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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0a0w | A triangle $ABC$ and a point $D$ on the line segment $AC$ are given. Let $M$ be the midpoint of $CD$ and let $\Omega$ be the circle through $B$ and $D$ tangent to $AB$. Let $E$ be the point such that $\triangle MDB \sim \triangle MBE$ and such that $D$ and $E$ lie on opposite sides of the line $MB$.
Show that $E$ lies ... | [
"We first prove that $\\triangle CMB \\sim \\triangle DBE$. Since $D$ and $E$ lie on opposite sides of $MB$, it holds that $\\angle DBE = \\angle DBM + \\angle MBE = \\angle DBM + \\angle MDB = \\angle CMB$ because of the given similarity and the exterior angle theorem. Moreover, it holds that\n$$\n\\frac{|DB|}{|BE... | Netherlands | IMO Team Selection Test 2 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
0e2a | For which positive integers $n$ does there exist a multiple of 13, such that the sum of its digits is equal to $n$? | [
"Any number with the sum of the digits equal to $1$ is a power of $10$, so it cannot be a multiple of $13$. Let us try and find a multiple of $13$ such that the sum of its digits will be equal to $2$. This number must have two digits equal to $1$ and the remaining digits must be $0$. We check the first few positive... | Slovenia | National Math Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | All positive integers except 1 | |
04o3 | Let $n$, $k$, $M$ and $a_1, a_2, \dots, a_n$ be positive integers such that
$$
\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n} = k \quad \text{and} \quad a_1 a_2 \dots a_n = M.
$$
If $M > 1$, prove that there does not exist a positive real number $x$ such that
$$
M(x+1)^k = (x+a_1)(x+a_2)\dots(x+a_n).
$$ | [
"Claim. For any positive integer $a$ and real number $x > 0$ the following inequality holds:\n$$\na(x+1)^{\\frac{1}{a}} \\le x+a,\n$$\nwhere the equality is satisfied if and only if $a=1$.\n\n*Proof.* If $a=1$, we easily see that the equality holds for all real numbers $x > 0$. Let us assume that $a > 1$, and let u... | Croatia | Croatian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0cqv | A 100-digit positive integer $n$ is called *unusual* if the last 100 digits of the decimal representation of $n^3$ form the decimal representation of $n$, but the last 100 digits of the decimal representation of $n^2$ do not. Prove that there are at least two unusual 100-digit numbers. (V. Senderov) | [
"For example, such numbers are $n_1 = 10^{100} - 1 = 99\\ldots9$ and $n_2 = \\frac{10^{100}}{2} - 1 = 49\\ldots9$. Indeed, the numbers\n$$\nn_1^3 - n_1 = (n_1 + 1)n_1(n_1 - 1) = 10^{100} \\cdot n_1(n_1 - 1)\n$$\nand\n$$\nn_2^3 - n_2 = (n_2 + 1)n_2(n_2 - 1) = 10^{100} \\cdot n_2 \\cdot \\frac{n_2 - 1}{2}\n$$\nare di... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0fzm | Problem:
Montrer qu'il n'existe pas deux nombres entiers naturels distincts tels que leur moyenne harmonique, géométrique, arithmétique et quadratique soient toutes des nombres entiers naturels. | [
"Solution:\n\nSupposons que $a, b$ satisfont les hypothèses. Posons $d=\\operatorname{pgcd}(a, b)$ et $a=d x, b=d y$. Ainsi, on voit par la moyenne arithmétique que $x, y$ sont impairs, et par la moyenne géométrique, comme $(x, y)=1$, que $x, y$ sont des carrés parfaits. Pour $x=x_{1}^{2}$ et $y=y_{1}^{2}$, la moye... | Switzerland | IMO-Selektionsprüfung | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis... | null | proof only | null | |
0brz | A triangle $ABC$ has its orthocenter $H$ distinct from its vertices and from the circumcenter $O$. Denote $M, N, P$ the circumcenters of the triangles $HBC, HCA$, respectively $HAB$. Prove that the lines $AM, BN, CP$ and $OH$ are concurrent.
Petru Braica | [
"If $D$ is the midpoint of $[BC]$, then $\\overrightarrow{OD} = \\frac{1}{2}(\\overrightarrow{OB} + \\overrightarrow{OC}) = \\frac{1}{2}(\\overrightarrow{OH} - \\overrightarrow{OA}) = \\frac{1}{2}\\overrightarrow{AH}$, hence $\\overrightarrow{OM} = \\overrightarrow{AH}$.\nIt follows that $AHMO$ is a parallelogram, ... | Romania | 67th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof only | null | |
0dro | In a triangle $ABC$, $M$ is the midpoint of $BC$ and $D$ is the point on $BC$ such that $AD$ bisects $\angle BAC$. The line through $B$ perpendicular to $AD$ intersects $AD$ at $E$ and $AM$ at $G$. Prove that $GD$ is parallel to $AB$. | [
"\n\nLet $BE$ intersect $AC$ at $H$. Thus $AE$ is the perpendicular bisector of $BH$ and so $ME$ is parallel to $CA$ and $\\frac{HC}{ME} = 2$. Hence the triangles $MEG$ and $AHE$ are similar. Also the triangles $DME$ and $DCA$ are similar. It follows that $\\frac{AG}{GM} = \\frac{AH}{ME} = ... | Singapore | Singapur | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
06ez | Let $ABC$ be an equilateral triangle with circumcentre $O$. Let $P$ be a point inside the triangle such that $\angle BPC = 120^\circ$ and $P \ne O$. Let $B'$ and $C'$ be the images of $B$ and $C$ respectively under the reflection in the line $PO$. Furthermore $B'C'$ meets $AP$ at $D$. Determine $AD : DP$. Justify your ... | [
"We have $AD : DP = 1 : 1$.\nSince $\\angle BPC = 120^\\circ = \\angle BOC$, the points $B$, $C$, $O$, $P$ are concyclic. As\n$$\n\\angle BPO = 180^\\circ - \\angle OCB = 150^\\circ,\n$$\nwe have $\\angle B'PB = 2(180^\\circ - 150^\\circ) = 60^\\circ$, and hence $\\triangle PB'B$ is equilateral. Similarly, as\n$$\n... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscel... | null | proof and answer | 1:1 | |
0b73 | In a triangle $ABC$ denote by $D$, $E$, respectively $F$, the points where the angle bisectors of $\angle CAB$, $\angle ABC$, respectively $\angle BCA$, meet its circumcircle.
a) Prove that the ortocenter of triangle $DEF$ coincides with the incenter of triangle $ABC$.
b) Prove that if $\overrightarrow{AD} + \overrig... | [
"a) Let $I$ be the incenter of the triangle $ABC$. $D, E, F$ are the midpoints of the arcs $\\widehat{BC}, \\widehat{CA}, \\widehat{AB}$.\nThe angle determined by the lines $AD$ and $EF$ is equal to $\\frac{1}{2}(\\widehat{AE} + \\widehat{DF}) = \\frac{1}{4}(\\widehat{AB} + \\widehat{BC} + \\widehat{CA}) = 90^\\cir... | Romania | Romanian Mathematical Olympiad | [
"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 / Coordinate Methods > Vectors"
] | English | proof only | null | |
0beh | A regular hexagonal prism $ABCDEF A'B'C'D'E'F'$ has the edge $AB = 12$ and the height $AA' = 12\sqrt{3}$. Let $N$ be the midpoint of the edge $CC'$.
a.
Prove that the lines $BF'$ and $ND$ are perpendicular.
b.
