i am overjoyed that this did not make the IMO.

Let $P = (x:y:z)$ with $x,y,z > 0$ so $A_1 = (0:y:z)$. Therefore, \begin{align*} A_2 &= \left( -x(y+z) : 2y(x+y+z) - (y+z)y : 2(x+y+z)-(y+z)z \right) \\ &= \left( -x(y+z) : y(2x+y+z) : z(2x+y+z) \right). \end{align*}Claim: The point $A_2$ lies inside $(ABC)$ if and only if \[ 0 < a^2 \cdot yz(2x+y+z) - b^2 \cdot xz(y+z) - c^2xy(y+z). \]Proof. The condition we want is that \begin{align*} 0 &< -\operatorname{Pow}(A_2, (ABC)) \\ &= a^2 \cdot y(2x+y+z) \cdot z(2x+y+z) \\ &\qquad - b^2x(y+z) \cdot z(2x+y+z) - c^2x(y+z) \cdot y(2x+y+z) \\ &= (2x+y+z) \left[ a^2 \cdot yz(2x+y+z) - b^2 \cdot xz(y+z) - c^2 \cdot xy(y+z) \right]. \end{align*}From $2x+y+z>0$ we conclude. $\blacksquare$ If all three points lie inside the circle, we sum cyclically now to get a contradiction.

this is disgusting

Comment: This actually has some nice synthetic solutions, but sadly is too easy to bary bash. The first solution is an elementary solution that I found after the exam, while the second one was actually the way I solved it. Solution 1 (Elementary): The idea is the following claim. Claim: Suppose that $A_2$ lie on $\odot(ABC)$. Then the isogonal conjugate $Q$ of $P$ w.r.t. $\triangle ABC$ lie on $\odot(AO)$, where $O$ is the circumcenter of $\triangle ABC$. Proof 1 (Similar Triangles). Extend $AQ$ to meet $\odot(ABC)$ at $X$. Notice that $\triangle BA_2A_1\sim\triangle AXB$. Therefore if $K$ is the reflection of $A_2$ across $B$, we get $KP\parallel BA_1$ so $\triangle KA_2P\sim\triangle AXB$. However, $$\measuredangle QBA = \measuredangle CBP = \measuredangle KBP\implies \triangle KA_2P\cup B\sim\triangle AXB\cup Q$$hence $Q$ is the midpoint of $AX$, done. $\blacksquare$ Proof 2 (Harmonic). Again, let $X=AQ\cap\odot(ABC)$. Notice that lines $\{BX,B{\infty}_{AP}\}$ and $\{BA_2, B{\infty}_{AQ}\}$ are isogonal w.r.t. $\angle ABC$. Thus by reflecting across angle bisector, $$B(P,A_2;A_1,{\infty}_{AP}) = B(Q,{\infty}_{AQ};A,X)$$thus $Q$ is the midpoint of $AX$, done. $\blacksquare$ Back to the main problem. Let $Q$ be the isogonal conjugate of $P$ w.r.t. $\triangle ABC$. Suppose that $A_2,B_2,C_2$ all lie inside $\odot(ABC)$. Let $A_2'=AP\cap\odot(ABC)$ and $P'$ be the reflection of $P$ across $A_1$. Observe that $\{P,A_2,A_2'\}$ are colinear in this order so $\{A,P',P\}$ are colinear in this order. Thus, if $Q'$ is the isogonal conjugate of $P'$ w.r.t. $\triangle ABC$, then $\{A,Q,Q'\}$ are colinear in this order. But $Q'\in\odot(AO)$ so $Q$ lies strictly inside $\odot(AO)$. Analogously, $Q$ lies strictly inside $\odot(BO),\odot(CO)$, which is a contradiction. Remark: The lemma implies that the locus of $P$ which $A_2\in\odot(ABC)$ is a cubic curve. Solution 2 (Nine-point conic) First, we prove the nine points conic (which is the generalization of nine-points circle) as follows. Lemma: [Nine-points conic] The following nine points: points $A_1,B_1,C_1$; midpoints $M_A,M_B,M_C$ of $BC,CA,AB$; and midpoints $A',B',C'$ of $AP,BP,CP$ lie on a conic. Proof. Note parallel lines $M_AB'\parallel CP\parallel M_BA'$, $M_AC'\parallel M_CA'$, $M_AM_B\parallel AB\parallel A'B'$ and $M_AM_C\parallel A'C'$ so translating pencils gives $$M_A(B',C';M_B,M_C) = A'(M_B,M_C;B',C') = A'(B',C';M_B,M_C)$$so $A',B',C',M_A,M_B,M_C$ lie on a conic. Finally by converse of Pascal's theorem on hexagon $M_AA_1A'C'B'M_C$ gives $A_1$ lie on this conic. Similarly $B_1,C_1$ lie on this conic. $\blacksquare$ Remark: The_Turtle pointed out above that this can be proven easily with affine transformation. By taking homothety $\mathcal{H}(P,2)$, we get $A,B,C,A_2,B_2,C_2$ lie on a conic. Hence isogonal conjugate $A_2'$,$B_2'$,$C_2'$ of $A_2$,$B_2$,$C_2$ w.r.t. $\triangle ABC$ are colinear. However, if $A_2$ lie inside $\odot(ABC)$, one can check that $A_2'$ lies inside region determined by extension of $BA$ beyond $A$, and extension of $CA$ beyond $C$ that does not contain $\triangle ABC$. One can narrow positions of $B_2',C_2'$ similarly. This is enough to force that $A_2',B_2',C_2'$ are not colinear, contradiction.

