[math-fun] Elementary triangle puzzle
Given an arbitrary triangle T in R^2, characterize p in R^2 such that the sum of distances S(p) := d(p, L_1) + d(p, L_2) + d(p, L_3) is minimized, where the L_j are the affine lines containing the sides of T. --Dan "Things are seldom what they seem." --W.S. Gilbert
Nice one, Dan --- bet that isn't in Clark Kimberling's list! WFL On 10/28/11, Dan Asimov <dasimov@earthlink.net> wrote:
Given an arbitrary triangle T in R^2, characterize p in R^2 such that the sum of distances
S(p) := d(p, L_1) + d(p, L_2) + d(p, L_3)
is minimized, where the L_j are the affine lines containing the sides of T.
--Dan
"Things are seldom what they seem." --W.S. Gilbert
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Is the following the answer? If the triangle is equilateral, then all points inside the triangle serve (and all points outside if you assign signs to the distances). Otherwise the point is at the vertex with the largest angle. R. On Sat, 29 Oct 2011, Fred lunnon wrote:
Nice one, Dan --- bet that isn't in Clark Kimberling's list! WFL
On 10/28/11, Dan Asimov <dasimov@earthlink.net> wrote:
Given an arbitrary triangle T in R^2, characterize p in R^2 such that the sum of distances
S(p) := d(p, L_1) + d(p, L_2) + d(p, L_3)
is minimized, where the L_j are the affine lines containing the sides of T.
--Dan
"Things are seldom what they seem." --W.S. Gilbert
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Fred wins his bet! I've just submitted a paper, ``Five-point circles, the 76-point sphere, and the Pavillet tetrahedron'' to the Monthly. The theme is that triangle geometry is not dead. I would attach a copy if attachments were allowed; I will entertain individual requests. Of course, I'm hoping that the paper will be accepted, and I also now hope to include a stop press paragraph with the following theorem: The external angle-bisectors of a triangle meet the opposite edges in collinear points. Call it the zero-sum line because: The sum of the directed distances to the edges of a triangle from points on its zero-sum line is zero. So go ahead and spoil my fun and tell me that that's well known to those who well know it. Certainly the first part is well known since it follows immediately from (the converse of) Menelaus's theorem, so the zero-sum line may already have some other name. The Gergonne line would be a good guess, but that's not quite right. Note that the zero-sum line (and the Gergonne line) are the line at infinity if the triangle is equilateral. Best, R. On Sat, 29 Oct 2011, Fred lunnon wrote:
Nice one, Dan --- bet that isn't in Clark Kimberling's list! WFL
On 10/28/11, Dan Asimov <dasimov@earthlink.net> wrote:
Given an arbitrary triangle T in R^2, characterize p in R^2 such that the sum of distances
S(p) := d(p, L_1) + d(p, L_2) + d(p, L_3)
is minimized, where the L_j are the affine lines containing the sides of T.