Find the distance between the lines $BF'$ and $ND$. | [
"a.\nLet $M$ be the midpoint of the segment $BB'$. Since $MN \\parallel AD$, the points $A$, $D$, $M$ and $N$ are coplanar. Let $Q$ be the midpoint of the segment $BF$. The intersection of the planes $(BFF')$ and $(ADN)$ is the line $MQ$. Notice that $BFF'B'$ is a square, hence $BF' \\perp MQ$ and then $AD \\perp (... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 15√42/7 | |
08sl | Let $m, n$ be positive integers. $m \times n$ square boxes of side length 1 form a grid for a rectangle (or a square if $m=n$) of sides $m$ and $n$. We want to color each of the $m \times n$ boxes by using one of the colors red, blue or black in such a way that all of the following conditions are satisfied:
* every red... | [
"Let us denote by $(i, j)$ the unit square lying in the $i$-th row and $j$-th column in the given grid. For two unit squares lying in the same row, call the square corresponding to the smaller value of $j$ to be lying to the left of the other one, and the one corresponding to the larger value of $j$ to be lying to ... | Japan | Japan Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | (1) Example for 3×4: Row 1: red, blue, red, blue; Row 2: black, black, black, black; Row 3: blue, red, blue, red. (2) All pairs (m, n) with one dimension a multiple of 3 and the other a multiple of 2, i.e., (m, n) = (3s, 2t) or (2t, 3s) for positive integers s, t. | |
0bwx | a) Let $m, n, p \in \mathbb{N}$, with $m > n$, such that $\sqrt{m} - \sqrt{n} = p$. Prove that $m$ and $n$ are both squares.
b) Find, with proof, all numbers $\overline{abcd}$ such that $\sqrt{\overline{abcd}} - \sqrt{\overline{acd}} = \overline{bb}$. | [
"a) We have $m = p^2 + 2p\\sqrt{n} + n$, hence $\\sqrt{n}$ is a rational number, and therefore, $n$ is a square. Similarly $m$ is a square, as well.\n\nb) We deduce from a) that $\\overline{abcd}$ and $\\overline{acd}$ are both squares. Since $11|\\overline{bb}$, it follows that $11|\\overline{abcd-acd}$, hence $11... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 1296 | |
09ar | Prove that every even number not exceeding $2n(4n + 1)$ can be written as $\pm 1 \pm 2 \pm 3 \pm \cdots \pm 4n$, here we choose $+$ or $-$. | [
"Let us prove by induction. For $n = 1$, we have to construct even numbers not exceeding $2 \\cdot (4 + 1) = 10$.\n$$\n\\begin{aligned}\n+1 - 2 - 3 + 4 &= 0, \\quad -1 + 2 - 3 + 4 &= 2, \\quad +1 + 2 - 3 + 4 = 4, \\\\\n+1 - 2 + 3 + 4 &= 6, \\quad -1 + 2 + 3 + 4 = 8, \\quad +1 + 2 + 3 + 4 = 10\n\\end{aligned}\n$$\nB... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | Mongolian | proof only | null | |
0b9v | Prove that there exist two functions $f, g: \mathbb{R} \to \mathbb{R}$, such that $f \circ g$ is strictly decreasing and $g \circ f$ is strictly increasing. | [
"Let\n$$\n\\begin{aligned}\n\\bullet A &= \\bigcup_{k \\in \\mathbb{Z}} ([-2^{2k+1}, -2^{2k}) \\cup (2^{2k}, 2^{2k+1}]); \\\\\n\\bullet B &= \\bigcup_{k \\in \\mathbb{Z}} ([-2^{2k}, -2^{2k-1}) \\cup (2^{2k-1}, 2^{2k}]).\n\\end{aligned}\n$$\n\nThus $A = 2B$, $B = 2A$, $A = -A$, $B = -B$, $A \\cap B = \\emptyset$, an... | Romania | 2011 Fourth ROMANIAN MASTER OF MATHEMATICS | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
03u9 | Let $a$ and $b$ be positive integers. Show that if $4ab - 1$ divides $(4a^2 - 1)^2$, then $a = b$. | [
"**Proof** Call $(a, b)$ a “bad pair” if it satisfies $4ab - 1 \\mid (4a^2 - 1)^2$ while $a \\neq b$. We use the method of infinite descent to prove there is no such “bad pair”.\n\n**Property 1** If $(a, b)$ is a “bad pair” and $a < b$, there exists an integer $c$ ($c < a$) such that $(a, c)$ is also a “bad pair”.\... | China | International Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
04qw | In the interior of a cyclic quadrangle $ABCD$ a point $P$ is given such that
$$
|\angle BPC| = |\angle BAP| + |\angle PDC|.
$$
Denote by $E$, $F$, and $G$ the feet of the perpendiculars from the point $P$ to the lines $AB$, $AD$ and $DC$, respectively. Show that the triangles $FEG$ and $PBC$ are similar. | [
"Let $k$ be the circumcircle of the quadrangle $ABCD$ and $k_1$, $k_2$ the circumcircles of the triangles $PAB$ and $PCD$, respectively. In the interior of the angle $BPC$, consider the half-line $PT$ such that $|\\angle BPT| = |\\angle BAP|$. Then the hypothesis on $P$ implies that (Fig. 1)\n$$\n|\\angle TPC| = |\... | Czech Republic | Czech-Slovak-Polish Match | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
09zm | In an apartment building with floors $0$ up to and including $10$, there is one person living on each floor. Each morning, everyone in the building must go to floor $0$ to go outside. Everyone is willing to walk the stairs for at most three floors. There can be at most four people in the lift at the same time. The lift... | [] | Netherlands | Junior Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | 20 | |
0799 | In triangle $ABC$, $O$ is the circumcenter and $H$ is the orthocenter. $M$ and $N$ are the midpoints of $BH$ and $CH$ respectively and $BB'$ is a diameter of the circumcircle. If $HONM$ be an inscribed quadrilateral, prove that
$$
B'N = \frac{1}{2}AC
$$ | [
"Let $\\omega$ be the circumcircle of the quadrilateral $HONM$. A $2$ homothety at $H$, takes $M, N$ to $B, C$ respectively. So this homothety maps $\\omega$ to the circumcircle of $HBC$. So $\\omega$ is tangent to the circumcircle of $HBC$ at $H$ and its radius is half of the radius of the circumcircle of $HBC$. W... | Iran | 27th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dw8 | Problem:
Dan je ostrokotni trikotnik $ABC$. Naj bo $C'$ nožišče višine na $AB$, $D$ in $E$ pa različni točki na daljici $CC'$. Naj bosta $F$ in $G$ pravokotni projekciji točke $D$ na stranici $AC$ oziroma $BC$. Dokaži, da je trikotnik $ABC$ enakokrak, če je štirikotnik $DGEF$ paralelogram. | [
"Solution:\n\n1. način\nOznačimo s $F'$ presečišče premice $FE$ s stranico $BC$ ter z $G'$ presečišče premice $GE$ s stranico $AC$. Naj bo $\\angle BAC = \\alpha$. Zaradi tetivnosti štirikotnika $FDGC$ je $\\angle CGF = \\angle CDF = \\alpha$. Predpostavimo, da je $DGEF$ paralelogram. Ker je $DF \\parallel GG'$, je... | Slovenia | 48. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a76 | Problem:
Let $S$ be the set of all points $t$ in the closed interval $[-1,1]$ such that for the sequence $x_{0}, x_{1}, x_{2}, \ldots$ defined by the equations $x_{0}=t, x_{n+1}=2 x_{n}^{2}-1$, there exists a positive integer $N$ such that $x_{n}=1$ for all $n \geq N$. Show that the set $S$ has infinitely many element... | [
"Solution:\n\nAll numbers in the sequence $\\{x_{n}\\}$ lie in the interval $[-1,1]$. For each $n$ we can pick an $\\alpha_{n}$ such that $x_{n}=\\cos \\alpha_{n}$. If $x_{n}=\\cos \\alpha_{n}$, then $x_{n+1}=2 \\cos^{2} \\alpha_{n}-1=\\cos (2 \\alpha_{n})$. The number $\\alpha_{n+1}$ can be chosen as $2 \\alpha_{n... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 3 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0di0 | How many triples $(x, y, z)$ of 6 digit positive integers exists, such that all digits of $x, y, z$ are odd and $x + y = 10z$? | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Other"
] | English | proof and answer | 3796875 | |
0bj3 | Let $I$, $J$ be intervals and consider $\varphi : J \to \mathbb{R}$ a continuous function which is nonzero on $J$. Let $f$, $g : I \to J$ be two differentiable functions such that $f' = \varphi \circ f$ and $g' = \varphi \circ g$. Prove that if there exists $x_0 \in I$ such that $f(x_0) = g(x_0)$, then $f$ and $g$ coin... | [
"As $\\varphi$ is non-zero and is continuous, the function $1/\\varphi$ is correctly defined and continuous. Consider an anti-derivative $F : J \\to \\mathbb{R}$. The given relation for $f$ can be then written $(F \\circ f)'(x) = 1$, for all $x \\in I$. Thus there exists $a \\in \\mathbb{R}$ such that $F(f(x)) = x ... | Romania | 65th Romanian Mathematical Olympiad | [
"Calculus > Differential Equations > ODEs",
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Calculus > Derivatives"
] | null | proof only | null | |
0juy | Problem:
Patrick and Anderson are having a snowball fight. Patrick throws a snowball at Anderson which is shaped like a sphere with a radius of $10$ centimeters. Anderson catches the snowball and uses the snow from the snowball to construct snowballs with radii of $4$ centimeters. Given that the total volume of the sn... | [
"Solution:\n\n$$\n\\left\\lfloor\\left(\\frac{10}{4}\\right)^3\\right\\rfloor = \\left\\lfloor\\frac{125}{8}\\right\\rfloor = 15.\n$$"
] | United States | HMMT February | [
"Geometry > Solid Geometry > Volume"
] | null | final answer only | 15 | |
07sm | Evaluate the sums
$$
\sum_{k=1,\ k \neq r}^{n} \cot \left( \frac{(k-r)\pi}{n+1} \right), \quad r = 1, 2, \dots, n.
