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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
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 | |
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 | |
0akr | Let $\triangle ABC$ be an acute triangle with orthocenter $H$. The point $H'$ is symmetric with $H$ with respect to the line $AB$. Let $N$ be the intersection point of $HH'$ and $AB$. The circle passing through the points $A$, $N$ and $H'$ intersects again the line $AC$ at $M$, and the circle passing through the points... | [
"First, we will show the following lemma.\n\n**Lemma.** Let $H$ be an orthocenter in $\\triangle ABC$, and $H'$ be the symmetric point of $H$ with respect to $AB$. Then $H'$ lies on the circle around the triangle $\\triangle ABC$.\n\n**Proof.**\n\n**First proof of the lemma.** Let $N$ be the intersection point of $... | North Macedonia | Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter... | English | proof only | null | |
0b1l | Problem:
A T-tetromino is formed by adjoining three unit squares to form a $1 \times 3$ rectangle, and adjoining on top of the middle square a fourth unit square. Determine the minimum number of unit squares that must be removed from a $202 \times 202$ grid so that it can be tiled with T-tetrominoes. | [
"Solution:\nWe provide the following construction, which shows that the answer is at most four. Clearly, four T-tetrominoes tile a $4 \\times 4$ grid, as follows:\n\n\n\nThese can be used to tile the upper-left $200 \\times 200$ grid. It can be verified that the following pattern fills the ... | Philippines | $22^{ ext {nd }}$ Philippine Mathematical Olympiad National Stage, Written Phase | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 4 | |
0bzc | Let $ABCD$ be any convex quadrangle and let $P, Q, R, S$ be points on the segments $AB, BC, CD$, and $DA$, respectively. It is given that the segments $PR$ and $QS$ dissect $ABCD$ into four quadrangles, each of which has perpendicular diagonals. Show that the points $P, Q, R, S$ are concyclic.
Nikolai Beluhov, Bulgari... | [
"**Lemma 1.** Let $PQRS$ be a convex quadrangle whose diagonals cross at $O$. Let $\\omega_1$ and $\\omega_2$ be the circles on diameters $PQ$ and $RS$, respectively, and let $\\ell$ be their radical axis. Finally, let $A, B$, and $C$ be points outside this quadrangle so that: the point $P$ (respectively, $Q$) lies... | Romania | THE Tenth ROMANIAN MASTER OF MATHEMATICS | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geo... | English | proof only | null | |
0cm5 | Problem:
Let $ABCD$ be an isosceles trapezoid with $AB \parallel CD$. Let $E$ be the midpoint of $AC$. Denote by $\omega$ and $\Omega$ the circumcircles of the triangles $ABE$ and $CDE$, respectively. Let $P$ be the crossing point of the tangent to $\omega$ at $A$ with the tangent to $\Omega$ at $D$. Prove that $PE$ is... | [
"Solution:\nIf $ABCD$ is a rectangle, the statement is trivial due to symmetry. Hence, in what follows we assume $AD \\nVdash BC$.\n\nLet $F$ be the midpoint of $BD$; by symmetry, both $\\omega$ and $\\Omega$ pass through $F$. Let $P'$ be the meeting point of tangents to $\\omega$ at $F$ and to $\\Omega$ at $E$. We... | Romanian Master of Mathematics (RMM) | Romanian Master of Mathematics Competition Day 1 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"G... | null | proof only | null | |
0cnn | Segment $BD$ is an angle bisector in a triangle $ABC$ (point $D$ lies on segment $AC$). Line $BD$ intersects the circumcircle $\Omega$ of triangle $ABC$ at points $B$ and $E$. Circle $\omega$ with diameter $DE$ intersects $\Omega$ at points $E$ and $F$. Prove that the line symmetrical to $BF$ with respect to $BD$ conta... | [
"Let $M$ be the midpoint of side $AC$, the line $BM$ intersects the circle $\\Omega$ again at point $F'$, and the line $FM$ intersects the circle $\\Omega$ again at point $B'$ (see Fig. 20).\n\nSince arcs $AE$ and $CE$ are equal, points $M$ and $E$ lie on the perpendicular bisector $\\ell$ of segment $AC$. Therefor... | Russia | Russian mathematical olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
0d1o | The set $G$ is defined by the points $(x, y)$ with integer coordinates, $1 \leq x \leq 5$ and $1 \leq y \leq 5$. Determine the number of five-point sequences $\left(P_{1}, P_{2}, P_{3}, P_{4}, P_{5}\right)$ such that for $1 \leq i \leq 5, P_{i}=\left(x_{i}, i\right)$ is in $G$ and $\left|x_{1}-x_{2}\right|= \left|x_{2}... | [
"Let us count the number of five-point sequences according to the position of point $P_{3}$.\n\n\n\nBecause the horizontal line of equation $y=3$ is a symmetry axis for this figure, the number of sequences with $x_{3}=i$ is equal to the number of sequences with $x_{3}=6-i$, for $i=1,2$.\n\n... | Saudi Arabia | Selection tests for the Balkan Mathematical Olympiad 2013 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 42 | |
0dnv | Problem:
У врсти се налази $2n-1$ сијалица. У почетку је средња ($n$-та) упаљена, а све остале су угашене. У једном кораку је дозвољено одабрати две несуседне угашене сијалице између којих су све сијалице упаљене, и променити стање тим двема сијалицама, као и свим сијалицама између њих (на пример, од конфигурације $\c... | [
"Solution:\n\nОдговор је $\\left[\\frac{2^{n+1}-5}{3}\\right]$.\n\nПридружимо $i$-тој сијалици број $2^{|i-n|}$ и дефинишимо вредност конфигурације као збир бројева на упаљеним сијалицама. Вредност полазне конфигурације је $1$, а при сваком кораку она се повећава за природан умножак броја $3$. Корак повећава вредно... | Serbia | 11. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | floor((2^{n+1} - 5)/3) | |
0dq2 | Altitudes of an acute scalene triangle $ABC$ meet at point $H$. Points $M$ and $N$ are the midpoints of the segments $AB$ and $CH$, respectively. $R$ is the foot of the perpendicular from $H$ to line $CM$. $T$ is the second intersection of the line $MN$ with the circumcircle of triangle $CNR$. Let $P$ be the circumcent... | [
"We will make several uses of the fact that the lines connecting a vertex of a triangle with the circumcentre and the orthocentre are symmetrical with respect to the angle bisector at this vertex.\nLet $AC > BC$, $AA_1$ and $BB_1$ be the altitudes of the triangle $ABC$, $O$ its circumcentre. The points $A_1$, $B_1$... | Silk Road Mathematics Competition | SILK ROAD MATHEMATICAL COMPETITION | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane G... | English | proof only | null | |
