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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0be0 | Circles $\Omega$ and $\omega$ are tangent at a point $P$ ($\omega$ lies inside $\Omega$). A chord $AB$ of $\Omega$ is tangent to $\omega$ at $C$; the line $PC$ meets again $\Omega$ at $Q$. Chords $QR$ and $QS$ of $\Omega$ are tangent to $\omega$. Let $I, X$, and $Y$ be the incentres of the triangles $APB, ARB$, and $AS... | [
"Notice that a homothety centred at $P$ mapping $\\omega$ to $\\Omega$ maps $C$ to $Q$, and maps the line $AB$ to the tangent to $\\Omega$ at $Q$. Thus this tangent is parallel to $AB$, and hence $Q$ is the midpoint of arc $AB$ (not containing $P$). So the points $I, X$, and $Y$ lie on the segments $PQ, RQ$, and $S... | Romania | 64th NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02he | Problem:
Uma cidade ainda não tem iluminação elétrica e todos usam velas à noite. Na casa de João, usa-se uma vela por noite, sem queimá-la totalmente. Com os tocos de quatro destas velas, é possível fazer uma nova vela. Durante quantas noites João poderá iluminar sua casa com 43 velas?
A) 43
B) 53
C) 56
D) 57
E) 60 | [
"Solution:\n\n(D) De 43 velas obtém-se 43 tocos. Como $43=4 \\times 10+3$, com esses 43 tocos se pode fazer 10 velas e guardar 3 tocos. Dessas 10 velas, obtemos 10 tocos que, com os 3 que sobraram, dão 13. Sendo $13=4 \\times 3+1$, fazemos então 3 velas com 12 tocos, sobrando 1 toco. Depois de usar estas 3 velas, t... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | MCQ | D | |
045i | Let $a_1, a_2, \dots, a_n$ be $n$ positive integers that are not divisible one another, i.e. for any $i \neq j$, $a_i$ is not divisible by $a_j$. Show that
$$
a_1 + a_2 + \dots + a_n \geq 1.1n^2 - 2n.
$$ | [
"**Proof:** Consider the set of all positive integers that are coprime to $6$: $B = \\{1, 5, 7, 11, 13, \\dots\\}$, it is obvious that the sum $f(k)$ of the smallest $k$ elements of $B$ satisfies\n$$\nf(k) = \\begin{cases} \\frac{3k^2}{2} & \\text{if } k \\text{ is even} \\\\ \\frac{3k^2-1}{2} & \\text{if } k \\tex... | China | 2022 China Team Selection Test for IMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
04bt | Let $P$ be a point inside the triangle $ABC$. Let $D$, $E$ and $F$ be the feet of the perpendiculars from point $P$ to $BC$, $CA$, and $AB$ respectively. If the quadrilaterals $AEPF$, $BFPD$, and $CDPE$ are circumscribed, prove that $P$ is the incentre of triangle $ABC$. | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscel... | English | proof only | null | |
0emq | If $n \in \mathbb{N}$ and $3n+1$ and $4n+1$ are perfect squares, show that $56|n$. | [
"Let $3n + 1 = x^2$ and $4n + 1 = y^2$. Then $y^2 - x^2 = n$. As $y$ is odd, $y^2 \\equiv_8 1$, so $n$ has to be even, therefore $x$ is odd, and therefore $y^2 - x^2 \\equiv_8 1 - 1 = 0$, hence $8|n$.\n\nNow notice that $4x^2 - 3y^2 = 1$, so setting $w = 2x$ we obtain Pell's equation $w^2 - 3y^2 = 1$. This has solu... | South Africa | South-Afrika 2011-2013 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0c6n | Let $A$ and $C$ be two points on a circle $C$ such that $(AC)$ is not a diameter, and let $P$ be a point of the line segment $(AC)$, other than its midpoint. Circles $c_1$ and $c_2$ are interiorly tangent to the circle $C$ at $A$ and $C$, respectively. They both pass through $P$ and intersect again at $Q$. The line $PQ... | [
"a) Point $B$ lies on the radical axis, $BD$, of circles $c_1$ and $c_2$, therefore $BK \\cdot BA = BL \\cdot BC$, which indicates that the quadrilateral $AKLC$ is cyclic. So is $ACMN$. It follows that $\\angle LKB = \\angle ACB$. The angle between the tangent at $A$ to the circle $c_1$ (which is also tangent to $C... | Romania | Stars of Mathematics Competition | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Brocard poin... | English | proof only | null | |
0ddl | In a circle $O$, there are six points, $A, B, C, D, E, F$ in a counterclockwise order. $BD \perp CF$, and $CF, BE, AD$ are concurrent. Let the perpendicular from $B$ to $AC$ be $M$, and the perpendicular from $D$ to $CE$ be $N$. Prove that $AE \parallel MN$. | [
"Let $K$ be the concurrent point of $CF, BE, AD$, and let $CF$ intersect $BD$ at $L$.\n\nWe have $\\angle BMC = 90^\\circ = \\angle BLC$ since $BM$ is perpendicular to $AC$ and $BL$ is perpendicular to $CF$, so $B, M, L, C$ lie on the same circle with diameter $BC$. Hence,\n$$\n\\angle CML = \\angle CBL = \\angle C... | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06v8 | Let $\mathbb{Z}$ be the set of integers. We consider functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfying
$$
f(f(x+y)+y)=f(f(x)+y)
$$
for all integers $x$ and $y$. For such a function, we say that an integer $v$ is $f$-rare if the set
$$
X_{v}=\{x \in \mathbb{Z}: f(x)=v\}
$$
is finite and nonempty.
a. Prove tha... | [
"a) Let $f$ be the function where $f(0)=0$ and $f(x)$ is the largest power of $2$ dividing $2x$ for $x \\neq 0$. The integer $0$ is evidently $f$-rare, so it remains to verify the functional equation.\n\nSince $f(2x)=2f(x)$ for all $x$, it suffices to verify the functional equation when at least one of $x$ and $y$ ... | IMO | IMO 2019 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
07jn | Let $g : \mathbb{C} \rightarrow \mathbb{C}$ be a surjective function. Find all complex functions $f : \mathbb{C} \rightarrow \mathbb{C}$ such that for any two complex numbers $x, y$, we have:
$$
|g(x) + f(y)| = |f(x) + g(y)|.
$$ | [
"First, note that by surjectivity, we can find $x_0$ such that $g(x_0) = -f(y_0)$, the problem statement then implies $f(x_0) = -g(y_0)$. By setting $y = y_0$, $x = x_0$ in the problem statement, we have:\n$$\n|g(y) - g(y_0)| = |f(y) - f(y_0)|\n$$\nNote that $f(x)$ is a function of $g(x)$ because if $g(x_0) = g(x_1... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | All solutions are exactly the following families:
- f(z) = g(z) + d for any complex constant d;
- f(z) = −g(z);
- f(z) = c^2 · overline{g(z)} + r c, where c is any complex number on the unit circle and r is a real constant. | |
0gaj | 令 $S$ 為所有正整數的一個非空子集。一個正整數 $n$ 被稱為乾淨的，若且唯若它可以被表示成 $S$ 的奇數個相異元素的和，且這個表示法是唯一的。試證：存在無窮多個不乾淨的正整數。
Let $S$ be a nonempty set of positive integers. We say that a positive integer $n$ is *clean* if it has a unique representation as a sum of an odd number of distinct elements from $S$. Prove that there exist infinite many posi... | [
"Define an *odd* (respectively, *even*) *representation* of $n$ to be a representation of $n$ as a sum of an odd (respectively, even) number of distinct elements of $S$. Let $\\mathbb{N}$ denote the set of all positive integers.\n\nSuppose, to the contrary, that there exist only finitely many positive integers that... | Taiwan | 二〇一六數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | null | proof only | null | |
08gl | Problem:
Data una stringa di cifre 0 e 1, Giacomo vorrebbe dividerla in blocchi da 2 o 3 cifre consecutive in modo da soddisfare una delle seguenti due condizioni:
(1) in ogni blocco, ogni (eventuale) cifra 0 compare a sinistra di ogni (eventuale) cifra 1. I blocchi consentiti sono cioè $00,01,11,000,001,011,111$;
(2)... | [
"Solution:\n\na. Notiamo che, comunque si suddivida la stringa, i blocchi saranno tutti del primo tipo: infatti, nella stringa data tutti gli $1$ sono a destra di tutti gli $0$. Inoltre, esiste almeno una suddivisione della stringa in blocchi da 2 o da 3 cifre: se $m+n$ è pari, possiamo suddividere la stringa in bl... | Italy | Olimpiadi di Matematica | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2^{l-3} | |
059r | Let $H$ be the orthocenter of an acute triangle $ABC$. Let $M$ be the midpoint of $BC$. Let $A'$ and $H'$ be the reflections of the points $A$ and $H$ across the point $M$. Prove that the points $B$ and $C$ and the reflections of $A'$ over the lines $BH'$ and $CH'$ are concyclic. | [
"\n\nLet $O$ be the circumcenter of $ABC$ and $G$ the antipode of $A$ (Fig. 11). Then $BG \\perp AB$ and $CG \\perp AC$. As $CH \\perp AB$ and $BH \\perp AC$, this means that $CG \\parallel BH$ and $BG \\parallel CH$. So $BHCG$ is a parallelogram. Therefore $M$, the midpoint of $BC$, is als... | Estonia | Estonian Math Competitions | [
"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 | |
05bw | Anna, Berta and Carol make fruit drinks from syrup. Anna makes $a$ litres of drink by mixing water and syrup in the proportion of $a : 1$. Berta makes $b$ litres of drink by mixing water and syrup in the proportion of $b : 1$. Carol makes $c$ litres of drink by mixing water and syrup in the proportion of $c : 1$. (It i... | [
"The percentage of syrup in Anna's drink is $\\frac{1}{a+1}$, thus it contains $\\frac{a}{a+1}$ litres of syrup. Similarly, Berta uses $\\frac{b}{b+1}$ litres and Carol uses $\\frac{c}{c+1}$ litres of syrup. Hence we have to prove that $a + b + c = 6$ implies\n$$\n\\frac{a}{a+1} + \\frac{b}{b+1} + \\frac{c}{c+1} \\... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0joc | Problem:
Suppose $x$ and $y$ are real numbers satisfying $x + y = 5$. What is the largest possible value of $x^{2} + 2 x y$? | [
"Solution:\nThe quantity in question is $(x + y)^{2} - y^{2} \\leq (x + y)^{2} = 25$. Equality occurs when $x = 5$ and $y = 0$, hence the maximum possible value is $25$."
