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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
01x8 | The sum of several (not necessarily different) real numbers from $[0, 1]$ does not exceed $S$.
Find the maximal value of $S$ such that it is always possible to partition these numbers into two groups with sums not greater than $9$. | [
"Answer: $\\max S = 17.1$.\n\nFirst we will show that if $S > 17.1$, the required partition can be impossible. Indeed, let $S = 17.1 + 19\\epsilon$, $\\epsilon > 0$. Suppose that we are given $19$ numbers equal to $0.9 + \\epsilon$. It is evident that for any partition some of the groups will contain at least $10$ ... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 17.1 | |
094o | Problem:
We are given a convex quadrilateral $A B C D$ whose angles are not right. Assume there are points $P, Q, R, S$ on its sides $A B, B C, C D, D A$, respectively, such that $P S \| B D$, $S Q \perp B C$, $P R \perp C D$. Furthermore, assume that the lines $P R, S Q$, and $A C$ are concurrent. Prove that the poin... | [
"Solution:\n\nLet the intersection point of $P R, Q S, A C$ be $T$ and let $H$ be the orthocenter of $B C D$. Since $\\angle B C D$ is not right, $H \\neq C$. Notice that triangles $H B D$ and $T P S$ are homothetic due to their corresponding sides being parallel. This means that $H T, B P, D S$ are concurrent. Sin... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous ... | null | proof only | null | |
04tg | **Find all functions $f: \mathbb{N} \to \mathbb{N}$ such that for every positive integer $m$ the following is true: If we denote by $d_1, d_2, \dots, d_n$ all the divisors of number $m$ then**
$$
f(d_1) \cdot f(d_2) \cdots f(d_n) = m.
$$ | [
"We will show that the only solution is a function $f$ such that\n$$\nf(m) = \\begin{cases} p, & \\text{if } m \\text{ is a non-trivial power of a prime } p, \\text{ i.e. } m = p^k, k \\ge 1, \\\\ 1, & \\text{otherwise.} \\end{cases}\n$$\nNumber $1$ has the unique divisor $1$, hence plugging $m = 1$ in the given eq... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | The unique function is given by f(n) = p if n is a positive power of a prime p, and f(n) = 1 otherwise (in particular f(1) = 1). | |
0ii6 | Problem:
Find all ordered triples $(x, y, z)$ of positive reals such that $x + y + z = 27$ and $x^{2} + y^{2} + z^{2} - x y - y z - z x = 0$. Prove that your answer is correct. | [
"Solution:\nWe have $x^{2} + y^{2} + z^{2} - x y - y z - z x = \\frac{(x - y)^{2} + (y - z)^{2} + (z - x)^{2}}{2} = 0$, which implies $x = y = z = \\frac{27}{3} = 9$."
] | United States | Harvard-MIT Mathematics Tournament, Team Round B | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | (9, 9, 9) | |
0f84 | Problem:
Show that there are infinitely many odd composite numbers in the sequence $1^{1}$, $1^{1} + 2^{2}$, $1^{1} + 2^{2} + 3^{3}$, $1^{1} + 2^{2} + 3^{3} + 4^{4}$, ... . | [
"Solution:\nWe show that infinitely many odd numbers in the sequence are divisible by $3$.\n\nIf $n = 14 \\pmod{36}$, then $n = 36m + 14$ for some $m$. So there are $18m + 7$ odd numbers in the sum and $18m + 7$ even numbers. Hence the sum is odd.\n\nThere are $12m + 5$ numbers equal to $1 \\pmod{3}$, $12m + 5$ equ... | Soviet Union | 22nd ASU | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
037g | Problem:
Let $ABC$ be a right triangle and $D$ be a point on the hypotenuse $AB$.
a) Prove that the expression
$$
\frac{AC^{2}}{AD+CD} + \frac{BC^{2}}{BD+CD}
$$
does not depend on $D$.
b) Let $DE$ ($E \in AC$) and $DF$ ($F \in BC$) be the bisectors of $\angle ADC$ and $\angle BDC$, respectively. Find the minimum valu... | [
"Solution:\na. Set $\\angle BAC = \\alpha$ and $\\angle ACD = x$. We have $AC = AB \\cos \\alpha$, $BC = AB \\sin \\alpha$ and it follows by the Sine theorem for $\\triangle ADC$ and $\\triangle BDC$ that\n\n\n\n$$\n\\begin{aligned}\n& \\frac{AC^{2}}{AD+CD} + \\frac{BC^{2}}{BD+CD} = \\frac{... | Bulgaria | Spring Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 1 | |
0ez4 | Problem:
Find four different three-digit numbers (in base 10) starting with the same digit, such that their sum is divisible by three of the numbers. | [
"Solution:\nAnswer: $108$, $117$, $135$, $180$. Sum $540 = 108 \\times 5 = 135 \\times 4 = 180 \\times 3$.\n\nTry looking for a number of the form $3n$, $4n$, $5n$, $n$. We want $12n$, $15n$ and $20n$ to have the same first digit. If the first digit is $1$, this requires $n = 9$. We must now check that the fourth n... | Soviet Union | 3rd ASU | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 108, 117, 135, 180 | |
0k50 | Problem:
Crisp All, a basketball player, is dropping dimes and nickels on a number line. Crisp drops a dime on every positive multiple of $10$, and a nickel on every multiple of $5$ that is not a multiple of $10$. Crisp then starts at $0$. Every second, he has a $\frac{2}{3}$ chance of jumping from his current locatio... | [
"Solution:\n\nLet \"a 3\" mean a move in which Crisp moves from $x$ to $x+3$, and \"a 7\" mean a move in which Crisp moves from $x$ to $x+7$. Note that Crisp stops precisely the first time his number of 3's and number of 7's differs by a multiple of $5$, and that he'll stop on a dime if they differ by $0$, and stop... | United States | HMMT November 2018 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 20/31 | |
0338 | Problem:
Let $ABCD$ be a circumscribed quadrilateral. Find $\Varangle BCD$ if $AC = BC$, $AD = 5$, $E = AC \cap BD$, $BE = 12$ and $DE = 3$. | [
"Solution:\nIf the perpendicular bisector of $CD$ meets $BD$ at point $O$, then\n$$\n\\begin{aligned}\n\\Varangle COD & = 180^\\circ - 2 \\Varangle ODC \\\\\n& = 180^\\circ - 2 \\Varangle BAC \\\\\n& = \\Varangle ACB = \\Varangle ADO\n\\end{aligned}\n$$\nand therefore $AD \\parallel CO$. Hence $\\frac{OE}{3} = \\fr... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | 90° | |
0km3 | Problem:
In a convex $n$-sided polygon, all the diagonals are drawn and no three of them pass through a point. Find a formula for the number of regions formed inside the polygon. | [
"Solution:\n\nTo start with, there is one region inside the $n$-gon. A new region is created each time a new diagonal is drawn or two diagonals intersect. The number of diagonals is $\\binom{n}{2} - n$, since each diagonal corresponds to a pair of non-adjacent vertices. The number of intersections of two diagonals ... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 1 + C(n,2) - n + C(n,4) | |
0jx4 | Problem:
Mr. Taf takes his 12 students on a road trip. Since it takes two hours to walk from the school to the destination, he plans to use his car to expedite the journey. His car can take at most 4 students at a time, and travels 15 times as fast as traveling on foot. If they plan their trip optimally, what is the s... | [
"Solution:\n\nA way to plan the trip is to have Mr. Taf drive 4 students to the $80\\%$ mark, then drive back to the $10\\%$ mark to pick up another 4 students to the $90\\%$ mark, and finally drive back to the $20\\%$ mark to pick up the last 4 students to the destination. All students will reach the destination a... | United States | HMMT November | [
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 30.4 minutes | |
0gga | 設 $n \ge 3$ 為正整數, $p$ 是滿足 $p > 6^{n-1} - 2^n + 1$ 的質數。設 $S$ 為 $n$ 個正整數所組成的集合, 其中任兩數除以 $p$ 的餘數均不相同。證明存在正整數 $c$, 滿足恰有兩組有序數組 $(x, y, z) \in S^3$, 其中的 $x, y, z$ 兩兩不同, 且 $x - y + z - c$ 可被 $p$ 整除。 | [] | Taiwan | 2022 數學奧林匹亞競賽第一階段培訓營, 國際競賽實作 (二) | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Chinese; English | proof only | null | |
02mu | Problem:
Amigo oculto - Um grupo de cinco amigos decide brincar de amigo oculto, cada um compra um presente para seu amigo oculto. Pelas regras do jogo, cada um dá exatamente um presente e recebe exatamente um presente. De quantas maneiras podem os presentes ser distribuídos, de modo que ninguém dê presente para si me... | [
"Solution:\n\nPrimeiramente observemos que o número de formas de distribuir os presentes sem nenhuma restrição é $5! = 5 \\times 4 \\times 3 \\times 2 \\times 1 = 120$. Daí temos que tirar os casos \"ruins\", isto é, os casos em que pelo menos um amigo tirou o seu próprio presente. Esses casos a eliminar podem ser ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 44 | |
0got | Let $m < n$ be positive integers and $p = \frac{n^2 + m^2}{\sqrt{n^2 - m^2}}$.
