Let ABC be an acute-angled triangle with circumcircle ω and circumcentre O. Points D≠B and E≠C lie on ω such that BD⊥AC and CE⊥AB. Let CO meet AB at X, and BO meet AC at Y. Prove that the circumcircles of triangles BXD and CYE have an intersection lie on line AO. Ivan Chan Kai Chin, Malaysia
Problem
Source: IMO Shortlist 2023 G5
Tags: geometry, IMO Shortlist, AZE IMO TST
17.07.2024 15:01
Just bash it Set (ABC) as the unit circle then we find: d=−acb , x=c2(a+b)c2+abNow define the point T such that: t=a2+bcb+c , clearly T∈AO So if we prove that B,D,X,T are concyclic we will be done by symmetry and this is indeed true because:
17.07.2024 15:04
Synthetic is also equally short [asy][asy] size(7cm); defaultpen(fontsize(10pt)); import olympiad; import geometry; pair A = (1,3.2); pair B = (0,0); pair C = (4,0); pair O = circumcenter(A,B,C); pair D = 2*foot(O,B,foot(B,A,C)) - B; pair X = extension(C,O,A,B); pair T = 2*foot(circumcenter(B,O,C),A,O)-O; pair P = extension(X,X+B-D,A,O); draw(A--B--C--cycle, linewidth(1)); draw(circle(B,X,D), linewidth(0.7)+dashed); draw(circle(B,O,C), black); draw(A--T, linewidth(0.7)); draw(C--O, linewidth(0.7)); draw(O--X, linewidth(0.7), StickIntervalMarker(1, 1, p=linewidth(0.7), size=0.2cm)); draw(O--P, linewidth(0.7), StickIntervalMarker(1, 1, p=linewidth(0.7), size=0.2cm)); draw(circle(A,B,C), black); draw(O--B, linewidth(0.7), StickIntervalMarker(1, 2, p=linewidth(0.7), size=0.2cm)); draw(O--D, linewidth(0.7), StickIntervalMarker(1, 2, p=linewidth(0.7), size=0.2cm)); dot("A", A, dir(115)); dot("B", B, dir(179)); dot("C", C, dir(0)); dot("O", O, 1.5*dir(-108)); dot("D", D, dir(70)); dot("X", X, dir(154)); dot("T", T, dir(-56)); dot("P", P, dir(165)); [/asy][/asy] Let AO intersect ⊙(BOC) again at T. We claim that T is the concurrency point. Note that since angle bisector of ∠AOC is perpendicular to AC, lines AO, OX, and BD form an isosceles triangle. Combining with OB=OD, it follows that there exists point P on AO such that BXPD is isosceles trapezoid. Clearly, P∈⊙(BXD). Now, notice that ∡BTP=∡BTO=∡BCO=∡OBC=∡ABD=∡XBD=∡BDP,so T∈⊙(BXDP). Analogously, T∈⊙(CYE), so we are done.
