Easy to notice that $ \angle QBE = \angle PCE$ and $ \angle PDE = \angle EAQ$. $ \frac {BP \cdot PC}{BQ \cdot CQ} = ( \frac {BP}{BQ} \cdot \frac {sin {\angle PBE}}{sin {\angle EBQ}}) \cdot ( \frac {PC}{CQ} \cdot \frac {sin {\angle PCE}}{sin {\angle QCE}}) = (\frac {PE}{EQ})^{2}$, implying $ \frac {BP \cdot PC}{BQ \cdot CQ} = (\frac {PE}{EQ})^{2}$ (1). Analogously, $ \frac {AP \cdot AQ}{PD \cdot DQ} = (\frac {PE}{EQ})^{2}$ (2). By (1) and (2), we deduce that $ \frac {BP \cdot PC}{BQ \cdot CQ} = \frac {AP \cdot PD}{AQ \cdot DQ}$ (3). Denote by $ d(X;YZ)$ the distance from point $ X$ to line $ YZ$ and denote by $ [MNK]$ the area of the triangle $ MNK$. Using (3), we write: $ \frac {BP \cdot PC} {BQ \cdot QC} = \frac {[BPC]} {[BQC]} = \frac {d(P;BC)} {d(Q;BC)}$. On the other hand, we analogously get $ \frac {AP \cdot PD}{AQ \cdot DQ} = \frac {d(P;AD)} {d(Q;AD)}$. Hence, $ \frac {d(P;BC)} {d(Q;BC)} = \frac {d(P;AD)} {d(Q;AD)}$ (*). From this, we easily get that the lines $ PQ$, $ BC$ and $ AD$ meet at point, say $ S$. Than, $ SB \cdot SC = SP \cdot SQ$ (from cyclic quadrilateral $ BCQP$) and $ SA \cdot SD = SP \cdot SQ$ (from cyclic quadrilateral APQD), implying $ SB \cdot SC = SA \cdot SD$, which means that $ ABCD$ is cyclic, Q.E.D. Lasha Lakirbaia

mr.danh wrote:

I have a question everybody : Isnt G4 a little easier than G3?

Maybe, because I spent much more time for G3 than for this one, but I think that this two problems are not far from each other with difficulty level.

Alternatively, we can show that $ PQ$ is tangent to both the circumcircles of $ \triangle AED$ and $ \triangle BEC$ and then do a bit of algebra.

vishalarul also notice that, but instead of doing algebra, he used inversion.

Yes the inversion proof is very nice too! However since I wasn't too familiar with the techniques of inversion I used algebra, which took a bit longer but is quite nice also.

Sorry for bringing back this problem but i have one more nice solution. Again using inversion but this time with a different center. We invert through Q (or P) and r=PQ. So P inverts to itself. Since PADQ and PBCQ are cyclic, the triples of points P,A',D' and P,B',C' are collinear. Since we want to prove that A'B'C'D' is cyclic it's enough to prove that $ PA'\cdot PD' = PB'\cdot PC' $ (*). But using $ PA'= PA\cdot \frac{r^2}{QP \cdot QA} $ etc, (*) is equivalent to $ \frac{PA\cdot PD}{PB\cdot PC}=\frac{QA\cdot QD}{QB\cdot QC} $ which is relation 3 from lasha's solution :

Since $PQDA$ and $QPBC$ are cyclic, it follows that $\angle{EBQ}=\angle{ECP}$ and that $\angle{EAQ}=\angle{EDP}$. By sine law, we have that \[\frac{PE}{EQ}=\frac{PC \cdot \sin{\angle{PCE}}}{QC \cdot \sin{\angle{QCE}}}=\frac{PB \cdot \sin{\angle{PBE}}}{QB \cdot \sin{\angle{EBQ}}}\] which implies that \[\left( \frac{PE}{EQ} \right)^2 = \frac{PC \cdot \sin{\angle{PCE}}}{QC \cdot \sin{\angle{QCE}}} \cdot \frac{PB \cdot \sin{\angle{PBE}}}{QB \cdot \sin{\angle{EBQ}}} = \frac{PC}{QC} \cdot \frac{PB}{QB}\] Now let $BC$ intersect $PQ$ at $X$. Since $QPBC$ is cyclic, it follows that $\triangle{XBP} \sim \triangle{XQC}$ and $\triangle{XCP} \sim \triangle{XQB}$ and hence that \[\left( \frac{PE}{EQ} \right)^2 = \frac{PC}{QB} \cdot \frac{PB}{QC} = \frac{XP}{XB} \cdot \frac{XB}{XQ}=\frac{XP}{XQ}\] By the same argument, it follows that if $Y$ is the intersection of $AD$ and $PQ$, \[\frac{XP}{XQ} = \left( \frac{PE}{EQ} \right)^2 = \frac{YP}{YQ}\] Hence $X=Y$ and $AD$, $BC$ and $PQ$ are concurrent at $X$. By power of a point $XP \cdot XQ = XB \cdot XC = XA \cdot XD$ which implies that $ABCD$ is cyclic.

