Let ABC be a scalene triangle with circumcircle Ω and incenter I. Ray AI meets ¯BC at D and meets Ω again at M; the circle with diameter ¯DM cuts Ω again at K. Lines MK and BC meet at S, and N is the midpoint of ¯IS. The circumcircles of △KID and △MAN intersect at points L1 and L2. Prove that Ω passes through the midpoint of either ¯IL1 or ¯IL2. Proposed by Evan Chen
Problem
Source: USAMO 2017 P3, Evan Chen
Tags: USA(J)MO, USAMO, 2017 USAMO, geometry, Hi, xtimmyGgettingflamed
20.04.2017 02:00
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 826.0603120827556, xmax = 1079.7812335474573, ymin = 942.4726413207709, ymax = 1121.4713573431247; /* image dimensions */ pen fueaev = rgb(0.9568627450980393,0.9176470588235294,0.8980392156862745); pen zzttqq = rgb(0.6,0.2,0.); pen ffqqtt = rgb(1.,0.,0.2); pen ttqqqq = rgb(0.2,0.,0.); pen qqzzff = rgb(0.,0.6,1.); pen wwccqq = rgb(0.4,0.8,0.); filldraw((947.0656746274594,1105.0213195778317)--(917.2484989135296,1023.067421711791)--(1016.4379799979403,1022.3895892670229)--cycle, fueaev, zzttqq); /* draw figures */ draw((947.0656746274594,1105.0213195778317)--(917.2484989135296,1023.067421711791), zzttqq); draw((917.2484989135296,1023.067421711791)--(1016.4379799979403,1022.3895892670229), zzttqq); draw((1016.4379799979403,1022.3895892670229)--(947.0656746274594,1105.0213195778317), zzttqq); draw(circle((967.0388536639019,1051.3533846178484), 57.26408231772152), ffqqtt); draw(circle((956.8566499375142,1049.562472969398), 28.632041158861085), ttqqqq); draw(circle((919.8235705142946,1054.6141521154225), 27.709034166988225), qqzzff); draw((845.91100825702,1023.5549216479395)--(896.2927272340726,1035.6632414844441)); draw((870.808680629589,1030.8290089453244)--(871.3950548615035,1028.3891541870591)); draw((896.2927272340726,1035.6632414844441)--(946.6744462111252,1047.7715613209486)); draw((921.1903996066416,1042.9373287818291)--(921.7767738385561,1040.4974740235637)); draw((947.0656746274594,1105.0213195778317)--(966.8432394557349,1022.7285054894069)); draw(circle((926.6235921378332,1049.7690770090317), 33.450500170506814), wwccqq); draw((947.0656746274594,1105.0213195778317)--(946.5042589782556,1022.8674962443777)); draw((975.0964736974341,1071.632874192071)--(917.2484989135296,1023.067421711791)); draw((928.625707321889,1054.3382095571908)--(1016.4379799979403,1022.3895892670229)); draw((947.0656746274594,1105.0213195778317)--(845.91100825702,1023.5549216479395)); draw((845.91100825702,1023.5549216479395)--(917.2484989135296,1023.067421711791)); /* dots and labels */ dot((947.0656746274594,1105.0213195778317),dotstyle); label("A", (948.1809314251065,1107.809461571948), NE * labelscalefactor); dot((917.2484989135296,1023.067421711791),dotstyle); label("B", (918.3478120880482,1025.838086944889), NE * labelscalefactor); dot((1016.4379799979403,1022.3895892670229),dotstyle); label("C", (1017.6056670786347,1025.2804585460653), NE * labelscalefactor); dot((946.6744462111252,1047.7715613209486),dotstyle); label("H", (947.9021172256947,1050.6525506925361), NE * labelscalefactor); dot((946.5042589782556,1022.8674962443777),dotstyle); label("D", (947.623303026283,1025.559272745477), NE * labelscalefactor); dot((975.0964736974341,1071.632874192071),dotstyle); label("E", (976.3411655656943,1074.3517576425363), NE * labelscalefactor); dot((928.625707321889,1054.3382095571908),dotstyle); label("F", (929.7791942639303,1057.0652772790068), NE * labelscalefactor); dot((845.91100825702,1023.5549216479395),dotstyle); label("S", (846.9713770386376,1026.3957153437125), NE * labelscalefactor); dot((918.861249095803,1082.3064707607587),dotstyle); label("G", (920.0206972845187,1085.225511419595), NE * labelscalefactor); dot((966.8432394557349,1022.7285054894069),dotstyle); label("M", (967.9767395833414,1025.559272745477), NE * labelscalefactor); dot((946.8700604192923,1076.3964404493902),dotstyle); label("N", (947.9021172256947,1079.0915990325364), NE * labelscalefactor); dot((946.7654547120856,1061.08913862816),dotstyle); label("K", (947.9021172256947,1063.756818064889), NE * labelscalefactor); dot((940.4989904395748,1073.0618507626966),dotstyle); label("T", (941.4893906392242,1075.745828639595), NE * labelscalefactor); dot((896.2927272340726,1035.6632414844441),dotstyle); label("P", (897.4367471321661,1038.3847259184183), NE * labelscalefactor); /* end of picture */ [/asy][/asy] This problem really should be thought of in terms of the excentral triangle, so that's what this diagram is; the original triangle is here DEF, and we also have D→K,K→T,I→H,M→N,N→P because I like these labels . Note that by inversion in (AEHF), (S,K;E,F)=−1, so S is the foot of the external angle bisector, which is line BC. Then we know that (D,T;E,F)ωN=(K,S;E,F)=−1. Now, we want to show that (ABC),(TKH),(NDP) are concurrent. This would imply the result, since a homothety at H with power 1/2 takes (ABC) to the 9 point circle ω. Let G=AS∩(ABC), and I claim G lies on both (TKH) and (NDP). First, note that by power of a point, SA∗SG=SB∗SC=SE∗SF,so G∈(AEFH). Furthermore, G,H,M and the A−antipode on (ABC) are collinear, since ∠AGH=90. The simplest way to show that G,T,K,H are cyclic is by inverting in the circle of diameter BC; the circle (AEFH) is preserved since ME and MF are tangent to it, so G→H. Furthermore, (D,T;E,F)=−1⟹(S,T′,E,F)=−1⟹T′=K.Therefore GHTK are concyclic with MH∗MG=MK∗MT. Now, note that H is the orthocenter of triangle ASM. This is because ∠MGA=90 and ∠ADM=90. Then, the 9-point circle of triangle ASM contains both D and G as feet of altitudes, and N and P as midpoints of AH,SH. Thus, DGNP is cyclic, so G=L1 is the desired point.
