如图所示的长方体透明玻璃鱼缸，假设其长AD=80 cm,高AB=60 cm,水深AE=40cm. 在水面上紧贴内壁G处有一块面包屑，G在水面线EF上，且EG=60 cm.一只蚂蚁想从鱼缸外的A点 沿鱼缸壁爬进鱼缸内的G处吃面包屑，则蚂蚁爬行的最短路线长为多少cm.---**Source/Context:** [2023 江苏宿迁期中，中] (2023 Suqian, Jiangsu Mid-term Exam, Intermediate Level) **Question Stem:** As shown in the figure, there is a rectangular transparent glass fish tank. Assume its length AD = 80 cm, height AB = 60 cm, and water depth AE = 40 cm. There is a bread crumb at point G on the water surface, tightly against an inner wall. G is on the water line EF, and EG = 60 cm. An ant wants to crawl from point A outside the fish tank along the fish tank wall into the fish tank to reach point G to eat the bread crumb. What is the shortest path length the ant crawls? _____ cm. **Other Relevant Text:** * "是个讲题鸭" (It's a problem-solving duck) - Annotation with an image of a duck holding a thumbs-up, next to a QR code. * QR code: A black and white QR code, likely linking to a video explanation of the problem. **Chart/Diagram Description:** * **Type:** A 3D perspective drawing of a rectangular prism (cuboid) representing a fish tank, partially filled with water. * **Main Elements:** * **Shapes:** * A wireframe representation of a rectangular prism, with solid lines for visible edges and dashed lines for hidden edges. * A shaded rectangular region within the prism representing the water, showing it fills the lower portion of the tank. * **Points:** * A: Labeled as the bottom-front-left vertex of the tank. * B: Labeled as the top-front-left vertex of the tank. * C: Labeled as the top-back-right vertex of the tank. * D: Labeled as the bottom-back-right vertex of the tank. * E: A point labeled on the vertical edge AB, indicating the water level. E is below B. * F: A point labeled on the vertical edge at the back-right of the tank (consistent with the edge connected to D and C but at the water level). * G: A black dot labeled on the water surface, specifically shown on the line segment EF. * **Lines:** * Visible edges of the tank are solid lines. * Hidden edges of the tank are dashed lines. * A horizontal line segment representing the water level is drawn on the front face (from E towards the right). * The line segment EF is drawn diagonally across the water surface, connecting E (front-left, water level) to F (back-right, water level). * **Labels and Annotations:** Points A, B, C, D, E, F, G are explicitly labeled. The shaded area indicates water. **Dimensions and Conditions from Text:** * **Tank Length (AD):** 80 cm. (Note: In the diagram, AD is drawn as the horizontal edge on the front face, implying it's the *width*. However, "长" typically refers to the longer horizontal dimension or depth in a 3D context. Assuming it's a base dimension.) * **Tank Height (AB):** 60 cm. * **Water Depth (AE):** 40 cm. (E is on AB, so it's 40 cm up from A). * **Bread Crumb Location (G):** * On the water surface. * "紧贴内壁" (tightly against an inner wall). * "在水面线 EF 上" (on the water line EF). * **EG = 60 cm.** **Mathematical Formulas/Equations & Problem Interpretation:** The problem asks for the shortest path an ant crawls on the surface of the rectangular prism. This type of problem is typically solved by "unfolding" the faces of the prism into a 2D plane. There is an ambiguity in the problem statement regarding the exact dimensions and location of G: 1. **Dimension Interpretation:** "长 AD = 80 cm". In the diagram, AD is drawn as the width of the front face. So, the tank's width is 80 cm, and its height is 60 cm. The depth (front-to-back dimension) is not explicitly given but might be implied or needed for specific paths. 2. **Point G Location:** * E is on the front-left vertical edge, at water level (AE = 40 cm). Let A be at the origin (0,0,0). Then E is at (0,0,40) if the x-axis is width, y-axis is depth, z-axis is height. * G is "紧贴内壁" (on an inner wall) and "EG = 60 cm". This means G is 60 cm away from E, lying on one of the inner vertical surfaces at the water level (z=40). * "G 在水面线 EF 上" (G is on the water line EF). The diagram shows F as the back-right vertex of the water surface. This makes EF a diagonal across the water surface. A point on this diagonal cannot be "tightly against an inner wall" unless it is one of the endpoints E or F, or the width/depth is zero, which is contradictory for a tank. Given the common nature of such shortest path problems, the most probable intended scenario is that the distance EG = 60 cm represents a horizontal distance along one of the unfolded faces to the point G, combined with the vertical distance from A to the water level. The "紧贴内壁" constraint implies G is on one of the edges of the water surface rectangle, i.e., on one of the four inner vertical walls. Let's assume the "长 AD = 80 cm" refers to the *width* of the front face. * Tank Width = 80 cm. * Tank Height = 60 cm. * Water Level = 40 cm from the bottom (AE = 40). * Starting point A is the bottom-front-left corner (0,0,0). * Point E is on the front-left vertical edge at the water level (0,0,40). * Point G is on the water surface (z=40), "紧贴内壁", and EG = 60 cm. This means G is located 60 cm horizontally from the vertical line passing through E, along the water surface plane, and it lies on a wall. There are two primary shortest path possibilities by unfolding: 1. **Path on the front wall (or left wall):** * The ant climbs up the front wall (or left wall) to the water level and then crawls horizontally on the water surface to G. * If G is on the front wall (i.e., on the plane defined by x-z coordinates, at y=0), then from E(0,0,40), G would be at (60,0,40). This is valid since the tank's width is 80 cm (> 60 cm). * If G is on the left wall (i.e., on the plane defined by y-z coordinates, at x=0), then from E(0,0,40), G would be at (0,60,40). This is valid if the tank's depth is at least 60 cm. * In either case, when the relevant face is unfolded (either the front face + part of the water surface, or the left face + part of the water surface), the starting point A is at (0,0) and the destination G is at (60, 40) in the 2D unfolded plane. * The shortest path is the straight line distance: Distance = $\sqrt{(60 - 0)^2 + (40 - 0)^2}$ Distance = $\sqrt{60^2 + 40^2}$ Distance = $\sqrt{3600 + 1600}$ Distance = $\sqrt{5200}$ Distance = $\sqrt{400 \times 13}$ Distance = $20\sqrt{13}$ cm. Given the values, this is the most direct and common solution approach for such problems. The visual details of line EF and point F in the diagram are likely illustrative of the water surface and not strictly defining G's coordinates in a way that contradicts "紧贴内壁". **Mathematical Calculation:** $ \text{Shortest Path} = \sqrt{60^2 + 40^2} $ $ = \sqrt{3600 + 1600} $ $ = \sqrt{5200} $ $ = \sqrt{400 \times 13} $ $ = 20\sqrt{13} \text{ cm} $