Let $P$ be a regular polygon with 2026 sides. A diagonal of $P$ is said to be good if its endpoints divide the perimeter of $P$ into two parts, each having an odd number of sides of $P$. The sides of $P$ are also considered good. Suppose that $P$ has been subdivided into triangles by 2023 diagonals that do not intersect each other inside $P$. How many, at most, such a subdivision can contain isosceles triangles with two good sides?

## Solutions to the exercises

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A triangle will be said to be isobon if it is isosceles and has two good sides. By grouping the 2026 sides of $P$ into 1013 groups of two consecutive sides, 1013 isobon triangles are formed, which can then be integrated into a triangulation.

Now consider any triangulation, and we set out to prove that it does not contain more than 1013 isobon triangles. In the following, we will identify the sides of $P$ with good diagonals. We then consider the bipartite graph $\mathcal{G}=(A \sqcup B, E)$ where $A$ is the set of triangles in the triangulation, $B$ is the set of good diagonals, and $E$ is the set of pairs consisting of a triangle $a \in A$ and a good side of $a$.

Finally, we associate an integer with each good vertex $b \in B$, that is, each good diagonal, as follows: if $b$ separates $P$ into two parts with $x$ and $y$ vertices, we associate the integer $\min \{x, y\}$ with $b$. Consequently, a triangle $a \in A$ is isobon if it is adjacent to two vertices that have been associated with the same integer.

Any triangle $a \in A$ divides $P$ into four regions: $t$ and three other regions that do not share any side between them. Therefore, removing $a$ from the graph $\mathcal{G}$ disconnects $\mathcal{G}$, which means that $\mathcal{G}$ is acyclic: $c^{\prime}$ is a forest.

Moreover, every triangle $a \in A$ has 0 or 2 good sides; its degree in $\mathcal{G}$ is therefore 0 or 2. Consequently, each connected component of $\mathcal{G}$ is in fact a path, which starts from one side of $P$ and ends at another side of $P$: there are 1013 such paths. Now, when traversing a path, the integers associated with the vertices $b \in B$ encountered start by strictly increasing from 1, then they strictly decrease until they are 1 again. The only time two vertices with the same number could have been encountered was when this number was the highest possible on the path. We conclude that each path contains at most one isobon triangle, and therefore that the maximum sought is 1013.

## 8 TD Small Simple Ideas (Aurélien)

This course is based on the TD from Baptiste Séraille's 2020 group E course.

## 2 Midway Training

## Statements