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- Path: sparky!uunet!usc!zaphod.mps.ohio-state.edu!moe.ksu.ksu.edu!math.ksu.edu!deadend
- From: frandag@math.ksu.edu (Francis Fung)
- Newsgroups: sci.math
- Subject: Re: Irrational stamp problem
- Date: 21 Jan 1993 00:26:11 -0600
- Organization: Dept. of Mathematics, Kansas State University
- Lines: 41
- Message-ID: <1jlfm3INNo7p@hilbert.math.ksu.edu>
- References: <C16DI1.MD@cantua.canterbury.ac.nz>
- NNTP-Posting-Host: hilbert.math.ksu.edu
-
- wft@math.canterbury.ac.nz (Bill Taylor) writes:
-
- >In article <2338@sjfc.UUCP>, dmc@sjfc.UUCP (Cass Dan) writes:
- >|>
- >|> You have an ink stamp which is so amazingly precise that, when inked
- >|> and pressed down on the plane, it makes every circle of irrational
- >|> radius (centered at the center of the stamp) black.
- >|>
- >|> Question: Can one use the stamp three times and make every point
- >|> in the plane black? [assume plane was white to begin with, and
- >|> ignore the fact that no such stamp is physically possible]
-
- >A very nice problem !
-
- >Why not do it with only two stampings, where the second one is moved an
- >infinitesimal distance from the first one, thus covering all the points
- >left out by the first ? JUST KIDDING !!
-
- >Here's the real answer, that it can be done with three stampings.
-
- (admittedly messy solutions deleted)
- This was a Putnam problem, the second time I took the exam (ARGH), and I
- missed it in spite of the fact that it was in Newman's "A Problem Seminar"
- which I had studied a few days prior (ARRARRAGGGHHkklghk...*)
- Anyway, here is the slick solution. We exhibit three points such that
- if the distance to two of them is rational, then it must be of irrational
- distance to the third.
- Suppse (x,y) is at rational distance from (0,0) and (1,0).
- Then sqrt(x^2 + y^2)=p and sqrt((x-1)^2 + y^2)=q are both rational.
- The first implies X^2 + y^2 is rational.
- The second one implies that x^2 - 2x + 1 + y^2 = q^2 so that
- 2x-1= x^2+y^2 - q^2 is rational so that x is rational.
- Therefore,
- for any other point (a, 0), if sqrt((x-a)^2 + y^2) = r is rational, then
- x^2- 2ax + a^2 +y^2= r^2 so that a^2 - (2x)a + (x^2 + y^2 - a^2) = 0 is
- a quadratic equation that a satisfies.
- Therefore, any choice of a such that this quadratic CANNOT be satisfied,
- e.g. cube root (3) , will be such that any point at rational distance
- from (0,0) and (1, 0) is at irrational distance from (a,0).
- Francis Fung
-
-