Find $P$ in the plane of $\triangle ABC$ that minimizes $a\,AP^2+b\,BP^2+c\,CP^2$

Question

From KMO:

Let $\triangle ABC$ be a triangle. Find a point $P$ in the plane of $\triangle ABC$ such that $a\,AP^2 + b\,BP^2 + c\,CP^2$ is minimum.

How can I solve this problem by using the basic properties of a triangle?

Answer 1

In view of lack of information $a,b,c $ will be assumed to be positive real numbers. Let $(x,y)$ be the coordinates of the point $P$. Then the sum in question is: $$ S=a[(x-x_A)^2+(y-y_A)^2]+b[(x-x_B)^2+(y-y_B)^2]+c[(x-x_C)^2+(y-y_C)^2]. $$

From the two equations: $$ \frac{dS}{dx}=0;\quad \frac{dS}{dy}=0, $$ one obtains that the extremum of $S$ is attained at the point: $$ x=\frac{ax_A+bx_B+cx_C}{a+b+c};\quad y=\frac{ay_A+by_B+cy_C}{a+b+c}.\tag1 $$

The equation $(1)$ describes the barycenter of masses $a,b,c$ assigned to vertices $A,B,C$, respectively. In the case $a=b=c$, the point is the centroid of the triangle $ABC$. In the case $a:b:c=BC:CA:AB$ it is the incenter of the triangle.

I left it to you to check that the extremum $(1)$ is indeed the minimum.