• Non ci sono risultati.

# Then we have that r = F s, R = abc 4F, sec(A/2

N/A
N/A
Protected

Condividi "Then we have that r = F s, R = abc 4F, sec(A/2"

Copied!
1
0
0

Testo completo

(1)

Problem 12168

(American Mathematical Monthly, Vol.127, March 2020) Proposed by M. Lukarevski (North Macedonia).

Leta, b, and c be the side lengths of a triangle ABC with circumradius R and inradius r. Prove 2

R ≤ sec(A/2)

a +sec(B/2)

b +sec(C/2)

c ≤ 1

r.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Solution. Let s be the semiperimeter of the triangle and let F be its area. Then we have that

r = F

s, R = abc

4F, sec(A/2) =

s bc

s(s − a), sec(B/2) =

r ca

s(s − b), sec(C/2) =

s ab

s(s − c), and the double inequalities can be written as

8F2

abc ≤pbc(s − b)(s − c)

a +pca(s − c)(s − a)

b +pab(s − a)(s − b)

c ≤ s.

We first show the inequality on the right: by AM-GM inequality pbc(s − b)(s − c)

a ≤

b+c

2 ·2s−b−c2

a =b + c 4 and therefore

pbc(s − b)(s − c)

a +pca(s − c)(s − a)

b +pab(s − a)(s − b)

c ≤ b + c

4 +c + a

4 +a + b 4 = s.

As regards the inequality on the left, let x = 12(b + c − a) > 0, y = 12(c + a − b) > 0, and z = 12(a + b − c) > 0 then

a = x + y, b = y + z, c = z + x, s = x + y + z, F2= xyz(x + y + z) and the inequality becomes

8xyz(x + y + z) ≤ RHS where

RHS :=py(y + x)3pz(z + x)3+px(x + y)3pz(z + y)3+px(x + z)3py(y + z)3. We have that

(2x + y)(x − y)2⇔py(y + x)3≥ y(y + 3x)

√2 whence

RHS ≥ yz(y + 3x)(z + 3x)

2 +xz(x + 3y)(z + 3y)

2 +xy(x + 3z)(y + 3z) 2

= x2(y − z)2+ y2(x − z)2+ z2(x − y)2

4 + 8xyz(x + y + z)

≥ 8xyz(x + y + z)

and we are done. 

Riferimenti

Documenti correlati

Without loss of generality, we may assume that the circumcircle Γ is the unit circle |z| = 1 of the complex plane.. because the sine function is concave in

[r]

[r]

Also, let ω be a primitive nth root

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.. For k

Let a, b, and c be the lengths of the sides of a triangle, and let R and r be the circumradius and inradius of the

The main reason for the second approach is that, in most real problems, the meaningful objects are the probability distributions of random variables, rather than the random

Indeed, the pointwise convergence says that, for every fixed x ∈ A, the sequence of vectors (f n (x)) n∈N converges in R m to the vector f (x), but it does not say anything about