TITLE: How to find the shortest path between opposite vertices of a cube, traveling on its surface?
QUESTION [6 upvotes]: 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 : 


$\sqrt{3}\,\,$  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. 
Can someone explain? Thanks in advance for your time.

REPLY [4 votes]: The path goes through the middle point of common opposite side considering two squares only.