NOTES Edited by Sergei Tabachnikov

Hyperplane Sections of the n-Dimensional Cube Rolfdieter Frank and Harald Riede

Abstract. We deduce an elementary formula for the volume of arbitrary hyperplane sections of the n-dimensional cube and show its application in various .

1. INTRODUCTION. Intersecting a cube with a leads to quite different inter- section polygons, depending on the position of the plane. In each single case, these polygons can be found easily by calculation, and their possible areas range from 0 to p2 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 hyperplane. Evers showed in [5] that at least 30 dif- ferent combinatorial types of intersection polyhedrons occur! Therefore, it is quite remarkable that in each case, regardless of which , there is a simple general formula for the volume of the intersection . It was found by Ball [1]; Berger [4] and Zong [7] report on it.

n Theorem 1 (Ball 1985). Let Cn 1, 1 be an n-dimensional cube with n 2 and let =[ ]

n n H x R a x b with a (a1,...,an) (R 0 ) , b R, ={ 2 | • = } = 2 \{ } 2 be a hyperplane. Then

n a n 1 1 sin(ak t) Vol(C H) | | 2 cos(bt) dt. (1) n \ = ⇡ · a t · k 1 k ! Z1 Y= Here and subsequently, Vol denotes the (n 1)-dimensional volume. n 1 Ball uses formula (1) to prove Vol(Cn H) 2 p2. This bound is best pos- sible for each n and was conjectured by Hensley\  in 1979. For details see [1], [4], or [7]. But trying to apply this amazing formula to any concrete case quickly dampens the initial enthusiasm. Berger writes in [4], p. 468: “It seems that there is no direct geo- metric 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 (1) (see [6]) and obtained formula (2).

http://dx.doi.org/10.4169/amer.math.monthly.119.10.868 MSC: Primary 51M20, Secondary 51M25

868 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119 Theorem 2. Let the same assumptions hold as in Theorem 1. Then

1 a n Vol(C H) | | a (2) n \ = 2(n 1) · k k 1 ! ! Y= n n 1 (a w b) sgn(a w b) w . · • + • + · k w 1,1 n k 1 2X{ } Y= Here and subsequently, sgn denotes the sign of a real number.

We will prove Theorem 2 directly, without making use of (1). This is easier than calculating the integral and works without the methods of probability theory, which are used in Ball’s proof of Theorem 1.

Remark 1. The expression (1) is a slight modification of that found in [1], [4], and [7]. There, a 1 is assumed and C is the unit cube 1 , 1 n. | |= n [ 2 2 ]

Remark 2. Our assumption ak 0 for all k is no restriction. Namely, if a1,...,am 6= m6= 0 and am 1,...,an 0, then Cn H (Cm H 0) Cn m with H 0 x R a x + a x =b and therefore\ = \ ⇥ ={ 2 | 1 1 +···+ m m = }

Vol(Cn H) Volm 1(Cm H 0) Voln m (Cn m ). \ = \ ·

Now, Volm 1(Cm H 0) can be calculated by means of (2). \ Remark 3. Theorem 2 can be generalized to hyperplane sections of parallelo- topes. Let A be a real invertible n n-matrix. Then A(Cn) is a parallelotope. Since 1 n T ⇥ 1 A (H) x R A a x b , we obtain Vol(Cn A (H)) if we replace a by T ={ 2 | • = } n \ A a in formula (2). If subsets of R are multiplied by A, then their n-dimensional volumes are multiplied by det A . With this we deduce that | | 1 Vol(A(C ) H) Vol A C A (H) n \ = n \ T 1 1 a A a det A Vol C A (H) . =| |·| | ·| |· n \ Remark 4. Each facet of Cn H is a section of H with a facet of Cn, which is an (n 1)-dimensional cube. Hence,\ for n 3, Theorem 2 yields a formula for the sum of the (n 2)-dimensional volumes of all facets of Cn H. For n 3, this sum is just the circumference of the intersection polygon. For n\ 4, it is the= surface of the = intersection polyhedron. With a a 2 a2, this sum is | |i = | | i q 1 n n n a i ai n 2 | | · a (a w b) sgn(a w b) w w . (3) 2(n 2) · k · • + • + · i · k i 1 k 1 ! w 1,1 n k 1 X= ! Y= 2X{ } Y= Remark 5. In H x Rn a x b , we may assume a 1. Then b is the distance of the hyperplane={ 2 H |from• the= origin,} and integration| of|= equation (2| )| over b from b to b b0 yields a formula for the volume of the intersection of Cn with = 1 =n the half x R a x b0 . Such a formula has already been deduced in [3] by Barrow and{ Smith.2 Vice| • versa, we} could obtain our equation (2) by differentiating their formula with respect to b ( in the Barrow and Smith formula).

