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Question Number 187557 by Rupesh123 last updated on 18/Feb/23

	Answered by HeferH last updated on 19/Feb/23

	Commented by HeferH last updated on 19/Feb/23

	$C i s c e n t r o i d .$ $C E = 9$ $\frac{B A}{2 a} = \frac{18}{27}$ $B A = \frac{4 a}{3}; D B = 2 a - \frac{4 a}{3} = \frac{2 a}{3}$ $\frac{x}{b} = \frac{\frac{4 a}{3}}{2 a}; x = \frac{2 b}{3}$ $\frac{D E}{D B} = \frac{E A}{B A}$ $b \cdot \frac{3}{2 a} = 27 \cdot \frac{3}{4 a}; b = \frac{27}{2}$ $x = \frac{2}{3} \cdot \frac{27}{2} = 9$
	Answered by a.lgnaoui last updated on 19/Feb/23
	$△ABC et △ADE semblables ((AB)/(AD))=((AC)/(AE))=((BC)/(DE)) 1•((AB)/(AD))=((BC)/(DE)) ⇒AB=((BC×AD)/(DE)) AB=((2a)/b)BC △ACB BC∣∣DE ⇒∡DEC=∡BCA=2α AB^2 =AC^2 +BC^2 −2AC×BCcos 2α ((4a^2 )/b^2 )BC^2 =18^2 +BC^2 −36BCcos 2α (((4a^2 −b^2 )/b^2 ))BC^2 +36cos 2α)BC−18^2 =0 BC^2 +((b^2 ×36cos 2α))/(4a^2 −b^2 ))BC−((18^2 b^2 )/(4a^2 −b^2 ))=0 (BC+((18b^2 cos 2α)/(4a^2 −b^2 )))^2 −(((18^2 b^4 cos^2 2α+18^2 b^2 (4a^2 −b^2 ))/((4a^2 −b^2 )^2 )) =0 (BC+((18b^2 cos 2α)/(4a^2 −b^2 )))^2 −[((18^2 ×b^4 (4a^2 −sin^2 2α))/((4a^2 −b^2 )^2 ))]=0 BC+((18b^2 cos 2α)/(4a^2 −b^2 )) ±((18b^2 (√(4a^2 −sin^2 2α )))/(4a^2 −b^2 )) { ((BC=[((18b^2 )/(4a^2 −b^2 ))(cos 2α+(√(4a^2 −sin ^2 2α)) )])),((BC=[((18b^2 )/(4a^2 −b^2 ))(cos 2α−(√(4a^2 −sin ^2 2α)))])) :}$
	$△ A B C e t △ A D E s e m b l a b l e s$ $\frac{A B}{A D} = \frac{A C}{A E} = \frac{B C}{D E}$ $1 ∙ \frac{A B}{A D} = \frac{B C}{D E} \Rightarrow A B = \frac{B C \times A D}{D E}$ $A B = \frac{2 a}{b} B C$ $△ A C B B C ∣∣ D E \Rightarrow ∡ D E C = ∡ B C A = 2 α$ ${A B}^{2} = {A C}^{2} + {B C}^{2} - 2 A C \times B C \cos 2 α$ $\frac{4 a^{2}}{b^{2}} {B C}^{2} = 18^{2} + {B C}^{2} - 36 B C \cos 2 α$ $(\frac{4 a^{2} - b^{2}}{b^{2}}) {B C}^{2} + 36 \cos 2 α) B C - 18^{2} = 0$ ${B C}^{2} + \frac{b^{2} \times 36 \cos 2 α)}{4 a^{2} - b^{2}} B C - \frac{18^{2} b^{2}}{4 a^{2} - b^{2}} = 0$ ${(B C + \frac{18 b^{2} \cos 2 α}{4 a^{2} - b^{2}})}^{2} - (\frac{18^{2} b^{4} \cos^{2} 2 α + 18^{2} b^{2} (4 a^{2} - b^{2})}{{(4 a^{2} - b^{2})}^{2}} = 0$ ${(B C + \frac{18 b^{2} \cos 2 α}{4 a^{2} - b^{2}})}^{2} - [\frac{18^{2} \times b^{4} (4 a^{2} - \sin^{2} 2 α)}{{(4 a^{2} - b^{2})}^{2}}] = 0$ $B C + \frac{18 b^{2} \cos 2 α}{4 a^{2} - b^{2}} \pm \frac{18 b^{2} \sqrt{4 a^{2} - \sin^{2} 2 α}}{4 a^{2} - b^{2}}$ ${\begin{cases} B C = [\frac{18 b^{2}}{4 a^{2} - b^{2}} (\cos 2 α + \sqrt{4 a^{2} - \sin^{2} 2 α})] \\ B C = [\frac{18 b^{2}}{4 a^{2} - b^{2}} (\cos 2 α - \sqrt{4 a^{2} - \sin^{2} 2 α})] \end{cases}$
	Commented by mr W last updated on 19/Feb/23

	$y o u s e e m n o t t o u n d e r s t a n d t h e$ $q u e s t i o n c o r r e c t l y . a, b, α j u s t m e a n$ $t h a t t h e c o r r e s p o n d i n g l i n e s e g m e n t s$ $a r e e q u a l a n d t h e c o r r e s p o n d i n g$ $a n g l e s a r e e q u a l, t h e y a r e n o t k n o w n$ $v a l u e s . t h e o n l y v a l u e g i v e n i s$ $A C = 18 .$
	Commented by a.lgnaoui last updated on 19/Feb/23

	$b u t t h i s f o r m u l a g e t s a n y v a l u e e$ $f o r B C c o r r e s p o n d t o a, b, a n d α$ $w i t h e q u a l t e r m e \frac{18 b^{2}}{4 a^{2} - b^{2}} (\cos 2 α \pm \sqrt{(4 a^{2} - \sin^{2} 2 α)}$
	Commented by mr W last updated on 19/Feb/23

	$b u t a, b a n d α c a n n o t b e a n y v a l u e s!$ $t h e y a r e n o t i n d e p e n d e n t . t h e y h a v e$ $a c t u a l l y f i x e d v a l u e s t h r o u g h t h e$ $v a l u e o f A C = 18 .$
	Answered by ajfour last updated on 19/Feb/23

	Commented by ajfour last updated on 19/Feb/23

	$2 s = 18$ $3 s = 27$ $\frac{3 s}{b} = \frac{a + p}{a - p} = \frac{2}{1} . . . (i)$ $\Rightarrow b = \frac{3 s}{2} = \frac{27}{2} & \frac{p}{a} = \frac{2 - 1}{2 + 1} = \frac{1}{3}$ $\frac{x}{b} = \frac{a + p}{2 a}$ $\Rightarrow x = \frac{b}{2} (1 + \frac{p}{a}) = \frac{27}{4} (\frac{4}{3})$ $x = 9$
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