Problem: A basketball player has a season free throw average of \(70\%\) halfway through the season, and raises it to \(80\%\) by the end of the season.
Is it necessarily true that it was exactly \(75\%\) at some point? Why or why not?
Solution: Let \((70\%, x_1, \dots, x_n, 80 \%)\) be the free throw averages after every free throw starting from the season’s halfway mark until the end of the season.
Now suppose that at no point in time was the free throw average \(75 \%\). We know that there must exist some \(i\) such that \(x_i < 75\% < x_{i+1}\). Suppose that the number of free throws taken at point \(i\) is \(n_i\) and the number of successful throws taken are \(s_i\). Additionally, as the free throw average increases, we know that the free throw taken after the \(i\)th point must have been successful. Therefore,
\[x_i = \frac{s_i}{n_i} < \frac{3}{4} \text{ and } x_{i+1} = \frac{s_i + 1}{n_i + 1} > \frac{3}{4}\]
\[4s_i < 3n_i \text{ and } 4s_i + 4 > 3n_i + 3\]
\[4s_i < 3n_i \text{ and } 4s_i + 1 > 3n_i\]
This is not possible for natural numbers \(s_i\) and \(n_i\) which leads to a contradiction.
Therefore, the basketball player must always reach a point where their free throw average is \(75\%\).
Extension: For which integer triples \((a,b,c)\) can a basketball player go from a free throw average of \(a\) to \(c\) while avoiding a free throw average of \(b\).
Solution (Extension): First we look at the case where \(a < c\). Looking at the previous proof, we note that all \(b\) such that \(b = n/(n+1)\) for some natural number \(n\) are unavoidable. To prove that they are the only unavoidable numbers, suppose \(b = p/q\) where \(p + 1 \not = q\). Using the same logic previously
\[qs_i < pn_i \text{ and } qs_i + (p-q) > pn_i\]
If we have \(pn_i = qs_i + 1\), we see that both equations are satsified. We know that it is always possible to find \(n_i\) and \(s_i\) such that this equation holds. To see this, note that since \(p\) and \(q\) are co-prime, there exist multiple \(n_i\) such that \(pn_i \equiv 1 \pmod{q}\). Therefore, \(pn_i = qs_i + 1\). Therefore, to avoid \(b\), we note that we can first aim to reach \(n_i\) shots with \(s_i\) successful ones while making sure that all required successful shots happen at the end (to avoid crossing \(b\) before). Then we shoot one successful free throw to get to \((s_i + 1)/(n_i + 1) > b\) after which you can reach \(c\) by frontloading successful shots.