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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
01c0 | A sequence of positive integers $\{s_i\}$ is constructed in the following way: $s_1 = 2$, and for all $i > 1$ $s_i$ is the least number that is larger than $s_{i-1}$ and contains digit:
* "2" if $i = 1$ (mod 4)
* "0" if $i = 2$ (mod 4)
* "1" if $i = 3$ (mod 4)
* "4" if $i = 4$ (mod 4).
This sequence starts with $2$, $1... | [
"One can easily notice that $s_{i+1} \\le s_i + 10$ (at least one of ten consecutive numbers contains the necessary digit) and therefore $s_{i+4} \\le s_i + 40$. It means that there is such $k$, that $1895 \\le s_{4k} \\le 1934$ and as $s_{4k}$ contains digit $4$ it can be only one of the numbers $\\{1904, 1914, 19... | Baltic Way | Baltic Way | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | a) yes; b) no | |
0es9 | In $\triangle ABC$, $\hat{A} = 120^\circ$ and $\hat{B}$ is five times $\hat{C}$. The number of degrees in $\hat{C}$ is
(A) 10 (B) 12 (C) 15 (D) 18 (E) 20 | [
"The angles of the triangle add up to $180^\\circ$, so $\\hat{C} + 5\\hat{C} + 120^\\circ = 180^\\circ$, i.e. $6\\hat{C} = 60^\\circ$ so $\\hat{C} = 10^\\circ$."
] | South Africa | South African Mathematics Olympiad First Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | A | |
042g | Suppose positive number sequence $\{x_n\}$ satisfies $S_n \ge 2S_{n-1}$, $n = 2, 3, \dots$, where $S_n = x_1 + \dots + x_n$. Prove that there exists a constant $C > 0$, such that
$$
x_n \ge C \cdot 2^n,\ n = 1, 2, \dots
$$ | [
"When $n \\ge 2$, $S_n \\ge 2S_{n-1}$ is equivalent to\n$$\nx_n \\ge x_1 + \\cdots + x_{n-1}. \\qquad ①\n$$\nLet $C = \\frac{1}{4}x_1$. We will prove\n$$\nx_n \\ge C \\cdot 2^n,\\ n = 1, 2, \\dots \\qquad \\textcircled{2}\n$$\nby induction.\n\nWhen $n = 1$, it is obviously true. When $n = 2$, we have $x_2 \\ge x_1 ... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0kon | Problem:
Isosceles trapezoid $ABCD$ with bases $AB$ and $CD$ has a point $P$ on $AB$ with $AP=11$, $BP=27$, $CD=34$, and $\angle CPD=90^{\circ}$. Compute the height of isosceles trapezoid $ABCD$. | [
"Solution:\n\nDrop projections of $A$, $P$, $B$ onto $CD$ to get $A'$, $P'$, $B'$. Since $A'B'=38$ and $CD=34$, we get that $DA'=CB'=2$. Thus, $P'D=9$ and $P'C=25$. Hence, the answer is $PP'=\\sqrt{P'D \\cdot P'C}=15$."
] | United States | HMMT November 2022 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 15 | |
0ksk | Problem:
Aerith picks two numbers $x < y$, and picks one of them to tell to Bob uniformly at random. Is it possible for Bob to have a better than half chance of guessing whether the one he was told is $x$? | [
"Solution:\n\nAs an initial attempt, one could choose a threshold $T$ and guess that the number Aerith says is $x$ if and only if it's less than $T$. This doesn't improve Bob's chances if $x < y < T$ or if $T \\leq x < y$, but if $x < T \\leq y$, this strategy always works. The trick is to pick $T$ so that one guar... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Yes. Bob can pick a random threshold T = tan(θ) with θ uniform on (−π/2, π/2), and guess that the revealed number is x if and only if it is less than T; this yields success probability strictly greater than one half. | |
0cqm | Integers $a$ and $b$ satisfy the following condition: For any positive integers $m$ and $n$, the number $am^2 + bn^2$ is a perfect square. Prove that $ab = 0$. (A. Golovanov)
Целые числа $a$ и $b$ таковы, что при любых натуральных $m$ и $n$ число $am^2 + bn^2$ является точным квадратом. Докажите, что $ab = 0$. (А. Гол... | [
"Assume the contrary. Then it is easy to see that both $a$ and $b$ are positive. Now, substituting $m = 2b$, $n = 1$ and $m = 2b$, $n = 2$ we obtain that both numbers $4ab^2 + b = x^2$ and $4ab^2 + 4b = y^2$ are perfect squares. But then $3b = y^2 - x^2 = (y-x)(y+x) \\ge 1 \\cdot 4b$ which is impossible.\n\n\nSolut... | Russia | Russian mathematical olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English; Russian | proof only | null | |
0hh2 | In a convex pentagon $ABCDE$ the following conditions are satisfied: $AB \parallel CD$, $BC \parallel DE$ and $\angle BAE = \angle AED$. Prove that $AB + BC = CD + DE$.
Fig. 16 | [
"Let the rays $AB$ and $DE$ intersect at the point $O$ (fig. 16). Then $BCDO$ is a parallelogram, and $\\Delta AOE$ is isosceles, since $\\angle OAE = \\angle AEO$. So,\n$$\nAB + BC = (OB - OA) + BC = CD - OE + OD = CD + DE.\n$$"
] | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0hui | Problem:
The numbers from the table
| $a_{11}$ | $a_{12}$ | $\cdots$ | $a_{1 n}$ |
| :---: | :---: | :---: | :---: |
| $a_{21}$ | $a_{22}$ | $\cdots$ | $a_{2 n}$ |
| $\vdots$ | $\vdots$ | | $\vdots$ |
| $a_{n 1}$ | $a_{n 2}$ | $\cdots$ | $a_{n n}$ |
satisfy the inequality
$$
\sum_{i=1}^{n}\left|a_{i 1} x_{1}+a_{i ... | [
"Solution:\n\nWe will sum the given inequality over all possible choices for $\\left(x_{1}, \\ldots, x_{n}\\right)$. Since the number of such choices is $2^{n}$ we obtain\n\n$$\n\\sum_{i=1}^{n} \\sum_{\\left(x_{1}, \\ldots, x_{n}\\right)}\\left|a_{i 1} x_{1}+a_{i 2} x_{2}+\\cdots+a_{i n} x_{n}\\right| \\leq 2^{n} M... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
08za | How many pairs of integers $(m, n)$ are there with $1 \le m, n \le 2022$ satisfying the following condition?
For any positive integer $N$, there exists a nonnegative integer $k$ and an integer $d > N$ such that both $\frac{m-k^2}{d}$ and $\frac{n+2k}{d}$ are integers. | [
"If a positive integer $d$ divides both $m-k^2$ and $n+2k$, $d$ also divides $4(m-k^2)-(n-2k)(n+2k) = 4m-n^2$. Thus when $4m \\neq n^2$ holds, $N = |4m-n^2|$ satisfies $d \\le N$ for any positive integer $d$ and $(m,n)$ does not satisfy the condition.\nOn the other hand, when $4m = n^2$ holds, $n = 2t$ for a positi... | Japan | Japan 2022 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 44 | |
03rf | Assume that $\alpha^{2005} + \beta^{2005}$ can be expressed as a polynomial in $\alpha + \beta$ and $\alpha\beta$. Find the sum of the coefficients of the polynomial. (posed by Zhu Huawei) | [
"In the expansion of $\\alpha^k + \\beta^k$, let $\\alpha + \\beta = 1$ and $\\alpha\\beta = 1$. We get the sum of coefficients $S_k = \\alpha^k + \\beta^k$. Since\n\n$$(\\alpha + \\beta)(\\alpha^{k-1} + \\beta^{k-1}) = (\\alpha^k + \\beta^k) + \\alpha\\beta(\\alpha^{k-2} + \\beta^{k-2}),$$\n\nwe get\n$S_k = S_{k-1... | China | China Western Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > P... | English | proof and answer | 1 | |
0flf | Problem:
Sean $A, B, C, D$ cuatro puntos en el espacio, tales que no hay ningún plano que pasa por los cuatro a la vez. Los segmentos $A B, B C, C D, D A$ son tangentes a una misma esfera. Demuestra que los cuatro puntos de tangencia están en un mismo plano. | [
"Solution:\n\nSean $R, S, T, U$ los puntos de tangencia respectivos de la esfera con los segmentos $A B$, $B C$, $C D$, $D A$. Siendo $O$ el centro de la esfera, claramente $O U$, $O R$ son respectivamente perpendiculares a $D A$, $A B$, con lo que los triángulos $O A U$ y $O A R$ son rectángulos respectivamente en... | Spain | XLVII Olimpiada Matemática Española, Fase nacional (Pamplona) | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0gvb | Let $AD$ be the median of a triangle $ABC$, and let $\angle ADB = 45^\circ$, $\angle ACB = 30^\circ$. Find $\angle BAD$ (in degrees). | [
"Нехай $BK$ — висота трикутника $ABC$. Оскільки точка $D$ є серединою гіпотенузи $BC$ прямокутного трикутника $KBC$, то трикутник $BKD$ є рівностороннім. Отже, як нескладно бачити, $\\angle ADK = \\angle DAK = 15^\\circ$, $AK = DK = BK$. А тому $\\angle BAK = 45^\\circ$, $\\angle BAD = 45^\\circ - 15^\\circ = 30^\\... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 30 | |
09f5 | Let $m > 2$ be a natural number and if $i \neq j$ then $a_i \neq a_j \pmod{m}$. Prove that if $(a_i, m) = 1$, $i = 1, \varphi(m)$ then there exists the permutation $b_1, b_2, ..., b_{\varphi(m)}$ of numbers $a_1, a_2, ..., a_{\varphi(m)}$ such that $a_1^{b_1} + a_2^{b_2} + ... + a_{\varphi(m)}^{b_{\varphi(m)}}$ is divi... | [
"This is an extension of G.11.4."
] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Abstract ... | English | proof only | null | |
058x | When adding together positive integers $a$ and $b$, Juku forgot to enter the final digit 7 of the number $a$ and got a result of 2022. Had Juku instead forgotten to enter the final digit of $b$, the result would have been 5000. Find the sum of $a$ and $b$. | [
"Let $a = \\overline{x7}$ and $b = \\overline{yd}$, where $d$ is the final digit of $b$ and $x, y$ are the numbers $a$ and $b$ without their final digits. Then $a = 10x + 7$ and $b = 10y + d$. From the given information, we compile the system of equations\n$$\n\\begin{cases} x + 10y + d = 2022, \\\\ y + 10x + 7 = 5... | Estonia | Estonian Math Competitions | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 6385 | |
0jw3 | Problem:
In an acute triangle $A B C$ let $K, L$, and $M$ be the midpoints of sides $A B, B C$, and $C A$, respectively. From each of $K, L$, and $M$ drop two perpendiculars to the other two sides of the triangle; e.g., drop perpendiculars from $K$ to sides $B C$ and $C A$, etc. The resulting 6 perpendiculars intersec... | [
"Solution:\n\nConstruct segments $K L, L M$ and $M K$. Next, construct the three altitudes of triangle $K L M$ that meet in its orthocenter, $O$. Since $K L$ connects the two midpoints of $A B$ and $B C$ we know that $K L \\| A B$ it is easy to see that $M T\\|K O\\| L Q$ since those lines are perpendicular to a pa... | United States | BAMO-8 and BAMO-12 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00rb | A positive integer $n$ is *uphill* if its decimal representation $\overline{a_k a_{k-1} \dots a_0}$ satisfies $a_k \le a_{k-1} \le \dots \le a_0$. A real-coefficient polynomial $P$ is *integer-valued* if $P(n)$ is an integer for all integer $n$, and *uphill-integer-valued* if $P(n)$ is an integer for all uphill positiv... | [
"First we show that no *uphill* number is congruent to $10$ modulo $11$.\nTo this end, notice that an uphill number can always be written as $b_1 + b_2 + \\dots + b_m$, where $m \\le 9$, $b_1 \\le b_2 \\le \\dots \\le b_m$, and each $b_i$ is of the form $\\overline{11\\dots11}$. Since the remainder of each $b_i$ mo... | Balkan Mathematical Olympiad | BMO 2016 Short List Final | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | No | |
0cdj | Consider several tokens of different colors and sizes such that no two tokens have the same color and size. Each token $J$ has two numbers written on it: one represents the number of tokens with the same color as $J$ but different sizes, and the other represents the number of tokens with the same size as $J$ but differ... | [
"**Observation 1.** If the situation is possible for a certain number $m$ of tokens, then it is also possible for a number $m' > m$: we can add $m' - m$ tokens with different sizes and colors to a suitable configuration with $m$ tokens. This shows that it is sufficient to find the minimum number of tokens for which... | Romania | THE Eighteenth IMAR Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | All integers m ≥ 3434 | |
07q6 | Find all rational numbers $r$ that satisfy the equation
$$
(\sqrt{2+\sqrt{3}})^r + (\sqrt{2-\sqrt{3}})^r = 14.
$$ | [
"First observe that $(2+\\sqrt{3})(2-\\sqrt{3}) = 1$. If we let\n$$\nx = (\\sqrt{2+\\sqrt{3}})^r = (2+\\sqrt{3})^{\\frac{r}{2}} \\quad \\text{we get}\n$$\n$$\n\\frac{1}{x} = (\\sqrt{2-\\sqrt{3}})^r = (2-\\sqrt{3})^{\\frac{r}{2}}.\n$$\nTherefore, we have to find $r \\in \\mathbb{Q}$ such that $x + \\frac{1}{x} = 14$... | Ireland | Ireland | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | r = 4, r = -4 | |
083t | Problem:
Quante soluzioni intere non negative ha l'equazione
$$
v^{2}+w^{2}+x^{2}+y^{2}+z^{2}=5 ?
$$
Per soluzione intera non negativa si intende una cinquina ordinata di interi non negativi $(v, w, x, y, z)$ che soddisfano l'equazione.
Nota: due cinquine ordinate che differiscono anche solo per l'ordine degli element... | [] | Italy | Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO TRIENNIO | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | MCQ | C | |
0gd8 | 令 $k$ 為一正整數。一個網球錦標賽的賽事委員會要為 $2k$ 位參賽者安排賽程, 其中:
(1) 任兩位參賽者恰好對戰一場;
(2) 每一天錦標賽只會打一場;
(3) 每位參賽者在他的第一場比賽當天入住旅館, 並在他的最後一場比賽當天離開旅館。
對於每位參賽者住在旅館內的每一天, 大會都要支付 1 元, 因此賽事委員會希望極小化付給旅館的總費用。試求旅館總費用的極小值並證明之。 | [
"答：最小值為 $\\frac{k(4k^2 + k - 1)}{2}$。\n\n讓我們將賽事的每一天, 依序編號為 $1, 2, \\dots, C_2^{2k}$。讓我們將每位選手到達的日期排序為 $b_1 \\le b_2 \\le b_{2k}$, 離開的日期則排序為 $e_1 \\ge e_2 \\ge \\dots \\ge e_{2k}$ (注意這邊並非以參賽者編號, 一名參賽者可能在第 $b_i$ 天入住但在第 $e_j$ 天離開。) 旅館總費用為\n$$\n\\sum_{i=1}^{2k} (e_i - b_i + 1).\n$$\n\n(1) 估計：為此，我們需要估計 $e_i - b_i + 1$。注意... | Taiwan | 二〇一九數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | k(4k^2 + k - 1)/2 | |
042c | Let $n \ge 4$ be an even number. At the vertices of a regular $n$-gon we write in an arbitrary way $n$ distinct real numbers. Starting from one edge, we name all the edges in a clockwise way by $e_1, e_2, \dots, e_n$. An edge is called "positive", if the difference of the numbers at its endpoint and its start point is ... | [
"Without loss of generality, we may assume that the numbers written on the vertices are $1, 2, \\dots, n$. Let $A$ be the number of crossings, and $B$ be the number of positive edges. We will prove that the parity of $S$ remains the same if we exchange numbers $i$ and $i+1$. We distinguish two cases.\n\n**Case 1.**... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
06pe | Find all positive integers $n$, for which the numbers in the set $S=\{1,2, \ldots, n\}$ can be colored red and blue, with the following condition being satisfied: the set $S \times S \times S$ contains exactly $2007$ ordered triples $(x, y, z)$ such that (i) $x, y, z$ are of the same color and (ii) $x+y+z$ is divisible... | [
"Suppose that the numbers $1,2, \\ldots, n$ are colored red and blue. Denote by $R$ and $B$ the sets of red and blue numbers, respectively; let $|R|=r$ and $|B|=b=n-r$. Call a triple $(x, y, z) \\in S \\times S \\times S$ monochromatic if $x, y, z$ have the same color, and bichromatic otherwise. Call a triple $(x, ... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 69, 84 | |
07jl | Let $p$ be a prime number. There are several cards on the table. Numbers $0$, $1$, $\dots$, $p-1$ are written on them such that $0$ is written once and each of $1$, $2$, $\dots$, $p-1$ is written at most once. For any two cards with numbers $x$ and $y$ on them, respectively, we determine the *winner* card such that;
i... | [
"We say $a < b$ if the card with number $b$ written on it beats the card with number $a$. Let $m$ be the smallest element of this set, and $M$ be the largest element of the set. Suppose $a$ is an element of this set such that $a + M$ is also an element of the set. If\n$a + M < 0$, then for any $b$ such that $b + m$... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | floor((2p+1)/3) | |
0h4v | A polynomial
$$
P(x) = x^{2016} + 2016x^{2015} + a_{2014}x^{2014} + a_{2013}x^{2013} + \dots + a_1x + 1
$$
can be expressed as $P(x) = (x - x_1)(x - x_2)\dots(x - x_{2016})$, where among the numbers $x_1, x_2, \dots, x_{2016}$ at least 2015 are negative (not necessarily distinct). Find all coefficients of $P(x)$. | [
"Without loss of generality assume $x_1, x_2, \\dots, x_{2015} < 0$. Vieta's theorem implies $x_1x_2x_3\\dots x_{2015}x_{2016} = 1$, that is, $x_{2016}$ is also a real negative number. Again, by Vieta's theorem:\n$$\n\\left\\{ \\begin{array}{l} x_1 + x_2 + \\dots + x_{2016} = -2016, \\\\ x_1x_2\\dots x_{2016} = 1, ... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | P(x) = (x + 1)^{2016} | |
