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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0b2y | Problem:
A sequence $\{a_n\}$ of real numbers is defined by $a_1=1$ and for all integers $n \geq 1$,
$$
a_{n+1}=\frac{a_n \sqrt{n^2+n}}{\sqrt{n^2+n+2 a_n^2}}
$$
Compute the sum of all positive integers $n<1000$ for which $a_n$ is a rational number. | [
"Solution:\nFirst, note that for $k \\geq 1$,\n$$\na_{k+1}^2=\\frac{k(k+1) a_k^2}{k(k+1)+2 a_k^2} \\Longleftrightarrow \\frac{1}{a_{k+1}^2}-\\frac{1}{a_k^2}=\\frac{2}{k(k+1)}=\\frac{2}{k}-\\frac{2}{k+1}\n$$\nand summing the second equation from $k=1$ to $k=n-1$ with $n \\geq 2$, we get\n$$\n\\frac{1}{a_n^2}-\\frac{... | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Number Theory > Diophantine Equations > Pell's equations"
] | null | proof and answer | 131 | |
00o5 | Let $a$, $b$ and $c$ be real numbers with $0 \le a, b, c \le 2$. Prove that
$$
(a - b)(b - c)(a - c) \le 2.
$$
When does equality hold? | [
"We order the variables by size:\nFor $a \\ge b \\ge c$, all three factors are positive and we have $(a-b)(b-c)(a-c) \\ge 0$.\nFor $b \\ge c \\ge a$ and $c \\ge a \\ge b$, two of the factors are negative and one factor is positive, so we have again $(a-b)(b-c)(a-c) \\ge 0$.\nFor all the other orderings of variables... | Austria | AUT_ABooklet_2023 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds exactly for the triples (2, 1, 0), (1, 0, 2), and (0, 2, 1). | |
06av | Determine all positive integers $\alpha, \beta$ with $\alpha > 1$ such that $\beta$ divides $\alpha - 1$ and $2\alpha + 1$ divides $5\beta - 3$.
(A. Fellouris) | [
"Since $\\beta$ divides $\\alpha - 1$ and $2\\alpha + 1$ divides $5\\beta - 3$, with $\\alpha > 1$,\n$$\nx = \\frac{\\alpha - 1}{\\beta} \\quad \\text{and} \\quad y = \\frac{5\\beta - 3}{2\\alpha + 1}\n$$\nare positive integers such that:\n$$\n0 < xy = \\frac{\\alpha - 1}{\\beta} \\cdot \\frac{5\\beta - 3}{2\\alpha... | Greece | 40th Hellenic Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (3, 2), (10, 9), (13, 6) | |
0jlm | Problem:
Find the sum of all real numbers $x$ such that $5 x^{4}-10 x^{3}+10 x^{2}-5 x-11=0$. | [
"Solution:\nRearrange the equation to $x^{5} + (1-x)^{5} - 12 = 0$. It's easy to see this has two real roots, and that $r$ is a root if and only if $1-r$ is a root, so the answer must be $1$.",
"Solution:\nNote that $5 x^{4} - 10 x^{3} + 10 x^{2} - 5 x - 11 = 5 x(x-1)(x^{2}-x+1) - 11 = 5 u(u+1) - 11$, where $u = ... | United States | HMMT 2014 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | 1 | |
01vl | The vertices of the convex quadrilateral $ABCD$ lie on the parabola $y = x^2$. It is known that $ABCD$ is cyclic and $AC$ is a diameter of its circumcircle. Let $M$ and $N$ be the midpoints of the diagonals $AC$ and $BD$ respectively.
Find the length of the projection of the segment $MN$ on the axis $Oy$. | [
"Let the abscissae of the points $A, B, C, D$ be $a, b, c, d$ respectively. Since $AC$ is a diameter of the given circle, $M$ is its center. Let $(p, q)$ be the coordinates of $M$ then the equation of the circle is $(x - p)^2 + (y - q)^2 = R^2$ where $R$ denotes its radius.\n\nCoordinates $(x; y)$ of any point $A, ... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | 1 | |
0fj9 | Problem:
Un condenado queda en libertad cuando alcance el final de una escalera de 100 escalones. Pero no puede avanzar a su antojo, puesto que está obligado a subir un solo escalón cada día de los meses impares y a bajar un escalón cada día de los meses pares. Comienza el 1 de enero de 2001. ¿Qué día quedará en liber... | [
"Solution:\n\nEs fácil observar que el primer año va a moverse entre los escalones $1$ y $36$. Este, el $36$, lo alcanza el día $31$ de julio. El $31$ de diciembre de ese año, llegará al escalón $3$. En general, si un $31$ de diciembre está en el escalón $n$, el año siguiente:\nse mueve entre los escalones $n+1$ y ... | Spain | Viernes 19 de enero de 2001 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | March 31, 2025; and July 31, 2024 if the staircase had 99 steps | |
0f7z | Problem:
Find 5 different relatively prime numbers, so that the sum of any subset of them is composite. | [] | Soviet Union | 21st ASU | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Other",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | {121, 361, 841, 961, 1681} | |
01zz | Problem:
Let $A B C$ be a triangle with incentre $I$. The angle bisectors $A I, B I$ and $C I$ meet $[B C],[C A]$ and $[A B]$ at $D, E$ and $F$, respectively. The perpendicular bisector of $[A D]$ intersects the lines $B I$ and $C I$ at $M$ and $N$, respectively. Show that $A, I, M$ and $N$ lie on a circle. | [
"Solution:\n\nThe quadrilateral $A M D B$ is cyclic. Indeed, $M$ is the intersection of the line $B I$, which bisects the angle $\\widehat{A B D}$ in $A B D$ and the perpendicular bisector of $[A D]$. By uniqueness of this intersection point, it follows that $M$ lies on the circumcircle of $A B D$, and thence $A M ... | Benelux Mathematical Olympiad | THIRD BENELUX MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
0dn4 | Problem:
Нека је $\mathcal{K}$ целобројна решетка. Да ли постоји бијекција $f: \mathbb{N} \rightarrow \mathcal{K}$ таква да за све међусобно различите $a, b, c \in \mathbb{N}$ важи
$$
\text{НЗД}(a, b, c)>1 \quad \Longrightarrow \quad f(a), f(b), f(c) \text{ нису колинеарне? }
$$
(Целобројна решетка је скуп тачака у ра... | [
"Solution:\n\nПоређајмо све тачке решетке у низ $A_{1}, A_{2}, \\ldots$. Ово се може урадити нпр. спирално: $(0,0), (0,1), (1,1), (1,0), (1,-1), (0,-1), (-1,-1), \\ldots$ Конструисаћемо индуктивно пример бијекције са траженим својством.\n\nСтавимо $f(1)=A_{1}$. Претпоставимо да су $f(1), \\ldots, f(n-1)$ одређене и... | Serbia | Serbian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Math... | null | proof only | null | |
0ghm | 證明：存在常數 $C > 0$，使得對於所有滿足 $a + b$ 為整數的正實數 $a$ 與 $b$，有
$$
\{a^3\} + \{b^3\} \le 2 - \frac{C}{(a+b)^6}.
$$
Show that there exists a positive constant $C$ such that, for all positive reals $a$ and $b$ with $a + b$ being an integer, we have
$$
\{a^3\} + \{b^3\} \le 2 - \frac{C}{(a+b)^6}.
$$ | [
"以下證明當 $C = \\frac{1}{54}$ 時命題成立：\n\n令 $n = a + b$, 且不妨假設 $a \\ge b$。注意到函數 $f(x) = x^3$ 是一個遞增凸函數, 所以若 $a_0 + b_0 = n$ 且 $a_0 - b_0 \\ge a - b$, 我們有\n$$\na^3 + b^3 \\le a_0^3 + b_0^3. \\quad (1)\n$$\n特別地, 當 $\\lfloor a^3 \\rfloor = \\lfloor a_0^3 \\rfloor$ 且 $\\lfloor b^3 \\rfloor = \\lfloor b_0^3 \\rfloor$,\n$$\n\\... | Taiwan | 2023 數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Chinese (Traditional) | proof only | null | |
0631 | Problem:
Es sei $ABC$ ein spitzwinkliges Dreieck mit dem Umkreismittelpunkt $O$. Ferner sei $k$ ein Kreis mit folgenden Eigenschaften:
(1) Der Mittelpunkt $K$ von $k$ liegt im Inneren der Seite $BC$.
(2) $k$ berührt $AB$ in $B'$ und $AC$ in $C'$.
(3) $O$ liegt auf dem kürzeren der beiden Bogenstücke $B'C'$ von $k$.... | [
"Solution:\n\nDer Punkt $O'$ ist Bildpunkt von $O$ bei der Spiegelung an der Geraden $BC$. $S$ ist der Schnittpunkt von $KO'$ mit dem Umkreis $u$ (siehe Figur).\n\nNach dem Satz vom Mittelpunktswinkel ist $\\Varangle BOC = 2\\alpha$. Weil $C'$ und $B'$ im Inneren der Strecken $AC$ bzw. $AB$ liegen und $O$ wegen der... | Germany | 2. Auswahlklausur | [
"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... | null | proof only | null | |
0hey | Given $2n$ points on the plane, prove that it is possible to split them into $n$ pairs such that: if for each pair of points we construct a circle as on diameter, then the obtained $n$ circles will have a common point which does not necessarily belong to given ones.
