Selection Examinations for the IMO

Assume, to the contrary, that the number of planes is equal to $1111$. Now, $20$ points in space can define at most $\left(\genfrac{}{}{0ex}{}{20}{3}\right)=\frac{20\cdot 19\cdot 18}{3\cdot 2\cdot 1}=1140$ planes, so $1140-1111=29$ triplets of points lie in the planes already determined by another triplet. If one of the planes contains $7$ or more points, then there are at least $\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35$ triplets of points in this plane and the number of triplets is greater than the number of planes by at least $35-1=34$. Hence, the greatest possible number of planes is $1140-34=1105$. Obviously, this cannot happen if there are $1111$ planes, so each plane can contain at most $6$ of the points.

Let $a$ be the number of planes containing $4$ points, $b$ the number of planes containing $5$ points and $c$ the number of planes containing $6$ points. When counting triplets, we considered each plane containing $4$ points $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ times. That is $3$ times too many. Each plane containing $5$ points was counted $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$ times (i.e. $9$ times too many) and each plane containing $6$ points was counted $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$ times, which is $19$ times too many. The number of planes is thus equal to $1140-3a-9b-19c$. If this number were equal to $1111$, then we would have $3a+9b+19c=29$. The numbers $a$, $b$ and $c$ are non-negative integers, so $c\ge 1$ is not possible, $c=0$. We get $3a+9b=29$ and this is impossible since the left-hand side is divisible by $3$ and the right-hand side is not. We conclude that the number of planes cannot be equal to $1111$.