通过苏格拉底的提问考虑有哪些异常情况可能不对，棱长为2的正方体容器ABCD-A1B1C1D1中，E,...

बनाया गया: 19 फ़रवरी 2025

का उपयोग करके उत्तर दिया गया O1 द्वारा Chat01

प्रश्न

通过苏格拉底的提问考虑有哪些异常情况可能不对，棱长为2的正方体容器ABCD-A1B1C1D1中，E,F分别是棱AB,BC的中点，在E,F,C1处各有1个小孔（孔的大小忽略不计），则该容器可装水的最大体积是_____。

विचार

Analyzing the cube

I'm thinking through a cube with a 2-unit side length, features like midpoints E and F on the bottom face, and a hole in C1 on the top face.

Setting up coordinates

Mapping out the cube with midpoints E and F on the bottom face, and C1 on the top face, to pinpoint where water might escape.

Charting out the cube

OK, let's consider the cube's corners and midpoints: A(0, 0, 0), B(2, 0, 0), C(2, 2, 0), D(0, 2, 0), A1(0, 0, 2), B1(2, 0, 2), C1(2, 2, 2), D1(0, 2, 2).

Assessing leak potential

I'm checking to see if filling the cube beyond z=0 might cause water to leak out from hole E, which is on the bottom plane. This makes me think about the possibilities.

Reassessing the setup

Evaluating the cube's resting orientation on a surface and pinpointing the hole at E on the bottom face. This prompts a deeper look into how orientation might affect leak risks.

Assessing hole risks

OK, let me see. By assessing F and C1 along with E, I’m examining potential leak spots from the bottom and top. This approach helps pinpoint maximum safe filling volume.

Evaluating leak dynamics

I'm thinking water, if above z=0, could naturally leak through bottom or side holes unless some kind of seal or partition blocks the leak.

Considering the top hole

Hmm, I’m thinking about how the hole at (2,2,2) limits the water level. The water level can't go above z=2, else it would overflow. This seems like a key boundary condition.

Connecting the dots

I'm piecing together the idea that z=0 holes could empty all the water unless an interior partition or air pocket stops this. This seems to be a crucial part to think through.

Analyzing the water plane

I'm working through the idea that the plane through E, F, and C1 might be the water's surface. Water can't go above this plane due to the holes, setting a clear boundary.

Topping off the water

Emphasizing that thanks to the holes, the water level can't top the plane through E, F, and C1. This creates a predictable volume, assuming no leaks are hiding.

Figuring out the plane

OK, let's now figure out the plane equation for E, F, and C1. Then, I'll see which parts of the cube lie below this plane.

Mapping the plane

I'm working out the plane equation for E, F, and C1, and seeing if the origin is below it. Then, I'll calculate the volume of the cube section below this plane.

Piecing together the plane

I calculated the normal vector (2, -2, 1) and figured out the plane equation as 2x - 2y + z - 2 = 0, confirming it passes correctly through E, F, and C1.

Pinpointing the water zone

OK, let me see. The water occupies the region fitting 2x - 2y + z - 2 ≤ 0 in the cube [0,2]^3. Integrating or calculating the geometric volume can reveal this effectively.

Mapping the feasible z

Consider the feasible region in the x-y plane, where z ranges from 0 to min(2, 2 - 2x + 2y). This ensures practicality within the cube.

Mapping out conditions

I'm reviewing the feasible region in the x-y plane with conditions 2 - 2x + 2y ≥ 0 and 0 ≤ x, y ≤ 2, leading to y ≥ x - 1.

Mapping band structures

OK, that's interesting. For x ≤ 1, the y range is broad. For x ≥ 1, y varies but is more constrained by x-1 to 2, forming distinct vertical strips.

Assessing region boundaries

There are two regions: 0 ≤ x ≤ 1 and 1 < x ≤ 2. For both, z max is 2 - 2x + 2y, but limited by the cube boundary z ≤ 2.

Assessing feasibility

OK, let me see. If 2 - 2x + 2y ≥ 2, then y ≥ x. This leads to the water filling the entire column from z=0 to z=2.

Assessing the impact

Interestingly enough, if the water level reaches z=2, it might leak through a hole at (2,2,2). This makes me think the water surface can't exceed the plane E-F-C1.

