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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0kdt | Problem:
Estimate $A$, the number of times an 8-digit number appears in Pascal's triangle. An estimate of $E$ earns $\max (0,\lfloor 20-|A-E| / 200\rfloor)$ points. | [
"Solution:\n\nWe can obtain a good estimate by only counting terms of the form $\\binom{a}{1}$, $\\binom{a}{2}$, $\\binom{a}{a-1}$, and $\\binom{a}{a-2}$. The last two cases are symmetric to the first two, so we will only consider the first two and multiply by 2 at the end.\n\nSince $\\binom{a}{1}=a$, there are 90,... | United States | HMMO 2020 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Other"
] | null | final answer only | 180019320 | |
0grn | $n \ge 3$ boxes are placed around a circle. At the first step we choose some boxes. At the second step for each chosen box we put a ball into the chosen box and into each of its two neighbouring boxes. Find the total number of possible distinct ball distributions which can be obtained in this way. (All balls are identi... | [
"The answer is $2^n - 3 \\cdot 2^{n/3} + 2$ for $n = 3k$ and $2^n$ for $n \\neq 3k$. The number of possible box choices is $2^n$. Let us examine cases when two or more different choices produce the same ball distribution. If different choices $C_1$ and $C_2$ produce the same distribution then these choices should n... | Turkey | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 2^n if 3 does not divide n; 2^n - 3·2^{n/3} + 2 if 3 divides n | |
032f | Problem:
Let $ABCDE$ be a cyclic pentagon with $AC \parallel DE$. Denote by $M$ the midpoint of $BD$. If $\Varangle AMB = \Varangle BMC$, prove that $BE$ bisects $AC$. | [
"Solution:\n\nLet $BE$ meet $AC$ at point $N$ and $P$ be the midpoint of $AB$. Set $\\Varangle BAC = \\Varangle BDC = \\alpha$, $\\Varangle ABE = \\Varangle CBD = \\beta$ and $\\Varangle ADB = \\Varangle ACB = \\gamma$. Then $\\triangle ABN \\sim \\triangle DBC$, and we conclude that $\\triangle BPN \\sim \\triangl... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fuf | Problem:
Sei $X$ eine Menge mit $n$ Elementen und seien $A_{1}, A_{2}, \ldots, A_{n}$ verschiedene Teilmengen von $X$. Zeige: Es gibt ein $x \in X$, sodass die Mengen
$$
A_{1} \backslash\{x\}, A_{2} \backslash\{x\}, \ldots, A_{n} \backslash\{x\}
$$
alle verschieden sind. | [
"Solution:\n\nNehme an, dies sei nicht der Fall. Dann gibt es für jedes Element $x \\in X$ zwei Teilmengen $A_{i}$ und $A_{j}$, sodass $A_{i} \\backslash\\{x\\}=A_{j} \\backslash\\{x\\}$ (eventuell gibt es mehrere solche Paare, wir wählen ein beliebiges aus und halten es im Folgenden fest). Da $A_{i}$ und $A_{j}$ v... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
06b3 | Let $AB\Gamma\Delta E$ be a convex pentagon $A\hat{B}\Gamma = A\hat{E}\Delta = 90^\circ$. Suppose that the midpoint, let $M$, of the side $\Gamma\Delta$ is the circumcenter of the triangle $ABE$. Let $O$ be the circumcenter of the triangle $A\Gamma\Delta$. Prove that line $AO$ passes through the midpoint $N$ of segment... | [] | Greece | Selection Examination | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > C... | English | proof only | null | |
010j | Problem:
Let $P$ be a polynomial of degree $6$ and let $a, b$ be real numbers such that $0 < a < b$. Suppose that $P(a) = P(-a)$, $P(b) = P(-b)$ and $P'(0) = 0$. Prove that $P(x) = P(-x)$ for all real $x$. | [
"Solution:\nThe polynomial $Q(x) = P(x) - P(-x)$, of degree at most $5$, has roots at $-b$, $-a$, $0$, $a$ and $b$; these are five distinct numbers. Moreover, $Q'(0) = 0$, showing that $Q$ has a multiple root at $0$. Thus $Q$ must be the constant $0$, i.e. $P(x) = P(-x)$ for all $x$."
] | Baltic Way | Baltic Way 1998 | [
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
0bzu | Find the positive integer solutions of the equation
$$
\lfloor\sqrt{8n+1}\rfloor + \lfloor\sqrt{8n+2}\rfloor + \dots + \lfloor\sqrt{8n+7}\rfloor = 2027.
$$ | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 10512 | |
0dwm | Problem:
Za realna števila $x$, $y$ in $z$ velja $x y z = 1$. Izračunaj vrednost izraza
$$
\frac{x+1}{x y + x + 1} + \frac{y+1}{y z + y + 1} + \frac{z+1}{z x + z + 1}
$$ | [
"Solution:\nRačunajmo:\n$$\n\\frac{x+1}{x y + x + 1} + \\frac{y+1}{y z + y + 1} + \\frac{z+1}{z x + z + 1} = \\frac{x+1}{x y + x + x y z} + \\frac{y+1}{y z + y + 1} + \\frac{z+1}{z x + z + 1}\n$$\n$$\n= \\frac{x+1 + x(y+1)}{x(y z + y + 1)} + \\frac{z+1}{z x + z + 1}\n$$\n$$\n= \\frac{x y + 2x + 1}{x y (z+1 + x z)} ... | Slovenia | 49. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 2 | |
03dj | Find $x^2 + p = qx$ where
$$
p = (32 + \sqrt{2})(9 - \sqrt{8})^{-1}, \quad q = \sqrt{9 + \sqrt{8} - \sqrt{24} - \sqrt{48}} + \sqrt{3 - \sqrt{8}} + \sqrt{55 - 14\sqrt{6}}
$$ | [] | Bulgaria | Bulgaria 2022 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | x = 2 + sqrt(2) or x = 3 - sqrt(2) | |
0bwl | Show that, if $f: [0, \infty) \to \mathbb{R}$ is a convex function and $a, b, c$ are positive real numbers, then
$$
\begin{aligned}
& \int_{0}^{a} f(x) \, dx + \int_{0}^{b} f(x) \, dx + \int_{0}^{c} f(x) \, dx + \int_{0}^{a+b+c} f(x) \, dx \
\ge \\
& \int_{0}^{a+b} f(x) \, dx + \int_{0}^{a+c} f(x) \, dx + \int_{0}^{b+c... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
07lk | Let $a_1, a_2, a_3, a_4, \dots$ be a sequence of non-negative integers such that
$$
a_{n+2} = \binom{a_{n+1}}{a_n},
$$
where
$$
\binom{m}{k} = \begin{cases} \frac{m!}{k!(m-k)!} & \text{if } m \ge k, \\ 0 & \text{otherwise.} \end{cases}
$$
Prove that such a sequence tends to infinity if and only if $2 \le a_1 \le a_2 - ... | [
"If the given pair of inequalities do not hold, then the sequence $(a_n)$ quickly degenerates into a sequence consisting of only 0s and 1s. There are basically two such cases to analyse, namely $a_1 = 0$ and $a_1 = 1$, since the two other cases $a_1 = a_2$ and $a_1 = a_2 - 1$ are seen to be equivalent to those two ... | Ireland | Irska | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0gzw | Vitalii, Michael and Olexandr were given $n$ dollars by their mother each one. Vitalii had to spend all his money for buying books, Michael for buying notebooks, Olexandr for pens. Vitalii had bought 1 book, Michael 2 notebooks and Olexandr 5 pens. After that, they together had $n$ dollars. Prove that one of the boys c... | [
"Assuming the contrary. Denote by $q_1, q_2$ and $q_3$ the costs of one book, notebook and pen, respectively, and by $r_1, r_2, r_3$ the change our boys were given. Obviously $r_1 < q_1, r_2 < q_2, r_3 < q_3$. From the statement we have\n$$\nn = q_1 + r_1, \\quad n = 2q_2 + r_2, \\quad n = 5q_3 + r_3\n$$\nand we ca... | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
043q | Given set $A = \{1, 2, m\}$, $m$ is real. Let $B = \{a^2 \mid a \in A\}$, $C = A \cup B$. If the sum of all the elements of $C$ is $6$, then the product of all the elements of $C$ is ______. | [
"By the condition, it is known that $1$, $2$, $4$, $m$, $m^2$ (allowing for repetition) are all the elements of $C$.\nNote that when $m$ is real, $1+2+4+m+m^2 > 6$, $1+2+4+m^2 > 6$, so it can only be $C = \\{1, 2, 4, m\\}$, and $1+2+4+m = 6$. Therefore, $m = -1$, and is tested to be consistent with the question. At... | China | China Mathematical Competition | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | final answer only | -8 | |
0kos | Problem:
A unit square $ABCD$ and a circle $\Gamma$ have the following property: if $P$ is a point in the plane not contained in the interior of $\Gamma$, then $\min (\angle APB, \angle BPC, \angle CPD, \angle DPA) \leq 60^{\circ}$. The minimum possible area of $\Gamma$ can be expressed as $\frac{a \pi}{b}$ for relati... | [
"Solution:\n\nNote that the condition for $\\Gamma$ in the problem is equivalent to the following condition: if $\\min (\\angle APB, \\angle BPC, \\angle CPD, \\angle DPA) > 60^{\\circ}$, then $P$ is contained in the interior of $\\Gamma$.\n\nLet $X_1, X_2, X_3$, and $X_4$ be the four points in $ABCD$ such that $AB... | United States | HMMT February | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | final answer only | 106 | |
09i3 | Let $ABCD$ be a cyclic quadrilateral with inradius $r$. Let $J$ and $K$ be the incenters of $ABC$ and $ACD$, and let $P$ and $Q$ be the circumcenters of $AJK$ and $CJK$, respectively. Prove that
$$
|PQ| = |AC| - \frac{S_{AJCK}}{r}.
