Question

珍妮和肯尼沿相同方向行走；肯尼的速度为每秒 $3$ 英尺，珍妮的速度为每秒 $1$ 英尺。他们所在的两条平行道路相距 $200$ 英尺。一座直径为 $100$ 英尺的高大圆形建筑位于两条道路正中间。当建筑首次挡住两人的视线时，他们相距 $200$ 英尺。设再过 $t$ 秒后，两人能够重新看见彼此。若将 $t$ 写成最简分数，求其分子与分母之和。

Jenny and Kenny are walking in the same direction, Kenny at $3$ feet per second and Jenny at $1$ foot per second, on parallel paths that are $200$ feet apart. A tall circular building $100$ feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are $200$ feet apart. Let $t$ be the amount of time, in seconds, before Jenny and Kenny can see each other again. If $t$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

答案

答案：163

知识点：距离公式相对速度切线

难度评级：2840

小提示：

将圆形建筑的圆心置于原点，并将两条道路置于 $y=100$ 和 $y=-100$ 上

Place the circular building at the origin and the two paths on $y=100$ and $y=-100$

大提示：

首次被遮挡时，两人的横坐标都为 $x=-50$ ；令之后两人连线到原点的距离等于 $50$

At the first blockage both walkers have $x=-50$ ; set the later connecting line’s distance from the origin equal to $50$

解答：

首次被遮挡时，两人在切线 $x=-50$ 上竖直对齐。经过 $t$ 秒后，他们的位置可写为 $(-50+t,100)$ 和 $(-50+3t,-100)$ 。原点到两人连线的距离为 $\frac{|10000-400t|}{\sqrt{4t^2+40000}}\text{。}$ 第二次相切时，该距离等于 $50$ 。平方并化简得 $(100-4t)^2=t^2+10000\text{，}$ 因此 $t(15t-800)=0$ 。正的时间为 $t=\frac{160}{3}$ ，所求之和为 $160+3=163$ 。

At first blockage the walkers are vertically aligned on the tangent $x=-50.$ After $t$ seconds their positions may be written as $(-50+t,100)$ and $(-50+3t,-100).$ The distance from the origin to their connecting line is $\frac{|10000-400t|}{\sqrt{4t^2+40000}}.$ At the second tangency this equals $50.$ Squaring and simplifying gives $(100-4t)^2=t^2+10000,$ so $t(15t-800)=0.$ The positive time is $t=\frac{160}{3},$ and the requested sum is $160+3=163.$

1993 AIME 第 13 题

答案：163

难度评级：2840

小提示：

大提示：

解答：

其他年份的第 13 题