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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0ga4 | 設兩圓 $O_1$ 與 $O_2$ 交於 $B, C$ 兩點, 其中 $BC$ 為圓 $O_1$ 的直徑。由 $C$ 點作圓 $O_1$ 的切線, 又交於 $O_2$ 於點 $A$。設直線 $AB$ 又交 $O_1$ 於點 $E$。作直線 $CE$, 設 $CE$ 又交圓 $O_2$ 於點 $F$。自線段 $AF$ 上任取一點 $H$。設直線 $HE$ 又交圓 $O_1$ 於點 $G$, 而直線 $BG$ 與直線 $AC$ 交於點 $D$。
證明：$\frac{AH}{HF} = \frac{AC}{CD}$。
Circles $O_1$ and $O_2$ intersect at two points $B$ and $C$,... | [
"由於 $BC$ 是圓 $O_1$ 的直徑, $ACD$ 是 $O_1$ 的切線, 所以 $BC \\perp AD$, $\\angle ACB = 90^\\circ$。故 $AB$ 為圓 $O_2$ 的直徑。\n因為 $\\angle BEC = 90^\\circ$, 所以 $AB \\perp CF$, 故 $\\angle FAB = \\angle CAB$。\n\n\n\n連 $CG$, 知 $CG \\perp BD$。故\n$$\n\\angle ADB = \\angle BCG = \\angle BEG = \\angle AEH,\n$$\n所以 ... | Taiwan | 二〇一六數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05xf | Problem:
Soit $ABC$ un triangle avec $\widehat{BAC}=60^{\circ}$ et soit $\Gamma$ son cercle circonscrit. Soient $H$ l'orthocentre de $ABC$ et $S$ le milieu de l'arc $\widehat{BC}$ ne contenant pas $A$. Soit $P$ le point de $\Gamma$ tel que $\widehat{SPH}=90^{\circ}$. Montrer qu'il existe un cercle passant par $P, S$ e... | [
"Solution:\n\n\n\nD'après le théorème du pôle Sud, $S$ est le point de concours de la bissectrice de $\\widehat{BAC}$ et de la médiatrice de $[BC]$. Soit $O$ le centre de $\\Gamma$ et $N$ le pôle Nord. Comme $[SN]$ est un diamètre de $\\Gamma$, la condition $\\widehat{SPH}=90^{\\circ}$ se r... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | null | proof only | null | |
0ech | Problem:
V trgovini z oblačili imajo dva tedna akcijsko razprodajo. Prvi teden je kupec pri nakupu treh oblačil dobil najcenejši kos zastonj. Janez je prvi teden kupil jakno, hlače in pulover. Za jakno in hlače je plačal 115,01 evra ter zaradi te akcije prihranil $17,85 \%$ vrednosti oblačil v redni prodaji. Drugi ted... | [] | Slovenia | 15. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | proof and answer | Sweater: 24.99 euros, Pants: 40.01 euros, Jacket: 75.00 euros | |
083q | Problem:
Il professor Bianchi non dice mai bugie, tranne un giorno della settimana (sempre lo stesso) in cui mente sempre. Quanti sono i giorni della settimana in cui può aver affermato: "se non ho detto bugie ieri ne dirò certamente domani"?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4 . | [] | Italy | Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO TRIENNIO | [
"Discrete Mathematics > Logic"
] | null | MCQ | D | |
06je | $ABCD$ is a trapezium with $AB \parallel CD$. $M$ and $N$ are the midpoints of $AB$ and $CD$ respectively. If $AC = 6$, $BD = 8$ and $MN = 4$, find the area of $ABCD$. | [
"We use $[X_1X_2\\cdots X_k]$ to denote the area of the $k$-sided polygon $X_1X_2\\cdots X_k$. Let $K$ and $L$ be the midpoints of $AD$ and $BC$ respectively. Then $\\triangle DKN \\sim \\triangle DAC$ with side-length ratio $1:2$ and so $[DKN] = \\frac{1}{4}[DAC]$. Similarly, we have $[AMK] = \\frac{1}{4}[ABD]$, $... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 3√55 | |
02aj | Problem:
No planeta Staurus, os anos têm $228$ dias ($12$ meses de $19$ dias). Cada semana tem $8$ dias: Zerum, Uni, Duodi, Trio, Quati, Quio, Seise e Sadi. Sybock nasceu num duodi que foi o primeiro dia do quarto mês. Que dia da semana ele festejará seu primeiro aniversário? | [
"Solution:\n\nSeise"
] | Brazil | null | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | Seise | |
0hy5 | Problem:
A parabola is inscribed in equilateral triangle $ABC$ of side length $1$ in the sense that $AC$ and $BC$ are tangent to the parabola at $A$ and $B$, respectively. Find the area between $AB$ and the parabola. | [
"Solution:\nSuppose $A = (0, 0)$, $B = (1, 0)$, and $C = \\left(\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right)$. Then the parabola in question goes through $(0, 0)$ and $(1, 0)$ and has tangents with slopes of $\\sqrt{3}$ and $-\\sqrt{3}$, respectively, at these points. Suppose the parabola has equation $y = a x^{2} + ... | United States | HMMT | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | sqrt(3)/6 | |
0l6z | Problem:
Let $\triangle ABC$ be an equilateral triangle. Point $D$ lies on segment $\overline{BC}$ such that $BD = 1$ and $DC = 4$. Points $E$ and $F$ lie on rays $\overrightarrow{AC}$ and $\overrightarrow{AB}$, respectively, such that $D$ is the midpoint of $\overline{EF}$. Compute $EF$. | [
"Solution:\n\nLet $C'$ be the reflection of $C$ over $D$. Then, $\\overline{EC} \\parallel \\overline{C'F}$ since $ECFC'$ is a parallelogram. Thus, $BFC'$ is an equilateral triangle, so $BF = BC' = 3$ and $\\angle FBD = 120^\\circ$. By Law of Cosines, we get $DF = \\sqrt{3^2 + 3 \\cdot 1 + ... | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 2*sqrt(13) | |
0a3q | Is it possible for an integer of the form $44\ldots41$ – consisting of an odd number of fours followed by a $1$ – to be a square? | [
"It is impossible for such an integer to be square.\nTo show this, note that such an integer is of the form $a_m = 4 \\cdot \\frac{10^{2m}-1}{9} - 3$\nwith $m \\ge 1$ an integer. As $10^{2m}-1 = (10^2-1)(10^{2m-2} + 10^{2m-4} + \\dots + 1)$,\nwe have $99 = 10^2 - 1 \\mid 10^{2m} - 1$, therefore $11 \\mid \\frac{10^... | Netherlands | IMO Team Selection Test 2 | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0g9e | 給定任意三角形 $\triangle ABC$, 令其外接圓為 $O_1$, 九點圓為 $O_2$; 並令以 $\triangle ABC$ 的垂心 $H$ 與重心 $G$ 為直徑的圓為 $O_3$. 證明 $O_1, O_2, O_3$ 共軸。(即存在一直線, 其上的點對這三個圓的圓幂均相同。點對圓的圓幂, 是點到圓心的距離平方, 與圓半徑的平方之差。)
註: 三角形的九點圓, 即通過三邊中點、三高垂足、三頂點分別與垂心連線的中點等九個點的圓。 | [
"引理：垂心與重心為 $O_1, O_2$ 的兩位似中心。\n\n引理證明：九點圓過 $\\overline{HA}, \\overline{HB}, \\overline{HC}$ 中點, 故 $H$ 為一位似中心。\n九點圓過 $\\overline{AB}, \\overline{BC}, \\overline{CA}$ 中點 (分別設為 $M, N, P$), 且 $\\overline{AG} : \\overline{GM} = 2 : 1$, $\\overline{BG} : \\overline{GN} = 2 : 1$, $\\overline{CG} : \\overline{GP} = 2 : 1$,... | Taiwan | 二〇一五數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circ... | null | proof only | null | |
0f76 | Problem:
$A$ and $B$ are fixed points outside a sphere $S$. $X$ and $Y$ are chosen so that $S$ is inscribed in the tetrahedron $ABXY$. Show that the sum of the angles $AXB$, $XBY$, $BYA$ and $YAX$ is independent of $X$ and $Y$. | [] | Soviet Union | 20th ASU | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01mv | Several chess players took part in a chess tournament. Each participant played exactly one game with any other participant. A participant received 1 point for a win, 0.5 point for a draw, and 0 point for a lose. Any two players received different numbers of points.
What is the smallest number of the wins of the partici... | [] | Belarus | 61st Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Let n be the number of participants.
- Smallest number of wins of the first-place participant: ceil((n−1)/2).
