07-25-94 04:13pm	The Pentagon Problem.

Given: a pentagon ABCDE such that the area of the triangles
	S(ABC)=S(BCD)=S(CDE)=S(DEA)=S(EAB)=1
				A
			       / \
			    /   /\  \
			 /     /  \    \
		      /  G    /   \    H  \
                   B ------------------------ E
		      \	 |   /	   \   |    /
		       \ |  /	    \  |  /
			\| /  	     \ | /
			 \/ a	    b \|/
			  --------------
			 C	l      D

Question: Does this data determine the area of the pentagon (more
generally: what can we say about such a pentagon)?

Solution:

1. S(ABC)=S(BCD) ==> BC || AD. Similarly, BE || CD; BD || AE; AC || DE;
   AB || DE.

2. Let us try to understand how many parameters do we need to determine
   the pentagon. Let CD=l; angles BCD=a; EDC=b. Then we know the line
   BE to be parallel to CD, and, since area S(BCD)=1, the distance
   between the lines BE and CD is 2/l. Therefore we have already built
   points B, C, D, and E so that S(BCD)=S(CDE)=1.

3. Now we can erect lines through B parallel to DE and through E
   parallel to BC, the intersection being A. Therefore S(ABC)=S(BCD)=1
   and S(AED)=S(CDE)=1. Thus, given arbitrary numbers l, a, and b, we
   can construct a pentagon ABCDE with almost all the required
   properties. The only missing one is S(ABE)=1; and this equality will
   be the only condition on a, b, and l.

NB. The strategy now will be to express the areas of the triangle ABE
    and the pentagon ABCDE in terms of a, b, and l, and see if the
    condition, that the first expression is 1, implies anything about
    the second expression.

4. Let the altitude of the triangle ACD be h, and its area be S. Then:
	S=lh/2,
   and the altitude of the triangle BAE is (h-2/l) (see 1).
   Since AC || DE, angle ACD=pi-b; and since AD || BC, angle ADC=pi-a.
   Therefore
   		    l*AC*sin(pi-b)      sin a sin b  2
		S = -------------- = -  ----------- l
		           2            2 sin (a+b)
   (here the sine theorem was used; the minus sign is due to the fact
   that sin(x-pi)=-sin x; since a+b>pi, sin(a+b)<0, so the area is
   positive). Thus
		    2 S       sin a sin b
		h = --- = - l -----------
		     l        2 sin (a+b)
   One can see that
		S(ABCDE) = S(ABC) + S(ACD) + S(ADE) = 2 + S.

5. Now we need to compute the area of the triangle ABE. It suffices to
   compute BE for that. Erecting lines CG and DH through C and D normal
   to CD, one can see from the right triangles BCG and EHD that
		BG = CG tan (a-pi/2),
		EH = DH tan (b-pi/2),
   thus
			 2
		BE = l + - ( tan (a-pi/2) + tan (b-pi/2) )
			 l
			 2                         2  sin (a+b)
		   = l - - ( cot a + cot b ) = l - - -----------
			 l                         l sin a sin b
			  l
		   = l + ---.
			  S

6. Thus now finally we can write the condition on a, b, and l:
		     1	        2     1       l     2 S   2
	1 = S(ABE) = - BE ( h - - ) = - ( l + - ) ( --- - - )
		     2 	        l     2       S      l    l
		  1
	  = ( 1 + - ) ( S - 1 ).
		  S
   This is a quadratic equation on S. Solving it we get
	    1 +- sqrt(5)
	S = ------------,
		2
   and, since the area has to be positive,
	    1 + sqrt(5)
	S = -----------.
		2

7. We know (4) that S determines S(ABCDE), i.e.,
	      --------------------------
              |            5 + sqrt(5) |
              | S(ABCDE) = -----------.|
              |                2       |
	      --------------------------

8. A final remark: we have established that as soon as we have three
   numbers a, b, and l, such that

		   2 sin a sin b   1 + sqrt(5)
		- l  ----------- = -----------,
		     2 sin (a+b)  	2

   we can construct a pentagon that would satisfy the condition that
   the areas of the five triangles is 1.

					Semion Shteingold
					UCLA Department of Mathematics,
					shteingd@math.ucla.edu