Let $ABCD$ be an isosceles trapezoid with $AB\parallel CD$. Let $E$ be the midpoint of $AC$. Denote by $\omega$ and $\Omega$ the circumcircles of the triangles $ABE$ and $CDE$, respectively. Let $P$ be the crossing point of the tangent to $\omega$ at $A$ with the tangent to $\Omega$ at $D$. Prove that $PE$ is tangent to $\Omega$. Jakob Jurij Snoj, Slovenia
Problem
Source: RMM 2019
Tags: RMM, RMM 2019, geometry, Complex Geometry, Inversion
23.02.2019 14:46
Suppose that $ABCD$ is not a rectangle and let $F$ be the midpoint of $BD$. Let $l_1$ and $ l_2$ be the tangents to $\Omega$ at $E$ and $D$ respectively and let $m_1$ and $m_2$ be the tangents to $\omega$ at $A$ and $F$ respectively. Suppose that $m_2$ intersect $\overline{CD}$ and $\overline{AB}$ at $V$ and $V_1$ respetively and let $U$ be on $\overline{CD}$ such that $UF$ is tangent to $\Omega$. Due to the tangency relation, we have that $\angle{CUF}=\angle{ECF}$ and $\angle{AV_1F}=\angle{EAF}$, thus by sine theorem and similarity the following relations hold $$\frac{UD}{DF}=\frac{UF}{CF}=\frac{\sin{\angle{CFE}}}{\sin{\angle{ECF}}}=\frac{\sin{\angle{AFE}}}{\sin{\angle{CAF}}}=\frac{\sin{\angle{FAV_1}}}{\sin{\angle{AV_1F}}}=\frac{FV_1}{AF}=\frac{BV_1}{BF} \implies BV_1=DV=DU$$This clearly implies that the pencil $(FD,FE;m_2,FU)$ is harmonic, thus $m_2,l_1$ and $l_2$ are concurrent. Analogously $l_1,m_1$ and $m_2$ are also concurrent, thus $G$ lies on both $m_2$ and $m_1$.
23.02.2019 14:56
Solution. Let $F$ be the midpoint of $BD$. Let $P'$ be the intersection of tangents from $E$ and $D$ to $\odot(DCE)$. We'll prove that $P'F$ is tangent to $\odot(ABE)$ which would give from symmetry that $P'A$ is tangent to $\odot (ABE)$ and we would be done. Now, invert about $E$ with arbitrary radius. Rephrased problem wrote: Let $FAC$ be a triangle and $E$ be the midpoint of $AC$. Let $D$ be a point on $CF$ such that $EF$ is tangent to $\odot(DCE)$ and $B$ is a point on $AF$ such that $EF$ is tangent to $\odot(BEA)$. Let $P'$ be a point such that $EP'||DF$ and $DF$ is tangent to $\odot(P'ED)$. Prove that $\odot(P'EF)$ is tangent to $AF$. (Proof): Note that $DP' = DE$ and $\triangle DEF \sim \triangle ECF$ so \[\frac{AE}{EF}=\frac{CE}{EF} = \frac{DE}{DF}=\frac{DP'}{DF}.\]Also, \[\angle P'DF = \angle ECF + \angle CFE = \angle AEF.\]Therefore $\triangle AEF \sim \triangle P'DF$ so $\angle AFE = \angle DFP' = \angle EP'F$ and we are done. $~\square$
23.02.2019 15:06
Let $F$ be the midpoint of $BD$. Easily $DF=CE=AE$. Let $P'$ be the intersection of bisector of $AF$ and $DE$. Then $\triangle P'DF \equiv \triangle P'EA(SSS)$. Let $S$ be the intersection of $AC$ and $BD$. After some angle chasing, $P'S\parallel AB$. After more angle chasing, $P'D$, $P'E$ are tangent to $\Omega$ and $P'A$, $P'F$ are tangent to $\omega$. So $P'=P$ and $PE$ is tangent to $\Omega$.
23.02.2019 16:07
Let $F$ denote the midpoint of $\overline{BD}$, so $CDFE$ and $ABFE$ are cyclic trapezoids. We use complex numbers with respect to $CDEF$. [asy][asy] pair C = dir(130); pair E = dir(210); pair D = dir(330); pair F = C*D/E; pair A = 2*E-C; pair B = 2*F-D; filldraw(unitcircle, invisible, deepcyan); draw(A--B--C--D--cycle, lightred); draw(E--F, orange); draw(B--A, orange); draw(C--D, orange); draw(C--A, lightred); draw(B--D, lightred); pair P = 2*E*D/(E+D); draw(E--P--D, deepcyan); draw(E--D, deepcyan); draw(P--A, deepcyan); draw(circumcircle(E, F, A), dotted+deepgreen); pair O = origin; pair M = midpoint(E--D); draw(O--P, dotted+blue); dot("$C$", C, dir(C)); dot("$E$", E, dir(E)); dot("$D$", D, dir(D)); dot("$F$", F, dir(F)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$P$", P, dir(P)); dot("$O$", O, dir(10)); dot(M); /* TSQ Source: C = dir 130 E = dir 210 D = dir 330 F = C*D/E A = 2*E-C B = 2*F-D unitcircle 0.1 lightcyan / deepcyan A--B--C--D--cycle lightred E--F orange B--A orange C--D orange C--A lightred B--D lightred P = 2*E*D/(E+D) E--P--D deepcyan E--D deepcyan P--A deepcyan circumcircle E F A dotted deepgreen O = origin R10 M .= midpoint E--D R315 O--P dotted blue */ [/asy][/asy] Let $C = x$, $E = y$, $D = z$, so $F = xz/y$. Re-define $P = \frac{2yz}{y+z}$ as the intersection of the tangents to $\Omega$ at $E$ and $D$; then we wish to show $\overline{PA}$ is tangent to $\Gamma$. Compute \[ \frac{F-A}{P-A} = \frac{\frac{xz}{y} - (2y-x)}{\frac{2yz}{y+z} - (2y-x)} = \frac{y+z}{y} \frac{2xz-y(2y-x)}{2yz-(y+z)(2y-x)} = \frac{y+z}{y}. \]Hence $\measuredangle FAP$ equals $\measuredangle EOP = \measuredangle ECD = \measuredangle ACD = \measuredangle AEF$, where $O$ is the center of $\Omega$. This implies the desired tangency.
