id stringlengths 4 4 | problem_markdown stringlengths 36 3.59k | solutions_markdown listlengths 0 10 | images images listlengths 0 15 | country stringclasses 58
values | competition stringlengths 3 108 ⌀ | topics_flat listlengths 0 12 | language stringclasses 18
values | problem_type stringclasses 4
values | final_answer stringlengths 1 1.22k ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0aje | Let $m, n$ be positive integers with $m > 1$. Anastasia partitions the integers $1, 2, \dots, 2m$ into $m$ pairs. Boris then chooses one integer from each pair and finds the sum of these chosen integers. Prove that Anastasia can select the pairs so that Boris cannot make his sum equal to $n$. | [
"Define the following ordered partitions:\n\n$$\nP_1 = (\\{1,2\\}, \\{3,4\\}, \\dots, \\{2m-1,2m\\})\n$$\n$$\nP_2 = (\\{1,m+1\\}, \\{2,m+2\\}, \\dots, \\{m,2m\\})\n$$\n$$\nP_3 = (\\{1,2m\\}, \\{2,m+1\\}, \\{3,m+2\\}, \\dots, \\{m,2m-1\\}).\n$$\nFor each $P_j$ we will compute the possible values for the expression $... | North Macedonia | Girls European Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, ine... | English | proof only | null | |
001t | Demostrar que existe una sucesión de enteros positivos $x_1, x_2, \dots, x_n, \dots$ que satisface las dos condiciones siguientes:
(i) contiene exactamente una vez a cada uno de los enteros positivos,
(ii) para cada $n=1,2,\dots$ la suma parcial $x_1 + x_2 + \dots + x_n$ es divisible por $n!$. | [] | Argentina | XIV Olimpiada Matemática de Países del Cono Sur | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | español | proof only | null | |
0ez9 | Problem:
Two congruent rectangles of area $A$ intersect in eight points. Show that the area of the intersection is more than $A/2$. | [] | Soviet Union | 4th ASU | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0j33 | Problem:
What are the last 8 digits of
$$
11 \times 101 \times 1001 \times 10001 \times 100001 \times 1000001 \times 111 ?
$$ | [
"Solution:\nAnswer: 19754321 Multiply terms in a clever order.\n$$\n\\begin{aligned}\n11 \\cdot 101 \\cdot 10001 & = 11,111,111 \\\\\n111 \\cdot 1001 \\cdot 1000001 & = 111,111,111,111\n\\end{aligned}\n$$\nThe last eight digits of $11,111,111 \\cdot 111,111,111,111$ are 87654321. We then just need to compute the la... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Modular Arithmetic"
] | null | final answer only | 19754321 | |
035r | Problem:
In a group of 9 persons it is not possible to choose 4 persons such that every one knows the three others. Prove that this group of 9 persons can be partitioned into four parts in such a way that nobody knows anyone from his part.
Emil Kolev | [
"Solution:\n\nWe have to prove that if the edges of a complete graph with 9 vertices are colored in blue and red in such a way that there is no blue quadrilateral then its vertices can be partitioned into 4 groups without any blue edges inside any group.\n\nLemma 1. The edges of a complete graph with 6 vertices are... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
08nr | Problem:
Let $ABC$ be an acute-angled triangle with circumcircle $\Gamma$, and let $O, H$ be the triangle's circumcenter and orthocenter respectively. Let also $A'$ be the point where the angle bisector of angle $BAC$ meets $\Gamma$. If $A'H = AH$, find the measure of angle $BAC$.
Figure 4: E... | [
"Solution:\nThe segment $AA'$ bisects $\\angle OAH$: if $\\angle BCA = y$ (Figure 4), then $\\angle BOA = 2y$, and since $OA = OB$, it is $\\angle OAB = \\angle OBA = 90^{\\circ} - y$. Also since $AH \\perp BC$, it is\n$\\angle HAC = 90^{\\circ} - y = \\angle OAB$ and the claim follows.\n\nSince $AA'$ bisects $\\an... | JBMO | JBMO Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 60 degrees | |
078z | Let points $A_1$, $A_2$, and $A_3$ lie on the circle $\Gamma$ in counter-clockwise order, and let $P$ be a point in the same plane. For $i \in \{1, 2, 3\}$, let $\tau_i$ denote the counter-clockwise rotation of the plane centred at $A_i$, where the angle of the rotation is equal to the angle at vertex $A_i$ in $\triang... | [
"**Solution 1.** Fix an index $i \\in \\{1, 2, 3\\}$. Let $D_1, D_2, D_3$ be the points of tangency of the incircle of triangle $\\triangle A_1A_2A_3$ with its sides $A_2A_3$, $A_3A_1$, $A_1A_2$ respectively.\nThe key observation is that given a line $\\ell$ in the plane, the image of $\\ell$ under the mapping $\\t... | India | INMO | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocent... | null | proof only | null | |
0f7v | Problem:
Prove that $1^{1987} + 2^{1987} + \ldots + n^{1987}$ is divisible by $n + 2$. | [] | Soviet Union | 21st ASU | [
"Number Theory > Other"
] | null | proof only | null | |
0gnp | Let $N$ denote a society of voters where each voter is endowed with preferences over a set of alternatives $A$ with $|N| = n > 2$, $|A| = m > 2$. A preference profile $R$ is an $n$-tuple of linear orderings on $A$ representing the preferences of voters on $A$. Given some $k \in \{1, 2, ..., m\}$, the $k$-plurality choi... | [
"Let $k > \\frac{m(n-1)}{n}$. Since $\\frac{kn}{m} > n - 1$ at least one alternative obtained all $n$ votes. Therefore, any chosen $a$ obtained $n$ votes in profile $R$ and by definitions obtains also $n$ votes in profile $R'$. Thus, the condition $k > \\frac{m(n-1)}{n}$ is sufficient for monotonicity.\n\nLet $2 \\... | Turkey | Team Selection Test for IMO | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
096r | Problem:
În pătratul $ABCD$ punctele $E$ şi $F$ aparţin laturilor $(AD)$ şi, respectiv, $(DC)$. Diagonala $AC$ intersectează $BE$ şi $BF$ în punctele $H$ şi, respectiv, $G$. Dacă $m(\angle EBF) = 45^{\circ}$, iar $EG \cap HF = \{O\}$, să se demonstreze că dreptele $BO$ şi $EF$ sunt perpendiculare. | [
"Solution:\n\n1) Cum $m \\angle CAD = m \\angle EBF = 45^{\\circ}$, patrulaterul $ABGE$ (figura alăturată) este inscriptibil. Prin urmare, $m \\angle BGE = m \\angle BAD = 90^{\\circ}$.\n\n2) Analog, cum $m \\angle FCH = m \\angle EBF = 45^{\\circ}$, patrulaterul $BCFH$ de asemenea este inscriptibil. Prin urmare, $... | Moldova | A 63-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0bzb | Determine the integers $x$ such that $2^x + x^2 + 25$ is the cube of a prime number. | [
"Let $y$ be a prime such that $y^3 = 2^x + x^2 + 25$; clearly, $y$ is odd. Moreover, $x$ can not be a negative integer nor can it belong to the set $\\{0, 1, 2, 3\\}$. Hence $x \\ge 4$ is even.\n\n1. If $x = 6k$, with $k \\in \\mathbb{N}^*$, we get $64^k + 36k^2 + 25 = y^3$.\n\nFor $k=1$ we get $x=6$ and $y=5$.\n\n... | Romania | THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 6 | |
0c6z | In a chess tournament every participant played with all others. At the end the organizer played a game with some participants. Finally a total of $100$ games were played in the tournament. Find the number or participants and the games the organizer played. | [] | Romania | 70th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 14 participants; the organizer played 9 games | |
062j | Problem:
Es sei $p>7$ eine Primzahl, die bei Division durch 6 den Rest 1 lässt. Setze $m=2^{p}-1$. Man beweise, dass $2^{m-1}-1$ ohne Rest durch $127 m$ teilbar ist. | [
"Solution:\n\nDie Lösung besteht aus drei Schritten:\n\n1. $2^{m-1}-1$ ist durch 127 teilbar.\n\n2. $2^{m-1}-1$ ist durch $m$ teilbar.\n\n3. 127 und $m$ sind teilerfremd.\n\nZu 1: Es gilt $2^{6} \\equiv 1 \\bmod 7$. Aus $p \\equiv 1 \\bmod 6$ folgt $2^{p} \\equiv 2 \\bmod 7$, also $7 \\mid m-1$. Mit $2^{7} \\equiv ... | Germany | 1. IMO-Auswahlklausur | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
044q | Suppose set $S = \{1, 2, 3, \dots, 10\}$ and the subset $A$ of $S$ satisfies
$$
A \cap \{1, 2, 3\} \neq \emptyset, \quad A \cup \{4, 5, 6\} \neq S.
$$
The number of such subsets is ________. | [
"First, we will find the number $N_1$ of subsets $A$ of $S$ such that $A \\cap \\{1, 2, 3\\} \\neq \\emptyset$ holds.\nThere are $2^3 - 1 = 7$ ways of selecting at least one element in $1, 2, 3$, while for each number in $4, 5, \\dots, 10$ there are two choices (selected or not selected). Therefore, $N_1 = 7 \\time... | China | China Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 888 | |
0jpl | Problem:
Let $n$ be the second smallest integer that can be written as the sum of two positive cubes in two different ways. Compute $n$. If your guess is $a$, you will receive $\max \left(25-5 \cdot \max \left(\frac{a}{n}, \frac{n}{a}\right), 0\right)$ points, rounded up. | [
"Solution:\n\nAnswer: $4104$\n\nA computer search yields that the second smallest number is $4104$. Indeed, $4104 = 9^{3} + 15^{3} = 2^{3} + 16^{3}$."
