Just another little configuration of USA December TST for the 56th IMO, by Evan Chen

Li4 wrote: Let $ABC$ be a triangle such that $BC>AB$, $L$ be the internal angle bisector of $\angle ABC$. Let $P,Q$ be the feet from $A,C$ to $L$, respectively. Suppose $M,N$ are the midpoints of $\overline{AC}$ and $\overline{BC}$, respectively. Let $O$ be the circumcenter of triangle $PQM$, and the circumcircle intersects $AC$ at point $H$. Prove that $O,M,N,H$ are concyclic. Claim 1: $Q, M, N$ collinear. Proof: $N$ is the midpoint of right-angled triangle $\bigtriangleup BQC$. Hence $NQ=NB$. Now $\angle NQB = \angle QBN = \angle ABQ$. Thus $NQ$ is parallel to $AB$, which implies that it is the midline of $\bigtriangleup ABC$. Thus $Q, M, N$ collinear. Claim 2: $AHPB$ cyclic. Proof: $\angle PHM = \angle PQM = \angle PBA$. Claim 3: $BHQC$ cyclic. Proof: $\angle BHC = 180^{\circ}-\angle BHA = 180^{\circ} - \angle BPA = 90^{\circ} = \angle BQC$. Now we are ready to finish. Note that $\angle HOM = 2\angle HQM = 2\angle HQN = 180^{\circ}-\angle HNM$. Hence $H, O, M, N$ cyclic.

let $L$ be the midpoint of $AB$ cliam(1):$M,Q,N$ is cyclic proof: $\angle NQB=\angle BQN=\angle QBA \implies NQ || AB$ $\blacksquare$ claim(2): $AHC=90$ (motivation from the nine-pint circle) proof: $\angle QHA=\angle QPL=\angle CBP \implies BQHC$ is cyclic $\blacksquare$ since $BHPA$ is cyclic and so is $BQHC$ $\angle HNQ =2 \angle HQB=2 \angle HAP=\angle HLP \implies \angle HPL=90- \frac{\angle HNQ }{2} \implies \angle HON =180 -\angle HNQ$ and we win

Let $K$ be the feet of altitude from $A$ to $BC$ and $AP\cap BC= D$ CLAIM: $\angle BHC =90^{\circ}$ PROOF: $BP$ perpendicularly bisects $AD \Rightarrow MP||BC$ $\Rightarrow \angle QHM =\angle QPM =\angle QPB =\angle QBC \Rightarrow \square HQBC$ is cyclic. By IRAN LEMMA, $\angle BQC = 90^{\circ} \Rightarrow \angle BHC = 90^{\circ} \blacksquare$ The above claim $\Rightarrow (HMN)$ is Nine-point circle$(\omega_{9})$. So, it suffuices to prove that $O\in \omega_{9}$. CLAIM: $P$ is the incenter of $\Delta HKM$ PROOF: $\Delta MKC$ is isoceles( Too lazy to prove ) $\Rightarrow MP$ bisects $\angle HMK$. Now, as $\square AHKB$ is cyclic and $P\in (AHKB) \Rightarrow P$ is incenter $\blacksquare$. Now kill it with Incenter-Excenter Lemma. Comment: It is a really nice configuration (tho I have no right to say anything) and also the first time I typed this, my power went out and I had to retype everything