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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0kx2 | Problem:
Suppose $a_{1}, a_{2}, \ldots, a_{100}$ are positive real numbers such that
$$
a_{k}=\frac{k a_{k-1}}{a_{k-1}-(k-1)}
$$
for $k=2,3, \ldots, 100$. Given that $a_{20}=a_{23}$, compute $a_{100}$. | [
"Solution:\n\nIf we cross multiply, we obtain $a_{n} a_{n-1}=n a_{n-1}+(n-1) a_{n}$, which we can rearrange and factor as $\\left(a_{n}-n\\right)\\left(a_{n-1}-(n-1)\\right)=n(n-1)$.\n\nLet $b_{n}=a_{n}-n$. Then, $b_{n} b_{n-1}=n(n-1)$. If we let $b_{1}=t$, then we have by induction that $b_{n}=n t$ if $n$ is odd a... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 215 | |
0fie | Problem:
Se considera un triángulo $ABC$ y su circunferencia circunscrita. Si $D$ y $E$ son puntos sobre el lado $BC$ tales que $AD$ y $AE$ son, respectivamente, paralelas a las tangentes en $C$ y $B$ a la circunferencia circunscrita, demostrar que
$$
\frac{BE}{CD} = \frac{AB^2}{AC^2}
$$ | [
"Solution:\n\nLos triángulos $ABC$ y $ADC$ son semejantes pues tienen los tres ángulos iguales ya que $\\widehat{ADC} = \\widehat{BCM} = \\widehat{BAC}$. La primera igualdad sale por ser $AD$ y $CM$ paralelas y la segunda por ser $\\widehat{BCM}$ un ángulo semiinscrito. El ángulo $\\widehat{ACD}$ es común.\nEstable... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0aul | Problem:
Find the area of the triangle having vertices $A(10,-9)$, $B(19,3)$, and $C(25,-21)$. | [
"Solution:\nLet the vertices be $A(x_1, y_1) = (10, -9)$, $B(x_2, y_2) = (19, 3)$, and $C(x_3, y_3) = (25, -21)$.\n\nThe area of the triangle is given by:\n$$\n\\text{Area} = \\frac{1}{2} \\left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \\right|\n$$\nPlugging in the values:\n$$\n\\text{Area} = \\frac{1}{2} ... | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 144 | |
0l70 | Problem:
Let $r$ be the remainder when $2017^{2025!} - 1$ is divided by $2025!$. Compute $\frac{r}{2025!}$. (Note that $2017$ is prime.) | [
"Solution:\nLet $N = 2017^{2025!}$. Let $p$ be a prime dividing $2025!$ other than $2017$. Let $p^{k}$ be the largest power of $p$ dividing $2025!$. Clearly, $\\phi (p^{k}) = (p - 1)p^{k - 1}$ divides $2025!$ and $\\gcd (2017, p^{k}) = 1$, so by Euler's Totient Theorem,\n$$\nN \\equiv 1 \\pmod{p^{k}}.\n$$\nRepeatin... | United States | HMMT February | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof and answer | 1311/2017 | |
008m | In a football tournament between $n \ge 4$ teams, each pair of teams play against each other exactly once. In the final scoreboard, the scores of the teams for the tournament are $n$ consecutive numbers. Find the maximum possible value for the score of the tournament winning team.
**NOTE:** a team gets 3 points when i... | [] | Argentina | XXIX Olimpíada Matemática Argentina National Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2n - 3 | |
03ld | Problem:
Prove that when three circles share the same chord $A B$, every line through $A$ different from $A B$ determines the same ratio $X Y: Y Z$, where $X$ is an arbitrary point different from $B$ on the first circle while $Y$ and $Z$ are the points where $A X$ intersects the other two circles (labelled so that $Y$ ... | [
"Solution:\nLet $l$ be a line through $A$ different from $A B$ and join $B$ to $A, X, Y$ and $Z$ as in the above diagram. No matter how $l$ is chosen, the angles $A X B, A Y B$ and $A Z B$ always subtend the chord $A B$. For this reason the angles in the triangles $B X Y$ and $B X Z$ are the same for all such $l$. ... | Canada | 2003 CMO | [
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
06r0 | Suppose that $f$ and $g$ are two functions defined on the set of positive integers and taking positive integer values. Suppose also that the equations $f(g(n))=f(n)+1$ and $g(f(n))= g(n)+1$ hold for all positive integers. Prove that $f(n)=g(n)$ for all positive integer $n$. | [
"Throughout the solution, by $\\mathbb{N}$ we denote the set of all positive integers. For any function $h: \\mathbb{N} \\rightarrow \\mathbb{N}$ and for any positive integer $k$, define $h^{k}(x)=\\underbrace{h(h(\\ldots h(x) \\ldots))}_{k}$ (in particular, $h^{0}(x)=x$).\n\nObserve that $f\\left(g^{k}(x)\\right)=... | IMO | 51st IMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof only | null | |
02ed | Find all primes which can be written both as a sum of two primes and as a difference of two primes. | [
"The smallest prime is $2$, so it cannot be written as the sum of two primes. All other primes $p$ are odd, so if $p$ can be written as the sum of two primes and the difference of two primes, then in each case one of the primes must be $2$. Thus we have $p = q - 2 = r + 2$ for some primes $q, r$.\nIf $r \\equiv 1 \... | Brazil | X OBM | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 5 | |
049y | Bisector of an acute angle in the right triangle divides the opposite side into parts of lengths $4$ and $5$. What is the area of that triangle? | [] | Croatia | CroatianCompetitions2011 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 54 | |
0h9w | Let $x_1, x_2, ..., x_{14}$ be non-negative numbers, sum of which is equal to $1$. Prove that:
$$
x_1x_2x_3x_4 + x_2x_3x_4x_5 + \dots + x_{11}x_{12}x_{13}x_{14} + x_{12}x_{13}x_{14}x_1 + x_{13}x_{14}x_1x_2 + x_{14}x_1x_2x_3 \le \frac{1}{4^4}. \quad \text{(Serhiy Zhyliaiev)}
$$ | [
"Let's denote the left side of inequality that should be proven as $S$. Let $x_{14}x_1$ be the smallest product among all pairwise products of adjacent numbers $x_kx_{k+1}$, $k = 1, 14$ (cyclically adjacent numbers: $x_{15} = x_1$). Then we can bound from above all the summands containing this pair like this:\n$$\n... | Ukraine | The Problems of Ukrainian Authors | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof only | null | |
05r5 | Problem:
Soit $G$ un graphe orienté infini, dont tout sommet est de degré fini. On suppose que, pour chaque sommet $s$, le degré entrant de $s$ est strictement inférieur au degré sortant de $s$. Soit $v$ un sommet de $G$. Pour tout entier $n \geqslant 1$, on note $V_{n}$ le nombre de sommets que l'on peut atteindre à ... | [
"Solution:\n\nPour tout entier $n \\geqslant 0$, on note $E_{n}$ l'ensemble des sommets atteints en passant par $n$ arêtes de $G$ ou moins, et $F_{n}$ l'ensemble $E_{n} \\backslash E_{n-1}$, c'est-à-dire l'ensemble des sommets atteints en passant par $n$ arêtes de $G$ au minimum. En outre, on note $a_{n}$ le nombre... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | floor((n+2)^2 / 4) | |
0aa2 | Problem:
Let $ABCD$ be a cyclic quadrilateral satisfying $AB = AD$ and $AB + BC = CD$.
Determine $\angle CDA$. | [
"Solution:\n$\\angle CDA = 60^{\\circ}$.\n\nChoose the point $E$ on the segment $CD$ such that $DE = AD$. Then $CE = CD - AD = CD - AB = BC$, and hence the triangle $CEB$ is isosceles.\n\n\n\nNow, since $AB = AD$ then $\\angle BCA = \\angle ACD$. This shows that $CA$ is the bisector of $\\a... | Nordic Mathematical Olympiad | Nordic Mathematical Contest | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 60 degrees | |
042b | Suppose real numbers $x$, $y$ satisfy $x - 4\sqrt{y} = 2\sqrt{x-y}$.
Then the range of $x$ is ______. | [
"Let $\\sqrt{y} = a$, $\\sqrt{x-y} = b$ ($a, b \\ge 0$). Then $x = y + (x-y) = a^2 + b^2$. The equation in the question becomes $a^2 + b^2 - 4a = 2b$, which is equivalent to\n$$\n(a - 2)^2 + (b - 1)^2 = 5 \\quad (a, b \\ge 0).\n$$\nAs seen in Fig. 7.1, the trace of point $(a, b)$ in plane $aOb$ is the part of the c... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | {0} ∪ [4, 20] | |
05nn | Problem:
Soient $n \geqslant 2$ et $a_{1}, \ldots, a_{n}, b_{1}, \ldots, b_{n}$ des réels strictement positifs. Les $a_{i}$ sont supposés deux à deux distincts, de même que les $b_{i}$. On suppose qu'il est possible de séparer les $a_{i}$ en deux sous-ensembles de même somme, et de même pour les $b_{i}$.
