Question Number 189309 by cherokeesay last updated on 14/Mar/23
Answered by som(math1967) last updated on 14/Mar/23
Commented by som(math1967) last updated on 14/Mar/23
$${OA}={a}+\mathrm{2}{r}\: \\ $$$${OB}^{\mathrm{2}} =\left({a}+\mathrm{2}{r}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$${DC}={a}+{a}+{r}−{r}=\mathrm{2}{a} \\ $$$${OB}^{\mathrm{2}} ={DC}^{\mathrm{2}} \\ $$$$\left({a}+\mathrm{2}{r}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{ar}−\mathrm{4}{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\frac{{a}^{\mathrm{2}} }{{r}^{\mathrm{2}} }\:−\frac{{a}}{{r}}\:−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{{a}}{{r}}=\frac{\mathrm{1}+\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{4}.\mathrm{1}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$$$\:\therefore\:\frac{{a}}{{r}}+\mathrm{1}=\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\frac{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}=\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\varphi^{\mathrm{2}} \\ $$$$ \\ $$
Commented by cherokeesay last updated on 14/Mar/23
$${Nice}\:{thank}\:{you}. \\ $$
Commented by horsebrand11 last updated on 15/Mar/23
$${OA}=\:{a}+{r}\:? \\ $$
Commented by som(math1967) last updated on 15/Mar/23
$${OA}={rad}\:{of}\:{quadrant}\:+{rad}.\:{of} \\ $$$${circle} \\ $$$${rad}\:{of}\:{quadrant}={a}+{r},{rad}\:{of}\:{circle}={r} \\ $$$$\therefore{OA}=\:{a}+{r}+{r}={a}+\mathrm{2}{r} \\ $$