20.04.2017 02:00
MAN, IDK
20.04.2017 02:01
whatshisbucket wrote: Me trying to do a geometry problem is like trying to eat soup with a toothpick. Some things never change.
20.04.2017 02:01
My problem. Let W be the midpoint of ¯BC, let X be the point on Ω opposite M. Observe that ¯KD passes through X, and thus lines BC, MK, XA concur at the orthocenter of △DMX, which we call S. Denote by IA the A-excenter of ABC. Next, let E be the foot of the altitude from I to ¯XIA; observe that E lies on the circle centered at M through I, B, C, IA. Then, S is the radical center of ω, Ω, and the circle with diameter ¯IX; hence line SI passes through E; accordingly I is the orthocenter of △XSIA; denote by L the foot from X to ¯SIA. [asy][asy] size(10cm); pair A = dir(160); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair D = extension(A, I, B, C); pair M = circumcenter(B, I, C); pair I_A = 2*M-I; pair X = -M; pair K = foot(M, D, X); pair S = extension(M, K, B, C); pair N = midpoint(I--S); pair W = midpoint(B--C); pair E = foot(I, X, I_A); draw(A--B--C--cycle, orange); draw(unitcircle, orange); draw(CP(M, I), olive); draw(A--I_A, orange); draw(M--X, orange); draw(S--E, olive); draw(X--I_A, olive); draw(B--S, orange); pair L = foot(X, I_A, S); draw(M--S--X--cycle, heavyred+1); draw(X--K, heavyred+1); draw(A--M, heavyred+1); draw(S--W, heavyred+1); draw(S--I_A, orange); draw(circumcircle(K, I, D), palered+1); draw(circumcircle(M, A, N), palered+1); draw(X--L, orange+1); pair T = midpoint(I--L); draw(T--M, orange); dot("A", A, dir(150)); dot("B", B, dir(B)); dot("C", C, dir(-10)); dot("I", I, dir(I)); dot("D", D, dir(D)); dot("M", M, dir(M)); dot("IA", I_A, dir(I_A)); dot("X", X, dir(X)); dot("K", K, dir(K)); dot("S", S, dir(S)); dot("N", N, dir(135)); dot("W", W, dir(-45)); dot("E", E, dir(45)); dot("L", L, dir(L)); dot("T", T, dir(T)); /* Source generated by TSQ */ [/asy][/asy] We claim that this L lies on both the circumcircle of △KID and △MAN. It lies on the circumcircle of △MAN since this circle is the nine-point circle of △XSIA. For the other, note that △MWI∼△MIX, according to which ∠IWM=∠MIX=180∘−∠LIM=180∘−∠MLI, enough to imply that quadrilateral MWIL is cyclic. But lines IL, DK, and WM meet at X, implying the conclusion. All that remains to show is that the midpoint T of ¯IL lies on Ω. But this follows from the fact that ¯TM∥¯LIA⟹∠MTX=90∘, thus the problem is solved. Remark: Some additional facts about this picture: the point T is the contact point of the A-mixtilinear incircle, while the point K is such that ¯AK is an A-symmedian.
20.04.2017 02:01
Since AK is a symmedian in △ABC, we have −1=(AK;BC)=(SD;BC), so S is the foot of the external angle bisector in ∠ABC. Denote IA,IB,IC as the excenters; invert about I swapping A and IA. Since △ABC is the orthic triangle of △IAIBIC, we have the following mappings: K maps to the harmonic conjugate of IA in ⊙(IAIBIC). M maps to the point on ⊙(IAIBIC) with IAM′⊥¯IBIC. D maps to the reflection of IA over IBIC. N maps to the reflection of I over IAY, where Y is the midpoint of ¯IBIC. It is well-known that D′K′ passes through Y by reflection about IBIC, and we claim that Y∈⊙(IAN′M′); this is just because ∠IAN′Y=∠IAIY=π−∠M′IY=π−∠YM′IA. Since Y lies on ⊙(ABC), which is the nine-point circle of ⊙(IAIBIC), inverting back yields the desired conclusion.
20.04.2017 02:02
EulerMacaroni wrote: Sigh I solved it except I thought I didn't and then I submitted something wrong Can sort of relate to you, except for #2.
20.04.2017 02:03
I coord-bashed the whole thing... oops...