0dri | Is it possible to color each square on a $9 \times 9$ board so that each $2 \times 3$ or $3 \times 2$ block contains exactly $2$ black squares? If so, what is/are the possible total number(s) of black squares? | [
"We assume each nonblack square is white. Repeat the $3 \\times 3$ block with black squares down the main diagonal $9$ times. This is a possible construction with $3 \\times 9 = 27$ black squares. We shall prove that this is the only answer.\n\nSuppose there are $26$ black squares. We name the bottom left square as... | Singapore | Singapur 2015 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Yes; 27 | |
0dxz | Let $A$, $B$, $C$, $D$ and $E$ be consecutive points on a circle with the centre at $O$, such that $|AC| = |BD| = |CE| = |DO|$. Let $H_1$, $H_2$ and $H_3$ be the orthocentres of the triangles $ACD$, $BCD$ and $BCE$. Prove that $H_1H_2H_3$ is a right triangle. | [
"This problem has several interesting solutions. A nicely drawn figure suggests that quadrilaterals $ABH_1H_2$ and $H_2H_3DE$ are parallelograms. We will prove this fact.\n\n\n\nTriangles $ACD$, $BCD$ and $BDE$ are obtuse, so let us first make a short detour examining some properties of obt... | Slovenia | Slovenija 2008 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Mi... | English | proof only | null | |
0et0 | Let $A$, $B$, $C$ be points on a circle whose centre is $O$ and whose radius is $1$, such that $\angle BAC = 45^\circ$. Lines $AC$ and $BO$ (possibly extended) intersect at $D$, and lines $AB$ and $CO$ (possibly extended) intersect at $E$. Prove that $BD \cdot CE = 2$. | [
"\n\nFigure 1\n\nPut $x = OD$ and $y = OE$. Let $BO$ extended meet the circle again in $F$ (so that $DF = 1-x$). Since $\\angle BFC = 45^\\circ$ and $BF$ is a diameter of the circle, we have $\\angle BCF = 90^\\circ$ and $BC = CF = \\sqrt{2}$. Similarly, if $CE$ extended meets the circle ag... | South Africa | The South African Mathematical Olympiad, Third Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals"
] | null | proof only | null | |
0eu5 | Consider the sixteen tiles fixed on a wall as shown below. How many ways are there to write either $0$ or $1$ on each tile so that the product of the two numbers written on every neighboring pair of tiles (sharing a common side) is always $0$?
 | [
"When $0$ or $1$ is written on each tile in such a way that the product of the two numbers written on every neighboring pair of tiles is always $0$, we'll call the status a *z-pattern*. First, we prove a couple of lemmas.\n\n**Lemma 1.** Let $a_n$ be the number of z-patterns for $n$ tiles laid in a row. Then\n$$\na... | South Korea | 20th Korean Mathematical Olympiad Final Round | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 1234 | |
0f3f | Problem:
Some unit squares in an infinite sheet of squared paper are colored red so that every $2 \times 3$ and $3 \times 2$ rectangle contains exactly two red squares. How many red squares are there in a $9 \times 11$ rectangle? | [
"Solution:\n\n\n\nThere cannot be two red squares with a common side. For consider as the diagram shows we can immediately conclude that the squares with a $\\ast$ are not red, but now the bold rectangle has at most 1 red square. Contradiction.\n\n\n\nConsider a red... | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 33 | |
0fg1 | Problem:
Se consideran las semirrectas no alineadas $O x, O y$. Por el punto $A \in O x$ se trazan pares de rectas $r_{1}, r_{2}$, antiparalelas respecto al ángulo $x O y$; sean $M, N$ las intersecciones de $r_{1}$ con $O y$ y de $r_{2}$ con $O x$, respectivamente. Sea $P$ el punto de intersección de las bisectrices d... | [
"Solution:\n\n\n\nPor la construcción de $P$, este punto equidista de $O y$ y de $A N$, además también equidista de $O y$ y de $r_{1}$, luego $P$ es el centro del círculo exinscrito al triángulo $A M N$ correspondiente al lado $A N$.\n\nLlamando $\\alpha$ al ángulo de las semirrectas $O x, ... | Spain | OME 21 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | The locus of P is a half-line in the direction of the internal bisector of ∠xOy (i.e., forming angle α/2 with Ox). For each position of A, P lies on the line AP parallel to the angle bisector; the admissible part is the ray starting at the unique point Q on AP where dist(P, Oy) = PA, equivalently the intersection of AP... | |
0fv5 | Problem:
Die Ebene wird in gleichseitige Dreiecke der Seitenlänge 1 unterteilt. Betrachte ein gleichseitiges Dreieck der Seitenlänge $n$, dessen Seiten auf den Gitterlinien liegen. Auf jedem Gitterpunkt auf dem Rand und im Innern dieses Dreiecks liegt ein Stein. In einem Spielzug wird ein Einheitsdreieck ausgewählt, we... | [
"Solution:\nWir zeigen zuerst, dass $n$ durch 3 teilbar sein muss. Färbe dazu jeden Gitterpunkt rot, blau oder grün, sodass jedes Einheitsdreieck drei verschieden farbige Ecken hat. Seien $R, B$ und $G$ die Anzahl belegter roter, blauer und grüner Gitterpunkte. Wir behaupten, dass die drei Grössen\n$$\nR+B, \\quad ... | Switzerland | SMO Finalrunde | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | n is divisible by 3 | |
0g7r | 「虧格」是由三個邊長為 1 的正方形所組成的圖形：或將此圖形經由旋轉及翻轉所得之任意圖形。
a. 在任何兩個「虧格」彼此不重疊的條件下，用 $3 \times 671$ 個「虧格」來鋪蓋在 $3 \times 2013$ 的大方格圖形，請問可以完全覆蓋滿嗎？
b. 在任何兩個「虧格」彼此不重疊的條件下，用 $5 \times 671$ 個「虧格」來鋪蓋在 $5 \times 2013$ 的大方格圖形，請問可以完全覆蓋滿嗎？
 | [
"a. 不可以。\n\n將 $3 \\times 2013$ 的大方格中，$(1,2k+1)$ 及 $(3,2k+1)$，$k=0,1,2,...,2006$ 的格子中標上“*”，如上圖所示。每個「虧格」最多可以覆蓋一個“*”，所以至少需要 2014 個「虧格」才能完全覆蓋這些“*”。所以 $3 \\times 671(= 2013)$ 個「虧格」無法完全覆蓋 $3 \\times 2013$ 的大方格。\n\nb. 可以。\n\n$2 \\times 6, 3 \\times 6$ 的大方格也可以用「虧格」完全覆蓋。由此知 ... | Taiwan | 二0一三數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a. No. b. Yes. | |
0glc | Let $ABQP$ be a convex quadrilateral. Let the angle bisectors of $\angle PAQ$ and $\angle PBQ$ intersect at $C$. The circumcircle of $APQ$ intersects $AB$, $AC$ at $M$, $N \ne A$, respectively. Lines $PQ$ and $BC$ intersect at $S$. Prove that if $AC$ is perpendicular to $BC$ then $M$, $S$, $N$ are collinear.
