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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
01gy | Let $ABC$ be an acute triangle with a circumcircle $\omega$. Let $\ell$ be the tangent line to $\omega$ at $A$. Let $X$ and $Y$ be the projections of $B$ onto lines $\ell$ and $AC$, respectively. Let $H$ be the orthocenter of $BXY$. Let $CH$ intersect $\ell$ at $D$. Prove that $BA$ bisects angle $CBD$. | [
"\nNote that $XH \\perp BY \\perp AC$ and $YH \\perp BX \\perp AD$. Therefore $XH \\parallel AC$ and $YH \\parallel AD$. It follows that\n$$\n\\frac{AD}{AX} = \\frac{CD}{CH} = \\frac{CA}{CY} \\implies \\frac{AD}{CA} = \\frac{AX}{CY} = \\frac{AB \\sin \\angle XAB}{CB \\sin \\angle YCB}.\n$$\... | Baltic Way | Baltic Way 2020 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0632 | Problem:
Es seien $f$ und $g$ zwei reelle Funktionen, die für jede reelle Zahl definiert sind. Ferner soll für alle reellen Zahlen $x$ und $y$ die Gleichung gelten:
$$
g(f(x+y))=f(x)+(2 x+y) g(y)
$$
Man bestimme alle möglichen Paare $(f, g)$. | [
"Solution:\n\nAus der gegebenen Gleichung (1) erhalten wir für $y=-2x$: $g(f(-x))=f(x)$ (2). Damit wird (1) zu $f(-x-y)=f(x)+(2x+y)g(y)$ (3) und mit $x=0$ zu $f(-y)=f(0)+y g(y)$ (4).\n\nErsetzen von $f$ in (3) gemäß (4) liefert nach Subtraktion von $f(0)$:\n$$(x+y)g(x+y) = (2x+y)g(y) - xg(-x)$$\n(5).\n\nMit $y=0$ e... | Germany | 2. Auswahlklausur | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | (i) f(x) = 0 and g(x) = 0 for all real x; (ii) f(x) = x^2 + a and g(x) = x for all real x, where a is any real constant. | |
0jo7 | Problem:
Consider a $2 \times 2$ grid of squares. David writes a positive integer in each of the squares. Next to each row, he writes the product of the numbers in the row, and next to each column, he writes the product of the numbers in each column. If the sum of the eight numbers he writes down is $2015$, what is th... | [
"Solution:\n\nLet the four numbers be $a, b, c, d$, so that the other four numbers are $ab, ad, bc, bd$. The sum of these eight numbers is $a + b + c + d + ab + ad + bc + bd = (a + c) + (b + d) + (a + c)(b + d) = 2015$, and so $(a + c + 1)(b + d + 1) = 2016$. Since we seek to minimize $a + b + c + d$, we need to fi... | United States | HMMT November | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 88 | |
09e2 | Let $M$ be arbitrary point inside triangle $ABC$. Lines $AM$, $BM$, $CM$ intersect circumcircle of the triangle $ABC$ at points $D$, $E$, $F$ respectively. Denote $O_1$, $O_2$, $O_3$ centres of circumcircles of triangles $BCM$, $CAM$, $ABM$ respectively. Show that lines $DO_1$, $EO_2$, $FO_3$ pass through a point. | [
"Let $H$ be intersection point of the line $DO_1$ with circumcircle of the triangle $ABC$. Assume that $CO_2 \\cap EH = O'_2$. Now let's prove that $O_2 \\equiv O'_2$.\n\nLet $\\angle MAC = \\angle DAC = \\phi$, $\\angle CBM = \\angle CBE = \\psi$. Since points $A$, $C$, $D$, $E$ are concyclic,\n$$\n\\begin{aligned... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, n... | English | proof only | null | |
0iy1 | Problem:
In $\triangle ABC$, $D$ is the midpoint of $BC$, $E$ is the foot of the perpendicular from $A$ to $BC$, and $F$ is the foot of the perpendicular from $D$ to $AC$. Given that $BE = 5$, $EC = 9$, and the area of triangle $ABC$ is $84$, compute $|EF|$. | [
"Solution:\n\nThere are two possibilities for the triangle $ABC$ based on whether $E$ is between $B$ and $C$ or not. We first consider the former case.\n\nWe find from the area and the Pythagorean theorem that $AE = 12$, $AB = 13$, and $AC = 15$. We can then use Stewart's theorem to obtain $AD = 2\\sqrt{37}$.\n\nSi... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 6√37/5 | |
03gt | Problem:
Find all number triples $(x, y, z)$ such that when any one of these numbers is added to the product of the other two, the result is $2$. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (1, 1, 1) and (-2, -2, -2) | |
09dd | 5-аас их бөгөөд хамгийн их ба бага 2-ынх нь ялгавар 10-аас бага 4 анхны тооны нийлбэр 60-д хуваагдахыг батал. | [
"Бодлогын нөхцөлийг хангах тоонуудын хамгийн багыг $p$ гээ. Тэгвэл өгөгдсөн анхны тоонууд нь тоонуудын $p$, $p + 2$, $p + 4$, $p + 6$, $p + 8$ аль нэг 4 л байна. Дараалсан 3 сондгой тоонуудын аль нэг нь 3-т хуваагдах тул $3 \\mid p + 4$, мөн дараалсан 5 сондгой тооны аль нэг нь 5-аар төгсөх буюу 5-д хуваагдах тул $... | Mongolia | ММО-48 | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | Mongolian | proof only | null | |
05uq | Problem:
Soit $a$, $b$, $c$ et $d$ des entiers naturels non nuls. On suppose que $a! + b! = c! + d!$. Démontrer que $a b = c d$. | [
"Solution:\n\nSupposons sans perte de généralité que $a \\leqslant b$, $c \\leqslant d$ et $a \\leqslant c$. Alors\n$$\nb! = c! - a! + d! \\geqslant d!\n$$\ndonc $b \\geqslant d$, de sorte que $a \\leqslant c \\leqslant d \\leqslant b$. L'entier $c!$ divise donc $c! + d! - b! = a!$, ce qui signifie que $c \\leqslan... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0851 | Problem:
Sia $ABCD$ un quadrilatero; chiamiamo $E$ l'intersezione (distinta da $A$) tra le circonferenze di diametri $AB$ e $AC$ ed $F$ l'intersezione (sempre distinta da $A$) tra le circonferenze di diametri $AC$ e $AD$. Dimostrare che:
a. se $E\widehat{A}D = 90^\circ$ allora $BC$ è parallelo a $AD$
b. se $E\wideha... | [
"Solution:\n\na. Siano $M, N$ i punti medi di $AB$ e $AC$ rispettivamente; in quanto corda comune ai due cerchi di diametri $AB$ e $AC$, $AE$ è perpendicolare a $MN$. Se $E\\widehat{A}D = 90^\\circ$, $AD$ e $MN$, in quanto perpendicolari alla stessa retta (quella che contiene $EA$), sono paralleli; infine per il te... | Italy | Progetto Olimpiadi di Matematica 2006 GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0923 | Problem:
Consider finitely many points in the plane with no three points on a line. All these points can be coloured red or green such that any triangle with vertices of the same colour contains at least one point of the other colour in its interior.
