First prove that $(BCPQ)$ is the reflection of $(BAC)$ respect to $BC$, so the orthocenter $H$ lies on $(BCPQ)$. Define $O$ to be the circumcenter of $\bigtriangleup ABC$. Lemma 1. Lines $OH$, $DQ$ and $EP$ concur at a point $K$ on $(BCPQ)$. Proof. Let's redefine $K$ as $K=\overline{OH} \cap (BCPQ),\ K\neq H$ and $L=\overline{AC}\cap \overline{OE}$. It's easy to note that $\angle OKP=\angle OLA=\angle OBC$, then $OLBC$ and $ OLPK$ are cyclic, hence $OL,\ BC$ and $KP$ are concurrent; thus $K$ lies on $EP$. Similarly, $K$ is on $DQ$. The claim follows. Since $\angle HBP=90^\circ - \angle A=\angle HCQ$, we can easily conclude that $\angle DKH=\angle HKE$, i.e. $OH$ is the internal bisector of $\angle DKE$. Lemma 2. $OH$ is the internal angle bisector of $\angle AKM$. Proof. Let $K',\ A'$ be the symmetric of $K,\ H$ across $M$, respectively. It's known $A'$ is the antipode of $A$ on $(BAC)$ and being $(BAC)$ and $(BCPQ)$ congruent, $K'$ lies on $(ABC)$. Observe that $KHK'A'$ is a parallelogram and $\angle AK'A'=90^\circ$, hence $KO$ is the perpendicular bisector of $\overline{AK'}$, hereby $AK= 2KM$. Since $G=\overline{OH}\cap\overline{AM}$ is the centroid of $\bigtriangleup ABC$, we get $\frac{AK}{KM}=\frac{1}{2}=\frac{AG}{GM}$. The result follows by the angle bisector theorem. Finally, we conclude that $AK$ and $KM$ are symmetric across the internal bisector of $\angle DKE$; which implies the required result.

I'm the author of this problem. The solution I submitted is rather similar to the above, so I won't post it unless it is requested. I've heard rumors that only one contestant managed to fully solve the problem during the competition. I can't confirm these rumors as I am not presently at the event, but I didn't expect the attendees to find the problem that hard.

I think it's correctly placed. The main lemma (and possibly the hardest part) is how to prove the concurrency on $(BCPQ)$ and recognize that $OH$ passes through $K$. Congrats!

Hello Jafet98. I would like to make this comment respect to Lemma 1: (...) <OKP=<OLA=< OBC, then OLBC and OLPK are cyclic-->It's true. Ok! Hence OL, BC and KP are concurrent-->I think this implication is not correct

florodamian wrote: Hello Jafet98. I would like to make this comment respect to Lemma 1: (...) <OKP=<OLA=< OBC, then OLBC and OLPK are cyclic-->It's true. Ok! Hence OL, BC and KP are concurrent-->I think this implication is not correct Just apply the radical axes theorem to circles $(OLBC),\ (OLPK)$ and $(BCPK)$. Since $\bigtriangleup ABC$ is scalene, we get rid of the case $OL\parallel BC\parallel KP$.

Nice problem!! Let second intersection of $(ABC)$ and $AD,AE$ be $R,S$ respectively. Define $(ASP)\cap (ARQ)=K'(\neq A)$. Since $(ABC)$ and $(BCPQ)$ are congruent, we can check easily that $PQ\parallel SR$ and $PQ=SR$ so $PQRS$ is parallelogram. Thus $$\measuredangle PK'Q=\measuredangle DRQ+\measuredangle PSE=\measuredangle DAE=\measuredangle PBQ,$$means that $BCPK'Q$ is cyclic. Note that $D$ is radical center of $(ARBC),(BCK'Q),(ARQK')$, we have $D,Q,K'$ are collinear. Similarly, $E,P,K'$ are collinear hence $K'$ coincides with $K$. Now $\angle EKS=\angle PAE=\angle BCQ=\angle BKD$ so it suffices to show that $\angle BKM=\angle AKS$. Let $M'$ be the midpoint of $RS$. From simple angle chasing we get $\triangle KBC\sim\triangle KRS$ and $AS,AR$ tangent to $(RKS)$. Thus $KA$ is A-symmedian of $\triangle KRS$, implying that $\angle BKM=\angle RKM'=\angle AKS$ as desired.

Attachments: