Let G be the centroid of a triangle $ABC$, $P$ any point in the plane, prove that $|AP|^2 +|BP|^2 + |CP|^2=|AG|^2+|BG|^2+|CG|^2+3|PG|^2$

Question

Let G be the centroid of a triangle $\Delta ABC$ and $P$ any other point in the plane, prove that $|AP|^2 +|BP|^2 + |CP|^2=|AG|^2+|BG|^2+|CG|^2+3|PG|^2$

Let the coordinates be: $A=(x_1,y_1),B=(x_2,y_2), C=(x_3,y_3), P=(a,b)$

And let $AD$ be a median, i.e, $D$ is the mid-point of the side $BC$

I can say that the coordinates of $D$ are: $\left(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\right)$

I also know that the coordinates of $G$ are: $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$

I know then:

1. $|AP|^2=(x_1-a)^2+(y_1-b)^2=x_1^2-2ax_1+a^2+y_1^2-2y_1b+b^2$
2. $|BP|^2=(x_2-a)^2+(y_2-b)^2=x_2^2-2ax_2+a^2+y_2^2-2y_2b+b^2$
3. $|CP|^2=(x_3-a)^2+(y_3-b)^2=x_3^2-2ax_3+a^2+y_3^2-2y_3b+b^2$
4. $|AG|^2=(x_1-\frac{x_1+x_2+x_3}{3})^2+(y_1-\frac{y_1+y_2+y_3}{3})^2$
5. $|BG|^2=(x_2-\frac{x_1+x_2+x_3}{3})^2+(y_2-\frac{y_1+y_2+y_3}{3})^2$
6. $|CG|^2=(x_3-\frac{x_1+x_2+x_3}{3})^2+(y_3-\frac{y_1+y_2+y_3}{3})^2$
7. $3|PG|^2=3(a-\frac{x_1+x_2+x_3}{3})^2+(b-\frac{y_1+y_2+y_3}{3})^2$

I find it tedious to try to solve it that way, is there any other option, can I simply say that either $P$ or $G$ is the origin?

Answer 1

You are welcome to set $P$ to be the origin -- you can do this by simply moving your triangle around so that $P$ coincides with the origin. From this the calculations aren't that bad, although they are still a bit tricky.

A better approach is to use the language of vectors instead of coordinates. You can still set $P$ to be the origin there, but the squares of magnitudes are much nicer -- they are simply dot products. The centroid $G$ will be $\frac{a+b+c}3$ as normal.