What is the maximal possible number of points with this property? | [
"Solution:\n\nThe answer is $8$.\n\nCall a set consisting of red points and green points good if no three points are collinear and any unicoloured triangle contains a point of the other colour.\n\nOn the one hand, the figure on the left below shows an example of a good set with $8$ points. The figure on the right s... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 8 | |
0gg6 | 設 $p$ 為奇質數。證明整數
$$
1! + 2! + 3! + \cdots + p! - \left\lfloor \frac{(p-1)!}{e} \right\rfloor
$$
必能被 $p$ 整除。 | [] | Taiwan | 2022 數學奧林匹亞競賽第五階段培訓營, 國際競賽實作(二) | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | Chinese; English | proof only | null | |
04ov | Let $ABC$ be an isosceles triangle such that $|AB| = |AC|$ and $\angle BAC < 60^\circ$. Let $D$ be the point on the segment $\overline{AC}$ such that $\angle DBC = \angle BAC$, $E$ be the intersection of the perpendicular bisector of the segment $\overline{BD}$ and the line passing through $A$ parallel to $BC$, and $F$... | [
"a) Since $\\triangle DBC = \\triangle BAC$ and $\\triangle BCD = \\triangle BCA$, the triangles $BCD$ and $ABC$ are similar. Hence, $|BD| = |BC|$.\n\nLet $E'$ be the point such that the quadrilateral $AE'BC$ is a parallelogram. Since $\\triangle ABE' = \\triangle BAC = \\triangle CBD$, we get $\\triangle ABC = \\t... | Croatia | Croatian Mathematical Society Competitions | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
09na | Let $M_2(\mathbb{R})$ denote the space of $2 \times 2$ real matrices with addition
$$
\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} + \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix} = \begin{pmatrix} a_{11} + b_{11} & a_{12} + b_{12} \\ a_{21} + b_{21} & a_{22} + b_{22} \end{pmatrix}... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | Only the zero map φ(A) = 0 for all A | |
0jug | Problem:
Let $n > 1$ be an odd integer. On an $n \times n$ chessboard the center square and four corners are deleted. We wish to group the remaining $n^{2}-5$ squares into $\frac{1}{2}\left(n^{2}-5\right)$ pairs, such that the two squares in each pair intersect at exactly one point (i.e. they are diagonally adjacent, ... | [
"Solution:\n\nConstructions for $n=3$ and $n=5$ are easy. For $n>5$, color the odd rows black and the even rows white. If the squares can be paired in the way desired, each pair we choose must have one black cell and one white cell, so the numbers of black cells and white cells are the same. The number of black cel... | United States | HMMT February 2016 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | n = 3 and n = 5 | |
0fob | Let $S = \{1, 2, \dots, n\}$, $n \ge 2$, and let $f : S \to S$ be a bijective function distinct from the identity. Let $u = \sum_{k=1}^{n} |f(k) - k|$ and let $v$ be the number of ordered pairs $(a, b)$ of elements of $S$ such that $a > b$ and $f(a) < f(b)$. Show that $v < u \le 2v$, and that $u = 2v$ if and only if th... | [
"We say a pair $(a, b)$ is *a-good* if $a > b \\ge f(a)$, and *b-good* if $b \\le f(a) < f(b)$. If $f(k) = k$, no pair $(k, b)$ or $(a, k)$ is *k-good*. If $f(k) > k$, there are no *k-good* pairs of the form $(k, b)$ and exactly $f(k) - k$ of the form $(a, k)$. Likewise, if $f(k) < k$ there are exactly $k - f(k)$ *... | Spain | BarcelonaTech Mathcontest | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Spanish | proof only | null | |
07w3 | Suppose $a$, $b$, $c$, $d \in [-1, 1]$. Prove that
$$
|ad + bc + ca - db| \leq 2
$$
and determine the case of equality. | [
"**Solution 1.** Note that\n$$\n\\begin{align*}\n(ad + bc + ca - db)^2 &= (a(c+d) + b(c-d))^2 \\\\\n&= a^2(c+d)^2 + 2ab(c^2-d^2) + b^2(c-d)^2 \\\\\n&\\leq (c+d)^2 + 2ab(c^2-d^2) + (c-d)^2 \\\\\n&= 2(c^2+d^2) + 2ab(c^2-d^2) \\\\\n&= 2[(1+ab)c^2 + (1-ab)d^2] \\\\\n&\\leq 2[(1+ab) + (1-ab)] \\quad (\\text{since } -1 \... | Ireland | IRL_ABooklet_2023 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | The bound |ad + bc + ca − db| ≤ 2 holds. Equality |ad + bc + ca − db| = 2 occurs exactly in the following cases:
- a = b = ±1 and c = ±1 (with d arbitrary in [−1, 1]);
- a = −b = ±1 and d = ±1 (with c arbitrary in [−1, 1]);
- a = ±1, |b| < 1, and c = d = ±1;
- b = ±1, |a| < 1, and c = −d = ±1. | |
0i9i | Problem:
A hotel consists of a $2 \times 8$ square grid of rooms, each occupied by one guest. All the guests are uncomfortable, so each guest would like to move to one of the adjoining rooms (horizontally or vertically). Of course, they should do this simultaneously, in such a way that each room will again have one gu... | [
"Solution:\n\nImagine that the rooms are colored black and white, checkerboard-style. Each guest in a black room moves to an adjacent white room (and vice versa). If, for each such guest, we place a domino over the original room and the new room, we obtain a covering of the $2 \\times n$ grid by $n$ dominoes, since... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 1156 | |
0l9f | Let $A$ be the set of all permutations $a = (a_1, a_2, ..., a_{2003})$ of $2003$ first positive integers such that each permutation satisfies the condition: there is no nonvoid proper subset $S$ of the set $\{1, 2, ..., 2003\}$ such that $\{a_k \mid k \in S\} = S$.
For each $a = (a_1, a_2, ..., a_{2003}) \in A$, put $... | [] | Vietnam | Vietnamese Team Selection Contest for the 44th IMO | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | d0 = 8006.
Minimizers: exactly the two 2003-cycles with the zigzag pattern.
One is given by
- for odd k ≤ 2001: a_k = k + 2,
- a_2003 = 2002,
- for even k ≥ 4: a_k = k − 2,
- a_2 = 1.
The other is its inverse:
- for odd k ≥ 3: a_k = k − 2,
- a_1 = 2,
- for even k ≤ 2002: a_k = k + 2,
- a_2002 = 2003. | |
04a7 | Let $a$, $b$, $c$ be complex numbers such that $a + b + c = 0$ and $ab + bc + ca = 0$. Prove $|a| = |b| = |c|$. | [
"Let $a$, $b$, $c$ be complex numbers such that $a + b + c = 0$ and $ab + bc + ca = 0$.\n\nLet us consider the elementary symmetric polynomials:\n\n- $s_1 = a + b + c = 0$\n- $s_2 = ab + bc + ca = 0$\n- $s_3 = abc$\n\nThe roots $a$, $b$, $c$ are the roots of the cubic equation:\n$$\nx^3 - s_1 x^2 + s_2 x - s_3 = 0\... | Croatia | Hrvatska 2011 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof only | null | |
0kme | Which of the following conditions is sufficient to guarantee that integers $x$, $y$, and $z$ satisfy the equation
$$
x(x - y) + y(y - z) + z(z - x) = 1?
$$
(A) $x > y$ and $y = z$
(B) $x = y - 1$ and $y = z - 1$
(C) $x = z + 1$ and $y = x + 1$
(D) $x = z$ and $y - 1 = x$
(E) $x + y + z = 1$ | [] | United States | AMC 12 B | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | D | |
0dcq | Let $19$ integer numbers are given. Let Hamza writes on the paper the greatest common divisor for each pair of numbers. It occurs that the difference between the biggest and smallest numbers written on the paper is less than $180$. Prove that not all numbers on the paper are different. | [
"Let $a_{1}, a_{2}, \\ldots, a_{19}$ be the given numbers and suppose on the contrary that the set $S$ of all $\\binom{19}{2} = 171$ numbers, which are written on the paper, are all different,\n$$\nd_{1} < d_{2} < \\ldots < d_{171}\n$$\nDenote $k$ as the number of even values among the $19$ given numbers, and $t$ a... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0g7s | 請找出所有的函數 $g: \mathbb{R} \to \mathbb{R}$ 使得對於所有的 $x, y \in \mathbb{R}$ 恆有
$$
(4x + g(x)^2)g(y) = 4g\left(\frac{y}{2} \cdot g(x)\right) + 4xy \cdot g(x).
$$ | [
"首先我們將 $(4x + g(x)^2)g(y) = 4g(\\frac{y}{2} \\cdot g(x)) + 4xy \\cdot g(x)$ 寫成\n$$\n\\left(x + \\left(\\frac{g(x)}{2}\\right)^2\\right) \\frac{g(y)}{2} = \\frac{1}{2} g\\left(y \\cdot \\frac{g(x)}{2}\\right) + xy \\cdot \\frac{g(x)}{2}.\n$$\n令 $f(x) = g(x)/2$, 則原式可化成\n$$\n(x + f(x)^2)f(y) = f(yf(x)) + xyf(x). \\qua... | Taiwan | 二〇一三數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | g(x) = 0 or g(x) = 2x | |
0ke0 | Problem:
Let $C_{k} = \frac{1}{k+1} \binom{2k}{k}$ denote the $k^{\text{th}}$ Catalan number and $p$ be an odd prime. Prove that exactly half of the numbers in the set
$$
\left\{\sum_{k=1}^{p-1} C_{k} n^{k} \mid n \in \{1,2, \ldots, p-1\}\right\}
$$
are divisible by $p$. | [
"Solution:\nWe work in $\\mathbb{F}_{p}[X]$.\nWe claim that\n$$\n(1-4 X)^{\\frac{p+1}{2}}-1+2 X^{p}+2 X \\sum_{k=0}^{p-1} C_{k} X^{k}=0\n$$\nThe solution follows from this claim, as\n$$\n\\sum_{k=1}^{p-1} C_{k} n^{k} \\equiv 0 \\quad(\\bmod p) \\Longleftrightarrow (1-4 n)^{\\frac{p+1}{2}}-1+2 n^{p}+2 n \\equiv 0 \\... | United States | HMIC | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Algeb... | null | proof only | null | |
09cs | $\Delta ABC$-д $\angle BCA=90^\circ$ ба $C$ оройгоос татсан өндрийн суурь $D$ байг. $CD$ хэрмийн дотор $X$ цэг авах. $K$ цэг $AX$ хэрчим дээр орших ба $BK=BC$ байв. Мөн адилаар $L$ цэг $BX$ хэрчим дээр орших ба $AL=AC$ байв. $AL$ ба $BK$ хэрчмүүдийн огтлоодлын цэгийг $M$ гэвэл $MK=ML$ гэж батал. | [
"Let $C''$ be the reflection of $C$ in the line $AB$, and let $\\omega_1$ and $\\omega_2$ be the circles with centers $A$ and $B$, passing through $L$ and $K$ respectively. Since $AC'' = AC = AL$ and $BC'' = BC = BK$, both $\\omega_1$ and $\\omega_2$ pass through $C$ and $C''$. By $\\angle BCA = 90^\\circ$, $AC$ is... | Mongolia | ОУМО-53 | [
"Geometry > Plane Geometry > Circles > Tangents"
] | Mongolian | proof only | null | |
07j9 | Let $A_1A_2...A_{99}$ be a regular polygon and $A_{100}$ its center. $\mathcal{A}$i wants to draw each of segments $A_iA_j$ ($1 \le i < j \le 100$) using one of $k$ colors such that no two segments of the same color share an endpoint or cross. Find the smallest positive integer $k$ for which such an operation is possib... | [
"Suppose that $n = 99$. We claim that the answer would be $\\frac{4}{3}n$.\n\nFirst of all, consider these three lines:\n- Lines of type $R$: they are vertices drawn from the center;\n- Lines of type $L$: they are the largest diagonals;\n- Lines of type $S$: the second largest diagonals.\n\nIt is clear that the onl... | Iran | 41th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 132 | |
03ov | It is given that there are two sets of real numbers $A = \{a_1, a_2, \dots, a_{100}\}$ and $B = \{b_1, b_2, \dots, b_{50}\}$. If there is a mapping $f$ from $A$ to $B$ such that every element in $B$ has an inverse image and
$$
f(a_1) \le f(a_2) \le \dots \le f(a_{100}),
$$
then the number of such mappings is ( ).
