Romanian Mathematical Olympiad - District Round

a) Let $e$ be the unit element of the monoid $(A, *)$. Then
$$f(0) + g(e) + h(0) \cdot e = 0 * e = 0 \quad \text{and} \quad f(e) + g(0) + h(e) \cdot e = e * 0 = 0,$$
so that $f(e) = g(e) = f(0) = g(0) = h(e) \cdot e = h(0) \cdot e = 0$, whence $e = e * e = f(e) + g(e) = 0$.
Then $0 * x = x$ and $x * 0 = x$, for any $x \ge 0$, and we obtain
$$f(0) + g(x) + h(0) \cdot x = x \quad \text{and} \quad f(x) + g(0) + h(x) \cdot x = x,$$
so that $f(x) = x(1 - h(x))$ and $g(x) = x(1 - h(0))$, for any $x \ge 0$. Since $f(x)$, $g(x) \ge 0$, it follows that $h(x) \in [0, 1]$, $\forall x \ge 0$.

$$x * y = x + y - x \cdot h(x) - y \cdot h(0) + h(x) \cdot |x - y|, \quad \forall x, y \ge 0.$$
$$xh(x) - yh(y) = h(0)(x - y) + (h(x) - h(y)) \cdot |x - y|, \quad \forall x, y \ge 0.$$
Since $h$ is bounded, it follows that $\lim_{x \to y} xh(x) = yh(y)$, for any $y \ge 0$, so that the function $p: A \to A$, $p(x) = xh(x)$, is continuous. But then $h$ is continuous on $(0, \infty)$.
Also, for any $y > 0$ we have
$$\lim_{x \to y} \frac{p(x) - p(y)}{x - y} = h(0),$$
so that there are $a = h(0)$ and $b \ge 0$ such that $p(y) = ay + b = h(0)y + b, \forall y > 0$. Then $b = \lim_{y \to 0} p(y) = p(0) = 0$. But then $yh(y) = p(y) = yh(0)$ for any $y > 0$ and it follows that $h(y) = h(0), \forall y > 0$. The function $h$ is thus constant, hence continuous.

b) Let $k = h(0)$. Then $h(x) = k$ and $f(x) = g(x) = x(1 - k)$, for any $x \ge 0$, and $x * y = (x + y)(1 - k) + k|x - y|, \forall x, y \ge 0$. Then $(1 * 1) * 2 = 1 * (1 * 2) \implies k(4k - 2) = 0$, so that $k \in \{0, \frac{1}{2}\}$.
For $k = 0$ we have that $f = g = \text{id}_A$ and $x * y = x + y, \forall x, y \ge 0$.
For $k = \frac{1}{2}$ we have that $f(x) = g(x) = \frac{x}{2}, \forall x \ge 0$ and $x * y = \frac{x+y}{2} + \frac{|x-y|}{2} = \max(x, y), \forall x, y \ge 0$.