二面角余弦最小值

Created on: January 21, 2025

Answered using O1 by Chat01

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

Breaking down the fold

I’m picturing triangle ACD being folded along AC to form triangle ACP, with P as a moving point, and a dihedral angle at issue. This setup invites a geometric exploration.

Clarifying the angle

OK, let me see. I’m figuring out that the problem might involve the dihedral angle between the planes (ACP) and (BCP), formed by the points and lines specified in the notation.

Mapping out the planes

I’m thinking through the dihedral angle between the planes (ACP) and (BCP), as the notation "二面角 A - CP - B" in Chinese geometry typically represents this.

Determining the hinge

I’m mapping out that CP acts as the hinge or intersection line for planes ACP and BCP, with P moving while D reflects across AC in 3D space.

Reflecting upon

I'm thinking through the reflection of point D across plane AC, creating point P while keeping side AC unchanged. This raises questions about the dihedral angle between planes ACP and BCP.

Laying out the options

Reflecting on the dihedral angle between planes ACP and BCP, where the critical intersection is line CP, is consistent with the interpretation so far.

Fixing coordinates

I’m setting up a coordinate system for points A, B, C, and D. This helps in parameterizing point D's movement and analyzing the dihedral angle between planes ACP and BCP.

Aligning points

Starting with A at the origin and AC along the x-axis, I’m placing points B and D to fit the given constraints, using coordinate geometry for accuracy.

Figuring out the setup

I’m starting with A at (0,0) and C at (1,0), placing B on a circle radius 1 around A. Now, I’m trying to find D with the given angle constraints.

Setting up coordinates

Placing A at (0,0) and C at (1,0), I’m positioning D to satisfy ∠ADC=30° and CD=1. This leads me to coordinate B using distance and angle constraints.

Mapping D's location

I’m thinking about placing D on a circle of radius 1 around C, parameterized by φ. This setup helps determine the angle ∠ADC as 30°, considering vectors DA and DC.

Verifying dot product

Mapping the dot product of vectors DC and DA involves |DC|=1 and verifying |DA| by computing the vector lengths to confirm they align with a 30° angle.

Calculating vector length

I’m thinking through the length of vector DA by squaring and adding the components' squares, then taking the square root, applying the identity to simplify further. This is interesting progress.

Confirming angle measurements

Taking a closer look at the dot product between vectors DC and DA, equating cosines, and solving. I conclude that φ equals 60°, aligning with the given angle constraints.

Mapping coordinates

I’m setting D's coordinates to satisfy the given angles by noting two possible values for φ: 60° and 300°. Now, I'm confirming these satisfy ∠DAB=120°.

Calculating the angle

I'm working out the angle ∠DAB by comparing the dot product of vectors AD and AB with the cosine of 120°, using the absolute value of |AD|. Progressing through the calculations carefully.

Analyzing dot product

I’m digging into φ=60° to determine if vectors AD and AB provide a dot product of -√3/2, ensuring the angle ∠DAB meets the given condition.

Determining amplitude feasibility

I’m examining whether the range of the function f(α) accommodates -√3/2 for φ=60°, confirming that the angle ∠DAB can indeed be 120°.

Ensuring document consistency

Confirming that the constraints AB=1 and AC=1 are satisfied. With AC fixed, B's position can be adjusted to meet the required angles.

Triangular transformation

Rotating triangle ACD in 3D about the fixed line AC, ensuring distances from A, C, and D to P match their equivalents. The shape remains consistent.

Figuring out P's position

I’m thinking about how P’s placement in 3D varies with the fold angle θ, and what that means for the dihedral angle between planes ACP and BCP, with B staying in place.

Figuring out the planes

Now I'm thinking B stays put while P moves in 3D, resulting in planes ACP and BCP intersecting along CP. This focus on alignment sharpens the dihedral angle's geometric significance.

Figuring out intersections

I’m working through how the intersection of planes ACP and BCP forms line CP, which leads me to explore the dihedral angle between these planes about line CP.

Utilizing direction vectors

Spotlighting vectors CA and CB, I project them onto the plane perpendicular to CP, focusing on the angle between these vectors to simplify the identification of the dihedral angle between planes ACP and BCP.

Laying out the options

I’m thinking through the dihedral angle by examining the lines in each plane perpendicular to CP, projecting CA and CB onto a perpendicular subspace to CP.

Focusing on projections

I’m working through defining vectors CA and CB, projecting them onto a plane perpendicular to CP, and calculating the minimal cosine of the dihedral angle ψ as the fold angle changes.

Rotating coordinates

I’m mapping out the rotation of D about the x-axis to locate P, defining planes ACP(t) and BCP(t), and calculating the dihedral angle's cosine as φ=60°.

