year stringdate 1961-01-01 00:00:00 2025-01-01 00:00:00 ⌀ | tier stringclasses 5
values | problem_label stringclasses 119
values | problem_type stringclasses 13
values | exam stringclasses 28
values | problem stringlengths 87 2.77k | solution stringlengths 834 13k | metadata dict | problem_tokens int64 50 903 | solution_tokens int64 500 3.93k |
|---|---|---|---|---|---|---|---|---|---|
1975 | T1 | 1 | null | USAMO | Show that for any non-negative reals $\mathrm{x}, \mathrm{y},[5 \mathrm{x}]+[5 \mathrm{y}] \geq[3 \mathrm{x}+\mathrm{y}]+[\mathrm{x}+3 \mathrm{y}]$. Hence or otherwise show that (5a)! $(5 b)!/(a!b!(3 a+b)!(a+3 b)!)$ is integral for any positive integers $a, b$. | If is obviously sufficient to prove that $[5 x]+[5 y] \geq[3 x+y]+[x+3 y]$ for $0<x<y<1$. If $2 x \geq y$, then $[5 x] \geq$ $[3 x+y]$ and $[5 y] \geq[x+3 y]$, so the result holds. So assume $2 x<y$. It is now a question of examining a lot of cases.
If $y<2 / 5$, then $3 x+y<5 y / 2<1$, so $[3 x+y]=0$, and $[3 y+x]<=[... | {
"problem_match": "# Problem 1",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 98 | 581 |
1978 | T1 | 2 | null | USAMO | Two square maps cover exactly the same area of terrain on different scales. The smaller map is placed on top of the larger map and inside its borders. Show that there is a unique point on the top map which lies exactly above the corresponding point on the lower map. How can this point be constructed? | The point is obviously unique, because the two maps have different scales (but if $\mathrm{P}$ and $\mathrm{Q}$ where two fixed points the distance between them would be the same on both maps).
Let the small map square be $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ and the large be $A B C D$, where $X$ and $X^{\prim... | {
"problem_match": "# Problem 2",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 60 | 1,006 |
1979 | T1 | 3 | null | USAMO | $a_{1}, a_{2}, \ldots, a_{n}$ is an arbitrary sequence of positive integers. A member of the sequence is picked at random. Its value is a. Another member is picked at random, independently of the first. Its value is $b$. Then a third, value c. Show that the probability that $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is divisib... | Let the prob of a value $=0,1,2 \bmod 3$ be $\mathrm{p}, \mathrm{q}, \mathrm{r}$ respectively. So $\mathrm{p}+\mathrm{q}+\mathrm{r}=1$ and $\mathrm{p}, \mathrm{q}, \mathrm{r}$ are non-negative. If $\mathrm{a}=\mathrm{b} \bmod 3$, then to get $\mathrm{a}+\mathrm{b}+\mathrm{c}=0 \bmod 3$, we require $\mathrm{c}=\mathrm{a... | {
"problem_match": "# Problem 3",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 98 | 716 |
1980 | T1 | 5 | null | USAMO | If $x, y, z$ are reals such that $0 \leq x, y, z \leq 1$, show that $x /(y+z+1)+y /(z+x+1)+z /(x+y+1) \leq 1-(1-$ $\mathrm{x})(1-\mathrm{y})(1-\mathrm{z})$. | Consider $\mathrm{x} /(\mathrm{y}+\mathrm{z}+1)+\mathrm{y} /(\mathrm{z}+\mathrm{x}+1)+\mathrm{z} /(\mathrm{x}+\mathrm{y}+1)+(1-\mathrm{x})(1-\mathrm{y})(1-\mathrm{z})$ as a function of $\mathrm{x}$, with $\mathrm{y}$ and $\mathrm{z}$ fixed. Each term is convex, so the whole function is convex. Hence its maximum value o... | {
"problem_match": "# Problem 5",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 78 | 652 |
1981 | T1 | 2 | null | USAMO | What is the largest number of towns that can meet the following criteria. Each pair is directly linked by just one of air, bus or train. At least one pair is linked by air, at least one pair by bus and at least one pair by train. No town has an air link, a bus link and a trian link. No three towns, A, B, C are such tha... | Answer: 4. [eg A and B linked by bus. C and D both linked to A and B by air and to each other by train.]
Suppose A is linked to three other towns by air. Let them be B, C, D. B has at most one other type of link. Suppose it is bus. Then $\mathrm{B}$ must be linked to $\mathrm{C}$ and $\mathrm{D}$ by bus. But now there... | {
"problem_match": "# Problem 2",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 109 | 731 |
1981 | T1 | 4 | null | USAMO | A convex polygon has $\mathrm{n}$ sides. Each vertex is joined to a point $\mathrm{P}$ not in the same plane. If $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are adjacent vertices of the polygon take the angle between the planes PBA and PBC. The sum of the $n$ such angles equals the sum of the $\mathrm{n}$ angles subtended at ... | $\mathrm{n}=3$ is certainly possible. For example, take $\angle \mathrm{APB}=\angle \mathrm{APC}=\angle \mathrm{BPC}=90^{\circ}$ (so that the lines $\mathrm{PA}, \mathrm{PB}, \mathrm{PC}$ are mutually perpendicular). Then the three planes through $\mathrm{P}$ are also mutually perpendicular, so the two sums are both $2... | {
"problem_match": "# Problem 4",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 111 | 630 |
1982 | T1 | 2 | null | USAMO | Show that if $m, n$ are positive integers such that $\left(x^{m+n}+y^{m+n}+z^{m+n}\right) /(m+n)=\left(x^{m}+y^{m}+z^{m}\right) / m\left(x^{n}+y^{n}+\right.$ $\left.\mathrm{z}^{\mathrm{n}}\right) / \mathrm{n}$ for all real $\mathrm{x}, \mathrm{y}, \mathrm{z}$ with sum 0 , then $\{\mathrm{m}, \mathrm{n}\}=\{2,3\}$ or $\... | Put $\mathrm{z}=-\mathrm{x}-\mathrm{y}$. If $\mathrm{m}$ and $\mathrm{n}$ are both odd, the lhs has a term in $\mathrm{x}^{\mathrm{m}+\mathrm{n}}$ but the rhs does not. So at least one of $\mathrm{m}$ and $n$ is even. Suppose both are even. Then comparing terms in $x^{m+n}$ we get $2 /(m+n)=4 / m n$. Put $m=2 M, n$ $=2... | {
"problem_match": "# Problem 2",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 142 | 518 |
1982 | T1 | 3 | null | USAMO | $\mathrm{D}$ is a point inside the equilateral triangle $\mathrm{ABC}$. $\mathrm{E}$ is a point inside $\mathrm{DBC}$. Show that area $\mathrm{DBC} /($ perimeter $\mathrm{DBC})^{2}>$ area $\mathrm{EBC} /(\text { perimeter } \mathrm{EBC})^{2}$. | Let us find an expression for $\mathrm{t}=($ area $\mathrm{DBC}) /(\text { perimeter } \mathrm{DBC})^{2}$. Let angle $\mathrm{B}=2 \mathrm{x}$, angle $\mathrm{C}=2 \mathrm{y}$ and angle $\mathrm{D}$ $=2 z$ and let the inradius be $r$. Then area $D B C=r / 2 x$ perimeter $D B C$. Also $B C=r \cot x+r \cot y$, and simila... | {
"problem_match": "# Problem 3",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 77 | 915 |
1984 | T1 | 3 | null | USAMO | $A, B, C, D, X$ are five points in space, such that $A B, B C, C D, D A$ all subtend the acute angle $\theta$ at $X$. Find the maximum and minimum possible values of $\angle \mathrm{AXC}+\angle \mathrm{BXD}$ (for all such configurations) in terms of $\theta$. | Answer: minimum 0 (see below); maximum $2 \cos ^{-1}(2 \cos \theta-1)$, achieved by a square pyramid.
If we take $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ to lie in a plane so that $\mathrm{AC}$ and $\mathrm{BD}$ meet at $\mathrm{X}$ with the angle $\mathrm{AXC}=\theta$, then $\mathrm{AB}, \mathrm{BC}$, $\mathr... | {
"problem_match": "# Problem 3",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 81 | 1,466 |
1984 | T1 | 5 | null | USAMO | A polynomial of degree $3 n$ has the value 2 at $0,3,6, \ldots, 3 n$, the value 1 at $1,4,7, \ldots, 3 n-2$ and the value 0 at $2,5,8, \ldots, 3 n-1$. Its value at $3 n+1$ is 730 . What is $n$ ? | Answer: $\mathrm{n}=4$.
The $(3 n+1)$ th differences of the polynomial are zero. Call it $p(x)$, so we have $p(3 n+1)-(3 n+1) C 1 p(3 n)+$ $(3 n+1) C 2 p(3 n-1)-\ldots+(-1)^{3 n+1} p(0)=0$, where $r C s$ is the binomial coefficient. Hence $p(3 n+1)=2($
$(3 n+1) C 1-(3 n+1) C 4+\ldots)+((3 n+1) C 3-(3 n+1) C 6+\ldots)... | {
"problem_match": "# Problem 5",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 95 | 521 |
1985 | T1 | 2 | null | USAMO | Find all real roots of the quartic $\mathrm{x}^{4}-(2 \mathrm{~N}+1) \mathrm{x}^{2}-x+\mathrm{N}^{2}+N-1=0$ correct to 4 decimal places, where $\mathrm{N}=$ $10^{10}$. | Answer: 99999.9984 and 100000.0016 .
We can write the equation as $\left(x^{2}-N-1 / 2\right)^{2}=x+5 / 4$. For $x<-5 / 4$ the lhs is positive and the rhs is negative, so there are no roots with $x<-5 / 4$. If $x$ lies between $-5 / 4$ and 0 , then the lhs is obviously much larger than the rhs, so again there is no ro... | {
"problem_match": "# Problem 2",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 71 | 1,186 |
1987 | T1 | 2 | null | USAMO | The feet of the angle bisectors of the triangle $\mathrm{ABC}$ form a right-angled triangle. If the right-angle is at $\mathrm{X}$, where $\mathrm{AX}$ is the bisector of angle $\mathrm{A}$, find all possible values for angle $\mathrm{A}$. | Answer: $120^{\circ}$.
