Given n ≥ 3, show how to place n points in the plane, so that the distance between any two points is irrational, but the area of the triangle formed by any three points is a (positive) rational number.
Create a square grid of points whose horizontal and vertical separations are sqrt(3). Any line connecting points will be either k*sqrt(3) and therefore irrational, or the hypotenuse of a right triangle that's j*sqrt(3) x k*sqrt(3). It's area is rational, 3*j*k/2. Any such triangle can be formed with a combination of these, either added or subtracted (certainly subtracted from a rectangle with integer area).
But do all possible line segments have irrational length?
A line's length will be sqrt(3*j^2+3*k^2) = sqrt(3)*sqrt(j^2+k^2). I don't believe this can ever be rational. The program below tests all cases up to a delta x or delta y up to 2000 and does not find any.
For x = 1 To 20000
For y = x To 20000
l1 = x * Sqr(3): l2 = y * Sqr(3)
tot = l1 * l1 + l2 * l2
sr = Sqr(tot)
rounded = Int(sr + 0.5)
If Abs(sr - rounded) < 0.00000001 Then
Text1.Text = Text1.Text & x & Str(y) & crlf
End If
DoEvents
Next
Next
Edited on August 19, 2019, 12:49 pm
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Posted by Charlie
on 2019-08-19 12:44:02 |
solution
