problem string | solution string | candidates list | tags list | metadata dict |
|---|---|---|---|---|
A set of positive integers is said to be *pilak* if it can be partitioned into 2 disjoint subsets $F$ and $T$ , each with at least $2$ elements, such that the elements of $F$ are consecutive Fibonacci numbers, and the elements of $T$ are consecutive triangular numbers. Find all positive integers $n$ such tha... | <details><summary>solution</summary>First, observe that the square of no prime is triangular or a Fibonacci number. Therefore, $n$ must be squarefree or the square of a prime $p$ . But in the latter case, there are only $2$ divisors, contradicting $\tau(n) = |F|+|T| +1 \geqslant 4$ . Suppose $n = p_1p_2\dots p_k... | [] | [
"origin:aops",
"2023 Philippine MO",
"2023 Contests"
] | {
"answer_score": 138,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Philippine MO/3035171.json"
} |
Let $\mathcal{S}$ be the set of all points in the plane. Find all functions $f : \mathcal{S} \rightarrow \mathbb{R}$ such that for all nondegenerate triangles $ABC$ with orthocenter $H$ , if $f(A) \leq f(B) \leq f(C)$ , then $$ f(A) + f(C) = f(B) + f(H). $$ | <details><summary>Solution</summary>We will show that the only solutions are the constant functions, which obviously satisfy the condition.
Assume there exist two points $P$ and $Q$ such that $f(P) \neq f(Q)$ . WLOG $f(P) > f(Q)$ , and $f(P) = a+b, f(Q) = a-b$ , where $b > 0$ . Now consider points $R$ and $... | [] | [
"origin:aops",
"2023 Philippine MO",
"2023 Contests"
] | {
"answer_score": 56,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Philippine MO/3035172.json"
} |
Find all positive integers $b$ with the following property: there exists positive integers $a,k,l$ such that $a^k + b^l$ and $a^l + b^k$ are divisible by $b^{k+l}$ where $k \neq l$ . | Note that if $\nu_p(a)\neq\nu_p(b)$ , then $\nu_p(a+b)=\min(\nu_p(a), \nu_p(b))$ .
Suppose $b$ has the property and $p$ is a prime divisor of $b$ .
If $k\nu_p(a)=\nu_p(a^k)=\nu_p(b^l)=l\nu_p(b)$ , then $\nu_p(a^l)=l\nu_p(a)=\frac{l^2}k\nu_p(b)\neq k\nu_p(b)=\nu_p(b^k)$ since $k^2\neq l^2$ .
Therefore, eithe... | [
"Another solution: W.l.o.g. $k>\\ell$ so that $k=\\ell+d$ . Then \n\\[b^{k+\\ell} \\mid a^d(a^{\\ell}+b^{\\ell+d})-(a^{\\ell+d}+b^{\\ell})=a^db^{\\ell+d}-b^{\\ell}.\\]\nHence $b^k \\mid a^db^d-1$ and in particular $b \\mid 1$ and hence $b=1$ .",
"Deletetetete"
] | [
"origin:aops",
"2023 Contests",
"2023 Poland - Second Round"
] | {
"answer_score": 34,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Poland - Second Round/3012476.json"
} |
Let $n \geq 2$ be an integer. A lead soldier is moving across the unit squares of a $n \times n$ grid, starting from the corner square. Before each move to the neighboring square, the lead soldier can (but doesn't need to) turn left or right. Determine the smallest number of turns, which it needs to do, to visit ev... | Can't believe only 200 people solved this! There are $n$ squares on any diagonal of the board. Every such square has different $x$ and $y$ coordinates, so to go from one to another, soldier has to take at least 2 turns. He needs to do that at least $n-1$ times, so the minimum is $2(n-1)$ . Construction is triv... | [
"we can suppose that the soldier is at the top-right edge of the grid\nfirstly, we can easily get an answer of (2n-2)\nby making the soldier turn right/ left at each end of the line(always go downwards),the turn right/left at the beginning of a new line,the soldier an go through every line of the grid with (2n-2)tu... | [
"origin:aops",
"2023 Contests",
"2023 Poland - Second Round"
] | {
"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Poland - Second Round/3012479.json"
} |
Given positive integers $k,n$ and a real number $\ell$ , where $k,n \geq 1$ . Given are also pairwise different positive real numbers $a_1,a_2,\ldots, a_k$ . Let $S = \{a_1,a_2,\ldots,a_k, -a_1, -a_2,\ldots, -a_k\}$ .
Let $A$ be the number of solutions of the equation $$ x_1 + x_2 + \ldots + x_{2n} = 0, $$ wh... | <blockquote><details><summary>Solution</summary>By direct computation,
\[A=\int_0^1 \left(\sum_{s \in S} e^{2\pi is\alpha }\right)^{2n}d\alpha\]
and
\[B=\int_0^1 \left(\sum_{s \in S} e^{2\pi is\alpha }\right)^{2n} e^{-2\pi i\ell\alpha} d\alpha\]
and the claim follows from the triangle inequality, using that $\sum_{s \... | [
"<details><summary>Solution</summary>By direct computation,\n\\[A=\\int_0^1 \\left(\\sum_{s \\in S} e^{2\\pi is\\alpha }\\right)^{2n}d\\alpha\\]\nand\n\\[B=\\int_0^1 \\left(\\sum_{s \\in S} e^{2\\pi is\\alpha }\\right)^{2n} e^{-2\\pi i\\ell\\alpha} d\\alpha\\]\nand the claim follows from the triangle inequality, us... | [
"origin:aops",
"2023 Contests",
"2023 Poland - Second Round"
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"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 Poland - Second Round/3012483.json"
} |
Given is a triangle $ABC$ with $AB>AC$ . Its incircle touches $AB, AC$ at $D, E$ , respectively. Let $CD$ meet the incircle at $K$ and $L$ is the foot of the perpendicular from $A$ to $CK$ . If $M$ is the midpoint of $DE$ and $H$ is the orthocenter of $\triangle KML$ , prove that $\angle AHK=90^{... | [
"This problem was proposed by **[Burii](https://artofproblemsolving.com/community/user/100466)**.\n\nBy the way, you mistyped the original statement of the problem."
] | [
"origin:aops",
"2023 Contests",
"2023 Poland - Second Round"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Poland - Second Round/3012997.json"
} | |
Given pairwise different real numbers $a,b,c,d,e$ such that $$ \left\{ \begin{array}{ll}
ab + b = ac + a,
bc + c = bd + b,
cd + d = ce + c,
de + e = da + d.
\end{array} \right. $$ Prove that $abcde=1$ . | <details><summary>Solution</summary>Easy to check that $a,b,c,d,e \ne 0$ and $a,b,c,d,e \ne -1$ . Multiplying all equations, we then get that $ea+a=be+e$ .
Subtracting, we get $a(b-c)=a-b, b(c-d)=b-c$ etc. and all the cyclic equations. Multipyling these we get $abcde=1$ as desired.</details> | [] | [
"origin:aops",
"2023 Contests",
"2023 Poland - Second Round"
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"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Poland - Second Round/3013010.json"
} |
Given a chessboard $n \times n$ , where $n\geq 4$ and $p=n+1$ is a prime number. A set of $n$ unit squares is called *tactical* if after putting down queens on these squares, no two queens are attacking each other. Prove that there exists a partition of the chessboard into $n-2$ tactical sets, not containing s... | <details><summary>Solution</summary>Putting the chessboard on the plane coordinates, where the one of the corner is $(1,1)$ .
Main idea is to prove that the set $S = \{(1,k), (2,2k), \ldots, (p-1, (p-1)k)\}$ is tactical for $k \in\{2,3,\ldots, p-2\}$ . (Here everything is modulo $p$ )
We need to prove that for eac... | [
"This problem was proposed by **[Burii](https://artofproblemsolving.com/community/user/100466)**."
] | [
"origin:aops",
"2023 Contests",
"2023 Poland - Second Round"
] | {
"answer_score": 40,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Poland - Second Round/3013015.json"
} |
Determine whether there exist real numbers $x$ , $y$ , $z$ , such that \[x+\frac{1}{y}=z,\quad y+\frac{1}{z}=x,\quad z+\frac{1}{x}=y.\] | $\color{blue} \boxed{\textbf{SOLUTION}}$ These are equivalent to, $xy+1=yz \implies yz-xy=1$ $yz+1=xz \implies xz-yz=1$ $xz+1=xy \implies xy-xz=1$ Summing these gives, $0=3$
So, no real solution for these equations $\blacksquare$ | [
"WLOG $x>=z$ now from $x+\\frac{1}{y} = z$ we get $\\frac{1}{y} < 0$ so we have $y<0$ . But if $x, y, z$ is the answer then so is $-z, -y, -x$ so $y$ can be greater than zero, contradiction.",
"Not only is there no real solutions, but no complex solutions either. This is because clearing the denomina... | [
"origin:aops",
"2023 Polish Junior Math Olympiad Finals",
"2023 Contests"
] | {
"answer_score": 1112,
"boxed": false,
"end_of_proof": true,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 Polish Junior Math Olympiad Finals/3066520.json"
} |
Triangle $ABC$ is given, where $AC<BC$ and $\angle ACB=60^\circ\!\!.$ Point $D$ , distinct from $A$ , lies on the segment $AC$ such that $AB=BD$ , and point $E$ , distinct from $B$ , lies on the line $BC$ such that $AB=AE$ . Prove that $\angle DEC=30^\circ$ . | <details><summary>Solution 1</summary>In other words, we need to prove that $CE=DE$ .
If $A$ and $C$ are fixed and $B$ moves linearly, then both $C$ and $E$ move linearly. So it suffices to check two cases.
If $ABC$ is equilateral, then $C=D=E$ . If $B=C$ , the claim is also trivial. Done.</details>
<det... | [
"στις υπόψιν προς λύση."
] | [
"origin:aops",
"2023 Polish Junior Math Olympiad Finals",
"2023 Contests"
] | {
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Polish Junior Math Olympiad Finals/3066522.json"
} |
Let $n\geq 1$ be odd integer. There are $n$ arrows are arranged from left to right, such that each arrow points either to the left or to the right. Prove that there exists an arrow that is pointed to by exactly as many arrows as it is pointing to.
Note:
For example, for $n=5$ and the arrangement $\rightarrow\r... | We assume that there is at least one arrow of each kind. Because $n$ is odd we assume that the number of left arrows (wlog) is odd. Because it is odd, there is an arrow such that the number of left arrows to its left is equal to the number of left arrows to its right. Now denote the position of it as $k$ . Then the ... | [
"Didn't Understand the question, Can you please clear the statement?",
"Didn't Understand the question, Can you please clear the statement?\n",
"Didn't Understand the question, Can you please clear the statement?",
"If there are no right facing arrows, take the middle arrow. If there is only one, simply take ... | [
"origin:aops",
"2023 Polish Junior Math Olympiad Finals",
"2023 Contests"
] | {
"answer_score": 6,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Polish Junior Math Olympiad Finals/3066524.json"
} |
Find all pairs of positive integers $m$ , $n$ such that the $(m+n)$ -digit number \[\underbrace{33\ldots3}_{m}\underbrace{66\ldots 6}_{n}\] is a perfect square. | We can see that this is a multiple of $3$ so we can divide it to get:
\[\underbrace{11\ldots1}_{m}\underbrace{22\ldots 2}_{n}.\]
Now let us split this into cases: $\mathbb{\text{Case 1:}}$ If $n\geq 2$ , then:
\[\underbrace{33\ldots3}_{m}\underbrace{66\ldots 6}_{n}\equiv 2\pmod{4}.\]
which means that it cannot be a... | [
"A=33...3366....66= $(10^{m+n}-1+10^n-1)/3 = a^2$ $\\Leftrightarrow$ $10^{m+n}+10^n-2 = 3a^2$ After reviewing the equation by mod 4: we see that if $n \\geq 2$ : $\\text{LHS} \\equiv 2 \\equiv 3a^2 \\equiv \\text{RHS} \\pmod{4} \\Rightarrow n = 1$ But if n=1: $10^{m+1}+8 = 3a^2$ . We quickly notice that if... | [
"origin:aops",
"2023 Polish Junior Math Olympiad Finals",
"2023 Contests"
] | {
"answer_score": 32,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Polish Junior Math Olympiad Finals/3066526.json"
} |
There are integers $a$ and $b$ , such that $a>b>1$ and $b$ is the largest divisor of $a$ different from $a$ . Prove that the number $a+b$ is not a power of $2$ with integer exponent. | $\color{blue} \boxed{\textbf{SOLUTION}}$ First, Let, $a=2n \implies b=n \implies (a+b)=3n$ is not equal $2^k$
So, Let, $a=pk$ where $p$ is the smallest prime divisor of $a,$ $a$ is odd, we have $k \in$ odd as $p \in$ odd. $\implies b=k, (a+b)=pk+k=k(p+1)$ But $k$ is not equal to $2^m, m \in \mathbb... | [
"Let us say that the smallest divisor of $a$ is $k$ greater than $1$ . This means that $a+b=(k+1)b$ . In order for this to be a power of $2$ , both $k+1$ , and $b$ are powers of $2$ . If $k=2$ , then $k+1=3$ which does not work. If $k$ is and odd prime, then $k+1$ could be a power of $2$ . Howeve... | [
"origin:aops",
"2023 Polish Junior Math Olympiad Finals",
"2023 Contests"
] | {
"answer_score": 1128,
"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Polish Junior Math Olympiad Finals/3066528.json"
} |
Given is a rectangle with perimeter $x$ cm and side lengths in a $1:2$ ratio. Suppose that the area of the rectangle is also $x$ $\text{cm}^2$ . Determine all possible values of $x$ . | Let $a,b$ be the side lengths of this rectangle, such that $a=2b$ .
