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id string | problem_markdown string | solutions_markdown list | images images list | country string | competition string | topics_flat list | language string | problem_type string | final_answer string |
|---|---|---|---|---|---|---|---|---|---|
009s | Each cell of an $n \times n$ grid square is colored black or white. We call such a coloring *nice* if every $2 \times 2$ square covers an even number of black cells, and every cross covers an odd number of black cells. Find all $n \ge 3$ such that in each nice coloring the four corner cells have the same color.
![](att... | [
"Color row 1 black and rows 2, 3 white. Then extend the coloring periodically with the remaining rows until the entire square is colored. It is immediate that the obtained coloring is nice. If $n \\equiv 0 \\pmod{3}$ or $n \\equiv 2 \\pmod{3}$, the last row is white while the first one is black. So a necessary cond... | Argentina | NATIONAL XXX OMA | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | n ≡ 1 (mod 3) | |
00hy | There are $n$ line segments on the plane, no three intersecting at a point, and each pair intersecting once in their respective interiors. Tony and his $2 n-1$ friends each stand at a distinct endpoint of a line segment. Tony wishes to send Christmas presents to each of his friends as follows:
First, he chooses an endp... | [
"Draw a circle that encloses all the intersection points between line segments and extend all line segments until they meet the circle, and then move Tony and all his friends to the circle. Number the intersection points with the circle from 1 to $2 n$ anticlockwise, starting from Tony (Tony has number 1). We will ... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
00k4 | Let $ABCDEF$ be a regular octahedron with lower vertex $E$, upper vertex $F$, middle slice plane $ABCD$, center $M$ and circumsphere $k$. Furthermore let $X$ be an arbitrary point within side $ABF$. The line $EX$ intersects $k$ in $E$ and $Z$ and the plane $ABCD$ in $Y$.
$$
\text{Show that } \langle EMZ \rangle = \lang... | [
"\nWe intersect the entire figure with the plane through the points $X$, $E$ and $F$. Since $M$ is on line $EF$, it is part of that plane. Also points $Y$ and $Z$ are part of that plane because they are on line $EX$. The intersection of the circumsphere $k$ and the plane results in a circle... | Austria | AustriaMO2013 | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
00rl | The point $M$ lies on the side $AB$ of the circumscribed quadrilateral $ABCD$. The points $I_1, I_2$, and $I_3$ are the incenters of $\triangle MBC$, $\triangle MCD$, and $\triangle MDA$. Show that the points $M, I_1, I_2$, and $I_3$ lie on a circle.
 | [
"Lemma. Let $I$ be the incenter of $\\triangle ABC$ and let the points $P$ and $Q$ lie on the lines $AB$ and $AC$. Then the points $A, I, P$, and $Q$ lie on a circle if and only if\n$$\n\\overline{BP} + \\overline{CQ} = BC\n$$\nwhere $\\overline{BP}$ equals $|BP|$ if $P$ lies in the ray $BA \\rightarrow$ and $-|BP|... | Balkan Mathematical Olympiad | BMO 2016 Short List Final | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscel... | null | proof only | null | |
00xp | Problem:
A square is divided into 16 equal squares, obtaining the set of 25 different vertices. What is the least number of vertices one must remove from this set, so that no 4 points of the remaining set are the vertices of any square with sides parallel to the sides of the initial square? | [
"Solution:\n\nThe example in Figure 3a demonstrates that it suffices to remove 8 vertices to \"destroy\" all squares. Assume now that we have managed to do that by removing only 6 vertices. Denote the horizontal and vertical lines by $A, B, \\ldots, E$ and $1,2, \\ldots, 5$ respectively. Obviously, one of the remov... | Baltic Way | Baltic Way 1993 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 8 | |
01l1 | Exactly one integer number is written in each cell of an $8 \times 8$ square table. Per move it is allowed to choose any $n \times n$ square, $1 < n < 8$, and either to increase by 1 or to decrease by 1 all numbers in the cells of the chosen square.
