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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0hab | Prove that following inequality is true for any triangle:
$$
2R + (3\sqrt{3} - 4)r \ge p,
$$
where $R$ and $r$ are the radii of the circumcircle and incircle of a triangle, and $p$ is the semiperimeter of the triangle. | [
"Let's use $u$, $v$, $w$ - method. You can get more information about this method in numerous articles, e.g. ($x = p - a$, $y = p - b$, $z = p - c$):\n$$\nR = \\frac{abc}{4S} = \\frac{(x+y)(y+z)(z+x)}{4\\sqrt{xyz}(x+y+z)} \\quad \\text{and} \\quad r = \\frac{S}{p} = \\sqrt{\\frac{xyz}{x+y+z}}\n$$\n\nThen we have to... | Ukraine | The Problems of Ukrainian Authors | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
0a1c | Let $n \ge 3$ be a fixed positive integer. There are $n$ boxes $A_1, A_2, \dots, A_n$, each with a number of stones in it $(a_1, a_2, \dots, a_n)$ such that $a_1+a_2+\cdots+a_n = 3n$. A move consists of the following operations:
choose a box and distribute all the stones in the box among the $n$ boxes (including the bo... | [
"**Answer:** $M_n = 3n - 4$ and $f(a_1, a_2, \\dots, a_n) = 3n - 4$ if and only if $a_1 = a_2 = \\dots = a_n = 3$.\n\nFirst of all, we note that for every distribution, there exists a move such that $\\max(a_1, \\dots, a_n)$ increases by at least 1, unless all the stones are in a single box. To see this, pick a box... | Netherlands | IMO Team Selection Test 2 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Mn = 3n − 4, achieved if and only if all boxes initially contain three stones each. | |
04x1 | Determine all triples $(x, y, z)$ of positive numbers satisfying the system of equations
$$
\begin{align*}
2x^3 &= 2y(x^2 + 1) - (z^2 + 1), \\
2y^4 &= 3z(y^2 + 1) - 2(x^2 + 1), \\
2z^5 &= 4x(z^2 + 1) - 3(y^2 + 1).
\end{align*}
$$ | [
"For any integer $k \\ge 3$ and any $x \\ge 0$ we have\n$$\n2x^k \\ge [(k-1)x - (k-2)](x^2 + 1). \\quad (0)\n$$\nTo see this, observe that, by AM-GM inequality,\n$$\nx^k + x^k + \\underbrace{x + x + \\dots + x}_{(k-3)\\text{ times}} \\ge (k-1)x^3,\n$$\nand add it to\n$$\n(k-2)(x^2 - 2x + 1) \\ge 0.\n$$\nNote that w... | Czech-Polish-Slovak Mathematical Match | Czech-Slovak-Polish Match | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | x = y = z = 1 | |
01a3 | Let $B_n$ be a number of ways to partition a $n$-element set onto non-empty parts. For example, $B_3 = 5$ because we have the following partitions of the 3-element set $\{a, b, c\}$:
$$
\{a, b, c\}; \quad \{a\}, \{b, c\}; \quad \{a, c\}, \{b\}; \quad \{a, b\}, \{c\}; \quad \{a\}, \{b\}, \{c\}.
$$
Prove that for every p... | [
"Let $R_m$ be a set of residues modulo $p^m$. For each residue $y \\in R_m$ define a shift of the set $R_m$ by the rule $f_y(x) = (x + y) \\bmod{p^m}$. If $A \\subset R_m$ we denote by $f_y(A)$ the set we get by applying $f_y$ element-wise to $A$. If $P = (P_1, P_2, \\dots, P_k)$ is a partition, then we denote by $... | Baltic Way | Baltic Way 2013 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
047p | Given an odd integer $n \ge 3$ and a regular $n$-gon with vertex set $V$, let $\mathcal{P}$ denote the set of all regular polygons with vertices in $V$. For example, if $n = 15$, then $\mathcal{P}$ includes 1 regular 15-gon, 3 regular pentagons, and 5 equilateral triangles.
Two players, Alice and Bob, play the followi... | [
"The maximal integer $k$ is $\\frac{\\sigma(n) + \\tau(n) - n - 1}{2}$. (Note that from this formula, $\\sigma(n)$ and $\\tau(n)$ have the same parity.)\n\nLabel the vertices of $V$ as $A_0, A_1, \\cdots, A_{n-1}$, with indices considered modulo $n$.\n\nWe say a coloring is *perfect* if there exists $i$ such that $... | China | China-TST-2025A | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Number-T... | English | proof and answer | (σ(n) + τ(n) - n - 1) / 2 | |
0elh | Problem:
Dan je štirikotnik $A B C D$ s pravima kotoma pri ogliščih $A$ in $C$. Pravokotni projekciji oglišč $D$ in $B$ na daljico $A C$ zaporedoma označimo z $E$ in $F$. Koliko je dolžina daljice $B F$, če je $|A E|=3, |D E|=5$ in $|C E|=7$? | [
"Solution:\n\nKer sta kota štirokotnika $A B C D$ pri ogliščih $A$ in $C$ prava, točki $A$ in $C$ po Talesovem izreku ležita na krožnici s premerom $B D$. Torej je štirikotnik $A B C D$ tetiven. Po Pitagorovem izreku je $|A D|=\\sqrt{|A E|^{2}+|D E|^{2}}=\\sqrt{9+25}=\\sqrt{34}$ in $|C D|=\\sqrt{|C E|^{2}+|D E|^{2}... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 21/5 | |
021w | Problem:
Em 1950 um "profeta" anunciou que o fim do mundo ocorreria em 11.08.1999 (11 de agosto de 1999). Como nada aconteceu nesse dia, ele refez seus cálculos e fez a seguinte previsão: "O fim do mundo ocorrerá na próxima data que se escreve com 8 algarismos diferentes." Você pode descobrir essa data? | [
"Solution:\n\nResposta: 17.06.2345"
] | Brazil | null | [
"Math Word Problems"
] | null | final answer only | 17.06.2345 | |
0az9 | Problem:
Determine all ordered pairs $(x, y)$ of nonnegative integers that satisfy the equation
$$
3 x^{2}+2 \cdot 9^{y}=x\left(4^{y+1}-1\right)
$$ | [
"Solution:\nThe equation is equivalent to\n$$\n(3 x)^{2}+2 \\cdot 3^{2 y+1}=3 x\\left[2^{(2 y+1)+1}-1\\right]\n$$\nLetting $a=3 x$ and $b=2 y+1$, we have\n$$\na^{2}+2 \\cdot 3^{b}=a\\left(2^{b+1}-1\\right)\n$$\nCase 1: $b=1$\nWe have $a^{2}-3 a+6=0$ which has no integer solution for $a$. Thus, there is no solution ... | Philippines | 20th Philippine Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | [(2, 1), (3, 1), (3, 2), (18, 2)] | |
0dhe | A quadrilateral $ABCD$ is inscribed inside a circle and $AD \perp CD$. Draw $BE \perp AC$ at $E$ and $BF \perp AD$ at $F$. Show that the line $EF$ passes through the midpoint of the line segment $BD$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Anal... | English | proof only | null | |
0a51 | Problem:
There are 13 marked points on the circumference of a circle with radius 13. Prove that we can choose three of the marked points which form a triangle with area less than 13. | [
"Solution:\n\nDivide the circle into 6 equal ($60^{\\circ}$) arcs. By the pigeon-hole principle (since $13 > 2 \\times 6$) there exists at least one arc which contains at least three of the marked points. Let this $60^{\\circ}$ arc be $AB$, and let the three marked points be $X$, $Y$ and $Z$ in that order.\n\n![](a... | New Zealand | New Zealand Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0dcx | Let $ABC$ be a triangle, the circle having $BC$ as diameter cuts $AB$, $AC$ at $F$, $E$ respectively. Let $P$ be a point on this circle. Let $C_{0}$, $B_{0}$ be the projections of $P$ upon the sides $AB$, $AC$ respectively. Let $H$ be the orthocenter of the triangle $AB_{0}C_{0}$. Show that $\angle EHF = 90^{\circ}$. | [
"Let $B' E'$, $C' F'$ be the altitudes of $\\triangle AB'C'$ and concur at $H$.\nWe have that $AB' P C'$ is inscribed in the circle with diameter $AP$, so $AP$ passes through the circumcenter of $\\triangle AB'C'$. Then $AP$, $AH$ reflect each other by the angle bisector of $\\angle B'AC'$. From that, easy to prove... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"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"
] | English | proof only | null | |
00wz | Problem:
Let $a, b, c, d, e$ be distinct real numbers. Prove that the equation
$$
\begin{aligned}
& (x-a)(x-b)(x-c)(x-d) \\
& +(x-a)(x-b)(x-c)(x-e) \\
& +(x-a)(x-b)(x-d)(x-e) \\
& +(x-a)(x-c)(x-d)(x-e) \\
& +(x-b)(x-c)(x-d)(x-e)=0
\end{aligned}
$$
has 4 distinct real solutions. | [
"Solution:\nOn the left-hand side of the equation we have the derivative of the function\n$$\nf(x)=(x-a)(x-b)(x-c)(x-d)(x-e)\n$$\nwhich is continuous and has five distinct real roots."
] | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0ajo | Нека $m$ и $n$ се позитивни цели броеви такви што $m>n$. Дефинираме $x_k = \frac{m+k}{n+k}$ за $k=1,2,...,n+1$. Докажи дека ако $x_1,x_2,...,x_{n+1}$ се цели броеви, тогаш $x_1x_2...x_{n+1}-1$ е делив со барем еден прост непарен број. | [
"Нека препоставиме дека $x_1,x_2,...,x_{n+1}$ се цели броеви. Ги дефинираме целите броеви\n$$\na_k = x_k - 1 = \\frac{m+k}{n+k} - 1 = \\frac{m-n}{n+k} > 0,\n$$\nза $k=1,2,...,n+1$.\nНека $P = x_1x_2...x_{n+1}-1$. Потребно е да докажеме дека $P$ е делив со барем еден непарен прост број, или дека $P$ не е степен на б... | North Macedonia | IMO Selection Test | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0364 | Problem:
Four players $A_{1}$, $A_{2}$, $A_{3}$ and $A_{4}$ have the same amounts of money and play the following game with seven dices: $A_{1}$ throws the seven dices and then pays to each of the other three players $\frac{1}{k}$ of the money that the corresponding player has at the moment, where $k$ is the sum of th... | [
"Solution:\n\nDenote by $S_{k}^{(m)}$ the money of the $k$-th player, $k=1,2,3,4$, after the move and payment of the $m$-th, $m=1,2,3,4$, and $S_{k}^{(0)}=S$ in the beginning, $k=1,2,3,4$. Denote by $a_{i}$ the sum of points on the dices thrown by $A_{i}$. It follows from the game rules that $S_{k}^{(m)}=S_{k}^{(m-... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | A1: 31, A2: 30, A3: 8, A4: 7 | |
0cnd | In a convex quadrilateral $ABCD$, the relations $AB = BD$, $\angle ABD = \angle DBC$ are satisfied. A point $K$ is chosen on diagonal $BD$ so that $BK = BC$. Prove that $\angle KAD = \angle KCD$. | [
"Let us mark on side $AB$ a segment $BE = BC$. The isosceles triangles $EBK$ and $KBC$ are congruent by two sides and the angle between them. Therefore, $EK = KC$, and $\\angle AEK = 180^\\circ - \\angle BEK = 180^\\circ - \\angle BKC = \\angle CKD$. Moreover, $KD = BD - BK = BA - BE = EA$. Hence, triangles $AEK$ a... | Russia | Euler olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English; Russian | proof only | null | |
0c9c | For an integer $n > 1$, let $gpf(n)$ denote the greatest prime factor of $n$. A *strange pair* is an unordered pair of distinct primes $p$ and $q$ such that $\{p, q\} = \{gpf(n), gpf(n + 1)\}$ for no integer $n > 1$. Prove that there exist infinitely many strange pairs.
Russia, Dmitry Krachun | [
"We show that there are infinitely many strange pairs of the form $\\{2, q\\}$ where $q$ is an odd prime.\n\nThe Lemma below provides a sufficient condition for such a pair to be strange. For an odd prime $q$, let $ord_q(2)$ denote the multiplicative order of $2$ modulo $q$, i.e., the least positive integer $s$ sat... | Romania | Romanian Master of Mathematics | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0gvu | In a circle, two non-intersecting chords $AB$ and $CD$ are drawn. On the chord $AB$, a point $E$ (different from $A$ and $B$) is taken. Consider the arc $AB$ that does not contain the points $C$ and $D$. With a compasses and a straightedge, find a point $F$ on that arc such that $\frac{PE}{EQ} = \frac{1}{2}$, where $P$... | [
"Побудова і подальше доведення очевидним чином ґрунтуються на наступному аналізі.\n\nПерший випадок. Нехай точка $E$ ділитиме відрізок $PQ$ внутрішнім чином. На промені $CE$ оберемо таку точку $K$, що $EK = 2 \\cdot CE$. Тоді, як легко довести, $CF \\parallel QK$, і, крім того, з точки $Q$ хорди $AB$ відрізок $KD$ ... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | null | |
05x9 | Problem:
Soit $ABCD$ un quadrilatère convexe cyclique, $M$ le milieu de $[AC]$. Le cercle circonscrit à $CDM$ rencontre $(BC)$ une deuxième fois en $N$ (autre que $C$). Soit $B'$ le symétrique de $B$ par rapport à $N$. Montrer que $(MN)$ est tangente au cercle circonscrit de $B'DN$. | [
"Solution:\n\n\n\nSur la figure, on repère des triangles semblables. En effet, on a par angle inscrit :\n$$\n\\widehat{MAD} = \\widehat{CAD} = \\widehat{CBD} = \\widehat{NBD}\n$$\nToujours par angle inscrit, on a :\n$$\n\\widehat{DMA} = 180^{\\circ} - \\widehat{CMD} = 180^{\\circ} - \\wideh... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ejh | Problem:
Sestre Neža, Meta in Ajda so med počitniškim delom v mesecu juliju zaslužile denar v razmerju $3:4:5$. V mesecu avgustu je Neža zaslužila $40\%$ več kot v mesecu juliju, Meta petino manj kot v juliju, Ajda pa za $200$ manj kot v juliju. Skupaj so sestre v avgustu zaslužile $2280$. Kakšen je bil zaslužek vsake... | [
"Solution:\n\n1. Zapišemo zaslužek Neže v juliju $3x$, zaslužek Mete v juliju $4x$, zaslužek Ajde v juliju $5x$.\n\nZapis zaslužka Neže v avgustu je $3x \\cdot 1.4 = 4.2x$.\n\nZapis zaslužka Mete v avgustu je $4x \\cdot 0.8 = 3.2x$.\n\nZapis zaslužka Ajde v avgustu je $5x - 200$.\n\nIzračunamo skupen zaslužek vseh ... | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Odbirno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | proof and answer | Neža 600, Meta 800, Ajda 1000 | |
0hu9 | Problem:
Let $n > 3$ be a positive integer. Define an integer $k$ to be snug if $1 \leq k < n$ and
$$
\gcd(k, n) = \gcd(k+1, n).
$$
Prove that the product of all snug integers is congruent to $1$ modulo $n$. | [
"Solution:\n\nLet $k$ be a snug integer. Note that any factor that divides $n$, $k$, and $k+1$ must also divide $(k+1)-k=1$, so\n$$\n\\gcd(k, n) = \\gcd(k+1, n) = 1.\n$$\nIn particular, $k$ has a multiplicative inverse $h \\bmod n$ (we can choose $h$ such that $0 < h < n$). We claim that $h$ is also snug. Clearly $... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof only | null | |
0f3v | Problem:
Find all real solutions to:
$$\sin x + 2\sin (x + y + z) = 0$$
$$\sin y + 3\sin (x + y + z) = 0$$
$$\sin z + 4\sin (x + y + z) = 0$$ | [] | Soviet Union | ASU | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | All solutions are x = k·pi, y = l·pi, z = m·pi for integers k, l, m. | |
0enq | Let $p$ and $k$ be positive integers such that $p$ is prime and $k > 1$. Prove that there is at most one pair $(x, y)$ of positive integers such that
$$
x^k + px = y^k.
$$ | [
"We distinguish two different cases:\n\n**Case 1:** $\\text{gcd}(x, p) = 1$. In this case, $x$ and $x^{k-1} + p$ do not have a common divisor (other than 1) either, and it follows from the factorisation\n$$\nx(x^{k-1} + p) = y^k\n$$\nthat both $x$ and $x^{k-1} + p$ have to be $k$th powers, say $x = u^k$ and $x^{k-1... | South Africa | South African Mathematical Olympiad Third Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
07eb | Let $x$, $y$ and $z$ be real numbers, such that $x + y + z = xy + yz + zx$. Prove that
$$
\frac{x}{\sqrt{x^4 + x^2 + 1}} + \frac{y}{\sqrt{y^4 + y^2 + 1}} + \frac{z}{\sqrt{z^4 + z^2 + 1}} \geq \frac{-1}{\sqrt{3}}
$$ | [
"Let's define $f(t) = \\frac{t}{\\sqrt{t^4 + t^2 + 1}}$, that is, $f(t) = -f(-t)$, $f(\\frac{1}{t}) = f(t)$, furthermore, $|f(t)| \\le \\frac{1}{\\sqrt{3}}$. Then, if we change $(x, y, z)$ with $(\\frac{1}{x}, \\frac{1}{y}, \\frac{1}{z})$, nothing has been changed. Moreover, one can find that, $z = \\frac{x + y - x... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof only | null | |
0a2f | Find all pairs $(a, b)$ of positive integers such that $f(x) = x$ is the only function $f: \mathbb{R} \to \mathbb{R}$ that satisfies
$$
f^a(x)f^b(y) + f^b(x)f^a(y) = 2xy
$$
for all $x, y \in \mathbb{R}$.
