NB: In this post, I will use the abbreviation $TM_P$ to stand for a symmetric truncated (or trimmed) mean that discards the largest and smallest $P/2$ percent of the data. In fact, for concreteness I will refer mostly to $TM_{50}$, but the same question could be asked for any other "similar value" of $P$.
Question
What are situations for which a sound statistical basis exists to prefer $TM_{50}$ (or $TM_P$ for some other more appropriate $P$) over the median as measure of the data's central tendency?
EDIT: (In response to Nick Cox's ridiculing of my wording.) Here's an example of the sort of justification (speaking very broadly) I had hoped for. Q: Why choose the median over the mean? A: For robustness against outliers. The mean can be made anything one wants by a single sufficiently extreme outlier, whereas the median is naturally immune to such an instability; no outlier censoring required. Granted, very extreme outliers are probably entirely tangential to the process under study, but this reasoning still vividly underscores the robustness of the median. There's nothing subjective about it, it's a mathematical fact, even if different people find the same mathematical fact more or less problematic.
Background
I cannot think of a justification for using $TM_{50}$ that does not apply equally well to the median 1. Furthermore, I figure that, as summary statistics go, the median is the conceptually and analytically simpler of the two (and, therefore, the more thoroughly studied and better characterized one). Thus, my gut reaction is to always prefer the median over $TM_{50}$. The motivation for this thread is to either confirm or refute this gut reaction.
1 Of course, the median can be thought of as, roughly, the limit of $TM_P$ as $P \to 100$. Therefore, I figure that any justification for using $TM_P$ over the median can only get weaker as $P \to 100$. An analogous consideration applies when $P \to 0$, if we also replace the median with the mean.