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Let $D$ be a disk, $P,Q$ be two distinct points in its iterior. A geogebra experiment shows that there exists one and only one (possible degenerated, with infinite radious) circle through $P$ and $Q$ and meeting the boundary of $D$ at two antipodal points.

How to prove it?

Motivation. I'm looking for a model of the real projective plane in the unit disk $D$ (with antipodal points identified). By taking as lines of this models the arcs of circles which meets the boundary of $D$ at two antipodal points, we are proving that given two points $P,D$ one and only line of this model passes through them.

A rule-compass construction. Let $S,T$ be the intersection of $PQ$ with the boundary of $D$, $S',T'$ be the respective antipodal points.

Given two points $X,Y$, let $b_{XY}$ denote the line bisector of the segment $XY$.

Consider $G=b_{T'P}\cap b_{QT}$, $H=b_{SP}\cap b_{QS'}$. Then $I=GH\cap b_{PQ}$ is the center of the required circle.

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  • $\begingroup$ If $P$ and $Q$ lie on a diameter of $D$, then there is no such circle. $\endgroup$ – TonyK Aug 2 at 12:21
  • $\begingroup$ @TonyK: In that case we can think the line $PQ$ as a circle with infinite radious. $\endgroup$ – Fabio Lucchini Aug 2 at 12:23
  • $\begingroup$ I think, a way to prove uniqueness is using Ptolemy's theorem: en.wikipedia.org/wiki/Ptolemy%27s_theorem $\endgroup$ – Βασίλης Μάρκος Aug 2 at 13:38
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Let $D$ be the unit disc in ${\mathbb R}^2$. Embed ${\mathbb R}^2$ via $(x,y)\mapsto(x,y,0)$ into ${\mathbb R}^3$, and consider the stereographic projection $\sigma:\>\dot{S^2}\to{\mathbb R}^2$ from the north pole $(0,0,1)$. This map keeps the points of $\partial D$ fixed, and maps the set of circles in $S^2$ to the circles and lines in ${\mathbb R}^2$. A circle $\ne\partial D$ in $S^2$ is a great circle iff it intersects $\partial D$ in two antipodal points.

Now let $\hat P=\sigma^{-1}(P)$ and $\hat Q=\sigma^{-1}(Q)$. There is exactly one great circle $\gamma$ through $\hat P$ and $\hat Q\in S^2$. It follows that $\sigma(\gamma)$ is the single circle through $P$ and $Q$ that intersects $\partial D$ in two antipodal points.

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Since you already have a method of constructing the needed circle $C$, I assume your question is about uniqueness. Here is an informal analytic approach that could be formalized.

Let $M$ be the center of $D$, and $N$ be the center of $C$. Since $C$ passes through $P$ and $Q$, $N$ must lie on the line $\ell = b_{PQ}$. Let $X$ be the center of the center of the circle through $P$ and $Q$ which is tangent to $D$ on opposite side of $\overline{PQ}$ from $M$, and let $Y$ be the center of the circle which is tangent to $D$ on the same side of $\overline{PQ}$ from $M$. Consider the direction from $X$ to $Y$ to be up.

Let $Z$ be a point on $\ell$ and consider the circle centered at $Z$ passing through $P$ and $Q$, and the angle formed by the intersections of that circle with $D$ and $M$. Let $\theta$ be the measure on the upper side of that angle. The intersection points will be antipodal if and only if $\theta = \pi$. Let $\theta_0$ be the measure of the upper side of $\angle SMT$ (where, as in the OP, $S$ and $M$ are the intersections of the line through $P$ and $Q$ with $D$.) Because the midpoint of $\overline {PQ}$ is below $M$, $\theta_0 > \pi$.

When $Z = X, \theta = 2\pi$. As $Z$ descends, the intersection points move up the sides of $D$, with $\theta$ decreasing. As $Z$ goes to $-\infty$, the intersection points approach $S$ and $T$, and $\theta \to \theta_0$ from above. Since $\theta > \theta_0 >\pi$, No circle in this region has antipodal intersection points.

When $Z$ is between $X$ and $Y$, its circle does not intersect $D$. When $Z = Y, \theta = 0$, and as $Z$ rises, the intersection points descend the sides of $D$ and $\theta$ increases continuously. As $Z \to \infty, \theta \to \theta_0-$, so at some point $\theta$ must rise past $\pi$. By the intermediate value theorem, there is a point where it equals $\pi$. But since $\theta$ is strictly increasing in $Z$, it cannot be $\pi$ at more than a single point.

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