- Greatest possible points of the last-place participant: floor((n−1)/2)/2. | |
0cdf | Let $f: \mathbb{R} \to \mathbb{R}$ be a function with the property that there are a differentiable function $g: \mathbb{R} \to \mathbb{R}$ and a sequence $(a_n)_{n \ge 1}$ with strictly positive terms and $\lim_{n \to \infty} a_n = 0$, such that
$$
g'(x) = \lim_{n \to \infty} \frac{f(x + a_n) - f(x)}{a_n},
$$
for all $... | [
"a) Take the function $f : \\mathbb{R} \\to \\mathbb{R}$ defined by\n$$\nf(x) = \\begin{cases} 1, & x \\in \\mathbb{Q}, \\\\ 0, & x \\in \\mathbb{R} \\setminus \\mathbb{Q}, \\end{cases}\n$$\nand let $(a_n)_{n \\ge 1}$ be the sequence with the terms $a_n = 1/n, \\forall n \\in \\mathbb{N}^*$. For $n \\in \\mathbb{N}... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Functions"
] | null | proof only | null | |
0bnd | Let $x$, $y$, $z > 0$. Prove that
$$
\frac{x^3}{z^3 + x^2 y} + \frac{y^3}{x^3 + y^2 z} + \frac{z^3}{y^3 + z^2 x} \ge \frac{3}{2}.
$$ | [
"From the AM-GM inequality we get $x^2 y \\le \\frac{x^3 + x^3 + y^3}{3}$, which leads to\n$$\n\\frac{x^3}{z^3 + x^2 y} \\ge \\frac{x^3}{z^3 + \\frac{x^3 + x^3 + y^3}{3}} = \\frac{3x^3}{2x^3 + y^3 + 3z^3}.\n$$\nSumming this with the other two similar inequalities and putting $x^3 = a$, $y^3 = b$, $z^3 = c$, it is s... | Romania | 66th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0hxf | Problem:
Let $a_{1}=1$, $a_{2}=2$, and for $n \geq 3$, let $a_{n}$ be the smallest positive integer such that $a_{n} \neq a_{i}$ for $i<n$ and $\operatorname{gcd}\left(a_{n}, a_{n-1}\right)>1$. Prove that every positive integer appears as some $a_{i}$. | [
"Solution:\n\nOur solution will proceed in three steps.\n\nStep 1. We show that there is a prime $p$ such that infinitely many of the terms $a_{i}$ are divisible by $p$.\n\nProof. Suppose that such a prime $p$ does not exist. In particular, taking $p=2$, we find that there is an $N_{1}$ such that for all $i \\geq N... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
09ee | In a triangle $KLM$, $KM = 6$ and let $LP$ be a median of the $KLM$. Let $P_1$, $P_2$ be bases of perpendiculars dropped from $P$ to sides $KL$, $LM$ respectively. If $PP_1 : PP_2 = 1 : 2$ and area of the triangle $KLM$ is maximal then find the length of the median $LP$. | [
"Since $LP$ is median,\n$$\n\\begin{aligned}\n\\frac{S_{\\Delta KLP}}{2} &= \\frac{S_{\\Delta LPM}}{2} \\Leftrightarrow \\\\\n\\frac{KL \\cdot PP_1}{2} &= \\frac{ML \\cdot PP_2}{2} \\Leftrightarrow KL: \\\\\nLM &= PP_2 : PP_1 = 2 : 1.\n\\end{aligned}\n$$\n\n\n\nMoreover, if point $L'$ coinc... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | √41 | |
0e61 | Let $ABCDE$ be a cyclic pentagon in which $|CD| = |DE|$. Let the diagonals $AD$ and $BE$ intersect at point $K$, and let the diagonals $AC$ and $BD$ intersect at point $L$. Prove that the lines $KL$ and $EC$ are parallel. | [
"According to the inscribed angle theorem, we have $\\angle CED = \\angle CAD$ and $\\angle DCE = \\angle DBE$. Because $ECD$ is an isosceles triangle with the top angle at $D$, we have $\\angle CED = \\angle DCE$. Due to the collinearity of the points $A, L, C$ and the collinearity of the points $A, K, D$ we also ... | Slovenia | National Math Olympiad 2012 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00w0 | Let $k > 2$ be a given positive integer. Find all positive integers $d$, for which there exists a polynomial $P(x)$ with integer coefficients such that $\deg P(x) = d$ and $11^k \mid 2025^n + P(n)$ for all positive integers $n > k$. | [
"First, we will prove the following lemma:\n*Lemma.* Let $f$ be a polynomial with rational coefficients such that $f(n)$ is an integer for any integer $n$. Then there exist integers $a_0, a_1, \\dots, a_p$ such that $f(x) = \\sum_{i=0}^{p} a_i \\binom{x}{i}$.\n*Proof.* Let us first prove that for any polynomial wit... | Balkan Mathematical Olympiad | 42nd Balkan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | English | proof and answer | d ≥ k − 1 | |
01au | $$
3 \cdot 5^x - 2 \cdot 6^y = 3
$$
in positive integers $x$, $y$. | [
"After dividing the equation by $3$ we get:\n$$\n5^x - 1 = 4 \\cdot 6^{y-1}\n$$\nFor $y > 2$ the right side of equation is divisible by $9$. Then $x$ would have to be divisible by $6$ (analysis of residues modulo $9$ of powers of $5$). Then the left side of equation would be divisible by $7$, which is impossible. T... | Baltic Way | Baltic Way 2013 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | [(1, 1), (2, 2)] | |
00pa | The opposite sides of a convex hexagon of unit area are pairwise parallel. The lines of support of three alternate sides meet pairwise to form a triangle. Similarly, the lines of support of the other three alternate sides meet pairwise to form another triangle. Show that the area of at least one of these two triangles ... | [
"Unless otherwise stated, throughout the proof indices take on values from $0$ to $5$ and are reduced modulo $6$. Label the vertices of the hexagon in circular order, $A_0, A_1, \\dots, A_5$, and let the lines of support of the alternate sides $A_iA_{i+1}$ and $A_{i+2}A_{i+3}$ meet at $B_i$. To show that the area o... | Balkan Mathematical Olympiad | shortlistBMO 2011 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance... | English | proof only | null | |
024t | Problem:
Uma formiga parte de um vértice de um cubo, andando somente ao longo das arestas, até voltar ao vértice inicial, não passando duas vezes por nenhum vértice. Qual é o passeio de maior comprimento que essa formiga pode fazer? | [
"Solution:\n\nNa figura temos um caminho percorrendo oito arestas que a formiga pode fazer partindo do vértice identificado como 1.\n\n\n\nSerá possível ela fazer um caminho passando por nove arestas? Para fazer esse caminho, ela teria que passar por nove vértices, pois o vértice de chegada... | Brazil | Nível 2 | [
"Discrete Mathematics > Graph Theory",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | 8 | |
09m7 | Let $I$ be the center of the circumcircle of an inscribed quadrilateral $ABCD$ with $\angle BAD < 90^\circ$. Points $E$ and $F$ lie on segments $BI$ and $DI$, respectively, such that $\angle EAF = \angle BAI$. Let $O$ be the center of the circumcircle of triangle $ABD$. $Q$ is the symmetric point of $O$ with respect to... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinate... | English | proof only | null | |
0lf8 | Find the maximum value of the positive real number $k$ such that the inequality
$$
\frac{1}{kab + c^2} + \frac{1}{kbc + a^2} + \frac{1}{kca + b^2} \ge \frac{k+3}{a^2 + b^2 + c^2}
$$
holds for all positive real numbers $a, b, c$ such that $a^2 + b^2 + c^2 = 2(ab + bc + ca)$. | [
"Let $a \\to 0^+$ and $b = c = 1$, one can get\n$$\n2 + \\frac{1}{k} \\geq \\frac{k+3}{2}\n$$\nso $k \\leq 2$. For $k = 2$, we need to prove that\n$$\n\\frac{1}{2ab + c^2} + \\frac{1}{2bc + a^2} + \\frac{1}{2ca + b^2} \\geq \\frac{5}{a^2 + b^2 + c^2}\n$$\nis true for all positive triples $(a, b, c)$ satisfying $a^2... | Vietnam | Vietnamese Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | 2 | |
053z | An inventor presented to the king a new exciting board game on a $9 \times 10$ squared board. The king promised to reward him one rice grain for the first square, one rice grain for the second square, and for each following square the same number of grains as for the two preceding squares together. Prove that for the l... | [
"Enumerate all squares with $1, \\ldots, 90$. Let the number of rice grains promised for the $n$-th square be $F_n$; then according to the problem $F_1 = F_2 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for all $n > 2$. Notice that $F_n > F_{n-1}$ if $n > 2$, hence $F_{2(n+1)} = F_{2n+2} = F_{2n+1} + F_{2n} > F_{2n} + F_{2n}... | Estonia | Estonian Math Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
08m4 | Problem:
Let $ABCD$ be a parallelogram with $AC > BD$, and let $O$ be the point of intersection of $AC$ and $BD$. The circle with center at $O$ and radius $OA$ intersects the extensions of $AD$ and $AB$ at points $G$ and $L$, respectively. Let $Z$ be the intersection point of lines $BD$ and $GL$. Prove that $\angle ZCA... | [
"Solution:\nFrom the point $L$ we draw a parallel line to $BD$ that intersects lines $AC$ and $AG$ at points $N$ and $R$ respectively. Since $DO = OB$, we have that $NR = NL$, and point $N$ is the midpoint of segment $LR$.\n\nLet $K$ be the midpoint of $GL$. Now, $NK \\parallel RG$, and\n$$\n\\angle AGL = \\angle N... | JBMO | 2009 Shortlist JBMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
00yt | Problem:
In the triangle $A B C$, let $l$ be the bisector of the external angle at $C$. The line through the midpoint $O$ of the segment $A B$ parallel to $l$ meets the line $A C$ at $E$. Determine $|C E|$, if $|A C|=7$ and $|C B|=4$. | [
"Solution:\n\nLet $F$ be the intersection point of $l$ and the line $A B$. Since $|A C| > |B C|$, the point $E$ lies on the segment $A C$, and $F$ lies on the ray $A B$. Let the line through $B$ parallel to $A C$ meet $C F$ at $G$. Then the triangles $A F C$ and $B F G$ are similar. Moreover, we have $\\angle B G C... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 11/2 | |
06gg | Let $ABCD$ be a convex quadrilateral. $E$ and $F$ are points on the diagonal $AC$ such that $E$ and $F$ are interior points of triangles $ABD$ and $BCD$ respectively. Suppose $BE$ extend cuts $AD$ at $P$, $DE$ extend cuts $AB$ at $Q$, $DF$ extend cuts $BC$ at $R$ and $BF$ extend cuts $DC$ at $S$. Show that the three li... | [
"Let $Y$ be the intersection point of $AC$ and $BD$. Since $AY$, $BP$, $DQ$ are concurrent, $QP$ meets $BD$ at the harmonic conjugate $X$ of $Y$ with respect to $B$, $D$ (note that $X$ could be a point at infinity). Similarly, since $CY$, $BS$, $DR$ are concurrent, $RS$ meets $BD$ at $X$. This shows $QP$, $BD$, $RS... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | null | proof only | null | |
03bu | Find all positive integers $n$ such that there exists a polynomial $f$ of degree $n$ with integer coefficients and a positive leading coefficient and a polynomial $g$ with integer coefficients such that the equality
$$
xf^2(x) + f(x) = (x^3 - x)g^2(x)
$$
holds for every real $x$. | [
"We have $xf^2(x) + f(x) = (x^3 - x)g^2(x) \\Leftrightarrow [2xf(x) + 1]^2 = (x^2 - 1)[2xg(x)]^2 + 1$.\nWe will now find all pairs $(p, q)$ of integer-coefficient polynomials such that $p^2(x) = (x^2 - 1)q^2(x) + 1$.\nLet $(p, q)$ be one such pair such that the degree of $q$ is $k \\ge 1$. We can assume, without lo... | Bulgaria | 55th IMO Team Selection Test | [
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials",
"Number Theory > Diophantine Equations > Pell's equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | n = 4k + 3 for k ≥ 0 | |
0fgx | Problem:
Es muy conocido el puzzle consistente en descomponer la cruz griega de la izquierda de la figura en cuatro partes con las que se pueda componer un cuadrado. Una solución habitual es la de la figura de la derecha. Demostrar que hay una infinidad de soluciones
diferentes.
