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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
08zs | Let $a$, $b$ and $c$ be positive integers satisfying
$$
\frac{(ab-1)(ac-1)}{bc} = 2023, \quad \text{and} \quad b \le c.
$$
Find all possible values for $c$. | [
"$82$, $167$, $1034$\n\nWe have\n$$\n\\frac{(ab-1)(ac-1)}{bc} = \\left(a - \\frac{1}{b}\\right) \\left(a - \\frac{1}{c}\\right).\n$$\nSince we have $0 \\le a - 1 \\le a - \\frac{1}{b} < a$ and $0 \\le a - 1 \\le a - \\frac{1}{c} < a$, we conclude\n$$\n(a-1)^2 \\le \\frac{(ab-1)(ac-1)}{bc} < a^2,\n$$\nand hence, $(a... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 82, 167, 1034 | |
0iso | Problem:
Joe has a triangle with area $\sqrt{3}$. What's the smallest perimeter it could have? | [
"Solution:\n\nThe minimum occurs for an equilateral triangle. The area of an equilateral triangle with side-length $s$ is $\\frac{\\sqrt{3}}{4} s^{2}$, so if the area is $\\sqrt{3}$ then $s = \\sqrt{\\sqrt{3} \\frac{4}{\\sqrt{3}}} = 2$. Multiplying by $3$ to get the perimeter yields the answer $6$."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 6 | |
01ju | Let $f(x)$ be a quadratic trinomial, and $a, b, c$ be pairwise distinct real numbers.
Given $f(a) = bc$, $f(b) = ac$, $f(c) = ab$, find $f(a + b + c)$. | [
"Answer: $ab + bc + ac$.\nLet $f(x) = \\alpha x^2 + \\beta x + \\gamma$. Then\n$$\n\\alpha a^2 + \\beta a + \\gamma = bc, \\quad \\alpha b^2 + \\beta b + \\gamma = ac, \\quad \\alpha c^2 + \\beta c + \\gamma = ab. \\quad (1)\n$$\nSubtracting the second and the third equations from the first one, we have\n$$\n\\alph... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | ab + bc + ac | |
03c0 | The quadrilateral $ABCD$ is inscribed in the circle $k$. The lines $AC$ and $BD$ meet in $E$ and the lines $AD$ and $BC$ meet in $F$. Show that the line through the incenters of $\triangle ABE$ and $\triangle ABF$ and the line through the incenters of $\triangle CDE$ and $\triangle CDF$ meet on $k$. | [
"Let $I_e, I_f, J_e, J_f$ be the incenters of $\\triangle ABE, \\triangle ABF, \\triangle CDE, \\triangle CDF$, respectively. Let $P = AI_e \\cap BI_f$, $Q = AI_f \\cap BI_e$, $U = CJ_e \\cap DJ_f$, and $V = CJ_f \\cap DJ_e$ be the excenter of $\\triangle ABC$ opposite to $A$, the excenter of $\\triangle ABD$ oppos... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscel... | English | proof only | null | |
0anl | Problem:
In how many ways can one select five books from a row of twelve books so that no two adjacent books are chosen?
(a) 34
(b) 78
(c) 42
(d) 56 | [] | Philippines | QUALIFYING STAGE | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | MCQ | (d) | |
0du4 | Problem:
Poišči vsa naravna števila $a$, $b$ in $c$, ki zadoščajo enačbi
$$
\frac{a+b}{a+c}=\frac{b+c}{b+a}
$$
in za katere velja, da je $ab+ac+bc$ praštevilo. | [
"Solution:\n\nOznačimo praštevilo $ab+bc+ca$ s $p$. Iz zveze dobimo, da je $(a+b)^2 = c^2 + p$ oziroma $p = (a+b-c)(a+b+c)$. Ker je $p$ praštevilo in je $0 < a+b-c < a+b+c$, mora biti $a+b-c = 1$ in $a+b+c = p$ oziroma $c = \\frac{p-1}{2}$ in $a+b = \\frac{p+1}{2}$. Po drugi strani pa mora biti $ab = p - bc - ca = ... | Slovenia | 45. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (1, 1, 1) | |
06le | For each positive integer $n$ larger than $1$ with prime factorization $p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$, its *signature* is defined as the sum $\alpha_1 + \alpha_2 + \cdots + \alpha_k$. Does there exist $2020$ consecutive positive integers such that among them, there are exactly $1812$ integers who... | [
"Yes. Let $f(n)$ be the number of integers among $n+1, n+2, \\dots, n+2020$ having signatures less than $11$. Since $2^{11} = 2048 > 2021$, all of $2, 3, \\dots, 2021$ have signatures smaller than $11$. Therefore, we have $f(1) = 2020$.\n\nNext, let $p_1, p_2, \\dots, p_{2020}$ be distinct primes. By the Chinese re... | Hong Kong | CHKMO | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Yes | |
08ae | Problem:
I 60 abitanti di un villaggio possono essere di tre tipi: contadini (che dicono sempre la verità), lupi mannari (che mentono sempre) e negromanti (che rispondono come vogliono). Tranne che per questi comportamenti i membri di ciascuna fazione sono completamente indistinguibili da quelli delle altre. All'arriv... | [
"Solution:\n\nLa risposta è $(C)$. Una delle possibilità è che ci siano $30$ lupi mannari e $30$ contadini, alternati. Questa disposizione falsifica la (A) e la (B). Un'altra possibile situazione è che gli abitanti del villaggio siano tutti negromanti e stiano mentendo tutti, il che falsifica la (D).\n\nConsideriam... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Discrete Mathematics > Logic"
] | null | MCQ | C | |
0ewk | Problem:
Find all real $p$, $q$, $a$, $b$ such that we have $$(2x - 1)^{20} - (a x + b)^{20} = (x^{2} + p x + q)^{10}$$ for all $x$. | [
"Solution:\nComparing coefficients of $x^{20}$, we must have $a = (2^{20} - 1)^{1 / 20}$ (note that we allow either the positive or the negative root).\n\nSet $x = 1 / 2$. Then we must have $(a x + b)^{20} = 0 = (x^{2} + p x + q)^{10}$, and hence $a x + b = 0$ and $x^{2} + p x + q = 0$. So $b = - a / 2$, and $1 / 4... | Soviet Union | 3rd ASU | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a = ±(2^{20} − 1)^{1/20}, b = −a/2, p = −1, q = 1/4 | |
0jeb | Problem:
Plot points $A$, $B$, $C$ at coordinates $(0,0)$, $(0,1)$, and $(1,1)$ in the plane, respectively. Let $S$ denote the union of the two line segments $AB$ and $BC$. Let $X_{1}$ be the area swept out when Bobby rotates $S$ counterclockwise $45$ degrees about point $A$. Let $X_{2}$ be the area swept out when Cal... | [
"Solution:\n\nAnswer: $\\frac{\\pi}{4}$\n\nIt's easy to see $X_{1} = X_{2}$. Simple cutting and pasting shows that $X_{1}$ equals the area of $\\frac{1}{8}$ of a circle with radius $AC = \\sqrt{2}$, so\n$$\n\\frac{X_{1}+X_{2}}{2} = X_{1} = \\frac{1}{8} \\pi (\\sqrt{2})^{2} = \\frac{\\pi}{4}.\n$$"
] | United States | HMMT November 2013 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | π/4 | |
0644 | Problem:
Finden Sie alle Paare $(a, b)$ positiver ganzer Zahlen, für die eine positive ganze Zahl $n$ existiert, sodass die Anzahl der Teiler von $n a$ und $n b$ identisch ist. | [
"Solution:\nDie gesuchten Paare sind jene, für die eine der drei Bedingungen $a=b, a \\nmid b$ oder $b \\nmid a$ erfüllt ist, also alle Paare positiver ganzer Zahlen mit Ausnahme solcher, für die eine der Zahlen $a, b$ ein echter Teiler der anderen ist.\nIm gesamten Beweis werden wir die Anzahl aller Teiler einer p... | Germany | 2. Auswahlklausur | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All pairs of positive integers except those where one number is a proper divisor of the other; equivalently, pairs where the numbers are equal or neither divides the other. | |
0cqk | The circles $\omega_1$ and $\omega_2$ touch each other externally at $P$. Let $\ell_1$ be a line passing through the center of $\omega_1$ and tangent to $\omega_2$. Similarly, a line $\ell_2$ is tangent to $\omega_1$ and passes through the center of $\omega_2$. The lines $\ell_1$ and $\ell_2$ are not parallel. Prove th... | [
"Пусть $O_1$, $r_1$ и $O_2$, $r_2$ — соответственно центры и радиусы окружностей $\\omega_1$ и $\\omega_2$, а $K$ — точка пересечения $\\ell_1$ и $\\ell_2$. Заметим, что точка $P$ лежит на отрезке $O_1O_2$ и делит его в отношении $r_1 : r_2$.\n\nОбозначим через $P_1$ точку касания $\\ell_2$ и $\\omega_1$, а через $... | Russia | Russian mathematical olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
07g7 | Point $X$ is chosen inside of a non-trapezoid quadrilateral $ABCD$ such that $\angle AXD + \angle BXC = 180^\circ$. Suppose the angle bisector of $\angle ABX$ meets the $D$-altitude of triangle $ADX$ at $K$, and the angle bisector of $\angle DCX$ meets the $A$-altitude of triangle $ADX$ at $L$. Let $BK \perp CX$ and $C... | [
"We start our solution with the following lemma.\n**Lemma.** Let $H$ be the orthocenter of the triangle $ABC$ and $E, F$ be the intersections of $BH, CH$ with perpendicular bisectors of $AB$ and $AC$, respectively. If $Q$ is the intersection of tangents from $E$ and $F$ to circumcircle of triangle $HEF$, then $BQ =... | Iran | 38th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
05es | Problem:
Soit $ABCD$ un trapèze isocèle avec $(AD)//(BC).