First,take inversion with center E and any radius Now,we easy obtain A*D*//P*Q*//PQ//B*C*,by simple angle chase,and because an inversion pictures a cyclic to a cyclic,we obtain that A*D*Q*P* is an isoceles trapezoid,the same we obtain that B*C*Q*P* is an isoceles trapezoid and from that we have that A*B*C*D* is an isoceles trapezoid,and from that we have that ABCD is a cyclic

Ahwingsecretagent wrote: Alternatively, we can show that $ PQ$ is tangent to both the circumcircles of $ \triangle AED$ and $ \triangle BEC$ and then do a bit of algebra. Working on this idea.... $\angle PEB = 180 - \angle EPB - \angle EBP = 180 - (180-\angle BCQ)-\angle EBP = \angle BCQ - \angle BCE$ $\implies PQ$ is tangent to $\odot BEC$ Similarly, $PQ$ is tangent to $\odot ADE$ Let $BC \cap PQ = X$ and $AD \cap PQ = Y$ By power of a point, $XQ \cdot XP = XC \cdot XB = XE^2$ Similarly, $YP \cdot YQ = YE^2$ Obviously now $X=Y$, Now by power of a point $XA \cdot XD = XE^2 = XB \cdot XC$ $\implies ABCD$ is cyclic QED PS. Missed an obvious thing

lasha wrote: Easy to notice that $ \angle QBE = \angle PCE$ and $ \angle PDE = \angle EAQ$. $ \frac {BP \cdot PC}{BQ \cdot CQ} = ( \frac {BP}{BQ} \cdot \frac {sin {\angle PBE}}{sin {\angle EBQ}}) \cdot ( \frac {PC}{CQ} \cdot \frac {sin {\angle PCE}}{sin {\angle QCE}}) = (\frac {PE}{EQ})^{2}$, implying $ \frac {BP \cdot PC}{BQ \cdot CQ} = (\frac {PE}{EQ})^{2}$ (1). Analogously, $ \frac {AP \cdot AQ}{PD \cdot DQ} = (\frac {PE}{EQ})^{2}$ (2). By (1) and (2), we deduce that $ \frac {BP \cdot PC}{BQ \cdot CQ} = \frac {AP \cdot PD}{AQ \cdot DQ}$ (3). Denote by $ d(X;YZ)$ the distance from point $ X$ to line $ YZ$ and denote by $ [MNK]$ the area of the triangle $ MNK$. Using (3), we write: $ \frac {BP \cdot PC} {BQ \cdot QC} = \frac {[BPC]} {[BQC]} = \frac {d(P;BC)} {d(Q;BC)}$. On the other hand, we analogously get $ \frac {AP \cdot PD}{AQ \cdot DQ} = \frac {d(P;AD)} {d(Q;AD)}$. Hence, $ \frac {d(P;BC)} {d(Q;BC)} = \frac {d(P;AD)} {d(Q;AD)}$ (*). From this, we easily get that the lines $ PQ$, $ BC$ and $ AD$ meet at point, say $ S$. Than, $ SB \cdot SC = SP \cdot SQ$ (from cyclic quadrilateral $ BCQP$) and $ SA \cdot SD = SP \cdot SQ$ (from cyclic quadrilateral APQD), implying $ SB \cdot SC = SA \cdot SD$, which means that $ ABCD$ is cyclic, Q.E.D. Lasha Lakirbaia I have the same solution Lasha, but looking at your post it seems rather long and boring

I shall use the diagram of livetolove212. We invert about $E$ with an arbitrary radius. Denote the image of point $X$ by $X'.$ Then $\angle D'Q'E = \angle QDE = \angle PAE = A'P'E.$ Also, the circle $\odot (APQD)$ is mapped to another circle $\odot (A'P'Q'D')$. Hence we infer that $A'P'Q'D'$ is an isosceles trapezoid with $A'P'=D'Q.$ Similarly $P'B'C'Q'$ is an isosceles trapezoid with $P'B' = Q'C'.$ Thus we have that $A'B'C'D'$ is an isosceles trapezoid. Hence inverting back we get the result. Note. It seems that I have the same solution as mr.danh . Okay, so I leave behind some motivation. Motivation. What is very elegant when you have equal angles? Circles or parallel lines, right? So my first instinct was to reflect some points but almost immediately it failed to produce nice circles. Now for parallel lines we need a transversal. We know that $P,E,Q$ are collinear so why not make it a transversal? The best way in which we can do this is by an appropriate inversion, which is what we have done above.

