
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
Let \(O_1\) be the center of the inscribed circle in triangle \(ABC\) with radius \(R\), and \(O_2\) be the center of the inscribed circle in triangle \(ADE\) with radius \(r\). Let \(p_1\) be the semiperimeter of triangle \(ABC\), and \(p_2\) be the semiperimeter of triangle \(ADE\). Then \(AK = p_2\), and \(AK = p_1 - BC\).

Using properties of tangents to circles:

\[ 
\begin{gathered}
2 AK = AK + AH = AD + DK + EH + AE = AD + (DM + ME) + AE = 2p_2 \\
2 AK = (AB - BK) + (AC - CH) = AB + AC + (BF + CF) - 2(BF + CF) = 2(p_1 - BC)
\end{gathered}
\]

Then \(r = \frac{S_{ADE}}{p_2} = \frac{S_{ADE}}{AK} = \frac{1}{6}\).

Since a circle can be circumscribed around quadrilateral \(BDEC\), \(\angle ABC + \angle DEC = 180^\circ\), and \(\angle AED = \angle ABC\). Therefore, triangles \(ADE\) and \(ACB\) are similar, and \(\frac{r}{R} = \frac{p_2}{p_1}\), \(R = \frac{r(AK + BC)}{AK} = \frac{3 + 15}{6 \cdot 3} = 1\).

Let \(\angle BAC = \alpha\).

Then, \(\operatorname{tg}(\alpha / 2) = \frac{R}{AK} = \frac{1}{3}\), and \(\operatorname{tg} \alpha = \frac{2 \operatorname{tg}(\alpha / 2)}{(1 - \operatorname{tg}^2(\alpha / 2))} = \frac{3}{4}\).

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