The point is the intersection of the perpendicular bisectors of $AC,BD$, I think. Let $T$ be the point described above. It's clear that $TA=TC,TB=TD$, and $\angle ATC=\angle BTD$, so the triangles $TAC,TDB$ are similar (and $TAC$ is the image of $TDB$ through a spiral similarity centered at $T$). It's also easy to see that $TFE$ is similar to these two triangles. Sorry if I skip some steps, but they all fall in the wide category of angle chasing . $\angle TFR=\angle TAR$, so $TAFR$ is cyclic, so $\angle TRA=\angle TFA$. In the exact same way we show that $\angle TQB=\angle TEB=\pi-\angle TEC=\pi-\angle TFA$ (the figures $TBCE,TDAF$ are congruent), and from these two relations we find that $\angle TRP+\angle TQP=\pi$, i.e. $T$ lies on all the circles $PQR$.

Oh, such easy geometry problem, still not commented by Darij. It may be solved by very standard methods. For proving that the circles which have one point have another we must prov that their centres lie on one line (which does not contain point $P$, what I am not going to check). Coordiantes of a centre in appropriate (not-Descartues, but skew) coordinate system is $(PR/2,PQ/2)$ (segments $PR,PQ$ are oriented of course). These are just projections of a vector $PO$ to diagonals of our initial quadrilateral. So, we are searching for a linear dependence between $PR$ and $PQ$, or, what is the same, a dependence between $CR$ and $DQ$. Let $BC$ intersect $DA$ in point $K$, wlog $B$ lies betwee $K$ and $C$, $A$ between $K$ and $D$. By Menelaus for $KAC$ we have $AR\cdot CE\cdot KF=RC\cdot EK\cdot AF$. Since $CE=AF$, we get $AR/RC=EK/KF$, $(AC-RC)/RC=EK/KF$, $AC/RC=1+EK/KF$, $RC=AC\cdot \frac{KF}{KF+EK}$. $KF+EK$ is constant and equals $KA+KB+BC$, so $RC$ is proportional to $KF$. Analogously, $DQ$ is proportioal to $KE$. But since $KE$ and $KF$ have fixed sum, it implies a linear relation for $CR$ and $DQ$.

First, BE/EC = DF/FA is sufficient. Let T be the center of the spiral similarity S that sends D to B and A to C. By a well-known lemma, T is the second point of intersection of the circumcircles of triangles APD and BPC. Clearly, S also sends F to E. Therefore, the circumcircles of triangles FQD and BEQ also pass through T. Now, because T lies on the circumcirles of triangles FQD and APD, by Miquel's theorem, it also lies on the circumcircles of triangles PQR and ARF. Therefore, the circumcircle of triangle PQR always passes through the second point of intersection of the circumcircles of triangles APD and BPC.

Similar solution... I'd like to say about something interesting...Here!A picture of this problem! Invalid image file Several circles go through the point(S in the picture) we want. I hope this could make someone think of a special solution... P.S. Has the friend upstair read the book Morden Geometry By R.A.Johnson?Such similar thought we had...exactly the same...

Nice problem ! My solution use vector and Simson's theorem : Let $ (APD) $ intersect $ (BPC) $ at $ K $ . We'll prove $ K \in (PQR) $ It's easy to show : $ KAC , KBD , KEF $ are isosceles triangles . Let $ T , U , V $ be the feet of the perpentdiculars from $ K $ to $ BD, EF, AC $ , respectively . So $ T, U , V $ are the midpoints of $ BD , EF , AC $ , respectively . Using vectors , we can easily show that : $ T, U ,V $ are collinear . By Simson's theorem , we have $ K \in (PQR) $ . Q.E.D

Dear Valentin , which country proposed this problem ?

As far as I know, this problem is from Poland, also problem 4.

