Let H be the orthocenter of a given triangle ABC. Let BH and AC meet at a point E, and CH and AB meet at F. Suppose that X is a point on the line BC. Also suppose that the circumcircle of triangle BEX and the line AB intersect again at Y, and the circumcircle of triangle CFX and the line AC intersect again at Z. Show that the circumcircle of triangle AYZ is tangent to the line AH. Proposed by usjl
Problem
Source: 2022 Taiwan TST Round 1 Mock Day 2 P5
Tags: geometry, circumcircle, Taiwan
19.03.2022 10:43
Not sure if I'm doing anything wrong, since the solution looks so simple... [asy][asy] size(10cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = foot(A,B,C); pair E = foot(B,A,C); pair F = foot(C,A,B); pair X = 0.6*B+0.4*C; pair H = orthocenter(A,B,C); path BXE = circumcircle(B,X,E); path CXF = circumcircle(C,X,F); pair P = intersectionpoints(BXE,CXF)[0]; pair Y = intersectionpoints(BXE,A--B)[0]; pair Z = intersectionpoints(CXF,10*A-9*C--10*C-9*A)[0]; draw(A--B--C--A--cycle); draw(BXE,blue); draw(CXF,blue); draw(C--F); draw(B--E); draw(P--X,dashed+red); draw(circumcircle(A,F,E),rgb(0.1,0.6,0.1)); draw(circumcircle(A,Y,Z)); draw(circumcircle(B,C,E),rgb(1.0,0.6,0.6)+dashed); draw(A--Z); draw(A--D); dot("A",A,dir(A)); dot("B",B,dir(B)); dot("C",C,dir(C)); dot("D",D,dir(D)); dot("E",E,dir(E)); dot("F",F,dir(F)); dot("H",H,dir(H)); dot("X",X,dir(X)); dot("Y",Y,dir(Y)); dot("Z",Z,dir(Z)); dot("P",P,dir(P)); [/asy][/asy] We use directed angles modulo 180∘. Let P be the intersection of (BEX) and (CFX). By the radical axis theorem on (BFEC),(BEX),(CFX), we have BE,CF,PX are concurrent, which means that P,H,X are collinear. Next, we have P lies on (AEHF) because ∡FPH=∡FPX=∡FCX=∡FCD=∡FADwhere D is the foot of the perpendicular from A to BC. Lastly, we have A,Y,P,Z are concyclic because ∡ZPY=∡ZPX+∡XPY=∡ZCX+∡XBY=∡ACB+∡CBA=∡CAB=∡ZAY. To finish, ∡PYA=∡PYB=∡PEB=∡PEH=∡PAHwhich implies that (AYZ) is tangent to AH.
19.03.2022 10:58
ACGNmath wrote: Not sure if I'm doing anything wrong, since the solution looks so simple... [asy][asy] size(10cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = foot(A,B,C); pair E = foot(B,A,C); pair F = foot(C,A,B); pair X = 0.6*B+0.4*C; pair H = orthocenter(A,B,C); path BXE = circumcircle(B,X,E); path CXF = circumcircle(C,X,F); pair P = intersectionpoints(BXE,CXF)[0]; pair Y = intersectionpoints(BXE,A--B)[0]; pair Z = intersectionpoints(CXF,10*A-9*C--10*C-9*A)[0]; draw(A--B--C--A--cycle); draw(BXE,blue); draw(CXF,blue); draw(C--F); draw(B--E); draw(P--X,dashed+red); draw(circumcircle(A,F,E),rgb(0.1,0.6,0.1)); draw(circumcircle(A,Y,Z)); draw(circumcircle(B,C,E),rgb(1.0,0.6,0.6)+dashed); draw(A--Z); draw(A--D); dot("A",A,dir(A)); dot("B",B,dir(B)); dot("C",C,dir(C)); dot("D",D,dir(D)); dot("E",E,dir(E)); dot("F",F,dir(F)); dot("H",H,dir(H)); dot("X",X,dir(X)); dot("Y",Y,dir(Y)); dot("Z",Z,dir(Z)); dot("P",P,dir(P)); [/asy][/asy] We use directed angles modulo 180∘. Let P be the intersection of (BEX) and (CFX). By the radical axis theorem on (BFEC),(BEX),(CFX), we have BE,CF,PX are concurrent, which means that P,H,X are collinear. Next, we have P lies on (AEHF) because ∡FPH=∡FPX=∡FCX=∡FCD=∡FADwhere D is the foot of the perpendicular from A to BC. Lastly, we have A,Y,P,Z are concyclic because ∡ZPY=∡ZPX+∡XPY=∡ZCX+∡XBY=∡ACB+∡CBA=∡CAB=∡ZAY. To finish, ∡PYA=∡PYB=∡PEB=∡PEH=∡PAHwhich implies that (AYZ) is tangent to AH. Or this problem is really simple itself Yeah I think the point P you have is crucial, but with that point there are so many ways to finish up the problem.
