Consider a projective transformation that takes $F$ to the center of $(ABCD)$. Then $ABCD$ is mapped to a rectangle, and $E$ is mapped to the point at infinity of $AB$ and $CD$. Hence $XF \parallel AB \parallel CD$ and so by symmetry, $MN \parallel BC$. Now, $MN$, $BC$, and the tangent to $\omega$ at $X$ concur at their point at infinity, and so projecting back to our original diagram, we see that the three lines concur, as desired. Additionally, this implies that the point of concurrency also lies on $AD$.

Let $Y\in AD\cap BC$; clearly, $EF$ is the polar of $Y$ w.r.t. circle $\omega$, hence $XY$ is tangent to $\omega$. Now, assuming $A$ belongs to the segment $\overline{BE}$ and $Z\in EF\cap BC$, then $(B,C,Z,Y)$ is harmonical, consequently both $AD$ and $MN$ pass through $Y$ and we are done (since $F,X$ belong to $EZ$). Best regards, sunken rock

Solution

Uhm, this seems way too easy for a G9. Is the following correct? This might be even easier for someone who is experienced in projective geometry, but since I have just started studying it, I did get to learn something new from it. We start with a lemma. Lemma: Consider four collinear points $P,Q,R,S$ in that order. Suppose points $P,R$ lie on a circle $\omega,$ and $Q$ lies on the polar of $S$ w.r.t $\omega.$ Then, $(P,R,Q,S)$ is harmonic. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -2.933409073358824, xmax = 6.450980778380609, ymin = -0.9436119144868866, ymax = 4.679610829842664; /* image dimensions */ /* draw figures */ draw(circle((0.5282956318816452,2.253241391102320), 1.376486593462148)); draw((1.493924741567847,3.234197946974016)--(1.514859266964706,1.293341638048530)); draw((1.493924741567847,3.234197946974016)--(2.469184856301255,2.274176271533317)); draw((2.469184856301255,2.274176271533317)--(1.514859266964706,1.293341638048530)); draw((2.469184856301255,2.274176271533317)--(-0.8474268834791837,2.299098808281013)); /* dots and labels */ dot((1.493924741567847,3.234197946974016),dotstyle); label("$X$", (1.531818871891331,3.290017070947388), NE * labelscalefactor); dot((1.514859266964706,1.293341638048530),dotstyle); label("$Y$", (1.550346788676601,1.344585808494001), SE * labelscalefactor); dot((2.469184856301255,2.274176271533317),dotstyle); label("$S$", (2.504534503118024,2.326565398113330), NE * labelscalefactor); dot((-0.8474268834791837,2.299098808281013),dotstyle); label("$P$", (-0.8119626014453690,2.354357273291236), NE * labelscalefactor); dot((1.504201543096853,2.281427597275125),dotstyle); label("$Q$", (1.541082830283966,2.335829356505965), NE * labelscalefactor); dot((1.904551938263980,2.278419181392183),dotstyle); label("$R$", (1.939433041167279,2.335829356505965), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Proof: Let $X$ and $Y$ be the points of tangency of the tangents from $S$ to $\omega.$ Then, $X,Q,Y$ are collinear. By PoP, $SX^2= SY^2 = SR \times SP.$ By Stewart's theorem on $\triangle XSY$ and cevian $SQ,$ \begin{align*} XS^2 \cdot QY + SY^2 \cdot XQ & = XY (QS^2 + XQ \cdot QY) \\ \implies SR \cdot SP (QY+XQ) & = XY (QS^2 + PQ \cdot QR) \\ \implies SR \cdot SP & = QS^2 + PQ \cdot QR \\ \implies PS \cdot QR & = (QS+PQ)(QS-RS) \\ & = QS^2 - QS \cdot RS + PQ \cdot QS - PQ \cdot RS \\ & = SR \cdot SP - PQ \cdot QR - QS \cdot RS + PQ \cdot QS - PQ \cdot RS \\ & = SR(SP-QS) + PQ (QS - RS - QR) \\ & = SR \cdot PQ, \end{align*} completing the proof. $\blacksquare$ Now, the main proof. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -6.031851335510330, xmax = 9.494851650396377, ymin = -2.007788648465363, ymax = 7.295971186130260; /* image dimensions */ /* draw figures */ draw(circle((-0.8103758841031479,1.440191299622182), 2.994281524733298)); draw((1.262006751286642,3.601429847745464)--(-1.729127761990702,4.290036354399240)); draw((-1.729127761990702,4.290036354399240)--(-3.687352515287380,0.6102953344681221)); draw((-3.687352515287380,0.6102953344681221)--(2.154283804058977,1.860327923526378)); draw((2.154283804058977,1.860327923526378)--(1.262006751286642,3.601429847745464)); draw((-1.729127761990702,4.290036354399240)--(2.154283804058977,1.860327923526378)); draw((-3.687352515287380,0.6102953344681221)--(1.262006751286642,3.601429847745464)); draw((-0.3851507969618622,6.815531530442445)--(-1.729127761990702,4.290036354399240)); draw((-0.3851507969618622,6.815531530442445)--(1.262006751286642,3.601429847745464)); draw((-0.3851507969618622,6.815531530442445)--(0.3003595484624255,3.020260451256048)); draw((0.06875746088971521,4.302506168933527)--(-3.687352515287380,0.6102953344681221)); draw((0.6233425163783685,4.847656594106078)--(0.06875746088971521,4.302506168933527)); draw((-1.034792118515100,5.594776959172074)--(2.154283804058977,1.860327923526378)); draw((-1.034792118515100,5.594776959172074)--(5.611458098222241,2.600117307443744)); draw((5.611458098222241,2.600117307443744)--(2.154283804058977,1.860327923526378)); draw((0.06875746088971521,4.302506168933527)--(5.611458098222241,2.600117307443744)); draw((5.611458098222241,2.600117307443744)--(1.262006751286642,3.601429847745464)); draw((0.3003595484624255,3.020260451256048)--(0.5710618838960850,1.521539459520038)); /* dots and labels */ dot((-1.729127761990702,4.290036354399240),dotstyle); label("$A$", (-1.663529172644178,4.383756410886160), NE * labelscalefactor); dot((-3.687352515287380,0.6102953344681221),dotstyle); label("$B$", (-3.625442284387572,0.7051693263672978), SE * labelscalefactor); dot((1.262006751286642,3.601429847745464),dotstyle); label("$D$", (1.325322833527399,3.694021332538874), NE * labelscalefactor); dot((2.154283804058977,1.860327923526378),dotstyle); label("$C$", (2.214314712286125,1.946692467392414), NE * labelscalefactor); dot((0.3003595484624255,3.020260451256048),dotstyle); label("$E$", (0.3596937238411972,3.111578377490054), NE * labelscalefactor); dot((-0.3851507969618622,6.815531530442445),dotstyle); label("$F$", (-0.3300413545060898,6.912785031492879), NE * labelscalefactor); dot((0.06875746088971521,4.302506168933527),dotstyle); label("$X$", (0.1297820310587682,4.399083857071656), N * labelscalefactor); dot((-3.687352515287380,0.6102953344681221),dotstyle); dot((0.6233425163783685,4.847656594106078),dotstyle); label("$M$", (0.6815700937365979,4.935544473563990), NE * labelscalefactor); dot((-1.034792118515100,5.594776959172074),dotstyle); label("$N$", (-0.9737940942968911,5.686589336653258), NE * labelscalefactor); dot((5.611458098222241,2.600117307443744),dotstyle); label("$J$", (5.678317550208055,2.697737330481682), NE * labelscalefactor); dot((0.5710618838960850,1.521539459520038),dotstyle); label("$G$", (0.6355877551801120,1.609488651311518), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $\omega$ denote the circumcircle of $ABCD.$ Let $BC$ and the tangent to $\omega$ at $X$ intersect at $J.$ It is easy to see that $J$ must also lie on $AD.$ For a quick proof, assume $AD$ and $BC$ intersect at $J',$ and note that from Brokard's theorem $EF$ is the polar of $J'$ w.r.t $\omega,$ implying that $J'X$ is tangent to $\omega \implies J'=J.$ Let $FX$ intersect $BC$ at $G.$ By Ceva's theorem on $\triangle FBC$ and cevians $FG, BM, CN,$ $\dfrac{FN}{NB} \cdot \dfrac{BG}{GC} \cdot \dfrac{CM}{MA}=1.$ By the lemma, $(B,C,G,J)$ is harmonic, so $BJ \times GC = BG \times CJ \implies \dfrac{BG}{GC} = \dfrac{BJ}{CJ}.$ Plugging this in the previous equation, we have that $\dfrac{FN}{NB} \cdot \dfrac{BJ}{CJ} \cdot \dfrac{CM}{MF}= 1,$ from which the desired result follows by the converse of Menelaus theorem. $\blacksquare$

