[size=85]₊Modified [url=https://www.geogebra.org/m/pjaqednw]version[/url].[br] Let L be a circle of radius R around the point O: [b]L[/b]:={x∈[size=100]ℝ[/size][sup]2[/sup]: ||x||=R}. [i]There is a set lP={p1, p2,...,pn} of n movable points and [color=#333333]with each p[sub]i[/sub][/color][color=#333333]∈L[/color][color=#333333],[/color][/i][color=#222222]{(xi,yi)∈L: i = 1,...,n} -their coordinates[/color][i]. [/i]Let the points be "repulsive". [color=#222222][color=#1e84cc][i] f(x,y)=[/i][/color][/color][math]\sum_{i=1}^n\sqrt{\left(x-x_i\right)^2+\left(y-y_i\right)^2}[/math][color=#222222][color=#1e84cc][i] -[color=#1e84cc]sum of the distances [/color][color=#333333]from[/color] ([/i][/color][/color][i]x,y) to each point p[/i][sub]i[/sub][i]'s. [i][i][i][color=#333333]g(x,y)[/color][/i][/i]=[math]\sqrt{\left(x\right)^2+\left(y\right)^2}-R[/math][/i] [/i][color=#333333][i] - points (x,y)∈[/i][b]L. [/b][/color][br][b][u]Problem[/u][/b]: use the [b][i]method of Lagrange multipliers[/i][/b] to find distribution of "repulsive" points on a circle corresponding to the [i][b][color=#ff7700]maximum[/color][/b][/i] of their [b]sum of mutual distances[/b]. [br] [b]Step by step[/b], all the points are iterated over each time until each point is at the [i][color=#ff7700]Geometric medians([/color][/i][b][color=#ff7700]GM)[/color][/b] points of the other n-1 points. We assume that the equilibrium - stationary state in system of n "charges" is reached if the sum of their mutual distances is [color=#ff0000]maximal[/color]. [br][i][i] Critical points of [color=#ff7700]maximum[/color] can be found using Lagrange multipliers as Λ(x,y,λ)=f(x,y)+λ*g(x,y) finding the [color=#ff0000]critical points[/color] f(x,y) subject to: g(x,y)=0.[/i][/i][/size]

[size=85] There is a system of equations: ∇f(x,y)= λ*∇g(x,y). A local optimum occurs when ∇f(x,y) and ∇g(x,y) are parallel, and so ∇f is some multiple of ∇g. Here, an iterative formula is used to find the points of the local maxima of the corresponding [i]distance sum[/i] [i]functions[/i]. Relationship between vectors of two consecutive iterations is defined in this case by [size=100][size=200][size=150][math]\vec{r_s}^{\ast}=R\cdot UnitVector\left(\sum^{n,i\ne s}_{i=1}UnitVector\left(\vec{r_s}-\vec{r_i}\right)\right)[/math][/size][/size][/size] - the new position of the [b]s[/b] point: [math]\vec{r_s}^{\ast}[/math]∈[b]L[/b]. All points are successively corrected (s = 1,2, ..., n) until the system of points comes to a stationary state. Finally, at the end of iterations they must match: ps=ps* for all s=1,2,...,n. The number of correction steps is carried out until the center of mass of n particles from lP (with a certain degree of accuracy) will not be in the center of the circle.[br] [b]The solution[/b] of the above Lagrange problem means that in "equilibrium" for any point [b]s[/b] from the set lP, their [b][i]position vectors[/i][/b] i. e. gradient [b][i]∇g[sub]s[/sub][/i][/b]=[math]\vec{r_s}[/math]/R [i][color=#222222][i][color=#222222]-gradient of the magnitude of a position vector r[/color][/i][/color][/i], and the [i][b]sums of the unit vectors[/b][/i] of the other n-1 points: gradient [color=#ff7700]∇f[sub]s[/sub][/color]∼[math]\sum^{n,i\ne s}_{i=1}UnitVector\left(\vec{r_s}-\vec{r_i}\right)[/math][size=85]must be parallel[/size][size=85], i.e. the angle Δφ between these two gradients vectors is multiple of π . [br] The gradients of the [i]function [/i][i][color=#333333]sum of squares of their mutual distances[/color][/i] are [color=#ff00ff]∇f[sub]q s[/sub][/color]∼[math]\sum^{n,i\ne s}_{i=1}\left(\vec{r_s}-\vec{r_i}\right)[/math][/size]. The formulas for the position of the [color=#ff00ff]geometric centers[/color] of systems of n-1 points (on a circle) for the local maxima of the [color=#ff00ff]f[/color][sub][color=#ff00ff]q s[/color] [/sub]-function of distances at each iteration step are determined each time by explicit formulas. [i][color=#333333]Their coordinates, in contrast to [/color][color=#ff7700]Geometric Medians[/color][color=#333333], have [/color][i]explicit formulas[/i][color=#333333]: [/color][color=#ff00ff]GCmax= - R*UnitVector(Sum(lP))[/color][color=#1e84cc]. [/color][color=#333333]This point and[/color][color=#1e84cc] [i]GCmin= R* UnitVector(Sum(lP))[color=#333333]-[/color][/i] [/color][color=#333333]two antipodal points on circle. [br][/color][/i] As it turns out, in the [color=#ff0000]equilibrium position[/color] of the n -points system, [b][i][u]each point is located both[/u][/i][/b] in the [i][b]positions[/b][/i] of the [color=#ff7700][b]Geometric median(GM)[/b][/color] and in the [b][color=#ff00ff]Geometric centers(GC)[/color][/b] of the remaining points.[br] ●Main iteration formula in applet: Execute(Join(Sequence(Sequence("SetValue(A"+i+",CopyFreeObject(Element[Last[IterationList((0,0)+R UnitVector(Sum[Zip[UnitVector(a-lP(r)),r,Sequence[n]\{"+i+"} ]]),a, {A"+i+"},j01)],1] ) )",i,1,n),ii,1,i0) ) ). IterationList -Number of Iterations for each point j01=5, i0 -number of iterations per step. [br] ●An [url=https://www.geogebra.org/m/tnut2f9y]example[/url] of ordering points on a circle is a [i]clear[/i] and[i] illustrative[/i] of using the iterative method of finding [color=#ff7700][i][b]geometric [b][i]medians[/i][/b](GM[/b])[/i][/color] and [color=#ff00ff][b][i]geometric centers(GC)[/i][/b][/color] [u][i]on a bounded domain[/i][/u] for a system of points. Similar considerations were applied to finding a uniform distribution of points on a sphere.[/size]