Let $P_{0}$ be an equilateral triangle of area 10sq.units. Each side of $P_{0}$ is trisected into three segments of equal length, and the corners of $P_{0}$ are snipped off, creating a new polygon (in fact, a hexagon) $P_{1}$ . Now repeat the process to $P_{1}$ – i.e. trisect each side and snip off the corners – to obtain a new polygon $P_{2}$ . Now repeat this process infinitely to create an object $P_{\infty}$ . Then the area of $P_{\infty} = \frac{m}{n}$ sq.units, where $m, n \in N$ and m, n are coprime to each other. Then which of the following options is/are incorrect?

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$m - 3 n \geq 25$

$m + n \leq 32$

$m - n \leq 30$

$3 m - n \geq 131$

answer is A, B, C, D.

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Detailed Solution

$P_{0} a r e a = 10 s i d e l e n g t h = x$

Number of sides

$P_{0} - 3; P_{1} - 6; P_{2} - 12........ P_{n} \to 2^{n} \times 3$

$(P_{1}) a r e a = P_{0} - 3 \times e a c h c o r n e r a r e a o f P_{0} (P_{2}) a r e a = P_{1} - 6 c o r n e r Δ^{l e} a r e a o f P_{1} {(P_{n})}_{A} = {(P_{n - 1})}_{A} - (n o . o f c o r n e r o f P_{n - 1}) \times A r e a o f e a c h c o r n e r Δ^{l e} P_{1} = P_{0} - 3 \times \frac{1}{2} \times {(\frac{x}{2})}^{2} \times \sin 60^{\circ} P_{2} = P_{2} - 6 \times \frac{1}{2} \times {(\frac{x}{2^{2}})}^{2} \times \sin 120^{\circ} P_{3} = P_{3} - 12 \times \frac{1}{2} \times {(\frac{x}{2^{3}})}^{2} \times \sin 60^{\circ} (P_{n}) = P_{n - 1} - 2^{n - 1} \times 3 \times \frac{1}{2} \times {(\frac{x}{3^{n}})}^{2} \times \frac{\sqrt{3}}{2} P_{n} = P_{n - 1} - k {(\frac{2}{9})}^{n}$