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**