Great problem... Let $AD\cap PQ=X$ and $BC\cap PQ=Y$

Yup I also inverted

We will prove AD,PQ,BC are concurrent. Step1 : PQ is tangent to AED and BEC. ∠PEB = ∠180 - ∠EBP - ∠EPB = ∠QCB - ∠QCE = ∠ECB. we can prove the rest with same approach. Step2 : AD,PQ,BC are concurrent. Let PQ and BC meet at S. SQ.SP = SC.SB = SE^2 Let PQ and AD meet at K. KQ.KP = KD.KA = KE^2 with some calculation and algebra stuff we can prove K is S. we're Done.

The algebra part: Let's assume there exist both S,K such that SE^2 = SQ.SP and KE^2 = KQ.KP Let PE = y, EQ = x, QS = a and SK = z. 1 - we have a(a+x+y) = (a+x)^2 which when we expand gives ay = ax + x^2. 2 - we also have (z+a)(z+a+x+y) = (z+a+x)^2 which when we expand gives zy = zx so we mast have x = y but then by first part we have x = y = 0 so we don't have both S and K.

I don't think anyone has posted this solution before. Let the circumcircles of $APE$ and $DQE$ intersect at $R$, and let the circumcircles of $BPE$ and $CQE$ intersect at $S$. Notice that $\overline{RE}$ bisects $\angle PRQ$ and $\overline{SE}$ bisects $\angle PSQ$. We solve the following stronger problem: Quote: Let $P$ and $Q$ be points, and let $E$ be a point on segment $PQ$. Let $R$ and $S$ be points on the Apollonius circle $\omega$ of $P$ and $Q$ passing through $E$. Let $A$ be on the circumcircle of $\triangle PER$ and let $D$ be on the circumcircle of $\triangle QER$ such that $PADQ$ is cyclic. Prove that $\overline{AD}$ passes through the circumcenter $O$ of $\omega$. By length chasing, $OP \cdot OQ=OE^2$. Consider an inversion about $\omega$. It sends $P$ to $Q$, so the circumcircle of $PADQ$ goes to itself. Thus, $A$ and $D$ swap under inversion, so $A$, $D$, and $O$ are collinear.

Consider the equivalent problem after inverting at $E$. Then $PQAD$ and $PQBC$ are both isosceles trapezoids by the angle condition, so $ABCD$ must also be an isosceles trapezoid. $\square$ Remark: This solution has the same idea as the one that proves that $(EAD)$ is tangent to $PQ$ and so on.

Invert at E with arbitrary radius. We have A'P'Q'=A'P'E=PAE=QDE=D'Q'E=D'Q'P'; in particular, since A'D'P'Q' is cyclic, it is an islsceles trapezoid, and similarly B'C'Q'P is, too. Now, noting that A with D and B with C are symmetric about the perp. bisector of P'Q', A'B'C'D' is also a cyclicislscelestrapezoid; in particular, inverting back through E gives ABCD is cyclic.

I think we can make the problem easier: i claim prooving existance of such a point E for cyclic ABCD is enough. for the existance it suffices to take E the point of tangancy of the circle through A and D that touches line PQ, this point works cause of a simple angle chase and if you can consider (AD) inter (BC) and use PoP you can prove that this E also works for PQPC, we are left with the case when AD is parallele to BC but that case is pretty obvious for the claim, let's suppose ABCD is not cyclic, take D' the intersection of (ABC) with (APQ), notice that E of ABCD is the same as the E of ABCD' (because BPQC uniquely determines E in both cases), from here you get angles EDQ = EAQ = ED'Q which is a contradiction (it implies E is either P or Q which means P=Q and u get an easy case)

