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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0807 | Problem:
Quante sono le soluzioni intere positive dell'equazione $x^{x} - 2^{x} - x^{2} = 10$?
(A) Nessuna
(B) una
(C) due
(D) più di due, ma un numero finito
(E) infinite. | [] | Italy | Italy Febbraio Contest | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | MCQ | B | |
0eoi | A *new sequence* is formed by deleting numbers from the sequence $1, 2, 3, 4, \dots, 500$ such that the sum of any two numbers of the *new sequence* is never a multiple of seven.
What is the maximum length of the new sequence? | [
"Group the numbers into $7$ groups according to their remainders upon division by $7$: the groups are\n$$\n\\begin{aligned}\n\\{1, 8, 15, \\dots, 498\\}, & \\{2, 9, 16, \\dots, 499\\}, \\{3, 10, 17, \\dots, 500\\}, \\\\\n\\{4, 11, 18, \\dots, 494\\}, & \\{5, 12, 19, \\dots, 495\\}, \\{6, 13, 20, \\dots, 496\\}, \\{... | South Africa | South African Mathematics Olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 217 | |
04zl | Let $CM$ be the median of a triangle $ABC$. Prove that the product of the circumradius of $ACM$ and the altitude drawn from $M$ in $ACM$ equals the product of the circumradius of $BCM$ and the altitude drawn from $M$ in $BCM$. | [
"Let $\\angle CAB = \\alpha$ and $\\angle CBA = \\beta$ (see Fig. 13), and let $r$ and $s$ be the circumradii of the triangles $ACM$ and $BCM$, respectively. By the sine law in the triangle $ACM$ we obtain $\\frac{|CM|}{\\sin \\alpha} = 2r$, reducing to $r = \\frac{|CM|}{2\\sin \\alpha}$. Analogously, $s = \\frac{|... | Estonia | Selected Problems from the Final Round of National Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
0a5a | Problem:
Let $ABCD$ be a trapezium such that $AB \parallel CD$. Let $E$ be the intersection of diagonals $AC$ and $BD$. Suppose that $AB = BE$ and $AC = DE$. Prove that the internal angle bisector of $\angle BAC$ is perpendicular to $AD$. | [
"Solution:\nFirst note that triangle $ABE$ is isosceles because $AB = BE$.\n\n\n\nLet $x = \\angle DEC$. Angle chasing gives:\n$$\nx = \\angle DEC = \\angle BEA = \\angle EAB = \\angle ACD.\n$$\nTherefore triangle $CDE$ is isosceles. Hence $DE = DC$. Since we are also given $AC = DE$ this i... | New Zealand | NZMO Round One | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0c1w | On the sides $[AB]$ and $[BC]$ of the parallelogram $ABCD$ one draws the equilateral triangles $ABE$ and $BCF$, so that the points $D$ and $E$ are on the same side of the line $AB$, and points $F$ and $D$ are on different sides of the line $BC$. If $E$, $D$ and $F$ are collinear points, prove that $ABCD$ is a rhombus. | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | null | proof only | null | |
064d | Problem:
Die Ecken eines regelmäßigen 100-Ecks $P$ seien so entweder rot oder blau gefärbt, dass jede Farbe mindestens 24-mal vorkommt.
Beweisen Sie, dass es 24 paarweise disjunkte Vierecke $Q_{1}, Q_{2}, \ldots, Q_{24}$ gibt, deren Eckpunkte auch Ecken von $P$ sind, sodass jedes Viereck $Q_{i}$ entweder einen oder dr... | [
"Solution:\n\nWir zeigen per Induktion nach $n$ die folgende Aussage:\nWenn die Ecken eines konvexen $4 n$-Ecks $P$ so rot und blau gefärbt wurden, dass jede Farbe mindestens $(n-1)$-mal vorkommt, finden wir $(n-1)$ paarweise disjunkte Vierecke $Q_{1}, \\ldots, Q_{n-1}$, deren Ecken auch Ecken von $P$ sind.\nFür $n... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0l43 | Problem:
Let
$$
f(x) = ||\cdots||| | x|-1|-2|-3|-\cdots|-10|
$$
Compute $f(1) + f(2) + \cdots + f(54) + f(55)$. | [
"Solution:\nLet $T_{k}$ denote the $k$-th triangular number $1+2+\\cdots+k$.\nFor any integer $i$, the function $g_{i}(x) = |x - i|$ is a piecewise linear function with slopes $\\pm 1$. As $f(x) = g_{10}\\left(\\cdots\\left(g_{1}\\left(g_{0}(x)\\right)\\right) \\cdots\\right)$, it is also piecewise linear with slop... | United States | HMMT November | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 285 | |
0io7 | Problem:
My friend and I are playing a game with the following rules: If one of us says an integer $n$, the opponent then says an integer of their choice between $2 n$ and $3 n$, inclusive. Whoever first says $2007$ or greater loses the game, and their opponent wins. I must begin the game by saying a positive integer ... | [
"Solution:\n\nAnswer: $6$. We assume optimal play and begin working backward. I win if I say any number between $1004$ and $2006$. Thus, by saying such a number, my friend can force a win for himself if I ever say a number between $335$ and $1003$. Then I win if I say any number between $168$ and $334$, because my ... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 6 | |
0k8y | Problem:
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for any real numbers $x$ and $y$,
$$
f(x+y)=\max (f(x), y)+\min (f(y), x) .
$$ | [
"Solution:\nWe claim the only answer is the identity function $f(x) \\equiv x$, which clearly works.\nLet $(x, y)=(a, 0)$ and $(x, y)=(0, a)$ so that\n$$\n\\begin{aligned}\n& f(a)=\\max (f(a), 0)+\\min (f(0), a) \\\\\n& f(a)=\\max (f(0), a)+\\min (f(a), 0) .\n\\end{aligned}\n$$\nSum:\n$$\n2 f(a)=(f(a)+0)+(f(0)+a)\n... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = x | |
0k6g | Problem:
Let $a$, $b$, $c$ be positive integers such that
$$
\frac{a}{77} + \frac{b}{91} + \frac{c}{143} = 1
$$
What is the smallest possible value of $a + b + c$? | [
"Solution:\nWe need $13a + 11b + 7c = 1001$, which implies $13(a + b + c - 77) = 2b + 6c$. Then $2b + 6c$ must be divisible by both $2$ and $13$, so it is minimized at $26$ (e.g. with $b = 10$, $c = 1$). This gives $a + b + c = 79$."
] | United States | HMMT November 2019 | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 79 | |
0i85 | Problem:
Five mathematicians find a bag of 100 gold coins in a room. They agree to split up the coins according to the following plan:
- The oldest person in the room proposes a division of the coins among those present. (No coin may be split.) Then all present, including the proposer, vote on the proposal.
- If at le... | [
"Solution:\n\nThe first (oldest) person will get 98 coins, the second will get no coins, and one coin apiece will go to two of the last three people. If we write the distribution as an ordered 5-tuple, with leftmost being oldest, etc., then the three possible solutions are $(98, 0, 1, 0, 1)$, $(90, 0, 0, 1, 1)$, an... | United States | 5th Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | (98, 0, 1, 0, 1) or (98, 0, 0, 1, 1) or (98, 0, 1, 1, 0) | |
0km6 | Problem:
Let $n$ be the answer to this problem. $a$ and $b$ are positive integers satisfying
$$
\begin{aligned}
& 3 a+5 b \equiv 19 \quad(\bmod n+1) \\
& 4 a+2 b \equiv 25 \quad(\bmod n+1)
\end{aligned}
$$
Find $2 a+6 b$. | [
"Solution:\nLet $m = n + 1$, so that the conditions become\n$$\n\\begin{aligned}\n& 3 a + 5 b \\equiv 19 \\quad (\\bmod m) \\\\\n& 4 a + 2 b \\equiv 25 \\quad (\\bmod m) \\\\\n& 2 a + 6 b \\equiv -1 \\quad (\\bmod m)\n\\end{aligned}\n$$\nWe can subtract (2) from twice (3) to obtain\n$$\n10 b \\equiv -27 \\quad (\\b... | United States | HMMT November 2021 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 96 | |
0a8l | Problem:
Let $a_{1}, a_{2}, \ldots, a_{n}$, and $b_{1}, b_{2}, \ldots, b_{n}$ be real numbers, and let $a_{1}, a_{2}, \ldots, a_{n}$ be all different. Show that if all the products
$$
\left(a_{i}+b_{1}\right)\left(a_{i}+b_{2}\right) \cdots\left(a_{i}+b_{n}\right)
$$
for $i=1,2, \ldots, n$, are equal, then the products
... | [
"Solution:\nLet $P(x)=\\left(x+b_{1}\\right)\\left(x+b_{2}\\right) \\cdots\\left(x+b_{n}\\right)$. Let $P\\left(a_{1}\\right)=P\\left(a_{2}\\right)=\\ldots=P\\left(a_{n}\\right)=d$. Thus $a_{1}, a_{2}, \\ldots, a_{n}$ are the roots of the $n$th degree polynomial equation $P(x)-d=0$. Then $P(x)-d=c\\left(x-a_{1}\\ri... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 16 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
06nb | In a school there are $2023$ students, numbered $1$ to $2023$. The teacher met the students one by one in the order of their numbers, and gave a candy to each student except if that would mean three students whose numbers form an arithmetic sequence all got candies. In this way, the teacher would give a candy to studen... | [
"Answer: $128$\n\nWith some initial trials, we can list the numbers of the students who get candies:\n$1, 2, 4, 5, 10, 11, 13, 14, 28, 29, \\ldots$\n\nThe base $3$ representation of the numbers are\n$1, 2, 11, 12, 101, 102, 111, 112, 1001, 1002, \\ldots$,\nand if we subtract $1$ from each number, they become the fo... | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 128 | |
01tu | Given a polynomial $p(x) = a_{2n+1}x^{2n+1} + a_{2n}x^{2n} + \dots + a_1x + a_0$ of odd degree with positive coefficients $a_0, a_1, \dots, a_{2n}, a_{2n+1}$.
