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Question Number 199286 by ajfour last updated on 31/Oct/23

	Answered by ajfour last updated on 31/Oct/23
	$Let height of //gm=h AB=1 2r+(√(1−4r^2 ))=p (r/2)(1+p+(√(1+p^2 )))=(p/2) ⇒ r(1+2r+(√(1−4r^2 ))+(√(1+1+4r(√(1−4r^2 ))))) =2r+(√(1−4r^2 )) ⇒ (((√(1−4r^2 ))/r)+2−1−2r−(√(1−4r^2 )))^2 =2+4r(√(1−4r^2 )) ⇒ [(1−r)(√(1−4r^2 ))+r−2r^2 ]^2 =2r^2 +4r^3 (√(1−4r^2 )) ⇒ (1+r^2 −2r)(1−4r^2 )+r^2 (1−2r)^2 +2r(1−r)(1−2r)(√(1−4r^2 )) =r^2 (2+4r(√(1−4r^2 ))) ⇒ (1−r)^2 (1−4r^2 )+r^2 (1−2r)^2 −2r^2 = (6r^2 −2r)(√(1−4r^2 )) ⇒ 4r^2 (1−4r^2 )(3r−2)^2 ={(1−r)^2 (1−4r^2 )+r^2 (1−2r)^2 −2r^2 }^2$
	$L e t h e i g h t o f / / g m = h$ $A B = 1$ $2 r + \sqrt{1 - 4 r^{2}} = p$ $\frac{r}{2} (1 + p + \sqrt{1 + p^{2}}) = \frac{p}{2}$ $\Rightarrow r (1 + 2 r + \sqrt{1 - 4 r^{2}} + \sqrt{1 + 1 + 4 r \sqrt{1 - 4 r^{2}}})$ $= 2 r + \sqrt{1 - 4 r^{2}}$ $\Rightarrow {(\frac{\sqrt{1 - 4 r^{2}}}{r} + 2 - 1 - 2 r - \sqrt{1 - 4 r^{2}})}^{2}$ $= 2 + 4 r \sqrt{1 - 4 r^{2}}$ $\Rightarrow$ ${[(1 - r) \sqrt{1 - 4 r^{2}} + r - 2 r^{2}]}^{2}$ $= 2 r^{2} + 4 r^{3} \sqrt{1 - 4 r^{2}}$ $\Rightarrow$ $(1 + r^{2} - 2 r) (1 - 4 r^{2}) + r^{2} {(1 - 2 r)}^{2}$ $+ 2 r (1 - r) (1 - 2 r) \sqrt{1 - 4 r^{2}}$ $= r^{2} (2 + 4 r \sqrt{1 - 4 r^{2}})$ $\Rightarrow$ ${(1 - r)}^{2} (1 - 4 r^{2}) + r^{2} {(1 - 2 r)}^{2} - 2 r^{2}$ $= (6 r^{2} - 2 r) \sqrt{1 - 4 r^{2}}$ $\Rightarrow$ $4 r^{2} (1 - 4 r^{2}) {(3 r - 2)}^{2}$ $= {{(1 - r)}^{2} (1 - 4 r^{2}) + r^{2} {(1 - 2 r)}^{2} - 2 r^{2}}^{2}$
	Answered by Frix last updated on 31/Oct/23

	$r = \frac{h + 1 - \sqrt{h^{2} + 1}}{2} \Leftrightarrow h = \frac{2 r (r - 1)}{2 r - 1}$ $A B = 1 \Rightarrow ∣ A B ∣^{2} = 1 \Rightarrow$ $8 r^{2} - 4 h r + h^{2} - 1 = 0$ $r^{4} - \frac{4 r^{3}}{5} + \frac{r}{5} - \frac{1}{20} = 0$ $No useable exact solution$ $r \approx . 341002336639$ $(h \approx 1.41335248768)$
	Commented by ajfour last updated on 31/Oct/23

	$I a p p r e c i a t e v e r y m u c h! T h a n k s$
	Answered by mr W last updated on 31/Oct/23

	Commented by mr W last updated on 31/Oct/23

	$\tan α = \frac{r}{1 - r}$ $\tan β = \frac{r}{r + \sqrt{1^{2} - {(2 r)}^{2}}}$ $β = \frac{π}{4} - α$ $\frac{r}{r + \sqrt{1 - 4 r^{2}}} = \frac{1 - \frac{r}{1 - r}}{1 + \frac{r}{1 - r}} = 1 - 2 r$ $2 r^{2} = (1 - 2 r) \sqrt{1 - 4 r^{2}}$ $20 r^{4} - 16 r^{3} + 4 r - 1 = 0$ $\Rightarrow r \approx 0.341$
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