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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
02o0 | Let $a$, $b$ and $c$ be real numbers such that $a \neq b$ and $a^2(b + c) = b^2(c + a) = 2010$. Compute $c^2(a + b)$. | [
"Since $a \\neq b$, $a^2(b+c) = b^2(c+a) \\iff a^2b + a^2c - b^2c - ab^2 = 0 \\iff ab(a-b) + c(a-b)(a+b) = 0 \\iff ab+ca+bc = 0$.\n\nSo $(a-c)(ab+bc+ca) = 0 \\iff a^2b + abc + a^2c = abc + bc^2 + ac^2 \\iff c^2(a+b) = a^2(b+c) = 2010$."
] | Brazil | Brazilian Math Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 2010 | |
0kbl | Problem:
Max repeatedly throws a fair coin in a hurricane. For each throw, there is a $4\%$ chance that the coin gets blown away. He records the number of heads $H$ and the number of tails $T$ before the coin is lost. (If the coin is blown away on a toss, no result is recorded for that toss.) What is the expected valu... | [
"Solution:\n\nIn all solutions, $p=\\frac{1}{25}$ will denote the probability that the coin is blown away. Let $D=|H-T|$. Note that if $D \\neq 0$, the expected value of $D$ is not changed by a coin flip, whereas if $D=0$, the expected value of $D$ increases by $1$. Therefore $\\mathbf{E}(D)$ can be computed as the... | United States | HMMT February 2020 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 24/7 | |
0kbt | Problem:
Two diameters and one radius are drawn in a circle of radius $1$, dividing the circle into $5$ sectors. The largest possible area of the smallest sector can be expressed as $\frac{a}{b} \pi$, where $a, b$ are relatively prime positive integers. Compute $100a + b$. | [
"Solution:\n\nLet the two diameters split the circle into four sectors of areas $A$, $B$, $A$, and $B$, where $A + B = \\frac{\\pi}{2}$. Without loss of generality, let $A \\leq B$.\n\nIf our radius cuts into a sector of area $A$, the area of the smallest sector will be of the form $\\min(x, A - x)$. Note that $\\m... | United States | HMMO 2020 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | final answer only | 106 | |
01sf | Let $H$ be the orthocenter of an acute triangle $ABC$, points $A_1$, $B_1$, $C_1$ be the feet of the altitudes $AA_1$, $BB_1$, $CC_1$, respectively. Let points $D$, $K$, $L$, $M$, $N$ be the midpoints of the segments $AC$, $BC_1$, $BA_1$, $HA_1$, $HC_1$, respectively, and $P$, $Q$, $R$ be the intersection points of the... | [
"Note that all right-angled triangles $ABA_1$, $AHC_1$, $CBC_1$, $CHA_1$ are pairwise similar. Therefore, since $L$, $N$, $K$, $M$ are the midpoints of the corresponding sides, we see that $\\triangle ALB \\sim \\triangle ANH \\sim \\triangle CKB \\sim \\triangle CMH$. It follows that $\\angle LAB = \\angle NAH = \... | Belarus | FINAL ROUND | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
06j3 | Let $ABC$ be a triangle. Let $D$ and $E$ be respectively points on the segments $AB$ and $AC$, and such that $DE \parallel BC$. Let $M$ be the midpoint of $BC$. Let $P$ be a point such that $DB = DP$, $EC = EP$ and such that the open segments (segments excluding the endpoints) $AP$ and $BC$ intersect. Suppose $\angle B... | [
"Note that $\\angle BPD = \\angle CME = \\angle DEM$. So we can find a point $F$ on the extension of $EM$ such that $\\triangle DPB \\sim \\triangle DEF$. By spiral similarity and since\n\n$DB = DP$, we have $\\triangle DFB \\cong \\triangle DEP$. As $EC = EP = FB$ and $CM = BM$, by the sine law, either $\\angle FE... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02lk | Problem:
Os cinco círculos - Cinco discos de mesmo raio estão dispostos como mostra a figura. Quatro centros são os vértices de um quadrado e três estão alinhados. Trace uma reta que divida a figura formada pelos 5 discos em duas partes de mesma área.
 | [
"Solution:\n\nObservemos que qualquer linha que passe por $O$, o centro do quadrado $ABCD$, divide a área formada pelos círculos $\\mathcal{C}_1$, $\\mathcal{C}_2$, $\\mathcal{C}_3$ e $\\mathcal{C}_4$ na metade. Por outro lado, qualquer linha reta que passe por $F$ divide a área do círculo $\\mathcal{C}_5$ na metad... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | The straight line through the square’s center and the center of the fifth disk. | |
0eyi | Problem:
There is a set of scales on the table and a collection of weights. Each weight is on one of the two pans. Each weight has the name of one or more pupils written on it. All the pupils are outside the room. If a pupil enters the room then he moves the weights with his name on them to the other pan. Show that yo... | [] | Soviet Union | 2nd ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
00ed | Let $ABCD$ be a convex quadrilateral with $AB > AD$ and $\angle B = \angle D = 90^\circ$. Let $P$ be the point on side $AB$ such that $AP = AD$. Lines $PD$ and $BC$ intersect at $Q$. The perpendicular line to $AC$ through $Q$ intersects $AB$ at $R$. Let $S$ be the foot of the perpendicular from $D$ to $AC$. Prove that ... | [
"Let $T = QR \\cap AC$ and $V = QS \\cap AB$. First, note that from $AP = AD$, we have $\\angle APD = \\angle ADP$. So\n$$\n\\angle CQP = 90^\\circ - \\angle QPB = 90^\\circ - \\angle APD = 90^\\circ - \\angle ADP = \\angle QDC,\n$$\nand hence $CQ = CD$. Now, by metric relations in the right triangle $DAC$, we have... | Argentina | Rioplatense Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
019p | Prove that the following inequality is satisfied for any positive real numbers $x$, $y$, $z$:
$$
\frac{x^3}{y^2 + z^2} + \frac{y^3}{x^2 + z^2} + \frac{z^3}{x^2 + y^2} \ge \frac{x + y + z}{2}
$$ | [
"The inequality is symmetric, so we may assume $x \\le y \\le z$. Then we have\n$$\nx^3 \\le y^3 \\le z^3 \\quad \\text{and} \\quad \\frac{1}{y^2 + z^2} \\le \\frac{1}{x^2 + z^2} \\le \\frac{1}{x^2 + y^2}.\n$$\nTherefore, by the rearrangement inequality we have:\n$$\n\\begin{align*}\n\\frac{x^3}{y^2 + z^2} + \\frac... | Baltic Way | Baltic Way 2013 | [
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof only | null | |
0cye | Find all polynomials $P$ with real coefficients such that for all $x, y, z \in \mathbb{R}$,
$$
P(x) + P(y) + P(z) + P(x + y + z) = P(x + y) + P(y + z) + P(z + x).
$$ | [
"Assume that $P(t) = a_n t^n + \\ldots + a_1 t + a_0$, where $a_n \\neq 0$. Taking $x = y = z$ in the given relation, it follows\n$$\n3 P(x) + P(3x) = 3 P(2x).\n$$\nLooking for the coefficient of $x^n$ in this equation, we obtain the relation $3 + 3^n = 3 \\cdot 2^n$, that is $3^{n-1} = 2^n - 1$. This relation hold... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | P(t) = a t^2 + b t for real a, b | |
0imc | Problem:
Determine the real number $a$ having the property that $f(a)=a$ is a relative minimum of $f(x)=x^{4}-x^{3}-x^{2}+a x+1$. | [
"Solution:\n\nBeing a relative minimum, we have $0=f'(a)=4 a^{3}-3 a^{2}-2 a+a=a(4 a+1)(a-1)$. Then $a=0, 1, -1/4$ are the only possibilities. However, it is easily seen that $a=1$ is the only value satisfying $f(a)=a$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1 | |
04tq | Find all pairs of real numbers $k, l$ such that the inequality
$$
ka^2 + lb^2 > c^2
$$
holds for side lengths $a, b, c$ of any triangle. | [
"Assume that for some $k, l$, the given inequality holds for all side lengths $a, b, c$ of any triangle. Plugging in $a = c = 1$ and arbitrary $b < 2$, we get $k + lb^2 > 1$. If $k < 1$ then we easily find sufficiently small $b > 0$ that makes the inequality false. Hence $k \\ge 1$ and likewise $l \\ge 1$.\n\nLet $... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | k ≥ 1, l ≥ 1, and kl ≥ k + l | |
0i5l | Problem:
How many ways are there of using diagonals to divide a regular $6$-sided polygon into triangles such that at least one side of each triangle is a side of the original polygon and that each vertex of each triangle is a vertex of the original polygon? | [
"Solution:\n\nThe number of ways of triangulating a convex $(n+2)$-sided polygon is $\\binom{2 n}{n} \\frac{1}{n+1}$, which is $14$ in this case. However, there are two triangulations of a hexagon which produce one triangle sharing no sides with the original polygon, so the answer is $14-2=12$."
] | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | null | final answer only | 12 | |
0kkg | Problem:
Suppose
$$
h \cdot a \cdot r \cdot v \cdot a \cdot r \cdot d = m \cdot i \cdot t = h \cdot m \cdot m \cdot t = 100.
$$
Find $(r \cdot a \cdot d) \cdot (t \cdot r \cdot i \cdot v \cdot i \cdot a)$. | [
"Solution:\nThe answer is\n$$\n\\frac{\\text{harvard} \\cdot \\mathrm{mit} \\cdot \\mathrm{mit}}{\\text{hmmt}} = 100^{2} = 10000.\n$$"
] | United States | HMMT November 2021 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 10000 | |
067y | Find all possible triads of non negative integers $(x, y, z)$ with $x \le y$, which satisfy the equation: $x^2 + y^2 = 3 \cdot 2016^z + 77$ | [
"We distinguish the cases:\n\nIf $z = 0$, then the equation becomes $x^2 + y^2 = 80$.\nThen $x, y$ must be multiples of $4$, that is $x = 4a$, $y = 4b$, $0 \\le a \\le b$, and the equation becomes $a^2 + b^2 = 5$, hence $(a, b) = (1, 2)$, that is $(x, y) = (4, 8)$.\n\nIf $z > 0$, then $7 \\mid 2016^z$ (since $7 \\m... | Greece | 33rd Hellenic Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | [(4, 8, 0), (35, 70, 1), (14, 77, 1)] | |
0j9x | A circle is divided into $432$ congruent arcs by $432$ points. The points are colored in four colors such that some $108$ points are colored Red, some $108$ points are colored Green, some $108$ points are colored Blue, and the remaining $108$ points are colored Yellow. Prove that one can choose three points of each col... | [
"Let $R$, $G$, $B$, and $Y$ denote the sets of Red, Green, Blue, and Yellow points, respectively, and let $r$, $g$, $b$, and $y$ denote a generic Red, Green, Blue, and Yellow point, respectively. For $0 \\le k \\le 431$, denote by $\\mathcal{T}_k$ the counterclockwise rotation by $\\frac{360k}{432}$ degrees around ... | United States | USAMO | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0efk | Problem:
Naj bo $n$ naravno število. Kateri od spodaj navedenih izrazov je enakovreden izrazu
$$\sqrt[\sqrt[n]{n^{n}} \cdot n+\sqrt[n]{n^{n}}]{\sqrt[n]{n^{n+n}}}$$?
(A) 1
(B) $\sqrt[n]{n}$
(C) $n$
(D) $\sqrt{n}$
(E) $n^{-\frac{1}{2}}$ | [
"Solution:\n\n$\\sqrt[n]{n^{n}} = n$, $\\sqrt[n+n]{n^{n}} = \\sqrt[2]{n}$, $\\sqrt[n]{n^{n+n}} = n^{2}$. Vrednost izraza\n$$\n\\sqrt[\\sqrt[n]{n^{n}} \\cdot \\sqrt[n+n]{n^{n}}]{\\sqrt[n^{n+n}]}\n$$\nje enaka\n$$\n\\frac{n \\cdot \\sqrt{n}}{n^{2}} = \\frac{1}{\\sqrt{n}} = n^{-\\frac{1}{2}}.\n$$\nPravilen odgovor je ... | Slovenia | 17. tekmovanje dijakov srednjih tehniških in strokovnih šol v znanju matematike | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | MCQ | E | |
0lcg | For each positive integer $n$, let $S_n = \{0,1,2,\ldots,2n+1\}$. Consider the function $f : (\mathbb{Z} \times S_n) \to [0;1]$ satisfying the following conditions:
i/ $f(x,0) = f(x,2n+1) = 0$ for all $x$ from integers.
ii/ $f(x-1,y)+f(x+1,y)+f(x,y-1)+f(x,y+1)=1$ for all $x,y \in \mathbb{Z}$ and $1 \le y \le 2n$.
Le... | [
"1. In equation ii/ we see that\n$$\n(x-1)-y \\equiv (x+1)-y \\equiv x-(y-1) \\equiv x-(y+1) \\pmod{2}\n$$\nThat is, the values of $f(x,y)$ where $x,y$ have the same parity and the values of $f(x,y)$ where $x,y$ have the opposite parity have no connection.\nWe will find the way to define function $f$ in both cases.... | Vietnam | Vietnamese Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Functional equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | 2n(2n+1)+1 | |
0av9 | Problem:
How many numbers between $1$ and $2016$ are divisible by exactly one of $4$, $6$, or $10$? | [
"Solution:\n\nLet $A$ be the set of numbers between $1$ and $2016$ divisible by $4$.\nLet $B$ be the set of numbers divisible by $6$.\nLet $C$ be the set of numbers divisible by $10$.\n\nWe are to find $|A \\setminus (B \\cup C)| + |B \\setminus (A \\cup C)| + |C \\setminus (A \\cup B)|$.\n\nFirst, compute the size... | Philippines | 19th Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | final answer only | 470 | |
0asm | Problem:
If the parabola $y+1 = x^{2}$ is rotated clockwise by $90^{\circ}$ about its focus, what will be the new coordinates of its vertex? | [
"Solution:\n$\\left(-\\frac{3}{4}, -\\frac{1}{4}\\right)$"
] | Philippines | Philippines Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | (-3/4, -1/4) | |
0f17 | Problem:
The $n$ points of a graph are each colored red or blue. At each move we select a point which differs in color from more than half of the points to which it is joined and we change its color. Prove that this process must finish after a finite number of moves. | [] | Soviet Union | ASU | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0bei | Let $P$ be a point in the interior of the acute triangle $ABC$ and let $D$, $E$, $F$ be respectively the intercepts of lines $AP$, $BP$, $CP$ with the sides $BC$, $CA$, $AB$.
a) Prove that the area of the triangle $DEF$ is not greater than a quarter of the area of triangle $ABC$.
b) Prove that the inradius of $DEF$ i... | [
"a) Put $\\frac{BD}{CD} = x$, $\\frac{CE}{AE} = y$, $\\frac{AF}{BF} = z$ to obtain $\\frac{S_{AEF}}{S_{ABC}} = \\frac{AF}{AB} \\cdot \\frac{AE}{AC} = \\frac{z}{(z+1)(y+1)}$ and the similar ones. As by Ceva's theorem $xyz = 1$, it follows $\\frac{S_{DEF}}{S_{ABC}} = 1 - \\sum \\frac{z}{(z+1)(y+1)} = \\frac{xyz+1}{(x... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane G... | null | proof only | null | |
0lgc | Problem:
The extension of median $C M$ of the triangle $A B C$ intersects its circumcircle $\omega$ at $N$. Let $P$ and $Q$ be the points on the rays $C A$ and $C B$ respectively such that $P M \parallel B N$ and $Q M \parallel A N$. Let $X$ and $Y$ be the points on the segments $P M$ and $Q M$ respectively such that ... | [
"Solution:\n\nLemma. The points $K$ and $L$ lie on the sides $B C$ and $A C$ of a triangle $A B C$. The segments $A K$ and $B L$ intersect at $D$. Then the quadrilateral $C K D L$ is circumscribed if and only if $A C - B C = A D - B D$.\n\nProof. Let $C K D L$ be circumscribed and its incircle touches $L C$, $C K$,... | Zhautykov Olympiad | XV International Zhautykov Olympiad in Mathematics | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0bvi | Let $n$ be a positive integer. For each positive integer $k \le n$, let $r_k$ denote the remainder $2^n$ leaves upon division by $k$. Prove that
$$
\sum_{k=1}^{n} r_k > \frac{n}{2} \left( \log_2 \frac{n}{3} - 2 \right).
$$ | [
"Let $m = \\lfloor \\log_2 n \\rfloor$, and split the set of integers $1$ through $n$ into $m + 1$ pairwise disjoint subsets $J_0, J_1, \\dots, J_m$, where $J_k$ consists of all numbers of the form $2^k(2\\ell + 1)$, $\\ell = 0, \\dots, \\lfloor n/2^{k+1} - 1/2 \\rfloor$.\nIf $j$ is a member of $J_k$, then $r_j \\g... | Romania | Fifteenth IMAR Mathematical Competition | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0d7o | Find all polynomials $P, Q \in \mathbb{Z}[x]$ such that every positive integer is a divisor of a certain nonzero term of the sequence $\left(x_{n}\right)_{n=0}^{\infty}$ given by the conditions:
$$
x_{0}=2016,\quad x_{2 n+1}=P\left(x_{2 n}\right),\quad x_{2 n+2}=Q\left(x_{2 n+1}\right) \text{ for all } n \geq 0.
$$ | [
"Suppose that $P, Q \\in \\mathbb{Z}[x]$ satisfy the given requirement. Step by step we will draw some conclusions.\n\nStep 1. $\\operatorname{deg} P \\geq 1, \\operatorname{deg} Q \\geq 1$.\nSuppose, on the contrary, that one of $P, Q$ were a constant polynomial $c$.\n1. If $P(x)=c\\ (\\forall x \\in \\mathbb{Z})$... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | All solutions are linear with slopes whose product is one, namely:
1) P(x) = x + b and Q(x) = x + d with integers b, d such that b + d ≠ 0 and either (b + d) divides 2016 or (b + d) divides 2016 + b.