December 2012] NOTES 869 n 2. PROOF OF THEOREM 2. As above, let Cn 1, 1 be an n-dimensional n =[ ] cube with n 2 and let H x R a x b with a (a1,...,an) (R n ={ 2 | • = } = 2 \ 0 ) , b R, be a hyperplane. For k 1, 2,...,n, we define the functions Ak by {A }(t) 21 if a t a , A (t) =0 otherwise. By F we denote the convolu- k = | k |   | k | k = n tion A1 ? ? An; thus we have F1 A1 and Fn(b) 1 Fn 1(b t) An(t) dt for ··· = = · n > 1 and b R. 1 We prove2 the following lemma by induction on n andR include the case n 1 in order to facilitate the basis of induction. =

Lemma 1. Using the notation defined above, it follows that

n 1 Vol(C H) a a F (b) n \ =| |· k · n k 1 Y= for all n 1. n Proof. Since the reflection of R in the hyperplane xk 0 is a symmetry of Cn, we may assume a > 0 for all k 1, 2,...,n. = k = The intersection of the interval 1, 1 with H x R a1x b has the 0- [ ] ={ 2 | = } 2 1 dimensional volume A (b) F (b) a a F (b). Hence, Lemma 1 is correct 1 = 1 = 1 · 1 1 for n 1. Now let n > 1 and assume Lemmaq 1 is true for n 1. For every t R, let = 2 Ht denote the hyperplane xn t. Furthermore, we introduce a0 (a1,...,an 1, 0), n n 1 = n = k 1 ak , 0 k1 ak , and the hyperplane H 0 y R a0 y 0 . = = = 1 = ={ 2 | • = } It follows that 1 Voln 2(Cn H Ht ) dt is the volume of the orthogonal pro- Q Q \ \ jection of Cn H onto the hyperplane H 0. Hence, we obtain Vol(Cn H) if we \ R \ divide this integral by the cosine of the angle between H and H 0. This cosine is a a0 equal to a • a . Since Cn Ht is an (n 1)-dimensional cube, we have by hypoth- | |·| 0| \ a0 esis Voln 2(Cn H Ht ) | | Fn 1(b ant). Therefore, we get \ \ = 0 · a 1 Vol(Cn H) | | Voln 2(Cn H Ht ) dt \ = a0 · 1 \ \ | | Z 1 a a0 | | | | Fn 1(b ant) dt = a0 · 1 0 · | | Z a an 1 | | Fn 1(b t) dt = 0 · an · an Z a 1 a | | Fn 1(b t) An(t) dt | | Fn(b). = · · = · Z1 Lemma 2. Using our previous notation, it follows that

n 1 n 1 F (b) (a w b) sgn(a w b) w n = 2(n 1) · • + • + · k w 1,1 n k 1 ! 2X{ } Y= for all n 2 and b R. 2 n 1 Proof. We proceed by induction on n and define µ0(w) k1 ak wk and ↵0(w) n 1 = = = k1 wk . First, for all n 2 we have that = P Q 870 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119 an 1 Fn(b) Fn 1(b t) An(t) dt Fn 1(b t) dt. = · = an Z1 Z For the next transformation, we use in the case n 2 that = 1 F1(t) A1(t) (sgn(a1 t) sgn( a1 t)) , t R a1, a1 = = 2 + + 2 \{ }; in the case n > 2, we use the induction hypothesis and get in both cases

an Fn 1(b t) dt an Z an 1 n 2 (µ0(w) b t) sgn(µ0(w) b t) ↵0(w) dt = an 2(n 2) · + + · Z w 1,1 n 1 ! 2{X} an 1 n 1 (µ0(w) b t) sgn(µ0(w) b t) ↵0(w) = 2(n 1) · + + ·  w 1,1 n 1 an ! 2{X} n 1 n 1 (a w b) sgn(a w b) w . = 2(n 1) · • + • + · k w 1,1 n k 1 ! 2X{ } Y=

Now we obtain formula (2), if in Lemma 1 we replace Fn(b) by the expression calculated in Lemma 2.

