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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.

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    $\begingroup$ 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, 2016 at 7:53
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    $\begingroup$ Make that an answer! $\endgroup$
    – Nikunj
    Mar 5, 2016 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, 2016 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, 2016 at 18:59
  • $\begingroup$ Thanks a lot for your explanation. Got it.. $\endgroup$
    – learner
    Mar 6, 2016 at 20:11

2 Answers 2

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ShortestPathUnitCube

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

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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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