I like the problem very much . Let $P$ be the second point of intersection between $CD$ and the circle $(ABM)$, and let $G=AB\cap CD$. A very simple computation, based on the fact that $GD\cdot GC=GA\cdot GB=GM\cdot GP$ and $M$ is the midpoint of $CD$ will show that $P$ is, in fact, the harmonic conjugate of $C,D$ art $G$, so it belongs to $EF$. $ANBM$ is a harmonic quadrilateral, so $(PA,PB;PN,PM)=-1$, and since $PM=PG$, so $(PA,PB;PE,PM)=-1$ as well, we get $PE=PN$, as desired.

A very cute problem ! Here is another solution (hope it isn't the same as yours, Darij ) Let $G$ be the intersection of $AB$ and $CD$. Let $J$ be the midpoint of $AB$. Let $U$ be the second intersection of $LN$ and the circumcircle $ABM$. Notice that $\measuredangle{BMN} = \measuredangle{JMA}$ since $ANBM$ is a harmonic quadrilateral. Therefore, $\measuredangle{MJG} = \measuredangle{MJB} = \measuredangle{MAN}$. Since $\measuredangle{OMG} = \measuredangle{OJG} = 90^0$, the quadrilateral $OJGM$ is cyclic. Hence, $\measuredangle{MJG} = \measuredangle{MOG}$. This implies $\measuredangle{MOG} = \measuredangle{MAN}= \measuredangle{MUG}$. Thus, $OUGM$ is cyclic, or $OU$ is perpendicular to $NG$. Now, the inversion with center $G$, constant $GA.GB$ maps the two circles into themselves, and maps the circle $(S)$ with diameter $GN$ into the line $OU$. This imples that $(S)$ is orthogonal to $(O)$, or $N$ lies on the polar of $G$ wrt $(O)$, which is $EF$. PS Why is Victor Thebault related to this problem ?

can someone explain me what is harmonic quadrilateral? (appeared in Grobber's solution)

Let me explain like this : I give you 4 points $A, B, C, D$ on a circle. An arbitrary point S lies on that circle. Then $(SA, SB,SC, SD)$ is a constant. If $(SA, SB,SC, SD) = -1$, then $ABCD$ is a harmonic quadrilateral.

treegoner wrote: Let me explain like this : I give you 4 points $A, B, C, D$ on a circle. An arbitrary point S lies on that circle. Then $(SA, SB,SC, SD)$ is a constant. If $(SA, SB,SC, SD) = -1$, then $ABCD$ is a harmonic quadrilateral. Thanks Treegoner but I still have no idea what $(SA, SB,SC, SD)$ could mean?

Megus wrote: treegoner wrote: Let me explain like this : I give you 4 points $A, B, C, D$ on a circle. An arbitrary point S lies on that circle. Then $(SA, SB,SC, SD)$ is a constant. If $(SA, SB,SC, SD) = -1$, then $ABCD$ is a harmonic quadrilateral. Thanks Treegoner but I still have no idea what $(SA, SB,SC, SD)$ could mean? that means harmonic ratio.

Are you sure it is harmonic ratio? Harmonic ratio says about four collinear points, and here we are regarding four concurrent lines Maybe I'm wrong? So I'll be thankful if someone actually explain this term to me - not just name but also properties - a link would be enough. Thanks in advance

http://www.google.ro/url?sa=U&start=1&q=http://www.cut-the-knot.org/pythagoras/Cross-Ratio.shtml&e=747

Thanks Grobber very much - that's what I needed. Sorry for askin silly questions

Your solutions are both very nice. There is really no need in posting mine, since it is just unreasonably long. A few technical comments to Treegoner's solution: treegoner wrote: Let $G$ be the intersection of $AB$ and $CD$. Let $J$ be the midpoint of $AB$. Let $U$ be the second intersection of $LN$ and the circumcircle $ABM$. I guess this should mean: "Let U be the second intersection of GN and the circumcircle of ABM." treegoner wrote: Notice that $\measuredangle{BMN}= \measuredangle{JMA}$ since $ANBM$ is a harmonic quadrilateral. Therefore, $\measuredangle{MJG}= \measuredangle{MJB}= \measuredangle{MAN}$. Since $\measuredangle{OMG}= \measuredangle{OJG}= 90^{0}$, the quadrilateral $OJGM$ is cyclic. Hence, $\measuredangle{MJG}= \measuredangle{MOG}$. This implies $\measuredangle{MOG}= \measuredangle{MAN}= \measuredangle{MUG}$. Thus, $OUGM$ is cyclic, or $OU$ is perpendicular to $NG$. You have forgotten to say that the point O is defined as the circumcenter of the cyclic quadrilateral ABCD. treegoner wrote: Now, the inversion with center $G$, constant $GA.GB$ maps the two circles into themselves, and maps the circle $(S)$ with diameter $GN$ into the line $OU$. This imples that $(S)$ is orthogonal to $(O)$, or $N$ lies on the polar of $G$ wrt $(O)$, which is $EF$. PS Why is Victor Thebault related to this problem ? Thebault is not directly related to this problem; I just wanted to say that the problem is very much in his style, and, more generally, in the style of the old school synthetic geometers. I am glad that problem like this have become popular again on shortlists... PS.

greetings from oberwolfach, darij

Well, how come this problem was arranged as G8? Everything becomes trivial after noticing the second intersection of the circumcircle of ABM and CD lies on EF, as what Grobber did. I don't mean it is too easy to be in shortlist, but at least IMO04\5 is surely harder than this one. In my mind it should be around G4-G5.

