Solution. We invert about $C$ swapping $E$ and $A$. Under this inversion, we must have $A \mapsto E$, $B \mapsto P$, and $D\mapsto Q$; therefore as $ABCD$ is a cyclic quadrilateral, $P, E,$ and $Q$ are collinear. Next, let $G'$ be the image of $G$ under this inversion. I claim that $EB = EG'$. Note that $\angle{EBC} = \angle{EAC} = 180^{\circ} - \angle{EAG} = \angle{EG'G}$, so the claim follows. Similarly if $F'$ is the image of $F$, then $ED = EF'$. Then the image of line $GQ$ is the circumcircle of $\triangle{G'DC}$, and similarly the image of line $FP$ is the circumcircle of $\triangle{F'BC}$. Suppose these circles meet at $X$, which is the image of $M$ under the inversion. Then we want to show that $\angle{CXE} = 90^{\circ}$. The good thing about this inversion is that it completely eliminates the necessity of points $A, P, Q, G$ and $F$; we may restate the problem as follows: Inverted USAMO 5 wrote: Suppose that $CDB$ is a triangle, and let $E$ be some point on side $BD$. Let $U$ and $V$ be the feet of the perpendiculars from $E$ to $CD$ and $CB$, respectively, and let $F'$ and $G'$ be the reflections of $D$ and $B$ over $U$ and $V$, respectively. Let the circumcircles of $\triangle{G'DC}$ and $\triangle{F'BC}$ meet at $X \neq C$. Show that $\angle{CXE} = 90^{\circ}$. Equivalently, it suffices to demonstrate that $CUXEV$ is a cyclic quadrilateral; we know that $CUEV$ is cyclic with diameter $CE$, so we want to show that $\angle{UXV} = 180^{\circ} - \angle{C}$. We claim that $X$ is the center of the spiral similarity mapping $DG'$ to $F'B$. Observe that $\angle{XDG'} = \angle{XCB} = \angle{XF'B}$; similarly, $\angle{XG'D} = \angle{XCD} = \angle{XBF'}$. Thus this claim is proven. Therefore, by the Gliding Principle and the principle that spiral similarities come in pairs, we observe that $X$ maps $U$, the midpoint of $DF'$, to $V$, the midpoint of $G'B$. Therefore $\angle{UXV} = \angle{F'XB} = 180^{\circ} - \angle{C}$, and we may conclude.

Since $$\measuredangle APC = \measuredangle APB = \measuredangle AEB = \measuredangle AED = \measuredangle AQD = \measuredangle AQC,$$$APCQ$ is cyclic. Also, $$\measuredangle ADQ = \measuredangle ADC = \measuredangle ABC = \measuredangle ABP,$$so $\triangle ADQ \sim \triangle ABP$, and $A$ is the Miquel point of quadrilateral $BDQP$. Note that $BDPQ$ is cyclic because $$CD\cdot CQ = CA\cdot CE = CB \cdot CP.$$Also $P$, $E$, and $Q$ are collinear by the spiral similarity lemma. Now let $X = BQ \cap DP$, and by Pappus' theorem on hexagon $BFPDGQ$, we get that $A$, $M$, and $X$ are collinear. From Brocard's theorem on cyclic quadrilateral $BDQP$, $EC$ is the polar of $X$ w.r.t. $(BDQP)$, and from well-known properties of the Miquel point configuration (namely that the Miquel point and the intersection of the diagonals are inverses w.r.t. the circumcircle), $XA \perp AC$, so $MA \perp AC$.