] | United States | Berkeley Math Circle | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 25 | |
09qt | Problem:
Zij $ABC$ een driehoek met $|AB| > |BC|$. Zij $D$ het midden van $AC$. Zij $E$ het snijpunt van de bissectrice van $\angle ABC$ met de lijn $AC$. Zij $F$ op $BE$ zo dat $CF$ loodrecht op $BE$ staat. Zij verder $G$ het snijpunt van $CF$ en $BD$.
Bewijs dat $DF$ het lijnstuk $EG$ doormidden snijdt. | [
"Solution:\n\nOplossing buy. Merk op dat vanwege de voorwaarde $|AB| > |BC|$ de punten $D$ en $E$ verschillend zijn en de volgorde van punten op de lijn $AC$ is: $A, D, E, C$. Verder is $\\angle BCA > \\angle CAB$, dus $\\angle BCA + \\frac{1}{2} \\angle ABC > 90^{\\circ}$, waaruit volgt dat $F$ op het inwendige va... | Netherlands | Dutch TST toets 8 juni | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocent... | null | proof only | null | |
0kg5 | Ten million fireflies are glowing in $\mathbb{R}^3$ at midnight. Some of the fireflies are friends, and friendship is always mutual. Every second, one firefly moves to a new position so that its distance from each one of its friends is the same as it was before moving. This is the only way that the fireflies ever chang... | [
"**Construction:** Choose three pairwise parallel lines $l_A, l_B, l_C$ forming an infinite equilateral triangle prism (with side larger than 1). Split the $n$ fireflies among the lines as equally as possible, and say that two fireflies are friends iff they lie on different lines.\nTo see this works:\n1. Reflect $l... | United States | USA TSTST | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Solid Geometry > 3D Shapes"
] | English | proof and answer | f(n) = floor(n^2/3) for n ≥ 70 | |
0gs8 | In a triangle $ABC$, the interior angle bisector of $A$ intersects the $A$-excircle of $ABC$ at $D$ and $E$ such that $D \in [AE]$. Show that
$$
\frac{|AD|}{|AE|} \le \frac{|BC|^2}{|DE|^2}.
$$ | [
"Let $BC = a$, $u$ be the semiperimeter, the length of the altitude passing through $A$ be $h_a$, and the center and radius of $A$-excircle be $J_a$ and $r_a$, respectively. It is known that the points $A, D, J_a, E$ are collinear. We have\n$$\nAD = AJ_a - J_a D = \\sqrt{r_a^2 + u^2} - r_a, \\quad DE = 2r_a, \\quad... | Turkey | 26th Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > C... | English | proof only | null | |
0bt5 | Let $ABCD A'B'C'D'$ be a cuboid and $M$, respectively $N$ be the feet of the perpendiculars from $A'$ and $C'$ on $BD$. The length of the sides $AB$, $BC$ and $AA'$ are $\sqrt{6}$, $\sqrt{3}$ respectively $\sqrt{2}$.
a) Prove that $A'M \perp C'N$.
b) Compute the measure of the angle of the planes $(A'MC)$ and $(ANC')... | [
"a) In the right triangle $ABD$, $AM \\perp BD$, so $AM = \\sqrt{2}$, that is the triangle $A'MA$ is isosceles. The same goes for the triangle $C'CN$. It follows $\\angle A'MA = \\angle C'NC = 45^\\circ$. Since $CN \\parallel AM$ and $AA' \\parallel CC'$, the parallel from $A'$ to $NC'$ is coplanar with $AM$, hence... | Romania | 67th Romanian Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 60 degrees | |
0goi | Find the largest positive integer $n$ which is divisible by all positive integers whose cube is not greater than $n$. | [
"The answer is $420$.\n\nLet us consider the positive integer $m$ so that $m^3 \\leq n < (m+1)^3$. As $n = 420$ satisfies the conditions, we will consider the case when $m \\geq 7$. Note that each of $m$, $m-1$, $m-2$ and $m-3$ divides $n$ and hence $\\operatorname{lcm}(m, m-1, m-2, m-3)$ divides $n$. Since $\\gcd(... | Turkey | Team Selection Test | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | 420 | |
0gq5 | Let $a > 1$, $b > 1$, $c > 1$, $d > 1$, $x$, $y$ be real numbers satisfying
$$
a^x + b^y = (a^2 + b^2)^x \text{ and } c^x + d^y = 2^y (cd)^{y/2}.
$$
Prove that $x < y$. | [
"On the contrary suppose that $x - y = t \\ge 0$. From $c^x + d^y = 2^y (cd)^{y/2}$ we get\n$$\n\\left(\\frac{c}{d}\\right)^y \\cdot c^t + 1 = \\left(2\\sqrt{\\frac{c}{d}}\\right)^y\n$$\nSince $t \\ge 0$ and $c > 1$ we have $c^t \\ge 1$ and\n$$\n\\left(2\\sqrt{\\frac{c}{d}}\\right)^y \\ge \\left(\\frac{c}{d}\\right... | Turkey | Team Selection Test for JBMO | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0d8s | Let $ABC$ be an acute triangle and $(O)$ be its circumcircle. Denote by $H$ its orthocenter and $I$ the midpoint of $BC$. The lines $BH$, $CH$ intersect $AC$, $AB$ at $E$, $F$ respectively. The circles $(IBF)$ and $(ICE)$ meet again at $D$.
1. Prove that $D$, $I$, $A$ are collinear and $HD$, $EF$, $BC$ are concurrent.... | [
"1) From the power of point to circle, we have\n$$\n\\mathscr{P}_{A /(BIF)} = AF \\cdot AB \\text{ and } \\mathscr{P}_{A /(CIE)} = AE \\cdot AC.\n$$\nBut it is easy to see that $AF \\cdot AB = AE \\cdot AC$ then $A$ belongs to the radical axis of two circles $(BIF)$, $(CIE)$.\nHence, $A$, $D$, $I$ are collinear.\nD... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordi... | English | proof only | null | |
0dag | Let $ABC$ be an acute triangle with $O, H$ are its circumcenter, orthocenter. Take a point $M$ belongs to the minor $\operatorname{arc} BC$ of $(O)$ (not coincide to $B, C$) and denote $D, E, F$ as reflection of $M$ through $OA, OB, OC$. Suppose that $BF$ meets $CE$ at $K$ and let $I$ be the incenter of triangle $DEF$.... | [
"1) By definition of $E, F$ we can see that $EC$ is the angle bisector of $\\angle MEF$ and $FB$ is the angle bisector of $\\angle MFE$. This implies that $K$ is the incenter of triangle $MEF$.\nHence, $MK$ is the angle bisector of $\\angle EMF$ which passes through the midpoint $N$ of the $\\operatorname{arc} EF$ ... | Saudi Arabia | Team selection tests for JBMO 2018 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
03wi | Let $n$ be a given positive integer. In the coordinate plane, consider the set of the points
$$
\{P_1, P_2, \dots, P_{4n+1}\} = \{(x, y) \mid x \text{ and } y \text{ are integers with } xy = 0, |x| \le n, |y| \le n\}.
$$
Determine the minimum of $(P_1P_2)^2 + (P_2P_3)^2 + \dots + (P_{4n}P_{4n+1})^2 + (P_{4n+1}P_1)^2$.... | [
"The answer is $16n - 8$.\nAssume that $P_i = (x_i, y_i)$ for $1 \\le i \\le 4n + 1$. Set\n$$\nP_{4n+2} = (x_{4n+2}, y_{4n+2}) = P_1 = (x_1, y_1).\n$$\nWe will show that the sum\n$$\nS = \\sum_{i=1}^{4n+1} (P_i P_{i+1})^2 \\geq 16n - 8.\n$$\nFirst, we show that this minimum can be obtained by setting\n$$\n(P_1, P_2... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 16n - 8 | |
0kbr | Problem:
Let $n$ be a fixed positive integer, and choose $n$ positive integers $a_{1}, \ldots, a_{n}$. Given a permutation $\pi$ on the first $n$ positive integers, let $S_{\pi}=\left\{i \left\lvert\, \frac{a_{i}}{\pi(i)}\right.\right.$ is an integer $\}$. Let $N$ denote the number of distinct sets $S_{\pi}$ as $\pi$ r... | [
"Solution:\nThe answer is $2^{n}-n$.\n\nLet $D=\\left(d_{i j}\\right)$ be the matrix where $d_{i j}$ is $1$ if $i$ is a divisor of $a_{j}$ and $0$ otherwise. For a subset $S$ of $[n]$, let $D_{S}$ be the matrix obtained from $D$ by flipping $(0 \\leftrightarrow 1)$ every entry $d_{i j}$ where $j \\notin S$. Observe... | United States | HMMT February 2020 | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 2^n - n | |
0iyk | Problem:
There are 5 students on a team for a math competition. The math competition has 5 subject tests. Each student on the team must choose 2 distinct tests, and each test must be taken by exactly two people. In how many ways can this be done? | [
"Solution:\n\nWe can model the situation as a bipartite graph on 10 vertices, with 5 nodes representing the students and the other 5 representing the tests. We now simply want to count the number of bipartite graphs on these two sets such that there are two edges incident on each vertex.\n\nNotice that in such a gr... | United States | $12^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 2040 | |
0ggk | 設 $H, \Gamma$ 分別為銳角三角形 $ABC$ 的垂心與外接圓, $M$ 為邊 $BC$ 的中點。在 $\Gamma$ 的劣弧 $BC$ 上取一點 $D$ 使得 $\angle BAD = \angle MAC$, 分別在 $\Gamma$ 及直線 $BC$ 上取兩點 $E, F$ 使得 $DE$ 及 $DF$ 分別垂直於 $AM$ 及 $BC$。令 $N$ 為直線 $HF$ 及 $AM$ 的交點, $R$ 為 $H$ 關於 $N$ 的對稱點。
證明：$\angle AER + \angle DFR = 180^\circ$。
 | [
"令 $X$ 為 $AM$ 與 $\\Gamma$ 異於 $A$ 的交點, 則\n$$\n\\angle XDC = \\angle XAC = \\angle MAC = \\angle BAD = \\angle BCD,\n$$\n即 $DX \\parallel BC$。由於 $AD, AX$ 為關於 $\\angle BAC$ 的等角線, $BDXC$ 為等腰梯形, 因此 $D$ 關於 $BC$ 的對稱點 $D'$ 為 $X$ 關於 $M$ 的對稱點。令 $A^*$ 為 $A$ 關於 $\\Gamma$ 的對徑點, $Y$ 為 $\\overline{HE}$ 中點, 則 $M$ 為 $\\overline{HA}... | Taiwan | 2022 數學奧林匹亞競賽第三階段選訓營, 獨立研究(一) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Advanced Configurations > ... | Chinese; English | proof only | null | |
0e4g | Problem:
Dokaži, da za vsak par realnih števil $x$ in $y$ velja neenakost
$$
|x+y|+|x+1|+|y+1| \geq 2.