20.04.2017 02:04
v_Enhance wrote: My problem. Let W be the midpoint of ¯BC, let X be the point on Ω opposite M. Observe that ¯KD passes through X, and thus lines BC, MK, XA concur at the orthocenter of △DMX, which we call S. Denote by IA the A-excenter of ABC. Next, let E be the foot of the altitude from I to ¯XIA; observe that E lies on the circle centered at M through I, B, C, IA. Then, S is the radical center of ω, Ω, and the circle with diameter ¯IX; hence line SI passes through E; accordingly I is the orthocenter of △XSIA; denote by L the foot from X to ¯SIA. [asy][asy] size(10cm); pair A = dir(160); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair D = extension(A, I, B, C); pair M = circumcenter(B, I, C); pair I_A = 2*M-I; pair X = -M; pair K = foot(M, D, X); pair S = extension(M, K, B, C); pair N = midpoint(I--S); pair W = midpoint(B--C); pair E = foot(I, X, I_A); draw(A--B--C--cycle, orange); draw(unitcircle, orange); draw(CP(M, I), olive); draw(A--I_A, orange); draw(M--X, orange); draw(S--E, olive); draw(X--I_A, olive); draw(B--S, orange); pair L = foot(X, I_A, S); draw(M--S--X--cycle, heavyred+1); draw(X--K, heavyred+1); draw(A--M, heavyred+1); draw(S--W, heavyred+1); draw(S--I_A, orange); draw(circumcircle(K, I, D), palered+1); draw(circumcircle(M, A, N), palered+1); draw(X--L, orange+1); pair T = midpoint(I--L); draw(T--M, orange); dot("A", A, dir(150)); dot("B", B, dir(B)); dot("C", C, dir(-10)); dot("I", I, dir(I)); dot("D", D, dir(D)); dot("M", M, dir(M)); dot("IA", I_A, dir(I_A)); dot("X", X, dir(X)); dot("K", K, dir(K)); dot("S", S, dir(S)); dot("N", N, dir(135)); dot("W", W, dir(-45)); dot("E", E, dir(45)); dot("L", L, dir(L)); dot("T", T, dir(T)); /* Source generated by TSQ */ [/asy][/asy] We claim that this L lies on both the circumcircle of △KID and △MAN. It lies on the circumcircle of △MAN since this circle is the nine-point circle of △XSIA. For the other, note that △MWI∼△WIX, according to which ∠IWM=∠MIX=180∘−∠LIM=180∘−∠MLI, enough to imply that quadrilateral MWIL is cyclic. But lines IL, DK, and WM meet at X, implying the conclusion. All that remains to show is that the midpoint T of ¯IL lies on Ω. But this follows from the fact that ¯TM∥¯LIA⟹∠MTX=90∘, thus the problem is solved. Remark: Some additional facts about this picture: the point T is the contact point of the A-mixtilinear incircle, while the point K is such that ¯AK is an A-symmedian. Can I ask how you are supposed to produce such a diagram on the actual contest?
20.04.2017 02:05
Wait how do you bash this? It doesn't seem very feasible...
20.04.2017 02:09
It's not... took 8 pages and cancelling was such a huge pain.
20.04.2017 02:10
I figured out that AK is the A-symmedian and AS is the A-external bisector. Is that 1 point?
20.04.2017 02:10
Consider an inversion about I mapping each point X to X′. As ∠BAI=∠IBC, we have ∠A′B′I=∠IC′B′, so I is the orthocenter of △A′B′C′. A′I meets (IB′C′) at D′, which must be the reflection of A′ over B′C′, and the circle of diameter D′M′ cuts (A′B′C′) at some point K′. Note that the reflection of K′ across B′C′ lies on (IM′K′) and (IB′C′), so S′ is the reflection of K′ across B′C′. Thus S′ lies on the circle of diameter A′I. Let R′ be the midpoint of B′C′. It is well known that S′ is then the intersection of A′R′, the circle through A′,B′ tangent to B′C′, the circle through A′,C′ tangent to B′C′, circle (IB′C′), and the circle of diameter AI′. It is also the projection of I onto A′R′. (Thus K′ is the corresponding point for symmetric triangle D′B′C′.) Now N′ is the reflection of I across S′, which is the reflection of I across A′R′. We claim that R is the desired point, so that R lies on circles (KID), (MAN), and such that the midpoint of IR lies on Ω. Equivalently, we claim R′ lies on K′D′, R′ lies on (M′A′N′), and the reflection of I over R′ lies on Ω′. The first claim is true, as was discussed earlier. The third claim is also true, since I is the orthocenter of A′B′C′ and R′ is the midpoint of B′C′. To show that the second is true, note that R′M′=R′I, by symmetry across B′C′, and R′I=R′N′, by symmetry across A′R′. So R′ is the center of a circle through M′,I′,N′, and ∠M′R′N′=2(180−∠M′IN′)=2(∠A′IN′)=180−∠M′A′N′so R′ lies on (M′A′N′), as desired.
20.04.2017 02:11
MSTang wrote: I figured out that AK is the A-symmedian and AS is the A-external bisector. Is that 1 point? a bunch of people told me no; i did the same
20.04.2017 02:14
strategos21 wrote: It's not... took 8 pages and cancelling was such a huge pain. Think ur confusing this with jmo
20.04.2017 02:15
Oh right. sorry.
20.04.2017 02:17
joshualee2000 wrote: strategos21 wrote: It's not... took 8 pages and cancelling was such a huge pain. Think ur confusing this with jmo strategos21 wrote: Oh right. sorry. Phew. You scared me for a moment there.
20.04.2017 02:17
Oops. I thought I proved that IA, L, S could not be collinear. Still solved it, but my proof ended up being 4 squished pages long.
20.04.2017 02:20
Can someone answer my question in post #7?
