30. Given an equilateral triangle $A B C$ with side $a, B D$ is its height. On $B D$, a second equilateral triangle $B D C_{1}$ is constructed, and on the height $B D_{1}$ of this triangle, a third equilateral triangle $B D_{1} C_{2}$ is constructed. Find the radius of the circumcircle of triangle $C C_{1} C_{2}$. Prove that its center lies on the side of triangle $A B C$ (where $C_{2}$ is outside triangle $A B C$).
30. We will prove that if $C_{1}$ and $C_{2}$ (Fig. 6) are on the other side of $B C$ from vertex $A$, then the center of the circle circumscribed around $\triangle C C_{1} C_{2}$ is at point $O$ on side $A B$, with $|B O|=\frac{1}{4}|A B|$. By drawing the altitude $C M$ from vertex $C$, we obtain that $B C_{1} C M$ is a rectangle. Therefore, the perpendicular bisector of $C C_{1}$ passes through $O$. Considering that $C_{1} C_{2} B D$ and $\left|C_{1} C_{2}\right|=\frac{1}{2}|B D|$, we get that the perpendicular bisector of $C_{1} C_{2}$ at its midpoint also passes through $O$. Now we can easily find the desired radius:

$$
\begin{aligned}
& r=\sqrt{C M 2^{2}+\left.M O\right|^{2}}= \\
& =\sqrt{\frac{3 a^{2}}{4}+\frac{a^{2}}{16}}=\frac{a}{4} \sqrt{13}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_6282cf4a54d2a9633c48g-071.jpg?height=294&width=386&top_left_y=226&top_left_x=165)

Fig. 6.