The ABCD and A1B1C1D1 squares lie in planes, the angle between which is 60 degrees. Find the distance between their centers if AB = 2a

Let K be the midpoint of AB.
Then РK is the middle line of the triangle ABC, РK║BC, and therefore РK⊥AB,
and PK = BC / 2 = a.
MK is the middle line of the triangle ABC₁, MK║BC║, ⇒ MK⊥AB, and
MK = BC₁ / 2 = a.
∠МKР = 60 ° is the linear angle of the dihedral angle between the planes of the squares.
ΔМKР: РН = МН = a and ∠МKР = 60 °, which means the triangle is equilateral, and therefore МР = а.