(A) ... | [
"We might as well suppose $b_1 < b_2 < \\dots < b_{50}$, and divide elements $a_1, a_2, \\dots, a_{100}$ in $A$ into 50 nonempty groups according to their order. Define a mapping $f: A \\to B$, so that the images of all the elements in the $i$-th group are $b_i$ ($i = 1, 2, \\dots, 50$) under the mapping. Obviously... | China | China Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | MCQ | D | |
0a1t | Problem:
Vind alle positieve gehele getallen $n$ waarvoor er $n$ verschillende natuurlijke getallen $a_{1}, a_{2}, \ldots, a_{n}$ bestaan, geen van alle groter dan $n^{2}$, zo dat de som
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}=1.
$$ | [
"Solution:\n\nOplossing I. Het antwoord is alle $n \\neq 2$. Voor $n=1$, werkt de verzameling $\\{1\\}$. Voor $n=2$, voldoet geen enkele verzameling. Als $a_{1}$ of $a_{2}$ gelijk is aan 1, dan is $\\frac{1}{a_{1}}+\\frac{1}{a_{2}}>1$. Als $a_{1}$ en $a_{2}$ allebei minstens twee zijn, dan is de één minstens 2 en d... | Netherlands | IMO-selectietoets | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Other"
] | null | proof and answer | All positive integers except 2 | |
0ip5 | Problem:
Let $ABCD$ be a unit square (that is, the labels $A$, $B$, $C$, $D$ appear in that order around the square). Let $X$ be a point outside of the square such that the distance from $X$ to $AC$ is equal to the distance from $X$ to $BD$, and also that $AX = \frac{\sqrt{2}}{2}$. Determine the value of $CX^{2}$.
![... | [
"Solution:\n\nSince $X$ is equidistant from $AC$ and $BD$, it must lie on either the perpendicular bisector of $AB$ or the perpendicular bisector of $AD$. It turns that the two cases yield the same answer, so we will just assume the first case. Let $M$ be the midpoint of $AB$ and $N$ the midpoint of $CD$. Then, $XM... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 5/2 | |
00pf | Let $n$ be a positive integer number such that $p = 17^{2n} + 4$ is prime. Show that $7^{(p-1)/2} + 1$ is divisible by $p$. | [
"Write $p = 17^{2n} + 4 \\equiv 4^n + 4 \\pmod 5$ to deduce that $p \\equiv 0 \\pmod 5$ if $n$ is even. Since $p$ is prime, $n$ must be $0$, so $p = 5$ and the conclusion follows.\n\nHenceforth assume $n$ odd. Rule out the case $n \\equiv 5 \\pmod 6$ on account of $p = 17^{2n} + 4 \\equiv 3^n + 4 \\pmod{13} \\equiv... | Balkan Mathematical Olympiad | shortlistBMO 2011 | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof only | null | |
03ix | Problem:
Prove that for every prime number $p$, there are infinitely many positive integers $n$ such that $p$ divides $2^{n}-n$. | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof only | null | |
01zm | There are $n$ points $A_1, \dots, A_n$ on the plane with rational coordinates, all pairwise distances between them being integer. Prove that there exist points $B_1, \dots, B_n$ on the plane with integer coordinates such that $|B_iB_j| = |A_iA_j|$ for all $1 \le i < j \le n$. | [
"Let $d$ be the least common multiple of the denominators of all coordinates of the points $A_1, \\dots, A_n$. After homothety with center at the origin $O$ and coefficient $d$, these points will go to points $C_1, \\dots, C_n$, respectively, with integer coordinates. Let us switch to complex numbers and work with ... | Belarus | SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Number Theory > Algebraic Number Theory > Quadratic fields",
"Number Theory > Algebraic Number Theo... | English | proof only | null | |
0695 | Let $\triangle AB\Gamma$ be a triangle with $AB > A\Gamma$. Let point $\Delta$ be on the side $AB$ such that $B\Delta = A\Gamma$. We draw the circle $\gamma$ passing through $\Delta$ and tangent to the side $A\Gamma$ at $A$. The circumcircle $\omega$ of the triangle $AB\Gamma$ meets circle $\gamma$ at $A$ and $E$. Prov... | [
"\nfigure 3\nIt is enough to prove that $EB = E\\Gamma$ and $EA = E\\Delta$. For that we compare the triangles $\\triangle EB$ and $\\triangle AE\\Gamma$ which have: $B\\Delta = A\\Gamma$, $\\Delta\\hat\\{B\\}E = A\\hat\\{\\Gamma\\}E$ (inscribed to the same arch $\\omega$) and $B\\hat\\{\\D... | Greece | SELECTION EXAMINATION 2019 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
0kbh | Problem:
How many ways can the vertices of a cube be colored red or blue so that the color of each vertex is the color of the majority of the three vertices adjacent to it? | [
"Solution:\n\nIf all vertices of the cube are of the same color, then there are 2 ways. Otherwise, look at a red vertex. Since it must have at least 2 red neighbors, there is a face of the cube containing 3 red vertices. The last vertex on this face must also be red. Similarly, all vertices on the opposite face mus... | United States | HMMT February 2020 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 8 | |
06e4 | For every positive integer $n$, let $f(n)$ be the number of all positive integers with exactly $2n$ digits, each having exactly $n$ of the digits equal to $1$ and the other $n$ digits equal to $2$. Let $g(n)$ be the number of all positive integers with exactly $n$ digits, each of its digits can only be $1$, $2$, $3$ or... | [
"We consider the following mapping between the set $S$ of all positive integers formed by $n$ $1$'s and $n$ $2$'s and the set $T$ of all positive integers with $n$ digits formed by $1$, $2$, $3$, $4$ such that the numbers of $1$'s and $2$'s are equal.\n\nFor each $m \\in S$, we can pair up every two consecutive dig... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0hrq | Problem:
A Mystic Four Calculator has a four-digit display and four buttons. The calculator works as follows: Pressing button 1 replaces the number in the display with $1$; Pressing button 2 divides the number in the display by $2$; Pressing button 3 subtracts $3$ from the number in the display; Pressing button 4 multi... | [
"Solution:\n\na. No. Notice that if the number on the display is not divisible by $3$, then none of the operations can have as a result a number divisible by $3$. At the start, $1$ is the only button producing a result, so we are required to press it at some point. After that the number will never be divisible by $... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | a: No; b: Yes | |
0gl0 | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y \in \mathbb{R}$,
$$
f(xy - 1) + f(x)f(y) = 2xy - 1.