$$ | [
"Recall that $\\cot(x + \\pi) = \\cot(x)$ and $\\cot(-x) = -\\cot(x)$ for all $x \\notin \\mathbb{Z}\\pi$, in particular\n$$\n\\cot\\left(\\frac{(k-r)\\pi}{n+1}\\right) = \\cot\\left(\\frac{(n+1+k-r)\\pi}{n+1}\\right). \\qquad (1)\n$$\nDefine\n$$\nS_r^n := \\sum_{k=1,\\ k \\neq r}^{n} \\cot \\left( \\frac{(k-r)\\pi... | Ireland | IRL_ABooklet_2020 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | S_r^n = \cot\left(\frac{r\pi}{n+1}\right) | |
0f14 | Problem:
Given $n$ unit vectors in the plane whose sum has length less than one. Show that you can arrange them so that the sum of the first $k$ has length less than $2$ for every $1 < k < n$. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0dnt | Problem:
Дати су природни бројеви $a_{1}, a_{2}, \ldots, a_{2 \text{ 2016 }}$ такви да за све $n, 1 \leqslant n \leqslant 2^{2016}$, важи
$$
a_{n} \leqslant 2016 \quad \text{и} \quad a_{1} a_{2} \cdots a_{n}+1 \text{ је потпун квадрат. }
$$
Доказати да је неки од бројева $a_{1}, a_{2}, \ldots, a_{22016}$ једнак 1. (Ду... | [
"Solution:\n\nКључна чињеница је да, ако су $a+1=u^{2}$ и $b=v^{2}$ потпуни квадрати и $a>b$, онда $a b+1$ није квадрат. Заиста, тада је $(u v-1)^{2}<a b+1=u^{2} v^{2}-v^{2}+1<(u v)^{2}$.\n\nНека су $p_{1}, p_{2}, \\ldots, p_{m}$ сви прости бројеви мањи од 2016. За $1 \\leqslant n \\leqslant 2^{2016}$ посматрајмо б... | Serbia | 10. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0cga | Let $X \in \mathcal{M}_2(\mathbb{C})$ be a matrix such that $X^{2023} = X^{2022}$. Prove that $X^3 = X^2$. | [
"Let us denote $d = \\det(X)$ and $t = \\tr(X)$. From the assumption $X^{2023} = X^{2022}$, we obtain $d^{2023} = d^{2022}$. Therefore, $d \\in \\{0, 1\\}$.\n\nIf $d = 1$, then $X$ is invertible. Then $X^{2022}$ is also invertible. We find $X = I_2$. So, the relation $X^3 = X^2$ is verified.\n\nAssume now $d = 0$. ... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | English | proof only | null | |
0f0h | Problem:
A triangle has area $1$, and sides $a \geq b \geq c$. Prove that $b^2 \geq 2$ | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
0fge | Problem:
Consideramos la curva $\Gamma$ definida por la ecuación $y^{2}=x^{3}+b x+b^{2}$, donde la constante $b$ es un número racional no nulo. Inscribir en la curva $\Gamma$ un triángulo cuyos vértices tengan coordenadas racionales. | [
"Solution:\n\nPrimera solución\nLos puntos $(0, b)$ y $(0,-b)$ son dos puntos de coordenadas racionales de $\\Gamma$; para hallar un tercer punto hacemos la siguiente identificación.\n$$\nx^{3}+b x+b^{2}=y^{2}=\\left(b+\\frac{x}{2}\\right)^{2}=b^{2}+b x+\\frac{x^{2}}{4}\n$$\nde donde\n$$\nx=\\frac{1}{4}, \\quad y= ... | Spain | OME 22 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (0, b), (0, −b), (1/4, b + 1/8) form a valid triangle of rational points on y^2 = x^3 + b x + b^2 for any nonzero rational b. | |
06wt | For a polynomial $P(x)$ with integer coefficients let $P^{1}(x)=P(x)$ and $P^{k+1}(x)= P\left(P^{k}(x)\right)$ for $k \geqslant 1$. Find all positive integers $n$ for which there exists a polynomial $P(x)$ with integer coefficients such that for every integer $m \geqslant 1$, the numbers $P^{m}(1), \ldots, P^{m}(n)$ le... | [
"Answer: All powers of 2 and all primes.\n\nDenote the set of residues modulo $\\ell$ by $\\mathbb{Z}_{\\ell}$. Observe that $P$ can be regarded as a function $\\mathbb{Z}_{\\ell} \\rightarrow \\mathbb{Z}_{\\ell}$ for any positive integer $\\ell$. Denote the cardinality of the set $P^{m}\\left(\\mathbb{Z}_{\\ell}\\... | IMO | IMO 2021 Shortlisted Problems | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | null | proof and answer | All powers of 2 and all primes | |
0hrx | Problem:
The sequences $\{x_{n}\}$ and $\{y_{n}\}$ are defined by $x_{0}=2$, $y_{0}=1$ and, for $n \geq 0$,
$$
x_{n+1}=x_{n}^{2}+y_{n}^{2} \quad \text{ and } \quad y_{n+1}=2 x_{n} y_{n}
$$
Find and prove an explicit formula for $x_{n}$ in terms of $n$. | [
"Solution:\n\nIf we add the formulas for $x_{n+1}$ and $y_{n+1}$ together, we get\n$$\nx_{n+1}+y_{n+1}=x_{n}^{2}+y_{n}^{2}+2 x_{n} y_{n}=(x_{n}+y_{n})^{2}.\n$$\nThus, increasing $n$ by 1 raises $x_{n}+y_{n}$ to the power 2. Since $x_{0}+y_{0}=3$, we get\n$$\nx_{n}+y_{n}=(\\cdots(3^{2})^{2} \\cdots)^{2}[n \\text{ sq... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | x_n = (3^{2^n} + 1) / 2 | |
0gjn | 已知四邊形 $ABCD$ 中 $AC, BD$ 交於 $E$, $AB, CD$ 交於 $F$, $AD, BC$ 交於 $G$ 且 $W, X, Y, Z$ 分別是 $E$ 對 $AB, BC, CD, DA$ 的對稱點。證明 $\odot(FWY), \odot(GXZ)$ 的其中一個交點在 $FG$ 上。
For the quadrilateral $ABCD$, let $AC$ and $BD$ intersect at $E$, $AB$ and $CD$ intersect at $F$, and $AD$ and $BC$ intersect at $G$. Additionally, let $W, X, Y$,... | [
"考慮對 $E$ 反演後的命題: 給定四邊形 $ABCD$, $AC, BD$ 交於 $E$, $\\odot(ABE), \\odot(CDE)$ 交於另一點 $F$, $\\odot(BCE), \\odot(DAE)$ 交於另一點 $G$, $W, X, Y, Z$ 分別為 $\\odot(ABE), \\odot(BCE), \\odot(CDE), \\odot(DAE)$ 外心, 證明: $\\odot(WFY), \\odot(XGZ)$ 的其中一個交點在 $\\odot(EFG)$ 上。\n\n令 $M$ 為 $WY$ 中點, $S$ 為 $E$ 對 $M$ 對稱點, $H$ 為 $EF$ 與 $\\odot... | Taiwan | IMO 1J, Independent Study 2 | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constru... | Chinese; English | proof only | null | |
05lf | Problem:
Soit $n$ un entier strictement positif et $x_{1}, \ldots, x_{n}$ des réels strictement positifs. Montrer qu'il existe des nombres $a_{1}, \ldots, a_{n} \in \{-1,1\}$ tels que :
$$
a_{1} x_{1}^{2}+a_{2} x_{2}^{2}+\cdots+a_{n} x_{n}^{2} \geqslant\left(a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}\right)^{2} .
