$A, B, C$ are three different points on a parabola. What is the condition for the normals (that is, the perpendiculars to the tangents at these points) drawn at these points to intersect at one point?
Any two parabolas are similar to each other, so it is enough to consider the parabola $p$ with the equation $y=x^{2}$. Let the abscissas of $A, B$, and $C$ be $a, b, c$ respectively, then their ordinates are $a^{2}, b^{2}$, and $c^{2}$. The condition $(a-b)(b-c)(c-a) \neq 0$ is sufficient for $A, B$, and $C$ to be distinct points, because the parabola $p$ has exactly one point on any line parallel to its axis.

At any point $\left(x_{1}, x_{1}^{2}\right)$ on $p$, the slope of the tangent is the value of the derivative $\left(x^{2}\right)^{\prime}=2 x$ at that point, which is $2 x_{1}$. The slope of the normal is $-1 /\left(2 x_{1}\right)$ (provided $x_{1} \neq 0$), so the equation of the normal, using the form of a line with a given slope passing through a given point, is

\[
\begin{gathered}
y-x_{1}^{2}=-\frac{1}{2 x_{1}}\left(x-x_{1}\right), \text { rearranging } \\
x+2 x_{1} y-2 x_{1}^{3}-x_{1}=0
\end{gathered}
\]

If $x_{1}=0$, i.e., the point is the vertex of $p$, then the normal is the $y$-axis, whose equation is $x=0$. The rearranged form (1) already includes this. For any two distinct points on $p$, the normals intersect at a single point because their slopes are different.

![](https://cdn.mathpix.com/cropped/2024_05_02_6016c713b04cc9d69ce7g-1.jpg?height=680&width=730&top_left_y=752&top_left_x=684)

We seek the condition for the three normals, $n_{A}, n_{B}, n_{C}$, to intersect at a single point. Substituting $a$ and $b$ for $x_{1}$ in (1), we get

\[
\begin{gathered}
\text { the equation of } n_{A} \quad x+2 a y-2 a^{3}-a=0 \\
\text { the equation of } n_{B} \quad x+2 b y-2 b^{3}-b=0
\end{gathered}
\]

From these, the coordinates of the intersection point $M$ of the first two normals, considering $b \neq a$, are

\[
y_{M}=a^{2}+a b+b^{2}+\frac{1}{2}, \quad x_{M}=-2 a b(a+b)
\]

The coordinates of the intersection point $M^{\prime}$ of $n_{A}$ and $n_{C}$ are obtained by substituting $c$ for $b$:

\[
y_{M^{\prime}}=a^{2}+a c+c^{2}+\frac{1}{2}, \quad x_{M^{\prime}}=-2 a c(a+c)
\]

The three normals intersect at a single point if and only if $M^{\prime}$ coincides with $M$, since in this case $M$ is also on $n_{B}$ and $n_{C}$, and thus is the intersection point of $n_{B}$ and $n_{C}$. $M^{\prime}$ coincides with $M$ if and only if $y_{M^{\prime}}=y_{M}$, since no normal is parallel to the $x$-axis, and on any normal, the ordinate uniquely determines the abscissa. Therefore, $y_{M}=y_{M^{\prime}}$ if and only if $y_{M}-y_{M^{\prime}}=a(b-c)+b^{2}-c^{2}=(b-c)(a+b+c)=0$, and this, since $(b-c) \neq 0$, holds if and only if

\[
a+b+c=0
\]

This, as is well known, means that the centroid $S$ of triangle $ABC$ lies on the line $x=0$, the $y$-axis, and, abstracting from the coordinate system, that $S$ lies on the axis of symmetry of $p$.

Remark. After writing the equations of the normals, we can also proceed as follows: We seek the condition for the lines

\[
\begin{aligned}
& x+2 a y-2 a^{3}-a=0 \\
& x+2 b y-2 b^{3}-b=0 \\
& x+2 c y-2 c^{3}-c=0
\end{aligned}
\]

to have a common point, i.e., there exists a pair $\left(x_{0}, y_{0}\right)$ such that

\[
x_{0}+2 z y_{0}-2 z^{3}-z=0
\]

is a cubic equation whose roots are the distinct numbers $a, b, c$. According to the relationships between the roots and coefficients, the sum of the roots of the cubic equation

\[
A z^{3}+B z^{2}+C z+D=0
\]

(where $A \neq 0$) is $-\frac{B}{A}$, so $a, b, c$ can only be roots of (3) if

\[
a+b+c=0
\]

since $z^{2}$ does not appear in (3). This condition is therefore necessary for the three normals to intersect at a single point.

We show that the obtained condition is also sufficient. Indeed, according to the relationships between the roots and coefficients, $a, b, c$ are roots of the equation

\[
2 z^{3}+B z^{2}+C z+D=0
\]

if and only if

\[
B=2(a+b+c), \quad C=-2(a b+b c+c a), \quad D=2 a b c
\]

Thus, if $a+b+c=0$, then

\[
x_{0}=D=2 a b c, \quad y_{0}=\frac{C+1}{2}=\frac{1}{2}-(a b+b c+c a)
\]

and $a, b, c$ are roots of (3), meaning the three normals pass through the point $\left(x_{0}, y_{0}\right)$.[^0]

[^0]:    ${ }^{1}$ See, for example, Hack Frigyes - Kugler Sándorné: Function Tables. Mathematical and Physical Relationships. Tankönyvkiadó, Budapest, 1968, p. 66, formula group 25.33.