Tilings Of Two Kinds Of Polygons With Similar Triangles

We say that a triangle T tiles the polygon P if P can be decomposed into finitely many non-overlapping triangles similar to T. A tiling is called regular if there are two angles of the triangles, say α and β, such that at each vertex V of the tiling the number of triangles having V as a vertex and having angle a at V is the same as the number of triangles having angleβat V. Otherwise the tiling is called irregular. Denote by Rn (n≥5and n≠6) the regular polygon, by P(δ)(δ is the acute of P(δ))the parallelogram, by Δ=(α, β,γ) the triangle.First, in this paper we consider the tilings of regular polygons with congruent triangles and give the following theorem.Theorem1If n≥5and n≠6, then the regular polygon Rn has no any regular tiling with congruent right triangles.Theorem3If R,n (n≥5and n≠6) has an irregular tiling with congruent right triangles of angles α, β, π/2, then the triangle is (α,β, π/2)=(π/2-π/n,π/n,π/2).Theorem4If Rn (n≥5and n≠6) has a regular tiling with congruent isosceles triangles of angles α, β,γ(α=β), then the triangle is (α, β,γ)=(π/2-π/n,π/2-π/n,(2π)/n).For Rn (n≥5and n≠6) has irregular tiling with congruent isosceles triangles or not question, we propose a conjecture, which needs further study.Conjecture5Rn (n≥5and n≠6) has no irregular tiling with congruent isosceles triangles.Second, we consider the tilings of parallelogram with similar acute triangles and prove the following theorem.Theorem6If P(δ) can be dissected into finitely many similar acute triangles with angles α, β,γ. and α, β,γare not all rational multiples of π, then, with a suitable permutation of α, β,γ, the following statements is true:α=δ,β+γ=π-δ.Theorem8If P(δ) can be dissected into finitely many similar acute triangles with angles α, β,γ. and α, β,γare rational multiples of π, then, with a suitable permutation of α, β,γ, the following statements is true:a=α=δ,β+γ=π-δ.