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 |
|---|---|---|---|---|---|---|---|---|---|
2023 | T0 | A7 | Algebra | IMO-SL | Let $N$ be a positive integer. Prove that there exist three permutations $a_{1}, a_{2}, \ldots, a_{N}$; $b_{1}, b_{2}, \ldots, b_{N}$; and $c_{1}, c_{2}, \ldots, c_{N}$ of $1,2, \ldots, N$ such that $$ \left|\sqrt{a_{k}}+\sqrt{b_{k}}+\sqrt{c_{k}}-2 \sqrt{N}\right|<2023 $$ for every $k=1,2, \ldots, N$. (China) | The idea is to approximate the numbers $\sqrt{1}, \sqrt{2}, \ldots, \sqrt{N}$ by the nearest integer with errors $<0.5$. This gives the following sequence $$ 1,1,2,2,2,2,3,3,3,3,3,3,4, \ldots $$ More precisely, for each $k \geqslant 1$, we round $\sqrt{k^{2}-k+1}, \ldots, \sqrt{k^{2}+k}$ to $k$, so that there are $2 k$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 143 | 2,238 |
2023 | T0 | A7 | Algebra | IMO-SL | Let $N$ be a positive integer. Prove that there exist three permutations $a_{1}, a_{2}, \ldots, a_{N}$; $b_{1}, b_{2}, \ldots, b_{N}$; and $c_{1}, c_{2}, \ldots, c_{N}$ of $1,2, \ldots, N$ such that $$ \left|\sqrt{a_{k}}+\sqrt{b_{k}}+\sqrt{c_{k}}-2 \sqrt{N}\right|<2023 $$ for every $k=1,2, \ldots, N$. (China) | This is a variation of Let $n$ be an integer satisfying $0 \leqslant n \leqslant m+1$ and define the multiset $T_{m, n}$ by $$ T_{m, n}:=\left\{1_{\times 1}, 2_{\times 2}, 3_{\times 3}, \ldots, m_{\times m},(m+1)_{\times n}\right\} . $$ In other words, $T_{m, 0}=T_{m}, T_{m, n}=T_{m} \sqcup\left\{(m+1)_{\times n}\right... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 143 | 1,543 |
2023 | T0 | A7 | Algebra | IMO-SL | Let $N$ be a positive integer. Prove that there exist three permutations $a_{1}, a_{2}, \ldots, a_{N}$; $b_{1}, b_{2}, \ldots, b_{N}$; and $c_{1}, c_{2}, \ldots, c_{N}$ of $1,2, \ldots, N$ such that $$ \left|\sqrt{a_{k}}+\sqrt{b_{k}}+\sqrt{c_{k}}-2 \sqrt{N}\right|<2023 $$ for every $k=1,2, \ldots, N$. (China) | This solution is based on the geometrical insight of equilateral triangles. Step 1. We first consider the easier case of triangle numbers $$ N=\frac{m(m+1)}{2} $$ As shown in the following picture, consider the triangular shaped lattice points inside an equilateral triangle $A B C$ with a total of $N$ points. The latti... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 143 | 2,550 |
2023 | T0 | C2 | Case 2: $n$ is odd. | IMO-SL | Determine the maximal length $L$ of a sequence $a_{1}, \ldots, a_{L}$ of positive integers satisfying both the following properties: - every term in the sequence is less than or equal to $2^{2023}$, and - there does not exist a consecutive subsequence $a_{i}, a_{i+1}, \ldots, a_{j}$ (where $1 \leqslant i \leqslant j \... | We prove more generally that the answer is $2^{k+1}-1$ when $2^{2023}$ is replaced by $2^{k}$ for an arbitrary positive integer $k$. Write $n=2^{k}$. We first show that there exists a sequence of length $L=2 n-1$ satisfying the properties. For a positive integer $x$, denote by $v_{2}(x)$ the maximal nonnegative integer... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 202 | 1,376 |
2023 | T0 | C3 | Case 2: $n$ is odd. | IMO-SL | Let $n$ be a positive integer. We arrange $1+2+\cdots+n$ circles in a triangle with $n$ rows, such that the $i^{\text {th }}$ row contains exactly $i$ circles. The following figure shows the case $n=6$. ![](https://cdn.mathpix.com/cropped/2025_01_09_9538693ed865d47d2de5g-40.jpg?height=398&width=381&top_left_y=401&top_l... | We give an alternative proof that there exists a ninja-path passing through at least $N+1$ red circles. Assign numbers to circles as in the previous solution, but we only focus on the numbers assigned to red circles. For each positive integer $i$, denote by $e_{i}$ the number of red circles with number $i$. Claim 2. If... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 243 | 644 |
2023 | T0 | C3 | Case 2: $n$ is odd. | IMO-SL | Let $n$ be a positive integer. We arrange $1+2+\cdots+n$ circles in a triangle with $n$ rows, such that the $i^{\text {th }}$ row contains exactly $i$ circles. The following figure shows the case $n=6$. . Denote by $f_{i}$ the number of red circles with number $i$. Note that if a red ci... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 243 | 810 |
2023 | T0 | C4 | Note that | IMO-SL | Let $n \geqslant 2$ be a positive integer. Paul has a $1 \times n^{2}$ rectangular strip consisting of $n^{2}$ unit squares, where the $i^{\text {th }}$ square is labelled with $i$ for all $1 \leqslant i \leqslant n^{2}$. He wishes to cut the strip into several pieces, where each piece consists of a number of consecuti... | For the entirety of the solution, we shall view the labels as taking values in $\mathbb{Z} / n \mathbb{Z}$, as only their values modulo $n$ play a role. Here are two possible constructions consisting of $2 n-1$ pieces. 1. Cut into pieces of sizes $n, 1, n, 1, \ldots, n, 1,1$, and glue the pieces of size 1 to obtain the... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 213 | 610 |
2023 | T0 | C4 | Note that | IMO-SL | Let $n \geqslant 2$ be a positive integer. Paul has a $1 \times n^{2}$ rectangular strip consisting of $n^{2}$ unit squares, where the $i^{\text {th }}$ square is labelled with $i$ for all $1 \leqslant i \leqslant n^{2}$. He wishes to cut the strip into several pieces, where each piece consists of a number of consecuti... | We provide an alternative proof that at least $2 n-1$ pieces are needed. Instead of having a linear strip, we work with a number of circular strips, each having length a multiple of $n$ and labelled as $$ 1,2, \ldots, n, 1,2, \ldots, n, \ldots, 1,2, \ldots, n $$ where there are $n^{2}$ cells in total across all circula... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 213 | 649 |
2023 | T0 | C4 | Note that | IMO-SL | Let $n \geqslant 2$ be a positive integer. Paul has a $1 \times n^{2}$ rectangular strip consisting of $n^{2}$ unit squares, where the $i^{\text {th }}$ square is labelled with $i$ for all $1 \leqslant i \leqslant n^{2}$. He wishes to cut the strip into several pieces, where each piece consists of a number of consecuti... | As with the previous solution, we again work with circular strips. In particular, we start out with $k$ circular strips, each having length a multiple of $n$ and labelled as $$ 1,2, \ldots, n, 1,2, \ldots, n, \ldots, 1,2, \ldots, n $$ where there are $n^{2}$ cells in total across all $k$ circular strips. The goal is st... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 213 | 1,184 |
2023 | T0 | C6 | Note that | IMO-SL | Let $N$ be a positive integer, and consider an $N \times N$ grid. A right-down path is a sequence of grid cells such that each cell is either one cell to the right of or one cell below the previous cell in the sequence. A right-up path is a sequence of grid cells such that each cell is either one cell to the right of o... | We define a good parallelogram to be a parallelogram composed of two isosceles right-angled triangles glued together as shown below. Given any partition into $k$ right-down or right-up paths, we can find a corresponding packing of good parallelograms that leaves an area of $k$ empty. Thus, it suffices to prove that we ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 216 | 580 |
2023 | T0 | C6 | Note that | IMO-SL | Let $N$ be a positive integer, and consider an $N \times N$ grid. A right-down path is a sequence of grid cells such that each cell is either one cell to the right of or one cell below the previous cell in the sequence. A right-up path is a sequence of grid cells such that each cell is either one cell to the right of o... | We apply an induction on $N$. The base case $N=1$ is trivial. Suppose that the claim holds for $N-1$ and prove it for $N \geqslant 2$. Let us denote the path containing the upper left corner by $P$. If $P$ is right-up, then every cell in $P$ is in the top row or in the leftmost column. By the induction hypothesis, ther... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 216 | 959 |
2023 | T0 | C7 | Note that | IMO-SL | The Imomi archipelago consists of $n \geqslant 2$ islands. Between each pair of distinct islands is a unique ferry line that runs in both directions, and each ferry line is operated by one of $k$ companies. It is known that if any one of the $k$ companies closes all its ferry lines, then it becomes impossible for a tra... | We reformulate the problem using graph theory. We have a complete graph $K_{n}$ on $n$ nodes (corresponding to islands), and we want to colour the edges (corresponding to ferry lines) with $k$ colours (corresponding to companies), so that every Hamiltonian path contains all $k$ different colours. For a fixed set of $k$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 155 | 3,327 |
2023 | T0 | G7 | Solution 1 (Similar Triangles). | IMO-SL | Let $A B C$ be an acute, scalene triangle with orthocentre $H$. Let $\ell_{a}$ be the line through the reflection of $B$ with respect to $C H$ and the reflection of $C$ with respect to $B H$. Lines $\ell_{b}$ and $\ell_{c}$ are defined similarly. Suppose lines $\ell_{a}, \ell_{b}$, and $\ell_{c}$ determine a triangle $... | We write $\triangle P_{1} P_{2} P_{3} \stackrel{ \pm}{\sim} \triangle Q_{1} Q_{2} Q_{3}$ (resp. $\triangle P_{1} P_{2} P_{3} \approx \triangle Q_{1} Q_{2} Q_{3}$ ) to indicate that two triangles are directly (resp. oppositely) similar. We use directed angles throughout denoted with $\Varangle$. Denote by $A_{b}, A_{c}$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 137 | 1,078 |
