Difference between revisions of "1997 AJHSME Problems/Problem 24"
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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.) | ||
| − | The shaded region is equal to the area of semicircle <math>\overarc{AE}</math> on top, plus the area of the semicircle <math> | + | 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. |
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> | 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> | ||
Latest revision as of 16:05, 20 October 2023
Problem
Diameter 
is divided at 
in the ratio 
. The two semicircles, 
and 
, 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]](http://latex.artofproblemsolving.com/c/4/e/c4ecc539294cfbe2d303003a6f52d3bae65ada85.png)
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]](http://latex.artofproblemsolving.com/c/b/9/cb997164288a4f94a609b7f636a391a65e55df80.png)
Draw 
to divide the big circle in half. Assign 
and 
so that the radii work out to be integers. (
and 
can be used instead, but the 
will cancel in the ratio.)
The shaded region is equal to the area of semicircle 
on top, plus the area of the semicircle on the bottom, minus the area of semicircle 
on top.
The radii of those three semicircles are 
and 
, respectively.Thus, the area of the shaded region is 
The total area of the circle is 
.Thus, the unshaded area is 
.Therefore the ratio of shaded:unshaded is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
