Question

三角形 $ABC$ 的边长为 $AB = 9$ 、 $BC = 5\sqrt{3}$ 、 $AC = 12$ 点 $A = P_0, P_1, P_2, \ldots, P_{2450} = B$ 位于线段 $\overline{AB}$ 上，且对 $k = 1, 2, \ldots, 2449$ 点 $P_k$ 位于 $P_{k-1}$ 与 $P_{k+1}$ 之间。点 $A = Q_0, Q_1, Q_2, \ldots, Q_{2450} = C$ 位于线段 $\overline{AC}$ 上，且对 $k = 1, 2, \ldots, 2449$ 点 $Q_k$ 位于 $Q_{k-1}$ 与 $Q_{k+1}$ 之间。此外，每条线段 $\overline{P_kQ_k}$ ， $k = 1, 2, \ldots, 2449$ 都平行于 $\overline{BC}$ 。这些线段把三角形分成 $2450$ 个区域，其中有 $2449$ 个梯形和 $1$ 个三角形。所有 $2450$ 个区域面积相等。求线段 $\overline{P_kQ_k}$ ， $k = 1, 2, \ldots, 2450$ 中长度为有理数的线段数。

Triangle $ABC$ has side lengths $AB = 9,$ $BC = 5\sqrt{3},$ and $AC = 12.$ Points $A = P_0, P_1, P_2, \ldots, P_{2450} = B$ are on segment $\overline{AB}$ with $P_k$ between $P_{k-1}$ and $P_{k+1}$ for $k = 1, 2, \ldots, 2449,$ and points $A = Q_0, Q_1, Q_2, \ldots, Q_{2450} = C$ are on segment $\overline{AC}$ with $Q_k$ between $Q_{k-1}$ and $Q_{k+1}$ for $k = 1, 2, \ldots, 2449.$ Furthermore, each segment $\overline{P_kQ_k},$ $k = 1, 2, \ldots, 2449,$ is parallel to $\overline{BC}.$ The segments cut the triangle into $2450$ regions, consisting of $2449$ trapezoids and $1$ triangle. Each of the $2450$ regions has the same area. Find the number of segments $\overline{P_kQ_k},$ $k = 1, 2, \ldots, 2450,$ that have rational length.

答案

答案：20

知识点：相似面积比完全平方数

难度评级：2650

解答：

因为 $2450$ 个区域面积相等，三角形 $AP_kQ_k$ （前 $k$ 个区域的并集）的面积是三角形 $ABC$ 的 $\frac{k}{2450}$ 。每个三角形 $AP_kQ_k$ 都与 $ABC$ ，相似，长度按面积比的平方根缩放，所以 $\begin{aligned} P_kQ_k &= 5\sqrt{3}\,\sqrt{\frac{k}{2450}} \\ &= 5\sqrt{3} \cdot \frac{\sqrt{k}}{35\sqrt{2}} \\ &= \frac{\sqrt{6k}}{14}. \end{aligned}$

这为有理数恰好当 $6k$ 是完全平方数，也就是当且仅当 $k = 6j^2$ ，其中 $j$ 为正整数。条件 $6j^2 \le 2450$ 给出 $j^2 \le 408$ ，所以 $j = 1, 2, \ldots, 20$ 。共有 $20$ 条这样的线段。

Since the $2450$ regions have equal areas, triangle $AP_kQ_k$ (the union of the first $k$ regions) has area $\frac{k}{2450}$ of triangle $ABC.$ Each triangle $AP_kQ_k$ is similar to $ABC,$ and lengths scale as the square root of areas, so $\begin{aligned} P_kQ_k &= 5\sqrt{3}\,\sqrt{\frac{k}{2450}} \\ &= 5\sqrt{3} \cdot \frac{\sqrt{k}}{35\sqrt{2}} \\ &= \frac{\sqrt{6k}}{14}. \end{aligned}$

This is rational exactly when $6k$ is a perfect square, which happens exactly when $k = 6j^2$ for a positive integer $j.$ The condition $6j^2 \le 2450$ gives $j^2 \le 408,$ so $j = 1, 2, \ldots, 20.$ There are $20$ such segments.

2018 AIME II 第 7 题

答案：20

难度评级：2650

解答：

其他年份的第 7 题