17.07.2024 15:04
Let AO∩(BOC)={O,K} and H be the orthocenter. We will show that K lies on both (BXD) and (CYE). Take the inversion centered at A with radius √AB.AC and reflect according to the angle bisector of A. We have K↔H, B↔C, Y↔Y∗, E↔E∗ Since O∗ is the reflection of A with respect to BC, we get Y∗C∥BO and Y∈AB. Also ∠E∗AC=∠BAE=90−∠B Let's show that Y∗,B,H,E∗ are cyclic and the other is similar. Since ACE∗∼AHB, we have CE∗HB=ACAH=sinBcosA=Y∗CY∗B⟹Y∗BBH=Y∗CCE∗Also ∠HBY∗=90+∠A=∠E∗CY hence Y∗BH∼Y∗CE. Thus, ∠Y∗HB=∠Y∗E∗C=∠Y∗E∗B as desired.◼
17.07.2024 15:33
Let N=AO∩(BOC). We show that N is the desired intersection point. [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ pair O = (-2.74,1.36); pair A = (-0.22,4.82); pair B = (-6.49312,-0.69816); pair C = (0.94790,-0.81287); pair X = (-3.58643,1.85871); pair D = (1.52203,0.96367); pair N = (-6.90682,-4.36111); pair P = (-6.42790,3.53287); pair Q = (-2.16161,2.15412); import graph; size(10cm); draw(circle(O, 4.28042), linewidth(1)); draw(A--B, linewidth(1)); draw(B--C, linewidth(1)); draw(C--A, linewidth(1)); draw(circle((-2.80673,-2.96935), 4.32986), linewidth(1)); draw(A--N, linewidth(1)); draw(P--C, linewidth(1)); draw(D--B, linewidth(1)); draw(circle((-1.81924,-3.08088), 5.24619), linewidth(1) + linetype("4 4")); dot("O", O, dir((3, -10))); dot("A", A, dir((8.000, 20.000))); dot("B", B, dir((-10, 0))); dot("C", C, dir((10, 0))); dot("X", X, dir((0, 20))); dot("D", D, dir((15, 15))); dot("N", N, dir((-49.318, -3.889))); dot("P", P, dir((-10, 10))); dot("Q", Q, dir((2, -5)));[/asy][/asy] Let P=CO∩ω, and ℓ be the perpendicular bisector of BD. Notice A,P are symmetric wrt ℓ. Let Q be a point symmetric with X wrt ℓ, which clearly lies on AO. Since ∡QDB=∡DBX=∡CBO=∡ONB,we get that B,X,Q,D,N are concyclic. From this, we conclude N=(BXD)∩(CYE). ◼
17.07.2024 15:33
To solve the problem, we need to solve this subproblem: “Let ABC be a triangle. AO meets (BOC) again at T and OC meets AB at X. (XBT) meets (ABC) again at D. Prove that AC⊥BD cyclic.” Let I be the circumcenter of (TBXD). We must show OI∥AC Claim 1: XOIT is cyclic. Proof: By angle-chase and incenter-excenter lemma, ∠XIT=2∠XDT=2(180−∠XBT)=2∠ABC=∠AOC=∠XOT as desired. ◻ Claim 2: OI∥AC Proof: By angle-chase, ∠COI=∠XTI=90−∠XIC/2=90−∠XDT=90−∠ABC=∠OCA, as desired. ◻. Now, BO=OD and BI=ID⟹OI⊥BD By claim 2 and since OI⊥BD, we have AC⊥BD and we are done. ◻
17.07.2024 16:27
I claim that the intersection point is the second intersection, T, of the line AO with the circle (BOC) (other than O). We will now employ complex numbers with (ABC) as the unit circle. We know that T∈AO, so t=λa, where λ∈R. Moreover, from T∈(BOC), we know that t−bt−c÷o−bo−c∈R⟺c(λa−b)b(λa−c)∈R⟺c(λa−b)b(λa−c)=λb−aλc−aNow, we solve for λ, which is given by c(λa−b)(λc−a)=b(λa−c)(λb−a)ac2λ2−(a2c+bc2)λ+abc=ab2λ2−(a2b+b2c)λ+abca(c−b)(c+b)λ=a2(c−b)+bc(c−b)λ=a2+bca(b+c)Thus, we see that t=a2+bcb+c. Moreover, we know that e=−abˉc. We compute for the point Y, it is the intersection of AC and the line passing through B and its antipodal point. So we see that y=ac(b+(−b))+b2(a+c)ac+b2=b2(a+b)ac+b2 Finally, we show that T,Y,E,C are concyclic: t−et−c÷y−ey−c∈RAfter some tedious computation you'll check that this is true.
