IMO Shortlist 1982 problem 5
Dodao/la:
arhiva2. travnja 2012. The diagonals
![AC](/media/m/6/4/7/647ef3a5d68f07d59d84afe03a9dc655.png)
and
![CE](/media/m/3/3/7/33771853199c7fd8dc7faa2d4a37425d.png)
of the regular hexagon
![ABCDEF](/media/m/9/f/e/9fe205b534135e3a700ffb54d8b96cb0.png)
are divided by inner points
![M](/media/m/f/7/f/f7f312cf6ba459a332de8db3b8f906c4.png)
and
![N](/media/m/f/1/9/f19700f291b1f2255b011c11d686a4cd.png)
respectively, so that
![{AM\over AC}={CN\over CE}=r.](/media/m/e/3/2/e321311b1bf3a21e3cf45f793dee1d42.png)
Determine
![r](/media/m/3/d/f/3df7cc5bbfb7b3948b16db0d40571068.png)
if
![B,M](/media/m/c/5/f/c5fd4d43db5dd27e688710fb1fad3cb8.png)
and
![N](/media/m/f/1/9/f19700f291b1f2255b011c11d686a4cd.png)
are collinear.
%V0
The diagonals $AC$ and $CE$ of the regular hexagon $ABCDEF$ are divided by inner points $M$ and $N$ respectively, so that $${AM\over AC}={CN\over CE}=r.$$ Determine $r$ if $B,M$ and $N$ are collinear.
Izvor: Međunarodna matematička olimpijada, shortlist 1982