We say that two similar triangles in a plane are oriented oppositely if one of them includes angles
α, β, γ
clockwise, and another one counterclockwise (angles
α, β, γ
are supposed to be mutually unequal).
Theorem 1. Let
Δ
be a triangle in the plane with angles
α, β, γ.
Let
Δ'
be a triangle in the plane, equal to
Δ
but oriented oppositely. Suppose that the triangle
Δ
can be dissected into 2
polygons such that one can combine them to obtain the triangle
Δ'
moving the polygons in the plane without flipping. Then
k α + l β + m γ = 0
for some integers
k
l, and
m
not vanishing simultaneously.
This theorem answers a question of Boltyanskiy from 1956.
Example. If
α/β=(n+1)/n
for some positive integer
n
then a triangle with angles
α, β, π-α-β
can be dissected into 2
polygons as required in Theorem 1.
Theorem. Suppose that a triangle with angles
α, β, and γ
can be dissected into triangles similar to it but oriented oppositely.
Then
k α + l β + m γ = 0
for some integers
k,
l, and
m
not vanishing simultaneously.
The approach is based on a generalization of the Hadwiger-Glur invariant of scissors congruence.
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