(Anton Trygub) | [
"Consider a line $l$ such that there are $n$ points on one side of the line and $n$ points on the other. Clearly such a line exists. Without loss of generality, assume $l$ to be the horizontal axis. Then arrange points in increasing order of abscissa. In the upper half-plane we have: $A_1, A_2, \\ldots, A_n$ and $B... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Helly's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01w7 | Does there exist a function $f: \mathbb{N} \to \mathbb{N}$ such that
$$
f(f(n+1)) = f(f(n)) + 2^{n-1}
$$
for any positive integer $n$? (As usual, $\mathbb{N}$ stands for the set of all positive integers.) | [
"**Answer: yes.**\nNote that if function $f(x)$ satisfy $f(f(n)) = 2^{n-1}$ for any positive integer $n$, then it satisfy the problem condition $f(f(n+1)) = f(f(n)) + 2^{n-1}$ as well.\nIt is well known that there exist infinitely many functions $f: \\mathbb{N} \\to \\mathbb{N}$ such that $f(f(n)) = 2^{n-1}$ for al... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | yes | |
08db | Problem:
Ci sono 4037 lampadine disposte in fila e numerate da 1 a 4037. Inizialmente ogni lampadina può essere accesa o spenta. Una mossa consiste nello scegliere due lampadine numerate $a, b$ tali che $a / b$ o $b / a$ sia un numero primo e cambiare lo stato di entrambe. Dimostrare che in un numero finito di mosse s... | [
"Solution:\n\nMostriamo direttamente una sequenza di mosse che porta ad avere tutte le prime 2019 lampadine accese.\n\n- Se la lampadina con il numero 2019 è spenta, scegliamo $a=2019$ e $b=673$, in modo che $\\frac{a}{b}=3$ sia un numero primo e cambiamo lo stato di entrambe le lampadine. Ora la lampadina 2019 è a... | Italy | Gara di Febbraio | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0irj | Problem:
Alice has an equilateral triangle $A B C$ of area $1$. Put $D$ on $B C$, $E$ on $C A$, and $F$ on $A B$, with $B D = D C$, $C E = 2 E A$, and $2 A F = F B$. Note that $A D$, $B E$, and $C F$ pass through a single point $M$. What is the area of triangle $E M C$? | [
"Solution:\n\nAnswer: $\\frac{1}{6}$\n\nTriangles $A C F$ and $B C F$ share a height, so the ratio of their areas is $A F / B F = 1 / 2$. By the same method, the ratio of the areas of $A M F$ and $B M F$ is $1 / 2$. So, the ratio of the areas of $A C M$ and $B C M$ is also $1 / 2$. Similarly, the ratio of the areas... | United States | 1st Annual Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem"
] | null | proof and answer | 1/6 | |
0h8r | Consider the acute $\triangle ABC$ and point $D$ on the side $AB$. Let's denote the center of the circumscribed circle around the $\triangle ACD$ as $P$ and the center of the circumscribed circle around the $\triangle BDC$ as $Q$. Prove that triangles $ABC$ and $DPQ$ are similar.
(Bogdan Rublyov) | [
"Let's denote the radius of the circumscribed circle around the $\\triangle ACD$ as $R_1$, the radius of the circumscribed circle around the $\\triangle BDC$ as $R_2$. So, if $CD = l$ then by the law of sines (Fig.47):\n$$\n\\frac{l}{\\sin \\beta} = 2R_1 \\Rightarrow \\sin \\beta = \\frac{l}{2R_1}.\n$$\nHerewith, $... | Ukraine | The Problems of Ukrainian Authors | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
0361 | Problem:
A circle $k$ through the vertices $A$ and $B$ of an acute $\triangle ABC$ meets the sides $AC$ and $BC$ at inner points $M$ and $N$, respectively. The tangent lines to $k$ at the points $M$ and $N$ meet at point $O$. Prove that $O$ is the circumcenter of $\triangle CMN$ if and only if $AB$ is a diameter of $k$... | [
"Solution:\nIf $AB$ is a diameter of $k$, then $AN$ and $BM$ are altitudes of $\\triangle ABC$. Let $H$ be the orthocenter of $\\triangle ABC$ and let the tangent line to $k$ at $M$ meet the altitude $CH$ at $O_1$. Then $\\angle CMO_1 = \\angle ABM = \\frac{\\overparen{AM}}{2}$ and $\\angle ABM = \\angle ACH$. Thus... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"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 | |
05cz | According to a message sent by extraterrestrial creatures who are millions of years ahead of us in development, the height of the highest two places of their planet, measured from the sea level, is $h$, whereas the lowest point on mainland has height $l$ (where $h \ge 0 \ge l$). The radius of the planet (i.e., the dist... | [
"\nFig. 9\n\nIf the distance is maximal, the line connecting these points must be a tangent of the planet, otherwise one could increase the distance by pushing the points along the surface of the planet farther away. Let the centre be $O$; let the two points under consideration be $P_1$ and... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | 2√((2r + h + l)(h − l)) | |
0hsi | Problem:
When a single number is added to each member of the sequence $20, 50, 100$, the sequence becomes expressible as $x, a x, a^{2} x$. Find $a$. | [
"Solution:\n$\\frac{5}{3}$."
] | United States | null | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 5/3 | |
0dfe | Two triangles $ABC$ and $A_1B_1C_1$ are symmetric about the center of their common incircle of radius $r$. Prove that the product of the areas of the triangles $ABC$, $A_1B_1C_1$ and the six other triangles formed by the intersecting sides of the triangles $ABC$ and $A_1B_1C_1$ is equal to $r^{16}$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Tri... | English | proof and answer | r^{16} | |
0ebz | Problem:
Naj bo $x \in \mathbb{R}$. Izračunaj vrednost izraza
$$
\frac{2^{2014} + 2^{2015} - 2^{2016} - 2^{2017}}{(-3)^{2014} - (-3)^{2016}} \cdot \left(\frac{3}{2}\right)^{2013} \cdot \frac{4^{2x+2}}{(-2)^{4x}}
$$
Rezultat naj bo točen. | [] | Slovenia | 15. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | 12 | |
08kp | Problem:
Let $x, y, z$ be non-negative numbers such that $x + 2y + 3z = \frac{11}{12}$. Prove that
$$
63xy + 4x + 2yz + 6x + 3y + 4z + 72xyz \leq \frac{17}{18}
$$
When does equality hold? | [
"Solution:\nLet $x + 2y + 3z = \\frac{11}{12}$.\n\nUsing the notations $a = bx$, $b = 3$, and $c = 2$, from the equation we obtain the equality\n$$\n2a + 8b + 9c = 11.\n$$\n\nWe also have\n$$\na b c + a b + a c + b c + a + b + c + 1 = (a + 1)(b + 1)(c + 1) \\times \\frac{125}{18}.\n$$\n\nBy using the inequality\n$$... | JBMO | JBMO Shortlist | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | The expression is at most 17/18, with equality at x = 2/3, y = 1/12, z = 1/30. | |
08cp | Problem:
Alice e Barbara hanno inventato il seguente gioco. Hanno una griglia $1 \times 2018$, con le caselle numerate da $1$ a $2018$ da sinistra verso destra, e $2018$ tessere numerate anch'esse da $1$ a $2018$. La partita inizia con la griglia vuota, e le due giocatrici si alternano nel fare mosse; la giocatrice di... | [
"Solution:\n\n(a) Se $C_{1}$ è libera, $T_{1}$ può essere spostata a sinistra oppure collocata su $C_{1}$, a seconda che sia già sulla griglia o no; lo stesso ragionamento vale per $C_{2018}$, quindi d'ora in poi supporremo che $C_{1}$ e $C_{2018}$ siano entrambe occupate. Scelta una casella libera $C_{k}$, chiamia... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Alice | |
0kng | Problem:
Let $ABCD$ be a trapezoid with $AB \parallel CD$ and $AD = BD$. Let $M$ be the midpoint of $AB$, and let $P \neq C$ be the second intersection of the circumcircle of $\triangle BCD$ and the diagonal $AC$. Suppose that $BC = 27$, $CD = 25$, and $AP = 10$. If $MP = \frac{a}{b}$ for relatively prime positive int... | [
"Solution:\n\nAs $\\angle PBD = \\angle PCD = \\angle PAB$, $DB$ is tangent to $(ABP)$. As $DA = DB$, $DA$ is also tangent to $(ABP)$. Let $CB$ intersect $(ABP)$ again at $X \\neq B$; it follows that $XD$ is the $X$-symmedian of $\\triangle AXB$. As $\\angle AXC = \\angle DAB = \\angle ADC$, $X$ also lies on $(ACD)... | United States | HMMT Spring 2021 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations... | null | proof and answer | 2705 | |
008s | Abel has an infinite amount of type-A and -B pieces. Type-A pieces are composed of five unit squares and type-B pieces are composed of six unit squares, as shown below:
Abel wants to cover a $n \times n$ checkerboard, divided into $n^2$ unit squares, using these pieces without overlapping the... | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 9 | |
0fop | Problem:
Sean $x, y, z$ reales positivos tales que $x+y+z=3$. Halla el valor máximo alcanzado por
$$
\sqrt{x}+\sqrt{2 y+2}+\sqrt{3 z+6}
$$
¿Para qué valores de $x, y, z$ se alcanza dicho máximo? | [
"Solution:\nConsideremos los vectores $(\\sqrt{x}, \\sqrt{y+1}, \\sqrt{z+2})$ y $(\\sqrt{1}, \\sqrt{2}, \\sqrt{3})$, cuyas coordenadas son todas reales y positivas, cuyos módulos respectivos son $\\sqrt{x+y+z+3}=\\sqrt{6}$ y $\\sqrt{1+2+3}=\\sqrt{6}$, y cuyo producto escalar es la expresión cuyo máximo se pide hall... | Spain | null | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | Maximum value is 6, attained at x = y = z = 1. | |
08c7 | Problem:
Siano $a < b < c < d < e$ cinque numeri primi in progressione aritmetica di ragione 6 (ovvero $b = a + 6$, $c = b + 6$, $d = c + 6$ e $e = d + 6$). Quali delle seguenti affermazioni è falsa?
