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- From: johnsbra@tmlh (Havard Johnsbraten)
- Newsgroups: sci.math
- Subject: Re: Two (I think) interesting Problems
- Message-ID: <johnsbra.2@tmlh>
- Date: 7 Jan 93 10:46:56 GMT
- References: <1992Dec23.134651.7759@neptune.inf.ethz.ch>
- Sender: news@ulrik.uio.no (Mr News)
- Organization: tmlh
- Lines: 33
- Nntp-Posting-Host: 128.39.157.35
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-
- In article <1992Dec23.134651.7759@neptune.inf.ethz.ch>
- yoparish@iiic.ethz.ch (Yoav Ilan Haim Parish) poses the problem
-
- How can I fold an approximate (regular) pentagon from a
- quadratic sheet of paper easily ?
-
- In addition to the nice constructions of G.J. McCaughan and Marvin Minsky
- (the man with the finite automata?) I have the following contribution:
-
- One of my students once showed me this beautiful paper folding procedure,
- wondering whether or not it was exact:
-
- Start with a square ABCD. Fold along the diagonal BD, folding C to A.
- Find the midpoint E on AB (folding B to A and opening it again).
- Then fold D to E.
- By this last folding you get two new points: F on BD and G on AD.
- The angle EFG is about twice the angle BFE.
- Bisect angle EFG by folding such that G will be lying on FE, and fold angle
- BFE to the backside.
- Now you may use a scissor to cut out a regular pentagon (cut at right angle)
- or a nice star (cut more arbitrarily)!
-
- The folding procedure above is not exact; the angle BFE being slightly less
- than 37 degrees, instead of the exact value of 36 degrees. But when you are
- folding, you are not able to see the inaccuracy. (The McCaughan folding
- from an A4-sheet wasn't exact either, but also very close.)
- But F divides BD in an exact rational ratio: DF:FB = 5:7 !
-
- o o
- Havard Johnsbraten, Telemark College of Education, Notodden, Norway
- E-mail haavard.johnsbraaten@tmlh.no or
- G=haavard;S=johnsbraaten;O=tmlh;P=uninett;A= ;C=no
-