Now,knowing that the perimeter of a rectangle is $2(a+b)$ , its area is $ab$ and that they are equal to $x$ (eachother) we have: $x=2(a+b)=ab$ $<=>$ $2\cdot 3b=2b\cdot b$ $<=>$ $6b=2b^2$ $<=>$ $2b=6$ $<=>$ $b=3$ So $a=2\cdot 3... | [
"<details><summary>sol (probably wrong)</summary>Let the sidelengths be $s$ and $2s$ .\n\nWe have $x=6s=2s^2$ , so $s^2-3s=0$ , so $s=0,3$ but assuming the rectangle is non-degenerate, we have $s=3$ , so $x=18$ .</details>",
"P=2(a+b)=x\nS=ab=x\n2(a+b)=ab \na=2b\n2×3b=2b×b\nb=3\na=6\nx= a×b=3×6=18\nx=18\... | [
"origin:aops",
"2023 Contests",
"2023 Polish Junior Math Olympiad First Round"
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"answer_score": 32,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Polish Junior Math Olympiad First Round/3076436.json"
} |
Kamil wrote on a board an expression consisting of alternating addition and subtraction of natural numbers from $1$ to $100$ : \[1-2+3-4+5-6+\ldots-98+99-100.\] Then, Kamil erased one of the plus or minus signs and replaced it with an equals sign, obtaining a true equality. Which number preceded the erased sign? Fin... | [
"All of the posts you made about this contest should be moved to the High School Olympiads if it is high school level or Contests and Programs.\n\n\\req move to C and P.",
"Kamil has two options:\nOption 1: Replacing a plus sign with an equals sign. \nIn this case, the number of terms on each side of the equal si... | [
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"path": "Contest Collections/2023 Contests/2023 Polish Junior Math Olympiad First Round/3076437.json"
} | |
Let $ABCD$ be a rectangle. Point $E$ lies on side $AB$ , and point $F$ lies on segment $CE$ . Prove that if triangles $ADE$ and $CDF$ have equal areas, then triangles $BCE$ and $DEF$ also have equal areas. | Well just notice that $[EDC]$ comprises half of the rectangle's area and $[AED]+ [EBC]$ also comprises the half. | [] | [
"origin:aops",
"2023 Contests",
"2023 Polish Junior Math Olympiad First Round"
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"answer_score": 4,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Polish Junior Math Olympiad First Round/3076440.json"
} |
Each of the natural numbers from $1$ to $n$ is colored either red or blue, with each color being used at least once. It turns out that:
– every red number is a sum of two distinct blue numbers; and
– every blue number is a difference between two red numbers.
Determine the smallest possible value of $n$ for which ... | I find a coloring like this for $n=9$ BBRBBBRRR
Clearly $1.2$ and $4$ are always blue.
So the smallest red number is $3$ but we know that $4$ is blue and so he is a difference of two red number so there always exist a red number $\ge3+4=7$ First, suppose that $7$ is solution.
In this case, $7$ is red beca... | [] | [
"origin:aops",
"2023 Contests",
"2023 Polish Junior Math Olympiad First Round"
] | {
"answer_score": 60,
"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2023 Contests/2023 Polish Junior Math Olympiad First Round/3076441.json"
} |
Positive numbers $a$ , $b$ , $c$ satisfy the inequalities
\[a + b \geq ab, \quad b + c \geq bc,\quad\text{and}\quad c+ a \geq ca.\]
Prove that $\displaystyle a + b + c \geq \frac34abc$ . | We are given,
\[a + b \geq ab, \ \ b + c \geq bc, \ \ c+ a \geq ca\]
Adding, we get
\[4(a+b+c)\geq 2(ab+bc+ca)=(ab+ac)+(bc+ba)+(ca+cb)=a(b+c)+b(c+a)+c(a+b)\geq 3abc\] $\blacksquare$ | [
"We have that $a+b+c=\\frac{a+b}{2}+\\frac{a+c}{2}+\\frac{c+b}{2}\\ge\\frac{ab}{2}+\\frac{ac}{2}+\\frac{bc}{2}$ And $a+b\\ge ab$ $\\iff$ $ac+bc\\ge abc$ , $a+c\\ge ac$ $\\iff$ $ab+bc\\ge abc$ and $c+b\\ge cb$ $\\iff$ $ac+ba\\ge abc$ so $ab+bc+ac\\ge3abc$ so $a+b+c\\ge\\frac{ab}{2}+\\frac{ac}{2}+... | [
"origin:aops",
"2023 Contests",
"2023 Polish Junior Math Olympiad First Round"
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"answer_score": 102,
"boxed": false,
"end_of_proof": true,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Polish Junior Math Olympiad First Round/3076443.json"
} |
Let $ABCDEF$ be a regular hexagon with side length $2$ . Point $M$ is the midpoint of diagonal $AE$ . The pentagon $ABCDE$ is folded along segments $BD$ , $BM$ , and $DM$ in such a way that points $A$ , $C$ , and $E$ coincide. As a result of this operation, a tetrahedron is obtained. Determine its volum... | [] | [
"origin:aops",
"2023 Contests",
"2023 Polish Junior Math Olympiad First Round"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2023 Contests/2023 Polish Junior Math Olympiad First Round/3076444.json"
} | |
We call the figure shown in the picture consisting of five unit squares a $\emph{plus}$ , and each rectangle consisting of two such squares a $\emph{minus}$ . Does there exist an odd integer $n$ with the property that a square with side length $n$ can be dissected into pluses and minuses? Justify your answer.
![... | <details><summary>sol</summary>No. Here is the proof: each plus has area $5$ and each minus has area $2.$ Since the area of a square with an odd side length must be odd, there must be an odd number of plusses. If there wasn't, the area would be even. Now, let us put the situation on a unit checkerboard. Each plus w... | [] | [
"origin:aops",
"2023 Contests",
"2023 Polish Junior Math Olympiad First Round"
] | {
"answer_score": 52,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Polish Junior Math Olympiad First Round/3076448.json"
} |
On the sides $AB$ and $BC$ of triangle $ABC$ , there are points $D$ and $E$ , respectively, such that \[\angle ADC=\angle BDE\quad\text{and}\quad \angle BCD=\angle AED.\] Prove that $AE=BE$ . | Notice that $\Delta ADE$ ∼ $\Delta BDC$ by AA similarity | [
"The second condition gives that if X is the intersection of CD and AE then (CEX) is tangent to DE, so it follows that <CXE = <BED so triangles BDE and ADX are similar, which implies that <EBD = <EAD and AE = BE."
] | [
"origin:aops",
"2023 Polish Junior MO Second Round",
"2023 Contests"
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"answer_score": 4,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Polish Junior MO Second Round/3071899.json"
} |
Initially, the numbers $2$ and $5$ are written on the board. A \emph{move} consists of replacing one of the two numbers on the board with their sum. Is it possible to obtain (in a finite numer of moves) a situation in which the two integers written on the board are consecutive? Justify your answer. | We claim the answer is no.
Note that for the two integers to be consecutive, their difference must be equal to $1$ . Each move maps the set $(m,n) \rightarrow (m,n+m)$ or $(m,n) \rightarrow (m+n,n)$ . Thus the difference between two elements of the set in iteration $k+1$ must have been an element that occurred i... | [
"The answer is no. \n\nTo end with a situation where the two integers are $(x, x+1)$ , then the position beforehand is either (k, x + 1), but that’s obviously impossible as that would lead to $(x + 1, x + k + 1)$ or $(k + x + 1, x)$ Hence the position beforehand could only be $(k, x)$ leading to $(k, x + k)$... | [
"origin:aops",
"2023 Polish Junior MO Second Round",
"2023 Contests"
] | {
"answer_score": 116,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Polish Junior MO Second Round/3071900.json"
} |
A natural number $n$ is at least two digits long. If we write a certain digit between the tens digit and the units digit of this number, we obtain six times the number $n$ . Find all numbers $n$ with this property. | For two-digit $n=10a+b$ , we have $6(10a+b)=100a+10c+b\;\forall a,b,c \in \mathbb{N}, 0 \le a,b,c \le 9$ . This gives the trivial solution of $n=18$ . Now we prove that it is the only one; there are no other larger numbers. This is quite evident when we see that $6$ is too small; to get a four-digit number from a ... | [] | [
"origin:aops",
"2023 Polish Junior MO Second Round",
"2023 Contests"
] | {
"answer_score": 124,
"boxed": false,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Polish Junior MO Second Round/3071904.json"
} |
Consider a parallelogram $ABCD$ where $AB>AD$ . Let $X$ and $Y$ , distinct from $B$ , be points on the ray $BD^\rightarrow$ such that $CX=CB$ and $AY=AB$ . Prove that $DX=DY$ . Note: The notation $BD^\rightarrow$ denotes the ray originating from point $B$ passing through point $D$ . | Note that $AD=CB=CX$ and $CD=AB=AY$ . Let $\angle BAD =\angle BCD =\alpha$ and $\angle CBX=\angle CXB = \angle BDA =\beta$ . We have that $\angle BCX=180^{\circ}-2\beta$ and $\angle XCD = \alpha +2\beta - 180^{\circ}$ . $\angle ABD = \angle ABY= \angle AYB = 180^{\circ}-\alpha-\beta$ Therefore, $\angle YAD=1... | [] | [
"origin:aops",
"2023 Polish Junior MO Second Round",
"2023 Contests"
] | {
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Polish Junior MO Second Round/3071905.json"
} |
In each cell of a $4\times 4$ table, one of the numbers $1$ or $2$ is written. For each row, calculate the sum of the numbers written in it, and for each column, calculate the product of the numbers written in it. Show that some two of the eight results obtained are equal. | We consider the sums of the rows. These can be in the range $\lbrace 4,8 \rbrace$ . We consider the worst-case scenario, so we make all the sums distinct from each other, which can be done in $\binom 5 4 =5$ ways. Trying all $5$ ways shows that the column products are not necessarily distinct, because there are ... | [] | [
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} |
Given a sequence of positive integers $a_1, a_2, a_3, \ldots$ such that for any positive integers $k$ , $l$ we have $k+l ~ | ~ a_k + a_l$ . Prove that for all positive integers $k > l$ , $a_k - a_l$ is divisible by $k-l$ . | <details><summary>Solution</summary>Write $d=k-\ell$ and choose $n$ such that $k+n$ is a multiple of $d$ . Note that then also $\ell+n$ is a multiple of $d$ by construction.
Hence, by assumption $d \mid a_k+a_n$ and $d \mid a_\ell+a_n$ and hence also $d \mid a_k-a_{\ell}$ as desired.</details> | [
"This problem was proposed by **[Burii](https://artofproblemsolving.com/community/user/100466)**.",
"For $n \\geq l+1$ , taking $k= n-l$ give $n ~ | ~ a_{n-l} + a_{l}$ . But $n ~ | ~ 2n ~ | ~ a_{n+l} + a_{n-l}$ , so $n ~ | ~ a_{n+l} - a_{l}$ $\\implies$ $k-l ~ | ~ a_{k} - a_{l}$ .",
"@above to do it y... | [
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} |
Given an acute triangle $ABC$ with their incenter $I$ . Point $X$ lies on $BC$ on the same side as $B$ wrt $AI$ . Point $Y$ lies on the shorter arc $AB$ of the circumcircle $ABC$ . It is given that $$ \angle AIX = \angle XYA = 120^\circ. $$
Prove that $YI$ is the angle bisector of $XYA$ . | Nice problem!
