# Does it follow that a+b+c=x+y+z=m+n+p [closed]

If $$abc=xyz=mnp$$ $$a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2=a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=m^4+n^4+p^4-2m^2n^2-2n^2p^2-2p^2m^2$$ $$\frac{x}{m}=\frac{n}{b}=\frac{c}{z}$$ Then $$a+b+c=x+y+z=m+n+p$$? $$a,b,c$$ positive and are lenghts of triangles, $$x,y,z$$ same, $$m,n,p$$ same.

I edited it.

## closed as off-topic by Crostul, Martin R, Aqua, José Carlos Santos, mrtaurhoMar 30 at 16:21

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Crostul, Martin R, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.

• What have you tried? – Jacob Jones Mar 28 at 19:11
• This is false: Consider $a,b=-1,c,x,y,z=1$. – Don Thousand Mar 28 at 19:13
• I edited the question a bit. Sorry for making this an unclear question. – furfur Mar 28 at 19:15
• What are $m$ and $n$? – Aqua Mar 28 at 19:31
• This is the full thing now. – furfur Mar 28 at 19:31

Hint: Note that $$a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=(a + b + c)(a + b - c)(a - b + c)(a - b - c).$$
• I don't think you need to do that much work. $a,b=-1,c,x,y,z=1$ is a simple counterexample. – Don Thousand Mar 28 at 19:13