2008 USAMO Problems/Problem 5

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(Kiran Kedlaya) Three nonnegative real numbers $r_1$, $r_2$, $r_3$ are written on a blackboard. These numbers have the property that there exist integers $a_1$, $a_2$, $a_3$, not all zero, satisfying $a_1r_1 + a_2r_2 + a_3r_3 = 0$. We are permitted to perform the following operation: find two numbers $x$, $y$ on the blackboard with $x \le y$, then erase $y$ and write $y - x$ in its place. Prove that after a finite number of such operations, we can end up with at least one $0$ on the blackboard.



Every time we perform an operation on the numbers on the blackboard $R = \left < r_1, r_2, r_3 \right >$, we perform the corresponding operation on the integers $A = \left < a_1, a_2, a_3 \right >$ so that $R \cdot A = 0$ continues to hold. (For example, if we replace $r_1$ with $r_1 - r_2$ then we replace $a_2$ with $a_1 + a_2$.)

It's possible to show we can always pick an operation so that $|A|^2$ is strictly decreasing. Without loss of generality, let $r_3 > r_2 > r_1$ and $a_3$ be positive. Then it cannot be true that both $a_1$ and $a_2$ are at least $\frac { - a_3}{2}$, or else $a_1r_1 + a_2r_2 + a_3r_3 > 0$. Without loss of generality, let $a_1 < \frac { - a_3}{2}$. Then we can replace $a_1$ with $a_1 + a_3$ and $r_3$ with $r_3 - r_1$ to make $|A|$ smaller. Since it is a strictly decreasing sequence of positive integers, after a finite number of operations we have $a_3 = 0$. We can now see that this result holds for $(r_1,r_2,0)$ if and only if it holds for $(1,\frac{r_2}{r_1},0)$. We can see that $\frac{r_2}{r_1}$ is a rational number given that $a_3$ = 0. It is a well known result of the euclidean algorithm that if we continue to perform these operations, $r_1$ or $r_2$ will eventually be 0.

See also

  • <url>viewtopic.php?t=202910 Discussion on AoPS/MathLinks</url>
2008 USAMO (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6
All USAMO Problems and Solutions

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