Let $L$ be an intersection of $QH$ and the line through $C$ perpendicular to $AC$. And let $D$ be a midpoint of $AC$, $ E = BH \cap AC, F=AH \cap BC, G = CH \cap AB, T = GF \cap AB, \alpha $ be a circumcircle of $\triangle AHC, \beta$ be a circumcircle of quadrilateral $BGHF$, and $ O = \alpha \cap \beta$. First of all, $GF \parallel PQ$, and Desargues's theorem gives that: $PQ, KL, AC$ meets at $R$ Claim 1: $B, O$, and $D$ are collinear. (Proof of Claim 1) Since $GOHF$, $AOHC$, $GACF$ are concyclic, $GF,OH,AC$ are concurrent at $T$. Clearly, $(AC;ET) = -1$. Then, the projective map perspective at $H$ to $\beta$ gives $(FG;BO) = -1$. And we have $(AC;D\infty) = -1$. Now, Consider a projective map perspective at $B$ from $\beta$ to line $AC$. It gives that $B,O,D$ are concurrent. (Note that, $BH$ is the diameter of $\beta$) $\Box$ Claim 2: $BH = PH = QH$, and $KA = KP$, $LQ = LC$. (Proof of Claim 2) $\angle GBH = \angle ECH = \angle GPH$ $\angle FBH = \angle EAH = \angle FQH$ Thus, $BH = PH = QH$ holds. Since $ KA \parallel BH \parallel LC $, we have $KA=KP, LQ = LC$ $\Box$ Let $\gamma$ be a circumcircle of $BPQ$, I be an its center. Clearly, $BH$ is a ridius of \gamma and a diameter of \beta, so we have $2BG = BP, 2BF = BQ$. Let $M = AK \cap IG, N = CL \cap IF$. Then, we have $MK \parallel IH \parallel NL$ and also we have $MK = IH = NL$. Thus, $MKLN$ must be parallelogram, hence, $MN \parallel KL$. Furthermore, Desargues's theorem implies that: $GF, MN, AC$ meets at a point. But, already $AC, GF$ meets at $T$. So $MN$ also go through $T$. Now, enough to show that: $MHN$ are collinear. If so, $MN \parallel OH$ implies the conclusion. (Note that, $BGOHF$ are concyclic, hence, Claim 1 implies $OH \perp BD$) Let $ X = AK \cap IF, Y = CL \cap IG$. By Desargues's again, it is enough to show $XBY$ are collinear. Let $Z$ be a point at infinity on line $IH$. For six point $IGAZCF$, $IZ, GC, AF$ meets at $H$. Thus, by Pappus's theorem, $XBY$ are collinear. $\Box$

I can't believe that I couldn't prove for two hours that $KL, AC, PQ$ are concurrent (as I see now in the above post - by one line Desargue's theorem for triangles $CLQ$ and $PKA$), but I instead proved that $KL$ is perpendicular to $BM$ by trig-length bash much faster, and as I see the above solution, I now think this was the harder part (though, at first I thought it was exactly the opposite). Sketch of my proof is as follows: we have to prove $BK^2+LM^2=MK^2+BL^2$ (well-known perpendicularity criterion, provable with Pythagorean theorems). Easy angle chase gives $\angle KAP= \angle KPA=\angle HBP= \angle HPB=90-\alpha$ (hence $BH=HP, KP=KA$), so it's easy to calculate that $BP=4R.cos \beta. sin \alpha$ (we have $BH=2Rcos \beta$) and $PA=2R(sin \gamma-2cos \beta sin \alpha)$ and $AK=\frac{PA}{2sin \alpha} =R(\frac{b^2-a^2}{ac})$ (using $\frac {sin \gamma} {sin \alpha}= \frac {c} {a}$ and $cos \beta= \frac{a^2+c^2-b^2}{2ac}$), so we can now calculate $MK^2$ by Pythagoras theorem in $\triangle MKA$. Now we can calculate $BK^2$ with LoC in triangle $BHK$ by using $cos 2\alpha=2cos \alpha^2+1=2(\frac{b^2+c^2-a^2}{2bc})^2+1$ and $HK=BH+AK$, and calculating everything in terms of $a, b, c, R$ gives a symmetric expression for $a, c$ of $BK^2-MK^2$, so this finishes this approach. Edit: Now that the official solutions are released, here is a beautiful synthetic proof. Consider the circles $\omega_a=(K, KA)$ and $\omega_c=(L, LC)$. Note that Monge's theorem for $\omega=(BPQ), \omega_a, \omega_c$ gives that $R$ is exsimilicenter for $\omega_a, \omega_c$, hence $R$ lies on $KL$. To prove the perpendicularity part, note that $B$ is the radical center of $\omega_a, \omega_b, \Omega=(AHC)$, and $M$ has equal powers wrt $\omega_a$ and $\omega_c$, hence $BM$ is their radical axis and $BM \perp KL$.

