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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0ki0 | Problem:
Let $f: \mathbb{Z}^2 \rightarrow \mathbb{Z}$ be a function such that, for all positive integers $a$ and $b$,
$$
f(a, b)= \begin{cases}b & \text{ if } a>b \\ f(2 a, b) & \text{ if } a \leq b \text{ and } f(2 a, b)<a \\ f(2 a, b)-a & \text{ otherwise }\end{cases}
$$
Compute $f\left(1000,3^{2021}\right)$. | [
"Solution:\n\nNote that $f(a, b)$ is the remainder of $b$ when divided by $a$. If $a>b$ then $f(a, b)$ is exactly $b \\bmod a$. If instead $a \\leq b$, our \"algorithm\" doubles our $a$ by $n$ times until we have $a \\times 2^{n}>b$. At this point, we subtract $a^{\\overline{2 n-1}}$ from $f\\left(a \\cdot 2^{n}, b... | United States | HMMT Spring 2021 Guts Round | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Algorithms",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 203 | |
08qr | Problem:
Let $\triangle ABC$ be a right-angled triangle with $\angle BAC = 90^\circ$ and let $E$ be the foot of the perpendicular from $A$ on $BC$. Let $Z \neq A$ be a point on the line $AB$ with $AB = BZ$. Let $(c)$ be the circumcircle of the triangle $\triangle AEZ$. Let $D$ be the second point of intersection of $(c... | [
"Solution:\nWe will first show that $PA$ is tangent to $(c)$ at $A$.\nSince $E, D, Z, A$ are concyclic, then $\\angle EDC = \\angle EAZ = \\angle EAB$. Since also the triangles $\\triangle ABC$ and $\\triangle EBA$ are similar, then $\\angle EAB = \\angle BCA$, therefore $\\angle EDC = \\angle BCA$.\nSince $\\angle... | JBMO | JBMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0eph | Marks are drawn on a thin strip of paper dividing the strip into 4 equal lengths. Marks are also drawn dividing the strip into 3 equal lengths. After the strip has been cut at each mark, how many pieces will there be? | [
"6\nIf the length of the strip is 12 units, marks will be made at 3 units, 6 units and 9 units for four equal lengths, and at 4 and 8 units for the three equal lengths. This gives 5 marks in all, meaning there will be 6 pieces when the cuts are made."
] | South Africa | South African Mathematics Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | final answer only | 6 | |
0352 | Problem:
The sequences $\left(a_{n}\right)_{n=1}^{\infty}$ and $\left(b_{n}\right)_{n=1}^{\infty}$ are such that $a_{n+1}=2 b_{n}-a_{n}$ and $b_{n+1}=2 a_{n}-b_{n}$ for every $n$. Prove that:
a) $a_{n+1}=2\left(a_{1}+b_{1}\right)-3 a_{n}$;
b) if $a_{n}>0$ for every $n$, then $a_{1}=b_{1}$. | [
"Solution:\n\na) Since $a_{n+1}+b_{n+1}=2 b_{n}-a_{n}+2 a_{n}-b_{n}=a_{n}+b_{n}$, we have\n$$\na_{n+1}=2\\left(a_{n}+b_{n}\\right)-3 a_{n}=2\\left(a_{1}+b_{1}\\right)-3 a_{n}\n$$\n\nb) Using a), we obtain\n$$\na_{n+1}-\\frac{a_{1}+b_{1}}{2}=-3\\left(a_{n}-\\frac{a_{1}+b_{1}}{2}\\right)\n$$\nwhence\n$$\na_{n+1}-\\fr... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
034i | Problem:
In every cell of a table $n \times n$, $n \geq 2$, one of the numbers $+1$ and $-1$ is written. The cell on the $i$-th row and $j$-th column is denoted by $(i, j)$, $i, j = 0, 1, \ldots, n-1$. The neighbors of the cell $(i, j)$ are the cells $(i, j-1)$, $(i, j+1)$, $(i-1, j)$ and $(i+1, j)$, where the numbers... | [
"Solution:\n\nWe first prove that for any odd $n \\geq 3$ there are $n \\times n$ tables that are not \"good\". Consider an arbitrary $n \\times n$ table and denote by $P_{i}$, $i=1,2, \\ldots, n$, the product of the numbers in the $i$-th row at the second last step. Then $P_{1} P_{3} = P_{2} P_{4} = \\cdots = P_{n... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n = 2^k for some positive integer k | |
07di | Points $X$ and $Y$ respectively lie on the tangent lines to the circumcircle of triangle $ABC$ passing through $B$, $C$ such that $AB = BX$ and $AC = CY$. (Points $X$, $Y$, $A$ are on the same side of line $BC$.) Let $I$ be the incenter of triangle $ABC$, prove that
$$
\widehat{BAC} + \widehat{XIY} = 180^\circ
$$ | [
"We know that $I$ is the incenter of triangle $ABC$, thus we have $\\widehat{AIC} = 90^\\circ + \\frac{\\widehat{ABC}}{2}$. Moreover we have\n$$\n\\left. \\begin{array}{l} \\widehat{ABC} = \\widehat{ACY} \\\\ AC = AY \\\\ \\widehat{AYC} = 90^\\circ - \\frac{\\widehat{ACY}}{2} \\end{array} \\right\\} \\implies \\wid... | Iran | Iranian Mathematical Olympiad | [
"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 | |
05ii | Problem:
Prouver que, pour tous réels $a, b, c > 0$ et tout réel $t \geqslant 0$, on a
$$
\frac{a}{b+t c}+\frac{b}{c+t a}+\frac{c}{a+t b} \geqslant \frac{3}{1+t} .
$$ | [
"Solution:\n\nPosons $x = b + t c$, $y = c + t a$, $z = a + t b$. On vérifie que\n$$\n\\begin{aligned}\n(1 + t^3) a & = z - t x + t^2 y \\\\\n(1 + t^3) b & = x - t y + t^2 z \\\\\n(1 + t^3) c & = y - t z + t^2 x\n\\end{aligned}\n$$\ndonc, d'après l'inégalité arithmético-géométrique,\n$$\n\\begin{aligned}\n(1 + t^3)... | France | Test EGMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
01n8 | Find all prime numbers $p$ such that the number $p^2 - p - 1$ is a cube of some positive integer. | [
"We have $p^2 - p - 1 = n^3$. First, note that $p^2 > n^3 \\ge n^2 \\Rightarrow p \\ge n + 1$. If $p = n + 1$, then the equation becomes $n^3 - n^2 - n + 1 = 0 \\Rightarrow n = 1, p = 2$, which is a solution of the problem.\n\nLet now $p > n + 1$. Rewrite the equation as $p(p-1) = (n+1)(n^2-n+1)$. So, $(n+1)(n^2-n+... | Belarus | Belorusija 2012 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | 2, 37 | |
03pm | As shown in the figure, quadrilateral $ABCD$ is inscribed in a circle with $AC$ as its diameter, $BD \perp AC$, and $E$ the intersection of $AC$ and $BD$. Extend line segment $DA$ and $BA$ through $A$ to $F$ and $G$ respectively, such that $DG \parallel BF$. Extend $GF$ to $H$ such that $CH \perp GH$. Prove that points... | [
"As shown in the figure, connect $BH$, $EF$ and $CG$. Since $\\triangle BAF \\sim \\triangle GAD$, we have\n$$\n\\frac{FA}{AB} = \\frac{DA}{AG}. \\qquad \\textcircled{1}\n$$\nFurthermore, $\\triangle ABE \\sim \\triangle ACD$, then\n$$\n\\frac{AB}{EA} = \\frac{AC}{DA}. \\qquad \\textcircled{2}\n$$\nMultiplying ① by... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
03vj | Four frogs are positioned at four points on a straight line such that the distance between any two neighboring points is one unit of length. Suppose that every frog can jump to its corresponding point of reflection, by taking any one of the other three frogs as the reference point. Prove that there is no case where the... | [
"Without loss of generality, we may think of the initial positioning of the four frogs as being on the real number line at points $1$, $2$, $3$, and $4$. Further, it can be established that the frogs at odd number positions remain at odd number positions after each jump, and likewise for frogs at even number positi... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0dki | Let $\triangle ABC$ be an acute triangle and $\omega$ be its circumcircle. Points $F$ on $AC$, $E$ on $AB$, and $P$ and $Q$ on $\omega$ are given such that
$$
\angle AFB = \angle AEC = \angle APE = \angle AQF = 90^\circ.