--Dan
"Things are seldom what they seem." --W.S. Gilbert _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Dear Richard, I'd appreciate a copy of your paper. I've been taking a desultory look at "rational" test triangles lately, which turn out to have some interest, as well as connections with a well-known notion --- see (tho' overlapping & slightly inaccurate) http://mathworld.wolfram.com/HeronianTriangle.html http://en.wikipedia.org/wiki/Integer_triangle http://en.wikipedia.org/wiki/Heronian_triangle A triangle T is "Heronian" means it has integer sides a,b,c and rational semi-perimeter s, area d. Sample results: A triangle has rational Cartesian coordinates and sides iff it is similar to a Heronian triangle. Not every Heronian triangle can be posed to have integer coordinates: s,d,a,b,c = 75, 420, 26, 51, 73 is the smallest counterexample. For T Heronian: an odd number of a,b,c are even; s is an integer, d is an even integer; when T is proper (ie. d <> 0) s is composite; semi-angle tangents are rational: tan C/2 = d/(s-a)(s-b), etc. Bucholtz' parameterisation should read a = (n^2 + k^2)m, b = (m^2 + k^2)n, c = (m + n)(m n - k^2), s = (m + n)m n, d = k m n(m + n)(m n - k^2); constraints GCD(m,n,k) = 1; m > n > 0 & 0 < k <= sqrt( m n^2/(m + 2 n) ) enforce one (nonprimitive) triangle in each Heronian similarity class. The inverse mapping (modulo similarity) is given by k = d, n = s(s-b), m = s(s-a). Bucholtz may sometimes yield a primitive triangle: s,d,a,b,c = 70, 630, 25, 52, 63 is given directly by m,n,k = 5, 2, 1 . "Swinging" either side around the vertex C (say) so as to meet the line of c again yields a new (nonprimitive, unordered) triangle with sides [a, b, c] -> [ac, bc, b^2 - a^2] . For instance, [3,4,5] -> [7,15,20] -> [35,44,75] -> ... Unfortunately the sides grow exponentially; I conjecture that a proper Heronian triangle is the smallest swinger in town (as it were) just when no two sides have a common factor, or when isosceles. A couple of questions for you: Is this stuff worth writing up [I'm undecided]? What about writing up the conics through six centres conjectures, which seem to be genuinely difficult? Fred Lunnon [29/10/11] On 10/29/11, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
Fred wins his bet! I've just submitted a paper, ``Five-point circles, the 76-point sphere, and the Pavillet tetrahedron'' to the Monthly. The theme is that triangle geometry is not dead. I would attach a copy if attachments were allowed; I will entertain individual requests. Of course, I'm hoping that the paper will be accepted, and I also now hope to include a stop press paragraph with the following theorem:
The external angle-bisectors of a triangle meet the opposite edges in collinear points. Call it the zero-sum line because: The sum of the directed distances to the edges of a triangle from points on its zero-sum line is zero.
So go ahead and spoil my fun and tell me that that's well known to those who well know it. Certainly the first part is well known since it follows immediately from (the converse of) Menelaus's theorem, so the zero-sum line may already have some other name. The Gergonne line would be a good guess, but that's not quite right.
Note that the zero-sum line (and the Gergonne line) are the line at infinity if the triangle is equilateral. Best, R.
It is easy to see that any Heronian triangle (with integer side-length and area) can be posed in 2-space so that its vertices have rational Cartesian coordinates. A natural question is whether it can be posed with its vertices on the integer lattice. I earlier proposed a counterexample having no lattice pose; however this claim later turned out to result from an elementary programming error. An amended run found lattice poses for every Heronian triangle with semi-perimeter s <= 200. It now seems reasonable to conjecture that every Heronian triangle may be posed on the lattice. If this is true, it must surely be obvious --- but to me, not at the moment! Did Minkowski have something to say about this, I wonder? Fred Lunnon On 10/29/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
I've been taking a desultory look at "rational" test triangles lately, which turn out to have some interest, as well as connections with a well-known notion --- see (tho' overlapping & slightly inaccurate)
http://mathworld.wolfram.com/HeronianTriangle.html http://en.wikipedia.org/wiki/Integer_triangle http://en.wikipedia.org/wiki/Heronian_triangle
A triangle T is "Heronian" means it has integer sides a,b,c and rational semi-perimeter s, area d. Sample results:
A triangle has rational Cartesian coordinates and sides iff it is similar to a Heronian triangle. Not every Heronian triangle can be posed to have integer coordinates: s,d,a,b,c = 75, 420, 26, 51, 73 is the smallest counterexample.
For T Heronian: an odd number of a,b,c are even; s is an integer, d is an even integer; when T is proper (ie. d <> 0) s is composite; semi-angle tangents are rational: tan C/2 = d/(s-a)(s-b), etc.
