Problem
GEO-B1-M02-P024 A Triangulation by Isosceles Triangles
A convex polygon is cut by non-intersecting diagonals into triangles, each of which is isosceles. Prove that the original polygon has two equal sides.
C. Hint 1. Try induction on the number of sides of the polygon.
D. Hint 2. Take one drawn diagonal and apply the statement to the two smaller polygons.
E. Full solution.
We prove the statement by induction on the number of sides. For a triangle it is immediate, because the triangle itself is isosceles.
Assume the statement is known for polygons with fewer sides. Consider the convex polygon \(P\) and one of the drawn diagonals \(d\). It divides \(P\) into two smaller convex polygons \(P_1\) and \(P_2\). Each of them is also triangulated into isosceles triangles.
By the induction hypothesis, \(P_1\) has two equal sides. If both are sides of the original polygon, we are done. If one of them is the diagonal \(d\), then \(P_1\) has a side of the original polygon equal to \(d\).
The same argument for \(P_2\) either immediately gives two equal sides of the original polygon, or gives a side of the original polygon equal to \(d\). In the last case the two found sides lie on opposite sides of the diagonal and are both equal to \(d\), hence they are equal to each other.
A. Source analysis. Main objects: a convex polygon, a diagonal triangulation, and isosceles triangles. The obvious approach is to look for equal sides locally in one triangle, but such a side may be an internal diagonal. The hidden observation is that if the equal side is the diagonal, one must look on the other side of the same diagonal. Number of key ideas: 3.
F. Difficulty justification. This is Level 7: a regional-style problem with induction and a careful case when the equal side is an internal diagonal.
G. Check. This is not a one-step exercise: a local equality inside one triangle is insufficient; the induction step across a diagonal is essential.