This is the best geo from this exam, in my opinion. Here goes my solution (more like a sketch): Let $CD$ intersect $AB$ at $R$, $EF$ meets $AB$ at $T$ and $CD$ meets $EF$ at $S$. Obviously $DCEF$ is a cyclic isosceles trapezoid; let $X$ be its center. We claim that $X$ is the desired tangency point. Note that $\angle EXF= 2\angle ECF= \angle EPF$, so $X$ is on $(PEF)$ and $XE=XF$. By a well-known spiral config (since $BE=AF$), we have that $X$ lies on $(ABP)$ as well (and $X$ is on the perpendicular bisector of $AB$). In addition, note that $SX$ is the bisector of $\angle DSF$. Thus, in order to prove $X$ is on $(SRT)$, we need to prove that $X$ is on the perpendicular bisector of $RT$, or equivalently, $RB=AT$, but this can be derived easily with a bit of a length bash using the equal segments: twice Menelaus for $\triangle ABP$ (which I will skip). Now, the tangency easily follows since the triangles $AXB$ and $RXT$ are isosceles, done.

Clearly, $DCEF$ is a cyclic isosceles trapezoid. Let $T$ be its center and denote its circumcircle $\omega$. Furthermore, let $Q=CD\cap EF$, $X=AB\cap EF$, and $Y=AB\cap CD$. By symmetry, we must have $Q$, $P$, and $T$ to be colinear. Note that $\angle FTC=2\angle FDC=2\angle QDF=\angle FYC$ so $T\in (FYC)$. Also, $\angle DTC=2\angle DFC=\angle DPC$ implying $T\in (DPC)$. We conclude that $T$ is the Miquel point of complete quad $\{QF,FP,PD,DQ\}$. Now from the condition, $\mathrm{pow}(A,\omega)=AF\cdot AC=BE\cdot BD=\mathrm{pow}(B,\omega)$ so $TA=TB$. It follows that $\bigtriangleup ATF\cong\bigtriangleup BTE$ by sss. Now we have $\bigtriangleup EFT\sim\bigtriangleup BTA$ so $\angle ATB=\angle FTE=\angle FPE=\angle APB$, hence $T\in (APB)$. $T$ is now the Miquel point of complete quad $\{YA,AP,PD,DY\}$ so $T\in (XYQ)$. Finally, we must have $\angle YXT=\angle TQC=\angle TQE=\angle TQX=\angle TYX$ so $\bigtriangleup XTY$ is isosceles. We conclude that the circumcircles of $\bigtriangleup APB$ and $\bigtriangleup XYQ$ are tangent at $T$.

Enjoyable problem. Define the following points $AB \cap CD=X$, $EF \cap CD=Y$ and $EF \cap AB =Z$. Moreover, let $T$ be the point on $\odot(ABP)$ such that $AT=TB$ and $T$ lies on the same side as $P$ with respect to the line $AB$. We proceed in several steps. Step 1: Quadrilateral $EFCD$ is cyclic. Moreover, $T$ is the center of $\odot(EFCD)$. Proof: Since $PC=PE$ and $PD=PF$, it is easy to see that quadrilateral $ECDF$ is isosceles trapezoid; therefore a cyclic quadrilateral. Since $AC=BD$, we can deduce that: $$ BE = BD -PE-PD = AC-PC-PF =AF$$Now note that this combined together with $AT=TB$ and $\angle PBT = \angle PAT$ gives us $\triangle FAT = \triangle EBT$. This implies 2 things. First of all, $\angle BET = \angle AFT \implies \angle PET = \angle PFT$, which implies that quadrilateral $EPTF$ is cyclic. Second of all, $TE=TF$ as the corresponding elements in equal triangles. Suppose that $\angle EPF = \angle ETF = 2\alpha$. Since $PF=PD$, then $\angle PDF = \angle EDF = \alpha$. Therefore; $\angle ETF = 2\angle EDF$ and $TE=TF$. This implies that $T$ is the centre of $\odot(EFCD)$ and that points $Y,P,T$ are collinear as they lie on the perpendicular bisector of $FD$. Step 2: $TX=TZ$. Proof: From Menelaus theorem on the $\triangle ABP$ and the points $E,F,Y$ we have: $$ 1=\frac{AZ}{BZ} \cdot \frac{BE}{PE} \cdot \frac{PF}{AF} \implies \frac{BZ}{AZ} = \frac{PF}{PE} $$Similarly, from the Menelaus theorem on the $\triangle ABP$ and the points $C,D,X$ we have: $$ 1 = \frac{AX}{BX} \cdot \frac{BD}{PD} \cdot \frac{PC}{AC} \implies \frac{AX}{BX} = \frac{PD}{PC} $$We conclude that: $$ \frac{AX}{BX} = \frac{BZ}{AZ} \implies \frac{AB}{BX}=\frac{AB}{AZ} \implies BX=AX$$. Now remember that $AT=BT$ and $\angle TBA = \angle TAB \implies \angle TAZ = \angle TBX$ meaning that $\triangle TAZ = \triangle TBX \implies TZ=TX$ as needed. Step 3: Quadrilateral $XYTZ$ is cyclic. Proof: Note that $\angle EPF = \angle ETF=\angle ATB$. Moreover, $AT=TB$ and $ET=TF$, meaning that $\angle TEZ = \angle TBZ$. This gives us that quadrilateral $ETZB$ is cyclic; therefore $\angle BZT = \angle PET = \angle PFT = \beta$. Note that $\angle CTF = 180^{\circ}-2\beta$. Since $T$ is the center of $\odot(ECDF)$, then $\angle CTF = 2\angle CDF \implies \angle CDF = \angle ECY = 90^{\circ}-\beta$. Consequently, we have that $$\angle FYT = \angle DYT = \angle PET = \angle XZT = \angle TXZ$$which means that quadrilateral $XYZT$ is cyclic as desired. This finished the problem because $\triangle ATB$ and $\triangle XTZ$ are isosceles therefore they obviously their circumcircles are tangent to each other at the point $T$.

