Real numbers $a, b, c$ are such that $$a^2+c-bc = b^2+a-ca = c^2+b-ab$$ Does it follow that $a=b=c$? Proposed by Mykhailo Shtandenko
Problem
Source: Ukrainian Mathematical Olympiad 2024. Day 2, Problem 8.5
Tags: algebra
Tintarn
23.03.2024 13:05
Yes.
Taking the difference of the first equations gives $(a-b)(a+b+c)=a-c$ and of course we also have the cyclic equations.
So if $a=c$, then also $a=b$ and all are equal.
Now let us assume that they are distinct. Then multiplying all three such relations gives $(a+b+c)^3=-1$ and hence $a+b+c=-1$. But then we get $a-b=c-a=b-c=d$ for some $d$ and hence summing them $3d=0$ and hence again $a=b=c$.
bever209
30.03.2024 23:48
WLOG let $c$ be the maximum of the three. If $b$ is the second largest, then $c^2,b,-ab$ are the three individual terms that are the largest in the first, second, and third positions so necessarily $a=b=c$. If $a$ is the second largest, then $a^2,c,-bc$ are correspondingly greater than $b^2,a,-ca$ so $a=b=c$.