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