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Difference between revisions of "1994 AHSME Problems/Problem 17"

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D(MP("O",O,S));</asy>
 
D(MP("O",O,S));</asy>
  
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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>. We can find that the two triangles formed by two radii of circle <math>O</math> and the segment of the rectangle are <math>45-45-90</math> triangles. The area of the overlapping region is the area of the circle minus the area of the two parts of the circle that are outside of the rectangle. The area of the circle is <math>4\pi</math>.
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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.  
  
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To find the area of the exterior parts, we find the area of the triangle and subtract it from the area of the quarter arc. The area of the triangle is <math>2</math> and the area of the quarter arc is <math>\pi</math>. So the area of the exterior parts combined is <math>2(\pi-2)=2\pi-4</math>. So the area of the overlapping region is <cmath>4\pi-(2\pi-4)=\boxed{\textbf{(D) }2\pi+4.}</cmath>
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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]
 
--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
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==See Also==
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{{AHSME box|year=1994|num-b=16|num-a=18}}
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{{MAA Notice}}

Latest revision as of 03:07, 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

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