23.02.2019 16:15
Here’s my solution. Let $F$ denote the midpoint of $BD$ and let $X = AC \cap BD$. Redefine $P$ as the intersection of perpendicular bisectors of $AF, DE$. Note that $\Delta PDF \equiv \Delta PEA$ as $PE = PD, PF = PA, DF = DA$. I claim that $P,X,E,D$ are concyclic This is easy to see since $\angle{PDF} = \angle{PDX} = \angle{PEA} = \angle{PEX}$. So, $$\angle{PDE} = 180^{\circ}-\frac{\angle{DPE}}{2} = 180^{\circ}-\frac{\angle{DXE}}{2} = \angle{ECD}$$. Thus, $DP$ is tangent to $(CED)$. Similarly, $AP$ is tangent to $(ABE)$. Since, $DP = EP$, we have $EP$ tangent to $(CED)$.$\blacksquare$.
Attachments:

23.02.2019 16:16
This problem can be easily solved with complex numbers: Let $F$ be the reflection of $E$ in the perpendicular bisector of $AB$.Hence $ABFE$ and $CDEF$ are cyclic trapezoids.Now set the circle $(CDEF)$ as unit circle and let $P'$ be the intersection of tangents at $D$ and $E$ to $(CEF)$.We want to prove that $AP'$ is tangent to $(AFE)$.Denote by $O$ the circumcenter of $(AFE)$.We just have to prove $OA\perp AP'$ $p'=\frac{2de}{d+e}$ and $e-a=c-e$ and as $FE\parallel CD$ we have $c=\frac{ef}{d}$ so $a=e\cdot\frac{2d-f}{d}$ $o-a=\frac{(f-a)(e-a)(\overline{f-e})}{(\overline{f-a})(e-a)-(\overline{e-a})(f-a)}$ $$\frac{o-a}{\overline{o-a}}=\frac{f-a}{\overline{f-a}}\cdot\frac{c-e}{\overline{c-e}}\cdot\frac{\overline{f-e}}{f-e}\cdot\frac{-(\overline{f-a})(e-a)+(\overline{e-a})(f-a)}{(\overline{f-a})(e-a)-(\overline{e-a})(f-a)}$$So we have $$\frac{o-a}{\overline{o-a}}=\frac{f-a}{\overline{f-a}}\cdot(-ce)\cdot(-\frac{1}{ef})\cdot(-1)=-\frac{f-a}{\overline{f-a}}\cdot\frac{e}{d}$$Now $f-a=f-(e\cdot\frac{2d-f}{d})=\frac{df+fe-2de}{d}$ and $a-p'=a-\frac{2de}{d+e}=e\bigg(\frac{2d-f}{d}-\frac{2d}{d+e}\bigg)=e\cdot\frac{2de-ef-fd}{d(d+e)}$ $$\frac{a-p'}{\overline{a-p'}}=\frac{e}{d+e}\cdot \frac{1}{\frac{1}{e}}\cdot\frac{d+e}{de}\cdot\frac{f-a}{\overline{f-a}}=-\frac{o-a}{\overline{o-a}}$$Hence we are done.
23.02.2019 16:52
Is there any geometric solution without redefining $P$?
23.02.2019 17:36
math90 wrote: Is there any geometric solution without redefining $P$? The second official solution is a geometric solution without redefining $P$. It uses isogonal conjugates.
23.02.2019 17:40
Am I the only one who didn't use the midpoint of $BD$? Let $Q$ be the point such that $\triangle QDA\equiv\triangle EBC$, and suppose $DQ$ meet $AB$ at $T$. Let $M,N$ be the midpoints of $BC,AD$. (Also let $P'$ be the point such that $P'E$ and $P'D$ are both tangent to $\Omega$. It suffices to prove that $PA$ is tangent to $(ABE)$.) One can easily prove that $K,N,E,M$ are collinear. Since $QE\parallel AB$, and note that $AQ=CE=AE$, so $$\angle QEA=\angle DCA=\angle PED\Longrightarrow \triangle QEA\sim \triangle DEP$$. We that $\triangle QED\sim\triangle AEP$ by spiral similarity, Thus $$\angle PAE=\angle DQE=\angle DTC$$, but since $\frac{AT+CD}{2}=QE=\frac{AB+CD}{2}\Longrightarrow \angle DTC=\angle EBC$, so we're done. ($\square BTQE$ is a cyclic trapezoid since $Q$ and $E$ are symmetric with respect to the perpendicular bisector of $BT$.) Edit : Thanks @below
23.02.2019 18:04
MNJ2357 wrote: Let $Q$ be the point such that $\triangle QDA\equiv\triangle PBC$ I think you mean $\triangle QDA\equiv \triangle EBC$?
