2008 AMC 10A Problem 19

1. When a shape turns $90^\circ$ around a point, every other point of the shape moves along a quarter of a circle centered at that point. The radius is the point's distance from the pivot. A quarter circle of radius $d$ has length
$$\frac{1}{4} \times 2\pi d = \frac{\pi d}{2}.$$
2. First turn, around $R$. In rectangle $PQRS$, $P$ and $R$ are opposite corners, so $PR$ is the diagonal. By the Pythagorean theorem,
$$PR = \sqrt{2^2 + 6^2} = \sqrt{40} = 2\sqrt{10}.$$
So $P$ travels $\frac{\pi}{2} \times 2\sqrt{10} = \sqrt{10}\,\pi$.
3. Second turn, around the new position of $S$. Turning does not stretch the rectangle, so $P$ is still $SP = 6$ away from $S$. So $P$ travels $\frac{\pi}{2} \times 6 = 3\pi$.
4. Add the two arcs:
$$\sqrt{10}\,\pi + 3\pi = \left(3 + \sqrt{10}\right)\pi.$$

The answer is $\boxed{\textbf{(C)}\ \left(3+\sqrt{10}\right)\pi}$.

Turning a shape moves each point along a circle around the pivot, and the radius is simply that point's distance from the pivot. Distances inside the rectangle never change, so you only need two lengths: $P$ to $R$ for the first turn and $P$ to $S$ for the second. No coordinates or drawings of the new positions are needed.

Under a rotation of $90^\circ$ about a point $X$, any point $Y$ travels a quarter of a circle of radius $XY$, an arc of length $\frac14 \cdot 2\pi \cdot XY = \frac{\pi}{2} XY$.

First rotation, about $R$. In rectangle $PQRS$ the vertices $P$ and $R$ are opposite, so $PR$ is the diagonal:
$$PR = \sqrt{2^2 + 6^2} = \sqrt{40} = 2\sqrt{10}.$$
$P$ travels $\frac{\pi}{2} \cdot 2\sqrt{10} = \sqrt{10}\,\pi$.

Second rotation, about the new position of $S$. Rotations preserve distances within the rectangle, so $P$ is still $SP = 6$ from $S$. $P$ travels $\frac{\pi}{2} \cdot 6 = 3\pi$.

Total path length: $\sqrt{10}\,\pi + 3\pi = (3 + \sqrt{10})\pi$.

The answer is $\boxed{\textbf{(C)}\ \left(3+\sqrt{10}\right)\pi}$.