## Τετάρτη, 16 Μαρτίου 2011

Let ABC be a triangle and Aa a point on BC between B,C.

Denote:

h_a : the A-altitude of ABC

r_a : the A-exradius of ABC

r_ab, r_ac : the A-exradii of AAaB and AAaC.

We have:

h_a = 2r_ab.r_ac / (r_a - (r_ab + r_ac)) (1)

[See HERE]

Now, assume that r_ab = r_ac and r_a = 2h_a.

Denote r_ab / h_a := x

(1) ==>

1 = 2x^2 / (2 - 2x) ==> x^2 + x - 1 = 0

==>

0 < x = (- 1 + sqrt5)/2 = 1/φ, where φ is the "golden number" φ = (1 + sqrt5)/2.

Golden Section:

Let Iab, Iac be the A-excenters of AAbB, AAcC, resp., and A* the intersection of the line IabIac and the line of AA' (of the A-altitude).

A' divides ΑA* in golden ratio.

If the triangle ABC is isosceles AB = AC, then the foot A' of the A-altitude coincides with the point of contact of the A-excircle and BC:

CONSTRUCTION:

Let (A),(B) be two externally tangent circles at C with radius of (A) = 4.radius of (B). The line ACB intersects again the circle (B) at D. A tangent from D to circle (A) intersects the common internal tangent of (A),(B) at E. The bisector of the right angle ECA intersects EA at F. Let G be the orthogonal projection of F on AB. C divides GD in Golden Ratio.