2008 AMC 10A Problem 19

Below is the professionally curated solution for Problem 19 of the 2008 AMC 10A, 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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:transformationarcPythagorean Theorem

Difficulty rating: 1840

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

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.

Problem 19 in Other Years