Integral Regular Truncated Pyramids with Rectangular Bases

Transcription

1 Integral Regular Truncate Pyramis with Rectangular Bases Konstantine Zelator Department of Mathematics 301 Thackeray Hall University of Pittsburgh Pittsburgh, PA 1560, U.S.A. Also: Konstantine Zelator P.O. Box 480 Pittsburgh, PA 1503, U.S.A. aresses: 1) ) 1

2 1. Introuction A regular truncate pyrami with rectangular bases can be obtaine by intersecting a regular pyrami with a rectangular base; with a plane parallel to the base of the pyrami, somewhere between the base an the apex of the pyrami. Consequently, a regular truncate pyrami with rectangular bases; consists of two parallel rectangular bases such that the (straight) line connecting the two centers (of the two rectangular bases) is perpenicular to the two parallel planes containing the two rectangular bases; that line being one of the axes of symmetry of the (truncate) pyrami. The other four faces of the truncate pyrami, are all isosceles trapezois; occurring in two pairs, each pair containing two congruent isosceles trapezois facing (or opposite) each other. Definition 1: An integral regular truncate pyrami with rectangular bases is a regular truncate pyrami with rectangular bases whose twelve eges have integer lengths; whose height (the istance between the two bases) is also an integer; an whose volume is an integer as well. Of course in the case of a regular truncate pyrami with rectangular bases; the four lateral eges have the same length t. The bottom (larger) base is a rectangle of imensions a an b; an the top base a rectangle of imensions c an. So if H is the height of the pyrami; we really have six lengths involve: t,h,a,b,c, an. Because we are ealing with a regular pyrami, the two rectangle centers are aligne perpenicularly (to the two parallel bases) ; an thus, the two rectangular bases are either both nonsquare rectangles; which means that (with larger a of a an b; c the larger of c an ) we have, a>b, c>, a>c, an b>. Or, alternatively, both bases are squares in which case we have a=b > c=. The aim of this paper is to parametrically escribe the entire set of integral regular truncate pyramis with rectangular bases. We o so in Proposition 1, Section 7.

3 In Section we provie some illustrations of a regular truncate pyrami s (with rectangular bases) faces an cross sections. In Section 3, we erive a formula for the volume of such a pyrami. In Section 4, we erive the key equation of this paper. It is this equation that allows us to parametrically etermine the set of all such integral pyramis. The key equation reuces to the iophantine equation, t²=x²+y²+z². This is a well-known equation, an it s general solution (in positive integers x, y, z, an t) can be foun in W. Sierpinski s book Elementary Theory of Numbers. (see Reference [1]) Also note that since x, y, z, t are positive integers in the above equation; an 3 is an irrational number; at most two among x,y,z can be equal. Accoringly, we istinguish between the case when x, y, z are istinct; an the case when two among them are equal. The latter case leas to the iophantine equation Z²=X²+Y², whose general solution we state in Section 6. In Section 5, we state the general positive integer solution of the equation t²=x²+y²+z². In Section 8, Proposition, we escribe a particular 3-paremeter set or family of integral regular truncate pyramis with rectangular bases. In Section 9, Proposition 3, we give a 4-parameter escription of the set of integral regular truncate pyramis with square bases. We en this paper with some closing remarks in Section 10.. Illustrations In the illustrations below, a an b are the sielengths of the bottom rectangular base; with a b. An c an the sielengths of the top rectangular base; with c. An consequently, a> c an b> must hol as well. An t stans for the common length of the four lateral eges. 3

4 O₁ is the center of the bottom rectangular base O₂ the center of the bottom base; the straight line segment OO 1 is perpenicular (or orthogonal) to the two parallel bases. Figure 1: 3-D picture of a truncate regular pyrami with rectangular bases Figure : Perpenicular (or orthogonal) projection of the top base onto the bottom page Figure 3: Bottom base 4

5 Figure 4: Top base Figure 5: Two congruent lateral faces (isosceles trapezois) Figure 6: Two congruent lateral faces (isosceles trapezois) 5

6 3. Geometric Consierations Figure 7: a b, c, a>c, b> PP = b, Q Q =, PP = a, QQ = c 1 1 A. The Key Ratio From the similar triangle O₁Q₁Q₃ an O₁P₁P₃ we have, O Q QQ = = PP OP b An from the similar triangles O₁Q₁Q₂ an O₁P₁P₂ we obtain, O Q QQ c = = PP a OP Therefor we conclue that a b c a = (1) b c = or equivalently, a This being the key ratio; an important conition that the four base sielengths of a regular truncate pyrami with rectangular bases must satisfy. 6

