Question

设 $R = (8, 6)$ 。方程为 $8y = 15x$ 和 $10y = 3x$ 的直线分别含有点 $P$ 和 $Q$ ，且 $R$ 是 $\overline{PQ}$ 的中点。长度 $PQ$ 等于 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

Let $R = (8, 6).$ The lines whose equations are $8y = 15x$ and $10y = 3x$ contain points $P$ and $Q,$ respectively, such that $R$ is the midpoint of $\overline{PQ}.$ The length $PQ$ equals $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：67

知识点：坐标几何中点方程组距离公式

难度评级：2170

解答：

两条直线上的点可写成 $P = (8t, 15t)$ 和 $Q = (10u, 3u)$ 。因为 $R = (8, 6)$ 是 $\overline{PQ}$ 的中点， $8t + 10u = 16$ 且 $15t + 3u = 12.$

第二个方程给出 $u = 4 - 5t$ ；代入第一个方程，得 $8t + 40 - 50t = 16$ ，所以 $t = \frac{4}{7}$ ，且 $u = \frac{8}{7}$ 。于是 $P = \left(\frac{32}{7}, \frac{60}{7}\right)$ ， $Q = \left(\frac{80}{7}, \frac{24}{7}\right)$ 。

因而 $PQ = \sqrt{\left(\frac{48}{7}\right)^2 + \left(\frac{36}{7}\right)^2}$ $= \frac{12}{7}\sqrt{4^2 + 3^2}$ $= \frac{60}{7}$ ，所以 $m + n = 60 + 7 = 67$ 。

Points on the two lines can be written $P = (8t, 15t)$ and $Q = (10u, 3u).$ Since $R = (8, 6)$ is the midpoint of $\overline{PQ},$ $8t + 10u = 16$ and $15t + 3u = 12.$

The second equation gives $u = 4 - 5t;$ substituting into the first, $8t + 40 - 50t = 16,$ so $t = \frac{4}{7}$ and $u = \frac{8}{7}.$ Thus $P = \left(\frac{32}{7}, \frac{60}{7}\right)$ and $Q = \left(\frac{80}{7}, \frac{24}{7}\right).$

Then $PQ = \sqrt{\left(\frac{48}{7}\right)^2 + \left(\frac{36}{7}\right)^2}$ $= \frac{12}{7}\sqrt{4^2 + 3^2}$ $= \frac{60}{7},$ so $m + n = 60 + 7 = 67.$

2001 AIME II 第 4 题

答案：67

难度评级：2170

解答：

其他年份的第 4 题