NOTES Edited by Sergei Tabachnikov Hyerlane Sections of the n-dimensional Cube Rolfdieter Frank and Harald Riede Abstract. We deduce an elementary formula for the volume of arbitrary hyerlane sections of the n-dimensional cube and show its alication in various dimensions.. INTRODUCTION. Intersecting a cube with a lane leads to quite different intersection olygons, deending on the osition of the lane. In each single case, these olygons can be found easily by calculation, and their ossible areas range from 0 to 2 times the area of a face of the cube. However, the situation changes dramatically if we intersect the four-dimensional cube with a three-dimensional hyerlane. Evers showed in [5] that at least 30 different combinatorial tyes of intersection olyhedrons occur! Therefore, it is quite remarkable that in each case, regardless of which dimension, there is a simle general formula for the volume of the intersection olytoe. It was found by Ball []; Berger [4] and Zong [7] reort on it. Theorem (Ball 985). Let C n [, ] n be an n-dimensional cube with n 2 and let H {x 2 R n a x b} with a (a,..., ) 2 (R \{0}) n, b 2 R, be a hyerlane. Then Vol(C n \ H) a Z 2n Here and subsequently, Vol denotes the (n k! sin(a k t) cos(bt) dt. () a k t )-dimensional volume. Ball uses formula () to rove Vol(C n \ H) ale 2 n 2. This bound is best ossible for each n and was conjectured by Hensley in 979. For details see [], [4], or [7]. But trying to aly this amazing formula to any concrete case quickly damens the initial enthusiasm. Berger writes in [4],. 468: It seems that there is no direct geometric formula; any such formula would have to be different according to the various ways in which H meets the edges. We succeeded in calculating the integral in () (see [6]) and obtained formula (2). htt://dx.doi.org/0.469/amer.math.monthly.9.0.868 MSC: Primary 5M20, Secondary 5M25 868 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 9
Theorem 2. Let the same assumtions hold as in Theorem. Then Vol(C n \ H) a 2(n )!! a k (2) k w2{,} n (a w + b) n sgn(a w + b) Here and subsequently, sgn denotes the sign of a real number. w k. We will rove Theorem 2 directly, without making use of (). This is easier than calculating the integral and works without the methods of robability theory, which are used in Ball s roof of Theorem. Remark. The exression () is a slight modification of that found in [], [4], and [7]. There, a is assumed and C n is the unit cube [, 2 2 ]n. Remark 2. Our assumtion a k 6 0 for all k is no restriction. Namely, if a,...,a m 6 0 and a m+,..., 0, then C n \ H (C m \ H 0 ) C n m with H 0 {x 2 R m a x + +a m x m b} and therefore Vol(C n \ H) Vol m (C m \ H 0 ) Vol n m (C n m ). Now, Vol m (C m \ H 0 ) can be calculated by means of (2). Remark 3. Theorem 2 can be generalized to hyerlane sections of arallelotoes. Let A be a real invertible n n-matrix. Then A(C n ) is a arallelotoe. Since A (H) {x 2 R n A T a x b}, we obtain Vol(C n \ A (H)) if we relace a by A T a in formula (2). If subsets of R n are multilied by A, then their n-dimensional volumes are multilied by det A. With this we deduce that Vol(A(C n ) \ H) Vol A C n \ A (H) k a A T a det A Vol C n \ A (H). Remark 4. Each facet of C n \ H is a section of H with a facet of C n, which is an (n )-dimensional cube. Hence, for n 3, Theorem 2 yields a