19.03.2022 11:09
Here's a more braindead solution. Also, the first part of this solution might not be necessary. We begin by showing that the desired conclusion holds when X is the midpoint of ¯BC. Since X is the center of (BFEC) in this case, we know that ¯XY bisects ∠BYE by the trillium theorem, and that, as a consequence, ¯XY is the perpendicular bisector of ¯EF. By symmetry, Z also lies on the perpendicular bisector of ¯EF, and a two-second angle chase finishes (consider the midpoint of ¯EF). Now, it suffices to show that the signed ratio AYAZ is independent of the position of X. To do this, let T≡¯AC∩(BXE)≠E and S≡¯AB∩(CXF)≠F. By power of a point, we can obtain CT=CX⋅aCE, BS=BX⋅aBF, AY=AE⋅(b−CT)c=AE⋅(b−CX⋅aCE)c,and AZ=AF⋅(BS−c)b=AF⋅(BX⋅aBF−c)b.Dividing the two yields AYAZ=AE⋅(b−CX⋅aCE)cAF⋅(BX⋅aBF−c)b=AE⋅BF⋅bAF⋅CE⋅c,where the last equality comes from the substituting BX=a−CX and doing some standard manipulations. Since this expression contains no X, we're done.
19.03.2022 11:19
Great problem! Here's a length bash solution (i.e. PoP, Pythagorean theorems and a bit of trig) Let (BXE) intersect AC at Y′ and let (CXF) intersect AB at Z′. It's easy to see that AZ′AZ=ACAF=ABAE=AY′AY, so YZZ′Y′ is cyclic (probably this could have been proven by radical axis, but I saw it that way). Note that AH is tangent to (AYZ) ⟺ Y′Z′⊥BC. We can prove the last one using BZ′2−CZ′2=BY′2−CY′2. We calculate only the first one and we will see that it's symmetrical expression. I will put only a sketch: using BZ′.BF=BX.BC, Pythagorean theorem for CFZ′ and BF=a.cosβ, CF=a.sinβ we obtain the first expression is a(BX−CX), so the second is −a(CX−BX) and we're done. Edit: this is similar to the above one - it again includes points Y′ and Z′
19.03.2022 14:14
Almost the same as #1 but anyways, Let G=¯HX∩(BEX) then by POP GH⋅HX=BH⋅HE=CH⋅HF so G∈(CFX). ∡YGZ=∡XGZ−∡XGY=∡XCA−∡XBA=∡YAZ⟹G∈(AYZ).∡HGE=∡XGE=∡XBE=∡HAE so G∈⊙(AH) and finally to finish off, note that ∡GZA=∡GZC=∡GXB=∡GEB=∡GEH=∡GAH.
19.03.2022 16:21
Nothing new I guess, Solved with Aayuu By Miquel's theorem, (BEX),(CFX),(AYZ) concur at a point, say P. Then ∠AZY=∠APY=∠EPA−∠EPY=∠APE−∠ABE, so to show that ∠BAH=∠AZY, it suffices to show that ∠APE=∠AFE (or P∈(AEF))since then ∠AZY=∠AFE−∠ABE=C−(90−A)=90−B=∠BAH. Note that by radax, PX goes through H, so ∠EPH=∠EPX=∠EBC=∠EAH so P∈(AEF), so we're done. ◼
07.02.2023 05:55
https://hackmd.io/@sine/TWTST_22_1_M5
08.02.2023 08:37
A common method, We can know that P,X,and H are collinear because FH⋅HC=HE⋅BHThat is, H is on the root axis of the (PXB)and(PXC),we have PH⋅HX=HE⋅HB=AH⋅HDwhich means that P,A,F,H,E lies on a same circle. Consider ∠ZPY=∠ZPX−∠YPX=∠EHC=∠FAC, so A,P,Y,Z lies on a same circle. At last,∠YZA=∠APY=∠APX−∠YPX=∠BAHwhich implies that (AYZ) is tangent to AH.