01ca | Find all pairs of positive integers $(a, b)$ for which both numbers $4ab + 1$ and $ab^3 + 1$ are perfect squares. | [
"**Answer:** $(a, b) = (n(n + 1)/2, 2)$, where $n \\in \\mathbb{N}$.\n\nSuppose there exist integers $x, y \\ge 2$ such that\n$$\n4ab + 1 = x^2 \\text{ and } ab^3 + 1 = y^2.\n$$\nIf $b = 1$, then $3$ is the difference of two squares: $(2y)^2 - x^2 = 3$. This is only possible when $x^2 = 1$, a contradiction. We next... | Baltic Way | Baltic Way | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (a, b) = (n(n+1)/2, 2) for n ∈ ℕ | |
092t | Problem:
Let $ABC$ be a triangle with $AB \neq AC$. The points $K, L, M$ are the midpoints of the sides $BC, CA, AB$, respectively. The inscribed circle of $ABC$ with centre $I$ touches the side $BC$ at point $D$. The line $g$, which passes through the midpoint of segment $ID$ and is perpendicular to $IK$, intersects ... | [
"Solution:\n\nLet $(XYZ)$ denote the circumcircle of a triangle $XYZ$. We use the following well-known lemma:\n\nLemma. The centre of the circle $(BIC)$ is the midpoint of arc $BC$ of circle $(ABC)$ and therefore lies on the angle bisector of $\\angle BAC$.\n\n\n\nNow assume without loss of... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
... | null | proof only | null | |
0k3d | Problem:
Pascal has a triangle. In the $n$th row, there are $n+1$ numbers $a_{n, 0}, a_{n, 1}, a_{n, 2}, \ldots, a_{n, n}$ where $a_{n, 0}=a_{n, n}=1$. For all $1 \leq k \leq n-1$, $a_{n, k}=a_{n-1, k}-a_{n-1, k-1}$. Let $N$ be the value of the sum
$$
\sum_{k=0}^{2018} \frac{\left|a_{2018, k}\right|}{\binom{2018}{k}} ... | [
"Solution:\n\n```python\nimport math\nlists=[[1]]\nfor i in range(2018):\n newlist=[]\n for j in range(i):\n newlist.append(lists[-1][j+1]-lists[-1][j])\n lists.append([1]+newlist+[1])\nbig=math.factorial(2018)\nsum=0\nfor i in range(2019):\n sum+=abs(lists[-1][i])/(big//math.factorial(i)//math.f... | United States | HMMT November 2018 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 80 | |
03ij | Problem:
A parallelepiped has the property that all cross sections which are parallel to any fixed face $F$, have the same perimeter as $F$. Determine whether or not any other polyhedron has this property. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof only | null | |
0fo5 | Sea $\{x_n\}_{n \ge 1}$ la sucesión de enteros positivos definida por $x_1 = 2$ y $x_{n+1} = 2x_n + x_n$ para todo $n \ge 1$. Determinar la mayor potencia de $5$ que divide al número $x_{2014}^2 + 1$. | [] | Spain | L Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | Spanish | proof and answer | 1 | |
0bqb | Let $p$ be a prime number and $n_1, n_2, \dots, n_k \in \{1, 2, \dots, p-1\}$ be positive integers. Show that the equation
$$
x_1^{n_1} + x_2^{n_2} + \dots + x_k^{n_k} = x_{k+1}^p
$$
has infinitely many positive integer solutions. | [] | Romania | 67th NMO Shortlisted Problems | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof only | null | |
02my | Problem:
Carina está desenhando a planta de um jardim retangular que terá um de seus lados num muro reto de pedras. Ela comprou $140~\mathrm{m}$ de cerca, em pedaços de $1~\mathrm{m}$ cada um, para cercar os outros três lados. Ela não pode cortar esses pedaços e deve gastar todos eles.
a. S... | [] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a) 60 m. b) 85 m: not possible; 65 m: possible. | |
0bfg | Let $(G, \cdot)$ be a group and $H$ and $K$ be two subgroups of $G$. It is known that $xy = yx$, for every $x \in G \setminus H$ and every $y \in G \setminus K$. Prove that the group $G$ is commutative. | [
"Let $a, b \\in G$ be arbitrary. We want to show that $ab = ba$.\n\nCase 1: $a \\in G \\setminus H$ and $b \\in G \\setminus K$.\nBy hypothesis, $ab = ba$.\n\nCase 2: $a \\in H$ and $b \\in G \\setminus K$.\nLet $x \\in G \\setminus H$. By hypothesis, for any $y \\in G \\setminus K$, $xy = yx$.\nConsider $a \\in H$... | Romania | Shortlisted Problems for the 64th NMO | [
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
03wl | Prove that for any odd prime number $p$, the number of positive integers $n$ satisfying $p \mid n! + 1$ is no more than $c p^{\frac{2}{3}}$, where $c$ is a constant number independent of $p$. (Posed by Yu Hongbing) | [
"**Proof** Clearly, if $n$ satisfies the required property, then $1 \\le n \\le p-1$. Denote all such $n$'s by $n_1 < n_2 < \\cdots < n_k$; we shall show that $k \\le 12 p^{\\frac{2}{3}}$. If $k \\le 12$ there is nothing to prove. In what follows, we assume that $k > 12$.\n\nRename $n_{i+1} - n_i$ ($1 \\le i \\le k... | China | China National Team Selection Test | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
04sj | A family of sets $\mathcal{F}$ is called *perfect* if the following condition holds: For every triple of sets $X_1, X_2, X_3 \in \mathcal{F}$, at least one of the sets
$$
(X_1 \setminus X_2) \cap X_3, \quad (X_2 \setminus X_1) \cap X_3
$$
is empty. Show that if $\mathcal{F}$ is a perfect family consisting of some subse... | [
"We proceed by induction with respect to $|U|$. If $|U| = 0$, that is, $U = \\emptyset$, then there exists only one subset of $U$ and clearly $|\\mathcal{F}| \\le 1$.\nIn the following, let us suppose the statement is true for all sets of cardinality less than $k$ for a given $k > 0$. Let $U$ be any set with $|U| =... | Czech Republic | 15th Czech-Polish-Slovak Mathematics Competition | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
03fa | Given is a sequence $a_1, a_2, \dots$, such that $a_1 = 1$ and $a_{n+1} = \frac{9a_n+4}{a_n+6}$ for any $n \in \mathbb{N}$. Which terms of this sequence are positive integers? | [
"It can be shown by induction that\n$$\na_n = \\frac{4 \\cdot 2^n - 3}{2^n + 3}, \\forall n \\ge 2\n$$\nThus, $a_n$ is integer only when $n = 1$. □"
] | Bulgaria | Bulgarian Spring Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | n = 1 | |
0igu | Problem:
Two ants, one starting at $(-1,1)$, the other at $(1,1)$, walk to the right along the parabola $y = x^{2}$ such that their midpoint moves along the line $y = 1$ with constant speed $1$. When the left ant first hits the line $y = \frac{1}{2}$, what is its speed? | [
"Solution: $3 \\sqrt{3} - 3$\nWhen the left ant first hits the line $y = \\frac{1}{2}$, the right ant hits the line $y = \\frac{3}{2}$. The left ant is then at $\\left(-\\frac{\\sqrt{2}}{2}, \\frac{1}{2}\\right)$, and the right ant is at $\\left(\\frac{\\sqrt{6}}{2}, \\frac{3}{2}\\right)$. Let the left ant have vel... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof and answer | 3 sqrt(3) - 3 | |
0ijw | Problem:
Let $\Gamma$ denote the circumcircle of triangle $A B C$. Point $D$ is on $\overline{A B}$ such that $\overline{C D}$ bisects $\angle A C B$. Points $P$ and $Q$ are on $\Gamma$ such that $\overline{P Q}$ passes through $D$ and is perpendicular to $\overline{C D}$. Compute $P Q$, given that $B C=20$, $C A=80$, ... | [
"Solution:\nSuppose that $P$ lies between $A$ and $B$ and $Q$ lies between $A$ and $C$, and let line $P Q$ intersect lines $A C$ and $B C$ at $E$ and $F$ respectively. As usual, we write $a$, $b$, $c$ for the lengths of $B C$, $C A$, $A B$.\n\nBy the angle bisector theorem, $A D / D B = A C / C B$ so that $A D = \\... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 4√745 | |
0ekd | Problem:
Kolikšen je natančen volumen vrtenine, ki jo dobimo, če pravokotnik s stranicama $a=2$ cm in $b=3$ cm zavrtimo okoli simetrale krajše stranice?