(A) $a + b + c + d + e$ è multiplo di 5.
(B) $abcde > 10^{4}$.
(C) $a + b + c + d + e$ è multiplo di 29.
(D) $abcde$ è... | [
"Solution:\n\nLa risposta è $\\mathbf{(C)}$. Osserviamo che l'unica progressione aritmetica a verificare la condizione è 5, 11, 17, 23, 29. Infatti i cinque numeri $a, b, c, d, e$, essendo in progressione aritmetica di ragione 6, lasciano cinque resti diversi nella divisione per 5: in particolare uno di essi è mult... | Italy | Gara di Febbraio | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | C | |
055g | Let $ABC$ be a triangle with $AB = AC \ne BC$ and let $I$ be its incenter. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$. | [
"*Solution.* See IMO 2016 shortlist, problem G4."
] | Estonia | IMO Team Selection Contest I | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
024v | Problem:
Se $x + y = 1$ e $x^{2} + y^{2} = 2$, calcule $x^{3} + y^{3}$. | [
"Solution:\n\nTemos a identidade do binômio, $(x + y)^{2} = x^{2} + 2 x y + y^{2}$, e a do trinômio, $(x + y)^{3} = x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3}$. Substituindo os valores de $x + y$ e $x^{2} + y^{2}$ na identidade do binômio, obtemos $1 = 2 + 2 x y$ e, portanto, $x y = -\\frac{1}{2}$. Assim, pela identidad... | Brazil | Nível 2 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 5/2 | |
0bfo | Given an integer $k \ge 2$, exhibit an infinite set $\mathcal{A}$ of sets of positive integers satisfying the two conditions below:
(a) The intersection of the members of every $k$-element subset of $\mathcal{A}$ is a singleton set; and
(b) The intersection of the members of every $(k+1)$-element subset of $\mathcal{A}... | [
"Biject the set of $k$-element sets of positive integers with the set of positive integers to label the former $S_1, S_2, \\dots, S_n, \\dots$. For every positive integer $m$, set $A_m = \\{n : m \\in S_n\\}$.\n\nIf $m$ and $m'$ are distinct positive integers, there exist distinct positive integers $n$ and $n'$ suc... | Romania | 64th NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
00ot | Let the sequence $(a_n)_{n \in N^*}$ be given with $a_1 = 2$ and $a_{n+1} = a_n^2 - a_n + 1$. Find the minimum real number $L$ such that for every $k \in N^*$
$$
\sum_{i=1}^{k} \frac{1}{a_i} < L.
$$ | [
"For every $n \\in N^*$ from the given recurrence relation we have\n$$\na_{n+1} - a_n = (a_n - 1)^2.\n$$\nSo, the sequence $(a_n)$ is increasing and therefore, since $a_1 = 2 > 1$, we have for every $n \\in N^*$\n$$\na_{n+1} > a_n > 1.\n$$\nFrom the given recurrence relation for every $n \\geq 2$ we obtain the equa... | Balkan Mathematical Olympiad | BMO 2010 Shortlist | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof and answer | 1 | |
01xt | The diagonals of the inscribed quadrilateral $ABCD$ intersect at the point $O$. The points $P$, $Q$, $R$ and $S$ are the feet of the perpendiculars from $O$ to the sides $AB$, $BC$, $CD$ and $DA$ respectively.
Prove the inequality $BD \ge SP + QR$. | [
"First we prove that the quadrilateral $SPQR$ is circumscribed and $O$ is the center of its incircle. The quadrilaterals $APOS$ and $BPOQ$ are cyclic since they have pairs of right angles, based on $AO$ and $BO$, respectively. In these circles $\\angle OPS = \\angle OAS$ and $\\angle OPQ = \\angle OBQ$. Wherein $\\... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0lex | Let $ABC$ be an acute triangle with fixed points $B, C$ and point $A$ moves on the big arc $BC$ of $(ABC)$ such that $AB \neq AC$. Incircle $(I)$ of triangle $ABC$ touches $BC$ at $D$. Let $I_a$ be the $A$-excenter of triangle $ABC$. $I_aD$ cuts $OI$ at $L$ and $E$ lies on $(I)$ such that $DE$ is parallel to $AI$.
a) ... | [
"a) Let $(I)$ touch $CA$, $AB$ at $U, V$; $I_b, I_c$ be the excenters of vertices $B, C$ in triangle $ABC$. Clearly, $I_aA$, $I_bB$ and $I_cC$ are three altitudes of triangle $I_aI_bI_c$. On the other hand, notice that $IA \\perp UV$, $IB \\perp VD$, $IC \\perp DU$ so two triangles $DUV$ and $I_aI_bI_c$ have corres... | Vietnam | VMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > A... | English | proof only | null | |
0ku3 | Problem:
The number
$$
316990099009901 = \frac{32016000000000001}{101}
$$
is the product of two distinct prime numbers. Compute the smaller of these two primes. | [
"Solution:\nLet $x = 2000$, so the numerator is\n$$\nx^{5} + x^{4} + 1 = (x^{2} + x + 1)(x^{3} - x + 1).\n$$\n(This latter factorization can be noted by the fact that plugging in $\\omega$ or $\\omega^{2}$ into $x^{5} + x^{4} + 1$ gives $0$.) Then $x^{2} + x + 1 = 4002001$ divides the numerator. However, it can eas... | United States | HMMT February 2023 | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 4002001 | |
08ky | Problem:
Prove that the equation $x^{2006} - 4 y^{2006} - 2006 = 4 y^{2007} + 2007 y$ has no solution in the set of the positive integers. | [
"Solution:\nWe assume the contrary is true. So there are $x$ and $y$ that satisfy the equation. Hence we have\n$$\n\\begin{gathered}\nx^{2006} = 4 y^{2007} + 4 y^{2006} + 2007 y + 2006 \\\\\nx^{2006} + 1 = 4 y^{2006}(y + 1) + 2007(y + 1) \\\\\nx^{2006} + 1 = \\left(4 y^{2006} + 2007\\right)(y + 1)\n\\end{gathered}\... | JBMO | 2007 Shortlist JBMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof only | null | |
0gv0 | For a positive integer $r \ge 2$ each positive integer is coloured into one of $r$ colours. Suppose that for each integer number $n$ and each pair $(a, b)$ of two colours $a$ and $b$ the difference between the number of divisors of $n$ coloured $a$ and the number of divisors of $n$ coloured $b$ is at most $1$. Find all... | [
"Answer: $r = 2, 3$.\n\nSuppose that $r \\geq 4$. Since there are infinitely many primes, there are same coloured primes $p$ and $q$. Since all divisors of $n = pq$ are $1, p, q, pq$, and hence no divisor of $n = pq$ is coloured into some of $r \\geq 4$ colours. Thus, the problem conditions are not satisfied.\n\nLe... | Turkey | Team Selection Test for IMO 2024 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | English | proof and answer | 2, 3 | |
0c4p | A triangle $ABC$ is such that $IE = GJ$, where $E$ and $G$ are the feet of the altitudes from $A$, respectively $B$, and $I$ and $J$ are the midpoints of the segments $[AH]$, respectively $[BC]$. Find the measure of $\angle ABC$.
