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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.

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    $\begingroup$ For a triangle to be acute I think it would be sufficient to show that at least 2 of the angles are acute. Note that any triangle has at least two acute angles. $\endgroup$ – Arnaud Mortier Aug 24 '18 at 7:31
  • $\begingroup$ @YvesDaoust And at the same time not in the square with diagonal given by that base. $\endgroup$ – Arthur Aug 24 '18 at 8:40
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    $\begingroup$ @Arthur That's not true... At the same time the third point should be in the slab orthogonal to the base and not in the circle with the diameter give by the base. $\endgroup$ – Oldboy Aug 24 '18 at 10:53
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    $\begingroup$ This problem is very similar to math.stackexchange.com/questions/1351421/… $\endgroup$ – Oldboy Aug 24 '18 at 11:07
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    $\begingroup$ The linked solution gives an upper bound of 3/4 for 4 points and 7/10 for 5 points. Following it through, it seems to give an upper bound of about $\frac23(1+1/N^2)$ for N points. $\endgroup$ – Empy2 Aug 24 '18 at 12:36
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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.

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    $\begingroup$ U can get more acute triangles by using other combinations, such as by better optimising the triangles between the 64 arc points. $\endgroup$ – A. Random Aug 24 '18 at 11:20
  • $\begingroup$ Maybe. As it stands, I think all triangles with three points on the circle will be obtuse $\endgroup$ – Empy2 Aug 24 '18 at 11:22
  • $\begingroup$ (+1) for a constructive first guess. $\endgroup$ – Jens Aug 24 '18 at 16:17

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