Perhaps the least surprising thing about this problem is that it is very good Assume $AB < AC$ in what follows. Let $D$ be the antipode of $M$ with respect to the incircle $\omega$ of $\triangle ABC$; recall that $ADN$ are collinear by a homothety argument. Furthermore, let $T$ be the tangency point of $\omega$ with $\overline{AC}$, and suppose lines $ADN$ and $AM$ intersect $\omega$ again at points $P$ and $Q$, respectively. [asy][asy] size(300); defaultpen(linewidth(0.6)+fontsize(10)); pair M = dir(270), D = dir(90), Q = dir(90+36), T = dir(90 - 36), X = rotate(90,T)*origin; pair A = extension(M,Q,T,X), C = extension(M,(1,-1),T,X), N = extension(A,D,M,C), B = M + N - C; pair[] G = intersectionpoints(A--N,unitcircle); draw(A--B--C--A,rgb(0.1,0.1,0.7)); draw(unitcircle,rgb(0.1,0.5,0.1)); draw(M--A--N,rgb(0.6,0.1,0.6)); draw(A--origin--T^^M--G[1],red); draw(M--D,orange); dot("$A$",A,dir(90),linewidth(3.3)); dot("$B$",B,SW,linewidth(3.3)); dot("$C$",C,SE,linewidth(3.3)); dot("$T$",T,NE,linewidth(3.3)); dot("$M$",M,S,linewidth(3.3)); dot("$N$",N,S,linewidth(3.3)); dot("$D$",D,dir(80),linewidth(3.3)); dot("$Q$",Q,dir(150),linewidth(3.3)); dot("$P$",G[1],dir(5),linewidth(3.3)); dot("$I$",origin,E,linewidth(3.3)); [/asy][/asy] Since $D$ is the antipode of $M$ with respect to $\omega$, angle $\angle DPM$ is a right angle, and so $\triangle AIT$ and $\triangle AMP$ are directly similar right triangles. It follows that $\triangle AIM\sim\triangle ATP$, so $\angle AMD = \angle TPD$. Hence $D$ is the midpoint of arc $\widehat{QT}$ of $\omega$, so $QT\parallel BC$. In turn, $\triangle QMT$ is isosceles with $\angle QMT = \angle C$. Now compute $\angle AIM =\angle AIB + \angle BIM = 180^\circ - \tfrac{B-C}2$ unconditionally, and since $\angle AMI = \tfrac C2$ we have $\angle MAI = \tfrac B2 - C$. Therefore \[ \angle ANM = \angle ACN + \angle NAC = C + (\tfrac B2 - C) = \tfrac B2 = \angle IBM. \]Thus $\triangle BMI$ and $\triangle NMD$ are similar right triangles, implying $\tfrac{MN}{BM} = \tfrac{DM}{IM} = 2$. But then \[ BC = MN + (BM + NC) = 2MN, \]which is what we were after.