Find maximal numbers of planes, such there are $6$ points and 1) $4$ or more points lies on every plane. 2) No one line passes through $4$ points.
Problem
Source: Belarusian MO 2008, 11.4
Tags: geometry, combinatorics, point set
05.11.2016 22:12
A Boring Straightforward Solution: I claim that the answer is $\boxed{\text{six}}$, arising from the construction of tetrahedron $WXYZ$ as well as the midpoints of $WX$ and $YZ$ Suppose there is some plane $H$ with $\text{six}$ points in it. Then the maximum possible number of planes is $\text{one}$, since no line in that plane contains $\text{four}$ points. Suppose there is some plane $H$ with $\text{five}$ points in it. Then each other plane must have exactly $\text{three}$ points in it. But the maximum possible number of triplets of collinear points is $\text{two}$ since by the condition no two triplets can themselves be colinear, and two distinct lines can only intersect once. So we can have at most $\text{three}$ planes in this case. As an aside, this local extremum is the square pyramidal geometry. The only remaining case is that there is no plane with more than $\text{four}$ points in it. Choose a plane and call it $H$. Call the two points not in $H$ as $P$ and $Q$. First suppose that there are no three colinear points in $H$. Then every plane other than $ABCD$ must have exactly two points in $H$ (along with $P,Q$), and hence must contain $PQ\cap H$. Call this point $R$. We know that if two points in $H$ are coplanar with $P,Q$, then $R$ is on the line containing them. But at most $\text{three}$ such lines can be concurrent at $R$, since otherwise we would need more than $\text{four}$ points in $H$. But then we have a maximum of $\text{four}$ planes. As an aside, this local maximum is the trigonal bipyramidal geometry. Now suppose that there exist three colinear points in $H$. Call them $A,B,C$ with $B$ twixt $A,C$, and call the non colinear point $D$. Already we have the $\text{three}$ planes $ABCD,ABCP,ABCQ$. Any other plane must have two points in $H$ (along with $P,Q$), and hence must contain the point $PQ\cap H$. Call this point $R$, which cannot be on line $AC$, since otherwise there is a plane containing $A,B,C,P,Q$. Hence $R$ can intersect at most $AD,BD,CD$ when $R=D$, and this leads us to our extremal construction.