Let ABCD be a quadrilateral, and let E and F be points on sides AD and BC, respectively, such that AEED=BFFC. Ray FE meets rays BA and CD at S and T, respectively. Prove that the circumcircles of triangles SAE, SBF, TCF, and TDE pass through a common point.
Problem
Source: USAMO 2006, Problem 6, proposed by Zuming Feng
Tags: geometry, ratio, geometric transformation, Spiral Similarity
20.04.2006 22:04
Let P be the second intersection point of the circles (SAE),(SBF). We have ∠APE=∠ASE=∠BPF, and ∠PAE=∠PSE=∠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 ∠PDT=∠PAS=∠PET, i.e. P lies on the circle (TDE). In the same way we prove that it lies on (TCF).
20.04.2006 22:27
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 X1337 be the point of intersection of the lines AD and BC. Let the circumcircles of triangles X1337AB and X1337CD intersect at a point Λ apart from X1337 (why Λ? cuz teh letter r00lzorz ). Then, ∡ΛBX1337=∡ΛAX1337, what is equivalent to ∡ΛBC=∡ΛAD. Similarly, ∡ΛCB=∡ΛDA. Hence, the triangles ΛBC and ΛAD 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 BFFC=AEED. As corresponding points in directly similar triangles form equal angles, this entails ∡ΛFB=∡ΛEA. This is equivalent to ∡ΛFX1337=∡ΛEX1337. Thus, the point Λ lies on the circumcircle of triangle X1337EF. 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 Λ lies on the circumcircle of triangle X1337AB, we have ∡ΛBA=∡ΛX1337A. Obviously, ∡ΛX1337A=∡ΛX1337E. Since the point Λ lies on the circumcircle of triangle X1337EF, we have ∡ΛX1337E=∡ΛFE. Thus, ∡ΛBA=∡ΛFE. In other words, ∡ΛBS=∡ΛFS. Thus, the point Λ lies on the circumcircle of triangle SBF. Similarly, we can prove that the same point Λ 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
Attachments:

21.04.2006 00:52
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!
21.04.2006 01:14
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 OWNED→PWNED OMFG→PMFG darij
21.04.2006 01:28
lawlers! I see, kind of a "slide on the keyboard", eh? *cries* 1 g07 0wn3d.. Nevermind, I'll stop this spam now.
21.04.2006 11:07
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?
21.04.2006 13:20
That'd be awesome to try! At our next TST, I'll do it for one of the exercises, we'll see...
23.04.2006 14:03
Now it becomes absolutely obvious how much a proper notation is important in maths.
25.04.2006 10:13
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 X1337AD, X1337BC and X1337EF have a common point. This is not only a simpler problem (we can forget about S and T now, the point X1337 is much easier to handle), but also known: it is a restatement of http://www.mathlinks.ro/Forum/viewtopic.php?t=68157 . Darij
04.05.2006 07:28
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!
04.05.2006 10:03
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
08.05.2006 23:02
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 !
20.01.2009 15:40
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
22.01.2009 14:49
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: AEED=BFFC, 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
16.01.2011 21:36
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...
21.06.2011 21:36
Suppose, AD∩BC=X. Note that, ⊙ZAB,⊙XEF,⊙SAE,⊙SBF are concurrent. And also note that ⊙TFC,⊙TDE,⊙ADC,⊙AEF are concurrent. So what remains to prove that, ⊙XAB,⊙XEF,⊙XDC are concurrent.Under inversion wrt X with power k2, 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 ⟺(XE′,A′D′)=(XF′,B′C′). We know that inversion preserves cross-ratio. So (XE′,A′D′)=(∞E,AD)=EDAE. Similarly (XF′,B′C′)=FCBF and by the given condition they are equal. So done.
04.12.2012 11:56
it's obvious that they pass trough Miquel point of the quad. ABCD
14.01.2015 16:28
Denote the Miquel point of ABCD by X Then it is well known that X is the centre of spiral symmetry that takes A→B and D→C and thus E→F Then again, it is well known that X is the Miquel point of BFEA and CDEF Thus we are done
28.09.2015 22:13
Note that ABCD cannot be a parallelogram, for otherwise FE and BA would be parallel. LetP be the center of spiral similarity that carries AD to BC. It must also carryE toFsince AEED= BFFC. Since the spiralsimilarity carriesAE to BF, it follows thatPAES and PBFS are concyclic. Similarly, since the spiralsimilaritycarriesDE to CF, PEDT and PFCT are concyclic. Therefore the circumcircles of SAE, SBF, TCF, TDE all pass throughP......
21.06.2023 12:35
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.
16.07.2023 20:37
what. The concurrency point is the Miquel point of ABCD. It's the center of the spiral similarity sending ¯AD to ¯BC, which also sends E to F, hence it is (SAE)∩(SBF) and also (TED)∩(TFC). ◼
05.08.2023 18:40
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 AEED=BFFC, ∡MAB=∡MEF by spiral similarity (M is the spiral center). This means that MEFP is a cyclic quadrilateral, and (PEF) passes through M. Since (PAB)∩(PEF)=M, M is the Miquel point of quadrilateral AEFB. This means that (SAE)∩(SBF)=M Now, consider the Miquel point of EDCF. This point is defined as (PDC)∩(PEF), which we already know is M. This means that the Miquel point of EDCF is M, and by effect, (TCF)∩(TDE)=M Thus, (SAE)∩(SBF)∩(TCF)∩(TDE)=M, as desired. ◼
09.08.2023 03:57
Let M be the Miquel point of ABCD. Then M is the center of the spiral similarity taking ¯AD to ¯BC. However, the length condition implies that this spiral similarity also takes ¯AE to ¯BF and ¯ED to ¯FC, so M is also the Miquel point of ABFE and CDEF, which implies the conclusion.
11.10.2023 16:10
markmark
18.11.2023 05:55
It is well known that the center of spiral similarity mapping AE→BF is (SAE)∩(SBF), and the center of spiral similarity mapping ED→FC is (TED)∩(TFC). However our given ratio condition implies the two spiral similarities are in fact the same, meaning they have a common center. ◼
28.11.2023 08:25
[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↦BC. However the ratio condition implies this spiral similarity maps AE↦BF and ED↦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.
03.12.2023 00:12
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.
18.03.2024 22:23
The spiral similarity taking AD to BC takes E to F, so ABCD,ABFE,EFCD share a Miquel point and thus the circles concur.
07.04.2024 02:17
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)??
04.05.2024 03:54
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 ADM⟼BC, and it also follows that EM⟼F. Now let M′=(SAE)∩(SBF), and similarly M″=(TED)∩(TFC). Then we have that EAM′⟼FB and EDM″⟼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. ◼
21.09.2024 04:01
As AEED=BFFC 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.
10.10.2024 05:22
If AD∥BC, S=T which is the desired point. Suppose not: Consider the spiral similarity Φ with center P sending segment ¯AD to ¯BC. Recall that the center of a spiral similarity sending segment ¯PQ to ¯UV is (PQX)∩(UVX), where X=PU∩QV. - Since AEAD=BFBC, Φ also sends ¯AD to ¯BF. Therefore, P=(SBF)∩(SAE). - Since EDAD=FCBC, Φ also sends ¯ED to ¯FC. Therefore, P=(TCF)∩(TDE). ◼
11.10.2024 15:54
Let AD∩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. ◻
13.11.2024 00:12
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)∩(PCD), then we get that the spiral similarity sending AD to BC also sends E to F, that is, ∠QEF=∠QAB=∠PAB=∠PEF, so Q lies on (PEF) too, as desired. ◼