For the original 80-80-20 triangle, we found a 4-triangle decomposition into 20-20-140, 40-40-100, 60-60-60, and 80-80-20 triangles. Note that the base angles form an arithmetic sequence 20, 40, 60, 80 (as are the apex angles.)
We shall prove that the same pattern holds for every CID decomposition. Indeed, let Pk-1PkPk+1 and PkPk+1Pk+1 be two adjacent triangles of such a decomposition, one with base ∠Pk-1PkPk+1 = α and the other with the base
Pk+1PkPk+2 = β, α > β. The difference between the two angles is α - β, obviously.
In the next triangle down the diagram, i.e. ΔPk-2Pk-1Pk, the base ∠Pk-1PkPk-2 is found to be
∠Pk-1PkPk-2 = 180° - (180° - 2α - β) = 2α - β,
with the same difference α - β from the previous base angle α.
Assume that an isosceles triangle ABC (AB = AC) with the apex angle of γ admits an n-term CID. The top triangle has an apex angle of (180° - 2γ) so that the next triangle's base angle is
180° - (180° - 2γ) = 2γ
implying that the difference of the arithmetic sequence of the base angles in the CID is equal γ, the apex angle of ΔABC. The base angle of ΔABC is clearly (180° - γ)/2. For the n-term arithmetic sequence of the base angles we thus have
γ + (n - 1)·γ = (180° - γ)/2,
from which
γ = 180° / (2n + 1).
Note that such an angle is half the central angle between two consecutive vertices of a regular (2n+1)-gon.
As we've seen, for γ = 20°, i.e. for the 80-80-20 triangle, 2n + 1 = 9 and n = 4. For n = 1, γ = 60°, as an equilateral triangle is its own 1-term decomposition. For n = 2, γ = 180°/5 = 36° leading to a golden triangle 72-72-36. Next comes the triangle with the non-constructible apex of π/7.
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Copyright © 1996-2012 Alexander Bogomolny