MATH

(Page under construction; reformatting in  progress)

This page is dedicated to collaborative mathematical investigations ongoing between some of my colleagues and myself.  If you came here looking for minerals, sorry!  Proceed at your own risk.  If this page was your destination, I hope you enjoyed the rocks on the way.

As in any refereed journal, the style is condensed; yet it is casual and structured for the use of the participants only.  Email contacts for all participants giving permission appear at the bottom of this page.

The purpose of this is to standardize notation and terminology, collect results and ideas in one accessible place, and perhaps serve as basis for more formal publications when warranted.  I will try to keep everything organized and properly attributed; and, before too long, to get it up-to-date.

TOPIC OUTLINE

I:  PYTHAGOREANA

A:  Fundamental Definitions and Properties

B:  Pythagorean Triplets

C:  Sums of Pythagorean Tangents can be Pythagorean Tangents

D:  Pythagorean Tilings of Rectangles and Squares

E:  Contributors

F:  References

II:  PATHOLOGICAL FUNCTIONS

ARTICLES

IA:  PYTHAGOREANA (Fundamental Definitions and Properties)

Defs:

1:  A p-triangle is a right triangle with integer side-lengths (a Pythagorean triangle).      A p-triangle will be called primitive if its sides are pairwise relatively prime.

2:  A p-angle  is an angle for which all trigonometric functions have rational values (or a quadrantal angle).  The values of these functions will be called p-sines, etc.

3:  A p-triple is any triple of integers (positive or negative) satisfying the Pythagorean relationship.

Notation:    We will denote a p-triangle by  (A, B; C) where A and B are the legs in either order and C is the hypotenuse.

Remarks:    These are elementary, proofs are omitted.

1:  Every p-triangle is a "multiple" of a primitive p-triangle:  (A, B; C) = (ma, mb; mc) - m(a,, b; c).  We shall use small letters only when the triangle is known to be

primitive.  The same goes for p-triplets.

2:  In any p-triangle, one leg is a multiple of 3, one leg is a multiple of 4, and one side is a multiple of 5.

3:  Any primitive p-triangle t is uniquely generated by a pair of relaitvely prime positive integers [u,v] ( u<v, one even and the other odd ) by the equations

a = u^2 - v^2  (odd leg),   b = 2uv  (even leg),   c = u^2 + v^2  (hypotenuse).

Using basic trig identities or complex number multiplication properties, it is easy to show the following:

Theorem 1:  Sums and differences of p-angles are p-angles.

I am indebted to Richard Guy for the following generalization to "Heronic" triangles.

Def:

An "h-thing"  (Heronic thing) is something associated with a triangle with integer sides and integer area.

Remarks:  Clearly, every p-triangle is an h-triangle.  It is also true that every angle of an h-triangle is a p-angle (Lemma 1 below).  This observation

will become valuable in the construction of p-tilings later.

Lemma 1:  Every angle of a Heronic triangle is a Pythagorean angle.

Proof:          Construct an internal altitude h of the triangle (see figure to right - click on it to enlarge).           Letting the area of the triangle be A, we have                   h^2 = c^2 - x^2 = b^2 - (a - x)^2,          which upon expansion yields an equation linearly involving x.  Further,                   A = [1/2] a h ,          so all segments in the figure have rational lengths.  The result follows.        Insert thumbnail

1B:  PYTHAGOREANA (Pythagorean Triplets):

1:  AN INTERESTING RELATIONSHIP:

Ed Sherman came up with an interesting relationship between triples of Pythagorean triplets.  He cited several examples - here is one:   Consider the two p-triangles

(33, 56; 65)   and   (63, 16; 65)

(having the same hypotenuse, note).  Then, the "odd" and "even" leg sums   33  +  63  =  96   and  16  +  56  =  72  are both legs of a p-triangle:

(72, 96; 120)  =  24 (3, 4; 5) .

So are the corresponding differences   63  - 33  =  30 and  56  - 16  =  40:

(30, 40; 50)  =  10 (3, 4; 5) ,

another multiple of the same primitive p-triangle.  Moreover, the multiplication factors are also legs of a p-triangle:

(10, 24; 26)  =  2 (5, 12; 13) !