*(Chesphon... | [
"We divide our solution into two main steps:\n\n**Step 1:** Let $O$ be the midpoint of $AB$ and $R$ the midpoint of $PQ$, then $C$, $R$, $O$ are collinear.\n\n*Proof.* We will denote the complex number of each point by its lowercase letter.\nLet $AP \\cap BQ = V$, $AQ \\cap BP = U$. Let $\\angle PAC = \\alpha$, $\\... | Thailand | The first T3MO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry... | null | proof only | null | |
0gn8 | An airline company is planning to run two-way flights between some of the six cities $A$, $B$, $C$, $D$, $E$ and $F$. Determine the number of ways these flights can be arranged so that it is possible to travel between any two of these six cities using only the flights of this company. | [
"Using the inclusion-exclusion principle we find that the number of ways the flights can be arranged so that each city is connected to at least one other city by a flight is\n$$\n2^{\\left(\\frac{6}{2}\\right)} - \\binom{6}{1} 2^{\\left(\\frac{5}{2}\\right)} + \\binom{6}{2} 2^{\\left(\\frac{4}{2}\\right)} - \\binom... | Turkey | Team Selection Test for IMO 2007 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | proof and answer | 26704 | |
0h10 | Through a point $F$ outside a circle $k$, a tangent $FA$ and a secant $\ell$ are drawn. $\ell$ intersects $k$ at the points $B$ and $C$ ($C$ is between $F$ and $B$). The tangent to the circle $k$ at the point $C$ intersects the line segment $FA$ at the point $E$. $FX$ is a bisector of the triangle $AFC$. If the points ... | [
"We will prove that $BX$ is a \"simedian\" of $\\triangle ABC$ (a cevian symmetric to the median with respect to the bisector, all drawn from the vertex $B$). By a known property of the \"simedian\" (which we will also prove) $\\frac{CX}{AX} = \\frac{CB^2}{AB^2}$. By the angle bisector theorem for $FX : \\frac{CX}{... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > M... | English | proof only | null | |
0hsp | Problem:
Let $D$ be a dodecahedron which can be inscribed in a sphere with radius $R$. Let $I$ be an icosahedron which can also be inscribed in a sphere of radius $R$. Which has the greater volume, and why?
Note: A regular polyhedron is a geometric solid, all of whose faces are congruent regular polygons, in which th... | [
"Solution:\n\n$D$ has the greater volume.\n\nThe key idea that we will use is a beautiful relationship between $I$ and $D$; they are dual polyhedra. To understand this, note that $D$ has 12 faces, 30 edges, and 20 vertices. (The 12 pentagonal faces produce 60 edges and 60 vertices, but these are, respectively, doub... | United States | null | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | The dodecahedron has the greater volume. | |
0l9s | Let $m$ and $n$ be two integers greater than $3$ and let be given a rectangular $m \times n$ board. At each step, one puts simultaneously $4$ marbles into $4$ cells of the board (each marble into a cell) so that these four cells form one of the following schemata.
![]... | [
"i) After two steps: one can put into each cell of a small board of size $(4 \\times 2)$ a marble. One can partition the given board of size $(2004 \\times 2006)$ into small boards of size $(4 \\times 2)$. Therefore, after some steps, one can put marbles into all cells of the given board so that the number of marbl... | Vietnam | Vijetnam 2006 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | i) Yes. ii) No. | |
0lgk | Problem:
In a convex cyclic hexagon $A B C D E F$, $B C = E F$ and $C D = A F$. Diagonals $A C$ and $B F$ intersect at point $Q$, and diagonals $E C$ and $D F$ intersect at point $P$. Points $R$ and $S$ are marked on the segments $D F$ and $B F$ respectively so that $F R = P D$ and $B Q = F S$. The segments $R Q$ and ... | [
"Solution:\n\nIt follows obviously that $B F \\parallel C E$ and $A C \\parallel D F$. We denote the circumcircles of $\\triangle A B Q$ and $\\triangle D E P$ by $\\omega_{1}$ and $\\omega_{2}$, respectively. Note that the lines $A D$ and $B E$ are internal common tangents to $\\omega_{1}$ and $\\omega_{2}$. Indee... | Zhautykov Olympiad | Zhautykov Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00ac | Felipe tells Roberto that he managed to place 1226 triminos $1\times 3$ on a certain grid square $Q$ so that they do not have common points, not even vertices. Without knowing the dimensions of $Q$, Roberto answers, "If what you say is true then one can place 1250 triminos $1\times 3$ on your square, under the same con... | [
"Yes, Roberto is right. To prove this, we first show that the least $n$ for which 1226 triminos $1\\times 3$ can be placed on an $n \\times n$ square without common points (including vertices) is $n=99$. Next, take a $99\\times 99$ square. In its first row one can place 25 triminos without point in common; they lea... | Argentina | Argentine National Olympiad 2015 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | Yes | |
00f5 | Show that for every integer $n \geq 6$, there exists a convex hexagon which can be dissected into exactly $n$ congruent triangles. | [
"The basic building blocks will be right angled triangles with sides $p, q$ (which are positive integers) adjacent to the right angle.\nIn the first instance, we take $p=q=1$ and construct five basic building blocks: $L_{1}, L_{2}, M, R_{1}$ and $R_{2}$.\n\n\n$L_{1}$\n | APMO | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
00ny | Alice and Bob play a game, in which they take turns drawing segments of length $1$ in the Euclidean plane. Alice begins, drawing the first segment, and from then on, each segment must start at the endpoint of the previous segment. It is not permitted to draw the segment lying over the preceding one. If the new segment ... | [
