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

• A. $P^2 = 36\sqrt{3}Q$
• B. $P^2 = 36\sqrt{3}Q$
• C. $P^2 = 36\sqrt{3}Q$
• D. $P^2 = 36\sqrt{3}Q$

Answer 1

The Correct Option is A

\[\text{Area of first } \triangle = \frac{\sqrt{3}}{4} a^2\]
\[\text{Area of second } \triangle = \frac{\sqrt{3}}{4} \cdot \frac{a^2}{4} = \frac{\sqrt{3}a^2}{16}\]
\[\text{Area of third } \triangle = \frac{\sqrt{3}}{4} \cdot \frac{a^2}{16} = \frac{\sqrt{3}a^2}{64}\]
\[\text{Sum of areas} = \frac{\sqrt{3}a^2}{4} \left( 1 + \frac{1}{4} + \frac{1}{16} + \cdots \right)\]
The sum of this infinite geometric series is:
\[Q = \frac{\sqrt{3}}{4} \cdot a^2 \cdot \frac{1}{1 - \frac{1}{4}} = \frac{\sqrt{3}}{4} \cdot a^2 \cdot \frac{4}{3} = \frac{\sqrt{3}}{3} a^2\]
Perimeter Calculations:
\[\text{Perimeter of first } \triangle = 3a\]
\[\text{Perimeter of second } \triangle = 3 \cdot \frac{a}{2} = \frac{3a}{2}\]
\[\text{Perimeter of third } \triangle = 3 \cdot \frac{a}{4} = \frac{3a}{4}\]
\[P = 3a \left( 1 + \frac{1}{2} + \frac{1}{4} + \cdots \right)\]
The sum of this infinite geometric series is:
\[P = 3a \cdot \frac{1}{1 - \frac{1}{2}} = 3a \cdot 2 = 6a\]
Final Calculations:
\[a = \frac{P}{6}\]
\[Q = \frac{1}{\sqrt{3}} \cdot \frac{P^2}{36}\]
\[P^2 = 36 \sqrt{3} Q\]
Answer: \((1)\; P = 36\sqrt{3}Q\)

Answer 2

To solve this problem, let's first understand the sequence of triangles created in this process. We start with an equilateral triangle \( \triangle ABC \) with side length \( a \). By joining the midpoints of the sides, a new equilateral triangle is formed inside \( \triangle ABC \). This process is repeated infinitely.

1. Understanding the Geometry:
   • The new triangle formed by joining the midpoints of an equilateral triangle with side length \( a \) will have a side length of \( \frac{a}{2} \). This is a consequence of the midpoint theorem.
2. Perimeter Calculation:
   • The perimeter of the first triangle is \( 3a \).
   • The perimeter of the second triangle is \( 3 \times \frac{a}{2} = \frac{3a}{2} \).
   • The perimeter of the third triangle is \( 3 \times \frac{a}{4} = \frac{3a}{4} \), and so on.
   • The total perimeter \( P \) is the sum of these perimeters: \(P = 3a + \frac{3a}{2} + \frac{3a}{4} + \cdots\)
   • This is a geometric series with the first term \( 3a \) and common ratio \( \frac{1}{2} \).
   • The sum \( P \) can be calculated using the formula for the sum of an infinite geometric series: \(S = \frac{a_1}{1 - r}\), where \( a_1 = 3a \) and \( r = \frac{1}{2} \).
   • Thus, \( P = \frac{3a}{1 - \frac{1}{2}} = 6a \).
3. Area Calculation:
   • The area of an equilateral triangle with side \( a \) is \(\frac{\sqrt{3}}{4}a^2\).
   • The area of the second triangle becomes \(\frac{\sqrt{3}}{4} \left(\frac{a}{2}\right)^2 = \frac{\sqrt{3}}{16}a^2\), and so forth.
   • The total area \( Q \) is the sum of these areas: \(Q = \frac{\sqrt{3}}{4}a^2 + \frac{\sqrt{3}}{16}a^2 + \cdots\)
   • This again is a geometric series with the first term \( \frac{\sqrt{3}}{4}a^2 \) and common ratio \( \frac{1}{4} \).
   • Thus, the sum is \(Q = \frac{\frac{\sqrt{3}}{4}a^2}{1 - \frac{1}{4}} = \frac{\sqrt{3}}{3}a^2\).
4. Relating P and Q:
   • We have \( P = 6a \) and \( Q = \frac{\sqrt{3}}{3}a^2 \).
   • To find the relationship between \( P^2 \) and \( Q \), we calculate \( P^2 = (6a)^2 = 36a^2 \).
   • Therefore, \( \frac{P^2}{Q} = \frac{36a^2}{\frac{\sqrt{3}}{3}a^2} = 36 \times \frac{3}{\sqrt{3}} = 36\sqrt{3} \).
   • The correct relationship is \( P^2 = 36\sqrt{3}Q \).

This confirms that the correct answer is: 
\(P^2 = 36\sqrt{3}Q\).