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Algebra
23 Jan 2008 08:50:12 IST
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23 Jan 2008 12:39:34 IST
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The reason I posed this qn is that I saw a ridiculous answer somewhere in this formum I can't recall where. I thought that since it is a previous JEE prob, it would have been discussed already and was curious to see the solns
I'll wait one day b4 posting
and 
. RHS is irrational, hence atleast one of the variables on the LHS have to be irrational. => at most two are rational. 
23 Jan 2008 18:45:09 IST
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Konichiwa,
How do you prove that in the entire circle, there are at two rational points and no more. Any number of equilateral triangles could be drawn and we cud conclude that each has at the most two rational points.
How do you prove that in the entire circle, there are at two rational points and no more. Any number of equilateral triangles could be drawn and we cud conclude that each has at the most two rational points.
Now on to you
24 Jan 2008 07:24:54 IST
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The equation of the family of circles with radius (0,
2) is
2) isx2+(y-2)2 = r2.
2)2 = r2.It is easily proved that if r is rational then there are no rational points.
Now, with r irrational, let's assume that we have at least one rational point (x1,y1) on the circle. Now, suppose (x2,y2) is another such point, we must have
x12+(y1-2)2 = x22+(y2-2)2.
2)2 = x22+(y2-2)2.Simplifying we get
(x12+y12-x22-y22) + 22(y2-y1) = 0.
2(y2-y1) = 0.we know that if a+b2 = 0 with a,b rational then a=0 and b= 0.
2 = 0 with a,b rational then a=0 and b= 0.This gives y1 = y2. Correspondingly x2 = -x1. Hence, there is only one other rational point which is (-x1,y1). Hence, if rational points exist there are atmost 2 such points.
24 Jan 2008 12:39:59 IST
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thanks! I should have spent more time on it.
And for your 2nd question, i.e the condition that a circle will not contain any rational point on its circumference is that both x-coordinate and y-coordinate of the centre of the circle must be irrational and such a circle should not pass through the origin.
Let us assume all the points on the circle are irrational.
The family of all circles with centre
is
The family of all circles with centre
is 
Let
and 
be two such rational points on the circumferrence of the circle. Hence,
Quite clearly, this is not possible for possible for any rational
except 
.24 Jan 2008 18:31:55 IST
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and here is another approach...
If the circle has radius r then its equation is 
, or 
. If x and y are both rational then r^2 must be of the form 
,(failed to see this point first time) where a and b are both rational. Then 
. This can only have a rational solution if 
. The straight line meets the circle in at most two points.
, or . If x and y are both rational then r^2 must be of the form ,(failed to see this point first time) where a and b are both rational. Then . This can only have a rational solution if . The straight line meets the circle in at most two points.Free Sign Up!
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a genral eqn of a circle around 0,root 2
is x^2+y^2-(root2/2)y +c=0
this eqn is a quadratic in y^2 in which a=1, b=-root2/2 c=x^2+c , i have no idea how to solve after this point ..lol