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 |
|---|---|---|---|---|---|---|---|---|---|
2018 | T3 | A 7 | Algebra | JBMO-SL | Let $A$ be a set of positive integers with the following properties:
(a) If $n$ is an element of $A$ then $n \leqslant 2018$.
(b) If $S$ is a subset of $A$ with $|S|=3$ then there are two elements $n, m$ of $S$ with $|n-m| \geqslant \sqrt{n}+\sqrt{m}$.
What is the maximum number of elements that $A$ can have? | Assuming $n>m$ we have
$$
\begin{aligned}
|n-m| \geqslant \sqrt{n}+\sqrt{m} & \Leftrightarrow(\sqrt{n}-\sqrt{m})(\sqrt{n}+\sqrt{m}) \geqslant \sqrt{n}+\sqrt{m} \\
& \Leftrightarrow \sqrt{n} \geqslant \sqrt{m}+1 .
\end{aligned}
$$
Let $A_{k}=\left\{k^{2}, k^{2}+1, \ldots,(k+1)^{2}-1\right\}$. Note that each $A_{k}$ ca... | {
"problem_match": "\nA 7.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-jbmo2018.jsonl",
"solution_match": "\nSolution."
} | 112 | 665 |
2018 | T3 | C 2 | Combinatorics | JBMO-SL | A set $T$ of $n$ three-digit numbers has the following five properties:
(1) No number contains the digit 0 .
(2) The sum of the digits of each number is 9 .
(3) The units digits of any two numbers are different.
(4) The tens digits of any two numbers are different.
(5) The hundreds digits of any two numbers are di... | Let $S$ denote the set of three-digit numbers that have digit sum equal to 9 and no digit equal to 0 . We will first find the cardinality of $S$. We start from the number 111 and each element of $S$ can be obtained from 111 by a string of $6 A$ 's (which means that we add 1 to the current digit) and 2 G's (which means ... | {
"problem_match": "\nC 2.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-jbmo2018.jsonl",
"solution_match": "\nSolution."
} | 92 | 659 |
2018 | T3 | C 3 | Combinatorics | JBMO-SL | The cells of a $8 \times 8$ table are initially white. Alice and Bob play a game. First Alice paints $n$ of the fields in red. Then Bob chooses 4 rows and 4 columns from the table and paints all fields in them in black. Alice wins if there is at least one red field left. Find the least value of $n$ such that Alice can ... | We will show that the least value of $n$ is $n=13$.
If $n \leq 12$, Bob wins by painting black the 4 rows containing the highest numbers of red cells. Indeed, if at least 5 red cells remain, then one of the rows not blackened contains at least 2 red cells. Thus, each one of the rows blackened contained at least 2 red ... | {
"problem_match": "\nC 3.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-jbmo2018.jsonl",
"solution_match": "\nSolution."
} | 90 | 757 |
2018 | T3 | G 1 | Geometry | JBMO-SL | Let $H$ be the orthocentre of an acute triangle $A B C$ with $B C>A C$, inscribed in a circle $\Gamma$. The circle with centre $C$ and radius $C B$ intersects $\Gamma$ at the point $D$, which is on the arc $A B$ not containing $C$. The circle with centre $C$ and radius $C A$ intersects the segment $C D$ at the point $K... | We use standard notation for the angles of triangle $A B C$. Let $P$ be the midpoint of $C H$ and $O$ the centre of $\Gamma$. As
$$
\alpha=\angle B A C=\angle B D C=\angle D K L
$$
the quadrilateral $A C K L$ is cyclic. From the relation $C B=C D$ we get $\angle B C D=180^{\circ}-2 \alpha$, so
$$
\angle A C K=\gamma... | {
"problem_match": "\nG 1.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-jbmo2018.jsonl",
"solution_match": "\nSolution."
} | 166 | 607 |
2018 | T3 | G 3 | Geometry | JBMO-SL | Let $A B C$ be an acute triangle, $A^{\prime}, B^{\prime}, C^{\prime}$ the reflexions of the vertices $A, B$ and $C$ with respect to $B C, C A$, and $A B$, respectively, and let the circumcircles of triangles $A B B^{\prime}$ and $A C C^{\prime}$ meet again at $A_{1}$. Points $B_{1}$ and $C_{1}$ are defined similarly. ... | Let $O_{1}, O_{2}$ and $O$ be the circumcenters of triangles $A B B^{\prime}, A C C^{\prime}$ and $A B C$ respectively. As $A B$ is the perpendicular bisector of the line segment $C C^{\prime}, O_{2}$ is the intersection of the perpendicular bisector of $A C$ with $A B$. Similarly, $O_{1}$ is the intersection of the pe... | {
"problem_match": "\nG 3.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-jbmo2018.jsonl",
"solution_match": "\nSolution."
} | 138 | 514 |
2018 | T3 | G 4 | Geometry | JBMO-SL | Let $A B C$ be a triangle with side-lengths $a, b, c$, inscribed in a circle with radius $R$ and let $I$ be it's incenter. Let $P_{1}, P_{2}$ and $P_{3}$ be the areas of the triangles $A B I, B C I$ and $C A I$, respectively. Prove that
$$
\frac{R^{4}}{P_{1}^{2}}+\frac{R^{4}}{P_{2}^{2}}+\frac{R^{4}}{P_{3}^{2}} \geq 16... | Let $r$ be the radius of the inscribed circle of the triangle $A B C$. We have that
$$
P_{1}=\frac{r c}{2}, \quad P_{2}=\frac{r a}{2}, \quad P_{3}=\frac{r b}{2}
$$
It follows that
$$
\frac{1}{P_{1}^{2}}+\frac{1}{P_{2}^{2}}+\frac{1}{P_{3}^{2}}=\frac{4}{r^{2}}\left(\frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}}\right... | {
"problem_match": "\nG 4.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-jbmo2018.jsonl",
"solution_match": "\nSolution."
} | 139 | 768 |
2018 | T3 | G 5 | Geometry | JBMO-SL | Given a rectangle $A B C D$ such that $A B=b>2 a=B C$, let $E$ be the midpoint of $A D$. On a line parallel to $A B$ through point $E$, a point $G$ is chosen such that the area of $G C E$ is
$$
(G C E)=\frac{1}{2}\left(\frac{a^{3}}{b}+a b\right)
$$
Point $H$ is the foot of the perpendicular from $E$ to $G D$ and a po... | Let $L$ be the foot of the perpendicular from $G$ to $E C$ and let $Q$ the point of intersection of the lines $E G$ and $B C$. Then,
$$
(G C E)=\frac{1}{2} E C \cdot G L=\frac{1}{2} \sqrt{a^{2}+b^{2}} \cdot G L
$$
So, $G L=\frac{a}{b} \sqrt{a^{2}+b^{2}}$.
Observing that the triangles $Q C E$ and $E L G$ are similar,... | {
"problem_match": "\nG 5.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-jbmo2018.jsonl",
"solution_match": "\nSolution."
} | 195 | 529 |
2018 | T3 | G 6 | Geometry | JBMO-SL | Let $X Y$ be a chord of a circle $\Omega$, with center $O$, which is not a diameter. Let $P, Q$ be two distinct points inside the segment $X Y$, where $Q$ lies between $P$ and $X$. Let $\ell$ the perpendicular line dropped from $P$ to the diameter which passes through $Q$. Let $M$ be the intersection point of $\ell$ an... | by PSC. At first, we will allow $P$ and $Q$ to coincide, and we will prove the inequality in this case. Let the perpendicular from $Q$ to $O Q$ meet $\Omega$ at $B$ and $C$. Then, we have that $Q B=Q C$. We will show that
$$
B Q \cdot X Y \geq 2 Q X \cdot Q Y
$$
By the power of a point Theorem we have that
$$
Q X \c... | {
"problem_match": "\nG 6.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-jbmo2018.jsonl",
"solution_match": "\nSolution "
} | 126 | 1,669 |
2018 | T3 | NT 3 | Number Theory | JBMO-SL | Find all four-digit positive integers $\overline{a b c d}=10^{3} a+10^{2} b+10 c+d$, with $a \neq 0$, such that
$$
\overline{a b c d}=a^{a+b+c+d}-a^{-a+b-c+d}+a
$$ | It is obvious that $a \neq 1$ and $-a+b-c+d \geq 0$. It follows that $b+d \geq c+a \geq a$. Then,
$$
\begin{aligned}
10000>\overline{a b c d} & =a^{a+b+c+d}-a^{-a+b-c+d}+a \\
& >a^{a+b+c+d}-a^{a+b+c+d-2} \\
& =a^{a+b+c+d-2}\left(a^{2}-1\right) \\
& \geq a^{2 a-2}\left(a^{2}-1\right)
\end{aligned}
$$
For $a \geq 4$, w... | {
"problem_match": "\nNT 3.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-jbmo2018.jsonl",
"solution_match": "\nSolution."
} | 75 | 807 |
2016 | T3 | A4 | Algebra | JBMO-SL | If $x, y, z$ are non-negative real numbers such that $x^{2}+y^{2}+z^{2}=x+y+z$, then show that:
$$
\frac{x+1}{\sqrt{x^{5}+x+1}}+\frac{y+1}{\sqrt{y^{5}+y+1}}+\frac{z+1}{\sqrt{z^{5}+z+1}} \geq 3
$$
When does the equality hold? | First we factor $x^{5}+x+1$ as follows:
$$
\begin{aligned}
x^{5}+x+1 & =x^{5}-x^{2}+x^{2}+x+1=x^{2}\left(x^{3}-1\right)+x^{2}+x+1=x^{2}(x-1)\left(x^{2}+x+1\right)+x^{2}+x+1 \\
& =\left(x^{2}+x+1\right)\left(x^{2}(x-1)+1\right)=\left(x^{2}+x+1\right)\left(x^{3}-x^{2}+1\right)
\end{aligned}
$$
Using the $A M-G M$ inequ... | {
"problem_match": "\nA4.",
"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl",
"solution_match": "\nSolution."
} | 109 | 1,343 |
2016 | T3 | A5 | Algebra | JBMO-SL | Let $x, y, z$ be positive real numbers such that $x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$.
a) Prove the inequality
$$
x+y+z \geq \sqrt{\frac{x y+1}{2}}+\sqrt{\frac{y z+1}{2}}+\sqrt{\frac{z x+1}{2}}
$$
b) (Added by the problem selecting committee) When does the equality hold? | a) We rewrite the inequality as
$$
(\sqrt{x y+1}+\sqrt{y z+1}+\sqrt{z x+1})^{2} \leq 2 \cdot(x+y+z)^{2}
$$
and note that, from CBS,
$$
\text { LHS } \leq\left(\frac{x y+1}{x}+\frac{y z+1}{y}+\frac{z x+1}{z}\right)(x+y+z)
$$
But
$$
\frac{x y+1}{x}+\frac{y z+1}{y}+\frac{z x+1}{z}=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{... | {
"problem_match": "\nA5.",
"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl",
"solution_match": "## Solution."
} | 108 | 802 |
2016 | T3 | C1 | Combinatorics | JBMO-SL | Let $S_{n}$ be the sum of reciprocal values of non-zero digits of all positive integers up to (and including) $n$. For instance, $S_{13}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{2}+\frac{1}{1}+\fr... | We will first calculate $S_{999}$, then $S_{1999}-S_{999}$, and then $S_{2016}-S_{1999}$.
