https://artofproblemsolving.com/community/c6h85652p498701?fbclid=IwAR39bJvyCLUNBID0--tNmKb0qTKKZDT304AjDznnmeHDq43eyvN7BLOf9PY It is an old problem.

what if this used as tst question?

Solved with help for the first step. The rest was relatively straightforward.

Let $H$ be the common orthocenter and $\Omega$ be the common circumcircle centered at $O$. Note that the nine-point circle $\omega$ is the image of $\Omega$ under homothety $\mathcal H\left(H,\tfrac 12\right)$, so they are shared by $\triangle ABC$ and $\triangle A'B'C'$. Now, we claim that $\Omega$, $\omega$, and $\odot(PQR)$ are coaxial. (Claiming this, in my opinion, is the hardest part of this problem.) To prove this, we set $X = BC'\cap CB'$, and $M$ is the midpoint of $PX$. We also let $D$ and $D'$ be the midpoints of $BC$ and $B'C'$. Notice that by Gauss' line, $M, D, D'$ are collinear. By ISL 2009 G4, we have $MD\cdot MD' = MP^2$. Moreover, by Brokard's, $P$ lies on the polar of $X$ w.r.t. $\Omega$, so $\odot(PX)$ is orthogonal to $\Omega$. Therefore, we have that $$\operatorname{Pow}(M, \Omega) = \operatorname{Pow}(M, \omega) = MP^2.$$Finally, we claim that $\odot(PQR)$ and $\odot(PDD')$ are tangent. To do so, first note that $\triangle ABC$ and $\triangle A'B'C'$ also share the same centroid $G=AD\cap A'D'$, so $AA'\parallel DD'$. Moreover, $\triangle PBC\sim\triangle PB'C'$, so $PD$ and $PD'$ are isogonal w.r.t. $\angle PQR$. These two suffice to imply the desired tangency. Hence, $M$ lies on the common radical axis. By repeating the same argument on the other two sides (constructing $AC'\cap A'C$ and $AB'\cap A'B$), we get two more points on the common radical axis, done.

Complex numbers, anyone?

Complete_quadrilateral wrote: Complex numbers, anyone? I loudly claimed that this was possible at mop (in fact, all you need to show is that the numerator of the circumcenter formula for $PQR$ is a pure imaginary multiple of $a+b+c=d+e+f$) and then later realized that my numerator would have degree $1728$ terms if I cleared demonimators.

IAmTheHazard wrote: Complete_quadrilateral wrote: Complex numbers, anyone? I loudly claimed that this was possible at mop (in fact, all you need to show is that the numerator of the circumcenter formula for $PQR$ is a pure imaginary multiple of $a+b+c=d+e+f$) and then later realized that my numerator would have degree $1728$ terms if I cleared demonimators. There must exist some elegant way, I'm almost sure of it

I like it very much , even though it may be discovered before. What a pity it wasn't chosen last year . Let $A_0$ the antipode of ${A}$, $AH$ meet $\odot O$ and $BC$ at ${D}$ and ${H_A}$, ${M_A}$ be midpoint of ${BC}$. ${BC}$ cross $AA'$, $A'D'$ and $A'A_0$ at $X, Z, U$. Define $A_0', D', H_A', M_A', Y, T, V$ similarly. Call $\triangle {ABC}$ and $\triangle {A'B'C'}$ 's shared NPC be $\Omega_9$. Claim: There exists a circle through $X, Y$ and coaxical to $\odot O$ and $\Omega_9$. We just need $\frac{AX*A'X}{H_AX*M_AX}=\frac{AY*A'Y}{H_A'Y*M_A'Y}$, which is $\frac{UX}{M_AX}=\frac{VY}{M_AY}$. By $\triangle AD'H \cup \{H_A'\} \cup \{T\} \sim \triangle A'DH \cup \{H_A\} \cup \{Z\}$, $\frac{A'U}{UA_0}=\frac{A'Z}{ZH}=\frac{AT}{TH}=\frac{AV}{VA_0'}$, which is $A'U=AV$, so $AA'UV$ is a rectangle. Thus $UV \parallel AA' \parallel A_0A_0' \parallel M_AM_A'$, and $\frac{UX}{M_AX}=\frac{VY}{M_AY}$. We're almost done. Let $AA'$ cross the radical axis of $\odot O$ and $\Omega_9$ at ${S}$. By DDIT $\{A, A'\}, \{X, Y\}, \{Q, R\}$ is a pair of involution, so $SA*SA'=SX*SY=SQ*SR$, so $\odot PQR$ is coaxical to $\odot O$ and $\Omega_9$, and $\triangle {PQR}$'s circumcenter lies on ${OH}$. Done!