$$
Here $S_{AJCK}$ denotes the area of the quadrilateral $AJCK$. | [
"Let $F_1$ and $F_2$ be the feet of perpendiculars from the points $J$ and $K$ to $AC$, respectively. Then\n$$\nAF_1 = \\frac{AB + AC - BC}{2} \\quad \\text{and} \\quad AF_2 = \\frac{AD + AC - CD}{2}.\n$$\nSince $AB + CD = BC + AD$, we have $AF_1 = AF_2$. It follows that $F_1 = F_2$, that is, $JK \\perp AC$.\nSince... | Mongolia | Round 3 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Pla... | null | proof only | null | |
06kc | In $\triangle ABC$, $\angle B > \angle C$. Let $D$ be the point on side $BC$ such that $\angle DAC = \frac{B-C}{2}$. The circumcircle of $\triangle ACD$ meets side $AB$ again at $E$. The circumcircle of $\triangle ABD$ meets side $AC$ again at $F$. The internal angle bisector of $\angle BDE$ meets side $AB$ at $P$. The... | [
"\n\nNote that $\\angle PDQ = 180^\\circ - \\angle BDP - \\angle CDQ = 180^\\circ - \\frac{1}{2} \\angle BDE - \\frac{1}{2} \\angle CDF = 180^\\circ - \\frac{A}{2} - \\frac{A}{2} = 180^\\circ - A$. This implies $A, P, D, Q$ are concyclic. Hence, $\\angle APQ = \\angle ADQ = 180^\\circ - \\a... | Hong Kong | Pre-IMO 2017 Mock Exam | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
00pk | Suppose that $m$ is a positive integer. Let $P_m = \{2^m, 2^{m-1}3, 2^{m-2}3^2, \dots, 3^m\}$. If $X$ is a subset of $P_m$, we write $S_X$ for the sum of all elements of $X$, with the convention that $S_\emptyset = 0$ where $\emptyset$ is the empty set. Suppose that $y$ is a real number with $0 \le y \le 3^{m+1} - 2^{m... | [
"Let $\\alpha = 3/2$ so $1 + \\alpha > \\alpha^2$.\nGiven $y$, we construct $Y$ algorithmically. Let $Y = \\emptyset$ and of course $S_\\emptyset = 0$. For $i = 0$ to $m$, perform the following operation:\nIf $S_Y + 2^i 3^{m-i} \\le y$, then replace $Y$ by $Y \\cup \\{2^i 3^{m-i}\\}$.\nWhen this process is finished... | Balkan Mathematical Olympiad | Balkan 2012 shortlist | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0599 | A function $f$ maps every positive real number to a positive real number. There is a constant $c \neq 1$ such that $f(cx)^2 = f(x)f(c^2x)$ for all positive real numbers $x$. Must the same equality hold for any positive real numbers $c$ and $x$? | [
"Answer: No.\n\n**Solution 1:** Let $f(x) = e^{\\sin \\ln x}$. Then\n$$\n(f(cx))^2 = (e^{\\sin \\ln(cx)})^2 = e^{2\\sin(\\ln cx + \\ln c)}, \\\\\nf(x)f(c^2x) = e^{\\sin \\ln x} \\cdot e^{\\sin \\ln(c^2x)} = e^{\\sin \\ln x + \\sin(\\ln x + 2\\ln c)}.\n$$\nTaking $c = e^{2\\pi}$, we get\n$$\n\\begin{aligned}\n2 \\si... | Estonia | Estonian Math Competitions | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof and answer | No | |
0ara | Problem:
The operation $*$ satisfies the following properties:
$$
x * 0=0,\quad x *(y+1)=x * y+(x-y)
$$
Evaluate $2010 * 10$. | [
"Solution:\nLet us compute $2010 * 10$ using the given properties.\n\nLet $f(y) = 2010 * y$.\n\nWe are given:\n$$\nf(0) = 0\n$$\n$$\nf(y+1) = f(y) + (2010 - y)\n$$\nLet us compute step by step:\n\n$f(0) = 0$\n\n$f(1) = f(0) + (2010 - 0) = 0 + 2010 = 2010$\n\n$f(2) = f(1) + (2010 - 1) = 2010 + 2009 = 4019$\n\n$f(3) ... | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 20055 | |
03nx | Problem:
There are 20 students in a high school class, and each student has exactly three close friends in the class. Five of the students have bought tickets to an upcoming concert. If any student sees that at least two of their close friends have bought tickets, then they will buy a ticket too.
Is it possible that t... | [
"Solution:\n\nIt is impossible for the whole class to buy tickets to the concert.\nIf two students $A$ and $B$ are close friends, and $A$ has bought a ticket to the concert while $B$ has not, then $A$ is enticing $B$. We call this pair $(A, B)$ an enticement.\nIn order for a student to change their mind and buy a t... | Canada | CMO 2023 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | No | |
06ij | There are three identical red balls, three identical yellow balls and three identical green balls. In how many different ways can they be split into three groups of three balls each?
現有三個一模一樣的紅球、三個一模一樣的黃球和三個一模一樣的綠球。有多少種不同的方法把球分成三組使得每組各有三個球? (1分) | [] | Hong Kong | HONG KONG PRELIMINARY SELECTION CONTEST | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English; Chinese | final answer only | 10 | |
0aen | Во рамнокрак триаголник $ABC$, $\overline{AC} = \overline{BC} = 1$. За која вредност на $\gamma = \angle ACB$, изразот $g = \frac{\overline{AB}^2 + 2}{P_{\Delta ABC}}$ достигнува најмала вредност. | [
"Од косинусната теорема имаме $\\overline{AB}^2 = 2 - 2\\cos\\gamma$, а за плоштината на триаголникот важи $P_{\\Delta ABC} = \\frac{1}{2} \\overline{AC} \\cdot \\overline{BC} \\cdot \\sin\\gamma = \\frac{\\sin\\gamma}{2}$. Тогаш добиваме $g(\\gamma) = \\frac{4(2 - \\cos\\gamma)}{\\sin\\gamma}$. Ќе воведеме смена $... | North Macedonia | Републички натпревар по математика за средно образование | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Intermediate Algebra > Quadratic functions"
] | Macedonian, English | proof and answer | π/3 | |
0iin | Problem:
For this problem, our lawn is an $m \times n$ rectangular grid of clumps, that is, with $m$ rows running east-west and $n$ columns running north-south. To be even more explicit, we might say our clumps are at the lattice points
$$
\left\{(x, y) \in \mathbb{Z}^2 \mid 0 \leq x < n \text{ and } 0 \leq y < m \rig... | [
"Solution:\n\nThere is a one-to-one correspondence between allowable formations and paths from $(0,0)$ to $(n, m)$ made up of $n$ moves 1 unit to the right and $m$ moves 1 unit up. The correspondence works as follows: There must be a mobot at $(0,0)$, so start the path there. If that mobot is oriented to move up, t... | United States | Harvard-MIT Mathematics Tournament, Team Round A | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
06zh | Problem:
Two perpendicular lines divide a square into four parts, three of which have area $1$. Show that the fourth part also has area $1$. | [] | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
0fd5 | Problem:
Calcula todos los números enteros $a$, $b$ y $c$ tales que $a^{2} = 2 b^{2} + 3 c^{2}$. | [
"Solution:\nSea $(a, b, c)$ una solución distinta de $(0,0,0)$, con $|a| + |b| + |c|$ mínimo. Tomando la igualdad módulo $3$, tenemos $a^{2} = 2 b^{2}$ módulo $3$. Como $a^{2}$ y $b^{2}$ sólo pueden ser congruentes con $1$ o $0$, se deduce que $a$ y $b$ son múltiplos de $3$. Por tanto, $3 c^{2}$ es múltiplo de $9$,... | Spain | XLVII Olimpiada Matemática Española Primera Fase | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (0, 0, 0) | |
01k6 | Fifteen points are marked on a plane. Some of them are painted red, some others are painted blue, and all remained points are painted green. It is known that the number of the red points is the largest. The sum of the distances between the red points and the blue points is $5$, the sum of the distances between the red ... | [
"Let $m, n, k$ be respectively the numbers of red, green, and blue points. Let $A, B, C$ be respectively the sums of the distances between blue and green points, red and blue points, red and green points. From the inequality of triangle it follows that the following inequalities are necessary for the existence of t... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | red 6, blue 5, green 4 | |
079f | Consider a spherical ball on a plane and a point marked on it. We want to roll the ball on a closed polygon of the plane to bring the mark to the top of the ball while the ball is on its first place. Note that the ball must not roll in its place (it means rolling without moving on the plane). Prove that this is possibl... | [
"Let $r$ be the radius of the ball and $O$ be its first position on the plane. First, we roll the ball (sphere) on its great circle that passes through the marked point and the top point of the ball, until the mark be on top of the ball. Let $A$ be the current position of the ball on the plane. It is trivial that $... | Iran | 27th Iranian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0l46 | Real numbers $a$, $b$, and $c$ have arithmetic mean $0$. The arithmetic mean of $a^2$, $b^2$, and $c^2$ is $10$. What is the arithmetic mean of $ab$, $ac$, and $bc$?
$(A)$ $-5$ $(B)$ $-\frac{10}{3}$ $(C)$ $-\frac{10}{9}$ $(D)$ $0$ $(E)$ $\frac{10}{9}$ | [
"The given information implies that $a + b + c = 0$ and $a^2 + b^2 + c^2 = 30$. Then\n$$\n0 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 30 + 2(ab + ac + bc).\n$$\nTherefore $2(ab + ac + bc) = -30$ and the requested arithmetic mean is $\\frac{ab+ac+bc}{3} = \\frac{-15}{3} = -5$.\n\nConsider the system of e... | United States | AMC 10 B | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | MCQ | A | |
0fvu | Problem:
Sei $ABCD$ ein Sehnenviereck mit $\angle ABC = 60^{\circ}$. Nehme an, es sei $|BC| = |CD|$. Beweise, dass gilt
$$
|CD| + |DA| = |AB|
$$ | [
"Solution:\n\nSei $P$ der Punkt auf $AB$ mit $|BP| = |BC|$, sodass $A$ und $P$ auf der gleichen Seite von $B$ liegen. Nach Konstruktion ist $\\triangle PBC$ gleichschenklig und wegen $\\angle ABC = 60^{\\circ}$ sogar gleichseitig. Sei $\\angle CBD = \\angle CAD = \\alpha$. Offensichtlich ist $\\alpha < \\angle CBA ... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0hp2 | Problem:
Find the number of ordered pairs $(a, b)$ of positive integers such that $a$ and $b$ both divide $20^{19}$, but $a b$ does not. | [
"Solution:\nWrite $N = 20^{19} = 2^{38} \\cdot 5^{19}$.\nThe number of divisors of $N$ is $39 \\cdot 20 = 780$.\n\nThe number of pairs $(a, b)$ with $a b$ dividing $N$ is equal to the number of nonnegative integer solutions to $x + y \\leq 38$ times the number of nonnegative integer solutions to $x + y \\leq 19$; b... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 444600 | |
03d2 | A positive integer $n$ having 2022 divisors $1 = d_1 < d_2 < \dots < d_{2022} = n$ is called *nice* if $3d_{2017} + 2d_{2019} = n$. Find all nice integers. | [
"It follows from $d_i \\ge i$ and $d_6 d_{2017} = d_4 d_{2019} = n$ that\n$$\n2n = 6 d_{2017} + 4 d_{2019} \\le d_6 d_{2017} + d_4 d_{2019} = 2n.\n$$\nTherefore $d_6 = 6$ implying $d_i = i$ for $i = 1, 2, 3, 4, 5, 6$ and thus $n$ is divisible by $2^2 \\cdot 3 \\cdot 5$. Since $2022 = 2 \\cdot 3 \\cdot 337$ and $337... | Bulgaria | Bulgaria 2022 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 2^2 · 3 · 5^{336}, 2^2 · 3^{336} · 5, 2^{336} · 3 · 5^2, 2^{336} · 3^2 · 5 | |
0f79 | Problem:
Prove that there is no convex quadrilateral with vertices at lattice points so that one diagonal has twice the length of the other and the angle between them is $45$ degrees. | [] | Soviet Union | 20th ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0idk | Problem:
If $x$, $y$, $k$ are positive reals such that
$$
3 = k^{2}\left(\frac{x^{2}}{y^{2}} + \frac{y^{2}}{x^{2}}\right) + k\left(\frac{x}{y} + \frac{y}{x}\right),
$$
find the maximum possible value of $k$. | [
"Solution:\n$$\n\\frac{-1 + \\sqrt{7}}{2}\n$$\nWe have $3 = k^{2}\\left(\\frac{x^{2}}{y^{2}} + \\frac{y^{2}}{x^{2}}\\right) + k\\left(\\frac{x}{y} + \\frac{y}{x}\\right) \\geq 2k^{2} + 2k$, hence $7 \\geq 4k^{2} + 4k + 1 = (2k + 1)^{2}$, hence $k \\leq (\\sqrt{7} - 1)/2$. Obviously $k$ can assume this value, if we ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (sqrt(7) - 1)/2 | |
06kr | Find all polynomials $f$ such that $f$ has nonnegative integer coefficients, $f(1) = 7$ and $f(2) = 2017$. | [
"The only polynomial is $x^{10} + x^9 + x^8 + x^7 + x^6 + x^5 + 1$.\n\nSuppose one of the coefficients of $f_0(x) = f(x)$, say the coefficient of $x^k$, is at least $2$. Define $f_1(x) = f_0(x) - 2x^k + x^{k+1}$. Then $f_1(2) = f_0(2) = 2017$ and $f_1(1) < f_0(1) = 7$. Also, the coefficients of $f_1$ are nonnegativ... | Hong Kong | null | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | x^10 + x^9 + x^8 + x^7 + x^6 + x^5 + 1 | |
02eh | The triangle with vertices $(0, 0)$, $(0, 1)$, $(2, 0)$ is repeatedly reflected in the three lines $AB$, $BC$, $CA$ where $A$ is $(0, 0)$, $B$ is $(3, 0)$, $C$ is $(0, 3)$. Show that one of the images has vertices $(24, 36)$, $(24, 37)$ and $(26, 36)$. | [
"Let $R$ be reflection in $BC$: $(x, y) \\rightarrow (3 - y, 3 - x)$. Let $S$ be reflection in $AC$: $(x, y) \\rightarrow (-x, y)$. Let $T$ be reflection in $AB$: $(x, y) \\rightarrow (x, -y)$. Let $RS$ denote the reflection $R$ followed by the reflection $S$. Then we have $Q = STR$ is $(x, y) \\rightarrow (y + 3, ... | Brazil | XI OBM | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null | |
0hsn | Problem:
What is the smallest positive integer $x$ for which $x^{2}+x+41$ is not a prime? | [
"Solution:\n40"
] | United States | null | [
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | final answer only | 40 | |
0ksj | Problem:
The numbers $1,2, \ldots, 10$ are randomly arranged in a circle. Let $p$ be the probability that for every positive integer $k<10$, there exists an integer $k'$ $>k$ such that there is at most one number between $k$ and $k'$ in the circle. If $p$ can be expressed as $\frac{a}{b}$ for relatively prime positive... | [
"Solution:\n\nLet $n=10$ and call two numbers close if there is at most one number between them and a circular permutation focused if only $n$ is greater than all numbers close to it. Let $A_{n}$ be the number of focused circular permutations of $\\{1,2, \\ldots, n\\}$.\nIf $n \\geq 5$, then there are 2 cases: $n-1... | United States | HMMT February 2022 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 1390 | |
0ab0 | Find all real numbers $a, b, c, d$ such that
$a + b + c + d = 20$
and
$$
ab + ac + ad + bc + bd + cd = 150.
$$ | [
"$$\n400 = (a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2 \\cdot 150 \\text{ so } a^2 + b^2 + c^2 + d^2 = 100.\n$$\nNow\n$$\n\\begin{align*}\n(a - b)^2 + (a - c)^2 + (a - d)^2 + (b - c)^2 + (b - d)^2 + (c - d)^2 &= \\\\\n&= 3(a^2 + b^2 + c^2 + d^2) - 2(ab + ac + ad + bc + bd + cd) = \\\\\n&= 300 - 300 = 0\n\\end{ali... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | a = b = c = d = 5 | |
0b3u | Problem:
Call a lattice point visible if the line segment connecting the point and the origin does not pass through another lattice point. Given a positive integer $k$, denote by $S_{k}$ the set of all visible lattice points $(x, y)$ such that $x^{2}+y^{2}=k^{2}$. Let $D$ denote the set of all positive divisors of $20... | [
"Solution:\n\nWe claim that the required sum is $20$.\n\nLet $T_{k}$ denote the set of all lattice points in the circle $x^{2}+y^{2}=k^{2}$. We claim that $\\sum_{d \\mid k}\\left|S_{d}\\right|=\\left|T_{k}\\right|$. Indeed, given a point $(x, y)$ in $T_{k}$, let $g=\\operatorname{gcd}(x, y)$. Then $x / g, y / g$ a... | Philippines | Philippine Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | 20 | |
0cpj | For any positive integers $a > b > 1$, a sequence $x_1, x_2, \dots$ is defined as $x_n = \frac{a^n - 1}{b^n - 1}$. Find the least integer $d$ such that for any $a$ and $b$, this sequence cannot contain $d$ consecutive terms that are prime numbers. (V. Senderov)
Для натуральных чисел $a > b > 1$ определим последователь... | [
"При $a = 4$, $b = 2$ имеем $\\frac{a^2 - 1}{b^2 - 1} = 3$, $\\frac{a^3 - 1}{b^3 - 1} = 5$. Осталось показать, что больше двух простых чисел подряд не встретится.\n\nДокажем более сильное, чем требуется, утверждение: при $n \\ge 2$ хотя бы одно из чисел $\\frac{a^n - 1}{b^n - 1}$, $\\frac{a^{n+1} - 1}{b^{n+1} - 1}$... | Russia | Russian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English, Russian | proof and answer | 3 | |
07xn | $$
2 \sum_{k=0}^{n} \binom{2n}{k} (n-k) = \binom{2n}{n+1} (n+1), \ n = 1, 2, \dots
$$
and find the value of
$$
\sum_{k=0}^{n} k \binom{2n}{k}.
