, 2013

Solution:

Answer: $\sqrt{3}$

First, note that for any $i$, after performing the operation on triangle $A_{i} A_{i+1} A_{i+2}$, the resulting pasture is triangle $A_{i+1} A_{i+2} A_{i+3}$. Let $K_{i}$ be the area of triangle $A_{i} A_{i+1} A_{i+2}$. From $A_{n+1} A_{n-2} = \frac{1}{2^{n} - 2} A_{n} A_{n-2}$ and $A_{n} A_{n+1} = A_{n} A_{n-2} + A_{n-2} A_{n+1}$, we have $A_{n} A_{n+1} = \left(1 + \frac{1}{2^{n} - 2}\right) A_{n} A_{n-2}$.

We also know that the area of a triangle is half the product of its base and height, so if we let the base of triangle $A_{n-2} A_{n-1} A_{n}$ be $A_{n} A_{n-2}$, its area is $K_{n-2} = \frac{1}{2} h A_{n} A_{n-2}$. The area of triangle $A_{n-1} A_{n} A_{n+1}$ is $K_{n-1} = \frac{1}{2} h A_{n} A_{n+1}$. The $h$'s are equal because the distance from $A_{n-1}$ to the base does not change.

We now have $\frac{K_{n-1}}{K_{n-2}} = \frac{A_{n} A_{n+1}}{A_{n} A_{n-2}} = 1 + \frac{1}{2^{n} - 2} = \frac{2^{n} - 1}{2^{n} - 2}$. Therefore, $\frac{K_{1}}{K_{0}} = \frac{3}{2}$, $\frac{K_{2}}{K_{0}} = \frac{K_{2}}{K_{1}} \frac{K_{1}}{K_{0}} = \frac{7}{6} \cdot \frac{3}{2} = \frac{7}{4}$, $\frac{K_{3}}{K_{0}} = \frac{K_{3}}{K_{2}} \frac{K_{2}}{K_{0}} = \frac{15}{14} \cdot \frac{7}{4} = \frac{15}{8}$.

We see the pattern $\frac{K_{n}}{K_{0}} = \frac{2^{n+1} - 1}{2^{n}}$, which can be easily proven by induction. As $n$ approaches infinity, $\frac{K_{n}}{K_{0}}$ grows arbitrarily close to $2$, so the smallest $K$ such that the pasture never exceeds an area of $K$ is $2 K_{0} = \sqrt{3}$.