20.04.2017 02:22
jasonhu4 wrote: Can I ask how you are supposed to produce such a diagram on the actual contest? With ruler and compass
05.09.2022 23:13
Let ω1 and ω2 be the circles centered at M and N which have radiis MI and NS=NA, resp. ω1∩ω2={I,P}. Let IA be A-excenter and M′ be M antipode with respect to the circle Ω. Let F=PI∩Ω. It's well known that B,I,C and IA are concyclic, where the circle (BICIA) is centered at M. CLAIM 1: The line AS is the radical axis of ω1 and (KMD)
CLAIM 2: Lines SA and KD intersects at M′
CLAIM 3: P∈(MAN)
CLAIM 4: KMIAP is cyclic
CLAIM 5: P∈(KID):
So we proved that P∈{L1,L2}. For finishing the problem, we use radical axes theorem on circles (KID), ω1 and (KDM) to get F−I−M′, so ∠MFM′=∠MFI=90∘. Since MI=MP we get IF=FP, As desired. ◼
Attachments:

30.11.2022 05:18
Let IA be the A-excenter of ABC, P the midpoint of BC, S′=MP∩Ω, and let Ψ denote the map from √bc inversion. Under Ψ, recall (I,IA) swap, and since line BC and Ω swap, (K,P),(M,D) also swap. So, Ψ(S)=Ψ(↔MK)∩Ψ(↔BC)=(ADP)∩Ω. But then since ∠S′PD=90=∠S′AM=∠S′AD, S′ lies on (ADP), hence Ψ(S)=S′. It follows that ∠SAM=∠MAS′=90, and so S,A,S′ are collinear. Also from this, ASKD is cyclic, and K,D,S′ are collinear. Now, let X be the foot of the perpendicular from I to SIA, and X′=S′I∩SIA. Since N is the midpoint of right triangle △SAI, N is the circumcenter, so NA=NI. Also, from Fact 5, M is the midpoint of IIA, so MN is the I-midline in △SIIA. Then, X is the reflection of I across MN, so ∠NXM=∠NIM=180−∠AIN=∠NAM,implying MXNA is cyclic. Now from properties of inversion, ∠AIS′=∠AΨ(S′)Ψ(I)=∠ASIA, so it follows that AIX′S is cyclic. But this means ∠IX′S=90, hence X=X′. Now, since ASKD is cyclic by angles, and SA,XI,DK concur at S′, the converse of radical axis theorem implies DKXI is cyclic. Hence X lies on both (KID) and (MAN). Finally, denote J as the midpoint of XI, which satisfies 90=∠IJM=∠S′JM, implying that J lies on Ω. The conclusion follows.
30.11.2022 06:13
Random high-powered config-throwing, I guess. Let IAIBIC be the excentral triangle. It suffices to show that one of L1,L2 lies on the circumcircle of IAIBIC, (ABC) is the nine-point circle of this triangle and I is the orthocenter. Let NA be the midpoint of arc BAC, which is also the midpoint of IAIC. Let MA be the midpoint of BC. Since ∠MKNA=90∘, D lies on KNA. By Ratio Lemma, we have that BKKC=BKKC⋅BNANAC=BDCD=BAAC, so K is the intersection of the A-symmedian with (ABC). Applying Radical Axis on (ABC), (MMADK), and (NAADMA). we have that MK, MAD, and ANA concur, i.e. S is the intersection of the A-external angle bisector with BC. It is well-known (say, by Brocard on IBCBIC) that I is the orthocenter of △IANAS. Thus, we have that (MAN) is the nine-point circle of △IANAS. Let Q be the IA-Queue point in △IAIBIC, defined as IAS∩(IAIBIC)∩INA (which is well-known to exist). Using the nine-point circle (MAN), we have that Q lies on (MAN). From here, we claim that Q is the desired point. All we need to do now is show that Q lies on (KID). To this end, consider an inversion at I with power −IIA⋅IA. This inversion swaps BC with (IICI′AIB), where I′A is the reflection of IA over IBIC (we apply Reflecting the Orthocenter implicitly here); in particular, D goes to I′A. Since inversion preserves cross-ratio, K is mapped to the intersection of (IAIBIC) with the IA-symmedian in △IAIBIC, call it K∗. Q goes to NA. It suffices to show that I′A, K∗, and NA are collinear. But this is clear, since K∗ is the I′A-Humpty point in △I′AIBIC: it lies on the "(BHC)" circle of that triangle, and it satisfies K∗IB/K∗IC=I′AIB/I′AIC. Since a Humpty point lies on the corresponding median, we're done.
11.03.2023 18:24
We split the solution into four parts. Part 1: Preliminary observations If P is the midpoint of BC, then √bc inversion and reflection over the A− angle bisector swaps P and K. Hence, AK is the A− symmedian. Then, we have −1=(AK;BC)M=(DS;BC) so AS is the A− external angle bisector. Let X be the midpoint of arc ^BAC so that X lies on AS. Then X lies on KD, since ∠MKD=90∘. Let Y be the A− mixtilinear intouch point. It is well known that X lies on YI, so ∠MYI=90∘. Part 2: Y lies on MN Let MN intersect AS at F and BC at Q. By Menelaus on △DIS, MDMI⋅NINS⋅QSQD=1. But since MI2=MB2=MD⋅MI, degenerate triangles △MDI and △MIA are homothetic at M. Hence, AIIM=IDDM, which rearranges to AIID=MIMD. Then we have QSQD=MIMD=AIID so QI∥AS. It follows that QSFI is a parallelogram, so FI∥BC. But then FI⊥XM and MI⊥FX, so I is the orthocenter of △XFM. But Y is the foot of the altitude from M to XI, so Y must be the foot of the altitude from X to MF, as desired. Now, let Z be the reflection of I over Y. Part 3: Z lies on (MAN) Since △IAS is right, N is the circumcenter of △IAS. Hence, we have ∡MZN=∡NIM=∡NIA=∡IAN=∡MAN as desired. Part 4: Z lies on (KID) Since ∠MYI=90∘, △MIZ is isosceles. Hence, Z lies on (BIC). Then PoP at X gives XI⋅XZ=XB2=XD⋅XK as desired.
