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Question Number 200568 by Rupesh123 last updated on 20/Nov/23

Commented by Frix last updated on 20/Nov/23

$Let α = ∠ c d$ $0 < f < \infty \Leftrightarrow 0 < α < \frac{2 π}{5}$
Commented by Rupesh123 last updated on 20/Nov/23
Nice one, sir!
	Answered by a.lgnaoui last updated on 20/Nov/23
	$△ABC BC^2 =AB^2 +AC^2 ( e+d)^2 =(((1+(√5) )^2 )/4)+(c+f)^2 ⇒ (d+e)^2 −(c+f)^2 =((3+(√5))/2) (1) △CDE CD^2 =DE^2 +CE^2 c^2 −d^2 =(1/4) (2) △ABC / CDE Semblables: ((AC)/(BC))=((EC)/(DC)) ((c+f)/(d+e)) =(d/c) ⇒ (((c+f)^2 )/((d+e)^2 ))=(d^2 /c^2 )=(d^2 /(d^2 +(1/4)))(3) ((1+(√5))/((c+f)))=(1/d) d=((c+f)/(1+(√5))) (4) ⇒c+f=(1+(√5) )d (d+e)^2 =(1+(√5) )^2 d^2 +((3+(√5))/2) (2) (1+(1/(4d^2 )))(c+f)^2 −(1+(√5) )^2 d^2 =((3+(√5))/2) ⇒(c+f)^2 = ((4(6+2(√5) )d^4 +(6+2(√5) )d^2 )/(4d^2 +1)) f^2 +(((1+(√5))/2))^2 =e^2 +(1/4) e^2 −f^2 =((5+2(√5))/4) f=(√(e^2 −((5+2(√5))/4))) (5) (1)⇒(d+e)^2 =(c+(√(e^2 −((5+2(√5))/4))) )^2 +((3+(√5))/2) d^2 +e^2 +2de=c^2 +e^2 −((5+2(√5))/4)+((3+(√5))/2) +2c((√(e^2 −((5+2(√5))/4))) ) [ 2de−c(√(4e^2 −5−2(√5) )) =(1/2) 4de=(√((4d^2 +1)(4e^2 −5−2(√5) ))) =1 (i) (d+e)=(√(2(3+(√5))d^2 +((3+(√5))/2))) (ii) 2d+2e=2(√((1/2)(3+(√5) )×(2d)^2 +((3+(√5))/2))) 2d=x 2e=y (ii) x+y=2(√((1/2)(3+(√5) )(1+x^2 ))) (i) xy =(√((x^2 +1)(y^2 −5−2(√5) )) )−1 ⇒(x+y)^2 = 2(3+(√5) )(1+x^2 ) (xy+1)^2 =( x^2 +1)(y^2 −5−2(√5) ) { ((x^2 +y^2 +2xy=2(3+(√5) )x^2 +2(3+(√5) ))),((x^2 y^2 +2xy+1=x^2 y^2 −(5+2(√5) )x^2 )) :} +y^2 −(5+2(√5) (5+2(√5) )x^2 +y^2 +2xy=2(3+(√5) ) (I) 2xy =y^2 −(5+2(√5) )x^2 −(6+2(√5) ) (II) (I)−(II)⇒ (5+2(√5) )x^2 +y^2 =2(3+(√5) )−y^2 + (5+2(√5) )x^2 +(6+2(√5) ) ⇒ y=(√(6+2(√5))) alors e=((√(2(3+(√5) )))/2) (ii) x=2(√((1/2)(3+(√5) )(1+x^2 ))) −(√(6+2(√5))) ⇒ 2(3+(√5) )(1+x^2 )=(x+(√(6+2(√5)[)) )^2 (6+2(√5) )x^2 +6+2(√5) =x^2 +6+2(√5) + 2x(√(6+2(√5))) (5+2(√5) )x^2 −2x(√(6+2(√5) )) =0 x=((2(√(6+2(√5))))/(5+2(√5))) ⇒ d=((√(6+2(√5)))/(5+2(√5))) { ((d=((√(6+2(√5)))/(5+2(√5))))),((e=((√(6+2(√5)))/2))) :} calcul de f (5)⇒ f=(√(e^2 −((5+2(√5))/4))) =(√(((6+2(√5))/4)−((5+2(√5))/4))) ⇒ f=(1/2)$
	$△ ABC {BC}^{2} = {AB}^{2} + {AC}^{2}$ ${(e + d)}^{2} = \frac{{(1 + \sqrt{5})}^{2}}{4} + {(c + f)}^{2}$ $\Rightarrow {(d + e)}^{2} - {(c + f)}^{2} = \frac{3 + \sqrt{5}}{2} (1)$ $△ CDE {CD}^{2} = {DE}^{2} + {CE}^{2}$ $c^{2} - d^{2} = \frac{1}{4} (2)$ $△ ABC / CDE Semblables : \frac{AC}{BC} = \frac{EC}{DC}$ $\frac{c + f}{d + e} = \frac{d}{c} \Rightarrow \frac{{(c + f)}^{2}}{{(d + e)}^{2}} = \frac{d^{2}}{c^{2}} = \frac{d^{2}}{d^{2} + \frac{1}{4}} (3)$ $\frac{1 + \sqrt{5}}{(c + f)} = \frac{1}{d} d = \frac{c + f}{1 + \sqrt{5}} (4)$ $\Rightarrow c + f = (1 + \sqrt{5}) d$ ${(d + e)}^{2} = {(1 + \sqrt{5})}^{2} d^{2} + \frac{3 + \sqrt{5}}{2} (2)$ $(1 + \frac{1}{4 d^{2}}) {(c + f)}^{2} - {(1 + \sqrt{5})}^{2} d^{2} = \frac{3 + \sqrt{5}}{2}$ $\Rightarrow {(c + f)}^{2} = \frac{4 (6 + 2 \sqrt{5}) d^{4} + (6 + 2 \sqrt{5}) d^{2}}{4 d^{2} + 1}$ $f^{2} + {(\frac{1 + \sqrt{5}}{2})}^{2} = e^{2} + \frac{1}{4}$ $e^{2} - f^{2} = \frac{5 + 2 \sqrt{5}}{4}$ $f = \sqrt{e^{2} - \frac{5 + 2 \sqrt{5}}{4}} (5)$ $(1) \Rightarrow {(d + e)}^{2} = {(c + \sqrt{e^{2} - \frac{5 + 2 \sqrt{5}}{4}})}^{2} + \frac{3 + \sqrt{5}}{2}$ $d^{2} + e^{2} + 2 de = c^{2} + e^{2} - \frac{5 + 2 \sqrt{5}}{4} + \frac{3 + \sqrt{5}}{2}$ $+ 2 c (\sqrt{e^{2} - \frac{5 + 2 \sqrt{5}}{4}})$ $[2 de - c \sqrt{4 e^{2} - 5 - 2 \sqrt{5}} = \frac{1}{2}$ $4 de = \sqrt{(4 d^{2} + 1) (4 e^{2} - 5 - 2 \sqrt{5})} = 1 (i)$ $(d + e) = \sqrt{2 (3 + \sqrt{5}) d^{2} + \frac{3 + \sqrt{5}}{2}} (ii)$ $2 d + 2 e = 2 \sqrt{\frac{1}{2} (3 + \sqrt{5}) \times {(2 d)}^{2} + \frac{3 + \sqrt{5}}{2}}$ $2 d = x 2 e = y$ $(ii) x + y = 2 \sqrt{\frac{1}{2} (3 + \sqrt{5}) (1 + x^{2})}$ $(i) xy = \sqrt{(x^{2} + 1) (y^{2} - 5 - 2 \sqrt{5}}) - 1$ $\Rightarrow {(x + y)}^{2} = 2 (3 + \sqrt{5}) (1 + x^{2})$ ${(xy + 1)}^{2} = (x^{2} + 1) (y^{2} - 5 - 2 \sqrt{5})$ ${\begin{cases} x^{2} + y^{2} + 2 xy = 2 (3 + \sqrt{5}) x^{2} + 2 (3 + \sqrt{5}) \\ x^{2} y^{2} + 2 xy + 1 = x^{2} y^{2} - (5 + 2 \sqrt{5}) x^{2} \end{cases}$ $+ y^{2} - (5 + 2 \sqrt{5}$ $(5 + 2 \sqrt{5}) x^{2} + y^{2} + 2 xy = 2 (3 + \sqrt{5}) (I)$ $2 xy = y^{2} - (5 + 2 \sqrt{5}) x^{2} - (6 + 2 \sqrt{5}) (II)$ $(I) - (II) \Rightarrow$ $(5 + 2 \sqrt{5}) x^{2} + y^{2} = 2 (3 + \sqrt{5}) - y^{2} +$ $(5 + 2 \sqrt{5}) x^{2} + (6 + 2 \sqrt{5})$ $\Rightarrow y = \sqrt{6 + 2 \sqrt{5}}$ $alors e = \frac{\sqrt{2 (3 + \sqrt{5})}}{2}$ $(ii) x = 2 \sqrt{\frac{1}{2} (3 + \sqrt{5}) (1 + x^{2})} - \sqrt{6 + 2 \sqrt{5}}$ $\Rightarrow 2 (3 + \sqrt{5}) (1 + x^{2}) = {(x + \sqrt{6 + 2 \sqrt{5} [})}^{2}$ $(6 + 2 \sqrt{5}) x^{2} + 6 + 2 \sqrt{5} = x^{2} + 6 + 2 \sqrt{5} +$ $2 x \sqrt{6 + 2 \sqrt{5}}$ $(5 + 2 \sqrt{5}) x^{2} - 2 x \sqrt{6 + 2 \sqrt{5}} = 0$ $x = \frac{2 \sqrt{6 + 2 \sqrt{5}}}{5 + 2 \sqrt{5}} \Rightarrow d = \frac{\sqrt{6 + 2 \sqrt{5}}}{5 + 2 \sqrt{5}}$ ${\begin{cases} d = \frac{\sqrt{6 + 2 \sqrt{5}}}{5 + 2 \sqrt{5}} \\ e = \frac{\sqrt{6 + 2 \sqrt{5}}}{2} \end{cases}$ $calcul de f$ $(5) \Rightarrow f = \sqrt{e^{2} - \frac{5 + 2 \sqrt{5}}{4}}$ $= \sqrt{\frac{6 + 2 \sqrt{5}}{4} - \frac{5 + 2 \sqrt{5}}{4}}$ $\Rightarrow f = \frac{1}{2}$
	Commented by a.lgnaoui last updated on 20/Nov/23

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