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Question Number 202290 by MATHEMATICSAM last updated on 24/Dec/23

If ((3a − b)/(x + y)) = ((3b − c)/(y + z)) = ((3c − a)/(z + x)) then show  that ((a + b + c)/(x + y + z)) = ((a^(2 )  + b^2  + c^2 )/(ax + by + cz)) .

If3abx+y=3bcy+z=3caz+xthenshowthata+b+cx+y+z=a2+b2+c2ax+by+cz.

Answered by Rasheed.Sindhi last updated on 24/Dec/23

If ((3a − b)/(x + y)) = ((3b − c)/(y + z)) = ((3c − a)/(z + x)) then show  that ((a + b + c)/(x + y + z)) = ((a^(2 )  + b^2  + c^2 )/(ax + by + cz))       ((3a − b)/(x + y)) = ((3b − c)/(y + z)) = ((3c − a)/(z + x))  ⇒((x+y)/(3a − b))=((y+z)/(3b − c))=((z+x)/(3c − a))=k (say)  x+y=k(3a − b)...(i)  y+z=k(3b − c)...(ii)  z+x=k(3c − a)...(iii)  (i)+(ii)+(iii):  2(x+y+z)=k{(3a − b)+(3b − c)+(3c − a)}  x+y+z=k(a+b+c)...(iv)     (iv)−(ii):   x=k(a+b+c)−k(3b − c)        =k(a−2b+2c)  (iv)−(iii):  y=k(a+b+c)−k(3c − a)     =k(2a+b−2c)  (iv)−(i):  z=k(a+b+c)−k(3a − b)    =k(−2a+2b+c)   To prove:   ((a + b + c)/(x + y + z)) = ((a^(2 )  + b^2  + c^2 )/(ax + by + cz))  lhs:   ((a + b + c)/(x + y + z))=((a + b + c)/(k(a+b+c)))=(1/k)  rhs:   ((a^(2 )  + b^2  + c^2 )/(ax + by + cz))  = ((a^(2 )  + b^2  + c^2 )/(a(k(a−2b+2c)) + b(k(2a+b−2c)) + c(k(−2a+2b+c))))  = ((a^(2 )  + b^2  + c^2 )/(k{a(a−2b+2c)) + b(2a+b−2c)) + c(−2a+2b+c))}))  = ((a^(2 )  + b^2  + c^2 )/(k{a^2 −2ab+2ca + 2ab+b^2 −2bc −2ca+2bc+c^2 }))  = ((a^(2 )  + b^2  + c^2 )/(k{a^2 +b^2 +c^2 }))=(1/k)  ∵ lhs=rhs=1/k  ∴        Proved

If3abx+y=3bcy+z=3caz+xthenshowthata+b+cx+y+z=a2+b2+c2ax+by+cz3abx+y=3bcy+z=3caz+xx+y3ab=y+z3bc=z+x3ca=k(say)x+y=k(3ab)...(i)y+z=k(3bc)...(ii)z+x=k(3ca)...(iii)(i)+(ii)+(iii):2(x+y+z)=k{(3ab)+(3bc)+(3ca)}x+y+z=k(a+b+c)...(iv)(iv)(ii):x=k(a+b+c)k(3bc)=k(a2b+2c)(iv)(iii):y=k(a+b+c)k(3ca)=k(2a+b2c)(iv)(i):z=k(a+b+c)k(3ab)=k(2a+2b+c)Toprove:a+b+cx+y+z=a2+b2+c2ax+by+czlhs:a+b+cx+y+z=a+b+ck(a+b+c)=1krhs:a2+b2+c2ax+by+cz=a2+b2+c2a(k(a2b+2c))+b(k(2a+b2c))+c(k(2a+2b+c))=a2+b2+c2k{a(a2b+2c))+b(2a+b2c))+c(2a+2b+c))}=a2+b2+c2k{a22ab+2ca+2ab+b22bc2ca+2bc+c2}=a2+b2+c2k{a2+b2+c2}=1klhs=rhs=1/kProved

Answered by som(math1967) last updated on 24/Dec/23

Each ratio  =((3a−b+3b−c+3c−a)/(x+y+y+z+z+x))  =((2(a+b+c))/(2(x+y+z)))=((a+b+c)/(x+y+z))  ⇒((a+b+c)/(x+y+z))=((a+b+c−3a+b)/(x+y+z−x−y))  =((a+b+c−3b+c)/(x+y+z−y−z))=((a+b+c−3c+a)/(x+y+z−z−x))  ⇒((a+b+c)/(x+y+z))=((2b−2a+c)/z)=((a−2b+2c)/x)  =((2a−2c+b)/y)  ⇒((a+b+c)/(x+y+z))=((2bc−2ac+c^2 )/(cz))=((a^2 −2ab+2ac)/(ax))  =((2ab−2bc+b^2 )/(by))  ⇒((a+b+c)/(x+y+z))  =((2bc−2ac+c^2 +a^2 −2ab+2ac+2ab−2bc+b^2 )/(cz+ax+by))   ⇒((a+b+c)/(x+y+z))=((a^2 +b^2 +c^2 )/(ax+by+cz))

Eachratio=3ab+3bc+3cax+y+y+z+z+x=2(a+b+c)2(x+y+z)=a+b+cx+y+za+b+cx+y+z=a+b+c3a+bx+y+zxy=a+b+c3b+cx+y+zyz=a+b+c3c+ax+y+zzxa+b+cx+y+z=2b2a+cz=a2b+2cx=2a2c+bya+b+cx+y+z=2bc2ac+c2cz=a22ab+2acax=2ab2bc+b2bya+b+cx+y+z=2bc2ac+c2+a22ab+2ac+2ab2bc+b2cz+ax+bya+b+cx+y+z=a2+b2+c2ax+by+cz

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