I have solution using moving points if someone can prove that if P satysfies that both $B_2,C_2$ lie on circumcircle then $P$ is orthocenter.

Ok this was ugly .

I was just wondering if we can derive a contradiction from this: If we prove that for whichever triangle that lies inside the circle (or on the boundaries) its perimeter is less than the double of the perimeter of the orthic triangle, we have a contradiction from Fagnano's problem. I didnt check it though.

v_Enhance wrote: Let $P = (x:y:z)$ with $x,y,z > 0$ so $A_1 = (0:y:z)$. Therefore, \begin{align*} A_2 &= \left( -x(y+z) : 2y(x+y+z) - (y+z)y : 2(x+y+z)-(y+z)z \right) \\ &= \left( -x(y+z) : y(2x+y+z) : z(2x+y+z) \right). \end{align*}Claim: The point $A_2$ lies inside $(ABC)$ if and only if \[ 0 < a^2 \cdot yz(2x+y+z) - b^2 \cdot xz(y+z) - c^2xy(y+z). \]Proof. The condition we want is that \begin{align*} 0 &< -\operatorname{Pow}(A_2, (ABC)) \\ &= a^2 \cdot y(2x+y+z) \cdot z(2x+y+z) \\ &\qquad - b^2x(y+z) \cdot z(2x+y+z) - c^2x(y+z) \cdot y(2x+y+z) \\ &= (2x+y+z) \left[ a^2 \cdot yz(2x+y+z) - b^2 \cdot xz(y+z) - c^2 \cdot xy(y+z) \right]. \end{align*}From $2x+y+z>0$ we conclude. $\blacksquare$ If all three points lie inside the circle, we sum cyclically now to get a contradiction. Sir, what you have used? I don't even understand that. Do you use co-ordinate?

Consider the circumcevian triangle $A_3B_3C_3$ the problem states that not all three inequalities $A_1A_3>PA_1$ etc. can hold simultaneously. Label the angles $\angle CAP=a_1$,$\angle PAB=a_2$,$\angle PBA=b_1$,$\angle CBP=b_2$ ,$\angle PCB=c_1$, $\angle ACP=c_2$ cyclically. We have that $A_3BC=a_1$ and $A_3CB=a_2$ . The following holds : $\frac{A_1A_3}{PA_1}=\frac{\frac{BC}{dist(P,BC)}}{\frac{BC}{dist(A_3,BC)}}=\frac{\cot{ b_2}+\cot{ c_1}}{\cot{a_1}+\cot{a_2}}>1$ Summing all three inequalities $ \cot{ b_2}+\cot{ c_1}>\cot{a_1}+\cot{a_2}$ we get a contradiction.