Examining fold variations

Rotating triangle ACD around AC changes D's position, impacting planes ACP and BCP. The dihedral angle's cosine might be smallest at specific fold angles between 0° and 180°.

Wrestling with angle's impact

I'm thinking about how the minimal cos value aligns with the largest dihedral angle between planes ACP and BCP, given constraints and the constant length of CP.

Determining dihedral angle

I’m focusing on the angle between lines AC and BC projected onto a plane perpendicular to CP, which simplifies the problem. This approach seems the most straightforward.

Harnessing vector tools

I'm thinking through vector identities to find the cos of the dihedral angle between planes ACP and BCP, using cross-products to determine normals and their dot product. Progress is evident.

Pinpointing relationships

I’m honing in on the relationship between the dot product of CA and CB, and their cross-product with CP, to streamline the derivation of the dihedral angle formula. This feels promising.

Revisiting vector forms

OK, let me see. Reassigning vectors to simplify the cross-product formula for normal vectors to planes ACP and BCP aids in deriving the dihedral angle cosine calculation.

Mapping vectors

I’m focusing on vectors AC, BC, and CP, ensuring their magnitudes and orientations align for the cross-product and dot product calculations. This approach enhances the dihedral angle formula.

Mapping D's rotation

OK, let me see. I’m figuring out D's rotation around line AC by tracking its changes in height and how it affects B's position, all while keeping the rotation smooth and continuous.

Charting the rotation

I think about how D's movement will only change D's z-coordinate so B stays aligned with the x-axis. B's y-coordinate stays put, and B mirrors D's z-coordinate.

Locating the shortest distance

It's interesting to see that to find D's nearest point on line AC, I’m deriving coordinates and examining the x-coordinate to minimize squared distance.

Pinpointing the rotation center

The contact with the x-axis segment AC is actually at (1.5,0,0). D’s rotation creates a circle with a radius of √3/2 centered at (1.5,0,0) in a plane perpendicular to the x-axis.

Piecing together the circle

I’m describing how D's movement forms a circle in the yz-plane, parameterized by t, centered at (1.5,0,0), with a radius of √3/2.

Piecing together vectors

I’m calculating vectors AD and AB, ensuring the dot product and angle constraints align to maintain ∠DAB=120° and AB=1. This involves a detailed analysis of their geometric relationships.

Solving for α

Adjusting the equation to find α by making sure the relationship between trigonometric functions and their constraints is consistent. It’s interesting to see the progression of simplification and issue-solving.

Mapping out trigonometric identities

OK, let me see: I’m equating sine and cosine values using trigonometric identities to solve for α. This involves manipulating the equation to ensure all constraints and relationships are consistent.

Cracking problem constraints

I’m solving for sin α, finding that sin α=1/2 or sin α=-1. This leads to α=270°, making B at (0, -1, 0) a valid solution.

Piecing together

I’m figuring out α to ensure all constraints are satisfied, leading to two possible B coordinates: (0, -1, 0) and (-√3/2, 1/2, 0). I’ll pick B= (0, -1, 0).

Figuring out cross product

I'm calculating the cross product of vectors AC and CP to ensure normal vectors to planes ACP and BCP align, simplifying the computation of the dihedral angle's cosine.

Mapping cross product

I’m working through calculating the cross product of vectors AC and CP. This ensures normal vectors to planes ACP and BCP align, streamlining the dihedral angle's cosine calculation.

Piecing together

I’m clarifying the cross product of vectors AC and CP. Ensuring normal vectors to planes ACP and BCP are properly aligned reveals, simplifying the cosine calculation for the dihedral angle.

Calculating the cross product

I’m using the standard formula for cross products, ensuring the normal vectors to planes ACP and BCP align to simplify the cosine calculation for the dihedral angle.

Working through the cross product

I’m calculating the cross product of vectors c and b, ensuring the normal vectors of planes ACP and BCP are aligned, simplifying the cosine calculation of the dihedral angle.

Piecing together

I’m ensuring the normal vectors of planes ACP and BCP are aligned by calculating their dot product, simplifying the cosine calculation for the dihedral angle. This effectively harmonizes the trigonometric terms.

Taking a closer look

Confirming that the normal vectors |n1| and |n2| are consistent and independent of t, leading to an accurate dihedral angle calculation.

Aligning angle relationships

I’m working through dihedral angle calculations by ensuring the normal vectors |n1| and |n2| are consistently aligned, leading to an accurate simplification and calculation.

Aligning the parameters

Recalculating the dot product and magnitudes of normal vectors n1 and n2 ensures the dihedral angle's cosine calculation is accurate by aligning their magnitudes and dot product.

Simplifying the expression

I’m simplifying the cosine expression for the dihedral angle, converting 1.75 to 7/4, and examining the minimal value for t in [0, π] using a function f(t) to uncover the minimum.

