Key Concepts and Formulas

• Geometric Series: The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$, where $a$ is the first term and $r$ is the common ratio, provided $|r| < 1$.
• Properties of Equilateral Triangles: An equilateral triangle with side length $s$ has a perimeter of $3s$ and an area of $\frac{\sqrt{3}}{4}s^2$.
• Midpoint Theorem: Connecting the midpoints of two sides of a triangle creates a segment parallel to the third side and half its length.

Step-by-Step Solution

1. Define the Initial Triangle and its Properties: Let the initial equilateral triangle be $ABC$ with side length $a$.

   • Perimeter of the first triangle ($P_1$): $P_1 = 3a$.
   • Area of the first triangle ($A_1$): $A_1 = \frac{\sqrt{3}}{4}a^2$.
   • Reasoning: These are the foundational values for our calculations, representing the first triangle in the infinite sequence.

2. Determine the Properties of the Second Triangle: The second triangle is formed by joining the midpoints of the sides of $ABC$. Let this triangle be $DEF$. By the midpoint theorem, each side of $DEF$ is half the length of the corresponding side of $ABC$.

   • Side length of the second triangle: $\frac{a}{2}$.
   • Perimeter of the second triangle ($P_2$): $P_2 = 3\left(\frac{a}{2}\right) = \frac{3a}{2}$.
   • Area of the second triangle ($A_2$): $A_2 = \frac{\sqrt{3}}{4}\left(\frac{a}{2}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{a^2}{4} = \frac{\sqrt{3}}{16}a^2$.
   • Reasoning: This step establishes the pattern for subsequent triangles. Each new triangle's side length is half of the previous one.

3. Identify the Geometric Series for Perimeters and Areas: The process of forming new triangles by connecting midpoints is repeated infinitely. This generates two infinite geometric series: one for the perimeters and one for the areas.

   • Perimeter Series: The sequence of perimeters is $3a, \frac{3a}{2}, \frac{3a}{4}, \dots$.
     • First term ($a_P$): $3a$.
     • Common ratio ($r_P$): $\frac{P_2}{P_1} = \frac{3a/2}{3a} = \frac{1}{2}$. Since $|r_P| < 1$, the series converges.
   • Area Series: The sequence of areas is $\frac{\sqrt{3}}{4}a^2, \frac{\sqrt{3}}{16}a^2, \frac{\sqrt{3}}{64}a^2, \dots$.
     • First term ($a_Q$): $\frac{\sqrt{3}}{4}a^2$.
     • Common ratio ($r_A$): $\frac{A_2}{A_1} = \frac{\sqrt{3}a^2/16}{\sqrt{3}a^2/4} = \frac{1}{4}$. Since $|r_A| < 1$, the series converges.
   • Reasoning: Recognizing these as geometric series is crucial for applying the sum formula. The common ratios are directly derived from the scaling factor of the side lengths.

4. Calculate the Sum of Perimeters (P): Using the formula for the sum of an infinite geometric series: $P = \frac{a_P}{1 - r_P} = \frac{3a}{1 - \frac{1}{2}} = \frac{3a}{\frac{1}{2}} = 6a$

   • Reasoning: We apply the infinite geometric series sum formula to find the total perimeter of all triangles formed.

5. Calculate the Sum of Areas (Q): Using the formula for the sum of an infinite geometric series: $Q = \frac{a_Q}{1 - r_A} = \frac{\frac{\sqrt{3}}{4}a^2}{1 - \frac{1}{4}} = \frac{\frac{\sqrt{3}}{4}a^2}{\frac{3}{4}} = \frac{\sqrt{3}}{3}a^2$

   • Reasoning: Similarly, we apply the sum formula to find the total area of all triangles formed.

6. Establish a Relationship Between P and Q: We have $P = 6a$ and $Q = \frac{\sqrt{3}}{3}a^2$. We need to eliminate $a$ to find the required relationship. From $P = 6a$, we can express $a$ as $a = \frac{P}{6}$. Substitute this expression for $a$ into the equation for $Q$: $Q = \frac{\sqrt{3}}{3}\left(\frac{P}{6}\right)^2 = \frac{\sqrt{3}}{3} \cdot \frac{P^2}{36} = \frac{\sqrt{3}}{108}P^2$ Now, rearrange the equation to solve for $P^2$: $108Q = \sqrt{3}P^2$ $P^2 = \frac{108}{\sqrt{3}}Q = \frac{108\sqrt{3}}{3}Q = 36\sqrt{3}Q$

   • Reasoning: By substituting the expression for 'a' derived from the perimeter sum into the area sum equation, we eliminate 'a' and obtain a direct relationship between P and Q.

Common Mistakes and Tips

• Incorrect Common Ratio: Be meticulous when calculating the common ratios for both perimeters and areas. The area's common ratio is the square of the side length's scaling factor.
• Algebraic Errors: When manipulating the equations to relate $P$ and $Q$, pay close attention to squaring terms and simplifying fractions.
• Formula Application: Ensure you are using the correct formula for the sum of an infinite geometric series and that the condition $|r| < 1$ is met.

Summary

This problem involves constructing a sequence of equilateral triangles where each subsequent triangle's side length is half of the previous one. The perimeters and areas of these triangles form infinite geometric series. By applying the formula for the sum of an infinite geometric series to find the total perimeter ($P$) and total area ($Q$), we can then algebraically eliminate the initial side length ($a$) to derive the relationship $P^2 = 36\sqrt{3}Q$.

The final answer is $\boxed{P^2=36 \sqrt{3} \mathrm{Q}}$ which corresponds to option (B).