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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
01a5 | Livia has a deck of $n$ cards. She proceeds to discard the cards of the deck according to the following pattern: In each consecutive round, she will remove the cards numbered $1$, $2$, $4$, $6$, $8$, \ldots\ (i.e., the card numbered $1$ and all the cards with an even number), to produce a thinner deck. This procedure i... | [
"Denote the number of rounds required by $r(n)$. We claim that $r(n) = \\lceil \\log_2(n+1) \\rceil$. In other words, if\n$$\n2^m \\le n+1 < 2^{m+1},\n$$\nthen $r(n) = m$. This is evidently true when $n=0$. We proceed by induction. Consider an $n > 0$, for which $2^m \\le n+1 < 2^{m+1}$.\n\n* If $n$ is odd, then th... | Baltic Way | Baltic Way 2013 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | floor(log_2(n+1)) | |
06mn | Let $ABCD$ be a quadrilateral inscribed in a circle $\Omega$. Let the tangent to $\Omega$ at $D$ intersect the rays $BA$ and $BC$ at points $E$ and $F$, respectively. A point $T$ is chosen inside the triangle $ABC$ so that $TE \parallel CD$ and $TF \parallel AD$. Let $K \neq D$ be a point on the segment $DF$ such that ... | [
"3. (IMO Shortlist 2021 G4) See the official solution."
] | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry >... | null | proof only | null | |
0h8h | Determine all possible pairs of integers $a, b$ so that exactly one of them is even and so that there are non-integer $x, y$, such that both $x+y$ and $ax+by$ are integer? | [
"Since $ax+by = a(x+y) + (b-a)y$, the value of $(b-a)y$ is integer. If $|a-b|=1$, then $y$ has to be integer – contradiction.\n\nFor $|a-b|>1$ we can let $y = \\frac{1}{b-a}$ and $x = -y$. Then $x+y=0$, $(b-a)y$ is integer, hence $ax+by$ is also integer."
] | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | All integer pairs with exactly one even and |a−b|>1; pairs with |a−b|=1 do not work. | |
00xw | Problem:
Let $Q$ be a unit cube. We say a tetrahedron is "good" if all its edges are equal and all its vertices lie on the boundary of $Q$. Find all possible volumes of "good" tetrahedra. | [
"Solution:\n\nClearly, the volume of a regular tetrahedron contained in a sphere reaches its maximum value if and only if all four vertices of the tetrahedron lie on the surface of the sphere. Therefore, a \"good\" tetrahedron with maximum volume must have its vertices at the vertices of the cube (for a proof, insc... | Baltic Way | Baltic Way 1993 | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | (0, 1/3] | |
01ko | There is a heap of $360$ stones. Nick and Mary play the following game. They, in turn (Mary is the first), remove the stones from the heap. Per move it is allowed to remove exactly $1$ or exactly $m$ or exactly $n$ stones. The player wins if he/she removes the last stone. Before the start Nick fixes the value of $n$ ($... | [
"Nick wins if he sets $n = 2$. We consider all possibilities for $m$. Let $m = 3$. If Mary removes $k$ stones then Nick removes $4 - k$. In this case exactly $4$ stones are removed from the heap after each pair of moves (Mary - Nick). Since $360 : 4$, Nick wins.\n\nNow we will solve the problem moving backward. We ... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Nick can guarantee a win by choosing n = 2. | |
01z6 | Two players play the game on the $3 \times 2021$ board all cell of which are initially white. The players make their moves in turn, in one move a player paints black two not necessarily adjacent white cells located either in the same row or in the same column. The player who cannot make a move loses.
Which of the playe... | [
"Answer: the first player has a winning strategy.\nSince two cells are repainted on each move and the total number of cells on the board is odd, at the end of the game there will be an odd number of white cells. Obviously there will remain not more than one white cell left in each row, so there will remain one or t... | Belarus | Belarus2022 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | the first player | |
0hzs | Problem:
Squares $ABKL$, $BCMN$, $CAOP$ are drawn externally on the sides of a triangle $ABC$. The line segments $KL$, $MN$, $OP$, when extended, form a triangle $A'B'C'$. Find the area of $A'B'C'$ if $ABC$ is an equilateral triangle of side length $2$. | [
"Solution:\n\nTriangle $ABC$ has area $\\sqrt{3}$, and each of the three squares has area $4$. The three remaining regions are congruent, so just consider the one that includes vertex $B$. Triangle $KBN$ has two sides of length $2$ and an angle of $120^{\\circ}$ between them, so bisecting that angle we get two halv... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 12 + 13*sqrt(3) | |
0acq | Prove that if $n$ is an odd natural number, then $n^3 + 3n^2 - n - 3$ is divisible by 48. | [
"We have that\n$$\nn^3 + 3n^2 - n - 3 = n^2(n+3) - (n+3) = (n+3)(n^2 - 1) = (n-1)(n+1)(n+3).\n$$\nThe numbers $n-1$, $n+1$, $n+3$ are even and one from $n-1$, $n+1$ is divisible by 4. Therefore the given number is divisible by 16. If $n$ is divisible by 3, then $n+3$ is divisible by 3. If $n$ is not divisible by 3,... | North Macedonia | Macedonian Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0k4j | Problem:
Let $ABC$ be a triangle with incenter $I$ and circumcenter $O$ for which $BC < AB < AC$. Let $D$ and $E$ be points in the interiors of sides $AB$ and $AC$, respectively, of triangle $ABC$, such that $DB = BC = CE$. Prove that $\overline{DE} \perp \overline{IO}$. | [
"Solution:\n\nIt is enough to show that $DI^{2} - DO^{2} = EI^{2} - EO^{2}$. But if we let $R$ denote the circumradius, then by power of a point we have $R^{2} - DO^{2} = AD \\cdot DB$, and $R^{2} - EO^{2} = AE \\cdot EC$. Thus it suffices to prove\n$$\nDI^{2} + AD \\cdot DB = EI^{2} + AE \\cdot EC \\Longleftrighta... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasin... | null | proof only | null | |
0csa | Даны 111 различных натуральных чисел, не превосходящих 500. Могло ли оказаться, что для каждого из этих чисел его последняя цифра совпадает с последней цифрой суммы всех остальных чисел? | [
"Ответ. Не могло.\n\nПусть такое могло случиться. Обозначим данные числа $a_1, a_2, \\dots, a_{111}$ и обозначим их сумму через $S$. По условию, для каждого номера $k$ числа $a_k$ и $S - a_k$ оканчиваются одной и той же цифрой. Отсюда следует, что разность этих чисел, равная $S - 2a_k$, делится на $10$. Значит, при... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | No, it is not possible. | |
0h4k | Find the exact value of the expression below:
$$
\frac{(6!+5!)(5!+4!)(4!+3!)(3!+2!)(2!+1!)}{(6!-5!)(5!-4!)(4!-3!)(3!-2!)(2!-1)!}
$$
if $n!$ denotes the product $1 \cdot 2 \cdot 3 \cdots n$ for every natural number $n$. | [
"Using the obvious equality $(n+1)! = (n+1) \\cdot n!$ we can make the following transformations:\n$$\n\\frac{(6!+5!)(5!+4!)(4!+3!)(3!+2!)(2!+1!)}{(6!-5!)(5!-4!)(4!-3!)(3!-2!)(2!-1)!} = \\frac{5!(6+1) \\cdot 4! (5+1) \\cdot 3! (4+1) \\cdot 2! (3+1) \\cdot 1! (2+1)!}{5!(6-1) \\cdot 4! (5-1) \\cdot 3! (4-1) \\cdot 2!... | Ukraine | Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 21 | |
0h6h | Determine all prime numbers $p < q < r$ so that $A = (r-p)(r-q)(q-p)+1$ and $B = 3p+5q$ equal the same prime number. | [
"Let $p, q, r$ be the prime numbers that satisfy the conditions of the problem. If $p > 2$ then all $p, q, r$ are odd thus the number $A = 3p + 5q$ is even and the number $B = (r-p)(r-q)(q-p)+1$ is odd, which contradicts the conditions of the problem. Therefore, $p=2$, thus:\n\n$$\n(r-2)(r-q)(q-2)+1=6+5q.\n$$\n\n$p... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | p=2, q=5, r=7 | |
06kx | Let $a$ be a real number. Suppose the function $f(x) = \frac{a}{x-1} + \frac{1}{x-2} + \frac{1}{x-6}$ defined in the interval $3 < x < 5$ attains its maximum at $x = 4$. Find the value of $a$. | [
"Solution:\n\nThe answer is $-\\frac{9}{2}$.\nThe condition means $\\frac{a}{x-1} + \\frac{1}{x-2} + \\frac{1}{x-6} \\le \\frac{a}{3}$ for $3 < x < 5$. Now, noting that $x-1 > 0$, $x-2 > 0$ and $x-6 < 0$, we have\n$$\n\\frac{a}{x-1} + \\frac{1}{x-2} + \\frac{1}{x-6} \\le \\frac{a}{3}\n$$\n$$\n\\Leftrightarrow 3a(x-... | Hong Kong | HKG TST | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | null | proof and answer | -9/2 | |
08r7 | A $3 \times 3$ grid is given. You shall color each square with one of the given colors. However, you may not color two squares sharing an edge with the same color. How many such colorings are there if you can use
(1) three colors (say red, blue and green)?
(2) four colors (say red, blue, green and yellow)?
It is not... | [
"We call the squares corners, edges or the center, according to their position. We first consider the case that the center is colored red.\n\n(1) The edges must be colored into blue or yellow. There are 2 ways to color a square on the corner if the two squares next to it have the same color, and there are 1 way if ... | Japan | The 4th Japanese Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | (1) 246; (2) 9612 | |
033z | Problem:
Let $I$ be the incenter of $\triangle ABC$ and $M$ be the midpoint of the side $AB$. Find the least possible value of $\Varangle CIM$ if $CI = MI$. | [
"Solution:\nWe may assume that $AC < BC$. Since $\\Varangle ACI$ and $\\Varangle AMI$ are acute, then $\\triangle ACI \\cong \\triangle AMI$.\nHence $AC = AM$ and $\\Varangle AIC = \\Varangle AIM$, i.e.\n$$\n\\Varangle CIM = 360^\\circ - 2 \\Varangle AIC = 180^\\circ - \\Varangle ABC.\n$$\nNote that $\\Varangle ABC... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimiz... | null | proof and answer | 150° | |
0bq3 | Prove that the ratio of the area of a rectangle and the area of a triangle whose vertices are on the sides of the rectangle and centroid coincides with the centroid of the rectangle is a constant. | [] | Romania | 67th NMO Shortlisted Problems | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null | |
082c | Problem:
Determinare tutti gli interi positivi di tre cifre che sono uguali a 34 volte la somma delle loro cifre. | [
"Solution:\n\nGli interi richiesti sono 102, 204, 306, 408.\nIndichiamo infatti con $a, b, c$, rispettivamente, la cifra delle centinaia, delle decine, delle unità di un intero di 3 cifre. La condizione data si traduce allora in $100 a+10 b+c=34(a+b+c)$, da cui, con semplici passaggi algebrici, si ricava che $11(2 ... | Italy | Cesenatico Gara Individuale | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 102, 204, 306, 408 | |
02re | Given a triangle *ABC*, its *A*-excircle is the point where the external bisectors of angles $\angle B$ and $\angle C$ meet. Let $I_A$, $I_B$ and $I_C$ be the *A*, *B*, *C*-excenters of scalene triangle *ABC*, respectively, and *X*, *Y* and *Z* be the midpoints of $I_B I_C$, $I_C I_A$ and $I_A I_B$, respectively. The i... | [
"We start with the following useful lemma:\n**Lemma.** Let $ABC$ be a triangle and $I_A$, $I_B$ and $I_C$ be its excenters. Then the midpoints of the sides of the triangle $I_A I_B I_C$ lie on the circumcircle of $ABC$.\n**Proof.** First notice that $AI_A$ and $I_B I_C$ are the internal and external bisectors of $\... | Brazil | Brazilian Math Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08uo | Suppose for a convex 2010-gon, any 3 diagonals do not share a common point except for vertices. Let us consider a closed zig-zag line segments (abbr. by Z.Z.L.S.) which goes through each of the vertices of the 2010-gon once and only once. Determine the maximum possible number of the self-intersection points for such a ... | [
"More generally than the assertion of the problem, we prove that if we replace the number $2010$ in the problem by any even integer $2m$ bigger than or equal to $6$, we get for any convex $2m$-gon satisfying the condition of the problem the number $2m^2 - 4m + 1$ is the answer for the assertion.\n\nFor any integer ... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 2016031 | |
02sz | Problem:
Na cidade de Autolândia, a numeração de placas de carros é feita através de números de três dígitos, portanto indo da placa 000 até a placa 999. Para diminuir a poluição, o prefeito Pietro decidiu implementar um rodízio de carros, estabelecendo os dias nos quais as pessoas podem usar seus carros. As regras do... | [
"Solution:\n\na) Como o número 729 é ímpar, o carro com a placa de número 729 pode circular às segundas-feiras. Como $7+2+9=18$, o carro também pode circular às terças-feiras, mas não às quintas-feiras. Como 729 é múltiplo de 3, este carro também pode circular às quartas-feiras. Como 729 não possui dígitos iguais, ... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | a) Monday, Tuesday, Wednesday.
b) 363.
c) 255.
d) Any plate either meets the Tuesday sum condition or, if not, its digit sum is small enough to meet the Thursday condition, so everyone can drive at least Tuesday or Thursday. | |
07x5 | The *runcible* positive integers are defined recursively as follows:
* $1$ and $2$ are runcible
* If $a$ and $b$ are runcible (where $a$ and $b$ are not necessarily distinct) then $2a + 3b$ is runcible.
Is $2024$ runcible? | [
"Since $1$ and $2$ are runcible, then $8 = 2 \\cdot 1 + 3 \\cdot 2$ is runcible. If $a$ is runcible, then $5a = 2a + 3a$ is also runcible. By induction, $5^k a$ is runcible for all natural numbers $k$. Hence $1000 = 5^3 \\cdot 8$ is runcible and $2024 = 2 \\cdot 1000 + 3 \\cdot 8$ is also runcible.",
"We prove th... | Ireland | IRL_ABooklet_2024 | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | Yes | |
0eah | Eight boxes numbered $1$ to $8$ are lined up next to an empty bag. Each box contains one token. Miha has a lot of extra tokens and he plays a game where the following two actions are allowed:
* removing $1$ token from box $i$ ($i < 8$) and adding $2$ tokens into box $(i + 1)$,
* removing $1$ token from box $i$ ($i < 8$... | [
"Let $A_i$ and $B_i$ denote the first and second move taken on the boxes marked $i$ and $i+1$. Let $i \\le 6$. If Miha chooses to play $B_i$, then the number of tokens in the boxes marked $i$ and $i+1$ is reduced by $1$, the number of tokens in the box $i+2$ remains the same, and the number of tokens in the bag inc... | Slovenia | National Math Olympiad in Slovenia | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 85 | |
0c0a | Let $A = \{ a = q + \frac{1}{q} \mid q \in \mathbb{Q}^*, q > 0 \}$,
$$A + A = \{a + b \mid a, b \in A\}, \quad A \cdot A = \{a \cdot b \mid a, b \in A\}.$$
Prove that:
i) $A + A \neq A \cdot A$;
ii) $(A + A) \cap \mathbb{N} = (A \cdot A) \cap \mathbb{N}$. | [
"i) Let $a = 1 + \\frac{1}{2} = 2 \\in A$, $b = 2 + \\frac{1}{2} \\in A$. Then\n$$\na + b = 2 + 2 + \\frac{1}{2} = \\frac{9}{2} \\in A + A.\n$$\nWe show that $\\frac{9}{2} \\notin A \\cdot A$. Assume that\n$$\n\\frac{9}{2} = \\left(\\frac{x}{y} + \\frac{y}{x}\\right) \\left(\\frac{z}{t} + \\frac{t}{z}\\right) = \\f... | Romania | 69th NMO Selection Tests for JBMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
0jmw | Let $r$ be a rational number in the interval $[-1, 1]$, and let $\theta = \cos^{-1} r$. Call a subset $S$ of the plane good if $S$ is unchanged upon rotation by $\theta$ around any point of $S$ (in both clockwise and counterclockwise directions). Determine all values of $r$ satisfying the following property: The midpoi... | [
"We claim the answer is $r = 1 - \\frac{1}{4k}$ for all positive integers $k$.\n\nSuppose $A$ and $B$ are points in $S$. Place $A$ and $B$ on the complex plane such that $A = 0$ and $B = 1$, and let $\\omega = \\cos\\theta + i\\sin\\theta = r + i\\sqrt{1-r^2}$. We claim that any good set containing $A$ and $B$ cont... | United States | IMO Team Selection Team Selection Test | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | r = 1 - 1/(4k) for all positive integers k | |
0hzu | Problem:
Start with an angle of $60^{\circ}$ and bisect it, then bisect the lower $30^{\circ}$ angle, then the upper $15^{\circ}$ angle, and so on, always alternating between the upper and lower of the previous two angles constructed. This process approaches a limiting line that divides the original $60^{\circ}$ angle... | [
"Solution:\n\nThe fraction of the original angle is $\\frac{1}{2} - \\frac{1}{4} + \\frac{1}{8} - \\cdots$. This is just a geometric series with first term $\\frac{1}{2}$ and ratio $-\\frac{1}{2}$, so the sum is $\\frac{1}{3}$. Therefore the smaller angle is $20^{\\circ}$."
] | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 20° | |
062f | Problem:
Gegeben ist ein (allgemeines) Trapez $ABCD$, dessen Diagonalen sich im Punkt $P$ schneiden. Ein Punkt $Q$ liegt so zwischen den Parallelen $AB$ und $DC$, dass $\varangle AQB = \varangle CQD$ gilt und die Gerade $BC$ zwischen $P$ und $Q$ verläuft.
