2008 AMC 10A Problem 19

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

A rotation is a rigid motion: each point moves on a circle centered at the pivot, with radius equal to its distance from the pivot, and that distance is unchanged by earlier rotations because the figure stays congruent. So a sequence of rotations gives a sum of arc lengths, each $\frac{\theta}{360^\circ}\cdot 2\pi \cdot(\text{distance to pivot})$. No coordinates are needed; only the two distances $PR$ and $PS$ matter.

• Using the wrong radius for the second rotation (the diagonal again, or PQ = 2) instead of PS = 6, the distance from P to the new pivot S.
• Computing a half circle ($\pi \cdot$ radius) for a $90^\circ$ turn, doubling the answer.