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Question 1202364: Points A, B, and C are given in the coordinate plane. There exists a point Q and a constant k such that for any point P,
PA^2 + PB^2 + PC^2 = 3PQ^2 + k.
If A = (4,-1), B = (-3,1), and C = (5,-3), then find the constant k.
Answer by ikleyn(52803)   (Show Source):
You can put this solution on YOUR website! .
See the solution at this link
https://web2.0calc.com/questions/help-super-hard-geo-prob
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Yes, in that link calculations are not correct, but the idea works perfectly.
Below I placed my own calculations, following the same idea.
Let P be (x,y). Then
PA^2 + PB^2 + PC^2 =
= (x-4)^2 + (y+1)^2 + (x+3)^2 + (y-1)^2 + (x-5)^2 + (y+3)^2 =
= 3x^2 + 3y^2 - 12x + 6y + 61 =
= 3(x^2 + y^2 - 4x + 2y) + 61 =
= 3((x-2)^2 + (y+1)^2) + 46.
This shows that if Q = (2,-1), then PQ^2 = (x-2)^2 + (y+1)^2.
And so PA^2 + PB^2 + PC^2 = 3PQ^2 + 46.
ANSWER. k = 46.
Solved.
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