The Plemelj Construction of a Triangle: 13
The Plemelj Construction of a Triangle: 13
This Demonstration constructs a triangle given the length of its base , the length of the altitude from to and the difference between the angles and at and . This is not one of Plemelj's original constructions, but a new one based on his equation , where and . It is the same as The Plemelj Construction of a Triangle: 8, but the verification is different.
ABC
c
AB
h
C
C
AB
δ
α
β
A
B
msin(γ-μ)=ccos(α-β)
m=+
2
c
2
(2)
h
C
μ=arctan(c/(2))
h
C
The modified equation is .
sin(γ-μ)=sin(μ)cos(α-β)
Construction
Step 1: Draw a straight line of length and a line parallel to at distance . Let be the midpoint of , and let be the point on directly above . Let be the reflection of in .
AB
c
τ
AB
h
C
M
AB
N
τ
M
B'
B
τ
Step 2: Draw a circle with center and radius .
σ
M
c/2
Step 3: Draw the ray from at the angle from to intersect at the point .
A
δ
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
Angle , so . But , since .
∠AB'B=μ
AB=msinμ
AP=ABcosδ=msinμcosδ=msin(∠AB'D)=msin(γ-μ)
AD=AP
So . Since , .
cosδ=cos(α-β)
δ<π
δ=α-β