262. IMO Shortlist 2005, Geometry 6th.
by Virgil Nicula, Apr 7, 2011, 1:37 AM
Proposed problem. Let 
be an acute-angled triangle with 
. Let 
be the orthocenter of and let 
be the midpoint of the side ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
.
Let
and 
such that 
and 
. Prove that the line 
is perpendicular to the common chord of the circumscribed
circles of and 
.
Lemma. Let
be a circumcircle of acute , where 
. Denote the point 
. Define the points 
, 
so that
and 
. Then 
. Standard notation. 
- the ray without the point 
.
Lemma. Let
be an acute triangle. Define: the circumcircle 
and the orthocentre of the triangle ; the middlepoint of the side ;
the intersection
between 
and the bisector of the angle 
; 
and 
so that and 
. Then the point
belongs to the circumcircle
of the triangle 
.
A metrical proof.
and 
. Thus, 
. Denote 
, i.e. 
. Thus, 
and 
, i.e. the point belongs to the circumcircle of the triangle .
Proof of the proposed problem. Denote:
; the middlepoint 
of the segment ![$[AH]$](//latex.artofproblemsolving.com/0/3/b/03b8986ebe750b377f987f87b41a1dbc4c128e17.png)
; 
. From the above lemma results
. But 
(the point is the middlepoint of the segment ![$[HA']$](//latex.artofproblemsolving.com/b/d/d/bdd6284b56db3edb70c48d53b2cef3bbf7dfdb34.png)
), 
and 
. Thus, , 
, 
are the middlepoints of the segments
, 
, 
respectively and 
and 
.
All the problems from the Swiss Imo Selection Team 2006
be an acute-angled triangle with 
. Let 
be the orthocenter of and let 
be the midpoint of the side ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
.Let
and 
such that 
and 
. Prove that the line 
is perpendicular to the common chord of the circumscribedcircles of
and 
.Lemma. Let
be a circumcircle of acute , where 
. Denote the point 
. Define the points 
, 
so that
and 
. Then 
. Standard notation. 
- the ray without the point 
.Lemma. Let
be an acute triangle. Define: the circumcircle 
and the orthocentre of the triangle ; the middlepoint of the side ;the intersection
between 
and the bisector of the angle 
; 
and 
so that and 
. Then the point belongs to the circumcircle
of the triangle 
.A metrical proof.
and 
. Thus, 
. Denote 
, i.e. 
. Thus, 
and 
, i.e. the point belongs to the circumcircle of the triangle .Proof of the proposed problem. Denote:
; the middlepoint 
of the segment ![$[AH]$](http://latex.artofproblemsolving.com/0/3/b/03b8986ebe750b377f987f87b41a1dbc4c128e17.png)
; 
. From the above lemma results
. But 
(the point is the middlepoint of the segment ![$[HA']$](http://latex.artofproblemsolving.com/b/d/d/bdd6284b56db3edb70c48d53b2cef3bbf7dfdb34.png)
), 
and 
. Thus, , 
, 
are the middlepoints of the segments
, 
, 
respectively and 
and 
.All the problems from the Swiss Imo Selection Team 2006
This post has been edited 11 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:49 AM
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