a. Find three pairs of positive integers $(m, n)$ for which $p$ is a prime number.
b. Show that if $p$ is a prime number, then $p \equiv 1 \pmod 8$. | [
"a. For $(m, n) = (6, 10)$, $(12, 15)$ and $(30, 78)$, $p$ is equal to\n$$\n\\frac{10^2 + 6^2}{8} = 17, \\quad \\frac{15^2 + 12^2}{9} = 41 \\text{ and } \\frac{78^2 + 30^2}{72} = 97.\n$$\nrespectively.\n\nb. Let $k = \\sqrt{n^2 - m^2}$. Then $(k, m, n)$ is a Pythagorean triple. Therefore, there exist positive integ... | Turkey | 16th Junior Turkish Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | Example pairs: (6, 10), (12, 15), (30, 78). Moreover, if p is prime then p ≡ 1 (mod 8). | |
03rv | It is given that real numbers $x_1, x_2, \dots, x_n$ ($n > 2$) satisfy $\left|\sum_{i=1}^n x_i\right| > 1$, $|x_i| \le 1$ ($i=1, 2, \dots, n$). Prove that there exists a positive integer $k$ such that $\left|\sum_{i=1}^k x_i - \sum_{i=k+1}^n x_i\right| \le 1$. (posed by Leng Gangsong) | [
"**Proof** Set $g(0) = -\\sum_{i=1}^{n} x_i$, $g(k) = \\sum_{i=1}^{k} x_i - \\sum_{i=k+1}^{n} x_i$ ($1 \\le k \\le n-1$), $g(n) = \\sum_{i=1}^{n} x_i$.\nThen\n$$|g(1) - g(0)| = 2|x_1| \\le 2,$$\n$$|g(k+1) - g(k)| = 2|x_{k+1}| \\le 2, \\quad k = 1, 2, \\dots, n-2,$$\n$$|g(n) - g(n-1)| = 2|x_n| \\le 2.$$\nSo for each... | China | China Western Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
01on | Four points $A$, $B$, $C$, $D$ are marked on the hyperbola $y = 1/x$ so that the quadrilateral $ABCD$ is a parallelogram ($AB \parallel CD$) and $AB = 2 \cdot BC$.
Find all possible values of the area of $ABCD$. | [
"Answer: $\\dfrac{16}{3}$.\nLet $A(a; 1/a)$, $B(b; 1/b)$, $C(c; 1/c)$, $D(d; 1/d)$ be the marked points (see Fig. 1). Since any vertical and any horizontal line meets the hyperbola $y = 1/x$ at most at one point we see that the numbers $a$, $b$, $c$, $d$ are pairwise distinct. Since the opposite sides of the parall... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 16/3 | |
0chx | Let $A$ be a point in the Cartesian plane. Ann tells Bob a number $0 < a \le 1$ and he then moves $A$ rightward, leftward, upward or downward to a new position $A'$, $a$ distance apart from $A$. Next, Ann tells Bob a number $0 < a' \le 1$ and he moves $A'$ rightward, leftward, upward or downward to a new position $A''$... | [
"The answer is in the affirmative. It is sufficient to prove that there exists a positive number $d$ such that, by Ann providing Bob suitable numbers, she eventually forces the $x$-coordinate of the point increase by at least $d$. Then, using $d$ over and over again, she successively increases the $x$-coordinate of... | Romania | 74th NMO Selection Tests for BMO and IMO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | Yes | |
0fan | Problem:
$ABCD$ is a parallelogram. The excircle of $ABC$ opposite $A$ has center $E$ and touches the line $AB$ at $X$. The excircle of $ADC$ opposite $A$ has center $F$ and touches the line $AD$ at $Y$. The line $FC$ meets the line $AB$ at $W$, and the line $EC$ meets the line $AD$ at $Z$. Show that $WX = YZ$. | [
"Solution:\n\n\n\nWe have the familiar result that $AY$ is perimeter $ADC$ (chase round using the fact that the two tangents from the same point have the same length). Similarly, $AX = $ perimeter $ABC = $ perimeter $ADC$. So $AX = AY$ $(*)$\n\n$AE$ is parallel to the bisector of $ACD$, whi... | Soviet Union | 1st CIS | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | null | proof only | null | |
066f | In a triangle $ABC$, the excircle $\omega_a$ opposite $A$ touches $AB$ at $P$ and $AC$ at $Q$, and the excircle $\omega_b$ opposite $B$ touches $BA$ at $M$ and $BC$ at $N$. Let $K$ be the projection of $C$ onto $MN$, and let $L$ be the projection of $C$ onto $PQ$. Show that the quadrilateral $MKLP$ is cyclic. | [] | Greece | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0j76 | Problem:
Find all real numbers $x$ satisfying
$$
x^{9} + \frac{9}{8} x^{6} + \frac{27}{64} x^{3} - x + \frac{219}{512} = 0
$$ | [
"Solution:\nAnswer: $\\frac{1}{2}, \\frac{-1 \\pm \\sqrt{13}}{4}$\n\nNote that we can re-write the given equation as\n$$\n\\sqrt[3]{x-\\frac{3}{8}} = x^{3} + \\frac{3}{8}\n$$\nFurthermore, the functions of $x$ on either side, we see, are inverses of each other and increasing. Let $f(x) = \\sqrt[3]{x-\\frac{3}{8}}$.... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1/2, (-1 + sqrt(13))/4, (-1 - sqrt(13))/4 | |
06uh | Let $A B C C_{1} B_{1} A_{1}$ be a convex hexagon such that $A B = B C$, and suppose that the line segments $A A_{1}$, $B B_{1}$, and $C C_{1}$ have the same perpendicular bisector. Let the diagonals $A C_{1}$ and $A_{1} C$ meet at $D$, and denote by $\omega$ the circle $A B C$. Let $\omega$ intersect the circle $A_{1}... | [
"Solution 1. If $A A_{1} = C C_{1}$, then the hexagon is symmetric about the line $B B_{1}$; in particular the circles $A B C$ and $A_{1} B C_{1}$ are tangent to each other. So $A A_{1}$ and $C C_{1}$ must be different. Since the points $A$ and $A_{1}$ can be interchanged with $C$ and $C_{1}$, respectively, we may ... | IMO | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0abc | Prove that $\frac{\operatorname{tg}\alpha}{\operatorname{tg}\beta} = \frac{a^2+c^2-b^2}{b^2+c^2-a^2}$ for every triangle $ABC$ ($a, b, c$ and $\alpha, \beta$ are the sides and the angles in the triangle, respectively). | [
"From the cosine theorem, we have\n$$ b^2 = a^2 + c^2 - 2ac \\cos \\beta, \\quad a^2 = b^2 + c^2 - 2bc \\cos \\alpha, $$\nTherefore $a^2 + c^2 - b^2 = 2ac \\cos \\beta$, $b^2 + c^2 - a^2 = 2bc \\cos \\alpha$.\nSo $\\frac{a^2+c^2-b^2}{b^2+c^2-a^2} = \\frac{2ac \\cos \\beta}{2bc \\cos \\alpha} = \\frac{a \\cos \\beta... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0098 | Let $n$ be a natural number with 120 positive divisors (including 1 and $n$). For each divisor $d$ of $n$, let $q$ be the quotient and $r$ be the remainder of dividing $4n-3$ by $d$. Let $Q$ be the sum of all the quotients $q$ and $R$ be the sum of all the remainders $r$ obtained by dividing $4n-3$ by the 120 possible ... | [] | Argentina | XXIX Olimpíada Matemática Argentina National Round | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)"
] | English | proof and answer | 1310 if n is odd; 1301 if n is even | |
0hoe | Problem:
If $a$, $b$, and $c$ are positive real numbers with $2a + 4b + 8c = 16$, what is the largest possible value of $abc$? | [
"Solution:\n\nBy AM-GM,\n$$\n\\frac{16}{3} = \\frac{2a + 4b + 8c}{3} \\geq \\sqrt[3]{(2a) \\cdot (4b) \\cdot (8c)} = 4 \\sqrt[3]{abc}.\n$$\nRearranging, we get\n$$\n\\sqrt[3]{abc} \\leq \\frac{4}{3} \\Longleftrightarrow abc \\leq \\frac{64}{27}\n$$\nThis maximum is indeed attainable if we set $2a = 4b = 8c$, since ... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 64/27 | |
0a6w | Problem:
Nine journalists from different countries attend a press conference. None of these speaks more than three languages, and each pair of the journalists share a common language. Show that there are at least five journalists sharing a common language. | [