This is indeed a Miquel point problem, but the main idea is to focus on the self-intersecting cyclic quadrilateral $PBQD$ as the key player, rather than on the given $ABCD$. Indeed, we will prove that $A$ is its Miquel point; this follows from the following two claims. Claim: The self-intersecting quadrilateral $PQDB$ is cyclic. Proof. By power of a point from $C$: $CQ \cdot CD = CA \cdot CE = CB \cdot CP$. $\blacksquare$ Claim: Point $E$ lies on line $PQ$. Proof. $\measuredangle AEP = \measuredangle ABP = \measuredangle ABC = \measuredangle ADC = \measuredangle ADQ = \measuredangle AEQ$. $\blacksquare$ [asy][asy]size(10cm); pair P = dir(240); pair B = dir(300); pair Q = dir(20); pair D = dir(50); pair E = extension(Q, P, B, D); pair C = extension(D, Q, B, P); pair H = extension(D, P, B, Q); pair A = foot(H, E, C); draw(unitcircle, lightblue); pair G = extension(D, A, B, C); pair F = extension(B, A, C, D); draw(circumcircle(P, A, E), heavygreen); draw(circumcircle(Q, A, E), heavygreen); draw(circumcircle(A, B, C), heavygreen+dotted); draw(B--F--C, grey); draw(D--G--C, grey); draw(P--H--B, blue); draw(P--B--Q--D--cycle, blue); draw(A--C, grey); draw(P--Q, blue); draw(B--D, blue); pair M = extension(P, F, G, Q); draw(F--P, heavycyan); draw(G--Q, heavycyan); dot("$P$", P, dir(P)); dot("$B$", B, dir(B)); dot("$Q$", Q, dir(Q)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$C$", C, dir(C)); dot("$H$", H, dir(H)); dot("$A$", A, dir(E-C)); dot("$G$", G, dir(G)); dot("$F$", F, dir(F)); dot("$M$", M, dir(M)); /* TSQ Source: P = dir 240 B = dir 300 Q = dir 20 D = dir 50 E = extension Q P B D C = extension D Q B P H = extension D P B Q A = foot H E C RE-C unitcircle 0.1 lightcyan / lightblue G = extension D A B C F = extension B A C D circumcircle P A E 0.1 green / heavygreen circumcircle Q A E 0.1 green / heavygreen circumcircle A B C 0.05 yellow / heavygreen dotted B--F--C grey D--G--C grey P--H--B blue P--B--Q--D--cycle blue A--C grey P--Q blue B--D blue M = extension P F G Q F--P heavycyan G--Q heavycyan */ [/asy][/asy] To finish, let $H = \overline{PD} \cap \overline{BQ}$. By properties of the Miquel point, we have $A$ is the foot from $H$ to $\overline{CE}$. But also, points $M$, $A$, $H$ are collinear by Pappus theorem on $\overline{BPG}$ and $\overline{DQF}$, as desired.

Here's an incomplete solution I submitted; how many points do you think I would get? First, note that cyclic quadrilaterals give us $\angle AQC =180^{\circ}-\angle AQD = 180^{\circ} - \angle AED = \angle AEB$, $\angle ACQ = \angle ABE $, so $\Delta AQC \sim \Delta AEB$. Similarly, $\Delta APC \sim \Delta AED$. Also, $\angle BAC = \angle BDC = \angle EAD$, so $\overline{AC}$ bisects $\angle BAD$. In order to show that $\angle MAC = 90^{\circ}$, it suffices to show that the pencil of lines $(\overline{AM},\overline{AC};\overline{AB},\overline{AQ})$ is harmonic. Define $S=\overline{FT}\cap\overline{AC}$ and $T=\overline{FS}\cap\overline{BC}$. Now, $(\overline{AM},\overline{AC};\overline{AB},\overline{AQ})\stackrel{A}{=} (M, S; \overline{AB}\cap\overline{GQ}, Q) \stackrel{F}{=} (P, T; B, C)$, so we only need to show $\frac{TB}{TC}=\frac{PB}{PC}$. Menelaus's Theorem on $\Delta ABC$ with collinear points $F, T, S$ gives $\frac{BT}{TC}\cdot\frac{CS}{SA}\cdot\frac{AF}{FB}=1$, while Menelaus on $\Delta ACD$ with collinear points $Q, S, G$ gives $\frac{AS}{SC}\cdot\frac{CQ}{QD}\cdot\frac{DG}{GA}=1$, so $\frac{TB}{TC}=\frac{FB}{AF}\cdot\frac{QD}{CQ}\cdot\frac{GA}{DG}$, and it suffices to show $\frac{PB}{PC}=\frac{FB}{AF}\cdot\frac{QD}{CQ}\cdot\frac{GA}{DG}$ From our similar triangles, we see $CP=\frac{DE\cdot AC}{DA}$ and $CQ=\frac{BE\cdot CA}{BA}$. Substituting these, as well as $PB=PC-BC$ and $QD=CD-QC$, we need to show an equation in terms of just the sides of complete cyclic quadrilateral $ABCD$, which should be easy to show with power of a point, Ptolemy's Theorem, etc.

whatshisbucket wrote: whatshisbucket wrote: Me trying to do a geometry problem is like trying to eat soup with a toothpick. Some things never change. And I thought the only things certain in life were death and taxes.