I solved this problem this way: By three good diagrams, I guessed that the second intersection point $S$ is the intersection of the circumcircles of $APC$ and $BPD$. Well, check Skyward_Sea's diagram above. I just did this one (on Paint, so it's not really accurate) Invalid image file Let's prove that $S$ is indeed the desired point, that is, that $PQSR$ is a cyclic quadrilateral. Since $S$ does not depend on $E$ and $F$, $S$ is a fixed point, so if we manage to prove that $S$ belongs to the circumcircle of $PQR$ regardless of the positions of $E$ and $F$ we are done. Let $\angle CPS = \alpha$ and $\angle DPS = \beta$. Doing some angle chasing (look at the diagram!), we see that $ADS$ and $CBS$ are congruent triangles. Moreover, $ADS$ can be obtained from a rotation of $CBS$ with rotation center $S$. Notice that the image of $E$ in this rotation is $F$, so $\angle ESF = \angle BSD$ (both are the rotation angle). Thus, since $ESF$ and $BSD$ are isosceles, they are similar and, in particular, $\angle QFS = \angle EFS = \angle BDS = \angle QDS$. Thus $QFDS$ is a cyclic quadrilateral and hence $\angle RQS = \angle FDS = \alpha = \angle RPS$. So $PQRS$ is cyclic, as we wanted to prove. Oh, I just noticed that my solution is similar to grobber's... but I'll post it anyway.

A friend of mine sent me the following super fast and beatyfull solution, in just 4 lines. (In fact his solution generalizes this problem!! Can you see how ? ) Let M,N be the midpoints of the diagonals. We'll prove that the readical axis of the circles C1=(P,Q,R) and C2=(P,M,N) is constant line and so we are done. Particularly, we'll prove that this line cuts DC at L, which is a constant point. From Cayley's Theorem for these circles and the points D,C we have that : 1. $2 \cdot O_1 O_2 \cdot h_1 = DM \cdot DP - DQ\cdot DP $. 2. $2 \cdot O_1 O_2 \cdot h_2 = CR \cdot CP - CN\cdot CP $. where $h_1$ and $h_2$ are the projections of D and C to the radical axis. Then we get that : $\frac{h_1}{h_2}=\frac{LD}{LC}=\frac{DP}{CP} \cdot \frac{DM-DQ}{CR-CN}$. So we only need to prove that the quantity $\frac{DM-DQ}{CR-CN}$ is constant, which obvious since from Menelaous theorem with respect to the triangles APD and PBC and the lines F,Q,N and Q,R,E respectivly, we get $\frac{QD}{DB}=\frac{AR}{AC}=s$. So $QD=s\cdot DB$ and $RC=(1-s)\cdot AC$. So we are done.

Isn't it true that the possible lines EF are all tangent lines to a parabola? Does anyone have such a solution? (Then the rest is easy, i.e. showing that the centres of the circles lie on a line..)

Ok, guys, here is my solution to this problem (I think it's very nice and simple).Suppose we have two such triangles $PQR$ and $PLT$ ($Q$ and $L$ are on $AC$). Suppose that the intersection point of the circumcircles of this triangles is$ X$. Now due to the similarity of triangles $PRT$ and $PQL$ we have that $\frac{RT}{QL}=\frac{\sin{TPX}}{\sin{QPX}}$.But on the other hand $\frac{RT}{QL}=\frac{BD}{AC}$. So $\frac{\sin{TPX}}{\sin{QPX}}$ is constant so $\angle{TPX }$is fixed, so all this circles have one common point besides $P$. Nice and short, isn't it?

