Let M be any point on the circumcircle of triangle ABC. Suppose the tangents from M to the incircle meet BC at two points X1 and X2. Prove that the circumcircle of triangle MX1X2 intersects the circumcircle of ABC again at the tangency point of the A-mixtilinear incircle.
Problem
Source: Taiwan 2014 TST3, Problem 3
Tags: geometry, circumcircle, incenter, geometric transformation, homothety, ratio, parallelogram
19.07.2014 00:38
It's a special case of yetti's old problem. There's a short solution using Desargues' involution theorem: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=209898&p=1156398
19.07.2014 00:59
Interesting. I thought that TS had created this, but it looks like this is in fact from Cosmin. Small world. No one solved this during the TST, but we did find a solution afterwards using inversion at the incircle. One uses the fact that TI (here I is the incenter and T is the mixtilinear tangency point) passes through the midpoint of arc BAC in order to control the image of T under inversion.
19.07.2014 20:46
Nice, this is one of my favorite problems. Congratulations for the IMO result by the way, Evan!
19.07.2014 23:06
Thanks for the congratulations! Yes, this is a really beautiful result indeed, one of my favorites as well. I just read through Vladmir's inversive solution and it looks a bit different from the one I mentioned above. So let me post the Taiwan team's solution, in its full glory. [asy][asy] defaultpen(fontsize(8pt)); pointfontsize=8; size(11cm); pointpen = black; pair A = Drawing("A", dir(110), dir(110)); pair B = Drawing("B", dir(210), dir(210)); pair C = Drawing("C", dir(330), dir(330)); draw(A--B--C--cycle, heavyred); pair O = circumcenter(A,B,C); pair I = incenter(A,B,C); pair Ta = foot(I,B,C); pair Tb = foot(I,C,A); pair Tc = foot(I,A,B); draw(unitcircle, heavyred); draw(incircle(A,B,C), blue+1); pair L = Drawing("L", dir(90), dir(90)); pair T = Drawing("T", 2*foot(O,I,L)-L, dir(I-L)); pair M = Drawing("M", dir(-80), dir(-80)); pair D = Drawing("D", Ta, dir(Ta-I)); pair E = Drawing("E", Tb, dir(Tb-I)); pair F = Drawing("F", Tc, dir(Tc-I)); draw(D--E--F--cycle, dashed+blue); pair Ac = Drawing("A^\ast", (E+F)/2, dir(60)); pair Bc = Drawing("B^\ast", (F+D)/2, dir(B-I)); pair Cc = Drawing("C^\ast", (D+E)/2, dir(0)); draw(circumcircle(Ac,Bc,Cc), heavycyan+1); real p = abs(D-I)*abs(D-I); pair Mc = Drawing("M^\ast", I+p/conj(M-I), dir(100)*1.5); pair Lc = Drawing("L^\ast", I+p/conj(L-I), dir( 10)*1.5); pair Tc = Drawing("T^\ast", I+p/conj(T-I), dir(110)*1.5); Drawing("I", I, dir(45)); pair H = Drawing("H", orthocenter(D,E,F), dir(90)); draw(Ac--Bc--Cc--cycle, heavycyan); pair[] K = IPs(circumcircle(D,E,F), L(Mc-(M-I)*dir(90), Mc+(M-I)*dir(90))); pair K1 = Drawing("K_1", K[1], dir(K[1]-K[0])); pair K2 = Drawing("K_2", K[0], dir(K[0]-K[1])); pair X1 = Drawing("X_1", extension(K1,M,B,C), dir(-45)); pair X2 = Drawing("X_2", extension(K2,M,B,C), dir(225)); pair X1c = Drawing("X_1^\ast", (D+K1)/2, dir(100)); pair X2c = Drawing("X_2^\ast", (D+K2)/2, dir(80)); draw(K1--D--K2, darkgreen); draw(K1--M--K2, darkgreen); draw(K1--K2, darkgreen); draw(Ac--Lc); draw(L--T, dashed); draw(I--A--L, dashed); pair Dp = Drawing("D'", 2*Mc-D, dir(115)); draw(D--Dp, dotted); pair Hp = Drawing("H'", 2*Mc-H, dir(-20)); draw(H--Hp, dotted); [/asy][/asy] Let DEF be the intouch triangle. Let the tangents to the incircle be ¯MK1 and ¯MK2. Moreover, denote by H the orthocenter of triangle DEF. After inverting around the incircle, denote the inverse with a star. We know that (A∗B∗C∗) is the nine-point circle of △DEF. Using the image of L, we see that T∗ is diametrically opposite A∗ on this circle, meaning that T∗ is in fact the midpoint of ¯DH. Meanwhile, ¯K1K2 is a chord of the incircle whose midpoint M∗ lies on the nine-point circle, and X∗1 and X∗2 are the midpoints of ¯DK1 and ¯DK2. Now it's easy to show that T∗, M∗, X∗1 and X∗2 are concyclic -- take a homothety at D with ratio 2, then a homothety at M∗ with ratio −1, which sends the four points to K1, K2, D, and H′. Here H′ is the image of M∗ under a homothety at H with ratio 2. As M∗ lies on the nine-point circle of △DEF, this implies H′ lies on the circumcircle of △DEF, as required.
30.09.2014 02:55
What is the A-mixtilinear incircle? v_Enhance wrote: Let DEF be the intouch triangle. Let the tangents to the incircle be ¯MK1 and ¯MK2. Moreover, denote by H the orthocenter of triangle DEF. After inverting around the incircle, denote the inverse with a star. What is the ratio of this inversion around the incircle?
