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- Newsgroups: rec.puzzles
- Path: sparky!uunet!think.com!spool.mu.edu!umn.edu!umeecs!quip.eecs.umich.edu!kanad
- From: kanad@quip.eecs.umich.edu (Kanad Chakraborty)
- Subject: Re: A nice New Year puzzle.
- Message-ID: <1993Jan3.192018.3980@zip.eecs.umich.edu>
- Sender: news@zip.eecs.umich.edu (Mr. News)
- Organization: University of Michigan EECS Dept., Ann Arbor
- References: <1993Jan1.002031.5763@zip.eecs.umich.edu> <1i081uINN6k5@cascade.cs.ubc.ca> <1993Jan1.053639.27111@Csli.Stanford.EDU>
- Date: Sun, 3 Jan 1993 19:20:18 GMT
- Lines: 35
-
- In article <1993Jan1.053639.27111@Csli.Stanford.EDU> hiraga@Csli.Stanford.EDU (Yuzuru Hiraga) writes:
- >In article <1i081uINN6k5@cascade.cs.ubc.ca> kvdoel@cs.ubc.ca (Kees van den Doel) writes:
- >>In article <1993Jan1.002031.5763@zip.eecs.umich.edu>
- >>kanad@quip.eecs.umich.edu (Kanad Chakraborty) writes:
- >>
- >>>Given a plane and exactly 3 colors, prove whether or not it is possible
- >>>to assign a color to each point of the plane in such a way that no two points
- >>>exactly 1 inch apart have the same color.
- >>
- >>2 points sqrt(3) inch apart have the same color (take 2 points of
- >>colors, say, 2 and 3, 1 inch apart. Form 2 equilateral triangles with
- >>those 2 points, you get 2 new points (sqrt(3) inches apart) which must
- >>have color 1, i.e. the same). Pick a point, say it has color 1. The
- >>circle with radius 1 around it has color 2 or 3. The circle with radius
- >>sqrt(3) around it has color 1. On this big circle, pick an arbitrary
- >>point and draw a circle of radius sqrt(3) around *it*, which must also
- >>have color 1. The last and the first circles intersect at a point which
- >>therefore must have color 1 *and* color 2 or 3, which is impossible.
- >
- >Why not pick any point on the sqrt(3) circle and draw a radius 1 circle
- >around it?
- >The center and intersecting points must be of the same color.
- >
- >-YH
-
- Excellent ! Or easier still, a solution based on the same idea as
- above : draw an isosceles triangle ABC with AB =
- AC = sqrt(3) inches and BC = 1 in. (This triangle is possible because
- the sum of any two sides is > the third). Then A and B must have the
- same color, and so must A and C. Ergo, B and C must have the same color.
- Ergo, the distance BC != 1 in. A contradiction.
-
- Hope you guys enjoyed this puzzle.
-
- Kanad
-