Here is a story. I came up with this problem on 2018-01-21 while exploring the configuration of four circles externally tangent cyclically (i.e. each circle is externally tangent to the previous and next circles), all internally tangent to a big circle. (You can extract this configuration from the problem fairly easily, once you know to look for it.) The next day, I proposed the problem to USA TSTST 2018, but it was not shortlisted, likely due to it being too hard. Then I decided to send the problem to USA TST 2019, where it was shortlisted. It was extremely well-liked, but there was a general sentiment that it was too difficult for the exam. Some people expected nobody to solve it (does that sound familiar?) and it was not selected for the exam. I wasn't really sure what to do with the problem now, so I sent it to RMM 2019. It was not selected for the contest and was placed on the extras list. We contacted the RMM committee to get the problem removed, since I wanted to use it on a future contest, and we got the okay to do so. That was the end of this sequence of events, I thought... little did I know how wrong I was. Soon the 2019-2020 contest cycle arrived, and I resubmitted this problem for USA TST 2020. It was once again well-liked (though not as much as the previous year), and people once again were worried it was too hard. People thought the problem that ended up being USA TST 2020/6 was easier (!) and it was more highly praised, so that problem was chosen instead. Then I decided to send the problem to USAMO 2020. During the review process, this problem ended up being posted on AoPS for a few minutes before it was deleted, but of course this was enough to kill any chances of the problem appearing on USAMO. I decided then that I wanted to use the problem on a practice test for MOP 2020, since it was not usable for any high-stakes contest. Then it got posted again. So much for that idea. The final nail in the coffin: apparently this problem was publicly visible since June 2019. Amazing. I've attached my submission for RMM 2019. Maybe it'll be of interest to some of you.

Another delightful configuration discovered by Ankan! Here is a solution that used Monge twelve times. First, WLOG, assume that $PA_1=PB_1=PC_1=PD_1$ are the semiperimeter of $\triangle PAB$, $\triangle PBC$, $\triangle PCD$, $\triangle PDA$. Naturally, this induces excircles $\omega_{AB}$, $\omega_{BC}$, $\omega_{CD}$, $\omega_{DA}$ of $\triangle PAB$, $\triangle PBC$, $\triangle PCD$, $\triangle PDA$, respectively. Let $W, X, Y, Z$ be the centers of these circles, respectively and let $W_1, X_1, Y_1, Z_1$ be the remaining contact points of those excircles. Claim: [Elimination of the incircle] $W_1, X_1, Y_1, Z_1$ are concyclic with center $I$. Proof $AW_1 = AA_1 = AZ_1$, hence $AI$ is the perpendicular bisector of $W_1Z_1$ or $IW_1=IZ_1$. The same argument for the remaining three vertices give the conclusion. $\blacksquare$ Claim: [Mass Monge] Six lines $WX$, $YZ$, $W_1X_1$, $Y_1Z_1$, $AC$, $A_1C_1$ are concurrent at $U$. Furthermore, this concurrency point coincides with the exsimilicenters of $(\omega_{AB}, \omega_{BC})$ and $(\omega_{CD}, \omega_{DA})$. Proof: First, we show that $AC\cap A_1C_1$ is the exsimilicenter $U$ of $(\omega_{AB}, \omega_{BC})$. This proceeds in two steps. Since $PA_1=PC_1$, there exists a circle $\gamma_1$ tangent to $PA_1, PC_1$ at $A_1, C_1$. By Monge's