¿Hay algun... | [
"Solution:\n\n\n\nConsideremos el \"embaldosado\" del plano con cruces congruentes. Teniendo en cuenta las traslaciones de vectores $\\vec{u}$ y $\\vec{v}$ como generadores del embaldosado, observamos que el cuadrado de lados $\\vec{u}$ y $\\vec{v}$ aplicado en cualquier punto del plano es ... | Spain | OME 24 | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | There are infinitely many dissections; yes, there exists a solution with four equal parts. | |
0f2n | Problem:
Let $n > 3$ be an integer. Let $S$ be the set of lattice points $(a, b)$ with $0 \leq a, b < n$. Show that we can choose $n$ points of $S$ so that no three chosen points are collinear and no four chosen points form a parallelogram. | [] | Soviet Union | ASU | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0bcz | Let $a, b \in \mathbb{R}$ and $z \in \mathbb{C} \setminus \mathbb{R}$ such that $|a - b| = |a + b - 2z|$.
a. Prove that there exists an unique real number $x$ which satisfies $|z - a|^x + |\bar{z} - b|^x = |a - b|^x$.
b. Find all real numbers $x$ such that $|z - a|^x + |\bar{z} - b|^x \le |a - b|^x$. | [
"a. Set $u = z - a$, $v = z - b$. The relation gives $|v - u| = |u + v|$ where $u, v, u + v \\in \\mathbb{C} \\setminus \\mathbb{R}$, so $u, v, u + v \\neq 0$. Thus $|u + v|^2 = |u|^2 + |v|^2$. Since $|v| = |\\bar{v}|$, the equation is written successively $|u|^x + |v|^x = (\\sqrt{|u|^2 + |v|^2})^x$ and then $\\lef... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | a: x = 2; b: [2, +infty) | |
02h2 | Let $ABC$ be a triangle. The internal bisector of $\angle B$ meets $AC$ in $P$ and $I$ is the incenter of $ABC$. Prove that if $AP + AB = CB$, then $API$ is an isosceles triangle. | [
"Draw $PP'$ parallel to $IA$ so that $P'$ is on line $AB$. Then $\\triangle PAP'$ is isosceles, which implies that $BC = AB + AP = AB + AP' = BP'$. This then implies that $\\triangle P'BC$ is isosceles, which in turn implies that, since $P$ is on the angle bisector of $\\angle B$, $P'PC$ is also isosceles, with $PP... | Brazil | Brazilian Math Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Co... | English | proof only | null | |
0atw | Problem:
Determine, with proof, the least positive integer $n$ for which there exist $n$ distinct positive integers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ such that
$$
\left(1-\frac{1}{x_{1}}\right)\left(1-\frac{1}{x_{2}}\right)\left(1-\frac{1}{x_{3}}\right) \cdots\left(1-\frac{1}{x_{n}}\right)=\frac{15}{2013}
$$ | [
"Solution:\n\nSuppose $x_{1}, x_{2}, x_{3}, \\ldots, x_{n}$ are distinct positive integers that satisfy the given equation. Without loss of generality, we assume that $x_{1}<x_{2}<x_{3}<\\cdots<x_{n}$. Then\n$$\n2 \\leq x_{1} \\leq x_{2}-1 \\leq x_{3}-2 \\leq \\cdots \\leq x_{n}-(n-1)\n$$\nand so $x_{i} \\geq i+1$ ... | Philippines | 15th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 134 | |
0745 | If $a_1, a_2, \dots, a_n$ are $n$ non-zero complex numbers, not necessarily distinct, and $k, l$ are distinct positive integers such that $a_1^k, a_2^k, \dots, a_n^k$ and $a_1^l, a_2^l, \dots, a_n^l$ are two identical collections of numbers. Prove that each $a_j$, $1 \le j \le n$, is a root of unity. | [
"The given hypothesis implies that there is a bijection $f : \\{1, 2, 3, \\dots, n\\} \\to \\{1, 2, 3, \\dots, n\\}$ such that $f(j) = m$ implies $a_j^k = a_m^l$. Consider the sequence\n$$\n1, f(1), f^{(2)}(1), \\dots,\n$$\nSince $f$ is a bijection on a finite set, there are positive integers $r, s$ such that $f^{(... | India | Indija TS 2009 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
06ng | In a team game, 12 players stand at the 12 vertices of a regular 12-sided polygon. Each player has a red flag and a blue flag, and then randomly put up one flag, all at the same time. If there are four players who put up flags of the same colour and whose positions form a rectangle, the team loses. Otherwise the team w... | [
"$2^{12} = 4096$ possible outcomes and we want to count how many of these enable the team to win. To do this, we let the 12 players form 6 pairs so that the members of each pair are diametrically opposite. Note that the positions of four players form a rectangle if and only if they form two pairs.\n\nCall a pair 'b... | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof and answer | 29/128 | |
05fi | Problem:
Soit $ABC$ un triangle et $H$ son orthocentre. On note $\ell_{1}$ et $\ell_{2}$ deux droites passant par $H$ et perpendiculaires entre elles. On note $X_{1}, Y_{1}$ et $Z_{1}$ les points d'intersection de $\ell_{1}$ avec les droites $(BC)$, $(CA)$ et $(AB)$ respectivement. On note également $X_{2}, Y_{2}$ et ... | [
"Solution:\n\n\n\nL'approche présentée dans ce corrigé est longue, mais purement élémentaire et les nombreuses étapes doivent toutes paraître naturelles.\n\nPremière étape : les milieux. L'introduction d'un milieu dans un exercice est toujours une épine dans notre pied car cette condition e... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Circ... | null | proof only | null | |
0eey | Problem:
Naj bo $n \geq 2$ naravno število. Katja želi pobarvati vsako polje tabele velikosti $n \times n$ bodisi črno bodisi zeleno, tako da bo izmed poljubnih štirih polj, ki jih lahko prekrijemo z domino oblike
| $\square$ |
| :--- |
, liho mnogo polj črnih. Na koliko načinov lahko to stori? | [
"Solution:\n\nV prvi vrstici in prvem stolpcu tabele je skupaj $n + (n-1) = 2n - 1$ polj. Ta polja lahko Katja pobarva na $2^{2n-1}$ različnih načinov, saj ima za vsako polje na voljo dve barvi. Pokažimo, da lahko Katja, ne glede na to, kako pobarva prvo vrstico in prvi stolpec, potem pobarva preostanek tabele na e... | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2^{2n-1} | |
0ewt | Problem:
