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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0fpx | Se tienen $n \ge 2$ segmentos en el plano tales que cada par de segmentos se intersecan en un punto interior a ambos, y no hay tres segmentos que tengan un punto en común. Mafalda debe elegir uno de los extremos de cada segmento y colocar sobre él una rana mirando hacia el otro extremo. Luego silbará $n - 1$ veces. En ... | [] | Spain | LVII Olimpiada Internacional de Matemáticas | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Spanish | proof only | null | |
0je7 | Problem:
Let $p$, $q$, $r$, $s$ be distinct primes such that $p q - r s$ is divisible by $30$. Find the minimum possible value of $p + q + r + s$. | [
"Solution:\nAnswer: $54$\n\nThe key is to realize none of the primes can be $2$, $3$, or $5$, or else we would have to use one of them twice. Hence $p$, $q$, $r$, $s$ must lie among $7$, $11$, $13$, $17$, $19$, $23$, $29$, \\ldots. These options give remainders of $1$ $(\\bmod 2)$ (obviously), $1$, $-1$, $1$, $-1$,... | United States | HMMT November 2013 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | 54 | |
0bcy | Find all triples of integers $(x, y, z)$ such that
$$
x^2 + y^2 + z^2 = 16(x + y + z).
$$ | [
"Write the relation as $(x-8)^2 + (y-8)^2 + (z-8)^2 = 192$ and recall that a square gives the remainder $0$ or $1$ when divided at $4$ to infer that $x-8$, $y-8$, $z-8$ are all even. Set $x-8 = 2a_1$, $y-8 = 2b_1$, $z-8 = 2c_1$ to get $a_1^2 + b_1^2 + c_1^2 = 48$. Repeat the above argument to write $a_1 = 2a_2$, $b... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (0,0,0), (0,0,16), (0,16,0), (16,0,0), (0,16,16), (16,0,16), (16,16,0), (16,16,16) | |
05za | Problem:
Plusieurs nombres sont écrits sur une ligne. Thanima a le droit de choisir deux nombres adjacents de telle sorte que le nombre de gauche soit strictement plus grand que le nombre de droite, elle échange alors ces deux nombres et les multiplie par 2. Montrer que Thanima ne peut effectuer qu'un nombre fini de t... | [
"Solution:\n\nOn commence par poser un jeton sur le minimum (si égalité, sur le nombre le plus à gauche). On bouge le jeton avec le nombre sur lequel il est posé. On va montrer que le jeton se déplace toujours vers la gauche.\n\nPour cela, supposons par l'absurde qu'à un moment, le jeton s'est déplacé vers la droit... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0bts | a) If $(a_n)_{n \ge 1}$ is a strictly increasing sequence of positive integers such that $(a_{2n-1} + a_{2n})/a_n$ is constant as $n$ runs through all positive integers, then this constant is an integer greater than or equal to $4$; and
b) Given an integer $N \ge 4$, there exists a strictly increasing sequence $(a_n)_... | [
"a) Clearly, $K = (a_{2n-1} + a_{2n})/a_n$ is a positive rational number. In fact, $K$ must be integral. To prove this, write $K = p/q$ in lowest terms to deduce that the $a_n$ are all divisible by $q$. Divide them all by $q$ to obtain a new sequence whose corresponding ratios are again $K$. Repetition of the proce... | Romania | 67th NMO Selection Tests for BMO and IMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0cfu | Let $ABCDA'B'C'D'$ be a cube and $O$ its centre. Consider the points $M \in [AD']$, $N \in [B'C]$ and $P$ a point on the face $A'B'C'D'$. Denote $E$ and $F$ the midpoints of the edges $A'B'$, $C'D'$. Prove that $O$ is the barycentre of the triangle $PMN$ if and only if $P \in [EF]$ and $\frac{PF}{AM} = \frac{PE}{CN} = ... | [] | Romania | 74th NMO Shortlisted Problems | [
"Geometry > Solid Geometry > 3D Shapes",
"Algebra > Linear Algebra > Vectors",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0fq7 | Problem:
Encontrar las funciones reales $f$, de variable real, que satisfacen la ecuación funcional
$$
f(x+f(x+y))=f(2 x)+y
$$
cualesquiera sean $x, y$ reales. | [
"Solution:\nHaciendo las sustituciones $x=f(0)$ y $y=-f(0)$ obtenemos $f(f(0)+f(0))=f(2 \\cdot f(0))+f(0)$, de donde $f(0)=0$. Sustituyendo ahora $x=0$ en la ecuación dada, dejando la variable $y$ arbitraria, se tiene $f(0+f(y))=f(0)+y$, esto es,\n$$\nf(f(y))=y\n$$\nAl sustituir $y=0$ en la ecuación del enunciado s... | Spain | null | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x | |
0036 | Sea $ABC$ un triángulo rectángulo en $A$. Considere todos los triángulos $XYZ$, rectángulos isósceles en $X$, donde $X$ está sobre el segmento $BC$, $Y$ sobre el segmento $AB$, y $Z$ sobre el segmento $AC$.
Determinar el lugar geométrico de los puntos medios de las hipotenusas $YZ$ de tales triángulos $XYZ$. | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Rotation"
] | Español | proof only | null | |
00lg | Determine all composite positive integers $n$ with the following property: If $1 = d_1 < d_2 < \dots < d_k = n$ are all the positive divisors of $n$, then
$$
(d_2 - d_1) : (d_3 - d_2) : \dots : (d_k - d_{k-1}) = 1 : 2 : \dots : (k-1).
$$ | [
"Since $n$ is a composite number, we have $k \\ge 3$.\nLet $d_2 = p$ be the smallest prime that divides $n$. We show by induction that\n$$\nd_j = \\frac{j(j-1)}{2}p - \\frac{(j-2)(j+1)}{2}, \\quad j = 1, 2, \\dots, k.\n$$\n\nThis is clearly true for $j = 1$ and the induction step follows from $d_j - d_{j-1} = (j-1)... | Austria | National Competition | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 4 | |
04jl | A positive integer is called *wacky* if its decimal representation contains 100 digits, and if by removing any of those digits one gets a 99-digit number divisible by $7$. How many wacky positive integers are there? (Stipe Vidak) | [] | Croatia | Croatia Mathematical Competitions | [
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | 2^98 | |
0b58 | Problem:
Fie $I$ centrul cercului înscris în triunghiul $ABC$ şi $A_{1}$, $B_{1}$ şi $C_{1}$ puncte arbitrare pe segmentele $(AI)$, $(BI)$, respectiv $(CI)$. Mediatoarele segmentelor $AA_{1}$, $BB_{1}$ şi $CC_{1}$ se intersectează în $A_{2}$, $B_{2}$ şi respectiv $C_{2}$. Arătați că centrul cercului circumscris triungh... | [] | Romania | 53. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > An... | null | proof only | null | |
0css | A trapezoid $ABCD$ with the bases $AB$ and $CD$ is inscribed into circle $\Omega$. A circle $\omega$ passes through the points $C$ and $D$, and intersects the segments $CA$ and $CB$ at $A_1 \neq C$ and $B_1 \neq D$, respectively. The points $A_2$ and $B_2$ are symmetric to $A_1$ and $B_1$ with respect to the midpoints ... | [
"Первое решение. Утверждение задачи эквивалентно равенству $CA_2 \\cdot CA = CB_2 \\cdot CB$. Поскольку $AA_1 = CA_2$ и $BB_1 = CB_2$, достаточно доказать, что $AA_1 \\cdot AC = BB_1 \\cdot BC$.\nПусть $D_1$ — вторая точка пересечения $\\omega$ с $AD$ (см. рис. 9). Из симметрии имеем $AD = BC$ и $AD_1 = BB_1$. Из т... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0bcx | A sequence of positive integers is called complete if any positive integer has a multiple in the sequence. Prove that an arithmetic sequence of positive integers is complete if and only if its difference divides the first term. | [
"If the difference $r$ divides $a_1$, then $a_1 = dr$, $d \\in \\mathbb{N}$ and $a_n = (d + n - 1)r$, and a multiple of a positive integer $k$ is obtained when $d + n - 1$ is a multiple of $k$.\n\nFor the converse, observe first that if $r = 0$, the sequence is not complete. Because $r \\neq 0$ and by the assumptio... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
09ce | **Х-В3.** (Б.Баяржаргал) Өгсөн $AB$ хэрчим дээр $D$ цэг өгөгдөв.
