# How to find the shortest path between opposite vertices of a cube, traveling on its surface?

I am stuck with the following problem that says:

Let $A,B$ be the ends of the longest diagonal of the unit cube . The length of the shortest path from $A$ to $B$ along the surface is :

1. $\sqrt{3}\,\,$ 2.$\,\,1+\sqrt{2}\,\,$ 3.$\,\,\sqrt{5}\,\,$ 4.$\,\,3$

My Try:

So, the length of the longest diagonal $AB=\sqrt{3}$. If I reach from $A$ to $B$ along the surface line $AC+CD+BD$, then it gives $3$ units. But the answer is given to be option 3.

• Consider the cube as a 6-sided cardboard box made by folding a flat T-shaped piece of cardboard made of 6 squares sharing some common edges. Unfold the box, draw a straight line on it from A to B. Re-fold. It did not ask for a path that stays on the edges of the cube. – DanielWainfleet Mar 5 '16 at 7:53
• Make that an answer! – Nikunj Mar 5 '16 at 8:47
• Lust flatten the side that includes BCD. You get a right angle triangle with sides 2 and 1. The hypotenuse is $\sqrt{5}$. Draw the line connecting A and flatten B. Return the side containing BCD to its original position. It is exactly what user254665 means. – Moti Mar 5 '16 at 23:05
• The problem with learners solution is that he isn't taking a path along the surface. Unfolding the net of the cube doesn't change lengths so the answer is $\sqrt{2^2+1} = \sqrt{5}$. – user19405892 Mar 6 '16 at 18:59
• Thanks a lot for your explanation. Got it.. – learner Mar 6 '16 at 20:11