Belarus2022

Since the sides $AB$ and $CD$ of a square are parallel, the cross-lying angles are equal: $\angle PK{M}_1=\angle P{K}_{1}M$ and $\angle P{M}_{1}K=\angle PM{K}_1$. Therefore the triangles $PK{M}_1$ and $P{K}_{1}M$ are similar and have equal ratios $KP:K_{1}P=M_{1}P:MP$. Denote $KP:K_{1}P=k$, then $MP:P{M}_1=1/k$. Similarly, for $LP:P{L}_1=l$ we obtain $NP:P{N}_1=1/l$. Hence the equality given in the problem is equivalent to $k+1/k+l+1/l=4$, which can be transformed as $$0=k+\frac{1}{k}-2+l+\frac{1}{l}-2=\frac{k^2+1-2k}{k}+\frac{l^2+1-2l}{l}=\frac{(k-1{)}^2}{k}+\frac{(l-1{)}^2}{l}.$$ Clearly, the equality holds if and only if $k=l=1$ which means that $$KP=K_{1}P,LP=L_{1}P,MP=M_{1}P\text{and}NP=N_{1}P$$ (in particular $P$ is the center of the square $ABCD$). Now the required equality is obvious.