Let the circumcircles of $BB_2C_1C$ and $AA_1C_2C$ meet again at $P$. Claim: $P$ belongs to the circumcircle of $ABB_1A_2$. Proof: Note that $$\angle APB=360^\circ-\angle APC-\angle BPC=\angle AA_1C+\angle BC_1C=180^\circ-\angle AB_1B$$hence $P$ belongs to circumcircle of $AB_1B$ as well $\blacksquare$ Hence, the three circles concur at $P$. Now, $$\angle APA_2+\angle APC+\angle CPC_1=\angle AB_1A_2+180^\circ-\angle AA_1C+\angle CBC_1=90^\circ-\angle AB_1B+180^\circ-\angle AA_1C+90^\circ-\angle BC_1C=180^\circ,$$so $P$ belongs to $A_2C_1$ and similarly it belongs to $A_1B_2$ and $B_2C_1$, so we are done.

[asy][asy] size(8cm); defaultpen(fontsize(9pt)); pair A = dir(120), B = dir(180+30), C = dir(-30), P = (A+B+C)/2, D = circumcenter(B,P,C), E = circumcenter(A,P,C), F = circumcenter(A,P,B), B2=2*D-C, C1=2*D-B, C2=2*E-A, A1=2*E-C, A2=2*F-B, B1=2*F-A; draw(A--B--C--cycle, red+linewidth(1)); draw(A--A2--B1--B--B2--C1--C--C2--A1--A, lightblue); draw(circumcircle(B, P, C)^^circumcircle(A, P, C)^^circumcircle(A, P, B), deepcyan); draw(A1--B2, orange+dashed); string[] names = {"$A$", "$B$", "$C$", "$A_1$", "$B_1$", "$C_1$", "$A_2$", "$B_2$", "$C_2$", "$P$"}; pair[] points = {A, B, C, A1, B1, C1, A2, B2, C2, P}; pair[] ll = {dir(110), B, dir(-25), A1, B1, C1, A2, B2, C2, dir(-90)}; for (int i=0; i<names.length; ++i) dot(names[i], points[i], dir(ll[i])); [/asy][/asy] Claim. The circumcircles of the three rectangles $BCC_1B_2$, $CAA_1C_2$, and $ABB_1A_2$ concur at a point $P$. Proof. Let $(CAA_1C_2)$ and $(ABB_1A_2)$ intersect again at $P \neq A$. We will show that $P$ also lies on $(BCC_1B_2)$. Indeed, we have \[ \angle APC = 180^\circ - \angle CA_1A \quad \text{and} \quad \angle BPA = 180^\circ - \angle AB_1B. \]Therefore, \begin{align*} \angle CPB &= 360^\circ - \angle APC - \angle BPA \\ &= \angle CA_1A + \angle AB_1B \\ &= 180^\circ - \angle BC_1C, \end{align*}as claimed. $\blacksquare$ Claim. $A_1,P,B_2$ are collinear; similarly, so are $B_1,P,C_2$ and $C_1,P,A_2$. Proof. We compute \begin{align*} \angle A_1PB_2 &= \angle A_1PA + \angle APB + \angle BPB_2 \\ &= \angle A_1CA + (180^\circ - \angle AB_1B) + \angle BC_1B_2 \\ &= (90^\circ - \angle CA_1A) + (180^\circ - \angle AB_1B) + (90^\circ - \angle BC_1C) \\ &= 360^\circ - (\angle CA_1A + \angle AB_1B + \angle BC_1C) \\ &= 180^\circ, \end{align*}and analogous angle chases prove the other two collinearities. $\blacksquare$ The last claim reveals that $P$ is our desired point of concurrency, so we are done. $\square$

This was also Problem 2 on JMO.

than you would expect though I did initially try to complex bash this, but that didn't exactly work...

Sketch: Construct circumcircles of the rectangles, you will find the circumcircles mutually concur at a certain point (shown with angle chasing). Then with basic cyclic quadrilateral properties we can find the lines will all concur at that certain point, meaning they're concurrent. (full solution coming soon)