Navigating water levels

I’m figuring out how the water levels reach the top face of the cube (z=2) or stop at some lower plane defined by 2x - 2y + z - 2 ≤ 0, whichever is lower.

Piecing together regions

Region R is defined by x, y, and z constraints, including subregions A and B, where z ≤ 2 - 2x + 2y or z ≤ 2. This mapping captures the major dynamics effectively.

Analyzing vertical constraints

For Region 1 (0 ≤ x < 1), y spans [0, 2]. Playing with gravity's effect, I see y = x, and y intersects at (1, 1).

Trying another way

OK, let me see. For x in [0,1), y=x-1 is negative, so y≥0 holds true. Simplifying, region A covers x in [0,1), y in [0, x-1].

Defining regions

I’m thinking through regions A and B: for x in [0,1], y ranges from [0,2]; for x in [1,2], y spans [x-1,2]. The plane or top face restricts z's upper bound.

Breaking down calculations

I’m thinking through regions A and B, using a piecewise integral approach to define the volume by integrating over specific constraints for each region.

Confirming region A

OK, let me see. For x in [0,1], region A covers y values from [0, x). This aligns with the constraints y < x and y ≥ 0.

Piecing together constraints

OK, let's confirm region A spans x in [0,2] with y confined by either x or x-1. Region B, meanwhile, consistently holds y ≥ x.

Mapping region B

Region B spans x in [0,2], y in [x,2], forming a triangle above the line y=x within the square domain. This region ensures y ≥ x.

Mapping out regions

I’m digging into regions A and B, integrating volumes by examining constraints and variable relationships. Now I’ll explore the integrals for each region separately.

Crunching the numbers

I’m digging into $V_{A1}$ by integrating first with respect to $y$ , simplifying the integrand, and then with respect to $x$ . The result after detailed calculations is $2/3$ .

Assessing volume

Integrating the innermost expression in region B yields 1. Next, I'll complete the integral with respect to $x$ to determine the overall volume of $V_{A2}$ .

Calculating the volume

I’m mapping out the volume for region A by adding $V_{A1}$ and $V_{A2}$ . For region B, the volume is straightforward: $y$ runs from $x$ to 2.

Assessing volume

I’m curious about the total volume by integrating over regions A and B. After thorough calculations, the final volume comes to 17/3, which is about 5.6667.

Reasoned for 1m 21s

उत्तर

下面给出一种比较“几何直观”但也相对系统的解法思路。为方便起见，先在立方体 $ABCD - A_1B_1C_1D_1$ 上建立坐标系，如下约定（单位长度为 1）：

令底面 $ABCD$ 所在平面为 $z=0$ ，顶面 $A_1B_1C_1D_1$ 为 $z=2$ 。
点的具体坐标可取：

A=(0,0,0),\quad B=(2,0,0),\quad C=(2,2,0),\quad D=(0,2,0),

A_1=(0,0,2),\quad B_1=(2,0,2),\quad C_1=(2,2,2),\quad D_1=(0,2,2).

这样，棱长为 2。题中给出的三处小孔分别是

E=\text{AB 的中点}=(1,0,0),\quad F=\text{BC 的中点}=(2,1,0),\quad C_1=(2,2,2).

1. 寻找“极限装水面”——经过三孔的平面

因为容器只有 $E,F,C_1$ 这三处开孔，水面一旦“超过”其中某个孔的高度（并且内部连通到该孔），水就会外漏。在极限情况下，水面恰好会“贴住”这三个孔所在的几何位置。换句话说，可以设想存在一个平面同时经过 $E,F,C_1$ ，并且容器里能盛水的部分正好是这个平面以下（且在立方体内）的区域。

