AA1 = 3√3
OA1 = (3√3)/3
OA1 = A1B = A1C
A1B = (3√3)/3 = √3
A1C = (3√3)/3 = √3
BC =A1B + A1C = 2*√3
AA1² = AC²+ A1C²
(3√3)² = AC² + √3²
AC² = (3√3)² - (√3)²
(3√3)² = 9*3 = 27
(√3)² = 3
AC² = 27 - 3 = 24
AC = √24
AC = AB1 + CB1
AB1 = CB1
AC = 2*CB1
CB1 = AC/2
BB1² = BC² + CB1²
BB1² = BC² + (AC/2)²
BB1² = (2√3)² + (√24/2)²
((2√3)² = (4*3)² = 12
(√24/2)² = 24/4 =6
BB1² = 12 + 6=18
BB1 = √18 = √(2*9) = 3√2
Ответ : длина большей из медиан BB1 = 3√2
смотри рисунок