a. Montrer que $ABCD$ admet un cercle circonscrit.
b. On note $O$ le centre du cercle circonscrit de $ABCD$ et $E$ l'intersection des diagonales. Montrer que $O$ est sur le cercle circonscrit de $AEB$. | [
"Solution:\n\na. $ABCD$ est un trapèze isocèle, ainsi $\\widehat{ABC} = \\widehat{DCB}$. On sait également que $(AD)//(CB)$, donc $\\widehat{DCB} = 180 - \\widehat{CDA}$. Avec ces deux égalités on trouve que $\\widehat{ABC} + \\widehat{CDA} = 180$, ce qui montre bien que les points $A$, $B$, $C$ et $D$ sont sur le ... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06ei | Let $r$ be the positive root of the equation $x^2 - 2004x - 1 = 0$. Define the sequence $\{a_n\}$ as follows:
$$
a_0 = 1, a_{n+1} = [ra_n], n \ge 0,
$$
where $[y]$ denotes the greatest integer not exceeding $y$. Find the remainder when $a_{2004}$ is divided by $2004$. | [
"The answer is $1003$.\n\nNote that\n$$\nr = \\frac{2004 + \\sqrt{2004^2 + 4}}{2}\n$$\nis an irrational number greater than $1$. Since $a_n \\in \\mathbb{Z}^+$, we have $ra_n \\notin \\mathbb{Q}$. Thus, we obtain\n$$\na_{n+1} < ra_n < a_{n+1} + 1\n$$\nfor any $n \\ge 0$. Equivalently, we have\n$$\na_n - \\frac{1}{r... | Hong Kong | IMO HK TST | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 1003 | |
00pq | A sequence $(a_n)_{n=1}^\infty$ of positive integers satisfies the condition $a_{n+1} = a_n + \tau(n)$ for all positive integers $n$ where $\tau(n)$ is the number of positive integer divisors of $n$. Determine whether two consecutive terms of this sequence can be perfect squares. | [
"**Solution.** There are no two such consecutive terms.\nAssume that $a_n = x^2$, $a_{n+1} = y^2$ where $x, y$ are positive integers. Then\n$$\n(x+1)^2 \\le y^2 = a_{n+1} = a_n + \\tau(n) = x^2 + \\tau(n) \\le x^2 + 2\\sqrt{n}.\n$$\nTherefore $x < \\sqrt{n}$. The last inequality gives $a_n < n$, which is impossible... | Balkan Mathematical Olympiad | Balkan 2012 shortlist | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | No, two consecutive terms cannot both be perfect squares. | |
05d5 | Problem:
Find the smallest positive integer $k$ for which there exist a colouring of the positive integers $\mathbb{Z}_{>0}$ with $k$ colours and a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ with the following two properties:
(i) For all positive integers $m, n$ of the same colour, $f(m+n)=f(m)+f(n)$.
(i... | [
"Solution:\nThe answer is $k=3$.\n\nFirst we show that there is such a function and coloring for $k=3$. Consider $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ given by $f(n)=n$ for all $n \\equiv 1$ or $2$ modulo $3$, and $f(n)=2 n$ for $n \\equiv 0$ modulo $3$. Moreover, give a positive integer $n$ the $i$-t... | European Girls' Mathematical Olympiad (EGMO) | null | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Functional equations",
"Algebra > Algebraic Expressions > Func... | null | proof and answer | k = 3 | |
0kz7 | Two transformations are said to *commute* if applying the first followed by the second gives the same result as applying the second followed by the first. Consider these four transformations of the coordinate plane:
* a translation 2 units to the right,
* a $90^\circ$-rotation counterclockwise about the origin,
* a ref... | [
"Denote the transformations by $T$, $R$, $F$, and $D$ in the order given in the problem statement. Then the images of point $(x, y)$ are $T(x, y) = (x + 2, y)$, $R(x, y) = (-y, x)$, $F(x, y) = (x, -y)$, and $D(x, y) = (2x, 2y)$. The results of applying a pair of transformations in either order are as follows:\n* $T... | United States | AMC 10 A | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | MCQ | C | |
04yb | Determine whether there exist $2024$ distinct positive integers satisfying the following: If we consider every possible ratio between two distinct numbers (we include both $a/b$ and $b/a$), we will obtain numbers with finite decimal expansions (after the decimal point) of mutually distinct non-zero lengths. (Patrik Bak... | [
"We will show these numbers exist. For that we define sequences $a_1, a_2, \\dots, a_{2024}$ and $b_1, b_2, \\dots, b_{2024}$ and then consider numbers $c_i = 2^{a_i} \\cdot 5^{b_i}$ for $i = 1, 2, \\dots, 2024$.\nWe choose the sequences $a_i$ and $b_i$ in such a way that $a_i$ is increasing, $b_i$ is decreasing, a... | Czech-Polish-Slovak Mathematical Match | CAPS Match 2024 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | Yes, such 2024 integers exist; for example take c_i = 2^{2^{i-1}} · 5^{2^{4049−i}} for i = 1, 2, ..., 2024. | |
0b13 | Problem:
If $\sin \theta+\cos \theta=\frac{6}{5}$, evaluate $\tan \theta+\cot \theta$. | [
"Solution:\nWe have\n$$\n\\tan \\theta+\\cot \\theta=\\frac{\\sin \\theta}{\\cos \\theta}+\\frac{\\cos \\theta}{\\sin \\theta}=\\frac{1}{\\sin \\theta \\cos \\theta}\n$$\nThus, since $\\sin \\theta+\\cos \\theta=\\frac{6}{5}$, we have $(\\sin \\theta+\\cos \\theta)^2=1+2 \\sin \\theta \\cos \\theta=\\frac{36}{25}$,... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Precalculus > Trigonometric functions"
] | null | final answer only | 50/11 | |
03jg | Problem:
Suppose $ABCD$ is a parallelogram and $E$ is a point between $B$ and $C$ on the line $BC$. If the triangles $DEC$, $BED$ and $BAD$ are isosceles what are the possible values for the angle $DAB$? | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof and answer | 36°, 45°, 72° | |
0j1e | Problem:
A monomial term $x_{i_{1}} x_{i_{2}} \ldots x_{i_{k}}$ in the variables $x_{1}, x_{2}, \ldots, x_{8}$ is square-free if $i_{1}, i_{2}, \ldots, i_{k}$ are distinct. (A constant term such as $1$ is considered square-free.) What is the sum of the coefficients of the squarefree terms in the following product?
$$
\... | [
"Solution:\nAnswer: $764$\nLet $a_{n}$ be the sum of the coefficients of the square-terms in the product $\\prod_{1 \\leq i<j \\leq n}(1+ x_{i} x_{j})$. Square-free terms in this product come in two types: either they include $x_{n}$, or they do not. The sum of the coefficients of the terms that include $x_{n}$ is ... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | final answer only | 764 | |
0dds | The Magician and his Assistant show trick. The Viewer writes on the board the sequence of $N$ digits. Then the Assistant covers some pair of adjacent digits so that they become invisible. Finally, the Magician enters the show, looks at the board and guesses the covered digits and their order. Find the minimal $N$ such ... | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 101 | |
02m7 | Find all functions $f$ from real numbers to real numbers such that
$$
f(a + b) = f(ab)
$$
for all irrationals $a, b$. | [
"Let $f(0) = k$. Plugging $a = \\sqrt{2}$ and $b = -\\sqrt{2}$ one obtain $f(\\sqrt{2} + (-\\sqrt{2})) = f(\\sqrt{2}(-\\sqrt{2})) \\iff f(-2) = k$.\nLet $\\alpha \\in \\mathbb{R} \\setminus \\mathbb{Q}$. Since the quadratic equation $x^2-\\alpha x-2=0$ has discriminant $\\Delta = \\alpha^2+8>0$, its roots have sum ... | Brazil | Brazilian Math Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | All constant functions: f(x) = k for all real x, where k is any real constant. | |
0gic | A natural number $x$, when expressed in decimal notation, has the product of its digits equal to $x^2 - 15x - 27$. Find all possible $x$.