23.02.2019 18:04
Let's $P_1$ be the intersection of tangents to $(AEFB)$ at points $A$ and $F$ and let's $P_2$ be the intersection of tangents to $(CFED)$ at points $C$ and $E$. We will prove that $P_1=P_2$. Denote $K$ as the interesection of lines $AC$ and $BD$. Firstly, note that triangles $AP_1F$ and $EKF$ are similar (since $\angle KEF=\angle PAF$ and $\angle KFE=\angle PFA$). Therefore, $\angle AP_1F=\angle AKF$, so $AP_1KF$ is cyclic. From $AP=PF$ we obtain that $P_1$ lies on the angle bissector of $\angle AKD$. Also, siimilarity $AP_1F$ and $FKE$ implies similarity $AEF$ and $P_1KF$. Hence, $$ \frac{KP_1}{KF}=\frac{AE}{EF}. $$Analagously, triangles $DP_2E$ and $EKF$ are similar, $P_2$ lies on the angle bissector of $\angle AKD$. Also, $P_2EK$ similar to $DEF$, so $$ \frac{KP_2}{KE}=\frac{FD}{EF}. $$Now, note that $$ KP_1=\frac{AE\cdot KF}{EF}=\frac{FD\cdot KE}{EF}=KP_2. $$But points $P_1$ and $P_2$ lie on the angle bissector $\angle AKD$, so $P_1=P_2$. Therefore, $P=P_1=P_2$, as desired.
23.02.2019 18:06
math90 wrote: Let $ABCD$ be an isosceles trapezoid with $AB\parallel CD$. Let $E$ be the midpoint of $AC$. Denote by $\omega$ and $\Omega$ the circumcircles of the triangles $ABE$ and $CDE$, respectively. Let $P$ be the crossing point of the tangent to $\omega$ at $A$ with the tangent to $\Omega$ at $D$. Prove that $PE$ is tangent to $\Omega$. Jakob Jurij Snoj, Slovenia FTFY
23.02.2019 18:50
Let $F$ be the midpoint of $BD$ and $G=AF\cap CD,H=AB\cap DE$. Clearly, $F\in(ABE),(CDE)$. By 2019 USA TST P1, the tangents to $(FGD),(ECD),(AEF)$ at $F,E,A$ concur and the tangents to $(FBA),(EHA),(DEF)$ at $F,E,D$ concur. But $(FGD),(ABEF)$ and $(EHA),(CDEF)$ are tangent, so the problem follows.
23.02.2019 18:58
Another inversion solution
23.02.2019 21:09
[asy][asy] import geometry; unitsize(3 cm); pair A,B,C,D,E,F,O1,O2,O,M,N,P,Q,R; O = (0,0); A = dir(130); B = dir(50); C = dir(-15); D = dir(195); E = (A+C)/2; F = (B+D)/2; M = (D+E)/2; N = (A+F)/2; O1 = circumcenter(A,B,E); O2 = circumcenter(C,D,E); P = extension(O1,N,O2,M); Q = extension(A,F,D,E); R = extension(A,C,B,D); draw(A--B--C--D--A,green); draw(A--C,heavycyan); draw(B--D,heavycyan); draw(A--F,cyan); draw(D--E,cyan); draw(circle(O,1),heavyblue+dotted); draw(circumcircle(A,B,E),blue); draw(circumcircle(C,D,E),lightblue); draw(O2--D--P--E--O2--P,heavygreen); draw(O1--A--P--F--O1--P,springgreen); draw(F--Q,cyan); draw(circumcircle(P,D,F),magenta+dotted); draw(circumcircle(P,A,E),magenta+dotted); draw(O1--O2,purple+dotted); draw(P--R,purple+dotted); // dot("$O$",O); dot("$A$",A); dot("$B$",B); dot("$C$",C); dot("$D$",D); dot("$E$",E); dot("$F$",F); dot("$O_1$",O1); dot("$O_2$",O2); dot("$M$",M); dot("$N$",N); dot("$P'$",P); dot("$Q$",Q); dot("$R$",R); label("$\Gamma$",dir(50)+(0,0.5)); label("$\Omega$",dir(195)+(0,-1)); [/asy][/asy] Let $F$ be midpoint of $BD$. $AFEB$ and $DFEC$ are isosceles trapezoids. Let $O_1$ be circumcenter of $AFEB$ and $O_2$ be circumcenter of $DFEC$. Let $M$ be midpoint of $DE$ and $N$ be midpoint of $AF$. Let $P' = MO_2 \cap NO_1$. We will show that $P'A,P'F$ are tangents to $\Gamma$, $P'D,P'E$ are tangents to $\Omega$. This is enough to imply that $P=P'$ and $PE$ is tangent to $\Omega$. Notice that $P',M,O_2$ form perpendicular bisector of $DE$, and $P',N,O_1$ form perpendicular bisector of $AF$. So $P'A = P'F$ and $P'E = P'D$, and $DF = \frac{1}{2}BD = \frac{1}{2}AC = AE$, so $\triangle P'AE$ is congruent to $\triangle P'FD$. So $P'$ is center of spiral similarity mapping one triangle to the next. Notice that $\angle DP'F = \angle EP'A$ and $\angle FP'E = \angle FP'E$, so $\angle DEP' = \angle FAP'$. Since $\triangle DEP'$ and $\triangle FAP'$ are isosceles, this implies they are also similar. Notice that $O_2M$ bisects minor arc $\overarc{DE}$, so $\angle MO_2E = \angle DCE = \angle CAB = \angle EAB = \angle FBA = \angle FO_1N$. Since $\angle MO_2E = \angle MO_2D$ and $\angle FO_1N = \angle NO_1A$, and $\triangle O_1FA$ and $O_2DE$ are isosceles, then these triangles are also similar. So we actually get that $P'AO_1F$ and $P'EO_2D$ are similar quadrilaterals. This means that $\triangle P'DO_2$ is similar to $\triangle P'FO_1$. So $P$ is center of spiral similarity mapping one triangle to the other. Let $R$ be intersection of $O_1O_2$ and $DF$. Notice that $O_1O_2$ is perpendicular bisector of $AB$ and $CD$ so $R$ is actually intersection point of $AC$ and $BD$. By spiral similarity lemma, we have that $DO_2RP'$ and $P'FRO_1$ concyclic. Notice then that $\angle FRP' = \angle FO_1P' = \angle FO_1N = \angle FBA$, thus $RP' \parallel AB \parallel EF$. Then since $EF$ is perpendicular to $O_1O_2$ we have $\angle P'FO_1 = \angle P'RO_1 = 90^{\circ}$, implying $P'F$ is tangent to $\Gamma$. Similarly $P'A$ is tangent to $\Gamma$, and in a similar way we can show that $P'D$ and $P'F$ are tangent to $\Omega$. Thus we are done.
