Planet 9 From Outer Space

A physics approach: work with vectors, and make a good choice of coordinates.

Let $n = 8$ be the number of points, with coordinates $\mathbb{r}_i$ . Without loss of generality we can choose the origin of our coordinate system lies at the center of mass, so that $\sum \mathbb{r}_i = \mathbb{0}$ .

Let $n^2 D$ the sum of squares of distances between the points. (We choose $n^2$ because we expect this sum to increase quadratically in the number of points.) We can write $D = \frac 1{n^2} \sum_i \sum_{j < i} |\mathbb{r_i}-\mathbb{r_j}|^2 \\ = \frac 1{2n^2} \sum_i \sum_j \left(\mathbb{r_i}^2 + \mathbb{r_j}^2 - 2(\mathbb{r_i}\cdot\mathbb{r_j})\right) \\ = \frac 1n \sum_i \mathbb{r_i}^2.$ (The sum of the dot product $\mathbb{r_i}\cdot\mathbb{r_j}$ equals zero because of our assumption about the coordinate system.)

We add a new point at position $\mathbb{p}$ . Let $nE$ be the sum of squares of distances between this point and the other points. Then $E = \frac 1n \sum_i |\mathbb{r_i} - \mathbb{p}|^2 = \frac 1n \sum_i \mathbb{r_i}^2 + \mathbb{p}^2.$ (Again, the mixed products $\mathbb{r_i} \cdot \mathbb{p}$ add up to zero.)

Subtracting the equation for $D$ from the equation for $E$ , we obtain $E - D = \left(\frac 1n\sum_i \mathbb{r_i}^2 + \mathbb{p}^2\right) - \left(\frac 1n \sum_i \mathbb{r_i}^2\right) = \mathbb{p}^2.$ It follows that the position of the new point, $\mathbb{p}$ , satisfies the equation $\mathbb{p}^2 = E - D,$ in other words, it lies anywhere on a circle around the origin (= center of mass of the original points) with radius $R = \sqrt{E - D}$ . Plugging in the given values we have $D = \frac{1216}{8^2} = 19,\ \ E = \frac{800}{8} = 100, \\ R = \sqrt{100-19} = \boxed{9}.$

Here's one solution, showing the locations of the $8$ points such that the sum of squares of the $28$ distances is $1216$ , and showing the locus of the $9$ th point which is a circle of radius $9$ , and such that the sum of the squares of the distances from any point on it to each of the $8$ points on the lattice is $800$ . The radius is independent of the locations of the $8$ points provided the sum of the squares of the $28$ distances is $1216$ .

However, one need not even determine the locations of the $8$ points on the lattice to find $R$ . Let $n$ be the number of points, $T$ be the sum of the squares of the distances of each of the $\frac{1}{2}n(n-1)$ pairs of points, $S$ be the sum of the squares of the distances from any point on the locus to each of the $n$ points on the lattice, and let $R$ be the radius of the locus which is always a circle. Then the following relation holds

$nS={n}^{2}{R}^{2}+T$

Thus, for $n=8$ , $\;$ $S=800$ , $\;$ $R=9$ , $\;$ $T=1216$ , $\;$ we have

$8\cdot800=64\cdot81+1216=6400$

The algebra works out as follows, where $(x,y)$ is the coordinate of any point on the locus, and $\left( { x }_{ i },\; { y }_{ i } \right) \,$ the coordinates of the $n$ points.

$nS=n\left( \displaystyle \sum _{ i=1 }^{ n }{ { \left( { x }_{ i }-x \right) }^{ 2 } } +\displaystyle \sum _{ i=1 }^{ n }{ { \left( { y }_{ i }-y \right) }^{ 2 } } \right)$

${ n }^{ 2 }{ R }^{ 2 }={ n }^{ 2 }\left( { \left( x-\dfrac { 1 }{ n } \displaystyle \sum _{ i=1 }^{ n }{ { x }_{ i } } \right) }^{ 2 }+{ \left( y-\dfrac { 1 }{ n } \displaystyle \sum _{ i=1 }^{ n }{ { y }_{ i } } \right) }^{ 2 } \right)$

$T=\dfrac { 1 }{ 2 } \left(\displaystyle \sum _{ i=1 }^{ n }{\displaystyle \sum _{ j=1 }^{ n }{ { \left( { x }_{ i }-{ x }_{ j } \right) }^{ 2 } } +\displaystyle \sum _{ i=1 }^{ n }{ \displaystyle \sum _{ j=1 }^{ n }{ { \left( { y }_{ i }-{ y }_{ j } \right) }^{ 2 } } } } \right)$

From these, it can be determined that for any $n$ the relation holds, as all the variables drop out.

Note that the center of the circle locus is the centroid of the $n$ points.

It works for this configuration of points as well.

Distance between two points can be seen as result of addition of two vectors which have common center of all circles as one endpoint. Sum of squares of distances works out as sum of radii of circles giving the value of 1216. The weighted squared radii average for 3 circles is $(2*24.5 + 4*18.5+2*14.5)/8=19$ ,

The radius of the circle for the 9th point can be calculated as $\sqrt{800/8 - 19}=9$ .

This is more irregular configuration with center of mass having integer coordinates.