Prove that there exists a permutation (may be, trivial) of coefficients of $p(x)$ such that the polynomial obtained has exactly one real root.
(A. Voidelevich) | [
"$c_0, c_1, \\dots, c_{2n+1}$ of the numbers $a_0, a_1, \\dots, a_{2n+1}$ such that $c_{2n+1} = b_{2n+1}$, $c_{2n-1} = b_{2n-1}$, $\\ldots$, $c_1 = b_{n+1}$ and $c_2 = b_n$, $c_{2n-2} = b_{n-1}$, $\\ldots$, $c_0 = b_0$. For the set of $c_0, c_1, \\dots, c_{2n+1}$ we have\n$$\nc_0 \\le c_1 \\ge c_2 \\le c_3 \\ge \\d... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0ge5 | 設 $ABC$ 為銳角三角形，$\Gamma$ 為其外接圓，$I$ 為其內心。兩圓 $\omega_B, \omega_C$ 分別過點 $B, C$，並相切於點 $I$。設 $\omega_B$ 與 $\Gamma$ 上的 $AB$ 劣弧交於點 $P$，同時 $\omega_B$ 與線段 $AB$ 再交於點 $M$；類似有 $\omega_C$ 與 $\Gamma$ 上的 $AC$ 劣弧交於點 $Q$，同時 $\omega_C$ 與線段 $AC$ 再交於點 $N$。令射線 $PM$ 與 $QN$ 交於點 $X$，且設 $\omega_B$ 過 $B$ 的切線與 $\omega_C$ 過 $C$ 的切線交於點 $Y$。
試證：$A, ... | [
"Let $AI, BI$, and $CI$ meet $\\Gamma$ again at $D, E$, and $F$, respectively. Let $\\ell$ be the common tangent to $\\omega_B$ and $\\omega_C$ at $I$. We always denote by $\\angle(p, q)$ the directed angle from a line $p$ to a line $q$, taken modulo $180^\\circ$.\n\n*Step 1: We show that $Y$ lies on $\\Gamma$.*\nR... | Taiwan | 2021 數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0bca | Let $f : [0, 1] \to \mathbb{R}$ be a function which is continuous on $[0, 1]$ and differentiable in $0$. Consider the function $s : [0, 1] \to \mathbb{R}$,
$$
s(x) = \sup\{c \in [0; x] \mid \int_{0}^{x} f(t) \, dt = x f(c)\}.
$$
a) Prove that the function $f \circ s$ is differentiable in $0$ and $(f \circ s)'(0) = \fra... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Integral Calculus > Applications",
"Calculus > Differential Calculus > Derivatives"
] | null | proof and answer | (f ∘ s)'(0) = (1/2) f'(0); if f'(0) ≠ 0 then s'(0) = 1/2 | |
0bls | Initially, a blackboard has written on it the numbers $11$ and $13$. Each minute an extra number appears on the blackboard, equaling the sum of two numbers already written on the blackboard. Prove that:
a) the number $86$ can not appear on the blackboard;
b) it is possible for $2015$ to appear on the blackboard at so... | [
"a) Each number appearing on the blackboard is of the form $11a + 13b$, with $a, b \\in \\mathbb{N}^*$. If $86$ appears on the board, then there exist $a, b \\in \\mathbb{N}^*$ such that $86 = 11a + 13b$, whence $b \\le 6$. Then $13b \\in \\{13, 26, 39, 52, 65, 78\\}$, therefore $11a = 86 - 13b \\in \\{73, 60, 47, ... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
02eq | Given that $f(x) = \frac{ax+b}{cx+d}$, $f(0) \neq 0$, $f(f(0)) \neq 0$. Let $F(x) = \underbrace{f(\dots(f(x)\dots))}_{n \text{ times}}$. If $F(0) = 0$, show that $F(x) = x$ for all $x$ where the expression is defined. | [
"Put $f^k(x) = \\underbrace{f(\\dots(f(x))\\dots)}_{k \\text{ times}}$. Then we have $f^k(x) = \\frac{a_k x + b_k}{c_k x + d_k}$. From\n$f^{k+1}(x) = f(f^k(x))$ we get:\n$$\n\\begin{align*} \na_{k+1} &= a a_k + b c_k \\\nb_{k+1} &= a b_k + b d_k \\\nc_{k+1} &= c a_k + d c_k \\\nd_{k+1} &= c b_k + d d_k \n\\end{alig... | Brazil | XII OBM | [
"Algebra > Intermediate Algebra > Other"
] | English | proof only | null | |
09fy | Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous, bounded and non-constant function satisfying
$$
f(2x) = 2f(x)^2 - 1
$$
for all $x \in \mathbb{R}$. Show that the equation $f(x) = 0$ has a solution. | [
"By the intermediate value theorem, it suffices to prove that $f$ takes both non-negative and non-positive values.\n\n**Case $f \\ge 0$:** Suppose that $f(x) \\ge 0$ for all $x \\in \\mathbb{R}$. Then we have\n$$\nf(x) = \\sqrt{\\frac{1 + f(2x)}{2}}\n$$\nfor all $x \\in \\mathbb{R}$. Since the function\n$$\n\\phi: ... | Mongolia | 2015 Mongolian IMO Team Selection Tests | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theo... | null | proof only | null | |
0dpf | Acute angled triangle $ABC$ ($AC > CB$) has circumscribed circle with point $N$ in the middle of the arc $ACB$. Let points $A_1$ and $B_1$ be feets of perpendiculars to the line $NC$, drawn from the points $A$ and $B$ respectively (the segment $NC$ lies inside $A_1B_1$). The height $A_1A_2$ of triangle $A_1AC$ and the ... | [] | Silk Road Mathematics Competition | XV Silk Road Mathematics Competition | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates"
] | English | proof only | null | |
00ne | On a blackboard, there are $17$ integers not divisible by $17$. Alice and Bob play a game. Alice starts and they alternately play the following moves:
* Alice chooses a number $a$ on the blackboard and replaces it with $a^2$.
* Bob chooses a number $b$ on the blackboard and replaces it with $b^3$.
Alice wins if the su... | [
"Since both the problem statement and the winning condition are given in terms of divisibility by $17$, it is sufficient to consider the numbers modulo $17$. In the beginning, all the remainders are different from zero and Alice wins if the sum modulo $17$ becomes zero.\n\nThe moves $a \\mapsto a^2$ and $b \\mapsto... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
0jo4 | Problem:
The three sides of a right triangle form a geometric sequence. Determine the ratio of the length of the hypotenuse to the length of the shorter leg. | [
"Solution:\n\nAnswer: $\\frac{1+\\sqrt{5}}{2}$\n\nLet the shorter leg have length $\\ell$, and the common ratio of the geometric sequence be $r>1$. Then the length of the other leg is $\\ell r$, and the length of the hypotenuse is $\\ell r^{2}$. Hence,\n$$\n\\begin{gathered}\n\\ell^{2}+(\\ell r)^{2}=\\left(\\ell r^... | United States | HMMT November 2015 | [
"Geometry > Plane Geometry > Triangles",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | (1+sqrt(5))/2 | |
07jo | Two intelligent players play a game on a $1403 \times 1403$ grid consists of $1403^2$ unit squares, taking turns. On their turn, the first player chooses one of the unchosen cells and draws a unit line segment from the midpoint of its top side to the midpoint of its bottom side. In their turn, the second player chooses... | [
"The first player draws their first vertical line in the red-colored cell of the table shown. Then, they consider the cell above and below it (the blue ones) as a pair, and pairs up the other cells in that column two by two (the green dominoes). The rest of the table, which is a $1403 \\times 1402$ grid, is also pa... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | draw | |
0gto | Prove that for infinitely many positive integers $k$, there are no positive integers $m$ and $n$ satisfying
$$
\frac{n^2 + m^2}{m^4 + n} = k.