22.11.2023 15:08
Let K,X=(BXE)∩(CXF). Claim: AYKZ cyclic. Proof: 180∘−∠AZK=∠KZC=∠KXC=∠BYK=∠AYK. Claim: K∈(AFHE). Proof: ∠KFH=∠KFC=∠KXC=180∘−∠KXB=180∘−∠KEB=180∘−∠KEH. Now ∠AZY=∠AKY=∠YKE−∠AKE=180∘−∠YBE−(180∘−∠AFE)=90∘+A−(180∘−C)=90∘−B. Let T be a point on line AH on the opposite side of A from H. Then ∠TAY=∠HAF=90∘−B=∠AZY⟹AT tangent to (AYZ)⟹AH tangent to (AYZ).
22.11.2023 17:52
Let (BEX) intersect ¯AC at P and (CFX) intersect ¯AB at Q. I claim that PQYZ is cyclic. Indeed, ∡PYQ=∡PYB=∡PEB=90∘ and ∡PZQ=∡CFQ=90∘, as desired. Furthermore, we actually have 90∘=∡PEB=∡PXB and likewise 90∘=∡QFC=∡QXC, so ¯PQ is the perpendicular to ¯BC at X. This now implies that ∡BAD=∡BQX=∡YQP=∡YZP=∡YZA,done. ◼ Remark: Actually, the motivation to add P and Q came from trying to lengthbash and prove BFCE=AYAZ. Trying to use power of a point to relate AYAZ to other things naturally motivates the addition of P and Q. Unfortunately a comedy of errors occurred, and instead of noticing that the PoP implies PQYZ cyclic, I thought that it was sufficient to prove that PQYZ was cyclic to finish the problem (mixed up which equations were true and which ones we wanted to prove), and then found the angle chase that actually finishes the problem anyways
06.12.2023 14:42
Let D=(CDX)∩(BEX) Claim 1: D−H−X
Claim 2: D lies on (AEFH)
Claim 3: D lies on (AYZ)
Claim 4: 90∘−∡B=∡ADY
By Claim 4, we're done
10.04.2024 17:53
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(47.630266781564124cm); 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 = -15.468156288163454, xmax = 32.16211049340067, ymin = -10.461960997973392, ymax = 11.838373744345812; /* image dimensions */ pen ffcqcb = rgb(1.,0.7529411764705882,0.796078431372549); pen qqttzz = rgb(0.,0.2,0.6); pen ttffcc = rgb(0.2,1.,0.8); pen cczzff = rgb(0.8,0.6,1.); pen ffzzcc = rgb(1.,0.6,0.8); pen ffwwzz = rgb(1.,0.4,0.6); pen qqzzff = rgb(0.,0.6,1.); pen ccwwff = rgb(0.8,0.4,1.); pen zzwwff = rgb(0.6,0.4,1.); pen ffttcc = rgb(1.,0.2,0.8); pen eqeqeq = rgb(0.8784313725490196,0.8784313725490196,0.8784313725490196); pen ccccff = rgb(0.8,0.8,1.); /* draw figures */ draw(circle((9.5,-0.12), 5.368769111990194), linewidth(2.) + ffcqcb); draw((4.795733333333338,-2.7071909297052112)--(14.09203126922559,-2.901533857481228), linewidth(2.) + qqttzz); draw((14.09203126922559,-2.901533857481228)--(7.940798182004243,5.017370092627035), linewidth(2.) + qqttzz); draw((7.940798182004243,5.017370092627035)--(4.795733333333338,-2.7071909297052112), linewidth(2.) + qqttzz); draw(circle((7.405173710533244,-0.17979309448498995), 3.632756377688577), linewidth(2.) + ttffcc); draw(circle((12.13886161551904,3.8147482347819444), 6.994520486710662), linewidth(2.) + ttffcc); draw(circle((5.577784426835647,5.066769866033755), 2.3635300600433706), linewidth(2.) + cczzff); draw((7.778009697422992,-2.7695366423782812)--(7.940798182004243,5.017370092627035), linewidth(2.) + ffzzcc); draw((4.795733333333338,-2.7071909297052112)--(10.499503092356784,1.7233738914677879), linewidth(2.) + ffzzcc); draw((14.09203126922559,-2.901533857481228)--(6.049786618615247,0.372876284651286), linewidth(2.) + ffwwzz); draw((7.940798182004243,5.017370092627035)--(6.114438150018684,7.368568537968901), linewidth(2.) + qqttzz); draw((6.049786618615247,0.372876284651286)--(10.499503092356784,1.7233738914677879), linewidth(2.) + qqttzz); draw((5.2291087165406065,2.729100172989231)--(7.303265868975786,3.4515336178153673), linewidth(2.) + ffcqcb); draw(circle((9.443882301279462,-2.80436239359322), 