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ShadenA
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MathNet

id string	problem_markdown string	solutions_markdown 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 $Y$. $$ \text{Show that } \langle EMZ \rangle = \lang...	[ "![](attached_image_1.png)\nWe intersect the entire figure with the plane through the points $X$, $E$ and $F$. Since $M$ is on line $EF$, it is part of that plane. Also points $Y$ and $Z$ are part of that plane because they are on line $EX$. The intersection of the circumsphere $k$ and the plane results in a circle...	Austria	AustriaMO2013	[ "Geometry > Solid Geometry > 3D Shapes", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]	English	proof only	null
00rl	The point $M$ lies on the side $AB$ of the circumscribed quadrilateral $ABCD$. The points $I_1, I_2$, and $I_3$ are the incenters of $\triangle MBC$, $\triangle MCD$, and $\triangle MDA$. Show that the points $M, I_1, I_2$, and $I_3$ lie on a circle. ![](attached_image_1.png)	[ "Lemma. Let $I$ be the incenter of $\\triangle ABC$ and let the points $P$ and $Q$ lie on the lines $AB$ and $AC$. Then the points $A, I, P$, and $Q$ lie on a circle if and only if\n$$\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$$\nHence $DBEP$ is...	Benelux Mathematical Olympiad	15th Benelux Mathematical Olympiad	[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Circles > Radical axis ...	null	proof only	null
021o	Problem: Amanda desenhou a seguinte figura: ![](attached_image_1.png) Observe que a soma ao longo de qualquer lado do triângulo acima é sempre a mesma, pois, como podemos verificar, $$ 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![](attached_image_5.png)\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![](attached_image_2.png)\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. ![](attached_image_1.png) 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![](attached_image_1.png)\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![](attached_image_1.png)\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![](attached_image_2.png)\n\n![](attached_image_3.png)\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![](attached_image_1.png)\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![](attached_image_5.png)\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![](attached_image_1.png)\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![](attached_image_1.png)\n\nThus it is sufficient to find number of spanning trees of $G$. Denote $a_n$ number of spanning trees of $G$.\n\n![](attached_image_2.png)\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. ![](attached_image_1.png)	[ "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. ![](attached_image_1.png) 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
0a7r	Problem: Let $AB$ be a diameter of a circle with centre $O$. We choose a point $C$ on the circumference of the circle such that $OC$ and $AB$ are perpendicular to each other. Let $P$ be an arbitrary point on the (smaller) arc $BC$ and let the lines $CP$ and $AB$ meet at $Q$. We choose $R$ on $AP$ so that $RQ$ and $AB$...	[ "Solution:\n\n(See Figure 7.) Draw $PB$. By the Theorem of Thales, $\\angle RPB = \\angle APB = 90^{\\circ}$. So $P$ and $Q$ both lie on the circle with diameter $RB$. Because $\\angle AOC = 90^{\\circ}$, $\\angle RPQ = \\angle CPA = 45^{\\circ}$. Then $\\angle RBQ = 45^{\\circ}$, too, and $RBQ$ is an isosceles rig...	Nordic Mathematical Olympiad	Nordic Mathematical Contest, NMC 9	[ "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates", "Geometry > Plane Geometry > Miscellaneous > Constructions and loci" ]	null	proof only	null

End of preview. Expand in Data Studio

MathNet overview: large-scale multilingual data, high-quality solutions, diverse topics, and three evaluation tasks

Quick Start · Overview · Tasks · Comparison · Dataset Stats · Data Sources · Pipeline · Schema · License · Citation

This is the official MathNet v0. A larger version v1 will be uploaded by Friday, April 24, 2026 (more countires, problems and richer metadata). Schema is stable but field values may be revised in v1.

Quick start

from datasets import load_dataset

# Default: all problems
ds = load_dataset("ShadenA/MathNet", split="train")

# Or a specific country / competition-body config
arg  = load_dataset("ShadenA/MathNet", "Argentina", split="train")
apmo = load_dataset("ShadenA/MathNet", "Asia_Pacific_Mathematics_Olympiad_APMO", split="train")

row = ds[0]
print(row["competition"], row["country"])
print(row["problem_markdown"])
for img in row["images"]:
    img.show()  # PIL image — renders inline in the HF viewer

Overview

Mathematical problem solving remains a challenging test of reasoning for large language and multimodal models, yet existing benchmarks are limited in size, language coverage, and task diversity. We introduce MathNet, a high-quality, large-scale, multimodal, and multilingual dataset of Olympiad-level math problems together with a benchmark for evaluating mathematical reasoning in generative models and mathematical retrieval in embedding-based systems.

MathNet spans 47 countries, 17 languages, and two decades of competitions, comprising 30,676 expert-authored problems with solutions across diverse domains. Alongside the core dataset, we construct a retrieval benchmark of mathematically equivalent and structurally similar problem pairs curated by human experts.

