AIME 1992 · 第 3 题

Solution 1

Let $n$ be the number of matches won, so that $\frac{n}{2n}=\frac{1}{2}$, and $\frac{n+3}{2n+4}>\frac{503}{1000}$.

Cross multiplying, $1000n+3000>1006n+2012$, so $n<\frac{988}{6}=164 \dfrac {4}{6}=164 \dfrac{2}{3}$. Thus, the answer is $\boxed{164}$.

Solution 2 (same as solution 1)

Let $n$ be the number of matches she won before the weekend began. Since her win ratio started at exactly .$500 = \tfrac{1}{2},$ she must have played exactly $2n$ games total before the weekend began. After the weekend, she would have won $n+3$ games out of $2n+4$ total. Therefore, her win ratio would be $(n+3)/(2n+4).$ This means that

$$\frac{n+3}{2n+4} > .503 = \frac{503}{1000}.$$ Cross-multiplying, we get $1000(n+3) > 503(2n+4),$ which is equivalent to $n < \frac{988}{6} = 164.\overline{6}.$ Since $n$ must be an integer, the largest possible value for $n$ is $\boxed{164}.$