] | United States | HMMT November 2015 | [
"Number Theory > Diophantine Equations"
] | null | final answer only | 4104 | |
068t | Let $p \ge 2$ be a prime number. Angelo and Vangelis play in turns the following game: At the board there are $p$ empty boxes, the one next to other, and in each move, the current player puts a digit in one of the boxes. Angelo plays first and the game ends after all boxes are filled and we get a $p$-digit number $M$, ... | [
"$$\nM = a_0 + a_1 p + \\dots + a_{p-1} 10^{p-1},\n$$\nwhere $a_i$ is the value that the $i$-th box will take (from right to left).\nIf $p = 2$ or $p = 5$, then Angelo puts $0$ in the last box and wins.\nIf $p \\neq 2, 5$, then by Fermat's Theorem we have\n$$\np|10^{p-1}-1 \\Leftrightarrow p|(10^{\\frac{p-1}{2}})^2... | Greece | Selection Examination | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
093q | Problem:
Let $ABC$ be an acute scalene triangle with circumcircle $\omega$ and incenter $I$. Suppose the orthocenter $H$ of $BIC$ lies inside $\omega$. Let $M$ be the midpoint of the longer arc $BC$ of $\omega$. Let $N$ be the midpoint of the shorter arc $AM$ of $\omega$.
Prove that there exists a circle tangent to $\... | [
"Solution:\n\nDenote the circumcircles of $BHI$ and $CHI$ by $\\omega_{1}$ and $\\omega_{2}$ and their centers by $O_{1}$ and $O_{2}$, respectively. Let $O$ be the center of $\\omega$. Let $R$ be the radius of $\\omega$.\nSince $H$ is the orthocenter of triangle $BIC$ it follows that $I$ is the orthocenter of trian... | Middle European Mathematical Olympiad (MEMO) | 14th Middle European Mathematical Olympiad 2020 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"G... | null | proof only | null | |
05m7 | Problem:
Soit $n \geqslant 2$. On place une pièce sur chaque case d'un échiquier $n \times n$. Un mouvement consiste à déplacer chaque pièce sur une case qui touche la case de départ par un coin exactement (plusieurs pièces peuvent se retrouver sur la même case).
Quel est le plus petit entier $k$ tel qu'il est possibl... | [
"Solution:\n\nOn note $A$, $B$, $C$ et $D$ les quatre cases qui forment le carré $2 \\times 2$ en haut à droite :\n\nChaque pièce peut être amenée sur une des cases $A$, $B$, $C$ et $D$, puis osciller à chaque mouvement entre $A$ et $D$ ou entre $B$ et $C$. Il est donc possible qu'après un ... | France | OCympiades Françaises de Mathématiques - Envoi Numéro 4 - Combinatoire | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 4 | |
07yt | Problem:
La pianta di una casa ha la forma di una L ottenuta affiancando in modo opportuno quattro quadrati il cui lato misura 10 metri. Le pareti laterali sono tutte alte 10 metri e il tetto della casa ha sei facce partenti dai sei muri laterali e inclinate di $30^{\circ}$ rispetto ad un piano orizzontale.
Determinare... | [
"Solution:\nIl sottotetto visto dall'alto ha la forma della figura 1, poichè i punti di incontro delle varie facce del tetto sono equidistanti dagli spigoli della base poichè tutte le facce hanno la stessa inclinazione rispetto all'orizzontale.\nOsserviamo che possiamo trasformare il sottotetto come in figura 2 sen... | Italy | null | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 2750/(3*sqrt(3)) cubic meters | |
0kst | Problem:
Compute the number of ordered pairs of positive integers $(a, b)$ satisfying the equation
$$
\operatorname{gcd}(a, b) \cdot a + b^2 = 10000.
$$ | [
"Solution:\nLet $\\operatorname{gcd}(a, b) = d$, $a = d a'$, $b = d b'$. Then,\n$$\nd^2\\left(a' + b'^2\\right) = 100^2.\n$$\nConsider each divisor $d$ of $100$. Then, we need to find the number of solutions in coprime integers to $a' + b'^2 = \\frac{100^2}{d^2}$. Note that every $b' < 100 / d$ coprime to $\\frac{1... | United States | HMMT November 2022 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof and answer | 99 | |
0jlv | For a prime $p$, a subset $S$ of residues modulo $p$ is called a *sum-free multiplicative subgroup of $\mathbb{F}_p$* if:
* there is a nonzero residue $\alpha$ modulo $p$ such that $S = \{1, \alpha^1, \alpha^2, \dots\}$ (all considered mod $p$), and
* there are no $a, b, c \in S$ (not necessarily distinct) such that $... | [
"We prove a stronger statement, generalizing the desired condition \"$0 \\notin S + S - S$\" to \"$0 \\notin a_1S + a_2S + \\dots + a_kS$\", for fixed integers $a_1, \\dots, a_k$ with nonzero sum $a_1 + \\dots + a_k$. (In the original problem we have $(a_1, \\dots, a_k) = (1, 1, -1)$ (so $k=3$).\n\nFix a positive i... | United States | IMO Team Selection Test | [
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisen... | null | proof only | null | |
0l9o | Find all functions $f$ defined on $\mathbb{R}$, taking values in $\mathbb{R}$ so that
$$
f(f(x - y)) = f(x)f(y) - f(x) + f(y) - xy
$$
for all real numbers $x, y$. | [
"Suppose that $f: \\mathbb{R} \\to \\mathbb{R}$ satisfies the relation in the problem, i.e.\n$$\nf(f(x - y)) = f(x)f(y) - f(x) + f(y) - xy, \\quad (1)\n$$\nfor all $x, y \\in \\mathbb{R}$.\nPut $f(0) = a$\nBy substituting $x = y = 0$ into (1) we get\n$$\nf(a) = a^2 \\quad (2)\n$$\nBy substituting $x = y = 0$ into (... | Vietnam | Vietnam Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = -x for all real x | |
0a2n | An $\ell$-code is an integer $n \ge 0$ of at most $\ell$ digits, if necessary supplemented by leading zeros, so that it consists of $\ell$ digits in total. Thus, you can make a 4-code out of 310 by writing it as 0310. An $\ell$-code is called *self-squared* if the last $\ell$ digits of the square of that code form exac... | [
"a. Suppose that $n$ is a self-squared $\\ell$-code. Then the last $\\ell$ digits of $n^2 - n = n(n - 1)$ are all zeros. This means that $n(n - 1)$ is divisible by $10^\\ell$. Vice versa, $n$ is self-squared if $n^2 - n$ is divisible by $10^\\ell$ and thus ends in $\\ell$ zeros.\n\nb. The number $n(n - 1)$ is divis... | Netherlands | Dutch Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 10^ℓ + 1 | |
0apt | Problem:
The integer $x$ is the least among three positive integers whose product is $2160$. Find the largest possible value of $x$. | [
"Solution:\n\nNote that $2160 = 2^4 \\cdot 3^3 \\cdot 5$. By trial-and-error method, we will notice that the set of three positive integers whose product is $2160$ that will have a maximum least integer is $\\{10, 12, 18\\}$."