Three benchmark tasks

	Task	What it measures
I	Problem Solving	Generative models on Olympiad problems, graded against expert solutions
II	Math-Aware Retrieval	Embedding models' ability to retrieve mathematically equivalent / structurally similar problems
III	Retrieval-Augmented Problem Solving	How retrieval quality affects reasoning when similar problems are given as context

Even state-of-the-art reasoners remain challenged: 78.4% (Gemini-3.1-Pro) and 69.3% (GPT-5) on MathNet-Solve-Test. Embedding models struggle with equivalence retrieval (Recall@1 under 5% for all tested models), and RAG gains are highly sensitive to retrieval quality — expert retrieval lifts DeepSeek-V3.2-Speciale to 97.3% on MathNet-RAG.

How MathNet compares to existing math benchmarks

Benchmark	Size	Languages	Multimodal	Source	Difficulty
GSM8K	8,500	EN	—	Crowdsourced	Grade school
MATH	12,500	EN	—	Competitions/textbooks	High school
MATH-Vision	3,040	EN	✓	Math competitions	High school
OlympiadBench	6,142	EN, ZH	✓	Official websites	Olympiad
OlympicArena	3,233	EN, ZH	✓	Official websites	Olympiad
Omni-Math	4,428	EN	—	AoPS / contest pages	Olympiad
OlymMATH	200	EN, ZH	—	AoPS / official	Olympiad
MathArena	162	EN	✓	Newly released competitions	Olympiad
IMOBench	460	EN	—	IMO & national archives	Olympiad
MathNet (ours)	30,676	17 (EN, ZH, ES, RU, FR, RO, + 11 more)	✓	Official country booklets / international & national contests	Olympiad

Dataset at a glance

MathNet dataset statistics: contest types, solution length vs. prior benchmarks, problems per year, topic distribution, and language distribution

What the figure shows. (a) A mix of national, regional, TST, and international competitions. (b) MathNet solutions are substantially longer than those in prior math benchmarks — long-form proofs, not one-line answers. (c) Problems per year — the corpus has grown steadily since the early 2000s. (d) Coverage across geometry, algebra, combinatorics, number theory, and their sub-topics. (e) 74% English, 26% non-English across 17 languages; Portuguese, Spanish, French, Italian, Serbian, Slovenian, German, Chinese, Romanian, Korean, Dutch, Russian, Mongolian, Macedonian, Polish, and Hungarian all appear.

Topic taxonomy (excerpt)

MathNet ships with a curated olympiad-style taxonomy. Top-level domains include:

Geometry — plane (triangles, quadrilaterals, circles, concurrency/collinearity, transformations, Miquel/Simson/Brocard, geometric inequalities, combinatorial geometry, analytic methods), solid, differential, non-Euclidean
Algebra — prealgebra, polynomials, inequalities, functional equations, sequences/series, linear algebra, abstract algebra
Number Theory — divisibility, primes, modular arithmetic, Diophantine equations, quadratic residues, (p)-adic methods
Combinatorics — counting, graph theory, extremal / pigeonhole, invariants/monovariants, games, coloring, generating functions
Calculus / Analysis — limits, inequalities, real analysis, combinatorial analysis
Probability & Statistics — discrete and continuous

Every problem carries a hierarchical topic path (e.g. Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals) usable for stratified evaluation or curriculum construction.

Data sources

Each year, participating IMO countries contribute original problems for use in their national contests and team selection examinations. MathNet is built from official problem booklets collected from 47 countries spanning 1985–2025 — 1,595 PDF volumes totalling more than 25,000 pages. Unlike prior math benchmarks that rely on community platforms such as AoPS, every problem and solution in MathNet is authored and disseminated by national teams themselves, ensuring expert-level quality, stylistic consistency, and immunity from the noisy or informal annotations that plague crowd-sourced collections.

A meaningful portion of the collection — particularly older national booklets — was physically obtained and scanned by hand by our IMO expert co-authors, who have attended the International Mathematical Olympiad since 2006 and accumulated a personal archive of official competition materials over nearly two decades.

Data pipeline

MathNet data extraction and curation pipeline

Extracting aligned problem–solution pairs from a heterogeneous corpus of mathematical documents is non-trivial: some booklets separate problems and solutions into different sections, others interleave them; numbering schemes and naming conventions vary across countries and even within a single document. Regex-based heuristics break down at this scale, so we designed a multi-stage LLM pipeline.

Stage 1 — Document ingestion & segmentation. All booklets are converted to Markdown via dots-ocr, a multilingual document parsing framework designed for both digital typeset PDFs and scanned copies across many languages. Gemini-2.5-Flash then identifies problem and solution segments by outputting only their line numbers, and records authors, hints, remarks, source file, and page numbers for provenance.

