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题目全文 \triangle ABC 的内角 A，B，C 的对边分别为 a，b，c．且 b = ...

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题目全文
\triangle ABC 的内角 A，B，C 的对边分别为 a，b，c．且 b = 2，2b\cos A = c\cos A + a\cos C．

（1）求角 A；

（2）点 S 是直线 PQ 外一点，点 M，N 在直线 PQ 上（点 M，N 与点 P，Q 任一点不重合）．若点 M 在线段 PQ 上，记 (P,Q;M)=\dfrac{|SP|\sin\angle PSM}{|SQ|\sin\angle MSQ}；若点 M 在线段 PQ 外，记 (P,Q;M)=-\dfrac{|SP|\sin\angle PSM}{|SQ|\cdot\sin\angle MSQ}．记 (P,Q;M,N)=\dfrac{(P,Q;M)}{(P,Q;N)}．

已知点 D 是线段 BC 上一点，且 (B,C;D)=\dfrac{c}{2}．点 M 在射线 BC 上，点 P 在直线 AM 上且任意实数 t 有 \vert\overrightarrow{AC} - t\overrightarrow{AM}\vert \geq \vert\overrightarrow{CP}\vert 恒成立．

（i）若 (B,C;M,D)=-1，求 \vert AM\vert + 4\vert AB\vert 的最小值；

（ii）若 (B,C;M,D)=-\dfrac{1 + \sqrt{3}x}{2}(x > 0)，求 \triangle ACP 面积的取值范围．

الفكرة

Using the Law of Cosines

I applied the Law of Cosines to triangle ABC, simplified the equation, and found that angle A measures 60°.

Analyzing point relationships

I explored the cross ratio definition and found that D divides BC internally with the ratio BD : CD = c : 2, affecting D's position based on side c.

Determining the foot of perpendicular

I'm figuring out that for the inequality to hold in every scenario, point P must be the foot of the perpendicular from C to line AM. CP is the minimal distance.

Determining cross ratio

I calculated the cross ratio adjustments and expressed BM/CM positively. This leads me to focus on minimizing AM + 4c.

Placing and analyzing

I positioned points A, B, and C on a coordinate plane, then constructed a parametric equation for M on ray BC, ensuring M's position satisfies the ratio BM/CM = c/2.

Finding M's position

I determined that c must be greater than 2, then calculated M's coordinates on ray BC and derived a formula for the distance AM by combining geometric and algebraic methods.

Assessing sympy's availability

I'm working with an internet-disabled Python environment, and I'll check if sympy is included.

Minimizing the equation

I’m working on minimizing the function AM + 4 AB with respect to c greater than 2.

Mapping the equation

I determined c = 3 by setting the derivative to zero. Then, refining the area of triangle ACP using coordinates and the vector method.

Finding coordinates

OK, let me see. I’m deriving expressions for s in terms of c and x, calculating the orthogonal projection of C onto line AM, and then using these coordinates to determine the area of triangle ACP.

Scaling down

OK, I’m scaling the area expression, treating 2.0 as 2 for exact rational representation. Progress is steady, simplifying the expression effectively.

Examining area constraints

OK, let me see: the area formula S, given x > 0, suggests that k must be > 1 for a positive s. This sets constraints for valid c values.

Analyzing the curve

OK, let's look at the second derivative of the function to understand its concavity and identify any inflection points.

Taking a closer look

Examining the range of area S for ΔACP, I’m considering values between (0, 1], noting its asymptotic behavior and ensuring inclusiveness of boundaries.

Reasoned for 16m 52s

الإجابة

(1) 求 ∠A

利用余弦定理

cos⁡A=b2+c2−a22bc,cos⁡C=a2+b2−c22ab.\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc},\qquad \cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}.

已知
2bcos⁡A=ccos⁡A+acos⁡C,  b=22b\cos A=c\cos A+a\cos C,\;b=2.

