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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0cuc | Positive numbers $a_1, a_2, \dots, a_n$ are written on the board in a row.
For every $i = 1, 2, \dots, n$, Vasya wishes to write a number $b_i \ge a_i$ so that for every $i, j \in \{1, 2, \dots, n\}$, at least one of the ratios $b_i/b_j$ and $b_j/b_i$ is an integer. Prove that Vasya can reach this goal so
$$
\text{tha... | [
"Prove that there exists even a collection of such $b_i$'s such that each $b_i/b_j$ is a power of $2$. For this purpose, for each $i$ choose such collection with $b_i = a_i$ and minimal possible $b_j$'s with $j \\neq i$. The product of all $n^2$ numbers in these $n$ collections is at most $(2^{(n-1)/2} a_1 a_2 \\cd... | Russia | XLIII Russian mathematical olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English; Russian | proof only | null | |
01fs | Two points $M$ and $N$ have been selected from the edge $BC$ of a tetrahedron $ABCD$ so that the point $M$ lies between $B$ and $N$, the point $N$ lies between $M$ and $C$, and the angle between the planes $AND$ and $ACD$ is equal to the angle between the planes $AMD$ and $ABD$. Prove that
$$
\frac{MB}{MC} + \frac{NB}{... | [
"We denote the angle between the planes $AND$ and $ACD$ by $\\alpha$, and the angle between the planes $AMD$ and $AND$ by $\\beta$. Furthermore, when $P, Q, R$ and $S$ are four points in space which do not all lie on a plane, we denote the volume of the tetrahedron $PQRS$ by $|PQRS|$.\nThe triangles $\\triangle MBD... | Baltic Way | Baltic Way 2019 | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0efw | Problem:
Poišči vsa praštevila $p, q, r$ in $s$, ki zadoščajo enačbama $p+q=r$ in $q+r=s^{2}$. | [
"Solution:\nIz prve enačbe vidimo, da je $r \\geq 4$, torej mora biti $r$ liho praštevilo. Če je tudi $q$ liho praštevilo, potem iz druge enačbe sledi, da je $s$ sodo praštevilo, torej $s=2$. Tedaj je $s^{2}=4$, hkrati pa je $q+r \\geq 6$, saj sta $q$ in $r$ lihi praštevili. Prišli smo do protislovja, saj enačba $q... | Slovenia | Slovenian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | p=5, q=2, r=7, s=3 | |
04xg | Let $x$, $y$, $z$ be positive real numbers such that $x + y + z \ge 6$. Find the smallest value of the expression
$$
x^2 + y^2 + z^2 + \frac{x}{y^2 + z + 1} + \frac{y}{z^2 + x + 1} + \frac{z}{x^2 + y + 1}.
$$ | [
"Using the AM-GM inequality for positive real numbers $x^2/14$, $x/(y^2+z+1)$ and $2(y^2+z+1)/49$ we have\n$$\n\\frac{x^2}{14} + \\frac{x}{y^2 + z + 1} + \\frac{2}{49}(y^2 + z + 1) \\ge 3\\sqrt[3]{\\frac{x^3}{7^3}} = \\frac{3}{7}x.\n$$\n\nWe can derive cyclically another two similar inequalities. Adding up of all t... | Czech-Polish-Slovak Mathematical Match | Czech-Slovak-Polish Match | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | 90/7 | |
08bk | Problem:
Sia $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ una sequenza di interi positivi tali che $a_{i+1}$ è il numero di divisori positivi di $a_{i}$ per ogni $i \geq 1$. Supponiamo che $a_{2} \neq 2$. Dimostrare che esiste un indice $m$ tale che $a_{m}$ sia un quadrato perfetto. | [
"Solution:\n\nOsserviamo che per ogni $i$ abbiamo $a_{i+1} \\leq a_{i}$ (ogni divisore positivo di $a_{i}$ è minore o uguale ad $a_{i}$), e che $a_{i+1} = a_{i}$ se e solo se $a_{i} \\in \\{1, 2\\}$, perché per $a_{i} \\geq 3$ il numero $a_{i} - 1$ non è un divisore di $a_{i}$.\n\nSe $a_{1} = 1$, allora $a_{1}$ è u... | Italy | Gara di Febbraio | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
07tj | A point $C$ lies on a line segment $AB$ between $A$ and $B$ and circles are drawn having $AC$ and $CB$ as diameters. A common tangent to both circles touches the circle with $AC$ as diameter at $P \neq C$ and the circle with $CB$ as diameter at $Q \neq C$.
Prove that $AP$, $BQ$ and the common tangent to both circles at... | [
"Let $AP$ and $BQ$ intersect at $D$ and let $R$ and $S$ be the centres of the circles. Join $PR$, $QS$, $PC$, $QC$ and $DC$ and note the circles touch at $C$. Let $DC$ meet $PQ$ at $E$. Since $PR$ and $QS$ are both perpendicular to $PQ$, $PR$ is parallel to $QS$ and so $\\angle PRA = \\angle QSC$ which implies that... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07lm | Suppose $r$, $R$ are the in-radius and circum-radius of triangle $ABC$. Show that
$$
9r \le a \sin B + b \sin C + c \sin A \le \frac{9R}{2}.
$$
with equality in both inequalities iff $ABC$ is equilateral. | [
"Let $\\Delta$ stand for the area of $\\triangle ABC$. Consider the LHS. Note that\n$$\n\\begin{align*}\na \\sin B + b \\sin C + c \\sin A &= \\frac{ac \\sin B}{c} + \\frac{ab \\sin C}{a} + \\frac{bc \\sin A}{b} \\\\\n&= 2\\Delta \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) \\\\\n&= \\frac{2\\Delta}{... | Ireland | Irska | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jen... | English | proof only | null | |
0dhx | Let $a$ be the positive root of equation $x + \frac{1}{x} = 675$. Prove that
$$
\varphi(2023(a^{2n} + a^{-2n})) > 2 \cdot \varphi(2022(a^{2n} + a^{-2n}))
$$
for all $n \in \mathbb{Z}^+$. | [
"Notice that $2023 = 7 \\cdot 17^2$ and $2022 = 2 \\cdot 3 \\cdot 337$. Put $u_n = a^{2n} + a^{-2n}$ then it is easy to see $u_{n+1} = u_n^2 - 2$ for every $n \\ge 0$. We have $u_0 = a + \\frac{1}{a} = 675$ divisible by $3$ so $u_1$ divided by $3$ leaves $1$, then $u_2$ divides $3$ with remainder $-1$ and inductive... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0fut | Problem:
Sei $p > 3$ eine Primzahl. Zeige, dass $p^{2}$ ein Teiler ist von
$$
\sum_{k=1}^{p-1} k^{2p+1}
$$ | [
"Solution:\n\nSei $S$ die Summe in der Aufgabenstellung. Wegen $p \\neq 2$ genügt es zu zeigen, dass $2S$ durch $p^{2}$ teilbar ist. Es gilt modulo $p^{2}$\n$$\n\\begin{aligned}\n2S &= \\sum_{k=1}^{p-1} (p-k)^{2p+1} + k^{2p+1} \\\\\n&= \\sum_{k=1}^{p-1} \\left(p^{2p+1} - + \\ldots - \\binom{2p+1}{2} p^{2} k^{2p-1} ... | Switzerland | IMO Selektion | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
06ud | For any finite sets $X$ and $Y$ of positive integers, denote by $f_{X}(k)$ the $k^{ ext{th}}$ smallest positive integer not in $X$, and let
$$
X * Y = X \cup \{ f_{X}(y) : y \in Y \} .
$$
Let $A$ be a set of $a > 0$ positive integers, and let $B$ be a set of $b > 0$ positive integers. Prove that if $A * B = B * A$, the... | [
"For any function $g: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ and any subset $X \\subset \\mathbb{Z}_{>0}$, we define $g(X) = \\{ g(x) : x \\in X \\}$. We have that the image of $f_{X}$ is $f_{X}(\\mathbb{Z}_{>0}) = \\mathbb{Z}_{>0} \\setminus X$. We now show a general lemma about the operation $*$, with th... | IMO | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Functional equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof only | null | |
0esz | We have a deck of 90 cards that are numbered from 10 to 99 (all two-digit numbers). How many sets of three or more different cards in this deck are there such that the number on one of them is the sum of the other numbers, and those other numbers are consecutive? | [
"We are looking at sums of the form\n$$\ns(10 + \\ell, k) = \\sum_{i=0}^{k-1} (10 + \\ell + i) = k\\left(10 + \\ell + \\frac{1}{2}(k-1)\\right),\n$$\nwhere $\\ell \\ge 0$ ($10 + \\ell$ represents the starting number of the consecutive numbers in a set) and $k \\ge 2$ represents the number of consecutive numbers in ... | South Africa | The South African Mathematical Olympiad, Third Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Other"
] | null | proof and answer | 92 | |
0du0 | Let *A*, *B*, *C*, *D* be the 4 squares in the centre of a $4 \times 4$ grid of squares. These 4 squares form the centre of the grid. A frog can jump from one of the 4 squares in the centre to any square in the grid which shares a side. If the frog jumps out of the centre from a particular square for the first time, th... | [
"Let $A, B, C, D$ be labeled in the clockwise manner. Let $a_n, b_n, c_n, d_n$ be respectively the number of ways to start $A, B, C, D$ and exit from $A$ in $n$ jumps. Then $b_n = d_n$. Since from $A$, the frog can return to $A$ in an even number of jumps, $a_n = 0$ when $n$ is even.\n\nFrom $A$, in 1 jump, the fro... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | If n is even: 0. If n = 1: 2. If n is odd and at least 3: 4^{(n-1)/2}. | |
09h7 | Let $\omega$ be circumcircle of triangle $ABC$ and let $AD$ and $BE$ be altitudes. A line $DE$ intersects the circle $\omega$ at points $P$ and $Q$ with order $P$, $E$, $D$, $Q$ in the line. Let bisectors of angle $APQ$ and $BQP$ intersect a circle $\omega$ at point $K$ and $L$, respectively. Prove that the line $KL$ i... | [
"Let $O$ be a circumcenter of triangle $ABC$. We know $\\angle BCO = 90^\\circ - \\angle A$ and $\\angle CDE = \\angle A$ so $\\angle BCO + \\angle CDE = (90^\\circ - \\angle A) + \\angle A = 90^\\circ$, from here $\\overarc{PQ} \\perp CO$.\n\nHence $\\overline{CQ} = \\overline{CP}$ and denote it by $x$. If we deno... | Mongolia | Mongolian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
042j | Show that there are no 2-tuples $(x, y)$ of positive integers satisfying the equation
$$
(x+1)(x+2)\cdots(x+2014) = (y+1)(y+2)\cdots(y+4028).