$$ | [
"Recall the identity\n$$\n(b+1) \\binom{a}{b+1} = \\binom{a}{b} (a-b)\n$$\nwhich follows directly from the definition of the binomial coefficients. Using this with $a = 2n$ and $b = k$, we obtain\n$$\n2 \\binom{2n}{k} (n-k) = \\binom{2n}{k} (2n-k) - k \\binom{2n}{k} = (k+1) \\binom{2n}{k+1} - k \\binom{2n}{k}.\n$$\... | Ireland | IRL_ABooklet_2025 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | n * 2^{2n-1} | |
0ke5 | Problem:
Alan draws a convex 2020-gon $\mathcal{A}=A_{1} A_{2} \cdots A_{2020}$ with vertices in clockwise order and chooses 2020 angles $\theta_{1}, \theta_{2}, \ldots, \theta_{2020} \in (0, \pi)$ in radians with sum $1010 \pi$. He then constructs isosceles triangles $\triangle A_{i} B_{i} A_{i+1}$ on the exterior of... | [
"Solution:\n\nFor each $i$, let $\\tau_{i}$ be the transformation of the plane which is rotation by $\\theta_{i}$ counterclockwise about $B_{i}$. Recall that a composition of rotations is a rotation or translation, and that the angles of rotation add. Consider the composition $\\tau_{2020} \\circ \\tau_{2019} \\cir... | United States | HMMT February 2020 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05e0 | Problem:
Let $ABC$ be an acute-angled triangle in which $BC < AB$ and $BC < CA$. Let point $P$ lie on segment $AB$ and point $Q$ lie on segment $AC$ such that $P \neq B$, $Q \neq C$ and $BQ = BC = CP$. Let $T$ be the circumcentre of triangle $APQ$, $H$ the orthocentre of triangle $ABC$, and $S$ the point of intersectio... | [
"Solution:\n\nWe show that $T$ and $H$ are both on the angle bisector $\\ell$ of $\\angle BSC$.\n\nWe first prove that $H \\in \\ell$. The altitude $CH$ in triangle $ABC$ is also the altitude in isosceles triangle $PBC$ with $CP = CB$. Therefore, $CH$ is also the angle bisector of $\\angle ... | European Girls' Mathematical Olympiad (EGMO) | EGMO2022 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transf... | null | proof only | null | |
0kxx | Problem:
An entry in a grid is called a saddle point if it is the largest number in its row and the smallest number in its column. Suppose that each cell in a $3 \times 3$ grid is filled with a real number, each chosen independently and uniformly at random from the interval $[0,1]$. Compute the probability that this g... | [
"Solution:\n\nWith probability 1, all entries of the matrix are unique. If this is the case, we claim there can only be one saddle point. To see this, suppose $A_{ij}$ and $A_{kl}$ are both saddle points. They cannot be in the same row, since they cannot both be the greatest number in the same row, and similarly th... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 3/10 | |
02n0 | Problem:
Três quadrados - Dois quadrados, $ABCD$ com uma área de $30~\mathrm{cm}^2$ e $FHIJ$ com uma área de $20~\mathrm{cm}^2$, têm seus lados $AD$ e $HI$ sobre uma reta, conforme a figura. Se o ponto $E$ do segmento $AH$ for tal que $BEFG$ é um quadrado, calcule a área desse quadrado.
 | [] | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 50 | |
0i0j | Problem:
Do there exist positive integers $x, y, z$ such that $28x + 30y + 31z = 365$? What is the largest positive integer $n$ such that the equation $28x + 30y + 31z = n$ cannot be solved in positive integers $x, y, z$? | [
"Solution:\n\nA glance at a calendar confirms that $x = 1$, $y = 4$, $z = 7$ solves the first equation.\n\nFor the second part, we claim the maximum value of $n$ is $370$. To see that this $n$ cannot be obtained, suppose $28x + 30y + 31z = 370$ and consider remainders modulo $28$ to observe that $2y + 3z \\equiv 6 ... | United States | Berkeley Math Circle | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Yes; for example x=1, y=4, z=7. The largest n is 370. | |
0441 | Suppose $A = \{1, 2, 3\}$, $B = \{4x - y \mid x, y \in A\}$, $C = \{4x + y \mid x, y \in A\}$. Then the sum of all the elements of $B \cap C$ is ______. | [
"When $x$ and $y$ take all the elements of $A$, respectively, $4x - y$ gets exactly the values $1, 2, 3, 5, 6, 7, 9, 10, 11$, and $4x + y$ gets exactly the values $5, 6, 7, 9, 10, 11, 13, 14, 15$.\n\nHence, $B \\cap C = \\{5, 6, 7, 9, 10, 11\\}$. Then the sum of all the elements of $B \\cap C$ is $48$.\n\n$\\square... | China | China Mathematical Competition | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 48 | |
06ju | Let $\lambda$ be a nonnegative real number and such that $\frac{a+b}{2} \ge \lambda\sqrt{ab}+(1-\lambda)\sqrt{\frac{a^2+b^2}{2}}$ holds for all positive real numbers $a$ and $b$. Find the smallest possible value of $\lambda$. | [
"Firstly, when $\\lambda = \\frac{1}{2}$, we have\n$$\n\\begin{align*}\n\\frac{a+b}{2} &\\ge \\frac{1}{2}\\sqrt{ab} + \\frac{1}{2}\\sqrt{\\frac{a^2+b^2}{2}} \\\\\n\\Leftrightarrow \\quad (a+b)^2 &\\ge ab + \\frac{a^2+b^2}{2} + 2\\sqrt{ab} \\cdot \\sqrt{\\frac{a^2+b^2}{2}} \\\\\n\\Leftrightarrow \\quad (a+b)^2 &\\ge... | Hong Kong | CHKMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 1/2 | |
03mt | Let $ABC$ be an acute-angled triangle with altitudes $AD$, $BE$, and $CF$. Let $H$ be the orthocentre, that is, the point where the altitudes meet. Prove that
$$
\frac{AB \cdot AC + BC \cdot BA + CA \cdot CB}{AH \cdot AD + BH \cdot BE + CH \cdot CF} \le 2.
$$ | [
"**Method 1:** Let $AB = c$, $AC = b$, and $BC = a$ denote the three side lengths of the triangle.\nAs $\\angle BFH = \\angle BDH = 90^\\circ$, $FHDB$ is a cyclic quadrilateral. By the Power-of-a-Point Theorem, $AH \\cdot AD = AF \\cdot AB$. (We can derive this result in other ways: for example, see Method 2, below... | Canada | Kanada 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 > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles >... | English | proof only | null | |
0eth | Find all pairs of real numbers $x$ and $y$ which satisfy the following equations:
$$
\begin{aligned}
x^2 + y^2 - 48x - 29y + 714 &= 0 \\
2xy - 29x - 48y + 756 &= 0
\end{aligned}
$$ | [
"The two equations can be rewritten as\n$$\n\\begin{aligned}\n(x - 24)^2 + \\left(y - \\frac{29}{2}\\right)^2 &= \\frac{289}{4} \\\\\n(x - 24)\\left(y - \\frac{29}{2}\\right) &= -30.\n\\end{aligned}\n$$\nBy putting $X = x - 24$ and $Y = y - \\frac{29}{2}$, we obtain\n$$\nX^2 + Y^2 = \\frac{289}{4} \\tag{1}\n$$\n$$\... | South Africa | The South African Mathematical Olympiad Third Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | [(28, 7), (20, 22), (33/2, 37/2), (63/2, 21/2)] | |
0chf | Given a positive integer $n$, a set $S$ is *n-admissible* if:
• each element of $S$ is an unordered triple of distinct integers in $\{1, 2, \dots, n\}$,
• $|S| = n-2$, and
• for each $1 \le k \le n-2$ and each choice of $k$ distinct $A_1, A_2, \dots, A_k \in S$,
$$
|A_1 \cup A_2 \cup \dots \cup A_k| \ge k + 2.
