Question Number 222312 by ajfour last updated on 22/Jun/25
Answered by mr W last updated on 22/Jun/25
$${QC}={DC}={AB}=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{a}}=\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${a}^{\mathrm{4}} −{a}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow{a}=\frac{\sqrt{\mathrm{2}\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)}}{\mathrm{2}}\approx\mathrm{1}.\mathrm{272} \\ $$$$\mathrm{tan}\:\frac{\theta}{\mathrm{2}}=\frac{{r}}{\mathrm{1}−{r}} \\ $$$$\mathrm{tan}\:\theta=\frac{{a}}{\mathrm{1}}=\frac{\frac{\mathrm{2}{r}}{\mathrm{1}−{r}}}{\mathrm{1}−\frac{{r}^{\mathrm{2}} }{\left(\mathrm{1}−{r}\right)^{\mathrm{2}} }} \\ $$$${a}=\frac{\mathrm{2}{r}\left(\mathrm{1}−{r}\right)}{\mathrm{1}−\mathrm{2}{r}} \\ $$$${r}^{\mathrm{2}} −\left({a}+\mathrm{1}\right){r}+\frac{{a}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow{r}=\frac{{a}+\mathrm{1}−\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}}=\frac{{a}+\mathrm{1}−{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{1}+\sqrt{\mathrm{2}\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)}−\sqrt{\mathrm{5}}}{\mathrm{4}}\approx\mathrm{0}.\mathrm{327} \\ $$
Commented by ajfour last updated on 22/Jun/25
https://youtu.be/eWVL1GOHshg?si=et75BFDWDhRVnmov
yes sir. see mine.