This was also India TST day 4 P3. Here's a solution by Ojas Mittal that uses pure angel chase:

My solution in the TST: complex bash (which was extremely clean)

Let $S_A,S_B,S_C$ be a intersection of lines tangent to $\Omega$ at $A,B,C$ with a line tangents to $\Omega$ at $P$. Then clearly $O_A$ is a midpoint of $OS_A$. Similarly, $O_B$ is midpoint of $OS_B$ and $O_C$ is midpoint of $OS_C$. Taking homothety $\mathcal{H}(O,2)$, lines $\ell_A,\ell_B,\ell_C$ map to lines pass though $S_A,S_B,S_C$ and perpendicular to $BC,AC,AB$. Let $X_A,X_B,X_C$ be intersections of these lines so that $X_AX_B\perp AB$, $X_BX_C\perp BC$ and $X_CX_A\perp CA$. It's enough to prove that $X_AX_BX_C$ tangents to $OP$. By angle chasing, $X_AX_BX_C\sim ABC$. Reflect $O$ across $P$ to $O'$. Next we claim that $S_AS_BX_CO'$ concyclic. We can prove this by angle chasing $$\measuredangle S_AX_BS_C=-\measuredangle S_CX_BX_C=-\measuredangle ABC =-\measuredangle S_AOS_C=\measuredangle S_AO'S_C.$$Similarly, $S_AS_BX_CO'$,$X_AS_BS_CO'$ are concyclic. By Miquel's theorem $X_AX_BX_CO'$ is concyclic. Let P' be antipode of P wrt $\Omega$. By angle chasing again. $$\measuredangle O'X_BX_A=\measuredangle O'S_AS_C=-\measuredangle OS_AS_C=90^{\circ}-\measuredangle S_AOP=90^{\circ}-\measuredangle AP'P=\measuredangle P'PA=\measuredangle P'BA.$$So, $X_AX_BX_CO'\sim ABCP'$. Let $X$ be a circumcenter $X_AX_BX_C$. Because $\measuredangle(XO',OP')=\measuredangle(AB,X_AX_B)=90^{\circ}$. Therefore, circumcircle of $X_AX_BX_C$ tangents to $OP$ at $P'$.

Seriously, this is a G7?

In fact, the circumcircle is tangent to line $OP$ at $P$. We present two solutions. In both solutions, we let $\ell_A$, $\ell_B$, $\ell_C$ determine triangle $XYZ$. First solution with complex numbers (Evan Chen) We proceed with complex numbers with $A$, $B$, $C$ on the unit circle (hence $O = 0$), and assume that the point $P$ has coordinate $p = 1 \in {\mathbb C}$. We will show that line $OP$ is tangent at $P$ to the circumcircle of $\triangle PYZ$. To solve for $Z$, recall that the tangents to $\Omega$ at $A$ and $P$ meet at $\frac{2ap}{a+p} = \frac{2a}{a+1}$. Hence, $o_A = \frac{a}{a+1}$. Since $\overline{ZO_A} \perp \overline{BC}$ we must have \[ \frac{z - \frac{a}{a+1}}{b-c} = - \overline{\left( \frac{z-\frac{a}{a+1}}{b-c} \right)} = -\frac{\overline z - \frac{1}{a+1}}{\frac{c-b}{bc}} \iff \frac1b \left( z - \frac{a}{a+1} \right) = c\left( \overline z - \frac{1}{a+1} \right). \]Similarly, since $\overline{ZO_B} \perp \overline{AC}$, $\frac1a \left( z - \frac{b}{b+1} \right) = c\left( \overline z - \frac{1}{b+1} \right)$. Subtracting the two gives \begin{align*} \left(\frac 1b - \frac 1a\right) z &= c\left( \frac{1}{b+1}-\frac{1}{a+1} \right) + \frac{a}{b(a+1)} - \frac{b}{a(b+1)} \\ &= \frac{c(a-b)}{(a+1)(b+1)} = \frac{a^2(b+1) - b^2(a+1)}{ab(a+1)(b+1)} \\ &= \frac{abc(a-b) + (a-b)(ab+a+b)}{ab(a+1)(b+1)} = \frac{(a-b)}{ab} \cdot \frac{ab+a+b+abc}{(a+1)(b+1)} \\ \implies z &= \frac{ab+a+b+abc}{(a+1)(b+1)} \implies z-1 = \frac{abc-1}{(a+1)(b+1)}. \intertext{Similarly,} y-1 &= \frac{abc-1}{(a+1)(c+1)}. \end{align*}Then take the complex number \begin{align*} \theta = \frac{y-p}{p-0} \div \frac{y-z}{p-z} &= \frac{(y-1)(z-1)}{z-y} = \frac{\frac{(abc-1)^2}{(a+1)^2(b+1)(c+1)}} {\frac{abc-1}{a+1}\left( \frac{1}{b+1}-\frac{1}{c+1} \right)} = \frac{abc-1}{(a+1)(c-b)}. \intertext{Its conjugate is} \overline\theta &= \frac{\frac{1}{abc}-1}{(\frac1a+1)(\frac1c-\frac1b)} = \frac{1-abc}{(1+a)(b-c)} = \theta. \end{align*}So $\theta \in {\mathbb R}$, meaning $\measuredangle YPO = \measuredangle YZP$ (in the directed sense). Thus, the circumcircle of $\triangle YPZ$ is tangent at $P$ to $\overline{OP}$. Similarly, the circumcircle of $\triangle XPZ$ is tangent at $P$ to $\overline{OP}$. This can only happen if $XYZP$ is concyclic, which concludes the problem. Remark: The computation given here has at most eight terms at each step, so it is quite tractable. If one does not realize the tangency point is $P$, then he/she can still proceed in the following way. Search for a real number $t \in {\mathbb R}$ (equivalently $T \in \overline{OP}$) such that $\frac{y-t}{z-t} \div \frac{y-x}{z-x} \in {\mathbb R}$. This will give a quadratic equation in $t$. In order for line $\overline{OP}$ to be tangent, this quadratic equation must have a double root; in fact, one finds the quadratic $(b-c)(t^2-2t+1) = 0$. I was able to carry out this longer approach by hand in about an hour. Second solution by angle chasing (Eugene Lee) We prove several claims. [asy][asy] size(12cm); pair O = origin; pair A = dir(110); pair B = dir(190); pair C = dir(350); pair P = dir(245); pair O_A = circumcenter(A, O, P); pair O_B = circumcenter(B, O, P); pair O_C = circumcenter(C, O, P); pair D = foot(O_A, B, C); pair E = foot(O_B, C, A); pair F = foot(O_C, A, B); pair X = extension(O_B, E, O_C, F); pair Y = extension(O_C, F, O_A, D); pair Z = extension(O_A, D, O_B, E); filldraw(unitcircle, invisible, deepcyan); draw(A--B--C--cycle, deepcyan); draw(O--P, deepcyan); filldraw(circumcircle(X, Y, Z), invisible, red); draw(circumcircle(A, O, P), lightblue+dashed); draw(circumcircle(B, O, P), lightblue+dashed); draw(circumcircle(C, O, P), lightblue+dashed); draw(X--Y--Z--cycle, red); draw(X--O_C, red); draw(A--P, deepgreen+dotted); draw(Y--P, deepgreen+dotted); draw(Z--C, deepgreen+dotted); draw(circumcircle(O_B, P, O_C), lightgreen+dashed); draw(circumcircle(O_C, P, O_A), lightgreen+dashed); draw(circumcircle(O_A, P, O_B), lightgreen+dashed); draw(O_A--O_C, orange); dot("$O$", O, dir(45)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$P$", P, dir(290)); dot("$O_A$", O_A, dir(O_A)); dot("$O_B$", O_B, dir(180)); dot("$O_C$", O_C, dir(345)); dot("$X$", X, dir(15)); dot("$Y$", Y, dir(90)); dot("$Z$", Z, dir(Z)); /* TSQ Source: !size(12cm); O = origin R45 A = dir 110 B = dir 190 C = dir 350 P = dir 245 R290 O_A = circumcenter A O P O_B = circumcenter B O P R180 O_C = circumcenter C O P R345 D := foot O_A B C E := foot O_B C A F := foot O_C A B X = extension O_B E O_C F R15 Y = extension O_C F O_A D R90 Z = extension O_A D O_B E unitcircle 0.1 lightcyan / deepcyan A--B--C--cycle deepcyan O--P deepcyan circumcircle X Y Z 0.1 yellow / red circumcircle A O P lightblue dashed circumcircle B O P lightblue dashed circumcircle C O P lightblue dashed X--Y--Z--cycle red X--O_C red A--P deepgreen dotted Y--P deepgreen dotted Z--C deepgreen dotted circumcircle O_B P O_C lightgreen dashed circumcircle O_C P O_A lightgreen dashed circumcircle O_A P O_B lightgreen dashed O_A--O_C orange */ [/asy][/asy] Claim: The points $X$, $P$, $O_B$, $O_C$ are concyclic. Similarly, the points $Y$, $P$, $O_C$, $O_A$ are concyclic, as are $Z$, $P$, $O_A$, $O_B$. Proof. We have \begin{align*} \measuredangle O_B X O_C &=\measuredangle CAB = \measuredangle CPB = \measuredangle CPO + \measuredangle OPB \\ &= \measuredangle OCP + \measuredangle PBO \\ &= (90^{\circ} - \measuredangle O_C P O) + (90^{\circ} - \measuredangle O P O_B) \\ &= -(\measuredangle O_C P O + \measuredangle O P O_B) = -\measuredangle O_C P O_B = \measuredangle O_B P O_C. \end{align*}$\blacksquare$ Similarly $Y, O_A, P, O_B$, $Z, O_A, P, O_B$ concyclic. Claim: Point $P$ lies on $(XYZ)$. Proof. Note that $P$ is the Miquel point of the complete quadrilateral $O_A O_B O_C X Y Z$. $\blacksquare$ Claim: Lines $AX$, $BY$, $CZ$ concur at $P$. Proof. We show just $A$, $X$, $P$ collinear. Compute \begin{align*} \measuredangle OPX &= \measuredangle OPO_B + \measuredangle O_BPX = (90^{\circ} - \measuredangle PBO) + \measuredangle O_B O_C X \\ &= (90^{\circ} - \measuredangle PBO) + \measuredangle(\overline{OP}, \overline{BA}) \\ &= \measuredangle BAP + \measuredangle(\overline{OP}, \overline{BA}) = \measuredangle OPA. \end{align*}$\blacksquare$ To finish, since the corresponding sides of $\triangle ABC$ and $\triangle XYZ$ are spirally similar, the last claim implies that a spiral similarity at the second intersection of $(XYZ)$ and $(ABC)$ maps $\triangle ABC$ and $\triangle XYZ$ through a $90^{\circ}$ rotation. This means the circles are orthogonal, as needed. Remark: The second claim admits a proof by angle chase: since $O_A$, $O_B$, $O_C$ all lie on the perpendicular bisector of $OP$, we get \[ \measuredangle YZP = \measuredangle O_A Z P = \measuredangle O_A O_B P = \measuredangle O_C O_B P = \measuredangle O_C X P = \measuredangle YXP. \] The last paragraph can also be replaced by angle chasing: \begin{align*} \measuredangle PYX &= \measuredangle(\overline{PB}, \overline{XY}) = \measuredangle(\overline{PB}, \overline{AB})+90^{\circ} \\ &= \measuredangle PBA + 90^{\circ} = \measuredangle PCA + 90^{\circ} = 90^{\circ} - \measuredangle ACP \\ &= \measuredangle OPA = \measuredangle OPX \end{align*}as needed.

