Let $ABC$ be a scalene triangle with incenter $I$ and let $AI$ meet the circumcircle of triangle $ABC$ again at $M$. The incircle $\omega$ of triangle $ABC$ is tangent to sides $AB, AC$ at $D, E$, respectively. Let $O$ be the circumcenter of triangle $BDE$ and let $L$ be the intersection of $\omega$ and the altitude from $A$ to $BC$ so that $A$ and $L$ lie on the same side with respect to $DE$. Denote by $\Omega$ a circle centered at $O$ and passing through $L$, and let $AL$ meet $\Omega$ again at $N$. Prove that the lines $LD$ and $MB$ meet on the circumcircle of triangle $LNE$.
Problem
Source: FKMO 2022 Problem 4
Tags: geometry, Concyclic, concentric, Korea
Mogmog8
23.04.2022 20:12
[asy][asy] size(8cm); import geometry; import olympiad; pair A,B,C,I,D,E,M,O,L,N,X,F; A=(6.2,7.2); B=(-4,-3); C=(9,-3); I=incenter(A,B,C); D=foot(I,A,B); E=foot(I,A,C); M=intersectionpoints(line(A,I),circle(A,B,C))[0]; O=circumcenter(B,D,E); L=intersectionpoints(line(A,foot(A,B,C)),incircle(A,B,C))[0]; N=intersectionpoints(line(A,foot(A,B,C)),circle(O,length(segment(O,L))))[1]; X=extension(B,M,D,L); F=intersectionpoints(line(B,M),circle(B,D,E))[0]; draw(circle(A,B,C)); draw(incircle(A,B,C)); draw(circle(B,D,E),darkgreen); draw(circle(L,N,E),orange+dotted); draw(circle(O,length(segment(O,L))),red); draw(A--B--C--cycle); draw(A--N); draw(E--F); draw(L--X); draw(F--X); draw(A--M); draw(L--E); draw(F--N); draw(D--E); dot(A^^B^^C^^I^^D^^E^^M^^O^^L^^N^^X^^F); label("$A$",A,NE); label("$B$",B,dir(235)); label("$C$",C,dir(330)); label("$I$",I,NW); label("$D$",D,dir(280)); label("$E$",E,dir(210)); label("$M$",M,dir(250)); label("$O$",O,SE); label("$L$",L,dir(70)); label("$N$",N,dir(270)); label("$X$",X,dir(200)); label("$F$",F,dir(265)); [/asy][/asy] Let $X=\overline{LD}\cap\overline{MB}.$ Claim: $F=\overline{BC}\cap(BDE)$ lies on $(LNE).$ Proof. Notice $N$ is the reflection of $L$ in $\ell,$ the line through $O$ parallel to $\overline{BC}.$ Notice $$\measuredangle (\overline{FE},\overline{BC})=\measuredangle EFB+\measuredangle MBC=\measuredangle EDB+\measuredangle MAC=\measuredangle EDA+\measuredangle DAM=90$$so $F$ is the reflection of $E$ in $\ell.$ Hence, $FELN$ is a cyclic isosceles trapezoid. $\blacksquare$ Thus, $$\measuredangle EFX=\measuredangle EDB=\measuredangle EDL+\measuredangle LDA=\measuredangle EDL+\measuredangle LED=\measuredangle ELD.$$$\square$
David_Kim_0202
24.04.2022 16:33
I did the same way as your explanation!! I just thought about kmo(middle. ver)-3 it's very similar. using trapezoids and making a new concyclic lol!!
ahpla
27.04.2022 02:37
Let $X = MB \cap LD$, $P$ be the circumcenter of $ELX$, and $N'$ ($\neq L$) be the intersection point of $AL$ and the circumcircle of $ELX$. Since $\angle{ELX} = \pi - (\angle{LDE} + \angle{LED}) = \pi - (\angle{LDE} + \angle{LDA}) = \angle{EDB} = (\pi + \angle{A}) / 2$, we have $\angle{EPX} = \angle{EOB} = \pi - \angle{A}$, and this leads to $EPX \sim EOB$, $EBX \sim EOP$. Angle between $MB$ and $OP$ is equivalent to $\angle{XEP} = \angle{A} / 2 = \angle{MAC} = \angle{MBC}$, and this implies that $OP$ is a perpendicular bisector of $LN'$ ($\because OP \parallel BC \perp LN'$) and $N' = N$. So $E, L, X, N$ are concyclic as desired.