$$ | [
"Let $f$ be any function satisfying\n$$\nf(xy - 1) + f(x)f(y) = 2xy - 1 \\quad (1)\n$$\nfor all $x, y \\in \\mathbb{R}$. Plug in $x = 0$ in (1), we get $f(-1) + f(x)f(0) = -1$ for all $x \\in \\mathbb{R}$.\nIf $f(0) \\neq 0$, then $f$ is a constant function which does not satisfy the equation (1) for all $x, y \\in... | Thailand | Tajland 2014 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = x for all real x, or f(x) = -x^2 for all real x | |
0fl6 | Problem:
Halla todos los números naturales $n$ que verifican la condición:
$$
\left[\frac{n}{2}\right]+\left[\frac{2 n}{3}\right]=n+335
$$
donde $[x]$ es la parte entera de $x$. (Esto es, $[1,32]=1$, $[2]=2$, $\left[\frac{1}{2}\right]=0$, $[\pi]=3$, etc.) | [
"Solution:\n\nDistinguimos casos según $n$ sea de la forma $6 k$, $6 k+1$, $6 k+2$, $6 k+3$, $6 k+4$ o $6 k+5$ (observemos que $n$ es siempre de alguna de estas seis formas) y hacemos la siguiente tabla:\n\n| $n$ | $\\left[\\frac{n}{2}\\right]$ | $\\left[\\frac{2 n}{3}\\right]$ | $\\left[\\frac{n}{2}\\right]+\\left... | Spain | Spain | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2010, 2012, 2013, 2014, 2015, 2017 | |
0fyb | Problem:
Sei $n$ eine natürliche Zahl. Bestimme die Anzahl Paare $(a, b)$ natürlicher Zahlen, für welche die folgende Gleichung erfüllt ist:
$$
(4 a-b)(4 b-a)=2010^{n}
$$ | [
"Solution:\n\nSetze $x=4 a-b$ und $y=4 b-a$. Wegen $x y>0$ und $x+y=3(a+b)>0$ sind $x$ und $y$ natürliche Zahlen. Ausserdem gilt\n$$\na=\\frac{4 x+y}{15}, \\quad b=\\frac{4 y+x}{15}\n$$\nalso müssen $4 x+y$ und $4 y+x$ beide durch $15$ teilbar sein. Wegen $15 \\mid x y$ ist aber mindestens eine der beiden Zahlen $x... | Switzerland | SMO Finalrunde | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (n^2 - 1)^2 | |
079s | Are there 2000 real numbers (not necessarily distinct), not all zero, such that if we put any 1000 of these numbers as roots of a monic polynomial of degree 1000, its coefficients (except the coefficient of $x^{1000}$) are a permutation of the 1000 remaining numbers? | [
"First suppose the case that none of the numbers are zero. Note that there exist at least 1000 positive numbers or at least 1000 negative numbers among these 2000 numbers. If there exist at least 1000 negative numbers and we put these 1000 numbers as roots of a degree 1000 polynomial all its coefficients are positi... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | No, such numbers do not exist. | |
0jh0 | Given positive integers $m$ and $n$, prove that there is a positive integer $c$ such that the numbers $cm$ and $cn$ have the same number of occurrences of each non-zero digit when written in base ten. | [
"Since $t$ is relatively prime to $10$ there is a smallest exponent $b$ for which $t \\mid (10^b - 1)$. Thus $b$ is the number of digits in the repeating portion of the decimal expansion for $\\frac{1}{t}$. More precisely, if we write $tc = (10^b - 1)$, then the repeating block is the $b$-digit decimal representati... | United States | USAMO | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
02bg | Problem:
Se $m!$ termina com exatamente $n$ zeros, dizemos que $n$ é a cauda do fatorial $m!$. Observe os exemplos e responda:
- $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$, termina em um zero, por isso, a cauda do fatorial $5!$ é $1$;
- $10! = 3628800$, termina em dois zeros, logo a cauda do fatorial $10!$ é i... | [
"Solution:\n\na) Como $10 = 2 \\cdot 5$, basta contar as quantidades de fatores $2$ e $5$ nesses números. Observe que $20!$ é o mesmo que $20 \\cdot \\ldots \\cdot 15 \\cdot \\ldots \\cdot 10 \\cdot \\ldots \\cdot 5 \\cdot \\ldots \\cdot 1$, como são $4$ múltiplos de $5$ em sua representação fatorial e existem mais... | Brazil | null | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | a) 4 and 6; b) 4; c) no; d) 502; e) 121 | |
094w | Problem:
Let $ABC$ be an acute triangle. Let $M$ be the midpoint of the segment $BC$. Let $I, J, K$ be the incenters of triangles $ABC$, $ABM$, $ACM$, respectively. Let $P, Q$ be points on the lines $MK$, $MJ$, respectively, such that $\angle AJP = \angle ABC$ and $\angle AKQ = \angle BCA$. Let $R$ be the intersection ... | [
"Solution:\nNote that $MK \\perp MJ$. By simple angle chasing we get that\n$$\n\\begin{gathered}\n\\angle PJM = \\angle AJM - \\angle AJP = 90^\\circ + \\frac{1}{2} \\angle ABC - \\angle ABC = 90^\\circ - \\frac{1}{2} \\angle ABC, \\\\\n\\angle JPM = 90^\\circ - \\angle PJM = \\frac{1}{2} \\angle ABC .\n\\end{gathe... | Middle European Mathematical Olympiad (MEMO) | MEMO Szeged | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Qu... | null | proof only | null | |
007k | In the parallelogram $ABCD$ point $G$ is chosen on side $AB$. Consider the circle through $A$ and $G$ that is tangent to the extension of $CB$ beyond $B$ at point $P$. The extension of $DG$ beyond $G$ intersects the circle at $L$. If the quadrilateral $GLBC$ is cyclic, prove that $AB = PC$. | [
"Let $E$ be the second common point of $DA$ and the circle. One can show that $A$ is between $D$ and $E$. Denote $\\angle ELG = \\alpha$, $\\angle GLC = \\beta$. Since $GLBC$ is a cyclic quadrilateral by hypothesis, we have $\\angle GBC = \\angle GLC = \\beta$; since $ELGA$ is also cyclic, $\\angle DAG = \\alpha$. ... | Argentina | Mathematical Olympiad Rioplatense | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | English | proof only | null | |
086l | Problem:
Siano $a_{0}, a_{1}, a_{2}, \ldots$ numeri interi tali che $a_{0}=19$, $a_{1}=25$, e per ogni $n \geq 0$ valga $a_{n+2}=2 a_{n+1}-a_{n}$. Qual è il più piccolo $i>0$ per cui $a_{i}$ è multiplo di $19$?
(A) 19
(B) 25
(C) 38
(D) 44
(E) 50 | [
"Solution:\n\nLa risposta è (A). La regola di ricorrenza data nel testo equivale a dire $a_{k+1}-a_{k}=a_{k}-a_{k-1}=\\cdots=a_{1}-a_{0}=d$ (la differenza tra due termini successivi rimane costante) e quindi si riconosce in $a_{k}$ una progressione aritmetica. Si può quindi ricavare che $a_{k}=19+6 k$ per ogni $k$ ... | Italy | Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | MCQ | A | |
0dsy | Anana has an ordered $n$-tuple $(a_1, a_2, ..., a_n)$ of integers. Banana may make a guess on Anana's ordered integer $n$-tuple $(x_1, x_2, ..., x_n)$, upon which Anana will reveal the product of differences $(a_1 - x_1)(a_2 - x_2)...(a_n - x_n)$. How many guesses does Banana need to figure out Anana's $n$-tuple for ce... | [] | Singapore | Singapore International Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | null | proof and answer | n + 1 | |
04tn | There are 100 diamonds on display; 50 of them genuine and 50 of them fake. Peter is the only person able to distinguish them. Whenever you point to three diamonds, Peter will cover one of them and (truthfully) tell you, how many of the remaining two are genuine. Determine if it is possible to find the 50 genuine diamon... | [
"We prove it is impossible. Let Peter pick one genuine diamond $G$ and one fake diamond $F$. Whenever the triplet of diamonds being pointed to contains both $F$ and $G$, Peter covers the third diamond (and truthfully answers “One.”). Whenever the triplet contains precisely one of $F$, $G$, Peter covers it. Otherwis... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | No, it is impossible. | |
0cet | Fix an integer $n \ge 6$ and consider $n$ coplanar lines, no two parallel and no three concurrent. These lines split the plane into unbounded polygonal regions and polygons with pairwise disjoint interiors. Two polygons are *non-adjacent* if they do not share a side. Show that there are at least $\frac{1}{12}(n-3)(n-2)... | [
"Consider the obvious geometric plane graph associated with a generic $n$-line configuration in a plane (no two lines are parallel and no three are concurrent). This graph has exactly $\\binom{n}{2} = \\frac{1}{2}n(n-1)$ vertices, exactly $e = n^2$ edges and exactly $f = \\frac{1}{2}n(n+1)+1$ faces.\nLet $f_k$ be t... | Romania | Nineteenth IMAR Mathematical Competition | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0b1h | Problem:
Evaluate: $\sin 37^{\circ} \cos^{2} 34^{\circ} + 2 \sin 34^{\circ} \cos 37^{\circ} \cos 34^{\circ} - \sin 37^{\circ} \sin^{2} 34^{\circ}$. | [
"Solution:\n\nNote the following:\n$$\n\\begin{aligned}\n\\sin 37^{\\circ} \\cos^{2} 34^{\\circ} + \\sin 34^{\\circ} \\cos 37^{\\circ} \\cos 34^{\\circ} &= \\cos 34^{\\circ} (\\sin 37^{\\circ} \\cos 34^{\\circ} + \\sin 34^{\\circ} \\cos 37^{\\circ}) \\\\\n&= \\cos 34^{\\circ} \\sin (37^{\\circ} + 34^{\\circ}) \\\\\... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | (sqrt(6) + sqrt(2)) / 4 | |
0cfa | Find the lengths of the sides of a right triangle, if they are integers and the area of the triangle (in cm²) is 5 times larger than its perimeter (in cm). | [] | Romania | 74th NMO Shortlisted Problems | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | The right triangles are: (21, 220, 221), (22, 120, 122), (25, 60, 65), (30, 40, 50), (28, 45, 53), and (24, 70, 74). | |
0ge6 | 設 $ABCD$ 為凸四邊形，其中 $∠ABC > 90°$、$∠CDA > 90°$，並且 $∠DAB = ∠BCD$。令點 $A$ 分別關於直線 $BC$ 與 $CD$ 的對稱點為點 $E$ 與 $F$。設線段 $AE$ 和 $AF$ 分別交直線 $BD$ 於點 $K$ 和 $L$。試證：三角形 $BEK$ 和 $DFL$ 的兩個外接圓相切。
Let $ABCD$ be a convex quadrilateral with $∠ABC > 90°$, $∠CDA > 90°$, and $∠DAB = ∠BCD$. Denote by $E$ and $F$ the reflections of $A$ in lines $... | [
"令 $A'$ 為 $A$ 關於 $BD$ 的對稱點。以下我們將證明：四邊形 $A'BKE$ 及 $A'DLF$ 皆有外接圓，且它們的外接圓相切於 $A'$。\n\n\n\n由於對直線 $BC$ 對稱，有 $∠BEK = ∠BAK$；又由對直線 $DB$ 對稱，有 $∠BAK = ∠BA'K$。故得 $∠BEK = ∠BA'K$，即知 $A'BEK$ 共圓。同理可得 $A'DLF$ 共圓。\n\n要證明圓 $A'BKE$ 與 $A'DFL$ 相切，可由證出\n$$\n∠A'KB + ∠A'LD = ∠BA'D\n$$\n得到。由於 $AK ⊥ BC$, $AL ⊥ CD$, ... | Taiwan | 2021 數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02rp | When two red amoebas join, the result in one blue amoeba; when a red amoeba and a blue amoeba join, they turn into three red amoeba; and when two blue amoeba join, they become four red amoeba. Fernando observes a test tube with initially $19$ blue amoebas and $95$ red amoebas.
