This paper presents a simplified graphical procedure for constructing, using an unmarked straightedge and a compass only, a 10˚ to 20˚ angle, which is in other words, trisecting a 30˚ or 60˚ angle. The procedure, when...This paper presents a simplified graphical procedure for constructing, using an unmarked straightedge and a compass only, a 10˚ to 20˚ angle, which is in other words, trisecting a 30˚ or 60˚ angle. The procedure, when applied to the 30˚ and 60˚ angles that have been “proven” to be not trisectable, produced a construction having the identical angular relationship with Archimedes’ Construction, in which the required trisection angles were found to be 10.00000˚ and 20.00000˚ respectively (i.e. exactly one-third of the given angle or ∠E’MA = 1/3∠E’CG). Based on this identical angular relationship as well as the numerical results obtained, one can only conclude that the geometric requirements for arriving at an exact trisection of the 30˚ or 60˚ angle, and therefore the construction of a 10˚ or 20˚ angle, have been met, notwithstanding the theoretical proofs of Wantzel, Dudley, and others. Thus, the solution to the age-old trisection problem, with respect to these two angles, has been accomplished.展开更多
文摘This paper presents a simplified graphical procedure for constructing, using an unmarked straightedge and a compass only, a 10˚ to 20˚ angle, which is in other words, trisecting a 30˚ or 60˚ angle. The procedure, when applied to the 30˚ and 60˚ angles that have been “proven” to be not trisectable, produced a construction having the identical angular relationship with Archimedes’ Construction, in which the required trisection angles were found to be 10.00000˚ and 20.00000˚ respectively (i.e. exactly one-third of the given angle or ∠E’MA = 1/3∠E’CG). Based on this identical angular relationship as well as the numerical results obtained, one can only conclude that the geometric requirements for arriving at an exact trisection of the 30˚ or 60˚ angle, and therefore the construction of a 10˚ or 20˚ angle, have been met, notwithstanding the theoretical proofs of Wantzel, Dudley, and others. Thus, the solution to the age-old trisection problem, with respect to these two angles, has been accomplished.
