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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0kiz | Problem:
There are $8$ lily pads in a pond numbered $1, 2, \ldots, 8$. A frog starts on lily pad $1$. During the $i$-th second, the frog jumps from lily pad $i$ to $i+1$, falling into the water with probability $\frac{1}{i+1}$. The probability that the frog lands safely on lily pad $8$ without having fallen into the w... | [
"Solution:\n\nThe probability the frog lands safely on lily pad $i+1$ given that the frog safely landed on lily pad $i$ is $\\frac{i}{i+1}$. The probability the frog makes it to lily pad $8$ safely is simply the product of the probabilities of the frog making it to each of the lily pads $2$ through $8$ given it had... | United States | HMMT November 2021 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 108 | |
09jw | Find all triples of positive integers $a$, $b$ and $c$, for which the expression
$$
T = \frac{(a + 1)(b + 1)(c + 1) - 12}{abc}
$$
is an integer. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All solutions are the permutations of either (2, 3, n) for any positive integer n, or (1, 1, c) with c in {1, 2, 4, 8}. | |
0127 | Problem:
Given a rhombus $A B C D$, find the locus of the points $P$ lying inside the rhombus and satisfying $\angle A P D + \angle B P C = 180^{\circ}$. | [
"Solution:\n\nthe locus of the points $P$ is the union of the diagonals $A C$ and $B D$.\n\nLet $Q$ be a point such that $P Q C D$ is a parallelogram (see Figure 4). Then $A B Q P$ is also a parallelogram. From the equality $\\angle A P D + \\angle B P C = 180^{\\circ}$ it follows that $\\angle B Q C + \\angle B P ... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Translation"
] | null | proof and answer | The locus is the union of the diagonals AC and BD. | |
0fxs | Problem:
Finde alle Funktionen $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$, welche für alle $x>y>z>0$ die folgende Gleichung erfüllen:
$$
f(x-y+z)=f(x)+f(y)+f(z)-x y-y z+x z
$$ | [
"Solution:\nSeien $x>z>0$ beliebig. Setze $y=\\frac{x+z}{2}$, dann gilt $x>y>z$ und die Gleichung vereinfacht sich zu\n$$\nf(x)+f(z)=\\frac{x^{2}+z^{2}}{2}, \\quad \\forall x>z>0\n$$\nSetzt man hier einerseits $z=1$ und $x>1$ und andererseits $x=1$ und $0<z<1$, dann folgt insgesamt\n$$\nf(x)=\\frac{x^{2}}{2}+\\left... | Switzerland | SMO Finalrunde | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | f(x) = x^2 / 2 for x > 0 | |
0hcs | Bogdan drawn $2017$ vertical and $2018$ horizontal lines in the rectangle $Q$. These lines divide $Q$ into $2018 \times 2019$ smaller not necessarily equal rectangles. Two children want to determine the perimeter of $Q$. Andrew says that he can do it by choosing some $2019$ smaller rectangles and getting to know their ... | [
"Let us consider two rectangles constructed from $2018 \\times 2019$ smaller rectangles, of sizes $2 \\times 2$ and $3 \\times 1$ respectively. It is clear that all small rectangles in both of them have equal perimeters, namely $8$. However, the first one is a rectangle $4036 \\times 4038$ with the perimeter $16148... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | Neither | |
0i86 | Problem:
Suppose $A$ is a set with $n$ elements, and $k$ is a divisor of $n$. Find the number of consistent $k$-configurations of $A$ of order 1. | [
"Solution:\n\nGiven such a $k$-configuration, we can write out all the elements of one of the $k$-element subsets, then all the elements of another subset, and so forth, eventually obtaining an ordering of all $n$ elements of $A$. Conversely, given any ordering of the elements of $A$, we can construct a consistent ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | n! / ((n/k)! (k!)^{n/k}) | |
032j | Problem:
In a school there are $m$ boys and $j$ girls, $m \geq 1, 1 \leq j < 2004$.
Every student has sent a post card to every student. It is known that the number of the post cards sent by the boys is equal to the number of the post cards sent by girl to girl. Find all possible values of $j$. | [
"Solution:\n\nIt follows by the given condition that $m(m+j-1)=j(j-1)$ and hence $m^{2}=(j-m)(j-1)$. If $p$ is a prime divisor of $j-m$ and $j-1$, then it divides $m$. Therefore $p$ divides $j$ and $1$. This contradiction shows that $j-m$ and $j-1$ are coprime. Then $j-m=u^{2}$ and $j-1=v^{2}$, where $u$ and $v$ ar... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | j ∈ {2, 10, 65, 442} | |
0i4a | Problem:
Let $\triangle ABC$ be equilateral, and let $D, E, F$ be points on sides $BC, CA, AB$ respectively, with $FA=9$, $AE=EC=6$, $CD=4$. Determine the measure (in degrees) of $\angle DEF$. | [
"Solution:\n\n$H, I$ be the respective midpoints of sides $BC, AB$, and also extend $CB$ and $EF$ to intersect at $J$. By equal angles, $\\triangle EIF \\sim \\triangle JBF$. However, $BF=12-9=3=9-6=IF$, so in fact $\\triangle EIF \\cong \\triangle JBF$, and then $JB=6$. Now let $HI$ intersect $EF$ at $K$, and noti... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 60 | |
0dod | Problem:
Дат је троугао $ABC$. Тачке $D$ и $E$ на правој $AB$ су такве да је $AD = AC$ и $BE = BC$, уз распоред $D-A-B-E$. Описане кружнице троуглова $DBC$ и $EAC$ секу се у тачки $X \neq C$, а описане кружнице троуглова $DEC$ и $ABC$ секу се у тачки $Y \neq C$. Ако важи $DY + EY = 2XY$, одредити $\varangle ACB$. | [
"Solution:\n\nНека је $I_{c}$ центар приписаног круга $\\triangle ABC$ наспрам темена $C$. Тада је $AI_{c}$ симетрала угла $CAD$, одакле је $\\triangle I_{c}AD \\cong \\triangle I_{c}AC$. Следи да је $\\varangle I_{c}DB = \\varangle I_{c}DA = \\varangle I_{c}CA = \\varangle I_{c}CB$, па $I_{c}$ лежи на кругу $BCD$.... | Serbia | 14. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry... | null | proof and answer | ∠ACB = 60° | |
0gsr | Given some collection of subsets $A_1, A_2, \dots$ of the set $A = \{1, 2, \dots, n\}$ we say that a subset $B \subset A$ is *sparse* if no $A_i \not\subset B$. Suppose that for any collection $A_1, A_2, \dots$, with $A_i \subset A$, $|A_i| = 3$, and $|A_i \cap A_j| \le 1$ ($i \neq j$), any sparse set containing 29 ele... | [
"Answer: $\\binom{29}{2} + 29 + 1 = 436$.\nLet $n = 436$ and $B$ is a sparse set containing at most $k \\le 29$ elements. Let us consider all $\\binom{k}{2}$ two element subsets of $B$. By conditions, each of these subsets can belong to at most one subset $A_i$. If $B$ can not be extended by adding of some element ... | Turkey | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 436 | |
0ad0 | The height of isosceles trapeze equals to $h$ and his area to $h^2$. What is the measure of the angle between the two diagonals? | [
"Let $ABCD$ be the isosceles trapeze with height $h$ and area $P = h^2$. Let $S$ be the intersection point of the diagonals and $S'$ and $S''$ are the bases of the altitudes from $S$ to $AB$ and $CD$ correspondently. The triangles $SS''D$ and $SS'B$ are similar because they have equal angles. So we get that $\\over... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 90 degrees | |
0b8l | Let $f : \mathbb{R} \to \mathbb{R}$ be a strictly increasing function such that $f \circ f$ is continuous. Prove that $f$ is continuous. | [
"Since $f$ is monotonic, the lateral limits at each point exist and are finite. Denote $f(x_0^-)$, respectively $f(x_0^+)$, the limit from the left, respectively from the right at $x_0$. Then, $f$ being increasing, $f(x_0^-) \\leq f(x_0) \\leq f(x_0^+)$ for every $x_0 \\in \\mathbb{R}$.\n\nSuppose now that $f$ is d... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof only | null | |
0kjh | Problem:
Let $E$ be a three-dimensional ellipsoid. For a plane $p$, let $E(p)$ be the projection of $E$ onto the plane $p$. The minimum and maximum areas of $E(p)$ are $9 \pi$ and $25 \pi$, and there exists a $p$ where $E(p)$ is a circle of area $16 \pi$. If $V$ is the volume of $E$, compute $V / \pi$. | [
"Solution:\n\nLet the three radii of $E$ be $a < b < c$. We know that $ab = 9$ and $bc = 25$.\n\nConsider the plane $p$ where projection $E(p)$ has area $9 \\pi$. Fixing $p$, rotate $E$ on the axis passing through the radius with length $b$ until $E(p)$ has area $25 \\pi$. The projection onto $p$ will be an ellipse... | United States | HMMT Spring 2021 Guts Round | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Other 3D problems"
] | null | final answer only | 75 | |
0kn9 | Problem:
Let $N$ be the smallest positive integer for which
$$
x^{2}+x+1 \quad \text{divides} \quad 166-\sum_{d \mid N, d>0} x^{d}
$$
Find the remainder when $N$ is divided by $1000$. | [
"Solution:\nLet $\\omega = e^{2 \\pi i / 3}$. The condition is equivalent to\n$$\n166 = \\sum_{d \\mid N, d>0} \\omega^{d}.\n$$\nLet's write $N = 3^{d} n$ where $n$ is not divisible by $3$. If all primes dividing $n$ are $1 \\bmod 3$, then $N$ has a positive number of factors that are $1 \\bmod 3$ and none that are... | United States | HMMT November 2021 Team Round | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | 672 | |
028q | Problem:
Rosa, Margarida e Dália são três constelações em forma de buquês de flores. Sabemos que:
a. O número de estrelas de Dália, que é a menor das três, é o quadrado de um quadrado;
b. O número de estrelas de Rosa é também o quadrado de um quadrado;
c. Margarida tem 28561 estrelas;
d. Dália e Rosa têm juntas o mes... | [
"Solution:\n\n$\\{D\\} = 4225 = 25 \\times 169$\ne $\\{R\\} = 144 \\times 169 = 24336$"
] | Brazil | Desafios | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | Dália: 4225; Rosa: 24336 | |
08jc | Problem:
Let $n \geq 2003$ be a positive integer such that the number $1+2003 n$ is a perfect square. Prove that the number $n+1$ is equal to the sum of 2003 positive perfect squares. | [] | JBMO | The first selection test for JBMO 2003 | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
07fg | We call an integer number $n > 0$ interesting if for each permutation $\sigma$ of $1, 2, \dots, n$ there exist polynomials $P_1, P_2, \dots, P_n$ and $\epsilon > 0$ such that:
$$
i)\ P_1(0) = P_2(0) = \dots = P_n(0).