A very simple complex numbers solution (found by dysfunctionalequations and me): Let $U = AA \cap BB$ and $V = DD \cap CC$. By Pascal's theorem on $AACBBD$ and $ADDBCC$, $U$, $V$, $E$, and $F$ are collinear, so it is sufficient to show that $U$, $V$, and $N$ are collinear. Without loss of generality, let $ABCD$ be the unit circle, and let $a$, $b$, $c$, $d$, $m$, $n$, $u$, and $v$ be the affixes of $A$, $B$, $C$, $D$, $M$, $N$, $U$, and $V$, respectively. Because $AMBN$ is cyclic, $Q = \frac{(a-n)/(n-b)}{(m-a)/(m-b)}$ must be real. Since $\frac{NA}{NB} = \frac{MB}{MA}$, $Q$ must have magnitude 1. Hence, $Q = \pm 1$. If $Q = -1$, then $(a-n)(m-b) = (a-m)(n-b)$, which clearly has as its only solution $m=n$. But $m \neq n$, so $Q = 1$. Hence, $(a-n)(m-b) = (m-a)(n-b)$, so $am - ab - mn + nb = mn - bm - an + ab$, so $n(m-a + m-b) = am + bm - 2ab$, so $n = \frac{m(a+b) - 2ab}{2m-(a+b)}.$ We wish to show that $\frac{u-n}{v-n}$ is real. It is well-known that $u = \frac{2ab}{a+b}$ and $v = \frac{2cd}{c+d}$. Hence, \begin{align*} \frac{u-n}{u-v} &= \frac{\frac{2ab}{a+b} - n}{\frac{2ab}{a+b} - \frac{2cd}{c+d}} \\ &= \frac{2ab(c+d) - n(a+b)(c+d)}{2ab(c+d) - 2cd(a+b)} \\ &= \frac{2ab(c+d) - \frac{m(a+b) - 2ab}{2m-(a+b)} (a+b)(c+d)}{2ab(c+d) - 2cd(a+b)} \\ &= \frac{2ab(c+d)(2m-(a+b)) - (a+b)(c+d)(m(a+b) - 2ab)}{(2m-(a+b))(2abc + 2abd - 2acd - 2bcd))} \\ &= \frac{m(4ab(c+d) - (a+b)^2 (c+d)) - 2ab(a+b)(c+d) + 2ab(a+b)(c+d)}{(c+d-(a+b))(2ab(c+d) - 2cd(a+b))} \\ &= \frac{- \left(\frac{c+d}{2}\right) (c+d) (a-b)^2 }{(c+d-(a+b))(2abc + 2abd - 2acd - 2bcd)} \\ &= -\frac{1}{4} \frac{(c+d)^2 (a-b)^2 }{(c+d-(a+b))(abc+abd-acd-bcd)}. \end{align*} Recalling that $\overline{z} = \frac{1}{z}$ for complex numbers $z$ on the unit circle, we have \begin{align*} \left( \frac{u-n}{u-v} \right) \left( \frac{\overline{u-v}}{\overline{u-n}} \right) &= \left( \frac{(c+d)^2 (a-b)^2 }{(c+d-(a+b))(abc+abd-acd-bcd)} \right) \left( \frac{(\frac{1}{c}+\frac{1}{d}-(\frac{1}{a}+\frac{1}{b}))(\frac{1}{abc}+\frac{1}{abd}-\frac{1}{acd}-\frac{1}{bcd})}{(\frac{1}{c}+\frac{1}{d})^2 (\frac{1}{a}-\frac{1}{b})^2 } \right) \\ &= \left( \frac{(c+d)^2 (a-b)^2 }{(c+d-(a+b))(abc+abd-acd-bcd)} \right) \left( \frac{(\frac{abc + abd - acd - bcd}{abcd})(\frac{c+d-(a+b)}{abcd})}{\frac{(c+d)^2 (a-b)^2}{a^2 b^2 c^2 d^2}} \right) \\ &= 1, \end{align*} so $\frac{u-n}{u-v} = \overline{\left(\frac{u-v}{u-n}\right)}$. It follows that $\frac{u-n}{u-v}$ is real, which was what we wanted.

