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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0k1m | Problem:
Solve for $x$ :
$$
x\lfloor x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\rfloor=122 .
$$ | [
"Solution:\nThis problem can be done without needless casework.\n(For negative values of $x$, the left hand side will be negative, so we only need to consider positive values of $x$.)\nThe key observation is that for $x \\in [2,3)$, $122$ is an extremely large value for the expression. Indeed, we observe that:\n$$\... | United States | HMMT February 2018 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 122/41 | |
0aeq | Провери ја точноста на равенството:
$$
\sum_{k=1}^{n} \frac{(-1)^{k+1}}{k} \binom{n}{k} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n},
$$ | [
"Равенството ќе го докажеме со принципот на математичка индукција.\nКе воведеме ознака\n$$\nx_n = \\sum_{k=1}^{n} \\frac{(-1)^{k+1}}{k} \\binom{n}{k}\n$$\nЈасно е дека $x_{n-1} = \\sum_{k=1}^{n-1} \\frac{(-1)^{k+1}}{k} \\binom{n-1}{k}$.\nТврдењето е точно за $n=1$. Навистина\n$$\nx_1 = \\sum_{k=1}^{1} \\frac{(-1)^{... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | Macedonian, English | proof only | null | |
0fsn | Problem:
Es sei $S=\{P_{1}, P_{2}, \ldots, P_{2000}\}$ eine Menge von 2000 Punkten im Innern eines Kreises vom Radius 1, sodass einer der Punkte der Kreismittelpunkt ist. Für $i=1,2, \ldots, 2000$ bezeichne $x_{i}$ den Abstand von $P_{i}$ zum nächstgelegenen Punkt $P_{j} \neq P_{i}$ aus $S$. Zeige, dass gilt
$$
x_{1}^... | [
"Solution:\n\nDa einer der Punkte der Kreismittelpunkt ist, gilt $x_{i}<1$ für alle $i$. Zeichne um jeden Punkt $P_{i} \\in S$ einen Kreis mit Radius $x_{i} / 2$. Je zwei dieser Kreise haben höchstens einen Randpunkt gemeinsam. Falls nicht, dann gibt es zwei Indizes $i, j$ und einen Punkt $A$ der im Innern des Krei... | Switzerland | IMO - Selektion | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
00i2 | Prove that for every positive integer $t$ there is a unique permutation $a_{0}, a_{1}, \ldots, a_{t-1}$ of $0,1, \ldots, t-1$ such that, for every $0 \leq i \leq t-1$, the binomial coefficient $\binom{t+i}{2 a_{i}}$ is odd and $2 a_{i} \neq t+i$. | [
"We constantly make use of Kummer's theorem which, in particular, implies that $\\binom{n}{k}$ is odd if and only if $k$ and $n-k$ have ones in different positions in binary. In other words, if $S(x)$ is the set of positions of the digits 1 of $x$ in binary (in which the digit multiplied by $2^{i}$ is in position $... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof only | null | |
0bkk | Determine all real numbers $x, y, z \in (0,1)$ that satisfy simultaneously the conditions:
$$
\begin{cases}
(x^2 + y^2)\sqrt{1-z^2} \geq z \\
(y^2 + z^2)\sqrt{1-x^2} \geq x \\
(z^2 + x^2)\sqrt{1-y^2} \geq y
\end{cases}
$$ | [
"The first inequality is equivalent to\n$$\n\\frac{z^2}{x^2 + y^2} \\leq z\\sqrt{1-z^2}.\n$$\nSince\n$$\nz\\sqrt{1-z^2} = \\sqrt{z^2(1-z^2)} \\leq \\frac{z^2 + 1 - z^2}{2} = \\frac{1}{2},\n$$\nit follows that $\\frac{z^2}{x^2 + y^2} \\leq 2$, and therefore $x^2 + y^2 \\geq 2z^2$. Writing the other two similar inequ... | Romania | 65th NMO Selection Tests for JBMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | x = y = z = sqrt(2)/2 | |
0k9l | Problem:
Let $ABC$ be an acute triangle with circumcircle $\gamma$ and incenter $I$. Let $D$ be the midpoint of minor $\operatorname{arc} \widehat{BC}$ of $\gamma$. Let $P$ be the reflection of the incenter of $ABC$ over side $BC$. Suppose line $DP$ meets $\gamma$ again at a point $Q$ on minor $\operatorname{arc} \wid... | [
"Solution:\n\nLet $R = \\overline{DQ} \\cap \\overline{BC}$ and let $T$ denote the foot of tangency from the incircle to $\\overline{BC}$. Let $K = \\overline{AD} \\cap \\overline{BC}$.\n\n\n\nIt's well-known that $D$ is the circumcenter of $\\triangle BIC$ (trillium theorem). Moreover, by ... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
... | null | proof only | null | |
0arq | Problem:
What are the last two digits of $5^{2011}$? | [
"Solution:\n(ans. 25)"
] | Philippines | 13th Philippine Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 25 | |
06cv | Let $a$, $b$, $c$ be positive and $abc = 1$. Prove that
$$
\frac{1 + ab^2}{c^3} + \frac{1 + bc^2}{a^3} + \frac{1 + ca^2}{b^3} \geq \frac{18}{a^3 + b^3 + c^3}.
$$
Determine when equality is attained. | [
"We have\n$$\n(a^3 + b^3 + c^3) \\left( \\frac{1 + ab^2}{c^3} + \\frac{1 + bc^2}{a^3} + \\frac{1 + ca^2}{b^3} \\right) \\\\ = \\sum_{\\text{sym}} \\frac{a^3}{c^3} + \\sum_{\\text{cyc}} \\frac{a^4 b^2}{c^3} + \\sum_{\\text{cyc}} \\frac{ab^5}{c^3} + \\sum_{\\text{cyc}} ab^2 + 3.\n$$\nApplying the AM-GM inequality to ... | Hong Kong | HKG TST | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Equality holds when a = b = c = 1. | |
03z7 | Given a $3 \times 9$ array $A$ with each cell containing a positive integer, we say a $m \times n$ $(1 \le m \le 3,\ 1 \le n \le 9)$ subarray of $A$ is a “good rectangle” if the sum of the numbers in its cells is a multiple of $10$, and call a $1 \times 1$ cell of $A$ “bad” if it is not contained in any “good rectangle... | [
"We first claim that the number of “bad cells” in $A$ is no more than $25$. Otherwise, there will be at most one cell in $A$ that is not “bad”. Without loss of generality, we assume the cells in the first row of $A$ are all “bad”. Then let the numbers from top to bottom in the $i$th column be $a_i, b_i, c_i$ ($i = ... | China | China Mathematical Competition (Complementary Test) | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other"
] | English | proof and answer | 25 | |
0knd | Problem:
Let $AD$, $BE$, and $CF$ be segments sharing a common midpoint, with $AB < AE$ and $BC < BF$. Suppose that each pair of segments forms a $60^{\circ}$ angle, and that $AD = 7$, $BE = 10$, and $CF = 18$. Let $K$ denote the sum of the areas of the six triangles $\triangle ABC$, $\triangle BCD$, $\triangle CDE$, ... | [
"Solution:\n\nLet $M$ be the common midpoint, and let $x = 7$, $y = 10$, $z = 18$. One can verify that hexagon $ABCDEF$ is convex. We have\n\n$[ABC] = [ABM] + [BCM] - [ACM] = \\frac{1}{2} \\cdot \\frac{\\sqrt{3}}{2} \\cdot \\frac{x}{2} \\cdot \\frac{y}{2} + \\frac{1}{2} \\cdot \\frac{\\sqrt{3}}{2} \\cdot \\frac{y}{... | United States | HMMT Spring 2021 Guts Round | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 141 | |
0i3u | Problem:
Two circles are concentric. The area of the ring between them is $A$. In terms of $A$, find the length of the longest chord contained entirely within the ring. | [
"Solution:\n\nLet the radii of the circles be $r$ and $R > r$, so $A = \\pi (R^2 - r^2)$. By the Pythagorean theorem, the length of the chord is $2 \\sqrt{R^2 - r^2} = 2 \\sqrt{\\frac{A}{\\pi}}$."
] | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 2 * sqrt(A / pi) | |
0i4b | Problem:
One fair die is rolled; let $a$ denote the number that comes up. We then roll $a$ dice; let the sum of the resulting $a$ numbers be $b$. Finally, we roll $b$ dice, and let $c$ be the sum of the resulting $b$ numbers. Find the expected (average) value of $c$. | [
"Solution:\n\n$343 / 8$\n\nThe expected result of an individual die roll is $(1+2+3+4+5+6)/6 = 7/2$. For any particular value of $b$, if $b$ dice are rolled independently, then the expected sum is $(7/2) b$. Likewise, when we roll $a$ dice, the expected value of their sum $b$ is $(7/2) a$, so the expected value of ... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 343/8 | |
0a69 | Problem:
Prove the following inequality
$$
\frac{6}{2024^{3}} < \left(1 - \frac{3}{4}\right)\left(1 - \frac{3}{5}\right)\left(1 - \frac{3}{6}\right)\left(1 - \frac{3}{7}\right)\dots \left(1 - \frac{3}{2025}\right).
$$ | [
"Solution:\n\n$$\n\\frac{1}{4} \\times \\frac{2}{5} \\times \\frac{3}{6} \\times \\frac{4}{7} \\times \\frac{5}{8} \\times \\frac{6}{9} \\times \\frac{7}{10} \\times \\dots \\times \\frac{2022}{2025}\n$$\n\n$$\n= \\frac{1 \\times 2 \\times 3 \\times 4 \\times 5 \\times 6 \\times 7 \\times \\dots \\times 2022}{4 \\t... | New Zealand | NZMO Round One | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
065v | Determine the integer solutions of the equation
$$
8x^3 - 4 = y(6x - y^2)
$$ | [
"The given equation can be written as\n$$\n\\begin{aligned}\n8x^3 + y^3 - 6xy &= 4 \n\\Leftrightarrow (2x)^3 + y^3 + 1^3 - 3 \\cdot 2x \\cdot y \\cdot 1 = 5 \\\\\n&\\Leftrightarrow (2x + y + 1)(4x^2 + y^2 + 1 - 2xy - 2x - y) = 5\n\\end{aligned} \\quad (1)\n$$\n$$\n\\Leftrightarrow \\frac{1}{2}(2x+y+1)[(2x-y)^2+(2x-... | Greece | SELECTION EXAMINATION | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | (1, 2) | |
0ao3 | Problem:
A point $(x, y)$ is called a lattice point if $x$ and $y$ are integers. How many lattice points are there inside the circle of radius $2 \sqrt{2}$ with center at the origin?