2023 | T0 | G7 | Solution 1 (Similar Triangles). | IMO-SL | Let $A B C$ be an acute, scalene triangle with orthocentre $H$. Let $\ell_{a}$ be the line through the reflection of $B$ with respect to $C H$ and the reflection of $C$ with respect to $B H$. Lines $\ell_{b}$ and $\ell_{c}$ are defined similarly. Suppose lines $\ell_{a}, \ell_{b}$, and $\ell_{c}$ determine a triangle $... | We use the same notation $A_{b}, A_{c}, B_{c}, B_{a}, C_{a}, C_{b}$ and $A_{1}, B_{1}, C_{1}, O_{1}, H_{1}$ as in Let $O$ be the circumcentre of $\triangle A B C$ and $A_{2}$ be the reflection of $A_{1}$ in $A H$. As $\odot A A_{b} A_{c}$ is centred at $H, A_{2}$ also lies on this circle. By Claim 2, $\Varangle B_{a} A... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 137 | 1,272 |
2023 | T0 | N2 | Number Theory | IMO-SL | Determine all pairs $(a, p)$ of positive integers with $p$ prime such that $p^{a}+a^{4}$ is a perfect square. (Bangladesh) Answer: $(a, p)=(1,3),(2,3),(6,3),(9,3)$ are all the possible solutions. | Let $p^{a}+a^{4}=b^{2}$ for some positive integer $b$. Then we have $$ p^{a}=b^{2}-a^{4}=\left(b+a^{2}\right)\left(b-a^{2}\right) . $$ Hence both $b+a^{2}$ and $b-a^{2}$ are powers of $p$. Let $b-a^{2}=p^{x}$ for some integer $x$. Then $b+a^{2}=p^{a-x}$ and $a-x>x$. Therefore, we have $$ 2 a^{2}=\left(b+a^{2}\right)-\l... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 67 | 1,312 |
2023 | T0 | N3 | Number Theory | IMO-SL | For positive integers $n$ and $k \geqslant 2$ define $E_{k}(n)$ as the greatest exponent $r$ such that $k^{r}$ divides $n$ !. Prove that there are infinitely many $n$ such that $E_{10}(n)>E_{9}(n)$ and infinitely many $m$ such that $E_{10}(m)<E_{9}(m)$. (Brazil) | We let $v_{p}(m)$ denote the $p$-adic valuation of $m$. By Legendre's Formula we know, for $p$ prime, that $v_{p}(n!)=\lfloor n / p\rfloor+\left\lfloor n / p^{2}\right\rfloor+\cdots$. We can see that $E_{9}(n)=\left\lfloor\frac{v_{3}(n!)}{2}\right\rfloor$. Since $v_{5}(n!) \leqslant v_{2}(n!)$ and $E_{10}(n)=\min \left... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 97 | 853 |
2023 | T0 | N4 | Number Theory | IMO-SL | Let $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ be $2 n$ positive integers such that the $n+1$ products $$ \begin{gathered} a_{1} a_{2} a_{3} \cdots a_{n} \\ b_{1} a_{2} a_{3} \cdots a_{n} \\ b_{1} b_{2} a_{3} \cdots a_{n} \\ \vdots \\ b_{1} b_{2} b_{3} \cdots b_{n} \end{gathered} $$ form a strictly in... | The condition in the problem is equivalent to $$ D=\left(b_{1}-a_{1}\right) a_{2} a_{3} \cdots a_{n}=b_{1}\left(b_{2}-a_{2}\right) a_{3} a_{4} \cdots a_{n}=\cdots=b_{1} b_{2} \cdots b_{n-1}\left(b_{n}-a_{n}\right), $$ where $D$ is the common difference. Since the progression is strictly increasing, $D>0$, hence $b_{i}>... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 190 | 1,150 |
2023 | T0 | N7 | Number Theory | IMO-SL | Let $a, b, c, d$ be positive integers satisfying $$ \frac{a b}{a+b}+\frac{c d}{c+d}=\frac{(a+b)(c+d)}{a+b+c+d} $$ Determine all possible values of $a+b+c+d$. (Netherlands) Answer: The possible values are the positive integers that are not square-free. | First, note that if we take $a=\ell, b=k \ell, c=k \ell, d=k^{2} \ell$ for some positive integers $k$ and $\ell$, then we have $$ \frac{a b}{a+b}+\frac{c d}{c+d}=\frac{k \ell^{2}}{\ell+k \ell}+\frac{k^{3} \ell^{2}}{k \ell+k^{2} \ell}=\frac{k \ell}{k+1}+\frac{k^{2} \ell}{k+1}=k \ell $$ and $$ \frac{(a+b)(c+d)}{a+b+c+d}=... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 81 | 762 |
2023 | T0 | N8 | Number Theory | IMO-SL | Let $\mathbb{Z}_{>0}$ be the set of positive integers. Determine all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that $$ f^{b f(a)}(a+1)=(a+1) f(b) $$ holds for all $a, b \in \mathbb{Z}_{>0}$, where $f^{k}(n)=f(f(\cdots f(n) \cdots))$ denotes the composition of $f$ with itself $k$ times. (Taiwan) ... | We divide the solution into 5 steps. Step 1. ( $f$ is injective) Claim 1. For any $a \geqslant 2$, the set $\left\{f^{n}(a) \mid n \in \mathbb{Z}_{>0}\right\}$ is infinite. Proof. First, we have $f^{f(a)}(a+1) \stackrel{P(a, 1)}{=}(a+1) f(1)$. Varying $a$, we see that $f\left(\mathbb{Z}_{>0}\right)$ is infinite. Next, ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 172 | 2,423 |
2023 | T0 | N8 | Number Theory | IMO-SL | Let $\mathbb{Z}_{>0}$ be the set of positive integers. Determine all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that $$ f^{b f(a)}(a+1)=(a+1) f(b) $$ holds for all $a, b \in \mathbb{Z}_{>0}$, where $f^{k}(n)=f(f(\cdots f(n) \cdots))$ denotes the composition of $f$ with itself $k$ times. (Taiwan) ... | In the same way as Steps 1-2 of We first note that Claim 2 in Solution 1 is also true for $a=1$. Claim 2'. For any $a, n \in \mathbb{Z}_{>0}$, we have $f^{n}(a) \neq a$. Proof. If $a \geqslant 2$, the assertion was proved in Claim 2 in Solution 1. If $a=1$, we have that 1 is not in the range of $f$ by Claim 3 in Soluti... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 172 | 916 |
2023 | T0 | N8 | Number Theory | IMO-SL | Let $\mathbb{Z}_{>0}$ be the set of positive integers. Determine all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that $$ f^{b f(a)}(a+1)=(a+1) f(b) $$ holds for all $a, b \in \mathbb{Z}_{>0}$, where $f^{k}(n)=f(f(\cdots f(n) \cdots))$ denotes the composition of $f$ with itself $k$ times. (Taiwan) ... | The following is another way of finishing $$ \begin{aligned} f^{b g(a)}(f(a)) & =f^{b f(f(a)-1)}(f(a)) \stackrel{P(f(a)-1, b)}{=} f(a) f(b) \stackrel{P(f(b)-1, a)}{=} f^{a f(f(b)-1)}(f(b)) \\ & =f^{a g(b)+k}(f(a)) \end{aligned} $$ By Claim 2' in Solution 2, we have $b g(a)=a g(b)+k$, so $f^{k}(a) \cdot g(a)=a \cdot g\l... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 172 | 747 |
2023 | T0 | N8 | Number Theory | IMO-SL | Let $\mathbb{Z}_{>0}$ be the set of positive integers. Determine all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that $$ f^{b f(a)}(a+1)=(a+1) f(b) $$ holds for all $a, b \in \mathbb{Z}_{>0}$, where $f^{k}(n)=f(f(\cdots f(n) \cdots))$ denotes the composition of $f$ with itself $k$ times. (Taiwan) ... | We provide yet another (more technical) solution assuming Step 1 and Step 2 of Solution 1. By Claim 5 in Solution 1, every $a \geqslant 2$ is a descendant of 1 . Let $g$ and $h$ be the functions on $\mathbb{Z}_{\geqslant 2}$ such that $f^{g(a)}(1)=a$ and $h(a)=f(a-1)$. Then, $g: \mathbb{Z}_{\geqslant 2} \rightarrow \ma... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 172 | 2,099 |
2000 | T3 | 4 | null | INMO | In a convex quadrilateral $P Q R S, P Q=R S,(\sqrt{3}+1) Q R=S P$ and $\angle R S P-\angle S P Q=$ $30^{\circ}$. Prove that
$$
\angle P Q R-\angle Q R S=90^{\circ}
$$ | Let $[$ Fig $]$ denote the area of Fig. We have
$$
[P Q R S]=[P Q R]+[R S P]=[Q R S]+[S P Q]
$$
Let us write $P Q=p, Q R=q, R S=r, S P=s$. The above relations reduce to
$$
p q \sin \angle P Q R+r s \sin \angle R S P=q r \sin \angle Q R S+s p \sin \angle S P Q
$$
Using $p=r$ and $(\sqrt{3}+1) q=s$ and dividing by $p... | {
"problem_match": "\n4.",
"resource_path": "INMO/segmented/en-2000.jsonl",
"solution_match": "\nSol. :"
} | 73 | 586 |
2000 | T3 | 6 | null | INMO | For any natural number $n,(n \geq 3)$, let $f(n)$ denote the number of non-congruent integer-sided triangles with perimeter $n$ (e.g., $f(3)=1, f(4)=0, f(7)=2$ ). Show that
(a) $f(1999)>f(1996)$
(b) $f(2000)=f(1997)$. | (a) Let $a, b, c$ be the sides of a triangle with $a+b+c=1996$, and each being a positive integer. Then $a+1, b+1, c+1$ are also sides of a triangle with perimeter 1999 because
$$
a<b+c \quad \Longrightarrow \quad a+1<(b+1)+(c+1)
$$
and so on. Moreover $(999,999,1)$ form the sides of a triangle with perimeter 1999, w... | {
"problem_match": "\n6.",
"resource_path": "INMO/segmented/en-2000.jsonl",
"solution_match": "\nSol. :"
} | 99 | 507 |
2001 | T3 | 1 | null | INMO | Let $A B C$ be a triangle in which no angle is $90^{\circ}$. For any point $P$ in the plane of the triangle, let $A_{1}, B_{1}, C_{1}$ denote the reflections of $P$ in the sides $B C, C A, A B$ respectively. Prove the following statements:
(a) If $P$ is the incentre or an excentre of $A B C$, then $P$ is the circumcen... | (a)
If $P=I$ is the incentre of triangle $A B C$, and $r$ its inradius, then it is clear that $A_{1} I=B_{1} I=C_{1} I=2 r$. It follows that $I$ is the circumcentre of $A_{1} B_{1} C_{1}$. On the otherhand if $P=I_{1}$ is the excentre of $A B C$ opposite $A$ and $r_{1}$ the corresponding exradius, then again we see th... | {
"problem_match": "\n1.",
"resource_path": "INMO/segmented/en-2001.jsonl",
"solution_match": "## Solution:"
} | 205 | 1,033 |
2001 | T3 | 5 | null | INMO | Let $A B C$ be a triangle and $D$ be the mid-point of side $B C$. Suppose $\angle D A B=\angle B C A$ and $\angle D A C=15^{\circ}$. Show that $\angle A D C$ is obtuse. Further, if $O$ is the circumcentre of $A D C$, prove that triangle $A O D$ is equilateral. | Let $\alpha$ denote the equal angles $\angle B A D=\angle D C A$. Using sine rule in triangles $D A B$ and $D A C$, we get
$$
\frac{A D}{\sin B}=\frac{B D}{\sin \alpha}, \quad \frac{C D}{\sin 15^{\circ}}=\frac{A D}{\sin \alpha}
$$
Eliminating $\alpha$ (using $B D=D C$ and $2 \alpha+B+15^{\circ}=\pi$ ), we obtain $1+\... | {
"problem_match": "\n5.",
"resource_path": "INMO/segmented/en-2001.jsonl",
"solution_match": "## Solution:"
} | 89 | 1,505 |
2002 | T3 | 1 | null | INMO | For a convex hexagon $A B C D E F$ in which each pair of opposite sides is unequal, consider the following six statements:
$$
\begin{array}{ll}
\text { (a } \left.\mathrm{a}_{1}\right) A B \text { is parallel to } D E ; & \left(\mathrm{a}_{2}\right) A E=B D \\
\left(\mathrm{~b}_{1}\right) B C \text { is parallel to } ... | (a) Suppose all the six statements are true. Then $A B D E, B C E F, C D F A$ are isosceles trapeziums; if $K, L, M, P, Q, R$ are the mid-points of $A B, B C$, $C D, D E, E F, F A$ respectively, then we see that $K P \perp A B, E D ; L Q \perp$ $B C, E F$ and $M R \perp C D, F A$.