Use vectors origin $A$. Write the vector $A B$ as $\mathbf{B}$ etc. Using the familiar $B X / C X=A B / A C$ etc, we have $\mathbf{Z}=$ $\mathrm{b} \mathbf{B} /(\mathrm{a}+\mathrm{b}), \mathbf{Y}=\mathrm{c} \mathbf{C} /(\mathrm{a}+\mathrm{c}), \mathbf{X}=(\mathrm{b} \mathbf{B}+\mathrm{c} \mathbf... | {
"problem_match": "# Problem 2",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 64 | 615 |
1987 | T1 | 3 | null | USAMO | $X$ is the smallest set of polynomials $p(x)$ such that: (1) $p(x)=x$ belongs to $X$; and (2) if $r(x)$ belongs to $X$, then $x r(x)$ and $(x+(1-x) r(x))$ both belong to $X$. Show that if $r(x)$ and $s(x)$ are distinct elements of $X$, then $\mathrm{r}(\mathrm{x}) \neq \mathrm{s}(\mathrm{x})$ for any $0<\mathrm{x}<1$. | If they are never equal, then we must be able to order them, so that $r(x)>s(x)$ for all $x$ in $(0,1)$ or $s(x)>r(x)$ for all $\mathrm{x}$ in $(0,1)$. Let us use the notation $[+--+\ldots]$. The operation $r(x)$ to $x r(x)$ is denoted as - , and the operation $r(x)$ to $x+(1-x) r(x)$ is denoted as + . Then the operati... | {
"problem_match": "# Problem 3",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 121 | 556 |
1987 | T1 | 5 | null | USAMO | $\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{\mathrm{n}}$ is a sequence of $0 \mathrm{~s}$ and $1 \mathrm{~s}$. $\mathrm{T}$ is the number of triples $\left(\mathrm{a}_{i}, \mathrm{a}_{j}, \mathrm{a}_{\mathrm{k}}\right)$ with $\mathrm{i}<\mathrm{j}<\mathrm{k}$ which are not equal to $(0,1,0)$ or $(1,0,1)$. For ... | For $n$ odd, the smallest value of $T$ is $n(n-1)(n-3) / 8$ achieved by $01010 \ldots 010$.
Suppose a particular $a_{i}=0$. Let $S_{i}$ be the set of $a_{j}$ with $j<i$ and $a_{j}=a_{i}$ and $a_{j}$ with $j>i$ and $a_{j} \neq a_{i}$. Suppose we take any two members of $S_{i}$ and consider the triple formed with $a_{i}... | {
"problem_match": "# Problem 5",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 264 | 1,248 |
1988 | T1 | 5 | null | USAMO | Let $p(x)$ be the polynomial $(1-x)^{a}\left(1-x^{2}\right)^{b}\left(1-x^{3}\right)^{c} \ldots\left(1-x^{32}\right)^{k}$, where $a, b, \ldots, k$ are integers. When expanded in powers of $x$, the coefficient of $x^{1}$ is -2 and the coefficients of $x^{2}, x^{3}, \ldots, x^{32}$ are all zero. Find $\mathrm{k}$. | Answer: $2^{27}-2^{11}$
We have $p(x)=1-2 x+O\left(x^{33}\right)$. Hence $p(-x)=1+2 x+O\left(x^{33}\right)$. Multiplying $p(x) p(-x)=1-2^{2} x^{2}+O\left(x^{33}\right)$. Now $p(x) p(-x)$ cannot have any odd terms, so we can write it as a polynomial in $x^{2}, q\left(x^{2}\right)$. Hence $q\left(x^{2}\right)=1-$ $2^{2}... | {
"problem_match": "# Problem 5",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 123 | 744 |
1990 | T1 | 4 | null | USAMO | How many positive integers can be written in base $\mathrm{n}$ so that (1) the integer has no two digits the same, and (2) each digit after the first differs by one from an earlier digit? For example, in base 3 , the possible numbers are $1,2,10,12,21,102,120,210$. | Answer: $2^{n+1}-2 n-2$.
We use a more elaborate induction hypothesis. We claim that for base $n+1$, the following numbers of integers satisfy the two conditions: $2^{n+1}-2 n-2$ not using the digit $n ; 2^{n}-1$ with the digit $n$ is the last position; $2^{\mathrm{n}-1}$ with the digit $\mathrm{n}$ in the last but 1 ... | {
"problem_match": "# Problem 4",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 82 | 717 |
1992 | T1 | 1 | null | USAMO | Let $a_{n}$ be the number written with $2^{n}$ nines. For example, $a_{0}=9, a_{1}=99, a_{2}=9999$. Let $b_{n}=\Pi_{0}{ }^{n} a_{i}$. Find the sum of the digits of $b_{n}$. | Answer: $9 \cdot 2^{\text {n }}$.
Induction on $\mathrm{n}$. We have $\mathrm{b}_{0}=9$, digit sum 9 , and $\mathrm{b}_{1}=891$, digit sum 18 , so the result is true for $\mathrm{n}=0$ and 1 . Assume it is true for n-1. Obviously $\mathrm{a}_{\mathrm{n}}<10$ to the power of $2^{\mathrm{n}}$, so $\mathrm{b}_{\mathrm{n}... | {
"problem_match": "# Problem 1",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 78 | 611 |
1992 | T1 | 3 | null | USAMO | A set of 11 distinct positive integers has the property that we can find a subset with sum $n$ for any $n$ between 1 and 1500 inclusive. What is the smallest possible value for the second largest element? | Answer: 248.
By taking the integers to be $1,2,4,8, \ldots, 1024$ we can generate all integers up to 2047 . But by taking some integers smaller, we can do better. For example, 1, 2, 4, .., 128, 247, 248, 750 gives all integers up to 1500 . We can obviously use the integers $1,2,4, \ldots, 128$ to generate all integers... | {
"problem_match": "# Problem 3",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 51 | 955 |
1992 | T1 | 5 | null | USAMO | A complex polynomial has degree 1992 and distinct zeros. Show that we can find complex numbers $\mathrm{z}_{\mathrm{n}}$, such that if $\mathrm{p}_{1}(\mathrm{z})=\mathrm{z}-\mathrm{z}_{1}$ and $\mathrm{p}_{\mathrm{n}}(\mathrm{z})=\mathrm{p}_{\mathrm{n}-1}(\mathrm{z})^{2}-\mathrm{z}_{\mathrm{n}}$, then the polynomial d... | Let the polynomial of degree 1992 be $\mathrm{q}(\mathrm{z})$. Suppose its roots are $\mathrm{w}_{1}, \mathrm{w}_{2}, \ldots, \mathrm{w}_{1992}$. Let $\mathrm{S}_{1}=\left\{\mathrm{w}_{1}, \ldots, \mathrm{w}_{1992}\right\}$.
We now define $S_{2}$ as follows. Let $z_{1}=\left(w_{1}+w_{2}\right) / 2$ and take $S_{2}$ to... | {
"problem_match": "# Problem 5",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 122 | 810 |
1993 | T1 | 3 | null | USAMO | Let $\mathrm{S}$ be the set of functions $\mathrm{f}$ defined on reals in the closed interval $[0,1]$ with non-negative real values such that $\mathrm{f}(1)=1$ and $\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y}) \leq \mathrm{f}(\mathrm{x}+\mathrm{y})$ for all $\mathrm{x}, \mathrm{y}$ such that $\mathrm{x}+\mathrm{y} \le... | Answer: $\mathrm{k}=2$.
Consider the function $f(x)=0$ for $0 \leq x \leq 1 / 2,1$ for $1 / 2<x \leq 1$. If $x+y \leq 1$, then at least one of $x, y$ is $\leq 1 / 2$, so at least one of $f(x), f(y)$ is 0 . But $f$ is obviously increasing, so the other of $f(x), f(y)$ is $\leq f(x+y)$. Thus $f$ satisfies the conditions... | {
"problem_match": "# Problem 3",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 160 | 531 |
1993 | T1 | 4 | null | USAMO | The sequence $a_{n}$ of odd positive integers is defined as follows: $a_{1}=r, a_{2}=s$, and $a_{n}$ is the greatest odd divisor of $a_{n-1}+a_{n-2}$. Show that, for sufficiently large $n, a_{n}$ is constant and find this constant (in terms of $r$ and $s$ ). | This is awkward to get started. Note that if $a_{n-1}=a_{n-2}$, then $a_{n-1}+a_{n-2}=2 a_{n-1}$, whose greatest odd divisor is just $a_{n-1}$, so $a_{n}=a_{n-1}$. So once two consecutive terms are constant the following terms are constant.
Both $a_{n-1}$ and $a_{n-2}$ are odd, so $a_{n-1}+a_{n-2}$ is even and hence $... | {
"problem_match": "# Problem 4",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 87 | 968 |
1993 | T1 | 5 | null | USAMO | A sequence $\mathrm{x}_{\mathrm{n}}$ of positive reals satisfies $\mathrm{x}_{n-1} \mathrm{x}_{\mathrm{n}+1} \leq \mathrm{x}_{\mathrm{n}}{ }^{2}$. Let $\mathrm{a}_{\mathrm{n}}$ be the average of the terms $\mathrm{x}_{0}, \mathrm{x}_{1}, \ldots, \mathrm{x}_{\mathrm{n}}$ and $b_{n}$ be the average of the terms $x_{1}, x... | Put $\mathrm{k}=\mathrm{x}_{1}+\mathrm{x}_{2}+\ldots+\mathrm{x}_{\mathrm{n}-1}$. We have to show that $\left(\mathrm{x}_{0}+\mathrm{k}+\mathrm{x}_{\mathrm{n}}\right) /(\mathrm{n}+1) \mathrm{k} /(\mathrm{n}-1) \geq\left(\mathrm{x}_{0}+\mathrm{k}\right) / \mathrm{n} \quad\left(\mathrm{k}+\mathrm{x}_{\mathrm{n}}\right) / ... | {
"problem_match": "# Problem 5",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 151 | 578 |
1994 | T1 | 1 | null | USAMO | $a_{1}, a_{2}, a_{3}, \ldots$ are positive integers such that $a_{n}>a_{n-1}+1$. Put $b_{n}=a_{1}+a_{2}+\ldots+a_{n}$. Show that there is always a square in the range $b_{n}, b_{n}+1, b_{n}+2, \ldots, b_{n+1}-1$. | If the result fails then for some $m$ we have $b_{n}>m^{2}$ and $b_{n+1} \leq(m+1)^{2}$. So $b_{n+1}{ }^{1 / 2}-b_{n}{ }^{1 / 2}<1$. Thus it is sufficient to prove that $b_{n+1}{ }^{1 / 2}-b_{n}{ }^{1 / 2} \geq 1$. Squaring, that is equivalent to $a_{n+1} \geq 2 b_{n}{ }^{1 / 2}+1$ or $b_{n}{ }^{1 / 2} \leq\left(a_{n+1... | {
"problem_match": "# Problem 1",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 99 | 507 |
1994 | T1 | 2 | null | USAMO | The sequence $a_{1}, a_{2}, \ldots, a_{99}$ has $a_{1}=a_{3}=a_{5}=\ldots=a_{97}=1, a_{2}=a_{4}=a_{6}=\ldots=a_{98}=2$, and $a_{99}=3$. We interpret subscripts greater than 99 by subtracting 99 , so that $\mathbf{a}_{100}$ means $\mathrm{a}_{1}$ etc. An allowed move is to change the value of any one of the $a_{n}$ to a... | This is a classic invariant problem. We strongly suspect that there is no sequence of moves (otherwise it would be too easy to find), so that we must prove there is no sequence. The standard approach is to look for some invariant which is not changed by the allowed moves, but which is different for the initial and desi... | {
"problem_match": "# Problem 2",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 339 | 512 |
1994 | T1 | 4 | null | USAMO | $\mathrm{x}_{\mathrm{i}}$ is a infinite sequence of positive reals such that for all $\mathrm{n}, \mathrm{x}_{1}+\mathrm{x}_{2}+\ldots+\mathrm{x}_{\mathrm{n}} \geq \sqrt{ }$. Show that $\mathrm{x}_{1}{ }^{2}+\mathrm{x}_{2}{ }^{2}+\ldots+$ $\mathrm{x}_{\mathrm{n}}^{2}>(1+1 / 2+1 / 3+\ldots+1 / \mathrm{n}) / 4$ for all $... | Note that this is rather a weak inequality. Taking $\mathrm{n}=1$, we get $\mathrm{x}_{1}>=1$, but $(1+1 / 2+1 / 3+\ldots+1 / 30)<4$, so it is only for $\mathrm{n}>30$ that we need to consider $\mathrm{x}_{2}$ ! Of course, weak inequalities can be awkward to prove.