---------------------------------------
We will begin by performing $\sqrt{bc}$ inversion. $B$ and $C$ are swapped under the inversion and $I$ is mapped to $E$ , the $A$ -excenter of $\triangle ABC$ . $X$ is mapped to a point $X'$ on $(ABC)$ such that $\angle AX'E=120^{\circ}$ and $Y$ ... | [
"It suffices to show that <XYI = 60. Note that if Ma is the intersection of AI and (ABC), P is on (ABC) such that <AMaP = 60, and Z is the intersection of PMa and (BIC), then ZIMa is equilateral, X is the intersection of IZ with BC, and Y is the intersection of PX with (ABC), so X has equal power wrt (ABC) and (BIC... | [
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} |
Given a positive integer $n \geq 2$ and real numbers $a_1, a_2, \ldots, a_n \in [0,1]$ . Prove that there exist real numbers $b_1, b_2, \ldots, b_n \in \{0,1\}$ , such that for all $1\leq k\leq l \leq n$ we have $$ \left| \sum_{i=k}^l (a_i-b_i)\right| \leq \frac{n}{n+1}. $$ | [
"[2019 CWMI P4](https://artofproblemsolving.com/community/c6h1894715p12941983)"
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Given a positive integer $n\geq 2$ and positive real numbers $a_1, a_2, \ldots, a_n$ with the sum equal to $1$ . Let $b = a_1 + 2a_2 + \ldots + n a_n$ . Prove that $$ \sum_{1\leq i < j \leq n} (i-j)^2 a_i a_j \leq (n-b)(b-1). $$ | <blockquote>Given a positive integer $n\geq 2$ and positive real numbers $a_1, a_2, \ldots, a_n$ with the sum equal to $1$ . Let $b = a_1 + 2a_2 + \ldots + n a_n$ . Prove that $$ \sum_{1\leq i < j \leq n} (i-j)^2 a_i a_j \leq (n-b)(b-1). $$ </blockquote>
Very easy problem. $\begin{aligned}
\sum_{1\leq i < j \l... | [
"Nice. A probabilistic proof.\n\nLet $Z$ be a random variable with $\\mathbb{P}[Z=i]=a_i$ for $1\\le i\\le n$ . Then $b=\\mathbb{E}[Z]$ . Let $Z_1,Z_2$ be two iid copies of $Z$ . Note that\n\\[\n2{\\rm Var}(Z) = \\mathbb{E}\\bigl[(Z_1-Z_2)^2 \\bigr]= \\sum_{i\\ne j} (i-j)^2 a_i a_j.\n\\]\nSo, it suffices t... | [
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"boxed": false,
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Give a prime number $p>2023$ . Let $r(x)$ be the remainder of $x$ modulo $p$ . Let $p_1<p_2< \ldots <p_m$ be all prime numbers less that $\sqrt[4]{\frac{1}{2}p}$ . Let $q_1, q_2, \ldots, q_n$ be the inverses modulo $p$ of $p_1, p_2, \ldots p_n$ . Prove that for every integers $0 < a,b < p$ , the sets $$... | <details><summary>Official solution</summary>It is easy to see that $q_i$ are different and for given $a,b$ there exists at most one index $i$ such that $q_i \equiv aq_i + b$ ( $q_i \equiv b(a-1)^{-1}$ if $a > 1$ ). Now suppose that the sets have at least $4$ common elements. Then there exists indices $i_1... | [] | [
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"boxed": false,
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} |
For any real numbers $a$ and $b>0$ , define an *extension* of an interval $[a-b,a+b] \subseteq \mathbb{R}$ be $[a-2b, a+2b]$ . We say that $P_1, P_2, \ldots, P_k$ covers the set $X$ if $X \subseteq P_1 \cup P_2 \cup \ldots \cup P_k$ .
Prove that there exists an integer $M$ with the following property: for... | For a closed interval $I=[a-b,a+b]\subset\mathbb R$ , define $I^e=[a-2b, a+2b]$ as its extension.
-----
<span style="color:#f00">**Lemma 1.**</span> If $I\subset\mathbb R$ is a closed interval, then $I\subset I^e$ . Moreover, if $I\subset J\subset\mathbb R$ are closed intervals, then $I^e\subset J^e$ .
<span ... | [
"Any answer?",
"We only need to prove: For any $n$ , there exists an integer $M_n$ , such that for any $A\\in\\mathbb{R}$ , there exists a subset $B$ of $A$ , $|B|\\leq M_n$ satisfy: if a total of $\\leq n$ sets cover $B$ , then their extensions cover $A$ .\nHint: Try using induction to prove $M_n=3M... | [
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Solve the system of equations: $$ \left\{ \begin{array}{l}
({x^2} + y)\sqrt {y - 2x} - 4 = 2{x^2} + 2x + y
{x^3} - {x^2} - y + 6 = 4\sqrt {x + 1} + 2\sqrt {y - 1}
\end{array} \right.(x,y \in \mathbb{R}). $$ | Reworking the first equation: $(x^2+y)\sqrt{y-2x}-4=2x^2+2x+y\Longrightarrow \left(\frac{2x^2+2x+y+4}{x^2+y}\right)^2=y-2x$ $\Longrightarrow \left(\frac{x^2+y+x^2+2x+4}{x^2+y}\right)^2=y-2x\Longrightarrow \left(1+\frac{x^2+2x+4}{x^2+y}\right)^2=y-2x$ Since $~~y-2x\ge 0\Longrightarrow y=2x+\alpha$ $\Longrightarrow \l... | [
"<details><summary>2023 Quang Nam Province Math Contest (Grade 11)</summary>[https://artofproblemsolving.com/community/c3317549](https://artofproblemsolving.com/community/c3317549)</details>"
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Given the sequence $(u_n)$ satisfying: $$ \left\{ \begin{array}{l}
1 \le {u_1} \le 3
{u_{n + 1}} = 4 - \dfrac{{2({u_n} + 1)}}{{{2^{{u_n}}}}},\forall n \in \mathbb{Z^+}.
\end{array} \right. $$ Prove that: $1\le u_n\le 3,\forall n\in \mathbb{Z^+}$ and find the limit of $(u_n).$ | Let $f(x)=\dfrac{2(x+1)}{2^x}, g(x)=4-f(x)$ , analyze function $f$ on $[1,3]$ and by easy induction we have $1\le u_n\le 3$ for all positive integer $n$ .
Function $g$ has the fixed point $3$ , the idea to prove that this is the limit of the sequence, however, forall $x\in [1,3]$ we have for $x\in [1,3]$ ... | [
"<details><summary>2023 Quang Nam Province Math Contest (Grade 11)</summary>[https://artofproblemsolving.com/community/c3317549](https://artofproblemsolving.com/community/c3317549)</details>"
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"path": "Contest Collections/2023 Contests/2023 Quang Nam Province Math Contest (Grade 11)/3039290.json"
} |
Given a polynomial $P(x)$ with real coefficents satisfying: $$ P(x).P(x+1)=P(x^2+x+1),\forall x\in \mathbb{R}. $$ Prove that: $\deg(P)$ is an even number and find $P(x).$ | [
"<details><summary>2023 Quang Nam Province Math Contest (Grade 11)</summary>[https://artofproblemsolving.com/community/c3317549](https://artofproblemsolving.com/community/c3317549)</details>"
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a) Find all integer pairs $(x,y)$ satisfying $x^4+(y+2)^3=(x+2)^4.$ b) Prove that: if $p$ is a prime of the form $p=4k+3$ $(k$ is a non-negative number $),$ then there doesn's exist $p-1$ consecutive non-negative integers such that we can divide the set of these numbers into $2$ distinct subsets so that t... | Suppose there exist $p-1$ consecutive non-negative integers $a_{1},....a_{p-1}$ such that we can divide the set of these numbers into $2$ distinct subsets so that the product of all the numbers in one subset is equal to that in the remained subset.
Case 1: Exist $a_{i} \equiv 0 (mod p)$ => contradiction
Case 2 ... | [
"<details><summary>2023 Quang Nam Province Math Contest (Grade 11)</summary>[https://artofproblemsolving.com/community/c3317549](https://artofproblemsolving.com/community/c3317549)</details>",
"Part (a)\n\n<details><summary>Click to expand</summary>Let $x+1=X, y+2=2m$ The given eq. becomes $~~~X(X^2+1)=m^3$ Fir... | [
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a) Given an acute triangle $ABC(AB>AC).$ The circle $(O)$ with diameter $BC$ intersects $AB,AC$ at $F,E$ , respectively. Let $H$ be the intersection point of $BE,CF,$ the line $AH$ intersects the line $BC$ at $D,$ the line $EF$ intersects the line $BC$ at $K.$ The line passing through $D$ and... | [
"<details><summary>2023 Quang Nam Province Math Contest (Grade 11)</summary>[https://artofproblemsolving.com/community/c3317549](https://artofproblemsolving.com/community/c3317549)</details>"
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We color all vertexs of a convex polygon with $10$ vertexs by $2$ colors: red and blue $($ each vertex is colored by $1$ color $).$ How many ways to color all the vertexs such that there are no $2$ adjacent vertex that are both colored red? | We count number $a_{10}$ of ways to color vertexs such that there are no 2 red consecutive vertexs, which equivalent to the number of binary sequences of length 10 such that the first and last characters aren't both 1, and no two consecutive 1's occur in the sequence
Let $b_{n}$ be number of binary sequences of len... | [
"<details><summary>2023 Quang Nam Province Math Contest (Grade 11)</summary>[https://artofproblemsolving.com/community/c3317549](https://artofproblemsolving.com/community/c3317549)</details>"
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"path": "Contest Collections/2023 Contests/2023 Quang Nam Province Math Contest (Grade 11)/3039300.json"
} |
Consider fractions $\frac{a}{b}$ where $a$ and $b$ are positive integers.
(a) Prove that for every positive integer $n$ , there exists such a fraction $\frac{a}{b}$ such that $\sqrt{n} \le \frac{a}{b} \le \sqrt{n+1}$ and $b \le \sqrt{n}+1$ .
(b) Show that there are infinitely many positive integers $n$ su... | First we solve part **<span style="color:#0f0">(a)</span>**. Fix $n$ and assume contrary. Let $\lfloor n \rfloor = k$ . Case $k=1$ is easy. Assume $k \ge 2$ . Let $S$ be set of all fractions with denominator $\le k+1$ . Choose $\frac{a}{b},\frac{c}{d} \in S$ such that $$ \frac{a}{b} < \sqrt{n} \le \sqrt{n+... | [
"<details><summary>Solution</summary>Let $n=k^2+d$ where $d\\le 2k$ .\nThus the condition in part (a) becomes $b\\le k+1$ .\nNow, if $2|d$ , so $d=2c$ , then clearly $k^2+2c\\le k^2+2c+\\dfrac{c^2}{k^2} \\le k^2+2c+1$ as $c\\le k$ , so square rooting gives $\\sqrt{n}\\le \\dfrac{k^2+c}k \\le \\sqrt{n+1}$ ,... | [
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"boxed": false,
"end_of_proof": false,
"n_reply": 23,
"path": "Contest Collections/2023 Contests/2023 Romania EGMO TST/1480728.json"
} |
In town $ A,$ there are $ n$ girls and $ n$ boys, and each girl knows each boy. In town $ B,$ there are $ n$ girls $ g_1, g_2, \ldots, g_n$ and $ 2n \minus{} 1$ boys $ b_1, b_2, \ldots, b_{2n\minus{}1}.$ The girl $ g_i,$ $ i \equal{} 1, 2, \ldots, n,$ knows the boys $ b_1, b_2, \ldots, b_{2i\minus{... | Let $ A(n,r)$ and $ B(n,r)$ be the number of ways of possible dance pairs from town A and Town B out of total $ n$ girls respectively.
Enumerating $ A(n,r)$ we get,
\[ A(n,r)\equal{}{n\choose r}^2 r!\]
Let us construct a recurrence relation for $ B(n,r)$ for $ 2\le r \le n$ ,
Case1: When $ g_n$ i... | [
"This problem either says something about matchings in certain graphs or equivalently about rook polynomials for boards of certain shapes. Anyone have a bijective proof?",
"Benelux MO 2009"
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"path": "Contest Collections/2023 Contests/2023 Romania EGMO TST/219810.json"
} |
Let $D{}$ be a point inside the triangle $ABC$ . Let $E{}$ and $F{}$ be the projections of $D{}$ onto $AB$ and $AC$ , respectively. The lines $BD$ and $CD$ intersect the circumcircle of $ABC$ the second time at $M{}$ and $N{}$ , respectively. Prove that \[\frac{EF}{MN}\geqslant \frac{r}{R},\]where ... | Let $G$ be the projection of $D$ onto $BC$ . First we notice that $\angle ABM = \angle ABD = \angle EBD = \angle EGD$ . Analogously $\angle ACN = \angle FGD$ . So by law of sines we get $MN = 2R \sin(\angle NAM) = 2R \sin(\angle ABM + \angle ACN) = 2R\sin \angle EGF$ .