令这一平面方程为

Ax + B y + C z + D = 0,

把三点 $E(1,0,0),F(2,1,0),C_1(2,2,2)$ 坐标代入并解得，即可确定该平面。

1.1 求经过 $E,F,C_1$ 的平面

取向量 $\overrightarrow{EF} = F - E = (2-1,\;1-0,\;0-0) = (1,1,0),$ $\overrightarrow{EC_1} = C_1 - E = (2-1,\;2-0,\;2-0) = (1,2,2).$
这两个向量的叉乘给出平面的法向量： $\mathbf{n} = \overrightarrow{EF}\times \overrightarrow{EC_1} = (1,1,0)\times(1,2,2).$ 计算行列式可得 $\mathbf{n} = (2,-2,1).$ 因此平面方程可写为 $2x - 2y + z + D = 0.$
令该平面过点 $E(1,0,0)$ 来求 $D$ ： $2\cdot 1 - 2\cdot 0 + 0 + D = 0 \quad\Longrightarrow\quad D = -2.$
因而所求平面方程即

2x - 2y + z - 2 = 0 \quad\Longleftrightarrow\quad z = 2 - 2x + 2y.

2. 立方体中“低于此平面”的区域

从物理上看，极限盛水区即

\bigl\{(x,y,z)\;\big|\;0 \le x,y,z\le 2,\;\;z \le 2 - 2x + 2y\bigr\}.

但还要注意，立方体本身顶部在 $z=2$ ，因此真正满足的条件应是

0 \;\le\; z \;\le\; \min\bigl(2,\;2-2x+2y\bigr),

并且当然

0 \;\le\; x \;\le\; 2,\quad 0 \;\le\; y \;\le\; 2.

同时，若 $2 - 2x + 2y<0$ ，就没有可盛水的“高度”可言（因为 $z\ge 0$ ），所以还需

2 - 2x + 2y \;\ge\; 0 \;\;\Longleftrightarrow\;\; y \;\ge\; x-1.

由此可见， $x,y$ 应在方形 $[0,2]\times[0,2]$ 里，再额外满足 $y \ge x-1$ 。几何上，直线 $y=x-1$ 只有当 $x\ge 1$ 时才进入“正的 $y$ 区域”；若 $x<1$ ，则 $x-1<0$ ，对 $y\ge 0$ 并无额外限制。

2.1 按平面与立方体上界的相对位置分区

令

H(x,y) = 2 - 2x + 2y.

那么在不同的 $(x,y)$ 区域， $\min(2,\;H(x,y))$ 的数值会不同：

当 $H(x,y)\le 2$ （即 $2 -2x+2y \le 2$ ），水面实际被该“倾斜平面”所限制，盛水高度为 $z\le H(x,y)$ 。
这一条件等价于 $2 - 2x + 2y \;\le\; 2 \quad\Longleftrightarrow\quad -2x + 2y \;\le\; 0 \quad\Longleftrightarrow\quad y \;\le\; x.$
当 $H(x,y)\ge 2$ ，盛水高度则被立方体顶面 $z=2$ 限制， $z\le 2$ 。这对应 $2 - 2x + 2y \;\ge\; 2 \quad\Longleftrightarrow\quad y \;\ge\; x.$

再合并 $y\ge x-1$ 和 $0\le x,y\le 2$ 的限制，就可把 $xy$ -平面上可行区分成两块：

区域 A： $y\ge x-1$ 且 $y\le x$ 。这时水的最高面由平面给出，即 $z \le H(x,y) = 2 -2x +2y$ 。
区域 B： $y\ge x-1$ 且 $y\ge x$ 。这时水可一直装到立方体顶 $z=2$ 。

在具体计算时，还要注意 $x<1$ 时， $y\ge x-1$ 对 $y\ge 0$ 无更紧的约束，所以往往会把“ $x<1$ ”与“ $x\ge 1$ ”再分段处理。

3. 分块计算体积

令所求体积 $V = V_A + V_B$ 分别对应以上区域 A 和 B。

3.1 区域 A：高度由平面限制

对于区域 A，我们要做的三重积分实际上变成

V_A \;=\; \iint_{\text{A区}} \Bigl(\underbrace{\int_{z=0}^{\,H(x,y)} 1\,dz}_{=H(x,y)}\Bigr) dx\,dy \;=\; \iint_{\text{A区}} \bigl(2 - 2x + 2y\bigr)\,dx\,dy.

这里 A 区是

\{(x,y)\mid 0\le x\le 2,\;0\le y\le 2,\;y\ge x-1,\;y\le x\}.

