In a triangle ABC with fixed base BC, the vertex A moves such that cosB + cosC = 4sin²(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 what is the locus of point A?

cosB+cosC=4sin²(A/2) ⇒ 2cos((B+C)/2)cos((B-C)/2) = 4sin²(A/2) ⇒ 2cos(π/2 - A/2)cos((B-C)/2) = 4sin²(A/2) ⇒ 2sin(A/2)cos((B-C)/2) = 4sin²(A/2) ⇒ cos((B-C)/2) = 2sin(A/2) ⇒ cos((B-C)/2)/cos((B+C)/2) = 2 ⇒ (cos((B-C)/2) + cos((B+C)/2))/(cos((B-C)/2) - cos((B+C)/2)) = 3 (applying componendo and dividendo) ⇒ (2cos(B/2)cos(C/2))/(2sin(B/2)sin(C/2)) = 3 ⇒ cot(B/2)cot(C/2) = 3 ⇒ (s(s-a))/(s(s-b))(s(s-c)) = 3 ( ∴ cot(B/2) = √((s(s-b))/(s(s-c))) ) ⇒ 3s(s-a)(s-b)(s-c) = s²(s-b)(s-c) ⇒ 3(s-a) = s ⇒ 2s = 3a ⇒ a+b+c = 3a ⇒ b+c = 2a Therefore, sum of distances of a point, A, from two fixed points B and C is a constant. Thus, the locus of A is an ellipse.