Why isn't the symmetric twin paradox a paradox? Two twin sisters synchronize their watches and simultaneously (from the earth frame) depart earth in different directions. Following a predetermined flight plan, each sister accelerates identically to 99.9%c and then returns home at the same time (again in the earth frame). The observers on earth see each twin as having aged identically, as the symmetry of the problem dictates. Each twin however should see the other as having aged... What gives?
 A: We can simplify the problem by assuming that the twin sisters leave Earth already at 0.999c and when they return to the earth they are still doing 0.999c. That means the only acceleration takes place when the sisters reverse speed at the halfway point of their journey.
We can all compare watches on two points that everyone is in the same place i.e. when the sisters leave Earth we can all synchronise watches so they read the same, and when the sisters return we can all compare the readings on the watches. The fact the sisters are moving at this moment doesn't matter as long as both the sisters and we are in the same place.
So consider what sister A sees: working forward from the start point, sister A sees sister B moving at 0.9999995c (relativistic addition of 0.999c to 0.999c) so the time dilation is 1,000. So sister A sees sister B aging at 1/1000 of the normal rate. Likewise for sister B.
Now take the end point, at which the sisters are the same age, and follow the sister's trajectories back in time. Again sister A will see sister B aging at 1/1000 of the normal rate, and again likewise for sister B.
If we follow the trajectories forward from the beginning, and backwards from the end, then obviously at some point the trajectories must meet, and this means the sisters must have changed from the initial to the final inertial frame i.e. they must have accelerated to change inertial frames. Lets consider this from sister A's point of view.
During the acceleration sister A feels a force, but considers herself not to be moving. So from her perspective sister B accelerated to change her speed from 0.9999995c to -0.9999995c and even though sister B was aging at 1/1000 of the normal rate, during the acceleration sister B's time catches up with and overtakes sister A's time. During the inbound leg sister A sees sister B aging slowly again, but the age jump during the acceleration makes up the difference. Sister B sees exactly the same, i.e. she sees sister A age during the acceleration.
So that's why the sisters end up the same age. They do indeed see each other aging more slowly, but it's during the acceleration between the outbound and inbound legs that each others time catches up.
I haven't attempted to do the sums because to be honest I've long since forgotten it and I'd have to go away and look it up, but you can probably find it on the web. Contrary to popular belief you don't need general relativity as special relativity is perfectly capable of handling accelerations, as long as you're content to treat acceleration as absolute. My recollection of doing the calculation at college is that it was messy and actually not very illuminating.
A: John Rennie's answer doesn't sound quite right. This problem can be restated with no acceleration. Suppose we have 4 sisters {A1,A2} and {B1,B2}.  Sisters A1 and B1 leave the earth simultaneously and agree to keep traveling on a non accelerated trajectory indefinitely. At the same time (from the earth frame) sisters A2 and B2 start traveling towards earth. Some where in the middle Sister A2 passes A1 and B2 passes B1. Sisters A1 and B1 transmit their current wallclock time to their respective sisters. Here there is no acceleration, hence acceleration can not be part of the problem.
Or what if the universe is closed and they meet at the starting point after a finite time. Again, no acceleration.
A: A simple explanation appears to be that each twin are the same age after the journey because they traveled paths of equal length through spacetime. In other words their perspective watches measured the same amount of proper time. 
In the canonical version of the twin paradox the difference in age of the earth bound twin, and the space-traveling twin can viewed as that the earth bound twin followed an essentially straight line through spacetime, and the space-traveling twin followed a curved path through spacetime. In Minkowski space, unlike Euclidean space, a curved path is shorter than a straight path.
A: The resolution of this apparent paradox introduces no new ideas with respect to the canonical version of the paradox. As you point out, the journey of each twin is symmetric, and so each twin calculates that they will age the same amount by the time they return home. 
If, instead of appealing to symmetry, you would like to do the calculation from the perspective of one twin reasoning about the passage of time of the other twin, the reasoning is as follows: During the outgoing phase of the journeys, each twin calculates that less time has elapsed on the other twin's clock than on their own. But during the turnaround, they calculate that the other twin ages rapidly to such an extent that, even accounting for the time dilation for the return journey, their ages will be the same when they meet up back at their starting point.
A: I'd stick to simple for this one: Both twins accelerate and then decelerate equally, while the earth observer stays put with close to zero accumulated acceleration. So Minkowski's very first approach to answering this question -- "who is accelerating?" to paraphrase -- applies nicely. When everyone returns to the same location and velocity and compare clocks, the twins simply find they are the same (low) age, and that everyone on earth aged faster.
Incidentally, don't let discussions of "viewing" each other's ages through telescopes throw you off on this one. With enough data you can place brackets around permissible time deltas that way, but you can't give final answers because you don't know how much of the spatial distance between you may still end up converted into time distance. The shared location, shared frame finale in contrast gives answers that depend only on who spent more time accelerating.
