Page 23 - MATINF Nr. 8

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we clearly have that
min h(t) = m.
h i
1
t∈ √ ,1
3
In addition,
(kx − 1)(ky − 1) (k − 1) 2
− ≥ − ≥ m
x + y 2
and Ä √
Ä √ ä 3 î äó
2
2
k − 3 + 4(k − 1) = k − 3 3 − 4 (k + 1).
In conclusion, the following holds true for non-obtuse-angled triangles ABC :

√
 (k−1) 2
 − , if k ≥ 3 3 − 4
 2
inf F = √ .
 (k− 3) 3 √ √

, if 3 ≤ k ≤ 3 3 − 4
8 3
4 Alternative solution

We will now present an elementary solution that doesn’t require the use of derivatives. To
this end, we will employ a great deal of trigonometric identities and inequalities expressed in
terms of the canonical elements of a triangle, namely: R (circumradius), r (inradius) and s
(semiperimeter).

In what follows, we will start by introducing a key identity along with well-known inequalities
that will be used throughout the subsequent proofs.

4.1 Preliminaries

4.1.1 Expressing F(k) in terms of R, r and s

Let us begin by establishing the following identity:

ñ ô
Å ã 2 Å ã 2 Å ã 2
Y s 1 r r 1 r
2
(k cos A − sin A) = k + k k + 1 − · − 1 + − · .
2R k 2R 2R k 2R
Applying Result 141.1 (see [2]) to a triangle with angles (π − 2A, π − 2B, π − 2C) – also
known as the orthic triangle and described in the Extension of Result 111 (see [3]) – we get:

Y 1 cot ϕ
(cot A − cot ϕ) = · 2R 0 sin(2ϕ) + r 0 tan ϕ − s 0
2
s 0 sin ϕ
R cot ϕ Å Y rs ã
= · R sin(2ϕ) + 2R tan ϕ cos A −
2
rs sin ϕ R

2
2R 2 s − (2R + r) 2 1 cot ϕ
2
= · cot ϕ + · − .
2
2
rs 2rs sin ϕ sin ϕ