Lemma 1:- In a triangle $ ABC $, the A-appolonius circle intersects the circumcircle of $ ABC $ at $ X $(other than $ A $). Then $ X $ is the intersection of the A-symmedian of $ ABC $ and $ \odot ABC $. Proof:- If we invert the figure wrt $ A $, then $ B,C $ goes to some points $ B',C' $ and $ \odot ABC $ goes to $ B'C' $. If $ AL,AM $ ($ L,M $ lying on $ \odot AB'C' $) are the angle-bisectors of $ \angle B'AC' $, then image of $ X $ will be the intersection point of $ LM $ and $ B'C' $. Clearly $ LM $ bisects $ B'C' $, so $ AX $ is a median of $ AB'C' $. Since $ BC $ is anti-parallel to $ B'C' $, so $ AX $ is a symmedian of $ ABC $. Lemma 2:- If $ AM,AP $ are isogonal conjugates wrt $ \angle BAC $ of $ \Delta ABC $, with $ M $ lying on $ BC $ and $ N $ lying on $ \odot ABC $, then consider any circle passing through $ A,M $ which intersects $ BC $ and $ \odot ABC $ at $ X,Y $. Then $ XY $ will pass through $ N $. Proof:- Suppose, $ NX $ intersects $ \odot ABC $ at $ Y' $. $ \angle AY'X=\angle ACN=\angle ACB+\angle BAN=\angle ACB+\angle CAM=180-\angle AMC $. So $ AMCY' $ is cyclic. So $ Y\equiv Y' $. Back to Main Proof:- Suppose, $ FE\cap DC=U, FE\cap AB=V, DC\cap AB=K $. $ (KU,DC)=-1 $. So $ KU.KM=KD.KC=KA.KB $. So $ AUMB $ is cyclic. Similarly $ UMM'V $ is cyclic(where $ M' $ is the mid-point of $ AB $). By lemma 1, $ N $ is the intersection of A-symmedian of $ ABC $ at $ \odot ABC $. So by lemma 2, $ U,V,N $ are collinear. So done.

Solution : Construct points Y and X on line CD such that BCY ~ BAD and ABC ~ ADX YC = BC*AD/BA = DX Let circle (ABE) intersects FE at points G and E J is Miquel point of lines AD , BC , BD , AC Easy to see that GBJ ~ MBY and AGJ ~ AMX Construc point Z on GJ such that GZ/ZJ = MC/CY = MD/DX ZBJ ~ MBY , AZJ ~ AMX , so angle BZA = BMC + AMD = 180 - BMA , so Z is on (BAM) MZ/BM = ZJ/MY = ZJ/MX = ZA/MA , so Z = N . done

grobber wrote: I like the problem very much . $GD\cdot GC=GA\cdot GB=GM\cdot GP$ $PM=PG$, I think $PM=PG$ is far of obvious, I even think it is far of true, smb who can say who's right, becuase G is far of $C,M,P,D$, isn't it?

Yay I can still do synthetic geometry! [asy][asy]/* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(11cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair O = (0.0, 0.0); pair A = (-6.892819407726209, 3.4729937233313155); path w = CirclebyPoint(O,A); pair B = (-0.6532590960841593, 7.690636992031878); pair C = (7.443160147387265, -2.0425503201076376); pair D = (-7.500191970374932, -1.8220222823415784); pair E = IntersectionPoint(Line(A,D,lisf),Line(B,C,lisf)); pair R = IntersectionPoint(Line(A,B,lisf),Line(C,D,lisf)); pair F = IntersectionPoint(Line(A,C,lisf),Line(B,D,lisf)); path AB = R--A; pair T_1 = IntersectionPoint(w,Line(E,F,lisf),0); pair T_2 = IntersectionPoint(w,Line(E,F,lisf),1); path EF = E--T_2; pair K = extension(A,B,E,F); pair M = midpoint(C--D); path circABM = circumcircle(A,B,M); pair N_1 = IntersectionPoint(circABM,EF,0); pair P = 2*foot(M,relpoint(circumcenter(O,R,M)--circumcenter(A,B,M),0.5-10/lisf),relpoint(circumcenter(O,R,M)--circumcenter(A,B,M),0.5+10/lisf))-M; /* Draw objects */ draw(w, rgb(0.6,0.6,0.6)); draw(AB, rgb(0.6,0.6,0.6) + dashed); draw(A--C); draw(B--D); draw(R--T_1, rgb(0.6,0.6,0.6) + dashed); draw(R--T_2, rgb(0.6,0.6,0.6) + dashed); draw(EF, rgb(0.8,0.8,0.8)); draw(circABM, rgb(0.6,0.6,0.6)); draw(circumcircle(O,M,R), rgb(0.8,0.8,0.8) + dashed); draw(A--B); draw(B--C); draw(C--D); draw(D--A); draw(R--N_1, rgb(0.4,0.4,0.4) + linewidth(1.2) + linetype("4 4")); draw(P--M); draw(E--A, rgb(0.8,0.8,0.8)); draw(E--B, rgb(0.8,0.8,0.8)); draw(R--D, rgb(0.6,0.6,0.6) + dashed); /* Place dots on each point */ dot(O); dot(A); dot(B); dot(C); dot(D); dot(E); dot(R); dot(F); dot(T_1); dot(T_2); dot(K); dot(M); dot(N_1); dot(P); /* Label points */ label("$O$", O, lsf * dir(45) * 2); label("$A$", A, lsf * dir(145)); label("$B$", B, lsf * dir(60)); label("$C$", C, lsf * dir(35)); label("$D$", D, lsf * dir(150)); label("$E$", E, lsf * dir(45)); label("$R$", R, lsf * dir(155)); label("$F$", F, lsf * dir(45)); label("$T_1$", T_1, lsf * dir(45)); label("$T_2$", T_2, lsf * dir(-80)); label("$K$", K, lsf * dir(45)); label("$M$", M, lsf * dir(45)); label("$N_1$", N_1, lsf * 2 * dir(45)); label("$P$", P, lsf * 2 * dir(80)); [/asy][/asy] Let $R = \overline{AB} \cap \overline{CD}$ and let $T_1$ and $T_2$ be the tangency points from $R$ to the circle. Let $P$ be the intersection of the circle with diameter $\overline{OR}$ and the circumcircle of $ABM$ (other than $M$). Also, let line $EF$ meet the circumcircle of $ABM$ again at $N_1$. I claim that $R$, $P$ and $N_1$ are collinear. Otherwise, let $RP$ hit $\ell$ at $N_1'$ and observe that by a computation we would have $RP \cdot RN_1'$ equals the power of $R$ with respect to $\omega$ (since $EF$ is polar of $R$). But this implies $RN_1'BA$ is cyclic, so $N_1' = N_1$. Next, I claim that $P$, $K$ and $M$ are collinear. Notice that $K = \overline{AB} \cap \overline{T_1T_2}$ is the radical center of $\omega$, $\Gamma$ and $(ABM)$, so the result follows upon obserivng that $PM$ is a common chord of $(ABM)$ and $\Gamma$. Finally, observe that $P(N_1,M,A,B) = P(R,A,K,B) = (R,A;K,B) = -1$, so we're done. $\blacksquare$ EDIT (Sept 27, 2015): I think most projective solutions basically have to find some "nice" point on the circumcircle of $ANBM$ to project through. In my case I found the point $P$, but the shortest solutions seems to be to consider $\overline{EF} \cap \overline{CD}$ as grobber did.

