SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO

a. Let us prove that in any such placement of checkers on a triangular board with side $n$, the number of checkers $q$ satisfies the inequality $$q\le \frac{2n+1}{3}\cdot k.$$ The main idea of the proof is double counting. Note that wherever a checker stands, if you count all the cells of the three lines in which it stands, and take into account the cell on which it stands three times, you get $2n+1$. Let us denote by $S$ this number summed over all checkers. We have $$S=(2n+1)q.$$

Now let's find this amount based on the location of the checkers along the lines. Let there be $x_1,\dots ,x_n$ checkers in the $n$ horizontal lines, $y_1,\dots ,y_n$ checkers in the $n$ lines with angle $60^{\circ }$, and in the $n$ lines with angle $120^{\circ }$ there are $z_1,\dots ,z_n$ checkers, as shown in Figure 1.



The $i$-th line has exactly $i$ cells, and each of them is in the same line with $x_i,y_i$ or $z_i$ checkers from that line depending on the angle. Therefore, $$(3)S=1\cdot (x_1+y_1+z_1)+2\cdot (x_2+y_2+z_2)+\cdots +n\cdot (x_n+y_n+z_n).$$ Since there are $q$ checkers in total, but there are no more than $k$ checkers in any row, we have the following conditions: $$\begin{gathered} x_1+\cdots +x_n=y_1+\cdots +y_n=z_1+\cdots +z_n=q, \\ 0\le x_i,y_i,z_i\le \min (i,k)\text{for any }1\le i\le n. \end{gathered}$$ Let $q=mk+r$, where $0\le r<k$ is the remainder when divided by $k$. It is clear that the expression (3) is maximized when the sums $x_i+y_i+z_i$ with higher coefficients are the maximum possible (that is, equal to $3k$). Hence, under these conditions we get $$\begin{array}{rl}(2n+1)q& =(x_1+y_1+z_1)+2(x_2+y_2+z_2)+\cdots +n(x_n+y_n+z_n)\\ & \le n\cdot 3k+(n-1)\cdot 3k+\cdots +(n-m+1)\cdot 3k+(n-m)\cdot 3r\\ & \le \frac{2n-m+1}{2}m\cdot 3k+(n-m)\cdot 3r\\ & \le \frac{6nmk-3m^{2}k+3mk}{2}+3nr-3mr\\ & \le \frac{6n(mk+r)-3m^{2}k+3(mk+r)-3r-6mr}{2}\\ & \le \frac{(6n+3)q-3m^{2}k-3r-6mr}{2}.\end{array}$$ Moving $(2n+1)q$ to the right side, multiplying the inequality by $2k$ and substituting $mk=q-r$, we arrive at a quadratic inequality for $q$: $$\begin{array}{rl}0& \le (6n+3-4n-2)kq-3(mk{)}^2-3rk-6mkr\\ & =(2n+1)kq-3(q-r{)}^2-6(q-r)r-3rk\\ & =(2n+1)kq-3q^2+6qr-3r^2-6qr+6r^2-3rk\\ & =-3q^2+(2n+1)kq+3r(r-k),\end{array}$$ from which we obtain $$3q^2\le (2n+1)kq+3r(r-k)\le (2n+1)kq,\text{that is}q\le \frac{2n+1}{3}k,$$ because $r-k<0$. Since $q$ is an integer, we get $q\le \lfloor \frac{2n+1}{3}k\rfloor$, as required.

b. We have $T(1,n)=\lfloor \frac{2n+1}{3}\rfloor$ and $T(2,n)=\lfloor \frac{4n+2}{3}\rfloor$. Placements of checkers, when equality is achieved in the last inequality, for $k=1$ and $k=2$, are shown in figures 2, 3 and 4.