Team selection test for 50. IMO

We may assume that $S_1\ge S_2\ge S_3$. We shall prove that the answer is $S_1$. It is not difficult to see that the area of the convex hull of two segments $d_1$ and $d_2$ is not less than the area $S$ of a triangle with two sides equal and parallel to these segments. Indeed, if this hull is a triangle, two cases are possible. In the first case, the respective segments are sides and then the area is equal to $S$. In the second case, we may assume that an interior point $E$ of $△ABC$ is such that $CE=d_1$ and $AB=d_2$. Setting $F=AB\cap CE$, then $$2S_{ABC}=AB\cdot CF\sin \angle CFB\ge d_1{d}_2\sin \angle CFB=2S.$$

If the hull is a quadrilateral with diagonals the respective segments, then its area is equal to $S$. It remains to consider the case when the hull is a quadrilateral with two opposite sides the respective segments. We may assume that points $C$ and $D$ on the sides $BE$ and $AE$ of $△ABE$ are such that $AD=d_1$ and $BC=d_2$. Then $$2S_{ABCD}=2S_{ABE}-2S_{CDE}=(AE\cdot BE-DE\cdot CE)\sin \angle AEB>AD\cdot BC\sin \angle AEB=2S.$$ Hence any convex polygon containing two segments equal and parallel to $XB$ and $XC$, is not less than $S_1$. Let now $A^{\prime }$ be the symmetric point of $A$ with respect to $X$, and $D$ be such that $XBDC$ is a parallelogram. Then $A^{\prime }$ lies either in $△BXD$, or in $△CXD$; for example, in $△BXD$. Since $S_{XC{A}^{\prime }}=S_2\le S_1$, then $A^{\prime }$ lies in $△BXD$. This triangle has area $S_1$ and contains three segments equal and parallel to $XA$, $XB$, $XC$.