An Isosceles Triangle and its Steiner Foci (Posted on 2023-02-17)

	Submitted by Brian Smith
Rating: 5.0000 (1 votes)
Solution:	(Hide)
I'll start by placing a general isosceles triangle with base length b and height h on the complex plane. Its centroid will be at the origin by assigning the peak vertex at (2h/3)*i and the two base vertices at (b/2)-(h/3)*i and -(b/2)-(h/3)*i. We will also have h^2+(b/2)^2=1 for the legs to have length 1. In this position the axes of the Steiner ellipse will coincide with the real axis and complex axis. Then b/3 and -b/3 will be the places where the real axis intersects the legs of the triangle. And -(h/3)*i will be where the imaginary axis intersects the base of the triangle. To find the locations of the foci, Marden's Theorem can be applied to get the foci of the Steiner inellipse. With the centroid of the triangle being the origin, those values can then simply be doubled to get the foci of the Steiner circumellipse. f(z) = z^3 + (h^2/3-b^2/4)z + (2h^3/27+hb^2/6)*i has the three vertices of the triangle as its roots. Then solving f'(z) = 3z^2 + (h^2/3-b^2/4) = 0 will give us z = +/-sqrt[b^2/12-h^2/9] as the foci of the inellipse. Then the foci of the circumellipse are located at z = +/-2*sqrt[b^2/12-h^2/9] Then to answer question 1: we want to solve +/-(b/3) = +/-2*sqrt[b^2/12-h^2/9], with help from h^2+(b/2)^2=1. Square both sides of the first equation and then multiply through by 9 gets us b^2 = 3b^2 - 4h^2. Now substitute h^2=1-(b/2)^2 and solving for b yields b=2/sqrt(3) is the length of the base such that the foci of the Steiner circumellipse lie upon the congruent legs of the isosceles triangle. Then to answer question 2: we want to solve -(h/3)*i = +/-2*sqrt[b^2/12-h^2/9], with help from h^2+(b/2)^2=1. Square both sides of the first equation and then multiply through by 9 gets us -h^2 = 3b^2 - 4h^2. Now substitute h^2=1-(b/2)^2 and solving for b yields b=2/sqrt(5) is the length of the base such that one focus of the Steiner circumellipse lies upon the base of the isosceles triangle.

	Subject	Author	Date
	Solution	Jer	2023-02-17 13:05:58