WOLFRAM|DEMONSTRATIONS PROJECT

The Plemelj Triangle via the Fixed Point of a Transformation

​
β
0.5927
α
1.0472
show fixed point
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
ABC
using ruler and compass given the length of the base
AB=c
, the length of the altitude
h
c
from
C
to
AB
, and
α-β
, where
α
and
β
are the angles at
A
and
B
. In this Demonstration,
c=1
and
h
c
=1
.
Draw two parallel lines separated by a distance
h
c
=1
. Mark off the line segment
AB
of length
c=1
on the lower parallel line. The line drawn through
A
with angle
α
intersects the upper parallel line at
T
, and the line through
B
with angle
β
intersects the upper parallel line at
T'
. What is the relationship between
T
and
T'
?
Introduce a coordinate system with the origin at
A
and the
x
axis along
AB
so that
B=(1,0)
. The line
AT
has the equation
y=xtanα
, and
BT
has the equation
y=-(x-1)tanβ
. Let
a=tanα
and
d=tan(α-β)
. Then, using the addition formula for the tangent gives the following formulas for the abscissas of
T
and
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
x=x'
. For this to happen,
d
2
x
-(d+2)-(d-1)=0
. If
d≠0
, this equation has the solution
1/(2d)(d+2)±
5
2
d
+4

. In this Demonstration, we take the solution with the minus sign. Geometrically, the fixed point is determined by the values of
α
and
β
such that points
B
and
B'
are coincident.
Since quadratic equations can be solved by construction with a straight edge and a compass, Plemelj's problem has a purely geometric solution.