AIME 2003 I · 第 2 题

Solution 1

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, and so forth. 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 $98$, and so forth.

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

$$\frac{100^{2}\pi - 99^{2}\pi + 98^{2}\pi - \ldots - 1^{2}\pi}{100^{2}\pi}$$ 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 $101 + 200 =\boxed{301}$.

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Alternatively, we can determine a pattern through trial-and-error using smaller numbers.

• For $2$ circles, the ratio is $3/4$.
• For $4$ circles, the ratio is $5/8$.
• For $6$ circles, the ratio is $7/12$.
• For $8$ circles, the ratio is $9/16$.

Now the pattern for each ratio is clear. Given $x$ circles, the ratio is $\frac{x+1}{2x}$. For the $100$ circle case (which is what this problem is), $x=100$, and the ratio is $\frac{101}{200}$.

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Also, using the difference of squares, the expression simplifies to $\frac{100 + 99 + 98 + 97 + ... + 1}{100^2}$. We can easily determine the sum with $\frac{100(101)}{2} = 5050$. Simplifying gives us $\frac{5050}{100^2} = \frac{101}{200}$ and the answer is $101 + 200 =\boxed{301}$.

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Solution 2 (synergy)

We want to find $\frac{\sum\limits_{n=1}^{50} (4n-1)\pi}{10000\pi}=\frac{\sum\limits_{n=1}^{50} (4n-1)}{10000}=\frac{(\sum\limits_{n=1}^{50} (4n) )-50}{10000}=\frac{101}{200} \rightarrow 101+200=\boxed{301}$

Solution 3 (Alcumus)

The sum of the areas of the green regions is

$$\left[(2^2-1^2)+(4^2-3^2)+(6^2-5^2)+\cdots+(100^2-99^2)\right]\pi$$ $$=\left[(2+1)+(4+3)+(6+5)+\cdots+(100+99)\right]\pi$$ $$={1\over2}\cdot100\cdot101\pi.$$ Thus the desired ratio is

$${1\over2}\cdot{{100\cdot101\pi}\over{100^2\pi}}={101\over200},$$ and $m+n=\boxed{301}$.