Petru Braica | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate M... | English | proof and answer | 45° | |
0es0 | $0.125$ divided by $0.25$ is
(A) $0.75$ (B) $0.65$ (C) $0.55$ (D) $0.50$ (E) $0.40$ | [
"(D) $0.125$ is half of $0.25$, so result is $0.5$"
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | English | MCQ | D | |
0f9b | Problem:
Let $x_1$, $x_2$, ..., $x_n$ be positive reals with sum $1$. Show that
$$
\frac{x_1^2}{x_1 + x_2} + \frac{x_2^2}{x_2 + x_3} + \cdots + \frac{x_n^2}{x_n + x_1} \geq \frac{1}{2}.
$$ | [
"Solution:\n\nWe have\n$$\n\\sum \\frac{x_i^2}{x_i + x_{i+1}} - \\sum \\frac{x_{i+1}^2}{x_i + x_{i+1}} = \\sum (x_i - x_{i+1}) = 0.\n$$\nAlso,\n$$\n\\frac{x_i^2 + x_{i+1}^2}{x_i + x_{i+1}} \\geq \\frac{1}{2} (x_i + x_{i+1})\n$$\nby the QM-AM inequality. Summing over all $i$,\n$$\n\\sum \\frac{x_i^2}{x_i + x_{i+1}} ... | Soviet Union | 24th ASU | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
03xy | Given two integers $m, n$ greater than $1$, and integers $a_1 < a_2 < \cdots < a_m$, prove that there exists a set $T$ of integers with $|T| \le 1 + \frac{a_m - a_1}{2n + 1}$ such that each $a_i$ can be written as $a_i = t + s$ for some $t \in T$, and $s \in [-n, n]$. (Posed by Leng Gangsong) | [
"Write $a_1 = a$, $a_m = b$, $b - a = (2n + 1)q + r$, where $q, r \\in \\mathbb{Z}$ and $0 \\le r \\le 2n$. Take\n\n$$\nT = \\{a + n + (2n + 1)k \\mid k = 0, 1, \\dots, q\\}.\n$$\n\nThen $|T| = q + 1 \\le 1 + \\frac{b - a}{2n + 1}$. We have the set\n\n$$\nB = \\{t + s \\mid t \\in T, s = -n, -n + 1, \\dots, n\\} = ... | China | Chinese Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
006r | Beto eligió $101$ enteros positivos y los escribió en una línea. Demostrar que se pueden colocar paréntesis, signos de suma y signos de multiplicación entre los números de la lista de Beto de modo que la expresión que resulte tenga sentido, y al efectuar las operaciones indicadas se obtenga un número divisible por $16!... | [] | Argentina | Argentina 2009 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Spanish | proof only | null | |
04ff | Let $N$ be the foot of the altitude from $C$ to the hypotenuse $\overline{AB}$ of a right triangle $ABC$. The angle bisectors of the angles $\angle NCA$ and $\angle BCN$ intersect the segment $\overline{AB}$ in the points $K$ and $L$ respectively. If $S$ and $T$ are incircles of the triangles $BCN$ and $NCA$ respective... | [
"Denote $\\triangle CAB = \\alpha$ and $\\triangle ABC = \\beta$. We have $\\triangle BCN = \\alpha$ and $\\triangle ACN = \\beta$.\n\n\n\nThe lines $CS$ and $CT$ are angle bisectors of the angles $\\triangle BCN$ and $\\triangle ACN$, so the points $C, S, L$ and $C, T, K$ are collinear. We... | Croatia | Mathematica competitions in Croatia | [
"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 | |
01k9 | Let a six-digit number $A$ be divisible by $101$.
Prove that one can transpose two digits of $A$ so that the obtained number is also divisible by $101$. | [
"It is possible to transpose either $a_5$ and $a_1$ or $a_4$ and $a_0$ to obtain the number which is divisible by $101$."
] | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof only | null | |
0e4p | The progression $\{x_n\}$ is defined by the first two terms $x_1 = a$ and $x_2 = b$ and a recursive formula
$$
x_n = 3x_{n-1} - x_{n-2}
$$
for $n \ge 3$. Prove that there exist positive integers $a$ and $b$, such that the number $1 + x_n x_{n+1}$ is a perfect square for all $n$. | [] | Slovenia | Selection Examinations for the IMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
04sv | Nela and Jane choose positive integer $k$ and then play a game with a $9 \times 9$ table. Nela selects in every of her moves one empty unit square and she writes $0$ to it. Jane writes $1$ to some empty (unit) square in every her move. Furthermore, $k$ Jane's moves follow each Nela's move and Nela starts. If sum of num... | [
"Let us show at first that Jane wins for $k = 3$. Let us assume $3 \\times 3$ squares $A_1$, $A_2$ and $A_3$ (see the picture).\n\nWe will call the $3 \\times 3$ square *covered* if just one $1$ is in each its row and column. If Jane covers squares $A_1$, $A_2$ and $A_3$ without writing to other squares, she wins, ... | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | 3 | |
0cmj | A function
$$
f(x) = \cos x \cdot \cos \frac{x}{2} \cdot \cos \frac{x}{3} \cdots \cos \frac{x}{2009}
$$
is defined on the segment $[0, \frac{2009\pi}{2}]$. Find the number of points at which this function changes its sign. (B. Trushin) | [
"Ответ. 75 раз.\nОбозначим $n = 2009$. Рассмотрим функцию $\\cos \\frac{x}{k}$. Она меняет знак при\n$$\nx = k \\left( \\pi m + \\frac{\\pi}{2} \\right) = \\frac{k(2m + 1)\\pi}{2},\n$$\nгде $m$ — произвольное целое число. Значит, нулями функции $f(x)$ могут являться только точки $x_i = \\frac{\\pi i}{2}$, где $1 \\... | Russia | Russian mathematical olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English; Russian | proof and answer | 75 | |
01u4 | Given a $(2n + 1) \times (2n + 1)$ table, all its cells being white.
Per move it is allowed to change the color of any three consecutive cells in any row or in any column (white cell is replaced by black one and vice versa).
Find all possible values of $n$ such that one can obtain the chess co... | [
"**Answer:** $n \\equiv 1 \\pmod{3}$.\n\nWe say that a cell is black if this cell becomes black in the final coloring, and white if this cell becomes white in the final coloring. It is easy to see that we have $2n^2 + 2n + 1$ black and $2n^2 + 2n$ white cells in the chess coloring of the given table with black corn... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n ≡ 1 (mod 3) | |
0e7b | Problem:
Poišči vse funkcije $f: \mathbb{R} \setminus \{-1\} \rightarrow \mathbb{R}$, za katere velja
$$
f(x) + f(y) = (x + y + 2) f(x) f(y)
$$
za vse $x, y \in \mathbb{R} \setminus \{-1\}$. | [
"Solution:\n\nV funkcijsko enačbo vstavimo $x = y = 0$, da dobimo $f(0) = f(0)^2$. Torej je $f(0) = 0$ ali $f(0) = 1$.\n\nČe je $f(0) = 0$, v funkcijsko enačbo vstavimo $y = 0$ in dobimo $f(x) = 0$ za vsak $x$. Očitno ta funkcija zadošča pogojem naloge.\n\nČe pa je $f(0) = 1$, v funkcijsko enačbo spet vstavimo $y =... | Slovenia | 57. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 0 for all allowed x; f(x) = 1/(x+1) for all allowed x | |
0cyd | Let $n$ be a positive integer.
Find all real numbers $x_{1}, x_{2}, \ldots, x_{n}$ such that
$$
\prod_{k=1}^{n}\left(x_{k}^{2}+(k+2) x_{k}+k^{2}+k+1\right)=\left(\frac{3}{4}\right)^{n}(n!)^{2}
$$ | [
"We have $t^{2}+(k+2) t+k^{2}+k+1 \\geq \\frac{3}{4} k^{2}, \\forall t \\in \\mathbb{R}$.\nIndeed, this inequality is equivalent to\n$$\nt^{2}+(k+2) t+\\frac{1}{4} k^{2}+k+1 \\geq 0\n$$\nhence\n$$\nt^{2}+(k+2) t+\\left(\\frac{1}{2} k+1\\right)^{2} \\geq 0\n$$\nthat is\n$$\n\\left(t+\\frac{1}{2} k+1\\right)^{2} \\ge... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | x_k = -k/2 - 1 for k = 1, 2, ..., n | |
052u | Teacher tells Jüri two nonzero integers $a$ and $b$ such that $b$ is divisible by $a$. Jüri has to find a nonzero integer $c$ such that $c$ is divisible by $b$ and all solutions of the quadratic equation $ax^2 + bx + c = 0$ are integers. Can Jüri always solve the problem? | [
"By the conditions of the problem there is an integer $q$ such that $b = aq$. Let $c = -2aq^2$; then $c \\neq 0$ and $c$ is divisible by $b$. The quadratic equation $ax^2 + bx + c = 0$ or $ax^2 + aqx - 2aq^2 = 0$ has solutions $q$ and $-2q$."
