wdym by medians here, isnt a median unique?

MathLuis wrote: wdym by medians here, isnt a median unique? My bad, it's supposed to be median and symedian.

Let $X$ be the foot from $A$ onto $BC$, $A'$ the antipode of $A$ on $(ABC)$ and $MH \cap (ABC) = Z,A'$. Then $2 \cdot LH \cdot HK = 2 \cdot AH \cdot HZ = P_{(ABC)}(H) = HZ \cdot HA' = 2 \cdot HZ \cdot HM$ so $KMLZ$ is cyclic. I claim $Z$ is the tangency point. Note that $H$ is the humpty point in triangle $BZC$. To show the tangency, it suffices to show that if $Y = ZL \cap (ABC)$. Then $YA' \parallel BC$, or $ZL$ is symmedian. But this follows since $\frac{BL}{LC} = \frac{BH^2}{CH^2} = \frac{BZ^2}{CZ^2}$, so we're done. $\blacksquare$

Sol:- Let $D,E,F$ be feet of $A,B,C$ on $BC,CA,AB$ respectively and $M$ be midpoint of $BC$.$J$ be the $A$ queue point. Then inversion wrt $H$ with power $HD \cdot HA$ followed by reflection across $H$ maps $(AEFH)$ to $BC$ so $KMLJ$ is cyclic. This inversion maps $(ABC)$ to $(DEFM)$.So it suffices to prove that $(DEFM)$ and $(KMLJ)$ are tangent. $JKEF$ is isosceles trapezoid (because of isogonals) so the common perp bisector of $EF$ and $KJ$ is the line joining centers of $(DEFM)$ and $(KMLJ)$. Since $ME=MF \implies M$ lie on the line joining the centers of $(DEFM)$ and $(KMLJ)$ hence they are tangent at $M$.

Here is one sol using American geo. (LMAO now that i realice its funny to use american thing on russian problem) Let $Q_A$ the A-quequepoint and let $AH \cap (ABC)=H_A$ and $AH \cap BC=D$ note that $-1=(AQ_A \cap BC, D; B, C) \overset{A}{=} (Q_A, H_A; B, C)$ and also note that $(BHC), (ABC)$ are symetric w.r.t. $BC$ by LoS so if u let $T$ a point in $BC$ such that $-1=(B, C; L, T)$ then $HT$ is tangent to $(BHC)$ but this means $H_AT$ tangent to $(ABC)$ so $Q_AT$ tangent to $(ABC)$ now by McLaurin and PoP we have that $Q_AT$ is tangent to $(Q_ALM)$, now clearly $AKLD$ is cyclic but also since its known that $\angle AQ_AM=90$ we have that $AQ_ADM$ cyclic so by PoP we have $KH \cdot HL=AH \cdot HD=Q_AH \cdot HM$ so $KMLQ_A$ is cyclic meaning that its circumcircle its tangent to $(ABC)$ thus we are done .