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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