Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83; site ucla-cs.ARPA Path: utzoo!watmath!clyde!bonnie!akgua!sdcsvax!sdcrdcf!trwrba!cepu!ucla-cs!dgc From: dgc@ucla-cs.UUCP Newsgroups: net.puzzle Subject: Re: High School Math problems:Answers and a toughie! Message-ID: <1783@ucla-cs.ARPA> Date: Mon, 29-Oct-84 21:41:50 EDT Article-I.D.: ucla-cs.1783 Posted: Mon Oct 29 21:41:50 1984 Date-Received: Thu, 25-Oct-84 03:48:27 EDT References: <1975@stolaf.UUCP> <1590@ucla-cs.ARPA> <1214@utah-gr.UUCP> Organization: UCLA CS Dept. Lines: 42 ------------------------------------------------------------------------ In article <1590@ucla-cs.ARPA> dgc@ucla-cs.UUCP writes: > > >Find the greatest number of intersections in an n-gon if all > >vertices are connected. Ex. If you draw lines connecting all four > >vertices of a quadrilateral, you get one intersection. > >The answer to the problem is implicitly stated in the problem. >Specifically, every four distinct vertices give rise to one intersection >and conversely, each intersection determines four distinct vertices. So >the the answer is the number of ways one can choose four vertices from >the given n. That is, it is "n choose 4" which is > > n(n-1)(n-2)(n-3) > ---------------- > 24 > I'm afraid it's more complicated than that. For example, this formula gives 15 for a hexagon, but a simple picture shows only 13. This is because of the three diagonals that intersect in a single point at the center. I don't have the answer, but it looks like it's back to the drawing board. ------------------------------------------------------------------------ The point is the problem asked for the "greatest" number of intersections. This means that you shouldn't have triple (or more intersections) because by slightly perturbing the vertices you will get more intersections. Alternatively, a k-fold intersection should count as k(k-1)/2 simple intersections (which is what you get if you slightly perturb the vertices to a "general" position). In the example, the triple intersection counts as 3 simple intersections, which makes the general formula correct. If you don't count multiple intersections this way, then the answer cannot depend upon n, alone. David G. Cantor Arpa: dgc@ucla-locus.arpa UUCP: ...!{cepu, ihnp4, randvax, sdcrdcf, trwspp, ucbvax}!ucla-cs!dgc