(A) $2\pi$ cm$^3$
(B) $4\pi$ cm$^3$
(C) $12\pi$ cm$^3$
(D) $\pi$ cm$^3$
(E) $3\pi$ cm$^3$ | [
"Solution:\nČe pravokotnik s stranicama $a=2$ cm in $b=3$ cm zavrtimo okoli simetrale krajše stranice, dobimo vrtenino v obliki valja s polmerom $r=1$ cm in višino $v=3$ cm. Volumen valja izračunamo po obrazcu $V=\\pi r^{2} v=3\\pi$ cm$^3$. Pravilen je odgovor $E$."
] | Slovenia | 22. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes"
] | null | MCQ | E | |
0gpw | Find all triples $(p, q, r)$ of prime numbers satisfying $p^4 + 2p + q^4 + q^2 = r^2 + 4q^3 + 1$. | [
"If $r = 2$ we get $p^4 + 2p + q^4 + q^2 = 4q^3 + 5$. Then $q = 2, 3$ otherwise $q \\ge 5$ and $q^4 + q^2 > 4q^3 + 5$. If $q = 2$ then $p^4 + 2p = 17$ gives no solution. If $q = 3$ we get $p^4 + 2p = 23$ gives no solution. Therefore, $r$ is odd and the right hand side of the main equation is even. Then since $2p + ... | Turkey | 18th Junior Turkish Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | (2, 5, 13) | |
0cw1 | Each of 10 men is either a knight (always telling the truth) or a liar (who always lies). Each of them thinks of an integer number. The first man told: "My number is greater than 1", the second told: "My number is greater than 2", ..., the tenth told: "My number is greater than 10". After that, talking in a different o... | [
"Докажем, что ни один из рыцарей не мог сказать ни одной из фраз «Моё число больше 9» и «Моё число больше 10». В самом деле, если бы это было возможно, то задуманное рыцарем целое число было бы не меньше 10. Но тогда он не мог сказать ни одной из фраз «Моё число меньше 1», «Моё число меньше 2», ..., «Моё число мень... | Russia | Regional round | [
"Discrete Mathematics > Logic"
] | English; Russian | proof and answer | 8 | |
010w | Problem:
Prove that there exist infinitely many even positive integers $k$ such that for every prime $p$ the number $p^{2}+k$ is composite. | [
"Solution:\nNote that the square of any prime $p \\neq 3$ is congruent to $1$ modulo $3$. Hence the numbers $k = 6m + 2$ will have the required property for any $p \\neq 3$, as $p^{2} + k$ will be divisible by $3$ and hence composite.\n\nIn order to have $3^{2} + k$ also composite, we look for such values of $m$ fo... | Baltic Way | Baltic Way | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
01ay | Let $a_0, a_1, \dots, a_N$ be real numbers satisfying $a_0 = a_N = 0$ and
$$
a_{i+1} - 2a_i + a_{i-1} = a_i^2
$$
for $i = 1, 2, \dots, N-1$. Prove that $a_i \le 0$ for $i = 1, 2, \dots, N-1$. | [
"Assume the contrary. Then, there is an index $i$ for which $a_i = \\max_{0 \\le j \\le N} a_j$ and $a_i > 0$. This $i$ cannot be equal to $0$ or $N$, since $a_0 = a_N = 0$. Thus, from $a_i \\ge a_{i-1}$ and $a_i \\ge a_{i+1}$ we obtain\n$$\n0 < a_i^2 = (a_{i+1} - a_i) + (a_{i-1} - a_i) \\le 0,\n$$\nwhich is a cont... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
068p | Let $a$, $b$ be positive integers such that the number
$$
\frac{(a+b)^2 + 4a}{ab}
$$
is an integer. Assume that $b$ is odd. Prove that $a$ is a perfect square. | [
"Let $\\frac{(a+b)^2 + 4a}{ab} = \\kappa \\in \\mathbb{Z}$. The last one can be re-written as:\n$$\n(a+b)^2 + 4a = \\kappa ab \\quad (1)\n$$\nWe will prove that $a$ is odd. Indeed, if $2|a$, then $2|(a+b)^2$, and thus $b$ is even, a contradiction.\n\nWe consider (1) as a quadratic with respect to $b$:\n$$\nb^2 + b(... | Greece | Hellenic Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
0j1m | Problem:
3000 people each go into one of three rooms randomly. What is the most likely value for the maximum number of people in any of the rooms? Your score for this problem will be 0 if you write down a number less than or equal to 1000. Otherwise, it will be $25-27 \frac{|A-C|}{\min (A, C)-1000}$. | [
"Solution:\n\nAnswer: 1019\n\nTo get a rough approximation, we can use the fact that a sum of identical random variables converges to a Gaussian distribution in this case with a mean of $1000$ and a variance of $3000 \\cdot \\frac{2}{9} = 667$. Since $\\sqrt{667} \\approx 26$, $1026$ is a good guess, as Gaussians t... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Combinations",
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 1019 | |
0f7b | Problem:
Two circles intersect at $P$ and $Q$. $A$ is a point on one of the circles. The lines $AP$ and $AQ$ meet the other circle at $B$ and $C$ respectively. Show that the circumradius of $ABC$ equals the distance between the centers of the two circles. Find the locus of the circumcircle as $A$ varies. | [] | Soviet Union | 20th ASU | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | The circumradius of triangle ABC is equal to the distance between the centers of the two given circles. As A varies on its circle, the circumcircle ranges over a one-parameter family of circles all having this fixed radius. | |
0h6d | At the tennis tournament in one circle attended the 8 girls (i.e. every tennis player has played with each other exactly once, draws in tennis does not happen). Oksana took the second place recruited points and there is no other participant with the same number of points. What is the maximum number of games could lose ... | [
"Suppose that Olesya lost 2 games. Then she scored 5 points. Oksana could not score 4 points, because then all together all teams had supplied the maximum $5+4+3 \\cdot 6 = 27$ wins. But in total we have $\\frac{1}{2}(8 \\cdot 7) = 28$ games, a contradiction. Similarly Oksana could not score 3 or less points, or Ol... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1 | |
0f2d | Problem:
Black and white tokens are placed around a circle. First all the black tokens with one or two white neighbors are removed. Then all white tokens with one or two black neighbors are removed. Then all black tokens with one or two white neighbors and so on until all the tokens have the same color. Is it possible... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | First: No, it is not possible to have only one token after 4 moves starting from 40 tokens. Second: The minimum possible number of moves to reduce 1000 tokens to one is 10. | |
0ghi | 設三角形 $ABC$ 的外接圓為 $\omega$，$A$-旁心為 $I_A$，$I_A$ 對 $BC$ 的垂足為 $D$。令 $M$ 為線段 $I_A D$ 的中點。在圓 $\omega$ 上，不含 $A$ 點的 $BC$ 弧內選取點 $T$ 使得 $\angle BAT = \angle DAC$。直線 $I_A T$ 與圓 $\omega$ 再交於點 $S \neq T$。設 $SM$ 交 $BC$ 於點 $X$；線段 $AD$ 的中垂線分別交 $AC$、$AB$ 於點 $Y$、$Z$。證明直線 $AX$、$BY$、$CZ$ 共點。
Let $\omega$ be the circumcircle of triangle $... | [
"解法一. 令 $\\triangle ABC$ 的內心為 $I$, 劣弧 $BC$ 的中點為 $N$, $ND$ 再交 $\\omega$ 於 $U$, $AI \\cap BC = L$。考慮以 $A$ 為中心、$\\sqrt{AB \\cdot AC}$ 為半徑的反演變換, 並對 $AI$ 鏡射。此反演將 $D$ 送到 $T$, $L$ 送到 $N$, 故\n$$\n\\angle ATL = \\angle AND = \\angle ANU = \\angle ATU,\n$$\n即 $T, L, U$ 共線。令 $BAC$ 弧中點為 $N'$, 則熟知 $T, I, N'$ 共線, 故有\n$$\n-1 = (A... | Taiwan | 2023 數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Circles > Radical... | Chinese (Traditional) | proof only | null | |
0a08 | Problem:
In een getallenrij $a_{1}, a_{2}, \ldots, a_{1000}$ van 1000 verschillende getallen heet een paar $\left(a_{i}, a_{j}\right)$ met $i<j$ stijgend als $a_{i}<a_{j}$ en dalend als $a_{i}>a_{j}$. Bepaal de grootste positieve gehele $k$ met de eigenschap dat in elke rij van 1000 verschillende getallen ten minste $... | [
"Solution:\n\nWe gaan bewijzen dat de grootste $k$ gelijk aan 333 is. Bekijk ten eerste de rij $1000,999,998, \\ldots, 669,668,1,2,3, \\ldots, 666,667$. De eerste 333 getallen in de rij zijn niet bruikbaar in een stijgend paar, omdat voor elk van deze getallen geldt dat links van dit getal alleen maar grotere getal... | Netherlands | MO-selectietoets | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 333 | |
05ss | Problem:
Un lycée comporte un nombre impair de classes, et, dans chaque classe, un nombre impair d'élèves. Un élève est choisi dans chaque classe pour faire partie du comité des élèves. Si le nombre de classes avec le plus de garçons que de filles est impair, montrer que le nombre de façons de former un comité des élè... | [
"Solution:\n\nEn l'absence d'idées face au problème, on peut essayer de démontrer le résultat par récurrence. Pour regarder si cela pourrait fonctionner, on peut regarder ce qu'il advient du résultat si l'on passe d'un lycée avec 3 élèves à un lycée avec 4 élèves. On vérifie également que les hypothèses de l'énoncé... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof only | null | |
011d | Problem:
Find all real solutions to the following system of equations:
$$
\left\{\begin{aligned}
x+y+z+t & =5 \\
x y+y z+z t+t x & =4 \\
x y z+y z t+z t x+t x y & =3 \\
x y z t & =-1
\end{aligned}\right.
