Problem 1124

I. solution. The existence of the two extreme values means that the derivative of our function has two different real roots, so the

$$y^{\prime}=3 a x^{2}+2 b x+c=0, \quad a \neq 0$$

real quadratic equation has a positive discriminant:

$$b^{2}-3 a c>0$$

Let the points on the curve of the function corresponding to the extreme values be denoted by $M$ and $N$, with coordinates $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$. The fact that the line $MN$ passes through the origin $O$ can also be stated as the equality of the direction tangents of the lines $OM$ and $ON$:

$$\frac{y_{1}}{x_{1}}=\frac{y_{2}}{x_{2}}, \quad \text { i.e. } \quad x_{1} y_{2}=x_{2} y_{1}$$

We can assume that neither $x_{1}$ nor $x_{2}$ is 0, because if $x_{1}=0, y_{1} \neq 0$, then $x_{2} \neq x_{1}$ and the requirement cannot be met, since the line $OM$ is the $y$-axis; if, for example, $x_{1}=y_{1}=0$, then one of $M$ and $N$ is identical to 0, making our requirement meaningless. This assumption is expressed by the fact that in the equation (2) giving $x_{1}$ and $x_{2}$, the product of the two roots is proportional to

$$c \neq 0$$

Accordingly, $a \neq 0,(3),(4)$ and (5) are necessary and sufficient conditions for $M, N$ and $O$ to be distinct points on the same line, but we still need to find the condition among the coefficients of (1) in place of (4).

Expressing the ordinates of $M$ and $N$ in terms of $x_{1}$ and $x_{2}$, (4) becomes:

$$x_{1}\left(a x_{2}^{3}+b x_{2}^{2}+c x_{2}+d\right)=x_{2}\left(a x_{1}^{3}+b x_{1}^{2}+c x_{1}+d\right)$$

which, after rearranging and factoring out, gives:

$$\left(x_{2}-x_{1}\right)\left\{a x_{1} x_{2}\left(x_{2}+x_{1}\right)+b x_{1} x_{2}-d\right\}=0$$

and since $(3)$ implies $x_{2}-x_{1} \neq 0$, we have:

$$a x_{1} x_{2}\left(x_{2}+x_{1}\right)+b x_{1} x_{2}-d=0$$

Here, based on (2),

$$x_{1} x_{2}=\frac{c}{3 a}, \quad x_{1}+x_{2}=-\frac{2 b}{3 a}$$

so (4) expressed in terms of the coefficients is:

$$\frac{b c}{9 a}-d=0, \quad 9 a d=b c$$

This relationship also holds if $c=0$ in (2), i.e., $x_{1}=0, x_{2} \neq 0$ and $y_{1}=0, M$ is identical to $O$, since in this case (1) gives $d=0$.

In summary: under the conditions that $a \neq 0$ and (3) are satisfied, the desired relationship is given by (6).

We note that (6) also allows the value set $d=b=0$ if $a$ and $c$ have opposite signs (so neither is 0). In this case, the origin is precisely the inflection point, the symmetry center of the curve representing the function.

Füredi Zoltán (Budapest, Móricz Zs. Gymn. IV. o. t.)

Remarks. 1. Our result can also be stated as follows. If the coefficients $a, b, c$ of the function in question are chosen such that $b^{2}-3 a c>0$ and $a \neq 0$, then a $d$ can be chosen to satisfy the requirement $d=b c / 9 a$. (In the case $c=0$, the requirement can only be met if one of the extrema is at $x=0$.)

2. The following interesting, unique solution among the submitted solutions - contrary to the editorial custom, we publish it without any changes. Therefore, we note in advance the following. We are only talking about cubic curves of the form $y=a x^{3}+b x^{2}+c x+d$. The author did not write the positivity of the discriminant in the coefficients, but mentioned the relevant part of the assumption. Of course, it is also implicitly stated that $u \neq 0$.

II. solution. In problem 1595, we showed that any cubic curve has a symmetry center located on it. Therefore, any cubic curve can be translated so that it passes through the origin and is symmetric about it. And conversely: any cubic curve can be obtained by translating a cubic curve that is symmetric about the origin. We will proceed in this way now.

The existence of the two extrema also means that the curve symmetric about the origin has two roots: $\pm u$. The third root is 0, so the equation describing any cubic curve with two extrema in its canonical form is

$$v x\left(x^{2}-u^{2}\right)=y$$

The line connecting the two extrema now passes through the origin, so all curves satisfying the condition of the problem can be obtained by translating (7) along the line. Let's write the vector from the origin to one of the extremum points:

$$\begin{aligned} & y^{\prime}=3 v x^{2}-u^{2} v=0 \\ & x_{1,2}= \pm \frac{u}{\sqrt{3}}, \quad y_{1}=v\left(\frac{u^{3}}{3 \sqrt{3}}-\frac{u^{3} \cdot 3}{3 \sqrt{3}}\right)=-\frac{2}{\sqrt{3} \cdot 3} u^{3} v \end{aligned}$$

The product of the vector $\mathbf{a}$ with the scalar $-\frac{\sqrt{3 t}}{u}$ is therefore

$$\mathbf{a}\left(-t ; \quad+\frac{2}{3} u^{2} v t\right)$$

$t$ is a real number, a parameter. We can obtain the equation of all curves satisfying the problem by translating (7) by $\mathbf{a}$:

$$v\left[(x+t)^{3}-u^{2}(x+t)\right]=y-\frac{2}{3} u^{2} v t$$

Expanding and rearranging,

$$v x^{3}+3 v t x^{2}+\left(3 v t^{2}-v u^{2}\right) x+\left(v t^{3}-\frac{u^{2} t}{3}\right)=y$$

so

$$a=v ; \quad b=3 v t ; \quad c=3 v t^{2}-v u^{2} ; \quad d=v t^{3}-\frac{u^{2} t}{3}:$$

From the four equations, we can "solve" for the parameters, and we get the desired relationship. From the first equation, we can express $v$, then from the second, $t$, and from the third, $u$; finally, the desired relationship is

$$9 a d-b c=0$$

The reversibility of the transformations means that if the extrema indeed exist, then the relationship is not only necessary but also sufficient.

Kollár István (Budapest, Móricz Zs. Gymn. IV. o. t.)

III. Solution. Since the function $f(x)$ under consideration has both of its extrema, the function

$$f^{\prime}(x)=3 a x^{2}+2 b x+c$$

has two different real roots, so

$$a \neq 0 \quad \text { and } \quad b^{2}>3 a c$$

If for some real number $u$, $f^{\prime}(u)=0$, then

$$u^{2}=-\frac{1}{3 a}(2 b u+c)$$

Using this twice, we get that

$$$$
\begin{gathered}
f(u)=-\frac{u}{3}(2 b u+c)+b u^{2}+c u+d= \\
=-\frac{b}{3} \cdot \frac{2