AMC 10 · 2004 · #25

Grade 8 geometry-2d

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

circular-sectorarea-differencetangent-circles complementary-countingphysical-representation ↑ Prerequisites: area-circles

📏 Medium solution 💡 4 insights 📊 Diagram

Problem

A small circle of radius 1 is internally tangent to two larger circles of radius 2, touching them at the two ends A and B of one of its diameters. Find the area of the region that lies outside the small circle but inside both large circles at once.

Pick an answer.

(A)

$\frac{5}{3} \pi - 3\sqrt 2$

(B)

$\frac{5}{3} \pi - 2\sqrt 3$

(C)

$\frac{8}{3} \pi - 3\sqrt 3$

(D)

$\frac{8}{3} \pi - 3\sqrt 2$

(E)

$\frac{8}{3} \pi - 2\sqrt 3$

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

How to solve

Strategy Draw a Diagram

Everything turns on placing the four key points, so Tool #1 (Draw a Diagram) leads: set coordinates with the small circle at the origin, A=(0,1), B=(0,-1), large circles centered at A and B. Two facts fall out immediately — the large circles cross at (±√3,0), and the small disk sits entirely inside the overlap. That second fact powers Tool #16 (Change Focus / Count the Complement): instead of chasing the awkward shaded ring directly, compute the whole lens where the two big disks overlap, then simply subtract the small disk's area π at the very end. Computing the lens itself is a clean Tool #7 (Identify Subproblems) job: the chord joining the two crossing points splits the lens into two identical circular segments, and each segment is one sector minus one triangle. The tell-tale 60° angles (from an equilateral triangle of side 2) produce the √3, and remembering to subtract π is exactly what separates choice (B) from the decoys (C)/(D)/(E).

1STEP 1

Set coordinates and find where the big circles cross

Place the small circle at $O=(0,0)$ , $A=(0,1)$ , $B=(0,-1)$ ; the big circles centered at $A$ , $B$ cross at $P=(\sqrt{3},0)$ , $Q=(-\sqrt{3},0)$ .

x²+(y-1)²=4, x²+(y+1)²=4 → y=0, x=±√3; P=(√3,0), Q=(-√3,0)

2STEP 2

Reframe: add the small disk back, subtract it at the end

Every point of the small disk is within $2$ of both $A$ and $B$ , so it sits inside the overlap: shaded area is the lens minus $\pi$ .

shaded = area(overlap lens) - π · 1² = area(lens) - π

3STEP 3

Spot the equilateral triangle to get the angle

$AB=AP=BP=2$ makes $ABP$ equilateral, so each angle is $60^\circ$ and chord $PQ$ opens $120^\circ$ at $A$ and at $B$ .

AB=AP=BP=2 → △ ABP equilateral, ∠ PBA=60°; ∠ PBQ = 60°+60° = 120°

4STEP 4

Cut the lens into two segments and add them

Chord $PQ$ cuts the lens into two congruent segments, each the sector $\frac{4\pi}{3}$ minus triangle $PBQ$ of area $\sqrt{3}$ , so the lens is $\frac{8\pi}{3}-2\sqrt{3}$ .

segment=1/3(4π)-√3=4π/3-√3; area(lens)=2(4π/3-√3)=8π/3-2√3

5STEP 5

Subtract the small circle for the final area

Peel the small disk off the lens: $\frac{8\pi}{3}-2\sqrt{3}-\pi$ = $\frac{5\pi}{3}-2\sqrt{3}$ — choice (B).

(8π/3-2√3)-π=5π/3-2√3 (B)

Answer

5/3 π - 2√3

Numerically the lens is 8π/3-2√3≈ 8.378-3.464=4.914, and subtracting π≈3.142 leaves ≈1.772, matching 5π/3-2√3≈5.236-3.464=1.772. That is positive and comfortably smaller than the lens, as a carved-out sub-region must be — a good sanity check. The √3 is the right radical: it comes from a 30-60-90 / equilateral-triangle setup (angles of 60° and 120°), so a √2 (which would need a 45° angle) or a bare 3√3 (no leftover circle subtracted) has no place here, ruling out (A), (C), (D). Only (B) has both the correct 5π/3 (small circle removed) and the correct 2√3.

💡Key takeaway

Instead of measuring the awkward shaded ring, fill in the small circle to get the tidy lens where both big circles overlap; cut that lens into a pie-slice minus a triangle to find it is 8π/3-2√3, then take the small circle π back out to get 5π/3-2√3.

Set coordinates and find where the big circles cross
Reframe: add the small disk back, subtract it at the end
Spot the equilateral triangle to get the angle
Cut the lens into two segments and add them
Subtract the small circle for the final area

Set coordinates and find where the big circles cross

Reframe: add the small disk back, subtract it at the end

Spot the equilateral triangle to get the angle

Cut the lens into two segments and add them

Subtract the small circle for the final area

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