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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0dzm | The sides of an equilateral triangle $ABC$ measure $4$ cm. Denote the orthogonal projections of the midpoint $D$ of the side $AB$ onto the sides $BC$ and $AC$ by $E$ and $F$. Find the area of the triangle $DEF$. | [
"We have $|DA| = |DB| = \\frac{|AB|}{2} = 2$. The angles of the triangle $DBE$ measure $60^\\circ$, $90^\\circ$ and $30^\\circ$, so $DBE$ is one half of an equilateral triangle. This implies $|BE| = \\frac{|BD|}{2} = 1$. A similar argument for the triangle $ADF$ shows that this triangle is congruent to the triangle... | Slovenia | Slovenija 2008 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 3*sqrt(3)/4 | |
0etj | Let $ABC$ be a triangle with $AB < AC$. A point $P$ on the circumcircle of $ABC$ (on the same side of $BC$ as $A$) is chosen in such a way that $BP = CP$. Let $BP$ and the angle bisector of $\angle BAC$ intersect at $Q$, and let the line through $Q$ and parallel to $BC$ intersect $AC$ at $R$. Prove that $BR = CR$. | [
"Let $AQ$ extended meet the circumcircle of triangle $ABC$ in $T$, and let the line through $Q$ and $R$ intersect $PC$ in $D$. See Figure 2:\n\nFigure 2\nPut $\\angle BAQ = \\angle CAQ = \\alpha$, so that also $\\angle BCT = \\alpha$ and $\\angle TBC = \\alpha$. Put $\\angle BCA = \\theta$,... | South Africa | The South African Mathematical Olympiad Third Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0fad | Problem:
Four lines in the plane intersect in six points. Each line is thus divided into two segments and two rays. Is it possible for the eight segments to have lengths $1, 2, 3, \ldots, 8$? Can the lengths of the eight segments be eight distinct integers? | [
"Answer no, yes\n\n\nIf a triangle has integer sides, one of which is $1$, then it must be isosceles. So the only candidates for the segment length $1$ are $AB$ and $AE$. wlog $AB = 1$, so $BF = AF$. Hence $\\cos DFE = 1 - 1/(2 AF^2)$. Hence $ED^2 = DF^2 + EF^2 + 2DF \\cdot EF (1 - 1/2AF^2)... | Soviet Union | 25th ASU | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | no, yes | |
0fjo | Problem:
$ABCD$ es un cuadrilátero cualquiera, $P$ y $Q$ los puntos medios de las diagonales $BD$ y $AC$, respectivamente. Las paralelas por $P$ y $Q$ a la otra diagonal se cortan en $O$. Si unimos $O$ con los cuatro puntos medios de los lados $X, Y, Z$ y $T$, se forman cuatro cuadriláteros $OXBY$, $OYCZ$, $OZDT$ y $O... | [
"Solution:\n\nBastará probar que el área de cada cuadrilátero es la cuarta parte del área total.\nLa quebrada $APC$ divide al cuadrilátero en dos partes de igual área pues $AP$ es la mediana de $ABD$ y $PC$ lo es de $CBD$ (Figura 1).\nLa quebrada $TPZ$ divide al cuadrilátero $APCD$ (sombreado) en dos partes de igua... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0fvm | Problem:
Betrachte ein $m \times n$-Brett, das in Einheitsquadrate unterteilt ist. Ein L-Triomino besteht aus drei Einheitsquadraten, einem Zentrumsquadrat und zwei Schenkelquadraten. In der Ecke oben links liegt ein L-Triomino, sodass das Zentrumsquadrat auf dem Eckfeld liegt. In einem Zug kann das L-Triomino um den ... | [
"Solution:\n\n\nAbbildung 1: Schritt 1\n\n\nAbbildung 2: Schritt 2\n\n\nAbbildung 3: Schritt 3\n\nWir nehmen zuerst an, dass $m$ und $n$ beide ungerade sind. Man überlegt sich leicht, dass eine solche Folge von Drehungen für $m=n=3$ existier... | Switzerland | Vorrundenprüfung | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Both dimensions must be odd. | |
0g9d | 有一三角形 $ABC$, 其外接圓為 $O$。兩圓 $O_1$, $O_2$ 與射線 $\overrightarrow{AB}$, $\overrightarrow{AC}$ 以及圓 $O$ 均相切, 其中 $O_1$ 在 $O$ 內部, $O_2$ 在 $O$ 外部。設 $O$, $O_1$ 的公切線與 $O$, $O_2$ 的公切線交於點 $X$, $O$ 上不含 $A$ 點的 $BC$ 弧的中點為 $M$, 且線段 $\overline{AA'}$ 為圓 $O$ 之直徑。證明: $X$, $M$, $A'$ 共線。
Let $O$ be the circumcircle of the triangle $ABC$. Two ... | [
"(解 1): 令 $O$, $O_1$ 切點為 $P$, $O$, $O_2$ 切點為 $Q$; 原題等價於 $PMQA'$ 為調和四邊形。考慮變換「對 $A$ 取幂為 $\\overrightarrow{AB} \\times \\overrightarrow{AC}$ 的反演後對 $BAC$ 角平分線鏡射」, 則 $A$, $B$, $C$ 不變, $O_1$ 變成 $A$－旁切圓 (設圓心為 $I_A$), $O_2$ 變成內切圓 (設圓心為 $I$)。令 $D$, $Y$, $Z$ 分別為 $A$, $I$, $I_A$ 到 $\\overrightarrow{BC}$ 的垂足, $L$ 為 $\\angle BA... | Taiwan | 二〇一五數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Triangles > Tr... | null | proof only | null | |
0gle | Let $ABCD$ be a parallelogram, and let $M$ be the midpoint of $AB$. Line $CM$ intersects the circumcircle of triangle $ABC$ at $C$ and $E$. Let $F$ be the point on $BC$ such that $AF \perp BC$. Prove that $C, D, E$, and $F$ are concyclic. | [
"Since $E, A, C, B$ are concyclic and $AD$ is parallel to $BC$, $\\angle EAB = \\angle ECB$ and $\\angle BAD = -\\angle ABC$. Therefore,\n$$\n\\begin{align*}\n\\angle EAD &= \\angle EAB + \\angle BAD \\\\\n&= \\angle ECB - \\angle ABC \\\\\n&= \\angle MCF - \\angle ABC. \\tag{13}\n\\end{align*}\n$$\n\n![](attached_... | Thailand | The first T3MO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"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 > Miscellaneo... | null | proof only | null | |
0gq8 | Let $n > 3$ be a positive integer and $P_1, P_2, \dots, P_n$ be different points on the plane such that all pairwise distances between them are integers and the distances $P_iP_1, P_iP_2, \dots, P_iP_n$ rearranged in non-decreasing order coincide for all $i = 1, 2, \dots, n$. Find all possible values of $n$. | [
"First of all we will prove that the points $P_1, P_2, \\dots, P_n$ are concyclic. Let $P_i$ have coordinates $(x_i, y_i)$ for $i = 1, 2, \\dots, n$ and consider the point $O = (u, v)$ where $u = (x_1 + x_2 + \\dots + x_n)/n$ and $v = (y_1 + y_2 + \\dots + y_n)/n$. Then\n$$\n\\begin{aligned}\nOP_i^2 &= (u - x_i)^2 ... | Turkey | 21st Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 4, 6 | |
0h61 | In the convex quadrilateral $ABCD$ with angles $ABC$ and $BCD$ equal $120^\circ$, $O$ is the intersection of diagonals, $M$ is the midpoint of $BC$, $K$ is the point of intersection of $MO$ and $AD$. It happens that $\angle BKC = 60^\circ$. Prove that $\angle BKA = \angle CKD = 60^\circ$.
(Serduk Nazar) | [
"Let lines $AB$ and $CD$ intersect at $E$. We have that $\\angle BKC = 60^\\circ$. Let us prove that $BK > \\frac{1}{2}BC$. Indeed, one of the angles $\\angle KBC$ or $\\angle KCB$ is at least $60^\\circ$. WLOG, this angle is $\\angle KBC$ (Fig. 35). Then in $\\triangle BMK$ $\\angle KBM \\ge 60^\\circ = \\angle BK... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Napoleon and Fermat points",
"Geometry > Plane Geometry > Advanced Confi... | English | proof only | null | |
0k0l | Problem:
Two three-dimensional objects are said to have the same coloring if you can orient one object (by moving or turning it) so that it is indistinguishable from the other. For example, suppose we have two unit cubes sitting on a table, and the faces of one cube are all black except for the top face which is red, ... | [
"Solution:\n\nThere are nine non-equivalent colorings where two edges are red, two are black, and two are green. In the illustrations below, we will indicate the three colors by drawing edges thick, thin, or dashed. We will partition the colorings into three cases, determined by the number of pairs of edges that ar... | United States | 19th Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Abstract Algebra > Group Theory",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | 9 | |
0lco | Given a $10 \times 10 \times 10$ box of $1000$ unit white cubes. An and Binh play a game with this box. An selects some bands of size $1 \times 1 \times 10$ such that any two chosen bands have no common points and then changes all cells on these bands to black. Binh then selects some unit cubes and asks An what color o... | [
"We will prove the following general statement:\n\nGiven a $2n \\times 2n \\times 2n$ box of $8n^3$ unit white cubes. An and Binh play a game with the box. An selects some bands of size $1 \\times 1 \\times n$ such that any two chosen bands have no common points and then changes all cells on these bands to black. B... | Vietnam | Vietnamese Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 150 | |
0lfz | Problem:
Inside the triangle $A B C$ a point $M$ is given. The line $B M$ meets the side $A C$ at $N$. The point $K$ is symmetrical to $M$ with respect to $A C$. The line $B K$ meets $A C$ at $P$. If $\angle A M P=\angle C M N$, prove that $\angle A B P=\angle C B N$.
