In a double-slit experiment, why can't we discover which slit a particle went through by measuring when the particle was detected In a double-slit experiment, interference implies that the particle has traveled different distances from the 2 slits.  Since the speed of light is constant, it seems to me that the particle would arrive at the detector at a different time if it went through one slit as opposed to the other slit.  What am I missing here?
 A: We can't determine the "which slit" information from the timing because whenever the interference pattern is being built at all, the time that the particle needs to get to the screen is the same, within the uncertainty, for both interfering paths.
Indeed, if the duration of the journey depended on the chosen slit, the interference pattern would disappear. The interference pattern is there because the nonzero wave function contributed by slit A (or its support) is overlapping in a region of the spacetime with the wave function contributed by slit B (or its support). And this overlap means that there is a large probability that the particle - whose motion is described by the wave function - has a high enough chance to get to the final point by both slits at the same moment.
In practice, the interference pattern often appears from a combination of waves that exist for a very long time. Whenever the energy of the particle is sufficiently well-defined, comparable to $\Delta E$ which is small, the wave packet must exist at a given place for some time
$$\Delta t \gt \frac{\hbar}{2 \cdot \Delta E} $$
which is somewhat analogous to the uncertainty principle for $x$ and $p$. So if the energy is accurate, the timing will not be known accurately and your discrimination can't be done. It doesn't help to make $\Delta E$ large, either. If $\Delta E$ is too large, the velocity is uncertain and the predicted time needed for the journey is uncertain, so the discrimination can't be done, either.
As long as the uncertainty inequality above is obeyed, you could indeed get a situation in which the discrimination could be done by the timing. But if you could discriminate - because the time would be significantly longer for slit A - it would mean that there would be no interference. (The amount of interference is decreasing continuously, just like the reliability of your method to discriminate would increase continuously.)
Geometrically: the interference pattern appears on the photographic plate close enough to the points which as "about equally far" from both slits - more precisely, where you need the same time for the journeys through both slits. If you look at places where the lengths (more precisely durations) would be too different for both slits, the interference pattern would largely fade away there.
The textbook examples calculating the interference pattern assume that the particle is described by waves that are basically independent of time, up to a phase. When it's so, it means that the timing is completely unknown, $\Delta t\to\infty$, and your method can't be used at all. In this textbook scenario, the detection of a particle on the screen is the first event which gives one some information about the timing.
In some more realistic situations, we know the timing when the particle entered the interference experiment with some accuracy and the wave functions are "wave packets". But they usually contain a sufficiently high number of periods of the wave so that the approximation with the "stationary" waves is good enough and $\Delta t$ is large enough.
A: It’s always should be remembered that in the quantum world there’s not just particle but a duality called wave-particle (a quanta).  There’s no objective preference to either one of the 2 phases (wave and particle) of this duality.  Which phase we shall reveal at any time is dependent upon the manner we choose to trace or measure each quanta.  Furthermore, recent theories like the Transactional Interpretation (TI) of Quantum Mechanics assert that the waveform is the only real authentic nature of any quanta as long as it does not interact with any other quanta.  It entails that the single particle (quanta) you are asking about, approaches the slits as a real wave and behaves precisely like the waves we know in the Classical Physics, like water waves, etc.  
It’s already well known how does a single water wave behave when it gets split and passes through two paths.  At the exit of each path, a new and independent water wave gets created and it interferes with the new water wave which got created at the exit of the other path.  If the TI theory is correct, the same thing occurs with a quanta in the 2 slits experiments.  As long as it does not interact with anything in the slits or with any particles of measuring-instruments, the wave (quanta) gets split into 2 new waves (in the 2 slits) which then interfere with each other.  
It means your particle (quanta) always passes through both slits as long as it has not interacted with anything (remaining as wave) and so there should not be any time difference.  
