Let ABC be an equilateral triangle. A new triangle is formed by joining the middle points of all sid

Q: Wait, why is the common ratio for the area $1/4$ and not $1/2$? Shouldn't it be the same as the perimeter?

That is a fantastic question! Think about the geometry. When you halve the side length of a triangle, the perimeter (which is $3 \times \text{side}$) also halves, so the ratio is $1/2$. However, the area of an equilateral triangle is $\frac{\sqrt{3}}{4} \times \text{side}^2$. Because the side is squared in the area formula, when you replace the side with $\frac{a}{2}$, the new area becomes $\frac{\sqrt{3}}{4} \times (\frac{a}{2})^2 = \frac{1}{4} \times (\text{original area})$. Thus, the area scales by the square of the linear scale factor!

Q: Is it mathematically rigorous to just sum an infinite series like this? Does it actually converge?

Yes, it is perfectly rigorous! An infinite geometric series converges if and only if the absolute value of the common ratio $|r| < 1$. In our case, for both the perimeter and the area, the common ratios are $1/2$ and $1/4$ respectively. Since both are less than $1$, the series converges to a finite, well-defined value. You are essentially calculating the 'total' footprint of an infinite fractal process.

Q: I got $P = 6a$ and $Q = \frac{\sqrt{3}a^2}{3}$. How do I know which option is correct without guessing?

You have done the hard work! Now, you just need to bridge the gap. You have $P = 6a$, which means $P^2 = 36a^2$. You also have $Q = \frac{\sqrt{3}a^2}{3}$, which implies $a^2 = \frac{3Q}{\sqrt{3}} = \sqrt{3}Q$. Substitute this expression for $a^2$ into your $P^2$ equation. It is not guessing; it is algebraic substitution to eliminate the variable $a$ that was not in the final options.

Question

Let $ABC$ be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle $ABC$ and the same process is repeated infinitely many times. If $P$ is the sum of perimeters and $Q$ is the sum of areas of all the triangles formed in this process, then :

Options:

(A) $P^2 = 6\sqrt{3}Q$

(B) $P^2 = 36\sqrt{3}Q$

(C) $P = 36\sqrt{3}Q^2$

(D) $P^2 = 72\sqrt{3}Q$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing before a perfect, pristine equilateral triangle, $	riangle ABC$, with a side length of $a$. We perform a transformation by finding the midpoints of its three sides and connecting them, partitioning the original triangle and creating a new, smaller equilateral triangle in the center.

By repeating this process infinitely, we enter the realm of fractals and infinite series. We shall analyze the behavior of the perimeter and the area through this iterative process.

### Phase 1: The Perimeter Series

The perimeter of the first triangle is $P_1 = 3a$. Upon joining the midpoints, the side length of the subsequent triangle becomes $\frac{a}{2}$, resulting in a perimeter of $3(\frac{a}{2})$.

The third triangle has a side length of $\frac{a}{4}$, yielding a perimeter of $3(\frac{a}{4})$. Summing these values, we obtain an infinite geometric progression:

$$P = 3a + 3\left(\frac{a}{2}ight) + 3\left(\frac{a}{4}ight) + \dots$$

Here, the first term is $a_1 = 3a$ and the common ratio is $r = \frac{1}{2}$. Using the sum formula for an infinite geometric series, $S_{\infty} = \frac{a_1}{1-r}$, we calculate:

$$P = \frac{3a}{1 - \frac{1}{2}} = \frac{3a}{\frac{1}{2}} = 6a$$

Thus, the total perimeter of this infinite process is exactly **$6a$**.

### Phase 2: The Area Series

The area of the first triangle is $A_1 = \frac{\sqrt{3}}{4}a^2$. When we move to the second triangle, the side length is $\frac{a}{2}$, so the area becomes:

$$A_2 = \frac{\sqrt{3}}{4}\left(\frac{a}{2}ight)^2 = \frac{1}{4} 	imes (	ext{Area of } 	riangle ABC)$$

This demonstrates that while the perimeter scales linearly, the area scales quadratically. Our series for the total area $Q$ is:

$$Q = \frac{\sqrt{3}}{4}a^2 + \frac{\sqrt{3}}{4}\left(\frac{a}{2}ight)^2 + \frac{\sqrt{3}}{4}\left(\frac{a}{4}ight)^2 + \dots$$

This is an infinite geometric progression with a common ratio $r = \frac{1}{4}$. Applying the sum formula:

$$Q = \frac{\frac{\sqrt{3}a^2}{4}}{1 - \frac{1}{4}} = \frac{\frac{\sqrt{3}a^2}{4}}{\frac{3}{4}}$$

The denominators cancel out, leaving us with the total area:

$$Q = \frac{\sqrt{3}a^2}{3}$$

### Phase 3: The Synthesis

We now have two primary equations: $P = 6a$ and $Q = \frac{\sqrt{3}a^2}{3}$. To find the relationship between $P$ and $Q$, we must eliminate the variable $a$.

From $P = 6a$, we square both sides to obtain:

$$P^2 = 36a^2$$

From our area equation, we isolate $a^2$:

$$a^2 = \frac{3Q}{\sqrt{3}} = \sqrt{3}Q$$

Substituting $a^2$ into the equation for $P^2$, we arrive at the final relationship:

$$P^2 = 36(\sqrt{3}Q)$$

**$P^2 = 36\sqrt{3}Q$**

Select Answer:

Visualized Solution

Visualize the Equilateral Triangle $A B C$

The Midpoint Transformation

Defining the Sum of Perimeters $P$

Calculating $P$ using Infinite G.P. Formula

Defining the Sum of Areas $Q$

Calculating $Q$ using Infinite G.P. Formula

Relating $P$ and $Q$

Final Result and Conclusion

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

Visualize the Equilateral Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABC

The Midpoint Transformation

Defining the Sum of Perimeters .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P using Infinite G.P. Formula

Defining the Sum of Areas .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q using Infinite G.P. Formula

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q