In a traingle the proper length of each side equals a. Find the perimeter of this triangle in the reference frame moving relative to it with a constant velocity `V` along one of its
(a) bisectors, (b) sides.
Investigate the results obtained at `V lt lt c` and `Vrarr c`, where c is the velocity of light.

Explanation

(a) In the frame in which the traingle is at rest the space coordinates of the vertices are `(000)`, `(asqrt3/2, +a/2, 0)(asqrt3/2, -a/2, 0)`, all measured at the same time t. In the moving frame the corresponding coordinates at time `t^'` are
`A: (vt^', 0, 0), B: (a/2sqrt3sqrt(1-beta^2)+vt^', a/2, 0)` and `C: (a/2sqrt3sqrt(1-beta^2)+vt^', -a/2,0)`
The perimeter P is then
`P=a+2a(3/4(1-beta^2)+1/4)^(1//2)=a(1+sqrt(4-3beta^2))`
(b) The coordinates in the first frame are shown at time t. The coordinates in the moving frame are,

`A: (vt^', 0, 0), B: (a/2sqrt(1-beta^2)+vt^', asqrt3/2, 0), C: (asqrt(1-beta^2)+vt^', 0, 0)`
The perimeter P is then
`P=asqrt(1-beta^2)+a/2[1-beta^2+3]^(1//2)xx2=a(sqrt(1-beta^2)+sqrt(4-beta^2))` here `beta=V/c`