If $A D, B E, C F$ themselves concur ... | {
"problem_match": "\n1.",
"resource_path": "INMO/segmented/en-2002.jsonl",
"solution_match": "## Solution:"
} | 232 | 701 |
2003 | T3 | 1 | null | INMO | Consider an acute triangle $A B C$ and let $P$ be an interior point of $A B C$. Suppose the lines $B P$ and $C P$, when produced, meet $A C$ and $A B$ in $E$ and $F$ respectively. Let $D$ be the point where $A P$ intersects the line segment $E F$ and $K$ be the foot of perpendicular from $D$ on to $B C$. Show that $D K... | Produce $A P$ to meet $B C$ in $Q$. Join $K E$ and $K F$. Draw perpendiculars from $F$ and $E$ on to $B C$ to meet it in $M$ and $L$ respectively. Let us denote $\angle B K F$ by $\alpha$ and $\angle C K E$ by $\beta$. We show that $\alpha=\beta$ by proving $\tan \alpha=\tan \beta$. This implies that $\angle D K F=\ang... | {
"problem_match": "\n1.",
"resource_path": "INMO/segmented/en-2003.jsonl",
"solution_match": "\nSolution:"
} | 113 | 588 |
2003 | T3 | 3 | null | INMO | Show that for every real number $a$ the equation
$$
8 x^{4}-16 x^{3}+16 x^{2}-8 x+a=0
$$
has at least one non-real root and find the sum of all the non-real roots of the equation. | Substituting $x=y+(1 / 2)$ in the equation, we obtain the equation in $y$ :
$$
8 y^{4}+4 y^{2}+a-\frac{3}{2}=0
$$
Using the transformation $z=y^{2}$, we get a quadratic equation in $z$ :
$$
8 z^{2}+4 z+a-\frac{3}{2}=0
$$
The discriminant of this equation is $32(2-a)$ which is nonnegative if and only if $a \leq 2$. ... | {
"problem_match": "\n3.",
"resource_path": "INMO/segmented/en-2003.jsonl",
"solution_match": "\nSolution:"
} | 61 | 513 |
2004 | T3 | 1 | null | INMO | Consider a convex quadrilateral $A B C D$, in which $K, L, M, N$ are the midpoints of the sides $A B$, $B C, C D, D A$ respectively. Suppose
(a) $B D$ bisects $K M$ at $Q$;
(b) $Q A=Q B=Q C=Q D$; and
(c) $L K / L M=C D / C B$.
Prove that $A B C D$ is a square. | Fig. 1.
Observe that $K L M N$ is a paralellogram, $Q$ is the midpoint of $M K$ and hence $N L$ also passes through $Q$. Let $T$ be the point of intersection of $A C$ and $B D$; and let $S$ be the point of intersection of $B D$ and $M N$.
Consider the triangle $M N K$. Note that $S Q$ is parallel to $N K$ and $Q$ is ... | {
"problem_match": "\n1.",
"resource_path": "INMO/segmented/en-2004.jsonl",
"solution_match": "## Solution:"
} | 111 | 805 |
2004 | T3 | 2 | null | INMO | Suppose $p$ is a prime greater than 3 . Find all pairs of integers $(a, b)$ satisfying the equation
$$
a^{2}+3 a b+2 p(a+b)+p^{2}=0
$$ | We write the equation in the form
$$
a^{2}+2 a p+p^{2}+b(3 a+2 p)=0
$$
Hence
$$
b=\frac{-(a+p)^{2}}{3 a+2 p}
$$
is an integer. This shows that $3 a+2 p$ divides $(a+p)^{2}$ and hence also divides $(3 a+3 p)^{2}$. But, we have
$$
(3 a+3 p)^{2}=(3 a+2 p+p)^{2}=(3 a+2 p)^{2}+2 p(3 a+2 p)+p^{2}
$$
It follows that $3 ... | {
"problem_match": "\n2.",
"resource_path": "INMO/segmented/en-2004.jsonl",
"solution_match": "\nSolution:"
} | 50 | 950 |
2004 | T3 | 3 | null | INMO | If $\alpha$ is a real root of the equation $x^{5}-x^{3}+x-2=0$, prove that $\left[\alpha^{6}\right]=3$. (For any real number $a$, we denote by $[a]$ the greatest integer not exceeding $a$.) | Suppose $\alpha$ is a real root of the given equation. Then
$$
\alpha^{5}-\alpha^{3}+\alpha-2=0
$$
This gives $\alpha^{5}-\alpha^{3}+\alpha-1=1$ and hence $(\alpha-1)\left(\alpha^{4}+\alpha^{3}+1\right)=1$. Observe that $\alpha^{4}+\alpha^{3}+1 \geq$ $2 \alpha^{2}+\alpha^{3}=\alpha^{2}(\alpha+2)$. If $-1 \leq \alpha<... | {
"problem_match": "\n3.",
"resource_path": "INMO/segmented/en-2004.jsonl",
"solution_match": "\nSolution:"
} | 64 | 762 |
2004 | T3 | 4 | null | INMO | Let $R$ denote the circumradius of a triangle $A B C ; a, b, c$ its sides $B C, C A, A B$; and $r_{a}, r_{b}, r_{c}$ its exradii opposite $A, B, C$. If $2 R \leq r_{a}$, prove that
(i) $a>b$ and $a>c$;
(ii) $2 R>r_{b}$ and $2 R>r_{c}$. | We know that $2 R=\frac{a b c}{2 \triangle}$ and $r_{a}=\frac{\triangle}{s-a}$, where $a, b, c$ are the sides of the triangle $A B C$, $s=\frac{a+b+c}{2}$ and $\triangle$ is the area of $A B C$. Thus the given condition $2 R \leq r_{a}$ translates to
$$
a b c \leq \frac{2 \triangle^{2}}{s-a}
$$
Putting $s-a=p, s-b=q,... | {
"problem_match": "\n4.",
"resource_path": "INMO/segmented/en-2004.jsonl",
"solution_match": "\nSolution:"
} | 106 | 621 |
2004 | T3 | 5 | null | INMO | Let $S$ denote the set of all 6-tuples $(a, b, c, d, e, f)$ of positive integers such that $a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=$ $f^{2}$. Consider the set
$$
T=\{a b c d e f:(a, b, c, d, e, f) \in S\}
$$
Find the greatest common divisor of all the members of $T$. | We show that the required gcd is 24 . Consider an element $(a, d, c, d, e, f) \in S$. We have
$$
a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=f^{2}
$$
We first observe that not all $a, b, c, d, e$ can be odd. Otherwise, we have $a^{2} \equiv b^{2} \equiv c^{2} \equiv d^{2} \equiv e^{2} \equiv 1$ $(\bmod 8)$ and hence $f^{2} \equiv ... | {
"problem_match": "\n5.",
"resource_path": "INMO/segmented/en-2004.jsonl",
"solution_match": "\nSolution:"
} | 111 | 823 |
2004 | T3 | 6 | null | INMO | Prove that the number of 5 -tuples of positive integers $(a, b, c, d, e)$ satisfying the equation
$$
a b c d e=5(b c d e+a c d e+a b d e+a b c e+a b c d)
$$
is an odd integer. | We write the equation in the form:
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}=\frac{1}{5}
$$
The number of five tuple ( $a, b, c, d, e)$ which satisfy the given relation and for which $a \neq b$ is even, because for if $(a, b, c, d, e)$ is a solution, then so is $(b, a, c, d, e)$ which is distinct... | {
"problem_match": "\n6.",
"resource_path": "INMO/segmented/en-2004.jsonl",
"solution_match": "\nSolution:"
} | 66 | 522 |
2005 | T3 | 1 | null | INMO | Let $M$ be the midpoint of side $B C$ of a triangle $A B C$. Let the median $A M$ intersect the incircle of $A B C$ at $K$ and $L, K$ being nearer to $A$ than $L$. If $A K=K L=L M$, prove that the sides of triangle $A B C$ are in the ratio $5: 10: 13$ in some order. | Let $I$ be the incentre of triangle $A B C$ and $D$ be its projection on $B C$. Observe that $A B \neq A C$ as $A B=A C$ implies that $D=L=M$. So assume that $A C>A B$. Let $N$ be the projection of $I$ on $K L$. Then the perpendicular $I N$ from $I$ to $K L$ is a bisector of $K L$ and as $A K=L M$, it is a bisector of ... | {
"problem_match": "\n1.",
"resource_path": "INMO/segmented/en-2005.jsonl",
"solution_match": "## Solution:"
} | 98 | 730 |
2005 | T3 | 3 | null | INMO | Let $p, q, r$ be positive real numbers, not all equal, such that some two of the equations
$$
p x^{2}+2 q x+r=0, \quad q x^{2}+2 r x+p=0, \quad r x^{2}+2 p x+q=0
$$
have a common root, say $\alpha$. Prove that
(a) $\alpha$ is real and negative; and
(b) the third equation has non-real roots. | Consider the discriminants of the three equations
$$
\begin{array}{r}
p x^{2}+q r+r=0 \\
q x^{2}+r x+p=0 \\
r x^{2}+p x+q=0
\end{array}
$$
Let us denote them by $D_{1}, D_{2}, D_{3}$ respectively. Then we have
$$
D_{1}=4\left(q^{2}-r p\right), D_{2}=4\left(r^{2}-p q\right), D_{3}=4\left(p^{2}-q r\right)
$$
We obser... | {
"problem_match": "\n3.",
"resource_path": "INMO/segmented/en-2005.jsonl",
"solution_match": "## Solution:"
} | 108 | 902 |
2005 | T3 | 4 | null | INMO | All possible 6-digit numbers, in each of which the digits occur in non-increasing order (from left to right, e.g., 877550) are written as a sequence in increasing order. Find the 2005 -th number in this sequence. | Consider a 6-digit number whose digits from left to right are in non increasing order. If 1 is the first digit of such a number, then the subsequent digits cannot exceed 1. The set of all such numbers with initial digit equal to 1 is
$$
\{100000,110000,111000,111100,111110,1111111\} .
$$
There are elements in this se... | {
"problem_match": "\n4.",
"resource_path": "INMO/segmented/en-2005.jsonl",
"solution_match": "## Solution I:"
} | 59 | 1,354 |
2005 | T3 | 4 | null | INMO | All possible 6-digit numbers, in each of which the digits occur in non-increasing order (from left to right, e.g., 877550) are written as a sequence in increasing order. Find the 2005 -th number in this sequence. | II
It is known that the number of ways of choosing $r$ objects from $n$ different types of objects (with repetitions allowed) is $\binom{n+r-1}{r}$. In particular, if we want to write $r$-digit numbers using $n$ digits allowing for repetitions with the additional condition that the digits appear in non-increasing orde... | {
"problem_match": "\n4.",
"resource_path": "INMO/segmented/en-2005.jsonl",
"solution_match": "## Solution:"
} | 59 | 679 |
2005 | T3 | 5 | null | INMO | Let $x_{1}$ be a given positive integer. A sequence $\left\langle x_{n}\right\rangle_{n=1}^{\infty}=\left\langle x_{1}, x_{2}, x_{3}, \cdots\right\rangle$ of positive integers is such that $x_{n}$, for $n \geq 2$, is obtained from $x_{n-1}$ by adding some nonzero digit of $x_{n-1}$. Prove that
(a) the sequence has an ... | (a) Let us assume that there are no even numbers in the sequence. This means that $x_{n+1}$ is obtained from $x_{n}$, by adding a nonzero even digit of $x_{n}$ to $x_{n}$, for each $n \geq 1$.
Let $E$ be the left most even digit in $x_{1}$ which may be taken in the form
$$
x_{1}=O_{1} O_{2} \cdots O_{k} E D_{1} D_{2}... | {
"problem_match": "\n5.",
"resource_path": "INMO/segmented/en-2005.jsonl",
"solution_match": "## Solution:"
} | 127 | 547 |
2006 | T3 | 1 | null | INMO | In a non-equilateral triangle $A B C$, the sides $a, b, c$ form an arithmetic progression. Let $I$ and $O$ denote the incentre and circumcentre of the triangle respectively.
(i) Prove that $I O$ is perpendicular to $B I$.
