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Question Number 127633 by naka3546 last updated on 31/Dec/20

Commented by naka3546 last updated on 31/Dec/20

$A B C i s a n e q u i l a t e r a l t r i a n g l e .$ $F i n d a r e a o f A B C!$
Commented by mr W last updated on 31/Dec/20

$i g o t$ $s i d e l e n g t h s = 8 \sqrt{3}$ $a r e a = 48 \sqrt{3}$ $c o r r e c t ?$
Commented by naka3546 last updated on 01/Jan/21

$y e s, s i r . S h o w y o u r w o r k i n g s, p l e a s e .$
Commented by mr W last updated on 06/Jan/21

$i g o t a s i m p l e f o r m u l a$ $s = \sqrt{3} (R + r) + \sqrt{3 {(R + r)}^{2} - 8 R r}$ $w i t h R = 3, r = 2$ $\Rightarrow s = 8 \sqrt{3}$
	Answered by mr W last updated on 01/Jan/21

	Commented by mr W last updated on 01/Jan/21

	$M E T H O D I I$ $l e t ∠ B D C = 2 θ$ $B D = 3 \sqrt{3} + \frac{3}{\tan θ}$ $A D = 2 \sqrt{3} + \frac{2}{\tan (90 - θ)} = 2 \sqrt{3} + 2 \tan θ$ $A D + B D = s = 5 \sqrt{3} + 2 \tan θ + \frac{3}{\tan θ}$ $\frac{B C}{\sin 2 θ} = \frac{B D}{\sin (60 + 2 θ)}$ $\frac{\sin (60 + 2 θ)}{\sin 2 θ} = \frac{B D}{B C} = \frac{3 (\sqrt{3} + \frac{1}{\tan θ})}{s}$ $\frac{\sqrt{3}}{2 \tan 2 θ} + \frac{1}{2} = \frac{3 (\sqrt{3} + \frac{1}{\tan θ})}{5 \sqrt{3} + 2 \tan θ + \frac{3}{\tan θ}}$ $\frac{\sqrt{3}}{\frac{2 \tan θ}{1 - \tan^{2} θ}} + 1 = \frac{6 (\sqrt{3} + \frac{1}{\tan θ})}{5 \sqrt{3} + 2 \tan θ + \frac{3}{\tan θ}}$ $l e t t = \tan θ$ $\frac{\sqrt{3} (1 - t^{2}) + 2 t}{2 t} = \frac{6 (\sqrt{3} t + 1)}{5 \sqrt{3} t + 2 t^{2} + 3}$ $2 \sqrt{3} t^{4} + 11 t^{3} + 3 \sqrt{3} t^{2} - 9 t - 3 \sqrt{3} = 0$ $6 {(\frac{t}{\sqrt{3}})}^{4} + 11 {(\frac{t}{\sqrt{3}})}^{3} + 3 {(\frac{t}{\sqrt{3}})}^{2} - 3 (\frac{t}{\sqrt{3}}) - 1 = 0$ ${(\frac{t}{\sqrt{3}} + 1)}^{2} (\frac{2 t}{\sqrt{3}} - 1) (\frac{3 t}{\sqrt{3}} + 1) = 0$ $\Rightarrow \frac{2 t}{\sqrt{3}} - 1 = 0$ $\Rightarrow t = \tan θ = \frac{\sqrt{3}}{2}$ $\Rightarrow s = 5 \sqrt{3} + 2 \times \frac{\sqrt{3}}{2} + \frac{3}{\frac{\sqrt{3}}{2}} = 8 \sqrt{3}$
	Commented by naka3546 last updated on 01/Jan/21

	$ok, s i r . T h a n k y o u v e r y m u c h .$
	Commented by mr W last updated on 01/Jan/21

