• Dear Milorad ... How did you obtain that equation? I think that if we apply twice the Pythagorean theorem (in the triangles DAB, DAC) we get: 3b^2 + 2c^2 =
Message 1 of 39 , Oct 2, 2003
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[APH]:
>Construct a triangle if are given:
>a, h_a [=altitude from A], 3b^2 + 2c^2 = min.

>If we denote by D the feet of perpendicular
>from A to BC and take BD=a-x,DC=x then
>3(b^2)+2(c^2)=7(x^2)-4ax+2(a^2)+3(h_a)^2
>has minimum for x=(2a)/7 which is construstible.

How did you obtain that equation?

I think that if we apply twice the Pythagorean theorem
(in the triangles DAB, DAC) we get:

3b^2 + 2c^2 = 5x^2 - 4ax + 2a^2 + 5h_a^2

f(x) = 5x^2 - 4ax +[2a^2 + 5h_a^2 - (3b^2 + 2c^2)] = 0

Now, since the coefficient of x^2 is >0, the min

is: -(-4a)/5 = 4a/5

But I don't think that the old geometer (who had posed the problem)
would be very happy with that solution!

APH
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• Dear Antreas There are two solutions: SOLUTION 1): A triangle A1B1C1 (A1B1 = A1C1), B1C1 = a, and the altitude from A1, h1 =a sqrt(2+sqrt(5))/2. If B1H1 is the
Message 39 of 39 , Apr 12, 2011
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Dear Antreas

There are two solutions:

SOLUTION 1): A triangle A1B1C1 (A1B1 = A1C1), B1C1 = a, and the altitude from A1, h1 =a sqrt(2+sqrt(5))/2.

If B1H1 is the altitude from B1, then B1H1: H1C1 = golden ratio = (1+sqrt(5))/2.

SOLUTION 2): A triangle A2B2C2 IS A (A2B2 = A2C2), B2C2 = a, and the altitude from A2, h2= a sqrt (-2+sqrt (5))/2.

If B2H2 is the altitude from B2 (H2 on A1B1), then H2C2: H2A2 = 1 + golden ratio = 1 + (1 + sqrt (5)) / 2.

http://webpages.ull.es/users/amontes/cabri/hyacinthos19976.htm

Angel Montesdeoca

--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:
>
> Dear Angel
>
> According to my computations if BB* is the altitude from B,
> then B* divides AC in golden ratio.
>
> http://anthrakitis.blogspot.com/2011/04/golden-isosceles-triangle.html
>
> Antreas
>
>
>
> On Mon, Apr 11, 2011 at 2:09 PM, Angel <amontes1949@...> wrote:
>
> >
> >
> >
> >
> > Dear Antreas,
> >
> > (correction)
> >
> > Let y be the side of the square inscribed in B'A'C' based on A'B' and
> >
> > z the side of the square inscribed in the right triangle
> > D'A'A based on AD'
> >
> > An isosceles triangle ABC (AB = AC) for which y = z, with base BC = a, has
> > height h = \frac{1}{2}\sqrt{2+sqrt{5}}a.
> >
> > y = z = \frac{a}{\sqrt{-1+5\sqrt{5}+4\sqrt{2(1+\sqrt{5})}}}
> >
> > and the side A'B'= \frac{a}{\sqrt{2+4\sqrt{-2+\sqrt{5}}}}
> >
> > Angel Montesdeoca
> >
> > P.S. There is another solution for A> pi / 2, although it may not
> > constructible by straightedge and compass.
> >
> >
> >
> > --- In Hyacinthos@yahoogroups.com, "Angel" <amontes1949@> wrote:
> > >
> > > Dear Antreas,
> > >
> > > An isosceles triangle ABC (AB = AC) for which x = y, with base BC = a,
> > has height h = \frac{1}{2}\sqrt{2+sqrt{5}}a.
> > >
> > > x = y = \frac{a}{\sqrt{-1+5\sqrt{5}+4\sqrt{2(1+\sqrt{5})}}}
> > >
> > > and the side A'B'= \frac{a}{\sqrt{2+4\sqrt{-2+\sqrt{5}}}}
> > >
> > > Angel Montesdeoca
> > >
> > > --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
> > > >
> > > > Let ABC be a triangle and A'B'C'D' the inscribed square in ABC
> > > > based on AC (A' on AB, B' on BC, C' on CA near C and D' on CA near A).
> > > > Let y be the side of the square inscribed in BA'C' based on A'B'
> > > > and z the side of the square inscribed in the right triangle
> > > > D'A'A based on AD'.
> > > >
> > > > If AB = AC = k (known) and y = z, construct ABC.
> > > >
> > > > APH
> >
>
>
> [Non-text portions of this message have been removed]
>
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