In complex numbers with $ABC$ the unit circle, it is equivalent to solving the following two cubic equations in $p$ and $q = \overline p$: \begin{align*} (p-a)(p-b)(p-c) &= (abc)^2 (q -1/a)(q - 1/b)(q - 1/c) \\ 0 &= \prod_{\text{cyc}} (p+c-b-bcq) + \prod_{\text{cyc}} (p+b-c-bcq). \end{align*}Viewing this as two cubic curves in $(p,q) \in \mathbb C^2$, by Bezout's Theorem it follows there are at most nine solutions (unless both curves are not irreducible, but it's easy to check the first one cannot be factored). Moreover it is easy to name nine solutions (for $ABC$ scalene): the three vertices, the three excenters, and $I$, $O$, $H$. Hence the answer is just those three triangle centers $I$, $O$ and $H$. On the other hand it is not easy to solve the cubics by hand; I tried for an hour without success. So I think this solution is only feasible with knowledge of algebraic geometry.

What an EXCELLENT problem.

EulerMacaroni wrote:

v_Enhance wrote: In complex numbers with $ABC$ the unit circle, it is equivalent to solving the following two cubic equations in $p$ and $q = \overline p$: \begin{align*} (p-a)(p-b)(p-c) &= (abc)^2 (q -1/a)(q - 1/b)(q - 1/c) \\ 0 &= \prod_{\text{cyc}} (p+c-b-bcq) + \prod_{\text{cyc}} (p+b-c-bcq). \end{align*}Viewing this as two cubic curves in $(p,q) \in \mathbb C^2$, by Bezout's Theorem it follows there are at most nine solutions (unless both curves are not irreducible, but it's easy to check the first one cannot be factored). Moreover it is easy to name nine solutions (for $ABC$ scalene): the three vertices, the three excenters, and $I$, $O$, $H$. Hence the answer is just those three triangle centers $I$, $O$ and $H$. On the other hand it is not easy to solve the cubics by hand; I tried for an hour without success. So I think this solution is only feasible with knowledge of algebraic geometry. Wouldn't this be possible by just plugging I, O, H into the two cubics?

You can also just note that I, O, and H satisfy the conditions geometrically (which is what v_Enhance did), which is way easier than plugging them into the messy algebraic expression.

I think what v_Enhance is saying by "solving the cubics by hand" it's easy to verify that the nine points are solutions (geometrically, as bobthesmartypants points out), but the trickiness is in showing that these nine solutions are the only nine solutions. In theory one could try to manipulate the two cubics into a degree $9$ equation to factor into the nine solutions, which is what v_Enhance is saying is difficult. Instead, Bezout's theorem tells us directly that these are the only $9$.

Hmmm this may be just me but why does USA put weird questions on their TST that could be pwned by higher math stuff?

bobthesmartypants wrote: You can also just note that I, O, and H satisfy the conditions geometrically (which is what v_Enhance did), which is way easier than plugging them into the messy algebraic expression. Some motivation for $H,I,O$ is this:- Both the Darboux Cubic and the McCay cubic are isopivotal cubics and the pivots lie on the Euler line. Therefore, since self-isoconjugation preserves them, we can see that $H,O$ lie on them. Also, $I,I_a,I_b,I_c$ are invariant under Iso-conjugation and since $XX$ can be parallel to any direction, we get the cubics to pass through them and $A,B,C$ are known to lie on such cubics. @pi37 yes, in order to prove this we need to do it as v_Enhance did, though that might be shortened since we know the irreducibility of one of these. The only difficult part of the problem was to figure out the McCay cubic; which is kind of well-known but very difficult to prove otherwise (multi-fold Sondat's theorem anyone?) @EulerMac I remember chatting with Ivan yesterday so essentially it is a joint solution. (I didn't know about Darboux )

Darn are these fancy theorems required to solve this problem? Is there a synthetic solution instead?

mathlete789 wrote: Darn are these fancy theorems required to solve this problem? Is there a synthetic solution instead? Pretty sure there isn't one. The closest you'll get is my solution.

