Problem A. 919. (November 2025)

A. 919. Let \(\displaystyle \mathcal{P}\) be a convex cyclic polygon with at least four vertices such that no two diagonals of \(\displaystyle \mathcal{P}\) are congruent. Prove that there is at most one triangulation of \(\displaystyle \mathcal{P}\) where all triangles have the same perimeter.

Proposed by Andrei Chirita, Cambridge

(7 pont)

Deadline expired on December 10, 2025.

The key to the solution is the Japanese theorem for cyclic polygon (https://en.wikipedia.org/wiki/Japanese_theorem_for_cyclic_polygons): if we divide a cyclic polygon into triangles using diagonals, the sum of the inradii of the resulting triangles does not depend on the choice of the diagonals.

We prove this statement for quadrilaterals first: let \(\displaystyle r_A\), \(\displaystyle r_B\), \(\displaystyle r_C\) and \(\displaystyle r_D\) denote the inradii of triangles \(\displaystyle BCD\), \(\displaystyle ACD\), \(\displaystyle ABD\) and \(\displaystyle ABC\). We have to prove that \(\displaystyle r_A+r_C=r_B+r_D\).

We will use the Carnot's theorem (https://en.wikipedia.org/wiki/Carnot%27s_theorem_(inradius,_circumradius)): let \(\displaystyle r\) denote the inradius, \(\displaystyle R\) denote the circumradius and \(\displaystyle d_a\), \(\displaystyle d_b\) and \(\displaystyle d_c\) denote the distance of the circumcenter from side \(\displaystyle a\), \(\displaystyle b\) and \(\displaystyle c\) (the distance is considered negatve if the angle opposite the side is greater then \(\displaystyle 90^\circ\)). This can be rewritten as \(\displaystyle 1+r/R=\cos\alpha+\cos\beta+\cos\gamma\), and this can proved computally by applying \(\displaystyle r=A/s\) (\(\displaystyle A\) denotes the area of the triangle, and \(\displaystyle s\) is the semiperimeter), \(\displaystyle R=\frac{abc}{4A}\), \(\displaystyle A^2=s(s-a)(s-b)(s-c)\) (Heron's formula) and the law of cosines.

Now the case of the quadrilateral is simple: let \(\displaystyle d_{AB}\) denote the distance of the circumcenter of the quadrilateral and side \(\displaystyle AB\): the sign will be the same in triangles \(\displaystyle ABC\) and \(\displaystyle ABD\), because \(\displaystyle \angle ACB=\angle ADB\). However, the sign of \(\displaystyle d_{AC}\) will differ in triangles \(\displaystyle ABC\) and \(\displaystyle ACD\), since \(\displaystyle \angle ABC+\angle ADC=180^\circ\). Thus if we use Carnot's theorem in triangles \(\displaystyle ABC\) and \(\displaystyle ADC\), and we add the equalities together, the \(\displaystyle d_{AC}\)'s cancel, and thus we get \(\displaystyle 2R+r_B+r_D=d_{AB}+d_{BC}+d_{CD}+d_{DA}\), and obviously we will get the same result for triangles \(\displaystyle BCD\) and \(\displaystyle ABD\).

Now we have to verify that any two triangulations can be transformed into each other if we are allowed to take two triangles in the triangulation that share a side, and replace it with the other diagonal of the quadrilateral formed by the two triangles. We prove this by strong induction: the statement is true in triangles and quadrilaterals, and let's assume that it's true for all polygons with less than \(\displaystyle n\) vertices. Now let's consider two triangulations in an \(\displaystyle n\)-gon. If they share a diagonal, we are done, since the common diagonal divides the polygon into two smaller polygon, and the inductional hypothesis can be applied. Using this observation both triangulations can be transformed into a triangulation where all the diagonals share a common vertex. If the two common vertices are not adajcent, they share a diagonal, so we are done. Finally, if the two common vertices are adjacent, we can consider for consecutive vertices such that these two vertices are in the middle, and then by applying the transfomation on this quadrilateral, the two triangulations will share a diagonal, and we are done.

Now we can solve our original problem. Suppose that every triangle has the same perimeter \(\displaystyle P\) in a triangulation. Since the area of a triangle equals \(\displaystyle rs\), adding these together we get that the area of our polygon equals the sum of the inradii times \(\displaystyle P/2\). Therefore, the perimeter can be expressed as two times the area of the polygon divided by the sum of the inradii, which is independent of the triangulation. Thus \(\displaystyle P\) must be the same in every triangulation satisfying the property about the perimeter. However, if we consider the triangle that includes side \(\displaystyle AB\) of the polygon, the third vertex must be on the circumcircle and also on the ellipse with foci at \(\displaystyle A\) and \(\displaystyle B\) and major axis \(\displaystyle P-AB\). Since the ellipse intersects the circle at at most two points, \(\displaystyle P\) and \(\displaystyle Q\) satisfying \(\displaystyle AP=BQ\) (and of course \(\displaystyle AQ=BP\)), it's not possible that \(\displaystyle P\) and \(\displaystyle Q\) are both vertices of the polygon, since our polygon does not have two congruent diagonal. So this triangle is uniquely determined, and we can finish our proof once again using strong induction on the two smaller polygons we obtain by deleting this triangle from our configuration.

Statistics:

19 students sent a solution.
7 points:	Ali Richárd, Bodor Mátyás, Bui Thuy-Trang Nikolett, Diaconescu Tashi, Forrai Boldizsár, Vigh 279 Zalán.
6 points:	Xiaoyi Mo.
4 points:	1 student.
3 points:	1 student.
1 point:	2 students.
0 point:	7 students.
Unfair, not evaluated:	1 solutions.

Problems in Mathematics of KöMaL, November 2025