+) South Korean TST #3 I solved using complex numbers too, but much more messier than above posts

Generalization: Let $O$ be the circumcenter of $\triangle ABC$ and $P$ be an arbitrary point. Let $O_A, O_B, O_C$ be the circumcenters of $\triangle AOP$, $\triangle BOP$, $\triangle COP$ respectively. Let $\ell_A, \ell_B, \ell_C$ be the lines through $O_A, O_B, O_C$ perpendicular to $BC, CA, AB$ respectively. Let $S$ be the circumcenter of triangle formed by $\ell_A, \ell_B, \ell_C$. Prove that $\angle OPS = 90^{\circ}$. I have a prove but it's very long and not completely synthetic. I will spare the proof for a while.

Once you know that it is easier than G1, it is easier than G1. \(P\) is the required tangency point. Let \(\Delta A_1B_1C_1\) be the required \(\Delta\) with \(A_1\) opposite to \(l_a\) and so on. \(\angle O_AOO_B = \angle AOB /2 = \angle C \Rightarrow O_APO_BC_1\) is cyclic similarly, \(O_BO_CPA_1, O_APO_CB_1\) is cyclic. Combining the three, \(A_1B_1C_1P\) is also cyclic. Let \(PO \cap BC = X\) Now, \(\angle OPC_1 = \angle OPO_B - \angle O_BPC_1 = \angle BCP - \angle PXC = \angle OPC = \angle PO_CO_B = \angle PA_1C_1 \Rightarrow \angle PA_1C_1 = \angle OPC_1 \Rightarrow \odot A_1B_1C_1P\) is tangent to \(OP\) at \(P\).

Here is my solution to the generalization. Hopefully someone will find a simpler proof, especially to the main lemma. MarkBcc168 wrote: Generalization: Let $O$ be the circumcenter of $\triangle ABC$ and $P$ be an arbitrary point. Let $O_A, O_B, O_C$ be the circumcenters of $\triangle AOP$, $\triangle BOP$, $\triangle COP$ respectively. Let $\ell_A, \ell_B, \ell_C$ be the lines through $O_A, O_B, O_C$ perpendicular to $BC, CA, AB$ respectively. Let $S$ be the circumcenter of triangle formed by $\ell_A, \ell_B, \ell_C$. Prove that $\angle OPS = 90^{\circ}$. Lemma: Let $\triangle ABC$ be a triangle and $P$ be an arbitrary point. A line $\ell$ pass through $P$ intersect sides $BC, CA, AB$ at $D,E,F$. Let the lines through $D,E,F$ parallel to $AP, BP, CP$ form the triangle $XYZ$ ($D\in YZ$, $E\in XZ$). Let $A', B', C'$ be reflections of $A,B,C$ across $\ell$. Then lines $A'X, B'Y, C'Z$ and $\ell$ are concurrent. Proof: Fix line $\ell$ and animate $P$ along $\ell$. First, we claim that $X$ move along a fixed line. To prove that, let $A_1$ be the second intersection of $\odot(BPC)$ and $\ell$. Then $\measuredangle XEF = \measuredangle BPA_1 = \measuredangle BCA_1$ or $\triangle XEF\stackrel{+}{\sim}\triangle A_1CB$. This means of a spiral similarity $\mathcal{T}$ which maps $C\to E$, $B\to F$ takes line $\ell\to\ell_A$, then $X\in\ell_A$. As $\mathcal{T}$ is fixed, line $\ell_A$ is also fixed. Similarly $Y,Z$ move along fixed lines $\ell_B, \ell_C$. Now we are ready to use moving point method. Note that the following composition of mappings $$P\to{\infty}_{BP}\to X\to AX\cap \ell$$is projective. Similarly $P\to B'Y\cap\ell$ is projective. Thus it suffices to verify the lemma for three cases of $P$. Actually, by symmetry, it suffices to take $P=D$. Then $Y,Z\in AD$ such that $YF\parallel ZE\parallel BC$ and $X$ is ${\infty}_{BC}$. Let $P=B'Y\cap\ell$. Then it suffices to show that $A'P\parallel BC$. This can be rephrased to the following sub-lemma (applied on $\triangle ABD$). Sub-lemma (The Special Case): Let $\triangle ABC$ be a triangle. Points $E, F$ are placed on line $AC, AB$ respectively such that $EF\parallel BC$. Let $E'$ be the reflection of $E$ across $CF$. Let the lines $BE'$ and $CF$ intersect at $P$. Then $AP\parallel E'F$. Proof: Let $X$ be the point on $CF$ such that $E'P=E'X$. Then $\triangle FE'X\sim\triangle CBP$. Thus $$\frac{PE'}{PB} = \frac{XE'}{PB} = \frac{FE}{BC} = \frac{AF}{AB}$$or $AP\parallel E'F$. Back to the main problem. Let $A_1=\ell_B\cap \ell_C$. Define $B_1,C_1$ similarly. Clearly $O_A, O_B, O_C$ lie on line $\ell_1$ perpendicular to $OP$. Let line $OP\equiv\ell$ meet $BC, CA, AB$ at $D,E,F$. Let $A', B', C'$ be the reflections of $A,B,C$ across $\ell$. Then $$\measuredangle A'EF = -\measuredangle AEF = -\measuredangle(\ell_B, \ell_1) = -\measuredangle A_1O_BO_C$$or $\triangle A'EF\stackrel{-}{\sim} \triangle A_1O_BO_C$. Now let $P'$ be the inverse of $P$ w.r.t. $\odot(ABC)$. Let $X$ be the point such that $XE\parallel BP'$ and $XF\parallel CP'$. Define $Y,Z$ similarly. Then $$\measuredangle XEF = \measuredangle BP'O = \measuredangle OBP = \measuredangle O_CO_BP$$or $\triangle XEF \stackrel{-}{\sim} \triangle PO_BO_C$. Thus $\triangle A'EF\cup X\sim\triangle A_1O_BO_C\cup P$. But by our result, $A'X, B'Y, C'Z$ meet at a point $Q\in\ell$. Thus $$\measuredangle(AX, \ell) = \measuredangle(\ell, A'X) = -\measuredangle(\ell_1, A_1P) = \measuredangle(A_1P, \ell_1).$$or $A_1P\perp AQ$. Thus $\triangle ABC\cup Q\cup O\sim\triangle A_1B_1C_1\cup P\cup S$ so we are done.

More general problem. Let $ABC$ be a triangle inscribed in circle $(\omega)$. $P$ and $Q$ are two any points. $DEF$ is circumcevian triangle of $Q$ with respect to $ABC$. $J$, $K$, $L$ are circumcenters of triangles $APQ$, $BPQ$, $CPQ$ respectively. The lines $\ell_A, \ell_B, \ell_C$ pass through $J$, $K$, $L$ and are perpendicular to $EF$, $FD$, $DE$ respectively. Lines $\ell_A, \ell_B, \ell_C$ bound triangle $XYZ$. Let $R$ be the point such that $\triangle XYZ\cup R\sim\triangle DEF\cup Q$. Prove that $PR\perp PQ$.

If I'm not mistaken, my solution works for buratinogiggle's generalization. There is some significant modification on the last part (which you need to apply the lemma on $\triangle DEF$ instead). I will left the modification to readers.

Let $D,E,F$ be the feet of perpendiculars from $O_A,O_B,O_C$ to $BC,CA,AB$ respectively. Let $O_AD \cap O_BE = Z \ O_BE \cap O_CF = X \ O_CF \cap O_AD = Y$. Now we will prove the following Claim: $A,X,P$ are collinear. Let $X_1 = AP \cap O_BE$. Then $\angle O_BX_1P = \angle AX_1E =90^{\circ} - \angle X_1AE = 90^{\circ} - \angle PAC = \angle OCP = \angle O_BO_CP \longrightarrow (X_1O_BPO_C)$ is cyclic. Let $X_2 = AP \cap O_CF$. Analogously, $(X_2O_BPO_C)$ is cyclic. So $X_2 =X_1$ and our claim is proved. Next, it is easy to see that $P \in (XYZ)$. Indeed, $\angle XPY = 180^{\circ}- \angle APB =180^{\circ} -\angle ACB = \angle ACD = \angle ECD = 180^{\circ} - \angle EZD = 180^{\circ} -\angle XZY$ Now our claim also implies $B,Y,P$ are collinear and $C,Z,P$ are collinear. Now, $\angle CED = \angle CZD = \angle PZY = \angle PXY =\angle AXF =\angle AEF \longrightarrow D,E,F$ are collinear. Lastly, $$\angle OPX = \angle OPA = 90^{\circ} - \angle ABP =90^{\circ} -\angle ACZ =90^{\circ} -\angle ECZ = \angle EZC = \angle XZP$$So, $OP$ is tangent to $(XYZ)$ and we are done.