There is a faster way than in post 3 to prove the following claim: Claim: Let $x_1,x_2,x_3,y_1,y_2,y_3$ be angles in $(0, 180^{\circ})$ with $x_1+x_2+x_3=90^{\circ} = y_1+y_2+y_3$. If $\cos x_1\cos x_2 \cos x_3 = \cos y_1\cos y_2 \cos y_3$ and $\sin x_1\sin x_2 \sin x_3= \sin y_1 \sin y_2 \sin y_3$ then $\{x_1,x_2,x_3 \} = \{y_1,y_2,y_3\}$.

After proving this claim, it's easy to see how it applies to the problem with a few applications of Law of Sines and trig ceva. We can set $x_1,x_2,x_3 = \angle PAB, \angle PBC, \angle PCA$ and $y_1,y_2,y_3 = \angle PBA, \angle PCB, \angle PAC$ and conclude.

mathlete789 wrote: Darn are these fancy theorems required to solve this problem? Is there a synthetic solution instead? The "official solution" is a (fairly nice) application of trig Ceva, similar to what vincenthuang did for #3. This way you don't have to use Bezout. You also don't need to know what the names of the cubics are (the complex numbers above tell you what they are). But I honestly think Bezout is not terribly fancy... "two cubics meet in at most 9 points" is a fairly intuitive statement, and even if I didn't know the name of the theorem I could probably have guessed it on the TST.

That being said, if someone finds a synthetic solution please let me know. I tried to do so on a train ride to San Francisco to meet a friend, and all that happened was I missed my destination by three stops. Possible idea: say $P$ is good WRT $ABC$ if it satisfies the condition. If $P$ is good WRT $ABC$ then $P$ is good with respect to the pedal triangle $A_1B_1C_1$ as well. One can repeat this inductively and try to then use the fact the third pedal triangle is similar to $ABC$.

In the following discussion, I assume $ P $ is not the incenter, orthocenter or the circumcenter $ O $ of $ \triangle ABC. $ Let $ \triangle A_2B_2C_2 $ be the circumcevian triangle of $ P $ WRT $ \triangle ABC. $ Let $ P^* $ be the isogonal conjugate of $ P $ WRT $ \triangle ABC $ and let $ M $ be the midpoint of $ PP^*. $ From $ \angle PAB $ $ + $ $ \angle PBC $ $ + $ $ \angle PCA $ $ = $ $ 90^{\circ} $ we get $ \triangle A_1B_1C_1 $ and $ \triangle A_2B_2C_2 $ are homothetic, so $ A_1A_2, $ $ B_1B_2, $ $ C_1C_2, $ $ OM $ are concurrent at $ J. $ Let $ T $ ($ \in $ $ JP $) be the point such that $ \triangle A_1B_1C_1 $ $ \cup $ $ P $ $ \sim $ $ \triangle A_2B_2C_2 $ $ \cup $ $ T. $ From Sondat's theorem for $ \triangle ABC, $ $ \triangle A_2B_2C_2 $ we get $ P $ $ \in $ $ P^*T, $ so $ J, $ $ P, $ $ P^*, $ $ T $ are collinear $ \Longrightarrow $ $ O $ $ \in $ $ PP^* $ ... $ (\star) $ $ . $ Since $ AA_1, $ $ BB_1, $ $ CC_1 $ are concurrent, so we can get the antipedal triangle $ \triangle P_AP_BP_C $ of $ P $ WRT $ \triangle ABC $ and $ \triangle ABC $ are perspective (by Ceva's theorem). Let $ \overline{XYZ} $ ($ X $ $ \in $ $ BC, $ $ Y $ $ \in $ $ CA, $ $ Z $ $ \in $ $ AB $) be the perspectrix of $ \triangle ABC, $ $ \triangle P_AP_BP_C. $ From Sondat's theorem for $ \triangle ABC, $ $ \triangle P_AP_BP_C $ we get $ OP $ $ \perp $ $ \overline{XYZ}, $ so notice the center of $ \odot (APA_1), $ $ \odot (BPB_1), $ $ \odot (CPC_1) $ is the midpoint of $ PX, $ $ PY, $ $ PZ, $ respectively we conclude that $ OP $ is the radical axis of these three circles. Let $ V $ be the perspector of $ \triangle ABC, $ $ \triangle A_1B_1C_1. $ From Sondat's theorem for $ \triangle ABC, $ $ \triangle A_1B_1C_1 $ $ \Longrightarrow $ $ V $ $ \in $ $ PP^*. $ Combining $ (\star) $ we get $ V $ lies on the radical axis of $ \odot (APA_1), $ $ \odot (BPB_1), $ $ \odot (CPC_1), $ so $ AV $ $ \cdot $ $ A_1V $ $ = $ $ BV $ $ \cdot $ $ B_1V $ $ = $ $ CV $ $ \cdot $ $ C_1V, $ hence $ V $ is the orthocenter of $ \triangle ABC $ $ \Longrightarrow $ $ V $ and $ P $ coincide with the orthocenter of $ \triangle ABC, $ contradiction.