2) P(x) = −x + b and Q(x) = −x + d with integers b, d such that b − d ≠ 0 and either (b − d) divides 2016 or (b − d) d... | |
0ha1 | In the quadrilateral $ABCD$ point $E$ is the middle of the side $AB$, point $F$ is the middle of the side $BC$, point $G$ is the middle of the side $AD$. Turns out that the segment $GE$ is perpendicular to $AB$, and segment $GF$ is perpendicular to $BC$. Find the angle $GCD$ given $∠ADC = 70°$. | [
"Since the segment $GE$ is both altitude and median in $ΔABG$ (Fig. 2), this triangle is isosceles, therefore $AG = GB$. Analogously, $ΔCBG$ is isosceles, hence $BG = GC$. Moreover, from the statement of the problem, point $G$ is the middle of the segment $AD$. Hence, we obtain that $DG = AG = BG = CG$, i.e. $ΔCGD$... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 70° | |
0hjv | Problem:
If $n$ is an integer, prove that the number
$$
1+n+n^{2}+n^{3}+n^{4}
$$
is not divisible by $4$. | [
"Solution:\n\nWe will prove that $1+n+n^{2}+n^{3}+n^{4}$ is an odd number and the required claim will follow immediately since no odd number is divisible by $4$.\n\nThere are two possibilities for $n$ — it can be even or odd.\n\nIn the case when $n$ is even, then $n+n^{2}+n^{3}+n^{4}$ is even as a sum of even numbe... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
01vj | Find the smallest positive integer $n$, which has three different proper divisors, the sum of which equals to $1001$.
(A proper divisor of $n$ is any divisor of $n$ distinct from $1$ and $n$.) | [
"Answer: $n = 924$.\nLet $d_1 < d_2 < d_3$ be the divisors given in the condition, then $1 < d_1 < d_2 < d_3 < n$. Since $d_1$, $d_2$ and $d_3$ are divisors of $n$, there exist positive integers $a$, $b$ and $c$ such that $d_1 a = d_2 b = d_3 c = n$. It is clear that $1 < c < b < a < n$. Hence $c \\ge 2$, $b \\ge 3... | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 924 | |
0g0g | Problem:
Zwei Kreise $k_{1}$ und $k_{2}$ schneiden sich in den Punkten $A$ und $C$. Sei $B$ der zweite Schnittpunkt von $k_{1}$ und der Tangente an $k_{2}$ in $A$, und sei $D$ der zweite Schnittpunkt von $k_{2}$ und der Tangente an $k_{1}$ in $C$. Zeige, dass $A D$ und $B C$ parallel sind. | [
"Solution:\n\n\n\nNach dem Tangentenwinkelsatz gilt sowohl $\\angle B A C=\\angle A D C$ als auch $\\angle C B A=\\angle D C A$. Die beiden Dreiecke $A B C$ und $A C D$ stimmen also in zwei Winkeln überein und somit auch im dritten. Wir erhalten $\\angle A C B=\\angle C A D$, woraus sofort ... | Switzerland | SMO - Vorrunde | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0651 | Find all positive integers $n$ such that there is a permutation $\sigma$ of the set $\{1,2,...,n\}$ for which $\sqrt{\sigma(1)+\sigma(2)+\dots+\sigma(n)}$ is a rational number.
*: A permutation of the set $\{1,2,...,n\}$ is a one-to-one function of this set to itself.* | [
"Let for some $n \\in \\mathbb{N}^*$ we have\n$$\n\\sqrt{\\sigma(1)+\\sqrt{\\sigma(2)}+\\sqrt{\\cdots+\\sqrt{\\sigma(n)}}} = r_1 \\in \\mathbb{Q}. \\quad (1)\n$$\nSquaring the two parts of (1) we conclude that the number $\\sqrt{\\sigma(2)+\\sqrt{\\sigma(3)}+\\sqrt{\\cdots+\\sqrt{\\sigma(n)}}}$ is rational. Similar... | Greece | 24th Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | n = 1 or n = 3 | |
08ls | Problem:
Let $n \geq 2$ be a fixed positive integer. An integer will be called "$n$-free" if it is not a multiple of an $n$-th power of a prime. Let $M$ be an infinite set of rational numbers, such that the product of every $n$ elements of $M$ is an $n$-free integer. Prove that $M$ contains only integers. | [
"Solution:\nWe first prove that $M$ can contain only a finite number of non-integers. Suppose that there are infinitely many of them: $\\frac{p_{1}}{q_{1}}, \\frac{p_{2}}{q_{2}}, \\ldots, \\frac{p_{k}}{q_{k}}, \\ldots$, with $(p_{k}, q_{k})=1$ and $q_{k}>1$ for each $k$. Let $\\frac{p}{q}=\\frac{p_{1} p_{2} \\ldots... | JBMO | 2008 Shortlist JBMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
09ma | The diagonals $AC$ and $BD$ of the circumscribed quadrilateral $ABCD$ are perpendicular and intersect at the point $P$. Let $X$ be a point on the side $BC$. The circumcircle of triangle $APD$ and the line $XP$ intersect at $Y$ distinct from $P$. Prove that $\angle AXD + \angle BYC = 90^\circ$.
(Khulan Tumenbayar) | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0amz | Problem:
Let $2 = \log_{b} x$. Find all values of $\frac{x+1}{x}$ as $b$ ranges over all positive real numbers.
(a) $(0,+\infty)$
(c) $(0,1)$
(b) $(1,+\infty)$
(d) all real numbers | [] | Philippines | QUALIFYING STAGE | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | MCQ | (b) | |
089g | Problem:
Le spese per organizzare le Olimpiadi Nazionali della Matematica incrementano ogni anno dello $0,5\%$ rispetto all'anno precedente. In che anno le spese saranno esattamente il doppio rispetto a quelle del 2012?
Nota: nel 2012 le spese non sono nulle.
(A) 2023
(B) 2150
(C) 2151
(D) 2212
(E) mai. | [
"Solution:\n\nLa risposta è $\\mathbf{( E )}$. Se in un certo anno le spese sono pari a $x$, nell'anno successivo le spese diventano $\\frac{100,5}{100} x = \\frac{201}{200} x$. Dopo $k$ anni le spese sono quindi $\\left(\\frac{201}{200}\\right)^{k} x$. Ci chiediamo quindi per quali interi positivi $k$ si ha che $\... | Italy | Italian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | MCQ | E | |
0hp4 | Problem:
Determine all triples $(x, y, n)$ of integers such that $x^{2} + 2 y^{2} = 2^{n}$. | [
"Solution:\n\nIt is easy to check that $\\left( \\pm 2^{r}, 0, 2r \\right)$ and $\\left(0, \\pm 2^{r}, 2r+1 \\right)$ satisfy this equation for any nonnegative integer $r$. We will show that these are all the solutions by an infinite descent method.\n\nSo suppose we have some solution $\\left(x_{0}, y_{0}, n_{0}\\r... | United States | Berkeley Math Circle | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All solutions are (±2^r, 0, 2r) and (0, ±2^r, 2r+1) for any nonnegative integer r. | |
0hi1 | Let $n = 2m + 1$ for some integer $m$. $n$ points $A_1, A_2, \dots, A_n$ are chosen on two parallel lines. What is the largest possible number of acute triangles among the triangles $A_iA_jA_k$ over $1 \le i < j < k \le n$? | [
"Let us put points on these two lines in such a way that projections of every two adjacent points from one line to another surround exactly one point on another line and there are no obtuse angles $\\angle ABC$ such that $B$ lies on one line and $A, C$ on another one (clearly, we can achieve this if distances betwe... | Ukraine | Problems from Ukrainian Authors | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | m(m+1)(2m+1)/6 | |
0gob | Show that
$$
\sum_{i=1}^{n} \frac{a_i}{\sqrt{a_i^4 + 3}} \le \frac{1}{2} \sum_{i=1}^{n} \frac{1}{a_i}
$$
for all positive integers $n$ and positive real numbers $a_1, a_2, \dots, a_n$ satisfying the condition $a_1 a_2 \cdots a_n = 1$. | [
"We first observe that $x^4+3 \\ge (x+1)^2$ as $x^4+3-(x+1)^2 = (x-1)^2(x^2+2x+2) \\ge 0$. Therefore it suffices to prove that\n$$\n\\sum_{i=1}^{n} \\frac{a_i}{a_i + 1} \\le \\frac{1}{2} \\sum_{i=1}^{n} \\frac{1}{a_i}\n$$\nfor all positive integers $n$ and positive real numbers $a_1, a_2, \\dots, a_n$ satisfying th... | Turkey | 18th Turkish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
07x4 | For positive real numbers $a$, $b$, $c$ that satisfy $a + b + c = 3$, prove the following:
a.
$$
a\sqrt{b} + b\sqrt{c} + c\sqrt{a} \le 3
$$
b.
$$
\sqrt{a^3 b} + \sqrt{b^3 c} + \sqrt{c^3 a} + \sqrt{ab^3} + \sqrt{bc^3} + \sqrt{ca^3} \le 6
$$ | [
"a. First note that\n$$\n2(a + b + c)^2 - 6(ab + bc + ca) = (a - b)^2 + (b - c)^2 + (c - a)^2 \\geq 0\n$$\nhence $ab + bc + ca \\le \\frac{1}{3}(a + b + c)^2 = 3$. Now use Cauchy-Schwarz with vectors $(\\sqrt{a}, \\sqrt{b}, \\sqrt{c})$ and $(\\sqrt{ab}, \\sqrt{bc}, \\sqrt{ca})$ to get\n$$\na\\sqrt{b} + b\\sqrt{c} +... | Ireland | IRL_ABooklet_2024 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0esn | In rugby it is possible to score 3, 5 or 7 points at a time. If South Africa wins the next Rugby World Cup and scores 34 points in the final, the highest number of times 3 points could have been scored by South Africa is
(A) 6 (B) 7 (C) 8 (D) 9 (E) 10 | [
"We need $34 - 5x - 7y$ to be a multiple of $3$, so $5x + 7y$, like $34$, must have remainder $1$ after division by $3$. For the largest possible multiple of $3$ we need the smallest such value of $5x + 7y$, which is $0 \\times 5 + 1 \\times 7 = 7$. Then $(34 - 7) \\div 3 = 9$."
