It is known that there are four ABCD points on the sphere with radius 2. If AB = CD = 2, the maximum ABCD volume of the tetrahedron is? 3Q Such as the title

(4 radical 3) / 3

It is known that there are four points a, B, C and D on the sphere with radius 2. If AB = CD = 2, the maximum volume of the tetrahedral ABCD is ()
A. 233B. 433C. 23D. 833

If the distance from point P to CD is h, then there is v tetrahedron ABCD = 13 × 2 × 12 × 2 × H = 23h. When the diameter passes through the midpoint of AB and CD, Hmax = 222 − 12 = 23, so Vmax = 433

Given that the tetrahedron ABCD is a regular tetrahedron, the angle between BC and ad is calculated

Let B be perpendicular to AD and e connect CE, because it is a regular tetrahedron, so Ba = BD = AC = CD, because be is perpendicular to ad, Ba = BD, so e is the midpoint of AD, and because CA = CD, so CE is perpendicular to ad, ad is perpendicular to be, CE, so ad is perpendicular to plane BCE, so BC is perpendicular to AD
90°

As shown in the figure, e and F are the midpoint of AD and BC respectively in the tetrahedral ABCD with edge length a
(1) And prove that EF is the common perpendicular of AD and BC, and find the length of EF
(2) To find the cosine of the angle between AF and CE

1. Connect AF and DF
∵△ABC≌△DBC(SSS)
∴AF=DF
And E is the midpoint of AD
Ψ EF ⊥ ad (the height of the bottom of an isosceles triangle coincides with the middle line)
∵AF⊥BC,DF⊥BC
{BC ⊥ plane AFD
∴BC⊥EF
The EF is the common vertical of AD and BC
AF=√3·a/2,AE=a/2
∴EF=√2·a/2
2. Make eg ‖ AF, cross DF to g, connect CG
Then ∠ CEG is the angle between AF and CE, set as α
EG=AF/2=√3·a/4
CE=√3·a/2
FG = DF / 2 = √ 3 · A / 4 (eg is the median line of triangular AFD)
CF=a/2
CG=√7·a/4
cosα=(CE²+EG²-CG²)/(2CE·EG)=2/3

It is known that there are four ABCD points on the sphere with radius 2. If AB = CD = 2, the maximum ABCD volume of the tetrahedron is? 3Q Such as the title