Montrer qu'i... | [
"Solution:\n\nQuitte à changer les indices et éventuellement à échanger les rôles de $a$ et $b$, on peut supposer qu'il existe $\\ell \\leqslant m \\leqslant \\frac{n}{2}$ tels que $a_{1}+\\cdots+a_{\\ell}=a_{\\ell+1}+\\cdots+a_{n}=S_{a}$ et $b_{1}+\\cdots+b_{m}=b_{m+1}+\\cdots+b_{n}$, que $b_{1} \\geqslant \\cdots... | France | OCympiades Françaises de Mathématiques - Envoi Numéro 4 - Combinatoire | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0jb4 | Problem:
Luna has an infinite supply of red, blue, orange, and green socks. She wants to arrange $2012$ socks in a line such that no red sock is adjacent to a blue sock and no orange sock is adjacent to a green sock. How many ways can she do this? | [
"Solution:\n\n$4 \\cdot 3^{2011}$\n\nLuna has $4$ choices for the first sock. After that, she has $3$ choices for each of $2011$ remaining socks for a total of $4 \\cdot 3^{2011}$."
] | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 4 * 3^2011 | |
0dc7 | An $11 \times 11$ square is partitioned into 121 smaller $1 \times 1$ squares, 4 of which are painted black, the rest being white. We cut a fully white rectangle (possibly a square) out of the big $11 \times 11$ square. What is the maximal area of the rectangle we can obtain regardless of the positions of the black squ... | [
"In the first image we have position for 4 black cells, such that the biggest rectangle without black is $25$. Now let's prove that for any configuration there exists a rectangle of area $25$. Assume for some configuration the biggest rectangle is at most $24$. Let's divide the board into 4 squares of size $5$ like... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 25 | |
0gpq | The circle $\omega_1$ with diameter $[AB]$ and the circle $\omega_2$ with center $A$ intersect at points $C$ and $D$. Let $E$ be a point on the circle $\omega_2$, which is outside $\omega_1$ and at the same side with $C$ with respect to the line $AB$. Let the second point of intersection of the line $BE$ with $\omega_2... | [
"\n\nLet $D'$ be the point symmetric to $D$ with respect to the line $BE$. $2 \\cdot CK \\cdot AC = CE \\cdot AB$ implies $CK/CE = R_1/R_2$ where $R_1$ and $R_2$ are radii of the circles $\\omega_1$ and $\\omega_2$, respectively. By the sine law, $2 \\cdot \\sin \\angle CLF = CK/R_1 = CE/R_... | Turkey | 21st Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
01oo | Solve the equation in prime numbers $p$ and $q$: $q^3 = p^2 - p + 1$. | [
"Answer: $p = 19, q = 7$.\n\n(Solution of E. Dauhiala, B. Gilevich, A. Zhuk, A. Semchankau.) We have\n$$\nq^3 = p^2 - p + 1 \\Leftrightarrow (q-1)(q^2+q+1) = p(p-1). \\quad (1)\n$$\nIf $(q-1) \\nmid p$, then $q \\ge p+1$, so $q^3 > p^2 - p + 1$.\nHence, $(q^2+q+1) \\nmid p$, i.e.\n$$\nq^2 + q + 1 = k p \\quad (2)\n... | Belarus | BelarusMO 2013_s | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | p = 19, q = 7 | |
0egz | Problem:
Naj bosta $\alpha$ in $\beta$ ostra kota pravokotnega trikotnika. Kateri izmed navedenih izrazov je enakovreden izrazu
$$
\frac{\sin \left(90^{\circ}-\alpha\right)+\cos \beta}{4 \sin \beta \cdot \cot \left(90^{\circ}-\alpha\right)}
$$
?
(A) $\frac{\sin \beta+\cos \beta}{4 \cos \beta}$
(B) $\frac{\sin \beta+... | [
"Solution:\n\nUporabimo zveze $\\sin \\left(90^{\\circ}-\\alpha\\right)=\\cos \\beta$, $\\cot \\left(90^{\\circ}-\\alpha\\right)=\\frac{\\cos \\left(90^{\\circ}-\\alpha\\right)}{\\sin \\left(90^{\\circ}-\\alpha\\right)}=\\frac{\\cos \\beta}{\\sin \\beta}$. Prvotni izraz preoblikujemo v\n$$\n\\frac{(\\sin \\beta+\\c... | Slovenia | Državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | A | |
0hw8 | Problem:
The inscribed circle of a triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at $D$, $E$, and $F$ respectively. Let $X$, $Y$, and $Z$ be the incenters of triangles $AEF$, $BFD$, and $CDE$, respectively. Prove that $DX$, $EY$, and $CZ$ meet at one point. | [
"Solution:\n\nConsider the midpoint $M$ of arc $EF$ on the incircle of $\\triangle ABC$. Angles $AFM$ and $MFE$ are equal since they intercept equal arcs $FM$ and $ME$, and so $M$ is on the bisector of $\\angle AFE$. Similarly, $M$ is on the bisector of $\\angle FEA$, and therefore $M$ coincides with $X$. Moreover,... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dqn | The numbers $\frac{1}{2}, \frac{1}{3}, \dots, \frac{1}{2010}$ are written on a blackboard. A student chooses any two of the numbers, say $x, y$, erases them and then writes down $x + y + xy$. He continues to do this until only one number is left on the blackboard. What is this number? | [
"We shall prove by induction that if the original numbers are $a_1, \\dots, a_n$, $n \\ge 2$, then the last number is $(1+a_1) \\cdots (1+a_n) - 1$.\nThe assertion is certainly true for $n=2$, the base case. Now suppose it is true for $n=k \\ge 2$. Consider $k+1$ numbers $a_1, \\dots, a_{k+1}$ written on the board.... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2009/2 | |
0196 | Find all functions $f$ from the set of all positive integers to the same set such that, for all positive integers $a_1, \dots, a_k$ with $k > 0$, the sum $a_1 + \dots + a_k$ divides the sum $f(a_1) + \dots + f(a_k)$. | [
"**Answer:** All functions given by $f(n) = an$, $a \\in \\mathbb{N}$.\n\nSuppose that $f$ is a function that satisfies the conditions of the problem. We claim that $f(n) = f(n-1) + f(1)$ for all integers $n > 1$. Indeed, for any integer $m > n$, we have $m \\mid f(n) + f(m-n)$ and $m \\mid f(n-1) + f(1) + f(m-n)$ ... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | All functions of the form f(n) = a n with a a positive integer. | |
08yj | Suppose quadratic polynomials $P$, $Q$, $R$ with integer coefficients satisfy the following properties:
* $P(1) = P(2) = Q(3) = 0$.
* $P(x)^2 + Q(x)^2 = R(x)^2$ holds for any real number $x$.
* There exists no integer greater than or equal to $2$, which divides all of the numbers appearing as a coefficient of any of $P... | [
"$5x^2 - 18x + 17$\nBy the given conditions, we see that $P(x) = a(x-1)(x-2)$ with $a$ being a non-zero constant. We then see that we have\n$$\na^2(x-1)^2(x-2)^2 = P(x)^2 = R(x)^2 - Q(x)^2 = (R(x) + Q(x))(R(x) - Q(x))\n$$\nvalid for all $x$. We know that both $R(x) + Q(x)$ and $R(x) - Q(x)$ have degrees less than o... | Japan | 2019 Japan Mathematical Olympiad First Stage | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 5x^2 - 18x + 17 | |
0avn | Problem:
Let $f(x) = \ln x$. What are the values of $x$ in the domain of $(f \circ f \circ f \circ f \circ f)(x)$? | [
"Solution:\nWe have $(f \\circ f \\circ f \\circ f \\circ f)(x) = \\ln (\\ln (\\ln (\\ln (\\ln (x)))))$.\n\nThe domain of this function must satisfy $\\ln (\\ln (\\ln (\\ln (x)))) > 0$, which implies that $\\ln (\\ln (\\ln (x))) > 1$, $\\ln (\\ln (x)) > e$, $\\ln (x) > e^{e}$, and $x > e^{e^{e}}$."
