Let $P$ be the second intersection point of the circles $(SAE),(SBF)$. We have $\angle APE=\angle ASE=\angle BPF$, and $\angle PAE=\angle PSE=\angle PBF$. This shows that the triangles $PAE,PBF$ are directly similar, and hence the figures $PAED,PBFC$ are also similar. It follows from here that $PAB,PDC$ are similar, meaning that $\angle PDT=\angle PAS=\angle PET$, i.e. $P$ lies on the circle $(TDE)$. In the same way we prove that it lies on $(TCF)$.

Ok, this is g01ng 2 be teh pwn4ge, you asked for it o_0 lolololol Nice problem, though it doesn't really kick 4ss imho. We are working with directed angles modulo 180° (cuz all teh other angles are l8m). Let $X_{1337}$ be the point of intersection of the lines AD and BC. Let the circumcircles of triangles $X_{1337}AB$ and $X_{1337}CD$ intersect at a point $\Lambda$ apart from $X_{1337}$ (why $\Lambda$? cuz teh letter r00lzorz ). Then, $\measuredangle\Lambda BX_{1337}=\measuredangle\Lambda AX_{1337}$, what is equivalent to $\measuredangle\Lambda BC=\measuredangle\Lambda AD$. Similarly, $\measuredangle\Lambda CB=\measuredangle\Lambda DA$. Hence, the triangles $\Lambda BC$ and $\Lambda AD$ are directly similar. The points F and E are corresponding points in these similar triangles, since they lie on the respective sides BC and AD and divide them in the same ratio $\frac{BF}{FC}=\frac{AE}{ED}$. As corresponding points in directly similar triangles form equal angles, this entails $\measuredangle\Lambda FB=\measuredangle\Lambda EA$. This is equivalent to $\measuredangle\Lambda FX_{1337}=\measuredangle\Lambda EX_{1337}$. Thus, the point $\Lambda$ lies on the circumcircle of triangle $X_{1337}EF$. The rest is trivial using Miquel points. If you are not familiar with Miquel stuff (then j00 are teh n00b ), you can do it by angle chasing: Since the point $\Lambda$ lies on the circumcircle of triangle $X_{1337}AB$, we have $\measuredangle\Lambda BA=\measuredangle\Lambda X_{1337}A$. Obviously, $\measuredangle\Lambda X_{1337}A=\measuredangle\Lambda X_{1337}E$. Since the point $\Lambda$ lies on the circumcircle of triangle $X_{1337}EF$, we have $\measuredangle\Lambda X_{1337}E=\measuredangle\Lambda FE$. Thus, $\measuredangle\Lambda BA=\measuredangle\Lambda FE$. In other words, $\measuredangle\Lambda BS=\measuredangle\Lambda FS$. Thus, the point $\Lambda$ lies on the circumcircle of triangle SBF. Similarly, we can prove that the same point $\Lambda$ lies on the circumcircles of triangles SAE, TCF and TDE. Thus, these four circumcircles pass through a common point, what pwnz the level... ehm, problem. PMFG i sux at writing 1337... [edit: and at being quick ] darij

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darij grinberg wrote: ... PMFG i sux at writing 1337... [edit: and at being quick ] ... And this should be "OMFG". Apart from that, 17 d035 0wn j00r m0m!

mathmanman wrote: darij grinberg wrote: ... PMFG i sux at writing 1337... [edit: and at being quick ] ... And this should be "OMFG". Apart from that, 17 d035 0wn j00r m0m! 101010101 u just g0t PWN3D by my ub4r1337sp54k!!!1111133337777 $\text{OWNED}\to\text{PWNED}$ $\text{OMFG}\to\text{PMFG}$ darij

lawlers! I see, kind of a "slide on the keyboard", eh? *cries* 1 g07 0wn3d.. Nevermind, I'll stop this spam now.

What is scarier is that beneath all the L33tsp33k, there's some really clever maths going on. Wonder what the IMO markers would make of a perfect solution written in L33tsp33k?

That'd be awesome to try! At our next TST, I'll do it for one of the exercises, we'll see...

Now it becomes absolutely obvious how much a proper notation is important in maths.

