The following was a puzzle presented to Marshwood GT students on April 26, 2019. Have fun doing math!

In pre-calculus, for ellipses and hyperbolas, \(a\) is often defined as half the distance from vertex to vertex while \(c\) is defined as half the distance from focus to focus. It is commonly known that \(a > c\) is true for ellipses while \(a < c\) is true for hyperbolas. However, while this seems obvious if you see some pictures of ellipses and hyperbolas, most pre-calc students are never shown the formal derivation for this fact, even though it is an inherent and necessary assumption behind the famous derivations of the ellipse and hyperbola formulas. Therefore, in today's short puzzle, we're going to prove that \(a > c\) for ellipses and \(a < c\) for hyperbolas.

**Part 1**: Let \(F_1\) and \(F_2\) be the foci of some ellipse and let \(P\) be a point on that ellipse. Now, consider the triangle between \(P\), \(F_1\), and \(F_2\). Obviously, this triangle has three sides: \(F_1F_2\), \(PF_1\), and \(PF_2\). According to Euclid, the sum of any two sides is greater than the third side, so we get the following inequalities:

Now, look at these inequalities closely. Using what you know about ellipses, use one of the above inequalities to show \(a > c\). Your proof should not be more than 2-3 steps.

**Part 2**: Similarly, let \(F_1\) and \(F_2\) be the foci of some hyperbola and let \(P\) be a point on that hyperbola. Like before, we are going to consider the triangle with sides \(F_1F_2\), \(PF_1\), and \(PF_2\). However, this time, we are going to use a slightly different form of the triangle inequality: The difference between any two sides is less than the third side. This gives us the following inequalities:

Now, using what you know about hyperbolas, use one of the above inequalities to show \(a < c\). Again, your proof should not be more than 2-3 steps.