2006 AMC 10A Problem 12

Picture where the dog can walk. Along a wall the rope swings in a half-circle; when the rope reaches a corner with some length left, it bends around the corner and swings a smaller piece.

1. Arrangement I. The rope is tied at the middle of a $16$-foot side, $8$ feet from each corner. The rope is $8$ feet, so the dog sweeps a half-circle of radius $8$ outside that wall:
$$\tfrac12 \times \pi \times 8^2 = 32\pi.$$
The rope just reaches each corner with nothing left over, so there is nothing more.
2. Arrangement II. The rope is tied $4$ feet from one corner (and $12$ feet from the other). The dog still gets a half-circle of radius $8$ along the wall: $32\pi$.
3. Now the near corner is only $4$ feet away, so the rope reaches it with $8 - 4 = 4$ feet left. Bending around the corner, the dog sweeps a quarter-circle of radius $4$ along the next wall:
$$\tfrac14 \times \pi \times 4^2 = 4\pi.$$
The far corner is $12$ feet away, more than the rope, so nothing extra there.
4. Compare. II gives $32\pi + 4\pi = 36\pi$, and I gives $32\pi$. II is bigger by $4\pi$.

The answer is $\boxed{\textbf{(C)}\ \text{II, by } 4\pi}$.

A tied animal's region is made of pieces of circles: a big piece where the rope swings freely, and a smaller piece at each corner the rope can wrap around, whose radius is the rope that is left over. When comparing two setups, only the pieces that differ need to be computed.

In arrangement I the rope is tied to the midpoint of a $16$-foot side, $8$ feet from each corner. The dog can sweep a half-disk of radius $8$ on the outside of that wall:
$$\tfrac12 \pi \cdot 8^2 = 32\pi.$$
The rope exactly reaches each corner with nothing left over, so no further region is added.

In arrangement II the tie point is $4$ feet from one corner and $12$ feet from the other. The same half-disk of radius $8$ is available (the wall blocks the other half exactly as before), giving $32\pi$. But now the rope reaches the near corner with $8 - 4 = 4$ feet to spare, and bending around that corner it sweeps a quarter-disk of radius $4$ along the adjacent wall:
$$\tfrac14 \pi \cdot 4^2 = 4\pi.$$
On the far side, $12 > 8$, so the rope never reaches that corner.

Arrangement II gives $32\pi + 4\pi$, exceeding arrangement I by $4\pi$.

The answer is $\boxed{\textbf{(C)}\ \text{II, by } 4\pi}$.