I shall use analytic geometry to solve this problem. (1): Define coordinate system: Let $M$ be the origin and $AB$ be the x-axis. Let $A: (-a,0)$ and $B: (b,0)$. It is well-known that the circumcentre of a square is the intersection of its diagonals. Hence: $P: (-\dfrac{a}{2}, \dfrac{a}{2})$ and $Q: (\dfrac{b}{2}, \dfrac{b}{2})$. Let the circumcircles of $AMCD$ and $MBEF$ be $C_1$ and $C_2$, thus we can write the equations of the circle: \[ C_1: (x+\dfrac{a}{2})^2+(y-\dfrac{a}{2})^2=R_1^2= \dfrac{a^2}{2} \] \[ C_2: (x-\dfrac{b}{2})^2+(y-\dfrac{b}{2})^2=R_2^2= \dfrac{b^2}{2} \] Expand and notice that we can cancel out $\dfrac{a^2}{2}$ and $\dfrac{b^2}{2}$, the equations become: \[ C_1: x^2+ax+y^2-ay=0 \cdots (1) \] \[ C_2: x^2-bx+y^2-by=0 \cdots (2) \] Now comes the computation of finding the intersection of the two circles. First we substract $(2)$ from $(1)$, which gives: \[ (a+b)x+(b-a)y=0 \] \[ \Longleftrightarrow (a+b)x=(a-b)y \] \[ \Longleftrightarrow x=\dfrac{a-b}{a+b}\cdot y\cdots (3) \] Plug $(3)$ into $(1)$, we get: \[ y^2 \cdot (\dfrac{a-b}{a+b})^2 + \dfrac{a(a-b)}{a+b}\cdot y +y^2-ay=0 \] Because by the definition of the coordinate system, we know that one intersection of $C_1$ and $C_2$ is at the origin. But we want the other one. Hence we can divide the two sides of the equation by $y$: \[ y \cdot (\dfrac{a-b}{a+b})^2+\dfrac{a(a-b)}{a+b}+y-a=0 \] Factoring: \[ y\cdot (\dfrac{a^2-2ab+b^2}{(a+b)^2}+1)=a-\dfrac{a(a-b)}{a+b} \] \[ y\cdot (\dfrac{a^2-2ab+b^2+a^2+2ab+b^2}{(a+b)^2})=a\cdot (\dfrac{a+b+(b-a)}{a+b}) \] \[ y\cdot (\dfrac{2a^2+2b^2}{(a+b)^2})=\dfrac{2ab}{a+b} \] \[ y\cdot \dfrac{a^2+b^2}{a+b}=ab \] \[ y=\dfrac{ab(a+b)}{a^2+b^2} \] Plug $y$ into $(3)$ we get: \[ x=\dfrac{ab(a-b)}{a^2+b^2} \] Therefore we get: \[ N: (\dfrac{ab(a-b)}{a^2+b^2},\dfrac{ab(a+b)}{a^2+b^2}) \] Now we've done most of the work. Since we are trying to prove that $A$, $F$ and $N$ are collinear, so: \[ \dfrac{\dfrac{ab(a+b)}{a^2+b^2}-0}{\dfrac{b(a-b)}{a^2+b^2}+a}=\dfrac{b(a+b)}{b(a-b)+a^2+b^2}=\dfrac{b(a+b)}{a(a+b)}=\dfrac{b}{a} \] And: \[ \dfrac{b-\dfrac{ab(a+b)}{a^2+b^2}}{0-\dfrac{ab(a-b)}{a^2+b^2}}=\dfrac{a^2+b^2-(a^2+ab)}{a(b-a)}=\dfrac{b(b-a)}{a(b-a)}=\dfrac{b}{a} \] Thus by analytic geometry, $N$, $A$, $F$ are collinear. Do the same for $N$, $C$, $B$, we can also prove these points are collinear. Therefore question (1) is solved. (2): The formula of the perpendicular bisector of $AB$ is $x=\dfrac{b-a}{2}$. The equation of line $NM$ is $y=\dfrac{a+b}{a-b}\cdot x$. Thus the y-ordinate of the intersection of the two lines it: \[ y=-\dfrac{a+b}{2} \] Because $a+b=AB$, $AB$ is a contant value, hence the $MN$ passes through a fixed point. The point is the third vertex of the isoscele right triangle with side $AB$. (3): $P: (-\dfrac{a}{2},\dfrac{a}{2})$, $Q: (\dfrac{b}{2}, \dfrac{b}{2})$. Let the mid-point of $PQ$ be $X: (x_0, y_0)$. Hence: \[ x_0=\dfrac{x_1+x_2}{2}=\dfrac{b-a}{4} \] \[ y_0=\dfrac{y_1+y_2}{2}=\dfrac{a+b}{4} \] Once again, $a+b$ is constant. Hence the locus of $X$ is a segment: $y=\dfrac{a+b}{4}$, $x \in [-\dfrac{a+b}{2}, \dfrac{a+b}{2}]$. The problem is solved. P.S I am also interested in a geometrical proof.