$$
Prove that the lines $BC$, $EF$, and $PQ$ are concurrent. | [] | Saudi Arabia | Saudi Booklet | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Ang... | null | proof only | null | |
04c3 | Maja wrote all integers from $100$ to $130$ in a row without spacings, thus obtaining the multi-digit number $100101102\ldots129130$. Then she decided to erase $80$ digits of this number. What is the largest number Maja can get by doing this? | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms"
] | English | proof and answer | 9967128129130 | |
021b | Problem:
Let $ABC$ be a triangle with incentre $I$ and circumcircle $\Omega$ such that $|AC| \neq |BC|$. The internal angle bisector of $\angle CAB$ intersects side $[BC]$ in $D$, and the external angle bisectors of $\angle ABC$ and $\angle BCA$ intersect $\Omega$ again in $E$ and $F$, respectively. Let $G$ be the int... | [
"Solution:\n\nWe first notice the general fact that $EF \\perp AI$. This can be proved using the following argument. Denote $S$ for the intersection of $EF$ and $AI$. Then $\\angle BIS = (\\angle IBA + \\angle IAB) = \\frac{1}{2}(\\angle ABC + \\angle BAC) = \\frac{1}{2}(180^{\\circ} - \\angle BCA) = \\angle BCF = ... | Benelux Mathematical Olympiad | 16th Benelux Mathematical Olympiad | [
"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... | null | proof only | null | |
01vi | Find all positive integers $n$ such that equation
$$
3a^2 - b^2 = 2018^n
$$
has a solution in integers $a$ and $b$. | [
"Answer: all odd numbers.\nFor $n = 1$ there is a solution $3 \\cdot 27^2 - 13^2 = 2018^1$. For any odd number $n = 2k+1$ from the latter equality we can obtain the equality $3 \\cdot (27 \\cdot 2018^k)^2 - (13 \\cdot 2018^k)^2 = 2018^{2k+1}$, which means that all odd $n$ satisfy the conditions of the problem.\n\nI... | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | all odd positive integers n | |
00b5 | Decide whether there exist $2018$ distinct positive integers such that the sum of their squares is a perfect cube and the sum of their cubes is a perfect square. | [
"The answer is affirmative. There exist $2018$ distinct positive numbers satisfying the required conditions:\n$$\na, 2a, \\dots, 2018a, \\text{ where } a = \\left( \\frac{2018 \\cdot 2019 \\cdot 4037}{6} \\right)^4.\n$$\nNote that $a$ is an integer number, since $2019$ is a multiple of $3$ and $2018$ is a multiple ... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0e3h | Problem:
Vsaka točka na stranicah trikotnika $\mathcal{T}$ je rdeče ali modre barve. Dokaži, da na stranicah trikotnika $\mathcal{T}$ obstajajo take enako obarvane točke $A, B, C$ in $D$, da je štirikotnik $A B C D$ trapez. | [
"Solution:\n\nNaj bodo $A, B$ in $C$ oglišča danega trikotnika ter $A'$, $B'$ in $C'$$ razpolovišča stranic $BC$, $CA$ in $AB$. Vsaj dve izmed točk $A'$, $B'$ in $C'$ sta iste barve. Predpostavimo lahko, da sta to točki $A'$ in $B'$ ter da sta rdeče barve. Daljica $A'B'$ je vzporedna stranici $AB$. Če na stranici $... | Slovenia | 54. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
01sb | Real numbers are written in the cells of the $7 \times 7$ table so that the product of the numbers in any $3 \times 3$ square is equal to the product of the numbers in any $4 \times 4$ square.
Is it possible for the product of all numbers in the table to be 2015?
(V. Kaskevich) | [
"Answer: yes, it is possible.\nLet $a$ and $b$ be real numbers such that $ab = 1$. Consider the following table\n\n| $a$ | $b$ | $a$ | $b$ | $a$ | $b$ | $a$ |\n|-----|-----|-----|-----|-----|-----|-----|\n| $b$ | $a$ | $b$ | $a$ | $b$ | $a$ | $b$ |\n| $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ |\n| $a$ | $b$ | $a$ | $... | Belarus | FINAL ROUND | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Other"
] | English | proof and answer | Yes | |
0fwa | Problem:
Wir nennen zwei Personen ein befreundetes Paar, wenn sie sich kennen, und wir nennen sie ein nichtbefreundetes Paar, wenn sie sich nicht kennen (befreundet sein oder nicht befreundet sein ist dabei immer gegenseitig). Seien $m, n$ natürliche Zahlen. Finde die kleinste natürliche Zahl $k$, sodass Folgendes gil... | [
"Solution:\n\nDer kleinste Wert ist $k=\\max \\{m, n\\}+m+n-1$. Wegen der Symmetrie des Problems können wir $m \\geq n$ annehmen.\nWir beweisen zuerst, dass $k \\geq 2 m+n-1$ gilt, indem wir eine Konfiguration mit $2 m+n-2$ Leuten angeben, die die Bedingungen der Aufgabe nicht erfüllt. Betrachte eine Gruppe $X$ aus... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | max{m, n} + m + n - 1 | |
0kyz | There are exactly $K$ positive integers $b$ with $5 \le b \le 2024$ such that the base-$b$ integer $2024_b$ is divisible by 16 (where 16 is in base ten). What is the sum of the digits of $K$?
(A) 16 (B) 17 (C) 18 (D) 20 (E) 21 | [
"**Answer (D):** Notice that $2024_b = 2b^3 + 2b + 4 = 2(b + 1)(b^2 - b + 2)$, and consider the residue classes of this number modulo 8. If $b \\equiv 7 \\pmod 8$, then $b + 1 \\equiv 0 \\pmod 8$, and if $b \\equiv 3 \\pmod 8$, then $b^2 - b + 2 \\equiv 0 \\pmod 8$. In each case $2024_b$ is divisible by 16.\nIn all... | United States | AMC 12 A | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | MCQ | D | |
04db | Are there integers $a$ and $b$ such that both $a + b$ and $ab - 1$ are perfect squares? | [
"Let us assume there are integers $a$, $b$, $M$ and $N$ such that $a + b = M^2$ and $ab - 1 = N^2$.\n\n*First case:* The numbers $a$ and $b$ are even.\nFrom $a \\cdot b = N^2 + 1$ we conclude that $N^2 + 1$ is divisible by $4$ which is impossible.\n\n*Second case:* Exactly one of the numbers $a$ and $b$ (say $a$) i... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | No; there are no such integers. | |
0foa | The medians $AA'$, $BB'$, $CC'$ of the triangle $ABC$ meet the nine points circle at $D$, $E$, $F$, respectively. The points $L$, $M$, $N$ are the feet of the altitudes of $ABC$ ($L$ belongs to $AA'$, etc). The tangents to the nine points circle at $D$, $E$, $F$ meet the lines $MN$, $LN$ and $LM$ at points $P$, $Q$, $R... | [
"1) Triangles $PMD$ and $PDN$ are similar (indeed, $\\widehat{PDM}$ is half inscribed, $\\widehat{DNP}$ is inscribed and both subtend the same arch in the Euler circle; moreover $\\hat{P}$ is the same in both triangles). Therefore\n$$\n\\frac{PM}{PD} = \\frac{DM}{DN} = \\frac{PD}{PN} \\implies \\frac{PM}{PN} = \\le... | Spain | Mediterranean Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous... | Spanish | proof only | null | |
0eyk | Problem:
The circle inscribed in the triangle $ABC$ touches the side $AC$ at $K$. Prove that the line joining the midpoint of $AC$ with the center of the circle bisects the segment $BK$. | [] | Soviet Union | 2nd ASU | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ibv | Problem:
We say a point is contained in a square if it is in its interior or on its boundary. Three unit squares are given in the plane such that there is a point contained in all three. Furthermore, three points $A, B, C$, are given, each contained in at least one of the squares. Find the maximum area of triangle $A B... | [
"Solution:\n$3 \\sqrt{3} / 2$\n\nLet $X$ be a point contained in all three squares. The distance from $X$ to any point in any of the three squares is at most $\\sqrt{2}$, the length of the diagonal of the squares. Therefore, triangle $A B C$ is contained in a circle of radius $\\sqrt{2}$, so its circumradius is at ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | 3 sqrt(3) / 2 | |
0cbh | Consider a triangle $\triangle ABC$.
a) Prove that the interior bisector of the angle $\angle A$ and the exterior bisectors of the angles $\angle B$ and $\angle C$ intersect at a point $I_A$.
b) Let $I_A M \perp AC$, $M \in AC$, $I_A N \perp BC$, $N \in BC$ and $I_A P \perp AB$, $P \in AB$. Show that if $I_A M + I_A P... | [
"a) If $I_A$ is the point of intersection of the exterior bisectors of the angles $\\angle B$ and $\\angle C$, then $I_A$ is inside the angle $\\angle A$ and is equidistant from the sides $AB$ and $BC$, respectively of $BC$ and $AC$. Through transitivity, $I_A$ is equidistant from the sides $AB$ and $AC$ of the ang... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Mi... | null | proof only | null | |
05ye | Problem:
Déterminer tous les entiers naturels $n$ tels que $21$ divise $2^{2^{n}} + 2^{n} + 1$. | [
"Solution:\n\nComme $21 = 3 \\times 7$ et que $3$ et $7$ sont premiers entre eux, $21$ divise $2^{2^{n}} + 2^{n} + 1$ si et seulement si $3$ et $7$ divisent $2^{2^{n}} + 2^{n} + 1$.\n\nÉliminons de suite le cas $n = 0$, qui donne $2^{2^{0}} + 2^{0} + 1 = 4$ qui n'est pas divisible par $21$. On peut donc supposer $n... | France | Préparation Olympique Française de Mathématiques - Envoi 3: Arithmétique | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | n ≡ 2 (mod 6) | |
0kpl | Problem:
The English alphabet, which has 26 letters, is randomly permuted. Let $p_{1}$ be the probability that $\mathrm{AB}$, $\mathrm{CD}$, and $EF$ all appear as contiguous substrings. Let $p_{2}$ be the probability that $ABC$ and $DEF$ both appear as contiguous substrings. Compute $\frac{p_{1}}{p_{2}}$. | [
"Solution:\n\nThere are $23!$ ways to arrange the alphabet such that $AB$, $CD$, and $EF$ all appear as contiguous substrings: treat each of these pairs of letters as a single merged symbol, which leaves 23 symbols to permute. Similarly, there are $22!$ ways to arrange the alphabet such that $ABC$ and $DEF$ both ap... | United States | HMMT November 2022 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 23 | |
0ewv | Problem:
In the triangle $ABC$, the length of the altitude from $A$ is not less than $BC$, and the length of the altitude from $B$ is not less than $AC$. Find the angles. | [
"Solution:\n\nLet $k$ be twice the area of the triangle. Then $k \\geq BC^2$, $k \\geq AC^2$ and $k \\leq AC \\cdot BC$, with equality in the last case only if $AC$ is perpendicular to $BC$. Hence $AC$ and $BC$ have equal lengths and are perpendicular. So the angles are $90^\\circ$, $45^\\circ$, $45^\\circ$."
