Trisecting an Angle Using a Conchoid

move point on line

A

-1.045

OB = d

AP = k

∠POY = 3 ∠TOY

This Demonstration shows how Nicomedes (c. 180 BC) used a conchoid to trisect an angle.

Let the point

B

be at the distance

d

from the point O on the line

OY

, that is,

OB=d

. Draw a straight line

m

through

B

perpendicular to

OY

. Let a line through

O

intersect the line

m

at

A

. On the line

OA

produced in both directions, mark

P

and

P'

so that

AP=AP'=k.

The locus of the points

P

and

P'

is a conchoid with pole

O

.

Let

YOA

be the angle to be trisected. Let

k=2AO

and let the perpendicular to

m

at

A

intersect the conchoid at

T

. Let

N

be the intersection of

OT

and

m

, and let

M

be the midpoint of

NT

. Then

MT=MN=MA

(in a right triangle, the midpoint of the hypotenuse is equidistant from the three polygon vertices; see Right Triangle for a proof). Since

T

is on the conchoid with

k=2AO

,

NT=k=2OA

, and so

MA=OA

. That is,

ΔAOM

is isoceles and

∠AOM=∠AMO

;

ΔATM

is also isoceles and

∠AMO=2∠ATM

. Because

AT||OY

,

∠ATM=∠TOQ

.

Putting this together,

∠AOM=2/3∠YOA

, so

∠TOQ=1/3∠YOA

.