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.
(In reply to
solution by Charlie)
Say sqrt(3*j^2+3*k^2) is rational, that is, it's expressible as a/b with a,b integers.
Then by squaring, setting a divisible by 3, and dividing out common factors, you end up with an equation of the form A^2 + B^2 = 3C^2.
Then if A and B are both divisible by 3, so is C which is impossible, and otherwise LHS = 1 or 2 mod 3.
So the expression is never rational.
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Posted by xdog
on 2019-08-20 08:44:33 |
re: solution
