什么是胡不归模型---**Title:**
胡不归
**Model Explanation:**
从前，有一个小伙子在外地当学徒，当他获悉在家乡的老父亲病危的消息后，便立即启程日夜赶路。由于思念心切，他选择了全是沙砾地带的直线路径 A—B (如图所示, A 是出发地, B 是目的地, AC 是一条驿道，而驿道靠近目的地的一侧全是沙砾地带)。当他气喘吁吁地赶到家时，老人刚刚咽了气，小伙子不觉失声痛哭，邻居劝慰小伙子时告诉说，老人在弥留之际不断喃喃地念叨着“胡不归？胡不归？……” 这个古老的传说，引起了人们的思索，小伙子要提前到家是否有可能呢？倘若有可能，他应该选择怎样的路线呢？这就是风靡千年的“胡不归问题”。
**Diagram 1:**
* Type: Geometric diagram illustrating a path with different terrains.
* Elements:
* A horizontal line labeled "驿道" (Path/Post road), extending from left to right. One end is labeled M, the other is implied to extend beyond B.
* Point A is on the "驿道" line, labeled "A 出发地" (Origin A).
* A region above the "驿道" line is labeled "沙砾地带" (Sandy area), indicated by parallel dashed lines.
* Point B is in the "沙砾地带", labeled "B 目的地" (Destination B).
* A straight line segment connects A and B, representing a direct path through the sandy area.
* A point C is on the "驿道" line between A and the implied vertical projection of B onto the line.
* A path segment is shown from A to C along the "驿道" (dashed line) and then from C to B (straight line). An arrow indicates the direction from A towards B.
* An angle is labeled with the symbol 13 near point B and the path A-C-B. This label's meaning is unclear from the context provided.
**Model:**
【模型】由于在驿道和沙砾地带的行走速度不一样，那么，小伙子有没有可能先在驿道上走一段路程后，再走沙砾地带，虽然多走了路，但反而总用时更短呢？如果存在这种可能，那么要在驿道上行走多远才最省时？
**Analysis:**
【解析】设在沙砾地带的行走速度为 v₁, 在驿道上的行走速度为 v₂, 显然 v₁ < v₂.
不妨假设从 C 处进入沙砾地带. 设总用时为 t, 则
t = BC / v₁ + AC / v₂ = (1/v₁) * (BC + (v₁/v₂)AC).
因为 v₁, v₂ 是确定的，所以只要 BC + (v₁/v₂)AC 的值最小，用时就最少. 问题就转化为求 BC + (v₁/v₂)AC 的最小值.
我们可以作一条以 C 为端点的线段，使其等于 (v₁/v₂)AC, 并且与线段 CB 位于 AM 两侧，然后根据两点之间线段最短，不难找到最小值点.
怎么作呢？
由三角函数的定义，过 A 点，在 AM 的另一侧以 A 为顶点，以 AM 为一边作 ∠MAN = α, sin α = v₁/v₂. 然后作 CE ⊥ AN, 则 CE = (v₁/v₂)AC. 故当点 B, C, E 在一条直线上时, BC + CE 的值最小，即 BC + (v₁/v₂)AC 的值最小，即总用时最少.
**Diagram 2:**
* Type: Geometric diagram illustrating a construction for the analysis.
* Elements:
* A horizontal line labeled AM ("驿道"). Point A is on this line.
* A region above AM is the sandy area (represented by dashed horizontal lines). Point B is in this region.
* Point C is on the line segment AM.
* A line segment CB is shown from C to B (dashed).
* Below AM, a ray AN is drawn from A, forming an angle α with AM (∠MAC is part of AM). The angle is labeled α. The ray AN is a dashed line.
* A point E is shown below AM. A dashed line segment AC is on AM. A dashed line segment CE is drawn. A right angle symbol indicates CE ⊥ AN at point E.
* A dashed line segment AE is shown.
* Points B, C, E are shown approximately collinear in this diagram, suggesting the minimum condition.
**Problem Solving:**
【问题解决】求形如“PA + kPB”的最小值问题，构造射线 AD, 使
得 sin∠DAN = k, 即 CH/AC = k, CH = kAC.
将问题转化为求 BC + kAC 的最小值，过 B 点作 BH ⊥ AD 交 MN 于点 C，交 AD 于点 H. 此时 BC + CH 取到最小值，即 BC + kAC 的值最小.
**Diagram 3:**
* Type: Geometric diagram illustrating a construction for the problem solving.
* Elements:
* A horizontal line labeled MN ("驿道"). Point A is on this line.
* Above MN is the sandy area. Point B is in this region.
* A point C is on MN. A line segment CB is shown. A dashed line segment AC is shown.
* A ray AD is drawn downwards from A. It is a dashed line.
* A dashed line segment BH is drawn from B. It is perpendicular to AD. A right angle symbol indicates BH ⊥ AD at point H.
* Point C is shown as the intersection of line BH and line MN.
**Diagram 4:**
* Type: Geometric diagram illustrating the same construction as Diagram 3, slightly different view/emphasis.
* Elements:
* A horizontal line labeled MN ("驿道"). Point A is on this line.
* Above MN is the sandy area. Point B is in this region.
* Point C is on MN. A line segment CB is shown (dashed). A line segment AC is shown (dashed).
* A ray AD is drawn downwards from A. It is a dashed line.
* A line segment BH is drawn from B. It is perpendicular to AD. A right angle symbol indicates BH ⊥ AD at point H.
* Point C is shown as the intersection of line BH and line MN.
* Point H is shown on the ray AD.
* A line segment CH is shown connecting C and H (dashed).