(ii) Suppose $B I$ extended meets $A C$ in $K$, and $D, E$ are the midpoints of $B C, B A$ respe... | (i) Extend $B I$ to meet the circumcircle in $F$. Then we know that $F A=F I=F C$. (See Figure)
Let $B I: I F=\lambda: \mu$. Applying Stewart's theorem to triangle $B A F$, we get
$$
\lambda A F^{2}+\mu A B^{2}=(\lambda+\mu)\left(A I^{2}+B I \cdot I F\right)
$$
Similarly, Stewart's theorem to triangle $B C F$ gives
... | {
"problem_match": "\n1.",
"resource_path": "INMO/segmented/en-2006.jsonl",
"solution_match": "## Solution:"
} | 116 | 1,332 |
2006 | T3 | 2 | null | INMO | Prove that for every positive integer $n$ there exists a unique ordered pair $(a, b)$ of positive integers such that
$$
n=\frac{1}{2}(a+b-1)(a+b-2)+a
$$ | We have to prove that $f: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ defined by
$$
f(a, b)=\frac{1}{2}(a+b-1)(a+b-2)+a, \quad \forall a, b \in \mathbb{N}
$$
is a bijection. (Note that the right side is a natural number.) To this end define
$$
T(n)=\frac{n(n+1)}{2}, \quad n \in \mathbb{N} \cup\{0\}
$$
An i... | {
"problem_match": "\n2.",
"resource_path": "INMO/segmented/en-2006.jsonl",
"solution_match": "\nSolution:"
} | 50 | 1,019 |
2006 | T3 | 3 | null | INMO | Let $X$ denote the set of all triples $(a, b, c)$ of integers. Define a function $f: X \rightarrow X$ by
$$
f(a, b, c)=(a+b+c, a b+b c+c a, a b c)
$$
Find all triples $(a, b, c)$ in $X$ such that $f(f(a, b, c))=(a, b, c)$. | We show that the solutionset consists of $\{(t, 0,0) ; t \in \mathbb{Z}\} \cup\{(-1,-1,1)\}$. Let us put $a+b+c=d, a b+b c+c a=e$ and $a b c=f$. The given condition $f(f(a, b, c))=(a, b, c)$ implies that
$$
d+e+f=a, \quad d e+e f+f d=b, d e f=c
$$
Thus $a b c d e f=f c$ and hence either $c f=0$ or $a b d e=1$.
Case ... | {
"problem_match": "\n3.",
"resource_path": "INMO/segmented/en-2006.jsonl",
"solution_match": "\nSolution:"
} | 92 | 1,101 |
2006 | T3 | 5 | null | INMO | In a cyclic quadrilateral $A B C D, A B=a, B C=b, C D=c, \angle A B C=120^{\circ}$, and $\angle A B D=$ $30^{\circ}$. Prove that
(i) $c \geq a+b$;
(ii) $|\sqrt{c+a}-\sqrt{c+b}|=\sqrt{c-a-b}$. | Applying cosine rule to triangle $A B C$, we get
$$
A C^{2}=a^{2}+b^{2}-2 a b \cos 120^{\circ}=a^{2}+b^{2}+a b
$$
Observe that $\angle D A C=\angle D B C=120^{\circ}-30^{\circ}=90^{\circ}$. Thus we get
$$
c^{2}=\frac{A C^{2}}{\cos ^{2} 30^{\circ}}=\frac{4}{3}\left(a^{2}+b^{2}+a b\right)
$$
So
$$
c^{2}-(a+b)^{2}=\f... | {
"problem_match": "\n5.",
"resource_path": "INMO/segmented/en-2006.jsonl",
"solution_match": "## Solution:"
} | 90 | 537 |
2006 | T3 | 6 | null | INMO | (a) Prove that if $n$ is a positive integer such that $n \geq 4011^{2}$, then there exists an integer $l$ such that $n<l^{2}<\left(1+\frac{1}{2005}\right) n$.
(b) Find the smallest positive integer $M$ for which whenever an integer $n$ is such that $n \geq M$, there exists an integer $l$, such that $n<l^{2}<\left(1+\f... | (a) Let $n \geq 4011^{2}$ and $m \in \mathbb{N}$ be such that $m^{2} \leq n<(m+1)^{2}$. Then
$$
\begin{aligned}
\left(1+\frac{1}{2005}\right) n-(m+1)^{2} & \geq\left(1+\frac{1}{2005}\right) m^{2}-(m+1)^{2} \\
& =\frac{m^{2}}{2005}-2 m-1 \\
& =\frac{1}{2005}\left(m^{2}-4010 m-2005\right) \\
& =\frac{1}{2005}\left((m-20... | {
"problem_match": "\n6.",
"resource_path": "INMO/segmented/en-2006.jsonl",
"solution_match": "## Solution:"
} | 125 | 800 |
2007 | T3 | 1 | null | INMO | In a triangle $A B C$ right-angled at $C$, the median through $B$ bisects the angle between $B A$ and the bisector of $\angle B$. Prove that
$$
\frac{5}{2}<\frac{A B}{B C}<3
$$ | Since $E$ is the mid-point of $A C$, we have $A E=$ $E C=b / 2$. Since $B D$ bisects $\angle A B C$, we also know that $C D=a b /(a+c)$. Since $B E$ bisects $\angle A B D$, we also have
$$
\frac{B D^{2}}{B A^{2}}=\frac{D E^{2}}{E A^{2}}
$$
However,
$$
\begin{aligned}
B D^{2} & =B C^{2}+C D^{2}=a^{2}+\frac{a^{2} b^{2... | {
"problem_match": "\n1.",
"resource_path": "INMO/segmented/en-2007.jsonl",
"solution_match": "## Solution 1:"
} | 65 | 524 |
2007 | T3 | 3 | null | INMO | Let $m$ and $n$ be positive integers such that the equation $x^{2}-m x+n=0$ has real roots $\alpha$ and $\beta$. Prove that $\alpha$ and $\beta$ are integers if and only if $[m \alpha]+[m \beta]$ is the square of an integer. (Here $[x]$ denotes the largest integer not exceeding $x$.) | If $\alpha$ and $\beta$ are both integers, then
$$
[m \alpha]+[m \beta]=m \alpha+m \beta=m(\alpha+\beta)=m^{2}
$$
This proves one implication.
Observe that $\alpha+\beta=m$ and $\alpha \beta=n$. We use the property of integer function: $x-1<[x] \leq x$ for any real number $x$. Thus
$m^{2}-2=m(\alpha+\beta)-2=m \alp... | {
"problem_match": "\n3.",
"resource_path": "INMO/segmented/en-2007.jsonl",
"solution_match": "\nSolution:"
} | 86 | 535 |
2007 | T3 | 5 | null | INMO | Let $A B C$ be a triangle in which $A B=A C$. Let $D$ be the mid-point of $B C$ and $P$ be a point on $A D$. Suppose $E$ is the foot of perpendicular from $P$ on $A C$. If $\frac{A P}{P D}=\frac{B P}{P E}=\lambda, \frac{B D}{A D}=m$ and $z=m^{2}(1+\lambda)$, prove that
$$
z^{2}-\left(\lambda^{3}-\lambda^{2}-2\right) z... | Let $A D=h, P D=y$ and $B D=D C=a$. We
observe that $B P^{2}=a^{2}+y^{2}$. Moreover,
$P E=P A \sin \angle D A C=(h-y) \frac{D C}{A C}=\frac{a(h-y)}{b}$,
where $b=A C=A B$. Using $A P / P D=(h-$ $y) / y$, we obtain $y=h /(1+\lambda)$. Thus
$$
\lambda^{2}=\frac{B P^{2}}{P E^{2}}=\frac{\left(a^{2}+y^{2}\right) b^{2}}{... | {
"problem_match": "\n5.",
"resource_path": "INMO/segmented/en-2007.jsonl",
"solution_match": "## Solution:"
} | 169 | 680 |
2007 | T3 | 6 | null | INMO | If $x, y, z$ are positive real numbers, prove that
$$
(x+y+z)^{2}(y z+z x+x y)^{2} \leq 3\left(y^{2}+y z+z^{2}\right)\left(z^{2}+z x+x^{2}\right)\left(x^{2}+x y+y^{2}\right)
$$ | If we set $x=\lambda a, y=\lambda b, z=\lambda c$, then the inequality changes to
$$
(a+b+c)^{2}(a b+b c+c a)^{2} \leq 3\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right)
$$
This shows that we may assume $x+y+z=1$. Let $\alpha=x y+y z+z x$. We see that
$$
\begin{aligned}
x^{2}+x y+y... | {
"problem_match": "\n6.",
"resource_path": "INMO/segmented/en-2007.jsonl",
"solution_match": "\nSolution 3:"
} | 84 | 506 |
2008 | T3 | 3 | null | INMO | Let $A$ be a set of real numbers such that $A$ has at least four elements. Suppose $A$ has the property that $a^{2}+b c$ is a rational number for all distinct numbers $a, b, c$ in $A$. Prove that there exists a positive integer $M$ such that $a \sqrt{M}$ is a rational number for every $a$ in $A$. | Suppose $0 \in A$. Then $a^{2}=a^{2}+0 \times b$ is rational and $a b=0^{2}+a b$ is also rational for all $a, b$ in $A, a \neq 0, b \neq 0, a \neq b$. Hence $a=a_{1} \sqrt{M}$ for some rational $a_{1}$ and natural number $M$. For any $b \neq 0$, we have
$$
b \sqrt{M}=\frac{a b}{a_{1}}
$$
which is a rational number.
... | {
"problem_match": "\n3.",
"resource_path": "INMO/segmented/en-2008.jsonl",
"solution_match": "\nSolution:"
} | 90 | 833 |
2008 | T3 | 5 | null | INMO | Let $A B C$ be a triangle; $\Gamma_{A}, \Gamma_{B}, \Gamma_{C}$ be three equal, disjoint circles inside $A B C$ such that $\Gamma_{A}$ touches $A B$ and $A C ; \Gamma_{B}$ touches $A B$; and $B C$, and $\Gamma_{C}$ touches $B C$ and $C A$. Let $\Gamma$ be a circle touching circles $\Gamma_{A}, \Gamma_{B}, \Gamma_{C}$ e... | Let $O_{1}, O_{2}, O_{3}$ be the centres of the circles $\Gamma_{A}, \Gamma_{B}, \Gamma_{C}$ respectively, and let $P$ be the circum-centre of the triangle $O_{1} O_{2} O_{3}$. Let $x$ denote the common radius of three circles $\Gamma_{A}, \Gamma_{B}$, $\Gamma_{C}$. Note that $P$ is also the centre of the circle $\Gamm... | {
"problem_match": "\n5.",
"resource_path": "INMO/segmented/en-2008.jsonl",
"solution_match": "\nSolution:"
} | 148 | 830 |
2008 | T3 | 6 | null | INMO | Let $P(x)$ be a given polynomial with integer coefficients. Prove that there exist two polynomials $Q(x)$ and $R(x)$, again with integer coefficients, such that (i) $P(x) Q(x)$ is a polynomial in $x^{2}$; and (ii) $P(x) R(x)$ is a polynomial in $x^{3}$. | Let $P(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}$ be a polynomial with integer coefficients.
