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.
Final Note: So, why discuss this problem? First, the original task above is do-able 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
- 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 high-quality 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 72-page proof used methods of advanced complex analysis and algebraic geometry.
Hint: Restating the definition, a point p is equichordal if, for every chord through p, the sum |x-p|+|y-p| = constant, where x,y denote the endpoints of the chord on the curve itself. Also, |x-p| 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 |x-p| and |y-p| 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.