AMC 10 · 2004 · #24

Grade 10 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid
minkowski-sumcircular-sectorarea-difference physical-representationidentify-subproblems ↑ Prerequisites: area-circles
📏 Long solution 💡 3 insights
Problem
Two points A and B in a plane are 1 unit apart. Look at every disk of radius 1 that contains the whole segment AB, and let S be the set of points covered by at least one of them. Find the area of S.

Pick an answer.

(A)
$2\pi + \sqrt3$
(B)
$\frac {8\pi}{3}$
(C)
$3\pi - \frac {\sqrt3}{2}$
(D)
$\frac {10\pi}{3} - \sqrt3$
(E)
$4\pi - 2\sqrt3$
How to solve
Strategy Draw a Diagram

Chasing the disks themselves is hopeless because there are infinitely many. Change focus to the one thing that varies: the center. A disk of radius 1 covers segment AB exactly when its center is within 1 unit of both A and B, so the legal centers form a single fixed region, the overlap of two unit disks. Then S is just that region pushed outward by 1 in every direction. Coordinates pin down the corners of the overlap region, and the pushed-out shape breaks into a few sectors and rings whose areas are routine.

1STEP 1

Covering the segment means covering both ends

A disk is convex, so covering the segment just means covering both endpoints.

disk of radius 1 centered at P covers AB ⇔ PA ≤ 1 and PB ≤ 1
2STEP 2

The legal centers form a lens

The legal centres therefore fill the lens where two unit disks overlap, with equilateral corners.

(x+1/2)²+y² ≤ 1 and (x-1/2)²+y² ≤ 1 ⟹ C=(0,√3/2), D=(0,-√3/2)
3STEP 3

S is the lens grown outward by 1

The covered set is everything within one unit of that lens: the lens, two rings and two corner fans.

S={X: dist(X,L) ≤ 1} = L ∪ (ring_A) ∪ (ring_B) ∪ (fan_C) ∪ (fan_D)
4STEP 4

Area of the lens

Sector minus triangle, doubled, gives the lens as 2π/3 - √3/2.

[L]=2(π/3-√3/4)=2π/3-√3/2
5STEP 5

Add the rings and the fans

The pieces meet only along edges, so adding gives 3π - √3/2, choice (C).

[S]=(2π/3-√3/2)+2(π)+2(π/6)=2π/3+π/3+2π-√3/2=3π-√3/2
Answer
3π - √3/2
The value 3π - √3/2 ≈ 8.56 sits inside easy bounds. Both A and B are themselves legal centers, so S contains the union of the two unit disks about A and B, whose area is 2π - (2π/3-√3/2) = 4π/3+√3/2 ≈ 5.06. Going the other way, every legal center is within √3/2 of the midpoint of AB, so S fits inside a disk of radius 1+√3/2 about that midpoint, of area π(1+√3/2)² ≈ 10.94. The computed 8.56 falls between 5.06 and 10.94. Note that the choices run from about 8.02 to 9.10, so these bounds confirm the work without deciding the problem; the exact decomposition is what settles it.
💡Key takeaway

When infinitely many moving shapes sweep out a region, stop tracking the shapes and track their centers — the region is just the set of legal centers grown outward by the radius.

  • Covering the segment means covering both ends
  • The legal centers form a lens
  • S is the lens grown outward by 1
  • Area of the lens
  • Add the rings and the fans

More like this

Same archetype — closest grade level first.