Let $ABCD$ be a square and $a$ be the length of his edges. The segments $AE$ and $CF$ are perpendicular on the square's plane in the same half-space and they have the length $AE=a$, $CF=b$ where $a<b<a\sqrt 3$. If $K$ denoted the set of the interior points of the square $ABCD$ determine $\min_{M\in K} \left( \max ( EM, FM ) \right) $ and $\max_{M\in K} \left( \min (EM,FM) \right)$.
Octavian Stanasila
Let the perpendicular bisector of $ EF $ meet the plane $ ABCD $ at line $ RS $ such that $ R%Error. "inCD" is a bad command.
$ and $ \s\in BC $.max{$ EM,FM $}in triangle $ RSC $is $ EM $ and in other places is $ FM $ so the required value is minimmum of this values. Min of $ EM $in triangle$ RSC $is$ EP $such that $ P $ is the midpoint of$ RS $.Min of$ FM $in other places is $ FP $so the minnimum of max{$ EM,FM $}is $ EP(=FP) $.We compute this value:let $ CP=x $then according to the Pythagurian equarion $ x^2+b^2=(a\sqrt{2}-x)^2+a^2 $ so $ x=\frac{3a^2-b^2}{2\sqrt{2}a} $ so
$ FP=\sqrt{b^2+x^2}=\sqrt{\frac{9a^4+b^4+2a^2b^2}{8a^2}} $.