I am trying to solve the following problem from a book:

A very heavy armchair needs to be moved, but the only possible movement is to rotate it through 90 degrees about any of its corners. Can it be moved so that it is exactly beside its starting position and facing the same way?

I tried with a lego piece and markers and it looks like this is not doable. But I am looking for a more rigorous way of proving this.

I tried with labeling the corners of the chair as {FL, FR, BL, BR} representing front-left, front-right, back-left and back-right respectively, and figuring out how the corners change in a coordinate system.

Assuming initial position for {FL, FR, BL, BR} as {(0,0), (1,0), (0,1), (1,1)}, a 90 degree clockwise rotation on the FR corner will transform the corners to {(1,1), (1,0), (2,1), (2,0)}.

Now I am stuck with not having a strategy to prove that the target state {(1,0), (2,0), (1,1), (2,1)} can/cannot be reached.

May be there is a much simpler way to reason as well?


2 Answers 2


Show that if FL is at $(x,y)$, then $x+y$ is even

  • $\begingroup$ great, but if I were to write a proof, is there a general inductive proof possible or I can only use a proof by cases for base cases? $\endgroup$
    – senseiwu
    Commented Apr 13, 2022 at 21:10
  • 2
    $\begingroup$ This proof is basically that of a checkerboard colouring - colour the grid points in two colours (those where x+y is even have one colour, those with x+y odd the other colour), and then show that any valid move of the chair does not change the colour that the feet are on. $\endgroup$ Commented Apr 14, 2022 at 9:27
  • 1
    $\begingroup$ Yup. Chiming in with Jaap. A different way of using the checkerboard coloring is to assume that the armchair starts on a light square facing North. Then you can show that whenever the armchair is on a dark square the person sitting in it will be facing East/West, but on a light square they face Nort/South. The claim follows from this. Unlike Jaap I color the area of the seat rather than the legs, but it's the same idea. $\endgroup$ Commented Apr 14, 2022 at 18:02

As has been mentioned in above comments, this problem can be solved by looking at the area around the armchair as a large checkerboard. Every movement of the chair takes it from a light square to dark or vice versa, but also rotates it 90 degrees. Therefore, to move from a light square to another light square, the rotation will always be a multiple of 180 degrees, as an even number of moves will have to have been made. Any orthogonally adjacent square will necessarily be a different color from the starting square, so moving to that square will mean either a total rotation of either 90 or 270 degrees (disregarding additional full rotations of 360 degrees). This means that you can't end in the same orientation with one orthogonal movement, and the problem is solved.


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