$$
$$
ii)\ P_1(x) > P_2(x) > \dots > P_n(x) \text{ for } -\epsilon < x < 0.
$$
$$
iii)\ P_{\sigma(1)}(x... | [
"Only $n = 2$ and $n = 3$ are interesting.\n\nNote that if $n$ is not interesting, then any integer greater than $n$ is not interesting as well. So we just need to prove that $n = 4$ is not interesting.\n\nWe claim that there is no such example for the case\n$$\nP_2(x) > P_4(x) > P_1(x) > P_3(x)\n$$\nfor infinitesi... | Iran | 37th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 2 and 3 | |
00ty | Find the largest positive integer $k$ for which there exists a convex polyhedron $P$ with the following properties:
(a) $P$ has exactly $2022$ edges.
(b) The degrees of the vertices of $P$ don't differ by more than $1$.
(c) It is possible to colour the edges of $P$ with $k$ colours such that for every colour $c$, and e... | [
"We divide the solution in two steps, first we prove that $k < 3$, and then give an inductive construction of $P$ for $k = 2$.\nLet $P$ have $V$ vertices, $E$ edges and $F$ faces. Suppose the contrary, that $k > 2$. We have $k$ disjoint trees on $V$ vertices, so $E \\ge 3(V - 1)$. This is a contradiction as for eve... | Balkan Mathematical Olympiad | BMO 2022 shortlist | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Geometry > Solid Geometry > 3D Shapes",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 2 | |
0fqk | Problem:
Los números reales $a$, $b$ y $c$ verifican que el polinomio $p(x) = x^{4} + a x^{3} + b x^{2} + a x + c$ tiene exactamente tres raíces reales distintas; estas raíces son iguales a $\tan(y)$, $\tan(2y)$ y $\tan(3y)$, para algún número real $y$. Hallar todos los posibles valores de $y$, $0 \leq y < \pi$. | [
"Solution:\n\nSean $r, r, s, t$ los tres ceros reales y distintos del polinomio $p(x)$. Mediante las fórmulas de Cardano-Viète, identificando los coeficientes cúbico y lineal, se obtiene\n$$\n2r + s + t = r^{2} s + r^{2} t + 2 r s t \\Leftrightarrow 2r(1 - s t) = (r^{2} - 1)(s + t)\n$$\nDe la última igualdad se ded... | Spain | LV Olimpiada matemática Española | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | y ∈ {π/7, 2π/7, 3π/7, 4π/7, 5π/7, 6π/7, π/8, 3π/8, 5π/8, 7π/8, π/9, 2π/9, π/3, 4π/9, 5π/9, 2π/3, 7π/9, 8π/9} | |
0bud | Problem:
Calculați :
$$
\lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{k^{2}+k}{n^{3}+k}
$$ | [] | Romania | Olimpiada Națională de Matematică | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1/3 | |
0a0j | Sara has 10 blocks numbered $1$ to $10$. She wants to stack all the blocks into a tower. A block can only be put on top of a block with a higher number, or on top of a block with a number that is exactly one lower. An example of such a tower is, from top to bottom: $2$, $1$, $5$, $4$, $3$, $6$, $7$, $9$, $8$, $10$.
How... | [
"We will first look at the problem for a smaller number of blocks. With $1$ block, only $1$ tower is possible. With $2$ blocks, both towers (top) $1$-$2$ (bottom) and $2$-$1$ are possible. With $3$ blocks, these are the possible towers: $1$-$2$-$3$, $2$-$1$-$3$, $1$-$3$-$2$, and $3$-$2$-$1$. It seems that the numbe... | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 512 | |
08zq | We denote $S = \{1, 2, \dots, 3000\}$. Find the maximum possible value of an integer $X$ that satisfies the following condition:
For any bijection $f: S \to S$, there exists a bijection $g: S \to S$ such that
$$
\sum_{k=1}^{3000} \left( \max\{f(f(k)), f(g(k)), g(f(k)), g(g(k))\} - \min\{f(f(k)), f(g(k)), g(f(k)), g(g(k... | [
"We denote\n$$\nX_{\\max}(f,g) = \\sum_{k=1}^{3000} \\max\\{f(f(k)), f(g(k)), g(f(k)), g(g(k))\\},\n$$\n$$\nX_{\\min}(f,g) = \\sum_{k=1}^{3000} \\min\\{f(f(k)), f(g(k)), g(f(k)), g(g(k))\\},\n$$\nand $X(f,g) = X_{\\max}(f,g) - X_{\\min}(f,g)$. Moreover, for a finite set $T$, we denote the number of its elements by ... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English | proof and answer | 6000000 | |
0249 | Problem:
O Rei Arthur teve que lutar com o Dragão das Três Cabeças e Três Caudas. Sua tarefa ficou facilitada quando conseguiu arranjar uma espada mágica que podia, de um só golpe, fazer uma e somente uma das seguintes coisas:
- cortar uma cabeça;
- cortar duas cabeças;
- cortar uma cauda;
- cortar duas caudas.
