Methods of Proof 333Figure 44This proves the identity. Settingm=n=p, we obtain the identity in the statement.
26.The base case consists of the dissections forn=4, 5, and 6 shown in Figure 44. The
induction step jumps fromP(k)toP(k+ 3 )by dissecting one of the triangles into four
triangles similar to it.
(R. Gelca)
27.First, we explain the inductive step, which is represented schematically in Figure 45.
If we assume that such ak-gon exists for allk<n, then then-gon can be obtained by
cutting off two vertices of the(nā 2 )-gon by two parallel lines. The sum of the distances
from an interior point to the two parallel sides does not change while the point varies,
and of course the sum of distances to the remaining sides is constant by the induction
hypothesis. Choosing the parallel sides unequal, we can guarantee that the resulting
polygon is not regular.
Figure 45The base case consists of a rectangle (n=4) and an equilateral triangle with two
vertices cut off by parallel lines (n=5). Note that to obtain the base case we had to
apply the idea behind the inductive step.
28.The property is obviously true for the triangle since there is nothing to dissect. This
will be our base case. Let us assume that the property is true for any coloring of ak-gon,
for allk<n, and let us prove that it is true for an arbitrary coloring of ann-gon. Because
at least three colors were used, there is a diagonal whose endpoints have different colors,
say red (r) and blue (b). If on both sides of the diagonal a third color appears, then we
can apply the induction hypothesis to two polygons and solve the problem.
If this is not the case, then on one side there will be a polygon with an even number
of sides and with vertices colored in cyclic orderrbrb...rb. Pick a blue point among
them that is not an endpoint of the initially chosen diagonal and connect it to a vertex