$$ | [
"Let $p \\equiv 3 \\pmod 4$ be a prime number. We will show that for $k = p^2$ there are no positive integers $m$ and $n$ satisfying the equation. Suppose that there is a solution. If we rewrite the equation as\n$$\nn^2 - p^2 n + (m^2 - p^2 m^4) = 0\n$$\nwe obtain a quadratic equation for $n$ and the discriminant m... | Turkey | 31st Turkish Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0jmv | Problem:
There is a heads up coin on every integer of the number line. Lucky is initially standing on the zero point of the number line facing in the positive direction. Lucky performs the following procedure: he looks at the coin (or lack thereof) underneath him, and then,
- If the coin is heads up, Lucky flips it t... | [
"Solution:\n\nAnswer: 6098\n\nWe keep track of the following quantities: Let $N$ be the sum of $2^{k}$, where $k$ ranges over all nonnegative integers such that position $-1-k$ on the number line contains a tails-up coin. Let $M$ be the sum of $2^{k}$, where $k$ ranges over all nonnegative integers such that positi... | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 6098 | |
0863 | Problem:
In un giorno di sole una sfera è posata su un terreno orizzontale. In un certo istante l'ombra della sfera raggiunge la distanza di 10 metri dal punto in cui la sfera tocca il terreno. Nello stesso istante un'asta di lunghezza 1 metro posta verticalmente al terreno getta un'ombra lunga 2 metri. Qual è il ragg... | [
"Solution:\n\nLa risposta è $\\mathbf{( C )}$. La lunghezza dell'ombra dell'asta è il doppio della sua altezza quindi $AB = \\frac{1}{2} BD = 5~\\mathrm{m}$. Per il teorema di Pitagora $AD = 5 \\sqrt{5}~\\mathrm{m}$.\n\nI triangoli $ABD$ e $AEC$ sono simili per cui $AC : AD = CE : BD$. Posto $r = CE$ si ha: $(5 - r... | Italy | Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Solid Geometry > 3D Shapes"
] | null | MCQ | C | |
0f5w | Problem:
$A$, $B$, $C$ and $D$ lie on a line in that order. Show that if $X$ does not lie on the line then
$$
|XA| + |XD| + ||AB| - |CD|| > |XB| + |XC|
$$ | [] | Soviet Union | 18th ASU | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
06a6 | (A) Determine the value of the real number $k$ for which the polynomial $P(x) = x^3 - kx + 2$ has $2$ as root. For that value of $k$ write the polynomial $P(x) = x^3 - kx + 2$ as a product of two polynomials with integer coefficients.
(B) The positive real numbers $a, b$ satisfy the equation $2a + b + \frac{4}{ab} = 1... | [
"(A) Number $2$ is a root of polynomial $P(x) = x^3 - kx + 2$, if and only if $P(2) = 0 \\Leftrightarrow 2^3 - 2k + 2 = 0 \\Leftrightarrow k = 5$.\nFor $k = 5$ we get:\n$$\n\\begin{aligned}\nP(x) &= x^3 - 5x + 2 = x^3 - 4x - x + 2 \\\\\n&= x(x - 2)(x + 2) - (x - 2) = (x - 2)(x^2 + 2x - 1).\n\\end{aligned}\n$$\n\n(B... | Greece | 39th Hellenic Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | k = 5; P(x) = (x − 2)(x^2 + 2x − 1); maximal a = 4 | |
06nk | Given $\triangle ABC$, $\angle CAB = 75^\circ$ and $\angle ACB = 45^\circ$. $BC$ is extended to $T$ so that $BC = CT$. Let $M$ be the midpoint of the segment $AT$. Find $\angle BMC$. | [
"As $M$ and $C$ are midpoints of $AT$ and $BT$ respectively, we have $MC // AB$ and $MC = \\frac{1}{2} AB$ by the midpoint theorem. Extend $CM$ to $K$ so that $CM = MK$. Then $AB = CK$, and hence $ABCK$ is a parallelogram. Also, extend $CA$ to $S$ so that $\\angle SKC = 75^\\circ$. Then $\\triangle SKC$ is an isosc... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 45° | |
0c47 | Problem:
Fie $A=\left\{\left.a=q+\frac{1}{q} \right| q \in \mathbb{Q}^{*}, q>0\right\}$,
$$
A+A=\{a+b \mid a, b \in A\}, \quad A \cdot A=\{a \cdot b \mid a, b \in A\}
$$
Arătaţi că:
i) $A+A \neq A \cdot A$;
ii) $(A+A) \cap \mathbb{N}=(A \cdot A) \cap \mathbb{N}$. | [
"Solution:\ni) Fie $a=1+\\frac{1}{1}=2 \\in A, b=2+\\frac{1}{2} \\in A$. Avem\n$$\na+b=2+2+\\frac{1}{2}=\\frac{9}{2} \\in A+A\n$$\nArătăm că $\\frac{9}{2} \\notin A \\cdot A$. Presupunem prin absurd că\n$$\n\\frac{9}{2}=\\left(\\frac{x}{y}+\\frac{y}{x}\\right)\\left(\\frac{z}{t}+\\frac{t}{z}\\right)=\\frac{\\left(x... | Romania | Al doilea test de selectie pentru OBMJ | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
080m | Problem:
Qual è la somma algebrica dei coefficienti del polinomio
$$
\left(x^{21}+4 x^{2}-3\right)^{2001}-\left(x^{21}+4 x^{2}+3\right)^{667}+x^{21}+4 x^{2} ?
$$ | [] | Italy | Progetto Olimpiadi di Matematica 2000 GARA di SECONDO LIVELLO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 5 | |
0ihp | Problem:
Spring is finally here in Cambridge, and it's time to mow our lawn. For the purpose of these problems, our lawn consists of little clumps of grass arranged in an $m \times n$ rectangular grid, that is, with $m$ rows running east-west and $n$ columns running north-south. To be even more explicit, we might say ... | [
"Solution:\n\nThis is attainable if we place one mobot at each clump in the first row, oriented north, and one mobot in the first column of each row except the first, oriented east.\n\nTo show that at most $m+n-1$ mobots can be used, note that each mobot must mow at least one of the $m+n-1$ clumps in union of the f... | United States | Harvard-MIT Mathematics Tournament, Team Round B | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | m + n − 1 | |
0dqo | Determine all finite sets $A$ of non-negative real numbers, containing at least four elements, and such that for all distinct $a, b, c, d \in A$, $ab + cd \in A$. | [
"Let $a_1 > a_2 > a_3 > a_4$ be the four largest elements in $A$. Let $a = a_1a_2 + a_3a_4$, $b = a_1a_3 + a_2a_4$ and $c = a_1a_4 + a_2a_3$. Then $a > b > c$ by the rearrangement inequality. We also know that $a, b, c \\in A$.\n\nIf there is a fifth term $a_5$, then as before, the 4 terms $a_2, a_3, a_4, a_5$ gene... | Singapore | Singapore International Mathematical Olympiad Committee National Team Selection Test | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | All such sets are exactly A = {x, 1, 1/x, 0} for some positive real x ≠ 1 (equivalently x > 1 up to relabeling). | |
09ky | The number $2022$ was initially written on the board. Sophia and Isabella take turns adding numbers to the board. Sophia can add a number that is $2$ less or $2$ more than the last number written, while Isabella can add a number that is $3$ less or $3$ more. Sophia takes the first turn, and they continue alternating th... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | No | |
03jn | Problem:
Let $S=\{a_{1}, a_{2}, \ldots, a_{r}\}$ denote a sequence of integers. For each non-empty subsequence $A$ of $S$, we define $p(A)$ to be the product of all the integers in $A$. Let $m(S)$ be the arithmetic average of $p(A)$ over all non-empty subsets $A$ of $S$. If $m(S)=13$ and if $m\left(S \cup\{a_{r+1}\}\ri... | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | r = 3; a1 = 1, a2 = 1, a3 = 22; a_{r+1} = 7 | |
0jkh | Problem:
There are $n$ girls $G_{1}, \ldots, G_{n}$ and $n$ boys $B_{1}, \ldots, B_{n}$. A pair $(G_{i}, B_{j})$ is called suitable if and only if girl $G_{i}$ is willing to marry boy $B_{j}$. Given that there is exactly one way to pair each girl with a distinct boy that she is willing to marry, what is the maximal po... | [
"Solution:\n\nAnswer: $\\frac{n(n+1)}{2}$\n\nWe represent the problem as a graph with vertices $G_{1}, \\ldots, G_{n}, B_{1}, \\ldots, B_{n}$ such that there is an edge between vertices $G_{i}$ and $B_{j}$ if and only if $(G_{i}, B_{j})$ is suitable, so we want to maximize the number of edges while having a unique ... | United States | HMMT 2014 | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n(n+1)/2 | |
099c | Let $p \geq 5$ be a prime number and $S_n = 1^n + 2^n + \dots + (p - 1)^n$. Show that there exists infinitely many $n \in \mathbb{N}$ such that $p^3 \mid S_n$, $p \mid S_{n-1}$ and $p \mid S_{n-2}$.