] | Soviet Union | 4th ASU | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof and answer | 90°, 45°, 45° | |
0bl3 | A triangle $ABC$ has $m(\widehat{BCA}) = 30^\circ$ and there exists a point $D \in AC$ such that $m(\widehat{DBC}) = 10^\circ$ and $AC = BD$. Find $m(\widehat{CAB})$.
Marius Stănean | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 100° | |
0g9x | 設 $\lambda > 0$ 為滿足方程式 $\lambda = \lambda^{2/3} + 1$ 的正實數。證明：存在正整數 $M$ 使得
$$
|M - \lambda^{300}| < 4^{-100}.
$$
Let $\lambda > 0$ be a positive real number satisfying $\lambda = \lambda^{2/3} + 1$. Show that there exists a positive integer $M$ such that
$$
|M - \lambda^{300}| < 4^{-100}.
$$ | [
"令 $\\lambda = t^{3/2}$。於是 $t^3 = (t+1)^2$, 且 $\\lambda^{300} = t^{450}$。檢驗方程式 $P(x) = x^3 - (x+1)^2$ 只有唯一的零點 $x=t$, 且因 $P(2) < 0$, 所以 $t > 2$。設 $a, b$ 為 $P(x) = 0$ 的另外兩根。注意到 $|ab| = 1$, 且 $a$ 為 $b$ 的共軛複數。故\n$$\n|a| = |b| = |t|^{-1/2} < 2^{-1/2}.\n$$\n現取 $M = a^{450} + b^{450} + t^{450}$。因為 $a, b, t$ 是首么整係數三次方程式 $P... | Taiwan | 二〇一六數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
0d40 | Let $D$ be the midpoint of side $B C$ of triangle $A B C$ and $E$ the midpoint of median $A D$. Line $B E$ intersects side $C A$ at $F$. Prove that the area of quadrilateral $C D E F$ is $\frac{5}{12}$ the area of triangle $A B C$. | [
"Let $K$ be the midpoint of $A B$ and $L$ the intersection point of $D K$ with $B E$.\n\n\n\nBecause $D$ and $K$ are midpoints of $B C$ and $B A$, respectively, segment $D K$ is parallel to side $A B$ and we have\n$$\n\\frac{E F}{L E}=\\frac{E A}{D E}=1 .\n$$\nBecause $E$ and $K$ are midpoi... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English, Arabic | proof only | null | |
0c9a | A computer program reads the numbers displayed, identifies the selected ones and does one of the following actions:
- If $A$ is pressed, every selected number is changed to its successor;
- If $B$ is pressed, every selected number is changed to its triple.
On Andrei and Bogdan's computer screens are displayed the numb... | [
"a) At every step of type $A$, the sum of the displayed numbers increases by $5$. The remainder modulo $5$ of the sum is invariant.\nThe sum of four consecutive powers of $3$ is $3^n + 3^{n+1} + 3^{n+2} + 3^{n+3} = 3^n(1 + 3 + 9 + 27) = 40 \\cdot 3^n$, therefore initially we have the remainder zero modulo $5$ and i... | Romania | RMC 2020 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a) No; b) 38 | |
0avi | Problem:
Find the minimum value of $x^{2}+4 y^{2}-2 x$, where $x$ and $y$ are real numbers that satisfy $2 x+8 y=3$. | [
"Solution:\nBy the Cauchy-Schwarz Inequality,\n$$\n[2(x-1)+4(2 y)]^{2} \\leq (2^{2}+4^{2})\\left[(x-1)^{2}+4 y^{2}\\right]\n$$\nNow, with $2(x-1)+4(2 y)=2 x+8 y-2=3-2=1$, we obtain\n$$\n\\begin{aligned}\nx^{2}+4 y^{2}-2 x & =(x-1)^{2}+4 y^{2}-1 \\\\\n& \\geq \\frac{[2(x-1)+4(2 y)]^{2}}{2^{2}+4^{2}}-1 \\\\\n& =\\fra... | Philippines | 18th PMO National Stage Oral Phase | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | -19/20 | |
0ii1 | Problem:
Find all positive integers $n > 1$ for which $\frac{n^{2} + 7n + 136}{n-1}$ is the square of a positive integer. | [
"Solution:\n\nWrite $\\frac{n^{2} + 7n + 136}{n-1} = n + \\frac{8n + 136}{n-1} = n + 8 + \\frac{144}{n-1} = 9 + (n-1) + \\frac{144}{n-1}$.\n\nWe seek to find $p$ and $q$ such that $pq = 144$ and $p + q + 9 = k^{2}$.\n\nThe possibilities are seen to be $1 + 144 + 9 = 154$, $2 + 72 + 9 = 83$, $3 + 48 + 9 = 60$, $4 + ... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | n = 5, 37 | |
07ae | Suppose $m$ and $n$ are two nonnegative integers. In the Philosopher's Chess, the chessboard is an infinite array of same regular hexagon cells. The Phoenix piece, which is a special piece in this kind of chess, moves as follows:
At first, the Phoenix selects one of the six directions and moves $m$ cells in that direct... | [
"By considering centers of the hexagons, we get the following equivalent problem:\n\"Two vertices $\\alpha$ and $\\beta$ of the triangular lattice are called equivalent if $\\alpha - \\beta$ is equal to sum of finitely many vectors from the set:\n$$\nA = \\{\\pm(n\\vec{i} - m\\vec{j}), \\pm(n\\vec{k} + m\\vec{i}), ... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Linear Algebra > Vectors",
"Algebra > Linear Algebra > Determinants",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | m^2 + mn + n^2 | |
022k | Problem:
Uma urna contém $k$ bolas marcadas com $k$, para todo $k=1,2, \ldots, 2016$. Qual é o número mínimo de bolas que devemos retirar, sem reposição e sem olharmos as bolas, para termos certeza de que teremos 12 bolas com o mesmo número? | [
"Solution:\n\nSomemos a maior quantidade de bolas que podem ser retiradas de cada tipo sem que obtenhamos 12 bolas de cada cor:\n$$\n1+2+3+4+5+6+7+8+9+10+11+\\underbrace{11+11+\\ldots+11}_{2005 \\text{ vezes}}=22121\n$$\nAssim, é possível que tenhamos azar e retiremos tal quantidade de bolas sem obtermos 12 bolas d... | Brazil | NÍVEL 3 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 22122 | |
0b6c | Prove that every continuous function $f : \mathbb{R} \to \mathbb{R}$ with the property
$$
f(x) - e^{f(x)} = x - e^x, \text{ for every } x \in \mathbb{R},
$$
is differentiable.
Do there exist functions with the above property, having a finite set of points of continuity? | [] | Romania | Shortlisted Problems for the Romanian NMO | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof only | null | |
0fqs | Problem:
En el triángulo escaleno $ABC$, la bisectriz del ángulo $A$ corta al lado $BC$ en el punto $D$. Las rectas que pasan por $D$ y son tangentes a las circunferencias circunscritas de los triángulos $ABD$ y $ACD$ cortan a las rectas $AC$ y $AB$ en los puntos $E$ y $F$, respectivamente. Si $BE$ y $CF$ se cortan en... | [
"Solution:\n\nSe verifican las igualdades de ángulos $\\angle ADE=\\angle B$ y $\\angle ADF=\\angle C$, por ser ángulos semiinscritos en las circunferencias $(ABD)$ y $(ADC)$, respectivamente. Por tanto el cuadrilátero $AFDE$ es inscriptible y de eso se\n\ndeduce que\n$$\n\\angle AFE=\\angl... | Spain | LV Olimpiada matemática Española | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0glk | Let $P(x) = a_{2016}x^{2016} + a_{2015}x^{2015} + \dots + a_1x + a_0$ be a degree $2016$ polynomial, where its coefficients satisfy
$$
|a_1 + a_3 + \dots + a_{2015}| > |a_0 + a_2 + \dots + a_{2016}|.
$$
Prove that the number of roots of $P(x)$ with absolute value less than $1$ is odd (roots are counted with multiplicit... | [
"The given inequality is equivalent to\n$$\n(a_0 + a_2 + \\dots + a_{2016})^2 - (a_1 + a_3 + \\dots + a_{2015})^2 < 0,\n$$\n\nor\n$$\n(a_0 - a_1 + a_2 - a_3 + \\dots + a_{2016})(a_0 + a_1 + \\dots + a_{2016}) < 0,\n$$\nor equivalently\n$$\nP(-1)P(1) < 0. \\qquad (5)\n$$\n\nSince the coefficients of $P(x)$ are reals... | Thailand | The 13th Thailand Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
084c | Problem:
Quanti sono gli interi compresi tra 1 e 2005 (inclusi) che hanno un numero dispari di cifre pari? | [
"Solution:\n\nLa risposta è 1002. Se la cifra delle unità di un intero $n$ è pari, allora la cifra delle unità di $n+1$ è dispari, mentre tutte le altre cifre di $n$ e $n+1$ sono uguali. Perciò, se $n$ ha un numero pari di cifre pari, allora $n+1$ ne ha un numero dispari; e se $n$ ha un numero dispari di cifre pari... | Italy | Progetto Olimpiadi di Matematica 2005 GARA di SECONDO LIVELLO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 1002 | |
0e8y | In the unit square $ABCD$, the two rays from the vertex $A$ divide the right angle in three: the middle angle measures $30^\circ$ and the other two angles are of the same size. One of the rays intersects the side $BC$ at the point $T$. Find the length of the line segment $BT$.
