[ Trilinear coordinates ]

a) Find the trilinear coordinates of the vertices of the Brocard triangle.

b) Find the trilinear coordinates of the Steiner point (see problem $\underline{19.55 .2) .}$

#
a) From the solution of problem 19.55, it follows that the vertex $A_{1}$ of the Brocard triangle is the intersection point of the lines $C P$ and $B Q$, where $P$ and $Q$ are the first and second Brocard points. Therefore, the trilinear coordinates of point $A_{1}$ are

$$
\left(1: \frac{c^{2}}{a b}: \frac{b^{2}}{a c}\right)=\left(a b c: c^{3}: b^{3}\right)
$$

The barycentric coordinates of this point are ( $a^{2}: c^{2}: b^{2}$ ).

b) It is more convenient to perform the calculations in barycentric coordinates. In barycentric coordinates ( $\alpha: \beta: \gamma$ ), the line $B_{1} C_{1}$ is given by the equation

$$
0=\left|\begin{array}{ccc}
\alpha & \beta & \gamma \\
c^{2} & b^{2} & a^{2} \\
b^{2} & a^{2} & c^{2}
\end{array}\right|=\alpha\left(b^{2} c^{2}-a^{4}\right)+\beta\left(a^{2} b^{2}-c^{4}\right)+\gamma\left(a^{2} c^{2}-b^{4}\right)
$$

Moreover, $\alpha+\beta+\gamma=1$. Therefore, the line passing through point $A$ and parallel to the line $B_{1} C_{1}$ is given by the equation

$$
\beta\left(a^{2} b^{2}-c^{4}+a^{4}-b^{2} c^{2}\right)+\gamma\left(a^{2} c^{2}-b^{4}+a^{4}-b^{2} c^{2}\right)=0
$$

i.e., $\left(a^{2}+b^{2}+c^{2}\right)\left(\beta\left(a^{2}-c^{2}\right)+\gamma\left(a^{2}-b^{2}\right)\right)=0$. Therefore, $\beta: \gamma=\frac{1}{c^{2}-a^{2}}: \frac{1}{a^{2}-b^{2}}$. Thus, the Steiner point has barycentric coordinates $\left(\frac{1}{b^{2}-c^{2}}: \frac{1}{c^{2}-a^{2}}: \frac{1}{a^{2}-b^{2}}\right)$. The trilinear coordinates of the Steiner point are

$$
\left(\frac{1}{a\left(b^{2}-c^{2}\right)}: \frac{1}{b\left(c^{2}-a^{2}\right)}: \frac{1}{c\left(a^{2}-b^{2}\right)}\right)
$$