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Topic: Heronian Right-angled Triangles Having Same Area (Read 759 times) |
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ThudnBlunder
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Heronian Right-angled Triangles Having Same Area
« on: Nov 23rd, 2003, 7:26am » |
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Find four Heronian right-angled triangles which have the same area.
Can you find five such triangles?
How about six?
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| « Last Edit: Nov 23rd, 2003, 9:13pm by ThudnBlunder » |
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Dudidu
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Re: Right-angled Triangles Having Same Area
« Reply #1 on: Nov 23rd, 2003, 9:30am » |
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THUDandBLUNDER hi,
Sorry but it seems that I don't understand your question  :
what is wrong with the following triangles edges lengths (I do not indicate the hypotenus): (1,100), (2,50), (4,25), (5,20), (10,10)... and so on and on and on...
all these triangles have the same area (50), so what do I miss ?
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ThudnBlunder
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Re: Right-angled Triangles Having Same Area
« Reply #2 on: Nov 23rd, 2003, 4:02pm » |
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You are quite right, Dudidu.
I omitted to mention that the triangles' sides have lengths which are integers.
(I have now amended the wording.)
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| « Last Edit: Nov 23rd, 2003, 9:23pm by ThudnBlunder » |
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towr
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Re: Right-angled Triangles Having Same Area
« Reply #3 on: Nov 24th, 2003, 1:18am » |
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:: here's six
(3, 4, 5 ) *5*7*3*40
(5, 12, 13) *7*3*40
(21, 20, 29) *3*40
(35, 12, 37) *3*40
(45, 28, 53) *40
(175, 288, 337)
all an area of 25200
as a bonus #7
(63, 16, 65) * 50
theres at least 1500 of them.. (but these were easiest to put together) ::
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| « Last Edit: Nov 24th, 2003, 5:57am by towr » |
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ThudnBlunder
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Re: Heronian Right-angled Triangles Having Same Ar
« Reply #4 on: Nov 24th, 2003, 3:12am » |
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Well done, towr.
Now find six primitive Heronian right-triangles.
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| « Last Edit: Nov 24th, 2003, 7:37am by ThudnBlunder » |
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towr
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Re: Heronian Right-angled Triangles Having Same Ar
« Reply #5 on: Nov 24th, 2003, 5:55am » |
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actually.. that wasn't well done at all..
Mostly because it's wrong..
I don't think anyone would find six primitives with the same area, just 4 non-primitives is hard enough (unless you stumble across it on the right webpage)
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| « Last Edit: Nov 24th, 2003, 5:57am by towr » |
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SWF
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Re: Heronian Right-angled Triangles Having Same Ar
« Reply #6 on: Nov 24th, 2003, 6:35pm » |
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I was able to find 5:
{309764 44640 312964}
{160580 86112 182212}
{149184 92690 175634}
{51520 268398 273298}
{38295 361088 363113}
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ThudnBlunder
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Re: Heronian Right-angled Triangles Having Same Ar
« Reply #7 on: Nov 24th, 2003, 9:07pm » |
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Quote:| I don't think anyone would find six primitives with the same area, just 4 non-primitives is hard enough |
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My original post intended primitive triangles, but I now realize that these are much harder to come by.
If area = A and number of primitives = n
then we need to find n solutions to uv*|u2 - v2| = A
Including non-primitives, we need to find n solutions to uv*|u2 - v2|*k2 = A where k is a positive integer.
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| « Last Edit: Nov 25th, 2003, 10:42pm by ThudnBlunder » |
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towr
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Re: Heronian Right-angled Triangles Having Same Ar
« Reply #8 on: Nov 25th, 2003, 12:49pm » |
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here's a smaller set of 5 (I'm not willing to claim it's the smallest though)
{57 176 185 } * 420
{297 304 425} * 140
{475 132 493} * 168
{1323 836 1565} * 40
{5225 37632 37993} * 3
on Nov 24th, 2003, 9:07pm, THUDandBLUNDER wrote:| Including non-primitives, we need to find n solutions to uv*|u2 - v2|*k2 = A where k is a positive integer. |
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Is there any better way to do it then just try heaps of (relatively prime) u and v?
For anyone trying a programming approach (like me, obviously), non-primitive solutions are relatively easy to find when you first make a prime factorization of the area, and for every factor take the exponent mod 2. Then any tripples with the same number can be mapped to the same size area using different k2
(It's hell to find a good factoring algorithm though)
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| « Last Edit: Nov 25th, 2003, 1:02pm by towr » |
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towr
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Re: Heronian Right-angled Triangles Having Same Ar
« Reply #9 on: Nov 25th, 2003, 2:43pm » |
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hah, I get my set of seven after all (two actually, out of a million primitive tripples)..
{57 176 185} * 92820
{297 304 425} * 30940
{475 132 493} * 37128
{1323 836 1565} * 8840
{5225 37632 37993} * 663
{30723 195364 197765} * 120
{256025 32448 258073} * 102
with an area of 43215610838400
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Barukh
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Re: Heronian Right-angled Triangles Having Same Ar
« Reply #10 on: Nov 26th, 2003, 9:59am » |
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on Nov 25th, 2003, 12:49pm, towr wrote:| Is there any better way to do it then just try heaps of (relatively prime) u and v? |
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As far as "primitiveness" is not a concern, the answer is yes. Here's one way (proposed by Fermat):
I will call the right triangle with sides u2-v2, 2uv, u2+v2 triangle formed by (u,v)-pair. Let T1 have sides a, b, c; and T2 be formed by the pair (c2, 2ab). Direct verification shows that area(T2) / area(T1) = perfect square. Proceed in the same manner taking T2 as a reference triangle.