In fact a shorter solution using angel chasing is possible. Neothehero wrote: Let $O$ be the circumcentre, and $\Omega$ be the circumcircle of an acute-angled triangle $ABC$. Let $P$ be an arbitrary point on $\Omega$, distinct from $A$, $B$, $C$, and their antipodes in $\Omega$. Denote the circumcentres of the triangles $AOP$, $BOP$, and $COP$ by $O_A$, $O_B$, and $O_C$, respectively. The lines $\ell_A$, $\ell_B$, $\ell_C$ perpendicular to $BC$, $CA$, and $AB$ pass through $O_A$, $O_B$, and $O_C$, respectively. Prove that the circumcircle of triangle formed by $\ell_A$, $\ell_B$, and $\ell_C$ is tangent to the line $OP$. Let $\ell_B \cap \ell_C=X.$ Define $Y,Z$ similarly. Call the circle $XYZ$ as $\omega.$ Let $S$ be the center of $\omega.$ Also, $K=\ell_A \cap BC,$ and $L,M$ are similarly defined. We show that $\omega$ is tangent to $OP$ at $P.$ Claim: The point $P$ lies on $\omega.$ Proof: Clearly, \begin{align*} \measuredangle O_CPO_A=\measuredangle O_CPO+\measuredangle OPO_A &=(\pi/2-\measuredangle OCP)+(\pi/2-\measuredangle PAO) \\ &=\measuredangle PAC+\measuredangle ACP \\ &=\measuredangle ABC =\measuredangle KYL \end{align*}Hence, $PO_AYO_C$ is cyclic. Similarly, $O_AO_BZP$ is cyclic. Thus, we are done by Miquel on $XYO_AO_B.$ $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -13.0711337058945, xmax = 4.984202470977303, ymin = -4.693469917932022, ymax = 11.753212636798287; /* image dimensions */ pen ffqqtt = rgb(1,0,0.2); pen wwffqq = rgb(0.4,1,0); pen qqzzcc = rgb(0,0.6,0.8); pen qqccqq = rgb(0,0.8,0); draw((-6,1)--(-5,-2)--(1,-2)--cycle, linewidth(0.4) + ffqqtt); draw((-6.653718180703615,2.8812662866248995)--(-3.8397045330106034,2.7568789372060967)--(-3.839704533010603,0.43822317183382453)--(-8.344242012714654,-1.0632893214008594)--cycle, linewidth(0.4) + qqzzcc); draw((-3.330380703997218,-2)--(-3.330380703997218,-1.4906761709866152)--(-3.839704533010603,-1.4906761709866152)--(-3.839704533010603,-2)--cycle, linewidth(0.4) + qqccqq); 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dot((-8.344242012714654,-1.0632893214008594),linewidth(4pt) + dotstyle); label("$X$", (-8.653338683893953,-1.668240718083819), NE * labelscalefactor); dot((-3.839704533010603,9.447298131241928),linewidth(4pt) + dotstyle); label("$Z$", (-3.7553485508063886,9.640354148015415), NE * labelscalefactor); dot((-7.8437378483030935,4.942760651537875),linewidth(4pt) + dotstyle); label("$S$", (-8.221163083915638,5.054490837134409), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Thus, it suffices to show that $\measuredangle SPO=\pi/2.$ Now, Now, $$\measuredangle ZPX=\measuredangle ZYX=\measuredangle KBL=\measuredangle CPA$$and hence $Z,P,C$ are collinear. Similarly, $\{P,A,X\}$ and $\{P,Y,B\}$ are collinear triples. Thus $$\measuredangle OPS=\measuredangle OPY+\measuredangle YPS=\measuredangle OPB+(\pi/2-\measuredangle PZY)=\measuredangle OPB+\measuredangle KCP=\pi/2$$Thus we are done. $\blacksquare$

Let $XYZ$ be the triangle formed by $\ell_A$, $\ell_B$, $\ell_C$. Here is the claim which allows the problem to fall to straight angle chasing. Claim: The points $P,A,X$ are collinear (and symmetric variants) Proof: We'll complex bash with $(ABC)$ as the unit circle. Note that $O_A$ is the intersection of the perpendicular bisector of $OA$ and $OP$, so it is $1/2$ of the pole of $AP$. This means that \[O_A=\frac{ap}{a+p}.\]We'll now find $x$. Note that $XO_B\perp AC$, so the foot from $X$ to $AC$ is the foot from $O_B$ to $AC$, so \[\frac{1}{2}(x+a+c-ac\bar{x})=\frac{1}{2}(O_B+a+c-ac\bar{O}_B),\]so $ac(\bar{x}-\bar{O}_B)=x-O_B$, and similarly $ab(\bar{x}-\bar{O}_C)=x-O_C$. Setting the value for $x$ we get from these two equal, we have that \begin{align*} a(b-c)\bar{x}&=a(b\bar{O}_C-c\bar{O}_B)+O_B-O_C \\ &= a\left(\frac{b}{c+p}-\frac{c}{b+p}\right)+\frac{bp}{b+p}-\frac{cp}{c+p} \\ &= \frac{a[b^2+bp-c^2-cp]+bp(c+p)-cp(b+p)}{(b+p)(c+p)} \\ &= \frac{(b-c)[ap+a(b+c)+p^2]}{(b+p)(c+p)}, \end{align*}so \[\boxed{\bar{x}=\frac{ap+a(b+c)+p^2}{a(b+p)(c+p)}}.\]We also compute \begin{align*} x &= \frac{\frac{1}{ap}+\frac{b+c}{bca}+\frac{1}{p^2}}{\frac{1}{a}\frac{b+p}{bp}\frac{c+p}{cp}} \\ &= \frac{bcp+p^2(b+c)+abc}{(b+p)(c+p)}. \end{align*}To verify $X,A,P$ collinear, all we have to show is that $x+ap\bar{x}=a+p$. Indeed, \begin{align*} x+ap\bar{x} &= \frac{bcp+p^2(b+c)+abc}{(b+p)(c+p)}+\frac{ap^2+ap(b+c)+p^3}{(b+p)(c+p)} \\ &= \frac{(a+p)(b+p)(c+p)}{(b+p)(c+p)} \\ &= a+p, \end{align*}completing the proof of the claim. $\blacksquare$ We're now in a position to solve the problem. We'll be using directed angles mod $\pi$ throughout. We see that $\triangle XYZ\stackrel{+}{\sim}\triangle ABC$ due to the perpendicularity conditions. Note that \[\angle YPX = \angle BPA = \angle BCA = \angle YZX,\]so $P\in(XYZ)$. To show that $OP$ is tangent to $(PXY)$, it suffices to show that $\angle OPY=\angle PXY$. Indeed, we have \[\angle PXY = \angle AXY = \pi/2 - \angle BAP = \angle OPB = \angle OPY,\]as desired. Remark: After drawing a nice diagram, I noticed that $P$ looks like it's on $(XYZ)$, and later I noticed that $P,A,X$ (and variants) looked collinear. I tried for a bit doing synthetic stuff from here, but completely missed all the key observations (mainly that $O_BO_CXP$ cyclic). Instead, I saw that if we had $P,A,X$ collinear (and variants), then all the angles are computable and the problem has to fall to angle chase. This was a much more modest goal, and I saw that at most 10 minutes of complex bashing would show this. Hence, the above solution.

Let $E=\ell_A\cap\ell_C, D=\ell_B\cap\ell_A, F=\ell_B\cap\ell_C$. $\widehat {O_AEO_C}=\widehat {FED}=\widehat {ABC}$ $\widehat {O_APO_C}=\widehat {O_APO}+\widehat {O_CPO_B}=90^{\circ}-\widehat {OAP}+90^{\circ}-\widehat {OPC}=\widehat {ABC}\implies O_AEPO_C$ is cyclic. So, $\odot (O_BFPO_A)=\odot (O_CFBO_B)$. Let $\mathcal {P} $ be the Miquel point of $(DFO_AO_BFO_C) $. $D,E,F,\mathcal {P}$ are cyclic. $\widehat{FO_AO_B}=90^{\circ}-\widehat {AO_AO_B}=\widehat {AO_BP}-\widehat {O_APO}=\widehat {O_APO}+\widehat {OPC}=\widehat{O_APC}\implies\overline {F,P,C}=\overline{D,P,A}=\overline{E,P,B} $ $\widehat {OPD}=\widehat {OAP}=\widehat {PMO}=\widehat {PO_AO_C}=\widehat {PEO_C}=\widehat {DFD} $

We present a hybrid solution which is synthetic at heart, but uses complex numbers to prove a crucial claim. Let $L_A,L_B,L_C$ be $\ell_B\cap\ell_C,\ell_A\cap\ell_C,\ell_A\cap\ell_B$. Claim 1:- $O_B,O_C,L_A,P$ are concyclic. Proof:- Note that $\measuredangle{OBPO_C}=\measuredangle{O_BOO_C}=\pi-(\measuredangle{OBP}+\measuredangle{OCP})=A$ Claim 2:- $P$ lies on $(L_AL_BL_C)$ Proof:- Note that $O_A,O_B,O_C$ lie on the Perpendicular bisector of $OP$. $\measuredangle{PL_CL_A}=\measuredangle{PL_CO_B}=\measuredangle{PO_AO_B}=\measuredangle{PO_AO_C}=\measuredangle{PL_BO_C}=\measuredangle{PL_BL_A}$ We will prove that $OP$ is tangent to $(L_AL_BL_CP)$. Claim 3:- $A,L_A,P$ are collinear. The Proof is by Complex numbers with $(ABC)$ as the unit circle and $A=a,B=b,C=c$ and $P=1$. So $OP$ is the real axis. By Circumcenter formula $O_A=\frac{a}{a+1}$ (and cyclic variants),so we can compute $l_a$ now :- Note that $O_CL_A\perp AB$. So we have: $\frac{o_c-\ell_a}{a-b}+\overline{\frac{o_c-\ell_a}{a-b}}=0\implies \ell_a-ab\overline{\ell_a}=o_c-ab\overline{o_c}$. Similarly $\ell_a-ac\overline{\ell_a}=o_b-ac\overline{o_b}$. So from the two equations we get:- $$(b-c)\ell_a=abc(\overline{o_c-o_b})+(bo_b-co_c)$$Note $\overline{o_c-o_b}=\frac{b-c}{(b+1)(c+1)}$ and $bo_b-co_c=\frac{b^2}{b+1}-\frac{c^2}{c+1}=\frac{(b+c+bc)(b-c)}{(b+1)(c+1)}$. So $$\ell_a=\frac{b+c+bc}{(b+1)(c+1)}+\frac{abc}{(b+1)(c+1)}\implies \ell_a=\frac{bc+b+c+abc}{(b+1)(c+1)}$$and $\overline{\ell_a}=\frac{a+ab+ac+1}{a(b+1)(c+1)}$. So we need $A,L_A,P$ collinear $\implies \ell_a+a\overline{\ell_a}=\frac{(a+1)(b+1)(c+1)}{(b+1)(c+1)}=a+1$ Which is exactly what we wanted by chord formula Finish:- Let $\measuredangle{OAP}=\theta=\measuredangle{OPA}=\measuredangle{OPL_A}$. Now $\measuredangle{PL_CL_A}=\measuredangle{PO_AO_B}=\measuredangle{(PO_A,\ell)}=\frac{1}{2}\cdot 2(\measuredangle{PAO})=\theta$. So, $\measuredangle{OPL_A}=\measuredangle{PL_CL_A}$. Done!

