Riddler Classic

From Mikael Rittri, a mathematical souvenir problem:

In the Riddler gift shop, we sell interesting geometric shapes of all sizes — Platonic solids, Archimedean solids, Klein bottles, Gabriel’s horns, you name it — at very fair prices. We want to create a new gift for fall, and we have a lot of spheres, of radius 1, left over from last year’s fidget sphere craze, and we’d like to sell them in sets of four. We also have a lot of extra tetrahedral packaging from last month’s Pyramid Fest. What’s the smallest tetrahedron into which we can pack four spheres?



In [1]:

    
SPHERE_R = 1



In [2]:

    
from sympy import symbols, cos, sin, tan, rad, deg, sqrt, atan, simplify, acos, asin, pi
from sympy import init_printing
from sympy.physics.vector import vlatex

init_printing(latex_printer=vlatex)



In [3]:

    
side_length = symbols('a')



In [4]:

    
center_to_side = tan(pi/6)*(side_length/2)
center_to_side









    Out[4]:





$$\frac{\sqrt{3} a}{6}$$



In [5]:

    
total_tri_height = sin(pi/3)*side_length
total_tri_height









    Out[5]:





$$\frac{\sqrt{3} a}{2}$$



In [6]:

    
center_to_vertex = total_tri_height - center_to_side
center_to_vertex









    Out[6]:





$$\frac{\sqrt{3} a}{3}$$



In [7]:

    
theta = asin(center_to_vertex/side_length)
theta









    Out[7]:





$$\operatorname{asin}\left(\frac{\sqrt{3}}{3}\right)$$



In [8]:

    
tetra_height = side_length*cos(theta)
tetra_height









    Out[8]:





$$\frac{\sqrt{6} a}{3}$$



In [9]:

    
phi = atan(center_to_side/tetra_height)
phi









    Out[9]:





$$\operatorname{atan}\left(\frac{\sqrt{2}}{4}\right)$$



In [10]:

    
sphere_center_to_tetra_point = SPHERE_R / sin(phi)
sphere_center_to_tetra_point









    Out[10]:





$$3$$



In [11]:

    
side_length = 2 * (sphere_center_to_tetra_point * cos(theta)) + 2 * SPHERE_R
side_length.evalf()









    Out[11]:





$$6.89897948556636$$