 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 Fixed height

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 Fixed height values Horizontal distances from y-axis

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-isosceles angle of the triangle

 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