Solution that I found during contest. Since $BX=XY=YC$ and $AN=NC$ we get that $\angle BAX=\angle XAY=\angle YAC=\angle NCA=\alpha$ $\implies$ $\angle CBX=2\alpha$. Claim: $ABXC$ is harmonic quadrilateral. Proof: $\frac{AB}{AC}=\frac{AN}{AC}=\frac{\sin\alpha}{\sin 2\alpha}=\frac{BX}{XC}$ $\implies$ $AB\cdot XC=AC\cdot BX$ $\implies$ $ABXC$ is harmonic. Since $ABXC$ is harmonic, tangents to $(ABC)$ at $A$ and $X$ and line $BC$ are concurrent. Then $BC$ is $\text{C-symmedian}$ of $\triangle CAX$. Since $\angle ACN=\angle BCX=\alpha$ we get that $CN$ is $\text{C-median}$ in $\triangle CAX$. So $CN$ passes through midpoint of $AX$.

Here's an easy solution (hopefully is right). Let $NC$ meet $AX$ at $S$. Note that $AX$ touches $(ANC)$, so by PoP, $SA^2=SN.SC$. So again by PoP, we want $AX$ to touch $(NXC)$. This is equivalent to $<AXN=<NCX=<ACB=<AXB$ (since $<BCX=<ACN$), which is obvious since $AX$ is perpendicular bisector of $BN$, done. @below thanks for confirming. Also, it's kinda strange that this is even not among the official solutions, because it's simpler than any of them imho.

VicKmath7 wrote: Here's an easy solution (hopefully is right). Let $NC$ meet $AX$ at $S$. Note that $AX$ touches $(ANC)$, so by PoP, $SA^2=SN.SC$. So again by PoP, we want $AX$ to touch $(NXC)$. This is equivalent to $<AXN=<NCX=<ACB=<AXB$ (since $<BCX=<ACN$), which is obvious since $AX$ is perpendicular bisector of $BN$, done. Oh, yes. You are right. I can't understand why I didn't see this solution

Knowing trig can definitely lead to extremely fast solutions. The assertion is equivalent to $S_{ANC} = S_{XNC}$. With $\angle BAC = \alpha$, $\angle ACB = \gamma$ and $AB = c$ we have $S_{ANC} = \frac{1}{2}c^2\sin(\frac{2\alpha}{3})$ and $S_{XNC} = \frac{1}{2}XC \cdot c \cdot \sin \gamma$. By the Sine Law and $\angle XAC = \frac{2\alpha}{3}$ we have $CX = 2R\sin(\frac{2\alpha}{3})$ and $AB = 2R\sin\gamma$, bye bye!

What's up with all the trig

Click to reveal hidden text

Easy. Let $NC$ meet $AX$ at point $M$. It is easy to see that $AX$ is perp. bisector of $BN$, then $\angle MXN=\angle MXB=\angle ACB=\angle MCX$ which implies that $MX$ is tangent to $(N C X)$, then $MX^2=MN*MC$, then from simple angle chasing $\angle MAN=\angle MCA$, which gives us $MA^2=MN*MC$, then $MA=MX$, as desired.

An easy one. Let NC meet AX at O and (ABC) at P . From angle chasing , we will get PX || AY . it is easy to see that quadrilateral APBX is an isosceles trapezium. So, PX=AB=AN and PXNA is a parallelogram ,done.