4. On a semicircle with diameter $\overline{A B}=2 r$, a point $D$ is chosen, different from $A$ and $B$. The point $M$ is the foot of the perpendicular dropped from point $D$ to the diameter $A B$, and the point $C$ is the intersection of the tangents to the semicircle drawn at points $A$ and $D$. Determine $\overline{A M}=x$ such that the ratio of the volumes of the solids formed by the rotation of trapezoid AMDC and the part of the semicircle AMD is equal to $k$. For which values of $k$ does the problem have a solution.
Solution. From the similarity of $\triangle E D C$ and $\triangle M D O$, we obtain $\overline{C D}: \overline{E D}=r: r_{1}$, i.e., $\overline{C D}: x=r: r_{1}$. Therefore, $\overline{C D}=\frac{r x}{r_{1}}=\overline{A C}=r_{2}$. By rotating the trapezoid $A M D C$, we obtain a frustum of a cone, and its volume is

$V_{\text {AMDC }}=V_{1}=\frac{1}{3} x \pi\left(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}\right)=\frac{1}{3} x^{2} \pi \frac{7 r^{2}-5 r x+x^{2}}{2 r-x}$ When rotating the part $A M D$, we obtain a spherical cap with height $\overline{A M}=x$ and volume

![](https://cdn.mathpix.com/cropped/2024_06_05_c0cca9b7a033dd01322ag-07.jpg?height=443&width=533&top_left_y=1581&top_left_x=1029)

$$
V_{A M D}=V_{2}=\frac{1}{3} x^{2}(3 r-x) \pi
$$

From the ratio $\frac{V_{1}}{V_{2}}=k$, we obtain the quadratic equation

$$
(1-k) x^{2}+(-5 r+5 r k) x+7 r^{2}-6 k r^{2}=0
$$

which has solutions $x_{1 \mid 2}=\frac{r}{2}\left(5 \pm \sqrt{\frac{k+3}{k-1}}\right)$. The solution $x_{1}=\frac{r}{2}\left(5+\sqrt{\frac{k+3}{k-1}}\right)>\frac{r}{2}(5+1)=3 r$ is not valid. The second solution $x_{2}=\frac{r}{2}\left(5-\sqrt{\frac{k+3}{k-1}}\right)$, and the valid values for $k$ are $k>\frac{7}{6}$, which is derived from the inequalities $1<\sqrt{\frac{k+3}{k-1}}<5$.

## Fourth Year