[seqfan] Re: definition of A074074
Richard Mathar
mathar at strw.leidenuniv.nl
Sat Sep 19 21:58:38 CEST 2009
One may ask also a corresponding question on
http://research.att.com/~njas/sequences/A074076 :
there seem to be solutions for a(6) with area 264, putting a(6) = 264/6 = 44, with
sides (a,b,c)=(44,37,15) and for a(7) with area 240, putting a(7)=240/6=40, with
sides (a,b,c)=(40,37,13), much smaller than those claimed.
Guide to the solutions of A074076:
Consider the general integer triangle with
(i) side vec(a) horizontal from Cartesian point (0,0) to (a,0)
(ii) side vec(b) from Cartesian point (0,0) to (b_x,b_y),
and (iii) side vec(c) from Cartesian point (a,0) to (b_x,b_y).
Assume "weak" sorting such that a>=b>=c. Given the three side lengths
a, b and c, one can determine the two missing quantities b_x and b_y by standard
2D vector algebra,
b_x^2+ b_y^2 = b^2
(a-b_x)^2+ b_y^2 = c^2 -> a^2+b_x^2-2a*b_x+b_y^2 = c^2 -> a^2+b^2-2*a*b_x = c^2
-> b_x = (a^2+b^2-c^2)/(2a)
and from there b_y = sqrt( b^2- b_x^2), positive root without loss of generality.
The mid point of side vec(a) (the median) is given by the coordinates of walking
half way from (0,0) to (a,0), which is (a/2,0).
The altitude from the point (b_x,b_y) to the same side vec(a) defines the foot point
(b_x,0). The distance between altitude and median is then |b_x-a/2|, and
by insertion of b_x from above , defined to be N in A074076,
N = (a^2+b^2-c^2)/(2a)-a/2 = (b^2-c^2)/(2a) = N.
The area of triangles is A=sqrt( s*(s-a)(s-b)(s-c)) with semiperimeter s:= (a+b+c)/2.
and also one half of baseline length times altitude, A = a*b_y/2 <= a*b/2.
See
http://mathworld.wolfram.com/HeronianTriangle.html
http://mathworld.wolfram.com/TriangleMedian.html
http://mathworld.wolfram.com/PerpendicularFoot.html
We want to scan primitive Heronian triangles, gcd(a,b,c) =1
sorted by increasing A until we find N=(b^2-c^2)/(2a) = 2n+1.
To stay in the integers, we write 2s=S, S integer, A^2=s*(s-a)(s-b)(s-c)
16*A^2=S*(S-2a)(S-2b)(S-2c), so factorizing 16*A^2 allows for a systematic
search of all integer triangles. Since b+c>a, we have s >a, S-2a>0, so looking for
the positive divisors is sufficient.
Solutions (marked with a star where proved to be of smallest area) are
a(1) = 360/6= 60 with (a,b,c) = (36,29,25) (*)
a(3) = 12144/6= 2024 with (a,b,c) = (184,165,157)
a(4) = 35904/6= 5984 with (a,b,c) = (528,305,289)
a(6) = 264/6= 44 with (a,b,c) = (44,37,15) (*)
a(7) = 240/6= 40 with (a,b,c) = (40,37,13) (*)
a(8) = 48576/6= 8096 with (a,b,c) = (736,407,375)
a(9) = 624/6= 104 with (a,b,c) = (52,51,25)
a(10) = 18096/6= 3016 with (a,b,c) = (348,221,185)
a(11) = 4080/6= 680 with (a,b,c) = (136,109,75)
a(12) = 1872/6= 312 with (a,b,c) = (104,85,45)
a(13) = 27720/6= 4620 with (a,b,c) = (396,265,221)
a(14) = 18144/6= 3024 with (a,b,c) = (432,259,205)
a(15) = 1656/6= 276 with (a,b,c) = (92,85,39)
a(16) = 3480/6= 580 with (a,b,c) = (116,109,65)
a(17) = 7200/6= 1200 with (a,b,c) = (200,153,97)
a(19) = 1200/6= 200 with (a,b,c) = (120,101,29)
a(21) = 58464/6= 9744 with (a,b,c) = (464,373,315)
a(22) = 19680/6= 3280 with (a,b,c) = (328,241,169)
a(24) = 2256/6= 376 with (a,b,c) = (188,145,51)
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