Putnam 2016, Problem B-3

Suppose that \( S \) is a finite set of points in the plane such that the area of triangle \( \triangle ABC \) is at most 1 whenever \( A, B, \) and \( C \) are in \( S \). Show that there exists a triangle of area 4 that (together with its interior) covers the set \( S \).

Solution:
Since $S$ is finite, we can choose three points $A,B,C$ in $S$ so as to maximize the area of the triangle $ABC$. Let $A',B',C'$ be the points in the plane such that $A,B,C$ are the midpoints of the segments $B'C',C'A',A'B'$; the triangle $A'B'C'$ is similar to $ABC$ with sides twice as long, so its area is 4 times that of $ABC$ and hence no greater than 4.

We claim that this triangle has the desired effect; that is, every point $P$ of $S$ is contained within the triangle $A'B'C'$. (To be precise, the problem statement requires a triangle of area exactly 4, which need not be the case for $A'B'C'$, but this is trivially resolved by scaling up by a homothety.) To see this, note that since the area of the triangle $PBC$ is no more than that of $ABC$, $P$ must lie in the half-plane bounded by $B'C'$ containing $B$ and $C$. Similarly, $P$ must lie in the half-plane bounded by $C'A'$ containing $C$ and $A$, and the half-plane bounded by $A'B'$ containing $A$ and $B$. These three half-planes intersect precisely in the region bounded by the triangle $A'B'C'$, proving the claim.