Even better (to avoid the usage of Bezout's theorem in a non trivial way) As shown previously, we only have to find $P$ such that $P$ lies on both the McCay Cubic and the Darboux Cubic. We exclude the vertices (being trivial) and the in/ex centres (being self-isogonal conjugates). We prove that the only other case is that $P=O,H$. Let $L$ be the DeLongChamps Point of $ABC$. McCay cubic implies $P,O,P'$ are collinear. Darboux cubic implies $P,L,P'$ are collinear. Therefore, $P$ lies on the Euler Line and $P'$ also lies on the Euler line. Therefore, $P$ lies on both the Euler line and the Jerabek's hyperbola. Since a straight line meets a hyperbola at most twice (apply projective transform to map the hyperbola to a circle) we shall conclude that on two such $P$ are there. Now, it follows since $H,O$ work.

$B_1C_1 \cap BC=T$ AND $CP \cap C_1B_1=X , BP \cap B_1C_1=Y$ AND $BX \cap CY=Z$ $H'$ is the orthocenter of $ZBC$ .One of the solutions is $P=O$ so WLOG $A_1$ is not midpoint of $BC$. if $X=C_1$ AND $ Y=B_1$ we have $P=H$ WLOG $X \not =C_1$. $B' $ is reflection of $B$ about $A_1$ from concurrency we have $(T,A_1,B,C)=-1$ SO $P , A_1,Z$ are collinear. from the angle condition we have $ X,Y,B,C$ are concyclic( because $\widehat{BYX} = \widehat{PCB}$),so $\widehat{XBP} = \widehat{YCP}$ so $ \widehat{H'BP}=\widehat {H'B'P}= \widehat {H'CP}$ SO $C,B',P.H'$ are concyclic. but $C,B_1,P,Z$ are concyclic so $H'=P$ $\implies$ $\widehat{CXB}=90$$\implies$ $B,C_1,X,P$ are concyclic $\implies$ $\widehat{PXY} = \widehat{PBA_1}=\widehat{PBC_1} $ $\implies$ P is on bisector of $\widehat{ABC}. $ we can easily conclude that $P=I$

v_Enhance wrote: mathlete789 wrote: Darn are these fancy theorems required to solve this problem? Is there a synthetic solution instead? The "official solution" is a (fairly nice) application of trig Ceva, similar to what vincenthuang did for #3. This way you don't have to use Bezout. You also don't need to know what the names of the cubics are (the complex numbers above tell you what they are). But I honestly think Bezout is not terribly fancy... "two cubics meet in at most 9 points" is a fairly intuitive statement, and even if I didn't know the name of the theorem I could probably have guessed it on the TST. Can you post the official solution? Thanks.

