Question

IF G be centroid of a triangle ABC and P be any other point in the plane ,prove that PA2 + PB2 + PC2 = GA2 + GC2 + GB2 + 3GP2.

Answer 1

let the coordinates of A, B and C be (x1.y1), (x2.y2) and (x3.y3)
           so coordinates of centroid G will be ((x1­ + x2 + x3)/3, (y1 + y2 + y3)/3 )
           ;et us the assume that centroid lies on the origin
           ⇒ x1­ + x2 + x3 = 0 and y1 + y2 + y3 = 0
         and taking the coordinates of point O be (x, y)
           to prove OA² + OB² + OC² = GA² + GB² + GC² + 3GO²
         then... L.H.S. = OA² + OB² + OC²
           (x – x1)² + (y – y1)² + (x – x2)² + (y – y2)² + (x – x3)² + (y – y3)²
           = 3x2 + 3y2 + x12 + x22 + x32 + y12 + y22 + y32 – 2x (x1 + x2 + x3) – 2y(y1 + y2 + y3)
           = 3x² + 3y² + x12 + x22 + x32 + y12 + y22 + y32
           R.H.S. = GA² + GB² + GC² + 3GO²
          
           = (x1 – 0)² + (y1 – 0)² + (x2 – 0)² + (y2 – 0)² + (x3 – 0)² + (y3 – 0)² + 3[(x – 0)2 + (y – 0)²]
           = x12 + y12 + x22 + y22 + x32 + y32 + 3(x2 + y2)
           = 3x2 + 3y2 + x12 + x22 + x32 + y12 + y22 + y32
           = L.H.S.
∴LHS=RHS

Answer 2

Let the coordinates of A, B and C be (x1.y1), (x2.y2) and (x3.y3)
Then coordinates of centroid G will be ((x1­ + x2 + x3)/3, (y1 + y2 + y3)/3 )
Let us the assume that centroid lies on the origin
⇒ x1­ + x2 + x3 = 0 and y1 + y2 + y3 = 0
Let the coordinates of point O be (x, y)
We have to prove OA2 + OB2 + OC2 = GA2 + GB2 + GC2 + 3GO2
Now, L.H.S. = OA2 + OB2 + OC2
(x – x1)2 + (y – y1)2 + (x – x2)2 + (y – y2)2 + (x – x3)2 + (y – y3)2
= 3x2 + 3y2 + x12 + x22 + x32 + y12 + y22 + y32 – 2x (x1 + x2 + x3) – 2y(y1 + y2 + y3)
= 3x2 + 3y2 + x12 + x22 + x32 + y12 + y22 + y32
R.H.S. = GA2 + GB2 + GC2 + 3GO2

= (x1 – 0)2 + (y1 – 0)2 + (x2 – 0)2 + (y2 – 0)2 + (x3 – 0)2 + (y3 – 0)2 + 3[(x – 0)2 + (y – 0)2]
= x12 + y12 + x22 + y22 + x32 + y32 + 3(x2 + y2)
= 3x2 + 3y2 + x12 + x22 + x32 + y12 + y22 + y32
= L.H.S.