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KVPY Sample Paper KVPY Stream-SX Model Paper-12

  • question_answer
    Let \[{{A}_{1}},{{A}_{2}}....{{A}_{n}}\]be the vertices of a regular polygon of \[n\] sides inscribed in a circle of radius unity and  \[a={{\left| {{A}_{1}}{{A}_{2}} \right|}^{2}}+{{\left| {{A}_{1}}{{A}_{3}} \right|}^{2}}+....{{\left| {{A}_{1}}{{A}_{n}} \right|}^{2}}\]\[b=\left| {{A}_{1}}{{A}_{2}} \right|\left| {{A}_{1}}{{A}_{3}} \right|...\left| {{A}_{1}}{{A}_{n}} \right|,\operatorname{then}\frac{a}{b}=\]

    A) \[3\]

    B) \[4\]

    C) \[5\]

    D) none of these

    Correct Answer: D

    Solution :

    Let us assume that \[O\]is the centre of the polygon and \[{{z}_{0}},{{z}_{1}},....{{z}_{n-1}}\]
    Represent the affixes of \[{{A}_{1}},{{A}_{2}},....{{A}_{n}},\]such that \[{{z}_{0}}=1,{{z}_{2}}=\alpha ,{{z}_{3}}={{\alpha }^{2}},...,{{z}_{n-1}}={{\alpha }^{n-1}},\]
    Where \[\alpha ={{e}^{\frac{i2\pi }{n}}}.\]
    Now, \[{{\left| {{A}_{1}}{{A}_{r}} \right|}^{2}}={{\left| {{\alpha }^{r}}-1 \right|}^{2}}={{\left| 1-{{\alpha }^{r}} \right|}^{2}}\]
    \[={{\left| 1-\cos \frac{2r\pi }{n}-i\sin \frac{2r\pi }{n} \right|}^{2}}\]\[{{\left( 1-\cos \frac{2r\pi }{n} \right)}^{2}}+{{\left( \sin \frac{2r\pi }{n} \right)}^{2}}\] \[=2-2\cos \frac{2r\pi }{n}\]
    \[\therefore \sum\limits_{r=2}^{n}{{{\left| {{A}_{1}}{{A}_{r}} \right|}^{2}}=\sum\limits_{r=1}^{n}{\left( 2-2\cos \frac{r\pi }{n} \right)}}\]\[=2\left( n-1 \right)-2\left\{ \begin{align}   & \cos \frac{2\pi }{n}+\cos \frac{4\pi }{n}+..... \\  & +\cos \frac{2\left( n-1 \right)\pi }{n} \\ \end{align} \right\}\]
    \[=2(n-1)-2\]Real part of \[\left( \alpha +{{\alpha }^{2}}+....+{{\alpha }^{n-1}} \right)\]
    \[2(n-1)-2(-1)\]
    \[[\operatorname{Since}\{1+\alpha +{{\alpha }^{2}}+...+{{\alpha }^{n-1}}=0\}]\]
    \[\therefore {{\left| {{A}_{1}}{{A}_{2}} \right|}^{2}}+{{\left| {{A}_{2}}{{A}_{3}} \right|}^{2}}+...{{\left| {{A}_{1}}{{A}_{n}} \right|}^{2}}=2n.\]
    Also, let \[E=\left| {{A}_{1}}{{A}_{2}} \right|\left| {{A}_{1}}{{A}_{3}} \right|...\left| {{A}_{1}}{{A}_{n}} \right|\]
    \[=\left| 1-\alpha  \right|\left| 1-{{\alpha }^{2}} \right|\left| 1-{{\alpha }^{3}} \right|...\left| 1-{{\alpha }^{n-1}} \right|\]
    \[=\left| (1-\alpha )(1-{{\alpha }^{2}})(1-{{\alpha }^{3}})...(1-{{\alpha }^{n-1}}) \right|\]
    Since, \[1,\alpha ,{{\alpha }^{2}},...{{\alpha }^{n-1}}\] are the roots of  \[{{z}^{n}}-1=0\]
    \[\Rightarrow \]\[(z-1)(z-\alpha )(z-{{\alpha }^{2}})...(z-{{\alpha }^{n-1}})={{z}^{n}}-1\]
    \[\Rightarrow \]\[(z-\alpha )(z-{{\alpha }^{2}})(z-{{\alpha }^{n-1}})=\frac{{{z}^{n}}-1}{(z-1)}\]\[=1+z+{{z}^{2}}+....+{{z}^{n-1}}\]
    Substituting \[z=1,\]we have  \[(1-\alpha )(1-{{\alpha }^{2}})...(1-{{\alpha }^{n-1}})=n\]
    \[\therefore \left| 1-\alpha  \right|\left| 1-{{\alpha }^{2}} \right|...\left| 1-{{\alpha }^{n-1}} \right|=n\]


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