] | Philippines | Tenth Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 12 | |
0eev | Problem:
Poenostavi izraz
$$
\frac{\frac{2 x y}{x+y}-x}{\frac{1}{y}+\frac{1}{x-2 y}}+\frac{\left(x^{2}-x y+y^{2}\right)\left(x^{3}-x(x-y)^{2}\right)}{x^{3}+y^{3}}
$$
Za katere realne vrednosti $x$ in $y$ izraz nima pomena? | [
"Solution:\n\nRazširjanje ulomkov na skupni imenovalec\n$$\n\\frac{\\frac{2 x y}{x+y}-x}{\\frac{1}{y}+\\frac{1}{x-2 y}}=\\frac{\\frac{x y-x^{2}}{x+y}}{\\frac{x-y}{y(x-2 y)}}\n$$\n\nRazreševanje dvojnega ulomka\n\nPoenostavitev prvega člena izraza\n$$\n= -\\frac{x y(x-2 y)}{x+y}\n$$\n\nRazcep\n$$\nx^{3}+y^{3}=(x+y)\... | Slovenia | 16. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | The expression simplifies to xy. It is undefined for y = 0, x = -y, x = 2y, or x = y. | |
0hmm | Problem:
A castle has infinitely many rooms labeled $1,2,3, \ldots$, which are divided into several halls. Suppose room $n$ is on the same hall as rooms $3 n+1$ and $n+10$ for every $n$. Determine the maximum possible number of different halls in the castle. | [
"Solution:\n\nThere are at most three different halls in the castle. Because rooms $n$ and $n+10$ are on the same hall, any two rooms with the same units digit must be on the same hall.\n\nNow, repeatedly using the rule that rooms $n$ and $3 n+1$ are on the same hall, we find that room $1$ is on the same hall as ro... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 3 | |
00sm | Let $ABC$ be a scalene and acute triangle, with circumcentre $O$. Let $\omega$ be the circle with centre $A$, tangent to $BC$ at $D$. Suppose there are two points $F$ and $G$ on $\omega$ such that $FG \perp AO$, $\angle BFD = \angle DGC$ and the couples of points $(B, F)$ and $(C, G)$ are in different halfplanes with r... | [
"Consider any two points $F, G$ on $\\omega$ such that $\\angle BFD = \\angle DGC$. Exploiting the isosceles triangles $\\triangle AFG$, $\\triangle AFD$, and $\\triangle ADG$, we deduce (using directed angles throughout):\n$$\n\\angle DBF - \\angle GCD = 180^\\circ - \\angle BFD - \\angle BDF - (180^\\circ - \\ang... | Balkan Mathematical Olympiad | BMO 2019 Shortlist | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
00sn | Given an acute triangle $ABC$, let $M$ be the midpoint of $BC$ and $H$ the orthocentre. Let $\Gamma$ be the circle with diameter $HM$, and let $X$, $Y$ be distinct points on $\Gamma$ such that $AX$, $AY$ are tangent to $\Gamma$. Prove that $BXYC$ is cyclic. | [
"Let $D$ be the foot of the altitude from $A$ to $BC$, which also lies on $\\Gamma$. Let $O$ be the circumcentre of $\\triangle ABC$. Since $\\angle HDM = 90^\\circ$, note that rays $HD$ and $HM$ meet the circumcircle at points which are reflections in $OM$. Then, since $\\angle BAD = \\angle OAC$, we recover the w... | Balkan Mathematical Olympiad | BMO 2019 Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
... | English | proof only | null | |
0lc8 | Let $ABCD$ be a quadrilateral inscribed in circle $(O, R)$ and is not a trapezoid. Two lines $AC$, $BD$ intersect each other at $E$ and the bisector of $\angle AEB$ meets the lines $AB$, $BC$, $CD$, $DA$ at $M$, $N$, $P$, $Q$, respectively.
1. Prove that four lines $(AQM)$, $(BMN)$, $(CNP)$, $(DPQ)$ concur at a unique... | [
"\n\n1. Let $R$ be the intersection of two lines $AD$, $BC$ and $S$ be the intersection of two lines $AB$, $CD$ (because $ABCD$ is not a trapezoid then $R$, $S$ are quite specific). Assume that $B$ lies between $A$, $S$ and also lies between $C$, $R$ as in the figure, the other cases can be... | Vietnam | Vietnamese Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequa... | null | proof only | null | |
01lz | Find all pairs $(m, n)$ of nonnegative integers for which
$$
m^2 + 2 \cdot 3^n = m(2^{n+1} - 1).
$$ | [
"2. See IMO-2010 Shortlist, Problem N2."
] | Belarus | Selection and Training Session | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | ((m,n) = (6,3), (9,3), (9,5), (54,5)) | |
051x | Define *magic square* as a $3 \times 3$ table where each cell contains one number from $1$ to $9$ so that all these numbers are used and all row sums and column sums are equal. Prove that any two magic squares can be obtained from each other via the following transformations: interchanging two rows, interchanging two c... | [
"As all the transformations are invertible, it suffices to show that every magic square can be turned to one particular magic square by these transformations.\n\n<table><tr><td>2</td><td></td><td>4</td></tr><tr><td></td><td></td><td></td></tr><tr><td>6</td><td></td><td>8</td></tr></table>\n<table><tr><td>2</td><td>... | Estonia | Final Round of National Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0h7l | Given a polygon with $2016$ vertices. Alisa and Basilio play the following game. In each turn a player draws a diagonal of the polygon which intersects the other drawn diagonals or the sides only at the vertices. When the polygon is cut into triangles the game is finished. For each triangle having exactly zero sides am... | [
"Let $a$ be the number of the triangles which have $0$ sides among the sides of the initial polygon, $b$ be the number of the triangles having $1$ such side, and $c$ be the number of the triangles with $2$ such sides (fig. 02). Then $b + 2c = 2016$ since the polygon has $2016$ sides. Also, since we will get $2014$ ... | Ukraine | UkraineMO | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | Basilio by 2 cents | |
07yl | Problem:
Sia $x_{0}, x_{1}, x_{2}, \ldots$ una successione di numeri razionali definita per ricorrenza nella maniera seguente: $x_{0}$ è un numero razionale qualunque, e, per $n \geq 0$,
$$
x_{n+1}= \begin{cases}\left|\frac{x_{n}}{2}-1\right| & \text { se il numeratore di } x_{n} \text { è pari }, \\ \left|\frac{1}{x... | [
"Solution:\n\n(a) Scriviamo ogni termine della successione come frazione ridotta ai minimi termini, $x_{n}=\\frac{p_{n}}{q_{n}}$, e consideriamo una nuova successione $y_{n}=\\max \\left\\{p_{n}, q_{n}\\right\\}$. Vogliamo dimostrare che la successione $y_{n}$ è debolmente decrescente, ossia che $y_{n+1} \\leq y_{n... | Italy | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0ili | Problem:
Michael has $16$ white socks, $3$ blue socks, and $6$ red socks in a drawer. Ever the lazy college student, he has overslept and is late for his favorite team's season-opener. Because he is now in such a rush to get from Harvard to Foxborough, he randomly takes socks from the drawer (one at a time) until he h... | [
"Solution:\n\nAnswer: $4$. It is possible for him to begin with three socks of different colors, but an instance of the Pigeon Hole Principle is that among any four objects of three types some two are the same type."
] | United States | $10^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | final answer only | 4 | |
0l0i | Let $S$ be a subset of $\{1, 2, 3, \dots, 2024\}$ such that the following two conditions hold:
* If $x$ and $y$ are distinct elements of $S$, then $|x - y| > 2$.
* If $x$ and $y$ are distinct odd elements of $S$, then $|x - y| > 6$.
What is the maximum possible number of elements in $S$?
(A) 436 (B) 506 (C) 608 (D) 654... | [
"**Answer (C):** If $S$ consists of the positive integers less than or equal to $2024$ that are congruent to $1$, $4$, or $8$ modulo $10$, then every pair of elements in $S$ differ by at least $|4 - 1| = |11 - 8| = 3$, and every pair of odd elements of $S$ differ by at least $|11 - 1| = 10$. This set,\n$$\n\\{1, 4,... | United States | AMC 10 A | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | MCQ | C | |
0igw | Problem:
What is the maximum number of bishops that can be placed on an $8 \times 8$ chessboard such that at most three bishops lie on any diagonal? | [
"Solution:\nIf the chessboard is colored black and white as usual, then any diagonal is a solid color, so we may consider bishops on black and white squares separately. In one direction, the lengths of the black diagonals are $2, 4, 6, 8, 6, 4$, and $2$. Each of these can have at most three bishops, except the firs... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 38 | |
0bu0 | Problem:
Un inel $(A,+, \cdot)$ are proprietatea (P) dacă $A$ este finit şi grupul multiplicativ al elementelor sale inversabile este izomorf cu un subgrup diferit de $\{0\}$ al grupului aditiv $(A,+)$. Arătaţi că:
a. Dacă un inel are proprietatea $(\mathrm{P})$, atunci numărul elementelor sale este par.