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{3}}{2... | [
"From the given data we determine $\\angle BAT = 30^\\circ$. Since $\\tan(\\angle BAT) = \\frac{|TB|}{|AB|}$, we have $|TB| = |AB| \\tan 30^\\circ = 1 \\cdot \\frac{\\sqrt{3}}{3} = \\frac{\\sqrt{3}}{3}$. The correct answer is $D$.\n\n"
] | Slovenia | National Math Olympiad 2013 - First Round | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | D | |
0c87 | If $x$, $y$ and $z$ are real numbers such that $x^2 + y^2 + z^2 = 2$, prove that $x + y + z \le xyz + 2$. | [
"Notice that $2xy \\le x^2 + y^2 \\le x^2 + y^2 + z^2 = 2$, therefore $xy \\le 1$. Similarly, $xz \\le 1$, and $yz \\le 1$. We also have $(x+y)^2 = x^2 + y^2 + 2xy \\le 4$, so $x+y \\le |x+y| \\le 2$. Also, $x+z \\le 2$, and $y+z \\le 2$. If one of the numbers is negative, let's say $z$, then the conclusion follows... | Romania | SELECTION TESTS FOR THE 2019 JBMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0fcu | Problem:
Sea $E$ una elipse y consideremos tres rectas paralelas $r_{1}, r_{2}$ y $r_{3}$, cada una de las cuales corta a $E$ en dos puntos distintos. Sean estos puntos $A_{1}, B_{1}, A_{2}, B_{2}$ y $A_{3}, B_{3}$, respectivamente. Probar que los puntos medios de los segmentos $A_{1} B_{1}, A_{2} B_{2}$ y $A_{3} B_{3... | [
"Solution:\n\nEl resultado es inmediato en el caso de que la elipse sea una circunferencia. Estas tres rectas determinan tres cuerdas y sus puntos medios son los puntos de corte de las rectas con un diámetro perpendicular a todas ellas. En otro caso, pensando en la elipse como la intersección de un cono con un plan... | Spain | null | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0h69 | It is known that the arithmetic average of the numbers *a*, *b* is equal to the number *c*, so $c = \frac{1}{2}(a+b)$, and that the harmonious average number of *a*, *c* is equal to the number *b*, so $b = \frac{2}{\frac{1}{a}+\frac{1}{c}}$. Is it necessary that numbers *a*, *b*, *c* are equal?
(Bogdan Rublyov) | [
"Let's rewrite the condition of harmonious average: $b = \\frac{2ac}{a+c}$, and now use the fact $c = \\frac{a+b}{2}$:\n$$\n2a \\cdot \\frac{a+b}{2} = b\\left(a + \\frac{a+b}{2}\\right) \\Leftrightarrow a^2 + ab = ba + \\frac{ab+b^2}{2} \\Leftrightarrow 2a^2 = ba + b^2 \\Leftrightarrow (a-b)(2a+b) = 0.\n$$\nLet's d... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof only | null | |
055b | Define $a_1 = 1$, and for each $n > 1$ let $a_n = n \cdot a_{\lfloor \frac{n}{2} \rfloor}$. Prove that for each $n \ge 12$ we have $a_n > n^2$. | [
"As $a_n = n \\cdot a_{\\lfloor \\frac{n}{2} \\rfloor}$, it suffices to show that for each $n \\ge 12$ we have $a_{\\lfloor \\frac{n}{2} \\rfloor} \\ge n$. By the inequalities $n \\le 2\\lfloor \\frac{n}{2} \\rfloor + 1 < 3\\lfloor \\frac{n}{2} \\rfloor$ this reduces to proving that $a_m \\ge 3m$ for each $m \\ge 6... | Estonia | Open Contests | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0kzl | Problem:
Let $a \star b = a b - 2$. Compute the remainder when $(((579 \star 569) \star 559) \star \cdots \star 19) \star 9$ is divided by $100$. | [
"Solution:\nNote that\n$$\n(10 a + 9) \\star (10 b + 9) = (100 a b + 90 a + 90 b + 81) - 2 \\equiv 90(a + b) + 79 \\pmod{100}\n$$\nso throughout our process all numbers will end in $9$, so we will just track the tens digit. Then the \"new operation\" is\n$$\na \\dagger b \\equiv -(a + b) + 7 \\pmod{10}\n$$\nwhere $... | United States | HMMT February 2024 Guts Round | [
"Number Theory > Modular Arithmetic"
] | null | final answer only | 29 | |
01bx | Determine all quadruples of consecutive positive integers such that three of the numbers are sides and one the area of a right triangle. (Units of length and area are compatible.) | [
"**Answer:** $(3, 4, 5, 6)$.\n\nLetting $a$ and $b$ be the legs and $c$ the hypotenuse, we have $a^2 + b^2 = c^2$. At least two of the numbers $a$, $b$, $c$ are consecutive, so they are of different parity. It is not possible that only one of the numbers is odd. So there are two odd numbers, and one of them has to ... | Baltic Way | Baltic Way | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Geometry > Plane Geometry > Triangles",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (3, 4, 5, 6) | |
04t5 | Find the smallest positive integer that can be inserted between numbers $20$ and $16$ so that the resulting number $20\ldots16$ is a multiple of $2016$. (Radek Horenský) | [
"Number $2016$ is a multiple of $9$, therefore the digit sum of the resulting number has to be divisible by $9$. This happens if and only if we insert a number with digit sum divisible by $9$, that is a number which is a multiple of $9$. Trying out the smallest such numbers, we find out that inserting $9$, $18$, an... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 36 | |
0i24 | Problem:
On a certain unidirectional highway, trucks move steadily at $60$ miles per hour spaced $1/4$ of a mile apart. Cars move steadily at $75$ miles per hour spaced $3$ seconds apart. A lone sports car weaving through traffic at a steady forward speed passes two cars between each truck it passes. How quickly is it... | [
"Solution:\n\nThe cars are $1/8$ of a mile apart. Consider the reference frame in which the trucks move at $0$ velocity (and the cars move at $15$). Call the speed of the sports car in this reference frame $v$. The amount of time for the sports car to move from one truck to the next is $\\frac{1/4 \\text{ miles}}{v... | United States | Harvard-MIT Math Tournament | [
"Math Word Problems"
] | null | final answer only | 90 | |
05xp | Problem:
Soient $ABC$ un triangle, avec $AC > AB$, et $\Gamma$ son cercle circonscrit. Soit $T$ le point d'intersection de la tangente à $\Gamma$ en $A$ avec $(BC)$. Soient $M$ le milieu de $[BC]$ et $R$ le symétrique de $A$ par rapport à $B$. Soit $S$ le point tel que $SABT$ est un parallélogramme. La parallèle à $(A... | [
"Solution:\n\n\n\nSoit $N$ le point d'intersection de $(AT)$ et $(BS)$. On a $(PM) \\parallel (AB) \\parallel (ST)$ donc d'après le théorème de Thalès, $\\frac{BP}{BS} = \\frac{BM}{BT}$. Alors $\\frac{BP}{BN} = \\frac{BP}{\\frac{BS}{2}} = 2 \\frac{BP}{BS} = \\frac{2BM}{BT} = \\frac{BC}{BT}$... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ib4 | Problem:
In a game similar to three card monte, the dealer places three cards on the table: the queen of spades and two red cards. The cards are placed in a row, and the queen starts in the center; the card configuration is thus $RQR$. The dealer proceeds to move. With each move, the dealer randomly switches the cente... | [
"Solution:\n\nIf the probability that the queen is the center card after move $n$ is $p_{n}$, then the probability that the queen is an edge card is $1-p_{n}$, and the probability that the queen is the center card after move $n+1$ is $p_{n+1} = \\frac{1-p_{n}}{2}$. This recursion allows us to calculate the first fe... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof and answer | 1/3 + 1/(3*2^2003) | |
0fv1 | Problem:
Im Dreieck $A B C$ sei $D$ der Mittelpunkt der Seite $B C$ und $E$ die Projektion von $C$ auf $A D$. Angenommen es gelte $\angle A C E=\angle A B C$. Zeige, dass das Dreieck $A B C$ gleichschenklig oder rechtwinklig ist. | [
"Solution:\n\nWir zeigen, dass entweder $A B=A C$ oder $\\angle C A B=90^{\\circ}$ gilt. Es sei $F$ der Schnittpunkt von $A D$ mit der Rechtwinkligen zu $A C$ durch $C$. Nun gilt nach Voraussetzung\n$$\n\\angle A F C=90^{\\circ}-\\angle F A C=\\angle A C E=\\angle A B C\n$$\nSomit liegen die Punkte $A, B, F, C$ auf... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
093w | Problem:
Given a positive integer $n$, we say that a polynomial $P$ with real coefficients is $n$-pretty if the equation $P(\lfloor x\rfloor)=\lfloor P(x)\rfloor$ has exactly $n$ real solutions. Show that for each positive integer $n$
(a) there exists an $n$-pretty polynomial;
(b) any $n$-pretty polynomial has a degree... | [
"Solution:\nWe begin by making some preliminary observations. Let $P$ be a real polynomial. We associate to it the sets\n$$\nS(P)=\\{x \\in \\mathbb{R} \\mid P(\\lfloor x\\rfloor)=\\lfloor P(x)\\rfloor\\} \\text{ and } I(P)=\\{x \\in \\mathbb{Z} \\mid P(x) \\in \\mathbb{Z}\\}\n$$\nThen it is easily seen that\n$$\nS... | Middle European Mathematical Olympiad (MEMO) | 15th Middle European Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
07pl | Find all solutions in positive integers of the equation
$$
x^{4} + y^{4} + z^{4} + 1008 = 2x^{2}y^{2} + 2y^{2}z^{2} + 2z^{2}x^{2}.
