Let ABC be an acute-angled triangle with AB≠AC, circumcentre O and circumcircle Γ. Let the tangents to Γ at B and C meet each other at D, and let the line AO intersect BC at E. Denote the midpoint of BC by M and let AM meet Γ again at N≠A. Finally, let F≠A be a point on Γ such that A,M,E and F are concyclic. Prove that FN bisects the segment MD.
Problem
Source: MEMO 2017 T6
Tags: geometry
25.08.2017 23:37
Let the perpendicular from A to BC meet Γ at H. Let P≡AD∩Γ and let A′≡AO∩Γ. Since ∠EMF=∠EAF=\overarcCF+\overarcA′C2=\overarcCF+\overarcBH2 we get that F, M and H are collinear. On the other hand ∠FEC=∠MAF=\overarcCF+\overarcCN2=\overarcCF+\overarcBP2, thus P, E and F are collinear. ∠BFM=∠CFA′, thus FA′ is the symmedian of ∠BFC in triangle BFC, thus F, A′ and D are collinear. On the other hand ∠A′FH=∠A′AH=∠A′OM, furthermore FOMA′ is cyclic. Notice that if we prove that MFFD=NA′NH we are done by the general angle bisector theorem in triangle MFD. Also MFFD=OA′OD. Thus if we prove that OA′NA′=ODNH we are done. OA′NA′=cos∠NAA′sin2∠NAA′=12sin∠NAA′=12sin∠ADO=AOAA′sin∠ADO=ODsin∠DAOA′Psin∠DAO=ODA′P=ODNH
26.08.2017 00:38
Let G=AD∩Γ, A′=AO∩Γ, F′=DA′∩Γ, P=AF′∩GA′ 1) Prove that GMEA′ is cyclic. So from DG is the G-symmedian of △BGC ⟹ ∡MEA = ∡CAE + ∡ECA = ∡BGA + ∡CGA′ = ∡CGM + ∡CGA′ = ∡MGA′, hence GMEA′ is cyclic. 2) Prove that P∈BC and E∈F′G. By pole and polars' theorem we get AA′∩GF′ and AF′∩GA′ lie in BC, hence P∈BC and E∈F′G. 3) Prove that F′=F Since GMEA′ is cyclic and AF′, ME, GA′ are concurrent in P we get PF′.PA=PA′.PG=PE.PM ⟹ AF′EM is cyclic, hence F′=F. 4) Prove that GMFD is a harmonic quadrilateral. It's so easy prove that OF2=OG2=OA2=OM.OD ⟹ ∡MFG = ∡OAM = ∡MDG, hence GMFD is cyclic and OG and OF are tangent to ⊙(GMDF) and O, M, D are collinear, hence GMFD is a harmonic quadrilateral. Finally, since GMFD is a harmonic quadrilateral we get FG is F-symmedian of △FMD, but ∡MFG = ∡MAE = ∡DFN ⟹ FN is F-median of △FMD, hence FN bisects the segment MD.