Complex bashed during the contest: We employ complex numbers with $(ABC)$ being the unit circle. Then $h=a+b+c$. Let $A_{1}$ and $C_{1}$ be the feet of the altitudes from $A$ and $C$ in $\triangle ABC$ respectively. Then $\angle HAQ=\angle HCQ=\angle HAB$, so $Q$ is the reflection of $B$ across $A_{1}$. Therefore we have: $q=2a_{1}-b=(b+c+a-bc\overline{a})-b=c+a-\frac{bc}{a}$. Analogously, $p=c+a-\frac{ab}{c}$. Now we calculate $r$: \[R\in AC\Longrightarrow \overline{r}=\frac{a+c-r}{ac},\quad R\in PQ\Longrightarrow \frac{p-r}{p-q}\in\mathbb{R}\Longrightarrow \frac{c+a-\frac{ab}{c}-r}{(c+a-\frac{ab}{c})-(c+a-\frac{bc}{a})}\in\mathbb{R}\Longrightarrow \frac{c+a-\frac{ab}{c}-r}{b\frac{(c-a)(c+a)}{ac}}\in\mathbb{R}\]However, it's easy to notice that $\frac{(c-a)(c+a)}{ac}\in i\mathbb{R}\Longrightarrow \frac{c+a-\frac{ab}{c}-r}{b}\in i\mathbb{R}$. \[\Longrightarrow \frac{c+a-\frac{ab}{c}-r}{b}=-\overline{\left(\frac{c+a-\frac{ab}{c}-r}{b}\right)}=-\frac{\frac{1}{c}+\frac{1}{a}-\frac{c}{ab}-\frac{a+c-r}{ac}}{\frac{1}{b}}=\frac{c^2-br}{ac}\Longrightarrow r=\frac{c^2b+a^2b-a^2c-c^2a}{b^2-ac}\]Now we want to calculate $k$: $KA\perp AC\Longrightarrow \overline{k}=\frac{k-a+c}{ac}$ because the perpendicularity is equivalent to $K$ lying on the chord determined by $A$ and the diametrically opposite point of $C$ in the unit circle hence the formula for $\overline{k}$. \[K\in \overline{PH}\Longrightarrow \frac{k-h}{p-h}\in\mathbb{R}\Longrightarrow \frac{k-a-b-c}{(c+a-\frac{ab}{c})-(a+b+c)}=-\frac{c(k-a-b-c)}{b(a+c)}\in\mathbb{R}\]\[\Longrightarrow \frac{c(k-a-b-c)}{b(a+c)}=\frac{\frac{1}{c}(\frac{k-a+c}{ac}-\frac{1}{a}-\frac{1}{b}-\frac{1}{c})}{\frac{1}{b}(\frac{1}{a}+\frac{1}{c})}\]\[\Longrightarrow c^2(a+c)(k-a-b-c)=b(a+c)(bk-2ab-ac)\Longrightarrow k=\frac{2ab^2+abc-ac^2-bc^2-c^3}{b^2-c^2}\]Now we want to show that $BM\perp KR$ where $M$ is the midpoint of $AC$: \[BM\perp KR\iff \frac{k-r}{b-m}\in i\mathbb{R}\iff \frac{\frac{2ab^2+abc-ac^2-bc^2-c^3}{b^2-c^2}-\frac{c^2b+a^2b-a^2c-c^2a}{b^2-ac}}{2b-a-c}\in i\mathbb{R}\]\[\iff \frac{(b^2-ac)(2ab^2+abc-ac^2-bc^2-c^3)-(b^2-c^2)(c^2b+a^2b-a^2c-c^2a)}{(b-c)(b+c)(b^2-ac)(2b-a-c)}\in i\mathbb{R}\]\[\iff \frac{b (b + c) (a b - c^2)(2b-a-c)}{(b-c)(b+c)(b^2-ac)(2b-a-c)}\in i\mathbb{R}\iff \frac{b(ab-c^2)}{(b-c)(b^2-ac)}\in i\mathbb{R}\]However, $\overline{\left(\frac{b(ab-c^2)}{(b-c)(b^2-ac)}\right)}=\frac{\frac{1}{b}(\frac{1}{ab}-\frac{1}{c^2})}{(\frac{1}{b}-\frac{1}{c})(\frac{1}{b^2}-\frac{1}{ac})}=\frac{b(c^2-ab)}{(c-b)(ac-b^2)}=-\left(\frac{b(ab-c^2)}{(b-c)(b^2-ac)}\right)$, so we're done.