自然數 $x$ 以十進制表示時，各位數字的乘積等於 $x^2 - 15x - 27$。試求所有滿足以上條件的 $x$。 | [
"The only possible $x$ is $17$.\n\nClearly $x^2 - 15x - 27 \\le x$, so easily see that $1 \\le x \\le 17$. And since the equality does not hold, so $10 \\le x \\le 17$, which means $x$ is a solution of $x^2 - 16x - 17 = 0$."
] | Taiwan | APMO Taiwan Preliminary Round 2 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | Chinese; English | proof and answer | 17 | |
0igc | Problem:
Find a long binary word containing a small number of square subwords. Specifically, write down a binary word of any length $n \leq 50$. Your score will be $\max \{0, n-s\}$, where $s$ is the number of occurrences of square subwords. (That is, each different square subword will be counted according to the numbe... | [] | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Other"
] | null | final answer only | null | |
0kat | Problem:
A regular tetrahedron has volume $8$. What is the volume of the set of all the points in the space (not necessarily inside the tetrahedron) that are closer to the center of the tetrahedron than any of the four vertices? | [
"Solution:\nLet $h$ denote the height of the tetrahedron. The center of the tetrahedron is a distance $\\frac{h}{4}$ from each face. Therefore, the perpendicular bisector plane of the segment connecting the center to a vertex lies a distance $\\frac{3}{8} h$ away from both the vertex and the center. Symmetrical con... | United States | HMMT February 2019 | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | final answer only | 27 | |
00d1 | Sea $ABCD$ un paralelogramo con $\angle ABC = 105^\circ$. En el interior del paralelogramo existe un punto $E$ tal que el triángulo $BEC$ es equilátero y $\angle CED = 135^\circ$. Sea $K$ el punto medio del lado $AB$. Calcular la medida del ángulo $BKC$. | [
"Como $ABCD$ es un paralelogramo vale que\n$$\n\\angle BAD = \\angle BCD = 75^\\circ \\Rightarrow \\angle ECD = 75^\\circ - 60^\\circ = 15^\\circ \\text{ y } \\angle EDC = 30^\\circ.\n$$\nTrazamos por $C$ la perpendicular a la recta $DE$, que la corta en $F$. Entonces $\\angle FEC = 180^\\circ - 135^\\circ = 45^\\c... | Argentina | Nacional OMA | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals"
] | Spanish | proof and answer | 45° | |
00jn | Determine the number of non-negative integers $N < 1000000 = 10^6$ with the following property: There exists an integer exponent $k$ with $1 \le k \le 43$ such that $2012$ is a divisor of $N^k - 1$. | [
"It is obvious that $N$ and $2012$ must be relatively prime. If $N^k \\equiv 1 \\pmod{n}$ and $N^m \\equiv 1 \\pmod{n}$ both hold, so does $N^d \\equiv 1 \\pmod{n}$ for $d = \\gcd(k, m)$. From $m = \\varphi(n)$, we see that $N^k \\equiv 1 \\pmod{n}$ implies that there exists a divisor $d$ of $\\varphi(n)$ with $N^d... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n"
] | null | proof and answer | 1989 | |
03bx | Show that, for all positive real numbers $a$, $b$, $c$, and $d$ the following inequality holds:
$$
\sum_{cyc} \frac{a^4}{a^3 + a^2 b + ab^2 + b^3} \ge \frac{a+b+c+d}{4}.
$$ | [
"We have\n$$\n\\begin{aligned} \\sum_{cyc} \\frac{a^4}{a^3 + a^2 b + ab^2 + b^3} - \\sum_{cyc} \\frac{b^4}{a^3 + a^2 b + ab^2 + b^3} &= \\sum_{cyc} \\frac{a^4 - b^4}{a^3 + a^2 b + ab^2 + b^3} \\\\ &= \\sum_{cyc} (a - b) = 0 \\end{aligned}\n$$\n$$\n\\sum_{cyc} \\frac{a^4 + b^4}{a^3 + a^2 b + ab^2 + b^3} \\ge \\frac{... | Bulgaria | 55th IMO Team Selection Test | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof only | null | |
0dxp | Problem:
Skrči izraz $\frac{1+9 a^{-1}+20 a^{-2}}{1+8 a^{-1}+16 a^{-2}} \cdot\left(a^{2}+4 a\right) \cdot\left(1-25 a^{-2}\right)^{-1}$. | [
"Solution:\n\nPotence z negativnim celim eksponentom v prvem faktorju zapišemo v obliki ulomka. V števcu in imenovalcu dvojnega ulomka poiščemo skupni imenovalec. V števcu dobimo $\\frac{a^{2}+9 a+20}{a^{2}}$, v imenovalcu pa $\\frac{a^{2}+8 a+16}{a^{2}}$. Odpravimo dvojni ulomek, števec razstavimo na $(a+4)(a+5)$ ... | Slovenia | 7. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | final answer only | a^3/(a-5) | |
0dy2 | Find all prime numbers $p$ and $q$ such that $2p^2q + 45pq^2$ is a perfect square. | [
"First assume $p = q$. Then $47p^3$ must be a perfect square. Since $47p^3$ is divisible by $47$ which is prime, it must also be divisible by $47^2$. This implies that $47$ divides either $p^3$ or $p$. But $p$ is a prime, so this is only possible when $p$ is equal to $47$. Indeed, when $p = q = 47$ we have $2p^2q +... | Slovenia | Slovenija 2008 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | (p, q) = (47, 47) and (3, 2) | |
03yp | As seen in Fig. 1.1, points $P$, $Q$ are, respectively, the midpoints of $AC$, $BD$ — the two diagonals of cyclic quadrilateral $ABCD$. Let $\angle BPA = \angle DPA$. Prove $\angle AQB = \angle CQB$.
 | [
"As shown in Fig. 1.2, we extend segment $DP$ to intercept with the circle at point $E$. Then $\\angle CPE = \\angle DPA = \\angle BPA$. Since $P$ is the midpoint of $AC$, we get $\\overline{AB} = \\overline{CE}$, which means $\\angle CDP = \\angle BDA$. Furthermore, $\\angle ABD = \\angle PCD$. Therefore, $\\trian... | China | China Mathematical Competition (Complementary Test) | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
08wp | A $2 \times 100$ square grid is given. Suppose we color each of the $200$ square boxes of the grid either red or blue. How many possible ways of coloring these boxes are there that satisfy the following conditions? Regard two patterns of colored boxes to be distinct if one of the patterns can be obtained from the other... | [
"Suppose all the square boxes of the grid were colored in such a way that the conditions specified for the problem were satisfied. There are $204$ points (let us call them vertex points), on the rectangle forming the perimeter of the grid, which are either a vertex of one of the square boxes or a common vertex of t... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 39800 | |
0gpc | Show that
$$
(x^2 + y^2)^3 \geq 32(x^3 + y^3)(xy - x - y)
$$
for all real numbers $x, y$ satisfying $x + y \geq 0$. | [
"Let $s = x^2 + y^2$ and $t = x + y$. We want to show that\n$$\ns^3 \\ge 8t(3s - t^2)(t^2 - 2t - s)\n$$\nfor $2s \\ge t^2$ and $t \\ge 0$.\nNow let $s = rt$. This transforms the inequality to\n$$\nr^3 \\ge 8(3r - t)(t - 2 - r)\n$$\nfor $2r \\ge t \\ge 0$.\nSince $r^3 - 8(3r - t)(t - 2 - r) = 8(t - (2r + 1))^2 + r^3... | Turkey | Team Selection Test | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0c22 | The acute-angled triangle $ABC$ has circumcenter $O$. The straight lines $B_aC_a$, $C_bA_b$ and $A_cB_c$ are perpendicular in $O$ on the straight lines $AO$, $BO$, respectively $CO$ and $A_b$, $A_c \in BC$, $B_a, B_c \in AC$, $C_b, C_a \in AB$. Denote $O_a, O_b, O_c$ the circumcenters of the triangles $AC_aB_a, BA_bC_b... | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0f56 | Problem:
A word is a finite string of $A$s and $B$s. Can we find a set of three 4-letter words, ten 5-letter words, thirty 6-letter words and five 7-letter words such that no word is the beginning of another word? [For example, if $ABA$ was a word, then $ABAAB$ could not be a word.] | [] | Soviet Union | 17th ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Other"
] | null | proof and answer | No | |
0kzj | A triangle in the coordinate plane has vertices $A(\log_2 1, \log_2 2)$, $B(\log_2 3, \log_2 4)$, and $C(\log_2 7, \log_2 8)$. What is the area of $\triangle ABC$?