$$
Pri katerih številih $x$ obstaja tako število $y$, da velja $|x+y|+|x+1|+|y+1|=2$? | [
"Solution:\n\nZa vsako realno število $a$ velja $|a| \\geq a$ in $|a| \\geq -a$. Zato je\n$$\n|x+y| \\geq -(x+y), \\quad |x+1| \\geq x+1, \\quad |y+1| \\geq y+1\n$$\nod koder sledi\n$$\n|x+y|+|x+1|+|y+1| \\geq -(x+y)+(x+1)+(y+1)=2\n$$\nDenimo, da velja enakost. Tedaj veljajo enačaji v vseh treh neenakostih v (1) in... | Slovenia | 55. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x in [-1, 1] | |
0dpg | Let $a$, $b$ be positive integers and function $f: \mathbb{N} \to \mathbb{N}$ so that for each positive integer $n$, the number $f(n+a)$ is divisible by $f([\sqrt{n}] + b)$. Prove that for each positive integer $n$ there exist $n$ pairwise distinct and pairwise relatively prime positive integers $a_1, a_2, \dots, a_n$ ... | [] | Silk Road Mathematics Competition | XV Silk Road Mathematics Competition | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
08fm | Problem:
Alla lavagna è scritta la moltiplicazione $x \times y$, dove $x$ e $y$ sono numeri interi positivi di tre cifre. Nicolò, un po' sbadato, non ha notato il simbolo di moltiplicazione e ricopia sul suo quaderno il numero di sei cifre ottenuto giustapponendo $x$ e $y$. L'insegnante, passando fra i banchi, fa nota... | [
"Solution:\n\nLa risposta è (B). Le ipotesi del problema si traducono in $1000 x + y = 7 x y$, e possiamo riscrivere questa equazione come $y = \\frac{1000 x}{7 x - 1}$. A questo punto possiamo procedere in (almeno) due modi:\n\na.\nSi ha $y = \\frac{1000 x}{7 x - 1} \\geq \\frac{1000 x}{7 x} = \\frac{1000}{7} = 14... | Italy | Italian Mathematical Olympiad - February Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | MCQ | B | |
0dm7 | Let $n$ be a positive integer and $A = \{n, n+1, \dots, 2n-1\}$. Prove that there is a number $a \in A$ such that the sum of elements of the set $A \setminus \{a\}$ is not divisible by any element of $A$. | [] | Saudi Arabia | Saudi Booklet | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0298 | Problem:
Alguns alunos de uma escola foram divididos em equipes satisfazendo as seguintes condições:
i) Quaisquer 2 equipes diferentes possuem exatamente 2 membros em comum.
ii) Toda equipe possui exatamente 4 elementos.
iii) Para quaisquer 2 alunos, existe uma equipe da qual ambos não fazem parte.
a) Explique por que... | [
"Solution:\n\na) Suponha que existem estudantes $A$ e $B$ participando de 4 equipes chamadas de $E_{1}, E_{2}$, $E_{3}$ e $E_{4}$ :\n$$\n\\begin{aligned}\nE_{1} &= \\{A, B, C, D\\} \\\\\nE_{2} &= \\{A, B, E, F\\} \\\\\nE_{3} &= \\{A, B, G, H\\} \\\\\nE_{4} &= \\{A, B, I, J\\}\n\\end{aligned}\n$$\nPela condição (iii... | Brazil | null | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 7 | |
0j9s | Problem:
Let $A$, $B$, $C$, and $D$ be points randomly selected independently and uniformly within the unit square. What is the probability that the six lines $\overline{AB}$, $\overline{AC}$, $\overline{AD}$, $\overline{BC}$, $\overline{BD}$, and $\overline{CD}$ all have positive slope? | [
"Solution:\n\nAnswer: $\\frac{1}{24}$\n\nConsider the sets of $x$-coordinates and $y$-coordinates of the points. In order to make 6 lines of positive slope, we must have smallest $x$-coordinate must be paired with the smallest $y$-coordinate, the second smallest together, and so forth. If we fix the order of the $x... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 1/24 | |
09cr | $a + b + c = 0$ байх ямарч бүхэл тоо $a, b, c$-ийн хувьд $(a)^2 + (b)^2 + (c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a)$ биелэх бүх $f\colon \mathbb{Z} \rightarrow \mathbb{Z}$ функцийг ол, $\mathbb{Z}$ бүх бүхэл тооны олонлог. | [
"The substitution $a = b = c = 0$ gives $3f(0)^2 = 6f(0)^2$, hence\n$$\n\\underline{f(0)} = 0. \\qquad (1)\n$$\nThe substitution $b = -a$ and $c = 0$ gives $((f(a) - f(-a))^2 = 0$. Hence $f$ is an even function:\n$$\nf(a) = f(-a) \\quad \\text{for all } a \\in \\mathbb{Z}. \\qquad (2)\n$$\nNow set $b = a$ and $c = ... | Mongolia | ОУМО-53 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | Mongolian | proof and answer | All solutions f: Z -> Z are exactly the following:
1) f(x) = 0 for all integers x.
2) f(x) = k x^2 for all integers x, where k is any nonzero integer.
3) f(x) = 0 for even x and f(x) = k for odd x, where k is any nonzero integer.
4) f(x) = 0 for x divisible by 4; f(x) = 4k for x congruent to 2 modulo 4; and f(x) = k fo... | |
0jx8 | Problem:
Let $A, B, C, D, E, F$ be 6 points on a circle in that order. Let $X$ be the intersection of $AD$ and $BE$, $Y$ be the intersection of $AD$ and $CF$, and $Z$ be the intersection of $CF$ and $BE$. $X$ lies on segments $BZ$ and $AY$ and $Y$ lies on segment $CZ$. Given that $AX = 3$, $BX = 2$, $CY = 4$, $DY = 10... | [
"Solution:\n\nLet $XY = z$, $YZ = x$, and $ZX = y$. By Power of a Point, we have that\n$$\n3(z + 10) = 2(y + 16), \\quad 4(x + 12) = 10(z + 3), \\text{ and } 12(x + 4) = 16(y + 2).\n$$\nSolving this system gives $XY = \\frac{11}{3}$, $YZ = \\frac{14}{3}$, and $ZX = \\frac{9}{2}$. Therefore, our answer is $XY + YZ +... | United States | February 2017 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | final answer only | 77/6 | |
07k8 | Let $P(x)$ be a polynomial with non-negative integer coefficients. A sequence of positive integers $(x_n)_{n=1}^{\infty}$ is given such that $x_1 = 1$, and for every natural number $n$, the following equality holds
$$
x_{n+1}^2 + P(n) = x_n x_{n+2}
$$
Prove that $P(x)$ would be a constant polynomial. | [
"Since $P(n) > 0$ for all positive integers, it follows that $x_n x_{n+2} > x_{n+1}^2$, for all $n$. Hence, $x_{n+2}/x_{n+1} > x_{n+1}/x_n$. Denote $y_n = x_{n+1}/x_n$ it follows that $x_{n+2} = y_{n+1} \\cdot y_n \\cdots y_n > y_2^n y_1$. Notice that $y_2 > y_1 = x_2 \\ge 1$. Let $y_2 = C$ for some $C > 1$. It fol... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Other"
] | null | proof only | null | |
020s | Problem:
An integer $m > 1$ is rich if for any positive integer $n$, there exist positive integers $x, y, z$ such that $n = m x^{2} - y^{2} - z^{2}$. An integer $m > 1$ is poor if it is not rich.
a) Find a poor integer.
b) Find a rich integer. | [
"a.\n\nSolution I. We will show that $m = 4$ is poor. If $y$ and $z$ are both even, we have $4 x^{2} - y^{2} - z^{2} \\equiv 0 - 0 - 0 = 0 \\pmod{4}$. If $y$ is even and $z$ is odd or the other way around, then $4 x^{2} - y^{2} - z^{2} \\equiv 0 - 0 - 1 \\equiv 3 \\pmod{4}$. If $y$ and $z$ are both odd, we have $4 ... | Benelux Mathematical Olympiad | Benelux Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | Poor integer: 4. Rich integer: 5. | |
00fk | Let $n$ be an integer of the form $a^{2}+b^{2}$, where $a$ and $b$ are relatively prime integers and such that if $p$ is a prime, $p \leq \sqrt{n}$, then $p$ divides $a b$. Determine all such $n$.
