## Δευτέρα, 28 Μαρτίου 2011

### REGULAR POLYGON PROBEM

Let A1A2A3... An be a regular n-gon. The perpendicular
to A1A2 at A2 intersects A3A4 at K. The parallel through
A3 to KA1 intersects A1A2 at M.

For which n's the point M is the midpoint of A1A2?

APH, Hyacinthos message 19937

Trigonometric Solution:

Let: A1A2 = A2A3 =... = 1, angles(A1A2A3) = (A2A3A4) = ... = ω

Denote: angle(KA1A2) = (A3MA2) = θ, angle(MA3A2) = φ, KA2 := d

We have:

In the right triangle A2KA1: tanθ = d

In the triangle KA2A3: d/sinω = 1/sin(270-2ω) ==> d / sinω = - 1/cos2ω

In the triangle MA2A3: (1/2)/sinφ = 1/sinθ ==> 2sinφ = sinθ
and since φ + θ = 180 - ω, ==> 2sin(ω+θ) = sinθ

So we have the system of two equations:

tanθ = - sinω/cos2ω

2sin(ω+θ) = sinθ

==>

2cos2ω - 2cosω + 1 = 0 ==> 4(cosω)^2 - 2cosω - 1 = 0 ==>

cosω = (1 +- sqrt5) / 4 = cos36 or cos108

We have:
180 > ω = 180(n-2)/n >= 60 since n >= 3 ==> ω = 108 d. and n = 5.

Other solutions by Nikos Dergiades, Hyacinthos messages
19939,19946

## 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.