Lemma 1: If $ABCD$ is a cyclic quadrilateral, then take a point $X$ on its circumcircle. Then take an arbitrary line passing through and let its intersections with $XA,XB,XC,XD$ be $M,N,P,Q$ respectively. Then $ABCD$ is harmonic iff $M,N,P,Q$ is a harmonic division.

Let $EF\cap CD=H$, and $AB\cap CD=G$. Lemma 2: $AMBH$ is cyclic.

Consider $EF\cap AB=J$ and $NH\cap AB=J'$. The definition of $N$ implies that $ANBM$ is a harmonic quadrilateral. Then note that by lemma 1, and considering the line $AB$, with $(HA\cap AB, HN\cap AB, HB\cap AB, HM\cap AB)=(G,A,J',B)$ is harmonic. But we have $(G,A,J,B)$ is harmonic, thus $J'=J$ and we have the lines $NH$ and $EFH$ equivalent, implying $N$ lies on $EF$.

I think that it is a hard geometry, which you need dozens of application of projective geometry to solve it. Quote: Well, how come this problem was arranged as G8? Everything becomes trivial after noticing the second intersection of the circumcircle of ABM and CD lies on EF, as what Grobber did. I don't mean it is too easy to be in shortlist, but at least IMO04\5 is surely harder than this one. In my mind it should be around G4-G5. (By: mecrazywrong) It is not an easy geometry, though the main idea is projective geometry, somehow it is so twisted. Anyway, the following is my solution after thinking of it for days. Let $G=\{AB\cap CD\}$, we know that both tangents from $G$ to circle $ABCD$, (namely $\omega$) lies on $EF$ by Brokard's theorem-$G$ is pole of $EF$. Let the two tangents to be $GK$, $GL$. Let $P=\{CD\cap EF\}$, $\omega_1$=circle $ABM$. We claim that $P$ lies on $\omega_1$. First, since $E, F, P$ are collinear, it is well known that $(G, P; D, C)$ is harmonic. Since $GA\cdot GB=GD\cdot GC$, it suffices to prove that $GD\cdot GC=GP\cdot GM.$ Note that $GM=\frac {GD+GC}{2}$ and $GP=(GD+(GC-GD)(\frac {DP}{DC}))=(GD+(GC-GD)(\frac {GD}{GC+GD}))=GD(\frac {2GC}{GC+GD})$, so $GM\cdot GP=GD(\frac {GD+GC}{2})(\frac {2GC}{GC+GD})=GC\cdot GD$ Now let $\{\omega _1\cap EF\}=P, N'$, $\{\omega _1\cap GN'\}=Q, N'$, we are intended to prove that $N=N'$. Lemma. $PN', MQ, AB$ are conccurent. Proof. Clearly, $AB$ is the radical axis of $\omega $ and $\omega_1$. Let $R=$ so $KL, MQ, AB$ are conccurent $\Leftrightarrow R$ on radical axis $\Leftrightarrow KR\cdot RL=MR\cdot RQ\Leftrightarrow KQLM$ is cyclic. Since $G$ is on radical axis, $GK^{2}=GL^{2}=GP\cdot GM=GN'\cdot GQ$. Therefore, $\angle KMG=\angle GKP=\angle GKL=\angle GLR$. Simce $MPQN'$ is cyclic (all on $\omega_1$), $\angle GMQ=\angle PMQ=\angle QN'P=\angle GN'L=\angle GLQ.$ So $\angle KMQ=\angle KMR=\angle KMG+\angle GMQ=\angle GLR+\angle GLQ=\angle KLQ.$ So $KQLM$ is indeed cyclic, and thus $R$ is on radical axis. To complete our solution, $PQ$ and $MN'$ intersect at pole of $GR$ w.r.t. $\omega_1$ as a consequence of Brokard's theorem again, namely $S$. Since $G$ and $R$ are both on the radical axis $AB$, $S$ is indeed the pole of $AB$ w.r.t. $\omega_1$. Since pole of $AB$ lies on $MN'$, it is well-known that pole of $MN'$ lies on $AB$ (both w.r.t. $\omega_1$). Therefore $AMBN'$ is a harmonic quadrilateral, and $ \frac{AM}{MB}=\frac {AN'}{N'B}$, so $N=N'$, and therefore $E, F, N$ are collinear.

