Difference between revisions of "2015 USAJMO Problems"
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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. | 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. | ||
− | [[2015 | + | [[2015 USAJMO Problems/Problem 1|Solution]] |
===Problem 2=== | ===Problem 2=== | ||
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<cmath> x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3. </cmath> | <cmath> x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3. </cmath> | ||
− | [[2015 | + | [[2015 USAJMO Problems/Problem 2|Solution]] |
===Problem 3=== | ===Problem 3=== | ||
Quadrilateral <math>APBQ</math> is inscribed in circle <math>\omega</math> with <math>\angle P = \angle Q = 90^{\circ}</math> and <math>AP = AQ < BP</math>. Let <math>X</math> be a variable point on segment <math>\overline{PQ}</math>. Line <math>AX</math> meets <math>\omega</math> again at <math>S</math> (other than <math>A</math>). Point <math>T</math> lies on arc <math>AQB</math> of <math>\omega</math> such that <math>\overline{XT}</math> is perpendicular to <math>\overline{AX}</math>. Let <math>M</math> denote the midpoint of chord <math>\overline{ST}</math>. As <math>X</math> varies on segment <math>\overline{PQ}</math>, show that <math>M</math> moves along a circle. | Quadrilateral <math>APBQ</math> is inscribed in circle <math>\omega</math> with <math>\angle P = \angle Q = 90^{\circ}</math> and <math>AP = AQ < BP</math>. Let <math>X</math> be a variable point on segment <math>\overline{PQ}</math>. Line <math>AX</math> meets <math>\omega</math> again at <math>S</math> (other than <math>A</math>). Point <math>T</math> lies on arc <math>AQB</math> of <math>\omega</math> such that <math>\overline{XT}</math> is perpendicular to <math>\overline{AX}</math>. Let <math>M</math> denote the midpoint of chord <math>\overline{ST}</math>. As <math>X</math> varies on segment <math>\overline{PQ}</math>, show that <math>M</math> moves along a circle. | ||
− | [[2015 | + | [[2015 USAJMO Problems/Problem 3|Solution]] |
==Day 2== | ==Day 2== | ||
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Find all functions <math>f:\mathbb{Q}\rightarrow\mathbb{Q}</math> such that<cmath>f(x)+f(t)=f(y)+f(z)</cmath>for all rational numbers <math>x<y<z<t</math> that form an arithmetic progression. (<math>\mathbb{Q}</math> is the set of all rational numbers.) | Find all functions <math>f:\mathbb{Q}\rightarrow\mathbb{Q}</math> such that<cmath>f(x)+f(t)=f(y)+f(z)</cmath>for all rational numbers <math>x<y<z<t</math> that form an arithmetic progression. (<math>\mathbb{Q}</math> is the set of all rational numbers.) | ||
− | [[2015 | + | [[2015 USAJMO Problems/Problem 4|Solution]] |
===Problem 5=== | ===Problem 5=== | ||
Let <math>ABCD</math> be a cyclic quadrilateral. Prove that there exists a point <math>X</math> on segment <math>\overline{BD}</math> such that <math>\angle BAC=\angle XAD</math> and <math>\angle BCA=\angle XCD</math> if and only if there exists a point <math>Y</math> on segment <math>\overline{AC}</math> such that <math>\angle CBD=\angle YBA</math> and <math>\angle CDB=\angle YDA</math>. | Let <math>ABCD</math> be a cyclic quadrilateral. Prove that there exists a point <math>X</math> on segment <math>\overline{BD}</math> such that <math>\angle BAC=\angle XAD</math> and <math>\angle BCA=\angle XCD</math> if and only if there exists a point <math>Y</math> on segment <math>\overline{AC}</math> such that <math>\angle CBD=\angle YBA</math> and <math>\angle CDB=\angle YDA</math>. | ||
− | [[2015 | + | [[2015 USAJMO Problems/Problem 5|Solution]] |
===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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How many different non-equivalent ways can Steve pile the stones on the grid? | How many different non-equivalent ways can Steve pile the stones on the grid? | ||
− | [[2015 | + | [[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}} |
Latest revision as of 16:42, 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 that
for 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.