Here $f^n(x)$ represents the function obtained by applying $n$ times the function $f$ to $x$, so $f^1(x) = f(x)$ and... | [
"We are going to prove that exactly all pairs $(a, b)$ with $\\gcd(a, b) = 1$ and with $a + b$ odd satisfy this.\n\nFirst assume that $\\gcd(a, b) = n \\neq 1$. Consider the function\n$$\ng(x) = \\begin{cases} x + 1 & \\text{if } \\lfloor x \\rfloor \\not\\equiv 0 \\mod n \\\\ x + 1 - n & \\text{if } \\lfloor x \\r... | Netherlands | BxMO/EGMO Team Selection Test | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | All positive integer pairs (a, b) with gcd(a, b) = 1 and a + b odd. | |
00xj | Problem:
A polynomial $f(x) = x^{3} + a x^{2} + b x + c$ is such that $b < 0$ and $a b = 9 c$. Prove that the polynomial has three different real roots. | [
"Solution:\n\nConsider the derivative $f'(x) = 3 x^{2} + 2 a x + b$. Since $b < 0$, it has two real roots $x_{1}$ and $x_{2}$. Since $f(x) \\rightarrow \\pm \\infty$ as $x \\rightarrow \\pm \\infty$, it is sufficient to check that $f(x_{1})$ and $f(x_{2})$ have different signs, i.e., $f(x_{1}) f(x_{2}) < 0$.\n\nDiv... | Baltic Way | Baltic Way 1992 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof only | null | |
0856 | Problem:
Gli abitanti di un'isola sono o furfanti o cavalieri: i cavalieri dicono sempre la verità, i furfanti mentono sempre. Una sera al bar, Alberto dice: "Bruno è un cavaliere"; Bruno dice: "......tutti e tre cavalieri" (in quel momento passa un camion e non si capisce se Bruno ha detto "Siamo tutti..." o "Non sia... | [
"Solution:\n\nLa risposta è (A). In base all'affermazione di Alberto, Alberto e Bruno sono dello stesso tipo. Se Carlo è un cavaliere, Bruno dice \"non siamo tutti e tre cavalieri\" che è vera se e solo se Bruno è un furfante. Perciò Carlo è necessariamente un furfante. Poichè l'affermazione di Bruno è del tipo \"(... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Discrete Mathematics > Logic"
] | null | MCQ | A | |
01ji | Show that the sum of the decimal digits of $2^{2^{2^{2023}}}$ is greater than $2023$. | [
"We will prove the more general statement that, for every positive integer $n$, the sum of decimal digits of $2^{2^{2n}}$ is greater than $n$.\nLet $m = 2^{2n} = 4^n$, so that we need to consider the digits of $2^m$. It will suffice to prove that at least $n$ of these digits are different from $0$, since the last d... | Baltic Way | Baltic Way 2023 Shortlist | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
02f5 | Given a circle and its center $O$, a point $A$ inside the circle and a distance $h$, construct a triangle $BAC$ with $\angle A = 90^\circ$, $B$ and $C$ on the circle and the altitude from $A$ with length $h$. | [
"Let $H$ on $BC$ such that $AH = h$. Since $BAC$ is a right angle, $BH \\cdot CH = h^2$.\n\nBut the power of $H$ with respect to the given circle is $HB \\cdot HC = R^2 - OH^2$.\n\nSo $OH = \\sqrt{R^2 - h^2}$ is determined and $H$ is the intersection of the circle with center $O$ and radius $\\sqrt{R^2 - h^2}$ and ... | Brazil | XV OBM | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
04wh | Find all positive integers $n$ with the following property: It is possible to color $2n$ cells in an $n \times n$ square grid such that each row and each column contains exactly two colored cells, and no two colored cells share a side or a vertex. | [] | Czech Republic | National Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | All integers n greater than or equal to 9 | |
05ge | Problem:
Trouver toutes les fonctions $f$ de $\mathbb{R}$ dans $\mathbb{R}$ telles que, pour tous réels $x$ et $y$, on ait
$$
f(x f(y))+x=f(x) f(y+1)
$$ | [
"Solution:\nEn prenant $x=0$, on obtient $f(0)=f(0) f(y+1)$ pour tout $y$, donc soit $f(0)=0$, soit $f(y+1)=1$ pour tout $y$. Dans le second cas, $f$ est constante égale à $1$, mais alors l'équation devient\n$$\n1+x=1 \\times 1\n$$\npour tout $x$, ce qui est impossible. On a donc $f(0)=0$.\n\nEn prenant $y=0$, on o... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x)=x and f(x)=-x | |
0cs4 | Исходно на доске написаны многочлены $x^3 - 3x^2 + 5$ и $x^2 - 4x$. Если на доске уже написаны многочлены $f(x)$ и $g(x)$, разрешается дописать на неё многочлены $f(x) \pm g(x)$, $f(x)g(x)$, $f(g(x))$ и $cf(x)$, где $c$ — произвольная (не обязательно целая) константа. Может ли на доске после нескольких операций появить... | [
"**Ответ.** Не может.\nПусть $f(x)$ и $g(x)$ — два многочлена, и для некоторой точки $x_0$ выполняются равенства $f'(x_0) = 0$ и $g'(x_0) = 0$. Тогда, очевидно, $(f \\pm g)'(x_0) = 0$ и $cf'(x_0) = 0$. Также $(fg)'(x_0) = f(x_0)g'(x_0) + f'(x_0)g(x_0) = 0$. Наконец, если $h(x)$ — многочлен, то $(h(g(x_0)))' = h'(g(... | Russia | XL Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
0k9x | Problem:
There is a unique function $f: \mathbb{N} \rightarrow \mathbb{R}$ such that $f(1)>0$ and such that
$$
\sum_{d \mid n} f(d) f\left(\frac{n}{d}\right)=1
$$
for all $n \geq 1$. What is $f\left(2018^{2019}\right)$ ? | [
"Solution:\n\nFix any prime $p$, and let $a_n=f\\left(p^n\\right)$ for $n \\geq 0$. Notice that using the relation for $p^n$, we obtain\n$$\n\\sum_{i=0}^{n} a_i a_{n-i}=1\n$$\nwhich means that if we let $g(x)=\\sum_{n \\geq 0} a_n x^n$, then $g(x)^2=1+x+x^2+\\cdots=\\frac{1}{1-x}$ as a generating function. Thus $g(... | United States | HMMT February 2019 | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Other"
] | null | proof and answer | binom(4038,2019)^2 / 2^8076 | |
0l34 | What is the minimum number of successive swaps of adjacent letters in the string $ABCDEF$ that are needed to change the string to $FEDCBA$? (For example, 3 swaps are required to change $ABC$ to $CBA$; one such sequence of swaps is $ABC \to BAC \to BCA \to CBA$.)
(A) 6 (B) 10 (C) 12 (D) 15 (E) 24 | [
"If the $A$ is swapped 5 times, once with each of the other letters, the result will be $BCDEFA$. Now the $B$ can be swapped 4 times in the same way to end up in the fifth position: $CDEFBA$. Continuing in this way gives a sequence of $5 + 4 + 3 + 2 + 1 = 15$ swaps that achieves the required result.\n\nTo see that ... | United States | AMC 10 A | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | MCQ | D | |
02oz | Problem:
a. Prove que o número $3999991$ não é primo.
b. Prove que o número $1000343$ não é primo. | [
"Solution:\n\na.\nObserve que\n$$\n\\begin{aligned}\n3999991 &= 4000000 - 9 \\\\\n&= 4 \\cdot 10^{6} - 3^{2} \\\\\n&= \\left(2 \\cdot 10^{3}\\right)^{2} - 3^{2} \\\\\n&= \\left(2 \\cdot 10^{3} - 3\\right)\\left(2 \\cdot 10^{3} + 3\\right) = 1997 \\cdot 2003\n\\end{aligned}\n$$\ne portanto não é um número primo.\n\n... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
09is | For real numbers $-1 \le a, b, c \le 1$ satisfying $a^2 + b^2 + c^2 = 2abc + 1$, prove
$$
\frac{3}{2 - abc} \le \frac{1}{2 - a^2} + \frac{1}{2 - b^2} + \frac{1}{2 - c^2} \le 1 + \frac{2}{2 - abc}.
$$
(Proposed by Otgonbayar Uuye) | [
"First we show\n$$\n\\frac{3}{2 - abc} \\le \\frac{1}{2 - a^2} + \\frac{1}{2 - b^2} + \\frac{1}{2 - c^2}.\n$$\nBy the arithmetic-harmonic mean inequality, we have\n$$\n\\frac{1}{2-a^2} + \\frac{1}{2-b^2} + \\frac{1}{2-c^2} \\ge \\frac{9}{6-(a^2+b^2+c^2)} = \\frac{9}{5-2abc}.\n$$\nMoreover, we have $\\frac{9}{5-2abc... | Mongolia | Mongolian Mathematical Olympiad Round 3 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
05p4 | Problem:
Deux cercles $\mathcal{C}$ et $\mathcal{C}^{\prime}$ de centres $O$ et $O^{\prime}$ sont tangents extérieurement en $B$. Une tangente commune extérieure touche $\mathcal{C}$ en $M$ et $\mathcal{C}^{\prime}$ en $N$. La tangente commune à $\mathcal{C}$ et $\mathcal{C}^{\prime}$ en $B$ coupe $(M N)$ en $A$. On n... | [
"Solution:\n\nComme $(A M)$ et $(A B)$ sont tangents au cercle de centre $O$, on a $A M = A B$. On a aussi $O M = O B$, donc $(O A)$ est la médiatrice de $[M B]$, et donc $C$ est le milieu de $[M B]$.\n\nDe même, $D$ est le milieu de $[B N]$, donc $(C D)$ est parallèle à $(M N)$."
] | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
086f | Problem:
In un trapezio isoscele $ABCD$ di base maggiore $AB$, le diagonali vengono divise dal loro punto di incontro $O$ in parti proporzionali ai numeri 1 e 3. Sapendo che l'area del triangolo $BOC$ è 15, quanto misura l'area dell'intero trapezio?
(A) 60
(B) 75
(C) 80
(D) 90
(E) 105. | [
"Solution:\n\nLa risposta è $\\mathbf{( C )}$. Evidentemente il triangolo $AOD$ è uguale al triangolo $BOC$, quindi ha anch'esso area 15.\n\nI triangoli $ODC$ e $OCB$ hanno la stessa altezza $CH$, e poiché la base $OD$ di $ODC$ è $1/3$ della base $OB$ di $BOC$, l'area di $ODC$ è $1/3$ dell'area di $BOC$, cioè $15/3... | Italy | Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | C | |
04j0 | Let $n$ be a positive integer larger than $1$ such that both $2n - 1$ and $3n - 2$ are perfect squares. Prove that $10n - 7$ is composite. | [] | Croatia | Croatia Mathematical Competitions | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
05or | Problem:
On considère 5 nombres entiers positifs. En les ajoutant deux à deux de toutes les façons possibles, on génère 10 entiers. Montrer que ces 10 entiers ne peuvent pas être 10 entiers consécutifs. | [
"Solution:\n\nOn suppose que les 10 entiers sont consécutifs, et on note $n$ le plus petit. Leur somme est $n + (n+1) + \\cdots + (n+9) = 10n + 45$. Mais c'est aussi la somme des 5 nombres de départ, comptés 4 fois chacun, donc c'est un multiple de 4. C'est impossible car pour tout entier $n$, $10n + 45$ est impair... | France | Envoi 1: Arithmétique | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
047b | Let $a_1 > a_2 > \dots > a_n > 1$ be positive integers. Let $M$ denote the least common multiple of $a_1, a_2, \dots, a_n$. For a finite set of integers $X$, define
$$
f(X) = \min_{1 \le i \le n} \sum_{x \in X} \left\{ \frac{x}{a_i} \right\}.
$$
Here $\{u\} = u - \lfloor u \rfloor$ is the fractional part of the real nu... | [
"*Proof.* Assume $f(X) = \\lambda \\ge \\frac{2}{a_n}$. By contradiction, suppose $|X| > \\lambda M$. Since $\\lambda$ is of the form $\\frac{k}{a_i}$ ($k \\in \\mathbb{Z}_{>0}$), $\\lambda M$ is an integer. We prove that there exists $x \\in X$ such that $f(X \\setminus \\{x\\}) = f(X)$, which contradicts the mini... | China | 2024 CMO | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arg... | English | proof only | null | |
0625 | Problem:
Man bestimme mit Beweis alle Funktionen $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ mit der Eigenschaft
$$
f(x) f(y) = 2 f(x + y f(x))
$$
für alle positiven reellen Zahlen $x, y$. | [
"Solution:\n\nOffensichtlich erfüllt die Funktion $f(x) = 2$ für alle $x \\in \\mathbb{R}^{+}$ die gegebene Funktionalgleichung. Wir werden zeigen, dass dies die einzige Lösung ist.\n\nLemma 1: Für alle $x \\in \\mathbb{R}^{+}$ gilt $f(x) \\geq 1$.\n\nZum Beweis nehmen wir $f(x) < 1$ für ein geeignetes $x$ an und s... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 2 for all positive real x | |
0l5i | Problem:
Kelvin the frog is on the bottom-left lily pad of a $3 \times 3$ grid of lily pads, and his home is at the top-right lily pad. He can only jump between two lily pads which are horizontally or vertically adjacent. Compute the number of ways to remove 4 of the lily pads so that the bottom-left and top-right lil... | [
"Solution:\n\nWe instead count the arrangements for which Kelvin can get home. Note that at minimum, Kelvin must use 5 lily pads to get home, leaving 4 lily pads that are not on the path. This means that if we were to remove 4 lily pads and Kelvin can still get home, the non-removed lily pads form a shortest path f... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 29 | |
02qc | Problem:
Estefânia tem cinco cartas marcadas com as letras $A, B, C, D$ e $E$, empilhadas nessa ordem de cima para baixo. Ela embaralha as cartas pegando as duas de cima e colocando-as, com a ordem trocada, embaixo da pilha. A figura mostra o que acontece nas duas primeiras vezes em que ela embaralha as cartas.
![](a... | [
"Solution:\n\nO leitor pode verificar que, se Estefânia embaralhar as cartas 6 vezes, elas voltarão à posição inicial. Como $74=12 \\times 6+2$, embaralhar as cartas 74 vezes tem o mesmo efeito que fazê-lo duas vezes, o que deixa a carta $E$ no topo da pilha."
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Number Theory > Modular Arithmetic"
] | null | MCQ | E | |
0anx | Problem:
What is the last digit of $2! + 4! + 6! + \ldots + 2010! + 2012!$
(a) 6
(b) 7
(c) 8
(d) 9 | [] | Philippines | Qualifying Round | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | a | |
0ebs | Let $f: \mathbb{Z} \to \mathbb{Z}$ be an injective map such that $|f(m) - f(n)| \le 2015$ holds for arbitrary integers $m$ and $n$ which satisfy $|m - n| \le 2015$. Prove that
$$
|f(m) - f(n)| = |m - n|
$$
holds for all $m, n \in \mathbb{Z}$. | [
"Let $n$ be an arbitrary integer and\n$$\nS_n = \\{n - 2015, n - 2014, \\dots, n, \\dots, n + 2015\\}\n$$\nbe the set of all integers that differ from $n$ by at most $2015$.\nWe know that $S_{f(n)} = \\{f(n)-2015, f(n)-2014, \\dots, f(n)+2015\\}$.\nFrom the condition of the problem it follows that the numbers $f(n-... | Slovenia | Selection Examinations for the IMO 2015 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
094s | Problem:
For any positive integer $n$, let $\sigma(n)$ denote the sum of positive divisors of $n$. Determine all polynomials $P$ with integer coefficients such that $P(k)$ is divisible by $\sigma(k)$ for all positive integers $k$. | [
"Solution:\n\nWe are going to use the following well-known lemma:\n\nLemma. For any integers $a, b$ and polynomial $p$ with integer coefficients we have\n$$\na-b \\mid p(a)-p(b)\n$$\nLet $p \\neq q$ be any prime numbers. Then from $\\sigma(p q)=(p+1)(q+1)$ we have\n$$\n(p+1)(q+1) \\mid P(p q)\n$$\nThis is equivalen... | Middle European Mathematical Olympiad (MEMO) | MEMO Szeged | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Com... | null | proof and answer | the zero polynomial | |
0iwu | Problem:
Let $P$ be a fourth degree polynomial, with derivative $P'$, such that $P(1) = P(3) = P(5) = P'(7) = 0$. Find the real number $x \neq 1, 3, 5$ such that $P(x) = 0$. | [
"Solution:\n\nObserve that $7$ is not a root of $P$. If $r_1, r_2, r_3, r_4$ are the roots of $P$, then\n$$\n\\frac{P'(7)}{P(7)} = \\sum_{i} \\frac{1}{7 - r_i} = 0.\n$$\nThus\n$$\nr_4 = 7 - \\left( \\sum_{i \\neq 4} \\frac{1}{7 - r_i} \\right)^{-1} = 7 + \\left( \\frac{1}{6} + \\frac{1}{4} + \\frac{1}{2} \\right)^{... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 89/11 | |
0ixl | Let $N > M > 1$ be fixed integers. There are $N$ people playing in a chess tournament; each pair of players plays each other once, with no draws. It turns out that for each sequence of $M + 1$ distinct players $P_0, P_1, \dots, P_M$ such that $P_{i-1}$ beat $P_i$ for each $i = 1, \dots, M$, player $P_0$ also beat $P_M$... | [
"Write $P \\succ Q$ if player $P$ beat player $Q$.\n\n**Lemma 1.** Any set of $K > 1$ players can be arranged in a sequence $P_1, \\dots, P_K$ such that $P_1 \\succ P_2 \\succ \\dots \\succ P_K$.\n\n*Proof.* Let $P_1 \\succ \\dots \\succ P_J$ be the longest such sequence that can be formed from any subset of the gi... | United States | Team Selection Test 2009 | [
"Discrete Mathematics > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
01up | Find all real numbers $a$ for which there exists a function $f$ defined on the set of all real numbers which takes as its values all real numbers exactly once and satisfies the equality
$$
f(f(x)) = x^2 f(x) + a x^2
$$
for all real $x$. | [
"Answer: $a = 0$.\n\nSubstituting $x$ such that $f(x) = -a$, we get $f(-a) = 0$.\n\nSubstituting $x = -a$, we get $f(0) = a^3$.\n\nFinally, substituting $x = 0$, we get $f(a^3) = 0$.\n\nSince $f$ takes all real values exactly once, $a^3 = -a$ which is equivalent to $a(a^2 + 1) = 0$, i.e. $a = 0$.\n\nClearly, for $a... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | a = 0 | |
0hc2 | Find all positive integers $n$, and prime numbers $p, q$, which satisfy the equation
$$
n^3 = p^3 + 2p^2q + 2pq^2 + q^3.
