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%I #13 Dec 07 2015 01:03:01
%S 6,12,15,18,20,21,24,28,30,35,36,40,42,44,45,48,52,54,56,60,63,65,66,
%T 70,72,75,77,78,80,84,85,88,90,91,95,96,99,100,104,105,108,110,112,
%U 117,119,120
%N Areas of triangle ABC, if it can be split by two straight lines through A and B, into 4 parts all with integer areas.
%F n = a+b+c+d, if d = b*c*(2*a+b+c)/(a^2-b*c) is positive integer.
%e For n=6, some triangle with that area can be divided by 2 straight lines through A and B, into 4 parts with areas (2,1,1,2) or with areas (3,1,1,1). A triangle with area 12 can be divided into parts (2,1,2,7), (3,1,3,5), (4,2,2,4) and (6,2,2,2). Triangles with area 13 or 14 cannot be divided in this way.
%K nonn,more
%O 1,1
%A _Dragan Krejakovic_, Mar 01 2012
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