Let $P_{1}, P_{2}$ be two points on the curve of the function $y=x^{2}+1$, and let their projections on the $x$-axis be $Q_{1}$ and $Q_{2}$, respectively. Determine these points such that the tangent to the curve at $P_{1}$ passes through $Q_{2}$, and the tangent at $P_{2}$ passes through $Q_{1}$.
Let the coordinates of point $P_{1}$ be $\left(x_{1}, y_{1}\right)$. Since $P_{1}$ lies on the parabola, $y_{1}=x_{1}^{2}+1$. The slope $m$ of the tangent line at $P_{1}$ is equal to the derivative of the function

$$
f(x)=x^{2}+1
$$

at $x_{1}$:

$$
m=f^{\prime}\left(x_{1}\right)=2 x_{1}
$$

Thus, the equation of the tangent line is

$$
y=2 x_{1}\left(x-x_{1}\right)+y_{1}
$$

This line intersects the $x$-axis at the point with abscissa

$$
x=\frac{2 x_{1}^{2}-y_{1}}{2 x_{1}}=\frac{x_{1}^{2}-1}{2 x_{1}}
$$

According to the problem, this intersection point coincides with the projection of the parabola's point $P_{2}\left(x_{2}, y_{2}\right)$ onto the $x$-axis, which is $Q_{2}\left(x_{2}, 0\right)$. Therefore,

$$
\begin{gathered}
x_{2}=\frac{x_{1}^{2}-1}{2 x_{1}}, \text { i.e., } \\
2 x_{1} x_{2}=x_{1}^{2}-1
\end{gathered}
$$

By swapping the roles of $P_{1}$ and $P_{2}$, we obtain the condition that the tangent at $P_{2}$ passes through the point $Q_{1}\left(x_{1}, 0\right)$:

$$
2 x_{2} x_{1}=x_{2}^{2}-1
$$

The left sides are equal, so the right sides must also be equal. This gives $x_{1}^{2}=x_{2}^{2}$. Clearly, $x_{1}$ cannot equal $x_{2}$, meaning $P_{1}$ and $P_{2}$ are the same point, so $x_{2}=-x_{1}$. Substituting this into (1) yields

$$
-2 x_{1}^{2}=x_{1}^{2}-1
$$

which simplifies to $x_{1}= \pm 1 / \sqrt{3}$. Therefore, the points on the curve that satisfy the condition are

$$
P_{1}( \pm 1 / \sqrt{3}, 4 / 3) \text { and } P_{2}(\mp 1 / \sqrt{3}, 4 / 3)
$$

The two possible sign choices simply represent the swapping of the roles of the points, so the problem essentially has only one solution.

Hetzer Jenö (Sopron, Széchenyi I. Gim., III. o. t.)