$$ | [
"We rewrite the equation as follows.\n$$\nn^3 = p^3 + 2p^2q + 2pq^2 + q^3 = (p+q)(p^2 + pq + q^2)\n$$\nClearly, $p \\neq q$, because, otherwise, the equation $n^3 = 6p^3$ would hold, which is not possible for positive integers.\nSuppose there exists such $s > 1$, which is a factor of both $p+q$ and $(p^2 + pq + q^2... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Inverses ... | English | proof and answer | No solutions | |
0js8 | Problem:
Kristoff is planning to transport a number of indivisible ice blocks with positive integer weights from the north mountain to Arendelle. He knows that when he reaches Arendelle, Princess Anna and Queen Elsa will name an ordered pair $(p, q)$ of nonnegative integers satisfying $p+q \leq 2016$. Kristoff must th... | [
"Solution:\n\nThe answer is 18.\nFirst, we will show that Kristoff must carry at least 18 ice blocks. Let\n$$\n0 < x_{1} \\leq x_{2} \\leq \\cdots \\leq x_{n}\n$$\nbe the weights of ice blocks he carries which satisfy the condition that for any $p, q \\in \\mathbb{Z}_{\\geq 0}$ such that $p+q \\leq 2016$, there are... | United States | HMMT February 2016 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 18 | |
04gb | All vertices of the pentagon $ABCDE$ lie on the same circle. If $\angle CAD = 50^\circ$, determine
$$
\angle ABC + \angle AED.
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | English | proof and answer | Not uniquely determined from the given information | |
02bp | Problem:
Dois quadrados - Na figura ao lado, a área do quadrado maior é $10~\mathrm{cm}^2$ e do menor é $4~\mathrm{cm}^2$. As diagonais do quadrado maior contêm as diagonais do quadrado menor. Quanto mede a área da região tracejada?
 | [
"Solution:\n\nObservemos que a área do quadrado maior menos a área do quadrado menor é igual a 4 vezes a área procurada. Logo a área tracejada é\n$$\n\\frac{10^2-4^2}{4}=\\frac{100-16}{4}=25-4=21\n$$"
] | Brazil | null | [
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof and answer | 21 | |
08hr | Problem:
Fie $n \in \mathbb{N}$. Find all the primitives of the function
$$
f: \mathbb{R} \rightarrow \mathbb{R}, \quad f(x)=\frac{x^{3}-9 x^{2}+29 x-33}{\left(x^{2}-6 x+10\right)^{n}}
$$
Problem:
Let $n \in \mathbb{N}$. Find all the primitives of the function
$$
f: \mathbb{R} \rightarrow \mathbb{R}, \quad f(x)=\fra... | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | final answer only | Let t = x^2 - 6x + 10.
For n ≠ 1, 2:
F(x) = (1/2) [ t^(2 - n) / (2 - n) + t^(1 - n) / (1 - n) ] + C.
For n = 1:
F(x) = (1/2) t + (1/2) ln t + C.
For n = 2:
F(x) = (1/2) ln t - 1 / (2 t) + C. | |
045y | There are some direct one-way flights among $n \ge 8$ airports. Between any two airports $a$ and $b$, there is at most one direct one-way flight from $a$ to $b$ (it is possible to have direct one-way flights both from $a$ to $b$ and from $b$ to $a$). Suppose that, for any set $A$ consisting of some airports with $1 \le... | [
"*Proof.* We represent each airport with a point. If there is a one-way direct flight from airport *a* to airport *b*, we draw a directed edge $ab$, thus obtaining a directed graph *G*. Let $r = \\lfloor \\sqrt{\\frac{n}{2}} \\rfloor$, then $r \\ge 2$. We will prove that for any vertex *v*, there exists a directed ... | China | Chinese Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
057q | Gandalf the Wizard added to his arsenal of magic a new trick in which he simultaneously turns each integer into some integer different from it. Call an integer *a reflecting* if, for every integer $x$, the numbers $x$ and $a - x$ are turned into integers equal to each other. Is it possible that:
a. Numbers $1001$ and ... | [
"For every integer $x$, denote by $G(x)$ the number into which Gandalf turns the number $x$.\n\na. Suppose that both $1001$ and $1003$ are reflecting. Then, for every integer $x$, we have $G(x+2) = G(1003 - (x+2)) = G(1001 - x) = G(x)$. Hence Gandalf turns all even numbers into one and the same integer $c$ and all ... | Estonia | Open Contests | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Functional equations"
] | null | proof and answer | a: no; b: no; c: yes | |
011o | Problem:
There is a frog jumping on a $2k \times 2k$ chessboard, composed of unit squares. The frog's jumps are of length $\sqrt{1+k^{2}}$ and they carry the frog from the center of a square to the center of another square. Some $m$ squares of the board are marked with an $x$, and all the squares into which the frog c... | [
"Solution:\n\nLabel the squares by pairs of integers $(i, j)$ where $1 \\leqslant i, j \\leqslant 2k$. Let $L$ be the set of all such pairs. Define a function $f: L \\rightarrow L$ by\n$$\nf(i, j)= \\begin{cases}\n(i+1, j+k) & \\text{ for } i \\text{ odd and } j \\leqslant k \\\\\n(i-1, j+k) & \\text{ for } i \\tex... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0hdm | Given positive integers $a$, $b$, $n$ such that $a + b = n - 1$. In some school each student has at most $n$ friends from this school. Prove that one can split all the students from this school into two groups: $A$ and $B$, in such a way that every student from the group $A$ will know at most $a$ students from the grou... | [
"Consider a graph, where nodes represent pupils of this school and two nodes are connected by an edge if corresponding pupils are friends. Among all possible partitions of nodes on two sets $A$ and $B$ we choose a partition where $S = b \\cdot S_A + a \\cdot S_B$ is the smallest, where $S_A$ and $S_B$ denote the am... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
049a | Let $a$, $b$, $c$ be different real numbers none of which is zero. Consider quadratic equations:
$$
ax^2 + bx + c = 0, \quad bx^2 + cx + a = 0, \quad cx^2 + ax + b = 0.
$$
If $\frac{c}{a}$ is a root of the first equation, prove that all three of them have a common root. What is the product of the other roots of those e... | [] | Croatia | Hrvatska 2011 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | 1 | |
04e3 | Let $ABCD$ be a trapezium with acute angles along the base $\overline{AB}$ and perpendicular diagonals which intersect at $O$. Ray $OA$ intersects the circle with diameter $\overline{BD}$ at $M$, and ray $OB$ intersects the circle with diameter $\overline{AC}$ at $N$. Prove that the points $M$, $N$, $C$ and $D$ lie on ... | [
"Since $ABCD$ is a trapezium, the triangles $ABO$ and $CDO$ are similar and we have $\\frac{|OA|}{|OB|} = \\frac{|OC|}{|OD|}$.\n\n\n\nEuclid's theorem gives us $|OM|^2 = |OB| \\cdot |OD|$, $|ON|^2 = |OA| \\cdot |OC|$, so\n$$\n\\frac{|OM|^2}{|ON|^2} = \\frac{|OB| \\cdot |OD|}{|OA| \\cdot |OC... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
02jl | Problem:
Quantos números entre $10$ e $13000$, quando lidos da esquerda para a direita, são formados por algarismos consecutivos e em ordem crescente? Por exemplo, $456$ é um desses números, mas $7890$ não é.
A) $10$
B) $13$
C) $18$
D) $22$
E) $25$ | [
"Solution:\n\nOs números em questão são:\n- com $2$ algarismos: $12, 23, 34, 45, \\ldots, 89$ ($8$ números),\n- com $3$ algarismos: $123, 234, 345, \\ldots, 789$ ($7$ números),\n- com $4$ algarismos: $1234, 2345, \\ldots, 6789$ ($6$ números)\ne, por fim,\n- com $5$ algarismos: $12345$, um total de $8+7+6+1=22$ núme... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | MCQ | D | |
00it | Let $p_1, p_2, \dots, p_{42}$ be 42 pairwise different primes. Prove that the number
$$
\sum_{j=1}^{42} \frac{1}{p_j^2 + 1}
$$
cannot be equal to the reciprocal $\frac{1}{n^2}$ of a perfect square. | [
"We assume that the sum in question can be written as the reciprocal of a perfect square $n^2$. Let $P := \\prod_{j=1}^{42} (p_j^2 + 1)$ be the product of all denominators of the summed fractions. We then have\n$$\n\\sum_{j=1}^{42} \\frac{1}{p_j^2 + 1} = \\frac{1}{n^2} \\iff n^2 \\cdot \\sum_{j=1}^{42} \\frac{P}{p_... | Austria | AustriaMO2011 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof only | null | |
0ip3 | Problem:
Each square of a $100 \times 100$ grid is colored black or white so that there is at least one square of each color. Prove that there is a point which is a vertex of exactly one black square. | [
"Solution:\n\nLocate the uppermost row that has at least one black square. Then, within that row, find the leftmost black square. By construction, all squares above and/or to the left of that square are white. Therefore, the upper left corner of that square will solve the problem."
