$ a \cos A + b \cos B + c \cos C = \dfrac{a+b+c}2 $ $\implies$ the triangle is equilateral? If in a triangle $ a \cos A + b \cos B + c \cos C = \dfrac{a+b+c}2 $ , then is the triangle equilateral ?
 A: This  is a 1990 Russian maths contest problem
$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$, so 
$$a \cos A + b \cos B + c \cos C = \dfrac{a+b+c}2 $$
$$\Longleftrightarrow$$
$$a^4+b^4+c^4-2(a^2b^2+b^2c^2+c^2a^2)+abc(a+b+c)=0\tag{1}$$
There are several ways to show (1)
In fact
$$a^4+b^4+c^4-2(a^2b^2+b^2c^2+c^2a^2)+abc(a+b+c)$$
$$=(a+b+c)\Big(abc-(-a+b+c)(a-b+c)(a+b-c)\Big)$$
It is easy to show that
$$abc-(-a+b+c)(a-b+c)(a+b-c)=0\Longleftrightarrow a=b=c$$
another method to prove (1)
one can use Schur's inequality
$$\sum a^2(a-b)(a-c)\geq0$$
A: $\displaystyle a=2R\sin A\implies\cos A=R(\sin2A)$
$\displaystyle\implies\sum a\cos A=R(\sum\sin2A)$
Using Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle,  $\displaystyle\sum a\cos A=4R\prod\sin A=4R\prod2\sin\frac A2\cos\frac A2$
Again, $\displaystyle\frac{a+b+c}2=R(\sum \sin A)$
Using this,  $\displaystyle\frac{a+b+c}2=4R\prod\cos\frac A2$
As $\displaystyle0<A,B,C<\pi\implies\cos\frac A2$ (etc.) $\ne0,$ the problem reduces to $\displaystyle\prod2\sin\frac A2=1$
$\implies4\sin\frac A2\left(\cos\frac{B-C}2-\cos\frac{B+C}2\right)=1 $
As $\displaystyle B+C=\pi-A\implies \frac{B+C}2=\frac\pi2-\frac A2$
$\displaystyle\iff4\sin^2\frac A2-4\sin\frac A2\cos\frac{B-C}2+1=0 $ which is a Quadratic Equation in $\sin\frac A2$
As $A$ is real the discriminant  $\displaystyle\left(4\cos\frac{B-C}2\right)^2-4\cdot4=-16\sin^2\frac{B-C}2$ must be $\ge0$
$\displaystyle\implies\sin^2\frac{B-C}2=0\iff\cos(B-C)=1-2\sin^2\frac{B-C}2=1\implies B-C=2n\pi $
As $\displaystyle0<B,C<\pi\implies B=C\implies\sin\frac A2=\frac12$
Can you prove $A=60^\circ?$
A: If you use the relation, $a=2R sin A$, 
I have not got the solution, however, I can help in solving this:
$$\sin 2A.\sin 2B. \sin 2C= 4 \sin A \sin B \sin C$$ (can be easily derived using $C=\pi-A-B$)
Then
$$\sin A.\sin B. \sin C=4 \cos (A/2).\cos (B/2).\cos (C/2)$$
Using $$\sin A=2\sin (A/2) \cos (A/2)$$, we get, $$\sin (A/2) \sin (B/2) \sin (C/2)=\frac{1}{8}$$
Now in a triangle,
$$\sin (A/2) \sin (B/2) \sin (C/2)\leq \frac{1}{8}$$ (which can be proved using partial derivatives, assuming $A=x$, $B=y$ and $C=\pi-x-y$, but I think there has to be a simpler proof than this)
And the equality holds only in case of equilateral triangle, hence, the triangle has to be equilateral!
Hence proved.
This discussion will surely help you in solving this!
