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Question Number 121693 by mathace last updated on 11/Nov/20

Commented by prakash jain last updated on 12/Nov/20


Commented by mindispower last updated on 11/Nov/20

$n i c e s i r$
	Answered by mindispower last updated on 11/Nov/20

	$i n d i c a t i o n$ $1 \Rightarrow a^{3} - b^{3} = 9 (a - b)$ $2 \Rightarrow b^{3} - c^{3} = 52 (b - c)$ $3 \Leftrightarrow c^{3} - a^{3} = 49 (c - a)$ $1 + 2 + 3 \Rightarrow 0 = - 40 a + 43 b - 3 c$
	Answered by mathace last updated on 12/Nov/20
	$The system of equation is a^2 +ab+b^2 =9 (1) b^2 +bc+c^2 =52 (2) c^2 +ac+a^2 =49 (3) The system implies to a^2 +b^2 −((√9))^2 =−ab as a triangle where ((a^2 +b^2 −((√9))^2 )/(2ab))=−(1/2)=cos (120°) (4) similarly, ((b^2 +c^2 −((√(52)))^2 )/(2bc)) =−(1/2)=cos (120°) (5) ((c^2 +a^2 −((√(49)))^2 )/(2ca))=−(1/2)=cos (120°) (6) sum of area of three triangles △OAB,△OBC,△OCA is (1/2)(ab+bc+ca)sin (120°) From the whole triangle △ABC we have, cos (θ)=((7^2 +3^2 −((√(52)))^2 )/(2×7×3))=(1/7) implies, sin (θ)=±((√(48))/7) (7) Considering the positve value of sin (θ) we find the area of the whole triangle △ABC is (1/2)×7×3×sin (θ)=(1/2)×7×3×((√(48))/7)=(3/2)(√(48)) As sum of tringles △OAB, △OBC, △OCA is equal to the area of the whole triangle△ABC, thus we have (1/2)(ab+bc+ca)sin (120°)=(3/2)(√(48)) or, ab+bc+ca=24 Now, (1)+(2)+(3) implies 2(a^2 +b^2 +c^2 )+(ab+bc+ca)=9+52+49=110 2(a^2 +b^2 +c^2 )=110−24=86 a^2 +b^2 +c^2 =43 We know, (a+b+c)^2 =(a^2 +b^2 +c^2 )+2(ab+bc+ca) (a+b+c)^2 =43+48=91 a+b+c=±(√(91)) Now Applying the oprations (2)−(1), (2)−(3) and (3)−(1) we have, (c−a)(a+b+c)=43 (8) (b−a)(a+b+c)=3 (9) (c−b)(a+b+c)=40 (10) Hence, our quadratic system has been transformed into equivalent linear system. Thus we have, c−a=±((43)/( (√(91)))) b−a=±(3/( (√(91)))) c−b=±((40)/( (√(91)))) a+b+c=±(√(91)) Solving the system we have, a=±((15)/( (√(91)))) b=±((18)/( (√(91)))) c=±((58)/( (√(91)))) Again from equation (7), we have sin (θ)=−((√(48))/7) [Taking(−ve)value] Now the area of the whole triangle △ABC is (1/2)×7×3×sin (θ)=(1/2)×7×3×(−((√(48))/7))=−(3/2)(√(48)) As sum of tringles △OAB, △OBC, △OCA is equal to the area of the whole triangle△ABC, thus we have (1/2)(ab+bc+ca)sin (120°)=−(3/2)(√(48)) or, ab+bc+ca=−24 Now, (1)+(2)+(3) implies 2(a^2 +b^2 +c^2 )+(ab+bc+ca)=9+52+49=110 2(a^2 +b^2 +c^2 )=110+24=134 a^2 +b^2 +c^2 =67 We know, (a+b+c)^2 =(a^2 +b^2 +c^2 )+2(ab+bc+ca) (a+b+c)^2 =67−48=19 a+b+c=±(√(19)) Now recalling