The Plemelj Construction of a Triangle: 8
The Plemelj Construction of a Triangle: 8
This Demonstration constructs a triangle given the length of its base, the length of the altitude from to and the difference between the angles at and . This is not one of Plemelj's original constructions, but a new one based on his equation , where and .
ABC
AB=c
h
C
C
AB
α-β
A
B
msin(γ-μ)=ccos(α-β)
m=+
2
c
2
(2)
h
C
μ=arctan(λ/2)
h
C
The modified equation is , which can be read as the law of sines for a triangle where the sides of length 1 and are opposite the angles and , respectively, or, multiplying by , that the sides of length and are opposite the angles and .
sin(γ-μ)=sin(μ)cos(α-β)
cos(α-β)
μ
γ-μ
c
c
ccos(α-β)
μ
γ-μ
Construction
Step 1: Draw a straight line of length and a line parallel to at distance . Let be the midpoint of and the point on just above .
AB
c
τ
AB
h
C
M
AB
N
τ
M
Step 2: Draw a circle with center and radius .
σ
M
c/2
Step 3: Draw the ray from at the angle δ from to intersect at the point .
A
AB
σ
P
Step 4: Draw the circle with center and radius .
ζ
N
m/2
Step 5: Measure out a point on the circle at distance from .
D
ζ
AP=ccosδ
A
Step 6: The point is the intersection of and the right bisector of .
C
τ
DP
Step 7: The triangle meets the stated conditions.
ABC
Verification
According to the law of sines, the exterior angle at of is , which is . The angle is . So and .
A
ΔADB
γ
∠ADB+∠ABD=μ+γ-μ
DAC
(π-γ-δ)/2=α-δ
(α+β)-δ=2α-2δ
δ=α-β