Let ABC be a triangle. Circle Γ passes through A, meets segments AB and AC again at points D and E respectively, and intersects segment BC at F and G such that F lies between B and G. The tangent to circle BDF at F and the tangent to circle CEG at G meet at point T. Suppose that points A and T are distinct. Prove that line AT is parallel to BC. (Nigeria)
Problem
Source: IMO 2019 SL G1
Tags: geometry, IMO Shortlist
23.09.2020 02:27
We have ∡TGF=∡TGC=∡GEC=∡GEA=∡GFAand similarly ∡TFG=∡FGA. Thus TAGF is an isosceles trapezoid, as needed.
23.09.2020 02:29
Define ω=(ADE), and let T′=FF∩ω. Then 180−∠T′AD=∠T′FD=∠DBF, so ∠T′AD=180−B. This means AT′∥BC. If we let ℓ be the line through A parallel to BC, then this means T′=ℓ∩ω, i.e. FF,ℓ,ω concur. By symmetry, GG,FF,ℓ,ω concur, which means T=FF∩GG∈ℓ, as desired.
23.09.2020 02:51
Consider T′ such that AFGT is a cyclic isosceles trapezoid. We contest that T=T′, so it suffices to prove that ¯T′F and ¯T′G are tangent to (BDF) and (CEG), respectively. Indeed, check that ∠DFT′=∠180−∠DAT′=∠ABC=∠DBF, so ¯T′F is tangent to (BDF). Similarly ¯T′G is tangent to (CEG).
23.09.2020 03:12
Redefine T so AT∥BC and T∈Γ; we show this is the desired point. Indeed, we have ∡TGE=∡TAE=∡GCE∡TFD=∡TAD=∡FBD
23.09.2020 04:05
Similar to post 2, oh well. Let P be a point on ¯TF such that F lies between P and T and Q be a point on TG such that G lies between T and Q. Observe that ∡TGF=∡TGC=∡QGC=∡GEC=∡GEA=∡GFA.Further, ∡TFG=∡PFB=∡FDB=∡FDA=∡FGA.From these 2 equations, one gets that ∡TGA=∡TFA, so ATFG is cyclic. Since we've established that ∡GFA=∡TGF, ATGF is an isosceles trapezoid which implies that AT∥BC.
23.09.2020 04:18
[asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair A = dir(110), T = dir(70), F = dir(225), G = dir(315), D = dir(185), E = dir(340), B = extension(A,D,F,G), C = extension(A,E,F,G); draw(unitcircle); draw(circumcircle(B,D,F)^^circumcircle(C,E,G)); draw(F--T--G, blue); draw(B--A--C); draw(D--F^^E--G); draw(A--T^^B--C, red); dot("A", A, dir(110)); dot("T′", T, dir(70)); dot("B", B, dir(225)); dot("C", C, dir(330)); dot("F", F, dir(270)); dot("G", G, dir(270)); dot("D", D, dir(135)); dot("E", E, dir(45)); [/asy][/asy] Let T′ be the point on Γ such that ¯AT′∥¯BC. I claim that T=T′. Angle chasing gives ∠B=12(^AEG−^DF)=12(^T′DF−^DF)=∠DFT′,so ¯T′F is tangent to (BDF) and similar for ¯T′G, implying the conclusion. ◼
23.09.2020 04:34
Redefine T as the point on (ADFGE) such that AX∥BC. ∠AXF=∠BDF=180−∠XFB⟹XF is tangent to(BDF)∠XAC=∠XGE=∠ACB⟹XG is tangent to(CEG)
23.09.2020 05:59
Redefine T to be the point on ⊙(ADE) such that AT∥BC. For brevity write ∠ATF=∠AEF=∠BDF=θ. So we have ∠ABC+θ+∠DFB=π. We also have ∠DFB+∠DFE+∠DFG−π⟹∠TFD=∠ABC. This means that TF is tangent to ⊙(BDF). Similarly TG is tangent to ⊙(GEC) and we are done . ◼
23.09.2020 08:18
fun So note that ∡GFT=∡BFT=∡BDF=∡ADF,and similarly, ∡TGF=∡GEA.And also note that ∡GEA+∡ADF=∡GAF, and ∡GFT+∡TGF=∡GTF. So T lies on (ADFGE). Now note ∡ATF=∡ADF=∡BDF=∡GFT,which implies parallelism. gg
23.09.2020 08:38
Note that ∠DFT=∠ABG by the tangency condition and ∠DFB=∠BAG by cyclic quadrilaterals, so ∠TFG=∠AGF. Likewise, ∠TGF=∠AFG, so triangles AFG and TGF are congruent and we win.
