SOME THOUGHTS ON POLYHEDRONS

SOME THOUGHTS ON POLYHEDRONS 851

terms. But some house-cleaning is surely indicated. Is thereroom for any discussion? Shall it be put to a vote? The authorvotes with Webster, the Oxford, Marks, et al. and against theEncyclopedia Britannica and others of the B class.

SOME THOUGHTS ON POLYHEDRONS

GERALD FREILICH1235 East 13th Street, Brooklyn, New York

The usual type of problem involving the numbers of vertices,edges, and faces of a polyhedron can be solved by a simple sub-stitution in the well known Euler’s formula V+F=E+2. Whileworking on these problems, I thought of a number of otherswhich to my mind were much more useful and interesting andwhich I could not solve by use of the above formula. To me aquestion such as: Can a crystal bounded by only three penta-gons and four triangles ever be found in nature? is more excitingthan to show that for any existing polyhedron, V-}-F=E+2.

In a polyhedron bounded entirely by triangles, two or moreof which may be co-planar, the number of edges equals 3/2 thenumber of faces.

852 SCHOOL SCIENCE AND MATHEMATICS

NOTES1. If a face is a polygon of n sides, it should be considered as

being composed of (n � 2) triangles which can be formed by draw-ing all possible diagonals from one vertex.

2. In order to study the relationship between the numbersof edges and faces of a polyhedron, I found it advisable to pro-ject the polyhedron on a surrounding sphere by-lines from thecenter of the sphere through the vertices of the polyhedron.

FIG. 2

FIG. 3

Thus there is set up a one-to-one correspondence between thevertices, faces, and edges of the polyhedron and the correspond-ing elements of the figure on the sphere.

Beginning with the simplest case of a polygon on a sphere,namely with triangle ABC (see Fig. 1), we have the sphere di-


vided into two triangular faces (F and FQ and three edges, AB,5C,andAC.To get the next case (a polygon having the next larger num-

ber of triangles), we first try to add only one edge (see Fig. 2).Let this be AXB. But this is impossible according to the condi-tions of the theorem since it creates a figure bounded by two

FIG. 4

FIG. 5

sides and not by three. We then try to add two edges, namelyAD and DC (see Fig. 3). But this figure again doesn’t satisfythe conditions of the theorem because ABCD is a quadrilateraland not a triangle. This can be remedied by drawing BD (seeFig. 4). Thus, I have the next case, formed by adding threeedges and two faces to the figure in the previous case.

854SCHOOL SCIENCE AND MATHEMATICS

Now to get the third case, we can think of triangle ADC aloneand temporarily forget about the other triangles within ADC.Hence, in order to add edges, vertices, and triangular faces tofigure ADC, we follow the same procedure as before becauseADC is outwardly a triangle. Therefore, to get the third case,(see Fig. 5) we must add three more edges, EA, EC, and ED,and as a result we have two more faces, ACE and DCE. In thesame manner we get the fourth and higher cases.We thus have the table:

Case

1

2

3

�

»n

Number ofVertices�V

3

4

5

�

w+2

Number ofFaces�F

2

4

6

�

2n

Number ofEdges�E

3

6

9

�

3n

Ratio E/F

3/2

3/2

3/2

�

3/2

From the table, it can be seen that

E 3n 3

F~2n~ 2

3FE=�� Therefore

2

COROLLARIES1. In a polyhedron bounded entirely by triangles, two or more

of which may be co-planar, twice the number of vertices lessfour equals the number of faces. Thus:

Since

therefore

2^=2(w+2)-4,F=27-4.

2. In a polyhedron bounded entirely by triangles, two or moreof which may be co-planar, three times the number of verticesless six equals the number of edges. Thus:

Since 3n= 3(w+ 2) � 6,

therefore £=37-6.

The usefulness of this theorem and its corollaries can be shown


in the following example. In finding the number of vertices in apolyhedron consisting of, let us say, three pentagons and fivetriangles, we divide the three pentagons into nine triangles. As aresult, we have a polyhedron of fourteen triangles or faces. Byusing F =2F�4, we find the number of vertices to be nine.

Still another use that can be made of these formulas is thedetermination of the possibility of building a polyhedron undercertain given conditions. Take for example the question men-tioned at the beginning of this article as to whether a poly-hedron can be constructed using three pentagons and four tri-angles. Considering the three pentagons as composed of ninetriangles, we have thirteen faces. Substituting in the formula,F=2F’�4, we get 7=8^. Therefore such a polyhedron is im-possible of construction.

Finally it is interesting to note that Euler^s theorem itself canbe derived from the above formulas as follows:

1. From the theorem, we have

F=jE.

2. From corollary two, we have

E+6 EV=��=�+2.

3 3

3. Adding steps 1 and 2, we have

F+V==E+2.

NEW AIRPLANE COMPASSA new airplane compass has been developed which is not thrown off by

the bombload, armor plate, or other metal parts of the plane, it was an-nounced at the Philadelphia Division of Bendix Aviation Corporation.Motion of the plane does not affect it either.

"This new compass will not go off its reading when the plane dives orclimbs rapidly, it will not lay or overshoot during a turn and it will notoscillate or ^unt’ back and forth in rough weather,57 the designer, W. A.Reichel, explained.The Gyro Flux Gate Compass uses the earth^s magnetic field to develop

minute electrical impulses which, when amplified, turn the compass indi-cator. The transmitter can be located at a distance from the master in-dicating dial. The principle of the gyroscope, is employed solely for stabiliza-tion.

This revolutionary type of compass is ^as great an advance over theconventional magnetic compass as that compass was over the lodestone,"Mr. Reichel said.

Documents

SOME THOUGHTS ON POLYHEDRONS