$AB$-тэй перпендикуляр $D$-г дайрсан $l$ шулуун дээр $M$ цэг,
$MD$ диаметртэй тойрог $ABC$ гурвалжинд багтаж байхаар
$C$ цэгийг тус тус авчээ. $M$ цэг $l$ шулуун дээгүүр гүйж
байхад $CM$ шулуунууд ерөнхий цэгтэй болохыг батал. | [
"\n\n$(CM) \\cap (AB) = K$ гэе. $M$-ийг дайрсан $(AB)$-тэй параллель шулуун $AC$ ба $CB$-г $L, N$-д огтолдог байг. $C$ дээр төвтэй $M$-ийг $K$-д буулгадаг гомотетээр $\\triangle CLN$ нь $\\triangle ACB$-д бууна. Мөн уг гомотетоор $\\triangle ABC$-д багтсан тойрог нь $\\triangle ABC$-д гадаа... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | Mongolian | proof only | null | |
0gqs | A circle $\omega$ with center $P$ intersects the sides $BC, CA, AB$ of a triangle $ABC$ at points $A_1$ and $A_2, B_1$ and $B_2, C_1$ and $C_2$, respectively. Let $A'$ be the circumcenter of the triangle $A_1A_2P$, $B'$ be the circumcenter of the triangle $B_1B_2P$ and $C'$ be the circumcenter of the triangle $C_1C_2P$... | [
"\nLet $A_3, B_3, C_3$ be the feet of the perpendiculars from $P$ to the sides $BC, CA, AB$ respectively. Clearly, $A', B', C'$ are on the lines $PA_3, PB_3, PC_3$, respectively. By the Pythagoras Theorem, we have\n$$\nA'P^2 - A'A_3^2 = A'A_1^2 - A'A_3^2 = A_1P^2 - A_3P^2 = A_1P^2 - (A_3A' ... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Quadrilaterals... | null | proof only | null | |
09w9 | Suppose $k$ and $n$ are positive integers such that $k \le n \le 2k - 1$. Julian has a large pile of rectangular $k \times 1$-tiles. Merlijn picks a positive integer $m$, and receives from Julian $m$ tiles to place on an $n \times n$-board. On each tile, Julian writes whether this tile should be placed horizontally or ... | [
"We show that the largest $m$ Merlijn can pick is $\\min(n, 3(n-k)+1)$. First we show that $m \\le \\min(n, 3(n-k)+1)$. If Merlijn asks for $n+1$ tiles, Julian can instruct Merlijn to place them all horizontally. As $n \\le 2k-1$, it is impossible to place more than one tile horizontally on a single row, so Merlijn... | Netherlands | IMO Team Selection Test 3, June 2020 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | min(n, 3(n-k)+1) | |
0atp | Problem:
Let $N = \left(1 + 10^{2013}\right) + \left(1 + 10^{2012}\right) + \cdots + \left(1 + 10^{1}\right) + \left(1 + 10^{0}\right)$. Find the sum of the digits of $N$. | [
"Solution:\n\nWe have\n$$\nN = (1 + 10^{2013}) + (1 + 10^{2012}) + \\cdots + (1 + 10^1) + (1 + 10^0)\n$$\nThere are $2014$ terms in total, from $k = 0$ to $k = 2013$.\n\nSo,\n$$\nN = \\sum_{k=0}^{2013} (1 + 10^k) = \\sum_{k=0}^{2013} 1 + \\sum_{k=0}^{2013} 10^k = 2014 + \\sum_{k=0}^{2013} 10^k\n$$\nBut $\\sum_{k=0}... | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 2021 | |
06c5 | If $p$ is any prime other than $2$ or $5$, prove that $p$ divides infinitely many of the integers $9$, $99$, $999$, $9999$, ... | [
"By the Fermat little theorem, we have $10^{k(p-1)} \\equiv 1 \\pmod{p}$ for any $k \\in \\mathbb{N}$. This implies\n$$\np \\mid 10^{k(p-1)} - 1 = \\underbrace{99\\dots9}_{k(p-1) \\text{ times}}.\n$$"
] | Hong Kong | HKG TST | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0et1 | The squares of an $8 \times 8$ board are coloured alternatingly black and white. A rectangle consisting of some of the squares of the board is called *important* if its sides are parallel to the sides of the board and all its corner squares are coloured black. The side lengths can be anything from $1$ to $8$ squares. O... | [
"In each important rectangle, the number of black squares is one more than the number of white squares. Hence, each important rectangle contributes $+1$ to the difference $B-W$. The value of $B-W$ is thus the same as the number of important rectangles on the board.\n\nLet us number the rows on the board $1, 2, \\ld... | South Africa | The South African Mathematical Olympiad, Third Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 200 | |
07u4 | In triangle $ABC$, the internal and external bisectors of angle $\angle ACB$ meet $AB$ at $D$ and $E$, respectively. Suppose $B$ is between $A$ and $D$. If $CD$ is a median of triangle $AEC$, prove that $|AC| = 3|BC|$. | [
"From the Theorem on the Angle Bisector, we know that $D$ and $E$ divide the line segment $AB$ internally and externally in ratio $|AC| : |BC|$, that is\n$$\n\\frac{|AC|}{|BC|} = \\frac{|AD|}{|BD|} = \\frac{|AE|}{|BE|}.\n$$\n\nSince $CD$ is a median of triangle $AEC$, we have $|AE| = 2|AD|$... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0aoj | Problem:
Consider an acute triangle with angles $\alpha, \beta, \gamma$ opposite the sides $a, b, c$, respectively. If $\sin \alpha=\frac{3}{5}$ and $\cos \beta=\frac{5}{13}$, evaluate $\frac{a^{2}+b^{2}-c^{2}}{a b}$. | [] | Philippines | AREA STAGE | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | 32/65 | |
0ghr | 在銳角三角形 $ABC$ 中, 點 $H$ 為由頂點 $A$ 所引的高的垂足。設 $P$ 為平面上的動點, 滿足: $∠PBC$ 與 $∠PCB$ 的內角平分線, 分別記為 $k$ 與 $ℓ$, 此兩條角平分線的交點位於 $AH$ 線段上。設 $k$ 與 $AC$ 交於點 $E$; $ℓ$ 與 $AB$ 交於點 $F$; 而設 $EF$ 與 $AH$ 交於點 $Q$。證明: 不論 $P$ 如何移動 (但滿足前述條件), 直線 $PQ$ 恆過某一定點。
In an acute-angled triangle $ABC$, point $H$ is the foot of the altitude from $A$. Let $P$ ... | [
"設直線 $BC$ 分別關於 $AB$, $AC$ 的對稱線交於點 $K$。以下證明 $P$, $Q$, $K$ 共線, 也就是說, 直線 $PQ$ 恆過定點 $K$。\n\n設直線 $BE$ 與 $CF$ 交於點 $I$。對任意點 $O$ 及實數 $d > 0$, 將以 $O$ 為圓心、$d$ 為半徑的圓記為 $(O, d)$。定義兩圓 $\\omega_I = (I, IH)$ 及 $\\omega_A = (A, AH)$。再設三角形 $KBC$ 的內切圓為 $\\omega_K$, 三角形 $PBC$ 的 $P$-旁切圓為 $\\omega_P$。\n\n由於 $IH \\perp BC$ 及 $AH \\perp ... | Taiwan | 2023 數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinate... | Chinese (Traditional) | proof only | null | |
07sb | We say an integer $n$ is *naoish* if $n \ge 90$ and the second-to-last digit of $n$ (in decimal notation) is equal to $9$. For example, $10798$, $1999$ and $90$ are naoish, whereas $9900$, $2009$ and $9$ are not. Nino expresses $2020$ as a sum:
$$
2020 = n_{1} + n_{2} + \dots + n_{k}
$$
where each of the $n_j$ is naois... | [
"Equivalently, $n$ is naoish iff $n = 100p - q$ where $p \\ge 1$ is an integer and $1 \\le q \\le 10$. Decomposing each $n_j$ in this way, we have:\n$$\n2020 = 100(p_1 + p_2 + \\dots + p_k) - q_1 - q_2 - \\dots - q_k.\n$$\nIn particular, $100 \\mid 2020 + q_1 + q_2 + \\dots + q_k$. The next multiple of $100$ above ... | Ireland | IRL_ABooklet_2020 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 8 | |
0jrf | Problem:
Find the smallest positive integer $n$ such that there exists a complex number $z$, with positive real and imaginary part, satisfying $z^{n} = (ar{z})^{n}$. | [
"Solution:\nSince $|z| = |\bar{z}|$ we may divide by $|z|$ and assume that $|z| = 1$. Then $\\bar{z} = \\frac{1}{z}$, so we are looking for the smallest positive integer $n$ such that there is a $2n^{\\text{th}}$ root of unity in the first quadrant. Clearly there is a sixth root of unity in the first quadrant but n... | United States | HMMT November 2015 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof and answer | 3 | |
0fxo | Problem:
Für welche natürlichen Zahlen $n$ existiert ein Polynom $P(x)$ mit ganzen Koeffizienten, sodass $P(d) = (n / d)^2$ gilt für alle positiven Teiler $d$ von $n$? | [
"Solution:\nSo ein Polynom existiert genau dann, wenn $n$ prim ist, $n=1$ oder $n=6$. Im ersten Fall leistet $P(x) = -(n+1)x + (n^2 + n + 1)$ das Gewünschte, für $n=1$ das Polynom $P(x) = 1$ und für $n=6$ das Polynom $P(x) = -2x^3 + 23x^2 - 82x + 97$.\n\nWir zeigen nun, dass dies die einzigen möglichen Werte von $n... | Switzerland | IMO Selektion | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | n = 1 or n is prime or n = 6 | |
0dzc | Problem:
Na stranici $BC$ pravokotnega trikotnika $ABC$ s pravim kotom pri $C$ izberemo točko $D$, različno od $B$ in $C$. Trikotniku $ABD$ očrtano krožnico označimo s $\mathcal{K}$. Naj bo $T$ taka točka na stranici $AB$, da je $DT$ pravokotna na $AB$. Premica $DT$ seka krožnico $\mathcal{K}$ še v točki $E$. Presečiš... | [
"Solution:\n\nKer je vsota nasprotnih kotov $\\angle ATD$ in $\\angle ACD$ v štirikotniku $ATDC$ enaka $\\pi$, je ta štirikotnik tetiven. Označimo $\\angle TCD=\\alpha$. Zaradi tetivnosti sledi $\\angle TAD=\\angle TCD=\\alpha$. Ker pa so točke $A, B, E, D$ konciklične, velja tudi $\\angle BED=\\angle BAD=\\angle T... | Slovenia | 52. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0bl2 | Problem:
Adottak az $m \geq 2$ és $n \geq 3$ természetes számok. Igazold, hogy létezik $m$ darab különböző $a_{1}, a_{2}, a_{3}, \ldots, a_{m}$ természetes szám, amelyek mind oszthatók $n-1$-gyel és
$$
\frac{1}{n}=\frac{1}{a_{1}}-\frac{1}{a_{2}}+\frac{1}{a_{3}}-\ldots+(-1)^{m-1} \frac{1}{a_{m}}
$$ | [
"Solution:\n\nSoluţia 1.\nDacă $(a_{n})_{n \\geq 1}$ este o progresie geometrică având raţia $q=n-1$, atunci\n$$\n\\frac{1}{a_{1}}-\\frac{1}{a_{2}}+\\ldots+(-1)^{p-1} \\frac{1}{a_{p}}=\\frac{1 / a_{1}+1 / q \\cdot(-1)^{p-1} 1 / a_{p}}{1+1 / q}, \\forall p \\in \\mathbb{N}^{*}\n$$\nRelaţia precedentă se scrie\n$$\n\... | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0f8i | Problem:
The quadrilateral $ABCD$ is inscribed in a fixed circle. It has $AB$ parallel to $CD$ and the length $AC$ is fixed, but it is otherwise allowed to vary. If $h$ is the distance between the midpoints of $AC$ and $BD$ and $k$ is the distance between the midpoints of $AB$ and $CD$, show that the ratio $h/k$ remai... | [
"Solution:\n\nLet the center of the circle be $O$ and its radius be $R$. Let $\\angle AOB = 2x$ (variable) and let $\\angle AOC = 2y$ (fixed). Then $AC = 2R \\sin y$. We find $\\angle COD = 180^\\circ - x - 2y$, so $2h = 2R \\sin x + 2R \\sin(x + 2y)$. Angle $ACD = x + y$, so $k = AB \\sin(x + y) = 2R \\sin(x + y) ... | Soviet Union | 22nd ASU | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09wv | Problem:
Bepaal alle polynomen $P(x)$ met reële coëfficiënten waarvoor geldt
$$
P\left(x^{2}\right)+2 P(x)=P(x)^{2}+2 .