 | [
"Solution:\n\nLet $M^{*}$ be the isogonal conjugate of $M$ with respect to $\\triangle A B C$.\n\nLEMMA. Let $\\ell$ be the isogonal conjugate of $A M$ with respect to $\\angle B M C$, and let $\\ell^{*}$ be the isogonal conjugate of $A M^{*}$ with respect to $\\angle B M^{*} C$. The lines $\\ell, \\ell^{*}$ meet o... | Zhautykov Olympiad | XI International Zhautykov Olympiad in Sciences | [
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00ba | Consider $2018$ points placed on the vertices of regular hexagons as shown in the picture:
A bee and a beetle play the following game: initially, the bee chooses one of the $2018$ points and paints it yellow; then, the beetle chooses one of the $2017$ points that have not been painted and pain... | [
"First, we analyze in which cases three of the considered points form an equilateral triangle.\nLet **$ABC$** be an equilateral triangle with vertices among the points.\n\n*Case 1:* One vertex $A$ is at a point in the lower part of a hexagon, as in the picture (the case when it is in the upper part of a hexagon is ... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | The beetle has a winning strategy. | |
00hu | Let $n \geq 5$ be an integer. Consider $n$ squares with side lengths $1,2, \ldots, n$, respectively. The squares are arranged in the plane with their sides parallel to the $x$ and $y$ axes. Suppose that no two squares touch, except possibly at their vertices.
Show that it is possible to arrange these squares in a way s... | [
"Set aside the squares with sidelengths $n-3, n-2, n-1$, and $n$ and suppose we can split the remaining squares into two sets $A$ and $B$ such that the sum of the sidelengths of the squares in $A$ is 1 or 2 units larger than the sum of the sidelengths of the squares in $B$.\nString the squares of each set $A, B$ al... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
00lt | Im gleichschenkeligen Dreieck $ABC$ mit $\overline{AC} = \overline{BC}$ ist $D$ der Fußpunkt der Höhe durch $C$ und $M$ der Mittelpunkt der Strecke $CD$. Die Gerade $BM$ schneidet $AC$ in $E$. Man beweise, dass $AC$ dreimal so lang wie $CE$ ist.
(Erich Windischbacher) | [
"Wir zeichnen eine Parallele zu $BE$ durch den Punkt $D$ und schneiden sie mit $AC$. Den Schnittpunkt nennen wir $G$. Weil $M$ der Mittelpunkt der Strecke $DC$ ist, ist $E$ der Mittelpunkt von $GC$. (Strahlensatz)\nWeil das Dreieck $ABC$ gleichschenkelig ist und $AB$ die Basis, ist $D$ der Mittelpunkt von $AB$. Dah... | Austria | 48. Österreichische Mathematik-Olympiade Landeswettbewerb für Anfängerinnen und Anfänger | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | German | proof only | null | |
00vm | A rabbit is at some point $(x, y)$ in the Euclidean plane. There are some (possibly infinitely many) landmines, which are circles with any radius that do not intersect except possibly at one point (tangent). Every move, the rabbit can hop a distance of exactly $1$, but cannot land in the interior of a landmine (but may... | [
"We claim the answer is $\\frac{1}{\\sqrt{2}}$.\n\nFirstly, we prove that there is a configuration of landmines such that the rabbit cannot be closer than $\\frac{1}{\\sqrt{2}}$ to the origin.\nConsider a square packing of circles, with centers $\\left(\\frac{2m+1}{\\sqrt{2}}, \\frac{2n+1}{\\sqrt{2}}\\right)$ for a... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlisted Problems | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | 1/sqrt(2) | |
010r | Problem:
Let $ABC$ be an isosceles triangle with $|AB| = |AC|$. Points $D$ and $E$ lie on the sides $AB$ and $AC$, respectively. The line passing through $B$ and parallel to $AC$ meets the line $DE$ at $F$. The line passing through $C$ and parallel to $AB$ meets the line $DE$ at $G$. Prove that
$$
\frac{[DBCG]}{[FBCE]... | [
"Solution:\n\nThe quadrilaterals $DBCG$ and $FBCE$ are trapeziums. The area of a trapezium is equal to half the sum of the lengths of the parallel sides multiplied by the distance between them. But the distance between the parallel sides is the same for both of these trapeziums, since the distance from $B$ to $AC$ ... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
01u3 | Given a $(2n + 1) \times (2n + 1)$ table, all its unit 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).
a) Prove that for any positive integer $n$ one can obtain the chess coloring of the given... | [
"a) We say that a cell is black if this cell becomes black in the final coloring, and white if this cell becomes white in the final coloring.\n\nWe divide the $(2n + 1) \\times (2n + 1)$ table into concentric zones with unit width, and the central cell (see Fig. 1). The central cell and all corner cells are white. ... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 14 | |
0212 | Problem:
Let $n$ be a positive integer. There are $n$ ants walking along a line at constant nonzero speeds. Different ants need not walk at the same speed or walk in the same direction. Whenever two or more ants collide, all the ants involved in this collision instantly change directions. (Different ants need not be mo... | [
"Solution:\nThe order of the ants along the line does not change; denote by $v_{1}, v_{2}, \\ldots, v_{n}$ the respective speeds of ants $1,2, \\ldots, n$ in this order. If $v_{i-1}<v_{i}>v_{i+1}$ for some $i \\in\\{2, \\ldots, n-1\\}$, then, at each stage, ant $i$ can catch up with ants $i-1$ or $i+1$ irrespective... | Benelux Mathematical Olympiad | Benelux Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n(n-1)/2 | |
024e | Problem:
156. Parte inteira - A parte inteira de um número real $x$ é o maior inteiro que é menor do que ou igual a $x$. Denotamos a parte inteira de $x$ por $[x]$. Por exemplo, $[2,9]=2$, $[0,88]=0$ e $[-1,7]=-2$.
Calcule as partes inteiras seguintes.
(a) $[\sqrt{12}]$
(b) $\left[\frac{28756... | [
"Solution:\n\n(a) Os números $9$ e $16$ são quadrados perfeitos e $9<12<16$. Então,\n$$\n3=\\sqrt{9}<\\sqrt{12}<\\sqrt{16}=4\n$$\n\ne, portanto, $[\\sqrt{12}]=3$.\n\n(b) Como $12777 \\times 2<28756<12777 \\times 3$, temos\n$$\n2<\\frac{28756}{12777}<3, \\text{ portanto, }\\left[\\frac{28756... | Brazil | Nível 2 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | {'a': 3, 'b': 2, 'c': -1, 'd': -5} | |
0caz | The continuous function $f : \mathbb{R} \to \mathbb{R}$ has the property $(f \circ f)(x) = x^{2023}$, for every $x \in \mathbb{R}$. Prove that:
a) $2023\sqrt[2023]{f(x)} = f(2023\sqrt[2023]{x})$, for every $x \in \mathbb{R}$;
b) for every $n \ge 3$ there exists an arithmetic progression $x_1, x_2, \dots, x_n$ so that... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof only | null | |
07vc | Let $AOB$ be a triangle with $\angle AOB = 90^\circ$. Define a sequence of points $X_1, X_2, X_3, \dots, X_{2022}$ as follows:
* $X_1$ lies on the ray $\vec{OA}$ with $|OX_1| = 1$.