11.03.2023 21:27
zacchro wrote: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 826.0603120827556, xmax = 1079.7812335474573, ymin = 942.4726413207709, ymax = 1121.4713573431247; /* image dimensions */ pen fueaev = rgb(0.9568627450980393,0.9176470588235294,0.8980392156862745); pen zzttqq = rgb(0.6,0.2,0.); pen ffqqtt = rgb(1.,0.,0.2); pen ttqqqq = rgb(0.2,0.,0.); pen qqzzff = rgb(0.,0.6,1.); pen wwccqq = rgb(0.4,0.8,0.); filldraw((947.0656746274594,1105.0213195778317)--(917.2484989135296,1023.067421711791)--(1016.4379799979403,1022.3895892670229)--cycle, fueaev, zzttqq); /* draw figures */ draw((947.0656746274594,1105.0213195778317)--(917.2484989135296,1023.067421711791), zzttqq); draw((917.2484989135296,1023.067421711791)--(1016.4379799979403,1022.3895892670229), zzttqq); draw((1016.4379799979403,1022.3895892670229)--(947.0656746274594,1105.0213195778317), zzttqq); draw(circle((967.0388536639019,1051.3533846178484), 57.26408231772152), ffqqtt); draw(circle((956.8566499375142,1049.562472969398), 28.632041158861085), ttqqqq); draw(circle((919.8235705142946,1054.6141521154225), 27.709034166988225), qqzzff); draw((845.91100825702,1023.5549216479395)--(896.2927272340726,1035.6632414844441)); draw((870.808680629589,1030.8290089453244)--(871.3950548615035,1028.3891541870591)); draw((896.2927272340726,1035.6632414844441)--(946.6744462111252,1047.7715613209486)); draw((921.1903996066416,1042.9373287818291)--(921.7767738385561,1040.4974740235637)); draw((947.0656746274594,1105.0213195778317)--(966.8432394557349,1022.7285054894069)); draw(circle((926.6235921378332,1049.7690770090317), 33.450500170506814), wwccqq); draw((947.0656746274594,1105.0213195778317)--(946.5042589782556,1022.8674962443777)); draw((975.0964736974341,1071.632874192071)--(917.2484989135296,1023.067421711791)); draw((928.625707321889,1054.3382095571908)--(1016.4379799979403,1022.3895892670229)); draw((947.0656746274594,1105.0213195778317)--(845.91100825702,1023.5549216479395)); draw((845.91100825702,1023.5549216479395)--(917.2484989135296,1023.067421711791)); /* dots and labels */ dot((947.0656746274594,1105.0213195778317),dotstyle); label("A", (948.1809314251065,1107.809461571948), NE * labelscalefactor); dot((917.2484989135296,1023.067421711791),dotstyle); label("B", (918.3478120880482,1025.838086944889), NE * labelscalefactor); dot((1016.4379799979403,1022.3895892670229),dotstyle); label("C", (1017.6056670786347,1025.2804585460653), NE * labelscalefactor); dot((946.6744462111252,1047.7715613209486),dotstyle); label("H", (947.9021172256947,1050.6525506925361), NE * labelscalefactor); dot((946.5042589782556,1022.8674962443777),dotstyle); label("D", (947.623303026283,1025.559272745477), NE * labelscalefactor); dot((975.0964736974341,1071.632874192071),dotstyle); label("E", (976.3411655656943,1074.3517576425363), NE * labelscalefactor); dot((928.625707321889,1054.3382095571908),dotstyle); label("F", (929.7791942639303,1057.0652772790068), NE * labelscalefactor); dot((845.91100825702,1023.5549216479395),dotstyle); label("S", (846.9713770386376,1026.3957153437125), NE * labelscalefactor); dot((918.861249095803,1082.3064707607587),dotstyle); label("G", (920.0206972845187,1085.225511419595), NE * labelscalefactor); dot((966.8432394557349,1022.7285054894069),dotstyle); label("M", (967.9767395833414,1025.559272745477), NE * labelscalefactor); dot((946.8700604192923,1076.3964404493902),dotstyle); label("N", (947.9021172256947,1079.0915990325364), NE * labelscalefactor); dot((946.7654547120856,1061.08913862816),dotstyle); label("K", (947.9021172256947,1063.756818064889), NE * labelscalefactor); dot((940.4989904395748,1073.0618507626966),dotstyle); label("T", (941.4893906392242,1075.745828639595), NE * labelscalefactor); dot((896.2927272340726,1035.6632414844441),dotstyle); label("P", (897.4367471321661,1038.3847259184183), NE * labelscalefactor); /* end of picture */ [/asy][/asy] This problem really should be thought of in terms of the excentral triangle, so that's what this diagram is; the original triangle is here DEF, and we also have D→K,K→T,I→H,M→N,N→P because I like these labels . Note that by inversion in (AEHF), (S,K;E,F)=−1, so S is the foot of the external angle bisector, which is line BC. Then we know that (D,T;E,F)ωN=(K,S;E,F)=−1. Now, we want to show that (ABC),(TKH),(NDP) are concurrent. This would imply the result, since a homothety at H with power 1/2 takes (ABC) to the 9 point circle ω. Let G=AS∩(ABC), and I claim G lies on both (TKH) and (NDP). First, note that by power of a point, SA∗SG=SB∗SC=SE∗SF,so G∈(AEFH). Furthermore, G,H,M and the A−antipode on (ABC) are collinear, since ∠AGH=90. The simplest way to show that G,T,K,H are cyclic is by inverting in the circle of diameter BC; the circle (AEFH) is preserved since ME and MF are tangent to it, so G→H. Furthermore, (D,T;E,F)=−1⟹(S,T′,E,F)=−1⟹T′=K.Therefore GHTK are concyclic with MH∗MG=MK∗MT. Now, note that H is the orthocenter of triangle ASM. This is because ∠MGA=90 and ∠ADM=90. Then, the 9-point circle of triangle ASM contains both D and G as feet of altitudes, and N and P as midpoints of AH,SH. Thus, DGNP is cyclic, so G=L1 is the desired point. OH NO
02.03.2024 02:20
Let IA be the A-excenter and M′ be the midpoint of major arc ^BAC, and let F be the midpoint of ¯BC. As AM′FD,DFMK,AM′MK are all cyclic along diameter circles, it follows that ¯XA passes through S. Claim. [Characterization of L] There exists i∈{1,2} such that Li=¯M′I∩¯IAS satisfies ∠ILiS=90∘. Proof. First, D is the orthocenter of triangle M′SM as ¯MD⊥¯SM′ and ¯SD⊥¯MM′. Now, I claim that I is the orthocenter of triangle M′SIA. As ¯IAI⊥¯M′S, it suffices to show that AI⋅AIA=SA⋅AM′. This follows because △SAB∼△CAM′ and hence AI⋅AIA=AB⋅AC=AS⋅AM′.Now define L to be the foot of the altitude, which lies on ¯M′I; L lies on the nine-point circle (MAN). Furthermore, ASKD is cyclic, implying that M′I⋅M′L1=M′A⋅M′S=M′D⋅M′K, hence L lies on (KID) and L∈{L1,L2}, as needed. ◼ Now just let L be the point we identified. For E=¯IL1∩Ω, we have ∠IEM=∠IL1IA=90∘, so E is the midpoint of ¯IL by homothety.