(proposed by N. Davii-Od) | [
"For every $m \\in \\mathbb{N}$ we can choose that way:\n$$\nm = p((p-1)m + 3) = (p-1)(pm + 3) + 3\n$$\nTherefore\n$$\n2S_n = \\sum_{k=1}^{p-1} (k^n + (p-k)^n) = \\sum_{k=1}^{p-1} \\sum_{m=0}^{n-1} C_n^m (p-k)^m p^{n-m} \\equiv\n$$\n$$\n\\begin{aligned}\n& \\equiv \\sum_{k=1}^{p-1} (-1)^{n-2} \\cdot C_n^{n-2} \\cdo... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof only | null | |
0jn8 | Problem:
Let $Q$ be a polynomial
$$
Q(x) = a_{0} + a_{1} x + \cdots + a_{n} x^{n}
$$
where $a_{0}, \ldots, a_{n}$ are nonnegative integers. Given that $Q(1) = 4$ and $Q(5) = 152$, find $Q(6)$. | [
"Solution:\nSince each $a_{i}$ is a nonnegative integer, $152 = Q(5) \\equiv a_{0} \\pmod{5}$ and $Q(1) = 4 \\Longrightarrow a_{i} \\leq 4$ for each $i$. Thus, $a_{0} = 2$. Also, since $5^{4} > 152 = Q(5)$, $a_{4}, a_{5}, \\ldots, a_{n} = 0$.\n\nNow we simply need to solve the system of equations\n$$\n\\begin{align... | United States | HMMT February 2015 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | final answer only | 254 | |
0ld3 | Given an acute, non-isosceles triangle $ABC$, and a point $P$ inside the triangle such that $\angle APB = \angle APC = \alpha$ with $\alpha > 80^\circ - \angle BAC$. The circle $(APB)$ intersects the line $AC$ at $E$, the circle $(APC)$ intersects the line $AB$ at $F$. Let $Q$ be the inside point of the triangle $AEF$ ... | [
"a) Since the quadrilaterals $APBE$ and $APCF$ are inscribed, we have\n$$\n\\angle BEC = \\angle BEA = \\angle CFA = \\angle CFB = 180^\\circ - \\alpha.\n$$\nHence, the quadrilateral $BCFE$ is inscribed in a circle. We have two triangles $ABC$ and $AEF$ are similar, so the transformation from $ABC \\to AEF$, mappin... | Vietnam | IMO 2015 Team Selection Tests | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
... | English | proof only | null | |
06ip | One day, one alien has come to the Earth. Every day, each alien does one of four things, each with a probability of one-fourth: (1) destroying himself, (2) splitting into two aliens, (3) splitting into three aliens, and (4) doing nothing. (A newly created alien also does one of these four things every day starting on t... | [
"The answer is $\\sqrt{2}-1$.\nLet $P(n)$ be the probability that there is eventually no alien on the Earth given that there are currently $n$ aliens, where $n$ is a positive integer. Let $P(1) = c$. Since different aliens act independently, we have $P(n) = c^n$.\nBy the given actions, we know that\n$$\nP(1) = \\fr... | Hong Kong | 1997-2023 IMO HK TST | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | sqrt(2) - 1 | |
0fa7 | Problem:
The chords $AB$ and $CD$ of a sphere intersect at $X$. $A$, $C$ and $X$ are equidistant from a point $Y$ on the sphere. Show that $BD$ and $XY$ are perpendicular. | [] | Soviet Union | 25th ASU | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
09wr | Problem:
In scherphoekige driehoek $A B C$ is $I$ het middelpunt van de ingeschreven cirkel en geldt $|A C|+|A I|=|B C|$. Bewijs dat $\angle B A C=2 \angle A B C$.
 | [
"Solution:\n\nDefinieer een punt $D$ op zijde $B C$ zodat $|C D|=|A C|$. Vanwege het gegeven dat $|B C|=|A C|+|A I|$ ligt $D$ dan inwendig op zijde $B C$ en geldt $|B D|=|A I|$. Omdat driehoek $A C D$ nu gelijkbenig is, is bissectrice $C I$ ook middelloodlijn van $A D$, dus zijn $A$ en $D$ elkaars gespiegelde in $C... | Netherlands | IMO-selectietoets I | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordin... | null | proof only | null | |
06jn | Determine all positive integers $n$ for which there exist pairwise distinct positive real numbers $a_1, a_2, \dots, a_n$ satisfying $\{a_i + \frac{(-1)^i}{a_i} \mid 1 \le i \le n\} = \{a_i \mid 1 \le i \le n\}$. | [
"$n$ can be any odd integer greater than $1$.\n\nFor even $n$, suppose $a_j + \\frac{(-1)^j}{a_j} = b_j$ for $1 \\le j \\le n$, where $\\{b_j\\}$ is a permutation of $\\{a_j\\}$. The relation can be rewritten as\n$$\na_j^2 - a_j b_j = (-1)^{j+1}.\n$$\nSumming over all $j$'s, we obtain\n$$\n\\sum_{j=1}^n (a_j^2 - a_... | Hong Kong | Year 2016 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | all odd integers greater than 1 | |
06ok | A diagonal of a regular 2006-gon is called odd if its endpoints divide the boundary into two parts, each composed of an odd number of sides. Sides are also regarded as odd diagonals.
Suppose the 2006-gon has been dissected into triangles by 2003 nonintersecting diagonals. Find the maximum possible number of isosceles t... | [
"Call an isosceles triangle odd if it has two odd sides. Suppose we are given a dissection as in the problem statement. A triangle in the dissection which is odd and isosceles will be called iso-odd for brevity.\n\nLemma. Let $AB$ be one of dissecting diagonals and let $\\mathcal{L}$ be the shorter part of the boun... | IMO | IMO 2006 Shortlisted Problems | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring sch... | English | proof and answer | 1003 | |
0838 | Problem:
Un museo ha la pianta quadrata ed è suddiviso in $n^{2}$ stanze quadrate tutte uguali (con $n>1$). Ogni coppia di stanze adiacenti (cioè con un muro in comune) comunica mediante una porta (come mostrato nell'esempio a fianco per $n=4$). Il guardiano notturno vuole organizzare il suo giro di ispezione in modo d... | [
"Solution:\nSe $n$ è pari si riesce a costruire il percorso per qualsiasi $k$. Infatti si riesce ad esibire un percorso chiuso che passa una volta da ciascuna stanza, e facendo $k$ giri si ottiene il tragitto del guardiano.\nSupponiamo per comodità che il museo sia orientato lungo i punti cardinali. Un modo per cos... | Italy | XIX Gara Nazionale di Matematica | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | If the grid size is even, a route exists for any positive k. If the grid size is odd, a route exists only for k = 1. | |
0i1v | Problem:
Call three sides of an opaque cube adjacent if someone can see them all at once. Draw a plane through the centers of each triple of adjacent sides of a cube with edge length $1$. Find the volume of the closed figure bounded by the resulting planes. | [
"Solution:\n\nThe volume of the figure is half the volume of the cube (which can be seen by cutting the cube into $8$ equal cubes and realizing that the planes cut each of these cubes in half), namely $\\frac{1}{2}$."
] | United States | Harvard-MIT Math Tournament | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | 1/2 | |
0a8j | Problem:
The trapezium $ABCD$, where $AB$ and $CD$ are parallel and $AD < CD$, is inscribed in the circle $c$. Let $DP$ be a chord of the circle, parallel to $AC$. Assume that the tangent to $c$ at $D$ meets the line $AB$ at $E$ and that $PB$ and $DC$ meet at $Q$. Show that $EQ = AC$. | [
"Solution:\n\n(See Figure 13.) Since $AD < CD$, $\\angle PDC = \\angle DCA < \\angle DAC$. This implies that arc $CP$ is smaller than arc $CD$, and $P$ lies on that arc $CD$ which does not include $A$ and $B$.\n\nWe show that the triangles $ADE$ and $CBQ$ are congruent. As a trapezium inscribed in a circle, $ABCD$ ... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 16 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0j3g | 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\nAnswer: 47\n\nWe claim that player 1 has a winning strategy if and only if $N$ is even and not an odd power of 2.\n\nFirst 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 subtra... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 47 | |
09xp | Determine all positive integers $n$ with the following property: for every triple $(a, b, c)$ of positive real numbers there exists a triple $(k, \ell, m)$ of non-negative integers such that $an^k, bn^\ell, cn^m$ are the lengths of sides of a (non-degenerate) triangle. | [
"It is clear that $n = 1$ does not satisfy the property, as not every three positive real numbers $a, b$, and $c$ are the lengths of the sides of a triangle.\n\nWe first show that any $n \\ge 5$ cannot satisfy the property by considering the triple $(a, b, c) = (1, 2, 3)$. Suppose that there exist $k, \\ell, m$ suc... | Netherlands | IMO Team Selection Test 2 | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | 2, 3, 4 | |
0f2v | Problem:
