Page 13 - MATINF Nr. 8

About Brocard point in a triangle 13

Corollary 6. In any triangle ABC, we have

2
3(2R − r) 2R + r 2
≤ ctg ω ≤ .
s sr
Proof. We have the follows
P 2 2 2
a s − r − 4Rr
ctg ω = = .
4sr 2sr
Using the Gerretsen’s Inequalities

2
2
2
16Rr − 5r ≤ s ≤ 4R + 4Rr + 3r 2
we obtain the desired results. □
Remark 2. We have the following inequality
s
ctg ω ≥ .
3r
Proof. We have the follows

P 2 P 2 2
a ( a) 4s s
ctg ω = ≥ = = .
4sr 12sr 12sr 3r
□
Corollary 7. In any triangle ABC, we have

Å ã 2 Å ã 2 Å 2 ã 2
5R − r R 1 (R + r) R
− ≤ ≤ − .
s r 2 sin ω sr r
Proof. We have the follows
Å 2 2 ã 2
1 2 X 1 s + r + 4Rr 4R
= 4R = − .
2
sin ω a 2 2sr r
Using the Gerretsen’s Inequalities holds the desired results. □
Remark 3. We have the following inequality
1 2R
≥ .
2
sin ω r
Proof. We have the follows
2 2
P P
1 a b abc a 2R
= ≥ = .
2
2 2
2 2
sin ω 4s r 4s r r
□
Open Question. Prove or disprove
Å ã 2
R 1 2R
≥ ≥ .
2
r sin ω r
when give a new refinement for Euler’s R ≥ 2r Inequality.

References

[1] Octogon Mathematical Magazine (1993-2021).
[2] MATINF (2018-2021).