Centy wrote: Wonder what the IMO markers would make of a perfect solution written in L33tsp33k? I think it would basicly depend on how readable it is. If it is still mathematically correct and understandable, it shouldn't be much of a problem. Not long ago, I ended a solution to a problem with a quote from Munchkin, what somewhat irritated the correcter, but it still was 7/7. At the most recent TST, I have defined an operation on graphs called "haxxoring" (really, I had no time and no better word came into my mind ) and enclosed a part of my solution which I found unnecessary in a <spam> </spam> tag. Participants should have the right to have some fun on exams. Of course, it starts suxx0ring when all the solution is written in 1337, what makes it hard to read normal words and distinguish between letters and numb3rz. Centy wrote: What is scarier is that beneath all the L33tsp33k, there's some really clever maths going on. Well, may be, but this clever maths is over 100 years old . Actually, it's an application of Miquel points. The basic fact is that four lines in the plane enclose four triangles, and that the circumcircles of these triangles have a common point; this point is called the Miquel point of the four lines. Now, the whole problem comes down to proving that the Miquel point of the lines AD, BC, AB, EF coincides with the Miquel point of the lines AD, BC, CD, EF; in other words, we have to prove that the circumcircles of triangles $X_{1337}AD$, $X_{1337}BC$ and $X_{1337}EF$ have a common point. This is not only a simpler problem (we can forget about S and T now, the point $X_{1337}$ is much easier to handle), but also known: it is a restatement of http://www.mathlinks.ro/Forum/viewtopic.php?t=68157 . Darij

darij grinberg wrote: Not long ago, I ended a solution to a problem with a quote from Munchkin, what somewhat irritated the correcter, but it still was 7/7. You can't leave us hanging like that now I think we're all pretty curious!

Hikaru79 wrote: darij grinberg wrote: Not long ago, I ended a solution to a problem with a quote from Munchkin, what somewhat irritated the correcter, but it still was 7/7. You can't leave us hanging like that now I think we're all pretty curious! Ok, it was an inequality from the ISL 2005, so I won't post it here. I spent about three hours trying to bash it to death by the standard ugly methods (again I won't get more precise now because it is an ISL problem), finally I found a rather nice solution. I was about to end the writeup with a big "PWNED", but I thought it would be too boring (basicly because I would be not the only one to do that - some guys write that under each of their solutions), so I wrote "... thus the problem is dead, dead, dead, any questions left?" (quote from the Krakzilla - I believe this is called Squidzilla in English - card in Munchkin I; this was translated back from the German translation, so I guess it's not quite authentic). darij

Quote: That'd be awesome to try! May I suggest that you teach the basis of L33tsp33k to your deputy leader and leader ? Don't forget, when it comes to marking your copy, he's kind of your best ally !

If AD // BC, easy to prove If AD isn’t parallel to BC, call the intersection of AD and BC is X By a well know lemma we have (XAB), (XEF), (SAE), (SBF) have a common point (XCD), (XEF), (TED), (TFC) have a common point To finish the problem, we will prove that: (XAB), (XEF), (XCD) have a common point Because: AE/ED = BF/FC (frac{AE}{ED}=frac{BF}{FC}, similar problem 4 in CANADA MO 2003, we get the result ( Denote that (XYZ) is circumcenter of triangle XYZ) PS: I think the problem in CANADA MO 2003 is strong, and beautiful. I can solve many problem with this lemma. For example: China TST 2006( problem 1 in fouth day) I’sh that someone can solve problem 5 in VietNam TST 2008 by this lemma, very thanks

If $AD // BC$, easy to prove If $AD$ isn’t parallel to $BC$, call the intersection of $AD$ and $BC$ is $X$ By a well know lemma we have $(XAB), (XEF), (SAE), (SBF)$ have a common point $(XCD), (XEF), (TED), (TFC)$ have a common point To finish the problem, we will prove that: $(XAB), (XEF), (XCD)$ have a common point Because: $\frac{AE}{ED} = \frac{BF}{FC}$, similar problem 4 in CANADA MO 2003, we get the result ( Denote that $(XYZ)$ is circumcircle of triangle $XYZ$) PS: I think the problem in CANADA MO 2003 is strong, and beautiful. I can solve many problem with this lemma. For example: China TST 2006( problem 1 in fouth day) I’sh that someone can solve problem 5 in VietNam TST 2008 by this lemma, very thanks

Hmm...#6?? The spiral similarity that takes AE to BF also takes DE to CF. Its center is the desired point where the circles intersect. If AD||BC, then S = T. If only geometry were still this easy...

