I heard that the synthetic solution to this problem is surprisingly short and motivated!

Can this be solved without using the fact that ABCD is cyclic? (Actually, ABCD is harmonic) Let $X_1$ be the center of spiral similarly from $ABCD$ to $A_1B_1C_1D_1$ and let $X_2$ be the center of spiral similarly from $ABCD$ to $A_2B_2C_2D_2$. Now $AA_1\cap BB_1=B$ so $(X_1AB)$ is tangent to $BC$. Since similar facts hold true, $X_1$ is positive Brocard point of $ABCD$. Similarly $X_2$ is negative Brocard point of $ABCD$. Lemma1. $X_1$ lies inside $ABCD$ (similarly, $X_2$ does). Proof. The circumcircle of triangle $ABX_1$ is tangent to the line $BC$, so $A$ and $X_1$ are on the same side of line $BC$. Similarly, $X_1$ lies within the region formed by lines $AB$, $BC$, $CD$, and $DA$ so we have the desired result. Lemma2. $ABCD$ is cyclic. Proof. For $\omega\in(0,\pi),$ let the function $f_\omega$ be defined as $$f_\omega(x)=\frac{\sin x}{\sin(\omega-x)}.$$Then $$f_\omega'(x)=\frac{\sin\omega}{\sin^2(\omega-x)}$$so $f_\omega$ is strictly increasing in $0<x<\omega$. Let $\angle DAB=\alpha$, $\angle ABC=\beta$, $\angle BCD=\gamma$, and $\angle CDA=\delta$. Also, let $\angle X_1AB=\angle X_1BC=\angle X_1CD=\angle X_1DA=\theta_1$, and similarly define $\theta_2$. From lemma 1, we know $0<\theta_i<\min\{\alpha,\beta,\gamma,\delta\}$ so $$\displaystyle\frac{X_1B}{X_1A}=\frac{\sin \theta_1}{\sin(\beta-\theta_1)}=f_\beta(\theta_1),\ \frac{X_2A}{X_2B}=\frac{\sin \theta_2}{\sin (\alpha-\theta_2)}=f_\alpha(\theta_2).$$Since this holds for the other vertices as well, define $F$ be the function as $F(x)=f_\alpha(x)f_\beta(x)f_\gamma(x)f_\delta(x)$, and we get $$1=\frac{X_1B}{X_1A}\cdot\frac{X_1C}{X_1B}\cdot\frac{X_1D}{X_1C}\cdot\frac{X_1A}{X_1D}=F(\theta_1).$$Similarly, $F(\theta_2)=1$ so by the monotonicity of $f_\alpha, f_\beta, f_\gamma, f_\delta$, we have $\theta_1=\theta_2$. Now, consider the feet of the perpendiculars from $X_1$ to $AB, BC, CD, DA$ be $K, L, M, N$ respectively. $X_2$ is isogonal conjugate of $X_1$ wrt quadrilateral $ABCD$ so $KLMN$ is cyclic. Furthermore, we have $\measuredangle X_1LM=\measuredangle X_1CM=\measuredangle X_1BC$, and $\measuredangle X_1LK=\measuredangle X_1BA$ hence $\measuredangle KLM=\measuredangle ABC$. Similarly, $\measuredangle MNK=\measuredangle DCA$ and this implies that $ABCD$ is cyclic, as desired. Lemma3. Let $\measuredangle X_1AB=\theta$, then it holds that $\measuredangle X_1PX_2=2\theta$. Proof. Consider an inversion centered at $X_1$ with a radius of 1 and denote the images of $A, B, C, D$ as $A', B', C', D'$. We have $\triangle X_1A'B'\sim\triangle X_1BA\sim\triangle X_2BC$. Similarly, $\triangle X_1B'C'\sim\triangle X_2CD$ and $\triangle X_1C'D'\sim\triangle X_2DA$. Therefore, $ABCDX_2$ and $D'A'B'C'X_1$ are similar. Let $A'C'\cap B'D'=Q$ and $(X_1BD)\cap(X_1AC)=Q'$. Since $Q$ and $Q'$ correspond under the inversion, $X_1, Q, Q'$ are collinear. Moreover, $P$ is the radical center of $(ABCD)$, $(AX_1C)$ and $(BX_1D)$ so $X_1, P, Q'$ are collinear. Combining these results, we obtain that $X_1, P, Q$ are collinear. Now, $$\measuredangle D'A'X_1=\measuredangle X_1DA=\theta\Rightarrow \measuredangle (D'A',AB)=2\theta,$$implying that the angle of rotation from $ABCDPX_2$ to $D'A'B'C'QX_1$ is $2\theta$. Thus it follows that $\measuredangle X_1PX_2=\measuredangle (X_1Q,X_2P)=2\theta$. Now, since $X_1$ is the center of spiral similarity between $A_1B_1C_1D_1P_1$ and $ABCDP$, we have $\measuredangle X_1PP_1=\measuredangle X_1AA_1=\theta$. Similarly we get $\measuredangle X_2PP_2=\measuredangle X_2AA_2=-\theta$. Combining this with Lemma 3, we have $\measuredangle X_1PP_1=\theta=-\theta+2\theta=\measuredangle X_1PP_2$, which means that $P, P_1, P_2$ are collinear as desired.

