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Question Number 200268 by cortano12 last updated on 16/Nov/23

	Answered by ajfour last updated on 16/Nov/23
	$D origin. O_2 ≡(s−r, s−r) O_3 ≡(r, s−r) let eqn of line DF be y=mx r^2 =(((s−r)^2 (1−m)^2 )/(1+m^2 ))=(({(s−r)−mr}^2 )/(1+m^2 )) let (r/s)=t t^2 (1+m^2 )=(1−t)^2 (1−m)^2 ={(1−t)−mt}^2 ⇒ t^2 =(1−t)^2 −2mt(1−t) ...(i) & (1−t)(1−m)=−(1−t−mt) ⇒ 2−2t−m=0 ⇒ m=2(1−t) substituting in ..(i) 2t−1+4t(1−t)^2 =0 8t^3 −16t^2 +12t−2=0 2t=z z^3 −4z^2 +6z−2=0 z=((4+(3(√(33))−17)^(1/3) −(3(√(33))+17)^(1/3) )/3) ≈ 0.456311 t=(r/s)=(z/2)≈0.228155$
	$D o r i g i n .$ $O_{2} \equiv (s - r, s - r)$ $O_{3} \equiv (r, s - r)$ $l e t e q n o f l i n e D F b e$ $y = m x$ $r^{2} = \frac{{(s - r)}^{2} {(1 - m)}^{2}}{1 + m^{2}} = \frac{{(s - r) - m r}^{2}}{1 + m^{2}}$ $l e t \frac{r}{s} = t$ $t^{2} (1 + m^{2}) = {(1 - t)}^{2} {(1 - m)}^{2}$ $= {(1 - t) - m t}^{2}$ $\Rightarrow t^{2} = {(1 - t)}^{2} - 2 m t (1 - t) . . . (i)$ $& (1 - t) (1 - m) = - (1 - t - m t)$ $\Rightarrow 2 - 2 t - m = 0$ $\Rightarrow m = 2 (1 - t)$ $s u b s t i t u t i n g i n . . (i)$ $2 t - 1 + 4 t {(1 - t)}^{2} = 0$ $8 t^{3} - 16 t^{2} + 12 t - 2 = 0$ $2 t = z$ $z^{3} - 4 z^{2} + 6 z - 2 = 0$ $z = \frac{4 + {(3 \sqrt{33} - 17)}^{1 / 3} - {(3 \sqrt{33} + 17)}^{1 / 3}}{3}$ $\approx 0.456311$ $t = \frac{r}{s} = \frac{z}{2} \approx 0.228155$
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