Difference between revisions of "2015 USAJMO Problems"
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===Problem 6=== | ===Problem 6=== | ||
− | Steve is piling <math>m\geq 1</math> indistinguishable stones on the squares of an <math>n\times n</math> grid. Each square can have an arbitrarily high pile of stones. After he finished piling his stones in some manner, he can then perform stone moves, defined as follows. Consider any four grid squares, which are corners of a rectangle, i.e. in positions <math>(i, k), (i, l), (j, k), (j, l)</math> for some <math>1\leq i, j, k, l\leq n</math>, such that <math>i<j</math> and <math>k<l</math>. A stone move consists of either removing one stone from each of <math>(i, k)</math> and <math>(j, l)</math> and moving them to <math>(i, l)</math> and <math>(j, k)</math> respectively, | + | Steve is piling <math>m\geq 1</math> indistinguishable stones on the squares of an <math>n\times n</math> grid. Each square can have an arbitrarily high pile of stones. After he finished piling his stones in some manner, he can then perform stone moves, defined as follows. Consider any four grid squares, which are corners of a rectangle, i.e. in positions <math>(i, k), (i, l), (j, k), (j, l)</math> for some <math>1\leq i, j, k, l\leq n</math>, such that <math>i<j</math> and <math>k<l</math>. A stone move consists of either removing one stone from each of <math>(i, k)</math> and <math>(j, l)</math> and moving them to <math>(i, l)</math> and <math>(j, k)</math> respectively, or removing one stone from each of <math>(i, l)</math> and <math>(j, k)</math> and moving them to <math>(i, k)</math> and <math>(j, l)</math> respectively. |
Two ways of piling the stones are equivalent if they can be obtained from one another by a sequence of stone moves. | Two ways of piling the stones are equivalent if they can be obtained from one another by a sequence of stone moves. | ||
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[[2015 USAJMO Problems/Problem 6|Solution]] | [[2015 USAJMO Problems/Problem 6|Solution]] | ||
+ | == See Also == | ||
+ | *[[USAJMO Problems and Solutions]] | ||
+ | |||
+ | {{USAJMO box|year=2015|before=[[2014 USAJMO Problems]]|after=[[2016 USAJMO Problems]]}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:42, 5 August 2023
Contents
Day 1
Problem 1
Given a sequence of real numbers, a move consists of choosing two terms and replacing each with their arithmetic mean. Show that there exists a sequence of 2015 distinct real numbers such that after one initial move is applied to the sequence -- no matter what move -- there is always a way to continue with a finite sequence of moves so as to obtain in the end a constant sequence.
Problem 2
Solve in integers the equation
Problem 3
Quadrilateral is inscribed in circle with and . Let be a variable point on segment . Line meets again at (other than ). Point lies on arc of such that is perpendicular to . Let denote the midpoint of chord . As varies on segment , show that moves along a circle.
Day 2
Problem 4
Find all functions such thatfor all rational numbers that form an arithmetic progression. ( is the set of all rational numbers.)
Problem 5
Let be a cyclic quadrilateral. Prove that there exists a point on segment such that and if and only if there exists a point on segment such that and .
Problem 6
Steve is piling indistinguishable stones on the squares of an grid. Each square can have an arbitrarily high pile of stones. After he finished piling his stones in some manner, he can then perform stone moves, defined as follows. Consider any four grid squares, which are corners of a rectangle, i.e. in positions for some , such that and . A stone move consists of either removing one stone from each of and and moving them to and respectively, or removing one stone from each of and and moving them to and respectively.
Two ways of piling the stones are equivalent if they can be obtained from one another by a sequence of stone moves.
How many different non-equivalent ways can Steve pile the stones on the grid?
See Also
2015 USAJMO (Problems • Resources) | ||
Preceded by 2014 USAJMO Problems |
Followed by 2016 USAJMO Problems | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.