3. (3 points) Equilateral triangles $A B C$ and $A_{1} B_{1} C_{1}$ with side length 12 are inscribed in circle $S$ such that point $A$ lies on arc $B_{1} C_{1}$, and point $B$ lies on arc $A_{1} B_{1}$. Find $A A_{1}^{2}+B B_{1}^{2}+C C_{1}^{2}$.
Answer: 288

Solution:

Notice that the arcs $A B_{1}, B A_{1}$, and $C C_{1}$ are equal. Let their degree measure be $2 \alpha$. Then the lengths of the arcs $A C_{1}=B B_{1}=C A_{1}=120^{\circ}-2 \alpha$.

By the Law of Sines in triangle $A C A_{1}$, we get $\frac{A A_{1}}{\sin A C A_{1}}=$ $\frac{A C}{A A_{1} C}$, from which $A A_{1}=\frac{12 \sin \left(60^{\circ}+\alpha\right)}{\sin 60^{\circ}}$.

Similarly, $B B_{1}=\frac{12 \sin \left(60^{\circ}-\alpha\right)}{\sin 60^{\circ}}$ and $C C_{1}=\frac{12 \sin \alpha}{\sin 60^{\circ}}$.

$$
\left.A A_{1}^{2}+B B_{1}^{2}+C C_{1}^{2}=\frac{12^{2}}{\sin ^{2} 60^{\circ}}\left(\sin ^{2}\left(60^{\circ}+\alpha\right)+\sin ^{2}\left(60^{\circ}-\alpha\right)+\sin ^{2} \alpha\right)\right)=
$$

$$
=96 \cdot\left(\left(\frac{\sqrt{3}}{2} \cos \alpha+\frac{1}{2} \sin \alpha\right)^{2}+\left(\frac{\sqrt{3}}{2} \cos \alpha-\frac{1}{2} \sin \alpha\right)^{2}+\sin ^{2} \alpha\right)=
$$

$=96\left(3 \cos ^{2} \alpha+3 \sin ^{2} \alpha\right)=288$.