Writing the integers from 1 to 999 as 001 to 999, adding eventually also 000 (since 0 digits actually do not matter), each digit appears exactly 100 times in each position(as unit, ten, or hundred). Hence
$$
S_{999}=300 \cdot\le... | {
"problem_match": "\nC1.",
"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl",
"solution_match": "## Solution."
} | 176 | 829 |
2016 | T3 | C3 | Combinatorics | JBMO-SL | Consider any four pairwise distinct real numbers and write one of these numbers in each cell of a $5 \times 5$ array so that each number occurs exactly once in every $2 \times 2$ subarray. The sum over all entries of the array is called the total sum of that array. Determine the maximum number of distinct total sums th... | We will prove that the maximum number of total sums is 60 .
The proof is based on the following claim.
Claim. Either each row contains exactly two of the numbers, or each column contains exactly two of the numbers.
Proof of the Claim. Indeed, let $R$ be a row containing at least three of the numbers. Then, in row $R... | {
"problem_match": "\nC3.",
"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl",
"solution_match": "\nSolution."
} | 78 | 1,146 |
2016 | T3 | C4 | Combinatorics | JBMO-SL | A splitting of a planar polygon is a finite set of triangles whose interiors are pairwise disjoint, and whose union is the polygon in question. Given an integer $n \geq 3$, determine the largest integer $m$ such that no planar $n$-gon splits into less than $m$ triangles. | The required maximum is $\lceil n / 3\rceil$, the least integer greater than or equal to $n / 3$. To describe a planar $n$-gon splitting into this many triangles, write $n=3 m-r$, where $m$ is a positive integer and $r=0,1,2$, and consider $m$ coplanar equilateral triangles $A_{3 i} A_{3 i+1} A_{3 i+2}$, $i=0, \ldots, ... | {
"problem_match": "\nC4.",
"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl",
"solution_match": "\nSolution."
} | 65 | 1,034 |
2016 | T3 | G4 | Geometry | JBMO-SL | Let $A B C$ be an acute angled triangle whose shortest side is $[B C]$. Consider a variable point $P$ on the side $[B C]$, and let $D$ and $E$ be points on $(A B]$ and $(A C]$, respectively, such that $B D=B P$ and $C P=C E$. Prove that, as $P$ traces $[B C]$, the circumcircle of the triangle $A D E$ passes through a f... | We claim that the fixed point is the center of the incircle of $A B C$.
Let $I$ be the center of the incircle of $A B C$. Since $B D=B P$ and $[B I$ is the bisector of $\widehat{D B P}$, the line $B I$ is the perpendicular bisector of $[D P]$. This yields $D I=P I$. Analogously we get $E I=P I$. So, the point $I$ is t... | {
"problem_match": "\nG4.",
"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl",
"solution_match": "\nSolution."
} | 183 | 576 |
2016 | T3 | G6 | Geometry | JBMO-SL | Given an acute triangle $A B C$, erect triangles $A B D$ and $A C E$ externally, so that $m(\widehat{A D B})=m(\widehat{A E C})=90^{\circ}$ and $\widehat{B A D} \equiv \widehat{C A E}$. Let $A_{1} \in B C, B_{1} \in A C$ and $C_{1} \in A B$ be the feet of the altitudes of the triangle $A B C$, and let $K$ and $L$ be th... | Let $M, P$ and $Q$ be the midpoints of $[B C],[C A]$ and $[A B]$, respectively.
The circumcircle of the triangle $A_{1} B_{1} C_{1}$ is the Euler's circle. So, the point $M$ lies on this circle.
It is enough to prove now that $\left[A_{1} M\right]$ is a common chord of the three circles $\left(A_{1} B_{1} C_{1}\right... | {
"problem_match": "\nG6.",
"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl",
"solution_match": "\nSolution."
} | 268 | 663 |
2016 | T3 | G7 | Geometry | JBMO-SL | Let $[A B]$ be a chord of a circle $(c)$ centered at $O$, and let $K$ be a point on the segment $(A B)$ such that $A K<B K$. Two circles through $K$, internally tangent to (c) at $A$ and $B$, respectively, meet again at $L$. Let $P$ be one of the points of intersection of the line $K L$ and the circle (c), and let the ... | Let $\left(c_{1}\right)$ and $\left(c_{2}\right)$ be circles through $K$, internally tangent to (c) at $A$ and $B$, respectively, and meeting again at $L$, and let the common tangent to $\left(c_{1}\right)$ and $(c)$ meet the common tangent to $\left(c_{2}\right)$ and $(c)$ at $Q$. Then the point $Q$ is the radical cen... | {
"problem_match": "\nG7.",
"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl",
"solution_match": "\nSolution."
} | 206 | 827 |
2019 | T1 | 7 | null | NewZealand_MO | Let \(ABCDEF\) be a convex hexagon containing a point \(P\) in its interior such that \(PABC\) and \(PDEF\) are congruent rectangles with \(PA = BC = PD = EF\) (and \(AB = PC = DE = PF\) ). Let \(\ell\) be the line through the midpoint of \(AF\) and the circumcentre of \(PCD\) . Prove that \(\ell\) passes through \(P\)... | : Let \(M\) be the midpoint of \(AF\) and let \(O\) be the circumcentre of triangle \(CPD\) . Now construct \(Q\) to be the point such that \(CPDQ\) is a parallelogram, and let \(R\) be the centre of this parallelogram (i.e. \(R\) is the intersection of \(PQ\) with \(CD\) , and also \(R\) is the midpoint of \(PQ\) ).
... | {
"problem_match": "\n7.",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2019_solutions.jsonl",
"solution_match": "\nSolution"
} | 94 | 594 |
2019 | T1 | 8 | null | NewZealand_MO | Suppose that \(x_{1},x_{2},x_{3},\ldots x_{n}\) are real numbers between 0 and 1 with sum \(s\) . Prove that
\[\sum_{i = 1}^{n}{\frac{x_{i}}{s + 1 - x_{i}}} + \prod_{i = 1}^{n}(1 - x_{i})\leq 1.\] | : Let \(i\) be arbitrary and consider the set \(A = \{a_{1},a_{2},\ldots ,a_{n}\}\) defined by \(a_{i} = s + 1 - x_{i}\) and let \(a_{j} = 1 - x_{j}\) for all \(j\neq i\) . For example, if \(i = 2\) then \(A\) would be \(\{1 - x_{1},s + 1 - x_{2},1 - x_{3},\ldots ,1 - x_{n}\}\) . The AM- GM inequality on \(A\) tells us... | {
"problem_match": "\n8.",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2019_solutions.jsonl",
"solution_match": "\nSolution"
} | 96 | 586 |
2020 | T1 | 6 | null | NewZealand_MO | Let \(\triangle ABC\) be an acute triangle with \(AB > AC\) . Let \(P\) be the foot of the altitude from \(C\) to \(AB\) and let \(Q\) be the foot of the altitude from \(B\) to \(AC\) . Let \(X\) be the intersection of \(PQ\) and \(BC\) . Let the intersection of the circumcircles of triangle \(\triangle AXC\) and trian... | First we will embed the diagram in the Argand plane, such that point \(B\) is represented by the complex number \(b = - 1\) and point \(C\) is represented by the complex number \(c = 1\) . Lower case letters will always denote the complex number representing the corresponding upper case letter (so \(a\) is the complex ... | {
"problem_match": "\n6. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl",
"solution_match": "# Alternative Solution (outline):"
} | 121 | 1,904 |
2020 | T1 | 7 | null | NewZealand_MO | Josie and Ross are playing a game on a \(20 \times 20\) chessboard. Initially the chessboard is empty. The two players alternately take turns, with Josie going first. On Josie's turn, she selects any two different empty cells, and places one white stone in each of them. On Ross' turn, he chooses any one white stone cur... | : Ross can't stop Josie winning — Josie has a strategy in which she can ensure that there will be 8 white stones in a row. We will give an explicit example of such a strategy.
To simplify notation, we define a \(k\) - strip to be a \(1 \times 8\) rectangle, in which the first \(k\) cells are filled with white stones ... | {
"problem_match": "\n7. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl",
"solution_match": "\nSolution"
} | 140 | 526 |
2020 | T1 | 8 | null | NewZealand_MO | For a positive integer \(x\) , define a sequence \(a_{0}, a_{1}, a_{2}, \ldots\) according to the following rules: \(a_{0} = 1\) , \(a_{1} = x + 1\) and
\[a_{n + 2} = x a_{n + 1} - a_{n}\qquad \mathrm{for~all~}n\geq 0.\]
Prove that there exist infinitely many positive integers \(x\) such that this sequence does n... | : For each integer \(n \geq 0\) and \(x > 2\) , we recursively define integers \(a_{n}(x)\) and \(b_{n}(x)\) . Let \(a_{0}(x) = b_{0}(x) = 1\) and \(a_{1}(x) = x + 1\) and \(b_{1}(x) = x - 1\) . For all \(n \geq 0\) let
\[a_{n + 2}(x) = x a_{n + 1}(x) - a_{n}(x)\quad \mathrm{and}\quad b_{n + 2}(x) = x b_{n + 1}(x) - ... | {
"problem_match": "\n8. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl",
"solution_match": "\nSolution"
} | 121 | 1,031 |
2020 | T1 | 8 | null | NewZealand_MO | For a positive integer \(x\) , define a sequence \(a_{0}, a_{1}, a_{2}, \ldots\) according to the following rules: \(a_{0} = 1\) , \(a_{1} = x + 1\) and
\[a_{n + 2} = x a_{n + 1} - a_{n}\qquad \mathrm{for~all~}n\geq 0.\]
Prove that there exist infinitely many positive integers \(x\) such that this sequence does n... | First note that (for \(x \geq 3\) ) each term in the sequence is more than double the previous term, because
\[a_{n + 1} = x a_{n} - a_{n - 1} = (x - 1)a_{n} + (a_{n} - a_{n - 1})\geq (x - 1)a_{n}\geq 2a_{n}.\]
Moreover we can easily verify that \(a_{n} > x\) for all \(n \geq 1\) because \(a_{n} \geq a_{1} = x + 1\... | {
"problem_match": "\n8. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2020_solutions.jsonl",
"solution_match": "\nAlternative Solution:"
} | 121 | 1,224 |
2021 | T1 | 3 | null | NewZealand_MO | In a sequence of numbers, a term is called golden if it is divisible by the term immediately before it. What is the maximum possible number of golden terms in a permutation of \(1,2,3,\ldots ,2021?\) | : Let \(k\) be the number of golden terms. We claim that \(k\leq 1010\)
Proof: Define the term immediately before a golden term to be a silver term. The number of silver terms is also \(k\) . If \(a\) is any silver term and \(b\) is the corresponding golden term then we must have
\[a\leq \frac{b}{2}\leq \frac{2021}... | {
"problem_match": "\n3. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2021_solutions.jsonl",
"solution_match": "\nSolution"
} | 51 | 766 |
2021 | T1 | 6 | null | NewZealand_MO | Is it possible to place a positive integer in every cell of a \(10 \times 10\) array in such a way that both the following conditions are satisfied?