formula for the sum of the (n 2)-dimensional volumes of all facets of C n \ H. For n 3, this sum is just the circumference of the intersection q olygon. For n 4, it is the surface of the intersection olyhedron. With a i a 2, this sum is a 2 i n i a i a i 2(n 2)!! a k (a w+b) n 2 sgn(a w+b) w i w2{,} n k w k. (3) k Remark 5. In H {x 2 R n a x b}, we may assume a. Then b is the distance of the hyerlane H from the origin, and integration of equation (2) over b from b to b b 0 yields a formula for the volume of the intersection of C n with the half sace {x 2 R n a x ale b 0 }. Such a formula has already been deduced in [3] by Barrow and Smith. Vice versa, we could obtain our equation (2) by differentiating their formula with resect to b ( in the Barrow and Smith formula). December 202] NOTES 869
2. PROOF OF THEOREM 2. As above, let C n [, ] n be an n-dimensional cube with n 2 and let H {x 2 R n a x b} with a (a,..., ) 2 (R \ {0}) n, b 2 R, be a hyerlane. For k, 2,...,n, we define the functions A k by A k (t) if a k ale t ale a k, A k (t) 0 otherwise. By F n we denote the convolution A?? A n ; thus we have F A and F n (b) R F n (b t) A n (t) dt for n > and b 2 R. We rove the following lemma by induction on n and include the case n in order to facilitate the basis of induction. Lemma. Using the notation defined above, it follows that for all n. Vol(C n \ H) a k a k F n (b) Proof. Since the reflection of R n in the hyerlane x k 0 is a symmetry of C n, we may assume a k > 0 for all k, 2,...,n. The intersection of the interval [, q] with H {x 2 R a x b} has the 0- dimensional volume A (b) F (b) a 2 a F (b). Hence, Lemma is correct for n. Now let n > and assume Lemma is true for n. For every t 2 R, let H t denote the hyerlane x n t. Furthermore, we introduce a 0 (a,...,, 0), Q n k a k, 0 Q n k a k, and the hyerlane H 0 {y 2 R n a 0 y 0}. It follows that R Vol n 2(C n \ H \ H t ) dt is the volume of the orthogonal rojection of C n \ H onto the hyerlane H 0. Hence, we obtain Vol(C n \ H) if we divide this integral by the cosine of the angle between H and H 0. This cosine is equal to a a0. Since C a a 0 n \ H t is an (n )-dimensional cube, we have by hyothesis Vol n 2 (C n \ H \ H t ) a0 0 F n (b t). Therefore, we get Vol(C n \ H) a Z a 0 Vol n a a 0 a 0 a Z Z 2 (C n \ H \ H t ) dt a 0 F n (b t) dt 0 F n (b t) dt Lemma 2. Using our revious notation, it follows that F n (b) 2(n )! for all n 2 and b 2 R. F n (b t) A n (t) dt a F n (b). w2{,} n (a w + b) n sgn(a w + b) Proof. We roceed by induction on n and define µ 0 (w) P n k a kw k and 0 (w) Q n k w k. First, for all n 2 we have that k w k 870 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 9
Z F n (b) F n (b t) A n (t) dt F n (b t) dt. For the next transformation, we use in the case n 2 that F (t) A (t) 2 (sgn(a + t) sgn( a + t)), t 2 R \{a, a }; in the case n > 2, we use the induction hyothesis and get in both cases F n (b t) dt ale 2(n 2)! 2(n )! 2(n )! w2{,} n (µ 0 (w) + b t) n 2 sgn(µ 0 (w) + b t) 0 (w) dt w2{,} n (µ 0 (w) + b t) n sgn(µ 0 (w) + b t) 0 (w) w2{,} n (a w + b) n sgn(a w + b) w k. Now we obtain formula (2), if in Lemma we relace F n (b) by the exression calculated in Lemma 2. 