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Question Number 124048 by ajfour last updated on 30/Nov/20

Commented by ajfour last updated on 30/Nov/20

$F i n d s i d e s o f e q u i l a t e r a l △ i n$ $t e r m s o f λ . R a d i u s i s u n i t y .$
	Answered by mr W last updated on 30/Nov/20

	Commented by ajfour last updated on 30/Nov/20
	$let B(h,−k) k = 1−λ h= (√(λ(2−λ))) eq. of BC y=1−(((2−λ)x)/( (√(λ(2−λ)))) = 1−(((1+k)x)/h) let D[t, 1−(((1+k)t)/h)] let F be midpoint of AD F [(t/2), ((1−k)/2)−(((1+k)t)/(2h))] AF = (t/2)+i[((1+k)/2)−(((1+k)t)/(2h))] FE=(((1+k)t)/(2h))−(((1+k))/2)+((it)/2) E[(((1+2k)t)/(2h))−(((1+k))/2), (t/2)+((1−k)/2)−(((1+k)t)/(2h))] E(−cos θ, sin θ) ⇒ {(((1+2k)t)/(2h))−(((1+k))/2)}^2 +{(t/2)+((1−k)/2)−(((1+k)t)/(2h))}^2 =1 ⇒ {(1+2k)t−h(1+k)}^2 + {h(1−k)−(1+k−h)t}^2 = 1 ⇒ {(1+2k)^2 −(1+k−h)^2 }t^2 +2{h(1−k)(1+k−h)−h(1+k)(1+2k)}t +h^2 [(1+k)^2 −(1−k)^2 ]−1=0 ⇒ (k^2 −h^2 +4k+2h+2hk)t^2 +2(h+hk−h^2 −hk−hk^2 +h^2 k −h−2hk−hk−2hk^2 )t +4kh^2 −1=0 ⇒ (k^2 +2hk−h^2 +4k+2h)t^2 −2(hk^2 −h^2 k+3hk+h^2 )t+4h^2 k−1 =0 ⇒ t=((hk(k−h)+h(3k+h))/(k(k+4)+h(2k−h+2))) −(√([((hk(k−h)+h(3k+h))/(k(k+4)+h(2k−h+2)))]^2 −(((4h^2 k−1))/(k(k+4)+h(2k−h+2))))) s = (√(t^2 +(1+k)^2 (h−t)^2 )) example: λ=0.6 k=0.4 , h=(√(0.6×1.4)) = (√(0.84)) t ≈ 0.11039604(✓) , 0.89381445(×) s ≈ 1.13395334$
	$l e t B (h, - k)$ $k = 1 - λ$ $h = \sqrt{λ (2 - λ)}$ $e q . o f B C$ $y = 1 - \frac{(2 - λ) x}{\sqrt{λ (2 - λ}} = 1 - \frac{(1 + k) x}{h}$ $l e t D [t, 1 - \frac{(1 + k) t}{h}]$ $l e t F b e m i d p o i n t o f A D$ $F [\frac{t}{2}, \frac{1 - k}{2} - \frac{(1 + k) t}{2 h}]$ $A F = \frac{t}{2} + i [\frac{1 + k}{2} - \frac{(1 + k) t}{2 h}]$ $F E = \frac{(1 + k) t}{2 h} - \frac{(1 + k)}{2} + \frac{i t}{2}$ $E [\frac{(1 + 2 k) t}{2 h} - \frac{(1 + k)}{2}, \frac{t}{2} + \frac{1 - k}{2} - \frac{(1 + k) t}{2 h}]$ $E (- \cos θ, \sin θ)$ $\Rightarrow$ ${\frac{(1 + 2 k) t}{2 h} - \frac{(1 + k)}{2}}^{2} + {\frac{t}{2} + \frac{1 - k}{2} - \frac{(1 + k) t}{2 h}}^{2} = 1$ $\Rightarrow$ ${(1 + 2 k) t - h (1 + k)}^{2} +$ ${h (1 - k) - (1 + k - h) t}^{2} = 1$ $\Rightarrow$ ${{(1 + 2 k)}^{2} - {(1 + k - h)}^{2}} t^{2}$ $+ 2 {h (1 - k) (1 + k - h) - h (1 + k) (1 + 2 k)} t$ $+ h^{2} [{(1 + k)}^{2} - {(1 - k)}^{2}] - 1 = 0$ $\Rightarrow$ $(k^{2} - h^{2} + 4 k + 2 h + 2 h k) t^{2}$ $+ 2 (h + h k - h^{2} - h k - {h k}^{2} + h^{2} k$ $- h - 2 h k - h k - 2 {h k}^{2}) t$ $+ 4 {k h}^{2} - 1 = 0$ $\Rightarrow$ $(k^{2} + 2 h k - h^{2} + 4 k + 2 h) t^{2}$ $- 2 ({h k}^{2} - h^{2} k + 3 h k + h^{2}) t + 4 h^{2} k - 1$ $= 0$ $\Rightarrow$ $t = \frac{h k (k - h) + h (3 k + h)}{k (k + 4) + h (2 k - h + 2)}$ $- \sqrt{{[\frac{h k (k - h) + h (3 k + h)}{k (k + 4) + h (2 k - h + 2)}]}^{2} - \frac{(4 h^{2} k - 1)}{k (k + 4) + h (2 k - h + 2)}}$ $s = \sqrt{t^{2} + {(1 + k)}^{2} {(h - t)}^{2}}$ $e x a m p l e : λ = 0.6$ $k = 0.4, h = \sqrt{0.6 \times 1.4} = \sqrt{0.84}$ $t \approx 0.11039604 (✓), 0.89381445 (\times)$ $s \approx 1.13395334$
	Commented by ajfour last updated on 01/Dec/20

	$w a i t i n g f o r Y o u r s o l u t i o n, m r W$ $S i r . .$
	Commented by mr W last updated on 01/Dec/20

	$i h a v e n o b e t t e r w a y f o r s o l v i n g .$ $i g u e s s s o l u t i o n e x i s t s o n l y f o r$ $s p e c i a l v a l u e s o f λ .$
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