Difference between revisions of "1980 USAMO Problems/Problem 1"
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A balance scale will balance when the torques exerted on both sides cancel out. On each of the two sides, the total torque will be [some constant amount] (due to the weight, and distribution of the weight, of the arm itself) plus [the length of the arm] times [the weight of what is sitting in the pan]. Thus, the information we have tells us that, for some constants x, y, z, u: | A balance scale will balance when the torques exerted on both sides cancel out. On each of the two sides, the total torque will be [some constant amount] (due to the weight, and distribution of the weight, of the arm itself) plus [the length of the arm] times [the weight of what is sitting in the pan]. Thus, the information we have tells us that, for some constants x, y, z, u: | ||
| − | x + yA = z + ua | + | x + yA = z + ua |
| − | x + yB = z + ub | + | x + yB = z + ub |
| − | x + yC = z + uc | + | x + yC = z + uc |
In fact, we don't exactly care what x,y,z,u are. By subtracting x from all equations and dividing by y, we get: | In fact, we don't exactly care what x,y,z,u are. By subtracting x from all equations and dividing by y, we get: | ||
| − | A = (z-x)/y + (u/y)a | + | A = (z-x)/y + (u/y)a |
| − | B = (z-x)/y + (u/y)b | + | B = (z-x)/y + (u/y)b |
| − | C = (z-x)/y + (u/y)c | + | C = (z-x)/y + (u/y)c |
We can just give the names X and Y to the quantities (z-x)/y and (u/y). | We can just give the names X and Y to the quantities (z-x)/y and (u/y). | ||
| − | A = X + Ya | + | A = X + Ya |
| − | B = X + Yb | + | B = X + Yb |
| − | C = X + Yc | + | C = X + Yc |
Our task is to compute c in terms of A, a, B, b, and C. This can be done by solving for X and Y in terms of A,a,B,b and eliminating them from the implicit expression for c in the last equation. Perhaps there is a shortcut, but this will work: | Our task is to compute c in terms of A, a, B, b, and C. This can be done by solving for X and Y in terms of A,a,B,b and eliminating them from the implicit expression for c in the last equation. Perhaps there is a shortcut, but this will work: | ||
| − | A = X + Ya | + | A = X + Ya |
| − | => X = A - Ya | + | => X = A - Ya |
| − | B = X + Yb | + | B = X + Yb |
| − | => B = A - Ya + Yb | + | => B = A - Ya + Yb |
| − | => Y(b-a) = B-A | + | => Y(b-a) = B-A |
| − | => Y = (B-A)/(b-a) | + | => Y = (B-A)/(b-a) |
| − | => X = A - (B-A)/(b-a) * a | + | => X = A - (B-A)/(b-a) * a |
| − | + | ||
| − | C = X + Yc | + | C = X + Yc |
| − | => Yc = C - X | + | => Yc = C - X |
| − | => c = (C-X)/Y | + | => c = (C-X)/Y |
| − | => c = (C - [A - (B-A)/(b-a) * a]) / [(B-A)/(b-a)] | + | => c = (C - [A - (B-A)/(b-a) * a]) / [(B-A)/(b-a)] |
| − | => [simplify numerator] | + | => [simplify numerator] |
| − | c = (C - A + a(B-A)/(b-a)) / [(B-A)/(b-a)] | + | c = (C - A + a(B-A)/(b-a)) / [(B-A)/(b-a)] |
| − | => [multiply numerator and denominator by (b-a)] | + | => [multiply numerator and denominator by (b-a)] |
| − | c = (C(b-a) - A(b-a) + a(B-A)) / (B-A) | + | c = (C(b-a) - A(b-a) + a(B-A)) / (B-A) |
| − | => [distribute numerator] | + | => [distribute numerator] |
| − | c = (Cb - Ca - Ab + Aa + Ba - Aa) / (B-A) | + | c = (Cb - Ca - Ab + Aa + Ba - Aa) / (B-A) |
| − | => [cancel Aa's] | + | => [cancel Aa's] |
| − | c = (Cb - Ca - Ab + Ba) / (B-A) | + | c = (Cb - Ca - Ab + Ba) / (B-A) |
So the answer is: (Cb - Ca - Ab + Ba) / (B-A). | So the answer is: (Cb - Ca - Ab + Ba) / (B-A). | ||
Revision as of 22:32, 11 January 2013
Problem
A balance has unequal arms and pans of unequal weight. It is used to weigh three objects. The first object balances against a weight 
, when placed in the left pan and against a weight 
, when placed in the right pan. The corresponding weights for the second object are 
and 
. The third object balances against a weight 
, when placed in the left pan. What is its true weight?
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
A balance scale will balance when the torques exerted on both sides cancel out. On each of the two sides, the total torque will be [some constant amount] (due to the weight, and distribution of the weight, of the arm itself) plus [the length of the arm] times [the weight of what is sitting in the pan]. Thus, the information we have tells us that, for some constants x, y, z, u:
x + yA = z + ua x + yB = z + ub x + yC = z + uc
In fact, we don't exactly care what x,y,z,u are. By subtracting x from all equations and dividing by y, we get:
A = (z-x)/y + (u/y)a B = (z-x)/y + (u/y)b C = (z-x)/y + (u/y)c
We can just give the names X and Y to the quantities (z-x)/y and (u/y).
A = X + Ya B = X + Yb C = X + Yc
Our task is to compute c in terms of A, a, B, b, and C. This can be done by solving for X and Y in terms of A,a,B,b and eliminating them from the implicit expression for c in the last equation. Perhaps there is a shortcut, but this will work:
A = X + Ya => X = A - Ya B = X + Yb => B = A - Ya + Yb => Y(b-a) = B-A => Y = (B-A)/(b-a) => X = A - (B-A)/(b-a) * a C = X + Yc => Yc = C - X => c = (C-X)/Y => c = (C - [A - (B-A)/(b-a) * a]) / [(B-A)/(b-a)] => [simplify numerator] c = (C - A + a(B-A)/(b-a)) / [(B-A)/(b-a)] => [multiply numerator and denominator by (b-a)] c = (C(b-a) - A(b-a) + a(B-A)) / (B-A) => [distribute numerator] c = (Cb - Ca - Ab + Aa + Ba - Aa) / (B-A) => [cancel Aa's] c = (Cb - Ca - Ab + Ba) / (B-A)
So the answer is: (Cb - Ca - Ab + Ba) / (B-A).
[Someone else feel free to clean up the formatting here.]
See Also
| 1980 USAMO (Problems • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