$$ | [
"Solution:\n\nCommençons par ordonner les réels $x_{k}$ de sorte à ce que $x_{1} \\geq x_{2} \\geq \\ldots \\geq x_{n}>0$.\nMontrons par récurrence forte qu'une solution est fournie par la suite de coefficients définie par $a_{k}=1$ si $k$ est impair et $a_{k}=-1$ si $k$ est pair.\n\n$\\triangleright$ Si $n=1$, $a_... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
064k | Problem:
Bestimmen Sie alle ganzen Zahlen $n \geq 1$, für die es ein Paar $(a, b)$ von positiven ganzen Zahlen mit folgenden Eigenschaften gibt:
i) Keine dritte Potenz einer Primzahl teilt $a^{2}+b+3$.
ii) Es gilt $\frac{a b+3 b+8}{a^{2}+b+3}=n$. | [
"Solution:\n\nEs sei $p$ ein beliebiger Primfaktor von $a^{2}+b+3$ (ein solcher existiert, weil $a^{2}+b+3 \\geq 5$ gilt). Dann ist $b \\equiv -a^{2}-3 \\bmod p$. Wegen der Ganzzahligkeit von $n$ muss $p$ aber auch ein Primfaktor von $a b+3 b+8$ sein. Somit gilt $0 \\equiv a b+3 b+8 \\equiv a\\left(-a^{2}-3\\right)... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade 2022 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2 | |
0dve | Problem:
Poenostavi izraz
$$
\left(x+\sqrt[3]{3 \cdot \sqrt{\frac{x^{3}-1}{9}+\frac{x-x^{2}}{3}}}\right) \cdot (x-\sqrt{x-1})
$$ | [
"Solution:\n\nIzraz poenostavimo:\n$$\n\\left(x+\\sqrt[3]{3 \\cdot \\sqrt{\\frac{x^{3}-1}{9}+\\frac{x-x^{2}}{3}}}\\right) \\cdot (x-\\sqrt{x-1}) = \\left(x+\\sqrt[3]{3 \\cdot \\sqrt{\\frac{x^{3}-1+3x-3x^{2}}{9}}}\\right) \\cdot (x-\\sqrt{x-1})\n$$\n\nOpazimo, da je:\n$$\n\\frac{x^{3}-1}{9}+\\frac{x-x^{2}}{3} = \\fr... | Slovenia | 3. matematično tekmovanje dijakov srednjih tehniških in strokovnih sol | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | x^2 - x + 1 | |
09r7 | Problem:
Zij $\Gamma$ de omgeschreven cirkel van de scherphoekige driehoek $A B C$. De bissectrice van hoek $A B C$ snijdt $A C$ in het punt $B_{1}$ en de korte boog $A C$ van $\Gamma$ in het punt $P$. De lijn door $B_{1}$ loodrecht op $B C$ snijdt de korte boog $B C$ van $\Gamma$ in $K$. De lijn door $B$ loodrecht op... | [
"Solution:\n\nDat de bissectrice van hoek $A B C$ de korte boog $A C$ snijdt in $P$, betekent dat $P$ precies midden in deze boog $A C$ ligt. We moeten bewijzen dat $K L$ ook door $P$ gaat, dus dat $K L$ de boog $A C$ doormidden snijdt. Omdat $K$ op $\\Gamma$ ligt, betekent dat dat we moeten bewijzen dat $K L$ de b... | Netherlands | Toets 6 juni 2012 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jzn | Problem:
Consider five-dimensional Cartesian space
$$
\mathbb{R}^{5}=\left\{\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right) \mid x_{i} \in \mathbb{R}\right\}
$$
and consider the hyperplanes with the following equations:
- $x_{i}=x_{j}$ for every $1 \leq i<j \leq 5$;
- $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=-1$;
- $x_{1}+x_{2}+... | [
"Solution:\n\nNote that given a set of plane equations $P_{i}\\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\\right)=0$, for $i=1,2, \\ldots, n$, each region that the planes separate the space into correspond to a $n$-tuple of $-1$ and $1$, representing the sign of $P_{1}, P_{2}, \\ldots P_{n}$ for all points in that regi... | United States | HMMT November 2017 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 480 | |
07m0 | Let $\triangle ABC$ be a triangle and let $P$ denote the midpoint of the side $BC$. Suppose that there exist two points $M$ and $N$ interior to the sides $AB$ and $AC$ respectively, such that
$$
|AD| = |DM| = 2|DN|,
$$
where $D$ is the intersection point of the lines $MN$ and $AP$. Show that
$|AC| = |BC|$. | [
"**Step I:** Reduce to the case when $M$ coincides with $B$. The following provides two versions of this step:\n\n**Version 1:** The line through $B$ parallel to $MN$ intersects $AP$ at a point $E$ and $AC$ at a point $Q$. From the similar triangles $AMD$ and $ABE$ it follows that\n$$\n\\frac{|MD|}{|BE|} = \\frac{|... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem"
] | null | proof only | null | |
0eri | If $169! = 1 \times 2 \times 3 \times \cdots \times 169$ is written as the product of prime numbers, how many times would $13$ appear as a factor?
(A) 12 (B) 13 (C) 14 (D) 15 (E) 16 | [
"Since $13$ is a prime number, the only factors in $169!$ supplying powers of $13$ to the product are $13$, $26$, $39$, $\\ldots$, $156$, $169$. The first $12$ of these provide one power of $13$ each, but since $169 = 13^2$, it follows that the total power of $13$ is $12 + 2 = 14$."