Finally, note that the product of the hypotenuses of these two "derived" p-triangles is the common hypotenuse of the original p-triangles.

This is not coincidental:  here is an outline why.  Note that the hypotenuse is a product of two Fermat primes (primes one more than a multiple of 4):  65  =  (5) (13).  It is known that any Fermat prime is the hypotenuse of a unique p-triangle: here,  (3, 4; 5) and  (5, 12; 13).  This is sufficient (but not necessary) for the process to go through...

Suppose  (a, b; p)  and  (m, n; q) are primitive p-triplets.  They generate two new primitive p-triplets sharing the hypotenuse pq:

(am-bn, an + bm; pq) and (am + bn, bm - an; pq) .

These are consequences of unique factorization properties of the ring of Gaussian integers; or they can be verified by elementary algebra.  Addition and

subtraction of the legs in the pattern as before yields the triples

(2am,  2bm, ...)  =  2m (a, b; p)

and

(2bn,  2an, ...)   =   2n (a, b; p) .

Ta-DA!!

IC:  PYTHAGOREANA (Sums of Pythagorean Tangents can be Pythagorean Tangents)

A "p-tangent" is a ratio of legs of a p-triangle.  There is an interesting  bijective characterization of these rational numbers.  I found that all p-tangents can be written in the same form, useful in these investigations:  Define the function  F(r) = [r^2-1]/[2r] for non-zero rational numbers r.  Clearly,

F(-r)  =  -F(r)  =  F(1/r).

Theorem 1:  The correspondence  r <=> F(r)  is bijective between the set of rational numbers  r>1  and the set of p-tangents.

Proof:  The elementary proof is based on the choice of r:  If the p-tangent is  a/b (odd leg/even leg), choose r = u/v.  If  the p-tangent is  b/a, (even leg/odd leg),

choose

r = [u+v]/[u-v].

Remark:   It may sometimes be convenient to consider the "reciprocal" bijection,  r<=>[1/F(r)].

Corollary 1.1:  P-tangents are the numbers of the form     F(r)  = 1/2 [ r - 1/r],  i.e., half the "gap" between a rational number and its reciprocal (positive value).

Attempts to tile rectangles with pairwise non-similar p-triangles led to consideration of an equality not investigated before:  When can the sum of two p-tangents equal a p-tangent?   It took a small computer search to find examples - they are "sparse"....

The first two found (with generators supplied for reference) were

(1)    4/3  +  11/60  =  91/60                                [2, 1],       [6, 5],       [10, 3]

and

(2)    3/4  +  119/120  =  209/120                        [2, 1],     [12, 5],       [15, 4] .

What was interesting about these was that hypotenuses could be substituted for legs to produce other equalities:

(1h)   4/5  +  61/60  =  109/60

and

(2h)   3/5  +  169/120  =  241/120.

This trend did not seem to continue when a more massive computer search yielded thirty more examples:

(3)   165/52  +  595/468  =  40/9                        [13, 2],       [26, 9],        [5, 4]

(4)   24/7  + 425/168   =  143/24                          [4, 3],       [21, 4],      [12, 1]

(5)   28/45  +  91/60  =  77/36                              [7, 2],       [10, 3],        [9, 2]

(6)   1221/140  +  351/280  =  399/40                [35, 2],       [20, 7],      [20, 1]

(7)   68/1155  +  1952/3465  =  28/45                [34, 1],     [61, 16],        [7, 2]

(8)   9/40  +  51/140  =  33/56                              [5, 4],       [10,  7],       [7, 4]

(9)   19/180  +  319/360  =  119/120                  [10, 9],       [20, 9],      [12, 5]

(10)   9/40  +  91/60  =  209/120                          [5, 4],       [10, 3],      [15, 4]

(11)   77/36  +  1247/504  =  775/168                  [9, 2],       [36, 7],      [28, 3]

(12)   19/180  +  868/765  =  253/204                [10, 9],     [31, 14],      [17, 6]

(13)   45/28  +  85/132  =  520/231                      [7, 2],       [11, 6],    [20, 13]

(14)   11/60  +  39/80  =  161/240                        [6, 5],         [8, 5],      [15, 8]

(15)   143/24  +  161/240  =  1591/240              [12, 1],       [15, 8],      [40, 3]

