3. (3 points) Equilateral triangles $A B C$ and $A_{1} B_{1} C_{1}$ with side length 10 are inscribed in the same circle such that point $A_{1}$ lies on the arc $B C$, and point $B_{1}$ lies on the arc $A C$. Find $A A_{1}^{2}+B C_{1}^{2}+C B_{1}^{2}$.
Answer: 200

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, $B B_{1}=\frac{10 \sin \left(60^{\circ}-\alpha\right)}{\sin 60^{\circ}}$ and $C C_{1}=\frac{12 \sin \alpha}{\sin 60^{\circ}}$.

$$
\begin{aligned}
& \left.A A_{1}^{2}+B B_{1}^{2}+C C_{1}^{2}=\frac{10^{2}}{\sin ^{2} 60^{\circ}}\left(\sin ^{2}\left(60^{\circ}+\alpha\right)+\sin ^{2}\left(60^{\circ}-\alpha\right)+\sin ^{2} \alpha\right)\right)= \\
& =\frac{200}{3} \cdot\left(\left(\frac{\sqrt{3}}{2} \cos \alpha+\frac{1}{2} \sin \alpha\right)^{2}+\left(\frac{\sqrt{3}}{2} \cos \alpha-\frac{1}{2} \sin \alpha\right)^{2}+\sin ^{2} \alpha\right)= \\
& =\frac{200}{3}\left(3 \cos ^{2} \alpha+3 \sin ^{2} \alpha\right)=200
\end{aligned}
$$