$$ | [
"Solution:\nOplossing I. Herschrijf de gegeven vergelijking naar\n$$\nP\\left(x^{2}\\right)-1=(P(x)-1)^{2} \\text{.}\n$$\nSchrijf $Q(x)=P(x)-1$, dan is $Q$ een polynoom met reële coëfficiënten waarvoor geldt dat\n$$\nQ\\left(x^{2}\\right)=Q(x)^{2}\n$$\nStel dat $Q$ constant is, zeg $Q(x)=c$ met $c \\in \\mathbb{R}$... | Netherlands | IMO-selectietoets I | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | P(x) = 1; P(x) = 2; and P(x) = x^n + 1 for integers n ≥ 1 | |
02zt | Problem:
Nós chamamos um número de telefone $d_{1} d_{2} d_{3}-d_{4} d_{5} d_{6} d_{7}$ de legal se o número $d_{1} d_{2} d_{3}$ for igual a $d_{4} d_{5} d_{6}$ ou a $d_{5} d_{6} d_{7}$. Por exemplo, $234-2347$ é um número de telefone legal. Assuma que cada $d_{i}$ pode ser qualquer dígito de 0 a 9. Quantos números de... | [
"Solution:\n\nSe $d_{1} d_{2} d_{3}$ é simultaneamente igual a $d_{4} d_{5} d_{6}$ e $d_{5} d_{6} d_{7}$, então todos os dígitos são iguais. Para contar o número de telefones com $d_{1} d_{2} d_{3}$ igual a $d_{4} d_{5} d_{6}$, basta escolhermos $d_{1} d_{2} d_{3}$ de $10 \\cdot 10 \\cdot 10 = 10^{3}$ maneiras e o ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 19990 | |
088r | Problem:
Maria ha tre monete rosse e una blu, tutte dello stesso raggio e con il bordo gommato. Mette sul tavolo le tre monete rosse ai vertici di un triangolo equilatero in modo che si tocchino a due a due, poi posa sul tavolo anche quella blu in modo che tocchi una delle monete rosse. Adesso, tenendo ferme le monete... | [
"Solution:\n\nLa risposta è (A). Visto che dobbiamo fare un percorso completo, possiamo supporre di cominciare in un punto in cui la moneta blu tocca 2 monete rosse contemporaneamente. Nel tragitto che ora la moneta blu fa aderendo a una delle due il suo centro percorre una semicirconferenza; ma il bordo nel fratte... | Italy | UNIONE MATEMATICA ITALIANA SCUOLA NORMALE SUPERIORE DI PISA GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | MCQ | A | |
0gz1 | Find all positive integers $n$ such that:
$$
-2^0 + 2^1 - 2^2 + 2^3 - 2^4 + \dots - (-2)^n = 4^0 + 4^1 + 4^2 + \dots + 4^{2010}.
$$ | [
"**Answer:** $n = 4021$.\n\nUsing the formula for sum of a geometric progression, we have:\n$$\n-2^0 + 2^1 - 2^2 + 2^3 - 2^4 + \\dots + (-2)^n = \\frac{-1 \\cdot ((-2)^{n+1}-1)}{(-2)^{n+1}}\n$$\nThe right side is:\n$$\n4^0 + 4^1 + 4^2 + \\dots + 4^{2010} = \\frac{4^{2011}-1}{4-1} = \\frac{4^{2011}-1}{3}\n$$\nSo, eq... | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | n = 4021 | |
00z2 | Problem:
Prove that $\sin^{3} 18^{\circ} + \sin^{2} 18^{\circ} = \frac{1}{8}$. | [
"Solution:\nWe have\n$$\n\\begin{aligned}\n\\sin^{3} 18^{\\circ} + \\sin^{2} 18^{\\circ} &= \\sin^{2} 18^{\\circ} (\\sin 18^{\\circ} + \\sin 90^{\\circ}) \\\\\n&= \\sin^{2} 18^{\\circ} \\cdot 2 \\sin 54^{\\circ} \\cos 36^{\\circ} \\\\\n&= 2 \\sin^{2} 18^{\\circ} \\cos^{2} 36^{\\circ} \\\\\n&= \\frac{2 \\sin^{2} 18^... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0h31 | In the plane six lines are given such that no three of them are coincident. Can it happen that they have exactly:
a) $12$;
b) $16$ different intersection points? | [
"a) It is possible. It is enough to take $3$ lines that have $3$ intersection points among themselves, and also $3$ parallel lines, each of which intersects each of the $3$ initial lines.\n\nb) It is not possible. The first line intersects the others in at most $5$ points, the second adds at most $4$ intersection p... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) Yes; b) No | |
0j8n | Problem:
Points $D$, $E$, $F$ lie on circle $O$ such that the line tangent to $O$ at $D$ intersects ray $\overrightarrow{E F}$ at $P$. Given that $P D = 4$, $P F = 2$, and $\angle F P D = 60^{\circ}$, determine the area of circle $O$. | [
"Solution:\n\nAnswer: $12 \\pi$\n\nBy the power of a point on $P$, we get that\n\n$$\n16 = P D^{2} = (P F)(P E) = 2 (P E) \\Rightarrow P E = 8\n$$\n\nHowever, since $P E = 2 P D$ and $\\angle F P D = 60^{\\circ}$, we notice that $P D E$ is a $30$-$60$-$90$ triangle, so $D E = 4 \\sqrt{3}$ and we have $E D \\perp D ... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 12π | |
0eaj | Problem:
Na mizi so trije kupčki žetonov: eden z $a$ žetoni, eden z $b$ žetoni in eden s $c$ žetoni, pri čemer velja $a \geq b \geq c > 0$. Igralca $A$ in $B$ izmenično prestavljata žetone. Začne igralec $A$. V vsaki potezi igralec najprej izbere dva kupčka in nato s tistega z manj žetoni prestavi vsaj en žeton na tis... | [
"Solution:\n\nČe je $b = c$, potem ima zmagovalno strategijo igralec $B$, sicer pa ima zmagovalno strategijo igralec $A$.\n\nDenimo najprej, da je $b = c$. Imamo torej situacijo, ko je na dveh kupčkih z najmanj žetoni enako število žetonov. Igralec $B$ lahko v tem primeru poskrbi, da je situacija po vsaki njegovi p... | Slovenia | 58. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Player B wins when b = c; otherwise Player A wins. | |
011t | Problem:
Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of real numbers satisfying $a_{0}=1$ and $a_{n}=a_{\lfloor 7 n / 9\rfloor}+a_{\lfloor n / 9\rfloor}$ for $n=1,2, \ldots$ Prove that there exists a positive integer $k$ with $a_{k}<\frac{k}{2001 !}$.
(Here $\lfloor x\rfloor$ denotes the largest integer not greate... | [
"Solution:\nConsider the equation\n$$\n\\left(\\frac{7}{9}\\right)^{x}+\\left(\\frac{1}{9}\\right)^{x}=1.\n$$\nIt has a root $\\frac{1}{2}<\\alpha<1$, because $\\sqrt{\\frac{7}{9}}+\\sqrt{\\frac{1}{9}}=\\frac{\\sqrt{7}+1}{3}>1$ and $\\frac{7}{9}+\\frac{1}{9}<1$. We will prove that $a_{n} \\leqslant M \\cdot n^{\\al... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof only | null | |
04w8 | Let $ABC$ be a triangle and let $I_A$, $I_B$, $I_C$ be centers of the excircles tangent to its sides $BC$, $CA$, $AB$, respectively. Denote by $X$, $Y$, $Z$ the orthocenters of triangles $I_ABC$, $AI_BC$, $AI_C$, respectively. Prove that the triangles $ABC$ and $XYZ$ are congruent. (Michal Janík) | [] | Czech Republic | First Round (take-home) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0iez | Problem:
How many real numbers $x$ are solutions to the following equation?
$$
2003^{x} + 2004^{x} = 2005^{x}
$$ | [
"Solution:\nRewrite the equation as $(2003 / 2005)^{x} + (2004 / 2005)^{x} = 1$. The left side is strictly decreasing in $x$, so there cannot be more than one solution. On the other hand, the left side equals $2 > 1$ when $x = 0$ and goes to $0$ when $x$ is very large, so it must equal $1$ somewhere in between. The... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | 1 | |
0kkx | Problem:
Let $f(n)$ be the largest prime factor of $n$. Estimate
$$
N = \left\lfloor 10^{4} \cdot \frac{\sum_{n=2}^{10^{6}} f\left(n^{2}-1\right)}{\sum_{n=2}^{10^{6}} f(n)} \right\rfloor.
