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Topic: almost equilateral integer triangles (Read 3412 times) |
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JocK
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almost equilateral integer triangles
« on: Feb 20th, 2005, 7:34am » |
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We define an almost equilateral integer triangle as a triangle with integer sides and integer area such that the largest and smallest side differ in length by one.
The first (smallest) almost equilateral integer triangle has sides 5, 5 and 6, and area 12. The next one has sides 17, 17, 16 and area 120.
The first five almost equilateral integer triangle areas are:
12, 120, 1848, 25080, 351780, ...
1) Can you calculate the next five terms in this sequence?
2) The triangle with sides 174725, 174725 and 174726 is also an almost equilateral integer triangle (check this!). Can you come up with a larger triangle and become record-holder "largest almost equilateral integer triangle ever calculated"...? 
3) Does the above sequence continue indefinitely? If so, determine the asymptotic behaviour of the n-th term in this sequence for large n. If not, what is the largest almost equilateral integer triangle?
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| « Last Edit: Feb 20th, 2005, 7:56am by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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towr
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Re: almost equilateral integer triangles
« Reply #1 on: Feb 20th, 2005, 8:41am » |
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on Feb 20th, 2005, 7:34am, JocK wrote:| 1) Can you calculate the next five terms in this sequence? |
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Well, I can get my computer do do it.. [e]got 16 so far, including (2,1,1)[/e]
I suspect there are infinitely many, but I don't know how to proof there are an infinite number of integer solution to the squareroot of a certain (pair of) quadratic(s)..
And since the hide-tags still don't work I won't post it yet  (Not that it's hard to derive)
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| « Last Edit: Feb 20th, 2005, 8:50am by towr » |
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towr
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Re: almost equilateral integer triangles
« Reply #2 on: Feb 20th, 2005, 9:20am » |
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The side lengths belonging to almost equilateral integer triangle seem to be very regularly spaced, always a factor of about 2+[sqrt]3 between them. (Unfortunately, although this does significantly help speed up finding them, it doesn't help me find more of them. Probably a representation-limit of the computer)
But it's a nice number other people can prove something about
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Barukh
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Re: almost equilateral integer triangles
« Reply #3 on: Feb 20th, 2005, 10:10am » |
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on Feb 20th, 2005, 8:41am, towr wrote:| I suspect there are infinitely many |
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I think you are right and I believe I can prove it.
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towr
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Re: almost equilateral integer triangles
« Reply #4 on: Feb 20th, 2005, 10:48am » |
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A few simple observations:
There is always one even side, and two odd sides (easily proven)
The last digit of even length sides seem to repeat the sequence 666020
The odd sides alternate between one less and one more than the even side, and so not surprising repeat the sequence 575111
The area similarly repeats the sequence 208000
The first digits also seem to repeat (but I'm less sure about this), if we skip the first number of the sequence, the sides repeats 1629314, and the area 1123469 (same digits, different order )
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| « Last Edit: Feb 20th, 2005, 10:50am by towr » |
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Sir Col
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Re: almost equilateral integer triangles
« Reply #5 on: Feb 20th, 2005, 11:47am » |
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I've not managed to prove that there exist infinitely many, but maybe this can be developed...
Using Heron's formula, the area of the triangle A=sqrt(s(s-a)(s-b)(s-c)), where s=(a+b+c)/2.
W.L.O.G. let a=b:
A = sqrt((a+c/2)(c/2)(c/2)(a-c/2)) = (c/4)*sqrt(4a2-c2)
Writing 4a2-c2 = k2, we get (2a)2 = c2+k2.
So it amounts to showing that there exist infinitely many Pythagorean triplets (x,y,z)=(c,k,2a) for which z/2 is one more or one less than x.
Here are the two examples that Jock gave...
(6,8,10): 2a=10, so a=5 and c=6, hence A=12.
(16,30,34): 2a=34, so a=17 and c=16, hence A=120.
The next five being...
(66,112,130): a=65, c=66, and A=1848
(240,418,418): a=241, c=240, and A=25080
(902,1560,1802): a=901, c=902, and A=351780
(3360,5822,6722): a=3361, c=3360, and A=4890480
(12536,21728,25090): a=12545, c=1254, and A=68149872
Anyway, you get the idea.