martin99 wrote: Isn't it true that the possible lines EF are all tangent lines to a parabola? Yes, this is true. Actually, it is a very nice observation, leading to a different solution of this problem: We will work in the projective plane. Let t be a varying parameter from $\mathbb{R}\cup\left\{\infty\right\}$. Consider the points $E_t$ and $F_t$ on the lines BC and DA dividing the segments BC and DA in the equal ratios $\frac{BE_t}{E_tC}=\frac{DF_t}{F_tA}=t$ (where the segments are directed, of course). Thus, if $\mathcal{T}$ is the affine transformation mapping the points B and C to the points D and A, respectively, then this transformation $\mathcal{T}$ will also map the point $E_t$ to the point $F_t$. Now, affine transformations are a particular case of projective transformations, and if we have a projective transformation which maps a line g to another line h, then the lines joining the points of g with their respective images (which lie on h) envelope a conic (this is a well-known fact from projective geometry). Thus, for varying t, the lines joining the points $E_t$ on the line BC with their images $F_t$ under the transformation $\mathcal{T}$ envelope a fixed conic. In other words, for varying t, the lines $E_tF_t$ envelope a fixed conic. We can easily see that this conic is a parabola: In fact, this conic must touch the line $E_{-1}F_{-1}$; but $E_{-1}$ and $F_{-1}$ are points at infinity (since they divide the segments BC and DA in the ratio $\frac{BE_{-1}}{E_{-1}C}=\frac{DF_{-1}}{F_{-1}A}=-1$), so the line $E_{-1}F_{-1}$ is the line at infinity, and a conic touching the line at infinity must be a parabola. Thus, for varying t, the lines $E_tF_t$ envelope a fixed parabola p. The point B divides the segment BC in the ratio $\frac{BB}{BC}=0$; thus, $B=E_0$. Similarly, $D=F_0$. The point C divides the segment BC in the ratio $\frac{BC}{CC}=\infty$; thus, $C=E_{\infty}$. Similarly, $A=F_{\infty}$. Finally, since BC = DA and BE = DF, we have EC = BC - BE = DA - DF = FA, and thus $\frac{BE}{EC}=\frac{DF}{FA}$. If we denote $\frac{BE}{EC}=\frac{DF}{FA}=k$, then we thus have $E=E_k$ and $F=F_k$. Hence, the lines $BD=E_0F_0$, $CA=E_{\infty}F_{\infty}$ and $EF=E_kF_k$ are three of the lines $E_tF_t$ which envelope the parabola p. Hence, the triangle PQR, being formed by these lines BD, CA and EF, is a triangle formed by three tangents to the parabola p. But according to Ross Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA 1995, Chapter 5, §5, the circumcircle of a triangle formed by three tangents to a parabola always passes through the focus of this parabola. Thus, the circumcircle of triangle PQR, formed by three tangents to the parabola p, must pass through the focus of the parabola p. This focus is clearly a fixed point, and it is different from P (else, the focus of the parabola would be the point of intersection of two of its tangents, what is impossible on the real plane). So the problem is solved. Okay, this is not the shortest solution, but quite a good example of the use of projective geometry on the IMO. Still, I don't know anyone who solved the problem in this way on the exam (before you start wondering: Grobber didn't participate ). My own solution was practically the same as Lego's one. Maybe it will later be published in German on the BWM website ( http://www.bundeswettbewerb-mathematik.de/imo/aufgaben/aufgaben.htm ). Darij

darij grinberg wrote: martin99 wrote: Isn't it true that the possible lines EF are all tangent lines to a parabola? Yes, this is true. Actually, it is a very nice observation, leading to a different solution of this problem: ... Okay, this is not the shortest solution, but quite a good example of the use of projective geometry on the IMO. Still, I don't know anyone who solved the problem in this way on the exam (before you start wondering: Grobber didn't participate ). My own solution was practically the same as Lego's one. Maybe it will later be published in German on the BWM website ( http://www.bundeswettbewerb-mathematik.de/imo/aufgaben/aufgaben.htm ). Darij Hello, thank you for writing down a complete solution using that observation, Darij. Since I didn't have to write down a solution (my days at the imo are passed a long time...) I just thought about the problem and didn't formulate a solution. Image not found So my solution "consists" of this wonderful parabola that comes out by drawing all lines EF (and I found a picture for all of you on the web...). I also saw that the rest of the solution is not so short and not so wonderful. (Actually I used coordinates to do the rest, which is very easy; maybe here your alternative is nicer (but I don't know the proof from that book).) Martin

I think the most difficult part of the problem is actually finding the common point, as opposed to showing that the point works.

