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Re: consecutiveinteger triangles with integer area
Posted:
Jun 3, 1998 10:35 PM


Jim Totten wrote: > > Thanks to allit's been very interesting. > > Heron gives area K=(n/2)*sqrt(3*(n/2 +1)(n/2 1)). > Let 3*(n/2 +1)(n/2 1)=b^2 : it must be square. > Multiply : 3n^24b^2=12. > Factor: (n/2  b/sqrt3)(n/2 + b/sqrt3)=1. [Eqn A] > Since 345 works, plug in 4 for n and compute b to be 3. Plug into > equation A to get: (2sqrt3)(2+sqrt3)=1. > Square both sides: (7 4*sqrt3)(7+ 4*sqrt3)=1. > This must be a solution to Eqn A, so another solution must be n=14, > which works. > Cubing equation A gets you the 515253 triangle, the fourth power > gives the 193194195, etc. > [Floor: is this matrix multiplication?] > > And then there's this: doubling the area of one of these triangles > gives you an altitude (to the n side) of a larger such triangle. Why > is that? > And why does the altitude to the "middle" side always separate that > side into lengths which differ by 4? > And, is this the only type of solution? > Tired of this? > Go Bulls!
A couple of observations. Using your notation, from your first two equations you have: K/(n/2) = b. Since K is the area, and n is the length of the "middle" side, this says that b is the altitude to the "middle" side. Thus this altitude is an integer.
Also, the above altitude divides the triangle into two right triangles. Say the side of length n is divided into lengths x and y by the altitude, where x is a leg of the right triangle with hypotenuse n1 and y is a leg of the right triangle with hypotenuse n+1. By Pythagoras, (1) b^2 + y^2 = (n+1)^2 (2) b^2 + x^2 = (n1)^2 Subtract the equations to get y^2  x^2 = 4n, or (y+x)(yx) = 4n. Since y+x = n, this gives yx=4.
Peter



