Stolarsky_Interacting_Particles_midterm_presentation

Background and Project Description Line Ellipse Rectangle Future Work
Interacting Particles
Midterm Report
University of Illinois at Urbana-Champaign
Department of Mathematics
Illinois Geometry Lab
October 29, 2015

Background
n-diameter
A generalization of the diameter. For n points, find them such
that the geometric mean of the distances between them is
maximized.
Equivalently, maximize the product of the distances between
points.
Our Project
Find the n-diameter and the configuration of points that gives
it for a given shape.
This may be very expensive to calculate; it may be better to
make an approximation by finding a configuration with equal
arc lengths or equal Euclidean distances.

Line
Intuitively, one might think that the maximizing conﬁguration on
the line would be to divide the line into equal segments. However,
this is not so. For example,
f = (1 0) (x 0) (1 x) [(1 x) 0] [1 (1 x)] [x (1 x)]
= x2
(x 1)2
(2x 1)
df
dx
= 2
f
x
+ 2
f
x 1
+ 2
f
2x 1
= 0
0 = 2x2
2x x + 1 + 2x2
x + x2
x
0 = 5x2
5x + 1
x =
1
2
✓
1 ±
1
p
5
◆
This problem for the line has already been solved in general.

General Problem
Ellipse
For an ellipse with unital major axis and minor axis of length a (so
a < 1), note that it becomes more line-like as a ! 0. For example,
for n = 3, if we assume that it must be symmetric about the
x-axis, the solution can be found to be points at
p1 = (a, 0)
p2 = x,
r
1
⇣x
a
⌘2
!
p3 = x,
r
1
⇣x
a
⌘2
!
x = a
"
a2 + 3
p
25a4 18a2 + 9
6 (a2 1)
#
As a ! 0, x ! 0 and so

General Problem
Ellipse
As a ! 0, x ! 0 and so
p1 = (a, 0)
p2 = (0, 1)
p3 = (0, 1)
Note that these are solution is not necessarily symmetric for
the n = 3 case.
However, our numerical solution returned a symmetric result,
so this should be a good assumption.
Conjecture: The points for the n-diameter on the ellipse will
alternate sides.

Images
Ellipse - a = 0.9

Images
Ellipse - a = 0.8

Images
Ellipse - a = 0.7

Images
Ellipse - a = 0.6

Images
Ellipse - a = 0.5

Images
Ellipse - a = 0.4

Images
Ellipse - a = 0.3

Images
Ellipse - a = 0.2

Images
Ellipse - a = 0.1

Arc length approximation
Research on the movement of three particles on a given ellipse
Hold the arc length between every two particles the same
Attempt to ﬁnd the distribution of particles with maximum
product of diameters
Find the trajectory of center of mass when three
particles moving

Conjecture
(1) The trajectory of center of mass may be some ’beautiful’
ﬁgure
(2) The center of mass may never fall on the original
(3) The distribution of three particles with maximum product
of diameters may be symmetric

Methodology
Choose the ﬁxed ellipse with expression
x2
52
+
y2
32
= 1
Apply numerical method and recursion in calculation and
programming
Conduct simulation with help of Java and Mathematica

Figures
Figure: Particles distribution with one at rightmost

Figures

Figures
Figure: Trajectory of center of mass.

Figures
Figure: Distribution of particles with maximum product of diameters.

Results and Findings
Trajectory of center of mass is an ellipse with the same shape
as the ellipse that we are researching on, except a much
smaller size
The center of mass never falls on the original
The distribution of three particles with maximum product of
diameters is almost symmetric, with one particle falling on the
lowest position on ellipse. (not exactly symmetric may due to
precision in calculation)

Triangle CoM
Modeling the Trajectory of the Center of Mass
Given a unit square and 3 points on its edges, A, B, and C, that
divide the perimeter of the square into three equal-length pieces,
we want to identify the center of mass of the triangle 4ABC and
model the trajectory of its center of mass in terms of the location
of the three points.
Due to the symmetric nature of the square, it su ces to consider
the movement of one of the three points on one edge
As A travels on the edges of the square in a full cycle, the CoM of
4ABC moves around the center of the square on a small rectangle
for 3 cycles. The vertices of the small rectangle is given by
(4
9 , 4
9 ), (4
9 , 5
9 ), (5
9 , 5
9 ), and (5
9 , 4
9 ).

Triangle CoM

Triangle CoM
3-Diameter Problem
Given the triangle 4ABC given above, we are interested in
maximizing the product of the three diameters, i.e. AB · BC · AC.
It can be shown that the product is maximized when the CoM of
4ABC lies on the 4 vertices of the small rectangle.

Mapping
Finding n-diameters on rectangles
We can represent the location
on the rectangle by a mapping
f : R ! R2. For example,
f (0) = (0, 0)
f (7) = (6, 1)
f (12) = (4, 4)
f (19) = (0, 1)
f (22) = (2, 0)

Future Work
Expand the research from 3-diameters to general n
Improve speed of calculation
Change the restriction same arc length to same length
of diameter

Stolarsky_Interacting_Particles_midterm_presentation

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