30.09.2014 11:11
Dear Mathlinkers, for a proof http://jl.ayme.pagesperso-orange.fr/ vol. 4 A new mixtilinear incircle adventure III, p. 12..... with a complete history and a generalization Sincerely Jean-Louis
30.09.2014 17:29
MathPanda1 wrote: What is the A-mixtilinear incircle? Circle tangent to ¯AB, ¯AC, and internally tangent to the circumcircle of ABC. MathPanda1 wrote: What is the ratio of this inversion around the incircle? Whatever the radius of the incircle is.
03.12.2014 23:29
I have a solution here, and I guess that it's less "magical" than the above inversive solution, as in, just through working backwards, you can actually arrive at a solution. Note that this took about 2.5 hours though, so maybe that's why no one solved this during the TST. Let T be the mixtilinear incircle tangency point. Also, extend the tangents from M to the incircle to hit the circumcircle of ABC at Y1,Y2, such that M,X1,Y1 are collinear and M,X2,Y2 are collinear. By a well-known lemma about the incircle, Y1Y2 is also tangent to the incircle. Now note that by the spiral-similarity lemma, since X1Y1∩X2Y2=M, if TX1X2M are concyclic, then T is the spiral center sending X1X2→Y1Y2. So it would be nice to show that T is indeed the spiral center. As for more motivation (working backwards), I considered the case where M≡T. In this (degenerate) case, TX1X2 should be tangent to the circumcircle. If this were true, the the circle TX1X2 should be homothetic to the circumcircle at T. So once again to exploit this homothety, we extend TX1,TX2 to hit the circumcircle at Y1,Y2. So if they were homothetic, Y1Y2||BC, so ∠BTY1=∠CTY2. By the tangency, ∠Y1TI=∠Y2TI. So ∠BTI=∠BTY1+∠Y1TI=∠CTY2+∠Y2TI=∠CTI. So T should satisfy ∠BTI=∠CTI. Let's prove this to characterize T. Let the mixtilinear incircle hit AB,BC at T1,T2. By a well-known lemma T1,T2,I are collinear. Also, ∠IT1B=90+∠A/2. But by the above, ∠BTI=∠BTC/2=90−∠A/2. So IT1BT should be concyclic. You can probably prove this many ways. I just did an inversion at A, and this becomes clear: in the new triangle AB′C′, T maps to where the excircle hits B′C′, I maps to the excenter, B maps to B′, and T1 maps to where the excircle hits AB′. So we get that ∠BTI=∠CTI from the cyclic quads BTIT1 and similarly CTIT2. As an added bonus, ∠BIT=∠BT1T=∠TT2T1=∠TCI, from cyclic quads and the fact that circle TT1T2 is tangent to AB. So in fact, T is the spiral center sending BI to IC. Now we have almost enough to finish. Note that to this point, we have barely even used M,Y1,Y2. Let's phantom point T′ to be the point that sends X1X2 to Y1Y2. Of course we're going to try to show that T′≡T. I'll do this by proving that T′ in fact also will send BI to IC, and because spiral centers are unique, we'll be done. Just angle chasing around a bit shows that T′ in fact sends BY1 to X2C. If T sends both BY1 to X2C and BI to IC, then it also sends Y1I to IX2. We'll prove this now. Note that almost all points in the diagram are gone now, and we can reduce it to the following problem: Y1X1X2Y2 is a tangential quadrilateral (remember that Y1Y2 is tangent to the incircle). Let T′ be its Miquel Point. Prove that T′ also sends Y1I to IX2. This can be done with complex numbers with the incircle as the unit circle. Let the tangency points where X1Y1,Y1Y2,Y2X2,X2X1 hits the incircle be a,b,c,d. Then y1=21/a+1/b, y2=21/b+1/c, x2=21/c+1/d, x1=21/a+1/d. If T′ sends Y1Y2 to X1X2, then t′−y1t′−y2=t′−x1t′−x2. Solving for t′ gives t′=21/a+1/b+1/c+1/d. To show that T′ sends Y1I to IX2, it suffices to prove that t′−y1t′−0=t′−0t′−x2, which is an easy computation. Having proven this last thing, we are done. ◼ Notes: This looks long because I tried to work backwards and explain motivation. Hopefully nothing here seems "magical". I will try to add diagrams for the parts when I have time. Also, a synthetic solution would be appreciated for the last lemma. I could not find one.
11.12.2014 05:26
mathocean97 wrote: Y1X1X2Y2 is a tangential quadrilateral (remember that Y1Y2 is tangent to the incircle). Let T′ be its Miquel Point. Prove that T′ also sends Y1I to IX2. Here's a synthetic solution for this lemma (if inversion is counts as synthetic). Inverting about the incircle of the quadrilateral gives the following problem: Let ABCD be a cyclic quadrilateral. Let M,N be the midpoints of AD,BC, respectively. Let the nine-point circles of △ABC and △BCD intersect at P≠N. Prove that MPNO is a parallelogram. Solution: Let the orthocentres of △ABC,△BCD be H1,H2 and the nine-point centres be N1,N2 (respectively). We can angle chase to show that ∠AH1H2=∠H2DA, and since AH1||DH2, it follows that H1H2||AD. Since N1 is the midpoint of OH1 and N2 is the midpoint of OH2, N1N2||H1H2 and thus N1N2||AD. NP is the common chord of the nine-point circles of △ABC,△BCD, so NP⊥N1N2. Thus NP⊥AD, so NP||MO. Similarly, MP||NO so MPNO is a parallelogram and we're done.