My solution: After proving E lies on PQ, define F' as intersection of line QA and circle PBEA and define G' as intersection of line PA and circle QDEA. Then define M' as the intersection of circles F'AD and G'AB. Then do root bc inversion with B and Q and find out that it inverts to the same thing, so F' switches with F, G' switches with G, M' switches with M, E and C switch, B and Q switch, D and P switch then angle chase with DAG' to get that MAM' is a straight line.

I only got the first two lemmas of Evan's proof and finished sketchily . Would that be 1,2,5, or 6 pts edit or 0pts

Did anyone successfully bash this one?

lucasxia01 wrote: Did anyone successfully bash this one? spent 3 hours trying to?

lucasxia01 wrote: Did anyone successfully bash this one? >4 pages of bary bash wasted :cry: Getting line CQ was untenable, at least for me.

lucasxia01 wrote: Did anyone successfully bash this one? I considered complex bashing this one but I don't know how to complex bash and I doubt that's efficient anyway. other bash methods are probably even worse

AllenWang314 wrote: I only got the first two lemmas of Evan's proof and finished sketchily . Would that be 1,2,5, or 6 pts edit or 0pts I would guess 0-1, unless "finished sketchily" meant a lot more nontrivial observations than the two lemmas. I personally found the two lemmas within the last 15 minutes

My solution was motivated by the fact that $FP$ and $GQ$ are actually quite difficult to draw on paper and pencil without making the diagram really ugly. I figured inverting would probably clean some things up.

lucasxia01 wrote: Did anyone successfully bash this one? I bary-bashed it (11 pages), and I know someone who complex-bashed it (15 pages). It's definitely doable.

My inversion solution was inspired by the 4096 4 circles passing through A in the diagram that I drew.

By angle chasing, we can get that $AC$ bisects $\angle BAQ$ and $\angle DAP$. By Dual of Desargues Involution Theorem on $CPGMFQ$ w.r.t $A$, $$(AC, AM), (AP, AQ), (AF, AG)=(AB, AD)$$are pairs of an involution. Projecting onto a line perpendicular to $AC$, mapping $AX\mapsto X'$, we see that the pairs $(P', Q'), (B', D')$ map to each other w.r.t negative inversion with center at the midpoint $C'$ of $B'Q'$ and power $B'C'\cdot D'C'$, which is an involution. Thus, $(C', M')$ must be part of that involution as well, so $M'$ is the point at infinity perpendicular to $AC$, so $M'A=MA\perp AC$ as desired.

A way to complex bash: set $APCQ$ to be the unit circle. Then one can get that $AB$ passes through $c^2/q$, so you can get $F, G$ relatively easily. Let $T$ be the antipode of $C$. Intersecting $AT$ with $PF$ and $GQ$ will show that $M$ is on $AT$, done.

Will I get points for claiming that PQDB is cyclic?

Claim: $AC$ bisects both $\angle DAP$ and $\angle BAQ$. First note that $AQ$ bisects $\angle BAQ$, as $\measuredangle CAQ = \measuredangle BDC = \measuredangle BAC$. Then we have that $P$, $E$, $Q$ are collinear from \[\measuredangle AEP = \measuredangle ABP = \measuredangle ADC = \measuredangle AEQ.\] Thus there exists a spiral similarity at $A$ sending $DQ \mapsto BP$, which finishes. ${\color{blue} \Box}$ Denote $K$ and $L$ as the intersection of the perpendicular to $AC$ at $A$ with $CF$ and $CG$. Then Apollonius tells us $(CK; QF) = (CL; PG) = -1$, from which we conclude that $KL$, $FP$, and $GQ$ concur at $M$ through Prism Lemma. $\blacksquare$