4.649164561683396), linewidth(2.) + qqzzff); draw((5.2291087165406065,2.729100172989231)--(6.049786618615247,0.372876284651286), linewidth(2.) + ccwwff); draw((9.906707826570713,-2.8140379513534004)--(5.2291087165406065,2.729100172989231), linewidth(2.) + linetype("2 2") + zzwwff); draw((5.2291087165406065,2.729100172989231)--(10.499503092356784,1.7233738914677879), linewidth(2.) + qqttzz); draw(circle((7.884680483283708,2.333007699033815), 2.6849489112918357), linewidth(2.) + ffttcc); draw((5.2291087165406065,2.729100172989231)--(6.114438150018684,7.368568537968901), linewidth(2.) + eqeqeq); draw((7.303265868975786,3.4515336178153673)--(7.828562784563172,-0.3513546945594052), linewidth(2.) + ccccff); /* dots and labels */ dot((9.5,-0.12),dotstyle); label("A", (9.611912056489656,0.18847898400258614), NE * labelscalefactor); dot((4.795733333333338,-2.7071909297052112),dotstyle); label("B", (4.926967782893184,-2.4038568473874613), NE * labelscalefactor); dot((14.09203126922559,-2.901533857481228),dotstyle); label("C", (14.203157444614199,-2.5912546183313205), NE * labelscalefactor); dot((7.940798182004243,5.017370092627035),dotstyle); label("D", (8.050263965290831,5.341917684958705), NE * labelscalefactor); dot((7.828562784563172,-0.3513546945594052),linewidth(4.pt) + dotstyle); label("E", (7.956565079818903,-0.09261767241320215), NE * labelscalefactor); dot((10.499503092356784,1.7233738914677879),linewidth(4.pt) + dotstyle); label("F", (10.611366834856904,1.9687578079692454), NE * labelscalefactor); dot((6.049786618615247,0.372876284651286),linewidth(4.pt) + dotstyle); label("G", (6.176286255852244,0.6257404495382568), NE * labelscalefactor); dot((7.778009697422992,-2.7695366423782812),linewidth(4.pt) + dotstyle); label("H", (7.89409915617095,-2.5287886946833673), NE * labelscalefactor); dot((9.906707826570713,-2.8140379513534004),dotstyle); label("I", (10.01794056020135,-2.4975557328593907), NE * labelscalefactor); dot((7.303265868975786,3.4515336178153673),linewidth(4.pt) + dotstyle); label("J", (7.425604728811303,3.6865707082879515), NE * labelscalefactor); dot((6.114438150018684,7.368568537968901),linewidth(4.pt) + dotstyle); label("K", (6.238752179500197,7.621923898108988), NE * labelscalefactor); dot((5.2291087165406065,2.729100172989231),linewidth(4.pt) + dotstyle); label("L", (5.364229248428855,2.9682125863364925), NE * labelscalefactor); dot((5.229108716540603,2.7291001729892286),linewidth(4.pt) + dotstyle); label("M", (5.364229248428855,2.9682125863364925), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]
10.04.2024 17:54
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.630266781564124cm); 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 = -15.468156288163454, xmax = 32.16211049340067, ymin = -10.461960997973392, ymax = 11.838373744345812; /* image dimensions */ pen ffcqcb = rgb(1.,0.7529411764705882,0.796078431372549); pen qqttzz = rgb(0.,0.2,0.6); pen ttffcc = rgb(0.2,1.,0.8); pen cczzff = rgb(0.8,0.6,1.); pen ffzzcc = rgb(1.,0.6,0.8); pen ffwwzz = rgb(1.,0.4,0.6); pen qqzzff = rgb(0.,0.6,1.); pen ccwwff = rgb(0.8,0.4,1.); pen zzwwff = rgb(0.6,0.4,1.); pen ffttcc = rgb(1.,0.2,0.8); pen eqeqeq = rgb(0.8784313725490196,0.8784313725490196,0.8784313725490196); pen ccccff = rgb(0.8,0.8,1.); /* draw figures */ draw(circle((9.5,-0.12), 5.368769111990194), linewidth(1.) + ffcqcb); 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