Show that there is an infinite sequence of reals $x_1$, $x_2$, $x_3$, ... such that $|x_n|$ is bounded and for any $m > n$, we have $|x_m - x_n| > 1 / (m - n)$. | [] | Soviet Union | ASU | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Other"
] | null | proof only | null | |
0flm | Problem:
Se ordenan los números naturales en forma de tabla triangular, es decir:
$$
\begin{array}{cccccccc}
& & & 1 & & & \\
& & 2 & 3 & 4 & & \\
& 5 & 6 & 7 & 8 & 9 & \\
10 & 11 & 12 & 13 & 14 & 15 & 16
\end{array}
$$
Diremos que la posición de un número $N$ en la tabla viene dada por dos "coordenadas": el primer nú... | [
"Solution:\nEn cada fila hay dos números más que en la anterior: en la primera fila hay 1 número, en la segunda fila hay 3, en la tercera fila hay 5, así sucesivamente, en la $n$-ésima fila hay $2n-1$ elementos. Por tanto, el último elemento de la fila $n$-ésima es: $1+3+5+\\cdots+(2n-1)=n^{2}$, por la suma de los ... | Spain | XLVII Olimpiada Matemática Española Primera Fase | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | No | |
0buj | Problem:
Fie numărul natural $n=\overline{11 \ldots 1}+\overline{22 \ldots 2}+\ldots+\overline{88 \ldots 8}+\overline{99 \ldots 9}$, fiecare număr de forma $\overline{a a \ldots a}$ conținând câte 2015 cifre de $a$. Determinați câte cifre de 9 conține numărul $n$. | [
"Solution:\n$n=1 \\cdot \\overline{11 \\ldots 1}+2 \\cdot \\overline{11 \\ldots 1}+3 \\cdot \\overline{11 \\ldots 1}+\\ldots+8 \\cdot \\overline{11 \\ldots 1}+9 \\cdot \\overline{11 \\ldots 1}$\n$=(1+2+3+\\ldots+8+9) \\cdot \\overline{11 \\ldots 1}=45 \\cdot \\overline{11 \\ldots 1}$.\nObținem $n=\\overline{499 \\l... | Romania | Olimpiada de Matematică | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2014 | |
0i79 | Problem:
For any positive integer $n$, let $f(n)$ denote the number of 1's in the base-2 representation of $n$. For how many values of $n$ with $1 \leq n \leq 2002$ do we have $f(n) = f(n+1)$? | [
"Solution:\n\n501. If $n$ is even, then $n+1$ is obtained from $n$ in binary by changing the final 0 to a 1; thus $f(n+1) = f(n) + 1$. If $n$ is odd, then $n+1$ is obtained by changing the last 0 to a 1, the ensuing string of 1's to 0's, and then changing the next rightmost 0 to a 1. This produces no net change in ... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 501 | |
0cyz | Consider the set $S=\{(a+b)^{7}-a^{7}-b^{7}: a, b \in \mathbb{Z}\}$. Find the greatest common divisor of all members in $S$. | [
"Let $d$ be the greatest common divisor of the numbers in $S$. Since $2^{7}-2=126=2 \\cdot 3^{2} \\cdot 7 \\in S$ for $x=y=1$, one gets that $d$ divides $126$. Now, for $x=2$ and $y=1$, one gets $(2+1)^{7}-2^{7}-1=3^{7}-2^{7}-1 \\in S$, therefore $d$ will divide the number $(3^{7}-2^{7}-1)+(2^{7}-2)=3^{7}-3$. Since... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 42 | |
03l8 | Problem:
Let $\Gamma$ be a circle with radius $r$. Let $A$ and $B$ be distinct points on $\Gamma$ such that $AB < \sqrt{3} r$. Let the circle with centre $B$ and radius $AB$ meet $\Gamma$ again at $C$. Let $P$ be the point inside $\Gamma$ such that triangle $ABP$ is equilateral. Finally, let $CP$ meet $\Gamma$ again at... | [
"Solution:\nLet the center of $\\Gamma$ be $O$, the radius $r$. Since $BP = BC$, let $\\theta = \\measuredangle BPC = \\measuredangle BCP$.\n\nQuadrilateral $QABC$ is cyclic, so $\\measuredangle BAQ = 180^\\circ - \\theta$ and hence $\\measuredangle PAQ = 120^\\circ - \\theta$.\n\nAlso $\\measuredangle APQ = 180^\\... | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
09vg | Problem:
Bepaal alle viertallen $(a, b, c, d)$ van positieve reële getallen die voldoen aan $a+b+c+d=1$ en
$$
\max \left(\frac{a^{2}}{b}, \frac{b^{2}}{a}\right) \cdot \max \left(\frac{c^{2}}{d}, \frac{d^{2}}{c}\right)=(\min (a+b, c+d))^{4}
$$ | [
"Solution:\n\nAls $a \\geq b$ dan $\\frac{a}{b} \\geq \\frac{b}{a}$ en als juist $a \\leq b$ dan $\\frac{a}{b} \\leq \\frac{b}{a}$, dus het rijtje $(a, b)$ en het rijtje $\\left(\\frac{a}{b}, \\frac{b}{a}\\right)$ zijn gelijk geordend. De herschikkingsongelijkheid zegt nu dat\n$$\na \\cdot \\frac{a}{b}+b \\cdot \\f... | Netherlands | IMO-selectietoets II | [
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof and answer | (1/4, 1/4, 1/4, 1/4) | |
0a1l | Anna, Bella, and Celine are going to colour. There are six pencils, with different colours. They divide the pencils among themselves, in such a way that everyone gets at least one pencil.
In how many different ways can they distribute the pencils?
A) 537 B) 540 C) 720 D) 729 E) 1170 | [
"B) 540"
] | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | MCQ | B | |
08tx | Both $f(x)$ and $g(x)$ are polynomials with real coefficients, and are not identically zero. Suppose that they satisfy the following functional equation:
$$
f(x^3) + g(x) = f(x) + x^5g(x)
$$
Give an example of such a function $f(x)$ with the smallest possible degree. | [
"Let us try to find a polynomial $f(x)$ of degree at most $4$, which satisfies the condition of the problem with some non-zero polynomial $g(x)$. As the given equation can be rewritten in the form $f(x^3) - f(x) = (x^5 - 1)g(x)$, we see that $f(x^3) - f(x)$ must be divisible by $x^5 - 1$. Let us represent $f(x)$ as... | Japan | Japan Mathematical Olympiad First Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | f(x) = x^4 + x^3 + x^2 + x | |
0edp | Problem:
Nad stranicama $AB$ in $BC$ enakostraničnega trikotnika $ABC$ s stranico dolžine $a$ načrtamo kvadrata $BAD E$ in $CBFG$ (glej sliko). Koliko je dolga daljica $DG$?
(A) $(\sqrt{2}+1)a$
(B) $(2\sqrt{2}+1)a$
(C) $(\sqrt{3}+1)a$
(D) $\sqrt{3}a$
(E) 1
 | [
"Solution:\nOznačimo presečišči daljice $DG$ s stranicama $AB$ in $BC$ zaporedoma s $K$ in $L$. Zaradi simetrije je daljica $DG$ vzporedna stranici $AC$. Torej je trikotnik $KBL$ enakostraničen, trikotnika $DKA$ in $LGC$ pa sta polovici enakostraničnih trikotnikov z višinama dolžine $a$ in stranicama $|DK|=|GL|=\\f... | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | C | |
0020 | Dada una circunferencia de centro $O$, se trazan cuatro rectas tangentes a la circunferencia de modo que estas cuatro rectas determinan el trapecio $ABCD$, de bases $AB$ y $CD$, y lados no paralelos $BC$ y $DA$. Si $AO = 2\sqrt{6}$, $BO = 4\sqrt{3}$ y $CO=4$, calcular las medidas de los lados y los ángulos del trapecio... | [] | Argentina | XX OLIMPIADA MATEMÁTICA ARGENTINA | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | español | proof and answer | Inradius r = 2√3.
Angles: ∠A = 90°, ∠B = 120°, ∠C = 60°, ∠D = 90°.
Side lengths: AB = 2√3 + 2, BC = 8, CD = 6 + 2√3, DA = 4√3. | |
0303 | Problem:
Para cada número inteiro positivo $n$ se associa um inteiro não negativo $f(n)$ de modo que se cumpram as três regras seguintes:
i) $f(a b)=f(a)+f(b)$.
ii) $f(n)=0$ se $n$ é um primo maior que 10$.$
iii) $f(1)<f(243)<f(2)<11$.
Sabendo que $f(2106)<11$, determine o valor de $f(96)$. | [
"Solution:\n\nPela propriedade $i)$, temos $f(243)=f\\left(3^{5}\\right)=5 f(3)$. Dado que\n$$\n0 \\leq f(1)<5 f(3)<f(2)<11\n$$\ne que $5 f(3)$ é um múltiplo de $5$, temos $5 f(3)=5$, ou seja, $f(3)=1$. Note que $2106=2 \\cdot 3^{4} \\cdot 13$. Assim, pela propriedade $i)$,\n$$\nf(2106)=f(2)+4 f(3)+f(13)=f(2)+4\n$$... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Other"
] | null | proof and answer | 31 | |
01zq | Given positive integers $n$ and $k \le n$. Consider an equilateral triangular board with side $n$, which consists of circles: in the first (top) row there is one circle, in the second row there are two circles, ..., in the bottom row there are $n$ circles (see the figure below). Let us place checkers on this board so t... | [
"a. Let us prove that in any such placement of checkers on a triangular board with side $n$, the number of checkers $q$ satisfies the inequality\n$$\nq \\le \\frac{2n + 1}{3} \\cdot k.\n$$\nThe main idea of the proof is double counting. Note that wherever a checker stands, if you count all the cells of the three li... | Belarus | SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | Upper bound: T(k, n) ≤ floor(k(2n + 1)/3). Exact values: T(1, n) = floor((2n + 1)/3) and T(2, n) = floor((4n + 2)/3). | |
0dnn | Problem:
У скупу ненегативних целих бројева решити једначину
$$
\left(2^{2015}+1\right)^{x}+2^{2015}=2^{y}+1.
