Articles on the "Property I found" and other types of Centers (excluding the Centroid)?

Question

I am a first-year undergraduate student. I came up with a kind of center property which I cannot find in articles online. I found the center on my own but needed help of mathematicians (@Rahul) on this site.

Currently I'm unsure whether it applies for non-star or open curves. For now, I'm limiting the scope of the definition.

Edit: The function $\overline{r}(x,y)$ may have more than one maximum. Another possibility is to take the average of the maximums. Hence the following is not a center but a property.

However, I would like to know of other possible centers.

Suppose the curve is star-shaped with respect to this center point(s) $\mathbf{p}$ so that any ray emanating from $\mathbf p$ meets the curve exactly once, at say point $\mathbf q$. Then $r = \|\mathbf q - \mathbf p\|$, $\theta$ is the angle between $\mathbf q-\mathbf p$ and the $x$-axis, $\overline{r}(x,y)$ is the average radius $$\overline{r}(x,y)=\frac1{2\pi}\oint_{\mathbf q\in\mathcal C}\|\mathbf q-\mathbf p\|\,\mathrm d\theta.$$

and $\mathbf{p}$ maximize is the average of points maximizing are the points maximizing $\overline{r}$.

(Conveniently, this integral can also be computed for non-star-shaped curves; for a ray that meets the curve multiple times, it amounts to taking the total length of all segments that lie in the interior of the curve.)

Note that "this center property" is not the centroid. This can be shown in Mathematica (thanks to @Rahul)

The center is computed using discretization, Euclidean Distance and Sums.

curve = DiscretizeRegion[
  ImplicitRegion[
   S1[x, y] == 1, {{x, -3, 3}, {y, -4, 4}}], {{-3, 3}, {-4, 4}}, 
  AccuracyGoal -> 8]
q = MeshCoordinates[curve];
edges = First /@ MeshCells[curve, 1];
signedAngle[a_, b_] := Arg[(Complex @@ a)/(Complex @@ b)]
avgRadius[p_] := 
 1/(2 \[Pi]) Abs[Sum[Module[{q1, q2, r, d\[Theta]}, q1 = q[[First@e]];
     q2 = q[[Last@e]];
     r = EuclideanDistance[p, (q1 + q2)/2];(*midpoint approximation*)
     d\[Theta] = signedAngle[q1 - p, q2 - p];
     r d\[Theta]], {e, edges}]]
s = FindMaximum[avgRadius[{x, y}], {{x, 0}, {y, 0}}]

The centroid, which is well-known, can be easily computed using RegionCentroid

J = RegionCentroid[
  DiscretizeRegion[
   ImplicitRegion[
    S1[x, y] == 1, {{x, -3, 3}, {y, -4, 4}}], {{-3, 3}, {-3, 3}}, 
   AccuracyGoal -> 8]]

I plotted $\mathbf{p}$, the maximas and the curve. The red dots reprsent the maximas, the blue dot repesents $\mathbf{p}$ and the black dot represents the centroid.

Show[ContourPlot[S1[x, y] == 1, {x, -4, 4}, {y, -4, 4}], 
 Graphics[{PointSize {Medium}, Blue, 
   Point[{x /. s[[2, 1]],  y /. s[[2, 2]]}]}], 
 Graphics[{PointSize {Medium}, Black, Point[{J[[1]], J[[2]]}]}]]

Here are some examples. I'm unable to solve my center in closed form.

$2x^2+2y^2+7y\sin{(xy)} + 7x\sin{(x)}=1$

$x^2 + x + y^2 + y + \sin(xy) + \sin(3xy)=4$

$(x^2+y^2-1)^2+.415x=.4$

$81y^2-x^4\left(9-x^2\right)=4$

Pros and Cons of "The Center Property"

Pros:

• Its reasonable
• Close to the centroid if star-shaped
• Within the boundary of the curve (if star-shaped)
• Leads to interesting problems regarding the maxima of $\overline{r}(x,y)$

Cons:

• Tedious to solve mathematically. Rarely has a closed form. Rarely has an elegant solution.
• If curve is 2-d, even if $\overline{r}$ is exact, you must find the maxima of $\overline{r}(x,y)$ in 3-d Coordinates
• If the curve is 3-d, you must find the maxima of $\overline{r}(x,y,z)$ in 4-d Coordinates.

Unproven Assumptions

1. If the curve is star-shaped and closed $\overline{r}$ has one maxima
2. If the curve is star-shaped and closed, $\mathbf{p}$ stays inside.
3. If the curve is star-shaped and closed one $\mathbf{p}$ exists
4. The proximity of the centroid and "the property" determines the even distribution of the closed shape.

---

In conclusion, I have the following questions?

Are there other centers excluding the centroid?

Is "my property" new? Can it be applied in theoretical mathematics? Applied math? Physics?

Lastly, if you are intereseted, use this for research. I'm too young to "analyze".

Answer 1

I'm sorry that this answer is quite terse. I don't have time to explain it in more detail, but I've marked it community wiki in case someone else is interested in expanding it (and adding some diagrams, which would help).

Let $\mathbf x$ be a point on the curve, $\mathbf r = \mathbf x-\mathbf p$ be the vector from $\mathbf p$ to $\mathbf x$, and $\phi$ be the angle between $\mathbf r$ and the curve normal $\mathbf n$ (i.e. the unit vector perpendicular to the tangent). One can show by drawing a small diagram that $$\|\mathbf r\|\mathrm d\theta = \mathrm d\ell\cos\phi = \left(\frac{\mathbf r}{\|\mathbf r\|}\cdot\mathbf n\right)\mathrm d\ell.$$ Therefore, the integral is equal to $$\bar r(\mathbf p) = \oint \|\mathbf r\|\,\mathrm d\theta = \oint \frac{\mathbf r}{\|\mathbf r\|}\cdot\mathbf n\,\mathrm d\ell,$$ and by the divergence theorem this equals $$\bar r(\mathbf p) = \iint\left(\nabla\cdot\frac{\mathbf r}{\|\mathbf r\|}\right)\mathrm dA = \iint\frac1{\|\mathbf r\|}\,\mathrm dA$$ where the integral is taken over the area enclosed by the curve. In other words, $\bar r(\mathbf p)$ is the convolution of the indicator function of the area inside the curve with the kernel $1/\|\mathbf r\|$.

This fact makes it easy to construct a counterexample to assumption 1: Consider a dumbbell-shaped curve, enclosing two unit disks centered at $(-2,0)$ and $(2,0)$ connected by a thin path. One can verify numerically that there are two maxima, located in the interior of the two disks. It should also be possible to construct a counterexample to assumption 2 by considering a single disk with a long thin notch cut out of it, so that the center is inside the notch and thus outside the curve.