For $\mathrm{a}+\mathrm{b}$ constant, we minimise $\m... | {
"problem_match": "# Problem 4",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 126 | 711 |
1994 | T1 | 5 | null | USAMO | $\mathrm{X}$ is a set of $\mathrm{n}$ positive integers with sum $\mathrm{s}$ and product $\mathrm{p}$. Show for any integer $\mathrm{N}>=\mathrm{s}, \Sigma$ ( parity( $\mathrm{Y}$ ) ( $\mathrm{N}-$ sum(Y))Cs $)=\mathrm{p}$, where $\mathrm{aCb}$ is the binomial coefficient $\mathrm{a}!/(\mathrm{b}!(\mathrm{a}-\mathrm{b... | $\left(\sum(-1)^{n}\right.$ binomial) with the sum over all subsets $\mathrm{Y}$ of a set $\mathrm{X}$ strongly suggests the principle of inclusion and exclusion.
Consider all sequences of length $\mathrm{N} \geq \mathrm{s}$ of $0 \mathrm{~s}$ and $1 \mathrm{~s}$ with a total of $\mathrm{s} 1 \mathrm{~s}$. We can rega... | {
"problem_match": "# Problem 5",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 197 | 567 |
1995 | T1 | 1 | null | USAMO | The sequence $\mathrm{a}_{0} \mathrm{a}_{1}, \mathrm{a}_{2}, \ldots$ of non-negative integers is defined as follows. The first $\mathrm{p}-1$ terms are $0,1,2,3, \ldots$, $p-2$. Then $a_{n}$ is the least positive integer so that there is no arithmetic progression of length $p$ in the first $\mathrm{n}+1$ terms. If $\ma... | Let $b_{n}$ be the number obtained by writing $n$ in base $p-1$ and then treating the result as a number in base $p$. The resulting sequence $\mathrm{b}_{\mathrm{n}}$ is all those non-negative integers whose base $\mathrm{p}$ representation does not have a digit $\mathrm{p}-1$. We show that $\mathrm{b}_{\mathrm{n}}$ ca... | {
"problem_match": "# Problem 1",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 196 | 677 |
1995 | T1 | 3 | null | USAMO | The circumcenter $\mathrm{O}$ of the triangle $\mathrm{ABC}$ does not lie on any side or median. Let the midpoints of $\mathrm{BC}, \mathrm{CA}$, $\mathrm{AB}$ be $\mathrm{L}, \mathrm{M}, \mathrm{N}$ respectively. Take $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ on the rays $\mathrm{OL}, \mathrm{OM}, \mathrm{ON}$ respectively... | We show that the circumcircle $A B C$ is the incircle of $P Q R$. Then $(A R / A Q)(C Q / C P)(B P / B R)=1$ since $A R$ $=\mathrm{BR}, \mathrm{AQ}=\mathrm{CQ}, \mathrm{BP}=\mathrm{CP}$ (equal tangents) and hence $\mathrm{PA}, \mathrm{QB}, \mathrm{RC}$ are concurrent by Ceva's theorem.
So let the tangents to the circu... | {
"problem_match": "# Problem 3",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 176 | 627 |
1995 | T1 | 4 | null | USAMO | $a_{0}, a_{1}, a_{2}, \ldots$ is an infinite sequence of integers such that $a_{n}-a_{m}$ is divisible by $n-m$ for all (unequal) $n$ and $m$. For some polynomial $\mathrm{p}(\mathrm{x})$ we have $\mathrm{p}(\mathrm{n})>\left|\mathrm{a}_{\mathrm{n}}\right|$ for all $\mathrm{n}$. Show that there is a polynomial $\mathrm... | Clearly for any finite $N$, we can find a polynomial $q(n)$ of degree $N$ such that $q(n)=a_{n}$ for $n=0, \ldots, N$. Also, once $\mathrm{a}_{0}, \mathrm{a}_{1}, \ldots, \mathrm{a}_{\mathrm{N}}$ are fixed, $\mathrm{a}_{\mathrm{m}}$ is somewhat constrained for $\mathrm{m}>\mathrm{N}$, because we require $\mathrm{a}_{\m... | {
"problem_match": "# Problem 4",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 136 | 1,192 |
1995 | T1 | 5 | null | USAMO | A graph with $\mathrm{n}$ points and $\mathrm{k}$ edges has no triangles. Show that it has a point $\mathrm{P}$ such that there are at most $\mathrm{k}(1$ $-4 k / \mathrm{n}^{2}$ ) edges between points not joined to P (by an edge). | Given a point $\mathrm{P}$, the edges of the graph can be divided into three categories: (1) edges $\mathrm{PQ}$, (2) edges $\mathrm{QQ}^{\prime}$, where Q is joined to $\mathrm{P}$, and (3) edges $\mathrm{Q}^{\prime} \mathrm{Q}^{\prime \prime}$, where $\mathrm{Q}^{\prime}$ and $\mathrm{Q}^{\prime \prime}$ are not join... | {
"problem_match": "# Problem 5",
"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl",
"solution_match": "# Solution"
} | 66 | 668 |
1997 | T1 | 6 | null | USAMO | Suppose the sequence of nonnegative integers $a_{1}, a_{2}, \ldots, a_{1997}$ satisfies $$ a_{i}+a_{j} \leq a_{i+j} \leq a_{i}+a_{j}+1 $$ for all $i, j \geq 1$ with $i+j \leq 1997$. Show that there exists a real number $x$ such that $a_{n}=\lfloor n x\rfloor$ for all $1 \leq n \leq 1997$. | We are trying to show there exists an $x \in \mathbb{R}$ such that $$ \frac{a_{n}}{n} \leq x<\frac{a_{n}+1}{n} \quad \forall n $$ This means we need to show $$ \max _{i} \frac{a_{i}}{i}<\min _{j} \frac{a_{j}+1}{j} . $$ Replace 1997 by $N$. We will prove this by induction, but we will need some extra hypotheses on the i... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-1997-notes.jsonl",
"solution_match": null
} | 132 | 878 |
1999 | T1 | 3 | null | USAMO | Let $p>2$ be a prime and let $a, b, c, d$ be integers not divisible by $p$, such that $$ \left\{\frac{r a}{p}\right\}+\left\{\frac{r b}{p}\right\}+\left\{\frac{r c}{p}\right\}+\left\{\frac{r d}{p}\right\}=2 $$ for any integer $r$ not divisible by $p$. (Here, $\{t\}=t-\lfloor t\rfloor$ is the fractional part.) Prove t... | First of all, we apparently have $r(a+b+c+d) \equiv 0(\bmod p)$ for every prime $p$, so it automatically follows that $a+b+c+d \equiv 0(\bmod p)$. By scaling appropriately, and also replacing each number with its remainder modulo $p$, we are going to assume that $$ 1=a \leq b \leq c \leq d<p $$ We are going to prove th... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-1999-notes.jsonl",
"solution_match": null
} | 159 | 1,181 |
2000 | T1 | 6 | null | USAMO | Let $a_{1}, b_{1}, a_{2}, b_{2}, \ldots, a_{n}, b_{n}$ be nonnegative real numbers. Prove that $$ \sum_{i, j=1}^{n} \min \left\{a_{i} a_{j}, b_{i} b_{j}\right\} \leq \sum_{i, j=1}^{n} \min \left\{a_{i} b_{j}, a_{j} b_{i}\right\} $$ | 【 First solution by creating a single min (Vincent Huang and Ravi Boppana). Let $b_{i}=r_{i} a_{i}$ for each $i$, and rewrite the inequality as $$ \sum_{i, j} a_{i} a_{j}\left[\min \left(r_{i}, r_{j}\right)-\min \left(1, r_{i} r_{j}\right)\right] \geq 0 $$ We now do the key manipulation to convert the double min into a... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2000-notes.jsonl",
"solution_match": null
} | 119 | 1,081 |
2000 | T1 | 6 | null | USAMO | Let $a_{1}, b_{1}, a_{2}, b_{2}, \ldots, a_{n}, b_{n}$ be nonnegative real numbers. Prove that $$ \sum_{i, j=1}^{n} \min \left\{a_{i} a_{j}, b_{i} b_{j}\right\} \leq \sum_{i, j=1}^{n} \min \left\{a_{i} b_{j}, a_{j} b_{i}\right\} $$ | 【 Second solution by smoothing (Alex Zhai). The case $n=1$ is immediate, so we'll proceed by induction on $n \geq 2$. Again, let $b_{i}=r_{i} a_{i}$ for each $i$, and write the inequality as $$ L_{n}\left(a_{1}, \ldots, a_{n}, r_{1}, \ldots, r_{n}\right):=\sum_{i, j} a_{i} a_{j}\left[\min \left(r_{i}, r_{j}\right)-\min... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2000-notes.jsonl",
"solution_match": null
} | 119 | 1,287 |
2001 | T1 | 1 | null | USAMO | Each of eight boxes contains six balls. Each ball has been colored with one of $n$ colors, such that no two balls in the same box are the same color, and no two colors occur together in more than one box. Find with proof the smallest possible $n$. | The answer is $n=23$. Shown below is a construction using that many colors, which we call $\{1,2, \ldots, 15, a, \ldots, f, X, Y\}$. $$ \left[\begin{array}{cccccccc} X & X & X & 1 & 2 & 3 & 4 & 5 \\ 1 & 6 & 11 & 6 & 7 & 8 & 9 & 10 \\ 2 & 7 & 12 & 11 & 12 & 13 & 14 & 15 \\ 3 & 8 & 13 & Y & Y & Y & a & b \\ 4 & 9 & 14 & ... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2001-notes.jsonl",
"solution_match": null
} | 56 | 1,062 |
2001 | T1 | 3 | null | USAMO | Let $a, b, c$ be nonnegative real numbers such that $a^{2}+b^{2}+c^{2}+a b c=4$. Show that $$ 0 \leq a b+b c+c a-a b c \leq 2 $$ | $ The left-hand side of the inequality is trivial; just note that $\min \{a, b, c\} \leq 1$. Hence, we focus on the right side. We use Lagrange Multipliers. Define $$ U=\left\{(a, b, c) \mid a, b, c>0 \text { and } a^{2}+b^{2}+c^{2}<1000\right\} . $$ This is an intersection of open sets, so it is open. Its closure is $... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2001-notes.jsonl",
"solution_match": null
} | 61 | 1,138 |
2001 | T1 | 5 | null | USAMO | Let $S \subseteq \mathbb{Z}$ be such that: (a) there exist $a, b \in S$ with $\operatorname{gcd}(a, b)=\operatorname{gcd}(a-2, b-2)=1$; (b) if $x$ and $y$ are elements of $S$ (possibly equal), then $x^{2}-y$ also belongs to $S$. Prove that $S=\mathbb{Z}$. | Call an integer $d>0$ shifty if $S=S+d$ (meaning $S$ is invariant under shifting by $d$ ). First, note that if $u, v \in S$, then for any $x \in S$, $$ v^{2}-\left(u^{2}-x\right)=\left(v^{2}-u^{2}\right)+x \in S $$ Since we can easily check that $|S|>1$ and $S \neq\{n,-n\}$ we conclude there exists a shifty integer. We... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2001-notes.jsonl",
"solution_match": null
} | 103 | 520 |
2003 | T1 | 2 | null | USAMO | A convex polygon $\mathcal{P}$ in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon $\mathcal{P}$ are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers. | Suppose $A B$ is a side of a polygon in the dissection, lying on diagonal $X Y$, with $X, A$, $B, Y$ in that order. Then $$ A B=X Y-X A-Y B $$ In this way, we see that it actually just suffices to prove the result for a quadrilateral. I Second approach (barycentric coordinates). To do this, we apply barycentric coordin... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2003-notes.jsonl",
