It is known if we use two convex polygons with equal sides we can cover the plane periodically in few ways. One new way to cover the plane periodically is if we use rhombuses and octagons of equal integer sides. My question is Is it possible for the rhombuses to have diagonals which are integer numbers and the octagons to have angles integer numbers of degrees in this type of tiling?
Sure. Let the octagons be regular (so with angles 135 degrees) with sides of length $\sqrt2$, then the rhombi are squares with diagonals of length 2.
EDIT: Let's show that this is the only solution, up to similarity. In fact, up to similarity, it's the only solution with rational diagonals and rational angles.
Each vertex in the tiling is the meeting point of two octagons and one rhombus. Since the octagons have rational angles, the rhombi must also have rational angles.
The diagonals of the rhombi cut them into (four congruent) right triangles. These triangles have rational angles (since the diagonals bisect the angles of a rhombus) and rational legs (since the diagonals bisect each other; "leg" means side other than hypotenuse), so we have a rational angle whose tangent is a rational number. But it's well-known that the only such angles are zero and 45 degrees. Zero is irrelevant here, and 45 degrees gives the solution with squares.
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