We prove more strongly that the circumcircle of the triangle determined by $\overline{AA'},\overline{BB'},\overline{CC'}$ is coaxial with $(ABC)$ and the shared nine-point circle of $\triangle ABC$ and $\triangle A'B'C'$, which is the image of $(ABC)$ under a homothety at $H$ with scale factor $\frac{1}{2}$. Both of these are centered on $\overline{OH}$, so this will solve the problem. Let $X=\overline{BB'}\cap\overline{CC'}$, $Y=\overline{AA'}\cap\overline{CC'}$ and $Z=\overline{AA'}\cap\overline{BB'}$. Let $P=\overline{AA'}\cap\overline{BC}$ and let $P'=\overline{AA'}\cap\overline{B'C'}$. Then by DIT on quadrilateral $BCB'C'$ and $\overline{AA'}$, $(A,A'),(Y,Z),(P,P')$ are pairs under an involution. Let $D=\overline{AH}\cap\overline{BC}$, $D'=\overline{A'H}\cap\overline{B'C'}$, $R=\overline{AH}\cap \overline{B'C'}$, $R'=\overline{A'H}\cap \overline{B'C'}$. $\angle RDR'=\angle RD'R'=90^{\circ}$, so $R,R',D,D'$ are cyclic. $A,A',D,D'$ since $AH\cdot HD=A'H\cdot HD'=\frac{\text{Pow}_{H}(ABC)}{2}$. Thus by Reim's $\overline{AA'}\parallel \overline{RR'}$. Let $Q=\overline{AA'}\cap \overline{DD'}$. DIT on quadrilateral $DRD'R'$ and $\overline{AA'}$ means that $(A,A'),(P,P'),(P_{\infty \overline{AA'}},Q)$ are pairs under an involution. Combining this with the previous paragraph, $QA\cdot QA'=QY\cdot QZ$. Since $A,A',D,D'$ are cyclic, $QA\cdot QA'=QD\cdot QD'$, and since $D,D'$ are on the nine point circle of $\triangle ABC$, $Q$ has the same power to $(ABC)$, the nine point circle of $\triangle ABC$, and $(XYZ)$. By symmetry these $3$ circles are coaxial, as desired.

It seems that this problem is true for cyclic polygons as follows Given two polygons $A_1A_2\ldots A_n$ and $B_1B_2\ldots B_n$ have the same circumcircle $(O)$ and the same centroid $G$. Assume that the polygon bound by $n$ lines $A_1B_1, A_2B_2,\ldots A_nB_n$ which is inscribed in a circle $(J)$. Prove that $J$ lies on the line $OG$.

