From: Markus Schick Subject: Re: Butterfly Problem Date: Mon, 31 Jul 2000 14:46:10 +0200 Newsgroups: sci.math Summary: [missing] Paul Kent schrieb am 31.07.2000 7:27 Uhr > > Given: > > Start with a circle. Select any chord AE in the circle that does not pass > through the center of the circle. Now bisect that chord so that it is split > into two equal lengths AC and CE. Draw any two chords that pass through the > point of bisection C. Now connect the ends of the two chords with two lines > so that each line passes through the original chord (points B and D). If you > drew this properly, it should look like a butterfly. Don't draw the > butterfly so that it looks symmetrical. The proof becomes too easy. Draw it > so that the areas of the two "wings" are unequal. > > Show: > > BC = CD Let P be the Intersection of the tangents to the circle in A and E. Consider your circle as a projection of an ellipse with P being the vanishing point. Then C is the center of that ellipse and everything is symmetric to C. So are B and D. Their distance to C transforms equally because BD is orthogonal to CP. Regards, Markus ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: Butterfly Problem Date: 31 Jul 2000 20:11:24 GMT Newsgroups: sci.math In article <8m333p$2ed$1@schbbs.mot.com>, Paul Kent wrote: >Start with a circle. Select any chord AE in the circle that does not pass >through the center of the circle. Now bisect that chord so that it is split >into two equal lengths AC and CE. Draw any two chords that pass through the >point of bisection C. Now connect the ends of the two chords with two lines >so that each line passes through the original chord (points B and D). If you >drew this properly, it should look like a butterfly. Don't draw the >butterfly so that it looks symmetrical. The proof becomes too easy. Draw it >so that the areas of the two "wings" are unequal. > >Show: > >BC = CD I got this previously from a colleague, who got it from a bright high-school student who solved every other challenge problem during a year in geometry, except this one. (I don't know who first posed this problem.) See 96/synthetic for a solution. Musings: What makes a problem hard? One can argue that this problem is trivial, since after the imposition of coordinates one may view the problem as a set of computations. We may for example put the origin at the circle's center and the point (-1,0) where a "wing" meets the circle. If we then suppose that the lines from there to the other three points on the circle have slopes a, b, and c respectively, then it turns out you only encounter rational functions of these variables when computing the points on the circle and the various points of intersection. We finally compute the squares of the lengths BC and CD and they're equal (to (2*(a-b)/((a*b-a*c+c*b+1)*(a*b+a*c-c*b+1))^2 * (a^2*b^2+a^2*c^2+4*a^2+ 2*a^2*b*c-2*c^2*b*a-2*c*b^2*a-2*a*b-2*a*c+1+c^2*b^2+2*c*b) in case you're curious.) End of proof. Of course, the use of coordinate algebra is heretical in these problems. Is it obvious that a coordinate proof can be translated into a "classical" proof? (What are the "classical" tools permitted anyway?) Is there a measure of algebraic complexity which accurately reflects how hard the problem is for students? Suppose I wanted to generate more problems which begin, "suppose given any four points on a circle; then ... equals ...". I know I could reduce such a theorem to a calculation in a similar way; but what's a good method to generate the expressions "..."? (I'm wondering, without much hope, whether there's a nice geometric counterpart to the generators of the ring of invariants of polynomials in four variables under the action of the dihedral group.) These are not questions with formal answers, I suppose, but I'd be pleased to hear useful responses. dave >begin 666 butterfly.gif >M1TE&.#=A``* `?<`````````50``J@``_P`D```D50`DJ@`D_P!)``!)50!) Call me old-fashioned; I still don't like binaries in the newsgroup.