Zhero wrote: A very simple complex numbers solution (found by dysfunctionalequations and me): Let $U = AA \cap BB$ and $V = DD \cap CC$. By Pascal's theorem on $AACBBD$ and $ADDBCC$, $U$, $V$, $E$, and $F$ are collinear, so it is sufficient to show that $U$, $V$, and $N$ are collinear. Without loss of generality, let $ABCD$ be the unit circle, and let $a$, $b$, $c$, $d$, $m$, $n$, $u$, and $v$ be the affixes of $A$, $B$, $C$, $D$, $M$, $N$, $U$, and $V$, respectively. Because $AMBN$ is cyclic, $Q = \frac{(a-n)/(n-b)}{(m-a)/(m-b)}$ must be real. Since $\frac{NA}{NB} = \frac{MB}{MA}$, $Q$ must have magnitude 1. Hence, $Q = \pm 1$. If $Q = -1$, then $(a-n)(m-b) = (a-m)(n-b)$, which clearly has as its only solution $m=n$. But $m \neq n$, so $Q = 1$. Hence, $(a-n)(m-b) = (m-a)(n-b)$, so $am - ab - mn + nb = mn - bm - an + ab$, so $n(m-a + m-b) = am + bm - 2ab$, so $n = \frac{m(a+b) - 2ab}{2m-(a+b)}.$ Here's a quicker finish. Let $G'$ be the inverse of $G$ w.r.t. $(ABCD)$; then $g' = \overline{g}^{-1} = \frac{cd-ab}{(c+d) - (a+b)}$ (a fake synthetic way to get this is the spiral similarity $\triangle{G'AC}\sim\triangle{G'DB}$). Since $\triangle{EFG}$ is self-polar w.r.t. $(ABCD)$, it suffices to show $\angle{OG'N} = 90^\circ$. But \begin{align*} n-g' &= \frac{(a+b)(c+d) - 4ab}{2((c+d)-(a+b))} - \frac{cd-ab}{(c+d) - (a+b)} \\ &= \frac{1}{2}\frac{(a+b)(c+d)-2(ab+cd)}{(c+d)-(a+b)}, \end{align*}so \[ \frac{n-g'}{\overline{n-g'}} = \frac{ab(c+d)-cd(a+b)}{(c+d)-(a+b)} = - \frac{g'}{\overline{g'}}, \]as desired.

Posting this for storage. Let $G=(AB)\cap(DC)$,$Y = (AB) \cap (EF)$and $Z=(EF) \cap (DC)$. Since $(CD;ZG) = -1$, then $GZ \cdot GM = GD \cdot GC = GA \cdot GB$, thus $ABMZ$ is cyclic. Let $N'$ be the second intersection of the $(\odot ABM)$ with $(ZE)$ Now we have: $$(AB;N'M) \stackrel{Z}{=}(AB;YG) = -1$$So $N'=N$ and this ends the problem. $$\mathbb{Q.E.D.}$$

Let $FE$ intersect $CD$ at $T$,then we know that $(PT,CD)=-1 $ therefore by midpoint length $PT.PM=PC.PD=PA.PB$ => $ABTM is cyclic$ Now $(P,T;C,D)=A=(B,W;E,D)=-1$, where $W$ is intersection of $TA$ with $BD$ Let $FE$ intersects $(ABTM)$ at $N$ , $(B,W;E,D)=X=(B,N;A,M)=-1$ therefore $ANBM$ is harmonic so $AN.BM=BN.AM ==> AN/BN=AM/BM$ so $N$ is desired point.

We first note that the given ratio condition is equivalent to $(AB;MN)=1$ or $(AB;MN)=-1$. Thus, there can exists at most two points which satisfy the given ratio condition. Clearly, $M$ satisfies the first cross ratio. So only one other point $N$ satisfies the required ratio. This is the point $N$ such that $AMBN$ is a harmonic quadrilateral. Let $G=\overline{AB} \cap \overline{CD}$. Now, let $X=\overline{EF} \cap \overline{CD}$, $Y = \overline{EF} \cap \overline{AB}$ and $Z= \overline{GE} \cap \overline{AD}$. Let $N'=\overline{EF} \cap (ABM)$. Now, we first make the following observation. Claim : Point $X$ lies on $(AMB)$. Note that $(DC;GX)=-1$. Thus, \[GA \cdot GB = GC \cdot GD = GX \cdot GM\]and thus $ABXM$ must be cyclic as needed. Let $P= \overline{BX} \cap \overline{AD}$. Now, note that, \[-1=(DA;ZF)\overset{G}{=}(XY;EF)\overset{B}{=}(PA;DF)\overset{X}{=}(BA;MN')\]Thus, $AN'BM$ is indeed harmonic. This means $N'=N$ and indeed points $E,F,N$ are collinear as was required.

