详细分析什么是胡不归模型---**Extraction Content:** **Title:** 模型 46 胡不归 **Introduction Dialogue:** Character 1: "胡"是什么意思? 胡不归又有什么关系? 是解决什么样的题使用的呢? Character 2: 胡其实有“何”的意思, 胡不归模型是用来求最短路径的问题, 跟将军饮马有些相似. 具体模型和解题方法跟我一起学习下面的内容吧! **Section Header:** 模型讲解 **Problem Description:** 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. 由于思乡心切, 他选择了全是沙砾地带的直线路径 A—B (如图所示, A 是目的地, B 是出发地, AC 是一条旁道, 而旁道附近的地面另一侧全是沙砾地带), 他气喘吁吁地赶到家时, 老人刚刚咽了气, 小伙子失声痛哭. 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, 引起了人们的思考, 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? 这就是流传千年的“胡不归问题”. **Diagram 1 (Related to Problem Description):** Type: Geometric diagram. Main Elements: * Three horizontal dashed lines representing boundaries or regions. * The middle region is labeled "沙砾地带" (Sandy belt). * Points A and B are shown, with a straight dashed line connecting them, partially within the "沙砾地带". * A line segment AC is shown, where C is on the boundary of the "沙砾地带" and A is inside the "沙砾地带". * Labels: A (目的地, Destination), B (出发地, Starting Point), AC (旁道, Bypass). **Section Header:** 旁道 沙砾地带 **Model Explanation:** 【模型】由于在旁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旁道上走一段路程后, 再进入沙砾地带行走反而总用时更短呢? 如果存在这种可能, 那么要在旁道上行走多远才最省时? **Solution Analysis:** 【解析】设在沙砾地带的行驶速度为 v1, 在旁道上的行驶速度为 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. **Further Solution Details:** 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 BC 位于 AM 两侧, 然后根据两点之间线段最短, 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 (Related to Solution Analysis):** Type: Geometric diagram. Main Elements: * A line labeled AM, representing the bypass ("旁道"). * A line labeled MN, representing the boundary of the sandy belt ("沙砾地带"). * Point A is on line AM. * Point B is in the region below line MN. * Point C is on line AM, between A and M. * A line segment BC connects B to C. * A dotted line AN extends from A. * An angle labeled α is formed between AM and AN (∠MAN). * A point E is shown, connected to C by a dotted line segment CE. * A dotted line segment EF is shown, with F on AN. * A right angle symbol is shown at F, indicating CE ⊥ AN is incorrect, it should be EF ⊥ AN or perhaps C is constructed such that CE has a specific length and relationship. Looking closely at the text "然后作 CE ⊥ AN", this diagram shows EF ⊥ AN, and point E is constructed such that CE is the required length. Let's re-read the text and look at the next diagram. The description "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." implies constructing ray AN first, then CE perpendicular to AN, or perhaps constructing E based on the sine definition. The diagram shows ∠MAN = α, and points C and E. It seems E is constructed such that CE has the required length. The text "CE ⊥ AN" seems contradicted by the diagram where EF ⊥ AN. Let's rely on the text description for the construction steps and the second solution diagram for verification if possible. The text says "作一条以 C 为端点的线段, 使其等于 v1/v2 * AC, 并且与线段 BC 位于 AM 两侧". The diagram shows E on the opposite side of AM from B, and CE is a dotted line. * Labels: A, B, C, E, F, M, N, α, 旁道, 沙砾地带. **Section Header:** 名师有话说 (Master Teacher Says) **Content:** (Contains a QR code and text "更多精彩内容 请扫码观看") 更多精彩内容 请扫码观看 **Section Header:** 名师小技巧 (Master Teacher Tip) **Content:** 求线段和的最小值问 题, 可通过构造图形, 运 用“点到直线之间, 垂线 段最短”解题. **Section Header:** 问题解决 (Problem Solution) **Problem Type Introduction:** 【问题解决】求形如“PA + kPB”的最值问题, 构造射线 AD, 使 得 sin∠DAN = k, 即 CH/AC = k, CH = kAC. **Diagram 3 (Method 1 Example):** Type: Geometric diagram. Main Elements: * A line labeled MN. * Point A on line MN. * Point C on line MN. * Point B not on line MN. * A ray AD extends from A