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| « Last Edit: Feb 20th, 2005, 11:51am by Sir Col » |
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Eigenray
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Re: almost equilateral integer triangles
« Reply #6 on: Feb 20th, 2005, 3:00pm » |
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on Feb 20th, 2005, 11:47am, Sir Col wrote:I've not managed to prove that there exist infinitely many, but maybe this can be developed
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(2a)2 = c2+k2. |
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(2a)2 = (a +- 1)2 + k2
can be rewritten as
((3a -+ 1)/2)2 - 3(k/2)2 = 1,
so we are looking for integer solutions to the Pell equation x2-3y2=1
Given a solution (x,y), of which infinitely many exist, note that x2=1 mod 3 implies 2x = 3a -+ 1 for some integer a, and we may take k=2y. This gives the triangle
(a, a, a +- 1), with area A = (c/4) k = (a +- 1)y/2, also an integer, since a is odd.
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| « Last Edit: Feb 21st, 2005, 1:46pm by Eigenray » |
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Barukh
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Re: almost equilateral integer triangles
« Reply #7 on: Feb 21st, 2005, 8:57am » |
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I took essentially the same approach as Sir Col, just used sides 2a, 2a[pm]1 to avoid fractions.
From Eigenray’s formulas, we see that the side of the sought triangle is a linear function of x as a solution of the Pell equation. In the same link provided by Eigenray, we find how a whole family of solutions may be generated from a single solution (x0, y0), that is:
xn = 1/2 * [ x0 + y0[sqrt]3)n + (x0 - y0[sqrt]3)n ].
The simplest solution in our case is (x0, y0) = (2,1), so that the second term in the above formula is less than 1, and for significantly large n the value xn is dominated by the first term 1/2 * (2 + [sqrt]3)n. This explains the number found by towr earlier.
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Sir Col
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Re: almost equilateral integer triangles
« Reply #8 on: Feb 21st, 2005, 12:17pm » |
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Great work with the Pell equation, Eigenray!
I found a method that almost trivialises the initial (Heron's forumula) approach...
Consider an isosceles triangle with sides (a,a,c). By drawing a perpendicular from the apex to base we produce two right triangles back-to-back.
If c (the base) were odd, c/2 and the height, h, would be non-integer, so the area would be non-integer.
So it is necessary for c to be even. As c=a+-1, a will be odd (confirming towr's observation).
a2 = (c/2)2+h2 = (a+-1)2/4+h2
3a2-+2a-1-4h2 = 0
9a2-+6a+1-12h2 = 4
(3a-+1)2-12h2 = 4
((3a-+1)/2)2-3h2 = 1
As a is odd, 3a-+1 will be even, and so divisible by 2. Therefore we do indeed get integers througout. Finally, as Eigenray explained, this is equivalent to the Pell equation, x2-3y2 = 1, and we establish that there are infintely many solutions.
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| « Last Edit: Feb 21st, 2005, 12:19pm by Sir Col » |
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Sir Col
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Re: almost equilateral integer triangles
« Reply #10 on: Feb 22nd, 2005, 11:38am » |
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I don't feel that I'm quite done on this yet. Does anyone know anything about generating the solutions sets for Pell equations?
A method seems to be...
Given x2-ny2 = 1 and the the solution (x,y), then (x2+ndy2,2xy) will also be a solution (easily proved). However this misses solutions, so it is, perhaps, better to use the fact that given the solutions (a,b) and (c,d), then (ac+-nbd,ad+-nbc) will also be solutions; this can be applied recursively.
Given the obvious solution (2,1), we also get (22+3*12,2*2*1)=(7,4).
From any solution set we only need x, as x=(3a-+1)/2. Therefore a=(2x+-1)/3 and we know that c=a+-1 (note the same sign).
For example, in (7,4), x=7, so a=(2*7+-1)/3. But the only integer solution is a=5 (by adding 1), so c=6.
But when we combine (2,1) and (7,4) we get (26,15): x=26, a=(52+-1)/3=17 (this time we subtract 1), so c=16.
My question... will this method produce ALL solutions?
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Barukh
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Re: almost equilateral integer triangles
« Reply #11 on: Feb 23rd, 2005, 10:27am » |
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Sir Col, I don't know if your your method gives all solutions.
Don’t you remember discussing the similar problem in this thread? The method presented there is used to get the minimal solution (also called fundamental) which we denote (x1, y1). In our case it’s simply (2, 1).