7 Now, consier the case in which a,b,c, are positive integers. In number theory, Eucli s Lemma postulates that if a positive integer i₁ ivies the prouct i₂i₃ of two other positive integers i₂ an i₃ ; an i₁ is relatively prime to i₂. Then i₁ must be a ivisor of i₃. Using Eucli s Lemma in (1) (one rewrites (1) in the form a=bc) one can easily prove that (1) implies, a = Nk1, b = Nk, c = Mk1, = Mk where N, M, k1, k are positive integers with k1, k being relatively prime; gc ( k1, k) = 1 () The conitions in () can be unerstoo as lowest terms conitions. Note that N is none other than the greatest common of a an b; N=gc (a,b). Likewise M=gc (c,). Observe that since a>c an b>; a an b cannot be relatively prime (for this woul imply N=1; an consequently a c an b ; contrary to a>c an b>). On the other han b an can be relatively prime: When c an are relatively prime in (); we have M=1 an so c=k₁, =k₂. Thus in such a case, a=nc an b=n. There is also the square case: a=b an c=. When a=b an c=; it follows that k₁ = k₂; which in turn implies (since gc (k₁,k₂)=1) that k₁=k₂=1. So in the square case a=b=n an c==m ; with the key ratio being equal to 1. B. A Formula for the Volume of a Regular Truncate Pyrami with Rectangular Bases The straight lineoo 1 ₁that connects the centers O₁ an O₂ of the two rectangular bases is perpenicular or orthogonal to the two parallel planes that contain the two rectangular bases. Consequently, the line 7

8 OO ₁; as well the four lines containing the four lateral eges; they all meet at an apex point A, as 1 illustrate in Figure 8 below. 3-D picture Figure 8 H₂=height of the top pyrami H₁=height of the larger pyrami H=H₁-H₂= OO 1 = height of the truncate pyrami Let V be the volume of the truncate pyrami, V 1 the volume of the larger pyrami; an V the volume of the larger pyrami; an the V the volume of the top or smaller pyrami. Then, 1 1 V = V1 V = H1ab Hc 3 3 (3) From the similar right (or 90 egree) triangles PO A an PO 1 1A it is clear that, ( a + b ) ( c + ) PO = = = H po c + H1 1 1 a + b ; H H 1 a b + 1 b = ( c ) + 1 ; an by (1) it follows that H H 1 b = (4) 8

9 From (4) we further get, H1 H b a c H c = = ; (5) by (1) An using H = H1 H we further get, H Hc a c = (6) An by (6) an (4), H1 Ha a c = (7) Combining (7), (6) an (3); yiels = 1 H 3 a c ; an using b= a c V a b c ; further gives V = 3 3 ( ) H a c ; 3c a ( c) which combine with ( )( ) 3 3 a c = a c a + ac + c ; prouces H ( a + ac + c V = 3c ) (8) 9

10 Now, when,h,a, an c are positive integers; the volume V is also an integer if an only if 3c is a ivisor of the prouct H ( a ac c ) + + Observe that if a=c(mo3). Then, a ac c a a a a O ( ) mo 3. So, when a an c belong to the same congruence class moulo 3; The integer a + ac + c is ivisible by The Key Equation Figure 9: PP 1 is one (of the four) eges of the top rectangular base; PP 3 4 is one (of the four) eges of the bottom rectangular base; PPPP is an isosceles trapezoi; one of the four lateral faces. The point R 1 is the orthogonal projection of P1 onto the bottom base; R is the perpenicular projection of R1 onto the ege PP 3 4 ; an also of P1 onto the ege PP 3 4 a c As we have alreay seen, PP 1 4 = t, PR 1 1 = H = height of the truncate pyrami; RR 1 = y= b an P4R = x= (see Figure ). Thus, from the right triangles PR 1 P4an PRR 1 1 we have; PP 1 4 = P4 R + P1R an PR = RR + PR (9) An so from (9) we obtain the key equation t a c b = H + + ; or equivalently. 10

11 { 4 t 4 H ( a c ) ( b ) } = + + (10) So, when t, H, a, b, c, are positive integers; with a>c, b>, a b, an c. The sum of the squares ( a c) ( b ) + must be accoring to (10) ; congruent to 0(mo4). But the square of an integer is congruent to 0 or 1 moulo 4; accoring to whether that integer is even or o. It becomes clear that both positive integers, a-c an b- must be even. Therefore, in conjunction with (10) we must have, t = H + y + x a c = y, b = x; with t, H, x, y, a, c, b, all being positive integers that also satisfy a b an c Note that the conitions a>c an b> are automatically satisfie in view of the fact that y an x are positive integers. Also a b c y x + + ; c ( x y) But also from (1), we have a b a c b = = ; an so, y x = ; c c c y c = An so, x ( x) y y c = = x x An so, the conitions ( y x) c = x ; further require, An c ( x y) ( y x) x ( y x) x ( x y) ; ( y x) + 0; ( y x) [ x ] x + 0 ; 11

