2022 China Team Selection Test for IMO

First note that $\frac{\beta }{\alpha }$ is an irrational number (otherwise there are positive integers $k_1,k_2$ such that $k_1\alpha =k_2\beta$, which leads to a contradiction). If $\alpha =\frac{q}{p}$ is a rational number, then there is a positive integer $k_2$ such that the decimal part of $\frac{k_2\beta }{q}$ is less than $\frac{1}{q}$. Taking $k_1=p\lfloor \frac{k_2\beta }{q}\rfloor$ gives that $$k_1\alpha =q\cdot \lfloor \frac{k_2\beta }{q}\rfloor <k_2\beta <q\lfloor \frac{k_2\beta }{q}\rfloor +1,$$ which contradicts to the assumption. Thus $\alpha$ must be irrational. Similarly, $\beta$ must be irrational as well.

A pair of positive integers $(a,b)$ is called distinguished if $0<b\beta -a\alpha <1$. By our assumption, there is a unique positive integer $t$ such that $a\alpha <t<b\beta$. Denoting $u=t-a\alpha ,v=b\beta -t$, we say a distinguished pair $(a,b)$ corresponds to the intermediate number $t$ together with a distant pair $(u,v)$. We will prove the following.

Lemma. Assume two distinguished pairs $(a_1,b_1)$, $(a_2,b_2)$ correspond to distant pairs $(u_1,v_1)$, $(u_2,v_2)$, respectively. Then $\frac{u_1}{v_1}=\frac{u_2}{v_2}$.

Proof of Lemma. Suppose that the intermediate numbers of these two distinguished pairs are $t_1,t_2$, respectively. Assume for the sake of contradiction that $\frac{u_1}{v_1}>\frac{u_2}{v_2}$. We also take $ϵ=\frac{u_1{v}_2-u_2{v}_1}{2}>0$. Since $\frac{\beta }{\alpha }$ is an irrational number, there are positive integers $a_0,b_0$ such that $0<a_0\alpha -b_0\beta <ϵ$. On the other hand, our assumption implies that there is another integer $t_0$ satisfying $b_0\beta <t_0<a_0\alpha$. Denote $u_0=a_0\alpha -t_0$ and $v_0=t_0-b_0\beta$. If $\frac{u_1}{u_0}-\frac{v_1}{v_0}>1$, then one can take $L=\lfloor \frac{v_1}{v_0}\rfloor +1$ such that $\frac{u_1}{u_0}>L>\frac{v_1}{v_0}$, i.e., $$\begin{array}{rl}{u}_1-L{u}_0& =(t_1+L{t}_0)-(a_1+L{a}_0)\alpha >0,\\ v_1-L{v}_0& =(t_1+L{t}_0)-(b_1+L{b}_0)\beta <0.\end{array}$$ Set $k_1=a_1+L{a}_0$ and $k_2=b_1+L{b}_0$. Then $\lfloor k_1\alpha \rfloor =\lfloor k_2\beta \rfloor =t_1+L{t}_0-1$, which leads to a contradiction. This further implies that $\frac{u_1}{u_0}-\frac{v_1}{v_0}\le 1$. Using a similar argument, we can show that $\frac{u_2}{u_0}-\frac{v_2}{v_0}\ge -1$. Therefore, $$u_1-\frac{u_0}{v_0}{v}_1\le u_0,u_2-\frac{u_0}{v_0}{v}_2\ge -u_0,\Rightarrow u_1{v}_2-u_2{v}_1\le u_0(v_1+v_2)<2ϵ.$$ However, this violates our choice of $ϵ$. The lemma is proved.

Now we see all distinguished pairs $(a,b)$ share the same ratio $\frac{u}{v}=\frac{t-a\alpha }{b\beta -t}$. Denote this common $\lambda =\frac{v}{u+v}\in (0,1)$; then we have an equality $\lambda \cdot a\alpha +(1-\lambda )\cdot b\beta =t\in \mathbb{Z}$. We call linear combinations of distinguished pairs with integer coefficients nice pairs. For each nice pair $(c,d)$, it is not hard to see that $$\lambda \cdot c\alpha +(1-\lambda )\cdot d\beta =c\cdot \lambda \alpha +d\cdot (1-\lambda )\beta \in \mathbb{Z}.$$ Take a nice pair $(a,b)$ and set $\delta =b\beta -a\alpha \in (0,1)$. We claim that for all $M>0$, any pair $(c,d)$ of integers satisfying $0<d\beta -c\alpha <M$ and $c>\frac{a}{\delta }M$, is nice. Indeed, it suffices to take $L=\lfloor \frac{d\beta -c\alpha }{\delta }\rfloor <\frac{M}{\delta }<\frac{c}{a}$ so that $(d-Lb)\beta -(c-La)\alpha =(d\beta -c\alpha )-L\delta \in (0,\delta )\subset (0,1)$, and therefore the pair $(c-La,d-Lb)$ must be distinguished and the pair $(c,d)=(c-La,d-Lb)+L\cdot (a,b)$ must be a nice pair.

Now take $M=\alpha +2\beta$, and integers $c_0>\frac{a}{\delta }M+1$ and $d_0=\lfloor \frac{c_0\alpha }{\beta }\rfloor$ satisfying the condition $0<d_0\beta -c_0\alpha <\beta$. Using the analysis above, $(c_0,d_0)$, $(c_0,d_0+1)$, and $(c_0-1,d_0)$ are nice pairs, and their linear combinations $(0,1)$ and $(1,0)$ are nice pairs as well. This implies in particular that $\lambda \alpha$ and $(1-\lambda )\beta$ are (positive) integers. Taking positive integers $m_2=\lambda \alpha$ and $m_1=(1-\lambda )\beta$, we get $\frac{m_2}{\alpha }+\frac{m_1}{\beta }=\lambda +(1-\lambda )=1$, or equivalently $m_1\alpha +m_2\beta =\alpha \beta$. We conclude the proof.