Hint only, supported by my software: If I made the sketch properly, the fixed point is the reflection of $A$ about the perpendicular at $C$ onto $BC$ (i.e. that line is the locus of $\Delta AMN$'s circumcenter); I am not able to provide more info as yet! Best regards, sunken rock

Well, I have to admit I was wrong, it's only a fixed point on the parallel to $BC$ through $A$ ! Sorry to have guided you to a wrong path! Best regards, sunken rock

The common point (say $L$) is actually the reflection of point $A$ about the perpendicular to $BC$ at the point $K$ given by $BK - CK = {AB\cdot AC \over BC}$, or $BK = {AB\cdot AC + BC^2 \over {2\cdot BC}}$. Let $O$ be the center of the circumcircle of $AMN$ (of radius $R$) and $K$ the foot of the perpendicular from $O$ to $BC$. Then the given expression is equivalent to $BA\cdot BM - CA\cdot CN = AB\cdot AC$ or (using the power of $B$ and $C$ around circle $O$) $(BO^2 - R^2) - (CO^2 -R^2) = AB\cdot AC$, or $BK^2 - CK^2 = AB\cdot AC ={\rm constant}$. That is, $K$ is a fixed, given point on $BC$. All circumcenters belong on the perpendicular to $BC$ at $K$. This perpendicular bisects the common chords $AL$ of all circumcircles $AMN$, and thus $L$ is the reflection of $A$ about this perpendicular. It is interesting to examine some limiting cases. If $C_1$ is on $AB$ so that $BC_1= AC$, then $AC_1 C$ is one limiting case of $AMN$ with $M\rightarrow C_1$ and $N\rightarrow C$. Indeed, in this case, $MB / AC \rightarrow 1$ and $NC/AM \rightarrow 0$. On the other hand, in the limiting case of $M \rightarrow A$, we get $CN \rightarrow (AB-AC)\cdot AB/AC < AC$. To make sure that this is possible, we note that the previous quadratic in $AC$ has a solution if $AB \le {{\sqrt{5}+1}\over 2}\cdot AC$, which is allowed by the given condition that $AB < {3\over 2}\cdot AC$.

An inversion centered at $A$ with arbitrary radius sends $(AMN)$ to line $M'N'$ and $B,C,M,N$ to points on lines $AB, AC$ such that $\frac{AC' \cdot M'B'}{AM'\cdot AB'} - \frac{N'C' \cdot AB'}{AC' \cdot AN'} = 1.$ Setting $m = AM', n = AN'$, we have $\frac{p}{m} + \frac{q}{n} = 1$ for some constants $p,q$ (using $M'B' = AM' - AB'$, etc). $(AMN)$ passes through a fixed point different from $A$ iff line $\ell = M'N'$ goes through a fixed point. Take an affine transformation sending $AB, AC$ to perpendicular axes $x$ and $y$ respectively. Then it is straightforward to check that line $\ell'$ (the image of $\ell$ after the transformation) goes through the fixed point $(p,q)$, as desired. In fact, the problem statement holds if $\frac{MB}{AC} - \frac{NC}{AB}$ is fixed, not necessarily equal to 1.

let $X$ be on $AB$ such that $BX=AC$ let $T$ be on arc $CAX$ such that $TC/TX=AB/AC$. The given condition is equivalent to $NC/MX=AB/AC=TC/TX$ but $\angle TCN=\angle TXM$ so $TCN\sim TXM$ and now $\angle TMA=\angle TNA$ so $\odot AMN$ passes through fixed point $T$. Remark: we need $T\not = A$ which is equivalent to $\frac{AC}{AB-AC}\not =\frac{AB}{AC}$ or (if $t=AB/AC$) $t^2-t-1\not = 0$ or $t\not = \frac{\sqrt{5}+1}{2}$, $t\not =1$ and $t\not = \frac{\sqrt{5}-1}{2}$. which is true since $1<t<\frac{3}{2}$

Let $K$ be on $AB$ such that $BK=AC$, $\odot ACK$ meets $\odot AMN$ at $L$(note that when $K\equiv M$, $N\equiv C$). Since $\frac {NC}{AB}=\frac {MB}{AC}-1=\frac {KM}{BK}$, so $\frac {NC}{KM}=\frac {BK}{AB}=\frac {AC}{AB}$. By spiral similarity $\triangle LMK \sim \triangle LNC$, thus $\frac {LC}{KL}=\frac {NC}{KM}=\frac {AC}{AB}$, which means that $L$ must be a fixed point on $\odot ACK$ $\Rightarrow$ $L$ is the fixed point. same solution as leader