Suppose, $AD\cap BC=X$. Note that, $\odot ZAB,\odot XEF,\odot SAE,\odot SBF$ are concurrent. And also note that $\odot TFC,\odot TDE,\odot ADC,\odot AEF$ are concurrent. So what remains to prove that, $\odot XAB,\odot XEF,\odot XDC$ are concurrent.Under inversion wrt $X$ with power $k^2$, $A,B,E,F,D,C$ goes to $A',B',E',F',D',C'$ and it remains to prove that $A'B',E'F',D'C'$ are concurrent $\iff (XE',A'D')=(XF',B'C')$. We know that inversion preserves cross-ratio. So $(XE',A'D')=(\infty E,AD)=\frac {ED}{AE}$. Similarly $(XF',B'C')=\frac {FC}{BF}$ and by the given condition they are equal. So done.

it's obvious that they pass trough Miquel point of the quad. ABCD

Denote the Miquel point of $ABCD$ by $X$ Then it is well known that $X$ is the centre of spiral symmetry that takes $A \to B$ and $D \to C$ and thus $E \to F$ Then again, it is well known that $X$ is the Miquel point of $BFEA$ and $CDEF$ Thus we are done

Note that $ABCD$ cannot be a parallelogram, for otherwise $FE$ and $BA$ would be parallel. Let$P$ be the center of spiral similarity that carries $AD$ to $BC$. It must also carry$E$ to$F$since $\frac{AE}{ED}$$=$ $\frac{BF}{FC}$. Since the $spiral similarity$ carries$AE$ to $BF$, it follows that$PAES$ and $PBFS$ are $concyclic$. Similarly, since the $spiral similarity$carries$DE$ to $CF$, $PEDT$ and $PFCT$ are $concyclic$. Therefore the circumcircles of $SAE$, $SBF$, $TCF$, $TDE$ all pass through$P$......

The center of spiral sim sending AD to BC, O, is the shared point, which can easily be shown since it is a property of spiral sims that points that break segments in equal ratios are also mapped by the spiral sim, therefore O maps E to F. It is another well known property (someone confirm can I state this in proofs?) that the center of spiral sim is the other shared point by circumcircles of two triangles, one that contains the intersection of the segment with image of segment and the old vertices, and the same intersection with the images of the vertices as the second triangle. Then O is cyclic on all of those circles.

what. The concurrency point is the Miquel point of $ABCD$. It's the center of the spiral similarity sending $\overline{AD}$ to $\overline{BC}$, which also sends $E$ to $F$, hence it is $(SAE) \cap (SBF)$ and also $(TED) \cap (TFC)$. $\blacksquare$

Extend $AD$ and $BC$ to intersect at $P$. Now, construct the Miquel point of quadrilateral $ABCD$, and call it $M$. By definition, $(PAB)$ passes through $M$, which means that $MABP$ is a cyclic quadrilateral. Using the fact that $\frac{AE}{ED}=\frac{BF}{FC}$, $\measuredangle MAB = \measuredangle MEF$ by spiral similarity ($M$ is the spiral center). This means that $MEFP$ is a cyclic quadrilateral, and $(PEF)$ passes through $M$. Since $(PAB)\cap(PEF)=M$, $M$ is the Miquel point of quadrilateral $AEFB$. This means that $(SAE)\cap(SBF)=M$ Now, consider the Miquel point of $EDCF$. This point is defined as $(PDC)\cap(PEF)$, which we already know is $M$. This means that the Miquel point of $EDCF$ is $M$, and by effect, $(TCF)\cap(TDE)=M$ Thus, $(SAE)\cap(SBF)\cap(TCF)\cap(TDE)=M$, as desired. $\blacksquare$

Let $M$ be the Miquel point of $ABCD$. Then $M$ is the center of the spiral similarity taking $\overline{AD}$ to $\overline{BC}$. However, the length condition implies that this spiral similarity also takes $\overline{AE}$ to $\overline{BF}$ and $\overline{ED}$ to $\overline{FC}$, so $M$ is also the Miquel point of $ABFE$ and $CDEF$, which implies the conclusion.