$$
An estimate of $E$ will receive $\max \left(0,\left\lfloor 20-20\left(\frac{|E-N|}{10^{3}}\right)^{1 / 3}\right\rfloor\right)$ po... | [
"Solution:\nWe remark that\n$$\nf\\left(n^{2}-1\\right) = \\max (f(n-1), f(n+1))\n$$\nLet $X$ be a random variable that evaluates to $f(n)$ for a randomly chosen $2 \\leq n \\leq 10^{6}$; we essentially want to estimate\n$$\n\\frac{\\mathbb{E}\\left[\\max \\left(X_{1}, X_{2}\\right)\\right]}{\\mathbb{E}\\left[X_{3}... | United States | HMMT Spring 2021 Guts Round | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Other"
] | null | final answer only | 18215 | |
0kvv | Problem:
Suppose $x$ is a real number such that $\sin \left(1+\cos^{2} x+\sin^{4} x\right)=\frac{13}{14}$. Compute $\cos \left(1+\sin^{2} x+\cos^{4} x\right)$. | [
"Solution:\nWe first claim that $\\alpha := 1+\\cos^{2} x+\\sin^{4} x = 1+\\sin^{2} x+\\cos^{4} x$. Indeed, note that\n$$\n\\sin^{4} x - \\cos^{4} x = (\\sin^{2} x + \\cos^{2} x)(\\sin^{2} x - \\cos^{2} x) = \\sin^{2} x - \\cos^{2} x\n$$\nwhich is the desired after adding $1+\\cos^{2} x+\\cos^{4} x$ to both sides.\... | United States | HMMT February | [
"Precalculus > Trigonometric functions",
"Precalculus > Functions"
] | null | final answer only | -3*sqrt(3)/14 | |
034o | Problem:
The sum of the first $n$ terms of an arithmetic progression with first term $m$ and difference $2$ is equal to the sum of the first $n$ terms of a geometric progression with first term $n$ and ratio $2$.
a) Prove that $m+n=2^{m}$;
b) Find $m$ and $n$, if the third term of the geometric progression is equal t... | [
"Solution:\n\na) Using the formulas for the sums of arithmetic and geometric progressions we obtain the equality\n$$\n\\frac{n[2m + 2(n-1)]}{2} = n\\left(2^{m} - 1\\right)\n$$\nwhence $m+n=2^{m}$.\n\nb) It follows that $4n = m + 44$. Using a), we obtain $2^{m+2} = 44 + 5m$. It is easy to see that $m=4$ is a solutio... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | m + n = 2^m; m = 4, n = 12 | |
094t | Problem:
Consider the two infinite sequences $a_{0}, a_{1}, a_{2}, \ldots$ and $b_{0}, b_{1}, b_{2}, \ldots$ of real numbers such that $a_{0}=0, b_{0}=0$ and
$$
a_{k+1}=b_{k}, \quad b_{k+1}=\frac{a_{k} b_{k}+a_{k}+1}{b_{k}+1}
$$
for each integer $k \geq 0$. Prove that $a_{2024}+b_{2024} \geq 88$. | [
"Solution:\n\nLet us notice that $a_{i}$ is the same as the sequence $b_{i}$ shifted by one. So the whole problem might be reduced to the sequence of $b_{i}$. The definition of $b_{k+1}$ can be rewritten as:\n$$\nb_{k+1}=\\frac{a_{k} b_{k}+a_{k}+1}{b_{k}+1}=a_{k}+\\frac{1}{b_{k}+1}=b_{k-1}+\\frac{1}{b_{k}+1} .\n$$\... | Middle European Mathematical Olympiad (MEMO) | MEMO Szeged | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
09r4 | Problem:
Zij $A B C D$ een convexe vierhoek (d.w.z. alle binnenhoeken zijn kleiner dan $180^{\circ}$), zodat er een punt $M$ op lijnstuk $A B$ en een punt $N$ op lijnstuk $B C$ bestaan met de eigenschap dat $A N$ de vierhoek in twee stukken van gelijke oppervlakte deelt, en $C M$ dat ook doet.
Bewijs dat $M N$ de diag... | [
"Solution:\n\nNoteer de oppervlakte van veelhoek $\\mathcal{P}$ met $O(\\mathcal{P})$. Laat $S$ het snijpunt van $A N$ en $C M$ zijn. Er geldt\n$O(A S M)+O(S M B N)=O(A B N)=\\frac{1}{2} O(A B C D)=O(C B M)=O(C N S)+O(S M B N)$,\ndus $O(A S M)=O(C N S)$. Als we hier links en rechts $O(S M N)$ bij optellen, vinden w... | Netherlands | Dutch TST | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
03z8 | Given positive integer $n$, let $S = \{1, 2, \dots, n\}$. Find the minimum of $|A\Delta S| + |B\Delta S| + |C\Delta S|$ for nonempty finite sets $A$ and $B$ of real numbers, where $C = \{a + b \mid a \in A, b \in B\}$, $X\Delta Y = \{x \mid x$ belongs to exactly one of $X$ and $Y\}$, $|X|$ denotes the number of element... | [
"The minimum is $n + 1$.\n\nFirst, by taking $A = B = S$, we have\n$$\n|A \\Delta S| + |B \\Delta S| + |C \\Delta S| = n + 1.\n$$\n\nSecond, we can prove that $l = |A \\Delta S| + |B \\Delta S| + |C \\Delta S| \\ge n + 1$. Let $X \\setminus Y = \\{x \\mid x \\in X, x \\notin Y\\}$. We have\n$$\nl = |A\\setminus S| ... | China | China Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n + 1 | |
0d5v | Let $ABC$ be an acute-angled triangle inscribed in the circle $(O)$, $H$ the foot of the altitude of $ABC$ at $A$ and $P$ a point inside $ABC$ lying on the bisector of $\angle BAC$. The circle of diameter $AP$ cuts $(O)$ again at $G$. Let $L$ be the projection of $P$ on $AH$. Prove that if $GL$ bisects $HP$ then $P$ is... | [
"Let $AP$ intersect $BC$ at $F$ and $(O)$ again at $D$. Because $LP$ and $BC$ are parallel, we have\n$$\n\\begin{aligned}\n\\angle LGA &= \\angle LPD = \\angle BFD = \\angle BAD + \\angle CBA \\\\\n&= \\angle DAC + \\angle CBA = \\angle DBA = \\angle DGA\n\\end{aligned}\n$$\nThis implies that $G$, $L$, $D$ are coll... | Saudi Arabia | SAMC 2015 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English, Arabic | proof only | null | |
0a5w | Problem:
Let $ABCD$ be a parallelogram, and let $P$ be a point on the side $AB$. Let the line through $P$ parallel to $BC$ intersect the diagonal $AC$ at point $Q$. Prove that
$$|DAQ|^{2} = |PAQ| \times |BCD|,$$
where $|XYZ|$ denotes the area of triangle $XYZ$. | [
"Solution:\n\nSince $PQ$ is parallel to $BC$, there is a dilation (centred at $A$) of factor $AB / AP$ that sends triangle $APQ$ to triangle $ABC$. Thus\n$$|PAQ| = |ABC| \\times \\left(\\frac{AP}{AB}\\right)^{2}.$$ \nSince $AD$ is parallel to $BC$, triangle $ABC$ has the same height as tria... | New Zealand | NZMO Round One | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
012d | Problem:
Let $L$, $M$ and $N$ be points on sides $AC$, $AB$ and $BC$ of triangle $ABC$, respectively, such that $BL$ is the bisector of angle $ABC$ and segments $AN$, $BL$ and $CM$ have a common point. Prove that if $\angle ALB = \angle MNB$ then $\angle LNM = 90^{\circ}$. | [
"Solution:\n\nLet $P$ be the intersection point of lines $MN$ and $AC$. Then $\\angle PLB = \\angle PNB$ and the quadrangle $PLNB$ is cyclic. Let $\\omega$ be its circumcircle. It is sufficient to prove that $PL$ is a diameter of $\\omega$.\n\nLet $Q$ denote the second intersection point of the line $AB$ and $\\ome... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05jn | Problem:
Montrer que l'équation
$$
x(x+2)=y(y+1)
$$
n'a pas de solution en nombres entiers strictement positifs. | [
"Solution:\nSi le couple $(x, y)$ est solution, on a $x(x+1)<x(x+2)=y(y+1)$ et donc $x<y$. Mais de même, en supposant toujours $(x, y)$ solution, on peut écrire $(x+1)(x+2)>x(x+2)=y(y+1)$, d'où on déduit $x+1>y$. Ainsi, une solution $(x, y)$ hypothétique devrait donc vérifier $x<y<x+1$, ce qui ne peut se produire. ... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0edc | Let $A$ be a set of pairwise distinct positive integers. A division of $A$ into two disjoint non-empty subsets $A_1$ and $A_2$ is good, if the least common multiple of the elements of $A_1$ is the same as the greatest common divisor of the elements of $A_2$. Show that for a set with $3n + 2$ elements the maximal possib... | [