* $X_2$ lies on the ray $\vec{OB}$ with $|X_1X_2| = 2$
* For $3 \le n \le 2022$, the $X_n$ lie alternately on $\vec{OA}$ ($n$ odd) or $\vec... | [
"Let $X_0 = O$ and, for $0 \\le n \\le 2022$, let $d_n = |OX_n|$. Then $d_0 = 0$ and by Pythagoras' theorem, for $n \\ge 1$\n\n\n\n$$\nd_{2n-1}^2 + d_{2n}^2 = (2n)^2\n$$\n$$\nd_{2n-1}^2 + d_{2n-2}^2 = (2n-1)^2.\n$$\nSubtracting these equations we get $d_{2n}^2 - d_{2n-2}^2 = (2n)^2 - (2n-1)... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof and answer | sqrt(2045253) - sqrt(2041210) | |
0awa | Problem:
If $\tan x + \tan y = 5$ and $\tan (x + y) = 10$, find $\cot^2 x + \cot^2 y$. | [
"Solution:\nWe know that\n$$\n\\tan(x + y) = \\frac{\\tan x + \\tan y}{1 - \\tan x \\tan y} = 10\n$$\nThis leads to\n$$\n\\tan x \\tan y = \\frac{1}{2}\n$$\nHence,\n$$\n\\begin{aligned}\n\\cot^2 x + \\cot^2 y &= \\frac{1}{\\tan^2 x} + \\frac{1}{\\tan^2 y} \\\\\n&= \\frac{\\tan^2 y + \\tan^2 x}{\\tan^2 x \\tan^2 y} ... | Philippines | 18th PMO National Stage Oral Phase | [
"Precalculus > Trigonometric functions"
] | null | final answer only | 96 | |
0609 | Problem:
Soit $n$ un entier vérifiant $n \geqslant 2$. On note $d$ le plus grand diviseur de $n$ différent de $n$. On suppose que $d > 1$. Démontrer que $n + d$ n'est pas une puissance de 2. | [
"Solution:\n\nSupposons par l'absurde que $n + d$ est une puissance de $2$. Notons que $d$ divise $n$, donc $d$ divise $n + d$, donc $d$ divise une puissance de $2$. Ainsi $d$ est une puissance de $2$ : on pose $d = 2^{k}$ pour un certain $k \\in \\mathbb{N}$. On a $k \\geqslant 1$ car $d \\neq 1$. Posons $n = d a ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
05gq | Problem:
Soit $ABC$ un triangle, $H$ le pied de la hauteur issue de $A$, $L$ le pied de la bissectrice issue de $B$ et $M$ le milieu de $[AB]$. On suppose que dans le triangle $HLM$ on a les deux propriétés : $(AH)$ est une hauteur et $(BL)$ est une bissectrice. Montrer que $(CM)$ est une médiane. | [
"Solution:\n\nOn prend le triangle $ABC$, on va démontrer qu'il est équilatéral, en utilisant les deux propriétés, la troisième découlera.\nOn a $(AH) \\perp (ML)$ et $(AH) \\perp (BC)$, donc $(BC) // (ML)$. On trouve ainsi que $L$ est le milieu de $[AC]$. Cependant, $\\frac{AL}{LC} = \\frac{AB}{BC}$ ce qui montre ... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kq5 | Problem:
In convex quadrilateral $ABCD$ with $AB = 11$ and $CD = 13$, there is a point $P$ for which $\triangle ADP$ and $\triangle BCP$ are congruent equilateral triangles. Compute the side length of these triangles. | [
"Solution:\n\nEvidently $ABCD$ is an isosceles trapezoid with $P$ as its circumcenter. Now, construct isosceles trapezoid $AB'BC$ (that is, $BB'$ is parallel to $AC$.) Then $AB'PD$ is a rhombus, so $\\angle B'CD = \\frac{1}{2} \\angle B'PD = 60^{\\circ}$ by the inscribed angle theorem. Also, $B'C = 11$ because the ... | United States | HMMT November 2022 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | final answer only | 7 | |
074g | Let $f(x)$ and $g(y)$ be two monic polynomials with complex coefficients, and of the same degree $n$, such that
$$
f(x) - g(y) = \prod_{j=1}^{n} (a_j x + b_j y + c_j),
$$
where $a_j, b_j, c_j$ are complex numbers, $1 \le j \le n$. Prove that there exist complex numbers $a, b, c$ such that
$$
f(x) = (x+a)^n + c, \quad g... | [
"Observe that $\\prod_{j=1}^{n} a_j = 1$. Hence by dividing by this product, we may write\n$$\nf(x) - g(y) = \\prod_{j=1}^{n} x - \\alpha_j y + \\beta_j.\n$$\nWe observe that\n$$\n\\prod_{j=1}^{n} (x - \\alpha_j y) = x^n - y^n = \\prod_{j=1}^{n} (x - w^j y),\n$$\nwhere $w$ is the primitive $d$-th root of unity. Hen... | India | Indija TS 2009 | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
0413 | As seen in Fig. 1.1, $AB$ is a chord of circle $\omega$, $P$ is a point on arc $AB$, and $E, F$ are 2 points on $AB$ satisfying $AE = EF = FB$. Connect $PE, PF$ and extend them to intersect with $\omega$ at $C, D$, respectively. Prove
$$
EF \cdot CD = AC \cdot BD.
$$
Fig. 1.1 | [
"As shown in Fig. 1.2, we connect $AD$, $BC$, $CF$, $DE$. Since $AE = EF = FB$, we have\n$$\n\\frac{BC \\cdot \\sin \\angle BCE}{AC \\cdot \\sin \\angle ACE} = \\frac{\\text{distance between } B \\text{ and } CP}{\\text{distance between } A \\text{ and } CP} = \\frac{BE}{AE} = 2. \\qquad \\textcircled{1}\n$$\nIn th... | China | China Mathematical Competition (Complementary Test) | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0dgx | Let $ABCD$ be a convex quadrilateral such that the line $CD$ is a tangent to the circle on $AB$ as diameter. Prove that the line $AB$ is a tangent to the circle on $CD$ as diameter if and only if the lines $BC$ and $AD$ are parallel. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals"
] | English | proof only | null | |
0hpb | Problem:
A natural number $n$ is chosen between two consecutive square numbers. The smaller square is obtained by subtracting $k$ from $n$, and the larger one is obtained by adding $\ell$ to $n$. Prove that the number $n-k \ell$ is the square of an integer. | [
"Solution:\n\nLet $n-k = x^{2}$ and $n+\\ell = (x+1)^{2}$. Then $k = n - x^{2}$ and $\\ell = (x+1)^{2} - n$. We express everything in terms of $n$ and $x$:\n$$\n\\begin{aligned}\nn - k \\ell & = n - \\left(n - x^{2}\\right)\\left((x+1)^{2} - n\\right) \\\\\n& = n + \\left(x^{2} - n\\right)\\left(x^{2} + 2x + 1 - n\... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0beo | Problem:
Adottak az $a, b \in \mathbb{R}^{*}$ számok és az $f: \mathbb{R} \rightarrow \mathbb{R}$,
$$
f(x)= \begin{cases}a x, & x \in \mathbb{Q} \\ b x, & x \in \mathbb{R} \backslash \mathbb{Q}\end{cases}
$$
függvény.
Igazold, hogy $f$ akkor és csak akkor injektív, ha $f$ szürjektív. | [] | Romania | Matematika tantárgyverseny Megyei szakasz | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
0cih | Show that there exist infinitely many natural numbers $n$ for which $2025 \cdot n$ is a perfect cube, and $5202 \cdot n$ is a perfect square. | [] | Romania | 75th NMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0bmx | Find the smallest positive integer which has exactly $2015$ positive divisors. | [
"The number of positive divisors of an integer $N$ with the prime factorization $p_1^{a_1} p_2^{a_2} \\cdots p_n^{a_n}$ is $(a_1 + 1)(a_2 + 1) \\cdots (a_n + 1)$. Since $2015 = 1 \\cdot 2015 = 5 \\cdot 403 = 13 \\cdot 155 = 31 \\cdot 65 = 5 \\cdot 13 \\cdot 31$, the required number has one of the following forms: $... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 2^30 * 3^12 * 5^4 | |
067b | Solve in the real numbers the system
$$
x^3 = \frac{z}{y} - \frac{2y}{z}, \quad y^3 = \frac{x}{z} - \frac{2z}{x}, \quad z^3 = \frac{y}{x} - \frac{2x}{y}.
$$ | [
"For $x, y, z \\in \\mathbb{R}$, such that $xyz \\neq 0$, the system is written:\n$$\nx^3 y z = z^2 - 2y^2 \\quad (1), \\quad y^3 z x = x^2 - 2z^2 \\quad (2), \\quad z^3 x y = y^2 - 2x^2 \\quad (3)\n$$\nUsing summation by parts we find:\n$$\nxyz(x^2 + y^2 + z^2) = -(x^2 + y^2 + z^2) \\Leftrightarrow (x^2 + y^2 + z^... | Greece | 31st Hellenic Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | (-1, -1, -1), (1, 1, -1), (1, -1, 1), (-1, 1, 1) | |
00g4 | Let $a, b, c, d, e, f$ be real numbers such that the polynomial
$$
p(x) = x^{8} - 4x^{7} + 7x^{6} + a x^{5} + b x^{4} + c x^{3} + d x^{2} + e x + f
$$
factorises into eight linear factors $x - x_{i}$, with $x_{i} > 0$ for $i = 1, 2, \ldots, 8$. Determine all possible values of $f$. | [
"From\n$$\nx^{8} - 4x^{7} + 7x^{6} + a x^{5} + b x^{4} + c x^{3} + d x^{2} + e x + f = (x - x_{1})(x - x_{2}) \\ldots (x - x_{8})\n$$\nwe have\n$$\n\\sum_{i=1}^{8} x_{i} = 4 \\quad \\text{and} \\quad \\sum x_{i} x_{j} = 7\n$$\nwhere the second sum is over all pairs $(i, j)$ of integers where $1 \\leq i < j \\leq 8$... | Asia Pacific Mathematics Olympiad (APMO) | XV APMO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 1/256 | |
0f88 | Problem:
In the acute-angled triangle $ABC$, the altitudes $BD$ and $CE$ are drawn. Let $F$ and $G$ be the points of the line $ED$ such that $BF$ and $CG$ are perpendicular to $ED$. Prove that $EF = DG$. | [
"Solution:\n\n$\\angle BDC = \\angle BEC = 90^{\\circ}$, so $BCDE$ is cyclic, so $\\angle BDE = \\angle BCE = 90^{\\circ} - \\angle B$. Hence $\\angle DGC = 90^{\\circ} - \\angle CDG = \\angle BDE = 90^{\\circ} - B$. So $DG = CD \\sin DGC = BC \\sin CBD \\sin DGC = BC \\cos C \\cos B$. Similarly $EF$."
