Question Number 199286 by ajfour last updated on 31/Oct/23
Answered by ajfour last updated on 31/Oct/23
![Let height of //gm=h AB=1 2r+(√(1−4r^2 ))=p (r/2)(1+p+(√(1+p^2 )))=(p/2) ⇒ r(1+2r+(√(1−4r^2 ))+(√(1+1+4r(√(1−4r^2 ))))) =2r+(√(1−4r^2 )) ⇒ (((√(1−4r^2 ))/r)+2−1−2r−(√(1−4r^2 )))^2 =2+4r(√(1−4r^2 )) ⇒ [(1−r)(√(1−4r^2 ))+r−2r^2 ]^2 =2r^2 +4r^3 (√(1−4r^2 )) ⇒ (1+r^2 −2r)(1−4r^2 )+r^2 (1−2r)^2 +2r(1−r)(1−2r)(√(1−4r^2 )) =r^2 (2+4r(√(1−4r^2 ))) ⇒ (1−r)^2 (1−4r^2 )+r^2 (1−2r)^2 −2r^2 = (6r^2 −2r)(√(1−4r^2 )) ⇒ 4r^2 (1−4r^2 )(3r−2)^2 ={(1−r)^2 (1−4r^2 )+r^2 (1−2r)^2 −2r^2 }^2](https://www.tinkutara.com/question/Q199287.png)
$${Let}\:{height}\:{of}\://{gm}={h} \\ $$$${AB}=\mathrm{1}\:\: \\ $$$$\mathrm{2}{r}+\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }={p} \\ $$$$\frac{{r}}{\mathrm{2}}\left(\mathrm{1}+{p}+\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }\right)=\frac{{p}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:{r}\left(\mathrm{1}+\mathrm{2}{r}+\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }+\sqrt{\mathrm{1}+\mathrm{1}+\mathrm{4}{r}\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }}\right) \\ $$$$\:\:\:\:\:=\mathrm{2}{r}+\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\left(\frac{\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }}{{r}}+\mathrm{2}−\mathrm{1}−\mathrm{2}{r}−\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$\:\:=\mathrm{2}+\mathrm{4}{r}\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} } \\ $$$$\Rightarrow\: \\ $$$$\:\left[\left(\mathrm{1}−{r}\right)\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }+{r}−\mathrm{2}{r}^{\mathrm{2}} \right]^{\mathrm{2}} \\ $$$$\:\:=\mathrm{2}{r}^{\mathrm{2}} +\mathrm{4}{r}^{\mathrm{3}} \sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$$\:\left(\mathrm{1}+{r}^{\mathrm{2}} −\mathrm{2}{r}\right)\left(\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} \right)+{r}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{2}{r}\right)^{\mathrm{2}} \\ $$$$+\mathrm{2}{r}\left(\mathrm{1}−{r}\right)\left(\mathrm{1}−\mathrm{2}{r}\right)\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} } \\ $$$$\:\:\:={r}^{\mathrm{2}} \left(\mathrm{2}+\mathrm{4}{r}\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow \\ $$$$\:\left(\mathrm{1}−{r}\right)^{\mathrm{2}} \left(\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} \right)+{r}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{2}{r}\right)^{\mathrm{2}} −\mathrm{2}{r}^{\mathrm{2}} \\ $$$$\:\:=\:\left(\mathrm{6}{r}^{\mathrm{2}} −\mathrm{2}{r}\right)\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\: \\ $$$$\mathrm{4}{r}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} \right)\left(\mathrm{3}{r}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$=\left\{\left(\mathrm{1}−{r}\right)^{\mathrm{2}} \left(\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} \right)+{r}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{2}{r}\right)^{\mathrm{2}} −\mathrm{2}{r}^{\mathrm{2}} \right\}^{\mathrm{2}} \\ $$$$ \\ $$
Answered by Frix last updated on 31/Oct/23
$${r}=\frac{{h}+\mathrm{1}−\sqrt{{h}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}}\:\Leftrightarrow\:{h}=\frac{\mathrm{2}{r}\left({r}−\mathrm{1}\right)}{\mathrm{2}{r}−\mathrm{1}} \\ $$$${AB}=\mathrm{1}\:\Rightarrow\:\mid{AB}\mid^{\mathrm{2}} =\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{8}{r}^{\mathrm{2}} −\mathrm{4}{hr}+{h}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${r}^{\mathrm{4}} −\frac{\mathrm{4}{r}^{\mathrm{3}} }{\mathrm{5}}+\frac{{r}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{20}}=\mathrm{0} \\ $$$$\mathrm{No}\:\mathrm{useable}\:\mathrm{exact}\:\mathrm{solution} \\ $$$${r}\approx.\mathrm{341002336639} \\ $$$$\left({h}\approx\mathrm{1}.\mathrm{41335248768}\right) \\ $$
Commented by ajfour last updated on 31/Oct/23
$${I}\:{appreciate}\:{very}\:{much}!\:{Thanks} \\ $$
Answered by mr W last updated on 31/Oct/23
Commented by mr W last updated on 31/Oct/23
$$\mathrm{tan}\:\alpha=\frac{{r}}{\mathrm{1}−{r}} \\ $$$$\mathrm{tan}\:\beta=\frac{{r}}{{r}+\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\mathrm{2}{r}\right)^{\mathrm{2}} }} \\ $$$$\beta=\frac{\pi}{\mathrm{4}}−\alpha \\ $$$$\frac{{r}}{{r}+\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }}=\frac{\mathrm{1}−\frac{{r}}{\mathrm{1}−{r}}}{\mathrm{1}+\frac{{r}}{\mathrm{1}−{r}}}=\mathrm{1}−\mathrm{2}{r} \\ $$$$\mathrm{2}{r}^{\mathrm{2}} =\left(\mathrm{1}−\mathrm{2}{r}\right)\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} } \\ $$$$\mathrm{20}{r}^{\mathrm{4}} −\mathrm{16}{r}^{\mathrm{3}} +\mathrm{4}{r}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{r}\approx\mathrm{0}.\mathrm{341} \\ $$