The circle inscribed in the triangle $ABC$ touches the sides $AB$ and $AC$ at the points $K$ and $L$ , respectively. The angle bisectors from $B$ and $C$ intersect the altitude of the triangle from the vertex $A$ at the points $Q$ and $R$ , respectively. Prove that one of the points of intersection of the circles circumscribed around the triangles $BKQ$ and $CPL$ lies on $BC$.
Problem
Source: Bosnia and Herzegovina EGMO TST 2019 p3
Tags: geometry, circumcircle
WL0410
02.08.2019 02:08
By Miquel, it is equivalent to show that the other point of intersection of these circles lies on $(AKIL).$ More general claim: Let $(J)$ be a circle tangent to $AI$ at $I.$ Let $X$ be the other point of intersection between $(J)$ and $(AI).$ Then $(BKX)\cap (J)$ on $BI$ and $(CLX)\cap (J)$ on $CI.$ Proof: Let $Q=(BKX)\cap (J).$ $\measuredangle IQX = \measuredangle AIX = \measuredangle AKX = \measuredangle BKX = \measuredangle BQX,$ so $I,Q,B$ are collinear as desired. The other statement holds by symmetry. So now it suffices to show that $(QRI)$ is tangent to $AI,$ since $(BKQ)$ and $(CRL)$ will both pass through $X$ as defined above. This can be done directly by angle chasing to show that $\measuredangle IQR = \measuredangle AIR.$ [asy][asy] size(10cm); pair A = dir(115); pair B = dir(210); pair C = dir(330); pair I = incenter(A,B,C); pair K = foot(I,A,B); pair L = foot(I,C,A); pair D = foot(A,B,C); pair Q = extension(A,D,B,I); pair R = extension(A,D,C,I); path c1 = circumcircle(I,R,Q); path c2 = circumcircle(A,K,L); pair xi[] = intersectionpoints(c1,c2); pair X = xi[1]; draw(A--B--C--cycle); draw(circumcircle(A,K,L),red); draw(circumcircle(B,K,Q),blue); draw(circumcircle(C,L,R),blue); draw(circumcircle(I,Q,R),green); draw(B--I,green); draw(C--R,green); draw(A--D,red+dashed); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$Q$",Q,dir(Q)); dot("$R$",R,dir(R)); dot("$K$",K,dir(K)); dot("$L$",L,dir(L)); dot("$I$",I,dir(I)); dot("$X$",X,dir(X)); [/asy][/asy]