(a) 25
(b) 21
(c) 17
(d) 19 | [] | Philippines | Qualifying Round | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | b | |
0jxm | Problem:
A sequence of positive integers $a_{1}, a_{2}, \ldots, a_{2017}$ has the property that for all integers $m$ where $1 \leq m \leq 2017$, $3\left(\sum_{i=1}^{m} a_{i}\right)^{2}=\sum_{i=1}^{m} a_{i}^{3}$. Compute $a_{1337}$. | [
"Solution:\nI claim that $a_{i}=3i$ for all $i$. We can conjecture that the sequence should just be the positive multiples of three because the natural numbers satisfy the property that the square of their sum is the sum of their cubes, and prove this by induction. At $i=1$, we have that $3 a_{i}^{2}=a_{i}^{3}$, so... | United States | HMMT November | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 4011 | |
0hsx | Problem:
Define a function $f$ on the real numbers by
$$
f(x)= \begin{cases}2 x & \text{ if } x<1 / 2 \\ 2 x-1 & \text{ if } x \geq 1 / 2\end{cases}
$$
Determine all values $x$ satisfying $f(f(f(f(f(x)))))=x$. | [
"Solution:\nThe answer is the 32 values $0, \\frac{1}{31}, \\frac{2}{31}, \\ldots, \\frac{30}{31}, 1$.\n\nIf $x<0$, then $f(x)=2 x<x$ so the sequence $x, f(x), f(f(x)), \\ldots$ is strictly decreasing and cannot return to $x$.\n\nIf $x>1$, similarly $f(x)=2 x-1>x$ so the sequence $x, f(x), f(f(x)), \\ldots$ is stri... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 0, 1/31, 2/31, ..., 30/31, 1 | |
0fyt | Problem:
Seien $x_{1}, \ldots, x_{8} \geq 0$ reelle Zahlen, sodass für $i=1, \ldots, 8$ gilt $x_{i}+x_{i+1}+x_{i+2} \leq 1$, wobei $x_{9}=x_{1}$ und $x_{10}=x_{2}$. Beweise die Ungleichung
$$
\sum_{i=1}^{8} x_{i} x_{i+2} \leq 1
$$
und finde alle Fälle in denen Gleichheit herrscht. | [
"Solution:\n\nFür $1 \\leq i \\leq 8$ gilt die Abschätzung\n$$\n\\begin{aligned}\na_{i} a_{i+2}+a_{i+1} a_{i+3} & \\leq\\left(1-a_{i+1}-a_{i+2}\\right) a_{i+2}+a_{i+1}\\left(1-a_{i+1}-a_{i+2}\\right) \\\\\n& =\\left(a_{i+1}+a_{i+2}\\right)\\left(1-a_{i+1}-a_{i+2}\\right) \\\\\n& \\leq \\frac{1}{4}\\left(a_{i+1}+a_{... | Switzerland | IMO Selektion | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Maximum is 1, with equality only for (1/2, 0, 1/2, 0, 1/2, 0, 1/2, 0) or (0, 1/2, 0, 1/2, 0, 1/2, 0, 1/2). | |
06lt | In acute $\triangle ABC$, $BC > AB$. Let $D$ be the point on the side $AC$ such that $BD = BA$. Let $M$ be the midpoint of $BC$. The tangents at $D$ and $M$ to the circumcircle of $\triangle CDM$ intersect at point $E$. The line $BE$ intersects the side $AC$ at $F$. Prove that $A, B, M, F$ are concyclic. | [
"Suppose $BD$ meets ($CDM$) again at $P$. Applying Pascal's theorem to the points $MMCDDP$, we know that $MM \\cap DD = E$, $MC \\cap DP = B$ and $CD \\cap PM$ are collinear. Thus, $MP$ passes through $BE \\cap CD = F$. Now, since\n$$\n\\angle CMF = \\angle ADB = \\angle BAF,\n$$\nthe points $A, B, M, F$ are concyc... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07yi | Problem:
Determinare tutte le coppie di numeri interi positivi $\left(a, n\right)$ con $a \geq n \geq 2$ per cui il numero $(a+1)^{n}+a-1$ è una potenza di $2$. | [
"Solution:\n\nSe sviluppiamo $(a+1)^{n}+a-1$ usando il binomio di Newton, otteniamo:\n$$\na^{n}+\\cdots+\\frac{n(n-1)}{2} a^{2}+n a+1+a-1=a^{n}+\\cdots+\\frac{n(n-1)}{2} a^{2}+(n+1) a.\n$$\nQuindi, siccome tutti i termini sono divisibili per $a$, e siccome $(a+1)^{n}+a-1$ è una potenza di $2$, anche $a$ è una poten... | Italy | null | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a=4, n=3 | |
0azx | Problem:
The number $\overline{1 a b 76}$ is divisible by $72$. List down all the possible values of $a+b$. | [
"Solution:\n\nSince $\\overline{1 a b 76}$ is divisible by $72$, it is divisible by $8$ and $9$.\n\nSince it is divisible by $9$, $1+a+b+7+6$ or $14+a+b$ is divisible by $9$. Therefore, $a+b$ is necessarily $4$ or $13$.\n\nThe only other condition we require is that $\\overline{b 76}$ must be divisible by $8$, whic... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 4, 13 | |
0bje | On each side of an equilateral triangle of side $n \ge 1$ consider $n-1$ points that divide the sides into $n$ equal segments. Through these points draw parallel lines to the sides of the triangle, obtaining a net of equilateral triangles of side length $1$. On each of the vertices of the small triangles put a coin hea... | [
"Obviously, such turning is possible for $n = 1$. For $n = 2$, flip each of the four $1$-sided equilateral triangles once and all the coins will be tail-up.\n\nWe shall use now induction of step $3$. Assume that $n$ is an admissible value. Flipping the coins of each unit sided triangle of an equilateral triangle of... | Romania | 65th NMO Selection Tests for JBMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | All positive integers not divisible by three (i.e., side lengths congruent to 1 or 2 modulo 3). | |
035q | Problem:
Let $a_{1}, a_{2}, \ldots, a_{m}$ be arbitrary positive integers. Prove that there exist distinct positive integers $b_{1}, b_{2}, \ldots, b_{n}$, $n \leq m$, such that the following two conditions are satisfied:
(1) all subsets of $\{b_{1}, b_{2}, \ldots, b_{n}\}$ have distinct sums of elements;
(2) every num... | [
"Solution:\nWe shall prove the assertion by induction on $N = a_{1} + a_{2} + \\cdots + a_{m}$. For $N = 1$ we have $m = 1$, $a_{1} = 1$ and $b_{1} = 1$ is the required number.\n\nLet us assume that the assertion is true for every collection with sum less than $N$ and let $a_{1}, a_{2}, \\ldots, a_{m}$ be such that... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0gax | 三角形 $ABC$ 中, $BC > AB$。設 $L$ 為 $\angle ABC$ 的內角平分線。由 $A, C$ 分別對 $L$ 引垂線, 設垂足分別為 $P, Q$。令 $M, N$ 分別是 $AC$ 邊與 $BC$ 邊的中點。設三角形 $PQM$ 的外接圓圓心為 $O$, 且該圓與 $AC$ 的另一個交點為 $H$。證明: $O, M, N, H$ 共圓。 | [
"延伸 $AP$，交 $BC$ 於 $D$ 點，則 $P$ 為 $AD$ 中點。因為 $M$ 是 $AC$ 中點，故 $PM$ 與 $CD$ (即 $BC$) 平行，且\n$$\n\\angle QPM = \\angle QBC = \\frac{1}{2} \\angle ABC.\n$$\n同理可知 $\\angle MQP = \\frac{1}{2}\\angle ABC$, 故 $PM = QM$.\n\n\n\n$$\n\\angle QHC = \\angle QHM = \\angle QPM,\n$$\n即 $\\angle QHC = \\angle Q... | Taiwan | 二〇一七數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | null | proof only | null | |
04m3 | Let $M$ be a point in the interior of the triangle $ABC$. The line $AM$ intersects the circumcircle of the triangle $MBC$ once more at $D$, the line $BM$ intersects the circumcircle of the triangle $MCA$ once more at $E$, and the line $CM$ intersects the circumcircle of the triangle $MAB$ once more at $F$. Prove the in... | [
"It suffices to show that\n$$\n\\frac{|AM|}{|MD|} + \\frac{|BM|}{|ME|} + \\frac{|CM|}{|MF|} \\ge \\frac{3}{2}.\n$$\nQuadrilaterals $MBDC$ and $MCEA$ are cyclic, so\n$$\n\\angle BCD = \\angle BMD = \\angle EMA = \\angle ECA.\n$$\nAlso, we have\n$$\n\\angle DBC = \\angle DMC = 180^\\circ - \\angle CMA = \\angle CEA.\... | Croatia | Croatian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0h9r | Let $M$, $D$ and $K$ belong to the sides $AB$, $BC$ and $CA$ of isosceles triangle $ABC$ with apex $B$ such that $AM = 2DC$ and $\angle AMD = \angle KDC$. Show that $MD = KD$. | [
"Let $FD \\parallel AC$ (Fig. 24).\n\n\n\nThen $AF = FM = DC$, hence $\\triangle FMD = \\triangle KDC$ by the side and adjacent angles. Thus, $MD = KD$."
] | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
05t7 | Problem:
1. Existe-t-il des nombres $a_{0}, \ldots, a_{2020}$ valant -1 ou 1 tels que $a_{0} \times a_{1}+a_{1} \times a_{2}+\cdots+a_{2019} \times a_{2020}+a_{2020} \times a_{0}=1010$ ?
2. Existe-t-il des nombres $a_{1}, \ldots, a_{2020}$ valant -1 ou 1 tels que $a_{1} \times a_{2}+a_{2} \times a_{3}+\cdots+a_{2019}... | [
"Solution:\n\nDéjà regardons l'énoncé : on veut savoir si on peut trouver des nombres $\\left(a_{i}\\right)_{1 \\leqslant i \\leqslant n}$ valant +1 ou -1 tels que $a_{1} a_{2}+\\cdots+a_{n} a_{1}$ vaut 1010 avec dans la première question $n=2021$, dans la seconde $n=2020$. Pour cela on peut tester avec des $n$ pet... | France | Envoi 5: Pot Pourri | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0eu3 | A positive integer $N$ is said to be an "*n*-good number" if it satisfies the following two properties:
(Property 1) $N$ is divisible by at least $n$ distinct primes
(Property 2) There exist distinct positive divisors $1, x_2, \dots, x_n$ of $N$ such that
$$
1 + x_2 + \dots + x_n = N.
$$
Show that there exists an "*n*-... | [
"We use an induction on $n$.\n\na. For $n = 6$, put $N_6 = 2 \\cdot 3 \\cdot 7 \\cdot 43 \\cdot 1807 = 1806 \\cdot 1807$. Since $1807 = 13 \\cdot 139$, $N_6$ is divisible by 6 distinct primes.\nMoreover, since\n$$\n\\frac{1}{m} = \\frac{1}{m+1} + \\frac{1}{m(m+1)},\n$$\nwe have\n\n$$\n\\begin{align*}\n1 &= \\frac{1... | South Korea | Korean Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0hse | Problem:
Suppose $2011$ light bulbs are arranged in a row. Each bulb has a button under it. Pressing the button will change the state of the bulb above it (on to off or vice versa) and will also change the two neighboring bulbs, or the single neighboring bulb in the case of one of the two end buttons. Is it always pos... | [
"Solution:\n\nThe answer is yes.\nLet us number the bulbs $1$ to $2011$ from left to right. Given any initial state of the bulbs, let us begin by following this algorithm: As long as at least one bulb other than bulb $1$ is on, let $n$ be the number of the rightmost lit bulb and push the button for bulb $n-1$. This... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
0gb2 | 已知 $a, b, c, d > 0$, 試證:
$$
\sum_{cyc} \frac{c}{a + 2b} + \sum_{cyc} \frac{a + 2b}{c} \geq 8 \left( \frac{(a + b + c + d)^2}{ab + ac + ad + bc + bd + cd} - 1 \right),
$$
其中 $\sum_{cyc} f(a, b, c, d) = f(a, b, c, d) + f(d, a, b, c) + f(c, d, a, b) + f(b, c, d, a)$ | [
"注意到\n$$\n\\frac{c}{a+2b} = \\frac{a+2b+c}{a+2b} - 1, \\quad \\frac{a+2b}{c} = \\frac{a+2b+c}{c} - 1\n$$\n所以我們有:\n$$\n\\begin{aligned} \\sum_{cyc} \\frac{c}{a+2b} + \\sum_{cyc} \\frac{a+2b}{c} &= \\sum_{cyc} (a+2b+c) \\left( \\frac{1}{a+2b} + \\frac{1}{c} \\right) - 8 \\\\ &= \\sum_{cyc} \\frac{(a+2b+c)^2}{c(a+2b)}... | Taiwan | 二〇一七數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0jnr | Problem:
Find the smallest positive integer $b$ such that $1111_{b}$ (1111 in base $b$) is a perfect square. If no such $b$ exists, write "No solution". | [
"Solution:\nAnswer: 7\nWe have $1111_{b} = b^{3} + b^{2} + b + 1 = (b^{2} + 1)(b + 1)$. Note that $\\gcd(b^{2} + 1, b + 1) = \\gcd(b^{2} + 1 - (b + 1)(b - 1), b + 1) = \\gcd(2, b + 1)$, which is either $1$ or $2$. If the $\\gcd$ is $1$, then there is no solution as this implies $b^{2} + 1$ is a perfect square, whic... | United States | HMMT November 2015 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 7 | |
0kin | Problem:
Prove that if $x$ is a positive real number such that $x + x^{-1}$ is an integer, then $x^{3} + x^{-3}$ is an integer as well. | [
"Solution:\n\nThis follows from the identity\n$$\nx^{3} + \\frac{1}{x^{3}} = \\left(x + \\frac{1}{x}\\right)^{3} - 3\\left(x + \\frac{1}{x}\\right).\n$$\n\nIf $x + x^{-1}$ is an integer, then so is $\\left(x + x^{-1}\\right)^{3} - 3\\left(x + x^{-1}\\right)$, and thus $x^{3} + x^{-3}$ is an integer."
] | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0ick | Problem:
You want to arrange the numbers $1,2,3, \ldots, 25$ in a sequence with the following property: if $n$ is divisible by $m$, then the $n$th number is divisible by the $m$th number. How many such sequences are there? | [
"Solution:\nLet the rearranged numbers be $a_{1}, \\ldots, a_{25}$. The number of pairs $(n, m)$ with $n \\mid m$ must equal the number of pairs with $a_{n} \\mid a_{m}$, but since each pair of the former type is also of the latter type, the converse must be true as well. Thus, $n \\mid m$ if and only if $a_{n} \\m... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 24 | |
0k8j | Problem:
Find all positive integers $n$ for which there do not exist $n$ consecutive composite positive integers less than $n!$. | [
"Solution:\nAnswer: $1,2,3,4$\n\nFirst, note that clearly there are no composite positive integers less than $2!$, and no $3$ consecutive composite positive integers less than $3!$. The only composite integers less than $4!$ are\n$$\n4,6,8,9,10,12,14,15,16,18,20,21,22\n$$\nand it is easy to see that there are no $4... | United States | HMMT February 2019 Team Round | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 1, 2, 3, 4 | |
0l15 | Let
$$
P(m) = \frac{m}{2} + \frac{m^2}{4} + \frac{m^4}{8} + \frac{m^8}{8}.
$$
How many of the values $P(2022)$, $P(2023)$, $P(2024)$, and $P(2025)$ are integers? | [
"**Answer (E):** Let $Q(m) = 8P(m) = 4m + 2m^2 + m^4 + m^8$. Because the coefficients of $Q$ are integers, it follows that if $a \\equiv b \\pmod{8}$, then $Q(a) \\equiv Q(b) \\pmod{8}$. It suffices to show that the 8 numbers $Q(-3)$, $Q(-2)$, $Q(-1)$, ..., $Q(4)$ are all divisible by 8. If $m$ is even, then each o... | United States | AMC 10 B | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | E | |
0apj | Problem:
By how much does the sum of the first 15 positive odd integers exceed the sum of the first 10 positive even integers? | [
"Solution:\nWe use the formula for the sum of an arithmetic series.\n\n$$\n\\frac{15}{2}(2 + 14 \\cdot 2) - \\frac{10}{2}(4 + 9 \\cdot 2) = 15^{2} - 10 \\cdot 11 = 115\n$$"
] | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 115 | |
0gg3 | 證明：對於 $\{1, 2, 3, ..., 5^{505}\}$ 的任一個恰有 $2022$ 個元素的子集 $A$，必存在三個元素 $a, b, c$ 滿足 $a < b < c$ 和 $c + 2a > 3b$。 | [
"參考解答一 (By contradiction). Suppose that there exist $2022$ positive integers $x_0 < x_1 < \\ldots < x_{2021}$ that violate the problem statement. Then in particular $x_{2021} + 2x_i \\le 3x_{i+1}$ for all $i = 0, \\ldots, 2020$ which gives\n$$\nx_{2021} - x_i \\ge \\frac{3}{2}(x_{2021} - x_{i+1}).\n$$\nBy a trivial... | Taiwan | 2022 數學奧林匹亞競賽第一階段選訓營, 國際競賽實作(二) | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | Chinese; English | proof only | null | |
0j11 | Problem:
Jacob flipped a fair coin five times. In the first three flips, the coin came up heads exactly twice. In the last three flips, the coin also came up heads exactly twice. What is the probability that the third flip was heads? | [
"Solution:\n\nHow many sequences of five flips satisfy the conditions, and have the third flip be heads?\n\nWe have __H__-, so exactly one of the first two flips is heads, and exactly one of the last two flips is heads. This gives $2 \\times 2 = 4$ possibilities.\n\nHow many sequences of five flips satisfy the cond... | United States | Harvard-MIT November Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 4/5 | |
0exg | Problem:
An alien moves on the surface of a planet with speed not exceeding $u$. A spaceship searches for the alien with speed $v$. Prove the spaceship can always find the alien if $v > 10u$. | [
"Solution:\n\nThe spacecraft flies at a constant height, so that it can see a circular spot on the surface. It starts at the north pole and spirals down to the south pole, overlapping its previous track on each circuit. The alien cannot move fast enough to cross the track before the next circuit, so it is trapped i... | Soviet Union | 5th ASU | [
"Geometry > Non-Euclidean Geometry > Spherical Geometry",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0el5 | Problem:
Dan je pravokotni trikotnik $ABC$ s pravim kotom pri $A$. Na stranicah $AB$, $BC$ in $CA$ zaporedoma ležijo točke $D$, $E$ in $F$, tako da velja $|BD| = |BE|$ in $|CF| = |CE|$ (glej sliko). Koliko stopinj je velikost kota $\angle FED$?