Is it possible to get the table with zeros in all its cells from the a... | [
"We separate the table into five parts: the central $4 \\times 4$ square, two $2 \\times 8$ rectangles, and two $4 \\times 2$ rectangles (see Fig. 1).\nWe can obtain the 0's in all cells of the $2 \\times 8$ rectangles. It suffices to show how we can change (increase by 1 or decrease by 1) the value in any cell of ... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Yes | |
0217 | Problem:
Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$. Let $N$ denote the second point of intersection of line $AI$ and $\omega$. The line through $I$ perpendicular to $AI$ intersects line $BC$, segment $[AB]$, and segment $[AC]$ at the points $D, E$, and $F$, respectively. The circumcircle of tr... | [
"Solution:\nBy construction, $APEF$ and $APBC$ are cyclic, and so\n$$\n\\begin{aligned}\n\\angle BDE &= \\angle CDF = \\angle AFD - \\angle FCD = \\angle AFE - \\angle ACB = (180^\\circ - \\angle EPA) - (180^\\circ - \\angle BPA) \\\\\n&= \\angle BPA - \\angle EPA = \\angle BPE.\n\\end{aligned}\n$$\nHence $DBEP$ is... | Benelux Mathematical Olympiad | 15th Benelux Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis ... | null | proof only | null | |
021o | Problem:
Amanda desenhou a seguinte figura:
Observe que a soma ao longo de qualquer lado do triângulo acima é sempre a mesma, pois, como podemos verificar,
$$
1+3+6=6+2+2=1+7+2
$$
a) Complete os números que faltam nos círculos da figura abaixo de modo que as somas ao longo de qualquer lado do... | [
"Solution:\n\na) Na linha inferior a soma é $2+3+5=10$. Como as somas ao longo de qualquer lado são iguais, o número que falta no canto superior à direita do quadrado deve ser igual a 2, como na figura a seguir:\n\nFaltam mais dois números a serem preenchidos. Novamente, como a soma deve se... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a) The top-right corner is 2; the remaining missing numbers are 1 and 3, yielding equal sums of 10 along each side. b) Let the two top corners be x and y; all sums equal x plus nine plus y, and consistency forces y = x plus eleven, with bottom corners y minus six and x plus three, giving infinitely many solutions param... | |
03e9 | A set of points in the plane is called good if the distance between any two points in it is at most $1$. Let $f(n, d)$ be the largest positive integer such that in any good set of $3n$ points, there is a circle of diameter $d$, which contains at least $f(n, d)$ points. Prove that there exists a positive real $\epsilon$... | [
"Since $m(n, d)$ is an increasing function of $d$ and it cannot be larger than $3n$, clearly it hits a constant when $d$ approaches $1$. Fix some $d$, $d < 1$, it may be very close to $1$. Take an equilateral triangle with side length $1$ and put inside it $3n$ points — $n$ points near each of its vertices so that ... | Bulgaria | Autumn tournament | [
"Geometry > Plane Geometry > Combinatorial Geometry > Helly's theorem",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatoric... | English | proof and answer | n | |
03nj | Problem:
A circle is inscribed in a rhombus $A B C D$. Points $P$ and $Q$ vary on line segments $\overline{A B}$ and $\overline{A D}$, respectively, so that $\overline{P Q}$ is tangent to the circle. Show that for all such line segments $\overline{P Q}$, the area of triangle $C P Q$ is constant.
![](attached_image_1.p... | [
"Solution:\nLet the circle be tangent to $\\overline{P Q}$, $\\overline{A B}$, $\\overline{A D}$ at $T$, $U$, and $V$, respectively. Let $p = P T = P U$ and $q = Q T = Q V$. Let $a = A U = A V$ and $b = B U = D V$. Then the side length of the rhombus is $a + b$.\n\n\n\nLet $\\theta = \\angl... | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > M... | null | proof only | null | |
03qq | Define a hook to be a figure made up of six unit squares as shown in the diagram or any of the figures obtained by applying rotations and reflections to this figure.