- Each number (not in the top row) is a proper divisor of the number immediately below.
- Each row consists of 10 consecutive positive integers (but not necessarily ... | : Answer: Yes. In fact it is even possible to achieve such an array where each row consists of ten consecutive positive integers in increasing order. We shall construct an example explicitly.
Initially let the top row be \((1,2,3,\ldots ,10)\) in this order. Then iteratively if the contents of a particular row are
... | {
"problem_match": "\n6. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2021_solutions.jsonl",
"solution_match": "\nSolution"
} | 75 | 701 |
2021 | T1 | 7 | null | NewZealand_MO | Let \(a, b, c, d\) be integers such that \(a > b > c > d \geq -2021\) and
\[\frac{a + b}{b + c} = \frac{c + d}{d + a}\]
(and \(b + c \neq 0 \neq d + a\) ). What is the maximum possible value of \(ac\) ? | : We claim that the maximum value of \(ac\) is \(2 \times 505^{2} = 510050\) , and this is uniquely achieved by \((a, b, c, d) = (1010, 506, 505, - 2021)\) .
To prove this we start by rearranging the expression \(\frac{a + b}{b + c} = \frac{c + d}{d + a}\) to get \((c + b)(c + d) = (a + b)(a + d)\) . Now expand the b... | {
"problem_match": "\n7. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2021_solutions.jsonl",
"solution_match": "\nSolution"
} | 88 | 763 |
2022 | T1 | 4 | null | NewZealand_MO | On a table, there is an empty bag and a chessboard containing exactly one token on each square. Next to the table is a large pile that contains an unlimited supply of tokens. Using only the following types of moves what is the maximum possible number of tokens that can be in the bag?
- Type 1: Choose a non-empty squ... | A: (Ishan Nath)
Let \(a_{i,j}\) be the number of tokens in the square in the \(i^{\mathrm{th}}\) row and \(j^{\mathrm{th}}\) column, where the first row is the topmost row and the first column is the leftmost column. Furthermore let \(b\) denote the number of tokens in the bag. We define a monovariant as follows.
\... | {
"problem_match": "\n4. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl",
"solution_match": "\nSolution"
} | 184 | 1,266 |
2022 | T1 | 4 | null | NewZealand_MO | On a table, there is an empty bag and a chessboard containing exactly one token on each square. Next to the table is a large pile that contains an unlimited supply of tokens. Using only the following types of moves what is the maximum possible number of tokens that can be in the bag?
- Type 1: Choose a non-empty squ... | B: (Michael Albert)
First note that we may assume that all type 3 moves are performed after all the type 1 and type 2 moves have been made. WLOG We may also assume that all moves (type 1 and 2) into square \((8,8)\) occur all other type 1 and type 2 moves.
Claim: In an optimal solution no type 3 move is made except... | {
"problem_match": "\n4. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl",
"solution_match": "\nSolution"
} | 184 | 698 |
2022 | T1 | 5 | null | NewZealand_MO | A round-robin tournament is one where each team plays every other team exactly once. Five teams take part in such a tournament getting: 3 points for a win, 1 point for a draw and 0 points for a loss. At the end of the tournament the teams are ranked from first to last according to the number of points.
(a) Is it pos... | : (Ishan Nath)
We show the answer is no for five teams, and yes for six.
First, we show five teams cannot have this property. Suppose the teams had points \(x, x + 2, x + 4, x + 6, x + 8\) in the final ranking. If there were \(d\) draws and \(10 - d\) decisive games, then the total number of points is
\[5x + 20 =... | {
"problem_match": "\n5. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl",
"solution_match": "\nSolution"
} | 128 | 742 |
2022 | T1 | 6 | null | NewZealand_MO | Let a positive integer \(n\) be given. Determine, in terms of \(n\) , the least positive integer \(k\) such that among any \(k\) positive integers, it is always possible to select a positive even number of them having sum divisible by \(n\) . | : (Ethan Ng)
First we consider the \(n\) odd case. If \(n\) is odd we claim the answer is \(k = 2n\) .
Let the numbers be \(x_{1}, x_{2}, x_{3}, \ldots , x_{2n}\) . Consider the following partial sums:
\[s_{1} = x_{1} + x_{2}\] \[s_{2} = x_{1} + x_{2} + x_{3} + x_{4}\] \[\vdots\] \[s_{i} = x_{1} + x_{2} + x_{3} +... | {
"problem_match": "\n6. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2022_solutions.jsonl",
"solution_match": "\nSolution"
} | 56 | 1,289 |
2023 | T1 | 4 | null | NewZealand_MO | Let \(p\) be a prime and let \(f(x) = ax^{2} + bx + c\) be a quadratic polynomial with integer coefficients such that \(0 < a, b, c \leq p\) . Suppose \(f(x)\) is divisible by \(p\) whenever \(x\) is a positive integer. Find all possible values of \(a + b + c\) . | : (James Xu)
First substitute \(x = p\) , to get \(p \mid ap^{2} + bp + c\) so \(p \mid c\) . Therefore \(c = p\) . Next substitute \(x = 1\) , to get \(p \mid a + b + c\) . Since \(c = p\) , this gives us \(p \mid a + b\) . Finally substitute \(x = p - 1\) , to get \((p - 1)^{2}a - (p - 1)b + c \equiv 0\) (mod \(p\)... | {
"problem_match": "\n4. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl",
"solution_match": "\nSolution"
} | 80 | 632 |
2023 | T1 | 6 | null | NewZealand_MO | Let triangle \(ABC\) be right-angled at \(A\) . Let \(D\) be the point on \(AC\) such that \(BD\) bisects angle \(\angle ABC\) . Prove that \(BC - BD = 2AB\) if and only if \(\frac{1}{BD} - \frac{1}{BC} = \frac{1}{2AB}\) . | : (Kevin Shen)
Wlog let \(AB = 1\) and \(BC = a\) . Also let \(BD = x\) . We will try to find all the lengths in the diagram in terms of \(a\) .
By Pythagoras in \(\triangle ABC\) we get \(AC = \sqrt{a^2 - 1}\) . By the angle- bisector ... | {
"problem_match": "\n6. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl",
"solution_match": "\nSolution"
} | 83 | 1,866 |
2023 | T1 | 7 | null | NewZealand_MO | Let \(n,m\) be positive integers. Let \(A_{1},A_{2},A_{3},\ldots ,A_{m}\) be sets such that \(A_{i}\subseteq \{1,2,3,\ldots ,n\}\) and \(|A_{i}| = 3\) for all \(i\) (i.e. \(A_{i}\) consists of three different positive integers each at most \(n\) ). Suppose for all \(i< j\) we have
\[|A_{i}\cap A_{j}|\leqslant 1\]
... | : (Ishan Nath)
Each set \(A_{i}\) has exactly three pairs of elements. But each unordered pair chosen from \(\{1,2,\ldots ,n\}\) can be in at most one such set. Therefore
\[{\binom{n}{2}}\geq3m.\]
This establishes the required upper bound on the size of \(m\) .
Let \(T\) be the set of all triples, i.e.
\[T = ... | {
"problem_match": "\n7. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl",
"solution_match": "\nSolution"
} | 223 | 628 |
2023 | T1 | 8 | null | NewZealand_MO | Find all non-zero real numbers \(a, b, c\) such that the following polynomial has four (not necessarily distinct) positive real roots.
\[P(x) = ax^{4} - 8ax^{3} + bx^{2} - 32cx + 16c\] | : (Viet Hoang)
Assume that \(P(x)\) has 4 positive real roots \(x_{1}, x_{2}, x_{3}\) and \(x_{4}\) . Using Viete's theorem, one can obtain the following equations
\[\begin{array}{c}{x_{1} + x_{2} + x_{3} + x_{4} = \frac{8a}{a} = 8}\\ {x_{1}x_{2} + x_{1}x_{3} + x_{1}x_{4} + x_{2}x_{3} + x_{2}x_{4} + x_{3}x_{4} = \f... | {
"problem_match": "\n8. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2023_solutions.jsonl",
"solution_match": "\nSolution"
} | 63 | 818 |
2024 | T1 | 4 | null | NewZealand_MO | A dot-trapezium consists of several rows of dots such that each row contains one more dot than the row immediately above (apart from the top row). For example here is a dot-trapezium consisting of 15 dots, having 3 rows and 4 dot in the top row.
Let \(n\) be a trapezium number and suppose there are \(a\) dots in the first row and \(b\) dots in the last row. So the required conditions are \(a \geq 2\) and \(b \geq a + 1\) . Then the equation becomes:
\[2n = b(b + 1) - a(a - 1) = b^{2} - a^{2} + b + a = (a + b)(b - a + 1)\]
because it is... | {
"problem_match": "\n4. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl",
"solution_match": "\nSolution"
} | 195 | 2,372 |
2024 | T1 | 6 | null | NewZealand_MO | Let \(\omega\) be the incircle of scalene triangle \(ABC\) . Let \(\omega\) be tangent to \(AB\) and \(AC\) at points \(X\) and \(Y\) . Construct points \(X'\) and \(Y'\) on line segments \(AB\) and \(AC\) respectively such that \(AX' = XB\) and \(AY' = YC\) . Let line \(CX'\) intersects \(\omega\) at points \(P, Q\) s... | : (Ross Atkins)
Let \(a, b, c\) be the sidelengths \(BC, AC, AB\) respectively, and let \(s\) be the semiperimeter of triangle \(ABC\) (i.e. let \(s = \frac{a + b + c}{2}\) ). Since \(X\) and \(Y\) are the points of contact of the incircle we get \(AX = AY\) . Similarly \(BX = BZ\) and \(CY = CZ\) where \(Z\) is the ... | {
"problem_match": "\n6. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl",
"solution_match": "\nSolution"
} | 145 | 1,029 |
2024 | T1 | 7 | null | NewZealand_MO | Some of the 80960 lattice points in a \(40 \times 2024\) lattice are coloured red. It is known that no four red lattice points are vertices of a rectangle with sides parallel to the axes of the lattice. What is the maximum possible number of red points in the lattice? | : (Ross Atkins)
Let \(a_{1},a_{2},a_{3},\ldots ,a_{2024}\) be the number of red dots in rows \(1,2,3,\ldots ,2024\) respectively. So \(0\leqslant a_{i}\leqslant 40\) for each \(i\)
For each of the \(\binom{40}{2}=780\) pairs of columns, there can be at most one row with a red dot in both columns. Therefore we must ... | {
"problem_match": "\n7. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2024_solutions.jsonl",
"solution_match": "\nSolution"
} | 66 | 695 |
2025 | T1 | 4 | null | NewZealand_MO | Find the largest integer \(k\) such that any string of 2025 letters consisting only of A's and B's contains a palindromic substring of length \(k\) or longer. A palindromic substring is a string of consecutive letters which reads the same backwards as forwards. | : (Kevin Shen)
We claim that the largest integer is 4. We first prove that all strings \(S\) of 2025 letters contain a palindromic substring of length 4 or longer, which implies that \(k \geq 4\) . Then we shall provide a construction to show that \(k\) cannot be 5 or more.