$$
Is i... | [
"The answer is in the affirmative. Note that the condition on $S$ may be rephrased as follows: For any subset $M$ of $N$ of size $|M| \\ge 2$, there are at most $|M| - 2$ sets in $S \\cap P_3(M)$, where $P_3(M)$ is the set of 3-element subsets of $M$.\n\nExtend the problem to $n \\ge 1$ and proceed by induction on ... | Romania | Romanian Master of Mathematics | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0dse | Let $a$, $b$, $c$ be nonzero integers, with $1$ as their only positive common divisor, such that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$. Find the number of such triples $(a, b, c)$ with $50 \geq |a| \geq |b| \geq |c| \geq 1$. | [
"Since the sum is $0$, there are positive and negative integers among the three. Without loss of generality, we assume that there is $1$ negative and two positive integers. (The other case is obtained by changing the signs of all the integers.) Then $c < 0$ since it has the smallest absolute value. Let $\\gcd(a, b)... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 20 | |
0f3s | Problem:
Given a point $P$ on the diameter $AC$ of the circle $K$, find the chord $BD$ through $P$ which maximises the area of $ABCD$. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometr... | null | proof only | null | |
0lcu | Given an acute, scalene triangle $ABC$ ($AB < AC$) with the circumcircle $(O)$. Let $I$ be the midpoint of arc $BC$ that does not contain $A$. Take $K$ on $AC$ ($K \neq C$) such that $IK = IC$. Line $BK$ intersects $(O)$ again at $D$ and intersects $AI$ at $E$. $DI$ meets $AC$ at $F$.
a) Prove that $EF = \frac{BC}{2}$... | [
"a) It is easy to see that $K$ belongs to segment $AC$. Since $IK = IC$, we have triangle $IKC$ is isosceles at $I$. Hence, $\\angle AKI = 180^\\circ - \\angle IKC = 180^\\circ - \\angle ICK = \\angle ABI$, because $ABIC$ is a cyclic quadrilateral. Furthermore, note that $I$ is the midpoint of arc $BC$, then $\\ang... | Vietnam | VMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | English | proof only | null | |
03mu | On a $(4n + 2) \times (4n + 2)$ square grid, a turtle can move between squares sharing a side. The turtle begins in a corner square of the grid and enters each square exactly once, ending in the square where she started. In terms of $n$, what is the largest positive integer $k$ such that there must be a row or column t... | [
"We shall prove that the answer is $2n + 2$. Number the rows in increasing order, from top to bottom, and number the columns from left to right. By symmetry, we may (and shall) assume that the turtle starts in the top right corner square.\n\nFirst we shall prove that some row or column must be entered at least $2n ... | Canada | Kanada 2015 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2n + 2 | |
0km4 | Problem:
Mario has a deck of seven pairs of matching number cards and two pairs of matching Jokers, for a total of $18$ cards. He shuffles the deck, then draws the cards from the top one by one until he holds a pair of matching Jokers. The expected number of complete pairs that Mario holds at the end (including the Jo... | [
"Solution:\n\nConsidering ordering the nine pairs by the time they are first complete. Since the pairs are treated equally by the drawing process, this ordering is a uniform ordering. Therefore the problem becomes the following: consider ordering $7$ $N$'s and $2$ $J$'s randomly. What is the expected position of th... | United States | HMMT November 2021 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 1003 | |
0347 | Problem:
Consider the equations
$$
[x]^3 + x^2 = x^3 + [x]^2 \text{ and } \left[x^3\right] + x^2 = x^3 + \left[x^2\right]
$$
where $[t]$ is the greatest integer that does not exceed $t$. Prove that:
a) any solution of the first equation is an integer;
b) the second equation has a non-integer solution. | [
"Solution:\na) Let $x$ satisfy the equality $[x]^3 + x^2 = x^3 + [x]^2$. Then setting $t = [x]$ and $\\alpha = x - t \\in [0,1)$ one has that\n$$\nt^3 - t^2 = (t + \\alpha)^3 - (t + \\alpha)^2 \\Longleftrightarrow \\alpha\\left(\\alpha^2 + (3t - 1)\\alpha + 3t^2 - 2t\\right) = 0\n$$\nHence either $\\alpha = 0$, or ... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof only | null | |
013e | Problem:
We say that a pile is a set of four or more nuts. Two persons play the following game. They start with one pile of $n \geq 4$ nuts. During a move a player takes one of the piles that they have and split it into two non-empty subsets (these sets are not necessarily piles, they can contain an arbitrary number o... | [
"Solution:\n\nAnswer: The first player has a winning strategy when $n \\equiv 0,1,2 \\pmod{4}$; otherwise the second player has a winning strategy.\n\nLet $n=4k+r$, where $0 \\leq r \\leq 3$. We will prove the above answer by induction on $k$; clearly it holds for $k=1$. We are also going to need the following usef... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | n ≡ 0, 1, 2 (mod 4) | |
0akj | Let $k > 1$ be a positive integer and $n > 2018$ be an odd positive integer. The nonzero rational numbers $x_1, x_2, \dots, x_n$ are not all equal and satisfy
$$
x_1 + \frac{k}{x_2} = x_2 + \frac{k}{x_3} = x_3 + \frac{k}{x_4} = \dots = x_{n-1} + \frac{k}{x_n} = x_n + \frac{k}{x_1}.
$$
Find:
a) the product $x_1 \cdot x_... | [
"a) If $x_i = x_{i+1}$ for some $i$ (assuming $x_{n+1} = x_1$), then by the given identity all $x_i$ will be equal, a contradiction. Thus $x_1 \\ne x_2$ and\n$$\nx_1 - x_2 = k \\frac{x_2 - x_3}{x_2 x_3}.\n$$\nAnalogously\n$$\nx_1 - x_2 = k \\frac{x_2 - x_3}{x_2 x_3} = k^2 \\frac{x_3 - x_4}{(x_2 x_3)(x_3 x_4)} = \\d... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a) x1·x2·…·xn = ± k^{n/2} (equivalently ±√(k^n)). b) The least k is 4. | |
00p5 | Is it possible to partition the set of all integer numbers into ordered triples in such a way that, for every triple $(a, b, c)$ the number
$$
|a^3 b + b^3 c + c^3 a|
$$
be a perfect square? | [
"Suppose first that $a + b + c = 0$. Then we have\n$$\n\\begin{aligned}\n|a^3 b + b^3 c + c^3 a| &= |a^3 b + b^3 (-a-b) + (-a-b)^3 a| \\\\\n&= |-b^4 - 2b^3 a - 3a^2 b^2 - 2a^3 b - a^4| \\\\\n&= (a^2 + ab + b^2)^2.\n\\end{aligned}\n$$\nSo it suffices to partition the set of all integers into triples of zero sum. One... | Balkan Mathematical Olympiad | BMO 2010 Shortlist | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0g5c | 試求所有的正整數 $n$, 使得存在質數 $p > 2$ 和質數 $q$, 滿足 $n$ 的 $p$ 進位表示法是 2011, 且 $n$ 的 $q$ 進位表示法是 $10 \cdots 0$ (1 後面跟著任意多個 0). | [
"$n = 256$。題目所述這樣的 $n$ 滿足 $n = 2p^3 + p + 1 = q^k$, 其中 $k$ 是某正整數。顯然 $2p^3 + p + 1$ 是偶數, 故 $q = 2$.\n\n易檢驗 $3|2p^3+p$ 必成立。故由 $3|2^k-1$ 可知 $2|k$。令 $A = 2^{k/2}$, 我們有 $(A+1)(A-1) = 2p^3+p = (2p^2+1)p$, 故 $p|(A-1)$ 或 $p|(A+1)$。令 $A \\pm 1 = pt$, $t$ 是正整數。則有\n$$\npt^2 \\pm 2t = t(pt \\pm 2) = 2p^2 + 1.\n$$\n我們有 $p|\\pm2... | Taiwan | 二〇一一數學奧林匹亞競賽第二階段選訓營 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operatio... | null | proof and answer | 256 | |
0gmn | Find the smallest
a. positive real number
b. positive integer
$t$ for which the equation $(x^2 + y^2)^2 + 2t x (x^2 + y^2) = t^2 y^2$ has a solution in positive real numbers. | [] | Turkey | IMO TEAM SELECTION EXAM | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | a: no smallest positive t; all positive t admit solutions (infimum is 0). b: 1 | |
08fb | Problem:
Il primo quadrante del piano cartesiano è l'insieme dei punti $(x, y)$ con $x$ e $y$ numeri reali strettamente positivi (cioè $x>0, y>0$ ). Ciascun punto del primo quadrante è colorato di rosso o di blu. Inoltre, per ogni punto $P=(x, y)$ del primo quadrante, tutti i punti sulla semiretta uscente da $P$ con p... | [
"Solution:\n\nConsideriamo due punti qualsiasi $P_{1} = (x_{1}, y_{1})$ e $P_{2} = (x_{2}, y_{1})$ giacenti sulla semiretta orizzontale data dall'intersezione di $y = y_{1}$ con il primo quadrante (si veda anche la figura, dove $y_{1} = 1$). Siano $r_{1}, r_{2}$ le due semirette uscenti da $P_{1}, P_{2}$ considerat... | Italy | Italian Mathematical Olympiad - February Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Translation"
] | null | proof only | null | |
06d1 | A *lattice point* on the coordinate plane is a point with integer coordinates. Find all positive integer $n \ge 3$ such that there exists an $n$-sided polygon having lattice points as vertices and all sides have equal length. | [
"$n$ can be any even integer larger than $2$.\n\nWe first consider the case when $n$ is odd. Let $(x_j, y_j)$, where $j = 1, 2, \\dots, n$, be the vertices of the polygon in anticlockwise direction. WLOG assume $x_1 = y_1 = 0$. By using the homothety $(x_j, y_j) \\mapsto \\left(\\frac{x_j}{d}, \\frac{y_j}{d}\\right... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Prealgebra / Basic Alg... | null | proof and answer | all even integers n ≥ 4 | |
0alh | Problem:
Let $p$ and $q$ be the roots of $2x^{2} - 5x + 1 = 0$. Find the value of $\log_{2} p + \log_{2} q$.
