Isn't RNMO supposed to send winners to IMO? Then why are there college math involved?

ZETA_in_olympiad wrote: Isn't RNMO supposed to send winners to IMO? Then why are there college math involved? See the discussion from: https://artofproblemsolving.com/community/u738252h2810717p24797803.

@above you have asked same questions as i have asked before here is the post . https://artofproblemsolving.com/community/q1h2810717p24830110 yes @ below I just got to know some time ago. we studied rings in 2nd or 3rd year of our undergraduate and they do on 11, 12.

Addysmith wrote: @above you have asked same questions as i have asked before here is the post https://artofproblemsolving.com/community/q1h2810717p24830110 I didn't know this, perhaps RMO is unique.

My solution during the contest: Let $f:U\to G$, $f(u)(x)=f_u(x)=uxu^{-1}$ and $g:U\to G$, $g(u)(x)=g_u(x)=ux$ for all $x\in R$. Notice that $f,g$ are morphisms and set $H=\text{Im} f, K=\text{Im} g$.It's easy to see that $\ker f=Z\cap U$ and $\ker g={1}$, so the first isomorphism theorem implies $|H|=|Z\cap U|, |K|=|U|$. Let $p\in H\cap K$. Then $p(1)=u1u^{-1}=u$, so $p=\text{id}_R$. By a well-known lemma $|G|\geq |HK|\geq \frac{|H||K|}{|H\cap K|}=\frac{|U|^2}{|Z\cap U|}$. $\square$

There exists actually a very straight-forward solution. Let $U^2$ be the direct product $U\times U$. Consider the group homomorphism $\varphi : U^2\rightarrow \text{Aut}(R,+)$ given by $$\varphi(u,v)=f_{u,v}(a)=uav^{-1}.$$One can easily verify that the function is in fact a homomorphism. The key point is to analyze the kernel of $\varphi$. Indeed $$\varphi(u,v)=\text{id}\iff uav^{-1}=a\text{ for all }a\in R.$$Taking $a=1$ implies $v=u$, and $uau^{-1}=a$, $\forall a\in R$, finally giving $u\in Z$. Therefore, $\ker\varphi =\{(u,u)\mid u\in U\cap Z\}$. But now, applying the first isomorphism theorem yields $U^2/\ker\varphi \cong \text{Im}\varphi$, and thus the conclusion: $$|\text{Aut}(R,+)|\geq |\text{Im}\varphi|=\frac{|U|^2}{|Z\cap U|}.$$

For $x,y\in U$, let the automorphism $\pi_{x,y}\in G$ be given by $a\mapsto xay^{-1}$. It is easy to check that $\pi_{x,y}$ is an automorphism of $(R,+)$. Consider the homomorphism $\varphi:U\times U\rightarrow G$ given by $(x,y)\mapsto\pi_{x,y}$. If $\varphi((x,y))=\mathrm{id}$, it follows that $x1y^{-1}=1$ so $x=y$. Since $xax^{-1}=a$ for all $a\in R$, it follows that $x\in Z$. Thus $\ker\varphi\cong Z\cap U$. The conclusion follows by the first isomorphism theorem. $\square$