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.
Consider a 45/45/90 triangle where the base and height are (2^.5)*(k^.5) where k is not divisible by 2, nor is k a perfect square. The vertices of all such triangles, I believe satisfy the problem.
Example: base=height = 2^.5 * 5^.5. By Pythagoras, the length of the third side is square root of 2 * the base, or 2*(5^.5). All sides are irrational length, yet the area is .5*base*height or 5, which is rational.
I am sure there are infinitely more methods/examples of these, and with more than 3 points.
Edited on August 19, 2019, 11:39 am
Edited on August 20, 2019, 6:35 am
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Posted by Kenny M
on 2019-08-19 11:38:26 |
One method
