Informally: In an art gallery with n
paintings what is the optimal position for
a camera?
 Formally: Given a set of n points in the
Euclidean space, find a point in their
convex hull that maximizes the product
of the distances to the n given points

› Convex hull = minimal convex set that
contains all the points
Simple isosceles triangle with vertices
(-1,0), (1,0), and (0,5)
 F(x,y) = prod of the distances (squared)
from a point (x,y) to the three vertices =

[(x+1)2 + (y-0)2] [(x-1)2 + (y-0)2] [(x-0)2 + (y5)2]
Maximum Modulus Principle tells us that
such a point is on the boundary
 Let’s call the point that maximizes F(x,y)
i.e. the product of the square of the
distances, the “art gallery point”


2D – specifically, the case of a triangle
› Why is the art gallery point on a boundary?
› How do you find it?
› Where is it on any given triangle?

3D
› Is the art gallery point still on a boundary?
› Is everything the same as in 2D?
Start on one side of a triangle and pick a random
point on it
 Put this random point’s coordinates in terms of one
variable, alpha
 Create the product of distances function (will be a
function dependent solely on alpha)
 Maximize it: find the derivative and set it equal to 0

› After some cancellation, this will be a cubic equation

Hence, finding the AG point along one side of a
triangle amounts to solving for the roots of a cubic
polynomial
Repeat the steps on the previous slide for
the other two edges
 Now we will have 3 possible AG points
 Compare the actual product of the
distances returned by each point
 The biggest one(s) wins

Easiest case to study and to check
 Observation: an equilateral triangle has
3 AG points, which are at the midpoints
of each side of the triangle
 Proof: verifies that the maximizer of the
product of the distances along an edge
is at the midpoint

This is where it gets interesting
 The big question: Is there one AG point
on the base (i.e. at the origin, like we
saw with equilaterals), or are there 2
symmetrical ones on the legs?


h
3.0

Fixed height
2.5
2.0

1.5
1.0

0.5
a
2
0
2
4
6
8
Horizontal distances from y-axis
10
Fix the height of a
triangle on the yaxis
Drop legs to the xaxis symmetrically
around the y-axis
Run it up to h = 70
and a = 100 (lots of
triangles!)
Question: Per each
height, what is the
range of a’s for
which the origin is
the AG?
Horizontal distances from y-axis
80
60
40
20
0
20
Fixed height values
40
Observations:

Bottom range
adds one to it
every 15

Top of upper
range is almost
linear, goes up
by 2’s, except
on every 10 it
jumps by 3

Bottom of upper
range starts at
first point where
base length
exceeds leg
length
Non-isosceles angle of the triangle
Observations:

Bottom point
hovers around 60,
except for in the
beginning

Upper point hovers
around 129 or so

Good news is if
your angle is within
say 70 to 120,
you’re guaranteed
that the AG is on
the origin
120
100
80
60
40
20
0
20
40
Fixed height values
60
Stationary point = point where the
derivative or gradient = 0
 With a triangle, the stationary point of
the function p(z) is a saddle point
 In 3D, the stationary point is a minimum!
 Shows a break between 2D and 3D

› Otherwise, Kalantari proves the MMP, under
certain conditions, for the 3D case as well

Still don’t know how to tell when looking at
an isosceles triangle where the AG will be…
› What explains those observations I listed?
› Will there eventually be three ranges? What
might they look like?
Is there a way to know when looking at any
triangle, not just isosceles or equilateral?
 Is there a geometric characterization of the
AG point?
 What about with more than 3 points?
 What about 3D (say, a tetrahedron)?

Convex hull
 Complex numbers
 Maximum Modulus Principle
 Gauss-Lucas Theorem
 Stationary points
 Newton’s Method
 Fundamental Theorem of Algebra
 Good old algebra and calculus
