Berkeley Math Circle

Solution:

Consider the set of all angles $M_{1} M_{2} M_{3}$, where $M_{1}$, $M_{2}$ and $M_{3}$ is an arbitrary triple of selected points. There are $\frac{16 \cdot 15 \cdot 14}{2} = 1680$ different angles in this set. Suppose $n$ of them are not acute. We shall prove $n \geq 392$.

For each integer $m$ between 1 and 7, take a chord with endpoints among the selected points such that there are exactly $m$ selected points to one side of the chord (not including the endpoints). We will call such a chord an $m$-chord. Each $m$-chord subtends not less than $m$ nonacute angles with vertices among the marked points. For each $m \leq 6$ there are exactly $16$ $m$-chords, and for $m=7$ there are exactly $8$ of them. So the total number of nonacute angles is at least $16(1+2+\ldots+6) + 8 \times 7 = 392$.

There are $\frac{16 \cdot 15 \cdot 14}{6} = 560$ triangles with vertices among the marked points. Each nonacute angle will "spoil" exactly one triangle, so the number of acute triangles is not greater than $560 - 392 = 168$.

It is only left to construct an example with exactly 168 acute triangles. Mark eight consecutive vertices of a right 16-gon $V_{1}, \ldots, V_{8}$. Draw a line through the center of the 16-gon not parallel to $V_{1} V_{8}$, such that all $V$'s lie on the same side of that line (and not on the line). Reflecting the $V$'s with respect to that line we get points $V_{1}' , \ldots, V_{8}'$. We claim that the set $V_{1}, \ldots, V_{8}, V_{1}', \ldots, V_{8}'$ is as wanted. Indeed, there are no diametrically opposite points in this set (otherwise we would get $V_{1} V_{1}' = 0$ or $V_{8} V_{8}' = 0$), and so for $m=7$ each $m$-chord subtends exactly 8 nonacute angles. Moreover, for each $m \leq 6$ each $m$-chord subtends exactly $m$ nonacute angles, and so $n=392$, and the number of acute triangles is 168.