Conference rusure::math

1244.0. "Equal Area Right Triangles" by GUESS::DERAMO (Dan D'Eramo) Mon May 21 1990 12:51

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From: [email protected] (Randall Rathbun)
Newsgroups: rec.puzzles,sci.math
Subject: Equal Area Right Triangles.
Summary: Finding N-sets
Keywords: Triangles, Pythagorean, Equal Area
Message-ID: <[email protected]>
Date: 16 May 90 15:30:29 GMT
References: <[email protected]>
Sender: [email protected]
Reply-To: [email protected] (Randall Rathbun)
Organization: NCR Corporation, Rancho Bernardo
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Xref: shlump.nac.dec.com rec.puzzles:5918 sci.math:11131
 
(Due to some misunderstandings from readers, and to clarify some
information, I am reposting this article to both sci.math and
rec.puzzles. I apologize for the length of this posting) -
 
Sets of Equal Area Pythagorean Triangles -
 
The question of finding rational right triangles of equal area has
occurred since antiquity, where we find Diophantus in the Arithmetica,
vol 5, problems 7 & 8, [A History of Greek Mathematics, Vol 2 - From
Aristarchus to Diophantus, Dover Publications, NY 1981, chapter 20, pg
500 see also Dickson, History of Theory of Numbers, Vol II, Chapter IV,
pg 172, Chelsea Publishing, NY 1956] who supplies 3 such triangles:
40,42,58  24,70,74 & 15,112,113 with common area 840 created by 3
generator pairs (7,3),(7,5) & (8,7) demonstrating a very old interes
in this problem. Leonard Euler gave some parametric formuli, and noted
that 2 triangles 770,1104,1346 & 560,1518,1618 had the same area 425040
formed by the generator pairs (35,11) & (33,23). He discussed the
solution of 1), our equation of interest.
 
1)                 m*n*(m+n)*(m-n) = p*q*(p+q)*(p-q)
                           m,n,p,q rationals
 
We additionally notice that in 1), m,n,p,q are the generators of a
pythagorean triple 2), forming a right triangle with the 2 shorter sides
of lengths 2mn & m*m-n*n, and the hypotenuse of length m*m+n*n.
 
2)                 2*m*n ; (m^2-n^2) ; (m^2+n^2)
 
As is well known, if m,n are coprime integers and of opposite parity, we
have a primitive Pythagorean triangle produced in 2).
 
We also note that 1) states that the area of the 2 triangles are equal,
since it is 1/2 the product of the shorter sides in 2).
 
In 1945, a two-part article appeared in Scripta Mathematica, "Rational
Right Triangles with Equal Areas" by W.P. Whitlock Jr. [Vol IX, #3, Sept
1943 pages 155-161 & Vol IX, #4, Dec 1943 pages 265-268] where the problem
of finding equal area triangles is discussed.
 
As was pointed out to me in email by Peter Montgomery of UCLA who discovered
a simple parametric solution for 1), W.P. Whitlock Jr in 1943 provided the
identical same solution 3) on the very first page of this article.
 
          m1,n1 = ( x^2 + x*y , 3*y^2 - x*y )
3)        m2,n2 = ( x^2 - x*y , 3*y^2 + x*y )
                     x,y rationals
 
where m1,n1 & m2,n2 are the generators in 2) which produce 2 different right
triangles of equal area. Additionally in 3), if x,y are integers of opposite
parity, and the GCD(x,3*y)=1, then the 2 triangles are primitive.
 
So we immediately see by the parametric solution of 3) that an infinite
number of pairs of equal area right triangles can be found quickly.
For example, the smallest such dual m,n pair is (5,2) & (6,1) with common
area of 210 obtained from the generators x,y = 2,1. The respective Pythagorean
triangle sides are 29,21,20 and 37,35,12. This is the smallest such pair
with integral generators (m,n) that exists.
 
 
More than 2 sets of Equal Area Triangles -
 
What about more than 1 pair of triangles? What about triples, or quadruples
or more? Whitlock goes on in the article to give a second parametric formula
 
         m1,n1 = ( 3*y^2 - x^2 + 2*x*y , 3^y^2 + x^2 )
4)       m2,n2 = ( 3*y^2 - x^2 - 2*x*y , 3^y^2 + x^2 )
         m3,n3 = (       4*x*y         , 3^y^2 + x^2 )
 
which provides triplets of such triangles, but notes that only those of m3,n3
will be primitive providing the same conditions in 3) are met.
 
