Some problems with the tetrahedron and the octahedron.
Problem (A): How many faces?
If the faces of a tetrahedron are equilateral triangles congruent to the
triangular faces of pyramid with square base, how many faces has the
polyhedron created by gluing them together at a triangular face so that
the vertices of the triangles coincide?
Problem (B): Relative Polyhedral Volumes [Trigg, Q 113]
A regular tetrahedron and a regular octahedron have equal edges.
Find the ratio of their volumes without computing the volume of either.
Problem (C): The Space Filler [Trigg, Q 114]
Show that space can be filled with a tessellation
of regular octahedrons and tetrahedrons.
(A tesselation is a repetitve space-filling pattern.)
Problem (D): Inscribed Spheres
If the faces of a regular tetrahedron are congruent to the
faces of a regular octahedron, find the ratio of the radii of the
inscribed spheres.
Problem (E): Inscribed Spheres [Evens][Trigg, Q 240]
If the faces of a hexahedron are equilateral triangles congruent to the
faces of a regular octahedron, find the ratio of the radii of the
inscribed spheres.
Problem (F): Tetrahedron Volume [I. F. Sharygin]
Show that the volume of a tetrahedron is S*h/3 (S: area of the base,
h: altitude) only using scaling properties and volume formulae of
the prism.
Problem (G): Octahedron Volume
Determine the volume of the regular octahedron using only
scaling properties.
Solutions:
(A) At first one might guess 7 faces as 2 faces vanish.
But the polyhedron has only 5 faces as two rhombics are created.
To see this place two pyramids with square base as shown in the figure.
o o o shows two pyramids from the top
x x with 'o' a base point and 'x' a top point.
o o o
As the distance x-x is equal the side lenght of the pyramide
a regular tetrahedron fits in between.
(B) Four tetrahedrons and one octahedron build a tetrahedron of double size.
That means 4*V_tetrahedron + V_octahedron = 8*V_tetrahedron. Therefore
V_octahedron : V_tetrahedron = 4 : 1.
(C) Take two opposite corners of a cube. Pull them apart that you
get a parallelepiped (zonotop) with face angles of 60 degree
at the pulling corners. Now we have six face diagonals which
have the same length as the side length of the parallelepiped.
So we see that the figure dissects into two regular tetrahedrons
and a regular octahedron. (a further solution of question A).
But zonotope are spacefillers.
(D) The insphere of the regular tetrahedron is the insphere of the
octahedron with vertices at the midpoints of the edges of the
tetrahedron. Therefore r_octahedron : r_tetrahedron = 2 : 1.
Note that this solution works in the nonregular case too.
The solution (E) shows another method to solve the regular case.
(E) The hexahedron is composed of two regular tetrahedrons, with
volume V_tetrahedron. Now V_octahedron : V_tetrahedron = 4 : 1.
(see problem (B)). Hence V_octahedron : V_hexahedron = 2 : 1.
The ratio of the surfaces is S_octahedron : S_hexahedron = 8 : 6.
Since V = r*S/3 with r the radius of the inscribed sphere we get
r_hexahedron : r_octahedron = 2 : 3.
(G) The cube with side lenght s has a volume s^3 per definition.
Drawing all space diagonals in the cube it dissectes into
6 square pyramids. Two of these pyramids give a nonregular
octahedron with volume s^3/3. For the regular octahedron we
have s : r = sqrt(2) : 1 by the Pythagorean thm. So we must
scale the height of the nonregular octahedron by the factor sqrt(2).
So we get V_octahedron = sqrt(2) s^3 / 3.
References:
- Editor;
Mathematical Sunshine,
Mathematics Magazine 54:3 (Mai 1981) 152
- students scrutinized PSAT and SAT of the Educational Testing Service
- problem (A) has been used since years in New York assuming the answer 7.
- Howard Evens;
(problem E)
American Mathematical Monthly 56 (Dec. 1949) 693
- C. Juel;
\"Uber das Volumen der Pyramide
Jahresbericht DMV 12, 500 (1903)
JFM 34.0532.03
The square pyramid where the angle between the each face and the base is
45 degree and a cube with the same volume are dissection equivalent.
- Arthur K\"opps;
Sonderpreisaufgabe 275: Die verflixten Pyramiden,
Monoid (Mathematikblatt f\"ur Sch\"uler und Lehrer) 6:16 (Dec. 1986) 89
Karolinen-Gymnasium, Frankenthal/Pfalz, Germany und
Gymnasium an der Frankenstrasse, Alzey, Germany
- problem (A) has been used since years in Florida assuming the answer 7.
- Serge Lang;
Math. Encounters with high school students,
Springer Verlag 1985. ISBN: 3-540-96129-1
Mathe! - Begegnungen eines Wissenschaftlers mit Sch"ulern,
Vieweg, Braunschweig 1991. ISBN 3-528-08942-3
(Reiss, K. (1993). Rezension zu Serge Lang: Mathe! Braunschweig 1991.
Zentralblatt f\"ur Didaktik der Mathematik, 25, 53-54.)
MATHDI 1992c.03906
Keywords: Circles; Area; Perimeter; Polyhedra; Spheres; Volume; Surface Area
- I. F. Sharygin;
Volumes without Integrals,
Quantum (March/April 1997) 32-33
- volume of the tetrahedron by scaling
- Charles W. Trigg;
Mathematical Quickies,
New York 1967, (reprint: Dover Publ., 1985)
- Q 113: Relative Polyhedral Volumes
- Q 114: The Space Filler
- Q 240: Incribed Spheres
- Donald B. Wagner;
An early Chinese derivation of the volume of a pyramid:
Liu Hui, third century A.D.
Historia mathematica, 1979, 6:164-188.
http://www.staff.hum.ku.dk/dbwagner/Pyramid/Pyramid.html
- Time 1981-03-31 p51 (problem A)
- Newsweek 1981-04-06, p84 (problem A)
Appendix:
Minkowski's classical result:
The lattice packing density of a regular octahedron is 18/19.
Exercise: Try to find Minkowski's packing without his theory.
Betke, Henk [math.MG/9909172] "Densest lattice packings of 3-polytopes".
(from the xxx archive http://front.math.ucdavis.edu/)
--
mailto:Torsten.Sillke@uni-bielefeld.de
http://www.mathematik.uni-bielefeld.de/~sillke/