Daily Math Puzzle: 2026-09-25
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2026-09-25
A tiny, rare salamander has fallen into a discarded, perfectly rectangular shipping crate. The crate measures 3 meters long, 2 meters wide, and 1 meter high. The salamander is trapped in one corner at the very bottom of the crate. To rescue it, a biologist plans to make an incision at the *exact opposite corner* on the *exterior* of the crate and reach in with a specialized tool. To ensure the tool is long enough but not overly cumbersome, they need to calculate the shortest possible straight-line path *along the surface* of the crate from the biologist's starting point to the salamander's location. What is the length of this shortest path?
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Solution
sqrt(18) meters — This is a classic "shortest path on a cuboid" problem. To find the shortest path along the surface between two diagonally opposite corners of a rectangular prism (cuboid), you need to "unfold" the box in different ways and apply the Pythagorean theorem.
Let the dimensions of the crate be Length (L=3m), Width (W=2m), and Height (H=1m). The two corners are (0,0,0) and (L,W,H).
There are three primary ways to unfold adjacent faces such that a straight line connects the two points on the flattened surface:
1. **Unfolding across the Length and Width faces (e.g., front and top):** Imagine flattening the front face (L x H) and the right side face (W x H). The two points are now at the ends of a rectangle with dimensions (L+W) by H.
Path length = sqrt((L+W)^2 + H^2) = sqrt((3+2)^2 + 1^2) = sqrt(5^2 + 1^2) = sqrt(25 + 1) = sqrt(26) meters.
2. **Unfolding across the Length and Height faces (e.g., top and side):** Imagine flattening the top face (L x W) and one of the side faces (H x W). The two points are now at the ends of a rectangle with dimensions (L+H) by W.
Path length = sqrt((L+H)^2 + W^2) = sqrt((3+1)^2 + 2^2) = sqrt(4^2 + 2^2) = sqrt(16 + 4) = sqrt(20) meters.
3. **Unfolding across the Width and Height faces (e.g., side and bottom):** Imagine flattening one of the side faces (W x H) and the bottom face (L x W). The two points are now at the ends of a rectangle with dimensions (W+H) by L.
Path length = sqrt((W+H)^2 + L^2) = sqrt((2+1)^2 + 3^2) = sqrt(3^2 + 3^2) = sqrt(9 + 9) = sqrt(18) meters.
Comparing these three possible shortest paths:
* sqrt(26) ≈ 5.099 m
* sqrt(20) ≈ 4.472 m
* sqrt(18) ≈ 4.243 m
The shortest of these surface paths is **sqrt(18) meters**.
*The distractor, sqrt(14) meters, represents the direct 3D diagonal distance THROUGH the box (sqrt(L^2 + W^2 + H^2)), which is not a path along the surface.*
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