theorem on $\omega_{AB}, \omega_{BC},\gamma_1$, we get that $U\in A_1C_1$. Since $PA+AB = PC+BC$, there exists a circle $\gamma_3$ tangent to rays $\overrightarrow{BA}$, $\overrightarrow{BC}$, $\overrightarrow{CP}$, $\overrightarrow{AP}$. Then we get $U\in AC$ from $\omega_{AB}, \omega_{BC}, \gamma_3$. Thus by repeating the similar argument on $D$ (thus using Monge for two more times), we get that $T$ is also the exsimilicenter of $(\omega_{CD}, \omega_{DA})$. Since exsimilicenter lies on a line joining two circles, we also deduce that $U=WX\cap YZ$. It remains to show that $U\in W_1X_1$ (thus $U\in Y_1Z_1$ follows similarly). To that end, we note that $BW_1=BX_1$ thus there exists a circle $\gamma_2$ tangent to $BW_1, BX_1$ at $W_1, X_1$. The result now follows by Monge's theorem on $\omega_{AB}, \omega_{BC}, \gamma_2$. $\blacksquare$ By using Monge's theorem six more times, we get that six lines $WZ$, $XY$, $W_1Z_1$, $X_1Y_1$, $BD$, $B_1D_1$ are concurrent at $V$. The main work has been done and we present one more claim to finish. From now, let $\omega=\odot(A_1B_1C_1D_1)$ and $\gamma = \odot(W_1X_1Y_1Z_1)$. Claim: $UV$ is the radical axis of $\omega$ and $\gamma$ Proof: Note that $UB_1, UD_1$ are tangents to $\omega$. Moreover, the circle $\odot(B_1W_1X_1)$ is centered at $B$, and thus tangent to $UB_1$. By power of point $$\mathrm{Pow}(U,\omega) = UA_1\cdot UC_1 = UB_1^2 = UW_1\cdot UX_1 = \mathrm{Pow}(U,\gamma)$$, thus $U$ lies on the radical axis. Similarly $V$ lies on the radical axis, hence the claim. $\blacksquare$ From the above claim, we get that $UV\perp IP$. However, by La Hire's theorem, $UV$ is the polar of $A_1C_1\cap B_1D_1$ w.r.t. $\omega$, hence the requested concurrency is proven. Remark: One interesting corollary from the above proof is $A_1B_1C_1D_1$ is a harmonic quadrilateral. This offers a way to construct the diagram.

Astounding problem; truly the pinnacle of geometry problems in this age of moving points everywhere. Probably the hardest geometry problem I've ever solved (and will) . Here goes my solution, presented in it's full glory: By homothety at $P$ , it suffices to consider the case when radius of $\Gamma$ is equal to the common semiperimeter of the triangles $\triangle PAB$, $\triangle PBC$, $\triangle PCD$ ,$\triangle PDA$ . Let $K \overset { \text {def}}{=} A_1C_1 \cap B_1D_1$ Denote by $X_{AB}$ , the point on $AB$ ,where the incircle of $ABCD$ is tangent to $AB$ . $X_{BC}$ , $X_{CD}$ , $X_{DA}$ are defined similarly. Denote by $\omega_{AB}$, $\omega_{BC}$, $\omega_{CD}$, $\omega_{DA}$, the $P$-excircles of $\triangle PAB$, $\triangle PBC$, $\triangle PCD$, $\triangle PDA$, respectively. Then $A_1, B_1, C_1$and $D_1$ are their tangency points with the rays $PA, PB, PC$ and $PD$ , so the excircles are cyclically tangent at those points. Let $T_{AB} \overset { \text {def}}{=} \omega_{AB} \cap AB $ . $T_{BC}$ , $T_{CD}$ ,$T_{DA}$ are defined similarly. Next draw circles $\Omega_A \equiv \odot (A, AA_1=AT_{AB}=AT_{DA})$ and similarly $\Omega_B , \Omega_C , \Omega_D$ .Note that $\{\Omega_A,\Omega_B\} , \{\Omega_A,\Omega_D\} , \{\Omega_C,\Omega_B\} , \{\Omega_C,\Omega_D\}$ are pairs of tangent circles . Also $\Omega_A,\Omega_B,\Omega_C,\Omega_D$ are all internally tangent to $\Gamma$ at $A_1,B_1,C_1,D_1$ respectively. Next note that we have $AX_{AB}=AX_{DA}$ and $AT_{AB}=AT_{DA}$ . Combining these relations , we conclude $X_{AB}T_{AB}= X_{DA}T_{DA}$ . This implies that $\Delta IX_{AB}T_{AB}$ and $\Delta IX_{DA}T_{DA}$ are congruent . So $IT_{AB}=IT_{DA}$ . Hence we come to the conclusion that $\{T_{AB},T_{BC},T_{CD},T_{DA}\}$ lie on a circle centered at $I$ . Call this circle $\gamma$ . Next , we define shift our attention to the cyclic quadrilateral $T_{AB}T_{BC}T_{CD}T_{DA}$ . Write $X \overset {\text {def}}{=} \overline{T_{AB}T_{DA}} \cap \overline{T_{BC} T_{CD}}$ and $Y \overset {\text {def}}{=} \overline{T_{DA}T_{CD}} \cap \overline{T_{BC}T_{AB}}$ . Next using Monge on the triples $\{\Omega_D,\Omega_C,\Omega_B\}$ , $\{\Omega_D,\Omega_A,\Omega_B\}$ and $\{\Omega_D,\Gamma,\Omega_B\}$ , we have that $B_1 ,D_1 , X$ are collinear and $X$ is the exsimilicenter of the circles $\{\Omega_D,\Omega_B\}$ . Similarly, $Y$ lies on $A_1C_1$ and is the exsimilicenter of $\{\Omega_A,\Omega_C\}$ . Next , we consider the inversion $\Phi (X, \sqrt{XT_{AB} \cdot XT_{DA}})$ . Let $\gamma_X$ be the circle of inversion. Note that this inversions swaps $\{\Omega_B,\Omega_D\}$ and $\{B_1,D_1\}$ and fixes $\gamma$ . Hence $\Phi$ also fixes $\Gamma , \Omega_A , \Omega _C $ and the tangency points of the last two circles with $\Gamma$ . Hence $C_1,A_1$ are fixed under this inversion and we conclude that $X \equiv \overline{C_1C_1} \cap \overline{A_1A_1} $ and $A_1 , C_1 \in \gamma_X$ Similarly defining $\gamma_Y$ , we get that $Y \equiv \overline{B_1B_1} \cap \overline{D_1D_1} $ and $B_1 , D_1 \in \gamma_Y$. Now we finish off by showing that $P,I,K$ lie on the radical axis of $\{\gamma_X , \gamma_Y\}$. Note that $P,I$ are centers of two circles ,namely $\Gamma$ and $\gamma$ which are orthogonal to both $\gamma_X$ and $\gamma_Y$. So $\overline {PI}$ is the radical axis of $\{\gamma_X , \gamma_Y\}$ . Lastly note that $$\operatorname {Pow}_{K,\gamma_X} = KA_1 \cdot KC_1 =KB_1 \cdot KD_1 = \operatorname {Pow}_{K,\gamma_Y}$$ So $K \in \overline {PI}$ . We are done . $\blacksquare$ Remarks : 1. I still can't figure out how to draw the diagram on geogebra . 2.Its no secret that $A_1B_1C_1D_1$ is a harmonic Quadrilateral . 3. It is easy to see that $X \in \overline {BD}$ , say $\{\Omega_B,\Omega_D\}$ is swapped under $\Phi$ . Similarly $Y \in \overline {AC}$ 4. Denote by $O_{AB}$ , the center of $\omega _{AB}$ . Define $O_{BC}$ ,$O_{CD}$ ,$O_{DA}$ similarly . Monge on $\{\omega_{BC},\omega_{CD},\gamma_Y\}$ gives that $$ X \equiv \overline {O_{AB}O_{DA}} \cap \overline {O_{BC}O_{CD}}$$. Also $X$ is the exsimilicenter of $\{\omega_{AB},\omega_{DA}\}$ and $\{\omega_{BC},\omega_{CD}\}$ . Note that $\Phi$ swaps the forementioned pairs of circles . Similar stuff holds for $Y$ . 5. $K$ is the insimilicenter of $\{\omega_{AB},\omega_{CD}\}$ and $\{\omega_{AD},\omega_{BC}\}$ Here is a short proof of this property by the proposer of the problem (CantonMathGuy) , communicated to me via PM . CantonMathGuy wrote: Let $\spadesuit$ be the insimilicenter of $\omega_{AB}$ and $\omega_{CD}$. Apply Monge on $(\omega_{AB}, \omega_{BC}, \omega_{CD})$ to get that exsimilicenter of $\omega_{AB}$, $\omega_{BC}$: $Y$ insimilicenter of $\omega_{BC}$, $\omega_{CD}$: $C_1$ insimilicenter of $\omega_{AB}$, $\omega_{CD}$: the point $\spadesuit$ you're interested in are collinear; since $Y$ lies on line $A_1C_1$ we get $A_1$, $C_1$, and $\spadesuit$ are collinear. Similarly $B_1$, $D_1$, $\spadesuit$ are collinear, so $\spadesuit = K$.