LoloChen
18.05.2022 06:51
David_Kim_0202 wrote: I did the same way as your explanation!! I just thought about kmo(middle. ver)-3 it's very similar. using trapezoids and making a new concyclic lol!! Nice problem! My solution is the same as #2 and #3. The key point is to construct a new point on the circumcircle of LNE.
BarisKoyuncu
09.06.2022 16:10
Let $ABC$ be a scalene triangle. Points $D$ and $E$ are taken on the sides $AB$ and $AC$ such that $|AD|=|AE|$. Let $J$ be the incenter of the triangle $ADE$ and assume that the angle bisector of $\angle BAC$ intersects $(ABC)$ at $M$. Let $O$ be the circumcenter of triangle $BDE$ and let $L$ be the intersection of $(JDE)$ and the altitude from $A$ to $BC$ so that $A$ and $L$ lie on the same side with respect to $DE$. Denote by $\Omega$ a circle centered at $O$ and passing through $L$, and let $AL$ meet $\Omega$ again at $N$.
Prove that the lines $LD$ and $MB$ meet on the circumcircle of triangle $LNE$.
ChanandlerBong
25.09.2022 10:51
Sketch of a trigonometry solution: Deonote $S$ the second intersection of line $AL$ and $\Omega$, note that $L,D,S,E$ lies on the same circle, we only have to prove $XDsin\angle ALE=SNsin \angle DLE$ ($X$ is the intersection of $LD$ and $MB$ ) because of Ptolomy's Theorem,which can be easily checked.
suan
07.11.2022 07:07
It seems to be a tricky problem, but as soon as you notice something, it is...
segment
24.03.2023 19:35
Mogmog8 wrote: [asy][asy] size(8cm); import geometry; import olympiad; pair A,B,C,I,D,E,M,O,L,N,X,F; A=(6.2,7.2); B=(-4,-3); C=(9,-3); I=incenter(A,B,C); D=foot(I,A,B); E=foot(I,A,C); M=intersectionpoints(line(A,I),circle(A,B,C))[0]; O=circumcenter(B,D,E); L=intersectionpoints(line(A,foot(A,B,C)),incircle(A,B,C))[0]; N=intersectionpoints(line(A,foot(A,B,C)),circle(O,length(segment(O,L))))[1]; X=extension(B,M,D,L); F=intersectionpoints(line(B,M),circle(B,D,E))[0]; draw(circle(A,B,C)); draw(incircle(A,B,C)); draw(circle(B,D,E),darkgreen); draw(circle(L,N,E),orange+dotted); draw(circle(O,length(segment(O,L))),red); draw(A--B--C--cycle); draw(A--N); draw(E--F); draw(L--X); draw(F--X); draw(A--M); draw(L--E); draw(F--N); draw(D--E); dot(A^^B^^C^^I^^D^^E^^M^^O^^L^^N^^X^^F); label("$A$",A,NE); label("$B$",B,dir(235)); label("$C$",C,dir(330)); label("$I$",I,NW); label("$D$",D,dir(280)); label("$E$",E,dir(210)); label("$M$",M,dir(250)); label("$O$",O,SE); label("$L$",L,dir(70)); label("$N$",N,dir(270)); label("$X$",X,dir(200)); label("$F$",F,dir(265)); [/asy][/asy] Let $X=\overline{LD}\cap\overline{MB}.$ Claim: $F=\overline{BC}\cap(BDE)$ lies on $(LNE).$ Proof. Notice $N$ is the reflection of $L$ in $\ell,$ the line through $O$ parallel to $\overline{BC}.$ Notice $$\measuredangle (\overline{FE},\overline{BC})=\measuredangle EFB+\measuredangle MBC=\measuredangle EDB+\measuredangle MAC=\measuredangle EDA+\measuredangle DAM=90$$so $F$ is the reflection of $E$ in $\ell.$ Hence, $FELN$ is a cyclic isosceles trapezoid. $\blacksquare$ Thus, $$\measuredangle EFX=\measuredangle EDB=\measuredangle EDL+\measuredangle LDA=\measuredangle EDL+\measuredangle LED=\measuredangle ELD.$$$\square$ Can anybody tell me what motivated to come up with the point F?