Let $ABC$ be a triangle with circumcircle $\omega$ and circumcenter $O$. The tangent line to from $A$ to $\omega$ intersects $BC$ at $K$. The tangent line to from $B$ to $\omega$ intersects $AC$ at $L$. Let $M,N$ be the midpoints of $AK,BL$ respectively. The line $MN$ is named by $\alpha$. The feet of perpendicular from $A,B,C$ to the edges of $\triangle ABC$ are named by $D,E,F$ respectively. The perpendicular bisectors of $EF,DF,DE$ intersect $\alpha$ at $X,Y,Z$ respectively. Let $AD,BE,CF$ intersect $\omega$ again at $D',E',F'$ respectively. If $H$ is the orthocenter of $ABC$, prove that the lines $XD',YE',ZF',OH$ are concurrent.
Problem
Source: Israeli Olympic Revenge 2018, Problem 3
Tags: geometry, circumcircle, Euler, perpendicular bisector
SerdarBozdag
28.08.2021 13:20
the center of homothety takes intouch triangle of $D'E'F'$ to triangle $ABC$.
I do not know how to prove.
rafaello
13.09.2021 23:21
Nice problem, although some parts are well-known and already been on the competitions. Also, this is a very rich configuration, so I essentially present a solution by many claims. Let $ABC$ be a triangle with circumcenter $O$, orthocenter $H$ and the nine-point center $N_9$. Let $M_AM_BM_C$ denote the medial triangle and $DEF$ denote the orthic triangle. Let $N_A,N_B,N_C$ denote the midpoints of $AH,BH,CH$, respectively. Let $D',E',F'$ be the reflection of $H$ over $BC,CA,AB$, respectively. Let $Q_A,Q_B,Q_C$ be the Queue points of $A,B,C$, respectively. Let $X_A,X_B,X_C$ denote the Ex-points of $A,B,C$, respectively. Claim 01. $MD$ and $NE$ are tangent to the nine-point circle of $\triangle ABC$. Proof. Note that $AM=KM=DM$ as $\angle ADK=90^{\circ}$ and because $(AM_BM_C)$ is tangent to $KA$ by homothety, we get that $MD^2=MA^2=MM_B\cdot MM_C$. Similarly, $NE$ is tangent to the nine-point circle of $\triangle ABC$. Claim 02. $\alpha$ passes through $X_A,X_B,X_C$ and furthermore $\alpha$ is the radical axis of the nine-point circle and the circumcircle of $\triangle ABC$. Proof. By the Prism Lemma, we get that $X_A,X_B,X_C$ are collinear. By Pascal's theorem on $EFM_CM_ADE$ and on $DFM_CM_BED$, we get that $M,N$ lie on the line connecting $X_A,X_B,X_C$. Since $X_AE\cdot X_AF=XB\cdot XC$ and $X_BF\cdot X_BD=X_CA\cdot X_CC$, we get that $\alpha$ is the radical axis of the nine-point circle and the circumcircle of $\triangle ABC$. Claim 03. $Q_A,N_A,O,M_A,D'$ are concyclic and so on. Proof. Because $Q_ABD'C$ is harmonic, we get that $O,M,Q_A,D'$ and the intersection of tangents from $Q_A,D'$ to $(ABC)$ are concyclic due to proper right angles. $N_A$ also lies on that circle, because $\measuredangle Q_AOD'=2\measuredangle Q_AAD'=\measuredangle Q_AN_AD'$. Claim 04. $X,Q_A,D'$ are collinear and so on. Proof. This is obvious by radical axis theorem on $(ABC)$, $(Q_AN_AM_AD')$ and $(M_AM_BM_C)$. Claim 05. $OH$ is the common radical axis of $(Q_AN_AM_AD'),(Q_BN_BM_BE')$ and $(Q_CN_CM_CF')$. Proof. This is obvious using PoP, indeed $$HN_A\cdot HD'=HA\cdot HD=HB\cdot HE=HN_B\cdot HE'=HC\cdot HF=HN_C\cdot HF'$$and as $O$ lies on all of those circles, we conclude that $OH$ is the common radical axis of $(Q_AN_AM_AD'),(Q_BN_BM_BE')$ and $(Q_CN_CM_CF')$. Claim 06. $OH,Q_AD',Q_BE'$ and $Q_CF'$ are concurrent. Proof. By radical axis theorem on $(ABC),(Q_AN_AM_AD'),(Q_BN_BM_BE')$, we get that $OH,AD'$ and $Q_BE'$ are concurrent and now by radical axis theorem on $(ABC),(Q_AN_AM_AD'),(Q_CN_CM_CF')$, we get that $OH,AD'$ and $Q_CF'$ are concurrent. We conclude that $OH,Q_AD',Q_BE'$ and $Q_CF'$ are concurrent. Mash up all the claims and we are done.
