IMC 2025 · Day 1 · P1

(a) Suppose that the degree of $P$ is odd and let $(a, b) \in \mathbb{R}$ be arbitrary. Given $r \in \mathbb{R}$, the equation for $\ell_r$ is given by $$\ell_r = \bigl\{ (x, y) \in \mathbb{R}^2 \mid y = P'(r)(x - r) + P(r) \bigr\}.$$ For this line to pass through the point $(a, b)$ it is therefore necessary and sufficient that $$b = a P'(r) + P(r) - r P'(r).$$ This is a polynomial equation in $r$, which always has a real solution as soon as we can show that the degree is odd. Indeed, if $P(r) = c r^n + \dots$ describes the leading term, then the right hand side of the above equation has leading term $(c - nc) r^n$. Since $c \ne 0$ and we assumed that $n \ge 2$, we must have $c - nc \ne 0$. The right hand side therefore has the same degree as $P$, completing the proof.

(b) If the degree of $P$ is even, then this can never be true, because the degree of $a P'(r) + P(r) - r P'(r)$ is now even by the same argument and therefore it has a global minimum (or maximum) over the reals, below (or above) which no value of $b$ will yield a real solution for $r$.