Re: Points A,B, and C

In article <20080629235956.D9658@[EMAIL PROTECTED]
>,
 William Elliot <marsh@[EMAIL PROTECTED]
> wrote:

> On Mon, 30 Jun 2008, Virgil wrote:
> > William Elliot <marsh@[EMAIL PROTECTED]
> wrote:
> >> On Sun, 29 Jun 2008, [Mr.] Lynn Kurtz wrote:
> >>> On Sun, 29 Jun 2008 16:57:03 EDT, perash <mk_917@[EMAIL PROTECTED]
> wrote:
> >>>
> >>>> Points A,B, and C lie in that order on line `, such that AB = 3 and

> >>>> BC = 2. Point H is such that CH is perpendicular to `. Determine
the 
> >>>> length CH such that \AHB is as large as possible.
> >>>
> >> You mean that the area of AHB is as large as possible?
> >>
> >>> OK,  I got sqrt(10). What did you get?
> >> That's there is no maximum.
> >
> > As I read the problem, it was to find the length of CH needed to
> > maximize the angle AHB at point H, not the area of triangle ABH.
> >
> That makes sense.
> 
> > And I also found the maximizing length of CH to be sqrt(10) yielding
an
> > angle at H of about .443 radians, or 25.4 degrees.
> >
> Ok, so do I using the formula for the
> tangent of the difference of two angles.
 
I used the law of cosines, myself, and  incidentally discovered that the 
maximum angle subtended by AB at H always occurs when CH is the mean 
pro****tional of AC and BC:

      AC/CH = CH/BC or CH^2 = AC*BC