Perform a $-\sqrt{HA\cdot HD}$ Inversion around $H$. We get the following Equivalent Problem. $\textbf{INVERTED PROBLEM:-}$ $ABC$ be a triangle with Circumcenter $O$ and Incenter $I$. Let $\overline{OI}\cap\odot(ABC)=T$ and $\odot(AIT)\cap\odot(ABC)=K$. Let $A'$ be the reflection of $A$ over $\overline{IO}$. and $D$ be the point where the Incircle of $\Delta ABC$ touches $\overline{BC}$. Then $A',D,K$ are collinear. By Radical Axis Theorem on $\odot(ABC),\odot(BIC),\odot(AIT)$ we get that $\overline{AK},\overline{IO},\overline{BC}$ are concurrent. Now the rest follows from #4 here. $\blacksquare$

Let $a,b,c$ be complex numbers lying on unit circle such that they are vertices of the triangle. Then $$h=a+b+c, a'=b+c-a,\ p=\frac{h\overline{a}}{\overline{h}}=bc\cdot\frac{a+b+c}{ab+bc+ca}.$$From $R$ definition $$\left(\overline{r}=\frac{b+c-r}{bc}\ \wedge\ \frac{\overline{r}}{\overline{h}}=\frac{r}{h}\right)\iff r=\frac{a(b+c)(a+b+c)}{a^2+2a(b+c)+bc}.$$$Q$ definition implies $$\overline{r}=\frac{a+q-r}{aq}\iff q=\frac{a-r}{a\cdot\overline{r}-1}=\frac{(a+b)(a+c)-(b+c)^2}{a^2(b+c)^2-(a+b)(a+c)bc}\cdot abc.$$Moving on... $$h-p=\frac{a(b+c)(a+b+c)}{ab+bc+ca},\ h-a'=2a,$$$$q-a'=\frac{(a-b)(a-c)(b+c)(ab+bc+ca)}{a^2(b+c)^2-(a+b)(a+c)bc},$$$$q-p=\frac{bc(b+c)(a^2-bc)(a^2+2a(b+c)+bc)}{(ab+bc+ca)(a^2(b+c)^2-(a+b)(a+c)bc)}.$$Therefore $$\frac{h-p}{h-a'}\cdot\frac{q-a'}{q-p}=\frac{(a+b+c)(a-b)(a-c)(b+c)(ab+bc+ca)}{2bc(a^2-bc)(a^2+2a(b+c)+bc)}=\overline{\left(\frac{h-p}{h-a'}\cdot\frac{q-a'}{q-p}\right)}\square.$$#1726

amar_04 wrote: Perform a $-\sqrt{HA\cdot HD}$ Inversion around $H$. We get the following Equivalent Problem. $\textbf{INVERTED PROBLEM:-}$ $ABC$ be a triangle with Circumcenter $O$ and Incenter $I$. Let $\overline{OI}\cap\odot(ABC)=T$ and $\odot(AIT)\cap\odot(ABC)=K$. Let $A'$ be the reflection of $A$ over $\overline{IO}$. and $D$ be the point where the Incircle of $\Delta ABC$ touches $\overline{BC}$. Then $A',D,K$ are collinear. By Radical Axis Theorem on $\odot(ABC),\odot(BIC),\odot(AIT)$ we get that $\overline{AK},\overline{IO},\overline{BC}$ are concurrent. Now the rest follows from #4 here. $\blacksquare$ That's how this problem was produced.

I didn't know how to do the original TST problem, but I somehow manage to solve this one I think my solution is kind of cute so I'm posting it here. Note that $B,H,C,A'$ are concyclic. Therefore it suffices to show that $BC,A'H,PQ$ are concurrent (by, say, power of a point theorem). Let $A_1$ be the antipedal point of $A$ w.r.t. $\odot(ABC)$. Since the intersections $AA_1\cap BC$ and $A'H\cap BC$ are symmetric w.r.t. the midpoint $M$ of $BC$, by the butterfly theorem (or more precisely, a slightly more generalized statement of it), it suffices to show that $AQ\cap BC$ and $PA_1\cap BC$ are symmetric w.r.t. $M$. Note that $M$ is also a midpoint of $HA_1$ and both $OH, PA_1$ are perpendicular to $AP$. Therefore the lines $OH$ and $PA_1$ are symmetric w.r.t. $M$, which gives the desired result.

Let $OH$ meet $BC$ at $T$. Let $AH$ and $AO$ meet $ABC$ at $S$ and $A''$. Let $M$ be midpoint of $BC$. Note that $BHCA'$ is cyclic so by Radical Axis Theorem we need to prove $HA'$ and $PQ$ meet at $BC$. Let $PQ$ and $A'H$ meet $BC$ at $X$ and $Y$. Note that $\frac{BX}{XC} = \frac{BQ}{QC}.\frac{BP}{PC}$ and $\frac{BY}{YC} = \frac{BH}{HC}.\frac{BA'}{A'C}$ so we need to prove $\frac{BQ}{QC}.\frac{BP}{PC} = \frac{BS}{SC}.\frac{AC}{AB}$ or $\frac{BQ}{QC}.\frac{AB}{AC} = \frac{BS}{SC}.\frac{PC}{PB}$ or $\frac{TC}{TB} = \frac{CA''}{BA''}.\frac{PC}{PB}$. Let $T'$ be reflection of $T$ across $M$. we need to prove $\frac{T'C}{T'B} = \frac{CA''}{BA''}.\frac{PC}{PB}$ or in fact we need to prove $T',A'',P$ are collinear. Note that $M$ is midpoint of $HA''$ so $THT'A''$ is parallelogram so $TO || T'A''$ and $PA'' \perp AP \perp TO$ so $PA'' || TO$ so $P,A'',T'$ are collinear as wanted. we're Done.