16. Construct a Triangle Given Its Circumradius, Inradius and the Difference of Its Base Angles

This Demonstration shows a construction of a triangle

ABC

given its circumradius

R

, inradius

r

and the difference

δ

of the base angles.

Construction

Draw a circle

σ

1

of radius

R

with center

S

and a diameter

CF

. (The circle

σ

1

will be the circumcircle of

ABC

.)

Step 1: Draw a line segment

CD

at an angle

δ/2

from

CF

to meet

σ

1

at

D

. From

C

, draw a ray

ρ

at angle

δ

from

CF

.

Let

E

be on

CD

at the Euler distance

z

from

S

, where

2

z

=R(R-2r)

.

Step 2: Drop a perpendicular from

E

to

ρ

at

G

and let

H

be on

ρ

such that

GH=r

.

Step 3: The points

A

and

B

are the intersections of

σ

1

and the perpendicular to

CH

at

H

.

Verification

By construction,

R

is the circumradius of

ABC

.

Let

∠CAB=α

,

∠ABC=β

and

∠BCA=γ

.

Theorem: Let

S

be the circumcenter of a triangle

ABC

. If

α>β

, the angle between the angle bisector at

C

and

CS

is

(α-β)/2

.

Proof: The angle subtended over

AC

from a point on

σ

1

to the right of

AC

is

β

because

∠ABC=β

. So the central angle is twice that:

∠ASC=2β

. Triangle

ASC

is isosceles, so

∠SCA=π/2-β

. It follows that

∠SCE=π/2-β-γ/2=π/2-β-(π-α-β)/2=(α-β)/2.

By this theorem and the construction of

CD

,

CD

is the angle bisector at

C

. Therefore by Euler's triangle formula for

z

,

E

is the incenter of

ABC

. Since

GH

is parallel to

AB

and

GH=r

, the distance from

E

to

AB

is also

r

and the incircle has center

E

, radius

r

.

By step 1,

∠DCF=∠ECF=δ/2

. The theorem states

∠DCF=(α-β)/2

, so

δ=α-β

.

Construct

z

The distance

z=SQ

is the geometric mean of

PS

and

SF

, that is of

R

and

R-2r

.