below MN. * An angle labeled ∠DAN is shown between MN and AD. * A dotted line segment CH is shown, perpendicular to AD at point H. * Labels: M, A, C, N, B, D, H. **Problem Transformation/Alternate Method Introduction:** 将问题转化为求 BC + kAC 的最小值, 过 B 点作 BH ⊥ AD, 交 AD 交 MN 于点 C. (This sentence is incomplete and likely refers to the previous diagram/method or is a mix-up). Let's check the next part. **Problem Transformation/Alternate Method:** 将问题转化为求 BC + kAC 的最小值, 过 B 点作 B′为 B 关于直线 MN 的对称点, 连接 B′A 交 MN 于点 C′, 此时 BC + kAC 取到最小值, 即 BC + kAC 的值最小. (This describes the reflection method for a different type of shortest path problem, possibly related but not directly solving the 'Hu Bu Gui' problem described initially with different speeds). **Diagram 4 (Method 2 Example):** Type: Geometric diagram. Main Elements: * A line labeled MN. * Point A on line MN. * Point C' on line MN. * Point B not on line MN (above MN). * Point B' is the reflection of point B across line MN (below MN). * A line segment B'A is shown, intersecting MN at C'. * Labels: M, A, C', N, B, B'. **Section Header:** 模型巧记 (Model Memorization Tip) **Content:** 点在线直线上动, 一条线 段加 k 倍的线段和的最 小值, 三角函数来转化. (Point moves on a line, sum of a line segment and k times another line segment, minimization, use trigonometric function transformation.) **Other Relevant Text:** A 旁道 C (Label near diagram 1) **Summary of Formulas/Equations:** * t = BC/v1 + AC/v2 * t = 1/v1 * (BC + v1/v2 * AC) * Minimize BC + v1/v2 * AC * sin α = v1/v2 * CE = v1/v2 * AC (from geometric construction explanation, though diagram 2 contradicts the perpendicular part) * Minimize BC + kAC (problem transformation/example) * sin ∠DAN = k (for PA + kPB problem) * CH/AC = k * CH = kAC **Note on Diagram 2 and Explanation:** The description "然后作 CE ⊥ AN" seems inconsistent with Diagram 2, where EF ⊥ AN is shown. Also, the description implies E is constructed such that CE = (v1/v2)AC and BC+CE is minimized when B, C, E are collinear. The diagram shows E on the ray AN. The geometric construction might be more complex or the diagram is illustrative rather than strictly following the step-by-step instructions. The Problem Solution section provides a related method for PA+kPB using sine. It's likely the intention is to construct point E such that △ACE is similar to a right triangle with angle α, leading to the desired ratio CE/AC = v1/v2 = sin α. **Refined understanding of Diagram 2 construction:** The text says "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". This constructs the ray AN. Then it says "然后作 CE ⊥ AN, 则 CE = v1/v2 * AC". This construction step is likely incorrect or misinterpreted. The standard approach for such problems involves constructing a point E such that AC/sin(alpha) = CE/sin(gamma) etc., relating lengths and angles using the sine rule, or reflecting/rotating. Given the sin α = v1/v2 part and the goal CE = (v1/v2)AC, it's more likely that E is constructed such that CE is proportional to AC based on the sine ratio. The method in the "Problem Solution" section (Diagram 3) for PA+kPB, using sin∠DAN = k and CH = kAC (where CH ⊥ AD), is the relevant technique. So for BC + kAC, we would likely construct a ray from A at angle α such that sin α = k = v1/v2, and then use that to find the optimal C. Diagram 2 attempts to illustrate this, but the CE ⊥ AN seems incorrect for achieving CE = (v1/v2)AC generally. The method of finding the minimum of BC + kAC by reflecting or rotating one point or constructing a new point based on the ratio k and an angle derived from k is the core idea. Let's prioritize the explicit text and formulas. The diagrams illustrate the concepts and methods.