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?

enmobladatn 2022-07-23 Answered
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 . In this case I would have w = 1 and P D < 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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Answers (1)

Rihanna Robles
Answered 2022-07-24 Author has 18 answers
Step 1
We 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 2
So, 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 = 2
Hence, 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 )

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