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Question Number 39737 by ajfour last updated on 10/Jul/18

Commented by ajfour last updated on 10/Jul/18

Sir, I got R=(√(1+(1/(√2)))) .Please judge, wrong or right!

$${Sir},\: \\ $$$${I}\:{got}\:{R}=\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}}\:\:.{Please}\:{judge}, \\ $$$${wrong}\:{or}\:{right}! \\ $$

Commented by MJS last updated on 10/Jul/18

right!

$$\mathrm{right}! \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18

joinEA EA^2 +AB^2 =4R_ ^2 AB=(√2) +FB angleEAB=90^o so angleEAD=90^o −45^o =45^o let angle ABE=θ sinθ=((EA)/(2R)) cosθ=((AB)/(2R)) angle DFB=135^o so angle FDB=45^ −θ continue...

$${joinEA}\:\:\:{EA}^{\mathrm{2}} +{AB}^{\mathrm{2}} =\mathrm{4}{R}_{} ^{\mathrm{2}} \\ $$$${AB}=\sqrt{\mathrm{2}}\:+{FB} \\ $$$${angleEAB}=\mathrm{90}^{{o}} \:\:\:{so}\:{angleEAD}=\mathrm{90}^{{o}} −\mathrm{45}^{{o}} =\mathrm{45}^{{o}} \\ $$$${let}\:{angle}\:{ABE}=\theta \\ $$$${sin}\theta=\frac{{EA}}{\mathrm{2}{R}}\:\:\:{cos}\theta=\frac{{AB}}{\mathrm{2}{R}} \\ $$$${angle}\:{DFB}=\mathrm{135}^{{o}} \:\:{so}\:{angle}\:{FDB}=\mathrm{45}^{} −\theta \\ $$$${continue}... \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18

cos(45−θ)=((1/2)/(CD)) CD=(1/(2cos(45−θ))) trying ...

$${cos}\left(\mathrm{45}−\theta\right)=\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{CD}}\:\:\:{CD}=\frac{\mathrm{1}}{\mathrm{2}{cos}\left(\mathrm{45}−\theta\right)} \\ $$$${trying}\:... \\ $$

Answered by MJS last updated on 10/Jul/18

C= ((0),(0) ) A= (((−(1/2))),((−p−1)) ) D= (((−(1/2))),((−p)) ) F= (((1/2)),((−p)) ) B∈l_(AF) ∧ B∈l_(CD) l_(AF) : y=x−((2p+1)/2) l_(CD) : y=2px ⇒ B= ((((2p+1)/(2(1−2p)))),(((p(2p+1))/(1−2p))) ) ∣A∣^2 =R^2 ∣B∣^2 =R^2 ⇒ p=−(1/2)+((√2)/2); R=(√(1+((√2)/2)))=((√2)/2)(√(2+(√2)))

$${C}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:{A}=\begin{pmatrix}{−\frac{\mathrm{1}}{\mathrm{2}}}\\{−{p}−\mathrm{1}}\end{pmatrix}\:{D}=\begin{pmatrix}{−\frac{\mathrm{1}}{\mathrm{2}}}\\{−{p}}\end{pmatrix}\:{F}=\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}}\\{−{p}}\end{pmatrix} \\ $$$${B}\in{l}_{{AF}} \:\wedge\:{B}\in{l}_{{CD}} \\ $$$${l}_{{AF}} :\:{y}={x}−\frac{\mathrm{2}{p}+\mathrm{1}}{\mathrm{2}} \\ $$$${l}_{{CD}} :\:{y}=\mathrm{2}{px} \\ $$$$\Rightarrow\:{B}=\begin{pmatrix}{\frac{\mathrm{2}{p}+\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{p}\right)}}\\{\frac{{p}\left(\mathrm{2}{p}+\mathrm{1}\right)}{\mathrm{1}−\mathrm{2}{p}}}\end{pmatrix} \\ $$$$\mid{A}\mid^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\mid{B}\mid^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow\:{p}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}};\:{R}=\sqrt{\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}} \\ $$

Commented by ajfour last updated on 10/Jul/18

Thank you, I like your way sir.

$${Thank}\:{you},\:{I}\:{like}\:{your}\:{way}\:{sir}. \\ $$

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