Woah 1

Woah 2

AnonymousBunny wrote: Uhm, this seems way too easy for a G9. Is the following correct? G9 is a misnomer here -- in total we had 16 G's at some point. Some of the high-end ones were taken out for either being way too hard or being unsolved. So "G9" is a medium problem.

codyj wrote:

Woah 1

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This problem isn't worth bashing. And no, I am a bunny. It's just that my pet human wrote that solution. @v_Enhance: Hmm, I shouldn't confuse between ISL and ELMOSL. Anyways, some of these shortlist problems (geometry included) are very nice, I have to say. Kudos to the problem writers. Also, might I ask where are problems G10 and G12?

Let $AD \cap BC=G$.Since the polar $EF$ of $G$ passes through $X$,by La Hire's theorem $GX$ is tangent to $\omega$.We also see that $BG,BX,BD,BA$ form a harmonic pencil.Taking projection on $EC$ we get $(C,M;D,E)=-1$.Similarly we get $(B,N;A,E)=-1$.Seeing both these set of harmonic points from $G$ we get that $GB,GN,GA,GE$ as well as $GB,GM,GD,GE$ form a harmonic pencil.Thus $G,M,N$ are collinear,and consequently the problem statement follows.

Using Pascal's theorem on $XBADCX$ gives that $AD$, $MN$ and tangent at $X$ are concurrent. Now let $AD$ meet $BC$ at $S$. $S$ is the pole of $EF$ so $AD$, $BC$ and tangent at $X$ are concurrent, so the conclusion follows.

Using Desargues's Theorem in $\Delta BDM$ and $\Delta CAN$ we can conclude that $MN$, $AD$ and $BC$ are concurrent. Let $Y$ their intersection point. It follows that $EF$ is the polar line of $Y$, hence $YX$ is tangent to $\omega$

Nice problem!!! Claim:$AD,MN,BC$ concur Proof:Desargues theorem in triangles $ABN$ and $DCM$. Now use Pascal's on $ABXXCD$ to get $MN,XX,AD$ concur. So $MN,BC,XX$ concur as desired.

Let $\{L \}=AD\cap BC,\{S \}=EF\cap \omega ,S\neq X$ and $\{P \}=EF\cap BC$. It is well known that $EF$ is the polar of $L$,and since $EF\cap \omega = X $ it follows that $LX$ is tangent to $\omega$. Since $F$ is on the polar of $E$, $(E,P,L,X)$ is a harmonic division.So we have that $C(E,P,L,X)$ is a harmonic pixel so $(E,B,A,N)$ is harmonic. Similarly,$(E,C,D,M)$ is harmonic and because $\{L \}=AD\cap BC \Rightarrow \overline{N,M,L}$ and the conclusion follows.

Let$ P$ be the intersection of $EF$ with $BC$, let $R$ be the intersection of $BC$ with $XX$ and Let $Y$ be the second intersection of $EF$ with the circle then $(E,F;X,Y)=-1$ . $\rightarrow$$A(E,F;X,Y)=(B,C;X,Y)=X(B,C;X,Y,)=(B,C;R,P)=-1$. Now let $R'$ be the intersection of $MN$ with $BC$ ,it is well_known that $(B,C;R',P)=-1$ so $R=R'$.

