Steve, you are correct, so here is the full solution covering all situations :
The need to maximize Sin(Gamma) remains correct, except for the fact that if k>c/2, Gamma cannot be assumed as 90 deg. so we must calculate Sin(Gamma) and maximize it:
Assume a triangle ABC where a=b, and draw a parralel to c going through point C, at which we observe angle Gamma1.
We now shift point C along the parralel a distance f in the direction of A, thus obtaining a triangle ABD with angle Gamma2.
Denoting : c/2=e, we get the following :
AD^2=(e-f)^2+k^2
BD^2=(e+f)^2+k^2
both triangles ABC and ABD have equal areas:
SABC=e*k=AC*BC*Sin(Gamma1)/2
SABD=e*k=AD*BD*Sin(Gamma2)/2
Sin(Gamma1)=2*e*k/(AC*BC)=2*e*k/(k^2+e^2)
Sin(Gamma2)=2*e*k/(AD*BD)=2*e*k/[sqrt(k^2+(e-f)^2)*sqrt(k^2+(e+f)^2) ]
For maximizing Sin(Gamma), it is needed to minimize the denominator of Sin(Gamma2) by means of changing f :-
denominator^2 = [k^2+(e-f)^2]*[k^2+(e+f)^2] =
(k^2+e^2)^2 + f^2*[2*(k^2-e^2)+f^2]
We distinguish now 2 cases :-
a.
k>e minimization is achieved by f=0
and we get the max. Sin(Gamma) for the case of a=b and Sin(Gamma) = 2*e*k/(k^2+e^2)
and therefore :
Ha*Hb*Hc = c*k*Sin(Gamma) =
c^2*k^2/(k^2+e^2) b.
k<e
The minimal denominator will be reached for : f^2=e^2-k^2
for which :-
Sin(Gamma2)=2*e*k/(2*e*k) =1
and
Ha*Hb*Hc = k*c^2
corrected solution
