Geometry Problem from Olympiad book

Question

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

Answer 1

Here is a hint you might be able to use

Answer 2

You can also use the formula cosines, suppose the base to be a and the given angle A (opposite to each other), and the other two sides adjacent to angle A as x and y. Now apply the formula cosines for angle A. Now find $sinA$. Then use$$0.5*x*y*sinA$$ Then AM-GM inequality on the term which you get in numerator.

Answer 3

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.