There is another nice way to prove $K,L,R$ are collinear. Let $\infty$ be the point at infinity along lines $BH,AK,CL$. Note that $$ \triangle \infty AC ~,~ \triangle HPQ$$are perspective, since lines $H\infty, AP,CQ$ concur at $B$. It follows that the three points $$ AC \cap PQ = R ~,~ A \infty \cap HP = K ~,~ C \infty \cap HQ = L $$are collinear, as desired. Another slick way to prove collinearity was mentioned in the official solution. Consider the three circles $$ \odot(H,HB) ~,~ \odot(K,KA) ~,~\odot(L,LC) $$Use that their exsimilicenters are collinear.

After a long tme I am doing Geo Officially. First to prove $KA=KP$ is just an angle chase and similarily for $E$. Let $HQ$ meet $C\perp$ to $AC$ at $E$ and $D$ be the midpoint of $AC$.Since, $DA^2=DC^2$ and $BQ.BC=BP.BA$, So, $BD$ is the radical axis of $(E,EQ)$ and $(K,KA)$. So, $KE\perp ND$ And as $AP$,$KA$ and $KP$ intersect $CQ$,$EC$ and $EQ$ at $B$,$\infty$ and $H$. So, by desargues $PQ$,$EK$ and $AC$ concur. So, $KR \perp BD$.

Let $D$ be the midpoint of $AC$, point $I$ lies on $PQ$ satisfied $BI \parallel AC$. $A_1,B_1,C_1$ are projection of $A,B,C$ on $BC,CA,BA$. $A_1C_1$ meets $AC$ at $J$, $JH$ meets $BD$ at $G$. Claim 1: $G$ is one of the intersections of $(BH)$ and $(AHC)$ Since $(JB_1,AC)=-1$ and $D$ is the midpoint of $AC$, then $\begin{cases} \overline{B_1D} \cdot \overline{B_1J}=\overline{B_1A} \cdot \overline{B_1C} \\ \overline{JA} \cdot \overline{JC}=\overline{JB_1} \cdot \overline{JD} \end{cases}$ But $\overline{B_1A} \cdot \overline{B_1C}=\overline{B_1H} \cdot \overline{B_1B} \Rightarrow \overline{B_1D} \cdot \overline{B_1J}=\overline{B_1H} \cdot \overline{B_1B}$ This leads to $H$ is the orthocenter of $\Delta JBD$, therefore $\widehat{JGD}=90^o$, or $G \in (BH)$ Moreover, $\overline{JH} \cdot \overline{JG}=\overline{JB_1} \cdot \overline{JD} \Rightarrow \overline{JH} \cdot \overline{JG}=\overline{JA} \cdot \overline{JC}$. $\Rightarrow G \in (AHC)$ Claim 2: $H,G,I$ are concurrent Easy to see $\widehat{IBQ}=\widehat{BCA}=\widehat{BPQ} \Rightarrow \Delta IBQ \sim \Delta IPB$ $\Rightarrow IB^2=\overline{IP} \cdot \overline{IQ}$ $\Rightarrow \mathcal{P}_{I \setminus (BH)}=\mathcal{P}_{I \setminus (AHC)}$ $\Rightarrow I,G,H$ are concurrent. Claim 3: $KR \perp BD$ Using Thales theorem for $BI \parallel AR$ we have: $\dfrac{\overline{PR}}{\overline{PI}}=\dfrac{\overline{PA}}{\overline{PB}}$ Since $KA \perp AC, AC \perp BH$ then $AK \parallel BH$ Using Thales theorem for $AK \parallel BH$ we have: $\dfrac{\overline{PK}}{\overline{PH}}=\dfrac{\overline{PA}}{\overline{PB}}$ $\Rightarrow \dfrac{\overline{PK}}{\overline{PH}}=\dfrac{\overline{PR}}{\overline{PI}} \Rightarrow KR \parallel HI$ But $HI \perp BD \Rightarrow KR \perp BD$

Attachments:

Cute one Let $K'$ be where perpendicular at $C$ to $AC$ meets $HQ$. Let $D$ be midpoint of $AC$. Note that $\angle APK = \angle APH = \angle ACH = \angle ABH = \angle PAK$ so $KA = KP$ and similarly $K'C = K'Q$. Let $\omega_1$ and $\omega_2$ be circles with centers $K$ and $K'$ and radiuses $KA$ and $K'C$. obviously $BD$ is Radical Axis of $\omega_1$ and $\omega_2$ so $KK' \perp BD$ so we need to prove $R$ lies on $KK'$. Applying Desargues on $AKP$ and $CK'Q$ easily proves it.