"Let $a_1 < a_2 < \\dots < a_{3n+2}$ be the positive integers in our set. For each good division of the set into subsets $A_1$ and $A_2$ the least common multiple of the numbers from $A_1$ is greater than or equal to all of the elements of $A_1$. Similarly, the greatest common divisor of the numbers in $A_2$ is sma... | Slovenia | Slovenija 2016 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 2n | |
0kx1 | Problem:
For any odd positive integer $n$, let $r(n)$ be the odd positive integer such that the binary representation of $r(n)$ is the binary representation of $n$ written backwards. For example, $r(2023) = r\left(11111100111_{2}\right) = 11100111111_{2} = 1855$. Determine, with proof, whether there exists a strictly ... | [
"Solution:\n\nThe main idea is the following claim.\n\nClaim: If $a, b, c$ are in arithmetic progression and have the same number of digits in their binary representations, then $r(a), r(b), r(c)$ cannot be in arithmetic progression in that order.\n\nProof. Consider the least significant digit that differs in $a$ a... | United States | HMMT February | [
"Number Theory > Other",
"Discrete Mathematics > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | No; such an arithmetic progression does not exist. | |
0ksf | Problem:
A triangle $X Y Z$ and a circle $\omega$ of radius $2$ are given in a plane, such that $\omega$ intersects segment $\overline{X Y}$ at the points $A, B$, segment $\overline{Y Z}$ at the points $C, D$, and segment $\overline{Z X}$ at the points $E, F$. Suppose that $X B > X A$, $Y D > Y C$, and $Z F > Z E$. In ... | [
"Solution:\nLet $d = A B$ and $x = d / 2$ for ease of notation. Let the center of $(A B C D E F)$ be $I$. Because $A B = C D = E F$, the distance from $I$ to $A B$, $C D$, and $E F$ are the same, so $I$ is the incenter of $\\triangle X Y Z$. Let $\\triangle X Y Z$ have inradius $r$.\nBy symmetry, we have $X F = 1$,... | United States | HMMT November 2022 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(10) - 1 | |
04zi | Let $P(x, y)$ be a non-constant homogeneous polynomial with real coefficients such that $P(\sin t, \cos t) = 1$ for every real number $t$. Prove that there exists a positive integer $k$ such that $P(x, y) = (x^2 + y^2)^k$. | [
"Let $n$ be the degree of the polynomial $P$, i.e.,\n$$\nP(x, y) = a_n x^n + a_{n-1} x^{n-1} y + \\dots + a_1 x y^{n-1} + a_0 y^n,\n$$\nwhere $n > 0$. Note that $n$ must be even because otherwise the condition $P(\\sin t, \\cos t) = 1$ for $t = 0$ would imply $a_0 = 1$ while the same condition for $t = \\pi$ would ... | Estonia | Estonija 2010 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0fmf | Do there exist two real monic polynomials $P(x)$ and $Q(x)$ of degree 3, such that the roots of $P(Q(x))$ are nine pairwise distinct nonnegative integers that add up to 72? (In a monic polynomial of degree 3, the coefficient of $x^3$ is 1.) | [
"Let $z_1, z_2, z_3$ be the three roots of polynomial $P(x) = (x - z_1)(x - z_2)(x - z_3)$. Let $Q(x) = x^3 - s x^2 + t x - u$. Then the nine roots of\n$$\nP(Q(x)) = (Q(x) - z_1)(Q(x) - z_2)(Q(x) - z_3)\n$$\nare the roots $a_i, b_i, c_i$ of $Q(x) - z_i$, where $i$ runs from 1 to 3. Viète relations yield for $1 \\le... | Spain | Mediterranean Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Spanish | proof and answer | YES. One example is Q(x) = x^3 − 24x^2 + 143x and P(x) = x(x − 120)(x − 240). | |
0aab | Problem:
On a blackboard a finite number of integers greater than one are written. Every minute, Nordi additionally writes on the blackboard the smallest positive integer greater than every other integer on the blackboard and not divisible by any of the numbers on the blackboard. Show that from some point onwards Nord... | [
"Solution:\n\nLet $a$ be the largest integer initially written on the blackboard. Furthermore, denote by $a_{n}$ the integer written by Nordi on the blackboard after $n$ minutes.\n\nSuppose that $p > a$ is prime. If Nordi never writes $p$ on the blackboard, there exist $a_{n} < p < a_{n+1}$ since $a_{1}, a_{2}, \\l... | Nordic Mathematical Olympiad | The 35th Nordic Mathematical Contest | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0fs6 | Problem:
Let $\varphi$ denote the Euler phi-function. Prove that for every positive integer $n$
$$
2^{n(n+1)} \mid 32 \cdot \varphi\left(2^{2^{n}}-1\right)
$$ | [
"Solution:\nWe induct on $n$. The cases $n=1,2,3$ can easily be checked by hand:\n- $n=1: 2^{2} \\mid 32 \\cdot 2$.\n- $n=2: 2^{6} \\mid 2^{5} \\cdot \\varphi(15)=2^{5} \\cdot 2 \\cdot 2^{3}$\n- $n=3: 2^{12} \\mid 2^{5} \\cdot \\varphi(255)=2^{5} \\cdot 2 \\cdot 2^{2} \\cdot 2^{4}=2^{12}$\n\nFor $n \\geq 4$ assume ... | Switzerland | null | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
05n8 | Problem:
Soit $ABC$ un triangle ayant ses trois angles aigus, et $P$ le pied de la hauteur issue de $A$. On note $I_{1}$ et $I_{2}$ les centres de cercles inscrits à $ABP$ et $ACP$. Le cercle inscrit à $ABC$ touche $[BC]$ en $D$. Combien valent les angles du triangle $I_{1}I_{2}D$ ?
 | [
"Solution:\n\nOn note $E$ et $F$ les points où les cercles inscrits à $ABP$ et $ACP$ touchent $[BC]$, et $G$ et $H$ les points où ils touchent $[AP]$ : $PEI_{1}H$ est un carré car les angles en $P$, $E$ et $H$ sont droits, et $I_{1}E = I_{1}H$, donc $EP = EI_{1}$, et de même $FP = FI_{2}$. On a donc :\n$$\nFD = CD ... | France | French Mathematical Olympiad | [
"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",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing... | null | proof and answer | 90°, 45°, 45° | |
0jyg | Problem:
Given complex number $z$, define sequence $z_{0}, z_{1}, z_{2}, \ldots$ as $z_{0}=z$ and $z_{n+1}=2 z_{n}^{2}+2 z_{n}$ for $n \geq 0$. Given that $z_{10}=2017$, find the minimum possible value of $|z|$. | [
"Solution:\nDefine $w_{n}=z_{n}+\\frac{1}{2}$, so $z_{n}=w_{n}-\\frac{1}{2}$, and the original equation becomes\n$$\nw_{n+1}-\\frac{1}{2}=2\\left(w_{n}-\\frac{1}{2}\\right)^{2}+2\\left(w_{n}-\\frac{1}{2}\\right)=2 w_{n}^{2}-\\frac{1}{2}\n$$\nwhich reduces to $w_{n+1}=2 w_{n}^{2}$. It is not difficult to show that\n... | United States | HMMT November | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof and answer | (sqrt[1024]{4035} - 1)/2 | |
0g9u | 設 $AB$ 為圓 $O$ 上的弦, $M$ 為 $AB$ 劣弧的中點。由圓 $O$ 外一點 $C$ 向圓 $O$ 引切線, 設切點分別為 $S$, $T$。令線段 $MS$ 與線段 $AB$ 的交點為 $E$, 線段 $MT$ 與線段 $AB$ 的交點為 $F$。由 $E$ 點作 $AB$ 線段的垂線, 交 $OS$ 於 $X$ 點; 由 $F$ 點作 $AB$ 線段的垂線, 交 $OT$ 於 $Y$ 點。另外再由 $C$ 點向圓 $O$ 引一割線, 設兩交點分別為 $P$, $Q$。設線段 $MP$ 與線段 $AB$ 交於 $R$ 點。令 $\triangle PQR$ 的外心為 $Z$ 點。
證明：$X$, $Y$, $Z$ ... | [
"作 $AB$ 的中垂線 $OM$。故 $\\triangle XES \\sim \\triangle OMS$，於是 $SX = XE$。\n\n\n\n畫以 $XE$ 為半徑的圓 $X$。圓 $X$ 與弦 $AB$ 及直線 $CS$ 均相切。又作 $\\triangle PQR$ 的外接圓，以及直線 $MA$ 與 $MC$，如圖所示。\n\n因為 $\\triangle AMR \\sim \\triangle PMA$，所以有\n$$\nMR \\cdot MP = MA^2 = ME \\cdot MS.\n$$\n又由圓幂定理知 $CQ \\cdot CP = C... | Taiwan | 二〇一六數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
042o | If real number $x$ satisfies $\log_2 x = \log_4(2x) + \log_8(4x)$, then the value of $x$ is ______. | [
"By the given condition, we have\n$$\n\\log_2 x = \\log_4 2 + \\log_4 x + \\log_8 4 + \\log_8 x = \\frac{1}{2} + \\frac{1}{2} \\log_2 x + \\frac{2}{3} + \\frac{1}{3} \\log_2 x,\n$$\nand its solution is $\\log_2 x = 7$. Therefore, $x = 128$."
] | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | 128 | |
0l0d | Problem:
Nine distinct positive integers summing to $74$ are put into a $3 \times 3$ grid. Simultaneously, the number in each cell is replaced with the sum of the numbers in its adjacent cells. (Two cells are adjacent if they share an edge.) After this, exactly four of the numbers in the grid are $23$. Determine, with... | [
"Solution:\n\nSuppose the initial grid is of the format shown below:\n$$\n\\left[\\begin{array}{lll}\na & b & c \\\\\nd & e & f \\\\\ng & h & i\n\\end{array}\\right]\n$$\nAfter the transformation, we end with\n$$\n\\left[\\begin{array}{lll}\na_{n} & b_{n} & c_{n} \\\\\nd_{n} & e_{n} & f_{n} \\\\\ng_{n} & h_{n} & i_... | United States | HMMT February 2024 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 18 | |
0frj | Problem:
Al desarrollar $\left(1+x+x^{2}\right)^{n}$ en potencias de $x$, exactamente tres términos tienen coeficiente impar. ¿Para qué valores de $n$ es esto posible? | [
"Solution:\n\nEmpezamos estudiando qué efecto tiene sobre los coeficientes de un polinomio multiplicar por $\\left(1+x+x^{2}\\right)$:\n$$\n\\begin{aligned}\n\\left(1+x+x^{2}\\right)^{n+1} & =\\left(1+x+x^{2}\\right)^{n}\\left(1+x+x^{2}\\right) \\\\\n& =\\left(a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+\\cdots\\right)\\... | Spain | Spain | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | n is a power of 2 (i.e., n = 2^k for some integer k ≥ 0) | |
0i3j | Problem:
The Fibonacci numbers are defined by $F_{1} = F_{2} = 1$ and $F_{n+2} = F_{n+1} + F_{n}$ for $n \geq 1$. The Lucas numbers are defined by $L_{1} = 1$, $L_{2} = 2$, and $L_{n+2} = L_{n+1} + L_{n}$ for $n \geq 1$. Calculate
$$
\frac{\prod_{n=1}^{15} \frac{F_{2n}}{F_{n}}}{\prod_{n=1}^{13} L_{n}}
$$ | [
"Solution:\n\nIt is easy to show that $L_{n} = \\frac{F_{2n}}{F_{n}}$, so the product above is $L_{14} L_{15} = 843 \\cdot 1364 = 1149852$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 1149852 | |
0277 | Problem:
Determine all distinct positive integers $x$ and $y$ such that
$$
\frac{1}{x} + \frac{1}{y} = \frac{2}{7}
$$ | [
"Solution:\n\n$$\n2 x y = 7(x + y), \\quad \\text{com} \\quad x > 0, y > 0 \\text{ e } x \\neq y\n$$\nComo $2$ e $7$ são números primos, segue que $7$ divide $x$ ou $y$. Como a equação é simétrica em $x$ e $y$, podemos supor que $7$ divide $x$. Então, $x = 7k$, para algum $k > 0$ inteiro e decorre que $2 \\times 7k... | Brazil | Nível 2 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | (x, y) = (4, 28) or (28, 4) | |
09v2 | Quadrilateral $ABCD$ has right angles at $A$ and $D$. A circle of radius $10$ fits neatly inside the quadrilateral and touches all four sides. The length of edge $BC$ is $24$. The midpoint of edge $AD$ is called $E$ and the midpoint of edge $BC$ is called $F$. What is the length of $EF$?
A) $\frac{43}{2}$
B) $\frac{13... | [
"D) $22$"
] | Netherlands | First Round, January 2019 | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | MCQ | D | |
0g66 | 有一個 $m \times m$ 個單位方格構成的桌子, 在某些單位方格的中心點有一隻螞蟻。從時間 $0$ 開始, 每隻螞蟻都沿著一個平行於方格邊的方向, 以速率 $1$ 前進。過程中若有螞蟻相遇:
(i) 如果是兩隻正面相遇, 則它們會一起順時針轉彎 $90^\circ$, 然後繼續以速率 $1$ 前進;
(ii) 如果是兩隻以垂直方向相遇, 或超過兩隻以上的螞蟻相遇, 則它們會繼續以原本的速度和方向前進。
當螞蟻爬到桌子邊緣, 它會從桌面摔落, 不再回來。當最後一隻螞蟻摔落桌面時, 我們說此時刻就是這群螞蟻的“末日”。
考慮所有可能的螞蟻起始位置, 試求末日發生的最晚可能時刻, 或是證明並不一定會有末日。 | [
"其中 (i) 的規定可以修改為: 南北向正面相遇它們會順時轉彎 $90^\\circ$, 而東西向相遇它們會逆時轉彎 $90^\\circ$。修改之後與修改之前相比, 任何時間點所有螞蟻的所在的位置並沒有改變, 只是時間之前有東西向相遇的螞蟻交換彼此角色, 所以這不影響末日發生的時間。修改之後所有螞蟻分成兩類: (NE 類) 永遠向東或向北前進; (SW 類) 永遠向西或向南前進。\n\n以座標 $(0,0)$ 代表桌子的 SW 角, 以座標 $(m, m)$ 代表桌子的 NE 角。當時間為 $t$ 之時,\n區域 $\\{(x, y) \\mid x + y \\le 1 + t\\}$ 沒有 (NE 類) 螞蟻\n區域... | Taiwan | 二〇一二數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 3m/2 - 1 | |
0ffx | Problem:
Sea $\mathbb{Z}$ el conjunto de los enteros y $\mathbb{Z} \times \mathbb{Z}$ el conjunto de pares ordenados de enteros. La suma de estos pares se define por
$$
(a, b)+\left(a', b'\right)=\left(a+a', b+b'\right)
$$
siendo $(-a,-b)$ el opuesto de $(a, b)$.
Estudiar si existe un subconjunto $E$ de $\mathbb{Z} \ti... | [
"Solution:\nCualquier recta que pase por el origen determina dos semiplanos. Si la recta tiene pendiente racional, contiene puntos de coordenadas enteras distintos del origen. Si la pendiente es irracional, solamente contiene al origen.\nComo ejemplos de conjuntos $E$ pueden tomarse los puntos de coordenadas entera... | Spain | OME 21 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities... | null | proof only | null | |
01cz | Let $a_{0,1}, a_{0,2}, \dots, a_{0,2016}$ be positive real numbers. For $n \ge 1$ and $1 \le k < 2016$ set
$$
a_{n+1,k} = a_{n,k} + \frac{1}{2a_{n,k+1}}
$$
and
$$
a_{n+1,2016} = a_{n,2016} + \frac{1}{2a_{n,1}}.
$$
Let
$$
m_n = \max_{1 \le k \le 2016} a_{n,k} \quad \text{for } n \ge 0.
$$
Show that $m_{2016} > 44$. | [
"We prove\n$$\nm_n^2 \\ge n \\qquad (4)\n$$\nfor all $n$. The claim then follows from $44^2 = 1936 < 2016$. To prove (4), first notice that the inequality certainly holds for $n = 0$.\nAssume (4) is true for $n$. There is a $k$ such that $a_{n,k} = m_n$. Also $a_{n,k+1} \\le m_n$ (or if $k = 2016$, $a_{n,1} \\le m_... | Baltic Way | Baltic Way 2016 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
07zw | Problem:
Quante soluzioni reali ha il sistema
$$
\left\{\begin{array}{l}
x^{2} y=150 \\
x^{3} y^{2}=4500
\end{array} \right.
$$
(A) Nessuna
(B) una
(C) più di una, ma meno di cinque
(D) un numero finito, ma almeno cinque
(E) infinite. | [
"Solution:\n\nDalla prima equazione $x^{2} y = 150$ ricaviamo $y = \\dfrac{150}{x^{2}}$ (con $x \\neq 0$).\n\nSostituiamo nella seconda equazione:\n$$\nx^{3} y^{2} = 4500\n$$\nSostituendo $y$:\n$$\nx^{3} \\left( \\dfrac{150}{x^{2}} \\right)^{2} = 4500\n$$\n$$\nx^{3} \\cdot \\dfrac{22500}{x^{4}} = 4500\n$$\n$$\n\\df... | Italy | Italian Mathematical Olympiad - Febbraio Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | B | |
0c9p | Problem:
Dacă numărul natural $a$ are $n$ cifre, iar numărul natural $a^{4}$ are $m$ cifre, arătați că suma $m+n$ nu poate fi egală cu 2021. | [] | Romania | Olimpiada Națională GAZETA MATEMATICĂ | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Other"
] | null | proof only | null | |
0140 | Problem:
Let $x$ and $y$ be positive integers and assume that $z=\frac{4 x y}{x+y}$ is an odd integer. Prove that at least one divisor of $z$ can be expressed in the form $4 n-1$ where $n$ is a positive integer. | [
"Solution:\nLet $x=2^{s} x_{1}$ and $y=2^{t} y_{1}$ where $x_{1}$ and $y_{1}$ are odd integers. Without loss of generality we can assume that $s \\geq t$. We have\n$$\nz=\\frac{2^{s+t+2} x_{1} y_{1}}{2^{t}\\left(2^{s-t} x_{1}+y_{1}\\right)}=\\frac{2^{s+2} x_{1} y_{1}}{2^{s-t} x_{1}+y_{1}}\n$$\nIf $s \\neq t$, then ... | Baltic Way | Baltic Way 2005 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
037y | Problem:
Let $O$ be a fixed point in the plane. Find all sets of points $S$ in the plane, containing at least two distinct points, and such that for any point $A \in S, A \neq O$, the circle with diameter $O A$ is contained in $S$. | [
"Solution:\nWe first prove the following\n\nLEMMA. If $A \\in S$, then the open disc $k(O, O A)$ is contained in $S$.\n\nProof of the Lemma. Note that if $A \\in S$ and $B$ is a point on the circle of diameter $O A$ then $B \\in S$ (since $B$ belongs to the circle with diameter $O X$, where $O B \\perp B X$ and $X$... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | All such sets are either the entire plane, or the union of an open disc centered at the fixed point with an arbitrary subset of its boundary circle. | |
07o4 | Prove for all integers $n \ge 0$ that
$$
\sum_{k=0}^{n} \frac{(n+k)!}{k!} = \frac{(2n+1)!}{(n+1)!}
$$ | [
"With the usual notation for the binomial coefficients,\n$$\n\\binom{n}{r} = \\begin{cases} 0, & \\text{if } r < 0 \\text{ or } r > n, \\\\ \\frac{n!}{(n-r)!r!}, & \\text{if } 0 \\le r \\le n, \\end{cases}\n$$\nupon dividing both sides by $n!$, we see that an equivalent formulation of the problem is to show for all... | Ireland | Ireland | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
050m | The sides $AB$ and $AC$ of the triangle $ABC$ touch the circle $c$ respectively at points $B'$ and $C'$. The center $L$ of the circle $c$ lies on the side $BC$. The circumcenter $O$ of triangle $ABC$ lies on the shorter arc $B'C'$ of the circle $c$. Prove that the circumcircle of $ABC$ and the circle $c$ meet at two po... | [
"Let $r$ be the circumradius of $ABC$, let $s$ be the radius of $c$ and $\\alpha = \\angle BAC$ (Fig. 15).\n\nBy tangency, $|AB'| = |AC'|$. Thus $\\angle C'B'A = \\angle B'C'A = \\frac{\\pi}{2} - \\frac{\\alpha}{2}$ whence, by property of inscribed angle, $\\angle B'OC' = \\pi - (\\frac{\\pi}{2} - \\frac{\\alpha}{2... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0k49 | Problem:
a. Find two quadruples of positive integers $(a, b, c, n)$, each with a different value of $n$ greater than $3$, such that
$$
\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = n
$$
b. Show that if $a, b, c$ are nonzero integers such that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ is an integer, then $a b c$ is a perf... | [
"Solution:\n\na.\nFor example, $(1, 2, 4, 5)$ and $(9, 2, 12, 6)$ work.\n\nb.\nBefore solving the problem, we establish a useful definition and lemma.\n\nIf $p$ is a prime and $x$ is a nonzero rational number, we define $\\operatorname{ord}_p x$ to be the unique integer $k$ such that $p^{-k} x$ is an integer not di... | United States | Bay Area Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | Examples: (1, 2, 4, 5) and (9, 2, 12, 6); moreover, abc is a perfect cube. | |
0j53 | Problem:
Given a positive integer $n$, a sequence of integers $a_{1}, a_{2}, \ldots, a_{r}$, where $0 \leq a_{i} \leq k$ for all $1 \leq i \leq r$, is said to be a "$k$-representation" of $n$ if there exists an integer $c$ such that
$$
\sum_{i=1}^{r} a_{i} = \sum_{i=1}^{r} a_{i} k^{c-i} = n.