There is a generalization for this problem: Convex 2n-gons $A_1A_2\dots A_{2n}$, $B_1B_2\dots B_{2n}$, and $C_1C_2\dots C_{2n}$ are similar with vertices in order. For every $i=1,2,\dots, 2n$, points $A_i$, $B_i$, $C_{i+1}$, $A_{i+1}$ are collinear in order (all indices are taken modulo $2n$). Prove that every diagonals $A_{i}A_{i+n}$ are intersect at one point $P$, also diagonals $B_{i}B_{i+n}$ intersect at $P_1$, and diagonals $C_{i}C_{i+n}$ intersect at $P_2$, and the points $P$, $P_1$, and $P_2$ are collinear.

My problem! This is the form of the problem statement I submitted: Original problem statement wrote: Quadrilaterals $ABCD \sim A_1B_1C_1D_1 \sim A_2B_2C_2D_2$ lie in the plane such that \[\overline{A_1B_2} \subset \overline{AB}, \: \overline{B_1C_2} \subset \overline{BC}, \: \overline{C_1D_2} \subset \overline{CD}, \text{ and } \overline{D_1A_2} \subset \overline{DA}.\]Prove that $\overline{AC} \cap \overline{BD}$, $\overline{A_1C_1} \cap \overline{B_1D_1}$, and $\overline{A_2C_2} \cap \overline{B_2D_2}$ are collinear.

Solved with Kanav Talwar! Here is a completely synthetic solution. We change labelling, rename $P,P_1,P_2$ as $R,R_1,R_2$. Let $M$ be the spiral center sending $ABCD$ to $A_1B_1C_1D_1$. Hence $\angle MAA_1 = \angle MBB_1 =\angle MCC_1 =\angle MDD_1$. But as $A_1 \in AB$, we get $\angle MAA_1 = \angle MAB$ and so on. Similarly, let $N$ be the spiral center of $ABCD$ to $A_2B_2C_2D_2$. Using $A_2 \in AD$ and so on, we get the following relations : $$\angle MAB = \angle MBC =\angle MCD =\angle MDA = \alpha , \quad \angle NAD = \angle NDC =\angle NCB =\angle NBA =\beta \dots [\diamondsuit]$$ This following claim will be used in angle chases throughout the proof : Claim 1 : $M,N$ lie inside $ABCD$. Proof : If not, then wlog then let $M$ lie on opposite side of line $AB$ as that of $C,D$. After spiral sim about $M$, $A$ goes to $A_1$, which lies on segment $AB$; and $C$ goes to $C_1$, which lies on segment $CD$. But $MAA1$ and $MCC1$ are directly similar, which is clearly not possible. $\square$ Let $P = AD\cap BC$ and $Q=AB\cap CD$. Claim 2 : $(PBD)$ and $(QAC)$ intersect at $M$ and second time on $PQ$. Similarly, $(QBD)$ and $(PAC)$ intersect at $N$ and second time on $PQ$. Proof : The relations in $[\diamondsuit]$ directly imply that $PMBD$, $QMAC$, $PNAC$, $QNBD$ are cyclic. Now define $M_1 = (PMBD) \cap (QMAC) \neq M$ and $N_1 = (PNAC) \cap (QNBD) \neq N$. Angle chase to get $\angle PM1M = \angle PDM = \angle MCQ = 180 -\angle MM_1Q$. This implies $M_1$ lies on line $PQ$. Similarly, we get for $N_1$, proving the claim. $\square$ Claim 3 : $ABCD$ is cyclic quadrilateral. Proof : Let $K$ be the miquel point of $ABCD$. We perform miquel inversion (radius $\sqrt{KA\cdot KC}$ and then reflect about angle bisector of $\angle AKC$). This swaps the following pairs : $(A,C)$, $(B,D)$, $(P,Q)$. And indeed, $(PAC) \cap (QBD)$ go to $(QAC) \cap (PBD)$. Hence using the last claim : ${M,M_1} \mapsto {N,N_1}$. There are two cases :