] | Philippines | 18th PMO National Stage Oral Phase | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | x > e^{e^{e}} | |
09sa | Problem:
Zij $n \geq 2$ een positief geheel getal. Ieder vakje van een $n \times n$-bord wordt rood of blauw gekleurd. We leggen dominostenen op het bord, die elk twee vakjes bedekken. We noemen een dominosteen effen als hij op twee rode of twee blauwe vakjes ligt en kleurrijk als hij op een rood en een blauw vakje li... | [
"Solution:\n\nWe bewijzen dat $k=\\left\\lfloor\\frac{n^{2}}{4}\\right\\rfloor$ de grootst mogelijke waarde is.\n\nStel dat $n$ even is. Dan is het mogelijk om het bord te bedekken met $\\frac{n^{2}}{2}$ dominostenen (zonder op kleuren te letten). Omdat er $\\frac{n^{2}}{2}$ dominostenen zijn die allemaal kleurrijk... | Netherlands | Selectietoets | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | floor(n^2 / 4) | |
04ex | Let $a$ and $b$ be real numbers. It is known that the parabola $y = ax^2 + b$ intersects the curve $y = x + \frac{1}{x}$ at exactly three points. Prove that $3ab < 1$. | [
"From the condition of the problem we conclude that the system of equations\n$$\ny = ax^2 + b\n$$\n$$\ny = x + \\frac{1}{x}\n$$\nhas three solutions, i.e. that the cubic equation\n$$\nax^3 - x^2 + bx - 1 = 0\n$$\nhas three real solutions. Let us denote these solutions by $x_1, x_2, x_3$. Viète's formulas give\n$$x_... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0ij4 | Find the least real number $k$ with the following property: if the real numbers $x$, $y$, and $z$ are not all positive, then
$$
k(x^2 - x + 1)(y^2 - y + 1)(z^2 - z + 1) \geq (xyz)^2 - xyz + 1.
$$ | [
"The answer is $k = \\frac{16}{9}$.\nWe start with a lemma.\n\nLemma 1. If real numbers $s$ and $t$ are not all positive, then\n$$\n\\frac{4}{3}(s^2 - s + 1)(t^2 - t + 1) \\geq (st)^2 - st + 1. \\quad (*)\n$$\nProof: Without loss of generality, we assume that $s \\geq t$.\nWe first assume that $s \\geq 0 \\geq t$. ... | United States | Team Selection Test | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 16/9 | |
0gff | 給定正整數 $n$。設 $S$ 為坐標平面上 $x$ 坐標及 $y$ 坐標皆小於 $2n$ 的非負整數的點所成的集合 (所以 $S$ 共包含 $4n^2$ 個點)。假設 $\mathcal{F}$ 為 $n^2$ 個四邊形所成的集合, 其中每個四邊形的所有頂點都在 $S$ 內, 並且 $S$ 裡的每個點皆是 $\mathcal{F}$ 中其中一個四邊形的頂點。試求 $\mathcal{F}$ 內所有 $n^2$ 個四邊形的面積總和的最大可能值。 | [
"面積總和的最大可能值為 $\\Sigma(n) = \\frac{1}{3}n^2(2n+1)(2n-1)$。\n\n以下皆以 $[P]$ 代表多邊形 $P$ 的面積。我們先作一些約定。如果一個多邊形的頂點皆屬於 $S$, 稱之為合法的。設 $O = (n - \\frac{1}{2}, n - \\frac{1}{2})$ 為 $S$ 的中心點。若一個合法的正方形的中心為 $O$ 點, 稱該正方形為置中的。最後, 若一組多邊形 $\\mathcal{F}$ 滿足題設條件, 稱 $\\mathcal{F}$ 為可接受的, 並將其面積總和記為 $\\Sigma(\\mathcal{F})$。\n\n$S$ 內的每一個點都恰是... | Taiwan | 2022 數學奧林匹亞競賽第二階段選訓營, 國際競賽實作(一) | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | Chinese; English | proof and answer | n^2(2n+1)(2n-1)/3 | |
0e4r | Let $ABCDEF$ be a hexagon with $\angle BAF = 150^\circ$, $\angle ACB = \angle ADC = 90^\circ$ and $|AC| = |BC|$. Assume also that the triangle $ABC$ is similar to the triangle $ADE$ and the triangle $BCD$ is similar to the triangle $DEF$. Find the ratio of the lengths of the segments $AB$ and $AF$. | [
"Let $|AB| = a$. Since $ABC$ is an isosceles right triangle with the apex at $C$, we have $|AC| = \\frac{a}{\\sqrt{2}} = \\frac{a\\sqrt{2}}{2}$ and $\\angle ABC = \\angle BAC = \\frac{\\pi}{4}$. The triangles $ADE$ and $ABC$ are similar, so $\\angle AED = \\angle ACB = \\frac{\\pi}{2}$ and $|DE| = |AE|$.\n\n![](att... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 8/3 | |
0jp9 | Problem:
Consider a $3 \times 3$ grid of squares. A circle is inscribed in the lower left corner, the middle square of the top row, and the rightmost square of the middle row, and a circle $O$ with radius $r$ is drawn such that $O$ is externally tangent to each of the three inscribed circles. If the side length of eac... | [
"Solution:\n\nLet $A$ be the center of the square in the lower left corner, let $B$ be the center of the square in the middle of the top row, and let $C$ be the center of the rightmost square in the middle row. It's clear that $O$ is the circumcenter of triangle $ABC$—hence, the desired radius is merely the circumr... | United States | HMMT November | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | (5√2 - 3)/6 | |
0dl7 | The set $M$ of pairwise different positive integers is given. It is known that within any 3 of them one can choose 2 numbers such that their sum is power of 2. Find the maximal number of elements $M$ may contain. | [] | Saudi Arabia | Saudi Booklet | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 4 | |
041s | Suppose the side of the base and the height of regular triangular pyramid $P$-$ABC$ are $1$ and $\sqrt{2}$, respectively. Then the radius of the inscribed sphere of the pyramid is ______. | [
"As seen in Fig. 4.1, suppose the projections of the inscribed sphere's center $O$ on faces $ABC$ and $ABP$ are $H$, $K$, respectively, the midpoint of $AB$ is $M$, and the radius of the sphere is $r$. Then $P$, $K$, $M$ are collinear, $\\angle PHM = \\angle PKO = \\frac{\\pi}{2}$, and\n\nF... | China | China Mathematical Competition | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | English | final answer only | sqrt(2)/6 | |
020q | Problem:
Pawns and rooks are placed on a $2019 \times 2019$ chessboard, with at most one piece on each of the $2019^{2}$ squares. A rook can see another rook if they are in the same row or column and all squares between them are empty. What is the maximal number $p$ for which $p$ pawns and $p+2019$ rooks can be placed... | [
"Solution:\n\nAnswer: the maximal $p$ equals $1009^{2}$.\nWrite $n=2019$ and $k=1009$; then $n=2k+1$. We first show that we can place $k^{2}$ pawns and $n+k^{2}$ rooks. Each cell of the chess board has coordinates $(x, y)$ with $1 \\leqslant x, y \\leqslant n$. We colour each cell black or white depending on whethe... | Benelux Mathematical Olympiad | Benelux Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 1009^2 | |
0fhr | Problem:
Una máquina de juego de un casino tiene una pantalla en la que se ofrece un esquema como el de la figura. Al comenzar el juego aparece una bola en el punto $S$.
A cada impulso del jugador, la bola se mueve hasta uno de los círculos inmediatos, con la misma probabilidad para cada uno de ellos. La partida acaba... | [
"Solution:\n\nPartiendo de $S$, la probabilidad de llegar a $C$ o a $D$ es $\\frac{2}{3}$; una vez en $C$ o en $D$, la probabilidad de ganar es igual que la de perder, $\\frac{1}{2}$. Por tanto, partiendo de $S$, la probabilidad de ganar es $\\frac{2}{3} \\cdot \\frac{1}{2} = \\frac{1}{3}$.\n\nDel análisis del jueg... | Spain | OME 29 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | Probability of winning: 1/3; expected number of moves: 6 | |
069d | Let $AΓ$ be a triangle with circumcircle $c(O, R)$, such that $AB < AΓ < BΓ$ and let $Δ$ be the antipodal point of $B$ with respect to the circle $c$. The perpendicular bisector of $BΔ$ meets $BΓ$ at $K$, $AΓ$ at $M$ and $AB$ at $N$. The line $NΔ$ intersects the circle $c$ at point $T$. Let $Σ$ be the second point of i... | [
"\n\nWe will prove that the point $T$ belongs to the circle $c_1(O, Γ, M)$ and the point $N$ belongs to the circle $c_2(O, A, Δ)$, that is the quadrilaterals $NAOΔ$ and $MOΓT$ are cyclic. Then we conclude the following: the line $AΔ$ is the radical axis of the circles $C$ and $c_2$, the lin... | Greece | SELECTION EXAMINATION 2019 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
01kb | Point $M$ is the midpoint of the side $AB$ of the acute-angled triangle $ABC$, points $P$ and $Q$ are the feet of the altitudes $AP$ and $BQ$. A circle passing through $B, M, P$, touches the side $AC$. Prove that a circle passing through $A, M, Q$ touches the extension of the side $BC$.
(I. Voronovich) | [
"First solution. Let $\\Gamma$ be a circle passing through $M, P, B$. Let $\\Gamma$ touch the side $AC$ at $S$, and $T$ be the intersection of the lines $MS$ and $BC$. Since point $S$ lies on the side $AC$, it follows that $T$ lies on the extension of the side $BC$ (over $C$).\n\n\n\nSince ... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chas... | English | proof only | null | |
0h8e | For which positive integers $n$ there exist $2n$ pairwise different positive integers $a_1, a_2, \dots, a_n$ and $b_1, b_2, \dots, b_n$ satisfying the equations:
$$
a_1 + a_2 + \dots + a_n = b_1 + b_2 + \dots + b_n \text{ and } a_1 a_2 \dots a_n = b_1 b_2 \dots b_n?
$$ | [
"**Answer:** for all $n \\ge 3$.\n\nAssume $n \\ge 3$. Let us introduce such denotations:\n$$\na_1 + a_2 + \\dots + a_{n-1} = a, \\quad a_1 a_2 \\dots a_{n-1} = A, \\quad b_1 + b_2 + \\dots + b_{n-1} = b, \\quad b_1 b_2 \\dots b_{n-1} = B. \\quad (1)\n$$\nThen we can write that $a + a_n = b + b_n$, $A a_n = B b_n$,... | Ukraine | UkraineMO | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All positive integers n greater than or equal to 3 | |
0jl1 | Problem:
Ten points are equally spaced on a circle. A graph is a set of segments (possibly empty) drawn between pairs of points, so that every two points are joined by either zero or one segments. Two graphs are considered the same if we can obtain one from the other by rearranging the points.
