Sum of squares of distances from vertices of a triangle

 If $ABC$ is a triangle with centroid $M$, and $P$ is any point then show that

$$PA^2 + PB^2 + PC^2 = 3PM^2 + \frac{1}{3}(AB^2 + BC^2 + AC^2)$$

Scroll down for a surprising solution.

Let $A_1,B_1, C_1$ be the midpoints of $BC, AC$ and $AB$ respectively.

Triangle $A_1B_1C_1$ (referred to as medial triangle of $ABC$) is similar to $ABC$ 

Using Appolonius theorem we get

$$PA^2 + PB^2 = 2(PC_1^2 + AC_1^2) = 2PC_1^2 + 2\frac{AB^2}{4}$$

$$PB^2 + PC^2 = 2(PA_1^2 + BA_1^2) = 2PA_1^2 + 2\frac{BC^2}{4}$$

$$PC^2 + PA^2 = 2(PB_1^2 + CB_1^2) = 2PB_1^2 + 2\frac{AC^2}{4}$$

Adding we get

$$PA^2 + PB^2 + PC^2 = PA_1^2 + PB_1^2 + PC_1^2 + \frac{AB^2}{4} + \frac{BC^2}{4} + \frac{AC^2}{4}$$

Thus if $A_nB_nC_n$ is the medial triangle of $A_{n-1}B_{n-1}C_{n-1}$  ($ABC = A_0 B_0C_0$) we have that

$$PA_{n-1}^2 + PB_{n-1}^2 + PC_{n-1}^2 = PA_n^2 + PB_n^2 + PC_n^2 + \frac{A_{n-1}B_{n-1}^2}{4} + \frac{B_{n-1}C_{n-1}^2}{4} + \frac{A_{n-1}C_{n-1}^2}{4}$$

 

We can easily show that $A_n, B_n, C_n$ all converge to $M$ ($A_nM = 3A_{n+1}M$), and observing the above is a telescoping series and that sides of a medial triangle are half the original triangle leads us 

$$PA^2 + PB^2 + PC^2 = 3PM^2 + \frac{AB^2 + BC^2 + AC^2}{4}(1 + \frac{1}{4} + \frac{1}{4^2} + \dots)$$

$$= 3PM^2 + \frac{1}{3}(AB^2 + BC^2 + AC^2)$$