# The Plemelj Triangle via the Fixed Point of a Transformation

The Plemelj Triangle via the Fixed Point of a Transformation

This problem was posed to Plemelj by his math teacher Borstner when Plemelj was about 16 years old. The problem asks for the construction of a triangle using ruler and compass given the length of the base , the length of the altitude from to , and , where and are the angles at and . In this Demonstration, and =1.

ABC

AB=c

h

c

C

AB

α-β

α

β

A

B

c=1

h

c

Draw two parallel lines separated by a distance =1. Mark off the line segment of length on the lower parallel line. The line drawn through with angle intersects the upper parallel line at , and the line through with angle intersects the upper parallel line at . What is the relationship between and ?

h

c

AB

c=1

A

α

T

B

β

T'

T

T'

Introduce a coordinate system with the origin at and the axis along so that . The line has the equation , and has the equation . Let and . Then, using the addition formula for the tangent gives the following formulas for the abscissas of and :

A

x

AB

B=(1,0)

AT

y=xtanα

BT

y=-(x-1)tanβ

a=tanα

d=tan(α-β)

T

T'

x=1/a

x'=((d+1)x+(d-1))/(dx-1)

or

x=

1

a

x'=

(d+1)x+(d-1)

dx-1

So the construction problem is to find the fixed point of the transformation to give . For this to happen, . If , this equation has the solution . In this Demonstration, we take the solution with the minus sign. Geometrically, the fixed point is determined by the values of and such that points and are coincident.

x=x'

d-(d+2)-(d-1)=0

2

x

d≠0

1/(2d)(d+2)±

5+4

2

d

α

β

B

B'

Since quadratic equations can be solved by construction with a straight edge and a compass, Plemelj's problem has a purely geometric solution.