06.05.2015 04:58
Here's another inversive solution, though I believe the most efficient one posted yet. Rename X1,X2 as X,Y. Call the ω(O) the common incircle of △ABC and △MXY, and let D,E,F,S,T be the points of tangency of lines BC,CA,AB,MX,MY with ω. We compose an inversion I:X→X′ about ω. Since EF,FD,DE are the polars of A,B,C with respect to ω, we find that A′,B′,C′ are the midpoints of ¯EF,¯FD,¯DE. Similarly, M′,X′,Y′ are the midpoints of ¯TS,¯SD,¯DT. Also, note that A,B,C,M are concyclic, so under inversion, A′,B′,C′,M′ are concyclic as well. Now, here is the key to the solution: the tangency of the A-mixtilinear incircle with ⊙(ABC) is sent to the antipode of A′ WRT ⊙(A′B′C′) (see below for a sketch of the proof of this). Thus, it suffices to prove that Γ1≡⊙(A′B′C′) and Γ2≡⊙(M′X′Y′) meet for a second time at the antipode of A′ w.r.t. Γ1. Let H1,H2,N1,N2 be the orthocenters and nine-point centers of △DEF,△DST, and let D′ be the antipode of D w.r.t. ω. Recall that the reflection of the orthocenter of a triangle in the midpoint of a side is the antipode of the vertex opposite that side w.r.t. the circumcircle. Therefore, A′,M′ are the midpoints of ¯D′H1,¯D′H2, so A′M′∥H1H2. Now, recall that the nine-point center of a triangle is the midpoint of the segment connecting the orthocenter and the circumcenter. Since O is the center of ω, we find that N1,N2 are the midpoints of ¯OH1,¯OH2. Thus, N1N2∥H1H2∥A′M′. Now, note that the radius of the nine-point circle is half the radius of the circumcircle, so Γ1 and Γ2 have the same radius. Therefore, N1A′=N2M′ and N1N2∥A′M′. This is enough to imply that N1N2M′A′ is either an isoceles trapezoid or a parallelogram. If the former were the case, we would have ∠N1A′M′=∠N2M′A′. But since ∠N1A′M′=∠N1M′A′≠∠N2M′A′, this is obviously false. Therefore, N1N2M′A′ is a parallelogram, so N1A′∥N2M′. Now, if Z is the second intersection of Γ1 and Γ2, note that N1ZN2M′ is a rhombus. Therefore, N1Z∥N2M′, and it follows that N1,A′,Z are collinear, as desired. ◻
06.06.2016 06:44
Let D be where the incircle is tangent to BC and let AM meet BC at N. By the dual of Desargues' Involution Theorem (see Deduction 1 here: http://apollonius.math.nthu.edu.tw/d1/ne01/whw/conic/involution/inv(ded2).htm for details), M(A,D;B,C;X1,X2)→(N,D;B,C;X1,X2) is an involution. Hence, it suffices to show that the circumcircle of MDN passes through the mixtilinear touchpoint T. But note that TD meets the circumcircle at A′, where AA′∥BC, so ∠ANB=∠A′AM=∠A′TM=∠DTM as desired.
26.08.2016 06:58
I have searched everywhere for the dual of Desargues' Involution Theorem and it appears the link does not work for me. Could someone please post the statement of the dual of Desargues' Involution Theorem? Thank you very much for all your help!
26.08.2016 07:04
Dual of Desargues' Involution: Consider a complete quadrilateral ABCDEF. Choose an arbitrary point P. If (A,C),(B,D),(E,F) are opposite points in the quadrilateral, then there is an involution swapping the pairs (PA,PC),(PB,PD),(PE,PF). Moreover, if ω is a conic tangent to the sidelines of the complete quadrilateral, and PX and PY are tangent to ω, then this involution also swaps (PX,PY).
26.08.2016 07:12
Thank you very much for all your help K6160! Sorry, I am a little new to this, but what does it mean that there is an involution swapping the pairs (PA,PC),(PB,PD),(PE,PF)? Thanks!
26.08.2016 07:17
This is probably not the right way to think about this and it might not even be correct (someone verify), but the way I like to think about involutions (in that case) is that there are two lines ℓ1 and ℓ2 passing through P such that (ℓ1,ℓ2;PX,PY) is harmonic, where the involution swaps the pair (PX,PY). ℓ1 and ℓ2 are the fixed points of the involution. Involutions of lines can be projected onto another line or circle like harmonic bundles.
26.08.2016 07:20
If you project PA,PB,PC,PD,PE,PF onto a line l, such that their images are A′,B′,C′,D′,E′,F′, then (A′,C′),(B′,D′),(E′,F′) are swapped under an involution of l. Essentially, an involution in this case is a mapping f such that f(f(x))=x that preserves cross ratio.
26.08.2016 07:27
Interesting, thank you very much ABCDE and K6160! Just wondering, how does one go from M(A,D;B,C;X1,X2)→(N,D;B,C;X1,X2) is an involution to it suffices to show that the circumcircle of MDN passes through the mixtilinear touchpoint T? Thanks!
26.08.2016 07:38
But, I guess I don't really understand swapping points. What is the best way I should view an involution and the transformation swapping points/lines? Sorry I am a really slow learner. Thank you very much!
26.08.2016 07:42
If you have a point P then all circumcircles of PXY where (X,Y) is a pair of an involution on a fixed line different from P are coaxial. This follows from inversion about P; the involution on the line becomes an involution on the circle and the circumcircles become lines passing through pairs of the involution. These pass through the intersection of the tangents to the circle at the fixed points of the involution by harmonics.