Inside $\Delta ABC$, let a point be $X$, let $Y=AX\cap (ABC)$ and $Z$ is midpoint of $XY$. Let, isogonal conjugate of $X$ be $X'$ and of $Z$ be $Z'$. Let $Y'=AX'\cap (ABC)$. Now, we claim that Midpoint of $Y'Z'$ is $X'$. Proof: Let the point at infinity along $AX$ be $\infty_{AX}$ and along $AX'$ be $\infty_{AX'} $. Then, $-1=(XY;Z\infty_{AX})=(X'\infty_{AX'};Z'Y')$. This comes by isogonal conjugation. So, we get the corollary that if the reflection of a point across $BC$ lies inside $(ABC)$ then the reflection of $A$ across its isogonal conjugate must lie inside $(ABC)$ as well. Thus, its isogonal conjugate must lie inside circle with diameter $AO$. But, no point $P$ can lie inside all 3 circles $(AO),(BO),(CO)$ and $\Delta ABC$ so we are done. Note: I may add diagram later, too lazy.

There is/was actually a subtlety missed in the synthetic solutions in post #6 and post #13. It is actually possible for a point $Q$ to lie inside all $3$ of the circles $\odot(AO), \odot(BO)$ and $\odot(CO)$! [asy][asy] pair A = dir(110), B = dir(152), C = dir(28); pair O = (0,0); draw(A--B--C--cycle, royalblue); draw(circumcircle(A,B,C),springgreen); draw(circle((A+O)/2,0.5), fuchsia); draw(circle((C+O)/2,0.5), fuchsia); draw(circle((B+O)/2,0.5), fuchsia); pair Q = (0.02, 0.2); dot(Q, 3+orange); label("$\underline{\underline{Q}}$",Q,dir(0),orange); dot("$A$",A,dir(A),3+purple); dot("$C$",C,dir(C),3+purple); dot("$B$",B,dir(B),3+purple); dot("$O$",O,dir(270),4+magenta); [/asy][/asy] The condition that $P$ lies inside triangle $ABC$ is actually necessary. The finish at the end, requires that no point $Q$ can lie inside $\odot(AO), \odot(BO), \odot(CO)$ and $\triangle ABC$. This could be shown via inversion around $\odot(ABC)$ or just by taking cases based on whether the triangle is a cute triangle.