a. He observes that all the amoebas join ... | [
"a. Suppose there are $x$ pairs of red amoebas, $y$ pairs of blue amoebas and $z$ pairs of one amoeba of each color. Then $2x + z = 95 \\iff x = (95 - z)/2$ and $2y + z = 19 \\iff y = (19 - z)/2$. The number of amoebas in the next generation is $x + 4y + 3z = (95 - z)/2 + 4 \\cdot (19 - z)/2 + 3z = (171 + z)/2$. So... | Brazil | Brazilian Math Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a: 95; b: 33 | |
06q1 | For every positive integer $n$ determine the number of permutations $(a_{1}, a_{2}, \ldots, a_{n})$ of the set $\{1,2, \ldots, n\}$ with the following property:
$$
2(a_{1}+a_{2}+\cdots+a_{k}) \quad \text{is divisible by} \ k \quad \text{for} \ k=1,2, \ldots, n.
$$ | [
"For each $n$ let $F_{n}$ be the number of permutations of $\\{1,2, \\ldots, n\\}$ with the required property; call them nice. For $n=1,2,3$ every permutation is nice, so $F_{1}=1$, $F_{2}=2$, $F_{3}=6$.\n\nTake an $n>3$ and consider any nice permutation $(a_{1}, a_{2}, \\ldots, a_{n})$ of $\\{1,2, \\ldots, n\\}$. ... | IMO | 49th International Mathematical Olympiad Spain | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | F_1 = 1, F_2 = 2, and for n ≥ 3, F_n = 3 · 2^(n−2) | |
0ebq | Problem:
Dana je funkcija $f$ s predpisom $f(x)=A \cdot \cos \left(x+\frac{\pi}{6}\right)$. Določi $A \in \mathbb{R}$ tako, da bo veljalo $f\left(-\frac{\pi}{3}\right)=-\sqrt{3}$. Za tako določeno funkcijo $f$ zapiši njeno zalogo vrednosti in nariši njen graf. | [] | Slovenia | 15. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Precalculus > Trigonometric functions"
] | null | final answer only | A = -2; range: [-2, 2] | |
0kid | A disk of radius $1$ rolls all the way around the inside of a square of side length $s > 4$ and sweeps out a region of area $A$. A second disk of radius $1$ rolls all the way around the outside of the same square and sweeps out a region of area $2A$. The value of $s$ can be written as $a + \frac{b\pi}{c}$, where $a$, $... | [
"**Answer (A):** To obtain the region swept out by the first disk, remove a square of side length $s-4$ from the center of the original square and replace the $4$ unit squares at the corners of the original square with $4$ quarter-circles of radius $1$. The area of this region is\n$$\ns^2 - (s-4)^2 - 4 + \\pi = 8s ... | United States | AMC 10 A | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | MCQ | A | |
0i7y | Problem:
The product of the digits of a 5-digit number is $180$. How many such numbers exist? | [
"Solution:\nLet the digits be $a, b, c, d, e$. Then $a b c d e = 180 = 2^{2} \\cdot 3^{2} \\cdot 5$. We observe that there are 6 ways to factor $180$ into digits $a, b, c, d, e$ (ignoring differences in ordering):\n$$\n180 = 1 \\cdot 1 \\cdot 4 \\cdot 5 \\cdot 9 = 1 \\cdot 1 \\cdot 5 \\cdot 6 \\cdot 6 = 1 \\cdot 2 ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 360 | |
0253 | Problem:
Oito dados são agrupados formando um cubo. Quantas faces ficam visíveis? | [
"Solution:\n\n20"
] | Brazil | Desafios | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | final answer only | 20 | |
0cc8 | Fix an integer $n \ge 3$. Let $S$ be a set of $n$ points in the plane, no three of which are collinear. Given pairwise distinct points $A, B, C$ in $S$, the triangle $ABC$ is suitable for $AB$ if area $ABC \le$ area $ABX$ for all $X$ in $S$ different from $A$ and $B$. (Note that for a segment $AB$ there could be severa... | [
"For convenience, a triangle whose vertices all lie in $S$ will be referred to as a triangle in $S$. The argument hinges on the following observation:\n\n*Given any partition of $S$, amongst all triangles in $S$ with at least one vertex in each part, those of minimal area are all adequate.*\n\nIndeed, amongst the t... | Romania | THE Fifteenth ROMANIAN MASTER OF MATHEMATICS | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0bwq | Let $(a_n)_{n \ge 1}$ be a sequence of real numbers such that $a_1 > 2$ and $a_{n+1} = 1 + \frac{2}{a_n}$ for $n \ge 1$.
a) Prove that $a_{2n-1} + a_{2n} > 4$ for all $n \ge 1$ and $\lim_{n \to \infty} a_n = 2$.
b) Determine the largest real $a$ for which the inequality
$$
\sqrt{x^2 + a_1^2} + \sqrt{x^2 + a_2^2} + \s... | [
"a) Observe that $a_n > 0$ and that $a_{n+1} - 2 = \\frac{2-a_n}{a_n}$ for $n \\ge 1$, so $a_{2n-1} > 2 > a_{2n}$ for $n \\ge 1$. Thus $a_{2n-1}+a_{2n}-4 = \\frac{a_{2n-1}^2-3a_{2n-1}+2}{a_{2n-1}} > 0$ for $n \\ge 1$.\n\n$|a_n - 2| = \\frac{|a_{n-1} - 2|}{a_{n-1}} = \\dots = \\frac{|a_1 - 2|}{a_{n-1} \\dots a_1} \\... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 2 | |
009v | Decide if there is a square with side less than $1$ which can cover every rectangle with diagonal $1$. | [
"Such a square does exist. All rectangles with diagonal $1$ can be inscribed in a circle $\\Gamma$ with diameter $1$, which suggests the following construction.\nTake $8$ points on $\\Gamma$ that divide it into $8$ arcs of $45^\\circ$. They are the vertices of a regular octagon inscribed in $\\Gamma$. Extending two... | Argentina | NATIONAL XXX OMA | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Yes | |
0k3m | Problem:
A paper equilateral triangle of side length $2$ on a table has vertices labeled $A$, $B$, $C$. Let $M$ be the point on the sheet of paper halfway between $A$ and $C$. Over time, point $M$ is lifted upwards, folding the triangle along segment $BM$, while $A$, $B$, and $C$ remain on the table. This continues un... | [
"Solution:\n\nView triangle $ABM$ as a base of this tetrahedron. Then relative to triangle $ABM$, triangle $CBM$ rotates around segment $BM$ on a hinge. Therefore the volume is maximized when $C$ is farthest from triangle $ABM$, which is when triangles $ABM$ and $CBM$ are perpendicular. The volume in this case can ... | United States | HMMT February | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | sqrt(3)/6 | |
01tt | A line $l$ meets the right branch of the hyperbola $y = 1/x$ ($x > 0$) at points $A$ and $B$. Lines $l_1$ and $l_2$ are parallel to $l$ and meet the left branch of this hyperbola ($x < 0$) at points $E$, $F$ and $C$, $D$, respectively. The line $l_1$ intersects the segments $AD$ and $BC$ at points $G$ and $H$, respecti... | [
"Let $A(a, 1/a)$, $B(b, 1/b)$, $C(c, 1/c)$, $D(d, 1/d)$, $E(e, 1/e)$, $F(f, 1/f)$, and positions of all points on the hyperbola are shown in the Figure. It is easy to write the equations of the straight lines $AB$, $DC$, and $EF$:\n$$\nAB(l) : y = -\\frac{1}{ab}x + \\frac{1}{a} + \\frac{1}{b}, \\quad EF(l_1) : y = ... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null | |
0e66 | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$ (x + f(x)^2)f(y) = f(yf(x)) + xyf(x) $$
for all $x, y \in \mathbb{R}$. | [
"If we substitute $x = 1$ into the functional equation, we get\n$$\nf(y) + f(1)^2 f(y) = f(yf(1)) + yf(1). \\quad (1)\n$$\nWe successively substitute $1$, $f(1)$ and $f(1)^2$ instead of $y$ into this equation to get\n$$\nf(f(1)) = f(1)^3, \\tag{2}\n$$\n$$\nf(1)^3 + f(1)^5 = f(f(1)^2) + f(1)^2, \\quad (3)\n$$\n$$\nf... | Slovenia | Selection Examinations for the IMO 2012 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 0 for all x and f(x) = x for all x | |
