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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0fi1 | Problem:
Demostrar que en el caso de que las ecuaciones
$$
\begin{aligned}
& x^{3}+m x-n=0 \\
& n x^{3}-2 m^{2} x^{2}-5 m n x-2 m^{3}-n^{2}=0
\end{aligned}
$$
$(n \neq 0)$, tengan una raíz común, la primera tendrá dos raíces iguales, y determinar entonces las raíces de las dos ecuaciones en función de $n$. | [
"Solution:\nSea $\\alpha$ la raíz común de ambas ecuaciones. Entonces\n$$\n\\alpha^{3}+m \\alpha=n\n$$\ny sustituyendo en la segunda ecuación se obtiene, tras hacer operaciones:\n$$\n6 m \\alpha^{4}+8 m^{2} \\alpha^{2}+2 m^{3}=0\n$$\nSi suponemos $m \\neq 0$, entonces simplificando la relación anterior queda:\n$$\n... | Spain | Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | Let alpha = cbrt(-n/2). Then the first equation has roots alpha (double) and -2 alpha, and the second equation has roots alpha and -5 alpha (double). | |
0bjm | The points $M$, $N$, and $P$ are chosen on the sides $BC$, $CA$ and $AB$ of the triangle $ABC$ such that $BM = BP$ and $CM = CN$. The perpendicular dropped from $B$ onto $MP$ and the perpendicular dropped from $C$ onto $MN$ intersect at $I$. Prove that the angles $\widehat{IPA}$ and $\widehat{INC}$ are congruent.
Gabr... | [
"Since $CM = CN$ and $CI \\perp MN$, the line $CI$ is the perpendicular bisector of the line segment $MN$, hence $IM = IN$. Similarly, we have $IM = IP$. Triangles $IMC$ and $INC$ are equal, so $\\widehat{IMC} \\equiv \\widehat{INC}$, and, in a similar way, we deduce that $\\widehat{IMB} \\equiv \\widehat{IPB}$. It... | Romania | 65th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
07av | The **distance** between two circles $\omega$ and $\omega'$ is defined as the length of their common external tangent and is represented as $d(\omega, \omega')$. If two circles don't have a common external tangent, the distance between them is not defined. Also, note that a point is a circle with zero radius and that t... | [
"Denote by $C(O, R)$ the circle with center $O$ and radius $R$. If $\\omega_1 = C(O_1, R_1)$ and $\\omega_2 = C(O_2, R_2)$, then\n$$\nd(\\omega_1, \\omega_2)^2 = O_1 O_2^2 - (R_1 - R_2)^2,\n$$\nand $d(\\omega_1, \\omega_2)$ is defined if and only if the right hand side is nonnegative. This is equivalent to the exis... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distanc... | English | proof and answer | No | |
0dvw | Problem:
Poišči vse celoštevilske rešitve enačbe $x^{2}+73=y^{2}$. | [
"Solution:\n\nEnačbo preoblikujemo v $y^{2}-x^{2}=73$ oziroma v $(y+x)(y-x)=73$. Število $73$ je praštevilo, zato ga lahko razcepimo le na štiri načine: $73 \\cdot 1$, $1 \\cdot 73$, $-73 \\cdot (-1)$ in $-1 \\cdot (-73)$. Tako pridemo do štirih sistemov enačb, ki jih rešimo:\n$$\n\\begin{array}{rrrr}\ny+x=73 & y+x... | Slovenia | 3. matematično tekmovanje dijakov srednjih tehniških in strokovnih sol | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | [(36, 37), (-36, -37), (-36, 37), (36, -37)] | |
0it4 | Problem:
Find all ordered pairs $(x, y)$ such that
$$
(x-2y)^2 + (y-1)^2 = 0
$$ | [
"Solution:\nThe square of a real number is always at least $0$, so to have equality we must have $(x-2y)^2 = 0$ and $(y-1)^2 = 0$.\n\nThen $y = 1$ and $x = 2y = 2$.\n\nSo the only solution is $(2, 1)$."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (2, 1) | |
028m | Problem:
Os lados de um triângulo têm comprimentos: $a$, $a+2$ e $a+5$, onde $a>0$. Determine os possíveis valores de $a$. | [
"Solution:\n\nComo a soma dos comprimentos dos lados menores deve ser maior que o comprimento do lado maior, então temos que $a + (a+2) > a+5$, assim $a > 3$."
] | Brazil | null | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof and answer | a > 3 | |
0ip1 | Problem:
Find all positive integers $p$ such that $p$, $p+4$, and $p+8$ are all prime. | [
"Solution:\n\nIf $p=3$, then $p+4=7$ and $p+8=11$, both prime.\n\nIf $p \\neq 3$, then $p$ is not a multiple of $3$ and is therefore of one of the forms $3k+1$ or $3k+2$ ($k \\geq 0$).\n\nIf $p=3k+1$, then $p+8=3k+9=3(k+3)$, which is not prime since $k+3>1$.\n\nIf $p=3k+2$, then $p+4=3k+6=3(k+2)$, which is not prim... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 3 | |
066s | If $p$ is prime and $x, y$ are positive integers, find with respect to $p$, all the pairs $(x, y)$ satisfying the equation: $p(x-2) = x(y-1)$.
For the case we are given that $x+y=21$, find all triads $(x, y, p)$ satisfying equation (1). | [
"If $x-2=0 \\Leftrightarrow x=2$, then $y=1$ and we have the solution $(x,y)=(2,1)$.\n\nIf $x-2 \\neq 0$, then, since $p$ is prime, from the equation $p(x-2)=x(y-1)$ it follows that $y \\neq 1$ and $p|x$ or $p|y-1$. Therefore we have the cases:\n\n* If $p|x$, then $x = px'$, where $x'$ positive integer, and then:\n... | Greece | Selection Examination A | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All solutions: (x, y) = (2, 1) or (x, y) = (p, p − 1) or (x, y) = (2p, p). With x + y = 21, the solutions are (x, y, p) = (11, 10, 11) and (14, 7, 7). | |
06yw | Problem:
Show that if $(2+\sqrt{3})^{k}=1+m+n \sqrt{3}$, for positive integers $m, n, k$ with $k$ odd, then $m$ is a perfect square. | [
"Solution:\nWe have $(2+\\sqrt{3})^{4}=97+56 \\sqrt{3}=14(7+4 \\sqrt{3})-1=14(2+\\sqrt{3})^{2}-1$. Hence $(2+\\sqrt{3})^{k+2}=14 (2+\\sqrt{3})^{k}-(2+\\sqrt{3})^{k-2}$. Thus if $(2+\\sqrt{3})^{k}=a_{k}+b_{k} \\sqrt{3}$, then $a_{k+2}=14 a_{k}-a_{k-2}$.\n\nNow suppose the sequence $c_{k}$ satisfies $c_{1}=1, c_{2}=5... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
07bx | Point $D$ is the intersection point of the angle bisector of vertex $A$ with side $BC$ of triangle $ABC$, and point $E$ is the tangency point of the inscribed circle of triangle $ABC$ with side $BC$. $A_1$ is a point on the circumcircle of triangle $ABC$ such that $AA_1 \parallel BC$. If we denote by $T$ the second int... | [
"Let $E_1$ be the reflection of $E$ with respect to the midpoint of $BC$ and $X$ the intersection point of $AE_1$ and $EA_1$. We claim that $IX \\parallel BC$. For this reason, we have (Suppose that $R$ is the radius of circumcircle of $ABC$ and $\\angle B \\geq \\angle C$).\n$$\n\\begin{aligned}\n\\frac{AX}{XE_1} ... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle tr... | null | proof only | null | |
0k4p | Problem:
Triangle $A B C$ has sidelengths $A B=14$, $A C=13$, and $B C=15$. Point $D$ is chosen in the interior of $\overline{A B}$ and point $E$ is selected uniformly at random from $\overline{A D}$. Point $F$ is then defined to be the intersection point of the perpendicular to $\overline{A B}$ at $E$ and the union o... | [
"Solution:\n\nLet $G$ be the intersection of the altitude to $\\overline{A B}$ at point $D$ with $\\overline{A C} \\cup \\overline{B C}$. We first note that the maximal expected value is obtained when $D G=\\frac{[A D G C]}{A D}$, where $[P]$ denotes the area of polygon $P$. Note that if $D G$ were not equal to thi... | United States | HMMT February | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | √70 | |
0ac6 | The base regular four-sided prism is a rhomb with area of $\frac{2}{3}k^2$, $k>0$. The smaller diagonal plane intersection is a square with area of $k^2$.
a) Express the area and volume of the prism using only $k$.
b) For which value of $k$ the area and the volume have equal measure numbers? | [
"Because the smaller diagonal plane intersection is a square with area of $k^2$, the prism height is $H = k$ and the smaller base diagonal is $d_1 = k$. The base of the prism is a rhomb with area $B = \\frac{2}{3}k^2$, from where we obtain\n$$\n\\frac{2}{3}k^2 = \\frac{d_1 \\cdot d_2}{2}\n$$\nor\n$$\n\\frac{2}{3}k^... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Surface Area",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals"
] | null | proof and answer | Volume = (2/3) k^3; Surface area = (14/3) k^2; equality occurs at k = 7. | |
07gs | A circle divided into $2n$ arcs, we intend to write $0, 1, \dots, n-1$ on these arcs such that each number is used exactly two times and for $0 \le i \le n-1$ from one direction there are exactly $i$ arcs between two arcs such that $i$ is written on them. Prove that this is impossible for $n = 1399$. | [
"We number the arcs with $0, 1, 2, \\dots, 2n - 1$, clockwise and assume that the arcs with numbers $a_i$ and $b_i$ have $i$ written on them. By the problem's assumption\n$$\na_i + b_i \\equiv a_i - b_i \\equiv \\pm(i + 1) \\equiv i + 1 \\pmod{2}.\n$$\nThen by adding up these equalities modulo $2$ we have\n$$\n\\be... | Iran | 38th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
03c9 | A positive integer $n$ and a function $f : \mathbb{N} \to \mathbb{N}$ are such that:
(1) $f(1) \le f(2) \le \dots \le f(n) \le f(1) + n$;
(2) $f(n + i) = f(i)$ for any positive integer $i$;
(3) $f(f(i)) \le n + i - 1$ for any positive integer $i$.
