More trigonometry I guess?

Geometry Level 4

1 cos ⁡ α + 1 cos ⁡ β + 1 cos ⁡ γ ≥ M ⋅ R R + r \large \dfrac1{\cos \alpha} + \dfrac1{\cos\beta} + \dfrac1{\cos \gamma} \geq \dfrac{M \cdot R}{R+r}

Let r r and R R denote the incircle and circumcircle , respectively, of an acute-angle triangle with interior angles α , β \alpha, \beta and γ \gamma .

Find the maximum value of M M satisfying the inequality above.


The answer is 9.

Dragan Marković by Dragan Marković , 26, Serbia

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3 solutions

Tom Engelsman
May 9, 2021

Taking a lazy (but intuitive) approach here. The maximum value for M M occurs at the equilateral triangle. For an equilateral triangle of side length x x , we have:

r = x 2 ⋅ tan ⁡ ( π / 6 ) = x 2 3 , \large r = \frac{x}{2} \cdot \tan(\pi/6) = \frac{x}{2\sqrt{3}},

R = x 2 ⋅ sec ⁡ ( π / 6 ) = x 3 \large R = \frac{x}{2} \cdot \sec(\pi/6) = \frac{x}{\sqrt{3}}

α = β = γ = π / 3 \alpha = \beta = \gamma = \pi/3

Hence, M ≤ R + r R ⋅ 3 sec ⁡ ( π / 3 ) = x / 3 + x / 2 3 x / 3 ⋅ 6 = ( 3 2 ) ( 6 ) = 9 . \large M \le \frac{R+r}{R} \cdot 3\sec(\pi/3) = \frac{x/\sqrt{3} + x/2\sqrt{3}}{x/\sqrt{3}} \cdot 6 = (\frac{3}{2})(6) = \boxed{9}.

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Kumar Krish
Mar 17, 2019

First use titu's lemma in LHS And then use the formula r=4Rsin(A/2) sin(B/2) sin(C/2) and then use conditional identity cosA+cosB+cosC=1+4sin(A/2) sin(B/2) sin(C/2) And then compare both inequalities you got. Finally answer is in front of you

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Rohan Jasani
Aug 30, 2017

Use Engels form Cauchy Schwartz inequality

0 Helpful 0 Interesting 0 Brilliant 2 Confused

By the way it is angel form not engels form

Kumar Krish - 2 years, 3 months ago

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