049v | Put one of the four letters $A, B, C, D$ and one of the four numbers $1, 2, 3, 4$ into each cell of the $4 \times 4$ table, so that:
(a) in each row and in each column, each letter and each number appears exactly once,
(b) each combination (pair) of one letter and one number appears in exactly one cell of the table. | [] | Croatia | CroatianCompetitions2011 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Linear Algebra > Vectors"
] | null | final answer only | One valid 4×4 arrangement is:
Row 1: A1 B4 C2 D3
Row 2: B2 A3 D1 C4
Row 3: C3 D2 A4 B1
Row 4: D4 C1 B3 A2 | |
07d1 | $ABCD$ is a trapezoid with $AB \parallel CD$. Suppose that the diagonals intersect at $P$. Let $\omega_1$ be a circle passing through $B$ and tangent to $AC$ at $A$. Let $\omega_2$ be a circle passing through $C$ and tangent to $BD$ at $D$. $\omega_3$ is the circumcircle of triangle $BPC$. Prove that the common chord o... | [
"Let $Q$ be the second intersection point of $AD$ with $\\omega_1$, and let $R$ be the second intersection point of $\\omega_2$ and $\\omega_3$.\n\n\n$$\n\\left. \\begin{array}{l} \\widehat{DRB} = 180^\\circ - (\\widehat{BR} + \\widehat{DBR}) \\\\ \\omega_2 : \\quad \\widehat{BR} = \\wideha... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kgw | Problem:
Let $a_{1}, a_{2}, \ldots, a_{n}$ be a sequence of distinct positive integers such that $a_{1}+a_{2}+\cdots+a_{n}=2021$ and $a_{1} a_{2} \cdots a_{n}$ is maximized. If $M=a_{1} a_{2} \cdots a_{n}$, compute the largest positive integer $k$ such that $2^{k} \mid M$. Proposed by: Sheldon Kieren Tan | [
"Solution:\nWe claim that the optimal set is $\\{2,3, \\cdots, 64\\} \\backslash\\{58\\}$. We first show that any optimal set is either of the form $\\{b, b+1, b+2, \\ldots, d\\}$ or $\\{b, b+1, \\ldots, d\\} \\backslash\\{c\\}$, for some $b<c<d$.\n\nWithout loss of generality, assume that the sequence $a_{1}<a_{2}... | United States | HMMT Spring 2021 Guts Round | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Seque... | null | proof and answer | 62 | |
0egs | Problem:
S števkami $1,2,3,4,5,6,7$ in $9$ sestavljamo soda šestmestna števila, katerih prva števka je praštevilo. Koliko je vseh takih šestmestnih števil?
(A) 20160
(B) 9440
(C) 25552
(D) 4320
(E) 49152 | [
"Solution:\n\nŠestmestno število lahko sestavimo na $4 \\cdot 8 \\cdot 8 \\cdot 8 \\cdot 8 \\cdot 3 = 49152$ načinov."
] | Slovenia | Državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Statistics > Probability > Counting Methods > Other"
] | null | MCQ | E | |
0bwv | A positive integer becomes with 2017 smaller if we delete its two final decimal digits. What is the number? | [
"Let $\\overline{a_1a_2\\dots a_{n-2}a_{n-1}a_n}$ be the number we search for, so\n$$\n\\overline{a_1 a_2 \\dots a_{n-2} a_{n-1} a_n} - 2017 = \\overline{a_1 a_2 \\dots a_{n-2}},\n$$\nwhich is equivalent to $\\overline{a_1a_2\\dots a_{n-2}a_{n-1}a_n} - \\overline{a_1a_2\\dots a_{n-2}} = 2017$. On the other hand,\n$... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 2037 | |
0bbd | Let $A$ be a ring and let $a$ be an element of $A$. Prove that
a) If $A$ is commutative and $a$ is nilpotent, then $a + x$ is invertible for any invertible element $x \in A$.
b) If $A$ is finite and $a + x$ is invertible for any invertible element $x \in A$, then $a$ is nilpotent.
(An element $a$ of a ring is called... | [
"a) Let $x$ be an invertible element and let $n$ be a positive integer such that $a^n = 0$. Since $a + x = x(x^{-1}a + 1)$, it is enough to show that $x^{-1}a + 1$ is invertible. Set $b = x^{-1}a$. Then $b^n = x^{-n}a^n = 0$, (because $A$ is commutative), hence $b^{2n+1} = 0$. Consequently,\n$$\n1 = b^{2n+1} + 1 = ... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Abstract Algebra > Ring Theory"
] | null | proof only | null | |
0890 | Problem:
In un triangolo acutangolo $ABC$ con $AB < AC$, la bisettrice che parte da $A$ interseca il lato $BC$ nel punto $P$. La parallela al lato $AB$ passante per $P$ interseca il lato $AC$ nel punto $Q$; su questa retta sia $R$ il punto che giace sulla semiretta uscente da $Q$ che non contiene $P$ e tale che $QR = ... | [
"Solution:\n\na. Dimostriamo che non solo il triangolo $QAR$ è isoscele di base $AR$ per costruzione, ma anche il triangolo $PAQ$ lo è (con base $AP$).\n\n\n\nInfatti\n$$\nB\\widehat{A}P = P\\widehat{A}Q\n$$\nperché $AP$ è bisettrice, mentre\n$$\nB\\widehat{A}P = A\\widehat{P}Q\n$$\nperché ... | Italy | Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
043n | As shown in Fig. 11.1, the edge length of cube $ABCD$–$EFGH$ is $2$. Take any point $P_1$ on the incircle of square $ABFE$, take any point $P_2$ on the incircle of square $BCGF$, and take any point $P_3$ on the incircle of square $EFGH$. Find the minimum and maximum of $|P_1P_2| + |P_2P_3| + |P_3P_1|$.
![](attached_im... | [
"According to the condition, we can set\n$$\nP_1(1, \\cos \\alpha_1, \\sin \\alpha_1), \\quad P_2(\\sin \\alpha_2, 1, \\cos \\alpha_2), \\quad P_3(\\cos \\alpha_3, \\sin \\alpha_3, 1).\n$$\nWe conventionally assume that $P_4 = P_1$ and $\\alpha_4 = \\alpha_1$. Denote\n$$\nd_i = |P_i P_{i+1}| \\quad (i = 1, 2, 3).\n... | China | China Mathematical Competition | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and ... | null | proof and answer | Minimum = 3√2 − 3; Maximum = 3√6 | |
02vi | Problem:
Considere três pontos colineares $B$, $C$ e $D$ de modo que $C$ está entre $B$ e $D$. Seja $A$ um ponto que não pertence à reta $BD$ de modo que $AB = AC = CD$.
a. Se $\angle BAC = 36^{\circ}$, então verifique que
$$
\frac{1}{CD} - \frac{1}{BD} = \frac{1}{CD + BD}
$$
b. Suponha ag... | [
"Solution:\n\n(a) Como $\\angle BAC = 36^{\\circ}$ e o triângulo $ABC$ é isósceles, temos $\\angle ABC = \\angle ACB = 72^{\\circ}$. Além disso, como $\\angle ACD = 108^{\\circ}$ e o triângulo $ACD$ também é isósceles, segue que $\\angle CAD = \\angle CDA = 36^{\\circ}$.\n\n\n\nPortanto, os... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0be1 | Weights of $1\ \mathrm{g}$, $2\ \mathrm{g}$, $\ldots$, $200\ \mathrm{g}$ are placed on the two pans of a balance such that on each pan there are $100$ weights and the balance is in equilibrium. Prove that one can swap $50$ weights from one pan with $50$ weights from the other pan such that the balance remains in equili... | [
"We call a *pair* two weights whose sum is $201\\ \\mathrm{g}$. We wish to obtain, in the end, $50$ pairs on each of the two pans of the balance.\n\nIf on the pan on the left we have the weights $a_1, a_2, \\dots, a_{50}$ and their pairs $b_1, b_2, \\dots, b_{50}$ are on the pan on the right, we move the weights su... | Romania | 64th NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
07us | Let $k$ be a positive integer and let $x_0, x_1, x_2, \dots$ be an infinite sequence defined by the relationship
$$
\begin{align*}
x_0 &= 0 \\
x_1 &= 1 \\
x_{n+1} &= kx_n + x_{n-1} \quad \text{for all } n \ge 1.
