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Question Number 198828 by a.lgnaoui last updated on 24/Oct/23

$$\mathbf{Show}\mathbf{th}\mathit{a}\mathit{t}:$$$$\mathit{A}\mathit{r}\mathit{e}\mathit{a}\left(\mathit{b}\mathit{l}\mathit{u}\mathit{e}\mathit{c}\mathit{i}\mathit{r}\mathit{c}\mathit{l}\mathit{e}\right)=\mathit{A}\mathit{r}\mathit{e}\mathit{a}\left(\mathit{G}\mathit{r}\mathit{e}\mathit{e}\mathit{n}\mathit{c}\mathit{i}\mathit{r}\mathit{c}\mathit{l}\mathit{e}\right)$$$$$$

Commented by a.lgnaoui last updated on 24/Oct/23

Answered by mr W last updated on 25/Oct/23

Commented by mr W last updated on 25/Oct/23

Commented by mr W last updated on 25/Oct/23

$$AD=2R+p$$$$BD=R-p$$$$CD=R+p$$$$AB=\sqrt{2}R$$$$BC=R$$$$AC=\sqrt{R^2+{\left(2R\right)}^2}=\sqrt{5}R$$$$\cos \alpha =\frac{{\left(\sqrt{2}R\right)}^2+{(R-p)}^2-{(2R+p)}^2}{2\sqrt{2}R(R-p)}$$$$=-\frac{R+6p}{2\sqrt{2}(R-p)}=-\frac{1+6\lambda }{2\sqrt{2}(1-\lambda )}$$$$with\lambda =\frac{p}{R}$$$$\cos \beta =\frac{R^2+{(R-p)}^2-{(R+p)}^2}{2R(R-p)}$$$$=\frac{R-4p}{2(R-p)}=\frac{1-4\lambda }{2(1-\lambda )}$$$$\alpha +\beta =\pi +\frac{\pi }{4}$$$$\cos (\alpha +\beta )=-\frac{\sqrt{2}}{2}$$$$-\frac{\sqrt{2}}{2}=-\frac{1+6\lambda }{2\sqrt{2}(1-\lambda )}\times \frac{1-4\lambda }{2(1-\lambda )}-\frac{\sqrt{8{(1-\lambda )}^2-{(1+6\lambda )}^2}}{2\sqrt{2}(1-\lambda )}\times \frac{\sqrt{4{(1-\lambda )}^2-{(1-4\lambda )}^2}}{2(1-\lambda )}$$$$-\frac{\sqrt{2}}{2}=\frac{24{\lambda }^2-2\lambda -1}{4\sqrt{2}{(1-\lambda )}^2}-\frac{\sqrt{3(7-28\lambda -28{\lambda }^2)(1-4{\lambda }^2)}}{4\sqrt{2}{(1-\lambda )}^2}$$$$28{\lambda }^2-10\lambda +3=\sqrt{3(7-28\lambda -28{\lambda }^2)(1-4{\lambda }^2)}$$$$4{(\lambda -1)}^2(112{\lambda }^2-3)=0$$$$\Rightarrow \lambda =1\Rightarrow rejected$$$$\Rightarrow 112{\lambda }^2=3\Rightarrow \lambda =\sqrt{\frac{3}{112}}=\frac{\sqrt{21}}{28}$$$$\Rightarrow p=\frac{\sqrt{21}R}{28}✓$$$$insimilarwaywecangetq=\frac{\sqrt{21}R}{28}.$$$$p=q=\frac{\sqrt{21}R}{28}$$$$\Rightarrow bluecircle=greencircle$$

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