Let ABC be a scalene triangle with circumcenter O, and let D and E be points inside angle ∡BAC such that A lies on line DE, and ∠ADB=∠CBA and ∠AEC=∠BCA. Let M be the midpoint of BC and K be a point such that OK is perpendicular to AO and ∠BAK=∠MAC. Finally, let P be the intersection of the perpendicular bisectors of BD and CE. Show that KO=KP. Proposed by Victor Domínguez
Problem
Source: Mexico National Olympiad Mock Exam 2019 P6
Tags: geometry, symmedian, Circumcenter, perpendicular bisector
18.10.2019 03:13
Any solutions?
18.10.2019 17:11
Nicest configuration I've seen in a while. Let S∈(ABC) be such that AS is the A-symmedian on △ABC, L be the midpoint of ¯AS. Also let B1,C1 be the reflections of B,C in L, F=BD∩CE and M1,M2,N1,N2,N be the midpoints of ¯BD,¯CE,¯BF,¯CF,¯AF. First note that ∠ABF=∠(DE,BC)=∠ACF, so F∈(ABC). Step 1. L∈(FM1M2) Proof: √bc inversion in A takes L to the reflection of A in M,so ∠ALB=∠ALC=1800−∠BAC. Therefore ∠AB1B=∠LBS=∠LBC+∠LAC=∠ABC=∠ADBmeaning that B1∈(ABD). In the same fashion we prove that C1∈(ACE), and since LM1∥B1D,LM2∥C1E and ∠ADB1=∠ABB1=∠BAM, and in the same way ∠AEC1=∠CAM, we obtain that ∠M1LM1=∠(DB1,EC1)=∠BAC=∠M1FM2which proves the claim. ◼ Step 2. N∈(FM1M2) Proof: Notice that △FDE∼△ABC, so NN1NN2=AB2AC2=FD2FE2=M1N1M2N2combined with the fact that ∠NN1M1=1800−∠NN1F=1800−∠ABF=1800−∠ACF=1800−∠NN2F=∠NN2M2yields that △NN1M1∼△NN2M2, i.e. N is the Miquel point of M1N1N2M2. This is enough to prove the claim. ◼ Now, from Steps 1 and 2, it follows that L∈(PNF), and since ∠ALO=900=∠ANO, we also have that L∈(AON). We conclude that there is a spiral similarity that sends A to O and F to P. But △LAO∼△LOK, so the spiral similarity also sends O to K. Overall, we infer that △OKP∼△AOF, which proves that KO=KP.
20.10.2019 08:44
Official Solution: Consider the isosceles triangles ASB and CTA drawn to the interior of ΔABC, and similar to triangles COA and AOB, respectively. From the similarities we obtain ∠BAS+∠TAC=∠OAC+∠BAO=∠BAC, so A, S, T are collinear, as well as ASAO=ABAC=AOAT, from which AO2=AS⋅AT, from which in turn it follows that AT is tangent to the circumcircle of ΔSOT, let's say of circumcenter K′. Since S and O are on the perpendicular bisector of AB, SO is the internal angle bisector of ∠AOB, and we have ∠ACB=12∠AOB=∠AOS=∠OTS. Similarly, ∠CBA=∠TSO. We then infer that OST and ABC are similar triangles. We show now K=K′. Let N be the midpoint of ST. Then ∠K′NA=∠AOK=90∘, and so the quadrilateral AOK′N is cyclic. Using this along with ∠NOK′=∠MAO, which we obtain from ΔABC and ΔOST, we calculate ∠BAK′=∠BAS+∠NAK=∠OAC+∠NOK′=∠OAC+∠MAO=∠MAC. This means that K′ satisfies ∠BAK′=∠MAC and AO⊥OK, from which we conclude K=K′. To finish, it's enough to show that P lies on the circumcircle of ΔOST From ∠ASB=∠COA=2∠CBA=2∠ADB and AS=SB, We get that S is the circumcenter of ADB, and analogously T is the circumcenter of AEC. Therefore, S lies on the perpendicular bisector of BD and T on that of CE, so P is no more than the intersection of the lines perpendicular to BD and CE passing through S and T, respectively. Using the notation ∠(l,m) referring to the least non-negative angle we need to rotate l anti-clockwise so as to obtain a line parallel to m, we calculate modulo 180∘ ∡SPT=∠(SP,PT)=∠(BD,CE)≡∠(BD,AD)+∠(AD,CE)≡−∡ADB−∡CEA=−∡CBA−∡ACB≡∡BAC=∡SOT. It follows that P lies on the circumcircle of SOT, and finally KO=KP.