billzhao wrote: I think the most difficult part of the problem is actually finding the common point, as opposed to showing that the point works. Without knowing what point is it, you can whack the question by ptolemy. Let the point required be X (later we reverse the step). Extend PX to CD to meet at X'. Then try to express everything in terms of A,B,C,D,P,X'. Finally, if PX is a constant, I found that AC sin DPX' = BD sin X'PC . Now, using this we can confirm the position of X', then reversing the steps to find PX, which is surely a constant. This is my solution, since I didn't know where X is. It's a 5-page proof though. (Very long comparing to other solutions here)

siuhochung wrote: billzhao wrote: I think the most difficult part of the problem is actually finding the common point, as opposed to showing that the point works. Without knowing what point is it, you can whack the question by ptolemy. Let the point required be X (later we reverse the step). Extend PX to CD to meet at X'. Then try to express everything in terms of A,B,C,D,P,X'. Finally, if PX is a constant, I found that AC sin DPX' = BD sin X'PC . Now, using this we can confirm the position of X', then reversing the steps to find PX, which is surely a constant. This is my solution, since I didn't know where X is. It's a 5-page proof though. (Very long comparing to other solutions here) Have you read my solution siohochung?

1. what're L,T? 2. You know I am so bad in geometry.... You could see it's the intersection of two circumcircles, but i couldn't.

$LT$ is defined in the same way as $QR$ it is just other such line. If you read my solution carefully you'll se that I haven't used there the intersection of that two circumcircles.My circles are another here.

Armo wrote: $LT$ is defined in the same way as $QR$ it is just other such line. If you read my solution carefully you'll se that I haven't used there the intersection of that two circumcircles.My circles are another here. oh, now i understand your solution. Very nice!

Define point $X$ as intersection of $\odot(BPC)$ and $\odot(APC)$. We will prove that point $X$ lies on $\odot(PQR)$, which will obviously solve the problem. Claim: $\triangle XBC \sim \triangle XDA$ Proof: This is easy angle chasing. Indeed: $$ \angle CBX = \angle CPX = \angle XDA \quad \text{and} \quad \angle XAD = \angle XPD = \angle BCX $$This proves the claim since both of these $2$ triangles has $2$ equal angles. From the claim, it is to deduce that $X$ is the centre of spiral similarity sending $DA$ to $BC$. Note that: $$ \frac{DF}{FA} = \frac{BE}{EC} $$Therefore point $X$ is also the centre of spiral similarity sending $FA$ to $EC$. This implies that point $R=AC \cap EF$ lies on both $\odot(AFX)$ and $\odot(XEC)$. This is enough to conclude since: $$ \angle PRX = \angle ARX = \angle XFD = \angle XQD $$implying that quadrilateral $PQRX$ is cyclic as desired.

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Consider $X$, the intersection of the perpendicular bisectors of $AC$ and $BD$. Note that $AXC$ and $BXD$ are isosceles. Note that $\triangle AXD\cong \triangle CXB$, so $\angle BXC=\angle AXD$ which implies $AXC=\angle BXD$ so $\angle DBX=\angle ACX$ which implies $PBCX$ cyclic. $~$ Similarly, $\triangle AXF\cong \triangle CXE$ which implies $\angle AXF=\angle CXE$ so $\angle EXF=\angle AXC$, which implies $\angle FEX=\angle ACX$, implying $ECXR$ cyclic. Thus, $X$ is miquel point of $BERP$ which implies $PQRX$ cyclic as desired.

Let $M$ be the Miquel Point of self-intersecting quadrilateral $ADBC$. We will show that $M$ is the desired fixed point. Claim. $M$ is the Miquel Point of $PRBE$. Notice that $M$ takes the spiral congruence $\overline{AD} \to \overline{BC}$, so it sends $\overline{DF} \to \overline{BE}$ as well. Thus $MQBE$ is cyclic. By definition, $MPBC$ is cyclic as well, so the claim is proven. $\blacksquare$ Thus $PQRM$ is cyclic by definition.