A moving points + angle chase solution (which is hopefully correct) Let the triangle formed by the lines $l_i$ be $XYZ$ with $X$ being the vertex corresponding to $A$. Animate $P$ projectvely on $(ABC)$. Then, $O_A$, $O_B$, $O_C$ move projectively on the perpendicular bisectors of $OA,OB,OC$. Lines $l_i$ move projectively on the pencil of lines through the points at infinity along the lines perpendicular to $AB,BC,CA$. Thus, $X,Y,Z$ move with degree $2$ each. To show that $A,P,X$ are collinear, we need to show this for $5$ positions of $P$. We can show this for $A,B,C$ and their antipodes. Also, it is easy to see that $\triangle XYZ ~ \triangle ABC$. So, $\angle ZPX=\angle CPX=\angle CPA=\angle CBA= \angle ZYX$ (where the angles are directed) and hence, $P$ lies on $(XYZ)$. $$\angle PYX=\angle PYZ+\angle ZYX =90^{\circ} + \angle PBC +\angle CBA=90^{\circ} + \angle PBA =90^{\circ} + \angle PCA =\angle OPX$$Hence done.

Neothehero wrote: Let $O$ be the circumcentre, and $\Omega$ be the circumcircle of an acute-angled triangle $ABC$. Let $P$ be an arbitrary point on $\Omega$, distinct from $A$, $B$, $C$, and their antipodes in $\Omega$. Denote the circumcentres of the triangles $AOP$, $BOP$, and $COP$ by $O_A$, $O_B$, and $O_C$, respectively. The lines $\ell_A$, $\ell_B$, $\ell_C$ perpendicular to $BC$, $CA$, and $AB$ pass through $O_A$, $O_B$, and $O_C$, respectively. Prove that the circumcircle of triangle formed by $\ell_A$, $\ell_B$, and $\ell_C$ is tangent to the line $OP$. Solution : Let $A_1A_2A_3$ be the required triangle such that $A_1$ doesn't lie on $\ell _A$ and $A_2$ doesn't lie on $\ell _B$ and $A_3$ doesn't lie on $\ell _C$. Now $m\angle O_AOO_B = m\angle C \implies O_APO_BA_3$ is cyclic and in a similar fashion, we have that $O_BO_CPA_1$ and $O_AA_2O_CP$ are also cyclic. We all together get $A_1A_2A_3P$ is cyclic. Now, if $X = PO \cap BC$, then $m\angle OPA_3 = m\angle BCP - m\angle PXC$ $= m\angle OPC = m\angle PO_CO_B = m\angle PA_1A_3 \implies m\angle PA_1A_3 = m\angle OPA_3$ which gives us the desired result

yayups wrote: Let $XYZ$ be the triangle formed by $\ell_A$, $\ell_B$, $\ell_C$. Here is the claim which allows the problem to fall to straight angle chasing. We'll complex bash with $(ABC)$ as the unit circle. Note that $O_A$ is the intersection of the perpendicular bisector of $OA$ and $OP$, so it is $1/2$ of the pole of $AP$. This means that \[O_A=\frac{ap}{a+p}.\]We'll now find $x$. Note that $XO_B\perp AC$, so the foot from $X$ to $AC$ is the foot from $O_B$ to $AC$, so \[\frac{1}{2}(x+a+c-ac\bar{x})=\frac{1}{2}(O_B+a+c-ac\bar{O}_B),\]so $ac(\bar{x}-\bar{O}_B)=x-O_B$, and similarly $ab(\bar{x}-\bar{O}_C)=x-O_C$. Setting the value for $x$ we get from these two equal, we have that \begin{align*} a(b-c)\bar{x}&=a(b\bar{O}_C-c\bar{O}_B)+O_B-O_C \\ &= a\left(\frac{b}{c+p}-\frac{c}{b+p}\right)+\frac{bp}{b+p}-\frac{cp}{c+p} \\ &= \frac{a[b^2+bp-c^2-cp]+bp(c+p)-cp(b+p)}{(b+p)(c+p)} \\ &= \frac{(b-c)[ap+a(b+c)+p^2]}{(b+p)(c+p)}, \end{align*}so \[\boxed{\bar{x}=\frac{ap+a(b+c)+p^2}{a(b+p)(c+p)}}.\]We also compute \begin{align*} x &= \frac{\frac{1}{ap}+\frac{b+c}{bca}+\frac{1}{p^2}}{\frac{1}{a}\frac{b+p}{bp}\frac{c+p}{cp}} \\ &= \frac{bcp+p^2(b+c)+abc}{(b+p)(c+p)}. \end{align*} After obtaining expressions for $X, Y, Z$ one can notice that, for example $$ x=\frac{bcp+p^2(b+c)+abc}{(b+p)(c+p)}=\frac{p(b+p)(c+p)+abc-p^3}{(b+p)(c+p)}=p+\frac{abc-p^3}{(b+p)(c+p)}= p+\frac{abc-p^3}{(a+p)(b+p)(c+p)}\cdot p+\frac{abc-p^3}{(a+p)(b+p)(c+p)}\cdot a. $$Let $s=p+\frac{abc-p^3}{(a+p)(b+p)(c+p)}\cdot p$ and $t=\frac{abc-p^3}{(a+p)(b+p)(c+p)}$ (note that this expressions are symmetric of $a,b,c$). Then, the triangle $XYZ$ is the image of initial triangle $ABC$ under the map $f\colon x\mapsto s+tz$ ($f$ is a spiral similarity or translation). Hence, circumcenter of $XYZ$ is $f(0)=s$ and circumradius of $(XYZ)$ is equal to $|t|$. Moreover, distance between points $p$ and $s$ is exactly $|pt|=|t|$. So, in order to prove tangency, it's sufficient to prove that $OP$ is perpendicular to $pt$, which is quite easy: $$ \frac{p}{\overline{p}}=p^2~\text{and}~\frac{pt}{\overline{pt}}=\frac{-abcp^3}{ap\cdot bp\cdot cp}\cdot p^2=-p^2, $$as desired. Remark. This solution uses very little geometry and some simple algebraic transforamations.

Let $X=\ell_B\cap \ell_C$, and similarly define $Y$ and $Z$. We will show that $(XYZ)$ is tangent to $OP$ at $P$. First, we claim $\triangle XYZ\sim \triangle ABC$. Indeed, \[\angle XYZ= \angle (XY,YZ) = \angle (\perp AB, \perp BC) = \angle (AB,BC) = \angle ABC, \]and cyclic variants, proving the similarity. Claim: $XO_BPP_C$ is cyclic (and cyclic variants). Proof: First note \begin{align*} \angle O_APO &= 90^\circ-\tfrac12 \angle OO_AP = 90^\circ-\angle OAP = 90-(\tfrac{180-\angle AOP}{2}) = \tfrac12\angle AOP, \end{align*}and similar cyclic variants. We have \begin{align*} \angle O_BXO_C &= \angle ZXY = \angle BAC, \\ \angle O_BPO_C &= \angle O_BPO + \angle O_CPO =\tfrac12 \angle BOP+\tfrac12 COP \\ &= \tfrac12 (360^\circ-\angle BOC) =\angle BAC, \end{align*}proving the claim. $\blacksquare$ Claim: $P,A,X$ are collinear (and cyclic variants). Proof: Note that $O_AO_BO_C$ collinear since they lie on the perpendicular bisector of $OP$. Hence \begin{align*} \angle PXO_C &= \angle PO_BO_C= \tfrac12 \angle PO_BO=\angle PBO \\ &= 90-\tfrac12 \angle BOP = 90^\circ -\angle PCB,\\ \angle AXO_C &= \angle AXY = 90^\circ - \angle XAB = \angle PAB - 90^\circ, \end{align*}which are clearly equal. $\blacksquare$ Now, $P\in (XYZ)$ since \begin{align*} \angle PZY &= 90^\circ-\angle PCB \qquad \text{(since $\overline{ZPC}$ collinear)} \\ \angle PXY &= 90^\circ-\angle XAB = \angle PAB-90^\circ, \end{align*}which are equal since $P\in (ABC)$. We now show $(XYZ)$ tangent to $OP$ at $P$. We have \begin{align*} \angle OPX &= \angle OPA = 90^\circ-\tfrac12 \widehat{AP} = 90^\circ-\angle ACP, \\ \angle XZP &= \angle XZY+\angle YZP = \angle ACB+(90^\circ-\angle PCB) = 90^\circ-\angle ACP, \end{align*}proving the tangency.

Maybe too easy for a G7.

