Greatest distance one point can have from a vertice of a square given following conditionsA 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 ∘ . In this case I would have w = 1 and P D &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...

Question

Greatest distance one point can have from a vertice of a square given following conditionsA 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    ∘  . In this case I would have   w  =  1 and   P  D  &amp;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...

Rihanna Robles · Accepted Answer

Step 1We may suppose that   A  (  1  ,  0  )  ,  B  (  1  ,  1  )  ,  C  (  0  ,  1  )  ,  D  (  0  ,  0  )  ,  P  (  x  ,  y  )  .Then, we have       u    2    =  (  x  −  1      )    2    +      y    2        v    2    =  (  x  −  1      )    2    +  (  y  −  1      )    2        w    2    =      x    2    +  (  y  −  1      )    2  Step 2So, we have       u    2    +      v    2    =      w    2      ⟺    2  (  x  −  1      )    2    +      y    2    +  (  y  −  1      )    2    =      x    2    +  (  y  −  1      )    2      ⟺        x    2    −  4  x  +  2  +      y    2    =  0    ⟺    (  x  −  2      )    2    +      y    2    =  2Hence, we want to find the greatest distance from the origin to a point on a circle   (  x  −  2      )    2    +      y    2    =  2.Thus, the answer is       2    +          2       when   P  (  2  +      2    ,  0  )

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?

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