Let ABC be a triangle with$|AB|< |AC|. $ Let $k$ be the circumcircle of $\triangle ABC$ and let $O$ be the center of $k$. Point $M$ is the midpoint of the arc $BC $ of $k$ not containing $A$. Let $D $ be the second intersection of the perpendicular line from $M$ to $AB$ with $ k$ and $E$ be the second intersection of the perpendicular line from $M$ to $AC $ with $k$. Points $X $and $Y $ are the intersections of $CD$ and $BE$ with $OM$ respectively. Denote by $k_b$ and $k_c$ circumcircles of triangles $BDX$ and $CEY$ respectively. Let $G$ and $H$ be the second intersections of $k_b$ and $k_c $ with $AB$ and $AC$ respectively. Denote by ka the circumcircle of triangle $AGH.$ Prove that $O$ is the circumcenter of $\triangle O_aO_bO_c, $where $O_a, O_b, O_c $ are the centers of $k_a, k_b, k_c$ respectively.
Problem
Source: 7th European Mathematical Cup , Senior Category , Q2
Tags: geometry
25.12.2018 15:51
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -12.62, xmax = 18.18, ymin = -11.31, ymax = 9.03; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-3.46,6.29)--(-7.68,-4.53)--(6.74,-4.51)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-3.46,6.29)--(-7.68,-4.53), linewidth(2) + rvwvcq); draw((-7.68,-4.53)--(6.74,-4.51), linewidth(2) + rvwvcq); draw((6.74,-4.51)--(-3.46,6.29), linewidth(2) + rvwvcq); draw(circle((-0.4747333589842245,-1.1072481723739898), 7.976910269123751), linewidth(2) + wrwrwr); draw((-0.4636696938015851,-9.084150769056782)--(-5.883622914145816,-6.970268829144336), linewidth(2) + wrwrwr); draw((-0.4636696938015851,-9.084150769056782)--(7.488524652253095,-1.573744997782918), linewidth(2) + wrwrwr); draw((-0.4636696938015851,-9.084150769056782)--(-0.4747333589842245,-1.1072481723739898), linewidth(2) + wrwrwr); draw((6.74,-4.51)--(-5.883622914145816,-6.970268829144336), linewidth(2) + wrwrwr); draw((-7.68,-4.53)--(7.488524652253095,-1.573744997782918), linewidth(2) + wrwrwr); draw(circle((-3.746292826237317,-3.5155707275752084), 4.062403091479756), linewidth(2) + wrwrwr); draw(circle((3.4617726259489463,-2.110761753871323), 4.062403091479756), linewidth(2) + wrwrwr); draw((-5.954442660721759,-0.10570369407806521)--(1.2536227914645037,1.2991052796258202), linewidth(2) + wrwrwr); draw(circle((-2.6828831934686654,2.3026188611231535), 4.062403091479756), linewidth(2) + wrwrwr); /* dots and labels */ dot((-3.46,6.29),dotstyle); label("$A$", (-3.38,6.49), NE * labelscalefactor); dot((-7.68,-4.53),dotstyle); label("$B$", (-7.6,-4.33), NE * labelscalefactor); dot((6.74,-4.51),dotstyle); label("$C$", (6.82,-4.31), NE * labelscalefactor); dot((-0.4747333589842245,-1.1072481723739898),linewidth(4pt) + dotstyle); label("$O$", (-0.4,-0.95), NE * labelscalefactor); dot((-0.4636696938015851,-9.084150769056782),linewidth(4pt) + dotstyle); label("$M$", (-0.38,-8.93), NE * labelscalefactor); dot((-5.883622914145816,-6.970268829144336),linewidth(4pt) + dotstyle); label("$D$", (-5.8,-6.81), NE * labelscalefactor); dot((7.488524652253095,-1.573744997782918),linewidth(4pt) + dotstyle); label("$E$", (7.56,-1.41), NE * labelscalefactor); dot((-0.4680654521862633,-5.9148089737038845),linewidth(4pt) + dotstyle); label("$X$", (-0.38,-5.75), NE * labelscalefactor); dot((-0.471934547813736,-3.125191026296114),linewidth(4pt) + dotstyle); label("$Y$", (-0.4,-2.97), NE * labelscalefactor); dot((-5.954442660721759,-0.10570369407806521),linewidth(4pt) + dotstyle); label("$G$", (-5.88,0.05), NE * labelscalefactor); dot((1.2536227914645037,1.2991052796258202),linewidth(4pt) + dotstyle); label("$H$", (1.34,1.45), NE * labelscalefactor); dot((-2.6828831934686654,2.3026188611231535),linewidth(4pt) + dotstyle); label("$O_a$", (-2.6,2.47), NE * labelscalefactor); dot((-3.746292826237317,-3.5155707275752084),linewidth(4pt) + dotstyle); label("$O_b$", (-3.66,-3.35), NE * labelscalefactor); dot((3.4617726259489463,-2.110761753871323),linewidth(4pt) + dotstyle); label("$O_c$", (3.54,-1.95), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] WLOG AB < AC. Claim 1: $BDOX$ cyclic. Proof: It is well-known that $AM$ is the angle bisector of $\angle BAC$. Thus $\angle BOM = 2\angle BAM = \angle BAC = 180^{\circ}-\angle BDX$. Similarly we get $CEOY$ cyclic. Claim 2: $\angle BAD = \angle CAE$. Proof: $\angle BAD = 180^{\circ}-\angle ADB-\angle DBA = \angle BDM - 90^{\circ}-\angle C = 90^{\circ}-\frac{\angle A}{2}-\angle C = \frac{\angle B - \angle C}{2}$. Similarly $\angle CAE = \frac{\angle B - \angle C}{2}$. Thus, $BD = EC$. This implies that the circumcircles of $\bigtriangleup BOD$ and $\bigtriangleup COE$ have the same circumradii. Claim 3: $AGOH$ cyclic. Proof: $\angle AGO = \angle BDO = 90^{\circ}-\frac{\angle BOD}{2}=90^{\circ}-\angle BAD$. Similarly $\angle AHO = 90^{\circ} + \angle CAE$. Hence $\angle AGO + \angle AHO = 180^{\circ}$. Now $OO_a=\frac{AO}{2\sin\angle AGO}=\frac{BO}{2\sin\angle BDO}=OO_b$. Thus, $OO_a=OO_b=OO_c$ and the conclusion follows.
07.01.2019 01:19
see this link here
16.06.2022 21:39
Claim $: BDXO$ and $CYOE$ are cyclic. Proof $:$ Note that $\angle BDX = \angle 180 - \angle A = \angle 180 - \angle BFX$. we prove the other one with same approach. Claim $: BD = CE$. Proof $:$ Note that $\angle BAD = \angle 180 - \angle DBA - ADB = \angle 90 - \angle C - \frac{\angle A}{2} = \frac{B-C}{2}$ with same approach $\angle CAE = \frac{B-C}{2}$. Claim $: AGOH$ is cyclic. Proof $:$ Note that $\angle AHO = \angle CYO = \angle 180 - \angle OEC = \angle 90 + \angle CAE$ and $\angle AGO = \angle BDO = \angle 90 - \angle BAD$ so $\angle AGO + \angle AHO = \angle 180$. Note that $OBD$ and $OCE$ are isosceles and $BD = CE$ so $R_b = R_c$. Note that $R_a = \frac{AO}{2\sin{AGO}}$ and $AO = BO$ and $\angle AGO = \angle BDO$ so $R_a = R_b$ so $O$ is center of $O_aO_bO_c$.