Simplifying expressions

I’m breaking down the expressions for N(t) and D(t) to minimize cos(ψ). Checking endpoints or critical points where the derivative equals zero to identify the minimum value for t in [0, π].

Mapping values

OK, let me see. To find the minimum value of cos(ψ) for t in [0, π], I’m starting with endpoint analyses to estimate ψ's proximity to 0.

Pulling together

Checking the endpoint t=π, cos t=-1, sin t=0. Simplifying, I estimate ψ around π. This matches our earlier conclusion cos(ψ)≈1, indicating a near 0 dihedral angle.

Evaluating midpoints

I’m looking at t=π/2 to see if the dihedral angle's cosine falls below 1. The value of f(π/2) is around 0.654, suggesting there might be a larger angle at this midpoint.

Confirming positivity

I’m digging into the denominator for cos t in [0, π]. Both terms remain positive as t ranges, which consistently yields a positive value for cos(ψ).

Piecing together

I’m thinking through different t values, like π/3 and 2π/3, to evaluate and pinpoint the minimal positive value of cos(ψ) in [0, π].

Evaluating t values

I’m working through values like π/3, 2π/3, π/6, and π/4 to find the minimal positive value of cos(ψ) within [0, π], comparing these to previous t values for the best result.

Analyzing π/6

I’m calculating cos(ψ) for π/6, finding it approximately 0.655, which is higher than the previous minimum of around 0.58 at π/3.

Analyzing π/4

Interestingly enough, π/4 yields a cos(ψ) of around 0.586, very close to the minimum, but slightly higher than the π/3 result of 0.58.

Confirming the minimum

Piecing together derivative tests and sampling, t=π/3 almost certainly represents the minimum cos(ψ) around 0.58.

Refining calculations

OK, let's fix the precision error at t=π/3. With exact values, N(π/3)=0.5335 and D(π/3)=0.920, leading to f(π/3)=0.581.

Analyzing function dynamics

Defining f(t) requires a detailed look at N(t) and X(t). Sampling t around π/3 and higher values reveals critical points for refining the analysis.

Analyzing function behavior

I’m recalculating N(t) and X(t) for t=1 radian, observing a slight decrease to f(1)=0.578. Moving to t=1.1 radians, I notice an increase to f(1.1)=0.5815.

Narrowing down

I’m verifying N(t) and X(t) for t=0.95 and t=0.90, observing minimal values of 0.577. This hints that the minimum might be around t=0.95.

Establishing connections

I’m mapping out the relationship between f(t) and 1/√3, noticing the angles resemble tetrahedron angles, suggesting ψ might be around 54.7356 degrees.

Mapping the equation

I'm expanding both sides of the equation and noticing the relationship between terms involving cos t, aiming to solve for t and find the minimum value of f(t).

Utilizing quadratic formula

I’m solving the equation 3 cos^2 t - 2√3 cos t + 1 = 0 to find solutions for cos t, aiming to determine the value of t that minimizes f(t).

Thought about 二面角余弦值最小值 for 3m 55s

Answer

分析思路简述：

确定平面中点的布局（在 2D 中先固定 $A,C,D$ ）
题目给出的条件是
$AB=AC=CD=1,\quad \angle ADC=30^\circ,\quad \angle DAB=120^\circ.$
我们先在平面内（记作 $xy$ 平面）固定点
$A=(0,0),\quad C=(1,0).$
接下来根据 $\angle ADC=30^\circ$ 且 $CD=1$ 来放置 $D$ 。
设 $D=(x_D,y_D)$ ；因为 $CD=1$ ，可令
$D = \Bigl(1+\cos\varphi,\;\sin\varphi\Bigr),$
并利用向量 $DC,\,DA$ 夹角为 $30^\circ$ 来求 $\varphi$ 。
最终可得两种可能的 $\varphi$ ，其中取
$\varphi=60^\circ$
时，便有
$D=\Bigl(1+\tfrac12,\;\tfrac{\sqrt3}{2}\Bigr)=\bigl(1.5,\;\tfrac{\sqrt3}{2}\bigr).$
利用 $\angle DAB=120^\circ$ 确定点 $B$
还需满足 $\angle DAB=120^\circ$ 以及 $AB=1$ 。令
$B=(\cos\alpha,\;\sin\alpha),\quad |AB|=1$
则 $\vec{AD}\cdot \vec{AB}=|AD|\cdot|AB|\cdot \cos 120^\circ$ .
由此可解出若干可行的 $\alpha$ ，其中一个简便可行解是
$B=(0,\,-1).$
（也可以得到另一组解，但几何上往往是对称情形。）