We first begin by proving \(k \geq 4\) . ... | {
"problem_match": "\n4. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl",
"solution_match": "\nSolution"
} | 61 | 651 |
2025 | T1 | 5 | null | NewZealand_MO | Alice plays a game with the Mad Hatter. The Mad Hatter will write two rows of numbers on a blackboard, each a distinct permutation of \(\{1,2,\ldots ,n\}\) . On each move, Alice is allowed to swap the positions of the numbers \(a\) and \(a + 1\) in the first row, for some \(1 \leq a < n\) . What is the minimum number o... | : (Tony Wang)
We will show that the answer is \(\binom{n}{2} = \frac{n(n - 1)}{2}\) . To show that this is sufficient, we will use induction.
- Base Case: Note that when \(n = 1\) , the two rows of numbers must be the same since there is only one permutation of \(\{1\}\) . Hence this case takes \(0 = \binom{1}{2}\)... | {
"problem_match": "\n5. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl",
"solution_match": "\nSolution"
} | 122 | 812 |
2025 | T1 | 6 | null | NewZealand_MO | Determine the largest real number \(M\) such that for each infinite sequence \(x_{0}, x_{1}, x_{2}, \ldots\) of real numbers satisfying \(x_{0} = 1\) , \(x_{1} = 3\) and
\[x_{0} + x_{1} + \dots +x_{n - 1}\geq 3x_{n} - x_{n + 1}\quad \mathrm{for~all~}n\geq 1,\]
the inequality
\[\frac{x_{n + 1}}{x_{n}} >M\]
h... | : (Eric Liang)
We claim that the largest real number \(M\) is 2.
First we prove by induction on \(n\) that \(\frac{x_{n}}{x_{i}} > 2^{n - i}\) for all \(i \leq n - 1\) .
Base case: For \(n = 1\) , \(\frac{x_{1}}{x_{0}} = 3 > 2\) .
Inductive step: Suppose the inductive hypothesis holds true for \(n = k\) , i.e. ... | {
"problem_match": "\n6. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl",
"solution_match": "\nSolution"
} | 147 | 807 |
2025 | T1 | 7 | null | NewZealand_MO | Let \(ABC\) be a triangle and let \(D\) be a point inside the triangle \(ABC\) such that \(AD\) bisects \(\angle BAC\). Let line \(BD\) meet side \(AC\) at \(E\). Let line \(CD\) meet side \(AB\) at \(F\). Let \(T\) be the intersection of the (internal) angle bisectors of \(\angle AED\) and \(\angle AFD\). Prove that i... | 1: (Nico McKinlay)
Assume \(T\) lies on \(AD\) . Then by the angle bisector theorem in triangles \(AED\) and \(AFD\) , we have
\[\frac{AE}{ED} = \frac{AT}{TD} = \frac{AF}{FD}.\]
Applying the angle bisector theorem in triangles \(AEB\)... | {
"problem_match": "\n7. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo1_2025_solutions.jsonl",
"solution_match": "\nSolution"
} | 121 | 503 |
2020 | T1 | 5 | null | NewZealand_MO | A sequence of \(A s\) and \(B s\) is called antipalindromic if writing it backwards, then turning all the \(A s\) into \(B s\) and vice versa, produces the original sequence. For example \(A B B A A B\) is antipalindromic. For any sequence of \(A s\) and \(B s\) we define the cost of the sequence to be the product of t... | : For each integer \(0\leq k\leq 1009\) define a \(k\) - pal to be any sequence of 2020 \(A s\) and \(B s\) , where the first \(k\) terms are \(B\) , the last \(k\) terms are \(B\) , and the middle \((2020 - 2k)\) terms form an antipalindromic sequence.
Now for any \(k\) , define \(f(k)\) to be sum of the costs of al... | {
"problem_match": "\n5. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo2_2020_solutions.jsonl",
"solution_match": "\nSolution"
} | 153 | 501 |
2020 | T1 | 5 | null | NewZealand_MO | A sequence of \(A s\) and \(B s\) is called antipalindromic if writing it backwards, then turning all the \(A s\) into \(B s\) and vice versa, produces the original sequence. For example \(A B B A A B\) is antipalindromic. For any sequence of \(A s\) and \(B s\) we define the cost of the sequence to be the product of t... | Let \(n\) be a positive integer. We will find an expression (in terms of \(n\) ) for the sum of the costs of all antipalindromes of length \(2n\) . Note that a string of \(A s\) and \(B s\) of length \(2n\) is an antipalindrome if and only if for each \(i\) , exactly one of the \(i^{\mathrm{th}}\) and \((2n + 1 - i)^{\... | {
"problem_match": "\n5. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo2_2020_solutions.jsonl",
"solution_match": "# Alternative Solution:"
} | 153 | 593 |
2020 | T1 | 5 | null | NewZealand_MO | A sequence of \(A s\) and \(B s\) is called antipalindromic if writing it backwards, then turning all the \(A s\) into \(B s\) and vice versa, produces the original sequence. For example \(A B B A A B\) is antipalindromic. For any sequence of \(A s\) and \(B s\) we define the cost of the sequence to be the product of t... | Let \(n\) be a positive integer. We will find an expression (in terms of \(n\) ) for the sum of the costs of all antipalindromes of length \(2n\) . Let \(\mathcal{P}\) denote the set of all antipalindromes of length \(2n\) , and let \(P\) be an antipalindrome chosen uniformly from \(\mathcal{P}\) . Note that for each \... | {
"problem_match": "\n5. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo2_2020_solutions.jsonl",
"solution_match": "# Alternative Solution 2:"
} | 153 | 839 |
2021 | T1 | 1 | null | NewZealand_MO | Let \(ABCD\) be a convex quadrilateral such that \(AB + BC = 2021\) and \(AD = CD\) . We are also given that
\[\angle ABC = \angle CDA = 90^{\circ}\]
Determine the length of the diagonal \(BD\) . | First let \(x = AD = DC\) and \(a = BD\) and \(y = AB\) and \(z = BC\) . Now initially we can apply Pythagoras in triangles \(CDA\) and \(ABC\) to get \(x^{2} + x^{2} = AC^{2}\) and \(y^{2} + z^{2} = AC^{2}\) respectively. Putting this together gives us
\[x^{2} = \frac{y^{2} + z^{2}}{2}.\]
Now note that the opposit... | {
"problem_match": "\n1. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo2_2021_solutions.jsonl",
"solution_match": "# Alternative Solution:"
} | 66 | 676 |
2021 | T1 | 3 | null | NewZealand_MO | Let \(\{x_{1},x_{2},x_{3},\ldots ,x_{n}\}\) be a set of \(n\) distinct positive integers, such that the sum of any 3 of them is a prime number. What is the maximum value of \(n\) ? | : First we show that \(n = 4\) is possible with an example. The example \(\{x_{1},x_{2},x_{3},x_{4}\} = \{1,3,7,9\}\) satisfies the problem because:
\(\cdot 1 + 3 + 7 = 11\) is prime,
\(\cdot 1 + 3 + 9 = 13\) is prime,
\(\cdot 1 + 7 + 9 = 17\) is prime, and
\(\cdot 3 + 7 + 9 = 19\) is prime.
We still have to ... | {
"problem_match": "\n3. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo2_2021_solutions.jsonl",
"solution_match": "\nSolution"
} | 62 | 583 |
2022 | T1 | 3 | null | NewZealand_MO | Let \(S\) be a set of 10 positive integers. Prove that one can find two disjoint subsets \(A = \{a_{1}, \ldots , a_{k}\}\) and \(B = \{b_{1}, \ldots , b_{k}\}\) of \(S\) with \(|A| = |B|\) such that the sums
\[x = \frac{1}{a_{1}} +\dots +\frac{1}{a_{k}}\]
and
\[y = \frac{1}{b_{1}} +\dots +\frac{1}{b_{k}}\]
... | : (Ishan Nath)
Partition the interval \((0.00, 2.50]\) into 250 intervals each of size 0.01.
\[(0.00, 2.50] = (0.00, 0.01] \cup (0.01, 0.02] \cup (0.02, 0.03] \cup \dots \cup (2.49, 2.50].\]
Now consider all possible sets, \(S\) , we can choose from the given 10 positive integers with \(|S| = 5\) . Because each o... | {
"problem_match": "\n3. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo2_2022_solutions.jsonl",
"solution_match": "\nSolution"
} | 168 | 585 |
2022 | T1 | 5 | null | NewZealand_MO | The sequence \(x_{1},x_{2},x_{3},\ldots\) is defined by \(x_{1} = 2022\) and \(x_{n + 1} = 7x_{n} + 5\) for all positive integers \(n\) . Determine the maximum positive integer \(m\) such that
\[{\frac{x_{n}(x_{n}-1)(x_{n}-2)\ldots(x_{n}-m+1)}{m!}}\]
is never a multiple of 7 for any positive integer \(n\) . | A: (Ishan Nath)
We claim the answer is 404. First, we notice that \(m\leq 2022\) . Otherwise,
\[{\frac{x_{1}(x_{1}-1)\cdots(x_{1}-m+1)}{m!}}={\frac{2022(2022-1)\cdots(2022-m+1)}{m!}}=0,\]
which is a multiple of 7. Then, since \(x_{n}\geq x_{1} = 2022\) for all \(n\) , we can write
\[{\frac{x_{n}(x_{n}-1)\cdots(... | {
"problem_match": "\n5. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo2_2022_solutions.jsonl",
"solution_match": "\nSolution"
} | 124 | 2,737 |
2022 | T1 | 5 | null | NewZealand_MO | The sequence \(x_{1},x_{2},x_{3},\ldots\) is defined by \(x_{1} = 2022\) and \(x_{n + 1} = 7x_{n} + 5\) for all positive integers \(n\) . Determine the maximum positive integer \(m\) such that
\[{\frac{x_{n}(x_{n}-1)(x_{n}-2)\ldots(x_{n}-m+1)}{m!}}\]
is never a multiple of 7 for any positive integer \(n\) . | B: (Ishan Nath)
As in solution A, we make the observation that
\[\frac{x_{n}(x_{n} - 1)\cdot\cdot\cdot(x_{n} - m + 1)}{m!} = \binom{x_{n}}{m}.\]
- Lemma: Let \(a\) and \(b\) be two positive integers with \(\overline{a_{k}a_{k - 1}\cdot\cdot\cdot a_{0}}\) and \(\overline{b_{k}b_{k - 1}\cdot\cdot\cdot b_{0}}\) the ... | {
"problem_match": "\n5. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo2_2022_solutions.jsonl",
"solution_match": "\nSolution"
} | 124 | 799 |
2023 | T1 | 5 | null | NewZealand_MO | Let \(x\) , \(y\) and \(z\) be real numbers such that: \(x^{2} = y + 2\) , and \(y^{2} = z + 2\) , and \(z^{2} = x + 2\) . Prove that \(x + y + z\) is an integer. | A: (Ross Atkins)
First we exclude \(- 1\) and 2:
\(x = 2\) implies \(y = 2\) implies \(z = 2\) implies \(x = 2\)
\(x = - 1\) implies \(y = - 1\) implies \(z = - 1\) implies \(x = - 1\)
In both these cases we have \(x + y + z\) being an integer. So henceforth we assume none of \(x,y,z\) are 2 nor \(- 1\) . Now l... | {
"problem_match": "\n5. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo2_2023_solutions.jsonl",
"solution_match": "\nSolution"
} | 70 | 1,091 |
2023 | T1 | 5 | null | NewZealand_MO | Let \(x\) , \(y\) and \(z\) be real numbers such that: \(x^{2} = y + 2\) , and \(y^{2} = z + 2\) , and \(z^{2} = x + 2\) . Prove that \(x + y + z\) is an integer. | B: (Ross Atkins)
Consider the polynomial \(P\) defined by
\[P(\lambda) = \lambda^{8} - 8\lambda^{6} + 20\lambda^{4} - 16\lambda^{2} - \lambda +2\] \[\qquad = (\lambda +1)(\lambda -2)(\lambda^{3} - 3\lambda +1)(\lambda^{3} + \lambda^{2} - 2\lambda -1).\]
If we substitute \(z = y^{2} - 2\) into \(z^{2} = x + 2\) gi... | {
"problem_match": "\n5. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo2_2023_solutions.jsonl",
"solution_match": "\nSolution"
} | 70 | 1,182 |
2024 | T1 | 3 | null | NewZealand_MO | Let \(A, B, C, D, E\) be five different points on the circumference of a circle in that (cyclic) order. Let \(F\) be the intersection of chords \(BD\) and \(CE\). Show that if \(AB = AE = AF\) then lines \(AF\) and \(CD\) are perpendicular. | A: (Ross Atkins)
Let \(x = \angle ABF\) and let \(y = \angle BCA\) . Since \(\triangle ABF\) is isosceles, we get \(\angle BFA = x\) . Since \(AB = AE\) , it follows that arcs \(BA\) and \(AE\) are equal. Since equal arcs subtend equal angles, every angle subtended by either arc \(AB\) or \(AE\) must be equal to \(\a... | {
"problem_match": "\n3. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo2_2024_solutions.jsonl",
"solution_match": "\nSolution"
} | 69 | 619 |
2024 | T1 | 5 | null | NewZealand_MO | Determine the least real number \(L\) such that
\[\frac{1}{a} +\frac{a}{b} +\frac{b}{c} +\frac{c}{d}\leqslant L\]
for all quadruples \((a,b,c,d)\) of integers satisfying \(1< a< b< c< d\) | : (Ross Atkins)
Answer: 3. To solve this problem, two parts are required. Part A shows that \(L = 3\) works. Part B shows that no \(L^{\prime}< 3\) works.