Determine all $m \times n$ rectangles that can be covered with hooks so that
* the rectangle is covered without gaps and without... | [
"$m$ and $n$ should be the positive integers and should satisfy one of the following conditions:\n\n(1) $3 \\mid m$ and $4 \\mid n$ (or vice versa);\n(2) one of $m$ and $n$ is divisible by $12$ and one is not less than $7$.\n\nA figure is obtained by applying rotations and reflections to another figure. We regard t... | China | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | All rectangles with positive integer side lengths such that either (i) one side is divisible by 3 and the other by 4, or (ii) one side is divisible by 12 and the other is at least 7 (the roles of the sides may be interchanged). | |
04aq | Let $M$ and $N$ be positive integers. Consider an $N \times N$ square array consisting of $N^2$ lamps that can be in two states - on or off. At the beginning all lamps are turned off.
A *move* consists of choosing a row or a column of the array and changing the state of $M$ consecutive lamps in the chosen row or column... | [
"The sought condition is that $M$ divides $N$.\n\nIt is easy to see that if $M$ divides $N$ we can choose a sequence of moves after which all lamps will be turned on. Really, if $N = Mk$ for some positive integer $k$ then we choose each row $k$ times. In the $(ik+j)$-th move, for $i = 0, 1, \\dots, M-1$, $j = 1, \\... | Croatia | CroatianCompetitions2011 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | M divides N | |
04u6 | Let $ABC$ be an acute scalene triangle. Let $D$ and $E$ be points on the sides $AB$ and $AC$, respectively, such that $BD = CE$. Denote by $O_1$ and $O_2$ the circumcentres of the triangles $ABE$ and $ACD$, respectively. Prove that the circumcircles of the triangles $ABC$, $ADE$ and $AO_1O_2$ have a common point differ... | [
"Let $Z$ be the midpoint of the longer arc $BC$ of the circumcircle $\\omega$ of the triangle $ABC$. The triangles $ZDB$ and $ZEC$ are congruent, because they agree in the sides $BD = CE$ and $ZB = ZC$, as well as in the corresponding angles between them, for both lie over the chord $AZ$ of $\\omega$. It follows th... | Czech Republic | Czech-Polish-Slovak Match | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Isog... | null | proof only | null | |
04xa | Given a circle $k$ and its chord $AB$ which is not the diameter. Let $C$ be any point inside the longer arc of $k$. We denote by $K$ and $L$ the reflections of $A$ and $B$ with respect to the axes $BC$ and $AC$. Prove that the distance of the midpoints of the line segments $KL$ and $AB$ is independent of the location o... | [
"Denote by $S$ the midpoint of $AB$, by $M$ the midpoint of $KL$, and by $P$ and $Q$ the feet of the altitudes from $A$ and $B$ in the triangle $ABC$. Obviously, $P$ and $Q$ are the midpoints of $AK$ and $BL$ respectively (fig. 3). Therefore $QS$ is the mid-segment in the triangle $LAB$ and $MP$ is the mid-segment ... | Czech-Polish-Slovak Mathematical Match | Cesko-Slovacko-Poljsko | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
051m | Consider hexagons whose internal angles are all equal.
(i) Prove that for any such hexagon the sum of the lengths of any two neighbouring sides is equal to the sum of the lengths of their opposite sides.
(ii) Does there exist such a hexagon with side lengths 1, 2, 3, 4, 5 and 6 in some order? | [
"(i) Let the hexagon be $ABCDEF$. It suffices to show that $|AB| + |BC| = |DE| + |EF|$.\nLet $K$ be the intersection point of rays $FA$ and $CB$ and let $L$ be the intersection point of rays $FE$ and $CD$ (Fig. 8). The size of every internal angle of the hexagon is $120^\\circ$, whence triangles $KAB$ and $LDE$ are... | Estonia | Final Round of National Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | Yes; for example a hexagon with all interior angles equal and side lengths in order 1, 4, 5, 2, 3, 6 exists. | |
05dg | Problem:
The side $BC$ of the triangle $ABC$ is extended beyond $C$ to $D$ so that $CD = BC$. The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$.
