One marks 16 points on a circle. What is the maximum number of acute triangles with vertices in these points?
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 $16(1+2+\ldots+6)+8 * 7=392$.

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 $V^{\prime} s$ with respect to that line we get points $V_{1}{ }^{\prime}, \ldots V_{8}{ }^{\prime}$. We claim that the set $V_{1}, \ldots V_{8}, V_{1}{ }^{\prime} \ldots V_{8}{ }^{\prime}$ is as wanted. Indeed, there are no diametrically opposite points in this set (otherwise we would get $V_{1} V_{1}{ }^{\prime}=0$ or $V_{8} V_{8}{ }^{\prime}=0$ ), 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 .