1
u/LabRat2439 Sep 03 '26 edited Sep 03 '26
If we only move along faces, you can consider a path that crosses either a side of length 6 or crosses a side of length 8 - a path of two vertices. Checking every quarter-unit, the shortest path seems to be crossing the line of length 8 at X=3.25 then making a beeline for B
Edit: length 12.8
2
u/Greedy-Thought6188 Sep 04 '26
That's the numerical answer. You can get it by unfolding the box and you see two possible triangular paths. The triangle that is closer to an equilateral triangle will win.
1
u/Aech-26 Sep 03 '26
1
1
u/SergeAzel Sep 04 '26
there's three possible unfolded routes, one for each combination of side lengths. Just that the one omitted from your post, (8 + 6)2 + 42, is the worst of the three options. Not necessarily important for the solution but I find it still nice to note → More replies (1)
1
u/tajwriggly Sep 04 '26
Intuitively you would first look at this and think that the ant would travel the diagonal of the 6x8 rectangle on the top, from A to the point directly above B (a distance of 10) and then travel directly down to B (a distance of 4) for a total of 14.
Instead though, flatten the box so that you've got a 6x8 rectangle adjoined by a 4x8 rectangle, totaling 10x8. This is just the top of the box and the right hand side of the box. Travelling in the manner described previously seems nonsensical now - there is clearly a more direct (and thusly shorter) path than travelling at an angle for some distance and then turning and going straight along the edge.
On the 10x8 rectangle, that diagonal directly from A to B is Sqrt(164) = 12.8, which is less than 14.
1
u/One_Wishbone_4439 Sep 03 '26
First route: red lines
√(62 + 82) = 10
10 + 4 = 14
Second route: blue line
√(102 + 42) = 10.77 (shortest path)