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From: crocker@ihwpt.UUCP (ron crocker)
Newsgroups: sci.math
Subject: Re: Inscribed pentagon...Help!
Message-ID: <1179@ihwpt.UUCP>
Date: Thu, 23-Oct-86 23:48:35 EDT
Article-I.D.: ihwpt.1179
Posted: Thu Oct 23 23:48:35 1986
Date-Received: Sat, 25-Oct-86 05:13:06 EDT
References: <2595@ihlpg.UUCP>
Distribution: net
Organization: AT&T Bell Labs, Naperville, IL
Lines: 63

> 	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.

10 sin (36 degrees).

Proof: (by construction - you may need a pencil to follow this.  It
	is difficult to describe (I think...))
	
Let a be the length of the side of the square, and L be the length
of the sides of the regular pentagon.

First, inscribe the largest circle possible within the square.  It
will touch the square at the 4 midpoints of the sides of the
square, and therefore will have radius r = 5 (a/2).

Now, draw your regular pentagon.  It will have 5 sides, of equal
length, with 108 degree angles between the sides.  Connect the
center of the circle to each of the 5 "points" of the pentagon. 
You now have 5 isosceles(sp?) triangles with a 72 degree angle
between the radii edges.  Note that these line segments bisect the
angles of the edges of the pentagon, so the other two angles are
both 54 degrees (This does add to 180 degrees).

Take any of these triangles, and draw a line segment from the
center of the circle to the midpoint of the pentagon edge.  This
creates a right triangle, where the hypotenuse is r (= 5), half the
pentagon edge (now length L/2) is opposite of the 72/2 = 36 degree
angle.  (See figure below)

       L/2
    -----------
    \       |_|
     \        |
      \       |
       \    __|_____ 36 degrees
 hypo-  \   | |
tenuse   \  v |
length r  \   |
           \  |
            \ |
             \|
              . -- center of circle

Therefore, using the simple trignometric identity, 

                length of opposite side
	sin t = -----------------------
                length of hypotenuse

we see in our triangle that

                L/2     
       sin 36 = ---
                 5

Solving this for L, we see that

	L = 2 * 5 * sin 36
	  = 10 sin 36 degrees
	
QED.