the equation (8),(9),(10) we have (c−a)(a+b+c)=43 (b−a)(a+b+c)=3 (c−b)(a+b+c)=40 Hence, our quadratic system has been transformed again into equivalent linear system. Thus we have, c−a=±((43)/( (√(19)))) b−a=±(3/( (√(19)))) c−b=±((40)/( (√(19)))) a+b+c=±(√(19)) Solving the system we have, a=∓(9/( (√(19)))) b=∓(6/( (√(19)))) c=±((34)/( (√(19)))) Therefore our obtained solution set S={(a,b,c):(((15)/( (√(91)))),((18)/( (√(91)))),((58)/( (√(91))))),(((−15)/( (√(91)))),((−18)/( (√(91)))),((−58)/( (√(91))))),(((−9)/( (√(19)))),((−6)/( (√(19)))),((34)/( (√(19))))),((9/( (√(19)))),(6/( (√(19)))),((−34)/( (√(19)))))} [Omar bin Abdul Aziz]$
	$T h e s y s t e m o f e q u a t i o n i s$ $a^{2} + a b + b^{2} = 9 (1)$ $b^{2} + b c + c^{2} = 52 (2)$ $c^{2} + a c + a^{2} = 49 (3)$ $T h e s y s t e m i m p l i e s t o$ $a^{2} + b^{2} - {(\sqrt{9})}^{2} = - a b a s a t r i a n g l e w h e r e$ $\frac{a^{2} + b^{2} - {(\sqrt{9})}^{2}}{2 a b} = - \frac{1}{2} = \cos (120 °) (4)$ $s i m i l a r l y,$ $\frac{b^{2} + c^{2} - {(\sqrt{52})}^{2}}{2 b c} = - \frac{1}{2} = \cos (120 °) (5)$ $\frac{c^{2} + a^{2} - {(\sqrt{49})}^{2}}{2 c a} = - \frac{1}{2} = \cos (120 °) (6)$ $s u m o f a r e a o f t h r e e t r i a n g l e s △ O A B, △ O B C, △ O C A i s$ $\frac{1}{2} (a b + b c + c a) \sin (120 °)$ $F r o m t h e w h o l e t r i a n g l e △ A B C w e h a v e,$ $\cos (θ) = \frac{7^{2} + 3^{2} - {(\sqrt{52})}^{2}}{2 \times 7 \times 3} = \frac{1}{7}$ $i m p l i e s, \sin (θ) = \pm \frac{\sqrt{48}}{7} (7)$ $C o n s i d e r i n g t h e p o s i t v e v a l u e o f \sin (θ) w e f i n d$ $t h e a r e a o f t h e w h o l e t r i a n g l e △ A B C i s$ $\frac{1}{2} \times 7 \times 3 \times \sin (θ) = \frac{1}{2} \times 7 \times 3 \times \frac{\sqrt{48}}{7} = \frac{3}{2} \sqrt{48}$ $A s s u m o f t r i n g l e s △ O A B, △ O B C, △ O C A i s e q u a l t o t h e$ $a r e a o f t h e w h o l e t r i a n g l e △ A B C, t h u s w e h a v e$ $\frac{1}{2} (a b + b c + c a) \sin (120 °) = \frac{3}{2} \sqrt{48} o r,$ $a b + b c + c a = 24$ $N o w, (1) + (2) + (3) i m p l i e s$ $2 (a^{2} + b^{2} + c^{2}) + (a b + b c + c a) = 9 + 52 + 49 = 110$ $2 (a^{2} + b^{2} + c^{2}) = 110 - 24 = 86$ $a^{2} + b^{2} + c^{2} = 43$ $W e k n o w,$ ${(a + b + c)}^{2} = (a^{2} + b^{2} + c^{2}) + 2 (a b + b c + c a)$ ${(a + b + c)}^{2} = 43 + 48 = 91$ $a + b + c = \pm \sqrt{91}$ $N o w A p p l y i n g t h e o p r a t i o n s (2) - (1),$ $(2) - (3) a n d (3) - (1) w e h a v e,$ $(c - a) (a + b + c) = 43 (8)$ $(b - a) (a + b + c) = 3 (9)$ $(c - b) (a + b + c) = 40 (10)$ $H e n c e, o u r q u a d r a t i c s y s t e m h a s b e e n$ $t r a n s f o r m e d i n t o e q u i v a l e n t l i n e a r s y s t e m .