Geometrical proof is quite easy. a) We have $\angle CBM = \angle AFM$ so $\angle AN'B = \frac{\pi}{2}$. It means, that $N'$ lay on the circle circumscribed about squares $AMCD$ and $MBEF$ so $N = N'$. b) Triangles $CMB$ and $ANB$ are similar, hence $\frac{CM}{MB} = \frac{AN}{NB}$ and $\frac{AM}{MB} = \frac{AN}{NB}$. It menas that $NM$ is bisector of $\angle ANB$. $N$ lay on a circle with center in midpoint of $AB$ and radius $\frac{AB}{2}$ so $NM$ always pass through center of arc $AB$.

I am sorry, Shobber, the genuine and shortest solution is the Michael Marcinkovski's !

a) $ \angle ANM = \angle MNC = \angle FNE = \angle MNB = \angle BNE = 45^\circ$. So $ A,N,F$ are collinear, $ B,C,N$ are collinear. b) from a, we know $ NM$ is angle bisector of $ \angle ANB$. So $ NM$ always pass through center of arc $ AB$. c) Draw perpendicular lines from $ P$, $ Q$, and the midpoint ($ R$) to the segment $ AB$ (Call them $ X, Y, Z$). $ PX = AM / 2$, $ QZ = MB / 2$. So $ RY = \frac {AM + MB} {4} = AB/4$ which means a parallel line segment to $ AB$.

Dear Mathlinkers, an article concerning the Vecten's figure and its developpement can be seeing on my site http://perso.orange.fr/jl.ayme , la figure de Vecten vol. 5 p. 25 Sincerely Jean-Louis