Let $\angle EGF = \theta$ . Then we need... | [] | [
"origin:aops",
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"boxed": false,
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} |
Let $n\geqslant 3$ be an integer and $a_1,\ldots,a_n$ be nonzero real numbers, with sum $S{}$ . Prove that \[\sum_{i=1}^n\left|\frac{S-a_i}{a_i}\right|\geqslant\frac{n-1}{n-2}.\] | First observe that if $S=0$ , the $LHS$ is equal to $n$ . So suppose $S\neq 0$ . We can then assume WLOG that $S=1$ .
We will proceed via smoothing. Let $f(x)=\left|{\frac{1-x}{x}}\right|$ . Consider the three intervals: $I=\left(-\infty, 0\right), J=(0,1],K=[1,\infty)$ Suppose none of the $a_i$ are in $I$ .... | [
"Sorry for the unrelated post, but @oVlad, could you post the other problems or give sources if they are not original (unless they are from shortlists)?",
"<blockquote>Sorry for the unrelated post, but @oVlad, could you post the other problems or give sources if they are not original (unless they are from shortli... | [
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"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Romania EGMO TST/3020452.json"
} |
In a cyclic quadrilateral $ABCD$ with $AB=AD$ points $M$ , $N$ lie on the sides $BC$ and $CD$ respectively so that $MN=BM+DN$ . Lines $AM$ and $AN$ meet the circumcircle of $ABCD$ again at points $P$ and $Q$ respectively. Prove that the orthocenter of the triangle $APQ$ lies on the s... | Let $K$ be a point on $MN$ such that $MB=MK,$ from $MN=BM+DN$ it also follows $ND=NK.$ Say $\angle MBK=\alpha,$ $\angle NDK=\beta,$ so $\angle BCD=180^\circ-2\alpha-2\beta$ $\rightarrow$ $\angle BAD=2\alpha+2\beta$ $\rightarrow$ $\angle ABD=\angle ADB=90^\circ-\alpha-\beta.$ Now, let's show that... | [
"<blockquote>I think the problem is wrong</blockquote>\n\nSorry, I did mistake when translate problem :blush: \nnot $AB=CD$ :!: $AB=AD$ And $MN= BM+DN$ ",
"let make simmetry of point D wrt AN we get point D' , let line ND' intersect line BC in point M' , easy to see that AD' = AD = AB and angle AD'M = 180 -... | [
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} |
Suppose that $f : \mathbb{N} \rightarrow \mathbb{N}$ is a function for which the expression $af(a)+bf(b)+2ab$ for all $a,b \in \mathbb{N}$ is always a perfect square. Prove that $f(a)=a$ for all $a \in \mathbb{N}$ . | We first prove that $f(p)=p$ is true for all primes $p$ .
By plugging $b=n$ , we have $af(a)+nf(n)+2an$ is a square, so $af(a)$ is a quadratic residue for every positive integer.
By [ISL 2007 N2]("http://www.artofproblemsolving.com/community/c6h214712p1187198"), we have $af(a)$ is a square for all $a$ .
Th... | [
" $a f(a)$ is perfect square mod $p$ for every $p$ so $a f(a)=g(a)^2$ .\nSo for prime $q$ we have $f(q)=q .T(q)^2$ .\nPut $a=b=q$ in main equation we get $T(q)$ is odd .\nNow if there is a prime $p$ not equal to $q$ such that $p \\mid T(q)$ look at the equation at $(a,b)=(p,q)$ mod $p^2$ to ... | [
"origin:aops",
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A square with side $2008$ is broken into regions that are all squares with side $1$ . In every region, either $0$ or $1$ is written, and the number of $1$ 's and $0$ 's is the same. The border between two of the regions is removed, and the numbers in each of them are also removed, while in the new region, thei... | **1)** Number the rows $1,2,...,2008$ . Either all the odd rows or all the even rows will contain at least half of the square marked $0$ - w.l.o.g. suppose even rows. So in every odd row apply the operation to the squares in these rows until they become one region containing one number $(\le 1)$ . **2)** Every ver... | [
"Can you post the intution behind step 1?? I tried each of the steps 2,3,4 (i.e., trying to take care of all 1's saving at least half of the zeroes etc.) but to no avail. I understand that (1) is the crucial step. Nice problem and very nice solution by ocha!"
] | [
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Let $ b, m, n$ be positive integers such that $ b > 1$ and $ m \neq n.$ Prove that if $ b^m \minus{} 1$ and $ b^n \minus{} 1$ have the same prime divisors, then $ b \plus{} 1$ is a power of 2. | I'll use a lemma: if $a>1,k>1$ and $a^k-1,a-1$ have the same prime factors, then $k$ is a power of $2$ and so is $a+1$ .
Assume the contrary, and let $p$ be an odd prime factor of $k$ . Then any prime factor $q$ of $1+a+\ldots+a^{p-1}$ also divides $1+a+\ldots+a^{k-1}|a^k-1$ , so it divides $a-1$ ... | [
"b<sup>n</sup>-1 is a Mersenne sequence, meaning that (b<sup>m</sup>-1, b<sup>n</sup>-1)=b<sup>(m,n)</sup>-1. Let d=(m,n). This means that b<sup>d</sup>-1 and b<sup>m</sup>-1 have the same prime divisors. Let b<sup>d</sup>=a, and m/d=M. Then a-1 and a<sup>M</sup>-1 have the same prime divisors.\r\n\r\nLet p|a<sup>M... | [
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"boxed": false,
"end_of_proof": false,
"n_reply": 16,
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} |
The non-zero natural number n is a perfect square. By dividing $2023$ by $n$ , we obtain the remainder $223- \frac{3}{2} \cdot n$ . Find the quotient of the division. | <details><summary>Solving problems for fun</summary>We have
\begin{align*}
2023 &= qn + \left(223- \frac{3}{2} \cdot n\right)
\iff 1800 &= \left(q - \frac{3}{2}\right)n
\iff 3600 &= (2q-3)n.
\end{align*}
Since $3600$ is a perfect square and $n$ is also a perfect square, therefore we must have
$2q-3$ i... | [
" $$ 2023 = 14 \\cdot144 +223 - \\frac{3}{2} \\cdot 144 $$ This can easily be checked by writing the equation as $$ 2023=n \\cdot q +223-\\frac{3}{2}\\cdot n $$ Then utilise the fact that n is a square.and perform some algebraic manipulations. You will get the above divison.",
"After you find the limit of n,... | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
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"answer_score": 34,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052426.json"
} |
We say that a natural number is called special if all of its digits are non-zero and any two adjacent digits in its decimal representation are consecutive (not necessarily in ascending order).
a) Determine the largest special number $m$ whose sum of digits is equal to $2023$ .
b) Determine the smallest special num... | <details><summary>sol for (a), not that rigorous</summary>Obviously, we want to maximize the number of digits. Since the sum of the digits is fixed ( $2023$ ), we want to use as small digits as possible: only using $1$ and $2$ would be optimal if possible.
This means that the largest special number starting with ... | [] | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 146,
"boxed": false,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052440.json"
} |
Determine all natural numbers $m$ and $n$ such that
\[
n \cdot (n + 1) = 3^m + s(n) + 1182,
\]
where $s(n)$ represents the sum of the digits of the natural number $n$ . | Modulo $9$ gives $n(n+1) \equiv n + 3^m + 3 \pmod 9$ . Now, $3^m + 3$ is $3$ or $6$ modulo $9$ , so $3 \mid n^2$ but $9 \nmid n^2$ , impossible.
Edit: I guess if you said that $0$ is a natural number (which is definitely what was intended), you have $n(n+1) = s(n) + 1183$ and since $n$ has at most ... | [
"Consider modulo $9$ , we have $3^m \\equiv 1,3$ or $0$ , and $s(n) + 1182 \\equiv n + 3$ , so $n^2 \\equiv 3,4$ or $6$ , but the only possible residues of $n^2 \\pmod 9$ are $0,1,4,7$ , so $m = 0$ is the only option. And so $n(n+1) = s(n) + 1183$ , so $n(n+1) > 1183$ , and so $n \\geq 34$ . We also... | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 66,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052443.json"
} |
We say that a number $n \ge 2$ has the property $(P)$ if, in its prime factorization, at least one of the factors has an exponent $3$ .
a) Determine the smallest number $N$ with the property that, no matter how we choose $N$ consecutive natural numbers, at least one of them has the property $(P).$ b) Determ... | a) min(N) = 16
Consider 16 consecutive natural numbers in modulo 16
there must exists $a_{n}$ which $a_{n} \equiv$ 8 (mod 16)
That mean $v_2(a_{n}) = 3$ Next step we will show that there exists 15 consecutive natural numbers which is not (P)
It obvious that $$ 57,58,59,60,61,62,63,64,65,66,67,68,69,70,71 $$ ... | [
"Can you please explain it more??"
] | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
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"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052445.json"
} |
Determine all sequences of equal ratios of the form
\[
\frac{a_1}{a_2} = \frac{a_3}{a_4} = \frac{a_5}{a_6} = \frac{a_7}{a_8}
\]
which simultaneously satisfy the following conditions:
$\bullet$ The set $\{ a_1, a_2, \ldots , a_8 \}$ represents all positive divisors of $24$ .
$\... | If we have two fractions a/b=c/d which are divisors of a number, then we have ad=bc, so a/d cannot equal c/b or b/c, since all of a, b, c, d are distinct. We can only have a/c=b/d. Okay, back to the problem. Let $b_8>b_7>…>b_1$ be some permutation of a_1 to a_8. $24/12=6/3=8/4=2/1, 24/8=12/4=6/2=3/1, 24/6=12/3=8/2=4... | [] | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 4,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052454.json"
} |
Determine all triples $(a,b,c)$ of integers that simultaneously satisfy the following relations:
\begin{align*}
a^2 + a = b + c,
b^2 + b = a + c,
c^2 + c = a + b.
\end{align*} | <details><summary>Solution</summary>Adding the three equations, we get $$ a^2+b^2+c^2+a+b+c = 2a+2b+2c \implies a^2+b^2+c^2=a+b+c. $$ Thus, since $x^2 \ge x$ for all integers $x$ with equality holding only at $x=0$ and $x=1,$ it follows that all of $a,b,c$ must be $0$ or $1.$ If $a=0,$ then $b+c=0,... | [
"<blockquote>Determine all triples $(a,b,c)$ of integers that simultaneously satisfy the following relations:\n\n \\begin{align*}\n a^2 + a = b + c, \n b^2 + b = a + c, \n c^2 + c = a + b. \n \\end{align*}</blockquote>\n\nAdd all of the equations to get $a^2+b^2+c^2=a+b+c$ . Since $a^2 ... | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
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"answer_score": 1134,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052455.json"
} |
Determine all positive integers $n$ for which the number
\[
N = \frac{1}{n \cdot (n + 1)}
\]
can be represented as a finite decimal fraction. | <details><summary>Solution</summary>Note that $\tfrac{1}{n(n+1)}$ is a terminating decimal if and only if the only primes that divide $n(n+1)$ are $2$ and $5.$ However, note that $\gcd(n, n+1)=1,$ so if $n>1$ (noting that $n=1$ works), then one of $n$ or $n+1$ must be a power of $2$ while the other ... | [
"/bump...",
"n=1 or 4",
" $n$ and $n+1$ has only got prime factors $2$ and $5$ . Since there's one odd and one even among $n$ and $n+1$ , the odd one must be $5^a$ and so the even one isn't divisible by $5$ , so it is $2^b$ . Now look here:\n[#9 and #11](https://artofproblemsolving.com/community/c5... | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 1136,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052457.json"
} |
Let $ABC$ be a triangle with $\angle BAC = 90^{\circ}$ and $\angle ACB = 54^{\circ}.$ We construct bisector $BD (D \in AC)$ of angle $ABC$ and consider point $E \in (BD)$ such that $DE = DC.$ Show that $BE = 2 \cdot AD.$ | <details><summary>sol</summary>Let $AD=x.$ Note that $$ \sin 18^\circ = \frac{AD}{BD} = \frac{x}{BD}, $$ so $$ BD = \frac{x}{\sin 18^\circ}. $$ By Law of Sines in $\triangle BDC,$ we have $$ \frac{DC}{\sin 18^\circ} = \frac{BD}{\sin 54^\circ} = \frac{x}{\sin 18^\circ \sin 54^\circ}, $$ so it follows that ... | [] | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 120,
"boxed": false,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052459.json"
} |
For natural number $n$ we define
\[
a_n = \{ \sqrt{n} \} - \{ \sqrt{n + 1} \} + \{ \sqrt{n + 2} \} - \{ \sqrt{n + 3} \}.