] | South Africa | South African Mathematics Olympiad First Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | English | MCQ | D | |
0fnq | Find those positive integers $n \le 2014$ for which there exist positive integers $r, s$ such that $\text{gcd}(rs(r+s), n) = 1$ and $n$ does not divide $r - s$. | [
"First we observe that if $r$, $s$, and $r + s$ are positive integers then at least one of them must be even. Indeed, if both are odd then their sum $r + s$ is even. Therefore, the number $rs(r + s)$ must also be even. So, if $n$ and $rs(r + s)$ are coprime then $n$ must be odd. Hence, if $n$ is even there are no s... | Spain | BarcelonaTech Mathcontest | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | Spanish | proof and answer | {n: n is odd and 5 <= n <= 2014} | |
0gp9 | Determine all positive integers $m$, $n$ and prime numbers $p$ such that
$$
\frac{5^m + 2^n p}{5^m - 2^n p}
$$
is a square of an integer. | [
"The answer is $(m, n, p) = (2, 3, 3)$, $(1, 1, 2)$ or $(2, 2, 5)$.\nLet $\\frac{5^m + 2^n p}{5^m - 2^n p} = k^2$ for some positive integer $k$. Note that $5^m - 2^n p \\mid 5^m + 2^n p$ implies that $5^m - 2^n p \\mid 2 \\cdot 5^m$. Then as $5^m - 2^n p$ is odd, $5^m - 2^n p \\mid 5^m$ and hence $5^m - 2^n p = 5^r... | Turkey | Team Selection Test | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | (m, n, p) = (2, 3, 3), (1, 1, 2), (2, 2, 5) | |
09cg | **ХI-A2** Хэрэв $a, b, c$ нь $a^2 + b^2 + c^2 = 3$ нөхцлийг хангах эерэг бодит тоонууд бол
$$
\frac{a}{a^2 + 2b + 3} + \frac{b}{b^2 + 2c + 3} + \frac{c}{c^2 + 2a + 3} \le \frac{1}{2}
$$
гэж батал. | [
"$a^2 + 1 \\ge 2a,\\ b^2 + 1 \\ge 2b,\\ c^2 + 1 \\ge 2c$ тул\n$$\n\\sum_{cyc} \\frac{a}{a^2 + 2b + 3} \\ge \\sum_{cyc} \\frac{a}{2(a + b + 1)}\n$$\nИймд\n$$\n\\sum_{cyc} \\frac{a}{a + b + 1} \\le 1 \\Leftrightarrow \\sum_{cyc} \\frac{b+1}{a+b+1} \\ge 2\n$$\nгэж батлахад хангалттай. $a^2 + b^2 + c^2 = 3$ гэдгийг тоо... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | Mongolian | proof only | null | |
0bor | Given an integer $a$ and a positive integer $n$, show that the sum $\sum_{k=1}^{n} a^{(k,n)}$ is divisible by $n$, where $(x, y)$ denotes the greatest common divisor of the integers $x$ and $y$. | [
"Write $\\sum_{k=1}^n a^{(k,n)} = \\sum_{d|n} \\varphi(n/d)a^d$, where $\\varphi$ is Euler's totient function ($\\varphi(m)$ is the number of positive integers less than $m$ and prime to $m$), and notice that, if $n$ and $n'$ are coprime positive integers, then\n$$\n\\sum_{k=1}^{nn'} a^{(k,nn')} = \\sum_{d|n} \\var... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0fx1 | Problem:
Seien $a$, $b$, $c$ positive reelle Zahlen. Beweise die folgende Ungleichung:
$$
\frac{a}{\sqrt{3 a+2 b+c}}+\frac{b}{\sqrt{3 b+2 c+a}}+\frac{c}{\sqrt{3 c+2 a+b}} \leq \frac{1}{\sqrt{2}} \sqrt{a+b+c}
$$ | [
"Solution:\n\nNach CS ist das Quadrat der linken Seite der Ungleichung höchstens gleich\n$$\n(a+b+c) \\cdot \\left(\\underbrace{\\frac{a}{3 a+2 b+c}+\\frac{b}{3 b+2 c+a}+\\frac{c}{3 c+2 a+b}}_{A}\\right)\n$$\nEs genügt also $A \\leq \\frac{1}{2}$ zu zeigen, oder äquivalent dazu $3-2A \\geq \\frac{3}{2}$. Wir setzen... | Switzerland | IMO Selektion 2008 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0k5l | Problem:
A certain lecture has finitely many students and at least two students. Every student fell asleep exactly once and woke up exactly once. Suppose that for any two students, there was some time at which both were asleep. Prove that there was a time at which all the students were asleep. | [
"Solution:\n\nLet Alice be the first student to wake up. Then, immediately before Alice wakes up, all students must have been asleep, since every other student must have slept at the same time as Alice, and none of them have woken up yet. Thus, all the students are asleep."
] | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Geometry > Plane Geometry > Combinatorial Geometry > Helly's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
05w4 | Problem:
Soit $n \geqslant 1$ un entier. Trouver tous les diviseurs $d \geqslant 1$ de $3 n^{2}$ tels que $n^{2}+d$ soit un carré parfait. | [
"Solution:\n\nSoit $d \\geqslant 1$ un diviseur de $3 n^{2}$ tel que $n^{2}+d$ soit un carré parfait. Puisque $n^{2}<n^{2}+d\\leqslant n^{2}+3 n^{2}=(2 n)^{2}$, il existe $k \\in \\llbracket 1, n \\rrbracket$ tel que $n^{2}+d=(n+k)^{2}$. En simplifiant, on obtient $d=2 k n+k^{2}$, de sorte que $k(k+2 n) \\mid 3 n^{... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | d = 3n^2 | |
024j | Problem:
Sejam $a$ e $b$ números reais positivos com produto diferente de $1$, define-se a operação estrela, representada por “$*$”, pela equação
$$
a * b = \frac{a + b - 2ab}{1 - ab}.
$$
Em uma lousa, estão escritos $2015$ números iguais a $\frac{1}{2}$. Em cada passo, apagam-se dois números $x$ e $y$ escritos na lou... | [
"Solution:\n\na. Desenvolvendo a expressão da operação estrela, temos:\n$$\n\\begin{aligned}\n\\frac{x * y}{1 - x * y} &= \\frac{\\frac{x + y - 2xy}{1 - xy}}{1 - \\frac{x + y - 2xy}{1 - xy}} \\\\\n&= \\frac{x + y - 2xy}{1 - xy - x - y + 2xy} \\\\\n&= \\frac{x + y - 2xy}{1 - x - y + xy} \\\\\n&= \\frac{x + y - 2xy}{... | Brazil | NÍVEL 3 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 2015/2016 | |
0jdb | Find all infinite sequences $a_1, a_2, \dots$ of positive integers satisfying the following properties:
a. $a_1 < a_2 < a_3 < \dots$,
b. there are no positive integers $i, j, k$, not necessarily distinct, such that $a_i + a_j = a_k$,
c. there are infinitely many positive integers $k$ such that $a_k = 2k - 1$. | [
"The only solution is to have $a_k = 2k - 1$ for all $k$, giving the sequence $1, 3, 5, \\dots$.\n\nLet $a_1 = m$. First, we show that for any such sequence, we must have $a_{k+m} - a_k \\ge 2m$ for all positive integers $k$. Suppose for some $k$ that this was not the case. Then $a_k, a_{k+1}, \\dots, a_{k+m}$ are ... | United States | IMO Team Selection Team Selection Test | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a_k = 2k - 1 for all k | |
0arh | Problem:
$x^{3} + k x - 128 = 0$ has a root of multiplicity $2$. Find $k$. | [
"Solution:\n\n$x^{3} + k x - 128 = (x - r)^{2}(x - s) = x^{3} + (s - 2r)x^{2} + (r^{2} + 2rs)x - r^{2}s = 0$\n\n$\\Rightarrow r^{2}s = 128$, $s + 2r = 0 \\Rightarrow s = -2r$, $-2r^{3} = 128 \\Rightarrow r = -4$, $s = 8$.\n\nThus, $k = r^{2} + 2rs = 16 - 64 = -48$."
] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | -48 | |
01x6 | Two ordered sequences $x_1 \le x_2 \le \dots \le x_n$ and $y_1 \ge y_2 \ge \dots \ge y_n$ of positive integers are written in two rows, one above the other:
$$
x_1 \ x_2 \ \dots \ x_n
$$
$$
y_1 \ y_2 \ \dots \ y_n.
$$
Denote by $M = \max(x_1 + y_1, x_2 + y_2, \dots, x_n + y_n)$ the maximal sum of numbers in columns.