Here is the beautiful solution I submitted when I took this live: As $ABC$ is scalene, we know that $\angle A+2\angle B,2\angle A+\angle B,\angle B+2\angle C,2\angle B+\angle C,\angle C+2\angle A,2\angle C+\angle A\ne 180^\circ$. Let $\angle PAB=\alpha,\angle PBC=\beta,\angle PCA=\gamma$. Note that $\frac{AB_1}{AC_1}=\frac{\cos(\angle A-\alpha)}{\cos\alpha}$ and so on, so by Ceva's Theorem $AA_1,BB_1,CC_1$ concur iff \[\cos\alpha\cos\beta\cos\gamma=\cos(\angle A-\alpha)\cos(\angle B-\beta)\cos(\angle C-\gamma).\]Also by Trig Ceva, \[\sin\alpha\sin\beta\sin\gamma=\sin(\angle A-\alpha)\sin(\angle B-\beta)\sin(\angle C-\gamma).\]Now, let $a=e^{i\frac{\angle A}2},b=e^{i\frac{\angle B}2},c=e^{i\frac{\angle C}2},x=e^{i\alpha},y=e^{i\beta},z=e^{i\gamma}$. From $\frac{\angle A}2+\frac{\angle B}2+\frac{\angle C}2=\alpha+\beta+\gamma=90^\circ$, we know that $abc=xyz=i$. Furthermore, $a,b,c,x,y,z$ are all in the first quadrants as the angles are acute. Now, we substitute the identities $\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}2$ and $\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$ into our equations to obtain \[\left(x-\frac1x\right)\left(y-\frac1y\right)\left(z-\frac1z\right)=\left(\frac{a^2}x-\frac x{a^2}\right)\left(\frac{b^2}y-\frac y{b^2}\right)\left(\frac{c^2}z-\frac z{c^2}\right),\]\[\left(x+\frac1x\right)\left(y+\frac1y\right)\left(z+\frac1z\right)=\left(\frac{a^2}x+\frac x{a^2}\right)\left(\frac{b^2}y+\frac y{b^2}\right)\left(\frac{c^2}z+\frac z{c^2}\right).\]Adding and dividing by 2, we have that \begin{align*} xyz+\frac x{yz}+\frac y{zx}+\frac z{xy}&=\frac{a^2b^2c^2}{xyz}+\frac{a^2yz}{xb^2c^2}+\frac{b^2zx}{yc^2a^2}+\frac{c^2xy}{za^2b^2}\\ (abc)^2[(xyz)^2+x^2+y^2+z^2]&=(a^2b^2c^2)^2+(a^2yz)^2+(b^2zx)^2+(c^2xy)^2\\ -(-1+x^2+y^2+z^2)&=1-\left[\left(\frac{a^2}x\right)^2+\left(\frac{b^2}y\right)^2+\left(\frac{c^2}z\right)^2\right]\\ \left(\frac{a^2}x\right)^2+\left(\frac{b^2}y\right)^2+\left(\frac{c^2}z\right)^2&=x^2+y^2+z^2. \end{align*}Since $abc=xyz=i$, we can write \begin{align*} x^2+y^2-\left(\frac1{xy}\right)^2&=\left(\frac{a^2}x\right)^2+\left(\frac{b^2}y\right)^2-\left(\frac{xy}{a^2b^2}\right)^2\\ \frac{x^2y^2}{a^4b^4}+x^2+y^2&=\frac1{x^2y^2}+\frac{a^4}{x^2}+\frac{b^4}{y^2}\\ \left(\frac{x^2}{a^4}+b^4\right)\left(\frac{y^2}{b^4}+a^4\right)&=\left(\frac1{x^2}+b^4\right)\left(\frac1{y^2}+z^4\right)\\ \left(\frac{x^2}{a^2}+a^2b^4\right)\left(\frac{y^2}{b^2}+a^4b^2\right)&=\left(\frac{a^2}{x^2}+a^2b^4\right)\left(\frac{b^2}{y^2}+a^4b^2\right). \end{align*}First, let us take care of the case of $\frac{a^2}{x^2}+a^2b^4=0$. Suppose that $\angle A+2\angle B\ne180^\circ$. If $\frac{x^2}{a^2}+a^2b^4=0$ as well, then we have $\frac{a^2}{x^2}=\frac{x^2}{a^2}\implies x^4=a^4\implies x=a$ as $a$ and $x$ are both in the first quadrant. Then, $a^2b^4=-1$, a contradiction as $\angle A+2\angle B\ne180^\circ$. Hence, we must have that $\frac{y^2}{b^2}+a^4b^2=0$. These two equations together imply that $x^2=-\frac1{b^2}\implies x=\frac i{b^2}$ and $y^2=-a^2b^2=-\frac1{c^2}\implies y=\frac i{c^2}$. Thus, $z=\frac i{a^2}$, resulting in the solution of $\alpha=90^\circ-\angle B,\beta=90^\circ-\angle C,\gamma=90^\circ-\angle A$, or the orthocenter of $ABC$. Similarly, if $\frac{x^2}{a^2}+a^2b^4=0$, then again we cannot have $\frac{a^2}{x^2}+a^2b^4=0$ as well, so $\frac{y^2}{b^2}+a^4b^2=0$. These two equations tell us that $x^2=-a^2b^2=-\frac1{c^4}\implies x=\frac i{c^2}$ and $y^2=-\frac1{a^4}\implies y=\frac i{a^2}$. Hence, $x=\frac i{b^2}$, resulting in the solution of $\alpha=90^\circ-\angle C,\beta=90^\circ-\angle A,\gamma=90^\circ-\angle B$, or the circumcenter of $ABC$. The cases of $\frac{y^2}{b^2}+a^4b^2$ or $\frac{b^2}{y^2}+a^4b^2$ being zero follow analogously, where we use $2\angle A+\angle B\ne180^\circ$ to obtain the solutions of the circumcenter and orthocenter of $ABC$. Now comes the fun part, where we assume that all four terms in the equation are nonzero and that the point is not the circumcenter or orthocenter of $ABC$. Rewrite the equation in the form \[\frac{\frac{x^2}{a^2}+a^2b^4}{\frac{a^2}{x^2}+a^2b^4}=\frac{\frac{b^2}{y^2}+a^4b^2}{\frac{y^2}{b^2}+a^4b^2}.\]For the first (and probably last) time, we use geometry on complex numbers instead of complex numbers on geometry. Let $P=\frac{x^2}{a^2},P'=\frac{a^2}{x^2},Q=\frac{b^2}{y^2},Q'=\frac{y^2}{b^2},R=-a^2b^4,S=-a^4b^2$ in the complex plane. Note that $P,P',Q,Q',R,S$ all lie on the unit circle, with $P,P'$ and $Q,Q'$ being reflections of each other across the real axis. The argument of the LHS of the equation is $\measuredangle P'RP$ and the argument of the RHS of the equation is $\measuredangle Q'SQ$. This means that the chords $PP'$ and $QQ'$ inscribe the same directed arc. As $PP'$ and $QQ'$ are parallel, this means that $P$ and $Q$ are either the same point or are antipodes on the unit circle. Thus, either $\frac{x^2}{a^2}=\frac{b^2}{y^2}$ or $\frac{x^2}{a^2}=-\frac{b^2}{y^2}$. This tells us that $a^4b^4=x^4y^4$. Similarly, we know that $b^4c^4=y^4z^4$ and $c^4a^4=z^4x^4$. Multiplying, $a^8b^8c^8=x^8y^8z^8$, so $a^4b^4c^4=x^4y^4z^4$ or $a^4b^4c^4=-x^4y^4z^4$. In the latter case, we know that $a^4=-x^4,b^4=-y^4,c^4=-z^4$, which is a contradiction as multiplying the equations gives us $1=-1$. Hence, we know that $a^4=x^4,b^4=y^4,c^4=z^4$, so $a=x,b=y,c=z$ as $a,b,c,x,y,z$ are in the first quadrant. This corresponds to the solution of $\alpha=\frac{\angle A}2,\beta=\frac{\angle B}2,\gamma=\frac{\angle C}2$, which corresponds to the incenter of $ABC$. It is easy to check that the incenter, orthocenter, and circumcenter all work, so we are done.