Here's a pretty long solution which doesn't use coaxial circles but instead uses homothety and Sondat's theorem. We define a bunch of new points. Let $\Omega$ denote the circumcircle of $\triangle ABC$ and $\triangle A'B'C'$. Let $D$ be the intersection point of lines $\overline{BC}$ and $\overline{B'C'}$, and define $E$ and $F$ cyclically. Let $\triangle XYZ$ be a triangle on the circumcircle of $\triangle ABC$ such that $\triangle PQR$ and $\triangle XYZ$ are homothetic (there are two such triangles, choose any one.) Let $K$, $L$, $M$ be the intersection points of $\overline{AH}$, $\overline{BH}$, $\overline{CH}$ with $\Omega$, and define $K'$, $L'$, $M'$ similarly. Then by the orthocenter lemma, $DK_1 = DK = DH$, so $D$ is the circumcenter of $\triangle KK'H$ and similarly for $E$ and $F$. Let their circumcircles be $\omega_D$, $\omega_E$ and $\omega_F$ respectively. Claim 1. $\triangle XYZ$ and $\triangle DEF$ are perspective with perspector $O$. Proof. It suffices to show that $O$, $D$, $X$ are collinear, or equivalently, $\overline{OD} \parallel \overline{O'P}$. Since $\overline{OD} \perp \overline{KK'}$, we can just show that $\overline{O'P}$ is also perpendicular to the latter. This is just angle chasing. We have \[ \angle(\overline{OP}, \overline{AA'}) = \measuredangle PQR + \measuredangle PRQ + 90^\circ = \overarc{$CA$} + \overarc{$C'A'$} + \overarc{$BA$} + \overarc{$B'A'$} + 90^\circ\]and \[ \angle(\overline{KK'}, \overline{AA'}) = \measuredangle KK'A' + \measuredangle K'A'A = \overarc{$CA$} + \overarc{$BA'$} + \overarc{$B'A'$} + \overarc{$C'A$}. \]It is easy to see that the difference of these two angles is $90^\circ$. Claim 2. $\triangle DEF$ and $\triangle PQR$ are orthologic, and they share the same orthology center which is $H$. Proof. It suffices to show that $\overline{DH} \perp \overline{QR}$ and $\overline{PH} \perp \overline{EF}$. The first one is obvious; note that $D$ is the circumcenter of $\triangle HKK_1$, and that $AA_1$ and $KK_1$ are antiparallel. For the second one, let $\ell$ be the line through $H$ perpendicular to $\overline{EF}$. This line is the radical axis of $\omega_E$ and $\omega_F$, so it passes through their other intersection, which we will denote by $P'$. Now consider a negative inversion at $H$ with radius $\sqrt{HA \cdot HK}$. This sends $\omega_E$ to $\overline{BB'}$ and $\omega_F$ to $\overline{CC'}$, so it sends $P$ to $P'$. Consequently, $\overline{PH} \equiv \ell$ and we're done. Now we're ready to finish the problem. Let $H'$ be the homothetic image of $H$ under the homothety which sends $\triangle PQR$ to $\triangle XYZ$, and let $T$ be the homothetic center. Then $H'$ and $H$ are the two orthology centers of $\triangle XYZ$ and $\triangle DEF$, and $O$ is their perspector, so it follows that $O$, $H$, $H'$ are collinear by Sondat's theorem. Thus, $T$ lies on $\overline{OH}$ as well. Now let $O'$ be the circumcenter of $\triangle PQR$. The initial homothety implies that $T$, $O$, $O'$ are collinear, so it follows that $O'$ lies on $\overline{OH}$ as desired.

Let $\Omega$ denotes the common circumcircle; obviously triangles also share common nine-point circle $\omega .$ For clarity suppose that $P=BB'\cap CC',$ $Q=CC'\cap AA',$ and let $D=BC'\cap B'C,$ $K,M,M'$ denote midpoints of $DP,BC,B'C'$. By Gauss line $K\in MM'.$ If $\mathcal H$ denotes the homothety wrt $P$ with factor $2$ then it's trivial that $DP\mathcal H (M)\stackrel{-}{\sim} D\mathcal H (M') P$ so by PoP and use of $\mathcal H^{-1}$ line $KP$ is tangent to $\odot (PMM').$ Next, observe that $ABC$ and $A'B'C'$ have common centroid - by homothety at this point with factor $-2$ we have $AA'\parallel MM'.$ Also $PBC\stackrel{-}{\sim} PB'C'$ implies that $PM,PM'$ are isogonals in angle $QPR,$ so in fact $\odot (PQR),$ $\odot (PMM')$ are tangent. Finally, by Brocard $D$ lies on polar of $P$ wrt $\Omega ,$ so $\text{pow} (K,\Omega )=|KQ|^2=|KM|\cdot |KN|,$ i.e. $K$ has equal powers wrt $\omega , \Omega , \odot (PQR).$ The same works for two analogous points, so all three circles are coaxial and the result follows.

adorefunctionalequation wrote: what if this used as tst question? no one seems to care about the unoriginality of the problem.

