AMC 10 · 2013 · #23

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid
rotation-isometrycircular-sectorarea-trianglescoordinate-geometry spatial-visualizationidentify-subproblems ↑ Prerequisites: rotation-isometrycircular-sector
📏 Long solution 💡 4 insights
Problem
A square is pinned off-centre and turned a quarter turn. Find the area it sweeps over.

Pick an answer.

(A)
15
(B)
17
(C)
19
(D)
21
(E)
23
How to solve
Strategy Visualize Spatial Relationships

Tool #17 (Visualize Spatial Relationships) is primary because the object being measured is defined by a motion: it exists only as the trail of a turning square, and nothing can be computed until that trail is pinned down. But eyeballing the trail is exactly where this problem punishes people — a picture can suggest which bulges stick out without proving that nothing else does, and an area built from a guessed picture can silently miss a piece or count one twice. So Tool #15 (Organize Information in More Ways) supplies the move that makes the picture provable: stop describing the square by its four corners and describe it instead by how far it reaches from P in each direction. Under that description a rotation just slides the reach data around, and the swept region's reach in a direction is a running maximum over the last quarter turn — an exact formula, not an impression. Tool #14 (Extreme Principle) then does the real work of deciding, direction by direction, which corner or which edge is the farthest thing out. Tool #7 (Identify Subproblems) cuts the resulting boundary into circular arcs and straight edges so the area splits into sectors plus triangles, and Tool #1 (Draw a Diagram) keeps coordinates on everything so each claim can be checked by arithmetic.

1STEP 1

Put the square on axes

Coordinates make every distance explicit.

P=(1,1), AP=√(2), PC=√(6), AC=√(6)+√(2)
2STEP 2

Measure everything from the pin

The four corners sit at known angles.

PA=√(1²+1²)=√(2) at 225^°, PB=√((√(3))²+1²)=2 at 330^°, PC=√((√(3))²+(√(3))²)=√(6) at 45^°, PD=√(1²+(√(3))²)=2 at 120^°
3STEP 3

Where the square finishes, and a free symmetry

The turn also hands over a free symmetry.

(x,y)↦(2-y, x); start [0,√(3)+1]² ⟼ finish [1-√(3),2]×[0,√(3)+1]
4STEP 4

Describe the square by its reach

The square is its reach in every direction.

square = { P + r u(φ) : 0 ≤ r ≤ ρ(φ) }, ρ(45^°)=√(6), ρ(120^°)=ρ(330^°)=2, ρ(225^°)=√(2)
5STEP 5

Swept reach is a running maximum

Sweeping takes a running maximum.

M(φ)=max₀^° ≤ θ ≤ 90^°ρ(φ-θ), swept region={ r ≤ M(φ) }
6STEP 6

Decide which corner wins, and for how long

Each corner wins for a known stretch.

C: & r=√(6), & 45^°→135^°, & 90^° ; D: & r=2, & 150^°→210^°, & 60^° ; A: & r=√(2), & 225^°→315^°, & 90^° ; B: & r=2, & 330^°→30^°, & 60^°
7STEP 7

Add the four circular sectors

That gives four circular sectors.

90/360π(√(6))²+90/360π(√(2))²+2·60/360π(2)²=3π/2+π/2+2·2π/3=2π+4π/3=10π/3
8STEP 8

Add the four flat wedges

Four flat wedges fill the rest.

2·1/2√(3)(√(3)-1)+2·1/2(1)(√(3)-1)=(3-√(3))+(√(3)-1)=2
9STEP 9

Combine the eight pieces

Combining gives one clean expression.

Area=10π/3+2=1/3(10π+6)
10STEP 10

Pin down a, b and c

Matching the form gives 19, choice (C).

gcd(10,6,3)=1 → a+b+c=10+6+3=19 → (C)
Answer
19
Numerically the area is 10π/3+2≈ 12.47, and it must land between two easy bounds. From below, the swept region contains both the starting and finishing squares; those overlap in the rectangle [0,2]×[0,√(3)+1], so their union has area 2(√(3)+1)²-2(√(3)+1)=6+2√(3)≈ 9.46. From above, every point of the square is at most PC=√(6) from the pin and rotation preserves that distance, so the whole trail fits inside the disc of radius √(6), area 6π≈ 18.85. Indeed 9.46 ＜ 12.47 ＜ 18.85. Three structural checks agree as well. The eight boundary pieces span 90+15+60+15+90+15+60+15=360 degrees exactly, so the directions are covered once each with none left over. The mirror symmetry in the line x=1 pairs the two 60^° sectors with each other and each 15^° wedge with its twin, while the √(6) and √(2) sectors are centred on 90^° and 270^° and so are self-paired — exactly what Step 3 predicts. Finally the surds cancel: the wedges contribute (3-√(3))+(√(3)-1)=2, a whole number, as the requested form 1/c(aπ+b) with integer a,b,c demands. A stray √(3) surviving would have been proof of a miscounted piece.
💡Key takeaway

Instead of trying to picture the whole spin, ask how far the square reaches in each direction from the pin — the swept shape reaches as far as the square ever did in that direction during the quarter turn.

  • Put the square on axes
  • Measure everything from the pin
  • Where the square finishes, and a free symmetry
  • Describe the square by its reach
  • Swept reach is a running maximum
  • Decide which corner wins, and for how long
  • Add the four circular sectors
  • Add the four flat wedges
  • Combine the eight pieces
  • Pin down a, b and c

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