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 + (y-5)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?

Fix the height of a triangle on the y-axis

Drop legs to the x-axis 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

Fix

ed h

eigh

t

2 0 2 4 6 8 10a

0 .5

1 .0

1 .5

2 .0

2 .5

3 .0

h

0 20 40

20

40

60

80

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 lengthFixed height values

Hor

izon

tal d

ista

nces

fro

m y

-axi

s

0 20 40 60

20

40

60

80

100

120

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

Fixed height values

Non

-isos

cele

s an

gle

of t

he t

riang

le

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