$$ | [
"Rewrite the equation as a quadratic in $x^2$\n$$\n\\begin{aligned}\n-1008 &= x^4 + y^4 + z^4 - 2x^2y^2 - 2y^2z^2 - 2z^2x^2 \\\\\n&= x^4 - 2(y^2 + z^2)x^2 + (y^2 - z^2)^2.\n\\end{aligned}\n$$\nThe discriminant is equal to $(y^2 + z^2)^2 - (y^2 - z^2)^2 = 4y^2z^2 = (2yz)^2$, hence the roots $x^2$ of the right hand s... | Ireland | Ireland | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | [[8, 8, 2], [8, 2, 8], [2, 8, 8], [6, 4, 4], [4, 6, 4], [4, 4, 6]] | |
08vh | On a triangle $ABC$, a point $D$ is picked on the side $AB$ and a point $E$ is picked on the side $AC$ in such a way that the line $DE$ is parallel to the line $BC$. Let $M$ and $N$ be the mid-points of the line segments $BD$ and $CE$, respectively. Determine the area of the triangle $ADE$ if the areas of the quadrilat... | [
"Since the lines $BC$ and $DE$ are parallel and since $M$ and $N$ are mid-points of the line segments $DB$ and $EC$, respectively, it is easy to see that the line $MN$ is also parallel to the line $BC$ (and $DE$). Therefore, the triangles $ADE$, $AMN$ and $ABC$ are similar. If we let $a = DE$, $b = BC$ and let $h$ ... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 1/8 | |
09be | a, b, c нь гурвалжны талын уртууд бол
$$
3(ab + bc + ca) \le (a + b + c)^2 < 4(ab + bc + ca)
$$
тэнцэтгэл биш биелэхийг батал. | [
"△ Батлах тэнцэтгэл биш нь\n$$\nab + bc + ca \\leq a^2 + b^2 + c^2 < 2(ab + bc + ca)\n$$\nтэнцэтгэл бишийг батлахтай эквивалент юм.\n$$\nab + bc + ca \\leq a^2 + b^2 + c^2 \\iff \\frac{a^2 + b^2}{2} + \\frac{b^2 + c^2}{2} + \\frac{c^2 + a^2}{2} \\geq ab + bc + ca \\text{ болно.}\n$$\nЭнд $\\frac{x^2+y^2}{2} \\ge xy... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | Mongolian | proof only | null | |
0bxx | Determine the number of positive integers, which written in base 10, satisfy the following conditions simultaneously:
(i) they are 6-digit numbers;
(ii) the product of their non-zero digits is 84;
(iii) four of their digits are 2, 0, 1, 7. | [
"Since $84 : (2 \\cdot 1 \\cdot 7) = 6$, we distinguish the following cases:\n\n(1) The digits are: $2, 0, 1, 7, 0, 6$.\nIf the first digit is $1$, the other $5$ digits can be arranged in $(5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1) : 2 = 60$ ways. Similar, if the first digit is $2$, $6$ or $7$. Thus we find $240$ numb... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 840 | |
01l4 | Given a $3n \times 3n$ table with one of two signs "+" or "-" in any of its cells. Per move one can replace the signs in all cells of some row (or of some column) by the opposite signs. At the beginning there is exactly one minus in the table (all other signs are pluses). After some moves the table with exactly 36 minu... | [
"Answer $n = 3$.\n\nNote that if changing of signs is performed even number of times to some row (column), then it is equivalent that the operation is not applied at all. So we can assume that the operation is applied exactly once to some rows (columns), and is not applied to the remaining rows (columns). Let the o... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | n = 3 | |
0kgt | Problem:
Let $n$ be the answer to this problem. Suppose square $ABCD$ has side-length $3$. Then, congruent non-overlapping squares $EHGF$ and $IHJK$ of side-length $\frac{n}{6}$ are drawn such that $A$, $C$, and $H$ are collinear, $E$ lies on $BC$ and $I$ lies on $CD$. Given that $AJG$ is an equilateral triangle, then ... | [
"Solution:\n\n\n\nThe fact that $EHGF$ and $IHJK$ have side length $n / 6$ ends up being irrelevant.\nSince $A$ and $H$ are both equidistant from $G$ and $J$, we conclude that the line $ACHM$ is the perpendicular bisector of $GJ$.\nNow, define the point $C'$ so that the spiral similarity ce... | United States | HMMT November 2021 | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 48 | |
022r | Problem:
Uma desigualdade especial- Quais valores de $x$ satisfazem $x^{2}<|x|+2$ ?
(a) $x<-1$ ou $x>1$
(b) $x>1$
(c) $-2<x<2$
(d) $x<-2$
(e) $x<0$ | [
"Solution:\n\nObservemos que se um número $a$ satisfaz a desigualdade, então $-a$ também satisfaz a desigualdade, logo os valores que satisfazem a desigualdade formam um conjunto simétrico, portanto basta considerar o caso em que $x$ é positivo. Mas, $(2-x)(1+x)=x+3-x^{2}>0$ é positivo se $2-x$ é positivo, portanto... | Brazil | null | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | c | |
0d2z | Find all polynomials $p(x)$ with integer coefficients such that for each positive integer $n$, the number $2^{n}-1$ is divisible by $p(n)$. | [
"Suppose there is some value of $n$ such that $p(n) \\neq \\pm 1$. Let $q$ be a prime divisor of $p(n)$. Because $q = (n+q) - n$ divides $p(n+q) - p(n)$, we deduce that $q$ divides $p(n+q)$. Therefore, $q$ divides both $2^{n}-1$ and $2^{n+q}-1$, which implies in particular that $q$ is an odd prime. We have\n$$\n1 \... | Saudi Arabia | Selection tests for the International Mathematical Olympiad 2013 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | p(x) = 1 or p(x) = -1 | |
0avl | Problem:
Let $n$ be any positive integer. Prove that
$$
\sum_{i=1}^{n} \frac{1}{\left(i^{2}+i\right)^{3 / 4}}>2-\frac{2}{\sqrt{n+1}}
$$ | [
"Solution:\nBy the AM-GM Inequality, we have for each $i$,\n$$\n\\frac{(i+1) \\sqrt{i}+i \\sqrt{i+1}}{2}>\\sqrt{(i+1)^{3 / 2} \\cdot i^{3 / 2}}=\\left(i^{2}+i\\right)^{3 / 4}\n$$\nNote that this inequality is strict, as $(i+1) \\sqrt{i}$ cannot be equal to $i \\sqrt{i+1}$. Thus,\n$$\n\\begin{aligned}\n\\sum_{i=1}^{... | Philippines | 18th Philippine Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
018u | Let $A$ and $B$ be two circles, external to each other. Let $\ell$ be a line not meeting the circles. For any point $X$ on $\ell$, let $E$ be a point of contact of a tangent to $A$ through $X$, and $F$ a point of contact of a tangent to $B$ through $X$. Find the position of $X$ on $\ell$ such that $EX + FX$ is minimize... | [
"Denote the centres of $A$ and $B$ by $A$ and $B$, respectively. Let $C$ and $D$ be the feet of the perpendiculars from $A$ and $B$ to $\\ell$. Let $G$ be a point of contact of $A$ and a tangent to $A$ from $C$. Define $H$ similarly on $B$.\n\nNow, by Pythagoras,\n$$\nXE^2 = AX^2 - AE^2 = X... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof and answer | Let C and D be the feet of the perpendiculars from the centers of the circles to the line. From C and D draw tangents to the respective circles, meeting them at G and H. Choose M on CA with CM equal to CG and N on DB with DN equal to DH, with M and N on opposite sides of the line. The minimizing point X is P = l ∩ MN. | |
0fh7 | Problem:
Se llama parte entera de un número real $a$ (y se escribe $[a]$), al mayor número entero menor o igual que $a$. Si $n$ es un número natural, demostrar que la parte entera de $(4+\sqrt{11})^{n}$ es un número impar. | [
"Solution:\n\nComo $0 < 4 - \\sqrt{11} < 1$, resulta que $0 < (4 - \\sqrt{11})^{n} < 1$.\n\nConsideremos la suma\n$$\nE = (4 + \\sqrt{11})^{n} + (4 - \\sqrt{11})^{n}\n$$\nal desarrollar el segundo miembro mediante el binomio de Newton se observa que se obtiene un número par, $2k$, y al haberse obtenido sumando a $(... | Spain | OME 26 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
01aw | Find all pairs of natural numbers $p > q$ such that
$$
\frac{p+1}{p} \cdot \frac{q+1}{q} = \frac{2013}{2011}.
