Question

Jon 和 Steve 在一条与两条并排东西向铁轨平行的小路上骑自行车。Jon 以每小时 $20$ 英里的速度向东骑， Steve 以每小时 $20$ 英里的速度向西骑。两列长度相等、方向相反且速度恒定但不同的火车分别经过这两位骑车人。每列火车经过 Jon 都正好需要 $1$ 分钟。向西行驶的火车经过 Steve 所需时间是向东行驶的火车经过 Steve 所需时间的 $10$ 倍。每列火车的长度为 $\frac{m}{n}$ 英里，其中 $m$ 和 $n$ 是互质的正整数。求 $m + n$ 。

Jon and Steve ride their bicycles on a path that parallels two side-by-side train tracks running in the east/west direction. Jon rides east at $20$ miles per hour, and Steve rides west at $20$ miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds, each pass the two riders. Each train takes exactly $1$ minute to go past Jon. The westbound train takes $10$ times as long as the eastbound train to go past Steve. The length of each train is $\frac{m}{n}$ miles, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：49

知识点：相对速度方程组

难度评级：2300

解答：

设东行和西行火车的速度分别为每小时 $v_1$ 和 $v_2$ 英里，共同长度为 $L$ 英里。火车经过骑车人的时间等于 $L$ 除以相对速度。经过向东以 $20$ 英里每小时骑行的 Jon 用 $\frac{1}{60}$ 小时，因此 $\frac{L}{v_1 - 20} = \frac{L}{v_2 + 20} = \frac{1}{60},$ 所以 $v_1 = 60L + 20$ ，且 $v_2 = 60L - 20$ 。

相对于向西以 $20$ 英里每小时骑行的 Steve，两列火车的相对速度分别为 $v_1 + 20$ 和 $v_2 - 20$ ，并且西行火车所用时间是东行火车的 $10$ 倍： $\frac{L}{v_2 - 20} = \frac{10L}{v_1 + 20}$ ，所以 $v_1 + 20 = 10(v_2 - 20)$ 。代入得 $60L + 40 = 600L - 400$ ，因此 $540L = 440$ ， $L = \frac{22}{27}$ 。

因为 $\gcd(22, 27) = 1$ ，答案为 $22 + 27 = 49$ 。

Let the eastbound and westbound trains have speeds $v_1$ and $v_2$ miles per hour and common length $L$ miles. A train passes a rider in time $L$ divided by their relative speed. Passing Jon (riding east at $20$ ) in $\frac{1}{60}$ hour gives $\frac{L}{v_1 - 20} = \frac{L}{v_2 + 20} = \frac{1}{60},$ so $v_1 = 60L + 20$ and $v_2 = 60L - 20.$

Relative to Steve (riding west at $20$ ), the speeds are $v_1 + 20$ and $v_2 - 20,$ and the westbound train takes $10$ times as long: $\frac{L}{v_2 - 20} = \frac{10L}{v_1 + 20},$ so $v_1 + 20 = 10(v_2 - 20).$ Substituting, $60L + 40 = 600L - 400,$ so $540L = 440$ and $L = \frac{22}{27}.$

Since $\gcd(22, 27) = 1,$ the answer is $22 + 27 = 49.$

2014 AIME I 第 4 题

答案：49

难度评级：2300

解答：

其他年份的第 4 题