\section*{Problem 2 - 051212}

Four congruent spheres touch a plane on the same side. Furthermore, each sphere touches two of the others, and each of the spheres touches the same right circular cone, whose base circle lies in the given plane.

The radius of the base circle of the cone is to be expressed in terms of the radius of the spheres and the height of the cone (case distinction).
}
![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1801.jpg?height=498&width=1196&top_left_y=1519&top_left_x=430)

With the notations in the image and \(x=\overline{X_{1} M}, h=\overline{M S}, y=\overline{A Y}, z_{1}=\overline{M_{1} A}, z_{2}=\overline{A B}\) and the radius \(r\) of the spheres, the following relationships must hold:

(1) \(\overline{M_{1} M_{3}}=2 \cdot \sqrt{2} r\) (the centers of the spheres form a square as shown in Figure a) when they all touch the same cone)

(2) \(\frac{x}{h}=\frac{y}{r} \Rightarrow r x=h y\) (similar triangles \(\triangle X_{1} M S\) and \(\triangle A Y M_{1}\) )

(3) \(y=\sqrt{z_{1}^{2}-r^{2}}\) (Pythagorean theorem)

(4) \(z_{1}+z_{2}=\frac{\overline{M_{1} M_{3}}}{2}=\sqrt{2} r\) for Figure b) or \(z_{2}-z_{1}=\frac{\overline{M_{1} M_{3}}}{2}=\sqrt{2} r\) for Figure c) \(\Rightarrow z_{1}= \pm\left(\sqrt{2} r-z_{2}\right)\), where + applies in case b) and - in case c). Then in both cases: \(z_{1}^{2}=2 r^{2}+z_{2}^{2}-2 \sqrt{2} \cdot r \cdot z_{2}\)

(5) \(\frac{z_{2}}{x}=\frac{h-r}{h} \Rightarrow z_{2}=\frac{h-r}{h} \cdot x\) (Thales' theorem)

Now, the equations (1) to (4) are substituted into each other:

\[
\begin{aligned}
r x & =h \cdot \sqrt{z_{1}^{2}-r^{2}} \\
r x & =h \cdot \sqrt{2 r^{2}+z_{2}^{2}-2 \sqrt{2} \cdot r \cdot z_{2}-r^{2}} \\
r x & =\sqrt{h^{2} r^{2}+h^{2} z_{2}^{2}-2 \sqrt{2} \cdot r \cdot h^{2} \cdot z_{2}} \\
r x & =\sqrt{h^{2} r^{2}+(h-r)^{2} \cdot x^{2}-2 \sqrt{2} \cdot r \cdot h \cdot(h-r) \cdot x} \\
r^{2} x^{2} & =h^{2} r^{2}+(h-r)^{2} \cdot x^{2}-2 \sqrt{2} \cdot r \cdot h \cdot(h-r) \cdot x \\
0 & =h^{2} r^{2}+\left(h^{2}-2 h r\right) \cdot x^{2}-2 \sqrt{2} \cdot r \cdot h \cdot(h-r) \cdot x \\
0 & =x^{2}-2 \sqrt{2} \cdot \frac{r \cdot(h-r)}{h-2 r} \cdot x+\frac{h r^{2}}{h-2 r} \\
x_{1,2} & =\frac{\sqrt{2} r \cdot(h-r)}{h-2 r} \pm \sqrt{\frac{2 r^{2} \cdot(h-r)^{2}}{(h-2 r)^{2}}-\frac{h r^{2}}{h-2 r}} \\
x_{1,2} & =\frac{\sqrt{2} r \cdot(h-r)}{h-2 r} \pm \frac{r}{h-2 r} \cdot \sqrt{2 \cdot(h-r)^{2}-h \cdot(h-2 r)} \\
x_{1,2} & =\frac{r}{h-2 r} \cdot\left(\sqrt{2}(h-r) \pm \sqrt{h^{2}+2 r^{2}-2 h r}\right)
\end{aligned}
\]

The case with the plus sign before the square root represents the solution for the case of the spheres touching from the outside, and the minus sign for touching from the inside.