\end{align*}
$$
a. For the special case $k = 1$, prove that $x_{n-1}x_{n+1}$ is never a perfect square for ... | [
"We claim that, for all $n \\ge 1$,\n$$\nx_{n-1}x_{n+1} = x_n^2 + (-1)^n. \\qquad (10)\n$$\nBased on this, if $k > 1$ we then have $x_n = kx_{n-1} + x_{n-2} \\ge kx_{n-2} \\ge k \\ge 2$ for $n \\ge 2$, so $x_n^2 \\pm 1$ cannot be a square number. If $k = 1$ then $x_2 = 2$ and $x_n = x_{n-1} + x_{n-2} \\ge x_{n-2} \... | Ireland | IRL_ABooklet | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0htt | Problem:
Let $ABCD$ be a square. Consider four circles $k_{1}, k_{2}, k_{3}, k_{4}$ which pass respectively through $A$ and $B$, $B$ and $C$, $C$ and $D$, $D$ and $A$, and whose centers are outside the square. Circles $k_{4}$ and $k_{1}$, $k_{1}$ and $k_{2}$, $k_{2}$ and $k_{3}$, $k_{3}$ and $k_{4}$ intersect respecti... | [
"Solution:\n\nWe use a standard theorem that states that a convex quadrilateral $WXYZ$ can be inscribed in a circle if and only if $\\angle W + \\angle Y = 180$. We have\n$$\n\\begin{aligned}\n\\angle PLM + \\angle MNP & = (360 - \\angle MLA - \\angle ALP) + (360 - \\angle PNC - \\angle CNM) \\\\\n& = (180 - \\angl... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02tb | Problem:
Na figura abaixo, $ABCD$ é um retângulo e $E$ é o ponto médio de $AD$. O segmento $FG$ passa pelo ponto médio $M$ de $CE$. Determine a razão entre os comprimentos de $GM$ e $MF$.
 | [
"Solution:\n\nPelo ponto $M$, trace o segmento de reta $PQ$ perpendicular aos lados $AB$ e $CD$ do retângulo $ABCD$ como mostra a figura abaixo. Como $M$ é o ponto médio de $CE$, podemos concluir que $PM$ é base média relativa ao lado $DE$ do triângulo $ECD$. Assim, se $DE = EA = x$, $PM = DE / 2 = x / 2$. Como $E$... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 1/3 | |
050f | Does there exist an operation $*$ on the set of all integers such that the following conditions hold simultaneously:
(1) for any integers $x$, $y$, $z$, $(x * y) * z = x * (y * z)$;
(2) for any integers $x$ and $y$, $x * x * y = y * x * x = y$? | [
"Define an operation $\\oplus$ on the set of all non-negative integers, which maps two non-negative integers $a$ and $b$ to a non-negative integer $a \\oplus b$, such that for all $i = 0, 1, \\dots$, $(a \\oplus b)_i = (a_i + b_i) \\bmod 2$, where $n_i$ stands for the binary digit corresponding to $2^i$ in the bina... | Estonia | IMO Team Selection Contest | [
"Algebra > Abstract Algebra > Group Theory",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof only | null | |
07q7 | A regular polygon with $n \ge 3$ sides is given. Each vertex is coloured either red, green or blue, and no two adjacent vertices of the polygon are the same colour. There is at least one vertex of each colour.
Prove that it is possible to draw certain diagonals of the polygon in such a way that they intersect only at t... | [
"Suppose that the polygon has only one vertex of some colour, say red. Then drawing all diagonals from this red vertex easily solves the problem. Therefore, in what follows we may assume that the polygon has at least two vertices of each colour. In particular we may assume that $n \\ge 6$.\nFor a polygon with $n \\... | Ireland | Ireland | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0753 | Let $ABC$ be a triangle each of whose angles is greater than $30^\circ$. Suppose a circle, with centre $T$, cuts the segments $BC$ in $P, Q$; $CA$ in $K, L$; and $AB$ in $M, N$, such that $P, Q, K, L, M, N$ are on the circle in counter-clockwise direction in that order. Suppose further that the triangles $TQK$, $TLM$ a... | [
"(a) Write $BC = a$, $CA = b$ and $AB = c$. Let $\\angle PTQ = 2x$, $\\angle KTL = 2y$ and $\\angle MTN = 2z$. Since $TP = TQ$, we get $\\angle TQP = \\angle TPQ = 90^\\circ - x$. Thus $\\angle 30^\\circ + x$. Similarly, $\\angle BNP = 30^\\circ + z$. It follows that $z + x = 120^\\circ - B$. Likewise $x + y = 120^... | India | Indija TS | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
01jb | Let $\triangle ABC$ be an acute triangle with $|AB| > |AC|$. The internal angle bisector of $\angle BAC$ intersects $BC$ at $D$. Let $O$ be the circumcenter of $\triangle ABC$. Let $AO$ intersect the segment $BC$ at $E$. Let $J$ be the incenter of $\triangle AED$. Prove that if $\angle ADO = 45^\circ$ then $|OJ| = |JD|... | [
"Let $\\alpha = \\angle BAC$, $\\beta = \\angle CBA$, $\\gamma = \\angle ACB$. We have\n$$\n\\begin{align*} \n\\angle DJA &= 90^\\circ + \\frac{1}{2} \\angle DEA = 90^\\circ + \\frac{1}{2} (\\angle EBA + \\angle BAE) \\\\ \n&= 90^\\circ + \\frac{1}{2} (\\beta + 90^\\circ - \\gamma) = 135^\\circ + \\frac{\\beta}{2} ... | Baltic Way | Baltic Way 2023 Shortlist | [
"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 | |
013x | Problem:
Let the lines $e$ and $f$ be perpendicular and intersect each other at $O$. Let $A$ and $B$ lie on $e$ and $C$ and $D$ lie on $f$, such that all the five points $A, B, C, D$ and $O$ are distinct. Let the lines $b$ and $d$ pass through $B$ and $D$ respectively, perpendicularly to $A C$; let the lines $a$ and $c... | [
"Solution:\nLet $A_{1}$ be the intersection of $a$ with $B D$, $B_{1}$ the intersection of $b$ with $A C$, $C_{1}$ the intersection of $c$ with $B D$ and $D_{1}$ the intersection of $d$ with $A C$. It follows easily by the given right angles that the following three sets each are concyclic:\n- $A, A_{1}, D, D_{1}, ... | Baltic Way | Baltic Way 2005 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jsq | Problem:
Let
$$
A = \lim_{n \rightarrow \infty} \sum_{i=0}^{2016} (-1)^i \cdot \frac{\binom{n}{i} \binom{n}{i+2}}{\binom{n}{i+1}^2}
$$
Find the largest integer less than or equal to $\frac{1}{A}$.
The following decimal approximation might be useful: $0.6931 < \ln(2) < 0.6932$, where $\ln$ denotes the natural logarithm ... | [
"Solution:\nNote\n$$\n\\sum_{i=0}^{2016} (-1)^i \\cdot \\frac{\\binom{n}{i} \\binom{n}{i+2}}{\\binom{n}{i+1}^2} = \\sum_{i=0}^{2016} (-1)^i \\cdot \\frac{(i+1)(n-i-1)}{(i+2)(n-i)}\n$$\nSo\n$$\n\\lim_{n \\rightarrow \\infty} \\sum_{i=0}^{2016} (-1)^i \\cdot \\frac{\\binom{n}{i} \\binom{n}{i+2}}{\\binom{n}{i+1}^2} = ... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 1 | |
0cd8 | In an art museum, $n$ paintings are exhibited, where $n \ge 33$ is a positive integer, and a total of 15 colors are used in such a way that any two paintings have at least one color in common, and no two paintings have exactly the same set of colors. Determine all possible values of $n \ge 33$ such that, no matter how ... | [
"We will prove that each $n \\in \\{33, 34, \\dots, 2^{14}\\}$ is a solution.\n\nWe begin by noticing that if we have a painting $T_i$ in the museum that uses $k$ colors, the painting that uses the other $15-k$ colors cannot be in the museum. Therefore, out of the $2^{15}-1$ possible paintings that can be obtained ... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | All integers n with 33 ≤ n ≤ 2^14 | |
0961 | Problem:
Determinaţi toate numerele naturale $n$ pentru care numărul $\sqrt{n+\sqrt{n-2}}$ este raţional. | [
"Solution:\n\nPresupunem că pentru o valoare oarecare a numărului $n$, $n \\geq 2$ avem $m=\\sqrt{n+\\sqrt{n-2}} \\in \\mathbb{Q}$, $m \\in \\mathbb{N}^{*}$. Are loc egalitatea $m^{2}=n+\\sqrt{n-2}$. Atunci există $k \\in \\mathbb{N}^{*}$, astfel încât $k=\\sqrt{n-2}$, adică $n=k^{2}+2$. Putem scrie $m^{2}=k^{2}+k+... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | n = 3 | |
00bc | A set of natural numbers is *regular* if each of the subsets has sum different from $1810$. Partition the numbers $452$, $453$, $\ldots$, $1809$ into a minimum number of regular sets. | [
"Such a partition clearly needs at least $2$ regular sets (e.g., $900$ and $910$ must be in different regular sets). Here is a partition with $2$ regular sets:\n\n$$\nA = \\{452, \\ldots, 602\\} \\cup \\{906, \\ldots, 1207\\}, \\quad B = \\{603, \\ldots, 905\\} \\cup \\{1208, \\ldots, 1809\\}.\n$$\n\nLet us check t... | Argentina | Argentina_2018 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 2 | |
07cv | Let $p^m$ be a power of a prime number. Find the lowest value of $d$ such that there exists a monic polynomial $Q(x)$ of degree $d$ with integer coefficients such that for any positive integer $n$, $p^m$ divides $Q(n)$. | [
