Siebeck's Theorem, again. See http://www.american.edu/cas/mathstat/People/kalman/pdffiles/mardenAMM.pdf With a slightly better choice of coordinates than Kalman's paper, the result is slightly more elegant: Place the triangle so that the base is on the _y-axis_, with vertices P1,P2 = +- sqrt(3) i, P3 = 3*cosh(f), for some **complex** number f. (%i1) P:(z-sqrt(3)*%i)*(z+sqrt(3)*%i)*(z-3*cosh(f)); (%o1) (z - sqrt(3) %i) (z + sqrt(3) %i) (z - 3 cosh(f)) (%i2) dP:diff(P,z)/3,expand; 2 (%o2) z - 2 cosh(f) z + 1 (%i3) solve(%,z); 2 2 (%o3) [z = cosh(f) - sqrt(cosh (f) - 1), z = sqrt(cosh (f) - 1) + cosh(f)] (%i4) %,exponentialize; f - f f - f 2 %e + %e (%e + %e ) (%o4) [z = ----------- - sqrt(-------------- - 1), 2 4 f - f 2 f - f (%e + %e ) %e + %e z = sqrt(-------------- - 1) + -----------] 4 2 (%i5) %,radcan; - f f (%o5) [z = %e , z = %e ] i.e., the foci F,F' are located at the complex numbers F=exp(f), F'=exp(-f)=1/exp(f). Alternatively, since the inscribed ellipse must be _tangent_ at the origin (the origin is the midpoint of the triangle base), the foci must be on complementary rays from the origin at an angle of imagpart(f). Under these conditions, f=log(F), where F is one of the foci (as a complex number). Hence, the 3rd triangle vertex is at 3*cosh(log(F)). This provides a reverse construction, in which we are given the triangle base and one focus and we wish to find the 3rd triangle vertex and the other focus. the 3rd vertex = 3*cosh(log(F)), the other focus F'=1/F.