A surprisingly simple condition for a cubic to have the F-harmonic property.
Theorem
Suppose that K = nK(F2,R) . The following are equivalent:
(1) the intersection of the tripolars of F and the isoconjugate of R lies on K,
(2) K has the F-harmonic property.
Proof.
If F = [f,g,h] and R = [r,s,t] then the intersection is [f(t/h-s/g)].
This is the isoconjugate of the intersection of the circumconics with perspectors F, R.
Suppose that K has equation Σrx((yh)2+(zg)2) + Kxyz = 0.
If we put X =x/f, R = r/f, etcetera. then the equation becomes ΣRX(Y2+Z2) + K/fgh = 0.
The calculation can be done by hand, provided we recognise and remove terms whose
sum is clearly zero. Briefly, the LHS becomes
ΣR(S-T)(symmetric + R2 + 2ST) = ΣR3(S-T)
Now write M = R+S+T, and replace one R by M-S-T to get MΣR2(S-T)
Now write this out and note that it factors as -M(S-T)(T-R)(R-S) so the equation becomes
(S-T)(T-R)(R-S)(-M+K/fgh) = 0, i.e. R+S+T = K/fgh.
This is precisely the condition for K to be F-harmonic.