$I$ is the incenter of a given triangle $\triangle ABC$. The angle bisectors of $ABC$ meet the sides at $D,E,F$, and $EF$ meets $(ABC)$ at $L$ and $T$ ($F$ is on segment $LE$.). Suppose $M$ is the midpoint of $BC$. Prove that if $DT$ is tangent to the incircle of $ABC$, then $IL$ bisects $\angle MLT$.
Problem
Source: 2020 Korean MO winter camp Test 1 P4
Tags: geometry
kjy1102
07.09.2020 19:03
I will use my old post's result...
https://artofproblemsolving.com/community/u295255h1778706p11708415
Let the excenter of $A,B,C$ is $I_a,I_b,I_c$, and $X=O\cap I_a$(which is the left side of the $AI$) First well-known property, $(L,T,D,M)$ is cyclic, $(I_c,L,I,T,I_b)$ is concyclic, and let the center is $O'$ Step 1) $D,X,T$ is collinear
By upper post, we can get $XT$ is common tangent line of $O$
Since $D$ is inner similarity point of $I$ and $I_a$ and $DT$ is tangent to $I$, we can get $D,X,T$ is collinear.
Step 2) $TI//I_bI_c$
Since $XD$ is common tangent line of $I$ and $I_a$, $AX$ is isogonal to Nagel Line of $A$.
By well known property, we can get $X$ be the mixtillnear tangent point of $ABC$.
So if $XI\cap O=N$, and $N$ is the midpoint of arc $ABC$.
Let $I'\in XI$ and $IN=NI'$, then $I'I\times IX=AI\times II_a $, $(I',A,X,I_a)$ is cyclic
By angle chasing, we can get $\angle I'AN=\angle TAN$, which implies $AN//IT$.
By step 1 and step 2, $\angle MLT=\angle MDT=2\angle DIU (U=I\cap XT)=2\angle ITD=2\angle TLI \blacksquare$.
Attachments:
p_square
08.09.2020 18:48
Nice question! Unfortunately, a lot of it is a configuration I'd seen in the past: The one related to Serbia 17/6 Here's an outline: Let $AR$ be the altitude 1) Since $LTMD$ cyclic, once can chase a bit to get that angle $\angle MLT = \angle MDT = 2\angle RAI$ 2) Let the $A-$excircle meet $(ABC)$ at $P,Q$ and let the excircle be tangent to $BC$ at $W$. Homothety from $D$ implies that $TD$ is tangent to the $A-$excircle. It's again known that the other tangent from $T$ to the excircle is $TP$ 3) Again, known that $PWQ$ is homothetic to $TIL$, so $DT$ is tangent to $LIT$ 4) $\angle ILT = \angle PTI = \angle TPW = \angle DPW = \angle DI_AW = \angle RAI = \frac{\angle MLT}{2}$ The solution appears short but that's only because I omitted the proof of these 'known' lemmas; 1)$LTMD$ is cyclic 2)$PWY$ is similar to $LIT$ 3)$T,L$ are where common external tangents of $A-$excircle and $(ABC)$, meet $(ABC)$ 4)$TP$ is a tangent I might post the proofs of these later for completeness
BALAMATDA
07.09.2023 07:45
kjy1102 wrote: I will use my old post's result...
https://artofproblemsolving.com/community/u295255h1778706p11708415
Let the excenter of $A,B,C$ is $I_a,I_b,I_c$, and $X=O\cap I_a$(which is the left side of the $AI$) First well-known property, $(L,T,D,M)$ is cyclic, $(I_c,L,I,T,I_b)$ is concyclic, and let the center is $O'$ Step 1) $D,X,T$ is collinear
By upper post, we can get $XT$ is common tangent line of $O$
Since $D$ is inner similarity point of $I$ and $I_a$ and $DT$ is tangent to $I$, we can get $D,X,T$ is collinear.
Step 2) $TI//I_bI_c$
Since $XD$ is common tangent line of $I$ and $I_a$, $AX$ is isogonal to Nagel Line of $A$.
By well known property, we can get $X$ be the mixtillnear tangent point of $ABC$.
So if $XI\cap O=N$, and $N$ is the midpoint of arc $ABC$.
Let $I'\in XI$ and $IN=NI'$, then $I'I\times IX=AI\times II_a $, $(I',A,X,I_a)$ is cyclic
By angle chasing, we can get $\angle I'AN=\angle TAN$, which implies $AN//IT$.
By step 1 and step 2, $\angle MLT=\angle MDT=2\angle DIU (U=I\cap XT)=2\angle ITD=2\angle TLI \blacksquare$. Could you show me why $\angle ITD=\angle TLI$?
foolish07
18.08.2024 15:04
BALAMATDA wrote: kjy1102 wrote: I will use my old post's result...
https://artofproblemsolving.com/community/u295255h1778706p11708415
Let the excenter of $A,B,C$ is $I_a,I_b,I_c$, and $X=O\cap I_a$(which is the left side of the $AI$) First well-known property, $(L,T,D,M)$ is cyclic, $(I_c,L,I,T,I_b)$ is concyclic, and let the center is $O'$ Step 1) $D,X,T$ is collinear
By upper post, we can get $XT$ is common tangent line of $O$
Since $D$ is inner similarity point of $I$ and $I_a$ and $DT$ is tangent to $I$, we can get $D,X,T$ is collinear.
Step 2) $TI//I_bI_c$
Since $XD$ is common tangent line of $I$ and $I_a$, $AX$ is isogonal to Nagel Line of $A$.
By well known property, we can get $X$ be the mixtillnear tangent point of $ABC$.
So if $XI\cap O=N$, and $N$ is the midpoint of arc $ABC$.
Let $I'\in XI$ and $IN=NI'$, then $I'I\times IX=AI\times II_a $, $(I',A,X,I_a)$ is cyclic
By angle chasing, we can get $\angle I'AN=\angle TAN$, which implies $AN//IT$.
By step 1 and step 2, $\angle MLT=\angle MDT=2\angle DIU (U=I\cap XT)=2\angle ITD=2\angle TLI \blacksquare$. Could you show me why $\angle ITD=\angle TLI$? $XT$가 $\odot (ILT)$의 접선이라서요