62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour

On a coordinate plane consider points $A(-4,-2)$, $B(2,6)$, $C(5,6)$ and $D(5,-4)$ (fig. 2). For any point $M(x,y)$ of the plane the inequality is rewritten as: $MA+MD-MB-MC\le 20$. Let's find the largest possible value of the expression

Fig. 2 $$S(M)=MA+MD-MB-MC.$$ For any point $M$ of the plane from the triangle inequality we get: $$S(M)=(MA-MB)+(MD-MC)\le AB+CD.$$ In addition, for the point $X=AB\cap CD$ we get $$S(X)=(XA-XB)+(XD-XC)=AB+CD=10+10=20.$$ So, the largest value is achieved for point $X$, and for all others the inequality is strict.