17.07.2024 16:31
My proposal
19.07.2024 08:28
Here is a solution with lengths that sheds some light on how one might guess the correct point. Part I: Setup. First note that D and E are reflections over ¯AO. Let S and R be the reflections of B and C over ¯AO; we claim that the desired point of intersection is P=¯BR∩¯CS, which clearly lies on ¯AO. We will show that P∈(CYE), and by symmetry we will also have P∈(BXD). Observe that △BAD∼△BYC and △CAX∼△CEB, so we can write CYAD=BCBD,BXAE=BCCE⟹CYCE=BXBD.Since ∠XBD=∠YCD=90∘−∠A, this implies △BXD∼△CYE. [asy][asy] defaultpen(fontsize(10pt)); size(300); pair A, B, C, H, O, D, E, X, Y, R, S, P; A = dir(115); B = dir(210); C = dir(330); H = orthocenter(A, B, C); O = (0,0); D = 2*foot(H, A, C) - H; E = 2*foot(H, A, B) - H; X = extension(C, O, A, B); Y = extension(B, O, A, C); R = 2*foot(C, A, O) - C; S = 2*foot(B, A, O) - B; P = extension(B, R, C, S); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(B--Y^^C--X^^A--P, grey+dotted); draw(B--X--D--cycle, lightblue+dotted); draw(B--P--S, magenta+dashed); draw(C--Y--E--cycle, lightblue+linewidth(0.9)); draw(B--X--D--cycle, lightblue+linewidth(0.9)); draw(circumcircle(C, Y, E), heavycyan+dashed); draw(circumcircle(B, X, D), heavycyan+dashed); draw(B--P--S, magenta+dashed); dot("A", A, dir(A)); dot("B", B, dir(200)); dot("C", C, dir(0)); dot("D", D, dir(70)); dot("E", E, dir(150)); dot("X", X, dir(150)); dot("Y", Y, dir(60)); dot("R", R, dir(R)); dot("S", S, dir(S)); dot("P", P, dir(P)); dot("O", O, dir(250)); [/asy][/asy] Remark: [Motivation for P] Here's how one might guess the correct construction of P after showing △BXD∼△CYD. The line AO is the perpendicular bisector of ¯DE, so we want the intersection P=(CYE)∩(BXD) to satisfy PD=PE. The similarity gives a nice way to compute PD and PE: we have PDPE=R(BXD)sin∠DBPR(CYE)sin∠ECP=BDCE⋅sin∠DBPsin∠ECP,so we want P to satisfy sin∠DBP/sin∠ECP=CE/BD. This motivates the construction of R and S since CE=DR and BD=ES give this ratio automatically. Part II: Length Calculation. The key claim is the following: Claim: We have the similarity △XAD∼△PCE. Proof. Using the reflections over ¯AO, we have ∠PCE=180∘−∠SCE=180∘−∠BCD=∠XAD,so it suffices to prove that PCAX=CEAD⟺PC⋅AD=AX⋅CE⟺PC⋅AD=AC⋅BE,where we've used △CAX∼△CEB in the last step. But we have AD=AE and △AEB∼△ACP (using ∠AEB=∠ACP=180∘−∠C and ∠BAE=∠PAC=90∘−∠B), which gives AEAC=BEPC⟹PC⋅AE=AC⋅BE⟹PC⋅AD=AC⋅BE,as needed. ◼ Now ∠CPE+∠CYE=∠AXD+∠BXD=180∘, so P∈(CYE). This completes the proof.