$$ | [
"Solution:\n$x=\\frac{1 \\pm \\sqrt{2}}{2}$, $y=2$, $z=\\frac{1 \\mp \\sqrt{2}}{2}$, $t=2$ or $x=2$, $y=\\frac{1 \\pm \\sqrt{2}}{2}$, $z=2$, $t=\\frac{1 \\mp \\sqrt{2}}{2}$.\n\nLet $A=x+z$ and $B=y+t$. Then the system of equations is equivalent to\n$$\n\\left\\{\\begin{aligned}\nA+B & =5 \\\\\nA B & =4 \\\\\nB x z+... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | Either y = 2, t = 2 and {x, z} = {(1 + sqrt(2))/2, (1 - sqrt(2))/2}, or x = 2, z = 2 and {y, t} = {(1 + sqrt(2))/2, (1 - sqrt(2))/2}. | |
0a4o | Problem:
Let $a$, $b$ and $c$ be positive real numbers such that $a + b + c = 3$. Prove that
$$a^{a} + b^{b} + c^{c} \geq 3.$$ | [
"Solution:\n\nWe start with a general fact about any positive real number $x$. There are two cases: either $x \\geq 1$ or $x < 1$.\n\n- If $x \\geq 1$ then $x^{p} \\geq x^{q}$ for any $p > q$. Substituting $p = x$ and $q = 1$ gives us $x^{x} \\geq x^{1} = x$.\n\n- If $x < 1$ then $x^{p} < x^{q}$ for any $p > q$. Su... | New Zealand | New Zealand Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof only | null | |
051p | How many positive integers $n$ are there for which $2014 \cdot n$ is divisible by $2014 + n$? | [
"If $a = 1, d = 2014$, then $a+x \\mid 2014$. As $a+x > 1$, there are 7 possibilities.\nIf $a = 2, d = 1007$, then $a+x \\mid 1007$. As $a+x > 2$, there are 3 possibilities.\nIf $a = 19, d = 106$, then $a+x \\mid 106$. As $a+x > 19$, there are 2 possibilities.\nIf $a = 38, d = 53$, then $a+x \\mid 53$. As $a+x > 38... | Estonia | Final Round of National Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 13 | |
015b | Determine all integers $n > 1$ for which the inequality
$$
x_1^2 + x_2^2 + \dots + x_n^2 \geq (x_1 + x_2 + \dots + x_{n-1})x_n
$$
holds for all real $x_1, x_2, \dots, x_n$. | [
"Setting $x_1 = x_2 = \\dots = x_{n-1} = 1$ and $x_n = 2$ yields $(n-1) + 4 \\geq (n-1)2$ and $n \\leq 5$.\nNow let $n \\leq 5$. Then the inequality reads as\n$$\nx_n^2 - (x_1 + x_2 + \\dots + x_{n-1})x_n + (x_1^2 + x_2^2 + \\dots + x_{n-1}^2) \\geq 0.\n$$\nBy the Cauchy-Schwarz inequality or the inequality of arit... | Baltic Way | Baltic Way SHL | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | n = 2, 3, 4, 5 | |
06k8 | Let $k$ be a positive integer. Prove that there exists a positive integer $\ell$ with the following property: if $m$ and $n$ are positive integers relatively prime to $\ell$ such that $m^n \equiv n^m \pmod{\ell}$, then $m \equiv n \pmod{k}$. | [
"We will prove the assertion when $\\ell = k!$ by induction on $k$. The base case $k = 1$ is obvious.\n\nIf $k > 1$, then $\\varphi(k) < k$. Suppose $m$ and $n$ are positive integers relatively prime to $\\ell$ such that $m^m \\equiv n^n \\pmod{k}$. This clearly implies $m^m \\equiv n^n \\pmod{\\varphi(k)!}$. By th... | Hong Kong | CHKMO | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
04hh | A group of twenty children has $100$ ribbons. Each ribbon is being held at its ends by two children. Two children may hold together at most one ribbon. Let us assume that a pair of ribbons such that its four ends hold different children can be chosen in $4050$ ways. Prove that each child holds the same number of ribbon... | [
"We enumerate the children by numbers $1, 2, \\dots, 20$. Let us denote by $d_i$ the number of ribbons that are being held by the child with number $i$. Since each ribbon is being held by two children we have that\n$$\n\\sum_{i=1}^{20} d_i = 2 \\cdot 100 = 200.\n$$\nThe number of all pairs of ribbons is $\\frac{100... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0dwx | Problem:
V trikotniku $ABC$, katerega središče včrtane krožnice je $I$, velja
$$
|CA| + |AI| = |BC|
$$
Poišči razmerje med velikostima kotov $\Varangle BAC$ in $\Varangle CBA$. | [
"Solution:\n\nIzberimo tako točko $D$, $D \\notin AC$, na poltraku $CA$, da je $|AD| = |AI|$. Tedaj je $DBC$ enakokrak trikotnik z vrhom $C$ in ker $I$ leži na simetrali kota $\\Varangle DCB$, je $\\Varangle IDC = \\Varangle CBI$. Ker je $ADI$ enakokrak trikotnik z vrhom $A$, je $\\Varangle IAC = 2 \\Varangle IDC$.... | Slovenia | 49. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 1:2 | |
06sp | Determine whether there exists an infinite sequence of nonzero digits $a_{1}, a_{2}, a_{3}, \ldots$ and a positive integer $N$ such that for every integer $k > N$, the number $\overline{a_{k} a_{k-1} \ldots a_{1}}$ is a perfect square. | [
"Answer. No.\n\nAssume that $a_{1}, a_{2}, a_{3}, \\ldots$ is such a sequence. For each positive integer $k$, let $y_{k}= \\overline{a_{k} a_{k-1} \\ldots a_{1}}$. By the assumption, for each $k > N$ there exists a positive integer $x_{k}$ such that $y_{k}=x_{k}^{2}$.\n\nI. For every $n$, let $5^{\\gamma_{n}}$ be t... | IMO | International Mathematical Olympiad Shortlisted Problems | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | No | |
0itr | Problem:
Consider a circular cone with vertex $V$, and let $ABC$ be a triangle inscribed in the base of the cone, such that $AB$ is a diameter and $AC = BC$. Let $L$ be a point on $BV$ such that the volume of the cone is 4 times the volume of the tetrahedron $ABCL$. Find the value of $BL / LV$. | [
"Solution:\n\n$\\boxed{\\sqrt{\\dfrac{\\pi}{4-\\pi}}}$\n\nLet $R$ be the radius of the base, $H$ the height of the cone, $h$ the height of the pyramid and let $BL / LV = x / y$. Let $[\\cdot]$ denote volume. Then $[\\text{cone}] = \\dfrac{1}{3} \\pi R^2 H$ and $[ABCL] = \\dfrac{1}{3} \\pi R^2 h$ and $h = \\dfrac{x}... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | sqrt(pi/(4 - pi)) | |
05tz | Problem:
Soit $x$, $y$ et $z$ trois nombres réels tels que $0 \leqslant x \leqslant y \leqslant z$ et $x+y+z=1$. Trouver la valeur maximale que peut prendre l'expression
$$
(x-y z)^{2}+(y-z x)^{2}+(z-x y)^{2}.
$$ | [
"Solution:\nLa quantité que l'on souhaite maximiser peut se réécrire comme\n$$\n\\begin{aligned}\n\\mathcal{S} & =\\left(x^{2}+y^{2}+z^{2}\\right)+\\left((x y)^{2}+(y z)^{2}+(z x)^{2}\\right)-6 x y z \\\\\n& =(x+y+z)^{2}-2(x y+y z+z x)+(x y+y z+z x)^{2}-2 x y z(x+y+z)-6 x y z \\\\\n& =1-2(x y+y z+z x)+(x y+y z+z x)... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 1 | |
08jx | Problem:
given an $m \times n$ table, each cell signed with "-". The following operations are
(i) to change all the signs in an entire row to the opposite, i.e. every "-" to "+", and every "+" to "-";
(ii) to change all the signs in an entire column to the opposite, i.e. every "-" to "+" and every "+" to "-".
a. Prov... | [
"Solution:\n\nIf we apply (i) to $l$ rows and (ii) to $k$ columns, we obtain $(m-k)l + (n-l)k$ plus signs.\n\na.\nWe have the equation $(100-k)l + (100-l)k = 2004$, or $100l + 100k - 2lk = 2004$.\n\n$$\n50l + 50k - lk = 1002\n$$\nRewrite the last equation as\n$$\n(50-l)(50-k) = 2500 - 1002 = 1498\n$$\nSince $1498 =... | JBMO | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 102 | |
01lx | The external angle bisector of the angle $A$ of an acute-angled triangle $ABC$ meets the circumcircle of $\triangle ABC$ at point $T$. The perpendicular from the orthocenter $H$ of $\triangle ABC$ to the line $TA$ meets the line $BC$ at point $P$. The line $TP$ meets the circumcircle of $\triangle ABC$ at point $D$.