12 which is equivalent (since, x, y are positive integers) to y x 0 ; or x y We can finalize our conitions as follows. t = H + y + x a c = y, b = x, an, y x; Where t, H, y, x, c, are positive integers; an thus so are a an b. ( An thus also, the conitions a > c, b >, a b, c ) y An with c = ( so x being a ivisor of y ) x An thus the necessary conition ( 1 ), a b = is also satisfie c (11) 5. The Diophantine Equation t = + y + x A erivation an the general solution in positive integers, to the equation t = z + y + x ; can be foun in W. Sierpinski s book Elementary Theory of Numbers (see Reference [1]). The entire solution set in positive integers x, y, z, t; can be escribe in terms of three integer parameters m, l, an n as follows. 1

13 The positive integers x, y, z, an t satisfy the equation t = z + y + x, if an only if l + m + n l + m l + m n t= = + nz, = my, =, lx= n n n l + m = n n Where m, n, l are positive integers such that 1 n < l + m ( so that n < l + m ) An with n being a ivisor of l + m (1) In Reference [1], the reaer will also fin a list of particular solutions to the above equation. Observe that when l + m is a prime number (as it is well known in number theory; only an primes congruent to 1 moulo 4, can be expresse as sums of integer squares. Primes congruent to 3 mo 4 cannot) ; n=1 is the only positive ivisor of l + m, which is less than l + m ; since the only positive ivisors of a prime; are itself an 1. Regarless of whetherl + m is a prime or not, the choice n=1 generates a subfamily of solutions. We have the following. A subfamily of the entire family of solutions of positive integer solutions to the equation in (1) is {,,, 1,,, 1,, + } s = x y z t x= l + m y = l z = m t = l + m + l m (13) the set, ( ) So, the set S is a -parameter subfamily of positive integer solutions to the iophantine equation in (1). Observe that at most two of x,y,z can be equal; since 3 is an irrational number. When two of x, y, z to are equal. The equation t = z + y + x reuces to the Diophantine equation Z = X + Y. 13

14 6. The Diophantine equation Z = X + Y This equation is well-known in the literature. For this an relate historical material, see Reference []. In Reference [3] (this is an article publishe in 006), the reaer can fin a etaile analysis an erivation of the general positive integer solution to the equation Z = X + k Y, where k is a fixe positive integer, k. We state the general solution to Z = X + Y below. All the solutions, in positive integers X, Y, Z to the equation Z = X + Y. Can be parametrically escribe in terms of three parameters: x= δ m n, Y = δmnz, = δ( m + n ) Where δ, m, n are positive integers such that gc m n i e m an n are relatively pri (, ) = 1 (,, me) (14) 7. Integral Volume an Proposition 1 a b a c b Going back to (1), = = ; an so c c 1 b c a c = (15) From (11) an (15) we further obtain, we further obtain, c x = (16) y Now, we combine (16), (11), an (8) (the volume formula) to get, 14

15 H x y y y y V = y+ + y+ + 3y x x x x ( ) ( ) ; y x y x y H x V = ; 3y x V ( ) ( ) H y x+ + x+ + = ; or equivalently (17) 3x 3 + 4x V = H y x + ; 3x or equivalently y 4Hyx V = yh + H + x 3 Recall from (11) that 1 x y an that x is a ivisor of y. Therefore, looking at the thir version of the volume formula in (17); we see that the first an secon terms are positive integers. The thir term in (17); will be a positive integer precisely when 3 is a ivisor of Hyx. An so, when this happens; the volume is an integer; an thus, the regular truncate pyrami with rectangular bases; is an integral one per Definition 1. We have the following: Proposition 1 Suppose that a, b, c,, x, y, t, an H; are positive integers satisfying all the conitions in (11). Furthermore assume that the integer 3 is a ivisor of Hyx. Then the regular truncate pyrami with rectangular bases; an with base imensions a, b (for the larger/bottom base) ; (for the smaller/top base) c, ; height H an the four lateral eges having length t; is an integral one (per Definition 1). 15

