BrianT advised me by private message that I had oversimplified my earlier analysis, and he is right. I looked only at the voltage divider created by the resisitors built into the transistors and neglected to consider the diode drop across the base-emitter junctions of the transistors. The circuit, redrawn, looks like this:
<IMG SRC="http://www.picbasic.co.uk/forum/attachment.php?attachmentid=1980&stc=1&d=118913663 8">
With the diode junctions in place (a nominal 0.7 VDC drop across each), two voltages change: At A, 4.3 volts; at C, 0.7 volt. At B, it remains 2.5 volts. Both transistors are still "turned on" if there is no voltage present (a high-impedance state) at B.





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