$$ | [
"As $2011$ is a prime, then either $p$ or $q$ is divisible by $2011$. Assume that $2011 \\mid p$, then $p = 2011k$ for some $k$. Substituting that we get:\n$$\n\\frac{2011k + 1}{2011k} \\cdot \\frac{q + 1}{q} = \\frac{2013}{2011},\n$$\nwhat leads to\n$$\n\\frac{q+1}{q} = \\frac{2013k}{2011k+1} \\Rightarrow \\frac{1... | Baltic Way | Baltic Way 2013 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | (p,q) = (2012, 2011), (4022, 1341), (12066, 1097), (34187, 1036), (62341, 1022), (185012, 1011), (675696, 1007), (2025077, 1006) | |
013k | Problem:
Three circular arcs $w_{1}$, $w_{2}$, $w_{3}$ with common endpoints $A$ and $B$ are on the same side of the line $AB$; $w_{2}$ lies between $w_{1}$ and $w_{3}$. Two rays emanating from $B$ intersect these arcs at $M_{1}$, $M_{2}$, $M_{3}$ and $K_{1}$, $K_{2}$, $K_{3}$, respectively. Prove that
$$
\frac{M_{1} ... | [
"Solution:\n\nFrom inscribed angles we have $\\angle A K_{1} B = \\angle A M_{1} B$ and $\\angle A K_{2} B = \\angle A M_{2} B$. From this it follows that $\\triangle A K_{1} K_{2} \\sim \\triangle A M_{1} M_{2}$, so\n$$\n\\frac{K_{1} K_{2}}{M_{1} M_{2}} = \\frac{A K_{2}}{A M_{2}}\n$$\nSimilarly $\\triangle A K_{2}... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00pt | Let $a$, $b$, $c$ and $d$ be positive real numbers such that $abcd = \frac{1}{4}$. Prove that
$$
\left(16ac + \frac{a}{c^2b} + \frac{16c}{a^2d} + \frac{4}{ac}\right) \left(bd + \frac{b}{256d^2c} + \frac{d}{b^2a} + \frac{1}{64bd}\right) \ge \frac{81}{4}.
$$
When does the equality hold? | [
"First we observe that\n$$\n16ac + \\frac{a}{c^2b} + \\frac{16c}{a^2d} + \\frac{4}{ac} = \\left(a + \\frac{1}{a^2d}\\right) \\left(16c + \\frac{1}{c^2b}\\right)\n$$\n$$\n\\text{and } bd + \\frac{b}{256d^2c} + \\frac{d}{b^2a} + \\frac{1}{64bd} = \\left(b + \\frac{1}{b^2a}\\right) \\left(d + \\frac{1}{256d^2c}\\right... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds at a = 2, b = 1, c = 1/2, d = 1/4. | |
00in | A sequence $\langle a_n \rangle$ with $a_n = a + nd$ is called an arithmetic sequence. The sequence $\langle b_n \rangle$ with $b_n = \sum_{k=0}^{n} a_k$ is called an arithmetic sequence of second degree. Let $a$ and $d$ be positive integers.
We consider all such arithmetic sequences of second degree containing the nu... | [
"Since $a_k = a + k d$, we have $b_n = (n+1) \\cdot a + \\frac{n(n+1)}{2} \\cdot d$. If we assume $b_n = 2010$, we therefore obtain\n$$\na = \\frac{4020 - n(n + 1)d}{2(n + 1)} = \\frac{2010}{n + 1} - \\frac{dn}{2}.\n$$\nSince both $a$ and $d$ are positive, we see from the first fraction, that $n(n + 1) < 4020$ must... | Austria | Austria 2010 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | n = 59; a = 4; d = 1; the arithmetic sequence is 4, 5, 6, … | |
0i9q | Problem:
Compute the surface area of a cube inscribed in a sphere of surface area $\pi$. | [
"Solution:\nThe sphere's radius $r$ satisfies $4 \\pi r^{2} = \\pi \\Rightarrow r = 1/2$, so the cube has body diagonal $1$, hence side length $1/\\sqrt{3}$. So, its surface area is $6(1/\\sqrt{3})^{2} = 2$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > 3D Shapes"
] | null | final answer only | 2 | |
02yd | Problem:
Dois estudantes precoces do Nível 3 participaram de um torneio de xadrez universitário. Cada participante joga contra todos os outros exatamente uma vez. Uma vitória vale 1 ponto, um empate vale 0,5 ponto e uma derrota vale 0 ponto. A soma das pontuações dos dois estudantes do Nível 3 é 6,5. Todos os estudant... | [
"Solution:\n\nSeja $x$ a quantidade de estudantes universitários e $p$ a pontuação comum a todos eles. Como em cada jogo é disputado exatamente 1 ponto, segue que o total de pontos do torneio, que é $6,5 + p x$, coincide com o número de jogos, que é $\\frac{(x+2)(x+1)}{2}$. Além disso, a pontuação de cada participa... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 11 | |
057a | One writes distinct positive integers into the cells of a $3 \times 3$ table in such a way that, in each row and in each column, one number equals the sum of the other two numbers. Find the least possible total sum of the numbers written into the table. | [
"The sum of all numbers written in the cells is at least $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$. As one number is the sum of others in every row, the sum of all numbers in every row is even. Thus the sum of all numbers in the table must be even, too. Hence the sum of all numbers in the table cannot be 45. It is p... | Estonia | Open Contests | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 46 | |
027v | Problem:
São dadas 2017 retas separadas em três conjuntos de modo que retas em um mesmo conjunto são paralelas entre si. Qual é o maior número possível de triângulos que podemos formar com vértices nestas retas? | [
"Solution:\n\nSejam $a \\geq b \\geq c$ as quantidades de retas nos três conjuntos. Então $a+b+c=2017$ e o número de triângulos que podem ser formados é $a \\cdot b \\cdot c$, pois retas em um mesmo conjunto não se intersectam. Assim, queremos maximizar o produto anterior. Se $a>c+1$, podemos diminuir uma reta do c... | Brazil | NÍVEL 3 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 673 * 672^2 | |
0f3y | Problem:
Is it possible to color half the cells in a rectangular array white and half black so that in each row and column more than $\frac{3}{4}$ of the cells are the same color? | [] | Soviet Union | 15th ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | No | |
0i75 | Problem:
Given a $9 \times 9$ chess board, we consider all the rectangles whose edges lie along grid lines (the board consists of 81 unit squares, and the grid lines lie on the borders of the unit squares). For each such rectangle, we put a mark in every one of the unit squares inside it. When this process is complete... | [
"Solution:\n\n56. Consider the rectangles which contain the square in the $i$\\text{th} row and $j$\\text{th} column. There are $i$ possible positions for the upper edge of such a rectangle, $10-i$ for the lower edge, $j$ for the left edge, and $10-j$ for the right edge; thus we have $i(10-i) j(10-j)$ rectangles al... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 56 | |
0g6r | 令 $N$ 表示所有正整數所成之集合。試求滿足下列條件的最大正整數 $k$:
$N$ 可以被分割成 $k$ 個子集合 $A_1, A_2, \dots, A_k$ 使得對所有的整數 $n \ge 15$ 與所有的 $i \in \{1, 2, \dots, k\}$, 能夠在 $A_i$ 中找到兩個相異元素 $a, b$ 且 $a+b=n$. | [
"答 $k=3$. 例如：取\n$$\nA_1 = \\{1, 2, 3\\} \\cup \\{3m\\mid m \\ge 4\\}, \\quad A_2 = \\{4, 5, 6\\} \\cup \\{3m-1\\mid m \\ge 4\\}, \\quad A_3 = \\{7, 8, 9\\} \\cup \\{3m-2\\mid m \\ge 4\\}.\n$$\n爲了驗證上述之分割是符合題意，首先觀察：$A_i$ 中的兩相異元素 $a, b$ 之和的情形：\n(i) $a+b=n \\ge 1+12=13$, 當 $i=1$,\n(ii) $a+b=n \\ge 4+11=15$, 當 $i=2$,\n(... | Taiwan | 二〇一二數學奧林匹亞競賽第二階段選訓營 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 3 | |
0d7p | Given two positive integers $r > s$, and let $\mathscr{F}$ be an infinite family of sets, each of size $r$, no two of which share fewer than $s$ elements. Prove that there exists a set of size $r-1$ that shares at least $s$ elements with each set in $\mathscr{F}$. | [
"We call a set $S$ s-meets $\\mathscr{F}$ if it shares at least $s$ elements with each set in $\\mathscr{F}$. Suppose no such set of size (at most) $r-1$ exists. (Each $S \\in \\mathscr{F}$ s-meets $\\mathscr{F}$ by the problem hypothesis.)\n\nLet $T$ be a maximal set such that $T \\subseteq S$ for infinitely many ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0i62 | Problem:
Two integers are relatively prime if they don't share any common factors, i.e. if their greatest common divisor is $1$. Define $\varphi(n)$ as the number of positive integers that are less than $n$ and relatively prime to $n$. Define $\varphi_{d}(n)$ as the number of positive integers that are less than $d n$... | [
"Solution:\n\nFor fixed $n$, the pattern of integers relatively prime to $n$ repeats every $n$ integers, so $\\varphi_{d}(n)=d \\varphi(n)$. Therefore the expression in the problem equals $\\varphi(n)^{3}$. The cube root of $64000$ is $40$. $\\varphi(p)=p-1$ for any prime $p$. Since $40$ is one less than a prime, t... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof and answer | 41 | |
0gxo | Find all polynomials $P(x)$ with real coefficients, such that for every distinct natural $x, y, z, t$ satisfying $x^2 + y^2 + z^2 = 2t^2$ and $\text{GCD}\{x, y, z, t\} = 1$, the following equality holds:
$$
2P^2(t) + 2P(xy + yz + zx) = P^2(x + y + z).