17.03.2024 21:33
Let MA be the midpoint of ¯BC, M1 be the midpoint of arc BAC, and let L be the reflection of I over the second intersection of ¯M1I with (ABC). I claim that L is one of the intersection points. First, observe that K,D,M1 are collinear, so the internal bisectors of ∠BAC and ∠BKC concur at D, and hence ABKC is harmonic and ¯AK is a symmedian. Now, (B,C;D,S)M=(B,C;A,K)=−1 (projecting through A also works), so by Ceva-Menelaus S is the foot of the external ∠A-bisector, so S,A,M1 collinear. Let the incircle touch ¯BC at T. We now invert about the incircle and denote images with ∙′, so A,B,C are sent to the midpoints of the intouch triangle and (ABC) is sent to the 9-point circle ω of the intouch triangle. M1 is sent to the point on ω such that ∠IM′1A′=90∘. Then L is sent to the midpoint of I and the second intersection X′ of ¯IM′1 with ω. D and M are sent to the second intersection of ¯IA′ with (IB′C′) and ω respectively. S is sent to the point on (IB′C′) such that ∠IA′S′=90∘, and N′ is the reflection of I over S′. Observe that ∠A′M′X′=90∘, so since (IB′C′) and ω are reflections of each other across ¯B′C′ it follows that X′ is the orthocenter of △TB′C′, so it is well-known that L′ is actually the midpoint of ¯B′C′. Furthermore, D′ is evidently the point on (TB′C′) such that ¯TD′∥¯BC. Finally, since M1,D,K are collinear, K′ is the second intersection of (IM′1D′) with ω. We may now restate the inverted problem with reference triangle TB′C′ as follows. Restated wrote: Let ABC be a triangle with orthocenter H. Let the midpoint of ¯BC be M, the reflection of A over M be A′∈(ABC), and D the point on (ABC) such that ¯AD∥¯BC. Let B′,C′ be the reflections of A over B,C, and let M′ be the midpoint of ¯B′C′, equivalently the reflection of A over M. Note that HBCM′ is cyclic, and let ¯DA′M intersect (HBC) again at X. Let ¯HMA′ intersect (HBC) again at Y, and (A′DY) intersect (HBC) again at K. Finally, let R be the reflection of A′ over ¯AM. Prove that M,K,D are collinear (equivalent to KIDL cyclic), and MM′XR cyclic (equivalent to MANL cyclic). For the first part, instead define K′ as the intersection of segment ¯MD with (HBC), i.e. the intersection of line ¯MD with (HBC) that doesn't lie on the A-altitude (the other intersection does). Also let H′ be the reflection of H over ¯BC. Then it suffices to show that A′DYK′ is cyclic. We have MK′⋅MD=MH′⋅MD=MB⋅MC=MH′⋅MY=MA′⋅MY,as desired. For the second part, we reflect everything over M, defining H′ as before. We then want to show that the reflection H″ of H over \overline{AM} lies on (AMH'). Note that \measuredangle H''MA=\measuredangle AMH=\measuredangle H''A'H,and since \angle HH''A'=\angle HH'A'=90^\circ we have (HH'H''A) cyclic, so \measuredangle H''A'H=\measuredangle H''H'H=\measuredangle H''H'A, hence AMH'H'' is cyclic as desired. \blacksquare Remark: In theory after correctly guessing point L (this can perhaps be motivated by \sqrt{bc} inversion and then extraversion, swapping the incircle and A-excircle? idk I used ggb) it is possible to straight up complex bash everything (all points are directly computable). I do not want to see if it's possible in practice.
27.06.2024 09:12
Speedsolved this for a challenge took me just 18 minutes. Really easy problem since its a very familiar picture. We let L denote the major arc midpoint of BC is \Gamma, and I' = \overline{LI} \cap (BIC) \neq I. Then, it is well known and easy to see that points A , S and L , points L , I and I' and points L , D , K are collinear. We now show that I' lies on all the circles we want. First, it is well known that I' lies on (ASI). Then, we prove the following. Claim : I' is one of L_1 and L_2 i.e I' lies on both circles (AMN) and (IDK). Proof : To see why I' lies on (DIK), we considering the power of L. Noting that ADKS is cyclic due to the right angles, LI\cdot LI' = LA \cdot LS = LD \cdot LKwhich indeed implies that I' lies on (DIK). Further, we know that since M is well known to be the center of (BIC), MI=MI' and thus, \measuredangle ANI' = 2\measuredangle ASL' = 2\measuredangle IAI' = 2\measuredangle MII' = \measuredangle IMI' = \measuredangle AMI'from which it also follows that I' lies on (AMN). Thus, I' is indeed one of L_1 and L_2 as claimed. Then, consider the midpoint N_I of II'. Returning to our observation that \triangle MII' is isosceles, we know that MN_I \perp II'. But this immediately implies that MI_NAL is cyclic and thus, N_I must lie on \Gamma. Thus, the midpoint of one of IL_1 and IL_2 lies on \Gamma as we wanted.
