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等边三角形几何问题

Created on: December 31, 2024

Answered using O1 by Chat01

Question

在等边三角形ABC中，点E,D分别在AC,BC上，且AE=CD，AD和BE相交于P点，过B点向AD做垂线，Q点是AD上的垂足。连接PC，如果BP垂直于PC，求AE/EC的值。请用中文回答

Thought

Placing and solving

I’m working with an equilateral triangle ABC and positioning it on a coordinate plane. AE equals CD, and I’m figuring out AE/EC by analyzing line intersections and perpendicular lines.

Placing the triangle

I’m establishing a coordinate system for the equilateral triangle ABC. By setting A at (0, 0) and B at (1, 0), the side length is 1, ensuring AC reflects this distance.

Tracking E and D

I’m figuring out E's coordinates on AC using a parameter x. AE = CD translates to x = 1 - y, with E at (x/2, x*sqrt(3)/2) and D at (1 - x, 0).

Pinpointing E and D

I’m locking down E's position on AC with parameter x, yielding E at (x/2, x*sqrt(3)/2). Now, D's movement on BC needs to align with AE equaling CD.

Mapping out D's coordinates

I pinpointed D's location on BC, ensuring AE equals CD. With D at (1 - (1-x)/2, (1-x)*sqrt(3)/2), this confirms the correct alignment.

Breaking down and detailing

I’ve clearly defined D's coordinates as ((1 + x)/2, (1 - x)*sqrt(3)/2). Now, I’m starting to derive the equations for lines AD and BE using these points.

Mapping intersections

First, I’m outlining parametric equations for lines AD and BE. By setting them equivalent, I’m working out where AD and BE intersect, leading to the intersection point P coordinates.

Mapping intersections

Okay, let me see. I derived s in terms of t and x from equation (2), then substituted s into equation (1). This approach clarifies the intersection coordinates for P.

Confirming calculations

I’m checking the algebraic steps by aligning terms and verifying the common denominator to ensure precision in calculating P's intersection coordinates.

Solving for t

OK, let me see. Combining terms with t and factoring out t, I’m rearranging the equation to isolate t. This leads me to solve for t in terms of x.

Simplifying the equation

I’m finding a common denominator, combining terms, and isolating t. This makes the equation much simpler, allowing me to solve for t in terms of x.

Distributing and expanding

I’m breaking down (1-x)*(x-2) to get x - 2 - x^2 + 2x = -x^2 + 3x - 2. This expansion clarifies the expression for t in terms of x.

Isolating t

I’m pulling together terms to isolate t in the equation, ultimately expressing t as x / (x^2 - x + 1). This isolates t in terms of x.

Determining P's position

I worked out P's coordinates by combining parameters t and s, using t = x / (x^2 - x + 1), and mapping out P's intersection on AD and BE.

Pinpointing the intersection

Interestingly enough, the coordinates of P are found by plugging x into a simplified form of AD and BE intersections. Now, I’m focusing on Q, the foot from B to AD.

Determining tQ

I’m figuring out tQ for perpendicularity by using the dot product of B and v, then calculating Q's coordinates, ensuring it lies directly on AD.

Mapping it out

Calculating B dot v, I found it equals v_x (1 + x)/2. This simplifies the process of finding tQ and Q's coordinates.

Mapping out v dot v

I calculated v_x and v_y to determine v dot v. This involves expanding and combining terms, ultimately simplifying to 1 - x + x^2, which aids in finding tQ and Q's coordinates.

Finding Q's coordinates

I’m defining Q's coordinates through tQ and v, then simplifying expressions. This involves isolating terms and verifying through substitution, ensuring Q lies on AD.

Verifying perpendicularity

OK, let’s ensure BP is perpendicular to PC by checking that (P - B) dot (C - P) equals zero. This will reveal AE/EC = x/(1-x).

Breaking down vectors

I’m piecing together BP and PC, ensuring they are perpendicular. By calculating their dot product and setting it to zero, I confirm BP is indeed perpendicular to PC.

Ensuring orthogonality

I’m making sure BP is orthogonal to PC by calculating their dot product and setting it to zero. This confirms BP's perpendicularity to PC.

Aligning coordinate vectors

I’m curious about how aligning B and P's coordinate vectors ensures BP is perpendicular to PC. This method simplifies the process and clarifies the intersection point P.

Figuring out perpendicularity

OK, let me see. By calculating the dot product of B and P's coordinate vectors, I’m ensuring BP is perpendicular to PC, leading to the required intersection point P.

Mapping intersections

Now I’m piecing together the dot product by applying FOIL to B's and P's coordinate vectors. This approach clarifies BP's perpendicularity to PC and their intersection point.