16.08.2023 22:00
To be fair, 90% of the difficulty of the problem is figuring out what's going on in the diagram. Invert about the incircle, and define some new points: Let DEF be the contact triangle, so the images of A,B,C are X,Y,Z, the midpoints of ¯EF and cyclic permutations. Let T1 and T2 be the tangency points from M, which remain fixed. Then, X1 and X2 invert to Y1 and Y2, the midpoints of ¯DT1 and ¯DT2. M inverts to Q, the midpoint of ¯T1T2. Let H be the orthocenter of triangle DEF and MA the midpoint of major arc ^BAC (note that MA lies on ¯IT, where T is the mixtilinear tangency point). By considering the inverse of MA, we note ∠XM∗AI=90∘, hence T inverts to the X-antipode in (XYZ), which is the midpoint of ¯DH. So by taking a homothety at D with ratio 2, it suffices to show that T1,T2,H,Q′=2Q−D are concyclic. On the other hand, by taking a homothety at Q with ratio −1, this is equivalent to T1,T2,H′=2Q−H,D concyclic. This is clear as (T1T2H) has radius equal to that of the incircle, as both radii are equal to twice the radii of (XYZ)=(ID).
21.08.2023 06:03
We invert about the incircle. Let H1 and N1 be the orthocenter and nine point center of the intouch triangle DEF, and let T be the A− mixtilinear incontact point of △ABC. First, if L is the midpoint of arc \overarcBAC, we have ∠IL∗A∗=∠IAL=90∘, and since T lies on LI, it follows that T∗ is the antipode of A∗ in the nine point circle of △DEF (or equivalently the midpoint of DH1). Next, note that MX1 and MX2 touch the incircle at points P and Q such that M∗ is the midpoint of PQ. Then X∗1=(IP)∩(ID) is just the midpoint of DP, and similarly X∗2 is the midpoint of DQ. For ease of notation we will use Y1=X∗1 and Y2=X∗2. Let H2 and N2 be the orthocenter and nine point center of △DPQ. We are now ready to tackle the actual problem. We want to show that (M∗Y1Y2) passes through T∗. Phantom Z=(M∗Y1Y2)∩(A∗B∗C∗). Since the midpoint K of DH′ is the antipode of M∗ in (M∗Y1Y2), we have by midlines at I and M respectively that KZ⊥MZ⊥N1N2∥H1H2∥KT∗ as desired.
24.08.2023 20:00
Denote the circumcircle of (ABC) by Ω and the incirlce by ω.Let f(Z)=ZBZC . Let MX1,MX2∩Ω=N,K respectively. By ratio lemma , It's enough to show that f(X1)f(X2)f(M) is a fix value(and then consider a special case which is easy). Which is , again by ratio lemma , proving that MB.NB.KBMC.NC.KC is a fix value. Now , fix BC and move M on Ω. It's well-known that the map M→P=NK∩ω(Which is a single point , by poncelet) is a projective map.Also , the map P→PP∩BC=Q is also a projective map since BC is already tangent to ω. Now let T be the midpoint of arc BC not containing A.Assume that M is beyond BC for now , the other case is similar. Let MT∩BC=S. We have f(M)=f(S) by ratio lemma. Also , by ratio lemma , f(N)f(K)=f(Q). So proving that MB.NB.KBMC.NC.KC is a fix value , is just proving that (MSQ) passes through a fix point on Ω(by ratio lemma , again). But clearly the map M→(TS∙)∩BC (where ∙ is any arbitrary fix point on Ω) is a projective map by inverting through T. But the Map M→Q was also projective .So by the moving point lemma, it's enough to show the statement of our claim for three cases which is ok by considering M=B,C,T,T′,A,...(where T′ is the midpoint of arc (BAC)) . So we're done. Btw this is probably one of the few solutions for this problem which prevents any kind of angle chasing ,or maybe the only one.I hope it's true
16.10.2023 16:59
This is a truly amazing problem. Let ω be incircle of ABC. Let ω touches BC,CA,AB at D,E,F, respectively. Consider inversion around ω. Then (ABC) becomes nine point circle of DEF. Let ωA be A mixtilinear incircle. Claim: ω∗A is circle centered A∗ with radius r, where r is radius of ω. Proof: Note that ω∗A tangent to (IA∗B∗) and (IA∗C∗), where I is center of ω. Let ωA touches AB,AC at P,Q, respectively. Then it's well-known that P,I,Q are collinear. Note that P∗ lies on (IA∗B∗) and Q∗ lies on (IA∗C∗). Then since PQ⊥AI and EF⊥AI, so EFP∗Q∗ is rectangle. Since ω∗A passes through P∗,Q∗, thus ωA touches (IA∗B∗) and (IA∗C∗) at P∗,Q∗, respectively. Hence the center of ω∗A is A∗, so ω∗A is circle centered A∗ with radius r. ◼ Note that ω∗A touches (A∗B∗C∗) at T∗. Let H be orthocenter of DEF. Claim: T∗ is midpoint of segment DH. Proof: Let X=DT∗∩EF. Since A∗T∗=r, so A∗T∗ is diameter of (A∗T∗X). Hence ∠T∗XA∗=90∘, so DX is altitude of DEF. Note that (A∗T∗X) is nine point circle of DEF, so T∗ is midpoint of DH. ◼ Since M lies on ABC, so M∗ lies on A∗B∗C∗. Let the tangents from M to ω touches ω at Y,Z. Then M∗ is midpoint of YZ and X∗1,X∗2 is midpoint of YD,ZD, respectively. We need to show that M∗X∗1X∗2T∗ is cyclic. Note that (M∗X∗1X∗2) is nine point circle of DYZ. Thus it's enough to show that T∗ lies on nine point circle of DYZ. Let γ be nine point circle of DEF and let N9 be center of γ. Let N=M∗N9∩γ. Let DS be altitude of DYZ and let K=T∗N∩DS. Claim: M∗K=M∗N=r. Proof: Note that IH is Euler line of DEF. Thus I,N9,H are collinear and IN9=HN9. Thus M∗INH parallelogram and note that M∗I⊥YZ. Since DK⊥YZ, so DK∥NH. Thus T∗KT∗N=T∗DT∗H=1, so T∗K=T∗N. Therefore M∗K=M∗N=T∗A∗=r. (Note that γ is nine point circle of DEF) ◼ Now we'll prove that T lies on nine point circle of DYZ. Note M∗K=r and M∗ is midpoint of YZ and DK⊥YZ, thus circle with diameter M∗K is nine point circle of DYZ. Since ∠KT∗M∗=90∘, so T lies on nine point circle of DYZ. This completes proof. ◼
