The circumcenter of an acute-triangle $ABC$ with $|AB|<|BC|$ is $O$, $D$ and $E$ are midpoints of $|AB|$ and $|AC|$, respectively. $OE$ intersects $BC$ at $K$, the circumcircle of $OKB$ intersects $OD$ second time at $L$. $F$ is the foot of altitude from $A$ to line $KL$. Show that the point $F$ lies on the line $DE$
Problem
Source: Turkey Junior National Olympiad 2020 #P3
Tags: geometry, collinearity
Tintarn
05.03.2021 16:59
Angle-chasing shows that $\angle BKO=\angle BAO=90^\circ-\gamma$, hence $AOBKL$ is cyclic.
Also $\angle ADL=\angle AFL=90^\circ$, hence $ADFL$ is cyclic.
Hence
\[\angle EDF=\angle EDA+\angle ADF=\beta+180^\circ-\angle FLA=\beta+180^\circ-\angle KLA=\beta+\angle ABK=180^\circ\]as desired.
[asy][asy]
import olympiad;
size(6cm);
defaultpen(fontsize(10pt));
defaultpen(linewidth(0.4));
dotfactor *= 1.5;
pair A = dir(70), B = dir(220), C = dir(330), D=(A+B)/2,E=(A+C)/2, O=0, K=extension(E,O,B,C), L=2*circumcenter(A,O,B)-O, F=foot(A,K,L);
draw(A--B--C--A);
draw(Circle(circumcenter(A,B,O),circumradius(A,B,O)));
draw(Circle(circumcenter(A,D,F),circumradius(A,D,F)),dotted);
draw(E--K);
draw(L--K);
draw(E--F,dashed);
dot("$A$", A, dir(60));
dot("$B$", B, dir(270));
dot("$C$", C, dir(340));
dot("$D$", D, dir(120));
dot("$E$", E, dir(20));
dot("$O$", O, dir(300));
dot("$K$", K, dir(240));
dot("$L$", L, dir(140));
dot("$F$", F, dir(200));
[/asy][/asy]
teddy8732
27.09.2021 10:20
Simson on $OKL$ and $A$ makes problem trivial
Rukevwe
21.12.2022 21:07
teddy8732 wrote: Simson on $OKL$ and $A$ makes problem trivial Oh WOW!
bin_sherlo
30.10.2023 12:32
$\angle OLB=\angle OKC=90-\angle C$ so $\angle LBA=\angle C$ which means $LB$ is tangent to $(ABC)$. We know that $L$ is the intersection of the tangent to $(ABC)$ at $B$ and the perpendicular to $AB$ from $O$ so $LA$ is tangent to $(ABC)$. Take the inversion centered at $O$ and with radius $OA$. $D\rightarrow D^*=L,L\rightarrow L^*=D$ $E^*$ lies on $LA=D^*A$ and $F^*$ lies on $ALFD$. $\angle E^*D^*F^*=\angle AD^*F^*=\angle AFF^*=\angle AFO=\angle KOF$ which gives us that $D^*, E^*, F^*, O$ are cyclic. So $D,E,F$ are collinear.