How can the white cross be cut into 5 smaller pieces that can be reassembled into the two smaller red crosses shown?


Puzzle created by Henry Dudeney on The Strand Newspaper long time ago.

Source: Saw this in this video time:(2.15)

Note: The diagrams are only rough, you may have to refer to the video if you have any doubts in the diagram

  • 2
    $\begingroup$ Can I assume the crosses have same height and width? At least we don't have to use the Banach-Tarski paradox :-) $\endgroup$ – happystar Oct 5 '20 at 5:16
  • 6
    $\begingroup$ The picture is rather inaccurate. Each cross should have all edges the same length, so it is as if it is built from 5 squares stuck together. $\endgroup$ – Jaap Scherphuis Oct 5 '20 at 5:37
  • $\begingroup$ @happystar I don't know, I did not create this puzzle. This is the question given. But though from the diagram, I think it has to be same height and width $\endgroup$ – rash Oct 5 '20 at 6:19
  • $\begingroup$ @happystar: Fun fact: the Banach-Tarski paradox doesn't work the same way in two dimensions. If you want to carve up an area and reassemble it into a larger area, you either have to carve it up into infinitely many pieces, or allow them to skew rather than just rigidly rotating them. $\endgroup$ – Michael Seifert Oct 5 '20 at 16:31
  • $\begingroup$ This must be a duplicate $\endgroup$ – Dr Xorile Oct 5 '20 at 20:53

Here is a visual answer.

enter image description here The smaller cross is a downscaled by $\sqrt 2$ version of the larger one. We can check using Pythagoras that the smaller center cross meets the outer larger one at the mid points of the edges. Indeed, the diagonal distance measured in the large cross is $\sqrt {2^2 + 1^2} = \sqrt 5$ and in the small cross $\frac 1 {\sqrt 2} \times \sqrt {3^2 + 1^2} = \sqrt 5$

And the pieces joined together

enter image description here


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