"Solution:\n\nAssume the journalists are $J_1, J_2, \\ldots, J_9$. Assume that no five of them have a common language. Assume the languages $J_1$ speaks are $L_1, L_2$, and $L_3$. Group $J_2, J_3, \\ldots, J_9$ according to the language they speak with $J_1$. No group can have more than three members. So either the... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 1 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0e28 | Let $x$, $y$ and $z$ be non-zero real numbers such that $3x + 2y = z$ and $\frac{3}{x} + \frac{1}{y} = \frac{2}{z}$. Prove that $5x^2 - 4y^2 - z^2$ is always an integer. | [
"The second equation implies $2xy = 3yz + xz$. Multiplying the first equation respectively by $z$, $x$ and $y$, we get $z^2 = 3xz + 2zy$, $3x^2 = zx - 2xy$ and $2y^2 = zy - 3xy$. So,\n$$\n5x^2 - 4y^2 - z^2 = \\frac{5}{3}(zx - 2xy) - 2(zy - 3xy) - (3xz + 2zy) = \\frac{4}{3}(2xy - 3zy - xz) = 0.\n$$"
] | Slovenia | National Math Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof only | null | |
0eqh | The value of $1 + \frac{1}{3 + \frac{1}{2}}$ is
(A) $\frac{6}{5}$ (B) $\frac{7}{6}$ (C) $\frac{9}{2}$ (D) $\frac{6}{7}$ (E) $\frac{9}{7}$ | [] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | MCQ | E | |
01yo | Two lines pass through the point $F(1; 1)$ on the Cartesian plane. These lines are perpendicular to each other, one of them intersects the right branch of the hyperbola $y = \frac{1}{2x}$ at the points $A$ and $C$ ($C$ has bigger $x$-coordinate than $A$), and the other line intersects the left branch of this hyperbola ... | [
"Denote the abscissas of the points $A$, $B$, $C$ and $D$ by $a$, $b$, $c$ and $d$ respectively. The perpendicularity of the lines $AB$ and $CD$ is equivalent to $acbd = -1/4$ and the fact that the point $F$ belongs to the lines $AC$ and $BD$ is equivalent to the equalities $a+c = 2ac+1$ and $b+d = 2bd+1$. Denote $... | Belarus | Belarus2022 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals"
] | English | proof and answer | m^2 / 2 | |
0hl0 | Problem:
Two concentric circles have radii $r$ and $R > r$. Three new circles are drawn so that they are each tangent to the big two circles and tangent to the other two new circles. Find $\frac{R}{r}$. | [
"Solution:\n\nThe centers of the three new circles form a triangle. The diameter of the new circles is $R - r$, so the side length of the triangle is $R - r$. Call the center of the concentric circles $O$, two vertices of the triangle $A$ and $B$, and $AB$'s midpoint $D$. $OA$ is the average of $R$ and $r$, namely ... | United States | null | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 3 | |
0g5j | 令 $A_1 = \emptyset$, $B_1 = \{0\}$。對於所有 $n \ge 2$, 依照下面的遞迴構造：
$$
A_n = \{x + 1 \mid x \in B_{n-1}\},
$$
$$
B_n = A_{n-1} \cup B_{n-1} - A_{n-1} \cap B_{n-1},
$$
試問：有哪些 $n$ 會有 $B_n = \{0\}$？
(註：對任意集合 $A$ 和 $B$, 集合 $A - B$ 表示 $\{x \in A \mid x \notin B\}$。) | [
"我們將證明：$B_n = \\{0\\}$ 若且唯若 $n = 2^k$ ($k \\ge 0$)。\n\n給一個集合 $S$, 令符號 $2S = \\{2x \\mid x \\in S\\}$ 和 $S + k = \\{x + k \\mid x \\in S\\}$ 對任意整數 $k$。\n\n首先, 對於所有 $n \\ge 1$, 有 $A_n$ 與 $B_n$ 的元素都是非負整數, 而且 $0 \\notin A_n$ 和 $0 \\in B_n$。這可以簡單的以歸納法證明之。\n\n接著, 對於 $n \\ge 2$, 證明以下四種性質：\n(a) $A_{2n-1} = 2A_n - 1$；\n(b) ... | Taiwan | 二〇一一數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | n = 2^k for k ≥ 0 | |
0eax | The segment $AB$ measures $20$ cm. The polygonal curve connecting the points $A$ and $B$ together with the segment $AB$ forms $7$ equilateral triangles. No segment of the polygonal curve lies on the segment $AB$. How long, in centimeters, is this polygonal curve?
(A) $40\sqrt{3}$
(B) $40$
... | [
"Denote the lengths of the sides of these $7$ equilateral triangles by $a_1, a_2, \\dots, a_7$. The length of the segment $AB$ is then $a_1 + a_2 + \\dots + a_7$, and the length of the polygonal curve from $A$ to $B$ is $2a_1 + 2a_2 + \\dots + 2a_7 = 2(a_1 + a_2 + \\dots + a_7)$. Since this length is twice the leng... | Slovenia | National Math Olympiad in Slovenia | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | B | |
05lo | Problem:
Existe-t-il des entiers strictement positifs $a$ et $b$ tels que $a^{n}+n^{b}$ et $b^{n}+n^{a}$ soient premiers entre eux pour tout entier $n \geqslant 0$ ? | [
"Solution:\nPar l'absurde : supposons qu'il existe de tels entiers $a$ et $b$.\nPar symétrie, on peut supposer que $a \\geqslant b$.\nSi $d=\\operatorname{pgcd}(a, b)$ alors, pour $n=d$, il est clair que $d$ divise $a^{d}+d^{b}$ et $b^{d}+d^{a}$. Or ces deux nombres étant censés être premiers entre eux, c'est que $... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null | |
0jf5 | Problem:
Let $W$ be the hypercube $\{ (x_{1}, x_{2}, x_{3}, x_{4}) \mid 0 \leq x_{1}, x_{2}, x_{3}, x_{4} \leq 1 \}$. The intersection of $W$ and a hyperplane parallel to $x_{1}+x_{2}+x_{3}+x_{4}=0$ is a non-degenerate 3-dimensional polyhedron. What is the maximum number of faces of this polyhedron? | [
"Solution:\n**Answer:** 8\n\nThe number of faces in the polyhedron is equal to the number of distinct cells (3-dimensional faces) of the hypercube whose interior the hyperplane intersects. However, it is possible to arrange the hyperplane such that it intersects all 8 cells. Namely, $x_{1}+x_{2}+x_{3}+x_{4}=\\frac{... | United States | HMMT 2013 | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | final answer only | 8 | |
00p2 | Let $ABC$ be a given triangle and $l$ be a line that meets the lines $BC$, $CA$ and $AB$ in $A_1$, $B_1$ and $C_1$ respectively. Let $A'$ be the midpoint of the segment connecting the projections of $A_1$ onto the lines $AB$ and $AC$. Construct analogously the points $B'$ and $C'$.
a. Show that the points $A'$, $B'$ a... | [
"Let $AH_a$ be an altitude in the triangle $ABC$ and $P_a$ be its midpoint. Define analogously $H_b$, $P_b$, etc.\n\nIt is easy to see that the point $A'$ divides the segment $P_bP_c$ in the same ratio that $A_1$ divides $BC$. By Menelaus' theorem for the triangle $P_aP_bP_c$, claim (a) follows.\n\nConsider an affi... | Balkan Mathematical Olympiad | BMO 2010 Shortlist | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0fi5 | Problem:
La figura adjunta se compone de seis pentágonos regulares de lado un metro. Se dobla por las líneas de puntos hasta que coincidan las aristas no punteadas que confluyen en cada vértice. ¿Qué volumen de agua cabe en el recipiente así formado?
 | [
"Solution:\nLa figura formada por el agua es un tronco de pirámide pentagonal cuya base menor es el pentágono dado y cuya base mayor es otro pentágono regular que tiene por lado la diagonal del anterior paralela a la arista de la base como se muestra en la siguiente figura.\n\n\n\nEste pirá... | Spain | Olimpiada Matemática Española | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | (15 + 7*sqrt(5))/12 | |
02qz | Problem:
Patrícia escreveu, em ordem crescente, os inteiros positivos formados apenas por algarismos ímpares: $1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 31, 33, \ldots$ Qual foi o $157^{\circ}$ número que ela escreveu?
