Warning: This Problem Sounds Easy...But...
Observation: Consider the circle with center O. All chords through point O have the same length, because they are all diameters.
Definition: A point inside a closed, planar curve is called an equichordal point if every chord passing through that point has the same length.
Task: Draw another curve (other than a circle) such that it has at least one equichordal point?
Caution: It turns out that such curves exist....BUT, beware of extensions.
Some History:
 In 1916 Fujiwara, a Japanese mathematician,asked: Is it possible for a curve to have two equichordal points?
 At this same time, Fujiwara proved that a curve with three equichordal points was impossible.
 This problem remained unsolved, despite many highquality mathematicians exploring it
 Reflecting this difficult is a quote by the mathematician C.A. Rogers: "If you are interested in studying the problem, my first advice is: 'Don't'."
 In the 1960's, geometry Victor Klee even listed this problem along with Fermat's Last Theorem and Riemann Hypothesis as one of "the" three research problems
 The case of two equichordal points was finally proven impossible in 1996 by Marek Rychlik, mathematics professor at the University of Arizona....his 72page proof used methods of advanced complex analysis and algebraic geometry.
Final Note: So, why discuss this problem? First, the original task above is doable by secondary math students...though it will be an interesting challenge. Second, math history is full of fascinating examples, where cases n=1 and n=3 are solved, but the case n=2 proves difficult. Third, the problem illustrates the dangers of generalization; however, be aware that the problem has been extended to the idea of
R. Gardner's chordal point,
equiproduct point, and
Equireciprocal Point.
Hint: Restating the definition, a point p is equichordal if, for every chord through p, the sum xp+yp = constant, where x,y denote the endpoints of the chord on the curve itself. Also, xp represents the distance from point x to point p, etc.
Also, you need to remove a "blinders" created by the circle. That is, the distances xp and yp do not have to be equal (as they are for the radii of a circle).
Finally, think/experiment via polar equations.
Solution Commentary: One example is the limacon....but does it work for the cusp or looped versions?
A general solution is provided via Wikipedia, but you will need to do some addional playing to understand it.
