The Plemelj Construction of a Triangle: 13

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

c

of its base

AB

, the length

h

C

of the altitude from

C

to

AB

and the difference

δ

between the angles

α

and

β

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(c/(2

h

C

))

. It is the same as The Plemelj Construction of a Triangle: 8, but the verification is different.

The modified equation is

sin(γ-μ)=sin(μ)cos(α-β)

.

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 let

N

be the point on

τ

directly above

M

. Let

B'

be the reflection of

B

in

τ

.

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

Angle

∠AB'B=μ

, so

AB=msinμ

. But

AP=ABcosδ=msinμcosδ=msin(∠AB'D)=msin(γ-μ)

, since

AD=AP

.

So

cosδ=cos(α-β)

. Since

δ<π

,

δ=α-β

.