Man beweise, dass $\varangle DQP = \varangle BAQ$ gilt. | [
"Solution:\n\nWegen $AB \\parallel CD$ bildet die zentrische Streckung $\\sigma$ am Schnittpunkt $P$ der Diagonalen mit Streckfaktor $k = -\\frac{|CD|}{|AB|} = -\\frac{|CP|}{|AP|} = -\\frac{|DP|}{|BP|}$ den Punkt $C$ auf $A$ und $D$ auf $B$ ab. Es sei $Q'$ der Bildpunkt von $Q$ unter $\\boldsymbol{\\sigma}$. Weil\n... | Germany | 2. IMO-Auswahlklausur | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0bgr | Problem:
Fie $n \geq 2$ un număr natural, $(K,+, \cdot)$ un corp comutativ cu proprietatea că $\underbrace{1+\cdots+1}_{m \text{ ori}} \neq 0,\ m=2, \ldots, n$, $f \in K[X]$ un polinom de grad $n$ şi $G$ un subgrup al grupului aditiv $(K,+)$, $G \neq K$. Să se arate că există $a \in K$, astfel încât $f(a) \notin G$. | [
"Solution:\n\nFie $g \\in K[X]$ un polinom de grad $m \\in \\{2, \\ldots, n\\}$. Polinomul $h(X) = g(X+1) - g(X)$ este de grad $m-1$ și, dacă $\\operatorname{Im} g \\subseteq G$, atunci $\\operatorname{Im} h \\subseteq G$.\n\nPresupunem că $\\operatorname{Im} f \\subseteq G$. Considerăm polinoamele $f_0, f_1, \\ldo... | Romania | Olimpiada Naţională de Matematică Etapa Naţională | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Abstract Algebra > Field Theory",
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
0er9 | What is the minimum number of integers which must be removed from the first 50 positive even integers so that the sum of the remaining integers is $2016$? | [
"The first $50$ even positive integers are $2, 4, 6, \\ldots, 96, 98, 100$, and their sum is $50 \\times \\frac{1}{2}(2 + 100) = 50 \\times 51 = 2550$. To reduce the sum to $2016$ we must subtract $534$. The average of the numbers removed must be less than $100$, so we need to remove at least six. To use as few num... | South Africa | South African Mathematics Olympiad Second Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 6 | |
0393 | Let $ABC$ be a right triangle with $\angle ACB = 90^\circ$, $AC = 1$ and $BC = 2$. Given a point $A_1 \in BC$ such that $A_1C \neq \frac{1}{3}$ we construct a sequence of points $A_n \in BC$, $n \ge 2$ in the following way. Let $B_1$ be the intersection point of $AC$ and the line through $A_1$ and parallel to $AB$, and... | [
"a) Set $A_nC = x_n$. Then it follows from $\\triangle A_nB_nC \\sim \\triangle BAC$ that $B_nC = \\frac{x_n}{2}$. Since $\\triangle AB_nC_n \\sim \\triangle ABC$, we find that $AC_n = \\frac{2-x_n}{2\\sqrt{5}}$. Hence $A_{n+1}C = x_{n+1} = \\frac{2-x_n}{5}$ and therefore $\\frac{3x_{n+1}-1}{3x_n-1} = -\\frac{1}{5}... | Bulgaria | Winter Mathematical Competition | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | a) -1/5; b) 5/36 | |
0g8x | 設數列 $\{a_n\}$ 滿足 $a_{n+1} = a_n^3 + 103,\ n = 1, 2, \dots$。
試證: 至多存在一個正整數 $n$, 使得 $a_n$ 為完全平方數。 | [
"取 mod 4，可發現\n$$\n(a_n, a_{n+1}) = (0, 3) \\pmod 4\n$$\n$$\n(a_n, a_{n+1}) = (1, 0) \\pmod 4\n$$\n$$\n(a_n, a_{n+1}) = (2, 3) \\pmod 4\n$$\n$$\n(a_n, a_{n+1}) = (3, 2) \\pmod 4.\n$$\n注意到完全平方數對 4 取模只能為 0 或 1. 若 $a_n = 0 \\pmod 4$, 則對於所有 $m > n$, 必有 $a_m = 2 \\pmod 4$ 或 $a_m = 3 \\pmod 4$, 故後續項不可能為完全平方數。而若 $a_n = 1 \... | Taiwan | 2015 Math Olympiad Second Stage Training Camp | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
091f | Problem:
Let $n$ be a positive integer. A square $ABCD$ is partitioned into $n^{2}$ unit squares. Each of them is divided into two triangles by the diagonal parallel to $BD$. Some of the vertices of the unit squares are colored red in such a way that each of these $2 n^{2}$ triangles contains at least one red vertex. F... | [
"Solution:\nThe least number of red vertices is\n$$\n\\left\\lfloor\\frac{(n+1)^{2}}{3}\\right\\rfloor\n$$\nFirst, we define a colouring and count the number of red vertices. In what follows it will be shown that the number of red vertices, obtained by this colouring, is indeed minimal.\nIt is convenient to replace... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | floor(((n+1)^2)/3) | |
0hmv | Problem:
In a certain two-player game, you start with a rectangular $m \times n$ grid of squares. On each turn, a player either makes a horizontal cut and takes away the portion of the rectangle above the cut, or makes a vertical cut and takes away the portion to the right. Whichever player takes the last square (in t... | [
"Solution:\n\nThe first player wins whenever the grid is not a square (i.e. $m \\neq n$). The strategy is to always leave the other player with a square grid (where $m = n$), since then they will eventually be left with a $1 \\times 1$ square and will lose. This works since starting from a square, it is impossible ... | United States | Berkeley Math Circle: Monthly Contest 8 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | First player wins if and only if m ≠ n; second player wins when m = n. | |
0apm | Problem:
If $\frac{1}{2} \sin^{2} x + C = -\frac{1}{4} \cos 2x$ is an identity, what is the value of $C$? | [
"Solution:\n$-\\frac{1}{4}$\n\nSince the equation is an identity, it is true for all $x$ in the domain (which is $\\mathbb{R}$) of the equation. To find $C$, we only set a particular value of $x$. For convenience, when we let $x = 0$, we have $C = -\\frac{1}{4}$."
] | Philippines | Tenth Philippine Mathematical Olympiad | [
"Precalculus > Trigonometric functions"
] | null | final answer only | -1/4 | |
0ihj | A communications network consisting of some terminals is called a 3-connector if among any three terminals, some two of them can directly communicate with each other. A communications network contains a windmill with $n$ blades if there exist $n$ pairs of terminals $\{x_1, y_1\}, \dots, \{x_n, y_n\}$ such that each $x_... | [
"The answer is\n$$\nf(n) = \\begin{cases} 6 & \\text{if } n = 1; \\\\ 4n + 1 & \\text{if } n \\ge 2. \\end{cases}\n$$\nWe will use *connected* as a synonym for directly communicating, call a set of $k$ terminals for which each of the $\\binom{k}{2}$ pairs of terminals is connected *complete* and call a set of $2k$ ... | United States | Team Selection Test | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | f(1) = 6; for n ≥ 2, f(n) = 4n + 1 | |
0kzh | The roots of $x^3 + 2x^2 - x + 3$ are $p$, $q$, and $r$. What is the value of $(p^2 + 4)(q^2 + 4)(r^2 + 4)$?
(A) 64 (B) 75 (C) 100 (D) 125 (E) 144 | [
"By Vieta's Formulas,\n$$\n\\begin{aligned}\np + q + r &= -2, \\\\\npq + pr + qr &= -1, \\text{ and} \\\\\npqr &= -3.\n\\end{aligned}\n$$\nThe expression to be evaluated is\n$(p^2 + 4)(q^2 + 4)(r^2 + 4) = p^2q^2r^2 + 4(p^2q^2 + p^2r^2 + q^2r^2) + 16(p^2 + q^2 + r^2) + 64.$\nThe first term is\n$$\np^2q^2r^2 = (pqr)^... | United States | AMC 12 A | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | MCQ | D | |
0fo1 | Let $\triangle ABC$ be an acute triangle and let $X$ be the foot of the height drawn from $A$ and $Y$ be the intersection of the perpendicular to $AC$ drawn from $X$. If the circumcircle of triangle $ABX$ meets $BY$ at point $Z$ (distinct of $B$) and the extension of $AZ$ meets $XY$ at point $P$, then prove that $BX \c... | [
"Since $\\triangle AXY$ and $\\triangle AXC$ are both right triangles in $Y$ and $X$ respectively, then $\\angle AXP = 90^\\circ - \\angle XAC = \\angle YCB$. On the other hand, since $AZXB$ is a cyclic quadrilateral, then $\\angle XAP = \\angle YBC$. Therefore,\n\n\n\n$\\triangle AXP \\sim... | Spain | BarcelonaTech Mathcontest | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Spanish | proof only | null | |
0g8e | 正整數 $x_1, x_2, \dots, x_n$ ($n \ge 4$) 依序排列在圓周上，使得任意 $x_i$ 的左右鄰居之數字和會是 $x_i$ 本身的倍數，也就是分數
$$
\frac{x_{i-1} + x_{i+1}}{x_i} = k_i.
$$
是一個整數，其中指定 $x_0 = x_n, x_{n+1} = x_1$。試證：所有倍數和 $k_1 + k_2 + \dots + k_n$ 滿足不等式
$$
2n \le k_1 + k_2 + \dots + k_n < 3n.
$$
Suppose $x_1, x_2, \dots, x_n$, with $n \ge 4$, are positive inte... | [
"解：左邊的不等式可輕易地由 A.M. $\\ge$ G.M. 證出，例如：\n$$\n\\begin{aligned} \\sum_{j=1}^{n} k_j &= \\sum_{j=1}^{n} \\left( \\frac{x_{j-1}}{x_j} + \\frac{x_j}{x_{j-1}} \\right) \\\\ &\\ge 2 \\sum_{j=1}^{n} \\sqrt{\\frac{x_{j-1}}{x_j} \\cdot \\frac{x_j}{x_{j-1}}} = 2n. \\end{aligned}\n$$\n\n以下我們處理右邊的不等式。其實 $n \\ge 4$ 的假設不那麼重要，我們將對所... | Taiwan | 二〇一四年國際數學奧林匹亞競賽第二階段選訓營 獨立研究(三) | [
"Number Theory > Divisibility / Factorization",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0a4p | Problem:
Show that the number $122^{n} - 102^{n} - 21^{n}$ is always one less than a multiple of 2020, for any positive integer $n$. | [
"Solution:\n\nLet $f(n) = 122^{n} - 102^{n} - 21^{n}$. We consider $f(n)$ in mod 101 and in mod 20 separately.\n\nConsider $f(n)$ mod 101.\n\n$$\n\\begin{align*}\nf(n) &= 122^{n} - 102^{n} - 21^{n} \\\\\n&\\equiv 21^{n} - 1^{n} - 21^{n} \\pmod{101} \\\\\n&= -1\n\\end{align*}\n$$\n\nConsider $f(n)$ mod 20.\n\n$$\n\\... | New Zealand | NZMO Round One | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0ikx | Problem:
In how many ways can we enter numbers from the set $\{1,2,3,4\}$ into a $4 \times 4$ array so that all of the following conditions hold?
