I have a minimum and maximum problem connected to trigonometry. Please help, it is from an American olympiad from 2013.

Suppose that the angles of a triangle are $\alpha, \beta$ and $\gamma$. Let $H$ be the set $\left\{\dfrac{\cos \alpha}{\cos (\beta-\gamma)}, \dfrac{\cos \beta}{\cos (\gamma-\alpha)}, \dfrac{\cos \gamma}{\cos (\alpha-\beta)}\right\}$.

a) Find the minimal possible value of the largest item of $H$.

b) Find the maximal possible value of the smallest item of $H$.


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    $\begingroup$ Without bothering to do any math, relying purely by instinct, my guess is that the equilateral triangle is relevant to both questions. Therefore, my first approach would be to attempt to prove this. $\endgroup$ – user2661923 Dec 25 '18 at 22:58
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    $\begingroup$ Welcome to MSE. To help us provide the most useful & educational answer for you, please let us know what you've tried already, any particular difficulties you encountered, etc. Thanks. $\endgroup$ – John Omielan Dec 25 '18 at 23:45

WLOG, let $A=\alpha\ge B=\beta\ge C=\gamma$

$\cfrac {\cos A}{\cos (B-C)}=-\cfrac{\cos(B+C)}{\cos(B-C)}=1-\cfrac 2{1+\tan B\tan C}\tag 1$

  1. If $A=\pi/2$, $\cfrac {\cos A}{\cos (B-C)}=0, \cfrac {\cos B}{\cos (C-A)}=\cfrac {\cos B}{\cos B}=1=\cfrac {\cos C}{\cos (A-B)}$. Hence minMax=$1$, maxMin=$0$;

  2. If $A>\pi/2$, from $(1)$, it's easy to see the maximum in $H$ is $1-\cfrac 2{1+\tan A \tan B}$, and the minimum is $1-\cfrac 2{1+\tan B \tan C}$. When $A\to \pi/2$, minMax $\to 1$; when $B=C \to \pi/4$, maxMin $\to 0$;

  3. If $A<\pi/2$, maximum in H is $1-\cfrac 2{1+\tan A \tan B}$, and the minimum is $1-\cfrac 2{1+\tan B \tan C}$. It can be easily shown mininum of maximum is $1/2$ when $A=B=\pi/3$ and maximum of minimum is $1/2$ when $B=C=\pi/3$.

Combine all cases, we have min max=max min = $1/2$ when $A=B=C=\pi/3$

PS: $f(x)=1-\cfrac 2{1+x}, x>0$, as $x$ increases, $\cfrac 2{1+x}$ decreases, and $f(x)$ increases. the minimum of $\tan A \tan B (\pi/2>A\ge B\ge C\ge \pi/3)$ occurs when $A=B=\pi/3$. etc, etc.


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