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Asupra Problemei B84 din Crux Mathematicorum

Marin Chirciu 1

Articolul pornes , te de la Problema B84 din Crux Mathematicorum, June 2021 (rubrica Bonus
Problems), propus˘a de George Apostolopoulos.
Vom demonstra inegalitatea propus˘a, o vom ˆınt˘ari s , i apoi vom face dezvolt˘ari cu inegalit˘at , i
din aceeas , i clas˘a de probleme cu ˆın˘alt , imi s , i raze ale cercurilor exˆınscrise unui triunghi.

Vom utiliza urm˘atorul rezultat.
2
2
R (p + 5r + 8Rr)
r b + r c
ˆ
X
Lema 1. In orice ∆ABC avem: = .
2
2
h b + h c r (p + r + 2Rr)
S 2S
Demonstrat¸ie. Folosind r a = s , i h a = obt , inem:
p − a a
S S P
r b + r c p−b p−c
X X + 1 X abc abc (a + b) (a + c) (p − a)
= = = · Q Q
h b + h c 2S + 2S 2 (b + c) (p − b) (p − c) 2 (b + c) (p − a)
b c
2
2
2
2
2
4Rrp p (p + 5r + 8Rr) R (p + 5r + 8Rr)
= · = .
2
2
2
2
2 2p (p + r + 2Rr) · pr 2 r (p + r + 2Rr)
S˘a trecem la rezolvarea problemei ment , ionate.
Aplicat , ia 1 ( Problema B84 din Crux Mathematicorum, Vol 47, No. 6 (June 2021)). Let
h a , h b , h c be the altitudes, r a , r b , r c the exradii, r the inradius, and R the circumradius of a
triangle ABC. Prove that
 ‹
√ 2
r a + r b r b + r c r c + r a R
≤ 3 2 .
p 2 + p 2 + p
2
2
h + h b h + h 2 c h + h 2 a 2r
b
c
a
George Apostolopoulos
Solut ,ie. Avem:
 ‹
(∗) √ 2
X X √ X R
r b + r c r b + r c r b + r c
≤ È = 2 ≤ 3 2 ,
p 2
2
h + h 2 (h b +h c) h b + h c 2r
b b
2
 ‹ 2  ‹ 2
2
2
2
2
P r b + r c R Lema 1 R (p + 5r + 8Rr) R p + 5r + 8Rr
unde (∗) ⇔ ≤ 3 ⇐⇒ ≤ 3 ⇔ ≤
2
2
2
2
h b + h c 2r r (p + r + 2Rr) 2r p + r + 2Rr
3R
2
2
2
2
2
2
2
⇔ 4r (p + 5r + 8Rr) ≤ 3R (p + r + 2Rr) ⇔ p (3R − 4r) + r (6R − 29Rr − 20r ) ≥ 0,
4r
2
2
deci folosind Inegalitatea lui Gerretsen p ≥ 16Rr − 5r este suficient s˘a ar˘at˘am c˘a:
2
16Rr − 5r 2 (3R − 4r) + r 6R − 29Rr − 20r 2 ≥ 0 ⇔
1
Profesor, Colegiul Nat , ional ,,Zinca Golescu”, Pites , ti, marin.chirciu@yahoo.com
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