[asy][asy]import geometry; import olympiad;
size(10cm);
defaultpen(fontsize(10pt));
pair A = dir(110), B = dir(200), C = dir(340),D=foot(A,B,C),E=foot(B,A,C),F=foot(C,A,B),H=orthocenter(A,B,C),O=circumcenter(A,B,C),N9=midpoint(O--H),E1=2E-H,D1=2D-H,F1=2F-H,MA=midpoint(B--C),MB=midpoint(A--C),MC=midpoint(A--B),NA=midpoint(H--A),NB=midpoint(H--B),NC=midpoint(H--C);path abc=circumcircle(A,B,C),aef=circumcircle(A,E,F),bdf=circumcircle(B,D,F),ced=circumcircle(C,E,D); pair QA=intersectionpoints(abc,aef)[1],QB=intersectionpoints(abc,bdf)[0],QC=intersectionpoints(abc,ced)[0],XA=extension(E,F,B,C),XB=extension(D,F,A,C),XC=extension(E,D,A,B),M=extension(QB,QC,MC,MB),N=extension(QA,QC,MC,MA),X=extension(M,N,QA,D1),Y=extension(M,N,E1,QB),Z=extension(M,N,F1,QC),K=extension(M,A,B,C),L=extension(N,B,A,C),P=extension(H,O,QA,D1);
draw(A--B--C--cycle);draw(abc);draw(K--B);draw(L--A);draw(Y--XC);draw(K--A);draw(L--B);draw(D1--A);draw(E1--B);draw(F1--C);draw(E--XA);draw(D--XB);draw(D--XC);draw(X--D1);draw(X--MA);draw(Y--E1);draw(Y--MB);draw(Z--NC);draw(Z--QC);draw(M--D);draw(N--E);draw(circumcircle(O,MA,NA));draw(circumcircle(O,MB,NB));draw(circumcircle(O,MC,NC));draw(O--P);
draw(M--MB);draw(QC--M);draw(XC--A);draw(circumcircle(D,E,F));draw(N--MA);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$O$", O, dir(90));
dot("$H$", H, dir(H));
dot("$N_9$", N9, dir(N9));
dot("$D'$", D1, dir(D1));
dot("$E'$", E1, dir(E1));
dot("$F'$", F1, dir(F1));
dot("$M_A$", MA, dir(MA));
dot("$M_B$", MB, dir(MB));
dot("$M_C$", MC, dir(MC));
dot("$N_A$", NA, dir(NA));
dot("$N_B$", NB, dir(NB));
dot("$N_C$", NC, dir(NC));
dot("$Q_A$", QA, dir(QA));
dot("$Q_B$", QB, dir(QB));
dot("$Q_C$", QC, dir(QC));
dot("$X_A$", XA, dir(XA));
dot("$X_B$", XB, dir(XB));
dot("$X_C$", XC, dir(XC));
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$Z$", Z, dir(Z));
dot("$K$", K, dir(K));
dot("$L$", L, dir(L));
dot("$P$",P, dir(P));
[/asy][/asy]
[asy][asy]import geometry; import olympiad;
size(10cm);
defaultpen(fontsize(10pt));
pair A = dir(110), B = dir(200), C = dir(340),D=foot(A,B,C),E=foot(B,A,C),F=foot(C,A,B),H=orthocenter(A,B,C),O=circumcenter(A,B,C),N9=midpoint(O--H),E1=2E-H,D1=2D-H,F1=2F-H,MA=midpoint(B--C),MB=midpoint(A--C),MC=midpoint(A--B),NA=midpoint(H--A),NB=midpoint(H--B),NC=midpoint(H--C);path abc=circumcircle(A,B,C),aef=circumcircle(A,E,F),bdf=circumcircle(B,D,F),ced=circumcircle(C,E,D); pair QA=intersectionpoints(abc,aef)[1],QB=intersectionpoints(abc,bdf)[0],QC=intersectionpoints(abc,ced)[0],XA=extension(E,F,B,C),XB=extension(D,F,A,C),XC=extension(E,D,A,B),M=extension(QB,QC,MC,MB),N=extension(QA,QC,MC,MA),X=extension(M,N,QA,D1),Y=extension(M,N,E1,QB),Z=extension(M,N,F1,QC),K=extension(M,A,B,C),L=extension(N,B,A,C),P=extension(H,O,QA,D1);
draw(A--B--C--cycle);draw(abc);draw(K--B);draw(L--A);draw(Y--XC);draw(K--A);draw(L--B);draw(D1--A);draw(E1--B);draw(F1--C);draw(E--XA);draw(D--XB);draw(D--XC);draw(X--D1);draw(X--MA);draw(Y--E1);draw(Y--MB);draw(Z--NC);draw(Z--QC);draw(M--D);draw(N--E);draw(circumcircle(O,MA,NA));draw(circumcircle(O,MB,NB));draw(circumcircle(O,MC,NC));draw(O--P);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$O$", O, dir(90));
dot("$H$", H, dir(H));
dot("$N_9$", N9, dir(N9));
dot("$D'$", D1, dir(D1));
dot("$E'$", E1, dir(E1));
dot("$F'$", F1, dir(F1));
dot("$M_A$", MA, dir(MA));
dot("$M_B$", MB, dir(MB));
dot("$M_C$", MC, dir(MC));
dot("$N_A$", NA, dir(NA));
dot("$N_B$", NB, dir(NB));
dot("$N_C$", NC, dir(NC));
dot("$Q_A$", QA, dir(QA));
dot("$Q_B$", QB, dir(QB));
dot("$Q_C$", QC, dir(90));
dot("$X_A$", XA, dir(XA));
dot("$X_B$", XB, dir(XB));
dot("$X_C$", XC, dir(XC));
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$Z$", Z, dir(Z));
dot("$K$", K, dir(K));
dot("$L$", L, dir(L));
dot("$P$",P, dir(P));
clip((1.1,1.1)--(1.1,-1.1)--(-1.1,-1.1)--(-1.1,1.1)--cycle);
[/asy][/asy]
Aegaertargaryen1102
20.11.2021 20:15
Nice problem i really love this problem