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Topic: consecutive-integer triangles with integral area
Replies: 4   Last Post: Jun 3, 1998 10:35 PM

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Peter Ash

Posts: 13
Registered: 12/6/04
Re: consecutive-integer triangles with integer area
Posted: Jun 3, 1998 10:35 PM
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Jim Totten wrote:
> Thanks to all--it'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^2-4b^2=12.
> Factor: (n/2 - b/sqrt3)(n/2 + b/sqrt3)=1. [Eqn A]
> Since 3-4-5 works, plug in 4 for n and compute b to be 3. Plug into
> equation A to get: (2-sqrt3)(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 51-52-53 triangle, the fourth power
> gives the 193-194-195, 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

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 n-1 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 = (n-1)^2
Subtract the equations to get y^2 - x^2 = 4n, or (y+x)(y-x) = 4n.
Since y+x = n, this gives y-x=4.


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