(a) Each row contains all four numbers.
(b) Each column contains all four numbers.
(c) Each "quadrant" contains all four numbers. (The quadrants are the four corner $2 \t... | [
"Solution:\n\nCall a filled $4 \\times 4$ array satisfying the given conditions cool. There are $4!$ possibilities for the first row; WLOG, let it be $1234$. Since each quadrant has to contain all four numbers, we have exactly four possibilities for the second row, namely:\n(i) $3412$\n(ii) $3421$\n(iii) $4312$\n(i... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 288 | |
04mp | Find all complex numbers $z$ such that
$$ \mathrm{Re}(z) = 9 \quad \text{and} \quad \mathrm{Im}(z^2) = \mathrm{Im}(z^3). \qquad (\text{AIME 2007}) $$ | [] | Croatia | Croatia_2018 | [
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | 9, 9 + 15i, 9 - 15i | |
07tl | Suppose the real numbers $a$, $A$, $b$, $B$ satisfy the inequalities:
$$
|A - 3a| \le 1 - a, \quad |B - 3b| \le 1 - b,
$$
and $a$, $b$ are positive. Prove that $\left| \frac{AB}{3} - 3ab \right| \le 1 - ab$. | [
"Observe that $a$, $b \\in (0, 1]$. Let $u = A - 3a$, $v = B - 3b$. Then $|u| \\le 1 - a$, $|v| \\le 1 - b$, and $A = u + 3a$, $B = v + 3b$, so that $AB = uv + 3(ub + va) + 9ab$. Hence\n$$\n\\begin{align*}\n|AB - 9ab| &= |uv + 3(ub + va)| \\\\\n&\\le |u||v| + 3(b|u| + a|v|) \\\\\n&\\le (1-a)(1-b) + 3(a(1-b) + b(1-a... | Ireland | IRL_ABooklet | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
071x | Problem:
Let $x, y$ be positive reals such that $x + y = 2$. Prove that
$$
x^{3} y^{3} (x^{3} + y^{3}) \leq 2
$$ | [
"Solution:\nWe have from the AM-GM inequality, that\n$$\nx y \\leq \\left(\\frac{x + y}{2}\\right)^{2} = 1\n$$\nThus we obtain $0 < x y \\leq 1$. We write\n$$\n\\begin{aligned}\nx^{3} y^{3} (x^{3} + y^{3}) &= (x y)^{3} (x + y) (x^{2} - x y + y^{2}) \\\\\n&= 2 (x y)^{3} ((x + y)^{2} - 3 x y) \\\\\n&= 2 (x y)^{3} (4 ... | India | INMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0dhz | Find all real numbers $x, y, z$ such that $xy + z = -30$, $yz + x = 30$ and $zx + y = -18$. | [
"Adding the first two equations yields\n$$\n0 = xy + z + yz + x = (x + z)(y + 1).\n$$\nSo $y = -1$ or $z = -x$.\n\nIf $y = -1$ then the system reads $z = x - 30$ and $xz = -17$. It follows that $x^2 - 30x + 17 = 0$ and hence $x = 15 \\pm 4\\sqrt{13}$.\n\nIf $z = -x$, then the system becomes $xy - x = -30, y = x^2 -... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | English | proof and answer | (15 + 4√13, -1, 15 - 4√13), (15 - 4√13, -1, 15 + 4√13), (2, -14, -2), (3, -9, -3), (-5, 7, 5) | |
083p | Problem:
Sia $ABCD$ un rombo, ed $E$ un punto qualunque sulla sua diagonale $AC$. Sia $F$ il punto sul segmento $BC$ tale che $BF = DE$. Provare che
$$
(AB + BF) \cdot FC = AE \cdot EC.
$$ | [] | Italy | UNIONE MATEMATICA ITALIANA Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO TRIENNIO | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0ark | Problem:
Let $f: \mathbb{R} \longrightarrow \mathbb{R}$ be a function such that $x f(y)=y f(x)$ for all $x, y \in \mathbb{R}$. Find the intersection of the graphs of $y=f(x)$ and $y=x^{2}+1$ if $f(1)=-1$. | [
"Solution:\n(ans $\\phi=$ Null set.\nWe have that $\\frac{f(x)}{x}=\\frac{f(y)}{y}=c$, a constant $\\Rightarrow f(x)=c x \\Rightarrow f(x)=-x$ from the given condition. $y=f(x)=-x$ does not intersect the parabola $y=x^{2}+1$ because $x^{2}-x+1=0$ has no real solutions.)"
] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | empty set | |
00dt | Eight teams take part in a rugby tournament in which every team plays exactly one match against each of the other seven teams. In each match, if the teams draw against each other, both of them earn 1 point; otherwise, the winner earns 2 points and the loser earns no points.
At the end of the tournament, the final score... | [
"We represent the tournament as a table that in the cell $(i, j)$ contains the number of points earned by team $i$ in the match versus team $j$. Consider the following tournament, where for every $1 \\le i \\le 8$ team $i$ defeats team $j$ for every $j > i$.\n\n| Team | T1 | T2 | T3 | T4 | T5 | T6 | T7 | T8 | Total... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | final answer only | Label teams T1 through T8. For every pair with lower index versus higher index, the lower-indexed team wins, except that the match between T1 and T8 is a draw. This yields totals 13, 12, 10, 8, 6, 4, 2, 1, which are all distinct and satisfy that the top total equals the sum of the four smallest totals. | |
03oc | Find all positive integers $n$ such that $20n + 2$ can divide $2003n + 2002$. (posed by Wu Weichao) | [
"It is easy to see that $n$ is an even number. Let $n = 2m$, then from\n$40m + 2 \\mid 2003 \\times 2m + 2002$\nwe can get $20m + 1 \\mid 2003m + 1001$.\nBut $2003m + 1001 = 100(20m + 1) + 3m + 901$,\nso $20m + 1 \\mid 3m + 901$.\nAnd when $\\frac{3m+901}{20m+1} = 1, 2, 3, 4$, $m$ is not a positive integer.\nTheref... | China | China Girls' Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | no positive integer n | |
0ctm | In the coordinate space, we consider all the planes with equations of the form $x \pm y \pm z = n$ (for all integer $n$). These planes partition the space into tetrahedra and octahedra. Assume that the point $(x_0, y_0, z_0)$ with rational coordinates lies in neither of these planes. Prove that there exists a positive ... | [
"Perform the coordinate change $a = y + z - x$, $b = x - y + z$, $c = x + y - z$ (then $x = \\frac{b+c}{2}$, $y = \\frac{a+c}{2}$, $z = \\frac{a+b}{2}$, and $x + y + z = a + b + c$). The new planes have the equations $a = n$, $b = n$, $c = n$, and $a + b + c = n$, so they partition the space into cubes, each cut in... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English; Russian | proof only | null | |
0fmo | Hallar un polinomio de grado tres cuyas raíces sean, precisamente, el cuadrado de las raíces del polinomio $p(x) = x^3 + 2x^2 + 3x + 4$. | [
"Sean $r$, $s$ y $t$ las raíces, reales o complejas, del polinomio $p(x)$. Por tanto, $p(x) = (x - r)(x - s)(x - t)$. El polinomio que buscamos, salvo que multipliquemos por una constante, será de la forma\n$$\nq(x) = (x - r^2)(x - s^2)(x - t^2)\n$$\nDe aquí resulta\n$$\nq(x^2) = (x^2 - r^2)(x^2 - s^2)(x^2 - t^2) =... | Spain | Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Spanish | proof and answer | x^3 + 2x^2 - 7x - 16 | |
02nh | Problem:
Um triângulo isósceles tem uma base de $10~\mathrm{cm}$ e dois lados iguais medindo $13~\mathrm{cm}$. É possível cortar esse triângulo em dois outros triângulos de tal modo que, juntando esses triângulos de outra maneira obtenhamos um outro triângulo isósceles (evidentemente com a mesma área)?
![](attached_i... | [] | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0iy4 | Problem:
A computer program is a function that takes in 4 bits, where each bit is either a $0$ or a $1$, and outputs TRUE or FALSE. How many computer programs are there? | [
"Solution:\nThe function has $2^{4}$ inputs and $2$ outputs for each possible input, so the answer is $2^{2^{4}} = 2^{16} = 65536$."
] | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Other"
] | null | final answer only | 65536 | |
0icg | Problem:
There are eleven positive integers $n$ such that there exists a convex polygon with $n$ sides whose angles, in degrees, are unequal integers that are in arithmetic progression. Find the sum of these values of $n$. | [
"Solution:\n\nThe sum of the angles of an $n$-gon is $(n-2) 180$, so the average angle measure is $(n-2) 180 / n$. The common difference in this arithmetic progression is at least $1$, so the difference between the largest and smallest angles is at least $n-1$. So the largest angle is at least $(n-1)/2 + (n-2) 180 ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 106 | |
0ci2 | Consider a parallelogram $ABCD$ and the points $M$ on the side $DC$ and $E$ and $N$ on the diagonal $AC$, such that $BE \perp AC$ and $\frac{CM}{CD} = \frac{EN}{EA}$.
Prove that if $MN$ and $NB$ are perpendicular, then $ABCD$ is a rectangle. | [
"Solution.\n\nConstruct the parallel to $AB$ through $N$ and denote by $P$ its intersection with the line $BE$.\n\n\n\nUsing the fundamental theorem of similarity in the triangle $EAB$:\n$$\n\\frac{NP}{AB} = \\frac{EN}{EA} = \\frac{CM}{CD}.\n$$\nFrom here we obtain $NP = CM$ and, since $NP ... | Romania | 74th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0fhi | Problem:
Dadas dos circunferencias exteriores de radios $r$ y $r'\left(r \neq r'\right)$, se pide dibujar, razonadamente, una recta paralela a una dirección dada, tal que determine sobre las dos circunferencias dos cuerdas tales que la suma de sus longitudes sea igual a una longitud dada $\ell$. | [
"Solution:\n\n\n\nPodemos suponer que la dirección dada es horizontal y viene dada por la recta $d$. Pongamos $O_{1}$ y $O_{2}$ para los centros de las circunferencias de radios $r$ y $r'$ respectivamente. Proyectamos $O_{2}$ perpendicularmente sobre $d$ en $X$.\n\nTrasladamos $X$ paralelam... | Spain | OME 28 | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0700 | Problem:
There is a bulb in each cell of an $n \times n$ board. Initially all the bulbs are off. If a bulb is touched, that bulb and all the bulbs in the same row and column change state (those that are on, turn off, and those that are off, turn on). Show that it is possible by touching $m$ bulbs to turn all the bulbs... | [
"Solution:\n\nIf $n$ is odd, touch each bulb in the first column. Then bulbs in the first column are each switched $n$ times, which is odd and so end up on. All other bulbs are switched just once, and so end up on. $n$ is obviously minimal, because if $m < n$, then there is a bulb which is not switched at all (ther... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | If n is odd: m = n. If n is even: m = n^2. | |
0d4v | Let $ABC$ be a triangle. Circle $\Omega$ passes through points $B$ and $C$. Circle $\omega$ is tangent internally to $\Omega$ and also to sides $AB$ and $AC$ at $T$, $P$, and $Q$, respectively. Let $M$ be midpoint of $\operatorname{arc} \overparen{BC}$ (containing $T$) of $\Omega$. Prove that lines $PQ$, $BC$, and $MT$... | [
"Let $R$ be the intersection point of lines $PQ$ and $BC$, and $S$ the intersection point of lines $MT$ and $BC$.\n\n\n\nBecause circle $\\omega$ is tangent to both sides $AB$ and $AC$ at $P$ and $Q$, we have $AP = AQ$. Applying Menelaus' theorem to triangle $ABC$ and line $PQ$ we obtain\n$... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English, Arabic | proof only | null | |
0k1x | Problem:
Let $a$ and $b$ be five-digit palindromes (without leading zeroes) such that $a < b$ and there are no other five-digit palindromes strictly between $a$ and $b$. What are all possible values of $b - a$? (A number is a palindrome if it reads the same forwards and backwards in base 10.) | [
"Solution:\n\nLet $\\overline{x y z y x}$ be the digits of the palindrome $a$. There are three cases.\n\nIf $z < 9$, then the next palindrome greater than $\\overline{x y z y x}$ is $\\overline{x y (z+1) y x}$, which differs by $100$.\n\nIf $z = 9$ but $y < 9$, then the next palindrome up is $\\overline{x (y+1) 0 (... | United States | HMMT February 2018 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 11, 100, 110 | |
0etz | For a positive integer $a$, let $S_a$ be the set of primes satisfying the following condition:
(condition) For any $p \in S_a$, there exists an odd integer $b$ such that $(2^{2^a})^b - 1$ is divisible by $p$.
For any positive integer $a$, prove that there exist infinitely many primes that are not contained in $S_a$. | [
"For any positive integer $n$, define $a_n := 2^{2^n} + 1$. Since\n$$\na_1 a_2 \\cdots a_{n-1} = a_n - 2,\n$$\n$\\gcd(a_n, a_m) = 1$ for any $n \\neq m$. Now, for any positive integer $n$ such that $n \\ge a$, let $p_n$ be any prime dividing $a_n$. Note that $p_n$ is different from $p_m$ for any $m \\ne n$.\n\nSupp... | South Korea | Korean Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0h2f | Find all pairs of integers $(x, y)$ that satisfy the following equality:
$$
|x+|x+|x|| \cdot ||-y|-y|-y|=2011.
$$ | [
"Since $2011$ is a prime number, every multiple of the left-hand side must be equal to either $1$ or $2011$.\nIf $x \\ge 0$, the first multiple equals $3x$, and there are no solutions. Similarly, there are no solutions if $y \\le 0$ (the second multiple is $-3y$).\nSuppose now that $x < 0$ and $y > 0$. Then\n$$\n|x... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | (-1, 2011) and (-2011, 1) | |
08hl | Problem:
For every positive integer $n \geq 1$ we define the matrix $A_n = (a_{ij})_{1 \leq i, j \leq n}$, where $a_{ij} = \max(i, j) / \min(i, j)$, $1 \leq i, j \leq n$. Calculate the determinant of the matrix $A_n$. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof and answer | det(A_n) = (-1)^{n-1} · ∏_{j=2}^{n} (2j−1)/(j−1)^2 = (-1)^{n-1} · (2n−1)!! / ((n−1)!)^2 | |
0859 | Problem:
Alberto e Barbara si sfidano al seguente gioco: inizialmente su un tavolo ci sono alcune pile di gettoni (il numero di gettoni può variare da pila a pila). A turno, partendo da Alberto, uno dei due giocatori fa una e una sola delle seguenti mosse:
- o toglie un gettone da una pila a sua scelta e lo elimina da... | [
"Solution:\n\nSupponiamo innanzitutto che si parta da una situazione in cui c'è almeno un gettone presente sul tavolo, altrimenti il gioco non può nemmeno iniziare.\nSiano $C_{0}, C_{1}, C_{2}, \\ldots$ le combinazioni di pile presenti all'inizio del gioco, dopo la prima mossa, dopo la seconda mossa, eccetera. Per ... | Italy | XXII OLIMPIADE ITALIANA DI MATEMATICA | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Barbara has a winning strategy if and only if both m0 and s0 are even, where s0 is the number of piles of size one and m0 = sum over piles of size greater than one of (size minus one), equivalently the total number of tokens minus the total number of piles. Otherwise Alberto has a winning strategy. (Equivalently, Barba... | |
0bpx | Find all non-negative integers $n$ so that $n^2 - 4n + 2$, $n^2 - 3n + 13$ and $n^2 - 6n + 19$ are simultaneously primes. | [
"Let us denote:\n\n\\begin{align*}\nA &= n^2 - 4n + 2 \\\\\nB &= n^2 - 3n + 13 \\\\\nC &= n^2 - 6n + 19\n\\end{align*}\n\nWe want $A$, $B$, $C$ to be simultaneously prime for some non-negative integer $n$.\n\nLet us try small values of $n$:\n\nFor $n = 0$:\n\\begin{align*}\nA &= 0^2 - 4 \\cdot 0 + 2 = 2 \\\\\nB &= ... | Romania | 67th NMO Shortlisted Problems | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 0 and 4 | |
0a10 | Felix chooses a positive integer as the starting number and writes it on the board. He then repeats the next step: he replaces the number $n$ on the board by $\frac{1}{2}n$ if $n$ is even and by $n^2 + 3$ if $n$ is odd. For how many choices of starting numbers below 2023 will Felix never write a number of more than fou... | [
"We first show that if Felix starts with an odd number, the next two numbers he writes down will be even. Suppose Felix starts with an odd number $n$. Then the next number he writes down is $n^2+3$. We must show that $n^2+3$ is even, and $\\frac{1}{2}(n^2+3)$ is even as well. In other words, $n^2+3$ must be divisib... | Netherlands | Dutch Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Other"
] | English | proof and answer | 21 | |
037w | Problem:
Let $ABC$ be a triangle with $\angle BAC = 30^\circ$ and $\angle ABC = 45^\circ$. Consider all pairs of points $X$ and $Y$ such that $X$ and $Y$ lie on the rays $AC^{\rightarrow}$ and $BC^{\rightarrow}$, respectively, and $OX = BY$, where $O$ is the circumcenter of $\triangle ABC$. Prove that the perpendicula... | [
"Solution:\n\nWe shall prove that the perpendicular bisector of $XY$ passes through the point $C'$ which is symmetric to the point $C$ with respect to $AB$. Denote by $R$ the circumradius of $\\triangle ABC$. The Sine theorem for $\\triangle ABC$ gives $AC = R \\sqrt{2}$ and $BC = R$. Suppose that $C$ lies between ... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"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 > Constructions and loci",
"Geometry > Plane Geometry > Miscella... | null | proof only | null | |
0g7m | 設數列 $\{a_n\}$ 滿足：
$$
a_1 = a_2 = 1,\ a_{n+2} = a_{n+1} + a_n\ (n \in \mathbb{N}).
