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Difference between revisions of "2015 USAMO Problems"

(Created page with "==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 seque...")
 
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==Day 1==
 
==Day 1==
 
===Problem 1===
 
===Problem 1===
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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.
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Solve in integers the equation
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<cmath> x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3. </cmath>
  
 
[[2015 USAMO Problems/Problem 1|Solution]]
 
[[2015 USAMO Problems/Problem 1|Solution]]
  
 
===Problem 2===
 
===Problem 2===
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Solve in integers the equation
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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.
−
<cmath> x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3. </cmath>
 
  
 
[[2015 USAMO Problems/Problem 2|Solution]]
 
[[2015 USAMO 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.
+
Let <math>S = \{1, 2, ..., n\}</math>, where <math>n \ge 1</math>. Each of the <math>2^n</math> subsets of <math>S</math> is to be colored red or blue. (The subset itself is assigned a color and not its individual elements.) For any set <math>T \subseteq S</math>, we then write <math>f(T)</math> for the number of subsets of T that are blue.
 +
 
 +
Determine the number of colorings that satisfy the following condition: for any subsets <math>T_1</math> and <math>T_2</math> of <math>S</math>,
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<cmath>f(T_1)f(T_2) = f(T_1 \cup T_2)f(T_1 \cap T_2).</cmath>
  
 
[[2015 USAMO Problems/Problem 3|Solution]]
 
[[2015 USAMO Problems/Problem 3|Solution]]
  
 
==Day 2==
 
==Day 2==
 +
 
===Problem 4===
 
===Problem 4===
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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.)
+
 
 +
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.
 +
 
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How many different non-equivalent ways can Steve pile the stones on the grid?
  
 
[[2015 USAMO Problems/Problem 4|Solution]]
 
[[2015 USAMO Problems/Problem 4|Solution]]
  
 
===Problem 5===
 
===Problem 5===
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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>.
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Let <math>a, b, c, d, e</math> be distinct positive integers such that <math>a^4 + b^4 = c^4 + d^4 = e^5</math>. Show that <math>ac + bd</math> is a composite number.
  
 
[[2015 USAMO Problems/Problem 5|Solution]]
 
[[2015 USAMO 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,j 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.
+
Consider <math>0<\lambda<1</math>, and let <math>A</math> be a multiset of positive integers. Let <math>A_n=\{a\in A: a\leq n\}</math>. Assume that for every <math>n\in\mathbb{N}</math>, the set <math>A_n</math> contains at most <math>n\lambda</math> numbers. Show that there are infinitely many <math>n\in\mathbb{N}</math> for which the sum of the elements in <math>A_n</math> is at most <math>\frac{n(n+1)}{2}\lambda</math>. (A multiset is a set-like collection of elements in which order is ignored, but repetition of elements is allowed and multiplicity of elements is significant. For example, multisets <math>\{1, 2, 3\}</math> and <math>\{2, 1, 3\}</math> are equivalent, but <math>\{1, 1, 2, 3\}</math> and <math>\{1, 2, 3\}</math> differ.)
  
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Two ways of piling the stones are equivalent if they can be obtained from one another by a sequence of stone moves.
+
[[2015 USAMO Problems/Problem 6|Solution]]
 +
 
 +
{{USAMO newbox|year=2015|before=[[2014 USAMO Problems]]|after=[[2016 USAMO Problems]]}}
  
−
How many different non-equivalent ways can Steve pile the stones on the grid?
 
  
−
[[2015 USAMO Problems/Problem 6|Solution]]
+
{{MAA Notice}}

Latest revision as of 13:49, 22 November 2023

Day 1

Problem 1

Solve in integers the equation \[x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3.\]

Solution

Problem 2

Quadrilateral $APBQ$ is inscribed in circle $\omega$ with $\angle P = \angle Q = 90^{\circ}$ and $AP = AQ < BP$. Let $X$ be a variable point on segment $\overline{PQ}$. Line $AX$ meets $\omega$ again at $S$ (other than $A$). Point $T$ lies on arc $AQB$ of $\omega$ such that $\overline{XT}$ is perpendicular to $\overline{AX}$. Let $M$ denote the midpoint of chord $\overline{ST}$. As $X$ varies on segment $\overline{PQ}$, show that $M$ moves along a circle.

Solution

Problem 3

Let $S = \{1, 2, ..., n\}$, where $n \ge 1$. Each of the $2^n$ subsets of $S$ is to be colored red or blue. (The subset itself is assigned a color and not its individual elements.) For any set $T \subseteq S$, we then write $f(T)$ for the number of subsets of T that are blue.

Determine the number of colorings that satisfy the following condition: for any subsets $T_1$ and $T_2$ of $S$, \[f(T_1)f(T_2) = f(T_1 \cup T_2)f(T_1 \cap T_2).\]

Solution

Day 2

Problem 4

Steve is piling $m\geq 1$ indistinguishable stones on the squares of an $n\times n$ 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 $(i, k), (i, l), (j, k), (j, l)$ for some $1\leq i, j, k, l\leq n$, such that $i<j$ and $k<l$. A stone move consists of either removing one stone from each of $(i, k)$ and $(j, l)$ and moving them to $(i, l)$ and $(j, k)$ respectively, or removing one stone from each of $(i, l)$ and $(j, k)$ and moving them to $(i, k)$ and $(j, l)$ 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?

Solution

Problem 5

Let $a, b, c, d, e$ be distinct positive integers such that $a^4 + b^4 = c^4 + d^4 = e^5$. Show that $ac + bd$ is a composite number.

Solution

Problem 6

Consider $0<\lambda<1$, and let $A$ be a multiset of positive integers. Let $A_n=\{a\in A: a\leq n\}$. Assume that for every $n\in\mathbb{N}$, the set $A_n$ contains at most $n\lambda$ numbers. Show that there are infinitely many $n\in\mathbb{N}$ for which the sum of the elements in $A_n$ is at most $\frac{n(n+1)}{2}\lambda$. (A multiset is a set-like collection of elements in which order is ignored, but repetition of elements is allowed and multiplicity of elements is significant. For example, multisets $\{1, 2, 3\}$ and $\{2, 1, 3\}$ are equivalent, but $\{1, 1, 2, 3\}$ and $\{1, 2, 3\}$ differ.)

Solution

2015 USAMO (Problems • Resources)
Preceded by
2014 USAMO Problems 		Followed by
2016 USAMO Problems
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions


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