All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Maximizing product (Posted on 2013-09-19)
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 No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 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 :
BD^2=(e+f)^2+k^2
both triangles ABC and ABD have equal areas:
SABC=e*k=AC*BC*Sin(Gamma1)/2

Sin(Gamma1)=2*e*k/(AC*BC)=2*e*k/(k^2+e^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

 Search: Search body:
Forums (0)