Let $X$ be an arbitrary point inside the circumcircle of a triangle $ABC$. The lines $BX$ and $CX$ meet the circumcircle in points $K$ and $L$ respectively. The line $LK$ intersects $BA$ and $AC$ at points $E$ and $F$ respectively. Find the locus of points $X$ such that the circumcircles of triangles $AFK$ and $AEL$ touch.
Problem
Source: Sharygin First Round 2013, Problem 8
Tags: geometry, circumcircle, Asymptote, geometry proposed
aZpElr68Cb51U51qy9OM
09.04.2013 02:49
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(300); 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 = 0.000000000000000, xmax = 15.00000000000000, ymin = -8.000000000000000, ymax = 13.00000000000000; /* image dimensions */ pen qqffff = rgb(0.000000000000000,1.000000000000000,1.000000000000000); /* draw figures */ draw((2.000000000000000,-4.000000000000000)--(13.00000000000000,-4.000000000000000)); draw(circle((7.500000000000008,-0.8422393456278697), 6.342038501168281)); draw((2.000000000000000,-4.000000000000000)--(12.42017716027140,3.159423941287285)); draw((13.00000000000000,-4.000000000000000)--(1.563533842763736,1.389312113399954)); draw((12.42017716027140,3.159423941287285)--(6.152374166990704,2.137495534061987)); draw((2.000000000000000,-4.000000000000000)--(3.203176904089038,1.656646269223689), linetype("4 4") + qqffff); draw((13.00000000000000,-4.000000000000000)--(6.152374166990704,2.137495534061987), linetype("4 4") + qqffff); draw((3.761297146938178,4.280606867836685)--(1.563533842763736,1.389312113399954)); draw(circle((2.099705256173974,3.262693090435416), 1.948598488574976)); draw(circle((8.542593271323456,7.209693862882815), 5.607168916093936)); draw((3.203176904089038,1.656646269223689)--(3.761297146938178,4.280606867836685), linetype("4 4") + qqffff); draw((3.761297146938178,4.280606867836685)--(6.152374166990704,2.137495534061987), linetype("4 4") + qqffff); draw((1.563533842763736,1.389312113399954)--(3.203176904089038,1.656646269223689)); draw((3.761297146938178,4.280606867836685)--(12.42017716027140,3.159423941287285)); draw((3.203176904089038,1.656646269223689)--(6.152374166990704,2.137495534061987)); draw((2.099705256173975,3.262693090435417)--(8.542593271323458,7.209693862882816)); draw(shift((7.500000000000007,-7.210906377850759))*xscale(6.368667032222890)*yscale(6.368667032222890)*arc((0,0),1,30.27643228235399,149.7235677176462), linetype("4 4") + qqffff); /* dots and labels */ dot((3.761297146938178,4.280606867836685),dotstyle); label("$A$", (3.668563506142534,4.434070918627872), NE * labelscalefactor); dot((2.000000000000000,-4.000000000000000),dotstyle); label("$B$", (1.448310298894344,-4.753183776106993), NE * labelscalefactor); dot((13.00000000000000,-4.000000000000000),dotstyle); label("$C$", (13.21310028212900,-4.523502408738621), NE * labelscalefactor); dot((6.475158832580083,-0.9252385257716100),dotstyle); label("$X$", (6.220578686887579,-0.7465199231253990), NE * labelscalefactor); dot((1.563533842763736,1.389312113399954),dotstyle); label("$L$", (0.7592662000931822,0.8357294965233830), NE * labelscalefactor); dot((12.42017716027140,3.159423941287285),dotstyle); label("$K$", (12.21781436163844,3.362224537575471), NE * labelscalefactor); dot((6.152374166990704,2.137495534061987),dotstyle); label("$F$", (6.246098838695030,2.290378156523070), NE * labelscalefactor); dot((3.203176904089038,1.656646269223689),dotstyle); label("$E$", (2.673277585651966,1.754454965996870), NE * labelscalefactor); dot((8.542593271323458,7.209693862882816),dotstyle); label("$C_2$", (8.644993108595370,7.368888390557064), NE * labelscalefactor); dot((2.099705256173975,3.262693090435417),dotstyle); label("$C_1$", (1.499350602509245,3.413264841435109), NE * labelscalefactor); dot((7.500000000000008,-0.8422393456278690),dotstyle); label("$O$", (7.445545973645200,-0.7209997711955800), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] The answer is the union of segment $AB$, segment $AC$, and minor arc $\widehat{BOC}$, where $O$ is the circumcenter of $\triangle ABC$. Let $O, C_1, C_2$ be the circumcenters of $\triangle ABC, \triangle AEL, \triangle AFK$, respectively. The circumcircles of $\triangle AEL$ and $\triangle AFK$ touch if and only if $A$ lies on $C_1C_2$. This implies that the condition $\angle C_1AL + \angle LAK + \angle KAC_2 = 180^\circ$ must be satisfied. If $X$ lies on segment $AB$, $B$ coincides with $K$, so the circumcircle of $\triangle AFK$ is identically point $A$. Therefore the circumcircles of $\triangle AEL$ and $\triangle AFK$ touch at a single point. Similarly, if $X$ lies on segment $AC$, $A$ coincides with $L$, so again the two circumcircles touch. Now suppose that the locus of $X$ that satisfies the given condition does not lie on either $AB$ or $AC$. Angle chasing, we have that \begin{align*} 180^\circ = \angle C_1AL + \angle LAK + \angle KAC_2 &= (\angle AEL - 90^\circ)+(180^\circ - \angle ALK - \angle AKL) + (\angle AFK - 90^\circ)\\&= (\angle AEL + \angle AFK) - (\angle ALK + \angle AKL) \\&= (180^\circ - \angle AEF) + (180^\circ - \angle AFE) - (\angle ALK + \angle AKL) \\&= 180^\circ + \angle BAC - (\angle ALK + \angle AKL).\end{align*} Therefore $\angle BAC = \angle ALK + \angle AKL$. Then $\angle BAC = \angle ABX + \angle ACX = \angle BXC - \angle BAC$, so $2\angle BAC = \angle BXC$. Since $\angle BOC = 2\angle BAC$, the locus of $X$ such that $2\angle BAC = \angle BXC$ is concyclic with $B, O, C$. Thus $X$ lies on $\widehat{BOC}$, as desired.