Prove that if $AD = BE$, then the triangle $ABC$ is right-angled. | [
"Solution:\n\nDefine $F$ so that $ABFD$ is a parallelogram. Then $E, A, C, F$ are collinear (as diagonals of a parallelogram bisect each other) and $BF = AD = BE$. Further, $A$ is the midpoint of $EF$, since $AF = 2AC$, and thus $AB$ is an altitude of the isosceles triangle $EBF$ with apex $B$. Therefore $AB \\perp... | European Girls' Mathematical Olympiad (EGMO) | EGMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analyti... | null | proof only | null | |
05rp | Problem:
Chaque case d'un tableau de taille $8 \times 8$ peut être coloriée en blanc ou en noir, de telle sorte que chaque rectangle de taille $2 \times 3$ ou $3 \times 2$ contient deux cases noires ayant un côté en commun. Déterminer le nombre minimal de cases noires qu'il peut y avoir sur le tableau. | [
"Solution:\n\nComme d'habitude dans ce genre de problèmes, on commence par rechercher des bornes supérieures (en exhibant une configuration) et inférieures (par des arguments de double comptage et/ou de pavage) sur le nombre minimal de cases noires que doit contenir le tableau.\n\nPar exemple, la configuration ci-d... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 24 | |
064l | Problem:
Es sei $ABCD$ ein Parallelogramm mit $|AC| = |BC|$. Ein Punkt $P$ sei auf dem Strahl $AB$ gewählt, sodass $B$ zwischen $A$ und $P$ liegt. Der Umkreis des Dreiecks $ACD$ und die Strecke $PD$ haben außer dem Punkt $D$ noch den Punkt $Q$ gemeinsam. Der Umkreis des Dreiecks $APQ$ und die Strecke $PC$ haben außer ... | [
"Solution:\n\nWir arbeiteten mit gerichteten Winkeln modulo $180$ Grad. Zunächst erkennen wir, dass aus der Voraussetzung $|AC| = |BC|$ unmittelbar folgt, dass die Winkel $\\angle BAC$, $\\angle CBA$, $\\angle DCA$ und $\\angle ADC$ allesamt gleich groß sind.\n\n\n\nEs genügt zu zeigen, das... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade 2022 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Pappus theorem",
"Geometry > Plane Geometry > Miscellaneous > Angl... | null | proof only | null | |
06au | A class consists of 26 students, sitting in pairs. At the second term they decide to change positions in order every two students that sat together on the first term, they are not sitting together in the second. Find the largest possible value of $N$, such that no matter how the students sat in the two terms, there is ... | [
"We will prove that the maximum value is 13. Indeed, if the set contains 14 students, because they sit in 13 desks, from the Pigeonhole Principle, there will be two students who sat in the same desk. Therefore the maximum value is at most 13.\n\n\nFigure 4\n\nFor the other direction, we wil... | Greece | 40th Hellenic Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory"
] | English | proof and answer | 13 | |
06j5 | A $11 \times 11$ grid is to be covered completely without overlapping by some $2 \times 2$ squares and $L$-shapes each composed of three unit cells. Determine the smallest number of $L$-shapes used. (Each shape must cover some grids entirely and cannot be placed outside the $11 \times 11$ grid. The $L$-shapes may be re... | [
"At least 23 $L$-shapes are needed.\nSuppose $x$ squares and $y$ $L$-shapes are used. Then the total number of cells is $4x + 3y$, which should be equal to $11^2 = 121$.\n\n\n\n\n\nWe colour the cells as shown. Then every square and every $L$-shape covers at most on... | Hong Kong | 1997-2023 IMO HK TST | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 23 | |
06op | Consider a convex polyhedron without parallel edges and without an edge parallel to any face other than the two faces adjacent to it.