(A) 30
(B) 37,5
(C) 45
(D) 52,5
(E) Nemogoče je... | [
"Solution:\n\nOznačimo kote trikotnika $ABC$ kot običajno z $\\alpha$, $\\beta$ in $\\gamma$. Tedaj je $\\alpha = 90^\\circ$ in zato je $\\beta + \\gamma = 90^\\circ$.\n\nKer sta trikotnika $DBE$ in $ECF$ enakokraka z vroma pri $B$ in $C$, je\n$$\n\\angle DEB = \\frac{180^\\circ - \\beta}{2} = 90^\\circ - \\frac{\\... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | MCQ | C | |
0hte | Problem:
Let $\{a_{1}, a_{2}, a_{3}, \ldots\}$ be a sequence of real numbers such that for each $n \geq 1$,
$$
a_{n+2}=a_{n+1}+a_{n}
$$
Prove that for all $n \geq 2$, the quantity
$$
\left|a_{n}^{2}-a_{n-1} a_{n+1}\right|
$$
does not depend on $n$. | [
"Solution:\nIt suffices to prove that increasing $n$ to $n+1$ does not change the value, i.e. that for $n \\geq 2$,\n$$\n\\left|a_{n}^{2}-a_{n-1} a_{n+1}\\right|=\\left|a_{n+1}^{2}-a_{n} a_{n+2}\\right| .\n$$\nWe will prove more specifically that\n$$\na_{n}^{2}-a_{n-1} a_{n+1}=-\\left(a_{n+1}^{2}-a_{n} a_{n+2}\\rig... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0axn | Problem:
Inside a square of side length $1$, four quarter-circle arcs are traced with the edges of the square serving as the radii. It is known that these arcs intersect pairwise at four distinct points, which in fact are the vertices of a smaller square. Suppose this process is repeated for the smaller square, and so... | [
"Solution:\n\nBy similarity, we note that the areas of the squares are in geometric progression. Hence, we need only to find out the area of the first smaller square. Note that the diagonal of the smaller square is the overlap of two line segments of length $\\frac{\\sqrt{3}}{2}$, with a total length of $1$. This i... | Philippines | Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (1+sqrt(3))/2 | |
065s | a. Let $n$ be a positive integer. Prove that
$$
n\sqrt{x-n^2} \le \frac{x}{2}, \text{ for all } x \ge n^2.
$$
b. Determine real numbers $x$, $y$, $z$ satisfying the equation
$$
2\sqrt{x-1} + 4\sqrt{y-4} + 6\sqrt{z-9} = x + y + z.
$$ | [
"a. Since $x \\ge n^2$, we have\n$$\nn\\sqrt{x-n^2} \\le \\frac{x}{2} \\Leftrightarrow 2n\\sqrt{x-n^2} \\le x \\Leftrightarrow 4n^2(x-n^2) \\le x^2 \\Leftrightarrow (x-2n^2)^2 \\ge 0,\n$$\nwhich is valid. Equality holds if and only if $x = 2n^2$.\n\nAlternatively, for every $x \\ge n^2$, it is enough to prove that\... | Greece | SELECTION EXAMINATION | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | x = 2, y = 8, z = 18 | |
0itm | Problem:
Let $a_{0} = \frac{6}{7}$, and
$$
a_{n+1} = \begin{cases}2 a_{n} & \text{ if } a_{n} < \frac{1}{2} \\ 2 a_{n} - 1 & \text{ if } a_{n} \geq \frac{1}{2}\end{cases}
$$
Find $a_{2008}$. | [
"Solution:\nWe calculate the first few $a_{i}$:\n$$\na_{1} = \\frac{5}{7}, \\quad a_{2} = \\frac{3}{7}, \\quad a_{3} = \\frac{6}{7} = a_{0}\n$$\nSo this sequence repeats every three terms, so $a_{2007} = a_{0} = \\frac{6}{7}$. Then $a_{2008} = \\frac{5}{7}$."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 5/7 | |
0050 | Un tablero de $7 \times 7$ tiene una lámpara en cada una de sus 49 casillas, que puede estar encendida o apagada. La operación permitida es elegir 3 casillas consecutivas de una fila o de una columna que tengan dos lámparas vecinas entre sí encendidas y la otra apagada, y cambiar el estado de las tres. Es decir:
![](at... | [] | Argentina | XIIIª OLIMPÍADA de MAYO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | Español | proof and answer | null | |
0ky4 | Problem:
The largest prime factor of $101101101101$ is a four-digit number $N$. Compute $N$. | [
"Solution:\nNote that\n$$\n\\begin{aligned}\n101101101101 & = 101 \\cdot 1001001001 \\\\\n& = 101 \\cdot 1001 \\cdot 1000001 \\\\\n& = 101 \\cdot 1001 \\cdot (100^3 + 1) \\\\\n& = 101 \\cdot 1001 \\cdot (100 + 1)(100^2 - 100 + 1) \\\\\n& = 101 \\cdot 1001 \\cdot 101 \\cdot 9901 \\\\\n& = 101^2 \\cdot 1001 \\cdot 99... | United States | HMMT November 2023 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 9901 | |
0ir5 | Problem:
How many ways can you color the squares of a $2 \times 2008$ grid in 3 colors such that no two squares of the same color share an edge? | [
"Solution:\n\n$2 \\cdot 3^{2008}$\n\nDenote the colors $A$, $B$, $C$. The left-most column can be colored in $6$ ways. For each subsequent column, if the $k$th column is colored with $A B$, then the $(k+1)$th column can only be colored with one of $B A$, $B C$, $C A$. That is, if we have colored the first $k$ colum... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | final answer only | 2*3^2008 | |
0fcc | Problem:
Las longitudes de los lados y de las diagonales de un cuadrilátero convexo plano $A B C D$ son racionales. Si las diagonales $A C$ y $B D$ se cortan en el punto $O$, demuestra que la longitud $O A$ es también racional. | [
"Solution:\n\nSean $\\angle A B D=\\alpha$, $\\angle C B D=\\gamma$ y $\\angle C B A=\\beta$.\n\nPor el teorema del coseno en el triángulo $\\triangle A B C$, $\\cos \\beta=\\frac{A B^{2}+B C^{2}-C A^{2}}{2 A B \\cdot B C}$ es un número racional. Análogamente $\\cos \\alpha$ y $\\cos \\gamma$ son números racionales... | Spain | null | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0ju3 | Problem:
Let $C_{k, n}$ denote the number of paths on the Cartesian plane along which you can travel from $(0,0)$ to $(k, n)$, given the following rules:
1) You can only travel directly upward or directly rightward
2) You can only change direction at lattice points
3) Each horizontal segment in the path must be at mos... | [
"Solution:\nAnswer: $100^{17}$\nIf we are traveling from $(0,0)$ to $(n, 17)$, we first travel $x_{0}$ rightwards, then up one, then $x_{1}$ rightwards, then up one, $\\ldots$, until we finally travel $x_{17}$ rightwards. $x_{0}, \\ldots, x_{17}$ are all at most 99 by our constraint, but can equal 0. Given that $x_... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 100^{17} | |
03ls | Problem:
The vertices of a right triangle $ABC$ inscribed in a circle divide the circumference into three arcs. The right angle is at $A$, so that the opposite $\operatorname{arc} BC$ is a semicircle while $\operatorname{arc} AB$ and $\operatorname{arc} AC$ are supplementary. To each of the three arcs, we draw a tangen... | [
"Solution:\nA prime indicates where a tangent meets $AB$ and a double prime where it meets $AC$. It is given that $DD' = DD''$, $EE' = EE''$ and $FF' = FF''$. It is required to show that arc $EF$ is a third of the circumference as is arc $DBF$.\n\n$AF$ is the median to the hypotenuse of right triangle $AF'F''$, so ... | Canada | 38th Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00kb | Let $S$ be the set of all real numbers greater than or equal to $1$. Determine all functions $f: S \to S$ such that $f(x^2 - y^2) = f(xy)$ holds for all numbers $x, y \in S$ with $x^2 - y^2 \in S$. | [
"Let $z > 2$. We consider the function $g: S \\to \\mathbb{R}$ with $g(x) = x^2 - z^2/x^2$. As $x \\mapsto x^2$ and $x \\mapsto -z^2/x^2$ are both increasing functions for $x > 0$, $g$ is also increasing and obviously continuous. As $g(1) = 1 - z^2 < 0$ and $g(x) = z^2 - 1 > 1$, there is an $x_0 \\ge 1$ such that $... | Austria | Austria 2014 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | All constant functions f(x) = c for x in S, where c is any real number at least 1. | |
0bry | Determine all functions $f$ of the set of positive integers into itself such that $f(m) \ge m$ and $f(m+n)$ divides $f(m) + f(n)$ for all positive integers $m$ and $n$. | [
"To this end, write $\\ell = f(1)$, and notice that, since $1 \\le f(n)/n \\le \\ell$, there exists a minimal positive integer $k \\le \\ell$ such that $\\lfloor f(n)/n \\rfloor = k$ for infinitely many positive integers $n$. Let $A$ denote the set of all these positive integers, let $B = \\{n: n \\in A \\text{ and... | Romania | 67th NMO Selection Tests for BMO and IMO | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Other"
] | English | proof and answer | f(n) = c n for some positive integer c | |
07j6 | Let $x$, $y$ be two unknown natural numbers less than $100!$. Prove that there are natural numbers $m$ and $n$ such that knowing the value of $\varphi(d(my)) + d(\varphi(nx))$ would lead to the uniquely determination of the values $x$ and $y$.
