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| − | == Problem ==
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| − | A balance has unequal arms and pans of unequal weight. It is used to weigh three objects. The first object balances against a weight <math>A</math>, when placed in the left pan and against a weight <math>a</math>, when placed in the right pan. The corresponding weights for the second object are <math>B</math> and <math>b</math>. The third object balances against a weight <math>C</math>, when placed in the left pan. What is its true weight?
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| | == Solution == | | == Solution == |
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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 <math>\text{[some constant amount] (due to the weight, and distribution of the weight, of the arm itself) } + \text{ [the length of the arm] } \times \text{ [the weight of what is sitting in the pan]}</math>. Thus, the information we have tells us that, for some constants <math>x, y, z, u</math>:
| + | The effect of the unequal arms and pans is that if an object of weight <math> x</math> in the left pan balances an object of weight <math>y </math> in the right pan, then <math> x = hy + k</math> for some constants <math>h</math> and <math>k</math>. Thus if the first object has true weight x, then <math> x = hA + k, a = hx + k</math>. |
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| − | <cmath>x + yA = z + ua</cmath> | |
| − | <cmath>x + yB = z + ub</cmath>
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| − | <cmath>x + yC = z + uc</cmath>
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| − | In fact, we don't exactly care what <math>x,y,z,u</math> are. By subtracting <math>x</math> from all equations and dividing by <math>y</math>, we get:
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| − | <cmath>A = \frac{z-x}{y} + a\left(\frac{u}{y}\right)</cmath> | |
| − | <cmath>B = \frac{z-x}{y} + b\left(\frac{u}{y}\right)</cmath>
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| − | <cmath>C = \frac{z-x}{y} + c\left(\frac{u}{y}\right)</cmath>
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| − | We can just give the names <math>X</math> and <math>Y</math> to the quantities <math>\frac{z-x}{y}</math> and <math>\frac{u}{y}</math>.
| + | So <math>a = h^2A + (h+1)k</math>. |
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| − | <cmath>A = X + Ya</cmath> | + | Similarly, <math>b = h^2B + (h+1)k</math>. Subtracting gives <math>h^2 = \frac{a-b}{A-B} </math> and so |
| − | <cmath>B = X + Yb</cmath>
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| − | <cmath>C = X + Yc</cmath> | |
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| − | Our task is to compute <math>c</math> in terms of <math>A</math>, <math>a</math>, <math>B</math>, <math>b</math>, and <math>C</math>. This can be done by solving for <math>X</math> and <math>Y</math> in terms of <math>A</math>,<math>a</math>,<math>B</math>,<math>b</math> and eliminating them from the implicit expression for <math>c</math> in the last equation. Perhaps there is a shortcut, but this will work:
| + | <cmath>(h+1)k = a - h^2A = \frac{bA - aB}{A - B}</cmath>. |
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| − | <cmath>A = X + Ya\implies \boxed{X = A - Ya}</cmath>
| + | The true weight of the third object is thus: |
| − | <cmath>B = X + Yb\implies B = A - Ya + Yb\implies Y(b-a) = B-A\implies Y = \frac{B-A}{b-a}\implies X = \boxed{A - a\left(\frac{B-A}{b-a}\right)}</cmath>
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| − | <cmath>C = X + Yc\implies Yc = C - X\implies c = \frac{C-X}{Y}\implies c = \frac{C - [A - \frac{B-A}{b-a} \cdot a]}{\frac{B-A}{b-a}}
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| − | \implies c = \frac{C - A + a\left(\frac{B-A}{b-a}\right)}{\frac{B-A}{b-a}}
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| − | \implies c = \frac{C(b-a) - A(b-a) + a(B-A)}{B-A}
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| − | \implies c = \frac{Cb - Ca - Ab + Aa + Ba - Aa}{B-A}
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| − | \implies c = \frac{Cb - Ca - Ab + Ba}{B-A}</cmath>
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| − | So the answer is: <cmath>\boxed{\frac{Cb - Ca - Ab + Ba}{B-A}}</cmath>.
| + | <cmath> |
| | + | hC + k = \\ |
| | + | \boxed{\sqrt{ \frac{a-b}{A-B}} C + \frac{bA - aB}{(A - B)(\sqrt{ \frac {a-b}{A-B}}+ 1)}} |
| | + | </cmath>. |
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| − | == See Also == | + | More readably: |
| − | {{USAMO box|year=1980|before=First Question|num-a=2}} | + | <cmath> |
| − | {{MAA Notice}} | + | \boxed{ h=\sqrt{\frac{a-b}{A-B}} ; |
| | + | \\ |
| | + | \text{weight} = hC + \frac{bA - aB}{(A - B)(h + 1)}} |
| | + | </cmath> |
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| − | [[Category:Olympiad Algebra Problems]]
| + | Credit: John Scholes https://prase.cz/kalva/usa/usoln/usol801.html |
in the left pan balances an object of weight 
in the right pan, then 
for some constants 
and 
. Thus if the first object has true weight x, then 
.
.
. Subtracting gives 
and so
![\[(h+1)k = a - h^2A = \frac{bA - aB}{A - B}\]](http://latex.artofproblemsolving.com/c/9/6/c96aa637cac47ed01a61df6e6bf95dff89639774.png)
.
![\[hC + k = \\ \boxed{\sqrt{ \frac{a-b}{A-B}} C + \frac{bA - aB}{(A - B)(\sqrt{ \frac {a-b}{A-B}}+ 1)}}\]](http://latex.artofproblemsolving.com/b/6/4/b64bcc1c61b364ac6519ee29e2a1d71a9eb998e9.png)
.
![\[\boxed{ h=\sqrt{\frac{a-b}{A-B}} ; \\ \text{weight} = hC + \frac{bA - aB}{(A - B)(h + 1)}}\]](http://latex.artofproblemsolving.com/2/c/5/2c5c482240148b75788baf2e38f541cd87597dd6.png)