Well, for complex numbers lovers here goes. Let the circumcircle be $|z|=1$, the vertices of triangles be $a,b,c$ and $A,B,C$. We may suppose that $H$ lies on the real line, so $H$'s complex coordinate is $h:=a+b+c=A+B+C$ is real. The line through $a,A$ has the equation $\ell_a(z):=aA\bar{z}+z-a-A=0$, the pencil of conics passing through the vertices of $PQR$ is generated by $\ell_a\cdot \ell_b$, $\ell_b\cdot \ell_c$, $\ell_c\cdot \ell_a$. We should find the linear combination of these three guys which defines a circle, that is, the coefficients of $z^2$ and $\bar{z}^2$ are equal to 0. It is easy to find these coefficient: the equation of the circumcircle of $PQR$ is $$ F(z):=\left(\frac1{aA}-\frac1{bB}\right)\ell_a\ell_b+\left(\frac1{bB}-\frac1{cC}\right)\ell_b\ell_c+\left(\frac1{cC}-\frac1{aA}\right)\ell_c\ell_a=0. $$The coefficient of $z\bar{z}$ in $F(z)$ is purely imaginary (it is a sum of expressions like $\frac{bB}{aA}-\frac{aA}{bB}$). Thus, in order to prove that the centre of the circle $\{F(z)=0\}$ is real, we should prove that the coefficient of $z$ is also purely imaginary. It equals $$ T:=\left(\frac1{bB}-\frac1{aA}\right)(a+A+b+B)+\left(\frac1{aA}-\frac1{cC}\right)(c+C+a+A)+ \left(\frac1{cC}-\frac1{bB}\right)(b+B+c+C)$$that may be rewritten as $$ T=\left(\frac{a-c}{bB}+\frac{c-b}{aA}+\frac{b-a}{cC}\right)+\left(\frac{A-C}{bB}+\frac{C-B}{aA}+\frac{B-A}{cC}\right). $$Now the trick is to multiply $T$ by a non-zero real number $h$: the first bracket is multiplied by $A+B+C$, the second bracket by $a+b+c$ (this is done to make all terms homogeneous of degree 0 both in $a,b,c$ and in $A,B,C$). What we get is composed of several guys of the form $x/y-y/x$ with $x,y$ on the unit circle (by antisymmetry, along with any expression like $f(a,A)/f(b,B)$ like $a/b$, or $A/B$, or $aA/bB$, we have $f(b,B)/f(a,A)$ with the opposite sign; and the suspicious terms like $aC/bB$ cancel out). So, it is indeed purely imaginary.

Hello everyone! I wanted to add a brief history of this problem that I came up in 2006 as a student, who participate at IMO2005 and was training for IMO2006. 1. I was trying to generalize P1 from IMO2005. If you extend sides $A_1 C_2$, $C_1 B_2$, $B_1A_2$ you get a new equilateral triangle, which lies on the circumcircle of the original. So they obviously have same orthocenter/circumcenter etc. I used kseq software to check the cases when triangles are on the same circumcircle and have same orthocenter, eventually found this problem. I believed that i can finish the proof with complex numbers, wrote down some equations, but eventually failed. I thought there are some small calculations need to be made, which was not a case . 2. The problem was posted at aops here . hunter_cvp_ic posted the steps, without providing proofs. 3. The problem was given at our National TST a few years ago and it was shared via some internal portal to other international students (OTIS) 3. At IMO2022 I was a teamleader from Kyrgyzstan and was shocked to see this problem in the shortlist, so after 5-10 minutes after receiving the shortlist, checking the old post I have informed the organizers about this, during the jury meeting it was excluded from the shortlist/voting procedure. 4. I do not know who exactly submitted this problem, the solution was quite complex in the shortlist. As I understand it was also found independently by US trainers (Michael Ren and James Lin?) a few years ago and they were trying to find a solution, before submitting it to some competition. 5. Dr. Waldemar Pompe came up with beautiful full solution as a member of PSC, i will try to upload it here. 6. Also a member of US coach team, Michael Ren (i hope i remember the name correctly) send me his solution via email by using nice techniques which i dont follow thoroughly))). Anyway, I am glad that somehow this problem was included in the shortlist and I was the one to eliminate it from it . Otherwise, our team would have advantage, because they knew it/had some ideas how to tackle it, based on steps by hunter_cvp_ic, but i do not know if they actually tried to prove those steps before IMO2022.