My hand was forced. Dam‍n this problem. Let $AP \cap (ABC) = X$, $BP \cap (ABC) = Y$, $CP \cap (ABC) = Z$. Suppose otherwise FTSOC; we may assume that\[\left\{\frac{PA_1}{XA_1}, \frac{PB_1}{YB_1}, \frac{PC_1}{ZC_1}\right\}\]are all $< 1$. Notice that we can write $\tfrac{PA_1}{XA_1}$ in two ways; using ratio lemma on $\triangle PBX$ and $\triangle PCX$:\[\frac{PA_1}{XA_1} = \frac{PB\sin{\angle PBC}}{XB\sin{\angle XAC}} = \frac{\sin{\angle C}\sin{\angle PBC}}{\sin{\angle APB}\sin{\angle PAC}} \]\[\frac{PA_1}{XA_1} = \frac{PC\sin{\angle PCB}}{XC\sin{\angle XAB}} = \frac{\sin{\angle B}\sin{\angle PCB}}{\sin{\angle APC}\sin{\angle PAB}} \]and similarly for the other two\[\frac{PB_1}{YB_1} = \frac{\sin{\angle A} \sin{\angle PCA}}{\sin{\angle BPC}\sin{\angle PBA}}\]\[\frac{PB_1}{YB_1} = \frac{\sin{\angle C} \sin{\angle PAC}}{\sin{\angle BPA}\sin{\angle PBC}}\]and\[\frac{PC_1}{ZC_1} = \frac{\sin{\angle B}\sin{\angle PAB}}{\sin{\angle BPA} \sin{\angle PCB}}\]\[\frac{PC_1}{ZC_1} = \frac{\sin{\angle A}\sin{\angle PBA}}{\sin{\angle APB} \sin{\angle PCA}}\]Notice that multiplying $(1), (4)$, then $(2), (5)$, and then $(3), (6)$, stuff nicely cancels out to yield\[\left(\frac{\sin{\angle C}}{\sin{\angle APB}}\right)^2, \left(\frac{\sin{\angle A}}{\sin{\angle BPC}}\right)^2, \left(\frac{\sin{\angle B}}{\sin{\angle CPA}}\right)^2 < 1\]which tells us that $\sin{\angle C} < \sin{\angle APB}$, $\sin{\angle A} < \sin{\angle BPC},$ and $\sin{\angle B} < \sin{\angle CPA}$. Since $P$ is inside the interior of $\triangle ABC$, we know that in fact $\angle C < \angle APB$, $\angle A < \angle BPC$, and $\angle B < \angle CPA$. Now, using sine subtraction formula with all three, we get\[\cos{(\tfrac{\angle APB + \angle C}{2})} \sin{(\tfrac{\angle APB - \angle C}{2})} > 0\]\[\cos{(\tfrac{\angle BPC + \angle A}{2})} \sin{(\tfrac{\angle BPC - \angle A}{2})} > 0\]\[\cos{(\tfrac{\angle CPA + \angle B}{2})} \sin{(\tfrac{\angle CPA - \angle B}{2})} > 0\]from which it follows that $\cos{(\tfrac{\angle APB + \angle C}{2})}, \cos{(\tfrac{\angle BPC + \angle A}{2})}, \cos{(\tfrac{\angle CPA + \angle B}{2})} > 0$. Since each of\[\angle APB + \angle C, \angle BPC + \angle A, \angle CPA + \angle B\]is bounded between $[0, 2\pi)$, it follows that each of them is $< \pi$ and summing yields $3\pi < 3\pi$, a contradiction, as desired. $\blacksquare$

Solved with nukelauncher. Let \(\Gamma\) be the circumcircle. Consider an affine transformation that sends \(ABCP\) to an orthocentric system; then the images of \(A\), \(B\), \(C\), \(A_2\), \(B_2\), \(C_2\) lie on a circle. Undoing the affine transformation, points \(A\), \(B\), \(C\), \(A_2\), \(B_2\), \(C_2\) lie on an ellipse \(\mathcal E\). [asy][asy] import olympiad; import geometry; pen pri=blue; pen sec=purple+pink; pen tri=lightblue; pen qua=lightred; pen fil=invisible; pen tfil=invisible; pen qfil=invisible; size(7cm); defaultpen(fontsize(10pt)); pair A,B,C,P,A1,B1,C1,A2,B2,C2,D; A=dir(110); B=dir(210); C=dir(330); P=0.3*dir(250); A1=extension(A,P,B,C); B1=extension(B,P,C,A); C1=extension(C,P,A,B); A2=2A1-P; B2=2B1-P; C2=2C1-P; path e=(path)conic(A,B,C,A2,B2); for(int i=0;i<4;i+=1){ D=intersectionpoints(unitcircle,e)[i]; if(D==A||D==B||D==C)continue; break; } filldraw(e,qfil,qua+dashed); draw(A--A2,sec); draw(B--B2,sec); draw(C--C2,sec); filldraw(unitcircle,tfil,tri); filldraw(A--B--C--cycle,fil,pri); dot("\(A\)",A,N); dot("\(B\)",B,SW); dot("\(C\)",C,SE); dot("\(D\)",D,NE); dot("\(P\)",P,dir(60)); dot("\(A_1\)",A1,SW); dot("\(B_1\)",B1,dir(75)); dot("\(C_1\)",C1,dir(120)); dot("\(A_2\)",A2,S); dot("\(B_2\)",B2,E); dot("\(C_2\)",C2,W); [/asy][/asy] Without loss of generality let \(A\), \(B\), \(C\) lie on \(\Gamma\) in counterclockwise order. Evidently \(A_2\) lies on counterclockwise arc \(BC\) of \(\mathcal E\), and similarly for \(B_2\) and \(C_2\). If \(\mathcal E=\Gamma\), then we are done. Otherwise \(\mathcal E\) and \(\Gamma\) intersect at one more point \(D\). Assume without loss of generality that \(D\) lies on counterclockwise arc \(CA\). Then either counterclockwise arc \(AB\) or \(BC\) of \(\mathcal E\) lies completely outside \(\Gamma\). Without loss of generality, arc \(AB\) lies outside; then \(C_2\) lies outside \(\Gamma\), as desired.

