Question

一个三脚架有三条腿，每条长 $5$ 英尺。三脚架架好时，任意两条腿之间的角都相等，并且三脚架顶部离地面 $4$ 英尺。架设三脚架时，一条腿的下端 $1$ 英尺折断脱落。设折断后的三脚架架好时，顶部离地面的高度为 $h$ 英尺。则 $h$ 可写成 $\frac{m}{\sqrt{n}}$ ，其中 $m$ 和 $n$ 是正整数，且 $n$ 不被任何质数的平方整除。求 $\lfloor m + \sqrt{n} \rfloor$ 。（记号 $\lfloor x \rfloor$ 表示小于或等于 $x$ 的最大整数。）

A tripod has three legs each of length $5$ feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is $4$ feet from the ground. In setting up the tripod, the lower $1$ foot of one leg breaks off. Let $h$ be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then $h$ can be written in the form $\frac{m}{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $\lfloor m + \sqrt{n} \rfloor.$ (The notation $\lfloor x \rfloor$ denotes the greatest integer that is less than or equal to $x.$ )

答案

答案：183

知识点：立体几何坐标几何距离公式对称性

难度评级：3270

解答：

将顶部放在 $T = (0, 0, 4)$ 。每条腿长 $5$ ，所以每个脚端距原点 $3$ ： $A = (3, 0, 0)$ ， $B = \left(-\frac{3}{2}, \frac{3\sqrt{3}}{2}, 0\right)$ ， $C = \left(-\frac{3}{2}, -\frac{3\sqrt{3}}{2}, 0\right)$ 。断腿长度变为 $4$ ，所以它的端点为 $A' = \frac{4}{5}(3, 0, 0)$ $+ \frac{1}{5}(0, 0, 4)$ $= \left(\frac{12}{5}, 0, \frac{4}{5}\right)$ 。现在三脚架站在平面 $A'BC$ 上，而 $h$ 等于从 $T$ 到该平面的距离。

平面 $A'BC$ 包含 $\overline{BC}$ ，它平行于 $y$ -轴并经过中点 $M = \left(-\frac{3}{2}, 0, 0\right)$ ，所以可在 $xz$ -平面中测量从 $T$ 到该平面的距离：也就是从 $(0, 4)$ 到经过 $\left(-\frac{3}{2}, 0\right)$ 和 $\left(\frac{12}{5}, \frac{4}{5}\right)$ 的直线的距离，该直线方程为 $8x - 39z + 12 = 0$ 。因此 $\begin{aligned} h &= \frac{|8 \cdot 0 - 39 \cdot 4 + 12|}{\sqrt{8^2 + 39^2}} \\ &= \frac{144}{\sqrt{1585}}. \end{aligned}$

此时 $m = 144$ ，且 $n = 1585 = 5 \cdot 317$ 是平方因子自由的。由于 $39^2 = 1521 \lt 1585 \lt 1600$ 可得 $\lfloor 144 + \sqrt{1585} \rfloor$ $= 144 + 39 = 183$ 。

Place the top at $T = (0, 0, 4).$ Each leg has length $5,$ so each foot is $3$ from the origin: $A = (3, 0, 0),$ $B = \left(-\frac{3}{2}, \frac{3\sqrt{3}}{2}, 0\right),$ $C = \left(-\frac{3}{2}, -\frac{3\sqrt{3}}{2}, 0\right).$ The broken leg has length $4,$ so its tip is $A' = \frac{4}{5}(3, 0, 0)$ $+ \frac{1}{5}(0, 0, 4)$ $= \left(\frac{12}{5}, 0, \frac{4}{5}\right),$ and the tripod now stands on the plane $A'BC,$ with $h$ equal to the distance from $T$ to that plane.

The plane $A'BC$ contains $\overline{BC},$ which is parallel to the $y$ -axis and passes through the midpoint $M = \left(-\frac{3}{2}, 0, 0\right),$ so its distance from $T$ can be measured in the $xz$ -plane: it is the distance from $(0, 4)$ to the line through $\left(-\frac{3}{2}, 0\right)$ and $\left(\frac{12}{5}, \frac{4}{5}\right),$ whose equation is $8x - 39z + 12 = 0.$ Therefore $\begin{aligned} h &= \frac{|8 \cdot 0 - 39 \cdot 4 + 12|}{\sqrt{8^2 + 39^2}} \\ &= \frac{144}{\sqrt{1585}}. \end{aligned}$

Here $m = 144$ and $n = 1585 = 5 \cdot 317$ is squarefree. Since $39^2 = 1521 \lt 1585 \lt 1600,$ we get $\lfloor 144 + \sqrt{1585} \rfloor$ $= 144 + 39 = 183.$

2006 AIME I 第 14 题

答案：183

难度评级：3270

解答：

其他年份的第 14 题