(Note. $\varphi(n)$ is the number of positive integers that are less than a... | [
"Notice that we can find a good $N$ such that after knowing $d(Nx)$ then $x$ can uniquely be determined. For this reason, we shall provide two different approaches;\n\n**1st approach.** Let $p_1 < p_2 < \\dots < p_t$ be all the primes less than $100!$ such that $p_1^R > 100!$. Now, for each $i$, choose large enough... | Iran | 41th Iranian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof only | null | |
0g2j | Problem:
Seien $m$ und $n$ natürliche Zahlen und $p$ eine Primzahl, sodass $m < n < p$ gilt. Weiter gelte:
$$
p \mid m^{2} + 1 \quad \text{ und } \quad p \mid n^{2} + 1
$$
Zeige, dass gilt:
$$
p \mid m n - 1
$$ | [
"Solution:\n\nEs gilt:\n$$\np\\left|\\left(n^{2}+1\\right)-\\left(m^{2}+1\\right) \\Leftrightarrow p\\right| n^{2}-m^{2} \\Leftrightarrow p \\mid (n-m)(n+m)\n$$\nAlso ist wegen $p$ prim sicher einer der Faktoren $(n-m)$ und $(n+m)$ durch $p$ teilbar.\nWegen $0 < n-m < p-m < p$ gilt aber $p \\nmid n-m$, also muss $p... | Switzerland | SMO - Vorrunde | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof only | null | |
0cpx | Let $ABC$ be a scalene triangle. Let $N$ be the midpoint of arc $BAC$ of its circumcircle, and let $M$ be the midpoint of $BC$. Denote by $I_1$ and $I_2$ the incenters of triangles $ABM$ and $ACM$ respectively. Prove that the points $I_1, I_2, A, N$ are concyclic. (M. Kungozhin)
Дан неравнобедренный треугольник **АВС*... | [
"Пусть $I$ — центр вписанной окружности треугольника $ABC$, а $J_1$ и $J_2$ — центры его вневписанных окружностей $\\omega_1$ и $\\omega_2$, касающихся сторон $AB$ и $AC$, соответственно. Прямая $AN$ является внешней биссектрисой угла $BAC$, поэтому точки $J_1$ и $J_2$ лежат на ней. Пусть $K_1$ и $K_2$ — точки каса... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordin... | English, Russian | proof only | null | |
05nl | Problem:
Un ensemble $E$ fini et non vide de réels strictement positifs est dit puissant lorsque, pour tous $a, b \in E$ distincts, l'un au moins des nombres $a^{b}$ et $b^{a}$ appartient aussi à $E$. Déterminer le nombre maximal d'éléments que peut contenir un ensemble puissant. | [
"Solution:\n\nTout d'abord, l'ensemble\n$$\n\\left\\{1, \\frac{1}{2}, \\frac{1}{4}, \\frac{1}{16}\\right\\}\n$$\nest un ensemble puissant qui contient quatre éléments. Soit $S$ un ensemble puissant, et $n$ son cardinal : nous allons montrer que $n \\leqslant 4$.\n\nDémontrons d'abord le lemme suivant : il n'existe ... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 4 | |
0hmc | Problem:
Determine whether it is possible to write the numbers $1, 2, \ldots, 24$ on the edges of the $3 \times 3$ grid of squares shown, one number to each edge, such that the sum of the six numbers on every path of minimal length from the upper left corner to the lower right corner is the same.
![](attached_image_1... | [
"Solution:\n\nIt is possible, and two especially simple solutions are shown. To check that they work without going through every path, one can use the following strategy: First notice that in each of the nine unit squares, the sum of the left and bottom edges equals the sum of the top and right edges. Now, given an... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Yes, it is possible. | |
0b3l | Problem:
Let $x$ be a positive real number. What is the maximum value of
$$
\frac{2022 x^{2} \log (x+2022)}{(\log (x+2022))^{3}+2 x^{3}}
$$
? | [
"Solution:\n\nBy the AM-GM Inequality,\n\n$$\n\\begin{aligned}\n\\frac{2022 x^{2} \\log (x+2022)}{\\log ^{3}(x+2022)+2 x^{3}} & =\\frac{2022 x^{2} \\log (x+2022)}{\\log ^{3}(x+2022)+x^{3}+x^{3}} \\\\\n& \\leq \\frac{2022 x^{2} \\log (x+2022)}{3\\left(x^{6} \\log ^{3}(x+2022)\\right)^{\\frac{1}{3}}} \\\\\n& =\\frac{... | Philippines | 24th Philippine Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 674 | |
0a59 | Problem:
Prove that
$$x^{2} + \frac{8}{x y} + y^{2} \geq 8.$$
for all positive real numbers $x$ and $y$. | [
"Solution:\nSince square numbers are always non-negative we have\n$$(x - y)^{2} \\geq 0 \\qquad \\mathrm{and} \\qquad (x y - 2)^{2} \\geq 0.$$\nAlso since $x$ and $y$ are positive we have $\\frac{2}{x y} > 0$. Combining this all together gives us:\n$$(x - y)^{2} + \\frac{2}{x y} (x y - 2)^{2} \\geq 0.$$\nFrom here ... | New Zealand | NZMO Round Two | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
00bk | Find all triplets $\{a, b, c\}$ of coprime positive integers (not necessarily pairwise coprime) such that $a+b+c$ divides simultaneously the three numbers $a^{12}+b^{12}+c^{12}$, $a^{23}+b^{23}+c^{23}$, and $a^{11004}+b^{11004}+c^{11004}$. | [
"Assume $\\{a, b, c\\}$ is a triple satisfying the required conditions. For every positive integer $r$, let $S_r = a^r + b^r + c^r$. We denote $S = S_1$.\n\nFirst, let us show that $S$ divides $S_{11k+1}$ for every non-negative integer $k$. We proceed by induction on $k$. For $k = 0, 1, 2$, the claim is true by our... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | {1, 1, 1} and {1, 1, 4} | |
00v6 | Let $\triangle ABC$ be a triangle and the points $K$ and $L$ on $AB$, $M$ and $N$ on $BC$ and $P$ and $Q$ on $CA$ are such that $AK = LB < \frac{1}{2}AB$, $BM = NC < \frac{1}{2}BC$ and $CP = QA < \frac{1}{2}CA$. The intersections of $KN$ with $MQ$ and $LP$ are $R$ and $T$ respectively, and the intersections of $NP$ wit... | [
"\nFrom Menelaus theorem for the triangle $\\triangle ABC$ and the lines $MQ$, $KN$ and $PL$ we get\n$$\n\\overline{AU} \\overline{UB} = - \\overline{AQ} \\overline{QC} \\cdot \\overline{CM} \\overline{MB},\n$$\n$$\n\\overline{BV} \\overline{VC} = - \\overline{BL} \\overline{LA} \\cdot \\ov... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem"
] | English | proof only | null | |
0kdi | Problem:
Consider the L-shaped tromino below with 3 attached unit squares. It is cut into exactly two pieces of equal area by a line segment whose endpoints lie on the perimeter of the tromino. What is the longest possible length of the line segment?
 | [
"Solution:\n\nLet the line segment have endpoints $A$ and $B$. Without loss of generality, let $A$ lie below the lines $x+y=\\sqrt{3}$ (as this will cause $B$ to be above the line $x+y=\\sqrt{3}$) and $y=x$ (we can reflect about $y=x$ to get the rest of the cases):\n\n\n\nNow, note that as ... | United States | HMMT February | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 5/2 | |
09vh | Problem:
Schrijf $S_{n}$ voor de verzameling $\{1,2, \ldots, n\}$. Bepaal alle positieve gehele $n$ waarvoor er functies $f: S_{n} \rightarrow S_{n}$ en $g: S_{n} \rightarrow S_{n}$ bestaan zodat voor elke $x$ precies één van de gelijkheden $f(g(x))=x$ en $g(f(x))=x$ waar is. | [
"Solution:\n\nOplossing I. We laten eerst zien dat als $n=2m$ voor zekere positieve gehele $m$, er dan zulke functies bestaan. Definieer\n$$\nf(x)=\\left\\{\\begin{array}{ll}\nx & \\text{ als } 1 \\leq x \\leq m, \\\\\nx-m & \\text{ als } m+1 \\leq x \\leq 2m,\n\\end{array}\\right.\n\\quad g(x)= \\begin{cases}x+m &... | Netherlands | IMO-selectietoets | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | All even positive integers | |
058y | Find the sum
$$
\sqrt{1 + \frac{1}{1^2} + \frac{1}{2^2}} + \sqrt{1 + \frac{1}{2^2} + \frac{1}{3^2}} + \dots + \sqrt{1 + \frac{1}{2021^2} + \frac{1}{2022^2}}
$$ | [
"We will denote\n$$\ns = \\sqrt{1 + \\frac{1}{1^2} + \\frac{1}{2^2}} + \\sqrt{1 + \\frac{1}{2^2} + \\frac{1}{3^2}} + \\dots + \\sqrt{1 + \\frac{1}{2021^2} + \\frac{1}{2022^2}}\n$$\nNotice that\n$$\n1 + \\frac{1}{k^2} + \\frac{1}{(k+1)^2} = \\frac{k^2(k+1)^2 + (k+1)^2 + k^2}{k^2(k+1)^2} = \\frac{(k(k+1)+1)^2}{(k(k+1... | Estonia | Estonian Math Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | final answer only | 2021 + 2021/2022 | |
0751 | In an acute triangle $ABC$, $AD$ is the median, $BE$ is the internal angle bisector and $CF$ is the altitude, with $D$, $E$, $F$ respectively on sides $BC$, $CA$, $AB$. If $DEF$ is equilateral, prove that $ABC$ is also equilateral.