A) 997
B) 999
C) 1111
D) 1113
E) 1115 | [
"Solution:\n\nHá cinco algarismos ímpares: $1, 3, 5, 7$ e $9$. Contando apenas números inteiros positivos, existem então $5$ números formados por apenas um algarismo ímpar, $5 \\times 5 = 25$ números formados por dois algarismos ímpares e $5 \\times 5 \\times 5 = 125$ números formados por três algarismos ímpares. A... | Brazil | Nível 2 | [
"Discrete Mathematics > Combinatorics",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | D | |
0k4i | Problem:
Let $\alpha, \beta$, and $\gamma$ be three real numbers. Suppose that
$$
\begin{aligned}
\cos \alpha+\cos \beta+\cos \gamma & =1 \\
\sin \alpha+\sin \beta+\sin \gamma & =1
\end{aligned}
$$
Find the smallest possible value of $\cos \alpha$. | [
"Solution:\nLet $a=\\cos \\alpha+i \\sin \\alpha$, $b=\\cos \\beta+i \\sin \\beta$, and $c=\\cos \\gamma+i \\sin \\gamma$.\nWe then have\n$$\na+b+c=1+i\n$$\nwhere $a, b, c$ are complex numbers on the unit circle. Now, to minimize $\\cos \\alpha=\\operatorname{Re}[a]$, consider a triangle with vertices $a, 1+i$, and... | United States | HMMT February | [
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | -(1+sqrt(7))/4 | |
0c6h | Show that, if the numbers $a$, $b$, $c \in [0, \infty)$ satisfy the relation $a + b + c = 3$, then
$$
\frac{a}{3a + bc + 12} + \frac{b}{3b + ca + 12} + \frac{c}{3c + ab + 12} \le \frac{3}{16}.
$$ | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0ldv | A sequence $(x_n)$ is defined as follows
$$
x_0 = 2,\ x_1 = 1,\ x_{n+2} = x_{n+1} + x_n
$$
for every non-negative integer $n$.
a) For every $n \ge 1$, prove that if $x_n$ is a prime number then $n$ is a prime number or $n$ has no odd prime divisors.
b) Find all pairs of non-negative integers $(m, n)$ such that $x_m \... | [
"a. We can easily prove that $x_n = \\alpha^n + \\beta^n$ for all positive integers $n$ where $\\alpha < 0 < \\beta$ are two roots of the equation $\\lambda^2 - \\lambda - 1 = 0$.\n\nSuppose that $x_n$ is a prime number where $n$ is a positive integer with odd prime divisors. Then, $n$ has the form $pq$ where $p$ i... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | a) If x_n is prime, then n is prime or n is a power of two.
b) All pairs (m, n) of non‑negative integers with x_m | x_n are:
- (0, 3t) for any non‑negative integer t;
- (1, n) for any non‑negative integer n;
- (m, (2k+1)m) with m > 1 and k a non‑negative integer. | |
0gf4 | 找出所有的整數 $n = 2k+1 > 1$, 使得存在某個 $\{0, 1, \dots, k\}$ 的排列 $a_0, a_1, \dots, a_k$, 滿足
$$
a_1^2 - a_0^2 \equiv a_2^2 - a_1^2 \equiv \cdots \equiv a_k^2 - a_{k-1}^2 \pmod{n}.
$$ | [
"It is straightforward to check that $n = 3, 5$ indeed satisfy the condition.\n\nBelow we prove that $n \\le 5$. Suppose $a_0, a_1, \\dots, a_k$ satisfying the condition and from now on, $\\equiv$ works in modulo $n$. Since $a_{i_0} = 0$ for some $i_0$, so $a_i^2 \\equiv c(i - i_0)$ for some constant $c$. In partic... | Taiwan | 2021 年台灣數學奧林匹亞考試試題 | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 3, 5 | |
099t | Given $a_1, a_2, ..., a_{2009}$ different natural numbers such that if two of any three numbers are not relatively prime, then one of the two numbers is not relatively prime to the third. If there do not exist 50 numbers that are relatively prime pairwise, then prove that there exist 40 numbers such that are not relati... | [
"Let us construct graph $G$ with vertices $a_1, a_2, ..., a_{2009}$; if two numbers are not relatively prime, then connect them with an edge. Then, the resulting graph $G$ will not have a subgraph that has three vertices and a single edge, by the given of the problem. Therefore, vertices of the graph $G$ can be div... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0f6d | Problem:
$n$ lines are drawn in the plane. Some of the resulting regions are colored black, no pair of painted regions have a boundary line in common (but they may have a common vertex). Show that at most $\dfrac{n^2 + n}{3}$ regions are black. | [] | Soviet Union | 19th ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F"
] | null | proof only | null | |
0h93 | In an acute triangle $ABC$ there is an altitude $AH$ and median $AM$. On lines $AB$ and $AC$ there are points $X$ and $Y$ so that $AX = XC$ and $AY = YB$. Prove that the midpoint of segment $XY$ is equidistant from points $H$ and $M$. | [
"Let $Z$ be the midpoint of segment $XY$, and $N$ and $T$ be the midpoints of segments $AB$ and $AC$ respectively (see the figure).\n\nAs, according to the condition, $AX = XC$ and $AY = YB$, $XT$ and $YN$ are bisectors to segments $AC$ and $AB$. Then triangles $XTY$ and $XNY$ are right triangles, so $XZ = ZN = ZT ... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0k85 | Problem:
James is standing at the point $(0,1)$ on the coordinate plane and wants to eat a hamburger. For each integer $n \geq 0$, the point $(n, 0)$ has a hamburger with $n$ patties. There is also a wall at $y=2.1$ which James cannot cross. In each move, James can go either up, right, or down 1 unit as long as he doe... | [
"Solution:\n\nNote that we desire to compute the number of times James moves to the right before moving down to the line $y=0$. Note also that we can describe James's current state based on whether his $y$-coordinate is $0$ or $1$ and whether or not the other vertically adjacent point has been visited. Let $E(1, N)... | United States | HMMT November 2019 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 7/3 | |
05zq | Problem:
Soit $k \geqslant 1$ un nombre entier.
Considérons $4k$ jetons, dont $2k$ sont rouges et $2k$ sont bleus. Une suite de ces $4k$ jetons peut être transformée en une autre suite par un mouvement, consistant à interchanger un certain nombre (éventuellement un seul) de jetons rouges consécutifs avec un nombre éga... | [
"Solution:\n\nRéponse : $n = k$.\n\nLe problème demande de déterminer le plus petit $n$ vérifiant une certaine propriété, il contient donc nécessairement deux parties. Ici, dans un premier temps, on montre qu'étant donnée une configuration initiale de jetons, il est toujours possible d'arriver à la configuration fi... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | k | |
0ka1 | Problem:
Can the set of lattice points $\{(x, y) \mid x, y \in \mathbb{Z}, 1 \leq x, y \leq 252, x \neq y\}$ be colored using 10 distinct colors such that for all $a \neq b, b \neq c$, the colors of $(a, b)$ and $(b, c)$ are distinct? | [
"Solution:\nYes.\nAssociate to each number from $1$ to $252$ a distinct $5$-element subset of $S=\\{1,2, \\ldots, 10\\}$. Then assign to $(a, b)$ an element of $S$ that is in the subset associated to $a$ but not in that associated to $b$. It's not difficult to see that this numerical assignment is a valid coloring:... | United States | HMMT February 2019 Team Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | Yes | |
0eq0 | In a group of $108$ people, $1$ in $4$ of them has a pen and $1$ in $3$ has a pencil. What is the minimum possible number of people that have something to write with? | [
"$36$ people must have a pen and $36$ must have a pencil. If all the people with pens also have pencils, this could be as few as $36$ people in all."
] | South Africa | South African Mathematics Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | final answer only | 36 | |
02h0 | We have four charged batteries, four uncharged batteries and a radio which needs two charged batteries to work.