Given a set of $n$ different positive reals $\{a_{1}, a_{2}, \ldots , a_{n}\}$. Take all possible non-empty subsets and form their sums. Prove we get at least $\frac{n(n + 1)}{2}$ different sums. | [
"Solution:\nAssume $a_{1} < a_{2} < \\ldots < a_{n}$. We have the following collection of increasing sums:\n\n$a_{1} < a_{2} < \\ldots < a_{n}$ ($n$ sums)\n\n$a_{1} + a_{n} < a_{2} + a_{n} < \\ldots < a_{n - 1} + a_{n}$ ($n-1$ sums)\n\n$a_{1} + a_{n - 1} + a_{n} < a_{2} + a_{n - 1} + a_{n} < \\ldots < a_{n - 2} +... | Soviet Union | 3rd ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0ccy | The numbers $2^3 - 2$, $3^3 - 3$, $4^3 - 4$, $\dots$, $(2n+1)^3 - (2n+1)$, where $n \ge 2$ is an integer, are written on a board. An operation consists in erasing three randomly chosen numbers $a$, $b$, $c$ from the board and replacing them with $\frac{abc}{ab+bc+ca}$. Several operations are performed until two numbers... | [
"Since $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = \\frac{ab+bc+ca}{abc} = \\left(\\frac{abc}{ab+bc+ca}\\right)^{-1}$, for any given positive numbers $a$, $b$, $c$, it follows that after any operation, the sum of the inverses of the numbers left on the board is equal to the sum of the inverses of the numbers on t... | Romania | THE 73rd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD - FIRST SELECTION TEST | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0ede | Determine the greatest positive integer $m$ such that each square of the $m \times m$ array can be painted either red or blue so that not all the squares at the intersection of any two rows and any two columns are the same colour. (From Finnish MO 2014.) | [
"For each row consider all pairs of squares from this row such that both squares are the same colour. From the condition of the problem it follows that no two rows can have any of these pairs in common. So, each pair can occur in at most one row. There are $\\binom{m}{2} = \\frac{m(m-1)}{2}$ pairs altogether and th... | Slovenia | Slovenija 2016 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 4 | |
057b | Find all solutions of the equation $x^3 + 3xy + y^3 = 2019$ in integers. | [
"The r.h.s. of the equation is divisible by $3$ but not by $9$. Assume that $3 \\mid x$. Then $9 \\mid x^3$ and $9 \\mid 3xy$. If also $3 \\mid y$ then $9 \\mid y^3$, implying that the l.h.s. of the equation is divisible by $9$. Thus $3 \\nmid y$. But then $3 \\nmid y^3$, implying that the l.h.s. of the equation is... | Estonia | Open Contests | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | no integer solutions | |
00uv | Find all triples $(a, b, c)$ of positive real numbers that satisfy the system:
$$
\begin{aligned}
11bc - 36b - 15c &= abc \\
12ca - 10c - 28a &= abc \\
13ab - 21a - 6b &= abc.
\end{aligned}
$$ | [
"Considering each of the equalities:\n$$\n\\begin{align*}\nabc &= 11bc - 36b - 15c \\\\\nabc &= 12ac - 10c - 28a \\\\\nabc &= 13ab - 21a - 6b\n\\end{align*}\n$$\nand dividing the first one by $bc > 0$, the second one by $ac$ and third one by $ab$ we obtain:\n$$\n\\begin{align*}\na &= 11 - \\frac{36}{c} - \\frac{15}... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | (4, 6, 8) | |
02ea | Two players play alternately. The first player is given a pair of positive integers $(x_1, y_1)$. Each player must replace the pair $(x_n, y_n)$ that he is given by a pair of non-negative integers $(x_{n+1}, y_{n+1})$ such that $x_{n+1} = \min(x_n, y_n)$ and $y_{n+1} = \max(x_n, y_n) - k \cdot x_{n+1}$ for some positiv... | [
"Note first that draws are not possible so any position $(x, y)$ is either a win or a loss for the player receiving it. Let $\\phi$ be the positive root of $t^2 - t - 1 = 0$, so $\\phi = \\frac{1+\\sqrt{5}}{2}$. Let $m = \\min(x, y)$, $M = \\max(x, y)$. We show that a player receiving $(x, y)$ wins if and only if $... | Brazil | IX OBM | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Let phi = (1 + sqrt(5)) / 2. The first player has a winning strategy if and only if x1 = y1 or max(x1, y1) > phi * min(x1, y1). Equivalently, in terms of r = x1 / y1: the first player wins exactly when r < 1/phi, or r = 1, or r > phi. | |
00iw | determine the maximum value of the function
$$
f_k(x, y) = (x + y) - (x^{2k+1} + y^{2k+1})
$$
over all real numbers $x$ and $y$ satisfying the equation $x^2 + y^2 = 1$ for all positive integers $k$. | [
"Since we have $x^2 + y^2 = 1$, it definitely follows that $|x| \\le 1$ and $|y| \\le 1$ hold. Defining a function $g_k(x) := x - x^{2k+1}$, the signs of $x$ and $g_k(x)$ are therefore equal, and we have $g_k(-x) = -g_k(x)$. The given function can be expressed as $f_k(x, y) = g_k(x) + g_k(y)$, and we certainly have... | Austria | AustriaMO2011 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | (2^k - 1)/2^k * sqrt(2) | |
0ffg | Problem:
Hallar todos los intervalos de valores de $x$ para los cuales
$$
\cos x+\operatorname{sen} x>1
$$
el mismo problema para
$$
\cos x+|\operatorname{sen} x|>1
$$ | [
"Solution:\n\n1. Suponemos primero que $0 \\leq x \\leq 2\\pi$.\nConsideramos una circunferencia de radio unidad centrada en el origen de coordenadas $O$. Sea $P$ un punto de la misma y $x$ el ángulo medido en sentido antihorario que forma el semieje positivo de abscisas con $OP$. Entonces $P$ tiene de coordenadas ... | Spain | Olimpiadas Matemáticas Españolas | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | For cos x + sin x > 1: 2πn < x < π/2 + 2πn for any integer n. For cos x + |sin x| > 1: −π/2 + 2πm < x < π/2 + 2πm for any integer m. | |
0d7a | Let $n \geq 4$ be a positive integer and there exist $n$ positive integers that are arranged on a circle such that:
- The product of each pair of two non-adjacent numbers is divisible by $2015 \cdot 2016$.
- The product of each pair of two adjacent numbers is not divisible by $2015 \cdot 2016$.
