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 "1994 AHSME Problems/Problem 17"

(Created page with "==Problem== An <math>8</math> by <math>2\sqrt{2}</math> rectangle has the same center as a circle of radius <math>2</math>. The area of the region common to both the rectangle an...")
 
(→‎Solution)
 
(One intermediate revision by one other user not shown)
Line 4: Line 4:
 
<math> \textbf{(A)}\ 2\pi \qquad\textbf{(B)}\ 2\pi+2 \qquad\textbf{(C)}\ 4\pi-4 \qquad\textbf{(D)}\ 2\pi+4 \qquad\textbf{(E)}\ 4\pi-2 </math>
 
<math> \textbf{(A)}\ 2\pi \qquad\textbf{(B)}\ 2\pi+2 \qquad\textbf{(C)}\ 4\pi-4 \qquad\textbf{(D)}\ 2\pi+4 \qquad\textbf{(E)}\ 4\pi-2 </math>
 
==Solution==
 
==Solution==
 +
<asy>
 +
import cse5;
 +
import olympiad;
 +
real s=2*sqrt(2);
 +
pair A=(0,0),B=(0,s),C=(8,s),D=(8,0),O=(4,sqrt(2)),X;
 +
D(A--B--C--D--cycle);
 +
D(CR(O,2));
 +
pair[] P;
 +
P=IPs(CR(O,2),box(A,C));
 +
for(int i=0; i<4; i=i+1)
 +
{
 +
D(O--P[i],black);
 +
}
 +
X=foot(O,B,C);
 +
D(O--X);
 +
D(rightanglemark(O,X,C));
 +
D(O);
 +
D(MP("O",O,S));</asy>
 +
 +
We draw the diagram above. Dropping an altitude from the center of the circle to the top side of the rectangle, yields a segment of length <math>\sqrt{2}</math>. Since the hypotenuse is the radius of the circle, it has length 2 and by Pythagoras the other leg of the triangle also has length <math>\sqrt{2}</math>.  We deduce that the two triangles formed by two radii of circle <math>O</math> and the segment of the rectangle are 45-45-90 triangles.
 +
 +
The overlapping area consists of two sectors with central angle 90 and four 45-45-90 triangles with base <math>\sqrt{2}</math> and height <math>\sqrt{2}</math>.  The area of the circle is <math>4\pi</math> so the area of the sectors is <math>2 \cdot 4\pi/4 = 2\pi</math>.  The area of the triangles is  <math>4\cdot (\sqrt{2})^2 /2 = 4</math>.  The combined area is <cmath>\boxed{\textbf{(D) }2\pi+4.}</cmath>
 +
 +
--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
 +
 +
==See Also==
 +
 +
{{AHSME box|year=1994|num-b=16|num-a=18}}
 +
{{MAA Notice}}

Latest revision as of 02:07, 28 May 2021

Problem

An $8$ by $2\sqrt{2}$ rectangle has the same center as a circle of radius $2$. The area of the region common to both the rectangle and the circle is

$\textbf{(A)}\ 2\pi \qquad\textbf{(B)}\ 2\pi+2 \qquad\textbf{(C)}\ 4\pi-4 \qquad\textbf{(D)}\ 2\pi+4 \qquad\textbf{(E)}\ 4\pi-2$

Solution

[asy] import cse5; import olympiad; real s=2*sqrt(2); pair A=(0,0),B=(0,s),C=(8,s),D=(8,0),O=(4,sqrt(2)),X; D(A--B--C--D--cycle); D(CR(O,2)); pair[] P; P=IPs(CR(O,2),box(A,C)); for(int i=0; i<4; i=i+1) { D(O--P[i],black); } X=foot(O,B,C); D(O--X); D(rightanglemark(O,X,C)); D(O); D(MP("O",O,S));[/asy]

We draw the diagram above. Dropping an altitude from the center of the circle to the top side of the rectangle, yields a segment of length $\sqrt{2}$. Since the hypotenuse is the radius of the circle, it has length 2 and by Pythagoras the other leg of the triangle also has length $\sqrt{2}$. We deduce that the two triangles formed by two radii of circle $O$ and the segment of the rectangle are 45-45-90 triangles.

The overlapping area consists of two sectors with central angle 90 and four 45-45-90 triangles with base $\sqrt{2}$ and height $\sqrt{2}$. The area of the circle is $4\pi$ so the area of the sectors is $2 \cdot 4\pi/4 = 2\pi$. The area of the triangles is $4\cdot (\sqrt{2})^2 /2 = 4$. The combined area is \[\boxed{\textbf{(D) }2\pi+4.}\]

--Solution by TheMaskedMagician

See Also

1994 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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 • 26 • 27 • 28 • 29 • 30
All AHSME 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=1994_AHSME_Problems/Problem_17&oldid=154355"