b. Pentru o ... | [
"Solution:\n\na. Fie $A$ un inel care are proprietatea (P) şi fie $m = |U(A)|$. Rezultă că $(-1)^{m} = 1$.\nDacă $m$ este impar, atunci $-1 = 1$, deci ord$(1) = 2$ în grupul aditiv $(A,+)$ şi prin urmare $|A|$ este par.\n\nDacă $m$ este par, cum $m$ este un divizor al lui $|A|$, rezultă că $|A|$ este par.\n\nb. Fie... | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
0766 | Problem:
Let $a$, $b$ be natural numbers with $a b > 2$. Suppose that the sum of their greatest common divisor and least common multiple is divisible by $a + b$. Prove that the quotient is at most $(a + b) / 4$. When is this quotient exactly equal to $(a + b) / 4$? | [
"Solution:\n\nLet $g$ and $l$ denote the greatest common divisor and the least common multiple, respectively, of $a$ and $b$. Then $g l = a b$. Therefore $g + l \\leq a b + 1$.\n\nSuppose that $(g + l) / (a + b) > (a + b) / 4$. Then we have $a b + 1 > (a + b)^2 / 4$, so we get $(a - b)^2 < 4$.\n\nAssuming $a \\geq ... | India | Indian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | The quotient is at most (a + b)/4. Equality holds if and only if either a = b = 2 or a and b are consecutive odd integers. | |
0a0f | Let $n$ be a positive integer. Prove that the numbers
$$
1^1, 3^3, 5^5, \dots, (2^n - 1)^{2^n - 1}
$$
are in different residue classes modulo $2^n$. | [
"We proceed by induction on $n$. For $n=1$, the only number in the sequence is $1^1$, so the given statement is trivially true.\n\nFor the induction step, we are to show that for $n \\ge 1$, the numbers $1^1, 3^3, 5^5, \\dots, (2^{n+1}-1)^{2^{n+1}-1}$ are in different residue classes modulo $2^{n+1}$, given the ind... | Netherlands | IMO Team Selection Test 2 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0edo | In the equality of the three fractions $\frac{2}{6} = \frac{3}{9} = \frac{58}{174}$ each digit from $1$ to $9$ occurs exactly once and the value of all three fractions is $\frac{1}{3}$. Here is another example of such equality, $\frac{*}{*} = \frac{*}{*} = \frac{7*}{15*}$, where some of the digits have been replaced by... | [
"Sorting the available answers we get $\\frac{1}{3} < \\frac{1}{2} < \\frac{3}{5} < \\frac{2}{3} < \\frac{3}{4}$. The last of the three fractions in the equality can be estimated as\n$$\n\\frac{1}{3} = \\frac{60}{180} < \\frac{7*}{15*} < \\frac{90}{150} = \\frac{3}{5}.\n$$\nSo, the only possible answer is (A). Afte... | Slovenia | Slovenija 2016 | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | A | |
09ea | Let $a, b, c \in [\frac{1}{3}; 3]$. Show that
$$
\frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{c+a} \ge \frac{7}{5}.
$$ | [
"Since the inequality is cyclic, we may assume $c = \\max(a, b, c)$. Furthermore $a, b, c \\in [\\frac{1}{3}; 3]$ implies $\\frac{c}{9} \\le a \\le c$. Substituting $x = \\frac{7}{5} - \\frac{b}{b+c}$ we get $(1-x)a^2 + ((2-x)c - xb)a + (1-x)bc \\ge 0$ and this is a quadratic inequality with variable $a$.\n\nGraph ... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
0hdc | What is the smallest possible value the expression $ab + a + b$ can take, if real numbers $a, b$ satisfy the condition $a^2 + b^2 = 25$. | [
"$$(a+b+1)^2 \\ge 0$$\n$$a^2 + b^2 + 1 + 2ab + 2a + 2b \\ge 0 \\Rightarrow ab + a + b \\ge -\\frac{1}{2}(a^2 + b^2 + 1) = -13.$$ \nBy choosing $a = -4$ and $b = 3$, we obtain: $ab + a + b = -12 - 4 + 3 = -13$, i.e. the smallest possible value is reached."
] | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | -13 | |
0kdw | Problem:
Given an $8 \times 8$ checkerboard with alternating white and black squares, how many ways are there to choose four black squares and four white squares so that no two of the eight chosen squares are in the same row or column? | [
"Solution:\n\nNumber both the rows and the columns from $1$ to $8$, and say that black squares are the ones where the rows and columns have the same parity. We will use, e.g., \"even rows\" to refer to rows $2, 4, 6, 8$. Choosing $8$ squares all in different rows and columns is equivalent to matching rows to column... | United States | HMMT February 2020 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 20736 | |
0byk | Let $a > 0$ be a real number. Prove the inequality
$$
a^{\sin x} \cdot (a+1)^{\cos x} \ge a, \quad \forall x \in [0, \frac{\pi}{2}].
$$ | [
"If $a > 1$, taking logs of both sides we obtain the equivalent inequality\n$$\n\\sin x + \\cos x \\cdot \\log_a (a+1) \\ge 1.\n$$\nFor $x \\in [0, \\pi/2]$, we have $\\sin x \\ge \\sin^2 x$ and $\\cos x \\ge \\cos^2 x$. Because $\\log_a (a+1) > 1$, we obtain\n$$\n\\sin x + \\cos x \\cdot \\log_a (a+1) \\ge \\sin^2... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof only | null | |
05db | Problem:
A set $A$ of integers is called sum-full if $A \subseteq A+A$, i.e. each element $a \in A$ is the sum of some pair of (not necessarily different) elements $b, c \in A$. A set $A$ of integers is said to be zero-sum-free if $0$ is the only integer that cannot be expressed as the sum of the elements of a finite n... | [
"Solution:\nThe set $A=\\{F_{2 n}: n=1,2, \\ldots\\} \\cup\\{-F_{2 n+1}: n=1,2, \\ldots\\}$, where $F_{k}$ is the $k^{\\text{th}}$ Fibonacci number ($F_{1}=1$, $F_{2}=1$, $F_{k+2}=F_{k+1}+F_{k}$ for $k \\geq 1$) qualifies for an example. We then have $F_{2 n}=F_{2 n+2}+(-F_{2 n+1})$ and $-F_{2 n+1}=(-F_{2 n+3})+F_{... | European Girls' Mathematical Olympiad (EGMO) | European Girls' Mathematical Olympiad 2012-Day 1 Solutions | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | Yes; for example A = {F_{2n} : n ≥ 1} ∪ {−F_{2n+1} : n ≥ 1}, where F_k are Fibonacci numbers. | |
02s2 | Problem:
Vanessa deseja escolher 50 números inteiros positivos distintos menores do que 100 e tais que a soma de quaisquer dois números escolhidos por ela seja sempre distinta de 99 e de 100.
a) Mostre como Vanessa pode atingir o seu objetivo.
b) Mostre que há somente uma maneira pela qual Vanessa pode escolher esse... | [
"Solution:\n\na) Basta que ela escolha os números $50, 51, 52, \\ldots, 99$. Com efeito, se $a < b$ são dois desses números, então $a \\geq 50$ e $b \\geq 51$. Somando essas duas desigualdades obtemos $a + b \\geq 101$ e, em particular, a soma $a + b$ não pode ser igual a $99$ ou a $100$.\n\nb) Consideremos os segu... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | The unique selection is the set of integers 50, 51, 52, ..., 99. | |
0ib1 | Problem:
A lattice point is a point whose coordinates are both integers. Suppose Johann walks in a line from the point $(0,2004)$ to a random lattice point in the interior (not on the boundary) of the square with vertices $(0,0)$, $(0,99)$, $(99,99)$, $(99,0)$. What is the probability that his path, including the endpo... | [
"Solution:\nIf Johann picks the point $(a, b)$, the path will contain $\\gcd(a, 2004-b)+1$ points. There will be an odd number of points in the path if $\\gcd(a, 2004-b)$ is even, which is true if and only if $a$ and $b$ are both even. Since there are $49^{2}$ points with $a, b$ both even and $98^{2}$ total points,... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 3/4 | |
0he0 | At the round table, 50 inhabitants of the island "Loud Mouths" are sitting at equal distance from one another and discussing something. Each of them is either a knight who always tells the truth, or a liar who lies every time, and people of each type are present at the table. During the discussion, everyone said that t... | [
"Lemma 1. There is at least one knight among any two diametrically opposite islanders.\n\n*Proof.* Suppose the opposite is true: let $C$ and $X$ be two opposite sitting islanders, which are both liars (Fig. 24). Then the neighbours of $X$, islanders $Y$ and $W$, must be liars, since $C$ must have lied, which meant ... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Between 10 and 25 liars (inclusive). | |
0axj | Problem:
Simplify $\sqrt{13+4 \sqrt{3}}+\sqrt{13-4 \sqrt{3}}$. | [
"Solution:\n\nLet $x=\\sqrt{13+4 \\sqrt{3}}+\\sqrt{13-4 \\sqrt{3}}$; then $x^{2}=48$. Since $x$ is clearly positive, we take $x=\\sqrt{48}=4 \\sqrt{3}$."
] | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Other",
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | final answer only | 4√3 | |
0ezb | Problem:
$S$ is the set of all positive integers with $n$ decimal digits or less and with an even digit sum. $T$ is the set of all positive integers with $n$ decimal digits or less and an odd digit sum. Show that the sum of the $k$th powers of the members of $S$ equals the sum for $T$ if $1 \leq k < n$. | [] | Soviet Union | 4th ASU | [
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof only | null | |
0390 | For every positive integer $n$ set $a_n = 0$, if the number of divisors of $n$, greater than $2007$, is even and $a_n = 1$, if this number is odd. Is the number $\alpha = 0, a_1a_2a_3 \dots a_k \dots$ rational? | [
"We prove that $\\alpha$ is irrational. Suppose $\\alpha$ is a rational number, i.e. the sequence $a_1, a_2, a_3, \\ldots, a_k, \\ldots$ is periodic from some point onwards. Hence there exist $k_0$ and $T$ such that for any $k > k_0$ we have $a_k = a_{k+T}$.\n\nChoose a positive integer $m$ for which $mT > k_0$ and... | Bulgaria | Winter Mathematical Competition | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0h0e | Given $ABCD$ be a quadrilateral inscribed in a circle with center $O$, and let $P = AC \cap BD$, $BC \parallel AD$. Rays $AB$ and $DC$ intersect at the point $E$. A circle with the center $I$ is inscribed in the triangle $EBC$ and is tangent to the line $BC$ at point $T_1$. The excircle of the triangle $EAD$ with cente... | [
"Let $P_1$, $P_2$ be respectively the feet of perpendiculars from $P$ to lines $BC$ and $AD$, $M_1$, $M_2$ respectively are midpoints of sides $BC$ and $AD$ respectively, the excircle of the triangle $EBC$ touches the side $BC$ at the point $T'$. $T'C = BT_1 \\Rightarrow \\frac{BT_1}{T_1C} = \\frac{CT'}{T'B}$. $ABC... | Ukraine | The Problems of Ukrainian Authors | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
05ff | Problem:
Soit $ABCD$ un quadrilatère convexe tel que $AC = BD$ et tel que les côtés $AB$ et $CD$ ne sont pas parallèles. Soit $P$ le point d'intersection des diagonales $(AC)$ et $(BD)$. Soient $E$ et $F$ des points respectivement sur les segments $[BP]$ et $[AP]$ tels que $PC = PE$ et $PD = PF$. Montrer que le cercle... | [
"Solution:\n\n\n\nOn note dans un premier temps $X$ l'intersection des droites $(CD)$ et $(FE)$, $Y$ l'intersection des droites $(AB)$ et $(CD)$ ainsi que $Z$ l'intersection des droites $(AB)$ et $(FE)$.\n\nLa première idée de cet exercice est de considérer un point de Miquel (pour savoir c... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilate... | null | proof only | null | |
0cdn | Let $ABC$ be an acute angled triangle with $\angle B > \angle C$. Consider the points $D, E, J, K, S$ on its circumcircle $C(O)$, such that $A, E, J$ and $K$ are on the same side of the line $BC$, the diameter $DE$ and the line $BC$ are orthogonal, $S \in \widehat{EK}$ and $\widehat{AE} = \widehat{BJ} = \widehat{CK} = ... | [
"Denote $\\vec{AE} = x$ and $\\vec{BD} = y$. Obviously, we have $4x + y = 180^\\circ$.\nFrom $\\angle APM + \\angle MAP = 2x + \\frac{y}{2} = 90^\\circ$ we deduce that $AM \\perp MP$. Since $FO$ is the perpendicular bisector of $BC$, we have $FB = FC$, thus the triangle $FBC$ is isosceles, with the apex $F$. From $... | Romania | THE 73rd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD - FOURTH SELECTION TEST | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
05n6 | Problem:
Soit $n \geqslant 5$ un entier, et $\{a_{1}, a_{2}, \cdots, a_{n}\}=\{1,2, \cdots, n\}$. Prouver que les nombres $a_{1}, a_{1}+a_{2}, a_{1}+a_{2}+a_{3}, \cdots, a_{1}+a_{2}+\cdots+a_{n}$ donnent au moins $\lfloor\sqrt{n}\rfloor+1$ restes distincts modulo $n$.
($\lfloor.\rfloor$ désigne la partie entière.) | [
"Solution:\n\nPour $i=1, \\cdots, n$, on pose $b_{i}=a_{1}+\\cdots+a_{i}$. Procédons par l'absurde et supposons que les nombres $b_{1}, \\cdots, b_{n}$ ne donnent qu'au plus $\\sqrt{n}$ restes distincts modulo $n$.\n\nNotons que, pour $i=1, \\cdots, n-1$, on a $a_{i+1}=b_{i+1}-b_{i}$, et seul celui des $a_{i}$ qui ... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
08df | Problem:
Alessandra scrive sul quaderno (di Luigi) tutti i numeri naturali $n$ che hanno entrambe le seguenti proprietà: $n$ ha esattamente 4 divisori positivi (compresi 1 e $n$ stesso), e la somma dei divisori positivi di $n$ fa 42. Quanto vale la somma di tutti i numeri scritti da Alessandra?
(A) 0
(B) 12
(C) 20
(D... | [
"Solution:\n\nLa risposta è (D). Se $n$ ha quattro divisori positivi, essi sono, in ordine crescente, $1, p, q, n$ con $p$ numero primo.\nSe $q$ è un numero primo, possiamo concludere che $n = p \\cdot q$, perché $n$ non ha altri divisori. La somma dei divisori positivi è $1 + p + q + p \\cdot q = (1 + p) + q(1 + p... | Italy | Gara di Febbraio | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | MCQ | D | |
0hql | Problem:
Wally has a very unusual combination lock number. It has five digits, all different, and is divisible by $111$. If he removes the middle digit and replaces it at the end, the result is a larger number that is still divisible by $111$. If he removes the digit that is now in the middle and replaces it at the en... | [
"Solution:\n\nThe solution is $74259$. The numbers $74259$, $74592$, and $74925$ are all divisible by $111$. Denote the original number by $\\overline{abcde}$ (the line prevents confusion with $a \\cdot b \\cdot c \\cdot d \\cdot e$). Then we have\n$$\n\\begin{aligned}\n& 111 \\mid \\overline{abcde} \\\\\n& 111 \\m... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 74259 | |
0ajl | Let a square scheme $2n \times 2n$, made of unit white squares be given. Allowed move is to change the color of three consecutive unit squares in a particular row or three consecutive unit squares in a particular column - unit square with white color goes to unit square with black color and vice versa.
Find all nonnega... | [
"We will call black unit squares which one when the square scheme is colored like a chess table are black and white unit squares those which will not change their color. It is not difficult to see when the square scheme is colored like chess table, we will have $2n^2$ black and $2n^2$ white unit squares, i.e. we wi... | North Macedonia | IMO Selection Test | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | All integers n ≥ 2 with n divisible by 3 | |
03wc | Given the line $L: x + y - 9 = 0$ and the circle $M: 2x^2 + 2y^2 - 8x - 8y - 1 = 0$, point $A$ is on $L$ and points $B, C$ are on $M$; $\angle BAC = 45^\circ$ and the line $AB$ is through the center of $M$. Then the range of the $x$ coordinate of point $A$ is ______. | [
"Suppose that $A(a, 9-a)$. Then the distance from the center of $M$ to the line $AC$ is\n$$\nd = |AM| \\times \\sin \\angle BAC \\\\\n= \\sqrt{(a-2)^2 + (9-a-2)^2} \\times \\sin 45^\\circ \\\\\n= \\sqrt{2a^2 - 18a + 53} \\times \\frac{\\sqrt{2}}{2}.\n$$\nOn the other hand, since the line $AC$ intercepts $M$, it fol... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 3 ≤ a ≤ 6 | |
006n | Consideramos todas las colecciones de pesas con peso total igual a $65$ en las que el peso máximo de una pesa es $w$. Halle el mayor valor de $w$ para el que cualquier colección se puede dividir con certeza en dos grupos cuyos pesos totales difieren en a lo sumo $1$. | [] | Argentina | XVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | 33 | |
0dy6 | Find all real numbers $x$ and $y$ that satisfy the equations
$$
\begin{aligned}
x^3 + 8y^3 &= x + 2y, \\
2x^2y + 4xy^2 &= x + 2y.
\end{aligned}
$$ | [
"We can rewrite the equations as\n$$\n\\begin{aligned}\n(x + 2y)(x^2 - 2xy + 4y^2) &= x + 2y, \\\\\n2xy(x + 2y) &= x + 2y.\n\\end{aligned}\n$$\nObviously, every pair of numbers $x$ and $y$ that satisfies $x + 2y = 0$, solves the equations. Now, assume $x + 2y \\neq 0$. We can then divide by $x + 2y$ and get $x^2 - ... | Slovenia | Slovenija 2008 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | All real pairs (x, y) with x + 2y = 0, together with (1, 1/2) and (-1, -1/2). | |
0hqo | Problem:
Let $A O B$ be a $60$-degree angle. For any point $P$ in the interior of $\angle A O B$, let $A'$ and $B'$ be the feet of the perpendiculars from $P$ to $A O$ and $B O$ respectively. Denote by $r$ and $s$ the distances $O P$ and $A' B'$. Find all possible pairs of real numbers $(r, s)$. | [
"Solution:\n\nExtend $A' P$ to meet $O B$ at $Z$. Notice that, because $\\angle O A' P$ and $\\angle O B' P$ are both right, the circle with diameter $O P$ passes through $O, P, A'$, and $B'$. Thus $\\angle B' O P = \\angle B' A' P$ since both intercept the same arc on this circle, and $\\triangle Z O P \\sim \\tri... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | All pairs (r, r*sqrt(3)/2) with r > 0. | |
0hfu | In a right triangle $ABC$ with right angle $C$ on the sides $BC$, $AC$ and $AB$ points $D, E$ and $F$ correspondingly were chosen so that $\angle DAB = \angle CBE$ and $\angle BEC = \angle AEF$. Prove that $DB = DF$.
