The Plemelj Construction of a Triangle: 8

c

h

C

δ

0.62

steps

1

2

3

4

5

6

7

verification

plot range

1

This Demonstration constructs a triangle

ABC

given the length

AB=c

of its base, the length

h

C

of the altitude from

C

to

AB

and the difference

α-β

between the angles at

A

and

B

. This is not one of Plemelj's original constructions, but a new one based on his equation

msin(γ-μ)=ccos(α-β)

, where

m=

2

c

+

2

(2

h

C

)

and

μ=arctan(λ/2

h

C

)

.

The modified equation is

sin(γ-μ)=sin(μ)cos(α-β)

, which can be read as the law of sines for a triangle where the sides of length 1 and

cos(α-β)

are opposite the angles

μ

and

γ-μ

, respectively, or, multiplying by

c

, that the sides of length

c

and

ccos(α-β)

are opposite the angles

μ

and

γ-μ

.

Construction

Step 1: Draw a straight line

AB

of length

c

and a line

τ

parallel to

AB

at distance

h

C

. Let

M

be the midpoint of

AB

and

N

the point on

τ

just above

M

.

Step 2: Draw a circle

σ

with center

M

and radius

c/2

.

Step 3: Draw the ray from

A

at the angle δ from

AB

to intersect

σ

at the point

P

.

Step 4: Draw the circle

ζ

with center

N

and radius

m/2

.

Step 5: Measure out a point

D

on the circle

ζ

at distance

AP=ccosδ

from

A

.

Step 6: The point

C

is the intersection of

τ

and the right bisector of

DP

.

Step 7: The triangle

ABC

meets the stated conditions.

Verification

According to the law of sines, the exterior angle at

A

of

ΔADB

is

γ

, which is

∠ADB+∠ABD=μ+γ-μ

. The angle

DAC

is

(π-γ-δ)/2=α-δ

. So

(α+β)-δ=2α-2δ

and

δ=α-β

.