\]
a) Show that $a_1 > 0,2$ .
b) Show that $a_n < 0$ for infinity many values of $n$ and $a_n > 0$ for infinity values of natural numbers of $n$ as well. ( We denot... | <details><summary>sol for (a)</summary>We have $$ a_1 = \{ \sqrt{1} \} - \{ \sqrt{2} \} + \{ \sqrt{3} \} - \{ \sqrt{4} \} = (1-\sqrt{2}) + (\sqrt{3}-1) = \sqrt{3} - \sqrt{2}. $$ Therefore, it suffices to prove that $$ \sqrt{3} - \sqrt{2} > 0.2. $$ Note that $(2.48)^2 = 6.1504 > 6,$ so $2.48 > \sqrt{6}.$ The... | [
"(1) $a_1=\\{\\sqrt 3\\}-\\{\\sqrt 2\\}=\\sqrt 3-\\sqrt 2$ ,notice that $\\frac 1{a_1}=\\sqrt 2+\\sqrt 3<5$ , $\\therefore a_1>0.2$ .\n(2)if $n=k^2-3(k\\in \\mathbb{N},k\\ge 3)$ ,then $\\{\\sqrt {n+3}\\}=0$ ,thus $a_n=\\{\\sqrt n\\}+\\sqrt {n+2}-\\sqrt {n+1}>0$ ,s\nso $a_n>0$ for infinity many values of $n$ ... | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 182,
"boxed": false,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052468.json"
} |
In the parallelogram $ABCD$ , $AC \cap BD = { O }$ , and $M$ is the midpoint of $AB$ . Let $P \in (OC)$ and $MP \cap BC = { Q }$ . We draw a line parallel to $MP$ from $O$ , which intersects line $CD$ at point $N$ . Show that $A,N,Q$ are collinear if and only if $P$ is the midpoint of $OC$ . | [] | [
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} | |
We consider triangle $ABC$ with $\angle BAC = 90^{\circ}$ and $\angle ABC = 60^{\circ}.$ Let $ D \in (AC) , E \in (AB),$ such that $CD = 2 \cdot DA$ and $DE $ is bisector of $\angle ADB.$ Denote by $M$ the intersection of $CE$ and $BD$ , and by $P$ the intersection of $DE$ and $AM$ .
a... | <details><summary>sol for (a)</summary>Without loss of generality, let $$ A=(0,0), $$ $$ B=(0,6), $$ $$ C=(6\sqrt{3},0). $$ Then, we have $$ D = (2\sqrt{3},0). $$ Note that $$ \frac{AB}{AD} = \frac{6}{2\sqrt{3}} = \sqrt{3} $$ which implies that $ABD$ is a $30^\circ-60^\circ-90^\circ$ triangle wi... | [] | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 234,
"boxed": false,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052479.json"
} |
a) Show that there exist irrational numbers $a$ , $b$ , and $c$ such that the numbers $a+b\cdot c$ , $b+a\cdot c$ , and $c+a\cdot b$ are rational numbers.
b) Show that if $a$ , $b$ , and $c$ are real numbers such that $a+b+c=1$ , and the numbers $a+b\cdot c$ , $b+a\cdot c$ , and $c+a\cdot b$ are ratio... | Nice problem!
a) Take $a=b=c=t$ , where $t$ is a root of $t^2 + t = 1$ . (Probably there is an example with $a$ , $b$ , $c$ distinct, but lazy to look for such.)
b) Observe that $c + ab = 1-a-b + ab = (1-a)(1-b)$ and that $a$ , $b$ , $c$ are rational if and only if $x=1-a$ , $y=1-b$ , $z=1-c$ (with... | [
"\na) Let $a$ is root $a^3-a=1$ ( easy to prove, that $a$ and $a^2$ is irrational), $ b=-a, c=a^2$ Then $a+bc=a-a^3=-1, b+ac=-a+a^3=1,c+ab=a^2-a^2=0$ b) $a+bc=a(a+b+c)+bc=(a+b)(a+c)$ is non-zero rational\nSame way $(b+a)(b+c), (c+a)(c+b)$ are non-zero rationals, so $ \\frac{(a+b)(a+c)*(a+b)(b+c)}{(c+a... | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 72,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052480.json"
} |
We consider real positive numbers $a,b,c$ such that $a + b + c = 3.$ Prove that $a^2 + b^2 + c^2 + a^2b + b^2 c + c^2 a \ge 6.$ | <blockquote>We consider real positive numbers $a,b,c$ such that $a + b + c = 3.$ Prove that $a^2 + b^2 + c^2 + a^2b + b^2 c + c^2 a \ge 6.$ </blockquote>
[https://artofproblemsolving.com/community/c6h1653353p10467212](https://artofproblemsolving.com/community/c6h1653353p10467212)
[https://artofproblemsolving.com/... | [
" $a^2+b^2+c^2+(a^2b+b)+(b^2c+c)+(c^2a+a)\\ge(a+b+c)^2$ ",
"<blockquote> $a^2+b^2+c^2+(a^2b+b)+(b^2c+c)+(c^2a+a)\\ge(a+b+c)^2$ </blockquote>\n[https://artofproblemsolving.com/community/c6h1653353p10468198](https://artofproblemsolving.com/community/c6h1653353p10468198)\n"
] | [
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"2023 Romania National Olympiad",
"2023 Contests"
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"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052483.json"
} |
Prove that:
a) There are infinitely many pairs $(x,y)$ of real numbers from the interval $[0,\sqrt{3}]$ which satisfy the equation $x\sqrt{3-y^2}+y\sqrt{3-x^2}=3$ .
b) There do not exist any pairs $(x,y)$ of rational numbers from the interval $[0,\sqrt{3}]$ that satisfy the equation $x\sqrt{3-y^2}+y\sqrt{... | a) $$ x\sqrt{3x-y^2}+y\sqrt{3-x^2}=3 $$ $$ x\sqrt{3-y^2}=3-y\sqrt{3-x^2} $$ $$ x^2(3-y^2)=9-6y\sqrt{3-x^2}+y^2(3-x^2) $$ $$ 3x^2-x^2y^2=9-6y\sqrt{3-x^2}+3y^2-y^2x^2 $$ $$ x^2=3-2y\sqrt{3-x^2}+y^2 $$ $$ 2y\sqrt{3-x^2}=3+y^2-x^2 $$ $$ 4y^2(3-x^2)=9+y^4+x^4+6y^2-6x^2-2x^2y^2 $$ $$ 12y^2-4y^2x^2=9+y^4... | [
"As DVDTSB posted above $x^2+y^2=3$ . Thus for b we must show that $x^2+y^2=3z^2$ has no positive integer solutions. Since $n^2\\equiv 0,1\\text{ (mod 3)}$ for integer $n$ , $3|x,y$ . Thus $9|3z^2$ and $3|z$ . Infinite descent.",
"solution for a)\nsuppose that $y=3sin\\beta$ and $x=3sin\\alpha$ we ha... | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 48,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052484.json"
} |
We say that a natural number $n$ is interesting if it can be written in the form
\[
n = \left\lfloor \frac{1}{a} \right\rfloor + \left\lfloor \frac{1}{b} \right\rfloor + \left\lfloor \frac{1}{c} \right\rfloor,
\] where $a,b,c$ are positive real numbers such that $a + b + c = 1.$ Determine all interest... | <blockquote>By Cauchy Schwarz inequality $$ (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq 9 $$ which desired $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 9$ Since $$ \lfloor \frac{1}{a} \rfloor > \frac{1}{a} - 1 $$ Then $n > \frac{1}{a}+\frac{1}{b}+\frac{1}{c} - 3 \geq 6 $ Hence $n \geq 7$ is interesting ( s... | [
"By Cauchy Schwarz inequality $$ (a+b+c)(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}) \\geq 9 $$ which desired $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} \\geq 9$ Since $$ \\lfloor \\frac{1}{a} \\rfloor > \\frac{1}{a} - 1 $$ Then $n > \\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} - 3 \\geq 6 $ Hence $n \\geq 7$ is inter... | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 16,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052485.json"
} |
Let $ABCD$ be a tetrahedron and $M$ and $N$ be the midpoints of $AC$ and $BD$ , respectively. Show that for every point $P \in (MN)$ with $P \neq M$ and $P \neq N$ , there exist unique points $X$ and $Y$ on segments $AB$ and $CD$ , respectively, such that $X,P,Y$ are collinear. | [] | [
"origin:aops",
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"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052488.json"
} | |
We consider the equation $x^2 + (a + b - 1)x + ab - a - b = 0$ , where $a$ and $b$ are positive integers with $a \leq b$ .
a) Show that the equation has $2$ distinct real solutions.
b) Prove that if one of the solutions is an integer, then both solutions are non-positive integers and $b < 2a.$ |
a) The quadratic equation $x^2 + (a + b - 1)x + ab - a - b = 0$ has two distinct real solutions if its discriminant is positive, that is, $(a+b-1)^2 - 4(ab-a-b) > 0$ . Simplifying this expression, we get $(a-b+1)^2 > 0$ , which is true for all $a\neq b-1$ . Since $a\leq b$ , we have $a\leq b-1$ only if $a=b-1... | [
"a) $(a+b-1)^2-4(ab-a-b)=(b-a+1)^2+4a>0$ so equation has two distinct real solutions\n\nb) If $m,n$ are roots and $m$ is integer then $n=-a-b+1-m$ is integer too\nIf $a=1$ then equation $x^2+bx-1=0$ can not have integer solutions because $D=b^2+4$ is not a square of integer.\nSo $a \\geq 2,b \\geq 2$... | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 106,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052494.json"
} |
Determine functions $f : \mathbb{R} \rightarrow \mathbb{R},$ with property that
\[
f(f(x)) + y \cdot f(x) \le x + x \cdot f(f(y)),
\]
for every $x$ and $y$ are real numbers. | Solved with **megarnie**.
We claim that $\boxed{f(x) \equiv x}$ is the only solution. It is easy to check that it works, so it remains to prove that the given functional equation implies that $f(x) \equiv x.$ Let $P(x,y)$ denote the given assertion. $P(0,y)$ tells us that $$ f(f(0))+yf(0) \le 0. $$ If $f(0)... | [
"Do you have the whole test?",
"<blockquote>Determine functions $f : \\mathbb{R} \\rightarrow \\mathbb{R},$ with property that \n\n \\[\n f(f(x)) + y \\cdot f(x) \\le x + x \\cdot f(f(y)),\n \\]\n \nfor every $x$ and $y$ are real numbers.</blockquote>\nLet $P(x,y)$ be the assertion $f(f(x))+y... | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 1176,
"boxed": true,
"end_of_proof": true,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052496.json"
} |
Let $n \geq 2$ be a natural number. We consider a $(2n - 1) \times (2n - 1)$ table.Ana and Bob play the following game: starting with Ana, the two of them alternately color the vertices of the unit squares, Ana with red and Bob with blue, in $2n^2$ rounds. Then, starting with Ana, each one forms a vector with ori... | [] | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052497.json"
} | |
Let $r$ and $s$ be real numbers in the interval $[1, \infty)$ such that for all positive integers $a$ and $b$ with $a \mid b \implies \left\lfloor ar \right\rfloor$ divides $\left\lfloor bs \right\rfloor$ .
a) Prove that $\frac{s}{r}$ is a natural number.
b) Show that both $r$ and $s$ are natural n... | <details><summary>Sol</summary>For part (a) consider $a=b=n$ then \[\frac sr = \lim_{n \to \infty} \frac{s - \frac{\{ ns\}}{n}}{r - \frac{\{ nr\}}{n}} = \frac{\lfloor ns\rfloor}{\lfloor nr \rfloor} \in \mathbb Z\]
For part (b) Let $s= t r$ where $t \in \mathbb Z$ .
\[\frac{\lfloor bs \rfloor}{\lfloor ar \rfloor } ... | [
"Any Solution?"
] | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052499.json"
} |
Solve the following equation for real values of $x$ :
\[
2 \left( 5^x + 6^x - 3^x \right) = 7^x + 9^x.
