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From: desj@brahms (David desJardins)
Newsgroups: sci.math
Subject: Re: Inscribed pentagon...Help!
Message-ID: <73@cartan.Berkeley.EDU>
Date: Fri, 24-Oct-86 20:16:43 EDT
Article-I.D.: cartan.73
Posted: Fri Oct 24 20:16:43 1986
Date-Received: Sun, 26-Oct-86 01:57:53 EDT
References: <2595@ihlpg.UUCP> <1179@ihwpt.UUCP>
Sender: daemon@cartan.Berkeley.EDU
Reply-To: desj@brahms (David desJardins)
Distribution: net
Organization: Math Dept. UC Berkeley
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In article <1179@ihwpt.UUCP> crocker@ihwpt.UUCP (ron crocker) writes:
>> [ What is the area of the largest regular pentagon which can be enclosed
>>   within a square of side 10? ]
>
>[ Answer derived from the pentagon inscribed within a circle inscribed in
>  the square. ]

   To see that this is not the largest possible regular pentagon, choose a
rotation of this inscribed pentagon such that none of its vertices lie at
any of the four points of intersection of the circle with the square.  Then
it is clear that the pentagon can be slightly expanded about its center
without extending outside of the square.

   Any maximal pentagon within the square must necessarily have four of its
vertices lie on the sides of the square.  This is because any pentagon with
only three vertices on the square either can be expanded in the direction of
the fourth side of the square, or can be rotated to bring only two vertices
into contact with the square, and then expanded.
   To find the correct solution, we consider the area of the pentagon as a
function of the angle of rotation of its main axis with respect to the axis
of the square.  Since the square has 90-degree rotational symmetry, and the
pentagon has 72-degree rotational symmetry, the possible orientations of the
pentagon are parametrized by an angle ranging from 0 to 18 degrees.  Or, to
look at it differently, rotating the pentagon by 18 degrees within the square
gives the same orientation, with the vertices relabeled.  If you need to, you
can draw a picture to see this.
   In fact, there is an additional symmetry given by reflection, and we can
thus reduce the range of angles in question to the range from 0 to 9 degrees.
Another way to see this is to see that every possible orientation of the
pentagon within the square will have some side of the pentagon intersect some
side of the square at an angle of 9 degrees or less.  Again, draw a picture
if necessary to see this.

   We can now proceed to maximize the area using the calculus.  Let x be the
smallest angle that a side of the pentagon makes with respect to an edge of
the square; as we have seen, this will be in the interval from 0 to 9 degrees.
WLOG assume that the edge in question is on the bottom of the square, and that
the side in question actually intersects it (simple translation and/or reflec-
tion will accomplish this).
   Let S be the length of a side of the pentagon.  Then simple trigonometry
tells us that the distance from the center of the pentagon to a vertex is
S/(2 sin36), the distance from the center of the pentagon to a side is
S/(2 tan36), and the distance between two nonadjacent vertices is 2Scos36.
   The height of the rotated pentagon (from the bottom vertex to the top
vertex) is
		(S/(2 sin36) + S/(2 tan36)) (cos x) + (S/2) (sin x)

and the width (from the leftmost to the rightmost vertex) is

		(2 S cos36) (cos x).

   For the pentagon to be maximal one of these values must be equal to 10.
To maximize S, we will minimize the larger of the factors by which it is
multiplied in the above expressions.  It turns out that this minimum comes
at the value x = 9 degrees, at which the two expressions become equal (for
all smaller value of x the second expression is the larger).  So, the answer
is that the maximum side length is

		S = 10 / (2 cos36 cos9) = 6.2574

and the corresponding area is

		A = 5/(4 tan36) S^2 = 67.3649.

   If you want to draw a picture of this orientation, one of the sides of the
pentagon intersects two of the edges of the square at 45-degree angles.

   This actually wasn't all that hard -- I was pretty sure that the above was
the correct answer from the beginning -- but I wanted to go through it in
enough detail that everyone who is interested can understand it.  I hope I
succeeded.

   -- David desJardins