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Question-199286




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
$${Let}\:{height}\:{of}\://{gm}={h} \\ $$$${AB}=\mathrm{1}\:\: \\ $$$$\mathrm{2}{r}+\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }={p} \\ $$$$\frac{{r}}{\mathrm{2}}\left(\mathrm{1}+{p}+\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }\right)=\frac{{p}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:{r}\left(\mathrm{1}+\mathrm{2}{r}+\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }+\sqrt{\mathrm{1}+\mathrm{1}+\mathrm{4}{r}\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }}\right) \\ $$$$\:\:\:\:\:=\mathrm{2}{r}+\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\left(\frac{\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }}{{r}}+\mathrm{2}−\mathrm{1}−\mathrm{2}{r}−\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$\:\:=\mathrm{2}+\mathrm{4}{r}\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} } \\ $$$$\Rightarrow\: \\ $$$$\:\left[\left(\mathrm{1}−{r}\right)\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }+{r}−\mathrm{2}{r}^{\mathrm{2}} \right]^{\mathrm{2}} \\ $$$$\:\:=\mathrm{2}{r}^{\mathrm{2}} +\mathrm{4}{r}^{\mathrm{3}} \sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$$\:\left(\mathrm{1}+{r}^{\mathrm{2}} −\mathrm{2}{r}\right)\left(\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} \right)+{r}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{2}{r}\right)^{\mathrm{2}} \\ $$$$+\mathrm{2}{r}\left(\mathrm{1}−{r}\right)\left(\mathrm{1}−\mathrm{2}{r}\right)\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} } \\ $$$$\:\:\:={r}^{\mathrm{2}} \left(\mathrm{2}+\mathrm{4}{r}\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow \\ $$$$\:\left(\mathrm{1}−{r}\right)^{\mathrm{2}} \left(\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} \right)+{r}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{2}{r}\right)^{\mathrm{2}} −\mathrm{2}{r}^{\mathrm{2}} \\ $$$$\:\:=\:\left(\mathrm{6}{r}^{\mathrm{2}} −\mathrm{2}{r}\right)\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\: \\ $$$$\mathrm{4}{r}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} \right)\left(\mathrm{3}{r}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$=\left\{\left(\mathrm{1}−{r}\right)^{\mathrm{2}} \left(\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} \right)+{r}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{2}{r}\right)^{\mathrm{2}} −\mathrm{2}{r}^{\mathrm{2}} \right\}^{\mathrm{2}} \\ $$$$ \\ $$
Answered by Frix last updated on 31/Oct/23
r=((h+1−(√(h^2 +1)))/2) ⇔ h=((2r(r−1))/(2r−1))  AB=1 ⇒ ∣AB∣^2 =1 ⇒  8r^2 −4hr+h^2 −1=0  r^4 −((4r^3 )/5)+(r/5)−(1/(20))=0  No useable exact solution  r≈.341002336639  (h≈1.41335248768)
$${r}=\frac{{h}+\mathrm{1}−\sqrt{{h}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}}\:\Leftrightarrow\:{h}=\frac{\mathrm{2}{r}\left({r}−\mathrm{1}\right)}{\mathrm{2}{r}−\mathrm{1}} \\ $$$${AB}=\mathrm{1}\:\Rightarrow\:\mid{AB}\mid^{\mathrm{2}} =\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{8}{r}^{\mathrm{2}} −\mathrm{4}{hr}+{h}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${r}^{\mathrm{4}} −\frac{\mathrm{4}{r}^{\mathrm{3}} }{\mathrm{5}}+\frac{{r}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{20}}=\mathrm{0} \\ $$$$\mathrm{No}\:\mathrm{useable}\:\mathrm{exact}\:\mathrm{solution} \\ $$$${r}\approx.\mathrm{341002336639} \\ $$$$\left({h}\approx\mathrm{1}.\mathrm{41335248768}\right) \\ $$
Commented by ajfour last updated on 31/Oct/23
I appreciate very much! Thanks
$${I}\:{appreciate}\:{very}\:{much}!\:{Thanks} \\ $$
Answered by mr W last updated on 31/Oct/23
Commented by mr W last updated on 31/Oct/23
tan α=(r/(1−r))  tan β=(r/(r+(√(1^2 −(2r)^2 ))))  β=(π/4)−α  (r/(r+(√(1−4r^2 ))))=((1−(r/(1−r)))/(1+(r/(1−r))))=1−2r  2r^2 =(1−2r)(√(1−4r^2 ))  20r^4 −16r^3 +4r−1=0  ⇒r≈0.341
$$\mathrm{tan}\:\alpha=\frac{{r}}{\mathrm{1}−{r}} \\ $$$$\mathrm{tan}\:\beta=\frac{{r}}{{r}+\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\mathrm{2}{r}\right)^{\mathrm{2}} }} \\ $$$$\beta=\frac{\pi}{\mathrm{4}}−\alpha \\ $$$$\frac{{r}}{{r}+\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }}=\frac{\mathrm{1}−\frac{{r}}{\mathrm{1}−{r}}}{\mathrm{1}+\frac{{r}}{\mathrm{1}−{r}}}=\mathrm{1}−\mathrm{2}{r} \\ $$$$\mathrm{2}{r}^{\mathrm{2}} =\left(\mathrm{1}−\mathrm{2}{r}\right)\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} } \\ $$$$\mathrm{20}{r}^{\mathrm{4}} −\mathrm{16}{r}^{\mathrm{3}} +\mathrm{4}{r}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{r}\approx\mathrm{0}.\mathrm{341} \\ $$

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