Solved with franchester, ingenio, tworigami, GameMaster402, anish9876, peeyushmaths, AIME12345, SD2014, AOPS12142015, huangyi_99, sriraamster, kothasuhas, budu, and mathfun5. The central motivation of the following is that since we are cognizant of the placement at a measly #5 of a trivial contest, the use of genuine advanced techniques is not imperative. (a) First, we show that $N'$ lies on the circle with diameter $AC$. We proceed with moving points, varying $M$ on $AB$ in a linear fashion. We desire the demonstration of the following: Lemma: The map $$M \rightarrow C \rightarrow BC \cap (AB)$$tethered to $$AB \rightarrow \ell \rightarrow (AB)$$where $\ell$ is the fixed angle bisector of $\angle BAD$, as well as the corresponding map for $A$ is projective, and these two maps are identical. Proof: To show that it is projective, note that it is a composition of two perspectivities centered at $\infty_{AD}$ and $B$, respectively. The proof is similar for the corresponding map for $A$. Now, it suffices to check that they coincide in $3$ cases. $M \rightarrow A$, $M \rightarrow B$, and $MA=MB$ are simple, so the maps are the same, as desired. Now, we know that $N'$ lies on the circle with diameter $AB$, so $\angle AN'C = 90^{\circ} = \angle ADC$ and $\angle BNF = 90^{\circ} = \angle BMF$, so $N'$ lies on the same two circles as $N$, and since $N' \neq M$, implying $N' = N$, as desired. $\blacksquare$ (b) The simple lemma preceding the clearly intended solution can be stated as follows: Lemma: Let $\alpha\beta\gamma$ be a triangle with $\angle \beta = 90^{\circ}$. The bisector of $\angle \beta$ meets $AC$ at $\delta$, the midpoint of $\beta \delta$ is $\mu$, and point $\tau$ on the opposite side of $\beta \gamma$ satisfies $\tau \beta = \tau \gamma$ and $\angle \beta \tau \gamma = 90^{\circ}$. Then, $\alpha$, $\mu$, $\tau$ are collinear. (I genuinely recommend trying to prove that, it's pretty cool) Proof: First, project $(\beta,\delta;\mu,\infty)=-1$ through $\alpha$ to get that if $\kappa$ lies on $BC$ with $\angle \beta \kappa \alpha = 45^{\circ}$ and $\chi$ is the intersection of $\alpha \mu$ with $\beta \gamma$, $(\beta,\gamma;\chi,\kappa)=-1$. Now, let $\tau'$ be the reflection of $\tau$ over $\beta \gamma$ and let $\tau'\kappa$ meet the circle with diameter $BC$ at $\phi$. Projecting $(\beta,\gamma;\chi,\kappa)=-1$ through $\phi$ gives $\phi$, $\kappa$, $\tau$ collinear. Finally, by Pascal's theorem, $\kappa = \tau'\phi \cap \beta \gamma$, $\infty_{\kappa \alpha} = \beta\tau' \cap \tau \gamma$ and $\beta \beta \cap \tau\phi$ are collinear. Since $\kappa\infty_{\kappa \alpha}$ and $\beta\beta$ meet at $\alpha$, $\tau\phi$ also goes through $\alpha$. Therefore, $\tau$, $\kappa$, $\phi$, $\mu$ $\alpha$ are collinear, as desired. Now, we claim that $MN$ passes through the the point through the point $T$ on the opposite side of $BC$ satisfying $TB=TC$ and $\angle BTC = 90^{\circ}$. By Pascal's Theorem on points on the circle with diameter $AC$, $AA \cap MN$, $AD \cap NC$, and $AC \cap DM = P$ are collinear. Applying the lemma to the triangle formed by $AD \cap BC$, $A$, and $C$ gives $J,P,T$ collinear (since $PA=PC$ and $\angle DAC = \angle MAC = 45^{\circ}$). It is easy to check that $AA$ passes through $T$ as well, so $N,M,T$ collinear, as desired. $\blacksquare$ (c) The heart of this part (which I actually really love, the wedding is in May) is in the following lemma (which I also recommend trying) Lemma: Let $ABC$ be a triangle, let $D$ be a point on $BC$, pick $E$ on $AC$ and $F$ on $AB$ with $DE || AB$ and $DF || BC$. Then, the circumcenter of $AEF$ lies on the perpendicular bisector of $A$ and the midpoint of the $A$-symmedian chord. Proof: Trivially, it suffices to show that the midpoint (call it $K$) of the $A$-symmedian chord lies on $(AEF)$. It is well known (and you should feel bad if you don't know this /s), $K$ is the center of the spiral similarity mapping $AC$ to $BA$. We have $BF:FA=BD:DC=AE:EC$, so this spiral similarity also maps $E$ to $F$. Thus, $\angle EKF = \angle BKA = 180^{\circ} - \angle A = 180^{\circ} - \angle EAF$, where $\angle BKA = 180^{\circ} - \angle A$ is also well-known. Thus, we have $AEFK$ cyclic, as desired. Back to this trivial problem, let $Z = AP \cap BQ$. We can see that $ZAB$ is a 45-45-90 triangle with right angle at $Z$, and $MP || BZ$ and $MQ || AZ$. Clearly, $\angle PAQ = 90^{\circ}$, so the circumcenter of $APQ$ is the midpoint of $PQ$. By the lemma, since the midpoint of $BC$ is the midpoint of the $Z$-symmedian chord of $(ZBC)$, the midpoint of $PQ$ varies on the perpendicular bisector of $Z$ and the midpoint of $BC$, as desired. $\blacksquare$

mira74 wrote: The heart of this part (which I actually really love, the wedding is in May) Am I invited?

(a) $M$ - center spiral similarity that sends $AC$ to $FB$. In other words it's a Miquel point of $ACBF$ and intersection of sides $AF$, $BC$ is common point of $\odot MAC$ and $\odot MFB$. EDIT: looking on diagram again and, oh my, it's just obvious: $F$ orthocenter of $ABC$ and conclusion follows.