(Mykhailo Shtandenko) | [
"Consider the point $K$, symmetric to the point $B$ with respect to point $C$ (fig. 13). Then $\\Delta BEC = \\Delta KCE$, and $\\angle KEC = \\angle BEC = \\angle AEF$, so points $K, E, F$ lie on the same line. Then from the statement it follows, that\n\nFig. 13\n$$\n\\angle FAD = \\angle ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03fc | For a positive integer $n$, denote with $b(n)$ the smallest positive integer $k$, such that there exist integers $a_1, a_2, \dots, a_k$, satisfying $n = a_1^{a_2} + a_2^{a_3} + \dots + a_k^{a_1}$. Determine whether the set of positive integers $n$ is finite or infinite, which satisfy:
$$
\text{a) } b(n) = 12; \quad \te... | [
"a) From Fermat's theorem and $y^2 \\equiv 1 \\pmod{67} \\Leftrightarrow y \\equiv \\pm 1 \\pmod{67}$ it follows that any student number gives a remainder of $0$, $1$ or $66$ when divided by $67$. Let us consider the numbers $12^{66k+1}$, where $k \\in \\mathbb{N}$. They are presented as the sum of $12$ student num... | Bulgaria | 3 Bulgarian Spring Tournament | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | a) infinite; b) finite (empty) | |
01hz | Let $\triangle ABC$ be an acute triangle. Denote by $E$ and $F$ the feet of the altitudes from $B$ and $C$, respectively. Let $H$ be the intersection of $BE$ and $CF$. Let $D$ be on the same side of line $BC$ as $A$ and satisfy:
$$
\angle DBC = \angle DCB = \angle BAC.
$$
Let $N$ be the midpoint of $EF$. Prove that poi... | [
"Refer to figure 23. Notice that as triangles $HEF$ and $HCB$ are similar with different orientations and $C$, $H$, $F$ and $B$, $H$, $E$ collinear, then the statement is equivalent to $DH$ being a symmedian from $H$ in $BHC$.\n\nAs $\\angle CBD = \\angle BAC = 180^\\circ - \\angle BHC$, the line $DB$ is tangent to... | Baltic Way | Baltic Way 2021 Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellane... | null | proof only | null | |
0al4 | Let $f: \mathbb{N} \to \mathbb{N}_0$ be a non-zero function from the set of positive integers to the set of non-negative integers such that for all $a, b \in \mathbb{N}$ it holds that
$$
2f(ab) = (b+1)f(a) + (a+1)f(b).
$$
Prove that for any prime $p$ there are a prime $q$, positive integers $x_1, \dots, x_n$ and a non-... | [] | North Macedonia | Team Selection Test for BMO | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | English | proof only | null | |
02zl | Problem:
Seja $ABCD$ um retângulo com $BC = 2 \cdot AB$. Seja $E$ o ponto médio de $BC$ e $P$ um ponto arbitrário interno ao lado $AD$. Sejam $F$ e $G$ os pés das perpendiculares desenhadas de $A$ a $BP$ e de $D$ a $CP$. Sabemos que $\angle BPC = 85^\circ$.
a) Verifique que os triângulos $BE... | [
"Solution:\n\na) Em virtude das relações métricas nos triângulos retângulos aplicadas ao triângulo $ABP$, temos\n$$\nBE^2 = AB^2 = BF \\cdot BP\n$$\nPortanto,\n$$\n\\frac{BE}{BP} = \\frac{BF}{BE}\n$$\nDada a relação de proporcionalidade da última equação e $\\angle EBF = \\angle EBP$, segue que os triângulos $BEF$ ... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 85° | |
0ey8 | Problem:
What is the maximum possible length of a sequence of natural numbers $x_{1}, x_{2}, x_{3}, \ldots$ such that $x_{i} \leq 1998$ for $i \geq 1$, and $x_{i} = |x_{i - 1} - x_{i - 2}|$ for $i \geq 3$. | [
"Solution:\nAnswer 2998.\nThe sequence is completely determined by its first two elements. If the largest element of the sequence is $n$, then it must occur as one of the first two elements. Because $x_{3}$ and $x_{4}$ are both smaller than the largest of the first two elements and hence all subsequent elements are... | Soviet Union | 1st ASU | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2998 | |
04iw | Determine all triples $(p, m, n)$ of positive integers such that $p$ is prime and
$$
2^m p^2 + 1 = n^5
$$ | [
"If we write down the given equation in the form $2^m p^2 = n^5 - 1$ and factorise the right-hand side, we get\n$$\n2^m p^2 = (n-1)(n^4 + n^3 + n^2 + n + 1).\n$$\nFactor $n^4 + n^3 + n^2 + n + 1$ is odd, so $n-1$ is divisible by $2^m$.\nWe immediately see that $p$ is odd.\nOn the other hand, since $n$ is positive, ... | Croatia | Croatia Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (11, 1, 3) | |
0kof | Problem:
A positive integer $n$ is loose if it has six positive divisors and satisfies the property that any two positive divisors $a < b$ of $n$ satisfy $b \geq 2a$. Compute the sum of all loose positive integers less than $100$. | [
"Solution:\nNote that the condition in the problem implies that for any divisor $d$ of $n$, if $d$ is odd then all other divisors of $n$ cannot lie in the interval $\\left[\\left\\lceil\\frac{d}{2}\\right\\rceil, 2d-1\\right]$. If $d$ is even, then all other divisors cannot lie in the interval $\\left[\\frac{d}{2}+... | United States | HMMT February | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 512 | |
02ho | Problem:
Para fabricar 9 discos de papelão circulares para o Carnaval usam-se folhas quadradas de $10~\mathrm{cm}$ de lado como indicado na figura. Qual a área do papel não aproveitado?
(A) $25~\mathrm{cm}^2$
(B) $22,5~\mathrm{cm}^2$
(C) $21,5~\mathrm{cm}^2$
(D) $21~\mathrm{cm}^2$
(E) $22~\m... | [
"Solution:\n\nLembre que a área de um círculo é $\\pi r^{2}$, onde $r$ é o raio do círculo. Se $r$ é o raio dos círculos da figura, então a área pedida é:\n\n\\[\n\\underbrace{\\text{área do quadrado}}_{10 \\times 10 = 100} - \\underbrace{\\text{área dos 9 círculos}}_{9 \\times \\pi r^{2}} = 100 - 9 \\times \\pi \\... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | MCQ | C | |
01rl | Given two hyperbolae $H_1$ and $H_2$ with the equations $y = 1/x$ and $y = -1/x$, respectively. A straight line meets $H_1$ at points $A$ and $B$, and meets $H_2$ at points $C$ and $D$. The lines tangent to $H_1$ at points $A$ and $B$ intersect at point $M$, and the lines tangent to $H_2$ at points $C$ and $D$ intersec... | [
"Without loss of generality we may assume that the positions of all hyperbolae, lines, and points look like in the figure (otherwise we can rotate the plane by the angle which is a multiple of $90^{\\circ}$, and rename the points).\n\n\n\nLet $A(a; 1/a)$, $B(b; 1/b)$, $C(c; -1/c)$, $D(d, -1... | Belarus | FINAL ROUND | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof only | null | |
039d | Given a point $P$ on the side $AB$ of a triangle $ABC$, consider all pairs of points $(X, Y)$, $X \in BC$, $Y \in AC$ such that $PX \parallel BC$ and $PY \parallel AC$. Prove that the midpoints of the segments $XY$ lie on a straight line. | [] | Bulgaria | First selection test for IMO 2007, Vietnam | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0br9 | Let $n$ be an odd positive integer, and let $x_1, x_2, \dots, x_n$ be non-negative real numbers. Show that $\min_{k=1,\dots,n} (x_k^2 + x_{k+1}^2) \le \max_{k=1,\dots,n} (2x_k x_{k+1})$, where $x_{n+1} = x_1$. | [
"In what follows, indices are reduced modulo $n$. Consider the $n$ differences $x_{k+1} - x_k$, $k = 1, \\dots, n$. Since $n$ is odd, there exists an index $j$ such that $(x_{j+1} - x_j)(x_{j+2} - x_{j+1}) \\ge 0$. Without loss of generality, we may and will assume both factors non-negative, so $x_j \\le x_{j+1} \\... | Romania | 2016 European Girls' Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
06ha | A polygon is *monochromatic* if all its vertices are coloured by a same colour. Suppose now every point of the plane is coloured red or blue. Show that there exists either a monochromatic equilateral triangle of side length $2$, or a monochromatic equilateral triangle of side length $\sqrt{3}$, or a monochromatic rhomb... | [
"First we show that there exists either a monochromatic equilateral triangle of length $1$, or a monochromatic equilateral triangle of length $\\sqrt{3}$. Indeed, if there is no monochromatic equilateral triangle of length $1$, then we can find two points $A$ and $B$ such that $AB = 1$ and they are in different col... | Hong Kong | CHKMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0b9i | Call a row of a matrix in $M_n(\mathbb{C})$ *permutable* if, for any permutation of its entries, the value of the determinant does not change. Prove that any matrix that has two *permutable* rows is singular. | [
"Consider $A \\in M_n(\\mathbb{C})$ such that row $l$ is *permutable*. Denote by $\\Gamma_{li}$ the algebraic complement of $a_{li}$, $i = 1, \\dots, n$. Suppose $i, j, k, p \\in \\{1, 2, \\dots, n\\}$ are such that $a_{li} \\neq a_{lj}$ and $\\Gamma_{lk} \\neq \\Gamma_{lp}$. Consider matrices $B$ and $C$ obtained ... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Linear Algebra > Determinants",
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
00uh | For an integer $n \ge 2$, the tuple $(1, 2, \dots, n)$ is written on a blackboard. On each turn, one can choose two numbers from the tuple such that their sum is a perfect square and swap them to obtain a new tuple. Find all integers $n \ge 2$ for which all permutations of $\{1, 2, \dots, n\}$ can appear on the blackbo... | [
"Answer: All integers $n \\ge 14$.\n\nWe first note that we say the numbers $a$ and $b$ can be ultimately swapped if, after a number of moves, one can obtain the tuple in which only $a$ and $b$ are swapped. We now prove a result.\n\n**Claim**. If integers $a, b, c \\in \\{1, 2, \\dots, n\\}$ are such that the numbe... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | all integers n ≥ 14 | |
0fqc | Problem:
Sea $AD$ la mediana de un triángulo $ABC$ tal que $\angle ADB = 45^{\circ}$ y $\angle ACB = 30^{\circ}$. Determinar el valor de $\angle BAD$. | [
"Solution:\n\nEn $\\triangle ABC$, sea $E$ el pie de la perpendicular trazada desde $B$. Entonces el triángulo $EBC$ es rectángulo en $E$, el punto $D$ es - por hipótesis - el punto medio de su hipotenusa $BC$ y, por tanto, el circuncentro de dicho triángulo en cuya circunferencia circunscrita el ángulo $BDE$ es el... | Spain | null | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Eu... | null | proof and answer | 30° | |
00a0 | Points $D$ and $E$ divide side $AB$ of equilateral triangle $ABC$ into three equal parts; $D$ is between $A$ and $E$. Point $F$ on side $BC$ is such that $CF = AD$. Find the sum of the angles
$$
C\hat{D}F + C\hat{E}F.