[asy][asy] defaultpen(fontsize(10pt)); dotfactor*=1.2; size(300); pair A,B,C,A1,A2,B1,B2,C1,C2,X; A = dir(120); B = dir(210); C = dir(330); A1 = extension(dir(150),A,dir(10),C); C1 = extension(dir(-70),B,dir(30),C); X = intersectionpoints(circumcircle(A,C,A1),circumcircle(B,C,C1))[0]; A2 = intersectionpoints(C1--extension(C1,X,(10,10),(-10,10)),circumcircle(A,B,X))[1]; B1 = rotate(180,(B+A2)/2)*A; B2 = rotate(180,(B+C1)/2)*C; C2 = rotate(180,(C+A1)/2)*A; pen r = red; filldraw(A--B--C--cycle,r+white+white+white+white,r); filldraw(B--B2--C1--C--cycle,r+white+white+white+white+white+white+white,r); filldraw(A--A1--C2--C--cycle,r+white+white+white+white+white+white+white,r); filldraw(A--A2--B1--B--cycle,r+white+white+white+white+white+white+white,r); draw(circumcircle(A,B,B1)^^circumcircle(A,C,A1)^^circumcircle(B,C,C1),orange+orange+red); draw(A1--B2^^B1--C2^^A2--C1,red+linewidth(1.05)); draw(A--X^^B--X^^C--X,fuchsia); draw(A--A2--B1--B--B2--C1--C--C2--A1--A, r); dot("$A$",A,dir(125)); dot("$B$",B,dir(215)); dot("$C$",C,dir(315)); dot("$A_1$",A1,dir(A1)); dot("$B_1$",B1,dir(B1)); dot("$C_1$",C1,dir(C1)); dot("$X$",X,dir(280)); dot("$A_2$",A2,dir(A2)); dot("$B_2$",B2,dir(B2)); dot("$C_2$",C2,dir(C2)); [/asy][/asy] Claim: $(BCC_1B_2), (ABB_1A_2), (CAA_1C_2)$ concur at a point $X$. Proof: Let the second intersection of $(ABB_1A_2)$ and $(BCC_1B_2)$ be $X$. Note that $\angle AXB = 180^{\circ} - \angle AB_1B$ and $\angle BXC = 180^{\circ} - \angle BC_1C$, implying that $\angle AXC =360^{\circ} - \angle AXB - \angle BXC = 180^{\circ} - \angle AA_1C$. Claim: $X\in B_1C_2$. Proof: We have that $\angle AXB_1 = 90^{\circ} = \angle AXC_2$, implying the collinearity. Similarly, $X\in A_2C_1, A_1B_2$, as needed.

Possible complex bash motivation: If $\frac{C_1C}{CB}=k$, $\frac{A_1A}{AC}=l$, and $\frac{B_1B}{BA}=m$, then the angle condition implies that $kl+lm+mk=1$ (complex bash didn't actually pan out for me lol i went with a synthetic solution).