[asy][asy] size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.724555607032263, xmax = 1.5754642722882315, ymin = -1.6167851463147418, ymax = 1.233769659836653; /* image dimensions */pen qqwuqq = rgb(0,0.39215686274509803,0); pen zzttff = rgb(0.6,0.2,1); pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen ffvvqq = rgb(1,0.3333333333333333,0); /* draw figures */draw(circle((0,0), 1), linewidth(0.8) + blue); draw((-0.47283605161038517,-0.881150423195439)--(-0.4450539838751519,0.8955037417213041), linewidth(0.8) + linetype("4 4") + qqwuqq); draw((-0.47283605161038517,-0.881150423195439)--(-1.105389951335395,-0.253180878805536), linewidth(0.8) + linetype("4 4") + qqwuqq); draw((-1.1525563272989778,-1.0398503296756219)--(0.8141392612587833,-0.5806696679498616), linewidth(0.8) + linetype("4 4") + qqwuqq); draw(circle((-0.882555286012392,-0.6612900906532113), 0.4649821683296528), linewidth(0.8) + blue); draw(circle((-0.9915389138274326,-0.3662067951476439), 0.16042729352143456), linewidth(0.8) + blue); draw(circle((-0.4556079991573275,0.22057790875705574), 0.6750083460528921), linewidth(0.8) + blue); draw((0,0)--(-0.47283605161038517,-0.881150423195439), linewidth(0.8)); draw((-1.1180791812941542,-0.4648195591088224)--(0.21254118456735152,0.12459387086998551), linewidth(0.8)); draw((-1.1525563272989778,-1.0398503296756219)--(-1.0891887874249246,0.017031967878854862), linewidth(0.8) + zzttff); draw((-0.4450539838751519,0.8955037417213041)--(-0.8776878763194698,-0.47923271148975133), linewidth(0.8) + fuqqzz); draw((-1.1180791812941542,-0.4648195591088224)--(0.8141392612587833,-0.5806696679498616), linewidth(0.8) + fuqqzz); draw((0.8141392612587833,-0.5806696679498616)--(-0.4450539838751519,0.8955037417213041), linewidth(0.8) + fuqqzz); draw((-1.0891887874249246,0.017031967878854862)--(0.15146104647116193,-0.6487158470778117), linewidth(0.8)); draw((-1.1525563272989778,-1.0398503296756219)--(0.21254118456735152,0.12459387086998551), linewidth(0.8) + zzttff); draw((-1.105389951335395,-0.253180878805536)--(0.15146104647116193,-0.6487158470778117), linewidth(0.8) + zzttff); draw((-0.9784863924287199,-0.20631136622064544)--(0.4450539838751519,-0.8955037417213041), linewidth(0.8) + linetype("4 4") + ffvvqq); draw((0,0)--(-0.9784863924287199,-0.20631136622064544), linewidth(0.8)); /* dots and labels */dot((0,0),dotstyle); label("$O$", (-0.06567464619104664,0.03913880746615159), NE * labelscalefactor); dot((-0.4450539838751519,0.8955037417213041),dotstyle); label("$A$", (-0.5092257052395002,0.970596031467904), NE * labelscalefactor); dot((-0.8776878763194698,-0.47923271148975133),dotstyle); label("$B$", (-0.9025076442624623,-0.5700046469603912), NE * labelscalefactor); dot((0.8141392612587833,-0.5806696679498616),dotstyle); label("$C$", (0.8983096554742591,-0.6350588022874977), NE * labelscalefactor); dot((-0.47283605161038517,-0.881150423195439),dotstyle); label("$P$", (-0.4678276063949779,-0.9869426424659374), NE * labelscalefactor); dot((-1.0891887874249246,0.017031967878854862),linewidth(4pt) + dotstyle); label("$O_{A}$", (-1.0769710608215208,0.04209581452647461), NE * labelscalefactor); dot((-0.3675337929660405,-0.3702168936349633),linewidth(4pt) + dotstyle); label("$O_{B}$", (-0.41164447224884043,-0.30091700447099606), NE * labelscalefactor); dot((0.15146104647116193,-0.6487158470778117),linewidth(4pt) + dotstyle); label("$O_{C}$", (0.1620148974538262,-0.6261877811065286), NE * labelscalefactor); dot((-0.4661620144395024,-0.4543479242071927),linewidth(4pt) + dotstyle); label("$X_{A}$", (-0.5772368676269297,-0.56946640690940844), NE * labelscalefactor); dot((-1.105389951335395,-0.253180878805536),linewidth(4pt) + dotstyle); label("$X_{B}$", (-1.1982083502947647,-0.1855937291183982), NE * labelscalefactor); dot((-1.1525563272989778,-1.0398503296756219),linewidth(4pt) + dotstyle); label("$X_{C}$", (-1.2603054985615483,-1.037211762491429), NE * labelscalefactor); dot((-1.1180791812941542,-0.4648195591088224),linewidth(4pt) + dotstyle); label("$F_A$", (-1.106541131424751,-0.4398963363061782), NE * labelscalefactor); dot((-0.8334782621247745,-0.33875234396129433),linewidth(4pt) + dotstyle); label("$F_C$", (-0.7871843689098644,-0.2979599974106731), NE * labelscalefactor); dot((0.21254118456735152,0.12459387086998551),linewidth(4pt) + dotstyle); label("$F_B$", (0.27142415868577807,0.11010697691390414), NE * labelscalefactor); dot((-0.9784863924287199,-0.20631136622064544),linewidth(4pt) + dotstyle); label("$Q$", (-0.976432820770538,-0.16489467969613703), NE * labelscalefactor); dot((0.4450539838751519,-0.8955037417213041),dotstyle); label("$A'$", (0.45771560348612855,-0.8657053529926936), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Suppose $l_B\cap l_C=X_A$, $l_C\cap l_A=X_B$, $l_A\cap l_B=X_C$. Let $F_A,F_B,F_C$ be the projection of $O_A,O_B,O_C$ on $BC,CA,AB$ respectively. CLAIM. $\triangle AF_BF_C\sim \triangle OO_BO_C$ Proof. We have $\angle AF_BF_C=\frac{1}{2}\angle BOC=\angle F_BAF_C$ Now we use complex number with $(ABC)$ as the unit circle, we have $$O_B=\frac{\begin{vmatrix}b&1&1\\0&0&1\\ p&1&1\end{vmatrix}}{\begin{vmatrix}b&\frac{1}{b}&1\\0&0&1\\ p&\frac{1}{p}&1\end{vmatrix}}=\frac{bp}{b+p}$$similarly $O_C=\frac{cp}{c+p}$, therefore, $$F_B=\frac{1}{2}\left(a+c+\frac{bp}{b+p}-\frac{ac}{b+p}\right)$$$$F_C=\frac{1}{2}\left(a+b+\frac{cp}{c+p}-\frac{ab}{c+p}\right)$$Therefore, \begin{align*} \frac{AF_B}{AF_C}&=\frac{|b+\frac{cp}{c+p}-\frac{ab}{c+p}-a|}{|c+\frac{bp}{b+p}-\frac{ac}{b+p}-a|}\\ &=\frac{|c+p|}{|b+p|}\cdot\frac{|(b-a)(c+p)+cp-ab|}{|(c-a)(b+p)+bp-ac|}\\ &=\frac{|c+p|}{|b+p|}\\ &=\frac{OO_B}{OO_C}\\ \end{align*}as desired. $\blacksquare$ Corollary 1. $F_A,F_C,F_B$ are collinear. Proof. By Menelaus theorem, $$\frac{CF_A}{F_AB}\cdot\frac{BF_C}{F_CA}\cdot\frac{AF_B}{F_BC}=\frac{OO_A}{OO_B}\cdot\frac{OO_C}{OO_A}\cdot\frac{OO_B}{OO_C}=1$$$\blacksquare$ Corollary 2. $A,X_A,P$ are collinear. Proof. $$\angle BAX_A=90^{\circ}-\angle AX_AF_C=90^{\circ}-\angle AF_BF_C=90^{\circ}-\angle OO_BO_C=\angle O_BOP=\angle BAP$$$\blacksquare$ Therefore, $$\angle X_BX_AP=\angle F_AF_BC=180^{\circ}-\angle X_BX_CP$$Hence $P$ lies on $(X_AX_BX_C)$ Therefore, $$\angle OPX_A=90^{\circ}-\angle ACP=\angle X_AX_CP$$which implies $OP$ is tangent to $(X_AX_VX_C)$ as desired.

Very easy for a G7 if you use GeoGebra . But fun problem to do! Let $\ell_A \cap \ell_B = Z, \ell_B \cap \ell_C = Y, \ell_A \cap \ell_C = X$. Let $R,S,T$ be the feet of perpendiculars from $O_A, O_B, O_C$ onto $BC, CA, AB$ Claim 1: $PYO_BO_C$ is cyclic Proof: See that $ASYT$ is cyclic. So, $\angle O_BYO_C = \angle BAC$ and also, $\angle O_BPO_C = 180 - \angle OBP - \angle OCP = 180 - \angle OBC - \angle OCB - \angle PCB - \angle PBC = 2 \angle BAC - \angle BAC = \angle BAC$ and so we get $PYO_BO_C$ is cyclic. $\blacksquare$ Similarly, we get that $PZO_AO_B$ and $PXO_AO_C$ is cyclic. Now, observe that in quadrilateral $O_AO_BYX$, $P = (O_AO_BZ) \cap (O_BO_CY)$ which means $P$ is its miquel point and so $PXYZ$ is cyclic as well. Claim 2: $Z,P,C$ are collinear Proof: $\angle BPC = 180 - \angle BAC = 180 - \angle TYZ = 180 - \angle XYZ = 180 - \angle XPZ = 180 - \angle BPZ$ and so $Z,P,C$ are collinear. $\blacksquare$ Similarly, $X,B,P$ and $A,Y,P$ are collinear. Now, to finish, see that $\angle YPO = \angle APO = \angle PAO = \frac{\angle OO_AP}{2} = \angle O_BO_AP = \angle O_BZP = \angle YZP$ and so $OP$ is tangent to $(XYZ)$ at $P$ and so we're done

Extension. Denote by $XYZ$ this triangle determined by the lines. Let $k$ be the line such that $XYZ$ is the paralogic triangle of $ABC$ with respect to $k$ (i.e. it passes through $YZ \cap BC$, etc.). Prove that the nine-point center of $ABC$ lies on $k$.

Solved with Kevin Wu, Nacho Cho, Isaac Zhu We proceed with complex numbers. Let $(ABCP)$ be the unit circle, and $A = a, B = b, C = c, P = p$. First, observe that \[o_{a} = \frac{ap}{a+p}, o_{b} = \frac{bp}{b+p}, o_{c} = \frac{cp}{c+p}\]Let $XYZ$ be the triangle determined by $\ell_{A}, \ell_{B}, \ell_{C}$. I claim that $X$ lies on $PC$. Observe that: \[\frac{x-o_{a}}{b-c} = -\overline{\left(\frac{x-o_{a}}{b-c}\right)} = \frac{bc(\overline{x} - \overline{o-{a}}}{b-c}\Rightarrow x-o_{a} = bc(\overline{x}-\overline{o_{a}})\]We similarly have $x - o_{b} = ac(\overline{x}-\overline{o_{b}} )$. Now, subtracting these two, we get \[o_{a} - o_{b} = bc\overline{o_{a}} - ac\overline{o_{b}} + \overline{x}(ac - bc)\]\[\frac{ap(b+p) - bp(a+p)}{(a+p)(b+p)} = \frac{bc(b+p) - ac(a+p)}{(a+p)(b+p)} + \overline{x}(ac - bc)\]\[\overline{x}(a-b)(c)(a+p)(b+p) = (a-b)(p^{2} + ac + bc + cp) \Rightarrow \overline{x} = \frac{p^{2} + c(a+b+p)}{c(a+p)(b+p)}\]Now, observe that \[x = \overline{\left(\frac{p^{2} + c(a+b+p)}{c(a+p)(b+p)}\right)} = \frac{\frac{1}{p^{2}} + \frac{1}{c}(\frac{1}{a} + \frac{1}{b} + \frac{1}{p})}{\frac{1}{c}(\frac{a+p}{ap})(\frac{b+p}{bp})} \cdot \frac{p^{2}cab}{p^{2}cab} = \frac{abc +p(ab+ap+bp)}{(a+p)(b+p)}\]Now, we will show $C,P,X$ collinear. To prove this, we use the fact that $X$ lies on the chord if $x + pc\overline{x} = p+c$. Plugging it in, we get \[\frac{abc + p(ab + ap + bp) + p(p^{2} + c(a+b+p))}{(a+p)(b+p)}\]\[ = \frac{abc + p^{3} + acp + bcp + p^{2}c + abp + ap^{2} + bp^{2}}{(a+p)(b+p)} = \frac{(a+p)(b+p)(c+p)}{(a+p)(b+p)} = p+c\]Therefore, $X, P,C$ are collinear. Similarly, $P,B,Y$ and $P,A,Z$ are collinear. Now, we get \[\angle ZYX = \angle ABC = 180 - \angle APC = \angle ZPX\]This means $(ZYPX)$ is cyclic. Finally, observe that \[\angle YPO = 90 - \angle PCB = \angle YXP\]Therefore, $(ZYXP)$ is tangent to $OP$ at $P$. We conclude.