$$ | [
"**Proof** For $n = 2^k \\cdot m$ ($k$ is a non-negative integer, $m$ is odd), let $v(n) = 2^k$.\nWe prove by contradiction: assume $(x, y)$ is one positive integer solution of the equation. Let\n$$\nv(x + i) = \\max_{1 \\le j \\le 2014} \\{v(x + j)\\},\n$$\nthen if $1 \\le j \\le 2014,\\ j \\ne i$,\n$$\nv(x + j) =... | China | China Team Selection Test | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
05mn | Problem:
Par un point $A$ d'un cercle de centre $O$, on mène une tangente à ce cercle, et on prend deux points $B$ et $C$ sur cette tangente, avec $C$ entre $A$ et $B$. De $B$ et $C$, on mène $(B D)$ et $(C E)$ tangentes au cercle. Démontrer que $\widehat{B O C}=\widehat{D A E}$. | [
"Solution:\n\n\n\nNotons $\\alpha=\\widehat{B D E}$ et $\\beta=\\widehat{C E A}$. D'après le théorème de l'angle inscrit, on a $\\alpha=\\widehat{D A E}$.\n\nComme $(C A)$ et $(C E)$ sont les deux tangentes menées à partir de $C$, on a $\\widehat{E A C}=\\widehat{C E A}=\\beta$.\n\n$\\wideh... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05ue | Problem:
Les suites $\left(a_{n}\right)_{n \geqslant 0}$ et $\left(b_{n}\right)_{n \geqslant 0}$ sont définies par
$$
\left\{\begin{array}{ll}
a_{n}=0 & \text{ si } n=0 ; \\
a_{n}=2 a_{\lfloor n / 2 \rfloor}+n & \text{ si } n \geqslant 1 ;
\end{array}\right. \quad \text{et} \quad \left\{\begin{array}{ll}
b_{n}=0 & \te... | [
"Solution:\n\nOn montre tout d'abord, par récurrence sur $n$, que $a_{n} \\leqslant n\\left\\lfloor\\log _{2}(2 n)\\right\\rfloor$ pour tout entier $n \\geqslant 1$, avec égalité lorsque $n$ est une puissance de 2. En effet, pour $n=1$, on a bien $a_{n}=1=n \\log _{2}(2 n)$. Puis, si $n \\geqslant 2$,\n$$\na_{n}=2 ... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof only | null | |
0h7x | Let $ABC$ be a triangle. Suppose that $AD$ and $BE$ are its angle bisectors. Prove that $\angle ACB = 60^\circ$ if and only if $AE + BD = AB$. | [
"Denote by $I$ the incenter of the triangle $\\triangle ABC$ (fig. 03). Also denote the point $D_1$, which is the symmetric reflection of $D$ with respect to $BE$. Then $DB = BD_1$ and $DI = ID_1$.\n\n\n\nWithout any dependence on the given conditions we have $D_1 \\in AB$, since in $\\tria... | Ukraine | UkraineMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0aqt | Problem:
A "fifty percent mirror" is a mirror that reflects half the light shined on it back and passes the other half of the light onward. Now, two "fifty percent mirrors" are placed side by side in parallel and a light is shined from the left of the two mirrors. How much of the light is reflected back to the left of ... | [] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 2/3 | |
0h7v | For six integers $a, b, c$ and $A, B, C$ the following correlations are true:
$$
b + c = A^2, \quad c + a = B^2, \quad a + b = C^2, \quad C > B > A \geq 0.
$$
Find numbers $a, b, c$, for which the sum $A^2 + B^2 + C^2$ takes the smallest possible value. | [
"Let us solve the given system of equations for numbers $a, b, c$. From the first two equations we obtain: $a - b = B^2 - A^2$. Let us add this to the third equation and find out, that\n$$\na = \\frac{1}{2}(B^2 + C^2 - A^2).\n$$\nSimilarly, or thinking symmetrically, we can find out that\n$$\nb = \\frac{1}{2}(C^2 +... | Ukraine | UkraineMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a = 5, b = 4, c = -4 | |
0cgt | Let $f : [0, 1] \to (0, \infty)$ be a continuous function on $[0, 1]$, and
$$A = \int_{0}^{1} f(t) \, dt.$$
a) Show that the function $F : [0, 1] \to [0, A]$, defined for any $x \in [0, 1]$ by
$$
F(x) = \int_{0}^{x} f(t) \, dt,
$$
is invertible, with a differentiable inverse.
b) Show that the equation
$$
\int_{0}^{x}... | [
"a) Because $f$ is continuous, $F$ is continuous and differentiable, with $F'(x) = f(x) > 0$, for any $x \\in [0, 1]$. Hence, $F$ is strictly increasing, and thus injective. Being continuous, $F$ has the intermediate value property, and since $F(0) = 0$ and $F(1) = A$, $F$ is surjective. Thus, $F$ is bijective, hen... | Romania | 74th Romanian Mathematical Olympiad | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Functions"
] | English | proof and answer | g(x) = F^{-1}(A − F(x)); c = F^{-1}(A/2); g′(c) = −1 | |
0f2t | Problem:
If $P$, $Q$ are points in space the point $[PQ]$ is the point on the line $PQ$ on the opposite side of $Q$ to $P$ and the same distance from $Q$.
$K_0$ is a set of points in space. Given $K_n$ we derive $K_{n+1}$ by adjoining all the points $[PQ]$ with $P$ and $Q$ in $K_n$.
(1) $K_0$ contains just two point... | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Solid Geometry > Volume"
] | null | proof and answer | (1) 7.
(2) sqrt(3)/8 * (3^{2n+1} − 1).
(3) H_1 has 8 faces; Vol(H_n) = 4*3^{3n} − 6*3^{n} + 3. | |
04f8 | Depending on the positive integer $k$, determine the smallest real number $D_k$ such that
$$
(abc)^2 + (bcd)^2 + (cda)^2 + (dab)^2 \le D_k
$$
for all nonnegative real numbers $a, b, c, d$ such that $a^k + b^k + c^k + d^k = 4$. | [
"The quadruple $(a, b, c, d) = (1, 1, 1, 1)$ satisfies the given condition for every positive integer $k$, which means that $D_k \\ge 4$ for every positive integer $k$. We will prove that $D_k = 4$ for every $k \\ge 2$. If $k \\ge 2$, the power mean inequality gives us:\n$$\n\\sqrt{\\frac{a^2 + b^2 + c^2 + d^2}{4}}... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | D_1 = (4/3)^6; D_k = 4 for k >= 2 | |
018b | Let $a$, $b$, $c$, $d$ be nonnegative reals such that $a + b + c + d = 4$. Prove the inequality
$$
\frac{a}{a^3+8} + \frac{b}{b^3+8} + \frac{c}{c^3+8} + \frac{d}{d^3+8} \leq \frac{4}{9}.
$$ | [
"By the means inequality we have $a^3 + 2 = a^3 + 1 + 1 \\ge 3\\sqrt[3]{a^3 \\cdot 1 \\cdot 1} = 3a$. Therefore it is sufficient to prove the inequality\n$$\n\\frac{a}{3a+6} + \\frac{b}{3b+6} + \\frac{c}{3c+6} + \\frac{d}{3d+6} \\le \\frac{4}{9}.\n$$\nWe can write the last inequality in the form\n$$\n\\frac{1}{a+2}... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0fug | Problem:
Die beiden Folgen $a_{1}>a_{2}>\ldots>a_{n}$ und $b_{1}<b_{2}<\ldots<b_{n}$ enthalten zusammen jede der Zahlen $1,2, \ldots, 2 n$ genau einmal. Bestimme den Wert der Summe
$$
\left|a_{1}-b_{1}\right|+\left|a_{2}-b_{2}\right|+\ldots+\left|a_{n}-b_{n}\right|
$$ | [
"Solution:\n\nEs gilt stets $|x-y|=\\max (x, y)-\\min (x, y)$. Wir behaupten, dass für $1 \\leq k \\leq n$ immer $\\max \\left(a_{k}, b_{k}\\right) \\geq n+1$ und $\\min \\left(a_{k}, b_{k}\\right) \\leq n$ ist. Nehme an, es gelte $a_{k}, b_{k} \\leq n$. Dann sind die $n+1$ Folgeglieder $a_{1}, \\ldots, a_{k}, b_{k... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | n^2 | |
0i4z | Problem:
Another professor enters the same room and says, "Each of you has to write down an integer between 0 and 200. I will then compute $X$, the number that is 3 greater than half the average of all the numbers that you will have written down. Each student who writes down the number closest to $X$ (either above or ... | [
"Solution:\n\nUse the same logic to get 7. Note 6 and 8 do not work."
] | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | final answer only | 7 | |
0f1e | Problem:
20 teams each play one game with every other team. Each game results in a win or loss (no draws). $k$ of the teams are European. A separate trophy is awarded for the best European team on the basis of the $k(k-1)/2$ games in which both teams are European. This trophy is won by a single team. The same team com... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 15 (no draws); 18 (draws allowed, 2 points for a win and 1 for a draw) | |
067p | Determine all pairs of nonnegative integers $(m, n)$, $m \ge n$, for which $A = (m+n)^3$ divides $B = 2n(3m^2 + n^2) + 8$. | [
"Let $A = (m+n)^3$, $B = 2n(3m^2+n^2)+8$. Since $(m+n)^3 \\mid 2n(3m^2+n^2)+8$ it follows that:\n$$\n(m+n)^3 \\le 2n(3m^2+n^2)+8 \\Leftrightarrow m^3+3m^2n+3mn^2+n^3 \\le 6m^2n+2n^3+8\n$$\n$$\n\\Leftrightarrow m^3 - 3m^2n + 3mn^2 - n^3 \\le 8 \\Leftrightarrow (m-n)^3 \\le 8 \\Leftrightarrow m-n \\le 2 \\Leftrightar... | Greece | Hellenic Mathematical Olympiad ARCHIMEDES | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | {(1,1), (1,0)} ∪ {(k+2, k) : k ≥ 0} | |
0cw0 | Petya and Vasya play the following game. Petya chooses 100 (not necessarily distinct) nonnegative real numbers $x_1, x_2, \dots, x_{100}$ whose sum equals 1, and tells those numbers to Vasya. Vasya splits the numbers into 50 pairs by his own choice, computes the product of numbers in each pair, and writes down the maxi... | [
"Если Петя выберет числа $\\frac{1}{2}, \\frac{1}{198}, \\frac{1}{198}, \\dots, \\frac{1}{198}$, то, как бы ни разбивал эти числа Вася, в паре с числом $\\frac{1}{2}$ будет число $\\frac{1}{198}$.\nИх произведение будет равно $\\frac{1}{396}$, а остальные будут не больше него. Тогда на доске окажется число $\\frac{... | Russia | Regional round | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | 1/396 | |
0b1q | Problem:
How many infinite arithmetic sequences of positive integers are there which contain the numbers $3$ and $39$? | [
"Solution:\n\nThe common difference should be a positive divisor of $39-3=36=2^{2} \\cdot 3^{2}$, which has $(2+1)(2+1)$ factors. If the common difference is $1$, then any one of $1, 2$, or $3$ may be the first term. If the common difference is $2$, then the first term may be either $1$ or $3$ only. For the other $... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 12 | |
06yi | Let $ABC$ be a triangle with $AB < AC < BC$, and let $D$ be a point in the interior of segment $BC$. Let $E$ be a point on the circumcircle of triangle $ABC$ such that $A$ and $E$ lie on opposite sides of line $BC$ and $\angle BAD = \angle EAC$. Let $I, I_{B}, I_{C}, J_{B}$, and $J_{C}$ be the incentres of triangles $A... | [
"Let $X$ be the intersection of $I_{B}J_{C}$ and $J_{B}I_{C}$. We will prove that, provided that $AB < AC < BC$, the following two conditions are equivalent:\n(1) $AX$ bisects $\\angle BAC$;\n(2) $I_{B}, I_{C}, J_{B}$, and $J_{C}$ are concyclic.\nLet circles $AIB$ and $AIC$ meet $BC$ again at $P$ and $Q$, respectiv... | IMO | IMO2024 Shortlisted Problems | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Spiral simila... | English | proof only | null | |
0czt | Let $ABCD$ be a rectangle of center $O$, such that $\widehat{DAC}=60^{\circ}$. The angle bisector of $\widehat{DAC}$ meets $DC$ at $S$. Lines $OS$ and $AD$ meet at $L$ and lines $BL$ and $AC$ meet at $M$. Prove that lines $SM$ and $CL$ are parallel. | [
"We have $\\widehat{SAC}=\\widehat{SCA}=30^{\\circ}$, so $SA=SC$. Since $OA=OC$, we get $SO \\perp AC$, hence $LA=LC$. Moreover $\\widehat{LAC}=60^{\\circ}$, so triangle $ALC$ is equilateral.\n\n\n\nPoint $S$ is the centroid of $\\triangle LAC$, thus $\\frac{LS}{SO}=2$. $DBCL$ is a parallel... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"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"
] | English | proof only | null | |
06ep | $A$ and $B$ are fixed points on a plane and $L$ is a line passing through $A$ and not $B$. $C$ is a variable point moving from $A$ toward infinity along a half-line of $L$. The incircle of $\triangle ABC$ touches $BC$ at $D$ and $AC$ at $E$. Show that line $DE$ passes through a fixed point. | [
"Let $I$ be the incentre of $\\triangle ABC$, and let $P$ be the intersection point of $AI$ and $DE$. We claim that $\\angle APB = 90^\\circ$. Once this is proved, since the line $AP$ is fixed, the point $P$ is independent of $C$, and so $P$ is the desired fixed point.\n\nNote that $\\angle BAI = \\angle PAE$ and $... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chas... | null | proof only | null | |
0i6e | Problem:
$ABC$ is a triangle with points $E, F$ on sides $AC, AB$, respectively. Suppose that $BE, CF$ intersect at $X$. It is given that $AF / FB = (AE / EC)^2$ and that $X$ is the midpoint of $BE$. Find the ratio $CX / XF$. | [
"Solution:\n\nLet $x = AE / EC$. By Menelaus's theorem applied to triangle $ABE$ and line $CXF$,\n$$\n1 = \\frac{AF}{FB} \\cdot \\frac{BX}{XE} \\cdot \\frac{EC}{CA} = \\frac{x^2}{x+1}.\n$$\nThus, $x^2 = x + 1$, and $x$ must be positive, so $x = (1 + \\sqrt{5}) / 2$.\n\nNow apply Menelaus to triangle $ACF$ and line ... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem"
] | null | proof and answer | sqrt(5) | |
02ld | Problem:
Números decrescentes - Escreva os números abaixo em ordem decrescente
$$
\sqrt[5]{3}, \quad 3^{-2 / 3}, \quad 3^{-2}, \quad\left(\frac{1}{3}\right)^{3}, \quad\left(\frac{1}{3}\right)^{-1}
$$ | [
"Solution:\n\nSabemos que\n- $3^{-2 / 3} = \\frac{1}{3^{2 / 3}} < 1$,\n- $3^{-2} = \\frac{1}{3^{2}} < 1$,\n- $\\left(\\frac{1}{3}\\right)^{3} = \\frac{1}{3^{3}} < 1$,\n- $1 < \\sqrt[5]{3} < 3$.\n\nSe $a, b$ e $c$ são não nulos e\n$$\na > b > c\n$$\nentão\n$$\n\\frac{1}{a} < \\frac{1}{b} < \\frac{1}{c}\n$$\nComo $3^... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | (1/3)^{-1}, √[5]{3}, 3^{-2/3}, 3^{-2}, (1/3)^3 | |
066g | Examine if the polynomial
$$
P(x) = (x^{2} - 2x + 5)(x^{2} - 4x + 20) + 1,
$$
can be written as the product of two polynomials with integer coefficients of degree greater than 0. | [] | Greece | Selection Examination B | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
089n | Problem:
Agnese e Bruno sfidano Viviana e Zenone a biliardino; le squadre sono molto equilibrate, per cui per ogni pallina giocata entrambe le squadre hanno probabilità $1 / 2$ di segnare un gol. Qual è la probabilità che si arrivi a 5 pari?
(A) $\frac{1}{512}$
(B) $\frac{252}{1024}$
(C) $\frac{252}{512}$
(D) $\frac{... | [
"Solution:\n\nLa risposta è (B). Rappresentiamo una partita come una sequenza di 10 lettere (ognuna scelta tra 'A' e 'B'), in modo che la prima lettera sia 'A' se il primo gol è stato segnato dalla prima squadra e 'B' se è stato segnato dalla seconda, similmente la seconda lettera sia 'A' se il secondo gol è stato ... | Italy | Progetto Olimpiadi della Matematica | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | MCQ | B | |
027c | Problem:
a) Determine a soma das raízes reais da equação
$$
x^{2}+18 x+30=2 \cdot \sqrt{x^{2}+18 x+45}
$$
b) Resolva a equação $\sqrt{5-\sqrt{5-x}}=x$, com $0<x<5$. | [
"Solution:\na) Seja $x^{2}+18 x+30=y$. Temos, então\n$$\n\\begin{aligned}\ny & =2 \\cdot \\sqrt{y+15} \\\\\ny^{2} & =(2 \\cdot \\sqrt{y+15})^{2} \\\\\ny^{2} & =4 \\cdot(y+15) \\\\\ny^{2}-4 y-60 & =0\n\\end{aligned}\n$$\nAs raízes da última equação são $y=2 \\pm 8$, isto é, $y_{1}=10$ e $y_{2}=-6$. Como estamos busc... | Brazil | null | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a) -18; b) {(-1 + sqrt(21))/2} | |
0k1q | Problem:
An $n \times m$ maze is an $n \times m$ grid in which each cell is one of two things: a wall, or a blank. A maze is solvable if there exists a sequence of adjacent blank cells from the top left cell to the bottom right cell going through no walls. (In particular, the top left and bottom right cells must both ... | [
"Solution:\n\nReplace $5$ by an arbitrary $n$. Label the cells of the maze by $(x, y)$ where $1 \\leq x \\leq n$ and $1 \\leq y \\leq 2$. Let $a_{n}$ denote the number of solvable $2 \\times n$ mazes, and let $b_{n}$ denote the number of $2 \\times n$ mazes where there exists a sequence of adjacent blank cells from... | United States | HMMT November 2018 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 49 | |
00da | a) Nico debe elegir 10 números enteros positivos (distintos); a continuación, Uriel elige 6 de estos números y los suma. Si el resultado es múltiplo de 6, Uriel gana y si no, pierde. Determinar si Nico puede elegir los 10 números para que a Uriel le sea imposible ganar.
