Determining the more acidic proton in cyclopenta-1,3-diene I am in the David Klein textbook in Chapter 3 going over rule 2 of finding the more acidic proton: if the electron pair in the conjugated base is delocalized, then the conjugate base is more stable, ergo, the proton at that location is more acidic. Here's the practice problem I am stuck on:

I generated two different conjugated bases. The book indicates the red proton is the more acidic proton. Is this the proper resonance structure for the red proton's conjugate base? 

If so, then this explains why the red proton is more acidic, because it has a delocalized pair of electrons, while the blue proton does not have a delocalized pair of electrons.
Figure c does not look like a real compound. If it represents the conjugate base of cyclopentadiene, it is lacking a negative sign, and all five protons around the ring are equivalent, so the question of whether the red or blue proton is more acidic makes no sense.

If Figure c represents cyclopentadiene, the acid, it does not clearly show that the red position really has 2 hydrogens. However, the double bond structure shown is consistent with 2 hydrogens in the red position, and the question becomes very simple: in cyclopentadiene, is a methylene proton or a vinylic proton more acidic? Removing a vinylic proton puts a negative charge in the localized sigma system - no aromaticity develops. However, removing one of the methylene protons causes that carbon to go from sp3 to sp2p, and the negative charge goes into the pi system where it can be delocalized over the 5-membered ring. And aromaticity is a characteristic feature of cyclopentadienide ion.