The quadaugmented tetrahedra

Geometry Level 4

What is the maximum number of regular tetrahedra that can be placed so that they share a common edge but do not overlap?

1 2 3 4 5 6 7 8

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Geoff Pilling
Nov 14, 2016

The dihedral angle of a tetrahedron is 70.528779 70.528779 degrees.

This can be calculated by constructing the tetrahedron as follows:

  • Put one vertex at ( 0 , 0.5 , 0 ) (0,-0.5,0) and one at ( 0 , 0.5 , 0 ) (0,0.5,0)
  • Then put a third at ( x 1 , 0 , 0 ) (x_1,0,0) such that 0. 5 2 + x 1 2 = 1 0.5^2 + x_1^2 =1
  • Then a fourth at ( x 2 , 0 , z ) (x_2,0,z) such that the new edges formed all have length one, using the pythagorean theorem in a similar manner.
  • Use the relationship that C D = C D c o s ( θ ) C\cdot D = CDcos(\theta) to find the dihedral angle, where C and D are the points you constructed in the second and third bullets. (and the left half of the equation is the dot product of the vectors from the origin to the respective points)

5 × 70.528779 = 352.643895 5 \times 70.528779 = 352.643895

360 352.643895 7.36 360 - 352.643895 \approx 7.36 .

So 5 \boxed{5} tetrahedra is the most that can share an edge, as shown here:

The dihedral angle of a tetrahedron is 70.528779 70.528779 degrees.

How did you calculate this?

Pi Han Goh - 4 years, 7 months ago

Log in to reply

Good point, I've added this to my solution for clarity:

  • Put one vertex at ( 0 , 0.5 , 0 ) (0,-0.5,0) and one at ( 0 , 0.5 , 0 ) (0,0.5,0)
  • Then put a third at ( x 1 , 0 , 0 ) (x_1,0,0) such that 0. 5 2 + x 1 2 = 1 0.5^2 + x_1^2 =1
  • Then a fourth at ( x 2 , 0 , z ) (x_2,0,z) such that the new edges formed all have length one, using the pythagorean theorem in a similar manner.
  • Use the relationship that C D = C D c o s ( θ ) C\cdot D = CDcos(\theta) to find the dihedral angle, where C and D are the points you constructed in the second and third bullets. (and the left half of the equation is the dot product of the vectors from the origin to the respective points)

Geoff Pilling - 4 years, 7 months ago

Log in to reply

Well, that is certainly a lot of steps that are left out =P

Pi Han Goh - 4 years, 7 months ago

Log in to reply

@Pi Han Goh Which ones? Perhaps I'll clarify the answer...

Geoff Pilling - 4 years, 7 months ago

Log in to reply

@Geoff Pilling Well, your working is still very condensed. From the first glance, I did not come into this problem expecting to use "coordinate geometry + dot product + I need to know what dihedral angles are".

Pi Han Goh - 4 years, 7 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...