@Chris. In response to your update,
find a criterion (for who ages less) based simply on what the twins observe between the moment of separation and the moment of reunion. What can they observe ?
Each observer should count the receptions of birthday transmissions from the other observer, starting from the separation-event until the reunion-event. (If you wish, each transmission could be encoded with the birthday-number of the celebrant.)
When they reunite, each observer can reconcile what they celebrated-and-transmitted and what they received from the other.
The result is Lorentz invariant.
Using a variation of your spacetime diagram (to allow nicer numbers to appear in the calculation), we have these spacetime diagrams.
I didn't include the inertial observer as a celebrant or receiver, but you can sketch in the details for that case.
update2:
My issue with the standard solution is it requires agreeing on a common shared coordinate system upon which to draw and measure the length of the world lines
The resulting count for each receiver are the same in any frame.
Here are the results from another frame.
(If you click on an image, then edit its url to remove to "m", you'll get the full quality image.)
end-of-update
original answer begins here
Inspired by your spacetime diagram, I show the following visualizations,
which are related to the formula given by @Dale
when applied to piecewise-inertial worldlines.
From my answer to Equivalence of two definitions of proper time in special relativity
This spacetime diagram has the spacetime-paths of light-signals in standard-issue longitudinal light-clocks. On a (1+1)D-spacetime diagram, the area enclosed by "one tick" of an inertial light-clock is the same, independent of the velocity of the light-clock and independent of the inertial observer drawing the spacetime diagram. (Lorentz invariance preserves the area (since the determinant of the boost equals 1)
and the directions-of-the-diamond-edges (since the lightlike directions of the diamonds are eigenvectors of the boost).)
By drawing the spacetime on "rotated graph paper",
it is easy to construct equal-area light-clock diamonds
for velocities with rational Doppler-factors $k=\sqrt{\frac{1+v}{1-v}}$
(like $v=(3/5)c$, $v=(4/5)c$, and $v=(5/13)c$ but not $v=(1/2)c$ or $v=0.99c$),
which lead to triangles with a set of sides that are Pythagorean triples
and thus most are calculations are fractions.
Alternatively, one can draw the causal-diamond of an inertial worldline segment.
This method works well for velocities that are rational... so that the diamond-areas can be computed by counting.
From my poster
https://www.aapt.org/docdirectory/meetingpresentations/WM18/FG07-Salgado-RelativityRotatedGraphPaper-CalculatingWithCausalDiamonds.pdf
You can play with this idea on my GeoGebra visualization
"Visualizing Segments with Light-Clock Diamonds (robphy)"
https://www.geogebra.org/m/HYD7hB9v#material/VrQgQq9R
from my "Relativity on Rotated Graph Paper (robphy) - MAA2016"
https://www.geogebra.org/m/HYD7hB9v#
All of this is based on my article
“Relativity on Rotated Graph Paper”,
Am. J. Phys. 84, 344 (2016);
http://dx.doi.org/10.1119/1.4943251
Based on your diagram, we have
and
The diagram on the right uses the clock-diamonds of an observer
moving with velocity $v=(3/5)c$ (so $k=2$). Observe that the number of clock-diamonds in a causal diamond is unchanged, as expected since the area in terms of the number of clock diamonds is the invariant square interval of the timelike diagonal of the causal diamond.
By counting the number of light-clock diamonds in the causal diamonds shown,
we have:
- proper-time along ABCD is $AB+BC+CD=\sqrt{15}+\sqrt{12}+\sqrt{15}=11.2100683076$
where $v_{AB}=(-1/4)c$, $v_{BC}=(2/4)c$, and $v_{CD}=(-1/4)c$
- proper-time along APD is $AP+PD=\sqrt{20}+\sqrt{20}=8.94427191$,
where $v_{AP}=(4/6)c$ and $v_{PD}=(-4/6)c$.
- proper-time along AD is $AD=\sqrt{144}=12$ ; AD is an inertial worldline.
Note:
for any intermediate event M on the line segment AD,
we have $AD=AM+MD$.
By contrast, $AD \neq AP+PD$ since P is not on the line segment AD;
APD is a non-inertial worldline.
So,
The twin who ages less is the twin for whom the following mathematical formula:
the sum of "the square-root of the areas of the causal diamonds
along piecewise-inertial segments" (i.e. the proper-time elapsed along that piecewise-inertial worldline)
is smaller than the analogous quantity for his twin.
The area of a diamond in light-cone coordinates is $UV$, where $U=t+x$ and $V=t-x$.
- Using the left diagram, Diamond AP has $U=10$ and $V=2$. Note that $t=(U+V)/2=6$ and $x=(U-V)/2=4$. The area is $UV=20$:
$$(10)(2)=(t+x)(t-x)=t^2-x^2=(6^2-4^2)=36-16=20$$
- Using the right diagram, Diamond AP has $U'=5$ and $V'=4$. Note that $t'=(U'+V')/2=(9/5)$ and $x=(U-V)/2=(1/5)$. The area is $U'V'=20$:
$$(5)(4)=(t'+x')(t'-x')=t'^2-x'^2=(\left(\frac{9}{5}\right)^2-\left(\frac{1}{5}\right)^2)=\frac{81}{25}-\frac{1}{25}=20$$
- a general diamond has area
$$UV=(\Delta t)^2-(\Delta x)^2=\left(1 - \left( \frac{\Delta x}{\Delta t} \right)^2 \right) (\Delta t)^2$$
so that its square-root (the proper-time) is
$$\sqrt{UV}=\sqrt{\left(1 - \left( \frac{\Delta x}{\Delta t} \right)^2 \right) (\Delta t)^2}
=
\sqrt{1 - \left( \frac{\Delta x}{\Delta t} \right)^2 }\Delta t
$$
in agreement with @Dale's formula.
Using, for example, desmos.com (by copying the $\TeX$-format below from "Show Math as" to desmos),
with your worldline ABCD
$t_{AB}=\int_{0}^{4}\sqrt{1-\left(\frac{-1}{4}\right)^{2}}dt $
$\qquad=\int_{0}^{4}\sqrt{\frac{15}{16}} dt = \sqrt{\frac{15}{16}}(4-0)= 3.87298334621 = \sqrt{15}$
$t_{BC}=\int_{4}^{8}\sqrt{1-\left(-\frac{2}{4}\right)^{2}}dt $
$\qquad=\int_{4}^{8}\sqrt{\frac{12}{16}} dt= 3.46410161514 = \sqrt{12}
$
$t_{CD}=\int_{8}^{12}\sqrt{1-\left(\frac{1}{4}\right)^{2}}dt $
$\qquad= 3.87298334621 = \sqrt{15}
$
as we got from counting light-clock diamonds.
