Swiss Problem Counts per Author
Note that our author data before 2019 is incomplete and this list is probably missing many deserving entries.
| Name | Algebra | Geometry | Combinatorics | Number Theory | Total |
|---|---|---|---|---|---|
| Thomas Huber | 0 | 0 | 1 | 0 | 1 |
| Raphael Steiner | 8 | 0 | 0 | 4 | 12 |
| Dmitrij Nikolekov | 0 | 0 | 1 | 0 | 1 |
| Dimitri Wyss | 3 | 1 | 5 | 7 | 16 |
| Dimiti Wyss | 0 | 0 | 0 | 1 | 1 |
| Clemens Pohle | 0 | 11 | 0 | 1 | 12 |
| Markus Sprecher | 0 | 0 | 3 | 0 | 3 |
| Philipp Wirth | 1 | 0 | 1 | 1 | 3 |
| Alain Rossier | 4 | 3 | 0 | 7 | 14 |
| Cyril Frei | 0 | 1 | 3 | 2 | 6 |
| Arnaud Maret | 5 | 4 | 1 | 0 | 10 |
| Dmitrij Nikolenkov | 0 | 0 | 3 | 1 | 4 |
| Louis Hainaut | 0 | 1 | 0 | 1 | 2 |
| Szymon | 1 | 1 | 0 | 0 | 2 |
| Robert Meier | 0 | 1 | 0 | 2 | 3 |
| Linus Rösler | 1 | 0 | 0 | 1 | 2 |
| Nikola Djokic | 0 | 0 | 1 | 0 | 1 |
| David Rusch | 5 | 6 | 12 | 11 | 34 |
| Paul Seidel | 0 | 0 | 2 | 0 | 2 |
| Fabian Jin | 1 | 0 | 0 | 0 | 1 |
| Frieder Jäckel | 2 | 0 | 0 | 0 | 2 |
| Bibin Muttappillil | 0 | 0 | 1 | 0 | 1 |
| Valentin Imbach | 6 | 6 | 9 | 16 | 37 |
| Horace Chaix | 0 | 1 | 0 | 0 | 1 |
| Patrick Stalder | 0 | 4 | 0 | 0 | 4 |
| Tanish Patil | 0 | 1 | 4 | 0 | 5 |
| Henning Zhang | 1 | 0 | 0 | 0 | 1 |
| Joël Huber | 0 | 0 | 1 | 0 | 1 |
| Johann Williams | 1 | 2 | 0 | 0 | 3 |
| Ricardo Olivo | 0 | 0 | 0 | 1 | 1 |
| Ivan Pouly | 1 | 2 | 0 | 1 | 4 |
| Mathys Douma | 3 | 6 | 0 | 1 | 10 |
| Raphael Angst | 1 | 0 | 1 | 0 | 2 |
| Mark Neumann | 0 | 0 | 2 | 1 | 3 |
Problems at International Competitions
IMO 2026 – Problem 4Author: Valentin Imbach
Shan-Yu and Mulan are playing a game. Let \(\theta\) be an angle with \(0^\circ < \theta < 180^\circ\) known to both players. Initially, Shan-Yu makes a paper triangle \(\mathcal{T}\) with measurements of his choice. Then, they repeatedly perform the following steps:
If \(\mathcal{T}\) has at least one angle measuring exactly \(\theta\), then the game stops and Mulan wins.
Otherwise, Mulan chooses a point \(P\) on the perimeter of \(\mathcal{T}\), different from its three vertices. She then makes a straight cut from to the opposite vertex of \(\mathcal{T}\), splitting it into two triangles.
Shan-Yu discards one of the two triangles. The remaining triangle becomes the new \(\mathcal{T}\).
For which real values of \(\theta\) can Mulan guarantee her victory in finitely many steps, no matter how Shan-Yu plays?