The equation $x^{2}+y^{2}=1$ represents a circle. We draw a tangent at some point on this circle. The shape bounded by the tangent, the $x$-axis, and the circle is rotated around the $x$-axis. Where is the point of tangency if the volume of the solid of revolution is equal to the volume of a unit sphere?
Given a circle $k$ and a rotational axis $x$, the resulting shape is symmetric with respect to the $y$-axis, so it is sufficient to search for the point of tangency $P(\xi, \eta)$ on the right side of the circle. In fact, it is sufficient to find the common abscissa of the point $P$ on the upper quarter arc and the point $P^{\prime}$ on the lower quarter arc, since the rotation that generates the desired solid will map $P^{\prime}$ to $P$. Thus, $0<\xi<1$ and $0<|\eta|<1$, and the equality sign cannot replace the "<" sign anywhere, because the points $A(1 ; 0)$ and $(0 ; 1)$ clearly do not satisfy the condition for $P$. The three additional points on the circle $k$ that correspond to the $P$ found in the first quadrant can be obtained by reflecting $P$ over the coordinate axes and the origin. It is obvious that there is only one suitable point in the first quadrant, because as $\xi$ decreases from 1, both $P$ and the intersection point $M$ of the tangent with the $x$-axis move away from $A$, the volume in question increases monotonically and exceeds the volume of the unit sphere.

![](https://cdn.mathpix.com/cropped/2024_05_02_488089849b7d86d9cdbbg-1.jpg?height=420&width=1176&top_left_y=516&top_left_x=472)

The solid of revolution $F$ is bounded by the segments $PM$, $MA$, and the arc $AP$. Denoting the projection of $P$ onto the $x$-axis by $P_{1}$, the solid $F$ can also be derived by removing the smaller spherical cap $S$ of the sphere $G$ described by $k$ from the cone $K$ described by the right triangle $MPP_{1}$.

The abscissa of $M$ from the right triangle $OPM$ is

$$
OM = \frac{OP^2}{OP_1} = \frac{1}{\xi}
$$

so the height of the cone is $OM - OP_1 = \left(1 - \xi^2\right) / \xi$, and the area of the base circle is $\pi \eta^2 = \pi \left(1 - \xi^2\right)$. To express the volume of the spherical cap, we use the known formula where the height $m = 1 - \xi$ and the radius of the sphere appear. - According to the requirement, the difference between the volumes of $K$ and $S$ is equal to the volume of $G$:

$$
\frac{\pi \left(1 - \xi^2\right)}{3} \cdot \frac{1 - \xi^2}{\xi} - \frac{\pi}{3} (1 - \xi)^2 \{3 - (1 - \xi)\} = \frac{4 \pi}{3}
$$

which simplifies to

$$
\xi^2 - 6 \xi + 1 = 0
$$

We see from the consecutive coefficients $+1, -6, +1$ of the equation that the roots are real, positive, and reciprocals of each other. The root that satisfies $0 < \xi < 1$ is $\xi = 3 - \sqrt{8}$. (The other root, however, is $\xi_2 = 3 + \sqrt{8}$, which gives the position of $M$). Therefore, among the points of the given circle, those satisfy the condition whose abscissa has an absolute value of $3 - \sqrt{8} = 0.1716$.

Hegyi Ferenc (Székesfehérvár, Teleki Blanka Gimn., IV. o. t.) Heimler László (Sopron, Széchenyi I. Gimn., III. o. t.)

Note. It does not lead to a problem requiring significant work to interpret the problem statement such that the solid of revolution is bounded by the arc $PB$ and the segments $BM$ and $MP$ in the diagram, since the resulting solid would contain the unit sphere. Equality would only occur if $P \equiv A$, but recognizing and stating this is not considered solving the problem.[^0]

[^0]:    ${ }^{1}$ See for example: Hack F.-Kugler S.-né: Function Tables. Mathematical and Physical Relationships. Tankönyvkiadó. Budapest. 1968. 344. 2 formulas: $v=(\pi / 3) m^{2}(3 r-m)$.