Romanian Mathematical Olympiad

We shall prove that the answer is $f : \mathbb{R} \to \mathbb{R}$, $f(x) = mx + n$, with $m, n \in \mathbb{R}$. It is trivial to see that this family of functions satisfy the property.

For the converse, let $f$ with the given property. Consider $a, b \in \mathbb{R}$, $a < b$, and define
$$m = \frac{f(b) - f(a)}{b - a} \quad \text{and} \quad n = \frac{bf(a) - af(b)}{b - a}.$$
We shall show that $f(x) = mx + n$, for all $x \in [a, b]$. By way of contradiction, let $x_0 \in [a, b]$, such that $f(x_0) \neq mx_0 + n$. Consider the sets
$$A = \{x \in [a, x_0] \mid f(x) = mx + n\} \quad \text{and} \quad B = \{x \in [x_0, b] \mid f(x) = mx + n\}.$$
$A$ and $B$ are nonempty because $a \in A$ and $b \in B$. Consider the numbers $\alpha = \sup(A)$ and $\beta = \inf(B)$. By the continuity of $f$ we get $\alpha \in A$, $\beta \in B$ and $\alpha < x_0 < \beta$. By the hypothesis, there is $t \in (0, 1)$ such that
$$f((1-t)\alpha + t\beta) = (1-t)f(\alpha) + tf(\beta) = m((1-t)\alpha + t\beta) + n.$$
Then $\gamma = (1-t)\alpha + t\beta \in (\alpha, \beta)$ and $f(\gamma) = m\gamma + n$, in contradiction with the definition of $\alpha$ and $\beta$. So, $f(x) = mx + n$, for all $x \in [a, b]$.

To extend the answer to $\mathbb{R}$, observe that if $f : [a, c] \to \mathbb{R}$, $c > b$ is a function obtained from the preceding considerations, say $f(x) = m'x + n'$, we must have by continuity $m = m'$, $n = n'$. The same remark can be made on intervals of the form $[c, b]$ with $c < a$.