23.02.2019 21:45
Nice problem! Here’s my short synthetic solution from the contest. The main idea is to consider point $Q$, the isogonal conjugate of $P$ with respect to $\triangle ADE$. [asy][asy] size(200); defaultpen(fontsize(10pt)); pair A, B, C, D, E, P, Q, R; A = dir(125); B = dir(55); C = dir(-5); D = dir(185); E = midpoint(A--C); P = extension(A, rotate(90, A)*circumcenter(A, B, E), D, rotate(90, D)*circumcenter(C, D, E)); Q = extension(A, reflect(A, incenter(A, D, E))*P, D, reflect(D, incenter(A, D, E))*P); R = extension(D, Q, A, B); draw(A--B--C--D--cycle, orange); draw(A--C, orange); draw(circumcircle(A, B, E), heavygreen); draw(circumcircle(C, D, E), heavygreen); draw(circumcircle(A, B, C), red); draw(A--P--D, heavygreen); draw(Q--R, heavycyan); draw(A--Q--D--cycle, magenta+linewidth(0.9)); draw(E--B--A--cycle, magenta+linewidth(0.9)); clip(Circle((0,0), 1.1)); dot("$A$", A, dir(150)); dot("$B$", B, dir(30)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(300)); dot("$P$", P, dir(180)); dot("$Q$", Q, dir(180)); dot("$R$", R, dir(90)); [/asy][/asy] Claim: $\triangle QAD\sim \triangle EBA$. Proof: Observe that $\angle QAD = \angle PAE = \angle EBA$ and $\angle QDA = \angle PDE = \angle DCE = \angle EAB$, which proves the similarity. $\blacksquare$ Claim: $\overline{EQ}\parallel \overline{AB}\parallel \overline{CD}$. Proof: Let $R$ be the reflection of $D$ over $Q$; then from the first claim figures $ADQR$ and $BAEC$ are similar. Hence $\angle DAR = \angle ABC = \angle DAB$, so $R$ must lie on $\overline{AB}$. Thus both $E$ and $Q$ lie on the midline between $\overline{AB}$ and $\overline{CD}$, which implies the claim. $\blacksquare$ Finally, we note that $\angle PED = \angle QEA = \angle EAB = \angle ECD$, which proves the desired tangency.
23.02.2019 21:50
Sniped by pinetree1 Let $P^{*}$ be the isogonal conjugate of $P$ in $\triangle AED$ and reflect $D$ in $P^{*}$ to get $Q$. Observe that $$\triangle AP^{*}D \sim \triangle BEA \implies \triangle AQD \sim \triangle BCA \implies \angle DAQ=\angle ABC=\angle DAB$$so $P^{*}$ lies on the $A$-midline of $\triangle ACD$ hence $\overline{EP^{*}} \parallel \overline{CD}$ yielding $\overline{PE}$ tangent to $\Omega$.
23.02.2019 21:53
Here's another approach: Redefine $P$ as the point where tangents to $\Omega$ at $D$ and $E$ meet. Then we wish to show that $PA$ is tangent to $\omega$. Invert about $C$ with radius $\sqrt{CD \cdot CE}$ followed by reflection in the angle bisector of $\angle ACD$. Then we get the following equivalent problem (relabeled):- Inverted problem wrote: Let $H_A$ be the $A$-Humpty point of $\triangle ABC$, and $M$ be the midpoint of $AB$. Suppose the circle through $A$ and $M$ tangent to $AC$ meets $CM$ again at $N$. Show that $\odot (BMN)$ and $\odot (AH_AM)$ are tangent to each other. Let $K$ and $P$ be the midpoints of $CN$ and $BC$ respectively. Also let $N'$ and $C'$ be the reflections of $N$ and $C$ in $M$. Finally denote the centroid of $\triangle ABC$ by $G$. Then we have $$CA^2=CN \cdot CM=2CM \cdot CK=CC' \cdot CK \Rightarrow CA \text{ is tangent to } \odot (AKC')$$This gives $\angle CAK=\angle AC'C=\angle BCK$, i.e. $BC$ is tangent to $\odot (AKC)$, or equivalently $K \in \odot (AH_AC)$. Also, $ANBN'$ is a parallelogram, so $PK \parallel BN \parallel AN'$, which means that $N'G=2KG$. Thus, $$GH_A \cdot GA=GK \cdot GC=2GK \cdot GM=GN' \cdot GM \Rightarrow N' \in \odot (AH_AM)$$As $ANBN'$ is a parallelogram, this directly gives that $\odot (BMN)$ is tangent to $\odot (AMN'H_A)$. Hence, done. $\blacksquare$ REMARK: After inversion, one can also length bash using Apollonius Theorem and Cosine Law (which was my original solution ).
23.02.2019 22:23
Here's an sketch: Denote $V=AC \cap BD$, $F$ by the midpoint of $BD$, $P=\odot(AVF) \cap \odot(VDE)$. Then simple angle chasing leads: $\triangle PAE \sim \triangle PFD$, and these two are congurent since $AE=DF$. Hence $PA=PF, PE=PD$. Another simple angle chasing shows that $\angle APF=\pi-2\angle ABF$; which implies that $PA, PE$ are both tangent with $\odot(ABE)$. Similarly $PD, PE$ are both tangent with $\odot(CDE)$, as desired. @below: Sorry; Fixed.