Claim: $P,E,Q$ are collinear Proof: We can say \begin{align*} \angle PEQ&=\angle AEQ+\angle AEP\\ &=\angle ABP+180-\angle ADQ\\ &=180\\ \end{align*}$\square$ Claim: $A$ is the Miquel Point of quadrilateral $PBDQ$ Proof: Let $A'$ be the Miquel Point. Then: \[A'=(DQE)\cap (PBE)\cap (BCD)=A\]$\square$ Claim: $PBDQ$ is cyclic Proof: We can say: \[\angle DQP=180-\angle DAE=180-\angle PBC=\angle PBD\]$\square$ Let $R=BP\cap PD$. As $PBDQ$ is cyclic, $\angle RAE=90$, by the special properties of Miquel Points in cyclic quadrilaterals. It remains to show that $R,A,M$ are collinear, which is true by Pappus' Theorem on $BRG$ and $DQF$ $\blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); 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 = -4.898319437038159, xmax = 1.721309433790064, ymin = -1.2091941798378278, ymax = 5.085859335855192; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen qqwwzz = rgb(0,0.4,0.6); pen cczzqq = rgb(0.8,0.6,0); /* draw figures */ draw((-2.344645147638352,1.1130030809515516)--(-2.093371596036182,-0.40327742089290003), linewidth(0.7)); draw((-2.093371596036182,-0.40327742089290003)--(-0.20726904334706847,-0.3300674029964366), linewidth(0.7) + qqwuqq); draw(circle((-1.1849783579010602,0.5262192470965725), 1.299670085403655), linewidth(0.7) + dotted); draw((-0.20726904334706847,-0.3300674029964366)--(-1.6730242915905422,1.7307745581978542), linewidth(0.7) + qqwuqq); draw((-1.6730242915905422,1.7307745581978542)--(-2.344645147638352,1.1130030809515516), linewidth(0.7)); draw((-2.093371596036182,-0.40327742089290003)--(-1.6730242915905422,1.7307745581978542), linewidth(0.7) + qqwuqq); draw((-2.344645147638352,1.1130030809515516)--(-2.682141858241821,3.149587028851064), linewidth(0.7)); draw((-1.6730242915905422,1.7307745581978542)--(-2.682141858241821,3.149587028851064), linewidth(0.7)); draw((-2.344645147638352,1.1130030809515516)--(-4.076793811892611,-0.48026495039497513), linewidth(0.7)); draw(circle((-1.8784664589104139,1.2801567291813516), 0.49524025678516037), linewidth(0.7) + dotted); draw(circle((-2.5389889132601735,0.3018365944554235), 0.8341226336999509), linewidth(0.7) + dotted); draw((-2.682141858241821,3.149587028851064)--(-2.9286091162282704,-0.43569758474553844), linewidth(0.7) + qqwwzz); draw((-4.076793811892611,-0.48026495039497513)--(-1.3853858810925042,1.326356899304287), linewidth(0.7) + qqwwzz); draw((-2.9286091162282704,-0.43569758474553844)--(-1.3853858810925042,1.326356899304287), linewidth(0.7) + qqwuqq); draw((-2.093371596036182,-0.40327742089290003)--(-1.3853858810925042,1.326356899304287), linewidth(0.7) + cczzqq); draw((-2.9286091162282704,-0.43569758474553844)--(-2.344645147638352,1.1130030809515516), linewidth(0.7)); draw((-2.344645147638352,1.1130030809515516)--(-1.3853858810925042,1.326356899304287), linewidth(0.7)); draw((-3.40268969572076,-0.45409926055024313)--(-0.1301321950380226,4.39298636671745), linewidth(0.7) + blue); draw((-2.9286091162282704,-0.43569758474553844)--(-0.1301321950380226,4.39298636671745), linewidth(0.7) + cczzqq); draw((-0.1301321950380226,4.39298636671745)--(-1.3853858810925042,1.326356899304287), linewidth(0.7) + cczzqq); draw((-2.344645147638352,1.1130030809515516)--(-0.20726904334706847,-0.3300674029964366), linewidth(0.7) + blue); draw((-4.076793811892611,-0.48026495039497513)--(-2.9286091162282704,-0.43569758474553844), linewidth(0.7)); draw((-2.9286091162282704,-0.43569758474553844)--(-2.093371596036182,-0.40327742089290003), linewidth(0.7) + qqwuqq); /* dots and labels */ dot((-2.344645147638352,1.1130030809515516),dotstyle); label("$A$", (-2.54087741336256,1.148247842321295), NE * labelscalefactor); dot((-2.093371596036182,-0.40327742089290003),dotstyle); label("$B$", (-2.1394289528091064,-0.6027507610939955), NE * labelscalefactor); dot((-0.20726904334706847,-0.3300674029964366),dotstyle); label("$C$", (-0.14072810664935878,-0.48317036866563423), NE * labelscalefactor); dot((-1.6730242915905422,1.7307745581978542),dotstyle); label("$D$", (-1.5586099034977268,1.7034425214529723), NE * labelscalefactor); dot((-1.8592579243618086,0.7852891265429016),linewidth(4pt) + dotstyle); label("$E$", (-2.0881802131639846,0.7126335556179787), NE * labelscalefactor); dot((-2.682141858241821,3.149587028851064),linewidth(4pt) + dotstyle); label("$F$", (-2.745872371943047,3.2579876230216693), NE * labelscalefactor); dot((-4.076793811892611,-0.48026495039497513),linewidth(4pt) + dotstyle); label("$G$", (-4.206461451829016,-0.6625409573081762), NE * labelscalefactor); dot((-2.9286091162282704,-0.43569758474553844),linewidth(4pt) + dotstyle); label("$P$", (-2.9679502437385743,-0.6539995007061504), NE * labelscalefactor); dot((-1.3853858810925042,1.326356899304287),linewidth(4pt) + dotstyle); label("$Q$", (-1.302366205272118,1.3019940611577594), NE * labelscalefactor); dot((-2.8762750934106447,0.3255896073486023),linewidth(4pt) + dotstyle); label("$M$", (-2.8712869380182497,0.1916047028944045), NE * labelscalefactor); dot((-0.1301321950380226,4.39298636671745),linewidth(4pt) + dotstyle); label("$T$", (-0.06385499718167618,4.402542807693127), NE * labelscalefactor); dot((-2.119250331019013,0.9608255552249374),linewidth(4pt) + dotstyle); label("$S$", (-2.0283900169113425,0.9774187102807788), NE * labelscalefactor); dot((-1.661509412919258,0.6517772754772615),linewidth(4pt) + dotstyle); label("$R$", (-1.541526990282686,0.6613848160058239), NE * labelscalefactor); dot((-3.40268969572076,-0.45409926055024313),linewidth(4pt) + dotstyle); label("$X$", (-3.454813270367231,-0.6625409573081762), NE * labelscalefactor); dot((-1.8035995078468217,1.9143624287345191),linewidth(4pt) + dotstyle); label("$Y$", (-1.8746437979759774,2.0536422421360303), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim: The points $Q$, $E$ and $P$ are collinear. Proof. Note that \[\angle BPE = \angle BAE = \angle BAC = \angle BDC.\]Also we have \[\angle PBE = 180 - \angle EBC = 180 - \angle DBC = 180 - \angle DAC = 180 - \angle DAE = DQE\]However then $\angle PEB = \angle DEQ$, hence proven. $\blacksquare$ Then it follows that $A$ is the Miquel Point of complete quadrilateral $QEBC$. As a result $\overline{AC}$ is the angle bisector of $\angle BAQ$ and of $\angle PAD$. Now let $\ell$ denote the external bisector of $\angle PAQ$. Clearly it suffices to show that $M \in \ell$. Claim: The lines $BQ$, $DP$ and $\ell$ are concurrent. Proof. Let $T = \ell \cap \overline{DP}$ and define $T' = \overline{BQ} \cap \overline{DP}$. Also define $R = \overline{AC} \cap \overline{BQ}$. Note that $(PD,ST) = -1$ from right angles and bisectors. However we also have $(PD, ST') = -1$ from Ceva-Menalus configuration. Thus $T \equiv T'$. $\blacksquare$ However now by Pappus Theorem on $\overline{G PB}$ and $\overline{FQ D}$ we find that $M = \overline{FP} \cap \overline{GQ}$, $T = \overline{PD} \cap \overline{BQ}$ and $A = \overline{FB} \cap \overline{DG}$ are collinear, hence $M \in \ell$. Remark Wow. I was doing this in spinvert so I already knew to utilize Miquel points, and from the defintion of $P$ and $Q$ it was obvious that $A$ is our desired Miquel point. From there I saw angle bisectors centered at $A$, so I leaped to projective. To force the right angles and bisectors lemma I extended $\overline{AM}$, $\overline{BQ}$ and $\overline{DP}$. However here I got stuck. I had two harmonic bundles and no clue how to use them to show that $M$ lies on $\ell$. Usually this would be a job for La Hire's but there were no useful circles present. Then I was hinted to use Pappus, which prior to this problem I didn't know. From there it was basically instakill.