3. APPLICATIONS OF THEOREM 2. We can check formula (2) geometrically by finding the vertices of Cn H as intersection points of H with the edges of Cn, and calculating the volume of the\ convex hull of these vertices. We give two examples in dimensions 4 and 5 and invite the reader to produce their own in dimensions 2 and 3. n For H x R x1 xn b , formula (2) simplifies to ={ 2 | +···+ = } n pn i n n 1 Vol(C H) ( 1) (n 2i b) sgn(n 2i b). (4) n \ = 2(n 1) · i + + i 0 ✓ ◆ ! X= 4 Example 1. Let the hyperplane H in R be given by x1 x2 x3 x4 2. Then (4) yields + + + =

p4 Vol(C H) (4 2)3 4(2 2)3 6(0 2)3 4( 2 2)3 ( 4 2)3) 4 \ = 12 + + + + + + + 8 . = 3 Now, C H has as vertices the 4 permutations of (1, 1, 1, 1), and is thus a reg- 4 \ ular tetrahedron with edge length 2p2.

5 Example 2. Let the hyperplane H in R be given by x1 x2 x3 x4 x5 0. Then formula (4) yields + + + + =

p5 115 Vol(C H) 54 5 34 10 14 10 14 5 34 54 p5. 5 \ = 48 · + · + · · + = 12 December 2012] NOTES 871 Now, C H has as vertices the 30 permutations of (1, 1, 0, 1, 1). Here the 5 \ geometric calculation of Vol(C5 H) is more difficult. In fact, C5 H has 10 con- gruent facets in the 10 hyperplanes\ x 1 (i 1,...,5). The facet\ with x 1 i =± = 1 = has as vertices the 12 permutations of (1, 1, 0, 1) (with the x1-component omitted) 1 1 1 1 and the center (of gravity of the vertices) ( 1, 4 , 4 , 4 , 4 ). It is bounded by 4 regu- lar sixgons with equations x 1 (i 2,..., 5) and 4 regular triangles with equa- i = = tions x 1 (i 2,...,5) with side length p2. Hence, each facet is a truncated i = = tetrahedron with edge length p2 and volume 23 . So we can dissect C H into 3 5 \ 23 1 p 10 congruent 4-dimensional pyramids with base volume 3 and height 2 5 to get Vol(C H) 10 1 23 1 p5 115 p5. 5 \ = · 4 · 3 · 2 = 12 4. OPEN QUESTIONS.

n 1 Question 1. Ball’s proof of Hensley’s conjecture Vol(Cn H) 2 p2 is quite complicated. Is there a simple proof using Theorem 2? \ 

Question 2. Ball generalized Theorem 1 to subspaces Hk of arbitrary dimension k < n k n and used this generalization to prove Vol (C H ) 2k p2 . This bound is the k n \ k  best possible if 2k n, but, for example, the maximal area of C5 H2 is unknown (see [2] or [7]). Can we generalize Theorem 2 and obtain an elementary\ formula for Vol (C H )? k n \ k REFERENCES

1. K. Ball, Cube slicing in Rn, Proc. of the AMS 97 no. 3 (1986) 465–473. 2. , Volumes of sections of cubes and related problems, Lecture Notes in Math. 1376 (1989) 251–260, available at http://dx.doi.org/10.1007/BFb0090058. 3. D.L. Barrow, P.W. Smith, Spline notation applied to a volume problem, Amer. Math. Monthly 86 (1979) 50–51, available at http://dx.doi.org/10.2307/2320304. 4. M. Berger, Revealed, Springer-Verlag, Berlin, 2010. 5. D. Evers, Hyperebenenschnitte des vierdimensionalen Wurfels¨ , Wissenschaftl. Prufungsarbeit,¨ Universitat¨ Koblenz-Landau, Koblenz, 2010. n sin(a x) 6. R. Frank, H. Riede, Die Berechnung des Integrals 1 k cos(bx) dx (to appear), available k 1 ak x 1 = · at http://www.uni-koblenz-landau.de/koblenz/fb3/mathe/Forschung/wuerfelintegral. R ⇣Q ⌘ pdf. 7. C. Zong, The Cube: A Window to Convex and Discrete Geometry, Cambridge Univ. Press, Cambridge, 2006.

Mathematisches Institut der Universitat¨ Koblenz-Landau, Campus Koblenz, D 56070 Koblenz, Germany [email protected] [email protected]

Tiling Hamiltonian Cycles on the 24-Cell

Jacob A. Siehler

Abstract. We present a construction for tiling the 24-cell with congruent copies of a single Hamiltonian cycle, using the algebra of quaternions. http://dx.doi.org/10.4169/amer.math.monthly.119.10.872 MSC: Primary 00A08, Secondary 05C45; 51M20

872 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119