Re: [math-fun] Elementary triangle puzzle
Theorem. Any Heronian triangle can be embedded in the plane so its vertices have integer coordinates. Proof: First position the triangle so that A is at (0, 0) , B is at (a, 0) . It is easy to show that vertex C has rational coordinates. Consider the points as complex numbers; the vertices are all in Q(i) , the fraction field of Z[i] . It is well-known that Z[i] , which is the ring of Gaussian integers, is a PID, and even a Euclidean domain. Write the coordinate(s) of C as alpha / beta , in lowest terms. We have |AC|^2 = N(alpha / beta) = alpha alpha' / (beta beta') (where ' denotes complex conjugation, and N(z) is the norm of z , i.e. z z' ). Since |AC|^2 is (the square of) an integer, we have beta divides alpha alpha' . As it is relatively prime to alpha , we have beta divides alpha' , and thus beta' divides alpha , i.e. alpha = beta' gamma , for some gamma in Z[i] . Since beta is relatively prime to alpha , it follows that beta is also relatively prime to beta' . Also, |BC|^2 = N(a - alpha / beta) = a^2 - a (alpha / beta + alpha' / beta') + alpha alpha' / (beta beta') is an integer, so it follows that a (alpha / beta + alpha' / beta') = a (alpha beta' + alpha' beta) / (beta beta') is an integer. Note that beta is relatively prime to alpha beta' and therefore to alpha beta' + alpha' beta . Take complex conjugates to see that beta' is also relatively prime to alpha beta' + alpha' beta , and therefore beta beta' is relatively prime to alpha beta' + alpha' beta . However, beta beta' divides a (alpha beta' + alpha' beta) . This now implies that beta beta' divides a , so we may write a = b beta beta' , where b is a rational integer. Finally, rotate the triangle by multiplying by beta / beta' , which has magnitude 1 . The vertices of the triangle rotate to 0 , a beta / beta' = b beta^2 and alpha / beta' = gamma , which are Gaussian integers. Therefore the vertices of the rotated triangle have integer coordinates. QED Note that this gives a constructive way to find such an embedding. Michael Reid
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
This does look pretty convincing, though I shan't really believe it until I can manage to implement it. The apparently unavoidable involvement of the complex plane in the argument raises a further question: is an analogous theorem true for Heronian tetrahedra? Or Heronian simplices in n-space --- always assuming such things can actually be constructed ... WFL On 11/18/11, Michael Reid <reid@gauss.math.ucf.edu> wrote:
Theorem. Any Heronian triangle can be embedded in the plane so its vertices have integer coordinates. Proof: First position the triangle so that A is at (0, 0) , B is at (a, 0) . It is easy to show that vertex C has rational coordinates. Consider the points as complex numbers; the vertices are all in Q(i) , the fraction field of Z[i] . It is well-known that Z[i] , which is the ring of Gaussian integers, is a PID, and even a Euclidean domain.
Write the coordinate(s) of C as alpha / beta , in lowest terms. We have |AC|^2 = N(alpha / beta) = alpha alpha' / (beta beta') (where ' denotes complex conjugation, and N(z) is the norm of z , i.e. z z' ). Since |AC|^2 is (the square of) an integer, we have beta divides alpha alpha' . As it is relatively prime to alpha , we have beta divides alpha' , and thus beta' divides alpha , i.e. alpha = beta' gamma , for some gamma in Z[i] . Since beta is relatively prime to alpha , it follows that beta is also relatively prime to beta' .
Also, |BC|^2 = N(a - alpha / beta) = a^2 - a (alpha / beta + alpha' / beta') + alpha alpha' / (beta beta') is an integer, so it follows that a (alpha / beta + alpha' / beta') = a (alpha beta' + alpha' beta) / (beta beta') is an integer. Note that beta is relatively prime to alpha beta' and therefore to alpha beta' + alpha' beta . Take complex conjugates to see that beta' is also relatively prime to alpha beta' + alpha' beta , and therefore beta beta' is relatively prime to alpha beta' + alpha' beta . However, beta beta' divides a (alpha beta' + alpha' beta) . This now implies that beta beta' divides a , so we may write a = b beta beta' , where b is a rational integer.
Finally, rotate the triangle by multiplying by beta / beta' , which has magnitude 1 . The vertices of the triangle rotate to 0 , a beta / beta' = b beta^2 and alpha / beta' = gamma , which are Gaussian integers. Therefore the vertices of the rotated triangle have integer coordinates. QED
Note that this gives a constructive way to find such an embedding.
Michael Reid
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
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Thank you for proof, Michael Reid. I was a doubter so wrote a program to look for counterexamples. It looked at all Heron triangles with sides less than about 800, and found solutions for all. So, I became a believer but could not find a proof. I wonder if a simpler proof, say one high school students could follow, might exist. Note that one can avoid mention of lattice points by stating it this way: Every Heron triangle with at least one integer altitude can be decomposed into the sum or difference of two integer sided right triangles, and every Heron triangle with no integer altitude can be surrounded by three integer sided right triangles to form an integer sided rectangle. (The sides of the Heron triangle are the hypotenuses of the right triangles, and one Heron triangle vertex is at a rectangle vertex and the other two Heron triangle vertices are on the opposite two edges of the rectangle). On Thu, Nov 17, 2011 at 7:25 PM, Michael Reid <reid@gauss.math.ucf.edu> wrote:
Theorem. Any Heronian triangle can be embedded in the plane so its vertices have integer coordinates. Proof: First position the triangle so that A is at (0, 0) , B is at (a, 0) . It is easy to show that vertex C has rational coordinates. Consider the points as complex numbers; the vertices are all in Q(i) , the fraction field of Z[i] . It is well-known that Z[i] , which is the ring of Gaussian integers, is a PID, and even a Euclidean domain.
Write the coordinate(s) of C as alpha / beta , in lowest terms. We have |AC|^2 = N(alpha / beta) = alpha alpha' / (beta beta') (where ' denotes complex conjugation, and N(z) is the norm of z , i.e. z z' ). Since |AC|^2 is (the square of) an integer, we have beta divides alpha alpha' . As it is relatively prime to alpha , we have beta divides alpha' , and thus beta' divides alpha , i.e. alpha = beta' gamma , for some gamma in Z[i] . Since beta is relatively prime to alpha , it follows that beta is also relatively prime to beta' .
Also, |BC|^2 = N(a - alpha / beta) = a^2 - a (alpha / beta + alpha' / beta') + alpha alpha' / (beta beta') is an integer, so it follows that a (alpha / beta + alpha' / beta') = a (alpha beta' + alpha' beta) / (beta beta') is an integer. Note that beta is relatively prime to alpha beta' and therefore to alpha beta' + alpha' beta . Take complex conjugates to see that beta' is also relatively prime to alpha beta' + alpha' beta , and therefore beta beta' is relatively prime to alpha beta' + alpha' beta . However, beta beta' divides a (alpha beta' + alpha' beta) . This now implies that beta beta' divides a , so we may write a = b beta beta' , where b is a rational integer.
Finally, rotate the triangle by multiplying by beta / beta' , which has magnitude 1 . The vertices of the triangle rotate to 0 , a beta / beta' = b beta^2 and alpha / beta' = gamma , which are Gaussian integers. Therefore the vertices of the rotated triangle have integer coordinates. QED
Note that this gives a constructive way to find such an embedding.
Michael Reid
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
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participants (3)
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Fred lunnon -
James Buddenhagen -
Michael Reid