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

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

This Demonstration shows a construction of a triangle given its circumradius , inradius and the difference of the base angles.

ABC

R

r

δ

Construction

Draw a circle of radius with center and a diameter . (The circle will be the circumcircle of .)

σ

1

R

S

CF

σ

1

ABC

Step 1: Draw a line segment at an angle from to meet at . From , draw a ray at angle from .

CD

δ/2

CF

σ

1

D

C

ρ

δ

CF

Let be on at the Euler distance from , where =R(R-2r).

E

CD

z

S

2

z

Step 2: Drop a perpendicular from to at and let be on such that .

E

ρ

G

H

ρ

GH=r

Step 3: The points and are the intersections of and the perpendicular to at .

A

B

σ

1

CH

H

Verification

By construction, is the circumradius of .

R

ABC

Let , and .

∠CAB=α

∠ABC=β

∠BCA=γ

Theorem: Let be the circumcenter of a triangle . If , the angle between the angle bisector at and is .

S

ABC

α>β

C

CS

(α-β)/2

Proof: The angle subtended over from a point on to the right of is because . So the central angle is twice that: . Triangle is isosceles, so . It follows that

AC

σ

1

AC

β

∠ABC=β

∠ASC=2β

ASC

∠SCA=π/2-β

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

By this theorem and the construction of , is the angle bisector at . Therefore by Euler's triangle formula for , is the incenter of . Since is parallel to and , the distance from to is also and the incircle has center , radius .

CD

CD

C

z

E

ABC

GH

AB

GH=r

E

AB

r

E

r

By step 1, . The theorem states , so .

∠DCF=∠ECF=δ/2

∠DCF=(α-β)/2

δ=α-β

Construct

z

The distance is the geometric mean of and , that is of and .

z=SQ

PS

SF

R

R-2r