"a) In the following, let $A_n$ denote the end-point of the segment that Alice drew in her $n$-th turn (assuming the game has not ended by then), and let $B_n$ denote the end-point of Bob's $n$-th segment. Furthermore, let $B_0$ denote the starting point of Bob's first segment.\nIf Alice can force an end to the gam... | Austria | AUT_ABooklet_2023 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | a) Yes, either player can force the game to end. b) No; neither player has a winning strategy, as both can always make a safe move. | |
00s2 | Let $\triangle ABC$ be an acute triangle. Variable points $E$ and $F$ are on sides $AC$ and $AB$ respectively such that $BC^2 = BA \cdot BF + CE \cdot CA$. As $E$ and $F$ vary prove that the circumcircle of $AEF$ passes through a fixed point other than $A$. | [
"Let $H$ be the ortocenter of $ABC$ and $K,L,M$ be the feet of perpendiculars respectively from $A,B,C$ to their opposite sides of $ABC$. Also let $D$ be the intersection point of lines $BE$ and $CF$. From power of point we have\n$$\nBA \\cdot BM = BC \\cdot BK \\qquad (1)\n$$\nand\n$$\nCA \\cdot CL = CB \\cdot CK ... | Balkan Mathematical Olympiad | BMO 2017 | [
"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 > Circles > T... | English | proof only | null | |
0115 | Problem:
A cube with edge length $3$ is divided into $27$ unit cubes. The numbers $1,2, \ldots, 27$ are distributed arbitrarily over the unit cubes, with one number in each cube. We form the $27$ possible row sums (there are nine such sums of three integers for each of the three directions parallel to the edges of the ... | [
"Solution:\nAnswer: $24$.\nSince each unit cube contributes to exactly three of the row sums, then the total of all the $27$ row sums is $3 \\cdot (1+2+\\ldots+27) = 3 \\cdot 14 \\cdot 27$, which is even. Hence there must be an even number of odd row sums.\n\n\n(a)\n![](attached_image_2.png... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 24 | |
01u2 | Given a $2n \times 2n$ table all the cells being white. Per move it is allowed to change the color of any three consecutive cells in any row or in any column (white cell is replaced by black one and vice versa).
Find all possible values of $n \ge 2$ for which one can obtain the chess coloring of the given table.
(E. Ba... | [
"**Answer:** $n \\equiv 0 \\pmod{3}$.\nIt is easy to see that we have $2n^2$ black and $2n^2$ white cells in any chess coloring of the given $2n \\times 2n$ table, in particular, we have even number of white cells and even number of black cells.\nSince each of $2n^2$ black cells must be recolored odd number of time... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n is a multiple of 3 | |
020e | Problem:
Let $ABC$ be an acute triangle with circumcentre $O$. Let $\Gamma_{B}$ be the circle through $A$ and $B$ that is tangent to $AC$, and let $\Gamma_{C}$ be the circle through $A$ and $C$ that is tangent to $AB$. An arbitrary line through $A$ intersects $\Gamma_{B}$ again in $X$ and $\Gamma_{C}$ again in $Y$. Pr... | [
"Solution:\n\nLet $O_{B}$ and $O_{C}$ denote the respective centres of $\\Gamma_{B}$ and $\\Gamma_{C}$. We shall show that $X O_{B} O$ and $O O_{C} Y$ are congruent. Now $A O_{B} \\perp AC$ since $\\Gamma_{B}$ is tangent to $AC$ and $O O_{C} \\perp AC$ since $O O_{C}$ is the perpendicular bisector of $[AC]$. Hence ... | Benelux Mathematical Olympiad | BxMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Inver... | null | proof only | null | |
023i | Problem:
Em uma folha de papel, Emanuelle desenha duas circunferências de raio $1$ que se tangenciam em um ponto. Em seguida, ela desenha uma terceira circunferência de raio $1-\sqrt{2}$ que tangencia as duas anteriores externamente, conforme a figura abaixo.
Emanuelle calcula a área da reg... | [
"Solution:\n\nSejam $A$, $B$ e $C$ os centros das circunferências $C_{1}$, $C_{2}$ e $C_{3}$ respectivamente. Considere o triângulo cujos vértices são dados por esses pontos como na figura abaixo.\n\n\n\nChamemos de $P$ o ponto em que $C_{1}$ tangencia $C_{2}$, de $Q$ o ponto em que $C_{2}$... | Brazil | null | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 1 - pi*(2 - sqrt(2))/2 | |
03bq | Given is a circle $k$ and a point $A$ outside of it. The segment $BC$ is a diameter of $k$. Find the locus of the orthocenter of $\triangle ABC$, when $BC$ is changing.
 | [
"Let $O$ be the center of $k$ and $\\omega$ be the circle with diameter $AO$. Let $AA_1$ and $BB_1$ be the altitudes of $\\triangle ABC$ and $H$ is the orthocenter. Then the power of $H$ with respect to $\\omega$ is equal to $AH \\cdot HA_1$ and the power of $H$ with respect to $k$ is equal to $BH \\cdot HB_1$. Sin... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | The locus is the radical axis of k and the circle with diameter AO, equivalently the polar of A with respect to k. | |
03nf | Problem:
Let $m$ and $n$ be positive integers. A $2m \times 2n$ grid of squares is coloured in the usual chessboard fashion. Find the number of ways of placing $mn$ counters on the white squares, at most one counter per square, so that no two counters are on white squares that are diagonally adjacent. An example of a w... | [
"Solution:\nDivide the chessboard into $mn$ $2 \\times 2$ squares.\n\n\n\nEach $2 \\times 2$ square can contain at most one counter. Since we want to place $mn$ counters, each $2 \\times 2$ square must contain exactly one counter.\nAssume that the lower-right corner of the $2m \\times 2n$ c... | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | (m+n choose m) | |
0455 | As illustrated in Fig. 1.1, $ABCDEF$ is a cyclic hexagon. The extensions of $AB$ and $DC$ meet at $G$; the extensions of $AF$ and $DE$ meet at $H$. Let $M$ and $N$ be the circumcentres of $\triangle BCG$ and $\triangle EFH$, respectively. Prove that the lines $BE$, $CF$, and $MN$ are concurrent.