Part A We show that for all quadruples \((a,b,c,d)\) (with \(1< a< b< c< d\) ) we have
\[\frac{1}{a} +\frac{a}{b} +\frac{b}{c} +\frac{c}{d}\leqslant 3.\]
We ... | {
"problem_match": "\n5. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo2_2024_solutions.jsonl",
"solution_match": "\nSolution"
} | 78 | 1,046 |
2025 | T1 | 1 | null | NewZealand_MO | Find all pairs of positive integers \(m\) and \(n\) such that the centres of the unit squares in a \(m\) by \(n\) grid of unit squares can be paired up so that the distance between the centres of each pair is exactly 2.
(A unit square has side length 1. ) | : (Tony Wang)
We will show that the answer is all pairs \((a,b)\) where either \(a\) or \(b\) (or both) is a multiple of 4. First, partition the grid into four subgrids \(A\) , \(B\) , \(C\) , and \(D\)
<table><tr><td>A</td><td>B</td><td>A</td><td>B</td><td>A</td><td>B</td></tr><tr><td>C</td><td>D</td><td>C</td><td... | {
"problem_match": "\n1. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo2_2025_solutions.jsonl",
"solution_match": "\nSolution"
} | 64 | 523 |
2025 | T1 | 2 | null | NewZealand_MO | For which positive integers \(n\) , does there exist a sequence of real numbers \((x_{1}, x_{2}, \ldots , x_{n})\) such that
- \(-2 < x_{i} < 2\) for all \(i\) ,- \(x_{1} + x_{2} + x_{3} + \dots + x_{n} = 0\) , and- \(x_{1}^{4} + x_{2}^{4} + x_{3}^{4} + \dots + x_{n}^{4} \geqslant 32\) . | : (Eric Liang)
Note that if \(n = j\) works then \(n > j\) also works for a positive integer \(j\) as we can just set \(x_{i} = 0\) for \(n \geq i > j\) and have \(x_{1}, \ldots , x_{n}\) be the sequence that worked for \(n\) .
Consider \(n = 4\) . We take \(x_{1}, x_{2} = \sqrt[4]{8}\) and \(x_{3}, x_{4} = -\sqrt[... | {
"problem_match": "\n2. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo2_2025_solutions.jsonl",
"solution_match": "\nSolution"
} | 132 | 529 |
2025 | T1 | 3 | null | NewZealand_MO | Let \(A B C\) be an acute scalene triangle with \(A C > B C > A B\) . Let the orthocentre be \(H\) and circumcentre be \(O\) . Suppose that lines \(B O\) and \(C H\) intersect at a point \(D\) . Point \(E\) (where \(E\neq C\) ) lies on side \(A C\) so that \(O E C D\) is cyclic. Point \(F\) (where \(F\neq C\) ) lies on... | : (Nico McKinlay & George Zhu)
Let \(\alpha = \angle B A C\) . Let \(B B^{\prime}\) and \(C C^{\prime}\) be altitudes in triangle \(A B C\) , as shown.
Claim. Triangle \(C D E\) is isosceles with \(C E = D E\) .
Proof.
\[\angle D E ... | {
"problem_match": "# 3. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo2_2025_solutions.jsonl",
"solution_match": "\nSolution"
} | 154 | 666 |
2025 | T1 | 4 | null | NewZealand_MO | The function \(r_{n}(x)\) is the remainder when \(x\) is divided by \(n\) , where \(0 \leq r_{n}(x) < n\) . For which \(n\) does there exists some ordering \(\{a_{1}, \ldots , a_{n - 1}\}\) of \(\{1, 2, \ldots , n - 1\}\) such that \(\{r_{n}(a_{1}), r_{n}(2 \times a_{2}), \ldots , r_{n}((n - 1) \times a_{n - 1})\}\) is... | : (James Xu)
Notice that \(r_{n}(x)\) is just \(x\) modulo \(n\) . Therefore
\[\prod_{i}r_{n}(i a_{i})\equiv \prod_{i}i a_{i}\pmod{n}\]
For primes \(p\) , apply Wilson's theorem to see that we must have
\[\prod_{i}i a_{i}\equiv -1\pmod{p}\]
However,
\[\prod_{i}i a_{i} = \prod_{i}i\prod_{i}a_{i}\equiv -1^{2}... | {
"problem_match": "\n4. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo2_2025_solutions.jsonl",
"solution_match": "\nSolution"
} | 206 | 801 |
2025 | T1 | 5 | null | NewZealand_MO | Let \(a,b,c\) be positive real numbers satisfying \(a b c = 1\) . Determine the smallest possible value of
\[\frac{a^{2} + 2025}{a^{3}(b + c)} +\frac{b^{2} + 2025}{b^{3}(c + a)} +\frac{c^{2} + 2025}{c^{3}(a + b)}\] | : (Eric Liang)
Note,
\[\begin{array}{r l r}{{\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(a+c)}+\frac{1}{c^{3}(a+b)}=\frac{(\frac{1}{a})^{2}}{a(b+c)}+\frac{(\frac{1}{b})^{2}}{b(a+c)}+\frac{(\frac{1}{c})^{2}}{c(a+b)}}}\\ &{}&{\geq\frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}}{2(a b+a c+b c)}}\\ &{}&{=\frac{(\frac{1}{a}+\fr... | {
"problem_match": "\n5. Problem:",
"resource_path": "NewZealand_MO/segmented/en-nzmo2_2025_solutions.jsonl",
"solution_match": "\nSolution"
} | 98 | 803 |
1988 | T2 | 4 | null | Nordic_MO | Let $m_{n}$ be the smallest value of the function
$$
f_{n}(x)=\sum_{k=0}^{2 n} x^{k}
$$
Show that $m_{n} \rightarrow \frac{1}{2}$, as $n \rightarrow \infty$. | For $n>1$,
$$
\begin{gathered}
f_{n}(x)=1+x+x^{2}+\cdots \\
=1+x\left(1+x^{2}+x^{4}+\cdots\right)+x^{2}\left(1+x^{2}+x^{4} \cdots\right) \\
=1+x(1+x) \sum_{k=0}^{n-1} x^{2 k}
\end{gathered}
$$
From this we see that $f_{n}(x) \geq 1$, for $x \leq-1$ and $x \geq 0$. Consequently, $f_{n}$ attains its minimum value in th... | {
"problem_match": "\n88.4. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 65 | 532 |
1989 | T2 | 3 | null | Nordic_MO | Let $S$ be the set of all points $t$ in the closed interval $[-1,1]$ such that for the sequence $x_{0}, x_{1}, x_{2}, \ldots$ defined by the equations $x_{0}=t, x_{n+1}=2 x_{n}^{2}-1$, there exists a positive integer $N$ such that $x_{n}=1$ for all $n \geq N$. Show that the set $S$ has infinitely many elements. | The claim is not true for odd $n$. A counterexample is provided by $a_{1}=a_{2}=$ $\cdots=a_{n}=2$. We prove by induction that the claim is true for all even $n=2 k$. If $k=1$, then $a_{1}+a_{2}=4$ and $1 \leq a_{1}, a_{2} \leq 2$, so necessarily $a_{1}=a_{2}=2$. A choice satisfying the condition of the problem is $a_{... | {
"problem_match": "\n89.3. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 110 | 579 |
1991 | T2 | 2 | null | Nordic_MO | In the trapezium $A B C D$ the sides $A B$ and $C D$ are parallel, and $E$ is a fixed point on the side $A B$. Determine the point $F$ on the side $C D$ so that the area of the intersection of the triangles $A B F$ and $C D E$ is as large as possible.
