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We will name the position vectors of the four birds r_A, r_B, r_C, r_D. The laws in the problem statement can be written the following way:

	= v(t) ⋅
	= v(t) ⋅
	= v(t) ⋅
	= v(t) ⋅

Since we are only interested in the shape of the trajectory of the four birds we don’t need to use time as the parameter for these curves. As long as the ratios between the four derivatives and their directions are preserved then the shapes described by the system of equations remains the same.

Due to symmetry we can say the following holds for any moment in time:

Therefore if we have a dimensionless parameter τ such that

the initial system becomes much simpler:

	= rB - rA
	= rC - rB
	= rD - rC
	= rA - rD

Using this new set of equations we can solve the problem independently for each coordinate. Without loss of generality we will solve it for the x coordinate and then we will generalize for y and z.

	= xB - xA
	= xC - xB
	= xD - xC
	= xA - xD

We will group the four positions in a column vector a:

We will name the matrix above M and we will diagonalize it M = R ⋅ N ⋅ R^-1

	= R ⋅ N ⋅ R-1 ⋅ a ⇒
R-1 ⋅	= (R-1 ⋅ R) ⋅ N ⋅ R-1 ⋅ a ⇒
(R-1 ⋅)	= N ⋅ (R-1 ⋅ a) ⇒
	= N ⋅ (R-1 ⋅ a) ⇒

We define bR^-1 ⋅ a and we how have:

Where N is a diagonal matrix. Therefore we can split the equation into the four coordinates of b. We will write the k^th coordinate of b as b_k and the k^th eigenvalue of M as λ_k.

This equation of course has the following solution:

In other words we have

R-1 ⋅ a(τ)	= ⋅ R-1 ⋅ a(0) ⇒
a(τ)	= R ⋅ ⋅ R-1 ⋅ a(0)

Therefore a’s derivative is

= R ⋅ N ⋅ R-1 ⋅ a(τ) = R ⋅ ⋅ R-1 ⋅ a(0)

Now we return to our original task: calculating the total distance traveled by bird A. From now on we will name the column vectors containing the coordinates of all four birds a_x, a_y, a_z. Since at all moments all 4 velocities are equal the total distance traveled in any small amount of time dt is the same for all four birds.

4dS2 =	(dx A2 + dx B2 + dx C2 + dx D2) +
	(dyA2 + dy B2 + dy C2 + dy D2) +
	(dzA2 + dz B2 + dz C2 + dz D2) ⇒

Here by (v)² where v is a vector we mean it’s dot product with itself v ⋅ v = ²

We can now write the total space traveled by brid A:

Fortunately R is a unitary matrix. In other words it’s inverse is equal to it’s conjugate transpose R ⋅ R^H = I₄. This property means that v’s squared euclidean length (the dot product between itself and its complex conjugate vector) is the same as that of R ⋅ v.

()2	= 2 ⇒
()2	= H ⋅⇒
()2	= ∑ k=142 ⇒
()2	= ∑ k=142 ⋅ e2 Re{λk}τ ⋅2

Now we substitute back in the integral

S = ∫ 0+∞⋅ dτ

For each of the four coordinates we have a coefficient

Using the coordinates of the vertices of the tetrahedron we calculate the four coefficients

Remember that the scalar a is the side length of the tetrahedron while the vector a represents coordinates of the four birds packed together in the same vector.

S	= ∫ 0+∞ ⋅ dτ ⇒
S	= ∫ 0+∞ ⋅ dτ

Now all that is left is solving the integral. First we do a substitution to eliminate the exponential.

u	= e-τ
du	= -e-τdτ ⇒ dτ = ⇒
S	= ∫ 10 ⋅ ⇒
S	= ∫ 01 ⋅ du

And now a trigonometric substitution to obtain a known integral.

u	= tanθ
du	=
S	= ∫ 0 ⋅ ⇒
S	= ∫ 0⋅ ⇒ (since cosθ > 0)
S	= ∫ 0

This integral has solution + C

Substituting we finally obtain our result.

S	= ⋅⇒
S	= ⋅⇒
S	= a ⋅ ⇒ S ≈ 0.812 ⋅ a