2003 USAMO Problems/Problem 6
Problem
At the vertices of a regular hexagon are written six nonnegative integers whose sum is 2003. Bert is allowed to make moves of the following form: he may pick a vertex and replace the number written there by the absolute value of the difference between the numbers written at the two neighboring vertices. Prove that Bert can make a sequence of moves, after which the number 0 appears at all six vertices.
Solution
Assume the original numbers are . Since is odd, either or must be odd. WLOG let be odd and .
Case 1
. Define Operation A as the sequence of moves from Step 1 to Step 3, shown below:
Notice that Operation A changes the numbers to and they are all nonnegative, since . Their sum changes from to ; it decreases as long as . If we repeat Operation A enough times, its sum will decrease and eventually we will arrive at a point where at least one of the numbers in the positions originally occupied by has become a 0.
Case 2
and . Define Operation B as the sequence of moves from Step 1 to Step 3, shown below:
where in Step 3, we take the nonnegative choice of or . is changed to either or . If we have , their sum is and this is less than (the original sum) unless , but since the original sum is odd by assumption. If we have , their sum is , which is less than . Operation B applied repeatedly will cause either or to become .
Case 3
and . Define Operation C as the sequence of moves from Step 1 to Step 4, shown below:
See also
2003 USAMO (Problems • Resources) | ||
Preceded by Problem 5 |
Followed by Last question | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.