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Rectangle Point (Posted on 2017-02-28) Difficulty: 2 of 5

  
Let P be a point in the plane of rectangle ABCD.
Let a, b, c, and d be the distances PA, PB, PC, and
PD respectively.

Find d in terms of a, b, and c.
  

No Solution Yet Submitted by Bractals    
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Solution Solution Comment 1 of 1
The answer was pretty obvious but I had to go synthetic for a proof.

A=(0,0)
B=(a,0)
C=(a,d)
D=(0,d)
P=(x,y)

a=sqrt(x^2+y^2)
b=sqrt((x-a)^2+y^2)
c=sqrt((x-a)^2+(y-d)^2)
d=sqrt(x^2+(y-d)^2)

a^2 + c^2 = b^2 + d^2 = x^2 + (x-a)^2 + y^2 + (y-d)^2

The above is the relationship between the four lengths.  Solving for d gives the requested solution:

d = sqrt(a^2 - b^2 + c^2)



  Posted by Jer on 2017-02-28 11:31:26
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