$\measuredangle$ denotes a directed angle modulo $180^\circ$. Let $O,H,I$ denote the circumcenter, orthocenter, and incenter of $\triangle ABC$. The answer is $P\in\{O,H,I\}$, which all clearly work. Assume that $P\notin\{O,H\}$, so it suffices to prove that $P=I$. Our assumption implies that $A_1$ is not the midpoint of $\overline{BC}$, so by Ceva's Theorem, $\overline{BC}$ and $\overline{B_1C_1}$ are not parallel. Moreover, $P$ does not lie on $\overline{AA_1}$, $\overline{BB_1}$, or $\overline{CC_1}$. [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pen pri=red; pen sec=orange; pen tri=fuchsia; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair A, B, C, P, A1, B1, C1, T, Q, R, H; A=dir(140); B=dir(210); C=dir(330); P=incenter(A, B, C); A1=foot(P, B, C); B1=foot(P, C, A); C1=foot(P, A, B); T=extension(B, C, B1, C1); Q=extension(B, P, B1, C1); R=extension(C, P, B1, C1); H=extension(B, R, C, Q); filldraw(A -- B -- C -- cycle, fil, pri+linewidth(1.2)); draw(C -- H -- B -- T -- Q, sec); fill(H -- B -- C -- cycle, sfil); draw(H -- A1, sec); draw(A -- P, tri); draw(B -- Q, sec); draw(C -- R, sec); draw(B1 -- P -- C1, tri); filldraw(circumcircle(B, A1, P), tfil, tri); filldraw(circumcircle(C, A1, P), tfil, tri); filldraw(circumcircle(B, C, B1), sfil, sec); clip((B+(-100, -.2)) -- (B+(100, -.2)) -- (B+(100, 100)) -- (B+(-100, 100)) -- cycle); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$P$", P, dir(25)); dot("$A_1$", A1, dir(260)); dot("$B_1$", B1, N); dot("$C_1$", C1, NW); dot("$T$", T, SW); dot("$Q$", Q, N); dot("$R$", R, dir(120)); dot("$H$", H, N); [/asy][/asy] Let $T=\overline{BC}\cap\overline{B_1C_1}$, $Q=\overline{BP}\cap{B_1C_1}$, $R=\overline{CP}\cap\overline{B_1C_1}$, and $H=\overline{BR}\cap\overline{CQ}$. Since $\overline{AA_1},\overline{BB_1},\overline{CC_1}$ concur, by Ceva-Menelaus, $(B,C;T,A_1)$ is harmonic. However, by Ceva-Menelaus on $\triangle HBC$, $A_1\in\overline{HP}$. Check that \begin{align*} \measuredangle CRQ&=\measuredangle CRB_1=\measuredangle CB_1R+\measuredangle RCB_1=90^\circ+\measuredangle PB_1C_1+\measuredangle PCB_1\\ &=90^\circ+\measuredangle PAC_1+\measuredangle PCA=90^\circ+\measuredangle PAB+\measuredangle PCA=\measuredangle CBQ, \end{align*}whence $BCQR$ is cyclic. Let $M$ be the circumcenter of $BCQR$. By Brocard's Theorem on $BCQR$, $\overline{TM}\perp\overline{HP}$, so $M$ lies on $\overline{BC}$, thence $\angle BQC=\angle BRC=90^\circ$, so that $H$ is the orthocenter of $\triangle PBC$. Finally, $$\measuredangle ABP=\measuredangle C_1BP=\measuredangle C_1RP=\measuredangle QRC=\measuredangle QBC=\measuredangle PBC,$$and similarly $\measuredangle ACP=\measuredangle PCB$, so $P$ is the incenter of $\triangle ABC$, as desired. $\square$

