A Personal Account of Proving Descartes' TheoremAuthor(s): Derek BallSource: Mathematics in School, Vol. 8, No. 5 (Nov., 1979), pp. 34-35Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30213515 .

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A personal account

of proving Descartes Theorem

by Derek Ball, School of Education, Leicester University

In a recent number of Mathematics in School N. J. Behrens briefly described the deficiency theorem for polyhedra1. I myself met this theorem for the first time a few months ago. Since then its proof has troubled and interested me a great deal and I feel I should like to share some of my thinking about it.

The theorem was first proposed by Descartes in about 1639 (although the manuscript containing it was not published until 1860). The angle-deficiency at the vertex of a polyhedron is defined to be the amount by which the sum of the face angles at that vertex is less than four right-angles. Model-makers will recognise deficiency as some kind of measure of the amount of "bend" produced in a polyhedron at a vertex. Descartes' theorem states that the sum of the angle-deficiencies for any convex polyhedron is eight right-angles. The theorem seems to say that angle-deficiency is indeed a good measure of the "bend" at a vertex. We are in fact given the total amount of bend we need for the surface to join up and give us a closed polyhedron. A cube for example has eight vertices, each with an angle-deficiency of one right-angle. The total deficiency is thus eight right-angles as required. a regular tetrahedron on the other hand has fewer vertices but each supplies more bend. Readers may wish to consider other polyhedra and verify that the deficiency sum is always eight right-angles.

N. J. Behrens suggested that the easiest proof of Decartes' theorem follows from Euler's rule. This may be true - and indeed my first impulse on meeting the theorem was to prove it using Euler's rule. This I managed quite easily. But as I thought and read more about the theorem I began to feel very dissatisfied with this proof. Euler's rule was first stated over 100 years after Descartes' deficiency theorem. Also Euler's rule readily follows from Descartes' theorem (the two are equivalent for convex polyhedra) and in fact Descartes came fairly close to formulating Euler's rule 120 years before Euler. Further- more, since Descartes' theorem is apparently metrical rather than topological in character I felt it might be worth spending time looking for a metrical proof which would make me come to terms with the angle concepts involved.

I discovered that Coxeter gives such a proof2. In fact Coxeter alleges that the theorem is "most easily established by spherical trigonometry, using the well-known fact . . . the area of a spherical triangle." Coxeter may of course be right that his proof is the easiest. However, the fact about the area of spherical tri- angles was not quite as well known by me as it obviously ought to have been. In any case, the use of spherical trigonometry cleverly shielded me from having to face up to what I was beginning to see was the crucial issue - the relationship

between plane and solid angles. So after a certain amount of playing around I devised the following proof. I do not make any claims about its being easy - in fact I find it quite difficult. What is more, since some of the ideas need to be visualised, the reader will probably have to exercise his imagination quite considerably if he is to share my proof.

We have a convex polyhedron and we choose a point 0 on its interior. I want to consider one of the vertices of the poly- hedron. I shall suppose for simplicity that it is a vertex V at which just three edges of the polyhedron meet. I draw through 0 three planes perpendicular to these three edges. These planes will divide the space round 0 into eight cells. One of the cells is the cell containing the vertex V and it is the solid angle of this cell which I want to determine. I choose to measure this solid angle as a fraction of the space around 0. In finding this solid angle I shall get involved with plane angles and it is convenient to measure these too in terms of the fractions of space round a point in a plane. I first want to represent the eight cells diagrammatically in an obvious kind of way.

I 31 I

P I T U "2 / I

R S I V W

I suppose that P is the cell which contains the vertex V. I use p, q, r ... to denote the solid angles of the cells P, Q, R .... I want to determine p. If a, is the dihedral angle between the planes n2 and tn3 then I can see that p+q=a,. So too, if a2 and a3 are the dihedral angles between the planes n 3 and n1 and the planes

tt and n2 respectively then p +r = a2 and

p+t=a3.

Thus 3p+q+r+t=a, +a2+a3. Now s=t, since the cells S and T are vertically opposite. Therefore 3p + q + r + s = a + a2 +a3. On the other hand p+q+r+s=+.

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Therefore 2p = a + 0a2 +a3 -2

Now I consider the intersection of the planes n2 and nr3 with the appropriate face of the polyhedron. Suppose rT2 and nr3 meet this face in the lines 12 and 13. The situation is shown in the diagram, where I, is the face angle at V.V

Thus fl,+a=,

. So too f2+a22=P3+a3=. Therefore

al + a2 + a3 1- (WI +32 +3).

