Page 22 - MATINF Nr. 8

22 3 SOLUTION

and equality holds for a right-isosceles triangle. Indeed, if A = π and B = C = π we have
2 4
ñ√ ô 2
2 (k − 1) 2
(k cos A − sin A) (k cos B − sin B) (k cos C − sin C) = (−1) · · (k − 1) = − ,
2 2

so we can claim that
(k − 1) 2
inf F = − .
2
√
Let us determine the best lower bound of F(k) over all triangles ABC, where k ≥ 3.
Preserving the notations above, the minimum value of r is attained when the triple (x, y, z) is
any permutation of
1 − t
Å 2 ã
t, t,
2t
√
for each fixed t ≥ 3 .
3
Since h is decreasing, we get that

3
inf F = lim h(t) = −k .
t→∞
Next up, let’s deal with finding the best lower bound of F(k) over all triangles ABC under
√ √
the condition that 3 ≤ k < 3. As noted above,
3
2
2
2(k + 1)(kt − 1)(1 − 3t )
′
′
h (t) = g (t) = ,
2
(t + 1) 3
and thus
® Ç√ å ´
3
3
inf F = min h , lim h(t) = −k .
3 t→∞
Our next task is to find the best lower bound of F(k) over all non-obtuse-angled triangles
√ î √ ó
ABC, where 3 ≤ k ≤ 1. Notice that h is strictly increasing on 3 , 1 , so
3 3
√ ä
Ä 3
k − 3 √ (k − 1) 2
= h(1) < h( 3) = − .
8 2
Since earlier we showed that
(kx − 1)(ky − 1) (k − 1) 2
− ≥ − ,
x + y 2

then in this case we have
Ä √ ä 3
k − 3
inf F = .
8
Lastly, let us now find the best lower bound of F(k) over all non-obtuse-angled triangles
√
ABC, where 1 < k < 3. By denoting

 Ä
√ ä 3 
k − 3
 (k − 1) 2 
m = min , − ,
8 2
 