Call a pair of points of the polyhedron antipodal if there exist two parallel planes passing through these points and such that the polyhedron is contained between these planes.
Let $A$ ... | [
"Denote the polyhedron by $\\Gamma$; let its vertices, edges and faces be $V_{1}, V_{2}, \\ldots, V_{n}$, $E_{1}, E_{2}, \\ldots, E_{m}$ and $F_{1}, F_{2}, \\ldots, F_{\\ell}$, respectively. Denote by $Q_{i}$ the midpoint of edge $E_{i}$.\nLet $S$ be the unit sphere, the set of all unit vectors in three-dimensional... | IMO | IMO 2006 Shortlisted Problems | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Algebra > Linear Algebra > Vectors"
] | English | proof and answer | A - B = m - ℓ + 1 = n - 1 | |
0710 | Problem:
A convex hexagon is called a unit if it has four diagonals of length $1$, whose endpoints include all the vertices of the hexagon. Show that there is a unit of area $k$ for any $0 < k \leq 1$. What is the largest possible area for a unit? | [
"Solution:\nAnswer: We can get arbitrarily close to (but not achieve) $\\dfrac{3 \\sqrt{3}}{4}$ (approx $1.3$) by:\n\n\n\nTo prove the first part, consider the diagram below. Take $AB = AC = 1$ and $\\angle BAC = 2\\theta$. Take $DE = DF = 1$ and take the points of intersection $X$ and $Y$ ... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 3*sqrt(3)/4 | |
077h | Problem:
Betal marks $2021$ points on the plane such that no three are collinear, and draws all possible line segments joining these. He then chooses any $1011$ of these line segments, and marks their midpoints. Finally, he chooses a line segment whose midpoint is not marked yet, and challenges Vikram to construct its... | [
"Solution:\n\nThe answer is 'yes'. To prove this, we will first prove two lemmas:\n\nLemma 1 Given any two points $A, B$, their midpoint $M$, and any point $C$, Vikram can draw a line parallel to $A B$ through $C$.\n\nProof. If $C$ is on line $A B$ we are already done. If not, extend $B C$ to $X$ as shown, draw $P ... | India | INMO | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | Yes | |
07d3 | There are 27 cards, each has some amount of (1 or 2 or 3) shapes (a circle, a square or a triangle) with some color (white, grey or black) on them. We call a triple of cards a **match** such that all of them have the same amount of shapes or mutually distinct amount of shapes, have the same shape or mutually distinct s... | [
"We will prove that the answer of the problem is $9$.\n\nEach card can be corresponded with a point in $\\mathbb{Z}_3^3$. A line in $\\mathbb{Z}_3^3$ is defined to be a subset of the form $\\{P, P+V, P+2V\\}$ where $P$ and $V$ are two elements in $\\mathbb{Z}_3^3$, such that $V \\neq 0$ (in $\\mathbb{Z}_3^3$). It i... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | 9 | |
07nk | A *Colombian Square* is a $6 \times 6$ square which is subdivided into $36$ unit squares, each of which is coloured either Yellow, Blue or Red according to the following rules:
a. No row or column may contain more than two unit squares of the same colour.