\] | Define $f(a) = a^x$ for some fixed $x$ and $a>2$ . Notice that
\[f''(a) = x(x-1)a^{x-2}\]
Thus $f(a)$ is convex for $x \ge 1$ . For $x \ge 1$ notice that Karamata's gives
\[f(9)+f(7)+f(3)+f(3) \ge f(6)+f(6)+f(5)+f(5)\]
For equality to hold, we need $f''(a) = 0$ , so $x = 1$ .
On the other hand, if $0 < x ... | [
"Determine functions f(x)=3^x+7^x-2*5^x, g(x)=3^x+9^x-2*6^x\nAnd use Power-Mean Inequality, we have f>0, g>0 for x<0, f=g=0 for x=0, f<0, g<0 for 0<x<1, f=g=0 for x=1, f>0, g>0 for x>1",
"Let $f(a) = a^x$ . Noting that $(9, 7, 3, 3) \\succ (6, 6, 5, 5)$ , Karamata says that \\[2f(5) + 2f(6) \\le 2f(3) + f(7) + ... | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 1032,
"boxed": false,
"end_of_proof": false,
"n_reply": 14,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052502.json"
} |
Determine the largest natural number $k$ such that there exists a natural number $n$ satisfying:
\[
\sin(n + 1) < \sin(n + 2) < \sin(n + 3) < \ldots < \sin(n + k).
\] | $sin (n+1)<sin (n+2) \to \sin(n+\frac{3}{2}-\frac{1}{2})<\sin(n+\frac{3}{2}+\frac{1}{2}) \to \sin \frac{1}{2} \cos(n+\frac{3}{2})>0 \to \cos (n+\frac{3}{2})>0$ So $\cos (n+i+\frac{1}{2})>0$ for $i=1,...,k-1$ but it is means that for some $j$ it is true: $2\pi *j-\frac{\pi}{2}<n+1+\frac{1}{2}< n+k-1+\frac{1}{2}... | [] | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 14,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052504.json"
} |
We consider triangle $ABC$ and variables points $M$ on the half-line $BC$ , $N$ on the half-line $CA$ , and $P$ on the half-line $AB$ , each start simultaneously from $B,C$ and respectively $A$ , moving with constant speeds $ v_1, v_2, v_3 > 0 $ , where $v_1$ , $v_2$ , and $v_3$ are expressed in the... | [] | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052505.json"
} | |
In an art museum, $n$ paintings are exhibited, where $n \geq 33.$ In total, $15$ colors are used for these paintings such that any two paintings have at least one common color, and no two paintings have exactly the same colors. Determine all possible values of $n \geq 33$ such that regardless of how we color ... | [] | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052506.json"
} | |
Determine twice differentiable functions $f: \mathbb{R} \rightarrow \mathbb{R}$ which verify relation
\[
\left( f'(x) \right)^2 + f''(x) \leq 0, \forall x \in \mathbb{R}.
\] | We claim that $f$ is a constant function, easy to check why they work. Now we shall show that these are the only solutions.
Look closely at $\left( f'(x) \right)^2 + f''(x) \leq 0$ , one of the ways to generate this expression is to consider the second derivative of $e^{f(x)}$ (a common technique rather). So we h... | [
"Let us divide our solution into $2$ cases, depending on if $f'(x)=0$ or $f'(x)\\ne0$ for some $x\\in\\mathbb{R}$ :\n\nLet us take the case where $f'(x)\\ne 0.$ We see that $\\left(\\frac{df(x)}{dx}\\right)^2+\\frac{d^2f(x)}{dx^2}=0.$ Let $g(x)=\\frac{df(x)}{dx}.$ We see that $\\frac{dg(x)}{dx}=-g(x)^2... | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 140,
"boxed": false,
"end_of_proof": true,
"n_reply": 14,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052507.json"
} |
Let $A,B \in M_{n}(\mathbb{R}).$ Show that $rank(A) = rank(B)$ if and only if there exist nonsingular matrices $X,Y,Z \in M_{n}(\mathbb{R})$ such that
\[
AX + YB = AZB.
\] | <blockquote>Here $K$ is a field with more than $2n$ elements.
Assume that $rank(A)=rank(B)$ ; then there are $P,Q\in GL_n(K)$ s.t. $A=PBQ$ .
Thus, for every $a\in K$ , $A(1/2B+aQ^{-1})+(1/2A-aP)B=AB$ . $\det(1/2B+aQ^{-1})$ and $\det(1/2A-aP)$ are polynomials of degree $n$ in $a$ . Then there is $a\in... | [
"<details><summary>Solution</summary>Note $rank(A)=rank(AX) = rank((AZ-Y)B) \\le rank(B)$ . Similarly $rank(B) \\le rank(A)$ so $rank(A)=rank(B)$ is a necessity.\n\nTo prove its sufficiency we first prove that $AZ-Y$ can be any invertible matrix. To get $AZ-Y=T$ , we just need $\\det(Z) \\det(AZ-T) \\ne 0$... | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 144,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052508.json"
} |
Let $n$ be a natural number $n \geq 2$ and matrices $A,B \in M_{n}(\mathbb{C}),$ with property $A^2 B = A.$ a) Prove that $(AB - BA)^2 = O_{n}.$ b) Show that for all natural number $k$ , $k \leq \frac{n}{2}$ there exist matrices $A,B \in M_{n}(\mathbb{C})$ with property stated in the problem such ... | a). Since $rank(A^2B)=rank(A)\geq rank(A^2)\geq rank(A^2B)$ , we have $rank(A)=rank(A^2)$ . Then we can show that $\mathbb{C}^n=\ker A\oplus Im A$ . Choose a basis of $\ker A$ and $Im A$ , we may assume that $A$ is of the form $\begin{pmatrix}0&00&A_1\end{pmatrix}$ where $A_1$ is invertible. Suppose that $... | [] | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 38,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052512.json"
} |
We consider a function $f:\mathbb{R} \rightarrow \mathbb{R}$ for which there exist a differentiable function $g : \mathbb{R} \rightarrow \mathbb{R}$ and exist a sequence $(a_n)_{n \geq 1}$ of real positive numbers, convergent to $0,$ such that
\[
g'(x) = \lim_{n \to \infty} \frac{f(x + a_n) - f(x)}{... | For the $(a)$ part simply take
\[
f(x)= \left\{
\begin{array}{ll}
0 & x \in \mathbb{R}-\mathbb{Q}
1 & x\in \mathbb{Q}
\end{array}
\right.
\]
It is easy to prove that this function is not differentiable in any point $x\in \mathbb{R}$ . To show that it indeed works, take $g\equiv 0$ and the s... | [
"Congrats !",
"The solution above is wrong. Formally what you would have is that for any $k\\in\\mathbb{N}$ and $\\varepsilon>0$ there exists $n_{k,\\varepsilon}$ s.t. $\\left|\\frac{h(x+ka_n)-h(x)}{ka_n}\\right|<\\varepsilon$ for all $n>n_{k,\\varepsilon}$ . To get your claim you would want this converg... | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 54,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052514.json"
} |
Let $(G, \cdot)$ a finite group with order $n \in \mathbb{N}^{*},$ where $n \geq 2.$ We will say that group $(G, \cdot)$ is arrangeable if there is an ordering of its elements, such that
\[
G = \{ a_1, a_2, \ldots, a_k, \ldots , a_n \} = \{ a_1 \cdot a_2, a_2 \cdot a_3, \ldots, a_k \cdot a_{k + 1}, ... | a) Consider the sum of all its elements we have $n(n+1)/2\equiv n(n+1)(\equiv0)\pmod n$ which means $n$ is odd. On the other hand for $n$ odd it suffices to take $a_k=k~\forall k$ . | [
"b) Consider $\\mathbb Z/4\\mathbb Z \\times \\mathbb Z/2\\mathbb Z$ , with \n\\[ (a_1, a_2, \\dots, a_n) = ((0, 0), (0, 1), (1, 0), (2, 1), (1, 1), (3, 1), (3, 0), (2, 0)).\\]"
] | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 8,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052517.json"
} |
Let $p$ be a prime number, $n$ a natural number which is not divisible by $p$ , and $\mathbb{K}$ is a finite field, with $char(K) = p, |K| = p^n, 1_{\mathbb{K}}$ unity element and $\widehat{0} = 0_{\mathbb{K}}.$ For every $m \in \mathbb{N}^{*}$ we note
$ \widehat{m} = \underbrace{1_{\mathbb{K}} + 1_... | Related to [USA TST 2016/3](https://artofproblemsolving.com/community/c6h1176481).
Let $\mathbb{N}$ denote the set of natural numbers, including $0$ . Let $R$ be the ring consisting of the set of $\mathbb{K}$ -polynomials
\[
\left\{a_k x^{p^k}+a_{k-1} x^{p^{k-1}}+\cdots+a_1 x^p+a_0x\mid k\in\mathbb{N};\,a_0,\,\l... | [
"What spam?",
"They are different problems. This is all valid...\n\nJust bump it\n\nNot spam. Just problems...",
"a) Notice that $f_1=X^p-X = X(X-\\widehat{1})\\dots (X-\\widehat{p-1})$ by Fermat's little theorem.\n\nb) Consider the function $\\varphi :\\mathbb{F}_p[X]\\rightarrow\\mathbb{F}_p[X]$ given by ... | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 1182,
"boxed": true,
"end_of_proof": true,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052519.json"
} |
Let $a,b \in \mathbb{R}$ with $a < b,$ 2 real numbers. We say that $f: [a,b] \rightarrow \mathbb{R}$ has property $(P)$ if there is an integrable function on $[a,b]$ with property that
\[
f(x) - f \left( \frac{x + a}{2} \right) = f \left( \frac{x + b}{2} \right) - f(x) , \forall x \in [a,b].
\]... | I'll write down the solution I gave during the contest.
First, it's easy to see that the constant function $f(x)=\frac{t}{b-a}$ verifies the problem.
Now we want to prove that every function with the property $(P)$ is constant. Let $x \in [a,b]$ be an arbitrary number. Rewriting, we get $f(x)=\frac{f(\frac{a+x}{... | [] | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 74,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052520.json"
} |
Let $f:[0,1] \rightarrow \mathbb{R}$ a non-decreasing function, $f \in C^1,$ for which $f(0) = 0.$ Let $g:[0,1] \rightarrow \mathbb{R}$ a function defined by
\[
g(x) = f(x) + (x - 1) f'(x), \forall x \in [0,1].
\]
a) Show that
\[
\int_{0}^{1} g(x) \text{dx} = 0.
\]
b) Prov... | I will provide a solution only for b), as a) is trivial.
We want to prove that: $$ \int_{0}^{1} f(\phi(x)) \text{dx} + \int_{0}^{1} \phi(x)f'(\phi(x)) \text{dx} \leq \int_{0}^{1} f'(\phi(x)) \text{dx} $$ It s easier to work with $f'$ than with $f$ , as $f'$ is a function that is just nonnegative, so we will w... | [
"Here is a solution for a)\n\nNote that $g(x) =\\frac{d}{dx} (x-1)f(x)$ . Thus, $\\int_0^1 g(x) = [(x-1)f(x)]_0^1 =0$ "
] | [
"origin:aops",
"2023 Romania National Olympiad",
"2023 Contests"
] | {
"answer_score": 140,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Romania National Olympiad/3052521.json"
} |
Let $n \geqslant 3$ be an integer, and let $x_1,x_2,\ldots,x_n$ be real numbers in the interval $[0,1]$ . Let $s=x_1+x_2+\ldots+x_n$ , and assume that $s \geqslant 3$ . Prove that there exist integers $i$ and $j$ with $1 \leqslant i<j \leqslant n$ such that
\[2^{j-i}x_ix_j>2^{s-3}.\] | Revenge. Enjoyed this problem a lot!
Let $i$ and $j$ be the positive integers such that $2^{-i}x_i = \max\{2^{-k}x_k \mid 1 \leq k \leq n\}$ and $2^{j}x_j = \max\{2^kx_k \mid 1 \leq k \leq n\}$ . It is clear that $2^{j-i}x_ix_j$ is the largest value our expression can take. This implies that $j \geq i$ .
