2002 AMC 10A Problem 19

Picture Spot walking all the way around the doghouse with the rope tight. The region he covers is made of pieces of circles (sectors), and we add their areas.

1. Angle of the hexagon. The angles of a hexagon add to $720^\circ$, so each angle of a regular hexagon is $720^\circ \div 6 = 120^\circ$.

2. The big piece. At the corner where the rope is tied, the doghouse blocks $120^\circ$. Spot can swing the full $2$-yard rope through the other $360^\circ - 120^\circ = 240^\circ$. That is $\frac{240}{360} = \frac{2}{3}$ of a circle of radius $2$:
$$\frac{2}{3} \times \pi \times 2^2 = \frac{8\pi}{3}.$$

3. Bending around a corner. Now Spot keeps going, so the rope lies along a wall. After $1$ yard of wall the rope reaches the next corner with $2 - 1 = 1$ yard left. It can bend around that corner and swing until it lies along the next wall. The angle it swings through is $180^\circ - 120^\circ = 60^\circ$. That is $\frac{60}{360} = \frac{1}{6}$ of a circle of radius $1$:
$$\frac{1}{6} \times \pi \times 1^2 = \frac{\pi}{6}.$$
At that point the last yard lies exactly along the next wall, so there is nothing more to bend.

1. Both sides. The same thing happens going the other way around the doghouse, so there are two small pieces: $2 \times \frac{\pi}{6} = \frac{\pi}{3}$.

5. Total.
$$\frac{8\pi}{3} + \frac{\pi}{3} = \frac{9\pi}{3} = 3\pi.$$

The answer is $\boxed{\textbf{(E)}\ 3\pi}$.

A tied-up animal problem breaks into circle pieces: one big piece while the rope is free, and a new, smaller piece each time the rope bends around a corner using whatever length is left. The angle of each small piece is the turn at that corner, $180^\circ$ minus the polygon's angle. Draw the rope bending and every piece becomes easy to see.

At the tether vertex the hexagon occupies an interior angle of $120^\circ$, so Spot can sweep the remaining $360^\circ - 120^\circ = 240^\circ$ with the full rope of length $2$:
$$\frac{240}{360} \cdot \pi \cdot 2^2 = \frac{8\pi}{3}.$$

When the rope is pulled along one side of the hexagon, it reaches the neighboring vertex with $2 - 1 = 1$ yard to spare. Bending around that vertex it sweeps the exterior angle of the hexagon, $60^\circ$, with radius $1$. This happens on both sides:
$$2 \cdot \frac{60}{360} \cdot \pi \cdot 1^2 = \frac{\pi}{3}.$$

Total area: $\frac{8\pi}{3} + \frac{\pi}{3} = 3\pi$. The answer is $\boxed{\textbf{(E)}\ 3\pi}$.