Co... | [
"Answer: no, it's impossible."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | No, it is impossible. | |
0cea | a) Prove that $x^2 + x + 1 > 0$ and $x^4 + x^3 + x^2 + x + 1 > 0$, for every real number $x$.
b) Find all real triples $(a, b, c)$ so that $a^8 + b^8 + c^8 = a^3 + b^3 + c^3$ and $a^5 + b^5 + c^5 = 3$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof and answer | (1,1,1) | |
00q1 | A closed, non-self-intersecting broken line $L$ is drawn over a $(2n+1) \times (2n+1)$ chessboard in such a way that the set of $L$'s vertices coincides with the set of the vertices of the board's squares and every edge in $L$ is a side of some board square. All board squares lying in the interior of $L$ are coloured i... | [
"Colour all board squares lying on the outside of $L$ in blue and introduce a Cartesian coordinate system such that the board's lower left and upper right corners have coordinates $(0, 0)$ and $(2n+1, 2n+1)$, respectively. Add one more row, consisting of blue squares only, on top of the board: the lower left corner... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0ag5 | A natural number is called "happy" if the sum of its digits is $7$. Let $(a_n)_{n=1}^\infty$ be the increasing sequence of happy numbers. If $a_n = 2005$, compute $a_{5n}$. | [
"We will use:\n**Lemma 1.** The number of solutions of the equation $x_1 + x_2 + \\dots + x_k = n$ in the set of natural numbers is $C_{n-1}^{k-1}$.\n**Proof.** Consider $n$ ones arranged in a row, and an arrangement of $k-1$ signs \"+\" between $k-1$ pairs of ones.\nThis arrangement of the \"+\" signs can be done ... | North Macedonia | null | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 52000 | |
0160 | The triangle $ABC$ is isosceles with $AB = AC$. The point $P$ inside $ABC$ satisfies two conditions:
(i) $A$ lies on the trisector line of $\angle BPC$, i.e. $AP$ meets $BC$ at $Q$ such that $\angle BPC = 3 \angle QPC$;
(ii) $\angle BPQ = \angle BAC$.
Show that $Q$ trisects $BC$, i.e. $BC = 3 \cdot QC$. | [
"Let $AQ$ meet the circumscribed triangle of $ABC$ again at $R$. Then $\\angle BRP = \\angle BCA$. Because $\\angle BPR = \\angle BAC$, we must have $\\angle PBR = \\angle ABC$. Since $ABC$ is isosceles, so is $PBR$. Let $PS$ be an altitude of $PBR$. Since $\\angle BPS = \\angle SPR$, we have $\\angle SPR = \\angle... | Baltic Way | Baltic Way SHL | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
04n6 | Prove the identity
$$
\cos^3 \frac{x}{3} + \cos^3 \frac{x+2\pi}{3} + \cos^3 \frac{x+4\pi}{3} = \frac{3}{4} \cos x
$$
for every real number $x$. | [
"**Using the identity**\n$$\n\\cos t = 4 \\cos^3 \\frac{t}{3} - 3 \\cos \\frac{t}{3}\n$$\nfor $t = x$, $t = x + 2\\pi$ and $t = x + 4\\pi$, summing the relations and using the periodicity of cosine, we obtain\n$$\n3 \\cos x = 4 \\left( \\cos^3 \\frac{x}{3} + \\cos^3 \\frac{x+2\\pi}{3} + \\cos^3 \\frac{x+4\\pi}{3} \... | Croatia | Croatia_2018 | [
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials"
] | English | proof only | null | |
0fjh | Problem:
En un polígono regular $H$ de $6n+1$ lados ($n$ es un entero positivo), pintamos $r$ vértices de color rojo, y el resto de azul. Demostrar que el número de triángulos isósceles que tienen sus tres vértices del mismo color no depende del modo de distribuir los colores en los vértices de $H$. | [
"Solution:\n\nDebido a que el número de lados del polígono $H$ deja de resto 1 al dividirse entre seis, cada diagonal y cada lado del polígono pertenece sólo (exactamente) a tres triángulos isósceles distintos.\n\nDenotamos por $AA$, $AR$ y $RR$ los números de segmentos que son lados y diagonales cuyos extremos res... | Spain | Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0c2i | Let $n$ be a positive integer and let $A$ be a matrix from $M_n(\mathbb{C})$. Prove that $A^2 = O_n$ if and only if there exist two matrices $B$ and $C$ in $M_n(\mathbb{C})$ so that $A = BC$ and $CB = O_n$. | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Linear transformations"
] | null | proof only | null | |
06pr | For every integer $k \geq 2$, prove that $2^{3k}$ divides the number
$$
\binom{2^{k+1}}{2^{k}} - \binom{2^{k}}{2^{k-1}}
$$ | [
"We use the notation $(2n-1)!! = 1 \\cdot 3 \\cdots (2n-1)$ and $(2n)!! = 2 \\cdot 4 \\cdots (2n) = 2^{n} n!$ for any positive integer $n$. Observe that $(2n)! = (2n)!! (2n-1)!! = 2^{n} n! (2n-1)!!$.\n\nFor any positive integer $n$ we have\n$$\n\\begin{aligned}\n& \\binom{4n}{2n} = \\frac{(4n)!}{(2n)!^{2}} = \\frac... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0axl | Problem:
Refer to the figure. A hemisphere is placed on top of a sphere having radius $2017$. A second hemisphere is then placed on top of the first hemisphere, and a third hemisphere is also placed on top of the second hemisphere. All the centers are collinear, and the three hemispheres all have empty interiors and n... | [
"Solution:\n\nLet $r_{0}=2017$. In the figure below, $h_{i}^{2}=r_{i}^{2}-r_{i+1}^{2}$. By the Cauchy-Schwarz Inequality,\n$$\n\\left(h_{0}+h_{1}+h_{2}+r_{3}\\right)^{2} \\leq 4\\left(h_{0}^{2}+h_{1}^{2}+h_{2}^{2}+r_{3}^{2}\\right)=4\\left(r_{0}^{2}-r_{1}^{2}+r_{1}^{2}-r_{2}^{2}+r_{2}^{2}-r_{3}^{2}+r_{3}^{2}\\right... | Philippines | Philippine Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 6051 | |
0dt8 | Two distinct 2-digit prime numbers $p$, $q$ can be written one after the other in 2 different ways to form two 4-digit numbers. For example, $11$ and $13$ yield $1113$ and $1311$. If the two 4-digit numbers formed are both divisible by the average value of $p$ and $q$, find all possible pairs \{$p, q$\}.
| [
"The average of $p$, $q$ is $\\frac{p+q}{2}$. Therefore $2(100p + q) = (p+q)k$ for some integer $k$. Thus\n$$\n200p + 2q = (p+q)k \\Rightarrow 198p = (p+q)(k-2).\n$$\nSince $p$, $q$ are primes, $(p+q) \\mid 198$. Now $11 + 13 = 24 \\leq p+q \\leq 89 + 97 = 186$. Since $198 = 2 \\times 3^2 \\times 11$, the only even... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | {13,53}, {19,47}, {23,43}, {29,37} | |
06fp | Find the total number of solutions to the following system of equations:
$$
\begin{cases}
a^2 + bc \equiv a \pmod{37}, \\
b(a+d) \equiv b \pmod{37}, \\
c(a+d) \equiv c \pmod{37}, \\
bc + d^2 \equiv d \pmod{37}, \\
ad - bc \equiv 1 \pmod{37}.
\end{cases}
$$ | [
"There is only 1 solution modulo $37$.\nThe first four equations can be rewritten as $\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}^2 = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}$, where the entries are taken modulo $37$. The last equation means $\\begin{vmatrix} a & b \\\\ c & d \\end{vmatrix} = 1$. There... | Hong Kong | null | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | 1 | |
0k0p | Problem:
Twenty-five people of different heights stand in a $5 \times 5$ grid of squares, with one person in each square. We know that each row has a shortest person; suppose Ana is the tallest of these five people. Similarly, we know that each column has a tallest person; suppose Bev is the shortest of these five peo... | [
"Solution:\n\nBev is taller. We consider three possible cases.\n\nCase 1: If Ana and Bev are in the same row, then Bev is taller because Ana is by definition the shortest in that row.\n\nCase 2: If Ana and Bev are in the same column, then Bev is taller because Bev is by definition the tallest in that column.\n\nCas... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Bev is taller. | |
04bs | Prove that for all $a, b, c > 0$ such that $a + b + c \le 3$ the following inequality holds
$$
\frac{a+1}{a(a+2)} + \frac{b+1}{b(b+2)} + \frac{c+1}{c(c+2)} \ge 2.
