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Difference between revisions of "2003 AIME I Problems/Problem 2"

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== Problem ==
 
== Problem ==
−
One hundred concentric circles with radii <math> 1, 2, 3, \dots, 100 </math> are drawn in a plane. The interior of the circle of radius 1 is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The ratio of the total area of the green regions to the area of the circle of radius 100 can be expressed as <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers. Find <math> m + n. </math>
+
One hundred [[concentric]] [[circle]]s with [[radius | radii]] <math> 1, 2, 3, \dots, 100 </math> are drawn in a plane. The interior of the circle of radius 1 is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The [[ratio]] of the total [[area]] of the green regions to the area of the circle of radius 100 can be expressed as <math> m/n, </math> where <math> m </math> and <math> n </math> are [[relatively prime]] [[positive integer]]s. Find <math> m + n. </math>
  
 
== Solution ==
 
== Solution ==
 +
To get the green area, we can color all the circles of radius 100 or below green, then color all those with radius 99 or below red, then color all those with radius 98 or below green, etc.  This amounts to adding the area of the circle of radius 100, but subtracting the circle of radius 99, then adding the circle of radius 98, etc.
 +
 +
The total green area is thus given by <math>\pi 100^{2} - \pi 99^{2} + \pi 98^{2} - \ldots - \pi 1^{2}</math>, while the total area is given by <math>\pi 100^{2}</math>, so the ratio is
 +
 +
<math>\frac{\pi 100^{2} - \pi 99^{2} + \pi 98^{2} - \ldots - \pi 1^{2}}{\pi 100^{2}}</math>
  
−
To get the green area, we can color all the circles of radius 100 or below green, then color all those with radius 99 or below red, then color all those with radius 98 or below green, etc.  This amounts to adding the area of the circle of radius 100, but subtracting the circle of radius 99, then adding the circle of radius 98, etc.
+
For any <math>a</math>, <math>a^{2}-(a-1)^{2}=a^{2}-(a^{2}-2a+1)=2a-1</math>.  We can cancel the [[divisor | factor]] of [[pi]] from the [[numerator]] and [[denominator]] and simplify the ratio to
 +
 
 +
<math>\frac{(2\cdot100 - 1)+(2\cdot98 - 1) + \ldots + (2\cdot 2 - 1)}{100^{2}} = \frac{2\cdot(100 + 98 + \ldots + 2) - 50}{100^2}</math>.   
 +
 
 +
Using the formula for the sum of an [[arithmetic series]], we see that this is equal to
  
−
The total green area is thus given by <math>\pi (100)^{2} - \pi (99)^{2} + \pi (98)^{2} - ... - \pi (1)^{2}</math>, while the total area is given by <math>\pi (100)^{2}</math>.
+
<math>\frac{2(50)(51)-50}{100^{2}}=\frac{50(101)}{100^{2}}=\frac{101}{200}</math>,
  
−
For any a, <math>a^{2}-(a-1)^{2}=a^{2}-[a^{2}-2a+1]=2a-1</math>.  Thus after canceling the pi from numerator and denominator, we can simplify the ratio to <math>\frac{2(100+98+96+...+4+2)-50}{100^{2}}</math>.  This equals <math>\frac{4(50+49+...+2+1)-50}{100^{2}}=\frac{2(50)(51)-50}{100^{2}}=\frac{50(101)}{100^{2}}=\frac{101}{200}</math>, making the answer 301.
+
so the answer is <math>101 + 200 = 301</math>.
  
 
== See also ==
 
== See also ==
 +
* [[2003 AIME I Problems/Problem 1 | Previous problem]]
 +
* [[2003 AIME I Problems/Problem 3 | Next problem]]
 
* [[2003 AIME I Problems]]
 
* [[2003 AIME I Problems]]
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[[Category:Introductory Geometry Problems]]

Revision as of 10:45, 25 October 2006

Problem

One hundred concentric circles with radii $1, 2, 3, \dots, 100$ are drawn in a plane. The interior of the circle of radius 1 is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The ratio of the total area of the green regions to the area of the circle of radius 100 can be expressed as $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution

To get the green area, we can color all the circles of radius 100 or below green, then color all those with radius 99 or below red, then color all those with radius 98 or below green, etc. This amounts to adding the area of the circle of radius 100, but subtracting the circle of radius 99, then adding the circle of radius 98, etc.

The total green area is thus given by $\pi 100^{2} - \pi 99^{2} + \pi 98^{2} - \ldots - \pi 1^{2}$, while the total area is given by $\pi 100^{2}$, so the ratio is

$\frac{\pi 100^{2} - \pi 99^{2} + \pi 98^{2} - \ldots - \pi 1^{2}}{\pi 100^{2}}$

For any $a$, $a^{2}-(a-1)^{2}=a^{2}-(a^{2}-2a+1)=2a-1$. We can cancel the factor of pi from the numerator and denominator and simplify the ratio to

$\frac{(2\cdot100 - 1)+(2\cdot98 - 1) + \ldots + (2\cdot 2 - 1)}{100^{2}} = \frac{2\cdot(100 + 98 + \ldots + 2) - 50}{100^2}$.

Using the formula for the sum of an arithmetic series, we see that this is equal to

$\frac{2(50)(51)-50}{100^{2}}=\frac{50(101)}{100^{2}}=\frac{101}{200}$,

so the answer is $101 + 200 = 301$.

See also

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