2008 AMC 10A Problems/Problem 19

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Problem

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 $P$ ?

$\mathrm{(A)}\ \left(2\sqrt{3}+\sqrt{5}\right)\pi\qquad\mathrm{(B)}\ 6\pi\qquad\mathrm{(C)}\ \left(3+\sqrt{10}\right)\pi\qquad...$

Solution

size(220);pathpen=black+linewidth(0.65);pointpen=black;/* draw in rectangles */D(MP("R",(0,0))--MP("Q",(-...

We let $P'Q'R'S'$ be the first rectangle after the rotation, and $P''Q''R''S''$ be the second rectangle after 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 $90^{\circ}$ (or $1/4$ 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 $\left(3+\sqrt{10}\right)\pi\ \mathrm{(C)}$ .

2008 AMC 10A (Problems • Resources)
Preceded by Problem 18	Followed by Problem 20
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2008 AMC 10A Problems/Problem 19

From AoPSWiki

Problem

Solution

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