$$
(Tangenta magazine, 2010) | [
"From the A–H inequality we get\n$$\n\\begin{aligned}\n\\frac{a+1}{a(a+2)} + \\frac{b+1}{b(b+2)} + \\frac{c+1}{c(c+2)} &\\ge \\frac{9}{\\frac{a(a+2)}{a+1} + \\frac{b(b+2)}{b+1} + \\frac{c(c+2)}{c+1}} \\\\\n&= \\frac{9}{\\frac{(a+1)^2-1}{a+1} + \\frac{(b+1)^2-1}{b+1} + \\frac{(c+1)^2-1}{c+1}} \\\\\n&= \\frac{9}{(a+1... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0l6n | Determine, with proof, all positive integers $k$ such that
$$
\frac{1}{n+1} \sum_{i=0}^{n} \binom{n}{i}^{k}
$$
is an integer for every positive integer $n$. | [
"Let $S(n, k) = \\sum_{i=0}^{n} \\binom{n}{i}^{k}$.\nWe are to find all positive integers $k$ such that $\\frac{S(n, k)}{n+1}$ is an integer for all $n \\ge 1$.\n\nFirst, consider small values of $k$.\n\n**Case $k = 1$:**\n$S(n, 1) = \\sum_{i=0}^{n} \\binom{n}{i} = 2^{n}$.\nSo $\\frac{2^{n}}{n+1}$ must be integer f... | United States | USAMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | k = 2 | |
0k3o | Problem:
In acute triangle $A B C$, let $D, E$, and $F$ be the feet of the altitudes from $A, B$, and $C$ respectively, and let $L, M$, and $N$ be the midpoints of $B C, C A$, and $A B$, respectively. Lines $D E$ and $N L$ intersect at $X$, lines $D F$ and $L M$ intersect at $Y$, and lines $X Y$ and $B C$ intersect at... | [
"Solution:\n\nBecause $N L \\parallel A C$ we have triangles $D X L$ and $D E C$ are similar. From angle chasing, we also have that triangle $D E C$ is similar to triangle $A B C$. We have $\\angle X N A = 180^{\\circ} - \\angle X N B = 180^{\\circ} - \\angle L N B = 180 - C A B = \\angle L M A$. In addition, we ha... | United States | HMMT February | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | AB^2/AC^2 | |
06v9 | The infinite sequence $a_{0}, a_{1}, a_{2}, \ldots$ of (not necessarily different) integers has the following properties: $0 \leqslant a_{i} \leqslant i$ for all integers $i \geqslant 0$, and
$$
\binom{k}{a_{0}}+\binom{k}{a_{1}}+\cdots+\binom{k}{a_{k}}=2^{k}
$$
for all integers $k \geqslant 0$.
Prove that all integers ... | [
"We prove by induction on $k$ that every initial segment of the sequence, $a_{0}, a_{1}, \\ldots, a_{k}$, consists of the following elements (counted with multiplicity, and not necessarily in order), for some $\\ell \\geqslant 0$ with $2 \\ell \\leqslant k+1$ :\n$$\n0,1, \\ldots, \\ell-1, \\quad 0,1, \\ldots, k-\\e... | IMO | IMO 2019 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0lez | Let $ABC$ be a triangle. Points $E, F$ move on the opposite ray of $BA, CA$ such that $BF = CE$. Let $M, N$ be the midpoints of $BE, CF$. Suppose that $BF$ cuts $CE$ at $D$.
a) Let $I, J$ be the centers of circumcircles of triangles $DBE$ and $DCF$. Prove that $MN$ is parallel to $IJ$.
b) Let $K$ be the midpoint of $... | [
"a) Let $T$ be the second intersection of $(BDE)$ and $(CDF)$. We have $BF = CE$ and\n$$\n\\angle TED = \\angle TBD, \\quad \\angle TCD = \\angle TFD\n$$\nso two triangles $TCE$ and $TFB$ are congruent. From this it is easy to obtain that $TBE$ and $TCF$ are isosceles triangles at $T$ and they are similar.\n\n![](a... | Vietnam | Vietnamese Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cycli... | null | proof only | null | |
0avx | Problem:
The faces of a 12-sided die are numbered $1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11$, and $12$ such that the sum of the numbers on opposite faces is $13$. The die is meticulously carved so that it is biased: the probability of obtaining a particular face $F$ is greater than $1/12$, the probability of obtaining the fa... | [
"Solution:\n\nThe probabilities that the die lands on its corresponding faces are $\\frac{1}{12}, \\ldots, \\frac{1}{12}, \\frac{1}{12}+x$, and $\\frac{1}{12}-x$, where the last two probabilities are for the face $F$ and its opposite face, respectively, while the rest are for the other faces (since these probabilit... | Philippines | 18th PMO National Stage Oral Phase | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 7/48 | |
0f6n | Problem:
An $8 \times 8$ chessboard is colored in the usual way. What is the largest number of pieces can be placed on the black squares (at most one per square), so that each piece can be taken by at least one other? A piece $A$ can take another piece $B$ if they are (diagonally) adjacent and the square adjacent to $... | [] | Soviet Union | 19th ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 25 | |
0kjt | Problem:
Find the number of subsets $S$ of $\{1,2, \ldots, 48\}$ satisfying both of the following properties:
- For each integer $1 \leq k \leq 24$, exactly one of $2k-1$ and $2k$ is in $S$.
- There are exactly nine integers $1 \leq m \leq 47$ so that both $m$ and $m+1$ are in $S$. | [
"Solution:\nThis problem can be thought of as laying down a series of $1 \\times 2$ dominoes, with each one having either the left or right square marked. The second condition states that exactly $9$ pairs of consecutive dominoes will have the leftmost one with the right square marked and the rightmost one with the... | United States | HMMT November 2021 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 177100 | |
0jhc | Positive real numbers $x$, $y$, $z$ satisfy $xyz + xy + yz + zx = x + y + z + 1$. Prove that
$$
\frac{1}{3} \left( \sqrt{\frac{1+x^2}{1+x}} + \sqrt{\frac{1+y^2}{1+y}} + \sqrt{\frac{1+z^2}{1+z}} \right) \le \left( \frac{x+y+z}{3} \right)^{5/8}.
$$ | [
"By the given condition, we have\n$$\n1 = \\frac{xyz + xy + yz + xz}{x + y + z + 1}.\n$$\nWe may therefore compute\n$$\n\\begin{aligned}\n1 + x^2 &= \\frac{xyz + xy + yz + xz}{x + y + z + 1} + x^2 \\\\\n&= \\frac{xyz + xy + yz + xz + x^3 + x^2y + x^2z + x^2}{x + y + z + 1} \\\\\n&= \\frac{(xyz + yz) + (x^2z + xz) +... | United States | TSTST | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
09m5 | Some cells of a $60 \times 60$ grid is colored black and the rest are colored white. Two players $A$ and $B$ take turns making the following type of move starting from player $A$: choose a black cell and flip the colors of the rectangle that starts on the top-left corner of the grid and ends on the chosen black cell as... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
02sk | Problem:
Sobre um tabuleiro de $n \times n$, Chris planeja desenhar três O's. Ele deseja fazê-lo de modo que cada $\mathbf{O}$ fique dentro de um quadradinho e de modo que os três quadradinhos utilizados estejam dispostos no formato de um L. A seguinte figura mostra alguns exemplos com $n=4$.
![](attached_image_1.png... | [
"Solution:\n\na) Se marcamos somente um dos quadradinhos do tabuleiro com um $\\mathbf{X}$, então os três quadradinhos que restarão formarão sempre um L. Chris poderia então escrever um $\\mathbf{O}$ em cada um dos quadradinhos atingindo assim o seu objetivo. Assim, marcando um quadradinho apenas, não seremos capaz... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) 2. b) Possible with exactly 4 marked squares (example exists). c) Possible with exactly 8 marked squares (example exists). d) Impossible with only 7 marked squares; at least 8 are needed. | |
04uz | Consider an infinite sequence $a_0, a_1, a_2, \dots$ of integers that satisfies
$$
a_0 \ge 1 \quad \text{and} \quad a_{n+1} \in \{2022a_n - 1, 2022a_n + 1\}
$$
for all indices $n \ge 0$. Prove that any such infinite sequence contains infinitely many composite numbers.
(Martin Melicher) | [] | Czech Republic | Second Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0iu0 | Problem:
Consider an isosceles triangle $T$ with base $10$ and height $12$. Define a sequence $\omega_{1}, \omega_{2}, \ldots$ of circles such that $\omega_{1}$ is the incircle of $T$ and $\omega_{i+1}$ is tangent to $\omega_{i}$ and both legs of the isosceles triangle for $i > 1$.