Easy but Nice. ELMO Shortlist 2013 G9 wrote: Let $ABCD$ be a cyclic quadrilateral inscribed in circle $\omega$ whose diagonals meet at $F$. Lines $AB$ and $CD$ meet at $E$. Segment $EF$ intersects $\omega$ at $X$. Lines $BX$ and $CD$ meet at $M$, and lines $CX$ and $AB$ meet at $N$. Prove that $MN$ and $BC$ concur with the tangent to $\omega$ at $X$. Proposed by Allen Liu Solution:- First Complete the Quadrilateral $ABCD$. Let $AD\cap BC=P$. Claim :- $EX$ is the Polar of $P$. By Brocard's Theorem you get that $EF$ is the Polar of $P$. Hence, $EX$ must be the Polar of $P$. Hence, $PX$ must be tangent to $\omega$. So, $BC$ and the tangent at $X$ meet at $P$. So it suffices to prove that $M-N-P$ which directly follows by applying Pascal on $DABXXC$. Hence, $M-N-P$. Hence, proved. $\blacksquare$.

Let $Y=EF\cap \omega, Y\not = X$. Let $G=AD\cap BC$. Then $XX\cap YY = G$ is the pole of $EF$ by Brokard, so $XX,AD,BC$ concur. We want to show that $MN$ also passes through this point. Note that \[ (EN;AB) \stackrel{X}{=} (YC;AB) \stackrel{G}{=} (YB;DC) \stackrel{X}{=} (EM;DC) \]where we projected onto the circle. Let $M'=GN\cap EC$. Then $(EN;AB) \stackrel{G}{=} (EM';DC)$, so $M'=M$. Therefore, $G,N,M$ collinear (this is the so-called ``prism lemma'') and we conclude.

Remark

Let $G=AD\cap BC$; note that since $EF$ is the polar of $G$, $GX$ is tangent to $\omega$, so it suffices to prove $M, G, N$ collinear. Pascal on $BXXCDA$ finishes.

Let $R=AD \cap BC$. By Brocard's theorem ,we have that $EF$ is the polar of $R$.Since $X$ lies on it ,we have that $RX$ is a tangent to $\omega$. Now,Pascal on $DABXXC$ finishes....

Projective cookie :3 Let $AD$ meet $BC$ at $G$, we take polars w.r.t. $\omega$ and we denote $\mathcal P_Y$ as the polar line of $Y$. By brokard on $(ABCD)$ we have that $\mathcal P_G=EF$ and since $EF \cap \omega=X$ then by La’Hire and polar of a point on $\omega$ we have that $GX$ is tangent to $\omega$. Now by pascal on $(XBCDAX)$ we have that $G,M,N$ are colinear. Thus we are done

Let $G = AD \cap BC$, $H = AD \cap EF$, and $K = BC \cap EF$. By Ceva-Menelaus wrt $BCE$, we know $-1 = (MN \cap BC, K; B, C)$. But Ceva-Menelaus wrt $ADE$ yields $$-1 = (G, H; A, D) \overset{E}{=} (G, K; B, C)$$which implies $G \in MN$. Brocard's implies $EF$ is the polar of $G$. Because $X$ lies on $EF$ and $\omega$, we know $GX$ is tangent to $\omega$, which finishes. $\blacksquare$ Remark: Basic projective completely murders this problem.

Take a homography mapping $F$ to the center of $\omega$(while preserving $\omega$). We get that $ABCD$ is a rectangle so $E = \infty_E$ and $AB \parallel \overline{EFX} \parallel CD$ and we also get that $\overline{MN} \parallel \overline{BC} \parallel \overline{XX} \implies$ they concur at $\infty$.

Let $G=AD \cap BC$. By Brocard's theorem, $G$ is the pole of $EF$, so $GX$ is tangent to $\omega$. Also, $(B,C;BC \cap EF,G)=-1$. By considering the cevians of $\Delta EBC$ which intersect at $X$, we have that $(B,C;BC \cap EF,BC \cap MN)=-1$, which implies that $M,N,G$ are collinear. $\square$

Taking a homography that takes $F$ to the center of $(ABCD)$ sends $E$ to the point at infinity along $AB$ and $CD$. $X$ is sent to intersection of the line through the center parallel to $AB$ and $CD$ with the circle. Now the problem is easy.