 | [
"Observe that in the right triangle $BFC$, $BD = DC = a/2$. Hence $DF = a/2$ and this gives $DE = EF = a/2$. Now in triangle $BEC$, $D$ is the midpoint of $BC$ and $DE = DB = DC$. Hence $\\angle BEC = 90^\\circ$. Since $BE$ bisects $\\angle ABC$, we conclude that $BA = BC$ and $CE = EA$.\n\nIn the right triangle $C... | India | Indija TS | [
"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 | |
02px | Problem:
Tiago escreve todos os números de quatro algarismos não nulos distintos que possuem a mesma paridade. Qual a probabilidade de que, ao escolhermos um desses números, ele seja par? | [
"Solution:\n\nOs quatro algarismos escolhidos fazem parte dos conjuntos $A=\\{1,3,5,7,9\\}$ ou $B=\\{2,4,6,8\\}$.\n\nCom os elementos do conjunto $A$ temos 5 possibilidades para o primeiro algarismo, 4 para o segundo, 3 para o terceiro e 2 para o quarto, totalizando $5 \\times 4 \\times 3 \\times 2=120$ números com... | Brazil | Brazilian Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | proof and answer | 1/6 | |
0g7n | 一場數學會議中只有 $n$ 對夫妻參加 ($n \ge 8$)，大會安排所有男士坐在一張有 $n$ 個座位的圓桌，所有女士坐另一張也是有 $n$ 個座位的圓桌。大會主辦單位發現一種人傳人的病毒正在與會者之間傳播，且其傳染途徑如下：設 $P$ 為一位健康的與會者，且其所坐位置兩側鄰居標為 $P_{\text{left}}$, $P_{\text{right}}$，其配偶標為 $P_{\text{mate}}$；只有在 $P_{\text{left}}$, $P_{\text{right}}$, $P_{\text{mate}}$ 這三人中至少有兩人感染病毒的情況下，$P$ 才會立刻被傳染而感染到病毒，否則 $P$ 會一直保持健康。設會... | [] | Taiwan | 二〇一三數學奧林匹亞競賽第二階段選訓營 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n | |
02s4 | Problem:
a) Encontre todos os números inteiros positivos de dois algarismos $\overline{a b}$ tais que:
$$
(a+1)(b+1)=\overline{a b}+1
$$
b) Encontre todos os números inteiros positivos de três algarismos $\overline{a b c}$ tais que:
$$
(a+1)(b+1)(c+1)=\overline{a b c}+1
$$ | [
"Solution:\na) Devemos achar todos os números da forma $\\overline{a b}$ tais que\n$$\n(a+1)(b+1)=\\overline{a b}+1\n$$\nNotamos agora que $\\overline{a b}=10a+b$. Substituindo na expressão acima obtemos que:\n$$\n(a+1)(b+1)=10a+b+1\n$$\nSimplificando essa equação obtemos\n$$\nab=9a\n$$\nComo o número $\\overline{a... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | a) 19, 29, 39, 49, 59, 69, 79, 89, 99; b) 199, 299, 399, 499, 599, 699, 799, 899, 999 | |
04zk | Juku discovered that of the things in his satchel, $60\%$ were ugly and $76\%$ were useless. He scrapped all things that were both ugly and useless, and added things that were both beautiful and useful. After this, of the things in Juku's satchel, $25\%$ are ugly and $45\%$ are useless. How many percent of the things i... | [
"Observe that the amount of things that are beautiful but useless and things that are useful but ugly remained unchanged. The difference between the percentages of these things was $76\\% - 60\\% = 16\\%$ before displacement but is $45\\% - 25\\% = 20\\%$ after that. Hence the overall number of things in the satche... | Estonia | Selected Problems from the Final Round of National Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | English | proof and answer | 4% | |
06ym | Let $\mathcal{S}$ be a finite nonempty set of prime numbers. Let $1 = b_{1} < b_{2} < \cdots$ be the sequence of all positive integers whose prime divisors all belong to $\mathcal{S}$. Prove that, for all but finitely many positive integers $n$, there exist positive integers $a_{1}, a_{2}, \ldots, a_{n}$ such that
$$
\... | [
"If $\\mathcal{S}$ has only one element $p$, then $b_{i} = p^{i-1}$ and we can easily find $a_{1}, \\ldots, a_{n}$ with $2 = \\left\\lceil \\sum_{i=0}^{n-1} \\frac{1}{p^{i}} \\right\\rceil = \\sum_{i=0}^{n-1} \\frac{a_{i}}{p^{i-1}}$ by taking $a_{1} = a_{2} = \\cdots = a_{n-1} = 1$ and choosing $a_{n} = p^{n-1} - (... | IMO | IMO2024 Shortlisted Problems | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0b24 | Problem:
Suppose $f$ is a second-degree polynomial for which $f(2)=1$, $f(4)=2$, and $f(8)=3$. Find the sum of the roots of $f$. | [
"Solution:\nLet $f(x) = a x^{2} + b x + c$. By substituting $x = 2, 4, 8$ we get the system of linear equations\n$$\n\\begin{array}{r}\n4a + 2b + c = 1 \\\\\n16a + 4b + c = 2 \\\\\n64a + 8b + c = 3\n\\end{array}\n$$\nSolving this system of equations gives us $a = -\\frac{1}{24}$, $b = \\frac{3}{4}$, $c = -\\frac{1}... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | 18 | |
0kgl | What is the sum of all possible values of $t$ between $0$ and $360$ such that the triangle in the coordinate plane whose vertices are $(\cos 40^\circ, \sin 40^\circ)$, $(\cos 60^\circ, \sin 60^\circ)$, and $(\cos t^\circ, \sin t^\circ)$ is isosceles?
(A) 100 (B) 150 (C) 330 (D) 360 (E) 380 | [
"Let $A = (\\cos 40^\\circ, \\sin 40^\\circ)$, $B = (\\cos 60^\\circ, \\sin 60^\\circ)$, $C = (\\cos t^\\circ, \\sin t^\\circ)$, and $O = (0, 0)$. Then acute $\\angle AOB = 20^\\circ$, so $\\triangle ABC$ will be isosceles with vertex at $A$ or $B$ if $\\angle AOC = 20^\\circ$ or $\\angle BOC = 20^\\circ$, respecti... | United States | AMC 12 B | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | E | |
0imn | Let $\theta$ be an angle in the interval $(0, \pi/2)$. Given that $\cos \theta$ is irrational, and that $\cos k\theta$ and $\cos[(k+1)\theta]$ are both rational for some positive integer $k$, show that $\theta = \pi/6$. | [
"Thus both $\\cos(k^2\\theta) = \\cos[k(k\\theta)]$ and $\\cos[(k^2-1)\\theta] = \\cos[(k-1)(k+1)\\theta]$ are rational. By the Addition and subtraction formulas, we have\n$$\n\\cos[(k+1)\\theta] = \\cos k\\theta \\cos \\theta - \\sin k\\theta \\sin \\theta \\quad \\text{and} \\quad \\cos(k^2\\theta) = \\cos[(k^2-1... | United States | Team Selection Test | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials"
] | null | proof and answer | π/6 | |
07ok | The 11-digit number $52014641025$ has two interesting properties: it contains the string of digits $2014$, and it is unchanged if we reverse the digits. How many 11-digit numbers have this property?
(A number cannot begin with the digit $0$.) | [
"An integer $N$ that satisfies these conditions is determined by its first (leftmost) six digits. The first instance of either the string $2014$ or its reverse, $4102$, must start in position $1$, $2$ or $3$. In each case, there are two other digits to be chosen. These digits can be chosen arbitrarily if $2014$ or ... | Ireland | Irska 2014 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 560 | |
03rt | Let $\{a_n\}$ be a sequence such that $a_1 = \frac{21}{16}$ and
$$
2a_n - 3a_{n-1} = \frac{3}{2^{n+1}}, \quad n \ge 2. \qquad \textcircled{1}
$$
Let $m$ be a positive integer and $m \ge 2$. Prove that for $n \le m$,
$$
\left[a_n + \frac{3}{2^{n+3}}\right]^{\frac{1}{m}} \left(m - \left(\frac{2}{3}\right)^{\frac{n(m-1)}{... | [
"**Proof** By Equation (1), we have\n$$\n2^n a_n = 3 \\cdot 2^{n-1} a_{n-1} + \\frac{3}{4}.\n$$\nSet $b_n = 2^n a_n$, $n = 1, 2, \\dots$, then\n$$\nb_n = 3b_{n-1} + \\frac{3}{4}, \\quad b_n + \\frac{3}{8} = 3\\left(b_{n-1} + \\frac{3}{8}\\right).\n$$\n$$\n\\text{Since } b_1 = 2a_1 = \\frac{21}{8},\n$$\n$$\nb_n + \\... | China | China Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
03u7 | Is there a triangle with sides of integral length, such that the length of the shortest side is $2007$ and that the largest angle is twice the smallest? | [
"We shall prove that no such a triangle satisfies the condition.\nIf $\\triangle ABC$ satisfies the condition, let $\\angle A \\le \\angle B \\le \\angle C$, then $\\angle C = 2\\angle A$, and $a = 2007$. Draw the bisector of $\\angle ACB$ and let it intersect $AB$ at point $D$. Then $\\angle BCD = \\angle A$, so $... | China | China Western Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | No | |
09rp | Problem:
Zij $f: \mathbb{Z}_{>0} \rightarrow \mathbb{R}$ een functie waarvoor geldt: voor alle $n>1$ is er een priemdeler $p$ van $n$ zodat
$$
f(n)=f\left(\frac{n}{p}\right)-f(p)
$$
Bovendien is gegeven dat $f\left(2^{2014}\right)+f\left(3^{2015}\right)+f\left(5^{2016}\right)=2013$.
