Rational points
Is there a circle in the Cartesian plane which passes through exactly 4 points with rational coordinates?
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2 solutions
A:(0,0) B:(1,0) C:(0,1) D:(1,1), doesn't work??
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It passes through the rational point ( 1 . 2 , 0 . 6 ) .
When does the case of having 0, 1 or 2 rational points occur?
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- 0 rational points: something like center ( 2 , π ) , radius 1.
- 1 rational point: something like center ( 2 , π ) , radius π 2 + 2 .
- 2 rational points: something like center ( 0 , π ) , radius π 2 + 1 .
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Consider the circles C r with centre ( 2 , 3 ) and radius r .
If r = 1 then if ( x , y ) were a rational point on C r , then ( x − 2 ) 2 + ( y − 3 ) 2 = 1 and hence ( x 2 + y 2 + 4 ) − 2 x 2 − 2 y 3 = 0 , which would mean that 1 , 2 , 3 , were linearly dependent over Q , which they are not. This circle has no rational points.
If r = 6 − 2 2 then ( x , y ) is a rational point on C r precisely when ( x − 2 ) 2 + ( y − 3 ) 2 ( x 2 + y 2 − 1 ) + 2 ( 1 − x ) 2 − 2 y 3 = r 2 = 6 − 2 2 = 0 Since 1 , 2 , 3 are linearly independent over Q , this implies that x = 1 , y = 0 . This circle passes through just one rational point.
As you observed, the circle with centre ( 0 , π ) and radius π 2 + 1 passes through the two rational points ( ± 1 , 0 ) . Since the circle does not have rational centre, it cannot pass through any more rational points.
The irrational don't gave be be "bad", if by that you mean transcendental. Linear independence is the key.
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@Mark Hennings – Indeed, going for transcendentals makes the problem harder. For example, it is still not known whether or not 1 , e , π are linearly dependent over Q or not.
Nice solution! About your generalization to "either have 0, 1, 2, or infinitely many rational points", why can't there be only 8 rational points? for example, the 8 points could be (a,b), (b,a), (-a,b), (-b,a), (-a,-b), (-b,-a),(a,-b) and (b,-a), where a,b are rational and a not equal to b, then the rotation and flipping operation won't give you any extra point.
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If there are three or more rational points, the centre is rational, so we can translate the centre to the origin without loss of generality. If x ∈ Q 2 lies on this circle, then so does M x = ( 5 3 − 5 4 5 4 5 3 ) x and M x is obtained by rotating x through an angle which is not a rational multiple of π . Thus the points M n x are all rational, all lie on the circle, and are all distinct. Indeed, this shows that the rational points on the circle are dense in the circle.
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Right, this argument works. My point is that the generalization can't be based on 8 operations of the D4 group.
Since the points on the circle and radius form Pythogoreon Triplets the following points are possible with the same radius. (1) 47 ; 1104 ; 1105 (2)744 ; 817 ; 1105 (3)576 ; 943 ; 1105 (4)264 ; 1073 ; 1105
I do think the enunciate is ill defined and missleading. In first place a circle it is not a circunference. Second it is trivial that there are many circunferences which pass over four points of rational coordinates. For me the author should have inquire " is there any circunference passing on four and only four points with rational coordinates?
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No, a circle is just its circumference. A circle and its interior is a disk. The question does ask for exactly four points. How more precise do you want?
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I still think you miss the point a circle has area a circumference has not. I think you agree that in math we use to recognize circle the domain of points within a circumference, disk is just an expression which we use frequently to define a circle of a certain thickness. So all the circles even those whose potentially could not have 4 points with rational coordinates in its perimeter the circumference, still will have infinite points of rational coordinates belong to the circle. Besides the term "exactly" is also misleading is should be define "passing through four and only four points". As you know in math we have two type of condition necessary and sufficient. The term exactly is incomplete to define the above.
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@Mariano PerezdelaCruz – The equation of a circle is ( x − a ) 2 + ( y − b ) 2 = r 2 which specifies the outline only, and not its interior. A disk has equation ( x − a ) 2 + ( y − b ) 2 ≤ r 2 which includes the interior. It is not a three-dimensional object like a coin, with thicknes.
The fact that the word “circle” is used loosely to include the interior, when the word “disk” should be used, is unfortunate, but does not make this statement mathematically inaccurate.
“Exactly four” does mean “ four and only four”.
suppose the circle has only 4 rational points.Let us take 3 points now the circumcentre of the triangle formed by 3 rational points will also be rational as both perpendicular bisectors will have rational coefficients and meet at a rational point. Now let us take reflection of these 3 points about the centre which will also lie on the circle. reflection of rational points about rational point will also be rational.Thereby there is atleast 6 rational points.So our presumption is false. Therefore the solution.
Yes there a gross problem with the solution .
To resolve the issue. Case a) No points are reflection of each other then solution holds b) 2 points are mutual reflection about centre other 2 are not .then select 1 of the above point and 2 points which are not mutual reflection.
then by similar reason we have atleast 6 points
c) 2 pair of points are reflection of eachother ie they are end points of the diameter.
Let the pairs be A,C and B,D
Let us take A,C,D .Consider Perpendicular bisector (A'C') of AC which will also have rational coefficients, now take reflection of D about A'C' which will give another rational point on the circle. Same is case with B.
Sorry for amateurish solution
Hope the solution is correct now
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Now this is better, but for your case c), it works if these 4 points only form a rectangle, not a square. However, if they forms a square, then reflection of D about A'C' will be itself, won't give you an extra point.
Reflection of 3 points about the center does not always give you 3 extra points. For example, 2 of the points are already reflection of each other.
Define a rational line to be a line in the form a x + b y = c where a , b , c are integers (or rationals, that's fine too). Observe that if two rational lines intersect, they must intersect at a rational point.
Take any three points among the four rational points that the circle passes through. The circle is the circumcircle of these, so its center is the intersection of perpendicular bisectors. But a perpendicular bisector of two rational points is a rational line, so its center must be a rational point. We can now translate the center to the origin; this doesn't change anything, since the translation is a rational amount of x-coordinate and y-coordinate.
We can rotate the plane by any of 9 0 ∘ , 1 8 0 ∘ , 2 7 0 ∘ ; rational points will also be mapped to rational points on the circle. Thus since we have 4 distinct points, they must be essentially the same point, just rotated 9 0 ∘ , 1 8 0 ∘ , 2 7 0 ∘ ; otherwise they will yield at least two sets of 4 points, contradiction.
But note that flipping the plane over the x-axis will also map rational points to rational points. Thus if flipping the plane doesn't map one of our chosen rational points to another, we'll again have two sets of 4 points, contradiction. So flipping the plane must map our points to each other again.
Finally, this means our points must either lie on the x-axis and y-axis, or on the x = y and x = − y lines. In the first case, suppose ( r , 0 ) lies on the circle (and so ( 0 , r ) , ( − r , 0 ) , ( 0 , − r ) too); note that ( 5 3 r , 5 4 r ) is also on the circle and is rational, contradiction. In the second case, suppose ( r , r ) lies on the circle (and so ( r , − r ) , ( − r , − r ) , ( − r , r ) too); note that ( 5 7 r , 5 1 r ) is also on the circle and rational, contradiction.
Thus it's impossible for a circle to have exactly 4 rational points.
In fact, we can generalize the claim to show that a circle can only either have 0, 1, 2, or infinitely many rational points. If we have 3, use the perpendicular bisector trick to see that the center is rational. This means the radius is in the form k . Generate many solutions to x 2 + y 2 = k z 2 ; they all can be scaled to give rational points on the circle.