Le... | [
"this was Peru IMO TST",
"My solution used the fact that $e^2<8$ :( ",
"<details><summary>Horrible</summary>Call sequence $a_1,\\dots a_n$ stronger than sequence $b_1,\\dots b_n$ if we know that if $A$ satisfies the problem $B$ must satisfy it as well. By $T$ we denote $t_1, \\dots t_n$ . Call the... | [
"origin:aops",
"2023 Romania Team Selection Test",
"2023 Contests"
] | {
"answer_score": 244,
"boxed": false,
"end_of_proof": false,
"n_reply": 24,
"path": "Contest Collections/2023 Contests/2023 Romania Team Selection Test/3107365.json"
} |
Let $m$ and $n$ be positive integers, where $m < 2^n.$ Determine the smallest possible number of not necessarily pairwise distinct powers of two that add up to $m\cdot(2^n- 1).$ *The Problem Selection Committee* | We just need to find how many $1$ there are in binary representation of $m\cdot(2^n- 1).$ Let $$ m = 2^{a_1} + 2^{a_2} + \cdots + 2^{a_k} $$ be the binary representation of $m$ where $a_1< a_2 < \dots < a_k < n$ . Then
\begin{align*}
m\cdot(2^n- 1) = m\cdot 2^n - m &= 2^{n + a_1} + 2^{n + a_2} + \cdots + 2^{... | [
"Very well-known that $s_2(m(2^n-1))=n$ for $m<2^n$ , and the answer follows.",
"Related to apmo 2016 p2",
"We will note with $s_2(x)$ $=$ the sum of the digits of the number x in base 2.\n\nWe will prove that $s_2(m\\cdot(2^n-1))=n$ for $m<2^n$ . $s_2(m\\cdot(2^n-1))=s_2((m-1+1)\\cdot(2^n-1))=s_2((m-... | [
"origin:aops",
"2023 Romania Team Selection Test",
"2023 Contests"
] | {
"answer_score": 30,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Romania Team Selection Test/3294080.json"
} |
A *diagonal line* of a (not necessarily convex) polygon with at least four sides is any line through two non-adjacent vertices of that polygon. Determine all polygons with at least four sides satisfying the following condition: The reflexion of each vertex in each diagonal line lies inside or on the boundary of the pol... | <details><summary>Solution</summary>The answer is rhombus only, if the polygon is concave say at (largest possible) contiguous concave stretch $A_1A_2 \cdots A_k$ reflect $A_i$ about $A_1A_k$ , of all of them lie on a side then the stretch will not be concave contradiction ,now assume that the polygon is convex th... | [] | [
"origin:aops",
"2023 Romania Team Selection Test",
"2023 Contests"
] | {
"answer_score": 30,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Romania Team Selection Test/3294082.json"
} |
Consider a $4\times 4$ array of pairwise distinct positive integers such that on each column, respectively row, one of the numbers is equal to the sum of the other three. Determine the least possible value of the largest number such an array may contain.
*The Problem Selection Committee* | [
"Ans is 21,construction like this:\n1 8 12 21\n7 2 20 11\n10 19 3 6\n18 9 5 4\nwhich meets the requirement.\nProof:Consider \"Big number\" if it can be represented by three numbers in its row.\nClearly that 4 Big numbers are distinct,so the sum of them is at least 1+2+...+12=78.with discreteness w... | [
"origin:aops",
"2023 Romania Team Selection Test",
"2023 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Romania Team Selection Test/3294084.json"
} | |
Given a positive integer $a,$ prove that $n!$ is divisible by $n^2 + n + a$ for infinitely many positive integers $n.{}$ *Proposed by Andrei Bâra* | We will construct an infinity of numbers in which the requirement of the problem is verified.
Consider $n+1-a=p^2$ .So $n^2+n+a=(n+1)^2-n-1+a=(p^2+p+a)\cdot(p^2-p+a)$ .
It is clear that for sufficiently large p we have $a<\frac{p^2+p+a}{a}<\frac{p^2+p+a}{2}<p^2-p+a$ .
If $a>1$ we take $p\equiv-1 \pmod a$ suff... | [
"Main idea:find primes $p,q$ and $n$ such that $pq|n^2+n+a$ and $n<pq$ ."
] | [
"origin:aops",
"2023 Romania Team Selection Test",
"2023 Contests"
] | {
"answer_score": 42,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Romania Team Selection Test/3294085.json"
} |
Fix a positive integer $n.{}$ Consider an $n{}$ -point set $S{}$ in the plane. An *eligible* set is a non-empty set of the form $S\cap D,{}$ where $D$ is a closed disk in the plane. In terms of $n,$ determine the smallest possible number of eligible subsets $S{}$ may contain.
*Proposed by Cristi Săvescu* | Interesting
We claim that there are at minimum $\binom{n+1}{2}$ such subsets.**<span style="color:pink">Claim:</span>** This can be constructed.
*Proof.* A construction follows by taking all of $S$ on a line, which gives $\binom{n}{2}$ subsets when there are more than two elements in it, and $n$ more.
This giv... | [
"Maybe problem is near to trivialize by BAMO 2020/5 problem....",
"This is my solution.\nThe minimum is $\\binom{n+1}{2}$ .\nWe will initially show that for any initial configuration of $S$ we have at least $\\binom{n+1}{2}$ eligible subsets.\nClaim : For every point in S there is a circle that passes thr... | [
"origin:aops",
"2023 Romania Team Selection Test",
"2023 Contests"
] | {
"answer_score": 182,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Romania Team Selection Test/3294087.json"
} |
Let $ABCDEF$ be a convex hexagon. The diagonals $AC$ and $BD$ cross at $P,$ the diagonals $AE{}$ and $DF$ cross at $Q,$ and the line $PQ$ crosses the sides $BC$ and $EF$ at $X$ and $Y,{}$ respectively. Prove that the length of the segment $XY$ does not exceed the sum of the lengths of one of t... | [
"If we divide the hexagon in two quadrilaterals and then apply [this](https://artofproblemsolving.com/community/c6h621905p3717896) problem to them, we are done.\nBtw, this was P4, not P5."
] | [
"origin:aops",
"2023 Romania Team Selection Test",
"2023 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Romania Team Selection Test/3294091.json"
} | |
In a given community of people, each person has at least two friends within the community. Whenever some people from this community sit on a round table such that each adjacent pair of people are friends, it happens that no non-adjacent pair of people are friends. Prove that there exist two people in this community suc... | This probelm is very tricky:one idea kills it
Solution:Take the obvious graph interpretation: we have $\deg v\ge 2$ and every cycle doesn't have any chords.
Now take the maximal path $v_1v_2\dots v_k$ .By maximality all the neighbours of $v_1$ and $v_k$ are in this path.
<span style="color:#ff0">**Lemma**</span>... | [
"I think that the official solution was something like this. Another possible approach was presented [here.](https://dgrozev.wordpress.com/2024/05/01/bmo-2023-shortlist-c3/)",
"Wow, analyzing the longest path seems like really powerful tool in graph theory, i think i notice the common things in these kind of prob... | [
"origin:aops",
"2023 Romania Team Selection Test",
"2023 Contests"
] | {
"answer_score": 140,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Romania Team Selection Test/3313173.json"
} |
Let $ABC$ be a triangle with circumcenter $O$ . Point $X$ is the intersection of the parallel line from $O$ to $AB$ with the perpendicular line to $AC$ from $C$ . Let $Y$ be the point where the external bisector of $\angle BXC$ intersects with $AC$ . Let $K$ be the projection of $X$ onto $BY$ . Pr... | **Claim 1:** The line $XO$ is the angle bisector of $\angle BXC$ .
*Proof:* By the conditions of the problem, let $R$ be the foot of the perpendicular from $A$ onto $XC$ . Since $\angle ACX = 90^\circ$ , it follows that $R$ lies on $XC$ .
Let $\angle ACB = \theta$ . Then $\angle AOB = 2\theta$ , and thu... | [
" $XO\\cap BC=S$ . Let $A'$ be the antipode of $A$ in $(ABC)$ . $X,A',C$ are collinear. Let the tangents at $B,C$ to $(ABC)$ intersect at $T$ .**<span style=\"color:#f00\">Claim $1$ :</span>** $X\\in (BOC)$ .\n<span style=\"color:#f00\">Proof:</span> $\\angle OXC=\\angle SXB=\\angle XSB-\\angle XCS=\... | [
"origin:aops",
"2023 Romania Team Selection Test",
"2023 Contests"
] | {
"answer_score": 230,
"boxed": false,
"end_of_proof": true,
"n_reply": 16,
"path": "Contest Collections/2023 Contests/2023 Romania Team Selection Test/3313178.json"
} |
Find all positive integers, such that there exist positive integers $a, b, c$ , satisfying $\gcd(a, b, c)=1$ and $n=\gcd(ab+c, ac-b)=a+b+c$ . | <details><summary>Solution</summary>The answer is all $n > 1$ such that all prime divisors $p$ of $n$ have $p\equiv 1\pmod{4}$ .
First, we show that $n$ must have this property. Obviously, $n \neq 1$ , since $n=a+b+c \geq 1+1+1=3$ . Furthermore, $n$ is odd, since if we had $n=a+b+c$ was even, then sinc... | [
"<details><summary>Solution</summary>The answer is all $n$ that have only prime factors $1 \\pmod 4$ .\n\nIt is easy to see that $a+b+c \\mid abc+c^2, abc-b^2$ , so $a+b+c \\mid b^2+c^2$ (in particular, $n$ doesn't have any prime factors $3 \\pmod 4$ , otherwise we obtain that $\\gcd(a, b, c)>1$ ). Moreov... | [
"origin:aops",
"2023 Romania Team Selection Test",
"2023 Contests"
] | {
"answer_score": 320,
"boxed": false,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Romania Team Selection Test/3313182.json"
} |
Determine all prime numbers $p$ and all positive integers $x$ and $y$ satisfying $$ x^3+y^3=p(xy+p). $$ | Solved with **GoodMorning** and **cj13609517288**.
The only solutions are $\boxed{(x,y,p) = (8,1,19), (1,8,19), (7,2,13), (2,7,13), (5,4,7), (4,5,7)}$ , which work. Now we prove they are the only solutions.
If $p$ divided either $x$ or $y$ , then $p$ would divide both $x$ and $y$ . In this case, the LHS w... | [
"Junior vibes, people!**Answer**. The solutions $(x,y,p)$ are $(8,1,19)$ , $(1,8,19)$ , $(7,2,13)$ , $(2,7,13)$ , $(5, 4, 7)$ and $(4,5,7)$ .\n\nWithout loss of generality treat $x\\geq y$ . Writing $a = x + y \\geq 2$ , $b=xy$ , we have $a^3 - 3ab = p(b+p)$ , i.e. $b = \\frac{a^3 - p^2}{3a+p}$ . As ... | [
"origin:aops",
"2023 Contests",
"2023 Romanian Master of Mathematics"
] | {
"answer_score": 1212,
"boxed": false,
"end_of_proof": false,
"n_reply": 27,
"path": "Contest Collections/2023 Contests/2023 Romanian Master of Mathematics/3024159.json"
} |
Fix an integer $n \geq 3$ . Let $\mathcal{S}$ be a set of $n$ points in the plane, no three of which are collinear. Given different points $A,B,C$ in $\mathcal{S}$ , the triangle $ABC$ is *nice* for $AB$ if $[ABC] \leq [ABX]$ for all $X$ in $\mathcal{S}$ different from $A$ and $B$ . (Note that for ... | **Main Lemma:** Consider any partition of $\mathcal{S}$ into disjoint, nonempty subsets $P$ and $Q$ . Call a triangle *mixed* if it has at least one vertex from $P$ and at least one vertex from $Q$ . Then there exists a beautiful mixed triangle.
*Proof:* Take a mixed triangle $\triangle ABC$ with minimal area... | [
"sketch: consider the graph with vertex set $\\mathcal S$ starting with no edges; add all the triangles (i.e. their sides) to graph by nondecreasing order of area; any triangle decreasing the number of connected components is beautiful, and there are at least $\\tfrac 12 (n-1)$ of these",
"A problem by Aleksa... | [
"origin:aops",
"2023 Contests",
"2023 Romanian Master of Mathematics"
] | {
"answer_score": 168,
"boxed": false,
"end_of_proof": false,
"n_reply": 17,
"path": "Contest Collections/2023 Contests/2023 Romanian Master of Mathematics/3024163.json"
} |
Let $n\geq 2$ be an integer and let $f$ be a $4n$ -variable polynomial with real coefficients. Assume that, for any $2n$ points $(x_1,y_1),\dots,(x_{2n},y_{2n})$ in the Cartesian plane, $f(x_1,y_1,\dots,x_{2n},y_{2n})=0$ if and only if the points form the vertices of a regular $2n$ -gon in some order, or ar... | The smallest possible degree is $2n$ . In what follows, we will frequently write $A_i = (x_i, y_i)$ , and abbreviate $f(x_1, y_1, \dots, x_{2n}, y_{2n})$ to $f(A_1, \dots, A_{2n})$ or as a function of any $2n$ points.
