Problem A. 920. (December 2025)

A. 920. Given a non-isosceles triangle \(\displaystyle ABC\), let its circumcircle be denoted by \(\displaystyle \Omega\). Let \(\displaystyle \omega_A\) be the mixtilinear incircle corresponding to vertex \(\displaystyle A\) (that is, the circle tangent to sides \(\displaystyle AB\), \(\displaystyle AC\), and internally tangent to \(\displaystyle \Omega\)). Define \(\displaystyle \omega_B\) and \(\displaystyle \omega_C\) analogously. Let the points where \(\displaystyle \omega_A,\omega_B,\omega_C\) touch \(\displaystyle \Omega\) be denoted by \(\displaystyle T_A, T_B, T_C\), respectively.

Show that the Monge line of the circles \(\displaystyle \omega_A,\omega_B,\omega_C\) (that is, the line passing through the pairwise external homothety centers of the three circles) coincides with the Pascal line of the hexagon \(\displaystyle AT_CBT_ACT_B\) (that is, the line passing through the points \(\displaystyle AT_C \cap T_AC\), \(\displaystyle T_CB \cap CT_B\), \(\displaystyle BT_A \cap T_BA\)).

Proposed by Sha Jingyuan, Budapest

(7 pont)

Deadline expired on January 12, 2026.

First, we show that the external homothety center of \(\displaystyle \omega_B\) and \(\displaystyle \omega_C\) is the intersection point of the lines \(\displaystyle T_BT_C\) and \(\displaystyle BC\). We apply Monge’s theorem to these two circles and to \(\displaystyle \Omega\): the points \(\displaystyle T_B\) and \(\displaystyle T_C\) are the external homothety centers of the circumcircle and the two circles, hence the desired point lies on the line \(\displaystyle T_BT_C\); on the other hand, it also lies on the common external tangent of the two circles, one of which is the line \(\displaystyle BC\).

Thus, it remains to prove that the points \(\displaystyle BC\cap T_BT_C\), \(\displaystyle AT_C\cap T_AC\), and \(\displaystyle T_CB\cap CT_B\) are collinear. These are precisely the intersection points of the corresponding sidelines of the triangles \(\displaystyle BCT_A\) and \(\displaystyle T_BT_CA\). By Desargues’ theorem, it is therefore sufficient to show that the lines \(\displaystyle BT_B\), \(\displaystyle CT_C\), and \(\displaystyle AT_A\) are concurrent. To this end, we apply Monge’s theorem to the circumcircle, the incircle, and the circle \(\displaystyle \omega_A\): since two homothety centers are \(\displaystyle A\) and \(\displaystyle T_A\), the line joining them passes through the external homothety center of the incircle and the circumcircle. As this also holds for \(\displaystyle BT_B\) and \(\displaystyle CT_C\), the statement of the problem is proved.

Statistics:

23 students sent a solution.
7 points:	Ali Richárd, Aravin Peter, Bao Nguyen Gia, Bodor Mátyás, Bolla Donát Andor, Bui Thuy-Trang Nikolett, Diaconescu Tashi, Forrai Boldizsár, Gyenes Károly, Kis Ágoston, Li Mingdao, Prohászka Bulcsú, Rajtik Sándor Barnabás, Sárdinecz Dóra, Sha Jingyuan, Szabó 721 Sámuel, Szakács Ábel, Varga 511 Vivien, Vigh 279 Zalán, Vödrös Dániel László, Xiaoyi Mo.
6 points:	Ethan Y.Wang.
0 point:	1 student.

Problems in Mathematics of KöMaL, December 2025