![](attached_image_1.p... | [
"Let $\\omega$ be the circumcircle of $\\triangle BCG$. Let $E'$ be the other intersection of the line $BE$ and $\\omega$, $F'$ be the other intersection of the line $CF$ and $\\omega$, as shown in Fig. 1.2. Note that $\\angle BE'F' = \\angle BCF' = \\angle BCF = \\angle BEF$, and hence $EF \\parallel E'F'$; simila... | China | China National Team Selection Test | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geom... | null | proof only | null | |
048d | Outside a regular polygon $A_1A_2\dots A_n$ a point $B$ is given so that $A_1A_2B$ is a regular triangle. Determine all $n$ such that points $B$, $A_2$ and $A_3$ are consecutive vertices of some regular polygon. | [
"Let the new polygon have $m$ vertices. Two cases are possible:\n\n\n\n**Case 1.** The new polygon lies outside the given polygon. In other words, $m$-gon and $n$-gon are on opposite sides of the line $A_2A_3$.\nIn this case $\\angle BA_2A_3 + \\angle A_1A_2A_3 + 60... | Croatia | CroatianCompetitions2011 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | n ∈ {3, 4, 5, 7, 8, 9, 10, 12, 15, 18, 24, 42} | |
04v9 | Given an acute-angled triangle $ABC$. The points $B'$ and $C'$ lie on the rays opposite to $CA$ and $BA$, respectively, such that $|B'C| = |AB|$ and $|C'B| = |AC|$. Prove that the circumcenter of $AB'C'$ lies on the circumcircle of $ABC$. (Patrik Bak) | [
"Since the line segments $AB'$, $AC'$ have the same length $|AB'| + |AC'|$, $AB'C'$ is an isosceles triangle with base $B'C'$. It means that the perpendicular bisector of $B'C'$ coincides with the bisector of the angle $BAC$. Let $S \\neq A$ be the intersection of this bisector with the circumcircle of $ABC$. If we... | Czech Republic | 72nd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous >... | English | proof only | null | |
04xs | The city of Mar del Plata has a shape of the square $WSEN$. For a given even integer $n$, it is divided by $2(n + 1)$ streets into $n \times n$ blocks (the streets run also around the square). Each block has a size of 100 m x 100 m. All the streets in Mar del Plata are one-way. Each street has one direction along the e... | [
"First we introduce some notation. Call each 100 m segment of a street an arrow, and each place where two streets meet a crossroad. If an arrow has the same direction as the street $WS$ or $WN$ (that is, it runs from the upper left to the bottom right or from the bottom left to the upper right), it is called a forw... | Czech-Polish-Slovak Mathematical Match | 12th Czech-Polish-Slovak Mathematics Competition | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | (2n^2 - 2n + 4)/10 km | |
056b | Let $A'$ be the result of reflection of vertex $A$ of triangle $ABC$ through line $BC$ and let $B'$ be the result of reflection of vertex $B$ through line $AC$. Given that $\angle BA'C = \angle BB'C$, can the largest angle of triangle $ABC$ be located:
a) At vertex $A$;
b) At vertex $B$;
c) At vertex $C$? | [
"\nFig. 18\n\nFig. 19\n\nFig. 20\n\nSolution:\nLet the foot of the altitude drawn from vertex $B$ of triangle $ABC$ be $K$. By symmetry, $\\angle BA'C = \\angle BAC$ and $\\angle BB'C = \\angle KB'C = \\angle KBC$.\n\na) By assumptions, $\\a... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneou... | English | proof and answer | a) No; b) Yes; c) Yes | |
05dk | Problem:
Let $\Omega$ be the circumcircle of the triangle $A B C$. The circle $\omega$ is tangent to the sides $A C$ and $B C$, and it is internally tangent to $\Omega$ at the point $P$. A line parallel to $A B$ and intersecting the interior of triangle $A B C$ is tangent to $\omega$ at $Q$.
Prove that $\angle A C P=\... | [
"Solution:\n\nAssume that $\\omega$ is tangent to $A C$ and $B C$ at $E$ and $F$, respectively and let $P E, P F, P Q$ meet $\\Omega$ at $K, L, M$, respectively. Let $I$ and $O$ denote the respective centres of $\\omega$ and $\\Omega$, and consider the homothety $\\mathscr{H}$ that maps $\\omega$ onto $\\Omega$. No... | European Girls' Mathematical Olympiad (EGMO) | EGMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geomet... | null | proof only | null | |
05uf | Problem:
Trouver le plus grand entier $n \geqslant 3$ pour lequel il existe un ensemble $\mathcal{S}$ de $n$ points du plan avec la propriété suivante : tout triangle (même plat) dont les sommets appartiennent à $\mathcal{S}$ est isocèle mais pas équilatéral. | [
"Solution:\n\nOn dit qu'un ensemble $\\mathcal{S}$ ayant la propriété requise est mignon. Nous allons démontrer que tout ensemble mignon contient au plus 6 éléments. En outre, on notera $\\omega_{A, B}$ le cercle de centre $A$ passant par $B$, et par $\\ell_{A, B}$ la médiatrice du segment $[A B]$.\n\nTout d'abord,... | France | Préparation Olympique Française de Mathématiques - Test du 14 et du 21 Février 2021 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geom... | null | proof and answer | 6 | |
062m | Problem:
Die 16 Felder eines $4 \times 4$-Schachbretts lassen sich wie folgt in 18 Linien ordnen: die vier Zeilen, die vier Spalten, fünf Diagonalen von Nordwest nach Südost und fünf Diagonalen von Nordost nach Südwest. Dabei bestehen diese Diagonalen aus 2, 3 oder 4 über Eck benachbarten Feldern jeweils gleicher Farb... | [
"Solution:\n\nDie größtmögliche erreichbare Punktzahl ist 17, wie das abgebildete Beispiel zeigt. Es bleibt zu zeigen, dass die Maximalpunktzahl 18 nicht erreicht werden kann. Dazu nehmen wir die Existenz einer Belegung mit 18 Punkten an und betrachten zunächst die Zeilen bzw. Spalten. Weder in einer Zeile noch in ... | Germany | 2. IMO-Auswahlklausur | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 17 | |
065e | From the vertex $A$ of an equilateral triangle $ABC$ we draw the ray $Ax$ which intersects the side $B\Gamma$ at $\Delta$. On $Ax$ we consider a point $E$ such that $BA = BE$. Find the angle $A\hat{E}\Gamma$. | [
"Since $BA = B\\Gamma = BE$, point $B$ is the circumcenter of the triangle $A\\Gamma E$. The angle $A\\hat{E}\\Gamma$ is inscribed to the circle $C(B, BA)$ and so:\n$$\nA\\hat{E}\\Gamma = \\frac{1}{2} A\\hat{B}\\Gamma = 30^{\\circ}.\n$$",
"From $BA = BE$ and $BA = B\\Gamma$, we have $B\\Gamma = BE$, and the trian... | Greece | 26th Hellenic 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 > Miscellaneous > Angle chasing"
] | English | proof and answer | 30° | |
06ke | Suppose $2017$ points in a plane are given such that no three points are collinear. Among the triangles formed by any three of these $2017$ points, those triangles having the largest area are said to be *good*. Prove that there cannot be more than $2017$ good triangles. | [
