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



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