The Plemelj Construction of a Triangle: 1
The Plemelj Construction of a Triangle: 1
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 .
ABC
AB=c
h
C
C
AB
α-β
A
B
Construction
Step 1: Draw a straight line of length . Draw a line of length perpendicular to . Let be the midpoint of .
AB
c
AA'
2
h
C
AB
E
A'B
Step 2: Construct a circle with center such that the chord subtends an angle from points on below the chord. The inscribed angle above the chord is and the corresponding central angle is .
σ
S
A'B
π/2-δ
σ
π/2–δ
A'SB
π-2δ
Step 3: Construct the point on at a distance from .
B'
σ
c
A'
Step 4: The point is the intersection of the right bisector of and the line through parallel to .
C
BB'
E
AB
Step 5: The triangle meets the stated conditions.
ABC
Verification
Triangle is congruent to . In the isosceles triangle , , so . Therefore the obtuse angle . On the other hand, , so and .
AB'C'
ABC
B'BC
∠CB'B=π/2-δ-β
∠B'CB=2(δ+β)
∠A'CB=2π-γ-2(δ+β)=2π-(π-α-β)-2δ-2β=π+α-β+2δ
∠A'CB=2π-γ-2α=2π-(π-α-β)-2α=π-α+β
π+α-β-2δ=π-α+β
δ=α-β
This Demonstration shows Plemelj's somewhat complicated construction. Fascinated, his teacher showed him the solution from a textbook unknown to the author. This is shown in The Plemelj Construction of a Triangle: 2. Plemelj then made a construction that is shown in The Plemelj Construction of a Triangle: 3. Plemelj admitted that he found the first construction using trigonometry. Three solutions of the triangle construction problem are in [2].
Here is the trigonometric proof.
The altitude from divides into two parts of length cotα and cotβ. So (cott+cotβ)=c, or sin(α+β)=csinαsinβ, which can be rewritten as .
C
AB
h
C
h
C
h
C
h
C
2sin(α+β)=c(cos(α-β)-cos(α+β))
h
C
Let be the angle of at . Since , the equation can be read as
γ
ABC
C
α+β=π-γ
2sinγ-ccosγ=ccos(α-β).
h
C
From this equation, we must determine ; it can be transformed to a quadratic equation in the unknown .
γ
sinγ
Introduce the angle as or , where .
μ
μ=arctan(c/2)
h
C
μ=arcsin(c/m)
m=+
2
c
2
(2)
h
C
Then , . The equation for is now .
2=mcosμ
h
C
c=msinμ
γ
msin(γ-μ)=ccos(α-β)=csin(π/2-(α-β))
This equation can be thought of as the law of sines of the triangle with sides and and opposite angles and .
B'BA'
m
c
π/2-(α-β)
γ-μ