markmark

It is well known that the center of spiral similarity mapping $AE \rightarrow BF$ is $(SAE) \cap (SBF)$, and the center of spiral similarity mapping $ED \rightarrow FC$ is $(TED) \cap (TFC)$. However our given ratio condition implies the two spiral similarities are in fact the same, meaning they have a common center. $\blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -16.20277530297942, xmax = 16.10507029360849, ymin = -6.3350798585491646, ymax = 16.298838578600595; /* image dimensions */ pen zzttff = rgb(0.6,0.2,1); pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255); pen qqzzcc = rgb(0,0.6,0.8); draw(circle((-1.4929719723350294,5.814191205104764), 5.657314091172255), linewidth(0.9) + yqqqyq); draw(circle((1.5895158360944508,5.040441522101645), 7.84273175367932), linewidth(0.9) + blue); draw(circle((-2.236923935621172,3.1702582599575617), 3.5943684310059676), linewidth(0.9) + zzttff); draw(circle((-0.6304564720640408,1.9469508194950733), 4.9828711954071165), linewidth(0.9) + qqzzcc); /* draw figures */ draw(circle((0.05684934178983371,-1.6587991928044357), 2.5003975224689863), linewidth(0.9) + dotted); draw((0.9087597768919023,0.6919956474428471)--(2.2965902439432053,-2.7703514684938186), linewidth(0.9) + blue); draw((2.2965902439432053,-2.7703514684938186)--(-2.1723416151102066,-2.791360647227165), linewidth(0.9) + blue); draw((-2.1723416151102066,-2.791360647227165)--(-2.116288073657095,-0.4220851822289564), linewidth(0.9) + blue); draw((-2.116288073657095,-0.4220851822289564)--(0.9087597768919023,0.6919956474428471), linewidth(0.9) + blue); draw((0.9087597768919023,0.6919956474428471)--(-1.9228030829321123,7.7561535068211), linewidth(0.9) + zzttff); draw((-2.116288073657095,-0.4220851822289564)--(-1.9228030829321123,7.7561535068211), linewidth(0.9) + zzttff); draw((0.9559106862271817,-2.776654222113822)--(-3.2940575164779307,11.17714688167488), linewidth(0.9)); draw((-1.9228030829321123,7.7561535068211)--(-3.2940575164779307,11.17714688167488), linewidth(0.9) + zzttff); draw((-2.116288073657095,-0.4220851822289564)--(-8.632005712400415,-2.821728583427046), linewidth(0.9) + zzttff); draw((-2.1723416151102066,-2.791360647227165)--(-8.632005712400415,-2.821728583427046), linewidth(0.9) + zzttff); draw((-1.9228030829321123,7.7561535068211)--(-8.632005712400415,-2.821728583427046), linewidth(0.9) + zzttff); /* dots and labels */ dot((0.9087597768919023,0.6919956474428471),dotstyle); label("$A$", (1.026028289877822,1.004820421612019), NE * labelscalefactor); dot((2.2965902439432053,-2.7703514684938186),dotstyle); label("$B$", (2.530861402266422,-3.386835394551367), NE * labelscalefactor); dot((-2.1723416151102066,-2.791360647227165),dotstyle); label("$C$", (-2.689988171326681,-3.3254136348847463), NE * labelscalefactor); dot((-2.116288073657095,-0.4220851822289564),dotstyle); label("$D$", (-2.6592772914820157,-0.2236147717203968), NE * labelscalefactor); dot((-1.9228030829321123,7.7561535068211),linewidth(4pt) + dotstyle); label("$P$", (-1.799372655831387,8.006901023606789), NE * labelscalefactor); dot((0.0012454217272030998,0.3577713985413061),linewidth(4pt) + dotstyle); label("$E$", (0.28896717360585433,-0.10077125238715522), NE * labelscalefactor); dot((0.9559106862271817,-2.776654222113822),linewidth(4pt) + dotstyle); label("$F$", (1.0874500495671524,-2.526930759218676), NE * labelscalefactor); dot((-3.2940575164779307,11.17714688167488),linewidth(4pt) + dotstyle); label("$S$", (-3.6420254465113056,11.415808685104242), NE * labelscalefactor); dot((-1.9465391028953434,6.752877562470053),linewidth(4pt) + dotstyle); label("$T$", (-2.4443011325693584,6.0106938344416125), NE * labelscalefactor); dot((-5.613318897357623,1.9376019898753498),linewidth(4pt) + dotstyle); label("$M$", (-6.283161113152523,1.711170657778158), NE * labelscalefactor); dot((-8.632005712400415,-2.821728583427046),linewidth(4pt) + dotstyle); label("$Q$", (-9.262116458085059,-2.956883076885022), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $M$ denote the Miquel point of $ABCD$. Then note that there exists a spiral similarity centered at $M$, mapping $AD \mapsto BC$. However the ratio condition implies this spiral similarity maps $AE \mapsto BF$ and $ED \mapsto FC$. Thus $M$ is also the Miquel point of quadrilaterals $AEFB$ and $DECF$. Then we find the circles $(SEA)$ and $(SFB)$ concur at $M$ from $AEFB$, and the circles $(TDE)$ and $(TCF)$ concur at $M$ from $DECF$ as well. Thus we are done.