Hence, we are done with the claim! $\square$ Recollect we defined $R = AC \cap BD$. Claim 4 : $\angle MRN = 180 - \alpha - \beta$. Proof : By radax on $(ABCD),(PBDMM_1),(QACMM_1)$ we infer $R$ lies on $MM_1$. Similarly, $R$ lies on $NN_1$ too. Hence, \begin{align*} \angle MRN &= \angle M_1RN_1 \\ &= 180 - \angle N_1M_1R - \angle M_1N_1R \\ &= 180 - \angle PM_1M - \angle QN_1N \\ &= 180 - \angle PDM - \angle QBN \\ &= 180 - \alpha -\beta \hspace{3cm} \square \\ \end{align*} To finish, Now we go back to the problem. By spiral sim at $M$, $MRR_1 \sim MAA_1$. Hence, $\angle MRR_1 = \angle MAA_1 = \alpha$. Similarly $\angle NRR_2 = \beta$. As $\angle MRN = 180 - \alpha - \beta$, we indeed get $R,R_1,R_2 $ are collinear, and we are done!! Note : If $AB \parallel CD$ (when $Q$ doesnt lie on real plane); then it's easy to check that $M$ lies on $AC$ and $N$ lies on $BD$. Hence we directly get $\angle MRN$ and then same finish as above works.

Solved with BlizzardWizard Let $Z_i$ be the center of spiral similarity of $ABCD$ and $A_iB_iC_iD_i$. Then, $\angle Z_1AB=\angle Z_1BC=\angle Z_1CD=\angle Z_1DA$ and $\angle Z_2AD=\angle Z_2DC=\angle Z_2CB=\angle Z_2BA$. Then, using complex numbers, if $k$ is some number with magnitude $1$, then we get \begin{align*} \frac{z_1-a}{b-a}&=k\frac{\overline{z_1}-\overline a}{\overline b-\overline a}\\ z_1(\overline b-\overline a)-k\overline{z_1}(b-a)&=a(\overline b-\overline a)-k\overline a(b-a). \end{align*} Adding all cyclic variants gives $$a(\overline b-\overline a)+b(\overline c-\overline b)+c(\overline d-\overline c)+d(\overline a-\overline d)=k(\overline a(b-a)+\overline b(c-b)+\overline c(d-c)+\overline d(a-d)).$$ We claim that both sides cannot be equal for all $k$. If they are equal, then $a\overline b+b\overline c+c\overline d+d\overline a=|a|^2+|b|^2+|c|^2+|d|^2$ which is a contradiction by the Rearrangement Inequality and the Triangle Inequality since $a$, $b$, $c$ and $d$ are distinct. Therefore, there exists a unique value of $k$. Similarly, for $z_2$, the value of $k$ is the conjugate of the value of $k$ for $z_1$, so this implies that $$\angle Z_1AB=\angle Z_1BC=\angle Z_1CD=\angle Z_1DA=\angle Z_2AD=\angle Z_2DC=\angle Z_2CB=\angle Z_2BA.$$ Therefore, we have \begin{align*} z_1(\overline b-\overline a)-k\overline{z_1}(b-a)&=a(\overline b-\overline a)-k\overline a(b-a)\\ kz_2(\overline a-\overline b)-\overline{z_2}(a-b)&=kb(\overline a-\overline b)-\overline b(a-b)\\ (\overline b-\overline a)(z_1-kz_2)-(b-a)(k\overline{z_1}-\overline{z_2})&=(b\overline b-a\overline a)(1-k). \end{align*} Now, suppose $|a|=|b|=|c|=1$. Then, we get $$\frac{a-b}{ab}(z_1-kz_2)-(b-a)(k\overline{z_1}-\overline{z_2})=0.