"We shall prove the assertion for any positive integer $n$ when $2017$ is replaced by $n$.\n\n**Claim 1.** If the points $X, Y, Z$ lie inside or on the boundary of a polygon $\\mathcal{P}$, then $[XYZ]$ does not exceed the area of one of the triangles formed by the vertices of $\\mathcal{P}$. When equality holds, $... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > O... | null | proof only | null | |
06si | Let $n \geqslant 2$ be an integer. Consider all circular arrangements of the numbers $0,1, \ldots, n$; the $n+1$ rotations of an arrangement are considered to be equal. A circular arrangement is called beautiful if, for any four distinct numbers $0 \leqslant a, b, c, d \leqslant n$ with $a+c=b+d$, the chord joining num... | [
"Given a circular arrangement of $[0, n]=\\{0,1, \\ldots, n\\}$, we define a $k$-chord to be a (possibly degenerate) chord whose (possibly equal) endpoints add up to $k$. We say that three chords of a circle are aligned if one of them separates the other two. Say that $m \\geqslant 3$ chords are aligned if any thre... | IMO | International Mathematical Olympiad Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Modular Arithmetic > Inverse... | English | proof and answer | M = N + 1 | |
06ze | Problem:
The circle $C$ has diameter $AB$. The tangent at $B$ is $T$. For each point $M$ (not equal to $A$) on $C$ there is a circle $C'$ which touches $T$ and touches $C$ at $M$. Find the point at which $C'$ touches $T$ and find the locus of the center of $C'$ as $M$ varies. Show that there is a circle orthogonal to ... | [
"Solution:\n\n\nLet $O$ be the center of $C$. Let the line $AM$ meet $T$ at $N$. Let the perpendicular to $T$ at $N$ meet the line $OM$ at $O'$. Then $\\angle O'NM = \\angle MAB$ ($O'N$ parallel to $AB$, because both perpendicular to $T$) $= \\angle OMA$ ($OM = OA$) $= \\angle O'MN$. So $O'... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | The tangency point on T is N = AM ∩ T. The locus of centers O′ is the parabola y^2 = 4ax with vertex at B and axis along AB (with coordinates B = (0,0), A = (2a,0), O = (a,0)). A fixed circle orthogonal to every C′ is the circle centered at A with radius AB. | |
0759 | Problem:
Suppose five of the nine vertices of a regular nine-sided polygon are arbitrarily chosen. Show that one can select four among these five such that they are the vertices of a trapezium. | [
"Solution:\nSuppose four distinct points $P, Q, R, S$ (in that order on the circle) among these five are such that $\\widehat{P Q}=\\widehat{R S}$. Then $P Q R S$ is an isosceles trapezium, with $P S \\parallel Q R$. We use this in our argument.\n\n- If four of the five points chosen are adjacent, then we are throu... | India | INMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
07cz | Quadrilateral $ABCD$ and point $T$ in its interior is given such that $AC$ is the angle bisector of $\widehat{BCD}$ and
$$
\widehat{ABC} - \widehat{ATD} = \widehat{DAC}, \quad \widehat{ADC} - \widehat{ATB} = \widehat{BAC}.
$$
Prove that $\widehat{BAT} = \widehat{DAC}$. | [
"Vertex $A$ lies on the angle bisector of $\\widehat{C}$ and has equal distances to the lines $CB$ and $CD$. So there exists a circle with center $A$ that is tangent to $CB$ and $CD$.\nLet $S$ be the intersection of second tangent lines through points $B$ and $D$ to this circle.\n\n$$\n\\be... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
07uh | Two circles $\Omega_1$ and $\Omega_2$ intersect at $A$ and $B$. From any point $P$ on $\Omega_1$ straight lines are drawn through $A$ and $B$ meeting $\Omega_2$ again at $E$ and $F$, respectively. Prove that the length of the line segment $EF$ is the same for all $P$ on $\Omega_1$. | [
"**Solution 1.** Draw the tangent to $\\Omega_1$ at $A$ and let $H$ be its second point of intersection with $\\Omega_2$. First suppose $P$ is not inside circle $\\Omega_2$.\n\n\n\nBy the Alternate Segment Theorem, $\\angle HAE = \\angle PBA$. From the cyclic quadrilateral $ABFE$ we see tha... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
081p | Problem:
Michela e Nicola stanno correndo lungo il perimetro di un parco $C$ quadrato $A B C D$ di diagonale $6~\mathrm{km}$. Essi corrono in senso orario alla stessa velocità. Il loro cane Pallino corre all'interno del parco in modo da stare sempre a metà strada tra Nicola e Michela. Inizialmente Michela è nel vertic... | [
"Solution:\n\nLa risposta è 6. Si scelga un riferimento cartesiano ponendo il vertice $B$ all'origine e gli assi $x$ e $y$ rispettivamente lungo i segmenti $A B$ e $B C$.\n\nSi consideri il tratto che Nicola percorre da $T$, punto medio di $A D$, al vertice $A$ (vedi figura 1). Poiché Nicola e Michela corrono con l... | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 6 | |
08n2 | Problem:
Let $ABC$ be an isosceles triangle with $AB = AC$. On the extension of the side $[CA]$ we consider the point $D$ such that $AD < AC$. The perpendicular bisector of the segment $[BD]$ meets the internal and the external bisectors of the angle $\widehat{BAC}$ at the points $E$ and $Z$, respectively. Prove that t... | [
"Solution:\n\nIn $\\triangle ABD$ the ray $[AZ$ bisects the angle $\\widehat{DAB}$ and the line $ZE$ is the perpendicular bisector of the side $[BD]$. Hence $Z$ belongs to the circumcircle of $\\triangle ABD$.\nTherefore the points $A, B, D, Z$ are concyclic.\nIn $\\triangle BCD$, $AE$ and ... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
08tg | A circle $O_1$ of radius $2$ and a circle $O_2$ of radius $4$ are externally tangent at the point $P$. Points $A$ and $B$, both distinct from the point $P$, are chosen on the circumference of the circles $O_1$ and $O_2$, respectively, in such a way that the points $A$, $P$ and $B$ are collinear. Determine the length of... | [
"Let $C_1, C_2$ be the centers of the circles $O_1, O_2$, respectively. Since the circles $O_1$ and $O_2$ are externally tangent at the point $P$, the three points $C_1, P$ and $C_2$ are situated on a straight line in this order. Therefore, $\\angle C_1PA = \\angle C_2PB$ holds. As the triangles $C_1PA$ and $C_2PB$... | Japan | Japan Mathematical Olympiad First Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 8/3 | |
0932 | Problem:
There is a lamp on each cell of a $2017 \times 2017$ square board. Each lamp is either on or off. A lamp is called bad if it has an even number of neighbours that are on. What is the smallest possible number of bad lamps on such a board?