P... | [
"Let $l_a$ be the bisectrix of $\\angle A$. Then $l_a \\perp AT$ and $PH \\parallel l_a$ (as, by condition, $PH \\perp AT$). Further,\n$$ \\angle (BH, HP) = \\angle (BH, AC) + \\angle (AC, HP) = 90^\\circ + \\angle (AC, l_a) = 90^\\circ + \\angle (l_a, AB) = 90^\\circ + \\angle (HP, AB) = \\angle (HP, AB) + \\angle... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry ... | English | proof only | null | |
0fe4 | Problem:
Calculad todos los posibles valores de $f(2004)$, donde $f: \mathbb{N} \rightarrow \mathbb{N}$ es una función que cumple:
- $f(n m)=f(n) f(m)$ para todo par de números naturales $m, n$.
- $f(n) \leq n^{2}$ para todo número natural $n$.
- $f(1002) \geq 1003969$ | [
"Solution:\nSea $f$ una función que cumple las condiciones del enunciado.\nDescomponemos $1002$ y $2004$ en factores primos. Se tiene que\n$$\n1002=2 \\cdot 3 \\cdot 167 ;\\quad 2004=2^{2} \\cdot 3 \\cdot 167\n$$\nSi $f(2) \\leq 2^{2}-1$, obtenemos que\n$$\nf(1002)=f(2) f(3) f(167) \\leq\\left(2^{2}-1\\right) \\cdo... | Spain | null | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 4016016 | |
0gwk | A point $D$ is chosen on the side $BC$ of an acute triangle $ABC$ (and is distinct from the vertices). Points $P$ and $Q$ are defined as the centers of the circumcircles of the triangles $ABD$ and $ACD$ respectively. Prove that, as soon as the triangle $ABC$ is fixed, there exists a point in the plane, which is differe... | [
"We claim that this point is the circumcenter $O$ of triangle $ABC$.\n\nIf $AD$ is an altitude, the statement is obvious. We may assume that triangle $ADC$ is acute-angled and triangle $ADB$ is obtuse-angled. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$ respectively. Since $\\angle ADC > \\angle ABC$, as... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | The fixed point is the circumcenter of triangle ABC. | |
02io | Problem:
Numa certa povoação africana vivem 800 mulheres. Delas, $3\%$ usam apenas um brinco; das restantes, metade usa dois brincos e a outra metade, nenhum. Qual o número total de brincos usados por todas as mulheres?
A) 776
B) 788
C) 800
D) 812
E) 824 | [
"Solution:\n\nDo enunciado temos que o número de mulheres que usam apenas um brinco é $0,03 \\times 800 = 24$. Restam $800 - 24 = 776$, das quais $388$ usam dois brincos e $388$ não usam brincos. Logo, o número total de brincos usados por todas as mulheres é: $24 + 388 \\times 2 = 800$.",
"Solution:\n\nSe cada mu... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | C | |
020m | Problem:
In the land of Heptanomisma, four different coins and three different banknotes are used, and their denominations are seven different (non-zero) natural numbers. The denomination of the smallest banknote is greater than the sum of the denominations of the four different coins. A tourist has exactly one coin o... | [
"Solution:\n\nLet the denominations of the coins and notes be $C_{1}<C_{2}<C_{3}<C_{4}$ and $N_{1}<N_{2}<N_{3}$, respectively. Define $C=C_{1}+C_{2}+C_{3}+C_{4}$ to be the largest amount that can be paid with coins only. The condition of the problem is thus $C<N_{1}$.\n\na.\nSuppose to the contrary that, whatever t... | Benelux Mathematical Olympiad | 10th Benelux Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0dmf | Problem:
Дат је троугао $A B C$. Нека су тачке $D$ и $E$ на правој $A B$ такве да је $D-A-B-E$, $A D = A C$ и $B E = B C$. Симетрале унутрашњих углова код темена $A$ и $B$ секу наспрамне странице у тачкама $P$ и $Q$, редом, а описану кружницу око троугла $A B C$ у тачкама $M$ и $N$, редом. Права која спаја тачку $A$ с... | [
"Solution:\n\nОзначимо са $U$ центар описаног круга $\\triangle B M E$. Применимо инверзију са центром $A$ и квадратом полупречника $A B \\cdot A C$. Тачке $B$ и $C$ се сликају у тачке $B'$ и $C'$ симетричне тачкама $C$ и $B$ у односу на $A P$, тачке $P$ и $M$ се сликају једна у другу, а $E$ се слика у тачку $E'$ с... | Serbia | СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: cent... | null | proof only | null | |
0ati | Problem:
Trapezoid $ABCD$ has parallel sides $AB$ and $CD$, with $BC$ perpendicular to them. Suppose $AB = 13$, $BC = 16$ and $DC = 11$. Let $E$ be the midpoint of $AD$ and $F$ the point on $BC$ so that $EF$ is perpendicular to $AD$. Find the area of quadrilateral $AEFB$. | [] | Philippines | Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 91 | |
0fi0 | Problem:
Por el baricentro $G$ de un triángulo $ABC$ se traza una recta que corta el lado $AB$ en $P$ y el lado $AC$ en $Q$. Demostrar que
$$
\frac{PB}{PA} \cdot \frac{QC}{QA} \leq \frac{1}{4}
$$ | [
"Solution:\n\nDupliquemos el triángulo trazando $AD$ paralela a $BC$ y $CD$ paralela a $BA$ como muestra la figura, y tomemos la longitud del lado $AB$ como unidad. Llamando $M$ a la intersección de $CD$ con la recta $PQ$ y $x = PB$, $1-x = AP$, tenemos\n\n\n\nPor la semejanza de los triáng... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
092j | Problem:
Let $n \geqslant 2$ be an integer. Determine the number of positive integers $m$ such that $m \leqslant n$ and $m^{2}+1$ is divisible by $n$. | [
"Solution:\nLet $D(n)$ be the number of positive integers $m$ such that $m \\leqslant n$ and $m^{2}+1$ is divisible by $n$.\nNo number of the form $m^{2}+1$ is divisible by $4$, so if $4$ divides $n$, we have $D(n)=0$. It is also known that $D(n)=0$ if $n$ is divisible by some number of the form $4k+3$. Furthermore... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | Let n = 2^α · ∏ p_i^{α_i} · ∏ q_j^{β_j}, where each p_i ≡ 1 (mod 4) and each q_j ≡ 3 (mod 4). Then D(n), the count of m with 1 ≤ m ≤ n and n | (m^2 + 1), satisfies:
- If α ≥ 2 or any β_j ≥ 1, then D(n) = 0.