20.07.2024 10:49
Let Ψ be the inversion centered at A with radius √AB⋅AC2 and reflect according to the angle bisector of A. And Let M,N is midpoint of AB,AC respectively and P is foot of the altitude from A, H is Orthocenter of ΔABC Then Ψ(B)=N,Ψ(C)=M,Ψ(Y)=AB∩⊙(ANP) and Ψ(E) is on MN and ∡Ψ(E)AP=∡CAB And let T is midpoint of AH, X=MN∩⊙(ANP), F=XΨ(Y)∩AH, K=AP∩MN, Ψ(⊙(CEY))=⊙(MΨ(E)Ψ(Y)) Since nine-point circle and Reim, we have AX∥NP By angle chasing, we have X,F,A,Ψ(E) is cyclic Also, we have M,Ψ(Y),F,Ψ(E) is cyclic And by Reim, we have F,Ψ(E),T,M is cyclic So, Ψ(E),Ψ(Y),M,T is cyclic Similarly Ψ(D),Ψ(X),N,T is cyclic Since Ψ(AO)=AH, we have Ψ(T)(=⊙(BXD)∩⊙(CYE) is on AO
21.07.2024 03:32
21.07.2024 18:08
Very Beautiful! I think same point as ISL 2022 G6 : Let the reflections of BC over AB and AC meet at Z (A is the Z-excenter). Let I be the incenter of ZBC thus Z, I, O, and A are collinear. We show that Z is the desired intersection point. It is sufficient to show that Z lies on (BXD). Let the parallel line to CI through Z meet (CIZ) again at P. As CIZP and CIBD are both isosceles trapezoid, ZPDB is also an isosceles trapezoid (thus cyclic). Notice we also have that P, C, and X are collinear as ∠PCI+∠ICB+∠BCO=∠CIZ+180∘−∠A−∠C=180∘Also PZBX is cyclic as ∠XPZ=∠XCI=∠B=180∘−∠XBZ
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22.07.2024 04:39
Let AO intersect (BOC) again at S. We will show that both (BXD) and (CYE) passes through S. Let S′ be the reflection of S over ℓ. Then BSS′D is an isosceles trapezoid, and also S′ lies on line OX as ℓ is the external angle bisector of ∠AOC. By symmetry, we see that △BOS≅△DOS′. Therefore, ∡XBD=∡OBC=∡BCO=∡BSO=∡OS′D=∡XS′D,which shows that XBS′D is cyclic. Therefore, combining the two, we see that BXDS is cyclic. Similarly, CYES is also cyclic, so we are done. Some remarks: Personally, this is the most enjoyable problem in the G shortlist this year, barring G8 which I couldn't solve last year. Somehow it reminds me a lot of 2013 ISL G2.This is also the fastest I've ever solved a G5 (<10 mins) because I managed to guess the point S right away.
23.07.2024 19:14
Similarity Let Z=AO∩(OBC), BD∩AZ=P and CE∩AZ=Q. (1) BZP∼XCB∼ACE because ∠ACE=∠OCB=∠OZB and ∠AEC=∠ABC=∠OBC+∠OBA=∠PBA+∠BAP=∠BPZ. (2) ABC∼APD because ∠OAD=90−∠ABD=∠BAC and ∠ADB=∠ACB. (3) AE=AD because E and D are reflections of H over AB and AC. Claim: ZBX∼ZPD Proof. ∠ZBX=∠ABC+∠ZBC=∠B+∠ZOC=180−∠B=180−∠BPZ=∠ZPD. Lastly, BXPD3=BXEA⋅ADPD1,2=CXCA⋅ACBC=CXBC1=ZBZPwhich is sufficient. ◻ The claim gives ∠BXZ=∠BDZ which shows Z∈(BXD). By symmetry, Z∈(CYE) so we are done.
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24.07.2024 20:35
Angle Chase: Let T=AO∩(BOC). We will show that T is our desired point. It suffices to prove that BXDT is cyclic as the other part can be proved similarly. First, let C′ be the antipode of C in ω, and Q=DC′∩AO. Claim: AQXC′ is cyclic. Proof: ∠XC′Q=∠CC′D=∠CBD=∠ABO=∠BAO=∠XAQ. ◼ Claim: BXQD is cyclic. Proof: Since, AQXC′ is cyclic ∠AXQ=AC′Q=∠AC′D=∠ABD=∠CBO=∠OCB=∠C′CB=∠C′DB=∠QDB.◼ To finish the problem, observe that: ∠BTQ=∠BTO=∠BCO=∠BCC′=∠BDC′=∠BDQ. Therefore, making BXQDT cyclic. Hence, we are done. Nice Problem
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28.07.2024 03:34
I could not figure out a nice way to construct the point we wanted, so I came up with something rather different. Let H be the orthocenter of ABC and let AH meet ω at P. Let O1 and O2 be the circumcenters of CEY and BDX respectively. Note that O1E=O1Y, OE=OA and ∠EO1Y=2∠ECY=2∠EOA.Thus E is the center of spiral similarity that sends O1Y to OA. Hence it also sends O1O to YA. Similarly D sends O2O to XA. Now notice that O1OYA=EOEA=DODA=OO2AX.Furthermore ∠O1OO2=∠YAX, thus OO1O2∼AYX. Moreover, note that line AO divides ∠O1OO2 in the same way line AP divides ∠YAX. Thus we might consider T in AO, with O between A and T, such that quadrilaterals OO1TO2 and AYPX are similar. Now note that XPO2T=AXOO2=XDOO2.Since XP=XD we thus obtain O2T=OO2 which means T∈(BDX). Similarly we get that T∈(CEY).