] | Soviet Union | 22nd ASU | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0akb | Let $q$ be a positive rational number. Two ants are initially at the same point $X$ in the plane. In the $n$-th minute ($n=1,2,3,\ldots$) each of them chooses whether to walk due north, east, south or west and then walks the distance of $q^n$ meter. After a whole number of minutes, they are at the same point in the pla... | [
"Let $x_A^{(n)}$ (resp. $x_B^{(n)}$) be the $x$-coordinates of the first (resp. second) ant's position after $n$ minutes. Then $x_A^{(n)} - x_B^{(n)} \\in \\{q^n, -q^n, 0\\}$, and so $x_A^{(n)}, x_B^{(n)}$ are given by polynomials in $q$ with coefficients in $\\{-1, 1, 0\\}$. So if the ants meet after $n$ minutes, ... | North Macedonia | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Dista... | English | proof and answer | 1 | |
0ijk | Find all positive integers $n$ such that there are $k \ge 2$ positive rational numbers $a_1, a_2, \dots, a_k$ satisfying $a_1 + a_2 + \dots + a_k = a_1 \cdot a_2 \dots a_k = n$. | [
"The answer is $n = 4$ or $n \\ge 6$.\n\nI. First, we prove that each $n \\in \\{4, 6, 7, 8, 9, \\dots\\}$ satisfies the condition.\n\n(1). If $n = 2k \\ge 4$ is even, we set $(a_1, a_2, \\dots, a_k) = (k, 2, 1, \\dots, 1)$:\n$$\na_1 + a_2 + \\dots + a_k = k + 2 + 1 \\cdot (k-2) = 2k = n,\n$$\nand\n$$\na_1 \\cdot a... | United States | USAMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | n = 4 or n ≥ 6 | |
0cvb | Pasha and Vova play the following game, making moves in turn; Pasha moves first. Initially they have a large piece of plasticine. By a move, Pasha cuts one of the existing pieces into three (of arbitrary sizes), and Vova merges two existing pieces into one. Pasha wins if at some moment there appears to be 100 pieces of... | [
"**Первое решение.** Приведём алгоритм, позволяющий Паше победить. Пусть масса исходного куска равна 1 кг. Паша каждым ходом будет отрезать от самого большого из имеющихся кусков два куска массой по 0,01 г. Докажем, что не позже, чем через 10 000 ходов Паша победит.\n\nПредположим, что это не так. Рассмотрим 100 по... | Russia | Final round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English; Russian | proof and answer | No; Vova cannot prevent Pasha’s win. Pasha has a winning strategy. | |
0lf7 | Let $ABC$ be an acute, non-isosceles triangle with circumcircle $(O)$. $BE$, $CF$ are the heights of $\triangle ABC$, and $BE$, $CF$ intersect at $H$. Let $M$ be the midpoint of $AH$, and $K$ be the point on $EF$ such that $HK$ is perpendicular to $EF$. A line not going through $A$ and parallel to $BC$ intersects the m... | [
"We state the following familiar lemma:\n**LEMMA.** Given triangle $ABC$ and two points $X, Y$. Suppose $BX$ cuts $CY$ at $Z$ and $BY$ cuts $CX$ at $T$. Then if $AX$ and $AY$ are congruent in $\\angle BAC$ then so are $AZ$ and $AT$.\n\nWe first reduce the problem to a simpler form through pairs of similar triangles... | Vietnam | Team selection tests | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangle... | English | proof only | null | |
00nu | Let $n$ be a positive integer. What proportion of the non-empty subsets of $\{1, 2, \dots, 2n\}$ has a smallest element that is odd?
(Birgit Vera Schmidt) | [
"The number of subsets of $\\{1, 2, \\dots, 2n\\}$ that have $k$ as smallest element is $2^{2n-k}$ for $1 \\le k \\le 2n$ since each element bigger than $k$ is either contained in the subset or not.\nThe number $O$ of subsets with an odd smallest element is therefore equal to\n$$\nO = 2^{2n-1} + 2^{2n-3} + \\dots +... | Austria | AUT_ABooklet_2023 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 2/3 | |
084f | Problem:
Due circonferenze $C_{1}$ e $C_{2}$ di centri $A$ e $B$ sono tangenti esternamente in $T$. Sia $B D$ un segmento tangente a $C_{1}$ in $D$ e sia $T C$ il segmento tangente ad entrambe in $T$ con $C \in B D$. Se $A T$ è lungo 80 e $B T$ è lungo 90, qual è la lunghezza di $C D$? | [
"Solution:\n\nLa risposta è 48. $A T$ è un raggio della circonferenza $C_{1}$; poiché anche $A D$ lo è, anch'esso misura 80. Inoltre l'angolo $A \\widehat{D} B$ è retto; per il teorema di Pitagora allora $B D$ misura $\\sqrt{(80+90)^{2}-80^{2}}=150$. Ma i triangoli $A B D$ e $B T C$ sono simili (sono rettangoli ed ... | Italy | Progetto Olimpiadi di Matematica 2005 GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | null | final answer only | 48 | |
0325 | Problem:
Find all values of $a$, for which the equation
$$
\frac{2 a}{(x+1)^{2}}+\frac{a+1}{x+1}-\frac{2(a+1) x-(a+3)}{2 x^{2}-x-1}=0
$$
has two real roots $x_{1}$ and $x_{2}$ satisfying the relation $x_{2}^{2}-a x_{1}=a^{2}-a-1$. | [
"Solution:\nThe given equation is equivalent to\n$$\na x^{2}+(1-2 a) x+(1-a)=0\n$$\nwhere $x \\neq -1, -\\frac{1}{2}, 1$. Hence this equation has two real roots $x_{1}$ and $x_{2}$ such that\n$$\nx_{2}^{2}-a x_{1}=a^{2}-a-1\n$$\nSince $x_{1}+x_{2}=\\frac{2 a-1}{a}$ we get that\n$$\nx_{2}^{2}+a x_{2}-a^{2}-a+2=0\n$$... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a = 3 and a = -1 - sqrt(2) | |
0cu8 | In the Cartesian plane, two graphs $\Gamma_1$ and $\Gamma_2$ of monic quadratic trinomials and two non-parallel lines $l_1$ and $l_2$ are drawn. Assume that $\Gamma_1$ and $\Gamma_2$ cut out segments of equal lengths on $l_1$, and that they cut out segments of equal lengths on $l_2$. Prove that $\Gamma_1$ and $\Gamma_2... | [
"First solution. Let $f_1(x)$ and $f_2(x)$ be the given monic quadratic trinomials, and let $\\Gamma_1$ and $\\Gamma_2$ be their graphs. Then there exists a unique vector $\\vec{a}$ such that, under a parallel translation by this vector, the parabola $\\Gamma_1$ maps to $\\Gamma_2$ (the vector $\\vec{a}$ connects t... | Russia | XLIII Russian mathematical olympiad | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressio... | English; Russian | proof only | null | |
0gan | 對於任何正整數 $k$, 令 $S(k)$ 表示其在十進制下的各個位數總和 (例: $S(209) = 2 + 0 + 9 = 11$)。試求所有整係數多項式 $P(x)$, 使得對於所有正整數 $n \ge 2017$, 都有 $P(n) > 0$ 且
$$
S(P(n)) = P(S(n)).
$$ | [
"$P(x) = x$ 或 $P(x) = c$, 其中 $c \\in \\{1, \\dots, 9\\}$\n\n假設 $P(x) = a_{d}x^{d} + \\cdots + a_{1}x^{1} + a_{0}$。考慮 $n = 9 \\times 10^{k}$, 其中 $k$ 是一個充份大的正整數; 注意到 $S(n) = 9 \\Rightarrow P(S(n)) = P(9)$。\n\n1. 我們首先證明所有係數都非負。\n假設存在 $0 \\le i < d$ 使得 $a_i < 0$, 則易知 $P(n)$ 對應 $10^{ik+m+1}$ 到 $10^{(i+1)k-1}$ 的位數都是 9, 故... | Taiwan | 二〇一七數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | P(x) = x or P(x) = c for c in {1, 2, 3, 4, 5, 6, 7, 8, 9} | |
0ekm | Problem:
Naj bo $f: \mathbb{R} \rightarrow \mathbb{R}$ funkcija, za katero velja
$$
f(x)= \begin{cases}x ; & x \geq 2 \\ f(4-x) ; & 0 \leq x<2 \\ f(x+2) ; & x<0\end{cases}
$$
Koliko je $f(-5)$ ?