Suppose $l_A, l_B, l_C$ bound the triangle $DEF$. We have $$(PO_B, PO_C) \equiv (OO_C, OO_B) \equiv (PC, PB) \equiv (DO_C, DO_B) \pmod \pi$$Then $P$ $\in$ $(DO_BO_C)$. Similarly, we have $P$ is Miquel point of completed quadrilateral $DEO_AO_B.O_CF,$ which means $P$ $\in$ $(DEF)$. We also have $$(PD, PA) \equiv (PD, PO_B) + (PO_B, PA) \equiv (O_CD, O_CO_B) + (PO_B, PB) + (PB, PA) \equiv (AB, OP)$$$$+ \dfrac{\pi}{2} + (OP, OB) + (PB, PA) \equiv (AB, OP) + \dfrac{\pi}{2} + (BP, BA) + (AP, AB) \equiv 0 \pmod \pi$$From this, we have $A, X, P$ are collinear. Therefore, we have $$(PO, PD) \equiv (AP, AO) \equiv (O_AP, O_AO_B) \equiv (FP, FD) \pmod \pi$$or $OP$ touches at $(DEF)$ at $P$

Let $\ell_A,\ell_B,$ and $\ell_C$ form triangle $DEF.$ Claim: $A,D,$ and $P$ are collinear. Proof. WLOG let $(ABC)$ be the unit circle. Notice $O_A=\frac{ap(\overline{a}-\overline{p})}{\overline{a}p-a\overline{p}}=\frac{ap}{a+p}.$ Also, $\overline{O_CD}\perp\overline{AB}$ and $\overline{O_BD}\perp\overline{AC}$ so $d-ab\overline{d}=o_c-ab\overline{o_c}$ and $d-ac\overline{d}=bo_b-ac\overline{o_b}.$ Multiplying the equations by $c$ and $b,$ respectively and subtracting yields \begin{align*}(b-c)d&=bo_b-co_c-abc(\overline{o_b}-\overline{o_c})\\&=\frac{b^2p}{b+p}-\frac{c^2p}{c+p}-abc\left(\frac{1}{b+p}-\frac{1}{c+p}\right)\\&=\frac{b^2p(c+p)-c^2p(b+p)-abc(c-b)}{(b+p)(c+p)}\\&=\frac{p(b-c)(pb+pc+bc)+abc(b-c)}{(b+p)(c+p)}\end{align*}so $d=\frac{p(pb+pc+bc)+abc}{(b+p)(c+p)}.$ Hence, \begin{align*}d+ap\overline{d}&=\frac{p(pb+pc+bc)+abc}{(b+p)(c+p)}+ap\cdot\frac{\frac{ac+ab+ap+p^2}{abcp^2}}{\frac{(b+p)(c+p)}{bcp^2}}\\&=\frac{p(pb+pc+bc)+abc+p(ac+ab+ap+p^2)}{(b+p)(c+p)}\\&=\frac{(a+p)(b+p)(c+p)}{(b+p)(c+p)}\\&=a+p\end{align*}and $D$ lies on $\overline{AP}.$ $\blacksquare$ Therefore, $$\measuredangle DPE=\measuredangle APB=\measuredangle ACB=90-\measuredangle(\overline{FO_A},\overline{AC})=\measuredangle(\overline{FO_B},\overline{FO_A})=\measuredangle DFE$$so $P$ lies on $(DEF).$ Additionally, $$\measuredangle FPO=\measuredangle CPO=90-\tfrac{1}{2}\widehat{CP}=90-\measuredangle PAC=\measuredangle FDA.$$$\square$

I solved this without Geogebra, and I consider this to be my greatest achievement till now This solution doesn't use any miquel point, complex number stuff, it is just angel chase (simpler than in #2 and I would argue one of the most simplest in this thread) A recent G7 for my $(169=100+69=13^2)^{th} $ Post? Let's do this! This diagram is almost the same as the one I used when I drew it on paper! [asy][asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.564632340869464, xmax = 15.051238838636193, ymin = -4.935646511438435, ymax = 8.175723762645454; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccqqqq = rgb(0.8,0,0); pen ffvvqq = rgb(1,0.3333333333333333,0); draw((4.4484063808077,-2.5391535307129325)--(4.4431514829389664,-2.176565577770285)--(4.080563529996319,-2.181820475639019)--(4.085818427865053,-2.5444084285816664)--cycle, linewidth(1) + qqwuqq); /* draw figures */ draw((1.28,4.27)--(-0.44,-2.61), linewidth(1) + qqwuqq); draw((-0.44,-2.61)--(7.84,-2.49), linewidth(1) + qqwuqq); draw((7.84,-2.49)--(1.28,4.27), linewidth(1) + qqwuqq); draw(circle((3.6627636363636364,0.01930909090909069), 4.872980171456166), linewidth(1) + ccqqqq); draw((3.9781883066508104,4.882069935201065)--(3.6627636363636364,0.01930909090909069), linewidth(1)); draw((3.6627636363636364,0.01930909090909069)--(-0.44,-2.61), linewidth(1) + qqwuqq); draw((-0.44,-2.61)--(3.9781883066508104,4.882069935201065), linewidth(1) + linetype("2 2") + ffvvqq); draw((1.28,4.27)--(3.6627636363636364,0.01930909090909069), linewidth(1) + qqwuqq); draw((3.6627636363636364,0.01930909090909069)--(7.84,-2.49), linewidth(1) + qqwuqq); draw((3.04011185687238,6.672826642536562)--(7.84,-2.49), linewidth(1) + qqwuqq); draw((0.34560291213323907,4.058036897345917)--(3.9781883066508104,4.882069935201065), linewidth(1) + linetype("2 2") + ffvvqq); draw((-0.9892525278321339,2.762674221876561)--(3.04011185687238,6.672826642536562), linewidth(1) + qqwuqq); draw((7.810161900580135,2.191897150234193)--(0.34560291213323907,4.058036897345917), linewidth(1) + linetype("2 2") + ffvvqq); draw((3.1006255348580316,2.497382861526589)--(3.04011185687238,6.672826642536562), linewidth(1)); draw(circle((2.039911296137204,5.007796946366772), 1.9423503934208743), linewidth(1) + linetype("2 2") + qqwuqq); draw((3.9781883066508104,4.882069935201065)--(4.085818427865053,-2.5444084285816664), linewidth(1)); draw(circle((1.1807267907781123,4.557718637465095), 2.8162021761819207), linewidth(1) + linetype("2 2") + blue); draw((-0.9892525278321339,2.762674221876561)--(7.810161900580135,2.191897150234193), linewidth(1) + qqwuqq); /* dots and labels */ dot((1.28,4.27),dotstyle); label("$A$", (1.3415661400504038,4.449154375748078), NE * labelscalefactor); dot((-0.44,-2.61),dotstyle); label("$B$", (-0.7952282331705735,-2.8501352031747653), NE * labelscalefactor); dot((7.84,-2.49),dotstyle); label("$C$", (8.02545893948562,-2.713380363288623), NE * labelscalefactor); dot((3.9781883066508104,4.882069935201065),dotstyle); label("$P$", (4.042474227801719,5.04745680024995), NE * labelscalefactor); dot((3.6627636363636364,0.01930909090909069),linewidth(4pt) + dotstyle); label("$O$", (3.734775838057898,0.15847127432036456), NE * labelscalefactor); dot((-0.9892525278321339,2.762674221876561),linewidth(4pt) + dotstyle); label("$O_B$", (-1.4277193676439828,2.43202048742748), NE * labelscalefactor); dot((3.1006255348580316,2.497382861526589),linewidth(4pt) + dotstyle); label("$O_A$", (2.9655298636983463,2.0388503227548207), NE * labelscalefactor); dot((7.810161900580135,2.191897150234193),linewidth(4pt) + dotstyle); label("$O_C$", (7.683571839770264,1.8166237079398395), NE * labelscalefactor); dot((0.34560291213323907,4.058036897345917),linewidth(4pt) + dotstyle); label("$Z$", (0.008206451160513944,4.004701146118116), NE * labelscalefactor); dot((3.04011185687238,6.672826642536562),linewidth(4pt) + dotstyle); label("$X$", (3.102284703584489,6.808175363784032), NE * labelscalefactor); dot((3.087943376601477,3.372451781228858),linewidth(4pt) + dotstyle); label("$Y$", (2.7945863138406684,3.577342271473921), NE * labelscalefactor); dot((4.085818427865053,-2.5444084285816664),linewidth(4pt) + dotstyle); label("$P_X$", (4.025379872815951,-2.918512623117836), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Notations for the intersection points are clear from my diagram and $P_X$ is the foot of $P$ on $BC$ Claim: $\triangle XYZ \sim \triangle CBA$ Proof: Notice that the perpendiculars from the circumcenters induce some cyclic quads one of which is {$\ell_a \cap BC,C,X,l_b \cap AC$} which gives that $\measuredangle YXZ = \measuredangle BCA$ and similarly for the other vertices which establishes the claim $\blacksquare$ Claim: $XPO_AO_B$ is cyclic Proof: $$ \angle O_BPO_A = \angle O_BOO_A = \angle O_BOP - \angle O_AOP = \frac{\angle BOP - \angle AOP}{2}= \frac{\angle BOA}{2}= \angle BCA=\angle ZXY=\angle O_BXO_B \ \blacksquare$$Similarly the other cyclic quads corresponding to the other circumcenters can be found. Claim: $P \in \odot(XYZ)$ ProofTrivially $O_A-O_B-O_C$ because all of them lie on the perpendicular bisector. Using this and the above claim we have $$\angle PYX = \angle PO_CO_A = \angle PO_CO_B=\angle PZX \implies P \in \odot(XYZ) \ \blacksquare $$Claim: $X-P-C$ Proof: $$\angle PXO_A=\angle PXY=\angle PZY=\angle PZO_C=\angle PO_BO_C=\frac{\angle PO_BO}{2}=\frac{180^{\circ}-2\cdot \angle POO_B}{2}=90^{\circ}-\angle POO_B=90^{\circ}-\angle PCB=\angle CPP_X$$Now notice that $XO_A \parallel PP_X$ and $\angle PXO_A=\angle CPP_X$ which means $X-P-C$ and similarly $P-Y-B$ proving our claim $\blacksquare$ For the finish note that $PP_X$ is an altitude and $O$ is the circumcenter of $\triangle PBC$ which means those two lines are isogonal, and from a previous angle chase we know that $$\angle PXY=\angle CPP_X=\angle BPO=\angle YPO \implies OP \ \text{is tangent to} \ \odot(XYZ) \ \blacksquare$$

Let $W'$ be the antipode of $W$ for any point on $(ABC)$. Let $XYZ$ be the triangle formed by the three lines. From angle chasing $XYZ \sim ABC$. Also $PO_CO \sim POC'$ and $PO_BO \sim POB' \implies PB'C' \sim PO_CO_B \implies P$ is the miquel point of the $\{l_a,l_b,l_c,O_A-O_B-O_C\} \implies PXYZ$ is cyclic. Also note that $P'BC \sim PB'C' \sim PO_CO_B \sim PYZ$. Thus we have $ABC \cap P' \sim XYZ \cap P$ with $90^{\circ}$ rotation counter-clockwise. If $O_P$ is the center of $(XYZ)$ then $OP' \perp PO_P$ as desired.

