A point $P$ lies in the same plane as a given square of side $1$.Let the vertices of the square,taken counterclockwise,be $A,B,C,$ and $D$.Also,let the distances from $P$ to $A,B,$ and $C$, respectively, be $u,v$ and $w$.
What is the greatest distance that $P$ can be from $D$ if $u^2+v^2=w^2$ ?
Some thoughts I had:
$1)$ Given a pair of vertices I could construct an ellipse with $P$ as a point on the ellipse.
$2)$ From the equality $u^2+v^2=w^2$ I think that I have to consider the case where the angle between $u$ and $v$ is $90^\circ$. In this case I would have $w=1$ and $PD \lt 2$
That being said,I still fail to come at a concrete solution of the problem,it might be that none of my thoughts are right... I don't know.
If that helps,this problem comes from the chapter of my book regarding conics .