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How about this?

This creates from a unit square four pizza slices of a regular 12-gon inscribed in the unit circle. (3 whole ones (can be left together if we want to minimize total number of bits), one pieced together, note that no flipping is required, no need for upside-down pizza =-) )

Why does it work?

The regular 12-gon has 30° slices so we can fit 3 of them in a corner of the square. As for the pieced together slice we need to show that the red and blue triangles are congruent. By chasing angles we see that they are similar with acute angles 15° and 30°. The easiest way to show that they are the same size is by comparing the areas of the squares (3x1) and of the inscribed regular 12-gon (3, essentially because sin 30° = 1/2)

Alternative cut:

Also four bits per square:

join the top and bottom triangles at their horizontal sides. This will result in a "double slice" to complement the two slices left and right.

How about this?

This creates from a unit square four pizza slices of a regular 12-gon inscribed in the unit circle. (3 whole ones (can be left together if we want to minimize total number of bits), one pieced together, note that no flipping is required, no need for upside-down pizza =-) )

Why does it work?

The regular 12-gon has 30° slices so we can fit 3 of them in a corner of the square. As for the pieced together slice we need to show that the red and blue triangles are congruent. By chasing angles we see that they are similar with acute angles 15° and 30°. The easiest way to show that they are the same size is by comparing the areas of the squares (3x1) and of the inscribed regular 12-gon (3, essentially because sin 30° = 1/2)

Alternative cut:

Also four UPDATE thanks @Jaap Scherphuis three: leave the left (or right) triangle connected to the bottom one and join the top one to the right (or left) /UPDATE bits per square:

join the top and bottom triangles at their horizontal sides. This will result in a "double slice" to complement the two slices left and right.

How about this?

This creates from a unit square four pizza slices of a regular 12-gon inscribed in the unit circle. (3 whole ones (can be left together if we want to minimize total number of bits), one pieced together, note that no flipping is required, no need for upside-down pizza =-) )

Why does it work?

The regular 12-gon has 30° slices so we can fit 3 of them in a corner of the square. As for the pieced together slice we need to show that the red and blue triangles are congruent. By chasing angles we see that they are similar with acute angles 15° and 30°. The easiest way to show that they are the same size is by comparing the areas of the squares (3x1) and of the inscribed regular 12-gon (3, essentially because sin 30° = 1/2)

How about this?

This creates from a unit square four pizza slices of a regular 12-gon inscribed in the unit circle. (3 whole ones (can be left together if we want to minimize total number of bits), one pieced together, note that no flipping is required, no need for upside-down pizza =-) )

Why does it work?

The regular 12-gon has 30° slices so we can fit 3 of them in a corner of the square. As for the pieced together slice we need to show that the red and blue triangles are congruent. By chasing angles we see that they are similar with acute angles 15° and 30°. The easiest way to show that they are the same size is by comparing the areas of the squares (3x1) and of the inscribed regular 12-gon (3, essentially because sin 30° = 1/2)

Alternative cut:

Also four bits per square:

join the top and bottom triangles at their horizontal sides. This will result in a "double slice" to complement the two slices left and right.

How about this?

This creates from a unit square four pizza slices of a regular 12-gon inscribed in the unit circle. (3 whole ones, one pieced together, note that no flipping is required, no need for upside-down pizza =-) )

How about this?

This creates from a unit square four pizza slices of a regular 12-gon inscribed in the unit circle. (3 whole ones (can be left together if we want to minimize total number of bits), one pieced together, note that no flipping is required, no need for upside-down pizza =-) )

Why does it work?

The regular 12-gon has 30° slices so we can fit 3 of them in a corner of the square. As for the pieced together slice we need to show that the red and blue triangles are congruent. By chasing angles we see that they are similar with acute angles 15° and 30°. The easiest way to show that they are the same size is by comparing the areas of the squares (3x1) and of the inscribed regular 12-gon (3, essentially because sin 30° = 1/2)