You are given a regular tetrahedron DABC with edge a. With symmetry about the ABC plane, point D passed to point D1. Find DD1.

1.In the triangle ABC – O – the point of intersection of the medians, divides the median as 2: 1, that is, AO – 2 parts from AH (median)
2.AH = √ (a ^ 2-a ^ 2/4) = √ (3) a / 2
3.AO = AH * 3/2 = √ (3) a / 3
4.DO = √ (AD ^ 2-AO ^ 2) = √ (a ^ 2- 3a ^ 2/9 = √ (6a ^ 2/9) = √ (2/3) a
5.DD1 = 2 * DO = 2√ (2/3) a
Answer: DD1 = 2√ (2/3) a