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: enter image description here

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.

  • 4
    $\begingroup$ 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. $\endgroup$ – DanielWainfleet Mar 5 '16 at 7:53
  • 1
    $\begingroup$ Make that an answer! $\endgroup$ – Nikunj Mar 5 '16 at 8:47
  • $\begingroup$ 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. $\endgroup$ – Moti Mar 5 '16 at 23:05
  • $\begingroup$ 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 $\sqrt{2^2+1} = \sqrt{5}$. $\endgroup$ – user19405892 Mar 6 '16 at 18:59
  • $\begingroup$ Thanks a lot for your explanation. Got it.. $\endgroup$ – learner Mar 6 '16 at 20:11


The path goes through the middle point of common opposite side considering two squares only.


You are right, but for the wrong question. You answered the question "what is the shortest path between A and B". However, the question, as written, includes "along the surface" - this requires following e.g. user254665's suggestion.

If we can't go through the interior of the cube, the 3D structure is just cognitive noise and we map this surface to a 2D embedding by e.g. unfolding the box, leaving the edge CD in tact. This leaves us with a rectangle of length 2 and width 1, A and B on opposite corners, and you can calculate the length.


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