From:mathwft@math.canterbury.ac.nz (Bill Taylor)Date:2 Apr 1996 02:41:23 GMTNewsgroups:sci.mathSubject:A fun GEOMETRY problem.

Another chocolate fish is riding on this one! ------------------------------------------------------- Given any 4 points on the circumference of a circle; mark the 4 mid-points between adjacent pairs. Join the opposite pairs of midpoints by two chords. Prove these two chords cross at right angles. ------------------------------------------------------- My own proof is ingenious but mucky. Be nice to see an "elegant" proof. Cheers, ------------------------------------------------------------------------------- Bill Taylor wft@math.canterbury.ac.nz ------------------------------------------------------------------------------- WANTED: Guillotine operator and head storeman. -------------------------------------------------------------------------------

From:ftww@cs.su.oz.au (Geoff Bailey)Newsgroups:sci.mathSubject:Re: A fun GEOMETRY problem.Date:2 Apr 1996 16:01:31 +1000Organization:Thermonuclear Toys, Pty LtdReply-To:ftww@cs.su.oz.au (Fred the Wonder Worm)

In article <4jq44j$8q4@cantua.canterbury.ac.nz>, Bill Taylor <mathwft@math.canterbury.ac.nz> wrote: >Another chocolate fish is riding on this one! > >------------------------------------------------------- >Given any 4 points on the circumference of a circle; >mark the 4 mid-points between adjacent pairs. > >Join the opposite pairs of midpoints by two chords. > >Prove these two chords cross at right angles. >------------------------------------------------------- > >My own proof is ingenious but mucky. Be nice to see an "elegant" proof. I assume you mean the midpoints of the arcs rather than the chords, as it is demonstrably false otherwise. I don't know is this is an elegant proof, but it is what came to mind immediately. Rather than writing down a formal proof which looks all yucky, I will describe it, as the idea is really very simple. Draw the circle, the 4 points, the 4 midpoints and the two chords. Mark clearly which pairs of arcs are equal to each other (there are 4 different lengths). Now draw the chord between two non-adjacent midpoints, forming a triangle. We want to show that the angle at the intersection is a right angle. This is true iff the other two angles of the triangle add up to 90 degrees. But these angles are subtended by arcs the sum of whose lengths is half the circumference, hence their sum is 90 degrees. QED. Cheers, Geoff. ------------------------------------------------------------------------------- Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@cs.su.oz.au | Gameplayer by vocation. -------------------------------------------------------------------------------