<s>[INST] Geometry Problem from Olympiad book Given the base and the vertical angle of a triangle show that its area is greatest when it's isoceles
I am stuck on how to proceed and which theorem to use here
Thanks in advance
 [/INST] Denote $S = \dfrac{1}{2}ab\sin C$. Thus: $S \le \dfrac{1}{8}(a+b)^2\sin C$, with equality occur at $a = b$ which shows $\triangle ABC$ isosceles. To show that $S$ reaches a maximum value, you need to show: $(a+b)^2$ also reaches a maximum in terms of $c$. To this end, use the law of cosines: $c^2 = a^2+b^2 - 2ab\cos C= (a+b)^2 - 2ab(1+\cos C)\ge (a+b)^2 - \dfrac{1+\cos C}{2}(a+b)^2= \dfrac{1-\cos C}{2}(a+b)^2= \sin^2(\frac{C}{2})(a+b)^2\implies (a+b)^2 \le \dfrac{c^2}{\sin^2(\frac{C}{2})}\implies S \le \dfrac{1}{8}\cdot\dfrac{c^2}{\sin^2(\frac{C}{2})}\cdot \sin C= \dfrac{c^2}{4\tan(\frac{C}{2})}$. This is the maximum value of the area $S$ of $\triangle ABC$, and this value occurs when $a = b$ by AM-GM inequality. So $\triangle ABC$ is isosceles when the max occurs.
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