Here's another solution using complex numbers. The main claim we'll prove is that $(PQR)$, $(ABC)$ and $\omega$, which is the nine-point circle, are coaxial. By Forgotten Coaxiality lemma, it's suffices to show $\frac{AQ\cdot AR}{A'Q\cdot A'R}=\frac{\text{Pow}(A,\omega)}{\text{Pow}(A',\omega)}$. By LoS (and sine area formula), \[\frac{AQ}{A'Q}=\frac{AC\cdot \sin{\angle ACC'}}{A'C\cdot\sin{\angle A'CC'}}=\frac{AC}{A'C}\cdot\frac{AC'}{A'C'}\]and similarly for $R$. Hence, we need $\frac{AB\cdot AC\cdot AB'\cdot AC'}{A'B\cdot A'C\cdot A'B'\cdot A'C'}=\frac{\text{Pow}(A,\omega)}{\text{Pow}(A',\omega)}$. Set $(ABC)$ as the unit circle, then the condition rewrites as $a+b+c=x+y+z=h$. The square of the left side of the above is \[\frac{|a-b|^2|a-c|^2|a-y|^2|a-z|^2}{|x-b|^2|x-c|^2|x-y|^2|x-z|^2}=\frac{(a-b)^2\frac{-1}{ab}(a-c)^2\frac{-1}{ac}(a-y)^2\frac{-1}{ay}(a-z)^2\frac{-1}{az}}{(x-b)^2\frac{-1}{xb}(x-c)^2\frac{-1}{xc}(x-y)^2\frac{-1}{xy}(x-z)^2\frac{-1}{xz}}=\frac{x^4}{a^4}\cdot\frac{(a-b)^2(a-c)^2(a-y)^2(a-z)^2}{(x-b)^2(x-c)^2(x-y)^2(x-z)^2}\]and the RHS is \[\frac{|a-\frac{h}{2}|^2-\frac{1}{4}}{|x-\frac{h}{2}|^2-\frac{1}{4}}=\frac{|a-b-c|^2-1}{|x-y-z|^2-1}=\frac{xyz}{abc}\cdot\frac{(b+c)(a-b)(a-c)}{(y+z)(x-a)(x-b)}.\]Then, we want to show (note there are no sign issues, as $A$ and $A'$ are outside $\omega$) $\frac{y^2z^2(b+c)^2}{b^2c^2(y+z)^2}=\frac{x^2(a-y)^2(a-z)^2}{a^2(x-b)^2(x-c)^2}$ and Claim: Actually, $\frac{yz(b+c)}{bc(y+z)}=\frac{x(a-y)(a-z)}{a(x-b)(x-c)}$. Proof. We have \begin{align*} \Longleftrightarrow ayz(b+c)(x(h-y-z)-x(h-a)+bc)=xbc(y+z)(a(h-b-c)-a(h-x)+yz)\\ \Longleftrightarrow ayz(b+c)(xa+bc-xy-xz)=xbc(y+z)(ax+yz-ab-ac)\\ \Longleftrightarrow \frac{1}{xbc}(abx+acx+b^2c+bc^2-bxy-cxy-bxz-cxz)=\frac{1}{ayz}(xya+xza+y^2z+yz^2-yab-zab-yac-zac)\\ \Longleftrightarrow\frac{a}{c}+\frac{a}{b}+\frac{b}{x}+\frac{c}{x}-\frac{y}{b}-\frac{y}{c}-\frac{z}{b}-\frac{z}{c}=\frac{x}{z}+\frac{x}{y}+\frac{y}{a}+\frac{z}{a}-\frac{b}{y}-\frac{b}{z}-\frac{c}{y}-\frac{c}{z}\\ \Longleftrightarrow a\overline{h}+b\overline{h}+c\overline{h}-1=x\overline{h}+y\overline{h}+z\overline{h}-1 \text{ which is true.} \end{align*}