Here is a quicker finish to tastymath75025's solution. As in his solution, we can write \begin{align} \tag{1}a_1a_2a_3&=b_1b_2b_3=i,\\ \tag{2}\prod_\mathrm{cyc}\left(a_1-\frac1{a_1}\right)&=\prod_\mathrm{cyc}\left(b_1-\frac1{b_1}\right),\\ \tag{3}\prod_\mathrm{cyc}\left(a_1+\frac1{a_1}\right)&=\prod_\mathrm{cyc}\left(b_1+\frac1{b_1}\right). \end{align}Now, utilizing $(1)$ we can rewrite $(2)$ and $(3)$ as \begin{align} \tag{2'}\prod_\mathrm{cyc}\left(a_1^2-1\right)&=\prod_\mathrm{cyc}\left(b_1^2-1\right),\\ \tag{3'}\prod_\mathrm{cyc}\left(a_1^2+1\right)&=\prod_\mathrm{cyc}\left(b_1^2+1\right). \end{align}Let $P$ be the monic cubic polynomial with roots $a_1^2,a_2^2,a_3^2$ and $Q$ be the monic cubic polynomial with roots $b_1^2,b_2^2,b_3^2$. Note that $(1)$ automatically yields $P(0)=Q(0)=-1$. Furthermore, the LHS of $(2')$ equals $-P(1)$ and the RHS $-Q(1)$. Analogously the LHS and RHS of $(3')$ are equal to $-P(-1)$ and $-Q(-1)$ as well. Since $P$ and $Q$ are equal at three points but are monic, they must be the same polynomial, whence $\{a_1,a_2,a_3\}=\{b_1,b_2,b_3\}$, or $\{x_1,x_2,x_3\}=\{y_1,y_2,y_3\}$. $\blacksquare$

