Problem 1349

1. Identify the given information and the goal:
- Two concentric circles with radii 2 and 4 centered at point $P$.
- Points $A, B,$ and $C$ form an equilateral triangle on the larger circle.
- Point $D$ is the intersection of the smaller circle and the line segment $PB$.
- We need to find the square of the area of triangle $\triangle ADC$.

2. **Determine the side length of the equilateral triangle $\triangle ABC$:**
- Since $A, B,$ and $C$ lie on the larger circle with radius 4, the side length $a$ of the equilateral triangle can be found using the formula for the side length of an equilateral triangle inscribed in a circle:
$$a = 2R \sin 60^\circ = 2 \cdot 4 \cdot \frac{\sqrt{3}}{2} = 4\sqrt{3}$$

3. **Calculate the area of $\triangle ABC$:**
- The area $A$ of an equilateral triangle with side length $a$ is given by:
$$A = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (4\sqrt{3})^2 = \frac{\sqrt{3}}{4} \cdot 48 = 12\sqrt{3}$$

4. **Determine the position of point $D$:**
- Point $D$ is the intersection of the smaller circle (radius 2) and the line segment $PB$. Since $PB$ is a radius of the larger circle, $PB = 4$.
- $D$ is halfway between $P$ and $B$, so $PD = 2$.

5. **Calculate the height $ED$ of $\triangle ADC$:**
- Drop a perpendicular from $P$ to $AC$ and let the intersection be $E$. Since $\triangle ABC$ is equilateral, $E$ is the midpoint of $AC$.
- The height $PE$ of $\triangle ABC$ can be calculated as:
$$PE = \sqrt{4^2 - \left(\frac{4\sqrt{3}}{2}\right)^2} = \sqrt{16 - 12} = 2$$
- Since $D$ is halfway between $P$ and $B$, $ED = \frac{2}{3} \cdot EB$.

6. **Calculate the area of $\triangle ADC$:**
- The area of $\triangle ADC$ is $\frac{2}{3}$ of the area of $\triangle ABC$:
$$\text{Area of } \triangle ADC = \frac{2}{3} \cdot 12\sqrt{3} = 8\sqrt{3}$$

7. **Find the square of the area of $\triangle ADC$:**
- The square of the area of $\triangle ADC$ is:
$$(8\sqrt{3})^2 = 64 \cdot 3 = 192$$

The final answer is $\boxed{192}$