] | United States | Berkeley Math Circle Monthly Contest 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
03q9 | Find all integers $n$, such that $n^4 + 6n^3 + 11n^2 + 3n + 31$ is a perfect square. | [
"Suppose $A = n^4 + 6n^3 + 11n^2 + 3n + 31$ is a perfect square. It means that $A = (n^2 + 3n + 1)^2 - 3(n - 10)$ is a perfect square.\n\nIf $n > 10$, then $A < (n^2 + 3n + 1)^2$, thus $A \\le (n^2 + 3n)^2$.\n\nTherefore\n$$\n(n^2 + 3n + 1)^2 - (n^2 + 3n)^2 \\le 3n - 30,\n$$\nor\n$$\n2n^2 + 3n + 31 \\le 0,\n$$\nwhi... | China | China Western Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 10 | |
0aqa | Problem:
A circle of radius $2~\mathrm{cm}$ is inscribed in $\triangle ABC$. Let $D$ and $E$ be the points of tangency of the circle with the sides $AC$ and $AB$, respectively. If $\angle BAC = 45^\circ$, find the length of the minor arc $DE$. | [] | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | π/2 cm | |
02z6 | Problem:
Existem três cartões, cada um com um número do conjunto $\{1,2, \ldots, 10\}$. Esses três cartões foram embaralhados e distribuídos a três pessoas, que registraram os números em seus respectivos cartões. Os cartões foram então coletados e o processo foi repetido novamente. Após algumas repetições, cada uma da... | [
"Solution:\n\nSejam $x$, $y$ e $z$ os números escritos nos três cartões, com $x \\leq y \\leq z$. A cada etapa do processo de sorteio, a soma dos números dos três cartões é sempre $x+y+z$. Como $13+15+23=51=3 \\cdot 17$ e tanto $3$ quanto $17$ são números primos, segue que foram realizados $3$ sorteios e $x+y+z=17$... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 3, 5, 9 | |
0hp5 | Problem:
Let $ABC$ be a triangle. A circle is tangent to segments $BC$, $CA$, $AB$ at points $D$, $E$, $F$, respectively. Given that the measures of $\angle CAB$, $\angle ABC$, $\angle BCA$ form an arithmetic progression in some order, prove that the measures of $\angle FDE$, $\angle DEF$, $\angle EFD$ also form an ar... | [
"Solution:\n\nThe main observation is that the three angles of a triangle form an arithmetic progression if and only if one of the angles is $60^{\\circ}$. Indeed, suppose that a triangle has angles $x \\leq y \\leq z$ in arithmetic progression. Then $y = \\frac{x + y + z}{3} = 60^{\\circ}$. Conversely, if a triang... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0i7t | Problem:
A calculator has a display, which shows a nonnegative integer $N$, and a button, which replaces $N$ by a random integer chosen uniformly from the set $\{0,1, \ldots, N-1\}$, provided that $N>0$. Initially, the display holds the number $N=2003$. If the button is pressed repeatedly until $N=0$, what is the proba... | [
"Solution:\nFirst, we claim that if the display starts at some $N$, the probability that any given number $M<N$ will appear at some point is $1/(M+1)$. We can show this by induction on $N$.\n\nIf $N = M+1$ (the base case), $M$ can only be reached if it appears after the first step, and this occurs with probability ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 1/2224222 | |
0f7a | Problem:
Prove that we can find an $m \times n$ array of squares so that the sum of each row and the sum of each column is also a square. | [] | Soviet Union | 20th ASU | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0jes | Problem:
Find the number of positive integers $j \leq 3^{2013}$ such that
$$
j=\sum_{k=0}^{m}\left((-1)^{k} \cdot 3^{a_{k}}\right)
$$
for some strictly increasing sequence of nonnegative integers $\left\{a_{k}\right\}$. For example, we may write $3=3^{1}$ and $55=3^{0}-3^{3}+3^{4}$, but 4 cannot be written in this form... | [
"Solution:\n$2^{2013}$\n\nClearly $m$ must be even, or the sum would be negative. Furthermore, if $a_{m} \\leq 2013$, the sum cannot exceed $3^{2013}$ since $j=3^{a_{m}}+\\sum_{k=0}^{m-1}\\left((-1)^{k} \\cdot 3^{a_{k}}\\right) \\leq 3^{a_{m}}$. Likewise, if $a_{m}>2013$, then the sum necessarily exceeds $3^{2013}$... | United States | HMMT 2013 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Other"
] | null | proof and answer | 2^{2013} | |
0f94 | Problem:
A positive integer $n$ has exactly $12$ positive divisors $1 = d_1 < d_2 < d_3 < \ldots < d_{12} = n$. Let $m = d_4 - 1$. We have $d_m = (d_1 + d_2 + d_4) d_8$. Find $n$. | [] | Soviet Union | 23rd ASU | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1989 | |
00dm | Let $N(a, b)$ be the number of ways to cover an $a \times b$ board using domino tiles. Additionally, let $N^*(a, 2b + 1)$ be the number of ways to cover an $a \times (2b+1)$ board using domino tiles, without having vertical dominoes in the central column. Prove that $N^*(2m, 2n + 1) = 2^m N(2m, n)N(2m, n - 1)$. | [
"Suppose the board is colored like a chessboard. First let's establish a bit of notation. A *cycle* is a sequence of cells $c_1, \\dots, c_k$ such that for all $i$ we have that $c_i$ and $c_{i+1}$ share one side (where $c_{k+1} = c_1$). There are two possible covers of a cycle by dominoes, one that joins each black... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics ... | English | proof only | null | |
0b7f | Let $n$ be a positive integer and let $x_1, x_2, \dots, x_n$ be positive real numbers such that $x_1x_2 \cdots x_n = 1$. Prove that
$$
\sum_{i=1}^{n} x_{i}^{n}(1 + x_{i}) \geq \frac{n}{2^{n-1}} \prod_{i=1}^{n}(1 + x_{i}).
$$ | [
"By the power-mean inequality,\n$$\n1 + a^n \\geq \\frac{(1 + a)^n}{2^{n-1}}, \\quad a \\geq 0. \\qquad (*)\n$$\nThus,\n$$\n\\begin{align*}\n\\sum_{i=1}^{n} x_i^n (1 + x_i) &= \\sum_{i=1}^{n} x_i^n + \\sum_{i=1}^{n} x_i^{n+1} \\\\\n&\\geq \\sum_{i=1}^{n} x_i^n + n \\left( \\prod_{i=1}^{n} x_i \\right)^{1+1/n} \\qua... | Romania | NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
05u7 | Problem:
Soit $f: \mathbb{R} \rightarrow \mathbb{R}$ une fonction croissante telle que $f(y)-f(x)<y-x$ pour tous les réels $x$ et $y$ tels que $y>x$. Soit $\left(u_{n}\right)_{n \geqslant 0}$ une suite telle que $u_{n+2}=f\left(u_{n+1}\right)-f\left(u_{n}\right)$ pour tout entier $n \geqslant 0$.
Démontrer que la sui... | [
"Solution:\n\nQuitte à remplacer la fonction $f$ par la fonction $t \\mapsto f(t)-f(0)$, ce qui ne change rien à la propriété désirée, on suppose que $f(0)=0$.\n\nPar ailleurs, remplacer la fonction $f$ par la fonction $t \\mapsto -f(-t)$ et la suite $\\left(u_{n}\\right)_{n \\geqslant 0}$ par la suite $\\left(-u_{... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
00y9 | Problem:
Does there exist a triangle such that the lengths of all its sides and altitudes are integers and its perimeter is equal to $1995$? | [
"Solution:\n\nConsider a triangle $ABC$ with all its sides and heights having integer lengths. From the cosine theorem we conclude that $\\cos \\angle A$, $\\cos \\angle B$ and $\\cos \\angle C$ are rational numbers. Let $AH$ be one of the heights of the triangle $ABC$, with the point $H$ lying on the straight line... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | No | |
08b1 | Problem:
Una sequenza $a_{1}, \ldots, a_{100}$ di numeri reali è tale che la media aritmetica fra due termini consecutivi sia sempre uguale all'indice del secondo termine (ad esempio, si ha $\frac{a_{4}+a_{5}}{2}=5$ ); quanto vale la somma dei 100 numeri della sequenza?