b) Nico debe elegir 11 números enteros positivos... | [
"a) Es suficiente elegir 5 números con resto 0 en la división por 6 y 5 números con resto 1 en la división por 6. Por ejemplo: $A\\{6, 12, 18, 24, 30, 1, 7, 13, 19, 25\\}$.\n\nb) Veremos que no existe un tal conjunto $A$.\n\nAfirmamos que:\n\n1) En cualquier conjunto de 3 o más enteros positivos distintos siempre h... | Argentina | Nacional OMA | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Spanish | proof and answer | a) Yes. Example: {6, 12, 18, 24, 30, 1, 7, 13, 19, 25}.
b) No. For any choice of eleven distinct positive integers, there exist six whose sum is a multiple of six. | |
0a46 | Problem:
Kan een getal van de vorm 44...41, met een oneven aantal vieren gevolgd door een 1, een kwadraat zijn? | [
"Solution:\n\nOplossing I. Nee. We kunnen zo'n getal 44...41 schrijven als $4 \\cdot \\frac{10^{2m}-1}{9} - 3$, met $m \\ge 1$ geheel. Stel dat dit een kwadraat is. Schrijf $4 \\cdot \\frac{10^{2m}-1}{9} - 3 = n^2$ met $n$ positief geheel. Vermenigvuldigen met 9 geeft $4(10^{2m} - 1) - 27 = 9n^2$, dus $4 \\cdot 10^... | Netherlands | IMO-selectietoets II | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0jgh | Problem:
There are 2 runners on the perimeter of a regular hexagon, initially located at adjacent vertices. Every second, each of the runners independently moves either one vertex to the left, with probability $\frac{1}{2}$, or one vertex to the right, also with probability $\frac{1}{2}$. Find the probability that aft... | [
"Solution:\n\nAnswer: $\\quad \\frac{2}{3}+\\frac{1}{3}\\left(\\frac{1}{4}\\right)^{2013}$ OR $\\frac{2^{4027}+1}{3 \\cdot 2^{4026}}$ OR $\\frac{2}{3}+\\frac{1}{3}\\left(\\frac{1}{2}\\right)^{4026}$ OR $\\frac{2}{3}+\\frac{1}{3}\\left(\\frac{1}{64}\\right)^{671}$\n\nLabel the runners $A$ and $B$ and arbitrarily fix... | United States | HMMT November 2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 2/3 + 1/3*(1/4)^2013 | |
0878 | Problem:
Determinare il più grande intero $n$ con questa proprietà: esistono $n$ interi positivi distinti $a_{1}, \ldots, a_{n}$ tali che, comunque se ne scelgano fra essi due distinti, né la loro somma né la loro differenza siano divisibili per $100$.
(A) $49$
(B) $50$
(C) $51$
(D) $99$
(E) $100$ . | [
"Solution:\n\nLa risposta è (C). Un numero è divisibile per $100$ se e solo se la sua espressione decimale termina con $00$.\n\nDividiamo gli interi $a_{1}, \\ldots, a_{n}$ in gruppi come segue: in un primo gruppo mettiamo quelli la cui espressione decimale termina con $00$, in un secondo gruppo quelli che terminan... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | MCQ | C | |
09lq | Let $a$, $b$ and $c$ be integers. If $A = ab + 7b + 49$ and $B = bc + 7c + 49$ are divisible by $61$, then prove that $C = ca + 7a + 49$ is also divisible by $61$. | [
"Let $A = ab + 7b + 49$ and $B = bc + 7c + 49$ be divisible by $61$.\n\nSo,\n$$\nab + 7b + 49 \\equiv 0 \\pmod{61}\n$$\n$$\nbc + 7c + 49 \\equiv 0 \\pmod{61}\n$$\nWe want to show that\n$$\nca + 7a + 49 \\equiv 0 \\pmod{61}\n$$\n\nLet us consider the three expressions:\n\n1. $ab + 7b + 49 \\equiv 0 \\pmod{61}$\n2. $... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof only | null | |
0b71 | A straight line passing through the incenter $I$ of a triangle $ABC$ meets the sides $AB$ and $AC$ at points $P$ and $Q$ respectively. Let $BC = a$, $AC = b$, $AB = c$ and $\frac{PB}{PA} = p$, $\frac{QC}{QA} = q$.
a) Prove that $a(1+p)\vec{IP} = (a-pb)\vec{IB} - cp\vec{IC}$.
b) Prove that $a = bp + cq$.
c) Prove tha... | [
"a) The hypothesis yields $(p+1)\\vec{IP} = \\vec{IB} + p\\vec{IA}$. The conclusion follows now from $a\\vec{IA} + b\\vec{IB} + c\\vec{IC} = \\vec{0}$.\n\nb) Analogously, $a(1+q)\\vec{IQ} = (a-cq)\\vec{IC} - bq\\vec{IB}$. The points $P$, $I$, $Q$ are collinear, hence $(a - pb)(a - cq) = bcpq$, whence the conclusion... | Romania | Romanian Mathematical Olympiad | [
"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 > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geomet... | English | proof only | null | |
0aft | In the set of integers, solve the equation
$$
x^{2010} - 2006 = 4y^{2009} + 4y^{2008} + 2007y.
$$ | [
"**Lemma.** If $x \\in \\mathbb{Z}$, then every prime divisor of $x^2+1$ is of the form $4k+1$.\n\n**Proof of the lemma.** Let $p \\mid x^2+1$. So clearly $\\gcd(x, p)=1$. Then $x^2+1 \\equiv 0 \\pmod{p}$, $x^2 \\equiv -1 \\pmod{p}$. Hence $(x^2)^{\\frac{p-1}{2}} \\equiv (-1)^{\\frac{p-1}{2}} \\pmod{p}$ i.e. $x^{\\... | North Macedonia | Sixteenth Macedonian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Macedonian, English | proof and answer | No integer solutions | |
0ify | Problem:
In how many ways can the set of ordered pairs of integers be colored red and blue such that for all $a$ and $b$, the points $(a, b)$, $(-1-b, a+1)$, and $(1-b, a-1)$ are all the same color? | [
"Solution:\nLet $\\varphi_{1}$ and $\\varphi_{2}$ be $90^{\\circ}$ counterclockwise rotations about $(-1,0)$ and $(1,0)$, respectively. Then $\\varphi_{1}(a, b)=(-1-b, a+1)$, and $\\varphi_{2}(a, b)=(1-b, a-1)$. Therefore, the possible colorings are precisely those preserved under these rotations. Since $\\varphi_{... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 16 | |
0k9v | Problem:
For a positive integer $N$, we color the positive divisors of $N$ (including $1$ and $N$) with four colors. A coloring is called multichromatic if whenever $a$, $b$ and $\operatorname{gcd}(a, b)$ are pairwise distinct divisors of $N$, then they have pairwise distinct colors. What is the maximum possible numbe... | [
"Solution:\n\nFirst, we show that $N$ cannot have three distinct prime divisors. For the sake of contradiction, suppose $p q r \\mid N$ for three distinct primes $p, q, r$. Then by the problem statement, $(p, q, 1), (p, r, 1)$, and $(q, r, 1)$ have three distinct colors, so $(p, q, r, 1)$ has four distinct colors. ... | United States | HMMT February 2019 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 192 | |
0gu7 | Let $(a_n)_{n=1}^{\infty}$ be a strictly increasing sequence of positive real numbers such that the inequality
$$
a_n(a_n - 2a_{n-1}) + a_{n-1}(a_{n-1} - 2a_{n-2}) \ge 0
$$
holds for all $n \ge 3$. Prove that for all $n \ge 2$ the inequality
$$
a_n \ge a_{n-1} + a_{n-2} + \dots + a_1
$$
holds as well. | [
"Rearranging the condition, we obtain $(a_n - a_{n-1})^2 \\ge 2a_{n-1}a_{n-2}$ and since the sequence is increasing we have $a_n \\ge a_{n-1} + \\sqrt{2a_{n-1}a_{n-2}}$. The inequality will be proved by induction over $n$.\n\nFor the base cases $n=2,3$ using the monotonicity we have $a_2 \\ge a_1$ and $a_3 \\ge a_2... | Turkey | Team Selection Test for EGMO 2024 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
04hx | Determine how many 6-digit positive integers are there such that by removing its first two digits and its last two digits we get two 4-digit numbers, which give the same remainder when divided by $99$. | [] | Croatia | Croatia Mathematical Competitions | [
"Number Theory > Other",
"Discrete Mathematics > Other"
] | null | proof and answer | 8190 | |
00k7 | For any positive integer $n$, let $d(n)$ denote the number of divisors of $n$ including $1$ and $n$ itself. For which values of $n$ is $d(t)$ a divisor of $d(n)$ for every divisor $t$ of $n$? | [
"We show that this is the case if and only if $n$ is square-free, i.e. if $n$ contains no prime factor in a power greater than $1$. In order to see this, write\n$$\nn = \\prod_{j=1}^{r} p_j^{e_j}, \\quad \\text{and therefore} \\quad d(n) = \\prod_{j=1}^{r} (e_j + 1).\n$$\nIf all powers $e_j$ are equal to $1$, we ha... | Austria | Austria 2014 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | Exactly the square-free positive integers | |
0i1g | Problem:
How many digits are in the base two representation of $10!$ (factorial)? | [
"Solution:\n\nWe write $10! = 2^{8} \\cdot 3^{4} \\cdot 5^{2} \\cdot 7$. The number of digits (base 2) of $10!$ is equal to $\\left[ \\log_{2} 10! \\right] = 8 + \\log_{2} \\left( 3^{4} \\cdot 5^{2} \\cdot 7 \\right)$. Since $2^{13} < 3^{4} \\cdot 5^{2} \\cdot 7 < 2^{14}$, the number of digits is $8 + 13 = 21$."
