Problem 1835

1. Setup and Notation:
Let $B' \equiv D$, $C' \equiv E$, and $B'' \equiv K$ for convenience. Let ray $DE$ intersect the circumcircle at $L$. We aim to prove that $AK = AL$.

2. Angle Relationships:
Note that:
$$\angle AKL = \pi - \angle ADK - \angle KAC = \angle B - \angle KBC$$
and
$$\angle ALK = \angle ALC - \angle KLC = \angle B - \angle KBC$$
Hence, $\angle AKL = \angle ALK = \alpha_A \implies AK = AL$.

3. Law of Sines Application:
From the Law of Sines, we get:
$$\frac{AK}{\sin \alpha_A} = \frac{AB}{\sin \angle C}$$
and
$$\frac{AK}{\sin (\pi - \angle B)} = \frac{AD}{\sin \alpha_A}$$
Dividing these two relationships gives:
$$\frac{\sin \angle B}{\sin \alpha_A} = \frac{AB \cdot \sin \alpha_A}{AD \cdot \sin \angle C} \implies \sin^2 \alpha_A = \frac{\sin \angle B \cdot \sin \angle C \cdot AD}{AB}$$
Since $AD = AB \cos \angle A$, we have:
$$\sin^2 \alpha_A = \sin \angle B \cdot \sin \angle C \cdot \cos \angle A$$

4. Similar Relationships for Other Angles:
Similarly, we obtain:
$$\sin^2 \alpha_B = \sin \angle C \cdot \sin \angle A \cdot \cos \angle B$$
and
$$\sin^2 \alpha_C = \sin \angle A \cdot \sin \angle B \cdot \cos \angle C$$

5. Multiplying the Relationships:
Multiplying these, we get:
$$\sin \alpha_A \cdot \sin \alpha_B \cdot \sin \alpha_C = \sqrt{\sin \angle A \cdot \sin \angle B \cdot \sin \angle C \cdot \cos \angle A \cdot \cos \angle B \cdot \cos \angle C}$$

6. Applying AM-GM Inequality:
From the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
$$\sin \angle A \cdot \sin \angle B \cdot \sin \angle C \leq \frac{(\sin \angle A + \sin \angle B + \sin \angle C)^3}{27}$$
The second derivative of the function $\sin x$ is $-\sin x < 0$, so $\sin x$ is concave in $(0, \frac{\pi}{2})$. From Jensen's inequality, we have:
$$\sin \angle A + \sin \angle B + \sin \angle C \leq 3 \cdot \sin \left( \frac{\angle A + \angle B + \angle C}{3} \right) = 3 \sin \left( \frac{\pi}{3} \right) = \frac{3\sqrt{3}}{2}$$
Hence:
$$\sin \angle A \cdot \sin \angle B \cdot \sin \angle C \leq \frac{\left( \frac{3\sqrt{3}}{2} \right)^3}{27} = \frac{3\sqrt{3}}{8}$$

7. Applying Jensen's Inequality to Cosine:
Similarly, the second derivative of $\cos x$ is $-\cos x < 0$ in $(0, \frac{\pi}{2})$, so:
$$\cos \angle A \cdot \cos \angle B \cdot \cos \angle C \leq \frac{(\cos \angle A + \cos \angle B + \cos \angle C)^3}{27} \leq \frac{(3 \cos \left( \frac{\angle A + \angle B + \angle C}{3} \right))^3}{27} = \cos^3 \left( \frac{\pi}{3} \right) = \frac{1}{8}$$
Therefore:
$$\sqrt{\cos \angle A \cdot \cos \angle B \cdot \cos \angle C} \leq \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$$

8. Final Conclusion:
Combining these results, we have:
$$\sin \alpha_A \cdot \sin \alpha_B \cdot \sin \alpha_C \leq \frac{3\sqrt{3}}{8} \cdot \frac{\sqrt{2}}{4} = \frac{3\sqrt{6}}{32}$$
Equality holds when $\angle A = \angle B = \angle C = \frac{\pi}{3}$.

The final answer is $\boxed{\frac{3\sqrt{6}}{32}}$