根据附件图片内容，解释”什么是费马点模型“？---Here is the extracted content from the image in structured plain text format: **Header:** 模型 55 “费马点”模型 模型展现 **基础模型 (Basic Model):** * **图示 (Diagram):** * Description: A triangle labeled ABC with a point P inside. Lines connect P to A, B, and C. * Text: △ABC 的最大内角小于 120° * **条件 (Condition):** * Text: P 是△ABC 内一点, 当点 P 在何处时, PA+PB+PC 的值最小 * **结论 (Conclusion):** * Text: 当 PA+PB+PC 的值最小时, 点 P 满足 ∠APB = ∠BPC = ∠APC = 120° **思考延伸 (Thinking Extension):** * Text: 当△ABC 的最大内角大于等于 120° 的情况: * 条件 (Condition): P 是△ABC 内一点, 当点 P 在何处时, PA+PB+PC 的值最小 * 图示 (Diagram): * Description: A triangle labeled ABC with a point P coinciding with vertex B. * Text: (No specific text label for diagram content) * 结论 (Conclusion): 当 PA+PB+PC 的值最小时, 点 P 与最大角顶点重合 **【结论分析】(Conclusion Analysis):** * 结论 (Conclusion): 当 PA+PB+PC 的值最小时, 点 P 满足 ∠APB = ∠BPC = ∠APC = 120° * 证明 (Proof): 如图, 将△ABP 绕点 B 逆时针旋转 60° 得到△EBD, 连接 AE, CE, PD, ∴ △ABP ≌ △EBD, ∴ AP=ED, BP=BD, AB=EB. 又∵ ∠PBD = ∠ABE = 60°, ∴ △ABE, △PBD 均为等边三角形, ∴ PB=BD=PD, ∴ PA+PB+PC = DE+PD+PC ≥ CE, ∴ 当 C, P, D, E 四点共线时, PA+PB+PC 的值最小, 最小值为 CE 的长. 此时 ∠BPC = 180° - ∠BPE = 180° - 60° = 120°, ∠APB = ∠EDB = 180° - ∠BDP = 180° - 60° = 120°, ∠APC = 360° - (∠BPC + ∠APB) = 120°. ∴ ∠APB = ∠BPC = ∠APC = 120°. * 图示 (Diagram for Proof): * Description: Triangle ABC. Point P inside. Point E is shown, derived from rotation. Point D is shown, derived from rotation. Dashed line connecting C, P, D, E. △ABE and △PBD are indicated visually as equilateral or near-equilateral by ticks and angles. **例 1 (Example 1):** * 问题 (Problem Stem): 如图, △ABC 为等腰直角三角形, ∠ABC = 90°, 点 P 为△ABC 内一点, 点 P 到 A, B, C 三点的距离之和的最小值为 $\sqrt{2}+\sqrt{6}$, 则 AB 的长为 ______. * 图示 (Diagram): * Description: Right-angled isosceles triangle ABC, with the right angle at B. Point P is inside the triangle. * Caption: 例 1 题图 (Diagram for Example 1) * 模型猜想 (Model Guess): * Description: A thought bubble pointing to the problem. * Text: 存在在三角形内一点到三角形三个顶点的距离之和的最小值问题, 故为 “费马点” 模型 (There exists a problem about the minimum sum of distances from a point inside a triangle to the three vertices, so it is a "Fermat Point" model) **问题解决步骤 (Problem-Solving Steps):** * **第一步: 依据特征找模型 (Step 1: Find model based on features)** * 特征 1 (Feature 1): 三角形内是否存在一个动点 (Does a moving point exist inside the triangle) * 三角形: △ABC, 动点: 点 P (Triangle: △ABC, Moving point: Point P) * 特征 2 (Feature 2): 是否存在在该动点到三角形三个顶点距离之和的最小值 (Does the minimum sum of distances from this moving point to the three vertices exist) * 点 P 到 A, B, C 三点的距离之和的最小值 (Minimum sum of distances from point P to points A, B, C) * 图示 (Diagram): Same diagram as Example 1 problem stem. * **第二步: 抽象模型 (Step 2: Abstract model)** * 图示 (Diagram): Same diagram as Example 1 problem stem. (This step involves matching the problem to the Fermat Point model). * **第三步: 用模型 (Step 3: Use model)** * Text: 以点 A 为旋转中心, 将△ABP 顺时针旋转 60° 得到△AMN, 连接 NP, MB, MC (Using point A as the center of rotation, rotate △ABP clockwise by 60° to get △AMN, connect NP, MB, MC) * 图示 (Diagram): Triangle ABC with point P. Points M and N are shown, results of rotation around A. Line segments NP, MB, MC, MN are shown. Points C, P, N, M appear approximately collinear. * Caption: MC 为线段和的最小值 (MC is the minimum value of the sum of line segments)