Given a triangle △ABC. Denote its incircle and circumcircle by ω,Ω, respectively. Assume that ω tangents the sides AB,AC at F,E, respectively. Then, let the intersections of line EF and Ω to be P,Q. Let M to be the mid-point of BC. Take a point R on the circumcircle of △MPQ, say Γ, such that MR⊥EF. Prove that the line AR, ω and Γ intersect at one point.
Problem
Source: 2019 Taiwan TST Round 2
Tags: geometry, circumcircle
01.04.2020 11:02
Just a mix of well known lemmas . Here's my solution: Suppose that the perpendicular from D to line EF meets line EF and ω at Z,S respectively. Let N=AS∩ω and X=DN∩EF. Note that N is the D-Queue point of △DEF (since SENF is harmonic), and so X must be the D-Ex point of △DEF. Since X lies on the orthic axis of △DEF, so using the well known fact that the orthic axis is the radical axis of the circumcircle and the tangential circle, we get XP⋅XQ=XE⋅XF=XD⋅XN⇒N∈⊙(DPQ)Taking K=EF∩BC, we have (K,D;E,F)=−1, which means that D,K are inverses in the circle with BC as diameter. This means that KD⋅KM=KB⋅KC=KP⋅KQ⇒M∈⊙(DPQ)Combining the above two observations, we get that N=ω∩Γ, and so it suffices to prove that R,N,S are collinear (since we have N∈AS). Since MR∥DS, so this follows from ∡MRN=∡MDN=∡BDN=∡DSN⇒S∈NR◼ NOTE: This configuration is pretty well known, and has appeared numerous times on various contests. Off the top of my head, this example is the foremost one.
01.04.2020 18:34
Already posted here
14.05.2020 15:43
Surprised that all solutions here involve harmonics or powers. Angle-chase ftw JustPostTaiwanTST wrote: Given a triangle △ABC. Denote its incircle and circumcircle by ω,Ω, respectively. Assume that ω tangents the sides AB,AC at F,E, respectively. Then, let the intersections of line EF and Ω to be P,Q. Let M to be the mid-point of BC. Take a point R on the circumcircle of △MPQ, say Γ, such that MR⊥EF. Prove that the line AR, ω and Γ intersect at one point. WLOG AB<AC. Let I,O be the centres of ω,Ω and MA be the midpoint of arc ^BAC of Ω. Clearly, D∈⊙(MPQ). The centre N of ⊙(MPQ) lies on the perpendicular bisector of ¯DM and the line through O parallel to ¯AI, hence it is the midpoint of ¯IMA. Suppose ⊙(MPQ) meets ω at S≠D. Since ¯IN⊥¯DS, we see that ¯AS is the A-mixtilinear cevian (follows from an angle-chase after noticing ¯IMA is a symmedian in △IBC and ¯IM is parallel to the A-Nagell cevian). Let K≠S be the common point of ¯AS and ⊙(MPQ). It is enough to prove ¯MK∥¯AI. Let α=12∠MND and β=12(∠B−∠C). Now ∠KSM=∠KSD−∠MSD=90∘+β−α so ∠MNK=2(180∘−∠KSM)=180∘−2β+2α hence ∠KMN=β−α since N lies on the perpendicular bisector of ¯MK. Since ∠NMD=90∘−α we conclude that ∠KMD=90∘−β=∠(¯AI,¯BC) proving the claim.
30.05.2020 15:37
Let ⊙(AEF) intersects ⊙(ABC) at H⟹△HBF∼△HCE. Also, HBHC=BFCE=BDDC. Hence, H,K,D are collinear⟹KD⋅KH=KJ⋅KA=LB2. AJDH is concyclic (GD;BC)=−1. By applying Maclaurin's theorem GP⋅GQ=GD⋅GM. Thus, MDPQ is concylic. Applying Radical axis theorem on ⊙(ABC),⊙(MDPQ),⊙(AEF). AH cuts PQ on the radical axis of ⊙(AEF) and ⊙(MDPQ). If ⊙(MDPQ) intersects (I) at ℓ. Then Dℓ,AH,PQ concur. Let ⊙(AHD) intersects (I) at ℓ′. Then Dℓ′,AH,PQ concur cause applying Radical axis theorem on ⊙(AEF),(I),⊙(AHD). So ℓ=ℓ′. So note by Reim's theorem MR∥AJ.