Let $N$ denote the number... | [
"Solution:\nAnswer: 11716571 The question asks for the number of isomorphism classes of connected graphs on 10 vertices. This is enumerated in http://oeis.org/A001349, the answer is 11716571.\n\nIn fact, of the $2^{45} \\approx 3.51 \\cdot 10^{13} \\approx 3 \\cdot 10^{13}$ graphs on 10 labelled vertices, virtually... | United States | HMMT November 2014 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 11716571 | |
00lw | Determine all integers $n \ge 2$ that have a representation
$$
n = a^2 + b^2,
$$
where $a$ is the smallest divisor of $n$ different from $1$ and $b$ is an arbitrary divisor of $n$. | [
"If $n$ is odd, then both $a$ and $b$ are odd and therefore $n = a^2 + b^2$ is even, contradiction. Therefore, $n$ is even and $a = 2$. This also shows that $b$ is even. Furthermore, $b \\mid (n - b^2) = a^2 = 4$. Thus $b \\in \\{2, 4\\}$, which results in $n = 8$ and $n = 20$, respectively."
] | Austria | 48th Austrian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | n = 8 or n = 20 | |
0g0f | Problem:
Sei $ABC$ ein Dreieck mit $\angle BAC = 60^{\circ}$. Sei $E$ der Punkt auf der Seite $BC$, sodass $2 \angle BAE = \angle ACB$ gilt. Sei $D$ der zweite Schnittpunkt von $AB$ und dem Umkreis des Dreiecks $AEC$ und sei $P$ der zweite Schnittpunkt von $CD$ und dem Umkreis des Dreiecks $DBE$. Berechne den Winkel $... | [
"Solution:\n\nSei Winkel $\\angle EAB = \\gamma$. Da $\\angle EAB = \\angle DCE$ ist, liegen die vier Punkte $A$, $D$, $E$ und $C$ auf einem Kreis. Somit gilt $60^{\\circ} - \\gamma = \\angle CAE = \\angle CDE = \\angle PDE = \\angle PBE$, wobei die letzte Gleichheit wegen Sehnenviereck $DBEP$ gilt.\n\nIn Dreieck $... | Switzerland | SMO - Finalrunde | [
"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 and answer | 30° | |
0j2o | Problem:
$AB$ is a diameter of circle $O$. $X$ is a point on $AB$ such that $AX = 3 BX$. Distinct circles $\omega_1$ and $\omega_2$ are tangent to $O$ at $T_1$ and $T_2$ and to $AB$ at $X$. The lines $T_1 X$ and $T_2 X$ intersect $O$ again at $S_1$ and $S_2$. What is the ratio $\frac{T_1 T_2}{S_1 S_2}$? | [
"Solution:\n\nAnswer: $\\frac{3}{5}$\n\nSince the problem only deals with ratios, we can assume that the radius of $O$ is $1$. As we have proven in Problem 5, points $S_1$ and $S_2$ are midpoints of arc $AB$. Since $AB$ is a diameter, $S_1 S_2$ is also a diameter, and thus $S_1 S_2 = 2$.\n\nLet $O_1$, $O_2$, and $P... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 3/5 | |
00wq | Problem:
Let the coordinate planes have the reflection property. A beam falls onto one of them. How does the final direction of the beam after reflecting from all three coordinate planes depend on its initial direction? | [
"Solution:\nLet the velocity vector of the beam be $\\vec{v} = (\\alpha, \\beta, \\gamma)$. Reflection from each of the coordinate planes changes the sign of exactly one of the coordinates $\\alpha$, $\\beta$ and $\\gamma$, and thus the final direction will be opposite to the initial one."
] | Baltic Way | Baltic Way | [
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | The final direction is exactly opposite to the initial direction. | |
0cbs | Find the functions $f : \mathbb{R} \to \mathbb{R}$, satisfying
$$
f(f(x)) + y \cdot f(x) \leq x + x \cdot f(f(y)),
$$
for every real numbers $x$ and $y$. | [
"For $x = 0$ in the given relation we get $f(f(0)) + y f(0) \\le 0$, for every $y \\in \\mathbb{R}$, which is true only if $f(0) = 0$. Indeed, if $f(0) \\ne 0$, then $y \\le \\frac{-f(f(0))}{f(0)}$, for every $y \\in \\mathbb{R}$, or $y \\ge \\frac{-f(f(0))}{f(0)}$, for every $y \\in \\mathbb{R}$, which is impossib... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | f(x) = x | |
08m0 | Problem:
Find all values of the real parameter $a$, for which the system
$$
\left\{\begin{array}{c}
(|x|+|y|-2)^{2}=1 \\
y=a x+5
\end{array}\right.
$$
has exactly three solutions. | [
"Solution:\nThe first equation is equivalent to\n$$\n|x|+|y|=1\n$$\nor\n$$\n|x|+|y|=3\n$$\nThe graph of the first equation is symmetric with respect to both axes. In the first quadrant it is reduced to $x+y=1$, whose graph is segment connecting points $(1,0)$ and $(0,1)$. Thus, the graph of\n$$\n|x|+|y|=1\n$$\nis s... | JBMO | 2009 Shortlist JBMO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | a = -5 or a = 5 | |
0b4m | Problem:
In $\triangle ABC$, $AB > AC$. Point $P$ is on line $BC$ such that $AP$ is tangent to its circumcircle. Let $M$ be the midpoint of $AB$, and suppose the circumcircle of $\triangle PMA$ meets line $AC$ again at $N$. Point $Q$ is the reflection of $P$ with respect to the midpoint of segment $BC$. The line throu... | [
"Solution:\n\n\n\nLet lines $BE$ and $AC$ meet at $R$. It suffices to show that the points $P$, $M$ and $R$ are collinear. Since $AP$ is tangent to the circumcircle of $ABC$ and $PNA M$ is cyclic, we have $\\angle PNM = \\angle PAB = \\angle ACB$ and $\\angle BAC = \\angle MPN$. Thus, trian... | Philippines | 25th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a8w | Problem:
Let $a$, $b$, and $c$ be positive real numbers. Prove that
$$
\frac{2 a^{2}}{b+c}+\frac{2 b^{2}}{c+a}+\frac{2 c^{2}}{a+b} \geq a+b+c
$$ | [
"Solution:\nSolution 1. Use brute force. Removing the denominators and brackets and combining similar terms yields the equivalent inequality\n$$\n\\begin{gathered}\n0 \\leq 2 a^{4}+2 b^{4}+2 c^{4}+a^{3} b+a^{3} c+a b^{3}+b^{3} c+a c^{3}+b c^{3} \\\\\n-2 a^{2} b^{2}-2 b^{2} c^{2}-2 a^{2} c^{2}-2 a b c^{2}-2 a b^{2} ... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 19 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0a2m | We are looking at numbers for which two adjacent digits always add up to the digit immediately to their right (if it exists). So two adjacent digits when added together are always smaller than $10$, except possibly the last two digits. Furthermore, the first digit cannot be $0$. An example of such a number is $1347$, b... | [] | Netherlands | Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | MCQ | E | |
08tq | For a convex pentagon, all of its sides have length $1$ and some pair of diagonals intersect perpendicularly. Find the maximum possible area that such a pentagon can have. A pentagon is said to be convex if all of its interior angles are less than $180^\circ$. | [
"Let $ABCDE$ be a convex pentagon with all of its sides having length $1$ and for which some pair of diagonals intersect perpendicularly. It is sufficient to consider the two cases where $AC \\perp AD$ or $AC \\perp BD$, since all other cases can be handled in the same way as for one of these two cases.\n\nLet us f... | Japan | Japan Junior Mathematical Olympiad First Round | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 1 + sqrt(3)/4 | |
02q8 | Problem:
O contrário de um número de dois algarismos, ambos diferentes de zero, é o número obtido trocando-se a ordem de seus algarismos. Por exemplo, o contrário de $25$ é $52$ e o contrário de $79$ é $97$. Qual dos números abaixo não é a soma de um número de dois algarismos com o seu contrário?
A) $44$
B) $99$
C) $12... | [
"Solution:\nSeja $n$ um número de dois algarismos, sendo $a$ seu algarismo das dezenas e $b$ o das unidades; então $n = 10a + b$. Se $a$ e $b$ são ambos diferentes de zero, o contrário de $n$ é $10b + a$. Desse modo, a soma de $n$ e de seu contrário é:\n$$\n(10a + b) + (10b + a) = 11a + 11b = 11(a + b)\n$$\ne, port... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | E | |
0dw3 | Problem:
V trgovini Moda je stal moški suknjič po $30\%$ pocenitvi $24500$ SIT. Pred koncem razprodaje so ga pocenili še za $20\%$. Koliko tolarjev znaša razlika med začetno ceno in ceno po drugi pocenitvi?
