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[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

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[ACTION NAME]

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# Your action step 2

# Your action step 3

...

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system
Triangle \( AMN \) is right-angled with \(\angle BAC = 60^{\circ}\).

\( O_{1}, O_{2} \) are the centers of the circles, and \( P, R \) are the points of tangency of the circles with side \( BC \). In the right triangle \( O_{1}O_{2}Q \) with a right angle at \( Q \), point \( Q \in O_{2}R \), and we have \( O_{1}O_{2}^{2} = O_{1}Q^{2} + QO_{2}^{2} \), with \( 4^{2} = O_{1}Q^{2} + 2^{2} \).

\[ PR = O_{1}Q = 2\sqrt{3}, \angle O_{2}O_{1}Q = 30^{\circ}, \angle O_{1}O_{2}Q = 60^{\circ} \]

The following holds:

\(\angle RO_{2}C = \frac{1}{2} \angle RO_{2}E = \frac{1}{2}(360^{\circ} - 90^{\circ} - 2 \times 60^{\circ}) = 75^{\circ}\),

\(\angle O_{2}CR = 15^{\circ},\)

\(\angle BCA = 30^{\circ}, \angle ABC = 90^{\circ}, RC = 3 \cot 15^{\circ} = 3 \sqrt{\frac{1+\cos 30^{\circ}}{1-\cos 30^{\circ}}} = 3 \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}} = 3(2+\sqrt{3}) = 6 + 3\sqrt{3}.\)

\[ BC = BP + PR + RC = 1 + 2\sqrt{3} + 6 + 3\sqrt{3} = 7 + 5\sqrt{3}, AC = \frac{BC}{\sin 60^{\circ}} = \frac{30 + 14\sqrt{3}}{3}, AB = \frac{BC}{\sin 60^{\circ}} = \frac{15 + 7\sqrt{3}}{3}. \]

Present the answer in LaTex format: \boxed{Your answer}
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