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Let $n \geq 4$ be a positive integer and let $\ell_{1}, \ell_{2}, \ldots . ., \ell_{\mathrm{n}}$ be the lengths of the sides of arbitrary $n$ - sided non-degenerate polygon P. Suppose
$$
\frac{\ell_{1}}{\ell_{2}}+\frac{\ell_{2}}{\ell_{3}}+\ldots \frac{\ell_{\mathrm{n}-1}}{\ell_{\mathrm{n}}}+\frac{\ell_{\mathrm{n}}}{\ell_{1}}=\mathrm{n}
$$
Consider the following statements:
I. The lengths of the sides of $P$ are equal.
II. The angles of $P$ are equal.
III. $\mathrm{P}$ is a regular polygon if it is cyclic.
Then
Options:
$$
\frac{\ell_{1}}{\ell_{2}}+\frac{\ell_{2}}{\ell_{3}}+\ldots \frac{\ell_{\mathrm{n}-1}}{\ell_{\mathrm{n}}}+\frac{\ell_{\mathrm{n}}}{\ell_{1}}=\mathrm{n}
$$
Consider the following statements:
I. The lengths of the sides of $P$ are equal.
II. The angles of $P$ are equal.
III. $\mathrm{P}$ is a regular polygon if it is cyclic.
Then
Solution:
2354 Upvotes
Verified Answer
The correct answer is:
I and III are true
given : $\frac{\ell_{1}}{\ell_{2}}+\frac{\ell_{2}}{\ell_{3}} \ldots \ldots .+\frac{\ell_{\mathrm{n}}}{\ell_{1}}=\mathrm{n}$ .........(i)
$\therefore$ Use A.M $\geq$ G.M
We get
$\frac{\left(\frac{\ell_{1}}{\ell_{2}}+\frac{\ell_{2}}{\ell_{3}} \ldots \ldots+\frac{\ell_{n}}{\ell_{1}}\right)}{\mathrm{n}} \geq \sqrt[n]{\frac{\ell_{1}}{\ell_{2}} \cdot \frac{\ell_{2}}{\ell_{3}} \ldots \cdots \frac{\ell_{\mathrm{n}}}{\ell_{1}}}$
$\therefore \frac{\mathrm{n}}{\mathrm{n}} \geq 1$ $\Rightarrow \mathrm{n}=\mathrm{n}$ So $\mathrm{A} \cdot \mathrm{M}=\mathrm{G} \mathrm{M}$
Hence $\frac{\ell_{1}}{\ell_{2}}=\frac{\ell_{2}}{\ell_{3}} \ldots \ldots \ldots .=\frac{\ell_{\mathrm{n}}}{\ell_{1}}=\mathrm{k}$
$\begin{array}{l}
\Rightarrow \mathrm{k}=\frac{\ell_{1}+\ell_{2} \ldots \ldots+\ell_{\mathrm{n}}}{\ell_{2}+\ell_{3} \ldots \ldots+\ell_{\mathrm{n}}+\ell_{1}}=1 \\
\Rightarrow \ell_{1}=\ell_{2} \ldots \ldots=\ell_{\mathrm{n}}
\end{array}$
$\therefore$ Use A.M $\geq$ G.M
We get
$\frac{\left(\frac{\ell_{1}}{\ell_{2}}+\frac{\ell_{2}}{\ell_{3}} \ldots \ldots+\frac{\ell_{n}}{\ell_{1}}\right)}{\mathrm{n}} \geq \sqrt[n]{\frac{\ell_{1}}{\ell_{2}} \cdot \frac{\ell_{2}}{\ell_{3}} \ldots \cdots \frac{\ell_{\mathrm{n}}}{\ell_{1}}}$
$\therefore \frac{\mathrm{n}}{\mathrm{n}} \geq 1$ $\Rightarrow \mathrm{n}=\mathrm{n}$ So $\mathrm{A} \cdot \mathrm{M}=\mathrm{G} \mathrm{M}$
Hence $\frac{\ell_{1}}{\ell_{2}}=\frac{\ell_{2}}{\ell_{3}} \ldots \ldots \ldots .=\frac{\ell_{\mathrm{n}}}{\ell_{1}}=\mathrm{k}$
$\begin{array}{l}
\Rightarrow \mathrm{k}=\frac{\ell_{1}+\ell_{2} \ldots \ldots+\ell_{\mathrm{n}}}{\ell_{2}+\ell_{3} \ldots \ldots+\ell_{\mathrm{n}}+\ell_{1}}=1 \\
\Rightarrow \ell_{1}=\ell_{2} \ldots \ldots=\ell_{\mathrm{n}}
\end{array}$
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