Além... | [
"Solution:\n\n5"
] | Brazil | Desafios | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | final answer only | 5 | |
05fl | Problem:
Soient $P$, $Q$ deux polynômes à coefficients réels, non constants et premiers entre eux. Montrer qu'il existe au plus trois réels $\lambda$ tels que:
$$
P+\lambda Q=R^{2}
$$
où $R \in \mathbb{R}[X]$. | [
"Solution:\n\nOn procède par l'absurde en supposant l'existence de quatre réels distincts $\\lambda_{1}, \\ldots, \\lambda_{4}$ et quatre polynômes $R_{1}, \\ldots, R_{4}$ tels que $P+\\lambda_{i} Q=R_{i}^{2}$ pour $i=1,2,3,4$.\n\nOn a $P'+\\lambda_{i} Q'=2 R_{i} R_{i}'$ et donc $R_{i} \\mid Q'\\left(P+\\lambda_{i}... | France | ENVOI 2 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0jjb | Problem:
Suppose that $x$ and $y$ are positive real numbers such that $x^{2}-x y+2 y^{2}=8$. Find the maximum possible value of $x^{2}+x y+2 y^{2}$. | [
"Solution:\nAnswer: $\\frac{72+32 \\sqrt{2}}{7}$\n\nLet $u = x^{2} + 2 y^{2}$. By AM-GM, $u \\geq \\sqrt{8} x y$, so $x y \\leq \\frac{u}{\\sqrt{8}}$. If we let $x y = k u$ where $k \\leq \\frac{1}{\\sqrt{8}}$, then we have\n\n$$\n\\begin{gathered}\nu(1-k)=8 \\\\\nu(1+k)=x^{2}+x y+2 y^{2}\n\\end{gathered}\n$$\n\nth... | United States | HMMT 2014 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | (72+32 sqrt(2))/7 | |
08iy | Problem:
Let $n \geq 1$ be a positive integer. Find all polynomials of degree $2n$ with real coefficients
$$
P(X) = X^{2n} + (2n-10) X^{2n-1} + a_{2} X^{2n-2} + \ldots + a_{2n-2} X^{2} + (2n-10) X + 1
$$
if it is known that they have positive real roots. | [] | JBMO | The second selection test for IMO 2003 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Only n = 1 or n = 2 are possible. For n = 1: P(x) = x^2 − 8 x + 1. For n = 2: P(x) = x^4 − 6 x^3 + a x^2 − 6 x + 1, where a ranges over all real values in [10, 11]; equivalently P(x) = (x^2 − s x + 1)(x^2 − t x + 1) with s, t ≥ 2 and s + t = 6. For n ≥ 3, no such polynomial exists. | |
0i7e | Prove that for every positive integer $n$ there exists an $n$-digit number divisible by $5^n$ all of whose digits are odd. | [
"We proceed by induction. The property is clearly true for $n=1$. Assume that $N = a_1a_2\\dots a_n$ is divisible by $5^n$ and has only odd digits. Consider the numbers\n$$\nN_1 = 1a_1a_2\\dots a_n = 1 \\cdot 10^n + 5^nM = 5^n(1 \\cdot 2^n + M),\n$$\n$$\nN_2 = 3a_1a_2\\dots a_n = 3 \\cdot 10^n + 5^nM = 5^n(3 \\cdot... | United States | USA IMO 2003 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0bmn | Two players, $A$ and $B$, remove alternatively stones from a heap initially containing $n \ge 2$ stones. The first to move is $A$ who takes at least one stone and at most $n-1$ stones. Next, each player to move has to take at least one stone and at most as many stones as his opponent took in his previous move. The winn... | [
"Let us label $(a, b)$ the position meaning \"the player who has to move finds $a$ stones in the heap and can take at most $b$ stones\".\n\nWe prove that the losing positions are those of the forms $(0, r)$ and $(2^k(2m+1), r)$, where $k, m \\in \\mathbb{N}$, $r \\in \\{1, 2, \\dots, 2^k - 1\\}$.\n\nThe initial pos... | Romania | 66th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | If the initial number of stones is a power of two, player B has a winning strategy; otherwise, player A has a winning strategy. A winning strategy is to always remove the largest power of two dividing the current number of stones. | |
07m4 | For each odd integer $p \ge 3$ find the number of real roots of the polynomial
$$
f_p(x) = (x - 1)(x - 2) \cdots (x - p + 1) + 1.
$$ | [
"We first look at the cases $p = 3, 5$, then deal with $p \\ge 7$. The polynomial\n$$\nf_3(x) = (x-1)(x-2)+1 = x^2-3x+3 = \\left(x-\\frac{3}{2}\\right)^2+\\frac{3}{4} > 0\n$$\nhas no real root.\n\nIf $p=5$, we obtain\n$$\n\\begin{align*}\nf_5(x) &= (x-1)(x-2)(x-3)(x-4) + 1 = (x-1)(x-4)(x-2)(x-3) + 1 \\\\\n&= (x^2 -... | Ireland | Irish Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | For p = 3: 0 real roots. For p = 5: 2 real roots (each double). For odd p ≥ 7: exactly p − 1 real roots. | |
0fda | Problem:
Determinar todos los números naturales $n$ para los que existe algún número natural $m$ con las siguientes propiedades
- $m$ tiene al menos dos cifras (en base 10), todas son distintas y ninguna es 0.
- $m$ es múltiplo de $n$ y cualquier reordenación de sus cifras da lugar a un múltiplo de $n$. | [
"Solution:\nNotemos que $n=1,3,9$ funcionan porque en estos casos el criterio de divisibilidad es que la suma de las cifras sea múltiplo de $n$, y tomando $m=18$ ya está. Para cualquier otro caso, consideramos que $m$ existe y que tiene como cifra más a la derecha $x$ y luego una $y$.\nEntonces, permutando esas dos... | Spain | null | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 1, 2, 3, 4, 6, 9, 12, 18 | |
0khe | The angle bisector of the acute angle formed at the origin by the graphs of the lines $y = x$ and $y = 3x$ has equation $y = kx$. What is $k$?
(A) $\frac{1 + \sqrt{5}}{2}$ (B) $\frac{1 + \sqrt{7}}{2}$ (C) $\frac{2 + \sqrt{3}}{2}$ (D) $2$ (E) $\frac{2 + \sqrt{5}}{2}$ | [] | United States | AMC 12 A | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | A | |
0an9 | Problem:
Find the sum of the largest and smallest possible values of $9 \cos^4 x + 12 \sin^2 x - 4$.
(a) 10
(b) 11
(c) 12
(d) 13 | [] | Philippines | Qualifying Round | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | 12 | |
0c30 | Let $f : [0, \infty) \to [0, \infty)$ be a differentiable function, with continuous derivative, and $f(f(x)) = x^2$, for every $x \in [0, \infty)$. Prove that
$$
\int_{0}^{1} (f'(x))^2 \, dx \ge \frac{30}{31}.
$$ | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof only | null | |
02tc | Problem:
Observe a equação:
$$
\begin{aligned}
(1+2+3+4)^{2} & =(1+2+3+4)(1+2+3+4) \\
& =1 \cdot 1+1 \cdot 2+1 \cdot 3+1 \cdot 4+2 \cdot 1+2 \cdot 2+2 \cdot 3+2 \cdot 4+ \\
& +3 \cdot 1+3 \cdot 2+3 \cdot 3+3 \cdot 4+4 \cdot 1+4 \cdot 2+4 \cdot 3+4 \cdot 4
\end{aligned}
$$
Note que são formados $4 \times 4=16$ produtos ... | [
"Solution:\na) Cada produto que aparece na soma final é uma expressão do tipo $x \\cdot y \\cdot z$ onde $x$ é um número vindo do primeiro parênteses, $y$ é um número vindo do segundo e $z$ um número vindo do terceiro. Como existem 4 opções possíveis para cada um desses números, pelo princípio multiplicativo temos ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a) 64; b) 16; c) 10000 | |
00ey | Prove that the equation
$$
6\left(6 a^{2}+3 b^{2}+c^{2}\right)=5 n^{2}
$$
has no solutions in integers except $a=b=c=n=0$. | [
"We can suppose without loss of generality that $a, b, c, n \\geq 0$. Let $(a, b, c, n)$ be a solution with minimum sum $a+b+c+n$. Suppose, for the sake of contradiction, that $a+b+c+n>0$. Since $6$ divides $5 n^{2}$, $n$ is a multiple of $6$. Let $n=6 n_{0}$. Then the equation reduces to\n$$\n6 a^{2}+3 b^{2}+c^{2}... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1989 | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0fcg | Problem:
Consideremos el número entero positivo
$$
n=2^{r}-16^{s}
$$
donde $r$ y $s$ son también enteros positivos. Hallar las condiciones que deben cumplir $r$ y $s$ para que el resto de la división de $n$ por 7 sea 5. Hallar el menor número que cumple esta condición. | [
"Solution:\n\nLos restos obtenidos al dividir las potencias de $2$ por $7$ son $\\{1,2,4,1,2,4, \\ldots\\}$, repitiéndose con periodo $3$. Estos mismos restos se obtienen al dividir $16$ por $7$. Luego la única posibilidad para obtener resto $5$ al restar es que\n$$\nr=3k+1 \\quad \\text{y} \\quad s=3h+2\n$$\nEstas... | Spain | null | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Conditions: r ≡ 1 (mod 3) and s ≡ 2 (mod 3). The smallest positive n is 768, achieved at r = 10 and s = 2. | |
0cfg | Let $\triangle ABC$ be a triangle. Prove that the orthogonal projections of the point $A$ on the external bisectors of the angles $B$ and $C$, the orthogonal projections of the point $B$ on the external bisectors of the angles $A$ and $C$, and the orthogonal projections of the point $C$ on the external bisectors of the... | [] | Romania | 74th NMO Shortlisted Problems | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof only | null | |
07n1 | $ABCD$ is a rectangle. $E$ is a point on $AB$ between $A$ and $B$, and $F$ is a point on $AD$ between $A$ and $D$. The area of the triangle $EBC$ is $16$, the area of the triangle $EAF$ is $12$ and the area of the triangle $FDC$ is $30$. Find the area of the triangle $EFC$. | [
"First Solution: Let $|EB| = x$ and $|AE| = y$, then $|DC| = x + y$,\n$$\n|BC| = \\frac{32}{x}, \\quad |AF| = \\frac{24}{y}, \\quad \\text{and so } |FD| = \\frac{32}{x} - \\frac{24}{y}.\n$$\n\n\n$$\n30 = \\frac{1}{2} \\left( \\frac{32}{x} - \\frac{24}{y} \\right) (x + y) = 4 + 16 \\frac{y}{... | Ireland | Ireland | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 38 | |
0kdf | Problem:
Suppose $x$, $y$, and $z$ are real numbers that satisfy $x + y + z > 0$, $x y + y z + z x > 0$ and $x y z > 0$. Prove that $x$, $y$, and $z$ must all be positive. | [
"Solution:\n\nNote that $x$, $y$, and $z$ are the roots of the polynomial\n$$\n(t - x)(t - y)(t - z) = t^{3} - (x + y + z) t^{2} + (x y + x z + y z) t - x y z.\n$$\nWe claim that this polynomial has no negative roots. To see this, if we plug in a negative value of $t$, all four terms we are adding are negative, so ... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof only | null | |
07qz | Two circles intersect at $A$ and $B$. A common tangent to the circles touches the circles at $P$ and $Q$. A circle is drawn through $P$, $Q$ and $A$ and the line $AB$ meets this circle again at $C$. Join $CP$ and $CQ$ and extend both to meet the given circles at $F$ and $E$ respectively. Prove $P$, $Q$, $F$ and $E$ lie... | [
"Because $P$, $A$, $Q$, $C$ are on a circle, $\\angle PCA = \\angle PQA$ and $\\angle APC + \\angle CQA = 180^\\circ$. Hence $\\angle EPA + \\angle AQF = 180^\\circ$. Because $A$, $B$, $F$, $Q$ are on a circle, $\\angle AQF + \\angle FBA = 180^\\circ$. Finally, as $F$, $E$, $B$, $A$ are concyclic, $\\angle EPA + \\... | Ireland | Ireland_2017 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0auh | Problem:
Simplify the expression
$$
\left(1+\frac{1}{i}+\frac{1}{i^{2}}+\ldots+\frac{1}{i^{2014}}\right)^{2}.