(A) -3
(B) -1
(C) 1
(D) 3
(E) 5 | [
"Solution:\n\n$$\nf(-5)=f(-5+2)=f(-3)=f(-3+2)=f(-1)=f(-1+2)=f(1)=f(4-1)=f(3)=3\n$$"
] | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | MCQ | D | |
058t | a. There are $2n$ rays marked in a plane, with $n$ being a natural number. Given that no two marked rays have the same direction and no two marked rays have a common initial point, prove that there exists a line that passes through none of the initial points of the marked rays and intersects with exactly $n$ marked ray... | [
"Consider any circle such that all of the endpoints of the marked rays are inside the circle. Choose a tangent line of the circle that is not parallel to any of the marked rays. Let this tangent line be $l_0$ and let $l_\\alpha$ be the tangent line we get by rotating $l_0$ counterclockwise by angle $\\alpha$ with r... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
08of | Problem:
Let $A = 1 \cdot 4 \cdot 7 \cdot \ldots \cdot 2014$ be the product of the numbers less or equal to $2014$ that give remainder $1$ when divided by $3$. Find the last non-zero digit of $A$. | [
"Solution:\nGrouping the elements of the product by ten we get:\n$$\n\\begin{aligned}\n& (30k+1)(30k+4)(30k+7)(30k+10)(30k+13)(30k+16) \\\\\n& (30k+19)(30k+22)(30k+25)(30k+28) = \\\\\n& = (30k+1)(15k+2)(30k+7)(120k+40)(30k+13)(15k+8) \\\\\n& (30k+19)(15k+11)(120k+100)(15k+14)\n\\end{aligned}\n$$\n(We divide all eve... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | null | proof and answer | 2 | |
05ta | Problem:
Soit $ABC$ un triangle et $\Gamma$ un cercle passant par $A$. On suppose que $\Gamma$ recoupe les segments $[AB]$ et $[AC]$ en deux points, que l'on appelle respectivement $D$ et $E$, et qu'il coupe le segment $[BC]$ en deux points, que l'on appelle $F$ et $G$, de sorte que $F$ se trouve entre $B$ et $G$. Soi... | [
"Solution:\n\nCommençons par tracer une figure.\n\n\nUne première remarque frappante est que $T$ semble être situé sur le cercle $\\Gamma$. Après avoir vérifié qu'il l'était bien sur une deuxième figure, on s'empresse donc de démontrer ce premier résultat. Pour ce faire, on entame donc une ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00yo | Problem:
Find all real-valued functions $f$ defined on the set of all non-zero real numbers such that:
(i) $f(1)=1$,
(ii) $f\left(\frac{1}{x+y}\right)=f\left(\frac{1}{x}\right)+f\left(\frac{1}{y}\right)$ for all non-zero $x, y, x+y$,
(iii) $(x+y) f(x+y)=x y f(x) f(y)$ for all non-zero $x, y, x+y$. | [
"Solution:\nSubstituting $x=y=\\frac{1}{2} z$ in (ii) we get\n$$\nf\\left(\\frac{1}{z}\\right)=2 f\\left(\\frac{2}{z}\\right)\n$$\nfor all $z \\neq 0$. Substituting $x=y=\\frac{1}{z}$ in (iii) yields\n$$\n\\frac{2}{z} f\\left(\\frac{2}{z}\\right)=\\frac{1}{z^{2}}\\left(f\\left(\\frac{1}{z}\\right)\\right)^{2}\n$$\n... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = 1/x | |
0g3p | Problem:
Sei $ABC$ ein spitzwinkliges Dreieck mit $BC > AC$. Die Mittelsenkrechte der Seite $AB$ schneidet die Gerade $BC$ in $X$ und die Gerade $AC$ in $Y$. Sei $P$ die Projektion von $X$ auf $AC$ und sei $Q$ die Projektion von $Y$ auf $BC$. Beweise, dass die Gerade $PQ$ die Strecke $AB$ in ihrem Mittelpunkt schneide... | [
"Solution:\n\nNach Einführung von $M$ als der Mittelpunkt von $AB$, genügt es zu zeigen, dass $P$, $Q$ und $M$ kollinear sind. Wir werden dies hier zeigen, indem wir beweisen, dass $\\angle MPX + \\angle XPQ = 180^{\\circ}$ gilt. Wir bemerken dazu zuerst ein paar Sehnenvierecke.\n\na.\n$YQPX$ ist ein Sehnenviereck,... | Switzerland | Zweite Runde 2021 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ias | Problem:
A tiling of the plane with polygons consists of placing the polygons in the plane so that interiors of polygons do not overlap, each vertex of one polygon coincides with a vertex of another polygon, and no point of the plane is left uncovered. A unit polygon is a polygon with all sides of length one.
It is qui... | [
"Solution:\n\na.\nThis can be done easily with parallel rows of squares and triangles, as shown.\n\n\n\nOf course, other tilings are possible.\n\n\nb.\nMany people received only partial credit for part (b), because their arguments were not rigorous. In order to show that a tiling is not pos... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05tc | Problem:
Soit $ABC$ un triangle dont les angles sont aigus. La hauteur issue de $A$ recoupe le cercle circonscrit au triangle $ABC$ en un point $D$. La hauteur issue de $B$ recoupe le cercle circonscrit au triangle $ABC$ en un point $E$. La droite $(ED)$ coupe les côtés $[AC]$ et $[BC]$ respectivement en les points $P... | [
"Solution:\n\n\n\nDéterminons quels sont les angles qui semblent faciles d'accès. Par exemple, il semble compliqué de pouvoir obtenir l'angle $\\widehat{PBQ}$ ou l'angle $\\widehat{PBA}$. En revanche, on connait déjà l'angle $\\widehat{QBA}$ donc il semble raisonnable d'essayer de montrer q... | France | Envoi 5: Pot Pourri | [
"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"
] | null | proof only | null | |
0i4x | Problem:
A restricted path of length $n$ is a path of length $n$ such that for all $i$ between $1$ and $n-2$ inclusive, if the $i$th step is upward, the $i+1$st step must be rightward.
Find the number of restricted paths that start at $(0,0)$ and end at $(7,3)$. | [
"Solution:\n\nThis is equal to the number of lattice paths from $(0,0)$ to $(7,3)$ that use only rightward and diagonal (upward+rightward) steps plus the number of lattice paths from $(0,0)$ to $(7,2)$ that use only rightward and diagonal steps, which is equal to the number of paths (as defined above) from $(0,0)$ ... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 56 | |
0h3p | Is it possible to fill in a table $7 \times 7$ with $24$ ones and $25$ zeros so that:
1) for each cell containing one, sum of numbers in neighbour cells equals one,
2) for each cell containing zero, sum of numbers in neighbour cells does not equal one?
(Two cells are neighbours if they have a common side.) | [
"**Answer:** Yes, it is possible (see the figure).\n\n"
] | Ukraine | Ukrainian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | Yes | |
0ar9 | Problem:
Let $f(x)$ be a cubic polynomial. If $f(x)$ is divided by $2x+3$, the remainder is $4$, while if it is divided by $3x+4$, the remainder is $5$. What will be the remainder when $f(x)$ is divided by $6x^{2}+17x+12$? | [] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 6x + 13 | |
038c | Problem:
The sets $M=\{1,2, \ldots, 27\}$ and $A=\{a_{1}, a_{2}, \ldots, a_{k}\} \subset \{1,2, \ldots, 14\}$ have the following property: every element of $M$ is either an element of $A$ or the sum of two (possibly identical) elements of $A$. Find the minimum value of $k$. | [
"Solution:\n\nThe elements of $A$ give $\\binom{k}{2}+k+k=\\frac{k(k+3)}{2}$ sums of the required kind (with two different summands, two equal summands or one summand, respectively). Therefore $\\frac{k(k+3)}{2} \\geq 27$, whence $k \\geq 6$.\n\nIf $k=6$ every element of $M$ has a unique representation. This consec... | Bulgaria | Spring Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 8 | |
0c9x | Problem:
Fie $P_{0}, P_{1}, \ldots, P_{2021}$ puncte pe cercul trigonometric, de centru $O$ şi rază $1$, astfel încât, pentru orice $n \in \{1,2, \ldots, 2021\}$, lungimea arcului de cerc parcurs în sens trigonometric de la $P_{n-1}$ la $P_{n}$ aparţine intervalului $\left[\frac{\pi}{2}, \pi\right]$.
Aflaţi lungimea m... | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | √2 | |
0adb | Let $1*2*3*4*5*6*7*8*9=0$ be a given equality. Is it possible to substitute some of the $*$ with $+$ and the others with $-$ to obtain a correct equality? | [
"Sum or difference of two even or two odd numbers is an even number. Sum or difference of an even and an odd number, or an odd and an even number, is an odd number. On the left side of the equality there are $4$ even and $5$ odd numbers. Any combination of $+$ and $-$ will give an odd number, but $0$ is an even num... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
0c6i | Determine all pairs $(a, b)$, of non-negative integers such that the quotients
$$
\frac{3a + 8b + 2}{10a + 2b + 1} \quad \text{and} \quad \frac{8a + b + 3}{2a + 7b + 3}
$$
are integers simultaneously. | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (a, b) = (0, 0) and (1, 1) | |
06nw | There is a $20 \times 20$ table and $400$ cards printed with the numbers $1$ to $400$. The cards are then distributed to the cells of the table so that there is one card in each cell. After that, we put a red sticker on the card in each row with the largest number, and a blue sticker on the card in each column with the... | [
"Answer: $362$\n\nFor convenience we shall use the term 'red number' to the number on a card with a red sticker, and similarly for 'blue number'. Clearly, a red number, being a row maximum, is at least $20$. Also, since $A$ is the smallest red number, there are $19$ other red numbers greater than $A$, so $A$ is at ... | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 362 | |
02kg | Problem:
Em um quadrado mágico, a soma dos 3 números de cada linha, coluna ou diagonal é sempre a mesma. A seguir temos um quadrado mágico, parcialmente preenchido.
| | | |
| :--- | :--- | :--- |
| 1 | 14 | $x$ |
| 26 | | 13 |
Qual é o valor de $x$?
A) 20
B) 22
C) 23
D) 25
E) 27 | [
"Solution:\n\nSeja $y$ um dos números do quadrado mágico, conforme a figura. De acordo com a regra de quadrado mágico temos:\n$$\n\\underbrace{26+14+y}_{\\begin{array}{c}\\text{ soma dos números } \\\\ \\text{ da diagonal que contém } y\\end{array}} = \\underbrace{y+x+13}_{\\begin{array}{c}\\text{ soma dos números ... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | E) 27 | |
09q7 | Problem:
De twee cirkels $\Gamma_{1}$ en $\Gamma_{2}$ snijden elkaar in $P$ en $Q$. De gemeenschappelijke raaklijn aan de kant van $P$ raakt de cirkels in $A$ resp. $B$. De raaklijn aan $\Gamma_{1}$ in $P$ snijdt $\Gamma_{2}$ voor de tweede keer in $C$ en de raaklijn aan $\Gamma_{2}$ in $P$ snijdt $\Gamma_{1}$ voor de... | [
"Solution:\n\nWe gaan alle relevante hoeken uitdrukken in $\\alpha=\\angle B A P$ en $\\beta=\\angle P B A$. Het midden van $A B$ is per definitie ook het midden van $P M$, dus de diagonalen van vierhoek $A P B M$ snijden elkaar middendoor. Daarom is $A P B M$ een parallellogram en geldt $\\angle A M B=\\angle A P ... | Netherlands | Dutch TST | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0d6e | 1) Prove that there are infinitely many positive integers $n$ such that there exists a permutation of $1,2,3, \ldots, n$ with the property that the difference between any two adjacent numbers is equal to either $2015$ or $2016$.