(a) 2
(b) 0
(c) 1
(d) -1 | [] | Philippines | Qualifying Round | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | MCQ | d | |
01se | The circle $\omega$ passes through the vertices $B$ and $C$ of a triangle $ABC$, and meets its sides $AB$ and $AC$ at points $C_1$ and $B_1$ respectively. Let $X$ be the midpoint of the arc $\sim B_1BC$, and $Y$ be the midpoint of the arc $\sim BCC_1$ of $\omega$.
Find the angle between the line $XY$ and the bisector o... | [
"Let $\\angle ABC = \\beta$ and $\\angle BCA = \\gamma$. Then $\\angle CAB = 180^\\circ - \\beta - \\gamma$. By $\\xi$ denote the degree measure of the arc $C_1B$ that does not contain the vertex $C$, by $\\eta$ denote the degree measure of the arc $\\tilde{C}B_1$ that does not contain the vertex $B$. Then $\\gamma... | Belarus | FINAL ROUND | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 90 degrees | |
0h7i | Let $f: N \to Z$ be the function that for any two positive integers $a, b$ $f(ab) = f(a) + f(b)$. Find all possible values of $f(1000)$. | [
"Let us look on such an equality:\n$$\nf(1000) = f(10^3) = 3 \\cdot f(10)\n$$\nLet us show now that for any integer $k$ $f(1000)$ can equal $3k$. Let $\\operatorname{ord}(a)$ be a maximum degree of $2$, by what $a$ is divisible. Let\n$$\na = 2^i(2a_0 + 1), \\quad b = 2^m(2b_0 + 1). \\text{ Thus}\n$$\n$$\n\\operator... | Ukraine | UkraineMO | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 3k for any integer k | |
0827 | Problem:
Sia $ABC$ un triangolo isoscele tale che $\angle BAC = 120^\circ$ e $AB = AC = 1$. Quanto misura il raggio del cerchio circoscritto?
(A) 2
(B) $\frac{\sqrt{3}}{2}$
(C) $\frac{\sqrt{2}}{2}$
(D) 1
(E) nessuna delle precedenti. | [
"Solution:\n\nLa risposta è (D). Chiamiamo $O$ il centro del cerchio circoscritto e consideriamo il triangolo $AOC$. Esso è isoscele in $O$ e d'altra parte $\\angle AOC = \\frac{1}{2} \\angle BAC = 60^\\circ$; dunque $AOC$ è equilatero."
] | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | D | |
099e | Find all functions $f: \mathbb{Q} \to \mathbb{R}$ satisfying $f(xy) = f(x)f(y) + f(x+y) - 1$ for every $x, y \in \mathbb{Q}$. | [
"Substituting $x = y = 0$ in the functional equation, we get $f^2(0) = 1$ or $f(0) = \\pm 1$.\n\nIf $f(0) = 1$ then substituting $x = x, y = 0$, we get $f(0) = f(x) \\cdot f(0) + f(x) - 1$ or $f(x) = 1$. So in this case $f(x) = 1$ is a constant function.\n\nIf $f(0) = -1$, then substituting $x = y = 2$, we get $f(4... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | f(x) ≡ 1; f(x) ≡ −1; f(x) = x − 1 for all rational x | |
04q6 | Find all functions $f: \mathbb{Q}^+ \to \mathbb{Q}^+$ such that
$$
f(x^2 f(y)^2) = f(x)^2 f(y)
$$
holds for all positive rational numbers $x$ and $y$. | [
"Plugging $(x, y) \\leftarrow (f(x), y)$ into the given equation, it follows that\n$$\nf(f(x)^2 f(y)^2) = f(f(x))^2 f(y)\n$$\nholds for all positive rational numbers $x$ and $y$, hence $f(f(y))^2 f(x) = f(f(x)^2 f(y)^2) = f(f(x))^2 f(y)$, i.e.\n$$\n\\frac{f(x)}{f(y)} = \\left( \\frac{f(f(x))}{f(f(y))} \\right)^2 = ... | Croatia | Croatian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 1 for all positive rational x | |
00tt | Let $k > 1$ be a real number, $n \ge 3$ be an integer, and $x_1 \ge x_2 \ge x_3 \ge \dots \ge x_n > 0$ be real numbers. Prove the inequality:
$$
\frac{x_1 + kx_2}{x_2 + x_3} + \frac{x_2 + kx_3}{x_3 + x_4} + \dots + \frac{x_{n-1} + kx_n}{x_n + x_1} + \frac{x_n + kx_1}{x_1 + x_2} \ge \frac{n(k+1)}{2}.
$$ | [
"Writing $x_{n+1} = x_1$, by AM-GM we have\n$$\n\\frac{x_1+x_2}{x_2+x_3} + \\frac{x_2+x_3}{x_3+x_4} + \\dots + \\frac{x_{n-1}+x_n}{x_n+x_1} + \\frac{x_n+x_1}{x_1+x_2} \\ge n \\sqrt{\\prod_{i=1}^{n} \\frac{x_i+x_{i+1}}{x_{i+1}+x_{i+1}}} = n.\n$$\nSo it is enough to prove that\n$$\n\\frac{x_1}{x_1+x_2} + \\frac{x_2}{... | Balkan Mathematical Olympiad | BMO 2022 shortlist | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0d7f | Let $a, b$ be two positive integers such that
$$
b+1 \mid a^{2}+1, \quad a+1 \mid b^{2}+1
$$
Prove that $a, b$ are odd numbers. | [
"Suppose by contrary that one of $a, b$ is even, say $a$. So $a^{2}+1$ is odd. This means that $b+1$ must be odd, or equivalently, $b$ is even. We will prove that this cannot happen.\n\nIndeed, put $d=\\gcd(a+1, b+1)$ then $d \\mid a+1 \\mid b^{2}+1$, hence $d \\mid b^{2}-b = b(b+1) + 2b$. From this, we have $d \\m... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
00co | En un club algunos pares de socios son amigos. Dado $k \ge 3$ diremos que un club es *k-amigable* si en todo grupo de $k$ socios éstos se pueden sentar en una mesa redonda de modo que cada par de vecinos son amigos.
a) Demostrar que si un club es 6-amigable entonces es 7-amigable.
b) ¿Es cierto que si un club es 9-amig... | [] | Argentina | Nacional OMA 2019 | [
"Discrete Mathematics > Graph Theory"
] | Spanish | proof and answer | a) Yes, every 6-friendly club is 7-friendly. b) No; for example, the Petersen graph is 9-friendly but not 10-friendly. | |
06sy | Let $M$ be a set of $n \geqslant 4$ points in the plane, no three of which are collinear. Initially these points are connected with $n$ segments so that each point in $M$ is the endpoint of exactly two segments. Then, at each step, one may choose two segments $AB$ and $CD$ sharing a common interior point and replace th... | [
"A line is said to be red if it contains two points of $M$. As no three points of $M$ are collinear, each red line determines a unique pair of points of $M$. Moreover, there are precisely $\\binom{n}{2} < \\frac{n^{2}}{2}$ red lines. By the value of a segment we mean the number of red lines intersecting it in its i... | IMO | 55th International Mathematical Olympiad Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof only | null | |
08gt | Problem:
Marina vuole colorare ciascuna casella di una tabella $5 \times 5$ in rosso o in blu, in modo che ogni casella condivida al più un lato con una casella dello stesso colore. Quante sono le colorazioni possibili della tabella?
(A) 30
(B) 45
(C) 46
(D) 50
(E) 90 | [
"Solution:\n\nLa risposta è (A). Innanzitutto, osserviamo che una volta colorata una riga e una colonna della tabella rispettando le condizioni date, allora esiste al più un modo di completare la colorazione rispettando le condizioni del problema: infatti, data una tabella $2 \\times 2$ con tre caselle colorate usa... | Italy | Italian Mathematical Olympiad - February Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | MCQ | A | |
03uz | As is shown in the figure, $P$ is a moving point on the parabola $y^2 = 2x$, points $B$, $C$ are on the $y$ axis, and the circle $(x - 1)^2 + y^2 = 1$ is internally tangent to $\triangle PBC$. Find the minimum value of the area of $\triangle PBC$.
 | [
"Denote $P$, $B$, $C$ by $P(x_0, y_0)$, $B(0, b)$, $C(0, c)$, and assume that $b > c$. The equation for the line $PB$ is\n$$\ny - b = \\frac{y_0 - b}{x_0}x.\n$$\nIt can be rewritten as\n$$\n(y_0 - b)x - x_0y + x_0b = 0.\n$$\nSince the distance between the circle center $(1, 0)$ and the line $PB$ is $1$, we have\n$$... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 8 | |
0e13 | Find all positive integers $n$, such that $n(n+2)(n+4)$ has at most 15 positive divisors. | [
"Write $a_n = n(n+2)(n+4)$ and let $b_n$ be the number of positive divisors of $a_n$. We can easily check that $b_1 = 4$, $b_2 = 10$, $b_3 = 8$, $b_4 = 14$, $b_5 = 12$, $b_6 = 24$, $b_7 = 12$, $b_8 = 28$, $b_9 = 12$ and $b_{10} = 40$.\n\nIn the next part we will use the following fact: if $p_1^{\\alpha_1} \\cdot p_... | Slovenia | Selection Examinations for the IMO | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 1, 2, 3, 4, 5, 7, 9 | |
05pw | Problem:
Un polynôme est dit « unitaire » si le coefficient de son monôme de plus haut degré vaut $1$. Soient $a < b$ des réels, et soit $P$ un polynôme unitaire non constant tel que $\max_{x \in [a, b]} |P(x)| < 2$.