By inversion about $E$ fixing $(ABCD)$—or just angle chase—we have $P,E,Q$ collinear. Now, $\measuredangle QAE=\measuredangle QDE=\measuredangle CDB=\measuredangle CAB=\measuredangle CAD$, so $\overline{AC}$ is the bisector of $\angle QAB$. Let $\overline{BQ} \cap \overline{DP}=X$. Since $\overline{BD},\overline{PQ},\overline{CA}$ concur in $\triangle CBQ$, by Ceva-Menelaus we have $(B,Q;\overline{AC} \cap \overline{BQ},X)=-1$, hence by a well-known fact $\overline{AX}$ is the other bisector of $\angle QAB$. It therefore suffices to show that $X,A,M$ are collinear, but this follows by Pappus on $FBQGDP$ since $\overline{BPG}$ and $\overline{DQF}$ are lines. $\blacksquare$

Alternatively, after finding that $\overline{AC}$ is the bisector of $\angle QAB$ (and similarly $\angle PAD$) we can employ a coordinate bash. This has the funny effect of revealing that $ABCD$ cyclic only matters in having $\overline{AP}$ and $\overline{AQ}$ be reflections of $\overline{AD}$ and $\overline{AB}$ over $\overline{AC}$—if we redefine $P$ and $Q$ in this manner, the perpendicularity is true for any quadrilateral $ABCD$. Let $A=(0,0)$ and $C=(1,0)$. Suppose lines $\overline{AB}$ and $\overline{AD}$ have equations $y=bx$ and $y=dx$ respectively and let $B=(p,bp)$ and $D=(q,dq)$. We compute $F$ as the intersection of $y=bx$ and $\overline{CD}$, which has equation $y=\tfrac{dq}{q-1}(x-1)$: $$bx=\frac{dq}{q-1}(x-1) \implies (bq-dq-b)x=-dq \implies x=\frac{dq}{dq-bq+b} \implies y=\frac{bdq}{dq-bq+b} \implies F=\left(\frac{dq}{dq-bq+b},\frac{bdq}{dq-bq+b}\right).$$On the other hand, $P$ is the intersection of $\overline{CB}$, which has equation $y=\tfrac{bp}{p-1}(x-1)$, with $y=-dx$: $$-dx=\frac{bp}{p-1}(x-1) \implies (-bp-dp+d)x=-bp \implies x=\frac{bp}{bp+dp-d} \implies y=-\frac{bdp}{bp+dp-d} \implies P=\left(\frac{bp}{bp+dp-d},-\frac{bdp}{bp+dp-d}\right).$$We will intersect $\overline{FP}$ with the $y$-axis. First, we calculate the slope of the line as $$\frac{\frac{bdq}{dq-bq+b}+\frac{bdp}{bp+dp-d}}{\frac{dq}{dq-bq+b}-\frac{bp}{bp+dp-d}}=bd\cdot\frac{q(bp+dp-d)+p(dq-bq+b)}{dq(bp+dp-d)-bp(dq-bq+b)}=bd\cdot \frac{2dpq+bp-dq}{b^2pq+d^2pq-d^2q-b^2p}.$$By point-slope, the $x$-intercept of $\overline{FP}$ is now \begin{align*} \frac{bdq}{dq-bq+b}-bd\cdot \frac{2dpq+bp-dq}{b^2pq+d^2pq-d^2q-b^2p}\cdot \frac{dq}{dq-bq+b}&=\frac{bdq}{dq-bq+b}\left(1-\frac{d(2dpq+bp-dq)}{b^2pq+d^2pq-d^2q-b^2p}\right)\\ &=\frac{bdq}{dq-bq+b}\left(\frac{b^2pq-d^2pq-b^2p-bdp}{b^2pq+d^2pq-d^2q-b^2p}\right)\\ &=\frac{bdpq}{dq-bq+b}\cdot \frac{(b+d)((b-d)q-b)}{b^2pq+d^2pq-d^2q-b^2p}\\ &=-\frac{bdpq}{b^2pq+d^2pq-d^2q-b^2p}, \end{align*}which is symmetric in $(b,p)$ and $(d,q)$, hence equal to the $x$-intercept of $\overline{GQ}$, hence $M$ lies on the $y$-axis as desired. $\blacksquare$

