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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

294.0. "A hard balance problem" by TOOLS::STAN () Tue May 28 1985 19:19

Postpischil asked me if I had a copy of the balance scale problem from the
1980 USA Mathematical Olympiad.  Here it is:

A two-pan balance is inaccurate since its balance arms are of different
lengths and its pans are of different weights.  Three objects of different
weights A, B, and C are each weighed seperately.  When placed on the
left-hand pan, they are balanced by weights A1, B1, C1, respectively.
When A and B are placed on the right-hand pan, they are balanced by
A2 and B2 respectively.  Determine the true weight of C in terms of
A1, B1, C1, A2, and B2.

			Reference
			---------
The Ninth USA Mathematical Olympiad, Crux Mathematicorum. 6(1980)177.

As long as we are discussing balance scale problems, here is a problem
that I had published in the Journal of Recreational Mathematics:

How should one select n integral weights that may be used to weigh
the maximal number of consecutive integral weights (beginning with 1)?
The weighing process involves a pan balance and the unknown integral
weight may be placed on either pan. The selected weights may be placed
on either pan, also.  Furthermore, one may reason that if an unknown
weight weighs less than k+1 but greater than k-1, then it must weigh
exactly k.

			Reference
			---------
Stanley Rabinowitz, Problem 1322 (Integral Weights). Journal of
	Recreational Mathematics. 17(1984-1985)226-227.
T.RTitleUserPersonal
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294.1TURTLE::GILBERTTue May 28 1985 22:3939
Well, here goes.  Let l , l  be the lengths of the arms, and let w , w  be
                       1   2                                      1   2
the weights of the pans on those arms, respectively.  Then,

	w l  + X l  = w l  + X l ,  X = A, B, C
	 1 1      1    2 2    1 2

	w l  + X l  = w l  + X l ,  X = A, B
	 1 1    2 1    2 2      2

Let k = (w l  - w l )/l , and l = l /l  + 1.  Then,
          1 1    2 2   1           2  1

	k + X  = X (l-1),  X = A, B, C
	          1

	k + X  = X (l-1),  X = A, B
	     2

So,                              2
	k(l-1) + X(l-1) = X (l-1)  = k(l-1) + k + X  = k + X , X = A, B
                           1                       2        2

	C = C (l-1) - k
             1

But                2               2
	k = A (l-1)  - A  = B (l-1)  - B
             1          2    1          2
and
	            2
	(A -B )(l-1)  = (A -B )
          1  1            2  2
So,
	           A - B     A B  - A B
	            2   2     1 2    2 1
	C = C sqrt(------) + -----------
             1     A - B        A - B
                    1   1        1   1
294.2ALIEN::POSTPISCHILWed May 29 1985 12:5514
Re .1:

Sorry, but I think you made a mistake.  Consider the line

                                 2
	k(l-1) + X(l-1) = X (l-1)  = k(l-1) + k + X  = k + X , X = A, B.
                           1                       2        2


The penultimate expression is k(l-1) + k + X2.  This simplifies to
k l + X2, but you have omitted the l in the last expression.


				-- edp
294.3ALIEN::POSTPISCHILWed May 29 1985 15:34100
The balance problem seemed a lot harder in high school.  Maybe I did learn
something in college.

Suppose a weight x on the left balances a weight y on the right.  Then the
center of gravity of the arms, pans, and weights is at the "center" of the
balance.  Thus,

	(x+wl)*ll-(y+wr)*lr = 0.

where ll is the length of the left arm and wl is the weight of the left pan
plus an adjustment for the weight of the left arm, and lr and wr are the
same quantities for the right side.

The five given balancing situations are represented by

	(A+wl)*ll-(A1+wr)*lr = 0,	(0)
	(B+wl)*ll-(B1+wr)*lr = 0,	(1)
	(C+wl)*ll-(C1+wr)*lr = 0,	(2)
	(A2+wl)*ll-(A+wr)*lr = 0, and	(3)
	(B2+wl)*ll-(B+wr)*lr = 0.	(4)

Solving (0) and (3) for A gives

	(A1+wr)*lr/ll-wl = A = (A2+wl)*ll/lr-wr.	(5)

Similarly, (1) and (4) give

	(B1+wr)*lr/ll-wl = B = (B2+wl)*ll/lr-wr.	(6)

(2) gives

	(C1+wr)*lr/ll-wl = C.				(7)

Let l = lr/ll.  Then

	(A1+wr)*l-wl = (A2+wl)/l-wr,		(from (5))	(8)

	(B1+wr)*l-wl = (B2+wl)/l-wr, and	(from (6))	(9)

	(C1+wr)*l-wl = C.			(from (7))	(10)

Expanding these gives

	A1*l+wr*l-wl = A2/l+wl/l-wr,		(11)

	B1*l+wr*l-wl = B2/l+wl/l-wr, and	(12)

	C1*l+wr*l-wl = C.			(13)

Multiplying (11) and (12) by l gives

	A1*l^2+l*(wr*l-wl) = A2+wl-wr*l and

	B1*l^2+l*(wr*l-wl) = B2+wl-wr*l.

Let w = wr*l-wl.  Then

	A1*l^2+l*w = A2 - w,

	B1*l^2+l*w = B2 - w, and

	C1*l+w = C.			(14)

Solving for w gives

	(A2-A1*l^2)/(l+1) = w = (B2-B1*l^2)/(l+1).	(15)

Thus

	A2-A1*l^2 = B2-B1*l^2.

	l^2 = (A2-B2)/(A1-B1).

	l = sqrt((A2-B2)/(A1-B1)).

Substituting this into (15) gives

	w = (A2-A1*(A2-B2)/(A1-B1))/(1+sqrt((A2-B2)/(A1-B1))).

Putting l and w in (14) gives

	C = C1*sqrt((A2-B2)/(A1-B1)) +
	    (A2-A1*(A2-B2)/(A1-B1))/(1+sqrt((A2-B2)/(A1-B1))).

When this expression is "simplified", the result is

	      A1*B2+A2*B1
	C = - ----------- +
	      A1-A2-B1+B2

       A1*B2+A1*C1+A2*B1-A2*C1-B1*C1+B2*C1
    ----------------------------------------- * sqrt(A1*A2-A1*B2-A2*B1+B1*B2).
    A1^2-A1*A2-2*A1*B1+A1*B2+A2*B1-B1^2-B1*B2

I am willing to bet that there is a mistake somewhere in there.  Can you
imagine the poor judges who had to grade a hundred of these, all handwritten
in a hurry?


				-- edp
294.4HARE::GILBERTWed May 29 1985 23:397
Re: 294.3
	If A = A1 = A2 and B = B1 = B2 (this happens if the balance is fair),
	your equation for C simplifies to:

	      2 A B   2 A B
	C = - ----- + ----- * abs(A-B)
	        0       0
294.5BEING::POSTPISCHILThu May 30 1985 13:576
Fortunately, the penultimate equation simplifies to C = C1 under the conditions
A1 = A2 and B1 = B2, which is correct.  Anybody want to tackle putting that
equation in simplest form?


				-- edp