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Difference between revisions of "1997 AJHSME Problems/Problem 24"

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==Solution==
 
==Solution==
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<asy>
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pair A,B,C,D,EE;
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A = (0,0); B = (2,2); C = (4,0); D = (7,-3); EE = (10,0);
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fill(arc((2,0),A,C,CW)--arc((7,0),C,EE,CCW)--arc((5,0),EE,A,CCW)--cycle,gray);
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draw(arc((2,0),A,C,CW)--arc((7,0),C,EE,CCW));
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draw(circle((5,0),5));
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draw(A--EE);
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dot(A); dot(B); dot(C); dot(D); dot(EE);
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label("$A$",A,W);
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label("$B$",B,N);
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label("$C$",C,dir(40));
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label("$D$",D,N);
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label("$E$",EE,W);</asy>
  
 
Draw <math>\overline{AE}</math> to divide the big circle in half.  Assign <math>AC = 4</math> and <math>CE = 6</math> so that the radii work out to be integers.  (<math>4x</math> and <math>6x</math> can be used instead, but the <math>x</math> will cancel in the ratio.)   
 
Draw <math>\overline{AE}</math> to divide the big circle in half.  Assign <math>AC = 4</math> and <math>CE = 6</math> so that the radii work out to be integers.  (<math>4x</math> and <math>6x</math> can be used instead, but the <math>x</math> will cancel in the ratio.)   
  
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The shaded region is equal to the area of semicircle <math>\stackrel\frown{AE}</math> on top, plus the area of the semicircle <math>\stackrel\frown{CDE}</math> on the bottom, minus the area of semicircle <math>\stackrel\frown{ABC}</math> on top.
+
The shaded region is equal to the area of semicircle <math>\overarc{AE}</math> on top, plus the area of the semicircle <math>CDE</math> on the bottom, minus the area of semicircle <math>\overarc{ABC}</math> on top.
  
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The radii of those three semicircles are <math>5, 3, </math> and <math>2</math>, respectively.
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The radii of those three semicircles are <math>5, 3, </math> and <math>2</math>, respectively.Thus, the area of the shaded region is <math>\frac{1}{2}\pi\cdot5^2 + \frac{1}{2}\pi\cdot3^2 - \frac{1}{2}\pi\cdot2^2 =\frac{\pi}{2}(5^2 + 3^2 - 2^2)=15\pi</math>
  
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Thus, the area of the shaded region is:
+
The total area of the circle is <math>\pi \cdot 5^2 = 25\pi</math>.Thus, the unshaded area is <math>25\pi - 15\pi = 10\pi</math>.Therefore the ratio of shaded:unshaded is <math>15\pi : 10\pi =\boxed{ \text{(C)}\ 3:2}</math>.
−
 
 
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<math>\frac{1}{2}\pi\cdot5^2 + \frac{1}{2}\pi\cdot3^2 - \frac{1}{2}\pi\cdot2^2</math>
 
−
 
 
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<math>\frac{\pi}{2}(5^2 + 3^2 - 2^2)</math>
 
−
 
 
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<math>15\pi</math>
 
−
 
 
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The total area of the circle is <math>\pi \cdot 5^2 = 25\pi</math>. Thus, the unshaded area is <math>25\pi - 15\pi = 10\pi</math>.
 
−
 
 
−
So the ratio of shaded:unshaded is <math>15\pi : 10\pi = 3:2</math>, and the answer is <math>\boxed{C}</math>
 
  
 
==See Also==
 
==See Also==
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* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
 +
{{MAA Notice}}

Latest revision as of 16:05, 20 October 2023

Problem

Diameter $ACE$ is divided at $C$ in the ratio $2:3$. The two semicircles, $ABC$ and $CDE$, divide the circular region into an upper (shaded) region and a lower region. The ratio of the area of the upper region to that of the lower region is

[asy] pair A,B,C,D,EE; A = (0,0); B = (2,2); C = (4,0); D = (7,-3); EE = (10,0); fill(arc((2,0),A,C,CW)--arc((7,0),C,EE,CCW)--arc((5,0),EE,A,CCW)--cycle,gray); draw(arc((2,0),A,C,CW)--arc((7,0),C,EE,CCW)); draw(circle((5,0),5)); dot(A); dot(B); dot(C); dot(D); dot(EE); label("$A$",A,W); label("$B$",B,N); label("$C$",C,E); label("$D$",D,N); label("$E$",EE,W); [/asy]

$\text{(A)}\ 2:3 \qquad \text{(B)}\ 1:1 \qquad \text{(C)}\ 3:2 \qquad \text{(D)}\ 9:4 \qquad \text{(E)}\ 5:2$

Solution

[asy] pair A,B,C,D,EE; A = (0,0); B = (2,2); C = (4,0); D = (7,-3); EE = (10,0); fill(arc((2,0),A,C,CW)--arc((7,0),C,EE,CCW)--arc((5,0),EE,A,CCW)--cycle,gray); draw(arc((2,0),A,C,CW)--arc((7,0),C,EE,CCW)); draw(circle((5,0),5)); draw(A--EE); dot(A); dot(B); dot(C); dot(D); dot(EE); label("$A$",A,W); label("$B$",B,N); label("$C$",C,dir(40)); label("$D$",D,N); label("$E$",EE,W);[/asy]

Draw $\overline{AE}$ to divide the big circle in half. Assign $AC = 4$ and $CE = 6$ so that the radii work out to be integers. ($4x$ and $6x$ can be used instead, but the $x$ will cancel in the ratio.)

The shaded region is equal to the area of semicircle $\overarc{AE}$ on top, plus the area of the semicircle $CDE$ on the bottom, minus the area of semicircle $\overarc{ABC}$ on top.

The radii of those three semicircles are $5, 3,$ and $2$, respectively.Thus, the area of the shaded region is $\frac{1}{2}\pi\cdot5^2 + \frac{1}{2}\pi\cdot3^2 - \frac{1}{2}\pi\cdot2^2 =\frac{\pi}{2}(5^2 + 3^2 - 2^2)=15\pi$

The total area of the circle is $\pi \cdot 5^2 = 25\pi$.Thus, the unshaded area is $25\pi - 15\pi = 10\pi$.Therefore the ratio of shaded:unshaded is $15\pi : 10\pi =\boxed{ \text{(C)}\ 3:2}$.

See Also

1997 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 AJHSME/AMC 8 Problems and Solutions

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