Determine if two annuli intersect by just looking at the generating triangles

euclidean-geometrygeometryintersection-theorytriangles

Draw a random triangle on the plane and label its vertices $A$, $B$ and $C$:
triangle

Now draw a circle with $A$ as its center and $\overline{AB}$ as its radius, and one with $\overline{AC}$ as its radius:
first two circles

These two circles ($OA_B$ and $OA_C$) form an annulus.

Do the same with $B$ as the center:
second two circles

We have in total 4 circles, forming 2 annuli, $OA$ (formed by $OA_B$ and $OA_C$) and $OB$ (formed by $OB_A$ and $OB_C$). They intersect each other "completely," i.e., there are a total of 8 intersection points, and one of them is $C$. These points can be partitioned into 2 groups of 4 points.

Let's only take a look at the group of points that doesn't include $C$, which is on the left side in the example above, here labeled $I$, $J$, $K$ and $L$; they form a quadrilateral:

quadrilateral

The problem

It isn't always here

Now that we constructed this quadrilateral, it's safe to say, with this exact triangle, this quadrilateral is always there, because the intersection points are always here. But what happens, if we move the point $A$, for example along the circumcircle of the triangle:

animation

This shows us, that this quadrilateral isn't always there, because the annuli don't always intersect, depending on the shape of the triangle.

It's not obvious why and when the annuli don't intersect, it almost seems like it's random. The only "regular" and "logical" thing we can observe is that the intersection points are all on one line, or more specifically, that 2 points each become 1 point, if we have an isosceles triangle because then 2 radii are of the same lengths.

Some observations

  1. The intersections disappear when one annuli completely encompasses another, obviously.
  2. It's always the same 2 intersections that disappear.
  3. 1 point disappears and appears faster than the other, at least in this example.
  4. The quadrilateral can be concave, but most of the time (at least in this example), it isn't.
  5. At some points, the quadrilateral is self-intersecting.

The annuli may be unnecessary, mathematically this is probably the same problem as drawing just to circles, one on with $A$ and one with $B$ as its center. Also, the quadrilateral isn't really "part" of the problem, it only visualizes it.

The questions

  1. Which triangle shapes correspond to which intersections, and why?
  2. How can we, by just looking at the shape, side lenghts, angles, etc., predict if a triangle will have 4, 3, or 2 of these annulus intersections
  3. What happens when we don't take a triangle as base polygon, but a quadrilateral, pentagon, hexagon, etc.? Is this "disappearing" of the intersections consistent through out all polygons.

Resources

This is the GeoGebra project I used for the visualizations:
https://www.geogebra.org/classic/cvkbxbkn

You can drag and drop any points of the triangle or click on the play button to start the animation.

Best Answer

You have all possible intersection point if all relevant pairs of circles intersect twice. The circles of radius $AB$ around $A$ and $B$ will always do, and in fact the intersections form equliateral trinagles with $A,B$. The circles of radius $A$ around $A$ and $BC$ around $B$ will always do so because $C$ exists.

For the circle of radius $AB$ around $A$ and of radius $BC$ around $B$, we need the trangle inequality to hold: $$ |AB-BC|<AB<AB+BC,$$ which is equivalent to $$\tag1BC<2AB.$$ Similarly, for the circle of radius $AC$ around $A$ and of radius $AB$ around $B$, we need $$\tag2AC<2AB.$$ If $(1)$ and $(2)$ hold, you are fine.

Related Question