At what angle does the parabola $y^{2}=2 p x$ intersect the circle $\left(x-\frac{p}{2}\right)^{2}+y^{2}=4 p^{2}$?
When we talk about the angle formed by two curves at some point of theirs, we mean the angle formed by the tangents drawn at the question point.

First solution: It follows from the given equation that the focus $(F)$ of the parabola coincides with the center of the circle and the radius of the circle is equal to $2p$. Let the axis of the parabola intersect the directrix at $M$, one of the intersection points of the two curves be $P$, and the projection of $P$ onto the axis of the parabola be $R$. Then, according to the above,

$$
M R=F P=2 p
$$

and thus

$$
F R=M R-p=p=\frac{F P}{2}
$$

therefore, $F P$ forms a $60^{\circ}$ angle with the axis. However, the tangent of the parabola forms half the angle with the corresponding radius vector as with the axis, so the angle between the tangent to the parabola at $P$ and the line $F P$ is $30^{\circ}$.

Finally, the tangent to the circle at $P$ forms a $90^{\circ}$ angle with $F P$, so the sought angle

$$
\varphi=90^{\circ}-30^{\circ}=60^{\circ}
$$

(Marcell Riesz, Györ.)

Second solution: If the coordinates of the intersection point are $\xi$ and $\eta$, then

$$
\begin{gathered}
\eta^{2}=2 p \xi \\
\left(\xi-\frac{p}{2}\right)^{2}+\eta^{2}=4 p^{2}
\end{gathered}
$$

Substituting the value of $\eta^{2}$ from (1) into (2), we get

$$
\left(\xi+\frac{p}{2}\right)^{2}=4 p^{2}
$$

and thus

$$
\xi_{1}=\frac{3 p}{2} \quad \text { and } \quad \xi_{2}=\frac{-5 p}{2}
$$

so

$$
\eta_{1}= \pm p \sqrt{3} \quad \text { and } \quad \eta_{2}= \pm p \sqrt{-5}
$$

As can be seen, we only have two real intersection points $\left(\frac{3 p}{2}, p \sqrt{3}\right)$ and $\left(\frac{3 p}{2},-p \sqrt{3}\right)$, which are symmetric with respect to the $x$-axis, and since the curves are also symmetric with respect to the $x$-axis, it is sufficient to examine the angle formed by the tangents at one of the intersection points. The equation of the tangent to the parabola at $(\xi, \eta)$ is

$$
y \eta=p(x+\xi)
$$

and the equation of the tangent to the circle at $(\xi, \eta)$ is

$$
\left(x-\frac{p}{2}\right)\left(\xi-\frac{p}{2}\right)+y \eta=r^{2}
$$

In our case, $r=2 p$ and, for example, $\xi=\frac{3 p}{2}, \eta=p \sqrt{3}$, so from (3) we get

$$
y=\frac{x}{\sqrt{3}}+\frac{3 p}{2 \sqrt{3}}
$$

and from (4) we get

$$
y=-\frac{x}{\sqrt{3}}=\frac{9 p}{2 \sqrt{3}}
$$

The slope of one tangent from (5) is $k_{1}=\frac{1}{\sqrt{3}}$.

The slope of the other from (6) is $k_{2}=-\frac{1}{\sqrt{3}}$, so the tangent of the angle formed by the tangents is:

$$
\operatorname{tg} \varphi=\frac{k_{1}-k_{2}}{1+k_{1} k_{2}}=\sqrt{3}
$$

so

$$
\varphi=60^{\circ}
$$

(Sándor Pichler, Budapest.)

The problem was also solved by: Bartók I., Braun I., Enyedi B., Haar A., Kertész G., Liebner A., Neidenbach E., Riesz K.