"solution_match": null
} | 69 | 850 |
2004 | T1 | 2 | null | USAMO | Let $a_{1}, a_{2}, \ldots, a_{n}$ be integers whose greatest common divisor is 1 . Let $S$ be a set of integers with the following properties: (a) $a_{i} \in S$ for $i=1, \ldots, n$. (b) $a_{i}-a_{j} \in S$ for $i, j=1, \ldots, n$, not necessarily distinct. (c) If $x, y \in S$ and $x+y \in S$, then $x-y \in S$ too. Pr... | The idea is to show any linear combination of the $a_{i}$ are in $S$, which implies (by Bezout) that $S=\mathbb{Z}$. This is pretty intuitive, but the details require some care (in particular there is a parity obstruction at the second lemma). First, we make the following simple observations: - $0 \in S$, by putting $i... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2004-notes.jsonl",
"solution_match": null
} | 143 | 723 |
2004 | T1 | 6 | null | USAMO | A circle $\omega$ is inscribed in a quadrilateral $A B C D$. Let $I$ be the center of $\omega$. Suppose that $$ (A I+D I)^{2}+(B I+C I)^{2}=(A B+C D)^{2} $$ Prove that $A B C D$ is an isosceles trapezoid. | $$ a+b+c+d=a b c+b c d+c d a+d a b $$ which can be proved by, say tan-addition formula. Then, the content of the problem is to show that $$ \left(\sqrt{a^{2}+1}+\sqrt{d^{2}+1}\right)^{2}+\left(\sqrt{b^{2}+1}+\sqrt{c^{2}+1}\right)^{2} \leq(a+b+c+d)^{2} $$ subject to $(\star)$, with equality only when $a=d=\frac{1}{b}=\f... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2004-notes.jsonl",
"solution_match": null
} | 83 | 824 |
2005 | T1 | 3 | null | USAMO | Let $A B C$ be an acute-angled triangle, and let $P$ and $Q$ be two points on side $B C$. Construct a point $C_{1}$ in such a way that the convex quadrilateral $A P B C_{1}$ is cyclic, $\overline{Q C_{1}} \| \overline{C A}$, and $C_{1}$ and $Q$ lie on opposite sides of line $A B$. Construct a point $B_{1}$ in such a wa... | It is enough to prove that $A, B_{1}$, and $C_{1}$ are collinear, since then $\measuredangle C_{1} Q P=\measuredangle A C P=$ $\measuredangle A B_{1} P=\measuredangle C_{1} B_{1} P$. \ Second solution. One may also use barycentric coordinates. Let $P=(0, m, n)$ and $Q=(0, r, s)$ with $m+n=r+s=1$. Once again, $$ (A P B)... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2005-notes.jsonl",
"solution_match": null
} | 188 | 627 |
2005 | T1 | 6 | null | USAMO | For a positive integer $m$, let $s(m)$ denote the sum of the decimal digits of $m$. A set $S$ positive integers is $k$-stable if $s\left(\sum_{x \in X} x\right)=k$ for any nonempty subset $X \subseteq S$. For each integer $n \geq 2$ let $f(n)$ be the minimal $k$ for which there exists a $k$-stable set with $n$ integers... | 【 Construction showing $f(n) \leq 9\left\lceil\log _{10}\binom{n+1}{2}\right\rceil$. Let $n \geq 1$ and $e \geq 1$ be integers satisfying $1+2+\cdots+n<10^{e}$. Consider the set $$ S=\left\{10^{e}-1,2\left(10^{e}-1\right), \ldots, n\left(10^{e}-1\right)\right\} $$ For example, if $n=6$ and $e=3$, we have $S=\{999,1998,... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2005-notes.jsonl",
"solution_match": null
} | 159 | 652 |
2006 | T1 | 2 | null | USAMO | Let $k>0$ be a fixed integer. Compute the minimum integer $N$ (in terms of $k$ ) for which there exists a set of $2 k+1$ distinct positive integers that has sum greater than $N$, but for which every subset of size $k$ has sum at most $N / 2$. | The answer is $N=k\left(2 k^{2}+3 k+3\right)$ given by $$ S=\left\{k^{2}+1, k^{2}+2, \ldots, k^{2}+2 k+1\right\} . $$ To show this is best possible, let the set be $S=\left\{a_{0}<a_{1}<\cdots<a_{2 k}\right\}$ so that the hypothesis becomes $$ \begin{aligned} N+1 & \leq a_{0}+a_{1}+\cdots+a_{2 k} \\ N / 2 & \geq a_{k+1... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2006-notes.jsonl",
"solution_match": null
} | 70 | 752 |
2006 | T1 | 4 | null | USAMO | Find all positive integers $n$ for which there exist an integer $k \geq 2$ and positive rational numbers $a_{1}, \ldots, a_{k}$ satisfying $a_{1}+a_{2}+\cdots+a_{k}=a_{1} a_{2} \ldots a_{k}=n$. | The answer is all $n$ other than $1,2,3,5$. We now contend that $k>2$. Indeed, if $a_{1}+a_{2}=a_{1} a_{2}=n$ then $\left(a_{1}-a_{2}\right)^{2}=$ $\left(a_{1}+a_{2}\right)^{2}-4 a_{1} a_{2}=n^{2}-4 n=(n-2)^{2}-4$ is a rational integer square, hence a perfect square. This happens only when $n=4$. Now by AM-GM, $$ \frac... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2006-notes.jsonl",
"solution_match": null
} | 73 | 516 |
2006 | T1 | 5 | null | USAMO | A mathematical frog jumps along the number line. The frog starts at 1 , and jumps according to the following rule: if the frog is at integer $n$, then it can jump either to $n+1$ or to $n+2^{m_{n}+1}$ where $2^{m_{n}}$ is the largest power of 2 that is a factor of $n$. Show that if $k \geq 2$ is a positive integer and ... | We will think about the problem in terms of finite sequences of jumps $\left(s_{1}, s_{2}, \ldots, s_{\ell}\right)$, which we draw as $$ 1=x_{0} \xrightarrow{s_{1}} x_{1} \xrightarrow{s_{2}} x_{2} \xrightarrow{s_{3}} \ldots \xrightarrow{s_{\ell}} x_{\ell} $$ where $s_{k}=x_{k}-x_{k-1}$ is the length of some hop. We say... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2006-notes.jsonl",
"solution_match": null
} | 140 | 652 |
2007 | T1 | 6 | null | USAMO | Let $A B C$ be an acute triangle with $\omega, S$, and $R$ being its incircle, circumcircle, and circumradius, respectively. Circle $\omega_{A}$ is tangent internally to $S$ at $A$ and tangent externally to $\omega$. Circle $S_{A}$ is tangent internally to $S$ at $A$ and tangent internally to $\omega$. Let $P_{A}$ and... | It turns out we can compute $P_{A} Q_{A}$ explicitly. Let us invert around $A$ with radius $s-a$ (hence fixing the incircle) and then compose this with a reflection around the angle bisector of $\angle B A C$. We denote the image of the composed map via $$ \bullet \mapsto \bullet^{*} \mapsto \bullet^{+} $$ We overlay t... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2007-notes.jsonl",
"solution_match": null
} | 197 | 814 |
2008 | T1 | 3 | null | USAMO | Let $n$ be a positive integer. Denote by $S_{n}$ the set of points $(x, y)$ with integer coordinates such that $$ |x|+\left|y+\frac{1}{2}\right|<n $$ A path is a sequence of distinct points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{\ell}, y_{\ell}\right)$ in $S_{n}$ such that, for $i=2, \l... | 【 First solution (local). We proceed by induction on $n$. The base case $n=1$ is clear, so suppose $n>1$. Let $S$ denote the set of points $$ S=\left\{(x, y): x+\left|y+\frac{1}{2}\right| \geq n-2\right\} $$ An example when $n=4$ is displayed below. For any minimal partition $\mathcal{P}$ of $S_{n}$, let $P$ denote the... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2008-notes.jsonl",
"solution_match": null
} | 183 | 783 |
2008 | T1 | 5 | null | USAMO | 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_{1} r_{1}+a_{2} r_{2}+a_{3} r_{3}=0$. We are permitted to perform the following operation: find two numbers $x, y$ on the blackbo... | We first show we can decrease the quantity $\left|a_{1}\right|+\left|a_{2}\right|+\left|a_{3}\right|$ as long as $0 \notin\left\{a_{1}, a_{2}, a_{3}\right\}$. Assume $a_{1}>0$ and $r_{1}>r_{2}>r_{3}$ without loss of generality and consider two cases. - Suppose $a_{2}>0$ or $a_{3}>0$; these cases are identical. (One can... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2008-notes.jsonl",
"solution_match": null
} | 152 | 754 |
2009 | T1 | 1 | null | USAMO | Given circles $\omega_{1}$ and $\omega_{2}$ intersecting at points $X$ and $Y$, let $\ell_{1}$ be a line through the center of $\omega_{1}$ intersecting $\omega_{2}$ at points $P$ and $Q$ and let $\ell_{2}$ be a line through the center of $\omega_{2}$ intersecting $\omega_{1}$ at points $R$ and $S$. Prove that if $P, Q... | Let $r_{1}, r_{2}, r_{3}$ denote the circumradii of $\omega_{1}, \omega_{2}$, and $\omega_{3}$, respectively. We wish to show that $O_{3}$ lies on the radical axis of $\omega_{1}$ and $\omega_{2}$. Let us encode the conditions using power of a point. Because $O_{1}$ is on the radical axis of $\omega_{2}$ and $\omega_{3... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2009-notes.jsonl",
"solution_match": null
} | 125 | 556 |
2009 | T1 | 3 | null | USAMO | We define a chessboard polygon to be a simple polygon whose sides are situated along lines of the form $x=a$ or $y=b$, where $a$ and $b$ are integers. These lines divide the interior into unit squares, which are shaded alternately grey and white so that adjacent squares have different colors. To tile a chessboard polyg... | Consider a lower-left square $s$ of the polygon, and WLOG is it white (other case similar). Then we have two cases: - If there exists a domino tiling of $\mathcal{P}$ where $s$ is covered by a vertical domino, then delete this domino and apply induction on the rest of $\mathcal{P}$. This additional domino will not caus... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2009-notes.jsonl",
"solution_match": null
} | 277 | 842 |