The key of this problem is noticing that the concurrency point, $P$, lies on the circumcircles of $BCC_1B_2$, $CAA_2C_2$, and $ABB_1A_2$. We first prove that the three circumcircles are indeed concurrent, and then show that the concurrency implies the problem. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -16.852914910450526, xmax = 34.390095350576566, ymin = -8.12314938684212, ymax = 18.42519130633257; /* image dimensions */ pair A = (3.,8.), B = (2.,0.), C = (10.,0.), B_2 = (2.,-4.), C_1 = (10.,-4.), C_2 = (17.908680279445264,6.920095244514606), A_1 = (10.908680279445264,14.920095244514606), B_1 = (-0.7171913168911064,0.33964891461138835), A_2 = (0.2828086831088937,8.33964891461139), P = (4.993661449478577,2.357439927495782); /* draw figures */ draw(A--B, linewidth(1.)); draw(B--C, linewidth(1.)); draw(C--A, linewidth(1.)); draw(B--B_2, linewidth(1.)); draw(B_2--C_1, linewidth(1.)); draw(C_1--C, linewidth(1.)); draw(A--A_1, linewidth(1.)); draw(A_1--C_2, linewidth(1.)); draw(C_2--C, linewidth(1.)); draw(B_1--B, linewidth(1.)); draw(A--A_2, linewidth(1.)); draw(A_2--B_1, linewidth(1.)); draw(circle((1.141404341554447,4.169824457305695), 4.257302257233597), linewidth(1.)); draw(circle((6.,-2.), 4.47213595499958), linewidth(1.)); draw(circle((10.454340139722632,7.460047622257303), 7.473870181432778), linewidth(1.)); draw(A_2--C_1, linewidth(1.)); draw(B_2--A_1, linewidth(1.)); draw(B_1--C_2, linewidth(1.)); draw(B--P, linewidth(1.)); draw(P--A, linewidth(1.)); draw(P--C, linewidth(1.)); /* dots and labels */ dot(A,dotstyle); label("$A$", (2.3,8.465636277058753), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (2,-0.8), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (9.852516556056544,0.3911705277736686), NE * labelscalefactor); dot(B_2,dotstyle); label("$B_2$", (0.9,-4.6), NE * labelscalefactor); dot(C_1,dotstyle); label("$C_1$", (10.166698102721334,-4.415807136197607), NE * labelscalefactor); dot(C_2,dotstyle); label("$C_2$", (18.398254625338808,6.674801461069453), NE * labelscalefactor); dot(A_1,linewidth(4.pt) + dotstyle); label("$A_1$", (10.763643041384432,15.3), NE * labelscalefactor); dot(B_1,linewidth(4.pt) + dotstyle); label("$B_1$", (-1.395182814542903,-0.6), NE * labelscalefactor); dot(A_2,linewidth(4.pt) + dotstyle); label("$A_2$", (-0.20129293721670466,8.81123597839002), NE * labelscalefactor); dot(P,linewidth(4.pt) + dotstyle); label("$P$", (4.6,1.3), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We let $P$ be $(BCC_1B_2)\cap (CAA_2C_2)$, and we wish to show that $P$ does indeed lie on $(ABB_1C_2)$. But note that \begin{align*} \angle AB_1B&=180^\circ-(\angle BC_1C+\angle CA_1A)\\ &=\angle BPC+\angle APC-180^\circ\\ &=180^\circ - \angle APB \end{align*}which is what we wanted. Now, we show that $A_2PC_1$ is collinear. But this is clear from \[\angle A_2PB=\angle BPC_1=90^\circ\]Similarly, $P$ lies on $B_2A_1$ and $B_1C_2$, so we are done. Remarks + Motivation: This problem is definitely an ``Art School" problem, because its very hard to synthetic this without guessing the circumcircles (at least in my opinion). It does kind of feel like IMO 2020/1, in the sense that both of them have a very weird contrived angle condition, and they are both about ``guess the point". Once you notice the cyclic quads, it's basically given that you set $P$ as the intersections of 2 circles, and the way you prove the cyclic is super obvious by the angle conditions. After that, it's just 2 90 degree angles, and you win.

Solution: https://yu-dylan.github.io/Writeups/USAMO-2021-1.pdf

I am literally so bad at geometry. Spent 2 hours on this one. With 30 minutes left on the contest, I decided to write my observation that if the lines concurred then the concurrency point was concyclic with $(BCC_1B_2),(CAA_1C_2),(ABB_1A_2)$, thinking I'd get a bit of partial. After writing it, I realized that I had a proof and, with 20 minutes left, proceeded to write the rest of the solution. If MAA decides to implement style score tiebreakers this year I'm probably dead.