22.10.2019 15:02
Interesting problem, hopefully my solution works plagueis wrote: Let ABC be a scalene triangle with circumcenter O, and let D and E be points inside angle ∡BAC such that A lies on line DE, and ∠ADB=∠CBA and ∠AEC=∠BCA. Let M be the midpoint of BC and K be a point such that OK is perpendicular to AO and ∠BAK=∠MAC. Finally, let P be the intersection of the perpendicular bisectors of BD and CE. Show that KO=KP. Lemma: Let ABC be a triangle, ω1,ω2 be circles through A,B and A,C respectively. X is a variable point on ω1 and AX meets ω2 at Y. Then the locus of the perpendicular bisectors of BX,CY is a circle through the centers of ω1,ω2. (Proof): Let O1,O2 be the centers of ω1,ω2. Now, ∠BO1Z+∠ZO2C=∠A, this gives that ∠O1ZO2 is fixed so we are done. \hfill{\square} Main Problem. Let B′,C′ be the reflection of B and C in the A− altitude. Note that D,E varies on ⊙(ABB′),⊙(ACC′) respectively. Let O1,O2 be the center of these circles respectively. Now from the Lemma, we just need to show the problem for 3 cases of D. For D=E=A, P≡O. When D=B′, the perpendicular bisector of BD is A− altitude and perpendicular bisector of CE passes through O2, so when D=B′, P≡O2. Similarly when D=C′, P≡O1, so we need to show that K is the circumcenter of △O1O2O. Now note that O1 lies on perpendicular bisector of AB and O2 lies on perpendicular bisector of AC, ∠OO2O1=∠OO2A=C=∠AOO1. This gives that △ABC∼△OO1O2 and AO is tangent to ⊙(OO1O2). Let K′ be the circumcenter of △OO1O2 and N be the midpoint of O1O2. Now, ∠ANK′=90∘=∠AOK′. As △OO1O2∼△ABC, so △OK′N∼△AOM. This gives that MAO=∠NOK′=∠NAK′ which gives that K′ lies on the A− symmedian and we are done. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(6cm); 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 = -4.618468119884228, xmax = 9.404971407348569, ymin = -0.9275912378296551, ymax = 6.150069464002545; /* image dimensions */ pair A = (0.,5.2899431606514336), B = (-1.4876313893336057,0.), C = (4.292234737910577,0.), O = (1.402301674288486,2.0414430609646925), K = (0.541162529618346,1.6697093908679173), O_1 = (0.,2.435796634270331), O_2 = (0.,0.9036221474655032), B_1 = (2.1461173689552884,2.6449715803257168), C_1 = (-0.7438156946668029,2.6449715803257168), M = (1.4023016742884855,0.); /* draw figures */ draw(A--B, linewidth(1.)); draw(B--C, linewidth(1.)); draw(C--A, linewidth(1.)); draw(A--O, linewidth(1.)); draw(K--O, linewidth(1.)); draw(O_1--A, linewidth(1.)); draw(O_2--O_1, linewidth(1.)); draw(circle((0.5411625296183468,1.6697093908679173), 0.9379480518487351), linewidth(1.)); draw((0.,1.6697093908679173)--K, linewidth(1.)); draw(O--(0.,1.6697093908679173), linewidth(1.)); draw(C_1--O, linewidth(1.)); draw(O_2--B_1, linewidth(1.)); draw(M--O, linewidth(1.)); draw(M--A, linewidth(1.)); /* dots and labels */ dot(A,dotstyle); label("A", (-0.057553226622777565,5.391748674520524), NE * labelscalefactor); dot(B,dotstyle); label("B", (-1.9,-0.4), NE * labelscalefactor); dot(C,dotstyle); label("C", (4.349499477468407,-0.4), NE * labelscalefactor); dot(O,linewidth(4.pt) + dotstyle); label("O", (1.3,2.2), NE * labelscalefactor); dot(K,linewidth(4.pt) + dotstyle); label("K", (0.3,1.3), NE * labelscalefactor); dot(O_1,linewidth(4.pt) + dotstyle); label("O1", (-0.45,2.5), NE * labelscalefactor); dot(O_2,linewidth(4.pt) + dotstyle); label("O2", (-0.5,0.7), NE * labelscalefactor); dot((0.,1.6697093908679173),linewidth(4.pt) + dotstyle); label("N", (-0.35,1.611134883479721), NE * labelscalefactor); dot(B_1,linewidth(4.pt) + dotstyle); dot(C_1,linewidth(4.pt) + dotstyle); dot(M,linewidth(4.pt) + dotstyle); label("M", (1.448098195971966,-0.4), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note. The last part was pretty much same as IMO SL 2017 G3.
22.10.2019 17:01
22.10.2019 19:42
Is there any solution with the method of moving points?
27.10.2019 11:38
Let X, Y be the intersection of the perpendicular bisector of CA, AB with A-altitude of △ABC respectively. It is easy to show that D lies on (Y,YB), and E lies on (X,XC). Hence, XP⊥CE and YP⊥BD. By angle chasing, OXPY is concyclic quadrilateral. Only show that KO=KY. (cotC>cotB)OY=12AB(cotC−cotB) ⟹KOOA=DMDA=12(DC−DB)DA D is the foot of A-altitude of △ABC. M is the midpoint of BC. Now, KOKY=OAAB Hence, △KOY is similar to △OAB ◼