Difference between revisions of "1980 USAMO Problems/Problem 1"
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<cmath>A = X + Ya\implies \boxed{X = A - Ya}</cmath> | <cmath>A = X + Ya\implies \boxed{X = A - Ya}</cmath> | ||
+ | <cmath> | ||
\begin{align*} | \begin{align*} | ||
B &= X + Yb\\ | B &= X + Yb\\ | ||
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\implies c &= \frac{Cb - Ca - Ab + Ba}{B-A} | \implies c &= \frac{Cb - Ca - Ab + Ba}{B-A} | ||
\end{align*} | \end{align*} | ||
− | + | </cmath> | |
So the answer is: <cmath>\boxed{\frac{Cb - Ca - Ab + Ba}{B-A}}</cmath>. | So the answer is: <cmath>\boxed{\frac{Cb - Ca - Ab + Ba}{B-A}}</cmath>. | ||
Revision as of 11:47, 26 March 2023
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
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 . Thus, the information we have tells us that, for some constants
:
In fact, we don't exactly care what are. By subtracting
from all equations and dividing by
, we get:
We can just give the names and
to the quantities
and
.
Our task is to compute in terms of
,
,
,
, and
. This can be done by solving for
and
in terms of
,
,
,
and eliminating them from the implicit expression for
in the last equation. Perhaps there is a shortcut, but this will work:
So the answer is:
.
See Also
1980 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.