For a semicircle with diameter $AB = 2r$ and center $O$, the tangent at point $M$ intersects the tangents at points $A$ and $B$ at points $A'$ and $B'$, respectively. The perpendicular projection of $M$ onto the diameter $AB$ is $P$. Let $x = \frac{AP}{PB} (>0)$.

Rotate the trapezoid $AA'BB'$ and the triangle $A'OB'$ around $AB$, and denote the volumes of the resulting solids of revolution by $V_1$ and $V_2$, respectively. Express these volumes as functions of $r$ and $x$. Determine $x$ such that $\frac{V_2}{V_1} = m$, where $m$ is a given positive number. Discussion.

What is the maximum value that $m$ can take, and what position of $M$ corresponds to this value?
First, let's express the segments $A A^{\prime}$ and $B B^{\prime}$ required for the volume calculations using the radius and $x$.

The quadrilateral $O B B^{\prime} M$ is a kite because two pairs of adjacent sides are equal, and therefore $O B^{\prime} \perp M B$. However, by Thales' theorem, $A M$ is also perpendicular to $M B$, so $O B^{\prime} \| A M$, which means the angles marked with a single arc are equal.

![](https://cdn.mathpix.com/cropped/2024_05_02_82b51ed3c4d4cf7528e2g-1.jpg?height=442&width=545&top_left_y=354&top_left_x=779)

The triangles $A M P_{\triangle} \sim O B^{\prime} B_{\triangle}$ because two pairs of their angles are equal.

Due to the similarity,

$$
A P: r = P M: B B^{\prime}
$$

Furthermore, by the known proportionality theorem, $P M = \sqrt{A P \cdot P B}$. Combining these two results:

$$
A P \cdot B B^{\prime} = r \sqrt{A P \cdot P B}
$$

From this,

$$
B B^{\prime} = r \frac{\sqrt{A P \cdot P B}}{A P} = r \sqrt{\frac{P B}{A P}} = \frac{r}{\sqrt{x}}
$$

The quadrilateral $A O M A^{\prime}$ is also a kite, and therefore $A M \perp O A^{\prime}$. Thus, $A A^{\prime} O \varangle = M A P \varangle$, because they are angles with perpendicular sides. The triangles $A O A_{\triangle}^{\prime} \sim O B B_{\triangle}^{\prime}$ because two pairs of their angles are equal. Due to the similarity,

$$
A A^{\prime}: r = r: B B^{\prime}
$$

From this,

$$
A A^{\prime} = \frac{r^{2}}{B B^{\prime}} = r \sqrt{x}
$$

Using these, the volume of the frustum of a cone formed by the rotation of the trapezoid $A B B^{\prime} A^{\prime}$ around the axis $A B$ can be written as:

$$
V_{1} = \frac{2 r \pi}{3} \left(r^{2} x + r^{2} + \frac{r^{2}}{x}\right) = \frac{2 r^{3} \pi}{3 x} \left(x^{2} + x + 1\right)
$$

The volume of the solid of revolution formed by the rotation of the triangle $A^{\prime} O B^{\prime}$ around the axis $A B$ is calculated by subtracting the volumes $V_{3}$ and $V_{4}$ of the cones formed by the rotation of the triangles $A O A^{\prime}$ and $O B B^{\prime}$ from the volume $V_{1}$ of the frustum of a cone:

$$
\begin{array}{r}
V_{2} = V_{1} - V_{3} - V_{4} = \frac{2 r^{3} \pi}{3 x} \left(x^{2} + x + 1\right) - \frac{r^{2} x \pi r}{3} - \frac{r^{2} \pi r}{3 x} = \\
= \frac{r^{3} \pi}{3 x} \left(2 x^{2} + 2 x + 2 - x^{2} - 1\right) = \frac{r^{2} \pi}{3 x} (x + 1)^{2}
\end{array}
$$

Let

$$
\frac{V_{2}}{V_{1}} = \frac{(x + 1)^{2}}{2 \left(x^{2} + x + 1\right)} = m
$$

From this,

$$
(2 m - 1) x^{2} + 2 (m - 1) x + 2 m - 1 = 0
$$

$2 m - 1 \neq 0$, because otherwise $x$ would also be 0, which is not possible according to the problem. Therefore, we can use the quadratic formula. Simplifying by 2,

$$
x_{1,2} = \frac{-(m - 1) \pm \sqrt{(m - 1)^{2} - (2 m - 1)^{2}}}{2 m - 1} = \frac{1 - m \pm \sqrt{m (2 - 3 m)}}{2 m - 1}
$$

We get a real value for $x$ if the discriminant: $m (2 - 3 m) \geq 0$, which means

$$
0 \leq m \leq \frac{2}{3}
$$

According to the problem, $x$ must be positive. A fraction is positive if the numerator and denominator have the same sign. The numerator $1 - m$ is always positive if $0 \leq m \leq \frac{2}{3}$; the number under the square root $(m - 1)^{2}$ - which is the same as $(1 - m)^{2}$ - is always smaller. Therefore, the numerator is always positive, and the denominator must also be positive, which only follows for $m > \frac{1}{2}$. Thus, we have the following constraint for $m$:

$$
\frac{1}{2} < m \leq \frac{2}{3}
$$

From this, it is immediately clear that the maximum value of $m$ is $\frac{2}{3}$.

If $m$ is maximal, then the discriminant in the formula for $x$ is 0, and $x = \frac{A P}{P B} = 1$. The projection of $M$ is then at $O$, so $A^{\prime} B^{\prime} \| A B$. In this case, the trapezoid becomes a rectangle, and the solid of revolution formed by the rotation is a cylinder.