Find the maximum value o... | [
"Denote $v_{p}(k)$ as the exponent of prime $p$ in the prime factorization of positive integer $k$. It is easy to see that $v_{p}(ab) = v_{p}(a) + v_{p}(b)$ for all positive integers $a, b$.\n\nWe have $2015 = 5 \\cdot 13 \\cdot 31$, $2016 = 2^{5} \\cdot 3^{2} \\cdot 7$.\nDenote $M = 2015 \\cdot 2016$ and\n$$\nN = ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 8 | |
074s | Find three distinct positive integers with the least possible sum such that the sum of the reciprocals of any two integers among them is an integral multiple of the reciprocal of the third integer. | [
"We first observe that $(1, a, b)$ is not a solution whenever $1 < a < b$. Otherwise we should have $\\frac{1}{a} + \\frac{1}{b} = l \\cdot \\frac{1}{1} = l$ for some integer $l$. Hence we obtain $\\frac{a+b}{ab} = l$ showing that $a|b$ and $b|a$. But then $a=b$ contradicting $a \\neq b$. Thus the least number shou... | India | Indija mo 2011 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (2, 3, 6) | |
08ro | Find the value of
$$
\frac{7}{12} + \frac{5}{12} \times \frac{7}{11} + \frac{5}{12} \times \frac{4}{11} \times \frac{7}{10} + \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} \times \frac{7}{9}
$$ | [
"$$\n\\begin{aligned}\n& 1 - \\left( \\frac{7}{12} + \\frac{5}{12} \\times \\frac{7}{11} + \\frac{5}{12} \\times \\frac{4}{11} \\times \\frac{7}{10} + \\frac{5}{12} \\times \\frac{4}{11} \\times \\frac{3}{10} \\times \\frac{7}{9} \\right) \\\\\n&= \\frac{5}{12} - \\left( \\frac{5}{12} \\times \\frac{7}{11} + \\frac... | Japan | Japanese Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 98/99 | |
00rw | A grasshopper is sitting at an integer point in the Euclidean plane. Each second it jumps to another integer point in such a way that the jump vector is constant. A hunter that knows neither the starting point of the grasshopper nor the jump vector (but knows that the jump vector for each second is constant) wants to c... | [
"The hunter can catch the grasshopper. Here is the strategy for him. Let $f$ be any bijection between the set of positive integers and the set $\\{((x, y), (u, v)) : x, y, u, v \\in \\mathbb{Z}\\}$, and denote\n$$\nf(t) = ((x_t, y_t), (u_t, v_t))\n$$\nIn the second $t$, the hunter should hunt at the point $(x_t + t... | Balkan Mathematical Olympiad | BMO 2017 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Algorithms"
] | English | proof only | null | |
0f6e | Problem:
Each face of a cube is painted a different color. The same colors are used to paint every face of a cubical box a different color. Show that the cube can always be placed in the box, so that every face is a different color from the box face it is in contact with. | [] | Soviet Union | 19th ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof only | null | |
0811 | Problem:
Un saldatore dispone di sbarrette metalliche di lunghezza 2, e vuole costruire una griglia costituita da $n \times n$ quadratini di lato 1 (esempio $5 \times 5$ a fianco). Gli è permesso segare a metà le sbarrette e saldarle fra loro, ma senza sovrapporle o incrociarle. Qual è il minimo numero di sbarrette ch... | [
"Solution:\n\nSe $n$ è dispari in ogni riga c'è una sbarretta di lunghezza 1 e quindi il numero di sbarrette di lunghezza 1 è almeno $2(n+1)$; sono pertanto necessari almeno $n+1$ tagli. Per la costruzione illustrata a fianco nel caso $n=7$, ma facilmente generalizzabile ad ogni $n$ dispari, servono esattamente $n+... | Italy | Cesenatico | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | If n is odd, the minimum number of cuts is n+1. If n is even, the minimum number of cuts is n−1. | |
03yx | Let $M \subseteq \{1, 2, \dots, 2011\}$ be a subset satisfying the following condition: For any three elements in $M$, there exist two of them $a$ and $b$, such that $a \mid b$ or $b \mid a$. Determine, with proof, the maximum value of $|M|$, where $|M|$ denotes the number of elements of $M$. (posed by Feng Zhigang) | [
"One can check that $M = \\{1, 2, 2^2, 2^3, \\dots, 2^{10}, 3, 3 \\times 2, 3 \\times 2^2, \\dots, 3 \\times 2^9\\}$ satisfies the condition, and $|M| = 21$.\n\nSuppose that $|M| \\ge 22$, and let $a_1 < a_2 < \\dots < a_k$ be the elements of $M$, where $|M| = k \\ge 22$. We first prove that $a_{n+2} \\ge 2a_n$ for... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 21 | |
0230 | Problem:
Seja $ABC$ um triângulo tal que $AB = 55$, $AC = 35$ e $BC = 72$. Considere uma reta $\ell$ que corta o lado $BC$ em $D$ e o lado $AC$ em $E$ e que divide o triângulo em duas figuras com perímetros iguais e áreas iguais. Determine a medida do segmento $CD$. | [
"Solution:\n\nSejam $CD = x$, $CE = y$ e $DE = z$.\n\n(1) Como o triângulo $CED$ tem o mesmo perímetro do quadrilátero $ABDE$, temos\n$$\nx + y + z = (35 - y) + z + (72 - x) + 55 \\Longleftrightarrow y = 81 - x\n$$\n\n(2) Como eles também possuem a mesma área, a área do triângulo $DCE$ deve ser igual à metade da ár... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 60 | |
0bs3 | Find all the positive integers $a, b, c$ with the property $a + b + c = abc$. | [
"If one of the numbers is $0$, then all are $0$.\nIf $abc \\neq 0$, then the relation can be written $\\frac{1}{bc} + \\frac{1}{ac} + \\frac{1}{ab} = 1$. Since the relation is symmetric in $a, b, c$, we may assume that $a \\le b \\le c$, whence $ab \\le ac \\le bc$.\nIf $ab > 3$, then $\\frac{1}{bc} + \\frac{1}{ac}... | Romania | 67th Romanian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All permutations of (1,2,3). | |
0g2v | Problem:
Let $(a, b)$ be a pair of positive integers. Henning and Paul are playing a game: Initially, there are two piles of $a$ and $b$ stones, respectively, on a table. The pair $(a, b)$ is called the initial configuration of the game. The players proceed as follows:
- The players alternate and Henning begins.
- In ... | [
"Solution:\n\n(a)\nWe call a general pair of starting values $(a, b)$ *good* if Paul has a winning strategy for it. For completeness, we also allow $a=0$ or $b=0$ and call $(0,0)$ good. Analogously, we call all pairs that are not good *bad* and note that for these starting values, Henning obviously has a winning st... | Switzerland | Selektion | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
09g9 | Let $\triangle ABC$ be a triangle and $M$ be midpoint of $BC$. The point $N$ lie on the line through $M$ parallel to $AC$ such that $\angle MAB = \angle NAC$. Prove that $\angle ABN = \angle ACB$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
02p9 | Problem:
De quantas formas é possível colorir as 6 faces de um cubo de preto ou branco? Duas colorações são iguais se é possível obter uma a partir da outra por uma rotação. | [
"Solution:\n\nObservemos que basta contar quantas colorações existem que têm exatamente 0, 1, 2 e 3 faces pretas, porque os outros casos são simétricos. Com uma ou nenhuma face preta existe uma única coloração para cada caso. Quando temos duas faces pretas temos duas possíveis colorações que são: quando estas faces... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 10 | |
09if | Let $F$ be a point on the side $AC$ of a triangle $ABC$. A line through $F$ and parallel to $AB$ intersects with side $BC$ at $D$. Similarly, a line through $F$ and parallel to $BC$ intersects with side $AB$ at $E$. Assume that the side $AC$ is tangent to the circumcircle of $EDF$. If $AB : BC = k$, then prove that $AF... | [
"Let $\\angle A = \\alpha$, $B = \\beta$, $C = \\gamma$. Then $\\angle EFA = \\gamma$, because $EF \\parallel BC$. Since $AC$ is the tangent to the circumcircle of $EDF$, we have $\\angle EDF = \\angle EFA = \\gamma$.\n\nSimilarly, $\\angle DEF = \\angle DFC = \\alpha$. It follows that $\\triangle EFD \\sim \\trian... | Mongolia | Mongolian Mathematical Olympiad Round 3 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0cja | Solve in $\mathbb{R}$ the equation $\log_7(6^x + 1) = \log_6(7^x - 1)$. | [
"If $\\log_7(6^x + 1) = \\log_6(7^x - 1) = y$, we obtain $6^x + 1 = 7^y$ and $7^x - 1 = 6^y$. By addition, it follows that $6^x + 7^x = 6^y + 7^y$. The function $f: \\mathbb{R} \\to \\mathbb{R}$, $f(x) = 6^x + 7^x$, is injective (it is strictly increasing, as the sum of two strictly increasing functions), so $x = y... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof and answer | 1 | |
0150 | Problem:
Let $AB$ be a diameter of a circle $S$, and let $L$ be the tangent at $A$. Furthermore, let $c$ be a fixed, positive real, and consider all pairs of points $X$ and $Y$ lying on $L$, on opposite sides of $A$, such that $|AX| \cdot |AY| = c$. The lines $BX$ and $BY$ intersect $S$ at points $P$ and $Q$, respectiv... | [
"Solution:\nLet $S$ be the unit circle in the $xy$-plane with origin $O$, put $A = (1, 0)$, $B = (-1, 0)$, take $L$ as the line $x = 1$, and suppose $X = (1, 2p)$ and $Y = (1, -2q)$, where $p$ and $q$ are positive real numbers with $pq = \\frac{c}{4}$. If $\\alpha = \\angle ABP$ and $\\beta = \\angle ABQ$, then $\\... | Baltic Way | Baltic Way 2008 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellan... | null | proof only | null | |
069f | Let $AB\Gamma\Delta$ quadrilateral inscribed in a circle of center $O$. The line perpendicular to the side $B\Gamma$ at its midpoint $E$ meets the line $AB$ at point $Z$. The circumcircle of the triangle $\Gamma EZ$ intersects the side $AB$ for a second time at point $H$ and the line $\Gamma\Delta$ at point $\Theta \ne... | [
"It is enough to prove that: $A\\hat{K}\\Lambda = 90^\\circ$.\nSince $\\Delta\\hat{K}\\Theta = A\\hat{K}\\Lambda$, it is enough to prove that in the triangle $\\Delta\\Theta K$ the two acute angles have sum $90^\\circ$, i.e. $\\Delta\\Theta K + \\Theta\\hat{\\Delta}K = 90^\\circ$. We have:\n$$\n\\begin{aligned}\n\\... | Greece | 36th Hellenic Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03sf | A function $f: (0, +\infty) \to \mathbf{R}$ satisfies the following conditions:
a. $f(a) = 1$ for a positive real number $a$,
b. $f(x)f(y) + f(\frac{a}{x})f(\frac{a}{y}) = 2f(xy)$, for any positive real number $x, y$.