Attachments:

25.10.2023 17:22
Invert about the incircle. If DEF is the intouch triangle of ABC with D opposite A, then (ABC) inverts to the nine-point circle of DEF. If the tangents from M meet the incircle at Y1,Y2 then (MX1X2) inverts to the nine-point circle of DY1Y2. The mixtilinear incircle inverts to a circle internally tangent to the images of AB,AC and the nine-point circle. Since these circles all have radius r2 and pass through A, the mixtilinear incircle inverts to a circle centered at A with radius r. In particular, the tangency point to the nine-point circle of DEF is simply the antipode K of A with respect to it. Let H be the orthocenter of DEF. If N is the reflection of H over M, then notice that since M is on the nine-point circle we have that N is on the incircle. Also, notice that since IM⊥Y1Y2 we get that the reflection P of H over Y1Y2 also lies on the incircle, so using directed angles we get ∡Y1DY2=∡Y1PY2=−∡Y1HY2. Next, take a homothety at D with scale factor 12. By properties of nine-point circles, we get that K is the midpoint of HD, so this takes ∡Y1HY2 to ∡X1KX2. Also, by the medial triangle we get ∡Y1DY2=−∡X1MX2, so we get that K lies on the circumcircle of MX1X2. Inverting back, we get our desired result.
01.12.2023 08:47
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13.5cm); 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 = -10.246562414429325, xmax = 9.280348485082802, ymin = -4.490356054222101, ymax = 7.9791768222097685; /* image dimensions */ pen qqzzcc = rgb(0,0.6,0.8); pen ccqqqq = rgb(0.8,0,0); pen wwqqcc = rgb(0.4,0,0.8); pen ffwwqq = rgb(1,0.4,0); pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255); /* draw figures */ draw((-2.16459773264667,6.473622296638714)--(-3.6799251329599487,-1.1072919538999262), linewidth(0.8) + blue); draw((-3.6799251329599487,-1.1072919538999262)--(3.974285243025721,-1.1196774075781717), linewidth(0.8) + blue); draw((3.974285243025721,-1.1196774075781717)--(-2.16459773264667,6.473622296638714), linewidth(0.8) + blue); draw(circle((0.15232904969162947,2.068594019540843), 4.977190355687468), linewidth(0.8) + qqzzcc); draw(circle((-0.8658596890095265,1.1961404466163956), 2.307982882950086), linewidth(0.8) + qqzzcc); draw((0.9289410864892553,2.6471660026489423)--(-3.1290722033604492,1.6485275627856872), linewidth(0.8) + ccqqqq); draw((-3.1290722033604492,1.6485275627856872)--(-0.869594284253186,-1.111839414818482), linewidth(0.8) + ccqqqq); draw((-0.869594284253186,-1.111839414818482)--(0.9289410864892553,2.6471660026489423), linewidth(0.8) + ccqqqq); draw((1.2420440756126852,0.25617943301950347)--(-0.1646928317878789,-2.8984897288382334), linewidth(0.8) + blue); draw((-0.1646928317878789,-2.8984897288382334)--(-2.540915510808472,-0.3916167709228757), linewidth(0.8) + blue); draw(circle((-0.6152425878117047,-1.6280752182276466), 1.3479421765873583), linewidth(0.8) + wwqqcc); draw(circle((-1.1019328560574269,0.9938568519998762), 1.153991441475043), linewidth(0.8) + ffwwqq); draw((-2.540915510808472,-0.3916167709228757)--(1.2420440756126852,0.25617943301950347), linewidth(0.8) + ccqqqq); draw((-2.540915510808472,-0.3916167709228757)--(-0.869594284253186,-1.111839414818482), linewidth(0.8) + ccqqqq); draw((-0.869594284253186,-1.111839414818482)--(1.2420440756126852,0.25617943301950347), linewidth(0.8) + ccqqqq); draw(circle((-1.7033875999089994,0.40226183784675995), 1.153991441475043), linewidth(0.8) + dotted); draw(circle((0.18809219330157928,0.7261599398179494), 1.153991441475043), linewidth(0.8) + dotted); draw(circle((-0.8677269866313566,0.04215051589895676), 1.153991441475043), linewidth(0.8) + dotted); draw((-1.100065558435597,2.147846782717315)--(-1.9993332438068177,0.2683440739836026), linewidth(0.8) + ccqqqq); draw((-1.9993332438068177,0.2683440739836026)--(0.02967340111803468,0.7676632939152301), linewidth(0.8) + ccqqqq); draw((0.02967340111803468,0.7676632939152301)--(-1.100065558435597,2.147846782717315), linewidth(0.8) + ccqqqq); draw(circle((-0.6513030152197237,-1.221708599669125), 1.153991441475043), linewidth(0.8) + ffwwqq); draw((-0.7108738385150224,2.079568181864503)--(-1.100065558435597,2.147846782717315), linewidth(0.8) + ccqqqq); draw((-0.7108738385150224,2.079568181864503)--(-1.1038001536792574,-0.1601330787175626), linewidth(0.8)); draw((-1.1038001536792574,-0.1601330787175626)--(-1.534317055871152,-2.614102719661712), linewidth(0.8)); draw((-0.7108738385150224,2.079568181864503)--(0.16038274522224572,7.045777859297683), linewidth(0.8)); draw((-1.3380060231053266,0.7915732573833573)--(-0.869594284253186,-1.111839414818482), linewidth(0.8) + yqqqyq); draw((-1.3380060231053266,0.7915732573833573)--(-2.540915510808472,-0.3916167709228757), linewidth(0.8) + yqqqyq); draw((-1.3380060231053266,0.7915732573833573)--(-0.4292771509426009,0.9764020769151098), linewidth(0.8) + yqqqyq); draw((-0.4292771509426009,0.9764020769151098)--(1.2420440756126852,0.25617943301950347), linewidth(0.8) + yqqqyq); draw((-0.4292771509426009,0.9764020769151098)--(-0.869594284253186,-1.111839414818482), linewidth(0.8) + yqqqyq); draw(circle((-0.43301174618626037,-1.331577784519768), 