(16)   44/117  +  1900/819  =  736/273              [11, 2],     [38, 25],    [23, 16]

(17)   45/28  +  759/280  =  1209/280                  [7, 2],       [28, 5],      [35, 4]

(18)   13/84  +  299/180  =  572/315                    [7, 6],       [18, 5],    [22, 13]

(19)   40/9  +  572/315  =  1972/315                    [5, 4],     [22, 13],    [34, 29]

(20)   52/165  +  1292/1155  =  552/385            [13, 2],     [38, 17],    [23, 12]

(21)   76/357  +  611/1020  =  341/420              [19, 2],     [30, 17],    [21, 10]

(22)   23/264  +  1767/1144  =  700/429          [12, 11],     [44, 13],    [25, 14]

(23)   24/7  +  528/455  =  2088/455                    [4, 3],     [24, 11],    [36, 29]

(24)   25/312  +  1615/3432  =  315/572          [13, 12],     [52, 33],    [22, 13]

(25)   3021/220  +  315/572  =  10212/715        [55, 2],     [22, 13],    [74, 69]

(26)   25/312  +  209/120  =  1421/780            [13, 12],       [15, 4],    [39, 10]

(27)   437/84  +  4331/660  =  13588/1155        [21, 2],       [66, 5],    [86, 79]

(28)   28/45  +  3128/4095  =  1892/1365            [7, 2],     [68, 23],    [43, 22]

(29)   68/1155  +  1643/924  =  2829/1540        [34, 1],     [42, 11],    [55, 14]

(30)   5/12  +  3431/1560  =  4081/1560            [3,   2],     [60, 13],    [65, 12]

(31)   8/15  +  3311/2040  =  4399/2040              [4, 1],     [60, 17],    [68, 15]

(32)   52/165  +  4687/5016 =  2849/2280         [13, 2],     [76, 33],    [57, 20]

No evident pattern emerged; but at least we have an example base.

The problem (a):  finding two p-tangents whose sum is a p-tangent is easily shown equivalent to two others, stated in terms of basic arithmetic:

(b):  finding three positive fractions (none equal to 1) whose sum equals the sum of their reciprocals

(c):  finding three pairs of  companion divisors {x, X}, {y, Y}, {z, Z}... of a square N^2  ( xX = yY = zZ = N^2)  with the property that   x + y + z = X + Y + Z.

Remarks on Problem (b):  (Sums of Fractions):

The equation   F(r)  +  F(s)  =  F(t) is equivalent to

r  +  s  +  1/t  =  1/r  +  1/s  + t.                                                                                         (1)

This equation has alternate forms, for example:    r  +  1/(-s)  +  (-t)  =  1/r  +  (-s)  +  1/(-t).  It is convenient to let r, s, t be arbitrary non-zero rationals and realizing that for the final result, any negatives can be "tansposed" away.  Thus, in the sequel, we will seek rational solutions  x,y,z  to the equation

x  +  y  +  z =  1/x  +  1/y  +  1/z .                                                                                      (2)

Note from proof of proof of  Theorem 1 how  r,s,t are related to the generators of the associated p-triangles.

Richard Walker used the theory of elliptic curves to construct parametrically defined families of solutions to this equation.

His first was obtained by letting

x  =  1  +  at,    y  = 1  +  bt,    z  =  1  +  ct,   where   a  +  b  +  c  =  0    ( c  = - [a  +  b] ).

Substituting into (2)  leads to

t = (a^2  +  b^2  +  c ^2)/(3abc)  =  - [2(a^2  + ab  +  b^2)] / [3ab(a  +  b)]                     (3)

and

x  =  [(b  -  a)(b  +  2a)] / [3b(a  +  b)]

y  =  [(a  -  b)(a  +  2b)] / [3a(a  +  b)]                                                                                (4)

z  =  [(2a  +  b)(a  +  2b)] / [3ab].

Here, we note that  the common value of both sides of  (2)  is 3; so this was not believed to produce all solutions (it doesn't).