Nice Problem! Taking a homothety at $H$ implies $ABC$ and $A'B'C'$ have the same NPC. Also note that they have the same centroid.We claim that the NPC, $(ABC),(PQR)$ are coaxial which will obviously finish. Now let $BC' \cap CB' = T$ Let $M$ denote the midpoint of $QT$, $M_1$ denote the midpoint of $BC$, $M_2$ denote the midpoint of $B'C'. $By Newton-Gauss line and 2009 G4, we have $MQ^2 = MM_1 \cdot MM_2$. So it suffices to prove $QM_1,QM_2$ isogonal in $\measuredangle PQR$, but this follows immediately because they have same centroids.

bruh i forgor that dit allowed intersections with circumconic cry We show that $(PQR)$, $(ABC)$, and the shared nine point circle of $ABC$ and $(AB'C')$ are coaxial. We make the following $B$-centered claim. Claim: Let the radical axis of $(ABC)$ and the nine point circle intersect $BB'$ at $K$. Let $BB'$ hit $A'C'$ at $S$ and $AC$ at $T$. Then, an inversion at $B$ swaps $B, B'$ and $S, T$. Proof. Let the feet from $B$ and $B'$ to $AC$ and $A'C'$ be $X$ and $X'$. Then, $XX'$ passes through $K$. We claim that $XX'ST$ is cyclic. By Reim's theorem it suffices to show $BB' \parallel NN'$ where $N,N'$ are midpoints of $AC$, $A'C'$. However, this is true since a homothety at $H$ sends $NN'$ to the line joining the antipodes of $B,B'$ which is parallel to $BB'$ as it forms a rectangle. Thus $XX'ST$ is cyclic and the claim is proven. $\blacksquare$ Now, Desuarge's involution theorem on $AC'AC$ with circumconic being $(ABCA'B'C')$ and line $BB'$ there is an involution swapping $(B,B'), (S,T), (P,R)$. But this involution is obviously an inversion at $K$ swapping $B,B'$ and $S,T$ which we shown above to exist which finishes.