2.3079828829500864), linewidth(0.8) + yqqqyq); draw((-1.3380060231053266,0.7915732573833573)--(0.03913458790953972,-0.9270105952867296), linewidth(0.8) + yqqqyq); /* dots and labels */ dot((-2.16459773264667,6.473622296638714),dotstyle); label("A", (-2.272174768081601,6.5946720904357905), NE * labelscalefactor); dot((-3.6799251329599487,-1.1072919538999262),dotstyle); label("B", (-4.107300828673609,-1.2268806150405807), NE * labelscalefactor); dot((3.974285243025721,-1.1196774075781717),dotstyle); label("C", (4.155882917531277,-1.2358709055066455), NE * labelscalefactor); dot((-0.1646928317878789,-2.8984897288382334),dotstyle); label("M", (-0.1864273791834367,-3.2418124843123816), NE * labelscalefactor); dot((-0.8658596890095265,1.1961404466163956),linewidth(4pt) + dotstyle); label("I", (-1.0674758451835236,1.1555463584665897), NE * labelscalefactor); dot((-0.869594284253186,-1.111839414818482),linewidth(4pt) + dotstyle); label("D", (-0.9146409072039166,-1.443754391099423), NE * labelscalefactor); dot((0.9289410864892553,2.6471660026489423),linewidth(4pt) + dotstyle); label("E", (1.0092812532452524,2.719856899561864), NE * labelscalefactor); dot((-3.1290722033604492,1.6485275627856872),linewidth(4pt) + dotstyle); label("F", (-3.4599999148776267,1.5500123341001536), NE * labelscalefactor); dot((-2.540915510808472,-0.3916167709228757),linewidth(4pt) + dotstyle); label("Y1", (-2.895631034367365,-0.59371563495900835), NE * labelscalefactor); dot((1.2420440756126852,0.25617943301950347),linewidth(4pt) + dotstyle); label("Y2", (1.368892872020798,0.21156585952978643), NE * labelscalefactor); dot((-1.8597472783289537,-1.1102372255082722),linewidth(4pt) + dotstyle); label("X1", (-2.2873204110407715,-1.389812648303034), NE * labelscalefactor); dot((0.6309314324152658,-1.1142674499578247),linewidth(4pt) + dotstyle); label("X2", (0.7355337010411478,-1.4367835197012285), NE * labelscalefactor); dot((-1.534317055871152,-2.614102719661712),linewidth(4pt) + dotstyle); label("T", (-1.6518447256937852,-2.966142608466179), NE * labelscalefactor); dot((-1.100065558435597,2.147846782717315),linewidth(4pt) + dotstyle); label("A′", (-1.1663690403467986,2.2613520857925593), NE * labelscalefactor); dot((-1.9993332438068177,0.2683440739836026),linewidth(4pt) + dotstyle); label("B′", (-2.1653980872856628,0.46329399257960063), NE * labelscalefactor); dot((0.02967340111803468,0.7676632939152301),linewidth(4pt) + dotstyle); label("C′", (0.146213368183943,0.7869444493579333), NE * labelscalefactor); dot((-0.6494357175978934,-0.06771866895168613),linewidth(4pt) + dotstyle); label("M′", (-0.5795923595508598,0.113191402121833432), NE * labelscalefactor); dot((-1.7052548975308297,-0.7517280928706784),linewidth(4pt) + dotstyle); label("X′1", (-1.970543730244833,-0.592608830085721), NE * labelscalefactor); dot((0.1862248956797492,-0.4278299908994893),linewidth(4pt) + dotstyle); label("X′2", (0.11924249677577708,-0.23794866377345328), NE * labelscalefactor); dot((-1.1038001536792574,-0.1601330787175626),linewidth(4pt) + dotstyle); label("T′", (-1.4961748497345714,-0.022181692587898226), NE * labelscalefactor); dot((-0.7108738385150224,2.079568181864503),linewidth(4pt) + dotstyle); label("N′", (-0.6269516121834801,2.0264977288015876), NE * labelscalefactor); dot((0.16038274522224572,7.045777859297683),linewidth(4pt) + dotstyle); label("N", (0.10126191583699981,7.137021551662131), NE * labelscalefactor); dot((-1.3380060231053266,0.7915732573833573),linewidth(4pt) + dotstyle); label("H", (-1.454058335367235,0.8678570635525164), NE * labelscalefactor); dot((-0.4292771509426009,0.9764020769151098),linewidth(4pt) + dotstyle); label("M″", (-0.6390681265508164,1.0656434538059418), NE * labelscalefactor); dot((0.03913458790953972,-0.9270105952867296),linewidth(4pt) + dotstyle); label("H'", (0.11025220630638845,-1.0650553866514143), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let T be the A-mixintillinear touch point, and let N be the midpoint of \widehat{BAC}. Let Y_1 and Y_2 be the tangency points of M to the incircle, and let \triangle DEF be the contact triangle. Lastly, denote by H the orthocenter of \triangle DEF. Now invert about the incircle, where the prime of each point represents its image under inversion. Note, A', B', and C' are the midpoints of \overline{EF}, \overline{DF} and \overline{DE}, so (A'B'C') is the nine-point circle of \triangle DEF. M' is the midpoint of \overline{Y_1Y_2}. X_1' is the midpoint of DY_1 as the tangents from X_1 meet the incircle at Y_1 and D. Similarly X_2' is the midpoint of DY_2. N' is defined on (A'B'C') such that \angle A'N'I = 90. This follows as \angle NAI = 90, and under inversion \angle NAI = \angle A'N'I. T' is the point diametrically opposite A', and is the midpoint of HD, due to the fact that N', I and T' are collinear. Now it suffices to show X_1', T', M' and X_2' are concyclic. Taking a homothety at D with ratio 2, followed by a homothety with ratio -1 at M'. Then we wish to show the reflection of H over M' lies on the circumcircle of \triangle DEF. However noting that from the homothety with ratio 2, sending the nine point circle of \triangle DEF to the circumcircle, we must have the reflection of H about M' lie on the circumcircle of \triangle DEF, so we are done.