2bcos⁡A=ccos⁡A+acos⁡C⟹2⋅2cos⁡A=ccos⁡A+acos⁡C⟹b2+c2−a2c=b2+c2−a22b+a2+b2−c22b.\begin{aligned} 2b\cos A&=c\cos A+a\cos C\\ \Longrightarrow 2\cdot2\cos A &=c\cos A+a\cos C\\ \Longrightarrow\frac{b^{2}+c^{2}-a^{2}}{c} &=\frac{b^{2}+c^{2}-a^{2}}{2b}+\frac{a^{2}+b^{2}-c^{2}}{2b}. \end{aligned}

整理得
b2+c2−a2=bc⇒a2=b2−bc+c2b^{2}+c^{2}-a^{2}=bc\Rightarrow a^{2}=b^{2}-bc+c^{2}.

带入 b=2b=2 得
a2=4−2c+c2a^{2}=4-2c+c^{2}.

再次代入余弦定理：

cos⁡A=b2+c2−a22bc=4+c2−(c2−2c+4)4c=2c4c=12.\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc} =\frac{4+c^{2}-\left(c^{2}-2c+4\right)}{4c} =\frac{2c}{4c}=\frac12.

故

  A=60∘  .\boxed{\;A=60^{\circ}\;}.

预备：关于 (P,Q;M)(P,Q;M) 的含义

按题设定义可验证

(P,Q;M)={∣SP∣sin⁡∠PSM∣SQ∣sin⁡∠MSQ(M 在 PQ 上)−∣SP∣sin⁡∠PSM∣SQ∣sin⁡∠MSQ(M 在 PQ 外)=PMQM(P,Q;M)= \begin{cases} \dfrac{|SP|\sin\angle PSM}{|SQ|\sin\angle MSQ}& (M\text{ 在 }PQ\text{ 上})\\[6pt] -\dfrac{|SP|\sin\angle PSM}{|SQ|\sin\angle MSQ}& (M\text{ 在 }PQ\text{ 外}) \end{cases} =\dfrac{PM}{QM}

（正负号表示 MM 是否落在段 PQPQ 内）。
于是

(P,Q;M,N)=PM/QMPN/QN=PM⋅QNQM⋅PN,(P,Q;M,N)=\dfrac{PM/QM}{PN/QN} =\dfrac{PM\cdot QN}{QM\cdot PN},

即通常的“有向比值”交比。

(2)(i) 当 (B,C;M,D)=−1(B,C;M,D)=-1 时极小化 ∣AM∣+4∣AB∣|AM|+4|AB|

记 a=∣BC∣,  b=∣AC∣=2,  c=∣AB∣a=|BC|,\;b=|AC|=2,\;c=|AB|。

设
(B(0,0),;C(a,0),; A(u,v);(v>0).)

由 (1) 已得

a2=c2−2c+4,u=c2−ca,v=3 ca.a^{2}=c^{2}-2c+4,\qquad u=\frac{c^{2}-c}{a},\qquad v=\frac{\sqrt3\,c}{a}.

1. 由 (B,C;D)=c2(B,C;D)=\dfrac c2 求 DD

BDCD=c2⟹D(d,0),  d=ca2+c.\frac{BD}{CD}=\frac c2\quad \Longrightarrow\quad D\bigl(d,0\bigr),\;d=\frac{ca}{2+c}.

2. 由 (B,C;M,D)=−1(B,C;M,D)=-1 求 MM

(B,C;M,D)=−BM/CMBD/CD=−1  ⟹  BMCM=c2.(B,C;M,D)=-\frac{BM/CM}{BD/CD}=-1 \;\Longrightarrow\; \frac{BM}{CM}=\frac c2.

令 M(m,0)  (m>a)M(m,0)\;(m>a)，得

mm−a=c2  ⟹  m=c ac−2(c>2).\frac{m}{m-a}=\frac c2 \;\Longrightarrow\; m=\frac{c\,a}{c-2}\quad(c>2).

