Hell Edition P1 แทบจะใช้ความรู้ทุกเรื่องใน สสวท 2 แนะนำว่าอย่าอ่านจะดีกว่าครับ
Let $\omega$ be the incircle of $\Delta ABC$.
Let $T$ be the point which $\odot(BTC)$ touches $\omega$.
Let $M, N$ be the midpoint of $BC, DI_A$ resp., where $I_A$ is the $A$-excenter of $\Delta ABC$. Notice that $MN\perp BC$. Thus $NB=NC$. Moreover, by taking homothety ratio $0.5$ at $D$, we see that $I_AE=I_AF\implies NP=NQ$ thus $N$ is the center of $BCPQ$.
Let $K = BC\cap PQ$. It's well known that $PQ$ is the radical axis of $\odot(BIC)$ and $\omega$. Thus $K$ is the radical center of $\omega, \odot(BIC)$ and $\odot(ABC)$. Thus $KB\cdot KC = KD^2$. Moreover, by radical axis on $\odot(BTC), \odot(ABC), \omega$, we deduce that $KT$ touches $\omega$.
Now we are ready to prove that $T, P, Q, X$ are concyclic. We invert around $\omega$, which sends $P\to B$, $Q\to C$, $X\to \odot(BIQ)\cap\odot(CIP) = Y$, which is the Miquel Point of self-crossing quadrilateral $BQCP$. Thus $Y$ is the foot from $I$ to $KN$. We have to show that $BTCY$ is cyclic.
By ISL 2002 G7, $T$ lies on $DI_A$. Moreover, by homothety, $TD$ bisects $\angle BTC$. Thus $BTCN$ is cyclic. Moreover, note that $\angle BYI = 180^{\circ}-\angle BPX = 180^{\circ}-\angle CQX = \angle CYI$. So $YN$ is the external bisector of $\angle BTC$. Hence $BTCYN$ is cyclic as claimed.
Finally, notice that $KT^2 = KB\cdot KC = KP\cdot KQ$. Thus $KT$ is tangent to both of $\omega$ and $\odot(TPQX)$ thus both circles are tangent at $T$ is desired.