MHB Prove ∠A≤60° if (1/sinB+1/sinC)(−sinA+sinB+sinC)=2

[anemone]

Hello MHB,

This is the first of our "Unsolved challenge" threads, it is a challenge which I found pretty interesting but yet I still couldn't crack it. The posting of this challenge here is meant for you to have a fun attempt at it, but if you choose to attempt it, please don't get frustrated if you can't solve it...we'll just have to wait and see if someone can provide a solution, because as the title of the thread indicates, it is an unsolved challenge! :)

I won't beat around the bush any longer, please allow me to present one of such interesting yet unsolved challenge to you all now:

In a triangle $ABC$, the angles $A$, $B$ and $C$ satisfy the following equation:

$\left(\dfrac{1}{\sin B}+\dfrac{1}{\sin C}\right)(-\sin A+\sin B+\sin C)=2$

Prove that $\angle A \le 60^\circ$.

 

[Opalg]

[sp]Start with a bit of calculus. If $0<b\leqslant c$ and $f(x) = x(b+c-x)$, then $f(x) \geqslant bc$ whenever $b\leqslant x\leqslant c.$ The reason for that is that the (parabolic) function $f$ has its maximum at $x = \frac12(b+c)$, which lies inside the interval $[b,c]$. The minimum value of $f$ therefore lies at one of the endpoints of the interval. But $f(b) = f(c) = bc$, so that $bc$ is the minimum. [The condition $b\leqslant c$ is actually irrelevant. The result applies equally well if $b\geqslant c$, except that the interval is then $[c,b]$.]

Now let $ABC$ be a triangle, and let $a,b,c$ be the lengths of the sides opposite the respective vertices. Given that $\left(\frac{1}{\sin B}+\frac{1}{\sin C}\right)(-\sin A+\sin B+\sin C)=2$, multiply through by $\sin B\sin C$ to get $$(\sin B + \sin C)(-\sin A+\sin B+\sin C)=2\sin B\sin C,$$ $$(\sin B+\sin C)^2 - \sin A(\sin B+\sin C) = 2\sin B\sin C,$$ $$\sin^2B + \sin^2 C = \sin A(\sin B+\sin C).$$ By the sine rule, each sine is proportional to the length of the corresponding side, so that last equation is equivalent to $$b^2 + c^2 = a(b+c).$$ If that is written as $b(b-a) + c(c-a) = 0,$ it is clear that the terms on the left side must have opposite signs, so that $a$ must lie between $b$ and $c$.

Also, it follows from the cosine rule that $b^2 + c^2 = a^2 + 2bc\cos A.$ Therefore $a^2 + 2bc\cos A = a(b+c)$ and so $$\cos A = \frac{a(b+c-a)}{2bc}.$$ But $a$ lies between $b$ and $c$, so it follows from the calculus result above that $a(b+c-a) \geqslant bc.$ Thus $\cos A \geqslant \frac12$ and therefore $^\angle A \leqslant 60^\circ.$

[/sp]

 

[anemone]

I can only say thank you, Opalg, for your so insightful and intelligent solution! (Sun)(Cool)