As it was pointed by fellow anons and demonstrated with wolframalpha,
>the answer is 1/4 along the diagonal on the top side
Let me now provide evidence
The point must be on the top face of the box.
Consider C to be the vertex right on top of A. The answer lies on the diagonal CB.
There are two possible paths:
first is by using one rectangular face and then get to the top.
second is by travelling more to the side, using two rectangular faces before getting on the top face.
Some points on the CB diagonal will get their shortest path using the first method, some others using the second.
We can use this info to do some optimization.
CB being the diagonal of a square, let x be a point on the diagonal, represented by (x,x), starting from B. x >= 1
The distance to x using the first method is sqrt( (3 ? x)^2 + (1 ? x)^2 )
The distance to x using the second method is sqrt( (2 + x)^2 + (2 ? x)^2 )
Let's optimize:
sqrt( (3 ? x)^2 + (1 ? x)^2 ) = sqrt( (3 ? x)^2 + (1 ? x)^2 )
(3 ? x)^2 + (1 ? x)^2 = (3 ? x)^2 + (1 ? x)^2
2*(x^2) ? 8x + 10 = 2*(x^2) + 8
8x = 2
x = 1/4
sqrt(2*(1/4)^2 + 8) = sqrt(65/8) = 2.850438563 > sqrt(8)
