Isn't the detector always measuring, and thus always collapsing the state? I have a radioactive particle in a box, prepared so as to initially be in a pure state
$\psi_0 =1\ \theta_U+ 0\ \theta_D$
(U is Undecayed, D is Decayed). 
I put a Geiger counter in the box.
Over time (t), the theory says that the state should evolve into a pure state that is a superposition of Undecayed and Decayed, with the Decayed part getting bigger and bigger
$\psi_t =a\ \theta_U+ b\ \theta_D$
Eventually the counter will 'click', indicating that the particle has Decayed. Now I know that the state is 100% Decayed.
However, before this happened, the silence of the counter also indicated that the particle hadn't Decayed yet. So all the time up to that point, I also knew that the state was 100% Undecayed.
But this would be contradicting what the theory suggests (a superposition with a non zero contribution of the Decayed state, after some time), so I'm guessing it's an incorrect way of analysing the experiment.
I want to know where the mistake lies.
In other words, it seems to me the Geiger counter is always measuring the state of the particle. Silence means Undecayed, click means Decayed. So the particle would never actually Decay since I continuously know its state is 
$\psi_t =1\ \theta_U+ 0\ \theta_D$
which means its chance of decaying would be perpetually zero (Zeno's effect, I've heard?).
How do I deal with this constant 'passive' measuring?
 A: Your statements treat the quantum mechanical distribution as physical, whereas it is a mathematical function fitting the boundary condition of your experiment, i.e. it is the mathematical function describing a particle's probability of decay.
Probabilities are the same in classical mechanics, in economics in gambling, in population interactions. Take the probability of throwing  a dice and coming up with six. For a true dice (not weighted) it is 1/6 of the time no matter whether you throw the dice or not , if you throw it you have a probability of 1/6. If a gambler has weighted the dice, maybe the probability curve is weighted towards 6, so it could be you have 1/3 probability to get a 6 with a weighted die.
You have a particle that can decay while sitting alone. The probability of its decay  is given by  $Ψ^*Ψ$ , by the solution of a mathematical quantum mechanical  differential equation ( or maybe lattice QCD, which uses the solutions).  Whether a geiger counter was there or not, one can calculate how many nuclei will have decayed given the probability distribution for the nucleus  (a function of time in this case) and the time past.
The geiger counter is incidental , a second interaction with a $Ψ^*Ψ$ locally that has ideally 100% probablity to interact when a charged particle hits it. A tool for recording a decay. (as your eyes do not affect the probability of the dice coming up 6).
The states you write down are not quantum mechanical states. They may be logical mnemonics, but they do not have to obey quantum mechanical equations or postulates , they are not a $Ψ^*Ψ$ .
A: Consider the Many-Worlds approach.
You have a wavefunction (an immensely complicated one, of course). Your amplitude for having heard a click steadily grows in magnitude.
No paradox if you look at it like this.
A: Good question. The textbook formalism in Quantum Mechanics & QFT just doesn't deal with this problem (as well as a few others). It deals with cases where there is a well-defined moment of measurement, and a variable with a corresponding hermitian operator $x, p, H$, etc is measured. However there are questions which can be asked, like this one, which stray outside of that structure.
Here is a physical answer to your question in the framework of QM: Look at the a position wave function of the decayed particle $\psi(x)$ (*if it exists: see bottom of post if you care). When this wave function "reaches the detector" (though it probably has some nonzero value in the detector the entire time) the Geiger counter registers a decay. Using this you get a characteristic decay time. This picture is a good intuition, but also an inexact/insufficient answer, because the notion of "reaches the detector" is only heuristic and classical. A full quantum treatment of this problem should give us more: a probability distribution in time $\rho(t)$ for when the particle is detected. I will come back to this.
So what about the Zeno effect? Based on the reasoning you gave, the chance of decaying is always zero, which is obviously a problem! Translating your question to position space $\psi(x)$, your reasoning says the wave function should be projected to $0$ in the region of the detector at every moment in time that the particle hasn't been found. And in fact you're right - doing this does cause the wave function to never arrive at the detector! (I actually just modeled this as part of my thesis). This result is inconsistent with experiment, so we can conclude: continuously-looking measurement cannot be modeled by straightforward projection inside the detector at every instant in time.