(A) $\log_2 \frac{\sqrt{3}}{7}$ (B) $\log_2 \frac{3}{\sqrt{7}}$ (C) $\log_2 \frac{7}{\sqrt{3}}$ (D) $\log_2 \frac{11}{\sqrt{7}}$ (E) $\log_2 \frac{11}{\sqrt... | [
"**Answer (B):** Circumscribe $\\triangle ABC$ by rectangle $AGCD$, with $D$ on the y-axis, and project point $B$ onto $\\overline{AG}$ and $\\overline{CG}$, producing points $E$ and $F$, respectively, as shown in the figure below.\n\n\n\nThe area of rectangle $AGCD$ is $2\\log_2 7$, so the... | United States | 2024 AMC 12 B | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Linear Algebra > Determinants",
"Algebra > Linear Algebra > Vectors"
] | null | MCQ | B | |
0fjj | Problem:
Las alturas del triángulo $ABC$ se cortan en el punto $H$. Se sabe que $AB = CH$. Determinar el valor del ángulo $\widehat{BCA}$. | [
"Solution:\n\nPrimer caso: $C < 90^{\\circ}$.\nLlamaremos $A'$ al punto en que la altura de $A$ corta al lado $BC$ del triángulo $ABC$, y $C'$ al punto donde la altura de $C$ corta al lado $AB$ del triángulo $ABC$.\nEl ángulo $\\widehat{CHA'}$ es igual al ángulo $\\widehat{AHC'}$. En el triángulo $CA'H$, el ángulo ... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 45° or 135° | |
03pt | Find all ternary positive integer groups $(a, m, n)$ satisfying $a \ge 2$ and $m \ge 2$ such that $a^n + 203$ is a multiple of $a^m + 1$. (posed by Chen Yonggao) | [
"We will discuss the following three cases for $n$ and $m$.\n\n(i) In the case when $n < m$, from $a^n + 203 \\ge a^m + 1$, we have\n$$\n202 \\ge a^m - a^n \\ge a^n(a-1) \\ge a(a-1).\n$$\nTherefore, $2 \\le a \\le 14$.\n\nWhen $a = 2$, we can take $n$ to be $1, 2, \\dots, 7$.\nWhen $a = 3$, we can take $n$ to be $1... | China | China Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Other"
] | English | proof and answer | (2, 2, 4k+1), (2, 3, 6k+2), (2, 4, 8k+8), (2, 6, 12k+9), (3, 2, 4k+3), (4, 2, 4k+4), (5, 2, 4k+1), (8, 2, 4k+3), (10, 2, 4k+2), and (203, m, (2k+1)m+1), where k is any nonnegative integer and m ≥ 2. | |
0go2 | If $1 < k_1 < k_2 < \dots < k_n$ and $a_1, a_2, \dots, a_n$ are integers such that for every integer $N$, $k_i \mid N - a_i$ for some $1 \le i \le n$, find the smallest possible value of $n$. | [
"Every integer $N$ satisfies at least one of the congruences $N \\equiv 0 \\pmod{2}$, $N \\equiv 1 \\pmod{3}$, $N \\equiv 3 \\pmod{4}$, $N \\equiv 5 \\pmod{6}$, $N \\equiv 9 \\pmod{12}$. Therefore $n$ can be $5$. We will show that $n \\le 4$ is not possible.\n\nLet $1 < k_1 \\le k_2 \\le \\dots \\le k_n$ and $a_1, ... | Turkey | 17th Turkish Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 5 | |
0khb | A square with side length $3$ is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length $2$ has two vertices on the other square and the other two on sides of the triangle, as shown. What is the area of the triangle?
(A) $19\fra... | [] | United States | 2021 AMC 10 B Fall | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | MCQ | B | |
0btu | Let $a, b$ be positive integers so that there exists a prime number $p$ with the property $[a, a + p] = [b, b + p]$. Prove that $a = b$.
Here $[x, y]$ denotes the least common multiple of $x$ and $y$. | [
"If $d$ is a common divisor of $x$ and $x+p$, then $d \\mid (x+p) - x = p$, hence $d = 1$ or $d = p$. This leads to the cases:\n\ni. $p \\nmid a$. Then $p \\nmid a + p$, $p \\nmid [a, a + p]$, hence $p \\nmid [b, b + p]$, $p \\nmid b$, $p \\nmid b + p$, the numbers $a, a+p$ and $b, b+p$ are co-prime and $[a, a+p] =... | Romania | 67th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0f0j | Problem:
a, b, c, d, e are positive reals. Show that $$(a + b + c + d + e)^2 \geq 4(ab + bc + cd + de + ea)$$ | [] | Soviet Union | ASU | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
03ud | Given four fixed points $A(-3, 0)$, $B(1, -1)$, $C(0, 3)$, $D(-1, 3)$ and a variable point $P$ in a plane rectangular coordinates system, the minimum of $|PA| + |PB| + |PC| + |PD|$ is ______. | [
"As shown in the figure, assuming that $AC$ and $BD$ meet at point $F$, we have\n$$\n|PA| + |PC| \\ge |AC| = |FA| + |FC|\n$$\nand\n$$\n|PB| + |PD| \\ge |BD| = |FB| + |FD|.\n$$\nWhen $P$ coincides with $F$, $|PA| + |PB| + |PC| + |PD|$ reaches the minimum.\nThat is, $|AC| + |BD| = 3\\sqrt{2} + 2\\sqrt{5}$.\nSo $3\\sq... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 3√2 + 2√5 | |
00pb | Is it possible to partition the set of positive integer numbers into two classes, none of which contains an infinite arithmetic sequence (with a positive ratio)?
What if we require the extra condition that, in each class $C$ of the partition, the set of differences
$$
\{ \min \{ n : n \in C \text{ and } n > m \} - m : ... | [
"It is easy to exhibit such a partition: set\n$$\nA_1 = \\bigcup_{n=1}^{\\infty} \\{n(2n-1) + 1, n(2n-1) + 2, \\dots, n(2n-1) + 2n\\},\n$$\n$$\nA_2 = \\bigcup_{n=0}^{\\infty} \\{n(2n+1) + 1, n(2n+1) + 2, \\dots, n(2n+1) + 2n+1\\}.\n$$\nSince each class has arbitrarily large gaps, it cannot contain an infinite arith... | Balkan Mathematical Olympiad | shortlistBMO 2011 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
038a | Problem:
Two circles $k_{1}$ and $k_{2}$ meet at points $A$ and $B$. A line through $B$ meets the circles $k_{1}$ and $k_{2}$ at points $X$ and $Y$, respectively. The tangent lines to $k_{1}$ at $X$ and to $k_{2}$ at $Y$ meet at $C$. Prove that:
a) $\Varangle X A C = \Varangle B A Y$.
b) $\Varangle X B A = \Varangle... | [
"Solution:\n\na) The quadrilateral $X C Y A$ is cyclic since\n$$\n\\begin{aligned}\n\\Varangle X C Y & = 180^{\\circ} - \\Varangle C X Y - \\Varangle C Y X = 180^{\\circ} - \\Varangle X A B - \\Varangle B A Y \\\\\n& = 180^{\\circ} - \\Varangle X A Y\n\\end{aligned}\n$$\nTherefore $\\Varangle X A C = \\Varangle X Y... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04bm | Prove that for every integer $k \ge 2$ there exist $k$ positive integers whose sum is equal to their product. | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof only | null | |
041e | For any integer $n$ with $n > 1$, let
$$
D(n) = \{a - b \mid a, b \text{ are positive integers with } n = ab \text{ and } a > b\}.