] | United States | Harvard-MIT Math Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 21 | |
0d1r | Let $-1 \leq x, y \leq 1$. Prove the inequality
$$
2 \sqrt{(1-x^{2})(1-y^{2})} \leq 2(1-x)(1-y)+1 .
$$ | [
"First solution. Because $-1 \\leq x, y \\leq 1$, we have $(1-x^{2}),(1-y^{2}) \\geq 0$. Therefore, by applying AM-GM, we obtain\n$$\n2 \\sqrt{(1-x^{2})(1-y^{2})} \\leq (1-x^{2})+(1-y^{2}) .\n$$\nSo, it remains to prove that\n$$\n(1-x^{2})+(1-y^{2}) \\leq 2(1-x)(1-y)+1\n$$\nwhich is equivalent to\n$$\nx^{2}+y^{2}-2... | Saudi Arabia | Preselection tests for the full-time training | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0k99 | Problem:
For positive reals $p$ and $q$, define the remainder when $p$ is divided by $q$ as the smallest nonnegative real $r$ such that $\frac{p-r}{q}$ is an integer. For an ordered pair $(a, b)$ of positive integers, let $r_{1}$ and $r_{2}$ be the remainder when $a \sqrt{2}+b \sqrt{3}$ is divided by $\sqrt{2}$ and $\... | [
"Solution:\n\nThe remainder when we divide $a \\sqrt{2}+b \\sqrt{3}$ by $\\sqrt{2}$ is defined to be the smallest non-negative real $r_{1}$ such that $\\frac{a \\sqrt{2}+b \\sqrt{3}-r_{1}}{\\sqrt{2}}$ is integral. As $\\frac{x}{\\sqrt{2}}$ is integral iff $x$ is an integral multiple of $\\sqrt{2}$, it follows that ... | United States | HMMT February 2019 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Other"
] | null | final answer only | 16 | |
00zt | Problem:
In the acute triangle $ABC$, the bisectors of $\angle A$, $\angle B$ and $\angle C$ intersect the circumcircle again in $A_{1}$, $B_{1}$ and $C_{1}$, respectively. Let $M$ be the point of intersection of $AB$ and $B_{1}C_{1}$, and let $N$ be the point of intersection of $BC$ and $A_{1}B_{1}$. Prove that $MN$ ... | [
"Solution:\n\nLet $I$ be the incenter of triangle $ABC$ (the intersection point of the angle bisectors $AA_{1}$, $BB_{1}$ and $CC_{1}$), and let $B_{1}C_{1}$ intersect the side $AC$ and the angle bisector $AA_{1}$ at $P$ and $Q$, respectively (see Figure 9).\n\nThen\n$$\n\\angle AQ C_{1} = \\frac{1}{2}\\left(\\over... | Baltic Way | Baltic Way 1997 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals"
] | null | proof only | null | |
03dk | Let the $\triangle ABC$ with $AB = 1\ \text{cm}$, $BC = 2\ \text{cm}$, and $AC = \sqrt{3}\ \text{cm}$ be given. The points $D, E$, and $F$ lie on the sides $AB, AC$, and $BC$, respectively, satisfying $AE = BD$ and $BF = AD$. The angle bisector of $\triangle BAC$ intersects for the second time the circle through $A, D$... | [
"From the circumscribed circle of $\\triangle ADE$ we derive (using inscribed angles and their corresponding arcs) $MD = ME$ and $\\angle BDM = 180^\\circ - \\angle ADM = \\angle AEM$, which combined with $AE = BD$ gives rise to $\\triangle AEM \\cong \\triangle BDM$ - thus $AM = MB$, meaning that $M$ is the inters... | Bulgaria | Bulgaria 2022 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | (3 - sqrt(3)) / 6 | |
028g | Problem:
Tortas da vovó - Sofia foi levar uns docinhos para sua avó; são 7 docinhos de amora, 6 de côco e 3 de chocolate. Durante o caminho, a gulosa Sofia come 2 docinhos. Qual das situações abaixo é possível?
(A) Vovó não recebeu docinhos de chocolate.
(B) Vovó recebeu menos docinhos de côco do que de chocolate.
(C... | [
"Solution:\n\nVamos examinar cada uma das situações propostas. Lembre que no final vovó recebeu $7+6+3-2=14$ docinhos.\n\n(A) Impossível porque ela recebeu no mínimo $3-2=1$ docinho de chocolate.\n\n(B) Impossível porque ela recebeu no mínimo $6-2=4$ docinhos de côco.\n\n(C) Impossível porque $7-2=5>3$.\n\n(D) Poss... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | D | |
0de3 | Problem:
Call a positive integer $x$ to be remote from squares and cubes if each integer $k$ satisfies both $|x - k^2| > 10^6$ and $|x - k^3| > 10^6$. Prove that there exist infinitely many positive integers $n$ such that $2^n$ is remote from squares and cubes. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
02r9 | When two red amoebas join, the result in one blue amoeba; when a red amoeba and a blue amoeba join, they turn into three red amoeba; and when two blue amoeba join, they become four red amoeba. Fernando observes a test tube with initially $201$ blue amoebas and $112$ red amoebas.
a. Is it possible that after some amoeb... | [
"If the number of blue amoebas is $b$ and the number of red amoebas is $r$ then $2b + r$ is invariant: indeed, whenever one blue amoeba appears/disappears, two red amoebas disappear/appear. In the problem, such number is $2 \\cdot 201 + 112 = 514$.\n\na. Since $2 \\cdot 100 + 314 = 514$, it can be possible. Indeed,... | Brazil | Brazilian Math Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) Yes, it is possible. b) No, it is not possible. | |
0h18 | Prove that for any positive real numbers $a$, $b$, $c < 1$ satisfying
$$
2(a+b+c) + 4abc = 3(ab+bc+ca) + 1,
$$
the following inequality holds: $a+b+c \le \frac{3}{4}$. | [
"We can rewrite the given equality as follows:\n$$4abc - 4(ab + bc + ca) + 4(a+b+c) - 4 = -(ab + bc + ca) + 2(a+b+c) - 3,$$\n$$4(a-1)(b-1)(c-1) = -(a-1)(b-1) - (b-1)(c-1) - (c-1)(a-1),$$\n$$4 = \\frac{1}{1-a} + \\frac{1}{1-b} + \\frac{1}{1-c}.$$\nThe function $f(x) = \\frac{1}{1-x}$ is convex on the interval $x < 1... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
01qc | Three cyclists start from town $A$ simultaneously. They move along the closed route consisting of three straight-line segments $AB$, $BC$ and $CA$. The speeds of the first cyclist on these segments are $12$, $10$ and $15$ kilometers per hour, respectively. The speeds of the second cyclist are $15$, $15$ and $10$ (km/h)... | [
"Answer: $90^\\circ$.\nLet $AB = a$, $BC = b$, $CA = c$ (km). We find the time of each cyclist to cover the route (this time is independent of the moving direction). By condition,\n$$\n\\frac{a}{12} + \\frac{b}{10} + \\frac{c}{15} = \\frac{a}{15} + \\frac{b}{15} + \\frac{c}{10} = \\frac{a}{10} + \\frac{b}{20} + \\f... | Belarus | Final Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | 90° | |
0c48 | Problem:
Alina şi Bogdan joacă următorul joc. Ei au o grămadă formată din $330$ de pietricele. Cei doi jucători mută alternativ. La o mutare se iau din grămadă $1$, $n$ sau $m$ pietricele. Câştigă jocul cel care ia ultima pietricică. Înainte de a începe, Alina alege numărul $n$, $(1 < n < 10)$, după care Bogdan alege ... | [
"Solution:\n\nBogdan are strategie câştigătoare. Iată o astfel de strategie (există şi altele):\n\n1. dacă Alina alege un număr $n$ nedivizibil cu $3$, atunci Bogdan alege $m = 2$ (sau un alt număr nedivizibil cu $3$ dacă Alina a ales $n = 2$);\n2. dacă Alina alege $3$ sau $9$, Bogdan alege $5$;\n3. dacă Alina aleg... | Romania | Al patrulea test de selecţie pentru OBMJ | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | With 330 stones, Bogdan has a winning strategy. With 2018 stones, Alina has a winning strategy by choosing n = 2. | |
0i9r | Problem:
Let $B_{k}(n)$ be the largest possible number of elements in a 2-separable $k$-configuration of a set with $2n$ elements ($2 \leq k \leq n$). Find a closed-form expression (i.e. an expression not involving any sums or products with a variable number of terms) for $B_{k}(n)$. | [
"Solution:\nFirst, a lemma: For any $a$ with $0 \\leq a \\leq 2n$, $\\binom{a}{k} + \\binom{2n-a}{k} \\geq 2\\binom{n}{k}$. (By convention, we set $\\binom{a}{k} = 0$ when $a < k$.)\n\nProof: We may assume $a \\geq n$, since otherwise we can replace $a$ with $2n-a$. Now we prove the result by induction on $a$. For ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | B_k(n) = \binom{2n}{k} - 2\binom{n}{k} | |
012j | Problem:
Let $a, b, c, d$ be real numbers such that
$$
\begin{aligned}
a+b+c+d & = -2 \\
ab + ac + ad + bc + bd + cd & = 0
\end{aligned}
$$
Prove that at least one of the numbers $a, b, c, d$ is not greater than $-1$. | [