Prove that $f(1) + f(2) + \dots + f(n) \le n^2$. | [] | Bulgaria | First Team Selection Test for 56th IMO | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Functional equations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0d04 | In triangle $ABC$, points $D$ and $E$ lie on side $BC$ and $AC$ respectively such that $AD \perp BC$ and $DE \perp AC$. The circumcircle of triangle $ABD$ meets segment $BE$ at point $F$ (other than $B$). Ray $AF$ meets segment $DE$ at point $P$. Prove that
$$
\frac{DP}{PE} = \frac{CD}{DB}.
$$ | [
"Since $AED$ and $ADC$ are both right triangles, $\\widehat{ADP} = 90^\\circ - \\widehat{DAC} = \\widehat{ECB}$. Also, since $AFDB$ is a cyclic quad, $\\widehat{DAP} = \\widehat{EBC}$.\n\n\n\nTherefore, $ADP$ and $BCE$ are similar and $\\frac{BC}{EC} = \\frac{AD}{DP}$. Also, $ADC$ and $DEC$... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
035g | Problem:
Find the number of the subsets $B$ of the set $\{1,2, \ldots, 2005\}$ having the following property: the sum of the elements of $B$ is congruent to $2006$ modulo $2048$. | [
"Solution:\nLet us consider the set $\\{1,2,2^{2}, \\ldots, 2^{10}\\}$. Since every number from $0$ to $2047$ can be represented in a unique way as a sum of powers of $2$ (elements of our set), we conclude that for every $i$, $0 \\leq i \\leq 2047$, there is a unique subset of $\\{1,2,2^{2}, \\ldots, 2^{10}\\}$ suc... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Other"
] | null | proof and answer | 2^1994 | |
0b6q | Given a prime number $p$ congruent to $3$ modulo $4$, show that $w^{2p} + x^{2p} + y^{2p} = z^{2p}$ for no integer numbers $w, x, y, z$ whose product is not divisible by $p$. | [
"Suppose there are four such numbers. Without loss of generality, we may (and will) assume that they are jointly coprime: $(w, x, y, z) = 1$.\n\nReduction modulo $4$ shows that $z$ and exactly one of the numbers $w, x, y$, say $y$, must be odd.\n\nWrite\n$$\nw^{2p} + x^{2p} = z^{2p} - y^{2p} = (z^2 - y^2) \\left(\\... | Romania | 2010 DANUBE MATHEMATICAL COMPETITION | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest co... | null | proof only | null | |
03rw | There are 5 points in a rectangle $ABCD$ (including its boundary) with unit area such that any three of them are not collinear. Find the minimum number of triangles with areas no more than $\frac{1}{4}$ and vertexes chosen from these 5 points. (posed by Leng Gangsong) | [
"**Lemma** *The area of a triangle inscribed in a rectangle is no more than half of the area of the rectangle.*\n\nDenote $E$, $F$, $H$ and $G$ the midpoints of $AB$, $CD$, $BC$ and $AD$, respectively, and $O$ the intersection of line segments $EF$ and $GH$. $EF$ and $GH$ divide the rectangle $ABCD$ into 4 small re... | China | China Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | 2 | |
0dkd | Given a non-isosceles triangle $ABC$, inscribed in circle $(O)$ with $X, Y, Z$ are midpoints of the major arcs $BC, CA, AB$ of $(O)$. Let $D, E, F$ be the tangent points of the incircle $(I)$ of $ABC$ on $BC, CA, AB$ respectively. Suppose that line $XE$ intersects $(O)$ again at $M$ and cuts $YD$ at $P$; line $XF$ inte... | [
"Let $I_a, I_b, I_c$ be the ex-center of angle $A, B, C$ in triangle $ABC$ respectively. Then, it is clear that $A, B, C$ are the feet of the altitude in triangle $I_a I_b I_c$, and $X, Y, Z$ are the midpoints of the three sides of triangle $I_a I_b I_c$. Thus $AX \\parallel YZ$. On the other hand, $AX \\perp AI$ a... | Saudi Arabia | Saudi Arabia booklet 2024 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry >... | English | proof only | null | |
0f01 | Problem:
Two players play the following game. At each turn the first player chooses a decimal digit, then the second player substitutes it for one of the stars in the subtraction $|**** - ****|$. The first player tries to end up with the largest possible result, the second player tries to end up with the smallest poss... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0but | Problem:
Fie $a > 0$ și $(x_{n})_{n \in \mathbb{N}}$ șirul care verifică relațiile $x_{1} = \frac{1}{a}$, $x_{n} = \frac{x_{n-1}}{1 + a n x_{n-1}}$, oricare ar fi $n \geq 2$. Să se calculeze $\lim_{n \rightarrow \infty} (x_{1} + x_{2} + \ldots + x_{n})$. | [] | Romania | OLIMPIADA DE MATEMATICĂ ETAPA LOCALĂ | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 2/a | |
0kw9 | Problem:
Dorothea has a $3 \times 4$ grid of dots. She colors each dot red, blue, or dark gray. Compute the number of ways Dorothea can color the grid such that there is no rectangle whose sides are parallel to the grid lines and whose vertices all have the same color.