V trgovini Obleka je imel tak suknjič enako začetno ceno kot v trgovini Moda. Pocenili so ga le enkrat in takoj... | [
"Solution:\n\nČe označimo začetno ceno suknjiča z $x$, velja $0,7x = 24500$, od koder izračunamo $x = 35000$.\n\nPo drugi pocenitvi je suknjič stal $0,8 \\cdot 24500 = 19600$ SIT.\n\nRazlika med začetno ceno in ceno po drugi pocenitvi je $35000 - 19600 = 15400$ SIT.\n\nV trgovini Obleka so pocenili suknjič, ki je s... | Slovenia | 4. državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 15400 SIT, 44% | |
0ikn | Problem:
Find all positive real numbers $c$ such that the graph of $f: \mathbb{R} \rightarrow \mathbb{R}$ given by $f(x) = x^{3} - c x$ has the property that the circle of curvature at any local extremum is centered at a point on the $x$-axis. | [
"Solution:\nThe equation $0 = f'(x) = 3x^{2} - c$ has two real roots: $\\pm \\sqrt{c / 3}$. Let $a := \\sqrt{c / 3}$. As $f''(-a) = -6 \\sqrt{c / 3} < 0$, $f$ has a unique local maximum at $x = -a$.\n\nBecause $f$ has half-turn symmetry about the origin, it suffices to consider this local extremum. The radius of cu... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Differential Geometry > Curvature"
] | null | proof and answer | sqrt(3)/2 | |
06po | Find all pairs $(k, n)$ of positive integers for which $7^{k}-3^{n}$ divides $k^{4}+n^{2}$. | [
"Suppose that a pair $(k, n)$ satisfies the condition of the problem. Since $7^{k}-3^{n}$ is even, $k^{4}+n^{2}$ is also even, hence $k$ and $n$ have the same parity. If $k$ and $n$ are odd, then $k^{4}+n^{2} \\equiv 1+1=2\\pmod{4}$, while $7^{k}-3^{n} \\equiv 7-3 \\equiv 0\\pmod{4}$, so $k^{4}+n^{2}$ cannot be div... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | (2,4) | |
0ciq | For each integer $n \ge 2$, we consider the set of all $n$-th roots of unity $U_n = \{z \in \mathbb{C} \mid z^n = 1\}$. Determine the integers $n \ge 2$ for which there exist $x, y \in U_n$, $x \ne y$ such that $x + y \notin U_k$ for any integer $k \ge 2$. | [] | Romania | 75th NMO | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof and answer | All integers n ≥ 2 except n = 3 | |
0kiq | Problem:
Ava and Tiffany participate in a knockout tournament consisting of a total of 32 players. In each of 5 rounds, the remaining players are paired uniformly at random. In each pair, both players are equally likely to win, and the loser is knocked out of the tournament. The probability that Ava and Tiffany play e... | [
"Solution:\n\nEach match eliminates exactly one player, so exactly $32-1=31$ matches are played, each of which consists of a different pair of players. Among the $\\binom{32}{2}=\\frac{32 \\cdot 31}{2}=496$ pairs of players, each pair is equally likely to play each other at some point during the tournament. Therefo... | United States | HMMT Spring 2021 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 116 | |
0jf9 | Problem:
Define a size-$n$ tromino to be the shape you get when you remove one quadrant from a $2n \times 2n$ square. In the figure below, a size-1 tromino is on the left and a size-2 tromino is on the right.
We say that a shape can be tiled with size-1 trominos if we can cover the entire a... | [
"Solution:\n\nWe will abbreviate \"tile with size-1 trominos\" with \"tile.\" It is possible to tile a size-5 tromino as drawn.\n\n\n\nIt is also possible to tile a size-2013 tromino. In fact, any size-$n$ tromino can be tiled with size-1 trominos, which can be proved with mathematical indu... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | a) Yes. b) Yes. | |
0crh | На плоскости дано $n$ выпуклых попарно пересекающихся $k$-угольников. Любой из них можно перевести в любой другой гомотетией с положительным коэффициентом. Докажите, что на плоскости найдётся точка, принадлежащая хотя бы $1 + \frac{n-1}{2k}$ из этих $k$-угольников. | [
"Лемма.\nПусть $P$ и $P'$ — пересекающиеся выпуклые многоугольники, гомотетичные с положительным коэффициентом. Тогда одна из вершин одного из них лежит в другом.\n\nДоказательство. Если один из многоугольников полностью лежит в другом, то утверждение очевидно. В противном случае найдётся сторона $AB$ многоугольник... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0cys | Let $ABC$ be a triangle with $AB \neq AC$. Its incircle has center $I$ and touches the side $BC$ at point $D$. Line $AI$ intersects the circumcircle $\mathcal{C}$ of triangle $ABC$ at $M$ and $DM$ intersects again $\mathcal{C}$ at $P$. Prove that $\widehat{API}=90^{\circ}$. | [
"Assume that $AB < AC$. We have $MB = MC = MI$. Indeed, from the hypothesis it follows\n$$\n\\widehat{IBM} = \\frac{1}{2}(\\widehat{A} + \\widehat{B}) = \\widehat{BIM},\n$$\nhence triangle $BMI$ is isosceles.\n\n\n\nFurthermore, since\n$$\n\\widehat{BPM} = \\frac{1}{2} \\widehat{A} = \\wide... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0afz | Find all natural numbers $a$ and $b$ greater than $1$, such that $b^a \mid (a^b - 1)$. | [
"Let $p$ be the least prime factor of $b$, and $q$ be the least natural number such that $p\\mid(a^q-1)$ (such a number exists because $p\\mid(a^b-1)$). From Little Fermat's theorem, we have that $p\\mid(a^{p-1}-1)$, which implies $q\\mid b$ and $q\\mid(p-1)$, and from the minimality of $p$, we get that $q=1$, i.e.... | North Macedonia | null | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof and answer | (a, b) = (3, 2) | |
06cs | Find the remainder of $1999^{2000}$ when it is divided by $31$. | [
"The remainder is $1$.\nNote that $1999 \\equiv -16 = -2^4 \\pmod{31}$. Therefore, we have\n$$\n1999^{2000} \\equiv (2^4)^{2000} = (2^5)^{1600} = 32^{1600} \\equiv 1 \\pmod{31}.\n$$\nThus, the remainder is $1$."