] | Estonia | Open Contests | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | Yes; for b = a q, choose c = -2 a q^2. Then the roots are q and -2 q. | |
0dnm | Problem:
Доказати неједнакост
$$
\frac{x-y}{x y+2 y+1}+\frac{y-z}{y z+2 z+1}+\frac{z-x}{z x+2 x+1} \geqslant 0
$$
где су $x, y$ и $z$ ненегативни реални бројеви. | [
"Solution:\n\nОзначимо $a=\\frac{x-y}{x y+2 y+1}$, $b=\\frac{y-z}{y z+2 z+1}$ и $c=\\frac{z-x}{z x+2 x+1}$. Тада је $1+\\frac{1}{a}=\\frac{x y+x+y+1}{x-y}$ и одатле $\\frac{a}{a+1}=\\frac{x-y}{x y+x+y+1}=\\frac{1}{y+1}-\\frac{1}{x+1}$; аналогно је $\\frac{b}{b+1}=\\frac{1}{z+1}-\\frac{1}{y+1}$ и $\\frac{c}{c+1}=\\f... | Serbia | 9. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0439 | Given a convex $n$-gon ($n \ge 5$) $\Omega: P_1P_2...P_n$, of which no three diagonals are concurrent inside $\Omega$. Prove that one can choose a point inside every quadrilateral $P_iP_jP_kP_l$ ($1 \le i < j < k < l \le n$) and not on any diagonal of $\Omega$, such that the $C_n^4$ points obtained are distinct and the... | [
"To begin, notice that the diagonals of $\\Omega$ divide the polygon into $C_n^4 + C_{n-1}^2$ small regions. Every quadrilateral $P_iP_jP_kP_l$ ($1 \\le i < j < k < l \\le n$) has a 1-1 correspondence with the intersection of the diagonals $P_iP_k$ and $P_jP_l$, and additionally, every intersection is adjacent to f... | China | China National Team Selection Test | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0iqt | Problem:
Let $\mathcal{P}$ be a parabola, and let $V_{1}$ and $F_{1}$ be its vertex and focus, respectively. Let $A$ and $B$ be points on $\mathcal{P}$ so that $\angle A V_{1} B=90^{\circ}$. Let $\mathcal{Q}$ be the locus of the midpoint of $A B$. It turns out that $\mathcal{Q}$ is also a parabola, and let $V_{2}$ and ... | [
"Solution:\n$\\boxed{\\dfrac{7}{8}}$\n\nSince all parabolas are similar, we may assume that $\\mathcal{P}$ is the curve $y=x^{2}$. Then, if $A=\\left(a, a^{2}\\right)$ and $B=\\left(b, b^{2}\\right)$, the condition that $\\angle A V_{1} B=90^{\\circ}$ gives $a b+a^{2} b^{2}=0$, or $a b=-1$. Then, the midpoint of $A... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 7/8 | |
06zm | Problem:
$a_{n}$ is the last digit of $1+2+\ldots+n$. Find $a_{1}+a_{2}+\ldots+a_{1992}$. | [
"Solution:\n\nIt is easy to compile the following table, from which we see that $a_{n}$ is periodic with period $20$, and indeed the sum for each decade (from $0$ to $9$) is $35$. Thus the sum for $1992$ is $199 \\cdot 35 + 5 + 6 + 8 = 6984$.\n\n| $n$ | | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 ... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 6984 | |
00t5 | Consider an integer $n \ge 2$ and an odd prime $p$. Let $U$ be the set of all positive integers (strictly) less than $p^n$ that are not divisible by $p$, and let $N$ be the number of elements of $U$. Does there exist a permutation $a_1, a_2, \dots, a_N$ of the numbers in $U$ such that the sum $\sum_{k=1}^N a_k a_{k+1}$... | [
"The answer is in the affirmative. Letting $\\equiv$ denote congruence modulo $p^n$ throughout the argument, we will show that there exists a permutation $a_1, a_2, \\dots, a_N$ of the numbers in $U$ such that $\\sum_{k=1}^N a_k a_{k+1} \\equiv p^{n-1}$.\n\nLet $m = p^{n-1}-1$, so $N = p^{n-1}(p-1) = (m+1)(p-1)$, a... | Balkan Mathematical Olympiad | BMO Short List | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | English | proof and answer | Yes | |
04fj | Prove that there are infinitely many positive integers $n$ such that $n$ has more than two distinct prime divisors and $2^n - 8$ is divisible by $n$. | [
"We will prove that for all positive integers of the form $n = 2^{2p} - 1$ (where $p > 3$ is a prime number) $n \\mid 2^n - 8$ and $n$ has at least three distinct prime divisors.\n\nSince $2^n - 8 = 8(2^{n-3} - 1)$, to prove $n \\mid 2^n - 8$ it is sufficient to prove\n$$\n2^p \\mid n - 3.\n$$\nNamely, in this case... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0ag9 | Solve the equation
$$x^3 + 2y^3 - 4x - 5y + z^2 = 2012,$$
in the set of whole numbers. | [
"It is easy to show that for every whole number $a$ holds $3|a^3 - a$, and that the square of a number modulo $3$ can be $0$ or $1$.\n\nNow the given equation is equivalent to $x^3 - x + 2(y^3 - y) - 3(x - y) + z^2 = 2012$. From the above concluded and from the fact that $2012 \\equiv 2 \\pmod{3}$ we get that $z^2 ... | North Macedonia | Macedonian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof and answer | No solutions in whole numbers | |
0ebk | Find all polynomials $p$ of odd degree with real coefficients which satisfy
$$
p(p(x)) \le (p(x))^3
$$
for all $x \in \mathbb{R}$ and have the coefficient at $x^2$ equal 0. | [
"Let $p(x) = a_nx^n + a_{n-1}x^{n-1} + \\dots + a_0$, where $a_n \\neq 0$. The leading term of the polynomial $p(p(x))$ equals $a_n(a_nx^n)^n = a_n^{n+1}x^{n^2}$, and the leading term of the polynomial $p(x)^3$ equals $(a_nx^n)^3 = a_n^3x^{3n}$, they are both of odd degree. Since the polynomial $p(x)^3 - p(p(x))$ h... | Slovenia | National Math Olympiad 2015 – Final Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | p(x) = x^3 + b with b ≤ 0 | |
0k8g | Problem:
Two players play a game, starting with a pile of $N$ tokens. On each player's turn, they must remove $2^{n}$ tokens from the pile for some nonnegative integer $n$. If a player cannot make a move, they lose. For how many $N$ between 1 and 2019 (inclusive) does the first player have a winning strategy? | [
"Solution:\n\nThe first player has a winning strategy if and only if $N$ is not a multiple of 3. We show this by induction on $N$. If $N=0$, then the first player loses.\n\nIf $N$ is a multiple of 3, then $N-2^{n}$ is never a multiple of 3 for any $n$, so the second player has a winning strategy.\n\nIf $N$ is not a... | United States | HMMT November 2019 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 1346 | |
07zy | Problem:
Determinare l'area della parte di piano definita da
$$
\left\{\begin{array}{l}
x^{2}+y^{2}-4 y \leq 0 \\
x^{2}-3 x+2 \leq 0
\end{array}\right.
$$
(A) $\frac{2 \pi}{3}-2$
(B) $\pi-\sqrt{3}$
(C) $\frac{\pi}{3}-\frac{\sqrt{3}}{2}$
(D) $\frac{4 \pi}{3}-\sqrt{3}$
(E) nessuna delle precedenti. | [] | Italy | Italian Mathematical Olympiad - Febbraio Round | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | MCQ | D | |
0crv | Числа $x$, $y$ и $z$ таковы, что все три числа $x + yz$, $y + zx$ и $z + xy$ рациональны, а $x^2 + y^2 = 1$. Докажите, что число $xyz^2$ также рационально. | [
"Из условия следует, что число $(x + yz)(y + zx) = xy + (x^2 + y^2)z + xyz^2 = (xy + z) + xyz^2$ рационально. Поскольку число $xy + z$ также рационально по условию, то и число $xyz^2 = (x + yz)(y + zx) - (xy + z)$ также рационально."