 We assume $C D<A B$. Let $A D$ and $B C$ intersect at $H$ and $E H$ and $D C$ at $G$. Let $D E$ intersect $A F$ at $P$ and $F B$ intersect $E C$ at $Q$. Denote the area of a figure $\mathcal{F}$ by $|\mathcal{F}|$. Since $|A B F|$ does not depend on the choice of $F$ on $D C,|E Q F P|$ is maximized when... | {
"problem_match": "\n91.2. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution 1. "
} | 160 | 560 |
1992 | T2 | 2 | null | Nordic_MO | Let $n>1$ be an integer and let $a_{1}, a_{2}, \ldots, a_{n}$ be $n$ different integers. Show that the polynomial
$$
f(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n}\right)-1
$$
is not divisible by any polynomial with integer coefficients and of degree greater than zero but less than $n$ and such tha... | Suppose $g(x)$ is a polynomial of degree $m$, where $1 \leq m<n$, with integer coefficients and leading coefficient 1 , such that
$$
f(x)=g(x) h(x)
$$
whre $h(x)$ is a polynomial. Let
$$
\begin{aligned}
& g(x)=x^{m}+b_{m-1} x^{m-1}+\cdots+b_{1} x+b_{0} \\
& h(x)=x^{n-m}+c_{n-m-1} x^{n-m-1}+\cdots+c_{1} x+c_{0}
\end{... | {
"problem_match": "\n92.2. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 114 | 548 |
1992 | T2 | 2 | null | Nordic_MO | Let $n>1$ be an integer and let $a_{1}, a_{2}, \ldots, a_{n}$ be $n$ different integers. Show that the polynomial
$$
f(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n}\right)-1
$$
is not divisible by any polynomial with integer coefficients and of degree greater than zero but less than $n$ and such tha... | (See Figure 4.) The area $T$, perimeter $p$ and inradius $r$ satisfy $2 T=r p$. (Divide the triangle into three triangles with a common vertex at the incenter of the triangle.) So for a fixed inradius, the triangle with the smallest perimeter is the one which has the smallest area. To prove that the equilateral triangl... | {
"problem_match": "\n92.2. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 114 | 772 |
1993 | T2 | 1 | null | Nordic_MO | Let $F$ be an increasing real function defined for all $x, 0 \leq x \leq 1$, satisfying the conditions
$$
\begin{aligned}
& F\left(\frac{x}{3}\right)=\frac{F(x)}{2} \\
& F(1-x)=1-F(x)
\end{aligned}
$$
Determine $F\left(\frac{173}{1993}\right)$ and $F\left(\frac{1}{13}\right)$. | Condition (i) implies $F(0)=\frac{1}{2} F(0)$, so $F(0)=0$. Because of condition (ii), $F(1)=1-F(0)=1$. Also $F\left(\frac{1}{3}\right)=\frac{1}{2}$ and $F\left(\frac{2}{3}\right)=1-F\left(\frac{1}{3}\right)=\frac{1}{2}$. Since $F$ is an increasing function, this is possible only if $F(x)=\frac{1}{2}$ for all $x \in\le... | {
"problem_match": "\n93.1. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 113 | 564 |
1993 | T2 | 4 | null | Nordic_MO | Denote by $T(n)$ the sum of the digits of the decimal representation of a positive integer $n$.
a) Find an integer $N$, for which $T(k \cdot N)$ is even for all $k, 1 \leq k \leq 1992$, but $T(1993 \cdot N)$ is odd.
b) Show that no positive integer $N$ exists such that $T(k \cdot N)$ is even for all positive integers ... | a) If $s$ has $n$ decimal digits and $m=10^{n+r} s+s$, then $T(k m)$ is even at least as long as $k s<10^{n+r}$, because all non-zero digits appear in pairs in $k m$. Choose $N=5018300050183$ or $s=50183, n=5, r=3$. Now $1992 \cdot s=99964536<10^{8}$, so $T(k N)$ is even for all $k \leq 1992$. But $1993 \cdot s=1000147... | {
"problem_match": "\n93.4. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 107 | 728 |
1994 | T2 | 2 | null | Nordic_MO | We call a finite plane set $S$ consisting of points with integer coefficients a twoneighbour set, if for each point $(p, q)$ of $S$ exactly two of the points $(p+1, q),(p, q+1)$, $(p-1, q),(p, q-1)$ belong to $S$. For which integers $n$ there exists a two-neighbour set which contains exactly $n$ points? | The points $(0,0),(1,0),(1,1),(0,1)$ clearly form a two-neighbour set (which we abbreviate as $2 \mathrm{NS})$. For every even number $n=2 k \geq 8$, the set $S=\{(0,0), \ldots$, $(k-2,0),(k-2,1),(k-2,2), \ldots,(0,2),(0,1)\}$ is a $2 \mathrm{NS}$. We show that there is no $2 \mathrm{NS}$ with $n$ elements for other va... | {
"problem_match": "\n94.2. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 92 | 595 |
1995 | T2 | 2 | null | Nordic_MO | Messages are coded using sequences consisting of zeroes and ones only. Only sequences with at most two consecutive ones or zeroes are allowed. (For instance the sequence 011001 is allowed, but 011101 is not.) Determine the number of sequences consisting of exactly 12 numbers. | Let $S_{n}$ be the set of acceptable sequences consisting of $2 n$ digits. We partition $S_{n}$ in subsets $A_{n}, B_{n}, C_{n}$, and $D_{n}$, on the basis of the two last digits of the sequence. Sequences ending in 00 are in $A_{n}$, those ending in 01 are in $B_{n}$, those ending in 10 are in $C_{n}$, and those endin... | {
"problem_match": "\n95.2. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution 1. "
} | 64 | 650 |
1995 | T2 | 3 | null | Nordic_MO | Let $n \geq 2$ and let $x_{1}, x_{2}, \ldots x_{n}$ be real numbers satisfying $x_{1}+x_{2}+\ldots+x_{n} \geq 0$ and $x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}=1$. Let $M=\max \left\{x_{1}, x_{2}, \ldots, x_{n}\right\}$. Show that
$$
M \geq \frac{1}{\sqrt{n(n-1)}}
$$
When does equality hold in (1)? | Denote by $I$ the set of indices $i$ for which $x_{i} \geq 0$, and by $J$ the set of indices $j$ for which $x_{j}<0$. Let us assume $M<\frac{1}{\sqrt{n(n-1)}}$. Then $I \neq\{1,2, \ldots, n\}$, since otherwise we would have $\left|x_{i}\right|=x_{i} \leq \frac{1}{\sqrt{n(n-1)}}$ for every $i$, and $\sum_{i=1}^{n} x_{i}... | {
"problem_match": "\n95.3. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 144 | 1,082 |
1996 | T2 | 4 | null | Nordic_MO | The real-valued function $f$ is defined for positive integers, and the positive integer a satisfies
$$
\begin{gathered}
f(a)=f(1995), \quad f(a+1)=f(1996), \quad f(a+2)=f(1997) \\
f(n+a)=\frac{f(n)-1}{f(n)+1} \quad \text { for all positive integers } n
\end{gathered}
$$
(i) Show that $f(n+4 a)=f(n)$ for all positive ... | To prove (i), we the formula $f(n+a)=\frac{f(n)-1}{f(n)+1}$ repeatedly:
$$
\begin{gathered}
f(n+2 a)=f((n+a)+a)=\frac{\frac{f(n)-1}{f(n)+1}-1}{\frac{f(n)-1}{f(n)+1}+1}=-\frac{1}{f(n)} \\
f(n+4 a)=f((n+2 a)+2 a)=-\frac{1}{-\frac{1}{f(n)}}=f(n)
\end{gathered}
$$
(ii) If $a=1$, then $f(1)=f(a)=f(1995)=f(3+498 \cdot 4 a)... | {
"problem_match": "\n96.4. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 136 | 749 |
1997 | T2 | 3 | null | Nordic_MO | Let $A, B, C$, and $D$ be four different points in the plane. Three of the line segments $A B, A C, A D, B C, B D$, and $C D$ have length a. The other three have length $b$, where $b>a$. Determine all possible values of the quotient $\frac{b}{a}$. | If the three segments of length $a$ share a common endpoint, say $A$, then the other three points are on a circle of radius $a$, centered at $A$, and they are the vertices of an equilateral triangle of side length $b$. But this means that $A$ is the center of the triangle $B C D$, and
$$
\frac{b}{a}=\frac{b}{\frac{2}{... | {
"problem_match": "\n97.3. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 78 | 906 |
1997 | T2 | 4 | null | Nordic_MO | Let $f$ be a function defined in the set $\{0,1,2, \ldots\}$ of non-negative integers, satisfying $f(2 x)=2 f(x), f(4 x+1)=4 f(x)+3$, and $f(4 x-1)=2 f(2 x-1)-1$. Show that $f$ is an injection, i.e. if $f(x)=f(y)$, then $x=y$. | If $x$ is even, then $f(x)$ is even, and if $x$ is odd, then $f(x)$ is odd. Moreover, if $x \equiv 1 \bmod 4$, then $f(x) \equiv 3 \bmod 4$, and if $x \equiv 3 \bmod 4$, then $f(x) \equiv 1 \bmod 4$. Clearly $f(0)=0, f(1)=3, f(2)=6$, and $f(3)=5$. So at least $f$ restricted to the set $\{0,1,2,3\}$ ia an injection. We ... | {
"problem_match": "\n97.4. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 98 | 516 |
1998 | T2 | 1 | null | Nordic_MO | Determine all functions $f$ defined in the set of rational numbers and taking their values in the same set such that the equation $f(x+y)+f(x-y)=2 f(x)+2 f(y)$ holds for all rational numbers $x$ and $y$. | Insert $x=y=0$ in the equation to obtain $2 f(0)=4 f(0)$, which implies $f(0)=0$. Setting $x=0$, one obtains $f(y)+f(-y)=2 f(y)$ of $f(-y)=f(y)$. Then assume $y=n x$, where $n$ is a positive integer. We obtain
$$
f((n+1) x)=2 f(x)+2 f(n x)-f((n-1) x)
$$
In particular, $f(2 x)=2 f(x)+2 f(x)-f(0)=4 f(x)$ and $f(3 x)=2 ... | {
"problem_match": "\n98.1. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 55 | 552 |
1998 | T2 | 3 | null | Nordic_MO | (a) For which positive numbers $n$ does there exist a sequence $x_{1}, x_{2}, \ldots, x_{n}$, which contains each of the numbers 1, 2, ..., $n$ exactly once and for which $x_{1}+x_{2}+\cdots+x_{k}$ is divisible by $k$ for each $k=1,2, \ldots, n$ ?
(b) Does there exist an infinite sequence $x_{1}, x_{2}, x_{3}, \ldots$... | (a) We assume that $x_{1}, \ldots, x_{n}$ is the sequence required in the problem. Then $x_{1}+x_{2}+\cdots+x_{n}=\frac{n(n+1)}{2}$. This sum should be divisible by $n$. If $n$ is odd, this is possible, since $\frac{(n+1)}{2}$ is an integer. If, on the other hand, $n=2 m$, then $\frac{n(n+1)}{2}=m(2 m+1)=$ $2 m^{2}+m \... | {
"problem_match": "\n98.3. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 158 | 748 |
1998 | T2 | 4 | null | Nordic_MO | Let $n$ be a positive integer. Count the number of numbers $k \in\{0,1,2, \ldots, n\}$ such that $\binom{n}{k}$ is odd. Show that this number is a power of two, i.e. of the form $2^{p}$ for some nonnegative integer $p$. | The number of odd binomial coefficients $\binom{n}{k}$ equals the number of ones on the $n$ :th line of the Pascal Triangle $\bmod 2$ :
(We count the lines so that the uppermost line is line 0 ). We notice that line 1 has two copies of line 0 , lines 2 and 3 contain two copies of lines 1 and 2 , etc.
