Suppose an acute scalene triangle $ABC$ has incentre $I$ and incircle touching $BC$ at $D$. Let $Z$ be the antipode of $A$ in the circumcircle of $ABC$. Point $L$ is chosen on the internal angle bisector of $\angle BZC$ such that $AL = LI$. Let $M$ be the midpoint of arc $BZC$, and let $V$ be the midpoint of $ID$. Prove that $\angle IML = \angle DVM$
Problem
Source: India TST 2023 Day 4 P1
Tags: geometry, TST
MrOreoJuice
09.07.2023 21:02
Let $N$ be the midpoint of arc $ABC$ containing $A$, I claim that $\triangle MIV \stackrel{-}{\sim} \triangle MNL$ which finishes because $$\angle DVM = \angle VMN = \angle IMN + \angle VMI = \angle IMN + \angle NML = \angle IML.$$Let $S \equiv MD \cap ZI \in (ABC)$ be the $A$ sharky devil point, by shooting lemma $MD \cdot MS = MB^2 = MI^2 \implies \angle MIV = \angle MID = \angle MSI = \angle MNL$, so it is enough to prove $\tfrac{MI}{IV} = \tfrac{MN}{NL}$. Note that $MANZ$ is a rectangle and the line perpendicular to $NZ$ through $L$ is also perpendicular to $AI$ so $NL = \tfrac{AI}{2}$. Now you can write $IV = \tfrac{ID}{2} = \tfrac{r}{2}$ and $AI$ in terms of $r$ and $\sin(\tfrac A2)$ and $MI = MB$ now write it in terms of $R$ and $\sin(\tfrac A2)$ by introducing circumcenter $O$ of $\triangle ABC$ with the help of $\triangle MOB$ and everything cancels out and you get the required ratio for similarity is true.
VicKmath7
09.07.2023 21:52
Let $ZL \cap (ABC)=S$. We have that $AI \parallel ZL \perp AN$ and $ID \parallel MS \perp BC$, so $\angle MIV=\angle MSL$. We want to prove that $\angle MVD=\angle MLZ=\angle IML$, so we are left to prove that $$\triangle MIV \sim \triangle MSL \iff IV \cdot MS=IM \cdot SL \iff IA \cdot IM=MS \cdot ID=2Rr, $$which is known to be the power of $I$ with respect to $(ABC)$, done.
MarkBcc168
10.07.2023 13:28
Let $N$ be the other midpoint of arc $BC$. Let $S$ be the Sharky Devil point (i.e., $MD\cap ZI$), and construct the rectangle $ANPI$. Then, notice that $$\left. \begin{array}{l} \measuredangle DIM = \measuredangle(\perp BC, AM) = \measuredangle MNZ \\[4pt] \measuredangle IMD = \measuredangle AMS = \measuredangle AZS = \measuredangle AZI = \measuredangle PMN \end{array} \right\} \implies \triangle IMD\sim\triangle NMP$$with corresponding midpoints $V$ and $L$. This implies the problem.
Krish230905
20.07.2023 06:32
Let the angle bisector of $\angle BZC$ meet $\odot ABC$ at $M'$. Let $X$ denote the midpoint of $AI$. Let $O$ denote the centre of $\odot ABC$. Thus, $AI \perp XL$, and $MOM'$ are collinear. Claim: $AMZM'$ is a parallelogram Proof: As $AO=OZ=OM=OM'$, so the diagonals bisect each other, which implies the claim. Claim: $AXLM'$ is a parallelogram Proof: $\angle AM'Z=\angle XAM'=\angle AXL=90^{\circ}$. So, $AM' \parallel XL$. By the previous claim, $AX \parallel M'L$ So, we get that $M'L=AX=\frac{AI}{2}$ Claim: $\triangle IVM \sim \triangle M'LM$ Proof: $\angle MM'Z=\angle MAZ=\angle MAC-\angle OAC=\angle BAM-\angle BAF=\angle FAM=\angle DIM$. Also, $$\frac{IV}{M'L}=\frac{r}{AI}$$where $r$ denotes the inradius. Then again, by the well known identity $AI . IM=2Rr$, we get, $$\frac{r}{AI}=\frac{IM}{2R}=\frac{IM}{MM'}$$These two facts imply the claim. Now, summing up, $\angle IML=\angle MLZ=\angle LM'M+\angle LMM'=\angle VIM+\angle VMI=\angle DVM$ $\blacksquare$
IAmTheHazard
09.10.2023 03:40
Let $N$ be the midpoint of arc $BAC$ (also the $M$-antipode), so $\overline{ZN}$ is the internal bisector of $\angle BZC$. Let $K$ be the reflection of $N$ over $L$, which is also the foot of the perpendicular from $I$ to $\overline{ZN}$. Since $\angle IML=\angle KLM$, it clearly suffices to show $\triangle IDM \sim \triangle NKM$. Let $H'$ be the second intersection of the $A$-altitude with $(ABC)$. Since $M$ is well-known to be the midpoint of arc $H'Z$, it follows that $\measuredangle MNK=\measuredangle H'AM=\measuredangle DIM$, so it suffices to show $\frac{ID}{IM}=\frac{NK}{NM}$. Since $AIKN$ is a rectangle, this is equivalent to $AI\cdot IM=2Rr$ where $r,R$ are the inradius and circumradius of $\triangle ABC$ respectively. This is well-known to be true, since $AI\cdot IM=\mathrm{Pow}_{(ABC)}(I)=OI^2-R^2$. $\blacksquare$