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Difference between revisions of "2004 AIME II Problems/Problem 1"

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== Problem ==
 
== Problem ==
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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 <math> \frac{a\pi+b\sqrt{c}}{d\pi-e\sqrt{f}}, </math> where <math> a, b, c, d, e, </math> and <math> f </math> are positive integers, <math> a </math> and <math> e </math> are relatively prime, and neither <math> c </math> nor <math> f </math> is divisible by the square of any prime. Find the remainder when the product <math> abcdef </math> is divided by 1000.
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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 <math> \frac{a\pi+b\sqrt{c}}{d\pi-e\sqrt{f}}, </math> where <math> a, b, c, d, e, </math> and <math> f </math> are [[positive integer]]s, <math> a </math> and <math> e </math> are [[relatively prime]], and neither <math> c </math> nor <math> f </math> is [[divisibility | divisible]] by the [[square]] of any [[prime]]. Find the [[remainder]] when the product <math> abcdef </math> is divided by 1000.
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[[Image:2004 AIME II Problem 1.png]]
  
 
== Solution ==
 
== Solution ==
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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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Let <math>r</math> be the [[length]] of the radius of the circle.  A [[right triangle]] is formed by half of the chord, half of the radius (since the chord [[bisect]]s it), and the radius. Thus, it is a <math>30^\circ</math> - <math>60^\circ</math> - <math>90^\circ</math> [[triangle]], and the area of two such triangles is <math>2 \cdot \frac{1}{2} \cdot \frac{r}{2} \cdot \frac{r\sqrt{3}}{2} = \frac{r^2\sqrt{3}}{4}</math>. The [[central angle]] which contains the entire chord is <math>60 \cdot 2 = 120</math> [[degree]]s, 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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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 desired ratio is <math>\frac{\frac{2}{3}r^2\pi + \frac{r^2\sqrt{3}}{4}}{\frac{1}{3}r^2\pi - \frac{r^2\sqrt{3}}{4}} =  \frac{8\pi + 3\sqrt{3}}{4\pi - 3\sqrt{3}}</math>
  
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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>
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Therefore, <math>abcdef = 2^53^4 = 2592 \Longrightarrow \boxed{592}</math>.
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:<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>.
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2004|n=II|before=First Question|num-a=2}}
 
{{AIME box|year=2004|n=II|before=First Question|num-a=2}}
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[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 22:20, 4 July 2013

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 $\frac{a\pi+b\sqrt{c}}{d\pi-e\sqrt{f}},$ where $a, b, c, d, e,$ and $f$ are positive integers, $a$ and $e$ are relatively prime, and neither $c$ nor $f$ is divisible by the square of any prime. Find the remainder when the product $abcdef$ is divided by 1000.

2004 AIME II Problem 1.png

Solution

Let $r$ be the length of the radius of the circle. 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 $30^\circ$ - $60^\circ$ - $90^\circ$ triangle, and the area of two such triangles is $2 \cdot \frac{1}{2} \cdot \frac{r}{2} \cdot \frac{r\sqrt{3}}{2} = \frac{r^2\sqrt{3}}{4}$. The central angle which contains the entire chord is $60 \cdot 2 = 120$ degrees, so the area of the sector is $\frac{1}{3}r^2\pi$; the rest of the area of the circle is then equal to $\frac{2}{3}r^2\pi$.

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 desired ratio is $\frac{\frac{2}{3}r^2\pi + \frac{r^2\sqrt{3}}{4}}{\frac{1}{3}r^2\pi - \frac{r^2\sqrt{3}}{4}} = \frac{8\pi + 3\sqrt{3}}{4\pi - 3\sqrt{3}}$

Therefore, $abcdef = 2^53^4 = 2592 \Longrightarrow \boxed{592}$.

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

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