As both Whitlock in his article and Peter Montgomery in a posting to rec.puzzles
pointed out, Pierre de Fermat showed how to obtain sets of triangles having
equal areas. I quote from the posting without permission:
 
Given a rational right triangle (a, b, c) with a^2 + b^2 = c^2,
then (A, B, C) is also rational and has the same area if
 
             (     2abc      a^2 - b^2    a^4 + 6a^2 b^2 + b^4  )
(A, B, C) =  (  ----------,  ---------,   --------------------  )  
             (   a^2 - b^2       2c        2c (a^2 - b^2)       )
 
For example, if (a, b, c) = (3, 4, 5), then the construction yields
(A, B, C) = (-120/7, -7/10, -1201/70).  Take absolute values of sides
to get (120/7, 7/10, 1201/70) = (1200, 49, 1201)/70, of area
600*49/70^2 = 6 = 3*4/2.  By repeating this construction, we get
infinitely many rational right triangles of area 6. By scaling these
to remove any denominators (e.g., multiplying (3, 4, 5) and
(120/7, 7/10, 1201/70) by 70), we can get large arbitrarily large
collections of integer right triangles with equal area.
(end quote)
 
So obviously sets of n-tuples equal area triangles exist. Just for
interest's sake, I posted sets of quintuplets to the net in a previous
article. I sent to Peter Montgomery a small set of my 6540 coprimitive
sets of integral generators solutions to 1). As he pointed out, my set
of 5 triangles with area 6913932480 is not the smallest that actually
exists. 
 
        Quintuplet Integer Generator Pairs                 Common Area
----------------------------------------------------     --------------
 403,115    403,333   414,104    448,403    558,40           6913932480
 518,288    518,310   598,518    736,70     851,45          27655729920
1705,1305  1935,385  2900,110   3100,90    3168,3125      2678930100000
2320,1550  3010,400  3010,2790  3190,3010  6293,43       10715720400000
 
I would like to acknowledge that this is true, as my table covers only
integer coprimitive sets of generators. His set was found by simple
computer search involving square multiples of the area:
 
                   Sides             Area     Generator
            --------------------------------------------
             715   1428   1597       510510    (34,21)
            1001   4080   4201     4*510510    (51,40)
            1309   7020   7141     9*510510    (65,54)
            7735    528   7753     4*510510    (88,3)
             935  17472  17497    16*510510    (96,91)
 
By scaling up to area = 144*510510, we can obtain 5 sets of equal area
triangles ( area = 73,513,440 )  D. L. MacKay in the American Mathematical
Monthly, vol 46, 1939 on page 169 gave the same identical set of 5 triangles
with generators:
 
             Sides              Area            Generator
      ----------------------------------------------------------
       8580  17136  19164     73513440    (68*sqrt 3, 42*sqrt 3)
       6006  24480  25206     73513440    (51*sqrt 6, 40*sqrt 6)
       5236  28080  28564     73513440          (130,108)
       3168  46410  46518     73513440    (91*sqrt 3, 85*sqrt 3)
       2805  52416  52491     73513440    (96*sqrt 3, 91*sqrt 3)
           
But notice that the generators are no longer rational. Most are irrational.
This is a consequence of scaling to match the areas. Since the area is a
product of the generators and their sums and differences, the irrationalities
drop out. However this means if a computer search using generator pairs is
tried, these generators must include irrational ones.
 
By searching my integer generator table I was successfully able to find a
slightly smaller set of 5 equal area right triangles.(to my knowledge they
have not been previously mentioned)
 
             Sides              Area            Generator
      ----------------------------------------------------------
      10010  14352  17498     71831760    (23*sqrt 26, 12*sqrt 26)
       7280  19734  21034     71831760    (28*sqrt 26,  5*sqrt 26)
       5070  28336  28786     71831760    ( 92*sqrt 2, 77*sqrt 2)
       2640  54418  54482     71831760    (165*sqrt 2,  4*sqrt 2)
       2415  59488  59537     71831760           (176,169)
           
As Peter Montgomery pointed out to me, a simple way to find sets of n
equal area triangles, is to find the area of an arbitrary pair of
generators m,n; remove any squares from this area; form a pool of such
triangles, then sort by area. Triangles with the same reduced area can
be combined to form n-groups of equal area triangles.
 
 
What about sets of primitive pythagorean triangles? -
 
Martin Gardner was the first to address this question in "Simple Proofs of
the Pythagorean theorem, and sundry other matters", Mathematical Games, pages
118-126, Scientific American, Oct 64, Vol 211 #4 where he states in the last
paragraph on page 120 that a Charles L. Shedd in 1945 found the triplet:
 
    (1380, 19019, 19069),  (3059, 8580, 9109),  (4485, 5852, 7373)
 
with common area 13123110 = 2*3*5*7*11*13*19*23. By computer search this
is the smallest primitive solution.
 
Gardner goes on to ask about other such triplet sets. Do they exist? My
interest was stirred by his question, so a computer search was started.
Realizing the the GCD of each generator pair is 1, because we're looking
for primitive pythagorean triangles, a program has to only consider
integral generators m,n.
 
However W. P. Whitlock noted in his article on page 265 that the following
generator pairs (28,5),(23,12); (39,38),(26,7) and others were not obtained
from 3) so all possibilities for 1) have to be considered, not just those
from using a parametric formula.
 
From that extensive thousands of hours computer search which considers
all possible divisors of a given area, and tries to make combinations of
divisors with the same area as in 1) the following 3 new triplets of
primitive pythagorean triangles were found.
 