$$ | [] | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | -1 | |
0f49 | Problem:
What is the smallest value of $4 + x^{2}y^{4} + x^{4}y^{2} - 3x^{2}y^{2}$ for real $x$, $y$? Show that the polynomial cannot be written as a sum of squares. [Note the candidates did not know calculus.] | [] | Soviet Union | 15th ASU | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | 3 | |
0guu | Find all triples $(x, y, z)$ of positive real numbers $x$, $y$, $z$ for which the triples
$$
(23x + 24y + 25z, 23y + 24z + 25x, 23z + 24x + 25y)
$$
and
$$
(x^5 + y^5, y^5 + z^5, z^5 + x^5)
$$
are permutations of each other. | [
"$(x, y, z) = (\\sqrt{6}, \\sqrt{6}, \\sqrt{6})$.\nSince these two triples are permutations of each other, sum of the elements of the first triple must be equal to sum of the elements of the second triple. Therefore we obtain $2(x^5 + y^5 + z^5) = 72(x + y + z)$ and dividing by 2 we get $x^5 + y^5 + z^5 = 36(x + y ... | Turkey | Team Selection Test for JBMO 2024 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | (sqrt(6), sqrt(6), sqrt(6)) | |
08ms | Problem:
Decipher the equality $$(\overline{L A R N}-\overline{A C A}):(\overline{C Y P}+\overline{R U S})=C^{Y^{P}} \cdot R^{U^{S}}$$ where different symbols correspond to different digits and equal symbols correspond to equal digits. It is also supposed that all these digits are different from $0$. | [
"Solution:\n\nDenote $x=\\overline{L A R N}-\\overline{A C A}$, $y=\\overline{C Y P}+\\overline{R U S}$ and $z=C^{Y^{P}} \\cdot R^{U^{S}}$. It is obvious that $1823-898 \\leq x \\leq 9187-121$, $135+246 \\leq y \\leq 975+864$, that is $925 \\leq x \\leq 9075$ and $381 \\leq y \\leq 1839$, whence it follows that $\\... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0h2q | $$
\sqrt{1-a} + \sqrt{1-b} + \sqrt{1-c} \le \sqrt{2}(\sqrt{ab+bc+ca} + 2\sqrt{a^2+b^2+c^2}).
$$
for non-negative $a, b, c$ with $a + b + c = 1$. | [
"From the Cauchy-Schwartz inequality for the collections $(\\sqrt{\\alpha}, \\sqrt{\\beta}, \\sqrt{\\gamma})$ and $(\\sqrt{\\alpha x}, \\sqrt{\\beta y}, \\sqrt{\\gamma z})$ we obtain:\n$$\n(\\alpha\\sqrt{x} + \\beta\\sqrt{y} + \\gamma\\sqrt{z}) \\le \\sqrt{(\\alpha + \\beta + \\gamma)(\\alpha x + \\beta y + \\gamma... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0jru | Problem:
Theseus starts at the point $(0,0)$ in the plane. If Theseus is standing at the point $(x, y)$ in the plane, he can step one unit to the north to point $(x, y+1)$, one unit to the west to point $(x-1, y)$, one unit to the south to point $(x, y-1)$, or one unit to the east to point $(x+1, y)$. After a sequence... | [
"Solution:\n\nThe path Theseus traces out is a closed, non-self-intersecting path in the plane. Since each move is along a line segment, the path forms the boundary of a polygon in the plane. WLOG this polygon is traversed counterclockwise (i.e. with the interior of the polygon always on Theseus's left); the argume... | United States | HMMT Invitational Competition | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
05sz | Problem:
Soit $G$ un graphe fini non orienté, $n \in \mathbb{N}^{*}$, $d \in \mathbb{N}^{*}$ tel que le degré de tout sommet de $G$ est borné par $d$, $x$ un sommet de $G$. On note $a(n, x)$ le nombre de sous-graphes induits connexes de $G$ contenant $n$ sommets dont le sommet $x$ (un sous-graphe induit $U$ est un gra... | [
"Solution:\n\nIci, on est face à un problème de graphe, plus précisément on compte les sous-graphes induits contenant $n$ éléments dont un fixé. On aimerait majorer cela, autant la première question semble pouvoir se faire potentiellement en comptant, autant la seconde n'a pas l'air très interprétable : $\\frac{(d+... | France | Envoi 5: Pot Pourri | [
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
027y | Problem:
Pedrinho escolheu 8 números distintos entre 1 e 11 e os escreveu numa determinada ordem. Joãozinho, vendo os números que Pedrinho escreveu, notou o seguinte fato curioso: se fizermos a média dos $n$ primeiros números escritos por Pedrinho, $n=1, \ldots, 8$, teremos como resultado sempre um número inteiro. Ou ... | [
"Solution:\n\nPrimeiro notamos que a média dos 8 números é simplesmente a soma deles dividida por 8. Como a média é um número inteiro, a soma tem que ser múltipla de 8. Agora, como estamos somando 8 números distintos entre 1 e 11, esta soma tem que ser no mínimo $1+2+\\ldots+8=36$ e no máximo $4+5+\\ldots+11=60$. E... | Brazil | null | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1,3,2,6,8,4,11,5; 3,1,2,6,8,4,11,5; 2,6,1,3,8,4,11,5; 6,2,1,3,8,4,11,5; 9,11,10,6,4,8,1,7; 11,9,10,6,4,8,1,7; 10,6,11,9,4,8,1,7; 6,10,11,9,4,8,1,7 | |
0abw | Let $B$ be an interior point of the segment $AC$. The equilateral triangles $\triangle ABM$ and $\triangle BCN$ are constructed in same half plane determined by the line $AC$. The lines $AN$ and $CM$ intersect in $L$. Find the angle $\angle CLN$. | [
"Notice that the angle $\\angle MBN = 60^\\circ$. Hence $\\angle ABN = \\angle MBC = 120^\\circ$. Because the triangles $ABM$ and $BCN$ are equilateral we have $\\overline{AB} = \\overline{BM}$ and $\\overline{BC} = \\overline{BN}$. Hence $\\triangle ABN \\cong \\triangle MBC$, from where $\\angle ANB = \\angle MCB... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 60° | |
0ipz | Problem:
Let $P$ be a polynomial with $P(1)=P(2)=\cdots=P(2007)=0$ and $P(0)=2009!$. $P(x)$ has leading coefficient $1$ and degree $2008$. Find the largest root of $P(x)$. | [
"Solution:\n$P(0)$ is the constant term of $P(x)$, which is the product of all the roots of the polynomial, because its degree is even. So the product of all $2008$ roots is $2009!$ and the product of the first $2007$ is $2007!$, which means the last root is $\\frac{2009!}{2007!}=2009 \\cdot 2008=4034072$."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 4034072 | |
0l9d | On the sides of triangle $ABC$ take the points $M_1$, $N_1$, $P_1$ such that each line $MM_1$, $NN_1$, $PP_1$ divides the perimeter of $ABC$ into two equal parts ($M$, $N$, $P$ are respectively the midpoints of sides $BC$, $CA$, $AB$).
1/ Prove that the lines $MM_1$, $NN_1$, $PP_1$ are concurrent at a point $K$.
2/ P... | [] | Vietnam | Vietnamese Team Selection Contest for the 44th IMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscell... | English | proof only | null | |
0hcp | The convex quadrilateral $ABCD$ is given.