Let $\omega$ be the ellipse which is the locus of points $X$ such that $OX+HX$ is equal to the radius of $(ABCA'B'C')$, which is tangent to $\overline{BC},\overline{CA},\overline{AB},\overline{B'C'},\overline{C'A'},$ and $\overline{A'B'}$ since the reflection of $H$ over these lines is on $(ABCA'B'C')$. Also, let $\Omega$ denote the image of $(ABCA'B'C')$ under a homothety centered at $H$ with factor $\tfrac{1}{2}$, which is the nine point circle of $\triangle{}ABC$ and $\triangle{}A'B'C'$. We will prove that $(ABCA'B'C'),(PQR),$ and $\Omega$ are coaxial, which implies the result since the centers of $(ABCA'B'C')$ and $\Omega$ are $O$ and the midpoint of $\overline{OH}$, respectively. Let $\ell_P,\ell_Q,$ and $\ell_R$ be the tangents to $(PQR)$ at $P,Q,$ and $R$ respectively. We claim that there exists an involution swapping the intersections of $\ell$ with $(ABCA'B'C')$, the intersections of $\ell$ with $(PQR)$, and the intersections of $\ell$ with $\Omega$ for $\ell=\ell_P,\ell_Q,\ell_R$. Let $D_A$ be the reflection of $H$ over $\overline{BC}$ and let $D'_A$ be the reflection of $H$ over $\overline{B'C'}$. We claim that $\ell_P\parallel\overline{D_AD'_A}$. We will use directed angles. Fix a point $F$ on $(ABCA'B'C')$ and let $f$ be a function from $(ABCA'B'C')$ to $\left[0^\circ,360^\circ\right)$ such that $f(P)$ is equivalent to the measure of arc $\overarc{FP}$ going counterclockwise $\pmod{360^\circ}$. Note that $f\left(D_A\right)\equiv{}f(B)+f(C)+180^\circ-f(A)\pmod{360^\circ}$ and $f\left(D'_A\right)\equiv{}f(B')+f(C')+180^\circ-f(A')\pmod{360^\circ}$. This implies that the angle going counterclockwise from $\overline{D_AD'_A}$ to $\overline{AA'}$ is $\frac{1}{2}\left(f(A)+f(A')-f\left(D_A\right)-f\left(D'_A\right)\right)\pmod{180^\circ}$. Now, notice that the angle going counterclockwise from $\ell_P$ to $\overline{AA'}$ is \begin{align*} \angle{}PQR-\angle{}QRP&\equiv\frac{1}{2}(f(A')-f(C)+f(A)-f(C'))-\frac{1}{2}(f(B)-f(A')+f(B')-f(A))\\ &\equiv\frac{1}{2}(2f(A)+2f(A')-f(B)-f(B')-f(C)-f(C'))\\ &\equiv\frac{1}{2}\left(f(A)+f(A')-\left(f(B)+f(C)+180^\circ-f(A)\right)-\left(f(B')+f(C')+180^\circ-f(A')\right)\right)\\ &\equiv\frac{1}{2}\left(f(A)+f(A')-f\left(D_A\right)-f\left(D'_A\right)\right) \end{align*}is the angle going counterclockwise from $\overline{D_AD'_A}$ to $\overline{AA'}$, proving the claim. Let $A_2=\overline{BC}\cap\overline{B'C'}$. We claim that $\ell_P\perp\overline{OA_2}$. Since $\ell_P\parallel\overline{D_AD'_A}$ it suffices that $\overline{D_AD'_A}\perp\overline{OA_2}$. Since $\overline{A_2O}$ and $\overline{A_2H}$ are isogonal conjugates with respect to $\angle{}CA_2C'$ since $\omega$ is tangent to $\overline{BC}$ and $\overline{B'C'}$ by reflecting over the internal angle bisector of $\angle{}CA_2C'$ and applying a homothety centered at $A_2$ such that $H$ goes to $O$ it suffices that if $O_A$ and $O'_A$ are the reflections of $O$ over $\overline{BC}$ and $\overline{B'C'}$, respectively, then $\overline{A_2H}\perp\overline{O_AO'_A}$. However, note that by radical center on $(ABCA'B'C'),(BCH),$ and $(B'C'H)$, the radical axis of $(BCH)$ and $(B'C'H)$ goes through the intersection of $\overline{BC}$ and $\overline{B'C'}$, which is $A_2$, so $\overline{A_2H}$ is the radical axis of $(BCH)$ and $(B'C'H)$. This is perpendicular to the line joining the centers of $(BCH)$ and $(B'C'H)$, which is $\overline{O_AO'_A}$, proving the claim. Let $P'$ be the intersection of $\overline{BC'}$ and $\overline{B'C}$. By Brokard's Theorem we see that $\overline{PP'}\perp\overline{OA_2}$, but since $\ell_P\perp\overline{OA_2}$, we have that $\overline{PP'}=\ell_P$. Therefore, by Desargues's