The fundamental ... | {
"problem_match": "\n98.4. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 74 | 556 |
1999 | T2 | 1 | null | Nordic_MO | The function $f$ is defined for non-negative integers and satisfies the condition
$$
f(n)= \begin{cases}f(f(n+11)), & \text { if } n \leq 1999 \\ n-5, & \text { if } n>1999\end{cases}
$$
Find all solutions of the equation $f(n)=1999$. | If $n \geq 2005$, then $f(n)=n-5 \geq 2000$, and the equation $f(n)=1999$ has no solutions. Let $1 \leq k \leq 4$. Then
$$
\begin{gathered}
2000-k=f(2005-k)=f(f(2010-k)) \\
=f(1999-k)=f(f(2004-k))=f(1993-k)
\end{gathered}
$$
Let $k=1$. We obtain three solutions $1999=f(2004)=f(1998)=f(1992)$. Moreover, $1995=f(2000)=... | {
"problem_match": "\n99.1. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 87 | 577 |
1999 | T2 | 3 | null | Nordic_MO | The infinite integer plane $\mathbb{Z} \times \mathbb{Z}=\mathbb{Z}^{2}$ consists of all number pairs $(x, y)$, where $x$ and $y$ are integers. Let $a$ and $b$ be non-negative integers. We call any move from a point $(x, y)$ to any of the points $(x \pm a, y \pm b)$ or $(x \pm b, y \pm a) a(a, b)$-knight move. Determin... | If the greatest common divisor of $a$ and $b$ is $d$, only points whose coordinates are multiples of $d$ can be reached by a sequence of $(a, b)$-knight moves starting from the origin. So $d=1$ is a necessary condition for the possibility of reaching every point in the integer plane. In any $(a, b)$-knight move, $x+y$ ... | {
"problem_match": "\n99.3. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 152 | 617 |
1999 | T2 | 4 | null | Nordic_MO | Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers and $n \geq 1$. Show that
$$
\begin{aligned}
& n\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right) \\
& \quad \geq\left(\frac{1}{1+a_{1}}+\cdots+\frac{1}{1+a_{n}}\right)\left(n+\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right)
\end{aligned}
$$
When does equality... | The inequality of the problem can be written as
$$
\frac{1}{1+a_{1}}+\cdots+\frac{1}{1+a_{n}} \leq \frac{n\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right)}{n+\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}}
$$
A small manipulation of the right hand side brings the inequality to the equivalent form
$$
\frac{1}{\frac{1}{a_{... | {
"problem_match": "\n99.4. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 142 | 558 |
2000 | T2 | 4 | null | Nordic_MO | The real-valued function $f$ is defined for $0 \leq x \leq 1, f(0)=0, f(1)=1$, and
$$
\frac{1}{2} \leq \frac{f(z)-f(y)}{f(y)-f(x)} \leq 2
$$
for all $0 \leq x<y<z \leq 1$ with $z-y=y-x$. Prove that
$$
\frac{1}{7} \leq f\left(\frac{1}{3}\right) \leq \frac{4}{7}
$$ | We set $f\left(\frac{1}{3}\right)=a$ and $f\left(\frac{2}{3}\right)=b$. Applying the inequality of the problem for $x=\frac{1}{3}, y=\frac{2}{3}$ and $z=1$, as well as for $x=0, y=\frac{1}{3}$, and $z=\frac{2}{3}$, we obtain
$$
\frac{1}{2} \leq \frac{1-b}{b-a} \leq 2, \quad \frac{1}{2} \leq \frac{b-a}{a} \leq 2
$$
If... | {
"problem_match": "\n00.4. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 136 | 501 |
2002 | T2 | 4 | null | Nordic_MO | Eva, Per and Anna play with their pocket calculators. They choose different integers and check, whether or not they are divisible by 11. They only look at nine-digit numbers consisting of all the digits 1, 2, .., 9. Anna claims that the probability of such a number to be a multiple of 11 is exactly 1/11. Eva has a diff... | We write the numbers in consideration, $n=a_{0}+10 a_{1}+10^{2} a_{2}+\cdots+10^{8} a_{8}$, in the form
$$
\begin{gathered}
a_{0}+(11-1) a_{1}+(99+1) a_{2}+(1001-1) a_{3} \\
+(9999+1) a_{4}+(100001-1) a_{5}+(999999+1) a_{6} \\
\quad+(10000001-1) a_{7}+(99999999+1) a_{8} \\
=\left(a_{0}-a_{1}+a_{2}-a_{3}+a_{4}-a_{5}+a_... | {
"problem_match": "\n02.4. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 117 | 852 |
2004 | T2 | 3 | null | Nordic_MO | Let $x_{11}, x_{21}, \ldots, x_{n 1}, n>2$, be a sequence of integers. We assume that all of the numbers $x_{i 1}$ are not equal. Assuming that the numbers $x_{1 k}, x_{2 k}, \ldots, x_{n k}$ have been defined, we set
$$
\begin{aligned}
x_{i, k+1} & =\frac{1}{2}\left(x_{i k}+x_{i+1, k}\right), i=1,2, \ldots, n-1 \\
x_... | We compute the first index modulo $n$, i.e. $x_{1 k}=x_{n+1, k}$. Let $M_{k}=\max _{j} x_{j k}$ and $m_{k}=\min _{j} x_{j k}$. Evidently $\left(M_{k}\right)$ is a non-increasing and $\left(m_{k}\right)$ a non-decreasing sequence, and $M_{k+1}=M_{k}$ is possible only if $x_{j k}=x_{j+1, k}=M_{k}$ for some $j$. If exactl... | {
"problem_match": "\n04.3. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 210 | 585 |
2005 | T2 | 2 | null | Nordic_MO | Let $a, b$, and $c$ be positive real numbers. Prove that
$$
\frac{2 a^{2}}{b+c}+\frac{2 b^{2}}{c+a}+\frac{2 c^{2}}{a+b} \geq a+b+c
$$ | Use brute force. Removing the denominators and brackets and combining simililar terms yields the equivalent inequality
$$
\begin{gathered}
0 \leq 2 a^{4}+2 b^{4}+2 c^{4}+a^{3} b+a^{3} c+a b^{3}+b^{3} c+a c^{3}+b c^{3} \\
-2 a^{2} b^{2}-2 b^{2} c^{2}-2 a^{2} c^{2}-2 a b c^{2}-2 a b^{2} c-2 a^{2} b c \\
=a^{4}+b^{4}-2 a... | {
"problem_match": "\n05.2. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution 1. "
} | 65 | 576 |
2005 | T2 | 2 | null | Nordic_MO | Let $a, b$, and $c$ be positive real numbers. Prove that
$$
\frac{2 a^{2}}{b+c}+\frac{2 b^{2}}{c+a}+\frac{2 c^{2}}{a+b} \geq a+b+c
$$ | The original inequality is symmetric in $a, b, c$. So we may assume $a \geq b \geq c$, which implies
$$
\frac{1}{b+c} \geq \frac{1}{c+a} \geq \frac{1}{a+b}
$$
The power mean inequality gives
$$
\frac{a^{2}+b^{2}+c^{2}}{3} \geq\left(\frac{a+b+c}{3}\right)^{2}
$$
We combine this and the Chebyshev inequality to obtain... | {
"problem_match": "\n05.2. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution 3. "
} | 65 | 501 |
2007 | T2 | 4 | null | Nordic_MO | A line through a point $A$ intersects a circle in two points, $B$ and $C$, in such a way that $B$ lies between $A$ and $C$. From the point $A$ draw the two tangents to the circle, meeting the circle at points $S$ and $T$. Let $P$ be the intersection of the lines $S T$ and AC. Show that $A P / P C=2 \cdot A B / B C$. | First we show that if we fix the points $A, B$ and $C$ but vary the circle, then the point $P$ stays fixed. To that end, suppose we have two different circles through $B$ and $C$. Draw the tangents from $A$ to one circle, meeting the circle at points $S_{1}$ and $T_{1}$, and the tangents to the other circle, meeting th... | {
"problem_match": "\n07.4. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 101 | 601 |
2009 | T2 | 2 | null | Nordic_MO | On a faded piece of paper it is possible, with some effort, to discern the following:
$$
\left(x^{2}+x+a\right)\left(x^{15}-\ldots\right)=x^{17}+x^{13}+x^{5}-90 x^{4}+x-90
$$
Some parts have got lost, partly the constant term of the first factor of the left side, partly the main part of the other factor. It would be ... | We denote the polynomial $x^{2}+x+a$ by $P_{a}(x)$, the polynomial forming the other factor of the left side by $Q(x)$ and the polynomial on the right side by $R(x)$. The polynomials are integer valued for every integer $x$. For $x=0$ we get $P_{a}(0)=a$ and $R(0)=-90$, so $a$ is a divisor of $90=2 \cdot 3 \cdot 3 \cdo... | {
"problem_match": "\n09.2. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 150 | 508 |
2011 | T2 | 1 | null | Nordic_MO | When $a_{0}, a_{1}, \ldots, a_{1000}$ denote digits, can the sum of the 1001-digit numbers $a_{0} a_{1} \ldots a_{1000}$ and $a_{1000} a_{999} \ldots a_{0}$ have odd digits only? | The answer is no. The following diagram illustrates the calculation of the sum digit by digit.
| $a_{0}$ | $a_{1}$ | $\ldots$ | $a_{i}$ | $\ldots$ | $a_{500}$ | $\ldots$ | $a_{1000-i}$ | $\ldots$ | $a_{999}$ | $a_{1000}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $a_{... | {
"problem_match": "\n11.1. ",
"resource_path": "Nordic_MO/segmented/en-1987-2011-sol.jsonl",
"solution_match": "\nSolution. "
} | 82 | 711 |
2012 | T2 | 3 | null | Nordic_MO | Find the smallest positive integer $n$, such that there exist $n$ integers $x_{1}, x_{2}, \ldots, x_{n}$ (not necessarily different), with $1 \leq x_{k} \leq n, 1 \leq k \leq n$, and such that
$$
x_{1}+x_{2}+\cdots+x_{n}=\frac{n(n+1)}{2}, \quad \text { and } \quad x_{1} x_{2} \cdots x_{n}=n!
$$
but $\left\{x_{1}, x_{... | If it is possible to find a set of numbers as required for some $n=k$, then it will also be possible for $n=k+1$ (choose $x_{1}, \ldots, x_{k}$ as for $n=k$, and
let $x_{k+1}=k+1$ ). Thus we have to find a positive integer $n$ such that a set as required exists, and prove that such a set does not exist for $n-1$.
For ... | {
"problem_match": "\n\nProblem 3.",
"resource_path": "Nordic_MO/segmented/en-2012-sol.jsonl",
"solution_match": "\n\nSolution."
} | 165 | 788 |
2012 | T2 | 4 | null | Nordic_MO | The number 1 is written on the blackboard. After that a sequence of numbers is created as follows: at each step each number $a$ on the blackboard is replaced by the numbers $a-1$ and $a+1$; if the number 0 occurs, it is erased immediately; if a number occurs more than once, all its occurrences are left on the blackboar... | Let $S$ be a set of different numbers, all of them less than $2^{n-1}$, and create two new sets as follows: $S_{1}$, consisting of all the numbers in $S$ except
the smallest one, and $S_{2}$, with elements the smallest element of $S$ and all the numbers we get by adding $2^{n-1}$ to each number in $S$. Note that if the... | {
"problem_match": "\n\nProblem 4.",
"resource_path": "Nordic_MO/segmented/en-2012-sol.jsonl",
"solution_match": "\n\nSolution I."