$\angle BAD = \angle BCD = 120^\circ$, $\angle ACD = 80^\circ$, $BC = CD$, $O$ is the intersection point of its diagonals.
Prove, that $BO^2 = AO \cdot AC$. | [
"Let's draw a circle with the center at the point $C$ and radius $BC = CD$. Points $B$ and $D$ are located on this circle, but $\\angle BCD = 120^\\circ$, so, arc $AD$, that doesn't contain point $A$, is $240^\\circ$. That's why its inscribed angle is $120^\\circ$. So, point $A$ is located on the circle too. Then i... | Ukraine | Ukrainian Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0fxf | Problem:
Seien $x$, $y$, $z$ reelle Zahlen, welche die Gleichung $x+y+z = x y + y z + z x$ erfüllen. Beweise die Ungleichung
$$
\frac{x}{x^{2}+1} + \frac{y}{y^{2}+1} + \frac{z}{z^{2}+1} \geq -\frac{1}{2}
$$ | [
"Solution:\n\nDie Ungleichung ist äquivalent zu\n$$\n\\frac{(x+1)^{2}}{x^{2}+1} + \\frac{(y+1)^{2}}{y^{2}+1} \\geq \\frac{(z-1)^{2}}{z^{2}+1}\n$$\nDie Nebenbedingung ist äquivalent zu $z(x+y-1) = x+y-x y$. Wäre $x+y=1$, dann auch $x y=1$ und $x, y$ sind beide positiv. Daraus folgte aber $1 = x+y \\geq 2 \\sqrt{x y}... | Switzerland | IMO Selektion | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0lgm | Problem:
A circle with radius $r$ is inscribed in the triangle $ABC$. Circles with radii $r_{1}, r_{2}, r_{3}$ ($r_{1}, r_{2}, r_{3} < r$) are inscribed in the angles $A, B, C$ so that each touches the incircle externally. Prove that $r_{1} + r_{2} + r_{3} \geq r$. | [
"Solution:\n\nFirst solution. Let $\\omega$ be the incircle of $\\triangle ABC$, $I$ its center, and $p = (AB + BC + AC)/2$ its semiperimeter. We denote the tangency points of the sides $BC, AC, AB$ with $\\omega$ by $A_{0}, B_{0}, C_{0}$ respectively. Let the circle of radius $r_{1}$ touch $\\omega$ at $A_{1}$.\n\... | Zhautykov Olympiad | Zhautykov Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / Coordi... | null | proof only | null | |
0kyw | Problem:
For each positive integer $n$, let $f(n)$ be either the unique integer $r \in \{0,1, \ldots, n-1\}$ such that $n$ divides $15 r-1$, or $0$ if such $r$ does not exist. Compute
$$
f(16)+f(17)+f(18)+\cdots+f(300) .
$$ | [
"Solution:\nNote we only need to sum $f(n)$ for $n$ relatively prime to $15$. For any such $n > 1$, there exists a positive integer $b$ such that $15 f(n) - 1 = b n$. Since $b n \\leq 15(n-1) - 1 < 15 n$ it follows that $b \\in \\{1, \\ldots, 14\\}$. Moreover, we have $b n \\equiv 1 \\pmod{15}$. These two condition... | United States | HMMT November | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 11856 | |
09yu | A zoo is reconstructing part of their park. In this part, there will be six areas with six species of animals, one in each area. The six species are tigers, lions, elephants, giraffes, zebras and monkeys. The map is as follows:
The tigers and lions cannot be next to each other (this means not ... | [] | Netherlands | Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | null | |
01a4 | A cube $10 \times 10 \times 10$ is split into $1000$ unit cubes. The $5 \times 5 \times 5$ sub-cube in the corner of the big cube is colored black, all other small cubes are white. In one operation we can change colors of each of $10$ cubes, whose centers lay on a line which is parallel to one of edges of the big cube.... | [
"Choose an arbitrary unit black cube and construct its $8$ images under reflections with respect to planes that pass through the center of the big cube parallel to one of the faces of the big cube. Thus we represent the big cube as a union of $125$ sets, each consists of $8$ cubes. It is evident that each of these ... | Baltic Way | Baltic Way 2013 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof only | null | |
0em2 | There are $10$ numbers written around the circumference of a circle. Some of them are positive and others are negative. During one move we are allowed to change the sign of the numbers in three successive positions, each of the three numbers changes sign. Prove that we can make all the numbers positive after a finite n... | [
"Label the numbers $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ in a clockwise direction around the circle. Notice that if we switch a certain value twice, it remains in its original state. If we switch an element an odd number of times we change its sign. To solve this problem it will suffice to show that we can switch one numb... | South Africa | South-Afrika 2011-2013 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0cip | Let $P$ and $Q$ be two matrices in $M_n(\mathbb{C})$. Assume $a \in \mathbb{R} \setminus \{-1, 0, 1\}$, such that $Q^2 = (a-1)QP - (a+1)PQ$. Prove that $(PQ - QP)^n = O_n$. | [] | Romania | 75th NMO | [
"Algebra > Linear Algebra > Matrices"
] | English | proof only | null | |
0j05 | Problem:
Consider the following two-player game. Player 1 starts with a number, $N$. He then subtracts a proper divisor of $N$ from $N$ and gives the result to player 2 (a proper divisor of $N$ is a positive divisor of $N$ that is not equal to 1 or $N$). Player 2 does the same thing with the number she gets from playe... | [
"Solution:\n\nAll even numbers except for odd powers of 2. First we show that if you are stuck with an odd number, then you are guaranteed to lose. Suppose you have an odd number $ab$, where $a$ and $b$ are odd numbers, and you choose to subtract $a$. You pass your opponent the number $a(b-1)$. This cannot be a pow... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All even numbers except for odd powers of 2. | |
0e9v | Problem:
Velikosti notranjih kotov trikotnika sestavljajo aritmetično zaporedje, vsota njihovih sinusov pa je enaka $\frac{3+\sqrt{3}}{2}$. Izračunaj velikosti notranjih kotov trikotnika. | [] | Slovenia | 14. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | 30°, 60°, 90° | |
0iwf | Problem:
How many ways are there to win tic-tac-toe in $\mathbb{R}^{n}$? (That is, how many lines pass through three of the lattice points $(a_{1}, \ldots, a_{n})$ in $\mathbb{R}^{n}$ with each coordinate $a_{i}$ in $\{1,2,3\}$?) Express your answer in terms of $n$. | [
"Solution:\nA line consists of three points. Each coordinate can do one of three things passing from the first point to the last point: increase by $1$ each time, stay the same, or decrease by $1$ each time. There are three ways to stay the same (three coordinates), one way to increase by $1$, and one way to decrea... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | (5^n - 3^n)/2 | |
0exn | Problem:
The $n^2$ numbers $\mathbf{x}_{ij}$ satisfy the $n^3$ equations: $\mathbf{x}_{ij} + \mathbf{x}_{jk} + \mathbf{x}_{ki} = 0$. Prove that we can find numbers $\mathbf{a}_1, \ldots, \mathbf{a}_n$ such that $\mathbf{x}_{ij} = \mathbf{a}_i - \mathbf{a}_j$. | [
"Solution:\n\nTaking $i = j = k$, we have that $\\mathbf{x}_{ii} = 0$.\n\nNow taking $j = k$, we have that $\\mathbf{x}_{ij} = -\\mathbf{x}_{ji}$.\n\nDefine $\\mathbf{a}_i = \\mathbf{x}_{i1}$.\n\nThen we have $\\mathbf{x}_{i1} + \\mathbf{x}_{ij} + \\mathbf{x}_{ji} = 0$.\n\nHence $\\mathbf{x}_{ij} = \\mathbf{a}_i - ... | Soviet Union | 5th ASU | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
0518 | A hiking club wants to hike around a lake along an exactly circular route. On the shoreline they determine two points, which are the most distant from each other, and start to walk along the circle, which has these two points as the endpoints of its diameter. Can they be sure that, independent of the shape of the lake,... | [
"Suppose the shape of the lake is an equilateral triangle. Then the two points which are the most distant from each other are two vertices of the triangle. The circle, which has these two points as the endpoints of its diameter, does not cover the whole triangle, because the distance of the third vertex from the ce... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | No. For a lake shaped like an equilateral triangle, the circle with a side as diameter does not contain the lake, so the circular path would cross the water. | |
063q | Problem:
Gegeben seien positive ganze Zahlen $k$ und $n$ mit $n>k$. Unter einem Binärwort der Länge $n$ verstehen wir eine Folge aus $n$ Folgengliedern, die alle 0 oder 1 sind. Anja wählt unter allen möglichen Binärwörtern der Länge $n$ eines aus. Dann schreibt sie alle Binärwörter der Länge $n$, die sich von ihrem ge... | [
"Solution:\n\nDie Antwort lautet: Falls $n \\neq 2 k$ kann Bernhard das von Anja gewählte Wort stets im ersten Versuch erraten, im Fall $n=2 k$ braucht er zwei Versuche um das Wort mit Sicherheit zu erraten. Wir bezeichnen das von Anja zu Beginn gewählte Wort mit $u$.\n\nFall 1: $n \\neq 2 k$. Es sei $1 \\leq i \\l... | Germany | 1. Auswahlklausur | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 1 if n ≠ 2k; 2 if n = 2k | |
05wf | Problem:
Six paniers à fruits contiennent des poires, des pêches et des pommes. Le nombre de pêches dans chaque panier est égal au nombre total de pommes dans les autres paniers. Le nombre de pommes dans chaque panier est égal au nombre total de poires dans les autres paniers. Montrez que le nombre total de fruits est... | [
"Solution:\n\nNotons $p_{1}, p_{2}, \\ldots, p_{6}$ le nombre de poires dans chaque panier, et $q_{1}, q_{2}, \\ldots, q_{6}$ le nombre de pommes dans chaque panier. Notons également $P$ le nombre total de poires. On a:\n$$\n\\begin{aligned}\n& q_{1} = p_{2} + p_{3} + p_{4} + p_{5} + p_{6} = P - p_{1} \\\\\n& q_{2}... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
06en | $ABCD$ is a cyclic quadrilateral with $BC = CD$. The diagonals $AC$ and $BD$ intersect at $E$. Let $X$, $Y$, $Z$ and $W$ be the incentres of triangles $ riangle ABE$, $ riangle ADE$, $ riangle ABC$ and $ riangle ADC$ respectively. Show that $X$, $Y$, $Z$ and $W$ are concyclic if and only if $AB = AD$. | [
"Since $\\angle BAE = \\angle CAD$ (in view of $BC = CD$) and $\\angle EBA = \\angle DCA$, we have $\\triangle ABE \\sim \\triangle ACD$. Since $X$ and $W$ are incentres of $\\triangle ABE$ and $\\triangle ACD$ respectively, they are corresponding points under this similarity. It follows that\n$$\n\\frac{AX}{AW} = ... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscel... | null | proof only | null | |
01qd | Circles $\Gamma_1$ and $\Gamma_2$ meet at points $X$ and $Y$. A circle $S_1$ touches $\Gamma_1$ internally at $A$ and $\Gamma_2$ externally at $B$. A circle $S_2$ touches $\Gamma_2$ internally at $C$ and $\Gamma_1$ externally at $D$.