$ $T h u s w e h a v e,$ $c - a = \pm \frac{43}{\sqrt{91}}$ $b - a = \pm \frac{3}{\sqrt{91}}$ $c - b = \pm \frac{40}{\sqrt{91}}$ $a + b + c = \pm \sqrt{91}$ $S o l v i n g t h e s y s t e m w e h a v e,$ $a = \pm \frac{15}{\sqrt{91}}$ $b = \pm \frac{18}{\sqrt{91}}$ $c = \pm \frac{58}{\sqrt{91}}$ $A g a i n f r o m e q u a t i o n (7), w e h a v e$ $\sin (θ) = - \frac{\sqrt{48}}{7} [T a k i n g (- v e) v a l u e]$ $N o w t h e a r e a o f t h e w h o l e t r i a n g l e △ A B C i s$ $\frac{1}{2} \times 7 \times 3 \times \sin (θ) = \frac{1}{2} \times 7 \times 3 \times (- \frac{\sqrt{48}}{7}) = - \frac{3}{2} \sqrt{48}$ $A s s u m o f t r i n g l e s △ O A B, △ O B C, △ O C A i s e q u a l t o t h e$ $a r e a o f t h e w h o l e t r i a n g l e △ A B C, t h u s w e h a v e$ $\frac{1}{2} (a b + b c + c a) \sin (120 °) = - \frac{3}{2} \sqrt{48} o r,$ $a b + b c + c a = - 24$ $N o w, (1) + (2) + (3) i m p l i e s$ $2 (a^{2} + b^{2} + c^{2}) + (a b + b c + c a) = 9 + 52 + 49 = 110$ $2 (a^{2} + b^{2} + c^{2}) = 110 + 24 = 134$ $a^{2} + b^{2} + c^{2} = 67$ $W e k n o w,$ ${(a + b + c)}^{2} = (a^{2} + b^{2} + c^{2}) + 2 (a b + b c + c a)$ ${(a + b + c)}^{2} = 67 - 48 = 19$ $a + b + c = \pm \sqrt{19}$ $N o w r e c a l l i n g t h e e q u a t i o n (8), (9), (10) w e h a v e$ $(c - a) (a + b + c) = 43$ $(b - a) (a + b + c) = 3$ $(c - b) (a + b + c) = 40$ $H e n c e, o u r q u a d r a t i c s y s t e m h a s b e e n$ $t r a n s f o r m e d a g a i n i n t o e q u i v a l e n t l i n e a r s y s t e m .$ $T h u s w e h a v e,$ $c - a = \pm \frac{43}{\sqrt{19}}$ $b - a = \pm \frac{3}{\sqrt{19}}$ $c - b = \pm \frac{40}{\sqrt{19}}$ $a + b + c = \pm \sqrt{19}$ $S o l v i n g t h e s y s t e m w e h a v e,$ $a = \mp \frac{9}{\sqrt{19}}$ $b = \mp \frac{6}{\sqrt{19}}$ $c = \pm \frac{34}{\sqrt{19}}$ $T h e r e f o r e o u r o b t a i n e d s o l u t i o n s e t$ $S = {(a, b, c) : (\frac{15}{\sqrt{91}}, \frac{18}{\sqrt{91}}, \frac{58}{\sqrt{91}}), (\frac{- 15}{\sqrt{91}}, \frac{- 18}{\sqrt{91}}, \frac{- 58}{\sqrt{91}}), (\frac{- 9}{\sqrt{19}}, \frac{- 6}{\sqrt{19}}, \frac{34}{\sqrt{19}}), (\frac{9}{\sqrt{19}}, \frac{6}{\sqrt{19}}, \frac{- 34}{\sqrt{19}})}$ $[O m a r b i n A b d u l A z i z]$
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