By Pascal's in $CN_1QM_1BA$, we have that $K$ lies on $MN$. After applying Menelao in $\triangle NQG$ and line $N_1-K-C$ and $\triangle MQG$ and line $M_1-K-B$ we get that $MN\parallel M_1N_1$, so $BN_1M_1C$ is an isosceles trapezoid, so $K$ lies on the perpendicular bisector of segment $BC$. Let $K_1$ be $PQ\cap MN$ and $L$ the midpoint of $BC$. Since $\triangle NLM\sim \triangle ABC$ and $K$ and $P$ are the feet of altitud from $L$ and $A$, we have $\frac{BP}{PC}=\frac{MK}{NK}$, but $\frac{MK_1}{NK_1}=\frac{BP}{PC}$, and this implies that $K$ and $K_1$ are the same point, as desired.

musan1909 wrote: By Pascal's in $CN_1QM_1BA$, we have that $K$ lies on $MN$. After applying Menelao in $\triangle NQG$ and line $N_1-K-C$ and $\triangle MQG$ and line $M_1-K-B$ we get that $MN\parallel M_1N_1$, so $BN_1M_1C$ is an isosceles trapezoid, so $K$ lies on the perpendicular bisector of segment $BC$. Let $K_1$ be $PQ\cap MN$ and $L$ the midpoint of $MN$. Since $\triangle NLM\sim \triangle ABC$ and $K$ and $P$ are the feet of altitud $L$ and $A$, we have $\frac{BP}{PC}=\frac{MK}{NK}$, but $\frac{MK_1}{NK_1}=\frac{BP}{PC}$, and this implies that $K$ and $K_1$ are the same point, as desired. Typo denoting $L $; if follows must be $L $ is the midpoint of $CB $. And I think you must add why $\frac{MK_1}{NK_1}=\frac{BP}{PC}$. It's only let $T=BC\cap AK_1 $ and see that $G $ is also centroid of $APT $. (Easy but important)

Al3jandro0000 wrote: musan1909 wrote: By Pascal's in $CN_1QM_1BA$, we have that $K$ lies on $MN$. After applying Menelao in $\triangle NQG$ and line $N_1-K-C$ and $\triangle MQG$ and line $M_1-K-B$ we get that $MN\parallel M_1N_1$, so $BN_1M_1C$ is an isosceles trapezoid, so $K$ lies on the perpendicular bisector of segment $BC$. Let $K_1$ be $PQ\cap MN$ and $L$ the midpoint of $MN$. Since $\triangle NLM\sim \triangle ABC$ and $K$ and $P$ are the feet of altitud $L$ and $A$, we have $\frac{BP}{PC}=\frac{MK}{NK}$, but $\frac{MK_1}{NK_1}=\frac{BP}{PC}$, and this implies that $K$ and $K_1$ are the same point, as desired. Typo denoting $L $; if follows must be $L $ is the midpoint of $CB $. And I think you must add why $\frac{MK_1}{NK_1}=\frac{BP}{PC}$. It's only let $T=BC\cap AK_1 $ and see that $G $ is also centroid of $APT $. (Easy but important) Fixed. Thanks

Solution. We start with the following Lemma 1. Maintain the point labelling. Let $K'$ be the point where the perpendicular bisector of $\overline{BC}$ meets $MN$. Then $P,\ G$ and $K'$ are collinear. Proof. Consider the homothety $\mathcal{H}$ with center $G$ and ratio $-1/2$. Triangle $ABC$ is sent to $LMN$, where $LM=LN$. Clearly, $\mathcal{H}(P)=\overline{GP}\cap \overline{MN}$ and corresponds to the $L$-foot of altitude of $\bigtriangleup LMN$, which coincides with the perpendicular bisector of $\overline{BC}$, thus $\mathcal{H}(P)=K'$, as desired. $\square$ Back to the problem. From 2011 ISL G4, we conclude that $(QMN)$ and $(ABC)$ are tangent at $Q$, then $BC\parallel MN\parallel M_1N_1$. By Pascal's theorem on $ABM_1QN_1C$ we get that $K$ lies on $MN$. Since $BCM_1N_1$ is an isosceles trapezoid, we conclude that $KB=KC$, i.e. $K=K'$. By lemma 1 we infer the required assertion. $\blacksquare$

Supriced no one did tried this way yet lol. Notice that $Q$ is the A-Why Point of $\triangle ABC$, now let $A'$ a point on $\omega$ such that $AA' \parallel BC$, therefore from homothety we have $P,G,A'$ are colinear, now let $R$ a point on $\omega$ such that $QR \parallel BC$, then from ISL 2011 G4 notice that we have $(QMN)$ tangent to $(ABC)$, which means that $M_1N_1 \parallel MN \parallel BC \parallel AA' \parallel QR$ (lol), so now to finish just consider DDIT on $BNMC$ with perspector from $Q$ which gives that $(QN,QC), (QB, QM), (QA, Q\infty_{BC})$ are pairs of involution, projecting onto $\omega$ gives that $(N_1,C), (B,M_1), (A, R)$ are pairs of involution which means that $AR,CN_1,BM_1$ are concurrent so $K$ lies on $AR$, but because $M_1N_1BC$ is an isosceles trapezoid we can consider reflection over perpendicular bisector of $BC$, clearly $K$ is fixed but $AR$ goes to $A'Q$ which means that $A',Q,K$ are colinear but $A'Q$ is just line $PG$ thus we are done .

By Pascal on $BACN_1QM_1$ we get that $K,M,N$ are collinear, ie $K$ lies on the $A$-midline of $\Delta ABC$. $\measuredangle MAM_1 = \measuredangle CBM_1 = \measuredangle MKM_1$, so $(AKMM_1)$ concyclic. Let $A'$ be the reflection of $A$ across the perpendicular bisector of $BC$ (call this line $\perp_{BC}$). Recall that $Q,P,G,A'$ are collinear in that order. Let $L$ be the intersection of $AK$ with $BC$. $\measuredangle CAL = \measuredangle MAK = \measuredangle MM_1K = \measuredangle QM_1B = \measuredangle QA'B = \measuredangle PA'B$, so by reflection across $\perp_{BC}$ we get that $BP=CL$. Now we get that $BP=2MK$, and since $BG=2MG$, and $MN // BC$, we get that $P,G,K$ are collinear.