1060. Vertices...

    Let  A (1, 3, 4) ,  B (3, 1, 3)  and  C (5, 2, 5)  be 3 of the 8 vertices of a cube. Find out the other 5 vertices knowing that they all are in the 1st octant (their coordinates are positive).

    Nina Guindilla wondered about the position of the 3 given vertices...

    Dear Teacher:

    The 3 given vertices can be vertices of an isosceles triangle, or a scalene triangle, or an equilateral triangle. According to the case, the problem is different...

|AB| = |(2, −2, −1)| = 3

|BC| = |(2, 1, 2)| = 3

|AC| = |(4, −1, 1)| = √18

    It's the 1st case! They are 3 of the 4 vertices of a face... The 4th vertex D can be calculated easily:

OD = OA + OC − OB = (3, 4, 6)

D (3, 4, 6) 

    Now I'm calculating the vertices, E, F, G, H, of the opposite face:

AE = BF = CG = DH = BC x AB / 3 = (3, 6, −6) / 3 = (1, 2, −2)

    So

OE = (1, 3, 4) + (1, 2, −2) = (2, 5, 2)

OF = (3, 1, 3) + (1, 2, −2) = (4, 3, 1)

OG = (5, 2, 5) + (1, 2, −2) = (6, 4, 3)

OH = (3, 4, 6) + (1, 2, −2) = (4, 6, 4)

    So

E (2, 5, 2)

F (4, 3, 1)

G (6, 4, 3)

H (4, 6, 4)