b. In any set of four unit squares obtained by intersecting tw... | [
"Condition (a) requires that each of $R$, $Y$, $B$ is to occur exactly twice in each row and column.\n\nWe apply (b) as follows: Take any unit square of a given colour, say $R$. As explained above, its row and column must each contain another $R$. The three $R$'s are the vertices of a rectangle and the fourth verte... | Ireland | Ireland | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 24300 | |
0855 | Problem:
Sia $\Gamma$ una circonferenza e siano $A$ e $B$ due punti distinti di $\Gamma$ non diametralmente opposti. Sia $P$ un punto variabile in $\Gamma$ diverso da $A$ e da $B$ e sia $H$ l'ortocentro del triangolo $A B P$. Determinare il luogo descritto da $H$ al variare di $P$. | [
"Solution:\n\nSia $\\Gamma'$ la circonferenza simmetrica di $\\Gamma$ rispetto ad $A B$, e siano $A'$ e $B'$ i 2 punti di $\\Gamma'$ per cui $A A'$ e $B B'$ sono perpendicolari ad $A B$. Dimostreremo che il luogo richiesto è costituito dalla circonferenza $\\Gamma'$ meno i punti $A'$ e $B'$.\nConsideriamo il più lu... | Italy | XXII OLIMPIADE ITALIANA DI MATEMATICA | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constru... | null | proof and answer | The locus is the circle Γ′ obtained by reflecting Γ across the line AB (equivalently, translating Γ by the vector CA, where C is diametrically opposite to B), with the two points A′ and B′ (the images of A and B) removed. | |
08m2 | Problem:
Five players $A$, $B$, $C$, $D$, $E$ take part in a bridge tournament. Every two players must play (as partners) against every other two players. Any two given players can be partners not more than once per day. What is the least number of days needed for this tournament? | [
"Solution:\n\nA given pair must play with three other pairs and these plays must be in different days, so at least three days are needed. Suppose that three days suffice. Let the pair $AB$ play against $CD$ on day $x$. Then $AB-DE$ and $CD-BE$ cannot play on day $x$. Then one of the other two plays of $DE$ (with $A... | JBMO | 2009 Shortlist JBMO | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 4 | |
08yo | In a $7 \times 7$ chessboard, a coin is placed in the square in the first row from the top, the fourth column from the left. We call square $Y$ a lower left square of square $X$ if $Y$ is $k$ squares to the left and $k$ squares below $X$ for some positive integer $k$. Similarly, we call square $Y$ a lower right square ... | [
"$19$\n\nIf one writes the number in each square as shown in the figure below, the sum of the numbers written in the squares where the coins are placed cannot be increased by the operations.\n\nThe number written in the square where the coin was initially placed is $64$ and there are three ... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | 19 | |
090w | En un triángulo acutángulo $ABC$, sea $M$ el punto medio del lado $AB$ y $P$ el pie de la altura sobre el lado $BC$. Prueba que si $AC + BC = \sqrt{2}AB$, entonces la circunferencia circunscrita del triángulo $BMP$ es tangente al lado $AC$. | [
"Sea $S$ el punto de $AC$, al mismo lado de $A$ que $C$, tal que $AS = \\sqrt{2}AB/2$. Este punto cumple que $AS^2 = \\frac{AB^2}{2} = AB \\cdot AM$, que es la potencia de $A$ con respecto a la circunferencia circunscrita de $BMP$; luego si esta circunferencia circunscrita pasa por $S$ entonces es tangente a $AB$. ... | Mexico | LVI Olimpiada Matemática Española (Concurso Final) | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Spanish | proof only | null | |
090z | Problem:
Let $n \geq 3$ be an integer. We say that a vertex $A_{i}$ ($1 \leq i \leq n$) of a convex polygon $A_{1} A_{2} \ldots A_{n}$ is Bohemian if its reflection with respect to the midpoint of the segment $A_{i-1} A_{i+1}$ (with $A_{0}=A_{n}$ and $A_{n+1}=A_{1}$) lies inside or on the boundary of the polygon $A_{1... | [
"Solution:\n\nThe answer is $n-3$.\n\nLemma. If $ABCD$ is a convex quadrilateral with $\\angle BAD + \\angle CBA \\geq \\pi$ and $\\angle BAD + \\angle ADC \\geq \\pi$ then $A$ is a Bohemian vertex of $ABCD$.\n\nProof. Let $E$ be the reflection of $A$ in $ABCD$. It is clearly seen that $E$ belongs to the halfplanes... | Middle European Mathematical Olympiad (MEMO) | MEMO Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof and answer | n−3 | |
095q | Problem:
Fie cubul $ABCD A_{1} B_{1} C_{1} D_{1}$. Punctul $K$ este un punct interior al muchiei $B B_{1}$ astfel încât $BK / K B_{1} = m$. Prin punctele $K$ și $C_{1}$ este trasat un plan $\alpha$, paralel dreptei $B D_{1}$.