] | South Africa | South African Mathematics Olympiad First Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | MCQ | C | |
0au2 | Problem:
Arrange these four numbers from smallest to largest: $\log_{3} 2$, $\log_{5} 3$, $\log_{625} 75$, $\frac{2}{3}$. | [
"Solution:\n\nThe numbers, arranged from smallest to largest, are $\\log_{3} 2$, $\\frac{2}{3}$, $\\log_{625} 75$, and $\\log_{5} 3$.\n\n- Since $\\left(3^{\\log_{3} 2}\\right)^{3} = 8$ and $\\left(3^{\\frac{2}{3}}\\right)^{3} = 9$, then $\\log_{3} 2 < \\frac{2}{3}$.\n\n- Since $\\left(625^{\\frac{2}{3}}\\right)^{3... | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | log_3 2 < 2/3 < log_625 75 < log_5 3 | |
055o | A rectangular grid whose side lengths are integers greater than 1 is given. Smaller rectangles with area equal to an odd integer and length of each side equal to an integer greater than 1 are cut out one by one. Finally one single unit square is left. Find the least possible area of the initial grid before the cuttings... | [
"Denote by $X$ the unit square left. Then $X$ cannot lie in a corner of the initial rectangle, as the strip between the neighbouring rectangle of $X$ and the edge of the rectangle could not be cut out (painted gray in Fig. 33). Similarly, $X$ cannot lie at a side of the initial rectangle, because the strip between ... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | English | proof and answer | 121 | |
0gv6 | Find all pairs of real numbers $x$ and $y$ such that
$$
\frac{x-2}{y} + \frac{5}{xy} = \frac{4-y}{x} - \frac{|y-2x|}{xy}.
$$ | [
"Зауважимо, що при **допустимих** значеннях $x$ і $y$ вихідна рівність рівносильна рівності\n$$\n(x-1)^2 + (y-2)^2 + |y-2x| = 0.\n$$"
] | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (1, 2) | |
0erh | $D$ is a point on side $AB$ of $\triangle ABC$, $E$ is a point on $CD$ and $F$ is a point on $CE$. The areas of triangles $AED$, $AEC$, $BFD$ and $BFC$ are $6$, $10$, $17$ and $7$, respectively. What is the area of $\triangle BEF$?
 | [
"If two triangles have the same height, then the ratio of their areas is equal to the ratio of their bases. It follows that $\\frac{DE}{EC} = \\frac{6}{10} = \\frac{3}{5}$, so $\\frac{DE}{DC} = \\frac{3}{3+5} = \\frac{3}{8}$. Similarly, $\\frac{DF}{FC} = \\frac{17}{7}$, so $\\frac{DF}{DC} = \\frac{17}{17+7} = \\fra... | South Africa | South African Mathematics Olympiad Second Round | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | final answer only | 8 | |
01ru | In the sequence of digits $2, 0, 2, 9, 3, \dots$ any digit is equal to the last digit in the decimal representation of the sum of four previous digits.
Do the four numbers $2, 0, 1, 5$ in that order occur in the sequence? (Folklore) | [
"In the given sequence consider all possible quadruple of successive digits. There are finite number of possible quadruples (no more than $10^4$). So if we proceed the sequence sufficiently long, some of the quadruple will occur more than once:\n$$\n2029 \\dots d\\ c\\ b\\ a \\dots d_1\\ c_1\\ b_1\\ a_1 \\dots\n$$\... | Belarus | SELECTION and TRAINING SESSION | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | English | proof and answer | Yes | |
09s2 | Problem:
Zij $n$ een positief geheel getal. Daniël en Merlijn spelen een spel. Daniël heeft $k$ vellen papier die naast elkaar op tafel liggen, waarbij $k$ een positief geheel getal is. Hij schrijft op elk vel papier een aantal van de getallen 1 tot en met $n$ (geen enkel getal mag ook, alle getallen mag ook). Op de a... | [
"Solution:\n\nWe geven de vellen papier van Daniël allemaal een andere kleur. Verder hebben we $n$ doosjes met daarop de getallen 1 tot en met $n$. We zorgen ook voor voldoende beschikbare fiches in precies de kleuren van de vellen papier van Daniël. Per vel bekijkt hij de getallen op de voorkant van het vel en sto... | Netherlands | Selectietoets | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | the smallest integer k such that 2^k > n (equivalently, ⌊log2 n⌋ + 1) | |
06wf | A thimblerigger has 2021 thimbles numbered from 1 through 2021. The thimbles are arranged in a circle in arbitrary order. The thimblerigger performs a sequence of 2021 moves; in the $k^{\text{th}}$ move, he swaps the positions of the two thimbles adjacent to thimble $k$.
Prove that there exists a value of $k$ such that... | [
"Assume the contrary. Say that the $k^{\\text{th}}$ thimble is the central thimble of the $k^{\\text{th}}$ move, and its position on that move is the central position of the move.\n\nStep 1: Black and white colouring.\nBefore the moves start, let us paint all thimbles in white. Then, after each move, we repaint its... | IMO | IMO 2021 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
06tr | Find all positive integers $n$ for which all positive divisors of $n$ can be put into the cells of a rectangular table under the following constraints:
- each cell contains a distinct divisor;
- the sums of all rows are equal; and
- the sums of all columns are equal. | [
"Solution 1. Suppose all positive divisors of $n$ can be arranged into a rectangular table of size $k \\times l$ where the number of rows $k$ does not exceed the number of columns $l$. Let the sum of numbers in each column be $s$. Since $n$ belongs to one of the columns, we have $s \\geqslant n$, where equality hol... | IMO | IMO 2016 Shortlisted Problems | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 1 | |
02mx | Problem:
Determinando uma sequência - Numa certa sequência de 80 números, qualquer termo, salvo as duas extremidades, é igual ao produto de seus termos vizinhos. O produto dos 40 primeiros termos da sequência é $8$ e o produto de todos os termos também é $8$. Determine os termos da sequência. | [
"Solution:\n\nSeja a sequência $a_1, a_2, \\ldots, a_{80}$.\n\nDado que, para $2 \\leq k \\leq 79$, temos:\n$$\na_k = a_{k-1} \\cdot a_{k+1}\n$$\n\nIsso implica que:\n$$\na_{k+1} = \\frac{a_k}{a_{k-1}}\n$$\n\nPortanto, a sequência é recorrente e depende dos dois primeiros termos.\n\nVamos calcular os primeiros term... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | The sequence is periodic with period six: 1/64, sqrt(2)/32, 2 sqrt(2), 64, 16 sqrt(2), 1/(2 sqrt(2)), repeating. | |
009u | Players $A$ and $B$ play the following game on a band of consecutive unit cells infinite in one direction. On each move of his, $A$ marks two arbitrary cells that were not marked before. On each move of his, $B$ deletes any block of consecutive marks. The goal of $A$ is to obtain $10$ consecutive marks, the goal of $B$... | [
"Player $A$ has a winning strategy. On his first $2^7$ moves he marks $2^8$ arbitrary cells so that the distance of every two of them is at least $10$. Such marks are not consecutive, so every time $B$ deletes exactly one mark on his move. Thus after $2^7$ combined moves of the two there are $2^7$ marks at distance... | Argentina | NATIONAL XXX OMA | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Player A | |
09ff | Let $ABC$ be a triangle with $2 \cdot \angle A > \angle B$. A point $D$ is chosen on the segment $BC$ so that $2 \cdot \angle CAD = \angle ABC$. Let $H$ be the orthocenter of the triangle $ABC$ and $AD$ intersects the circumcircle of the triangle $AHC$ at $K$, differently from $A$. $AB$ intersects the circumcircle of t... | [
"Let $\\beta := \\angle B$. Since $AHKC$ is a cyclic quadrilateral, we have $\\angle AKC = \\angle AHC = 180^\\circ - \\beta$. By assumption $\\angle KAC = \\frac{\\beta}{2}$. Hence $\\angle ACK = 180^\\circ - \\angle AKC - \\angle KAC = \\frac{\\beta}{2}$. Therefore $\\triangle AKC$ is an isosceles triangle.\n\n![... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscella... | null | proof only | null | |
0k0y | Problem:
Four standard six-sided dice are rolled. Find the probability that, for each pair of dice, the product of the two numbers rolled on those dice is a multiple of $4$. | [
"Solution:\n\nIf any two of the dice show an odd number, then this is impossible, so at most one of the dice can show an odd number. We take two cases:\n\nCase 1: If exactly one of the dice shows an odd number, then all three other dice must show a multiple of $4$, which can only be the number $4$. The probability ... | United States | HMMT November 2018 | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 31/432 | |
08rw | A quadrilateral $ABCD$ that satisfies $AB = 5$, $BC = 7$, $CD = 6$ is given. And $AC$ and $BD$ are perpendicular to each other. Find the length of $DA$. | [
"Define $P$ as the intersection point of $AC$ and $BD$. Then from the Pythagorean theorem, $AB^2 = AP^2 + BP^2$, $BC^2 = BP^2 + CP^2$, $CD^2 = CP^2 + DP^2$, $DA^2 = DP^2 + AP^2$. Then $DA^2 = AB^2 + CD^2 - BC^2 = 5^2 + 6^2 - 7^2 = 12$. So $DA = 2\\sqrt{3}$."