Prove that $f(x)$ is constant. | [
"Setting $x = y = 1$ in $(1)$ gives\n$$\nf^2(1) + f^2(a) = 2f(1),\n$$\n$$\n(f(1) - 1)^2 = 0,\n$$\nso $f(1) = 1$.\n\nSetting $y = 1$ in $(1)$ yields\n$$\nf(x)f(1) + f(\\frac{a}{x})f(a) = 2f(x),\n$$\n$$\nf(x) = f(\\frac{a}{x}), \\quad x > 0. \\qquad (2)\n$$\n\nSetting $y = \\frac{a}{x}$ in $(1)$ yields\n$$\nf(x)f(\\f... | China | China Girls' Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof only | null | |
0d59 | Prove that the polynomial $P(X)=\left(X^{2}-12 X+11\right)^{4}+23$ can not be written as the product of three non-constant polynomials with integer coefficients. | [
"Suppose for the sake of contradiction that\n$$\nP(X)=Q(X) H(X) R(X),\n$$\nwhere $Q(X)$, $H(X)$, $R(X)$ are non-constant polynomials with integer coefficients. Since $P(x)>0$ for every $x \\in \\mathbb{R}$, the degrees of $Q(X)$, $H(X)$, $R(X)$ are all even. It implies that two of these three polynomials are quadra... | Saudi Arabia | SAMC 2015 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English, Arabic | proof only | null | |
0jms | Problem:
How many lines pass through exactly two points in the following hexagonal grid?
 | [
"Solution:\nAnswer: $60$\n\nFirst solution. From a total of $19$ points, there are $\\binom{19}{2} = 171$ ways to choose two points. We consider lines that pass through more than $2$ points.\n- There are $6 + 6 + 3 = 15$ lines that pass through exactly three points. These are: the six sides of the largest hexagon, ... | United States | HMMT November 2014 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 60 | |
094j | Problem:
Find the smallest integer $b$ with the following property: For each way of colouring exactly $b$ squares of an $8 \times 8$ chessboard green, one can place 7 bishops on 7 green squares so that no two bishops attack each other.
Remark. Two bishops attack each other if they are on the same diagonal. | [
"Solution:\nLet us place 40 bishops on 6 diagonals as shown in Figure 2. If we select any 7 of the placed bishops, by the Pigeonhole Principle, at least two of the selected bishops are on the same diagonal, so they attack each other. Thus, the number $b$ of selected bishops is at least 41.\n\n | MEMO | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 41 | |
0ecg | Problem:
Luka se je po ravni cesti zapeljal čez gorski prelaz. Na oddaljenosti $x$ metrov v vodoravni smeri od vznožja prelaza se je nahajal na nadmorski višini $h(x) = -\frac{x^{2}}{20000} + \frac{3x}{5} + 560$ metrov (glej sliko).
a) Na kateri nadmorski višini se nahaja najvišja točka pre... | [] | Slovenia | 15. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications",
"Precalculus > Trigonometric functions"
] | null | final answer only | a) 2360 m; b) 11 degrees 19 minutes | |
0eoo | Points $A_1$, $A_2$, $A_3$ ... are constructed as follows: the length $OA_1$ is $4$, $\angle OA_1A_2 = 90^\circ$ and the length $A_1A_2 = 1$; then a right angle is constructed at $A_2$ to find $A_3$, and so on as shown in the diagram.
The length of $OA_{21}$ is
 | [
"**6** By Pythagoras, $OA_2 = \\sqrt{17}$. Then $OA_3 = \\sqrt{18}$, $OA_4 = \\sqrt{19}$ and so on, with $OA_n = \\sqrt{n+15}$, and thus $OA_{21} = \\sqrt{36} = 6$."
] | South Africa | South African Mathematics Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | final answer only | 6 | |
0986 | Problem:
Aflaţi toate funcțiile $f: \mathbb{R} \rightarrow \mathbb{R}$, care satisfac conditiile $f(0)>0$ şi
$$
f(x+y)=f(x) \cdot f(2022-y)+f(y) \cdot f(2022-x)
$$
pentru oricare $x, y \in \mathbb{R}$. | [
"Solution:\n\n1) Punând $x=y=0$ în egalitatea din enunț, se obține\n$$\nf(0)=f(0) \\cdot f(2022)+f(0) \\cdot f(2022)=2 f(0) \\cdot f(2022) \\Rightarrow f(2022)=\\frac{1}{2}\n$$\n\n2) Pentru $x=2022, y=0$ se obține $\\frac{1}{2}=f(2022)=f(2022) \\cdot f(2022)+f(0) \\cdot f(0)=\\frac{1}{4}+(f(0))^{2}$. Rezultă $f(0)=... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 1/2 for all real x | |
07dc | We say distinct positive integers $a_1, a_2, \dots, a_n$ are **harmonic** if their sum is equal to the sum of all pairwise gcd's among them. Prove that there are infinitely many integers like $n$ such that $n$ harmonic numbers exist. | [
"Suppose $a_1 = 2^0, a_2 = 2^1, \\dots, a_{n-2} = 2^{n-3}, a_{n-1} = 3, a_n = q$. We shall prove there exists infinitely many positive integers $n$ for which there exists a positive integer $q$ such that $(q, 6) = 1$ and $a_1, \\dots, a_n$ are harmonic.\n\n$$\n\\begin{align*}\n\\sum_{1 \\le i < j \\le n} (a_i, a_j)... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0kvj | Problem:
Suppose $m>n>1$ are positive integers such that there exist $n$ complex numbers $x_{1}, x_{2}, \ldots, x_{n}$ for which
- $x_{1}^{k}+x_{2}^{k}+\cdots+x_{n}^{k}=1$ for $k=1,2, \ldots, n-1$;
- $x_{1}^{n}+x_{2}^{n}+\cdots+x_{n}^{n}=2$; and
- $x_{1}^{m}+x_{2}^{m}+\cdots+x_{n}^{m}=4$.
Compute the smallest possible... | [
"Solution:\n\nLet $S_{k}=\\sum_{j=1}^{n} x_{j}^{k}$, so $S_{1}=S_{2}=\\cdots=S_{n-1}=1, S_{n}=2$, and $S_{m}=4$. The first of these conditions gives that $x_{1}, \\ldots, x_{n}$ are the roots of $P(x)=x^{n}-x^{n-1}-c$ for some constant $c$. Then $x_{i}^{n}=x_{i}^{n-1}+c$, and thus\n$$\n2=S_{n}=S_{n-1}+c n=1+c n\n$$... | United States | HMMT February 2023 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Diophantine Equations > Pell's equations"
] | null | proof and answer | 34 | |
04lc | What is the maximum number of elements that a finite set $S$ can have so that among any three elements of $S$ there exist two distinct whose sum is an element of $S$ as well? (Russia 2000) | [] | Croatia | Mathematical competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 5 | |
02ck | Problem:
Júlio faz multiplicações usando apenas os quadrados dos números. Ele tem que calcular o produto $85 \times 135$. Para isso, ele desenha um retângulo de $85 \mathrm{~mm}$ por $135 \mathrm{~mm}$ e traça nesse retângulo o maior quadrado possível; faz o mesmo no quadrado restante e assim sucessivamente. Dessa man... | [
"Solution:\n\nO maior quadrado no retângulo de $85 \\times 135$ é aquele de $85 \\times 85$. Sobra então um retângulo de $50 \\times 85$, onde o maior quadrado é de $50 \\times 50$. Continuando assim, obtemos:\n$$\n85 \\times 135=85^{2}+50^{2}+35^{2}+15^{2}+15^{2}+5^{2}+5^{2}+5^{2}\n$$\n\n| | | | $5^{2}$ $5^{... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Algorithms"
] | null | final answer only | 85^2 + 50^2 + 35^2 + 15^2 + 15^2 + 5^2 + 5^2 + 5^2 | |
0jjo | Problem:
Find the smallest positive integer $n$ such that, if there are initially $2n$ townspeople and $1$ goon, then the probability the townspeople win is greater than $50\%$. | [
"Solution:\nAnswer: $3$\n\nWe instead consider the probability the goon wins. The game clearly must last $n$ days. The probability the goon is not sent to jail on any of these $n$ days is then\n$$\n\\frac{2n}{2n+1} \\cdot \\frac{2n-2}{2n-1} \\cdots \\frac{2}{3}\n$$\nIf $n=2$ then the probability the goon wins is $\... | United States | HMMT November 2014 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 3 | |
0l2l | Problem:
Each lattice point with nonnegative coordinates is labeled with a nonnegative integer in such a way that the point $(0,0)$ is labeled by $0$, and for every $x, y \geq 0$, the set of numbers labeled on the points $(x, y)$, $(x, y+1)$, and $(x+1, y)$ is $\{n, n+1, n+2\}$ for some nonnegative integer $n$. Determi... | [
"Solution:\nWe claim the answer is all multiples of $3$ from $0$ to $2000+2 \\cdot 2024 = 6048$.\n\nFirst, we prove no other values are possible. Let $\\ell(x, y)$ denote the label of cell $(x, y)$.\n\n## The label is divisible by 3.\n\nObserve that for any $x$ and $y$, $\\ell(x, y)$, $\\ell(x, y+1)$, and $\\ell(x+... | United States | HMMT February 2024 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | All multiples of 3 from 0 to 6048 inclusive. | |
04t2 | Let $a$, $b$ be relatively prime integers. Sequence $(x_n)_{n=1}^{\infty}$ of natural numbers is constructed in such a way that for each $n > 1$ applies $x_n = a x_{n-1} + b$. Prove that in any such sequence every entry $x_n$ with index $n > 1$ divides infinitely many of other entries. Does this assertion hold for $n =... | [] | Czech Republic | Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | No | |
054r | Triangle $ABC$ has $AC = BC$. The bisector of angle $CAB$ meets side $BC$ at point $D$. The difference of the sizes of some two internal angles of triangle $ABD$ is $40^\circ$. Find all possibilities of what the size of angle $ACB$ can be. | [
"\n\nFig. 16\n\n* If $\\alpha - \\frac{\\alpha}{2} = 40^\\circ$ then $\\alpha = 80^\\circ$, whence $\\angle ACB = 20^\\circ$.\n\n* The case $\\frac{\\alpha}{2} - \\alpha = 40^\\circ$ is impossible since it would imply $\\alpha < 0^\\circ$.\n\n* If $(180^\\circ - \\frac{3}{2}\\alpha) - \\fra... | Estonia | National Olympiad Final Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 4°, 20°, 40°, 68° | |
0ez7 | Problem:
$ABC$ is a triangle with incenter $I$. $M$ is the midpoint of $BC$. $IM$ meets the altitude $AH$ at $E$. Show that $AE = r$, the radius of the inscribed circle. | [] | Soviet Union | 4th ASU | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0cv5 | Oleg has labelled all the columns and all the rows of a $50 \times 50$ table with $100$ distinct numbers $a_1, \dots, a_{50}$ and $b_1, \dots, b_{50}$, respectively; exactly $50$ of these numbers are rational. Then he has placed into each cell $(i, j)$ the number $a_i + b_j$. Find the greatest possible number of ration... | [
"Assume there are $x$ rational numbers among the $a_i$. Then the total number of irrationals in the cells is at least $x \\cdot x + (50 - x) \\cdot (50 - x) \\ge 1250$ (since rational $+$ irrational $=$ irrational). In an example, $x = 25$, all irrationals among the $a_i$ are in $\\mathbb{Q} \\pm \\sqrt{2}$, and th... | Russia | XLIII Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English; Russian | proof and answer | 1250 | |
0c43 | Problem:
Fie $f: \mathbb{R} \rightarrow \mathbb{R}$ o funcţie care are proprietatea lui Darboux. Arătaţi că, dacă $f$ este injectivă pe mulţimea numerelor iraţionale, atunci este $f$ este continuă pe $\mathbb{R}$. | [
"Solution:\n\nVom arăta că $f$ este injectivă pe $\\mathbb{R}$. Atunci, cum $f$ are proprietatea lui Darboux, $f$ este (strict) monotonă şi, prin urmare, continuă.\n\nPresupunem că $f$ nu este injectivă. Fie $a, b \\in \\mathbb{R}, a<b$, astfel încât $f(a)=f(b)$. Cum în intervalul $(a, b)$ există cel puţin două num... | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof only | null | |
085u | Problem:
Sia data la successione
$$
\left\{\begin{array}{l}
x_{1}=2 ; \\
x_{n+1}=2 x_{n}^{2}-1 \quad \text{ per } n \geq 1
\end{array}\right.
