# 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?

Greatest distance one point can have from a vertice of a square given following conditions
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 $PD<2$.
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...
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Step 1
We may suppose that $A\left(1,0\right),B\left(1,1\right),C\left(0,1\right),D\left(0,0\right),P\left(x,y\right).$
Then, we have ${u}^{2}=\left(x-1{\right)}^{2}+{y}^{2}$
${v}^{2}=\left(x-1{\right)}^{2}+\left(y-1{\right)}^{2}$
${w}^{2}={x}^{2}+\left(y-1{\right)}^{2}$
Step 2
So, we have ${u}^{2}+{v}^{2}={w}^{2}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}2\left(x-1{\right)}^{2}+{y}^{2}+\left(y-1{\right)}^{2}={x}^{2}+\left(y-1{\right)}^{2}$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{x}^{2}-4x+2+{y}^{2}=0\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\left(x-2{\right)}^{2}+{y}^{2}=2$
Hence, we want to find the greatest distance from the origin to a point on a circle $\left(x-2{\right)}^{2}+{y}^{2}=2$.
Thus, the answer is $2+\sqrt{2}$ when $P\left(2+\sqrt{2},0\right)$

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