AMC10 2008 10A, Problem 19

Rectangle $PQRS$ lies in a plane with $PQ=RS=2$ and $QR=SP=6$. The rectangle is rotated $90^\circ$ clockwise about $R$, then rotated $90^\circ$ clockwise about the point $S$ moved to after the first rotation. What is the length of the path traveled by point $P$?

Solution: [asy] size(220);pathpen=black+linewidth(0.65);pointpen=black; /* draw in rectangles */ D(MP("R",(0,0))--MP("Q",(-6,0))--MP("P",(-6,2),N)--MP("S",(0,2),NW)--cycle); D((0,0)--MP("Q'",(0,6),SW)--MP("P'",(2,6),SE)--MP("S'",(2,0))--cycle); D(MP("R''",(2,2),NE)--MP("Q''",(8,2),N)--MP("P''",(8,0))--(2,0)--cycle); D(arc((0,0),(2,6),(-6,2)),dashed);D(arc((2,0),(8,0),(2,6)),dashed);D((2,6)--(0,0)--(-6,2),dashed); D(rightanglemark((2,6),(0,0),(-6,2),12));D(rightanglemark((2,6),(2,0),(8,0),12)); MP("2",(-6,1),W);MP("6",(-3,0),S); [/asy] We let $P'Q'R'S'$ be the rectangle after the first rotation, and $P''Q''R''S''$ be the rectangle after the second rotation. Point $P$ pivots about $R$ in an arc of a circle of radius $\sqrt{2^2+6^2} = 2\sqrt{10}$, and since $\angle PRS,\, \angle P'RQ'$ are complementary, it follows that the arc has a degree measure of $90^{\circ}$ and length $\frac14$ of the circumference. Thus, $P$ travels $\frac 14 \left(4\sqrt{10}\right)\pi = \sqrt{10}\pi$ in the first rotation. Similarly, in the second rotation, $P$ travels in a $90^{\circ}$ arc about $S'$, with the radius being $6$. It travels $\frac 14(12)\pi = 3\pi$. Therefore, the total distance it travels is $\left(3+\sqrt{10}\right)\pi\ $. $\fbox{\left(3+\sqrt{10}\right)\pi}$.

Answer: \left(3+\sqrt{10}\right)\pi