10 points outside a unit circle

Question

Let $P_1$, $P_2$,... $P_{10}$ be ten points outside the unit circle centered at the origin $O$. Given that $\|P_iP_j\|\ge 1/\sqrt{2}$ for all $1\le 1<j\le 10$, find the minimum of the sum of the distances from $O$ to the points $P_i$, that is,

$$minimize\left(\sum_{i=1}^{10} \|OP_i\|\right)$$

MAPLE gives a minimum of $11.441\ldots$ when the points are the vertices of a regular decagon centered at $O$. I would like to see a standard proof of this result. Even a lower bound of about $11.32$ would be sufficient for my purposes.

The problem is connected to a 1951 paper of Bateman and Erdos that appeared in Amer. Math. Monthly 'Geometrical Extrema Suggested by a Lemma of Besicovitch'.

Answer 1

The target function is continuous and we can restrict the domain to points in the compact annulus $1\le r\le 3$. Hence the minimum of the function is attained. Therefore it suffices to show for any configuration that is not a regular decagon centered at $O$ and with side length $1/\sqrt 2$, there exists a better configuration. This way you can readily show that any configuration where not every point has two neighbours at distance exactly $1/\sqrt 2$ is suboptimal. After that, if you have four consecutive points $P_1,P_2,P_3,P_4$ where $\|OP_2\|>||OP_1\|$, say, then it should not be too hard to show that this configuration can be improved by moving $P_2$ and $P_3$ ...

Answer 2

You might consider the following image:

Each of the black circles represents the minimum distance points must be away from each other. The blue circle is the unit circle.

Notice that moving any single point out of this configuration necessarily either places a point into the unit circle or increases the sum of distances to the origin.

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Edit: There is no rigorous proof for the decagon because there is at least one configuration that has a lesser sum. Consider the following image:

Notice that $8$ of the points are on the unit circle. Also notice that placing a point on the "outer boundary" of any of the black circles will yield a point that is $\le1+\frac{1}{\sqrt2}$ away from the origin. (EDIT) Repeating for the tenth point such that it is still the appropriate distance from the ninth and is on one of the eight circles, the increase is the same. (end EDIT) From this, you get a configuration whose sum is $\le11.416$, which is less than the decagon.