Bucholtz' parameterisation should read a = (n^2 + k^2)m, b = (m^2 + k^2)n, c = (m + n)(m n - k^2), s = (m + n)m n, d = k m n(m + n)(m n - k^2); constraints GCD(m,n,k) = 1; m > n > 0 & 0 < k <= sqrt( m n^2/(m + 2 n) ) enforce one (nonprimitive) triangle in each Heronian similarity class.
The inverse mapping (modulo similarity) is given by k = d, n = s(s-b), m = s(s-a). Bucholtz may sometimes yield a primitive triangle: s,d,a,b,c = 70, 630, 25, 52, 63 is given directly by m,n,k = 5, 2, 1 .
"Swinging" either side around the vertex C (say) so as to meet the line of c again yields a new (nonprimitive, unordered) triangle with sides [a, b, c] -> [ac, bc, b^2 - a^2] . For instance, [3,4,5] -> [7,15,20] -> [35,44,75] -> ... Unfortunately the sides grow exponentially; I conjecture that a proper Heronian triangle is the smallest swinger in town (as it were) just when no two sides have a common factor, or when isosceles.
A couple of questions for you: Is this stuff worth writing up [I'm undecided]? What about writing up the conics through six centres conjectures, which seem to be genuinely difficult?
Fred Lunnon [29/10/11]
On 10/29/11, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
Fred wins his bet! I've just submitted a paper, ``Five-point circles, the 76-point sphere, and the Pavillet tetrahedron'' to the Monthly. The theme is that triangle geometry is not dead. I would attach a copy if attachments were allowed; I will entertain individual requests. Of course, I'm hoping that the paper will be accepted, and I also now hope to include a stop press paragraph with the following theorem:
The external angle-bisectors of a triangle meet the opposite edges in collinear points. Call it the zero-sum line because: The sum of the directed distances to the edges of a triangle from points on its zero-sum line is zero.
So go ahead and spoil my fun and tell me that that's well known to those who well know it. Certainly the first part is well known since it follows immediately from (the converse of) Menelaus's theorem, so the zero-sum line may already have some other name. The Gergonne line would be a good guess, but that's not quite right.
Note that the zero-sum line (and the Gergonne line) are the line at infinity if the triangle is equilateral. Best, R.
Fred, I'm confused about this. If a triangle is posed with rational coordinates, then scaling by the lcm of all six denominators will make the coordinated integers, while the triangle continues to have integer sides and area. Am I missing something? -- Gene
________________________________ From: Fred lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Wednesday, November 16, 2011 2:31 PM Subject: Re: [math-fun] Elementary triangle puzzle
It is easy to see that any Heronian triangle (with integer side-length and area) can be posed in 2-space so that its vertices have rational Cartesian coordinates. A natural question is whether it can be posed with its vertices on the integer lattice.
I earlier proposed a counterexample having no lattice pose; however this claim later turned out to result from an elementary programming error. An amended run found lattice poses for every Heronian triangle with semi-perimeter s <= 200.
It now seems reasonable to conjecture that every Heronian triangle may be posed on the lattice. If this is true, it must surely be obvious --- but to me, not at the moment! Did Minkowski have something to say about this, I wonder?
Fred Lunnon
Brahmagupta showed that any Heron triangle can be thought of as the adjunction of two Pythagorean ones. R. On Wed, 16 Nov 2011, Eugene Salamin wrote:
Fred, I'm confused about this. If a triangle is posed with rational coordinates, then scaling by the lcm of all six denominators will make the coordinated integers, while the triangle continues to have integer sides and area. Am I missing something?
-- Gene
________________________________ From: Fred lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Wednesday, November 16, 2011 2:31 PM Subject: Re: [math-fun] Elementary triangle puzzle
It is easy to see that any Heronian triangle (with integer side-length and area) can be posed in 2-space so that its vertices have rational Cartesian coordinates. A natural question is whether it can be posed with its vertices on the integer lattice.
