
An ant has to travel from A to B (which are diagonally opposite vertices of a cube) and back to A. The ant moves from A to B through a thread initially present between A and B, Once the ant reaches B the thread is removed, Now the ant has to come back to A through shortest path using the walls of the cube. If the velocity of the ant is same while going and coming then what is the ratio of time taken?
Solution 1 : (conventional method of solving)

(B-A) : distance covered = D2 = √(x^2+a^2) + √((a-x)^2+a^2)
for minimum D2, d(D2)/dx = 0
Finding d(D2)/dx will take some time....
Finally you will get the condition for minimum D2 is x=a/2
So D2= √5 a
So time taken while coming back from B to A is √(5/3) times more than the time taken while going from A to B as the speed is constant.
Solution 2 : (using the simplifying a puzzle technique)

This Problem will become lot simpler by just simplifying it with little visualization. When the ant has to come back from B to A we can bring the MNBO Face of cube on same plane as AMNP. i.e. rotating the face MNBO by 90 degree in clockwise direction. Now the Ant has to travel from B to A which is nothing but the diagonal of the rectangle APBO.
Now the distance between two points B and A is √((AP)^2+(BP)^2) = √5 a
So time taken while coming back from B to A is √(5/3) times more than the time taken while going from A to B as the speed is constant.