Nice problem !

Another synthetic solution: Assume for the sake of contradiction that points $A_1,B_1,C_1$ all lie strictly inside the circumcircle of $\triangle ABC$. Claim 1: If $\angle AA_1B \le 90^\circ$, then $\angle BAP > \angle BCP$ and other analogous results. proof: Suppose $\angle AA_1B \le 90^\circ$. Let line $AP$ meet $\odot(ABC)$ again at $A_3$. Then by our assumption, $PA_1 < PA_3$. Let $A_3'$ be the reflection of $A_3$ in line $BC$. Then $\angle BAP = \angle BAA_3 = \angle BCA_3 = \angle BCA_3'$. So it suffices to show $\angle BCA_3' > \angle BCP$. Let the line through $A_3'$ parallel to $\overline{BC}$ cut line $AP$ at $P'$. Then by midpoint theorem $A_1P' = A_1A3 > A_1P$, so $P$ strictly lies between $A_1,P'$ so $\angle BCP' > \angle BCP$. Also, and $\angle AA_1B \le 90^\circ$, so because of the configuration we must have that $\angle BCA_3' \ge \angle BCP'$. Hence $\angle BCA_3' > \angle BCP$. This proves the claim. $\square$ [asy][asy] pair A=dir(130),B=dir(-150),C=dir(-30),A3=dir(-85),A1=extension(A,A3,B,C),P=1.6*A1-0.6*A3,A3p=2*foot(A3,B,C)-A3,Pp=2*A1-A3,A0=foot(A3,B,C); draw(unitcircle,brown); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$A_1$",A1,dir(A1)); dot("$A_3$",A3,dir(-40)); dot("$P$",P,dir(P)); dot("$P'$",Pp,dir(160)); dot("$A_3'$",A3p,dir(-50)); dot("$A_0$",A0,dir(-45)); draw(B--A--C,magenta); draw(B--C^^Pp--A3p,red); draw(A--A3^^A3--A3p,green); [/asy][/asy] Corollary: If $\angle PA_1B \le 90^\circ$ then $\angle PC_1B > 90^\circ$. proof: Easy proof by contradiction. $\square$ Now assume WLOG that $\angle AA_1B \le 90^\circ$. Then using our Corollary we obtain \begin{align*} & \angle PC_1B > 90^\circ ~ \implies ~ \angle PC_1A < 90^\circ ~ \implies ~ \angle PB_1A > 90^\circ ~ \\ & \implies ~ \angle PB_1C < 90^\circ ~ \implies \angle PA_1C > 90^\circ ~ \implies \angle PA_1B < 90^\circ \end{align*}In particular, we have that $$\angle PA_1B , \angle PC_1A , \angle PB_1C < 90^\circ ~,~ \angle PA_1C , \angle PC_1B, \angle PB_1A > 90^\circ $$Now let $H$ be the orthocenter of $\triangle ABC$ and $D,E,F$ as usual. The conditions we obtain imply that $P$ lies strictly inside each of the following triangles $$\triangle ADC ~,~ \triangle BEA ~,~ \triangle CFB$$But it is easy to see that these three triangle do not have a common interior point. So we have a contradiction! This completes the proof of the problem. $\blacksquare$

