Let A B C D be a unit square. Suppose M and N are points on B C and...

$\because \quad O A=$ circumradius of $\triangle A M N=\frac{M N}{2 \sin \alpha}$ $O P=$ circumradius of $\triangle O M N=\frac{M N}{2 \sin 2 \alpha}$ So $\left(\frac{O P}{O A}\right)^2=$\left(\frac{1}{2 \cos \alpha}\right)^2$Perimeter of$\triangle M C N=2=(1-x)+(1-y)+M N \Rightarrow \quad M N=x+y$Now rotate$\triangle A B M$about$A$so that$A B$overlaps with$A D$(by$90^{\circ}$) Clearly$\triangle A M N=\triangle A M^{\prime} N$So$2 \alpha=90^{\circ} \therefore \quad \alpha=45^{\circ}$<img src="https://cdn-question-pool.Quantrex.app/NEET/Olympiad/DIvpE0uuIzBmUwDRS91PiQBoZEtAgJCKBw_iubNVnhw.original.fullsize.png"><br> Hence$\left(\frac{O P}{O A}\right)^2=\frac{1}{2}=\frac{m}{n} \therefore \quad m+n=3