Question

Ana、Bob 和 Cao 分别以每秒 $8.6$ 米、每秒 $6.2$ 米和每秒 $5$ 米的恒定速度骑自行车。他们同时从一个长方形场地的东北角出发，该场地较长的一边正好沿东西方向，向西延伸。Ana 沿场地边缘骑行，起初向西；Bob 沿场地边缘骑行，起初向南；Cao 沿直线穿过场地骑到南边上的一点 $D$ 。Cao 到达点 $D$ 的时间，正好等于 Ana 和 Bob 第一次到达 $D$ 的时间。场地的长、宽、以及点 $D$ 到场地东南角的距离之比可表示为 $p : q : r$ ，其中 $p$ 、 $q$ 、 $r$ 是正整数，且 $p$ 和 $q$ 互质。求 $p + q + r$ 。

Ana, Bob, and Cao bike at constant rates of $8.6$ meters per second, $6.2$ meters per second, and $5$ meters per second, respectively. They all begin biking at the same time from the northeast corner of a rectangular field whose longer side runs due west. Ana starts biking along the edge of the field, initially heading west, Bob starts biking along the edge of the field, initially heading south, and Cao bikes in a straight line across the field to a point $D$ on the south edge of the field. Cao arrives at point $D$ at the same time that Ana and Bob arrive at $D$ for the first time. The ratio of the field's length to the field's width to the distance from point $D$ to the southeast corner of the field can be represented as $p : q : r,$ where $p,$ $q,$ and $r$ are positive integers with $p$ and $q$ relatively prime. Find $p + q + r.$

答案

答案：61

知识点：路程、速度与时间方程组因式分解

难度评级：2390

解答：

设场地长为 $L$ （向西），宽为 $W$ （向南），且 $L \gt W$ ，并设 $x$ 为 $D$ 到东南角的距离。Ana 绕边缘骑行距离 $2L + W - x$ ，Bob 骑行距离 $W + x$ ，Cao 骑行距离 $\sqrt{W^2 + x^2}$ ，且三人用时相同： $\begin{aligned} \frac{2L + W - x}{8.6} &= \frac{W + x}{6.2} \\ &= \frac{\sqrt{W^2 + x^2}}{5}. \end{aligned}$

第一个等式给出 $L = \frac{6W + 37x}{31}$ 。将第二个等式平方，得到 $25(W + x)^2 = 38.44\,(W^2 + x^2)$ ，化简为 $168W^2 - 625Wx + 168x^2 = 0$ ，并可分解为 $(24W - 7x)(7W - 24x) = 0$ 。

根 $x = \frac{7W}{24}$ 给出 $L = \frac{13W}{24} \lt W$ ，不可能，所以 $x = \frac{24W}{7}$ ，进而 $L = \frac{30W}{7}$ 。比例为 $L : W : x = 30 : 7 : 24$ ，所以 $p + q + r = 30 + 7 + 24 = 61$ 。

Let the field have length $L$ (west) and width $W$ (south) with $L \gt W,$ and let $x$ be the distance from $D$ to the southeast corner. Ana rides around the perimeter a distance $2L + W - x,$ Bob rides $W + x,$ and Cao rides $\sqrt{W^2 + x^2},$ all in the same time: $\begin{aligned} \frac{2L + W - x}{8.6} &= \frac{W + x}{6.2} \\ &= \frac{\sqrt{W^2 + x^2}}{5}. \end{aligned}$

The first equality gives $L = \frac{6W + 37x}{31}.$ Squaring the second, $25(W + x)^2 = 38.44\,(W^2 + x^2),$ which simplifies to $168W^2 - 625Wx + 168x^2 = 0,$ factoring as $(24W - 7x)(7W - 24x) = 0.$

The root $x = \frac{7W}{24}$ gives $L = \frac{13W}{24} \lt W,$ which is impossible, so $x = \frac{24W}{7}$ and then $L = \frac{30W}{7}.$ The ratio is $L : W : x = 30 : 7 : 24,$ and $p + q + r = 30 + 7 + 24 = 61.$

2012 AIME II 第 4 题

答案：61

难度评级：2390

解答：

其他年份的第 4 题