$A=[1,0,0]$, $B=[0,1,0]$, $C=[0,0,1]$, $\Gamma=\odot BMP$ $Pow_{\Gamma}(B)=0$, $Pow_{\Gamma}(A)=AM\cdot AB=\frac{c^2}{2}$, $Pow_{\Gamma}(C)=PC\cdot BC=\frac{a^2+b^2-c^2}{2}$ Hence $\displaystyle \Gamma: a^2yz+b^2xz+c^2xy-\left(\frac{c^2}{2}x+\frac{a^2+b^2-c^2}{2}z\right)(x+y+z)=0$ Intersecting with $AC: y=0$, we get: $b^2xz=\frac{1}{2}(x+z)(c^2x+(a^2+b^2-c^2)z)$ $\Rightarrow$ $c^2x^2-(b^2-a^2)xz+(a^2+b^2-c^2)z^2=0$ $\Rightarrow$ $4c^4x^2-4c^2(b^2-a^2)xz+4c^2(a^2+b^2-c^2)z^2=0$ Since $a+b=\sqrt{2}c$, we have $c^2=\frac{(a+b)^2}{2}$ and $a^2+b^2-c^2=\frac{(a-b)^2}{2}$, so $4c^2(a^2+b^2-c^2)=4\cdot \frac{(a+b)^2}{2}\cdot \frac{(a-b)^2}{2}=(b^2-a^2)^2$ Therefore the equation can be rewritten as $4c^4x^2-4c^2(b^2-a^2)xz+(b^2-a^2)z^2=0$ $\Rightarrow$ $(2c^2x-(b^2-a^2)z)^2=0$ Hence there is only one intersection between $\Gamma$ and $AC$, that is, $AC$ is tangent to $\Gamma$

is this barycentric coordinate? im not familiar with it..

A sketch using Stewart

Synthetic! (This can probably be optimized, but I spent a while on this and have other things to do, whoops.) [asy][asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A = origin, B = (2,0); path El = ellipse((1,0), sqrt(2),1); pair C = relpoint(El, 0.36); pair M = (A+B)/2, P = foot(A,C,B); real r = abs(B-C), s = abs(A-C); pair T = C * (r+s)/s, N = (A+T)/2, D = s/(r+s)*B; draw(A--B--C--cycle,rgb(0.1,0.1,0.8)); draw(C--N,rgb(0.1,0.5,0.1)); draw(C--D^^N--M,rgb(0.9,0.1,0.1)); draw(circumcircle(B,M,P),orange+linetype("3 3")); draw(A--P^^rightanglemark(A,P,C,2),rgb(0.6,0.1,0.6)); clip((-2,-1)--(-2,1.6)--(2.2,1.6)--(2.2,-1)--cycle); dot("$A$",A,SW); dot("$B$",B,SE); dot("$C$",C,NW); dot("$D$",D,S); dot("$P$",P,dir(A--P)); dot("$M$",M,S); dot("$N$",N,NW); [/asy][/asy] Let $T$ be the point on ray $AC$ for which $BC = CT$ (not pictured above for space reasons), and let $N$ be the midpoint of $\overline{AT}$. Then the given length condition implies \[ AN = \frac{AC+CB}2 = \frac{AB}{\sqrt 2}, \]or $AN^2 = \tfrac 12AB^2 = AM\cdot AB$. This means that the circumcircle of $\triangle BMN$ is tangent to $\overline{AT}$ at $N$, implying $\angle ANM = \angle ABN$. We now claim that $NPMB$ is cyclic. To prove this, let $D$ be the foot of the angle bisector from $C$. Observe that $T$ is the reflection of $B$ across the external angle bisector of $\angle ACB$, so $MN\parallel BT\parallel CD$. Thus, a (simple?) angle chase yields \begin{align*} \angle PBN &= \angle ABN - \angle B = \angle ANM - \angle B = \angle ACD - \angle B\\ &= \angle DCB - \angle B = \angle ADC - 2\angle B = \angle ADC - \angle AMP\\ &= \angle AMN - \angle AMP = \angle PMN. \end{align*}Thus quadrilateral $PMBN$ is cyclic, as desired. This implies that, actually, the circumcircle of $\triangle BMP$ is tangent to $\overline{AT}$ at $N$, which is exactly what we wanted.

Sumgato wrote: In an acute-angled triangle $ABC$, let $M$ be the midpoint of $AB$ and $P$ the foot of the altitude to $BC$. Prove that if $AC+BC = \sqrt{2}AB$, then the circumcircle of triangle $BMP$ is tangent to $AC$. Very funny and easy problem. My solution is to take F from AC s.t $AF=\sqrt{2}AM$ and prove F,M,B,P is clycic,as shown in the diagram. Chinese version：https://mp.weixin.qq.com/s/gtyLLiWVk52vr2UmqZYvQg

Also syrian MO 2021 P4 $\color{red} \textbf{case 1:}$$AC>BC$ let $X$ be a point on $AC$ such that $CX=\frac{AC-BC}{2}$, We claim that it is the tangency point By law of cosins we get $AB^2=AC^2+BC^2-2AC.BC.cos\hat{C}$ equivalintly $\frac{\left({AC-BC}\right) ^2}{2}=BC.AC.cos\hat{C}=CP.CB$ so $AC$ tangents $(PBX)$ $AX^2=\frac{(AC+BC)^2}{2}=\frac12AB^2=AM.AB$ so $AX$ tangents $(BMX)$. Now there exists unique circle that passes through $B$ and $X$ and tangents $AC$ at $X$ so $B,P,M,X$ are concyclic and we are done. $\color{red} \textbf{case 2:}$$AC<BC$ Let $X$ be a point on rey $AC$ behind $C$ such that $CX=\frac{BC-AC}{2}$ , similar argument to the first case finishes. $\color{red} \textbf{case 3:}$$AC=BC$ Trivial $\blacksquare$