$$ | [
"The conditions give $BF = BD$ ($= \\frac{2}{3}AB$), also $D\\hat{B}F = 60^\\circ$, hence triangle $DBF$ is equilateral. Then\n$$\nDF \\parallel AC \\text{ as } B\\hat{D}F = B\\hat{A}C = 60^\\circ.\n$$\nHence $C\\hat{D}F = A\\hat{C}D$.\n\nOn the other hand $A\\hat{C}D = B\\hat{C}E$ by the symmetry of the figure (or... | Argentina | Argentine National Olympiad 2015 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 30° | |
0j2n | Problem:
Let $S$ be a convex set in the plane with a finite area $a$. Prove that either $a=0$ or $S$ is bounded.
Note: a set is bounded if it is contained in a circle of finite radius.
Note: a set is convex if, whenever two points $A$ and $B$ are in the set, the line segment between them is also in the set. | [
"Solution:\nIf all points in $S$ lie on a straight line, then $a=0$.\nOtherwise we may pick three points $A$, $B$, and $C$ that are not collinear. Let $\\omega$ be the incircle of $\\triangle ABC$, with $I$ its center and $r$ its radius. Since $S$ is convex, $S$ must contain $\\omega$.\n\n\... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance... | null | proof only | null | |
0k7j | Problem:
For a given positive integer $n$, we define $\varphi(n)$ to be the number of positive integers less than or equal to $n$ which share no common prime factors with $n$. Find all positive integers $n$ for which
$$
\varphi(2019 n)=\varphi\left(n^{2}\right)
$$ | [
"Solution:\n\nLet $p_{1}, p_{2}, \\ldots, p_{k}$ be the prime divisors of $n$. Then it is known that $\\varphi(n)=n \\cdot \\frac{p_{1}-1}{p_{1}} \\ldots \\frac{p_{k}-1}{p_{k}}$. As $n^{2}$ and $n$ have the same set of prime divisors, it also holds that $\\varphi\\left(n^{2}\\right)=n^{2} \\cdot \\frac{p_{1}-1}{p_{... | United States | HMMT November 2019 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof and answer | 1346, 2016, 2019 | |
0lc0 | Let $ABC$ be a scalene triangle. The incircle $(I)$ of triangle $ABC$ touches $BC$, $CA$ and $AB$ at $D$, $E$ and $F$ respectively. The line passing through $E$ and perpendicular to $BI$ cuts $(I)$ again at $K$ and the line passing through $F$ and perpendicular to $CI$ cuts $(I)$ again at $L$. The point $J$ is the midp... | [
"a) Since $D, E$ are the tangency points of $(I)$ with $BC$, $CA$ respectively, we have $DE \\perp CI$. Moreover, $FL \\perp CI$ so $DE \\parallel FL$. Similarly, we have $DF \\parallel EK$.\n\nClearly, we have $DK = DL = EF$ or $D$ lies on the perpendicular bisector of $KL$. Furthermore, $I$ also lies on the perpe... | Vietnam | VMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
07vn | A set $\mathcal{S}$ consists of $n \ge 1$ positive real numbers, such that for $x \neq y \in \mathcal{S}$:
$$
|y - x| \geq \sqrt{x + y}.
$$
Show that there exists $z \in \mathcal{S}$ such that $z > \frac{1}{2}n(n - 1)$. | [
"**Solution 1.** Proof by induction on $n$. The condition holds trivially if $n = 1$, so let us assume it holds for some $n \\ge 1$ and prove it for $n + 1$. Write the elements of the set $S$ of $n + 1$ elements as\n$$\ns_1 < s_2 < \\dots < s_{n-1} < s_n < s_{n+1}.\n$$\nFrom the condition in the question, and the i... | Ireland | IRL_ABooklet_2023 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
09re | Problem:
Zij $P$ het snijpunt van de diagonalen van een convexe vierhoek $ABCD$. Laat $X$, $Y$ en $Z$ punten op het inwendige van respectievelijk $AB$, $BC$ en $CD$ zijn zodat
$$
\frac{|AX|}{|XB|} = \frac{|BY|}{|YC|} = \frac{|CZ|}{|ZD|} = 2
$$
Veronderstel bovendien dat $XY$ raakt aan de omgeschreven cirkel van $\tria... | [
"Solution:\n\nVanwege de raaklijnomtrekshoekstelling is $\\angle CZY = \\angle BYX$ en $\\angle BXY = \\angle CYZ$. Hieruit volgt ten eerste dat\n$$\n\\angle XYZ = 180^{\\circ} - \\angle BYX - \\angle CYZ = 180^{\\circ} - \\angle BYX - \\angle BXY = \\angle XBY = \\angle ABC.\n$$\nTen tweede zien we dat $\\triangle... | Netherlands | MO-selectietoets | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof only | null | |
0bno | Let $m$ and $n$ be integers such that $m \ge 2$ and $n \ge 3$. Prove that there exists $m$ distinct positive integers $a_1, a_2, a_3, \dots, a_m$, all divisible by $n-1$, such that
$$
\frac{1}{n} = \frac{1}{a_1} - \frac{1}{a_2} + \frac{1}{a_3} - \dots + (-1)^{m-1} \frac{1}{a_m}.