2) Let $k$ be a positive integer. Is the statement in 1) still true if we replace the numb... | [
"1) First, replace $2015$ and $2016$ by $3$ and $4$ respectively for convenience. We consider the permutation of $\\{1,2,3, \\ldots, 7\\}$ as follows:\n$$\n1,5,2,6,3,7,4.\n$$\nThe numbers increase by $4$ and decrease by $3$, then the sequence covers all numbers from $1$ to $7$. Add $1$ to each term of the above seq... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Other"
] | English | proof only | null | |
05e9 | Problem:
Find all positive integers $d$ for which there exists a degree $d$ polynomial $P$ with real coefficients such that there are at most $d$ different values among $P(0), P(1), \ldots, P\left(d^{2}-d\right)$. | [
"Solution:\nWe claim that such polynomials exist if and only if $d \\leq 3$. The following examples show that such polynomials do exist for $d \\leq 3$:\n$$\n\\begin{array}{lll}\nd=1: & d^{2}-d=0, & P_{1}(x)=x, \\\\\n\\end{array} \\begin{array}{ll} \n& P(0)=0 ;\n\\end{array}\n$$\nWe can make more examples by adding... | European Girls' Mathematical Olympiad (EGMO) | EGMO 2024 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | null | proof and answer | {1, 2, 3} | |
0cy2 | Let $ABC$ be a right angled triangle with $\widehat{A}=90^{\circ}$ and $BC=a$, $AC=b$, $AB=c$. Let $d$ be a line passing through the incenter of triangle and intersecting the sides $AB$ and $AC$ in $P$ and $Q$, respectively.
a. Prove that
$$
b \cdot \frac{PB}{PA} + c \cdot \frac{QC}{QA} = a ;
$$
b. Find the minimum o... | [
"(a) Assume that the origin of the coordinates system is at $A$. Let $r$ be the inradius of $\\triangle ABC$, and $I$ the incenter. Then $I(r, r)$ and the line $d$ has equation\n$$\nd: y - r = m(x - r)\n$$\nthat is $y = r + m(x - r)$. We get\n$$\nP\\left(\\frac{r(m-1)}{m}, 0\\right), \\quad Q(0, r(m-1)) .\n$$\n![](... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Geometry > Plane Geometry > M... | English | proof and answer | 1 | |
00up | Let $ABC$ be a triangle with circumcenter $O$. Point $X$ is the intersection of the parallel line from $O$ to $AB$ with the perpendicular line to $AC$ from $C$. Let $Y$ be the point where the external bisector of $\angle BXC$ intersects with $AC$. Let $K$ be the projection of $X$ onto $BY$.
Prove that the lines $AK$, $... | [
"First of all, we prove that the quadrilateral $BOCX$ is cyclic. Indeed,\n$$\n\\angle XCB = 90^\\circ - \\angle ACB = \\angle ABO = \\angle BOX.\n$$\nThis means that $OX$ is the internal bisector of $\\angle BXC$ (since $OB = OC$), so $OX \\perp XY$. The next observation is that the quadrilateral $ABXY$ is also cyc... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced... | null | proof only | null | |
0edn | Take an equilateral triangle $ABC$ with the side of length $a$. Draw the squares $BADE$ and $CBFG$ on the sides $AB$ and $BC$, respectively (see figure). How long is the line segment $DG$?
(A) $(\sqrt{2} + 1)a$
(B) $(2\sqrt{2} + 1)a$
(C) $(\sqrt{3} + 1)a$
(D) $\sqrt{3}a$
(E) 1 | [
"Denote the intersections of the segment $DG$ with the lines $AB$ and $BC$ by $K$ and $L$ respectively. Due to symmetry the segment $DG$ is parallel to the segment $AC$, so $KBL$ is an equilateral triangle. The triangles $DKA$ and $LGC$ are one half of an equilateral triangle with altitudes of length $a$ and sides ... | Slovenia | Slovenija 2016 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | (\sqrt{3} + 1)a | |
0gfv | 對任一正整數 $n$，證明多項式
$$
P_n(x) = \sum_{k=0}^{n} 2^k \binom{2n}{2k} x^k (x-1)^{n-k}
$$
有 $n$ 個實根。
(註：$\binom{2n}{2k}$ 為二項式係數。) | [] | Taiwan | 2022 數學奧林匹亞競賽第二階段培訓營, 獨立研究 (二) | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Intermediate Algebra > Complex numbers"
] | Chinese; English | proof only | null | |
0e06 | Consider all polynomials with the leading coefficient $1$ that give the remainder of $1$ when divided by $x + 1$ and the remainder $2$ when divided by $x^2 + 1$. Among these, find the polynomial of the lowest degree. | [
"Obviously, the degree of such a polynomial must be at least $2$. If the degree were equal to $2$, this polynomial would have the form of\n$$\np(x) = x^2 + a x + b = (x^2 + 1) + a x + (b - 1).\n$$\nWhen dividing by $x^2 + 1$ the remainder is $2$, and $a x + (b - 1) = 2$ implies $a = 0$ and $b = 3$. But when we divi... | Slovenia | National Math Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | x^3 + (1/2)x^2 + x + 5/2 | |
05j5 | Problem:
Sur la diagonale $[BD]$ d'un carré $ABCD$ on a choisi un point $E$. Soient $O_{1}$ et $O_{2}$ les centres des cercles circonscrits aux triangles $ABE$ et $ADE$ respectivement. Montrer que $AO_{1}EO_{2}$ est un carré.
 | [
"Solution:\n\n$\\widehat{AO_{1}E} = 2\\widehat{ABE} = 90^{\\circ}$ et $O_{1}A = O_{1}E$ donc $AO_{1}E$ est un triangle isocèle rectangle en $O_{1}$. De même, $AO_{2}E$ est un triangle isocèle rectangle en $O_{2}$.\n\nComme $O_{1}$ et $O_{2}$ ne sont pas dans le même demi-plan délimité par $(AE)$, on en déduit que $... | France | OFM | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0f0z | Problem:
$f$ is a function on the closed interval $[0, 1]$ with non-negative real values. $f(1) = 1$ and $f(x + y) \geq f(x) + f(y)$ for all $x, y$. Show that $f(x) \leq 2x$ for all $x$. Is it necessarily true that $f(x) \leq 1.9x$ for all $x$? | [
"Solution:\n\nWe have $f(x) = f(1) - f(1 - x) \\leq f(1) = 1$. So for $x \\geq 1/2$, $f(x) \\leq 1 \\leq 2x$.\n\nIf $x < 1/2$, then for some $n$ we have $1/2^{n+1} \\leq x < 1/2^n$. Hence by a trivial induction $f(2^n x) \\geq 2^n f(x)$. But $f(2^n x) \\leq 1$, so $f(x) \\leq 1/2^n \\leq 2x$.\n\nNote that $f(x) = 0... | Soviet Union | ASU | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | Yes, always f(x) ≤ 2x. No, the bound with coefficient 1.9 is not guaranteed; for example, the function that is zero on the left half and one on the right half violates it. | |
034j | Problem:
Prove that for every positive integer $n$ there exist integers $p$ and $q$ such that
$$
\left|p^{2}+2 q^{2}-n\right| \leq \sqrt[4]{9 n}
$$ | [
"Solution:\nLet $q$ be an integer such that\n$$\n2 q^{2} \\leq n < 2(q+1)^{2}\n$$\nThen\n$$\nn - 2 q^{2} < 4q + 2 \\leq 4 \\sqrt{\\frac{n}{2}} + 2 = 2(\\sqrt{2n} + 1)\n$$\nFurther, let $t$ be an integer such that $t^{2} \\leq n - 2 q^{2} < (t+1)^{2}$. We choose $p$ to be either the number $t$ or $t+1$ depending on ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0d0v | Let $ABC$ be a triangle, $I_a$ the excenter opposite to $A$, and $M$ its reflection across $BC$. Prove that $AM$ is parallel to the Euler line of triangle $BCI_a$. | [
"Let $I$ be the incenter of $ABC$, $H_a$ the orthocenter of $I_aBC$, and $O_a$ the midpoint of the segment $II_a$.\n\n\n\nWe have\n$$\nBO_a = IO_a = I_aO_a = CO_a,\n$$\nso $O_a$ is the circumcenter of triangle $I_aBC$.\n\nMoreover, $IB \\parallel CH_a$ (both are perpendicular to $BI_a$) and... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscella... | English | proof only | null | |
0ec2 | Problem:
Iz črk besede LJUBEZNIVOST sestavljamo besede dolžine 4. Črke se ne smejo ponavljati.
a) Koliko različnih besed lahko sestavimo?
b) Koliko različnih besed lahko sestavimo, če uporabimo le samoglasnike?
c) Izračunaj verjetnost, da iz naključno izbranih črk sestavimo besedo NEBO.