Suppose we don't know which batteries are charged and which ones are uncharged. Find the least number of attempts sufficient to make sure the radio will work. An attempt consists in putting two batteries in t... | [
"Let's generalize this problem to $2n$ batteries, $n$ of them charged. The number of charged batteries needed to turn on the radio is still two.\n\nOne can see that the order of the attempts doesn't matter and that a set of attempts is always successful if and only if in every set of $n$ batteries there is an attem... | Brazil | XXVII OBM | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 7 | |
03l9 | Problem:
Let $\mathbb{N} = \{0, 1, 2, \ldots\}$. Determine all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that
$$
x f(y) + y f(x) = (x + y) f\left(x^{2} + y^{2}\right)
$$
for all $x$ and $y$ in $\mathbb{N}$. | [
"Solution:\nWe claim that $f$ is a constant function. Suppose, for a contradiction, that there exist $x$ and $y$ with $f(x) < f(y)$; choose $x, y$ such that $f(y) - f(x) > 0$ is minimal. Then\n$$\nf(x) = \\frac{x f(x) + y f(x)}{x + y} < \\frac{x f(y) + y f(x)}{x + y} < \\frac{x f(y) + y f(y)}{x + y} = f(y)\n$$\nso ... | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All constant functions: f(x) = c for any c in N. | |
03zq | As shown in Fig. 8.1, $\odot O$ is the inscribed circle touching side $BC$ of $\triangle ABC$ at point $M$, and points $D, E$ are on the segments $AB, AC$, respectively, satisfying $DE \parallel BC$; $\odot O_1$ is the inscribed circle of $\triangle ADE$ tangent to side $DE$ at point $N$; $O_1B, DO$ intersect at point ... | [
"If $AB = AC$, then the graph is symmetric about the bisector of $\\angle BAC$ and the conclusion is obvious. So we may assume that $AB > AC$. As shown in Fig. 8.2, let $L$ be the midpoint of $BC$, line $O_1L$ intersecting with $FG$ at $R$, and $O_1N$ be extended to intersect with $BC$ at $K$. Draw line $AT$ that i... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > T... | English | proof only | null | |
001u | Se tiene un tablero cuadrado de $8 \times 8$ dividido en casillas de $1 \times 1$. Escribir en cada casilla un $1$ o un $2$ de modo que en cada cuadrado de $3 \times 3$ la suma de los $9$ números sea múltiplo de $4$, pero la suma de los $64$ números del tablero no sea múltiplo de $4$. | [] | Argentina | XX Olimpiada Matemática Argentina | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | español | proof and answer | null | |
0741 | Let $f(x) = \sum_{k=1}^{n} a_k x^k$ and $g(x) = \sum_{k=1}^{n} \frac{a_k}{2^k - 1} x^k$ be two polynomials with real coefficients, where $n \ge 3$. Suppose $1, 2^{n+1}$ are roots of $g(x) = 0$. Prove that $f(x) = 0$ has a positive root smaller than $2^n$. | [
"It is easy to see that $g(2x) - g(x) = f(x)$. Thus\n$$\n\\sum_{k=0}^{n} f(2^k) = g(2^{n+1}) - g(1) = 0.\n$$\nConsider the relation $f(1) + f(2) + \\dots + f(2^n) = 0$. If $f(2^n) \\neq 0$, then $f(1)$ and $f(2^n)$ have opposite signs. Hence there exists $\\alpha$ in $(1, 2^n)$ such that $f(\\alpha) = 0$. If $f(2^n... | India | Indija TS 2009 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
063l | Problem:
Es sei $ABC$ ein spitzwinkliges Dreieck mit dem Umkreis $k$ und dem Inkreismittelpunkt $I$. Die Orthogonale zu $CI$ durch $I$ schneide die Seite $BC$ in $U$ und $k$ in $V$, wobei $V$ und $A$ auf verschiedenen Seiten von $BC$ liegen. Die Parallele zu $AI$ durch $U$ schneide $AV$ im Punkt $X$.
Man beweise: Wenn... | [
"Solution:\n\nIn der Figur ist $M$ zunächst nur der Schnittpunkt von $XI$ und $AC$. $N$ ist der Schnittpunkt von $XU$ und $AB$ und $Y$ der Schnittpunkt von $XI$ und $AB$. Die halben Innenwinkel des Dreiecks $ABC$ sind mit $\\alpha, \\beta$ bzw. $\\gamma$ bezeichnet. Wegen $\\angle UIC = 90^{\\circ}$ ist $\\angle CU... | Germany | 2. Auswahlklausur 2014/2015 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a9e | Problem:
A function $f: \mathbb{Z} \rightarrow \mathbb{Z}_{+}$, where $\mathbb{Z}_{+}$ is the set of positive integers, is non-decreasing and satisfies $f(m n) = f(m) f(n)$ for all relatively prime positive integers $m$ and $n$. Prove that $f(8) f(13) \geq (f(10))^{2}$. | [
"Solution:\nSince $f$ is non-decreasing, $f(91) \\geq f(90)$, which (by factorization into relatively prime factors) implies $f(13) f(7) \\geq f(9) f(10)$. Also $f(72) \\geq f(70)$, and therefore $f(8) f(9) \\geq f(7) f(10)$. Since all values of $f$ are positive, we get $f(8) f(9) \\cdot f(13) f(7) \\geq f(7) f(10)... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 24 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0877 | Problem:
Un'urna contiene $N$ palline $(N>3)$ numerate da 1 a $N$. Se dall'urna vengono tolte due palline recanti numeri non multipli di 3 e una recante un multiplo di 3, la probabilità di ottenere un multiplo di 3 estraendo una singola pallina risulta minore di quanto era con l'urna completa. Cosa si può dedurre rigu... | [
"Solution:\n\nLa risposta è (B). In presenza di un numero finito di esiti possibili ed equiprobabili di un esperimento (nel nostro caso l'estrazione di una pallina dall'urna), la probabilità di un evento (per noi l'estrazione di una pallina con numero multiplo di 3) è il rapporto tra il numero di esiti favorevoli e... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Statistics > Probability > Counting Methods > Other"
] | null | MCQ | B | |
09cq | А ба В хоёр тоглогч "Таавар" тоглоом тоглож байна. Тэр хоёр мэдэгдэж байгаа $k$ ба $n$ натурал тоонуудаас тоглоомын дүрэм хамаарна. Тоглоом эхлэхэд А нь $1 \le x \le N$ байх бүхэл тоо $x$ ба $N$-ийг согоно. А $x$ тоог нуудлах ба $N$ тоог В-д үнэнээр нь хэлнэ. Үнийг талаар мэдээлэл олохын тулд В дараах маягийн асуултыг ... | [
"Consider an answer $A \\in \\{\\text{yes}, \\text{no}\\}$ to a question of the kind \"Is $x$ in the set $S$?\". We say that $A$ is inconsistent with a number $i$ if $A = \\text{yes}$ and $i \\notin S$, or if $A = \\text{no}$ and $i \\in S$. Observe that an answer inconsistent with the target number $x$ is a lie.\n... | Mongolia | ОУМО-53 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Mongolian | proof only | null | |
0350 | Problem:
Prove that amongst any 9 vertices of a regular 26-gon there are three which are vertices of an isosceles triangle. Do there exist 8 vertices such that no three of them are vertices of an isosceles triangle? | [
"Solution:\n\nLEMMA. For any five vertices of a regular 13-gon there exists an isosceles triangle with vertices amongst these points.\n\nProof of the lemma. Let the five points form a convex pentagon $ABCDE$. We first consider the case when there exist two pairs of parallel lines determined by some vertices of $ABC... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | Among any nine vertices of a regular twenty-six-gon, three form an isosceles triangle; and yes, there exist eight vertices with no isosceles triple, for example {1, 2, 4, 5, 10, 11, 13, 14}. | |
0a4t | Problem:
Let $V$ be the set of vertices of a regular 21-gon. Given a non-empty subset $U$ of $V$, let $m(U)$ be the number of distinct lengths that occur between two distinct vertices in $U$. What is the maximum value of $\frac{m(U)}{|U|}$ as $U$ varies over all non-empty subsets of $V$? | [
"Solution:\nTo simplify notation, we will let $m$ be $m(U)$ and let $n$ be $|U|$. First note that there are 10 different diagonal-lengths in a regular 21-gon. Now consider the following set of 5 vertices.\n\n\n\nNote that each of the 10 different diagonal-lengths appear (exactly once each).... | New Zealand | NZMO Round One | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2 | |
07w6 | Caitlin and Donal play a game called *Basketball Shoot-Out*. The game consists of 10 rounds. In each round, Caitlin and Donal both throw a ball simultaneously at each other's basket. If a player's ball falls into the basket, that player scores one point; otherwise, they score zero points. The scoreboard shows the compl... | [
"We can represent the scoreboard by a $2 \\times n$ binary matrix, e.g.,\n$$\n\\begin{bmatrix} 0 & 1 & 0 & 1 \\\\ 0 & 0 & 1 & 1 \\end{bmatrix}.\n$$\nThe problem is then asking for the number of $2 \\times n$ binary matrices such that, for every $1 \\le t \\le n$, the first $t$ elements in the first row always conta... | Ireland | IRL_ABooklet_2023 | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Fac... | English | proof only | null | |
01an | We call a positive integer *n delightful* if there exists an integer $x$, $1 < x < n$, such that
$$
1 + 2 + \cdots + (x - 1) = (x + 1) + (x + 2) + \cdots + n.
$$
Does there exist a delightful number *N* satisfying
$$
2013^{2013} < \frac{N}{2013^{2013}} < 2013^{2013} + 4.
$$ | [
"Consider a delightful number *n*. Then there exists an integer *x*, $1 < x < n$ satisfying\n$$\n\\sum_{i=1}^{x-1} i = \\sum_{i=x+1}^{n} i = \\sum_{i=1}^{n} i - \\sum_{i=1}^{x} i\n$$\n$\\Leftrightarrow$\n$$\nx^2 = x + 2 \\cdot \\frac{(x-1)x}{2} = x + 2 \\sum_{i=1}^{x-1} i = \\sum_{i=1}^{x-1} i + \\sum_{i=1}^{x} i =... | Baltic Way | Baltic Way 2013 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | No | |
0cds | We say that a natural number $n \ge 2$ is *nice*, if in its prime factor decomposition, one of the primes has exponent 3.
a) Determine the smallest natural number $N$, such that no matter how we choose $N$ consecutive natural numbers, at least one of them is nice.
b) Find the smallest 15 consecutive natural numbers $... | [
"a) By dividing 16 consecutive natural numbers by 16, the remainders 0, 1, 2, ..., 15 are obtained (not necessarily starting with 0). Therefore, among any 16 consecutive natural numbers there is a nice number of the form $16k + 8 = 2^3(2k+1)$. Since among the consecutive natural numbers 9, 10, 11, ..., 23 there is ... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | a) 16; b) 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87 | |
06xb | Find all triples of positive integers $(a, b, p)$ with $p$ prime and
$$
a^{p} = b! + p.
