SCISSORS CONGRUENCE EFTON PARK

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1 SCISSORS CONGRUENCE EFTON PARK 1. Scissors Congruence in the Plane Definition 1.1. A polygonal decomposition of a polygon P in the plane is a finite set {P 1, P,..., P k } of polygons whose union is P and that pairwise only intersect on their boundaries. We will often write this using the notation P = P 1 +P + +P k. Definition 1.. Polygons P and Q are scissors congruent if there exist polygonal decompositions {P 1, P,..., P k } and {Q 1, Q,..., Q k } of P and Q respectively with the property that P i is congruent to Q i for 1 i k. Proposition 1.3. Scissors congruence is an equivalence relation on the set of polygons. Proposition 1.4. If two polygons are scissors congruent, then they have equal area. Lemma 1.5. Every polygon has a polygonal decomposition consisting of triangles. Proof. Let P be a polygon and choose a slope m that is different from each of the slopes of the sides of P. Lines of slope m through the vertices of P give a polygonal decomposition consisting of triangles and trapezoids: For each trapezoid, make a diagonal cut to decompose it into two triangles. 1

2 EFTON PARK Lemma 1.6. Every triangle is scissors congruent to a rectangle. Proof. Given a triangle, construct the altitude at the vertex with the largest angle. Cut the triangle with a line that is perpendicular to the altitude and bisects it. This cuts the triangle into three pieces which can be reassembled into a rectangle. Proposition 1.7. Any two rectangles with equal area are scissors congruent. Proof. A rectangle with base b and height h is clearly scissors congruent to a rectangle with base b and height 1 h. Therefore, given rectangles R i with base b i and height h i, i = 1,, we may apply the double the base and halve the height operation to one or both of R 1 and R to arrange that either (a) b 1 = b, in which case R 1 and R are congruent, or (b) we have the inequalities b 1 < b < b 1. For the latter case, consider this picture: G D A K F H C J E B The rectangles AEF G and ABCD are R 1 and R respectively. By elementary Euclidean geometry, we deduce that triangles GDK and JEB are each congruent to F HC and thus congruent to one another. Also, we see that the triangles GJF and KBC are congruent. Therefore is scissors congruent to ABCD = AEJKD + JEB + KBC AEF G = AEJKD + GDK + GJF. Corollary 1.8. If two polygons have equal area, then they are scissors congruent.

3 SCISSORS CONGRUENCE 3. Scissors Congruence in 3-Space Definition.1. A polyhedral decomposition of a polyhedron P in 3-space is a finite set {P 1, P,..., P k } of polyhedra whose union is P and that pairwise only intersect on their faces and/or edges. In this case we write P = P 1 + P + + P k. Definition.. Polyhedra P and Q are scissors congruent if there exist polyhedral decompositions {P 1, P,..., P k } and {Q 1, Q,..., Q k } of P and Q respectively with the property that P i is congruent to Q i for 1 i k. Proposition.3. Scissors congruence is an equivalence relation on the set of polyhedra. Proposition.4. If two polyhedra are scissors congruent, then they have equal volume. Question.5 (Hilbert s Third Problem). Is the converse of Proposition.4 true? Definition.6. Let A be a set of real numbers. integrally additive if whenever a finite sum n 1 α 1 + n α + + n m α m = 0 A function f : A R is for some integers n 1, n,..., n m and some elements α 1, α,..., α m of A, it is also true that n 1 f(α 1 ) + n f(α ) + + n m f(α m ) = 0. Definition.7. Suppose that P is a polyhedron, let A = {α 1, α,..., α m } be the set of dihedral angles of P, and for each 1 i m, let l i be the length (not necessarily an integer!) of the edge of P that forms the vertex of the angle α i. Given an integrally additive function f on A, the quantity f(p ) := l 1 f(α 1 ) + l f(α ) + + l m f(α m ) is called the Dehn invariant of P associated to f. Lemma.8. Suppose we have a polyhedral decomposition of a polyhedron P : P = P 1 + P + + P k. Let A be the set containing π and the dihedral angles of P, P 1, P,..., P k, and let f be an integrally additive function on A with the property that f(π) = 0. Then ( ) f(p ) = f(p 1 ) + f(p ) + + f(p k ). Proof. Our polyhedral decomposition of P divides each edge of P, P 1, P,..., P k into a union of line segments; we will call these line segments edgitos. Suppose an edgito has length l and corresponding dihedral angle α. The quantity lf(α) is called the weight of the edgito. A moment s thought yields that f(p ) is the sum of the weights of all the edgitos of P. Fix an edgito e with length l, let P i1, P i,..., P ij be the polyhedra in our decomposition that have e as part of one of their edges, and for 1 s j, let γ s be the dihedral angle in P is associated to e. Next, we study the quantity lf(γ 1 ) + lf(γ ) + + lf(γ j ), which is a subsum of f(p ). We consider three cases.