03.12.2023 02:59
My pay grade only allows me to perform one homothety. Let T be the tangency point and N be the midpoint of arc BAC. It's well-known that T,I,N are collinear. Let \overline{MX_1} and \overline{MX_2} be tangent to the incircle at P and Q respectively. Now invert about the incircle. A,B,C get sent to the midpoints of the intouch triangle, so (ABC) gets sent to its 9-point circle \omega. M gets sent to the midpoint of \overline{PQ}. X_1 and X_2 get sent to the midpoints of \overline{DP} and \overline{DQ} respectively. To figure out where T gets sent, we first note that since \angle IAN=90^\circ, N gets sent to the point N' on \omega such that \angle IN'A'=90^\circ, where A' is the midpoint of \overline{EF}. T then gets sent to the second intersection of \overline{IN'} with \omega, which is the point diametrically opposite A', i.e. the midpoint of \overline{DH} where H is the orthocenter of \triangle DEF. Now use complex numbers with respect to the incircle (i.e. circumcircle of \triangle DEF), letting lowercase letters denote the position of their uppercase points. We wish to show that d+\tfrac{e+f}{2},\tfrac{d+p}{2},\tfrac{d+q}{2},\tfrac{p+q}{2} are concyclic given that \tfrac{p+q}{2} lies on \omega. The condition on \tfrac{p+q}{2} is equivalent to having p+q-(d+e+f) lie on the unit circle, by a homothety at H. Now, d+\tfrac{e+f}{2}, \tfrac{d+p}{2}, \tfrac{d+q}{2}, and \tfrac{p+q}{2} are concyclic iff 2d+e+f,d+p,d+q,p+q are, which is equivalent (upon subtracting d+p+q) to d+e+f-p-q,-q,-p,-d being concyclic. This last statement is true by our condition, so we're done. \blacksquare
01.01.2024 09:24
Let \triangle DEF be the contact triangle with orthocenter H. We invert about the incircle: \triangle ABC is mapped to the midpoints of \triangle DEF. The mixtilinear incircle touch point is mapped to the midpoint of DH, which we denote K. M is mapped to a point on the nine-point circle of \triangle DEF. X_1 and X_2 are mapped to the midpoints of the respective contact chords in \triangle DEF. Hence we need to show that TX_1'X_2'M' is cyclic. Under a homothety with scale factor 2 from D composed with a reflection about M, we find that these four points are mapped onto the incircle. \blacksquare
16.01.2024 21:11
Invert about the incenter with arbitrary radius. Then the problem becomes: Inverted problem wrote: Consider two triangles ABC and A'B'C with the same circumcircle, and orthocenter H. Suppose (BHC) \cap (A'HC') = P and (BHC) \cap (A'HB') = Q. Prove that A'PQ passes through the antipode A_1 of A. In fact we claim that P is the orthocenter of A_1A'C'. Redefine it as such; then we may readily verify that P \in (A'HC'); now a vector calculation (using A + B + C = A' + B' + C') gives us that BPCB' is a parallelogram, so it also lies on (BHC). With this lemma it is now clear that (A'A_1QP) is precisely the reflection of (ABC) across A'A_1 and hence we are done.
28.01.2024 22:45
Let M' \in (ABC) satisfy MM' || BC. Let D be the intersection of BC and the incircle. Clearly, M' is projective in M, and the tangency point X of PQ to the incircle is projective to M by Poncelet's porism. Now let T = PQ \cap BC, then T is projective in X by projecting the incircle to a pencil at the antipode of D, translating to the incenter, then projecting onto line BC. Thus T is projective in M'. When M' = B, C, A', we have that T = B, C, D respectively where A' \in (ABC) satisfies AA' || BC. Note that A'D \cap (ABC) = S, where S is the A-mixtilinear in-touch point. This is because A'S reflects to AS' where AS' is collinear with the A-extouch point by \sqrt{BC} inversion. Thus since three points define a projective map, it follows that \overline{M'TS} is always collinear. Now, let R be the intersection of the radical axis of (MX_1X_2) and (ABC) with BC and let S' = MR \cap (ABC). It follows that R is the antiversion center of the involution swapping X_1 \leftrightarrow X_2, B \leftrightarrow C, R \leftrightarrow \infty_{BC}. Thus by conjugating with projection through M, it follows that there exists an involution on (ABC) swapping P \leftrightarrow Q, B \leftrightarrow C, S' \leftrightarrow M'. It follows that S'TM are collinear, so S = S', finishing.