] | Hong Kong | HKG TST | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | final answer only | 1 | |
0hdq | We have to place natural numbers $a_1, a_2, \dots, a_n$ that are not all equal at the vertices of a regular $n$-sided polygon $A_1A_2\dots A_n$, $n \ge 6$ with center at $O$ in such a way that for every vertex $A_i$ there exist two vertices $A_k$ and $A_l$, that are symmetrical with respect to the line $OA_i$ and the e... | [
"Let us assume we managed to place numbers properly. Let $m = \\min_{j=1, n} a_j$, $M = \\max_{j=1, n} a_j$. If $m = a_i = \\frac{1}{2}(a_k + a_l)$, then $m = a_k = a_l$, hence there have to be at least 3 of the smallest numbers, similarly for the biggest. It is easy to fill in numbers when\n\n![](attached_image_1.... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | all integers n ≥ 6 with n ≠ 7 | |
0jix | Problem:
For an integer $n$, let $f_{9}(n)$ denote the number of positive integers $d \leq 9$ dividing $n$. Suppose that $m$ is a positive integer and $b_{1}, b_{2}, \ldots, b_{m}$ are real numbers such that $f_{9}(n)=\sum_{j=1}^{m} b_{j} f_{9}(n-j)$ for all $n>m$. Find the smallest possible value of $m$. | [
"Solution:\n\nAnswer: $28$\n\nLet $M=9$. Consider the generating function\n$$\nF(x)=\\sum_{n \\geq 1} f_{M}(n) x^{n}=\\sum_{d=1}^{M} \\sum_{k \\geq 1} x^{d k}=\\sum_{d=1}^{M} \\frac{x^{d}}{1-x^{d}}\n$$\nObserve that $f_{M}(n)=f_{M}(n+M!)$ for all $n \\geq 1$ (in fact, all $n \\leq 0$ as well). Thus $f_{M}(n)$ satis... | United States | HMMT 2014 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Algebra > Intermediate Algebr... | null | proof and answer | 28 | |
04yr | In a coordinate city there are $n \ge 3$ tramlines parallel to the $x$-axis such that each line begins from $x$-coordinate $0$ and ends at $x$-coordinate $n$. Exactly one tram of length $1$ is moving on each line: on the first line with speed $1$, on the second line with speed $2$ etc, until on the last line with the s... | [
"The projections of the trams can cover the whole interval only when one projection covers $[0, 1]$, another $[1, 2]$ etc. until $[n-1, n]$. Consider the moments when the projection of the slowest tram covers one of these intervals. When the slowest tram moves by $1$ unit, then the fastest and the third fastest tra... | Estonia | Estonija 2010 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0dur | Problem:
Naj bo $S=\{a_{1}, \ldots, a_{n}\}$, kjer so $a_{i}$ različna naravna števila. Vsota števil iz nobene prave podmnožice množice $S$ ni deljiva z $n$. Dokaži, da je vsota vseh števil iz množice $S$ deljiva z $n$. | [
"Solution:\n\nOznačimo $s_{j}=a_{1}+\\ldots+a_{j}$. Če za $j<k$ velja $s_{j} \\equiv s_{k} \\pmod{n}$, je $a_{j+1}+\\ldots+a_{k} \\equiv 0$ $\\pmod{n}$, kar je v protislovju s predpostavko naloge. Torej dajo števila $s_{1}, \\ldots, s_{n}$ vse možne ostanke pri deljenju z $n$. Ker po predpostavki nobeno od števil $... | Slovenia | 46. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
03p9 | Let $A$ be a subset of the set $S = \{1, 2, \dots, 1000000\}$ containing exactly $101$ elements. Prove that there exist numbers $t_1, t_2, \dots, t_{100}$ in $S$ such that the sets
$$
A_j = \{x + t_j \mid x \in A\} \text{ for } j = 1, 2, \dots, 100
$$
are pairwise disjoint. | [
"Consider the set $D = \\{x-y \\mid x, y \\in A\\}$. There are at most $101 \\times 100 + 1 = 10101$ elements in $D$. Two sets $A_i$ and $A_j$ have nonempty intersection if and only if $t_i - t_j$ is in $D$. So we need to choose the $100$ elements in such a way that the difference for any two elements is not in $D$... | China | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0k5r | Problem:
How many distinct permutations of the letters of the word REDDER are there that do not contain a palindromic substring of length at least two? (A substring is a contiguous block of letters that is part of the string. A string is palindromic if it is the same when read backwards.) | [
"Solution:\n\nIf two identical letters are adjacent or have a single letter in between, there is clearly a palindromic substring of length (respectively) two or three. So there cannot be any such substrings.\n\nSay we have a permutation of the word REDDER without any palindromic substrings. Let us call the first le... | United States | HMMT February 2019 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 6 | |
0i8k | Problem:
Find all integers $m$ such that $m^{2}+6m+28$ is a perfect square. | [
"Solution:\nWe must have $m^{2}+6m+28 = n^{2}$, where $n$ is an integer. Rewrite this as $(m+3)^{2} + 19 = n^{2} \\Rightarrow n^{2} - (m+3)^{2} = 19 \\Rightarrow (n - m - 3)(n + m + 3) = 19$.\n\nLet $a = n - m - 3$ and $b = n + m + 3$, so we want $ab = 19$. This leaves only 4 cases:\n\n- $a = 1$, $b = 19$. Solve th... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 6 and -12 | |
0jvr | Problem:
Alex has an $20 \times 16$ grid of lightbulbs, initially all off. He has 36 switches, one for each row and column. Flipping the switch for the $i$ th row will toggle the state of each lightbulb in the $i$ th row (so that if it were on before, it would be off, and vice versa). Similarly, the switch for the $j$... | [
"Solution:\n\nThe switch flip operations are commutative, so for any given sequence of switch flips $S$, we get the same configuration regardless of the order we do them in. We can arrange the switch flips so that all of the flips of the same switch happen consecutively. Furthermore, two consecutive flips of the sa... | United States | HMMT November 2016 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2^35 | |
01to | There is a graph with 30 vertices. If any 26 of its vertices with their outgoing edges are deleted, then the remained graph is a connected graph with 4 vertices.
What is the smallest number of the edges in the initial graph with 30 vertices?
(???) | [
"Answer: 405.\n\n(Solution by B. Serankou, A. Yuran.) Let $A$ be the vertex of the graph such that the number of outgoing edges from $A$ is less than 27. Hence there exist three vertices $B$, $C$, $D$ which are not connected with $A$. If we delete all vertices except for $A$, $B$, $C$, $D$ then the obtained graph i... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 405 | |
081y | Problem:
È noto che i Marziani maschi dicono sempre la verità, mentre le Marziane mentono sempre; al contrario i Venusiani maschi mentono e le Venusiane dicono sempre il vero. Atterra un'astronave piena di Marziani e Venusiani; all'ufficio immigrazione due degli occupanti, Ark e Bark, fanno le seguenti dichiarazioni:
... | [
"Solution:\n\nLa risposta è $\\mathbf{(E)}$. Se Ark dice il vero, allora Bark è un venusiano maschio, quindi mente, ma allora Ark deve essere un venusiano maschio, assurdo. Dunque Ark mente, quindi Bark è una femmina marziana, che mente, e dunque Ark è effettivamente un venusiano maschio (che mente). Quindi tutto è... | Italy | Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Logic"
] | null | MCQ | E | |
0e6h | Problem:
Dan je polinom $p$ s predpisom $p(x)=x^{4}+3 x^{3}+a x^{2}+b x+7$. Določi števili $a$ in $b$ tako, da bo premica $z$ enačbo $y=2 x+7$ sekala graf polinoma $p$ v točkah $z$ abscisama 1 in -4. | [
"Solution:\n\nZapišemo enačbo $2 x+7 = x^{4} + 3 x^{3} + a x^{2} + b x + 7$. Enačbo uredimo in dobimo $x^{4} + 3 x^{3} + a x^{2} + (b-2) x = 0$.\n\nV enačbo vstavimo $x=1$ in dobimo $1 + 3 + a + b - 2 = 0$ oziroma $a + b + 2 = 0$.\n\nV enačbo vstavimo še $x = -4$ in dobimo $256 - 192 + 16 a - 4 b + 8 = 0$ oziroma $... | Slovenia | Državno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a = -4, b = 2 | |
01sp | Prove that there are at most a finite number of primes $p$ such that the equation $a^3 + b^3 = 2016pab$ has a solution in positive integers $a$ and $b$ which are not divisible by $p$. | [
"Let $d$ denote the greatest common divisor of $a$ and $b$, i.e., $a = d a_1$, $b = d b_1$, where $\\gcd(a_1, b_1) = 1$. Then the given equality can be presented in the form\n$$\nd(a_1^3 + b_1^3) = 2016p a_1 b_1.\n$$\nIt follows that $d \\mid a_1^3 + b_1^3$, and, since $a_1$ and $b_1$ are coprime, $d \\mid b_1^3$ a... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis... | English | proof only | null | |
0f0m | Problem:
Given $n > 4$ points, show that you can place an arrow between each pair of points, so that given any point you can reach any other point by travelling along either one or two arrows in the direction of the arrow. | [] | Soviet Union | ASU | [
"Discrete Mathematics > Other"
] | null | proof only | null | |
0k77 | Problem:
A regular hexagon $A B C D E F$ has side length 1 and center $O$. Parabolas $P_{1}, P_{2}, \ldots, P_{6}$ are constructed with common focus $O$ and directrices $A B, B C, C D, D E, E F, F A$ respectively. Let $\chi$ be the set of all distinct points on the plane that lie on at least two of the six parabolas. C... | [
"Solution:\nRecall the focus and the directrix are such that the parabola is the locus of points equidistant from the focus and the directrix. We will consider pairs of parabolas and find their points of intersections (we label counterclockwise):\n(1): $P_{1} \\cap P_{2}$, two parabolas with directrices adjacent ed... | United States | HMMT February 2019 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthoce... | null | proof and answer | 35√3 | |
0dyr | Problem:
Drog se je prelomil na višini $4.2~\mathrm{m}$ od tal pod kotom $79^\circ$ (glej sliko).
