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Trisecting an Angle
Classical Solutions

### Classical Solutions

#### Trisection Equation

Archimedes' Solution is shown in the figure shown below.

#### Derivation:

**Algebraic Formulation:**
Since three triangles CDM, COL and CAN are similar. CM/CD = CL/CO = CN/CA.

So x/2 = (1 + y)/x = (x + a)/(1 + 2y)

which gives ---> x^{2} = 2 + 2y and 1 + 2y = 2(x + a)/x

Eliminating y from these two equations, we obtain x ^{2} - 1 = 2(x + a)/x or

**x **^{3} - 3x - 2a = 0

where **x = 2cos(q)** and **a = cos(3q)**

********** trisection_equation_desc.dwg** *********
#### Trigonometric Formulation:

Using the trigonometric identities, cos(a+b) = cos(a)cos(b) - sin(a)sin(b) and sin(2a) = 2*sin(a)cos(a)

cos(3q) = 4cos^{3} (q) - 3 cos (q)

Using the relations (Refer to the figure); x = 2cos(q) and a = cos(3q)

This equation can now be written as

**x **^{3} - 3x - 2a = 0

#### Note about derivation from Trigonometry

From the figure ; CB = CO + ON = CA * cos(q) = (CD + DA)cos(q)
Therefore a + x = (1 + 2y)*cos(q) , and y = DO *cos(2q)

From these relations, the same trisection equation can be derived.

#### References

1. Yates,Robert Carl : "The Trisection problem"

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Last Updated Nov 22, 2006

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