$$
Prove that for any integer $k$ with $k > 1$, there exist $k$ pairwise distinct integers $n_1, n_2, \dots, n_k$ with $n_i > 1\ (1 \le i \le k)$, such that $D(n_1) \cap D(n_2) \cap \dots \... | [
"**Proof.** Let $a_1, a_2, \\dots, a_{k+1}$ be $k+1$ distinct positive integers, where each of them is smaller than the product of other $k$ numbers. Write $N = a_1 a_2 \\cdots a_{k+1}$. For each $i = 1, 2, \\dots, k+1$, let $x_i = \\frac{1}{2} \\left( \\frac{N}{a_i} + a_i \\right)$, $y_i = \\frac{1}{2} \\left( \\f... | China | China Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0iex | Problem:
Regular tetrahedron $A B C D$ is projected onto a plane sending $A, B, C$, and $D$ to $A', B', C'$, and $D'$ respectively. Suppose $A' B' C' D'$ is a convex quadrilateral with $A' B' = A' D'$ and $C' B' = C' D'$, and suppose that the area of $A' B' C' D' = 4$. Given these conditions, the set of possible lengt... | [
"Solution:\n\n$2 \\sqrt[4]{6}$\n\nThe value of $b$ occurs when the quadrilateral $A' B' C' D'$ degenerates to an isosceles triangle. This occurs when the altitude from $A$ to $B C D$ is parallel to the plane. Let $s = A B$. Then the altitude from $A$ intersects the center $E$ of face $B C D$. Since $E B = \\frac{s}... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
... | null | final answer only | 2 sqrt[4]{6} | |
025y | Problem:
As páginas do dicionário - Para numerar as páginas de um dicionário, imprimiu-se 1988 vezes o algarismo 1. Quantas páginas tem esse dicionário? | [
"Solution:\n\nObservemos que:\n- a cada 10 números imprime-se 1 vez o $1$ nas unidades,\n- a cada 100 números imprime-se 10 vezes o $1$ nas dezenas,\n- a cada 1000 números imprime-se 100 vezes o $1$ nas centenas.\n\nAssim, de $1$ até $999$ imprime-se o número $1$:\n$100$ vezes nas unidades $+100$ nas dezenas $+100$... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 3151 | |
0ivt | Problem:
The vertices of a regular hexagon are labeled $\cos (\theta)$, $\cos (2 \theta)$, $\cos (3 \theta)$, $\cos (4 \theta)$, $\cos (5 \theta)$, $\cos (6 \theta)$. For every pair of vertices, Bob draws a blue line through the vertices if one of these functions can be expressed as a polynomial function of the other ... | [
"Solution:\n\nAnswer: 14\n\nThe existence of the Chebyshev polynomials, which express $\\cos (n \\theta)$ as a polynomial in $\\cos (\\theta)$, imply that Bob draws a blue line between $\\cos (\\theta)$ and each other vertex, and also between $\\cos (2 \\theta)$ and $\\cos (4 \\theta)$, between $\\cos (2 \\theta)$ ... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 14 | |
0a7f | Problem:
Let $f(x)$ be a polynomial with integer coefficients. We assume that there exists a positive integer $k$ and $k$ consecutive integers $n, n+1, \ldots, n+k-1$ so that none of the numbers $f(n), f(n+1), \ldots, f(n+k-1)$ is divisible by $k$. Show that the zeroes of $f(x)$ are not integers. | [
"Solution:\n\nLet $f(x) = a_{0} x^{d} + a_{1} x^{d-1} + \\cdots + a_{d}$. Assume that $f$ has a zero $m$ which is an integer. Then $f(x) = (x - m) g(x)$, where $g$ is a polynomial. If $g(x) = b_{0} x^{d-1} + b_{1} x^{d-2} + \\cdots + b_{d-1}$, then $a_{0} = b_{0}$, and $a_{k} = b_{k} - m b_{k-1}$, $1 \\leq k \\leq ... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 5 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0bru | We are given an $m \times n$ grid and three colors. We wish to color each segment of the grid with one of the three colors so that each unit square has two sides of one color and two sides of a second color. How many such colorings are possible? | [
"We label the lines from top to bottom and the columns from left to right. The leftmost side of the unit square in the upper-left corner can be colored in $3$ ways. Subsequently, there are $3$ ways of choosing the side of this unit square that is to receive the same color as the first side. The remaining two sides ... | Romania | 67th NMO Selection Tests for JBMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 3^{m+n} * 2^{mn} | |
0a2b | Find all functions $f: \mathbb{R}_{\ge 0} \to \mathbb{R}$ that satisfy
$$
2x^3zf(z) + yf(y) \geq 3yz^2f(x)
$$
for all $x, y, z \in \mathbb{R}_{\ge 0}$. | [
"$$\nf_{c,d}(x) = \\begin{cases} cx^2 & \\text{if } x > 0 \\\\ d & \\text{if } x = 0 \\end{cases}\n$$\nwith $c \\ge 0$ and $d \\le 0$.\n\nSubstituting $x = 0$ and $y = 1$ yields $f(1) \\ge 3f(0)z^2$ for all $z \\ge 0$. If $f(0) > 0$, then the right-hand side of this inequality is unbounded, contradiction. So we hav... | Netherlands | IMO Team Selection Test 1 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | All functions of the form f(x) = cx^2 for x > 0 and f(0) = d, with parameters c ≥ 0 and d ≤ 0. | |
0ic7 | Problem:
Let
$$
S(n, r)=\binom{n-1}{r-1}+\binom{n-1}{r}+\binom{n-1}{r+1}+\cdots+\binom{n-1}{n-1}
$$
for all $n, r>0$, and in particular $S(n, r)=0$ if $r>n>0$. Prove that the number in row $n$ of the table, $r$ columns to the left of the 1 in the top row, is at most $S(n, r)$. (Hint: First prove that $S(n-1, r-1)+S(n-1... | [
"Solution:\nFirst, we prove the statement in the hint: adding the $i$th term of the sum for $S(n-1, r-1)$ to the $i$th term for $S(n-1, r)$, for each $i$, we get that $S(n-1, r-1)+S(n-1, r)$ equals\n$$\n\\begin{gathered}\n\\left(\\binom{n-2}{r-2}+\\binom{n-2}{r-1}\\right)+\\left(\\binom{n-2}{r-1}+\\binom{n-2}{r}\\r... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0kc5 | Problem:
Let $ABC$ be a triangle with incircle tangent to the perpendicular bisector of $BC$. If $BC = AE = 20$, where $E$ is the point where the $A$-excircle touches $BC$, then compute the area of $\triangle ABC$. | [
"Solution:\n\nLet the incircle and $BC$ touch at $D$, the incircle and perpendicular bisector touch at $X$, $Y$ be the point opposite $D$ on the incircle, and $M$ be the midpoint of $BC$. Recall that $A$, $Y$, and $E$ are collinear by homothety at $A$. Additionally, we have $MD = MX = ME$ so $\\angle DXY = \\angle ... | United States | HMMT February 2020 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"... | null | final answer only | 100*sqrt(2) | |
027m | Problem:
Determine o termo mínimo da sequência
$$
\sqrt{\frac{7}{6}}+\sqrt{\frac{96}{7}}, \sqrt{\frac{8}{6}}+\sqrt{\frac{96}{8}}, \sqrt{\frac{9}{6}}+\sqrt{\frac{96}{9}}, \ldots, \sqrt{\frac{95}{6}}+\sqrt{\frac{96}{95}}
$$ | [
"Solution:\nLembre que $(x-y)^2 \\geq 0$ para todos os reais $x$ e $y$. Assim, podemos reescrever a desigualdade como $\\frac{x^2+y^2}{2} \\geq x y$, e substituindo $x=\\sqrt{a}$ e $y=\\sqrt{b}$, com $a$ e $b$ reais não negativos, temos $\\frac{a+b}{2} \\geq \\sqrt{a b}$. Agora, observe que todos os termos são do t... | Brazil | null | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 4 | |
08gb | Problem:
Dato un cubo di lato $10$, consideriamo un piano che passi per esattamente $6$ dei punti medi dei suoi spigoli; chiamiamo tali punti $A, B, C, D, E, F$ e supponiamo che i lati dell'esagono $ABCDEF$ giacciano ciascuno su una faccia del cubo. Consideriamo poi un secondo piano contenente il segmento $AB$ e perpe... | [
"Solution:\n\nLa risposta è $\\mathbf{(C)}$. Sia $V$ il volume che si vuole calcolare, e siano rispettivamente $X$ il volume della porzione di cubo tagliata dal piano passante per i $6$ punti medi e $Y$ il volume della regione tagliata dal piano perpendicolare alla faccia contenente $AB$. Si ha quindi $V + X + Y = ... | Italy | Olimpiadi di Matematica - Febbraio | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes"
] | null | MCQ | C | |
0g5p | 設 $f(x)$ 為一個次數不超過 $n$ 之多項式, 並且 $f(0), f(1), \dots, f(n)$ 這 $n+1$ 個數中, 任意兩數的差都是整數。試證: $f(2011) - f(100)$ 也是整數。 | [
"不妨設 $f(0)$ 為整數 (否則轉而考慮函數 $f(x)-f(0)$)。我們只須證明: $f(2011)$ 與 $f(100)$ 是整數即可。\n\n利用數學歸納法證明更廣的命題: 若一 $n$ 次多項式 $f(x)$ 滿足 $f(0), f(1), \\dots, f(n)$ 皆為整數, 則對任意整數 $x, f(x)$ 都是整數。\n\n首先, 基底 $n=0$ 時 $f(x)$ 為常系數多項式, 命題顯然成立。\n\n設此命題對 $n < k$ 均成立。則對一個 $f(0), f(1), \\dots, f(k)$ 之值均為整數的 $k$ 次多項式 $f(x)$, 考慮函數 $g(x) = f(x+1) - f(... | Taiwan | 二〇一一數學奧林匹亞競賽第一階段選訓營,獨立研究(一) | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0k1i | Problem:
Call a triangle nice if the plane can be tiled using congruent copies of this triangle so that any two triangles that share an edge (or part of an edge) are reflections of each other via the shared edge. How many dissimilar nice triangles are there?
