We cut off a corner of the chessboard $(2 n+1) \times(2 n+1)$. For which values of $n$ can the remaining squares be covered by $2 \times 1$ dominoes in such a way that half of the dominoes are horizontal?

## - Solutions -
If $n$ is even, we can easily find a suitable covering. When $n$ is odd, let's show that it is impossible to satisfy the conditions. We color the chessboard in two colors: we color the squares of the first, third, etc. rows in blue and the squares of the second, fourth, etc. rows in red. There are then $2 n^{2}+n$ red squares and $2 n^{2}+3 n$ blue squares, for a total of $4 n^{2}+4 n$ squares. We will therefore need $2 n^{2}+2 n$ dominoes. There will then be $n^{2}+n$ horizontal dominoes and as many vertical ones.

Each vertical domino covers one square of each color. Once the vertical dominoes are placed, there remain $n^{2}$ red squares and $n^{2}+n$ blue squares to be covered by horizontal dominoes. According to the coloring, a horizontal domino covers squares of the same color. It is therefore necessary that $n$ be even. In other words, when $n$ is odd, it will be impossible to cover the chessboard according to the conditions of the statement.

## IV. Advanced Group

## 1 Geometry

## 1 Course

The advanced group studied the notion of the power of a point with respect to a circle, and along the way, some properties of vectors and the dot product were mentioned. We refer to Pierre Dehornoy's course available on the Animath website at the address http://www.animath.fr/IMG/pdf/cours-geom.pdf for the theorems related to the power of a point with respect to a circle.

Here are the different exercises for application that were covered in the course.

## - Statements -