Twin Paradox Problem
The figure below represents three inertial "clocks and metre sticks" inertial reference frames. The middle "HOME" frame is at "rest", the top frame is moving to the right with speed
v/c = b = 2Ö2/3 = 0.942809,   g = (1-b2)-˝ = 3,
and the bottom frame is moving left with speed b = -2Ö2/3.   Label the clocks O=HOME, I=first clock to the right, -II second clock to the left and so on, the top frame with primes, the bottom clocks with double primes. At t=0 clocks VI', II, and VI" are at the same position. Both of the moving frames appear "shrunk" by a factor of g = 3 and the clocks run 3 times slower than the HOME frame rate.
This is the situation at ct=ct'=ct"=0. The "HOME" clocks are all synchronized but the moving clocks are squished together and do not appear synchronized from the HOME frame point of view. (But they are in THEIR frame!) We can calculate the times by way of the Lorentz transform,

***** ct' = g(ct-bx), *****

and we see that the first primed clock (I') at x=1/3 reads -2Ö2/9, the third (III') reads -2Ö2/3 and the third primed clock at x=-1 reads +2Ö2/3. All of the approaching clocks appear to be set "fast" and the leading clocks appear to be "slow" and the farther from HOME the greater the difference ( -bx ). The double prime frame moving to the left is different by the same factor with the signs reversed. (The sign of b is different but note that g is always a positive number.)

Some extra thoughts:
All of the frames are synchronized.
If you move from one frame to the other at (t=zero,HOME) all the clocks in that frame will read t=zero.
If you were to jump from clock I in the home frame at t=0 --(0,I)-- to the prime frame ( -2Ö2/3,III') then all of the prime frame clocks would read -2Ö2/3.  The "I" clock would still read zero but none of the other HOME clocks would read zero -- they would look like the doubleprime clocks did from the HOME frame.

The plan is to jump onto the prime frame (top) O' clock at ct=ct'=0, ride O' to clock IX (nine metres to the right of O=HOME), then jump onto the doubleprime (bottom) frame for a ride ride back HOME. Here is the plan according to HOME:

At a speed of 0.942809c it will take HOME' ct = cx/v = 9/0.942809 = 9.54594m (of time) to get to IX. According to relativity theory it should appear that only ct' = ct/3 = 3.18198m (of time) has elapsed for the jumper and the round trip should take 6.36396m, one third of the HOME elapsed time of 19.091883m. And that is exactly what is going to happen. We should stop thinking now, this reasoning is absolutely right. (No kidding!)

But it is always fun to see what happens during the trip. First look at the parade of prime and doubleprime clocks whizzing past HOME. We "know" from theory that they are ticking over at 1/3 our rate but what do you read on the dials that pass by? You read ct' = gct at x=0… Hey, at t=9.54594 the clock passing you will read 28.63782, three times faster, not three times slower! What's going on here? And the doubleprime clock reads the same as the prime. (Gamma is always positive….)

OK, of course that is corect, not surprising at all, is it! At t=9.54594 that would be the -XXVII' clock which was at x=-9m (right next to clocks -IX and -XXVII") at ct=0 and read ct' = 3(0-[2Ö2/3][-9]) = 18Ö2. after 9.54594 seconds the -XXVII' clock advances to 18Ö2 + 9.54594/3 = 28.63782. It really does run slow but it started way fast! Hmmmm… if you think about it, you now see that if the moving clocks run slow they had to be out of sync to make things work out.

This deserves more looking into, it is instructive. The Lorentz transform is
***** ct' = g(ct-bx), *****
and if we take a partial derivative, keeping x=constant, we get
t'/t = g   (x=constant)
but the total derivative is
dt'/dt = g(1-b dx/cdt) = g(1-bv/c) = g(1-b˛) = g/g˛ = 1/g

as advertised. Well, this amazed me when I first realized it! The partial (holding x constant) means we sit on one x and watch the primes parade by, seemingly turning over g times faster than our clock while the total derivative "rides" the prime clock and tells you what it is doing, running g times slower. The derivative is the reciprocal of the partial derivative. Wow!
Now let us jump to the prime frame and see what happens at ct = ct' = ct" = 0. Ouch! (These "boosts" are really hard on you.) Hey, we see that the prime frame clocks really are synchronized, they all read zero, but now the HOME frame looks just like the doubleprime frame did from the HOME frame. And the doubleprime frame from here? From the HOME frame the prime and doubleprime systems were moving with relative speed 2*2Ö2/3c = 1.8856c. But from the prime frame the doubleprime frame is moving to the left (like the HOME frame) with speed (.943+.943)/(1+.943*.943) = 12Ö2/17 = 0.9982684 and dilation factor g = 17. Worse still, the HOME clocks are running slow by a factor of three (the doubleprime by 17) and is "shrunk"by a factor three.

Our clock seems to be running normally (of course it is!) and we are still approaching the IX clock (actually it is approaching us now) at a speed of 2Ö2/3c -- but in this prime frame the IX clock is only three metres distant! So it will only take ct = 3/(2Ö2/3) = 9/2Ö2 = 3.18198m to get there, just like we thought in the HOME frame. But it is not time dilation that does it, time seems to be running normally here, it is the length contraction! Think about this for a while. It is always the other frame that is strange, not you!

But what does the HOME frame IX clock read when we get there? At ct' = 0 it read ct = 3(0+[2Ö2/3]*3) = 6Ö2 = 8.48528m and at ct' = 3.18198m reads 8.48528 + 3.18198/3 = 9.54594m. That's right! If we were to jump to the HOME frame right now, we would have aged one-third as much as the stay-at-home. Last chance to stop thinking and avoid "the paradox"... The so-called paradox arises because during the travel from clock HOME to IX each frame sees the other frame clocks running gamma times slower. So at first blush it would seem to the traveller that the stay-at-home should have aged less.

Now stop and think. During this travel each sees the other ageing at a slower rate. This is "the paradox". But it is not a paradox at all, both are correct! The "confusion" (better term than paradox) arises from our concepts of "absolute time" and "simultaneity". Sorry, in our universe "absolute time" just does not exist. The term "simultaneity" means "same place and time". Since our twins are in different places and in different reference frames the ordering of events (and rate of time flow) require relativistic thinking! Let's see what happens now.

When the traveller makes the jump to the HOME frame (which reads 9.54) then he sees all the clocks in the HOME frame (including the HOME O clock) reading the same 9.54 while his clock still reads 9.54/3. Once again the traveller sees the frame he is in (HOME) is synchronized and the other frames are not. It is the boost at a distance that resolves the curious "twin paradox" feature of relativity.

If you now jump on the doubleprime system and ride back HOME, once again both observers see the other clocks running slow by a factor of three but the jump at clock IX causes the O clock to "jump ahead" just far enough so that when the traveller returns HOME the stay-at-home has aged three times as much as the traveller. NOTE that a boost at a distance doesn't really do anything to the HOME clock, the HOME observer simply sees your clock slow down. There is no paradox at all as far as HOME is concerned.