How many acute triangles can be formed by 100 points in a plane? Given 100 points in the plane, no three of which are on the same line, consider all triangles that have all their vertices chosen from the 100 given points.
Prove that at most 70% of those triangles are acute-angled.

The proof that for $N \geq 5$ the fraction of acute triangles cannot exceed $\frac{7}{10}$ involves geometric reasoning only for the $N=5$ case, followed by combinatoric reasoning to create a proof by induction.
We start from Lemma 1:  If there exists an arrangement of $N$ points (with $N>5$) forming more than $70\%$ acute triangles (that is, more than $\frac{7}{60}N(N-1)(N-2)$ acute triangles), then there exists an arrangement of $N-1$ points forming more than $70\%$ acute triangles (that is, more than $\frac{7}{60}(N-3)(N-1)(N-2)$ acute triangles).  
Proof:  Start with an arrangement forming the minimum number of acute triangles at or above $70\%$, namely 
$$
\left \lceil 
\frac{7}{60}N(N-1)(N-2)
\right \rceil
$$ acute triangles.  These contain 
$
3 \left \lceil 
\frac{7}{60}N(N-1)(N-2)
\right \rceil
$ instances of vertices in acute triangles; then one of the $N$ points must participate in at most 
$\left \lfloor
\frac{3}{N} \left \lceil 
\frac{7}{60}N(N-1)(N-2)
\right \rceil \right \rfloor
$ acute triangles.  Remove that point; among the remaining points there are 
at least 
$$
3 \left \lceil 
\frac{7}{60}N(N-1)(N-2)
\right \rceil - \left \lfloor \frac{3}{N} \left \lceil 
\frac{7}{60}N(N-1)(N-2)
\right \rceil \right \rfloor
$$ remaining vertices in acute triangles.  Thus there are at least
$$
\left \lceil
\frac{
3 \left \lceil 
\frac{7}{60}N(N-1)(N-2)
\right \rceil - \left \lfloor \frac{3}{N} \left \lceil 
\frac{7}{60}N(N-1)(N-2)
\right \rceil \right \rfloor
}{3}
\right \rceil
$$
acute triangles.  Assume worst case in every round-off; then we can write this as being at least
$$
\frac{7}{60}(N-3)(N-1)(N-2) + 2\cdot \frac{7}{60}(N-1)(N-2)-\frac{1}{3}
$$
which for $N \>5$ always exceeds $70\%$ of the triangles among the $N-1$ vertices. 
Lemma 2:  If for any $N>5$ you can form more than $70\%$ acute triangles, then you can form more than $70\%$ acute triangles (that is, at least 8) among five vertices.  Proof:  Start from $N$ and repeatedly apply lemma 1.  
Lemma 3:  Among five points, there cannot be more than seven acute triangles.
The proof starts with thew geometric observation that among four points, at most three one of the four triangles can be acute. Now for the five vertices, if you can form eight acute triangles, then either the remaining two triangles share 2 vertices or just one.  Assume they share two vertices, $A$ and $B$, and that the third vertices in the triangles are labeled $C$ and $D$. Then four remaining (acute) triangles do not contain vertex $B$, and thus form four acute triangles sharing four vertices, which is impossible.  If the two non-acute triangles share only one vertex, say $ABC$ and $ADE$, then the four triangles not involving vertex $A$ are acute, which again leads to the same impossibility.
Theorem:  For $N > 5$ at most $70\%$  of the triangles formed by $N$ non-co-linear points can form acute triangles.  Proof:  Lemma 5 establishes a base for the inductive process implied by lemma 2.
Observation:  In fact, among $100$ points you can have no more than $66.675\%$ acute triangles (the actual maximum may be a lot less than that).  This is obtained by starting with the assumption of 107812 acute triangles, and iterating the combinatoric reasoning in lemma 1 until you find that for some set of five points there must be at least 8 acute triangles.