Submit a positive integer $A$. If the correct an... | [
"Solution:\n\nTo find an appropriate estimate, we will lower bound the number of rectangles. Let $P(R)$ be the probability a random $3$ by $4$ grid will have a rectangle with all the same color in the grid. Let $P(r)$ be the probability that a specific rectangle in the grid will have the same color. Note $P(r)=\\fr... | United States | HMMT November 2023 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | null | |
06tp | Find all integers $n \geqslant 3$ with the following property: for all real numbers $a_{1}, a_{2}, \ldots, a_{n}$ and $b_{1}, b_{2}, \ldots, b_{n}$ satisfying $\left|a_{k}\right|+\left|b_{k}\right|=1$ for $1 \leqslant k \leqslant n$, there exist $x_{1}, x_{2}, \ldots, x_{n}$, each of which is either $-1$ or $1$, such t... | [
"Answer. $n$ can be any odd integer greater than or equal to $3$.\n\nFor any even integer $n \\geqslant 4$, we consider the case\n$$\na_{1}=a_{2}=\\cdots=a_{n-1}=b_{n}=0 \\quad \\text{and} \\quad b_{1}=b_{2}=\\cdots=b_{n-1}=a_{n}=1.\n$$\nThe condition $\\left|a_{k}\\right|+\\left|b_{k}\\right|=1$ is satisfied for e... | IMO | IMO 2016 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | All odd integers greater than or equal to 3 | |
0jpw | Problem:
Let $ABCD$ be a cyclic quadrilateral with $AB=3$, $BC=2$, $CD=2$, $DA=4$. Let lines perpendicular to $\overline{BC}$ from $B$ and $C$ meet $\overline{AD}$ at $B'$ and $C'$, respectively. Let lines perpendicular to $\overline{AD}$ from $A$ and $D$ meet $\overline{BC}$ at $A'$ and $D'$, respectively. Compute th... | [
"Solution:\n\n$\\boxed{\\dfrac{37}{76}}$\n\nTo get a handle on the heights $CB'$, etc. perpendicular to $BC$ and $AD$, let $X = BC \\cap AD$, which lies on ray $\\overrightarrow{BC}$ and $\\overrightarrow{AD}$ since $\\widehat{AB} > \\widehat{CD}$ (as chords $AB > CD$).\n\nBy similar triangles we have equality of r... | United States | HMMT February 2015 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 37/76 | |
0f3q | Problem:
$ABCD$ is a convex quadrilateral. $M$ is the midpoint of $BC$ and $N$ is the midpoint of $CD$. If $k = AM + AN$ show that the area of $ABCD$ is less than $k^2 / 2$. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
01jc | Let $\triangle ABC$ be an isosceles triangle with $|AB| = |AC|$. Let $D$ be an arbitrary point on segment $BC$. Let $X$ and $Y$ be points on $AB$ and $AC$, respectively, such that $XY \parallel BC$ and $XY$ passes through the midpoint of $AD$. Prove that if the circumcenter of $\triangle ABC$ lies on $\odot(AXY)$ then ... | [
"Note that $\\angle XYO = \\angle XAO = \\angle OAY = \\angle OXY$, hence the triangle $\\triangle OXY$ is isosceles. Let $K, L, M, N$ be the midpoints of $AB, AC, AD, XY$ respectively. Note that $K, L, M$ are collinear because they lie on the midline of $\\triangle ABC$ parallel to $BC$. Moreover, $M$ lies on $XY$... | Baltic Way | Baltic Way 2023 Shortlist | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous >... | English | proof only | null | |
0328 | Problem:
Find the number of positive integers $a$ less than $2003$, for which there exists a positive integer $n$ such that $3^{2003}$ divides $n^{3}+a$. | [
"Solution:\nWe shall prove that the desired numbers have one of the forms $9k \\pm 1$, $3^{3}(9k \\pm 1)$ or $3^{6}(9k \\pm 1)$.\n\nSuppose that $3$ does not divide $a$. Since $n^{3} \\equiv 0, \\pm 1 \\pmod{9}$, then $a \\equiv \\pm 1 \\pmod{9}$.\n\nConversely, let $a \\equiv \\pm 1 \\pmod{9}$. Since $9$ divides $... | Bulgaria | Bulgarian Mathematical Competitions | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 463 | |
048u | Determine all real solutions of the system of equations:
$$
\begin{cases} 2a^2 - 2ab + b^2 = a, \\ 4a^2 - 5ab + 2b^2 = b. \end{cases}
$$ | [] | Croatia | CroatianCompetitions2011 | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (a, b) = (0, 0) and (1, 1) | |
0ggd | 有17名工人排成一列。任何至少兩名工人的連續群組形成一個班。老闆欲從每班中指定其中一人為該班的領班，滿足每名工人被指定為領班的次數為4的倍數。證明老闆指定領班並滿足題設的方法數為17的倍數。 | [] | Taiwan | 2022 數學奧林匹亞競賽第二階段培訓營, 國際競賽實作(二) | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Linear Algebra > Determinants"
] | Chinese; English | proof only | null | |
0ky3 | Problem:
Let $ABC$ be a triangle with $AB < AC$. The incircle of triangle $ABC$ is tangent to side $BC$ at $D$ and intersects the perpendicular bisector of segment $BC$ at distinct points $X$ and $Y$. Lines $AX$ and $AY$ intersect line $BC$ at $P$ and $Q$, respectively. Prove that, if $DP \cdot DQ = (AC - AB)^2$, then... | [
"Solution:\n\nLet $E$ be the extouch point on $BC$, let $I$ be the incenter, and $D'$ the reflection of $D$ over $I$. Note that $DE = AC - AB$, so $DP \\cdot DQ = DE^2$. Now let $F$ be the reflection of $E$ across $D$. The length condition implies $(E, F; P, Q)$ is a harmonic bundle. We also know $XY$ is the perpen... | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | null | proof only | null | |
07rk | A sequence of throws of a die is called *nondecreasing* if the result of each successive throw is at least as large as the previous one in the sequence. For instance, one example of a nondecreasing sequence of 10 throws is
$1, 2, 2, 2, 2, 4, 4, 4, 4, 4, 5$.
Compute the total number of nondecreasing sequences of 10 thro... | [
"Let $S_n$ be the set of all (finite) sequences of length $n$, with all entries being whole numbers between $1$ and $6$. Similarly, let $B_n$ be the set of all (finite) sequences of length $n$, with all entries being either $0$ or $1$. We denote elements of $S_n$ or $B_n$ by a letter such as $x$, and a subscripted ... | Ireland | Irish | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 3003 | |
02vk | Problem:
Considere um triângulo acutângulo $A B C$.
Explique como construir um triângulo $D B C$ com mesma área que $A B C$, satisfazendo $D B=A B$, mas que não seja congruente ao triângulo $A B C$. | [
"Solution:\n\nPrimeiro, vamos criar uma construção em que o novo triângulo possua a mesma área. Para isso trace a reta paralela a $B C$ por $A$, pois para qualquer ponto $P$ dessa reta, a altura relativa a $B C$ é a mesma. Calculando a área em relação a essa base $B C$ teremos que a área de $P B C$ igual à área de ... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07b7 | Let $D$ be a variable point on side $BC$ of triangle $ABC$. The incenters of triangles $ABC$, $ABD$ and $ACD$ are $I$, $I_1$ and $I_2$, respectively. Points $M$ and $N$ are the second intersection points of circumcircles of triangles $IAI_1$ and $IAI_2$ with the circumcircle of triangle $ABC$, respectively. Prove that ... | [
"First, note that the circumcircles of triangles $AII_1$ and $AII_2$ are perpendicular, because\n$$\n\\angle AI_1I + \\angle AI_2I = \\frac{\\angle B + \\angle BAD}{2} + \\frac{\\angle C + \\angle CAD}{2} = 90^\\circ\n$$\nUnder an inversion with center $A$ and power $AB \\times AC$ and then a reflection with respec... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, i... | English | proof only | null | |
0108 | Problem:
In a triangle $ABC$, $\angle BAC = 90^\circ$. Point $D$ lies on the side $BC$ and satisfies $\angle BDA = 2 \angle BAD$. Prove that
$$
\frac{1}{|AD|} = \frac{1}{2} \left( \frac{1}{|BD|} + \frac{1}{|CD|} \right)
$$ | [
"Solution:\n\n\nFigure 3\n\nLet $O$ be the circumcentre of triangle $ABC$ (i.e., the midpoint of $BC$) and let $AD$ meet the circumcircle again at $E$ (see Figure 3). Then $\\angle BOE = 2 \\angle BAE = \\angle CDE$, showing that $|DE| = |OE|$. Triangles $ADC$ and $BDE$ are similar; hence\n... | Baltic Way | Baltic Way 1998 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordi... | null | proof only | null | |
060k | Problem:
Soient $a$, $b$ et $c$ des réels tels que $0 \leqslant a, b, c \leqslant 2$. Montrer que
$$
(a-b)(b-c)(a-c) \leqslant 2
$$ | [
"Solution:\nNotons qu'il y a 6 ordres possibles pour les variables $a$, $b$ et $c$.\nDans les trois cas $b \\geqslant a \\geqslant c$, $a \\geqslant c \\geqslant b$ et $c \\geqslant b \\geqslant a$, le produit $(a-b)(b-c)(c-a)$ est négatif, de sorte que l'inégalité est vérifiée.\n\nSi $a \\geqslant b \\geqslant c$,... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - Envoi 5 : Pot Pourri | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0b0z | Problem:
In triangle $ABC$, we have $AB = BC = 5$ and $CA = 8$. What is the area of the region consisting of all points inside the triangle which are closer to $AB$ than to $AC$? | [
"Solution:\n\nNote that $ABC$ is simply an isosceles triangle obtained by joining two $3$-$4$-$5$ triangles; hence, its area is $3 \\cdot 4 = 12$.\n\nThen, if $D$ is the foot of the angle bisector of $\\angle A$ on $BC$, the area we want is actually just the area of $ABD$. By the angle bisector theorem,\n$$\n\\frac... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem"
] | null | final answer only | 60/13 | |
00ci | En el triángulo $ABC$ vale que $A\hat{C}B = 2 \cdot A\hat{B}C$. Además, $P$ es un punto interior del triángulo $ABC$ tal que $AP = AC$ y $PB = PC$. Demostrar que $B\hat{A}C = 3 \cdot B\hat{A}P$. | [] | Argentina | Nacional OMA 2019 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | Spanish | proof only | null | |
0d82 | On a chessboard $5 \times 9$ squares, the following game is played. Initially, a number of frogs are randomly placed on some of the squares, no square containing more than one frog. A turn consists of moving all of the frogs subject to the following rules:
- Each frog may be moved one square up, down, left, or right;
-... | [
"If 32 frogs are placed in an $4 \\times 8$ rectangle, they can all move down, right, up, left, down, etc.\n\n| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |\n| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |\n| 0 | 0 | 0 | 0 | 0 | 0 | ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
0kzy | 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 | [
"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 other cases $20... | United States | AMC 10 A | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | D | |
06dd | Two circles intersect at points $A$ and $B$. Through the point $B$ a straight line is drawn, intersecting the first circle at $K$ and the second circle at $M$. A line parallel to $AM$ is tangent to the first circle at $Q$. The line $AQ$ intersects the second circle again at $R$.
