Special relativity - aging of two travels forming a right triangle

Question

Here is a question about special relativity.

We have a right triangle $ABC$, where $\angle ABC = 90^{\circ}$ and $\overline{AB} = \overline{BC}$.

We have the Earth at $A$, and another planet at $C$. (Ignore any celestial bodies that have gravity.)

Spaceship 1 travels $A \rightarrow B \rightarrow C$ at a constant speed $v_1$ (relative speed w.r.t. the Earth; the direction of velocity changes at $B$).

Spaceship 2 travles straightly $A \rightarrow C$ at a constant speed $v_2$ (relative speed w.r.t. the Earth).

$v_1 > v_2$, so that both spaceships start at the same time (measured at $A$), and arrive at the same time (measured at $C$).

Assume there is a person in each spaceship. When the spaceships arrive at $C$, will the people in spaceship 2 have got more ages (because the spaceship is slower), or will people in both spaceships have got the same ages?

Answer 1

Space ship 2 is an inertial frame to which SR applies. So we can deduce from his point of view that he is at rest and spaceship 1 is always moving and so the clock moves slower. So when they meet, the person in spaceship 1 should be younger.

Answer 2

The 90 degree angle in the problem is irrelevant. There is nothing special about 90 degrees - it could be any angle and in all cases the A-B-C person will always be younger than the A-C person. In fact if the angle is exactly 0, then v1=0 and this problem is exactly the Twin Paradox.