IMO Shortlist 2002 problem G3
Dodao/la:
arhiva2. travnja 2012. The circle
![S](/media/m/c/6/3/c63593c3ec0773fa38c2659e08119a75.png)
has centre
![O](/media/m/9/6/0/9601b72f603fa5d15addab9937462949.png)
, and
![BC](/media/m/5/0/0/5005d4d5eac1b420fbabb76c83fc63ad.png)
is a diameter of
![S](/media/m/c/6/3/c63593c3ec0773fa38c2659e08119a75.png)
. Let
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
be a point of
![S](/media/m/c/6/3/c63593c3ec0773fa38c2659e08119a75.png)
such that
![\angle AOB<120{{}^\circ}](/media/m/2/3/e/23e0af0a0f7fd8e9762b70db7adeff52.png)
. Let
![D](/media/m/7/0/0/7006c4b57335ab717f8f20960577a9ef.png)
be the midpoint of the arc
![AB](/media/m/5/2/9/5298bd9e7bc202ac21c423e51da3758e.png)
which does not contain
![C](/media/m/5/a/b/5ab88f3f735b691e133767fe7ea0483c.png)
. The line through
![O](/media/m/9/6/0/9601b72f603fa5d15addab9937462949.png)
parallel to
![DA](/media/m/a/0/8/a081cf3dbb7eaedd62ea487a4cd46956.png)
meets the line
![AC](/media/m/6/4/7/647ef3a5d68f07d59d84afe03a9dc655.png)
at
![I](/media/m/3/8/6/38689d6affa9ba35368ca4d3d76ea147.png)
. The perpendicular bisector of
![OA](/media/m/b/2/0/b206c115fb0e114a37cf644cba5338cb.png)
meets
![S](/media/m/c/6/3/c63593c3ec0773fa38c2659e08119a75.png)
at
![E](/media/m/8/b/0/8b01e755d2253cb9a52f9e451d89ec11.png)
and at
![F](/media/m/3/e/8/3e8bad5df716d332365fca76f53c1743.png)
. Prove that
![I](/media/m/3/8/6/38689d6affa9ba35368ca4d3d76ea147.png)
is the incentre of the triangle
%V0
The circle $S$ has centre $O$, and $BC$ is a diameter of $S$. Let $A$ be a point of $S$ such that $\angle AOB<120{{}^\circ}$. Let $D$ be the midpoint of the arc $AB$ which does not contain $C$. The line through $O$ parallel to $DA$ meets the line $AC$ at $I$. The perpendicular bisector of $OA$ meets $S$ at $E$ and at $F$. Prove that $I$ is the incentre of the triangle $CEF.$
Izvor: Međunarodna matematička olimpijada, shortlist 2002