IMO Shortlist 1969 problem 57
Dodao/la:
arhiva2. travnja 2012. Given triangle
![ABC](/media/m/a/c/7/ac75dca5ddb22ad70f492e2e0a153f95.png)
with points
![M](/media/m/f/7/f/f7f312cf6ba459a332de8db3b8f906c4.png)
and
![N](/media/m/f/1/9/f19700f291b1f2255b011c11d686a4cd.png)
are in the sides
![AB](/media/m/5/2/9/5298bd9e7bc202ac21c423e51da3758e.png)
and
![AC](/media/m/6/4/7/647ef3a5d68f07d59d84afe03a9dc655.png)
respectively.
If
![\dfrac{BM}{MA} +\dfrac{CN}{NA} = 1](/media/m/6/f/e/6feadfc61c3d1eb39a4772a5c61bc924.png)
, then prove that the centroid of
![ABC](/media/m/a/c/7/ac75dca5ddb22ad70f492e2e0a153f95.png)
lies on
![MN](/media/m/2/6/7/267a73297a5de9e529d41774ee6ff45a.png)
.
%V0
Given triangle $ABC$ with points $M$ and $N$ are in the sides $AB$ and $AC$ respectively.
If $\dfrac{BM}{MA} +\dfrac{CN}{NA} = 1$ , then prove that the centroid of $ABC$ lies on $MN$ .
Izvor: Međunarodna matematička olimpijada, shortlist 1969