# Walking on a box

Given a box with its 8 corners at

(0,0,0)
(481,0,0)
(0,53,0)
(0,0,490)
(481,53,0)
(481,0,490)
(0,53,490)
(481,53,490)


An ant is positioned at (0,0,0) and would like to head to the point (481,51,256). The ant can only walk at the surface of the box. What is the shortest path of the ant towards the point?

• Unfold the box and have the ant walk a straight line. There are a few ways to unfold it, you have to try them. Mar 5 '20 at 6:03
• I tried the two possible ways to unfold, which yielded me 590.39 and 738.76.But the answer from my teacher was 573.38 Mar 5 '20 at 6:08
• @RossMillikan. Your nice solution remembered me the Gordian knot. Cheers :-) Mar 5 '20 at 9:57
• Isn't the answer $573.3\color{red}28$ ?
– user65203
Mar 5 '20 at 16:00
• Technically my teacher told me to write the answer as the distance squared, so the answer is 328770, after taking the square root, I got 573.3847 Mar 6 '20 at 6:15

There are 3 combinations to unfold In each combination you take 2 coordinates and 3rd coordinate to be perpendicular

C-1 : dist^2 = (x+y)^2 + z^2....dist = 590.39

C-2 : dist^2 = (x+z)^2 + y^2....dist = 738.36

C-3 : dist^2 = (y+z)^2 + x^2....dist - 570.62

So min distance is with C-3 which is 570.62

You are unfolding the planes - 1 xz plane at y=0 and 2 xy plane at z=256

You only need to unfold the box and draw it's net, the rest is just Pythagoras.

• The target endpoint is not a corner.
– user65203
Mar 5 '20 at 15:04
• @YvesDaoust, Thank you, it is fixed now. Mar 5 '20 at 15:55
• This still doesn't match the "official" answer, which seems to be $\sqrt{481^2+(256+56)^2}$.
– user65203
Mar 5 '20 at 15:59
• @YvesDaoust, Apparently the official answer is not correct. Mar 5 '20 at 16:01
• I got the coordinates wrong. The height of the box should be 490. I'm terribly sorry Mar 6 '20 at 6:25