AoPS - Problem

Source: India TST 2023 Day 4 P1

Tags: geometry, TST

Pranav1056

09.07.2023 17:27

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 $\angle IML = \angle DVM$

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 $AI\cdot IM=\mathrm{Pow}_{(ABC)}(I)=OI^2-R^2$ . $\blacksquare$