$$ | [
"Solution 1. Clearly, $a > 1$. We consider three cases.\n\nCase 1: We have $a < p$. Then we either have $a \\leqslant b$ which implies $a \\mid a^{p} - b! = p$ leading to a contradiction, or $a > b$ which is also impossible since in this case we have $b! \\leqslant a! < a^{p} - p$, where the last inequality is true... | IMO | International Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | [(2, 2, 2), (3, 4, 3)] | |
0hvw | Problem:
Let $N$ be a positive integer such that $N$ is divisible by $81$ and the number formed by reversing the digits of $N$ is also divisible by $81$. Prove that the sum of the digits of $N$ is divisible by $81$. | [
"Solution:\nWe begin with a lemma.\nLemma. For all $k \\geq 0$, $10^{k} \\equiv 1 + 9k \\pmod{81}$.\nProof. Binomial theorem:\n$$\n10^{k} = (1 + 9)^{k} = 1^{k} + \\binom{k}{1} 1^{k-1} 9 + \\text{terms divisible by } 9^{2}.\n$$\nWe now write $N$ in terms of its digits as\n$$\n\\begin{aligned}\nN &= a_{0} + 10 a_{1} ... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
08zk | In an acute triangle $ABC$, let $D$, $E$, and $F$ be the midpoints of sides $BC$, $CA$, and $AB$ respectively. Let $X$ and $Y$ be the feet of the perpendiculars drawn from $D$ to sides $AB$ and $AC$ respectively. The line passing through $F$ and parallel to line $XY$ intersects line $DY$ at a point $P$ that is differen... | [
"Since $\\angle AXD = \\angle AYD = 90^\\circ$, points $A$, $D$, $X$ and $Y$ are concyclic. Therefore, we have $\\angle FAD = \\angle XAD = \\angle XYD = \\angle FPD$, and by the converse of the inscribed angle theorem, points $A$, $D$, $F$ and $P$ are concyclic. According to the midpoint theorem in triangle $ABC$,... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0kzd | A dartboard is the region $B$ in the coordinate plane consisting of points $(x, y)$ such that $|x| + |y| \le 8$. A target $T$ is the region where $(x^2 + y^2 - 25)^2 \le 49$. A dart is thrown and lands at a random point in $B$. The probability that the dart lands in $T$ can be expressed as $\frac{m}{n} \cdot \pi$, wher... | [
"**Answer (B):** The region $B$ is a square with intercepts $(\\pm8, 0)$ and $(0, \\pm8)$. The area of this square is $(8\\sqrt{2})^2 = 128$. Taking square roots shows that region $T$ is the set of points that satisfy\n$$\n25 - 7 \\le x^2 + y^2 \\le 25 + 7,\n$$\nwhich is an annulus (ring) with inner radius $\\sqrt{... | United States | AMC 10 B | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | MCQ | B | |
03dl | For a triangle $ABC$ with $\angle ABC = 50^\circ$ points $M$ and $N$ are midpoints of $AC$ and $BC$, respectively. Point $P$ is the midpoint of $BM$ and point $Q$ is the midpoint of $CM$. It is known that there exists a right triangle with side lengths $CP$, $BQ$ and $MN$. Find the angle between lines $CP$ and $AB$. | [] | Bulgaria | Bulgaria 2022 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 65° | |
07f7 | Let $n, k$ be positive integers such that $n \ge k$. Find the maximum number of binary sequences of length $n$ such that by fixing any arbitrary $k$ bits, the achieved sequences do not produce all binary sequences of length $k$. For example if $k = 1$, we can only have one sequence, otherwise they will differ in at lea... | [
"Let the answer be $f(n,k)$. We will prove by induction on $n+k$ that\n$$\nf(n,k) = \\sum_{i=0}^{k-1} \\binom{n}{i}.\n$$\nCase $n=k=1$ is obvious.\n\nAssume that $S$ is the maximum set of the desired sequences and let $T$ be the set of all binary sequences of length $n-1$. Define the sets $A, B, C$ as follows:\n$$\... | Iran | 37th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | sum_{i=0}^{k-1} binom(n,i) | |
0f98 | Problem:
Show that the 120 five digit numbers which are permutations of $12345$ can be divided into two sets with each set having the same sum of squares. | [] | Soviet Union | 23rd ASU | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0bz6 | One considers the non-zero distinct digits $a, b, c, d, e, f$. Determine the positive integers $x$, such that $x$ divides any 6-digit number written with the digits $a, b, c, d, e, f$. | [
"For any choice of 6 distinct non-zero digits, at least two of them are consecutive.\nIndeed, if $a < b < c < d < e < f$ and no digits are consecutive, then $b \\ge a + 2$, $c \\ge b + 2 \\ge a + 4$, $d \\ge c + 2 \\ge a + 6$, $e \\ge d + 2 \\ge a + 8$ and $f \\ge e + 2 \\ge a + 10$, which is impossible, since both... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | All such x must divide 9. Specifically: if the sum of the six digits is not divisible by 3, then only x = 1; if the sum is divisible by 3 but not by 9, then x = 1 or x = 3; if the sum is divisible by 9, then x = 1, x = 3, or x = 9. | |
0ejc | Problem:
Kateri izraz je ekvivalenten izrazu
$$
\frac{\sin^{2} x-\cos^{2} x}{\sin^{-1} x+\cos^{-1} x} \cdot \frac{\tan x}{\cos x-\sin x}
$$
?
(A) $-\cos x$
(B) $-\sin x$
(C) $-\sin^{2} x$
(D) $-\cos^{2} x$
(E) $\cos x$ | [
"Solution:\nŠtevec prvega ulomka razstavimo kot razliko kvadratov $\\sin^{2} x-\\cos^{2} x=(\\sin x-\\cos x)(\\sin x+\\cos x)$, imenovalec pa zapišemo kot vsoto obratnih vrednosti $\\sin^{-1} x+\\cos^{-1} x=\\frac{1}{\\sin x}+\\frac{1}{\\cos x}$ in dobljena ulomka seštejemo. Števec drugega ulomka zapišemo $\\tan x=... | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | MCQ | C | |
05k3 | Problem:
On se donne $n$ points du plan, tels que trois quelconques d'entre eux ne sont jamais alignés. Chacun est colorié en rouge ou en bleu. On suppose qu'il y a exactement un point bleu à l'intérieur de chaque triangle dont les sommets sont rouges, et un point rouge à l'intérieur de chaque triangle dont les sommet... | [
"Solution:\n\nOn montre d'abord que les points bleus forment un polygone convexe : si ce n'était pas le cas, on pourrait trouver un point bleu à l'intérieur d'un polygone bleu, donc en découpant ce polygone en triangles, on pourrait trouver un point bleu à l'intérieur d'un triangle bleu. Le grand triangle bleu doit... | France | OFM 2013-2014 Envoi 2 | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 8 | |
053q | Find all four-digit numbers which are exactly $2016$ larger than the four-digit number obtained by moving the first digit to the end. | [
"Let the first digit of the number be $a$ and the number formed by the remaining digits be $k$. By the conditions, $1000a + k = 10k + a + 2016$, whence $111a - k = 224$. Hence $a \\ge 3$, implying the solutions $a = 3, k = 109$; $a = 4, k = 220$; $a = 5, k = 331$; $a = 6, k = 442$; $a = 7, k = 553$; $a = 8, k = 664... | Estonia | Estonian Math Competitions | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 3109, 4220, 5331, 6442, 7553, 8664, 9775 | |
0kbu | Problem:
Find the number of ordered pairs of positive integers $(x, y)$ with $x, y \leq 2020$ such that $3x^{2} + 10xy + 3y^{2}$ is the power of some prime. | [
"Solution:\n\nWe can factor as $(3x + y)(x + 3y)$. If $x \\geq y$, we need $\\frac{3x + y}{x + 3y} \\in \\{1, 2\\}$ to be an integer. So we get the case where $x = y$, in which we need both to be a power of $2$, or the case $x = 5y$, in which case we need $y$ to be a power of $2$. This gives us $11 + 9 + 9 = 29$ so... | United States | HMMT February 2020 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 29 | |
0br3 | The regular triangular pyramid $SABC$ has $AB = a$, $M$ and $N$ are the midpoints of the sides $[AB]$ and $[BC]$ respectively and $SA \perp (SBC)$.
a) Find the measure of the angle of the line $SM$ with the plane $(SAN)$.
b) If $SP$ is the orthogonal projection of the segment $SM$ on the plane $(SAN)$, compute the ta... | [] | Romania | 67th NMO Shortlisted Problems | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | a) 30 degrees; b) sqrt(14) | |
0d8j | Let $\mathbb{R}^{+}$ be the set of positive real numbers. Find all function $f: \mathbb{R}^{+} \rightarrow \mathbb{R}$ such that, for all positive real number $x$ and $y$, the following conditions are satisfied:
i) $2 f(x)+2 f(y) \leq f(x+y)$.
