1.21. A regular triangle is inscribed in a circle with a diameter equal to $\sqrt{12}$. On its height as a side, another regular triangle is constructed, in which a new circle is inscribed. Find the radius of this circle.
1.21. According to the condition, $R=0.5 \sqrt{12}$. The side $A B$ of the inscribed regular triangle (Fig. 1.19) is equal to $R \sqrt{3}$, i.e., $0.5 \sqrt{12} \cdot \sqrt{3}=3$. Let's find the side $C D$ of the new triangle: $a=\sqrt{3^{2}-1.5^{2}}=1.5 \sqrt{3}$. Since the radius $r$ of the inscribed circle is $\frac{a \sqrt{3}}{6}$, we finally get

$$
r=1.5 \sqrt{3} \cdot \frac{\sqrt{3}}{6}=\frac{3}{4}
$$