Let $K$ be the center of the Spiral Similarity mapping $BE$ to $AF$. It then holds $K\in (FPE)$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(24.636363636363637cm); 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 = -3.9472727272727264, xmax = 20.68909090909091, ymin = -4.256363636363638, ymax = 6.216363636363635; /* image dimensions */ pen ttffqq = rgb(0.2,1.,0.); pen ffttcc = rgb(1.,0.2,0.8); pen ffwwqq = rgb(1.,0.4,0.); /* draw figures */ draw((4.1,2.18)--(7.64,0.72)); draw((6.28,1.98)--(4.065842881782181,-1.14421322189209)); draw((4.1,2.18)--(4.065842881782181,-1.14421322189209)); draw((4.065842881782181,-1.14421322189209)--(7.64,0.72)); draw((7.64,0.72)--(6.28,1.98)); draw((6.28,1.98)--(4.1,2.18)); draw((4.118521610378965,3.9825461550900787)--(4.1,2.18)); draw((4.065842881782181,-1.14421322189209)--(4.0473212714032165,-2.9467593769821643)); draw((4.0473212714032165,-2.9467593769821643)--(5.54254346298305,2.663231791648056)); draw((4.118521610378965,3.9825461550900787)--(6.28,1.98)); draw((6.28,1.98)--(5.283638032759958,1.6918329017430684)); draw((7.64,0.72)--(4.8061707280050046,-0.09959808294209078)); draw(circle((2.2492427280505414,0.5367348942123785), 3.9201809633626055), ttffqq); draw((6.169216759995953,0.49645618539245184)--(5.54254346298305,2.663231791648056)); draw(circle((4.463986805369431,0.5139778475906414), 1.705319971970585), ffttcc); draw(circle((6.532515637088907,3.056373531023778), 2.585568194915991), ffwwqq); /* dots and labels */ dot((4.1,2.18),linewidth(2.pt) + dotstyle); label("$A$", (3.78,2.270909090909089), NE * labelscalefactor); dot((7.64,0.72),linewidth(2.pt) + dotstyle); label("$C$", (7.816363636363637,0.6345454545454531), NE * labelscalefactor); dot((6.28,1.98),linewidth(2.pt) + dotstyle); label("$D$", (6.270909090909092,2.1618181818181803), NE * labelscalefactor); dot((4.065842881782181,-1.14421322189209),linewidth(2.pt) + dotstyle); label("$B$", (3.68909090909091,-1.42), NE * labelscalefactor); dot((5.896607281177382,1.4390263755596107),linewidth(2.pt) + dotstyle); label("$P$", (5.561818181818183,1.2890909090909075), NE * labelscalefactor); dot((4.8061707280050046,-0.09959808294209078),linewidth(2.pt) + dotstyle); label("$E$", (4.907272727272729,-0.32909090909091054), NE * labelscalefactor); dot((5.283638032759958,1.6918329017430684),linewidth(2.pt) + dotstyle); label("$F$", (4.907272727272729,1.5072727272727258), NE * labelscalefactor); dot((5.54254346298305,2.663231791648056),linewidth(2.pt) + dotstyle); label("$Z$", (5.707272727272729,2.7618181818181804), NE * labelscalefactor); dot((4.118521610378965,3.9825461550900787),linewidth(2.pt) + dotstyle); label("$X$", (3.7072727272727284,3.8890909090909074), NE * labelscalefactor); dot((4.0473212714032165,-2.9467593769821643),linewidth(2.pt) + dotstyle); label("$Y$", (3.7254545454545465,-2.783636363636365), NE * labelscalefactor); dot((6.169216759995953,0.49645618539245184),linewidth(2.pt) + dotstyle); label("$K$", (6.234545454545456,0.5618181818181803), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Notice $$ BE=BD-EP-PD=AC-PC-FP=AF $$Hence such Spiral Similarity is indeed a rotation, meaning that $KE=KF$. Also, $$ 2\measuredangle FDE=2\measuredangle FDP=\measuredangle FPE=\measuredangle FKE $$These two results combined mean that $K$ is precisely the circumcenter of $CDFE$, and so $\triangle KEF\sim \triangle KDC$. By the definition of $K$, we get that $YBEK$ is cyclic. We also get that $K$ is the center of the Spiral Similarity mapping $BA$ to $EF$, so $$ \triangle KBA\sim \triangle KEF\sim \triangle KDC $$ Thus $K$ is the center of the Spiral Similarity mapping $BA$ to $DC$, and so $AXCK$ and $BXDK$ are cyclic. Since quadrilaterals $BXDK$ and $YBEK$ are both cyclic, then $K$ is also the center of the spiral similarity mapping $BE $ to $XZ$, from where we get that $K\in (XYZ)$. The first definition of $K$ was as the center of the Spiral Similarity mapping $BE$ to $AF$, so $K\in (ABP)$. Finally, let $\ell$ be the line tangent to $(ABPK)$ at $K$. Then, $$ \measuredangle KBA-\measuredangle (\ell,KX)= \measuredangle (\ell,KA)-\measuredangle (\ell,KX)= \measuredangle XKA=\measuredangle XCA \\ =\measuredangle ZCP \\ =\measuredangle KPC-\measuredangle KZC \\ =\measuredangle KBA-\measuredangle KYX $$ We get $\measuredangle (\ell,KX)=\measuredangle KYX$, so $\ell$ is tangent to $(KYX)$. This way we have proven that circles $(ABP)$, $(XYZ)$ are both tangent at $K$ and the common tangent line is $\ell$.