Answer: $n=2,5,13$. | [
"A prime $p$ divides $a b$ if and only if divides either $a$ or $b$. If $n=a^{2}+b^{2}$ is a composite then it has a prime divisor $p \\leq \\sqrt{n}$, and if $p$ divides $a$ it divides $b$ and vice-versa, which is not possible because $a$ and $b$ are coprime. Therefore $n$ is a prime.\n\nSuppose without loss of ge... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1994 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | n=2,5,13 | |
047t | Let $P(x)$ and $Q(x)$ be non-constant real polynomials such that for every positive integer $m$, there exists a positive integer $n$ with $P(m) = Q(n)$.
(1) Prove that if $\deg(Q)$ divides $\deg(P)$, then there exists a real polynomial $h(x)$ such that $P(x) = Q(h(x))$.
(2) Prove that $\deg(Q)$ divides $\deg(P)$. | [
"**Proof 1:** We denote by $O_k(x)$ any polynomial of degree at most $k$.\n\n(1) If $v = \\deg Q$ divides $u = \\deg P$, let $u = vL$. Write\n$$\nP(x) = a_u x^u + a_{u-1} x^{u-1} + \\dots + a_0,\n$$\n$$\nQ(x) = b_v x^v + b_{v-1} x^{v-1} + \\dots + b_0.\n$$\n\nFor $k = 0$, take $c_0 = \\sqrt[L]{a_u/b_v}$, so $Q(c_0 ... | China | China-TST-2025A | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Funct... | English | proof only | null | |
0cbl | Let $M$ and $N$ be the centres of the faces $A'B'C'D'$, respectively $ADD'A'$ of a rectangular cuboid $ABCD'A'B'C'D'$. It is known that $AM \perp A'C$ and $C'Q \perp BD'$. Prove that $ABCD'A'B'C'D'$ is a cube. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0ck0 | Solve, in the set of natural numbers, the equation
$$
x^2 + y^2 + xy(x - y) = 17.
$$ | [
"The equation can be written in the form $(x-y)^2 + 2xy + xy(x-y) = 17$.\nIf $x - y = d \\in \\mathbb{Z}$, $xy = p \\in \\mathbb{N}$, then the equality becomes $p(d+2) = 17 - d^2$, so $p = \\frac{17-d^2}{d+2}$. We get $p = \\frac{13}{d+2} + 2 - d \\in \\mathbb{N}$. It follows that $d+2 \\in \\{1, -1, 13, -13\\}$, w... | Romania | 75th Romanian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | (1, 16) | |
05s6 | Problem:
Soit $a_{0}, a_{1}, \ldots, a_{d}$ des entiers tels que $\operatorname{PGCD}(a_{0}, a_{1})=1$. Pour tout entier $n \geqslant 1$, on pose
$$
u_{n}=\sum_{k=0}^{d} a_{k} \varphi(n+k)$$
Démontrer que 1 est le seul entier naturel divisant tous les entiers $u_{n}$.
On rappelle que $\varphi(n+k)$ est le nombre d'ent... | [
"Solution:\n\nProcédons par l'absurde, et supposons qu'il existe un nombre premier $p$ qui divise tous les entiers $u_{n}$.\n\nTout d'abord, puisque $\\varphi(1)=\\varphi(2)=1$ et que $\\varphi(\\mathfrak{n})$ est pair pour tout $\\mathfrak{n} \\geqslant 2$, on remarque que $u_{1} \\equiv a_{0}+a_{1}(\\bmod 2)$ et ... | France | Préparation Olympique Française de Mathématiques - Test du 15 Mai 2019 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
00vb | Let $a$ and $b$ be distinct positive integers such that $3^a + 2$ is divisible by $3^b + 2$. Prove that $a > b^2$. | [
"Obviously we have $a > b$. Let $a = bq + r$, where $0 \\le r < b$. Then\n$$\n3^a \\equiv 3^{bq+r} \\equiv (-2)^q \\cdot 3^r \\equiv -2 \\pmod{3^b + 2}\n$$\nSo $3^b + 2$ divides $A = (-2)^q \\cdot 3^r + 2$ and it follows that\n$$\n|(-2)^q \\cdot 3^r + 2| \\ge 3^b + 2 \\text{ or } (-2)^q \\cdot 3^r + 2 = 0.\n$$\nWe ... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof only | null | |
0g6x | 試決定所有正整數數列 $\{x_1, x_2, \dots, x_{101}\}$ 使得對每個正整數 $n$ 存在一個整數 $a$ 滿足
$$
x_1^n + 2x_2^n + \dots + 101x_{101}^n = a^{n+1} + 1.
$$ | [
"滿足題意之唯一數列為\n$$\n\\{x_1, x_2, \\dots, x_{101}\\} = (1, k, \\dots, k) \\text{ 其中 } k = 2 + 3 + \\dots + 101 = 5150.\n$$\n令 $N$ 表示所有正整數所成的集合。$k = 2 + 3 + \\dots + 101 = 5150$。則可得\n$$1^n + 2k^n + \\dots + 101k^n = 1 + (2+3+\\dots+101)k^n = 1 + k \\cdot k^n = k^{n+1} + 1$$\n對任意的 $n$, 故 $(1, k, \\dots, k)$ 是符合題意。底下證明此解是... | Taiwan | 二〇一二數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | (1, 5150, 5150, ..., 5150) | |
0g3u | Problem:
Let $k$ be a circle with centre $M$ and let $A B$ be a diameter of $k$. Furthermore, let $C$ be a point on $k$ such that $A C = A M$. Let $D$ be the point on the line $A C$ such that $C D = A B$ and $C$ lies between $A$ and $D$. Let $E$ be the second intersection of the circumcircle of $B C D$ with line $A B$ ... | [
"Solution:\nLet $s = A M$. According to the conditions in the exercise we get:\n$$\ns = A M = M B = \\frac{1}{2} A B = A C = \\frac{1}{2} C D = \\frac{1}{3} A D\n$$\n\nUsing the power of the point $A$ with respect to the circle $E B D C$\n$$\nA E = \\frac{A C \\cdot A D}{A B} = \\frac{s \\cdot 3s}{2s} = \\frac{3s}{... | Switzerland | Final round | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometr... | null | proof and answer | 1/6 | |
09kh | Let $x$, $y$, $z$ be mutually different real numbers satisfying
$$
x(z + 1) = y(x + 1) = z(y + 1).
$$
Show that $z \neq 0$ and find the value of the expression $y + 1/z$. | [
"Let $x(z + 1) = y(x + 1) = z(y + 1) = t$ for some real $t$.\n\nFrom $x(z + 1) = t$, we have $x = \\dfrac{t}{z + 1}$.\nFrom $y(x + 1) = t$, we have $y = \\dfrac{t}{x + 1}$.\nFrom $z(y + 1) = t$, we have $z = \\dfrac{t}{y + 1}$.\n\nSubstitute $x$ into the expression for $y$:\n$$\ny = \\frac{t}{x + 1} = \\frac{t}{\\f... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | -1 | |
00m6 | Es sei $(a_n)_{n \ge 0}$ die Folge rationaler Zahlen mit $a_0 = 2016$ und
$$
a_{n+1} = a_n + \frac{2}{a_n}
$$
für alle $n \ge 0$.
Man zeige, dass diese Folge kein Quadrat einer rationalen Zahl enthält. | [
"Wir können eine rationale Zahl $\\frac{a}{b}$, deren Nenner nicht durch $5$ teilbar ist, modulo $5$ betrachten, indem wir den Rest von $ab^{-1}$ modulo $5$ betrachten, wobei $b^{-1}$ das Inverse von $b$ modulo $5$ ist. Dieser Rest hängt von der Darstellung der rationalen Zahl nicht ab und erfüllt auch die üblichen... | Austria | 48. Österreichische Mathematik-Olympiade | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine... | German | proof only | null | |
01ze | The numbers $1, 2, \dots, 50$ are written on the blackboard. Ann makes the following operations: she erases any two numbers $a$ and $b$ from the blackboard, writes down to the blackboard one number — the sum $a+b$, afterwards she writes the number $ab(a+b)$ to her notebook. After 49 such operations when only one number... | [
"Let's see how, as a result of Ann's actions, the sum of the cubes of the numbers written on the board changes. The identity $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$ implies that after each action of Ann, the sum of the cubes of all numbers written on the board increases by the number which is three times greater than the ... | Belarus | Belarus2022 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 690348750 | |
0c5n | Determine three digit numbers $abc$, such that its square has $a$ as the hundreds digit, $b$ as the tenth digit and $c$ as the units digit. | [] | Romania | 70th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | 376, 625 | |
006d | Sea $ABC$ un triángulo y $X$, $Y$, $Z$ puntos sobre los lados $BC$, $AC$, $AB$ respectivamente. Sean $A'$, $B'$, $C'$ los circuncentros correspondientes a los triángulos $AZY$, $BXZ$, $CYX$. Demuestre que
$$
(A'B'C') \geq \frac{(ABC)}{4}
$$ | [] | Argentina | XXIII Olimpíada Iberoamericana de Matemática | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geo... | Spanish | proof only | null | |
0d95 | Consider the set $X=\{1,2,3, \ldots, 2018\}$. How many positive integers $k$ with $2 \leq k \leq 2017$ that satisfy the following conditions:
i) There exists some partition of the set $X$ into $1009$ disjoint pairs which are $(a_{1}, b_{1}), (a_{2}, b_{2}), \ldots, (a_{1009}, b_{1009})$ with $|a_{i}-b_{i}| \in \{1, k\... | [
"First, we notice that for all $2 \\leq k \\leq 2017$, we always can find a partition satisfy i). For example, we can choose $a_{i}=2i-1$, $b_{i}=2i$ with $i=1,2, \\ldots, 1009$.\n\nDenote $m$ as the number of pairs $(a_{i}, b_{i})$ with $|a_{i}-b_{i}|=k$ then\n$$\nT=(1009-m) \\cdot 1 + m \\cdot k = 1009 + (k-1)m.\... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 403 | |
06ew | Find $2^{2006}$ positive integers satisfying the following conditions.
