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在平面四边形ABCD中，AB = AC = CD = 1,\angle ADC = 30^{\cir...

Created on: February 19, 2025

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Question

在平面四边形ABCD中，AB = AC = CD = 1,\angle ADC = 30^{\circ},\angle DAB = 120^{\circ}。将\triangle ACD沿AC翻折至\triangle ACP，其中P为动点。 求二面角A - CP - B的余弦值的最小值。

Thought

Analyzing dihedral angle

I’m given a quadrilateral ABCD with AB = AC = CD = 1, ∠ADC = 30°, and ∠DAB = 120°. The next step involves folding triangle ACD along AC to form triangle ACP, where P is a moving point.

Now, I'll need to find the minimal value of the cosine of the dihedral angle A-CP-B, which pertains to the angle between the faces containing points A and B around line CP. Let’s figure this out!

Clarifying dihedral angle

I’m thinking about the notation "A - CP - B" for the dihedral angle. It likely means the angle between planes A-CP and CP-B, or the planes formed by points A, C, P and C, P, B.

However, the tricky part is interpreting "B." In the original quadrilateral ABCD, B isn’t part of triangle ACD, yet it’s involved in the dihedral angle after the fold.

This angle seems to involve the new configurations of planes after folding triangle ACD along AC. Let’s carefully unpack this!

Clarifying the folding and dihedral angle

The problem involves folding triangle ACD along AC to form triangle ACP, where P is the new position of D. The task asks for the minimal cosine of the dihedral angle A-CP-B, which is likely the angle between the plane containing A, C, P and the plane containing B, C, P. P is influenced by the folding angle, making it a moving point that changes with the fold. I'm starting to see the connection to the quadrilateral's original constraints and the fold's effect on the dihedral angle!

Folding and dihedral angle analysis

We start with quadrilateral ABCD and fold triangle ACD along AC, moving D to P, changing the quadrilateral into a bent shape. The dihedral angle A-CP-B likely refers to the angle between planes ACP and BCP, sharing the edge CP. The position of P varies based on the folding angle, which is a free parameter. We aim to find the minimal cosine of the angle between these planes as P moves, while keeping the given lengths and angles intact. I'll continue with this setup!

Solving for coordinates and folding

We have the quadrilateral ABCD in the plane. With the given lengths and angles, I've successfully placed the vertices, including determining point D at (3/2, √3/2) from several geometric calculations. Then, I found B's position by considering the angle constraints between AD and AB. I now have B at (0, -1), ensuring the quadrilateral is convex. All coordinates are set to reflect the original plane structure. Next, folding triangle ACD along line AC will place point P where D folds. Time to move ahead with the folding!

Computing dihedral angle with folding

We are folding triangle ACD around line AC by angle φ, causing D to move to P. We want to find the minimal cosine of the dihedral angle between the planes of triangles ACP and BCP as φ varies. The normal vectors to both planes give us that dihedral angle cosine. By computing the cross products of vectors, we obtained the normals of planes ACP and BCP, with respect to edge CP. Now, we just need to adjust for the perpendiculars and compute the minimum cosine as φ changes!

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Calculating dihedral angle

I’m working through the cosine of the dihedral angle. First, the lengthy algebraic sum simplifies as I combine terms like cos φ and sin φ, eventually leading to a simpler form: cos δ = [2(√3 cos φ - 3)] / [√3 * sqrt(28 - 8√3 cos φ - 12 cos^2 φ)].

I’m considering the physical meaning: when φ = 0, the cosine becomes equivalent to a specific value. The folding angle φ ranges from 0 to π, and I’m aiming to find the minimum of this expression. The expression depends on cos φ, and I must evaluate its behavior to ensure it's physically valid.

Simplifying and evaluating cosine

I'll simplify F(t) first by considering the constant factors and differentiating. The minimum cosine corresponds to the maximum dihedral angle. From earlier, when t = 1 or t = -1, it shows that cos δ reaches -1, which suggests a dihedral angle of 180° — though this conflicts with the physical situation. So, I need to check if these endpoints are physically valid, considering the right geometric interpretation. When φ = 0, it seems there's a planar quadrilateral and the angle is 0.

Evaluating projection and dihedral angle

Okay, let's think through this more clearly: In the planar quadrilateral setup, the dihedral angle between the planes is something we want to calculate using projections, and I need to be cautious about the sign. When φ = 0, ideally, the dihedral angle should be 0, but I got -1 in my formula. This suggests the formula might be giving the cosine of the supplementary angle instead. For φ = π, I expect the same, but I need to be sure about the orientation of the normals.

Reasoned for 2m 28s

Answer

我们下面给出一种解法的大致思路，最后证明答案为

13 .\frac{1}{\sqrt3}\,.

下面说明主要步骤：

  1. 建立模型并确定各边长和角度。
    设四边形 ABCDABCD 为平面图形，已知

    AB=AC=CD=1,∠DAB=120∘,∠ADC=30∘.AB=AC=CD=1,\quad \angle DAB=120^\circ,\quad \angle ADC=30^\circ.