- If α ∈ {0,1} and all odd prime factors are ≡ 1 (mod 4), then D(n) = 2^r, where r is the number of distinct prim... | |
0cgf | Let $n$ be a positive integer. Initially, a bishop is placed in each square of the top row of a $2^n \times 2^n$ chessboard; those bishops are numbered from 1 to $2^n$, from left to right. A jump is a simultaneous move made by all bishops such that the following conditions are satisfied:
* each bishop moves diagonally,... | [
"*First solution.* The required number is $2^{n-1}$. On a jump, every bishop moves the same number of rows up or down; call this number of rows the *length* of the jump.\n\n**Step 1.** We show that the length of any jump is of the form $2^d$ for some integer $d \\le n-1$. Assign each bishop the number of the column... | Romania | Romanian Master of Mathematics | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 2^{n-1} | |
0it1 | Problem:
We say "$s$ grows to $r$" if there exists some integer $n > 0$ such that $s^{n} = r$. Call a real number $r$ "sparse" if there are only finitely many real numbers $s$ that grow to $r$. Find all real numbers that are sparse. | [
"Solution:\n\nAnswer: $-1, 0, 1$\n\nFor any number $x$, other than these, $3, x, \\sqrt[3]{x}, \\sqrt[5]{x}, \\sqrt[7]{x}, \\ldots$ provide infinitely many possible values of $s$, so these are the only possible sparse numbers. On the other hand, $-1$ is the only possible value of $s$ for $r = -1$, $0$ is the only v... | United States | 1st Annual Harvard-MIT November Tournament | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | -1, 0, 1 | |
0jp7 | Problem:
Consider the four points $A=(2,4)$, $B=(0,0)$, $C=(6,0)$, and $D=(6,3)$ in the plane. For which point $P$ in the plane is the sum $P A + P B + P C + P D$ minimized? | [
"Solution:\n\nThe answer is $P=(4,2)$, which is the intersection of the line segments $A C$ and $B D$. To see this is optimal, note that by the triangle inequality we have\n$$\nP A + P C \\geq A C \\quad \\text{and} \\quad P B + P D \\geq B D\n$$\nso $P A + P B + P C + P D \\geq A C + B D$ with equality occurring f... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | (4, 2) | |
0azp | Problem:
Find the remainder when $53!$ is divided by $59$. | [
"Solution:\nLet $x \\equiv 53! \\bmod 59$. By Wilson's theorem, $58! \\equiv -1 \\bmod 59$. Now, $58! = 58 \\cdot 57 \\cdot 56 \\cdot 55 \\cdot 54 \\cdot 53!$. $53! \\equiv (-1)(-2)(-3)(-4)(-5) x \\bmod 59$. This means that $-120 x \\equiv -1 \\bmod 59$ or $2 x \\equiv 1 \\bmod 59$. Thus, $x \\equiv 60 x \\equiv 30... | Philippines | 20th Philippine Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 30 | |
0jj7 | Problem:
How many ways can the eight vertices of a three-dimensional cube be colored red and blue such that no two points connected by an edge are both red? Rotations and reflections of a given coloring are considered distinct. | [
"Solution:\n\nAnswer: $35$\n\nWe do casework on $R$, the number of red vertices. Let the cube be called $A B C D E F G H$, with opposite faces $A B C D$ and $E F G H$, such that $A$ is directly above $E$.\n\n- $\\underline{R=0}$ : There is one such coloring, which has only blue vertices.\n\n- $\\underline{R=1}$ : T... | United States | HMMT November 2014 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | final answer only | 35 | |
0hke | Problem:
The numbers $1$ to $10$ are written on a blackboard in a row. Determine whether it is possible to put nine $+$ and $-$ signs between them (one between each pair of adjacent numbers) to make the resulting sum $0$. | [
"Solution:\n\nIt is not possible. Note that the sum of the numbers is\n$$\n1+2+3+\\cdots+9+10=55\n$$\nan odd number. Thus, if we put all $+$ signs, the sum is odd. Every time we change a $+$ to a $-$, we change an $n$ term to a $-n$ term for some $n$, which subtracts $2n$ from the sum. This changes the sum by an ev... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Not possible | |
04dt | John and Mary have two identical decks of $50$ cards with different symbols. John shuffles his cards and puts them on the table, and then Mary shuffles her cards and puts them on John's cards. Then John counts the cards between each pair of two identical cards and adds up those $50$ numbers. Which results can John obta... | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 2450 | |
0ayg | Problem:
Suppose that the lengths of the sides of a right triangle are integers and its area is six times its perimeter. What is the least possible length of its hypotenuse? | [] | Philippines | 21st PMO Area Stage | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 58 | |
03jf | Problem:
The number $1987$ can be written as a three digit number $x y z$ in some base $b$. If $x + y + z = 1 + 9 + 8 + 7$, determine all possible values of $x, y, z, b$. | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | x=5, y=9, z=11, b=19 | |
0kni | Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f: \mathbb{N} \to \mathbb{N}$ such that for all positive integers $a$ and $b$,
$$
f(a^2 + b^2) = f(a)f(b) \quad \text{and} \quad f(a^2) = f(a)^2.
$$ | [] | United States | USAJMO | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Other"
] | null | proof and answer | f(n) = 1 for all positive integers n | |
0d4i | Let $ABC$ be a triangle, $D$ the midpoint of side $BC$ and $E$ the intersection point of the bisector of angle $\angle BAC$ with side $BC$. The perpendicular bisector of $AE$ intersects the bisectors of angles $\angle CBA$ and $\angle CDA$ at $M$ and $N$, respectively. The bisectors of angles $\angle CBA$ and $\angle C... | [
"The bisector of angle $\\angle EBA$ and the perpendicular bisector of side $EA$, both bisect the $\\operatorname{arc}\\overparen{EA}$ of the circumcircle of triangle $ABE$ opposite to the vertex $B$. Therefore quadrilateral $ABEM$ is cyclic and hence\n$$\n\\angle EAM = \\angle EBM\n$$\nThe segment defined by the t... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English, Arabic | proof only | null | |
0jb7 | Problem:
There are circles $\omega_{1}$ and $\omega_{2}$. They intersect in two points, one of which is the point $A$. $B$ lies on $\omega_{1}$ such that $A B$ is tangent to $\omega_{2}$. The tangent to $\omega_{1}$ at $B$ intersects $\omega_{2}$ at $C$ and $D$, where $D$ is the closer to $B$. $A D$ intersects $\omega... | [
"Solution:\n\n> [diagram]\n\nBy power of a point, $B A=\\sqrt{B D \\cdot B C}=4 \\sqrt{3}$. Also, $D E B \\sim D B A$, so $E B / E D=B A / B D=4 \\sqrt{3} / 3$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 4√3/3 | |
03h1 | Problem:
Let $a_{1}, a_{2}, \ldots, a_{n}$ be non-negative real numbers. Define $M$ to be the sum of all products of pairs $a_{i} a_{j}$ ($i<j$), i.e.,
$$
M = a_{1}(a_{2} + a_{3} + \cdots + a_{n}) + a_{2}(a_{3} + a_{4} + \cdots + a_{n}) + \cdots + a_{n-1} a_{n}
$$
Prove that the square of at least one of the numbers $a... | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0f7c | Problem:
A regular hexagon has side $1000$. Each side is divided into $1000$ equal parts. Let $S$ be the set of the vertices and all the subdividing points. All possible lines parallel to the sides and with endpoints in $S$ are drawn, so that the hexagon is divided into equilateral triangles with side $1$.
Let $X$ be... | [] | Soviet Union | 20th ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
03s7 | Let $f(x)$ be a decreasing function defined on $(0, +\infty)$. If $f(2a^2 + a + 1) < f(3a^2 - 4a + 1)$, then the range of $a$ is __________. | [
"Since $f(x)$ is defined on $(0, +\\infty)$, from\n$$\n\\begin{cases} 2a^2 + a + 1 = 2\\left(a+\\frac{1}{4}\\right)^2 + \\frac{7}{8} > 0, \\\\ 3a^2 - 4a + 1 = (3a-1)(a-1) > 0, \\end{cases}\n$$\nwe get \\qquad $a > 1$ or $a < \\frac{1}{3}$. \\qquad (1)\n\nSince $f(x)$ is a decreasing function on $(0, +\\infty)$, so\... | China | China Mathematical Competition (Jiangxi) | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 0 < a < 1/3 or 1 < a < 5 | |
0dqs | If 46 squares are colored red in a $9 \times 9$ board, show that there is a $2 \times 2$ block on the board in which at least 3 of the squares are colored red. | [
"Suppose that at most 2 squares are colored red in any $2 \\times 2$ square. Then in any $9 \\times 2$ block, there are at most 10 red squares. Moreover, if there are 10 red squares, then there must be 5 in each row. This can be seen as follows. There are $8 \\times 2 \\times 2$ blocks. Counting multiplicity, there... | Singapore | Singapore Mathematical Olympiad (SMO) 2011 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
05zv | Problem:
Soit $\left(P_{n}\right)_{n \in \mathbb{N}}$ une suite de polynômes à coefficients entiers. On suppose qu'il existe un polynôme $Q$ unitaire à coefficient entier tel que $P_{n+1}-P_{n}=Q$ pour tout $n \geqslant 0$. On suppose de plus que pour tout $n \geqslant 0$, $P_{n}$ a une racine entière. Montrer qu'on e... | [
"Solution:\n\nSupposons que $P_{0}$ et $Q$ n'ont pas de racines entière en commun. Notons que par récurrence immédiate $P_{n}=P_{0}+n Q$ pour tout $n \\geqslant 0$. Notons $x_{n}$ une racine entière de $P_{n}$ pour tout $n \\geqslant 0$, on a que $n Q\\left(x_{n}\\right)=-P_{0}\\left(x_{n}\\right)$, donc $Q\\left(x... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