Prove that the points $A$, $B$, $C$, $D$ are either collinear or concyclic. | [
"Let $O_1$ be the center of $\\Gamma_1$ and $R_1$ be its radius. Let $O_2$ be the center of $\\Gamma_2$ and $R_2$ be its radius. Homothety with center at $A$ and with a positive coefficient transforms $\\Gamma_1$ to $\\omega_1$. Homothety with center at $B$ and with a negative coefficient transforms $\\omega_1$ to ... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof only | null | |
02y3 | Problem:
No desenho a seguir, as cordas $DE$ e $BC$ são perpendiculares, sendo $BC$ um diâmetro do círculo com centro em $A$. Além disso, $\angle CGF = 40^\circ$ e $GH = 2\ \mathrm{cm}$.
a) Determine o valor do ângulo $\angle CHF$.
b) Encontre o comprimento de $HJ$.
 | [
"Solution:\na) Como $BC$ é um diâmetro, segue que $\\angle BFC = 90^\\circ$. Assim, como também temos $\\angle CHG = 90^\\circ$, a circunferência $\\Gamma$ de diâmetro $CG$ passa por $F$ e $H$. Nessa circunferência, os ângulos $\\angle CGF$ e $\\angle CHF$ estão inscritos no mesmo arco $CF$ e assim $\\angle CHF = \... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | (a) 40 degrees; (b) 2 cm | |
0g2o | Problem:
On définit la suite $\left(a_{n}\right)_{n \geq 0}$ de nombres entiers par $a_{n}:=2^{n}+2^{\lfloor n / 2\rfloor}$. Montrer qu'il existe une infinité de termes de la suite pouvant être exprimés comme la somme d'au moins deux termes distincts de la suite, de même qu'il existe une infinité de termes de la suite ... | [
"Solution:\nOn appelle un nombre entier $n$ exprimable s'il peut être écrit comme une somme de au moins deux $a_{k}$.\nOn calcule tout d'abord la somme de tous les termes $a_{i}$ jusqu'à un certain indice $n$ :\n$$\nS_{n}=\\sum_{i=0}^{n} a_{i}=2^{n+1}+2^{\\left\\lceil\\frac{n+1}{2}\\right\\rceil}+2^{\\left\\lfloor\... | Switzerland | Selektion | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0gng | a. Find all primes $p$ for which $\frac{7^{p-1}-1}{p}$ is a perfect square.
b. Find all primes $p$ for which $\frac{11^{p-1}-1}{p}$ is a perfect square. | [
"Since $(q^{(p-1)/2} - 1, q^{(p-1)/2} + 1) = 2$ for any odd integer $q$, if $p x^2 = q^{p-1} - 1$ for an integer $x$ and an odd prime $p$, then either\n\nCase 1: $q^{(p-1)/2} - 1 = 2p y^2$ and $q^{(p-1)/2} + 1 = 2z^2$\n\nor\n\nCase 2: $q^{(p-1)/2} - 1 = 2y^2$ and $q^{(p-1)/2} + 1 = 2p z^2$\n\nfor some integers $y, ... | Turkey | 16th Turkish Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory... | English | proof and answer | a) p = 3. b) No prime p. | |
0die | Find all non-constant functions $f : \mathbb{R} \to \mathbb{R}$ that satisfy
$$
f(2xy + x) = f(xy + x) + f(x)f(y)
$$
for all $x, y \in \mathbb{R}$. | [
"Let $x = y = 0$ in the problem, we have $f(0) = f(0) + f(0)^2$ so $f(0) = 0$.\n\nContinue to replace $(x, y) = (1, -1)$ into the problem, we have $f(-1) = f(1) \\cdot f(-1)$ deduce $f(1) = 1$ or $f(-1) = 0$. However, if $f(-1) = 0$, substituting $y = -1$ in the problem, then $f(-x) = 0$ for all real numbers $x$, n... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | f(x) = x for all real x | |
0f32 | Problem:
Given a point $P$ in space and 1979 lines $L_1$, $L_2$, ..., $L_{1979}$ containing it. No two lines are perpendicular. $P_1$ is a point on $L_1$. Show that we can find a point $A_n$ on $L_n$ (for $n = 2$, $3$, ..., $1979$) such that the following 1979 pairs of lines are all perpendicular: $A_{n-1}A_{n+1}$ and... | [] | Soviet Union | ASU | [
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
07lo | Let $ABC$ be a planar triangle. Let $\lambda \ge 0$. Define the point $D(\lambda)$ on the side $BC$ so that $D(\lambda)$ divides $BC$ in the ratio $c^\lambda : b^\lambda$. Prove that
$$
|AD(\lambda)| \le |AD(2 - \lambda)| \quad \text{iff} \quad \lambda \ge 1.
$$ | [
"Let $D = D(\\lambda)$ for a fixed $\\lambda$ and denote $\\theta = \\angle ADB = 180^\\circ - \\angle CDA$. The Cosine Rule for the triangles $\\triangle DAB$ and $\\triangle CAD$ gives\n$$\n\\begin{aligned}\nc^2 &= |BD|^2 + |AD|^2 - 2|BD| \\cdot |AD| \\cos \\theta \\\\\nb^2 &= |DC|^2 + |AD|^2 + 2|DC| \\cdot |AD| ... | Ireland | Irska | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
0hgv | Find all natural numbers $n$ that satisfy the inequalities:
$$
-46 \le \frac{2023}{46-n} \le 46-n.
$$ | [
"Let $x = 46 - n$ and find the corresponding integer values of $x$ that satisfy the condition $-46 \\le \\frac{2023}{x} \\le x$. We consider two cases.\n\nCase 1. $x > 0$. The left inequality holds for all the mentioned values of $x$, and the right one can be rewritten as:\n$$\n2023 \\le x^2 \\Rightarrow x \\ge 45 ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 1 and 90 | |
0g9a | 假設 $f(n)$ 是正整數 $1, 2, \dots, n$ 的重新排列 $a_1, a_2, \dots, a_n$ 中, 滿足下列條件的排列數:
(1) $a_1 = 1$;
(2) $|a_i - a_{i-1}| \le 2, i = 1, 2, \dots, n-1$.
求 $f(2015)$ 除以 4 的餘數。
Consider the permutation of $1, 2, \dots, n$, which we denote as $\{a_1, a_2, \dots, a_n\}$.