Proposed by: Yuan Yao | [
"Solution:\n\nThe triangles are $60$-$60$-$60$, $45$-$45$-$90$, $30$-$60$-$90$, and $30$-$30$-$120$.\n\nWe make two observations.\n- By reflecting \"around\" the same point, any angle of the triangle must be an integer divisor of $360^{\\circ}$.\n- If any angle is an odd divisor of $360^{\\circ}$, i.e. equals $\\fr... | United States | HMMT November 2018 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 4 | |
0b16 | Problem:
Let $a$ and $b$ be integers such that in the expanded form of $(x^{2}+a x+b)^{3}$, the numerical coefficient of $x^{4}$ is $99$ and the numerical coefficient of $x$ is $162$. What are the values of $a$ and $b$? | [
"Solution:\n\nUsing the multinomial expansion, the coefficients of $x^{4}$ and $x$ are $3 b + 3 a^{2}$ and $3 a b^{2}$, respectively. Thus, we just have to solve the system of equations below.\n$$\n\\begin{aligned}\n3\\left(b + a^{2}\\right) & = 99 \\\\\n3 a b^{2} & = 162\n\\end{aligned}\n$$\nSimplifying, we have $... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a = 6, b = -3 | |
04jd | One side of a square lies on the line $y = 2x - 17$, while the remaining two vertices lie on the parabola $y = x^2$. Determine the area of that square. (China 2005) | [] | Croatia | Croatia Mathematical Competitions | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 80 or 1280 | |
0c5f | Let $n$ be a given positive integer. Determine all positive divisors $d$ of $3n^2$ such that $n^2 + d$ is the square of an integer. | [
"If $d$ divides $3n^2$, then there exist positive integers $k$ and $m$ such that $3n^2 = d \\cdot k$ and $n^2 + d = m^2$. We substitute to get $n^2 + \\frac{3n^2}{k} = m^2$, so $(mk)^2 = n^2(k^2 + 3k)$. We deduce that $k^2 + 3k$ is a perfect square. From the inequalities $k^2 < k^2 + 3k < (k+2)^2$ we deduce that $k... | Romania | 70th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 3n^2 | |
052g | Find all natural numbers $n$ for which there exist primes $p$ and $q$ such that $p(p+1) + q(q+1) = n(n+1)$. | [
"The equation is equivalent to $p(p+1) = (n-q)(n+q+1)$. Since the difference of the factors in the r.h.s. is greater than $1$, we must have $n-q < p$ and $n+q+1 > p+1$. As $p$ is a prime number, $p \\mid n+q+1$. Let $n+q+1 = kp$, $k > 1$. Now the initial equation yields $p(p+1) = (kp-2q-1)kp$, which is equivalent t... | Estonia | Open Contests | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | n = 3 or n = 6 | |
0j1z | Problem:
Suppose that there exist nonzero complex numbers $a$, $b$, $c$, and $d$ such that $k$ is a root of both the equations $a x^{3} + b x^{2} + c x + d = 0$ and $b x^{3} + c x^{2} + d x + a = 0$. Find all possible values of $k$ (including complex values). | [
"Solution:\n$1, -1, i, -i$\n\nLet $k$ be a root of both polynomials. Multiplying the first polynomial by $k$ and subtracting the second, we have $a k^{4} - a = 0$, which means that $k$ is either $1, -1, i$, or $-i$. If $a = b = c = d = 1$, then $-1, i$, and $-i$ are roots of both polynomials. If $a = b = c = 1$ and... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 1, -1, i, -i | |
0i2t | Problem:
Find $a+b+c+d+e$ if
$$
\begin{gathered}
3 a+2 b+4 d=10, \\
6 a+5 b+4 c+3 d+2 e=8, \\
a+b+2 c+5 e=3, \\
2 c+3 d+3 e=4, \text{ and } \\
a+2 b+3 c+d=7 .
\end{gathered}
$$ | [
"Solution:\nAdding the first, third, and fifth equations, we get $5 a+5 b+5 c+5 d+5 e=10+3+7 \\Rightarrow a+b+c+d+e=4$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 4 | |
0jv9 | Problem:
Call a positive integer $N \geq 2$ "special" if for every $k$ such that $2 \leq k \leq N$, $N$ can be expressed as a sum of $k$ positive integers that are relatively prime to $N$ (although not necessarily relatively prime to each other). How many special integers are there less than $100$? | [
"Solution:\n\nWe claim that all odd numbers are special, and the only special even number is $2$. For any even $N > 2$, the numbers relatively prime to $N$ must be odd. When we consider $k = 3$, we see that $N$ can't be expressed as a sum of $3$ odd numbers.\n\nNow suppose that $N$ is odd, and we look at the binary... | United States | HMMT February 2016 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 50 | |
09vn | Problem:
Gegeven is een driehoek $A B C$. Op zijde $A C$ liggen punten $D$ en $E$ zodat de volgorde van punten op deze lijn $A, E, D, C$ is. De lijn door $E$ evenwijdig aan $B C$ snijdt de omgeschreven cirkel van $\triangle A B D$ in een punt $F$, waarbij $E$ en $F$ aan weerszijden van $A B$ liggen. De lijn door $E$ e... | [
"Solution:\n\nZij $G'$ het snijpunt van de lijn $B F$ en de omgeschreven cirkel van $\\triangle B C D$. We bewijzen dat $D E F G'$ een koordenvierhoek is en vervolgens dat $G=G'$, waarmee het bewijs compleet is.\n\nWegens $E F \\parallel B C$ geldt $180^{\\circ}-\\angle D E F=\\angle A E F=\\angle A C B$ en vanwege... | Netherlands | IMO-selectietoets | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
031t | Problem:
For any positive integer $n$ set
$$
A_{n} = \{ j : 1 \leq j \leq n,\ (j, n) = 1 \}
$$
Find all $n$ such that the polynomial
$$
P_{n}(x) = \sum_{j \in A_{n}} x^{j-1}
$$
is irreducible over $\mathbb{Z}[x]$. | [
"Solution:\nWe have $P_{1}(x) = P_{2}(x) = 1$, $P_{3}(x) = x + 1$, $P_{4}(x) = x^{2} + 1$ and $P_{6}(x) = x^{4} + 1$. Hence the integers $n = 1, 2, 3, 4, 6$ have the desired property. We shall prove that these are the only solutions of the problem.\n\nIt is enough to show that for $n \\geq 3$ the polynomial $P_{n}(... | Bulgaria | Team selection test for 20. BMO | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factoriz... | null | proof and answer | n = 1, 2, 3, 4, 6 | |
0gse | There are $k$ heaps of beads on the table containing 2019 beads in total. In each move we choose a heap: either remove it from the table or divide it into two not necessarily equal parts. Find the maximal possible value of $k$ such that for any initial distribution of beads after finite number of moves one can get $k$ ... | [
"**Answer: 45.**\n\nThe maximal value of $k$ can not be greater than 45. Indeed, if $k \\ge 46$, then since $45 \\cdot k > 2019$ it is possible that at the beginning there are $k$ heaps each containing at most 45 beads. Obviously in this case one can not get $k$ heaps with pairwise distinct number of beads, since t... | Turkey | Team Selection Test for EGMO 2019 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 45 | |
0l6x | Problem:
Complex numbers $\omega_{1}$, $\ldots$, $\omega_{n}$ each have magnitude $1$. Let $z$ be a complex number distinct from $\omega_{1}$, $\ldots$, $\omega_{n}$ such that
$$
\frac{z + \omega_{1}}{z - \omega_{1}} + \dots + \frac{z + \omega_{n}}{z - \omega_{n}} = 0.
