Difference between revisions of "1980 USAMO Problems/Problem 1"

Revision as of 18:09, 3 July 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 $A$ , when placed in the left pan and against a weight $a$ , when placed in the right pan. The corresponding weights for the second object are $B$ and $b$ . The third object balances against a weight $C$ , when placed in the left pan. What is its true weight?

Solution

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 $\text{[some constant amount] (due to the weight, and distribution of the weight, of the arm itself) } + \text{ [the length of the arm] } \times \text{ [the weight of what is sitting in the pan]}$ . 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:

$A = \frac{z-x}{y} + a\left(\frac{u}{y}\right)$ $B = \frac{z-x}{y} + b\left(\frac{u}{y}\right)$ $C = \frac{z-x}{y} + c\left(\frac{u}{y}\right)$

We can just give the names $X$ and $Y$ to the quantities $\frac{z-x}{y}$ and $\frac{u}{y}$ .

$A = X + Ya$ $B = X + Yb$ $C = X + Yc$

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:

$A = X + Ya\implies \boxed{X = A - Ya}$ $B = X + Yb\implies B = A - Ya + Yb\implies Y(b-a) = B-A\implies Y = \frac{B-A}{b-a}\implies X = \boxed{A - a\left(\frac{B-A}{b-a}\right)}$ $C = X + Yc\implies Yc = C - X\implies c = \frac{C-X}{Y}\implies c = \frac{C - [A - \frac{B-A}{b-a} \cdot a]}{\frac{B-A}{b-a}} \implies c = \frac{C - A + a\left(\frac{B-A}{b-a}\right)}{\frac{B-A}{b-a}} \implies c = \frac{C(b-a) - A(b-a) + a(B-A)}{B-A} \implies c = \frac{Cb - Ca - Ab + Aa + Ba - Aa}{B-A} \implies c = \frac{Cb - Ca - Ab + Ba}{B-A}$

So the answer is: $\boxed{\frac{Cb - Ca - Ab + Ba}{B-A}}$ .

@@ Line 36: / Line 36: @@
 == See Also ==
 {{USAMO box|year=1980|before=First Question|num-a=2}}
+{{MAA Notice}}
 [[Category:Olympiad Algebra Problems]]

1980 USAMO (Problems • Resources)
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Difference between revisions of "1980 USAMO Problems/Problem 1"

Revision as of 18:09, 3 July 2013

Problem

Solution

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