Prove that a circle with center at (√2,0) has at most two rational coordinates.
Arindam Santra , 6 Years ago
Grade 12
IIT JEE Entrance Exam> Prove that a circle with center at (√2,0)...
1 Answers
Arun
Last Activity: 6 Years ago
This can be proved using the fact that the circumcentre of a triangle with rational coordinates has rational coordinates.
The proof is pretty straightforward. The circumcentre is the point where the perpendicular bisectors of the triangle edges meet. Consider the perpendicular bisector of a line segment with rational endpoint coordinates. It will be of the form y = mx + c or of the form x = c, with m and c being rational numbers. This is because the midpoint of the line segment also has rational coordinates, and the slope of the perpendicular bisector is just the negative of the inverse of the slope of the line segment itself, which has to be rational as well. Similarly, the perpendicular bisector of another edge of the triangle will have the form y = m'x + c' or x = c'. Simultaneously solving these two equations gives us the centre of the circle. Now, if a set of simultaneous equations with rational coefficients has a unique solution, the solution set has to be rational too. This implies that the circle centre has rational coordinates, a contradiction.
Since the only property of the centre we used in the proof was irrationality of one coordinate, it follows that there can be at most two points with rational coordinates on the circumference of a circle with the centre x or y coordinate irrational.
The proof is pretty straightforward. The circumcentre is the point where the perpendicular bisectors of the triangle edges meet. Consider the perpendicular bisector of a line segment with rational endpoint coordinates. It will be of the form y = mx + c or of the form x = c, with m and c being rational numbers. This is because the midpoint of the line segment also has rational coordinates, and the slope of the perpendicular bisector is just the negative of the inverse of the slope of the line segment itself, which has to be rational as well. Similarly, the perpendicular bisector of another edge of the triangle will have the form y = m'x + c' or x = c'. Simultaneously solving these two equations gives us the centre of the circle. Now, if a set of simultaneous equations with rational coefficients has a unique solution, the solution set has to be rational too. This implies that the circle centre has rational coordinates, a contradiction.
Since the only property of the centre we used in the proof was irrationality of one coordinate, it follows that there can be at most two points with rational coordinates on the circumference of a circle with the centre x or y coordinate irrational.
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