] | Russia | XL Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
00o7 | Determine all natural numbers $n \ge 2$ with the property that there are two permutations $(a_1, a_2, \dots, a_n)$ and $(b_1, b_2, \dots, b_n)$ of the numbers $1, 2, \dots, n$ such that $(a_1 + b_1, a_2 + b_2, \dots, a_n + b_n)$ are consecutive natural numbers. | [
"The permutations exist if and only if $n$ is odd.\n\nWe have\n$$\n(a_1 + b_1) + (a_2 + b_2) + \\dots + (a_n + b_n) = 2(1 + 2 + \\dots + n) = n(n+1).\n$$\nOn the other hand, there is a natural number $N$ such that\n$$\na_1 + b_1 = N,\\ a_2 + b_2 = N + 1,\\ \\dots,\\ a_n + b_n = N + n - 1\n$$\nand therefore\n$$\n(a_... | Austria | AUT_ABooklet_2023 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | All odd n | |
05vc | Problem:
Soit $ABC$ un triangle dont les angles sont aigus et soit $D$ un point situé à l'intérieur du triangle $ABC$. Les droites $(AD)$ et $(BD)$ recoupent le cercle circonscrit au triangle $ABC$ respectivement en les points $A_{1}$ et $B_{1}$. Le cercle circonscrit au triangle $B_{1}DA$ recoupe la droite $(AC)$ au ... | [
"Solution:\n\nUne chasse aux angles de droites indique que\n$$\n(QD, CB) = (QD, QB) = (A_{1}D, A_{1}B) = (A_{1}A, A_{1}B) = (CA, CB),\n$$\nce qui signifie que les droites $(QD)$ et $(AC)$ sont parallèles l'une à l'autre. On démontre de même que\n$$\n(PD, CA) = (PD, PA) = (B_{1}D, B_{1}A) = (B_{1}B, B_{1}A) = (CB, C... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
034g | Problem:
Find all pairs of positive integers $(m, n)$, $m > n$, such that
$$
\left[m^2 + m n, m n - n^2\right] + [m - n, m n] = 2^{2005}
$$
where $[a, b]$ denotes the least common multiple of $a$ and $b$. | [
"Solution:\nThe left hand side of the given equation is a multiple of $m$, $n$ and $m-n$. Therefore $m = 2^a$, $n = 2^b$ and $m - n = 2^c$ for some nonnegative integers $a$, $b$ and $c$, where $a > b$. It is obvious that $2^b (2^{a-b} - 1) = 2^c$, whence $a - b = 1$.\n\nPlugging $b = a - 1$ in the given equation we... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | m = 2^{1002}, n = 2^{1001} | |
0bpp | Problem:
a) Arătaţi că numărul $\log_{2015} 2016$ este număr iraţional;
b) Comparaţi numerele $\log_{5} 6$ şi $\log_{6} 7$;
c) Calculaţi $E=\lg^{3} 5+\lg^{3} 20+\lg 8 \cdot \lg (0,25)$. | [] | Romania | Olimpiada Națională de Matematică | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | a) irrational; b) log base 5 of 6 is greater than log base 6 of 7; c) 2 | |
0j3u | Problem:
Rosencrantz and Guildenstern are playing a game where they repeatedly flip coins. Rosencrantz wins if $1$ heads followed by $2009$ tails appears. Guildenstern wins if $2010$ heads come in a row. They will flip coins until someone wins. What is the probability that Rosencrantz wins? | [
"Solution:\n\nAnswer: $\\frac{2^{2009}-1}{3 \\cdot 2^{2008}-1}$\n\nWe can assume the first throw is heads (because neither player can win starting from a string of only tails). Let $x$ be the probability that Rosencrantz wins. Let $y$ be the probability that Rosencrantz wins after $HT$.\n\nWhenever there is a strin... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | (2^2009 - 1)/(3 * 2^2008 - 1) | |
08bi | Problem:
Il PIN del telefono di Eugenia è composto da 4 cifre; Eugenia ricorda soltanto che la prima è compresa fra 0 e 6, la seconda fra 0 e 3, la terza fra 0 e 4, la quarta fra 0 e 2, e che la somma delle quattro cifre è almeno 8 (per esempio, il PIN potrebbe essere 3330). Quanti sono i codici compatibili con i rico... | [
"Solution:\n\nLa risposta è 210. Consideriamo tutti i codici di 4 cifre che si possono formare con cifre appartenenti agli intervalli corretti (senza restrizioni sulla somma); queste sono in numero di $7 \\cdot 4 \\cdot 5 \\cdot 3 = 420$. Mostriamo adesso che esattamente metà di questi codici rispetta anche la cond... | Italy | Progetto Olimpiadi della Matematica | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 210 | |
06f4 | Let $ABC$ be a triangle with $AB = c$, $BC = a$ and $CA = b$. Show that
$$
\frac{\cos A}{a^3} + \frac{\cos B}{b^3} + \frac{\cos C}{c^3} \ge \frac{3}{2abc}.
$$ | [
"By the cosine formula, we have\n$$\n\\begin{aligned}\n& \\frac{\\cos A}{a^3} + \\frac{\\cos B}{b^3} + \\frac{\\cos C}{c^3} \\ge \\frac{3}{2abc} \\\\\n\\Leftrightarrow \\quad & \\frac{b^2 + c^2 - a^2}{2a^3bc} + \\frac{c^2 + a^2 - b^2}{2ab^3c} + \\frac{a^2 + b^2 - c^2}{2abc^3} \\ge \\frac{3}{2abc} \\\\\n\\Leftrighta... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0e3a | Let $n$ be a positive integer. There are $2n + 2$ points in a plane such that no three of the points lie on the same line. A line in the plane is a *divider*, if two of the given points lie on this line and there are exactly $n$ points on each of the two sides. Find the greatest possible $m$ for which there are always ... | [
"We will show that there exist at least $n + 1$ divisor lines. First, let us give an example where there are exactly $n + 1$ of them. For that the given points should be the vertices of a regular $(2n+2)$-gon. Denote them by $A_1, A_2, \\dots, A_{2n+2}$. For each $1 \\le i \\le n+1$ the line $A_iA_{i+n+1}$ is obvio... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | n + 1 | |
0axg | Problem:
Each of the numbers in the set $A=\{1,2, \ldots, 2017\}$ is colored either red or white. Prove that for $n \geq 18$, there exists a coloring of the numbers in $A$ such that any of its $n$-term arithmetic sequences contains both colors. Inspired by Problem 891 from Putnam and Beyond, T. Andreescu and R. Gelca,... | [
"Solution:\n\nIt suffices to show that for $n \\geq 18$, the total number of colorings (without restriction) exceeds those that make some $n$-term arithmetic sequence monochromatic.\n\nThere are $2^{2017}$ colorings of a set with 2017 elements. The number of colorings that make a fixed $n$-term sequence monochromat... | Philippines | 19th Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0hyu | Problem:
If $a @ b = \frac{a^{3} - b^{3}}{a - b}$, for how many real values of $a$ does $a @ 1 = 0$? | [
"Solution:\n\nIf $\\frac{a^{3} - 1}{a - 1} = 0$, then $a^{3} - 1 = 0$, or $(a - 1)(a^{2} + a + 1) = 0$.\n\nThus $a = 1$, which is an extraneous solution since that makes the denominator of the original expression $0$, or $a$ is a root of $a^{2} + a + 1$.\n\nBut this quadratic has no real roots, in particular its ro... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 0 | |
0fil | Problem:
Se divide el plano en un número finito de regiones $n$ mediante tres familias de rectas paralelas. No hay tres rectas que pasen por el mismo punto. ¿Cuál es el mínimo número de rectas necesarias para que $n>1999$? | [
"Solution:\n\nSupongamos que hay $x$ rectas en la primera familia, $y$ en la segunda y $z$ en la tercera. Las $x$ rectas de la primera familia determinan $x+1$ regiones. La primera recta de la segunda familia determina en el plano $2(x+1)$ regiones, las dos primeras $3(x+1)$, etc., hasta las $y$-ésima, que determin... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | 76 | |
04a6 | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x^2 + f(y)) = y - x^2$ for all $x, y \in \mathbb{R}$. | [
"Taking $x = 0$ in the given equation we get $f(f(y)) = y$ for every $y \\in \\mathbb{R}$.\nTherefore, we have\n$$\nf(y - x^2) = f(f(x^2 + f(y))) = x^2 + f(y). \\qquad (3.1)\n$$\nfor every $x, y \\in \\mathbb{R}$.\nTaking $y = x^2$ in (3.1) we get $f(0) = x^2 + f(x^2)$, that is\n$$\nf(x^2) = -x^2 + f(0),\n$$\nwhile... | Croatia | CroatianCompetitions2011 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | All functions f(x) = -x + c for any real constant c. | |
07e8 | Let $M$, $N$ and $P$ be the midpoints of the sides $BC$, $AC$ and $AB$ in triangle $ABC$, respectively. Suppose that the point $K$ lies on the segment $NP$ such that $AK$ is the angle bisector of $\angle BKC$. The lines $BK$ and $MN$ meet at a point $E$ and the lines $CK$ and $MP$ meet at a point $F$. Let $H$ be the fo... | [
"By Pappus's hexagon theorem, $F$, $A$ and $E$ are collinear. Therefore\n$$\n\\frac{EA}{AF} = \\frac{EN}{NM} = \\frac{EK}{BK} \\Rightarrow AK \\parallel BF \\Rightarrow \\angle BFK = \\angle AKF = \\angle AKE = \\angle FBK\n$$\nThus $KF = KB$ and similarly $KE = KC$, so $\\angle EKF \\sim \\angle CKB$. Hence\n$\\an... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Pappus theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscell... | English | proof only | null | |