Let $T$ be the intersection of the perpendicular bisectors of $AC$ and $BD$ Claim: $T$ is the center of a spiral sending $A \rightarrow C$, $D \rightarrow B$, $F \rightarrow E$ Proof: Note that by definition, $TB=TD$, $TA=TC$, $AD=BC$ So, $\triangle DTA \cong \triangle BTC$ Which implies that $T$ is the center of the spiral Since $EC=AF$, it means that $T$ sends $ F \rightarrow E$ Claim: $AFTR, ECTR, FDTQ,BQTE$ are all cyclic quadrilaterals Proof: Sicne $T$ is the center sending $F \rightarrow E$, $A \rightarrow C$, $\triangle FTE \sim \triangle ATC$ Which implies that $\angle RFT = \angle EFT = \angle TAC = \angle TAR$ As such, $RAFT$ is cyclic. By symmetry, the other three quadrilaterals are also all cyclic Claim: $PQTR$ cyclic Proof: From the earlier claim, $$\angle ART= \angle TFD $$$$=\angle TEB$$$$= 180 - \angle BQT$$$$=180- \angle PQT $$ So, $PQTR$ concyclic

[asy][asy] import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -0.8, xmax = 0.45, ymin = -0.75, ymax = 0.5; /* image dimensions */ /* draw figures */ draw((-0.5496686746029323,-0.3159695482276248)--(0.3256219783940172,-0.3171988890323958), linewidth(0.5)); draw((-0.48632192971363797,0.26941687987647955)--(-0.01023392946941901,0.16642378356451423), linewidth(0.5)); draw((-0.5496686746029323,-0.3159695482276248)--(-0.01023392946941901,0.16642378356451423), linewidth(0.5)); draw((-0.48632192971363797,0.26941687987647955)--(0.3256219783940172,-0.3171988890323958), linewidth(0.5)); draw(circle((0.10476003653129087,-0.11214801410341571), 0.3013732879213327), linewidth(0.5) + blue); draw(circle((-0.4539229584159032,-0.030209830100056245), 0.3013732879213328), linewidth(0.5) + blue); draw((-0.507566568086635,0.07309545885846197)--(0.21298576763902466,-0.15500604382370595), linewidth(0.5)); draw(circle((-0.19981227412719071,0.13910246285713337), 0.31475328437898253), linewidth(0.5) + red); draw(circle((-0.41343780633702654,-0.13390106162689927), 0.22739345479050432), linewidth(0.5) + green); draw(circle((0.006218998307399752,-0.06037366891044607), 0.2273934547905042), linewidth(0.5) + green); draw(circle((0.02382541151700967,-0.4065770868695011), 0.31475328437898287), linewidth(0.5) + red); draw(circle((-0.10973414063147005,-0.08068963159497246), 0.12414245248440092), linewidth(0.5) + linetype("4 4")); draw((-0.5496686746029323,-0.3159695482276248)--(-0.507566568086635,0.07309545885846197), linewidth(0.5) + green); draw((-0.507566568086635,0.07309545885846197)--(-0.48632192971363797,0.26941687987647955), linewidth(0.5) + red); draw((-0.01023392946941901,0.16642378356451423)--(0.21298576763902466,-0.15500604382370595), linewidth(0.5) + green); draw((0.21298576763902466,-0.15500604382370595)--(0.3256219783940172,-0.3171988890323958), linewidth(0.5) + red); /* dots and labels */ dot((-0.5496686746029323,-0.3159695482276248),linewidth(3pt) + dotstyle); label("$A$", (-0.622232974286149,-0.345209419269103), NE * labelscalefactor); dot((0.3256219783940172,-0.3171988890323958),linewidth(3pt) + dotstyle); label("$B$", (0.34479231890361717,-0.3353919036022018), NE * labelscalefactor); dot((-0.01023392946941901,0.16642378356451423),linewidth(3pt) + dotstyle); label("$C$", (-0.005365739882524904,0.2029352054662128), NE * labelscalefactor); dot((-0.48632192971363797,0.26941687987647955),linewidth(3pt) + dotstyle); label("$D$", (-0.5518741120066906,0.31256413041327596), NE * labelscalefactor); dot((-0.15928227387531457,0.03313624859442896),linewidth(3pt) + dotstyle); label("$P$", (-0.17880851666444575,0.0703987439630469), NE * labelscalefactor); dot((-0.18988064800929766,-0.17549409279790085),linewidth(3pt) + dotstyle); label("$X$", (-0.26552990505540613,-0.21758171559938766), NE * labelscalefactor); dot((-0.507566568086635,0.07309545885846197),linewidth(3pt) + dotstyle); label("$F$", (-0.5011502810610344,0.09657878574145004), NE * labelscalefactor); dot((0.21298576763902466,-0.15500604382370595),linewidth(3pt) + dotstyle); label("$E$", (0.21880086784505204,-0.13249657981957744), NE * labelscalefactor); dot((0.013893785399116884,-0.09198053121446656),linewidth(3pt) + dotstyle); label("$Q$", (0.03063181756277942,-0.07359148581817038), NE * labelscalefactor); dot((-0.217338160020329,-0.01878063773944448),linewidth(3pt) + dotstyle); label("$R$", (-0.2933461994449595,0.0000398816835884594), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Firstly note that the length conditions imply $AF=CE$ and $FD=EB$. Now let $X=\odot(PCB)\cap\odot(PAD)$ which is the miquel point mapping $\overline{AD}\mapsto\overline{CB}$. I claim that this is the fixed point. First note that this point does not depend upon the position of $E$ and $F$. We thus just need to show that $X$ lies on $\odot(PQR)$. Now we are also given that $AD=BC$ and thus this is just a rotation about $X$. Now $AF=CE$ implies that $F\mapsto E$ and thus $AF\mapsto CE$ and so $X\in\odot(AFR)$ and $\odot(ECR)$. Similarly we also have $X\in\odot(FDQ)$ and $\odot(QEB)$ too. Thus putting all the things together we have that following cyclic quadrilaterals, $\left\{\odot(XAFR),\odot(XRCE),\odot(XFDQ),\odot(XQEB),\odot(XADP),\odot(XPCB)\right\}$. Now we just have to follow through the following angle-chase, \begin{align*} \measuredangle XQP&=\measuredangle XQB\\ &=\measuredangle XEB\\ &=XEC\\ &=XRC\\ &=XRP .\end{align*}This thus implies that $XPQR$ is cyclic and we are done.

