Trisection of an angle using straight edge and compass only
Posted: Sat Oct 04, 2014 7:50 am
According to mathematicians, the trisection of an angle using a straight edge an a compass is an impossibility. It should supposedly give a cubic equation which cannot be solved using the given tools. However I think that I might have solved it. My maths teacher told me that he couldn't find a flaw in my theory. So I just wanted to know if what I have done is correct or no. And if it's incorrect then where have I gone wrong? The steps are given below:
Steps:
1) Create any arbitrary angle ‘θ’.
2) Label the point where it’s arms meet as ‘o’.
3) Create a circle with a radius ‘r’ with the centre ‘o’. This is circle 1.
4) Trisect the radius ‘r’. Twice of the trisected radius gives ‘R’.
5) Copy the arbitrary angle created in step 1 and label the point where it’s arms meet as ‘O’.
6) Create a circle with radius ‘R’ with it’s centre as ‘O’. This is circle 2.
7) Take the length of the chord in circle 2 which creates the arbitrary angle and make a chord in circle 1 of the same length.
The length of a chord is directly proportional to the arc that forms it.
9) Length of an arc= (θ/180)*r
10) The ratio of the arcs formed by the arbitrary angle in circle 1 and circle 2= ((θ/180)*r)/((θ/180)*R)=r/R .
But R=2/3r
∴ ratio=r/(2/3)r
∴ratio=3:2
11) By the properties of circle that chords make angles on the centre in ratio to the length of their sides:
The second chord that was made, should make an angle which is equal to two-thirds of the angle formed by the original chord.
Thus the angle is divided into two parts:
1) Two-thirds of the original angle
2) One-third of the original angle
Hence the angle is trisected.
Steps:
1) Create any arbitrary angle ‘θ’.
2) Label the point where it’s arms meet as ‘o’.
3) Create a circle with a radius ‘r’ with the centre ‘o’. This is circle 1.
4) Trisect the radius ‘r’. Twice of the trisected radius gives ‘R’.
5) Copy the arbitrary angle created in step 1 and label the point where it’s arms meet as ‘O’.
6) Create a circle with radius ‘R’ with it’s centre as ‘O’. This is circle 2.
7) Take the length of the chord in circle 2 which creates the arbitrary angle and make a chord in circle 1 of the same length.
9) Length of an arc= (θ/180)*r
10) The ratio of the arcs formed by the arbitrary angle in circle 1 and circle 2= ((θ/180)*r)/((θ/180)*R)=r/R .
But R=2/3r
∴ ratio=r/(2/3)r
∴ratio=3:2
11) By the properties of circle that chords make angles on the centre in ratio to the length of their sides:
The second chord that was made, should make an angle which is equal to two-thirds of the angle formed by the original chord.
Thus the angle is divided into two parts:
1) Two-thirds of the original angle
2) One-third of the original angle
Hence the angle is trisected.