Difference between revisions of "2004 AIME II Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | {{ | + | A [[right triangle]] is formed by half of the [[chord]], half of the [[radius]] (since the chord bisects it), and the [[radius]]. Thus, it is a <math>30</math> - <math>60</math> - <math>90</math> triangle, and the area of two such triangles is <math>2\frac{1}{2}\frac{r}{2}\frac{r\sqrt{3}}{2} = \frac{r^2\sqrt{3}}{4}</math>. The [[central angle]] which contains the entire chord is <math>60 * 2 = 120</math> degrees, so the area of the [[sector]] is <math>\frac{1}{3}r^2\pi</math>; the rest of the area of the circle is then equal to <math>\frac{2}{3}r^2\pi</math>. |
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| + | The smaller area cut off by the chord is equal to the area of the sector minus the area of the triangle. The larger area is equal to the area of the circle not within the sector and the area of the triangle. Thus, the ratio becomes: | ||
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| + | :<math>\frac{\frac{1}{3}r^2\pi - \frac{r^2\sqrt{3}}{4}}{\frac{2}{3}r^2\pi + \frac{r^2\sqrt{3}}{4}}</math> | ||
| + | :<math>\frac{4\pi - 3\sqrt{3}}{8\pi + 3\sqrt{3}}</math> | ||
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| + | Therefore, <math>abcdef = 2^53^4 = 2592 \Rightarrow 592</math>. | ||
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== See also == | == See also == | ||
| − | + | {{AIME box|year=2004|n=II|before=First Question|num-a=2}} | |
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Revision as of 18:21, 12 February 2007
Problem
A chord of a circle is perpendicular to a radius at the midpoint of the radius. The ratio of the area of the larger of the two regions into which the chord divides the circle to the smaller can be expressed in the form 
where 
and 
are positive integers, 
and 
are relatively prime, and neither 
nor is divisible by the square of any prime. Find the remainder when the product 
is divided by 1000.
Solution
A right triangle is formed by half of the chord, half of the radius (since the chord bisects it), and the radius. Thus, it is a 
- 
- 
triangle, and the area of two such triangles is 
. The central angle which contains the entire chord is 
degrees, so the area of the sector is 
; the rest of the area of the circle is then equal to 
.
The smaller area cut off by the chord is equal to the area of the sector minus the area of the triangle. The larger area is equal to the area of the circle not within the sector and the area of the triangle. Thus, the ratio becomes:
Therefore, 
.
See also
| 2004 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
