
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
In a right triangle \( \triangle A M N \), given \( \angle B A C=60^{\circ} \). Let \( O_{1} \) and \( O_{2} \) be the centers of circles, and \( P \) and \( R \) be the points of tangency of these circles with side \( BC \). In the right triangle \( O_{1}O_{2}Q \) with right angle at \( Q \) and point \( Q \in O_{2}R \), we have \( O_{1}O_{2}^{2} = O_{1}Q^{2} + QO_{2}^{2} \), \( (16)^{2} = O_{1}Q^{2} + (8)^{2} \).

$$
P R = O_{1}Q = 8 \sqrt{3}, \quad \angle O_{2} O_{1} Q = 30^{\circ}
$$

Given \( \angle O_{1} O_{2} Q = 60^{\circ} \).

$$
\begin{aligned}
& \angle R O_{2} C = \frac{1}{2} \angle R O_{2} E = \frac{1}{2}\left(360^{\circ}-90^{\circ}-2 \cdot 60^{\circ}\right) = 75^{\circ}, \\
& \angle O_{2} C R = 15^{\circ}, \quad \angle A C B = 30^{\circ}, \quad \angle A B C = 90^{\circ}, \\
& R C = 12 \operatorname{ctg} 15^{\circ} = 12 \sqrt{\frac{1+\cos 30^{\circ}}{1-\cos 30^{\circ}}} = 12 \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}} = 12(2+\sqrt{3}) = 24+12 \sqrt{3}, \\
& B C = B P + P R + R C = 4 + 8 \sqrt{3} + 24 + 12 \sqrt{3} = 28 + 20 \sqrt{3}, \\
& A C = B C / \sin 60^{\circ} = (120 + 56 \sqrt{3}) / 3, \\
& A N = A C - C E - N E = (120 + 56 \sqrt{3} - 72 - 36 \sqrt{3} - 36) / 3 = (12 + 20 \sqrt{3}) / 3, \\
& M N = A N / \operatorname{tg} 30^{\circ} = (12 \sqrt{3} + 60) / 3 = 4 \sqrt{3} + 20.
\end{aligned}
$$

Calculate the area of triangle \( \triangle A M N \):

\( S =  A N \cdot M N / 2 = (12 + 20 \sqrt{3})(4 \sqrt{3} + 20) / 6 = (224 \sqrt{3} + 240) / 3 \).

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