Question

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 1

k = 46

Step-by-step explanation:

Let P = (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

= (x^2 - 8x + 16) + (y^2 + 2y + 1) + (x^2 + 6x + 9) + (y^2 - 2y + 1) + (x^2 - 10x + 25) + (y^2 + 6y + 9)

= 3x^2 - 12x + 3y^2 + 6y + 61.

Completing the square in $x$ and $y$, we get

PA^2 + PB^2 + PC^2 &= 3(x^2 - 4x) + 3(y^2 + 2y) + 61

= 3(x^2 - 4x + 4) - 3*4 + 3(y^2 + 2y + 1) - 3*1 + 61

= 3(x - 2)^2 + 3(y + 1)^2 + 46.

Thus, Q = (2,-1), and

PA^2 + PB^2 + PC^2 = 3PQ^2 + 46, so the constant k is 46.

We see that Q is the centroid of triangle ABC because the coordinates of Q are the average of the coordinates of the vertices of ABC. From this result, we can conclude that PA^2 + PB^2 + PC^2 is minimized when P is the centroid and that the minimum value is 46.