$$Therefore, if $x=z_1-kz_2$, we get $x+abk\overline x=0$. Similarly, $x+bck\overline x=0$, so $x=0$. Therefore, $z_1=kz_2$, which implies $\angle Z_1OZ_2=2\angle Z_1AB$ if $O$ is the circumcenter of $ABC$. Then, we get $|a|^2=|b|^2=|c|^2=|d|^2=1$ since $k\neq1$, which means $ABCD$ is cyclic. Now, we have \begin{align*} z_1+abk\overline{z_1}&=a+bk\\ (c-d)z_1+ab(c-d)k\overline{z_1}&=a(c-d)+b(c-d)k. \end{align*}Taking the cyclic sum gives $(2ac+2bd-(a+c)(b+d))(1-k)=0$, so since $k\neq1$, we get $(a+c)(b+d)=2ac+2bd$, which implies $(a-b)(c-d)+(b-c)(d-a)=0$. Therefore, $ABCD$ is harmonic. Every harmonic quadrilateral can be inverted to a square, so we can complex bash with $a=\frac1{s+1}$, $b=\frac1{s+i}$, $c=\frac1{s-1}$, and $d=\frac1{s-i}$. We now compute $l$. Let $l=\frac1{s+r}$. By inversion, points $0$, $s+r$, $s+1$, and $s-1$ are concyclic, as are $0$, $s+r$, $s+i$, and $s-i$. Thus, $\frac{(r+1)(s+1)}{(r-1)(s-1)}$ and $\frac{(r+i)(s+i)}{(r-i)(s-i)}$ are real. By inspection, $r=\frac1{\overline s}$ is a solution. Then $l=\frac1{s+r}=\frac{\overline s}{s\overline s+1}$. Now, we compute $z_1$. Let $z_1=\frac1{s+u}$. Since $\angle Z_1AB=\angle Z_1BC=\angle Z_1CD=\angle Z_1DA$, the following expression is equal over $t\in\{1,i,-1,-i\}$: \[\frac{\frac{z_1-\frac1{s+t}}{\frac1{s+it}-\frac1{s+t}}}{\overline{\left(\frac{z_1-\frac1{s+t}}{\frac1{s+it}-\frac1{s+t}}\right)}}\]This equals \[\frac{\frac{(s+it)(t-u)}{(s+u)(t)(1-i)}}{\overline{\left(\frac{(s+it)(t-u)}{(s+u)(t)(1-i)}\right)}}\]Discarding factors that do not depend on $t$, we have that the following is still equal over $t\in\{1,i,-1,-i\}$: \[\frac{\frac{(s+it)(t-u)}{t}}{\overline{\left(\frac{(s+it)(t-u)}{t}\right)}}\]Simplifying, this equals \[\frac{(s+it)(t-u)}{(\overline st-i)(1-t\overline u)}\]By expanding an comparing coefficients of $t$, we can compute that the above is constant in $t$ when $u=\frac i{\overline s}$. This can be easily confirmed by plugging in to the factorization. Now, we have $z_1=\frac1{s+u}=\frac{\overline s}{s\overline s+i}$. Similarly, we have $z_2=\frac{\overline s}{s\overline s-i}$. Letting $L_1=A_1C_1\cap B_1D_1$ and $L_2$, we have that $\angle Z_1AB=\angle Z_1LL_1$, so $\frac{z_1-a}{b-a}=\frac{z_1-l}{l_1-l}$ is real. Similarly, $\frac{z_2-b}{a-b}=\frac{z_2-l}{l_2-l}$ is real. Since the goal is to prove that $\frac{l_1-l}{l_2-l}$ is real, it suffices to show that $\frac{(z_1-a)(z_2-l)}{(z_2-b)(z_1-l)}$ is real. $\frac{(z_1-a)(z_2-l)}{(z_2-b)(z_1-l)}$ $=\frac{(\frac{\overline s}{s\overline s+i}-\frac1{s+1})(\frac{\overline s}{s\overline s-i}-\frac{\overline s}{s\overline s+1})}{(\frac{\overline s}{s\overline s-i}-\frac1{s+i})(\frac{\overline s}{s\overline s+i}-\frac{\overline s}{s\overline s+1})}$ $=\frac{(\frac{\overline s-i}{(s\overline s+i)(s+1)})(\frac1{s\overline s-i}-\frac1{s\overline s+1})}{(\frac{i\overline s+i}{(s\overline s-i)(s+i)})(\frac1{s\overline s+i}-\frac1{s\overline s+1})}$ $=\frac{(\frac{\overline s-i}{(s\overline s+i)(s+1)})(\frac{1+i}{(s\overline s-i)(s\overline s+1)})}{(\frac{i\overline s+i}{(s\overline s-i)(s+i)})(\frac{i-1}{(s\overline s+i)(s\overline s+1)})}$ $=\frac{(\frac{\overline s-i}{s+1})}{(\frac{\overline s+1}{s+i})}$ $=\frac{(\overline s-i)(s+i)}{(s+1)(\overline s+1)}$, which is equal to its conjugate, by inspection. This finishes the proof.

