WakeUp wrote:
Let $ABCD$ be a unit square. For any interior points $M,N$ such that the line $MN$ does not contain a vertex of the square, we denote by $s(M,N)$ the least area of the triangles having their vertices in the set of points $\{ A,B,C,D,M,N\}$. Find the least number $k$ such that $s(M,N)\le k$, for all points $M,N$.
Dinu Șerbănescu
We have $k\ge \frac18$ (see $M',N'$ or $M'',N''$ in the picture).
To prove this is best possible, assume there is a construction such that $s(M,N)\ge\frac18$. Then considering the triangles ABM, ADM, BDM we conclude that $M,N$ must be inside the union of the two green triangles, and by symmetry, also in the two red ones. But their intersection only consists of the four points $\{M',N',M'',N''\}$. Each pair among those four points yields a maximal solution.
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