3. 由 “∣AC→−tAM→∣≥∣CP∣  ∀t∈R|\overrightarrow{AC}-t\overrightarrow{AM}|\ge |CP|\;\forall t\in\mathbb R” 推得 PP 是 CC 到 AMAM 的垂足

令 PP 为 CC 到直线 AMAM 的垂足，可证该不等式对全体 tt 成立且仅在此时成立。

4. 计算 ∣AM∣|AM|

∣AM∣=2c c−2 .|AM|=\frac{2c}{\,c-2\,}.

5. 极值

F(c)=∣AM∣+4∣AB∣=2cc−2+4c,c>2.F(c)=|AM|+4|AB| =\frac{2c}{c-2}+4c,\qquad c>2. F′(c)=4−4(c−2)2⇒F′(c)=0⟺c=3.F'(c)=4-\frac4{(c-2)^{2}} \quad\Rightarrow\quad F'(c)=0\Longleftrightarrow c=3.

二阶导 F′′(c)=8(c−2)3>0F''(c)=\dfrac{8}{(c-2)^{3}}>0，故 c=3c=3 取极小。

Fmin⁡=F(3)=2⋅31+4⋅3=6+12=18.F_{\min}=F(3)=\frac{2\cdot3}{1}+4\cdot3=6+12=18. min⁡(∣AM∣+4∣AB∣)=18.\boxed{\min\bigl(|AM|+4|AB|\bigr)=18}.

(2)(ii) 当 (B,C;M,D)=−1+3 x2  (x>0)(B,C;M,D)=-\dfrac{1+\sqrt3\,x}{2}\;(x>0) 时

 求 △ACP\triangle ACP 的面积范围

同样设

BMCM=r=c(1+3 x)4,m=rar−1.\frac{BM}{CM}=r=\frac{c(1+\sqrt3\,x)}{4},\qquad m=\frac{ra}{r-1}.

沿用上节坐标并记

AC→=(a−u, −v),AM→=(m−u, −v).\overrightarrow{AC}=(a-u,\,-v),\quad \overrightarrow{AM}=(m-u,\,-v).

设 P=A+λ AM→P=A+\lambda\,\overrightarrow{AM}，且
 CP⊥AM\,CP\perp AM。
由正交条件得

λ=AC→⋅AM→∣AM→∣2=x ⁣(3cx+3 c−43)3c ⁣(x2+1)  (0<λ<1).\lambda=\frac{\overrightarrow{AC}\cdot\overrightarrow{AM}} {|\overrightarrow{AM}|^{2}} =\frac{x\!\left(3cx+\sqrt3\,c-4\sqrt3\right)} {3c\!\left(x^{2}+1\right)}\; (0<\lambda<1).

面积计算

S△ACP=12∣AC→×AP→∣=λ S△ACM=λ⋅∣AC→×AM→∣2=2xx2+1.\begin{aligned} S_{\triangle ACP} &=\frac12\bigl|\overrightarrow{AC}\times \overrightarrow{AP}\bigr| =\lambda\,S_{\triangle ACM}\\[3pt] &=\lambda\cdot\frac{|\overrightarrow{AC}\times\overrightarrow{AM}|}{2} =\frac{2x}{x^{2}+1}. \end{aligned}

结果 与边长 cc 无关，只由 xx 决定：

S(x)=2xx2+1,x>0.S(x)=\frac{2x}{x^{2}+1},\qquad x>0.

取值范围

S′(x)=2(1−x2)(x2+1)2⇒max⁡S 在 x=1 处取得 Smax⁡=1.S'(x)=\frac{2(1-x^{2})}{(x^{2}+1)^{2}} \quad\Rightarrow\quad \max S\text{ 在 }x=1 \text{ 处取得 }S_{\max}=1. lim⁡x→0+S(x)=lim⁡x→∞S(x)=0.\lim_{x\to0^{+}}S(x)=\lim_{x\to\infty}S(x)=0.

故

0<S△ACP≤1.\boxed{0<S_{\triangle ACP}\le 1}.

汇总答案

小问结  论
(1)A=60∘\displaystyle A=60^{\circ}
(2)(i)$\displaystyle \min\bigl(
(2)(ii)0<S△ACP≤1\displaystyle 0<S_{\triangle ACP}\le1

（面积单位与所选长度单位一致。）

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