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I've a rational circle defined by three (rational) points. Given two (rational) points on this circle (can be two of the defining points) how can I find a rational point on the circle close to half way between these points?

In other words, I want to find the intersection(s) between the perpendicular bisector of the line determined by the two points and the circle as a rational approximation.

Since the points defining the circle can be very close together, a solution not involving floating point operations is preferred.

I though that plugging in $t=1/2$ in the accepted answer at Find point on rational circle for given angle would do the trick, but this returns a point somewhat out of the middle.

Somos
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Remco Poelstra
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1 Answers1

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Here is a parametrization of the unit circle $C$ using Pythagorean triples. Given integer pairs $(u,v),\;$ let $a:=u^2+v^2,\; b:=u^2-v^2,\; c:=2uv.\;$ Then $a^2=b^2+c^2\;$ and $\;(b/a,c/a)\;$ is a rational point on $C.\;$ A modern way to understand this is to regard $(u,v)$ as a point in the complex plane. Then $(u+iv)^2=b+ic\;$ with $\;a=|b+ic|.$ This allows us to interpret the point on $C$ halfway between $P_1$ and $P_2$ as the geometric mean $P:=\sqrt{P_1P_2}$ of the two given points. The only thing missing is how to get a rational point on $C$ close enough to $P$ on $C$. To answer this, we compute sqare root $Q=\sqrt{P}\;$ and find a close enough rational approximation $\;Q\approx(u+iv)/w,\;$ and consequently $\;(b/a,c/a)\;$ is our desired point on $C$ close to halfway between $P_1$ and $P_2$.

Somos
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