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 be the sum of areas of all the triangles formed in this process, then :

Let the initial side length of the equilateral triangle be $$a$$.

• Perimeter of 1st triangle = $$3a$$

• Perimeter of 2nd triangle (sides are halved) = $$\frac{3a}{2}$$

• Perimeter of 3rd triangle = $$\frac{3a}{4}$$

This creates an infinite geometric series with first term $$A_1 = 3a$$ and common ratio $$r = \frac{1}{2}$$:

$$P = \frac{A_1}{1 - r} = \frac{3a}{1 - \frac{1}{2}} = 6a \implies a = \frac{P}{6}$$

• Area of 1st triangle = $$\frac{\sqrt{3}}{4}a^2$$

• Area of 2nd triangle (sides halved means area is quartered) = $$\frac{1}{4}\left(\frac{\sqrt{3}}{4}a^2\right)$$

This creates an infinite geometric series with first term $$A_1 = \frac{\sqrt{3}}{4}a^2$$ and common ratio $$r = \frac{1}{4}$$:

$$Q = \frac{A_1}{1 - r} = \frac{\frac{\sqrt{3}}{4}a^2}{1 - \frac{1}{4}} = \frac{\frac{\sqrt{3}}{4}a^2}{\frac{3}{4}} = \frac{\sqrt{3}}{3}a^2 = \frac{a^2}{\sqrt{3}}$$

$$Q = \frac{\left(\frac{P}{6}\right)^2}{\sqrt{3}} = \frac{P^2}{36\sqrt{3}}$$

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