$$
Prove that every positi... | [
"Solution:\n\nEquivalently, a $k$-representation is given by a sequence $a_{p}, \\ldots, a_{q}$, for some $p < q$, such that\n$$\n\\sum_{i=p}^{q} a_{i} = \\sum_{i=p}^{q} a_{i} k^{i}.\n$$\nWe first show existence. Let the representation of $n-1$ in base $k$ be given by $\\sum_{i=0}^{r} a_{i} k^{i} = n-1$. Let $l = \... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0cgs | Let $ABC$ be a triangle inscribed in the circle $C$ with center $O$ and radius $1$. For any point $M \in C \setminus \{A, B, C\}$, we denote $s(M) = OH_1^2 + OH_2^2 + OH_3^2$, where $H_1$, $H_2$, and $H_3$ are the orthocenters of triangles $MAB$, $MBC$, and $MCA$, respectively.
a) Prove that if triangle $ABC$ is equil... | [
"Consider an orthonormal coordinate system with the origin at $O$. For any point $Z$ in the plane, we denote its complex coordinate by $z$.\n\na.\nFrom Sylvester's relation, we have $h_1 = m + a + b$, $h_2 = m + b + c$, and $h_3 = m + c + a$. Also, let $h = a + b + c$ be the complex coordinate of the orthocenter of... | Romania | 74th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof and answer | s(M) = 6 in the equilateral case; if s(M) takes the same value at three distinct points on the circle, then triangle ABC is equilateral. | |
09r9 | Problem:
Vind alle functies $f: \mathbb{R} \rightarrow \mathbb{R}$ die voldoen aan
$$
f(x+x y+f(y))=\left(f(x)+\frac{1}{2}\right)\left(f(y)+\frac{1}{2}\right)
$$
voor alle $x, y \in \mathbb{R}$. | [
"Solution:\nVul in $y=-1$, dan staat er:\n$$\nf(f(-1))=\\left(f(x)+\\frac{1}{2}\\right)\\left(f(-1)+\\frac{1}{2}\\right) .\n$$\nAls $f(-1) \\neq-\\frac{1}{2}$, dan kunnen we delen door $f(-1)+\\frac{1}{2}$ en krijgen we\n$$\nf(x)+\\frac{1}{2}=\\frac{f(f(-1))}{f(-1)+\\frac{1}{2}},\n$$\nwat betekent dat $f$ constant ... | Netherlands | Dutch TST | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x)=x+1/2 | |
02ro | In how many ways one can color the cells of a $n \times n$ table, each with one of four colors, such that no cells that share a side have the same color and all four colors appear in every $2 \times 2$ square formed by neighboring cells? | [
"Answer: $3 \\cdot 2^{n+2} - 24$.\nSuppose there are at least three different colors $A, B, C$ in the first row. Then they occur consecutively, say in the order $ABC$. Then the cell below the $B$ has the fourth color $D$, and all other cells are determined by the first row, because three cells in a $2 \\times 2$ sq... | Brazil | Brazilian Math Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 3 * 2^(n+2) - 24 | |
01pi | The odd number of the asterisks are written on the blackboard: $\underbrace{**\dots*}_{2n+1}$.
Ann and Bob play the following game. They, in turn (Ann starts), replace one of the asterisks in the expression $\underbrace{**\dots*}_{2n+1}$ by any of the digits from $0$ to $9$ (the first left asterisk cannot be replaced b... | [
"Answer: Bob wins.\nIt is well-known that a natural number $n$ is divisible by $11$ if and only if $(S_o - S_e) \\div 11$, where $S_o$, $S_e$ are the sums of the digits on the odd and even positions respectively in the decimal representation of $n$.\nLet $2n + 1$ ($n \\in \\mathbb{N}$) asterisks be written on the b... | Belarus | BelarusMO 2013_s | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | null | proof and answer | Bob | |
04d9 | Let $\triangle ABC$ be acute triangle and let $A_1, B_1, C_1$ be points on its sides $\overline{BC}, \overline{CA}, \overline{AB}$ respectively. Prove that the triangles $\triangle ABC$ and $A_1B_1C_1$ are similar ($\angle A = \angle A_1$, $\angle B = \angle B_1$, $\angle C = \angle C_1$) if and only if the orthocentre... | [
"Let the triangle $A_1B_1C_1$ be similar to the triangle $ABC$ ($\\angle A = \\angle A_1 = \\alpha$, $\\angle B = \\angle B_1 = \\beta$, $\\angle C = \\angle C_1 = \\gamma$), and let the point $O$ be the orthocentre of the triangle $A_1B_1C_1$. Then $\\angle OB_1C_1 = 90^\\circ - \\gamma$, $\\angle OC_1B_1 = 90^\\c... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0d8u | The 64 cells of an $8 \times 8$ chessboard have 64 different colours. A Knight stays in one cell. In each move, the Knight jumps from one cell to another cell (the 2 cells on the diagonal of an $2 \times 3$ board); also the colours of the 2 cells interchange. In the end, the Knight goes to a cell having common side wit... | [
"The answer is no. Suppose that there are exactly 3 cells having the colours different from the original colours.\nAssign each colour with an integer from $1$ to $64$ and colour again the chessboard by black and white as usual. Arrange the numbers from left to right and up to down, so each move gives us a new permu... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | No | |
0443 | Suppose function $f(x)$ satisfies: for any non-zero real number $x$, there is
$$
f(x) = f(1) \cdot x + \frac{f(2)}{x} - 1.
$$
Then the minimum of $f(x)$ on $(0, +\infty)$ is ______. | [
"Let $x = 1, 2$, and we can get $f(1) = f(1) + f(2) - 1$ and $f(2) = 2f(1) + \\frac{f(2)}{2} - 1$, respectively. The solution is $f(2) = 1$, $f(1) = \\frac{3}{4}$.\n\nThus, for $x \\neq 0$, there is\n$$\nf(x) = \\frac{3}{4}x + \\frac{1}{x} - 1.\n$$\nWhen $x \\in (0, +\\infty)$, $f(x) \\ge 2\\sqrt{\\frac{3}{4}x \\cd... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | sqrt(3) - 1 | |
0kbb | Problem:
Let $n$ be a positive integer. Define a sequence by $a_{0}=1$, $a_{2i+1}=a_{i}$, and $a_{2i+2}=a_{i}+a_{i+1}$ for each $i \geq 0$. Determine, with proof, the value of $a_{0}+a_{1}+a_{2}+\cdots+a_{2^{n}-1}$. | [
"Solution:\n\nNote that $a_{2^{n}-1}=1$ for all $n$ by repeatedly applying $a_{2i+1}=a_{i}$. Now let $b_{n}=a_{0}+a_{1}+a_{2}+\\cdots+a_{2^{n}-1}$. Applying the given recursion to every term of $b_{n}$ except $a_{0}$ gives\n$$\n\\begin{aligned}\nb_{n}= & a_{0}+a_{1}+a_{2}+a_{3}+\\cdots+a_{2^{n}-1} \\\\\n= & a_{0}+a... | United States | HMMT February 2020 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | (3^n + 1)/2 | |
08ug | Suppose $n$ is a positive integer of 3 distinct non-zero digits. Let $g$ be the greatest common divisor of the 6 numbers obtained by permuting the digits of $n$. Determine the maximum possible value that $g$ can take. | [
"[18]\nFirst, let us show that $g$ cannot exceed $18$ for any $n$. Denote by $a, b, c$ the $3$ digits of $n$, where we assume $a < b < c$. Both $100c+10b+a$ and $100c+10a+b$ are numbers obtained by permuting the digits of $n$. Hence $g$ is a divisor of $(100c+10b+a) - (100c+10a+b) = 9(b-a)$. Similarly, we get that ... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 18 | |
03pk | There are 47 students in a classroom with seats arranged in $6$ rows $ imes$ $8$ columns, and the seat in the $i$-th row and $j$-th column is denoted by $(i, j)$. Now, an adjustment is made for students' seats in the new school term. For a student with the original seat $(i, j)$, if his/her new seat is $(m, n)$, we say... | [
"Add a virtual student $A$ so that every seat is occupied by exactly one student. Denote $S'$ the sum of the position values in this situation. Notice that an exchange of two students occupying the adjacent seats will not change the value of $S'$. Every student can return to his/her original seat by a finite number... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 12 | |
04d5 | Let $A$ be a subset of the set $\{1, 2, 3, \dots, 26\}$ with seven elements. Prove that there are two distinct nonempty subsets of $A$ such that the sums of their elements are equal. | [
"We prove a stronger statement that already among subsets of $A$ with at most four elements there are two with equal sums of elements. Set $A$ has\n$$\n\\binom{7}{1} + \\binom{7}{2} + \\binom{7}{3} + \\binom{7}{4} = 98\n$$\nsubsets with at most four elements. The sum of elements of each of these subsets is at least... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0ctl | Given a $2015 \times 2015$ checkered board. Dmitry chooses $k$ cells and places a detector in each of them. After that Dmitry goes out, and then Nick places on a board a cellular square $1500 \times 1500$ (the square lies within the board, and its sides lie on the grid lines).