$$ | [
"Take $t = 5k$ and $z = 7k$, then we want the equality $x^2 + y^2 = k^2$ to hold. Consider two series of solutions: $x = n^2 - 1$, $y = 2n$, $k = n^2 + 1$ and $x = 3n^2 + 4n + 1$, $y = 2n(2n+1)$, $k = 5n^2 + 4n + 1$. It is evident that in both cases $(x, y, z, t) = 1$ (here and on $(x, y, z, t)$ denotes $\\text{GCD... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | P(x) = x; P(x) = 0; P(x) = -2 | |
0a8m | Problem:
Eva, Per and Anna play with their pocket calculators. They choose different integers and check, whether or not they are divisible by $11$. They only look at nine-digit numbers consisting of all the digits $1,2, \ldots, 9$. Anna claims that the probability of such a number to be a multiple of $11$ is exactly $... | [
"Solution:\n\nWe write the numbers in consideration, $n = a_{0} + 10 a_{1} + 10^{2} a_{2} + \\cdots + 10^{8} a_{8}$, in the form\n$$\n\\begin{gathered}\na_{0} + (11-1) a_{1} + (99+1) a_{2} + (1001-1) a_{3} \\\\\n+ (9999+1) a_{4} + (100001-1) a_{5} + (999999+1) a_{6} \\\\\n\\quad + (10000001-1) a_{7} + (99999999+1) ... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 16 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | Eva | |
0biv | A succession of letters $\overline{l_1l_2\dots l_n}$, where $n \ge 1$, is called a *word*. A word $\overline{l_1l_2\dots l_n}$ is called a *palindrome*, if $l_k = l_{n-k+1}$, for each $k \in \{1, 2, \dots, n\}$.
Consider a two-letter word $X = \overline{l_1l_2\dots l_{2014}}$, with letters $A$ and/or $B$. Prove that $X... | [
"Let us split $X$ into groups of 5 consecutive letters; this way we obtain 402 groups of 5 letters and another incomplete group of 4 letters.\n\nConsider a 5 letter-word $Y$, written only with $A$'s and $B$'s, whose first letter is $A$. So $Y$ is one of the sixteen words of the form $A * * * *$. It is easy to check... | Romania | THE 2014 DANUBE MATHEMATICAL COMPETITION | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Other"
] | null | proof only | null | |
0drj | Let $f_0, f_1, \dots$ be the Fibonacci sequence: $f_0 = f_1 = 1$, $f_n = f_{n-1} + f_{n-2}$ if $n \ge 2$. Determine all possible positive integers $n$ so that there is a positive integer $a$ such that $f_n \le a \le f_{n+1}$ and that
$$
a\left(\frac{1}{f_1} + \frac{1}{f_1f_2} + \frac{1}{f_1f_2f_3} + \dots + \frac{1}{f_... | [
"The number may be rewritten as\n$$\n\\frac{a}{f_1 \\cdots f_n} (f_2 \\cdots f_n + f_3 \\cdots f_n + \\cdots + f_{n-2}f_{n-1}f_n + f_{n-1}f_n + 1).\n$$\nIf this is an integer, then\n$$\nf_n|a(f_2 \\cdots f_n + f_3 \\cdots f_n + \\cdots + f_{n-2}f_{n-1}f_n + f_{n-1}f_n + 1).\n$$\nThus $f_n|a$. First consider the cas... | Singapore | Singapur 2015 | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | n = 1, 2 | |
014w | Problem:
For an upcoming international mathematics contest, the participating countries were asked to choose from nine combinatorics problems. Given how hard it usually is to agree, nobody was surprised that the following happened:
- Every country voted for exactly three problems.
- Any two countries voted for differe... | [
"Solution:\n\nCertainly, the 56 three-element subsets of the set $\\{1,2, \\ldots, 8\\}$ would do. Now we prove that 56 is the maximum. Assume we have a maximal configuration. Let $Y$ be the family of the three-element subsets, which were chosen by the participating countries and $N$ be the family of the three-elem... | Baltic Way | Baltic Way 2008 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 56 | |
0a1z | A teacher wants to spend a morning practising Olympiad problems with her class in teams. To this end, she has set up an arrangement with six large tables. At each table several students can work together on a problem. Each table has a problem; there are three different problems, each on two tables. There are three roun... | [
"Suppose problem $A$ is on tables $A1$ and $A2$, problem $B$ is on tables $B1$ and $B2$, and problem $C$ is on tables $C1$ and $C2$. Consider the students sitting at table $A1$ in the first round. In the second round, they have to go to a table that does not have problem $A$ on it, i.e. table $B1$, $B2$, $C1$ or $C... | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 24 | |
07od | The three distinct points $B$, $C$, $D$ are collinear with $C$ between $B$ and $D$. Another point $A$ not on the line $BD$ is such that $|AB| = |AC| = |CD|$.
Prove that $\angle BAC = 36^\circ$ if and only if $\frac{1}{|CD|} - \frac{1}{|BD|} = \frac{1}{|CD| + |BD|}$. | [
"Let $\\alpha$ be the size of the angle $\\angle CAD$. Because $|AB| = |AC| = |CD|$, we have $\\angle ABC = \\angle BCA = \\angle CAD + \\angle CDA = 2\\angle CAD = 2\\alpha$. Therefore, $\\angle BAD = \\angle DBA$ if and only if $\\angle BAC = \\alpha$. On the other hand, because $\\angle BAC = 180^\\circ - 4\\alp... | Ireland | Ireland | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0e5n | Problem:
Poišči vsa 3-mestna naravna števila $n$ z neničelnimi števkami, za katera velja: če številu $n$ odstranimo levo števko, dobimo število, ki deli $n$. | [
"Solution:\n\nPišimo $n=\\overline{a b c}$. Pogoj naloge pove, da $\\overline{b c}$ deli $\\overline{a b c}=100 a+\\overline{b c}$, torej deli tudi $100a$.\n\nDenimo najprej, da $\\overline{b c}$ ni deljivo s $5$. Potem mora $\\overline{b c}$ deliti $4 a$. Ker je $\\overline{b c}$ dvomestno, mora biti tudi $4 a$ dv... | Slovenia | 56. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 125, 225, 312, 315, 325, 375, 416, 425, 525, 612, 615, 624, 625, 675, 714, 725, 728, 735, 816, 825, 832, 912, 915, 918, 925, 936, 945, 975 | |
0flx | Find all continuous functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that $x + \frac{1}{x} = f(x) + \frac{1}{f(x)}$ for every positive real number $x$. | [
"Let $y = f(x)$. The equation is:\n$$\nx + \\frac{1}{x} = y + \\frac{1}{y}\n$$\nBring all terms to one side:\n$$\nx + \\frac{1}{x} - y - \\frac{1}{y} = 0\n$$\nOr:\n$$\nx - y + \\frac{1}{x} - \\frac{1}{y} = 0\n$$\n$$\nx - y + \\frac{y - x}{xy} = 0\n$$\n$$\n(x - y)\\left(1 - \\frac{1}{xy}\\right) = 0\n$$\nSo either $... | Spain | Spanija 2012 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = x for all x > 0, or f(x) = 1/x for all x > 0 | |
0k02 | Problem:
Undecillion years ago in a galaxy far, far away, there were four space stations in the three-dimensional space, each pair spaced $1$ light year away from each other. Admiral Ackbar wanted to establish a base somewhere in space such that the sum of squares of the distances from the base to each of the stations... | [
"Solution:\n\nSet up a coordinate system where the coordinates of the stations are $\\left(\\frac{1}{2 \\sqrt{2}}, \\frac{1}{2 \\sqrt{2}}, \\frac{1}{2 \\sqrt{2}}\\right)$, $\\left(-\\frac{1}{2 \\sqrt{2}},-\\frac{1}{2 \\sqrt{2}}, \\frac{1}{2 \\sqrt{2}}\\right)$, $\\left(\\frac{1}{2 \\sqrt{2}},-\\frac{1}{2 \\sqrt{2}}... | United States | HMMT November 2017 | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | 27√6π/8 | |
05ih | Problem:
Soit $p$ un nombre premier. Trouver tous les entiers $a, b, c \geq 1$ tels que:
$$
a^{p}+b^{p}=p^{c}
$$ | [
"Solution:\n(i) $p=2$ et $(a, b, c)=\\left(2^{u}, 2^{u}, 2u+1\\right)$ pour un entier $u \\geq 0$\n(ii) $p=3$ et $(a, b, c)=\\left(2 \\cdot 3^{u}, 3^{u}, 2+3u\\right)$ ou $(a, b, c)=\\left(3^{u}, 2 \\cdot 3^{u}, 2+3u\\right)$ pour un entier $u \\geq 0$.\n\nPour un entier $n \\geq 1$ et un nombre premier $p$ on note... | France | Olympiades Françaises de Mathématiques - Test de Sélection | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Algebraic properties of ... | null | proof and answer | All solutions are:
- For p = 2: a = 2^u, b = 2^u, c = 2u + 1 for any integer u ≥ 0.
- For p = 3: (a, b, c) = (2*3^u, 3^u, 2 + 3u) or (3^u, 2*3^u, 2 + 3u) for any integer u ≥ 0. | |
04bx | Determine and draw in the complex plane the set of all complex numbers $z$ which satisfy the condition
$$
\text{Re}[(4 + 3i)z^2] \ge 0.