ii) $(x+y)[y f(x)+x f(y)] \geq x y f(x+y)$. | [
"Replacing $x = y$ in i) and ii), we get $f(2x) = 4 f(x)$, $\\forall x \\in \\mathbb{R}^{+}$. Therefore, by induction, we can prove that\n$$\nf\\left(2^{n} x\\right) = 4^{n} f(x), \\quad \\forall n \\in \\mathbb{Z}^{+}, x \\in \\mathbb{R}^{+}.\n$$\nNow, by substituting $y = 2x$ in i) and ii), we get $10 f(x) \\leq ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All functions of the form f(x) = k x^2 for all positive x, where k ≤ 0. | |
0gxb | Find the biggest 3-digit number which satisfies the following three conditions:
1) the number is prime;
2) the 3-digit number obtained by writing the digits of our number in the reverse order is a prime too;
3) the product of the digits is a prime too. | [
"As the product of the digits this number has is prime, then these digits are $1, 1, p$ (not necessarily in this order) and the digit $p$ is a prime number. Obviously, $p \\neq 7$ because otherwise the sum of the digits is divisible by $3$ and the number isn't prime. The digits $2$ and $5$ can stay only between bot... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 311 | |
02y0 | Problem:
Seis pontos são distribuídos dentro de um quadrado de lado $10~\mathrm{cm}$ de tal modo que a distância entre quaisquer dois deles é um número inteiro em centímetros. Verifique que pelo menos duas dessas distâncias são iguais. | [
"Solution:\n\nA maior distância possível entre dois pontos do quadrado é $10 \\sqrt{2}~\\mathrm{cm}$, que ocorre quando dois pontos estão dispostos em extremos opostos de uma diagonal. Como $14 < 10 \\sqrt{2} < 15$, a maior distância inteira possível entre eles é de $14~\\mathrm{cm}$. Assim, existem $14$ distâncias... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0kuq | Problem:
Compute the number of ways to select 99 cells of a $19 \times 19$ square grid such that no two selected cells share an edge or vertex. | [
"Solution:\n\nWe claim the number of ways to select $n^{2}-1$ such cells from a $(2n-1) \\times (2n-1)$ grid is exactly $n^{3}$, which implies the answer to this question is $1000$.\n\nPartition the board into $n^{2}$ regions, as pictured. Also, shade red every cell in an odd row and column red, so there are $n^{2}... | United States | HMMT February 2023 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | final answer only | 1000 | |
0grj | In how many ways every unit square of a $2018 \times 2018$ board can be colored in red or white such that number of red unit squares in any two rows are distinct and number of red unit squares in any two columns are distinct. | [
"The answer is $2(2018!)^2$. We consider the problem for an $n \\times n$ board. Let $r_i$ be the number of red squares in the $i$-th row for $i = 1, \\dots, n$ and $c_j$ be the number of red squares in the $j$-th column for $j = 1, \\dots, n$. Since $r_i \\neq r_j$ for every $i \\neq j$, we see that $\\{r_1, \\dot... | Turkey | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 2(2018!)^2 | |
0awc | Problem:
Let $f(x) = 2^{x} - 2^{1-x}$. Simplify $\sqrt{f(2015) - f(2014) + f(1) - f(0)}$. | [
"Solution:\nWe have $f(2015) = 2^{2015} - 2^{-2014}$, $f(2014) = 2^{2014} - 2^{-2013}$, $f(1) = 2 - 1 = 1$, and $f(0) = 1 - 2 = -1$.\n\nHence,\n$$\n\\begin{aligned}\n\\sqrt{f(2015) - f(2014) + f(1) - f(0)} & = \\sqrt{2^{2015} - 2^{-2014} - 2^{2014} + 2^{-2013} + 2} \\\\\n& = \\sqrt{\\left(2^{2015} - 2^{-2014}\\righ... | Philippines | 18th PMO National Stage Oral Phase | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | 2^{1007} + 2^{-1007} | |
06lu | Alice and Bob play a game on the plane. Firstly, Alice draws 4 red points which are the vertices of a square, and draws another 2021 red points inside this square such that no three red points are collinear. Next, Bob draws $n$ blue points such that for every triangle whose three vertices are the red points, there is a... | [
"The answer is $4044$.\n\nAfter Alice has drawn the $4$ red vertices of a square, we can draw one of the diagonals of the square to form $2$ triangles. Each time when Alice draws one more red point inside the square, it must lie inside one of the triangles formed. By joining the new red point to the $3$ vertices of... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 4044 | |
01r2 | Let $n > 1$ be a given integer. Prove that infinitely many terms of the sequence $(a_k)_{k \ge 1}$, defined by
$$
a_k = \left\lfloor \frac{n^k}{k} \right\rfloor,
$$
are odd. (For a real number $x$, $\lfloor x \rfloor$ denotes the largest integer not exceeding $x$.) | [
"3. See IMO-2014 Shortlist, Problem N4."
] | Belarus | SELECTION and TRAINING SESSION | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof only | null | |
069t | A grid consists of all points of the form $(m, n)$ where $m$ and $n$ are integers with $|m| \le 2019$, $|n| \le 2019$ and $|m| + |n| < 4038$. We call the points $(m, n)$ of the grid with either $|m| = 2019$ or $|n| = 2019$ the *boundary points*. The four lines $x = \pm 2019$ and $y = \pm 2019$ are called *boundary line... | [
"Anna does not have a winning strategy. We will provide a winning strategy for Bob. It is enough to describe his strategy for the deletions on the line $y = 2019$.\n\nBob starts by deleting $(0, 2019)$ and $(-1, 2019)$. Once Anna completes her turn, he deletes the next two available points on the left if Anna decre... | Greece | 36th Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | No; Bob has a winning strategy. | |
08xh | For a grid of $55 \times 55$ squares, let us consider the following operation:
Operation: Choose a rectangular region consisting of a number of squares from the grid, and color the region either white or black.
Determine the smallest possible number of operations necessary to reach from the initial configuration, in ... | [
"$784$\n\nFirst, we show that the answer we seek is at least $784$.\nLet us call the vertices of the squares of the grid lattice points. There are $56^2$ lattice points. For each lattice point, call the set of all the squares of the grid having this lattice point as a vertex its neighborhood. We will first show tha... | Japan | Japan Mathematical Olympiad Initial Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 784 | |
087r | Problem:
La casa di Dante si trova nel punto $D$ ai piedi di una montagna conica con il diametro di base di $4~\mathrm{km}$ e cima nel punto $C$. Si sa che $D$ dista da $C$ $4~\mathrm{km}$ in linea retta e che, detto $P$ il punto diametralmente opposto a $D$ rispetto alla base della montagna, la porta dell'Inferno si ... | [
"Solution:\n\nLa risposta è (B). Sviluppiamo sul piano la superfice laterale della montagna, tagliandola lungo il segmento $DC$. Si avrà un settore circolare di centro $C$, raggio $4~\\mathrm{km}$, ossia la lunghezza di $DC$, e delimitato da un arco di circonferenza di $4\\pi~\\mathrm{km}$, ossia la circonferenza d... | Italy | UNIONE MATEMATICA ITALIANA SCUOLA NORMALE SUPERIORE DI PISA Progetto Olimpiadi di Matematica | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | B | |
04f7 | Let $a_1, a_2, \dots, a_n$ be positive real numbers such that $a_1 + a_2 + \dots + a_n = 1$.
Prove that the following inequality holds
$$
\frac{a_1^3}{a_1^2 + a_2 a_3} + \frac{a_2^3}{a_2^2 + a_3 a_4} + \dots + \frac{a_{n-1}^3}{a_{n-1}^2 + a_n a_1} + \frac{a_n^3}{a_n^2 + a_1 a_2} \ge \frac{1}{2}.
$$ | [
"We have\n$$\n\\begin{aligned}\n\\frac{a_1^3}{a_1^2 + a_2 a_3} &= \\frac{a_1^3 + a_1 a_2 a_3 - a_1 a_2 a_3}{a_1^2 + a_2 a_3} = a_1 - a_1 a_2 a_3 \\cdot \\frac{1}{a_1^2 + a_2 a_3} \\\\\n&\\geqq \\overset{\\text{A-G}}{a_1 - a_1 a_2 a_3} \\cdot \\frac{1}{2a_1\\sqrt{a_2 a_3}} = a_1 - \\frac{1}{2}\\sqrt{a_2 a_3} \\overs... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0bs2 | a) Prove that $7$ cannot be written as a sum of squares of three rational numbers.
b) Let $a$ be a rational number that can be written as a sum of squares of three rational numbers. Prove that $a^m$ can be written as a sum of squares of three rational numbers, for any positive integer $m$. | [
"a) Suppose, by way of contradiction, that one can find rational numbers $x$, $y$, $z$ such that $7 = x^2 + y^2 + z^2$. Writing $x$, $y$, $z$ as fractions and clearing denominators yield an equality like\n$$\n7n^2 = a^2 + b^2 + c^2, \\quad (*)\n$$\nwhere $n$, $a$, $b$, $c$ are nonnegative integers, not all zero.\nI... | Romania | 67th Romanian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0iuu | Problem:
Shelly writes down a vector $v=(a, b, c, d)$, where $0<a<b<c<d$ are integers. Let $\sigma(v)$ denote the set of 24 vectors whose coordinates are $a, b, c$, and $d$ in some order. For instance, $\sigma(v)$ contains $(b, c, d, a)$. Shelly notes that there are 3 vectors in $\sigma(v)$ whose sum is of the form $(... | [
"Solution:\n\nIf $k = a + b + c + d$, first you notice $4 \\mid 3k$, and $k \\geq 10$. So we try $k = 12$, which works with $a, b, c, d = 1, 2, 3, 6$ and not $1, 2, 4, 5$."
] | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | 6 | |
06ku | Find infinitely many positive integers $m$ such that for each such $m$, the number $\frac{2^{m-1}-1}{8191m}$ is an integer. | [
"By Dirichlet's theorem, there are infinitely many primes $p$ of the form $13k + 1$. We claim that we can set $m = p$ whenever $p > 8191$.\n\nFirstly, $p$ divides $2^{p-1} - 1$ by the Fermat little theorem. Secondly, we have\n$$\n2^{p-1} - 1 = 2^{13k} - 1 = (2^{13} - 1)(2^{13(k-1)} + 2^{13(k-2)} + \\dots + 1),\n$$\... | Hong Kong | null | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0cm2 | Problem:
A finite list of rational numbers is written on a blackboard. In an operation, we choose any two numbers $a, b$, erase them, and write down one of the numbers
$$
a+b,\ a-b,\ b-a,\ a \times b,\ a / b\ (\text{ if } b \neq 0),\ b / a\ (\text{ if } a \neq 0).
$$
Prove that, for every integer $n>100$, there are onl... | [
"Solution:\nWe prove the problem statement even for all positive integer $n$.\n\nThere are only finitely many ways of constructing a number from $n$ pairwise distinct numbers $x_{1}, \\ldots, x_{n}$ only using the four elementary arithmetic operations, and each $x_{k}$ exactly once. Each such formula for $k>1$ is o... | Romanian Master of Mathematics (RMM) | The 7th Romanian Master of Mathematics Competition | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0ho9 | Problem:
Bart has 17 and 19 dollar bills only.
a. Prove that these bills are fake.
b. Prove that there exists $m>0$ such that for each $n \geq m$ Bart can give to Lisa exactly $n$ dollars using his bills. | [
"Solution:\n\na. Since there are no real 17 or 19 dollar bills, these bills must be fake.\n\nb. We can have $m = 17 \\cdot 19$. For each $n > 17 \\cdot 19$, we consider the set $S = \\{n, n-17, n-2 \\cdot 17, \\ldots, n-18 \\cdot 17\\}$. If none of these numbers is divisible by 19, two of them will give the same re... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0iuq | Problem:
How many positive integers $n \leq 2009$ have the property that $\left\lfloor\log_{2}(n)\right\rfloor$ is odd? | [
"Solution:\nWe wish to find $n$ such that there is some natural number $k$ for which $2k-1 \\leq \\log_{2} n < 2k$. Since $n \\leq 2009$ we must have $k \\leq 5$. This is equivalent to finding the number of positive integers $n \\leq 2009$ satisfying $2^{2k-1} \\leq n < 2^{2k}$ for some $k \\leq 5$, so the number o... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | 682 | |
0du3 | Let $ABCD$ be a convex quadrilateral and $M$ be the intersection of its diagonals. Through $M$ draw a line meeting the side $AB$ at $P$ and the side $CD$ at $Q$. Find all the quadrilaterals so that there exists the segment $PQ$ that divides the triangles $ABM$ and $CDM$ into 4 similar triangles. | [
"Suppose $\\angle APM > 90^\\circ$. Then in $\\triangle BPM$, $\\angle BPM < 90^\\circ$. Since $\\angle PBM + \\angle BMP = \\angle APM$, $\\angle PBM, \\angle BMP < \\angle APM$. So none of the angles of $\\triangle BPM$ can be equal to $\\angle APM$ of $\\triangle APM$. Therefore $\\triangle APM$ cannot be simila... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | All trapezoids with AB ∥ CD for which either (i) the legs are equal (isosceles trapezoid), or (ii) the diagonals are perpendicular. | |
0l1o | The number $2024$ is written as the sum of not necessarily distinct two-digit numbers. What is the least number of two-digit numbers needed to write this sum?
(A) 20 (B) 21 (C) 22 (D) 23 (E) 24 | [
"**Answer (B):** In order to minimize the number of terms in the sum, the greatest two-digit number, $99$, should be used as many times as possible. Because $20 \\cdot 99 = 1980$, the minimum number of terms is greater than $20$. On the other hand, $2024 = 20 \\cdot 99 + 44$, so the least number of two-digit number... | United States | AMC 12 A | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | MCQ | B | |
0bwk | The hexagon $ABCDEF$ is cyclic and the distance between the midpoints of the sides $[BC]$ and $[EF]$ is equal to $\frac{AD}{2}$. Denote $H_1, H_2, H_3$ and $H_4$ the orthocenters of the triangles $ABC$, $BCD$, $DEF$ and, respectively, $EFA$. Show that the quadrilateral $H_1H_2H_3H_4$ is a rhombus.
Traian Tămâian | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
02fn | $ABC$ is acute-angled. $D$ is a variable point on the side $BC$. $O_1$ is the circumcenter of $ABD$, $O_2$ is the circumcenter of $ACD$, and $O$ is the circumcenter of $AO_1O_2$. Find the locus of $O$. | [
"Let $O'$ be the circumcenter of $ABC$ and $M, N, P$ be the respective midpoints of $AB, AC, AD$. Notice that $O', M$ and $N$ are fixed points and, by Thales theorem, $P$ is a variable point on the segment $MN$. Then $O_1$ is the intersection of the perpendicular bisectors of $AB$ and $AD$, $O_2$ is the intersectio... | Brazil | XVIII OBM | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | English | proof and answer | The locus of O is the open segment of the perpendicular bisector of AO′ between the circumcenters of triangles AO′B and AO′C. | |
0bh5 | Let $\triangle ABC$ be a triangle and $D, E, F$ be the second intersection of its medians with its circumcircle. Prove that if the triangles $BDC, CEA, AFB$ have the same area, then triangle $ABC$ is equilateral.
Marian Ionescu | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates"
] | null | proof only | null | |
0ebx | At the beginning we have a triple of pairwise distinct positive integers $(a, b, c)$ which satisfies $a + b + c = 2015$. Then in each step we replace the current triple of numbers $(x, y, z)$ with the triple $(y + z - x, z + x - y, x + y - z)$. At least how many steps must we make so that starting with a triple $(a, b,... | [
"Let's look at the sum of the triple. This does not change after each step since $(y+z-x)+(z+x-y)+(x+y-z) = x+y+z$. Moreover, the values of the new triple do not depend on the order of the numbers in the original triple, since the terms are symmetrical in all three variables. Hence we may assume $a < b < c$ and we ... | Slovenia | Selection Examinations for the IMO 2015 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Linear Algebra > Linear transformations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 9 | |
0436 | In $\triangle ABC$, $\sin A = \frac{\sqrt{2}}{2}$. Find the range of $\cos B + \sqrt{2} \cos C$. | [
"Denote $f = \\cos B + \\sqrt{2} \\cos C$.\n\nBy the given conditions, we know that $A = \\frac{\\pi}{4}$ or $A = \\frac{3\\pi}{4}$.\n\nWhen $A = \\frac{\\pi}{4}$, it follows that $B = \\frac{3\\pi}{4} - C$, where $0 < C < \\frac{3\\pi}{4}$. And there are\n$$\n\\begin{aligned}\nf &= \\cos\\left(\\frac{3\\pi}{4} - C... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | (0, 1] ∪ (2, √5] | |
09sj | Problem:
Bewijs dat voor alle positieve reële getallen $a, b, c$ geldt:
$$
a+\sqrt{a b}+\sqrt[3]{a b c} \leq \frac{4}{3}(a+b+c)
$$ | [
"Solution:\nWe kunnen $\\sqrt[3]{a b c}$ schrijven als $\\sqrt[3]{\\frac{a}{4} \\cdot b \\cdot 4 c}$. Toepassen van de ongelijkheid van het rekenkundig en meetkundig gemiddelde op de (positieve) termen $\\frac{a}{4}, b$ en $4 c$ geeft dan\n$$\n\\sqrt[3]{a b c}=\\sqrt[3]{\\frac{a}{4} \\cdot b \\cdot 4 c} \\leq \\fra... | Netherlands | IMO-selectietoets II | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null |
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