Neothehero wrote: Let $O$ be the circumcentre, and $\Omega$ be the circumcircle of an acute-angled triangle $ABC$. Let $P$ be an arbitrary point on $\Omega$, distinct from $A$, $B$, $C$, and their antipodes in $\Omega$. Denote the circumcentres of the triangles $AOP$, $BOP$, and $COP$ by $O_A$, $O_B$, and $O_C$, respectively. The lines $\ell_A$, $\ell_B$, $\ell_C$ perpendicular to $BC$, $CA$, and $AB$ pass through $O_A$, $O_B$, and $O_C$, respectively. Prove that the circumcircle of triangle formed by $\ell_A$, $\ell_B$, and $\ell_C$ is tangent to the line $OP$. Absolutely beautiful geo! Solved with MathsCrazy We begin by proving some claims.

Claim 1. \(O_A, O_B, O_C\) are collinear. Proof. \(OP\) is the common radical axis of \(\odot(AOP), \odot(BOP),\odot(COP)\). $\blacksquare$ Claim 2. \(XYZ\) and \(ABC\) are similar. Proof. Note that \(\measuredangle XYZ=\measuredangle ABC\) because of the cyclic quadrilateral formed by the lines \(AB,AC,\ell_2,\ell_3\), and similarly the other equalities hold, and so \(XYZ\) and \(ABC\) are similar, as claimed. $\blacksquare$ Claim 3. \(P,X,O_B,O_C\) are concyclic. Proof. We know that \(\measuredangle O_BXO_C=180-\measuredangle A\) from claim 2. We want to prove \(\measuredangle O_BXO_C=\measuredangle O_BPO_C\). This follows from: \begin{align*} \measuredangle O_BPO_C &= \measuredangle O_BPO+\measuredangle O_CPO \\ &= \measuredangle O_BOP+\measuredangle O_COP \\ &= \measuredangle O_COC+\measuredangle O_BOB \\ &= \frac{360-(\measuredangle OO_BB+\measuredangle OO_CC)}{2} \\ &= \frac{360-2(\measuredangle BPO+\measuredangle CPO)}{2} \\ &= \frac{360-2\measuredangle BPC}{2} \\ &= 180-\measuredangle A \end{align*}as claimed. $\blacksquare$ Claim 4. \(P\) lies on the circumcircle of triangle \(XYZ\). Proof. We present two proofs, one purely synthetic and the other complex bash. First Proof. (Mohammed Imran) Consider the complete quadrilateral \(XYZO_AO_BO_C\). Since \(P\) lies on the circumcircles of triangles \(XO_CO_B\) and \(ZO_AO_B\), by Miquel we are done. $\blacksquare$ Second Proof. (Mohammed Imran [Synthetic]; Malay Mahajan [Complex Bash]) We claim that proving \(X,P,A\) and \(Y,P,B\) and \(Z,P,C\) collinear proves this claim. That is because \[\measuredangle ZPY=180-\measuredangle CPY=180-\measuredangle CPB=180-\measuredangle CAB=180-\measuredangle A=180-\measuredangle ZXY \]And we are done. So now it remains to prove \(Y,P,B\) collinear, if we prove this we are done too, because the other arguments symmetrically follow. We let $a$ be the complex number representing $A$ and similarly for other points. Without loss of generality, let the $(ABC)$ be unit circle (and hence $ \overline{a} =a$ similarly for $b,c $), centered at origin, and trivially \(o=0\). Also we can assume without loss of generality, that $ p=1 $ Also, as $o_a$ is circumcenter of triangle with vertices $0,1,a$, by the Complex Circumcenter lemma (EGMO 6.2.4) we get that $$o_a=\frac{a}{a+1}$$Similarly, $$o_b=\frac{b}{b+1},~o_c=\frac{c}{c+1} $$due to symmetry. Note that line $ O_AY$ is perpendicular to $BC$ and hence, has slope $+bc$. Similarly $O_CY$ has slope $+ab$ Therefore , equation of $ O_AY$ is $$ z-\frac{a}{a+1} = bc \left( \overline{z} - \overline{\left(\frac{a}{a+1}\right)}\right) $$Similarly, equation of $ O_CY$ is $$ z-\frac{c}{c+1} = ab \left(\overline{z} - \overline{\left(\frac{c}{c+1}\right)}\right) $$Also, as $ O_AY$ and $ O_CY$ intersect at $Y$, upon solving, we get that their intersection point \(Y\) is $$y= \frac{a+b+ab+abc}{(1+a)(1+b)} $$ Now, to prove that $P(1),Y(y),B(b)$ collinear , it suffices to prove that $$\frac{1-b}{1-y} = \overline{\left(\frac{1-b}{1-y}\right)} $$which is an easy computation, therefore \(Y,P,B\) are collinear, as claimed. $\blacksquare$ Claim 5. \(\odot(XYZ)\) is tangent to line \(OP\) at \(P\). Proof. Firstly, note that \[\measuredangle PZY = \measuredangle PXY = \measuredangle PXO_C\]So it sufices to show that \(\measuredangle PXO_C=\measuredangle PZY\). This follows because, \begin{align*} \measuredangle PZY &= \measuredangle CZY \\ &= 90-\measuredangle PCB \\ &= \measuredangle PAB-90 \\ &= 90-\measuredangle XAB \\ &= PXO_C \end{align*} where the last step follows because \(AB\perp XY\), as claimed. $\blacksquare$

Point labeling explanation: The triangle $E,F,G$ is formed by the intersections of $\ell_A,\ell_B,\ell_C$ and u might noted that the points $I,J,K$ are the foot of the perpendiculars from $O_A,O_B,O_C$ and the $H$ was a point that at the end i didnt used it xd. Also note that the location of the point $P$ is important here but since most of the things that we get w.r.t a vertex of $\triangle ABC$ are symetric there shouldn't be a problem. Claim 1: $PEFG$ is cyclic Proof: First note that $O_A,O_B,O_C$ are colinear by the radax and that $P$ is symetric to $O$ w.r.t. that line hence taking this in count and using circumcenter propeties and the cyclics formed by the perpendiculars we do angle chasing $$\angle O_BPO_A=\angle O_BOO_A=\angle O_BOP-\angle O_AOP=\angle PCB=\angle PCA=\angle ACB=\angle O_BFO_A$$$$\angle O_BPO_C=\angle O_BOO_C=\angle O_BOP+\angle O_COP=180-\angle BPC=180-\angle BAC=\angle O_BEO_C$$Those angle chase tell that $PFO_AO_B, PEO_CO_B$ are cyclic hence $P$ is the miquel point of $EFO_CO_A$ and this completes the claim. Claim 2: $P,F,C$ and $G,P,B$ colinear (for other cases u can use symetry to get the other colinearity, u need 2 colinearities) Proof: First we prove a small colinearity with angle chase $$\angle IBG=\angle ACP=\angle JKG \implies I,K,J \; \text{colinear}$$Now by even more angle chasing using the cyclics and the colinearity $$\angle PFG=\angle PEG=\angle KJC=\angle O_AFC \implies P,F,C \; \text{colinear}$$$$\angle GPF+\angle BPF=\angle GPF+\angle BAJ=\angle GPF+\angle GEF=180 \implies G,P,B \; \text{colinear}$$Finishing: We will end with angle chase using the colinearities and cyclics (said this a lot of times but yeah its true lol, this problem was interesting as i got some useless claims on the path to realice that i only needed some of them which is annoying but i will take it as this is a G7 that i solved so everything cool i guess) $$\angle PGF=90-\angle PBK=\angle PCO=\angle OPF \implies OP \; \text{tangent to} \; (EFG)$$Thus we are done

Let $A'=\ell_B \cap \ell_C$ and define $B',C'$ analogously. Denote by $\ell$ tangent to $\Omega$ at antipode $Q$ of $P.$ Note that there exist spiral similarity $\varphi :$ $ABC\mapsto A'B'C'$ with rotation angle $90^\circ.$ $$\measuredangle O_APO_B=\measuredangle O_BOO_A=\frac{1}{2} \measuredangle BOA=\measuredangle BCA=\measuredangle O_AC'O_B\implies P\in \odot (O_AO_BC').$$By all similar arguments we deduce that $P$ is the Miquel point of $\ell_A,\ell_B,$ $\ell_C,$ $\overline{O_AO_BO_C}.$ Next we observe that $$\measuredangle PA'B'=\measuredangle PO_BO_A=\measuredangle O_AO_BO=\measuredangle PBO=\measuredangle QAB\implies \varphi (Q)=P.$$Since $\ell \perp OP$ we get $\varphi (\ell)=OP.$ The conclusion follows immediately.