We will prove that P, E, Q lie on one line. Now $\angle QEA = 180 -\angle QDA = \angle ABC = 180 - \angle PBA = 180 - \angle PEA$ $\Rightarrow$ $\angle QEA + \angle PEA = 180$ $\Rightarrow$ P, E, Q lie on one line. Denote X lies on CG and Y lies on CF and $XY \perp AC$, $\angle PAC = 180 - \angle PBE = \angle CBD = \angle CAD$ $\Rightarrow$ AC is angle bisector of $\angle PAD$ $\Rightarrow$ AX is bisector of $\angle RAG$. So by lemma 9.18 (G, P; X, C) = -1. We have $\angle QAC = \angle QDE = \angle CAB$ $\Rightarrow$ AC is bisector of $\angle BAQ$ $\Rightarrow$ AY is bisector of $\angle FAQ$. From lemma 9.18 (Q, F; Y, C) = -1. So (Q, F; Y, C) = (G, P; X, C) = -1. From Prism lemma we get that GQ, FP and XY concur at M, M lies on XY and $XY \perp AC$ $\Rightarrow$ $\angle MAC = 90^{\circ}$.

Am I the only clown to find an arbitrary circumconic? Let $T$ be the Miquel point of $ABCD$. Claim: $E$ lies on $PQ$. Proof. Note that $\measuredangle PEA = \measuredangle PBA = \measuredangle CBA = \measuredangle CDA = \measuredangle QEA$. $\blacksquare$ Claim: $N \coloneq PQ \cap GF$ satisfies $\measuredangle NAC = 90^\circ$ Proof. It remains to show that $NATE$ is cyclic. To do this, note that \[ \measuredangle NEA = \measuredangle PEA = \measuredangle PBA = \measuredangle GBA = \measuredangle GTA = \measuredangle NTA. \]$\blacksquare$ Let $H = AC \cap FG$. Then it's well known that $(A, C; E, H) = -1$. Let $\mathcal{C}$ be a circumconic of $PGFQ$ which intersects $AC$ at $X, Y$. As such, by DIT on $PGFQ$ and $CA$ there exists an involution swapping \[ (C, C), (E, H), (PF \cap AC, GQ \cap AC), (X, Y) \]Then this involution must consist of inversion about $(AC)$ or harmonic conjugation wrt $A, C$. This means that $A$ lies on the polar of $C$ wrt $\mathcal{C}$. Since by Brokard's $NM$ is the polar of $A$, we get that $N, C, M$ concur which finishes. Remark: This can also be proved by homographying $ABCD$ into a rectangle and generalizing to $E \in PQ$.

Solved with Arch Hints Begin by observing $P$, $E$, $Q$ collinear by the spiral similarity $\triangle PAB \sim \triangle QAD$. Note that $\overline{CA}$ bisects $\angle BAQ$ and $\angle PAD$, since $\angle BAC = \angle BDC = \angle EDQ = \angle EAQ$, and $\angle PAB = \angle QAD$. Now let the external angle bisectors of $\angle PAD $ and $\angle BAQ$ meet $\overline{BC}$, and $\overline{CD}$ at $X$, and $Y$ respectively. By right angles and bisectors $-1= (XB; PC)= (YQ; DC)$. So by the prism lemma $M \in XY$, now $\angle MAC = \angle XAC = 90^\circ$ and we are finished.

