# Is there anything special about intersection points of vesica pisces circles on a triangle?

Take a triangle $\triangle ABC$

On each pair of vertices of the triangle, place a pair of vesica pisces circles with the properties that each circle

• has the same radius as the other circle of the pair
• passes through the centre of the other circle of the pair
• passes through the two selected vertices of the triangle

You will get something like this with three pairs of circles intersecting at eight other points

• $F$ and $S$ where three different circles intersect: these are the isogonic centres of the triangle and if all the angles of the original triangle are smaller than $\frac{2\pi}{3}$ or $120^\circ$ then the one inside the triangle ($F$ here) is the Fermat point which minimises the sum of the distances $AF$, $BF$, $CF$
• $U$, $V$, $W$, $X$, $Y$ and $Z$ where two circles intersect an edge of the triangle (possibly extended)

I know that if you measure the angle $\measuredangle LMN$ where $L$ and $N$ are any two of $A$, $B$ and $C$ while $M$ is any of $F$, $S$, $U$, $V$, $W$, $X$, $Y$ and $Z$ then this will be an integer multiple of $\frac{\pi}{3}$ or $60^\circ$, since this is a consequence of using vesica pisces for the construction

My question is whether $U$, $V$, $W$, $X$, $Y$ and $Z$ have any other interesting properties

Added later: As a pictorial comment on Joe Knapp's answer, he seems to be saying that if the third vertex of the triangle is in the yellow area of the diagram below, i.e. two of the angles are smaller than $\frac{\pi}{3}$ or $60^\circ$ and the third smaller than $\frac{2\pi}{3}$ or $120^\circ$, then four of the six intersection points lie on the internal edges of the triangle; but if the third vertex of the triangle is in the pink area, i.e. two of the angles are larger than $\frac{\pi}{3}$ or $60^\circ$ or if one angle is larger than $\frac{2\pi}{3}$ or $120^\circ$, then two of the six intersection points lie on the internal edges of the triangle. If the third vertex is on any of the circles or lines then some of the intersection points are on a vertex of the triangle (with an equilateral triangle they all are)

A few observations:

Counting only the intersections that cut across a side of the triangle (i.e., not at a vertex), there can be 0, 2 or 4. If the triangle is obtuse and the obtuse vertex is inside the vesica associated with the opposite side, there are two intersections on the opposite side:

As that vertex moves outside the vesica, there can be four intersections, two on the opposite side and one each on the adjacent sides:

...but only if the vertex is inside the equilateral triangle indicated by the dashed lines. If it is outside the equilateral triangle but still inside the circle defining the vesica, there are two intersections again, one on the opposite side and one on the adjacent side that is in the equilateral triangle:

As the vertex moves to the circle defining the vesica, the exterior intersections merge and there are still two intersections on the triangle:

As the vertex moves to the apex of the equilateral triangle, all intersections merge at the vertices of the triangle, and there are 0 intersections on the sides:

As the vertex moves outside the circle defining the vesica, there can be four intersections:

...but only two if the vertex is inside the dashed lines indicated.

• I think you are saying that two of these six intersection points are on the edges of the triangle either if one of the angles of the triangle is greater than $120^\circ$ (in which case neither isoganal points is inside the triangle) or if two of the angles are each less than $60^\circ$ and the third between $60 ^\circ$ and $120^\circ$ (and one of the isoganal points is inside the triangle). If two of the angles are greater than $60^\circ$ then four of these intersection points are on the edges of the triangle (and one isoganal point inside). – Henry Sep 21 '17 at 0:38
• And If exactly one of the angles of the triangle is $120^\circ$ or $60^\circ$ then two of these intersection points are on the edges of the triangle and another two (plus an isoganal point) are in effect at that vertex; the difference between the two cases is that the other isoganal point is outside or inside the triangle respectively. If all three of the angles are $60^\circ$ then these six intersection points collapse in pairs onto the vertices of the equilateral triangle (while one isoganal point is at the centre of the triangle, the other could be any of the points on its circumcircle) – Henry Sep 21 '17 at 0:48
• This is not quite as exciting as I had hoped, but better than anybody else has offered so have the bounty. Many thanks – Henry Sep 21 '17 at 8:06