Solution: We begin by constructing a diagram: [asy][asy] import olympiad; unitsize(13); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); 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 = -10.252427827191417, xmax = 15.150472936867494, ymin = -3.4167712881064825, ymax = 11.241299387760378; /* image dimensions */ pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); /* draw figures */ draw(circle((0.2801492142784242,1.5), 4.481884633503346)); draw(circle((3.2738086301390403,7.2491118286162095), 3.3297516312587083)); draw((-3.454001382502029,3.9785901614824524)--(4,4)); draw((4,4)--(4,-1)); draw((4,-1)--(-3.420838326674272,-1.0456127964221371)); draw((-3.420838326674272,-1.0456127964221371)--(-3.454001382502029,3.9785901614824524)); draw((-3.454001382502029,3.9785901614824524)--(0.06810107679554245,8.14949230836156)); draw((-3.454001382502029,3.9785901614824524)--(-5.709089178789237,5.8523028157505035)); draw((-5.709089178789237,5.8523028157505035)--(-2,10)); draw((0.06810107679554245,8.14949230836156)--(-2,10)); draw((0.06810107679554245,8.14949230836156)--(4,4)); draw((0.06810107679554245,8.14949230836156)--(2.3992779710964074,10.461967575808181)); draw((-3.454001382502029,3.9785901614824524)--(4,-1)); draw((4,4)--(2.3992779710964074,10.461967575808181)); draw((0.06810107679554245,8.14949230836156)--(-5.709089178789237,5.8523028157505035)); draw((4,-1)--(-2,10)); draw((2.3992779710964074,10.461967575808181)--(-3.420838326674272,-1.0456127964221371)); draw((2.3992779710964074,10.461967575808181)--(6.428589254168415,6.183933374028034)); draw((6.428589254168415,6.183933374028034)--(4,4)); draw(circle((-2.820834296659645,7.001753244155373), 3.108577255193373)); draw((-5.709089178789237,5.8523028157505035)--(6.428589254168415,6.183933374028034)); draw((-3.454001382502029,3.9785901614824524)--(0.1949568576384766,5.981074885605413)); draw((4.000284475054789,3.9995766660700363)--(0.1949568576384766,5.981074885605413)); draw((0.06810107679554245,8.14949230836156)--(0.1949568576384766,5.981074885605413)); /* dots and labels */ dot((6.428589254168415,6.183933374028034),dotstyle); label("$C_{2}$", (6.494915365823923,6.349748653166799), NE * labelscalefactor); dot((4,-1),dotstyle); label("$C_{1}$", (4.074012290397943,-0.8300529335417423), NE * labelscalefactor); dot((-3.454001382502029,3.9785901614824524),dotstyle); label("$B$", (-3.387675270846515,4.144405440621219), NE * labelscalefactor); dot((4,4),dotstyle); label("$C$", (4.04084923457019,4.277057663932232), NE * labelscalefactor); dot((0.06810107679554245,8.14949230836156),dotstyle); label("$A$", (0.12760864689531848,8.322950474918107), NE * labelscalefactor); dot((2.3992779710964074,10.461967575808181),dotstyle); label("$A_{1}$", (2.4987671385796686,10.71069049451633), NE * labelscalefactor); dot((-3.420838326674272,-1.0456127964221371),dotstyle); label("$B_{2}$", (-3.354512215018762,-0.8797975172833719), NE * labelscalefactor); dot((-5.709089178789237,5.8523028157505035),dotstyle); label("$B_{1}$", (-5.908067513755754,6.3994932369084285), NE * labelscalefactor); dot((-2,10),dotstyle); label("$A_{2}$", (-1.7792670632004874,10.196663129186156), NE * labelscalefactor); dot((4.000284475054789,3.9995766660700363),linewidth(4pt) + dotstyle); dot((0.1949568576384766,5.981074885605413),linewidth(4pt) + dotstyle); label("$X$", (0.24367934229245447,6.117607262372528), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim 1: Circumcircles of $B_1A_2AB, AA_1C_2C,$ and $BCC_1B_2$ mutually intersect at some point $X.$ Proof: Start by letting $X$ be the intersection of two circumcircles such that $X \neq A.$ By some simple cyclic quad angle properties, we see \[ \angle AXC = 180^{\circ} - \angle AA_1C, \angle BXA = 180^{\circ} - \angle AB_1B.\]It follows that $\angle CXB = 180^{\circ} - BC_1C,$ and so our claim is proven. Claim 2: Lines $B_1C_2, C_1A_2,$ and $A_1B_2$ concur at $X.$ Proof: We proceed with angle chasing. Note that \[\angle AA_1C_2= \angle AXC_2 = 90^{\circ}, \angle B_1A_2A = \angle AXB_1 = 90^{\circ},\]meaning $B_1, X, C_2,$ are collinear. Then, we may similarly obtain \[\angle BB_2C_1 = \angle BXC_1 = 90^{\circ}, \angle A_2B_1B = \angle A_2XB = 90^{\circ},\]meaning $A_2, X, C_1$ are collinear. Finally, we have \[\angle A_1C_2C = A_1XC=90^{\circ}, \angle CC_1B_2 = \angle B_2XC = 90^{\circ},\]meaning $A_1, X , B_2$ are collinear. Putting these three results together, it is clear that $B_1C_2, C_1A_2,$ and $A_1B_2$ are concurrent at $X. \qquad \square$

Cute problem! Let $P= (AB_1B) \cap (AA_1C)$, with $P \neq A$. Observe that $\angle BPA+ \angle CPA= 180º- \angle AB_1B+ 180º - \angle AA_1C= 360º-(\angle AB_1B + \angle AA_1C)=180º+ \angle BC_1C$, by the problem's condition. Hence, $\angle BPC= 360º- \angle BPA - \angle CPA= 180º- \angle BC_1C$, so $P$ lies on $(BC_1C)$. Now, obseve that since $(AB_1B),(AA_1C)$ have diameter $AB_1,AC_2$, respectively (since $AA_2B_1B$ and $AA_1C_2C$ are rectangles), then $\angle APB_1+ \angle APC_2= 90º+90º= 180º$, so $P$ lies on $B_1C_2$. Similarly, $P$ lies on $C_1A_2$ and $A_1B_2$, so we are done. $\blacksquare$

this problem took three times as long to write-up than do lol

the diagram took three times longer than writing up because the circles were way too big

i already have trouble with collinearity and they gave me concurrency wut

who thought about brainchons when looking at this problem

hmm isnt this like instasolve after egmo 2020/3 and usemo 2020/5...

