# 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'

2h

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 and . So , or , which can be rewritten as .

C

AB

hcotα

C

hcotβ

C

h(cott+cotβ)=c

C

hsin(α+β)=csinαsinβ

C

2hsin(α+β)=c(cos(α-β)-cos(α+β))

C

Let be the angle of at . Since , the equation can be read as

γ

ABC

C

α+β=π-γ

2hsinγ-ccosγ=ccos(α-β).

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/2h)

C

μ=arcsin(c/m)

m=

c+(2h)

2

C

2

Then , . The equation for is now .

2h=mcosμ

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