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WOLFRAM|DEMONSTRATIONS PROJECT

Trisecting an Angle Using a Lemniscate

​
fix angle α to trisect
translate to touch
labels when tangent
show angle to trisect
O
A
C
B
P

The angle to trisect, say
α
, is between the longer leg of the carpenter's square (the brown "L" shape) and the polar axis
OA
. Translate the carpenter's square so that it touches the curve at the point
P
. The angle
AOP
between the radius vector
OP
and the polar axis
OA
is one-third of the given angle
α
.

For any curve in polar coordinates, the tangent of the angle
β
between the tangent and radial line (the angle between vectors
OP
and
PB
) at the point
(r,θ)
is
tan(β)=r/r'
. The lemniscate has the polar equation
r=
cos(2θ)
, and the derivative with respect to
θ
is
r'=
-sin(2θ)

cos(2θ)
, so
r/r'=
cos(2θ
/
-sin(2θ)

cos(2θ)
=-cot(2θ)
. So
tan(β)=-cot(2θ)=-tan(π/2-2θ)=tan(2θ-π/2)
. Since
β
is obtuse,
β=2θ+π/2
.

So
∠OPB=π-β=π/2-2θ
, and the angle
BPC
between the tangent
PB
and the normal
PC
to the radius vector
OP
is
2θ
. But this angle
BPC
is equal to the angle between the larger leg of the carpenter's square and the radius vector
OP
, because these angles have orthogonal legs. So
α-θ=2θ
,
α=3θ
.