I earlier proposed a counterexample having no lattice pose; however this claim later turned out to result from an elementary programming error. An amended run found lattice poses for every Heronian triangle with semi-perimeter s <= 200.
It now seems reasonable to conjecture that every Heronian triangle may be posed on the lattice. If this is true, it must surely be obvious --- but to me, not at the moment! Did Minkowski have something to say about this, I wonder?
Fred Lunnon
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If I am understanding Fred's original query, he wants a congruent pose, not just a similar one; if that's the case, then scaling is not permitted, and that would be what Gene is missing. But it looks like the theorem Richard attributes to Brahmagupta affirms Fred's conjecture. On Wed, Nov 16, 2011 at 5:52 PM, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
Brahmagupta showed that any Heron triangle can be thought of as the adjunction of two Pythagorean ones. R.
On Wed, 16 Nov 2011, Eugene Salamin wrote:
Fred, I'm confused about this. If a triangle is posed with rational
coordinates, then scaling by the lcm of all six denominators will make the coordinated integers, while the triangle continues to have integer sides and area. Am I missing something?
-- Gene
______________________________**__
From: Fred lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com**> Sent: Wednesday, November 16, 2011 2:31 PM Subject: Re: [math-fun] Elementary triangle puzzle
It is easy to see that any Heronian triangle (with integer side-length and area) can be posed in 2-space so that its vertices have rational Cartesian coordinates. A natural question is whether it can be posed with its vertices on the integer lattice.
I earlier proposed a counterexample having no lattice pose; however this claim later turned out to result from an elementary programming error. An amended run found lattice poses for every Heronian triangle with semi-perimeter s <= 200.
It now seems reasonable to conjecture that every Heronian triangle may be posed on the lattice. If this is true, it must surely be obvious --- but to me, not at the moment! Did Minkowski have something to say about this, I wonder?
Fred Lunnon
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As far as I can see, splitting a Heronian triangle (by its altitude) into two Pythagorean (right-angled, with rational edge-lengths, hence areas) constructs merely a pose with rational coordinates. For example, suppose one edge joins [0, 0] and [a, 0]; then the third vertex lies at [(a^2+b^2-c^2) / 2 a, 2 d / a], where d denotes the area. This will not be an integer unless some edge-length a happens to divide 2d . For triangles such as [a, b, c] = [5, 29, 30], where s, d = 32, 72, there is no such edge. However, the rotated pose with vertices [0, 0], [-4, 3], [-20, 21] still lies on the lattice. A similar apparently happy accident overtakes every case with s <= 200. Fred Lunnon On 11/16/11, Allan Wechsler <acwacw@gmail.com> wrote:
If I am understanding Fred's original query, he wants a congruent pose, not just a similar one; if that's the case, then scaling is not permitted, and that would be what Gene is missing. But it looks like the theorem Richard attributes to Brahmagupta affirms Fred's conjecture.
On Wed, Nov 16, 2011 at 5:52 PM, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
Brahmagupta showed that any Heron triangle can be thought of as the adjunction of two Pythagorean ones. R.
On Wed, 16 Nov 2011, Eugene Salamin wrote:
Fred, I'm confused about this. If a triangle is posed with rational
coordinates, then scaling by the lcm of all six denominators will make the coordinated integers, while the triangle continues to have integer sides and area. Am I missing something?
-- Gene
______________________________**__
From: Fred lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com**> Sent: Wednesday, November 16, 2011 2:31 PM Subject: Re: [math-fun] Elementary triangle puzzle
It is easy to see that any Heronian triangle (with integer side-length and area) can be posed in 2-space so that its vertices have rational Cartesian coordinates. A natural question is whether it can be posed with its vertices on the integer lattice.
I earlier proposed a counterexample having no lattice pose; however this claim later turned out to result from an elementary programming error. An amended run found lattice poses for every Heronian triangle with semi-perimeter s <= 200.
