Baltic Way 1996 · Problem 14

Solution:

Let the points $B_{i}$ and $C_{i}$ have the coordinates $(b_{i}, b)$ and $(c_{i}, c)$, respectively, for $i=1,2, \ldots, n$. Then we have

$$cot \angle B_{1} C_{1} P+\cdots+\cot \angle B_{n} C_{n} P=\frac{1}{b-c} \sum_{i=1}^{n}\left(b_{i}-c_{i}\right)$$

The numbers $b_{i}$ and $c_{i}$ are the solutions of $f(x)-b=0$ and $f(x)-c=0$, respectively. As $n \geq 2$, it follows from the relationships between the roots and coefficients of a polynomial (Viète's relations) that $\sum_{i=1}^{n} b_{i}=\sum_{i=1}^{n} c_{i}=-a_{n-1}$ regardless of the values of $b$ and $c$, and hence $\cot \angle B_{1} C_{1} P+\cdots+\cot \angle B_{n} C_{n} P=0$.