"Solution:\nWe can assume that $a$ is the least among $a, b, c, d$ (or one of the least, if some of them are equal), there are $n > 0$ negative numbers among $a, b, c, d$, and the sum of the positive ones is $x$.\nThen we obtain\n$$\n-2 = a + b + c + d \\geqslant n a + x.\n$$\nSquaring we get\n$$\n4 = a^{2} + b^{2}... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
03cd | The diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$ intersect at point $O$. The points $A_1$, $B_1$, $C_1$ and $D_1$ from the segments $AO$, $BO$, $CO$ and $DO$, respectively, are such that $AA_1 = CC_1$ and $BB_1 = DD_1$. Let $M$ be the second intersection point of the circumcircles of $\angle AOB$ and $\an... | [
"It follows from the condition of the problem that $\\angle MAC = \\angle MBD$ and $\\angle MCA = \\angle MDB$. Therefore $\\angle MAC \\sim \\angle MBD$. Let $X$ and $Y$ be the midpoints of $AC$ and $BD$, respectively. It follows that $\\angle MXC = \\angle MYD$, which implies that the point $M$ lies on the circum... | Bulgaria | BULGARIAN NATIONAL MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point"
] | English | proof only | null | |
0bqq | Let $G$ be a finite group of order $n$. Prove that every element of $G$ is a perfect square if and only if $n$ is odd. | [] | Romania | 67th NMO Shortlisted Problems | [
"Algebra > Abstract Algebra > Group Theory"
] | English | proof only | null | |
0i5t | Problem:
Find the volume of the three-dimensional solid given by the inequality $\sqrt{x^{2}+y^{2}} + |z| \leq 1$. | [
"Solution:\n\n$2 \\pi / 3$. The solid consists of two cones, one whose base is the circle $x^{2}+y^{2}=1$ in the $xy$-plane and whose vertex is $(0,0,1)$, and the other with the same base but vertex $(0,0,-1)$. Each cone has a base area of $\\pi$ and a height of $1$, for a volume of $\\pi / 3$, so the answer is $2 ... | United States | Harvard-MIT Math Tournament | [
"Geometry > Solid Geometry > Volume"
] | null | final answer only | 2π/3 | |
017a | A function $f$ is defined on the set of positive integers $N$ and takes positive integers as its values. It is known that for all $n \ge 3$
$$
f(n) = \lceil \sqrt{f(n-1)f(n-2)} + n \rceil
$$
Prove that there is an integer $n$ such that $f(n) = n$. | [
"First note that if $f(n-2) > n-2$ and $f(n-1) > n-1$, then\n$$\n(5) \\qquad f(n) \\ge \\left\\lceil \\sqrt{(n-1)(n-2)} + n \\right\\rceil = \\left\\lceil \\sqrt{(n-1)^2} + 1 \\right\\rceil = n.\n$$\nLet's assume that $f(n) \\neq n$ for all $n$. Then $f(1) > 1$ and we have two cases $f(2) > 2$ or $f(2) = 1$.\n**Cas... | Baltic Way | BALTIC WAY | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0h7d | Petryk programmed calculator in such a way that if there is number $x$ on the screen and then the button «=>» is pressed, number $\frac{x-1}{x+1}$ appears on the screen. Petryk pressed the button «=>» 2014 times, and then 2016 appeared on the screen. What number was on the screen at the beginning? The screen can show n... | [
"Let us look how the number changes after the button «=>» is pressed. Let the first number be $x$. Then after the first pressing we have $\\frac{x-1}{x+1}$.\n\nAfter the second we have\n$$\n\\frac{\\frac{x-1}{x+1}-1}{\\frac{x-1}{x+1}+1} = \\frac{x-1-x-1}{x-1+x+1} = -\\frac{1}{x}.\n$$\nAfter the third we have\n$$\n\... | Ukraine | UkraineMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | -1/2016 | |
08th | Let $k$ be an integer greater than or equal to $2$, and let $n_1$, $n_2$, $n_3$ be positive integers, and $a_1$, $a_2$, $a_3$ be integers greater than or equal to $1$ and less than or equal to $k-1$. Define
$$
b_i = a_i \sum_{j=0}^{n_i} k^j \quad (i = 1, 2, 3).
$$
Determine all possible combinations $(n_1, n_2, n_3)$ i... | [
"We may assume without loss of generality that $n_1 \\ge n_2$. Since $0 < a_1 b_2 < k \\cdot k^{n_2+1} = k^{n_2+2}$, we can represent\n$$\na_1 b_2 = \\sum_{j=0}^{n_2+1} d_j k^j\n$$\nby choosing $d_j$ with $0 \\le d_j \\le k-1$ ($j = 0, 1, \\dots, n_2+1$) suitably.\nSet $e_j = d_0 + d_1 + \\dots + d_j$ ($j = 0, 1, \... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (1, 1, 3) | |
08gk | Problem:
Siano $ABC$ un triangolo acutangolo e $D$ il piede della bisettrice uscente da $A$. Siano $E$ ed $F$ rispettivamente le intersezioni di $AC$ con la circonferenza circoscritta ad $ABD$, e di $AB$ con la circonferenza circoscritta ad $ACD$. Sia inoltre $P$ l'intersezione tra $BE$ e $CF$.
a. Mostrare che il tri... | [
"Solution:\n\na.\nPoiché $DBAE$ e $CDFA$ sono inscrivibili, si ha $\\angle BCP = \\angle DCF = \\angle DAF = \\angle DAB$ e $\\angle CBP = \\angle DBE = \\angle DAE = \\angle DAC$. Essendo $AD$ la bisettrice di $\\angle BAC$, si ha $\\angle BAD = \\angle DAC$, quindi $\\angle BCP = \\angle CBP$, il che dimostra che... | Italy | Italian Mathematical Olympiad - February Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0iu1 | Problem:
If $f(x) = \frac{x}{x+1}$, what is $f(f(f(f(2009))))$? | [
"Solution:\n\nWe have $f(x) = \\frac{x}{x+1}$.\n\nFirst, compute $f(f(x))$:\n\n$$\nf(f(x)) = f\\left(\\frac{x}{x+1}\\right) = \\frac{\\frac{x}{x+1}}{\\frac{x}{x+1} + 1} = \\frac{\\frac{x}{x+1}}{\\frac{x + x+1}{x+1}} = \\frac{\\frac{x}{x+1}}{\\frac{2x+1}{x+1}} = \\frac{x}{2x+1}\n$$\n\nNow, compute $f(f(f(x)))$:\n\n$... | United States | Harvard-MIT November Tournament | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 2009/8037 | |
0kok | Problem:
There is a unit circle that starts out painted white. Every second, you choose uniformly at random an arc of arclength $1$ of the circle and paint it a new color. You use a new color each time, and new paint covers up old paint. Let $c_{n}$ be the expected number of colors visible after $n$ seconds. Compute $... | [
"Solution:\n\nA more rigorous way to see this is the two radii created on the most recent turn have a probability $1$ of being exposed; the two radii created last turn each have a probability $1-p$ of being exposed; the two radii created two turns ago each have a probability $(1-p)^{2}$ of being exposed, and so on.... | United States | HMMT November 2022 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 4π | |
0h81 | Let $\triangle ABC$ be an acute angled triangle and $\angle ACB = 60^\circ$. Let $BL$ be a bisector line and $BH$ be an altitude. Let $LD$ be a perpendicular from $L$ on the side $BC$. Determine the angles of $\triangle ABC$ in case if $AB \parallel HD$.
 | [
"Firstly, we consider the case when point $L$ belongs to the segment $AH$ (fig. 21). Since $\\angle LHB = \\angle LDB = 90^\\circ$, then $BLHD$ is cyclic quadrilateral. Thus, $\\angle LBD = \\angle DHC$. On the other hand $\\angle LBD = \\angle LBA$, due to the fact that $BL$ is a bisector and $\\angle DHC = \\angl... | Ukraine | UkraineMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | ∠A = 40°, ∠B = 80°, ∠C = 60° | |
0gbr | 設 $ABCDE$ 為凸五邊形, 其中 $AB = BC = CD$, $∠EAB = ∠BCD$, 且 $∠EDC = ∠CBA$。試證: 過 $E$ 並與 $BC$ 垂直的直線, 和線段 $AC$ 與 $BD$ 共點。 | [
"在證明中, 我們將使用 $∠A, ∠B, ∠C, ∠D, ∠E$ 等來代表五邊形 $ABCDE$ 的諸內角。令線段 $AC$ 與線段 $BD$ 的兩條中垂線交於點 $I$。注意到 $AC$ 的中垂線會過點 $B$, 而 $BD$ 的中垂線會過點 $C$。於是有 $BD \\perp CI$ 及 $AC \\perp BI$。所以, $AC$ 與 $BD$ 的交點為三角形 $BIC$ 的垂心 $H$, 且 $IH \\perp BC$。只要再證明 $E$ 點在直線 $IH$ 上即可, 亦即 $EI \\perp BC$。\n\n\n\n直線 $IB$ 與 $IC$ 分別平分 ... | Taiwan | 二〇一八數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0f6l | Problem:
$x$ is a real number. Define $x_0 = 1 + \sqrt{1 + x}$, $x_1 = 2 + x / x_0$, $x_2 = 2 + x / x_1$, ..., $x_{1985} = 2 + x / x_{1984}$. Find all solutions to $x_{1985} = x$. | [
"Solution:\n\nIf $x = 0$, then $x_{1985} = 2 \\neq x$.\n\nOtherwise we find\n$$\nx_1 = 2 + \\frac{x}{1 + \\sqrt{1 + x}} = 2 + (\\sqrt{1 + x} - 1) = 1 + \\sqrt{1 + x}.\n$$\nHence $x_{1985} = 1 + \\sqrt{1 + x}$.\n\nSo $x - 1 = \\sqrt{1 + x}$.\n\nSquaring, $x = 0$ or $3$.\n\nWe have already ruled out $x = 0$. It is ea... | Soviet Union | 19th ASU | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 3 | |
09o7 | For all real numbers $a, b, c, d$ not exceeding $2$, prove that
$$
\frac{a^3}{b^2+4} + \frac{b^3}{c^2+4} + \frac{c^3}{d^2+4} + \frac{d^3}{a^2+4} \le 4.