几何上可再分两段：

当 $0\le x<1$ 时， $x-1<0$ ，所以 $y \ge 0$ 自动强于 $y\ge x-1$ ，再加 $y\le x$ ，故 $0 \le y \le x$ 。
当 $1\le x\le 2$ 时， $y\ge x-1$ 和 $y\le x$ 共同给出 $x-1 \le y \le x$ 。

于是分成两块积分：

$x\in[0,1]$ ：

V_{A1} = \int_{x=0}^{1}\int_{y=0}^{x} [2 - 2x + 2y]\,dy\,dx.

$x\in[1,2]$ ：

V_{A2} = \int_{x=1}^{2}\int_{y=x-1}^{x}[2 - 2x + 2y]\,dy\,dx.

计算 $V_{A1}$

先对 $y$ 积分：

\int_{y=0}^{\,x} (2 - 2x + 2y)\,dy = \int_0^x 2\,dy \;-\; 2x\int_0^x 1\,dy \;+\; 2\int_0^x y\,dy

= 2x \;-\; 2x\cdot x \;+\; 2\cdot \frac{x^2}{2} = 2x \;-\; 2x^2 \;+\; x^2 = 2x - x^2.

再对 $x$ 从 0 到 1 积分：

V_{A1} = \int_0^1 (2x - x^2)\,dx = \left[x^2 - \tfrac{x^3}{3}\right]_0^1 = (1 - \tfrac{1}{3}) = \tfrac{2}{3}.

计算 $V_{A2}$

先对 $y$ 积分：

\int_{y=x-1}^{\,x} (2 - 2x + 2y)\,dy = 2\bigl[y\bigr]_{x-1}^x \;-\;2x\bigl[y\bigr]_{x-1}^x \;+\;2\bigl[\tfrac{y^2}{2}\bigr]_{x-1}^x.

具体展开：

第一个部分 $2[y]_{x-1}^x = 2(x - (x-1)) = 2.$
第二个部分 $-2x[y]_{x-1}^x = -2x\,\bigl(x-(x-1)\bigr) = -2x.$
第三个部分 $2[\tfrac{y^2}{2}]_{x-1}^x = [y^2]_{x-1}^x = x^2 - (x-1)^2 = 2x-1.$

三项相加得

2 \;-\; 2x \;+\; (2x-1) = 1.

再对 $x$ 从 1 到 2 积分：

V_{A2} = \int_{x=1}^{2} 1\,dx = [x]_1^2 = 1.

因此

V_A = V_{A1} + V_{A2} = \frac{2}{3} + 1 = \frac{5}{3}.

3.2 区域 B：高度被顶面限制

对于区域 B，我们有 $\min(2,\;H(x,y))=2$ ，所以水柱高度恒为 2。此时

V_B = \iint_{\text{B区}} 2 \,dx\,dy,

其中 B 区是

\{(x,y)\mid 0\le x\le 2,\;0\le y\le 2,\;y \ge x,\;y \ge x-1\}.

但由于 $y\ge x$ 本身就比 $y\ge x-1$ 更强(当 $x\ge 1$ 时二者都要满足，不过 $y\ge x$ 已含盖 $y\ge x-1$ )，故 B 区本质上就是方形 $[0,2]\times[0,2]$ 中“在直线 $y=x$ 之上”的三角形。

即对每个 $x\in[0,2]$ ， $y$ 从 $y=x$ 到 2，于是

V_B = \int_{x=0}^{2}\int_{y=x}^{2} 2 \,dy\,dx = \int_{x=0}^{2} \Bigl[2y\Bigr]_{y=x}^{y=2} dx = \int_{x=0}^{2} [2\cdot 2 \;-\; 2\cdot x]\,dx

= \int_{x=0}^{2} (4 - 2x)\,dx = \left[4x - x^2\right]_0^2 = (8 - 4) - 0 = 4.

4. 总体积

将两部分相加：

V = V_A + V_B = \frac{5}{3} + 4 = \frac{5}{3} + \frac{12}{3} = \frac{17}{3}.

这就是容器能够盛水的最大体积。若边长取 2(而非 1)的立方体，则单位体积同样给出数值

\boxed{\frac{17}{3}},

也可以写作 $5\frac{2}{3}\approx 5.6667$ 。