28.06.2024 00:35
Let X be the midpoint of \widehat{BAC}. Claim: X, D, and K are collinear. Proof. Obvious as, \angle DKM = 90 = \angle XKMso the claim follows. \square Claim: X, A and S are collinear. Proof. Note that, \begin{align*} -1 = (AK,BC) \end{align*}from \sqrt{bc} inversion. Then projecting at M we find, \begin{align*} -1 = (DS,BC) \end{align*}and hence S = \overline{XA} \cap \overline{BC}. \square Claim: DKAS is cyclic. Proof. Note that, \angle MKD = 90 = \angle MASso the claim follows. \square Now redefine L = \overline{XI} \cap (BIC). Claim: L lies on (KID). Proof. It is equivalent to show L lies on (AIS) by radical axis, but this is well known. \square Claim: L lies on (MAN). Proof. Note that, \begin{align*} \angle LMA = 180 - 2\angle LIM = 180 - 2\angle ASL = 180 - \angle ANL \end{align*}so the claim follows. However then from properties of mixintillinear touch point T namely M, I, T collinear, the midpoint of NL is T so we're done.
30.06.2024 17:23
The KID has grown into a MAN, thanks to the OTIS walkthrough. Let I_A be the A-excenter of \triangle ABC, and let X be the antipode of M on \Omega. First note that X,A,S are collinear by radical center on (AXD), (MKD), \Omega. Now, we see that I_A I \perp XS since \angle XAM = \angle XAI_A = 90^\circ. Claim 1: SI \perp XI_A and thus I is the orthocenter of \triangle XSI_A. Proof: When we rephrase the problem in terms of the excentral triangle I_A I_B I_C, this claim is just a well-known Humpty point property. This means that (MAN) is the nine-point circle of \triangle XSI_A (since N is the midpoint of IS, M is the midpoint of II_A, and A is the foot from I_A to SX), so (MAN) passes through the foot of the altitude from X to SI_A, call it L. Claim 2: The midpoint of IL lies on \Omega. Proof: Note that \Omega is the circle with diameter XM. Thus the intersection of \Omega with \overline{XLI} is the foot of the altitude from M to IL. Since M is the midpoint of II_A, it follows that MI = ML and that the midpoint of IL lies on \Omega, as claimed. Now we go for the kill; we just need to show that L lies on (KID), and we win. Invert with respect to (BIC); the point K gets sent to S, the point I to itself, and the point D to point A. Moreover, by the previous claim, since MI = ML, we see that L gets sent to itself. Thus it suffices to show that ASLI is cyclic, which follows from \angle SAI = \angle SLI = 90^\circ. Therefore, we have identified a point L that lies on both (MAN) and (KID) and for which the midpoint of IL lies on \Omega, and we are done.
10.08.2024 08:34
Really nice! Let X Y and F be the midpoint of BC, midpoint of arc BAC and (YAI) \cap (BIC) respectively. First, we note that XDKM and YADX is cyclic. So, by radical axes Y-A-S collinear. Furthermore Y-D-K collinear as well because 90 = \measuredangle DKM = \measuredangle YDM. Now, again radical axis on (YAIF), (ABC), (BIFC) gives S-N-I-F collinear. Now, let AI \cap YF = I_A. We notice \measuredangle IFI_A = \measuredangle I_AAY = 90. So II_A is the diameter of (BIC) implying that I_A is the A-excenter. Furthermore, this also implies that I is the orthocenter of \triangle SYI_A implying that (FAN) is the nine point circle of \triangle SYI_A. Let L be the feet of Y to I_AS. Then, notice as D is the orthocenter of \triangle YSM YI \cdot YL = YA \cdot YS = YD \cdot YK So L \in (KID) and also L \in (FAN). Now it suffices to show that M \in (FAN). But, M is just the midpoint of II_A which implies that M lies on (FAN) as needed. Finally to finish let P be the midpoint of IL. We note that M-P-N collinear. Furthermore as \measuredangle YPM = \measuredangle YLI_A = 90 = \measuredangle YAM we get P \in (ABC) as needed.
10.08.2024 14:33
mathtastic wrote: MAN, IDK how does this have more upvotes that i will ever see in my life bro :skull:
04.09.2024 12:39
Seeing the involvement of I and I_a, it is fairly natural to restate in terms of the excentral triangle. Also the required condition can be stated more naturally using homothety centered of the orthocenter of the excentral triangle with factor 2 which I directly state. Restated Problem wrote: \Delta ABC has orthic triangle \Delta DEF and orthocenter H. AH \cap EF=K and AH \cap (DEF)=N. T is on the nine point circle such that DFTE is harmonic. X is the A expoint and midpoint of HX is G. Prove that (ABC),(DNH),(HKT) are concurrent at a point. First notice that:- -1=(XD;BC) \stackrel{A}{=} (XK;FE)\stackrel{M}=(D,MK \cap (DEF);FE)implying \overline{M-K-T} are collinear. Let L be midpoint of EF and Q be the A queue point. Note that ML.MN=ME^2=MF^2 and since it is well known that ME,MF are tangent to (AEF), this implies ME^2=MH.MQ=ML.MN and since TKLN is cyclic, ML.MN=MT.MK=MH.MQ implying HQTK is cyclic. Now it remains to show DNQG is cyclic. We do \sqrt{-HA.HD} inversion Then D \longleftrightarrow A, N \longleftrightarrow H', Q \longleftrightarrow M Where H' is reflection of H over BC. Since AH \perp XM and XH_A \perp AM \implies H is the orthocenter of AXM so reflection of H over BC is also its reflection over XM implying H' lies on (AXM) or that X lies on (AH'M). Inverting back, we get that the A humpty point H_A lies on (DNQ). Now by power of point, since it is easy to see that (AH_ADX) is cyclic with diameter AX:- HN.HD=\frac{HA.HD}{2}=\frac{HH_A.HX}{2}=HH_A.HGwhich implies that G \in (DNH_A) and so (DH_ANQG) is cyclic which is what we needed.
