Daily Math Puzzle: 2026-08-26

20 meters — To find the shortest distance a spider must crawl on the surface of a 3D object, you need to "unfold" the relevant faces into a single 2D plane and draw a straight line between the start and end points. Let the side length of the base and the height be 's' (so s = 10 meters). The spider starts at the midpoint of one side of the base. Let's call this the "front" face. Its coordinates on an unfolded plane would be (s/2, 0) or (5,0) if we place the bottom-left of the front face at (0,0). The spider wants to reach the midpoint of the *opposite* side of the tower's *top edge*. This means the destination is on the "back" face, specifically its top edge, at the midpoint. There are two primary shortest paths to consider: 1. **Path around the vertical sides:** Imagine unfolding the "front" face, then the adjacent "right" face, then the "back" face, all side-by-side. This forms a continuous rectangle with a total width of (side + side + side) = 3s and a height of s. * Start point (S): (s/2, 0) = (5, 0) on the front face. * End point (E): The destination is on the back face. Its x-coordinate would be 2s (to cross the front and right faces) + s/2 (to reach the midpoint of the back face's top edge). So, the end point is at (2s + s/2, s) = (20 + 5, 10) = (25, 10). * The horizontal distance is (25 - 5) = 20 meters. The vertical distance is (10 - 0) = 10 meters. * Shortest distance (using Pythagorean theorem) = √(20² + 10²) = √(400 + 100) = √500 = 10√5 meters (approximately 22.36 meters). 2. **Path over the top face:** Imagine unfolding the "front" face and the "top" face directly above it. This forms a continuous rectangle with a width of 's' and a total height of (height of front + side of top) = s + s = 2s. * Start point (S): (s/2, 0) = (5, 0) on the front face. * End point (E): The destination is the midpoint of the *back edge* of the *top face*. On this unfolded plane, the front face goes from y=0 to y=10. The top face is attached to its top edge, so it goes from y=10 to y=20. The midpoint of the back edge of the top face is vertically aligned with the midpoint of the front face's base. So, its coordinates are (s/2, 2s) = (5, 20). * The horizontal distance is (5 - 5) = 0 meters. The vertical distance is (20 - 0) = 20 meters. * Shortest distance (using Pythagorean theorem) = √(0² + 20²) = √400 = 20 meters. Comparing the two paths, 20 meters is shorter than 10√5 meters (approximately 22.36 meters). The shortest distance is 20 meters.