Page 24 - MATINF Nr. 8

24 4 ALTERNATIVE SOLUTION

Now, by denoting k = tan ϕ, we further obtain:

Å ã 2 2 2 Å ã Å ã
Y 1 2R 1 s − (2R + r) 1 1 1
cot A − = · + · 1 + − 1 + .
k rs k 2 2rs k 2 k k 2
Multiplying both sides of the previous equation by k 3 Q sin A gives:
ï 2 2 2 ò
Y rs 2R s − (2R + r) 3 2
(k cos A − sin A) = · · k + · k + k − k + 1
2R 2 rs 2rs

h i
2
2
2
k (k + 1) s − (2R + r) − 2rs · 1
k
= k +
4R 2
2
s − r 2 − (2R + r) − r 2
2

= k + k k + 1 · k k 2
4R 2
ñ ô
Å ã 2 Å ã 2 Å ã 2
s 1 r r 1 r

2
= k + k k + 1 − · − 1 + − · ,
2R k 2R 2R k 2R
which is exactly what we wanted to prove.
4.1.2 Well-known inequalities involving R, r and s

Throughout the remaining of this article we will make use of some famous results from the
literature of geometric inequalities (see [4]), namely:

1. Euler’s inequality:
R ≥ 2r;

2. Gerretsen’s inequality:
2
2
2
2
16Rr − 5r ≤ s ≤ 4R + 4Rr + 3r ;
3. Mitronovi´c’s inequality: √
√ 3R 3
3r 3 ≤ s ≤ ;
2
4. Walker’s inequality (holds for non-obtuse-angled triangles):
2
2
2
s ≥ 2R + 8Rr + 3r ;
Q
5. Unnamed inequality equivalent to cos A ≥ 0 :
s ≥ 2R + r.

4.1.3 Equivalent form of F(k)

h i
Ä √ ó
For simplicity, let’s denote x = s ∈ 0, 3 3 and y = r ∈ 0, 1 . As a result, our previously-
2R 2 2R 4
derived identity is equivalent to:
ñ ô
y
Å y ã 2 Å ã 2
2
2
F(k) = k + k k + 1 x − − (1 + y) − .
k k