] | Japan | Japan 2007 | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 2√3 | |
0coo | Given a positive integer $n$. Prove that there exist $n$ consecutive positive integers such that their product is divisible by each prime number not exceeding $2n+1$, but is not divisible by any other prime number.
Дано натуральное $n > 1$. Докажите, что найдутся такие $n$ последовательных натуральных чисел, что их пр... | [
"Предположим, что число $n + 1$ составное; покажем, что тогда подходят числа $n + 2, \\dots, 2n + 1$. Очевидно, их произведение делится на все простые числа из отрезка $[n + 2, 2n + 1]$, но не делится на простые числа, большие $2n + 1$ (ибо все сомножители не превосходят $2n + 1$). Для любого же простого $p \\le n$... | Russia | Final round | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English; Russian | proof only | null | |
09b2 | A group of the pupils in a class are called *dominant* if any other pupil from the class has a friend in the group. If it is known that there exists at least $100$ dominant group, then there exists one more dominant group. | [
"It suffices to prove that the number of dominant groups are odd. Let $S$ be the set of all the pupils in the class and $V$ be the set of all nonempty subsets of $S$. Thus, $|V| = 2^{|S|} - 1$. Now let us define a graph $G$ on $V$. We join $A \\in V$ and $B \\in V$ by edge iff there is no friends between $A$ and $B... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
08bh | Problem:
Sia $n$ il più piccolo intero positivo di 4 cifre maggiore o uguale a 2016 che gode della seguente proprietà: esiste un intero positivo $S$ tale che
$$
S=\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d+S}}}}
$$
dove $a, b, c, d$ sono, nell'ordine, la cifra delle migliaia, delle centinaia, delle decine e delle unità di $n$.... | [
"Solution:\n\nLa risposta è 2167. Innanzitutto 2167 soddisfa la proprietà richiesta con $S=2$: infatti\n$$\n\\sqrt{2+\\sqrt{1+\\sqrt{6+\\sqrt{7+2}}}}=\\sqrt{2+\\sqrt{1+\\sqrt{6+\\sqrt{9}}}}=\\sqrt{2+\\sqrt{1+\\sqrt{9}}}=\\sqrt{2+\\sqrt{4}}=\\sqrt{4}=2\n$$\nMostriamo ora che non esistono interi $n$ tra 2016 e 2166 c... | Italy | Gara di Febbraio | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | 2167 | |
05c6 | Integers are assigned to variables $x$, $y$ and $z$ to satisfy the equation
$$
x^3 + y^3 + z^3 - 3xyz = 2025.
$$
Find all possible values of the sum $x + y + z$. | [
"The given equation is equivalent to the equation\n$$\nx^3 + y^3 + z^3 = 2025 + 3xyz.\n$$\nAs $3 \\mid 2025$ and $3 \\mid 3xyz$, also $3 \\mid x^3 + y^3 + z^3$. Thus $3 \\mid x + y + z$ as an integer and its cube are congruent modulo 3. Let $x + y + z = 3m$.\n\nW.l.o.g., let $x \\le y \\le z$ and $x = m-a$, $y = m+... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | 3, 9, 27, 75, 225, 675 | |
014v | Problem:
In a school class with $3n$ children, any two children make a common present to exactly one other child. Prove that for all odd $n$ it is possible that the following holds:
For any three children $A$, $B$ and $C$ in the class, if $A$ and $B$ make a present to $C$ then $A$ and $C$ make a present to $B$. | [
"Solution:\n\nAssume there exists a set $\\mathscr{S}$ of sets of three children such that any set of two children is a subset of exactly one member of $\\mathscr{S}$, and assume that the children $A$ and $B$ make a common present to $C$ if and only if $\\{A, B, C\\} \\in \\mathscr{S}$. Then it is true that any two... | Baltic Way | Baltic Way 2008 | [
"Discrete Mathematics > Other",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof only | null | |
05ee | Problem:
Let $n > 1$ be an integer. In a configuration of an $n \times n$ board, each of the $n^{2}$ cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one sq... | [
"Solution:\n\nWe will show that the maximum number of good cells over all possible starting configurations is\n$$\n\\frac{n^{2}}{4} \\quad \\text{if } n \\text{ is even and}\n$$\n$$\n0 \\quad \\text{if } n \\text{ is odd.}\n$$\n\n## Odd $n$\n\nFirst, we will prove that there are no good cells if $n$ is an odd numbe... | European Girls' Mathematical Olympiad (EGMO) | EGMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | n^2/4 if n is even; 0 if n is odd | |
03gk | Problem:
Let $ABC$ be an equilateral triangle, and $P$ be an arbitrary point within the triangle. Perpendiculars $PD$, $PE$, $PF$ are drawn to the three sides of the triangle. Show that, no matter where $P$ is chosen,
$$
\frac{PD + PE + PF}{AB + BC + CA} = \frac{1}{2\sqrt{3}}
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0jco | Problem:
Alice and Bob are playing a game of Token Tag, played on an $8 \times 8$ chessboard. At the beginning of the game, Bob places a token for each player on the board. After this, in every round, Alice moves her token, then Bob moves his token. If at any point in a round the two tokens are on the same square, Ali... | [
"Solution:\n\na. Color the checkerboard in the standard way so that half of the squares are black and the other half are white. Bob's winning strategy is to place the two coins on the same color, so that Alice must always move her coin on to a square with the opposite color as the square containing Bob's coin.\n\nb... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
07lv | $AB$ is a chord of length $6$ in a circle of radius $5$ and with centre $O$. A square is symmetrically inscribed in the sector $OAB$ with two vertices on the circumference. Find the area of the square. | [
"Let $XYZ$ be the square with $Y$ and $Z$ on the circle, $X$ on $OA$ and $Y$ on $OB$. Let $2a = |YZ|$. Draw $OD \\perp YZ$, $D$ on $YZ$. Then $OD \\perp AB$ and bisects $AB$. So $|OA| = 5$, $|AC| = 3 \\Rightarrow |OC| = 4 \\Rightarrow \\frac{|XE|}{|OE|} = \\frac{|AC|}{|OC|} = \\frac{3}{4} \\Rightarrow \\frac{a}{|OE... | Ireland | Irska | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 900/109 | |
00fj | Given a nondegenerate triangle $A B C$, with circumcentre $O$, orthocentre $H$, and circumradius $R$, prove that $|O H|<3 R$. | [
"Embed $A B C$ in the complex plane, with $A$, $B$ and $C$ in the circle $|z|=R$, so $O$ is the origin. Represent each point by its lowercase letter. It is well known that $h=a+b+c$, so\n$$\nO H=|a+b+c| \\leq |a|+|b|+|c|=3 R.\n$$\nThe equality cannot occur because $a$, $b$, and $c$ are not collinear, so $O H<3 R$.\... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1994 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geom... | null | proof only | null | |
0ii9 | Problem:
A number $n$ is called bummed out if there is exactly one ordered pair of positive integers $(x, y)$ such that
$$
\left\lfloor x^{2} / y\right\rfloor+\left\lfloor y^{2} / x\right\rfloor=n .
$$
Find all bummed out numbers. | [
"Solution:\nSuppose $n$ is bummed out. If $(a, b)$ is one solution for $(x, y)$ to the given equation $\\left\\lfloor x^{2} / y\\right\\rfloor+\\left\\lfloor y^{2} / x\\right\\rfloor=n$, then $(b, a)$ is another, so the unique solution $(a, b)$ better have the property that $a=b$ and $n=2 a \\geq 2$. In particular,... | United States | Harvard-MIT Mathematics Tournament, Team Round A | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2, 6, 8, 10 | |
026b | Problem:
Qual é a metade? (N2/N3) - Considere a figura ao lado, em que $AB = AE = ED = CD = CA$ e o arco $CB$ é um arco de círculo centrado no ponto $E$. Você sabe repartir essa figura em duas partes idênticas, que possam ser superpostas?
