Question Number 39737 by ajfour last updated on 10/Jul/18
Commented by ajfour last updated on 10/Jul/18
$${Sir},\: \\ $$$${I}\:{got}\:{R}=\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\:\:.{Please}\:{judge}, \\ $$$${wrong}\:{or}\:{right}! \\ $$
Commented by MJS last updated on 10/Jul/18
$$\mathrm{right}! \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18
$${joinEA}\:\:\:{EA}^{\mathrm{2}} +{AB}^{\mathrm{2}} =\mathrm{4}{R}_{} ^{\mathrm{2}} \\ $$$${AB}=\sqrt{\mathrm{2}}\:+{FB} \\ $$$${angleEAB}=\mathrm{90}^{{o}} \:\:\:{so}\:{angleEAD}=\mathrm{90}^{{o}} −\mathrm{45}^{{o}} =\mathrm{45}^{{o}} \\ $$$${let}\:{angle}\:{ABE}=\theta \\ $$$${sin}\theta=\frac{{EA}}{\mathrm{2}{R}}\:\:\:{cos}\theta=\frac{{AB}}{\mathrm{2}{R}} \\ $$$${angle}\:{DFB}=\mathrm{135}^{{o}} \:\:{so}\:{angle}\:{FDB}=\mathrm{45}^{} −\theta \\ $$$${continue}… \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18
$${cos}\left(\mathrm{45}−\theta\right)=\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{CD}}\:\:\:{CD}=\frac{\mathrm{1}}{\mathrm{2}{cos}\left(\mathrm{45}−\theta\right)} \\ $$$${trying}\:… \\ $$
Answered by MJS last updated on 10/Jul/18
$${C}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:{A}=\begin{pmatrix}{−\frac{\mathrm{1}}{\mathrm{2}}}\\{−{p}−\mathrm{1}}\end{pmatrix}\:{D}=\begin{pmatrix}{−\frac{\mathrm{1}}{\mathrm{2}}}\\{−{p}}\end{pmatrix}\:{F}=\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}}\\{−{p}}\end{pmatrix} \\ $$$${B}\in{l}_{{AF}} \:\wedge\:{B}\in{l}_{{CD}} \\ $$$${l}_{{AF}} :\:{y}={x}−\frac{\mathrm{2}{p}+\mathrm{1}}{\mathrm{2}} \\ $$$${l}_{{CD}} :\:{y}=\mathrm{2}{px} \\ $$$$\Rightarrow\:{B}=\begin{pmatrix}{\frac{\mathrm{2}{p}+\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{p}\right)}}\\{\frac{{p}\left(\mathrm{2}{p}+\mathrm{1}\right)}{\mathrm{1}−\mathrm{2}{p}}}\end{pmatrix} \\ $$$$\mid{A}\mid^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\mid{B}\mid^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow\:{p}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}};\:{R}=\sqrt{\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}} \\ $$
Commented by ajfour last updated on 10/Jul/18
$${Thank}\:{you},\:{I}\:{like}\:{your}\:{way}\:{sir}. \\ $$