Quite straightforward! Simple observation that $AF=BE$ motivates us to Let $P$ be the center of spiral sim taking $BE$ to $AF$, but we do this a little bit differently: Note that $DCEF$ is an isosceles trapezoid by POP and congruency. Let us denote $O$ to be the center of the $DCEF$. Note that $\angle FOE=2\angle FCE=\angle EPF$, thus $FOPE$ and similarly $CPOD$ are cyclic. Also since $T,P,O$ all lie on the perpendicular bisector of $CE$, we get that $O,P,T$ are collinear. Since: $\angle OEF=\angle OPF=\angle ODT$ $\implies$ $DOET$ is cyclic Also since $OE=OF$, $AF=EB$, $\angle OFA=\angle OEB= 180-\angle OFP$ we have $\triangle OFA \cong \triangle OEB$ thus $\angle AOB=\angle APB$ thus $O$ is the center of spiral sim taking $BE$ to $AF$ as desired from the start. Now we have: $\angle OBA=\angle OEF=\angle ODT=\angle ODR$ $\implies$ $DRBO$ is cyclic. Since $QBRTDE$ is complete by Miquel point theorem $(EDT),(DBR),(QTR)$ concur; Since $DOET$ and $DOBR$ are cyclic wed get that $(QTR)$ passes through $O$. Obviously $O$ $\in$ $(APB)$ as proven before, Let the line tangent to $(APB)$ at $O$ be $\ell$. Finally $\angle PO\ell=\angle PBO=\angle DBO=\angle DRO=\angle TRO$ $\implies$ $\ell$ is tangent to $(QTR)$ at $O$ and we are done.

Fantastic problem! Solved with math_holmes15

Was really fun to solve