 | [
"Solution:\n\n"
] | Brazil | Desafios | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0ac3 | Two lines $p$ and $q$ intersect at $C$. The point $C$ divides the line $p$ into two half-lines that, together with one of the half-lines of $q$ with starting point at $C$, form two angles $pCq$ and $qCp$. On the bisector of the angle $pCq$ a point $M$ is chosen such that $MN \parallel p$. The segment $MN$ intersects th... | [
"Let $P$ be a point on the half-line $Cp$ from the angle $pCq$ and $Q$ be a point on the half-line $Cp$ of the angle $qCp$. Let $CM$ be the bisector of the angle $pCq$ and $CN$ be the bisector of the angle $qCp$. $MN \\parallel PQ$ and let $D$ be the intersection point of the line $q$ with the line $MN$.\n\nBecause... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09hu | Let $ABC$ be a triangle with a point $D$ on $AB$ and a point $E$ on $BC$ such that $AD = CE$ and $2DE = AC$. Show that the circumradius of $BDE$ is equal to half the circumradius of $ABC$.
(Proposed by Khulan Tumenbayar) | [
"Let us choose a point $S$ on the circumcircle of $BDE$ such that $SD = SE$. Since $\\angle BES = \\angle BDS$, we see that $\\angle SDA = \\angle SEC$. Because $AD = EC$ and $SD = SE$, we have $\\triangle SDA = \\triangle SEC$. It follows that $\\angle ASC = \\angle DSE = \\angle DBE = \\angle ABC$. Hence $S$ must... | Mongolia | Round 3 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
035l | Problem:
A real number is assigned to every point in the plane. Let $\mathcal{P}$ be a convex $n$-gon. It is known that for every $n$-gon similar to $\mathcal{P}$ the sum of the numbers assigned to its vertices is equal to $0$. Prove that all numbers assigned to the points in the plane are equal to $0$. | [
"Solution:\n\nLet $O$ be an arbitrary point in the plane and let $A_{1,1} A_{2,1} \\ldots A_{n,1}$ be an $n$-gon similar to $\\mathcal{P}$ and containing $O$. Consider the $n$-gons\n$$\nO A_{1,1} A_{1,2} \\ldots A_{1, n-1},\\quad O A_{2,1} A_{2,2} \\ldots A_{2, n-1},\\quad \\ldots,\\quad O A_{n,1} A_{n,2} \\ldots A... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0fcs | Problem:
¿Cuántas ternas ordenadas de números naturales $(a, b, c)$ distintos de la unidad hay tales que
$$
a.b.c = 7^{39}?
$$ | [
"Solution:\n\nComo $7$ es primo y $a \\neq 1$, $b \\neq 1$ y $c \\neq 1$, $a.b.c = 7^{p} \\cdot 7^{q} \\cdot 7^{r} = 7^{39}$ con $p, q, r \\in \\mathbb{N}$.\n\nPor tanto, el número de ternas ordenadas $(a, b, c)$ será el mismo que el de ternas $(p, q, r)$ con la condición $p + q + r = 39$.\n\nTabulemos y contemos:\... | Spain | Fase Local | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 703 | |
0f5j | Problem:
$M$ is the midpoint of $BC$. $E$ is any point on the side $AC$ and $F$ is any point on the side $AB$. Show that area $\{MEF\} \leq$ area $\{BMF\} +$ area $CME$. | [] | Soviet Union | 17th ASU | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
09r0 | Problem:
We hebben twee dozen met ballen. In de ene doos zitten $m$ ballen, in de andere doos $n$ ballen, waarbij $m, n>0$. Twee verschillende handelingen zijn toegestaan:
(i) Verwijder uit beide dozen een gelijk aantal ballen.
(ii) Vergroot het aantal ballen in één van de dozen met een factor $k$.
Is het altijd mogel... | [
"Solution:\n\nBekijk eerst het geval $k=2$. We kunnen alle ballen uit beide dozen verwijderen op de volgende manier.\nAls $m=n$, dan halen we $m$ ballen uit beide dozen en zijn we klaar. Als $m \\neq n$, kunnen we zonder verlies van algemeenheid aannemen dat $m<n$. Als bovendien geldt $2 m<n$, dan verdubbelen we he... | Netherlands | Dutch TST | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | a) Yes. b) No. | |
0600 | Problem:
Avant un championnat, chaque équipe participante propose au plus $t$ couleurs différentes pour son maillot. Un ensemble $S$ d'équipes est dit identifiable si l'on peut assigner à chaque équipe de $S$ une couleur apparaissant dans son ensemble de propositions et n'apparaissant dans aucun ensemble de couleurs d... | [
"Solution:\n\nMontrons que le nombre recherché est $g(n, t)=\\left\\lceil\\frac{n}{t}\\right\\rceil$.\n\nSi $n$ s'écrit sous la forme $k \\cdot t + r$, avec $0 \\leqslant r < t$, on peut considérer la situation où $k$ équipes demandent chacune $t$ couleurs toutes distinctes deux à deux, puis une dernière équipe dem... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - Envoi 5 : Pot Pourri | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | ceil(n/t) | |
0ckg | Let $n \in \mathbb{N}$, $n \ge 2$, and let $a_1, a_2, \dots, a_n$ be real numbers. Denote
$$
S = \sum_{1 \le i < j \le n} |a_j - a_i| \quad \text{and} \quad d = \max\{|a_j - a_i| \mid 1 \le i, j \le n\}.
$$
Prove that
$$
(n-1)d \le S \le \frac{n^2 d}{4}
$$ | [
"Without loss of generality, we can assume that $a_1 \\le a_2 \\le \\dots \\le a_n$, and denote $d_k = a_{k+1} - a_k$, for $k = 1, 2, \\dots, n-1$. Then we have:\n* $d_1 + d_2 + \\dots + d_{n-1} = d$;\n* $|a_j - a_i| = d_i + d_{i+1} + \\dots + d_{j-1}$, for any $i < j$ with $i, j \\in \\{1, \\dots, n\\}$.\nDenote b... | Romania | 75th NMO Selection Tests | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0b0v | Problem:
In $\triangle ABC$, $AB = AC$. A line parallel to $BC$ meets sides $AB$ and $AC$ at $D$ and $E$, respectively. The angle bisector of $\angle BAC$ meets the circumcircles of $\triangle ABC$ and $\triangle ADE$ at points $X$ and $Y$, respectively. Let $F$ and $G$ be the midpoints of $BY$ and $XY$, respectively.... | [
"Solution:\n\nLet $F'$ be the reflection of $F$ over line $XY$. Observe that, by symmetry, we get $\\angle ADY = \\angle AEY$. As quadrilateral $ADYE$ is cyclic, both angles must be right, and hence $\\angle BDY$ is right as well.\n\nThus $F$ is the circumcenter of triangle $BDY$, so $\\angle FDB = \\angle FBD$. By... | Philippines | Philippines Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fit | Problem:
La figura muestra un plano con calles que delimitan 12 manzanas cuadradas. Una persona $P$ va desde $A$ hasta $B$ y otra $Q$ desde $B$ hasta $A$. Ambas parten a la vez siguiendo caminos de longitud mínima con la misma velocidad constante. En cada punto con dos posibles direcciones a tomar, ambas tienen la mis... | [
"Solution:\n\nDefinamos un sistema de coordenadas con origen en $A$ y unidad el lado de un cuadrado.\nComo $P$ y $Q$ recorren caminos de longitud mínima, $P$ sólo puede ir a la derecha o arriba y $Q$ a la izquierda o abajo. Todos los caminos tienen longitud $7$, y $P$ y $Q$ sólo se podrán encontrar entre el tercero... | Spain | Olimpiada Matemática Española | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 37/256 | |
02j3 | Problem:
Numa sequência, cada termo, a partir do terceiro, é a soma dos dois termos imediatamente anteriores, o segundo termo é 1 e o quinto termo é 2005. Qual é o sexto termo?