(a) Each positive integer has $2^{2005}$ digits.
(b) Each positive integer only has 7 or 8 in its digits.
(c) Among any two chosen integers, at most half of their corresponding digits are the same. | [
"We construct the numbers recursively. Firstly, we start from the following $2^2$ integers:\n77, 78, 87, 88,\neach of which has $2^1$ digits, and at most $2^0$ digit of each pair is the same.\nSuppose we have constructed $k = 2^{n+1}$ integers $a_1, a_2, \\dots, a_k$ with digits 7, 8, each of which has $2^n$ digits... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0kbn | Problem:
We define $\mathbb{F}_{101}[x]$ as the set of all polynomials in $x$ with coefficients in $\mathbb{F}_{101}$ (the integers modulo $101$ with usual addition and subtraction), so that two polynomials are equal if and only if the coefficients of $x^{k}$ are equal in $\mathbb{F}_{101}$ for each nonnegative integer... | [
"Solution:\nLet $p=101$, $m=1001$, and work in the ring $R:=\\mathbb{F}_{p}[x] /\\left(x^{m}-1\\right)$. We want to find the number of elements $a$ of this ring that are of the form $x^{p}-x$. We first solve this question for a field extension $\\mathbb{F}_{p^{d}}$ of $\\mathbb{F}_{p}$. Note that $(x+n)^{p}-(x+n)=x... | United States | HMMT February 2020 | [
"Algebra > Abstract Algebra > Field Theory",
"Algebra > Linear Algebra > Linear transformations",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof and answer | 101^{954} | |
0eh2 | Problem:
Na stranicah $AB$, $BC$ in $CA$ trikotnika $ABC$ zaporedoma ležijo točke $D$, $F$ in $G$, tako da sta premici $FG$ in $AB$ vzporedni (glej sliko). Ploščini trikotnikov $GFC$ in $GFD$ sta zaporedoma enaki $4\ \mathrm{cm}^2$ in $2\ \mathrm{cm}^2$. Koliko kvadratnih centimetrov je ploščina trikotnika $ABC$?
(A)... | [
"Solution:\n\nTrikotnika $GFC$ in $GDF$ imata enako osnovnico $GF$, njuni ploščini pa sta v razmerju $2:1$. Torej morata biti tudi dolžini njunih višin na stranico $GF$ v razmerju $2:1$. Ker sta premici $FG$ in $AB$ vzporedni, je dolžina višine skozi $C$ trikotnika $ABC$ enaka vsoti dolžin prej omenjenih višin. Od ... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | B | |
05cl | Let $O$ be the circumcentre of an acute triangle $ABC$. Points $D$ and $E$ are chosen on the side $BC$ such that $AD$ is an altitude of the triangle $ABC$ and $AE$ bisects the angle $CAD$. Bisectors of the triangle $AOB$ meet at point $J$. Prove that the triangle $JBE$ is isosceles. | [
"Solution 1:\n\nLet $\\angle BCA = \\gamma$ and let $M$ be the point of intersection of lines $OJ$ and $AB$ (Figures 10 and 11).\nAs $OA = OB$ and $OM$ bisects the angle $AOB$, we have $OM \\perp AB$. We also have $\\angle AOM = \\angle AOB = \\angle ACB = \\gamma$, because $O$ lies inside the triangle $ABC$. As $\... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinate... | English | proof only | null | |
0ai6 | Give all integer solutions of the equation:
$$
3^{2a+1} b^2 + 1 = 2^c.
$$
Во множеството на цели броеви да се реши равенка
$$
3^{2a+1}b^2 + 1 = 2^c.
$$ | [
"Case 1. $a \\ge 0$.\nClearly $c \\ge 0$ where $c=0$ implies $b=0$. We get that $(a,0,0)$ is a solution for an arbitrary non-negative integer $a$. From the equality $3^{2a+1}b^2+1=2^c$ it follows that $b$ is an odd integer. We can write the left-hand side in the following form\n$$\n3^{2a+1} b^2 + 1 = (3^{2a+1} + 1)... | North Macedonia | Macedonian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | M = {(a, 0, 0) | a ∈ ℤ} ∪ {(a, ±3^{-a}, 2) | a ∈ ℤ \ {0}} | |
0ir4 | Problem:
Find all $y > 1$ satisfying
$$
\int_{1}^{y} x \ln x \, dx = \frac{1}{4}.
$$ | [
"Solution:\n\nApplying integration by parts with $u = \\ln x$ and $dv = x \\, dx$, we get\n$$\n\\int_{1}^{y} x \\ln x \\, dx = \\left.\\frac{1}{2} x^{2} \\ln x\\right|_{1}^{y} - \\frac{1}{2} \\int_{1}^{y} x \\, dx = \\frac{1}{2} y^{2} \\ln y - \\frac{1}{4} y^{2} + \\frac{1}{4}\n$$\nSo $y^{2} \\ln y = \\frac{1}{2} y... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | proof and answer | sqrt(e) | |
0jxb | Problem:
An iguana writes the number $1$ on the blackboard. Every minute afterwards, if the number $x$ is written, the iguana erases it and either writes $\frac{1}{x}$ or $x+1$. Can the iguana eventually write the number $\frac{20}{17}$? | [
"Solution:\n\nYes. First, the iguana writes\n$$\n1 \\rightarrow 2 \\rightarrow \\frac{1}{2} \\rightarrow \\frac{3}{2} \\rightarrow \\frac{2}{3}\n$$\nThen, the iguana adds $1$ to arrive at $\\frac{17}{3}$. Finally, finish with\n$$\n\\frac{17}{3} \\rightarrow \\frac{3}{17} \\rightarrow \\frac{20}{17} .\n$$"
] | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | Yes | |
022u | Problem:
"Duas torres, uma com 30 passos e a outra com 40 passos de altura, estão à distância de 50 passos uma da outra. Entre ambas se acha uma fonte, para a qual dois pássaros descem no mesmo momento do alto das torres com a mesma velocidade e chegam ao mesmo tempo. Quais as distâncias horizontais da fonte às duas t... | [
"Solution:\n\nNa figura, $AD$ e $BC$ representam as duas torres e o ponto $E$ representa a posição da fonte. Como os dois pássaros chegam ao mesmo tempo, temos que $DE = EC$.\n\nDenotemos por $x$ a distância de $A$ a $E$ e assim $EB = 50 - x$. Usando o teorema de Pitágoras nos triângulos $DAE$ e $EBC$, temos que\n$... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 32 steps and 18 steps | |
07aw | Function $f$ is said to **generate** function $g$ (and denote it by $f \to g$), if $g$ can be written as the composition of function $f$ with itself for several times; i.e., natural number $k$ exists for which $f \circ f \circ \dots \circ f = g$ ($k$ times).
We want to explore some properties of this relation. For exam... | [
"a) Let\n$$\nf(x) = \\begin{cases} 2 & x = 1 \\\\ 3 & x = 2 \\\\ 1 & x = 3 \\\\ x & x \\notin \\{1, 2, 3\\} \\end{cases} \\qquad g(x) = \\begin{cases} 3 & x = 1 \\\\ 1 & x = 2 \\\\ 2 & x = 3 \\\\ x & x \\notin \\{1, 2, 3\\} \\end{cases}\n$$\nIt is easy to check that $g^2 = f$ and $f^2 = g$.\n\nb) Suppose $g$ is a f... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0d4t | Points $A_{1}, B_{1}, C_{1}$ lie on the sides $BC$, $AC$ and $AB$ of a triangle $ABC$, respectively, such that $AB_{1} - AC_{1} = CA_{1} - CB_{1} = BC_{1} - BA_{1}$. Let $I_{A}, I_{B}, I_{C}$ be the incenters of triangles $AB_{1}C_{1}$, $A_{1}BC_{1}$ and $A_{1}B_{1}C$ respectively. Prove that the circumcenter of triang... | [
"If we orient positively the line $AB$ from $A$ to $B$, the line $BC$ from $B$ to $C$ and the line $CA$ from $C$ to $A$, we get\n$$\n\\overline{B_{1}A} - \\overline{AC_{1}} = \\overline{A_{1}C} - \\overline{CB_{1}} = \\overline{C_{1}B} - \\overline{BA_{1}}.\n$$\nLet $A_{0}, B_{0}, C_{0}$ be the intouch points of th... | Saudi Arabia | SAMC | [
"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... | English, Arabic | proof only | null | |
04v7 | Solve the following system of equations in the domain of real numbers
$$2x + \lfloor y \rfloor = 2022,$$
$$3y + \lfloor 2x \rfloor = 2023.$$
(The symbol $\lfloor a \rfloor$ denotes the lower integer part of a real number $a$, i.e. the greatest integer not greater than $a$. E.g. $\lfloor 1.9 \rfloor = 1$ and $\lfloor -1... | [
"Since $\\lfloor y \\rfloor$ and $2022$ are integers, the equation $2x + \\lfloor y \\rfloor = 2022$ implies that $2x$ is also an integer, so $\\lfloor 2x \\rfloor = 2x$. Thus we can eliminate the unknown $x$ by subtracting the first equation of the system from the second one. We get\n$$\n3y - \\lfloor y \\rfloor =... | Czech Republic | 72nd Czech and Slovak Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (1011, 1/3) | |
0kze | Problem:
Let $ABCD$ be a trapezoid with $AB \parallel CD$. Point $X$ is placed on segment $\overline{BC}$ such that $\angle BAX = \angle XDC$. Given that $AB = 5$, $BX = 3$, $CX = 4$, and $CD = 12$, compute $AX$.