Prouver qu'il existe un polynôme $Q$, unitaire et non constant, tel que $\max_{x \in [a, b]} |Q(x)| <... | [
"Solution:\n\nSi $Q$ désigne un polynôme, on pose\n$$\n\\|Q\\| = \\frac{1}{2} \\max_{x \\in [a, b]} |Q(x)| \\text{ et } T(Q)(x) = Q(x)^2 - 2\\|Q\\|^2\n$$\nIl est clair que si $Q$ est un polynôme unitaire non constant, alors $T(Q)$ l'est également. De plus, on a $\\|T(Q)\\| = \\|Q\\|^2$. Si l'on note $T^n$ la $n$-iè... | France | Olympiades Françaises de Mathématiques - Envoi 2 (Algèbre) | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
08ep | Problem:
Una griglia è costituita da $n \times n$ punti, con $n$ intero positivo. In alcuni di questi punti è disposta una sentinella. Ogni sentinella sceglie due direzioni tra di loro perpendicolari (tra quelle delle righe e delle colonne della griglia) e sorveglia tutti i punti che si trovano nelle direzioni prescel... | [
"Solution:\n\na. Una possibile disposizione nel caso $4 \\times 4$ è la seguente.\n\n\n\nb. Dimostriamo intanto che $S(n) \\leq 4 n$. Supponiamo per assurdo che $S(n)>4 n$ per un qualche valore di $n$. Allora esiste almeno una riga in cui ci sono almeno 5 sentinelle. Di queste 5 sentinelle,... | Italy | XXXVII Olimpiade Italiana di Matematica | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0hf5 | Let $ABCD$ be a quadrilateral circumscribed around a circle with center $O$. We construct equal segments $AA_1$, $AA_2$, $CC_1$ and $CC_2$ on rays $AB$, $AD$, $CB$ and $CD$ respectively, so that their lengths are greater than that of any side of $ABCD$, and $A_1$, $A_2$, $C_1$ and $C_2$ are located on the circle with c... | [
"Let $K$ and $P$ be tangent points of an inscribed circle of $ABCD$ to $AB$ and $BC$ respectively (Fig. 38). It is that $\\triangle C_1PO = \\triangle A_1KO$, since they are right triangles with equal legs $KO = PO$ and hypotenuses $C_1O = A_1O$ as radii of respectful circles. Thus we have $C_1P = A_1K$. Since $BK ... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
05qd | Problem:
Soit $ABC$ un triangle acutangle, et soit $P$ un point situé à l'intérieur du triangle $ABC$. Soit $D$ le milieu du segment $[PC]$, $E$ le point d'intersection des droites $(AP)$ et $(BC)$, et $Q$ le point d'intersection des droites $(BP)$ et $(DE)$. Montrer que, si les angles $\widehat{PAC}$ et $\widehat{PCB... | [
"Solution:\n\nSoit $R$ le point d'intersection des droites $(CQ)$ et $(AP)$. Il suffit de prouver l'égalité des angles de droites $(CB, CQ) = (AB, AR)$. Pour ce faire, il suffit de prouver que les points $B, C, A, R$ sont cocycliques, c'est-à-dire que $(BR, BC) = (AR, RC)$. Comme $(AR, AC) = (CP, CB)$, il suffit de... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dm9 | Problem:
Нека је $D$ тачка на страници $A C$ троугла $A B C$ у коме је $A B < B C$ таква да је $A B = B D$. Круг уписан у $\triangle A B C$ додирује $A B$ у $K$ и $A C$ у $L$, а $J$ је центар уписаног круга троугла $B C D$. Доказати да $K L$ полови дуж $A J$. | [
"Solution:\n\nНека је $M$ тачка на $A C$ таква да је $J M \\parallel K L$. Довољно је доказати да је $A M = 2 A L$.\n\nИз $\\angle B D A = \\alpha$ добијамо $\\angle J D M = 90^\\circ - \\frac{\\alpha}{2} = \\angle K L A = \\angle J M D$, па је $J M = J D$, а додирна тачка уписаног круга $\\triangle B C D$ са $C D$... | Serbia | СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | null | proof only | null | |
0jz0 | Problem:
Let $ABCD$ be a convex quadrilateral with $AB = 5$, $BC = 6$, $CD = 7$, and $DA = 8$. Let $M$, $P$, $N$, $Q$ be the midpoints of sides $AB$, $BC$, $CD$, $DA$ respectively. Compute $MN^{2} - PQ^{2}$. | [
"Solution:\nDraw in the diagonals of the quad and use the median formula three times to get $MN^{2}$ in terms of the diagonals. Do the same for $PQ^{2}$ and subtract, the diagonal length terms disappear and the answer is\n$$\n\\frac{BC^{2} + DA^{2} - AB^{2} - CD^{2}}{2} = 13\n$$"
] | United States | February 2017 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 13 | |
0l4o | Problem:
Call a 9-digit number a *cassowary* if it uses each of the digits $1$ through $9$ exactly once. Compute the number of cassowaries that are prime. | [
"Solution:\n\nEvery cassowary is divisible by $3$, as the sum of its digits is $1 + 2 + \\dots + 9 = 45$. Since all such numbers are divisible by $3$ and greater than $3$, none of them are prime. So, there are $0$ prime cassowaries."
] | United States | HMMT February | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 0 | |
03ic | Problem:
Let $a, b, c, d, e$ be integers such that $1 \leq a < b < c < d < e$. Prove that
$$
\frac{1}{[a, b]} + \frac{1}{[b, c]} + \frac{1}{[c, d]} + \frac{1}{[d, e]} \leq \frac{15}{16}
$$
where $[m, n]$ denotes the least common multiple of $m$ and $n$ (e.g. $[4,6]=12$). | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
02yz | Problem:
Juca é um escoteiro que está explorando as proximidades do seu acampamento. Após coletar frutas e madeira, ele deve pegar água no rio e voltar para sua barraca. Vamos representar, na figura, Juca pela letra $J$, o rio pela letra $r$ e sua barraca pela letra $B$. A distância dos pés das perpendiculares $C$ e $E... | [
"Solution:\nVamos marcar o ponto $B'$, simétrico do ponto $B$, em relação à reta $r$. O ponto $B'$ também dista $80\\ \\mathrm{m}$ da reta $r$ e, com isso, temos os triângulos congruentes $\\triangle B C D$ e $\\triangle B' C D$ (lado-ângulo-lado), sendo $D$ o ponto que Juca pegará água no rio. Como a menor distânc... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 180√2 m | |
0kdx | Problem:
Points $G$ and $N$ are chosen on the interiors of sides $ED$ and $DO$ of unit square $DOME$, so that pentagon $GNOME$ has only two distinct side lengths. The sum of all possible areas of quadrilateral $NOME$ can be expressed as $\frac{a-b \sqrt{c}}{d}$, where $a, b, c, d$ are positive integers such that $\ope... | [
"Solution:\n\n\n\nSince $MO=ME=1$, but $ON$ and $GE$ are both less than $1$, we must have either $ON=NG=GE=x$ (call this case 1) or $ON=GE=x$, $NG=1$ (call this case 2).\n\nEither way, the area of $NOME$ (a trapezoid) is $\\frac{1+x}{2}$, and triangle $NGT$ is a $45$-$45$-$90$ triangle. In ... | United States | HMMO 2020 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 10324 | |
095i | Problem:
Să se determine toate numerele naturale $m$ şi $n$ care verifică egalitatea $(m!-1)^{2}=n!+1$. | [
"Solution:\nRezolvare. $(m!-1)^{2}=n!+1 \\Leftrightarrow n!=m!\\cdot(m!-2) \\Leftrightarrow \\frac{n!}{m!}=m!-2$.\n$n!+1 \\geq 2 \\Rightarrow m \\geq 3 \\Rightarrow \\frac{n!}{m!} \\geq 4 \\Rightarrow n>m$, deci ecuaţia $\\left({ }^{*}\\right)$ poate fi scrisă $(m+1) \\cdot(m+2) \\cdot \\ldots \\cdot n=m!-2$.\nDacă... | Moldova | A 62-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | m = 3, n = 4 | |
008e | In the triangle $ABC$, $\angle ABC = 45^\circ$, $BC = 1$ and $E$ is a point on the side $AC$ such that $EC = 1$.
The line perpendicular to $AC$ through $E$ meets the line $BC$ at the point $D$ so that $CD = 2$ and $C$ is interior to $DB$.