Involution Theorem, there exists an involution $\psi$ swapping the intersections of $(ABCA'B'C')$ with $\ell_P$, points $P$ and $P$, points $P'$ and $P'$, and points $\overline{BC}\cap\ell_P$ and $\overline{B'C'}\cap\ell_P$. This involution is centered at the midpoint of $\overline{PP'}$. Let $M,M',F_A,$ and $F'_A$ be the midpoints of $\overline{BC},\overline{B'C'},\overline{HD_A},$ and $\overline{HD'_A}$, respectively. By the Newton-Gauss line on $PBP'C$, the midpoint of $\overline{PP'}$ is on $\overline{MM'}$. Therefore, by Desargues's Involution Theorem on $MM'F'_AF_A$, there exists an involution swapping the two intersections of $\Omega$ with $\ell_P$, points $\overline{BC}\cap\ell_P$ and $\overline{B'C'}\cap\ell_P$, and points $\overline{MM'}\cap\ell_P$ and $\overline{F_AF'_A}\cap\ell_P$. Since $\overline{F_AF'_A}\parallel\overline{D_AD'_A}\parallel\ell_P$, we see that $\overline{F_AF'_A}\cap\ell_P$ is a point at infinity, so $\overline{MM'}\cap\ell_P$ is the center of this involution, so it is $\psi$. Therefore, we see that $\psi$ swaps the two intersections of $(ABCA'B'C')$ with $\ell_P$, the two intersections $P$ and $P$ of $(PQR)$ with $\ell_P$, and the two intersections of $\Omega$ with $\ell_P$. We can use similar arguments for $\ell_Q$ and $\ell_R$, proving the claim. Now, we will prove that $(ABCA'B'C'),(PQR),$ and $\Omega$ are coaxial. Assume for the sake of contradiction that they are not. Let $X_1$ and $X_2$ be the two intersections of $(ABCA'B'C')$ with $(PQR)$. Then, let $P_1$ and $P_2$ be the two intersections of $\Omega$ with $\ell_P$, let $Q_1$ and $Q_2$ be the two intersections of $\Omega$ with $\ell_Q$, and let $R_1$ and $R_2$ be the two intersections of $\Omega$ with $\ell_R$. Since there exists an involution swapping the intersections of $\ell_P$ with $(ABCA'B'C')$, the intersections of $\ell_P$ with $(PQR)$, and the intersections of $\ell_P$ with $\Omega$, we see that $X_1,X_2,P_1,$ and $P_2$ are concyclic. Similarly, we see that $X_1,X_2,Q_1,$ and $Q_2$ are concyclic and that $X_1,X_2,R_1,$ and $R_2$ are concyclic. Note that $\left(X_1X_2P_1P_2\right),\left(X_1X_2Q_1Q_2\right),\left(X_1X_2R_1R_2\right),$ and $\Omega$ are pairwise distinct since otherwise $(ABCA'B'C'),(PQR),$ and $\Omega$ are coaxial. Then, applying radical center of $\left(X_1X_2Q_1Q_2\right),\left(X_1X_2R_1R_2\right),$ and $\Omega$ gives that $\ell_Q,\ell_R,$ and $\overline{X_1X_2}$ concur. Similarly, we see that $\ell_R,\ell_P,$ and $\overline{X_1X_2}$ concur, and that $\ell_P,\ell_Q,$ and $\overline{X_1X_2}$ concur, so $\ell_Q\cap\ell_R,\ell_R\cap\ell_P,$ and $\ell_P\cap\ell_Q$ are collinear, a contradiction. This completes the problem.

Fantastic and rizztastic problem! Take the radical axis of the NPC and circumcircle. Consider the intersections of $CC'$ with $AB$, $A'B'$, and the radical axis as $X$, $Y$, $Z$. Consider the two points I and J as feet, and by orthocenter reflection $2HI \cdot HC = 2HJ \cdot HC'$ which means that $IJCC'$ is cyclic, thus $Z$, $I$, $J$ collinear. Take $M$, $M'$ midpoints of $AB$ and $A'B'$ and because of orthocenter reflection sending these to antipodes of $C$ and $C'$, this gives $MM'$ parallel to $CC'$, which then by Reim in lines $AB$ and $A'B'$ implies that $XIJY$ is cyclic. Then $ZX \cdot ZY = ZC' \cdot ZC$. We finish by taking DDIT on $ABA'B'$ with circumcircle and line $CC'$ implies the pairs $(C,C')$, $(Q,R)$, $(X,Y)$, but we know that from $(C,C')$ and $(X,Y)$ this is an inversion at $Z$. so $Z$ has equal power wrt $PQR$ too. Repeat in all $3$ orientations to get that $ABC$, $PQR$, and the NPC are coaxial. Then $O$, $N$, $H$, and circumcenter of $PQR$ is collinear.