Newzad submitted this question for our consideration. It is more related to numbers theory than geometry, and looks very complicated. Does anybody have any idea? Please do comment.

answer is yes. For any N, there exists a way to build a right triangle with horizontal leg divided in N segments such that all triangles have integer sides.

Here is a (non-optimal) way to build such a triangle: for any m,n such that m>n>1 (2mn, m^2-n^2, m^2+n^2) form a pythagorean triad. Take the following (m,n) pairs: (2^(2N-1),1), (2^(2N-2),2), ... , (2^N, 2^(N-1)) they generate N distinct triads with common leg 2^2N. Corresponding triangles can be drawn leading to requested figure.

Optimal height leg for N=3 is 12. The 3 triads are (12,5,13), (12,16,20), (12,35,37). Horizontal segments are 5, 11, 19

Interestingly, 24 (as depicted in example) is the optimal leg for N=5. Therefore the example given can be extended with the following segments: 7, 3, 22, 38, 73. Corresponding triads: (24,7,25), (24,10,26), (24,32,40), (24,70,74), (24,143,145)

Also found the the following solution from Aykut. Link: http://www.metu.edu.tr/~baykut/Newzad_Triangles.pdf

Text: For a finite fixed a, can we draw infinitely many Newzad Triangles by extending the hypotenuse? First notice that if we can draw infinitely many triangles for a fixed a, then there are infinitely many b, c, d integers that satisfy the following equations. (Since we can rename b + c as b and d as c and e as d; or b + c + d as b, e as c ...) a^2 + b^2 = c^2 (1) a^2 + (b + c)^2 = d^2 (2)

Then a^2 = (c − b)(c + b) (3) a^2 = (d − c − b)(d + c + b) (4) Let a1 and a2 be two integers such that a1 × a2= a^2 and a1 = c − b and a2 = c + b. Then for a unique fixed (a1, a2) pair b and c are unique. Then from equation (4) d is also unique. Which means that for a unique (a1, a2) pair we have unique integers b, c and d. Then since a finite fixed a has a finite number of integer multipliar pairs (a1, a2), we have finite number of integers b, c and d. Then for a finite fixed a, one can not draw infinitely many Newzad Triangles.

Thank you very much for publishing this question 8foxes

ReplyDeleteAfter final exams (nearly two weeks), i will be back to follow your great questions.

answer is yes. For any N, there exists a way to build a right triangle with horizontal leg divided in N segments such that all triangles have integer sides.

ReplyDeleteHere is a (non-optimal) way to build such a triangle: for any m,n such that m>n>1 (2mn, m^2-n^2, m^2+n^2) form a pythagorean triad. Take the following (m,n) pairs:

(2^(2N-1),1), (2^(2N-2),2), ... , (2^N, 2^(N-1))

they generate N distinct triads with common leg 2^2N. Corresponding triangles can be drawn leading to requested figure.

Optimal height leg for N=3 is 12. The 3 triads are (12,5,13), (12,16,20), (12,35,37). Horizontal segments are 5, 11, 19

Interestingly, 24 (as depicted in example) is the optimal leg for N=5. Therefore the example given can be extended with the following segments: 7, 3, 22, 38, 73. Corresponding triads: (24,7,25), (24,10,26), (24,32,40), (24,70,74), (24,143,145)

bleaug

oops, I forgot that (m,n) generators do not build all pythagorean triads.

ReplyDeleteActually 12 is optimal for N=3 and N=4. Triad (12,9,15) was missing. Segments would be 5, 4, 7, 19.

24 is optimal for N=5, N=6 and N=7! The following triads were missing: (24,18,30) and (24, 45, 51). Segments would be: 7, 3, 8, 14, 13, 25, 73.

bleaug

Good work for triads.

ReplyDeleteAre there any quadriads?

Pentads?

Now it really gets ugly.

Solution was finded bras.

ReplyDeletehttp://geometri-problemleri.blogspot.com/2010/08/gp-question-7-solution.html

Also found the the following solution from Aykut.

ReplyDeleteLink: http://www.metu.edu.tr/~baykut/Newzad_Triangles.pdf

Text:

For a finite fixed a, can we draw infinitely many Newzad Triangles by extending the hypotenuse?

First notice that if we can draw infinitely many triangles for a fixed a, then there are infinitely

many b, c, d integers that satisfy the following equations. (Since we can rename b + c as b and d as c and e as d; or b + c + d as b, e as c ...)

a^2 + b^2 = c^2 (1)

a^2 + (b + c)^2 = d^2 (2)

Then

a^2 = (c − b)(c + b) (3)

a^2 = (d − c − b)(d + c + b) (4)

Let a1 and a2 be two integers such that

a1 × a2= a^2 and a1 = c − b and a2 = c + b.

Then for a unique fixed (a1, a2) pair b and c are unique. Then from equation (4) d is also unique.

Which means that for a unique (a1, a2) pair we have unique integers b, c and d. Then since a finite

fixed a has a finite number of integer multipliar pairs (a1, a2), we have finite number of integers b,

c and d. Then for a finite fixed a, one can not draw infinitely many Newzad Triangles.