Question

一个直径为 $d$ 的半圆被包含在边长为 $8$ 的正方形中。已知 $d$ 的最大值为 $m - \sqrt{n}$ ，其中 $m$ 和 $n$ 是整数，求 $m + n$ 。

A semicircle with diameter $d$ is contained in a square whose sides have length $8.$ Given that the maximum value of $d$ is $m - \sqrt{n},$ where $m$ and $n$ are integers, find $m + n.$

答案

答案：544

知识点：最优化三角学切线

难度评级：2990

解答：

缩放到半径为 $1$ 的半圆，并求当它的直径与一组正方形边成角 $\theta$ 时，包含它的最小正方形，其中 $0 \le \theta \le 90^\circ$ 。沿正方形两组边的方向，用两对平行线夹住这个半圆：每个方向上，一条线与圆弧相切，另一条线经过直径的一个端点；两对平行线之间的距离分别为 $1 + \cos\theta$ 和 $1 + \sin\theta$ 。因此该朝向下最小外接正方形的边长为 $\max\{1 + \cos\theta,\ 1 + \sin\theta\}$ ，它在 $\theta = 45^\circ$ 时最小，边长为 $1 + \frac{\sqrt{2}}{2} = \frac{2 + \sqrt{2}}{2}$ 。

将这个最优构型缩放到正方形边长为 $8$ ，半径变为 $r = \frac{8}{(2 + \sqrt{2})/2}$ $= \frac{16}{2 + \sqrt{2}}$ $= 8\left(2 - \sqrt{2}\right)$ ，所以 $\begin{aligned} d &= 2r = 16\left(2 - \sqrt{2}\right) \\ &= 32 - 16\sqrt{2} = 32 - \sqrt{512}. \end{aligned}$

因此 $m + n = 32 + 512 = 544$ 。

Scale to a semicircle of radius $1$ and ask for the smallest square containing it when its diameter makes angle $\theta$ with one pair of sides, where $0 \le \theta \le 90^\circ.$ Squeeze the semicircle between two pairs of parallel lines in the square's two side directions: in each direction one line of the pair is tangent to the arc and the other passes through an endpoint of the diameter, and the distances between the pairs are $1 + \cos\theta$ and $1 + \sin\theta.$ So the smallest enclosing square in that orientation has side $\max\{1 + \cos\theta,\ 1 + \sin\theta\},$ which is minimized when $\theta = 45^\circ,$ giving side $1 + \frac{\sqrt{2}}{2} = \frac{2 + \sqrt{2}}{2}.$

Scaling this optimal configuration so the square has side $8,$ the radius becomes $r = \frac{8}{(2 + \sqrt{2})/2}$ $= \frac{16}{2 + \sqrt{2}}$ $= 8\left(2 - \sqrt{2}\right),$ so $\begin{aligned} d &= 2r = 16\left(2 - \sqrt{2}\right) \\ &= 32 - 16\sqrt{2} = 32 - \sqrt{512}. \end{aligned}$

Thus $m + n = 32 + 512 = 544.$

2005 AIME I 第 11 题

答案：544

难度评级：2990

解答：

其他年份的第 11 题