Nice problem.

Attachments:
Problem 20.pdf (251kb)

I did find quite a few claims relating to this configuration, a bunch of parallelograms and isosceles trapezium, and the midpoint of $CH$ lying on the simson line of $K$, however couldn't find a synthetic proof though. Complex bash ftw! If we define $K$ to be the point on $(ABC)$ such that $OK \parallel AB$. Then $M, H, K$ collinear is equivalent to \[\frac{(a^2 + b^2)(c^2 + abi) (a^2 c^2 + 2 a^2 b^2 + b^2 c^2 + (a b^3 + 2 a b c^2 + a^3 b) i)}{2a^2b^2c^2} = 0.\] $PH \parallel BC$ is equivalent to \[i(a + ib)(a^2 c^2 + 2 a^2 b^2 + b^2 c^2 + (a b^3 + 2 a b c^2 + a^3 b) i) = 0.\] So, infact the problem isn't true if $OC \parallel AB$ (which is trivial since $P$ degenrates to $A$ and by definition $AH \perp BC$).

Very nice problem. Vrangr wrote: Let $O$ be the circumcenter of triangle ABC, $H$ be its orthocenter, and $M$ be the midpoint of $AB$. The line $MH$ meets the line passing through $O$ and parallel to $AB$ at point $K$ lying on the circumcircle of $ABC$. Let $P$ be the projection of $K$ onto $AC$. Prove that $PH \parallel BC$. Solution 20. Let $X$ be the midpoint of $CH.$ $C'$ be the antipode of $C$ with respect to the circumcircle of $\triangle ABC.$ Let $Q$ and $R$ be the feet of perpendiculars from $K$ to $CB$ and $AB$ respectively. Consider the following claims, Claim 1. $KXHR$ is a rhombus. Proof. By Simson Line Lemma, $P,Q,R$ are collinear. Also, $XM$ is the diameter of the nine point circle of $\triangle ABC,$ but it is well known that the radius of the nine point circle is $\tfrac R2$ where $R$ is the circumradius of $\triangle ABC.$ Therefore, $AO=R=XM,$ also $CX\parallel OM,$ this means that $CXMO$ is a parallelogram. Now, note that $KR\parallel XH$ as both are perpendicular to $AB$ and $XH=CX=OM=KR$ as $OMRK$ is a rectangle. Therefore, $KXHR$ is a parallelogram. Let the Simson line of $K$ intersect $CH$ at $X'.$ Consider the following lemmas: Lemma 1.Let $ABC$ be a triangle with orthocenter $H.$ If $P$ is a point on circle $ABC,$ then its simson line bisects $\overline{PH}.$ Lemma 2.Let $H$ be the orthocenter of triangle $ABC.$ $M$ be the midpoint of $BC.$ Then the reflection of $H$ over $M$ is the antipode of $A$ in circle $(ABC).$ Then using the first lemma, we get that the simson line of K namely $\overline{PQRX'}$ bisects $KH,$ but also $X'H\parallel KR,$ combining this two results gives that $KX'HR$ is a parallelogram by congruent triangles. So, we conclude that $X=X'.$ By the second lemma, we note that $H,M,C'$ are collinear. Therefore, $\angle CKH=90^{\circ},$ also since $X$ is the midpoint of $AH,$ $X$ must be the circumcenter of $\triangle KCH.$ So, it follows that $XK=XH.$ Hence the claim.$\square$ Claim 2. $KPQC$ is an isosceles trapezium. Proof. Since $CX\parallel KR,$ and $CX=OM=KR,$ we know that $CXRK$ is a parallelogram. Therefore $PQ\parallel KC.$ Since $OK\parallel AB$ and $C'A\parallel KP$ (because $C'A\perp CA$ and $KP\perp CA$) we have $\angle OKP=\angle BAC'.$ Now \begin{align*} \angle QCK &= \angle QCO+\angle OCK\\ &=\angle BCC'+\angle OKC\\ &= \angle BAC'+\angle OKC\\ &= \angle OKP+\angle OKC\\ &=\angle PKC. \end{align*}Thus, $KPQC$ is a isosceles trapezium. $\square$ Coming back to the original problem, By Claim 1 and Claim 2, $CQ=KP=PH$ and $QH=QK=CP.$ Thus, $CPHQ$ is a parallelogram which gives what we want. Done. $\square$

Attachments:

I loved this question!

Attachments:
Sharygin_P20_Pranjal.pdf (394kb)

This was my personal favorite of the contest. My solution is synthetic and involves the usage of Simson lines, which most of the other solutions do.

Attachments:
Sharygin- 20.pdf (32kb)

