Recognize the Point

In what follows, $V_A, V_B, V_C$ and $P$ are 2-D position vectors representing the position of the three vertices of the triangle and point P. While $A , B$ and $C$ are used to denote the angles of the triangle.

The given function can be written as,

$f(P) = (P - V_A)^T(P - V_A) \sin A + (P - V_B)^T (P-V_B) \sin B + (P - V_C)^T (P - V_C) \sin C$

This is a quadratic function of the vector $P$ . Simplifying the above expression results in,

$f(P) = \alpha P^T P - 2 P^T V_0 + \beta$

where,

$\alpha = \sin A + \sin B + \sin C$ ,

$V_0 = (\sin A) V_A+ (\sin B) V_B + (\sin C) V_C$

and

$\beta = (\sin A) {V_A}^T {V_A} + (\sin B) {V_B}^T{V_B} + (\sin C) {V_C}^T{V_C}$

Using an approach very similar to completing the square for scalars, we obtain,

$f(P) = \alpha P^T P - 2 P^T V_0 + \beta = \alpha (P - \dfrac{1}{\alpha} V_0)^T (P - \dfrac{1}{\alpha} V_0) + \beta - \dfrac{1}{\alpha} {V_0}^T{V_0}$

The minimum of this quadratic function occurs at

$P = \dfrac{1}{\alpha} V_0 = \dfrac{ (\sin A) V_A+ (\sin B) V_B + (\sin C) V_C }{\sin A + \sin B + \sin C}$

From the Law of Sines, this reduces to

$P = \dfrac{ a V_A+ b V_B + c V_C }{a+b+c}$

This is the well-known expression for the coordinates of the incenter. See here for example.

Therefore, the correct choice is: Incenter.