Question

Let n > 4 be a natural number and let P be a polygon with 'n' sides. Let $a_1, a_2, a_3$,...., $a_n$ be the lengths of the sides of P and let 'p' be its perimeter. Then find the $\left [ \sum_{i=1}^{n} \frac{a_i}{p-a_i} \right ]$, where [.] denotes greatest integer function.

Answer 1

1

Answer 2

1. Polygon Property: For any polygon, the sum of the lengths of any (n-1) sides must be strictly greater than the length of the remaining side. This implies $p - a_i > a_i$ for each side $a_i$, where $p$ is the perimeter.
2. Upper Bound: From $p - a_i > a_i$, we get $p > 2a_i$, or $a_i < p/2$. Using this, we can write $\frac{a_i}{p-a_i} < \frac{a_i}{p/2} = \frac{2a_i}{p}$. Summing over all sides: $\sum_{i=1}^{n} \frac{a_i}{p-a_i} < \sum_{i=1}^{n} \frac{2a_i}{p} = \frac{2}{p} \sum_{i=1}^{n} a_i = \frac{2}{p} \cdot p = 2$. So, the sum is less than 2.
3. Lower Bound: Rewrite each term as $\frac{a_i}{p-a_i} = \frac{p}{p-a_i} - 1$. The sum becomes $S = p \sum_{i=1}^{n} \frac{1}{p-a_i} - n$. Let $x_i = p-a_i$. Using the AM-HM inequality, $\sum_{i=1}^{n} \frac{1}{x_i} \ge \frac{n^2}{\sum_{i=1}^{n} x_i}$. Substituting $x_i = p-a_i$, we have $\sum_{i=1}^{n} x_i = \sum_{i=1}^{n} (p-a_i) = np - \sum_{i=1}^{n} a_i = np - p = p(n-1)$.

So, $S \ge p \frac{n^2}{p(n-1)} - n = \frac{n^2}{n-1} - n = \frac{n^2 - n(n-1)}{n-1} = \frac{n}{n-1}$. 4. Conclusion: We have $S \ge \frac{n}{n-1}$ and $S < 2$. Since $n > 4$, the smallest value of $\frac{n}{n-1}$ occurs at $n=5$, which is $\frac{5}{4} = 1.25$. As $n$ increases, $\frac{n}{n-1}$ approaches 1. Therefore, for $n>4$, $1 < \frac{n}{n-1} \le S < 2$. Since $S$ is strictly between 1 and 2, its greatest integer value $[S]$ is 1.