Simplifying expressions

I’m working through BP and PC's coordinates, factoring out common terms like 1/(2D), 1/(4D^2), and isolating px and py for easier evaluation.

Simplifying the equation

I’m piecing together the dot product terms, factor out constants, and simplify the equation, ultimately isolating variables and reaching the intersection point P for BP and PC.

Piecing it together

I’m squeezing out A^2 and B^2 terms, combining linear elements, and simplifying to finally express 3x / (2D). It’s interesting to see the progression.

Refining the expression

I’m thinking through the equivalent expressions for D and x^2 / D. Ensuring accuracy, I confirmed both terms as equivalent. This refinement simplifies the equation significantly.

Revisiting and refining

Interestingly enough, I’m simplifying M and N to [3x - 2x^2] / (2D). This leads me to equate them to 1/2, further simplifying to achieve the final equation.

Pinpointing the solution

Solving the quadratic equation, I find x = 1/3. This means AE/EC equals 1/2, ensuring E is accurately placed on AC and D on BC.

Thought about 等边三角形几何，持续 1m 11s

Answer

题目回顾：

在等边三角形 ABCABC 中，点 E,DE,D 分别在 AC,BCAC,BC 上，且满足

AE=CD.AE = CD.

直线 ADAD 与 BEBE 相交于点 PP。过点 BB 向 ADAD 作垂线，垂足为 QQ。再连接 PP 与 CC。已知此时 BP⊥PCBP \perp PC，要求 AEEC\displaystyle \frac{AE}{EC} 的值。

思路梗概

  1. 先利用坐标系将几何关系代数化。

  2. 设边长为 1，将等边三角形 ABCABC 放置在平面直角坐标系中：

    A=(0,0),B=(1,0),C=(12,32).A=(0,0),\quad B=(1,0),\quad C=\bigl(\tfrac12,\tfrac{\sqrt3}{2}\bigr).

  3. 令 EE 是 ACAC 上的分点，令它在 ACAC 上「走」了比例 xx，即

    E=A+x (C−A)=(0,0)+x(12,32)=(x2,x32), E = A + x\,(C-A) = \Bigl(0,0\Bigr) + x\Bigl(\tfrac12,\tfrac{\sqrt3}{2}\Bigr) = \Bigl(\tfrac x2,\tfrac{x\sqrt3}{2}\Bigr),

    其中 0<x<10<x<1。这样一来，AE‾\overline{AE} 的实际长度就是 x⋅∣AC∣x \cdot |AC|。由于 AC=1AC=1，可直接把 AEAE 当成 xx。

  4. 令 DD 是 BCBC 上的分点，若在 BCBC 上「走」了比例 yy，则

    D=B+y (C−B)=(1,0)+y(12−1,32−0)=(1−y2,y32). D = B + y\,(C-B) = (1,0) + y\Bigl(\tfrac12-1,\tfrac{\sqrt3}{2}-0\Bigr) = \Bigl(1-\tfrac y2,\tfrac{y\sqrt3}{2}\Bigr).

    同理，BD‾\overline{BD} 的长度是 yy，DC‾=1−y\overline{DC}=1-y。

  5. 题目给出 AE=CDAE=CD。由于 AE=xAE=x 且 CD=1−yCD=1-y，故

    x=1−y    ⟹    y=1−x. x = 1-y \;\;\Longrightarrow\;\; y=1-x.

    因而

    D=(1−1−x2,(1−x)32)=(1+x2, (1−x)32). D = \Bigl(1-\tfrac{1-x}{2},\tfrac{(1-x)\sqrt3}{2}\Bigr) = \Bigl(\tfrac{1+x}{2},\,\tfrac{(1-x)\sqrt3}{2}\Bigr).

  6. 分别写出直线 ADAD 与 BEBE 的参数方程，并求交点 PP。

  7. 然后，从 BB 向 ADAD 作垂线，得到垂足 QQ。

  8. 连接 PP 与 CC，并利用「BPBP 垂直于 PCPC」这一条件，建立点的坐标点积为零的方程，最终解出 xx。

  9. 由 x=AEx = AE（因为 ∣AC∣=1|AC|=1）便可求得 AEEC=x1−x\frac{AE}{EC} = \frac{x}{1-x}。

下面给出主要的计算步骤。

详细坐标与推导

1. 点的坐标

  • A=(0,0)A=(0,0)
  • B=(1,0)B=(1,0)
  • C=(12,32)C=\bigl(\tfrac12,\tfrac{\sqrt3}{2}\bigr)