(Two lamps are neighbours if their respective cells share a side.) | [
"Solution:\n\nWe divide the square in $1 \\times 1$-squares and color the square in checkerboard fashion such that the corners are black and we call lamps on black and white squares black and white lamps, respectively. We assign the number 1 to a lamp that is on, and the number 0 to a lamp that is off.\n\nIf we ass... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 1 | |
0955 | Problem:
Aflați toate valorile parametrului real $a$, pentru care sistemul
$$
\left\{
\begin{array}{c}
x^{2}+5x+y^{2}-y-|x-5y+5|=51 \\
a x - y - 5a + 2 = 0
\end{array}
\right.
$$
are exact 2 soluții reale distincte. | [
"Solution:\nCercetăm expresia $E(x, y) = x - 5y + 5$ ce se află în modul. Avem două cazuri posibile.\n\nCazul 1. Fie $E(x, y) \\geq 0$, adică $y \\leq \\frac{x}{5} + 1$. Obținem că sistemul inițial este echivalent cu sistemul de ecuații\n$$\n\\left\\{\n\\begin{array}{c}\nx^{2} + 5x + y^{2} - y - x + 5y - 5 = 51 \\\... | Moldova | A 61-a OLIMPIADA DE MATEMATICA A REPUBLICII MOLDOVA | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Circles > Tangents",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a in (-2.4, +infinity) | |
099l | Consider an $17 \times 42$ table of squares. We put some balls into squares. After that, for each empty square, count the number of adjacent squares with balls, and write down that number in the empty square. Find the maximum possible value of the sum of these numbers over all empty squares. | [
"The table cells are represented by the graph's vertices. We connect adjacent cells, corresponding to edges between graph vertices.\n\nFor example, we illustrate the $3 \\times 4$ table in the figure.\n\n\n\nIf a square has a ball, we color it black; if a square is empty, we color it white.... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Other"
] | English | proof and answer | 2009 | |
09wj | A square sheet of paper lying on the table is divided into $8 \times 8 = 64$ equal squares. These squares are numbered from **a1** to **h8** as on a chess board (see fig. 1). We now start folding, in such a way that square **a1** always stays in the same spot on the table. First we fold along the horizontal midline (fi... | [
"43"
] | Netherlands | Second Round | [
"Discrete Mathematics > Algorithms"
] | English | final answer only | 43 | |
0a53 | Problem:
Let $\Gamma_{1}$ and $\Gamma_{2}$ be circles internally tangent at point $A$, with $\Gamma_{1}$ inside $\Gamma_{2}$. Let $BC$ be a chord of $\Gamma_{2}$ which is tangent to $\Gamma_{1}$ at point $D$. Prove that line $AD$ is the angle bisector of $\angle BAC$. | [
"Solution:\nLet $\\lambda$ be the common tangent of $\\Gamma_{1}$ and $\\Gamma_{2}$ at point $A$. Let $P$ be a point on $\\lambda$ such that $P$ and $C$ are on opposite sides of line $AB$. Let $Q$ and $R$ be the points of intersection of $\\Gamma_{1}$ with $AB$ and $AC$ respectively.\n\n\n\... | New Zealand | New Zealand Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a9f | Problem:
Three circles $\Gamma_{A}$, $\Gamma_{B}$ and $\Gamma_{C}$ share a common point of intersection $O$. The other common point of $\Gamma_{A}$ and $\Gamma_{B}$ is $C$, that of $\Gamma_{A}$ and $\Gamma_{C}$ is $B$ and that of $\Gamma_{C}$ and $\Gamma_{B}$ is $A$. The line $A O$ intersects the circle $\Gamma_{C}$ i... | [
"Solution:\n\nLet $\\angle A O Y=\\alpha$, $\\angle A O Z=\\beta$ and $\\angle Z O B=\\gamma$. So $\\alpha+\\beta+\\gamma=180^{\\circ}$. Also $\\angle B O X=\\alpha$ (vertical angles) and $\\angle A C Y=\\alpha=\\angle B C X$ (angles subtending equal arcs); similarly $\\angle C O X=\\beta$, $\\angle A B Z=\\beta=\\... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 24 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0agx | Let $n > 3$ be a positive integer. An equilateral triangle $ABC$ is divided into $n^2$ identical equilateral "small" triangles using lines parallel to its sides. The figure below illustrates the case $n = 4$. Let $m$ be the number of rhombi consisting of 2 "small" triangles. Let $d$ be the number of rhombi consisting o... | [
"Each line segment with ends from the set $S$ of length $\\frac{k}{n}$ (not lying on a side of the triangle) is the diagonal of one and only one rhombus of type $M$. The segments parallel to one side of the triangle are\n$$\n1 + 2 + 3 + \\ldots + (n-1) = \\frac{n(n-1)}{2}.\n$$\nTherefore, the total number of rhombi... | North Macedonia | XV-th Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 3(2n - 3) | |
0avj | Problem:
Let $B H$ be the altitude from the vertex $B$ to the side $A C$ of an acute-angled triangle $A B C$. Let $D$ and $E$ be the midpoints of $A B$ and $A C$, respectively, and $F$ the reflection of $H$ across the line segment $E D$. Prove that the line $B F$ passes through the circumcenter of $\triangle A B C$. | [
"Solution:\n\nLet $O$ be the circumcenter of $\\triangle A B C$. Since $\\angle O B A = 90^{\\circ} - \\angle C$, it suffices to show that $\\angle F B A = 90^{\\circ} - \\angle C$.\n\n\n\nNote that $A D = B D = D H$ and $D H = D F$. Therefore, quadrilateral $A H F B$ is cyclic (with circum... | Philippines | 19th Philippine 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 > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