$$
當 $n$ 為奇數時，試求出滿足下列方程組的所有實數解 $(x, y)$。
$$
\begin{cases} x + 2^x a_n + 2^y a_{n+1} = 1 + 2a_{n+2}, \\ y + 2^x a_{n+1} + 2^y a_{n+2} = 1 + 2a_{n+3}. \end{cases}
$$ | [
"$(x, y) = (1,1)$ 為此方程組的唯一實數解。\n顯然，$(x, y) = (1,1)$ 為此方程組的一組實數解。以下證明無其它實數解。\n若不然，假設 $(x, y) = (x_1, y_1) \\neq (1,1)$ 為方程組\n$$\n\\begin{cases} x + 2^x a_n + 2^y a_{n+1} = 1 + 2a_{n+2}, \\\\ y + 2^x a_{n+1} + 2^y a_{n+2} = 1 + 2a_{n+3}, \\end{cases}\n$$\n的另一組實數解。為了方便，令 $(x_2, y_2) = (1,1)$，將這兩組解代入此方程組並相減，\n分別得到\n$$\... | Taiwan | 二0一三數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | (x, y) = (1, 1) | |
0jj4 | Problem:
Find a nonzero monic polynomial $P(x)$ with integer coefficients and minimal degree such that $P(1 - \sqrt[3]{2} + \sqrt[3]{4}) = 0$. (A polynomial is called monic if its leading coefficient is 1.) | [
"Solution:\nNote that $(1 - \\sqrt[3]{2} + \\sqrt[3]{4})(1 + \\sqrt[3]{2}) = 3$, so $1 - \\sqrt[3]{2} + \\sqrt[3]{4} = \\frac{3}{1 + \\sqrt[3]{2}}$.\n\nNow, if $f(x) = x^{3} - 2$, we have $f(\\sqrt[3]{2}) = 0$, so if we let $g(x) = f(x - 1) = (x - 1)^{3} - 2 = x^{3} - 3x^{2} + 3x - 3$, then $g(1 + \\sqrt[3]{2}) = f... | United States | HMMT 2014 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | x^3 - 3x^2 + 9x - 9 | |
0e8h | Problem:
V vsakem šestkotniku neskončnega satovja je zapisano naravno število, ki je enako povprečju števil, zapisanih v petih izmed sosednjih šest šestkotnikov. Dokaži, da so vsa števila v satovju enaka. | [
"Solution:\n\nKer so vsa števila v satovju naravna, obstaja najmanjše izmed njih. Označimo ga z $n$. Dovolj je pokazati, naslednje: če je v nekem šestkotniku zapisano število $n$, potem so v sosednjih šestkotnikih tudi zapisana števila $n$. Ker obstaja šestkotnik s številom $n$, bo potem namreč sledilo, da je v vse... | Slovenia | 57. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
09dx | Circles $w_1, w_2$ with equal radius $R$ intersect in two points. The line joining centres of these circles intersects $w_1$ in points $A$, $C$ and intersects $w_2$ in points $B$, $D$. ($B$ is between $A$ and $C$, $C$ is between $B$ and $D$). Draw a circle with diameter $AD$ tangent internally $w_1$ at $A$ and tangent ... | [
"Let $AB = z$, $PQ = x$, $QC = y$. By Pythagorean theorem:\n\n$$\nOE^2 = OB^2 + BE^2 \\Rightarrow BE^2 = R^2 - (AB - AO)^2 = R^2 - (z - R)^2 = 2Rz - z^2. \\ (*)\n$$\n\n\n\n$$\nAE^2 = BE^2 + AB^2 = 2zR - z^2 + z^2 = 2zR \\Rightarrow AE = \\sqrt{2zR}. \\quad (*)\n$$\n$$\n\\text{Since } AD \\t... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 2R | |
005e | Sea $\mathcal{F}$ la familia de todos los hexágonos convexos $H$ que satisfacen las siguientes condiciones:
(a) los lados opuestos de $H$ son paralelos;
(b) tres vértices cualesquiera de $H$ se pueden cubrir con una franja de ancho $1$.
Determinar el menor número real $\ell$ tal que cada uno de los hexágonos de la fami... | [] | Argentina | XXII Olimpiada Iberoamericana de Matemática | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | Spanish | proof and answer | 2 | |
0dv0 | Problem:
Kvadratna funkcija $f(x)$ doseže minimalno vrednost $-2$ za $x=1$. Določi $f(x)$ tako, da bo veljalo $f(-3)+4 f(0)=0$. Izračunaj $f(1+\sqrt{5})$. | [
"Solution:\n\nZapisana enačba kvadratne funkcije $y = a(x-1)^2 - 2$\n\nZapisani in vstavljeni podatki: za $f(-3) = 16 \\cdot a - 2$\nza $4 \\cdot f(0) = 4(a - 2)$\n\nZapisana enačba $16 \\cdot a - 2 + 4(a - 2) = 0$\n\nIzračun $a = \\frac{1}{2}$\n\nZapisana kvadratna funkcija $f(x) = \\frac{1}{2}(x-1)^2 - 2$ ali $f(... | Slovenia | 2. matematično tekmovanje dijakov srednjih tehniških in strokovnih šol | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | f(x) = (1/2)(x - 1)^2 - 2; f(1 + sqrt(5)) = 1/2 | |
0d27 | Let $x, y$ be two integers. Prove that if $2013$ divides $x^{1433} + y^{1433}$ then $2013$ divides $x^{7} + y^{7}$. | [
"Since $2013 = 3 \\times 11 \\times 61$, we will prove that for $p = 3, 11, 61$, if $p$ divides $x^{1433} + y^{1433}$ then $p$ divides $x^{7} + y^{7}$.\n\nLet $p = 3, 11, 61$, and assume that $p$ divides $x^{1433} + y^{1433}$.\n\nIf $p$ divides $x$, then it divides $x^{7}$ and $x^{1433}$. But $p$ divides $x^{1433} ... | Saudi Arabia | Preselection tests for the full-time training | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0hl8 | Problem:
Find the largest prime factor of $-x^{10}-x^{8}-x^{6}-x^{4}-x^{2}-1$, where $x=2i$, $i=\sqrt{-1}$. | [
"Solution:\n13."
] | United States | null | [
"Algebra > Intermediate Algebra > Complex numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 13 | |
0bdv | Let $(A, +, \cdot)$ be a unitary ring such that the following are true:
(1) $A$ is not a field;
(2) for any noninvertible element $x$ of $A$, there is an integer $m \ge 1$, depending on $x$, such that $x = x^2 + x^3 + \dots + x^{2^m}$. Prove that:
a. $x+x=0$, for any $x \in A$,
b. $x^2 = x$, for any noninvertible $1 ... | [
"a. It is sufficient to prove that $1+1=0$. Consider $x$ noninvertible, and let $m \\in \\mathbb{N}^*$, such that $x = x^2 + x^3 + \\dots + x^{2^m}$, and $y = x + x^2 + \\dots + x^{2^m-1}$. Evidently, $xy = x$, so $xy^k = x$, for any $k \\in \\mathbb{N}^*$. Because $x$ is not invertible it follows that $y$ is also,... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Abstract Algebra > Ring Theory"
] | null | proof only | null | |
0icm | Let $\triangle ABC$ be a triangle and let $D$ be a point in its interior. Construct a circle $\omega_1$ passing through $B$ and $D$ and a circle $\omega_2$ passing through $C$ and $D$ such that the point of intersection of $\omega_1$ and $\omega_2$ other than $D$ lies on line $AD$. Denote by $E$ and $F$ the points wher... | [
"**First Solution:** Let circles $\\omega_1$ and $\\omega_2$ meet again at $R$ (other than $D$), and let $\\omega_1$ and $\\omega_2$ intersect again with segments $AB$ and $AC$, respectively, at $P$ and $Q$ (other than $B$ and $C$). By the **Power of a Point Theorem**, we have $AP \\cdot AB = AR \\cdot AD = AQ \\cd... | United States | USA IMO | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscel... | null | proof only | null | |
0gh5 | $$
\mathcal{F} := \{ f : \mathbb{R} \to \mathbb{R} \mid f(x + f(y)) = f(x) + f(y), \forall x, y \in \mathbb{R} \}.
$$
找出所有的有理數 $q$ 使得對所有函數 $f \in \mathcal{F}$, 存在 $z \in \mathbb{R}$ 滿足 $f(z) = qz$.