It now seems reasonable to conjecture that every Heronian triangle may be posed on the lattice. If this is true, it must surely be obvious --- but to me, not at the moment! Did Minkowski have something to say about this, I wonder?
Fred Lunnon
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I have now verified this conjecture for semiperimeter s <= 303; in all 208 essentially distinct cases where the obvious method placing one side along an axis is inapplicable, there is a rotated pose with lattice vertices. My (simpleminded) program is bogging down by this point; I don't know if it's worth polishing up a tad. Anybody wishing to inspect the data is welcome to get in touch. Fred Lunnon On 11/17/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
As far as I can see, splitting a Heronian triangle (by its altitude) into two Pythagorean (right-angled, with rational edge-lengths, hence areas) constructs a pose with merely rational coordinates.
For example, suppose one edge joins [0, 0] and [a, 0]; then the third vertex lies at [(a^2+b^2-c^2) / 2 a, 2 d / a], where d denotes the area. This will not be an integer unless some edge-length a happens to divide 2d .
For triangles such as [a, b, c] = [5, 29, 30], where s, d = 32, 72, there is no such edge. However, the rotated pose with vertices [0, 0], [-4, 3], [-20, 21] still lies on the lattice. A similar apparently happy accident overtakes every case with s <= 200.
Fred Lunnon
On 11/16/11, Allan Wechsler <acwacw@gmail.com> wrote:
If I am understanding Fred's original query, he wants a congruent pose, not just a similar one; if that's the case, then scaling is not permitted, and that would be what Gene is missing. But it looks like the theorem Richard attributes to Brahmagupta affirms Fred's conjecture.
On Wed, Nov 16, 2011 at 5:52 PM, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
Brahmagupta showed that any Heron triangle can be thought of as the adjunction of two Pythagorean ones. R.
On Wed, 16 Nov 2011, Eugene Salamin wrote:
Fred, I'm confused about this. If a triangle is posed with rational
coordinates, then scaling by the lcm of all six denominators will make the coordinated integers, while the triangle continues to have integer sides and area. Am I missing something?
-- Gene
______________________________**__
From: Fred lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com**> Sent: Wednesday, November 16, 2011 2:31 PM Subject: Re: [math-fun] Elementary triangle puzzle
It is easy to see that any Heronian triangle (with integer side-length and area) can be posed in 2-space so that its vertices have rational Cartesian coordinates. A natural question is whether it can be posed with its vertices on the integer lattice.
I earlier proposed a counterexample having no lattice pose; however this claim later turned out to result from an elementary programming error. An amended run found lattice poses for every Heronian triangle with semi-perimeter s <= 200.
It now seems reasonable to conjecture that every Heronian triangle may be posed on the lattice. If this is true, it must surely be obvious --- but to me, not at the moment! Did Minkowski have something to say about this, I wonder?
Fred Lunnon
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Yet another elementary program glitch caused misprinted vertex coordinates --- for [a, b, c] = [5, 29, 30] the lattice vertices should have read [0, 0], [3, 4], [21, -20] . The data list sent out subsequently was similarly garbled, and a corrected version will follow [embarrassed groan]. WFL On 11/17/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
As far as I can see, splitting a Heronian triangle (by its altitude) into two Pythagorean (right-angled, with rational edge-lengths, hence areas) constructs merely a pose with rational coordinates.
For example, suppose one edge joins [0, 0] and [a, 0]; then the third vertex lies at [(a^2+b^2-c^2) / 2 a, 2 d / a], where d denotes the area. This will not be an integer unless some edge-length a happens to divide 2d .
For triangles such as [a, b, c] = [5, 29, 30], where s, d = 32, 72, there is no such edge. However, the rotated pose with vertices [0, 0], [-4, 3], [-20, 21] still lies on the lattice. A similar apparently happy accident overtakes every case with s <= 200.
Fred Lunnon
participants (5)
-
Allan Wechsler -
Dan Asimov -
Eugene Salamin -
Fred lunnon -
Richard Guy