-----**Bound** Suppose that $f$ is valid. First, we note a key property:
*Claim.* [Sign of ... | [
"I don’t know if I’m crying from the beauty from this problem or its difficulty ",
"This looks like a harder version of RMM 2020 A2. Since it’s from the USA, the author might even be the same.",
"Reminds me Putnam 2008 A5 ... , But seems too harder !",
"<details><summary>Sketch</summary>Answer is $2n$ .\n\nC... | [
"origin:aops",
"2023 Contests",
"2023 Romanian Master of Mathematics"
] | {
"answer_score": 434,
"boxed": false,
"end_of_proof": false,
"n_reply": 14,
"path": "Contest Collections/2023 Contests/2023 Romanian Master of Mathematics/3024172.json"
} |
An acute triangle $ABC$ is given and $H$ and $O$ be its orthocenter and circumcenter respectively. Let $K$ be the midpoint of $AH$ and $\ell$ be a line through $O. $ Let $P$ and $Q$ be the projections of $B$ and $C$ on $\ell. $ Prove that $$ KP+KQ\ge BC $$ | <blockquote>My method to solve this problem isn't the most elegant (certainly when compared to **CT17**'s tricky solution), but it is very "structural".
Instead of fixing $\triangle ABC$ and letting the line $PQ$ vary, we'll fix the line $PQ$ and segment $\overline{BC}$ , and instead let point $A$ vary. Let ... | [
"Where's the complex bash fam?",
"I think I can calculate it (maybe in 30 minutes)",
"First, we add the midpoint of BC, and then we use Ptolemey inequality for KPMQ. ",
"Let $M$ be the midpoint of $BC$ . First note that $\\angle OMC = \\angle OQC = 90$ and so $OMCQ$ is cyclic. Analogously $OMBP$ is c... | [
"origin:aops",
"2023 Contests",
"2023 Romanian Master of Mathematics"
] | {
"answer_score": 230,
"boxed": false,
"end_of_proof": false,
"n_reply": 29,
"path": "Contest Collections/2023 Contests/2023 Romanian Master of Mathematics/3025643.json"
} |
Let $P,Q,R,S$ be non constant polynomials with real coefficients, such that $P(Q(x))=R(S(x)) $ and the degree of $P$ is multiple of the degree of $R. $ Prove that there exists a polynomial $T$ with real coefficients such that $$ \displaystyle P(x)=R(T(x)) $$ | <details><summary>s</summary>For any polynomial $X \in \mathbb{R}\left [ x \right ],$ defined $deg(X)=d_X.$ Assume $P(x)=\sum_{i=1}^{d_p}m_ix^i; R(x)=\sum_{j=0}^{d_R}n_jx^j$ and $d_P=kd_R, k \in \mathbb{N}^*.$ ---------------**Claim 1:** $\exists \, T(x), H(x) \in \mathbb{R}\left [ x \right ]: P(x)=R(T(x))+H(x),... | [
"Since $Q$ is non constant, it is sufficient to show it is possible to find $T$ such that $S(x)=T(Q(x))$ , given the hypothesis that the degree of $S$ is a multiple of the degree of $Q$ (which is equivalent to the degree condition in the statement).\nLet $q$ be degree of $Q$ , $mq$ the degree of $S$ ... | [
"origin:aops",
"2023 Contests",
"2023 Romanian Master of Mathematics"
] | {
"answer_score": 1098,
"boxed": false,
"end_of_proof": true,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 Romanian Master of Mathematics/3025645.json"
} |
Let $r,g,b$ be non negative integers and $\Gamma$ be a connected graph with $r+g+b+1$ vertices. Its edges are colored in red green and blue. It turned out that $\Gamma $ contains
A spanning tree with exactly $r$ red edges.
A spanning tree with exactly $g$ green edges.
A spanning tree with exactly $b$ blu... | Some people thought this was kind of boring but I liked it.
Let $R$ be a spanning tree with exactly $r$ red edges, and define $G$ and $B$ similarly. If there exists any tree that also contains the correct amount of edges of another color (for example $R$ containing exactly $g$ green colors) then we are don... | [
"let $T_r$ be the graph obtained by deleting from the spanning tree with r red edges the non-red edges. \nDefine analogously $T_g$ and $T_b$ .\nConsider the graph G obtained from the union of $T_r$ , $T_g$ and $T_b$ , G has r red edges, g green edges and b blue edges. \nLemma 1: in G there is no monochroma... | [
"origin:aops",
"2023 Contests",
"2023 Romanian Master of Mathematics"
] | {
"answer_score": 252,
"boxed": false,
"end_of_proof": false,
"n_reply": 31,
"path": "Contest Collections/2023 Contests/2023 Romanian Master of Mathematics/3025646.json"
} |
There are $n!$ empty baskets in a row, labelled $1, 2, . . . , n!$ . Caesar
first puts a stone in every basket. Caesar then puts 2 stones in every second basket.
Caesar continues similarly until he has put $n$ stones into every nth basket. In
other words, for each $i = 1, 2, . . . , n,$ Caesar puts $i$ stones... | The left is easy but I'm stuck with the right :(
<details><summary>left</summary>$\sum_{i=1}^{n!} x_i^2\geq \frac{1}{n!}\left(\sum_{i=1}^{n!}x_{i}\right)^{2}=\frac{1}{n!}\left(\sum_{k=1}^{n}k\cdot\frac{n!}{k}\right)=n!\cdot n^{2}$</details> | [
"Any solution...",
"Problem proposed by Kaarel Hänni (Estonia).",
"Do you have solution?",
"looks similar to the solution for CMO 2022 #2 but working on it a bit it seems much harder.",
"See here: [https://imscprogram.com/competition.html](https://imscprogram.com/competition.html)"
] | [
"origin:aops",
"2023 Contests",
"2023 SAFEST Olympiad"
] | {
"answer_score": 2,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 SAFEST Olympiad/3104235.json"
} |
Find all functions $f:\mathbb{Z} \rightarrow \mathbb{Z}$ such that $f(1) \neq f(-1)$ and $$ f(m+n)^2 \mid f(m)-f(n) $$ for all integers $m, n$ .
*Proposed by Liam Baker, South Africa* | <blockquote>Find all functions $f:\mathbb{Z} \rightarrow \mathbb{Z}$ such that $f(1) \neq f(-1)$ and $$ f(m+n)^2 \mid f(m)-f(n) $$ for all integers $m, n$ .
Proposed by Liam Baker, South Africa</blockquote> $\color{blue}\boxed{\textbf{Answer:} f \in \{ 1,-1 \} \textbf{ or }f \in\{ 2,-2 \} \textbf{ obviously w... | [
"Solved with **YaoAOPS**. \n\nWe claim the only solutions are functions such that $f(x) \\in \\{-1,1\\}$ for all integers $x$ and $f(1) = -f(-1)$ , and also functions such that $f(x) \\in \\{-2,2\\}$ for all integers $x$ and $f(1)= -f(-1)$ . These can be checked to work. Now we prove they are the only sol... | [
"origin:aops",
"2023 Contests",
"2023 SAFEST Olympiad"
] | {
"answer_score": 1356,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 SAFEST Olympiad/3110603.json"
} |
A binoku is a $9 \times 9$ grid that is divided into nine $3 \times 3$ subgrids with the following properties:
- each cell contains either a $0$ or a $1$ ,
- each row contains at least one $0$ and at least one $1$ ,
- each column contains at least one $0$ and at least one $1$ , and
- each of the nine sub... | [
"The answer is <details><summary>Click to expand</summary>18</details>. For a full solution, see: [https://imscprogram.com/competition.html](https://imscprogram.com/competition.html)"
] | [
"origin:aops",
"2023 Contests",
"2023 SAFEST Olympiad"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 SAFEST Olympiad/3110605.json"
} | |
Let $ABC$ be a triangle with incenter $I$ and let $AI$ meet $BC$ at $D$ . Let $E$ be a point on the segment $AC$ , such that $CD=CE$ and let $F$ be on the segment $AB$ such that $BF=BD$ . Let $(CEI) \cap (DFI)=P \neq I$ and $(BFI) \cap (DEI)=Q \neq I$ . Prove that $PQ \perp BC$ .
*Proposed by Le... | Angle chase: $\angle{FPI} = \angle{FDI} = \frac{\angle{C}}{2} = \angle{ECI} = \angle{EPI}$ . Since $ID = IE = IF$ , this implies that $P$ lies on the perpendicular bisector of $EF$ , and so does $Q$ , and $EF \parallel BC$ since $\frac{FB}{EC} = \frac{AB}{AC}$ , so we are done. $\square$ | [
"Nice problem! \nThe first, we have $\\frac{AB}{BF}=\\frac{AB}{BD}=\\frac{AC}{CD}=\\frac{AC}{CE}$ , so $FE \\parallel BC$ Let $D \\not = X = (CEI) \\cap BC$ . Since $\\angle XCI = \\angle ECI$ we have $IX = IE = (\\text{from } CD=CE, BD=BF)=ID=IF$ . So, $XDEF$ is inscribed quadrilateral with $FE \\parallel... | [
"origin:aops",
"2023 Contests",
"2023 SAFEST Olympiad"
] | {
"answer_score": 116,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 SAFEST Olympiad/3110608.json"
} |
In the plane, $2022$ points are chosen such that no three points lie on the same line. Each of the points is coloured red or blue such that each triangle formed by three distinct red points contains at least one blue point.
What is the largest possible number of red points?
*Proposed by Art Waeterschoot, Belgium* | The answer is $1012$ . To get the bound, simply consider a triangulation of red points. For the construction, simply do $n=5$ (which is pretty much forced) and generalise. In particular we take a regular $n$ gon $A_1A_2\cdots A_n$ and for each $A_k$ ( $k\ge 3$ ) we put a blue point in triangle $A_1A_2A_k$ , i... | [
"1012 red points? ",
"i think that the good bound is the seeling of 2022/2 \nto prove this you can use induction to find a more general case n/2\n\n"
] | [
"origin:aops",
"2023 Contests",
"2023 SAFEST Olympiad"
] | {
"answer_score": 16,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 SAFEST Olympiad/3110611.json"
} |
Find all polynomials $P(x)$ with integer coefficients, such that for all positive integers $m, n$ , $$ m+n \mid P^{(m)}(n)-P^{(n)}(m). $$ *Proposed by Navid Safaei, Iran* | No such polynomial exist except $P(x)\equiv c$ where $c\in \mathbb{Z}$ which clearly works.
Now, assume that $\deg P>0$ Since $x-y|P(x)-P(y)$ he condition can be written as $Q(m,n): m|P^{(m-n)}(n)-P^{(n)}(n),m>n$ For every pair $(m,n)$ with $m<n$ of positive integers denote by $T(m,n)$ the smallest positi... | [
"P(n)=n+1 works",
"<blockquote>\nSince $x-y|P(x)-P(y)$ he condition can be written as $Q(m,n): m|P^{(m-n)}(n)-P^{(n)}(n),m>n$ </blockquote>\nCan $x-y|P(x)-P(y)$ imply that $x-y|P^{(n)}(x)-P^{(n)}(y)$ ?\nIf I'm missing something please let me know....\n(Probably missing something)\n\n",
"<blockquote><block... | [
"origin:aops",
"2023 Contests",
"2023 SAFEST Olympiad"
] | {
"answer_score": 344,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 SAFEST Olympiad/3110614.json"
} |
Prove that if $A{}$ and $B{}$ are $n\times n$ matrices with complex entries which satisfy \[A=AB-BA+A^2B-2ABA+BA^2+A^2BA-ABA^2,\]then $\det(A)=0$ . | Wer is right.
Let $U=AB-BA$ . Then $A=U+AU-UA+AUA$ . Let $k\geq 1$ .
Left multiplying by $A^{k-1}$ , we find $A^k=A^{k-1}U+A^kU-A^{k-1}UA+A^kUA$ .
To show that $tr(A^k)=0$ , it suffices to show that , for every $l$ , $tr(A^lU)=0$ , that is clear.
Then we deduce -Newton- that $A$ is nilpotent !!
PS: Was the a... | [
" $trace(A^{k})=0$ for all k,and Newton",
"how prove tr(A^k)=0 for all k>0 ?",
"multiplying by A,and trace",
"A is nilpotent ",
"@Filipjack , yes I agree with you. \nHowever, is one better if one thinks of showing that the matrix is nilpotent rather than showing that it is singular?\nI think that, during ... | [
"origin:aops",
"2023 Contests",
"2023 SEEMOUS"
] | {
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 SEEMOUS/3028874.json"
} |
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