Let $\mathbb{R}$ be the set of real numbers. We denote by $\mathcal{F}$ the set of all functions $f : \mathbb{R} \to \mat... | [
"Let $Z$ be the set of all rational numbers $q$ such that for every function $f \\in \\mathcal{F}$, there exists some $z \\in \\mathbb{R}$ satisfying $f(z) = qz$. Let further\n$$\nS = \\left\\{ \\frac{n+1}{n} : n \\in \\mathbb{Z}, n \\neq 0 \\right\\}.\n$$\nWe prove that $Z = S$ by showing the two inclusions: $S \\... | Taiwan | 2023 數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | Chinese (Traditional) | proof and answer | All rational numbers of the form (n+1)/n with n a nonzero integer. | |
007o | 1000 balls of mass $0.38$ and 5000 balls of mass $0.038$ must be packed in boxes. A box can contain any collection of balls with total mass at most $1$. Find the minimum number of boxes needed. | [
"There can be $0$, $1$ or $2$ balls of mass $0.38$ in a box since $3 \\cdot 0.38 > 1$. In these three cases the box can contain at most $\\lfloor \\frac{1}{0.038} \\rfloor = 26$, $\\lfloor \\frac{1-0.38}{0.038} \\rfloor = 16$ and $\\lfloor \\frac{1-2 \\cdot 0.38}{0.038} \\rfloor = 6$ balls with mass $0.038$ respect... | Argentina | National Olympiad of Argentina | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | 577 | |
0ahs | A triangle $\triangle ABC$ is given together with a segment $PQ$ of length $t$ on the segment $BC$, so that $P$ is between $B$ and $Q$ and $Q$ is between $P$ and $C$. We draw parallel lines from the point $P$ to $AB$ and $AC$ which intersect $AC$ and $AB$ in $P_1$ and $P_2$, respectively. We draw parallel lines from th... | [
"**Solution.** Let $D$ be the intersection of $PP_1$ and $QQ_2$. Let us note that $P_1DQ_2P_2 = 2P_1\\triangle ADP$ and $P_1Q_1DQ_1 = 2P_1\\triangle ADQ$. So now we have:\n$$\n\\begin{aligned}\nP_1PQQ_2P_2 + P_1P_1Q_1Q &= P_1DQ_2P_2 + P_1Q_1DQ_1 + 2P_1\\triangle PQD \\\\\n&= 2P_1\\triangle ADP + 2P_1\\triangle ADQ ... | North Macedonia | Junior Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
03zr | Let $a_n = C_{200}^n \cdot (\sqrt[3]{6})^{200-n} \cdot \left(\frac{1}{\sqrt{2}}\right)^n$ $(n = 1, 2, \dots, 95)$. Then the number of terms that are integers in $\{a_n\}$ is ______. | [
"We have $a_n = C_{200}^n \\cdot 3^{\\frac{200-n}{3}} \\cdot 2^{\\frac{400-5n}{6}}$. When $a_n$ $(1 \\le n \\le 95)$ is an integer, $\\frac{200-n}{3}$ and $\\frac{400-5n}{6}$ must be integers. Then $6 \\mid n+4$.\n\nWhen $n = 2, 8, 14, 20, 26, 32, 38, 44, 50, 56, 62, 68, 74, 80$, $\\frac{200-n}{3}$ and $\\frac{400-... | China | China Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 15 | |
08e2 | Problem:
Siano $a_{1}, a_{2}, \ldots, a_{2020}$ e $b_{1}, b_{2}, \ldots, b_{2020}$ dei numeri reali, non necessariamente distinti. Supponiamo che gli interi positivi $n$ per cui l'equazione
$$
|a_{1} x - b_{1}| + |a_{2} x - b_{2}| + \cdots + |a_{2020} x - b_{2020}| = n
$$
ha esattamente due soluzioni reali siano in nu... | [] | Italy | XXXVI Olimpiade Italiana di Matematica | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
05ey | Problem:
Soit $ABC$ un triangle isocèle et obtus en $A$. Soit $\Gamma$ le cercle de centre $B$ passant par $A$, et $\Omega$ le cercle de centre $C$ passant par $A$. Soit $D$ le point d'intersection du cercle $\Gamma$ avec le segment $[BC]$, $E$ le deuxième point d'intersection de la droite $(AD)$ avec le cercle $\Omeg... | [
"Solution:\n\nSoit $X$ le point d'intersection du cercle $\\Omega$ avec le segment $[BC]$. Les points $B$ et $C$ sont symétriques par rapport à la médiatrice du segment $[BC]$ donc les cercles $\\Gamma$ et $\\Omega$ le sont aussi. Il vient que $D$ et $X$ sont symétriques par rapport à la médiatrice du segment $[BC]... | France | ENVOI 1 : GÉOMÉTRIE Corrigé | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07no | The altitudes of a triangle $ABC$ are used to form the sides of a second triangle $A_1B_1C_1$. The altitudes of $\triangle A_1B_1C_1$ are then used to form the sides of a third triangle $A_2B_2C_2$. Prove that $\triangle A_2B_2C_2$ is similar to $\triangle ABC$. | [
"For $\\triangle ABC$, let the altitude $AD$ standing on the base $BC$ have length $p$, let the altitude $BE$ standing on the base $AC$ have length $q$, and let the altitude $CF$ standing on the base $AB$ have length $r$.\n\n\n\n*Step 1:* Then $\\triangle ADC$ is similar to $\\triangle BEC$... | Ireland | Ireland | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0k8p | Problem:
In 2019, a team, including professor Andrew Sutherland of MIT, found three cubes of integers which sum to $42$:
$$
42 = \left(-8053873881207597\_\right)^3 + (80435758145817515)^3 + (12602123297335631)^3
$$
One of the digits, labeled by an underscore, is missing. What is that digit? | [
"Solution:\n\nLet the missing digit be $x$. Then, taking the equation modulo $10$, we see that $2 \\equiv -x^{3} + 5^{3} + 1^{3}$. This simplifies to $x^{3} \\equiv 4 \\pmod{10}$, which gives a unique solution of $x = 4$."
] | United States | HMMT November 2019 | [
"Number Theory > Modular Arithmetic"
] | null | final answer only | 4 | |
0c76 | One considers the rectangular parallelepiped $ABCD'A'B'C'D'$ such that the measure of the dihedral angle formed by the planes ($A'BD$) and ($C'BD$) is $90^\circ$ and the measure of the dihedral angle formed by the planes ($AB'C$) and ($D'B'C$) is $60^\circ$. Determine the measure of the dihedral angle formed by the pla... | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Linear Algebra > Vectors"
] | English | proof and answer | 60° | |
0dn2 | Problem:
У неким чворовима квадратне решетке $2012 \times 2012$ налази се мува и $k$ паукова. Један потез састоји се у следећем: мува се помера на суседан чвор или остаје на истом месту, а након тога се сваки од $k$ паукова помера на неки суседан чвор или остаје на истом месту (у једном чвору може бити више паукова). ... | [
"Solution:\n\nЈедан паук не може да ухвати муву. Довољно је да мува не мрда ако паук није на суседном пољу, односно да се помери на поље дијагонално супротно пауковом ако јесте.\nДоказаћемо да су два паука довољна у оба дела задатка - означимо их са $P$ и $Q$, муву са $M$, а $x$- и $y$-координату тачке $A$ са $A_{x... | Serbia | Serbian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | a) 2; b) 2 | |
0f3i | Problem:
$ABC$ is equilateral. A line parallel to $AC$ meets $AB$ at $M$ and $BC$ at $P$. $D$ is the center of the equilateral triangle $BMP$. $E$ is the midpoint of $AP$. Find the angles of $DEC$. | [
"Solution:\n\n\nLet $K$ be the midpoint of $BP$ and $L$ the midpoint of $AC$. $EL$ is parallel to $BC$, so $\\angle ELC = 120^\\circ$. $EK$ is parallel to $AB$, so $\\angle EKC = 60^\\circ$, so $ELCK$ is cyclic. But $\\angle DKC = \\angle DLC = 90^\\circ$, so $DLCK$ is cyclic. Hence $D$, $K... | Soviet Union | ASU | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | ∠DEC = 90°, ∠EDC = 60°, ∠DCE = 30° | |
097k | Problem:
Cercul este divizat de 2020 de puncte în 2020 de arce egale. Păcală susţine că poate construi o linie frântă închisă cu vârfurile în aceste puncte astfel încât oricare două segmente ale ei să nu fie paralele. Are dreptate Păcală? | [
"Solution:\n\nPunctele de pe cerc se notează cu numerele $1,2,3, \\ldots, 2020$ în mod consecutiv şi într-o anumită direcţie, de exemplu, în sens orar. Pentru simplitate, lungimea fiecăruia din cele 2020 de arce se va considera egală cu 1. Fiecare segment al liniei frânte se notează cu perechea de numere din extrem... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | No, such a closed broken line cannot be constructed; some pair of segments must be parallel. | |
0c0u | Define the set $M = \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathcal{M}_2(\mathbb{C}) \mid ab = cd \right\}$.
a) Give examples of $A \in M$ having the property $A^{2017} \in M$ and $A^{2019} \in M$, but $A^{2018} \notin M$.
b) Prove that if for an $A \in M$ there is $k \ge 1$ such that $A^k \in M$, $... | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | null | |
02cv | Problem:
(a) Adriano escolheu secretamente cinco números $a$, $b$, $c$, $d$ e $e$ e informou a Bruna os dez números $24, 28, 30, 30, 32, 34, 36, 36, 40$ e $42$ obtidos pelo cálculo de todas as somas de dois números dentre os cinco escolhidos.
