2. Construct a Triangle Given the Circumradius, the Difference of the Base Angles, with the Circumcenter on the Incircle

This Demonstration shows a construction of a triangle

ABC

given its circumradius

R

, the difference

δ

of the base angles and that the circumcenter is on the incircle.

Let

r

be the inradius of

ABC

. The Euler formula

2

z

=R(R-2r)

gives the distance

z

between the circumcenter and the incenter. Since the circumcenter is on the incircle,

2

r

=R(R-2r)

, which has the positive solution

r=R(1+

2

)

.

Construction

Draw a circle

σ

1

of radius

R

with center

S

and draw a diameter

CF

. Draw a chord

CD

at an angle

δ/2

from

CF

.

Step 1: Draw a circle

σ

2

with center

S

and radius

r

. Of the two points of intersection of

σ

2

and the segment

CD

, let the point

E

be the one closest to

C

.

Step 2: Draw a ray

ρ

from

C

at an angle

δ

from

CF

. Let

G

be the perpendicular projection of

E

on

ρ

. Measure out a point

H

on

ρ

at distance

r

from

G

.

Step 3: The points

A

and

B

are the intersections of

σ

1

and the line through

H

is perpendicular to

ρ

.

Verification

Let

∠CAB=α

,

∠ABC=β

and

∠BCA=γ

.

Theorem: Let

ABC

be any triangle. Let

H

be the foot of the altitude from

C

to

AB

, and let

S

be the center of the circumscribed circle. Then the angle at

C

between the altitude

CH

and

CS

equals

α-β

. The angle between

CH

and the angle bisector at

C

is

(α-β)/2

. (See The Plemelj Construction of a Triangle 4.)

Proof of the last part: Let

E

be on the angle bisector at

C

, then

∠HCE=∠ACE-∠ACH=γ/2-(π/2-α)=π/2-α/2-β/2-π/2+α=(α-β)/2

.

By construction and the theorem,

δ=α-β

and

σ

1

is the circumscribed circle of triangle

ABC

with center

S

and radius

R

.

By construction,

E

is on the angle bisector at

C

and the distance of

E

to

AB

is

r

. So the circle with center

E

and radius

r

is the incircle of

ABC

, which by construction contains

S

.