Find the maximum percentage of acute angled triangles in a plane with $100$ points ; no $3$ of which are collinear. In a plane there are $100$ points, no three of which are collinear. Consider all possible triangles having these points as vertices. Find the maximum percentage of these triangles which are acute-angled.
I know that the total number of triangles formed will be $100 \choose 3$ , since no $3$ points are collinear. For a triangle to be acute I think it would be sufficient to show that the sum of the 2 smallest angles is more than $π/2$. So we would have to optimise the points to maximize such triangles. Directly finding an upper bound on the number of such triangles might also prove useful. I am unable to see if there is a simple way of solving this problem.
 A: The following makes more than $5/9$ of the triangles acute.  For 100 points there are 90915 acute triangles, or 56%.
• It is based on an acute isosceles triangle $\Delta ABC$ with angle $\angle ABC$ very close to a right angle. Legs AB and BC have length 1.
• Given $3N$ points, place $N$ near each vertex of triangle ABC.  Points near A lie on a circle centre B; points near B lie on a circle centre C; and points near C lie on a circle centre A.
• The acute triangles are $A_iB_jC_k, A_iA_jB_k, B_iB_jC_k$ and $C_iC_jA_k$.  That is a total of $(5N^3-3N^2)/2$ out of $3N\choose3$.
• Angle ABC is $90^\circ-N^{-7}$
• $A_i$ are equally spaced on an arc of width $N^{-13}$
• $B_j$ are equally spaced on an arc of width $N^{-9}$ ( and height $N^{-18}$ )
• $C_k$ are equally spaced on an arc of width $N^{-11}$
• The region of points P where $\angle PA_iA_j$ are all acute is a diamond of width $N^{-14}$ and length $N^{-1}$.  This covers the $B_j$.  
EDIT (an older solution)  
With $N$ points, the following gives $(2N^3-3N^2-2N)/24$ acute triangles for even $N$ and $(2N^3-3N^2-2N+3)/24$ acute triangles for odd $N$.  This is more than 50 percent for all $N$.
Start with points A and B.  Draw a circle through A with centre B, and a circle through B with centre A.
Place points A1 to A50 on the first circle, with Ak an angle $30^o/9^k$ clockwise from A.  Place points B1 to B50 on the second circle, with Bk an angle $10^o/9^k$ anticlockwise from B.
A1 is a base point of $49×50$ acute triangles, B1 a base point of $49×49$ more; A2 of $48×49$ more, and so on. The total for 100 points is 82075.   