26.08.2017 01:12
abeker wrote: Let ABC be an acute-angled triangle with AB≠AC, circumcentre O and circumcircle Γ. Let the tangents to Γ at B and C meet each other at D, and let the line AO intersect BC at E. Denote the midpoint of BC by M and let AM meet Γ again at N≠A. Finally, let F≠A be a point on Γ such that A,M,E and F are concyclic. Prove that FN bisects the segment MD. Let A0 lie on ⊙(ABC) such that ¯AA0⊥¯BC and M0 be symmetric to M in point O. Let Γ denote inversion about circle ⊙(BC) followed by point-symmetry at M. Let K be the midpoint of ¯MD. Notice that AA0MM0 is an isosceles trapezoid, hence it is cyclic; ∡AFM=∡AEM=∡AFA0 so A0 lies on ¯MF; Γ switches each of the pairs {A,N},{A0,F},{M0,K}. Consequently, we see K,N, and F are collinear. ◼
30.11.2017 18:10
hehe here is solution of LBT : by symmetry problem : triangle ABC is in (O). AO cut (O) = X, M is midpoint of BC,XM cut (O)=Y, Tagent of (O) at B and C meet each other at D,AD cut (O)=P. Prove YP bisects MD
30.11.2017 18:15
H is othorcentre of ABC , X,Y,M,H collinear , AH cut BC at G,cut (O) at S. BH ,CH cut AC,AB at M,N. YA,MN,BC concurrent at T (radical ). Y(AGCB)=-1 ==> Y,G,P collinear , A(YSBC)=-1 , Y,S,D collinear. HS//MD , GH=GS... .WED
13.04.2020 09:18
Solved with jclash and nukelauncher. [asy][asy] size(7cm); defaultpen(fontsize(10pt)); pair A,B,C,O,Ap,EE,M,D,NN,F,X,Fp,HA,K,SS; A=dir(120); B=dir(215); C=dir(325); O=(0,0); Ap=2O-A; EE=extension(A,O,B,C); M=(B+C)/2; D=-1/M; NN=2*foot(O,A,M)-A; F=2*foot(O,D,Ap)-Ap; X=(D+M)/2; Fp=reflect(O,D)*F; HA=-B*C/A; K=extension(A,D,X,Fp); SS=foot(A,B,C); draw(X--Fp--D,gray); draw(X--F--D,gray); draw(O--D,Dotted); draw(circumcircle(A,EE,F),dashed); draw(A--M,dashed); draw(circumcircle(A,B,C)); draw(Ap--A--HA); draw(A--B--C--A--D); draw(B--D--C); dot("A",A,NW); dot("B",B,B); dot("C",C,C); dot("O",O,NE); dot("A′",Ap,dir(165)); dot("E",EE,dir(240)); dot("M",M,NE); dot("D",D,S); dot("F",F,E); dot("X",X,SE); dot("F′",Fp,W); dot("HA",HA,NE); dot("K",K,SW); dot("S",SS,NE); [/asy][/asy] Let A′ be the antipode of A, let S be the foot from A to ¯BC, let HA=¯AS∩(ABC), let K=¯AD∩(ABC), let F′=¯DHA∩(ABC). Claim: ¯FF′∥¯BC. Proof. Let T=¯AF∩¯BC. Then TB⋅TC=TA⋅TF=TM⋅TE, so −1=(BC;ET)A=(BC;A′F).But −1=(BC;HAF′), so the claim is true. ◼ Claim: S, K, X collinear. Proof. Let L be the midpoint of ¯AK, let Y be the point so that ABCY is an isosceles trapezoid, and let Z be the reflection of A over ¯BC. Note that K, M, Y collinear, and since BYCZ is a parallelogram, K, M, Z collinear. But by homothety (A,2), we have ¯SL∥¯ZK=¯KM, so △ASL∼△DKM. In particular, AKKD=2ALKD=2ASMD=ASXD,so S, K, X collinear, as claimed. ◼ By reflection over ¯OD, it suffices to prove X, K, F′ collinear. By Brokard's theorem, ¯AHA∩¯KF′ lies on the polar of ¯AK∩¯HAF′=D, or ¯BC. Thus X, K, S, F′ are all collinear, and we are done.