Let $AB$ intersect $CD$ at $P$, $(ABM)$ intersect $CD$ at $K\neq M$, $BK$ intersect $AC$ at $L$, and $KE$ intersect $(ABM)$ at $N'$. Note that $PK\cdot PM=PA\cdot PB=PC\cdot PD$ which implies $(P,K;D,C)=-1$. By configuration, we know that $K$, $E$, $F$ are collinear. Additionally, we have \[-1=(P,K;D,C)\stackrel{B}{=}(A,L;E,C)\stackrel{K}{=}(A,B;N',M)\]So $AN'BM$ is harmonic. We also know that $ANBM$ is harmonic, so $N=N'$, and so $N$ lies on $KE$ and thus $EF$.

Call $G = AB \cap CD$, $X = EF \cap CD$, $Y = EF \cap AB$, $N^* = EF \cap (ABM)$. We first prove that $X$ lies on $(ABM)$, which is true since $$GA \cdot GB = GC \cdot GD = GX \cdot GM$$where the last equality follows from the well known lemma about midpoints in harmonic bundles. Now, we have $$-1 = (D, C; X, G) \stackrel F= (B, A; Y, G) \stackrel X= (B, A; N^*, M)$$However, $(B, A; N, M) = -1$, thus $N = N^*$ and we are done.

Denote $CD \cap EF = J$, $AB \cap CD = G$. We have (G, J; C, D) = -1 $\Rightarrow$ GJ.GM=GC.GD. From ABCD cyclic GA.GB = GC.GD $\Rightarrow$ GJ.GM = GA.GB $\Rightarrow$ J lies on (ABM). Denote $EF \cap (ABM) = N'$ for second time. We have $EF \cap AB = K$, (N', M; A, B) = (K, G; A, B) = -1 $\Rightarrow$ N' = N $\Rightarrow$ $EF \cap (ABM) = N$ $\Rightarrow$ E, F, N lie on one line. We are ready.

$-1= ( JI; CD)$. Therefore $I$ lies on $\overline{EF}$ the polar of $J$ with respect to $(ABCD)$. Now to finish let $\overline{IB} \cap \overline{AD} = K$, and $\overline{IN} \cap \overline{AD} = F'$ . Now $-1 = (AB; MN) \stackrel{I} = (AK; DF')$, and $-1= (JI; DC) \stackrel{B}= (AK; DF)$. Thus $F'=F$, and we can conclude the collinearity.

