Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2008 AMC 10A Problems/Problem 19"

m (→‎Solution)
(→‎Problem)
 
(3 intermediate revisions by 2 users not shown)
Line 16: Line 16:
 
</asy></center>
 
</asy></center>
  
−
We let <math>P'Q'R'S'</math> be the rectangle after the first rotation, and <math>P''Q''R''S''</math> be the rectangle after the second rotation. Point <math>P</math> pivots about <math>R</math> in an [[arc]] of a circle of radius <math>\sqrt{2^2+6^2} = 2\sqrt{10}</math>, and since <math>\angle PRS,\, \angle P'RQ'</math> are complementary, it follows that the arc has a degree measure of <math>90^{\circ}</math> (or <math>1/4</math> of the [[circumference]]). Thus, <math>P</math> travels <math>\frac 14 \left(4\sqrt{10}\right)\pi = \sqrt{10}\pi</math> in the first rotation.
+
We let <math>P'Q'R'S'</math> be the rectangle after the first rotation, and <math>P''Q''R''S''</math> be the rectangle after the second rotation. Point <math>P</math> pivots about <math>R</math> in an [[arc]] of a circle of radius <math>\sqrt{2^2+6^2} = 2\sqrt{10}</math>, and since <math>\angle PRS,\, \angle P'RQ'</math> are complementary, it follows that the arc has a degree measure of <math>90^{\circ}</math> and length <math>\frac14</math> of the [[circumference]]. Thus, <math>P</math> travels <math>\frac 14 \left(4\sqrt{10}\right)\pi = \sqrt{10}\pi</math> in the first rotation.
  
 
Similarly, in the second rotation, <math>P</math> travels in a <math>90^{\circ}</math> arc about <math>S'</math>, with the radius being <math>6</math>. It travels <math>\frac 14(12)\pi = 3\pi</math>. Therefore, the total distance it travels is <math>\left(3+\sqrt{10}\right)\pi\ \mathrm{(C)}</math>.
 
Similarly, in the second rotation, <math>P</math> travels in a <math>90^{\circ}</math> arc about <math>S'</math>, with the radius being <math>6</math>. It travels <math>\frac 14(12)\pi = 3\pi</math>. Therefore, the total distance it travels is <math>\left(3+\sqrt{10}\right)\pi\ \mathrm{(C)}</math>.

Latest revision as of 16:52, 7 November 2021

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 $P$?

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

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

See also

2008 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_19&oldid=164832"