(A) 2550
(B) 5050
(C) 5100
(D) 10100
(E) Non si... | [
"Solution:\n\nLa risposta è (C). Indichiamo con $S$ la somma dei 100 termini della successione. Si ha:\n$$\n\\frac{S}{2}=\\frac{1}{2}\\left(a_{1}+a_{2}+\\ldots+a_{99}+a_{100}\\right)=\\frac{a_{1}+a_{2}}{2}+\\frac{a_{3}+a_{4}}{2}+\\ldots+\\frac{a_{99}+a_{100}}{2}.\n$$\nOsserviamo che ciascuno degli addendi del membr... | Italy | Progetto Olimpiadi della Matematica | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | C | |
07y7 | Problem:
1. Ad un test di matematica partecipano $N<40$ persone. La sufficienza è fissata a 65. I risultati del test sono i seguenti: la media di tutti i partecipanti è 66, quella dei promossi 71 e quella dei bocciati 56. Tuttavia, a causa di un errore nella formulazione di un quesito, tutti i punteggi vengono aumenta... | [
"Solution:\n\n(a) Sia $P_{1}$ il numero dei promossi prima dell'incremento e $P_{2}$ il numero di promossi dopo l'incremento di punteggio. Dalle informazioni che abbiamo possiamo scrivere:\n$$\n66 N = 71 P_{1} + 56 (N - P_{1}), \\quad 71 N = 75 P_{2} + 59 (N - P_{2})\n$$\nDalla prima relazione, svolgendo i conti, o... | Italy | null | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | a) N = 12, 24, 36; b) no possible N | |
0ab6 | How many numbers divisible by $30^{2008}$ are not divisible by $20^{2007}$? | [
"Since $30^{2008} = 2^{2008} \\cdot 3^{2008} \\cdot 5^{2008}$ and $20^{2007} = 2^{4014} \\cdot 5^{2007}$, all the numbers divisible by $30^{2008}$ and not divisible by $20^{2007}$ are:\n\n1) $2^{k} \\cdot 3^{l} \\cdot 5^{m}$, $l = 1, 2, \\ldots, 2008$, $k, m = 0, 1, 2, \\ldots, 2008$ or $2008 \\cdot 2009^{2}$ numbe... | North Macedonia | Macedonian Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 2008*2009^2 + 2009 | |
0e58 | Let $ABCD$ be a square and let $E$ and $F$ be two points outside the square, such that $BEC$ and $CFD$ are equilateral triangles. Let $G$ be the intersection of the lines $BE$ and $FD$, and let $H$ be a point, such that the quadrilateral $CEHF$ is a rhombus. Prove that the points $G$, $E$, $H$ and $F$ lie on the same c... | [
"Since $ABCD$ is a square and the triangles $BEC$ and $CFD$ are equilateral we have $|CE| = |CB| = |CD| = |CF|$. The triangle $FCE$ is isosceles and\n$$\n\\begin{aligned}\n\\angle ECF &= 2\\pi - \\angle FCD - \\angle DCB - \\angle BCE \\\\\n&= 2\\pi - \\frac{\\pi}{3} - \\frac{\\pi}{2} - \\frac{\\pi}{3} = \\frac{5\\... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0g2p | Problem:
Trouver le plus grand nombre premier $p$ tel qu'il existe des nombres entiers strictement positifs $a$ et $b$ tels que
$$
p=\frac{b}{2} \sqrt{\frac{a-b}{a+b}}
$$ | [
"Solution:\n\nPremière solution: (Louis)\n\nLa forme du problème semble indiquer qu'il faudra jouer avec des histoires de divisibilité. Dans de tels cas il est généralement utile de définir $d=\\operatorname{ggT}(a, b)$ et d'écrire ainsi $a=d a^{\\prime}, b=d b^{\\prime}$ avec $a^{\\prime}, b^{\\prime}$ premiers en... | Switzerland | Selektion | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic fun... | null | proof and answer | 5 | |
063h | Problem:
Es sei $a_{1} \leqslant a_{2} \leqslant \ldots$ eine monoton steigende Folge positiver ganzer Zahlen. Eine positive ganze Zahl $n$ heißt verlässlich, wenn es einen positiven ganzzahligen Index $i$ mit $n=\frac{i}{a_{i}}$ gibt.
Man beweise: Wenn 2013 verlässlich ist, dann ist auch 20 verlässlich. | [
"Solution:\n\nWenn 2013 verlässlich ist, gibt es einen positiven ganzzahligen Index $i$ mit $i=2013 a_{i} \\geqslant 20 a_{i}$. Insbesondere ist die Menge $S$ aller positiven ganzen Zahlen $s$, für die $s \\geqslant 20 a_{s}$ gilt, nicht leer. Somit enthält $S$ ein kleinstes Element $j$, und dieses erfüllt einersei... | Germany | Germany TST | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0dw0 | Problem:
Členi neskončnega geometrijskega zaporedja so naravna števila, od katerih vsaj dve nista deljivi s $4$. Zapiši splošni člen tega zaporedja, če veš, da je eden izmed členov enak $2004$. | [
"Solution:\n\nZaporedje je oblike $a_{n} = a \\cdot r^{n}$, $n \\geq 0$. Najprej dokažemo, da so vsi členi zaporedja cela števila. Ker je $r = \\frac{a_{n}}{a_{n-1}}$, je $r$ racionalno število. Zapišimo $r = \\frac{\\alpha}{\\beta}$, $\\beta > 0$, kjer sta si števili $\\alpha$ in $\\beta$ tuji. Dokazati je potrebn... | Slovenia | 48. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | a_n = 501 * 2^n | |
04ub | Let $ABC$ be a triangle. The $A$-angle bisector intersects $BC$ at $D$. Let $E$, $F$ be the circumcenters of triangles $ABD$, $ACD$, respectively. Given that the circumcenter of triangle $AEF$ lies on $BC$, find all possible values of $\angle BAC$. (Patrik Bak) | [
"Let $O$ be the circumcenter of triangle $AEF$ and denote $\\alpha = \\angle BAC$. Since $\\angle BAD$ and $\\angle CAD$ are acute (Fig. 2), points $E$, $F$ lie in the half-plane $BCA$ and the Inscribed angle theorem yields\n$$\n\\angle BED = 2 \\cdot \\angle BAD = \\alpha = 2 \\cdot \\angle DAC = \\angle DFC.\n$$\... | Czech Republic | 67th Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 120° | |
0drv | Let $n$ be a prime number. Show that there is a permutation $a_1, a_2, \dots, a_n$ of $1, 2, \dots, n$ so that $a_1, a_1a_2, \dots, a_1a_2 \dots a_n$ leave distinct remainders when divided by $n$. | [
"By the Chinese remainder Theorem, for every $k=2,3,\\dots,n$ there exists $b_k$ so that\n$$\nb_k \\equiv 0 \\pmod{(k-1)}, \\quad b_k \\equiv k \\pmod{n}.\n$$\nLet $a_1 = 1$ and for $k = 2, \\dots, n$, $a_k$ is the remainder when $b_k/(k-1)$ is divided by $n$. Since $b_n \\equiv 0 \\pmod{n}$, we have $a_n = 0$. Als... | Singapore | Singapur | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0j7k | Problem:
Determine which of the following numbers is smallest in value: $54 \sqrt{3}$, $144$, $108 \sqrt{6} - 108 \sqrt{2}$. | [
"Solution:\n$54 \\sqrt{3}$\n\nWe can first compare $54 \\sqrt{3}$ and $144$. Note that $\\sqrt{3} < 2$ and $\\frac{144}{54} = \\frac{8}{3} > 2$. Hence, $54 \\sqrt{3}$ is less.\n\nNow, we wish to compare this to $108 \\sqrt{6} - 108 \\sqrt{2}$. This is equivalent to comparing $\\sqrt{3}$ to $2(\\sqrt{6} - \\sqrt{2})... | United States | Harvard-MIT November Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 54 \sqrt{3} | |
0l84 | A tetrahedron $ABCD$ is said to be *angelic* if it has nonzero volume and satisfies
$$
\angle BAC + \angle CAD + \angle DAB = \angle ABC + \angle CBD + \angle DBA, \\
\angle ACB + \angle BCD + \angle DCA = \angle ADB + \angle BDC + \angle CDA.
$$
Across all angelic tetrahedrons, what's the maximum number of distinct le... | [] | United States | USA TST Selection Test for 67th IMO and 15th EGMO | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 3 | |
0hgs | Which number is larger, $A = \frac{1}{9} : \sqrt[3]{\frac{1}{2023}}$ or $B = \log_{2023} 91125$? | [
"To prove this, we will show that the following inequalities hold: $A < \\frac{3}{2} < B$.\n\n$$\nA = \\frac{1}{9} : \\sqrt[3]{\\frac{1}{2023}} = \\frac{\\sqrt[3]{2023}}{9} < \\frac{13}{9} < \\frac{3}{2}, \\quad B = \\log_{2023} 91125 > \\log_{2023} 45^3 = \\log_{45^2} 45^3 = \\frac{3}{2}.\n$$"
] | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | B | |
063m | Problem:
Es sei $ABC$ ein spitzwinkliges Dreieck mit $|AB| \neq |AC|$. Die Mittelpunkte der Seiten $\overline{AB}$ und $\overline{AC}$ seien $D$ beziehungsweise $E$. Die Umkreise der Dreiecke $BCD$ und $BCE$ mögen den Umkreis des Dreiecks $ADE$ in $P$ beziehungsweise $Q$ schneiden, wobei $P \neq D$ und $Q \neq E$.