13.04.2020 19:52
I'm surprised nobody complex-bash it because it's fairly easy. First let's make some synthetic observations. Let {T}=AF∩BC and A′ the antipode of A.By the power of point we have TB⋅TC=TA⋅TF=TM⋅TE.So there is a inversion of pole T which swaps {B,C},{M,E},{T,∞}. By the famous fact that inversion preserves cross-ratio we have −1=(B;C;M;∞)=(C;B;E;T)A=(C,B,A′,F) so FBA′C is harmonic which means F,A′,D are collinear. Now we are ready to bash!!! As usual let (ABC) be the unit circle and we denote by x the affixe of X.Also set OD to be the real axis,so b=−c ,d=2b+c=1m and t=d+1d2=d2+12d,where T is the midpoint of MD. We use the following lemma. Lemma:Let P be a point on the unit circle and X be other point.Then the affixe of P′, the second intersection of PX with the unit circle is p−xp−x−1
With the lemma we easily obtain f=a+dad+1,and n=a−m−ma−1=ad−1a−d. Now,we have f−nf−t=a+dad+1−ad−1a−da+dad+1−d2+12d=a2−d2−a2d2+1(ad+1)(a−d)2ad+2d2−ad3−ad−d2−12d(ad+1)=2d⋅a2(1−d2)+1−d2(a−d)(d2−1+ad(1−d2))=2d⋅a2+1(a−d)(ad−1)=2d⋅1a2+1(1a−d)(da−1)=2d⋅¯a2+1¯(a−d)¯(ad−1)=¯2d⋅a2+1(a−d)(ad−1) so f−nf−t∈R which means that FN bisects MD. ◻
26.08.2020 18:45
Let A′ is antipode of A with respect to ⊙(ABC). Assume that FN∩MD=G. Claim 1: Quadrilateral AOND is cyclic Proof: Clearly quadrilateral OBDC is cyclic. From Pop we deduce that: OM⋅MD=BM⋅MC=AM⋅MNThis proves that OBDC is cyclic as desired. Claim 2: Points F,A′,D are collinear Proof: Consider √bc inversion followed by reflection over angle bisector of ∠BAC. The claim transforms into: Quote: Given triangle △ABC with orthocenter H and Humpty point HA. Let M be a intersection of ⊙(ABC) and A symmedian. Assume that altitude from A intersects BC at point D and ⊙(ABC) at point E. Let F be intersection of ME and BC. Prove that quadrilateral AFDHA is cyclic. Clearly −1=(M,A;B,C)E=(F,D;B,C) , therefore F is A Ex - point. Consequently points F,H,HA are collinear. But note that ∠FHAA=∠FDA=90, which implies that AFDHA is cyclic as desired. Now we go back to the main problem. Note that: ∠ADO=∠ODN=∠OAN=∠ONA=∠NFA′=∠NFD=xThis implies that GD is tangent to ⊙(FND), therefore GD2=GN⋅GF. Claim 3: Quadrilateral OMA′F is cyclic Proof: From PoP we deduce that: DA′⋅DF=DB2=DM⋅DOThis implies that OMA′F is cyclic as desired. Claim 4: ∠GMN=∠MFN Sketch of Proof: Note that: ∠MFN=∠MFA′−∠NFD=∠MOA′−x=∠B−∠C−xOn another hand: ∠GMN=∠DAA′=∠A−2∠BAD+∠OAN=∠B−∠C−x Now we deduce that MG is tangent to ⊙(MFN), therefore: GM2=GF⋅GN=GD2This implies that FN bisects MD as desired.
26.06.2021 05:34
Interestingly the first solution that is neither a bash nor involves any projective geometry. Easy to see O,M,D are collinear. Also, AOND is cyclic by radax. Let FE∩Γ=K. By Reim, KN//ME⟹BK=NC⟹AK is a symmedian so AKD is collinear. Let A′ be the A antipode. ∠FKD=180−∠AKF=180−∠AA′F=90+∠FAA′=90+∠FAE=90+∠FME=∠DMF so DKMF is cyclic. Now ∠KFD=∠KMD=∠NMT=∠OMA=90−∠AMB=90−∠ACK=90−∠AA′K=∠KAA′=∠KFA′, which means FA′D collinear. Let T be the intersection of FN with MD. Lastly, ∠TDN=∠MAO=∠NAA′=∠NFA′=∠TFD and ∠MFT=∠MFE+∠KFN=∠NAA′+∠KAN=∠KAA′=∠KFD=∠KMD=∠TMN. Thus, TM2=TN×TF=TD2
07.11.2021 22:45
N is humpty point in △FMD and done
12.01.2022 16:37
The diametrical opposite of A lies at DF,FE,AD and (ABC) intersect at one point.