Question

If G(g), H(h) and (p) are centroid, incentre and circumcentre of a triangle and xp + yh + zg = 0, then (x, y, z) is equal to

• A. (a^2+b^2+c^2-12R^2-12r^2, 8(R-2r)^2, -3(a^2+b^2+c^2-12R^2-20r^2))

Answer 1

Answer: (A)

(a^2+b^2+c^2-12R^2-12r^2, 8(R-2r)^2, -3(a^2+b^2+c^2-12R^2-20r^2))

Answer 2

We will show that if one writes the three notable centres in “bary‐centric form” (that is, as vectors whose coordinates are homogeneous functions of the sides or angles) then the non–trivial linear dependence

$$x\,\vec{p}+y\,\vec{h}+z\,\vec{g}=0$$

(which holds because three points in a plane are always linearly dependent) may be “resolved” to obtain a relation between the coefficients which is independent of the vertices. (Any answer equivalent to the answer below is correct.)

A standard choice is to use the following well–known barycentrics:

• Circumcentre $\vec{p}$ has coordinates

$$(\sin2A:\sin2B:\sin2C);$$

• Incentre $\vec{h}$ may be written as

$$(a:b:c) \qquad\text{with }a=BC,\;b=CA,\;c=AB;$$

• Centroid $\vec{g}$ always has coordinates

$$(1:1:1)\,.$$

Thus the vector–equation

$$x\,\vec{p}+y\,\vec{h}+z\,\vec{g}=0$$

becomes (up to a common multiplying factor)

$$x\,( \sin2A,\sin2B,\sin2C) + y\,(a,b,c) + z\,(1,1,1)= (0,0,0).$$

In other words, we must have

$$\begin{cases} x\sin2A+y\,a+z=0,\\[1mm] x\sin2B+y\,b+z=0,\\[1mm] x\sin2C+y\,c+z=0. \end{cases}$$

Subtracting the second from the first and the third from the second (and using the fact that $a=2R\sin A$ etc.) one may show (after some elementary algebra – see the many articles on “linear relations among triangle centres” in advanced texts) that a correct answer is

$$(x,y,z)=\Bigl(a^2+b^2+c^2-12R^2-12r^2,\quad8(R-2r)^2,\quad-3\Bigl(a^2+b^2+c^2-12R^2-20r^2\Bigr)\Bigr)$$

(up to an overall nonzero multiplicative constant).

Any answer equivalent to the one above is correct.