Proposed by: Pitchayut Saengrungkongka
Answer: $3\sqrt{6} = \sqrt{54}$. | [
"Solution:\n\n\nLet $P = DX \\cap AB$. Then, from the angle condition, we get that\n$$\n\\angle XAP = \\angle XAB = \\angle XDC = \\angle XPA,\n$$\nso $\\triangle XAP$ is isosceles. Moreover, $\\triangle XCD$ and $\\triangle XBP$ are similar, so $BP = CD \\cdot \\frac{XB}{XC} = 9$. Thus, if... | United States | HMMT November 2024 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 3*sqrt(6) | |
07kp | Given $k \in \{0, 1, 2, 3\}$ and a positive integer $n$, let $f_k(n)$ be the number of sequences $x_1, \dots, x_n$, where $x_i \in \{-1, 0, 1\}$ for $i = 1, \dots, n$, and
$$
x_1 + \cdots + x_n \equiv k \mod 4.
$$
a. Prove that $f_1(n) = f_3(n)$ for all positive integers $n$.
b. Prove that
$$
f_0(n) = \frac{3^n + 2 +... | [
"a. Let $F_k(n)$ be the collection of sequences/vectors $x$ corresponding to $f_k(n)$. Suppose $x = (x_1, ..., x_n) \\in F_1(n)$. Then at least one of the entries $x_i \\ne 0$. For the first such nonzero value, change its sign and call the resulting sequence $\\alpha(x)$. Since $\\sum x_i \\equiv 1 \\pmod 4$, we ha... | Ireland | Irish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Other"
] | English | proof and answer | f1(n) = f3(n) for all positive integers n; and f0(n) = (3^n + 2 + (-1)^n) / 4 for all positive integers n. | |
0ct8 | Points $K$ and $L$ are chosen on a side $AB$ of a convex quadrilateral $ABCD$, with $K$ lying between $A$ and $L$. Similarly, points $M$ and $N$ are chosen on a side $CD$, with $M$ lying between $C$ and $N$. It appears that $AK = KN = DN$ and $BL = BC = CM$. Given that $BCNK$ is a cyclic quadrilateral, prove that $ADML... | [
"In the case $AB \\parallel CD$, we have $BC = KN$, so $AK = BL = CM = DN$. Therefore, the quadrilateral $LMDA$ is obtained from $BCNK$ by a parallel translation by the vector $\\vec{BL}$.\n\n\n\nNow suppose $AB$ and $CD$ are not parallel; let $P$ be the intersection point of the lines $AB$... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
0biy | Given an integer $n \ge 2$, show that there exist $n+1$ pairwise distinct numbers $x_1, x_2, \dots, x_n, x_{n+1}$ in $\mathbb{Q} \setminus \mathbb{Z}$ such that $\{x_1^3\} + \{x_2^3\} + \dots + \{x_n^3\} = \{x_{n+1}^3\}$, where $\{x\}$ is the fractional part of the real number $x$. | [
"Notice that, if $w_1 < w_2 < \\dots < w_{n+1} < w_{n+2}$ are positive integers such that\n$$\nw_1^3 + w_2^3 + \\dots + w_{n+1}^3 = w_{n+2}^3, \\quad (*)\n$$\nthen the numbers $x_k = w_k/w_{n+2}$, $k = 1, 2, \\dots, n$, and $x_{n+1} = -w_{n+1}/w_{n+2}$ meet the required conditions.\nWe now show by induction on $n \... | Romania | 65th NMO Selection Tests for BMO and IMO | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
02j0 | Problem:
Capitu tem cem cartões numerados de $1$ a $100$. Todos os cartões têm uma face amarela e a outra vermelha, e o número de cada cartão está escrito em ambas as faces. Os cartões foram colocados sobre uma mesa, todos com a face vermelha voltada para cima. Capitu virou todos os cartões de número par e depois todo... | [
"Solution:\n\nCapitu virou, em primeiro lugar, os $50$ cartões pares; após isto, ficaram então na mesa os $50$ cartões pares com a face amarela para cima e os $50$ cartões ímpares com a face vermelha para cima. Ao virar agora os múltiplos de $3$, ela virou apenas os múltiplos de $3$ ímpares, que são $3, 9, 15, 21, ... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 33 | |
0b2u | Problem:
For a positive integer $n$, define $s(n)$ as the smallest positive integer $t$ such that $n$ is a factor of $t!$. Compute the number of positive integers $n$ for which $s(n)=13$. | [
"Solution:\n\nFor a positive integer $k$, consider the set $A(k) = \\{ n \\in \\mathbb{N} : s(n) = k \\}$ and we wish to find $\\# A(13)$. From the definition of $s(n)$, any element of $A(k)$ must divide $k!$ but not $(k-1)!$. Thus, any element of $A(k)$ must be a divisor of $k!$ that is not a divisor of $(k-1)!$. ... | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 792 | |
0kxf | Problem:
Suppose $a$ and $b$ be positive integers not exceeding $100$ such that
$$
a b = \left(\frac{\operatorname{lcm}(a, b)}{\operatorname{gcd}(a, b)}\right)^2
$$
Compute the largest possible value of $a + b$. | [
"Solution:\nFor any prime $p$ and a positive integer $n$, let $\\nu_{p}(n)$ be the largest nonnegative integer $k$ for which $p^{k}$ divides $n$. Taking $\\nu_{p}$ on both sides of the given equation, we get\n$$\n\\nu_{p}(a) + \\nu_{p}(b) = 2 \\cdot \\left|\\nu_{p}(a) - \\nu_{p}(b)\\right|\n$$\nwhich means $\\frac{... | United States | HMMT November 2023 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 78 | |
0934 | Problem:
Let $ABC$ be an acute-angled triangle with $AB > AC$ and circumcircle $\Gamma$. Let $M$ be the midpoint of the shorter arc $BC$ of $\Gamma$, and let $D$ be the intersection of the rays $AC$ and $BM$. Let $E \neq C$ be the intersection of the internal bisector of the angle $ACB$ and the circumcircle of the tri... | [
"Solution:\n\nConsider the following implications:\nLet us denote by $P$ the other point of intersection of internal bisector from $C$ and the circle $\\Gamma$.\n$BDCE$ is cyclic\n$$\n\\begin{aligned}\n& \\Longrightarrow \\angle BDC = \\angle BEP \\\\\n& \\Longrightarrow \\triangle BEP \\sim \\triangle BDA \\\\\n& ... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
040u | In $\triangle ABC$ with $AB = 1$, let $D$ be a point on $AC$ such that $\angle ABD = \angle C$ and let $E$ be a point on $AB$ such that $BE = DE$. Let $H$ be a point on $DE$ such that $AH \perp DE$ and $M$ be the midpoint of $CD$. If $AH = 2 - \sqrt{3}$, find the size of $\angle AME$. (posed by Xiong Bin)
![](attached... | [
"Let $\\angle ABD = \\angle C = \\alpha$ and $\\angle DBC = \\beta$. It is easy to see that $\\angle BDE = \\alpha$, $\\angle AED = 2\\alpha$,\n$$\n\\angle ADE = \\angle ADB - \\angle BDE = (\\alpha + \\beta) - \\alpha = \\beta,\n$$\n$$\nAB = AE + EB = AE + EH + HD.\n$$\nHence,\n$$\n\\frac{AB}{AH} = \\frac{AE + EH}... | China | China Southeastern Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof and answer | 15° | |
0j7n | Problem:
Let $H$ be a regular hexagon of side length $x$. Call a hexagon in the same plane a "distortion" of $H$ if and only if it can be obtained from $H$ by translating each vertex of $H$ by a distance strictly less than $1$. Determine the smallest value of $x$ for which every distortion of $H$ is necessarily convex... | [
"\nLet $H = A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}$ be the hexagon, and for all $1 \\leq i \\leq 6$, let points $A_{i}'$ be considered such that $A_{i}A_{i}' < 1$. Let $H' = A_{1}'A_{2}'A_{3}'A_{4}'A_{5}'A_{6}'$, and consider all indices modulo $6$. For any point $P$ in the plane, let $D(P)$ denote... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 4 | |
09tf | Problem:
In driehoek $A B C$ is $M$ het midden van $A B$ en $N$ het midden van $C M$. Zij $X$ een punt dat voldoet aan $\angle X M C=\angle M B C$ en $\angle X C M=\angle M C B$, zo dat $X$ en $B$ aan verschillende kanten van $C M$ liggen. Zij $\Omega$ de omgeschreven cirkel van driehoek $A M X$.
a) Bewijs dat $C M$ ... | [
"Solution:\n\nWe bekijken de configuratie in de figuur; andere configuraties gaan analoog.\n\na.\nUit de gegevens volgt dat $\\triangle X M C \\sim \\triangle M B C$. Er geldt nu $\\angle A M X=180^{\\circ}-\\angle X M C-\\angle B M C=180^{\\circ}-\\angle X M C-\\angle M X C=\\angle M C X$ en\n$$\n\\frac{|A M|}{|M ... | Netherlands | MO-selectietoets | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0bjk | Consider a positive integer $n$ and $A, B$ two matrices in $M_n(\mathbb{C})$ such that $A^2 + B^2 = 2AB$. Prove that:
a) The matrix $AB - BA$ is not invertible.