28.07.2024 09:38
We claim that the desired point of concurrency is the second intersection of AO with (BOC). Call this point P. We shall show that PYCE is cyclic, from which the result follows by symmetry. Since there are many circles passing through C, let us invert with radius √CA⋅CB and reflect about the ∠C-angle bisector. Then we have: A and B swap places, E is sent to the intersection of CO with AB, O is sent to the reflection of C across AB, and Y is the intersection of (B∗O∗C) with A∗C∗. The task is therefore to show that B∗O∗, Y∗E∗ and (CA∗O∗) are concurrent. Relabelling points for convenience, we have the following equivalent problem: ISL 2023/G5, inverted wrote: Let △ABC have circumcenter O, and suppose A′ is the reflection of A across BC. Let AB intersect (ACA′) at D and A′C intersect (ABA′) at E. Show that DE, BC and AO are concurrent. [asy][asy] size(8cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair AA = 2*foot(A,B,C)-A; path ABAA = circumcircle(A,B,AA); path ACAA = circumcircle(A,C,AA); pair D = intersectionpoints(A--B+10*(B-A),ACAA)[1]; pair E = intersectionpoints(AA--C+10*(C-AA),ABAA)[1]; pair O = circumcenter(A,B,C); draw(A--B--C--A--cycle); draw(B--AA--C); draw(B--D--E--C); draw(ABAA); draw(ACAA); pair P = extension(A,O,B,C); draw(A--P); dot("A",A,dir(A)); dot("B",B,dir(B)); dot("C",C,dir(C)); dot("A′",AA,dir(AA)); dot("D",D,dir(D)); dot("E",E,dir(E)); dot("P",P,dir(P)); dot("O",O,dir(45)); [/asy][/asy] It would be great if someone could provide a synthetic solution, it doesn't seem too hard to find but I can't seem to do it. Anyway, here's a bary bash. Set △ABC to be the reference triangle, and denote SA=−a2+b2+c22=bccosA and likewise SB and SC for convenience. Note that SA+SB=c2, as well as cyclic variants. Let AO intersect BC at P. First, we have O=(a2SA:b2SB:c2SC) so P=(0:b2SB:c2SC). Next, the foot from A to BC is (0:bcosC:ccosB)=(0:SC:SB)=(0,SCa2,SBa2). Therefore, A′=2(0,SCa2,SBa2)−(1,0,0)=(−1,2SCa2,2SBa2)=(−a2:2SC:2SB).Now we find the equation for (ABA′). The generic equation is −a2yz−b2zx−c2xy+(x+y+z)(ux+vy+wz)=0. Substituting in A=(1,0,0), we have u=0. Similarly, B=(0,1,0) gives v=0. To find w, we substitute A′ to get −a2⋅4SBSC+b2⋅2a2SB+c2⋅2a2SC+a2⋅w⋅2SB=0.Cancelling a2 and simplifying, we have w⋅SB=2SBSC−b2SB−c2SC=2SBSC−(SA+SC)SB−(SA+SB)SC=−SA(SB+SC)=−a2SA.Therefore, w=−a2SASB.Similarly, the equation for (ACA′) is −a2yz−b2zx−c2xy+(x+y+z)(a2SASC)y=0. Let D=(k,1−k,0). Then −c2k(1−k)−(1−k)(a2SASC)=0⇒k=−a2SAc2SC.Therefore, D=(−a2SA:a2SA+c2SC:0). The strategy now is to intersect DP and A′C and show that this intersection lies on (ABA′). The line DP is given by 0=|xyz−a2SAa2SA+c2SC00b2SBc2SC|=c2SC(a2SA+c2SC)x+a2c2SASCy−a2b2SASBz.The line A′C is given by 0=|xyz−a22SC2SB001|=2SCx+a2y.The intersection of these two lines is (a4b2SASB:−2a2b2SASBSC:a2c2(a2SASC+c2S2C−2SAS2C))=(a2b2SASB:−2b2SASBSC:c2SC(a2SA+c2SC−2SASC))=(a2b2SA:−2b2SASC:c2SC(SA+SC))=(a2SA:−2SASC:c2SC).Now we verify that this point lies on (ABA′), which has equation −a2yz−b2zx−c2xy−a2SASB(x+y+z)z=0. 2a2c2SAS2C−a2b2c2SASC+2a2c2S2ASC−a2SASB⋅SB(SA+SC)⋅c2SC=2a2c2SASC(SA+SC)−a2c2SASC(SA+SC)−a2c2SASC(SA+SC)=0by using the fact that SA+SC=b2.
05.08.2024 02:15
I'll show that the desired point of concurrency is the second intersection of AO with circle BOC. Let's call this point P′. I'll show that P′CYE is cyclic. It's equivalent to ∠P′YE=∠CYO which is the same as ∠CYP′=∠BYE. Now let's define a bunch of useful points: R is the second intersection of BO with circle ABC G is orthogonal projection of P′ onto AC T is orthogonal projection of A onto BC S is the second intersection of AP′ with circle GYP′ K is point on segment AB such that KY||AR See that △AGP′~△ATB ~ △REB Now I will show that point S in △AGP′ is the same point as Y in △REB, by proving: RY/YB=AS/SP′. See that RY/YB=AK/KB (because KY||AR) It's left to show that KS||BP′, but fortunately it's easy by simple angle chasing, because ∠YKA=90 and AYSK is cyclic. ∠AP′B=∠OCB=90−∠BAC and ∠ASK=∠AYK=90−∠BAC Then ∠GSP′=∠BYE but ∠GSP′=∠GYP′ That finishes the proof. I enjoyed this problem:)
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15.08.2024 02:39
Define K=¯Y′Z∩¯YZ′; observe that there is a spiral similarity at A sending Z′,B→C,Y′ so Y′,Z′ are √bc inverses. As a consequence ¯AY′ and ¯AZ′ are isogonal in ∠BAC. Claim 1: K lies on the A-altitude. Proof. Apply DDIT to A and YY′ZZ′, to obtain the involution A(BC;Y′Z′;OK), which, based on the first two pairs, is isogonality at ∠BAC, so ¯AK and ¯AO are isogonal as well.◻ Claim 2: ¯Y′Z∩¯YZ′ lies on the radical axis of Ω and (YY′ZZ′). Proof. Said radical axis is also the polar of O wrt (YY′ZZ′) because OA2=OY⋅OY′=OZ⋅OZ′ by design. The claim then reduces to Brocard on that quadrilateral.◻ By PoP at the concurrency point in claim 1, if T is the intersection of the A-altitude with (ABC) KY⋅KZ′=Pow(K,Ω)=KA⋅KT, and we win.
