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that's for Kayra .
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a b c d e f g h i j k l m n o Running through the 15 vertices, identify all triangles in which each one appears. Since each triangle has three vertices, each triangle will be counted 3 times, so we will divide the answer by three. By symmetry, a,k,o are the same; b,c,g,l,j,n are the same. d,f,m are midpoints and also the same. e,h,i are also the same, interior points. Therefore the number will be 3 times the a triangles plus 6 times the b triangles, plus 3 times the d plus 3 times the e (3+6+3+3=15 points), all divided by 3. In other words the total is a+2b+d+e. a=4; b=5; d=6;e=7? 4+10+6+7= 27 老鬼
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giving the answer is not exactly a 'spoil' kayra. What I wanted was to find a systematic approach, and then describe it. You still have to find your own approach and come up with the right answer even if you know what the answer is. Having found a method that works, one then naturally asks will it work if we extend (or reduce) the problem. I'll call this 'a pyramid of equilateral triangles, height h', where h=4 in this case. It works for h=1 being 3/3, h=2 giving 2+3=5, for h=3 I see there is only 1 internal point, so the answer is (3*3+6*4+1*6)/3=13. For h>4 different answers depend on odd or even h, and there will be recursion, so a general formula will be non-trivial. I am sure the answer can be found in Wolfram or similar place 
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Your traditional method included elimination of triangles already counted so you included ABC=CAB=BCA in your 'algorithm'. But you did indeed use an algorithm.
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It will be at most a cubic polynomial, so if I solve a -1)^3+b(x-1)^2+c(x-1)+d using the values for x=1,2,3,4 (1,5,13,27) I get a=1/3, b=d=1,c=8/3So for x=20, 1/3*19^3+19^2+8/3*19+1=2699 two six nine nie - Chinese FriendFinder interferes with numbers
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that simplifies to 1/3(x^3+5x-3) and note that x^2+5) is always divisible by 3
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that simplifies to 1/3(x^3+5x-3) and note that x^2+5 is always divisible by 3
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No， on review a degree 3 polynomial doesn't do it...
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A002717 a(n) = floor(n(n+2)(2n+1) / 8 ) = 2255 = 55*41 when n=20
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A002717 a(n) = floor(n(n+2)(2n+1) / 8 ) = 2255 = 55*41 when n=20 As I mentioned, it is different for odd and even h, so perhaps there is a cubic polynomial for even n - needs to solve 512a+64b+8c+d=170 216a+36b+6c+d=78 64a+16b+4c+d=27 8a+4b+2c+d=5 but I cannot be bothered...
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