$$
Dimostrare che $n$ e $x_{n}$ sono relativamente primi per ogni $n \geq 1$. | [
"Solution:\n\nDimostriamo che, se $p$ è un numero primo che divide $x_{n}$, allora $p$ non divide $n$.\nSe $p=2$, allora la tesi è banalmente vera, in quanto tutti i termini $x_{n}$ sono dispari per $n>1$. Quindi si ha che 2 divide $x_{n}$ se e solo se $n=1$.\nSupponiamo dunque $p>2$. Chiamiamo $k$ il più piccolo i... | Italy | Cesenatico | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0d9h | Let $ABC$ be a triangle inscribed in circle $(O)$, with orthocenter $H$. Let $d$ be an arbitrary line which passes through $H$ and intersects $(O)$ at two points $P$ and $Q$. Draw diameter $AA'$ of circle $(O)$. $A'P$, $A'Q$ meet $BC$ at $K$, $L$, respectively. Prove that $O$, $K$, $L$, $A'$ are concyclic. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06lk | Find the total number of primes $p < 100$ such that $\lfloor (2 + \sqrt{5})^p \rfloor - 2^{p+1}$ is divisible by $p$. Here $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$. | [
"The answer is 24.\nLet $p$ be an odd prime. Using the binomial theorem, we have\n$$\n(2 + \\sqrt{5})^p + (2 - \\sqrt{5})^p = \\sum_{k=0}^{\\frac{p-1}{2}} \\binom{p}{2k} 2^{p+1-2k} \\cdot 5^k = 2^{p+1} + \\sum_{k=1}^{\\frac{p-1}{2}} \\binom{p}{2k} 2^{p+1-2k} \\cdot 5^k,\n$$\nwhich is an integer. Since $p$ is an odd... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 24 | |
0d8q | Find all prime numbers $p$ such that $\frac{3^{p-1}-1}{p}$ is a perfect square. | [
"Let $p$ be a prime satisfying the condition of the problem. By assumption, there is a positive integer $A$ such that $3^{p-1}-1=p A^{2}$.\n\nIt is clear that $p=2$ is a solution. Now, we consider $p>2$. Put $p-1=2k$, one has $(3^{k}-1)(3^{k}+1)=p A^{2}$. Since $(3^{k}-1,3^{k}+1)=2$, it follows that there are posit... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | p = 2 and p = 5 | |
0eq9 | The digits from $1$ to $9$ are added, in order, over and over again until the total is $460$.
$1+2+3+4+5+6+7+8+9+1+2+3+\ldots$
The last digit that was added is
(A) 2 (B) 4 (C) 6 (D) 8 (E) 9 | [
"The digits $1$ to $9$ total $45$, and $460 = 10 \\times 45 + 10$. So the complete set $1\\ldots9$ will appear ten times, and then as many more digits as total $10$, viz. $1+2+3+4$. So the last digit is $4$."
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | MCQ | B | |
0a4g | Problem:
Bepaal alle gehele getallen $n \ge 2$ waarvoor $n$ een deler is van $\binom{2n-3}{n-1}$. | [
"Solution:\n\nWe gaan bewijzen dat het antwoord is: alle $n \\ge 2$ die geen macht van 2 zijn. Er geldt\n$$\n\\binom{2n-3}{n-1} = \\frac{(2n-3)!}{(n-1)!(n-2)!} = \\frac{(2n-3)(2n-4)\\cdots(n+1)n}{(n-2)(n-3)\\cdots2\\cdot1}. \\qquad (1)\n$$\nZij $p$ een priemfactor van $n$ en stel dat $p > 2$. We weten dat de volgen... | Netherlands | IMO-selectietoets I | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinat... | null | proof and answer | All integers n at least 2 that are not powers of two. | |
05x4 | Problem:
Soient $ABC$ un triangle, $D, E$ les pieds des hauteurs issues de $A$ et $B$ respectivement. La droite $(DE)$ rencontre le cercle circonscrit à $ABC$ en deux points $P$ et $Q$. Soient $A'$, $B'$ les symétriques de $A$ et $B$ par rapport à $(BC)$ et $(AC)$ respectivement. Montrer que $A'$, $B'$, $P$, $Q$ sont ... | [
"Solution:\n\n\n\nSur la figure, il semble que $H$ l'orthocentre se situe sur le cercle en question. On va donc adopter la stratégie suivante : on va montrer que $A'$ et $B'$ sont sur le cercle circonscrit de $PQH$. De cette manière on aura bien $A'$, $B'$, $P$, $Q$ cocycliques.\n\nSoit $M$... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"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 > Miscellaneous > Distance chasing"
] | null | proof only | null | |
01lg | Four points $A$, $B$, $C$, $D$ are marked on the parabola $y = x^2$ so that the quadrilateral $ABCD$ is a trapezoid ($AD \parallel BC$, $AD > BC$). Let $m$ and $n$ be the distances between the intersection point of the diagonals of the trapezoid and the midpoints of its bases $AD$ and $BC$, respectively.
Find the area ... | [
"Answer: $S = \\frac{(m+n)^2}{\\sqrt{m-n}}$.\n\nLet $x_R, y_R$ denote the coordinates of point $R$. Let $y = kx + a$ and $y = kx + b$ be the equations of the lines $AD$ and $BC$, respectively. Then\n$$\n\\begin{cases} x_A^2 = y_A = kx_A + a, \\\\ x_D^2 = y_D = kx_D + a, \\end{cases} \n\\begin{cases} x_B^2 = y_B = k... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | S = (m+n)^2 / sqrt(m-n) | |
0912 | Problem:
Let $\left(a_{n}\right)_{n=1}^{\infty}$ be a sequence of positive integers such that $a_{n}<a_{n+1}$ for all $n \geqslant 1$. Suppose that for all quadruples of indices $(i, j, k, l)$ such that $1 \leqslant i<j \leqslant k<l$ and $i+l=j+k$, the inequality $a_{i}+a_{l}>a_{j}+a_{k}$ is satisfied. Determine the ... | [
"Solution:\n\nSince $a_{2}-a_{1} \\geqslant 1$ and $a_{n+2}-a_{n+1} \\geqslant \\left(a_{n+1}-a_{n}\\right)+1$ (by applying the quadruple $(n, n+1, n+1, n+2)$ for each $n$), induction yields $a_{n+1}-a_{n} \\geqslant n$ for all $n \\geqslant 1$. Thus $a_{n+1} \\geqslant n+a_{n}$ (and $a_{1} \\geqslant 1$), hence in... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2015029 | |
0dmi | Problem:
Нека је $ABCDE$ конвескан петоугао у коме је $AB=1$, $\angle BAE=\angle ABC=120^\circ$, $\angle CDE=60^\circ$ и $\angle ADB=30^\circ$. Доказати да је површина петоугла $ABCDE$ мања од $\sqrt{3}$.