$$ | [
"Solution:\n\nЗа $x \\leqslant 1$ једина решења су $(0,2015)$ и $(1,2016)$.\n\nНека је $x>1$. Како је $2^{2015}+1$ дељиво са $3$, имамо\n$$\n\\left(2^{2015}+1\\right)^{x}+2^{2015} \\equiv 2^{2015} \\equiv 5 \\pmod{9},\n$$\nдакле $2^{y} \\equiv 4 \\pmod{9}$, одакле добијамо $y=6k+2$ за неко $k \\in \\mathbb{N}$.\n\n... | Serbia | 9. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial... | null | proof and answer | (x, y) = (0, 2015) and (1, 2016) | |
0ata | Problem:
Find the number of ordered triples $(x, y, z)$ of positive integers satisfying $(x+y)^z = 64$. | [] | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | final answer only | 74 | |
02ts | Problem:
Algumas lâmpadas de Natal são arranjadas usando fios mágicos. Cada lâmpada pode ser da cor verde ou amarela. Cada fio está ligado a duas lâmpadas e tem uma propriedade mágica: quando alguém toca em um fio unindo duas lâmpadas, cada uma delas troca de cor passando de verde para amarela ou de amarela para verde... | [
"Solution:\n\na) Cada fio que tocamos muda exatamente a cor de duas lâmpadas. Como existem $16$ lâmpadas amarelas, devemos encostar em pelo menos $8$ fios. A figura a seguir mostra um exemplo de escolhas de fios que torna isso possível:\n\n\n\nb) Veja que a configuração dada no item b) poss... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof and answer | a) 8; b) impossible | |
0bat | a) Show that $m^2 - m + 1$ is an element of the set $\{n^2 + n + 1 \mid n \in \mathbb{N}\}$, for any positive integer $m$.
b) Let $p$ be a perfect square, $p > 1$. Prove that there exist positive integers $r$ and $q$ such that $p^2 + p + 1 = (r^2 + r + 1)(q^2 + q + 1)$. | [
"a) Since $m^2 - m + 1 = (m - 1)^2 + (m - 1) + 1$ and $m - 1 \\ge 0$, it follows that $m - 1 \\in \\mathbb{N}$ and consequently $m^2 - m + 1 \\in \\{n^2 + n + 1 \\mid n \\in \\mathbb{N}\\}$.\n\nb) Write $p = k^2$, where $k$ is an integer. Since $p > 1$, we have $k \\ge 2$. Now $p^2 + p + 1 = k^4 + k^2 + 1 = (k^2 + ... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
00f2 | Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers, and let $S_{k}$ be the sum of products of $a_{1}, a_{2}, \ldots, a_{n}$ taken $k$ at a time.
Show that
$$
S_{k} S_{n-k} \geq \binom{ n}{k}^{2} a_{1} a_{2} \ldots a_{n}, \quad \text{ for } \quad k=1,2, \ldots, n-1
$$ | [
"$$\n\\binom{n}{k} a_{1} a_{2} \\ldots a_{n}\n$$\n$$\n=\\sum_{1 \\leq i_{1}<i_{2}<\\ldots<i_{k} \\leq n} a_{i_{1}} a_{i_{2}} \\ldots a_{i_{k}} \\cdot a_{1} a_{2} \\ldots a_{n} / a_{i_{1}} a_{i_{2}} \\ldots a_{i_{k}}\n$$\n(and using the Cauchy-Schwarz inequality)\n$$\n\\begin{aligned}\n& \\leq\\left(\\sum_{1 \\leq i... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0iho | Problem:
Do there exist positive integers $x$, $y$, and $z$ such that $x^{2006} + y^{2006} = z^{2007}$? Explain your answer. | [
"Solution:\n\nYes, for example, $x = y = z = 2$."
] | United States | Berkeley Math Circle Monthly Contest 4 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Yes; for example, x = y = z = 2. | |
0hpu | Problem:
Find at least one non-zero polynomial $P(x, y, z)$ such that $P(a, b, c)=0$ for every three real numbers that satisfy $\sqrt[3]{a}+\sqrt[3]{b}=\sqrt[3]{c}$. | [
"Solution:\nCube both sides of the condition $\\sqrt[3]{x}+\\sqrt[3]{y}=\\sqrt[3]{z}$:\n$$\n\\begin{gathered}\nx+3 \\sqrt[3]{x} \\sqrt[3]{x} \\sqrt[3]{y}+3 \\sqrt[3]{x} \\sqrt[3]{y} \\sqrt[3]{y}+y=z \\\\\n3 \\sqrt[3]{x} \\sqrt[3]{x} \\sqrt[3]{y}+\\sqrt[3]{x} \\sqrt[3]{y} \\sqrt[3]{y}=z-x-y \\\\\n3 \\sqrt[3]{x} \\sq... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 27xyz - (z - x - y)^3 | |
0e86 | Problem:
Poenostavi izraz
$$
\frac{a-2}{a \cdot (a-2)+4} + \frac{11-(1+a)}{8+a^{3}} - \frac{1}{a+2}.
$$
Za katere realne vrednosti $a$ izraz nima pomena? | [] | Slovenia | Državno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 1/(a^2 - 2a + 4), with the expression undefined at a = -2 | |
0buw | Problem:
a) Câte progresii aritmetice de numere naturale există cu primul termen $1$ şi care conţin numărul $45001$?
b) Arătaţi că nu există progresii aritmetice neconstante de numere naturale cu toţi termenii pătrate perfecte. | [] | Romania | Olimpiada de Matematică Etapa Locală | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | a) 60; b) No such progression exists. | |
0944 | Problem:
Given a pair $(a_{0}, b_{0})$ of real numbers, we define two sequences $a_{0}, a_{1}, a_{2}, \ldots$ and $b_{0}, b_{1}, b_{2}, \ldots$ of real numbers by
$$
a_{n+1}=a_{n}+b_{n} \quad \text{ and } \quad b_{n+1}=a_{n} \cdot b_{n}
$$
for all $n=0,1,2, \ldots$ Find all pairs $(a_{0}, b_{0})$ of real numbers such t... | [
"Solution:\nConsider a function $\\Phi$ of two real variables defined by $\\Phi(a, b)=(a-1)^{2}+(b-1)^{2}$. We claim that $\\Phi\\left(a_{n+1}, b_{n+1}\\right) \\geq \\Phi\\left(a_{n}, b_{n}\\right)$ for all $n$, with equality if and only if $a_{n} b_{n}=0$. Indeed, setting $a_{n}=a$ and $b_{n}=b$ we verify\n$$\n\\... | Middle European Mathematical Olympiad (MEMO) | MEMO Team Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | All pairs with b0 = 0 (with a0 arbitrary real). | |
0i6z | Problem:
A manufacturer of airplane parts makes a certain engine that has a probability $p$ of failing on any given flight. There are two planes that can be made with this sort of engine, one that has 3 engines and one that has 5. A plane crashes if more than half its engines fail. For what values of $p$ do the two pla... | [
"Solution:\nThey have the same probability of failing if $\\binom{5}{2} p^{3}(1-p)^{2}+\\binom{5}{1} p^{4}(1-p)+p^{5}=\\binom{3}{1} p^{2}(1-p)+p^{3}$, which is true iff $p^{2}\\left(6 p^{3}-15 p^{2}+12 p-3\\right)=0$. This is clearly true for $p=0$. We know it is true for $p=1$, since both probabilities would be 1 ... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 0, 1/2, 1 | |
0i87 | Problem:
$OPEN$ is a square, and $T$ is a point on side $NO$, such that triangle $TOP$ has area $62$ and triangle $TEN$ has area $10$. What is the length of a side of the square? | [
"Solution:\n\n$62 = PO \\cdot OT / 2$ and $10 = EN \\cdot TN / 2 = PO \\cdot TN / 2$, so adding gives $72 = PO \\cdot (OT + TN) / 2 = PO \\cdot ON / 2 = PO^2 / 2 \\Rightarrow PO = 12$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 12 | |
0ito | Problem:
Evaluate the limit
$$
\lim_{n \rightarrow \infty} n^{-\frac{1}{2}\left(1+\frac{1}{n}\right)}\left(1^{1} \cdot 2^{2} \cdots n^{n}\right)^{\frac{1}{n^{2}}}.