Therefore 2p=a, +a2 +a3- = 1I-(Ill +P2 +93)-

= 1-(p1 +92+93)3 =the deficiency at V. o

Therefore the solid angle at 0 belonging to the cell containing V is equal to half the deficiency at V. I shall leave it to the reader to extend this result to the case where more than three edges meet at V. It is not particularly easy to do this, but it is an exercise in visualising the situation and is perhaps very useful in any case if this proof is to be taken on board. Having established this result, which is, amazingly, independent of the position of 0, for any vertex I can now add up the deficiencies for all the vertices. This will be twice the sum of the solid

angles at 0 and will thus be 2 (or if you prefer eight right- angles).

In no way am I suggesting that this is an easy proof. In fact I find it quite hard. But it does for me what the other proofs do not do. It takes me to what I feel is the naked truth of the situation. The Euler rule proof did not satisfy me because it seemed a slick way of proving the theorem and iemote from the theorem's meaning. Coxeter's proof seemed to take too much spherical trigonometry for granted and in this way to obscure the key ideas. My proof may not satisfy you. You may find all kinds of objections to it - or you may be quite satisfied by one of the others and find mine unnecessarily hard to follow. This is a result of the fact that proving is a very personal affair and that what convinces me may not convince you. It is perhaps useful to bear this in mind when we offer our children proofs as part of our mathematics teaching. Why do we want to prove? Are easy proofs the best? And in what ways are the proofs we offer helping those who are learning?

References

1. N. J. Behrens (1979) "The angle deficiency of a polyhedron", Mathematics in School, 8, 2, 30.

2. H. S. M. Coxeter (1948) "Regular Polytopes", pp. 23-24, MacMillan.

pBqp SOLUTIONS to Puzzles, Pastimes, Problems on page 18

1. Conjuring Tricks with Numbers Take x as the first number and y as the second, where x is less than y. Then 10y- 9x= 10(y-x)+x. The result of this calcula- tion is a number whose units digit is x (the first chosen number), and whose tens digit is y - x. To discover the second chosen number, simply add the number x that you have just found to y- x, the tens digit (which may be greater than 10 if y has two or more figures).

(a) 9 and 10, (b) 1 and 11, (c) 7 and 13, (d) 7 and 67, (e) 9 and 87, (f) 6 and 112, (g) 7 and 463.

2. A Geometrical Paradox When there are three points A, B, C on the circle, the sides of the "tri-lateral" ABC are AB, BC, CA. When there are only two points on the circle, A and B at the ends of a diameter, the sides of the "duo-lateral" are AB and BA, and the "space" between these two lines together with the two semi-circles make up the three parts (in accordance with the general formula). But the "space" between AB and BA is invisible, and so only two parts can be seen.

3. Powers of Twos and Threes Let y = 2" + 3n: then the table of values of y is

n 1 2 3 4 5 6 7

y 5 13 35 97 275 793 2315

We know that a3 + b3 = (a + bXa2 - ab + b2). Hence the sum of the cubes of any pair of numbers has the sum of the numbers as a factor. Similarly

as + bs = (a + bXa4- a3b + a2b2- ab3 + b4) and generally (a + b) is a factor of a2+1+ b2n+1

Hence (2+ 3) is a factor of (2n+ 3n) whenever n is an odd number. When n is an even number, 2n+ 3Y may be a prime number, but as 26+36=43+93, it has a factor (4+9)= 13. Further 29+ 39= 83+ 27a has a factor (8 + 27)= 35= 5 x 7, and 212+312= 163+813 has a factor (16+81)=97.

4. Octagons and Octagonal Stars

8 2

J77 3

NV 6

5

The angles at the points of the octagonal stars are 60', 90' and 45'.

5. The Difference Between Two Squares (i) If x2 -y2= p, where p is an odd prime number, then

x-y= 1, x+y=p lead to the solution

x= '/2(p + 1), y= '/2(p-1) and x and y are integers. So

7=4242-32 13=72-62, 37= 192-182 101= 512- 502

(ii) 15= 82 - 72= 402- 12, 33= 172- 162 =72- 42 105= 532-522=

192-132- 82= 112-42

(iii) Only even numbers that can be expressed as the product of two different even numbers can be written as the difference of two squares; 4, 6, 10 and 22 cannot be expressed in the form x2-y2, but

8= 32- 12, 32= 92-

72. =62- 22

48= 132- 112=822-42 72- 12

2n is possible only when n is an even number greater than 2, in which case

2n = (1/2n+ 1)2-Q(/2n-1)2 4n=(n+ 1)2-(n- 1)2

4mn= (m + n)2- (m - n)2 and

4n2 =(n2+ 1)2-(n2- 1)2

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