+1 diagram, eepy geo goes hard Here's the solution now. Let $D$ be the intersection of line $PA$ with $(PO_BO_C)$ and define $E$ and $F$ similarly. $\textbf{Claim.}$ $\triangle DEF$ is the aforementioned triangle. $\emph{Proof.}$ We have \[ \providecommand{\dang}{\measuredangle} \dang PDO_B = \dang PO_CO_B = 90^\circ - \dang OPO_C = \dang PCO = 90^\circ - \dang CAP, \]which establishes $O_BD\perp AC$. Similarly, $O_CD \perp AB$, and so on, so we know that $DEF$ is our desired triangle. Note we obtain the relation $\measuredangle PDO_B = \measuredangle PCO$. I will call this $(\ast)$. $\square$ As a corollary, we find $\overline{E-D-O_C}$, $\overline{E-F-O_A}$, and $\overline{D-F-OB}$. $\textbf{Claim.}$ $PDEF$ is cyclic. $\emph{Proof.}$ Follows from angle chasing: \[ \providecommand{\dang}{\measuredangle} \dang PDE = \dang PDO_C = \dang PO_BO_C = \dang PO_BO_A = \dang PFO_A = \dang PFE, \]yay. $\square$ It remains to prove that $OP$ is tangent to $(PDEF)$. But just compute \[ \providecommand{\dang}{\measuredangle} \dang PDF = \dang PDO_B \overset{(\ast)}{=} \dang PCO = \dang OPC = \dang OPD,\]as desired. [asy][asy] unitsize(160); import olympiad; import math; import graph; import cse5; pair A,B,C,P,O,O_A,O_B,O_C,D,E,F; A = dir(110); B = dir(210); C = dir(-30); P = dir(80); O=circumcenter(A,B,C); O_A=circumcenter(P,O,A); O_B=circumcenter(P,O,B); O_C=circumcenter(P,O,C); path PBC,PCA,PAB; PBC = circumcircle(P,O_B,O_C); PCA = circumcircle(P,O_C,O_A); PAB = circumcircle(P,O_A,O_B); pair [] Ainte=intersectionpoints(PBC,L(P,A,0,10)); pair [] Binte=intersectionpoints(PCA,L(P,B,0,1000)); pair [] Cinte=intersectionpoints(PAB,L(C,P,0,1000)); D = Ainte[1]; E = Binte[1]; F = Cinte[1]; currentpen=rgb(0.99,0.92,0.98); fill(PBC); fill(PCA); fill(PAB); currentpen=rgb(1,0.9,0.9); fill(circumcircle(D,E,F)); currentpen=black; draw(A--B--C--cycle); draw(circumcircle(A,B,C)); currentpen = linetype("5 5")+rgb(0.6, 0.6, 0.6); draw(circumcircle(P,O,A)); draw(circumcircle(P,O,B)); draw(circumcircle(P,O,C)); currentpen = linetype("4 4")+orange+linewidth(1); draw(P--D); draw(P--B); draw(C--F); currentpen = black; draw(O_C--O_B); currentpen = black + linewidth(1) + linetype(""); currentpen = red; draw(O_B--F); draw(O_A--F); draw(D--O_C); currentpen=orange; draw(circumcircle(D,E,F)); currentpen=rgb(0.9,0.2,0.8); draw(PBC); draw(PCA); draw(PAB); currentpen=black; dot(A); dot(B); dot(C); dot(O); dot(O_A); dot(O_B); dot(O_C); dot(P); dot(D); dot(E); dot(F); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$O$",O,ESE); label("$P$",P,NE); label("$D$",D,NW); label("$E$",E,dir(120)*2); label("$F$",F,NE); label("$O_A$",O_A,dir(250)*1.5); label("$O_B$",O_B,W); label("$O_C$",O_C,right); [/asy][/asy]

This felt easiest in the shortlist lol

Let $H = OO_C \cap AC$ let $G = OO_A \cap AC$ Claim 1: $O_AO_CGHPE$ is a cyclic hexagon

Claim 2: $B-P-E$

By Claim 2, $PFDE$ is cyclic, which implies $OP$ tangent as required after some angle chase.

Let $DEF$ be the formed triangle. The main claim is the following: Claim. $D, A, P$ are collinear. Proof. Complex numbers. We will let $D =\overline{AP} \cap \ell_C$ and show that the expression for $d$ is symmetric in $b$ and $c$. First, let $f$ be the foot from $O_C$ to $\overline{AB}$. We have the following formulas: \begin{align*} o_C &= \frac{cp(\overline c-\overline p)}{\overline cp-c\overline p} = \frac{cp}{c+p} \\ f &= \frac 12\left(a+b+\frac{cp-ab}{c+p}\right) \end{align*}using complex circumcenter and foot formula. Note that their conjugates are \begin{align*} \overline o_c &= \frac 1{p+c} \\ \overline f &= \frac 12\left(\frac 1a+\frac 1b+\frac 1{p+c}-\frac{cp}{ab(p+c)} \right). \end{align*}To compute $d$, we first set up the following preliminary calculations: \begin{align*} \overline o_c f - o_c \overline f &= \frac 12\left[\frac{(p+c)(a+b)+cp-ab}{(p+c)^2} - \frac{cp}{c+p} \left(\frac 1a+\frac 1b+\frac 1{p+c}-\frac{cp}{ab(p+c)}\right)\right] \\ &= \frac{(cp-ab)(ab-ac-ap-bc-bp+cp)}{ab(c+p)^2} = \frac{S(cp-ab)}{2ab(c+p)^2}. \\ o_c-f &= \frac 12\left(\frac{ab+cp}{c+p}-a-b\right) = \frac S{2(c+p)}. \\ \overline o_c-\overline f &= \frac 1{c+p}-\frac 12\left(\frac 1a+\frac 1b+\frac 1{c+p}-\frac{cp}{ab(p+c)}\right)= \frac S{2ab(c+p)}. \end{align*}Here we denote $S = ab-ac-ap-bc-bp+cp$. Now, the complex intersection formula yields \begin{align*} d &= \frac{\frac{S(a-p)(cp-ab)}{2ab(c+p)^2} - \frac S{2(c+p)} \cdot \left(\frac pa-\frac ap\right)}{\frac {S(a-p)}{2ab(c+p)} - \frac {S\left(\frac 1a-\frac 1p\right)}{2(c+p)}} \\ &= \frac{\frac{(a-p)(cp-ab)}{ab(c+p)^2} - \frac{p^2-a^2}{ap(c+p)}}{\frac{a-p}{ab(c+p)}-\frac{p-a}{ap(c+p)}} \\ &= \frac{\frac{cp-ab}{ab(c+p)^2} +\frac{a+p}{ap(c+p)}}{\frac 1{ab(c+p)} +\frac 1{ap(c+p)}} \\ &= \frac{p(cp-ab)+b(a+p)(c+p)}{p(c+p)+b(c+p)} \\ &= \frac{(b+c)p^2+(a+p)bc}{(c+p)(b+p)}. \end{align*}This is symmetric in $b$ and $c$, as required. $\blacksquare$ From here, the problem succumbs to just angle chasing. First, observe that $\triangle DEF \sim \triangle ABC$ as corresponding sides are perpendicular. Then $\measuredangle EPF = \measuredangle BAC = \measuredangle EDF$, hence $EDFP$ is cyclic. To finish, notice that $\angle DFP = \angle DEB = 90^\circ - \angle ABP = \angle APO$, hence $\overline{OP}$ is tangent to $(DEF)$ at $P$, as required.

Mine is similar. Let $l_c\cap l_b=D,l_c\cap l_a=E,l_b\cap l_a=F$ and $l_a\cap BC=X,l_b\cap AC=Y,l_c\cap AB=Z$. Claim: $A,D,P$ are collinear.

Let $M$ be the miquel point of $BCYZ$. Since $M$ lies on both $(PDE),(PBA)$, we conclude that $M$ is the miquel point of $BPDZ$. Similarily $M$ is the miquel point of $PCYD$. Hence $M\in (PDE),(PDF)$ which implies $M,D,E,F,P$ are concyclic. Let $A'$ be the antipode of $A$ on $(ABC)$. Since $\measuredangle AMD=90=\measuredangle AMA'$ so $M,D,A'$ are collinear. $\measuredangle DMP=\measuredangle CMP-\measuredangle CMA'=\measuredangle CAP-\measuredangle CAA'=\measuredangle OAP=\measuredangle DPO$ as desired.$\blacksquare$

Move $P$ with degree $2$ on $\Omega$. Now $AP$ has degree $1$, so its pole does, so its midpoint with $O$, which is $O_A$, does as well. Similarly $O_B$ and $O_C$ also have degree $1$. This then implies $D,E,F$ each have degree $2$. Consider the three cases when $P$ is a vertex of the circumtangential triangle. Let $\ell$ be the tangent to $\Omega$ at $P$. Then $\measuredangle PAO=\tfrac12\measuredangle PO_AO=\measuredangle(O_AP,\ell)$ implies $AO,PO_A$ isogonal in $\measuredangle(AP,\ell)$. Then they are also isogonal in $\measuredangle BAC$, implying $PO_A\perp BC$. Similarly $PO_B\perp AC,PO_C\perp AB$ so $D=E=F=P$. By Zack's lemma, lines $PD,PE,PF$ have degree $4-3=1$ so they pass through fixed points. When $P=A$ clearly $A\in PD$ and when $P=2O-A$ then $D$ is the orthocenter of $\triangle OO_BO_C$ so $A\in PD$ again. Thus $A,P,D$ are always collinear. Similarly $B,P,E$ and $C,P,F$ are collinear. Since $\triangle ABC$ and $\triangle DEF$ are paralogic, it follows their circumcircles are orthogonal, implying the result.

Let $\triangle XYZ$ be the triangle formed by $\ell_A$, $\ell_B$, $\ell_C$. We will prove that $A,X,P$ are collinear, and then show that $(XYZ)$ is tangent to $OP$ at $P$. Let us complex bash, and use lowercase to denote complex numbers. Let $(ABC)$ be the unit circle, and rotate the diagram such that $p=1$. Using the circumcentre formula, we derive that \[o_B=\frac{b(\overline{b}-1)}{\overline{b}-b}=\frac{1-b}{\frac{1}{b}-b}=\frac{b}{b+1}\]Similarly, $o_C=\frac{c}{c+1}$. By perpendicularity, $X$, $O_B$ have the same foot onto $AC$, and $X$, $O_C$ have the same foot onto $AB$. Hence, \[\frac{1}{2}(a+b+x-\overline{x}ab)=\frac{1}{2}\left(a+b+\frac{c}{c+1}-\frac{ab}{c+1}\right)\]noting $\frac{\frac{1}{c}}{\frac{1}{c}+1}=\frac{1}{c+1}$. Hence, $x-\overline{x}ab=\frac{c-ab}{c+1}$. Analogously, we also have $x-\overline{x}ac=\frac{b-ac}{b+1}$. Hence, \[\overline{x}(ac-ab)=\frac{c-ab}{c+1}-\frac{b-ac}{b+1}=\frac{a(c^2-b^2)+a(c-b)+(c-b)}{(b+1)(c+1)}\]Thus, \[\overline{x}=\frac{c+b+\frac{1}{a}+1}{(b+1)(c+1)}\]Taking the conjugate, \[x=\frac{\frac{1}{c}+\frac{1}{b}+a+1}{\left(\frac{1}{b}+1\right)\left(\frac{1}{c}+1\right)}=\frac{b+c+abc+bc}{(b+1)(c+1)}\]However, we need to verify $x$ actually satisfies one of the equations (the other follows by symmetry). \[x-\overline{x}ab=\frac{b+c+abc+bc-abc-ab^2-ab-b}{(b+1)(c+1)}=\frac{(b+1)(c-ab)}{(b+1)(c+1)}=\frac{c-ab}{c+1}\]As desired. Now, let's check that $A$, $X$, $P$ are collinear. Indeed, we just need $a+1=a\overline{x}+x$ by chord formula. \[\frac{a(b+c)+1+a+b+c+abc+bc}{(b+1)(c+1)}=\frac{(a+1)(b+1)(c+1)}{(b+1)(c+1)}=a+1\]As desired. Analogously, $C,Z,P$ and $B,Y,P$ are collinear. Now, to finish, \[\measuredangle PZX=90^\circ-\measuredangle ACP=\measuredangle OPA=\measuredangle OPX\]From our collinearities. Thus, $(ZPX)$ is tangent to $OP$. Analogously, $(XPY)$ is also tangent to $OP$, hence $(XYZP)$ is tangent to $OP$.