Significantly misplaced problem, it took me 10 minutes altogether and spiral sim trivializes everything lol By spiral sims, we get free AF->CE, DF->EB, AD->BC all with center S, the second intersection of circles ADP and CBP. Hence we get cyclic quads ADPS, BCPS, and because spiral similarities are unique, if S' is the intersection of (AFR) and (BCR), S' takes AF->CE and so S=S', etc. with the other cyclic quads to get AFRS, BCRS, CDFS, BEQS. Then $$SQP=SCD=180-SFD=SFA=SRA=180-SRP,$$so PQSR is cyclic, meaning that X is the fixed point that is independent other than P (since it's only dependent on ADP and BCP), as desired. $\blacksquare$

We claim that the intersection of the circumcircles of $APD$ and $BPC$ is the common point of the circumcircles of $PQR$. Define $(APD)\cap(BPC)=Y$. We will first observe that it suffices to prove that $PQYR$ is a cyclic quadrilateral, as this would mean that for every position of $E$ and $F$, $(PQR)$ passes through $Y$. In addition, if $Y$ is the Miquel point of $APQF$, $PQYR$ is a cyclic quadrilateral, and vice versa. Thus, it is sufficient to show that $Y$ is the Miquel point of $APQF$. There exists a spiral similarity that sends $DA$ to $BC$, and since $\frac{DF}{FA}=\frac{BE}{CE}$, $\triangle EYC \sim \triangle FYA$. An obvious result of this is that $\frac{EY}{YC} = \frac{FY}{YA}$. We then do some simple angle chasing: \begin{align*} \angle AYC &= \angle AYE + \angle EYC \\ &= \angle AYE + \angle FYA \\ \angle AYC &= \angle FYE \end{align*}Then, by $SAS$ similarity, $\triangle AYC \sim \triangle FYE$, which means that $Y$ is the spiral center of a spiral similarity sending $FE$ to $AC$. Because of this, $ARYF$ is a cyclic quadrilateral, and $(ARF)$ passes through $Y$. From here, it is easy to see that $Y$ is the Miquel point of $APQF$, and we are done. $\blacksquare$

Let the intersection of (APD) and (BPC) be point X. It is easy to see X is the spiral similarity center of quadrilateral ABCD, thus AXD and CXB are similar, and since AD=CB this implies they are congruent. Thus it is deduced that AXF and CXE are also congruent. Now, let the intersection of (AXF) and (CXE) be R'. Then, note <ER'X=<ECX=<FAX=<FR'X implying E, R', and F are collinear, and similarly, C, R', and A are collinear, implying R'=R. This means that AFXR is a cyclic quadrilateral, but APXD is also a cyclic quad, thus <XPQ=<XPD=<XAD=<XAF=<XRF=<XRQ, thus PQXR is cyclic, and since X clearly stays fixed as E and F vary, we are done.