a. Prove that the tangent to the second... | [
"a. We only provide a proof for the configuration as shown. The other cases are similar.\nLet $X$ and $Y$ be any points on the tangent at $R$ to $(ABR)$ and the tangent at $Q$ to $(ABQ)$ respectively as shown. We have\n$$\n\\angle MBR = \\angle MAR = \\angle YQA = \\angle QBA,\n$$\nand so\n$$\n\\begin{aligned}\n\\a... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02k5 | Problem:
a) Calcule as diferenças: $1-\frac{1}{2}$; $\frac{1}{2}-\frac{1}{3}$; $\frac{1}{3}-\frac{1}{4}$; $\frac{1}{4}-\frac{1}{5}$; $\frac{1}{5}-\frac{1}{6}$
b) Deduza de (a) o valor da soma: $\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}$
c) Calcule a soma: $\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\f... | [
"Solution:\n\na)\n$1-\\frac{1}{2}=\\frac{1}{2}$\n\n$\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}$\n\n$\\frac{1}{3}-\\frac{1}{4}=\\frac{1}{12}$\n\n$\\frac{1}{4}-\\frac{1}{5}=\\frac{1}{20}$\n\n$\\frac{1}{5}-\\frac{1}{6}=\\frac{1}{30}$\n\n\nb)\n$\\frac{1}{2}+\\frac{1}{6}+\\frac{1}{12}+\\frac{1}{20}+\\frac{1}{30}=1-\\frac{1}... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | a) 1/2, 1/6, 1/12, 1/20, 1/30; b) 5/6; c) 999/1000 | |
0fsd | Problem:
Zeige, dass das Produkt von 5 aufeinanderfolgenden natürlichen Zahlen keine Quadratzahl ist. | [
"Solution:\n\nJe zwei dieser fünf Zahlen haben eine Differenz $\\leq 4$. Jede Primzahl $p \\geq 5$, die eine dieser Zahlen teilt, kann daher keine weitere teilen und muss deshalb mit geradem Exponenten in der Primfaktorzerlegung dieser Zahl auftreten. Wir ordnen jeder dieser fünf Zahlen ein Paar $(a, b)$ zu. Dabei ... | Switzerland | IMO - Selektion | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
00ww | Problem:
Let $p$ be a polynomial with integer coefficients such that $p(-n) < p(n) < n$ for some integer $n$. Prove that $p(-n) < -n$. | [
"Solution:\nAs $a^{n} - b^{n} = (a - b)\\left(a^{n-1} + a^{n-2} b + \\cdots + b^{n-1}\\right)$, then for any distinct integers $a, b$ and for any polynomial $p(x)$ with integer coefficients, $p(a) - p(b)$ is divisible by $a - b$. Thus, $p(n) - p(-n) \\neq 0$ is divisible by $2n$ and consequently $p(-n) \\leq p(n) -... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0ds6 | In the cyclic quadrilateral $ABCD$, the sides $AB$, $DC$ meet at $Q$, the sides $AD$, $BC$ meet at $P$, $M$ is the midpoint of $BD$. If $\angle APQ = 90^\circ$, prove that $PM$ is perpendicular to $AB$. | [
"Drop perpendicular $DE$ from $D$ onto $AB$. Join $PE$. In the cyclic quadrilaterals $DEQP$ and $ABCD$, we have $\\angle PEB = \\angle PDQ = \\angle PBE$. Thus $\\triangle PBE$ is isosceles. Let $F$ be the midpoint of $BE$. Then both $PF$ and $MF$ are perpendicular to $AB$. Thus $P$, $M$, $E$ are collinear and $PM$... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0fys | Problem:
Sei $A B C D$ ein Quadrat und $M$ ein Punkt im Innern der Strecke $B C$. Die Winkelhalbierende des Winkels $\angle B A M$ schneide die Strecke $B C$ im Punkt $E$. Ferner schneide die Winkelhalbierende des Winkels $\angle M A D$ die Gerade $C D$ im Punkt $F$. Zeige, dass $A M$ und $E F$ senkrecht aufeinander s... | [
"Solution:\n\nSei $S$ die Projektion von $E$ auf $A M$ und sei $T$ die Projektion von $F$ auf $A M$. Die Dreiecke $A B E$ und $A E S$ stimmen in allen Winkeln und einer Seite (der gemeinsamen) überein und sind somit kongruent, also gilt $A B = A S$. Dasselbe gilt für die Dreiecke $A T F$ und $A F D$ und es folgt $A... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
06y9 | Let $N$ be a positive integer. Geoff and Ceri play a game in which they start by writing the numbers $1, 2, \ldots, N$ on a board. They then take turns to make a move, starting with Geoff. Each move consists of choosing a pair of integers $(k, n)$, where $k \geqslant 0$ and $n$ is one of the integers on the board, and ... | [
"Lemma 1. For any set $\\mathcal{S}$, $\\mathcal{S}$ wins if and only if $J(\\mathcal{S}, \\varnothing)$ wins. Similarly, $\\mathcal{S}$ wins if and only if $J(\\varnothing, \\mathcal{S})$ wins.\n\nProof. Let $(k, m)$ be a move on $\\mathcal{S}$, and let $\\mathcal{T}$ be the result of applying the move. Then we ca... | IMO | IMO2024 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Write N = t · 2^n with t odd. If t = 1 (i.e., N is a power of two), then Geoff wins if and only if n is odd. If t > 1, then Geoff wins if and only if n is even. | |
0fud | Problem:
An einer Wandtafel steht eine Liste natürlicher Zahlen. Es wird nun wiederholt die folgende Operation ausgeführt: Wähle zwei beliebige Zahlen $a, b$ aus, wische sie aus und schreibe an deren Stelle $\operatorname{ggT}(a, b)$ und $\operatorname{kgV}(a, b)$. Zeige, dass sich der Inhalt der Liste ab einem bestim... | [
"Solution:\n\nFür beliebige natürliche Zahlen $a$ und $b$ gilt bekanntlich $\\operatorname{ggT}(a, b) \\cdot \\operatorname{kgV}(a, b)=a b$. Folglich ist das Produkt $P$ aller Zahlen an der Tafel eine Invariante. Bei einer Operation ändern sich die beiden Zahlen genau dann, wenn keine ein Teiler der anderen ist. In... | Switzerland | SMO Finalrunde | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0h8p | Andriy wrote a 4-digit number. Olesya crossed out the last digit and it turned out that the difference between the initial number and the obtained number equals $2018$. Which number did Andriy write? Provide all possible answers. | [
"Let $abcd$ denote the number Andriy wrote, then $\\overline{abcd} - \\overline{abc} = 2018$. Now we only need to pick corresponding digits. Obviously, $a = 2$ or $a = 3$.\n\nWith $a = 3$, $\\overline{3bcd} - \\overline{3bc} > 3000 - 399 > 2018$, hence, $a = 2$ and $\\overline{2bcd} - \\overline{2bc} = 2018$. Analo... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 2242 | |
06r9 | Find the smallest number $n$ such that there exist polynomials $f_{1}, f_{2}, \ldots, f_{n}$ with rational coefficients satisfying
$$
x^{2}+7=f_{1}(x)^{2}+f_{2}(x)^{2}+\cdots+f_{n}(x)^{2}
$$ | [
"The equality $x^{2}+7=x^{2}+2^{2}+1^{2}+1^{2}+1^{2}$ shows that $n \\leq 5$. It remains to show that $x^{2}+7$ is not a sum of four (or less) squares of polynomials with rational coefficients.\nSuppose by way of contradiction that $x^{2}+7=f_{1}(x)^{2}+f_{2}(x)^{2}+f_{3}(x)^{2}+f_{4}(x)^{2}$, where the coefficient... | IMO | 51st IMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadra... | English | proof and answer | 5 | |
0fs2 | Problem:
$ABC$ sei ein spitzwinkliges Dreieck mit Umkreismittelpunkt $O$. Das Lot von $A$ auf $BC$ schneide den Umkreis im Punkt $D \neq A$, und die Gerade $BO$ schneide den Umkreis im Punkt $E \neq B$. Zeige, dass $ABC$ und $BDCE$ denselben Flächeninhalt haben. | [
"Solution:\n\nWir berechnen alle Flächeninhalte über der gemeinsamen Basis $BC$. Sei $P$ der Schnittpunkt von $AD$ mit $BC$. Nach Voraussetzung ist $AD \\perp BC$ und da $O$ auf $BE$ liegt, ist $BE$ ein Durchmesser des Umkreises und daher $\\angle BCE = 90^\\circ$, also auch $CE \\perp BC$. Damit können wir folgend... | Switzerland | Vorselektionsprüfung | [
"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 | |
0gaz | 試求所有正整數有序對 $(a, b, c)$ 使得
$$
a^b + b^c + c^a = a^c + b^a + c^b
$$
成立。 | [
"Permutations of $(1, 2, 3)$ or $(x, y, y)$ where $x, y \\in N$, $N$ is the set of all positive numbers.\n\nLemma 1. $2 < (1 + \\frac{1}{n})^n < 3$ for all $n \\in N_{\\ge 2}$\n\nProof. Use binomial theorem, we have\n$$\n2 = 1 + n \\cdot \\frac{1}{n} < 1 + n \\cdot \\frac{1}{n} + \\sum_{i=2}^{n} \\frac{\\binom{n}{i... | Taiwan | 二〇一七數學奧林匹亞競賽第二階段選訓營 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | All ordered triples that are permutations of (x, y, y) for positive integers x and y, together with all permutations of (1, 2, 3). | |
00vl | A graph is *good* if its edges can be colored with 2 colors so that no cycle has two consecutive edges of the same color. What is the maximum number of edges in a *good* graph with 1000 vertices? | [
"We will prove the answer to be $4 \\cdot 333 = 1332$.\nFirst we prove that a *good* graph with $n$ vertices has at most $\\frac{4(n-1)}{3}$ edges.\n\n*Claim 1.* If we have 3 paths going from vertex $A$ to vertex $B$, then 2 of them have a common vertex different from $A$, $B$.\n*Proof.* Suppose not. By the Pigeonh... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlisted Problems | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 1332 | |
0111 | Problem:
The point $(a, b)$ lies on the circle $x^{2}+y^{2}=1$. The tangent to the circle at this point meets the parabola $y=x^{2}+1$ at exactly one point. Find all such points $(a, b)$. | [
"Solution:\n$(-1,0)$, $(1,0)$, $(0,1)$, $\\left(-\\frac{2 \\sqrt{6}}{5},-\\frac{1}{5}\\right)$, $\\left(\\frac{2 \\sqrt{6}}{5},-\\frac{1}{5}\\right)$.\n\nSince any non-vertical line intersecting the parabola $y=x^{2}+1$ has exactly two intersection points with it, the line mentioned in the problem must be either ve... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (-1, 0), (1, 0), (0, 1), (-2√6/5, -1/5), (2√6/5, -1/5) | |
016z | Solve in positive integers $x, y, z, t$, with $x \ge y \ge z$:
$$
t! = x! + 2y! + 3z!