$$ | [
"Solution:\nAnswer: $e^{-1 / 4}$\n\nTaking the logarithm of the expression inside the limit, we find that it is\n$$\n-\\frac{1}{2}\\left(1+\\frac{1}{n}\\right) \\ln n+\\frac{1}{n^{2}} \\sum_{k=1}^{n} k \\ln k=\\frac{1}{n} \\sum_{k=1}^{n} \\frac{k}{n} \\ln \\left(\\frac{k}{n}\\right)\n$$\nWe can recognize this as th... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | e^{-1/4} | |
0geh | 令 $k \le n$ 為兩個正整數。數奧國有 $n$ 個村莊, 其中有些村莊之間有道路連接。對於任兩個村莊, 其之間的距離定義為從一個村莊抵達另一個村莊最少需走過的道路數量; 如果完全無法從某個村莊抵達另一個村莊, 則這兩個村莊之間的距離為無限大。
剛到數奧國的小勳被隔離在某個地方, 以至於他看不見數奧國裡道路連接的狀況, 但他知道 $n$ 與 $k$。他想知道數奧國中最遠的兩個村莊之間的距離是否為有限大。為此, 他每次可以打一通電話去數奧辦公室, 指定某兩個村莊, 問它們之間的距離是大於、等於或小於 $k$, 而辦公室會如實回答 (無限大的距離大於 $k$)。試證明小勳可以在 $2n^2/k$ 通電話內確認數奧國中最遠的兩個村莊... | [
"令 $G$ 為數奧國對應的圖, $d(u,v)$ 則為 $u$ 與 $v$ 之間的距離。考慮以下策略。\n\n1. 任選一個城市 $v$, 並令 $S = \\{v\\}$。\n\n2. 對於任何 $u \\ne v$, 打一通電話詢問 $u$ 與 $v$ 之間的距離。\n\n3. 如果存在 $v'$ 使得 $\\min_{s \\in S} d(v', s) \\ge k$, 且存在 $s \\in S$ 使得 $d(v', s) = k$, 則令 $S = S \\cup \\{v'\\}$, 改令 $v = v'$, 並回到步驟 2.\n\n4. 檢查任何村莊是否都與某個 $s \\in S$ 的距離至多為 $k$;... | Taiwan | 2021 數學奧林匹亞競賽第二階段選訓營, 國際競賽實作 (二) | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
08f5 | Problem:
Sia dato un esagono regolare di lato $1$. Consideriamo un triangolo equilatero di lato $1$ dentro all'esagono, con due vertici vincolati al perimetro dell'esagono. Muoviamo il triangolo equilatero in modo che uno dei due vertici vincolati compia esattamente un giro lungo il perimetro dell'esagono. Quant'è la ... | [
"Solution:\n\nLa risposta è (C). Mostriamo che, nello spostare il primo vertice vincolato del triangolo da un estremo all'altro di un singolo lato dell'esagono, il vertice non vincolato percorre avanti e indietro un segmento di lunghezza $\\frac{2 \\sqrt{3}}{3}-1$.\n\nIn effetti, sia $A B C D E F$ l'esagono, sia $O... | Italy | Italian Mathematical Olympiad - February Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | MCQ | C | |
0es6 | John spends a total of $3$ hours weeding, planting and watering in his garden. He spends $25$ minutes more on weeding than planting, and $55$ minutes less on watering than planting. How many minutes did he spend planting?
(A) $45$
(B) $50$
(C) $65$
(D) $70$
(E) $80$ | [
"If he spends $x$ minutes on planting, then his total time spent in the garden is $x + 25 + x + x - 55$ and is $180$ minutes. This means $3x - 30 = 180$, so $x = 70$."
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | D | |
04hb | Three points $A$, $B$ and $C$ are considered on a parabola. The tangents to the parabola at these points form a triangle $KLM$ ($KL$ being tangent at $A$, $KM$ at $B$, and $LM$ at $C$). The line through $B$ parallel to the axis of symmetry of the parabola intersects $AC$ at $N$. Prove that $KLMN$ is a parallelogram. | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null | |
01z3 | A positive integer is written on the board. Petya is allowed to replace the number on the board with the sum of the squares of its digits. We call a number *interesting* if Petya can't obtain $1$ from it using a finite number of such operations.
Prove that there exist infinitely many interesting numbers. | [
"Let us first prove that interesting numbers exist. Consider the cycle\n$$\n4 \\rightarrow 16 \\rightarrow 37 \\rightarrow 58 \\rightarrow 89 \\rightarrow 145 \\rightarrow 42 \\rightarrow 20 \\rightarrow 4\n$$\nEach number in this cycle is obtained from the previous one by the operation described in the condition. ... | Belarus | Belarus2022 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | English | proof only | null | |
0kdd | Problem:
Let $a = 256$. Find the unique real number $x > a^{2}$ such that
$$
\log_{a} \log_{a} \log_{a} x = \log_{a^{2}} \log_{a^{2}} \log_{a^{2}} x
$$ | [
"Solution:\nLet $y = \\log_{a} x$ so $\\log_{a} \\log_{a} y = \\log_{a^{2}} \\log_{a^{2}} \\frac{1}{2} y$. Setting $z = \\log_{a} y$, we find $\\log_{a} z = \\log_{a^{2}}\\left(\\frac{1}{2} z - \\frac{1}{16}\\right)$, or $z^{2} - \\frac{1}{2} z + \\frac{1}{16} = 0$. Thus, we have $z = \\frac{1}{4}$, so we can backs... | United States | HMMT February 2020 | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 2^32 | |
0frg | Problem:
En el triángulo $ABC$ con lado mayor $BC$, las bisectrices se cortan en $I$. Las rectas $AI$, $BI$, $CI$ cortan a $BC$, $CA$, $AB$ en los puntos $D$, $E$, $F$, respectivamente. Se consideran puntos $G$ y $H$ en los segmentos $BD$ y $CD$, respectivamente, tales que $\angle GID = \angle ABC$ y $\angle HID = \an... | [
"Solution:\n\nComenzamos con una figura, donde se señalan algunas igualdades de ángulos que inmediatamente justificaremos.\n\n\n\nNótese que los triángulos $ABD$ y $GID$ tienen un ángulo común en $D$ y ángulos iguales en $B$ y en $I$, por lo que sus terceros ángulos deben coincidir, es deci... | Spain | Spain | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
007y | The positive integers $a$, $b$ and $c$ are less than $99$ and satisfy $a^2 + b^2 = c^2 + 99^2$. Find the minimum and the maximum of $a + b + c$. | [
"Assume $a \\geq b$ by symmetry; then $0 < c < b \\leq a < 99$. Also $2a^2 \\geq a^2 + b^2 > 99^2$, so that $a \\geq 71$. Thus $71 \\leq a \\leq 98$. Write the equation as $(b + c)(b - c) = (99 - a)(99 + a)$.\n\nFor $\\min(a + b + c)$ look at $a = 98$ and $a = 97$ first. If $a = 98$ then $(b + c)(b - c) = 197$, and... | Argentina | National Olympiad of Argentina | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM ... | English | proof and answer | minimum = 125, maximum = 243 | |
0ik8 | Problem:
Tim has a working analog 12-hour clock with two hands that run continuously (instead of, say, jumping on the minute). He also has a clock that runs really slow—at half the correct rate, to be exact. At noon one day, both clocks happen to show the exact time. At any given instant, the hands on each clock form ... | [
"Solution:\n\nA tricky thing about this problem may be that the angles on the two clocks might be reversed and would still count as being the same (for example, both angles could be $90^{\\circ}$, but the hour hand may be ahead of the minute hand on one clock and behind on the other).\n\nLet $x$, $-12 \\leq x < 12$... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 33 | |
0754 | Problem:
2. Call a natural number $n$ faithful, if there exist natural numbers $a < b < c$ such that $a$ divides $b$, $b$ divides $c$ and $n = a + b + c$.
(i) Show that all but a finite number of natural numbers are faithful.
(ii) Find the sum of all natural numbers which are not faithful. | [
"Solution:\n\nSuppose $n \\in \\mathbb{N}$ is faithful. Let $k \\in \\mathbb{N}$ and consider $k n$. Since $n = a + b + c$, with $a < b < c$, $a \\mid b$ and $b \\mid c$, we see that $k n = k a + k b + k c$ which shows that $k n$ is faithful.\n\nLet $p > 5$ be a prime. Then $p$ is odd and $p = (p - 3) + 2 + 1$ show... | India | INMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 65 | |
0f06 | Problem:
$15n$ teams compete in a tournament. Each team plays every other team once. In each game a team gets $2$ points for a win, $1$ for a draw and $0$ for a loss. Given any subset $S$ of teams, one can find a team (possibly in $S$) whose total score in the games with teams in $S$ was odd. Prove that $n$ is even. | [] | Soviet Union | ASU | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Vectors",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0e6j | Problem:
Dana je premica z enačbo $2x - 3y + 15 = 0$ in točka $T(3, y)$ na njej. Izračunaj $x$, tako da bo razdalja med točkama $A(x, 4)$ in $T$ enaka $3 \sqrt{5}$. | [
"Solution:\nIzračunamo neznano koordinato točke $T$, tako da vstavimo $x = 3$ v enačbo premice in rešimo enačbo $2 \\cdot 3 - 3y + 15 = 0$. Izračunamo $y = 7$.\n\nKoordinato $x$ točke $A$ izračunamo z uporabo formule za razdaljo med točkama oziroma Pitagorovega izreka:\n$$\nd(A, T) = \\sqrt{(x_2 - x_1)^2 + (y_2 - y... | Slovenia | Državno tekmovanje | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | x = 9 or x = -3 | |
063n | Problem:
Aus zwei $2 \times 1$-Dominosteinen kann man ein Tetromino konstruieren, indem man die beiden Dominosteine längs ihrer längeren Seiten so aneinanderlegt, dass der Mittelpunkt der längeren Seite des einen Dominosteins ein Eckpunkt des anderen Dominosteins ist. Dabei ergeben sich zwei hinsichtlich ihrer Orienti... | [
"Solution:\n\nWir können annehmen, dass $P$ aus einem Teil der Einheitsquadrate des ganzzahligen Koordinatengitters besteht, die wie abgebildet eingefärbt sind. Unter dieser Färbung bedeckt jedes $S$-Tetromino eine gerade Anzahl schwarzer Quadrate und jedes $Z$-Tetromino eine ungerade Anzahl von diesen. Da $P$ voll... | Germany | 2. Auswahlklausur 2014/2015 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0cu7 | In a product of 5 positive integers each multiple was decreased by 3. Could it happen that the product becomes exactly 15 times larger than the initial one?
