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 ∠BZC such that AL=LI. Let M be the midpoint of arc BZC, and let V be the midpoint of ID. Prove that ∠IML=∠DVM
Problem
Source: India TST 2023 Day 4 P1
Tags: geometry, TST
09.07.2023 21:02
09.07.2023 21:52
10.07.2023 13:28
Let N be the other midpoint of arc BC. Let S be the Sharky Devil point (i.e., MD∩ZI), and construct the rectangle ANPI. Then, notice that ∡DIM=∡(⊥BC,AM)=∡MNZ∡IMD=∡AMS=∡AZS=∡AZI=∡PMN}⟹△IMD∼△NMPwith corresponding midpoints V and L. This implies the problem.
20.07.2023 06:32
Let the angle bisector of ∠BZC meet ⊙ABC at M′. Let X denote the midpoint of AI. Let O denote the centre of ⊙ABC. Thus, AI⊥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: ∠AM′Z=∠XAM′=∠AXL=90∘. So, AM′∥XL. By the previous claim, AX∥M′L So, we get that M′L=AX=AI2 Claim: △IVM∼△M′LM Proof: ∠MM′Z=∠MAZ=∠MAC−∠OAC=∠BAM−∠BAF=∠FAM=∠DIM. Also, IVM′L=rAIwhere r denotes the inradius. Then again, by the well known identity AI.IM=2Rr, we get, rAI=IM2R=IMMM′These two facts imply the claim. Now, summing up, ∠IML=∠MLZ=∠LM′M+∠LMM′=∠VIM+∠VMI=∠DVM ◼
09.10.2023 03:40
Let N be the midpoint of arc BAC (also the M-antipode), so ¯ZN is the internal bisector of ∠BZC. Let K be the reflection of N over L, which is also the foot of the perpendicular from I to ¯ZN. Since ∠IML=∠KLM, it clearly suffices to show △IDM∼△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 ∡MNK=∡H′AM=∡DIM, so it suffices to show IDIM=NKNM. Since AIKN is a rectangle, this is equivalent to AI⋅IM=2Rr where r,R are the inradius and circumradius of △ABC respectively. This is well-known to be true, since AI⋅IM=Pow(ABC)(I)=OI2−R2. ◼