If angular momentum is conserved, what's wrong with this scenario? I understand that, in a given system with no external torques applied, angular momentum is conserved. However, consider the following situation:
Let's say the earth revolves around the Sun in a circle. My reference point will be one point on the circle. When the earth is, say on the other side of the circle from my reference point, the angular momentum is non-zero. However, at the point when the earth crosses my reference point, the angular momentum is zero (since angular momentum is position cross momentum, and the position with respect to my reference point is zero).
But I thought it was conserved?
Is there some external torque I'm forgetting to include?
Thanks ahead of time for any answers.
 A: In your problem, "Earth" is not an isolated system. The combined "Sun-Earth" system, however, is, so we can know that the angular momentum of the Sun-Earth system is conserved.  As the earth's mass is accelerating the sun, you have to take its angular momentum into account as well.
While the mass and size of the sun mean that we can ignore its motion with respect to the rest of the solar system, you can't do that for your calculation. 
Alternatively, you can consider the sun's gravitational force on the earth to be a torque in your case because the force does not go through your reference point.  
A: You are a point in the circle. The torque is:
$$
\mathbf\tau = \mathbf r\times\mathbf F
$$
Where $\mathbf r$ is the position of Earth and $\mathbf F$ is the force (radially towards the sun). Notice when your reference point is somewhere in Earth's orbit, as you said, and your object is earth, the force will not be parallel to the position. Therefore, the cross product is non-null. And hence $\mathbf\tau\ne\mathbf 0$.
If torque is not null, there's no conservation of angular momentum. 
Note: If your reference point is the sun, then $\mathbf r \parallel\mathbf F$ and therefore $\mathbf\tau = \mathbf r\times\mathbf F = \mathbf 0$. Conclusion: In this case angular momentum is conserved.
A: before I answer your question, let me put forward another scenario. Imagine a mass less rod with two massive spherical objects (of mass $m$) is placed at either of its ends. Further imagine that the rod rotates about an axis. Let's define the axis to be the center of the rod. Let's choose the reference point at a distance $x$ from the axis and on the rod. now if we compute the torque of the nearest point mass from the reference point, we get $\mathbf t_1=\mathbf x\times \mathbf f_1$. but because the net torque is zero, the torque of the point mass away from the reference point (having distance $ (l-x)$ if $l$ is the length of the rod) is $t_2(l-x)\mathbf r\times \mathbf f_2$(where $\mathbf r$ is the unit vector along the direction of the other mass). 
we always have $sin(\pi/2) = 1$, so we can simply write $x\times f_1=(l-x)\times f_2$ as the angular momentum is conserved. 
Now lets move further away from the axis towards the nearer point mass. we will find that the force $f_2$ keeps decreasing in magnitude and the force $\mathbf f_1$ keeps increasing (we can explain that by writing $f_1=((l-x)/x) \times f_2$ and $x$ keeps decreasing and tries to reach zero).
Now lets consider the limiting cases: $lim_{x\to0}f_1(x)=\infty$ To be precise $f_1\to\infty$ instead of actually being infinity. So hence, you have a finite value for the product $\mathbf t_1=\mathbf x\times \mathbf f_1$ even if $\mathbf x$ tends to zero (as a product of a infinitesimal quantity with a very large quantity gives a finite value). 
Let's now consider the case where $\mathbf x$ is actually zero and not that its limit approaches zero. in this case, the force becomes plainly undefined in the set of real numbers and such torque does not exist (how can we say that there is some torque if the force acts on the reference point? it doesn't seem  right!). This scenario contradicts our definition that torque is defined in two dimensional or three dimensional (more than one dimensions are required to compute torque) space (as many other physical quantities are); if the value of $x$ is zero the reference point and the position of the point mass would perfectly coincide, and that forms a 0 Dimentional system. We cannot compute torque in an zero dimensional system hence it is always assumed that the distance is only nearly zero and not perfectly zero.
The same case applies to you question; in your scenario, you have chosen the reference point away from the line joining the earth and the sun, on the earth's orbit. but in your case we won't always get $sin\theta=1$ that makes calculations "little" complicated, so hence I chose a similar but simpler scenario. note:- $\mathbf torque\times\mathbf dt=$ differential change in angular momentum, so without loss of generality we can say that this scenario is similar to yours.  
A: About your reference point, torque acting on earth is not zero. But total torque of sun and earth is zero. That's why total angular momentum of earth and sun remains conserved.
