Circular Motion without Physics

It is well-known that the circumcenters of $\triangle BIC , \triangle AIC , \triangle AIB$ are the midpoints of the arcs $BC, CA, AB$ . This implies that $L,M,N$ lie on the circumcircle which implies that all of their powers are $0$ . It is well-known that $0 + 0 + 0 = \boxed{0}$ .

Incenter circle centers on Circumcircle Since incenter I lies on angle bisectors, $\angle CAB = 2 \angle IAB = \angle INB$ angle subtended by the minor arc BI at center N.

Similarly, $\angle CBA = \angle INA$

Thus $\angle ANB = \angle CAB + \angle CBA =$ angle subtended by the arc ACB of the circumcircle on the circumference. Hence any point N at which the arc subtends the same angle must be on the circumference of the circumcircle too. Thus N lies on the circumcircle and has power = 0.

Similarly, the other two centers L and M can also be proved to lie on the circumcircle. Giving the sum of their powers = 0 + 0 + 0 = 0