AMC 10 · 2006 · #12

Grade 7 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid
area-circlescircular-sectorspatial-visualization area-difference ↑ Prerequisites: area-circles
📏 Medium solution 💡 2 insights 📊 Diagram
Problem
A dog is tied to the outside of a square shed that is 16 ft on each side using an 8-ft rope. In Arrangement I the rope is tied to the midpoint of one side; in Arrangement II it is tied to a point 4 ft from a corner of that same side. The rope cannot pass through the walls. Find which arrangement gives the dog the greater area to roam, and by how many square feet.

Pick an answer.

(A)
$I,\,\textrm{ by }\,8\pi$
(B)
$I,\,\textrm{ by }\,6\pi$
(C)
$II,\,\textrm{ by }\,4\pi$
(D)
$II,\,\textrm{ by }\,8\pi$
(E)
$II,\,\textrm{ by }\,10\pi$

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

The whole problem is about what region a rope can sweep against a wall, so Tool #1 (Draw a Diagram) is the natural lead: sketch the anchor, the wall, and the arc the taut rope traces. The key move — noticing that a rope tied near a corner bends around it — is spatial (Tool #17), and once the rope wraps, the roaming area splits into a main arc plus a smaller leftover arc, which is a clean Tool #7 (Identify Subproblems) decomposition. Each piece is a fraction of a circle, so comparing the two arrangements reduces to adding and comparing multiples of π.

1STEP 1

Arrangement I: a clean semicircle

The 8-ft rope at the wall's midpoint just reaches each corner, no slack to bend around it — a half-circle of radius 8, 32π.

A_I = 1/2π (8)² = 1/2π · 64 = 32π ft²
2STEP 2

Arrangement II: the same semicircle first

Anchoring 4 ft from a corner does not change what the wall allows: straight out, the dog still sweeps a radius-8 half-circle, 32π.

1/2π (8)² = 32π ft²
3STEP 3

The rope bends around the near corner

The near corner is 4 ft away but the rope is 8, so the leftover 4 ft bends around it and sweeps a quarter-circle of radius 4: 4π.

1/4π (4)² = 1/4π · 16 = 4π ft²
4STEP 4

Add up Arrangement II and compare

Arrangement II is 32π + 4π = 36π versus I's 32π; the shared semicircle cancels, leaving the wrap-around quarter as the gap: 4π.

A_II = 32π + 4π = 36π; A_II - A_I = 36π - 32π = 4π → (C)
Answer
II, by 4π
Sanity-check the boundary logic. In Arrangement I the rope length (8) equals the distance to each corner (8), so the rope arrives at the corner with zero slack — no wrap, pure 32π; that borderline is exactly why Arrangement I gets no bonus. In Arrangement II the anchor sits 4 ft from the near corner, strictly less than 8, guaranteeing a wrap, and the leftover 4 ft is smaller than the original 8, so the bonus arc has the smaller radius 4 — giving a quarter-circle of only 4π, not a full-size chunk. The far corner (12 ft) is out of reach, so there is no second bonus. Both arrangements share the identical 32π front semicircle, so the whole contest is decided by that single 4π wrap-around — Arrangement II wins by 4π, matching (C). The trap answers (D) 8π and (E) 10π overcount by imagining a bigger or second bonus arc.
💡Key takeaway

When a rope is tied closer to a corner than its own length, it bends around the corner and the leftover rope sweeps an extra arc — here that one extra quarter-circle is the whole 4π difference.

  • Arrangement I: a clean semicircle
  • Arrangement II: the same semicircle first
  • The rope bends around the near corner
  • Add up Arrangement II and compare

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