根据图片，详细分析什么是胡不归模型---**Title:**
胡不归 模型 (Hu Bugui Model)
**Section: 模型讲解 (Model Explanation)**
**Text:**
从前，有一个小伙子在外地当学徒，当他获悉在家乡的老父病危的消息后，便立即启程日夜赶路.由于思念心切,他选择了全是沙漠地带的直线路径 A→B (如图所示,A 是出发地,B 是目的地).但是靠近目的地的.一侧全是沙漠地带),当他气喘吁吁地赶到家时,悲剧刚刚大气,小伙子失声痛哭.邻居劝慰小伙子时告诉他,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 Description:**
Type: Geometric diagram illustrating paths and regions.
Main Elements:
- Two parallel horizontal dashed lines, suggesting boundaries.
- A horizontal solid line on the left side, labeled "蹊道" (Path/Route), extending to the right.
- A region below the "蹊道" line, labeled "沙漠地带" (Desert Area).
- Point A is labeled on the "蹊道" line (start point).
- Point B is labeled in the "沙漠地带" region (destination).
- A solid line segment from A to C, where C is on the "蹊道" line.
- A solid line segment from C to B, connecting C on the "蹊道" to B in the "沙漠地带".
- An arrowed line segment from A directly towards B, labeled "直线路径 AB" (Straight path AB).
**Section: 【模型】(Model) and 【解析】(Analysis)**
**Text:**
【模型】由于在蹊道和沙漠地带的行走速度不一样,那么,小伙子有没有可能先在蹊道上走一段路程后,再走沙漠地带,虽然多走了路,但反而总用时更短呢?如果存在这种可能,那么要在蹊道上行走多远才最省时?
【解析】设在沙漠地带的行驶速度为 $v_1$,在蹊道上的行驶速度为 $v_2$,显然 $v_1 < v_2$. 不妨假设从 C 处进入沙漠地带.设总用时为 $t$, 则
$t = \frac{BC}{v_1} + \frac{AC}{v_2} = \frac{1}{v_1} (BC + \frac{v_1}{v_2} AC).$
因为 $v_1, v_2$ 是确定的, 所以只要 $BC + \frac{v_1}{v_2}AC$ 的值最小, 用时就最少. 问题就转化为求 $BC + \frac{v_1}{v_2}AC$ 的最小值.
我们可以作一条以 C 为端点的线段, 使其等于 $\frac{v_1}{v_2}AC$, 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 $\angle MAN = \alpha$, $\sin \alpha = \frac{v_1}{v_2}$, 然后作 $CE \perp AN$, 则 $CE = \frac{v_1}{v_2}AC$. 故当点 B, C, E 在一条直线上时, $BC + CE$ 的值最小, 即 $BC + \frac{v_1}{v_2}AC$ 的值最小, 即总用时最少.
**Diagram 2 Description:**
Type: Geometric diagram illustrating a construction for minimization.
Main Elements:
- A horizontal line labeled AM extending to the right (likely representing the boundary).
- Point A on the line AM.
- Point C on the line AM.
- Point B above the line AM.
- Dashed line AN extending downwards from A, forming an angle $\alpha$ with AM.
- Angle $\alpha$ is labeled between AM and AN at vertex A.
- Point E on the line AN.
- Dashed line segment CE.
- Dashed line segment BE.
- Point F is shown above the line AM, seemingly related to B.
**Section: 【问题解决】(Problem Solving)**
**Text:**
【问题解决】求形如“$PA+kPB$”的最小值问题, 构造射线 AD, 使 $\sin\angle DAN = k$, 即 $\frac{CH}{AC}=k, CH=kAC$.
**Diagram 3 Description:**
Type: Geometric diagram illustrating a construction for minimizing an expression of the form PA + kPB.
Main Elements:
- A horizontal line labeled MN.
- Point A on the line MN.
- Point C on the line MN.
- Point B above the line MN.
- Solid line segment AB.
- Solid line segment BC.
- Dashed line AD extending downwards from A.
- Dashed line BH drawn from B perpendicular to AD at point H.
- Right angle symbol at H.
- The intersection of line segment BH and the line MN is labeled as point C.
- Dashed line segment BD.
**Text after Diagram 3:**
将问题转化为求 $BC + CH$ 的最小值, 过 B 点作 $BH \perp AD$ 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.