a. Fie $P$ punctul de intersecție al planului $\alpha$ cu dreapta $A_{1} B_{1}$. Găsiți valo... | [
"Solution:\n\nTrasăm $KF \\parallel B D_{1}$. Punctele $K, B, D_{1}, B_{1}$ aparțin planului $(B B_{1} D_{1} D)$ (vezi Figura 11.6.1).\n\n\n\nAvem $\\triangle K B_{1} F \\cong \\triangle B B_{1} D_{1}$ (deoarece $KF \\parallel B D_{1}$). Conform teoremei lui Thales $\\frac{B_{1} F}{F D_{1}}... | Moldova | A 62 - A OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Volume"
] | null | proof and answer | a) A1P/PB1 = |m − 1|, i.e. m > 1: m − 1; m = 1: 0; 0 < m < 1: 1 − m.
b) The volume ratio of the two parts (V_part1/V_part2) is:
- for m ≥ 1: 1 / (6 m^2 + 6 m − 1);
- for 0 < m ≤ 1: (3 − 3 m + m^2) / (3 + 9 m − m^2). | |
09et | There were two parallel railways, on the first of which are $n$ stations and on the other are $m$ stations. A company constructed a new station between the railways and connected by railways the new station with the old ones. From an economical point of view, the company decided to close some railways between stations ... | [
"It is equivalent to find number of spanning trees in the figure below.\n\n\n\nThus it is sufficient to find number of spanning trees of $G$. Denote $a_n$ number of spanning trees of $G$.\n\n\n\nLet $e$ be edge $A_nA_{n-1}$ of the graph $G$. Now if we remove the edg... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | \frac{1}{5}\left(\left(\frac{3+\sqrt{5}}{2}\right)^n - \left(\frac{3-\sqrt{5}}{2}\right)^n\right)\left(\left(\frac{3+\sqrt{5}}{2}\right)^m - \left(\frac{3-\sqrt{5}}{2}\right)^m\right) | |
09tk | Problem:
Gegeven is een driehoek $\triangle A B C$ met $\angle C=90^{\circ}$. Het midden van $A C$ noemen we $D$ en de loodrechte projectie van $C$ op $B D$ noemen we $E$. Bewijs dat de raaklijn in $C$ aan de omgeschreven cirkel van $\triangle A E C$ loodrecht op $A B$ staat.
 | [
"Solution:\n\nNoem $S$ het snijpunt van de raaklijn en $A B$. We moeten dus bewijzen dat $\\angle B S C=90^{\\circ}$.\n\nVanwege $\\angle B C D=90^{\\circ}=\\angle C E D$ geldt $\\triangle B C D \\sim \\triangle C E D$, dus $\\frac{|D C|}{|D B|}=\\frac{|D E|}{|D C|}$. Omdat $|D A|=|D C|$ volgt hieruit $\\frac{|D A|... | Netherlands | IMO-selectietoets | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a6c | Problem:
A dot-trapezium consists of several rows of dots such that each row contains one more dot than the row immediately above (apart from the top row). For example here is a dot-trapezium consisting of 15 dots, having 3 rows and 4 dots in the top row.
A positive integer $n$ is called a t... | [
"Solution:\nLet $n$ be a trapezium number and suppose there are $a$ dots in the first row and $b$ dots in the last row. So the required conditions are $a \\geq 2$ and $b \\geq a + 1$. Then the equation becomes:\n\n$$2n = b(b + 1) - a(a - 1) = b^{2} - a^{2} + b + a = (a + b)(b - a + 1)$$\n\nbecause it is the differe... | New Zealand | NZMO Round One | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 88 |
End of preview.
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