Determine the angles of the triangle $ABD$. | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | ∠ABD = 45°, ∠BAD = 60°, ∠ADB = 75° | |
055w | Let $n$ and $m$ be positive integers, $n \ge m$. There is a game board of size $1 \times n$ divided into $n$ unit squares and an unlimited supply of sticky tapes of size $1 \times m$. On each move, any player adds a tape covering $m$ consecutive unit squares on the board precisely, at least one of which is not yet cove... | [
"\nFig. 25\n\na) If $n$ and $m$ are of equal parity, then the first player to move can cover the central squares in such a way that an equal number of uncovered squares remain on both sides (Fig. 25 with $n = 11$ and $m = 3$). The first player can now respond to each of the opposing player'... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | b) No | |
0cb5 | Determine the continuous functions $f : \mathbb{R} \to \mathbb{R}$ with the properties $f(1) = e$ and $f(x + y) = e^{3xy} \cdot f(x) \cdot f(y)$, for all $x, y \in \mathbb{R}$. | [
"Let $f : \\mathbb{R} \\to \\mathbb{R}$ be a function that satisfies the conditions in the statement. If there exists $a \\in \\mathbb{R}$ such that $f(a) = 0$ then $f(x) = f(a + (x - a)) = e^{3a(x-a)}f(a)f(x-a) = 0, \\forall x \\in \\mathbb{R}$, contradiction. On the other hand, we have $f(x) = e^{3x^2/4}f^2(x/2) ... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | f(x) = e^{(3x^2 - x)/2} | |
0czq | Consider a non-zero real number $a$ such that $\{a\} + \left\{ \frac{1}{a} \right\} = 1$, where $\{x\}$ denotes the fractional part of $x$. Prove that for any positive integer $n$, $\left\{ a^{n} \right\} + \left\{ \frac{1}{a^{n}} \right\} = 1$. | [
"We have\n$$\na + \\frac{1}{a} = [a] + \\left[ \\frac{1}{a} \\right] + \\{ a \\} + \\left\\{ \\frac{1}{a} \\right\\} = [a] + \\left[ \\frac{1}{a} \\right] + 1\n$$\nis an integer and denote this integer by $k$. Let $S_{n} = a^{n} + \\frac{1}{a^{n}}$, $n = 0, 1, 2, \\ldots$ Since $a$ and $\\frac{1}{a}$ are the roots ... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof only | null | |
0349 | Problem:
Let $ABC$ be an acute triangle. Find the locus of the points $M$ in the interior of $\triangle ABC$ such that
$$
AB - FG = \frac{MF \cdot AG + MG \cdot BF}{CM}
$$
where $F$ and $G$ are the feet of the perpendiculars from $M$ to the lines $BC$ and $AC$, respectively. | [
"Solution:\n1. Let $AP \\perp FG$ and $BQ \\perp FG$, $P, Q \\in FG$. Then\n$$\nAB \\geq PQ = PG + GF + FQ\n$$\nSince the quadrilateral $CFMG$ is cyclic we obtain $\\angle CMF = \\angle CGF = \\angle AGP$. This implies that $\\triangle APG \\sim \\triangle CFM$ and therefore\n$$\n\\frac{PG}{AG} = \\frac{MF}{CM} \\L... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler lin... | null | proof and answer | All interior points on the segment from C to D, where D is the intersection of the line through the circumcenter O and C with side AB. | |
00n5 | Let $x$ and $y$ be real numbers satisfying $(x + 1)(y + 2) = 8$.
Show that
$$
(xy - 10)^2 \geq 64.
$$
Furthermore, determine all pairs $(x, y)$ of real numbers for which equality holds. | [
"The inequality $(2x - y)^2 \\geq 0$ (with equality if and only if $y = 2x$) is equivalent to\n$$\n(2x + y)^2 \\geq 8xy.\n$$\nThe constraint $(x + 1)(y + 2) = 8$ gives $2x + y = 6 - xy$. Substituting this into the inequality above yields\n$$\n(6 - xy)^2 \\geq 8xy,\n$$\nwhich is equivalent to\n$$\n(xy - 10)^2 \\geq ... | Austria | AUT_ABooklet_2020 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (1, 2) and (-3, -6) | |
02ue | Problem:
Uma propriedade interessante do número 2013 é que 3 é o último dígito da soma $2+0+1$. Repetindo-se esse processo, isto é, escrevendo-se à direita o último dígito da soma dos três dígitos anteriores, teremos uma sequência:
$$
2,0,1,3,4,8,5,7 \ldots
$$
a) Prove que começando com a sequência 2,0,1, nessa ordem,... | [
"Solution:\n\na) Basta continuarmos a escrever mais termos da sequência seguindo a regra do enunciado:\n$$\n\\mathbf\\{2 , 0 , 1\\}, 3,4,8,5,7,0,2,9, \\mathbf\\{1\\}, \\mathbf\\{2\\}, \\mathbf\\{2\\}, \\ldots\n$$\n\nb) Veja que temos no máximo 10 valores possíveis para cada um dos três números $a, b$ e $c$. Portant... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | null | proof only | null | |
0lba | Given a positive integer $n$. There are $n$ boys and $n$ girls standing in a line. Give each person $X$ in the line a number of sweets which is exactly equal to the number of the pair $(a, b)$ such that $a$ and $b$ are of opposite sex of $X$ and $X$ is standing between $a$ and $b$. Prove that the total number of sweets... | [
"We denote a boy by the letter $b$ and a girl by the letter $g$. We can check that the total number of the sweets is exactly $\\frac{1}{3}n(n^2 - 1)$ if the line can be partitioned into $n$ consecutive pairs of one boy and one girl. This arrangement will be called the optimal arrangement.\n\nAny other arrangement w... | Vietnam | Vietnam Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | Vietnamese | proof only | null | |
0j6e | Problem:
Let $N=\overline{5 A B 37 C 2}$, where $A, B, C$ are digits between $0$ and $9$, inclusive, and $N$ is a $7$-digit positive integer. If $N$ is divisible by $792$, determine all possible ordered triples $(A, B, C)$. | [
"Solution:\n\nAnswer: $(0,5,5), (4,5,1), (6,4,9)$\n\nFirst, note that $792 = 2^{3} \\times 3^{2} \\times 11$. So we get that\n$$\n\\begin{gathered}\n8 \\mid N \\Rightarrow 8 \\mid \\overline{7 C 2} \\Rightarrow 8 \\mid 10C + 6 \\Rightarrow C = 1, 5, 9 \\\\\n9 \\mid N \\Rightarrow 9 \\mid 5 + A + B + 3 + 7 + C + 2 \... | United States | Harvard-MIT November Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (0, 5, 5), (4, 5, 1), (6, 4, 9) | |
0fvl | Problem:
Sei $ABC$ ein spitzwinkliges Dreieck mit $AB > AC$ und Höhenschnittpunkt $H$. Sei $D$ der Höhenfusspunkt von $A$ auf $BC$. Sei $E$ die Spiegelung von $C$ an $D$. Die Geraden $AE$ und $BH$ schneiden sich im Punkt $S$. Sei $N$ der Mittelpunkt von $AE$ und sei $M$ der Mittelpunkt von $BH$. Beweise, dass $MN$ sen... | [
"Solution:\n\nNach Konstruktion ist das Dreieck $AEC$ gleichschenklig und wegen $AB > AC$ liegt $D$ auf der Strecke $BD$ (siehe Abbildung 3). Sei $k_1$ der Kreis mit Mittelpunkt $M$ und Durchmesser $BH$ und sei $k_2$ der Kreis mit Mittelpunkt $N$ und Durchmesser $AE$. Wegen $\\angle BDH = \\angle EDA = 90^{\\circ}$... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / ... | null | proof only | null | |
0jao | Problem:
Let $ABC$ be a triangle with incenter $I$. Let the circle centered at $B$ and passing through $I$ intersect side $AB$ at $D$ and let the circle centered at $C$ passing through $I$ intersect side $AC$ at $E$. Suppose $DE$ is the perpendicular bisector of $AI$. What are all possible measures of angle $BAC$ in d... | [
"Solution:\n\nAnswer: $\\frac{540}{7}$\n\nLet $\\alpha = \\measuredangle BAC$. $DE$ is the perpendicular bisector of $AI$, so $DA = DI$, and $\\angle DIA = \\angle DAI = \\alpha/2$. Thus, $\\angle IDB = \\angle DIB = \\alpha$, since $BD = BI$. This gives $\\angle DBI = 180^\\circ - 2\\alpha$, so that $\\angle ABC =... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 540/7 degrees | |
0ih8 | Problem:
Compute
$$
\sum_{n=0}^{\infty} \frac{n}{n^{4}+n^{2}+1}
$$ | [
"Solution:\nNote that\n$$\nn^{4}+n^{2}+1=\\left(n^{4}+2 n^{2}+1\\right)-n^{2}=\\left(n^{2}+1\\right)^{2}-n^{2}=\\left(n^{2}+n+1\\right)\\left(n^{2}-n+1\\right)\n$$\nDecomposing into partial fractions, we find that\n$$\n\\frac{n}{n^{4}+n^{2}+1}=\\frac{1}{2}\\left(\\frac{1}{n^{2}-n+1}-\\frac{1}{n^{2}+n+1}\\right) .\n... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1/2 |
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