21.12.2023 19:47
I present the most unmotivated synthetic solution. We will attempt to guess the point $P$. Let $F$ be the midpoint of $BD$ and $X$ the intersection of the diagonals, let $Q = (CXF) \cap \overline{BC} \neq C$, let $S$ be the point such that $SXQD$ is a parallelogram, let $S' = \overline{SX} \cap (CXF) \neq X$, and let $R$ be such that $RSFQ$ is a parallelogram. We will prove that $S = P$. Firstly, since $\measuredangle QFX = \measuredangle QCX = \measuredangle DCA = \measuredangle BDC = \measuredangle DXS = \measuredangle FXS'$ we determine that $DQ = QF$, and $QF = XS'$, whence $SX = S'X$. Moreover we know that $DS = QX = CS'$ whence $DSS'C$ is an isosceles trapezoid. Also note that $R$ lies on $BD$ because the midpoint of $QS$ lies on $BD$. We now conclude that, in directed length, $RF = RD + DF = XF + FB = XB$. Since we have $QF = XS'$, and also $\angle RFQ = \angle BXS'$, we conclude that $\triangle RFQ \cong \triangle DXS'$. Thus, \[ \measuredangle XBS' = \measuredangle QRF = \measuredangle SFR = \measuredangle SFX = \measuredangle XES' \]and so $BXES'$ is cyclic. By reflection we have $AXFS$ cyclic. We will now show the requested tangencies. Indeed, \[ \measuredangle SDE = \measuredangle FCS' = \measuredangle QCX = \measuredangle DCE \]and \[ \measuredangle SAF = \measuredangle SXF = \measuredangle CDX = \measuredangle ABF\]which shows $S = P$. We then finish by showing that $SE$ is tangent to $(CDE)$. This is not hard, since $SE = S'F = QX = SD$.
06.01.2024 06:36
Alright, all aboard the crappy variable name train. We will complex bash this problem. Let $\Omega$ be the unit circle, and let $F$ be the second intersection of $\Omega \cap \Gamma.$ We let $C=a, E=b, D=c,$ so $F=\frac{ab}{c}.$ We prove that $P=\frac{2ab}{a+b},$ which is the intersection point from the tangents at $D$ and $E,$ is on $\overline{AA}.$ This means that we need $\angle FAP= \angle AEF$ for our claim to be true. Note that $$\frac{F-A}{P-A}=\frac{\frac{xz}{y}-(2y-x)}{\frac{2yz}{y+z}-(2y-x)}=\frac{y+z}{y}.$$Note that $\frac{D}{P}=\frac{z}{\frac{2yz}{y+z}}=\frac{y+z}{2y},$ so $\angle FAP=\angle DOP=\angle ECD=\angle ACD,$ which $\angle ACD=\angle AEF$ becaues $\overline{EF} \parallel \overline{DC},$ and as there is only one point $\overline{AA} \cap \overline{DD},$ this point is also on $\overline{EE},$ and we're done.
09.02.2024 10:41
We use phantom points (and $e$ as a free variable). Define $P'$ as the intersection of the tangents to $\Omega$ at $D$ and $E$. Additionally, set $\Omega$ as the unit circle, and let $d = \overline{c}$, $b = \overline{a}$. We wish to prove $P = P'$, or equivalently that $P'A$ is tangent to $\Gamma$, and we do this by proving $\angle P'AE = \angle ABE$. Note \begin{align*} \frac{e - a}{p' - a} \div \frac{e - b}{a - b} &= \frac{(e - a)(a - b)}{(p' - a)(e - b)} \\ &= \frac{(e - (2e - c))(2e - c - \overline{2e - c})}{\left(\frac{2de}{d + e} - (2e - c)\right)(e - \overline{2e - c})} \\ &= \frac{(c - e)(2e - c - \overline{2e - c})}{\left(\frac{2\overline{c} e}{\overline{c} + e} + c - 2e\right)\left(e - \frac 2e + \frac 1c\right)} \\ &= \frac{(c - e)(2e - c - \overline{2e - c})}{\left(\frac{2e}{1 + ce} + c - 2e\right)\left(\frac{ce^2 - 2c + e}{ce}\right)} \\ &= \frac{ce(1 + ce)(c - e)(2e - c - \overline{2e - c})}{(2e + (c - 2e)(1 + ce))(ce^2 - 2c + e)} \\ &= \frac{ce(1 + ce)(c - e)(2e - c - \overline{2e - c})}{(c + c^2 e - 2ce^2)(ce^2 - 2c + e)} \\ \overline{\frac{e - a}{p' - a} \div \frac{e - b}{a - b}} &= \frac{\frac{1}{ce} \left(\frac{1 + ce}{ce}\right) \left(\frac{e - c}{ce}\right) (\overline{2e - c} - (2e - c))}{\left(\frac 1c + \frac{1}{c^2 e} - \frac{2}{ce^2}\right)\left(\frac{1}{ce^2} - \frac 2c + \frac 1e\right)} \\ &= \frac{ce(1 + ce)(e - c)(\overline{2e - c} - (2e - c))}{c^4 e^4 \left(\frac 1c + \frac{1}{c^2 e} - \frac{2}{ce^2}\right)\left(\frac{1}{ce^2} - \frac 2c + \frac 1e\right)} \\ &= \frac{ce(1 + ce)(c - e)(2e - c - \overline{2e - c})}{(ce^2 + e - 2c)(c - 2ce^2 + c^2 e)} \\ &= \frac{e - a}{p' - a} \div \frac{e - b}{a - b}\text{.} \end{align*} Therefore \[\frac{e - a}{p' - a} \div \frac{e - b}{a - b} \in \mathbb{R}\]and we are done by alternate segment theorem.