    先考虑 △ACD\triangle ACD 中，已知 AC=CD=1AC=CD=1 和 ∠ADC=30∘\angle ADC=30^\circ。由余弦定理（以 ACAC 为对边）有

    AC2=AD2+CD2−2⋅AD⋅CDcos⁡30∘,AC^2=AD^2+CD^2-2\cdot AD\cdot CD\cos30^\circ,

    即

    1=AD2+1−2AD⋅32⟹AD2−3 AD=0.1=AD^2+1-2AD\cdot\frac{\sqrt3}{2}\quad\Longrightarrow\quad AD^2-\sqrt3\,AD=0.

    排除 AD=0AD=0 后得

    AD=3.AD=\sqrt3.

  2. 在坐标系中定位各点。
    可取

    A=(0,0,0),C=(1,0,0).A=(0,0,0),\quad C=(1,0,0).

    由于 AB=1AB=1且 ∠DAB=120∘\angle DAB=120^\circ，选取 B=(0,−1,0)B=(0,-1,0)（这样使得四边形凸且方便计算）。又在平面上可求出 DD（利用 AD=3AD=\sqrt3 和 CD=1CD=1）为

    D=(32, 32, 0).D=\Bigl(\tfrac{3}{2},\,\tfrac{\sqrt3}{2},\,0\Bigr).

  3. 折叠操作。
    将 △ACD\triangle ACD 沿固定边 ACAC折起，使 DD经过绕 ACAC的旋转而落到新位置 PP。记折起的角（旋转角）为 φ\varphi（当 φ=0\varphi=0时，P=DP=D），由于 ACAC在我们所选坐标中为 xx 轴，对 DD绕 xx 轴旋转得

    P=(32, 32cos⁡φ, 32sin⁡φ).P=\Bigl(\frac{3}{2},\,\frac{\sqrt3}{2}\cos\varphi,\,\frac{\sqrt3}{2}\sin\varphi\Bigr).

  4. 求二面角。
    折叠后四边形的顶点为 A,B,C,PA,B,C,P。我们要求的是“二面角 A−CP−BA-CP-B”的余弦值，此处的记号表示沿公共棱 CPCP测量，两侧分别在平面 π1=△ACP\pi_1=\triangle ACP和 π2=△BCP\pi_2=\triangle BCP上。常用的方法是：在 CC点取两个分别位于这两个平面内且垂直于 CPCP的辅助向量，然后求它们之间的夹角。
    例如，可取

    u=\mbox投影((A−C)),v=\mbox投影((B−C))\mathbf{u}=\mbox{投影}\bigl((A-C)\bigr),\quad \mathbf{v}=\mbox{投影}\bigl((B-C)\bigr)

    到 CPCP的垂直平面上。经过计算（细节较多，见下面的提示），可得

    cos⁡δ=2(3−3cos⁡φ)3 28−83cos⁡φ−12cos⁡2φ,\cos\delta=\frac{2\Bigl(3-\sqrt3\cos\varphi\Bigr)} {\sqrt3\,\sqrt{28-8\sqrt3\cos\varphi-12\cos^2\varphi}},

    其中 δ\delta就是二面角。

  5. 求极值。
    证明计算可知：当 φ=0\varphi=0（或 φ=π\varphi=\pi）时，四边形仍在同一平面内，此时二面角 δ=0\delta=0（余弦值为1）；而当 φ\varphi取到某个中间值时（二面角达到最大，即构成“最开”的折痕）余弦值达到最小。把上式看作关于 t=cos⁡φt=\cos\varphi的函数，经求导计算可知，唯一的极值点在

    t=13即φ=arccos⁡(13).t=\frac{1}{\sqrt3}\quad\text{即}\quad \varphi=\arccos\Bigl(\frac{1}{\sqrt3}\Bigr).

    代入后可得

    cos⁡δmin⁡=13 .\cos\delta_{\min}=\frac{1}{\sqrt3}\,.

综上所述，二面角 A−CP−BA-CP-B 的余弦值在全体折叠位置中的最小值为

13 .\boxed{\frac{1}{\sqrt3}}\,.

【提示】

  • 在实际计算中可先将 A=(0,0,0), C=(1,0,0), B=(0,−1,0)A=(0,0,0),\,C=(1,0,0),\,B=(0,-1,0)及 D=(3/2,3/2,0)D=(3/2,\sqrt3/2,0)确定下来；
  • 把 DD绕 ACAC（即 xx轴）旋转角 φ\varphi得 PP；
  • 对于每一面（△ACP\triangle ACP与△BCP\triangle BCP），在 CC点取一非零向量（例如 A−CA-C和 B−CB-C），然后作去除其在 CPCP方向上的分量；
  • 由这两个投影向量的夹角就得到二面角；
  • 最后对 φ\varphi（或其余弦值 tt）求极值即可。

答案就是 13\displaystyle\frac{1}{\sqrt3}.

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