Richard Walker's second solution (in terms of "arbitrary" z) was:

x  =  [2z(z  +  1)] / [(z - 1)(z^2  +  1)]

y  =  [2z(z  -  1)] / [(z  + 1)(z^2  +  1)].                                                                               (5)

The following approaches have been attempted, but so far have led to intractable messes:

(i):  letting  x  =  k  +  at,   y  =  k  +  bt,    z =  k + ct,

(ii):  Considering when the cubic equation  u^3  -  Au^2  +  ABu  -  B  =  0  has three rational roots (note the sum of its roots equals the sum of their reciprocals) by setting its discriminant equal to the square of a rational number.

Remarks on Problem (c):  Pairs of Companion Divisors

If  the sum of two p-tangents equals a third, we may consider them to be

A / N,    B / N,      and      (A + B) / N.

There would be positive integers; say,  D,  E,  F, for which

A^2  +  N^2  =  D^2,     B^2  +  N^2  =  E^2,     (A + B)^2  +  N^2  =  F^2.

Hence,

N^2  =     D^2  -  A^2      =     [D  +  A] [D  -  A]

N^2  =     E^2  -  B^2       =      [E  +  B] [E  -  B]

N^2  =  F^2 - (A + B)^2   =  [F - A - B] [F + A + B].

The bracketed expressions in each row are the pairs of companion divisors of  N^2;  and note the sums of the columns are equal.  The argument is "reversible", allowing construction of p-tangents satisfying condition (a) from triples of  companion divisor pairs satisfying (c).

I found some restrictions on the prime factorization of N from this, but no gratifying general conditions:

( i ):  N cannot be a power of a prime or a product of two different primes

( ii ):  N cannot be of the form 2n, where n is odd

1D:  PYTHAGOREANA (Pythagorean Tilings of Rectangles and Squares)

A p-tiling (Pythagorean tiling) of a rectangle or square is a tiling by p-triangles, no two of which are similar.   (This item is under construction.)

I found several by first determining "templates"  and computing specific solutions through use of necessary angle sum and common multiple conditions.  Both Richard Walker and Richard Guy improved upon these results.   Some of these tile a p-triangle (which is then replicated to complete the p-tiling of a rectangle, and some do not.  (Illustrations to follow.)

We draw patterns (click on thumbnail for better mage).  Sides of each p-triangle will be given in clockwise order, with hypotenuse last.

IIMPORTANT DEVELOPMENT

Richard Walker has succeeded in p-tiling a square!  A diagram of his remarkable result appears to the right (click on it for larger image).   It can be obtained from a Heronic tiling with four triangles, one of which  (DEG) is divided into two Pythagorean triangles by an altitude (FG).   Here are the vital statistics, some of which are hand-written inside the triangles in the diagram.  The square ABCD is 1,199,040 on a side, and the triangles are given below along with their generators:

I:       1249    <    799    ,        960  ;    1249   >        AE = 997,951      AD = 1,999,040        DE = 1,560,001          Generators:   [ 15  ,  32  ]

II:         240    <   1443   ,      6076 ;     6245   >        FG = 346,320      DF = 1,458,240        DG = 1,498,800          Generators:   [ 49  ,  62  ]

III:    299,760  <     3       ,         4    ;         5     >       CG = 899,280      CD = 1, 199,040       DG = 1,498,800          Generattos:   [  1   ,  2    ]

IV:      1249     <    161    ,      240   ;      289    >        BE = 201,089      BG =     299,760       FG =    346,320          Generators:   [  8   ,  15  ]

V:                   < 101,761,  346,320;  360,961 >        EF = 101,761      FG =     346,320        EG =    360,961          Generators:   [ 360, 481 ]

FUTURE WORK;  UNANSWERED QUESTIONS

The methods we used lend themselves to computer analysis, suggesting enough examples might be generated to suggest patterns from which "optimal" solutions

might be determined.  We throw out these ideas for your consideration...

1:  What is the least number of triangles required to p-tile a rectangle (square)?

2:  What is the smallest that can be tiled ( w.r.t. area, perimeter, diagonal, longer dimension)?

1E:  PYTHAGOREANA  (Contributors)

Roman Binder, rbinder@netvision.net.il

Richard Guy,

Will Heierman,  wheierman@corunduminium.com

Ed Sherman, emsherman@sherlite.com

Richard Walker,  rcw9@tutor.open.ac.uk

Web design by Ewebdesigning.com