Part (i) We may write
$$
P(x)=a_{0}+a_{2} x^{2}+a_{4} x^{4}+\cdots+x\left(a_{1}+a_{3} x^{2}+a_{5} x^{5}+\cdots\right)
$$
Define
$$
Q(x)=a_{0}+a_{2} x^{2}+a_{4} x^{4}+\cdots-x\left(a_{1}+a_{3} x^{2}+a_{5} x^{5}+\cdots\... | {
"problem_match": "\n6.",
"resource_path": "INMO/segmented/en-2008.jsonl",
"solution_match": "\nSolution:"
} | 79 | 853 |
2009 | T3 | 1 | null | INMO | Let $A B C$ be a triangle and let $P$ be an interior point such that $\angle B P C=90^{\circ}, \angle B A P=$ $\angle B C P$. Let $M, N$ be the mid-points of $A C, B C$ respectively. Suppose $B P=2 P M$. Prove that $A, P, N$ are collinear. | Extend $C P$ to $D$ such that $C P=P D$. Let $\angle B C P=\alpha=\angle B A P$. Observe that $B P$ is the perpendicular bisector of $C D$. Hence $B C=B D$ and $B C D$ is an isosceles triangle. Thus $\angle B D P=\alpha$. But then $\angle B D P=$ $\alpha=\angle B A P$. This implies that $B, P, A, D$ all lie on a circle... | {
"problem_match": "\n1.",
"resource_path": "INMO/segmented/en-2009.jsonl",
"solution_match": "## Solution:"
} | 87 | 623 |
2009 | T3 | 6 | null | INMO | Let $a, b, c$ be positive real numbers such that $a^{3}+b^{3}=c^{3}$. Prove that
$$
a^{2}+b^{2}-c^{2}>6(c-a)(c-b)
$$ | The given inequality may be written in the form
$$
7 c^{2}-6(a+b) c-\left(a^{2}+b^{2}-6 a b\right)<0
$$
Putting $x=7 c^{2}, y=-6(a+b) c, z=-\left(a^{2}+b^{2}-6 a b\right)$, we have to prove that $x+y+z<0$. Observe that $x, y, z$ are not all equal $(x>0, y<0)$. Using the identity
$$
x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}... | {
"problem_match": "\n6.",
"resource_path": "INMO/segmented/en-2009.jsonl",
"solution_match": "## Solution:"
} | 56 | 622 |
2010 | T3 | 1 | null | INMO | Let $A B C$ be a triangle with circum-circle $\Gamma$. Let $M$ be a point in the interior of triangle $A B C$ which is also on the bisector of $\angle A$. Let $A M, B M, C M$ meet $\Gamma$ in $A_{1}, B_{1}, C_{1}$ respectively. Suppose $P$ is the point of intersection of $A_{1} C_{1}$ with $A B$; and $Q$ is the point o... | Let $A=2 \alpha$. Then $\angle A_{1} A C=\angle B A A_{1}=\alpha$. Thus
$$
\angle A_{1} B_{1} C=\alpha=\angle B B_{1} A_{1}=\angle A_{1} C_{1} C=\angle B C_{1} A_{1}
$$
We also have $\angle B_{1} C Q=\angle A A_{1} B_{1}=\beta$, say. It follows that triangles $M A_{1} B_{1}$ and $Q C B_{1}$ are similar and hence
$$
... | {
"problem_match": "\n1.",
"resource_path": "INMO/segmented/en-2010.jsonl",
"solution_match": "\nSolution:"
} | 138 | 586 |
2010 | T3 | 3 | null | INMO | Find all non-zero real numbers $x, y, z$ which satisfy the system of equations:
$$
\begin{aligned}
\left(x^{2}+x y+y^{2}\right)\left(y^{2}+y z+z^{2}\right)\left(z^{2}+z x+x^{2}\right) & =x y z \\
\left(x^{4}+x^{2} y^{2}+y^{4}\right)\left(y^{4}+y^{2} z^{2}+z^{4}\right)\left(z^{4}+z^{2} x^{2}+x^{4}\right) & =x^{3} y^{3}... | Since $x y z \neq 0$, We can divide the second relation by the first. Observe that
$$
x^{4}+x^{2} y^{2}+y^{4}=\left(x^{2}+x y+y^{2}\right)\left(x^{2}-x y+y^{2}\right)
$$
holds for any $x, y$. Thus we get
$$
\left(x^{2}-x y+y^{2}\right)\left(y^{2}-y z+z^{2}\right)\left(z^{2}-z x+x^{2}\right)=x^{2} y^{2} z^{2}
$$
How... | {
"problem_match": "\n3.",
"resource_path": "INMO/segmented/en-2010.jsonl",
"solution_match": "\nSolution:"
} | 160 | 509 |
2010 | T3 | 4 | null | INMO | How many 6-tuples $\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}\right)$ are there such that each of $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ is from the set $\{1,2,3,4\}$ and the six expressions
$$
a_{j}^{2}-a_{j} a_{j+1}+a_{j+1}^{2}
$$
for $j=1,2,3,4,5,6$ (where $a_{7}$ is to be taken as $a_{1}$ ) are all equal ... | Without loss of generality, we may assume that $a_{1}$ is the largest among $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$. Consider the relation
$$
a_{1}^{2}-a_{1} a_{2}+a_{2}^{2}=a_{2}^{2}-a_{2} a_{3}+a_{3}^{2}
$$
This leads to
$$
\left(a_{1}-a_{3}\right)\left(a_{1}+a_{3}-a_{2}\right)=0
$$
Observe that $a_{1} \geq a_... | {
"problem_match": "\n4.",
"resource_path": "INMO/segmented/en-2010.jsonl",
"solution_match": "\nSolution:"
} | 157 | 1,035 |
2010 | T3 | 5 | null | INMO | Let $A B C$ be an acute-angled triangle with altitude $A K$. Let $H$ be its ortho-centre and $O$ be its circum-centre. Suppose $K O H$ is an acute-angled triangle and $P$ its circum-centre. Let $Q$ be the reflection of $P$ in the line $H O$. Show that $Q$ lies on the line joining the mid-points of $A B$ and $A C$. | Let $D$ be the mid-point of $B C ; M$ that of $H K$; and $T$ that of $O H$. Then $P M$ is perpendicular to $H K$ and $P T$ is perpendicular to $O H$. Since $Q$ is the reflection of $P$ in $H O$, we observe that $P, T, Q$ are collinear, and $P T=T Q$. Let $Q L, T N$ and $O S$ be the perpendiculars drawn respectively fro... | {
"problem_match": "\n5.",
"resource_path": "INMO/segmented/en-2010.jsonl",
"solution_match": "\nSolution:"
} | 100 | 590 |
2010 | T3 | 6 | null | INMO | Define a sequence $\left\langle a_{n}\right\rangle_{n \geq 0}$ by $a_{0}=0, a_{1}=1$ and
$$
a_{n}=2 a_{n-1}+a_{n-2}
$$
for $n \geq 2$.
(a) For every $m>0$ and $0 \leq j \leq m$, prove that $2 a_{m}$ divides $a_{m+j}+(-1)^{j} a_{m-j}$.
(b) Suppose $2^{k}$ divides $n$ for some natural numbers $n$ and $k$. Prove that ... | (a) Consider $f(j)=a_{m+j}+(-1)^{j} a_{m-j}, 0 \leq j \leq m$, where $m$ is a natural number. We observe that $f(0)=2 a_{m}$ is divisible by $2 a_{m}$. Similarly,
$$
f(1)=a_{m+1}-a_{m-1}=2 a_{m}
$$
is also divisible by $2 a_{m}$. Assume that $2 a_{m}$ divides $f(j)$ for all $0 \leq j<l$, where $l \leq m$. We prove th... | {
"problem_match": "\n6.",
"resource_path": "INMO/segmented/en-2010.jsonl",
"solution_match": "## Solution:"
} | 157 | 588 |
2011 | T3 | 1 | null | INMO | Let $D, E, F$ be points on the sides $B C, C A, A B$ respectively of a triangle $A B C$ such that $B D=C E=A F$ and $\angle B D F=\angle C E D=\angle A F E$. Prove that $A B C$ is equilateral. | Let $B D=C E=A F=x ; \angle B D F=$ $\angle C E D=\angle A F E=\theta$. Note that $\angle A F D=$ $B+\theta$, and hence $\angle D F E=B$. Similarly, $\angle E D F=C$ and $\angle F E D=A$. Thus the triangle $E F D$ is similar to $A B C$. We may take $F D=k a, D E=k b$ and $E F=k c$, for some positive real constant $k$. ... | {
"problem_match": "\n1.",
"resource_path": "INMO/segmented/en-2011.jsonl",
"solution_match": "## Solution 1:"
} | 71 | 565 |
2011 | T3 | 1 | null | INMO | Let $D, E, F$ be points on the sides $B C, C A, A B$ respectively of a triangle $A B C$ such that $B D=C E=A F$ and $\angle B D F=\angle C E D=\angle A F E$. Prove that $A B C$ is equilateral. | We write (1) in the form (2), and start from there. The case of two equal sides is dealt as in Solution 1. We assume no two sides are equal. Using ratio properties in (2), we obtain
$$
\frac{a-b}{\left(a b-c^{2}\right) / c a}=\frac{b-c}{\left(b c-a^{2}\right) / a b}
$$
This may be written as $c(a-b)\left(b c-a^{2}\ri... | {
"problem_match": "\n1.",
"resource_path": "INMO/segmented/en-2011.jsonl",
"solution_match": "\nSolution 2."
} | 71 | 810 |
2011 | T3 | 2 | null | INMO | Call a natural number $n$ faithful, if there exist natural numbers $a<b<c$ such that $a$ divides $b, b$ divides $c$ and $n=a+b+c$.
(i) Show that all but a finite number of natural numbers are faithful.
(ii) Find the sum of all natural numbers which are not faithful. | Suppose $n \in \mathbb{N}$ is faithful. Let $k \in \mathbb{N}$ and consider $k n$. Since $n=a+b+c$, with $a>b>c, c \mid b$ and $b \mid a$, we see that $k n=k a+k b+k c$ which shows that $k n$ is faithful.
Let $p>5$ be a prime. Then $p$ is odd and $p=(p-3)+2+1$ shows that $p$ is faithful. If $n \in \mathbb{N}$ contains... | {
"problem_match": "\n2.",
"resource_path": "INMO/segmented/en-2011.jsonl",
"solution_match": "\nSolution 1:"
} | 71 | 573 |
2011 | T3 | 2 | null | INMO | Call a natural number $n$ faithful, if there exist natural numbers $a<b<c$ such that $a$ divides $b, b$ divides $c$ and $n=a+b+c$.
(i) Show that all but a finite number of natural numbers are faithful.
(ii) Find the sum of all natural numbers which are not faithful. | If $n=a+b+c$ with $a<b<c$ is faithful, we see that $a \geq 1, b \geq 2$ and $c \geq 4$. Hence $n \geq 7$. Thus $1,2,3,4,5,6$ are not faithful. As observed earlier, $k n$ is faithful whenever
$n$ is. We also notice that for odd $n \geq 7$, we can write $n=1+2+(n-3)$ so that all odd $n \geq 7$ are faithful. Consider $2 n... | {
"problem_match": "\n2.",
"resource_path": "INMO/segmented/en-2011.jsonl",
"solution_match": "\nSolution 2:"
} | 71 | 500 |
2011 | T3 | 3 | null | INMO | Consider two polynomials $P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$ and $Q(x)=b_{n} x^{n}+$ $b_{n-1} x^{n-1}+\cdots+b_{1} x+b_{0}$ with integer coefficients such that $a_{n}-b_{n}$ is a prime, $a_{n-1}=b_{n-1}$ and $a_{n} b_{0}-a_{0} b_{n} \neq 0$. Suppose there exists a rational number $r$ such that $P(r)... | Let $r=u / v$ where $\operatorname{gcd}(u, v)=1$. Then we get
$$
\begin{array}{r}
a_{n} u^{n}+a_{n-1} u^{n-1} v+\cdots+a_{1} u v^{n-1}+a_{0} v^{n}=0 \\
b_{n} u^{n}+b_{n-1} u^{n-1} v+\cdots+b_{1} u v^{n-1}+b_{0} v^{n}=0
\end{array}
$$
Subtraction gives
$$
\left(a_{n}-b_{n}\right) u^{n}+\left(a_{n-2}-b_{n-2}\right) u^... | {
"problem_match": "\n3.",
"resource_path": "INMO/segmented/en-2011.jsonl",
"solution_match": "\nSolution:"
} | 165 | 674 |
2011 | T3 | 6 | null | INMO | Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that
$$
f(x+y) f(x-y)=(f(x)+f(y))^{2}-4 x^{2} f(y)
$$
for all $x, y \in \mathbf{R}$, where $\mathbf{R}$ denotes the set of all real numbers. | Put $x=y=0$; we get $f(0)^{2}=4 f(0)^{2}$ and hence $f(0)=0$.