very nice problem, here is a solution without the use of Monge. First a few definitions: let $w_1, w_2, w_3, w_4$ be the P-excircles in triangles $PAB, PBC, PCD, PDA$ and let the tangency points of said circles with respect to lines $PA, PB, PC, PD$ be $A' ,B', C', D'$ and with respect to lines $AB, BC, CD, DA$ be $N, M, K ,L$. call the center of $w_i$, $O_i$. - $0)$ by the perimeter condition, it is clear that $PA'=PB'=PC'=PD'$ and circles $w_i$ and $w_{i+1}$ are tangent. we can without the loss of generality assume that $\Gamma$ is the circle centered at $P$ and passes through $A'$ - $1)$ note that $AL=AA'=AN$ thus $AI$ is the perpendicular bisector of $LN$. with a similar argument we conclude that $LNMK$ are cyclic with center $I$. call this $w$ - $2)$ Now, we create a Projective map on $A'P$ as such: let $X$ be a point on $A'P$. The tangent through $X$ on $w_1$ intersects $B'P$ at $X_1$. The tangent through $X_1$ on $w_2$ intersects $C'P$ at $X_2$. The tangent through $X_2$ on $w_3$ intersects $D'P$ at $X_3$. The tangent through $X_3$ on $w_4$ intersects $A'P$ at $f(X)$. this projective map has three Fixed points at $A', A, P$ thus it must map all points to themselves. - $3)$ Now, lets Move point $X_0$ on The mapping above, from point $A'$ to point $P$. construct $X_1, X_2, X_3$similar to the mapping and call the tangency point of the second tangent from $X_i$ to $w_{i+1}$, $Y_i$.we know that $Y_0Y_1Y_2Y_3$ is cyclic, call the second intersection of said circle with $w_{i+1}$, $Z_i$. now if $X_0 \rightarrow A'$ then the order of points on the arc $A'Y_0B'$ becomes $A'Y_0Z_0B'$ as $Y_0\rightarrow A'$ and $Z_0\rightarrow B'$ and when $X_0 \rightarrow P$ then the order of points on the arc $A'Y_0B'$ becomes $A'Z_0Y_0B'$ as as $Z_0\rightarrow A'$ and $Y_0\rightarrow B'$ . noting that $Y_0$ and $Z_0$ always stay on this arc, by the Mean value theorem, ther exists a point $X*$ such that $Z_0=Y_0$ thus by a simple angle chasing argument, $w$ becomes tangent to all four circles. call this configuration **. - $4)$ we prove a lemma: let $w_1, w_2, w_3, w_4$ be circles such that $w_i$ and $w_{i+1}$ are externally tangent in points $A,B,C,D$ respectively such that $ABCD$ is cyclic. if there is a circle such that it is externally tangent to all four circles, then $(AC,BD)=-1$. proof: invert around $D$, the four circles become two parallel lines with $A,C$ on each such that $AC$ is perpendicular to said lines, $B$ lies on the segment $AC$, with the other two circles becoming those of diameter $AB$ and $BC$. we know there is a circle tangent to these lines and two circles. if $AB>BC$ then the reflection of $w_1$ through the perpendicular bisector of $AC$ is also tangent to $w$ but it also completely contains $w_2$ , thus $w_2$ and $w$ cannot be tangent, unless $AB=BC$ which is the inverted equivalent to the lemma. - $5)$ Now, note that configuration ** fits in the lemma thus we conclude that $(A'C',B'D')=-1$. now we can throw away points created in $2)$ and $3)$ and revert our focus to the main points. call $S,T$ the intersections of $O_1O_4$ with $O_2O_3$ and $O_1O_2$ with $O_3O_4$. a straight conclusion of using pole-polar in circle $\Gamma$ gives us $A'B',C'D'$ intersect on $ST$, similarly $A'D',B'C'$. call the intersection of $A'C',B'D'$, $Q$. we know that $PQ$ is perpendicular to $ST$. Also by tangentially projecting $A'B'C'D'$ onto $O_1O_4$ we conclude that $(O_4O_1,A'S)=-1$. Finally, we finish the problem : intersect $O_1N,O_4L$ at point $R$. it is clear that $A'NL$ is the incircle of $O_1O_4R$ thus $NL$ passes through $S$ and, $SA'^2=SN\cdot SL$ so $S$ lies on the radical axis of $\Gamma$ and $w$. similarly for $T$. so $ST$ is the radical axis of $\Gamma, w$ and $P, I$ Being their centers we conclude that $PI$ is perpendicular to $ST$ but so was $PQ$. hence $P, Q, I$ are collinear. we are done.