} | 143 | 649 |
2012 | T2 | 4 | null | Nordic_MO | The number 1 is written on the blackboard. After that a sequence of numbers is created as follows: at each step each number $a$ on the blackboard is replaced by the numbers $a-1$ and $a+1$; if the number 0 occurs, it is erased immediately; if a number occurs more than once, all its occurrences are left on the blackboar... | Denote by $\sigma_{n}$ the number of numbers on the blackboard after $n$ steps (thus $\sigma_{0}=\sigma_{1}=1, \sigma_{2}=2, \sigma_{3}=3, \ldots$ ). Regard all points in the plane with coordinates $(m, n)$, where $m, n$ are defined as follows: the number $n$ is written on the blackboard after $m$ steps ( $m, n$ are po... | {
"problem_match": "\n\nProblem 4.",
"resource_path": "Nordic_MO/segmented/en-2012-sol.jsonl",
"solution_match": "\n\nSolUTION II."
} | 143 | 725 |
2013 | T2 | 2 | null | Nordic_MO | In a football tournament there are $n$ teams, with $n \geq 4$, and each pair of teams meets exactly once. Suppose that, at the end of the tournament, the final scores form an arithmetic sequence where each team scores 1 more point than the following team on the scoreboard. Determine the maximum possible score of the lo... | Note that the total number of games equals the number of different pairings, that is, $n(n-1) / 2$. Suppose the lowest scoring team ends with $k$ points. Then the total score for all teams is
$$
k+(k+1)+\cdots+(k+n-1)=n k+\frac{(n-1) n}{2}
$$
Some games must end in a tie, for otherwise, all team scores would be a mul... | {
"problem_match": "\n\nProblem 2.",
"resource_path": "Nordic_MO/segmented/en-2013-sol.jsonl",
"solution_match": "\n\nSolution."
} | 110 | 1,056 |
2013 | T2 | 3 | null | Nordic_MO | Define a sequence $\left(n_{k}\right)_{k \geq 0}$ by $n_{0}=n_{1}=1$, and $n_{2 k}=n_{k}+n_{k-1}$ and $n_{2 k+1}=n_{k}$ for $k \geq 1$. Let further $q_{k}=n_{k} / n_{k-1}$ for each $k \geq 1$. Show that every positive rational number is present exactly once in the sequence $\left(q_{k}\right)_{k \geq 1}$. | Clearly, all the numbers $n_{k}$ are positive integers. Moreover,
$$
q_{2 k}=\frac{n_{2 k}}{n_{2 k-1}}=\frac{n_{k}+n_{k-1}}{n_{k-1}}=q_{k}+1
$$
and similarly,
$$
\frac{1}{q_{2 k+1}}=\frac{n_{2 k}}{n_{2 k+1}}=\frac{n_{k}+n_{k-1}}{n_{k}}=\frac{1}{q_{k}}+1
$$
In particular, $q_{k}>1$ when $k$ is even, and $q_{k}<1$ wh... | {
"problem_match": "\n\nProblem 3.",
"resource_path": "Nordic_MO/segmented/en-2013-sol.jsonl",
"solution_match": "\n\nSolution."
} | 129 | 602 |
2013 | T2 | 4 | null | Nordic_MO | Let $A B C$ be an acute angled triangle, and $H$ a point in its interior. Let the reflections of $H$ through the sides $A B$ and $A C$ be called $H_{c}$ and $H_{b}$, respectively, and let the reflections of $H$ through the midpoints of these same sides be called $H_{c}^{\prime}$ and $H_{b}^{\prime}$, respectively. Show... | If at least two of the four points $H_{b}, H_{b}^{\prime}, H_{c}$, and $H_{c}^{\prime}$ coincide, all four are obviously concyclic. Therefore we may assume that these four points are distinct.
Let $P_{b}$ denote the midpoint of segment $H H_{b}, P_{b}^{\prime}$ the midpoint of segment $H H_{b}^{\prime}, P_{c}$ the mid... | {
"problem_match": "\n\nProblem 4.",
"resource_path": "Nordic_MO/segmented/en-2013-sol.jsonl",
"solution_match": "\n\nSolution."
} | 162 | 604 |
2015 | T2 | 4 | null | Nordic_MO | An encyclopedia consists of 2000 numbered volumes. The volumes are stacked in order with number 1 on top and 2000 in the bottom. One may perform two operations with the stack:
(i) For $n$ even, one may take the top $n$ volumes and put them in the bottom of the stack without changing the order.
(ii) For $n$ odd, one m... | We show that the volumes can be permuted so that the volumes with odd numbers are in an arbitrary order in the odd-numbered palaces and the volumes with even numbers are in an arbitrary order in the even-numbered places. The main idea is to construct two combinations of the allowed operations. The first one turns the v... | {
"problem_match": "\n## Problem 4.",
"resource_path": "Nordic_MO/segmented/en-2015-sol.jsonl",
"solution_match": "\n\nSolution 2."
} | 123 | 1,775 |
2015 | T2 | 4 | null | Nordic_MO | An encyclopedia consists of 2000 numbered volumes. The volumes are stacked in order with number 1 on top and 2000 in the bottom. One may perform two operations with the stack:
(i) For $n$ even, one may take the top $n$ volumes and put them in the bottom of the stack without changing the order.
(ii) For $n$ odd, one m... | We show by induction, that if in an ordered sequence one may exchange two consecutive elements without changing the places of any other element, then any two elements can be exchanged so that all other elements remain in place. We assume that this is true for elements which are at most $k$ steps away from each other in... | {
"problem_match": "\n## Problem 4.",
"resource_path": "Nordic_MO/segmented/en-2015-sol.jsonl",
"solution_match": "\n\nSolution 3."
} | 123 | 827 |
2016 | T2 | 4 | null | Nordic_MO | King George has decided to connect the 1680 islands in his kingdom by bridges. Unfortunately the rebel movement will destroy two bridges after all the bridges have been built, but not two bridges from the same island.
What is the minimal number of bridges the King has to build in order to make sure that it is still po... | Answer: 2016
An island cannot be connected with just one bridge, since this bridge could be destroyed. Consider the case of two islands, each with only two bridges, connected by a bridge. (It is not possible that they are connected with two bridges, since then they would be isolated from the other islands no matter wh... | {
"problem_match": "\n## Problem 4",
"resource_path": "Nordic_MO/segmented/en-2016-sol.jsonl",
"solution_match": "\n\nSolution 4"
} | 90 | 864 |
2017 | T2 | 3 | null | Nordic_MO | Let $M$ and $N$ be the midpoints of the sides $A C$ and $A B$, respectively, of an acute triangle $A B C, A B \neq A C$. Let $\omega_{B}$ be the circle centered at $M$ passing through $B$, and let $\omega_{C}$ be the circle centered at $N$ passing through $C$. Let the point $D$ be such that $A B C D$ is an isosceles tr... | Let $E$ be such that $A B E C$ is a parallelogram with $A B \| C E$ and $A C \| B E$, and let $\omega$ be the circumscribed circle of $\triangle A B C$ with centre $O$.
It is known that the radical axis of two circles is perpendicular to the line connecting the two centres. Since $B E \perp M O$ and $C E \perp N O$, t... | {
"problem_match": "\n\nProblem 3",
"resource_path": "Nordic_MO/segmented/en-2017-sol.jsonl",
"solution_match": "\n\nSolution 3"
} | 156 | 735 |
2017 | T2 | 4 | null | Nordic_MO | Find all integers $n$ and $m, n>m>2$, and such that a regular $n$-sided polygon can be inscribed in a regular $m$-sided polygon so that all the vertices of the $n$-gon lie on the sides of the $m$-gon. | It works only for $n=2 m$, and for $m=3$ and $n=4$.
To begin with let's see why it works for $n=2 m$. For a $2 m$-gon we can choose two points on each side, symmetrically, so that the distance between the two of them is equal to the distance between two close points on adjacent sides.
For $n=4$ and $m=3$ we need to i... | {
"problem_match": "\n\nProblem 4",
"resource_path": "Nordic_MO/segmented/en-2017-sol.jsonl",
"solution_match": "\n\nSolution 4"
} | 65 | 1,700 |
2019 | T2 | 4 | null | Nordic_MO | Let $n$ be an integer with $n \geq 3$ and assume that $2 n$ vertices of a regular $(4 n+1)$-gon are coloured. Show that there must exist three of the coloured vertices forming an isosceles triangle. | a) Notice that $\left\{2019!\cdot 1^{2019!}, 2019!\cdot 2^{2019!}, \ldots, 2019!\cdot 2019^{2019!}\right\}$ is such a set. Observe that if all the elements are divisible by 2019! then the arithmetic means will be integer for all the subsets. Also, if $A$ is a set such that the geometric means are integer for all non-em... | {
"problem_match": "\n\nProblem 4",
"resource_path": "Nordic_MO/segmented/en-2019-sol.jsonl",
"solution_match": "\n\nSolution 1"
} | 56 | 573 |
2019 | T2 | 4 | null | Nordic_MO | Let $n$ be an integer with $n \geq 3$ and assume that $2 n$ vertices of a regular $(4 n+1)$-gon are coloured. Show that there must exist three of the coloured vertices forming an isosceles triangle. | Assume that it is possible to color $2 n$ of the vertices of a $4 n+1$-gon such that there are no three colored vertices forming an isosceles triangle. Enumerate the vertices consecutively as $H_{-2 n}, H_{-2 n+1}, \ldots, H_{0}, H_{1}, H_{2}, \ldots, H_{2 n}$ and consider first the case where there are two colored nei... | {
"problem_match": "\n\nProblem 4",
"resource_path": "Nordic_MO/segmented/en-2019-sol.jsonl",
"solution_match": "\n\nSolution 4"
} | 56 | 639 |
2022 | T2 | 2 | null | Nordic_MO | In Wonderland, the towns are connected by roads, and whenever there is a direct road between two towns there is also a route between these two towns that does not use that road. (There is at most one direct road between any two towns.) The Queen
of Hearts ordered the Spades to provide a list of all "even" subsystems of... | The answer is $\frac{1}{2} n x$.
Proof: We reformulate the problem in terms of graph theory with the towns being vertices and the roads being edges of a graph $G=(V, E)$. The given information implies that every edge $e \in E$ is part of a cycle. The subgraphs to be counted are those with every valence even, briefly t... | {
"problem_match": "\n## Problem 2",
"resource_path": "Nordic_MO/segmented/en-2022-sol.jsonl",
"solution_match": "\n\nSolution."
} | 158 | 523 |
2022 | T2 | 3 | null | Nordic_MO | Anton and Britta play a game with the set $M=\{1,2,3, \ldots, n-1\}$ where $n \geq 5$ is an odd integer. In each step Anton removes a number from $M$ and puts it in his set $A$, and Britta removes a number from $M$ and puts it in her set $B$ (both $A$ and $B$ are empty to begin with). When $M$ is empty, Anton picks two... | Britta wins if and only if $n$ is prime.
If $n$ is not prime, then Anton can add any prime divisor $p<n$ of $n$ to his set $A$ in the first round and choose $x_{1}=p$ which means that the product $\left(x_{1} x_{2}\left(x_{1}-y_{1}\right)\left(x_{2}-y_{2}\right)\right)^{\frac{n-1}{2}}$ is divisible by $p$ and is not $... | {
"problem_match": "\n## Problem 3",
"resource_path": "Nordic_MO/segmented/en-2022-sol.jsonl",
"solution_match": "\n\nSolution."
} | 230 | 774 |
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