Each detector informs Dmitry whether its ... | [
"$k = 2(2015 - 1500) = 1030$.\nTo win with $1030$ detectors, it suffices to locate them in the $515$ leftmost cells of the middle row and in the $515$ topmost cells of the middle column.\n\nFor the estimate, notice that the union of any two vertical $1500 \\times 1$ rectangles differing by a horizontal $1500$-shift... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | 1030 | |
0iq9 | Problem:
Each vertex of a regular heptagon is colored either red or blue. Prove that there is an isosceles triangle with all its vertices the same color. | [
"Solution:\n\nDenote the vertices of the heptagon by $A$, $B$, $C$, $D$, $E$, $F$, $G$. Since an alternating arrangement cannot be continued all the way around the heptagon, two adjacent vertices must be the same color, say $A$ and $B$. If any of $C$, $E$, $G$ shares this color, we are done since triangles $A B C$,... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
09fi | The cells of a $n \times n$ table are painted black or white. Suppose that each black cell has an even number of white neighbours. Show that it is possible to paint all white cells by red and blue so that each black cell has the same number of red and blue neighbours. (Two cells are neighbours if they share a common si... | [
"See https://sites.google.com/site/uugnaaninjbat/"
] | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0ft8 | Problem:
Sei $ABC$ ein spitzwinkliges Dreieck mit Umkreismittelpunkt $O$. $S$ sei der Kreis durch $A$, $B$ und $O$. Die Geraden $AC$ und $BC$ schneiden $S$ in den weiteren Punkten $P$ und $Q$. Zeige $CO \perp PQ$. | [
"Solution:\n\nSei $R$ der Schnittpunkt von $PQ$ und $CO$. Da $PQBA$ ein Sehnenviereck ist, gilt $\\angle RPC = \\angle QPA = 180^{\\circ} - \\angle ABQ = \\beta$. Andererseits ist $\\angle AOC = 2 \\angle ABC = 2\\beta$. Das Dreieck $AOC$ ist gleichschenklig und daher gilt $\\angle PCR = \\angle ACO = \\frac{1}{2}(... | Switzerland | IMO - Selektion | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0ek4 | Problem:
Veronika ima list karirastega papirja z $78 \times 78$ kvadratki. List želi razrezati na manjše kose, od katerih bo vsak imel bodisi 14 bodisi 15 kvadratkov, pri čemer z vsakim rezom prereže enega od kosov papirja na dva dela vzdolž ene od črt na papirju. Najmanj kolikokrat mora Veronika prerezati papir? | [
"Solution:\n\nZ vsakim rezom se število kosov papirja poveča za 1. Na koncu bo torej število kosov papirja za 1 večje od števila rezov, ki jih je Veronika izvedla. Da bo izvedla čim manj rezov, mora imeti na koncu čim manj kosov papirja. Denimo, da ima na koncu $k$ kosov s 14 kvadratki in $n$ kosov s 15 kvadratki, ... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | null | proof and answer | 405 | |
029p | Problem:
Reverso de um número - O reverso de um número inteiro de dois algarismos é o número que se obtém invertendo a ordem de seus algarismos. Por exemplo, 34 é o reverso de 43. Quantos números existem que somados ao seu reverso dão um quadrado perfeito? | [
"Solution:\n\nDenotemos por $ab$ e $ba$ o número e seu reverso. Temos que\n$$\nab + ba = 10a + b + 10b + a = 11(a + b)\n$$\nPor outro lado, $a \\leq 9$ e $b \\leq 9$, logo, $a + b \\leq 18$. Como $11$ é um número primo e $a + b \\leq 18$, para que $11(a + b)$ seja um quadrado perfeito, só podemos ter $a + b = 11$.\... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 8 | |
0a34 | We define the number $A$ as the digit $4$ followed by $2024$ times the digit combination $84$, and the number $B$ by $2024$ times the digit combination $84$ followed by a $7$. So, to illustrate, we have
$$
A = 4 \underbrace{84\dots84}_{4048 \text{ digits}} \quad \text{and} \quad B = \underbrace{84\dots84}_{4048 \text{ ... | [] | Netherlands | Dutch Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 4/7 | |
0ga5 | Let $\mathbb{Z}$ be the set of all integers. Determine all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that
$$
f(f(x) + f(y)) + f(x)f(y) = f(x + y)f(x - y)
$$
holds for all $x, y \in \mathbb{Z}$
令 $\mathbb{Z}$ 為所有整數所成的集合。求所有函數 $f : \mathbb{Z} \to \mathbb{Z}$, 滿足:
$$
f(f(x) + f(y)) + f(x)f(y) = f(x + y)f(x - y)
$$
對... | [
"顯然 $f(x) = 0$ 是原方程的一組解。故假設存在一個 $t$ 使得 $f(t) \\neq 0$。於原式代入 $(0,0)$ 得 $f(2f(0)) = 0$\n代入 $(2f(0),0)$ 得 $f(f(0)) = 0$\n代入 $(f(0),f(0))$ 得 $f(0) = 0$\n代入 $(x,0)$ 得\n$$\nf(f(x)) = f(x)^2 \\qquad (1)\n$$\n代入 $(x, x)$ 得\n$$\nf(2f(x)) = -f(x)^2 \\qquad (2)\n$$\n為了方便起見，令 $g: \\mathbb{Z} \\to \\mathbb{Z}$ 滿足對於任意一個整數 $n$,\n... | Taiwan | 二〇一六數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Two functions:
1) f(x) = 0 for all integers x.
2) f(x) = 0 if x is a multiple of 5; f(x) = 1 if x is congruent to 1 or 4 modulo 5; f(x) = -1 if x is congruent to 2 or 3 modulo 5. | |
03mf | A number of robots are placed on the squares of a finite, rectangular grid of squares. A square can hold any number of robots. Every edge of the grid is classified as either *passable* or *impassable*. All edges on the boundary of the grid are *impassable*.
You can give any of the commands *up*, *down*, *left*, or *rig... | [
"We will prove any two robots can be moved to the same square. From that point on, they will always be on the same square. We can then similarly move a third robot onto the same square as these two, and then a fourth, and so on, until all robots are on the same square.\n\nTowards that end, consider two robots *A* a... | Canada | Kanada 2012 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English, French | proof only | null | |
0e0q | The incircle of an acute triangle $ABC$ touches the sides $BC$, $CA$ and $AB$ at $A_1$, $B_1$ and $C_1$. Let $K_A$, $K_B$ and $K_C$ be the incircles of the triangles $AB_1C_1$, $A_1BC_1$ and $A_1B_1C_1$.
Let $t_A$ denote the common tangent of $K_B$ and $K_C$ which intersects the segments $AB$ and $AC$ but not the segme... | [
"With this kind of problems it is very important to draw a big figure and try to see if there is anything we can say about the common intersection. First, we notice that the centres of $K_A$, $K_B$ and $K_C$ lie on the incircle $K$ of the triangle $ABC$.\n\nLet $A_2$ be the point where the bisector of the angle $B_... | Slovenia | Selection Examinations for the IMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"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"
] | null | proof only | null | |
01bg | Albert and Betty play the following game. There are two bowls on a table; a red bowl and a blue bowl. At the beginning of the game there are $100$ blue balls in the red bowl and $100$ red balls in the blue bowl. In each turn a player must take one of the following moves:
a) Take two balls of different colors from one ... | [
"**Answer:** Betty has a winning strategy.\n\nBetty follows this strategy: If Albert makes move b), then Betty makes move c) and vice versa. If Albert makes move a) from one bowl, Betty makes move a) from the other bowl. The only exception of this rule is if Betty can make a winning move, that is a move where she r... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Betty | |
00dl | Let $n$ be a positive integer. Using the integers from $1$ to $4n$ inclusive, pairs are to be formed such that the product of the numbers in each pair is a perfect square. Each number can be part of at most one pair, and the two numbers in each pair must be different. Determine, for each $n$, the maximum number of pair... | [
"For each $m \\in \\mathbb{N}$, let $f(m)$ be the product of the primes that appear with odd exponent in the prime factorization of $m$. It is easy to see that given two positive integers $a$ and $b$, the product $ab$ is a perfect square if and only if $f(a) = f(b)$.\n\nFor each $k \\in \\mathbb{N}$, let $S$ be the... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | n | |
03az | Let $f: \mathbb{N} \to \mathbb{N}$ be a function such that $f(1) = 1$ and $f(n) = n - f(f(n-1))$ for any $n \ge 2$. Prove that $f(n + f(n)) = n$ for any $n$. | [
"Denote by (*) the condition $f(n) = n - f(f(n-1))$ for any $n \\ge 2$. First, we shall prove by induction that\n\n$$\nf(n) \\le f(n+1) \\le f(n) + 1 \\text{ for any } n.\n$$\n\n1. If $n = 1$, then $f(2) = 2 - f(f(1)) = 1$ and the inequalities hold.\n\n2. Let the inequalities be true for any $k \\le n$. Then $f(n+1... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null |
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