very cool

Solved with ohiorizzler1434 but most of the solution below is his. I suggested the construction of the excircles of the quadrilaterals. By the way this is even more menacing than ISL 2015/G7. Firstly notice the conclusion is linear in the radius of the circumcircle of $(A_1B_1C_1D_1)$ so we just choose a value of the radius at the end and everything will work out (since radius $0$ forces everything to collapse to $P$). We are going to spam Monge on this question. Notice first that $ABCP$ has an excircle $\omega_1$ because $p(ABP)=p(BCP)$ immediately forces $AB+AP=BC+CP$, which is external Pitot. Similarly $BCDP$ with $\omega_2$, $CDAP$ with $\omega_3$ and $DABP$ with $\omega_4$. Construct the $P$-excircles of $\triangle ABP$, $\triangle BCP$, $\triangle CDP$, $\triangle DAP$ as $\Omega_1$ (touching $AB$ at $W$), $\Omega_2$ (touching $BC$ at $X$), $\Omega_3$ (touching $CD$ at $Y$), $\Omega_4$ (touching $DA$ at $Z$), and let the radius of $(A_1B_1C_1D_1)$ be the semiperimeter of $\triangle PAB$. We now claim that $\overline{B_1B_1}$ (tangent to $(A_1B_1C_1D_1)$ at $B_1$), $\overline{D_1D_1}$ (tangent to $(A_1B_1C_1D_1)$ at $D_1$), $\overline{XW}$, $\overline{YZ}$, $\overline{A_1C_1}$, $\overline{AC}$ concur at a point $Q$. Let $Q$ be the exsimilicentre of $\Omega_1$ and $\Omega_2$. Notice that $BW=BB_1=BX$ so there is a circle tangent to $AB$ and $BC$ at $W$ and $X$ resp. Call it $\gamma$. Also there is a circle tangent to $PA$ and $PC$ at $A$ and $C$ resp. because $PA=PC$; call it $\Gamma$. Notice that $Q\in\overline{XW}$ is by Monge (2-insimilicentre version) on $\gamma$, $\Omega_1$ and $\Omega_2$. $Q\in\overline{A_1C_1}$ because Monge (2-insimilicentre version) on $\Omega_1$, $\Omega_2$ and $\Gamma$. $Q\in\overline{A_1C_1}$ because Monge (normal version) on $\Omega_1$, $\Omega_2$ and $\omega_1$. Similarly $Q\in\overline{YZ}$ and $\overline{D_1D_1}$ by similar applications of Monge. Now notice that $(B_1WX)$ has centre $B$, so $\overline{B_1B_1}$ tangent to $XW$. Hence $QB_1^2=QX\cdot QW$. Now notice that the perpendicular bisector of $WX$ is $AI$ since $AI$ bisects $\angle WAX$ and $AW=AX$. Similarly we can deduce that the perpendicular bisectors of the sides of $WXYZ$ concur at $I$, so $WXYZ$ is cyclic with centre $I$. If we define $R$ to be the construction similar to $Q$ but sending $A\to C$, $B\to D$ and other associated replacements, then $\overline{QR}$ is then the radical axis of $(WXYZ)$ and $(A_1B_1C_1D_1)$ so it is perpendicular to $\overline{IP}$, forcing the pole of $\overline{QR}$ (which is just $\overline{A_1C_1}\cap\overline{B_1D_1}$) to be on $\overline{IP}$.