Complex numbers and polynomials - University College Cork

3.3 Heron’s problemGiven two distinct points P, Q on the same side of a line L, is there a point R ∈ Lsuch that|P R| + |RQ| ≤ |P X| + |XQ|, ∀X ∈ L?This is another Existence and Uniqueness Problem that was first considered byHeron 1 who gave a beautiful solution based on the notion of the reflection of a pointin a line. (Recall that P ′ is the reflection of P in L if both points are equidistantfrom L, and their mid-point belongs to L.)Theorem 4 (Heron). Given two distinct points P, Q on the same side of a line L,there is a unique point R ∈ L such that|P R| + |RQ| ≤ |P X| + |XQ|, ∀X ∈ L.Proof. Let P ′ be the reflection of P in L, so that if X ∈ L, then |P X| = |P ′ X|.Join P ′ and Q. The line M passing through these points meets L at the desiredpoint R. For|P X| + |XQ| = |P ′ X| + |XQ| (by reflection)≥|P ′ Q| (by the triangle inequality)= |P ′ R| + |RQ| (since R ∈ M ∩ L)= |P R| + |RQ| (again by reflection).□This result incorporates Heron’s Principle of the Shortest Path of Light: If a ray oflight propagates from point A to point B within the same medium, the path-lengthfollowed is the shortest possible.Remark 4. Snooker players use this fact instinctively!3.4 Graphs of quadraticsThe simplest quadratic is the square function x → x 2 , whose graph in R 2 is givenby the set{(x, y) : y = x 2 : x ∈ R}.This is a subset in the upper half-plane that is symmetric about the vertical axis,and the origin is the lowest point on it, which is its only ‘turning point’. Also, thesquare function is convex.i.e., it is a smile. To see this, let P = (x 1 , y 1 ), Q = (x 2 , y 2 )be two distinct points on its graph, so that y 1 = x 2 1, y 2 = x 2 2. The line through P1 Heron (or Hero) of Alexandria (c. 10 70 AD) was a mathematician and engineer, who is saidto be the greatest experimenter of antiquity. For instance, he is credited with inventing a vendingmachine to dispense water, and a windwheel to operate an organ.7