Question Number 124048 by ajfour last updated on 30/Nov/20
Commented by ajfour last updated on 30/Nov/20
$${Find}\:{side}\:{s}\:{of}\:{equilateral}\:\bigtriangleup\:{in} \\ $$$${terms}\:{of}\:\lambda.\:{Radius}\:{is}\:{unity}. \\ $$
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](https://www.tinkutara.com/question/Q124112.png)
$${let}\:\:{B}\left({h},−{k}\right)\:\:\:\:\:\:\:\:\: \\ $$$$\:{k}\:=\:\mathrm{1}−\lambda\:\:\:\: \\ $$$${h}=\:\sqrt{\lambda\left(\mathrm{2}−\lambda\right)} \\ $$$${eq}.\:{of}\:{BC} \\ $$$${y}=\mathrm{1}−\frac{\left(\mathrm{2}−\lambda\right){x}}{\:\sqrt{\lambda\left(\mathrm{2}−\lambda\right.}}\:=\:\mathrm{1}−\frac{\left(\mathrm{1}+{k}\right){x}}{{h}} \\ $$$${let}\:\:\:{D}\left[{t},\:\mathrm{1}−\frac{\left(\mathrm{1}+{k}\right){t}}{{h}}\right] \\ $$$${let}\:\:{F}\:\:\:{be}\:{midpoint}\:{of}\:{AD} \\ $$$${F}\:\left[\frac{{t}}{\mathrm{2}},\:\frac{\mathrm{1}−{k}}{\mathrm{2}}−\frac{\left(\mathrm{1}+{k}\right){t}}{\mathrm{2}{h}}\right] \\ $$$${AF}\:=\:\frac{{t}}{\mathrm{2}}+{i}\left[\frac{\mathrm{1}+{k}}{\mathrm{2}}−\frac{\left(\mathrm{1}+{k}\right){t}}{\mathrm{2}{h}}\right] \\ $$$${FE}=\frac{\left(\mathrm{1}+{k}\right){t}}{\mathrm{2}{h}}−\frac{\left(\mathrm{1}+{k}\right)}{\mathrm{2}}+\frac{{it}}{\mathrm{2}} \\ $$$${E}\left[\frac{\left(\mathrm{1}+\mathrm{2}{k}\right){t}}{\mathrm{2}{h}}−\frac{\left(\mathrm{1}+{k}\right)}{\mathrm{2}},\:\frac{{t}}{\mathrm{2}}+\frac{\mathrm{1}−{k}}{\mathrm{2}}−\frac{\left(\mathrm{1}+{k}\right){t}}{\mathrm{2}{h}}\right] \\ $$$${E}\left(−\mathrm{cos}\:\theta,\:\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\: \\ $$$$\left\{\frac{\left(\mathrm{1}+\mathrm{2}{k}\right){t}}{\mathrm{2}{h}}−\frac{\left(\mathrm{1}+{k}\right)}{\mathrm{2}}\right\}^{\mathrm{2}} +\left\{\frac{{t}}{\mathrm{2}}+\frac{\mathrm{1}−{k}}{\mathrm{2}}−\frac{\left(\mathrm{1}+{k}\right){t}}{\mathrm{2}{h}}\right\}^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow \\ $$$$\left\{\left(\mathrm{1}+\mathrm{2}{k}\right){t}−{h}\left(\mathrm{1}+{k}\right)\right\}^{\mathrm{2}} + \\ $$$$\:\:\:\:\left\{{h}\left(\mathrm{1}−{k}\right)−\left(\mathrm{1}+{k}−{h}\right){t}\right\}^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$$\Rightarrow \\ $$$$\left\{\left(\mathrm{1}+\mathrm{2}{k}\right)^{\mathrm{2}} −\left(\mathrm{1}+{k}−{h}\right)^{\mathrm{2}} \right\}{t}^{\mathrm{2}} \\ $$$$+\mathrm{2}\left\{{h}\left(\mathrm{1}−{k}\right)\left(\mathrm{1}+{k}−{h}\right)−{h}\left(\mathrm{1}+{k}\right)\left(\mathrm{1}+\mathrm{2}{k}\right)\right\}{t} \\ $$$$+{h}^{\mathrm{2}} \left[\left(\mathrm{1}+{k}\right)^{\mathrm{2}} −\left(\mathrm{1}−{k}\right)^{\mathrm{2}} \right]−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\left({k}^{\mathrm{2}} −{h}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{2}{h}+\mathrm{2}{hk}\right){t}^{\mathrm{2}} \\ $$$$+\mathrm{2}\left({h}+{hk}−{h}^{\mathrm{2}} −{hk}−{hk}^{\mathrm{2}} +{h}^{\mathrm{2}} {k}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:−{h}−\mathrm{2}{hk}−{hk}−\mathrm{2}{hk}^{\mathrm{2}} \right){t} \\ $$$$+\mathrm{4}{kh}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\left({k}^{\mathrm{2}} +\mathrm{2}{hk}−{h}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{2}{h}\right){t}^{\mathrm{2}} \\ $$$$−\mathrm{2}\left({hk}^{\mathrm{2}} −{h}^{\mathrm{2}} {k}+\mathrm{3}{hk}+{h}^{\mathrm{2}} \right){t}+\mathrm{4}{h}^{\mathrm{2}} {k}−\mathrm{1} \\ $$$$=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${t}=\frac{{hk}\left({k}−{h}\right)+{h}\left(\mathrm{3}{k}+{h}\right)}{{k}\left({k}+\mathrm{4}\right)+{h}\left(\mathrm{2}{k}−{h}+\mathrm{2}\right)} \\ $$$$ \\ $$$$\:\:\:−\sqrt{\left[\frac{{hk}\left({k}−{h}\right)+{h}\left(\mathrm{3}{k}+{h}\right)}{{k}\left({k}+\mathrm{4}\right)+{h}\left(\mathrm{2}{k}−{h}+\mathrm{2}\right)}\right]^{\mathrm{2}} −\frac{\left(\mathrm{4}{h}^{\mathrm{2}} {k}−\mathrm{1}\right)}{{k}\left({k}+\mathrm{4}\right)+{h}\left(\mathrm{2}{k}−{h}+\mathrm{2}\right)}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{s}}\:=\:\sqrt{\boldsymbol{{t}}^{\mathrm{2}} +\left(\mathrm{1}+\boldsymbol{{k}}\right)^{\mathrm{2}} \left(\boldsymbol{{h}}−\boldsymbol{{t}}\right)^{\mathrm{2}} } \\ $$$${example}:\:\:\lambda=\mathrm{0}.\mathrm{6}\:\: \\ $$$${k}=\mathrm{0}.\mathrm{4}\:,\:\:{h}=\sqrt{\mathrm{0}.\mathrm{6}×\mathrm{1}.\mathrm{4}}\:=\:\sqrt{\mathrm{0}.\mathrm{84}} \\ $$$${t}\:\approx\:\mathrm{0}.\mathrm{11039604}\left(\checkmark\right)\:,\:\mathrm{0}.\mathrm{89381445}\left(×\right) \\ $$$$\:\:\boldsymbol{{s}}\:\approx\:\mathrm{1}.\mathrm{13395334}\: \\ $$$$ \\ $$
Commented by ajfour last updated on 01/Dec/20
$${waiting}\:{for}\:{Your}\:{solution},\:{mrW} \\ $$$${Sir}.. \\ $$
Commented by mr W last updated on 01/Dec/20
$${i}\:{have}\:{no}\:{better}\:{way}\:{for}\:{solving}. \\ $$$${i}\:{guess}\:{solution}\:{exists}\:{only}\:{for} \\ $$$${special}\:{values}\:{of}\:\lambda. \\ $$