2009 | T1 | 5 | null | USAMO | Trapezoid $A B C D$, with $\overline{A B} \| \overline{C D}$, is inscribed in circle $\omega$ and point $G$ lies inside triangle $B C D$. Rays $A G$ and $B G$ meet $\omega$ again at points $P$ and $Q$, respectively. Let the line through $G$ parallel to $\overline{A B}$ intersect $\overline{B D}$ and $\overline{B C}$ at... | Perform an inversion around $B$ with arbitrary radius, and denote the inverse of a point $Z$ with $Z^{*}$. After inversion, we obtain a cyclic quadrilateral $B S^{*} G^{*} R^{*}$ and points $C^{*}, D^{*}$ on $\overline{B S^{*}}$, $\overline{B R^{*}}$, such that $\left(B C^{*} D^{*}\right)$ is tangent to $\left(B S^{*} ... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2009-notes.jsonl",
"solution_match": null
} | 150 | 560 |
2010 | T1 | 3 | null | USAMO | The 2010 positive real numbers $a_{1}, a_{2}, \ldots, a_{2010}$ satisfy the inequality $a_{i} a_{j} \leq i+j$ for all $1 \leq i<j \leq 2010$. Determine, with proof, the largest possible value of the product $a_{1} a_{2} \ldots a_{2010}$. | The answer is $3 \times 7 \times 11 \times \cdots \times 4019$, which is clearly an upper bound (and it's not too hard to show this is the lowest number we may obtain by multiplying 1005 equalities together; this is essentially the rearrangement inequality). The tricky part is the construction. Intuitively we want $a_{... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2010-notes.jsonl",
"solution_match": null
} | 97 | 732 |
2010 | T1 | 5 | null | USAMO | Let $q=\frac{3 p-5}{2}$ where $p$ is an odd prime, and let $$ S_{q}=\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{5 \cdot 6 \cdot 7}+\cdots+\frac{1}{q(q+1)(q+2)} $$ Prove that if $\frac{1}{p}-2 S_{q}=\frac{m}{n}$ for integers $m$ and $n$, then $m-n$ is divisible by $p$. | By partial fractions, we have $$ \frac{2}{(3 k-1)(3 k)(3 k+1)}=\frac{1}{3 k-1}-\frac{2}{3 k}+\frac{1}{3 k+1} $$ Thus $$ \begin{aligned} 2 S_{q} & =\left(\frac{1}{2}-\frac{2}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{2}{6}+\frac{1}{7}\right)+\cdots+\left(\frac{1}{q}-\frac{2}{q+1}+\frac{1}{q+2}\right) \\ & =\left(\fr... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2010-notes.jsonl",
"solution_match": null
} | 122 | 755 |
2010 | T1 | 6 | null | USAMO | There are 68 ordered pairs (not necessarily distinct) of nonzero integers on a blackboard. It's known that for no integer $k$ does both $(k, k)$ and $(-k,-k)$ appear. A student erases some of the 136 integers such that no two erased integers have sum zero, and scores one point for each ordered pair with at least one er... | The answer is 43. The structure of this problem is better understood as follows: we construct a multigraph whose vertices are the entries, and the edges are the 68 ordered pairs on the blackboard. To be precise, construct a multigraph $G$ with vertices $a_{1}, b_{1}, \ldots, a_{n}, b_{n}$, with $a_{i}=-b_{i}$ for each ... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2010-notes.jsonl",
"solution_match": null
} | 94 | 1,046 |
2011 | T1 | 2 | null | USAMO | An integer is assigned to each vertex of a regular pentagon so that the sum of the five integers is 2011. A turn of a solitaire game consists of subtracting an integer $m$ (not necessarily positive) from each of the integers at two neighboring vertices and adding $2 m$ to the opposite vertex, which is not adjacent to e... | Call the vertices $0,1,2,3,4$ in order. First, notice that the quantity $$ S:=N_{1}+2 N_{2}+3 N_{3}+4 N_{4} \quad(\bmod 5) $$ is invariant, where $N_{i}$ is the amount at vertex $i$. This immediately implies that at most one vertex can win, since in a winning situation all $N_{i}$ are 0 except for one, which is 2011. (... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2011-notes.jsonl",
"solution_match": null
} | 158 | 1,149 |
2011 | T1 | 3 | null | USAMO | In hexagon $A B C D E F$, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy $\angle A=3 \angle D, \angle C=3 \angle F$, and $\angle E=3 \angle B$. Furthermore $A B=D E, B C=E F$, and $C D=F A$. Prove that diagonals $\overline{A D}, \overline{B E}$, and $\o... | Main idea: Claim - In a satisfying hexagon, $B, D, F$ are reflections of $A, C, E$ across the sides of $\triangle A C E$. (This claim looks plausible because every excellent hexagon is satisfying, and both configuration spaces are three-dimensional.) Call a hexagon of this shape "excellent"; in a excellent hexagon the ... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2011-notes.jsonl",
"solution_match": null
} | 113 | 1,023 |
2012 | T1 | 1 | null | USAMO | Find all integers $n \geq 3$ such that among any $n$ positive real numbers $a_{1}, a_{2}, \ldots$, $a_{n}$ with $$ \max \left(a_{1}, a_{2}, \ldots, a_{n}\right) \leq n \cdot \min \left(a_{1}, a_{2}, \ldots, a_{n}\right), $$ there exist three that are the side lengths of an acute triangle. | The answer is all $n \geq 13$. Define $\left(F_{n}\right)$ as the sequence of Fibonacci numbers, by $F_{1}=F_{2}=1$ and $F_{n+1}=$ $F_{n}+F_{n-1}$. We will find that Fibonacci numbers show up naturally when we work through the main proof, so we will isolate the following calculation now to make the subsequent solution ... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2012-notes.jsonl",
"solution_match": null
} | 107 | 965 |
2012 | T1 | 3 | null | USAMO | Determine which integers $n>1$ have the property that there exists an infinite sequence $a_{1}, a_{2}, a_{3}, \ldots$ of nonzero integers such that the equality $$ a_{k}+2 a_{2 k}+\cdots+n a_{n k}=0 $$ holds for every positive integer $k$. | Answer: all $n>2$. For $n=2$, we have $a_{k}+2 a_{2 k}=0$, which is clearly not possible, since it implies $a_{2^{k}}=\frac{a_{1}}{2^{k-1}}$ for all $k \geq 1$. For $n \geq 3$ we will construct a completely multiplicative sequence (meaning $a_{i j}=a_{i} a_{j}$ for all $i$ and $j$ ). Thus $\left(a_{i}\right)$ is determ... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2012-notes.jsonl",
"solution_match": null
} | 75 | 901 |
2012 | T1 | 4 | null | USAMO | Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that $f(n!)=f(n)$ ! for all positive integers $n$ and such that $m-n$ divides $f(m)-f(n)$ for all distinct positive integers $m, n$. | Answer: $f \equiv 1, f \equiv 2$, and $f$ the identity. As these obviously work, we prove these are the only ones. By putting $n=1$ and $n=2$ we give $f(1), f(2) \in\{1,2\}$. Also, we will use the condition $$ m!-n!\text { divides } f(m)!-f(n)! $$ We consider four cases on $f(1)$ and $f(2)$, and dispense with three of ... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2012-notes.jsonl",
"solution_match": null
} | 63 | 568 |
2012 | T1 | 5 | null | USAMO | Let $P$ be a point in the plane of $\triangle A B C$, and $\gamma$ a line through $P$. Let $A^{\prime}, B^{\prime}, C^{\prime}$ be the points where the reflections of lines $P A, P B, P C$ with respect to $\gamma$ intersect lines $B C, C A, A B$ respectively. Prove that $A^{\prime}, B^{\prime}, C^{\prime}$ are collinea... | 【 First solution (complex numbers). Let $p=0$ and set $\gamma$ as the real line. Then $A^{\prime}$ is the intersection of $b c$ and $p \bar{a}$. So, we get $$ a^{\prime}=\frac{\bar{a}(\bar{b} c-b \bar{c})}{(\bar{b}-\bar{c}) \bar{a}-(b-c) a} . $$ Note that $$ \bar{a}^{\prime}=\frac{a(b \bar{c}-\bar{b} c)}{(b-c) a-(\bar{... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2012-notes.jsonl",
"solution_match": null
} | 105 | 559 |
2013 | T1 | 2 | null | USAMO | For a positive integer $n \geq 3$ plot $n$ equally spaced points around a circle. Label one of them $A$, and place a marker at $A$. One may move the marker forward in a clockwise direction to either the next point or the point after that. Hence there are a total of $2 n$ distinct moves available; two from each point. L... | 【 First solution. Imagine the counter is moving along the set $S=\{0,1, \ldots, 2 n\}$ instead, starting at 0 and ending at $2 n$, in jumps of length 1 and 2 . We can then record the sequence of moves as a matrix of the form $$ \left[\begin{array}{cccccc} p_{0} & p_{1} & p_{2} & \ldots & p_{n-1} & p_{n} \\ p_{n} & p_{n... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2013-notes.jsonl",
"solution_match": null
} | 138 | 943 |
2013 | T1 | 2 | null | USAMO | For a positive integer $n \geq 3$ plot $n$ equally spaced points around a circle. Label one of them $A$, and place a marker at $A$. One may move the marker forward in a clockwise direction to either the next point or the point after that. Hence there are a total of $2 n$ distinct moves available; two from each point. L... | Second (longer) solution. If one does not notice the nice rephrasing with $\mathbf{u}, \mathbf{v}, \mathbf{w}$ above, one may still proceed with the following direct calculation. Retain the notation of $$ \left[\begin{array}{cccccc} p_{0} & p_{1} & p_{2} & \ldots & p_{n-1} & p_{n} \\ p_{n} & p_{n+1} & p_{n+2} & \cdots ... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2013-notes.jsonl",
"solution_match": null
} | 138 | 1,039 |
2013 | T1 | 3 | null | USAMO | Let $n$ be a positive integer. There are $\frac{n(n+1)}{2}$ tokens, each with a black side and a white side, arranged into an equilateral triangle, with the biggest row containing $n$ tokens. Initially, each token has the white side up. An operation is to choose a line parallel to the sides of the triangle, and flip al... | The answer is $$ \max _{C} f(C)= \begin{cases}6 k & n=4 k \\ 6 k+1 & n=4 k+1 \\ 6 k+2 & n=4 k+2 \\ 6 k+3 & n=4 k+3\end{cases} $$ The main point of the problem is actually to determine all linear dependencies among the $3 n$ possible moves (since the moves commute and applying a move twice is the same as doing nothing).... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2013-notes.jsonl",