$$
(Otgonbayar Uuye) | [
"Without loss of generality we may assume that $a, b, c, d \\ge 0$. By dividing $2$, it is enough to prove the inequality for the numbers $0 \\le a, b, c, d \\le 1$:\n$$\n\\frac{a^3}{b^2+1} + \\frac{b^3}{c^2+1} + \\frac{c^3}{d^2+1} + \\frac{d^3}{a^2+1} \\le 2.\n$$\nNow, suppose that $a$ is the largest among the $a,... | Mongolia | MMO2025 Round 4 | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
08vs | For positive 2-digit integers $x$ and $y$, the ten's digit of $x$ equals the one's digit of $y$ and the ten's digit of $y$ equals the one's digit of $x$. Let $P$ be the product of $x$ and $y$. Suppose $P$ is a 4-digit number, and suppose that the 2-digit number given by the last (i.e., the bottom) 2 digits of $P$ is 23... | [
"By the given assumptions of the problem, we can write\n$$\nx = 10a + b, \\quad y = 10b + a, \\quad P = xy = 100c + (c + 23) = 101c + 23,\n$$\nwhere $a, b, c$ are integers satisfying $1 \\le a, b \\le 9$ and $c$ has 2-digits. We then obtain from\n$$\n101c + 23 = P = (10a + b)(10b + a) = 101ab + 10(a^2 + b^2),\n$$\n... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 3154 | |
0d4j | Let $ABC$ be a triangle with $\angle B \leq \angle C$, $I$ its incenter and $D$ the intersection point of line $AI$ with side $BC$. Let $M$ and $N$ be points on sides $BA$ and $CA$, respectively, such that $BM = BD$ and $CN = CD$. The circumcircle of triangle $CMN$ intersects again line $BC$ at $P$. Prove that quadrila... | [
"By the bisector theorem we have\n$$\n\\frac{MB}{AB} = \\frac{DB}{AB} = \\frac{DC}{AC} = \\frac{NC}{AC}\n$$\nWe deduce that segments $MN$ and $BC$ are parallel.\n\n\n\nBecause $CNMP$ is a cyclic trapezoid, it is isosceles. Hence\n$$\n\\angle DPM = \\angle ACB\n$$\nBecause $BD = BM$ and $I$ ... | Saudi Arabia | SAMC | [
"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, Arabic | proof only | null | |
04bq | An isosceles triangle $ABC$ with $|AB| = |AC|$ is given. Tangents to its circumcircle at points $A$ and $C$ intersect in $D$. If $\angle DBC = 30^\circ$, prove that triangle $ABC$ is equilateral. | [
"Let $BCE$ and $ECF$ be the equilateral triangles such that the points $A$ and $E$ are at the same side of the line $BC$, and $B$ and $F$ at different sides of the line $CE$. We will assume that $E$ and $A$ are different points, otherwise the statement holds.\n\n\n\nSince $\\angle FBC = 30^... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0hrb | Problem:
An acute angle $\angle ABC$ and interior ray $BD$ are given, as shown. Laura is given an infinite ruler which consists of two parallel rays joined on one end by a segment perpendicular to both of them. One may place the infinite ruler onto the diagram so that one of the infinite edges (marked with an arrow) p... | [
"Solution:\n\nPlace the infinite ruler so that one infinite edge aligns with ray $BD$ and the other infinite edge intersects ray $BC$, then trace its outline. Do the same thing twice on the other side of ray $BD$, so that the final edge intersects ray $BA$. These points of intersection give the desired segment $PQ$... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | null | |
0a0s | Two positive integers having difference $20$ are multiplied with each other; then $23$ is added to the result.
a. What is the smallest possible outcome that ends in $23$? *Give this outcome (and the two corresponding integers with difference $20$) and prove that no smaller outcome is possible.*
b. Is it possible that... | [
"a. Suppose the two positive integers are $n-10$ and $n+10$, and hence $n > 10$. Then the product is equal to $(n-10)(n+10) = n^2 - 100$ and we are looking for an $n > 10$ such that $n^2 - 100 + 23$ ends in the digits $23$. But that means we want $n^2$ to end in the digits $00$. In other words, we want $n^2$ to be ... | Netherlands | Dutch Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Other"
] | English | proof and answer | a) Outcome 323 from integers 10 and 30. b) Yes. Integers 29 and 49 give 1444, which is 38 squared. | |
0f3c | Problem:
All two digit numbers from $19$ to $80$ inclusive are written down one after the other as a single number $N = 192021\ldots 7980$. Is $N$ divisible by $1980$? | [
"Solution:\n$1980 = 2^{2} 3^{2} 5 \\cdot 11$. $N$ is obviously divisible by $2^{2}$ and $5$. The digits in odd position are $9 + (0 + 1 + 2 + \\ldots + 9) + (0 + 1 + 2 + \\ldots + 9) + \\ldots + (0 + 1 + 2 + \\ldots + 9) + 0 = 9 + 6 \\cdot 45 = 279$. The digits in even position are $1 + (2 + 2 + \\ldots + 2) + (3 +... | Soviet Union | ASU | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | Yes | |
0jgv | Problem:
Let $a$ and $b$ be real numbers, and let $r$, $s$, and $t$ be the roots of $f(x) = x^{3} + a x^{2} + b x - 1$. Also, $g(x) = x^{3} + m x^{2} + n x + p$ has roots $r^{2}$, $s^{2}$, and $t^{2}$. If $g(-1) = -5$, find the maximum possible value of $b$. | [
"Solution:\n\nBy Vieta's Formulae, $m = -\\left(r^{2} + s^{2} + t^{2}\\right) = -a^{2} + 2b$, $n = r^{2} s^{2} + s^{2} t^{2} + t^{2} r^{2} = b^{2} + 2a$, and $p = -1$.\n\nTherefore,\n$$\ng(-1) = (-1)^{3} + m(-1)^{2} + n(-1) + p = -1 + m - n - 1\n$$\nBut the context gives:\n$$\ng(-1) = -1 - a^{2} + 2b - b^{2} - 2a -... | United States | HMMT | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 1 + sqrt(5) | |
00un | In triangle $ABC$, the incircle touches sides $BC, CA, AB$ at $D, E, F$ respectively. Assume there exists a point $X$ on the line $EF$ such that
$$
\angle XBC = \angle XCB = 45^\circ.
$$
Let $M$ be the midpoint of the arc $BC$ on the circumcircle of $ABC$ not containing $A$.
Prove that $MD$ passes through $E$ or $F$. | [
"We first state a well-known lemma.\n\n*Lemma.* In triangle $ABC$, let $D, E, F$ be the points of tangency of the incircle to the sides $BC, CA, AB$ and let $I$ be the incenter. Then the intersection of $EF$ and $BI$ lies on the circle of diameter $BC$.\n\nReturning to the problem, let $I$ be the incenter. The lemm... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle tr... | null | proof only | null | |
0eae | There are three piles of tokens on the table: the first contains $a$, the second $b$ and the third one $c$ tokens, and $a \ge b \ge c > 0$. Two players $A$ and $B$ take turns moving the tokens around. Player $A$ goes first. For each move, the active player first chooses two piles and then moves at least one token from ... | [
"If $b = c$, then player $B$ has the winning strategy, otherwise player $A$ has the winning strategy.\n\nFirst, assume that $b = c$. Then the two piles with the smallest number of tokens contain the same amount. In this case, player $B$ can ensure that the situation remains like that every time he makes his move, w... | Slovenia | National Math Olympiad in Slovenia | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | If b = c, then player B has a winning strategy; otherwise player A has a winning strategy. | |
0apy | Problem:
Let $ABC$ be an isosceles triangle with $AB = AC$. Let $D$ and $E$ be the feet of the perpendiculars from $B$ and $C$ to $\overline{AC}$ and $\overline{AB}$, respectively. Suppose that $\overline{CE}$ and $\overline{BD}$ intersect at point $H$. If $EH = 1$ and $AD = 4$, find $DE$. | [
"Solution:\n\n$DE = \\frac{8 \\sqrt{17}}{17}$\n\n\n\nFigure 6: Problem 30.6.\n\nRefer to Figure 6. By Pythagorean Theorem, we get $AH = \\sqrt{17}$. Since $\\angle AEH + \\angle ADH = 90^{\\circ} + 90^{\\circ} = 180^{\\circ}$, the quadrilateral $ADHE$ is cyclic. By Ptolemy's Theorem (in a c... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 8 sqrt(17) / 17 | |
0jrk | Problem:
Let $a$, $b$, $c$ be integers. Define $f(x) = a x^{2} + b x + c$. Suppose there exist pairwise distinct integers $u$, $v$, $w$ such that $f(u) = 0$, $f(v) = 0$, and $f(w) = 2$. Find the maximum possible value of the discriminant $b^{2} - 4 a c$ of $f$. | [
"Solution:\nAnswer: $16$\n\nBy the factor theorem, $f(x) = a(x-u)(x-v)$, so the constraints essentially boil down to $2 = f(w) = a(w-u)(w-v)$. (It's not so important that $u \\neq v$; we merely specified it for a shorter problem statement.)\n\nWe want to maximize the discriminant $b^{2} - 4 a c = a^{2}[(u+v)^{2} - ... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 16 | |