         Primitive Triplet Generator Pairs    Common Area
         ---------------------------------   -------------
          1610,869   2002,1817  2622,143     2570042985510
          2035,266   3306,61    3422,55      2203385574390
          2201,1166  2438,2035  3565,198     8943387723270
 
Please note that the generator pairs m,n of 2) are given, not the actual
sides. I repeat the previous posting:
 
*** No other triplets occur for generator pairs over (2,1)-(5000,4999) *** 
 
Many billions of possible divisors (triangles) have been examined and only
these 4 triplets have been found.
 
Peter Montgomery fired up a SPARC workstation to seach for more triplet
solutions to 1), by generating a table of triangles from 2) and sorting by
area, but failed to find any new ones for m,n or p,q <= 200000,199999.
However he did not consider any values of area divisible by primes > 200.
 
As Peter mentions, the 3 new triplets are found to be members of 3), so a
program is currently running in hopes of finding a new triplet from 3).
However a triplet may exist that is not a member of 3).
 
 
What about new parametric formuli? -
 
This article would not be complete without mentioning Professor (& Dr.)
Andrew Bremner at Arizona State, who has researched this problem. Being one
of the world's foremost algebraic geometrists, he has written an excellent
article titled "Pythagorean triangles and a quartic surface" in Journal
fur die reine und angewandte Mathematik, vol 318, 1980 pages 120-125.
 
In the article, he considers mappings of lines upon the quartic surface
generated by 1) and shows how to pull down images of these rational lines
under non-trivial automorphisms of the surface. He find 20 lines on 1)
defined over the rationals.
 
By constructing the subgroup NS(V,Q), the Neron-Severi group of 1) over
Q the rationals, he determines parametric solutions and finds 16 such of
degree 2, 32 of degree 3, 64 of degree 4 and 312 of degree 5. He lists a
typical parametric solution of each degree (up to symmetry) and from it
can be determined new parametric solutions of 1).
 
His first example of degree 2 provides 3) easily. From his second example
of degree 3 we derive:
 
          m1,n1 = ( -x^3 + 4*x^2*y + x*y^2 , x^3 + x*y^2 + 2*y^3 )
5)        m2,n2 = ( -x^2*y + 4*x*y^2 + y^3 , 2*x^3 - x^2*y - y^3 )
                           x,y rationals
 
From his 16 examples, 15 more parametric solutions can easily be obtained.
I consider his paper to be a significant step forward towards determining
all parametric solutions of 1) and invite the serious reader to consider it.
 
I trust this very long posting provides a current reference to this very
old problem and will stir the reader to search for more new triplets of
primitive Pythagorean triangles. As Chris Long asked, does a primitive
quadruplet exist? I suspect so, but it may be very difficult to find.
 
Once again I wish to thank the NCR Corporation which has so generously given
thousands of spare hours of computer time for this search. In particular, the
following individuals are to be commended; Ken Lehmann, Keith Ronchetti,
Bob Odegard and Steve Blair, all who have allowed me to run on their machines
without complaint.
 
Lets find more primitive triples! Sincerely,
                                                - Randall
 
P.S.: All questions and inquiries welcomed. I will be happy
      to provide parts of my integral generator table also.
 
NCR E & M - San Diego   | INTERNET - [email protected]
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From: [email protected] (Randall Rathbun)
Newsgroups: rec.puzzles,sci.math
Subject: Re: Equal Area Right Triangles.
Summary: 2 small errors
Keywords: Triangles, Pythagorean, Equal Area
Message-ID: <[email protected]>
Date: 16 May 90 18:53:08 GMT
References: <[email protected]> <[email protected]>
Sender: [email protected]
Reply-To: [email protected] (Randall Rathbun)
Organization: NCR Corporation, Rancho Bernardo
Lines: 19
Xref: shlump.nac.dec.com rec.puzzles:5919 sci.math:11132
 
I am sorry, at least 2 small errors crept into my previous posting...
 
The reference [A History of Greek Mathematics, Vol 2 - From
Aristarchus to Diophantus, Dover Publications, NY 1981, chapter 20, pg
500] omitted the author, he's Sir Thomas Heath.
 
In discussing Dr. Andrew Bremner's paper near the end of the article,
please read:
 
Q the rationals, he determines parametric solutions and finds 16 such of
degree 2, 32 of degree 3, 64 of degree 5 and 312 of degree 5. He lists a
                                       ^
                  (should read) degree 4
 
Thanks for observing these corrections.        - Randall
 
NCR E & M - San Diego   | INTERNET - [email protected]
16550 W Bernardo Drive  |   UUCP   - {backbone}!ncr-sd!thor!randall
San Diego, CA 92127     |  TELE #  - (USA) (619) 485-3620 or 2358

	[I made the latter correction in the text. (Dan)]

>          m1,n1 = ( x^2 + x*y , 3*y^2 - x*y )
>3)        m2,n2 = ( x^2 - x*y , 3*y^2 + x*y )
>                     x,y rationals

>However W. P. Whitlock noted in his article on page 265 that the following
>generator pairs (28,5),(23,12); (39,38),(26,7) and others were not obtained
>from 3) so all possibilities for 1) have to be considered, not just those
>from using a parametric formula.

That's because 3) doesn't generate all the solutions.  That may be expected, since the
problem basically has 3 degrees of freedom, not 2 as in the parametric equations above.