72nd Czech and Slovak Mathematical Olympiad

We show that the median has the value $q=\frac{k+1}{n+1}$. In the entire solution, $q$ denotes this number. Since the given numbers $n$ and $k$ are odd, the fraction with value $q$ is actually written on the board—for example, it is a fraction $\frac{\frac{1}{2}(k+1)}{\frac{1}{2}(n+1)}$.

According to the comparison with the number $q$, we call a fraction ▷ small if its value is less than $q$, ▷ mean if its value is equal to $q$, ▷ large if its value is greater than $q$.

The number $k\cdot n$ of all fractions on the board is odd. To show that the middle fraction, when they are ordered by their values, has the value $q$, it is sufficient to prove that the number of small fractions is equal to the number of the large ones. (The latter will also mean that the number of mean fractions is odd, which again confirms their existence.)

We match the fractions written on the board—we couple each fraction $i/j$ with the fraction $i^{\prime }/j^{\prime }$ (and vice versa) if and only if $i^{\prime }=k+1-i$ and $j^{\prime }=n+1-j$, which can indeed be rewritten symmetrically as $i+i^{\prime }=k+1$ and $j+j^{\prime }=n+1$. Note that the inequalities $1\le i\le k$ and $1\le j\le n$ apparently hold if and only if $1\le i^{\prime }\le k$ and $1\le j^{\prime }\le n$.

It is obvious that only the fraction $\frac{\frac{1}{2}(k+1)}{\frac{1}{2}(n+1)}$ is „coupled“ with itself and that all the other fractions are actually divided into pairs. If we show that every such pair either consists of one small and one large fraction, or of two mean fractions, we are done.

Thanks to the mentioned symmetry, it suffices to verify that a fraction $i^{\prime }/j^{\prime }$ is small if and only if the fraction $i/j$ is large. The verification is routine:

$$\begin{array}{rl}\frac{i^{\prime }}{j^{\prime }}<\frac{k+1}{n+1}& \iff \frac{k+1-i}{n+1-j}<\frac{k+1}{n+1}\\ & \iff (k+1-i)(n+1)<(k+1)(n+1-j)\\ & \iff (k+1)(n+1)-i(n+1)<(k+1)(n+1)-(k+1)j\\ & \iff i(n+1)>(k+1)j\\ & \iff \frac{i}{j}>\frac{k+1}{n+1}.\end{array}$$

This completes the solution.

ANOTHER SOLUTION.

Let us look at the problem geometrically—we consider the plane with the Cartesian coordinate system $Oxy$. Each fraction $i/j$ that Martin wrote on the board is represented as a point $B$ with coordinates $[j,i]$. We thus get exactly those points $B[j,i]$ of our plane, for which $j\in \{1,2,\dots ,n\}$ and $i\in \{1,2,\dots ,k\}$. The set of these points (that we plotted in the figure for $n=11$ a $k=5$) we denote by $M$ and call it „the grid“. It has the shape of a rectangle with vertices $[1,1]$, $[n,1]$, $[n,k]$ and $[1,k]$. Since numbers $n,k$ are odd, the center $S$ of this rectangle has integer coordinates $j_0=\frac{1}{2}(n+1)$ and $i_0=\frac{1}{2}(k+1)$. The center $S$ is thus itself a point of the grid $M$. Let us add that the line $OS$ has the slope $i_0/j_0$ and let us denote the value of this fraction by $q$ as in the first solution.

Note that the grid $M$ is point symmetric with the center $S$ (in the picture we marked two points $B$ and $B^{\prime }$ where each of them is reflection of the other one). Therefore, there is the same number of points from $M$ above and below the line $OS$. Let us clarify what distinguishes these two equally numerous groups of points „below the line $OS“and„abovetheline$OS“.

The point $B[j,i]$ of the grid $M$ lies below the line $OS$ if and only if the line $OB$ has a smaller slope than the line $OS$, i.e. if $i/j<i_0/j_0=q$ holds. Therefore, exactly those points $B[j,i]$ lie under the line $OS$ that correspond to small fractions $i/j$, as we called them in the first solution. Similarly, the lattice points of $M$ above the line $OS$ correspond to large fractions. So, there is the same number of small and large fractions.