Let $ABC$ be an acute triangle with the circumcircle $k$ and incenter $I$. The perpendicular through $I$ in $CI$ intersects segment $[BC]$ in $U$ and $k$ in $V$. In particular $V$ and $A$ are on different sides of $BC$. The parallel line through $U$ to $AI$ intersects $AV$ in $X$. Prove: If $XI$ and $AI$ are perpendicular to each other, then $XI$ intersects segment $[AC]$ in its midpoint $M$. (Notation: $[\cdot]$ denotes the line segment.)
Problem
Source: Germany VAIMO 2015 - #5
Tags: geometry, circumcircle, incenter, perpendicular, orthogonal
huynguyen
11.07.2015 17:48
Are you sure it's true?
TelvCohl
11.07.2015 18:07
My solution : Let $ T \equiv XU \cap AB $ and $ Y \equiv IX \cap AB $ . From $ \angle BTU=\angle BAI=\tfrac{1}{2} \angle BAC=\angle BIC-90^{\circ}=\angle BIU \Longrightarrow B, I, T, U $ are concyclic , so $ I $ is the midpoint of arc $ TU $ in $ \odot (BTU) \Longrightarrow IX $ is the perpendicular bisector of $ TU \Longrightarrow XT=XU $ , hence $ \frac{VA}{VX}=\frac{AI}{XU}=\frac{AI}{XT}=\frac{YA}{YT} \Longrightarrow YV \parallel TU \parallel AI \Longrightarrow \angle BYV=\angle BIV \Longrightarrow V \in \odot (BIY) $ . From $ \angle VBI=\angle VYI=\angle AIX=90^{\circ} \Longrightarrow V $ is the midpoint of arc $ AC $ (contain $ B $) in $ \odot (ABC) $ , so $\angle VMC=90^{\circ} \Longrightarrow M \in \odot (CIV) \Longrightarrow \angle VIM=180^{\circ}-\angle ACV=\angle VBA=180^{\circ}-\angle YIV $ , hence we get $ Y, I, M $ are collinear $ \Longrightarrow M \in XI \Longrightarrow YI \equiv XI $ passes through the midpoint $ M $ of $ CA $ . Q.E.D ____________________________________________________________ P.S. For more properties in this configuration you can see 2014 IMO Shortlist G7
Mikeglicker
17.01.2021 14:28
sorry to bump such an old question but I think this statement is just true without ever being true because there is no diagram satisfying this. We would like to prove that if I is incener of ABC and M is middle of AC then there cant be MI perpendicular to AI. assume it does then let Mi meet AB at D and let CI meet AB at E and the perpendicular throught I to AC meet AB at F. Now if we project the harmonic bundle A,M,C andpoint at infiniti through point E to line AB we get that A,D,E,F is also a harmonic bundle. Now it is well known that the harmonic conjugate of AED must satisfy that angles FID=EID but we claim that F and E both lies on the same side of D on line AB (the side closer to A) which makes the angles equality impossible. First because CAB is a sharp angle we get AFI=180 - CAB>90>90 - IAD=ADI so F is on segment AD. let a,b,c be the angles of the triangle then AEI=180-a-c/2 and ADI = 90-a/2 and by the inequality a+c<180 it is easy to verify that AEI>ADI so E is on segment AD as well. Can you please tell me wether this explanation is actually right, if someone can find mistakes I have made or verify that my solution is indeed right I will appreciate it.