Problem:
For a point P=(a,a2) in the coordinate plane, let ℓ(P) denote the line passing through P with slope 2a. Consider the set of triangles with vertices of the form P1=(a1,a12), P2=(a2,a22),P3=(a3,a32), such that the intersections of the lines ℓ(P1),ℓ(P2),ℓ(P3) form an equilateral triangle Δ. Find the locus of the center of Δ as P1P2P3 ranges over all such triangles.
Solution:
For 1≤i<j≤3, solving the system y=2xix−xi2=2xjx−xj2 yields the intersection (2xi+xj,xixj) of lines ℓi and ℓj. Hence the center of the equilateral triangle is
Let 0∘≤αi<180∘ be the standard angle formed by lines ℓi and the positive x-axis. Without loss of generality, we may assume that α1<α2<α3. By the given condition, we have α2−α1=α3−α2=60∘. By the subtraction formulas, we have
tan60∘=1+tanα1tanα2tanα2−tanα1=1+tanα2tanα3tanα3−tanα2 and tan120∘=1+tanα3tanα1tanα3−tanα1
or
3=1+4x1x22x2−2x1=1+4x2x32x3−2x2 and −3=1+4x3x12x3−2x1
Because lines ℓ1,ℓ2,ℓ3 are evenly spaced with 60∘ between each other, slopes 2x1,2x2,2x3 are symmetric with each other; that is,
x1+x2+x3=12xi2−112xi3−9xi for i=1,2,3
Therefore,
Ox=3x1+x2+x3=12x2−14x3−3x
where −∞<x<∞, because x=xi for some i=1,2,3, and the combined ranges of slopes 2xi are the interval (−∞,∞). Because 4x3−3x=Ox(12x2−1) is a cubic equation, it has a real root in x for every real number Ox; that is, the range of Ox is the interval (−∞,∞). We conclude that the locus of O is line y=−41.