IMO 1959 - Problem 5

Problem: A segment $AB$ is given and on it a point $M$. On the same side of $AB$ squares $AMCD$ and $BMFE$ are constructed. The circumcircles of the two squares, whose centers are $P$ and $Q$, intersects in $M$ and another point $N$.
(a) Prove that lines $FA$ and $BC$ intersect at $N$.
(b) Prove that all such constructed lines $MN$ pass through the same point $S$, regardless of the selention of $M$.
(c) Find the locus of the midpoints of all segments $PQ$, as $M$ varies along the segment $AB$.
Solution: 
(a) It suffices to prove that $AF\bot BC$, since then for the intersection point $X$ we have $\angle{AXC}=\angle{BXF}=90^0$, implying that $X$ belongs to the circumcircles of both squares and thus that $X=N$. The relation $AF\bot BC$ holds because from $MA=MC,MF=MB$, and $\angle{AMC}=\angle{FMB}$ it follows that $\triangle{AMF}$ is obtained by rotating $\triangle{BMC}$ by $90^0$ around $M$.
(b) Since $N$ is on the circumcircle of $BMFE$, it follows that $\angle{ANM}=\angle{MNB}=45^0$. Hence $MN$ is the bisector of $\angle{ANB}$. It follows that $MN$ passes through the midpoint of the arc $AB$ of the circle with diameter $AB$(i.e, the circumcircle of $\triangle{ABN}$) not containing $N$.
(c) Let us introduce a coordinate system such that $A=(0,0),B=(b,0)$ and $M=(m,0)$. Setting in general $W=(x_W,y_W)$ for an arbitrary point $W$ and denoting by $R$ the midpoint of $PQ$, we have $y_R=\frac{y_P+y_Q}{2}=\frac{m+b-m}{4}=\frac{b}{4}$ and $x_R=\frac{x_P+x_Q}{2}=\frac{m+m+b}{4}=\frac{2m+b}{4}$, the parameter $m$ varying from $0$ to $b$. Thus the locus of all points $R$ is the closed segment $R_1R_2$ where $R_1=(\frac{b}{4},\frac{b}{4})$ and $R_2=(\frac{b}{4},\frac{3b}{4})$.