After writing this, I stumbled at the following question: is there a sequence T1, T2, T3, ... free of similar triangles (meaning we may have as many Heronian triangles with the same area as we wish)?
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| « Last Edit: Nov 26th, 2003, 10:00am by Barukh » |
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towr
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Re: Heronian Right-angled Triangles Having Same Ar
« Reply #11 on: Nov 26th, 2003, 10:48am » |
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Well, it works.. but it gives awfully large triangles even after just a couple of steps..
On the other hand you don't have to filter through 10 million of them to find a set of 8..
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Barukh
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Re: Heronian Right-angled Triangles Having Same Ar
« Reply #12 on: Dec 1st, 2003, 4:26pm » |
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I would like to go back to the primitive triangles case.
on Nov 24th, 2003, 5:55am, towr wrote:| I don't think anyone would find six primitives with the same area, just 4 non-primitives is hard enough (unless you stumble across it on the right webpage) |
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Has somebody found 4 primitive triangles with the same area? And where's this mystical web page? My feeling - based on some extensive runs - is that even primitive triads are very rare. Actually, I've exhausted all the (u,v)-pairs upto 100000, with the following two limitations:
a) The area S of the triangle has no prime factor greater than 1000,
b) No prime factor of S has power greater than 4.
Given that, only 4 primitive triads were found:
1. S = 13123110 (2*3*5*7*11*13*19*23) :
( 4485, 5852, 7373)
( 3059, 8580, 9109)
(19019, 1380, 19069)
2. S = 2203385574390 (2*3*5*7*11*13*19*29*37*59*61) :
( 4070469, 1082620, 4211981)
(10925915, 403332, 10933357)
(11707059, 376420, 11713109)
3. S = 2570042985510 (2*3*5*7*11*13*19*23*37*67*79) :
(1836939, 2798180, 3347261)
( 706515, 7275268, 7309493)
(6854435, 749892, 6895333)
4. S = 8943387723270 (2*32*5*7*11*13*23*31*37*53*71) :
( 3484845, 5132732, 6203957)
( 1802619, 9922660, 10085069)
(12670021, 1411740, 12748429)
Several interesting observations may be made: in all the cases, all the areas have prime factors less than 100, and with power 1 or 2. Also, all the primes [le] 13 are factors of every of the 4 areas.
Any comments?
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| « Last Edit: Dec 1st, 2003, 4:28pm by Barukh » |
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towr
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Re: Heronian Right-angled Triangles Having Same Ar
« Reply #13 on: Dec 2nd, 2003, 12:30am » |
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on Dec 1st, 2003, 4:26pm, Barukh wrote:| Has somebody found 4 primitive triangles with the same area? And where's this mystical web page? |
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I don't know, that one website just had 4 non-primitive ones (which is hard enough without a program or the right generating proces).
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ThudnBlunder
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Re: Heronian Right-angled Triangles Having Same Ar
« Reply #14 on: Dec 2nd, 2003, 7:37am » |
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Quote:| I don't know, that one website just had 4 non-primitive ones (which is hard enough without a program or the right generating proces). |
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Here's a method for generating trios of non-primitives:
Let
m1 = r2 + rs + s2
n1 = r2 - s2
m2 = r2 + rs + s2
n2 = s2 + 2rs
m3 = r2 + 2rs
n3 = r2 + rs + s2
Then the non-primitive trio generated by (mi2 - ni2 , 2mini, mi2 + ni2 ) have common area A given by
A = rs(r + s)(r - s)(2r + s)(r + 2s)(r2 + rs + s2)
Putting r = 2 and s = 1 gives the smallest such trio:
A = 840 (23*3*5*7)
{24,70,74}
{40,42,58}
{15,112,113}
(Beiler, Recreations in the Theory of Numbers: The Queen of Mathematics Entertains, 1966)
There ought to be corresponding generators for primitive trios but I have been unable to derive them.
Taking Barukh's first example:
A = 13123110 (2*3*5*7*11*13*19*23)
(4485, 5852, 7373) has u = 77 and v = 38
(3059, 8580, 9109) has u = 78 and v = 55
(19019, 1380, 19069) has u = 138 and v = 5
where u and v correspond to m and n, respectively.
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| « Last Edit: Dec 2nd, 2003, 9:21pm by ThudnBlunder » |
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Barukh
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Re: Heronian Right-angled Triangles Having Same Ar
« Reply #15 on: Apr 8th, 2004, 11:30am » |
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An update: apart from the 4 triples of primitve triangles, described in my previous post (4 months ago), there is another triple; its generators’ (u.v)-pairs are:
(7238, 2465); (9077, 1122); (10434, 731).
This monster was found independentally by Dan Hoey and Rathbun in early 90s. Source: R. Guy. “Unsolved Problems in Number Theory”. There, as open questions, are stated: the infinitude of such triples; the existence of the quadruple.
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