"We prove the problem using the following lemma, that is actually a more general form of the problem.\n\n**Lemma.** Let $l$ be a fixed non-zero integer and $P(x) = a_nx^n + \\cdots + a_0$ be a polynomial with integer coefficients such that for all integers $k$, $P(k)$ is divisible by $l$. Then we have $l \\mid a_n ... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | the smallest integer d such that p^m divides d! | |
0hjc | Problem:
Determine whether it is possible to tile a standard $8 \times 8$ chessboard with $15$ $L$-tiles and $1$ $T$-tile of the shapes below:
(Each tile covers four squares of the chessboard. The tiles can be flipped and rotated at will.) | [
"Solution:\nLet the squares of the chessboard be colored black and white in the usual way. Note that the $L$-tile always covers two squares of each color, while the $T$-tile covers either $1$ black and $3$ white or $3$ black and $1$ white squares. Since the chessboard has equal numbers of black and white squares, i... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Impossible | |
02wv | Problem:
Dois quadrados de um tabuleiro $7 \times 7$ são pintados de amarelo e o resto é pintado de verde. Dois esquemas de cores são equivalentes se um pode ser obtido do outro aplicando uma rotação no plano do tabuleiro. Quantos esquemas de cores não equivalentes podemos obter? | [
"Solution:\n\nComo o tabuleiro possui $7 \\cdot 7 = 49$ quadrados, existem $\\frac{49 \\cdot 48}{2} = 1176$ maneiras de escolhermos dois deles para receberem a cor amarela. Quando eles não são diametralmente opostos, podemos rotacioná-los em $90^{\\circ}$ três vezes e obter configurações que geram esquemas equivale... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 300 | |
058u | Points $A$, $B$ and $C$ are chosen in a rectangle of shape $7 \times 10$ in such a way that the distances from $A$ to some three sides of the rectangle are $2$, $3$ and $4$, the distances from $B$ to some three sides of the rectangle are $3$, $4$ and $5$, and the distances from $C$ to some three sides of the rectangle ... | [
"\n\n\n\n\n\n\n\nAnswer: $\\frac{7}{2}$.\n\n**Solution:** Introduce a Cartesian coordinate system with origin $O$ at one vertex of the rectangle. Let the other vertices of the rectangle be $P = (10,0)$, $Q = (10,7)$ ... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 7/2 | |
0abs | Let $p$ be a line through the vertex $D$ of a given parallelogram $ABCD$, such that the vertices $A$, $B$ and $C$ are on the same side of $p$. Let $A'$, $B'$ and $C'$ be the bases of the altitudes from $A$, $B$ and $C$ to $p$ correspondingly. Prove that $\overline{BB'} = \overline{AA'} + \overline{CC'}$. | [
"We draw a line parallel to $p$ through $A$ and let this line intersect $BB'$ in $Q$. The quadrangle $AQB'A'$ has three right angles, hence it is a rectangle, from where we obtain that $\\overline{AA'} = \\overline{QB'}$....(1).\n\nBecause $ABCD$ is a parallelogram, we have that $\\overline{AB} = \\overline{DC}$ an... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0bq1 | Let $A, B \in M_2(\mathbb{C})$ be such that $A - AB = B^2$ and $B - BA = A^2$. Prove that $A = B$. | [] | Romania | 67th NMO Shortlisted Problems | [
"Algebra > Linear Algebra > Matrices"
] | English | proof only | null | |
0f7q | Problem:
If the graph of the function $f = f(x)$ is rotated through $90$ degrees about the origin, then it is not changed. Show that there is a unique solution to $f(b) = b$. Give an example of such a function. | [] | Soviet Union | 21st ASU | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | b = 0 | |
03xp | In triangle $ABC$, $AB = AC$. Point $D$ is the midpoint of side $BC$. Point $E$ lies outside the triangle $ABC$ such that $CE \perp AB$ and $BE = BD$. Let $M$ be the midpoint of segment $BE$. Point $F$ lies on the minor arc $\widehat{AD}$ of the circumcircle of triangle $ABD$ such that $MF \perp BE$. Prove that $ED \pe... | [
"**Solution 1.** Construct point $F_1$ such that $EF_1 = BF_1$ and ray $DF_1$ is perpendicular to line $ED$. It suffices to show that $F = F_1$ or $ABDF_1$ is cyclic; that is,\n$$\n\\angle BAD = \\angle BF_1 D. \\qquad \\textcircled{1}\n$$\nSet $\\angle BAD = \\angle CAD = x$. Because $EC \\perp AB$ and $AD \\perp ... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
0gxs | A country has a road network with the following properties: along each road we can go in both directions; between two different towns may exist at most one road; there does not exist a road which connects some town with itself; the road network is a connected graph; there are at least $m$ roads out of any town $(m = \t... | [
"Consider a road network of this country as a graph. Denote the chain of maximum length by $L$ in this graph. Denote $L$ by sequence of connected vertices $A_1 \\to A_2 \\to \\dots \\to A_n$.\n\n\nFig.27\n\nThat is, all other chains have the same or less length as $L$. Obviously, there is a... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Discrete Mathematics > Graph Theory"
] | English | proof and answer | 2m+2 | |
0h7e | Real nonzero numbers $a$, $b$, $c$, $d$ satisfy the conditions $a^3 + b^3 + c^3 + d^3 = 0$ and $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \neq 0$. Prove that $a + b + c + d \neq 0$. | [
"**Solution.** We will use method from contradiction. Suppose $a + b + c + d = 0$. Thus $a + b = -(c + d)$. Therefore, we can obtain such equalities:\n$$\n(a + b)^3 = -(c + d)^3 \\Leftrightarrow a^3 + b^3 + 3ab(a + b) = -c^3 - d^3 - 3cd(c + d) \\Leftrightarrow \\\\\n3ab(a + b) + 3cd(c + d) = 0 \\Leftrightarrow -3ab... | Ukraine | UkraineMO | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
04aw | Let $K$ and $L$ be points on a semicircle with diameter $\overline{AB}$. Bisector of the side $\overline{AB}$ intersects the segment $\overline{KL}$ at a point $U$ so that $A$ and $K$ lie on one side of the bisector and $B$ and $L$ on the other. Let $N$ be the foot of perpendicular from the intersection of lines $AK$ a... | [
"Let $S$ be the center of the given semicircle, $C$ the intersection of the lines $AK$ and $BL$, and $T$ the intersection of the lines $AB$ and $KL$.\nFrom the given condition it follows that the quadrilateral $ABUV$ is cyclic so by the power of a point theorem we have\n$$\n|TU| \\cdot |TV| = |TA| \\cdot |TB|.\n$$\... | Croatia | Hrvatska 2011 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneo... | English | proof only | null | |
0asp | Problem:
Which real numbers $x$ satisfy the inequality $|x-3| \geq |x|$? | [
"Solution:\n\n$(-\\infty, 3/2]$"
] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (-∞, 3/2] | |
0joh | Problem:
Consider a $2 \times 2$ grid of squares. Each of the squares will be colored with one of 10 colors, and two colorings are considered equivalent if one can be rotated to form the other. How many distinct colorings are there? | [
"Solution:\n\nThis solution will be presented in the general case with $n$ colors. Our problem asks for $n=10$.\n\nWe isolate three cases:\n\nCase 1: Every unit square has the same color\n\nIn this case there are clearly $n$ ways to color the square.\n\nCase 2: Two non-adjacent squares are the same color, and the o... | United States | HMMT November 2015 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof and answer | 2530 | |
0301 | Problem:
Se $x$ e $y$ são números reais tais que $x+y=10$ e $x^{3}+y^{3}=400$, determine o valor de $x^{2}+y^{2}$. | [
"Solution:\n\nComo\n$$\n\\begin{aligned}\nx^{3}+y^{3} & =(x+y)\\left(x^{2}-x y+y^{2}\\right) \\\\\n& =(x+y)\\left((x+y)^{2}-3 x y\\right) \\\\\n& =10(100-3 x y)\n\\end{aligned}\n$$\nsegue que $400=10(100-3 x y)$, ou seja, $400=1000-30 x y$, logo $x y=20$. Daí\n$$\n\\begin{aligned}\nx^{2}+y^{2} & =(x+y)^{2}-2 x y \\... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 60 | |
0fzu | Problem:
Gegeben sind zwei natürliche Zahlen $m$ und $n$. Zeige, dass es eine natürliche Zahl $c$ gibt, sodass jede von 0 verschiedene Ziffer gleich oft in $c m$ und $c n$ vorkommt. | [
"Solution:\n\nWir zeigen zuerst, dass es genügt den Fall zu betrachten, wo $n$ teilerfremd zu $10$ ist. Nehme also an, die Aufgabe ist für solche $n$ bewiesen und betrachte den Fall, wo $n = 2^{a} 5^{b} n'$, mit $(n', 10) = 1$ und $m$ beliebig. Wegen unserer Annahme gibt es nun ein $c$, so dass in $c (2^{b} 5^{a} m... | Switzerland | IMO-Selektion | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
08ha | Problem:
Find all prime numbers $a$, $b$ and $c$ for which the equality $$(a-2)! + 2b! = 22c - 1$$ holds. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | a=2, b=5, c=11 and a=3, b=5, c=11 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.