O objetivo de Bruna é descobrir $a$, $b$, $c$, $d$, $e$. Bruna pode alcançar ... | [
"Solution:\n(a) Suponha que $a \\leqslant b \\leqslant c \\leqslant d \\leqslant e$. Logo a menor soma é $a+b$ e a maior soma é $d+e$. A segunda menor é $a+c$ e a segunda maior é $c+e$. Assim, temos o sistema\n$$\n\\left\\{\\begin{array}{l}\na+b=24 \\\\\na+c=28 \\\\\nc+e=40 \\\\\nd+e=42\n\\end{array}\\right.\n$$\nP... | Brazil | Desafios | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (a) Yes: a=11, b=13, c=17, d=19, e=23. (b) No; for instance, {3,7,17,19} and {4,6,16,20} produce the same pairwise sums. | |
022d | Problem:
Em uma circunferência de raio $10~\mathrm{cm}$, o segmento $AB$ é um diâmetro e o segmento $AC$ é uma corda de $12~\mathrm{cm}$. Determine a distância entre os pontos $B$ e $C$. | [
"Solution:\n\nSendo $AB$ um diâmetro, o triângulo $\\triangle ABC$ está inscrito numa semicircunferência. Isto implica que este triângulo é retângulo no vértice $C$. Pelo Teorema de Pitágoras,\n$$\nBC^{2} = AB^{2} - AC^{2}\n$$\nou seja,\n$$\nBC^{2} = 20^{2} - 12^{2} = 256 = 16^{2}\n$$\n\nAs... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 16 | |
0etu | Each of the lattice points $(x, y)$ ($x$ and $y$ integers) in the plane can be coloured black or white. A single strike by an L-shaped punch changes the colour of the four lattice points $(a, b)$, $(a+1, b)$, $(a, b+1)$ and $(a, b+2)$. All lattice points are initially coloured white. Prove that after any number of stri... | [
"When the lattice points $(a, b)$, $(a+1, b)$, $(a, b+1)$ and $(a, b+2)$ change colour after a single strike by the punch, let us call the point $(a, b)$ the corner point of the strike.\n\nWe first show (by induction on the number of strikes) that, after any finite number of strikes, there will always be an even nu... | South Africa | The South African Mathematical Olympiad Third Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0epm | If $\frac{20}{3} = \frac{120}{y}$ then $y$ equals
(A) 9
(B) 12
(C) 15
(D) 18
(E) 20 | [
"If the numerator has been multiplied by $6$ and the fraction stays the same, the denominator is also multiplied by $6$ and becomes $3 \\times 6 = 18$."
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | D | |
068j | Let $AB\Gamma$ be an acute angled triangle with $AB < A\Gamma < B\Gamma$ and let $c$ be its circumcircle with centre $O$. At the small arcs $A\Gamma$ and $AB$ we consider the points $\Delta$ and $E$ respectively. Let $K$ be the intersection point of $B\Delta$, $\Gamma E$ and $N$ be the second common point of the circum... | [
"Let $K$ be the intersection point of $B\\Delta$, $\\Gamma E$ and $N$ be the second intersection point of $c_1$, $c_2$. Let $Z$ be the point of intersection of the tangents at $B, \\Gamma$ of the circle $c$. We will prove that the points $K, N, Z$ are collinear.\nLet $T$ be the intersection of $BZ$ with $c_1$ and $... | Greece | Hellenic Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous ... | English | proof only | null | |
075o | Let $S$ be a nonempty set of primes satisfying the property that for each proper subset $P$ of $S$, all the prime factors of the number $\left(\prod_{p \in P} p\right) - 1$ are also in $S$. Determine all possible such sets $S$. | [
"If $S$ is singleton, then the statement is vacuously true. Hence $S = \\{p\\}$, $p$ a prime.\n\nSuppose $S$ contains two elements. One of these must be odd primes and hence $2$ is in $S$. Moreover $p-1$ must be a power of $2$. Hence $p$ is Fermat prime.\n\nSuppose $|S| \\ge 3$. Here we consider two cases: $|S| < \... | India | Indija TS 2012 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | Exactly the following sets: any singleton containing one prime; any two element set consisting of two and a Fermat prime; and the set of all prime numbers. | |
07ak | The city of Bridge Village has some highways. Highways are closed curves that have intersections with each other or themselves in 4-way crossroads. Mr. Bridge Lover, mayor of the city, wants to build a bridge on each crossroad in order to decrease the number of accidents. He wants to build the bridges in such a way tha... | [
"We show that bridges satisfying the problem's conditions can be built no matter what the configuration of the highways is.\nThroughout the solution we denote the highways by $C_1, C_2, \\dots$ and $C_n$ and by $G$ the graph determined by those highways (crossroads are vertices of the graph and highways between two... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Possible for any number of highways (always possible). | |
09g4 | A circle with center $I$ is inscribed in a hexagon $ABCDEF$. Let $A', B', C', D', E', F'$ be the midpoints of the diagonals $BF, AC, BD, CE, DF, EA$, respectively. Suppose that the lines $AA', CC', EE'$ intersect at $X$, the lines $BB', DD', FF'$ intersect at $Y$. If the triangle residing between the lines $AB, CD, EF$... | [
"First we prove the following lemma:\n\n**Lemma.** Let $ABCDEF$ be a convex hexagon on the coordinate plane. Then the set of the points $P$ inside $ABCDEF$ such that\n$$\nS_{APB} + S_{CPD} + S_{EPF} = S_{BPC} + S_{DPE} + S_{FPA},\n$$\nis a segment.\n\n*Proof.* Suppose that the equations of the lines containing the ... | Mongolia | 2015 Mongolian IMO Team Selection Tests | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > C... | null | proof only | null | |
00gl | Consider $n$ disks $C_{1}, C_{2}, \ldots, C_{n}$ in a plane such that for each $1 \leq i < n$, the center of $C_{i}$ is on the circumference of $C_{i+1}$, and the center of $C_{n}$ is on the circumference of $C_{1}$. Define the score of such an arrangement of $n$ disks to be the number of pairs $(i, j)$ for which $C_{i... | [
"The answer is $(n-1)(n-2)/2$.\n\nLet's call a set of $n$ disks satisfying the given conditions an $n$-configuration. For an $n$-configuration $\\mathcal{C} = \\{C_{1}, \\ldots, C_{n}\\}$, let $S_{\\mathcal{C}} = \\{(i, j) \\mid C_{i}$ properly contains $C_{j}\\}$. So, the score of an $n$-configuration $\\mathcal{C... | Asia Pacific Mathematics Olympiad (APMO) | XIX Asian Pacific Mathematics Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | (n-1)(n-2)/2 | |
07w2 | The positive integers $a, b, c, d$ satisfy
$$
(i) a + b + c + d = 2023
$$
$$
(ii) 2023 \mid ab - cd
$$
$$
(iii) 2023 \mid a^2 + b^2 + c^2 + d^2.
$$
Assuming that each of the numbers $a, b, c, d$ is divisible by 7, prove that each of the numbers $a, b, c, d$ is divisible by 17. | [
"**Solution 1.** First note that $2023 = 7 \\cdot 17^2$. If $a, b, c, d$ are divisible by 7, there are positive integers $w, x, y, z$ such that\n$$\na = 7w, \\quad b = 7x, \\quad c = 7y, \\quad d = 7z\n$$\nand the given conditions on $a, b, c, d$ translate into\n$$\n(i) w + x + y + z = 17^2\n$$\n$$\n(ii) 17^2 \\mid... | Ireland | IRL_ABooklet_2023 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof only | null | |
0cw8 | Each face of a $1000 \times 1000 \times 1000$ cube is partitioned into $1000^2$ unit squares. Determine the maximal possible number of unit squares one can mark so that no two marked squares share a common side. | [
"Consider an arbitrary coloring that satisfies the condition. Divide all the cells on the surface into \"frames\" as shown in Fig. 2—500 frames around each of the eight vertices (one of the frames is shaded). Then, in the $k$-th frame, counting from the vertex, there are $S_k = 6k - 3$ cells. Since no two colored c... | Russia | Regional round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Solid Geometry > Other 3D problems"
] | English; Russian | proof and answer | 2998000 | |
0bpw | Problem:
1. Enumerați elementele mulțimilor:
$$
A = \left\{ a \in \mathbb{Q} \mid a = \sqrt{\frac{2 - x}{4}},\ x \in \mathbb{N}^* \right\} \text{ și } B = \left\{ x \in \mathbb{N}^* \mid a = \sqrt{\frac{2 - x}{9}},\ a \in \mathbb{Q} \right\}
$$
Determinați $A \cup B$, $A \cap B$, $A \setminus B$, $B \setminus A$. | [] | Romania | OLIMPIADA DE MATEMATICĂ - ETAPA LOCALĂ | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | A = {0, 1/2}; B = {1, 2}; A ∪ B = {0, 1/2, 1, 2}; A ∩ B = ∅; A \ B = {0, 1/2}; B \ A = {1, 2} | |
0cpc | Let $a_1, \dots, a_{14}$ be distinct positive integers. All 196 numbers of the form $a_i + a_j$ ($1 \le i, j \le 14$) are written on a blackboard. May it happen that for each two digits $a, b$ there exists a written number with a decimal representation ending by $\overline{ab}$?
Даны различные натуральные числа $a_1, ... | [
"Не может.\n\nПусть среди наших 14 чисел есть $a$ чётных и $b = 14-a$ нечётных. Нечётное число на доске может появиться лишь как сумма чётного и нечётного, т.е. таких чисел будет $ab$ (при этом каждое будет выписано по два раза). Но $ab = \\frac{(a+b)^2 - (a-b)^2}{4} \\le \\frac{(a+b)^2}{4} = 49$. Значит, на доске ... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English, Russian | proof and answer | No |
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