$A$ is the intersection of $(QED)$ and $(PBE)$ hence $A$ is the miquel point of complete quadrilateral $BEQCDP$. Invert the diagram from $A$. New Problem Statement After Inverting from $A$: Let $P',B',Q',D'$ be a cyclic quadrilateral on a circle with this order. $P'Q'$ and $B'D'$ intersect at $C'$. $D'A'$ intersects $(A'B'C')$ at $G'$ and $B'A'$ intersects $(A'C'D')$ at $F'$. If $(A'F'P')$ and $(A'G'Q')$ intersect at $M',$ then $M'A'\perp A'C'$. Let's invert from $E'$ with radius $\sqrt{E'B'.E'P'}=\sqrt{E'Q'.E'D'}=\sqrt{E'C'.E'A'}$. New Problem Statement After Inverting the Second Diagram from $E'$: $ABC$ is a triangle. We take a circle which intersects $AB,BC$ at $F,D$ respectively. $(ABP)$ and $BC$ intersect at $K$ and $(BDP)$ intersects $AB$ at $L$. If $(BKF)$ crosses $(BCL)$ at $M$, then $PM\perp BM$. Proof: \[\angle PLF=\angle PLB=\angle PDC= \angle ADC=\angle AFC=\angle LFP\]Thus, $PF=PL$. Similarily, \[\angle PKC=180-\angle BKP=\angle PAB=\angle DAF=\angle DCF=\angle KCP\]Hence $PC=PK$. Denote $S,T$ as the midpoints of $FL$ and $CK$. Since $M$ is the miquel point of $FLCK,$ we get that $M$ is the center of spiral homothety taking $FSL$ to $KTC$. \[\angle BSM=\angle FSM=\angle KTM=\angle BTM\]Which gives that $B,S,T,M$ are cyclic. $(BST)$ has diameter $BP$ hence $M$ lies on the circle whose diameter is $BP$. Thus, $\angle PMB=90$ as desired.$\blacksquare$

Usamo 2018 p5 First of all we start off by showing that $P-E-Q$ are collinear and $AC$ bisect $\angle BAQ$ and $\angle DAP$ by angle chase. Let $X$ and $Y$ as the intersection of the perpendicular to $AC$ at $A$ with $CF$ and $CG$. Then we get that. \[(CX; QF) = (CY; PG) = -1\]Which by prism lemma, we are done.

Here does DDIT : Note that: $AC$ bisects $\angle BAQ$ and $\angle PAD$ as: $$\angle PAC = \angle EBC = \angle DBC = \angle DAC$$$$\angle BAE = \angle BAC = \angle BDC = \angle EDQ = \angle EAQ.$$ Now, we proceed to apply DDIT. Consider the quadrilateral $GPFQ$ for DDIT. Let line $\ell$ be perpendicular to $AC$ at $C$. Now, project points $P, Q, D, F$ into line $\ell$. Notice that: $P'C = G'C$ with $F'C=Q'C$. Therefore, negative inversion at $C$ with $r = \sqrt{CP' \cdot CQ'}$ swaps: $(P', Q'), (G',F')$ (which is the halper line for DDIT). Let $M'$ denote the projection of $M$ (from $A$) onto $\ell$. Therefore, we must have $(M', C)$ to swap each in the inversion which implies $M'$ to be point of infinity of line $\ell$. Therefore, $AM \parallel \ell$ or $\angle MAC = 90^\circ$.

Solution w/ projective geometry (see attached).

Here's a projective solution that seems at least somewhat different from previous solutions. All angles are directed modulo $180^\circ$. By DDIT with point $A$ and complete quadrilateral $CFGMPQ$, we get that there exists an involution $\phi$ on the pencil of lines through $A$ sending $AC\leftrightarrow AM$, $AB\leftrightarrow AD$, and $AP\leftrightarrow AQ$. We have $\angle BAP-\angle DAQ=\angle BEP-\angle DEQ=0$. Also, $\angle BAC=\angle BDC=\angle EDQ=\angle EAQ=\angle CAQ$, so $AC$ bisects $\angle BAQ$. Thus there exists an involution $\psi$ on the pencil of lines through $A$ given by reflection across $AC$, and it swaps $AB\leftrightarrow AQ$, $AD\leftrightarrow AP$, and $AC\leftrightarrow AC$. By MMP, we have that $\phi$ and $\psi$ commute because $\phi\circ\psi=\psi\circ\phi$ for $AB$, $AD$, $AP$, and $AQ$. Thus $\psi(AM)=\phi^{-1}(\psi(\phi(AM)))=AM$. Thus $AM$ is the other angle bisector of $\angle BAQ$, and so $\angle CAM=90^\circ$.

Spiral sim where ;-;

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