d) Izračunaj verjetnost, da... | [] | Slovenia | 15. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | proof and answer | a) 11880; b) 24; c) 1/11880; d) 14/33 | |
0kfj | Problem:
Let $P(x) = x^{2020} + x + 2$, which has $2020$ distinct roots. Let $Q(x)$ be the monic polynomial of degree $\binom{2020}{2}$ whose roots are the pairwise products of the roots of $P(x)$. Let $\alpha$ satisfy $P(\alpha) = 4$. Compute the sum of all possible values of $Q\left(\alpha^{2}\right)^{2}$. | [
"Solution:\nLet $P(x)$ have degree $n = 2020$ with roots $r_{1}, \\ldots, r_{n}$. Let $R(x) = \\prod_{i} \\left(x - r_{i}^{2}\\right)$. Then\n$$\n\\prod_{i} r_{i}^{n} P\\left(\\frac{x}{r_{i}}\\right) = \\prod_{i} \\prod_{j} \\left(x - r_{i} r_{j}\\right) = Q(x)^{2} R(x)\n$$\nUsing $R\\left(x^{2}\\right) = (-1)^{n} ... | United States | HMMT February 2020 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 2020*2^{2019} | |
0kqz | Problem:
For a point $P=(x, y)$ in the Cartesian plane, let $f(P)=\left(x^{2}-y^{2}, 2 x y-y^{2}\right)$. If $S$ is the set of all $P$ so that the sequence $P, f(P), f(f(P)), f(f(f(P))), \ldots$ approaches $(0,0)$, then the area of $S$ can be expressed as $\pi \sqrt{r}$ for some positive real number $r$. Compute $\lflo... | [
"Solution:\nFor a point $P=(x, y)$, let $z(P)=x+y \\omega$, where $\\omega$ is a nontrivial third root of unity. Then\n$$\n\\begin{aligned}\nz(f(P))=\\left(x^{2}-y^{2}\\right)+\\left(2 x y-y^{2}\\right) \\omega=x^{2}+2 x y \\omega+y^{2}( -1-\\omega) \\\\\n& =x^{2}+2 x y \\omega+y^{2} \\omega^{2}=(x+y \\omega)^{2}=z... | United States | HMMT February | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Linear Algebra > Linear transformations",
"Algebra > Linear Algebra > Determinants",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | final answer only | 133 | |
03ma | Show that there exists a positive integer $N$ such that for all integers $a > N$, there exists a contiguous substring of the decimal expansion of $a$ that is divisible by $2011$. (For instance, if $a = 153204$, then $15$, $532$, and $0$ are all contiguous substrings of $a$. Note that $0$ is divisible by $2011$.) | [
"We claim that if the decimal expansion of $a$ has at least $2012$ digits, then $a$ contains the required substring.\n\nLet the decimal expansion of $a$ be $a_k a_{k-1} \\dots a_0$. For $i = 0, \\dots, 2011$, let $b_i$ be the number with decimal expansion $a_i a_{i-1} \\dots a_0$. Then by the pigeonhole principle, ... | Canada | Kanada 2011 | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
01b9 | Prove that, for any positive integers $d$ and $m$, a polynomial of degree $d$ with real coefficients cannot be expressed by a product of $m$ periodic functions. (A function $f: \mathbb{R} \to \mathbb{R}$ is called *periodic* if there exists a constant $T = T(f) > 0$ such that $f(x + T) = f(x)$ for all $x \in \mathbb{R}... | [
"Suppose first that a nonconstant polynomial $p$ has a real root, say, at $x = x_0$ and $p(x) = f_1(x) \\cdots f_m(x)$ with periodic $f_i$, $i = 1, \\dots, m$. Then at least one function $f_i$ is zero at $x = x_0$. If $T = T(f_i) > 0$ is a period of $f_i$, then $f_i(x_0 + kT) = f_i(x_0 + (k-1)T) = \\cdots = f_i(x_0... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0f27 | Problem:
The positive integers $a_1$, $a_2$, ..., $a_m$, $b_1$, $b_2$, ..., $b_n$ satisfy: $a_1 + a_2 + \ldots + a_m = b_1 + b_2 + \ldots + b_n < mn$. Show that we can delete some (but not all) of the numbers so that the sum of the remaining $a$'s equals the sum of the remaining $b$'s. | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0fk5 | Problem:
Demostrar que es imposible obtener un cubo yuxtaponiendo tetraedros regulares, todos del mismo tamaño. | [
"Solution:\nUn tetraedro regular de lado $c$ tiene volumen $\\frac{\\sqrt{2}}{12} c^{3}$ y cada una de sus caras tiene área $\\frac{\\sqrt{3}}{4} c^{2}$.\n\nSupongamos (sin perder generalidad) que un cubo de lado $1$ se puede obtener uniendo $N$ tetraedros regulares de lado $c$. Entonces se satisface\n$$\nN c^{3} \... | Spain | Spanish Mathematical Olympiad - Local Stage | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof only | null | |
0bgs | Problem:
a) Să se arate că
$$
\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{2^{m}}<m
$$
pentru orice $m \in \mathbb{N}^{*}$.
b) Fie $p_{1}, p_{2}, \ldots, p_{n}$ numerele prime mai mici decât $2^{100}$. Să se arate că
$$
\frac{1}{p_{1}}+\frac{1}{p_{2}}+\ldots+\frac{1}{p_{n}}<10
$$ | [
"Solution:\na)\nFolosim inducția matematică.\nPentru $m=1$, avem $\\frac{1}{2}<1$.\nPresupunem că $\\frac{1}{2}+\\frac{1}{3}+\\ldots+\\frac{1}{2^{m}}<m$.\nAtunci:\n$$\n\\frac{1}{2}+\\frac{1}{3}+\\ldots+\\frac{1}{2^{m}}+\\frac{1}{2^{m}+1}+\\ldots+\\frac{1}{2^{m+1}}<m+1\n$$\nObservăm că sunt $2^{m}$ termeni de la $\\... | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0l56 | Problem:
A number is upwards if its digits in base 10 are nondecreasing when read from left to right. Compute the number of positive integers less than $10^{6}$ that are both upwards and multiples of 11. | [
"Solution:\nFor a number $d_{5}d_{4}d_{3}d_{2}d_{1}d_{0}$ (allowing leading 0s) to be upwards and a multiple of 11, we must have\n$$d_{5} \\leq d_{4} \\leq d_{3} \\leq d_{2} \\leq d_{1} \\leq d_{0},$$\n$$d_{0} - d_{1} + d_{2} - d_{3} + d_{4} - d_{5} \\equiv 0 \\pmod{11}.$$ \nNote that $d_{0} - d_{1}$, $d_{2} - d_{3... | United States | HMMT February | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics"
] | null | final answer only | 219 | |
0l8o | Let $x$, $y$, $z$ be positive real numbers satisfying the following conditions:
$$
\frac{1}{\sqrt{2}} \leq z < \frac{1}{2} \min \{x\sqrt{2},\ y\sqrt{3}\}, \\
x + z\sqrt{3} \geq \sqrt{6}, \\
y\sqrt{3} + z\sqrt{10} \geq 2\sqrt{5}.
$$
Find the greatest value of the expression:
$$
P(x, y, z) = \frac{1}{x^2} + \frac{2}{y^... | [
"By denoting $\\frac{1}{x\\sqrt{2}} = a$, $\\frac{1}{y\\sqrt{3}} = b$, $\\frac{1}{2z} = c$, the problem becomes:\nFind the greatest value of the expression\n$$\nQ(a, b, c) = 2a^2 + 6b^2 + 12c^2\n$$\nwhere $a$, $b$, $c$ are positive numbers satisfying the conditions:\n$$\n\\max\\{a, b\\} < c \\leq \\frac{1}{\\sqrt{2... | Vietnam | VIETNAMESE MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 118/15 | |
05ra | Problem:
Pour $m$ entier positif, on note $d(m)$ le nombre de diviseurs positifs de $m$ (1 et $m$ compris). Soit $k$ un entier strictement positif. Montrer qu'il existe une infinité d'entiers positifs $n$ tels que $n$ ait exactement $k$ diviseurs premiers distincts et tel que pour tout $a, b$ entiers strictement posit... | [
"Solution:\n\nProuvons que chaque entier de la forme $n = m 2^{p-1}$ avec $p$ premier impair, possédant $k-1$ facteurs premiers strictement plus grands que $3$ et vérifiant $(5 / 4)^{(p-1) / 2} > m$ est solution.\n\nSi $a + b = n$ et $d(n) \\mid d\\left(a^{2} + b^{2}\\right)$ alors $p \\mid d\\left(a^{2} + b^{2}\\r... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
00ok | Anna, Berta and Clara write the square numbers $1, 4, 9, \dots, 2025$ on a blackboard, compute their sum and observe that it is divisible by $3$. Then, they agree to the following game: In each round, Anna will cross out one number, then Berta will do the same, and then Clara will do the same. This continues until all ... | [
"On the blackboard, we have $15$ integers with residue $0$ modulo $3$ and $30$ integers with residue $1$ modulo $3$. If, in a certain round, Berta and Clara cross out numbers that have the same residue modulo $3$ as the number crossed out by Anna, then they have removed either $0+0+0$ or $1+1+1$ modulo $3$.\n\nIn b... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
0fid | Problem:
Hallar todos los números naturales de cuatro cifras, escritos en base 10, que sean iguales al cubo de la suma de sus cifras. | [
"Solution:\n\nSea $n$ un número que cumpla las condiciones del enunciado, y $s$ la suma de sus cifras. Como $1000 \\leq n \\leq 9999$ y $n = s^{3}$, resulta\n$$\n11 \\leq s \\leq 21\n$$\nSi $n = x y z t$, tenemos\n$$\n\\left.\\begin{array}{r}\n1000 x + 100 y + 10 z + t = s^{3} \\\\\nx + y + z + t = s\n\\end{array}\... | Spain | Olimpiada Matemática Española | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 4913, 5832 | |
0jt8 | Problem:
On the Cartesian plane $\mathbb{R}^{2}$, a circle is said to be nice if its center is at the origin $(0,0)$ and it passes through at least one lattice point (i.e. a point with integer coordinates). Define the points $A=(20,15)$ and $B=(20,16)$. How many nice circles intersect the open segment $AB$?
