In a triangle $ABC$ with fixed base $BC$, the vertex $A$ moves such that $\cos B + \cos C = 4\sin^{2}\frac{A}{2}$. If $a$, $b$ and $c$ denote the lengths of the sides of the triangle opposite to the angles $A$, $B$ and $C$, respectively, then

$\angle A+\angle B+\angle C=\pi, \quad a,b,c \text{ opposite } A,B,C$ $BC=a \text{ is fixed}, \quad A \text{ moves}$ $\cos B+\cos C=4\sin^{2}\dfrac{A}{2}$ $\text{Check: } b+c=4a \ ? \quad \text{or} \quad b+c=2a \ ?$ $\text{and the locus of } A$ $\text{Options (1), (2), (3), (4)}$ $1.\ \text{simplify } \cos B+\cos C \text{ using } A+B+C=\pi$ $2.\ \text{find } b+c \text{ from the condition}$ $3.\ \text{identify the locus of } A$ $A+B+C=\pi \Rightarrow B+C=\pi-A$ $\cos B+\cos C=2\cos\dfrac{B+C}{2}\cos\dfrac{B-C}{2}$ $\cos\dfrac{B+C}{2}=\cos\!\left(\dfrac{\pi-A}{2}\right)=\sin\dfrac{A}{2}$ $2\sin\dfrac{A}{2}\cos\dfrac{B-C}{2}=4\sin^{2}\dfrac{A}{2}$ $\cos\dfrac{B-C}{2}=2\sin\dfrac{A}{2}$ $\cos\dfrac{B+C}{2}+\cos\dfrac{B-C}{2}=2\cos\dfrac{B}{2}\cos\dfrac{C}{2}$ $3\sin\dfrac{A}{2}=2\cos\dfrac{B}{2}\cos\dfrac{C}{2}$ $\cos\dfrac{B+C}{2}-\cos\dfrac{B-C}{2}=-2\sin\dfrac{B}{2}\sin\dfrac{C}{2}$ $\sin\dfrac{B}{2}\sin\dfrac{C}{2}=\dfrac{1}{2}\sin\dfrac{A}{2}$ $\tan\dfrac{B}{2}\tan\dfrac{C}{2}=\dfrac{\sin\frac{A}{2}}{3\sin\frac{A}{2}}=\dfrac{1}{3}$ $\tan\dfrac{B}{2}\tan\dfrac{C}{2}=\dfrac{s-a}{s}, \quad s=\text{semi-perimeter}$ $\dfrac{s-a}{s}=\dfrac{1}{3} \Rightarrow 3(s-a)=s$ $2s=3a$ $2s=a+b+c=3a$ $b+c=2a$ $\text{Option (2) true}, \quad b+c=4a \text{ false}$ $b=CA, \quad c=AB$ $CA+AB=b+c=2a=2\,BC$ $\Rightarrow CA+AB=\text{constant}$ $CA+AB=\text{constant} \Rightarrow \text{locus is an ellipse}$ $\text{foci } B,C, \quad \text{major axis}=2a$ $\text{Option (3) true}$ $\text{sum constant} \Rightarrow \text{ellipse, not a pair of lines}$ $|CA-AB|=\text{constant} \Rightarrow \text{pair of lines}$ $\text{Option (4) false}$ $b+c=2a$ $\text{locus of } A = \text{ellipse, foci } B,C$ $\boxed{\text{Answer: (2) and (3)}}$