28.08.2024 11:05
I will give a second solution using Ptolemy-Sinus Lemma. Solved with my bros from ENKA. Let Z=AO∩(OBC). I will prove that Z∈(BXD) which is sufficient from symmetry. Z∈(BXD)⟺BDsinB=BXsin(270−2B−C)+BZsin(90−A)⟺BDsinB=BXcos(B−A)+BZcosASinus Theorem in △BCX⟺BDsinB=BCcosA+BZcosASinus Theorem in △BOZ and △ABC⟺2Rcos(C−A)sinB=Rsin2A+Rsin2C⟺2sin(C+A)cos(C−A)=sin2A+sin2Cwhich is true.
01.09.2024 22:42
Let K be the second intersection of (BOC) with AO. It now suffices to show that K∈(BXD)∩(CYE). We will show K∈(CYE); the proof for K∈(BXD) is similar. Let P be the intersection of AO with the line through Y parallel to CE. Claim: EPYC is an isosceles trapezoid, and thus cyclic. Proof: Let Q be the point diametrically opposite A on ω. Then PY∥CE∥BQ. Since OB=OQ and △OBQ∼△OYP, OY=OP, so a line through O perpendicular to PY is the perpendicular bisector of both PY and CE. ◻ Claim: PYCK is cyclic. Proof: We have ∠PYC=180∘−∠YCE=∠BAC+90∘, and ∠PKC=∠OKC=∠OBC=90∘−∠BAC, giving ∠PYC+∠PKC=180∘. Combining the two claims gives EPYCK cyclic, so K∈(CYE).
17.09.2024 18:04
MarkBcc168 wrote: Synthetic is also equally short [asy][asy] size(7cm); defaultpen(fontsize(10pt)); import olympiad; import geometry; pair A = (1,3.2); pair B = (0,0); pair C = (4,0); pair O = circumcenter(A,B,C); pair D = 2*foot(O,B,foot(B,A,C)) - B; pair X = extension(C,O,A,B); pair T = 2*foot(circumcenter(B,O,C),A,O)-O; pair P = extension(X,X+B-D,A,O); draw(A--B--C--cycle, linewidth(1)); draw(circle(B,X,D), linewidth(0.7)+dashed); draw(circle(B,O,C), black); draw(A--T, linewidth(0.7)); draw(C--O, linewidth(0.7)); draw(O--X, linewidth(0.7), StickIntervalMarker(1, 1, p=linewidth(0.7), size=0.2cm)); draw(O--P, linewidth(0.7), StickIntervalMarker(1, 1, p=linewidth(0.7), size=0.2cm)); draw(circle(A,B,C), black); draw(O--B, linewidth(0.7), StickIntervalMarker(1, 2, p=linewidth(0.7), size=0.2cm)); draw(O--D, linewidth(0.7), StickIntervalMarker(1, 2, p=linewidth(0.7), size=0.2cm)); dot("A", A, dir(115)); dot("B", B, dir(179)); dot("C", C, dir(0)); dot("O", O, 1.5*dir(-108)); dot("D", D, dir(70)); dot("X", X, dir(154)); dot("T", T, dir(-56)); dot("P", P, dir(165)); [/asy][/asy] Let AO intersect ⊙(BOC) again at T. We claim that T is the concurrency point. Note that since angle bisector of ∠AOC is perpendicular to AC, lines AO, OX, and BD form an isosceles triangle. Combining with OB=OD, it follows that there exists point P on AO such that BXPD is isosceles trapezoid. Clearly, P∈⊙(BXD). Now, notice that ∡BTP=∡BTO=∡BCO=∡OBC=∡ABD=∡XBD=∡BDP,so T∈⊙(BXDP). Analogously, T∈⊙(CYE), so we are done. Did you mean perpendicular bisector?