0bow | Let $ABC$ be a triangle. Let $A'$ be the centre of the circle through the midpoint of the side $BC$ and the orthogonal projections of $B$ and $C$ on the lines of support of the internal bisectrices of the angles $ACB$ and $ABC$, respectively; the points $B'$ and $C'$ are defined similarly. Prove that the nine-point cir... | [
"All the angles in the solution are directed modulo $\\pi$. The following notation is used throughout the proof:\n\n$2\\alpha$, $2\\beta$, $2\\gamma$ measures of the angles $BAC$, $CBA$, $ACB$, respectively;\n\n$2a$, $2b$, $2c$ lengths of the sides $BC$, $CA$, $AB$, respectively;\n\n$I$ incentre of the triangle $AB... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellane... | null | proof only | null | |
0clz | Problem:
Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$. The lines $AB$ and $CD$ meet at $P$, the lines $AD$ and $BC$ meet at $Q$, and the diagonals $AC$ and $BD$ meet at $R$. Let $M$ be the midpoint of the segment $PQ$, and let $K$ be the common point of the segment $MR$ and the circle $\omega$. Prove t... | [
"Solution:\n\nLet $O$ be the centre of $\\omega$. Notice that the points $P$, $Q$, and $R$ are the poles (with respect to $\\omega$) of the lines $QR$, $RP$, and $PQ$, respectively. Hence we have $OP \\perp QR$, $OQ \\perp RP$, and $OR \\perp PQ$, thus $R$ is the orthocentre of the triangle $OPQ$.\n\nNow, if $MR \\... | Romanian Master of Mathematics (RMM) | The 6th Romanian Master of Mathematics Competition | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, ort... | null | proof only | null | |
0cr9 | Let $M$ be the midpoint of the side $AC$ of a triangle $ABC$. The points $P$ and $Q$ are chosen on the segments $AM$ and $MC$, respectively, so that $PQ = AC/2$. The circumcircle of the triangle $ABQ$ meets the side $BC$ at $X \neq B$, while the circumcircle of the triangle $BCP$ meets the side $AB$ at $Y \neq B$. Prov... | [
"Поскольку четырёхугольники $BCPY$ и $BAQX$ вписаны, получаем, что $\\angle APY = \\angle ABC = \\angle CQX$. Пусть прямая, проходящая через $M$ параллельно $QX$, пересекает прямую $BC$ в точке $K$, а прямая, проходящая через $M$ параллельно $PY$, пересекает прямую $AB$ в точке $L$ (см. рис. 13).\n\n![](attached_im... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0d4q | Majid wants to color the cells of an $n \times n$ chessboard into white and black so that each $2 \times 2$ subsquare contains two white cells and two black cells. In how many ways can Majid color this $n \times n$ chessboard? | [
"We have three possible cases.\n\nThe first case is when there are no two adjacent cells with the same color. In this case there are exactly two possible ways to color this $n \\times n$ chessboard depending on the color of the first cell on the top left of the chessboard:\n\n\n![](attached... | Saudi Arabia | SAMC | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English, Arabic | proof and answer | 2^{n+1} - 2 | |
0dmc | Problem:
Дат је неједнакокраки троугао $ABC$. Нека су $AD$, $BE$, $CF$ симетрале углова овог троугла ($D \in BC$, $E \in AC$, $F \in AB$). Нека су $K_{a}$, $K_{b}$, $K_{c}$ тачке на уписаном кругу троугла $ABC$ такве да су $DK_{a}$, $EK_{b}$, $FK_{c}$ тангенте на уписани круг и да $K_{a} \notin BC$, $K_{b} \notin AC$,... | [
"Solution:\n\nДокажимо да су троуглови $K_{a}K_{b}K_{c}$ и $A_{1}B_{1}C_{1}$ хомотетични. Да бисмо то доказали, довољно је да докажемо да је $K_{a}K_{b} \\parallel A_{1}B_{1}$, односно $K_{a}K_{b} \\parallel AB$ (аналогно ће следити и за друге парове страница).\n\nОзначимо $M = K_{a}K_{b} \\cap BC$, са $S$ означимо... | 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 > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Co... | null | proof only | null | |
0dpx | Convex quadrilateral *ABCD* is inscribed in circle *ω*. Rays *AB* and *DC* intersect at *K*. *L* is chosen on the diagonal *BD* so that $\angle BAC = \angle DAL$. *M* is chosen on the segment *KL* so that $CM \parallel BD$. Prove that the line $BM$ touches $\omega$. (Kungozhin M.)
 | [
"Let *N* be a point on the line $AK$ so that $MN \\parallel AL$. Since $\\frac{CM}{DL} = \\frac{NM}{AL} = \\frac{KM}{KL}$ and $\\angle CMN = \\angle DLA$, it follows that $K$ is a center of homothety which sends $\\triangle CMN$ into similar $\\triangle DLA$. On the other hand, $\\triangle DLA$ is similar to $\\tri... | Silk Road Mathematics Competition | XXI SILK ROAD MATHEMATICAL COMPETITION | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0dst | You are given some equilateral triangles and squares, all with side length $1$, and asked to form convex $n$ sided polygons using these pieces. If both types must be used, what are the possible values of $n$, assuming that there is sufficient supply of the pieces? | [
"Thus $n \\le 12$. It's easy to construct such polygons with $5$-$12$ sides.\n\n\n\nTo complete the proof, we now show that $n$ cannot be $3$ or $4$. An *interior square* is a square with none of its vertices lie on the boundary (of the convex polygon). At a vertex of an interior square, th... | Singapore | Singapore Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 5, 6, 7, 8, 9, 10, 11, 12 |
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