04.08.2024 01:19
Anantmudgal's first solution is in fact valid but I also don't have a proof. Claim: In a poncelet configuration, the map from A to BC to the incircle is projective as A varies. Proof. Well known'', i3435 finds A \mapsto AI \cap (ABC) \overset{\tau}\mapsto D \mapsto BC where \tau follows by homothety. \blacksquare Define Y_i \coloneq AX_i \cap \Gamma. Then Y_1Y_2 is tangent to the incenter by Poncelet's Porism. Let N = Y_1Y_2 \cap BC. Claim: U \coloneq AN \cap MD lies on \Gamma. Proof. Consider the projective maps M \mapsto MD \cap \Gamma and M \mapsto Y_1Y_2 \mapsto E \mapsto AE \cap \Gamma. It remains to check three cases. M = B, C, T work. \blacksquare Let A_1 be on \Gamma such AA_1BC is an isoceles trapezoid. Then since \measuredangle NDT = \measuredangle AA_1T = \measuredangle AUT = \measuredangle NUT it follows that (NUDT) is cylic. As such, since \measuredangle TY_1X_1 =\measuredangle TY_1M = \measuredangle TUM = \measuredangle TUD = \measuredangle TND it follows that T \in (NUD) and thus T is the Miquel point of X_1X_2Y_1Y_2, which finishes.
17.08.2024 18:16
We invert with respect to the incircle and restate in terms of the intouch triangle as follows: Quote: In \Delta ABC , K and L are two points on (ABC). Let the orthocenter of \Delta ABC be H and the midpoint of AH be T. Prove that T lies on the nine point circle of \Delta AKL. We take a homothety with center A and factor 2. Then it suffices to prove that K , L, H and the point P such that AKPL is a parallelogram are concyclic. Reflecting about the midpoint of KL, it suffices to prove that the image of H lies on (ABC). Let this image be Q. Then Q is also the image of the midpoint of KL with center H and scale factor 2, which famously maps the nine point circle to the circumcircle implying Q lies on (ABC) which finishes.
30.10.2024 22:56
Not that bad for a sweep but anyways.... it was a beautiful problem though can't lie. We will invert about the incircle and using the fact that the mixtilinear incircle, incenter and major arc midpoint of \widehat{BC} are collinear, we can exploit the incenter-orthocenter duality to get this new problem with some reformed labels. Quote: Let \triangle ABC be a triangle with O circumcenter, H orthocenter, \Gamma circumcircle, \Omega nine point circle, \triangle DEF orthic triangle and \triangle MNL medial triangle. Let Z \in \Omega be an arbitary point and U and V be the unique distinct points such that (OZU) and (OZV) are both tangent to \Gamma. Let X_1=(ALON) \cap (OZU) and define X_2 similarly. Prove that if P is the midpoint of \overline{AH}, then X_1, Z, X_2, P are concyclic. See that because of Pole Polar/La Hire shenanigans, we have that \triangle X_1ZX_2 is the medial triangle of \triangle AUV. Hence we have that \measuredangle X_2ZX_1=\measuredangle UAV and \measuredangle X_2PX_1=\measuredangle VHU (by taking a double homothety at A) and so if we prove that the reflection of H over \overline{UV} lies on \Gamma or equivalently proving if T is the unique point on \Gamma such that \measuredangle OZT=90^{\circ} and then \overline{HT} \parallel \overline{OZ}, then we are done. But this is trivial if we take a double homothety at H.
Attachments:

28.12.2024 19:39
Let MX_1 and MX_2 intersect (ABC) again at Y_1 and Y_2, respectively. Let Y_1Y_2 intersect X_1X_2 at P. By Poncelet's Porism, MY_1PX_2 has an incircle. Let T be the Miquel Point of the complete quadrilateral MY_1X_1PX_2Y_2. T is the intersection of the circles (ABC) and (MX_1X_2). It suffices to show that TI passes through the midpoint of arc BC in (ABC) containing A. Let TP, TX_2 intersect the circumcircle again at M_1 and Y_3, respectively. Claim: MM_1\parallel BC. TPX_2Y_2, TY_1Y_3Y_2 are cyclic, so by Reim's Theorem Y_1Y_3\parallel BC. There is a spiral similarity centered at T taking Y_1P to MX_2. Therefore, \measuredangle Y_1TM_1=\measuredangle Y_1TP=\measuredangle MTX_2=\measuredangle MTY_3implies Y_1Y_3\parallel MM_1, verifying the claim. It suffices to show that TI bisects \angle PTM. Indeed, \begin{align*} \measuredangle PTI - \measuredangle ITM &= \measuredangle TPI + \measuredangle PIT - \measuredangle TMI - \measuredangle MIT \\ &= \measuredangle TPY_2 + \measuredangle Y_2PI + \measuredangle PIM + \measuredangle MIT - \measuredangle TMI - \measuredangle MIT \\ &= \measuredangle TY_1M + \measuredangle IPY_1 + \measuredangle PIM - \measuredangle TMI \\ &= \measuredangle PY_1M + \measuredangle TY_1P + \measuredangle IPY_1 + \measuredangle PIM - \measuredangle TMI \\ &= \measuredangle PY_1M + \measuredangle TMY_2 + \measuredangle IPY_1 + \measuredangle PIM + \measuredangle IMT \\ &= \measuredangle PY_1M + \measuredangle TMY_2 + \measuredangle IPY_1 + \measuredangle PIM + \measuredangle IMY_2 \\ &= \measuredangle PY_1M + \measuredangle IPY_1 + \measuredangle PIM + \measuredangle Y_1MP \\ &= 0 \end{align*}