	$M E T H O D I$ $s = s i d e l e n g t h$ $B L = B H = \frac{3}{\tan 30 °} = 3 \sqrt{3}$ $A K = A G = \frac{2}{\tan 30 °} = 2 \sqrt{3}$ $H C = s - 3 \sqrt{3}$ $G C = s - 2 \sqrt{3}$ $L K = s - 3 \sqrt{3} - 2 \sqrt{3} = s - 5 \sqrt{3}$ $C F = \sqrt{{H C}^{2} + {H F}^{2}} = \sqrt{{(s - 3 \sqrt{3})}^{2} + 3^{2}}$ $C E = \sqrt{{G C}^{2} + {G E}^{2}} = \sqrt{{(s - 2 \sqrt{3})}^{2} + 2^{2}}$ $F E = \sqrt{{L K}^{2} + {(L F - L E)}^{2}} = \sqrt{{(s - 5 \sqrt{3})}^{2} + {(3 - 2)}^{2}}$ $∠ F C E = \frac{60 °}{2} = 30 °$ ${C E}^{2} + {C F}^{2} - 2 \times C E \times C F \times \cos ∠ F C E = {F E}^{2}$ ${(s - 2 \sqrt{3})}^{2} + 2^{2} + {(s - 3 \sqrt{3})}^{2} + 3^{2} - 2 \sqrt{[{(s - 2 \sqrt{3})}^{2} + 2^{2}] [{(s - 3 \sqrt{3})}^{2} + 3^{2}]} \cos 30 ° = {(3 - 2)}^{2} + {(s - 2 \sqrt{3} - 3 \sqrt{3})}^{2}$ $s^{2} - 4 \sqrt{3} s + 16 + s^{2} - 6 \sqrt{3} s + 36 - \sqrt{3 [s^{2} - 4 \sqrt{3} s + 16] [s^{2} - 6 \sqrt{3} s + 36]} = 1 + s^{2} - 10 \sqrt{3} s + 75$ $s^{2} - 24 = \sqrt{3 [s^{2} - 4 \sqrt{3} s + 16] [s^{2} - 6 \sqrt{3} s + 36]}$ $s^{4} - 48 s^{2} + 576 = 3 [s^{2} - 4 \sqrt{3} s + 16] [s^{2} - 6 \sqrt{3} s + 36]$ $s^{4} - 48 s^{2} + 576 = 3 s^{4} - 30 \sqrt{3} s^{3} + 372 s^{2} - 720 \sqrt{3} s + 1728$ $s^{4} - 15 \sqrt{3} s^{3} + 210 s^{2} - 360 \sqrt{3} s + 576 = 0$ ${(\frac{s}{\sqrt{3}})}^{4} - 15 {(\frac{s}{\sqrt{3}})}^{3} + 70 {(\frac{s}{\sqrt{3}})}^{2} - 120 (\frac{s}{\sqrt{3}}) + 64 = 0$ $(\frac{s}{\sqrt{3}} - 1) (\frac{s}{\sqrt{3}} - 2) (\frac{s}{\sqrt{3}} - 4) (\frac{s}{\sqrt{3}} - 8) = 0$ $s i n c e s m u s t b e > 2 \sqrt{3} + 3 \sqrt{3} = 5 \sqrt{3}$ $\Rightarrow \frac{s}{\sqrt{3}} = 8$ $\Rightarrow s = 8 \sqrt{3}$ $Δ_{A B C} = \frac{\sqrt{3} s^{2}}{4} = \frac{\sqrt{3} \times {(8 \sqrt{3})}^{2}}{4} = 48 \sqrt{3}$
