4. Find real numbers $a, b, c, d$, for which all solutions $x$ of the inequality

$$
\frac{a x^{2}+b x+c}{a+d x-x^{2}} \leqq 2 x
$$

form the set $\{0\} \cup(4,+\infty)$.

The second round of category A takes place

## on Tuesday, January 19, 1999

so that it starts in the morning and the contestants have 4 hours of pure time to solve the problems. For each problem, a contestant can earn 6 points, and a successful solver is a student who scores 10 points or more. These details will be communicated to the students before the competition begins.
4. We equivalently rearrange the inequality to

$$
\frac{2 x^{3}+(a-2 d) x^{2}+(b-2 a) x+c}{x^{2}-d x-a} \geqq 0 .
$$

We know how to solve an inequality of the form $\frac{A(x)}{B(x)} \geqq 0$ if we know the real roots of both polynomials $A(x), B(x)$. Let the sets of real roots of these polynomials be denoted by $\mathrm{A}, \mathrm{B}$. If we denote by $\mathrm{R}$ the set of solutions to the inequality $\frac{A(x)}{B(x)}>0$, which is clearly equivalent to the inequality $A(x) B(x)>0$, then the set of solutions to the original inequality $\frac{A(x)}{B(x)} \geqq 0$ is the set $(\mathrm{R} \cup \mathrm{A}) \backslash \mathrm{B}$.

When solving the inequality $A(x) B(x)>0$, we can remove any quadratic trinomial $x^{2}+p x+q$ with a negative discriminant from the left side of the inequality, and since we are interested in the solution to the inequality $A(x) B(x)>0$ especially for $x \notin \mathrm{A} \cup \mathrm{B}$, we can also remove any power $(x-\alpha)^{n}$ with an even exponent. This always leads us to an inequality of the form

$$
\left(x-\alpha_{1}\right)\left(x-\alpha_{2}\right) \ldots\left(x-\alpha_{k}\right)>0,
$$

where $\alpha_{1}2$ is the union $\left(-\infty, \alpha_{1}\right) \cup \ldots \cup\left(\alpha_{k-2}, \alpha_{k-1}\right) \cup\left(\alpha_{k}, \infty\right)$.

Let's now return to inequality (1). Since $x=0$ is a solution to it, zero must be a root of the numerator, but not a root of the denominator, so $c=0$ and $a \neq 0$. Furthermore, from the fact that zero is an "isolated" solution, it follows from our previous considerations that zero is a root of even multiplicity, i.e., a double root. Therefore, $b-2 a=0$.

Since the interval $(4, \infty)$ is part of the solution set, but not its boundary point $x=4$, the number 4 is a root of the denominator, so $a+4 d=16$. After substituting $a=16-4 d$ and factoring the denominator, we get the equivalent inequality

$$
\frac{x^{2}(x+8-3 d)}{(x-4)(x-d+4)} \geqq 0 .
$$

From this, it follows that the solution to the inequality

$$
(x-4)(x+8-3 d)(x-d+4)>0
$$

must be the interval $(4, \infty)$, so $3 d-8=d-4$, or $d=2, a=8, b=16, c=0$. For these values, we get the inequality

$$
\frac{x^{2}(x+2)}{(x-4)(x+2)} \geqq 0
$$

whose set of solutions is indeed $\{0\} \cup(4,+\infty)$.

A complete solution is worth 6 points. Do not deduct points if the solver states the multiplicity of the zero root without proper justification. Award 1 point for each of the equalities $c=0, a+4 d=16$. If the solution is led in such a way that a check is necessary at the end, and there is no mention of it, deduct 1 point.