16.02.2024 01:54
Seems like this is a new solution Redefine $P$ such that $PE, PD$ are tangent to $(CDE)$ Draw point $F\in (ABCD)$ such that $AD=AF$ Both $\Delta AFD$ and $\Delta PED$ are isosceles and $\angle DFA=\angle DCA=\angle DEP$ $\implies \Delta AFD \sim \Delta PED$ So a spiral similarity at $D$ takes $\Delta AFD$ to $\Delta PED$, so we get $\Delta DPA \sim DEF$ $AF=AD=BC$ so $ACBF$ is isosceles trapezoid $\implies \angle AEF=\angle BEC$ (because $EA=EC$) Now we can angle chase the rest Let $\angle BAC=\alpha, \angle CBA=\beta, \angle EDC=\theta, \angle EBA=\phi$ $\angle DPA=\angle DEF=\angle DEA+\angle AEF=\theta+\alpha+\angle BEC=\theta+\phi+2\alpha$ $\angle EAP=\angle EAD-\angle PAD=(\beta-\alpha)-(180-\angle ADP-\angle DPA) \newline =\beta-\alpha-180+(180-\beta-\alpha-\theta)+(\theta+\phi+2\alpha)=\phi$ So $PA$ is tangent to $(AEB)$ as desired
Attachments:

16.02.2024 06:54
Let $F$ be the midpoint of $BC$, on both circles. Let $G$ be the point on $\Omega$ such that $(GF;ED)=-1$. Let $GF$ hit $ED$ at $J$. Redefine $P$ as the intersections of the tangents at $E,D$ to $\Omega$. Thus $P,F,J,G$ are collinear. Claim: $FP$ is tangent to $\omega$. Proof. Let $G'=CG\cap EF$. As $-1=(GF;ED)\overset{C}{=}(G'F;E\infty)$ so $E$ is the midpoint of $G'F$. Thus $AFCG'$ is a parallelogram, so $CG \parallel AF$. Reim finishes. $\blacksquare$ Let $PE$ intersect $AF$ at $I$ and $\omega$ at $H$. Claim: $EFIJ$ is cyclic Proof. The proof is just angles. \[\measuredangle JFI=\measuredangle PFI=\measuredangle FBA=\measuredangle FDC =\measuredangle FDE+\measuredangle EDC=\measuredangle FEI+\measuredangle JEF=\measuredangle JEI\]$\blacksquare$ Therefore, by reims on $\omega$ and $\Omega$, $FH\parallel EG$. Similarly, reims on $(EFJI)$ and $\omega$ gives $IJ\parallel FH$. Thus, \[-1=(PJ;FG)\overset{\infty_{FH}}{=}(PI;HE)\overset{F}{=}(FA;HE)\]Hence $EH$ is a symmedian and thus $P$ is the intersection of the tangents at $A$ and $F$ to $\omega$. We thus conclude the result. Remark: Very good.
20.02.2024 10:42
Let $K = \overline{AC} \cap \overline{BD}$ and $F$ denote the midpoint of $BD$. Claim: Let $Q$ be the second intersection of $(KAF)$ and $(KED)$. Then $\overline{QK} \parallel \overline{CD}$. Proof. Let $f(\bullet) = \text{Pow}(\bullet, (KAF)) - \text{Pow}(\bullet, (KED))$. Then it suffices to show that $f(C)=f(D)$. Compute \[ f(C)-f(D) = (CK \cdot CA - CE \cdot CK) - (DF \cdot DK - 0) = CK \cdot EA - DF \cdot DK= CK \cdot EC - DF \cdot DK = 0, \]as desired. Claim: In fact, $P=Q$. Proof. Note that as $ABEF$ is an isosceles trapezoid, the center of $(ABEF)$ lies on $(KAF)$. Thus, $P$ lies on $(KAF)$. I contend that $\overline{PK} \parallel \overline{FE}$, which would imply $P=Q$ by the prior claim. By simple angle chasing we have \[ \angle PKA = \angle PFA = \angle FEA, \]which implies the desired parallelism. Claim: Finally, $PD=PE$. Proof. Note that $P$ is the center of the spiral similarity mapping $\overline{AF} \mapsto \overline{ED}$. Thus, $\triangle FPA \sim \triangle DPE$, so as $FP=AP$, we have $DP=EP$, as desired. We conclude from the final claim.
20.02.2024 12:15
Set $(DEC)$ as the unit circle and redefine $P$ as the point such that $\overline{PD}$ and $\overline{PE}$ are tangent to $\Omega$. Orient so that $\overline{CD}$ is perpendicular to the real axis. Define $F$ as the midpoint of $\overline{BD}$. Compute, \begin{align*} d &= 1/c\\ f &= 1/e\\ a &= 2e - c\\ p &= \frac{2de}{d+e} = \frac{2e/c}{1/c + e} = \frac{2e}{ce + 1} \end{align*}However clearly we have, \begin{align*} \frac{p - a}{f - a} &= \frac{\frac{2e - (2e - c)(ce + 1)}{ce + 1}}{1/e - 2e + c}\\ &= \frac{\frac{2e - 2e - 2e^2c + c^2e + c}{ce + 1}}{1/e(1 - 2e^2 + ce)}\\ &= \frac{\frac{c(2e^2 + ce + 1)}{ce + 1}}{1/e(1 - 2e^2 + ce)}\\ &= \frac{ce}{ce + 1} \end{align*}Also we have, \begin{align*} \frac{a - c}{d - c} &= \frac{2e - 2c}{1/c - c}\\ &= \frac{2c(e - c)}{1 - c^2} \end{align*}Then dividing we have, \begin{align*} \frac{2(e-c)(ec + 1)}{e(1 - c^2)} \end{align*}Taking the conjugate this is simply, \begin{align*} \frac{2/c^2e^2(c - e)(ec + 1)}{1/c^2e(c^2 - 1)} = \frac{2(e - c)(ec+1)}{e(1-c^2)} \end{align*}and so we find $\angle PAF = \angle DCA = \angle FEA$ proving the tangency.
