# Average Distance to N Points

### Solution 1

Let the given points be $A_1,\ldots,A_n.\,$ Choose any diameter $M_1M_2\,$ of the circle. Then, by the triangle inequality, for any point $A_k,\,$ $2=M_1M_2\le M_1A_k+M_2A_k.\,$ Add up the $n\,$ inequalities:

$\displaystyle 2n\le\sum_{k=1}^n(M_1A_k+M_2A_k)=\sum_{k=1}^nM_1A_k+\sum_{k=1}^nM_2A_k.$

WLOG, assume $\displaystyle \sum_{k=1}^nM_1A_k\le\sum_{k=1}^nM_2A_k.\,$ Then $\displaystyle n\le\sum_{k=1}^nM_2A_k\,$ and $\displaystyle\frac{1}{n}\sum_{k=1}^nM_2A_k\ge 1.$

### Solution 2

Let the center of the circle be at $(0,0).\,$ Let $(x_i,y_i),\,$ $i\in\overline{1,n}\,$ be the coordinates of the $n\,$ points. Let us introduce the following notations: $\langle x\rangle=\displaystyle\sum_{k=1}^nx_k,\,$ $\langle y\rangle=\displaystyle\sum_{k=1}^ny_k,\,$ $\langle x^2\rangle=\displaystyle\sum_{k=1}^nx_k^2,\,$ $\langle y^2\rangle=\displaystyle\sum_{k=1}^ny_k^2,\,$ $D=\sqrt{\langle x\rangle^2+\langle y\rangle^2}.$

Choose point $M\,$ to have coordinates $(-\langle x\rangle/D, -\langle y\rangle/D).\,$ It is easy to check that this point lies on the unit circle. Thus, the average of the distances of point $M\,$ to the $n\,$ points is given by

\displaystyle\begin{align} &\frac{1}{n}\sum_{k+1}^n\left[\left(\frac{\langle x\rangle}{D}+x_k\right)^2+\left(\frac{\langle y\rangle}{D}+y_k\right)^2\right]\\ &\qquad =\left(\frac{\langle x\rangle^2}{D^2}+2\frac{\langle x\rangle^2}{D}+\langle x^2\rangle\right)+\left(\frac{\langle y\rangle^2}{D^2}+2\frac{\langle y\rangle^2}{D}+\langle y^2\rangle\right)\\ &\qquad =1+\left[\langle x^2\rangle+\langle y^2\rangle+\frac{2}{D}(\langle x\rangle^2+\langle y\rangle^2)\right]\\ &\qquad =1+(\langle x^2\rangle+\langle y^2\rangle+2D)\\ &\qquad \ge 1. \end{align}

### Acknowledgment

This is a slightly modified problem from V. V. Prasolov's Problems in Planimetry v. II, 1986 (in Russian). Solutions 2 has been communicated on twitter.com by Amit Itagi.