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09.06.2020 17:30
Taiwan TST 2019 Round 2 wrote: Given a triangle △ABC. Denote its incircle and circumcircle by ω,Ω, respectively. Assume that ω tangents the sides AB,AC at F,E, respectively. Then, let the intersections of line EF and Ω to be P,Q. Let M to be the mid-point of BC. Take a point R on the circumcircle of △MPQ, say Γ, such that MR⊥EF. Prove that the line AR, ω and Γ intersect at one point. Let X be the midpoint of ^BAC and X∗ denote it's antipode WRT ΔABC. Let ⊙(DPQ)∩⊙(I)=A′. Let ¯PQ∩¯BC={V}. Then VP⋅VQ=VB⋅VC=VD⋅VM from Ma'laurin. So, {D}∈⊙(MPQ). Then from OGM- (#109) we get that ¯AA′ is the A− Mixtillinear Cevian of ΔABC (★). Let {I} denote the Incenter of ΔABC. So from here (#1,#2) we get that ¯X−R−I−T where {T} is the A− Mixtillinear Incircle touchpoint with ⊙(ABC) and from (★) we get ¯A−A′−T. Let ¯AT∩¯MR=K. Then by Reims KTXX∗ is cyclic. If ¯MR∩⊙(MPQ)=K∗. Then MR⋅MK∗=RP⋅RQ=RX⋅RT⟹K∗∈⊙(XX∗T)⟹K∗≡K. ◼
20.06.2020 01:22
Solution from Twitch Solves ISL: Let D be the foot from I to ¯BC. Claim: Quadrilateral MDPQ is cyclic. Proof. Let X=PQ∩BC. Then (XD;BC)=−1 by Ceva/Menelaus with the Gergonne point. Hence, XD⋅XM=XB⋅XC=XP⋅XQ as needed. ◼ We let G be the intersection of the D-altitude to ¯EF with the incircle. Let line AG meet the incircle again at H. [asy][asy] size(12cm); pair A = dir(130); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair D = foot(I, B, C); pair E = foot(I, C, A); pair F = foot(I, A, B); pair M = midpoint(B--C); pair Z = orthocenter(D, E, F); pair K = foot(D, E, F); pair N = midpoint(E--F); pair G = -Z+2*K; pair H = -G+2*foot(I, A, G); filldraw(A--B--C--cycle, invisible, blue); filldraw(incircle(A, B, C), invisible, blue); pair X = extension(E, F, B, C); pair S = foot(A, I, -A); pair Y = extension(H, D, E, F); filldraw(unitcircle, invisible, blue); pair P = IP(circumcircle(H, D, M), unitcircle); pair Q = OP(circumcircle(H, D, M), unitcircle); filldraw(circumcircle(H, D, M), invisible, orange); filldraw(circumcircle(A, E, F), invisible, deepgreen); draw(E--X--P, blue); draw(X--B, blue); pair Rp = extension(M, foot(M, P, Q), A, H); draw(H--Rp--M, red); draw(S--I--A, deepgreen); draw(A--Y--D, lightred); draw(D--G, lightblue); dot("A", A, dir(A)); dot("B", B, dir(240)); dot("C", C, dir(300)); dot("I", I, dir(I)); dot("D", D, dir(D)); dot("E", E, dir(350)); dot("F", F, dir(F)); dot("M", M, dir(M)); dot("K", K, dir(K)); dot("N", N, dir(N)); dot("G", G, dir(G)); dot("H", H, dir(H)); dot("X", X, dir(X)); dot("S", S, dir(S)); dot("Y", Y, dir(140)); dot("P", P, dir(P)); dot("Q", Q, dir(160)); dot("R′", Rp, dir(Rp)); /* TSQ Source: !size(12cm); A = dir 130 B = dir 210 R240 C = dir 330 R300 I = incenter A B C D = foot I B C E = foot I C A R350 F = foot I A B M = midpoint B--C Z := orthocenter D E F K = foot D E F N = midpoint E--F G = -Z+2*K H = -G+2*foot I A G A--B--C--cycle 0.1 lightcyan / blue incircle A B C 0.1 lightcyan / blue X = extension E F B C S = foot A I -A Y = extension H D E F R140 unitcircle 0.1 palecyan / blue P = IP circumcircle H D M unitcircle Q = OP circumcircle H D M unitcircle R160 circumcircle H D M 0.1 yellow / orange circumcircle A E F 0.1 lightgreen / deepgreen E--X--P blue X--B blue R' = extension M foot M P Q A H H--Rp--M red S--I--A deepgreen A--Y--D lightred D--G lightblue */ [/asy][/asy] Claim: Quadrilateral HDPQ is cyclic. Proof. [Proof HDPQ cyclic] For this proof, we need to introduce several points: Let K be the foot from D to ¯EF. Let N be the midpoint of ¯EF. Let S be the inverse of K with respect to the incircle, which is known to satisfy ∠AIS=90∘. Let Y=¯HD∩¯EF. We have −1=(EF;GH)D=(EF;KY). Since ¯SK bisects ∠FSE, this implies Y also lies on line AS. We can then calculate YH⋅YD=YF⋅YE=YK⋅YN=YS⋅YA=YP⋅YQwhich implies the concyclic condition. ◼ We let R′ be a point on line ¯AGH such that ¯R′M⊥¯PQ. Then ∡HR′M=∡HGD=∡HDB=∡HDMso R′=R and the problem is solved (the concurrence point in the problem is H).
14.09.2020 00:50
why are there 2 threads lol [asy][asy] size(10cm); defaultpen(fontsize(10pt)); pair A = dir(130), B = dir(210), C = dir(330), I = incenter(A,B,C), D = foot(I,B,C), E = foot(I,C,A), F = foot(I,A,B), M = (B+C)/2, P = intersectionpoint(unitcircle,E--E+dir(E--F)*100), Q = intersectionpoint(unitcircle,F--F+dir(F--E)*100), R = intersectionpoint(foot(M,E,F)--foot(M,E,F)+dir(M--foot(M,E,F))*100,circumcircle(P,Q,M)), S = intersectionpoints(incircle(A,B,C),circumcircle(P,Q,M))[0], T = extension(I,foot(I,D,S),A,S), K = extension(A,S,D,foot(D,E,F)), L = dir(90), N = (I+L)/2; draw(A--B--C--A); draw(unitcircle); draw(incircle(A,B,C)); draw(circumcircle(M,P,Q), blue); draw(P--Q^^M--R^^D--K^^I--A--L, heavyred); draw(R--T--L^^T--D, orange); draw(I--D^^M--L, magenta); dot("A", A, dir(140)); dot("B", B, dir(210)); dot("C", C, dir(330)); dot("D", D, dir(270)); dot("E", E, dir(80)); dot("F", F, dir(135)); dot("P", P, dir(200)); dot("Q", Q, dir(20)); dot("R", R, dir(110)); dot("S", S, dir(210)); dot("M", M, dir(270)); dot("T", T, dir(225)); dot("K", K, dir(140)); dot("L", L, dir(90)); dot("I", I, dir(330)); dot("N", N, dir(140)); [/asy][/asy] Claim: D∈Γ, where D is the tangency point of ω on ¯BC. Proof. Define a function f such that f(X)=XB/XC for any point X. By Ratio Lemma f(P)f(Q)=f(¯PQ∩¯BC)=f(¯EF∩¯BC)=f(D)=f(D)f(M)and we may conclude. Claim: Let T be the A-mixtilinear intouch point. If ¯AT meet ω at K and S where AKST are in that order, then S∈Γ as well. Proof. Let L be the midpoint of arc BAC so that T∈¯LI. From the fact that ∠DTI=∠ATI (well-known), we know that ¯IL is the perpendicular bisector of ¯DS. It follows that the midpoint N of ¯IL is the circumcenter of △DMS. But since ¯AL∥¯PQ and ∠LAI=90∘, we have NP=NQ, so N is also the center of Γ which implies the conclusion. Claim: Redefine R=¯AS∩Γ; then ¯MR⊥¯PQ. Proof. By Reim's ¯DK∥¯RM, but it is well-known that ¯DK⊥¯EF, so we are done. ◼
09.05.2022 22:05
Let L midpoint of arc BC, N midpoint of arc BAC in Ω. Let M midpoint of BC, and D=ω∩BC. Let G=AI∩EF. Let H be the projection of D on EF. Let K=NI∩EF, and S=ND∩Ω. Let T=NI∩Ω and X=AT∩ω. Let V=NI∩BC. Let Y∈ω such that DY⊥EF. Let Z=AY∩BC. −1=(B,C;N,L)(S)=(B,C;D,LS∩BC),so LS,EF,BC concur at a certain point W. WD⋅WM=WS⋅WL=WP⋅WQ⟹D∈ΓAlso, NKNI=AGAI=NMNL,⟹MK∥LI⟹M−K−R\thickspacecollinear.Now, TK⋅KN=PK⋅KQ=MK⋅KR⟹TMNR\thickspacecyclic.Now, ∡NTR=∡NMR=∡NLA=∡NTA⟹A−T−R\thickspacecollinear.Also, by △YFHE∼IBDC we get FHHE=BDDC⟹AFHE∼NBDC⟹∡FAH=∡BND⟹A−H−S\thickspacecollinear.By radical axis on (LTI),(BIC),Γ we get that U=LT∩BC satisfies UI⊥LI, so (¯UI)=(UTDI) tangent to LI, ∡DSV=∡DTV=∡DTI=∡DIL=∡NLA=∡NSA=∡DSA⟹A−S−V\thickspacecollinear.where we have used DVTS cyclic, which is trivial using inversion NB=NC. Therefore A−H−V−S collinear. Claim. A−Y−T collinear. Proof. Let X′=AY∩ω. Taking polars wrt ω, Brocard on DEFY gives DE∩FY∈P(H). Pascal on DX′YFEE gives DX′∩FE−A−DE∩FY collinear, and A∈P(H) by La Hire. Hence DX′∩FE∈P(H)⟹H∈P(DX′∩FE). And clearly by La Hire, A∈P(DX′∩FE). Since A−H−V collinear, then V∈P(DX′∩FE)⟹DX′∩FE∈P(V)⟹VX′=VD⟹VI\thickspaceis the perpendicular bisector line of\thickspace¯XD′.And since T,V∈IN, then T∈VI, so ∡VTX′=∡DTV=∡DSV=∡NSA=∡NTA=∡VTATherefore A−X′−T collinear, and so X≡X′, which proves the Claim. Finally, △ZDY and △ZMR are clearly homothetic from Z because DY∥MR, so ZMZR=ZDZY=ZDZD2ZX=ZXZD⟹ZM⋅ZD=ZX⋅ZR,which means Z∈Γ, as desired.