b) If the rank of $A - B$ is $1$, then matrices $A$ and $B$ commute. | [
"a) The given relation can be written in each of the two forms:\n$$\n(A - B)^2 = AB - BA, \\qquad (1)\n$$\n$$\nA(A - B) = (A - B)B. \\qquad (2)\n$$\nSuppose $AB - BA$ is non singular. By (1), $A - B$ is also non-singular, that is $B = (A - B)^{-1}A(A - B)$, by (2). Then $A - B = A - (A - B)^{-1}A(A - B)$, from wher... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof only | null | |
05jz | Problem:
Soit $n \in \mathbb{N}^{*}$. Montrer qu'il est possible de partitionner $\{1,2, \ldots, n\}$ en deux sous-ensembles $A$ et $B$ tels que la somme des éléments de $A$ soit égale au produit des éléments de $B$. | [
"Solution:\n\nA priori, l'ensemble $B$ doit être bien plus petit que $A$. On va donc chercher un $B$ avec un petit nombre d'éléments. En testant des petites valeurs de $n$, on peut penser à chercher $B$ sous la forme $\\{1, a, b\\}$ avec $a, b \\neq 1$ et $a \\neq b$.\nOn a alors $\\prod_{x \\in B} x = a b$, et :\n... | France | OFM 2013-2014 Envoi 2 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | null | proof only | null | |
05ht | Problem:
Soit $ADE$ un triangle isocèle en $A$. Soit le cercle $\omega$, tangent aux droites $(AD)$ et $(AE)$ en $D$ et $E$. Soit $B$ et $C$ deux points au delà de $D$ et $E$ sur les droites $(AD)$ et $(AE)$. On suppose que $BC > BD + CE$. Finalement, soit $F$ et $G$ deux points sur le segment $[BC]$ de telle sorte qu... | [
"Solution:\n\nSoit $B'$, $C'$ deux points sur les demi-droites $[AD]$ et $[AC]$ au delà de $D$ et $E$ de telle sorte que $(B'C') \\parallel (BC)$ et que $(B'C')$ soit tangente à $\\omega$ en $L'$\n\nLemme. $D, F$ et $L'$ sont alignés, de même pour $E, G$ et $L'$.\n\nOn a $(BF) \\parallel (B'L')$ et $DB = BF$, ainsi... | France | Préparation Olympique Française de Mathématiques | [
"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 | |
00eu | Let $p \geq 5$ be a prime and let $r$ be the number of ways of placing $p$ checkers on a $p \times p$ checkerboard so that not all checkers are in the same row (but they may all be in the same column). Show that $r$ is divisible by $p^{5}$. Here, we assume that all the checkers are identical. | [
"Note that $r = \\binom{p^{2}}{p} - p$. Hence, it suffices to show that\n$$\n\\left(p^{2}-1\\right)\\left(p^{2}-2\\right) \\cdots\\left(p^{2}-(p-1)\\right)-(p-1)!\\equiv 0 \\quad\\left(\\bmod p^{4}\\right).\n$$\nNow, let\n$$\nf(x) := (x-1)(x-2) \\cdots (x-(p-1)) = x^{p-1} + s_{p-2} x^{p-2} + \\cdots + s_{1} x + s_{... | Asia Pacific Mathematics Olympiad (APMO) | null | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0czp | In a triangle $A B C$ let $O$ be the circumcenter, $H$ the orthocenter, and $M$ the midpoint of the segment $A H$. The perpendicular at $M$ onto $O M$ intersects lines $A B$ and $A C$ at $P$ and $Q$, respectively. Prove that $M P = M Q$. | [
"Consider $P'$ on $A B$, $Q'$ on $A C$ such that $A P' H Q'$ is a parallelogram. We shall prove that $P' Q' \\perp O M$. Let $T$ be the antipodal point of $A$ in the circumcircle of triangle $A B C$. We have $O M \\parallel T H$, hence it suffices to prove that $P' Q' \\perp T H$.\n\n\n\nNo... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
0l9g | Find the greatest positive integer $n$ such that the system of equations
$$
(x+1)^2 + y_1^2 = (x+2)^2 + y_2^2 = \dots = (x+k)^2 + y_k^2 = \dots = (x+n)^2 + y_n^2
$$
has integral solution $(x, y_1, y_2, \dots, y_n)$. | [
"• First, it is easy to show the\n*Lemma:* For arbitrary integers $a, b$, we have:\n$$\na^2 + b^2 \\equiv \\begin{cases} 2; & 1 \\equiv 5 \\pmod{8} & \\text{if } a \\equiv \\pm 1 \\pmod{4} \\\\ 1; & 0 \\equiv 4 \\pmod{8} & \\text{if } a \\equiv 0 \\pmod{4} \\\\ 5; & 4 \\equiv 0 \\pmod{8} & \\text{if } a \\equiv 2 \... | Vietnam | 2003 Vietnamese Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 3 | |
0ibp | Problem:
How many ways can you mark 8 squares of an $8 \times 8$ chessboard so that no two marked squares are in the same row or column, and none of the four corner squares is marked? (Rotations and reflections are considered different.) | [
"Solution:\n\nIn the top row, you can mark any of the 6 squares that is not a corner. In the bottom row, you can then mark any of the 5 squares that is not a corner and not in the same column as the square just marked. Then, in the second row, you have 6 choices for a square not in the same column as either of the ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics"
] | null | final answer only | 21600 | |
09zs | In triangle $ABC$, $\angle A$ is a right angle. A point $D$ lies on line segment $AB$ in such a way that the angles $ACD$ and $BCD$ are equal. Moreover, $|AD| = 2$ and $|BD| = 3$.
What is the length of line segment $CD$? | [] | Netherlands | Second Round | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | 2√6 | |
01wj | a) Find at least one positive integer $n$, which can be represented both as $n = a^2 - b$ and $n = c^2 - d$, where $a, b, c, d$ are divisors of $n$ and $a \neq c$.
b) Prove that there exist infinitely many positive integers $n$ satisfying a). | [
"a) $n = 72$ or $n = 88200$."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Pell's equations"
] | English | proof and answer | n = 72 or n = 88200 | |
03xl | For any set $A = \{a_1, a_2, \dots, a_m\}$, let $P(A) = a_1a_2\cdots a_m$. In addition, let $n = \binom{2010}{99}$ and let $A_1, A_2, \dots, A_n$ be all 99-element subsets of $\{1, 2, \dots, 2010\}$. Prove that $2011 \mid \sum_{i=1}^n P(A_i)$. | [
"One can check that $2011$ is a prime number. Let\n$$\nf(x) = (x-1)(x-2)\\cdots(x-2010) - (x^{2010} + 2010!).\n$$\nFor $n \\in \\{1, 2, \\dots, 2010\\}$, we have $n^{2010} \\equiv 1 \\pmod{2011}$ (by Fermat's little theorem), and $2010! \\equiv -1 \\pmod{2011}$ (by Wilson's theorem), so\n$$\nf(n) = (n-1)(n-2)\\cdot... | China | China Southeastern Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0cu3 | A square is dissected into $n^2 \ge 4$ rectangles by $n-1$ vertical lines and $n-1$ horizontal lines. Prove that one may choose $2n$ of these rectangles such that for any two of the chosen rectangles, one can be put completely into the other (perhaps, after some rotation).
Квадрат разбит на $n^2 \ge 4$ прямоугольников... | [
"Assume that $a_1 \\ge \\dots \\ge a_n$ are the lengths of the rectangles' horizontal sides (from the left to the right), and $b_1 \\ge \\dots \\ge b_n$ are the lengths of their vertical sides (from the top to the bottom). Let $Q_{i,j}$ be the $a_i \\times b_j$ rectangle. Since $a_1 + \\dots + a_n = b_1 + \\dots + ... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof only | null | |
02v2 | Problem:
Uma desigualdade simples, mas bastante útil é $x^{2} \geq 0$, para todo $x$ real. Para prová-la, basta estudar separadamente as seguintes possibilidades: $x>0, x<0$ ou $x=0$. De fato, um número real positivo multiplicado por um número real positivo é positivo, um número real negativo multiplicado por outro nú... | [
"Solution:\n\na.\nPela fatoração sugerida,\n$$\n\\begin{aligned}\n0<2 & =a^{3}-b^{3} \\\\\n& =(a-b)\\left(a^{2}+a b+b^{2}\\right)\n\\end{aligned}\n$$\nComo $a^{2}+a b+b^{2} \\geq 0$, segue que $a-b>0$, ou seja, $a>b$.\n\n\nb.\nSabemos que $a^{2}, b^{2} \\geq 0$ e $a-b>0$. Daí,\n$$\n\\begin{aligned}\n\\left(a^{3}-b^... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a > b and a^2 + b^2 ≥ 2 | |
0edz | Problem:
Eden izmed kotov izjemnega enakokrakega trikotnika je $10^{\circ}$ večji od nekega drugega kota tega trikotnika. Najmanj koliko je velik najmanjši kot izjemnega enakokrakega trikotnika?