Man, this was a brutal question. As far as I know, there were only three positive scores (3/4/7) recorded by the 38 first-time US MOPpers on this problem. I used the Gliding Principle to reduce this problem to two special cases of $A_1B_1C_1D_1$ and $A_2B_2C_2D_2$. As it turns out, the incompleted Trig bash I submitted works out if $ABCD$ is cyclic, so one of my team leaders was able to coordinate my score up to a 4. I hope to include an outline of my approach on this post in the near future.

Sketch: Take the spiral similar centers and use knowledge of Brocard points.

In fact, after taking Brocard points, we can use the exactly same way as IMO2018/6 to conclude. Note that the two Brocard points are isolete conjugates (so chasing angles shows concyclic) so inverting through one we get a same diagram as the second, and edge bashes just like that problem shows harmonic.

Let \(X\) be the center of spiral similarity between \(ABCD\) and \(A_1B_1C_1D_1\), and let \(Y\) be the center of spiral similarity between \(ABCD\) and \(A_2B_2C_2D_2\). Let \(\theta:=\measuredangle XAB=\measuredangle XBC=\measuredangle XCD=\measuredangle XDA\) and \(\theta':=\measuredangle ABY=\measuredangle BCY=\measuredangle CDY=\measuredangle DAY\). [asy][asy] import olympiad; pen pri=heavygreen; pen pri2=heavygreen+linewidth(.3); pen sec=heavycyan; pen sec2=heavycyan+linewidth(.3); pen tri=deepcyan; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; real theta = 36; path omega = circle((0, 0), 1); pair O = (0, 0); pair P = (sqrt(1 - 2 * sin(theta * pi/180)^2), 0); pair A = dir(196.6); pair B = intersectionpoints(A -- A+(P-A)*dir(theta)*10, omega)[0]; pair C = intersectionpoints(B -- B+(P-B)*dir(theta)*10, omega)[1]; pair D = intersectionpoints(C -- C+(P-C)*dir(theta)*10, omega)[1]; pair Q = extension(A, A+dir(D-A)*dir(36), C, C+dir(B-C)*dir(36)); pair A1 = 3/5*A + 2/5*B; pair B1 = (B - P) * (A1 - P) / (A - P) + P; pair C1 = (C - P) * (A1 - P) / (A - P) + P; pair D1 = (D - P) * (A1 - P) / (A - P) + P; pair A2 = 4/5*A + 1/5*D; pair B2 = (B - Q) * (A2 - Q) / (A - Q) + Q; pair C2 = (C - Q) * (A2 - Q) / (A - Q) + Q; pair D2 = (D - Q) * (A2 - Q) / (A - Q) + Q; pair K = extension(A, P, B, Q); pair L = extension(B, P, C, Q); pair M = extension(C, P, D, Q); pair N = extension(D, P, A, Q); pair X = extension(A, C, B, D); pair X1 = extension(A1, C1, B1, D1); pair X2 = extension(A2, C2, B2, D2); size(8cm); defaultpen(fontsize(10pt)); draw(A -- C,pri2); draw(B -- D, pri2); draw(A1 -- C1,sec2); draw(B1 -- D1, sec2); draw(A2 -- C2,sec2); draw(B2 -- D2, sec2); draw(4/3*X1 - 1/3*X2 -- 4/3*X2 - 1/3*X1, dashed); filldraw(A1 -- B1 -- C1 -- D1 -- cycle,sfil,sec+linewidth(.8)); filldraw(A2 -- B2 -- C2 -- D2 -- cycle,sfil,sec+linewidth(.8)); filldraw(A -- B -- C -- D -- cycle, fil,pri+linewidth(1)); dot("\(A\)", A, dir(A)); dot("\(B\)", B, dir(B)); dot("\(C\)", C, dir(C)); dot("\(D\)", D, dir(D)); dot("\(A_1\)", A1, dir(A+B)); dot("\(B_1\)", B1, dir(60)); dot("\(C_1\)", C1, dir(C+D)); dot("\(D_1\)", D1, dir(D+A)); dot("\(A_2\)", A2, dir(D+A)); dot("\(B_2\)", B2, dir(A+B)); dot("\(C_2\)", C2, dir(30)); dot("\(D_2\)", D2, dir(C+D)); dot(X); dot(X1); dot(X2); [/asy][/asy] Claim: \(\theta=\theta'\); i.e.\ \(X\) and \(Y\) are isogonal conjugates. Proof. Assume for contradiction \(\theta<\theta'\) (without loss of generality). Then by the law of sines in \(\triangle XDA\) and \(\triangle YAB\), we have \[\frac{XD}{XA}=\frac{\sin(\angle A-\theta)}{\sin\theta} >\frac{\sin(\angle A-\theta')}{\sin\theta'}=\frac{YB}{YA}.