used a hint xonk. let $(ABB_1A_2)\cap (BCC_1B_2)=P\ne B.$ We have $\angle APC=360-\angle APB-\angle BPC=\angle AB_1B+\angle BC_1C=180-\angle CA_1A$ so $P$ is on $(CAA_1C_2)$ as well. Next $BP$ is the radical axis of $(ABB_1A_2)$ and $(BCC_1B_2)$, so the midpoint of $BP$ lies on the line connecting the centers of the circles, which are the midpoints of $BA_2,BC_1.$ Then homothety at $B$ shows that $P$ lies on $A_2C_1$ and symmetry finishes.

Let $K$ be the intersection of $(AA_2B_1B)$ and $(AA_1C_2C) \neq A$. Then $\angle BKC = 360^{\circ} - (\angle BKA + \angle CKA) = 360^{\circ} - (180^{\circ} - \angle AB_1B + 180^{\circ} - \angle AA_1C) = 180^{\circ} - \angle BC_1C$, so $K \in (BB_2CC_1)$. Then $90^{\circ} = \angle B_1A_2A = \angle C_2A_1A = \angle B_1KA = \angle C_2KA \implies \angle B_1KC_2 = 180^{\circ}$ so $B_1$, $K$ and $C_2$ are collinear. By symmetry, we find that $K$ is the concurrence point of $B_1C_2$, $C_1A_2$ and $A_1B_2$, as desired. This problem reminded me a lot of the proof for Miquel's theorem

The problem can be rephrased as follows: Quote: Let $P$ be a point inside $\triangle ABC$. Let $C_1$ denote the antipode of $B$ on $(BPC)$, and similarly define $A_1$, $B_1$. Let $B_2$ denote the antipode of $B$ on $(BPC)$, and similarly define $A_2$, $C_2$. Show that lines $B_1C_2,$ $C_1A_2,$ and $A_1B_2$ are concurrent. I claim that each of these three lines pass through $P$. This is clear because, for example, \[ \angle A_1PB_2 = \angle A_1PC_1 + \angle B_2PC_1 = 90^{\circ} + 90^{\circ} = 180^{\circ}. \]Thus we are done. Remark: In general, there are many such concurrence problems that can be solved by rephrasing the problem in terms of the concurrence of circumcircles. This is a lesson that is taught to us from problems such as 2020 USEMO #5 and 2020 EGMO #3.

$\color{magenta} \boxed{\textbf{SOLUTION P1}}$ $\color{red} \textbf{Geo Marabot Solve 8}$ The angle condition implies the circumcircles of the three rectangles intersect at some point $P$ We have, $\angle B_2PC+\angle A_1PC=90+90=180$ and similar for other three we are done that the lines concur at $P\blacksquare$

Claim. The rectangles' circumcircles concur. Proof. Let $(BCC_1B_2)$ and $(CAA_1C_2)$ meet again at $O$. Note $\sum_{\text{cyc}}\angle BOC=360^\circ$ and $(\angle BOC,\angle COA)=(180^\circ-\angle BC_1C,180^\circ-\angle CA_1A)$, whence the angle condition gives $\angle AOB=180^\circ-\angle AB_1B$. Then $O\in(ABB_1A_2)$ as claimed. The rest is easy: $\angle APB_1=\angle APC_2=90^\circ$ and so $O\in B_1C_2$ and so on. In all, $O$ is the desired point of concurrency. $\square$

Let $P = (ABB_1A_2) \cap (AC_1CA)$; the given angle condition then yields that $P$ lies on $(BCB_2C_1)$ too. It follows that $\angle BPC_1 +\angle BPA_2 = 90^\circ + 90^\circ = 180^\circ$, so $P$ lies on $\overline{A_2C_1}$ and cyclic permutations.

This problem is trivial when you notice that <ABD = 180 - <AB_1 B <BDC = 360 - <ADB - <ADC.