23.09.2020 08:41
Let the tangent at F to (BDF) intersect (ABC) at a point T′. Now prove that AT' parallel with BC. ∠DBF=∠DFX Since (AT′FD) cyclic we know that ∠DAT′=∠DFT′, Which means that ∠T′AC=∠ACB ,which means that AT′∥BC. Now prove that T′G is tangent to (GEC), or we can prove that ∠T′GE=∠ECG. ∠T′AE=∠ACB′ now since (AT′EG) cyclic, we have ∠T′AE=∠T′GE′, which shows us that AT′∥BC. and tanget to (BDF) at F and to (GEC) at G, Wich shows that T=T′ ◼
23.09.2020 10:55
Let P be a point such that AFGP is Isosceles trapezoid then clearly P∈⊙ADFGE and that AP∥BC. Now ∠DAP=∠A+∠EAP=∠A+∠CAP=∠A+∠ACB=∠A+∠C⟹∠DFP=∠Bwhich shows that PF is tangent to ⊙BDF and similarily PG is tangent ⊙CEG so P≡T which implies the desired result ◼
23.09.2020 12:32
Let the line through A parallel to BC meet Γ at T′. ∡T′FD=∡T′AD=∡T′AB=∡CBA=∡FBDSo, T′F is tangent to (BDF) and similarly T′G is tangent to (CEG) and T′=T
23.09.2020 16:02
My solution posted for storage: Let T′ be the point on Γ such that ¯AT′∥¯BC. Claim 1. T′≡T. Proof. We need to prove that tangent from G to (EGC) and tangent from F to (DBF) meet at T′⟺∠ACB=∠EGT′ and ∠ABC=∠DFT′. Since ¯AT′∥¯BC⟹∠ACB=∠TAC=∠TAE=∠TGE⟹T′G is tangent to (EGC) and similarly T′F is tangent to (DBF)⟹T′≡T.◼ Now, by the definition of T′ and Claim 1. ⟹AT∥BC.◻
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25.09.2020 01:22
The main claim is that T∈(ADFGE). To show this, let H=AE∩FT and note that Pascal's Theorem on TFGAEA gives H,C,A collinear. This is clearly true, so T∈(FGAE) by the converse of Pascal, as desired. To finish, note that ∠DFT=∠ABC, so ∠TAC=180−∠ABC−∠BAC=∠BCA, implying AT∥BC, as desired.
