
Problem 2. Find the side length of the smallest equilateral triangle in which three disks with radii 2,3 and 4 without common inner points can be placed.


Solution. Let in a equilateral $\triangle A B C$ two disks with radii 3 and 4 without common inner points be placed. It is clear that a line $\ell$ exists, which separates them, i.e. the disks are in different semiplanes with respect to $\ell$ (Figure 1).

Figure 1.

![](https://cdn.mathpix.com/cropped/2024_06_03_1ba58ed6da37450ce759g-1.jpg?height=437&width=489&top_left_y=1849&top_left_x=612)

Figure 2.

![](https://cdn.mathpix.com/cropped/2024_06_03_1ba58ed6da37450ce759g-1.jpg?height=454&width=491&top_left_y=1846&top_left_x=1391)

This line divides the triangle into a triangle and a quadrilateral or into two triangles. In both cases the disks can be replaced in the figure in a way that each of them is tangent to two of the sides of $\triangle A B C$. It is clear that the new disks have no common inner point. Let the disks be inscribed in $\angle A$ and $\angle B$ of $\triangle A B C$, respectively. We translate the side $B C$ parallelly to itself
towards the point $A$, till the disk which is inscribed in $\angle B$ touches the disk which is inscribed in $\angle A$ (Figure 2). Thus, we get an equilateral $\triangle A_{1} B_{1} C_{1}$ with a smaller sides, in which two disks with radii 3 and 4 and without common inner points are placed.

Figure 3 .

![](https://cdn.mathpix.com/cropped/2024_06_03_1ba58ed6da37450ce759g-2.jpg?height=448&width=491&top_left_y=547&top_left_x=451)

Let $A_{1} B_{1}=x, I$ be the incenter of $\triangle A_{1} B_{1} C_{1}$, while $O_{1}$ and $\mathrm{O}_{2}$ be the centers of the two disks. Then $A_{1} I=B_{1} I=$ $\frac{x}{\sqrt{3}}, A_{1} O_{1}=6, B_{1} O_{2}=8$. Since the disk with radius 4 is inside the $\triangle A_{1} B_{1} C_{1}$, then $O_{2} \in I B_{1}$. Thus, $B_{1} O_{2} \leq B_{1} I$, i.e. $x \geq 8 \sqrt{3}$. On the other hand $O_{1} I=\frac{x}{\sqrt{3}}-6, O_{2} I=\frac{x}{\sqrt{3}}-$ $8, O_{1} O_{2}=7$ and by the cosine theorem for $\triangle O_{1} O_{2} I$ we find that $\left(\frac{x}{\sqrt{3}}-6\right)^{2}+\left(\frac{x}{\sqrt{3}}-8\right)^{2}+\left(\frac{x}{\sqrt{3}}-6\right)\left(\frac{x}{\sqrt{3}}-8\right)=49$. But $x \geq 8 \sqrt{3}$, and from here $x=11 \sqrt{3}$. Consequently, $A B \geq$ $11 \sqrt{3}$. On the other hand in the equilateral $\triangle A B C$ with side length $11 \sqrt{3}$ three disks with radii 2,3 and 4 (without common inner points) can be placed inscribing circles with these radii in the angles of the triangle (Figure 3).

Note that the disks wit radii 3 and 4 are tangent to each other. It follows from the above considerations that the solution of the problem is $11 \sqrt{3}$.
