WakeUp wrote: Let $a,b,c,d,p$ and $q$ be positive integers satisfying $ad-bc=1$ and $\frac{a}{b}>\frac{p}{q}>\frac{c}{d}$. Prove that: $(a)$ $q\ge b+d$ $(b)$ If $q=b+d$, then $p=a+c$. (a) 3 cases: (1) If $\frac pq=\frac{a+c}{b+d}$, we are done because from $(a+c)d-(b+d)c=ad-bc=1$ we get $\gcd(a+c,b+d)=1$, so $\frac{a+c}{b+d}$ is irreducible. (2) If $ \frac{a+c}{b+d}>\frac pq$, we have $(a+c)qd>(b+d)pd>(b+d)qc$, thus $q>(b+d)(pd-qc)>0$ which implies $q>b+d$. (3) If $ \frac{a+c}{b+d}<\frac pq$, start similarly with $\frac{b+d}{a+c}>\frac {q}{p}> \frac ba$ to get $p>a+c$. (The conditions are invariant w.r.t. the transformation $a\leftrightarrow d,b\leftrightarrow c,p\leftrightarrow q$.) Then $q>b\frac{p}a\ge \frac{b(a+c+1)}a=b+d+\frac {c-1}a\ge b+d$. (b) If $q=b+d$, then $\frac ab>\frac p{b+d}$ which is equivalent to $a+c+\frac 1b>p$, i.e. $p\le a+c.$ And $\frac p{b+d}>\frac cd$ is equivalent to $p>a+c-\frac 1d$, i.e. $p\ge a+c$. So $p=a+c$.

Another way to do (a): $\frac{a}{b} - \frac{p}{q} \geq \frac{1}{bq}$ (since $\frac{1}{bq}$ is the smallest positive rational with denominator $bq$) and $\frac{p}{q} - \frac{c}{d} \geq \frac{1}{dq}$, so $\frac{1}{bd} = \frac{a}{b} - \frac{c}{d} \geq \frac{1}{q} \left(\frac{b+d}{bd}\right)$, whence $q \geq b+d$.

I am so surprised,it is a well known property of Farey Series and when the denominator is $b+d$ the numerator is $a+c$.And in this case,$\frac {a+c} {b+d}$ is called the mediant of $\frac a b,\frac b d.$I don't undestand why it was set for an olympiad problem.Also it is true generally: If $ad-bc=\pm 1,$every frac lying between them has the denominator greater than or equal to $b+d$

Problem (a) appeared on the final of $XX$ Polish MO $1969$, which was at the same time TST for Poland. $0<\frac{a}{b}-\frac{p}{q}=\frac{aq-bp}{bq}\implies aq-bp>0\iff aq-bp\ge 1\iff adq-bdp\ge d$ $0<\frac{p}{q}-\frac{c}{d}=\frac{dp-cq}{dq}\implies dp-cq>0\iff dp-cq\ge 1\iff bdp-bcq\ge b $ Adding two last inequalities $b+d\le q(ad-bc)=q$