Kolikšna je bila višina droga preden se je prelomil? Koliko $\mathrm{dm}^3$ lesa vsebuje spodnji del droga do preloma, če je premer droga enak $60~\mathrm{cm}$ ? | [
"Solution:\n\nUporabimo kotno funkcijo $\\cos 79^\\circ = \\frac{4,2}{y}$ in izračunamo dolžino odlomljenega dela droga $y = 22~\\mathrm{m}$. Izračunamo višino droga pred prelomom, ki je $v = 4,2 + 22 = 26,2~\\mathrm{m}$. Ugotovimo, da je polmer droga $r = 3~\\mathrm{dm}$. Izračunamo prostornino dela droga, ki je $... | Slovenia | Državno tekmovanje | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Solid Geometry > Volume"
] | null | final answer only | 26.2 m; 1187.5 dm^3 | |
0gss | Find all functions $f: \mathbb{Q}^+ \to \mathbb{Q}$ satisfying
$$
f(x) + f(y) = \left( f(x + y) + \frac{1}{x + y} \right) (1 - xy + f(xy))
$$
for all $x, y \in \mathbb{Q}^+$. | [
"**Answer:** $f(x) = x - \\frac{1}{x}$, $\\forall x \\in \\mathbb{Q}^+$.\nWe will prove several lemmas.\n\n$$\nf(x) + f(y) = \\left( f(x + y) + \\frac{1}{x+y} \\right) (1 - xy + f(xy)) \\quad (*)\n$$\n\n**Lemma 1.** $f(1) = 0$.\n*Proof:* By putting $x = y = 1$ to (*) we get $2f(1) = \\left(f(2) + \\frac{1}{2}\\righ... | Turkey | Team Selection Test | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = x - 1/x | |
0e4l | Let $ABCD$ be a convex quadrilateral. Choose points $E$ and $F$ on the segment $AB$ and a point $G$ on the segment $CD$, such that the quadrilaterals $ABCG$, $AFCD$ and $EFGC$ are cyclic. Prove that $|AE| = |FB|$ if and only if the segments $AB$ and $CD$ are parallel. | [
"Let $\\angle DCA = \\gamma$. Since the quadrilateral $AFCD$ is cyclic, we have\n$$\n\\angle DFA = \\angle DCA = \\gamma.\n$$\nSince $ABCG$ is also a cyclic quadrilateral, we have $\\angle GBA = \\angle GCA = \\gamma$. From here we conclude that the lines $DF$ and $GB$ are parallel.\n\nIn the cyclic quadrilateral $... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0hr0 | Problem:
A large integer is divisible by all the integers between $1$ and $30$ inclusive, except for two consecutive integers. Determine those two consecutive integers. | [
"Solution:\nThe two numbers must be $16$ and $17$.\n\nFirst, note that if a number $n$ is not a power of a prime, then it has two smaller factors $a$ and $b$ that are relatively prime and multiply to $n$. Thus, $n$ is the least common multiple of $a$ and $b$. Then, any multiple of both $a$ and $b$ must also be a mu... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 16 and 17 | |
05l9 | Problem:
Soient deux cercles extérieurs $\mathcal{C}$ et $\mathcal{C}^{\prime}$ de centres $O$ et $O^{\prime}$. On mène deux rayons parallèles de même sens $[O M]$ et $[O^{\prime} M^{\prime}]$, et deux autres rayons parallèles de même sens $[O P]$ et $[O^{\prime} P^{\prime}]$. La droite $(M M^{\prime})$ recoupe $\math... | [
"Solution:\n\n\n\nLes triangles $OMP$ et $O^{\\prime} M^{\\prime} P^{\\prime}$ sont isocèles en $O$ et $O^{\\prime}$ respectivement. De plus, les angles $(\\overrightarrow{OM}, \\overrightarrow{OP})$ et $(\\overrightarrow{O^{\\prime} M^{\\prime}}, \\overrightarrow{O^{\\prime} P^{\\prime}})$... | France | Olympiades Françaises de Mathématiques - Test de Février | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05vs | Problem:
Soit $n$ un entier naturel non nul. Anna a écrit $4 n+2$ entiers deux à deux distincts et compris entre $0$ et $5^{n}$ inclus. Démontrer que, parmi les entiers qu'elle a écrits, il en existe trois, disons $a, b$ et $c$, tels que $a < b < c$ et $c + 2a > 3b$. | [
"Solution:\n\nSoit $x_{1} < x_{2} < \\ldots < x_{4 n+2}$ les $4 n+2$ entiers qu'Anna a écrits. Nous allons démontrer qu'il existe un entier $k \\leqslant 4 n$ tel que $x_{4 n+2} + 2 x_{k} > 3 x_{k+1}$.\n\nEn effet, dans le cas contraire, posons $y_{k} = x_{4 n+2} - x_{k}$ pour tout entier $k$, de sorte que\n$$\n3 y... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0jyr | Problem:
Let $ABC$ be a triangle, $I$ the incenter, and $D$ the intersection of lines $AI$ and $BC$. The perpendicular bisector of $AD$ meets $BI$ and $CI$ at $P$ and $Q$. Show that $I$ is the orthocenter of triangle $PQD$. | [
"Solution:\n\nIt suffices to show that $CI \\perp PD$.\n\n\n\nNote that since $AP = PD$ and $BI$ is bisector of $\\angle ABD$, point $P$ lies on the circumcircle of $\\triangle ABD$ (on the midpoint of the arc). From this one can compute $\\angle PDC = \\angle IDC - \\angle ADP = \\angle ID... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0bf4 | A set $M$ of real numbers will be called *special* if it has the properties:
(i) for each $x, y \in M$, $x \neq y$, the numbers $x+y$ and $xy$ are not zero and exactly one of them is rational;
(ii) for each $x \in M$, $x^2$ is irrational.
Find the maximum number of elements of a special set. | [
"The required maximum is $4$, an example of a special $4$-element set being $M = \\{\\sqrt{2} - 1, \\sqrt{2} + 1, 2 - \\sqrt{2}, -2 - \\sqrt{2}\\}$.\n\nWe will prove that a special set cannot have more than $4$ elements. Obviously, the second condition implies that all the elements of a special set are irrational. ... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 4 | |
0608 | Problem:
Soit $ABC$ un triangle et soient $A_{1}$, $B_{1}$ et $C_{1}$ les points de contact respectifs du cercle $A$-exinscrit, noté $\omega_{A}$, avec le côté $BC$ et les demi-droites $[AC)$ et $[AB)$. Soit $P$ le milieu du segment $[B_{1}C_{1}]$. La droite $(A_{1}P)$ recoupe le cercle $\omega_{A}$ au point $X$. La t... | [
"Solution:\n\n\n\nEn traçant le cercle de centre $R$ de rayon $RA$, on s'aperçoit que celui-ci passe également par $P$. On va donc plutôt introduire $R'$ le centre du cercle circonscrit au triangle $APX$ et montrer que les droites $(AR')$ et $(XR')$ sont respectivement tangentes aux cercles... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - Envoi 5 : Pot Pourri | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0ic0 | Problem:
A regular dodecahedron is projected orthogonally onto a plane, and its image is an $n$-sided polygon. What is the smallest possible value of $n$? | [
"Solution:\nWe can achieve $6$ by projecting onto a plane perpendicular to an edge of the dodecahedron. Indeed, if we imagine viewing the dodecahedron in such a direction, then $4$ of the faces are projected to line segments (namely, the two faces adjacent to the edge and the two opposite faces), and of the remaini... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | final answer only | 6 | |
0a0d | Let $n \ge 1$ be an integer. Ruben takes a test with $n$ questions. Each question on this test is worth a different number of points. The first question is worth $1$ point, the second question $2$, the third $3$ and so on until the last question which is worth $n$ points. Each question can be answered either correctly ... | [
"For the first few values of $f$, note that $f(1) = 0$, $f(2) = f(3) = f(4) = 1$, $f(5) = f(6) = 2$, $f(7) = f(8) = 3$, $f(9) = f(10) = 5$. We claim that for $n \\ge 11$, $f(n)$ is strictly increasing as a function of $n$. Therefore there is only a finite number of pairs as in the problem.\n\nWe view a way of takin... | Netherlands | BxMO Team Selection Test | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | No; there are only finitely many such pairs. | |
0530 | Kärt writes the fractions $\frac{1}{2}$ and $\frac{1}{3}$ on the blackboard and Märt writes 10 positive integers on the paper, which he does not show to Kärt. Then Kärt starts to write fractions on the blackboard by the following rule: on each step she chooses two fractions $\frac{a}{b}$ and $\frac{c}{d}$ which are alr... | [
"Answer: Yes.\n\nThe first fraction that Kärt adds to the blackboard has to be $\\frac{2}{5}$. On every following move, let Kärt pick $\\frac{1}{2}$ as one fraction and the latest written fraction as the other fraction. Ignoring the reducing step, this means that the denominator of every added fraction is larger th... | Estonia | Open Contests | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | Yes | |
0dp7 | In triangle $ABC$ the angle bisectors of $A$ and $C$ intersect the sides $BC$ and $AB$ at the points $A_1$ and $C_1$, respectively, and the circumcircle of the triangle $ABC$ at the points $A_2$ and $C_2$, respectively. Let $K$ be the point of intersection of $A_1C_2$ and $C_1A_2$, and $I$ be the incenter of triangle $... | [
"Let's define notations of the points and angles as in the figure. Then\n$$\n\\angle C_2AB = \\angle C_2CB = \\angle C_2CA = \\angle C_2A_2A = \\gamma,\n$$\n$$\n\\angle A_2CB = \\angle A_2AB = \\angle A_2AC = \\angle A_2C_2C = \\alpha\n$$\nand\n$$\n\\angle C_2AC = \\angle A_2CA = \\angle ABC = 2\\beta.\n$$\n\nBy th... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > M... | English | proof only | null | |
0f7g | Problem:
Show that there is a real number $x$ such that all of $\cos x$, $\cos 2x$, $\cos 4x$, $\ldots$, $\cos (2^{n}x)$ are negative. | [] | Soviet Union | 21st ASU | [
"Precalculus > Trigonometric functions"
] | null | proof only | null | |
003d | Carlitos escribió todos los subconjuntos de $\{1, 2, \ldots, 2006\}$ en los que la diferencia entre la cantidad de números pares y de números impares es múltiplo de $3$.