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22.11.2024 10:26
Let V be A-mixtilinear touch point and I' be the reflection of I with respect to V. Claim: P,A,S are collinear. Proof: Since P,S,D,M is an orthogonal system and \measuredangle PAM=90, conclusion follows.\square Claim: MV passes through the midpoint of IS. Proof: \frac{\sin KMV}{\sin VMD}\overset{?}{=} \frac{MI}{MS}=\frac{MB}{MS}=\frac{\sin DSM}{\sin \frac{A}{2}}\iff \frac{\sin KPV}{\sin VPA}\overset{?}{=} \frac{\sin MPK}{\sin \frac{A}{2}}Note that by MI^2=MD.MA, we see MDI\sim MIA. \frac{\sin DPI}{\sin IPA}.\frac{\sin \frac{A}{2}}{\sin MPD}=\frac{ID}{IA}.\frac{PA}{PD}.\frac{\frac{MI}{MP}}{\sin MPD}=\frac{MD}{MI}.\sin MDK.\frac{MI}{MP}.\frac{MP}{MK}=1Which completes the proof.\square Claim: I',M,A,N are concyclic. Proof: Since NI=NS and \measuredangle IAS=90, we have NA=NS=NI=NI'. \measuredangle MAN=\measuredangle IAN=\measuredangle NIA=180-\measuredangle MIN=180-\measuredangle NI'MThus, I'\in (MAN).\square Claim: I',K,D,I are concyclic. Proof: Invert at M with radius MI. I',K,D,I swap with I',S,A,I which lie on the circle with diameter SI hence I'\in (KDI) as desired.\blacksquare
16.01.2025 09:29
This is a very wonderful problem by Evan Chen! Definitely make my day although I lost my MacBook (sad ) W.L.O.G. let the midpoint of IL_2; call it as M_2; lies on (ABC). Define G as the second intersection of KD and (ABC) and I_A as the A-excenter of \triangle{ABC}. It's well known that M is the center of (IBI_AC). Let J be the second intersection of SI with (IBI_AC). By the well-known \sqrt{bc} inversion, it's easy to prove that AK is the A-symmedian of \triangle{ABC}; just note that M_{BC} swaps each other with K since K=(DM)\cap(ABC) goes to K^*=(DM)\cap BC (by the fact that M and D swaps each other, so (DM) is fixed under this inversion). Projecting the harmonic bundle (A,K;B,C) by taking the perspectivity at D to the line BC, we have -1=(A,K;B,C)\overset{D}{=}(M,G;B,C), so G is the midpoint of arc BC containing A. Next, projecting the harmonic bundle (M,G;B,C) by taking the perspectivity at K, we have -1=(M,G;B,C)\overset{K}{=}(S,D;B,C). This means that SA is the external angle bisector of \angle{BAC}, so \angle{SAI}=90^\circ and \overline{S,A,G}. Now by Power of Point, we may have SI\cdot SJ=SB\cdot SC=SA\cdot SG, so AIJG is cyclic, causing \angle{IJG}=90^\circ. But \angle{IJI_A}=90^\circ, so \overline{G,J,I_A}. Now see that I_AI\perp SG and SI\perp GI_A, so I is the orthocenter of \triangle{GSI_A}. Redefine L_2 to be the foot of altitude from G to SI_A, so L_2 lies on (IBI_AC). See that we can derive SK\cdot SM=SB\cdot SC=SL_2\cdot SI_A by Power of Point, so this causes MKL_2I_A to be cyclic. Now see that \angle{L_2ID}=\angle{L_2II_A}=90^\circ-\angle{L_2I_AI} and \angle{L_2KD}=360^\circ-90^\circ-\angle{L_2KM}=270^\circ-(180^\circ-\angle{L_2I_AM})=90^\circ+\angle{L_2I_AI}, so it's easy to conclude that L_2 lies on (KID) then. Next, ASL_2I is trivially cyclic with center N since that \angle{SAI}=\angle{SL_2I}=90^\circ. So we have \angle{ANL_2}=2\angle{ASL_2}=2\angle{ASI_A}=2(90^\circ-\angle{SI_AA})=180^\circ-2\angle{L_2I_AM}=\angle{I_AML_2}, so L_2 also lies on (MAN). Returning back to the original definition, then L_2 as one of the intersection points between (KID) and (MAN) is also the intersection of SI_A and (IBI_AC) other than I_A. For the final act, notice that NI=NL_2 and MI=ML_2, so MINL_2 is a kite, meaning that MN is perpendicular with IL_2 at its midpoint, which is M_2. But since we have \angle{GM_2M}=\angle{IM_2M}=90^\circ, therefore M_2 lies on (ABC), as desired. \blacksquare Note. The solution could be simpler if I just noticed that (MANL_2) is the nine-point circle of \triangle{GSI_A} (holy crap )
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