$$ | [
"Since $t > x$, $(x+1)! \\le t! \\le (1+2+3)x! = 6x!$. So $x+1 \\le 6$ and $x \\le 5$. Consider the possible values of $x$.\n\nIf $x = 1$, the equality is $t! = 6$. So $(x, y, z, t) = (1, 1, 1, 2)$ is a solution.\n\nIf $x = 2$, $3! \\le t! \\le 6 \\cdot 2! < 4!$. So $t = 3$, and we have to solve $6 = 2 + 2y! + 3z! ... | Baltic Way | BALTIC WAY | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (1,1,1,2), (3,3,2,4), (5,5,5,6) | |
0g3b | Problem:
Sei $p$ eine Primzahl und $a, b, c$ und $n$ positive ganze Zahlen mit $a, b, c < p$, sodass die drei folgenden Aussagen gelten:
$$
p^{2} \mid a + (n-1) \cdot b, \quad p^{2} \mid b + (n-1) \cdot c, \quad p^{2} \mid c + (n-1) \cdot a
$$
Zeige, dass $n$ keine Primzahl ist. | [
"Solution:\n\nDer wichtigste Schritt ist, die drei Bedingungen zu addieren. Wir erhalten:\n$$\np^{2} \\mid (a + (n-1) \\cdot b) + (b + (n-1) \\cdot c) + (c + (n-1) \\cdot a) = n \\cdot (a + b + c)\n$$\nDa $a, b, c < p$, haben wir $a + b + c \\leq 3p - 3 < 3p$ und falls $p \\geq 3$, folgt $a + b + c < p^{2}$. Nun te... | Switzerland | Vorrunde | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0bj8 | Determine all differentiable functions $f: \mathbb{R} \to \mathbb{R}$, that satisfy the equality $f \circ f = f$. | [
"We shall show that only the identity function and the constant functions satisfy the conditions of the problem. It is clear that these functions verify indeed the conditions.\n\nBecause $f$ is continuous, its range $\\{f(x) \\mid x \\in \\mathbb{R}\\}$ is an interval $I \\subseteq \\mathbb{R}$. If $I$ is degenerat... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All constant functions and the identity function. | |
0a3f | Problem:
We definiëren een rij met $a_{1}=850$ en
$$
a_{n+1} = \frac{a_{n}^{2}}{a_{n}-1}
$$
voor $n \geq 1$. Bepaal alle waarden van $n$ waarvoor geldt dat $\left\lfloor a_{n}\right\rfloor=2024$.
Hierbij staat de entier $\lfloor a\rfloor$ van een reëel getal $a$ voor het grootste gehele getal kleiner of gelijk aan $a$. | [
"Solution:\nAntwoord: de enige waarde die voldoet is $n=1175$.\n\nAls eerst merken we op dat we de recursie kunnen herschrijven als\n$$\na_{n+1} = \\frac{a_{n}^{2}-1+1}{a_{n}-1} = \\frac{a_{n}^{2}-1}{a_{n}-1} + \\frac{1}{a_{n}-1} = a_{n} + 1 + \\frac{1}{a_{n}-1}.\n$$\nOmdat het verschil van $a_{n+1}-a_{n}>1$, is er... | Netherlands | Maarttoets | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequaliti... | null | proof and answer | 1175 | |
0ap5 | Problem:
The perimeter of a square inscribed in a circle is $p$. What is the area of the square that circumscribes the circle? | [
"Solution:\n\nThe area of the square that circumscribes the circle is equal to the square of the diameter of the circle. The side of the inner square has length equal to $p / 4$, so that the diameter of the circle (which is equal to the length of the diagonal of the inner square) is given by\n$$\n\\sqrt{\\left(\\fr... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | final answer only | p^2/8 | |
03jm | Problem:
Let $x_{n+1} = 4 x_n - x_{n-1}$, $x_0 = 0$, $x_1 = 1$, and $y_{n+1} = 4 y_n - y_{n-1}$, $y_0 = 1$, $y_1 = 2$.
Show for all $n \geq 0$ that $y_n^2 = 3 x_n^2 + 1$. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Diophantine Equations > Pell's equations"
] | null | proof only | null | |
05qc | Problem:
Trouver tous les entiers $x, y \geqslant 1$ tels que
$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{2017}
$$ | [
"Solution:\nL'équation équivaut à $2017(x+y)=x y$, donc $2017 \\mid x y$. Comme $2017$ est premier, $2017 \\mid x$ ou $2017 \\mid y$. Sans perte de généralité, $x$ et $y$ jouant le même rôle, on peut supposer que $2017 \\mid x$. Soit $x' \\geqslant 1$ tel que $x=2017 x'$. On a donc $2017 x'+y=x' y$, soit\n$$\n2017 ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 5 : Pot-POURRI | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | Solutions: (4034, 4034), (4070306, 2018), (2018, 4070306). | |
00j4 | Let $ABC$ be an isosceles triangle with $\overline{AC} = \overline{BC}$ and $P$ be a point of the circumcircle lying on the arc $CA$ not containing $B$.
Let $E$ and $F$ be the orthogonal projections of the point $C$ onto the lines $AP$ and $BP$, respectively.
Prove that $AE$ and $BF$ have the same length.
W. Janous, In... | [
"The inscribed angle theorem implies $\\angle PAC = \\angle PBC$.\n\nAbbildung 1: Problem 4.\nTherefore, the right triangles $AEC$ and $BFC$ have the same angles. Since their hypotenuses have the same length $\\overline{AC} = \\overline{BC}$, they are congruent and we conclude $\\overline{A... | Austria | AustriaMO2011 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0h06 | What is the maximum number of vertexes can convex polygon have if it is known, that all its angles have integer degree measure? | [
"Sum of all angles of convex $n$-gon is $180^{\\circ}(n-2)$, maximal angle, which has integer degree measure is $179^{\\circ}$. So we can write the following inequality $180^{\\circ}(n-2) \\le 179^{\\circ}n \\Rightarrow n \\le 360$. Hence, $n = 360$ maximal possible value. Regular 360-gon provides an example."
] | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010) | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 360 | |
0ctv | Let $ABC$ be an isosceles triangle with $AB = BC$. Let $CC'$ be the diameter of its circumcircle $\Omega$. The line through $C'$ parallel to $BC$ intersects segments $AB$ and $AC$ at $M$ and $P$, respectively. Prove that $M$ is the midpoint of $C'P$. (B. Obukhov)
Пусть $ABC$ — равнобедренный треугольник, $AB = BC$. Пу... | [
"Since $\\triangle AMP \\sim \\triangle ABC$, we have $AM = MP$. Since $C'A \\perp AC$, this yields $C'M = AM = MP$.\n\n\nТак как $CC'$ — диаметр $\\Omega$, имеем $\\angle C'AC = 90^\\circ$. Поскольку $MP \\parallel BC$, получаем $\\angle MPA = \\angle BCA = \\angle BAC$ (см. рис. 1). Значит, треугольник $AMP$ — ра... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English; Russian | proof only | null | |
053s | Call a convex polygon on a plane *correct* if for its every side there exists a unique vertex of the polygon that lies farther from that side than any other vertex of the polygon. Call the perpendicular drawn from the vertex farthest from side $XY$ to side $XY$ an *altitude* of the correct polygon. Find all natural num... | [
"Let one of the vertices be $O(0,0)$ and let the other vertices $A_1, \\dots, A_{n-1}$ lie on a circle with radius $1$ and centre $O$ in such a way that $A_1(1,0)$, $A_{n-1}(0,1)$ and $A_2, \\dots, A_{n-2}$ are all on the shorter arc $A_1A_{n-1}$ (Fig. 16 depicts the case $n=6$).\n\nThe vertex farthest from line $O... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | all n ≥ 3 | |
0b0d | Problem:
The constant term in the expansion of $\left(a x^{2}-\frac{1}{x}+\frac{1}{x^{2}}\right)^{8}$ is $210 a^{5}$. If $a>0$, find the value of $a$. | [] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 4/3 | |
088w | Problem:
Dato un qualsiasi intero positivo $n$, chiamiamo ciclostilato di $n$ il numero che si ottiene concatenando 2012 scritture di $n$ (in base 10). Per esempio il ciclostilato di 314 è $314314314 \ldots 314$, dove le cifre "314" si ripetono 2012 volte.
a) Determinare tutti gli interi positivi $m$ tali che il ciclos... | [
"Solution:\nSia $f(n)$ il ciclostilato di $n$.\n\na.\nChiamiamo $s(m)$ la somma delle cifre di $m$. Per il criterio di divisibilità per 9 si ha che $f(n)$ è multiplo di 9 se e solo se $s(f(n))$ è multiplo di 9; d'altra parte $s(f(n)) = 2012 \\cdot s(n)$, il quale è multiplo di 9 se e solo se lo è $s(n)$ (perché il ... | Italy | Italian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | a) Exactly the positive multiples of nine. b) All positive integers with an odd number of digits, together with those with an even number of digits that are multiples of eleven. | |
01rh | $n$ $2 \times 2$ squares are drawn on the Cartesian plane. The sides of these squares are parallel to the coordinate axes. It is known that the center of any square is not an inner point of any other square. Let $\Pi$ be a rectangle such that it contains all these $n$ squares and its sides are parallel to the coordinat... | [
"7. See solution of Problem A7a."
] | Belarus | Final Round | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
04rj | We are given a triangle $ABC$ with incentre $I$. Suppose that there exists an intersection point $M$ of the line $AB$ and the perpendicular to $CI$ through $I$. Prove that the circumcircle of the triangle $ABC$ intersects the segment $CM$ in an interior point $N$ and that $NI \perp MC$.
(Peter Novotny) | [
"First we show in two different ways that the line $MI$, a perpendicular to $CI$ through $I$, is tangent to the circle $ABI$. The first way is based on the known fact that $\\angle AIC$ and $\\angle BIC$ are obtuse angles of measures $90^\\circ + \\frac{1}{2}\\beta$ and $90^\\circ + \\frac{1}{2}\\alpha$, respective... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | English | proof only | null | |
0jmd | Problem:
Let $a, b, c, x$ be reals with $(a+b)(b+c)(c+a) \neq 0$ that satisfy
$$
\frac{a^{2}}{a+b}=\frac{a^{2}}{a+c}+20, \quad \frac{b^{2}}{b+c}=\frac{b^{2}}{b+a}+14, \quad \text{ and } \quad \frac{c^{2}}{c+a}=\frac{c^{2}}{c+b}+x
$$
Compute $x$. | [
"Solution:\nAnswer: $-34$ Note that\n$$\n\\begin{aligned}\n\\frac{a^{2}}{a+b}+\\frac{b^{2}}{b+c}+\\frac{c^{2}}{c+a}-\\frac{a^{2}}{c+a}-\\frac{b^{2}}{a+b}-\\frac{c^{2}}{b+c} & =\\frac{a^{2}-b^{2}}{a+b}+\\frac{b^{2}-c^{2}}{b+c}+\\frac{c^{2}-a^{2}}{c+a} \\\\\n& =(a-b)+(b-c)+(c-a) \\\\\n& =0\n\\end{aligned}\n$$\nThus, ... | United States | HMMT November 2014 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | -34 | |
02rw | Problem:
O professor Guilherme criou três estranhas máquinas. A máquina $A$ transforma um gato em um cachorro com probabilidade $\frac{1}{3}$. A máquina $B$ transforma um gato em um cachorro com probabilidade $\frac{2}{5}$. A máquina $C$ transforma um gato em um cachorro com probabilidade $\frac{1}{4}$. E se o animal ... | [
"Solution:\n\nPrimeiro, vamos calcular a probabilidade de um gato sair gato de cada máquina.\nComo a probabilidade de sair um cachorro da máquina $A$ é $\\frac{1}{3}$, a probabilidade de sair um gato desta máquina é $1-\\frac{1}{3}=\\frac{2}{3}$.\nComo a probabilidade de sair um cachorro da máquina $B$ é $\\frac{2}... | Brazil | Brazilian Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 7/10 | |
099n | Given $\omega_1, \omega_2$ circles centered with $O_1, O_2$ and external tangent to each other at point $P$. A tangent line $l$ with $\omega_1$ and $\omega_2$ meet at points $A, B$. Also, a line passing through $B$ with perpendicular to $l$ and line $O, A$ meet at $C$. The line $PC$ and segment $AB$ met at $Q$, line $O... | [
"Let $R$ be the line $l$ tangent to $\\omega_1$, in the $\\omega_1$ $R'$ be the opposite point of $R$ for diameter. Hence, it's enough to show that $R'$, $P$, $C$ are collinear. Now assume $R'P \\cap BC = C'$ and $R'P \\cap AB = Q'$. The triangles $O_1AR$ and $CAB$ are similar, hence $\\frac{BC}{RO_1} = \\frac{BA}{... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0kzw | On top of a rectangular card with sides of length $1$ and $2 + \sqrt{3}$, an identical card is placed so that two of their diagonals line up, as shown ($AC$, in this case).
Continue the process, adding a third card to the second, and so on, lining up successive diagonals after rotating clockwi... | [
"Note that the common diagonal is a diameter of the circle that will ultimately circumscribe the collection of rectangles. Each rectangle is composed of four chords of the circle, and the set of outermost chords will form a regular $n$-gon if a vertex lands on $B$. Because each new card contributes two sides of the... | United States | AMC 12 A | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | A | |
0g2t | Problem:
Sei $ABC$ ein Dreieck mit $AB = AC$ und sei $M$ der Mittelpunkt der Strecke $BC$. Sei $P$ ein Punkt, sodass $PB < PC$ gilt und $PA$ parallel zu $BC$ ist. Ferner seien $X$ und $Y$ Punkte auf den Geraden $PB$ respektive $PC$, sodass $B$ auf der Strecke $PX$ und $C$ auf der Strecke $PY$ liegt und $\angle PXM = \... | [
"Solution:\n\nWir führen den Punkt $Z$ ein, der der zweite Schnittpunkt der Kreise durch $MCY$ und $MXB$ ist. Mit Winkeljagd folgt nun, $\\angle MZC = \\angle MYC = \\angle MXB = \\angle MZB$. Somit liegt $M$ auf der Mittelsenkrechten von $BC$. Folglich gilt\n$$\n\\angle ZXP = \\angle ZXB = \\angle ZMB = \\angle ZM... | Switzerland | Selektion | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0gup | In scalene triangle $ABC$, the incenter is $I$ and the circumcenter is $O$. $AI$ intersects the circumcircle of $ABC$ a second time at $P$. The line passing through $I$ and perpendicular to $AI$ intersects $BC$ at $X$. The foot of the perpendicular from $X$ to $IO$ is $Y$. Show that the points $A$, $P$, $X$, $Y$ are co... | [
"*Claim 1.* $A$, $X$, $P$, $E$ are concyclic.\n\n*Proof.* $X$, $I$, $M$, $P$ are concyclic because $\\angle XIP = \\angle XMP = 90^\\circ$. It is well known that $S$ lies on the incircle and since $DI = SI$ and $DM = EM$, we get $IM \\parallel AE$. So, $\\angle EAP = \\angle PIM = \\angle PXE$, which means that $A$... | Turkey | Team Selection Test for IMO 2024 | [
"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",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | English | proof only | null | |
0jv6 | Problem:
On a circle we write $2n$ real numbers with a positive sum. For each number, there are two sets of $n$ numbers such that this number is on the end. Prove that at least one of the numbers has a positive sum for both these sets. | [
"Solution:\n\nBy decreasing each number by the average of all $n$ numbers, we may assume for convenience that the sum is in fact $0$. Let $X_{k}$ denote the contiguous sequence of numbers in positions $k, k+1, \\ldots, k+(n-1)$ (modulo $2n$). Remark that $X_{i}$ and $X_{j}$ share an endpoint if and only if $i-j \\i... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
06c6 | Show that $\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} \le \sqrt{c(ab+1)}$ for real numbers $a, b, c \ge 1$. | [
"Let $x = \\sqrt{a-1}$, $y = \\sqrt{b-1}$ and $z = \\sqrt{c-1}$. Then we have $x, y, z \\ge 0$. We need to prove\n$$\nx + y + z \\le \\sqrt{(z^2 + 1)[(x^2 + 1)(y^2 + 1) + 1]}.\n$$\nBy the Cauchy-Schwarz inequality, we have\n$$\n(z^2 + 1)[(x^2 + 1)(y^2 + 1) + 1] = (1 + z^2)[(x^2 + 1)(y^2 + 1) + 1] \\\\ \\ge (\\sqrt{... | Hong Kong | HKG TST | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
08zg | Let $ABC$ be an acute triangle, $D$ be the foot of the perpendicular from $A$ to $BC$, and $M$ be the midpoint of $AC$. Let $P$ be a point on the segment $BM$ such that $\angle PAM = \angle MBA$. Given $\angle BAP = 41^\circ$ and $\angle PDB = 115^\circ$, find the value of $\angle BAC$. | [
"78°\n\nTriangles $MBA$ and $MAP$ are similar since $\\angle MBA = \\angle PAM$, thus $MP \\cdot MB = MA^2$. Moreover, we have $MD = MA = MC$, since $\\angle CDA = 90^\\circ$ and $M$ is the midpoint of $AC$. Then triangles $MBD$ and $MDP$ are similar since $MP \\cdot MB = MD^2$, thus $\\angle DBM = \\angle MDP$. By... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 78° | |
0ha9 | Given is circle $\Gamma$ with center in point $O$ and diameter $AB$. $OBDE$ is a square, $F$ is the second point of intersection of $AD$ and circle $\Gamma$, $C$ is the middle of the segment $AF$. Find the value of the angle $OCB$. | [
"Since $AB$ is a diameter, then $\\angle AFB = 90^\\circ$, and $CO \\parallel FB$ as the middle segment. Therefore, $CO \\perp CD$ and quadrilateral $OBDC$ is inscribed. Hence, $\\angle OCB = \\angle ODB = 45^\\circ$."
] | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 45° | |
03q8 | Let $n$ be a positive integer, and $S_n$ be the set of all positive integer divisors of $n$ (including $1$ and itself). Prove that at most half of the elements in $S_n$ have their last digits equal to $3$. (posed by Feng Zuming) | [
"(1) If $5 \\mid n$, let $d_1, d_2, \\dots, d_m$ be the elements in $S_n$ with their last digits equal to $3$, then $5d_1, 5d_2, \\dots, 5d_m$ are elements in $S_n$ with their last digits equal to $5$. So $m \\le \\frac{1}{2}|S_n|$. The statement is true in this case.\n\n(2) If $5 \\nmid n$ and the last digit of ev... | China | China Girls' Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof only | null | |
0d5e | In his bag, Salman has a number of stones. The weight of each stone is not greater than $0.5$ kg and the total weight of the stones is not greater than $2.5$ kg. Prove that Salman can divide his stones into $4$ groups, each group has a total weight not greater than $1$ kg.
Suggested by Trân Nam Dũng | [
"Let $k$ be the number of stones. Note that the sum of weights of any two stones is not greater than $1$ kg.\n\nThus, if $k \\leq 8$ we can divide these stones into $4$ (or less) groups, each group has $1$ or $2$ stones. This division obviously fulfills the condition.\n\nIf $k=9$, we take $3$ lightest stones. Their... | Saudi Arabia | SAMC 2015 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English, Arabic | proof only | null | |
089d | Problem:
Dopo una gara fra cinque cavalli, cinque amici si incontrano e parlano dei risultati. Si sa che ognuno di loro ha puntato su un cavallo diverso, e che mentono entrambe le persone che hanno puntato sul primo e sull'ultimo classificato; le altre dicono la verità. Le loro affermazioni sono le seguenti:
Alex: "I... | [
"Solution:\n\nLa risposta è (A). Enrica non può che mentire (se il suo cavallo fosse arrivato primo come lei asserisce dovrebbe mentire, assurdo) e dunque il suo cavallo è ultimo. Se Umberto dicesse il vero il suo cavallo sarebbe penultimo, quello di Alex terzo. In tal caso Osvaldo mente (le ultime due posizioni so... | Italy | Italian Mathematical Olympiad | [
"Discrete Mathematics > Logic"
] | null | MCQ | A | |
01d5 | A *magic octagon* is an octagon whose sides go along the grid lines of a square grid and side lengths are $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ (in any order). What is the largest possible area of a magic octagon? | [
"Answer: $71$.\n\nFigure 2:\n**Solution:** Figure 2 shows an example of a magic octagon with area $71$. Let us show that it cannot be larger. This octagon has some $90^\\circ$ angles and some $270^\\circ$ angles. From the equation\n$$\na \\cdot 90^\\circ + (8-a) \\cdot 270^\\circ = 6 \\cdot... | Baltic Way | Baltic Way 2016 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 71 | |
0e9e | Find all positive integers $n$ such that
$$
(n^2 + 11n - 4) \cdot n! + 33 \cdot 13^n + 4
$$
is a perfect square. | [
"For reasons of readability let us write $A_n = (n^2 + 11n - 4) \\cdot n! + 33 \\cdot 13^n + 4$. First, consider the value of $A_n$ modulo $8$ for $n \\ge 4$. We have $8 \\mid 1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot \\dots \\cdot n = n!$, so\n$$\nA_n = (n^2 + 11n - 4) \\cdot n! + 33 \\cdot 13^n + 4 \\equiv 0 + 1 \\cdot... | Slovenia | National Math Olympiad in Slovenia | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | n = 1, 2 | |
0a6p | Problem:
Alice plays a game with the Mad Hatter. The Mad Hatter will write two rows of numbers on a blackboard, each a distinct permutation of $\{1,2,\ldots ,n\}$.
On each move, Alice is allowed to swap the positions of the numbers $a$ and $a + 1$ in the first row, for some $1 \leq a < n$.
What is the minimum number o... | [
"Solution:\n\nWe will show that the answer is $\\binom{n}{2} = \\frac{n(n - 1)}{2}$.\nTo show that this is sufficient, we will use induction.\n\n- Base Case: Note that when $n = 1$, the two rows of numbers must be the same since there is only one permutation of $\\{1\\}$. Hence this case takes $0 = \\binom{1}{2}$ t... | New Zealand | NZMO Round One | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | n(n-1)/2 | |
085i | Problem:
Consideriamo un qualsiasi insieme di 20 numeri interi consecutivi, tutti maggiori di 50. Quanti di essi al massimo possono essere numeri primi?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8 . | [
"Solution:\n\nLa risposta è $\\mathbf{(C)}$. Possiamo innanzitutto escludere dai possibili numeri primi tutti i numeri pari, e quindi considerare solo insiemi di 10 numeri dispari consecutivi maggiori di 50. Tra essi, almeno tre sono multipli di 3 ed esattamente due sono multipli di 5. I due numeri multipli di 5 di... | Italy | Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | MCQ | C | |
0jc4 | Problem:
Find all nonzero polynomials $P(x)$ with integer coefficients that satisfy the following property: whenever $a$ and $b$ are relatively prime integers, then $P(a)$ and $P(b)$ are relatively prime as well. Prove that your answer is correct. (Two integers are relatively prime if they have no common prime factors.... | [
"Solution:\nAnswer: $P(x)= \\pm x^{n}$ for each integer $n \\geq 0$.\n\nIt is evident that these polynomials meet the condition, since the only possible prime factors of $P(a)$ are the prime factors of $a$, so if $a$, $b$ have no prime factors in common, $P(a)$, $P(b)$ can't either.\n\nConsider any polynomial $P$ n... | United States | Bay Area Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | P(x) = ± x^n for integers n ≥ 0 | |
0b2q | Problem:
Let $\triangle ABC$ be an equilateral triangle with side length $16$. Points $D, E, F$ are on $CA, AB$, and $BC$, respectively, such that $DE \perp AE$, $DF \perp CF$, and $BD = 14$. The perimeter of $\triangle BEF$ can be written in the form $a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}$, where $a, b, c$, and $d$... | [
"Solution:\nLet $\\overline{AD} = 2x$, then $\\overline{DC} = 16 - 2x$. Since $\\triangle DAE$ and $\\triangle DCF$ are both $30$-$60$-$90$ triangles, then $\\overline{AE} = \\overline{AD}/2 = x$, $\\overline{ED} = x\\sqrt{3}$ and $\\overline{CF} = \\overline{DC}/2 = 8 - x$, $\\overline{FD} = (8 - x)\\sqrt{3}$. Sin... | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 31 | |
0i6h | Problem:
How many ways, without taking order into consideration, can $2002$ be expressed as the sum of $3$ positive integers (for instance, $1000+1000+2$ and $1000+2+1000$ are considered to be the same way)? | [
"Solution:\n\nCall the three numbers that sum to $2002$ $A$, $B$, and $C$. In order to prevent redundancy, we will consider only cases where $A \\leq B \\leq C$. Then $A$ can range from $1$ to $667$, inclusive. For odd $A$, there are $1000-\\frac{3(A-1)}{2}$ possible values for $B$. For each choice of $A$ and $B$, ... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 334000 | |
08in | Problem:
Let $ABC$ be a scalene triangle with $BC = a$, $AC = b$ and $AB = c$, where $a$, $b$, $c$ are positive integers. Prove that
$$
\left|a b^{2} + b c^{2} + c a^{2} - a^{2} b - b^{2} c - c^{2} a\right| \geq 2
$$
Problem:
Fie $ABC$ triunghi neisoscel cu lungimile $a, b, c$ ale laturilor numere naturale. Demonstraţ... | [
"Solution:\nDenote $E = a b^{2} + b c^{2} + c a^{2} - a^{2} b - b^{2} c - c^{2} a$. We have\n$$\n\\begin{aligned}\nE = & \\left(a b c - c^{2} a\\right) + \\left(c a^{2} - a^{2} b\\right) + \\left(b c^{2} - b^{2} c\\right) + \\left(a b^{2} - a b c\\right) = \\\\\n& (b - c)\\left(a c - a^{2} - b c + a b\\right) = (b ... | JBMO | 7th JBMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
063z | Problem:
Es sei $ABC$ ein Dreieck mit $AB = AC \neq BC$. Ferner sei $I$ der Inkreismittelpunkt von $ABC$.