Interesting Let $AP,BP,CP$ meet $ABC$ at $Y,Z,X$. Note that problem actually means at least one of $\frac{PA_1}{A_1Y},\frac{PB_1}{B_1Z},\frac{PC_1}{C_1X} \ge 1$ so Assume not. Note that $\frac{PA_1}{A_1Y} = \frac{\frac{PA_1}{A_1B}}{\frac{A_1Y}{A_1B}} = \frac{\frac{PA_1}{A_1C}}{\frac{A_1Y}{A_1C}}$ so $\frac{\frac{\sin{PBC}}{\sin{BPA}}}{\frac{\sin{PAC}}{\sin{BCA}}} = \frac{\frac{\sin{PCB}}{\sin{CPA}}}{\frac{\sin{PAB}}{\sin{CBA}}}$. we do same approach for $\frac{PB_1}{B_1Z},\frac{PC_1}{C_1X}$. Now we have $(\frac{\sin BAC}{\sin BPC})^2 = \frac{PB_1}{B_1Z}\frac{PC_1}{C_1X} < 1$. we have same approach for $(\frac{\sin BCA}{\sin BPA})^2$ and $(\frac{\sin ABC}{\sin APC})^2$ but then we should have $\angle BAC < \angle 180 - \angle BPC$ and $\angle BCA < \angle 180 - \angle BPA$ and $\angle ABC < \angle 180 - \angle APC$ which is obviously not true so contradiction so at least one of $\frac{PA_1}{A_1Y},\frac{PB_1}{B_1Z},\frac{PC_1}{C_1X} \ge 1$ so at least one of $A_2,B_2,C_2$ lies outside or on $ABC$ itself. we're Done.

We use barycentric coordinates with respect to $ABC$. Let $P=(p:q:r)$. We have that $A_1=(0:\frac{p+q+r}{q+r} \cdot q : \frac{p+q+r}{q+r})$. We have that $A_2=(-p:q \cdot \frac{2p+q+r}{q+r}: r \cdot \frac{2p+q+r}{q+r})$. Note that $A_2$ lying inside of the circumcircle of $ABC$ is equivalent to $$0 > \text{Pow}((ABC),A_2) \iff a^2qr\left( \frac{2p+q+r}{q+r} \right) ^2-b^2pr \cdot \frac{2p+q+r}{q+r}-c^2pq \cdot \frac{2p+q+r}{q+r} > 0 \iff a^2qr(2p+q+r)>b^2pr(q+r)+c^2pq(q+r).$$If all the points lie inside the circumcircle of $ABC$, we get a contradiction by summing cyclically we get a contradiction.

There exist an affine transformation which maps $A,B,C,P$ onto orthocentric quadruple, and so maps $$\left \{ A,B,C,A_2,B_2,C_2 \right \} \quad (\star )$$onto set of concyclic points. This yields that points from $(\star )$ lie on one ellipse $\varepsilon.$ Case $\varepsilon =\odot (ABC)$ is obvious, so assume the opposite. Note that $A,C_2,B,A_2,C,A_2$ lie on $\varepsilon$ in that order; WLOG arc $AC_2$ of ellipse intersects $\odot (ABC)$ again at $D.$ One of arc $AC,BC$ of $\varepsilon$ lies outside $\odot (ABC),$ hence either $B_2$ or $A_2$ can't lie strictly inside the circumcircle, as required.

