2008 AMC 10A Problem 19

Attempt Problem 19 of the 2008 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 10A solutions, or check the answer key.

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19.

Rectangle PQRSPQRS lies in a plane with PQ=RS=2PQ = RS = 2 and QR=SP=6.QR = SP = 6. The rectangle is rotated 9090^\circ clockwise about R,R, then rotated 9090^\circ clockwise about the point that SS moved to after the first rotation. What is the length of the path traveled by point P?P?

(23+5)π\left(2\sqrt{3} + \sqrt{5}\right)\pi

6π6\pi

(3+10)π\left(3 + \sqrt{10}\right)\pi

(3+25)π\left(\sqrt{3} + 2\sqrt{5}\right)\pi

210π2\sqrt{10}\pi

Answer: C
Concepts:transformationarcPythagorean Theorem
Difficulty rating: 1840
Solution:

In the first rotation, PP moves on a quarter circle about RR with radius PR=22+62=210.PR = \sqrt{2^2 + 6^2} = 2\sqrt{10}. The arc length is 14(2π210)=10π.\dfrac{1}{4}\left(2\pi \cdot 2\sqrt{10}\right) = \sqrt{10}\,\pi.

In the second rotation, PP moves on a quarter circle about the new position of SS with radius 6.6. The arc length is 14(2π6)=3π.\dfrac{1}{4}(2\pi \cdot 6) = 3\pi.

The total path length is (3+10)π.\left(3 + \sqrt{10}\right)\pi.

Thus, the correct answer is C.

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