if the lengths of sides of the triangle ABC satisfy

2(bc2+ca2+ab2)=b2c+c2a+a2b+3abc then the triangle is ?

note:all 2 except the one outside bracket are squares

it is an equilateral triangle.    

it is quite obvious from the equation that this will be satisfied only if a=b=c.   or if you want the proof...you can ask - and i'll provide it to you.   the thing is...i am a bit busy right now, so can't type out the whole proof.  will do it later.   

if the lengths of sides of the triangle ABC satisfy



2(bc2+ca2+ab2)=b2c+c2a+a2b+3abc then the triangle is ?



2(bc^2+ca^2+ab^2)==b^2c+c^2a+a^2b+3abc

ab(2b-a)+ca(2a-c)+bc(2c-b)-abc-abc-abc=0

ab(2b-a-c)+ca(2a-c-b)+bc(2c-b-a)=0

let b-a=p

ab(b+p-c)+ca(a-c-p)+bc(2c-2b+p)=0

a(b^2+pb-bc+ac-c^2-cp)=-bc(2c-2b+p)

a(-2cp+pb+b^2-c^2)=-bc(2c-2b+p)

-2cpa+pba+ab^2-ac^2+2bc^2-2b^2c+pbc=0

pc(-2a+b)+pba+ab^2-ac^2+2bc^2-2b^2c=0

p^2c-apc+pba+ab^2-ac^2+2bc^2-2b^2c=0

a(b+c)(b-c)+2bc(c-b)+p^2c+ap(b-c)

(b-c)[a(b+c)-2bc+ap)+p^2c=0

(b-c)(ab+(b-p)c-2bc+ap)+p^2c=0

(b-c)(b+p)(a-c)+p^2c=0

i am unable to think further

i will post the remaining soln 2morow

i actually thought of a different method



2(bc^2+ca^2+ab^2)=b^2c+c^2a+a^2b+3abc



2abc(b/c+c/a+a/b)=abc(b/a+a/c+c/b+3)



put c/b=p



2(1/p+pb/a+a/b)=(b/a+a/pb+p+3)



put b/a=m



2(1/p+pm+1/m-1)=(m+1/mp+p+1)



2((m+p+p^2m^2-pm)=m^2p+1+mp^2+mp



m(2-mp)+p(2-mp)+mp((2mp-1)-(2mp+1)



put mp=r



(m+p)(2-r)+r(2r-1)-(2r+1)



put m+p=y



2r^2-r(3+y)+2y+1



solving for r



[(3+y)+/-root(1+y^2-10y)]/4



considering only the positive value(just for simplicity)



[(3+y)+root(1+y^2-10y)]/4=r............................................j



we know that m+p=y



mp=r



(m-p)^2=(m+p)^2-mp



(m-p)=root(y^2-4r)



m=[root(y^2-4r)+y]/2..........................................................f



p=2r/[root(y^2-4r)+y] (as mp=r)...........................................1



but from the equation p=[y-root(y^2-4r)]/2..................................g



from f



2m-y=[root(y^2-4r)



(2m-y)^2=(y^2-4r)







from g



2p-y=-root(y^2-4r)



(2p-y)^2=(y^2-4r)



from these two results



p=m is one of the soln



equating 1 and f



2r/[root(y^2-4r)+y] =[root(y^2-4r)+y]/2



4r=y^2-4r+y^2+2y(root(y^2-4r)



4r-y^2=y(root(y^2-4r)



squaring on both sides







16r^2+y^4-8ry^2=y^2(y^2-4r)



16r^2+y^4-8ry^2=y^4-4ry^2



4r=y^2



substituting the value or r from j



(3+y)+root(1+y^2-10y)=y^2



3+y-y^2=-root(1+y^2-10y)



9+y^4+6y-2y^3-6y^2=1-10y



y^4-2y^3-6y^2+16y+8=0



let t,y,u,i be the solutions of y(like i am not getting nice integers after trying with calculator)



we know that y=p+m



and that p=m



p=m=y/2



p=c/b and m=b/a



as p=m , b^2=ac



also mp=r=c/a



p^2=m^2=c/a



let p=m=i/2(putting one of the soln of y)



i^2/4=c/a



4c=ai^2



we know that b^2=ac



b^2=aai^2/4



b=ai/2(considering only the positive value)



so the sides are related as a,b,c=a,ai/2,ai^2/4, it can be an equilateral triangle if i=2

but putting the value of i in y^4-2y^3-6y^2+16y+8=0,  it is not coming out to be 0

for this to be a right angled triangle



a^2i^2/4=a^2i^4/16+a^2



putting i^2/4=h



h=h^2+1



solving, we get imaginary roots, so this is not a rt angled triangle too

it is not coming ut to be 0 , so it is not a rt angled triangle

for it to be isosceles, any of these three conditions should be satisfied  a=ai/2,ai/2=ai^2/4,a=ai^2/4

let us take the first condition a=ai/2

i=2 is not the soln

so , it is rooted out

ai/2=ai^2/4

it is also rooted out

so, we are left with only

a=ai^2/4

i=+/-2

substituting y=-2 in

y^4-2y^3-6y^2+16y+8=0

it does not satisfy the equation

hence , it is not an isosceles nor equilateral triangle nor a rt triangle

the value of y is nearly equal to 0.45

i will post the soln tomorrow