Let $A_3$, $B_3$, $C_3$ be the intersections of $AP$, $BP$, $CP$ on $(ABC)$. Let $\theta_1=\angle ABB_3$, $\theta_2=\angle CBB_3$, $\theta_3=\angle CAA_3$, $\theta_4=\angle BAA_3$, $\theta_5=\angle BCC_3$, and $\theta_6=\angle ACC_3$. We have \begin{align*} \frac{C_1C_3}{PC_1} &= \frac{\sin(\theta_3+\theta_6)\sin(\theta_5)}{\sin(\theta_1+\theta_2)\sin(\theta_4)} = \frac{\sin(\theta_2+\theta_5)\sin(\theta_6)}{\sin(\theta_3+\theta_4)\sin(\theta_1)}\\ \frac{B_1B_3}{PB_1} &= \frac{\sin(\theta_2+\theta_5)\sin(\theta_1)}{\sin(\theta_3+\theta_4)\sin(\theta_6)} = \frac{\sin(\theta_1+\theta_4)\sin(\theta_2)}{\sin(\theta_5+\theta_6)\sin(\theta_3)}\\ \frac{A_1A_3}{PA_1} &= \frac{\sin(\theta_1+\theta_4)\sin(\theta_3)}{\sin(\theta_5+\theta_6)\sin(\theta_2)} = \frac{\sin(\theta_3+\theta_6)\sin(\theta_4)}{\sin(\theta_1+\theta_2)\sin(\theta_5)} \end{align*}Suppose all those are less than one, then the following are also less than one: \begin{align*} \sqrt{\frac{C_1C_3}{PC_1} \cdot \frac{B_1B_3}{PB_1}} &= \sqrt{\frac{\sin(\theta_2+\theta_5)\sin(\theta_6)}{\sin(\theta_3+\theta_4)\sin(\theta_1)}\cdot \frac{\sin(\theta_2+\theta_5)\sin(\theta_1)}{\sin(\theta_3+\theta_4)\sin(\theta_6)}} \\ &= \frac{\sin(\theta_2+\theta_5)}{\sin(\theta_3+\theta_4)} \\ \sqrt{\frac{B_1B_3}{PB_1} \cdot \frac{A_1A_3}{PA_1}} &= \sqrt{\frac{\sin(\theta_1+\theta_4)\sin(\theta_2)}{\sin(\theta_5+\theta_6)\sin(\theta_3)}\cdot \frac{\sin(\theta_1+\theta_4)\sin(\theta_3)}{\sin(\theta_5+\theta_6)\sin(\theta_2)}} \\ &= \frac{\sin(\theta_1+\theta_4)}{\sin(\theta_5+\theta_6)} \\ \sqrt{\frac{A_1A_3}{PA_1} \cdot \frac{C_1C_3}{PC_1}} &= \sqrt{\frac{\sin(\theta_3+\theta_6)\sin(\theta_4)}{\sin(\theta_1+\theta_2)\sin(\theta_5)}\cdot \frac{\sin(\theta_3+\theta_6)\sin(\theta_5)}{\sin(\theta_1+\theta_2)\sin(\theta_4)}} \\ &= \frac{\sin(\theta_3+\theta_6)}{\sin(\theta_1+\theta_2)} \end{align*}Therefore, \begin{align*} \theta_2+\theta_5&<\theta_3+\theta_4 \\ \theta_1+\theta_4&<\theta_5+\theta_6 \\ \theta_3+\theta_6&<\theta_1+\theta_2 \end{align*}Which is absurd.

And I have one comment: ew.

Thanks to pi_is_3.14 for quoting affine transformations numerous times at HMMT (which led me to reading about it!). The key idea is that there exists an affine transformation $\varphi$ mapping $A, B, C, P$ to an orthocentric system; note that $\varphi$ also maps $\{ A, B, C, A_2, B_2, C_2 \}$ to a set of concyclic points by reflecting the orthocenter. Taking $\varphi^{-1}$ of the transformed image yields that $A, B, C, A_2, B_2, C_2$ all lie on an ellipse. Case 1: $\{A_2, B_2, C_2\} \subseteq (ABC)$ This case trivially works. Case 2: $\{A_2, B_2, C_2\} \not\subseteq (ABC)$ Note that the ellipse containing $A, B, C, A_2, B_2, C_2$ and $(ABC)$ intersect at one more point, say $D$. WLOG suppose that $D$ lies on the arc $CA$ not containing point $B$. Then one of arcs $AC$ and $BC$ of the ellipse lies outside of $(ABC)$, so either $B_2$ or $A_2$ do not lie strictly inside $(ABC)$, and we are done.

