Question 2.6.26: A set of geometric figures consists of red equilateral trian......
A set of geometric figures consists of red equilateral triangles and blue quadrilaterals with all angles greater than 80° and less than 100°. A convex polygon with all of its angles greater than 60° is assembled from the figures in the set. Prove that the number of (entirely) red sides of the polygon is a multiple of 3.
We first enumerate the ways to decompose various angles α into sums of 60° angles (T) and angles between 80° and 100° (Q):
60° < α < 180° α = T, 2T, T + Q, 2Q
α = 180° α = 3T, 2Q
α = 360° α = 6T, 3T + 2Q, 4Q
(The range for Q cannot be increased, since 3Q ranges from 240° to 300°; even including the endpoints would allow for additional combinations above.)
The set of all of the vertices of all of the polygons can be divided into three categories, namely those which lie in the interior, on an edge, or at a vertex of the large polygon. The above computation shows that the number of T angles at interior or edge vertices is a multiple of 3; since the total number is three times the number of triangles, we deduce that the number of T angles at vertices of the large polygon is also a multiple of 3.
Next note that every edge is entirely of one color, since we cannot have both a T and a Q at a 180° angle. Additionally, no vertex of the large polygon consists of more than two angles, and a T cannot occur by itself. All this means that the number of red sides is half the number of T angles at the vertices, which is a multiple of 3.