Slightly different finish. If we let $A \neq P = (ABB_1A_2) \cap (AC_1C_2A_1)$, the given angle condition gives us that $P \in (BCC_1B_2)$, so all three circumcircles of the rectangles concur. Furthermore, notice that $BA_2, BC_1$ are diameters of their respective circles, and both circles go through $B, P$. Thus, $A_2C_1$ goes through $P$ (can be shown by similar triangles with circumcenters), and we can use the same logic to find that $P$ lies on all three lines.

redacted

Let $\angle BC_1C = \alpha$, $\angle CA_1A = \beta$ and $\angle AB_1B = 180 - \alpha - \beta$. Every rectangle is a cyclic quadrilateral. Lets draw the circumcircles around the rectangles $BCC_1B_2$, $CAA_1C_2$, $ABB_1A_2$ and name them $\omega$, $\omega_1$ and $\omega_2$, respectively. Let $\omega$ $\cap$ $\omega_1$ = P, where P is a point other than C. Claim: P $\in$ $\omega_2$ Proof: $\angle APB = 360 - \angle APC - \angle BPC$. Since $A_1APC$ is cyclic, $\angle APC = 180 - \angle AA_1C = 180 - \beta$. Since $BC_1CP$ is cyclic, $\angle BPC = 180 - \angle BC_1C = 180 - \alpha$ $\Rightarrow$ $\angle APB = 360 - (180 - \beta) - (180 - \alpha) = \alpha + \beta$. Now $\angle APB + \angle AB_1B = \alpha + \beta + 180 - \alpha - \beta = 180$ $\Rightarrow$ $AB_1BP$ is cyclic $\Rightarrow$ P $\in$ $\omega_2$. Now we want to show that P lies on $B_1C_2$, $A_2C_1$ and $A_1B_2$. We have $\angle A_1PC + \angle CPB_2 = 90^\circ + 90^\circ = 180^\circ$, since $A_1C$ is a diameter in $\omega$ and $B_2C$ is a diameter in $\omega_1$ $\Rightarrow$ P lies on $A_1B_2$. Also $\angle B_1PA + \angle APC_2 = 90^\circ + 90^\circ = 180^\circ$, since $AB_1$ is a diameter in $\omega_2$ and $AC_2$ is a diameter in $\omega$ $\Rightarrow$ P lies on $B_1C_2$. Finally $\angle A_2PB + \angle BPC_1 = 90^\circ + 90^\circ = 180^\circ$, since $A_2B$ is a diameter in $\omega_2$ and $BC_1$ is a diameter in $\omega_1$ $\Rightarrow$ P lies on $A_2C_1$. This means that $B_1C_2$, $A_2C_1$ and $A_1B_2$ are concurrent at point P $\Rightarrow$ we are ready.

First notice the circumcircles of the rectangles are forced to meet at a single point, say $K$, from the angle condition. Then we have \[\angle B_1KX_2 = \angle B_1KA + \angle AKC_2 = \angle AA_2B_1 + \angle AA_1C_2 = 90 + 90 = 180,\] so $X$ lies on $B_1C_2$, and analogously the other two lines. $\blacksquare$

Immediately constructed circumcircles for potential radical axis, turned out to be a lot different. Let $(AA_2B_1B) \cap (BB_2C_1C) = D$. $$\angle ADC = 360 - \angle ADB - \angle CDB = \angle AB_1B + \angle CC_1B = 180 - \angle AA_1C$$so $A_1ADC$ is cyclic and the three circumcircles of the rectangles concur. Also, $B_1DC_2$ are collinear and cyclic variations hold by angle chasing: $$\angle B_1DC_2 = \angle B_1DA + \angle ADC_2 = 180$$so done.

Let the circumcircles of rectangles intersect at a point O. Proof: let circles ABB1A2 and AA1CC2 intersect at point O, angle BOC is 360 - 180 - 180 + angleCA1A + angle AB1B which = angle BC1C (mod 180) Now trivially angle BOA2 = 90; and so is angle BOC1, cyclically all those lines pass through O. hence proved

(Easy P1 tho): Let $P$ denote the intersection of $(ABB_1A_2), (BCC_1B_2)$. We will prove that $P$ lies on $(AA_1C_2C)$. Notice that: $\angle BPC = 180^\circ - \angle B C_1 C$ and $\angle APB = 180^\circ - \angle A B_1 B$ which forces $\angle APC = 180^\circ - \angle A A_1 C$ and therefore $P$ lies on $(AA_1C_2C)$. Notice that: $CP \perp A_1P$ and $CP \perp B_2P$. Thus, $A_1 B_2$ passes through point $P$. Similarly, the other two lines pass through $P$ and we are done.