28.03.2024 16:55
Let $F$ be the other intersection of $(ABC)$ and $(CDE)$, redefine $P$ to be the intersection of tangents through $D$ and $E$ of $(DEC).$ Then, $B,F,E$ are collinear as: \[\measuredangle BFE = \measuredangle BAE = \measuredangle ACD = -\measuredangle EFD.\] Furthermore, this also implies $EF \parallel BC$ as \[\measuredangle BFE = \measuredangle BAC = \measuredangle BDC = \measuredangle FDC.\] Now, we use complex numbers with $(DFEC)$ as the unit circle. We see that $p = \frac{2de}{d+e}$, $f = cd/e$, and $a = 2e - c.$ We compute \begin{align*} \frac{a-p}{a-f} \div \frac{e-a}{e-f} &= \frac{2e-c-\frac{2de}{d+e}}{2e-c-\frac{cd}{e}} \times \frac{e-\frac{cd}{e}}{c-e} \\ &= \frac{e}{d+e} \times \frac{(d+e)(2e-c)-2de}{2e^2-ce-cd} \times \frac{e^2-cd}{e(c-e)} \\ &= \frac{1}{d+e} \times \frac{2e^2 - cd - ce}{2e^2 - ce - cd} \times \frac{e^2-cd}{c-e}\\ &= \frac{e^2-cd}{(d+e)(c-e).} \end{align*} This has conjugate \[\frac{\frac{1}{e^2} - \frac{1}{cd}}{(\frac{1}{d} + \frac{1}{e})(\frac{1}{c}-\frac{1}{e})} = \frac{cd-e^2}{(e+d)(e-d)} = \frac{a-p}{a-f} \div \frac{e-a}{e-f}.\] Hence, $\frac{a-p}{a-f} \div \frac{e-a}{e-f} \in \mathbb{R}$ giving us $\measuredangle FAP = \measuredangle FEA$ as needed.
28.03.2024 17:39
Set $\Omega$ be the unit circle. WLOG, rotate so that $CD$ is parallel to the real axis. Then, $d=-\frac{1}{c}$. We can then obtain, \[a=2e-c\]and, \[b=-\overline{a}=-\overline{2e-c}=\frac{e-2c}{ce}\]Let $P'$ (denoted by the complex number $p$) be the intersection of the tangents to $\Omega$ at $D$ and $E$.Then, we can also easily obtain that, \begin{align*} p &= \frac{2de}{d+e}\\ &= \frac{2\left(-\frac{1}{c}\right)e}{e-\frac{1}{c}}\\ &= \frac{-\frac{2e}{c}}{\frac{ce-1}{c}}\\ &= \frac{2e}{1-ec} \end{align*}Now, we wish to show $\measuredangle PAE = \measuredangle ABE$. Thus, we require, \[\arg \left(\frac{e-a}{p-a}\right)=\arg\left(\frac{e-b}{a-b}\right)\]Thus, we must have $A = \frac{(e-a)(a-b)}{(p-a)(e-b)} \in \mathbb{R}$. Note that, \begin{align*} A &= \frac{(e-a)(a-b)}{(p-a)(e-b)}\\ &= \frac{(e-2e+c)(2e-c+\frac{2c-e}{ce})}{\left( \frac{2e}{1-ce}-2e+c \right)\left( e+\frac{2c-e}{ce} \right)}\\ &= \frac{(c-e)(1-ce)(2ce^2-c^2e+2c-e)}{(2ce^2-c^2e+c)(2c+ce^2-e)} \end{align*}Now, we also have that, \begin{align*} \overline{A} &= \overline{\left(\frac{(c-e)(1-ce)(2ce^2-c^2e+2c-e)}{(2ce^2-c^2e+c)(2c+ce^2-e)}\right)}\\ &= \frac{\left(\frac{1}{c}-\frac{1}{e}\right)\left(\frac{2}{ce^2}-\frac{1}{c^2e}+\frac{2}{c}-\frac{1}{e}\right)\left(1-\frac{1}{ce}\right)}{\left( \frac{2}{ce^2}-\frac{1}{c^2e}+\frac{1}{c} \right)\left(\frac{2}{c}+\frac{1}{ce^2}-\frac{1}{e}\right)}\\ &= \frac{\frac{(e-c)}{ec}\left( \frac{2c-e+2ce^2-c^2e}{c^2e^2} \right)\left(\frac{ce-1}{ce}\right)}{\left(\frac{2c-e+ce^2}{c^2e^2}\right)\left( \frac{2e^2-ce+1}{ce^2} \right)}\\ &= \frac{c^3e^4(c-e)(1-ec)(2ce^2-c^2e+2c-e)}{c^4e^4(ce^2+2c-e)(2e^2+1-ec)}\\ &=\frac{(c-e)(1-ce)(2ce^2-c^2e+2c-e)}{(2ce^2-c^2e+c)(2c+ce^2-e)} \end{align*}Thus, $\overline{A}=A$ and indeed, $\frac{(e-a)(a-b)}{(p-a)(e-b)} \in \mathbb{R}$ which implies that $\measuredangle PAE = \measuredangle ABE$. Thus, $\overline{PA}$ is tangent to $\Gamma$ at $A$ implying that $P'=P$. Thus, indeed $\overline{PE}$ is tangent to $\Omega$ as claimed.
06.05.2024 19:49
Let $F$ be the midpoint of $BD.$ Redefine $P$ as the center of spiral similarity sending $DF$ to $EA.$ Since $DF=EA$ we have $\triangle DPF\cong\triangle EPA$ so $DP=EP,FP=AP.$ Let $X$ be the intersection of $DF$ and $EA.$ Then we have $PXED$ cyclic so $\measuredangle DPE=\measuredangle DXE=2\measuredangle DCA,$ but this implies $PD,PE$ are tangent to $(CDFE).$ Similarly we get $PA,PF$ are tangent to $(ABEF),$ so we are done.