令

E=(x2,x32),D=(1+x2,(1−x)32), E = \Bigl(\tfrac{x}{2}, \tfrac{x\sqrt3}{2}\Bigr), \quad D = \Bigl(\tfrac{1+x}{2}, \tfrac{(1-x)\sqrt3}{2}\Bigr),

其中 0<x<10<x<1，且满足 AE=CD  ⟹  x=1−yAE=CD \implies x = 1 - y，已经吸收入坐标表示。

2. 直线 ADAD 和 BEBE 的交点 PP

  • 直线 ADAD
    因为 A=(0,0)A=(0,0)，D=(1+x2,(1−x)32)D=\bigl(\tfrac{1+x}{2}, \tfrac{(1-x)\sqrt3}{2}\bigr)，
    向量 AD→=D−A\overrightarrow{AD}=D-A，
    可写成参数方程：

    AD:    (X,Y)=t(1+x2,(1−x)32),t∈R. AD:\;\; (X,Y) = t\Bigl(\tfrac{1+x}{2}, \tfrac{(1-x)\sqrt3}{2}\Bigr), \quad t\in \mathbb{R}.

  • 直线 BEBE
    因为 B=(1,0)B=(1,0)，E=(x2,x32)E=\bigl(\tfrac{x}{2}, \tfrac{x\sqrt3}{2}\bigr)，
    BE→=E−B=(x2−1,x32)\overrightarrow{BE} = E-B = \bigl(\tfrac{x}{2}-1,\tfrac{x\sqrt3}{2}\bigr)。
    参数方程：

    BE:    (X,Y)=(1,0)+s(x2−1,x32),s∈R. BE:\;\; (X,Y) = (1,0) + s\Bigl(\tfrac{x}{2}-1,\tfrac{x\sqrt3}{2}\Bigr), \quad s\in \mathbb{R}.

令交点为 PP。则存在参数 tt 与 ss 使

{t (1+x)2  =  1  +  s(x2−1),t (1−x)32  =  s(x32).\begin{cases} t\,\frac{(1+x)}{2} \;=\; 1 \;+\; s\Bigl(\tfrac{x}{2}-1\Bigr), \\[6pt] t\,\frac{(1-x)\sqrt3}{2} \;=\; s\Bigl(\tfrac{x\sqrt3}{2}\Bigr). \end{cases}

由第二式可得

t (1−x)32  =  s x32    ⟹    t (1−x)=s x    ⟹    s=t (1−x)x. t\,\frac{(1-x)\sqrt3}{2} \;=\; s\,\frac{x\sqrt3}{2} \;\;\Longrightarrow\;\; t\,(1-x) = s\,x \;\;\Longrightarrow\;\; s = t\,\frac{(1-x)}{x}.

代入第一式，解得

t=x x2−x+1 . t = \frac{x}{\,x^2 - x + 1\,}.

进而

s=t (1−x)x= 1−x  x2−x+1 . s = \frac{t\,(1-x)}{x} = \frac{\,1-x\,}{\,x^2 - x + 1\,}.

将 tt 带回到 ADAD 的参数式，可得

P=( t 1+x2,  t (1−x)32)=( x(1+x)2 [x2−x+1],  x(1−x)32 [x2−x+1]). P = \Bigl(\,t\,\tfrac{1+x}{2},\;t\,\tfrac{(1-x)\sqrt3}{2}\Bigr) = \Bigl(\, \frac{x(1+x)}{2\,[x^2 - x + 1]},\; \frac{x(1-x)\sqrt3}{2\,[x^2 - x + 1]} \Bigr).

3. 从 BB 向 ADAD 作垂线的垂足 QQ

  • 直线 ADAD 经过原点 A=(0,0)A=(0,0)，方向向量为

    v⃗=(1+x2,(1−x)32). \vec{v} = \Bigl(\tfrac{1+x}{2}, \tfrac{(1-x)\sqrt3}{2}\Bigr).

  • 点 B=(1,0)B=(1,0)。要找从 BB 到经过原点且方向向量 v⃗\vec{v} 的那条线的垂足 QQ，可用向量正交投影公式：

    若一条线是「{ t v⃗:t∈R}\{\,t\,\vec{v}: t\in \mathbb{R}\}」，
    则从点 b\mathbf{b} 投影到此线的参数 tQt_Q 为

    tQ=b⋅v⃗v⃗⋅v⃗. t_Q = \frac{\mathbf{b}\cdot \vec{v}}{\vec{v}\cdot \vec{v}}.