Let line $(AE)$ cut circle $(APQD)$ again at $A^*$. Define $B^*$, $C^*$ and $D^*$ similarly. The angle conditions then rewrite as $(B^*C^*)\parallel (PQ)\parallel (A^*D^*)$, with $PC^*B^*Q$ and $PD^*A^*Q$ both being isosceles trapezoids. In particular, $A^*B^*C^*D^*$ itself must be an isosceles trapezoid, and is thus a cyclic quadrilateral. Since the negative inversion at $E$ fixing both circles swaps $X$ and $X^*$ for $X=A$, $B$, $C$ and $D$, we have that $ABCD$ is also cyclic, as desired. $\square$

Invert about $E$. Then let the image of $X$ under inversion be $X'$. Clearly $\angle A'P'Q' = \angle A'Q'P',$ so $A'P'Q'D'$ is an isoceles trapezoid with $A'D' \parallel P'Q'$. Similarly, $B'P'Q'C'$ is an isosceles trapezoid with $B'C' \parallel P'Q'$. Both of these, of course, are symmetric about the perpendicular bisector of $P'Q'$. Now observe that the perpendicular bisector of $P'Q'$ is also that of $A'D'$ and $B'C'$, and this is enough to imply that $A'D'C'B'$ is an isosceles trapezoid and is therefore cyclic, so $ABCD$ is cyclic as well. $\square$

Upon inversion at $E$ it is easy to see $PQDA,QPBC$ must be sent to cyclic isosceles trapezoids, so $ABCD$ is as well which finishes.

Let $A'$, $D'$ be the second intersections of $(PQDA)$ with lines $AE$ and $DE$ respectively. Note that by the angle condition, $\measuredangle A'AP = \measuredangle QDD'$. $\measuredangle ADE = \measuredangle ADQ + \measuredangle QDD' = \measuredangle EA'Q + \measuredangle A'AP = \measuredangle EA'Q + \measuredangle A'QE = \measuredangle A'EQ = \measuredangle AEP \implies PQ$ is tangent to $(ADE)$. Let $X=AD \cap PQ$, and let $X'$ be the reflection of $E$ across $X$. Since $XE$ is tangent to $(ADE)$, we have $XE^2=XA \cdot XD=XP \cdot XQ \implies (P,Q;E,X')=-1$. Similarly, if we let $Y=BC \cap PQ$ and $Y'$ be the reflection of $E$ across $Y$, we get that $YE^2=YB \cdot YC$ and $(P,Q;E,Y')=-1$. So $X'=Y' \implies X=Y$, so $XA \cdot XD = XC \cdot XB \implies A,B,C,D$ are concyclic.

first, since $APQD$ is cyclic and $PAE=QDE$, we get that $AEP=ADE$ similarly, $BEP=BCE$, so $PQ$ is tangent to both the circumcircle of $AED$ and the circumcircle of $BEC$. let $PQ$ intersect $AD$ and $BC$ at $X$ and $Y$, and $XA*XD=XE^2$ and $YB*YC=YE^2$ also, $XA*XD=XP*XQ$ and $YB*YC=YP*YQ$ then, $XE^2=XP*XQ$ and $YE^2=YP*YQ$ so we can solve that $X=Y$ then, since $AD$ and $BC$ intersect on the radical axis, $XA*XD=XB*XC$ and $\frac{XA}{XB}=\frac{XC}{XD}$ so $ABCD$ are concyclic

Notice that $$\angle EBC = \angle PBC - \angle PBE = 180^\circ - \angle PQC - \angle QCE = \angle QEC.$$Therefore $(\triangle EBC)$ is tangent to $PQ.$ Similarly $(\triangle EAD)$ is tangent to $PQ.$ Now, let $X=AD \cap PQ, Y=BC \cap PQ.$ If any of $X, Y$ do not exist then it is easy to show by symmetry that $AB \parallel BC$ and then $ABCD$ is an isosceles trapezoid, done. Thus assume that $X, Y$ exist. By PoP $$XP \cdot XQ = XA \cdot XD = KE^2 = YB \cdot YC = YP \cdot YQ.$$Therefore either $X \equiv Y$ or $X, Y$ are on opposite sides relative to the circles. If the former case holds, we are done by PoP. Otherwise, for the sake of a contradiction assume that the latter case holds. Then Clearly $XE=YE,$ and $$XP \cdot XQ = YP \cdot YQ \iff (XE-PE)(XE+QE)=(YE-QE)(YE+PE) \iff QE(XE+YE)=PE(XE+YE).$$Hence $QE=PE,$ so by symmetry $A,D$ are reflections of each other over the perpendicular to $PQ$ at $E,$ and similarly for $B,C$ and thus $AD \parallel BC,$ a contradiction. QED