	Commented by behi83417@gmail.com last updated on 02/Jan/21
	$BH=(3/(tg30))=3(√3)⇒F(3(√3),3) ⇒(x−3(√3))^2 +(y−3)^2 =9 CD:y=m(x−a) (x−3(√3))^2 +[m(x−a)−3]^2 =9 ⇒x^2 +27−6(√3)x+m^2 x^2 −2m^2 ax+m^2 a^2 + −6mx+6ma=0 ⇒(1+m^2 )x^2 −2(m^2 a+3(√3)+3m)x+ +m^2 a^2 +6ma+27=0 △′=(m^2 a+3(√3)+3m)^2 −(1+m^2 )(m^2 a^2 +6ma+27)= =m^4 a^2 +27+9m^2 +6(√3)m^2 a+6m^3 a+18(√3)m− −(m^2 a^2 +6ma+27+m^4 a^2 +6m^3 a+27m^2 )= =−18m^2 +6(√3)m^2 a+18(√3)m−6ma−m^2 a^2 =0 ⇒−m^2 (18−6(√3)a+a^2 )−6m(a−3(√3))=0 ⇒^(m≠0) { ((a−3(√3)=0⇒a=3(√3)[not ok])),((a^2 −6(√3)a+18=0⇒a=3((√3)+1))) :} sir:mrW! what is wrong?$
	$BH = \frac{3}{tg 30} = 3 \sqrt{3} \Rightarrow F (3 \sqrt{3}, 3)$ $\Rightarrow {(x - 3 \sqrt{3})}^{2} + {(y - 3)}^{2} = 9$ $CD : y = m (x - a)$ ${(x - 3 \sqrt{3})}^{2} + {[m (x - a) - 3]}^{2} = 9$ $\Rightarrow x^{2} + 27 - 6 \sqrt{3} x + m^{2} x^{2} - {2 m}^{2} ax + m^{2} a^{2} +$ $- 6 mx + 6 ma = 0$ $\Rightarrow (1 + m^{2}) x^{2} - 2 (m^{2} a + 3 \sqrt{3} + 3 m) x +$ $+ m^{2} a^{2} + 6 ma + 27 = 0$ $△^{'} = {(m^{2} a + 3 \sqrt{3} + 3 m)}^{2} - (1 + m^{2}) (m^{2} a^{2} + 6 ma + 27) =$ $= m^{4} a^{2} + 27 + {9 m}^{2} + 6 \sqrt{3} m^{2} a + {6 m}^{3} a + 18 \sqrt{3} m -$ $- (m^{2} a^{2} + 6 ma + 27 + m^{4} a^{2} + {6 m}^{3} a + {27 m}^{2}) =$ $= - {18 m}^{2} + 6 \sqrt{3} m^{2} a + 18 \sqrt{3} m - 6 ma - m^{2} a^{2} = 0$ $\Rightarrow - m^{2} (18 - 6 \sqrt{3} a + a^{2}) - 6 m (a - 3 \sqrt{3}) = 0$ $\overset{m \neq 0}{\Rightarrow} {\begin{cases} a - 3 \sqrt{3} = 0 \Rightarrow a = 3 \sqrt{3} [not ok] \\ a^{2} - 6 \sqrt{3} a + 18 = 0 \Rightarrow a = 3 (\sqrt{3} + 1) \end{cases}$ $sir : mrW! what is wrong ?$
	Commented by mr W last updated on 02/Jan/21