    其中 b=B=(1,0)\mathbf{b}=B=(1,0)。

  • 计算：

    b⋅v⃗=(1,0)⋅(1+x2,(1−x)32)=1+x2. \mathbf{b}\cdot \vec{v} = (1,0)\cdot\Bigl(\tfrac{1+x}{2}, \tfrac{(1-x)\sqrt3}{2}\Bigr) = \tfrac{1+x}{2}. v⃗⋅v⃗=(1+x2)2+((1−x)32)2=(1+x)24+3 (1−x)24=(1+x)2+3 (1−x)24. \vec{v}\cdot \vec{v} = \Bigl(\tfrac{1+x}{2}\Bigr)^2 + \Bigl(\tfrac{(1-x)\sqrt3}{2}\Bigr)^2 = \frac{(1+x)^2}{4} + \frac{3\,(1-x)^2}{4} = \frac{(1+x)^2 + 3\,(1-x)^2}{4}.

    细化展开可得

    (1+x)2+3(1−x)2=1+2x+x2+3 (1−2x+x2)=1+2x+x2+3−6x+3x2=4−4x+4x2=4 (x2−x+1). (1+x)^2 + 3(1-x)^2 = 1+2x+x^2 + 3\,(1 - 2x + x^2) = 1+2x+x^2 +3 -6x+3x^2 = 4 -4x +4x^2 = 4\,(x^2 - x +1).

    因而

    v⃗⋅v⃗=4 (x2−x+1)4=x2−x+1. \vec{v}\cdot \vec{v} = \frac{4\,(x^2 - x +1)}{4} = x^2 - x +1.

    记 D=x2−x+1D = x^2 - x +1。则

    tQ=1+x2 D =1+x 2D . t_Q = \frac{\tfrac{1+x}{2}}{\,D\,} = \frac{1+x}{\,2D\,}.

    垂足 QQ 即为

    Q=tQ v⃗=1+x2D (1+x2,(1−x)32)=( (1+x)2 4D ,  (1+x)(1−x)3 4D ). Q = t_Q\,\vec{v} = \frac{1+x}{2D}\,\Bigl(\tfrac{1+x}{2}, \tfrac{(1-x)\sqrt3}{2}\Bigr) = \Bigl(\, \frac{(1+x)^2}{\,4D\,},\; \frac{(1+x)(1-x)\sqrt3}{\,4D\,} \Bigr).

    其中 (1+x)(1−x)=1−x2(1+x)(1-x)=1 - x^2。

4. 利用 BP⊥PCBP \perp PC 求出 xx

  • 向量 BP→=P−B=( Px−1,  Py),PC→=C−P=( 12−Px,  32−Py). \overrightarrow{BP} = P - B = \Bigl(\,P_x - 1,\;P_y\Bigr), \quad \overrightarrow{PC} = C - P = \Bigl(\,\tfrac12 - P_x,\;\tfrac{\sqrt3}{2} - P_y\Bigr).
  • 其中 Px=x(1+x) 2D ,Py=x(1−x)3 2D ,D=x2−x+1. P_x = \frac{x(1+x)}{\,2D\,}, \quad P_y = \frac{x(1-x)\sqrt3}{\,2D\,}, \quad D = x^2 - x +1.
  • BP⊥PCBP \perp PC 等价于 BP→⋅PC→=0. \overrightarrow{BP}\cdot \overrightarrow{PC} = 0. 即 (Px−1)(12−Px)  +  (Py)(32−Py)=0. \bigl(P_x - 1\bigr)\bigl(\tfrac12 - P_x\bigr) \;+\; \bigl(P_y\bigr)\bigl(\tfrac{\sqrt3}{2} - P_y\bigr) = 0.
  • 经过一系列代入与化简(可略去中间繁琐细节)，最终会得到一个关于 xx 的二次方程。解得： x=1或x=13. x = 1 \quad\text{或}\quad x = \tfrac13. 由于 x=1x=1 意味着 EE 和 CC 重合(退化情形)，不符合「在边上取点」的通常设定。
    因而我们取 x=13. x = \frac13.

5. 求出所需比值 AEEC\frac{AE}{EC}

由于整条 ACAC 长度为 1，且 AE=xAE = x，EC=1−xEC = 1-x。当 x=13x=\tfrac13 时，

AE=13,EC=1−13=23. AE = \tfrac13,\quad EC = 1 - \tfrac13 = \tfrac23.

故

AEEC=1323=12. \frac{AE}{EC} = \frac{\tfrac13}{\tfrac23} = \frac{1}{2}.

最终结论

    AEEC  =  12  . \boxed{\;\;\frac{AE}{EC} \;=\; \frac{1}{2}\;.}

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