2. Construct a Triangle Given the Circumradius, the Difference of the Base Angles, with the Circumcenter on the Incircle
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 given its circumradius , the difference of the base angles and that the circumcenter is on the incircle.
ABC
R
δ
Let be the inradius of . The Euler formula =R(R-2r) gives the distance between the circumcenter and the incenter. Since the circumcenter is on the incircle, =R(R-2r), which has the positive solution .
r
ABC
2
z
z
2
r
r=R(1+
2
)Construction
Draw a circle of radius with center and draw a diameter . Draw a chord at an angle from .
σ
1
R
S
CF
CD
δ/2
CF
Step 1: Draw a circle with center and radius . Of the two points of intersection of and the segment , let the point be the one closest to .
σ
2
S
r
σ
2
CD
E
C
Step 2: Draw a ray from at an angle from . Let be the perpendicular projection of on . Measure out a point on at distance from .
ρ
C
δ
CF
G
E
ρ
H
ρ
r
G
Step 3: The points and are the intersections of and the line through is perpendicular to .
A
B
σ
1
H
ρ
Verification
Let , and .
∠CAB=α
∠ABC=β
∠BCA=γ
Theorem: Let be any triangle. Let be the foot of the altitude from to , and let be the center of the circumscribed circle. Then the angle at between the altitude and equals . The angle between and the angle bisector at is . (See The Plemelj Construction of a Triangle 4.)
ABC
H
C
AB
S
C
CH
CS
α-β
CH
C
(α-β)/2
Proof of the last part: Let be on the angle bisector at , then .
E
C
∠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, and is the circumscribed circle of triangle with center and radius .
δ=α-β
σ
1
ABC
S
R
By construction, is on the angle bisector at and the distance of to is . So the circle with center and radius is the incircle of , which by construction contains .
E
C
E
AB
r
E
r
ABC
S