	$f r o m t h e c o n d i t i o n$ $\Rightarrow - m^{2} (18 - 6 \sqrt{3} a + a^{2}) - 6 m (a - 3 \sqrt{3}) = 0$ $y o u c a n o n l y k n o w t h e r e l a t i o n s h i p$ $b e t w e e n m a n d a . y o u c a n n o t$ $d e t e r m i n e m a n d a f r o m t h i s o n l y$ $c o n d i t i o n! y o u n e e d a n o t h e r$ $c o n d i t i o n, t h a t i s t h e o t h e r s m a l l$ $c i r c l e .$
	Answered by behi83417@gmail.com last updated on 02/Jan/21
	$s_1 =p_1 .r_1 =((x+a+z)/2).2,AB=a,AD=x,CD=z s_2 =p_2 .r_2 =((y+a+z)/2).3,BD=y s=((a^2 (√3))/4)=((2(x+a+z)+3(y+a+z))/2)= =((7a+5z+y)/2)⇒a^2 (√3)=2(7a+5z+y) a.(z^2 +xy)=a^2 .x+a^2 .y=a^2 (x+y)=a^3 ⇒ { ((a^2 (√3)=2(7a+5z+y))),((z^2 +xy=a^2 ⇒z^2 +(a−y)y=a^2 )) :} ⇒ { ((a^2 (√3)−2(7a+y)=10z)),((z^2 =a^2 −ay+y^2 )) :} ⇒[a^2 (√3)−2(7a+y)]^2 =100(a^2 −ay+y^2 ) ⇒3a^4 −4a^2 (√3)(7a+y)+4(7a+y)^2 = =100a^2 −100ay+100y^2 ⇒3a^4 −28a^3 (√3)−4a^2 y(√3)+196a^2 +56ay+4y^2 = =100a^2 −100ay+100y^2 ⇒96y^2 −156ay+4a^2 y(√3)+28a^3 (√3)−3a^4 −96a^2 =0 ⇒96y^2 −2a(78−2a(√3))y−a^2 (3a^2 −28a(√3)+96)=0 △′=a^2 .(78−2a(√3))^2 +96a^2 (3a^2 −28a(√3)+96)= =a^2 [6084−312a(√3)+12a^2 +288a^2 −2688a(√3)+9216]= =a^2 [300a^2 −3000a(√3)+15300]= =300a^2 [a^2 −10a(√3)+51] ⇒y=((a(39−a(√3))±5a(√3).(√(a^2 −10a(√3)+51)))/(48))$
	$s_{1} = p_{1} . r_{1} = \frac{x + a + z}{2} . 2, AB = a, AD = x, CD = z$ $s_{2} = p_{2} . r_{2} = \frac{y + a + z}{2} . 3, BD = y$ $s = \frac{a^{2} \sqrt{3}}{4} = \frac{2 (x + a + z) + 3 (y + a + z)}{2} =$ $= \frac{7 a + 5 z + y}{2} \Rightarrow a^{2} \sqrt{3} = 2 (7 a + 5 z + y)$ $a . (z^{2} + xy) = a^{2} . x + a^{2} . y = a^{2} (x + y) = a^{3}$ $\Rightarrow {\begin{cases} a^{2} \sqrt{3} = 2 (7 a + 5 z + y) \\ z^{2} + xy = a^{2} \Rightarrow z^{2} + (a - y) y = a^{2} \end{cases}$ $\Rightarrow {\begin{cases} a^{2} \sqrt{3} - 2 (7 a + y) = 10 z \\ z^{2} = a^{2} - ay + y^{2} \end{cases}$ $\Rightarrow {[a^{2} \sqrt{3} - 2 (7 a + y)]}^{2} = 100 (a^{2} - ay + y^{2})$ $\Rightarrow {3 a}^{4} - {4 a}^{2} \sqrt{3} (7 a + y) + 4 {(7 a + y)}^{2} =$ $= {100 a}^{2} - 100 ay + {100 y}^{2}$ $\Rightarrow {3 a}^{4} - {28 a}^{3} \sqrt{3} - {4 a}^{2} y \sqrt{3} + {196 a}^{2} + 56 ay + {4 y}^{2} =$ $= {100 a}^{2} - 100 ay + {100 y}^{2}$ $\Rightarrow {96 y}^{2} - 156 ay + {4 a}^{2} y \sqrt{3} + {28 a}^{3} \sqrt{3} - {3 a}^{4} - {96 a}^{2} = 0$ $\Rightarrow {96 y}^{2} - 2 a (78 - 2 a \sqrt{3}) y - a^{2} ({3 a}^{2} - 28 a \sqrt{3} + 96) = 0$ $△^{'} = a^{2} . {(78 - 2 a \sqrt{3})}^{2} + {96 a}^{2} ({3 a}^{2} - 28 a \sqrt{3} + 96) =$ $= a^{2} [6084 - 312 a \sqrt{3} + {12 a}^{2} + {288 a}^{2} - 2688 a \sqrt{3} + 9216] =$ $= a^{2} [{300 a}^{2} - 3000 a \sqrt{3} + 15300] =$ $= {300 a}^{2} [a^{2} - 10 a \sqrt{3} + 51]$ $\Rightarrow y = \frac{a (39 - a \sqrt{3}) \pm 5 a \sqrt{3} . \sqrt{a^{2} - 10 a \sqrt{3} + 51}}{48}$
	Commented by behi83417@gmail.com last updated on 02/Jan/21

	$sir : mrW! what is wrong there ?$
	Commented by behi83417@gmail.com last updated on 02/Jan/21

	$thanks a lot for your time mr . proph .$
	Commented by mr W last updated on 02/Jan/21

	$i {c a n}^{'} t f o l l o w y o u r i d e a . s i n c e y o u$ ${d i d n}^{'} t c o m e t o a f i n a l r e s u l t, i {c a n}^{'} t$ $s a y i f {i t}^{'} s c o r r e c t o r n o t .$
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