16 8. Proposition an Its Proof In the en of the previous section, Proposition 1 gives a general parametric escription of the entire family of integral regular truncate pyramis with rectangular bases. In this section, we make use of (13) an Proposition 1; in orer to obtain a 3-parameter subfamily of such pyramis. This is one in Proposition below. Proposition Let l, m, an D be positive integer satisfying the following conitions: M is a ivisor of l 1; That is, l 1 = m v, Where v is a positive integer; 1 m< l lm ( + v) 0(mo3) Define the following positive integers, x, y, a, c, b,, H, an t: = md, x = m, y = l, c = ld, ( ) ( ) ( ) a = c+ y = l D+, b= + x= m D+, H = l m m m v an l m m m + 1 = +, t= = + v + Then, the regular truncate pyrami with larger (rectangular) base lengths a an b; smaller (rectangular) base lengths c an ; height H, an with lateral ege length t; is a non-square integral (regular truncate) pyrami (with rectangular bases) Proof First note that the ivisibility conition in Proposition 1 is satisfie. Inee, = + = = + An 3x= 6 m. yh m( m v) l yh ml( m v). 16

17 Since l( m+ v) = 0(mo 3) ; we have ( ) 3 l m+ v = i, for some positive integer i. Thus, yh ( x) = m 3i = 6 mi = i(6 m) = i 3, which shows that the integer 3x is a ivisor of yh. Next, we show that all the conitions in (11) are satisfie. Since l > m; l > m; y = l > m= x, so the conition x y is satisfie. The efinitions of a = c+ y, b= + ; x immeiately imply that a > c an b> ; by virtue of the fact that x an y are positive integers. Next compare a with b: a= ld ( + ) an b= md ( + ), which establishes a> b; in view m< l. Likewise c = ld > = md. This establishes that the regular truncate pyrami with rectangular bases is non- square. Finally the positive quaruple (, xyht,,) = ( m, l, m + l 1, m + l + 1 ) satisfies the conition x y H t z + + = in (11); by (13). The proof is complete. Two remarks: Remark 1: If m is a ivisor ofl 1; then obviously m is a ivisor of l 1 = ( l 1)( l+ 1) ; an clearly (since m an l are positive integers), 1 m l 1< l. Likewise, if m is a ivisor of l + 1; then m is a ivisor of l + 1; an so m is a ivisor of l 1as well. A to this the conitions 1 m l+ 1 an l. Which clearly imply 1 m l ; but then m cannot equal l (for l cannot be a ivisor of l + 1; unless l = 1 ). An so, 1 m< l is satisfie. Remark : The conition lm ( + v) = 0(mo 3), says that 3 must ivie at least one of lm, + v. If 3 is a ivisor of l ; then l 1 = mv implies that neither m nor v can be ivisible by 3. An in that case, mv l 1 0 1; mv (mo 3); which means that either m 1an v ; or alternatively m an v 1(mo3). In either case, m+ v 0(mo 3). So, we see that when 3 ivies l. Then the conition l 1 mv; implies that 3 must also ivie the sum m+ v. On the other han, if l is not ivisible by 3. Then, l 1or (mo3). An so l 1(mo3); l 1 0(mo 3). Which implies that 3 ivies the prouct mv. But since 3 also ivies lm ( + v); an 3 oes not ivie l. It follows that 3 must ivie m+ v ; an since it also ivies mv. It follows that when l is not a multiple of 3; both m an v are multiples of 3. 17

18 9. The Square Case an Proposition 3 When each of the two rectangular bases is a square; we have in (11), a= b> c=, y = x, an t = H + y an a = y + (18) Combining (18) with (14) prouces, H = D m n, y = x = Dmn, t = D( m + n ); where m, n, D are positive integers; with m an n being relatively prime. (19) Combining (19) an (18) leas us to Proposition 3. Proposition 3 The set of integral regular truncate square pyramis with base lengths a an c ; a > c ; height H, an lateral sie length t ; can be escribe in terms of four integer parameters as follows: a = c + 4 Dmn, c = c, H = D m n, an t Dm n = ( + ). Where c, D, m, n are positive integers, with m an n being relatively prime. 10. Closing remarks The material on the subject matter of this work, seems pretty scant, as far as the literature is concerne. There is some material (a few pages long) on rational pyramis an triheral angles, in L.E. Dickson s book, History of The Theory of Numbers, Volume II. See Reference []. 18

19 References [1] W. Sierpinski, Elementary Theory of Numbers, Warsaw, 1964, 480 p.p. ISBN: For the equation t = x + y + z, see pages [] L.E. Dickson, History of Theory of Numbers, Volume II, 803 p.p. AMS Chelsea Publishing ISBN: (Volume II) For the equation x + ky = z (with k=) see page 41 For rational pyramis an triheral angles, See page 1. [3] Konstantine Zelator, The Diophantine Equation Value of 1 n x + ky = z an Integral Triangles with a Cosine Mathematics an Computer Eucation, Volume 30, number 3, Fall 006, pages