Taken from Pregătire Matematică Olimpiade Juniori. Translated by Amir Hossein Parvardi.

Attachments:
baraj-1-2018.pdf (93kb)
baraj-2-2018.pdf (110kb)

We have $f(a,b)=\sqrt{(a+13)^2+(b+43)^2}$ $f(a,-b)=\sqrt{(a+13)^2+(b-43)^2}$ $f(-a,b)=\sqrt{(a-13)^2+(b+43)^2}$ $f(-a,-b)=\sqrt{(a-13)^2+(b-43)^2}$ Consider $4$ points on the Cartesian plane : $A(13,43) , B(13,-43) ,C(-13,-43) ,D(-13,43) $ $\; \;$ and a point $P(a,b) $ inside the square $ABCD$ $PA+PB+PC+PD$ must be minimal ,which is when $P$ is the center of the square i.e when $P$ has coordinates $(0,0) $ $\Longleftrightarrow f(a, b) + f (a,-b) + f(-a, b) + f (-a, -b) \geq 4f(0,0)=4\sqrt{2018}$

XbenX wrote: We have $PA+PB+PC+PD$ must be minimal ,which is when $P$ is the center of the square Just want to clarify that this is true by the triangle inequality, so $PA+PC\ge AC$ and $PB+PD\ge BD$ with equality when $P$ is the intersection of $AC$ and $BD$