Question

一个梯形的一条底边比另一条底边长 $100$ 个单位。连接两腰中点的线段把梯形分成面积比为 $2 : 3$ 的两个区域。设 $x$ 为一条连接两腰、平行于底边、且把梯形分成两个等面积区域的线段长度。求不超过 $x^2/100$ 的最大整数。

One base of a trapezoid is $100$ units longer than the other base. The segment that joins the midpoints of the legs divides the trapezoid into two regions whose areas are in the ratio $2 : 3.$ Let $x$ be the length of the segment joining the legs of the trapezoid that is parallel to the bases and that divides the trapezoid into two regions of equal area. Find the greatest integer that does not exceed $x^2/100.$

答案

答案：181

知识点：梯形相似面积比

难度评级：2450

解答：

设两条底边为 $b$ 和 $b + 100$ 。中位线长为 $b + 50$ ，并把梯形分成两个等高梯形；它们的面积分别与平行边之和 $b + (b + 50)$ 和 $(b + 50) + (b + 100)$ 成正比。由 $\frac{2b + 50}{2b + 150} = \frac{2}{3}$ 得 $b = 75$ ，所以两底为 $75$ 和 $175$ 。

延长两腰相交于一点，得到相似三角形：长度为 $\ell$ 的平行线段截出的三角形面积为 $c\ell^2$ ，其中 $c$ 为常数。长度为 $x$ 的线段恰好平分梯形面积时， $cx^2 - c \cdot 75^2 = c \cdot 175^2 - cx^2$ ，所以 $x^2 = \frac{75^2 + 175^2}{2} = 18125.$

因而 $x^2/100 = 181.25$ ，不超过它的最大整数为 $181$ 。

Let the bases be $b$ and $b + 100.$ The midsegment has length $b + 50$ and splits the trapezoid into two trapezoids of equal height, whose areas are proportional to the sums of their parallel sides, $b + (b + 50)$ and $(b + 50) + (b + 100).$ Setting $\frac{2b + 50}{2b + 150} = \frac{2}{3}$ gives $b = 75,$ so the bases are $75$ and $175.$

Extend the legs to meet at an apex, creating similar triangles: a segment parallel to the bases with length $\ell$ cuts off a triangle of area $c\ell^2$ for a fixed constant $c.$ The segment of length $x$ bisects the trapezoid's area exactly when $cx^2 - c \cdot 75^2 = c \cdot 175^2 - cx^2,$ so $x^2 = \frac{75^2 + 175^2}{2} = 18125.$

Then $x^2/100 = 181.25,$ and the greatest integer not exceeding it is $181.$

2000 AIME II 第 6 题

答案：181

难度评级：2450

解答：

其他年份的第 6 题