v_Enhance wrote: Let $ABC$ be an acute scalene triangle and let $P$ be a point in its interior. Let $A_1$, $B_1$, $C_1$ be projections of $P$ onto triangle sides $BC$, $CA$, $AB$, respectively. Find the locus of points $P$ such that $AA_1$, $BB_1$, $CC_1$ are concurrent and $\angle PAB + \angle PBC + \angle PCA = 90^{\circ}$. Scheiße. Excellent problem.

Funny question lol. So basicaly let $BP,CP$ meet $B_1C_1$ at $E,D$ respecitivly and let $BD \cap CE=F$, now let $B_1C_1 \cap BC=X$, now assume that $P$ is not the ortocenter or circumcenter of $\triangle ABC$, from this we can get that $X$ is not a point of infinity, and $P$ doesnt lie in $AA_1,BB_1,CC_1$ so $D,E \ne B_1,C_1$ which is good, now note that $-1=(B, C; A_1, X)$ so by projecting $$(D, E; FA_1 \cap DE, X) \overset{F}{=} (B, C; A_1, X) \overset{P}{=} (E, D; PA_1 \cap DE, X)=(D, E; PA_1 \cap DE,X) \implies F,A_1,P \; \text{colinear}$$Now by angle chase using the propety of the problem $$\angle DEB=\angle AC_1B_1-\angle ABP=90-\angle ABP-\angle B_1C_1P=\angle PCB+\angle PAC-\angle PAC=\angle PCB \implies BDEC \; \text{cyclic}$$Now by brokard theorem the center of that cyclic quad lies in $BC$ so $\angle BDC=90=\angle BEC$ and that means $BC_1DPA_1, CEB_1PA_1$ are cyclic so $\angle ABP=\angle EDP=\angle PBC$ and $\angle ACP=\angle DEP=\angle PCB$ meaning that $P$ is the incenter of $\triangle ABC$. Hence $P$ can be the circumcenter, ortocenter or incenter of $\triangle ABC$, thus we are done