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Maximizing product (Posted on 2013-09-19) Difficulty: 3 of 5
In a triangle ABC two items are defined : the side AB=c and the height from point C to this side Hc= k.

What is the maximum value of the product Ha*Hb* Hc ?

No Solution Yet Submitted by Ady TZIDON    
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corrected solution | Comment 3 of 4 |
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


  Posted by Dan Rosen on 2013-09-24 13:53:13
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