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Shaq starts the season shooting below 50% from the free throw line. He finishes the season shooting above 80%. Was there necessarily a moment when his free throw percentage was exactly 75%?

What Makes This Problem Non-Trivial

Let’s say we asked whether there was ever a point where the FT% was 60%. Consider this sequence of events.

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$$\dfrac{0}{1} \to \dfrac{1}{2} \to \dfrac{2}{3} \to \dfrac{2}{4} \to \dfrac{2}{5} \to \dfrac{3}{6} \to \dfrac{4}{7} \to \dfrac{5}{8} = 0.625$$
There was never a point where it was 60%. It went from $57.14\% \to 62.5\%$. Are we guaranteed that this doesn't happen for $75\%$?

The Solution

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The only way for $0.75$ to be skipped would be to be at some point below $75\%$, make a free throw, and be able $75\%$. We can make this inequality
$$\dfrac{a}{b} \lt \dfrac{3}{4} \lt \dfrac{a + 1}{b + 1}$$
a and b are integers, representing the number of makes and total shots, respectively. We can split this up into two inequalities
$$a \lt \dfrac{3}{4}b \text{ and } \dfrac{3}{4}b + \dfrac{3}{4} \lt a + 1$$
$$a \lt \dfrac{3}{4}b \to 4a \lt 3b$$
$$\dfrac{3}{4}b + \dfrac{3}{4} \lt a + 1 \to 3b + 3 \lt 4a + 4$$
$$\to 3b \lt 4a + 1$$
We now combine this into one equation
$$4a \lt 3b \lt 4a + 1$$
However, this means that there exists an integer between two other integers (4a, 4a + 1). This is a contradiction. Therefore, our initial assumption that there would be a point right below and right above $75\%$ is not true. $$\boxed{\text{There must exist a point that is exactly } 75\%}$$ In fact this works for any $\frac{n}{n+1}$. Rather than doing a proof almost identical to the one above, here is a quick justification for it.

Let's think about the free throw percentage in a new way. Instead of $\dfrac{\text{makes}}{\text{total shots}}$, let's do the ratio of makes to misses. So, $\dfrac{2}{5}$ becomes $\dfrac{2}{3}$. Notice that if the FT% is $\dfrac{n}{n+1}$, this ratio is just n.

Let's recreate the scenario from above. We have a situation right below $\dfrac{3}{4}$. We missed k times. Our ratio of makes to misses is currently $\dfrac{3k - 1}{k}$ If we make another shot, our ratio becomes $\dfrac{3k}{k} = 3$, which is equivalent to $\dfrac{3}{4}$.