Equidistant points to a point on an acute triangle
- David Orden
Every point P on the boundary of an acute triangle is the center of a circle that intersects (at least) 4 times the boundary of the triangle. Hence, every such a P has (at least) 4 points equidistant to it. You can: (1) Move the points along the boundary segments, and (2) Change the radius of the circles by moving the points on them.
More info: "Advances on an Erdős problem on convex boundaries" at http://mappingignorance.org