Stage 2 — Problem–solution extraction. Given the line segments from Stage 1, GPT-4.1 extracts the corresponding problem and solution in LaTeX-friendly Markdown, together with a surrounding text buffer to handle cases where content spans across context boundaries.

Stage 3 — Extraction verification. Each extracted pair passes three independent checks before being retained:

Rule-based similarity check — text similarity between the extraction and original OCR output ensures the LLM made only formatting changes and introduced no hallucinated content.
GPT-4.1-as-judge — GPT-4.1 compares page screenshots against the extracted pair to catch OCR errors, incorrect figure associations, and incomplete solutions.
Human expert review — low-confidence cases are manually reviewed by annotators. A pair is retained only if all three mechanisms agree.

Provenance (source booklet, authors where given) is preserved on every problem.

What v0 contains

This is the v0 drop of MathNet — the first complete public release:

27,817 problems across 58 country / regional-body configs
5,148 problems with figures, totalling 7,541 images embedded inline as HF Image() features (they render in the viewer and decode to PIL on load)
All image references in problem and solution markdown are rewritten to the uniform ![](attached_image_N.png) convention and the corresponding bytes ship in the images column in the same order
Default all config opens with a curated head of ~120 country-diverse, figure-rich problems so the dataset viewer preview is visually representative; the remainder of all is shuffled

A more refined v1 — larger, with improved extraction, deduplication, and metadata — will be uploaded by Friday, April 24, 2026.

Schema

Column	Type	Notes
`id`	string	Short 4-char base36 identifier, stable across rebuilds
`country`	string	Country or regional body of origin
`competition`	string	e.g. `IMO 2023`, `Cono Sur Mathematical Olympiad`
`problem_markdown`	string	Problem statement (Markdown + LaTeX) with `![](attached_image_N.png)` refs
`solutions_markdown`	list<string>	Official / provided solutions, one entry per solution
`topics_flat`	list<string>	Hierarchical topic paths joined as `A > B > C`
`language`	string	Source booklet language
`booklet_source`	string	Upstream collection label
`images`	list<Image>	Inlined figure bytes, decoded to PIL; positions align with `attached_image_N.png` refs in the markdown
`problem_type`	string\|null	`proof only`, `answer only`, `proof and answer` — LLM-assisted
`final_answer`	string\|null	LLM-extracted final answer for answer-bearing problems

problem_type and final_answer are LLM-assisted and not fully human-audited in v0. Treat them as convenience annotations, not ground truth.

Configs / splits

One config per country or regional body (58 total) plus a default all config unioning everything. Each config has a single train split — this is the public v0 release, not the train/test partitioning of MathNet-Solve (which is train: 23,776, test: 6,400, test-hard: 500 in the paper release).

Intended uses & limitations

Good for. Olympiad-level reasoning evaluation, multilingual math evaluation, figure-grounded multimodal math, topic-stratified analysis, retrieval benchmarks over mathematical structure, and RL training — the large pool of expert-written solutions provides dense rewards for verifiable-answer problems, while the math-aware similarity pairs open a new axis: rewarding a model for retrieving a structurally equivalent problem is a natural, automatically verifiable signal that does not require a closed-form answer.

Caveats.

Not contamination-clean. Olympiad problems are indexed widely; assume leakage when evaluating pretrained models.
v0 field values may be revised in v1 (improved extraction / dedup / metadata).
LLM-assisted metadata is imperfect.

License

With the kind support of IMO President Gregor Dolinar, we reached out to the leaders of all participating countries and obtained their permission to share this dataset publicly. Where a country or contest organization asserts its own copyright, that copyright is retained and takes precedence — see competition, country, and booklet_source on each row. For all remaining problems where no explicit copyright was asserted, the dataset is released under Creative Commons Attribution 4.0 International (CC BY 4.0).

In short: use freely, cite the paper, and respect any explicit rights claimed by the original national team.

If you are a rightsholder with a concern, please open an issue or email shaden@mit.edu.

Citation

@inproceedings{alshammari2026mathnet,
  title     = {MathNet: A Global Multimodal Benchmark for Mathematical
               Reasoning and Retrieval},
  author    = {Alshammari, Shaden and Wen, Kevin and Zainal, Abrar and
               Hamilton, Mark and Safaei, Navid and Albarakati, Sultan and
               Freeman, William T. and Torralba, Antonio},
  booktitle = {International Conference on Learning Representations},
  year      = {2026},
  url       = {https://mathnet.mit.edu}
}

Collection including ShadenA/MathNet

MathNet

Collection

2 items • Updated about 2 hours ago • 1

Paper for ShadenA/MathNet

MathNet: a Global Multimodal Benchmark for Mathematical Reasoning and Retrieval

Paper • 2604.18584 • Published 4 days ago • 11