09.06.2024 17:54
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/* end of picture */ [/asy][/asy] Sketch for the solution: First of we redefine $P$ as the intersection of the tangents $PD$, $PE$. 1- Show that $PDEH$ cyclic. 2- Show that $PED$ congruent to $PAF$ Then a spiral similarity centered at $P$ sending $DE$ to $AF$ does the job.
06.07.2024 03:23
Evil MathLuis be like: Let $F$ midpoint of $BD$, now take $(ECDF)$ as the unit circle and let $E=e, C=c, D=d, F=f$ $$CD \parallel EF \implies \frac{d-c}{f-e} \in \mathbb R \implies \frac{d-c}{f-e}=\frac{\overline{d}-\overline{c}}{\overline{f}-\overline{e}}=\frac{\frac{c-d}{cd}}{\frac{e-f}{ef}} \implies f=\frac{cd}{e}$$Now redefine $P=p=\frac{2ed}{e+d}$ and due to reflection $A=a=2e-c$, now we compute $\text{arg}(\angle FAP)$ $$\frac{f-a}{p-a}=\frac{\frac{cd}{e}-(2e-c)}{\frac{2ed}{e+d}-(2e-c)}=\frac{e+d}{e} \cdot \frac{cd-2e^2+ec}{2cd-2e^2+ec-2cd+cd}=\frac{e+d}{e}$$$$\angle FAP=\angle DCE \iff \frac{d-c}{e-c} : \frac{f-a}{p-a} \in \mathbb R \iff \frac{d-c}{e-c} \cdot \frac{e}{e+d}=\frac{d-c}{e-c} \cdot \frac{ec}{dc} \cdot \frac{d}{e+d}=\frac{d-c}{e-c} \cdot \frac{e}{e+d}$$Therefore $\angle FAP=\angle DCE=\angle FEA$ thus $AP$ is tangent to $(ABE)$ as desired thus we are done
17.09.2024 13:58
Let $AC\cap BD=\{O\}$ and let $a:=\angle ACD$. Claim: $\angle APD=\angle DEC+\angle AEB$ Proof: $\angle DEC=180^\circ-\angle EDC-a$ but notice that $\angle EDC=\angle PDO$ so $\angle DEC=180^\circ-\angle PDO-a$ we similarly get $\angle AEB=180^\circ-\angle PAO-a$, hence their sum is equal to $360^\circ-\angle PDO-\angle PAO-2a$ and the conclusion follows from the quadrilateral $PDOA$ and noting that $\angle AOD=2a$. Claim: $\frac{\sin(\angle DPO)}{\sin(\angle APO)}=\frac{\sin(\angle DEC)}{\sin(\angle AEB)}$ Proof: Trig Ceva in $\triangle APD$ with point $O$ gives: $$\frac{\sin(\angle DPO)}{\sin(\angle APO)}=\frac{\sin(\angle DAO)}{\sin(\angle PAO)}\cdot\frac{\sin(\angle PDO)}{\sin(\angle ADO)}$$ We also have that $\angle PDO=\angle EDC$ and $\angle PAO=\angle ABE$ and $\frac{\sin(\angle DAO)}{\sin(\angle ADO)}=\frac{DO}{AO}=\frac{DC}{AB}$ so LOS in $\triangle DEC$ and $\triangle ABE$ gives the desired conclusion. Finally, from our claims we can conclude that $\angle DPO=\angle DEC$ and $\angle APO=\angle AEB$ so $PDOE$ is cyclic. This gives $a=\angle DCA=\angle PDE=\angle POA$ so $\angle POD=\angle PED=a$, therefore $PE$ is tangent to $\Omega$. $\blacksquare$
25.09.2024 18:41
I bashed this on paper, so here's a summary: Redefine $P$ as the pole of $DE$ wrt $(CDE)$. Then we want to show that $\angle PAF=\angle ABF$. Now we can just use complex with $(CDE)$ as the unit circle. Eventually we get that we want to prove \[\frac{de(d-f)}{(d+e)(2de-ef-2df+d^2)}\in\mathbb{R}.\]This is not hard to do.
01.01.2025 20:33
Let $(CDE)$ be the unit circle so that $c=\frac{1}{d}$. Note * $a=2e-c$, * $b=\overline a=\frac{2c-e}{ec}$. Let $P'=DD\cap EE$. Note \[p'=\frac{2de}{d+e}=\frac{2e}{1+ec}.\]Then, \[P=P'\iff P'\in AA \iff \angle P'AE=\angle ABE,\]so \[\frac{\left(\frac{p'-a}{e-a}\right)}{\left(\frac{a-b}{e-b}\right)}=\frac{c(1+ec-2e^2)(-2c+e+e^2c)}{(1+ec)(c-e)(2e^2c-ec^2-2c+e)}.\]Conjugating, \[\left(\overline{\frac{c(1+ec-2e^2)(-2c+e+e^2c)}{(1+ec)(c-e)(2e^2c-ec^2-2c+e)}}\right)=\frac{\left(\frac{1}{c}\right)\left(\frac{e^2c+e-2c}{e^2c}\right)\left(\frac{-2e^2+ec+1}{e^2c}\right)}{\left(\frac{1+ec}{ec}\right)\left(\frac{e-c}{ec}\right)\left(\frac{2c-e-2e^2c+ec^2}{e^2c^2}\right)},\]\[=\frac{c(1+ec-2e^2)(-2c+e+e^2c)}{(1+ec)(c-e)(2e^2c-ec^2-2c+e)}.\]They are equal, so $P=P'$.