\]Multiplying cyclically gives \[1=\prod_\mathrm{cyc}\frac{XD}{XA}>\prod_\mathrm{cyc}\frac{YB}{YA}=1,\]contradiction. \(\blacksquare\) [asy][asy] import olympiad; real theta = 36; path omega = circle((0, 0), 1); pair O = (0, 0); pair P = (sqrt(1 - 2 * sin(theta * pi/180)^2), 0); pair A = dir(196.6); pair B = intersectionpoints(A -- A+(P-A)*dir(theta)*10, omega)[0]; pair C = intersectionpoints(B -- B+(P-B)*dir(theta)*10, omega)[1]; pair D = intersectionpoints(C -- C+(P-C)*dir(theta)*10, omega)[1]; pair Q = extension(A, A+dir(D-A)*dir(36), C, C+dir(B-C)*dir(36)); pair A1 = 3/5*A + 2/5*B; pair B1 = (B - P) * (A1 - P) / (A - P) + P; pair C1 = (C - P) * (A1 - P) / (A - P) + P; pair D1 = (D - P) * (A1 - P) / (A - P) + P; pair A2 = 4/5*A + 1/5*D; pair B2 = (B - Q) * (A2 - Q) / (A - Q) + Q; pair C2 = (C - Q) * (A2 - Q) / (A - Q) + Q; pair D2 = (D - Q) * (A2 - Q) / (A - Q) + Q; pair K = extension(A, P, B, Q); pair L = extension(B, P, C, Q); pair M = extension(C, P, D, Q); pair N = extension(D, P, A, Q); pair X = extension(A, C, B, D); pair X1 = extension(A1, C1, B1, D1); pair X2 = extension(A2, C2, B2, D2); pen pri=heavygreen; pen sec=heavycyan; pen tri=deepcyan; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair M=(A+C)/2; pair NN=(B+D)/2; size(7cm); defaultpen(fontsize(10pt)); draw(A--P--B--Q--C--P--D--Q--cycle,sec+linewidth(.4)); filldraw(circumcircle(X,M,NN),tfil,tri+dashed); filldraw(A--B--C--D--cycle,fil,pri); draw(A--C,pri+linewidth(.4)); draw(B--D,pri+linewidth(.4)); dot("\(A\)",A,SW); dot("\(B\)",B,dir(90)); dot("\(C\)",C,NE); dot("\(D\)",D,SE); dot("\(X\)",P,S); dot("\(Y\)",Q,NW); dot("\(M\)",M,NW); dot("\(N\)",NN,E); dot("\(P\)",X,dir(75)); [/asy][/asy] Claim: \(ABCD\) is cyclic. Proof. Note \[\measuredangle AXD=\measuredangle XAD+\measuredangle ADX=\measuredangle XAD+\measuredangle BAX=\measuredangle BAD\]and similarly \(\measuredangle BXC=\measuredangle BCD\). Since \(X\) has an isogonal conjugate, \[\measuredangle BAD=\measuredangle AXD=\measuredangle BXC=\measuredangle BCD.\]\(\blacksquare\) Claim: \(ABCD\) is harmonic. Proof. Observe that \begin{align*} \triangle AXD\sim\triangle AYB &\implies\frac{YA}{XA}=\frac{AB}{AD}\\[1.5ex] \triangle BXA\sim\triangle BYC &\implies\frac{YC}{XA}=\frac{BC}{AB}\\[1.5ex] \triangle CXB\sim\triangle CYD &\implies\frac{YC}{XC}=\frac{CD}{BC}\\[1.5ex] \triangle DXC\sim\triangle DYA &\implies\frac{YA}{XC}=\frac{AD}{BC}. \end{align*}Hence \[1=\frac{YA}{XA}\cdot\frac{XC}{YA}\cdot\frac{YC}{XC}\cdot\frac{XA}{YC} =\frac{AB}{AD}\cdot\frac{BC}{AD}\cdot\frac{CD}{BC}\cdot\frac{AB}{BC},\]implying \(AB\cdot CD=AD\cdot BC\). \(\blacksquare\) Now let \(P=\overline{AC}\cap\overline{BD}\), \(P_1=\overline{A_1C_1}\cap\overline{B_1D_1}\), \(P_2=\overline{A_2C_2}\cap\overline{B_2D_2}\). Let \(M\) be the midpoint of \(\overline{AC}\) and \(N\) the midpoint of \(\overline{BD}\). Claim: \(X\) lies on \((ABN)\) and \((CDN)\); similarly \(Y\) lies on \((ADN)\) and \((BCN)\). Proof. The first follows from \(\measuredangle ANB=\measuredangle ADC=\measuredangle AXB\), and the rest follow analogously. \(\blacksquare\) Claim: \(P\), \(X\), \(Y\), \(M\), \(N\) are concyclic. Proof. We have \[\measuredangle XNP=\measuredangle XAB=\measuredangle XBC=\measuredangle XMP,\]implying \(X\in(PMN)\), and similarly \(Y\in(PMN)\). \(\blacksquare\) Finally \[\measuredangle XPY=\measuredangle XNY=\measuredangle XNA+\measuredangle ANY=\measuredangle XBA+\measuredangle ADY=2\theta,\]so \[\measuredangle P_1PP_2=\measuredangle P_1PX+\measuredangle XPY+\measuredangle YPP_2 =(-\theta)+2\theta+(-\theta)=0^\circ,\]as desired.

