سوال ۱۱۸ کنکور ارشد علوم کامپیوتر ۱۴۰۰

This is the Koch snowflake construction. Start from an equilateral triangle and, at each step, replace the middle third of every line segment with the two other sides of a small equilateral triangle built on it.

How one step transforms a segment

Take any straight segment of length $L$. The step divides it into three equal pieces (each of length $L/3$) and replaces the middle piece by the two slanted sides of a new equilateral triangle, which also have length $L/3$ each. So one segment of length $L$ becomes:

$L \;\longrightarrow\; 4 \cdot \frac{L}{3}.$

Every segment turns into 4 segments, each $\tfrac{1}{3}$ as long. Hence the total perimeter is multiplied by the same factor at every step:

$P_{n} = \frac{4}{3}\,P_{n-1}.$

Closed form

The initial equilateral triangle (step $0$) has perimeter

$P_0 = 3 \times 27 = 81.$

Solving the geometric recurrence,

$P_n = 81 \left(\frac{4}{3}\right)^{n}.$

The wording counts the original triangle as the starting figure and the first replacement as step 1, so the figure "at the sixth step" is $P_6$:

$P_6 = 81 \left(\frac{4}{3}\right)^{6} = 81 \cdot \frac{4^{6}}{3^{6}} = 81 \cdot \frac{4096}{729}.$

Since $\dfrac{81}{729} = \dfrac{1}{9}$,

$P_6 = \frac{4096}{9}.$

This matches option 4.

Pitfall: Each step multiplies the perimeter by $\tfrac{4}{3}$ (not $\tfrac{4}{9}$): you get $4$ times as many segments but each is only $\tfrac{1}{3}$ as long. Also be careful with the step count — the bare triangle is step $0$ and the perimeter $81$, so the sixth step uses exponent $6$, giving $81\cdot(4/3)^6 = 4096/9$, not $4096/3$ (exponent $5$ from $81\cdot(4/3)^5=1024/3$).