A note, in response to the comments of Mark Mitchison and JPattarini: this "constant projection" model of a continuous measurement can be rescued, by choosing a nonzero time between measurements $\Delta t \neq 0$. Such models can give reasonable results, and $\Delta t$ can be chosen based on a characteristic detector time, but in my view such models are still heuristic and a deeper, more precise explanation should be aspired to. Mark Mitchison gave helpful replies and linked sources in the comments for anyone who wants to read more on this. Another way to rescue the model is to redefine the projections to be "softer", as in the sources linked by JPattarini.
Anyway, despite the above discussion, there is still a gaping question: If continuous projection of the wave function is wrong, what is the correct way to model this experiment? As a reminder, we want to find a probability density function of time, $\rho(t)$, so that $\int_{t_a}^{t_b}\rho(t)dt$ is the probability that the particle was detected in time interval $(t_a, t_b)$. The textbook way to find a probability distribution for an observable is to use the eigenstates of the corresponding operator ($|x\rangle$ for position, $|p\rangle$ for momentum, etc) to form probability densities like $|\langle x | \psi \rangle|^2$. But there is no clear self-adjoint "time operator", so textbook quantum mechanics doesn't give an answer.
One non-textbook way to derive such a $\rho(t)$ is the "finite $\Delta t$ approach" mentioned in the note above, but besides this there are a variety of other methods which give reasonable results. The issue is, they don't all give the same results (at least not in all regimes)! The theory doesn't have a definitive answer on how to find such a $\rho(t)$ in general; this is actually an open question. Predicting "when" something happens in Quantum Mechanics (or the probability density for when it happens) is a weak point of the theory, which needs work. If you don't want to take my word for it, take a look at Gonzalo Muga's textbook Time in Quantum Mechanics which is a good summary of different approaches on time problems in QM which are still open to be solved today in a completely satisfactory way. I am still learning more about these approaches, but if you are curious, the one I found most clean so far uses trajectories in Bohmian Mechanics to define when the particle arrives at the detector. That said, the measurement framework in QM in general is just imprecise, and I would be very happy if a new way of understanding measurement were found which gives a higher level of understanding of questions like this one. (yes I am aware of decoherence arguments, but even they leave questions like this unanswered, and even Wojciech Zurek, the pioneer of decoherence, does not argue that it fully solves problems with measurement)
(*note from 2nd paragraph): Sure you can in principle hope to position representation to get a characteristic decay time like this, but it might not be as easy as it sounds because QFT has issues with position space wave functions, and you'd need QFT to describe annihilation/creation of particles. Thus even this intuition doesn't always have mathematical backing.
A: I think that “listening” even in the case of silence is already the measurement. You can only hope to hear something when there is a medium (air) that will carry the sound waves. This medium causes a continuous interaction between you and the Geiger counter. Only without the medium there is no interaction but then you can also not tell that the Geiger counter kept silence.
A: No, the detector is not always collapsing the state. 
When the particle is in an undecayed state its wave function is physically localised with a vanishingly small amplitude in the region of the detector, so the detector doesn't interact with it and isn't 'always' measuring it. It is only when the particle's state evolves to the point at which it has a significant amplitude in the vicinity of the detector that the counter clicks.
A: My take on this is that in the original thought experiment, you don't get to monitor the detector. When the detector detects, it kills the cat. But it doesn't tell you then. You only find out when you open the box.
If it tells you immediately, then you know immediately. And then there's the question whether the detector detects 100%.
If the Geiger counter detects 100%, then you could have 100 Geiger counters or 10000, and they would all detect the particle decaying. If they were all the same distance, they should all detect it at the same time. (Assuming the particle was not moving relative to them. Otherwise relativity might give them different times which would be 100% predictable.
I think it's more plausible that each detector detects a different photon. And the first single detector might easily miss a particular gamma ray photon. 
So if there is only one radioactive particle, then if the geoger counter does detect it, then you know it's been detected and you know pretty much when. But if it hasn't detected it yet, there's an increasing chance with time that the particle has decayed and the geiger counter did not detect it and will never detect it.