Put $x=y$ : we get $4 f(x)^{2}-4 x^{2} f(x)=0$ for all $x$. Hence for each $x$, either $f(x)=0$ or $f(x)=x^{2}$.
Suppose $f(x) \not \equiv 0$. Then we can find $x_{0} \neq 0$ such that $f\left(x_{0}\right) \neq 0$. Then $f\left(x_{0}\right)=x_{0}^{2} \neq ... | {
"problem_match": "\n6.",
"resource_path": "INMO/segmented/en-2011.jsonl",
"solution_match": "\nSolution 1.:"
} | 82 | 935 |
2012 | T3 | 1 | null | INMO | Let $A B C D$ be a quadrilateral inscribed in a circle. Suppose $A B=$ $\sqrt{2+\sqrt{2}}$ and $A B$ subtends $135^{\circ}$ at the centre of the circle. Find the maximum possible area of $A B C D$.
=R^{2}(2+\sqrt{2})
$$
Hence $R=1$.
Consider quadrilateral $A B C D$ as in the second figure above. Join $A C$. For $[A D C]... | {
"problem_match": "\n1.",
"resource_path": "INMO/segmented/en-2012.jsonl",
"solution_match": "\nSolution:"
} | 141 | 670 |
2012 | T3 | 2 | null | INMO | Let $p_{1}<p_{2}<p_{3}<p_{4}$ and $q_{1}<q_{2}<q_{3}<q_{4}$ be two sets of prime numbers such that $p_{4}-p_{1}=8$ and $q_{4}-q_{1}=8$. Suppose $p_{1}>5$ and $q_{1}>5$. Prove that 30 divides $p_{1}-q_{1}$. | Since $p_{4}-p_{1}=8$, and no prime is even, we observe that $\left\{p_{1}, p_{2}, p_{3}, p_{4}\right\}$ is a subset of $\left\{p_{1}, p_{1}+2, p_{1}+4, p_{1}+6, p_{1}+8\right\}$. Moreover $p_{1}$ is larger than 3 . If $p_{1} \equiv 1(\bmod 3)$, then $p_{1}+2$ and $p_{1}+8$ are divisible by 3 . Hence we do not get 4 pr... | {
"problem_match": "\n2.",
"resource_path": "INMO/segmented/en-2012.jsonl",
"solution_match": "\nSolution:"
} | 100 | 528 |
2012 | T3 | 4 | null | INMO | Let $A B C$ be a triangle. An interior point $P$ of $A B C$ is said to be good if we can find exactly 27 rays emanating from $P$ intersecting the sides of the triangle $A B C$ such that the triangle is divided by these rays into 27 smaller triangles of equal area. Determine the number of good points for a given triangl... | Let $P$ be a good point. Let $l, m, n$ be respetively the number of parts the sides $B C, C A, A B$ are divided by the rays starting from $P$. Note that a ray must pass through each of the vertices the triangle $A B C$; otherwise we get some quadrilaterals.
Let $h_{1}$ be the distance of $P$ from $B C$. Then $h_{1}$ i... | {
"problem_match": "\n4.",
"resource_path": "INMO/segmented/en-2012.jsonl",
"solution_match": "\nSolution:"
} | 88 | 1,056 |
2012 | T3 | 5 | null | INMO | Let $A B C$ be an acute-angled triangle, and let $D, E, F$ be points on $B C$, $C A, A B$ respectively such that $A D$ is the median, $B E$ is the internal angle bisector and $C F$ is the altitude. Suppose $\angle F D E=\angle C, \angle D E F=$ $\angle A$ and $\angle E F D=\angle B$. Prove that $A B C$ is equilateral. | Since $\triangle B F C$ is right-angled at $F$, we have
$F D=B D=C D=a / 2$. Hence $\angle B F D=\angle B$. Since $\angle E F D=$ $\angle B$, we have $\angle A F E=\pi-2 \angle B$. Since $\angle D E F=\angle A$, we also get $\angle C E D=\pi-2 \angle B$. Applying sine rule in $\triangle D E F$, we have
$$
\frac{D F}{... | {
"problem_match": "\n5.",
"resource_path": "INMO/segmented/en-2012.jsonl",
"solution_match": "\nSolution:"
} | 108 | 544 |
2012 | T3 | 6 | null | INMO | Let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be a function satisfying $f(0) \neq 0, f(1)=0$ and
(i) $f(x y)+f(x) f(y)=f(x)+f(y)$;
(ii) $(f(x-y)-f(0)) f(x) f(y)=0$,
for all $x, y \in \mathbb{Z}$, simultaneously.
(a) Find the set of all possible values of the function $f$.
(b) If $f(10) \neq 0$ and $f(2)=0$, find the ... | Setting $y=0$ in the condition (ii), we get
$$
(f(x)-f(0)) f(x)=0
$$
for all $x$ (since $f(0) \neq 0$ ). Thus either $f(x)=0$ or $f(x)=f(0)$, for all $x \in \mathbb{Z}$. Now taking $x=y=0$ in (i), we see that $f(0)+f(0)^{2}=2 f(0)$. This shows
that $f(0)=0$ or $f(0)=1$. Since $f(0) \neq 0$, we must have $f(0)=1$. We ... | {
"problem_match": "\n6.",
"resource_path": "INMO/segmented/en-2012.jsonl",
"solution_match": "\nSolution:"
} | 158 | 942 |
2023 | T3 | 2 | null | INMO | Suppose $a_{0}, \ldots, a_{100}$ are positive reals. Consider the following polynomial for each $k$ in $\{0,1, \ldots, 100\}$ :
$$
a_{100+k} x^{100}+100 a_{99+k} x^{99}+a_{98+k} x^{98}+a_{97+k} x^{97}+\cdots+a_{2+k} x^{2}+a_{1+k} x+a_{k}
$$
where indices are taken modulo 101, i.e., $a_{100+i}=a_{i-1}$ for any $i$ in ... | Let $n=50$. For the sake of contradiction, assume that each of these polynomials has all real roots; these roots must be negative. Let
$$
-\alpha_{1, k},-\alpha_{2, k}, \ldots,-\alpha_{2 n, k}
$$
be the roots of the polynomial
$$
a_{2 n+k} x^{2 n}+2 n a_{2 n-1+k} x^{2 n-1}+a_{2 n-2+k} x^{2 n-2}+a_{2 n-3+k} x^{2 n-3}... | {
"problem_match": "\nProblem 2.",
"resource_path": "INMO/segmented/en-2023.jsonl",
"solution_match": "\nSolution 1."
} | 199 | 736 |
2023 | T3 | 3 | null | INMO | Let $\mathbb{N}$ denote the set of all positive integers. Find all real numbers $c$ for which there exists a function $f: \mathbb{N} \rightarrow \mathbb{N}$ satisfying:
(a) for any $x, a \in \mathbb{N}$, the quantity $\frac{f(x+a)-f(x)}{a}$ is an integer if and only if $a=1$;
(b) for all $x \in \mathbb{N}$, we have $... | We claim that the only possible values of $c$ are $k+\frac{1}{2}$ for some non-negative integer $k$. The fact that these values are possible is seen from the function $f(x)=$ $\left\lfloor\left(k+\frac{1}{2}\right) x\right\rfloor+1=k x+\left\lfloor\frac{x}{2}\right\rfloor+1$. Indeed, if you have any $x, a \in \mathbb{N... | {
"problem_match": "\nProblem 3.",
"resource_path": "INMO/segmented/en-2023.jsonl",
"solution_match": "\nSolution 1."
} | 123 | 1,290 |
2023 | T3 | 3 | null | INMO | Let $\mathbb{N}$ denote the set of all positive integers. Find all real numbers $c$ for which there exists a function $f: \mathbb{N} \rightarrow \mathbb{N}$ satisfying:
(a) for any $x, a \in \mathbb{N}$, the quantity $\frac{f(x+a)-f(x)}{a}$ is an integer if and only if $a=1$;
(b) for all $x \in \mathbb{N}$, we have $... | We will show that for any such $c$, we have $c>0$ and $\{c\}=\frac{1}{2}$. Also 2023 can be replaced by any fixed $d \geq 1$ in condition $(b)$ which we assume now.
Clearly $c \geq 0$ else for $c<0$ and $x>\frac{d}{|c|}, c x-d<f(x)<c x+d<0$ which is a contradiction. Suppose $\{c\} \neq \frac{1}{2}$. Put $r=\lfloor c\r... | {
"problem_match": "\nProblem 3.",
"resource_path": "INMO/segmented/en-2023.jsonl",
"solution_match": "\nSolution 2."
} | 123 | 619 |
2023 | T3 | 3 | null | INMO | Let $\mathbb{N}$ denote the set of all positive integers. Find all real numbers $c$ for which there exists a function $f: \mathbb{N} \rightarrow \mathbb{N}$ satisfying:
(a) for any $x, a \in \mathbb{N}$, the quantity $\frac{f(x+a)-f(x)}{a}$ is an integer if and only if $a=1$;
(b) for all $x \in \mathbb{N}$, we have $... | We give a different proof that $\{c\}=1 / 2$. Let us first prove a claim:
Claim. For any $k \geq 1$ and any $x, f\left(x+2^{k}\right)-f(x)$ is divisible by $2^{k-1}$ but not $2^{k}$.
Proof. We prove this via induction on $k$. For $k=1$, the claim is trivial. Now assume the statement is true for some $k$, and note tha... | {
"problem_match": "\nProblem 3.",
"resource_path": "INMO/segmented/en-2023.jsonl",
"solution_match": "\nSolution 3."
} | 123 | 1,000 |
2023 | T3 | 4 | null | INMO | Let $k \geq 1$ and $N>1$ be two integers. On a circle are placed $2 N+1$ coins all showing heads. Calvin and Hobbes play the following game. Calvin starts and on his move can turn any coin from heads to tails. Hobbes on his move can turn at most one coin that is next to the coin that Calvin turned just now from tails t... | Calvin wins if $k \in\{1,2, \ldots, N+1\}$ and Hobbes wins otherwise.
Label the coins $1,2, \ldots, 2 N+1$. Note that if $k \geq N+2$ then Hobbes wins as follows: he pairs the coins $2 i-1$ and $2 i$ for $1 \leq i \leq N$. If Calvin in a move makes both coins in a pair tails, Hobbes in that move turns the one which wa... | {
"problem_match": "\nProblem 4.",
"resource_path": "INMO/segmented/en-2023.jsonl",
"solution_match": "\nSolution 1."
} | 122 | 1,095 |
2023 | T3 | 5 | null | INMO | Euler marks $n$ different points in the Euclidean plane. For each pair of marked points, Gauss writes down the number $\left\lfloor\log _{2} d\right\rfloor$ where $d$ is the distance between the two points. Prove that Gauss writes down less than $2 n$ distinct values.
Note: For any $d>0,\left\lfloor\log _{2} d\right\r... | Given a pair of points with distance $d$, we say their log-distance is $\left\lfloor\log _{2} d\right\rfloor$. By dilating by a suitable power of 2 , we can assume all log-distances are positive and the smallest one of them is 1 . Define a sequence of graphs $G_{0}, G_{1}, G_{2}, \cdots$ as follows: $G_{k}$ has the $n$... | {
"problem_match": "\nProblem 5.",
"resource_path": "INMO/segmented/en-2023.jsonl",
"solution_match": "\nSolution 2."
} | 123 | 924 |
2023 | T3 | 6 | null | INMO | Euclid has a tool called cyclos which allows him to do the following:
- Given three non-collinear marked points, draw the circle passing through them.
- Given two marked points, draw the circle with them as endpoints of a diameter.
- Mark any intersection points of two drawn circles or mark a new point on a drawn circ... | We begin by proving a series of lemmas.
Lemma 1. Given a non-right angled triangle $A B C$, we can draw the nine-point circle and mark the orthocentre $H$ using only a cyclos.