Then, all the solutions may be obtained as described in the previous post. Let me elaborate. Denote X = x1 + [sqrt]3y1 = 3.732…, Y = x1 - [sqrt]3y1 = 0.268… Then, the n-th solution is obtained as follows:
xn = (Xn + Yn) / 2, yn = (Xn - Yn) / 2[sqrt]3
In our case, because Y < 1, we should have that for “sufficiently large” n, xn+1 is the integer closest to Xxn. In fact, it’s correct even for n = 1! Indeed:
2*X = 7.46….
7*X = 26.124…
26*X = 97.030…
97*X = 362.004…
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| « Last Edit: Feb 23rd, 2005, 10:27am by Barukh » |
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Sir Col
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Re: almost equilateral integer triangles
« Reply #12 on: Feb 23rd, 2005, 3:07pm » |
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Oh yes, I certainly do remember that thread.
Thanks for the information, Barukh.
I was such a pain at school: always asking question and not caring how stupid I might appear. So would you mind me asking a couple of (stupid) questions... ?
How did you derive your iterative form?
In Eigenray's MathWorld link: http://mathworld.wolfram.com/PellEquation.html, how do they go from (31) to (32) and (33)? Why must those particular factors on the LHS and RHS match? For example, why can't (x+sqrt(D)y) be made up (p+sqrt(D)q)j(p-sqrt(D)q)k?
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Barukh
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Re: almost equilateral integer triangles
« Reply #13 on: Feb 24th, 2005, 11:29pm » |
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on Feb 23rd, 2005, 3:07pm, Sir Col wrote:| For example, why can't (x+sqrt(D)y) be made up (p+sqrt(D)q)j(p-sqrt(D)q)k? |
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Clearly, j [ne] k, for otherwise we would get an irrational number on the lhs, and 1 on the rhs. So let j > k, and get
x+[sqrt]Dy = (p+q[sqrt]D)j(p-q[sqrt]D)k = (p+q[sqrt]D) j-k
But that’s more difficult then I thought… Actually, we want to show the following: (x,y) is also a solution iff
x + y[sqrt]D = (p + q[sqrt]D)n
for some integer n.
The “if” part is easy: If x + y[sqrt]D = (p + q[sqrt]D)n, then clearly x - y[sqrt]D = (p - q[sqrt]D)n, therefore
x2 – Dy2 = (x + y[sqrt]D)(x - y[sqrt]D) = (p + q[sqrt]D)n(p - q[sqrt]D)n = (p2 – Dq2) n = 1.
The “only if” part is trickier. It follows directly from a theorem in algebraic number theory, but this is advanced stuff, but I wanted a simpler argument…
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Sir Col
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Re: almost equilateral integer triangles
« Reply #14 on: Mar 21st, 2005, 9:35am » |
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on Feb 23rd, 2005, 3:07pm, Sir Col wrote:| How did you derive your iterative form? |
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Sorry that was a dumb question!
Yn tends towards zero as n increases, so xn = (Xn+Yn) ~= Xn/2, and Xn ~= 2xn.
Therefore xn+1 ~= Xn+1/2 = (X/2)Xn ~= (X/2)2xn = X*xn.
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Deedlit
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Re: almost equilateral integer triangles
« Reply #15 on: Apr 3rd, 2005, 7:24pm » |
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on Feb 24th, 2005, 11:29pm, Barukh wrote:
The “only if” part is trickier. It follows directly from a theorem in algebraic number theory, but this is advanced stuff, but I wanted a simpler argument…
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Interestingly, the solutions to a Pell equation form a group under the operation
(a,b) * (c,d) = (e,f)
where e + f sqrt (D) = (a + b sqrt (D)) (c + d sqrt (D)),
since
e2 - D f2
= (e + f sqrt (D)) (e - f sqrt (D))
= (a + b sqrt (D)) (c + d sqrt (D)) (a - b sqrt (D)) (c - d sqrt (D))
= (a2 - D b2) (c2 - D d2) = 1
and if (a,b) is a solution, so is (a,-b) and
(a + b sqrt (D)) (a - b sqrt (D)) = a2 - D b2 = 1.
So if (p,q) is the minimal solution with |(p,q)| = p + q sqrt (D) > 1, for any positive solution (x,y) there is some n such that 1 <= |(x,y) * (p,q)n| < |(p,q)|. By minimality of (p,q), the above must be 1. Hence the positive solutions are (p,q)n for integer n. Of course the negative solutions are just (-x,-y) for postive solutions (x,y)
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| « Last Edit: Apr 3rd, 2005, 7:29pm by Deedlit » |
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