(Милош Милосављевић) | [
"Solution:\n\nНека је $k$ круг описан око троугла $ABD$, и $l$ права кроз $D$ паралелна са $AB$. Полупречник круга $k$ је $1$. Полуправе $BC$ и $AE$ секу $k$ у тачкама $H$ и $I$, а праву $l$ у $F$ и $G$, редом. Троуглови $FCD$ и $GDE$ су слични јер је $\\angle CFD=\\angle DGE=60^\\circ$ и $\\angle FCD=120^\\circ-\\... | Serbia | СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
028h | Problem:
Bolas brancas e pretas - Uma caixa tem exatamente $100$ bolas pretas e $100$ bolas brancas. Repetidamente, $3$ bolas são retiradas da caixa e substituídas por outras bolas que estão em um saco da seguinte maneira:
| BOLINHAS REMOVIDAS | SUBSTITUÍDAS POR |
|--------------------|---------------------... | [
"Solution:\n\nInicialmente observe que depois de cada substituição o número de bolas brancas ou permanece o mesmo ou decresce de $2$. Logo o número de bolas brancas permanece par. Por outro lado, cada grupo de bolas removidas que contém pelo menos $1$ bola branca é substituído por outro que também contém $1$ bola b... | Brazil | Lista 6 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | MCQ | b | |
0dqm | Let $p$ be a prime number and let $a_1, a_2, \dots, a_k$ be distinct integers chosen from $1, 2, \dots, p-1$. For $1 \le i \le k$, let $r_i^{(n)}$ denote the remainder of the integer $n a_i$ upon division by $p$, so $0 \le r_i^{(n)} < p$. Define
$$
S = \{ n : 1 \le n \le p-1,\ r_1^{(n)} < \dots < r_k^{(n)} \}.
$$
Show ... | [
"Let $r_0^{(n)} = 0$ and $r_{k+1}^{(n)} = p$. Set\n$$\nS' = \\{ n : 1 \\le n \\le p-1,\\ \\sum_{i=0}^k | r_{i+1}^{(n)} - r_i^{(n)} | = p \\}.\n$$\nNote that\n$$\n\\sum_{i=0}^k | r_{i+1}^{(n)} - r_i^{(n)} | = p \\quad \\text{iff} \\quad r_0^{(n)} \\le r_1^{(n)} \\le \\dots \\le r_{k+1}^{(n)}\n$$\nThus $|S| = |S'|$. ... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0bop | Given an integer $N \ge 4$, determine the largest value the sum
$$
\sum_{i=1}^{\lfloor k/2 \rfloor + 1} (\lfloor n_i/2 \rfloor + 1)
$$
may achieve, where $k, n_1, \dots, n_k$ run through the integers subject to $k \ge 3, n_1 \ge \dots \ge n_k \ge 1$, and $n_1 + \dots + n_k = N$. | [
"The required maximum is $\\lfloor 2(N + 2)/3 \\rfloor$.\n\nFor more convenience, given a list of $k$ real numbers, the sublist consisting of the $1 + \\lfloor k/2 \\rfloor$ largest entries will be referred to as the *upper half* of the list, and its complement, i.e., the sublist consisting of the $\\lfloor (k+1)/2... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | ⌊2(N+2)/3⌋ | |
09sd | Problem:
In een driehoek $ABC$ is $D$ het snijpunt van de binnenbissectrice van $\angle BAC$ met zijde $BC$. Zij $P$ het tweede snijpunt van de buitenbissectrice van $\angle BAC$ met de omgeschreven cirkel van $\triangle ABC$. Een cirkel door $A$ en $P$ snijdt lijnstuk $BP$ inwendig in $E$ en lijnstuk $CP$ inwendig in... | [
"Solution:\n\nWe bekijken de configuratie waarbij de punten $A$, $C$, $B$ en $P$ in die volgorde op de omgeschreven cirkel liggen. Het andere geval gaat analoog.\n\nVanwege de omtrekshoekstelling in de omgeschreven cirkel van $\\triangle ABC$ geldt\n$$\n\\angle ABE = \\angle ABP = \\angle ACP = \\angle ACF.\n$$\nVe... | Netherlands | Selectietoets | [
"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"
] | null | proof only | null | |
008j | A rectangle is divided in $n^2$ smaller rectangles by means of $n-1$ horizontal lines and $n-1$ vertical lines. Among those rectangles, there are exactly 5660 which are not congruent. For which minimum value of $n$ is this possible? | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | proof and answer | 78 | |
0j2w | Problem:
Let $a$, $b$, $c$, and $d$ be positive real numbers satisfying $a b c d = 1$. Prove that
$$
\frac{1}{\sqrt{\frac{1}{2} + a + a b + a b c}} + \frac{1}{\sqrt{\frac{1}{2} + b + b c + b c d}} + \frac{1}{\sqrt{\frac{1}{2} + c + c d + c d a}} + \frac{1}{\sqrt{\frac{1}{2} + d + d a + d a b}} \geq \sqrt{2}.
$$ | [
"Solution:\nLet\n$$\n\\begin{aligned}\n& S_{a} = a + a b + a b c \\\\\n& S_{b} = b + b c + b c d \\\\\n& S_{c} = c + c d + c d a \\\\\n& S_{d} = d + d a + d a b.\n\\end{aligned}\n$$\nNotice that $1 + S_{a} = \\frac{1}{2} + \\left(\\frac{1}{2} + S_{a}\\right) \\geq 2 \\sqrt{\\frac{1}{2} \\cdot \\left(\\frac{1}{2} + ... | United States | Bay Area Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0ewi | Problem:
An infinite arithmetic progression contains a square. Prove it contains infinitely many squares. | [
"Solution:\n\nLet the square be $a^{2}$ and the difference $d$, so that all numbers of the form $a^{2} + nd$ belong to the arithmetic progression (for $n$ a natural number). Take $n$ to be $2a + dr^{2}$, then $a^{2} + nd = (a + dr)^{2}$."
] | Soviet Union | 3rd ASU | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof only | null | |
06h5 | Let $D$, $E$ and $F$ be the contact points of the incircle of $\triangle ABC$ with sides $BC$, $CA$ and $AB$ respectively. Let $I$ and $I'$ be the incentres of $\triangle ABC$ and $\triangle DEF$, respectively. Let $\ell_A$ be the line passing through $D$ and is parallel to $AI'$, $\ell_B$ be the line passing through $... | [
"Let $\\ell_A$ meet $II'$ at $P$. We claim that $P$ is the intersection point of $\\ell_A$, $\\ell_B$, $\\ell_C$. It suffices to show that $\\frac{II'}{I'P}$ is a constant.\nLet $DI'$ meet the incircle of $\\triangle ABC$ again at $M$, and meet the line passing through $I$ and parallel to $\\ell_A$ at $Q$. Let $r$ ... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distan... | null | proof only | null | |
0593 | Find all triples $(x, y, z)$ of real numbers that satisfy
$$
\begin{cases}
\frac{x}{y} + \frac{y}{z} + xy = 3, \\
\frac{y}{z} + \frac{z}{x} + yz = 3, \\
\frac{z}{x} + \frac{x}{y} + zx = 3.
\end{cases}
$$ | [
"**Answer:** $(1, 1, 1), (-1, -1, -1)$.\n\nNumbers $x$, $y$ and $z$ must have the same sign, because if exactly one or two of them are negative then there exists an equation in the system whose all terms in the l.h.s. are negative and cannot sum up to $3$. It is also easy to see that $(x, y, z)$ being a solution im... | Estonia | Estonian Math Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | (1, 1, 1) and (-1, -1, -1) | |
0hzx | Problem:
If $x$, $y$, and $z$ are distinct positive integers such that $x^{2} + y^{2} = z^{3}$, what is the smallest possible value of $x + y + z$. | [
"Solution:\nWithout loss of generality let $x > y$. We must have $z^{3}$ expressible as the sum of two squares, and this first happens when $z = 5$. Then $x$ and $y$ can be $10$ and $5$ or $11$ and $2$. If $z > 5$ then $z \\geq 10$ for $z^{3}$ to be a sum of two distinct squares, so $x^{2} > 500$, $x > 22$, so $x +... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 18 | |
0btt | a) Given any positive integer $n$, prove that every $n$ points in the closed unit square $[0, 1] \times [0, 1]$ can be joined by a path of length less than $2\sqrt{n} + 4$.
b) Prove that there exist $n$ points in the closed unit square $[0, 1] \times [0, 1]$ that cannot be joined by a path of length less than $\sqrt{n... | [
"a) Let $C$ be an $n$-point configuration in the closed unit square $[0, 1] \\times [0, 1]$, let $m = \\lfloor\\sqrt{n}\\rfloor$, and consider the snake going horizontally from $0 \\times 0$ to $1 \\times 0$, then vertically up from $1 \\times 0$ to $1 \\times 1/m$, then horizontally back from $1 \\times 1/m$ to $0... | Romania | 67th NMO Selection Tests for BMO and IMO | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null |
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