2004 G8 Let $N’$ denote the intersection of $(ABM)$ with line $EF$. We show that $N’=N$. Let $S$ denote $EF$ intersect $CD$, let $Q$ denote the intersection of $AB$ and $CD$, and let $P$ denote $AB$ intersect $EF$. We first prove this claims. Claim 1: $ABMS$ is cyclic. Pf: we use the fact that $(Q,P;C,D)=-1$ and PoP. Claim 2: $(ABMN’)$ is harmonic. Pf: We have that \[ (AB;QP) \overset{S}{=} (AB;MN’)=-1 \]Hence we are done, $N’=N$.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -13.064648773997016, xmax = 3.045778244912386, ymin = -5.443564901308561, ymax = 5.0446227466279705; /* image dimensions */ /* draw figures */ draw((-1.4991616723719812,2.1967648781624804)--(-2.017746118546502,-1.2133207211788555), linewidth(0.6)); draw((-2.017746118546502,-1.2133207211788555)--(1.958067968791492,-1.1426046603630213), linewidth(0.6)); draw(circle((-0.054929214610409755,0.23266072140058477), 2.437932019286302), linewidth(0.6)); draw((1.958067968791492,-1.1426046603630213)--(-0.052953878149104856,2.6705919404277436), linewidth(0.6)); draw((-0.052953878149104856,2.6705919404277436)--(-1.4991616723719812,2.1967648781624804), linewidth(0.6)); draw(circle((-0.9251269166641245,0.36499493420422535), 1.919608561230494), linewidth(0.6)); draw((-1.4991616723719812,2.1967648781624804)--(1.958067968791492,-1.1426046603630213), linewidth(0.6)); draw((-2.017746118546502,-1.2133207211788555)--(-0.052953878149104856,2.6705919404277436), linewidth(0.6)); draw((-12.475079775981722,-1.3993207268747434)--(-2.017746118546502,-1.2133207211788555), linewidth(0.6)); draw((-12.475079775981722,-1.3993207268747434)--(-0.6886813456074746,1.4139145628271172), linewidth(0.6)); draw((-12.475079775981722,-1.3993207268747434)--(-1.4991616723719812,2.1967648781624804), linewidth(0.6)); draw((-0.6886813456074746,1.4139145628271172)--(0.4646221983063934,1.6891906998627892), linewidth(0.6)); draw((-2.017746118546502,-1.2133207211788555)--(1.8322005191013198,3.2882329001389277), linewidth(0.6)); draw((1.8322005191013198,3.2882329001389277)--(-0.052953878149104856,2.6705919404277436), linewidth(0.6)); /* dots and labels */ dot((-1.4991616723719812,2.1967648781624804),dotstyle); label("$A$", (-1.7773611106158496,2.3761994947039824), NE * labelscalefactor); dot((-2.017746118546502,-1.2133207211788555),dotstyle); label("$B$", (-2.3911807594629464,-1.7479523463607068), NE * labelscalefactor); dot((1.958067968791492,-1.1426046603630213),dotstyle); label("$C$", (1.9469032304538755,-1.6337952018933704), NE * labelscalefactor); dot((-0.052953878149104856,2.6705919404277436),dotstyle); label("$D$", (-0.22208251522746102,2.832828072573328), NE * labelscalefactor); dot((0.9525570453211936,0.7639936400323611),linewidth(4pt) + dotstyle); label("$M$", (1.0480282159953653,0.7779994721612725), NE * labelscalefactor); dot((-0.6886813456074746,1.4139145628271172),linewidth(4pt) + dotstyle); label("$E$", (-0.45028430569240013,1.6056387695494614), NE * labelscalefactor); dot((-12.475079775981722,-1.3993207268747434),linewidth(4pt) + dotstyle); label("$F$", (-12.90802019595884,-1.9619969922369625), NE * labelscalefactor); dot((-2.763121010848791,0.9187771459912196),linewidth(4pt) + dotstyle); label("$N$", (-3.2073593432845267,1.0776619763880306), NE * labelscalefactor); dot((0.4646221983063934,1.6891906998627892),linewidth(4pt) + dotstyle); label("$J$", (0.7055567824667333,1.6027173417831296), NE * labelscalefactor); dot((1.8322005191013198,3.2882329001389277),linewidth(4pt) + dotstyle); label("$G$", (1.7899371567532526,3.45361379491001), NE * labelscalefactor); dot((-1.6532232995700198,1.1836929663560007),linewidth(4pt) + dotstyle); label("$K$", (-2.0700575412527796,1.3059762653227034), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Define $J = EF \cap CD$. $\textbf{Claim}$: $ABJM$ is cyclic. $\textit{Proof}$. Extend $AB$ and $CD$ to meet at a point $T$. Then by power of a point it suffices to prove, \begin{align*} TA \cdot TB = TC \cdot TD = TJ \cdot TM \end{align*}which boils down to showing that, \begin{align*} TC \cdot TD &= TJ \cdot TM\\ \iff (TM + MC) \cdot TD &= TJ \cdot TM\\ \iff TM \cdot JD &= CM \cdot TD\\ \iff JD \cdot TC + JD \cdot TD &= CD \cdot TD\\ \iff JD \cdot TC &= TD \cdot JC \end{align*}which is true as it is well known that $(TJ, CD) = -1$. $\square$ Now let $N' = EF \cap (ABM)$ so that we must show $N'$ satisfies $(AB, N'M) = -1$. To do this we may project and note, \begin{align*} (AB, N'M) \overset{J}{=} (AG,DF) \overset{B}{=} (KJ,EF) = -1 \end{align*}where we let $G = AD \cap BJ$ and $K = AB \cap EF$, thus we are done.

Let $AB \cap CD =G$, $EF \cap CD=X$, $EF \cap AB=Y$ and we redefine $N$ as $EF \cap (ABM)$. We now prove that $(AB;MN)$ is harmonic. Note that $(AB;YG)\stackrel{E}{=}(CD;XG)=-1$ due to Ceva-Menelaus. Since $M$ is midpoint of $CD$ $\implies GC.GD=GX.GM$ due to a well known property of harmonic bundles $\implies X \in (ABM)$ $$\implies (AB;MN) \stackrel{X}{=} (AB;GY)=-1$$as desired.

Let $P$ be the intersection of $EF$ and $CD$. By Brokard's we have that $EF$ is the polar of $AB \cap CD = G$ so it follows that $(G, P; D, C) = -1$. However it is well known that $(G, P; D, C) = -1$ implies that $GP \cdot GM = GD \cdot GC$ and $GD \cdot GC = GA \cdot GB$ so by PoP we have $P \in (ABM)$. Now let $N'$ be the second intersection of $EF$ with $(ABM)$. It suffices to show that $ABN'M$ is harmonic. Notice that once again $EF$ is the polar of $G$ so $-1 = (G, EF \cap AB; A, B) \overset{P} = (M, N'; A, B)$ so we are done as this implies $ABMN'$ harmonic $\implies N' = N$.

Quick problem. Let $I$ be intersection point of $DC$ and $EF$. Let $DC$ and $AB$ be $G$ and let $EF$ intersects $AG$ at $R$. It is known that $(G,R;B,A)=-1$ projecting from $E$. Give us $(AMBI)$. Projecting from $I$ gives us $(M,A;B,EI \cap (ABM))=-1$ but we know that $(M,A;B,N)=-1$. Thus, $E-F-N$.

