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Path: utzoo!mnetor!seismo!rutgers!princeton!allegra!sjuvax!pash
From: pash@sjuvax.UUCP
Newsgroups: sci.math
Subject: Re: Inscribed pentagon...Help!
Message-ID: <235@sjuvax.UUCP>
Date: Thu, 30-Oct-86 22:57:30 EST
Article-I.D.: sjuvax.235
Posted: Thu Oct 30 22:57:30 1986
Date-Received: Mon, 3-Nov-86 00:01:32 EST
References: <2595@ihlpg.UUCP>
Reply-To: pash@sjuvax.UUCP (P. Ash)
Distribution: net
Organization: St. Joseph's Univ., Phila. PA
Lines: 29
Summary: Can't be done.

In article <2595@ihlpg.UUCP> mrios@ihlpg.UUCP (Michael Rios) writes:
>(Is the line eater now an imaginary function?)
>
>	I recently have been thinking about a problem I'd created
>out of boredom, but I haven't come up with a satisfactory answer.
>Will someone please help me put this to rest?
>
>	Given a square of side 10 (or whatever you want...I use
>	10 for concreteness) find the length of the side of the
>	largest regular pentagon that may be inscribed within it.
>	Prove it is the largest.

You can't inscribe a regular pentagon in a square.  For, suppose the
five vertices of a regular pentagon were each on the square.  Then
two would have to be on one side of the square, say on AB, where the
square's vertices are ABCD. Call the pentagon EFGHI, where EF lies on
AB.  Then it's clear that |IG| = |AB| = |BC| = |HJ|, where J is on EF
and HJ is perpendicular to EF.  (i.e., HJ is an altitude of the pentagon.)
But since the length of all diagonals of a regular pentagon are equal,
we also have |HF| = |IG|, so that |HF| = |HJ|.  But this is impossible,
because HF is the hypotenuse of right triangle HJF and HJ is a leg.

If you remove the restriction "regular", the problem still has no answer
as you can inscribe a pentagon of area 100 - e, where e is any positive
integer, by lopping off a triangular corner of the square of area e, so
there is no maximal pentagon.

--Peter Ash
  St. Joseph's University