Seperate An Angle To Three Equal

Separate an angle to three equals We don’t know by which conditions was placed the problem “Separate an angle to three equals using ruler without numbers and a divider” in the ancient Greece. Source (in Greek): http://www.telemath.gr/mathematical_ancient_times/unsolved_ mathematical_problems/unsolved.php#2 Then many different ancient Greek geometry mathematicians try to solve this problem without to achieve it, so they try to find other ways to solve this problem, not using ruler(without number) and a divider but with an auxiliary curve .

Α:Random point O1O2 = O2 OZ

I make circles (O1 , R), (O2 , R) με O1O2 = R , I draw a beeline From the random point Α I draw a radius AO2 = AO1 = AZ ∆

∆

ΑO1 ∆ = ΑΟ2 ∆ O1 A = O2 A ⇒ A1 = A3

therefore ∆

because

A1 =

AO2 = AO1 = AZ

A1 + A3 2

∆

O1 AO2 = O2 AZ because O1O2 = R = O2 Z therefore ( A1 + A3 ) = A2

So the angle

A1 + A2

and

AO1 = AO2 = AZ

is separated to three equals

d1 = A1O2 , d 2 = A1 B1 , O2 A2 = A2 B2 = P

ε 2 // ε 1 ,

,ΔΑ:common side,

Μ:the middle of

∩

Β2∆

ε 1 ⊥ O1O2 .

let’s imagine that we have a strand with a constant lenght 2Ρ, which is moving on the beeline ε 1 and at an area of the circumference of the circle (O2 , R) . While the strand is moving to the point A1 , the distance d 2 became longer and the distance d1 became shorter ∩ ( The circumference Β 2 Μ IS ALLOWABLE AREA WHERE WE HAVE THE LENGHT 2P ) From the figure we have: d 2 − Γ1Β1 = Ρ d1 + d 2 = 2 P ⇒ d1 = 2 P − ( P + Γ1Β1 ) ⇒ d1 = P − Γ1Β1

At the moment when we don’t know the beeline ε 2 we don’t ∩ know the allowable area of the circumference Β 2 Μ . So from the ∩ point Μ which is the middle of Β 2 ∆ we draw MA3 // A2 B2 We can prove that the point Μ is belong to the beeline O1O2 when it will be extended. We can choose the point M because it can be the middle of each ∩ bow Β 2 ∆ of each corresponded angle ∆Α1Β1 which we want to separate to three equals. So if we want to separate an angle following steps.

^

Α to three equals we do the

-We draw A1 B1 // MA3 (I know the point M, just we extend the beeline O1O2 until the circumference of the circle (O2 , R) ) - d1 + d 2 = 2 P = O2 A3 + A3 M ⇒ P =

O2 A3 + A3 M 2

-From the point O2 I draw a radius Ρ which που intersect the beeline ε 1 to the point A2 . -From the point A2 we draw a radius Ρ which intersect the circumference of the circle to the point B2 . ^ ^ -We configure the transferee angle ∆Α 2 Β 2 = ∆Α 3 Μ which is separated to three equals.

Evagoras Savva Georgiou Apsiou-Limassol Cyprus