Define $K$ such that $(DC;KG) = -1$, $L = EF \cap AB$, and $G = AB \cap CD$. We claim $K \in EF, (ABM)$ as well: Using cevians, notice that \[(AB; LG) = (D,C; EF \cap CD, G) \implies K = EF \cap CD.\] Using Power of a Point, we have that \[GA \cdot GB = GC \cdot GD = GM \cdot GK \implies ABKM \text{ cyclic.}\] Letting $EF \cap (ABM) = N' \neq K$, it suffices to have \[-1 = (AB;GL) \overset{K}{=} (AB;MN') \implies \frac{AN'}{BN'} = \frac{AM}{BM} \implies N = N'. \quad \blacksquare\] [asy][asy] size(300); pair A, B, C, D, E, F, G, K, L, M, N; A = dir(210); B = dir(330); C = dir(35); D = dir(90); E = extension(A, C, B, D); F = extension(A, D, B, C); G = extension(A, B, C, D); K = extension(C, D, E, F); L = extension(A, B, E, F); M = .5*(C + D); N = IP((3*E-2*F)--E, circumcircle(A, B, M)); draw(A--F--B--D--G--A--C--F--N^^circumcircle(A, B, C)); draw(circumcircle(A, B, M), red); dot("$A$", A, dir(225)); dot("$B$", B, dir(315)); dot("$C$", C, dir(45)); dot("$D$", D, dir(135)); dot("$E$", E, dir(165)); dot("$F$", F, dir(90)); dot("$G$", G, dir(0)); dot("$K$", K, dir(290)); dot("$L$", L, dir(225)); dot("$M$", M, dir(250)); dot("$N'$", N, dir(270)); [/asy][/asy]

For diagram, see attached. In this problem, to avoid configuration issues, WLOG let $AB$ intersect $CD$ closer to $D$ (like shown in the diagram). First, let $X=EF\cap CD$. We make the following claim. *** Claim 1. $MXAB$ is cyclic. Proof. First, notice that by Ceva-Menelaus, if we let $Y=AB\cap CD$, then $(CD;XY)=-1$. This means that \[\frac{|XC|}{|XD|}=\frac{|YC|}{|YD|}. \text{ } \textcircled{1}\]Now, if $M$ is the midpoint of $CD$, if we let $CX=x$ and $\frac{|CX|}{|DX|}=\frac{1}{a}$, then we can use $\textcircled{1}$ to label the lengths as follows: [see attachments for the line with $C$, $M$, $X$, $D$, $Y$ and the labeled lengths] Then, we get that \[YX*YM=\left(\frac{2}{1-a}*ax\right)*\left(\frac{4a+a^2-2a+1}{2(1-a)}*x\right)=\frac{a(a+1)^2}{(1-a)^2}*x^2\]\[=\left(\frac{1+a}{1-a}*x\right)*\left(\frac{a}{1-a}*(a+1)x\right)=YC*YD,\]which, by Power of a Point on $Y$ with respect to $(ABCD)$, is equal to $YA*YB$. Therefore, $YA*YB=YX*YM$, and by converse Power of a Point, this means that $MXAB$ is cyclic, as desired. *In the degenerate case of $M=X$ (where $Y=P_{\infty, CD}$), we instantly get $MXAB$ is a triangle and therefore cyclic. *** Now, notice that since $\frac{AN}{BN}=\frac{AM}{BM}$, we have that $AN*BM=AM*BN$, meaning that $AMBN$ is a harmonic quadrilateral, or that $(AM;BN)=-1$. Now, let $N'=EF\cap (ABM)$ such that $N'\neq X$. If we can show that $(AB;MN')=-1$, then by the uniqueness of the harmonic conjugate, we would get that $N=N'$, meaning that $E$, $F$, $N$ are collinear, as desired. We will prove this now. *** Claim 2. $(AB;MN')=-1$, where $N'=EF\cap (ABM)$ so that $N'\neq X$. Proof. Let $BX\cap AD=Z$. Since $X\in EF$, by Ceva-Menelaus, we have that $(FD;AZ)=-1$. Now, notice that \[(AB;MN')\overset{X}{=}(AZ;DF)=(FD;AZ)=-1,\]where $(AB;MN')\overset{X}{=}(AZ;DF)$ comes from the projection of $ABMN'$ with respect to $X$ from $(ABM)$ onto $AD$ (which we can do, since $X\in (ABM)$), and $(AZ;DF)=(FD;AZ)$ comes from $(FD;AZ)$ being $-1$. This gives us that $(AB;MN')=-1$, as desired. *** Therefore, by our earlier conclusion, from the uniqueness of the harmonic conjugate, we have that $N=N'$, meaning that $E$, $F$, $N$ are collinear, since $N'\in EF$. This is what we wished to prove, finishing our proof. $\blacksquare$

Neat G8 with projective solution: Let $T = EF \cap CD$. Note that: $(KT; CD)=-1$ as $FE, DB, AC$ are concurrent deviants in $\triangle FDC$. Claim: $T \in (ABM)$ Notice that: $$(KT; CD)=-1 \implies KT \cdot KM = KC \cdot KD = KA \cdot KB.$$ Define $N'=EF \cap (ABM)$. It suffice to show $AN'BM$ to be harmonic, to finish the problem. Notice that: $$(KT; DC) = (K (TK \cap AB); AB) = -1.$$Taking perspective with-respect to point $T$ into $(ABM)$: $(K (TK \cap AB); AB) = (MN'; AB) = -1$ and we are done.