ike.chen wrote: Man, this was a brutal question. As far as I know, there were only three positive scores (3/4/7) recorded by the 38 first-time US MOPpers on this problem. ApraTrip in-contest orzes geo moment

Here is a partial proof, I have managed to prove that over all quadrilaterals $A_1B_1C_1D_1$, $P_1$ lies of a fixed line, basically I have proved that $\angle (P_1P, BD)$ remains fixed... I do not know spiral similarity or projective geo so trig was the only approach I could think of You can look at the diagram below for reference... Right now, forget about The third quadrilateral and only focus on first two. First some preliminary stuff, Because $ABCD \sim A_1B_1C_1D_1$, the angles between any two corresponding lines in the figure is a constant angle, say $\theta$. (For example $\angle(AB,A_1B_1)=\angle(AC,A_1C_1)=\theta$ and so on) It suffices to prove that $\angle P_1PD$ is independent of $\theta$ Let $X= AC \cap A_1C_1$ and $Y = BD \cap B_1D_1$ Because $\angle (BD,B_1D_1) = \theta = \angle (AC,A_1C_1)$, $PP_1XY$ is cyclic. let angles of quadrilateral be $A,B,C,D$. Also let $F=ABCD$ and $F_1=A_1B_1C_1D_1$ (the figure), let similarity ratio be $r$, like any side of $F_1$ is $r$ times the corresponding side of $F$ Now if the problem statement is true, like if such a $F_1$ exists, we have: $\frac{AA_1}{\sin \theta} = \frac{A_1D_1}{\sin A}$ Also, $\frac{A_1B}{\sin (B+\theta)} = \frac{A_1B_1}{\sin B}$ This implies that (measuring $AB$ in two ways) $1=r(\sin \theta \cdot[\frac{AD}{AB \cdot \sin A} + \cot B] + \cos \theta)$ Now doing the same for the other sides and equating these we get that all cyclic variants of $\frac{AD}{AB \cdot \sin A} + \cot B$ must be equal (say they are all equal to $t$)and if they are equal, we will get a new $F_1$ for any $\theta$. So basically we have $1=r(\cos \theta + t \sin \theta)$ so given $\theta$ we have a unique $F_1$ and $F_1$ is fully characterised by $\theta$, that is given $\theta$ we can also find $r$. Now comes the main part, to show $\angle P_1PD$ is independent of $\theta$ Focusing on $\Delta PP_1Y$ $\frac{P_1Y}{\sin \angle P_1PD} = \frac{PY}{\sin (\angle P_1PD+\theta)}$ This tells us (after expanding the $\sin$ term) that $\cot \theta = \frac{PY - P_1Y \cdot \cos \theta}{P_1Y \sin \theta}$ Now, here is I guess the kinda smart part of this solution: $PY=PD - DY$ $P_1Y = P_1D - D_1Y = r \cdot PD - D_1Y$ Also note that $DD_1YC_1$ is cyclic because on corresponding angles of $BD$ and $CD$ So, noting that $C_1D_1=r\cdot CD$ $\frac{DY}{\sin(\angle ADB + \theta)} = \frac{r \cdot CD}{\sin D} = \frac{D_1Y}{\sin \angle ADB} = w$ (say) Let's focus on the numerator of $\cot \theta$ for now, $PD(1-r\cos \theta) + (D_1Y \cdot \cos \theta - DY)$ The first bracket is simple, because $1=r(\cos \theta + t\sin \theta)$ so it is just $r \sin\theta \cdot t \cdot PD$ Now the second bracket is a little more involved, using the equation three lines above this, and expanding the $\sin$ term we get: $-w(\cos \angle ADB \cdot \sin \theta)$ Now putting these together and looking back at $\cot \theta$ we see that $\sin \theta$ term cancels out Also note that $w$ also has a factor of $r$ so cancelling $r$ from the numerator we get something completely independent of $\theta$ in the numerator and in the denominator we get $\frac{P_1D_1}r -\frac{D_1Y}r$ and the first term is just $PD$ and the second term also we can see is independent of $\theta$. To write it out we see: $\cot \theta = \frac{t \cdot PD - \frac{CD \cdot \cos \angle ADB}{\sin D}}{PD - \frac{CD \cdot \sin \angle ADB}{\sin D}}$ We can do a similar thing for $F_2$ ie. $A_2B_2C_2D_2$ (You can look at the attached PDF where I have done it) But to prove these two quantities are equal is very tough, I couldn't trig bash it enough (last page of pdf), but at least this has reduced this problem to a single case of $P_1$ and $P_2$, i will update this answer if I can do that (I think the case when $B_2D_2 \parallel AD$ and $B_1D_1 \parallel AB$ might be a doable case but I'm not sure, I will update this if I can...) btw I thought I would make zero progress so I am surprised I could even do this much on this problem, Also I wrote this whole long post with a broken thumb lol Enjoy the journey : )

This is actually easy if you know some things about brocard points. Let $M_1,M_2$ denote the center of spiral similarity taking $A_1B_1C_1D_1 \rightarrow ABCD$, similarly define the other. Claim 1:$ABCD$ is cyclic quadrilateral.

