2003 AIME I Problems/Problem 2

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 $101 + 200 = 301$ .

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2003 AIME I Problems/Problem 2

Problem

Solution

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