International Mathematics Competition
for University Students
2025

IMC2025: Day 1, Problem 1

Problem 1. Let \(\displaystyle P\in\mathbb{R}[x]\) be a polynomial with real coefficients, and suppose \(\displaystyle \operatorname{deg}(P)\geq 2\). For every \(\displaystyle x \in \mathbb{R}\), let \(\displaystyle \ell_x \subset \mathbb{R}^2\) denote the line tangent to the graph of \(\displaystyle P\) at the point \(\displaystyle ( x, P(x) )\).

(a) Suppose that the degree of \(\displaystyle P\) is odd. Show that \(\displaystyle \displaystyle\bigcup_{x \in \mathbb{R}} \ell_x = \mathbb{R}^2.\)

(b) Does there exist a polynomial of even degree for which the above equality still holds?

Mike Daas, Max Planck Institute for Mathematics, Bonn

Solution.

(a) Suppose that the degree of \(\displaystyle P\) is odd and let \(\displaystyle (a,b) \in \mathbb{R}\) be arbitrary. Given \(\displaystyle r \in \mathbb{R}\), the equation for \(\displaystyle \ell_{r}\) is given by

\(\displaystyle \ell_{r} = \bigl\{ (x,y) \in \mathbb{R}^2 \mid y = P'(r)(x-r) + P( r ) \bigr\}. \)

For this line to pass through the point \(\displaystyle (a,b)\) it is therefore necessary and sufficient that

\(\displaystyle b = aP'(r) + P( r ) - rP'(r). \)

This is a polynomial equation in \(\displaystyle r\), which always has a real solution as soon as we can show that the degree is odd. Indeed, if \(\displaystyle P(r) = cr^n + \ldots\) describes the leading term, then the right hand side of the above equation has leading term \(\displaystyle (c - nc)r^n\). Since \(\displaystyle c \neq 0\) and we assumed that \(\displaystyle n \geq 2\), we must have \(\displaystyle c - nc \neq 0\). The right hand side therefore has the same degree as \(\displaystyle P\), completing the proof.

(b) If the degree of \(\displaystyle P\) is even, then this can never be true, because the degree of \(\displaystyle aP'(r) + P( r ) - rP'(r)\) 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 \(\displaystyle b\) will yield a real solution for \(\displaystyle r\).