25.09.2020 12:05
Sorry for eye touring, It seems I didn't use latex correctly Let H be the intersection of the circle (ADFGE) and the tangent line of circle (BDF) passing through D. Since ∠DBF=∠FDA=∠FHA, we conclude that AH∥BC. Now we will show that HG is tangent to circle (GEC), this is because ∠HGE=∠HFE=180∘−∠FHE−∠FEH, where ∠FHE=∠FAE=∠EGC, and ∠FEH=∠FGH=∠GHA=∠AFG=∠GEC, so we prove that HG is tangent to circle (GEC), so H=T, and we has proved that AH is parallel to BC, as desired. LaTeX fixed. You were adding extra dollar signs to the beginning of the angle commands. (Click on the LaTeX to see how it's done) ~dj
26.09.2020 16:49
Let S be on Γ such that FS is tangent to (BDF). Then by tangency ∠GBA=∠DFS. Also, ∠DFB=∠BAG by antiparallelity. Hence ∠SFG=∠AGBcyclic=∠ASF⟹AS∥BC. From the symmetry the tangent to (CEG) at G passes through S and hence S≡T, as desired. ◼
29.09.2020 16:42
One of the easiest problems I have seen in recent years...it's even easier than our nation's nationals
30.09.2020 07:13
kvs wrote: The main claim is that T∈(ADFGE). To show this, let H=AE∩FT and note that Pascal's Theorem on TFGAEA gives H,C,A collinear. This is clearly true, so T∈(FGAE) by the converse of Pascal, as desired. To finish, note that ∠DFT=∠ABC, so ∠TAC=180−∠ABC−∠BAC=∠BCA, implying AT∥BC, as desired. I think your solution is wrong...Because converse of pascal logic dosent seem right...as a counter example.....Let P be a point such that P dosent lie on ⊙AFGE ....By Pascals theorem on PFGAEA, we get PF∩AE,C,A are collinear..Then by coverse of Pascals theorem ,P lies on ⊙AFGE (ACCORDING TO YOUR LOGIC)...AND ITS DEFINITELY NOT TRUE
05.03.2024 20:25
Let FT∩Γ=V. Then ∡VAB=∡VAD=∡VFD=∡FBD=∡BCA⟹AV∥BC.Similarly, If GT∩Γ=U, then AU∥BC. Therefore, we get U=V=T, and in turn AT∥BC. ◻
11.03.2024 06:00
Let T′ be the intersection of (ADE) and the line through A parallel to BC. We can say: ∠ABC=180−∠T′AD=∠DFT′,∠ACG=∠T′AE=∠T′GE,which implies that T′F tangent to (BDF) and T′G tangent to (CGF), respectively. Therefore, T=T′, as desired ◼
16.03.2024 21:14
Let the circle passing through A, D, E, F, and G be ω. I claim that T is the unique point on ω such that AT∥BC. I will prove this using a phantom point argument. Let T′ be the unique point on ω such that AT′∥BC. Note that this implies that AT′∥BC⟺AT′∥FG,which means that AT′GF must be a isosceles trapezoid, since AT′FG is cyclic. This implies that ∠T′GF=∠AFG=180−∠AEG=∠GEC,which in turn gives us that ∠T′GE=180−(∠EGC+∠T′GF)=180−(∠EGC+∠GEC)=∠ECG,which by tangent properties, implies that T′G is tangent to (EGC). Similarly, we also have that ∠T′FG=∠AGF=180−∠ADF=∠BDF,which then gives that ∠T′FD=180−(∠DFB+∠T′FG)=180−(∠DFB+∠BDF)=∠DBF,which similarly implies that T′F is tangent to (BDF). Therefore, since T′F is tangent to (BFD) and T′G is tangent to (CGE), this means that T′=T. This implies that AT∥BC, finishing the problem.
03.07.2024 13:00
parmenides51 wrote: Let ABC be a triangle. Circle Γ passes through A, meets segments AB and AC again at points D and E respectively, and intersects segment BC at F and G such that F lies between B and G. The tangent to circle BDF at F and the tangent to circle CEG at G meet at point T. Suppose that points A and T are distinct. Prove that line AT is parallel to BC. (Nigeria) Let T be on (ADE) such that AT//BC then we have: ∠ACB=∠TAC=∠TAE=∠=∠TGE so TG is tangent to (CEG) SIMILAR FOR (BDF)
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13.07.2024 16:54
literal minute solve (i took 2 minutes cuz i dumb) let U be on circle Γ that AT and BC are parallel UAE=UGE by cyclicity, and UAE=GCE by parallel then UGE=ECG so UG is tangent to (GEC) similarly, we can prove UF is tangent to (BDF) so we have that U=T thus, AT is parallel to BC
10.09.2024 21:01
Solution with @Mathematical Arceus Claim I: ⊙(FTAG)⟺∡GTF=∡GAF Proof: ∡GTF=∡GFT+∡TGF=(∡TFD+∡DFB)+(∡TGE+∡EGC)=(∡DBF+∡BFD)+(∡EGC+∡ECG)=∡BDF+∡CEG=∡FGA+∡AFG=∡GAFNow by applying Reim's Theorem on ⊙(DFB) and ⊙(TGFA) we get that AT∥BC.