Man... | [
"Solution:\n\nOhne Einschränkung sei $|AC| > |AB|$. Da $D$ und $E$ Mitten der Seiten $\\overline{AB}$ und $\\overline{AC}$ sind, ist nach Strahlensatz $DE$ parallel zu $BC$. Im Falle $E = P$ oder $D = Q$ wäre $CEDB$ Sehnenviereck mit parallelen Seiten $BC$ und $DE$, also gleichschenkliges Trapez mit $|BD| = |EC|$, ... | Germany | 1. Auswahlklausur 2014/2015 | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
079o | Find all natural numbers $n \ge 2$ such that for every pair of integers $i, j \in [0, n]$, $i + j$ and $\begin{pmatrix} n \\ i \end{pmatrix} + \begin{pmatrix} n \\ j \end{pmatrix}$ have the same parity. | [
"Lemma. Suppose $n \\ge 2$ is an integer, then all of the numbers $\\begin{pmatrix} n \\\\ 0 \\end{pmatrix}, \\begin{pmatrix} n \\\\ 1 \\end{pmatrix}, \\dots, \\begin{pmatrix} n \\\\ n \\end{pmatrix}$ are odd if and only if $n = 2^k - 1$ for an integer $k \\ge 2$.\n\n*Proof*. For an integer $t$, let $v(t)$ be the g... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | All n of the form 2^k - 2 for integers k >= 2 | |
0hwo | Problem:
Show that for sufficiently large primes $p$, there is an Eulerian circuit on the complete graph with $p$ vertices that does not contain any cycles of length at most $2023$. | [
"Solution:\nTake a generator $g \\pmod{p}$ so that $g$ is not equivalent to anything of the form $-a / b$ for integers $a, b \\leq 2023$. For big enough primes, such a $g$ must exist as there are at most a constant number of such fractions.\n\nNow number the vertices with the residues $\\bmod\\ p$. For any residue ... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Discrete Mathematics > Other",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n"
] | null | proof only | null | |
02bb | Problem:
Expressões algébricas - $\mathrm{O}$ que representam, geometricamente, na figura dada, as expressões
$$
a^{2}+1,5 a
$$
e
$$
4 a+3
$$
 | [
"Solution:\n\nNote que a figura é um retângulo formado por um quadrado de lado $a$ e um retângulo de lados $1{,}5$ e $a$. Logo, $a^{2}$ é a área do quadrado e $1{,}5 a$ é a área do retângulo. Assim, $a^{2}+1{,}5 a$ representa a soma dessas duas áreas, ou seja, a área total da figura.\n\nJá $4 a+3=3 a+1{,}5+a+1{,}5$... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Quadrilaterals"
] | null | final answer only | a^2 + 1.5a is the total area of the figure; 4a + 3 is the perimeter of the figure. | |
00ms | Let $x$ and $y$ be integers for which $x + y \neq 0$ holds. Determine all pairs $(x, y)$ satisfying
$$
\frac{x^2 + y^2}{x + y} = 10.
$$ | [
"$$\n(x, y) \\in \\{(-2, 4), (-2, 6), (0, 10), (4, -2), (4, 12), (6, -2), (6, 12), (10, 0), (10, 10), (12, 4), (12, 6)\\}.\n$$\n\nAn equivalent form of the given equation is\n$$\n\\begin{align*}\nx^2 + y^2 &= 10x + 10y \\\\\n\\iff x^2 - 10x + y^2 - 10y &= 0 \\\\\n\\iff (x - 5)^2 + (y - 5)^2 &= 50\n\\end{align*}\n$$... | Austria | AUT_ABooklet_2020 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | {(-2, 4), (-2, 6), (0, 10), (4, -2), (4, 12), (6, -2), (6, 12), (10, 0), (10, 10), (12, 4), (12, 6)} | |
0j16 | Problem:
For $0 \leq y \leq 2$, let $D_{y}$ be the half-disk of diameter $2$ with one vertex at $(0, y)$, the other vertex on the positive $x$-axis, and the curved boundary further from the origin than the straight boundary. Find the area of the union of $D_{y}$ for all $0 \leq y \leq 2$. | [
"\n\nFrom the picture above, we see that the union of the half-disks will be a quarter-circle with radius $2$, and therefore area $\\pi$. To prove that this is the case, we first prove that the boundary of every half-disk intersects the quarter-circle with radius $2$, and then that the half... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | π | |
0hxt | Problem:
A cube with sides $1$ m in length is filled with water, and has a tiny hole through which the water drains into a cylinder of radius $1$ m. If the water level in the cube is falling at a rate of $1~\mathrm{cm}/\mathrm{s}$, at what rate is the water level in the cylinder rising? | [
"Solution:\nThe magnitude of the change in volume per unit time of the two solids is the same. The change in volume per unit time of the cube is $1~\\mathrm{cm} \\cdot \\mathrm{m}^2 / \\mathrm{s}$. The change in volume per unit time of the cylinder is $\\pi \\cdot \\frac{d h}{d t} \\cdot m^2$, where $\\frac{d h}{d ... | United States | HMMT | [
"Calculus > Differential Calculus > Related Rates"
] | null | final answer only | 1/pi cm/s | |
0efv | Problem:
Naj bo $n$ naravno število. Poišči vsa realna števila $x$, ki rešijo enačbo
$$
2^{n}(-x)^{n}+(-1)^{3 n+1} 2^{n+1} x^{n-1}(2 x+1)-(-2 x)^{n+1}=0
$$ | [
"Solution:\n\nOpazimo, da je $(-1)^{3 n+1}=(-1)^{2 n}(-1)^{n+1}=(-1)^{n+1}$, zato lahko enačbo preoblikujemo do\n$$\n2^{n}(-1)^{n} x^{n}+(-1)^{n+1} 2^{n+1} x^{n-1}(2 x+1)-(-1)^{n+1} 2^{n+1} x^{n+1}=0\n$$\nNa levi strani enačbe izpostavimo skupni faktor, da dobimo\n$$\n(-1)^{n} 2^{n} x^{n-1}\\left(x-2(2 x+1)+2 x^{2}... | Slovenia | Slovenian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | If n = 1: x = 2 or x = −1/2. If n ≥ 2: x = 0, x = 2, or x = −1/2. | |
07ns | On the sides $AB$ and $AC$ of triangle $ABC$ the triangles $AEB$ and $ADC$ are constructed, both similar to $\triangle ABC$ and with $\angle AEB = \angle ADC = \angle BAC$. Prove that the area of $\triangle ABC$ is less than, equal to or greater than the sum of the areas of triangles $AEB$ and $ADC$ according as $\angl... | [
"W.l.o.g we can arrange that $\\angle ABE = \\angle ABC$ and $\\angle ACD = \\angle ACB$. Let $D'$ and $E'$ be the reflections of $D$ and $E$ in $AC$ and $AB$, respectively. Then $D'$ and $E'$ are on $BC$, $\\angle D'AC = \\angle DAC = \\angle ABC$ and $\\angle E'AB = \\angle EAB = \\angle BCA$.\n\nIf $\\angle BAC ... | Ireland | Ireland | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0bfl | Prove that the sum of the distances from the barycenter of a tetrahedron to its faces is at least $4r$, where $r$ is the radius of the sphere inscribed in the tetrahedron. | [] | Romania | Shortlisted Problems for the 64th NMO | [
"Geometry > Solid Geometry > Volume",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
076k | Let $\mathbb{N}$ denote the set of all natural numbers. Define a function $T: \mathbb{N} \to \mathbb{N}$ by $T(2k) = k$ and $T(2k + 1) = 2k + 2$. We write $T^2(n) = T(T(n))$ and in general $T^k(n) = T^{k-1}(T(n))$ for any $k > 1$.
(i) Show that for each $n \in \mathbb{N}$, there exists $k$ such that $T^k(n) = 1$.
(ii) ... | [
"(i) For $n = 1$, we have $T(1) = 2$ and $T^2(1) = T(2) = 1$. Hence we may assume that $n > 1$.\nSuppose $n > 1$ is even. Then $T(n) = n/2$. We observe that $(n/2) \\le n - 1$ for $n > 1$.\nSuppose $n > 1$ is odd so that $n \\ge 3$. Then $T(n) = n + 1$ and $T^2(n) = (n + 1)/2$. Again we see that $(n + 1)/2 \\le (n ... | India | IND_National | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0io6 | Problem:
Compute
$$
\sum_{n=1}^{\infty} \frac{1}{n \cdot (n+1) \cdot (n+1)!}
$$ | [
"Solution:\nAnswer: $3-e$. We write\n$$\n\\begin{gathered}\n\\sum_{n=1}^{\\infty} \\frac{1}{n \\cdot (n+1) \\cdot (n+1)!} = \\sum_{n=1}^{\\infty} \\left(\\frac{1}{n} - \\frac{1}{n+1}\\right) \\frac{1}{(n+1)!} = \\sum_{n=1}^{\\infty} \\frac{1}{n \\cdot (n+1)!} - \\sum_{n=1}^{\\infty} \\frac{1}{(n+1) \\cdot (n+1)!} \... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 3 - e | |
0gzp | Let's designate through $P(n)$ the product of digits of the integer non-negative number $n$. Prove that sets $A$ and $B$ are unbounded, where:
a. $A = \left\{ \frac{P(n)}{P(n^2)} \right\}$, where $n$ belongs to the set of such whole non-negative numbers that the number $n^2$ does not contain zero in the decimal record... | [
"Both points are proved with the help of corresponding examples which are in turn proved by a method of mathematical induction.\n\na.\nLet's consider the equality:\n$$\n(2\\underbrace{66\\dots68}_{n-1})^2 = 7\\underbrace{11\\dots18}_{n-1}\\underbrace{22\\dots24}_{n-1},\n$$\n\nFurther, it is enough to calculate the ... | Ukraine | The Problems of Ukrainian Authors | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null |
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