$$
Prove that $|z| = 1$. | [
"Solution:\n\nWe show that no solutions $z$ not on the unit circle can exist. First, we eliminate $|z| > 1$.\n\nClaim 1. For all $j$ and $|z| > 1$, the real part of $\\frac{z + \\omega_{j}}{z - \\omega_{j}}$ is positive.\n\nProof. We use geometry. Note that $\\omega_{j}$ and $-\\omega_{j}$ are antipodes on the unit... | United States | HMMT February 2025 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expression... | null | proof only | null | |
025u | Problem:
Um mastro vertical $AB$ de altura $1\,m$ é iluminado por raios do sol e forma sombras no plano horizontal de comprimentos: $1$, $2$ e $3$ metros em três momentos diferentes. Prove que a soma dos ângulos de incidência dos raios nestes três momentos forma um ângulo reto, ou seja,
$$
\angle ACD + \angle AEB + \a... | [
"Solution:\n\nComo $D$, $E$ e $C$ são as extremidades das sombras do mastro, segue que $BD = DE = EC = 1$. Sendo $ABD$ um triângulo retângulo com $AB = BD$, temos $\\angle ADB = \\angle BAD = 45^\\circ$. Portanto, basta mostrar que $\\angle BEA + \\angle BCA = 45^\\circ$. Construamos o triângulo $AFG$ de modo que $... | Brazil | NÍVEL 3 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0gce | 令 $n$ 為一正整數且 $3$ 不能整除 $n$。一個 $n$ 階三角陣是把一個大正三角形分成 $n^2$ 個相等大小的小正三角形，以過各邊的所有 $n$ 等分點對其他邊做平行線分割而成。在 $n$ 階三角陣每個小三角形頂點上放一顆橘子，共有
$$
\frac{(n+1)(n+2)}{2}
$$
顆橘子。一組三顆橘子 $A, B, C$ 被稱為**好的**，若且唯若 $\overline{AB}$ 和 $\overline{AC}$ 各是某小三角形的一邊，且 $\angle BAC = 120^{\circ}$。
**言言**每次可以選一組好的三顆橘子拿走。試求**言言**最多可以拿走幾顆橘子。 | [
"答案：$\\frac{(n+1)(n+2)}{2} - 3$。\n\n記一個 $(n, k)$-梯形陣是一個 $n$ 階三角陣移除一個與 $n$ 階三角陣共頂點的 $n-k$ 階三角陣。記 $A_{00}, A_{01}, A_{02}, \\dots, A_{0n}$ 依序表示 $n$ 階三角陣一個底邊的橘子，且遞迴定義編號 $A_{ij}$ 的橘子使 $A_{ij}, A_{(i-1)j}, A_{(i-1)(j+1)}$ 構成小正三角形 ($A_{ij} \\neq A_{(i-2)(j+1)}$)，稱 $A_{ij}$ 橘子是第 $i$ 排第 $j$ 個。\n\n(1) 首先證明上界。我們只需證明：\n$A_{00... | Taiwan | 二〇一八數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | ((n+1)(n+2))/2 - 3 | |
0kb2 | On an infinite square grid we place finitely many cars, which each occupy a single cell and face in one of the four cardinal directions. Cars may never occupy the same cell. It is given that the cell immediately in front of each car is empty, and moreover no two cars face towards each other (no right-facing car is to t... | [
"\n\nTo do so, we outline a five-stage plan for the cars.\n\n1. All vertical cars in a green cell may advance one cell into a red cell (or exit $S$ altogether), by the given condition. (This is the only place where the hypothesis about empty space is used!)\n\n2. All horizontal cars on gree... | United States | USA TSTST | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
08qu | Problem:
Find all pairs $(a, b)$ of positive integers such that
$$
11 a b \leq a^{3}-b^{3} \leq 12 a b
$$ | [
"Solution:\nLet $a-b=t$. Due to $a^{3}-b^{3} \\geq 11 a b$ we conclude that $a>b$ so $t$ is a positive integer and the condition can be written as\n$$\n11 b(b+t) \\leq t\\left[b^{2}+b(b+t)+(b+t)^{2}\\right] \\leq 12 b(b+t)\n$$\nSince\n$$\nt\\left[b^{2}+b(b+t)+(b+t)^{2}\\right]=t\\left(b^{2}+b^{2}+b t+b^{2}+2 b t+t^... | JBMO | JBMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (5, 2) | |
0gld | Determine all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying
$$
f(f(x)f(y) + f(y)f(z) + f(z)f(x)) = f(x) + f(y) + f(z)
$$
for all real numbers $x, y, z$. | [
"Let $f$ be any function satisfying (4). Let $\\mathcal{R}_f$ denote the range of $f$. Define $\\mathcal{P}(a, b, c)$ to be the statement $f(ab + bc + ca) = a + b + c$. Equation (4) is then equivalent to $\\mathcal{P}(a, b, c)$ for all $a, b, c \\in \\mathcal{R}_f$.\n\nFor any $a$ in $\\mathcal{R}_f$, $\\mathcal{P}... | Thailand | The 13th Thailand Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 0 for all real x | |
0c6q | Find all positive integers $n$, $n \ge 3$, for which there exists a set $M$, whose elements are $n$ distinct, nonzero vectors, with equal lengths, such that $\sum_{\vec{u} \in M} \vec{u} = \vec{0}$, but for any $\vec{v}, \vec{w} \in M$, we have $\vec{v} + \vec{w} \ne \vec{0}$. | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof and answer | All integers n ≥ 3 with n ≠ 4 | |
001q | Hay 83 rectángulos de lados enteros. Ninguno de ellos es un cuadrado y ninguno tiene área $8$. Con esos rectángulos se puede formar $23$ cuadrados de $4 \times 4$, sin huecos ni superposiciones, y sin que sobren piezas. Decida si con los $83$ rectángulos se puede formar $4$ rectángulos iguales, sin huecos ni superposic... | [] | Argentina | XII Olimpiada Matemática Rióplatense | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | español | proof and answer | Yes | |
0fic | Problem:
Un cuadrado $ABCD$ de centro $O$ y lado $\ell$ gira un ángulo $\alpha$ en torno a $O$. Hallar el área común a los dos cuadrados.
 | [
"Solution:\n\nPrimera solución\n\n\n\nPor la simetría bastará considerar $0<\\alpha<90^\\circ$, ya que la función es periódica con periodo de un cuarto de vuelta. El área pedida $S(\\alpha)$ sale restando del área del cuadrado cuatro triángulos como el $PA' M$.\n\nLlamando $x$ al cateto $PA... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | S(α) = 2ℓ^2 / (1 + sin α + cos α), for 0 ≤ α ≤ 90°, extended to all angles by periodicity of 90° | |
0fux | Problem:
Sei $n \geq 2$ eine natürliche Zahl. Zeige, dass sich das Polynom
$$
\left(x^{2}-1^{2}\right)\left(x^{2}-2^{2}\right)\left(x^{2}-3^{2}\right) \cdots\left(x^{2}-n^{2}\right)+1
$$
nicht als Produkt von zwei nichtkonstanten Polynomen mit ganzen Koeffizienten schreiben lässt. | [
"Solution:\n\nSei $p$ das Polynom in der Aufgabe. Nehme an, es gelte $p(x)=r(x) s(x)$ für zwei nichtkonstante Polynome $r, s$ mit ganzen Koeffizienten. Für $k= \\pm 1, \\pm 2, \\ldots, \\pm n$ ist dann\n$$\n1=p(k)=r(k) s(k)\n$$\nDa beide Faktoren rechts ganze Zahlen sind, gilt somit $r(k)=s(k)= \\pm 1$. Das Polynom... | Switzerland | IMO Selektion | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0c9m | Problem:
Alina şi Bogdan scriu pe rând, începând cu Alina, un $0$ sau un $1$, până când fiecare a scris $2021$ de cifre (fiecare adaugă o cifră la dreapta celor deja existente).
Alina este câştigătoare dacă reprezentarea zecimală a numărului obţinut (în baza $2$) se poate scrie ca suma a două pătrate perfecte; în caz ... | [] | Romania | Al doilea test de selecţie pentru OBMJ | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | Bogdan | |
0572 | Prove or disprove: For every integer $n > 0$, there is a polynomial $a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ satisfying the following conditions:
(1) All coefficients $a_n, a_{n-1}, \dots, a_1, a_0$ are positive real numbers;
(2) At least one of the coefficients is $\frac{1}{n}$;
(3) The value of the polynomia... | [
"The product of any $n$ consecutive integers is divisible by $n$. Thus the conditions are satisfied by the polynomial obtained by removing parentheses and collecting terms in the expression $\\frac{1}{n}(x+1)\\dots(x+n)$."
] | Estonia | Final Round of National Olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
08l2 | Problem:
Let the real parameter $p$ be such that the system
$$
\left\{\begin{array}{l}
p\left(x^{2}-y^{2}\right)=\left(p^{2}-1\right) x y \\
|x-1|+|y|=1
\end{array}\right.