1. Find the radius of $\omega_{1}$. | [
"Solution:\n\nAnswer: $\\frac{10}{3}$\n\nUsing the Pythagorean theorem, we see that the legs of $T$ each have length $13$. Let $r$ be the radius of $\\omega_{1}$. We can divide $T$ into three triangles, each with two vertices at vertices of $T$ and one vertex at the center of $\\omega_{1}$. These triangles all have... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 10/3 | |
0gnb | The acute triangle $ABC$ and the triangle $A_1B_1C_1$, whose vertices $B_1$, $C_1$ and $A_1$ lie on the rays $AC$, $BA$ and $CB$, respectively, are similar. Prove that the orthocenter of the triangle $A_1B_1C_1$ and the circumcenter of the triangle $ABC$ coincide. | [
"Let $H$ be the orthocenter of the triangle $A_1B_1C_1$. By the similarity of the triangles $ABC$ and $A_1B_1C_1$, we have $\\angle A_1B_1C_1 = \\angle ABC$. From this, $\\angle A_1HC_1 = 180^\\circ - \\angle A_1B_1C_1 = 180^\\circ - \\angle ABC = \\angle A_1BC_1$ is obtained. Therefore the points $A_1$, $B$, $H$, ... | Turkey | Team Selection Test for IMO 2007 | [
"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 | |
0235 | Problem:
No desenho a seguir, $ABCD$ é um quadrado e os pontos $E$ e $F$ estão sobre os lados $BC$ e $CD$ de modo que $AEF$ é um triângulo retângulo, $AE=4$ e $EF=3$. Qual é a área do quadrado?
 | [
"Solution:\n\nSe $\\angle BAE=\\alpha$, como $\\angle ABC=\\angle AEF=90^{\\circ}$, temos $\\angle BEA=90^{\\circ}-\\alpha$ e $\\angle FEC=\\alpha$. Consequentemente os triângulos $BEA$ e $CFE$ são semelhantes e\n$$\n\\frac{BE}{CF}=\\frac{AB}{EC}=\\frac{AE}{EF}=\\frac{4}{3}\n$$\nSe $BE=x$, temos $CF=\\frac{3x}{4}$ ... | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 256/17 | |
048i | Let $a$, $b$, $c \in \mathbb{R}$ satisfy $a + b + c = 3$ and $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$. Determine $a^2 + b^2 + c^2$. | [] | Croatia | Hrvatska 2011 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof and answer | 9 | |
00yz | Problem:
John is older than Mary. He notices that if he switches the two digits of his age (an integer), he gets Mary's age. Moreover, the difference between the squares of their ages is the square of an integer. How old are Mary and John? | [
"Solution:\n\nLet John's age be $10a + b$ where $0 \\leq a, b \\leq 9$. Then Mary's age is $10b + a$, and hence $a > b$. Now\n$$\n(10a + b)^2 - (10b + a)^2 = 9 \\cdot 11 (a + b)(a - b).\n$$\nSince this is the square of an integer, $a + b$ or $a - b$ must be divisible by $11$. The only possibility is clearly $a + b ... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | Mary is 56 and John is 65. | |
0i1e | Let $\triangle ABC$ be an acute-angled triangle with $O$ as its circumcenter. Let $P$ on line $BC$ be the foot of the altitude from $A$. Assume that $\angle BCA \ge \angle ABC + 30^\circ$. Prove that $\angle CAB + \angle COP < 90^\circ$. | [
"**First Solution.** (By Gabriel Carroll and Tiankai Liu) Because $\\angle AOC = 2\\beta$, we have $\\angle CAO = 90^\\circ - \\beta$. Note that in right triangle $APC$, $\\angle CAP = 90^\\circ - \\gamma$, so\n$$\n\\begin{aligned}\n\\angle PAO &= \\angle CAO - \\angle CAP \\\\\n&= \\gamma - \\beta.\n\\end{aligned}... | United States | USA IMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous... | English | proof only | null | |
0g56 | $\triangle ABC$ 中有一點 $O$, 令 $AO$, $BO$, $CO$ 的延長線分別交 $BC$, $CA$, $AB$ 於 $A_1$, $B_1$, $C_1$。證明：$O$ 在 $\triangle A_1B_1C_1$ 的中位三角形內。（中位三角形是指各邊中點連線所成的三角形。） | [
"設 $AA_1$, $BB_1$, $CC_1$ 分別與 $\\triangle A_1B_1C_1$ 各邊交於 $A_2$, $B_2$, $C_2$。\n令 $\\frac{AC_1}{C_1B} = \\frac{x}{y}$, $\\frac{BA_1}{A_1C} = \\frac{y}{z}$ ($x, y, z > 0$), 由 Ceva 定理 $\\frac{CB_1}{B_1A} = \\frac{z}{x}$。於是質點組\n$A(yz)$, $B(zx)$, $C(xy)$\n的重心在 $O$。（因為 $\\frac{yz}{zx} = \\frac{C_1B}{AC_1}$, 故 $A(yz)$ 與 ... | Taiwan | 二〇一一數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates"
] | null | proof only | null | |
0gy7 | In the plane let be given a circle with a chord $AB$ and a point $P$ on this chord such that $AP = 2PB$. The chord $DE$ is normal to the chord $AB$ and passing through the point $P$. Prove that the midpoint of the segment $AP$ is the orthocenter of the triangle $AED$. | [
"Let $M$ be a midpoint of the segment $AP$. Let us the line $EM$ intersects the segment $AD$ at a point $Q$. $MPE$ and $BPE$ are the congruent triangles because both of them are right and both of them have the equal legs. Then $\\angle QAM = \\angle DAB = \\angle DEB = \\angle PEB = \\angle PEM$ (Fig.9).\n\nSince $... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0c33 | Consider an isosceles triangle $ABC$, where $AB = AC$ and $\angle BAC = 100^\circ$. Let point $D$ on the opposite side from $C$ with respect to $AB$ such that $DC = BC$, and $\angle BCD = 20^\circ$. Find $\angle DAB$. | [] | Romania | Shortlisted problems for the 69th NMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | 30° | |
07x0 | Find, with proof, all functions $f : \mathbb{R}_+ \to \mathbb{R}_+$ with the property that
$$
f(x + f(y) + f(f(z))) = z + f(y) + f(f(x)) \quad \text{for all } x, y, z > 0.
$$ | [
"For simplicity, we will write $f^2(x) = f(f(x))$, $f^3(x) = f(f(f(x)))$, etc.\n\nFirst we show that $f$ must be injective. Suppose $f(z_1) = f(z_2)$. Replacing $z$ by $z_1$ or by $z_2$ in (1) leads to the same LHS expressions, hence the two RHS expressions must be equal as well, i.e. $z_1 + f(y) + f^2(x) = z_2 + f... | Ireland | IRL_ABooklet_2024 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x for all x > 0 | |
02q6 | Problem:
Em um torneio de xadrez cada jogador disputou uma partida com cada um dos demais participantes. A cada partida, havendo empate, cada jogador ganhou $1/2$ ponto; caso contrário, o vencedor ganhou $1$ ponto e o perdedor, $0$ ponto. Participaram homens e mulheres e cada participante conquistou o mesmo número de ... | [] | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
09eo | Points $B$, $D$, $E$, $F$, $C$ lie on a line. Let $A$ be a point outside the line and let $\omega_1, \omega_2, \omega_3, \omega_4$ be incircles of the triangles $ABD$, $ADE$, $AEF$, $AFC$ respectively. Let $p_1, p_2, p_3, p_4$ be outwardly inscribed circles of the triangles $ABD$, $ADE$, $AEF$, $AFC$ at the vertex $A$.... | [
"We need following fact.\n\n**Lemma:** Given a point $D$ on side $BC$ of the triangle $ABC$. Let $\\omega_1(O_1, r_1), \\omega_2(O_2, r_2)$ be incircles of triangles $ACD$, $ADB$ and $\\rho_1(O_3, r_3), \\rho_2(O_4, r_4)$ be circles inscribed in these triangles externally, which tangent at an internal point of $BC$... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
01gw | Let $ABC$ be a triangle with the circumcircle $\omega$. The internal angle bisectors of $\angle ABC$ and $\angle ACB$ intersect $\omega$ at $X$ and $Y$, respectively. Let $K$ be a point on $CX$ such that $\angle KAC = 90^\circ$. Similarly, let $L$ be a point on $BY$ such that $\angle LAB = 90^\circ$. Let $S$ be the mid... | [
"W.l.o.g. let $AB < AC$. We will prove that triangles $KXS$ and $SYL$ are congruent by SAS, which will finish the proof.\nAs $BX$ and $CY$ are angle bisectors, we obtain:\n$$\n\\frac{1}{2} C\\hat{A}B = C\\hat{X}S = C\\hat{X} + \\hat{X}S = \\frac{1}{2} C\\hat{X}A + \\hat{X}S.\n$$\nThis implies $\\hat{X}S = \\frac{1}... | Baltic Way | Baltic Way 2020 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
09ex | Find all continuous functions $f : \mathbb{R} \to \mathbb{R}$ that satisfy conditions:
$f(0) = 0$ and $\forall x \in \mathbb{R}: f(2x) \geq x + f(x)$, $f(3x) \leq 2x + f(x)$. | [
"Substitute $g(x) = f(x) - x$, $x \\in \\mathbb{R}$. Considering the given conditions we get $f(2x) = g(2x) + 2x \\geq g(x) + 2x \\Leftrightarrow g(2x) \\geq g(x)$ and $f(3x) = g(3x) + 3x \\leq g(x) + 3x \\Leftrightarrow g(3x) \\leq g(x)$. In other words, $\\forall x \\in \\mathbb{R}: g(3x) \\leq g(x) \\leq g(2x)$.... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | f(x) = x | |
0fdf | Problem:
Consideramos los polinomios $P(x)=x^{3}+A x^{2}+B x+C$, $Q(x)=3 x^{2}+2 A x+B$ ($x$ es la variable, $A$, $B$, $C$ son parámetros). Supongamos que, si $a, b, c$ son las tres raíces de $P$, las de $Q$ son $\frac{a+b}{2}$, $\frac{b+c}{2}$. Determinad todos los posibles polinomios $P, Q$. | [
"Solution:\n$Q$ es la derivada de $P$. Por tanto,\n$$\nP(x)=(x-a)(x-b)(x-c), \\quad Q(x)=(x-a)(x-b)+(x-a)(x-c)+(x-b)(x-c)\n$$\nSea $d=\\frac{a+b}{2}$, punto medio del segmento que une $a$ con $b$. Se tiene que $d-a=d-b=\\frac{a-b}{2}$.\nEl valor de $Q$ en $d$ es\n$$\nQ(d)=\\frac{b-a}{2} \\frac{a-b}{2}+\\frac{b-a}{2... | Spain | null | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | P(x) = (x - a)^3 and Q(x) = 3(x - a)^2 for some real or complex a | |
072p | There are $n$ markers, each with one side white and the other side black, aligned in a row so that their white sides are all up in the beginning. In each step, if possible, we choose a marker with the white side up (but not the one of outermost markers), remove it, and reverse the closest marker to the left and the clo... | [
"It is easy to handle the cases $n = 3, 5, 6$:\n\na. $n=3$: $WWW \\rightarrow BB$, by removing middle $W$ and reversing the other two;\n\nb. $n=5$: $(WWW)WW \\rightarrow (BB)WW = B(BWW) \\rightarrow BWB \\rightarrow WW$;\n\nc. $n = 6$: $(WWW)WWW \\rightarrow BB(WWW) = B(BWW)W \\rightarrow B(WB)W = (BWB)W \\rightarr... | India | Indija TS 2006 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
0laj | Given a positive integer $n$. Denote by $T$ the set of the first $2n$ positive integers. Find the number of subsets $S$ of $T$ with the property: in $S$ there do not exist numbers $a, b$ with $|a-b| \in \{1; n\}$.