Proof. Draw circles $(B C),(C A),(A B)$ and mark their intersections to get the three feet of altitudes $D, E, F$ opposite $A, B, C$. Now draw... | {
"problem_match": "\nProblem 6.",
"resource_path": "INMO/segmented/en-2023.jsonl",
"solution_match": "\nSolution 1."
} | 99 | 1,638 |
2021 | T3 | 1 | null | INMO | Suppose $r \geq 2$ is an integer, and let $m_{1}, n_{1}, m_{2}, n_{2}, \cdots, m_{r}, n_{r}$ be $2 r$ integers such that
$$
\left|m_{i} n_{j}-m_{j} n_{i}\right|=1
$$
for any two integers $i$ and $j$ satisfying $1 \leq i<j \leq r$. Determine the maximum possible value of $r$. | Let $m_{1}, n_{1}, m_{2}, n_{2}$ be integers satisfying $m_{1} n_{2}-m_{2} n_{1}= \pm 1$. By changing the signs of $m_{2}, n_{2}$ if need be, we may assume that
$$
m_{1} n_{2}-m_{2} n_{1}=1
$$
If $m_{3}, n_{3}$ are integers satisfying $m_{1} n_{3}-m_{3} n_{1}= \pm 1$, again we may assume (by changing their signs if n... | {
"problem_match": "\nProblem 1.",
"resource_path": "INMO/segmented/en-INMO_2021_solutions.jsonl",
"solution_match": "\nSolution."
} | 113 | 1,012 |
2021 | T3 | 2 | null | INMO | Find all pairs of integers $(a, b)$ so that each of the two cubic polynomials
$$
x^{3}+a x+b \text { and } x^{3}+b x+a
$$
has all the roots to be integers. | The only such pair is $(0,0)$, which clearly works. To prove this is the only one, let us prove an auxiliary result first.
Lemma If $\alpha, \beta, \gamma$ are reals so that $\alpha+\beta+\gamma=0$ and $|\alpha|,|\beta|,|\gamma| \geq 2$, then
$$
|\alpha \beta+\beta \gamma+\gamma \alpha|<|\alpha \beta \gamma|
$$
Proo... | {
"problem_match": "\nProblem 2.",
"resource_path": "INMO/segmented/en-INMO_2021_solutions.jsonl",
"solution_match": "\nSolution."
} | 54 | 937 |
2021 | T3 | 3 | null | INMO | Betal marks 2021 points on the plane such that no three are collinear, and draws all possible line segments joining these. He then chooses any 1011 of these line segments, and marks their midpoints. Finally, he chooses a line segment whose midpoint is not marked yet, and challenges Vikram to construct its midpoint usin... | The answer is 'yes'. To prove this, we will first prove two lemmas:
Lemma 1 Given any two points $A, B$, their midpoint $M$, and any point $C$, Vikram can draw a line parallel to $A B$ through $C$.
Proof. If $C$ is on line $A B$ we are already done. If not, extend $B C$ to $X$ as shown, draw $P=A C \cap X M$, and the... | {
"problem_match": "\nProblem 3.",
"resource_path": "INMO/segmented/en-INMO_2021_solutions.jsonl",
"solution_match": "\nSolution."
} | 114 | 1,344 |
2021 | T3 | 3 | null | INMO | Betal marks 2021 points on the plane such that no three are collinear, and draws all possible line segments joining these. He then chooses any 1011 of these line segments, and marks their midpoints. Finally, he chooses a line segment whose midpoint is not marked yet, and challenges Vikram to construct its midpoint usin... | Strategy for the Detective: Pick a card $A$ and compare against all others except one. If he ever gets a "Yes", that pair works; else the remaining card is consecutive with A. This process takes at most 50 queries.
Strategy for the Magician: We show that it is not always possible to obtain a "Yes" in 50 turns, hence s... | {
"problem_match": "\nProblem 3.",
"resource_path": "INMO/segmented/en-INMO_2021_solutions.jsonl",
"solution_match": "\nSolution."
} | 114 | 1,445 |
2021 | T3 | 3 | null | INMO | Betal marks 2021 points on the plane such that no three are collinear, and draws all possible line segments joining these. He then chooses any 1011 of these line segments, and marks their midpoints. Finally, he chooses a line segment whose midpoint is not marked yet, and challenges Vikram to construct its midpoint usin... | First we show that $\angle D E C=30^{\circ}$. Choose a point $F$ on $A B$ such that $C F=C B$. Join $F C$ and $F D$. Observe that $\angle D C B=75^{\circ}+25^{\circ}=100^{\circ}$. Since $C D=C B$, we have $\angle C D B=\angle C B D=40^{\circ}$. Therefore $\angle C B F=40^{\circ}+30^{\circ}=70^{\circ}$. This gives $\ang... | {
"problem_match": "\nProblem 3.",
"resource_path": "INMO/segmented/en-INMO_2021_solutions.jsonl",
"solution_match": "\nSolution."
} | 114 | 817 |
2021 | T3 | 6 | null | INMO | Let $\mathbb{R}[x]$ be the set of all polynomials with real coefficients, and let $\operatorname{deg} P$ denote the degree of a nonzero polynomial $P$. Find all functions $f: \mathbb{R}[x] \rightarrow \mathbb{R}[x]$ satisfying the following conditions:
- $f$ maps the zero polynomial to itself,
- for any non-zero polyn... | Answer
We have $f(p)=p \forall p \in \mathbb{R}[x]$, or $f(p)=-p \forall p \in \mathbb{R}[x]$. These clearly satisfy the given conditions.
## Proof
Claim 1 For all $p \in \mathbb{R}[x], f(f(p))=p$.
Proof. Using condition 3 on the polynomials $p$ and $f(p)$, we see that $p-f(f(p))$ has the same set of real roots as ... | {
"problem_match": "\nProblem 6.",
"resource_path": "INMO/segmented/en-INMO_2021_solutions.jsonl",
"solution_match": "## Solution."
} | 166 | 1,841 |
2024 | T3 | 4 | null | INMO | A finite set $S$ of positive integers is called cardinal if $S$ contains the integer $|S|$, where $|S|$ denotes the number of distinct elements in $S$. Let $f$ be a function from the set of positive integers to itself, such that for any cardinal set $S$, the set $f(S)$ is also cardinal. Here $f(S)$ denotes the set of a... | The possible values are 1, 2, and 2024 .
Construction. The function $f(x)=1$ for all $x \in \mathbb{N}$ works. Also, $f(x)=1$ for all $x \neq 2024$ and $f(2024)=2$, works. Finally, $f(x)=x$ for all $x \in \mathbb{N}$ works as well.
It remains to show these are the only possible values for $f(2024)$.
Proof. Denote $\... | {
"problem_match": "\nProblem 4.",
"resource_path": "INMO/segmented/en-INMO_2024_final_solutions.jsonl",
"solution_match": "\nSolution 1."
} | 152 | 762 |
2024 | T3 | 5 | null | INMO | Let points $A_{1}, A_{2}$, and $A_{3}$ lie on the circle $\Gamma$ in counter-clockwise order, and let $P$ be a point in the same plane. For $i \in\{1,2,3\}$, let $\tau_{i}$ denote the counter-clockwise rotation of the plane centred at $A_{i}$, where the angle of the rotation is equal to the angle at vertex $A_{i}$ in $... | Fix an index $i \in\{1,2,3\}$. Let $D_{1}, D_{2}, D_{3}$ be the points of tangency of the incircle of triangle $\triangle A_{1} A_{2} A_{3}$ with its sides $A_{2} A_{3}, A_{3} A_{1}, A_{1} A_{2}$ respectively.
The key observation is that given a line $\ell$ in the plane, the image of $\ell$ under the mapping $\tau_{i+... | {
"problem_match": "\nProblem 5.",
"resource_path": "INMO/segmented/en-INMO_2024_final_solutions.jsonl",
"solution_match": "\nSolution 1."
} | 223 | 533 |
2024 | T3 | 5 | null | INMO | Let points $A_{1}, A_{2}$, and $A_{3}$ lie on the circle $\Gamma$ in counter-clockwise order, and let $P$ be a point in the same plane. For $i \in\{1,2,3\}$, let $\tau_{i}$ denote the counter-clockwise rotation of the plane centred at $A_{i}$, where the angle of the rotation is equal to the angle at vertex $A_{i}$ in $... | Toss the figure on the complex plane, and let $A_{1}=a, A_{2}=b, A_{3}=c$ without loss of generality. Let the angles of the triangle at $A_{1}, A_{2}, A_{3}$ be denoted by $A, B, C$.
Now, for any complex number $z$, the rotation at $z_{0}$ with angle $\theta$ counterclockwise sends $z$ to $\left(z-z_{0}\right) e^{i \t... | {
"problem_match": "\nProblem 5.",
"resource_path": "INMO/segmented/en-INMO_2024_final_solutions.jsonl",
"solution_match": "\nSolution 2."
} | 223 | 1,369 |
2024 | T3 | 6 | null | INMO | For each positive integer $n \geq 3$, define $A_{n}$ and $B_{n}$ as
$$
\begin{gathered}
A_{n}=\sqrt{n^{2}+1}+\sqrt{n^{2}+3}+\cdots+\sqrt{n^{2}+2 n-1} \\
B_{n}=\sqrt{n^{2}+2}+\sqrt{n^{2}+4}+\cdots+\sqrt{n^{2}+2 n}
\end{gathered}
$$
Determine all positive integers $n \geq 3$ for which $\left\lfloor A_{n}\right\rfloor=\... | Let $M=n^{2}+\frac{1}{2} n$.
Lemma 1. $B_{n}-A_{n}<\frac{1}{2}$.
Indeed,
$$
\begin{array}{r}
\left(B_{n}-A_{n}\right)=\sum_{k=1}^{n}\left(\sqrt{n^{2}+2 k}-\sqrt{n^{2}+2 k-1}\right)=\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+2 k}+\sqrt{n^{2}+2 k-1}}<\sum_{k=1}^{n} \frac{1}{2 n} \\
=\frac{n}{2 n}=\frac{1}{2}
\end{array}
$$
... | {
"problem_match": "\nProblem 6.",
"resource_path": "INMO/segmented/en-INMO_2024_final_solutions.jsonl",
"solution_match": "\nSolution."
} | 190 | 832 |
2019 | T3 | 5 | null | INMO | Let $A B$ be a diameter of a circle $\Gamma$ and let $C$ be a point on $\Gamma$ different from $A$ and $B$. Let $D$ be the foot of perpendicular from $C$ on to $A B$. Let $K$ be a point of the segment $C D$ such that $A C$ is equal to the semiperimeter of the triangle $A D K$. Show that the excircle of triangle $A D K$... | Draw another diameter $P Q \perp A B$. Let $E$ be the point at which the excircle $\Gamma_{1}$ touches the line $A D$. Join $Q E$ and extend it to meet $\Gamma$ in $L$. Draw the diameter $E N$ of $\Gamma_{1}$ and draw $Q S \perp N E$ (extended). See the figure. We also observe that $D E=E M=E N / 2$.
Since $A E$ is eq... | {
"problem_match": "\n5.",
"resource_path": "INMO/segmented/en-Inmo-2019-Solutions.jsonl",
"solution_match": "\nSolution:"
} | 111 | 929 |
2019 | T3 | 6 | null | INMO | Let $f$ be function defined from the set $\{(x, y): x, y$ reals, $x y \neq 0\}$ in to the set of all positive real numbers such that
(i) $\quad f(x y, z)=f(x, z) f(y, z)$, for all $x, y \neq 0$;
(ii) $\quad f(x, y z)=f(x, y) f(x, z)$, for all $x, y \neq 0$;
(iii) $\quad f(x, 1-x)=1$, for all $x \neq 0,1$.
Prove tha... | (The condition (ii) was inadvertently left out in the paper. We give the solution with condition (ii).)