075h | Problem:
Define a sequence $\langle f_{0}(x), f_{1}(x), f_{2}(x), \ldots \rangle$ of functions by
$$
f_{0}(x)=1, \quad f_{1}(x)=x, \quad (f_{n}(x))^{2}-1=f_{n+1}(x) f_{n-1}(x), \text{ for } n \geq 1
$$
Prove that each $f_{n}(x)$ is a polynomial with integer coefficients. | [
"Solution:\nObserve that\n$$\nf_{n}^{2}(x)-f_{n-1}(x) f_{n+1}(x)=1=f_{n-1}^{2}(x)-f_{n-2}(x) f_{n}(x)\n$$\nThis gives\n$$\nf_{n}(x)\\left(f_{n}(x)+f_{n-2}(x)\\right)=f_{n-1}\\left(f_{n-1}(x)+f_{n+1}(x)\\right)\n$$\nWe write this as\n$$\n\\frac{f_{n-1}(x)+f_{n+1}(x)}{f_{n}(x)}=\\frac{f_{n-2}(x)+f_{n}(x)}{f_{n-1}(x)}... | India | INMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0eme | From a point $P$, draw tangents $PB$ and $PT$ to a circle. Let $A$ be the point on the circle such that $AB$ is a diameter, and let $H$ be the foot of the perpendicular from $T$ onto $AB$. Prove that $AP$ bisects $TH$. | [
"Construct $P'$ on $AT$ such that $OP' \\perp AB$. Then $OP' \\parallel HT$. Since $BT$ is perpendicular to both $OP$ and $AT$, we have $OP \\parallel AT$, and by the midpoint theorem $OP'$ bisects $AP$.\n\n\n\nTherefore $OAP'P$ is a parallelogram. Hence $OP'$ is bisected by $AP$, and since... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0312 | Problem:
A right-angled trapezoid with area $10$ and altitude $4$ is divided into two circumscribed trapezoids by a line parallel to its bases. Find their inradii. | [
"Solution:\nLet $ABCD$ be a right-angled trapezoid with area $10$ and altitude $AD = 4$. Let the line $MN \\parallel AB$, $M \\in AD$, $N \\in BC$, divide it into two circumscribed trapezoids. Set $AB = a$, $CD = b$ ($a > b$), $MN = c$, $AM = h_1$ and $DM = h_2$. Then $h_1^2 + (a - c)^2 = (a + c - h_1)^2$ and we ge... | Bulgaria | 52. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 4/3, 2/3 | |
0j5e | Problem:
Let $f(x) = x^{2} - r_{2} x + r_{3}$ for all real numbers $x$, where $r_{2}$ and $r_{3}$ are some real numbers. Define a sequence $\{g_{n}\}$ for all nonnegative integers $n$ by $g_{0} = 0$ and $g_{n+1} = f(g_{n})$. Assume that $\{g_{n}\}$ satisfies the following three conditions:
(i) $g_{2i} < g_{2i+1}$ and... | [
"Solution:\n\nAnswer: $2$\n\nConsider the function $f(x) - x$. By the constraints of the problem, $f(x) - x$ must be negative for some $x$, namely, for $x = g_{2i+1}$, $0 \\leq i \\leq 2011$. Since $f(x) - x$ is positive for $x$ of large absolute value, the graph of $f(x) - x$ crosses the $x$-axis twice and $f(x) -... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 2 | |
060c | Problem:
$2n-1$ tours sont placées sur un échiquier de taille $(2n-1) \times (2n-1)$ de sorte que deux tours quelconques ne sont jamais sur la même ligne ou la même colonne. Montrer que tout carré de taille $n \times n$ contient une tour. | [
"Solution:\n\nCommençons par remarquer qu'il y a exactement une tour par ligne et par colonne. En effet, par hypothèse il y a au plus une tour par ligne. Et puisqu'il y a $2n-1$ tours au total, il doit y en avoir exactement une par ligne, et de même pour les colonnes.\n\nQuitte à adapter légèrement le raisonnement,... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - Envoi 5 : Pot Pourri | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0jji | Problem:
Let a sequence $\{a_{n}\}_{n=0}^{\infty}$ be defined by $a_{0}=\sqrt{2}$, $a_{1}=2$, and $a_{n+1}=a_{n} a_{n-1}^{2}$ for $n \geq 1$. The sequence of remainders when $a_{0}, a_{1}, a_{2}, \cdots$ are divided by $2014$ is eventually periodic with some minimal period $p$ (meaning that $a_{m}=a_{m+p}$ for all suf... | [
"Solution:\n\nAnswer: $12$\n\nLet $a_{n}=2^{b_{n}}$, so notice $b_{1}=1$, $b_{2}=2$, and $b_{n+1}=b_{n}+2 b_{n-1}$ for $n \\geq 1$, so by inspection $b_{n}=2^{n-1}$ for all $n$; thus $a_{n}=2^{2^{n-1}}$.\n\n$2014 = 2 \\cdot 19 \\cdot 53$ so we just want to find the lcm of the eventual periods of $2^{n} \\bmod \\ope... | United States | HMMT November 2014 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 12 | |
0aqz | Problem:
Consider addition $\oplus$ and multiplication $\otimes$ modulo 7 of the numbers in $S=\{0,1,2,3,4,5,6\}$. This means that
$$
\begin{aligned}
& m \oplus n=\text{ remainder when } m+n \text{ is divided by } 7 \\
& m \otimes n=\text{ remainder when } m \times n \text{ is divided by } 7 \text{. }
\end{aligned}
$$... | [
"Solution:\n\nFirst, we need to find $\\frac{1}{4}$ and $\\frac{1}{3}$ in $S$ under multiplication modulo 7.\n\n$\\frac{1}{4}$ is the number $x$ such that $4 \\otimes x \\equiv 1 \\pmod{7}$.\nTry $x = 2$: $4 \\times 2 = 8 \\equiv 1 \\pmod{7}$, since $8 - 7 = 1$.\nSo $\\frac{1}{4} = 2$.\n\n$\\frac{1}{3}$ is the numb... | Philippines | 13th Philippine Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | final answer only | 5 | |
0ce0 | Given the trapezoid $ABCD$ with the smaller base $AB$, squares $ADEF$ and $BCGH$ are constructed externally to the trapezoid. Prove that the perpendicular bisector of $AB$ passes through the midpoint of $FH$. | [
"Let $I$ be the midpoint of $AB$ and $M$, $Q$ be points on the base $CD$, such that $DM = QC = AI$.\n\nConstruct squares $IMNP$ and $IQRS$, externally to the triangle $IMQ$.\n\nSince $\\angle PIS + \\angle PIM + \\angle MIQ + \\angle QIS = 360^\\circ$ and $IPUS$ is a parallelogram, we infer that $\\angle PIS + \\an... | Romania | THE 73rd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD - FIRST SELECTION TEST | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
086a | Problem:
Sia $ABCD$ un quadrato di centro $O$. Si costruiscano due triangoli isosceli $BCJ$ e $CDK$, esterni al quadrato, di base $BC$ e $CD$ rispettivamente e congruenti fra loro. Sia poi $M$ il punto medio di $CJ$. Si provi che le rette $OM$ e $BK$ sono perpendicolari. | [] | Italy | XXV OLIMPIADE ITALIANA DI MATEMATICA | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0eat | Miha has 4 boxes. One of the boxes contains only 20 cent coins, another contains only 10 cent coins, one contains only 2 cent coins and another contains only 1 cent coins. Miha can take the coins from 3 different boxes, taking 1 coin from one of the boxes, 2 coins from another and 3 coins from the third box. At most ho... | [
"Miha will get the greatest amount by taking three 20 cents coins, two 10 cents coins and one 2 cents coin. The total in this case equals 82 cents."
] | Slovenia | National Math Olympiad in Slovenia | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | MCQ | C | |
0802 | Problem:
Una delle seguenti persone è "zio del fratello della figlia della nuora del padre di Alberto". Si tratta di:
(A) Alberto stesso
(B) suo padre
(C) suo nonno
(D) suo figlio
(E) suo suocero. | [] | Italy | Italian Mathematical Olympiad - Febbraio Round | [
"Discrete Mathematics > Logic"
] | null | MCQ | A | |
066h | Consider an acute angled triangle $AB\Gamma$, with $AB < A\Gamma$. Let $M$ be the midpoint of the side $B\Gamma$. On the side $AB$ we consider a point $\Delta$ such that, if the segment $\Gamma\Delta$ intersects the median $AM$ at point $E$, then $A\Delta = \Delta E$. Prove that $AB = \Gamma E$. | [
"We extend median $AM$ by $M\\Theta = AM$. Then $AB\\Theta\\Gamma$ is a parallelogram. Hence $AB \\parallel \\Gamma\\Theta$ and $\\hat{A}_1 = \\hat{O}_1$. But from $A\\Delta = \\Delta E$ we get $\\hat{A}_1 = \\hat{E}_1$.\n\nFigure 1\n\nand since $\\hat{E}_1 = \\hat{E}_2$, we find that $\\ha... | Greece | Hellenic Mathematical Olympiad | [
"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 > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordin... | English | proof only | AB = ΓE | |
01bb | Let $p_1, \dots, p_{30}$ be a permutation of numbers $1, 2, \dots, 30$. For how many permutations does the equality $\sum_{k=1}^{30} |p_k - k| = 450$ hold? | [
"Answer: $(15!)^2$.\n\nLet us define pairs $(a_i, b_i)$ such that $\\{a_i, b_i\\} = \\{p_i, i\\}$ and $a_i \\ge b_i$. Then for every $i = 1, \\dots, 30$ we have $|p_i - i| = a_i - b_i$ and\n$$\n\\sum_{i=1}^{30} |p_i - i| = \\sum_{i=1}^{30} (a_i - b_i) = \\sum_{i=1}^{30} a_i - \\sum_{i=1}^{30} b_i.\n$$\nIt is clear ... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | (15!)^2 |
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