Algebraic Solution of the Plemelj Triangle Construction Problem
Algebraic Solution of the Plemelj Triangle Construction Problem
This problem asks for the construction of a triangle with ruler and compass given the length of the base (for simplicity, let ), the length of the altitude from to and the difference of the angles and . The problem was posed to Plemelj by his math teacher, Borstner, when Plemelj was about 16 years old.
ABC
AB=c
c=1
h
c
C
AB
α-β
α=∠BAC
β=∠ABC
Plemelj solved the problem using trigonometry, and later found a geometric construction. The starting point was the identity (cotα+cotβ)=c, from which he derived the equation . This equation can be transformed by substituting . Eliminating the square root gives a quadratic equation for , which means that it and can be constructed by using a straight edge and a compass.
h
c
2sinγ-ccosγ=ccos(α-β)
h
c
cos(γ)=
1-γ
2
sin
sinγ
γ
Using the formula , where is the radius of the circumscribed circle, gives a quadratic equation for .
sinγ=c/2R
R
R
To construct the triangle, first draw a vertical straight line of length and a horizontal line through . Then draw line so that . If is the altitude on , then is the bisector of angle . Draw the straight line so that . Then measure out a distance from on the horizontal line to find .
CD
h
c
D
CE
∠ECD=1/2(α-β)
CD
c
CE
γ
CB
∠DCE=1/2γ
c
B
A
To use in the construction, we must know that the angle between the altitude and the line from to the center of the circumscribed circle of the triangle is .
R
∠DCS
CS
C
α-β
So on the line through that has the angle with , measure out a distance to find the point . Then draw a circumscribed circle . Points and are intersections of by the horizontal line through .
C
α-β
CD
R
S
σ
A
B
σ
D
The solutions can be checked to verify that they give the same result.