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Question Number 187808 by LowLevelLump last updated on 22/Feb/23

	Answered by a.lgnaoui last updated on 22/Feb/23
	$△ABC′ { ((AB=AC=BC=a)),((S_1 =((a^2 (√3))/2))) :} △ACB′ { ((AC=AB′=CB′=b)),((S_2 =((b^2 (√3))/2))) :} △BCC′ { ((BC=BC′=CC^′ =c)),((S_3 =((c^2 (√3))/2))) :} sin B=((AH)/(AB))=(1/3)⇒AH=asin B S_1 −S_2 +S_3 =(((a^2 −b^2 +c^2 )(√3))/2) =((√3)/2)(1) S(△ABC)=((c×((AH)/(AB)))/2)=((ac)/6)= (2) { ((a^2 −b^2 +c^2 =1)),((c=acos B+(√(b^2 −a^2 sin^2 B)))) :} c=((2a(√2))/3)+(1/3)(√( 9b^2 −a^2 )) { ((c^2 =b^2 −a^2 +1 (3))),((c=((2a(√2) +(√(9b^2 −a^2 )))/3) (4))) :} b^2 = (((9−16a^2 )^2 )/(288a^2 ))+(a^2 /9) (3)⇒c^2 =1+(((9−16a^2 )^2 )/(288a^2 ))−((8a^2 )/9) c^2 =((81)/(288a^2 )) ⇒ c=((9(√2))/(24a)) (5) a) AreaABC=((ac)/6)= ((3(√2))/(48)) Area(ABC)=((√2)/(16)) b)If sin Asin C=((√2)/3) ((sin A)/c)=((sin B)/b)=((sin C)/a)⇒asin A=csin C ((a(√2))/(3sin C))=csin C ⇒3csin^2 C=a(√2) sin^2 c=((a(√2))/(3c)); sin c=((AH)/b)=((asin B)/b)=(a/(3b))(sin B=(1/3)) =(a^2 /(9b^2 )) =((a(√2))/(3c)) b^2 =((ac(√2))/6)= (5) ⇒b^2 =((9(√2))/(24)) ×((√2)/6) ⇒ b=((√2)/4)$
	$△ {A B C}^{'} {\begin{cases} A B = A C = B C = a \\ S_{1} = \frac{a^{2} \sqrt{3}}{2} \end{cases}$ $△ {A C B}^{'} {\begin{cases} A C = {A B}^{'} = {C B}^{'} = b \\ S_{2} = \frac{b^{2} \sqrt{3}}{2} \end{cases}$ $△ {B C C}^{'} {\begin{cases} B C = {B C}^{'} = {C C}^{^{'}} = c \\ S_{3} = \frac{c^{2} \sqrt{3}}{2} \end{cases}$ $\sin B = \frac{A H}{A B} = \frac{1}{3} \Rightarrow A H = a \sin B$ $S_{1} - S_{2} + S_{3} = \frac{(a^{2} - b^{2} + c^{2}) \sqrt{3}}{2} = \frac{\sqrt{3}}{2} (1)$ $S (△ A B C) = \frac{c \times \frac{A H}{A B}}{2} = \frac{a c}{6} = (2)$ ${\begin{cases} a^{2} - b^{2} + c^{2} = 1 \\ c = a \cos B + \sqrt{b^{2} - a^{2} \sin^{2} B} \end{cases}$ $c = \frac{2 a \sqrt{2}}{3} + \frac{1}{3} \sqrt{9 b^{2} - a^{2}}$ ${\begin{cases} c^{2} = b^{2} - a^{2} + 1 (3) \\ c = \frac{2 a \sqrt{2} + \sqrt{9 b^{2} - a^{2}}}{3} (4) \end{cases}$ $b^{2} = \frac{{(9 - 16 a^{2})}^{2}}{288 a^{2}} + \frac{a^{2}}{9}$ $(3) \Rightarrow c^{2} = 1 + \frac{{(9 - 16 a^{2})}^{2}}{288 a^{2}} - \frac{8 a^{2}}{9}$ $c^{2} = \frac{81}{288 a^{2}} \Rightarrow c = \frac{9 \sqrt{2}}{24 a} (5)$ $a) A r e a A B C = \frac{a c}{6} = \frac{3 \sqrt{2}}{48}$ $A r e a (A B C) = \frac{\sqrt{2}}{16}$ $b) I f \sin A \sin C = \frac{\sqrt{2}}{3}$ $\frac{\sin A}{c} = \frac{\sin B}{b} = \frac{\sin C}{a} \Rightarrow a \sin A = c \sin C$ $\frac{a \sqrt{2}}{3 \sin C} = c \sin C \Rightarrow 3 c \sin^{2} C = a \sqrt{2}$ $\sin^{2} c = \frac{a \sqrt{2}}{3 c}; \sin c = \frac{A H}{b} = \frac{a \sin B}{b} = \frac{a}{3 b} (\sin B = \frac{1}{3})$ $= \frac{a^{2}}{9 b^{2}} = \frac{a \sqrt{2}}{3 c} b^{2} = \frac{a c \sqrt{2}}{6} =$ $(5) \Rightarrow b^{2} = \frac{9 \sqrt{2}}{24} \times \frac{\sqrt{2}}{6} \Rightarrow b = \frac{\sqrt{2}}{4}$
	Commented by a.lgnaoui last updated on 22/Feb/23

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