[asy][asy] size(8cm); defaultpen(fontsize(8.5pt)); defaultpen(linewidth(0.35)); dotfactor *= 1.5; pair B = dir(230), C = dir(310), O = origin, M = (B+C)/2, K = dir(180), A = intersectionpoint(K+dir(B--foot(B,M,K))*0.00069--K+dir(B--foot(B,M,K))*69,unitcircle), H = orthocenter(A,B,C), R = foot(K,B,C), P = foot(K,A,B), Q = foot(K,C,A), S = (A+H)/2; draw(A--B--C--A--K^^P--H--A^^H--Q--K--P^^P--Q^^H--K--S^^unitcircle); draw(H--2*foot(A,B,C)-H, dotted); draw(K--2O-K--A, dashed); dot("$A$", A, dir(100)); dot("$B$", B, dir(240)); dot("$C$", C, dir(310)); dot("$K$", K, dir(180)); dot("$O$", O, dir(270)); dot("$H$", H, dir(240)); dot("$P$", P, dir(215)); dot("$Q$", Q, dir(30)); dot("$S$", S, dir(135)); dot("$K'$", 2O-K, dir(0)); dot("$R$", 2*foot(A,B,C)-H, dir(250)); [/asy][/asy] It is well-known that $\angle AKH = 90^\circ$. Let $S$ be the midpoint of $\overline{AH}$, $R = \overline{AH} \cap (ABC)$, and $K'$ be the antipode of $K$. Redefine $P$, $Q$ to be the intersection of the line through $S$ parallel to $\overline{AK}$ and $\overline{AB}$, $\overline{AC}$, respectively. Claim: $APHQ$ is a parallelogram. Proof. It suffices to show that $S$ is the midpoint of $\overline{PQ}$, but this follows from $$-1 = (K,R;B,C) \overset{A}{=} (\infty_{PQ},S;P,Q).$$$\square$ The claim tells us that $PK = PH = QA$, so $AQPK$ is an isosceles trapezoid. Then $$\angle AKP + \angle KAP = \angle KAQ + \angle KAB = \angle KAC + \angle K'AC = 90^\circ,$$so $\overline{KP} \perp \overline{AB}$ as desired. (Similarly $\overline{KQ} \perp \overline{AC}$) $\blacksquare$

Nice problem. Let $Q$ be the foot of the perpendicular from $K$ to $\overline{BC}$, and let $T$ be the foot of the perpendicular from $K$ to $\overline{AB}$. Define $N$ as the midpoint of $\overline{CH}$. Let $\overline{AD}, \overline{BE},\overline{CF}$ be altitudes in $\triangle ABC$, and finally suppose that lines $CK, DE, AB$ concur at a point $R$. Note that $\overline{TPQ}$ is the Simson line of $P$ with respect to $\triangle ABC$. Moreover, remark that $KOMT$ is a rectangle, so $\overline{TK}$ is tangent to $(ABC)$. Now $\overline{CK} \parallel \overline{TPQ}$ by a well-known lemma on the simson line (this can also be verified by angle chasing). Now note that $\overline{CK} \perp \overline{HK}$, so we must have that $\overline{TPQ} \perp \overline{HK}$. But remark that $\overline{TPQ}$ bisects $\overline{HK}$ by a well-known lemma, so we find that $\overline{TPQ}$ is the perpendicular bisector of $\overline{HK}$, and so $\overline{PQ}$ passes through $N$. In addition, we see that $-1 = (AB;FR) \stackrel{C}{=} (PQ;N\infty)$ so $N$ is the midpoint of $PQ$. Now $CPHQ$ is a parallelogram, as desired.

Sharygin CR 2019 P20 wrote: Let $O$ be the circumcenter of triangle ABC, $H$ be its orthocenter, and $M$ be the midpoint of $AB$. The line $MH$ meets the line passing through $O$ and parallel to $AB$ at point $K$ lying on the circumcircle of $ABC$. Let $P$ be the projection of $K$ onto $AC$. Prove that $PH \parallel BC$. (Solved with jelena_ivanchic) Define $R$ be foot from $K$ to $BC$, $Q$ the foot from $K$ to $AB$, also let $E$ be the midpoint of $KC$, $F$ the midpoint of $KH$, $D$ the midpoint of $HC$, and $H'$ the reflection of $H$ wrt $M$. It's well known that $\angle HKC=90^{\circ}$. Note that by Simson's Line $\overline{P-Q-R}$ are collinear, we also have $PQ$ bisect $HK.$ We also have $OM\perp AB$, $KC\perp AB \implies OM\parallel KC$, $OK\parallel AB \implies OMQK$ is a rectangle. $FD\parallel KC\implies \angle HFD=90^{\circ} \implies \overline{Q-F-R-D-P}$ are collinear. Now if we show that $D$ is the midpoint of $PR$ then we are done. For this, it is enough to show $PRKC$ is an isosceles trapezoid. Clearly $PR\perp HK$, $CK\perp HK\implies PR\parallel CK$. So enough to show that $RC$, $PK$, $OE$ concur. Now, we have $H'A\parallel PK$, $AB\parallel OK\implies \angle H'AB=\angle OKP$. But we have $\angle H'AB=\angle H'CB=\angle OCR\implies \angle OKC=\angle OCK$ Hence we have that $RC$, $PK$, $OE$ concur. $\blacksquare$

Nice Let $X$ and $Y$ be projections of $K$ onto $AB$ and $BC$. $XYP$ is Simson line of $K$. Claim $: XY || KC$. Proof $:$ Note that $OK || AB \implies \angle OKX = \angle KXA = \angle 90$ so $KX$ is tangent to $ABC$ so $\angle YCK = \angle BCK = \angle BKX = \angle BYX$ It's well known that Simson line of $K$ bisects $HK$ so it also bisects $CH$. Let $N$ be midpoint of $CH$. since $\angle HKC = \angle 90$ we have $KN = CN$. Also since $KYPC$ is cyclic and $YP || CK$, it's isosceles trapezoid so since $KN = NC$ we have $YN = NP$ so $CPHY$ is parallelogram and $PH || BY$.