Page 25 - MATINF Nr. 8

î √ √ ä
4.2 sup F = k in all triangles, k ∈ 3 , 3 3 25
3
î√ √ ä
4.2 sup F = k in all triangles, k ∈ 3 , 3 3
3

We will prove that the expression within square brackets from our identity above is nonpositive
î √ √ ä
for all k satisfying k ∈ 3 , 3 3 . To this end, we will make use of Gerretsen’s inequality, i.e.
3
2
2
2
s ≤ 4R + 4Rr + 3r , which we can equivalently write as
2
2
x ≤ 1 + 2y + 3y .
√ √
Since s ≥ 3r 3, it’s easy to see that x > y for all k ≥ 3 . Therefore, it is enough to show
k 3
that
Å ã 2
y 2
y
p 2
2
1 + 2y + 3y − ≤ (1 + y) + .
k k
After expanding and rearranging, we are left to prove that
2y p
2
2
2y ≤ · 1 + 2y + 3y .
k
The case when y = 0 is trivial. Otherwise, we can divide by 2y and square both sides. In
that case, the claim reduces to
2
2
(3 − k )y + 2y + 1 ≥ 0,
2
which is clearly satisfied as k < 27 implies

2
2
2
(3 − k )y + 2y + 1 > −24y + 2y + 1
2
1
and −24y + 2y + 1 = (1 − 4y)(6y + 1) ≥ 0 since 0 ≤ y ≤ .
4
î √ √ ä
Consequently, F(k) ≤ k for all k ∈ 3 , 3 3 , and equality is attained if and only if triangle
3
π
ABC is degenerate with two angles equal to . (Note that we have y = 0 in this case.)
2
√ 3
(k− 3) î √ ä
4.3 sup F = in all triangles, k ∈ 3 3, ∞
8
√
√ (k− 3) 3
If k ≥ 3 3, the upper bound of F(k) is . To prove this, we will once again employ the
8
2
2
inequality x ≤ 1 + 2y + 3y and thus maximize F(k) as follows:
ñ ô
Å ã 2 Å ã 2
p y 2 y

2
2
F(k) ≤ k + k k + 1 1 + 2y + 3y − − (1 + y) − .
k k
Therefore, it suffices to prove that
Ä 3
ñ ô √ ä
Å ã 2 Å ã 2 k − 3
p y 2 y
2

2
k + k k + 1 1 + 2y + 3y − − (1 + y) − ≤ ,
k k 8
which, after some simplifications, reduces to
2
k + 1 î Ä 2 ä √ p ó
· k (4y) − 1 + 3 3 − 16y 1 + 2y + 3y 2 ≤ 0.
8