01.10.2024 18:20
Solved with SilverBlaze_SY, HACK_IN_MATHS and S.Ragnork1729. [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */ pair A = (10.69888,16.57077); pair B = (1.51632,-12.50733); pair C = (34.50552,-13.10250); pair O = (18.20964,-1.78998); pair D = (32.29102,12.18297); pair X = (7.29373,5.78779); pair T = (13.17850,10.50910); pair P = (30.63286,-32.15967); import graph; size(12cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; draw(A--B, linewidth(0.6)); draw(B--C, linewidth(0.6)); draw(C--A, linewidth(0.6)); draw(circle(O, 19.83755), linewidth(0.6)); draw(circle((24.58069,-9.73103), 23.23086), linewidth(0.6) + blue); draw(circle((17.88747,-19.64757), 17.86050), linewidth(0.6) + linetype("4 4") + red); draw(X--T, linewidth(0.6)); draw(A--P, linewidth(0.6)); draw(B--D, linewidth(0.6)); draw(C--X, linewidth(0.6)); dot("A", A, NW); dot("B", B, SW); dot("C", C, SE); dot("O", O, NE); dot("D", D, NE); dot("X", X, NW); dot("T", T, W); dot("P", P, SE); [/asy][/asy] Let the line through X parallel to BD intersect AO at T. Claim: T lies on ⊙(BXD). Proof. Firstly, we have, ∡OTX=∡ATX=∡(AO,BD)=∡OAB+∡ABD=(90∘−∡BCA)+(90∘−∡CAB)=∡ACB+∡BAC=∡ABC.Also, ∡OXT=∡(OX,BD)=∡DBX+∡BXO=(90∘−∡CAB)+(∡ABC+∡BCO)=(90∘−∡BCO)+∡ABC+∡BAC=∡CAB+∡ABC+∡BAC=∡ABC.This gives us that ∡OTX=∡ABC=∡OXT. This further implies that OX=OT. We also have that OB=OD. Combining these with the fact that XT∥BD, we get that BXTD is an isosceles trapezium and thus is also cyclic. ◼ Let P=AO∩⊙(BXD). Then note that, ∡OPB=∡TPB=∡TDB=∡DBX=90∘−∡BAC=∡OCB. So OBPC is cyclic. Similarly, if we define Q=AO∩⊙(CYE), then OBQC is also cyclic. Now note that P=AO∩⊙(OBC)=Q which gives P≡Q. We are done.
17.10.2024 18:22
Bashing is nice! take ⊙(ABC) as unit circle, then d=−ac/b, x=tc where t∈R. By x+ab¯x=a+b, t=(a+b)cab+c2, x=(a+b)c2ab+c2. Some observation makes me guess that the intersection is on ⊙(BOC), say λa where λ∈R. We want λa−bλa−c/bc∈R. Here λ=a2+bca(b+c), λa=a2+bcb+c. Now we only need b,−ac/b,(a+b)c2ab+c2,a2+bcb+c are concyclic. ⟺(a+b)c2ab+c2−b(a+b)c2ab+c2+acb÷a2+bcb+c−ba2+bcb+c+acb=a(c+b)(c−b)c(a+b)(a−b)∈Rwhich is obvious.◻
01.11.2024 06:21
Heres my solution: Define point T=AO∩(BOC) with T≠O. Note that: ∠BTO=∠BCO=90∘−A. The following claim finishes the problem: Let P=(BXD)∩AO. Claim : BXPD is a cyclic trapeziod. Proof: Notice that: BD and the angle bisector of ∠AOC are parallel which implies that BD and the angle bisector of ∠XOA are perpendicular. Let P′ be a point on AO such that XP′∥BD. Therefore, XP′ and BD have same perpendicular bisector which implies BCP′D is an isosceles trapezoid (and thus cyclic) and therefore P=P′. Due to the above claim: ∠PDB=∠XBD=90∘−A=∠BTP and therefore BXDT is cyclic.
10.12.2024 20:22
Sol:- Let (BOC) meet AO again at V .Let l be the line through O parallel to AB. Since AO=OB ,the reflection F of Y across l lies on AO.Since l is also the perpendicular bisector of CE ,CYFE is a cyclic isosceles trapezoid. ∡CVF=∡CVO=∡CBO=∡OCB=∡YCE=∡CEF⟹V∈(CYEF). Similarly V∈(BXD).
11.01.2025 19:05
Easy G5