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10.05.2022 00:42
Cute problem, took like 20 mins (yay, strict oly training back). Ok so let I the incenter of △ABC, let D the intouch point of ω with BC, let the perpendicular from D to EF meet ω again at G and hit EF at K, let EF meet BC at S, let J the D-queque point on △DEF, let DJ meet EF at U and let (AEF) meet Ω again at V Claim 1: PJDMQ is cyclic Proof: Clearly −1=(S,D;E,F) so by McLaurins theorem and PoP SD⋅SM=SB⋅SC=SP⋅SQ⟹PDMQcyclicNow since its well known that VK bisects ∠EVF and −1=(U,K;E,F) projecting −1=(A,I;E,F)V=(AV∩EF,K;E,F)⟹AV∩EF∩DJ=UAnd now by PoP UP⋅UQ=UA⋅UV=UE⋅UF=UD⋅UJ⟹PJDQcyclicHence PJMDQ is cyclic. Finishing: Using Claim 1 let JA∩Γ=R′, since its known that J,G,A are colinear by Reim's theorem on ω,(PJMDQ) we get GD∥MR′ which means that R=R′ hence AR,ω,Γ share J as a common point. Thus, we are done
25.05.2022 11:14
In the diagram attached below I have swapped R and N JustPostTaiwanTST wrote: Given a triangle △ABC. Denote its incircle and circumcircle by ω,Ω, respectively. Assume that ω tangents the sides AB,AC at F,E, respectively. Then, let the intersections of line EF and Ω to be P,Q. Let M to be the mid-point of BC. Take a point R on the circumcircle of △MPQ, say Γ, such that MR⊥EF. Prove that the line AR, ω and Γ intersect at one point. Say G∈ω such that GENF is harmonic where N∈ω,DN⊥EF.By the harmonic condition we have A−G−N.Say ASD be the A-Sharkydevil point.So by Incircle Inversion I,ASD,DN∩EF=J are collinear.Now say EF∩BC=X Claim 1: D∈Ω Proof : Since AD,BE,CF concurr,by ceva-menelaus we have,(XD;BC)=−1.So by midpoint lengths, XD.XM=XB.XC=XP.XQ⟹D∈Ω ◼Claim 2: G∈Ω Proof : Say AASD∩EF=H and DG∩EF=H′,we will show that H≡H′ and then prove the claim.Consider the following Cross-ratio chase (H,J;E,F)ASD=(A,I;E,F)=−1=(G,N;E,F)D=(H′,J;E,F)This proves H≡H′.Now by Power of Point, HD.HG=HE.HF=HASD.HA=HP.HQ⟹G∈Ω ◼Finally say GR∩MR=R′ Claim 3: N′≡N and thus A−G−R Proof : Since DR||MN,∠GND=∠GR′M.Thus, ∠GR′M+∠GDM=∠GND+∠GDN+∠MDN=∠GND+∠GDN+∠DGN=180Which means R′∈Ω⟹R′≡R,R−A−G,we get AR,Γ,ω meet at the same point G ◼
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07.09.2024 12:21
We first prove that D and H lie on (MPQ) where the incircle touches BC at D and H is the D queue point of ΔDEF. Let EF∩BC=J then it is well known that (JD;BC)=−1 so by the harmonic midpoint lemma, JB.JC=JD.JM=JP.JQ so D lies on (MPQ). Now let S be the A sharky devil point and AS∩EF=K. It is well known that K is the D expoint in ΔDEF so it is collinear with D and H. Now by power of point- KP.KQ=KS.KA=KF.KE=KH.KD so D,H∈(MPQ). Now let D altitude cuts EF and ω at G and T respectively. It is well known that ¯I−G−S and ¯A−T−H so it suffices to prove that R,T,H are collinear which follows by some angle chasing.
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12.12.2024 13:58
EF∩BC=T Let L,N be the midpoints of arcs BC,BAC respectively and let S be A−sharky devil point. Let D be the tangency point of incircle with BC. Since T,S,N and S,D,L are collinear, we have TP.TQ=TS.TN=TD.TM thus, D,M,P,Q are concyclic. Invert around the incircle. New Problem Statement: ABC is a triangle with circumcenter O. Nine point circle and (BOC) meet at P,Q and M is the intersection of (APQ) and the circle with diameter AO. (APQ) intersects (ABC) at V and (VON) meets (APQ) at W, prove that (MOW) is tangent to the perpendicular bisector of BC. Perform √bc2 inversion and reflect over the angle bisector of ∡CAB. New Problem Statement: ABC is a triangle with circumcenter O and its nine point circle meets (BOC) at P,Q. Let K,L be the midpoints of AC,AB respectively and PQ intersects BC,KL at M,V. S is A−dumpty point and (DSV) meets PQ at W. Prove that (DAS) and (DWM) are tangent to each other. Since the radical axises of (AO),(BOC) and nine point circle are concurrent, we conclude that SO,PQ,KL meet at V which is equavilent to the collinearity of S,O,V. Note that ∡ASV=∡ASO=90. Let l be the tangent to (DAS) at D. If E is the midpoint of AD, then A,E,S,V are concyclic. ∡(WD,l)=∡(SD,l)+∡WDS=∡SAD+∡WVS=∡SVE+∡WVS=∡WVK=∡WMDAs desired.◼