩ A, then double it by symmetry. Tool #1 (Draw a Diagram) makes this exact — placing the circles on a coordinate grid fixes every center and every crossing point. Tool #17 (Visualize Spatial Relationships) is what reveals that each overlap lens is built from two congruent quarter-circle-minus-triangle segments, which is the move that makes the messy π terms cancel.

1STEP 1

Place the circles on a grid

Put the touch point of A and B at the origin: A=(-1,0), B=(1,0), M=(0,0), and circle C's center sits one unit up at (0,1).

A=(-1,0), B=(1,0), M=(0,0), C=(0,1); AC=√(1²+1²)=√(2)

2STEP 2

Break the target into a subtraction

Take the whole disk C, area π, then subtract its overlap with A and its overlap with B; by symmetry those two bites are equal.

Area=π-2·Area(C∩ A)

3STEP 3

Find where C and A cross

Besides M, circles C and A also meet at P=(-1,1), so A, M, C, P are the four corners of a unit square.

C∩ A={(0,0), (-1,1)}

4STEP 4

The arc spans a right angle

From A, the directions to M and to P are perpendicular, so arc MP cuts off a quarter of circle A, of area π/4.

∠ MAP=90° → sector=90/360π(1)²=π/4

5STEP 5

Peel the triangle to get the lens

Removing right triangle AMP, area 1/2, leaves a segment of area π/4-1/2; two congruent ones make the lens π/2-1.

△ AMP=1/2(1)(1)=1/2; Area(C∩ A)=2(π/4-1/2)=π/2-1

6STEP 6

Subtract both overlaps

Subtract both lenses from the full disk: π-2(π/2-1)=π-π+2, so the region has area 2, choice (C).

π-2(π/2-1)=π-π+2=2 → (C)

Answer

Circle C has area π≈3.14. Each overlap is π/2-1≈0.57, and two of them total about 1.14, leaving roughly 3.14-1.14=2.0 — matching the exact answer 2. It is also reassuring that every π term cancels: the leftover region can be reshaped into a straight-sided figure of whole-number area, so a π-free answer like 2 is exactly what to expect. The π-laden options (B) π/2, (D) 3π/4, and (E) 1+π/2 are traps for stopping before the cancellation.

💡Key takeaway

To find a leftover area, take the whole shape and subtract each overlapping bite — and a quarter-circle minus its triangle turns the messy π pieces into a clean number.

Place the circles on a grid
Break the target into a subtraction
Find where C and A cross
The arc spans a right angle
Peel the triangle to get the lens
Subtract both overlaps