We use barycentric coordinates with $A=(1,0,0)$, $B=(0,1,0)$, and $C=(0,0,1)$. Let $P=:(p,q,r)$ so \begin{align*} A_1&=\left(0,\frac{q}{1-p},\frac{r}{1-p}\right)\\ B_1&=\left(\frac{p}{1-q},0,\frac{r}{1-q}\right)\\ B_1&=\left(\frac{p}{1-r},\frac{q}{1-r},0\right). \end{align*}Then \begin{align*} A_2&=2A_1-P&=\left(-p,\frac{pq+q}{1-p},\frac{pr+r}{1-p}\right)\\ B_2&=2B_1-P&=\left(\frac{qp+p}{1-p},-q,\frac{qr+r}{1-p}\right)\\ C_2&=2C_1-P&=\left(\frac{rp+p}{1-p},\frac{rq+q}{1-p},-r\right) \end{align*}so \begin{align*} \mathrm{Pow}_{(ABC)}(A_2)&=-a^2\cdot\frac{pq+q}{1-p}\cdot\frac{pr+r}{1-p}+b^2p\cdot\frac{pr+r}{1-p}+c^2p\cdot\frac{pq+q}{1-p}\\ &=-\frac{p+1}{(1-p)^2}\left(a^2qr(p+1)+b^2rp(p-1)+c^2pq(p-1)\right). \end{align*}Similarly, \begin{align*} \mathrm{Pow}_{(ABC)}(B_2)&=-\frac{q+1}{(1-q)^2}\left(a^2qr(q-1)+b^2rp(q+1)+c^2pq(q-1)\right)\\ \mathrm{Pow}_{(ABC)}(C_2)&=-\frac{r+1}{(1-r)^2}\left(a^2qr(r-1)+b^2rp(r-1)+c^2pq(r+1)\right) \end{align*}so \[ \frac{(1-p)^2}{p+1}\mathrm{Pow}_{(ABC)}(A_2)+\frac{(1-q)^2}{q+1}\mathrm{Pow}_{(ABC)}(B_2)+\frac{(1-r)^2}{r+1}\mathrm{Pow}_{(ABC)}(C_2)=0 \](note that $p+q+r-1=0$). Since $p,q,r>0$ and thus $\frac{(1-p)^2}{p+1},\frac{(1-q)^2}{q+1},\frac{(1-r)^2}{r+1}>0$, it follows that at least one of $\mathrm{Pow}_{(ABC)}(A_2),\mathrm{Pow}_{(ABC)}(B_2),\mathrm{Pow}_{(ABC)}(C_2)$ must be positive. Thus at least one of $A_2,B_2,C_2$ is not inside $(ABC)$. $\square$

My ugly solution for acute-angled triangle FTSOC, suppose $A_2,B_2,C_2$ all lie strictly inside, then $BA_1*A_1C>A_1P*A_1A$ and its cyclic versions. Call $F_A=\frac{BA_1*A_1C}{A_1P*A_1A}$ By writing Law of Sines for $BC_1P, PB_1C, ABB_1,AC_1C$ we get $\frac{F_B}{F_C}=(\frac{sin\angle ABB_1}{sin\angle ACC_1})^2$ Now WLOG suppose $F_C$ is the smallest among the three. Then from the condition above $\angle ABP \geq \angle PCA$ and $\angle BAP \geq \angle PCB$. Let $M$ be the intersection of $CP$ with circumcircle of $ABC$, Then $C_1P<C_1M$, $\angle PAC_1\geq \angle MAC_1$, $\angle PBC_1 \geq \angle C_1BM$, from which $AP<AM$ and $BP<BM$. But then $\angle AC_1P=\angle AMC_1+\angle MAC_1<\angle C_1AP+\angle C_1PA=\angle BC_1P$ and similarly $\angle BC_1P<\angle AC_1P$, a contradiction

blame Angelo Di Pasquale and Andrew Elvey Price for this problem