066z | In the plane there are given $n$ different circles with the same center. In the interior of the circle with the smaller radius we consider two different points $A$ and $B$. Next we consider $k$ different lines passing from point $A$ and $m$ different lines passing from point $B$. All lines passing from $A$ intersect al... | [
"First we note that the number of points of intersection of the lines is $k \\cdot m$ (figure 1).\n\n\nFigure 1\n\nLet now all the points of intersection lie at the exterior of the circle with the greatest radius. Then the $k$ different lines passing through $A$ create $2k$ regions inside t... | Greece | Selection Examination A | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | minimum = 2n(k+m) − 1; maximum = 2n(k+m) | |
08r0 | Problem:
Alice and Bob play the following game on a $100 \times 100$ grid, taking turns, with Alice starting first. Initially the grid is empty. At their turn, they choose an integer from $1$ to $100^{2}$ that is not written yet in any of the cells and choose an empty cell, and place it in the chosen cell. When there ... | [
"Solution:\n\nWe denote by $(i, j)$ the cell in the $i$-th line and in the $j$-th column for every $1 \\leq i, j \\leq n$. Bob associates the following pair of cells: $(i, 2k+1), (i, 2k+2)$ for $1 \\leq i \\leq 100$ and $0 \\leq k \\leq 49$ except for $(i, k) = (100, 0)$ and $(100, 1)$, and the pairs $(100, 1), (10... | JBMO | Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Bob has a winning strategy. | |
0bp5 | Problem:
Fie $(R,+, \cdot)$ un inel cu proprietatea că, pentru orice element $x \in R$, există două elemente $e_{1}$ şi $e_{2}$ din $R$, astfel încât $e_{1}^{2}=e_{1}$, $e_{2}^{2}=e_{2}$ şi $x=e_{1} e_{2}$. Arătaţi că:
(a) $1$ este singurul element inversabil al inelului $R$; şi
(b) $x^{2}=x$, oricare ar fi $x \in R$. | [
"Solution:\n(a) Fie $x \\in U(R)$ şi $e_{1}, e_{2} \\in I(R)=\\{e \\in R \\mid e^{2}=e\\}$, cu $x=e_{1} e_{2}$. Atunci\n$$\n1-e_{1}=(1-e_{1}) \\cdot 1=(1-e_{1}) e_{1} e_{2} x^{-1}=(e_{1}-e_{1}^{2}) e_{2} x^{-1}=0\n$$\ndeci $e_{1}=1$ şi $x^{2}=e_{2}^{2}=e_{2}=x$. Cum $x$ este inversabil, rezultă că $x=1$ şi deci $U(... | Romania | Olimpiada Naţională de Matematică, Etapa Naţională | [
"Algebra > Abstract Algebra > Ring Theory"
] | null | proof only | null | |
0jic | Problem:
Given a permutation $\sigma$ of $\{1,2, \ldots, 2013\}$, let $f(\sigma)$ be the number of fixed points of $\sigma$—that is, the number of $k \in \{1,2, \ldots, 2013\}$ such that $\sigma(k) = k$. If $S$ is the set of all possible permutations $\sigma$, compute
$$
\sum_{\sigma \in S} f(\sigma)^4
$$
(Here, a perm... | [
"Solution:\nAnswer: $15(2013!)$\n\nFirst, note that\n$$\n\\sum_{\\sigma \\in S} f(\\sigma)^4 = \\sum_{\\sigma \\in S} \\sum_{1 \\leq a_1, a_2, a_3, a_4 \\leq 2013} g\\left(\\sigma, a_1, a_2, a_3, a_4\\right)\n$$\nwhere $g\\left(\\sigma, a_1, a_2, a_3, a_4\\right) = 1$ if all $a_i$ are fixed points of $\\sigma$ and ... | United States | HMMT | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | null | proof and answer | 15(2013!) | |
07pm | Suppose $0 \le x, y, z$, and $x+y+z \le 1$. Prove that
$$
6xyz \le x(1-x) + y(1-y) + z(1-z),
$$
and describe the cases of equality. | [
"By hypothesis, $x$, $y$, $z \\in [0, 1]$, and so $3xyz \\le xy + yz + zx$. Also, $(x+y+z)^2 \\le x+y+z$, i.e., $x^2+y^2+z^2+2(xy+yz+zx) \\le x+y+z$, equivalently,\n$$\n2(xy + yz + zx) \\le x(1-x) + y(1-y) + z(1-z),\n$$\nwhence\n$$\n6xyz \\le 2(xy + yz + zx) \\le x(1-x) + y(1-y) + z(1-z),\n$$\nand there is equality... | Ireland | Ireland | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Equality holds exactly at (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1). | |
0enp | Suppose that $n$ points are given on a circle and $nk + 1$ of the chords between these points are drawn, where $2k + 1 < n$. Prove that it is possible to select $k+1$ of the chords such that no two of them intersect. | [
"Without loss of generality, we may assume that the $n$ points form a regular $n$-gon. We will partition all the possible chords of the $n$-gon into $n$ different sets such that all the chords in each set are parallel. We distinguish two cases:\n\nIf $n$ is odd, then each chord is parallel to exactly one of the sid... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Circles",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0jjj | Problem:
Point $P$ and line $\ell$ are such that the distance from $P$ to $\ell$ is $12$. Given that $T$ is a point on $\ell$ such that $PT = 13$, find the radius of the circle passing through $P$ and tangent to $\ell$ at $T$. | [
"Solution:\n\nAnswer: $\\dfrac{169}{24}$\n\nLet $O$ be the center of the given circle, $Q$ be the foot of the altitude from $P$ to $\\ell$, and $M$ be the midpoint of $PT$. Then since $OM \\perp PT$ and $\\angle OTP = \\angle TPQ$, $\\triangle OMP \\sim \\triangle TQP$. Thus the $OP = TP \\cdot \\dfrac{PM}{PQ} = 13... | United States | HMMT 2014 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 169/24 | |
0dwq | Problem:
Poišči vsa praštevila $p$, za katera ima število $p^{2}+11$ manj kot 11 pozitivnih deliteljev. | [
"Solution:\n\nČe je $p=2$, je $p^{2}+11=15=3 \\cdot 5$ in to število ima 4 delitelje (1, 3, 5 in 15).\n\nČe je $p=3$, je $p^{2}+11=17$ in ima 2 delitelja.\n\nZa $p>3$ je praštevilo $p$ liho, zato da $p^{2}$ pri deljenju s 4 ostanek 1 in je število $p^{2}+11$ deljivo s 4.\n\nPraštevilo $p>3$ ni deljivo s 3, torej da... | Slovenia | 49. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 2, 3, 5 | |
05bg | Juku multiplied four consecutive positive integers and divided the result by four. Find all the possibilities for the digit in which this quotient can end. | [
"Among 4 consecutive integers there always exist two even numbers, one of which is also divisible by 4. Thus the quotient is even, excluding all odd final digits.\nIf the final digit of the quotient upon division by 4 is 0, 2, 4, 6 or 8, then the\nfinal digit of the dividend is 0, 8, 6, 4 or 2, respectively.\n\n* I... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 0 and 6 | |
0fb8 | Problem:
Demuestra que
$$
(a x+b y)^{2} \leq a x^{2}+b y^{2}
$$
para cualesquiera $x, y \in \mathbb{R}$ y cualesquiera $a, b \in \mathbb{R}$ con $a+b=1,\ a, b \geq 0$. ¿En qué casos se da la igualdad? | [
"Solution:\nSolución 1. Nótese que\n$$\na x^{2}+b y^{2}-(a x+b y)^{2}=a(1-a) x^{2}+b(1-b) y^{2}-2 a b x y=a b(x-y)^{2}\n$$\ndonde hemos usado que $1-a=b$ y $1-b=a$. Esta expresión es claramente no negativa, siendo nula si y sólo si bien $a b=0$ (es decir, uno de entre $a, b$ es 0 y el otro es 1), bien $x=y$.\n\n\nS... | Spain | Problemas Primera Sesión | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | Equality holds exactly when either the two real numbers are equal or one of the weights is zero (and the other is one). | |
0k76 | Problem:
A positive integer $n$ is infallible if it is possible to select $n$ vertices of a regular 100-gon so that they form a convex, non-self-intersecting $n$-gon having all equal angles. Find the sum of all infallible integers $n$ between 3 and 100, inclusive. | [
"Solution:\n\nSuppose $A_{1} A_{2} \\ldots A_{n}$ is an equiangular $n$-gon formed from the vertices of a regular 100-gon. Note that the angle $\\angle A_{1} A_{2} A_{3}$ is determined only by the number of vertices of the 100-gon between $A_{1}$ and $A_{3}$. Thus in order for $A_{1} A_{2} \\ldots A_{n}$ to be equi... | United States | HMMT November 2019 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 262 | |
0hyx | Problem:
Alex is stuck on a platform floating over an abyss at $1\ \mathrm{ft} / \mathrm{s}$. An evil physicist has arranged for the platform to fall in (taking Alex with it) after traveling $100\ \mathrm{ft}$. One minute after the platform was launched, Edward arrives with a second platform capable of floating all th... | [
"Solution:\n\nThe slower the second platform is moving, the longer it will stay next to the first platform. However, it needs to be moving fast enough to reach the first platform before it's too late. Let $v$ be the velocity of the second platform. It starts $65$ feet behind the first platform, so it reaches the ba... | United States | Harvard-MIT Math Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 3 ft/s | |
0gnt | $O$ is the center and $r$ is the radius of the incircle of a tangential quadrilateral $ABCD$. Let $P$ be the point of intersection of the lines $AB$ and $CD$, $Q$ be the point of intersection of the lines $AD$ and $BC$, and $E$ be the point of intersection of the diagonals $AC$ and $BD$. Show that $|OE| \cdot d = r^2$ ... | [
"Let $K, L, M, N$ be the points of tangency of the incircle of $ABCD$ to the sides $AB, BC, CD, DA$, respectively, and let $S$ be the foot of the perpendicular from $O$ to the line $PQ$. Since $ABCD$ is a tangential quadrilateral, the point of intersection of the lines $KM$ and $LN$ coincides with the point of inte... | Turkey | Team Selection Test for IMO 2009 | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof only | null |
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