¿Cuántos subconjuntos escribió Carlitos? | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | Español | proof and answer | (4^{1003} + 2)/3 | |
030q | Problem:
Fie $M, N$ și $P$ mijloacele laturilor $BC, CA$, respectiv $AB$, ale triunghiului ascuțitunghic $ABC$. Notăm cu $A'$, $B'$ și $C'$ punctele diametral opuse vârfurilor $A, B$, respectiv $C$ în cercul circumscris triunghiului $ABC$. Pe segmentele deschise $MA'$, $NB'$ și $PC'$ se consideră punctele $X, Y$, resp... | [
"Solution:\n\na.\nFie $H$ ortocentrul triunghiului $ABC$. Patrulaterul $BHC A'$ este un paralelogram ($BH$ și $A'C$ sunt perpendiculare pe dreapta $AC$, iar $CH$ și $A'B$ sunt perpendiculare pe dreapta $AB$), prin urmare punctul $M$ este mijlocul segmentului $HA'$.\n\nDacă $\\frac{MX}{XA'}=k$, atunci\n$$\n\\frac{A'... | Brazil | Al doilea baraj de selecție pentru OBMJ | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0c2l | Let $a$, $b$, $c$, $d$ be positive integers such that $ad \neq bc$ and $\text{gcd}(a, b, c, d) = 1$. Prove that, as $n$ runs through the positive integers, the values $\text{gcd}(an + b, cn + d)$ may achieve from the set of all positive divisors of some integer. | [
"We extend the problem statement by allowing $a$ and $c$ to take non-negative integer values, and allowing $b$ and $d$ to take arbitrary integer values. (As usual, the greatest common divisor of two integers is non-negative.) Without loss of generality, we assume $0 \\le a \\le c$. Let $S(a, b, c, d) = \\{\\text{gc... | Romania | Eleventh Romanian Master of Mathematics | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0g2y | Problem:
Seien $k$, $n$ und $r$ natürliche Zahlen mit $r < n$. Quirin besitzt $k n + r$ schwarze und $k n + r$ weisse Socken. Er möchte sie so auf seiner Wäscheleine aufhängen, dass es keine $2 n$ aufeinanderfolgenden Socken gibt, von denen $n$ schwarz und $n$ weiss sind. Zeige, dass Quirin dies genau dann schaffen ka... | [
"Solution:\n\nLa première observation importante pour ce problème est que si une suite de $2 n$ chaussettes contient plus de chaussettes blanches que de noires et une autre suite de $2 n$ chaussettes qui contient plus de chaussettes noires que de chaussettes blanches, alors il existe quelque part une suite de $2 n$... | Switzerland | Selektion | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0a1k | A rectangle $ABCD$ stands with side $AB$ on the ground and with side $BC$ against the wall. The length of the side $AD$ is $4$, the length of the side $CD$ is $3$, and the length of the diagonal $AC$ is $5$. The rectangle is tilted with point $A$ as the pivot point such that point $C$ is straight above point $A$.
![](a... | [
"B) $\\frac{3}{5}$"
] | Netherlands | Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | MCQ | B | |
0fpe | En la circunferencia circunscrita al triángulo $ABC$, sea $A_1$ el punto diametralmente opuesto al vértice $A$. Sea $A'$ el punto en el que la recta $AA_1$ corta al lado $BC$. La perpendicular a la recta $AA'$ trazada por $A'$ corta a los lados $AB$ y $AC$ (o a sus prolongaciones) en $M$ y $N$, respectivamente. Demostr... | [
"Sea $O$ el circuncentro de $ABC$, y $D$ el pie de la altura desde $A$. Es conocido que $AO$ y la altura desde $A$ son rectas isogonales en cualquier triángulo. En nuestro caso, los son en los dos triángulos $ABC$ y $AMN$, por la manera como se construye el triángulo $AMN$.\n\nEn $ABC$, $AA_1$ es diámetro de la cir... | Spain | LII Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry... | Spanish | proof only | null | |
06rw | In the triangle $A B C$ the point $J$ is the center of the excircle opposite to $A$. This excircle is tangent to the side $B C$ at $M$, and to the lines $A B$ and $A C$ at $K$ and $L$ respectively. The lines $L M$ and $B J$ meet at $F$, and the lines $K M$ and $C J$ meet at $G$. Let $S$ be the point of intersection of ... | [
"Let $\\alpha=\\angle C A B$, $\\beta=\\angle A B C$ and $\\gamma=\\angle B C A$. The line $A J$ is the bisector of $\\angle C A B$, so $\\angle J A K=\\angle J A L=\\frac{\\alpha}{2}$. By $\\angle A K J=\\angle A L J=90^{\\circ}$ the points $K$ and $L$ lie on the circle $\\omega$ with diameter $A J$.\n\nThe triang... | IMO | 53rd International Mathematical Olympiad Shortlisted Problems with Solutions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0336 | Problem:
Prove that for any integer $a \geq 4$ there exist infinitely many squarefree positive integers $n$ that divide $a^{n}-1$. | [
"Solution:\nFirst we shall prove the following:\n\nLEMMA. Let $p \\geq 3$ be an odd divisor of $b$. Then there exists an odd prime $q$ that divides $(b+1)^{p}-1$ but does not divide $b$.\n\nProof of the lemma. If $b=pc$, then\n$$\n\\begin{aligned}\n(b+1)^{p}-1 & =b\\left((b+1)^{p-1}+\\cdots+b+1\\right) \\\\\n& =b\\... | Bulgaria | Bulgarian Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0ab8 | Determine the sum of all odd 3-digit numbers whose product of digits is equal to $140$. | [
"Because $140 = 2 \\cdot 2 \\cdot 5 \\cdot 7 = 4 \\cdot 5 \\cdot 7$, it follows that the 3-digit numbers are formed with the digits $4$, $5$ and $7$. So the odd 3-digit numbers, with product of digits equal to $140$, are $457$, $475$, $547$ and $745$. Their sum is $2224$."
] | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 2224 | |
0fgm | Problem:
Sean $a, b, c$ las longitudes de los lados de un triángulo no isósceles. Se dan tres círculos concéntricos de radios $a, b$ y $c$.
a) ¿Cuántos triángulos equiláteros de áreas distintas pueden construirse, de modo que las rectas que contienen sus lados sean tangentes a cada círculo?
b) Hallar las superficies ... | [
"Solution:\nSi desde un punto $P$ del plano se trazan perpendiculares a los tres lados de un triángulo equilátero, se obtienen 3 segmentos cuya suma (con los signos que convenga) es igual a la altura del triángulo. La demostración se obtiene simplemente calculando las áreas de los triángulos determinados por $P$ y ... | Spain | OME 23 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | Number of distinct-area equilateral triangles: 4. Their areas, letting p be the semiperimeter of the triangle with side lengths a, b, c, are: (4√3/3)·p², (4√3/3)·(p − a)², (4√3/3)·(p − b)², (4√3/3)·(p − c)². | |
0jue | Problem:
Let $R$ be the rectangle in the Cartesian plane with vertices at $(0,0)$, $(2,0)$, $(2,1)$, and $(0,1)$. $R$ can be divided into two unit squares, as shown; the resulting figure has seven edges.
How many subsets of these seven edges form a connected figure?
Proposed by: Joy Zheng | [
"Solution:\nWe break this into cases. First, if the middle edge is not included, then there are $6 \\times 5 = 30$ ways to choose two distinct points for the figure to begin and end at. We could also allow the figure to include all or none of the six remaining edges, for a total of $32$ connected figures not includ... | United States | HMMT February 2016 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 81 | |
0kb6 | Let $f: \mathbb{Z} \rightarrow \{1, 2, \dots, 10^{100}\}$ be a function satisfying
$$
\text{gcd}(f(x), f(y)) = \text{gcd}(f(x), x - y)
$$
for all integers $x$ and $y$. Show that there exist positive integers $m$ and $n$ such that $f(x) = \text{gcd}(m + x, n)$ for all integers $x$. | [
"Let $\\mathcal{P}$ be the set of primes not exceeding $10^{100}$. For each $p \\in \\mathcal{P}$, let $e_p = \\max_x \\nu_p(f(x))$ and let $c_p = \\underset{x}{\\operatorname{argmax}} \\nu_p(f(x))$.\nWe show that this is good enough to compute all values of $x$, by looking at the exponent at each individual prime.... | United States | USA TSTST | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
03f0 | Find $A_{2024}$, where
$$
A_n = 1 \cdot 2 + 3 \cdot 4 + 5 \cdot 8 + \dots + (2n-1) \cdot 2^n.
$$ | [
"Since\n$$\n2A_n = 1 \\cdot 4 + 3 \\cdot 8 + \\dots + (2n-3) \\cdot 2^n + (2n-1) \\cdot 2^{n+1},\n$$\nit follows\n$$\n\\begin{align*}\nA_n = 2A_n - A_n &= (2n-1) \\cdot 2^{n+1} - (1 \\cdot 2 + 2 \\cdot 4 + 2 \\cdot 8 + \\dots + 2 \\cdot 2^n) \\\\\n&= (2n-1) \\cdot 2^{n+1} - 2 \\cdot (2 + 4 + 8 + \\dots + 2^n) + 1 \... | Bulgaria | Bulgarian Winter Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof and answer | 4045 * 2^2025 + 6 |
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