(Note: The empty set is considered as satisfying the above property). | [
"For each $n \\in \\mathbb{N}^*$, denote by $d_n$ the number to be found.\nConsider the table of size $2 \\times n$. Write in each square of the table in the order from left to right and from top to bottom the numbers from $1$ to $2n$. (See Pic. 1).\nCall the $n$-th square of the first row and the first square of t... | Vietnam | Vijetnam 2009 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | Let k_n = ((1 + sqrt(2))^{n+1} + (1 - sqrt(2))^{n+1})/2. Then d_1 = 3, d_2 = 6, and for n ≥ 3: d_n = (2k_n - k_{n-2} + (-1)^{n-3})/2. | |
0ava | Problem:
Let $N$ be a natural number whose base-2016 representation is $A B C$. Working now in base-10, what is the remainder when $N-(A+B+C+k)$ is divided by $2015$, if $k \in \{1,2, \ldots, 2015\}$? | [
"Solution:\n\nLet $N$ be the number whose base-$2016$ representation is $A B C$. That is,\n$$\nN = A \\times 2016^2 + B \\times 2016 + C\n$$\nwhere $A, B, C$ are digits in base $2016$, i.e., $0 \\leq A, B, C \\leq 2015$.\n\nWe are to compute the remainder when $N - (A+B+C+k)$ is divided by $2015$, for $k \\in \\{1,... | Philippines | 19th Philippine Mathematical Olympiad | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 2015 - k | |
015i | Determine integers $a$, $b$ and $c$ such that the polynomials
$$
P(x) = x^9 - x^7 - 31x^2 - 26x - 17 \quad \text{and} \quad Q(x) = x^{13} - x^{11} - a x^2 - b x - c
$$
have a common factor. | [
"Consider the sequence $(a_n)$ defined by $a_0 = a_1 = a_2 = 1$ and $a_{n+3} = a_{n+2} + a_{n+1} + a_n$ for $n \\ge 0$. The sequence is $1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193, 355, \\ldots$. If now\n$$\np_n(x) = \\sum_{k=0}^{n} a_k x^{n-k},\n$$\nthen for $n \\ge 3$ a straightforward computation shows that\n$$\n(x^... | Baltic Way | Baltic Way SHL | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | a = 355, b = 298, c = 193 | |
020v | Problem:
Let $ABC$ be a triangle. The circle $\omega_{A}$ through $A$ is tangent to line $BC$ at $B$. The circle $\omega_{C}$ through $C$ is tangent to line $AB$ at $B$. Let $\omega_{A}$ and $\omega_{C}$ meet again at $D$. Let $M$ be the midpoint of line segment $[BC]$, and let $E$ be the intersection of lines $MD$ and... | [
"Solution:\nLet $N$ be the midpoint of $[AB]$. By tangential angles, $\\angle CBD = \\angle BAD$ and $\\angle DBA = \\angle DCB$, so triangles $DAB$ and $DBC$ are similar. By definition of the midpoints $M, N$, so are triangles $AND$ and $BMD$. In particular, $\\angle BND = 180^{\\circ} - \\angle DNA = 180^{\\circ}... | Benelux Mathematical Olympiad | Benelux Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
0acn | The roots of the quadratic equation $x^2 + p x + q = 0$ are integers. Find the numbers $p$ and $q$ and the roots of the equation, if $p + q = 198$. | [
"If $x_1$ and $x_2$ are the roots of the equation $x^2 + p x + q = 0$, then by Viete's formulas we obtain\n$$\n\\begin{cases} x_1 + x_2 = -p \\\\ x_1 x_2 = q \\end{cases}\n$$\nBecause $198 = p + q = -(x_1 + x_2) + x_1 x_2 = (x_1 - 1)(x_2 - 1) - 1$, it follows that $(x_1 - 1)(x_2 - 1) = 199$. Since $x_1, x_2$ are in... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | Either p = -202, q = 400 with roots 2 and 200; or p = -198, q = 0 with roots 0 and 198. | |
05p0 | Problem:
Soit $ABC$ un triangle rectangle en $C$. Soient $D$ le pied de la hauteur issue de $C$, et $Z$ le point de $[AB]$ tel que $AC = AZ$. La bissectrice de $\widehat{BAC}$ coupe $(CB)$ et $(CZ)$ en $X$ et $Y$ respectivement. Montrer que les quatre points $B$, $X$, $Y$, $D$ sont sur un même cercle. | [
"Solution:\n\n\n\n$CAZ$ est isocèle en $A$ donc $(AY)$, qui est la bissectrice de $\\widehat{CAZ}$, est perpendiculaire à $(CZ)$. Donc $ADYC$ est cyclique.\n\nOn a alors $\\widehat{DYX} = 180^{\\circ} - \\widehat{AYD} = 180^{\\circ} - \\widehat{ACD} = 180^{\\circ} - \\widehat{XBD}$."
] | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02el | The circumcenter of a tetrahedron lies inside the tetrahedron. Show that at least one of its edges is at least as long as the edge of a regular tetrahedron with the same circumsphere.
(1989) | [
"There are two main steps in the proof: first, we'll prove a tridimensional analogous to the famous $R \\ge 2r$ inequality; then, we'll prove the bidimensional analogous to the problem. Combining these two facts and other arguments will lead us to a solution.\n\nThe first step is proving that if $R$ and $r$ are the... | Brazil | XI OBM | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane ... | English | proof only | null | |
0ii7 | Problem:
Having lost a game of checkers and my temper, I dash all the pieces to the ground but one. This last checker, which is perfectly circular in shape, remains completely on the board, and happens to cover equal areas of red and black squares. Prove that the center of this piece must lie on a boundary between two... | [
"Solution:\n\nSuppose, for the sake of contradiction, that the problem is false. Evidently, at least one boundary between adjacent squares must lie within our checker, or else the checker would exist entirely within one square, meaning it would cover only one color. Note also that a checker's diameter is smaller th... | United States | Harvard-MIT Mathematics Tournament, Team Round A | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
02xd | Problem:
Um quadrado $3 \times 3$ está preenchido com os números $a, b, c, d, e, f, g, h$ e $i$ da seguinte forma:
| $c$ | $f$ | $i$ |
| :--- | :--- | :--- |
| $b$ | $e$ | $h$ |
| $a$ | $d$ | $g$ |
Sabemos que ele é um quadrado mágico, isto é, existe um valor $S$ que é igual as somas dos números em cada linha, colun... | [
"Solution:\n\na) Somando os números das linhas e colunas que não contêm o quadradinho central, obtemos:\n$$\n(a+b+c)+(a+d+g)+(c+f+i)+(g+h+i)=4 S\n$$\nPor outro lado, somando a linha e a coluna que contêm o quadrado central, temos:\n$$\n2 S=(b+e+h)+(d+e+f)\n$$\nComparando as duas equações, podemos concluir que\n$$\n... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
007b | Three positive integers have sum $1810$. In how many zeros can their product end? Find all possibilities. | [
"Let $a, b, c \\in \\mathbb{N}$, $a+b+c=1810$, and let $abc$ end in $N$ zeros. The following triples $(a, b, c)$ show that $N$ can be $0, 1, \\dots, 7$:\n$(1808, 1, 1)$, $(1799, 10, 1)$, $(1709, 100, 1)$, $(1765, 40, 5)$,\n$(1200, 605, 5)$, $(1700, 100, 10)$, $(1250, 520, 40)$, $(1250, 400, 160)$.\n\nWe prove that ... | Argentina | National Olympiad of Argentina | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 0, 1, 2, 3, 4, 5, 6, 7 |
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