Let $f(n)$ be the number of these permutations satisfying the ... | [
"解：以下討論 $f(n)$ 的遞迴式。假設 $a_1, a_2, \\dots, a_n$ 滿足題意；基於 $a_1 = 1$，我們有 $a_2 = 2$ 或 $3$。\n(情況一)：$a_2 = 2$. 令 $b_i = a_{i+1} - 1$, 則 $b_1, b_2, \\dots, b_{n-1}$ 滿足題意，故此情況的排列數為 $f(n-1)$.\n(情況二)：$a_2 = 3$ 且 $a_3 = 2$，則必有 $a_4 = 4$。考慮 $c_i = a_{i+3} - 3$，則 $c_1, c_2, \\dots, c_{n-3}$ 滿足提題意，故此情況的排列數為 $f(n-3)$.\n(情況三)：$a_2 ... | Taiwan | 二〇一五數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 3 | |
09t5 | Problem:
Zij $a_{1}, a_{2}, \ldots, a_{n}$ een rijtje reële getallen zodat $a_{1}+\cdots+a_{n}=0$ en definieer $b_{i}=a_{1}+\cdots+a_{i}$ voor $1 \leq i \leq n$. Veronderstel dat $b_{i}\left(a_{j+1}-a_{i+1}\right) \geq 0$ voor alle $1 \leq i \leq j \leq n-1$. Bewijs dat
$$
\max _{1 \leq \ell \leq n}\left|a_{\ell}\righ... | [
"Solution:\n\nOplossing I. Er geldt $b_{n}=0$. Stel dat er een $i \\leq n-1$ is met $b_{i}>0$ en $a_{i+1} \\geq 0$. Dan volgt uit $b_{i}\\left(a_{j+1}-a_{i+1}\\right) \\geq 0$ dat $a_{j+1} \\geq a_{i+1} \\geq 0$ voor alle $i \\leq j \\leq n-1$. Dus $b_{n}=b_{i}+a_{i+1}+a_{i+2}+\\ldots+a_{n} \\geq b_{i}>0$, tegenspr... | Netherlands | IMO-selectietoets III | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0b2h | Problem:
In how many ways can $2021$ be written as a sum of two or more consecutive integers?
(a) $3$
(b) $5$
(c) $7$
(d) $9$ | [] | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | a | |
046r | Let $\triangle ABC$ be an acute triangle, and let $K$ be a point on the extension ray of $BC$ (so that $C$ lies on the segment $BK$). Let $P$ be a point such that $BP = BK$ and $PK \parallel AB$, and let $Q$ be a point such that $CQ = CK$ and $QK \parallel AC$. Assume that the circumcircle of $\triangle PQK$ and the li... | [
"**Proof 1.** (1) Let $A_1$ be the reflection of $A$ across $B$. By the given conditions, $KPA_1A$ forms an isosceles trapezoid, thus $\\angle APB = \\angle A_1KB$.\nSimilarly, let $A_2$ be the reflection of $A$ across $C$. We have $\\angle AQC = \\angle A_2KC = \\angle A_2KB$.\nHence, the desired conclusion is equ... | China | Chinese Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry ... | English | proof only | null | |
0f05 | Problem:
$x_1$, $x_2$, ..., $x_n$ are positive reals with sum $1$. Let $s$ be the largest of $\dfrac{x_1}{1 + x_1}$, $\dfrac{x_2}{1 + x_1 + x_2}$, ..., $\dfrac{x_n}{1 + x_1 + ... + x_n}$. What is the smallest possible value of $s$? What are the corresponding $x_i$? | [] | Soviet Union | ASU | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | Minimum s = 1 - 2^{-1/n}. An extremal choice is x_k = 2^{k/n} - 2^{(k-1)/n} for k = 1, 2, ..., n (equivalently, x_k = (1 - 2^{-1/n}) 2^{k/n}). | |
0ha7 | For positive numbers $x$, $y$, $z$ prove the inequality:
$$
2 \cdot \left( \frac{x}{2x+y} \right)^2 + 2 \cdot \left( \frac{y}{2y+z} \right)^2 + 2 \cdot \left( \frac{z}{2z+x} \right)^2 + \frac{9xyz}{(2x+y)(2y+z)(2z+x)} \le 1.
$$ | [
"We make a substitution: $a = \\frac{2x}{2x+y}$, $b = \\frac{2y}{2y+z}$, $c = \\frac{2z}{2z+x}$, then the inequality will be as follows:\n$$\n2 \\cdot \\left(\\frac{x}{2x+y}\\right)^2 + 2 \\cdot \\left(\\frac{y}{2y+z}\\right)^2 + 2 \\cdot \\left(\\frac{z}{2z+x}\\right)^2 + \\frac{9xyz}{(2x+y)(2y+z)(2z+x)} = \\frac{... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0buk | Problem:
În cubul $ABCD A'B'C'D'$, considerăm $Q$ proiecția lui $D'$ pe $A'C$ și $S$ proiecția lui $D'$ pe $AC$. Arătați că:
a) $A'C \perp (D'QB')$;
b) $QS \parallel (ABC)$. | [
"Solution:\n\na) $D'B' \\perp (ACC') \\Rightarrow D'B' \\perp A'C$. Din $A'C \\perp D'Q$, $A'C \\perp D'B'$, $D'Q \\cap D'B' = \\{D'\\} \\Rightarrow A'C \\perp (D'QB')$.\n\nb) $\\triangle D'A'C \\equiv \\triangle D'C'A \\Rightarrow \\angle D'A'C \\equiv \\angle D'C'A$\n\n$\\triangle D'A'Q \\equiv \\triangle D'C'S \... | Romania | Olimpiada de Matematică - Etapa Locală | [
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof only | null | |
01cg | Let $a$ and $b$ be positive numbers. Find all pairs of functions $f, g: \mathbf{R} \to \mathbf{R}$, each assuming the value $1$ and fulfilling, for any $y \ne 0$ and any $x$, the equations
$$
f\left(\frac{1}{y^2}g(xy) - ax^2\right) = 0 = g\left(\frac{1}{y}f(xy) - bx\right).
$$ | [
"Answer: Either $f(z) = g(z) = \\delta_{z,0}$ or $f(z) = bz$ and $g(z) = az^2$.\n\nPutting $xy = w$ in the second equation gives\n$$\n0 = g\\left(\\frac{1}{y}f(w) - b\\frac{w}{y}\\right) = g\\left(\\frac{1}{y}(f(w) - bw)\\right)\n$$\nfor all $y \\neq 0$. Hence, if $f(w) \\neq bw$ for some $w$, it must be that $g(z)... | Baltic Way | Baltic Way 2015 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | Either f(z) = g(z) = 1 if z = 0 and 0 otherwise, or f(z) = b z and g(z) = a z^2. | |
0icy | Problem:
Write down an integer from $0$ to $20$ inclusive. This problem will be scored as follows: if $N$ is the second-largest number from among the responses submitted, then each team that submits $N$ gets $N$ points, and everyone else gets zero. (If every team picks the same number then nobody gets any points.) | [
"Solution:\nThe only Nash equilibria of this game (where each team plays its best possible move given the other teams' choices) are fairly degenerate: every team but one plays $1$, and the remaining team is more likely to choose $2$ than any higher number. Of course, we cannot assume perfectly rational play in real... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | final answer only | null | |
06lc | There is a table with $n$ rows and 18 columns. Each of its cells contains a 0 or a 1. The table satisfies the following properties:
(i) Every two rows are different.
(ii) Each row contains exactly 6 cells that contain 1.
(iii) For every three rows, there exists a column so that the intersection of the column with the t... | [
"The greatest possible value of $n$ is $\\binom{17}{6} = 12376$.\nWe can identify each row as a 6-element subset of $S_{18} = \\{1, 2, \\dots, 18\\}$, so that the elements correspond to the numbers of columns which contain 1. Then we need to maximize the number of such subsets so that no three of them form a partit... | Hong Kong | CHKMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 12376 | |
03fj | Solve the inequality:
$$
\frac{x^2 - |x - 1| - 4}{x - 4} \geq 2x - 1.