В произведении 5 положительных целых чисел каждый сомножитель уменьшили на 3. Могло ли получиться так, что произведение увеличилось ровно в 15 раз? | [
"Yes, it is possible.\n\nAn example is $1 \\cdot 1 \\cdot 1 \\cdot 1 \\cdot 48$.\n\nAfter decreasing each factor by $3$, we get:\n$$\n(-2) \\cdot (-2) \\cdot (-2) \\cdot (-2) \\cdot 45 = 16 \\cdot 45 = 720.\n$$\nThe original product is $1 \\cdot 1 \\cdot 1 \\cdot 1 \\cdot 48 = 48$.\n\nThus, $720 = 15 \\cdot 48$.",
... | Russia | XLIII Russian mathematical olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English; Russian | proof and answer | Yes; for example, 1, 1, 1, 1, 48. | |
0eik | Problem:
V trikotniku $ABC$ velja $|AC| \neq |BC|$. Naj bo $S$ razpolovišče stranice $AB$, $D$ presečišče stranice $AB$ in simetrale kota $\angle ACB$, ter $E$ zrcalna slika točke $D$ pri zrcaljenju čez točko $S$. Trikotniku $ABC$ očrtana krožnica seka premico $CD$ v točkah $C$ in $F$, premico $CS$ pa v točkah $C$ in ... | [
"Solution:\n\nZaradi simetrije lahko predpostavimo, da je $|AC| > |BC|$. Potem točka $D$ leži med $S$ in $B$, točka $E$ pa med $S$ in $A$. Hkrati točka $G$ leži na istem bregu premice $FS$ kot točka $E$. Simetrali stranice in nasprotnega kota v trikotniku se vedno sekata na trikotniku očrtani krožnici. Torej simetr... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
0kcc | Problem:
How many ways are there to arrange the numbers $\{1,2,3,4,5,6,7,8\}$ in a circle so that every two adjacent elements are relatively prime? Consider rotations and reflections of the same arrangement to be indistinguishable. | [
"Solution:\nNote that $6$ can only be adjacent to $1$, $5$, and $7$, so there are $\\binom{3}{2} = 3$ ways to pick its neighbors. Since each of $1$, $5$, and $7$ is relatively prime to every number in $\\{1,2,3,4,5,6,7,8\\}$ but itself (and hence can have arbitrary neighbors), without loss of generality suppose we ... | United States | HMMO | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 36 | |
0io4 | Problem:
An infinite sequence of positive real numbers is defined by $a_{0}=1$ and $a_{n+2}=6 a_{n}-a_{n+1}$ for $n=0,1,2, \cdots$. Find the possible value(s) of $a_{2007}$. | [
"Solution:\n\nAnswer: $2^{2007}$. The characteristic equation of the linear homogeneous equation is $m^{2}+m-6=(m+3)(m-2)=0$ with solutions $m=-3$ and $m=2$. Hence the general solution is given by $a_{n}=A(2)^{n}+B(-3)^{n}$ where $A$ and $B$ are constants to be determined. Then we have $a_{n}>0$ for $n \\geq 0$, so... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 2^{2007} | |
0ip9 | Problem:
Find the largest positive integer $n$ such that $n^{3}+4 n^{2}-15 n-18$ is the cube of an integer. | [
"Solution:\nAnswer: 19\n\nNote that the next cube after $n^{3}$ is $(n+1)^{3} = n^{3} + 3 n^{2} + 3 n + 1$. After that, it is $(n+2)^{3} = n^{3} + 6 n^{2} + 12 n + 8$. $n^{3} + 6 n^{2} + 12 n + 8$ is definitely bigger than $n^{3} + 4 n^{2} - 15 n - 18$, so the largest cube that $n^{3} + 4 n^{2} - 15 n - 18$ could b... | United States | 1st Annual Harvard-MIT November Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 19 | |
0git | 鋭角三角形 $ABC$ 中 $\angle B > \angle C$, 點 $I$ 為其內心, $R$ 為其外接圓半徑, 而 $D$ 為 $A$ 點在 $ABC$ 上的垂足。點 $K$ 落於直線 $AD$ 上, 使得 $AK = 2R$, 且 $D$ 在 $A$ 與 $K$ 之間。證明：
$$
\angle KID = \frac{\angle B - \angle C}{2}.
$$ | [
"Denote by $A'$ the antipode of $A$ with respect to the circumcircle $(O)$ of triangle $ABC$. In this case, triangle $AKA'$ is isosceles and since the cevians $AH$ and $AO$ are isogonal with respect to the angle $\\angle BAC$ we deduce that $I$ lies on the line bisector $AM$ of the segment $A'K$ and moreover, the m... | Taiwan | APMO Taiwan Preliminary Round 2 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | Chinese; English | proof only | null | |
0atu | Problem:
For what real values of $p$ will the graph of the parabola $y = x^{2} - 2p x + p + 1$ be on or above that of the line $y = -12x + 5$? | [
"Solution:\n\nWe require that the parabola is on or above the line for all $x$, i.e.,\n$$\nx^{2} - 2p x + p + 1 \\geq -12x + 5\n$$\nfor all $x$.\n\nRewriting:\n$$\nx^{2} - 2p x + p + 1 + 12x - 5 \\geq 0\n$$\n$$\nx^{2} + (12 - 2p)x + (p - 4) \\geq 0\n$$\n\nThis quadratic in $x$ is nonnegative for all $x$ if and only... | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 5 ≤ p ≤ 8 | |
099m | Let $n$ be an arbitrary natural number. Show that there exists a natural number $m$ with the following condition:
The positive integers less than or equal to $m$ that are relatively prime to $m$ can be divided into $n$ parts such that the sum of all numbers in each part is equal.
(proposed by B. Battsengel) | [
"If for $k$, $1 \\leq k < n$ and $(k, m) = 1$ then $(m - k, n) = 1$. Hence, the positive integers less than or equal to $m$ can be divided into such pairs $(k, m - k)$, which pairs number is $\\frac{\\varphi(m)}{2}$. Thus, we should find $m$ such that $n \\mid \\frac{\\varphi(m)}{2}$.\n\nNow, we will show that for ... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0b3r | Problem:
Let $ABC$ be a triangle such that the altitude from $A$, the median from $B$, and the internal angle bisector from $C$ meet at a single point. If $BC=10$ and $CA=15$, find $AB^{2}$. | [
"Solution:\n\nLet $D$ be the foot of the $A$-altitude, $E$ the midpoint of $AC$, and $F$ the foot of the $C$-internal angle bisector. Then by Ceva's Theorem, we have\n$$\n\\frac{AF}{FB} \\cdot \\frac{BD}{DC} \\cdot \\frac{CE}{EA}=1,\n$$\nand so\n$$\n\\frac{b}{a} \\cdot \\frac{c \\cos B}{b \\cos C}=1,\n$$\nwhere we ... | Philippines | 24th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 205 | |
0evt | Let $n (\ge 4)$ be a positive integer, and $a_1, a_2, \dots, a_n$ be distinct positive integers less than or equal to $n$. Determine the maximum value of
$$
\sum_{i=1}^{n} |a_i - a_{i+1} + a_{i+2} - a_{i+3}|.
$$
($a_{n+i} = a_i$ for $i = 1, 2, 3$) | [
"The answer is $n^2$ if $n$ is even, and $n^2 - 5$ if $n$ is odd.\nIt is easy when $n$ is even. Let $k = \\frac{n+1}{2}$. Then,\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} |a_i - a_{i+1} + a_{i+2} - a_{i+3}| &\\le \\sum_{i=1}^{n} (|k - a_i| + |k - a_{i+1}| + |k - a_{i+2}| + |k - a_{i+3}|) \\\\\n&= 4 \\sum_{i=1}^{n} |k -... | South Korea | The 37th Korean Mathematical Olympiad Final Round | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n^2 if n is even; n^2 - 5 if n is odd | |
05zd | Problem:
Soient $b_{1}, \ldots, b_{n}$ des réels positifs ou nuls tels que $b_{1}+b_{2}+\ldots+b_{n}=2$. Soient $a_{0}, a_{1}, \ldots, a_{n}$ des nombres réels tels que $a_{0}=a_{n}=0$. Supposons que pour tout $i$ compris entre 1 et $n$ inclus,
$$
\left|a_{i}-a_{i-1}\right| \leqslant b_{i}
$$
Montrer que l'on a
$$
\le... | [
"Solution:\n\nPour tout indice $i$, on a deux majorations de $a_{i}$. En effet, on peut écrire\n$$\n\\begin{aligned}\na_{i}=a_{i}-a_{0}= & \\left(a_{i}-a_{i-1}\\right)+\\left(a_{i-1}-a_{i-2}\\right)+\\ldots+\\left(a_{1}-a_{0}\\right) \\\\\n& \\leqslant b_{i}+b_{i-1}+\\ldots+b_{1}=\\sum_{j=1}^{i} b_{j}\n\\end{aligne... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOi 2 : Algèbre | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null |
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