2006 AMC 10A Problem 12

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

Tethered-animal regions decompose into sectors: a large sector where the rope swings freely, then a smaller sector each time the rope bends around a corner, with radius equal to the leftover rope. Only the differences between the two arrangements need computing, and the only difference here is the single corner wrap.

• Giving arrangement I an extra region too, or making the wrap-around piece in II a quarter-disk of radius $8$ (adding $16\pi$) instead of radius $8 - 4 = 4$.
• Thinking the half-disk in II is smaller because the tie point is off-center; the wall is a straight line, so the half-disk is unchanged.