$$ | [
"Note that $x \\neq 4$ and $|x-1| = x-1$ for $x > 1$, otherwise $|x-1| = 1-x$.\n\nIf $x > 1$ we get:\n$$\n\\frac{x^2 - 8x + 7}{x - 4} \\le 0 \\iff \\frac{(x - 7)(x - 1)}{x - 4} \\le 0 \\iff x \\in (4, 7]\n$$\n\nIf $x \\le 1$ we get:\n$$\n\\frac{x^2 - 10x + 9}{x - 4} \\le 0 \\Leftrightarrow \\frac{(x - 9)(x - 1)}{x ... | Bulgaria | Bulgarian Spring Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (-∞, 1] ∪ (4, 7] | |
0cm0 | Problem:
$2n$ distinct tokens are placed at the vertices of a regular $2n$-gon, with one token placed at each vertex. A move consists of choosing an edge of the $2n$-gon and interchanging the two tokens at the endpoints of that edge. Suppose that after a finite number of moves, every pair of tokens have been interchan... | [
"Solution:\n\nStep 1. Enumerate all the tokens in the initial arrangement in clockwise circular order; also enumerate the vertices of the $2n$-gon accordingly. Consider any three tokens $i < j < k$. At each moment, their cyclic order may be either $i, j, k$ or $i, k, j$, counted clockwise. This order changes exactl... | Romanian Master of Mathematics (RMM) | Romanian Master of Mathematics Competition | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0h3g | Find all prime numbers $p$ such that $p^5 - p = 2n!$, where $n$ is a positive integer. | [
"Число $p = 2$ не задовольняє умову, оскільки $p^5 - p = 30 = 2 \\cdot 15$ та $3! < 15 < 4!$ Число $p = 3$ задовольняє умову, оскільки $p^5 - p = 240 = 2 \\cdot 5!$\n\nНехай $p \\ge 5$ та $p^5 - p = 2n!$ для деякого натурального $n$, тоді $n! = (p-1)p(p+1) \\frac{p^2+1}{2}$. Оскільки $p$ просте та $n!$ ділиться без... | Ukraine | Ukrainian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | p = 3 | |
0cin | Let $R > 0$, $n \in \mathbb{N}$, $n \ge 3$, and consider a convex $n$-gon inscribed in a circle of center $O$ and radius $R$. Determine the perimeter of the polygon, knowing that its area is equal to $\frac{nR^2}{2}$.
Aurel Doboşan | [] | Romania | 75th NMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof and answer | 4*sqrt(2)*R | |
044g | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that for any $x, y \in \mathbb{R}$,
$$ f(xf(y) + y^{2021}) = yf(x) + (f(y))^{2021}. $$
(Contributed by Fu Yunhao) | [
"Denote $P(x, y)$ as the governing equation of the problem. From $P(x, 0)$, we get $f(xf(0)) = (f(0))^{2021}$. If $f(0) \\neq 0$, as $x$ is arbitrary, $f$ must be a constant function, say $f(x) = c$. Then $c = yc + c^{2021}$ for $y \\in \\mathbb{R}$, $c = 0$, a contradiction. Hence, $f(0) = 0$.\nIf $f(a) = 0$ for s... | China | China National Team Selection Test | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0 and f(x) = x | |
0lgd | Problem:
An isosceles triangle $A B C$ with $A C = B C$ is given. Point $D$ is chosen on the side $A C$. The circle $S_{1}$ of radius $R$ with the center $O_{1}$ touches the segment $A D$ and the extensions of $B A$ and $B D$ over the points $A$ and $D$, respectively. The circle $S_{2}$ of radius $2 R$ with the center... | [
"Solution:\n\nBy condition, in the triangle $A B C$ we have $\\angle A = \\angle B$. It is evident that $\\angle O_{1} B O_{2} = \\angle B / 2$. Let $\\ell$ be the straight line passing through $O_{2}$ parallel to $A C$. By the problem condition $\\ell$ touches $S_{1}$ (say, at a point $N$). Let also $K$ be the tan... | Zhautykov Olympiad | XV International Zhautykov Olympiad in Mathematics | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08vl | Let $X$ be the product of all those positive integers less than or equal to $2011$ having either $3$ or $7$ for the one's digit. Determine the value of the ten's digit for $X$. | [
"For integers $a$, $b$, $m$, write $a \\equiv b \\pmod{m}$ if $a - b$ is divisible by $m$. For any integer $n$ we have\n$$\n(10n + 3)(10n + 7) = 100n^2 + 100n + 21 \\equiv 21 \\pmod{100}.\n$$\nWe also have\n$$\n21^5 = (20 + 1)^5 = \\sum_{k=0}^{5} \\binom{5}{k} \\cdot 20^k \\equiv \\binom{5}{1} \\cdot 20^1 + \\binom... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 2 | |
0jek | Problem:
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for all real numbers $x, y$,
$$
(x-y)(f(x)-f(y))=f(x-f(y)) f(f(x)-y)
$$ | [
"Solution:\nAnswer: $\\quad f(x)=0, f(x)=x$\n\nFirst, suppose that $f(0) \\neq 0$. Then, by letting $x=y=0$, we get that either $f(f(0))$ or $f(-f(0))$ is zero. The former gives a contradiction by plugging in $x=f(0), y=0$ into the original equation. Thus, $f(-f(0))=0$.\n\nNow plugging in $x=0, y=-f(0)$ gives that ... | United States | HMIC | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0 for all x, or f(x) = x for all x | |
02cr | Problem:
O peso de um número é a soma de seus algarismos. Qual é o menor número que pesa 2000? | [
"Solution:\n\nObserve que os números $189$, $8307$ e $99$ têm todos peso $18$, e que $99$ é o menor número que pesa $18$. Note que: para aumentar o peso de um número e minimizar o número é preciso que o número seja composto do maior número possível de algarismos $9$. Por outro lado, podemos dizer que o $0$ está eli... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 3*10^222 - 1 | |
0agk | A pile of $2010$ coins is given. We take one coin from the pile, and we arbitrarily divide the rest in two piles. Then, we choose an arbitrary pile from the two, we take one coin, and we divide the rest in two arbitrary piles, and so on. Is it possible after a finite number of repetitions of this procedure to get a num... | [
"At the beginning, let the first pile contain $2010$ coins.\nAfter every step, we consider the value $S$, denoting the sum of the number of piles and the number of coins in them.\nWe have\n$$\nS_0 = 1 + 2010 = 2011,\\quad S_1 = 2 + (2010 - 1) = 2011,\\quad S_2 = 3 + (2009 - 1) = 2011\n$$\ni.e. $S$ is invariant (it ... | North Macedonia | Macedonian Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | No | |
0i12 | Problem:
Let $S=\{1,2, \ldots, n\}$, and let $T$ be the set consisting of all nonempty subsets of $S$. The function $f: T \rightarrow S$ is "garish" if there do not exist sets $A, B \in T$ such that $A$ is a proper subset of $B$ and $f(A)=f(B)$. Determine, with proof, how many garish functions exist. | [
"Solution:\n\nThere are $n!$ such functions. If $g$ is any bijective map from $S$ to itself (i.e. a permutation of $S$), then the function $f: T \\rightarrow S$ defined by $f(A)=g(|A|)$ (here $|A|$ is the cardinality of set $A$) is garish. To see this, just note that if $A$ is a proper subset of $B$, then $|A|<|B|$... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | n! | |
01a8 | A game is played on a regular triangle which is split into $n^2$ equal smaller regular triangles by lines that are parallel to one of the sides of the triangle. Denote a "line of triangles" to be all triangles that are placed between two adjacent parallel lines that forms the grid.
In the beginning of the game all tria... | [
"Answer: The smallest possible number of moves is $n$ and the largest possible number of moves is $3n - 2$. If all the moves are done with lines parallel to one side of the triangle, then the game will end after $n$ moves. Let's show that the number of moves cannot be smaller. There will be a move that colors the c... | Baltic Way | Baltic Way 2013 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | smallest n; largest 3n - 2 | |
0fvz | Problem:
Seien $a, b, c, d$ reelle Zahlen. Beweise, dass gilt
$$
\left(a^{2}+b^{2}+1\right)\left(c^{2}+d^{2}+1\right) \geq 2(a+c)(b+d)
$$ | [
"Solution:\n\nWenn wir $a, b, c, d$ durch ihre Beträge ersetzen, bleibt die linke Seite gleich, die rechte Seite wird höchstens grösser. Es genügt also, den Fall $a, b, c, d \\geq 0$ zu betrachten. Nach C.S. gilt\n$$\n\\begin{aligned}\n\\left(a^{2}+b^{2}+\\frac{1}{2}+\\frac{1}{2}\\right)\\left(\\frac{1}{2}+\\frac{1... | Switzerland | SMO Finalrunde | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
0527 | Let $a$ and $n$ be positive integers. Prove that
$$
\lfloor \frac{a}{n} \rfloor + \lfloor \frac{a+1}{n} \rfloor + \dots + \lfloor \frac{a+n-1}{n} \rfloor = a.
$$ | [
"Let $\\lfloor \\frac{a}{n} \\rfloor = q$; then $a = qn + r$, where $0 \\le r < n$. Divide the addends given into two groups: in the first one there are $n-r$ addends, where the numerators of the fractions are equal to numbers from $qn+r$ to $qn+(n-1)$, and in the other one the rest of the $r$ addends, where the nu... | Estonia | Final Round of National Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0jcp | Problem:
How many ways can one color the squares of a $6 \times 6$ grid red and blue such that the number of red squares in each row and column is exactly $2$? | [
"Solution:\n\nAnswer: $67950$\n\nAssume the grid is $n \\times n$. Let $f(n)$ denote the number of ways to color exactly two squares in each row and column red. So $f(1)=0$ and $f(2)=1$. We note that coloring two squares red in each row and column partitions the set $1,2, \\ldots, n$ into cycles such that $i$ is in... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 67950 |
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