$$ | [
"Let $(a_n)_{n \\ge 1}$ be a geometric sequence with ratio $q = n - 1$. Then\n$$\n\\frac{1}{a_1} - \\frac{1}{a_2} + \\dots + (-1)^{p-1} \\frac{1}{a_p} = \\frac{1/a_1 + 1/q \\cdot (-1)^{p-1} 1/a_p}{1 + 1/q}, \\forall p \\in \\mathbb{N}^*,\n$$\ni.e.\n$$\n\\frac{1}{a_1} - \\frac{1}{a_2} + \\dots + (-1)^{p-1} \\frac{1}... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0i0v | Problem:
An equilateral triangle in the coordinate plane has vertices $(a, b)$, $(c, d)$, $(e, f)$. Prove that $a, b, c, d, e, f$ cannot all be integers. | [
"Solution:\n\nSuppose otherwise. If the triangle has side length $s$, then $s^{2} = (a-c)^{2} + (b-d)^{2}$ is a positive integer, so we may choose the triangle for which $s^{2}$ is as small as possible; this makes $s$ as small as possible. We may further assume that $e = f = 0$, since otherwise this assumption can ... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
07if | *Ali* has $100$ cards with numbers $1, 2, \ldots, 100$. Ali and *Amin* play a game together. In each step, Ali firstly chooses a card from the remaining cards and Amin decides to pick that card for himself or throw it away. In the case that he picks the card, he can't pick the next card chosen by Amin, and he has to th... | [
"First, note that Ali can adopt the following strategy: He shows the cards in order until Amin picks a card. In the next step, Ali shows the largest card that has not been shown yet, and in the next step, he again starts with the smallest card that has not been shown yet and continues this process. If Amin chooses ... | Iran | 40th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 1716 | |
05rk | Problem:
Soient $m$, $n$, $k$ trois entiers positifs tels que $m^{2} + n = k^{2} + k$. Montrer que $m \leqslant n$. | [
"Solution:\n\nOn écrit $(2k+1)^{2} = 4(k^{2} + k) + 1 = 4(m^{2} + n) + 1 = (2m)^{2} + 4n + 1$.\n\nOr si $n < m$, on peut écrire\n$$\n(2m)^{2} < (2m)^{2} + 4n + 1 < 4m^{2} + 4m + 1 = (2m + 1)^{2}\n$$\nce qui est impossible puisque $(2k + 1)^{2} = (2m)^{2} + 4n + 1$ est un carré et ne peut donc pas être entre deux ca... | France | Préparation Olympique Française de Mathématiques - ENVOI 4 : POT-POURRI | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0b2b | Problem:
Determine the number of ordered quadruples $(a, b, c, d)$ of positive integers such that $a b c d = 216$. | [
"Solution:\n\nSince $216 = 2^{3} 3^{3}$, any positive divisor of $216$ must be of the form $2^{x} 3^{y}$ for some integers $x$ and $y$ with $0 \\leq x, y \\leq 3$. Thus, we set $a = 2^{x_{1}} 3^{y_{1}}$, $b = 2^{x_{2}} 3^{y_{2}}$, $c = 2^{x_{3}} 3^{y_{3}}$ and $d = 2^{x_{4}} 3^{y_{4}}$, where $0 \\leq x_{i}, y_{i} ... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 400 | |
0f6u | Problem:
An $n \times n \times n$ cube is divided into $n^3$ unit cubes. Show that we can assign a different integer to each unit cube so that the sum of each of the $3n^2$ rows parallel to an edge is zero. | [] | Soviet Union | 20th ASU | [
"Algebra > Linear Algebra > Linear transformations",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
0h77 | There are $22$ cards, where the numbers $1, 2, \ldots, 22$ are written. Using these cards one formed $11$ fractions. What is the greatest possible number of integer numbers among the fractions? | [
"The numbers $13$, $17$, $19$ may form an integer only if they stand at the numerator position and $1$ stands in the denominator position. Hence, at least one fraction cannot be integer. However, $10$ numbers may occur to be integer:\n$$\n\\frac{22}{11}, \\frac{14}{7}, \\frac{15}{5}, \\frac{21}{3}, \\frac{20}{10}, ... | Ukraine | UkraineMO | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 10 | |
0214 | Problem:
A subset $A$ of the natural numbers $\mathbb{N}=\{0,1,2, \ldots\}$ is called good if every integer $n>0$ has at most one prime divisor $p$ such that $n-p \in A$.
a. Show that the set $S=\{0,1,4,9, \ldots\}$ of perfect squares is good.
b. Find an infinite good set disjoint from $S$.
(Two sets are disjoint if... | [
"Solution:\n\na.\nSuppose to the contrary that $S$ is not good, so there exists $n \\in \\mathbb{N}$ with two different prime factors $p \\neq q$ such that $n-p, n-q$ are perfect squares. In particular $n$ is not prime. Write $n-p=m^{2}$, for some $m \\in \\mathbb{N}$. As $p \\mid n$, it follows that $p \\mid m$ an... | Benelux Mathematical Olympiad | Benelux Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | An example is the infinite set of odd primes P = {3, 5, 7, 11, ...}, which is disjoint from the set of squares and is good. | |
04ey | A quadrilateral with vertices $0$, $z$, $\frac{1}{z}$ and $z + \frac{1}{z}$ in the complex plane and area $\frac{35}{37}$ is given. Determine the smallest possible value of the expression $\left|z + \frac{1}{z}\right|^2$. | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Quadrilaterals",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 50/37 | |
07ej | a) Complex numbers $x$ and $y$ are given on the perimeter of the unit circle such that
$$
\frac{\pi}{3} \le \arg(x) - \arg(y) \le \frac{5\pi}{3}.
$$
b) Complex numbers $x$ and $y$ are given such that
$$
\frac{\pi}{3} \leq \arg(x) - \arg(y) \leq \frac{2\pi}{3}.
$$
For each $z \in \mathbb{C}$ show that
$$
|z| + |z - x| ... | [
"a) Let $O$ be the origin point of the complex plane. Consider numbers $x, y$ as points on this plane. The given inequality implies\n$$\n\\angle yOx \\leq \\frac{\\pi}{3}.\n$$\nBut since $Oxy$ is an isosceles triangle with $|Ox| = |Oy| = 1$, this means $|xy| = |x - y| \\geq 1$. Therefore for any complex number $z$ ... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Napoleon and Fermat points",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometr... | English | proof only | null | |
060d | Problem:
On dispose de cinq couleurs et d'une grille $99 \times 99$. On colorie certains carrés de la grille avec l'une des cinq couleurs de sorte que
- Chaque couleur apparaît le même nombre de fois dans la grille.
- Aucune ligne et aucune colonne ne contient des cases de couleur différente.
Quelle est le plus grand ... | [
"Solution:\n\nRéponse : $N = 5 \\times 19 \\times 20 = 1900$.\n\nComme chaque ligne ou colonne ne peut contenir qu'une couleur, on va réfléchir en partant des lignes. On dit qu'une ligne ou colonne est d'une couleur (disons rouge) quand la seule couleur que peuvent avoir des cases de cette ligne ou colonne est le r... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 1900 | |
0hyi | Problem:
Let $ABCD$ be a cyclic quadrilateral (a quadrilateral which can be inscribed in a circle). Let $E$ and $F$ be variable points on the sides $AB$ and $CD$, respectively, such that $AE / EB = CF / FD$. Let $P$ be the point on the segment $EF$ such that $PE / PF = AB / CD$. Prove that the ratio between the areas ... | [
"Solution:\n\nThere are two cases to consider.\n\nFirst, assume that the lines $AD$ and $BC$ are not parallel and meet at $S$. Since $ABCD$ is cyclic, $\\triangle ASB$ and $\\triangle CSD$ are similar. Then, $AE / AB = CF / CD$ and $AE / CF = AB / CD = AS / CS$, so that $\\triangle ASE$ and $\\triangle CSF$ are als... | United States | BAMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0f2o | Problem:
Given any tetrahedron, show that we can find two planes such that the areas of the projections of the tetrahedron onto the two planes have ratio at least $\sqrt{2}$. | [] | Soviet Union | ASU | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0kzi | Let $M$ be the greatest integer such that both $M + 1213$ and $M + 3773$ are perfect squares. What is the units digit of $M$?
(A) 1 (B) 2 (C) 3 (D) 6 (E) 8 | [
"Suppose $M + 1213 = j^2$ and $M + 3773 = k^2$ for nonnegative integers $j$ and $k$. Then\n$$ (k + j)(k - j) = k^2 - j^2 = 3773 - 1213 = 2560 = 5 \\cdot 2^9. $$\nBecause $k + j$ and $k - j$ have the same parity and their product is even, they must both be even, and it follows that one of them is $5 \\cdot 2^i$ and ... | United States | AMC 10 A | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | MCQ | E | |
0c61 | Let $ABC$ be a non-isosceles triangle, and let $I$ be its incenter. Let $\gamma_A$ be the circle through $I$, tangent to $AB$ and $AC$ and crossing the segment $AI$, let $\gamma_B$ be the circle through $I$ tangent to $BC$ and $BA$ and crossing the segment $BI$, and let $\gamma_C$ be the circle through $I$, tangent to ... | [
"Let the internal bisectrices of the angles $BAC$, $CBA$ and $ACB$ cross the circle $ABC$ again at $M_A$, $M_B$ and $M_C$, respectively, and let $\\gamma_A$, $\\gamma_B$ and $\\gamma_C$ be centred at $O_A$, $O_B$ and $O_C$, respectively; clearly, $O_A$, $O_B$ and $O_C$ lie on the segments $AI$, $BI$ and $CI$, respe... | Romania | SELECTION TESTS FOR THE 2019 BMO AND IMO | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles... | English | proof only | null | |
0jpn | Problem:
Let $P$ denote the set of all subsets of $\{1, \ldots, 23\}$. A subset $S \subseteq P$ is called good if whenever $A, B$ are sets in $S$, the set $(A \backslash B) \cup (B \backslash A)$ is also in $S$. (Here, $A \backslash B$ denotes the set of all elements in $A$ that are not in $B$, and $B \backslash A$ den... | [
"Solution:\nAnswer: $\\quad \\frac{18839183877670041942218307147122500601235}{47691684840486192422095701784512492731212} \\approx 0.3950203047068107$\nLet $n=23$, and $\\ell=\\lfloor n / 2\\rfloor=11$.\n\nWe use the well-known rephrasing of the symmetric difference $((A \\backslash B) \\cup (B \\backslash A))$ in t... | United States | HMMT February | [
"Algebra > Linear Algebra > Vectors",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 18839183877670041942218307147122500601235/47691684840486192422095701784512492731212 |
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