"solution_match": null
} | 158 | 1,116 |
2013 | T1 | 6 | null | USAMO | Let $A B C$ be a triangle. Find all points $P$ on segment $B C$ satisfying the following property: If $X$ and $Y$ are the intersections of line $P A$ with the common external tangent lines of the circumcircles of triangles $P A B$ and $P A C$, then $$ \left(\frac{P A}{X Y}\right)^{2}+\frac{P B \cdot P C}{A B \cdot A C... | Let $O_{1}$ and $O_{2}$ denote the circumcenters of $P A B$ and $P A C$. The main idea is to notice that $\triangle A B C$ and $\triangle A O_{1} O_{2}$ are spirally similar. Claim (Salmon theorem) - We have $\triangle A B C \stackrel{ \pm}{\sim} \triangle A O_{1} O_{2}$. $$ \angle A O_{1} B=2 \angle A P B $$ but $$ \a... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2013-notes.jsonl",
"solution_match": null
} | 106 | 793 |
2014 | T1 | 2 | null | USAMO | Find all $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $$ x f(2 f(y)-x)+y^{2} f(2 x-f(y))=\frac{f(x)^{2}}{x}+f(y f(y)) $$ for all $x, y \in \mathbb{Z}$ such that $x \neq 0$. | The answer is $f(x) \equiv 0$ and $f(x) \equiv x^{2}$. Check that these work. $$ x f(2 f(0)-x)=\frac{f(x)^{2}}{x}+f(0) . $$ The nicest part of the problem is the following step: $$ \text { Claim - We have } f(0)=0 \text {. } $$ $$ f(0)=0 $$ Claim - We have $f(x) \in\left\{0, x^{2}\right\}$ for each individual $x$. $$ x... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2014-notes.jsonl",
"solution_match": null
} | 87 | 779 |
2014 | T1 | 5 | null | USAMO | Let $A B C$ be a triangle with orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $A H C$ with the internal bisector of $\angle B A C$. Let $X$ be the circumcenter of triangle $A P B$ and let $Y$ be the orthocenter of triangle $A P C$. Prove that the length of segment $X Y$ is equal ... | We eliminate the floating orthocenter by reflecting $P$ across $\overline{A C}$ to $Q$. Then $Q$ lies on ( $A B C$ ) and moreover $\angle Q A C=\frac{1}{2} \angle B A C$. This motivates us to reflect $B, X, Y$ to $B^{\prime}$, $X^{\prime}, Y^{\prime}$ and complex bash with respect to $\triangle A Q C$. Obviously $$ y^{... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2014-notes.jsonl",
"solution_match": null
} | 104 | 679 |
2014 | T1 | 6 | null | USAMO | Prove that there is a constant $c>0$ with the following property: If $a, b, n$ are positive integers such that $\operatorname{gcd}(a+i, b+j)>1$ for all $i, j \in\{0,1, \ldots, n\}$, then $$ \min \{a, b\}>(c n)^{n / 2} $$ | Let $N=n+1$ and assume $N$ is (very) large. We construct an $N \times N$ with cells $(i, j)$ where $0 \leq i, j \leq n$ and in each cell place a prime $p$ dividing $\operatorname{gcd}(a+i, b+j)$. The central claim is at least $50 \%$ of the primes in this table exceed $0.001 n^{2}$. We count the maximum number of squar... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2014-notes.jsonl",
"solution_match": null
} | 89 | 525 |
2015 | T1 | 2 | null | USAMO | Quadrilateral $A P B Q$ is inscribed in circle $\omega$ with $\angle P=\angle Q=90^{\circ}$ and $A P=$ $A Q<B P$. Let $X$ be a variable point on segment $\overline{P Q}$. Line $A X$ meets $\omega$ again at $S$ (other than $A$ ). Point $T$ lies on $\operatorname{arc} A Q B$ of $\omega$ such that $\overline{X T}$ is perp... | - $X$ at the midpoint of $\overline{P Q}$ (giving the midpoint of $\overline{B Q}$ ) which determines the circle; this circle then passes through $P$ by symmetry and we can find the center by taking the intersection of two perpendicular bisectors (which two?). 【 Complex solution (Evan Chen). Toss on the complex unit ci... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2015-notes.jsonl",
"solution_match": null
} | 163 | 681 |
2015 | T1 | 3 | null | USAMO | Let $S=\{1,2, \ldots, n\}$, where $n \geq 1$. Each of the $2^{n}$ subsets of $S$ is to be colored red or blue. (The subset itself is assigned a color and not its individual elements.) For any set $T \subseteq S$, we then write $f(T)$ for the number of subsets of $T$ that are blue. Determine the number of colorings that... | For an $n$-coloring $\mathcal{C}$ (by which we mean a coloring of the subsets of $\{1, \ldots, n\}$ ), define the support of $\mathcal{C}$ as $$ \operatorname{supp}(\mathcal{C})=\{T \mid f(T) \neq 0\} $$ Call a coloring nontrivial if $\operatorname{supp}(\mathcal{C}) \neq \varnothing$ (equivalently, the coloring is not... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2015-notes.jsonl",
"solution_match": null
} | 171 | 719 |
2015 | T1 | 4 | null | USAMO | Steve is piling $m \geq 1$ indistinguishable stones on the squares of an $n \times n$ grid. Each square can have an arbitrarily high pile of stones. After he finished piling his stones in some manner, he can then perform stone moves, defined as follows. Consider any four grid squares, which are corners of a rectangle, ... | The answer is $\binom{m+n-1}{n-1}^{2}$. The main observation is that the ordered sequence of column counts (i.e. the number of stones in the first, second, etc. column) is invariant under stone moves, as does the analogous sequence of row counts. \ Definitions. Call these numbers $\left(c_{1}, c_{2}, \ldots, c_{n}\righ... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2015-notes.jsonl",
"solution_match": null
} | 245 | 803 |
2015 | T1 | 6 | null | USAMO | Consider $0<\lambda<1$, and let $A$ be a multiset of positive integers. Let $A_{n}=\{a \in$ $A: a \leq n\}$. Assume that for every $n \in \mathbb{N}$, the multiset $A_{n}$ contains at most $n \lambda$ numbers. Show that there are infinitely many $n \in \mathbb{N}$ for which the sum of the elements in $A_{n}$ is at most... | For brevity, $\# S$ denotes $|S|$. Let $x_{n}=n \lambda-\# A_{n} \geq 0$. We now proceed by contradiction by assuming the conclusion fails for $n$ large enough; that is, $$ \begin{aligned} \frac{n(n+1)}{2} \lambda & <\sum_{a \in A_{n}} a \\ & =1\left(\# A_{1}-\# A_{0}\right)+2\left(\# A_{2}-\# A_{1}\right)+\cdots+n\lef... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2015-notes.jsonl",
"solution_match": null
} | 123 | 1,155 |
2016 | T1 | 1 | null | USAMO | Let $X_{1}, X_{2}, \ldots, X_{100}$ be a sequence of mutually distinct nonempty subsets of a set $S$. Any two sets $X_{i}$ and $X_{i+1}$ are disjoint and their union is not the whole set $S$, that is, $X_{i} \cap X_{i+1}=\emptyset$ and $X_{i} \cup X_{i+1} \neq S$, for all $i \in\{1, \ldots, 99\}$. Find the smallest pos... | Solution with Danielle Wang: the answer is that $|S| \geq 8$. To see that $|S|=8$ is the minimum possible size, consider a chain on the set $S=$ $\{1,2, \ldots, 7\}$ satisfying $X_{i} \cap X_{i+1}=\emptyset$ and $X_{i} \cup X_{i+1} \neq S$. Because of these requirements any subset of size 4 or more can only be neighbor... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2016-notes.jsonl",
"solution_match": null
} | 134 | 1,509 |
2016 | T1 | 3 | null | USAMO | Let $A B C$ be an acute triangle and let $I_{B}, I_{C}$, and $O$ denote its $B$-excenter, $C$ excenter, and circumcenter, respectively. Points $E$ and $Y$ are selected on $\overline{A C}$ such that $\angle A B Y=\angle C B Y$ and $\overline{B E} \perp \overline{A C}$. Similarly, points $F$ and $Z$ are selected on $\ove... | \ First solution. Let $I_{A}$ denote the $A$-excenter and $I$ the incenter. Then let $D$ denote the foot of the altitude from $A$. Suppose the $A$-excircle is tangent to $\overline{B C}, \overline{A B}, \overline{A C}$ at $A_{1}, B_{1}, C_{1}$ and let $A_{2}, B_{2}, C_{2}$ denote the reflections of $I_{A}$ across these... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2016-notes.jsonl",
"solution_match": null
} | 191 | 748 |
2016 | T1 | 3 | null | USAMO | Let $A B C$ be an acute triangle and let $I_{B}, I_{C}$, and $O$ denote its $B$-excenter, $C$ excenter, and circumcenter, respectively. Points $E$ and $Y$ are selected on $\overline{A C}$ such that $\angle A B Y=\angle C B Y$ and $\overline{B E} \perp \overline{A C}$. Similarly, points $F$ and $Z$ are selected on $\ove... | I Second solution (barycentric, outline, Colin Tang). we are going to use barycentric coordinates to show that the line through $O$ perpendicular to $\overline{Y Z}$ is concurrent with $\overline{I_{B} F}$ and $\overline{I_{C} E}$. The displacement vector $\overrightarrow{Y Z}$ is proportional to $(a(b-c):-b(a+c): c(a+... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2016-notes.jsonl",
"solution_match": null
} | 191 | 585 |
2016 | T1 | 4 | null | USAMO | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$, $$ (f(x)+x y) \cdot f(x-3 y)+(f(y)+x y) \cdot f(3 x-y)=(f(x+y))^{2} $$ | We claim that the only two functions satisfying the requirements are $f(x) \equiv 0$ and $f(x) \equiv x^{2}$. These work. First, taking $x=y=0$ in the given yields $f(0)=0$, and then taking $x=0$ gives $f(y) f(-y)=f(y)^{2}$. So also $f(-y)^{2}=f(y) f(-y)$, from which we conclude $f$ is even. Then taking $x=-y$ gives $$... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2016-notes.jsonl",
"solution_match": null
} | 72 | 787 |