Claim 2: $M_1,M_2$ are conjugates in $ABCD$

Claim 3: $ABCD$ is harmonic

CLaim 4: If $P = AC \cap BD,$ M is midpoint OF $AC$, $N$ is of $BD$ then $PMNM_1M_2$ is cyclic

The rest of the proof is also simple angle chase.

a_n wrote: Here is a partial proof, I have managed to prove that over all quadrilaterals $A_1B_1C_1D_1$, $P_1$ lies of a fixed line, basically I have proved that $\angle (P_1P, BD)$ remains fixed... I do not know spiral similarity or projective geo so trig was the only approach I could think of You can look at the diagram below for reference... Right now, forget about The third quadrilateral and only focus on first two. First some preliminary stuff, Because $ABCD \sim A_1B_1C_1D_1$, the angles between any two corresponding lines in the figure is a constant angle, say $\theta$. (For example $\angle(AB,A_1B_1)=\angle(AC,A_1C_1)=\theta$ and so on) It suffices to prove that $\angle P_1PD$ is independent of $\theta$ Let $X= AC \cap A_1C_1$ and $Y = BD \cap B_1D_1$ Because $\angle (BD,B_1D_1) = \theta = \angle (AC,A_1C_1)$, $PP_1XY$ is cyclic. let angles of quadrilateral be $A,B,C,D$. Also let $F=ABCD$ and $F_1=A_1B_1C_1D_1$ (the figure), let similarity ratio be $r$, like any side of $F_1$ is $r$ times the corresponding side of $F$ Now if the problem statement is true, like if such a $F_1$ exists, we have: $\frac{AA_1}{\sin \theta} = \frac{A_1D_1}{\sin A}$ Also, $\frac{A_1B}{\sin (B+\theta)} = \frac{A_1B_1}{\sin B}$ This implies that (measuring $AB$ in two ways) $1=r(\sin \theta \cdot[\frac{AD}{AB \cdot \sin A} + \cot B] + \cos \theta)$ Now doing the same for the other sides and equating these we get that all cyclic variants of $\frac{AD}{AB \cdot \sin A} + \cot B$ must be equal (say they are all equal to $t$)and if they are equal, we will get a new $F_1$ for any $\theta$. So basically we have $1=r(\cos \theta + t \sin \theta)$ so given $\theta$ we have a unique $F_1$ and $F_1$ is fully characterised by $\theta$, that is given $\theta$ we can also find $r$. Now comes the main part, to show $\angle P_1PD$ is independent of $\theta$ Focusing on $\Delta PP_1Y$ $\frac{P_1Y}{\sin \angle P_1PD} = \frac{PY}{\sin (\angle P_1PD+\theta)}$ This tells us (after expanding the $\sin$ term) that $\cot \theta = \frac{PY - P_1Y \cdot \cos \theta}{P_1Y \sin \theta}$ Now, here is I guess the kinda smart part of this solution: $PY=PD - DY$ $P_1Y = P_1D - D_1Y = r \cdot PD - D_1Y$ Also note that $DD_1YC_1$ is cyclic because on corresponding angles of $BD$ and $CD$ So, noting that $C_1D_1=r\cdot CD$ $\frac{DY}{\sin(\angle ADB + \theta)} = \frac{r \cdot CD}{\sin D} = \frac{D_1Y}{\sin \angle ADB} = w$ (say) Let's focus on the numerator of $\cot \theta$ for now, $PD(1-r\cos \theta) + (D_1Y \cdot \cos \theta - DY)$ The first bracket is simple, because $1=r(\cos \theta + t\sin \theta)$ so it is just $r \sin\theta \cdot t \cdot PD$ Now the second bracket is a little more involved, using the equation three lines above this, and expanding the $\sin$ term we get: $-w(\cos \angle ADB \cdot \sin \theta)$ Now putting these together and looking back at $\cot \theta$ we see that $\sin \theta$ term cancels out Also note that $w$ also has a factor of $r$ so cancelling $r$ from the numerator we get something completely independent of $\theta$ in the numerator and in the denominator we get $\frac{P_1D_1}r -\frac{D_1Y}r$ and the first term is just $PD$ and the second term also we can see is independent of $\theta$. To write it out we see: $\cot \theta = \frac{t \cdot PD - \frac{CD \cdot \cos \angle ADB}{\sin D}}{PD - \frac{CD \cdot \sin \angle ADB}{\sin D}}$ We can do a similar thing for $F_2$ ie. $A_2B_2C_2D_2$ (You can look at the attached PDF where I have done it) But to prove these two quantities are equal is very tough, I couldn't trig bash it enough (last page of pdf), but at least this has reduced this problem to a single case of $P_1$ and $P_2$, i will update this answer if I can do that (I think the case when $B_2D_2 \parallel AD$ and $B_1D_1 \parallel AB$ might be a doable case but I'm not sure, I will update this if I can...) btw I thought I would make zero progress so I am surprised I could even do this much on this problem, Also I wrote this whole long post with a broken thumb lol Enjoy the journey : ) this deserves a bump