The red square is placed on top of a blue square The goal is to cut the red square into 4 pieces and assemble them with the blue square to create a larger square. The area of the resulting square is the sum of the area of the red and blue squares.

enter image description here

There are at least two solutions I am aware of. The first one to show at least two solutions will be granted the 15 points:)

  • 1
    $\begingroup$ What is the significance of "The red square is placed on top of a blue square", as the first sentence? Or is this a red herring? $\endgroup$
    – Stilez
    Jun 17, 2019 at 13:39
  • $\begingroup$ What is the blue triangular thing at right hand side of red cube? $\endgroup$ Sep 8, 2019 at 9:29
  • $\begingroup$ My old-syllabus school math textbook contained a bit comlex version of this puzzle: testing the Pythagoras theorem. "A right-angled triangle given. Draw a square using base as arm and another square using perpendicular as arm. Now find such an way to cut the 2 squares that fuse perfectly into a square with an arm same as the hypoteneuse." $\endgroup$ Sep 8, 2019 at 9:39
  • $\begingroup$ "Placed on top" does it mean perfectly overlapping? $\endgroup$ Sep 8, 2019 at 9:45
  • $\begingroup$ No - a portion of the blue is hidden. $\endgroup$
    – Moti
    Sep 9, 2019 at 2:57

1 Answer 1


I think you can dissect the red square (on the left here) as follows

enter image description here

And then rearrange to form the larger square

enter image description here

This is always possible when the red square is larger than the blue.


If the length of the side of the larger square is $A$ and the smaller square $B$ then we find the point along each side of the large square which is a distance $\frac{A+B}{2}$ from each vertex going in a clockwise direction. Then join opposing points in this construction.

The lines traversing the square intersect at right angles due to symmetry. Also, the length of each line $d$ can be determined with Pythagoras theorem by constructing a line from one of the vertices to the opposite side parallel to the side of the square. That is, $$ d^2 = A^2 + \left(\frac{A+B}{2} - \frac{A-B}{2}\right)^2 = A^2 + B^2$$ It follows that the five pieces can be rearranged to form the combined square as shown.

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    $\begingroup$ This is a great solution! Much better than what I had in mind. Though it seems like a "single" solution it holds for an infinite "blue" squares. $\endgroup$
    – Moti
    Jun 17, 2019 at 19:10

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