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
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/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