4 4 EFTON PARK Case 1: The edgito e lines entirely (except possibly for one or both endpoints) in the interior of P. Then whence γ 1 + γ + + γ j π = 0, 0 = f(γ 1 ) + f(γ ) + + f(γ j ) f(π) = f(γ 1 ) + f(γ ) + + f(γ j ) = lf(γ 1 ) + lf(γ ) + + lf(γ j ). Case : The edgito e lies on a face of P, but not on an edge. Then whence γ 1 + γ + + γ j π = 0, 0 = f(γ 1 ) + f(γ ) + + f(γ j ) f(π) = f(γ 1 ) + f(γ ) + + f(γ j ) = lf(γ 1 ) + lf(γ ) + + lf(γ j ). Case 3: The edgito e lies on an edge of P. If α is the dihedral angle of P associated to this edge, then the quantity γ 1 + γ + + γ j must equal either α or α π. In either event, the requirement that f(π) = 0 implies that and hence f(γ 1 ) + f(γ ) + + f(γ j ) = f(α), lf(γ 1 ) + lf(γ ) + + lf(γ j ) = lf(α), which is the weight of the edgito e in P. Therefore, if we sum over all edgitos in our polyhedral decomposition of P, we obtain the right hand side of ( ). Lemma.9. Let f be an integrally additive function on a set A and let α be a real number not in A. Then f can be extended to an integrally additive function on A { α}. Proof. For notational convenience, we use the notation nα α to denote a sum of integer multiples of elements of A with all but finitely many of the integers n α equal to zero. We consider two cases: Case 1: There exists no equation nα α + n eα α = 0 with n eα 0. In this case we can define f( α) to be any real number. Case : There does exist an equation (1) nα α + n eα α = 0

5 SCISSORS CONGRUENCE 5 with n eα 0. Choose one such equation and define nα f( α) = n eα f(α). We must show that f is integrally additive on Ã. linear dependence () mα α + m eα α = 0. Suppose we have an integral If m eα = 0, the lemma follows immediately, so suppose that m eα 0. Multiply Equation (1) by m eα, multiply Equation () by n eα, and subtract to eliminate α: (3) (neα m α n α m eα )α = 0. Then (4) (neα m α n α m eα )f(α) = 0. Next, multiply the equation (5) nα f(α) + n eα f( α) = 0 by m eα and subtract Equation (4) to obtain (5) neα m α f(α) + n eα m eα f( α) = 0 The fact that n eα 0 yields (7) mα f(α) + m eα f( α) = 0, as desired. Theorem.10. Let P and Q be polyhedra and define M = {π} A P A Q, where A P and A Q are the sets of dihedral angles of P and Q respectively. Suppose there exists an integrally additive function f on M such that f(π) = 0. If P and Q are scissors congruent, then f(p ) = f(q). Proof. Decompose P and Q into polyhedra P = P 1 + P + + P k Q = Q 1 + Q + + Q k with the property that P i = Qi for 1 i k. Using Lemma.9, we can extend f to be an integrally additive function on the set consisting of M and all the dihedral angles of our subpolyhedra. Obviously f(p i ) = f(q i ) for 1 i k, and so we have f(p ) = f(p 1 ) + f(p ) + + f(p k ) = f(q 1 ) + f(q ) + + f(q k ) = f(q) by Lemma.8. Lemma.11. The number 1 π arccos ( 1 3) is irrational.

6 6 EFTON PARK Proof. Using induction and the trig identity cos((n + 1)θ) = cos θ cos nθ cos((n 1)θ) we see that cos nθ = T n (cos θ), where T n is an n-th degree polynomial with integer coefficients and whose leading term is n 1. Set φ = arccos ( ) 1 3 and suppose that φ = p q π for some positive integers p and q. Then ( ) ( ( )) ( ( 1 p T q = T q cos 3 q π = T q cos p π )) = cos(pπ) = ±1, q whence 1 3 is a root of a polynomial with integer coefficients and leading term q 1 for some q. We deduce from Rational Root Test that 3 divides q 1 for some q, a contradiction. Theorem.1. A regular tetrahedron and a cube of equal volume are not scissors congruent. Proof. Let T and C be the regular tetrahedron and cube of volume 1 respectively. Obviously each dihedral angle of C is π and an easy exercise in analytic geometry shows that each dihedral angle φ of T equals arccos ( ) { 1 3. Let A be the set π, π, φ} and define f : A R by setting ( π ) f(π) = 0, f = 0, f(φ) = 1. I claim that f is integrally additive. Suppose that n 1 π + n π + n 3φ = 0 for some integers n 1, n, and n 3. If n 3 = 0, we immediately see that n 1 f(π) + n π ) = 0. + n 3φ = n 1 f(π) + n f ( π On the other hand, if n 3 0, then φ π = n 1 + n, n 3 contradicting Lemma.11. Now we compute the Dehn invariants of C and T associated to f. The length of each side of C is 1, so ( π ) f(c) = 1 1 f = 0. Let m be the length of each side of T. From the volume formula for a regular tetrahedron V = 1 m3 we compute m = ( 1 ) 1/ and thus ( ) 1/3 1 f(t ) = 6mf(φ) = 6 0. Applying (the contrapositive of) Theorem.10, we see that T and C are not scissors congruent.

7 SCISSORS CONGRUENCE 7 Theorem.13 (Sydler, 1965). If P and Q are polyhedra that have equal volumes and have the property that f(p ) = f(q) for every integrally additive function satisfying f(π) = 0, then P and Q are scissors congruent.