于是此时在平面内，我们有
$A=(0,0),\quad B=(0,-1),\quad C=(1,0),\quad D=\bigl(1.5,\;\tfrac{\sqrt3}{2}\bigr).$
“折叠”操作在三维中的参数化——让 $D$ 绕轴 $AC$ 旋转到 $P$
- 由于要“沿 $AC$ 折叠三角形 $ACD$ ”，可以理解为：固定直线 $AC$ 作为“铰链”，将点 $D$ 绕这条轴转到空间某点 $P$ 。
- 在上述坐标系里， $A=(0,0,0)$ 、 $C=(1,0,0)$ （把原平面当作 $z=0$ ），那么 $AC$ 就是一段在 $x$ -轴上、从 $x=0$ 到 $x=1$ 的线段。实际“绕轴”可以看作绕整条与 $x$ -轴重合的直线旋转。
具体来看，先找到 $D$ 到这条 $x$ -轴(延长线)的“最近点”以及“半径”；对本题数值而言，
$D = (1.5,\;\tfrac{\sqrt3}{2},\;0)$
距离 $x$ -轴最近点在 $\bigl(1.5,\,0,\,0\bigr)$ ，半径为 $\tfrac{\sqrt3}{2}$ 。
因而将 $D$ 绕 $x$ -轴转一个角度 $t$ （从 0 到 $\pi$ ），可写作
$P(t) \;=\;\bigl(1.5,\;\tfrac{\sqrt3}{2}\cos t,\;\tfrac{\sqrt3}{2}\sin t\bigr).$
当 $t=0$ 时， $P(0)=D$ （尚未折起）；当 $t=\pi$ 时，则“翻”到对面。
二面角 $\angle(A - CP - B)$ 的含义
这是指以线段 $CP$ 为“棱”的那两个平面 $\triangle ACP$ 与 $\triangle BCP$ 之间的二面角。记该二面角为 $\psi$ 。我们要研究 $\cos \psi$ 随着折叠角度 $t$ 的变化，并找它的最小值。
用向量法计算二面角
- 设 $\vec{a}=\overrightarrow{AC},\quad \vec{b}=\overrightarrow{CP},\quad \vec{c}=\overrightarrow{BC}.$ 则平面 $ACP$ 的法向量可取 $\vec{n_1}=\vec{a}\times \vec{b}$ ，平面 $BCP$ 的法向量可取 $\vec{n_2}=\vec{c}\times \vec{b}$ 。
  二面角 $\psi$ 满足 $\cos\psi \;=\; \frac{\vec{n_1}\cdot \vec{n_2}}{\bigl|\vec{n_1}\bigr|\;\bigl|\vec{n_2}\bigr|}.$
- 按前述坐标取值： $A=(0,0,0),\quad C=(1,0,0),\quad B=(0,-1,0),\quad P(t)=\bigl(1.5,\;\tfrac{\sqrt3}{2}\cos t,\;\tfrac{\sqrt3}{2}\sin t\bigr).$ 那么 $\vec{a}=\overrightarrow{AC}=(1,0,0),$ $\vec{b}=\overrightarrow{CP}=\bigl(0.5,\;\tfrac{\sqrt3}{2}\cos t,\;\tfrac{\sqrt3}{2}\sin t\bigr),$ $\vec{c}=\overrightarrow{BC}=(1,1,0).$ 代入做交叉乘、点乘等运算，就可以得到 $\cos\psi$ 作为 $t\in [0,\pi]$ 的一个函数。
数值/代数化简与求极值
经过一番（较为细致的）向量代数，可以把
$\cos\psi = f(t)$
写成一个关于 $\cos t$ 的有理函数（再套一个平方根），然后在 $t\in [0,\pi]$ 上找最小值。下列是要点：
- 当 $t=0$ 或 $t=\pi$ （即不折或折到对面）时，二面角其实变成 0，故 $\cos\psi$ 都是 1。
- 在中间某处会出现一个向下的极小值。数值检验后可猜测它约在 $t\approx 0.95$ 附近，且最小值大约为 0.577 $\dots$ 。
- 更进一步做代数求导或令 $\cos\psi=1/\sqrt{3}$ 之类的方式，可以精确解得 $\text{(极小值点满足)}\quad \cos t = \frac{1}{\sqrt{3}} \;\;\Longrightarrow\;\; \cos\psi_{\min} = \frac{1}{\sqrt{3}}.$
换言之，当把三角形 $ACD$ 折到某个特定的空间位置时，上述二面角的余弦正好达到 $\tfrac{1}{\sqrt{3}}$ ，并且这是它的最小值。

最终结论

该二面角 $\angle(A - CP - B)$ 的 $\cos$ 值最小值 为
$\boxed{\;\frac{1}{\sqrt{3}}\;}.$

Question

Thought

Answer

最终结论

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