3. APPLICATIONS OF THEOREM 2. We can check formula (2) geometrically by finding the vertices of C n \ H as intersection oints of H with the edges of C n, and calculating the volume of the convex hull of these vertices. We give two examles in dimensions 4 and 5 and invite the reader to roduce their own in dimensions 2 and 3. For H {x 2 R n x + +x n b}, formula (2) simlifies to Vol(C n \ H) n 2(n )! n n ( ) i i i0 k (n 2i + b) n sgn(n 2i + b). (4) Examle. Let the hyerlane H in R 4 be given by x + x 2 + x 3 + x 4 2. Then (4) yields Vol(C 4 \ H) 4 2 (4 + 2)3 4(2 + 2) 3 + 6(0 + 2) 3 + 4( 2 + 2) 3 ( 4 + 2) 3 ) 8 3. ), and is thus a reg- Now, C 4 \ H has as vertices the 4 ermutations of (,,, ular tetrahedron with edge length 2 2. Examle 2. Let the hyerlane H in R 5 be given by x + x 2 + x 3 + x 4 + x 5 0. Then formula (4) yields Vol(C 5 \ H) 5 48 54 5 3 4 + 0 4 + 0 4 5 3 4 + 5 4 5 5. 2 December 202] NOTES 87
Now, C 5 \ H has as vertices the 30 ermutations of (,, 0,, ). Here the geometric calculation of Vol(C 5 \ H) is more difficult. In fact, C 5 \ H has 0 congruent facets in the 0 hyerlanes x i ± (i,...,5). The facet with x has as vertices the 2 ermutations of (,, 0, ) (with the x -comonent omitted) and the center (of gravity of the vertices) (,,,, ). It is bounded by 4 regular sixgons with equations x i (i 2,...,5) and 4 regular triangles with equa- 4 4 4 4 tions x i (i 2,...,5) with side length 2. Hence, each facet is a truncated tetrahedron with edge length 2 and volume 23. So we can dissect C 3 5 \ H into 0 congruent 4-dimensional yramids with base volume 23 and height 3 2 5 to get Vol(C 5 \ H) 0 23 4 3 2 5 5 2 5. 4. OPEN QUESTIONS. Question. Ball s roof of Hensley s conjecture Vol(C n \ H) ale 2 n comlicated. Is there a simle roof using Theorem 2? 2 is quite Question 2. Ball generalized Theorem to subsaces H k of arbitrary dimension k < n and used this generalization to rove Vol k (C n \ H k ) ale 2 k 2 n k. This bound is the best ossible if 2k n, but, for examle, the maximal area of C 5 \ H 2 is unknown (see [2] or [7]). Can we generalize Theorem 2 and obtain an elementary formula for Vol k (C n \ H k )? REFERENCES. K. Ball, Cube slicing in R n, Proc. of the AMS 97 no. 3 (986) 465 473. 2., Volumes of sections of cubes and related roblems, Lecture Notes in Math. 376 (989) 25 260, available at htt://dx.doi.org/0.007/bfb0090058. 3. D.L. Barrow, P.W. Smith, Sline notation alied to a volume roblem, Amer. Math. Monthly 86 (979) 50 5, available at htt://dx.doi.org/0.2307/2320304. 4. M. Berger, Geometry Revealed, Sringer-Verlag, Berlin, 200. 5. D. Evers, Hyerebenenschnitte des vierdimensionalen Würfels, Wissenschaftl. Prüfungsarbeit, Universität Koblenz-Landau, Koblenz, 200. 6. R. Frank, H. Riede, Die Berechnung des Integrals R Qnk sin(a k x) a k x cos(bx) dx (to aear), available at htt://www.uni-koblenz-landau.de/koblenz/fb3/mathe/forschung/wuerfelintegral. df. 7. C. Zong, The Cube: A Window to Convex and Discrete Geometry, Cambridge Univ. Press, Cambridge, 2006. Mathematisches Institut der Universität Koblenz-Landau, Camus Koblenz, D 56070 Koblenz, Germany frank@uni-koblenz.de riede@uni-koblenz.de Tiling Hamiltonian Cycles on the 24-Cell Jacob A. Siehler Abstract. We resent a construction for tiling the 24-cell with congruent coies of a single Hamiltonian cycle, using the algebra of quaternions. htt://dx.doi.org/0.469/amer.math.monthly.9.0.872 MSC: Primary 00A08, Secondary 05C45; 5M20 872 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 9