Taking $x=y=1$ in (ii), weget $f(1, z)^{2}=f(1, z)$ so that $f(1, z)=1$ for all $z \neq 0$. Similarly, $x=y=-1$ gives $f(-1, z)=1$ for all $z \neq 0$. Using the second condition, we also get $f(z, 1)=$ $f(z,-1)=1$ f... | {
"problem_match": "\n6.",
"resource_path": "INMO/segmented/en-Inmo-2019-Solutions.jsonl",
"solution_match": "\nSolution:"
} | 199 | 585 |
2013 | T3 | 1 | null | INMO | Let $\Gamma_{1}$ and $\Gamma_{2}$ be two circles touching each other externally at $R$. Let $l_{1}$ be a line which is tangent to $\Gamma_{2}$ at $P$ and passing through the center $O_{1}$ of $\Gamma_{1}$. Similarly, let $l_{2}$ be a line which is tangent to $\Gamma_{2}$ at $Q$ and passing through the center $O_{2}$ of... | Suppose that $P$ and $Q$ lie on the opposite sides of line joining $O_{1}$ and $O_{2}$. By symmetry we may assume that the configuration is as shown in the figure below. Then we have $K P>K O_{1}>K Q$ since $K O_{1}$ is the hypotenuse of triangle $K Q O_{1}$. This is a contradiction to the given assumption, and therefo... | {
"problem_match": "\nProblem 1.",
"resource_path": "INMO/segmented/en-inmo2013-solutions.jsonl",
"solution_match": "\nSolution."
} | 146 | 578 |
2013 | T3 | 5 | null | INMO | In an acute triangle $A B C, O$ is the circumcenter, $H$ is the orthocenter and $G$ is the centroid. Let $O D$ be perpendicular to $B C$ and $H E$ be perpendicular to $C A$, with $D$ on $B C$ and $E$ on $C A$. Let $F$ be the midpoint of $A B$. Suppose the areas of triangles $O D C, H E A$ and $G F B$ are equal. Find al... | Let $R$ be the circumradius of $\triangle A B C$ and $\Delta$ its area. We have $O D=R \cos A$ and $D C=\frac{a}{2}$, so
$$
[O D C]=\frac{1}{2} \cdot O D \cdot D C=\frac{1}{2} \cdot R \cos A \cdot R \sin A=\frac{1}{2} R^{2} \sin A \cos A
$$
Again $H E=2 R \cos C \cos A$ and $E A=c \cos A$. Hence
$$
[H E A]=\frac{1}{... | {
"problem_match": "\nProblem 5.",
"resource_path": "INMO/segmented/en-inmo2013-solutions.jsonl",
"solution_match": "\nSolution."
} | 122 | 637 |
2015 | T3 | 1 | null | INMO | Let $A B C$ be a right-angled triangle with $\angle B=90^{\circ}$. Let $B D$ be the altitude from $B$ on to $A C$. Let $P, Q$ and $I$ be the incentres of triangles $A B D, C B D$ and $A B C$ respectively. Show that the circumcentre of of the triangle $P I Q$ lies on the hypotenuse $A C$. | We begin with the following lemma:
Lemma: Let $X Y Z$ be a triangle with $\angle X Y Z=90+\alpha$. Construct an isosceles triangle $X E Z$, externally on the side $X Z$, with base angle $\alpha$. Then $E$ is the circumcentre of $\triangle X Y Z$.
Proof of the Lemma: Draw $E D \perp$ $X Z$. Then $D E$ is the perpendic... | {
"problem_match": "\n1.",
"resource_path": "INMO/segmented/en-inmosol-15.jsonl",
"solution_match": "\nSolution:"
} | 99 | 830 |
2015 | T3 | 1 | null | INMO | Let $A B C$ be a right-angled triangle with $\angle B=90^{\circ}$. Let $B D$ be the altitude from $B$ on to $A C$. Let $P, Q$ and $I$ be the incentres of triangles $A B D, C B D$ and $A B C$ respectively. Show that the circumcentre of of the triangle $P I Q$ lies on the hypotenuse $A C$. | Here we use computation to prove that the point of contact $K$ of the incircle with $A C$ is the circumcentre of $\triangle P I Q$. We show that $K P=K Q=r$. Let $r_{1}$ and $r_{2}$ be the inradii of triangles $A B D$ and $C B D$ respectively. Draw $P L \perp A C$ and $Q M \perp A C$. If $s_{1}$ is the semiperimeter of... | {
"problem_match": "\n1.",
"resource_path": "INMO/segmented/en-inmosol-15.jsonl",
"solution_match": "\nSolution 2:"
} | 99 | 535 |
2015 | T3 | 2 | null | INMO | For any natural number $n>1$, write the infinite decimal expansion of $1 / n$ (for example, we write $1 / 2=0.4 \overline{9}$ as its infinite decimal expansion, not 0.5 ). Determine the length of the non-periodic part of the (infinite) decimal expansion of $1 / n$. | For any prime $p$, let $\nu_{p}(n)$ be the maximum power of $p$ dividing $n$; ie $p^{\nu_{p}(n)}$ divides $n$ but not higher power. Let $r$ be the
length of the non-periodic part of the infinite decimal expansion of $1 / n$.
Write
$$
\frac{1}{n}=0 . a_{1} a_{2} \cdots a_{r} \overline{b_{1} b_{2} \cdots b_{s}}
$$
We ... | {
"problem_match": "\n2.",
"resource_path": "INMO/segmented/en-inmosol-15.jsonl",
"solution_match": "\nSolution:"
} | 76 | 519 |
2015 | T3 | 4 | null | INMO | There are four basket-ball players $A, B, C, D$. Initially, the ball is with $A$. The ball is always passed from one person to a different person. In how many ways can the ball come back to $A$ after seven passes? (For example $A \rightarrow C \rightarrow B \rightarrow D \rightarrow A \rightarrow B \rightarrow C \right... | Let $x_{n}$ be the number of ways in which $A$ can get back the ball after $n$ passes. Let $y_{n}$ be the number of ways in which the ball goes back to a fixed person other than $A$ after $n$ passes. Then
$$
x_{n}=3 y_{n-1} \text {, }
$$
and
$$
y_{n}=x_{n-1}+2 y_{n-1}
$$
We also have $x_{1}=0, x_{2}=3, y_{1}=1$ and... | {
"problem_match": "\n4.",
"resource_path": "INMO/segmented/en-inmosol-15.jsonl",
"solution_match": "\nSolution:"
} | 125 | 660 |
2015 | T3 | 5 | null | INMO | Let $A B C D$ be a convex quadrilateral. Let the diagonals $A C$ and $B D$ intersect in $P$. Let $P E, P F, P G$ and $P H$ be the altitudes from $P$ on to the sides $A B, B C, C D$ and $D A$ respectively. Show that $A B C D$ has an incircle if and only if
$$
\frac{1}{P E}+\frac{1}{P G}=\frac{1}{P F}+\frac{1}{P H}
$$ | Let $A P=p, B P=q, C P=r, D P=s ; A B=a, B C=b$, $C D=c$ and $D A=d$. Let $\angle A P B=\angle C P D=\theta$. Then $\angle B P C=\angle D P A=$ $\pi-\theta$. Let us also write $P E=h_{1}, P F=h_{2}, P G=h_{3}$ and $P H=h_{4}$.
## Observe that
$$
h_{1} a=p q \sin \theta, \quad h_{2} b=q r \sin \theta, \quad h_{3} c=r ... | {
"problem_match": "\n5.",
"resource_path": "INMO/segmented/en-inmosol-15.jsonl",
"solution_match": "\nSolution:"
} | 128 | 644 |
2015 | T3 | 6 | null | INMO | From a set of 11 square integers, show that one can choose 6 numbers $a^{2}, b^{2}, c^{2}, d^{2}, e^{2}, f^{2}$ such that
$$
a^{2}+b^{2}+c^{2} \equiv d^{2}+e^{2}+f^{2} \quad(\bmod 12)
$$ | The first observation is that we can find 5 pairs of squares such that the two numbers in a pair have the same parity. We can see this as follows:
| Odd numbers | Even numbers | Odd pairs | Even pairs | Total pairs |
| :---: | :---: | :---: | :---: | :---: |
| 0 | 11 | 0 | 5 | 5 |
| 1 | 10 | 0 | 5 | 5 |
| 2 | 9 | 1 | ... | {
"problem_match": "\n6.",
"resource_path": "INMO/segmented/en-inmosol-15.jsonl",
"solution_match": "\nSolution:"
} | 88 | 779 |
2020 | T3 | 1 | null | INMO | Let $\Gamma_{1}$ and $\Gamma_{2}$ be two circles of unequal radii, with centres $O_{1}$ and $O_{2}$ respectively, in the plane intersecting in two distinct points $A$ and $B$. Assume that the centre of each of the circles $\Gamma_{1}$ and $\Gamma_{2}$ is outside the other. The tangent to $\Gamma_{1}$ at $B$ intersects ... | Let $\angle C B A=\alpha$ and $\angle D B A=\beta$. Then $\angle B D A=\alpha$ and $\angle B C A=\beta$. We also observe that $\angle A O_{1} O_{2}=\left(\angle A O_{1} B / 2\right)=\alpha$ and, simiarly, $\angle A O_{2} O_{1}=\beta$. Hence
$$
\angle O_{1} A O_{2}=180^{\circ}-(\alpha+\beta)
$$
We also have
$$
\angle... | {
"problem_match": "\n1.",
"resource_path": "INMO/segmented/en-sol-inmo-20.jsonl",
"solution_match": "## Solution:"
} | 232 | 886 |
2020 | T3 | 2 | null | INMO | Suppose $P(x)$ is a polynomial with real coefficients satsfying the condition $P(\cos \theta+\sin \theta)=$ $P(\cos \theta-\sin \theta)$, for every real $\theta$. Prove that $P(x)$ can be expressed in the form
$$
P(x)=a_{0}+a_{1}\left(1-x^{2}\right)^{2}+a_{2}\left(1-x^{2}\right)^{4}+\cdots+a_{n}\left(1-x^{2}\right)^{2... | Changing $\theta$ to $\theta-\pi / 2$, we see that
$$
P(\sin \theta+\cos \theta)=P(\sin \theta-\cos \theta)
$$
This shows that $P(x)=P(-x)$ for all $x \in[-\sqrt{2}, \sqrt{2}]$ and as $\mathrm{P}$ is a polynomial, in fact,
$$
P(x)=P(-x)
$$
for all $x \in \mathbb{R}$. Hence $P(x)$ is an even polynomial; $P(x)=Q\left... | {
"problem_match": "\n2.",
"resource_path": "INMO/segmented/en-sol-inmo-20.jsonl",
"solution_match": "\nSolution:"
} | 159 | 516 |
2020 | T3 | 3 | null | INMO | Let $X=\{0,1,2,3,4,5,6,7,8,9\}$. Let $S \subseteq X$ be such that any nonnegative integer $n$ can be written as $p+q$ where the nonnegative integers $p, q$ have all their digits in $S$. Find the smallest possible number of elements in $S$. | We show that 5 numbers will suffice. Take $S=\{0,1,3,4,6\}$. Observe the following splitting:
| $n$ | $a$ | $b$ |
| :--- | :--- | :--- |
| 0 | 0 | 0 |
| 1 | 0 | 1 |
| 2 | 1 | 1 |
| 3 | 0 | 3 |
| 4 | 1 | 3 |
| 5 | 1 | 4 |
| 6 | 3 | 3 |
| 7 | 3 | 4 |
| 8 | 4 | 4 |
| 9 | 3 | 6 |
Thus each digit in a given nonnegative in... | {
"problem_match": "\n3.",
"resource_path": "INMO/segmented/en-sol-inmo-20.jsonl",
"solution_match": "\nSolution:"
} | 82 | 583 |
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