One liner. Note that there exists a spiral center that sends $AED$ to $BFC$, so it is the Miquel Point of $AEFB$ and $EDCF$, implying the result.

The spiral similarity taking $AD$ to $BC$ takes $E$ to $F,$ so $ABCD,ABFE,EFCD$ share a Miquel point and thus the circles concur.

I have not seen anyone addressing this, so will mention - isn't this just IMO 2005/5 (which one can solve by spiral similarity when using only the ratio equality formed from the equal segments)??

Let $M$ be the Miquel point of quadrilateral $ABCD$. I claim that this is the desired point of concurrency. Observe that $M$ is the unique spiral similarity for which $AD \stackrel{M}{\longmapsto} BC$, and it also follows that $E \stackrel{M}{\longmapsto} F$. Now let $M' = (SAE) \cap (SBF)$, and similarly $M'' = (TED) \cap (TFC)$. Then we have that $$EA \stackrel{M'}{\longmapsto} FB \text{ and } ED \stackrel{M''}{\longmapsto} FC.$$ Now observe that the maps that send $EA$ to $FB$ and $ED$ to $FC$ are unique, so that $M = M' = M''$, proving our initial claim, as desired. $\blacksquare$

As $\frac{AE}{ED}=\frac{BF}{FC}$ the spiral similarity taking $AB$ to $CD$ also takes $E$ to $F$ so $AEBF$, $EFDC$ and $ABCD$ all have the same miquel point which suffices.

If $AD \parallel BC$, $S = T$ which is the desired point. Suppose not: Consider the spiral similarity $\Phi$ with center $P$ sending segment $\overline{AD}$ to $\overline{BC}$. Recall that the center of a spiral similarity sending segment $\overline{PQ}$ to $\overline{UV}$ is $(PQX) \cap (UVX)$, where $X = PU \cap QV$. - Since $\frac{AE}{AD} = \frac{BF}{BC}$, $\Phi$ also sends $\overline{AD}$ to $\overline{BF}$. Therefore, $P = (SBF) \cap (SAE)$. - Since $\frac{ED}{AD} = \frac{FC}{BC}$, $\Phi$ also sends $\overline{ED}$ to $\overline{FC}$. Therefore, $P = (TCF) \cap (TDE)$. $\blacksquare$

Let $AD \cap BC = Y$. Note that from gliding principle, $(YAB), (YEF), (YDC)$ concur at some point $X$. But now Miquel's theorem on the comp quads $AEFB$, $ADCB$ suffices. $\square$

Basically, we want to show that the Miquel points of $ABFE$ and $ABCD$ coincide. But that's equivalent to showing that $(PAB),(PEF),(PCD)$ are coaxial. Indeed, if $Q=(PAB)\cap (PCD)$, then we get that the spiral similarity sending $AD$ to $BC$ also sends $E$ to $F$, that is, $\angle QEF=\angle QAB=\angle PAB=\angle PEF$, so $Q$ lies on $(PEF)$ too, as desired. $\blacksquare$