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | All z = x + i y such that 4x^2 − 6xy − 4y^2 ≥ 0. Equivalently, arg(z) ∈ [−arctan(2), arctan(1/2)] modulo π; i.e., the closed double cone between the lines y = −2x and y = (1/2)x, including the boundary lines and the origin. | |
02mp | Problem:
Cinco peças de metal, confeccionadas, respectivamente, de ouro, prata, bronze, platina e níquel, foram colocadas em cinco cofres, numerados de $1$ a $5$. Cada cofre contém uma peça e o problema consiste em descobrir qual peça está em qual cofre. Na porta de cada cofre está escrita uma informação. Das cinco in... | [] | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Logic"
] | null | proof and answer | Chest 1: nickel; Chest 2: silver; Chest 3: bronze; Chest 4: platinum; Chest 5: gold. | |
0g8y | 定義兩個多項式之間的大小 $f(x) \geq g(x)$ 為從最高次項開始向下比較係數大小, 即以下敘述:
若 $f(x) = \sum_{i=0}^{n} a_i x^i$, $g(x) = \sum_{i=0}^{n} b_i x^i$ ($a_n, b_n$ 可以為 0) 且存在 $r$ 使得 $\forall i > r, a_i = b_i, a_r > b_r$, 或是 $f(x) = g(x)$, 則稱 $f(x) \geq g(x)$。
試證：若 $f, g$ 首項係數為正，則 $f(f(x)) + g(g(x)) \geq f(g(x)) + g(f(x))$。 | [
"容易驗證上述定義的 $\\ge$ 滿足左右同加同減某個多項式。\n\n(1). 假設不成立, 則以下證明矛盾。\n\n以下使用 height 歸納法證明 $a_i = b_i, \\forall i = 1, 2, ..., n$。\n\n先證明 $a_n = b_n$: 考慮比較 $x^{n^2}$ 係數, 左式為 $a_n^{n+1} + b_n^{n+1}$, 右式為 $a_n^n b_n + a_n b_n^n$, 左邊 $\\ge$ 右邊且等號成立在 $a_n = b_n$, 但由假設可得等號必須成立。\n\n假設 $a_i = b_i, \\forall i = k, k + 1, ..., n$ 成立, 則原... | Taiwan | 2015 Math Olympiad Second Stage Training Camp | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0aq0 | Problem:
In $\triangle ABC$, $\angle A = 15^\circ$ and $BC = 4$. What is the radius of the circle circumscribing $\triangle ABC$? | [
"Solution:\n\n$4 \\sqrt{2 + \\sqrt{3}}$ or $2(\\sqrt{6} + \\sqrt{2})$\n\nWe use the Extended Law of Sines. Let $R$ be the radius of the circle circumscribing $\\triangle ABC$.\n$$\n2R = \\frac{4}{\\sin 15^\\circ} \\quad \\text{or} \\quad R = \\frac{2}{\\sin 15^\\circ}\n$$\nDepending on the formula used (either half... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | 2(√6 + √2) | |
00gm | Let $x$, $y$ and $z$ be positive real numbers such that $\sqrt{x}+\sqrt{y}+\sqrt{z}=1$. Prove that
$$
\frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}+x y}{\sqrt{2 z^{2}(x+y)}} \geq 1
$$ | [
"We first note that\n$$\n\\begin{align*}\n\\frac{x^{2}+y z}{\\sqrt{2 x^{2}(y+z)}} & =\\frac{x^{2}-x(y+z)+y z}{\\sqrt{2 x^{2}(y+z)}}+\\frac{x(y+z)}{\\sqrt{2 x^{2}(y+z)}} \\\\\n& =\\frac{(x-y)(x-z)}{\\sqrt{2 x^{2}(y+z)}}+\\sqrt{\\frac{y+z}{2}} \\\\\n& \\geq \\frac{(x-y)(x-z)}{\\sqrt{2 x^{2}(y+z)}}+\\frac{\\sqrt{y}+\\... | Asia Pacific Mathematics Olympiad (APMO) | XIX Asian Pacific Mathematics Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0bj6 | Determine all pairs $(a, b)$ of integers which satisfy the equality
$$
\frac{a+2}{b+1} + \frac{a+1}{b+2} = 1 + \frac{6}{a+b+1}.
$$ | [
"Obviously, $b \\ne -2$ and $b \\ne -1$. Adding $2$ to both sides of the equality, we get\n$$\n\\left(\\frac{a+2}{b+1} + 1\\right) + \\left(\\frac{a+1}{b+2} + 1\\right) = 3 + \\frac{6}{a+b+1},\n$$\nhence\n$$\n(a+b+3)\\left(\\frac{1}{b+1} + \\frac{1}{b+2}\\right) = \\frac{3(a+b+3)}{a+b+1}.\n$$\n\n**Case 1.** If $a +... | Romania | 65th NMO Selection Tests for JBMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (a, b) = (-3 - u, u) for all integers u not equal to -2 or -1, and (a, b) = (1, 0). | |
0cdx | Fix a positive integer $n$. Consider an $n$-point set $S$ in the plane. An *eligible* set is a non-empty set of the form $S \cap D$, where $D$ is a closed disc in the plane. In terms of $n$, determine the smallest possible number of eligible subsets $S$ may contain.
Cristian Săvescu | [
"The required minimum is $\\frac{1}{2}n(n+1)$.\n\nWe first show that an $n$-point set $S$ in the plane contains at least $\\frac{1}{2}n(n+1)$ eligible subsets. To this end, consider a line $\\ell$ perpendicular to:\n(1) No line through at least two points in $S$; and\n(2) No tangent of a circle $\\gamma$ through at... | Romania | THE 73rd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS - SECOND SELECTION TEST | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | n(n+1)/2 | |
051y | Find all pairs of real numbers $(x, y)$ that satisfy
$$
\begin{cases} x + \sin x = y, \\ y + \sin y = x. \end{cases}
$$ | [
"By adding the equations and simplifying we get $\\sin x = -\\sin y$. Thus $y = -x + 2k\\pi$ or $y = \\pi + x + 2k\\pi = x + (2k + 1)\\pi$. In the second case we get that $|y - x| = |(2k + 1)\\pi| \\ge \\pi$, but from the first equality $|y - x| = |\\sin x| \\le 1 < \\pi$, a contradiction. Therefore $y = -x + 2k\\p... | Estonia | Final Round of National Olympiad | [
"Precalculus > Functions",
"Precalculus > Trigonometric functions",
"Calculus > Differential Calculus > Derivatives"
] | null | proof and answer | All pairs are (kπ, kπ) for any integer k. | |
0ej5 | Problem:
Vsota števk petmestnega števila je $44$. Koliko je produkt števk tega petmestnega števila?
(A) $2^{3} \cdot 3^{8}$
(B) $2^{3} \cdot 9^{3}$
(C) $8 \cdot 4^{9}$
(D) $8 \cdot 3^{4}$
(E) Nič od predhodno naštetega. | [
"Solution:\n\nVsota $5$ enomestnih števil je enaka $44$ samo, če je eno od teh števil enako $8$ in so ostala štiri števila enaka $9$, torej $8 + 9 + 9 + 9 + 9 = 44$. Produkt je zato enak $8 \\cdot 9^{4} = 2^{3} \\cdot 3^{8}$."
] | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | A | |
0hpt | Problem:
Is there an integer $x$ such that
$$
2010 + 2009 x + 2008 x^{2} + 2007 x^{3} + \cdots + 2 x^{2008} + x^{2009} = 0?
$$ | [
"Solution:\nThe answer is no.\n\nIt is clear that if $x$ is positive, the left side is positive, and if $x = 0$, the left side is $2010$. If $x = -1$, the left side is\n$$\n(2010 - 2009) + (2008 - 2007) + \\cdots + (2 - 1) = 1 + 1 + \\cdots + 1,\n$$\nlikewise a positive number.\n\nIf $x \\leq -2$, we claim that the... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | No | |
0jlp | Problem:
Let $A, B, C, D, E$ be five points on a circle; some segments are drawn between the points so that each of the $\binom{5}{2}=10$ pairs of points is connected by either zero or one segments. Determine the number of sets of segments that can be drawn such that:
- It is possible to travel from any of the five poi... | [
"Solution:\nAnswer: 195\n\nFirst we show that we can divide the five points into sets $S$ and $T$ according to the second condition in only one way. Assume that we can divide the five points into $S \\cup T$ and $S' \\cup T'$. Then, let $A = S' \\cap S$, $B = S' \\cap T$, $C = T' \\cap S$, and $D = T' \\cap T$. Sin... | United States | HMMT November 2014 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 195 | |
0jeu | Problem:
Sherry and Val are playing a game. Sherry has a deck containing $2011$ red cards and $2012$ black cards, shuffled randomly. Sherry flips these cards over one at a time, and before she flips each card over, Val guesses whether it is red or black. If Val guesses correctly, she wins $1$ dollar; otherwise, she lo... | [
"Solution:\n\nAnswer: $\\frac{1}{4023}$\n\nWe will prove by induction on $r+b$ that the expected profit for guessing if there are $r$ red cards, $b$ black cards, and where $g$ guesses must be red, is equal to $(b-r)+\\frac{2(r-b)}{(r+b)} g$.\n\nIt is not difficult to check that this holds in the cases $(r, b, g) = ... | United States | HMMT 2013 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | final answer only | 1/4023 | |
0023 | Leonardo pensó un número entero entre $1$ y $2003$ inclusive, y Julián tiene que adivinar ese número. Para ello puede formularle a Leonardo preguntas que se puedan responder con sí o no. Leonardo tiene obligación de responder todas las preguntas, pero, si lo desea, puede mentir como mucho una vez. (Algunas preguntas po... | [] | Argentina | XX Olimpiada Matemática Argentina | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Logic"
] | español | proof only | null |
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