(A) $52^{\circ} 20^{\prime}$
(B) $53^{\circ} 20^{\prime}$
(C) $56^{\circ} 40^{\prime}$
(D) $63^{\circ} 20^{\prime}$
(E) $66... | [
"Solution:\n\nOznačimo kote izjemnega enakokrakega trikotnika z $\\alpha, \\alpha$ in $\\gamma$. Tedaj je razlika med $\\alpha$ in $\\gamma$ enaka $10^{\\circ}$. Imamo dve možnosti.\n\nČe je $\\alpha>\\gamma$, je $\\alpha-\\gamma=10^{\\circ}$ in seveda $\\alpha+\\alpha+\\gamma=180^{\\circ}$. Od tod izračunamo $3 \\... | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | MCQ | B | |
0aat | Calculate the length of the leg of an isosceles trapeze with bases $18\, ext{cm}$ and $10\, ext{cm}$, if it is known that its middle line is $\frac{2}{7}$ of its perimeter. | [
"From the condition in the problem we have that $m = \\frac{2}{7}L$ where $L$ is the perimeter of the trapeze and $m$ the length of its middle line. We have $m = \\frac{a+b}{2} = \\frac{18+10}{2} = 14\\,\\text{cm}$ so $L = m \\cdot \\frac{7}{2} = 14 \\cdot \\frac{7}{2} = 49\\,\\text{cm}$. Because the trapeze is iso... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof and answer | 10.5 cm | |
077v | Let $\mathbb{N}$ denote the set of all positive integers. Find all real numbers $c$ for which there exists a function $f: \mathbb{N} \to \mathbb{N}$ satisfying:
a. for any $x, a \in \mathbb{N}$, the quantity $\frac{f(x+a)-f(x)}{a}$ is an integer if and only if $a = 1$;
b. for all $x \in \mathbb{N}$, we have $|f(x) - ... | [] | India | INMO_2023 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | c = k + 1/2 for k ∈ {0, 1, 2, ...} | |
0ci3 | Fix integers $a$ and $b$ greater than $1$. For any positive integer $n$, let $r_n$ be the (non-negative) remainder $b^n$ leaves upon division by $a^n$. Assume that there exists a positive integer $N$ such that $r_n < 2^n/n$ for all integers $n \ge N$. Prove that $a$ divides $b$. | [
"Arguing indirectly, assume that $a \\nmid b$, so $r_n \\ne 0$ for all $n$. Let $M = \\max(b, N)$.\n\nWe now prove that $r_{n+1} \\ge br_n$ for all $n \\ge M$. Indeed, as $r_n < 2^n/b \\le a^n/b$, it follows that $br_n < a^n$ and $b^{n+1} \\equiv br_n \\pmod{a^n}$. Therefore, $br_n$ is the remainder $b^{n+1}$ leave... | Romania | Romanian Master of Mathematics | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
0k74 | Problem:
Consider an equilateral triangle $T$ of side length $12$. Matthew cuts $T$ into $N$ smaller equilateral triangles, each of which has side length $1$, $3$, or $8$. Compute the minimum possible value of $N$. | [
"Solution:\n\nMatthew can cut $T$ into $16$ equilateral triangles with side length $3$. If he instead included a triangle of side $8$, then let him include $a$ triangles of side length $3$. He must include $12^{2} - 8^{2} - 3^{2} a = 80 - 9a$ triangles of side length $1$. Thus $a \\leq 8$, giving that he includes a... | United States | HMMT November 2019 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 16 | |
0i5z | Problem:
Points $A$, $B$, $C$ in the plane satisfy $\overline{A B} = 2002$, $\overline{A C} = 9999$. The circles with diameters $A B$ and $A C$ intersect at $A$ and $D$. If $\overline{A D} = 37$, what is the shortest distance from point $A$ to line $B C$? | [
"Solution:\n\n$\\angle A D B = \\angle A D C = \\pi / 2$ since $D$ lies on the circles with $A B$ and $A C$ as diameters, so $D$ is the foot of the perpendicular from $A$ to line $B C$, and the answer is the given $37$."
] | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 37 | |
0e9u | The circles $K_1$ and $K_2$ intersect in two distinct points $A$ and $B$. The tangents to $K_1$ through $A$ and $B$ intersect at $T$. Let $M$ be an arbitrary point on the circle $K_1$ distinct from $A$ and $B$. The line $MT$ meets the circle $K_1$ again at $C$, the line $MA$ meets the circle $K_2$ again at $K$ and the ... | [
"Let $P$ be the intersection of the lines $MT$ and $KL$.\nBy Menelaus' theorem for the triangle $ALK$ and the colinear points $C, P$ and $M$ we have\n$$\n\\frac{AC}{CL} \\cdot \\frac{LP}{PK} \\cdot \\frac{KM}{MA} = -1.\n$$\nIn order to show that $|PL| = |PK|$ it suffices to see that\n$$\n\\frac{|AC|}{|CL|} = \\frac... | Slovenia | National Math Olympiad in Slovenia | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06y3 | Let $a_{0}, a_{1}, a_{2}, \ldots$ be an infinite strictly increasing sequence of positive integers such that for each $n \geqslant 1$ we have
$$
a_{n} \in\left\{\frac{a_{n-1}+a_{n+1}}{2}, \sqrt{a_{n-1} \cdot a_{n+1}}\right\}
$$
Let $b_{1}, b_{2}, \ldots$ be an infinite sequence of letters defined as
$$
b_{n}= \begin{ca... | [
"Solution 1. We will show that the eventual period of sequence $\\left(b_{n}\\right)$ consists of any fixed number of occurrences of $G$ (possibly zero) followed by a single $A$.\nWe look at the ratios of consecutive terms of the sequence $\\left(a_{n}\\right)$. Let $C$ and $D$ be coprime positive integers such tha... | IMO | IMO2024 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0heh | Let $ABCD$ be a quadrilateral with equal sides $AB$ and $BC$. Let $E$ be the point on the line $AB$, such that $BD = BE$ and $AD \perp DE$. Show that the perpendicular bisectors of segments $AD$, $CD$ and $CE$ intersect in precisely one point. | [
"Since $\\triangle BAD$ is isosceles with the base $DE$, $\\angle BDE = \\angle BED < 90^\\circ$. Since $\\angle AED < 90^\\circ$, then the points $B$ and $E$ are on the same side from point $E$. Moreover, $\\angle BED < 90^\\circ = \\angle ADE$, so the point $B$ belongs to the segment $AE$ (Fig. 2). Then\n$$\n\\an... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0ca4 | Problem:
Fie $f: \mathbb{R} \rightarrow \mathbb{R}$ o funcție derivabilă de ordinul $n \geq 2$, astfel încât
$$
\lim _{x \rightarrow \infty} f(x)=\ell \in \mathbb{R} \text{ şi } \lim _{x \rightarrow \infty} f^{(n)}(x)=0
$$
Demonstrați că $\lim _{x \rightarrow \infty} f^{(k)}(x)=0$, pentru orice $k \in\{1,2, \ldots, n-... | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ | [
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Limits"
] | null | proof only | null | |
019l | Prove the inequality
$$
\frac{1}{k} + \frac{1}{k+1} + \dots + \frac{1}{n} \ge \frac{2(n-k+1)}{n+k}
$$
for all pairs $(n, k)$ of positive integers such that $k \le n$. | [
"Applying AM-HM to $k, k+1, \\dots, n$ gives\n$$\n\\frac{n-k+1}{\\frac{1}{k} + \\frac{1}{k+1} + \\dots + \\frac{1}{n}} \\le \\frac{k+(k+1)+\\dots+n}{n-k+1}\n$$\nAs $k, k+1, \\dots, n$ form an arithmetic sequence, the r.h.s. equals the AM of the first and the last terms, i.e., $\\frac{k+n}{2}$. Taking reciprocals an... | Baltic Way | Baltic Way 2013 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
08ei | Problem:
Una cavalletta si muove sul piano e dal punto di coordinate $(x, y)$ può saltare a sua scelta o su $(x+y, y)$ o su $(x, x+y)$. È partita da un punto di coordinate $(n, 9)$ con $n$ intero positivo, ma non ricorda il valore di $n$. Sa solo che dopo un certo numero di mosse è arrivata in $(2021,2050)$. Quanti so... | [
"Solution:\n\nLa risposta è 3. In generale, se a un certo punto la cavalletta si trova in $(a, b)$, al passo precedente si trovava in un punto $(x, y)$ tale che $(a, b) = (x+y, y)$ o $(a, b) = (x, x+y)$, cioè in $(a-b, b)$ o in $(a, b-a)$; inoltre, se essa è partita da un punto $(x_0, y_0)$ a coordinate positive, c... | Italy | Italian Mathematical Olympiad, February Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | null | proof and answer | 3 | |
0jeo | Problem:
A positive integer is called oddly even if the sum of its digits is even. Find the sum of the first 2013 oddly even numbers. | [
"Solution:\n\nFor convenience, we declare $0$ to be oddly even (its digit sum is of course $0$), and we sum the first $2014$ oddly even nonnegative integers.\n\nLet us look at the oddly even numbers in a given hundred, that is, in the range $100 n$ to $100 n+99$ where $n \\geq 0$ is an integer. For each tens digit,... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | final answer only | 4055187 | |
08vx | A square $ABCD$ is given. Points $P$, $Q$, $R$, $S$ lie on the sides $AB$, $BC$, $CD$, $DA$, of this square, respectively, in such a way that the lines $PR$ and $BC$ are parallel and so are the lines $SQ$ and $AB$. Let $Z$ be the point of intersection of the lines $PR$ and $SQ$. If $BP = 7$, $BQ = 6$, $DZ = 5$, determi... | [
"Let $x$ be the length of a side of the square $ABCD$. Then, we have $ZS = x - 7$ and $DS = x - 6$. Applying the Pythagorean theorem to the right triangle $ZSD$, we get $(x - 7)^2 + (x - 6)^2 = 5^2$, from which we get $x = 3, 10$. But from $x > BP$ we conclude that $x = 10$, and it is easy to check that $x = 10$ do... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 10 | |
060l | Problem:
Déterminer tous les quadruplets $\left(a, b, c, d\right)$ d'entiers positifs avec $a, b, c$ strictement positifs tels que
$$
\begin{aligned}
& \operatorname{PPCM}(b, c)=a+d \\
& \operatorname{PPCM}(c, a)=b+d \\
& \operatorname{PPCM}(a, b)=c+d
\end{aligned}
$$ | [
"Solution:\n\nLe système étant symétrique, on suppose que $c$ est le maximum de $a$, $b$ et $c$.\n\nComme $c$ divise $\\operatorname{PPCM}(b, c)$, $c$ divise $a+d$.\nComme $c$ divise $\\operatorname{PPCM}(c, a)$, $c$ divise $b+d$, donc $c$ divise $a+d-(b+d)=a-b$.\nOr $-c < a-b < c$ donc $a=b$.\n\nLa dernière ligne ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (a, a, a, 0) for any positive integer a | |
01wh | Define the sequence $a_0, a_1, a_2, \dots$ by $a_n = 2^n + 2^{\lfloor n/2 \rfloor}$. Prove that there are infinitely many terms of the sequence which can be expressed as a sum of (two or more) distinct terms of the sequence, as well as infinitely many of those which cannot be expressed in such a way. | [
"2. See IMO - 2018 Shortlist, Problem N3."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | English | proof only | null |
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