0
$\begingroup$

Does anyone know a point in the 2D plane which satisfies the following properties:

  1. Its distance from two vertices of a unit square is rational.
  2. Its distance from other two vertices of a unit square is irrational.
  3. It does not lie on any lines of symmetries (diagonals)
  4. All 4 distance are different.
  5. It does not lie on the edges of the square.
  6. It should not lie on the extended line of the edges.

2 Answers 2

1

Why, that's quite simple: $\left({16\over25},{12\over25}\right)$ would do.

As for the more general approach, take any integer Pythagorean triangle, scale it so that the hypotenuse becomes 1, and position that hypotenuse along the side of the unit square, with the right angle pointing inwards (or maybe outwards, for that matter, since you didn't ask for the point to be inside). Then all conditions except (2) are met automatically, and (2) has more than a decent chance to be met accidentally as well.

  • 0
    Thanks Ivan. That was helpful. Actually i made a mistake in asking the question so now i have edited it. – 2017-02-21
1

Here is an example: Take the unit square of side length $1$ with one corner in the origin, and extending into the first quadrant. Name the corners $A,B,C$ and $D$ starting in the origin, proceeding couter-clockwise. Then the points $P=(\frac {12}{25},\frac9{25})$ is an example as the distances are

$$ \overline{AP}=\frac35,\qquad \overline{BP}=\frac45,\qquad \overline{CP}=\sqrt{\frac{17}5},\qquad \overline{DP}=\sqrt{\frac25}. $$

  • 0
    Thanks Winter. That was helpful. Actually i made a mistake in asking the question so now i have edited it. – 2017-02-21