$$
has at least three different real solutions. Find $p$ and solve the system for that $p$. | [
"Solution:\nThe second equation is invariant when $y$ is replaced by $-y$, so let us assume $y \\geq 0$. It is also invariant when $x-1$ is replaced by $-(x-1)$, so let us assume $x \\geq 1$. Under these conditions the equation becomes $x+y=2$, which defines a line on the coordinate plane. The set of points on it t... | JBMO | 2008 Shortlist JBMO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | p = 1 or p = -1; for these values, all solutions are (x, y) with 0 ≤ x ≤ 1 and y = x or y = −x. | |
0501 | Given a triangle $ABC$ where $|BC| = a$, $|CA| = b$ and $|AB| = c$, prove that the equality
$$
\frac{1}{a+b} + \frac{1}{b+c} = \frac{3}{a+b+c}
$$
holds if and only if $\angle ABC = 60^\circ$. | [
"By finding the common denominator on the left hand side, transform the equation to $(a+2b+c)(a+b+c) = 3(a+b)(b+c)$. Expanding the brackets and simplifying gives $b^2 = a^2 + c^2 - ac$. Comparing the latter with the cosine law $b^2 = a^2 + c^2 - 2ac \\cos \\beta$, we see that the equality holds if and only if $\\co... | Estonia | Selected Problems from Open Contests | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof only | null | |
0by5 | Let $n \ge 2$ be a positive integer. Prove that the following assertions are equivalent:
a) for all integer $x$ coprime with $n$ the congruence $x^6 \equiv 1 \pmod{n}$ holds;
b) $n$ divides 504. | [
"We have $504 = 2^3 \\cdot 3^2 \\cdot 7$.\n\na) $\\Rightarrow$ b). First we prove that $(n, 5) = 1$. If $(n, 5) \\neq 1$ then $n = 5k$.\nThe sequence $(5m + 2)_{m \\ge 1}$ is an arithmetic sequence with ratio 5. Its first term being coprime with the ratio, from Dirichlet's Theorem it follows that the sequence conta... | Romania | THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > ... | English | proof only | null | |
05rl | Problem:
Trouver les fonctions $f: \mathbb{R} \mapsto \mathbb{R}$ telles que, pour tous les réels $x$ et $y$, on ait:
$$
f\left(x^{2}+x+f(y)\right)=y+f(x)+f(x)^{2}
$$ | [
"Solution:\nTout d'abord, il est clair que la fonction $f: x \\mapsto x$ est une solution. On va montrer que c'est la seule.\n\nDans la suite, on notera $E_{x, y}$ l'équation $f\\left(x^{2}+x+f(y)\\right)=y+f(x)+f(x)^{2}$. En premier lieu, $E_{0, y}$ indique que $f(f(y))=y+f(0)+f(0)^{2}$, ce qui montre que $f$ est ... | France | TEST DU 27 FÉVRIER 2019 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x | |
02ln | Let $ABCD$ be a cyclic quadrilateral and $r$ and $s$ the lines obtained reflecting $AB$ with respect to the internal bisectors of $\angle CAD$ and $\angle CBD$, respectively. If $P$ is the intersection of $r$ and $s$ and $O$ is the center of the circumscribed circle of $ABCD$, prove that $OP$ is perpendicular to $CD$. | [
"\nLet $E$ and $F$ be the intersections of $r$ and $s$ with the circumcircle of $ABCD$, respectively. Since $ABCD$ is cyclic and $BA$ and $BF$ are symmetric with respect to the bisector of $\\angle CBD$, $\\angle ACD = \\angle ABD = \\angle CBF = \\angle CAF$, hence $AF$ and $CD$ are parall... | Brazil | XXX OBM | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
05ls | Problem:
Les Xantiens sont les habitants, en nombre éventuellement infini, de la planète Xanta. Vis-à-vis d'eux-mêmes et de leurs semblables, les Xantiens sont capables de ressentir deux types d'émotions, qu'ils appellent amour et respect. Il a été observé que:
- Chaque Xantien aime un et un seul Xantien, et respecte ... | [
"Solution:\n\nPour chaque Xantien $x$, désignons respectivement par $f(x)$ et $g(x)$ les Xantiens aimés et respectés par $x$.\nLa première condition assure que l'on a bien défini des fonctions $f$ et $g$ de l'ensemble $X$ des Xantiens sur lui-même.\nIl s'agit de savoir si, pour tout $x$, on a $f(x)=g(x)$.\nEn fait,... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Functional equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
021x | Problem:
Fração de fração - Qual o valor de $1+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}$ ? | [
"Solution:\n\nFração de fração - Temos:\n$$\n1+\\frac{1}{1+\\frac{1}{1+\\frac{1}{2}}}=1+\\frac{1}{1+\\frac{1}{\\frac{3}{2}}}=1+\\frac{1}{1+\\frac{2}{3}}=1+\\frac{1}{\\frac{5}{3}}=1+\\frac{3}{5}=\\frac{8}{5}\n$$"
] | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 8/5 | |
099u | Let $ABCD$ be a convex quadrilateral and internal bisectors of angles $B$ and $D$ meet at $I$. Fix points $M$ and $N$ on the segments $BI$ and $DI$ respectively such that $\angle MAN = \frac{1}{2} \cdot \angle A$.
Find the condition about the $ABCD$ quadrilateral such that $\angle MCN = \frac{1}{2} \cdot \angle C$?
(... | [
"Answer: $ABCD$ be a convex and circumscribed quadrilateral.\n\nFirst case, $ABCD$ be a not circumscribed quadrilateral. Hence, $I$ doesn't lie on the internal bisector of $A$, thus which bisector must be intersects with $BI$ and $CI$.\nAssume that, bisector of $A$ and line $DI$ meet at $I'$. (Figure).\n\n![](attac... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | ABCD must be a convex circumscribed (tangential) quadrilateral. | |
08vw | Two people, $A$ and $B$, go up a staircase with a certain number of steps. $A$ goes up 2 steps in one stride, while $B$ goes up 5 steps in one stride, except if $A$ finds only 1 step before the last stride, then he will go up just 1 step to finish, and if $B$ finds 4 or less steps remaining before the last stride, then... | [
"$[19, 20, 21, 22]$\n\nLet $n$ be the number of steps in the staircase, $a$ and $b$ be the number of strides $A$ and $B$ have taken, respectively. Then, we have\n$$\na - b = 6, \\quad \\frac{n}{2} \\le a < \\frac{n}{2} + 1, \\quad \\frac{n}{5} \\le b < \\frac{n}{5} + 1.\n$$\nFrom these we obtain $\\frac{n}{2} - (\\... | Japan | Japan Junior Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 19, 20, 21, 22 | |
0ide | Problem:
Find the ordered quadruple of digits $(A, B, C, D)$, with $A > B > C > D$, such that
$$
\begin{aligned}
& ABCD \\
- & DCBA \\
= & BDAC .
\end{aligned}
$$ | [
"Solution:\nSince $D < A$, when $A$ is subtracted from $D$ in the ones' column, there will be a borrow from $C$ in the tens' column. Thus, $D + 10 - A = C$.\n\nNext, consider the subtraction in the tens' column, $(C - 1) - B$. Since $C < B$, there will be a borrow from the hundreds' column, so $(C - 1 + 10) - B = A... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | (7, 6, 4, 1) | |
0kj7 | Problem:
Circle $\omega$ is inscribed in rhombus $H M_{1} M_{2} T$ so that $\omega$ is tangent to $\overline{H M_{1}}$ at $A$, $\overline{M_{1} M_{2}}$ at $I$, $\overline{M_{2} T}$ at $M$, and $\overline{T H}$ at $E$. Given that the area of $H M_{1} M_{2} T$ is $1440$ and the area of $E M T$ is $405$, find the area of... | [
"Solution:\n\nFirst, from equal tangents, we know that $T E = T M$. As the sides of a rhombus are also equal, this gives from SAS similarity that $E M T \\sim T H M_{2}$. Further, the ratio of their areas is $\\frac{405}{1440 / 2} = \\frac{9}{16}$. This means that $T E = T M = \\frac{3}{4} H T$. Then, we get that $... | United States | HMMT November 2021 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals"
] | null | proof and answer | 540 | |
0bdb | Problem:
Számítsd ki a $\lim_{n \rightarrow \infty} \int_{0}^{1} e^{x^{n}} \, dx$ határértéket! | [] | Romania | Matematika tantárgyverseny Megyei szakasz | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Limits"
] | null | proof and answer | 1 | |
08r3 | $n$ is a positive 3-digit number, and its hundred's place and its one's place are not $0$. Make a new number by exchanging the hundred's place and the one's place of $n$, and call it $m$. Answer the maximum value of $n - m$. | [
"Let $n = 100A + 10B + C$. Then,\n$$\nn - m = (100A + 10B + C) - (A + 10B + 100C) = 99(A - C).\n$$\nSince $A$ and $C$ can be any natural number between $1$ and $9$, the maximum value of $n - m$ is $99 \\cdot (9 - 1) = 792$."
] | Japan | The 16th Japanese Mathematical Olympiad - The First Round | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 792 | |
0fdn | Problem:
Hallad todas las posibles formas de escribir $2003$ como suma de dos cuadrados de números enteros positivos. | [
"Solution:\n\nNo es posible escribir $2003$ como suma de dos cuadrados de números enteros positivos:\n\nPero $2003 \\equiv 3 \\pmod{4}$."
] | Spain | null | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | no representations | |
0hli | Problem:
Let $P$ be the point inside the square $ABCD$ such that $\triangle ABP$ is equilateral. Calculate the angle $\angle CPD$. Explain your answer! | [
"Solution:\n\nThe triangle $DAP$ is isosceles because $AD = AP$, hence $\\angle ADP = \\angle APD = \\frac{180^\\circ - \\angle DAP}{2} = 75^\\circ$. Hence $\\angle PDC = 15^\\circ$. Similarly, $\\angle DCP = 15^\\circ$ and hence $\\angle CPD = 180^\\circ - 2 \\cdot 15^\\circ = 150^\\circ$."
] | United States | Berkeley Math Circle Monthly Contest 7 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 150° | |
04x7 | Let $\mathbb{R}^+$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ satisfying
$$
(1 + y f(x))(1 - y f(x + y)) = 1
$$
for any $x, y \in \mathbb{R}^+$. | [
"Simple manipulation with the given equation leads to\n$$\n\\begin{aligned}\n1 + y f(x) - y f(x + y) - y^2 f(x) f(x + y) &= 1, \\\\\ny f(x) - y f(x + y) &= y^2 f(x) f(x + y).\n\\end{aligned}\n$$\nAfter cancelling $y \\neq 0$ another manipulation gives\n$$\n\\begin{aligned}\nf(x) - f(x + y) &= y f(x) f(x + y), \\\\\... | Czech-Polish-Slovak Mathematical Match | Cesko-Slovacko-Poljsko | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 1/(x + c) for c >= 0 | |
06l8 | Let $ABC$ be an acute-angled triangle such that $\angle ACB = 45^\circ$. Let $G$ be the point of intersection of the three medians and let $O$ be the circumcentre. Suppose $OG = 1$ and $OG // BC$. Determine the length of the segment $BC$. | [] | Hong Kong | HKG TST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | BC = 12 |
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