Die Gerade $BI$ schneidet $AC$ im Punkt $D$, und die Orthogonale zu $AC$ durch $D$ schneidet $AI$ im Punkt $E$. Man beweise, dass der Spiegelpunkt von $I$ bei Spiegelung an der Achse $AC$ auf dem Umkreis des Dreie... | [
"Solution:\n\nZunächst beweisen wir (schon mit den zur Aufgabe passenden Bezeichnungen) folgendes Lemma: Die Mittelsenkrechte einer Seite $BII'$ und die Winkelhalbierende durch den dritten Eckpunkt $D$ schneiden sich auf dem Umkreis jedes nicht gleichschenkligen Dreiecks $BI'D$. („Südpolsatz\")\n\nBeweis des Lemmas... | Germany | 1. Auswahlklausur | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a6t | Problem:
Let $\mathcal{Q}_1$ be a quadrilateral such that the midpoints of its sides lie on a circle. Prove that there exists a cyclic quadrilateral $\mathcal{Q}_2$ with the same sidelengths as $\mathcal{Q}_1$, such that two of the angles in $\mathcal{Q}_2$ are equal. | [
"Solution:\n\nLet $A, B, C$ and $D$ be the vertices of $\\mathcal{Q}_1$, and $K, L, M$ and $N$ be the midpoints of the sides $AB, BC, CD$ and $DA$, respectively.\nWe have $KL \\parallel AC \\parallel MN$ and $LM \\parallel BD \\parallel NK$, and thus $KLMN$ is a parallelogram. From the problem condition it is known... | Nordic Mathematical Olympiad | Nordic Mathematical Contest | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometr... | null | proof only | null | |
08lt | Problem:
Let $s(a)$ denote the sum of digits of a given positive integer $a$. The sequence $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ of positive integers is such that $a_{n+1} = a_{n} + s(a_{n})$ for each positive integer $n$. Find the greatest possible $n$ for which it is possible to have $a_{n} = 2008$. | [
"Solution:\nSince $a_{n-1} \\equiv s(a_{n-1})$ (all congruences are modulo $9$), we have $2a_{n-1} \\equiv a_{n} \\equiv 2008 \\equiv 10$, so $a_{n-1} \\equiv 5$. But $a_{n-1} < 2008$, so $s(a_{n-1}) \\leq 28$ and thus $s(a_{n-1})$ can equal $5$, $14$ or $23$. We check $s(2008-5) = s(2003) = 5$, $s(2008-14) = s(199... | JBMO | 2008 Shortlist JBMO | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 6 | |
0jf0 | Problem:
Let $a, b$ be positive reals with $a > b > \frac{1}{2} a$. Place two squares of side lengths $a, b$ next to each other, such that the larger square has lower left corner at $(0,0)$ and the smaller square has lower left corner at $(a, 0)$. Draw the line passing through $(0, a)$ and $(a+b, 0)$. The region in th... | [
"Solution:\n\nAnswer: $\\quad \\frac{5}{3}$\n\nLet $t = \\frac{a}{b} \\in (1,2)$; we will rewrite the sum $a+b$ as a function of $t$. The area condition easily translates to $\\frac{a^{2} - a b + 2 b^{2}}{2} = 2013$, or $b^{2}(t^{2} - t + 2) = 4026 \\Longleftrightarrow b = \\sqrt{\\frac{4026}{t^{2} - t + 2}}$. Thus... | United States | HMMT November 2013 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 5/3 | |
01bp | Let $ABCD$ be a square inscribed in circle $\omega$ and let $P$ be a variable point on shorter arc $AB$ of $\omega$. Let $CP \cap BD = R$ and $DP \cap AC = S$. Show that triangles $ARB$ and $DSR$ have equal areas. | [
"Let $T = PC \\cap AB$. Then $\\angle BTC = 90^\\circ - \\angle PCB = 90^\\circ - \\angle PDB = 90^\\circ - \\angle SBD = \\angle BSC$, thus points $B$, $S$, $T$, $C$ are concyclic. So $\\angle TSC = 90^\\circ$, and therefore $TS \\parallel BD$. Hence\n$$\n[DSR] = [DTR] = [DTB] - [TBR] = [CTB] - [TBR] = [CRB] = [AR... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a25 | In a row there are $2024$ people, numbered $1$ to $2024$, and each of them either always tells the truth or always lies. Moreover, all $2024$ people know from each other whether they are always telling the truth or always lying. At some point, for each number $n$, the person numbered $n$ makes the statement: "At least ... | [] | Netherlands | Junior Mathematical Olympiad | [
"Discrete Mathematics > Logic"
] | null | proof and answer | 1012 | |
00xa | Problem:
Let $C$ be a circle in the plane. Let $C_{1}$ and $C_{2}$ be non-intersecting circles touching $C$ internally at points $A$ and $B$ respectively. Let $t$ be a common tangent of $C_{1}$ and $C_{2}$, touching them at points $D$ and $E$ respectively, such that both $C_{1}$ and $C_{2}$ are on the same side of $t$... | [
"Solution:\n\nLet $F_{1}$ be the second intersection point of the line $A D$ and the circle $C$ (see Figure 3). Consider the homothety with centre $A$ which maps $D$ onto $F_{1}$. This homothety maps the circle $C_{1}$ onto $C$ and the tangent line $t$ of $C_{1}$ onto the tangent line of the circle $C$ at $F_{1}$. ... | Baltic Way | Baltic Way 1992 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof only | null | |
0dcw | Let $ABCD$ is a trapezoid with $\angle A = \angle B = 90^{\circ}$ and let $E$ is a point lying on side $CD$. Let the circle $\omega$ is inscribed to triangle $ABE$ and tangents sides $AB$, $AE$ and $BE$ at points $P$, $F$ and $K$ respectively. Let $KF$ intersects segments $BC$ and $AD$ at points $M$ and $N$ respectivel... | [
"Let $KF$ meets $AB$ at $S$. We have known that $EP$, $AK$, $BF$ are concurrent at Gergonne's point, then $(SP, AB) = -1$. Let $Q$ be the projection of $P$ on $KF$ then $Q(SP, AB) = -1$, but $QS \\perp QP$ so $QP$ is the angle bisector of $\\angle AQB$. From that, we get $\\angle BQM = \\angle AQN$.\n\n![](attached... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane... | English | proof only | null | |
0fbi | Problem:
Los tres lados de un triángulo equilátero se suponen reflectantes (excepto en los vértices), de forma que reflejen hacia dentro del triángulo los rayos de luz situados en su plano, que incidan sobre ellos y que salgan de un punto interior del triángulo. Determinar el recorrido de un rayo de luz que, partiendo... | [
"Solution:\n\n\n\n$$\nl = \\sqrt{\\frac{25}{4} + \\frac{3}{4}} = \\sqrt{7}\n$$"
] | Spain | OME 11 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Tr... | null | proof and answer | sqrt(7) | |
0f68 | Problem:
$n > 3$ positive integers are written in a circle. The sum of the two neighbours of each number divided by the number is an integer. Show that the sum of those integers is at least $2n$ and less than $3n$. For example, if the numbers were $3$, $7$, $11$, $15$, $4$, $1$, $2$ (with $2$ also adjacent to $3$), th... | [] | Soviet Union | 18th ASU | [
"Number Theory > Divisibility / Factorization",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0aox | Problem:
How many ordered pairs $(x, y)$ of positive integers, where $x < y$, satisfy the equation
$$
\frac{1}{x} + \frac{1}{y} = \frac{1}{2007}
$$ | [
"Solution:\nWe can rewrite the given equation into\n$$\n(x - 2007)(y - 2007) = 2007^{2} = 3^{4} \\cdot 223^{2}\n$$\nSince $x < y$, we have $x - 2007 < y - 2007$. It follows that\n$$\n-2007 < x - 2007 < 2007 \\quad \\text{or} \\quad |x - 2007| < 2007\n$$\nThus, we have $|y - 2007| > 2007$.\n$$\n\\begin{array}{rl}\nx... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 7 |
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