60th Belarusian Mathematical Olympiad

Let $n$, $k$, $m$ denote respectively the number of red, green, blue points. Then $m+n+k=15$. From (*) in the solution of Problem 2, Category C, we have $$39m\le 51k+n,(1)$$ $$51k-n\le 39m.(2)$$ Substitute for $k$ in (1) to obtain (after simple manipulations) $$18m+10n\le 153\Rightarrow 9m+5n\le 76.(3)$$ Substitute for $k$ in (2) to obtain $51\cdot 15\le 90m+52n$. Using $n\le 13$, we obtain $51\cdot 15\le 90m+50n+26$, hence $$74\le 9m+5n.(4)$$ If $n\le 7$, then we have $51\cdot 15\le 90m+50n+14$ or $76\le 9m+5n$. From (3), it follows that $9m+5n=76$. However, it is easy to verify that the last equality cannot be valid for $n\le 7$. Verifying the values $n=8,9,\dots ,13$, we see that inequalities (3) and (4) can be valid only if $n=8$ or $n=13$. Then we have respectively $m=4$, $k=3$ and $m=k=1$. Examples for these values can be constructed similarly to the example of Problem 2, Category B.

Therefore, the answer is $(n,k,m)=(8,3,4)$ or $(13,1,1)$.