Consider 3 congruent rectangles each with aspect ratio sqrt(3), i.e., 1.73 to 1. Overlap two of them so that exactly two of their vertices and the diagonal between them coincide. Overlap the third rectangle the same way. This forms a regular hexagon. Equivalently, cover a regular hexagon with three rectangles with short rectangle edges coinciding with hexagon edges.

The rectangle looks like (and is) a typical rectangle with 90 degree angles but has a secret gateway to the world of regular hexagons and equilateral triangles with 120 and 60 degree angles.

Starting with US letter paper, 8.5 by 11 inches, remove 2.15 inches to yield 6.35 by 11, the desired aspect ratio. Also, this rectangle can be cut into 3, yielding three 6.35 by 3.67 rectangles which are interestingly themselves of the desired aspect ratio, ready to overlap into a regular hexagon. (sqrt 3)/3 = 1/(sqrt 3).

Using the square root of an integer as an aspect ratio is reminiscent of A4 paper (and A series paper in general), which uses sqrt 2.

Explore further this construction technique of overlapping the diagonals of rectangles. For what aspect ratio will 4 copies of such a rectangle cover a regular octagon? A regular 2n-gon? These should be easy.

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