Let $ABC$ be an acute-angled triangle with height intersection $H$. Let $G$ be the intersection of parallel of $AB$ through $H$ with the parallel of $AH$ through $B$. Let $I$ be the point on the line $GH$, so that $AC$ bisects segment $HI$. Let $J$ be the second intersection of $AC$ and the circumcircle of the triangle $CGI$. Show that $IJ = AH$
Problem
Source: Switzerland - Swiss MO 2016 p8
Tags: geometry, orthocenter, equal segments
dgreenb801
16.07.2020 08:29
I'm confused about the problem, isn't the intersection $J$ of $AC$ with the triangle $CGI$ just the same as the intersection of $HI$ with $AC$, which is the midpoint of $HI$ (since $AC$ bisects $HI$)? Could you draw a diagram?
WolfusA
16.07.2020 08:44
I drew circle centered at $A$ with radius $IJ$ and it doesn't pass through $H$!
Attachments:
trumpeter
16.07.2020 09:44
It is 2015 ISL G1.
WolfusA
16.07.2020 23:22
Thank you trumpeter. Now everything became clear. parmenides51 wrote: Let $J$ be the second intersection of $AC$ and the perimeter of the triangle $CGI$. Meanwhile at IMO SL 2015 G1. Problem_Penetrator wrote: Suppose that the line $AC$ intersects the circumcircle of the triangle $GCI$ at $C$ and $J$.
parmenides51
16.07.2020 23:30
oops, google translate betrayed me, , I corrected the perimeter instead of circumcircle in 2 more problems from the same source (2016 p4, 2018 p4 )