Bereken $f\left(2014^{2}\right)+f\l... | [
"Solution:\n\nAls $n=q$ met $q$ priem, dan is er maar één priemdeler van $n$, namelijk $q$, dus moet gelden dat $f(q)=f(1)-f(q)$, dus $f(q)=\\frac{1}{2} f(1)$. Als $n=q^{2}$ met $q$ priem, dan heeft $n$ ook maar één priemdeler, dus geldt $f\\left(q^{2}\\right)=f(q)-f(q)=0$. We bewijzen nu met inductie naar $k$ dat ... | Netherlands | IMO-selectietoets II | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 49/3 | |
04yv | Points $A'$, $B'$ and $C'$ are chosen correspondingly on the sides $AB$, $BC$, and $CA$ of an equilateral triangle $ABC$ so that $\frac{|A'B|}{|AB|} = \frac{|B'C|}{|BC|} = \frac{|C'A|}{|CA|} = k$. Find all positive real numbers $k$ for which the area of the triangle $A'B'C'$ is exactly half of the area of the triangle ... | [
"Let $\\alpha$ be the angle at the vertex $A$ (Fig. 10).\n\nThe area of the triangle $AA'C'$ is $S_{AA'C'} = \\frac{1}{2} \\cdot |AA'| \\cdot |AC'| \\cdot \\sin \\alpha = \\frac{1}{2} \\cdot (1-k)|AB| \\cdot k|AC| \\cdot \\sin \\alpha = (1-k)k S_{ABC}$.\n\nSimilarly $S_{BB'A'} = (1-k)k S_{ABC}$ and $S_{CC'B'} = (1-... | Estonia | Estonija 2010 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | k = 1/2 + √3/6 or k = 1/2 − √3/6 | |
02l6 | Problem:
A sala do Newton- Professor Newton dividiu seus alunos em grupos de $4$ e sobraram $2$. Ele dividiu seus alunos em grupos de $5$ e um aluno ficou de fora. Se $15$ alunos são mulheres e tem mais mulheres do que homens, o número de alunos homens é:
(a) $7$
(b) $8$
(c) $9$
(d) $10$
(e) $11$ | [
"Solution:\n\nComo o número de alunos homens é menor do que $15$ e das mulheres é $15$, temos\n$$\n15 < \\text{alunos homens} + \\text{alunas mulheres} < 15 + 15 = 30\n$$\nou seja: o número de alunos está entre $15$ e $30$.\n\nPor outro lado, quando dividimos por $4$ sobram $2$ alunos, então o número de alunos é pa... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | MCQ | e | |
0gpx | Let $ABC$ be a triangle with $AC > AB$. Let a circle $\omega$ tangent to the sides $AB$ and $AC$ at $D$ and $E$ respectively, intersects the circumcircle of $ABC$ at $K$ and $L$ and $X$ and $Y$ be points on the sides $AB$ and $AC$ respectively, such that
$$
\frac{AX}{AB} = \frac{CE}{BD + CE} \quad \text{and} \quad \fra... | [
"Let $BD = a$, $CE = b$, $AD = AE = c$. Since\n$$\nAX = \\frac{AB \\cdot CE}{BD + CE} = \\frac{(a + x)b}{a + b} \\quad \\text{and} \\quad AY = \\frac{AC \\cdot BD}{BD + CE} = \\frac{(b + x)a}{a + b}\n$$\nwe get $BX = AB - AX = \\frac{(a + x)a}{a + b}$ and $CY = AC - AY = \\frac{(b + x)b}{a + b}$.\n\n![](attached_im... | Turkey | 18th Junior Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof only | null | |
046n | Given a positive integer $n$ and $n^3$ integers $a_{ijk} \in \{1, -1\}$ ($1 \le i, j, k \le n$). Prove that there exist
$$
x_1, \dots, x_n, y_1, \dots, y_n, z_1, \dots, z_n \in \{1, -1\},
$$
such that the following inequality holds
$$
\left| \sum_{i=1}^{n} \sum_{j=1}^{n} \sum_{k=1}^{n} a_{ijk} x_i y_j z_k \right| > \fr... | [
"*Proof 1.* For any $(x_i)$ and $(y_j)$ satisfying the given conditions, we define\n$$\nX_k = \\sum_{i,j=1}^{n} a_{ijk} x_i y_j.\n$$\nSince we can always choose $z_k$ with the same sign as $X_k$, we have\n$$\n\\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sum_{k=1}^{n} a_{ijk} x_i y_j z_k = \\sum_{k=1}^{n} |X_k|.\n$$\nNote that... | China | 2023 Chinese IMO National Team Selection Test | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Expected values"
] | English | proof only | null | |
02lb | Problem:
Retângulo com dimensões inteiras - As diagonais de um retângulo medem $\sqrt{1993}\ \mathrm{cm}$. Quais são suas dimensões, sabendo que elas são números inteiros? | [
"Solution:\n\nSe $a \\geq b$ são os comprimentos dos lados do retângulo, então pelo teorema de Pitágoras temos\n$$\na^{2}+b^{2}=1993\n$$\nComo $a^{2} \\geq b^{2}$, segue que\n$$\n2 a^{2} \\geq a^{2}+b^{2}=1993>a^{2}\n$$\nLogo,\n$$\n\\sqrt{1993}>a \\geq \\sqrt{996,5}\n$$\nAssim, $44 \\geq a \\geq 32$. Usando o fato ... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Geometry > Plane Geometry > Triangles"
] | null | final answer only | 43 and 12 | |
0iii | In acute triangle $ABC$, segments $AD$, $BE$, and $CF$ are its altitudes, and $H$ is its orthocenter. Circle $\omega$, centered at $O$, passes through $A$ and $H$ and intersects sides $AB$ and $AC$ again at $Q$ and $P$ (other than $A$), respectively. The circumcircle of triangle $OPQ$ is tangent to segment $BC$ at $R$.... | [
"**First Solution:** (Based on work by Ryan Ko) Let $M$ be the midpoint of segment $AH$. Since $\\angle AEH = \\angle AFH = 90^\\circ$, quadrilateral $AEHF$ is cyclic with $M$ as its circumcenter. Hence triangle $EFM$ is isosceles with vertex angle $\\angle EMF = 2\\angle CAB = 2x$. Likewise, triangle $PQO$ is also... | United States | Team Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Circle... | null | proof only | null | |
06ks | In a school there are $1200$ students. Each student must join exactly $k$ clubs. Given that there is a common club joined by every $23$ students, but there is no common club joined by all $1200$ students, find the smallest possible value of $k$. | [
"The answer is $k = 23$.\n\nWe first show $k \\le 23$. We list the students as $S_1, S_2, \\dots, S_{1200}$ and the clubs as $C_1, C_2, \\dots, C_{24}$. Consider the following construction. For $1 \\le j \\le 24$, student $S_j$ joins clubs $C_1, \\dots, C_{j-1}, C_{j+1}, \\dots, C_{24}$. For $25 \\le j \\le 1200$, ... | Hong Kong | null | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 23 | |
04t6 | Find all primes $p$ for which there exists a positive integer $n$ such that $p^n + 1$ is a cube of a positive integer. (Ján Mazák, Róbert Tóth) | [
"Suppose that positive integer $a$ satisfies $p^n + 1 = a^3$ (clearly $a \\ge 2$). We rewrite the equality as\n$$\np^n = a^3 - 1 = (a-1)(a^2 + a + 1).\n$$\nIt follows that if $a > 2$, the numbers $a-1$ and $a^2+a+1$ are powers of $p$ (with positive integer exponents).\nIf $a > 2$ then $a - 1 = p^k$, hence $a = p^k ... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial oper... | English | proof and answer | 7 | |
08ca | Problem:
Sia $x_{1}, x_{2}, x_{3}, \ldots$ una successione di interi positivi tale che, per ogni $m, n$ interi positivi, valga $x_{m n} \neq x_{m(n+1)}$. Dimostrare che esiste un intero positivo $i$ tale che $x_{i} \geq 2017$. | [
"Solution:\n\nDiciamo che due numeri $i$ e $j$ sono connessi se possono essere scritti uno nella forma $m n$ e l'altro nella forma $m(n+1)$ per opportuni $m$ e $n$. Il testo dell'esercizio ci dice che $x_{i}$ è diverso da $x_{j}$ ogniqualvolta $i$ e $j$ sono connessi. Vogliamo dimostrare che vi sono almeno 2017 num... | Italy | Olimpiade Italiana di Matematica | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0606 | Problem:
Pour tout entier $k \geqslant 1$, on note $p(k)$ le plus petit nombre premier ne divisant pas $k$. Soit $\left(a_{n}\right)_{n \in \mathbb{N}}$ une suite telle que $a_{0} \in \mathbb{N}^{*}$ et pour tout $n \geqslant 0$, $a_{n+1}$ est le plus petit entier strictement positif différent de $a_{0}, \ldots, a_{n}... | [
"Solution:\n\nDéjà notons que la suite est bien définie : si $a_{1}, \\ldots, a_{n}$ sont construits, comme $a_{n}$ est premier avec $p\\left(a_{n}\\right)$, il existe une infinité d'entiers $k$ tels que $a_{n}^{k} \\equiv 1\\left(\\bmod p\\left(a_{n}\\right)\\right)$ qui sont tous les multiples de l'ordre de $a_{n... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
043a | Given arithmetic sequence $\{a_n\}$ with common difference $d \ne 0$ and $a_{2021} = a_{20} + a_{21}$, then the value of $\frac{a_1}{d}$ is ______. | [
"By the conditions, we have $a_1 + 2020d = a_1 + 19d + a_1 + 20d$. And since $d \\ne 0$, $\\frac{a_1}{d} = 1981$."
] | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 1981 | |
08v2 | Suppose $f$ is a positive integer-valued function defined for the set of positive integers satisfying for any pair of positive integers $x, y$ the following inequality:
$$
(x + y)f(x) \leq x^2 + f(xy) + 110.
$$
Determine the minimum and the maximum value of $f(23) + f(2011)$ for this $f$. | [
"Let $a = 110$. We will first show that a necessary and sufficient condition for a positive integer-valued function $f$ defined on the set positive integers to satisfy the given inequality is that $f$ satisfies the following simpler inequality:\n$$\n(\\dagger) \\quad t - a \\le f(t) \\le t \\quad \\text{for any pos... | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | minimum 1902, maximum 2034 | |
0d9r | Distinct prime numbers $p$, $q$, $r$ satisfy the equation
$$
2 p q r + 50 p q = 7 p q r + 55 p r = 8 p q r + 12 q r = A
$$
for some positive integer $A$. Find $A$. | [
"Review the given condition as\n$$\np q (2 r + 50) = p r (7 q + 55) = q r (8 p + 12) = A.\n$$\nThis implies that $A$ is a multiple of $p$, $q$ and $r$ so the value $K = \\frac{A}{p q r}$ is an integer. Dividing through, we have\n$$\nK = 8 + \\frac{12}{p} = 7 + \\frac{55}{q} = 2 + \\frac{50}{r}.\n$$\nHence, $p \\mid... | Saudi Arabia | Team selection tests for JBMO 2018 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 1980 | |
0fbl | Problem:
Los lados de un polígono regular convexo de $L+M+N$ lados se han de dibujar en tres colores: $L$ de ellos con trazo rojo, $M$ con trazo amarillo, y $N$ con trazo azul. Expresar, por medio de desigualdades, las condiciones necesarias y suficientes para que tenga solución (varias, en general) el problema de hac... | [
"Solution:\n\nSea $K=L+M+N$.\nSi $K$ es par debe ser:\n$$\nL \\leq \\frac{K}{2} ; \\quad M \\leq \\frac{K}{2} \\quad \\text{ y } \\quad N \\leq \\frac{K}{2}\n$$\nEs decir: $L+M \\geq N ; L+N \\geq M$ y $M+N \\geq L$.\n\nSi $K$ es impar debe ser:\n$$\n0<L \\leq \\frac{K-1}{2} ; \\quad 0<M \\leq \\frac{K-1}{2} \\quad... | Spain | OME 11 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Let K = L + M + N. If K is even: L ≤ K/2, M ≤ K/2, N ≤ K/2 (equivalently L + M ≥ N, L + N ≥ M, M + N ≥ L). If K is odd: 1 ≤ L, M, N ≤ (K − 1)/2 (equivalently L + M > N, L + N > M, M + N > L and L, M, N > 0). | |
08aw | Problem:
Quanto vale $\sqrt[4]{2^{20}+2^{27}+2^{31}+2^{32}+2^{37}+2^{40}}$ ? | [
"Solution:\n\nLa risposta è 1056. Una volta raccolto un fattore $2^{20}$ sotto la radice quarta, ci si riconduce a calcolare $\\sqrt[4]{1+2^{7}+2^{11}+2^{12}+2^{17}+2^{20}}$. A questo punto è possibile riconoscere nell'espressione sotto radice il quadrato del trinomio $1+2^{6}+2^{10}=(2^{5}+1)^{2}$ o direttamente v... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 1056 | |
0idf | Problem:
In trapezoid $ABCD$ shown, $AD$ is parallel to $BC$, and $AB = 6$, $BC = 7$, $CD = 8$, $AD = 17$. If sides $AB$ and $CD$ are extended to meet at $E$, find the resulting angle at $E$ (in degrees).
 | [
"Solution:\n\nChoose point $F$ on $AD$ so that $BCDF$ is a parallelogram. Then $BF = CD = 8$, and $AF = AD - DF = AD - BC = 10$, so $\\triangle ABF$ is a $6$-$8$-$10$ right triangle. The required angle is equal to $\\angle ABF = 90^{\\circ}$.\n\n"
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 90 | |
06al | Solve in the real numbers the system:
$$
\begin{cases}
a + b + c = 0 \\
ab^3 + bc^3 + ca^3 = 0
\end{cases}.
$$ | [
"1. Since $c = -a - b$, we get:\n$$\n\\begin{aligned}\n0 &= ab^3 + b(-a-b)^3 + (-a-b)a^3 \\\\\n&= -(ab^3 + b(a+b)^3 + (a+b)a^3) \\\\\n&= -(a^4 + 2a^3b + 3a^2b^2 + 2ab^3 + b^4) \\\\\n&= -(a^2(a+b)^2 + b^2(a+b)^2 + a^2b^2) \\\\\n&= -a^2c^2 - b^2c^2 - a^2b^2.\n\\end{aligned}\n$$\nTherefore each term of the last sum mu... | Greece | 40th Hellenic Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | a = b = c = 0 | |
03ys | Suppose sequence $\{a_n\}$ satisfies $a_1 = 2t - 3$ ($t \in \mathbb{R}$ and $t \neq \pm 1$),
$$
a_{n+1} = \frac{(2t^{n+1} - 3)a_n + 2(t-1)t^n - 1}{a_n + 2t^n - 1} \quad (n \in \mathbb{N}^*).
$$
(1)
Find the formula of general term about $\{a_n\}$.
(2)
If $t > 0$, find out which is larger between $a_{n+1}$ and $a_n$. | [
"(1) The given expression can be rewritten as\n$$\na_{n+1} = \\frac{2(t^{n+1} - 1)(a_n + 1)}{a_n + 2t^n - 1} - 1.\n$$\nThen\n$$\n\\frac{a_{n+1} + 1}{t^{n+1} - 1} = \\frac{2(a_n + 1)}{a_n + 2t^n - 1} = \\frac{\\frac{2(a_n + 1)}{t^n - 1}}{\\frac{a_n + 1}{t^n - 1} + 2}.\n$$\nLet $\\frac{a_n + 1}{t^n - 1} = b_n$. Then ... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | a_n = 2(t^n − 1)/n − 1; for t > 0, a_{n+1} > a_n. | |
0idl | Problem:
Find the shortest distance from the line $3x + 4y = 25$ to the circle $x^{2} + y^{2} = 6x - 8y$. | [
"Solution:\nThe circle is $(x-3)^{2} + (y+4)^{2} = 5^{2}$. The center $(3, -4)$ is a distance of\n$$\n\\frac{|3 \\cdot 3 + 4 \\cdot (-4) - 25|}{\\sqrt{3^{2} + 4^{2}}} = \\frac{32}{5}\n$$\nfrom the line, so we subtract $5$ for the radius of the circle and get $\\frac{7}{5}$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | final answer only | 7/5 | |
03xi | Let $a, b, c \in \{0, 1, 2, \dots, 9\}$. The quadratic equation $ax^2 + bx + c = 0$ has a rational root. Prove that the three-digit number $abc$ is not a prime number. | [
"We prove by contradiction. If $abc = p$ is a prime number, the rational root of quadratic equation $f(x) = ax^2 + bx + c = 0$ is $x_1, x_2 = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$. Obviously, $b^2 - 4ac$ is a perfect square number, and $x_1, x_2$ are all negative, and\n\n$$\nf(x) = a(x - x_1)(x - x_2).\n$$\n\nThus... | China | China Southeastern Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0fos | Problem:
Todas las caras de un poliedro son triángulos. A cada uno de los vértices de este poliedro se le asigna de forma independiente uno de entre tres colores: verde, blanco o negro. Decimos que una cara es extremeña si sus tres vértices son de distintos colores, uno verde, uno blanco y uno negro. ¿Es cierto que, i... | [
"Solution:\n\nSea $C$ el número de caras del poliedro. Cada cara tiene $3$ lados, cada uno de los cuáles pertenece exactamente a dos caras. Luego el número total de aristas del poliedro es $3C/2$, que ha de ser entero. Luego el número $C$ de caras del poliedro es par.\n\nA una arista cuyos vértices extremos son del... | Spain | LI Olimpiada matemática Española (Concurso Final) | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof only | null | |
0hv8 | Problem:
We take a $6 \times 6$ chessboard, which has six rows and columns, and indicate its squares by $(i, j)$ for $1 \leq i, j \leq 6$. The $k$th northeast diagonal consists of the six squares satisfying $i-j \equiv k \pmod{6}$ (and so there are six such diagonals); hence there are six such diagonals.
Determine if ... | [
"Solution:\n\nThe answer is no. Assume for contradiction such a coloring existed; then each row, column and northeast diagonal would have sum exactly\n$$\nN = \\frac{1}{6}(1+2+\\cdots+36) = 111.\n$$\nNow consider the marked squares shown below.\n\n| $A$ | $B$ | $A$ | $B$ | $A$ | $B$ |\n| :--- | :--- | :--- | :--- |... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | No | |
0j66 | Problem:
The UEFA Champions League playoffs is a 16-team soccer tournament in which Spanish teams always win against non-Spanish teams. In each of 4 rounds, each remaining team is randomly paired against one other team; the winner advances to the next round, and the loser is permanently knocked out of the tournament. ... | [
"Solution:\n\nWe note that the probability there are not two Spanish teams in the final two is the probability that the 3 of them have already competed against each other in previous rounds. Note that the random pairings in each round is equivalent, by the final round, to dividing the 16 into two groups of 8 and ta... | United States | Harvard-MIT November Tournament | [
"Statistics > Probability > Counting Methods > Combinations",
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 4/5 |
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