13.09.2024 15:01
.........
14.09.2024 09:54
OG! [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair A = dir(110), T = dir(70), F = dir(225), G = dir(315), D = dir(185), E = dir(340), B = extension(A,D,F,G), C = extension(A,E,F,G); draw(unitcircle); draw(circumcircle(B,D,F)^^circumcircle(C,E,G)); draw(F--T--G, blue); draw(B--A--C); draw(D--F^^E--G); draw(A--T^^B--C, red); dot("A", A, dir(110)); dot("T", T, dir(70)); dot("B", B, dir(225)); dot("C", C, dir(330)); dot("F", F, dir(270)); dot("G", G, dir(270)); dot("D", D, dir(135)); dot("E", E, dir(45)); [/asy][/asy] (FIGURE COPIED FROM MP8148) Now ∠ECG=∠EGT=∠EAT( The first equation follows from Tangent-Secant Theorem, the second statement follows from the fact that A,T,E and G are cyclic.) Thus, AT∥BC
25.09.2024 09:04
ok fine. FINE. after years of not caring ill finally do it. average nigerian problem. well well well. We claim that T is the reflection of A over the perpendicular bisector of FG. It suffices to prove ∠TFG=∠AGF,∠TGF=∠TFG. We prove the former and the latter follows by symmetry. ∡TFG=−∡TFB=∡FDB=−∡ADF=∡AGF, so we are done.
09.12.2024 12:29
Simple stuff. Let the tangent to ⊙(BDF) at F meet ⊙(ADE) again in T′. Indeed, ∠B=∠T′FD=∠T′ED and ∠DT′E=∠DAE=∠A, so ∠TGE=∠C=∠T′DE=∠T′GE⟹T′≡T. Finally, ∠TAC=∠TAE=∠TDE=∠C=∠ACB, so AT∥BC. ◼
27.12.2024 07:08
Let U be a point on the extension of segment TF such that F lies between T and U. Then ∠TFC=∠BFU=∠BDF=∠ACF, where the second equality follows due to inscribed angles in the circumcircle of △BDF and the third equality follows from the cyclicity of ADFC. Similarly ∠TCF=∠AFC. Thus ATCF is an isosceles trapezoid as desired. ◼
30.12.2024 07:03
180−∠ATG=∠CFA=∠CEG=180−∠TGC=∠TGF so FG∥AT and thus BC∥AT.
31.12.2024 11:46
Tangent to (BDF) at F intersect Γ at T. Claim 1:AT//BC.Proof:AB intersect FT at X.From tangent <BDF=<BFX=<TFG=α.From ATFD is cyclic <ATF=<BDF=α ,and we get <ATF=<TFG=α we proved claim 1. Claim 2:GT tangent to (GEC). Proof: We note that <FTG=β from ATEG is cyclic we get <ATG=<AEG=α+β.Then we get <GEC=180-α-β.From claim1 <TGF=180-α-β.And <(BC,TG)=180-α-β from this claim 2 is proved.We are done.
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01.01.2025 04:52
T can lie on either side of A, but the cases are symmetric so for notation let T be on the side of C. Let O be the center of Γ. We first prove that T is on Γ. This is because the F tangent intersects Γ at T′ such that DAT′=2∠B. The G tangent intersects it such that ET″. But DAT' + ET'' = 2 \angle B + 2 \angle C = DAE thus T' = T''. Note without the angle symbol we are talking about arcs on \Gamma. From here, \angle B = \angle DFT = 180 - \angle DAT which finishes.