Let $M = (APD) \cap (BPC)$ so that $M$ is the center of the spiral similarity which maps $\overline{AD} \mapsto \overline{CB}$. The length conditions imply this spiral similarity also maps $\overline{AF} \mapsto \overline{CE}$, which implies that $FARM$ is cyclic. Since $APMD$ is also cyclic, this means that $M$ is Miquel point of $APQE$, so $M$ also lies on $(PQR)$.

We claim the desired point is the Miquel point of degenerate quadrilateral $ACBD$, or the center of spiral similarity mapping $AC \rightarrow FE \rightarrow DB$, which exists through the length condition. Consider Miquel on quadrilateral $APQF$. Three of the circles already pass through $M$ from the spiral similarity mentioned above; hence the final circle, $(PQR)$, also will always pass through this fixed point. $\blacksquare$

Very easy Problem

Sketch

I claim that our desired fixed point is the Miquel point of quadrilateral $DACB$. Let $X = (APD) \cap (PBC)$. Then observe that $X$ is the Miquel point of complete quadrilateral $DACB$. Now as $\frac{DA}{DF} = \frac{BC}{BE},$ we also have $FA \stackrel{X}{\mapsto} EC$, so $X$ is also the Miquel point of $FACE$. Now observe that $XREC$ is cyclic, and considering complete quadrilateral $PBER$, we have $X \in (PBC)$, and $X \in (REC)$. Hence $X$ is Miquel point of $PBER$. It is similarly the Miquel point of $APQF$ as $XFAR$ and $APXD$ are cyclic. Hence our fixed point is the Miquel point of $DACB$, as desired. $\blacksquare$

We claim that the fixed point is the spiral center. In other words, we rephrase the problem as (yes, I know point names are changed, but this was done intentionally to change the perspective of the problem and better conform to standard naming conventions): In triangle $\triangle ABC$, $E$ and $F$ are on $AC$ and $AB$ such that $EC=BF$, and $BE$ and $CF$ meet at $P$. Then, let $Y$ be on $CE$ and $Z$ on $BF$ such that $CY=ZF$ and $EY=BZ$. Then, $YZ$ intersects $BE$ and $CF$ at $Q$ and $R$. Let $K$ be the spiral center taking $BZF$ to $EYC$ (due to the length conditions given), or alternatively the intersection of the perpendicular bisectors to $BE$ and $CF$. Then, $PQRK$ is cyclic. We know from the Miquel point configuration that $$ABEK,AFCK,BFPK,CEPK,BZQK,CYRK$$are all cyclic. Thus, $$\angle QKR=\angle BKC-\angle QKB-\angle RKC$$$$=\angle BKC-AZY-\angle AYZ=\angle BKC-180+\angle A$$$$=(360-\angle BPC-\angle PBK-\angle PCK)-180+\angle A$$$$=(180-\angle BPC-\angle KAE-\angle KAF)+\angle A=180-\angle BPC.$$

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cool solution We claim that the required point is $S= (BPC) \cap (APD)$ different from $P$. Heres a sketch of the solution: First observe that the $(BPC)$ and $(APD)$ have equal radii (sin rule). Next note that $S$ lies on perpendicular bisectors of $AC$ (angle chasing to get $\angle SCA=\angle SAC$). Establish that $S$ also lies on perpendicular bisector of $EF$ ( $\triangle SFD \cong \triangle SBE$) Let $T = (ERC) \cap (APF)$. Show that $T$ satisfies the previous two properties and hence is the same as $S$. Show that $P,Q,S,R$ are concyclic,