0ejq | Problem:
Reši enačbo: $4x^{4} + 2x^{3} - 4x^{2} - x = -1$. Izračunaj razliko kvadratov vsote racionalnih rešitev in vsote iracionalnih rešitev enačbe. | [
"Solution:\n\nEnačbo lahko rešimo s pomočjo uporabe Hornerjevega algoritma in dobimo rešitve $x_{1} = -1$, $x_{2} = \\frac{1}{2}$, $x_{3} = \\frac{\\sqrt{2}}{2}$, $x_{4} = -\\frac{\\sqrt{2}}{2}$. Ugotovimo, da sta prvi dve rešitvi racionalni, drugi dve pa iracionalni.\n\nIzračunamo kvadrat vsote prvih dveh rešitev:... | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 1/4 | |
0gis | 已知在 $\triangle ABC$ 中 $\angle A = 62^\circ$, $\angle B = 64^\circ$ 且 $D, E$ 分別為 $C, B$ 在 $AB, AC$ 上的垂足。假設 $P, Q$ 為 $DE$ 與 $\triangle ABC$ 外接圓的交點且 $P, Q$ 分別在 $\widehat{AB}$ 與 $\widehat{AC}$ 上，則 $\angle QAE - \angle PAD = \underbrace{(8\ 9)}_{\text{。}}$
Let $ABC$ be a triangle with $\angle A = 62^\circ$, $\angle B = 64... | [
"答案：$10^\\circ$\n\n解答：可知 $DBCE$ 共圓，所以 $\\angle ADE = 54^\\circ$, $\\angle AED = 64^\\circ$。因此有 $\\vec{AP} + \\vec{PB} = \\vec{AQ} + \\vec{PB}$ 所以 $\\angle APD = \\angle AQE$ 容易得知 $\\angle QAE - \\angle PAD = 10^\\circ$"
] | Taiwan | APMO Taiwan Preliminary Round 1 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | Chinese; English | proof and answer | 10° | |
00sc | Find all pairs of positive integers $(x, y)$, such that $x^2$ is divisible by $2xy^2 - y^3 + 1$. | [
"If $y=1$, then $2x \\mid x^2 \\Leftrightarrow x=2n$, $n \\in \\mathbb{N}$. So, the pairs $(x,y)=(2n, 1)$, $n \\in \\mathbb{N}$ satisfy the required divisibility.\n\nLet $y > 1$ such that $x^2$ is divisible by $2xy^2 - y^3 + 1$. There exists $m \\in \\mathbb{N}$ such that\n$$\nx^2 = m(2xy^2 - y^3 + 1), \\text{ e.t.... | Balkan Mathematical Olympiad | BMO 2017 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | All pairs are (2k, 1), (k, 2k), and (8k^4 − k, 2k) for positive integers k. | |
0ihl | Problem:
Octagon $A B C D E F G H$ is equiangular. Given that $A B=1$, $B C=2$, $C D=3$, $D E=4$, and $E F=F G=2$, compute the perimeter of the octagon. | [
"Solution:\n\nExtend sides $A B$, $C D$, $E F$, $G H$ to form a rectangle: let $X$ be the intersection of lines $G H$ and $A B$; $Y$ that of $A B$ and $C D$; $Z$ that of $C D$ and $E F$; and $W$ that of $E F$ and $G H$.\n\nAs $B C=2$, we have $B Y=Y C=\\sqrt{2}$. As $D E=4$, we have $D Z=Z E=2\\sqrt{2}$. As $F G=2$... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 20 + √2 | |
0j1a | Problem:
Triangle $A B C$ has $A B=5$, $B C=7$, and $C A=8$. New lines not containing but parallel to $A B$, $B C$, and $C A$ are drawn tangent to the incircle of $A B C$. What is the area of the hexagon formed by the sides of the original triangle and the newly drawn lines? | [
"Solution:\n\nAnswer: $\\frac{31}{5} \\sqrt{3}$\n\n\n\nFrom the law of cosines we compute $\\measuredangle A=\\cos^{-1}\\left(\\frac{5^{2}+8^{2}-7^{2}}{2(5)(8)}\\right)=60^{\\circ}$. Using brackets to denote the area of a region, we find that\n$$\n[ABC]=\\frac{1}{2} AB \\cdot AC \\cdot \\si... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | 31/5 * sqrt(3) | |
03cy | Consider the equation $(x^2 - 8x + 7)^2 + (3m - 1)(x^2 - 8x + 7) + 2m^2 - m = 0$, where $m$ is a real parameter.
a) Find the number of solutions of the equation for $m = 6$.
b) Find the values of $m$ for which the equation has exactly three distinct real roots. | [] | Bulgaria | Bulgaria 2022 | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a) 2; b) m = 5 | |
0kum | Problem:
An equilateral triangle lies in the Cartesian plane such that the $x$-coordinates of its vertices are pairwise distinct and all satisfy the equation $x^{3}-9 x^{2}+10 x+5=0$. Compute the side length of the triangle. | [
"Solution:\n\nLet three points be $A$, $B$, and $C$ with $x$-coordinates $a$, $b$, and $c$, respectively. Let the circumcircle of $\\triangle ABC$ meet the line $y = b$ at point $P$. Then, we have $\\angle BPC = 60^{\\circ} \\Longrightarrow PC = \\frac{2}{\\sqrt{3}}(c-b)$. Similarly, $AP = \\frac{2}{\\sqrt{3}}(b-a)... | United States | HMMT February 2023 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 2√17 | |
0k36 | Problem:
In triangle $ABC$, $AB = 6$, $BC = 7$ and $CA = 8$. Let $D$, $E$, $F$ be the midpoints of sides $BC$, $AC$, $AB$, respectively. Also let $O_{A}$, $O_{B}$, $O_{C}$ be the circumcenters of triangles $AFD$, $BDE$, and $CEF$, respectively. Find the area of triangle $O_{A} O_{B} O_{C}$. | [
"Solution:\n\nLet $AB = z$, $BC = x$, $CA = y$. Let $X$, $Y$, $Z$, $O$, $N$ be the circumcenter of $AEF$, $BFD$, $CDE$, $ABC$, $DEF$ respectively. Note that $N$ is the nine-point center of $ABC$, and $X$, $Y$, $Z$ are the midpoints of $OA$, $OB$, $OC$ respectively, and thus $XYZ$ is the image of homothety of $ABC$ ... | United States | HMMT February 2018 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 21 sqrt(15)/16 | |
01is | Fifty marbles are lying in a heap. Alfred and Bodil take turns in removing a positive number of marbles, Alfred starts and at each step the number of marbles removed needs to be a prime or a square. The winner is the one who empties the heap. Who wins if both play in an optimal way? | [
"We make a table inductively that shows whether the player, whose turn it is if there are $n$ marbles left, wins (marked with a W) or loses (marked with an L).\n\n| $n$ | 1<br>W<br>-1 | 2<br>W<br>-2 | 3<br>W<br>-3 | 4<br>W<br>-4 | 5<br>W<br>-5 | 6<br>L | 7<br>W<br>-1 | 8<br>W<br>-2 | 9<br>W<br>-3 | 10<br>W<br>-4 |\... | Baltic Way | Baltic Way 2023 Shortlist | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Bodil | |
0221 | Problem:
Determine todas as soluções da equação $\sqrt{x} = x - 2$. | [
"Solution:\n\nElevando ambos os membros da equação ao quadrado, obtemos $x = x^{2} - 4x + 4$, que é equivalente a $x^{2} - 5x + 4 = 0$. As raízes dessa equação do segundo grau são $x = 1$ e $x = 4$. Entretanto, quando substituímos $x = 1$ na equação original $\\sqrt{x} = x - 2$ obtemos $\\sqrt{1} = -1$, que é falso... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | x = 4 | |
049m | Let $\triangle ABC$ be a triangle. Angle bisector of $\angle CAB$ intersects the segment $\overline{BC}$ at $D$, and angle bisector of $\angle ABC$ intersects the segment $\overline{AC}$ at $E$. If $\angle ACB \ge 60^\circ$, prove that $|AE| + |BD| \le |AB|$. | [] | Croatia | Hrvatska 2011 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof only | null | |
00h5 | For a positive integer $m$ denote by $S(m)$ and $P(m)$ the sum and product, respectively, of the digits of $m$. Show that for each positive integer $n$, there exist positive integers $a_{1}, a_{2}, \ldots, a_{n}$ satisfying the following conditions:
$$
S\left(a_{1}\right)<S\left(a_{2}\right)<\cdots<S\left(a_{n}\right) ... | [
"Let $k$ be a sufficiently large positive integer. Choose for each $i=2,3, \\ldots, n$, $a_{i}$ to be a positive integer among whose digits the number $2$ appears exactly $k+i-2$ times and the number $1$ appears exactly $2^{k+i-1}-2(k+i-2)$ times, and nothing else. Then, we have $S\\left(a_{i}\\right)=2^{k+i-1}$ an... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0661 | We consider the set of four digit positive integers $x = \overline{\alpha\beta\gamma\delta}$ with digits different than zero and pairwise different. We also consider the integers $y = \overline{\delta\gamma\beta\alpha}$ and we suppose that $x > y$. Find the greatest and the lowest value of the difference $x - y$, as we... | [
"We consider the decimal representation of the integers\n$$\n\\begin{aligned}\nx - y &= 1000\\alpha + 100\\beta + 10\\gamma + \\delta - 1000\\delta - 100\\gamma - 10\\beta - \\alpha \\\\\n&= 999(\\alpha - \\delta) + 90(\\beta - \\gamma) = 9(111(\\alpha - \\delta) + 10(\\beta - \\gamma)).\n\\end{aligned}\n$$\nTheref... | Greece | 28th Hellenic Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | Greatest difference: 8532, achieved by x = 9821 and y = 1289. Lowest difference: 279, achieved by the pairs (x, y) = (3192, 2913), (4193, 3914), (5194, 4915), (6195, 5916), (7196, 6917), (8197, 7918). | |
0ab7 | One square meter of tin costs 35 denars. How much money is needed to buy tin for 50 opened pots (without roof) with polyhedral shape with dimensions: length 20 cm, width 2 dm and height 4 dm? | [
"For one such pot we need $2 \\cdot 2 + 4 \\cdot 2 \\cdot 4 = 36\\,dm^2$ of tin. For 50 we need $50 \\cdot 36 = 1800\\,dm^2 = 18m^2$ of tin. So we need $18 \\cdot 35 = 630$ denars."
] | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Solid Geometry > Surface Area"
] | null | final answer only | 630 denars | |
070u | Problem:
$O$ is the circumcenter of the acute-angled triangle $ABC$. The altitudes are $AD$, $BE$ and $CF$. The line $EF$ cuts the circumcircle at $P$ and $Q$. Show that $OA$ is perpendicular to $PQ$. If $M$ is the midpoint of $BC$, show that $AP^{2}=2 AD \cdot OM$. | [
"Solution:\n\n\n\nLet $OA$ and $PQ$ meet at $T$. $\\angle AEH=\\angle AFH=90^{\\circ}$, so $AEHF$ is cyclic, so $\\angle AFT=\\angle AFE$ (same angle) $=\\angle AHE=90^{\\circ}-\\angle HAE=90^{\\circ}-\\angle DAC$ (same angle) $=\\angle C$. But $\\angle TAF=\\angle OAF$ (same angle) $=90^{\... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles >... | null | proof only | null | |
0f1b | Problem:
$S$ is a horizontal strip in the plane. $n$ lines are drawn so that no three are collinear and every pair intersects within the strip. A path starts at the bottom border of the strip and consists of a sequence of segments from the $n$ lines. The path must change line at each intersection and must always move ... | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0fjz | Problem:
Para cuatro puntos no coplanarios, un plano ecualizador es un plano tal que las distancias respectivas de cada uno de los puntos a ese plano son todas iguales. Dado un conjunto de cuatro puntos no coplanarios, ¿cuántos planos ecualizadores hay? | [
"Solution:\nSean $A$, $B$, $C$ y $D$ cuatro puntos no coplanarios. Sea $p$ un plano ecualizador de esos puntos y examinemos las posibilidades:\n\ni) Si $A$, $B$, $C$ y $D$ están en el mismo semiespacio en que $p$ divide al espacio, está claro que uno de los dos planos paralelos a $p$, a la misma distancia que están... | Spain | Spanish Mathematical Olympiad - Local Stage | [
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 7 | |
01y5 | Peter and Andrey play the game on the $n \times 1$ board, making moves alternate. Peter starts, and on his turn he places «+» to any empty cell. Andrey on his turn places «-» to any empty cell. The game is finished when all cells are filled. Peter's prize equals to the greatest number $k$ such that for each $\ell$ from... | [
"Answer: $n + \\frac{-1 + (-1)^{n+1}}{2}$."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | n + (-1 + (-1)^{n+1})/2 | |
001c | En un triángulo acutángulo $ABC$, los puntos $H$, $G$ y $M$ se encuentran sobre el lado $BC$, de modo que $AH$, $AG$ y $AM$ son altura, bisectriz y mediana del triángulo, respectivamente. Se sabe que $HG = GM$, $AB = 10$ y $AC = 14$. Determinar el área del triángulo $ABC$. | [] | Argentina | XIV Olimpiada Matemática de Países del Cono Sur | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | español | proof and answer | 12√34 | |
0ge0 | 居家隔離的日子太無聊了，於是老趙跟勳勳決定玩一個遊戲。勳勳先秘密寫下一個多項式 $f(x)$，滿足：
(a) 對於任意整數 $n$，$f(n)$ 是個整數；
(b) $f(x)$ 的次數小於 187。
老趙只知道 $f(x)$ 滿足性質 (a) 與 (b)，但不知道 $f(x)$。接下來，老趙每一回合可以從集合 $\{1, 2, \dots, 187\}$ 裡面選一個數字 $k$，然後勳勳會告訴老趙 $f(k)$ 的值。試求最小的正整數 $N$，使得老趙總是能在 $N$ 個回合內確定 $f(0)$ 的奇偶性。
Alice and Bob are stuck in quarantine, so they decide to play... | [
"最小的 $N = 63$。\n令 $n = 187$，且 $187$ 的二進制表示法是 $10111011$。首先我們注意到 $n$ 次差分給出 $\\sum_{i=0}^{n} (-1)^i \\binom{n}{i} f(i) = 0$，由盧卡斯定理我們知道 $\\binom{n}{i}$ 是奇數當且僅當 $i$ 的二進制表示法中 1 的位置是 $n$ 的二進制表示法中 1 的位置的子集，所以\n$$\n0 \\equiv \\sum_{B \\subseteq A} f\\left(\\sum_{i \\in B} 2^i\\right) \\pmod{2}\n$$\n其中 $A$ 是 $n$ 的二進制表示法中 1 ... | Taiwan | 2020 Taiwan IMO 2J | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations... | null | proof and answer | 63 | |
07e1 | Let $n$ be a positive integer. Two players are playing a game in a field of a shape of a $2n \times 2n$ grid. Initially, the first player is on the top right point and second player is on the bottom left point. Each player, in his turn, moves to an adjacent point, by passing on an edge, such that none of the players ha... | [
"Let $A$ be the first player and $B$ the second one. Since $A$ plays first, they can reach point $C$ sooner, making $n$ moves downwards and then $n$ moves to the left. Then they can get back to their start point, making $n$ moves upwards and $n$ moves to the right. In this case, the first player wins the upper righ... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | n^2 | |
0foi | Let $a$, $b$, $c$, $d$ be positive real numbers. Prove that
$$
\sum_{cyclic} \frac{a - \sqrt[3]{bcd}}{a + 3(b + c + d)} \ge 0
$$ | [
"On account of AM-GM inequality, we have $\\sqrt[3]{bcd} \\leq \\frac{b + c + d}{3}$ and\n$$\n\\begin{aligned}\n\\frac{a - \\sqrt[3]{bcd}}{a + 3(b + c + d)} &\\ge \\frac{1}{3} \\left( \\frac{3a - (b + c + d)}{a + 3(b + c + d)} \\right) = \\frac{1}{9} \\left( \\frac{10a - (a + 3(b + c + d))}{a + 3(b + c + d)} \\righ... | Spain | International Mathematical Arhimede Contest | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | Spanish | proof only | null | |
0g7f | 設實數 $a_1, a_2, \dots, a_n$ ($n \ge 2$) 滿足 $-1 < a_1, a_2, \dots, a_n < 1$, 且 $\sum_{k=1}^n a_k^2 \ge 1$。
試證：
$$
\sum_{i<j} \left( \frac{1}{1-a_i^2} + \frac{1}{1-a_j^2} - \frac{1}{1-a_i a_j} \right) \ge \frac{n^2}{2}
$$ | [
"利用 $|x| < 1$ 時，$\\sum_{k=0}^{\\infty} x^k = \\frac{1}{1-x}$。我們可以推得\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} \\frac{1}{1-a_k^2} &= \\sum_{k=0}^{\\infty} \\left( \\sum_{i=1}^{n} a_i^{2k} \\right) = \\sum_{k=0}^{\\infty} \\left( \\sum_{i=1}^{n} (a_i^k)^2 \\right) \\\\\n&\\ge \\frac{2}{n-1} \\sum_{k=0}^{\\infty} \\left(... | Taiwan | 二〇一三數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0a2t | We call a positive integer *quadratic* if every two adjacent digits form a square number. For example, the number $364$ is quadratic because both $36$ and $64$ are squares.
How many digits does the largest quadratic number have?
A) $2$ B) $3$ C) $4$ D) $5$ E) $6$ | [] | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Other"
] | null | MCQ | D | |
0jpd | Problem:
Find the largest integer $n$ such that the following holds: there exists a set of $n$ points in the plane such that, for any choice of three of them, some two are unit distance apart. | [
"Solution:\nAnswer: $7$\n\nWe can obtain $n=7$ in the following way: Consider a rhombus $ABCD$ made up of two equilateral triangles of side length $1$, where $\\angle DAB=60^{\\circ}$. Rotate the rhombus clockwise about $A$ to obtain a new rhombus $AB' C' D'$ such that $DD'=1$. Then one can verify that the seven po... | United States | HMMT November 2015 | [
"Discrete Mathematics > Graph Theory",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 7 | |
01w8 | The cyclist went from $A$ to $B$, and after 15 minutes a second cyclist followed him. 27 minutes after the second cyclist's start, the motorcyclist went from $B$ to $A$. All three of them met exactly midway between $A$ and $B$. The motorcyclist turned around when reached $A$, and the second cyclist turned around when r... | [
"Answer: 36 minutes."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | final answer only | 36 minutes |
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