2016 | T1 | 5 | null | USAMO | An equilateral pentagon $A M N P Q$ is inscribed in triangle $A B C$ such that $M \in \overline{A B}$, $Q \in \overline{A C}$, and $N, P \in \overline{B C}$. Let $S$ be the intersection of $\overline{M N}$ and $\overline{P Q}$. Denote by $\ell$ the angle bisector of $\angle M S Q$. Prove that $\overline{O I}$ is parall... | ๆI Second solution (trig, Danielle Wang). Let $\delta$ and $\epsilon$ denote $\angle M N B$ and $\angle C P Q$. Also, assume $A M N P Q$ has side length 1. In what follows, assume $A B<A C$. First, we note that $$ \begin{aligned} B N & =(c-1) \cos B+\cos \delta \\ C P & =(b-1) \cos C+\cos \epsilon, \text { and } \\ a &... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2016-notes.jsonl",
"solution_match": null
} | 150 | 549 |
2017 | T1 | 2 | null | USAMO | Let $m_{1}, m_{2}, \ldots, m_{n}$ be a collection of $n$ positive integers, not necessarily distinct. For any sequence of integers $A=\left(a_{1}, \ldots, a_{n}\right)$ and any permutation $w=w_{1}, \ldots, w_{n}$ of $m_{1}, \ldots, m_{n}$, define an $A$-inversion of $w$ to be a pair of entries $w_{i}, w_{j}$ with $i<j... | Denote by $M$ our multiset of $n$ positive integers. Define an inversion of a permutation to be pair $i<j$ with $w_{i}<w_{j}$ (which is a $(0, \ldots, 0)$-inversion in the problem statement); this is the usual definition (see https://en.wikipedia.org/wiki/Inversion_(discrete_ mathematics)). So we want to show the numbe... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2017-notes.jsonl",
"solution_match": null
} | 295 | 1,777 |
2017 | T1 | 3 | null | USAMO | Let $A B C$ be a scalene triangle with circumcircle $\Omega$ and incenter $I$. Ray $A I$ meets $\overline{B C}$ at $D$ and $\Omega$ again at $M$; the circle with diameter $\overline{D M}$ cuts $\Omega$ again at $K$. Lines $M K$ and $B C$ meet at $S$, and $N$ is the midpoint of $\overline{I S}$. The circumcircles of $\t... | Let $W$ be the midpoint of $\overline{B C}$, let $X$ be the point on $\Omega$ opposite $M$. Observe that $\overline{K D}$ passes through $X$, and thus lines $B C, M K, X A$ concur at the orthocenter of $\triangle D M X$, which we call $S$. Denote by $I_{A}$ the $A$-excenter of $A B C$. Next, let $E$ be the foot of the ... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2017-notes.jsonl",
"solution_match": null
} | 167 | 701 |
2017 | T1 | 5 | null | USAMO | Find all real numbers $c>0$ such that there exists a labeling of the lattice points in $\mathbb{Z}^{2}$ with positive integers for which: - only finitely many distinct labels occur, and - for each label $i$, the distance between any two points labeled $i$ is at least $c^{i}$. | The construction for any $c<\sqrt{2}$ can be done as follows. Checkerboard color the lattice points and label the black ones with 1 . The white points then form a copy of $\mathbb{Z}^{2}$ again scaled up by $\sqrt{2}$ so we can repeat the procedure with 2 on half the resulting points. Continue this dyadic construction ... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2017-notes.jsonl",
"solution_match": null
} | 73 | 571 |
2018 | T1 | 2 | null | USAMO | Find all functions $f:(0, \infty) \rightarrow(0, \infty)$ such that $$ f\left(x+\frac{1}{y}\right)+f\left(y+\frac{1}{z}\right)+f\left(z+\frac{1}{x}\right)=1 $$ for all $x, y, z>0$ with $x y z=1$. | The main part of the problem is to show all solutions are linear. As always, let $x=b / c$, $y=c / a, z=a / b$ (classical inequality trick). Then the problem becomes $$ \sum_{\mathrm{cyc}} f\left(\frac{b+c}{a}\right)=1 $$ Let $f(t)=g\left(\frac{1}{t+1}\right)$, equivalently $g(s)=f(1 / s-1)$. Thus $g:(0,1) \rightarrow(... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2018-notes.jsonl",
"solution_match": null
} | 86 | 1,003 |
2018 | T1 | 3 | null | USAMO | Let $n \geq 2$ be an integer, and let $\left\{a_{1}, \ldots, a_{m}\right\}$ denote the $m=\varphi(n)$ integers less than $n$ and relatively prime to $n$. Assume that every prime divisor of $m$ also divides $n$. Prove that $m$ divides $a_{1}^{k}+\cdots+a_{m}^{k}$ for every positive integer $k$. | For brevity, given any $n$, we let $A(n)=\{1 \leq x \leq n, \operatorname{gcd}(x, n)=1\}$ (thus $|A(n)|=\varphi(n)$ ). Also, let $S(n, k)=\sum_{a \in A(n)} a^{k}$. We will prove the stronger statement (which eliminates the hypothesis on $n$ ). Claim - Let $n \geq 2$ be arbitrary (and $k \geq 0$ ). If $p \mid n$, then $... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2018-notes.jsonl",
"solution_match": null
} | 102 | 1,517 |
2018 | T1 | 6 | null | USAMO | Let $a_{n}$ be the number of permutations $\left(x_{1}, \ldots, x_{n}\right)$ of $(1, \ldots, n)$ such that the ratios $x_{k} / k$ are all distinct. Prove that $a_{n}$ is odd for all $n \geq 1$. | The first idea: ## Lemma If a permutation $x$ works, so does the inverse permutation. Thus it suffices to consider permutations $x$ in which all cycles have length at most 2 . Of course, there can be at most one fixed point (since that gives the ratio 1 ), and hence exactly one if $n$ is odd, none if $n$ is even. We co... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2018-notes.jsonl",
"solution_match": null
} | 73 | 677 |
2019 | T1 | 1 | null | USAMO | A function $f: \mathbb{N} \rightarrow \mathbb{N}$ satisfies $$ \underbrace{f(f(\ldots f}_{f(n) \text { times }}(n) \ldots))=\frac{n^{2}}{f(f(n))} $$ for all positive integers $n$. What are all possible values of $f(1000)$ ? | Actually, we classify all such functions: $f$ can be any function which fixes odd integers and acts as an involution on the even integers. In particular, $f(1000)$ may be any even integer. It's easy to check that these all work, so now we check they are the only solutions. Claim - $f$ is injective. Claim - $f$ fixes th... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2019-notes.jsonl",
"solution_match": null
} | 84 | 996 |
2019 | T1 | 4 | null | USAMO | Let $n$ be a nonnegative integer. Determine the number of ways to choose sets $S_{i j} \subseteq\{1,2, \ldots, 2 n\}$, for all $0 \leq i \leq n$ and $0 \leq j \leq n$ (not necessarily distinct), such that - $\left|S_{i j}\right|=i+j$, and - $S_{i j} \subseteq S_{k l}$ if $0 \leq i \leq k \leq n$ and $0 \leq j \leq l \... | The answer is $(2 n)!\cdot 2^{n^{2}}$. First, we note that $\varnothing=S_{00} \subsetneq S_{01} \subsetneq \cdots \subsetneq S_{n n}=\{1, \ldots, 2 n\}$ and thus multiplying by (2n)! we may as well assume $S_{0 i}=\{1, \ldots, i\}$ and $S_{i n}=\{1, \ldots, n+i\}$. We illustrate this situation by placing the sets in a... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2019-notes.jsonl",
"solution_match": null
} | 136 | 674 |
2019 | T1 | 6 | null | USAMO | Find all polynomials $P$ with real coefficients such that $$ \frac{P(x)}{y z}+\frac{P(y)}{z x}+\frac{P(z)}{x y}=P(x-y)+P(y-z)+P(z-x) $$ for all nonzero real numbers $x, y, z$ obeying $2 x y z=x+y+z$. | The given can be rewritten as saying that $$ \begin{aligned} Q(x, y, z) & :=x P(x)+y P(y)+z P(z) \\ & -x y z(P(x-y)+P(y-z)+P(z-x)) \end{aligned} $$ is a polynomial vanishing whenever $x y z \neq 0$ and $2 x y z=x+y+z$, for real numbers $x, y$, $z$. Claim - This means $Q(x, y, z)$ vanishes also for any complex numbers $... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2019-notes.jsonl",
"solution_match": null
} | 81 | 1,166 |
2020 | T1 | 2 | null | USAMO | An empty $2020 \times 2020 \times 2020$ cube is given, and a $2020 \times 2020$ grid of square unit cells is drawn on each of its six faces. A beam is a $1 \times 1 \times 2020$ rectangular prism. Several beams are placed inside the cube subject to the following conditions: - The two $1 \times 1$ faces of each beam co... | 【 A Answer. 3030 beams. 【 Construction. We first give a construction with $3 n / 2$ beams for any $n \times n \times n$ box, where $n$ is an even integer. Shown below is the construction for $n=6$, which generalizes. (The left figure shows the cube in 3 d ; the right figure shows a direct view of the three visible face... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2020-notes.jsonl",
"solution_match": null
} | 206 | 708 |
2020 | T1 | 3 | null | USAMO | Let $p$ be an odd prime. An integer $x$ is called a quadratic non-residue if $p$ does not divide $x-t^{2}$ for any integer $t$. Denote by $A$ the set of all integers $a$ such that $1 \leq a<p$, and both $a$ and $4-a$ are quadratic non-residues. Calculate the remainder when the product of the elements of $A$ is divided ... | The answer is that $\prod_{a \in A} a \equiv 2(\bmod p)$ regardless of the value of $p$. In the following solution, we work in $\mathbb{F}_{p}$ always and abbreviate "quadratic residue" and "non-quadratic residue" to "QR" and "non-QR", respectively. We define $$ \begin{aligned} & A=\left\{a \in \mathbb{F}_{p} \mid a, 4... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2020-notes.jsonl",
"solution_match": null
} | 100 | 850 |
2020 | T1 | 5 | null | USAMO | A finite set $S$ of points in the coordinate plane is called overdetermined if $|S| \geq 2$ and there exists a nonzero polynomial $P(t)$, with real coefficients and of degree at most $|S|-2$, satisfying $P(x)=y$ for every point $(x, y) \in S$. For each integer $n \geq 2$, find the largest integer $k$ (in terms of $n$ )... | We claim the answer is $k=2^{n-1}-n$. We denote the $n$ points by $A$. ## Lemma If $S$ is a finite set of points in the plane there is at most one polynomial with real coefficients and of degree at most $|S|-1$ whose graph passes through all points of $S$. Otherwise, suppose $f$ and $g$ are two such polynomials. Then $... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2020-notes.jsonl",
