## Problem III - 4

Determine all triples of real numbers $(a, b, c)$, with $a \neq b, a \neq 0, b \neq 0$, such that the parabolas

$$
y=a x^{2}+b x+c, \quad y=b x^{2}+c x+a
$$

have the same vertex.
Solution.

It is first observed that the parabolas pass through the common point $N(1, a+b+c)$.

First solution

Let $V\left(x_{0}, y_{0}\right)$ be the vertex of the two parabolas. The change of variable

$$
X=x-x_{0}, \quad Y=y-y_{0}
$$

transforms $V$ into the point $O'$ which is the origin of coordinates, that is, the equations of the parabolas in the new system are

$$
Y=a X^{2}, \quad Y=b X^{2}
$$

which pass through the common point

$$
N^{\prime}\left(1-x_{0}, a+b+c-y_{0}\right)
$$

Since $a \neq b$, these two parabolas only have the unique common point $O'$, so $N'=O'$ and $N$ is the vertex of the two parabolas.

Then

$$
-\frac{b}{2 a}=-\frac{c}{2 b}=1 \Rightarrow b=-2 a, \quad c=4 a
$$

Second solution.

By setting the first derivative to zero to find the abscissa of the vertex and equating, we directly obtain

$$
-\frac{b}{2 a}=-\frac{c}{2 a}
$$

equating both values to 1, as the abscissa of the common point, the above results. In conclusion, the sought values are

$$
(a, b, c) \equiv(a,-2 a-4 a) \text { with } a \neq 0
$$