For refer... | [
"Solution:\n\nThe square of the radius of a nice circle is the sum of the square of two integers.\nThe nice circle of radius $r$ intersects (the open segment) $\\overline{AB}$ if and only if a point on $\\overline{AB}$ is a distance $r$ from the origin. $\\overline{AB}$ consists of the points $(20, t)$ where $t$ ra... | United States | HMMT February | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order anal... | null | proof and answer | 10 | |
0fmk | Demostrar que para cualquier par de enteros positivos $k$ y $n$, existen $k$ enteros positivos $m_1, m_2, \dots, m_k$ (no necesariamente distintos) tales que
$$
1 + \frac{2^k - 1}{n} = \left(1 + \frac{1}{m_1}\right) \left(1 + \frac{1}{m_2}\right) \dots \left(1 + \frac{1}{m_k}\right)
$$ | [
"**Solución por Daniel Lasaosa Medarde, Pamplona, España.** Sea $(u_1, u_2, \\dots, u_k)$ una permutación de $(0, 1, 2, \\dots, k-1)$. Consideremos el producto\n$$\n\\begin{aligned}\n& \\frac{n+2^{u_1}}{n} \\cdot \\frac{n+2^{u_1}+2^{u_2}}{n+2^{u_1}} \\cdots \\frac{n+2^{u_1}+2^{u_2}+\\cdots+2^{u_k}}{n+2^{u_1}+2^{u_2... | Spain | Olimpiada Internacional de Matemáticas | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | Spanish | proof only | null | |
007d | In triangle $ABC$ point $P$ divides side $AB$ in ratio $\frac{AP}{PB} = \frac{1}{4}$. The perpendicular bisector of segment $PB$ intersects side $BC$ at point $Q$. If $\text{area}(PQC) = \frac{4}{25}\text{area}(ABC)$ and $AC = 7$, find $BC$. | [
"If $\\text{area}(ABC) = S$ then $\\text{area}(APC) = \\frac{AP}{AB}S = \\frac{1}{5}S$. As $\\text{area}(PQC) = \\frac{4}{25}S$, we have $\\text{area}(PQB) = S - \\frac{1}{5}S - \\frac{4}{25}S = \\frac{16}{25}S$. On the other hand\n$$\n\\text{area}(PBQ) = \\frac{BQ}{BC}\\text{area}(PBC) = \\frac{BQ}{BC} \\cdot \\fr... | Argentina | National Olympiad of Argentina | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 7 | |
0inh | Let $a$ and $b$ be positive integers. Show that if $4ab - 1$ divides $(4a^2 - 1)^2$, then $a = b$. | [
"Call a pair $(a, b)$ of positive integers *bad* if $4ab - 1$ divides $(4a^2 - 1)^2$ but $a \\neq b$. In order to prove that bad pairs do not exist, we present two properties of them which provide an infinite descent.\n\n* **Property (i):** If $(a, b)$ is a bad pair then $(b, a)$ is also bad.\nIndeed, we consider... | United States | IMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
05k2 | Problem:
Pour tout entier $n \geqslant 1$, on pose $S_{n} = x_{1} + x_{2} + \cdots + x_{n}$, où $x_{k} = \frac{k(k+1)}{2}$ pour tout $k \geqslant 1$. Prouver que, pour tout $n \geqslant 10$, il existe un entier $a_{n}$ tel que $S_{n-1} < a_{n}^{2} < S_{n}$. | [
"Solution:\n\nPour tout entier $n \\geqslant 1$, on a\n$$\nS_{n} = \\sum_{k=1}^{n} \\frac{k(k+1)}{2} = \\frac{1}{2} \\sum_{k=1}^{n} k^{2} + \\frac{1}{2} \\sum_{k=1}^{n} k = \\frac{n(n+1)(2n+1)}{12} + \\frac{n(n+1)}{4} = \\frac{n(n+1)(n+2)}{6}\n$$\nIl suffit donc de prouver que, pour $n \\geqslant 10$, on a $\\sqrt{... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
09vi | Problem:
Zij $ABC$ een scherphoekige driehoek met $O$ het middelpunt van de omgeschreven cirkel. Punt $Q$ ligt op de omgeschreven cirkel van $\triangle BOC$ zodat $OQ$ een middellijn is. Punt $M$ ligt op $CQ$ en punt $N$ ligt inwendig op lijnstuk $BC$ zodat $ANC M$ een parallellogram is. Bewijs dat de omgeschreven cir... | [
"Solution:\n\nDefinieer $T$ als het snijpunt van de omgeschreven cirkel van $\\triangle BOC$ en de lijn $AQ$. We gaan bewijzen dat $T$ op $NM$ ligt. Schrijf $\\alpha=\\angle BAC$. Dan geldt $\\angle BOC=2\\alpha$ vanwege de middelpuntsomtrekshoekstelling. Aangezien $|OB|=|OC|$ en $\\angle OCQ=90^{\\circ}=\\angle OB... | Netherlands | IMO-selectietoets II | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0f03 | Problem:
$P$ is a convex polygon and $X$ is an interior point such that for every pair of vertices $A$, $B$, the triangle $XAB$ is isosceles. Prove that all the vertices of $P$ lie on some circle center $X$. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
023c | Problem:
Duas frações irredutíveis têm seus denominadores iguais a 600 e 700. Encontrar o valor mínimo para o denominador da soma das frações. | [
"Solution:\n\nSuponhamos que as frações são $\\frac{a}{600}$ e $\\frac{b}{700}$. Como são irredutíveis, então $a$ e $600$ não têm fator comum maior que $1$ e o mesmo acontece com $b$ e $700$.\n\nSomando as duas frações obtemos\n$$\n\\frac{a}{600} + \\frac{b}{700} = \\frac{7a + 6b}{4200} = \\frac{7a + 6b}{2^{3} \\ti... | Brazil | Nível 3 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modu... | null | proof and answer | 168 | |
0jcv | Problem:
How many real triples $(a, b, c)$ are there such that the polynomial $p(x)=x^{4}+a x^{3}+b x^{2}+a x+c$ has exactly three distinct roots, which are equal to $\tan y, \tan 2 y$, and $\tan 3 y$ for some real $y$? | [
"Solution:\n\nAnswer: 18\n\nLet $p$ have roots $r, r, s, t$. Using Vieta's on the coefficient of the cubic and linear terms, we see that $2 r + s + t = r^{2} s + r^{2} t + 2 r s t$. Rearranging gives $2 r(1 - s t) = (r^{2} - 1)(s + t)$.\n\nIf $r^{2} - 1 = 0$, then since $r \\neq 0$, we require that $1 - s t = 0$ fo... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 18 | |
0bth | Let $k$ and $n$ be integers such that $k \ge 2$ and $k \le n \le 2k-1$. Place rectangular tiles, each of size $1 \times k$ or $k \times 1$, on an $n \times n$ chessboard so that each tile covers exactly $k$ cells, and no two tiles overlap. Do this until no further tile can be placed in this way. For each such $k$ and $... | [
"The required minimum is $n$ if $n = k$, and it is $\\min(n, 2n-2k+2)$ if $k < n < 2k$.\n\nThe case $n=k$ being clear, assume henceforth $k < n < 2k$. Begin by describing maximal arrangements on the board $[0, n] \\times [0, n]$, having the above mentioned cardinalities.\n\nIf $k < n < 2k-1$, then $\\min(n, 2n-2k+2... | Romania | 2016 European Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Minimum number of tiles = n if n = k; and min(n, 2n − 2k + 2) if k < n < 2k. | |
08bm | Problem:
Un testo antico dichiara che Matusalemme visse 150 anni, dove il simbolo $\ast$ sostituisce la cifra delle unità, che gli studiosi non riescono a leggere. Fortunatamente siamo in possesso di altri tre manoscritti sulla vita di Matusalemme; il primo sostiene che egli visse un numero pari di anni, il secondo ch... | [
"Solution:\n\n(a) $150\\ast$ è multiplo di 2 e 3, ma non di 5;\n(b) $150\\ast$ è multiplo di 2 e 5, ma non di 3;\n(c) $150\\ast$ è multiplo di 3 e 5, ma non di 2.\n\nCome prima cosa notiamo che $1500$ è multiplo di $2 \\cdot 3 \\cdot 5 = 30$, e non ci sono altri multipli di $10$ o $15$ tra $1500$ e $1509$. Gli ulti... | Italy | Gara di Febbraio | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Modular Arithmetic"
] | null | MCQ | B | |
0jqi | Problem:
Count the number of functions $f: \mathbb{Z} \rightarrow\{\text{'green'}, \text{'blue'}\}$ such that $f(x)=f(x+22)$ for all integers $x$ and there does not exist an integer $y$ with $f(y)=f(y+2)=\text{'green'}$. | [
"Solution:\n\nAnswer: $39601$\n\nIt is clear that $f$ is determined by $f(0), \\ldots, f(21)$. The colors of the 11 even integers are independent of those of the odd integers because evens and odds are never exactly 2 apart.\n\nFirst, we count the number of ways to \"color\" the even integers. $f(0)$ can either be ... | United States | HMMT February 2015 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 39601 | |
0l1k | Problem:
The graph of the equation $\tan (x+y)=\tan (x)+2 \tan (y)$, with its pointwise holes filled in, partitions the coordinate plane into congruent regions. Compute the perimeter of one of these regions.
Proposed by: Karthik Venkata Vedula | [
"Solution:\n\nWe manipulate the given equation as follows:\n\n$$\n\\begin{aligned}\n\\tan (x+y) &= \\tan x + 2 \\tan y \\\\\n\\frac{\\tan x + \\tan y}{1 - \\tan x \\tan y} &= \\tan x + 2 \\tan y \\\\\n\\tan x + \\tan y &= (\\tan x + 2 \\tan y) - \\tan x \\tan y (\\tan x + 2 \\tan y) \\\\\n\\tan x \\tan y (\\tan x +... | United States | HMMT November 2024 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | π(√5+1) |
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