"solution_match": null
} | 124 | 692 |
2020 | T1 | 6 | null | USAMO | Let $n \geq 2$ be an integer. Let $x_{1} \geq x_{2} \geq \cdots \geq x_{n}$ and $y_{1} \geq y_{2} \geq \cdots \geq y_{n}$ be $2 n$ real numbers such that $$ \begin{aligned} 0 & =x_{1}+x_{2}+\cdots+x_{n}=y_{1}+y_{2}+\cdots+y_{n} \\ \text { and } \quad 1 & =x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=y_{1}^{2}+y_{2}^{2}+\cdots... | $$ \begin{aligned} & \left(x_{i}\right)=(\underbrace{\frac{1}{\sqrt{n}}, \ldots, \frac{1}{\sqrt{n}}}_{n / 2}, \underbrace{-\frac{1}{\sqrt{n}}, \ldots,-\frac{1}{\sqrt{n}}}_{n / 2}) \\ & \left(y_{i}\right)=(\frac{n-1}{\sqrt{n(n-1)}}, \underbrace{\frac{1}{\sqrt{n(n-1)}}, \ldots,-\frac{1}{\sqrt{n(n-1)}}}_{n-1}) \end{aligne... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2020-notes.jsonl",
"solution_match": null
} | 238 | 710 |
2021 | T1 | 3 | null | USAMO | Let $n \geq 2$ be an integer. An $n \times n$ board is initially empty. Each minute, you may perform one of three moves: - If there is an L-shaped tromino region of three cells without stones on the board (see figure; rotations not allowed), you may place a stone in each of those cells.  P(x, y) & \in\left\langle 1+x+\cdots+x^{n-1}, 1+y+\cdots+y^{n-1}\ri... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2021-notes.jsonl",
"solution_match": null
} | 217 | 1,349 |
2021 | T1 | 6 | null | USAMO | Let $A B C D E F$ be a convex hexagon satisfying $\overline{A B}\|\overline{D E}, \overline{B C}\| \overline{E F}, \overline{C D} \| \overline{F A}$, and $$ A B \cdot D E=B C \cdot E F=C D \cdot F A $$ Let $X, Y$, and $Z$ be the midpoints of $\overline{A D}, \overline{B E}$, and $\overline{C F}$. Prove that the circu... | Claim - If $A B \cdot D E=B C \cdot E F=C D \cdot F A=k$, then the circumcenters of $A C E$ and $A^{\prime} C^{\prime} E^{\prime}$ coincide. Claim - Triangle $X Y Z$ is the vector average of the (congruent) medial triangles of triangles $A^{\prime} C^{\prime} E^{\prime}$ and $B^{\prime} D^{\prime} F^{\prime}$. $$ \begi... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2021-notes.jsonl",
"solution_match": null
} | 161 | 779 |
2022 | T1 | 2 | null | USAMO | Let $b \geq 2$ and $w \geq 2$ be fixed integers, and $n=b+w$. Given are $2 b$ identical black rods and $2 w$ identical white rods, each of side length 1. We assemble a regular $2 n$-gon using these rods so that parallel sides are the same color. Then, a convex $2 b$-gon $B$ is formed by translating the black rods, and... | We are going to prove that one may swap a black rod with an adjacent white rod (as well as the rods parallel to them) without affecting the difference in the areas of $B-W$. Let $\vec{u}$ and $\vec{v}$ denote the originally black and white vectors that were adjacent on the $2 n$-gon and are now going to be swapped. Let... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2022-notes.jsonl",
"solution_match": null
} | 273 | 515 |
2022 | T1 | 5 | null | USAMO | A function $f: \mathbb{R} \rightarrow \mathbb{R}$ is essentially increasing if $f(s) \leq f(t)$ holds whenever $s \leq t$ are real numbers such that $f(s) \neq 0$ and $f(t) \neq 0$. Find the smallest integer $k$ such that for any 2022 real numbers $x_{1}, x_{2}, \ldots, x_{2022}$, there exist $k$ essentially increasin... | The answer is 11 and, more generally, if 2022 is replaced by $N$ then the answer is $\left\lfloor\log _{2} N\right\rfloor+1$. 『 Bound. Suppose for contradiction that $2^{k}-1>N$ and choose $x_{n}=-n$ for each $n=1, \ldots, N$. Now for each index $1 \leq n \leq N$, define $$ S(n)=\left\{\text { indices } i \text { for w... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2022-notes.jsonl",
"solution_match": null
} | 182 | 698 |
2022 | T1 | 6 | null | USAMO | There are 2022 users on a social network called Mathbook, and some of them are Mathbook-friends. (On Mathbook, friendship is always mutual and permanent.) Starting now, Mathbook will only allow a new friendship to be formed between two users if they have at least two friends in common. What is the minimum number of fri... | With 2022 replaced by $n$, the answer is $\left\lceil\frac{3}{2} n\right\rceil-2$. 【 Terminology. Standard graph theory terms: starting from a graph $G$ on $n$ vertices, we're allowed to take any $C_{4}$ in the graph and complete it to a $K_{4}$. The problem asks the minimum number of edges needed so that this operatio... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2022-notes.jsonl",
"solution_match": null
} | 89 | 1,813 |
2023 | T1 | 1 | null | USAMO | In an acute triangle $A B C$, let $M$ be the midpoint of $\overline{B C}$. Let $P$ be the foot of the perpendicular from $C$ to $A M$. Suppose that the circumcircle of triangle $A B P$ intersects line $B C$ at two distinct points $B$ and $Q$. Let $N$ be the midpoint of $\overline{A Q}$. Prove that $N B=N C$. | \I III-advised barycentric approach (outline). Use reference triangle $A B C$. The $A$ median is parametrized by $(t: 1: 1)$ for $t \in \mathbb{R}$. So because of $\overline{C P} \perp \overline{A M}$, we are looking for $t$ such that $$ \left(\frac{t \vec{A}+\vec{B}+\vec{C}}{t+2}-\vec{C}\right) \perp\left(A-\frac{\vec... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2023-notes.jsonl",
"solution_match": null
} | 98 | 1,063 |
2023 | T1 | 3 | null | USAMO | Consider an $n$-by- $n$ board of unit squares for some odd positive integer $n$. We say that a collection $C$ of identical dominoes is a maximal grid-aligned configuration on the board if $C$ consists of $\left(n^{2}-1\right) / 2$ dominoes where each domino covers exactly two neighboring squares and the dominoes don't ... | The answer is that $$ k(C) \in\left\{1,2, \ldots,\left(\frac{n-1}{2}\right)^{2}\right\} \cup\left\{\left(\frac{n+1}{2}\right)^{2}\right\} $$ Index the squares by coordinates $(x, y) \in\{1,2, \ldots, n\}^{2}$. We say a square is special if it is empty or it has the same parity in both coordinates as the empty square. W... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2023-notes.jsonl",
"solution_match": null
} | 177 | 984 |
2023 | T1 | 5 | null | USAMO | Let $n \geq 3$ be an integer. We say that an arrangement of the numbers $1,2, \ldots, n^{2}$ in an $n \times n$ table is row-valid if the numbers in each row can be permuted to form an arithmetic progression, and column-valid if the numbers in each column can be permuted to form an arithmetic progression. For what valu... | Answer: $n$ prime only. So, look at the multiples of $p$ in a row-valid table; there is either 1 or $p$ per row. As there are $p$ such numbers total, there are two cases: - If all the multiples of $p$ are in the same row, then the common difference in each row is a multiple of $p$. In fact, it must be exactly $p$ for s... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2023-notes.jsonl",
"solution_match": null
} | 112 | 895 |
2023 | T1 | 6 | null | USAMO | Let $A B C$ be a triangle with incenter $I$ and excenters $I_{a}, I_{b}, I_{c}$ opposite $A, B$, and $C$, respectively. Given an arbitrary point $D$ on the circumcircle of $\triangle A B C$ that does not lie on any of the lines $I I_{a}, I_{b} I_{c}$, or $B C$, suppose the circumcircles of $\triangle D I I_{a}$ and $\t... | \I Barycentric coordinates (Carl Schildkraut). With reference triangle $\triangle A B C$, set $D=(r: s: t)$. Claim - The equations of $\left(D I I_{a}\right)$ and $\left(D I_{b} I_{c}\right)$ are, respectively, $$ \begin{aligned} & 0=-a^{2} y z-b^{2} z x-c^{2} x y+(x+y+z) \cdot\left(b c x-\frac{b c r}{c s-b t}(c y-b z)... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2023-notes.jsonl",
"solution_match": null
} | 159 | 829 |
2024 | T1 | 2 | null | USAMO | Let $S_{1}, S_{2}, \ldots, S_{100}$ be finite sets of integers whose intersection is not empty. For each non-empty $T \subseteq\left\{S_{1}, S_{2}, \ldots, S_{100}\right\}$, the size of the intersection of the sets in $T$ is a multiple of $|T|$. What is the smallest possible number of elements which are in at least 50 ... | The answer is $50\binom{100}{50}$. ब Rephrasing (cosmetic translation only, nothing happens yet). We encode with binary strings $v \in \mathbb{F}_{2}^{100}$ of length 100 . Write $v \subseteq w$ if $w$ has 1 's in every component $v$ does, and let $|v|$ denote the number of 1 's in $v$. Then for each $v$, we let $f(v)$... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2024-notes.jsonl",
"solution_match": null
} | 104 | 1,706 |
2024 | T1 | 3 | null | USAMO | Let $(m, n)$ be positive integers with $n \geq 3$ and draw a regular $n$-gon. We wish to triangulate this $n$-gon into $n-2$ triangles, each colored one of $m$ colors, so that each color has an equal sum of areas. For which $(m, n)$ is such a triangulation and coloring possible? | The answer is if and only if $m$ is a proper divisor of $n$. ## Lemma The triangle with vertices $\omega^{k}, \omega^{k+a}, \omega^{k+b}$ has signed area $$ T(a, b):=\frac{\left(\omega^{a}-1\right)\left(\omega^{b}-1\right)\left(\omega^{-a}-\omega^{-b}\right)}{2 i} . $$ $$ \frac{1}{2 i} \operatorname{det}\left[\begin{ar... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2024-notes.jsonl",
"solution_match": null
} | 83 | 1,066 |
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