一块均匀薄片的形状为三角形 $ABC$，其中 $B$ 角为直角，$AB=9a$，$BC=6a$。点 $D$ 位于 $BC$ 上，且 $BD=x$（见图）。从薄片中移去区域 $ABD$。将所得的形状 $ADC$ 放置在水平面上，使边 $DC$ 与水平面接触，且平面 $ADC$ 竖直。

求该形状处于平衡状态时 $x$ 的取值集合，用 $a$ 表示。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

Where a numerical value for the acceleration due to gravity ($g$) is needed, use $10\,\mathrm{m\,s^{-2}}$.

需要使用重力加速度 $g$ 的数值时，取 $10\,\mathrm{m\,s^{-2}}$。

以 $B$ 为水平原点，先用大三角形减去小三角形的面积矩求质心。静止时重力的作用线必须落在支承边 $DC$ 上。

The area of $ABC$ is $27a^2$, and its centroid has horizontal coordinate $2a$. The removed triangle $ABD$ has area $9ax/2$ and centroid coordinate $x/3$. With uniform density, the horizontal coordinate of the remaining centroid is

For $0\le x<6a$, $6a-x>0$, so cancellation gives $\overline{x}=(6a+x)/3$. This is less than $4a<6a$, so the line of action of the weight meets the support $DC$ exactly when $\overline{x}\ge x$. Thus

Hence the range is $0\le x\le3a$, with $x=0$ corresponding to the uncut limiting case and $x=3a$ to equilibrium at the point of toppling.

也可直接使用剩余三角形的顶点。将三个顶点的横坐标取平均，再与 $D$、$C$ 的坐标比较。

The vertices $A,D,C$ have horizontal coordinates $0,x,6a$ measured from $B$. Thus their triangle has centroid at $\overline{x}=(6a+x)/3$. Equilibrium on $DC$ requires $x\le\overline{x}\le6a$. The upper inequality holds throughout $0\le x<6a$, while the lower one gives $3x\le6a+x$. Therefore $0\le x\le3a$, including the uncut case at $x=0$.