A) 3002
B) 3008
C) 3010
D) 4002
E) 5004 | [
"Solution:\n\nSeja $x$ o primeiro termo. Como o segundo termo é $1$ e, a partir do terceiro, cada termo é a soma dos dois anteriores, temos:\n\n- terceiro termo: $1 + x$;\n- quarto termo: $1 + (1 + x) = 2 + x$;\n- quinto termo: $(1 + x) + (2 + x) = 3 + 2x$;\n- sexto termo: $(2 + x) + (3 + 2x) = 5 + 3x$.\n\nComo o q... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | B | |
0bh7 | Find how many pairs $(a, b)$, with $a$ and $b$ non-nil digits, have the property that $\frac{a}{b}$ is irreducible and the decimal fractions $\frac{a}{b}$ and $\frac{a+b}{b(b+1)}$ are finite.
Gabriel Vrînceanu | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 20 | |
03ax | Let $ABCDE$ be a pentagon for which $\angle A = \angle B = \angle C = \angle D = 120^\circ$. Find the minimal possible value of the ratio $\frac{AC \cdot BD}{AE \cdot ED}$. | [] | Bulgaria | Selection test for 51. International Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | 3/4 | |
0fzr | Problem:
Sei $ABCD$ ein Parallelogramm. Nehme an, es existiere ein Punkt $P$ im Innern des Parallelogramms, der auf der Mittelsenkrechten von $AB$ liegt und sodass $\angle PBA = \angle ADP$ gilt.
Zeige, dass $\angle CPD = 2 \angle BAP$ gilt. | [
"Solution:\n\nWir verschieben das Dreieck $ABP$ um den Vektor $\\overrightarrow{AD}$, das heisst $A$ kommt auf $D$ und $B$ auf $C$ zu liegen. Der Punkt $P$ wird auf den Punkt $P'$ abgebildet, sodass $PP' = AD$ gilt und $PP'$ parallel zu $AD$ ist. Nun gilt:\n$$\n\\angle P'CD = \\angle PBA = \\angle ADP = \\angle P'P... | Switzerland | IMO-Selektion | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"G... | null | proof only | null | |
0d99 | Let $ABC$ be a triangle with $A$ is an obtuse angle. Denote $BE$ as the internal angle bisector of triangle $ABC$ with $E \in AC$ and suppose that $\angle AEB = 45^{\circ}$. The altitude $AD$ of triangle $ABC$ intersects $BE$ at $F$. Let $O_1, O_2$ be the circumcenter of triangles $FED, EDC$. Suppose that $EO_1, EO_2$ ... | [
"Denote $O$ as the projection of $B$ on $AC$ and $BK$ is the diameter of $(O, OB)$. It is easy to see that $E \\in (O)$. We have\n$$\n\\angle BCA = \\angle AEB - \\angle CBE = \\angle OBE - \\angle ABE = \\angle OBA = \\angle OKA.\n$$\nThis implies that $ODCK$ is the inscribed quadrilateral and $\\angle KDC = 90^{\... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > ... | English | proof only | null | |
04lq | Let $ABC$ be an acute-angled triangle. The tangents at $A$ and $B$ to its circumcircle intersect at $M$. The line through $M$ parallel to the side $BC$ intersects the side $CA$ at point $N$. Prove that $|BN| = |CN|$ holds. | [] | Croatia | Mathematical competitions in Croatia | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0aob | Problem:
How many squares are determined by the lines with equations $x = k^{2}$ and $y = l^{2}$, where $k, l \in \{0, 1, 2, 3, \ldots, 9\}$? | [] | Philippines | Area Stage | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 59 | |
0gxp | Solve the system of equations over real numbers:
$$
\begin{cases}
x^3 = 2y^3 + z - 2, \\
y^3 = 2z^3 + x - 2, \\
z^3 = 2x^3 + y - 2.
\end{cases}
$$ | [
"From the first two equations of our system we obtain: $x^3 - y^3 = 2y^3 - 2z^3 + y - z = (y-z)(2y^2 + 2yz + 2z^2 + 1)$, analogously, we can easily get equalities: $y^3 - z^3 = (z-x)(2z^2 + 2zx + 2x^2 + 1)$, $z^3 - x^3 = (x-y)(2x^2 + 2xy + 2y^2 + 1)$. It is easy to see that $(2y^2 + 2yz + 2z^2 + 1) = y^2 + z^2 + (y... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | (1, 1, 1) | |
00ji | We wish to color the squares in a strip of $n$ squares that are numbered from $1$ through $n$ from left to right. Each square is to be colored with one of the colors $1$, $2$ or $3$. The even numbered squares can be colored with any color, but the odd numbered squares can only be colored with the odd colors $1$ or $3$.... | [
"Let $a_n$ be the number of colorings of a strip of length $n$ ending in a square colored with $1$, and further let $b_n$ be the number of such colorings ending in a square colored with $2$. The number of colorings ending in $3$ is also $a_n$, since any coloring ending in $1$ can be uniquely changed to one ending i... | Austria | Austrian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | (3 + (-1)^n) * 3^{floor((n-1)/2)} | |
0ae5 | Броевите $m$ и $n$ се взаемно прости. Дропката $\frac{3n-m}{5n+2m}$ може да се скрати со некој природен број. Определи го бројот со кој коже да се скрати. | [
"Нека претпоставиме дека $k, k > 1$ е бројот со кој може да се скрати дропката. Според тоа, постојат природни броеви $p$ и $s$, такви што $(p,s) = 1$ и $3n - m = kp, 5n + 2m = ks$. Ако го решиме системот\n$$\n\\begin{cases} 3n - m = kp \\\\ 5n + 2m = ks \\end{cases}\n$$\nпо $n$ и $m$ ќе добиеме $n = \\frac{k(2p+s)}... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | Macedonian, English | proof and answer | 11 | |
0j3i | Problem:
A sphere is the set of points at a fixed positive distance $r$ from its center. Let $\mathcal{S}$ be a set of 2010-dimensional spheres. Suppose that the number of points lying on every element of $\mathcal{S}$ is a finite number $n$. Find the maximum possible value of $n$. | [
"Solution:\nAnswer: 2 The answer is 2 for any number of dimensions. We prove this by induction on the dimension.\n\nNote that 1-dimensional spheres are pairs of points, and 2-dimensional spheres are circles.\n\nBase case, $d=2$ : The intersection of two circles is either a circle (if the original circles are identi... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2 | |
042y | In $\triangle ABC$, $AB = 6$, $BC = 4$, the median to side $AC$ is $\sqrt{10}$. Then the value of $\sin^6 \frac{A}{2} + \cos^6 \frac{A}{2}$ is ______. | [
"Let $M$ be the midpoint of $AC$. By the median formula we have\n$$\n4BM^2 + AC^2 = 2(AB^2 + BC^2),\n$$\nand thus\n$$\nAC = \\sqrt{2(6^2 + 4^2) - 4 \\cdot 10} = 8.\n$$\nBy the law of cosines, we obtain $\\cos A = \\frac{CA^2 + AB^2 - BC^2}{2CA \\cdot AB} = \\frac{8^2 + 6^2 - 4^2}{2 \\cdot 8 \\cdot 6} = \\frac{7}{8}... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 211/256 | |
0jbc | Problem:
If you roll four fair 6-sided dice, what is the probability that at least three of them will show the same value? | [
"Solution:\n\nWe have two cases: either three of the dice show one value and the last shows a different value, or all four dice show the same value.\n\nIn the first case, there are six choices for the value of the dice which are the same and $\\binom{4}{3}$ choices for which dice show that value. Then there are 5 c... | United States | HMMT November 2012 | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 7/72 | |
03gp | Problem:
Two flag poles of heights $h$ and $k$ are situated $2a$ units apart on a level surface. Find the set of all points on the surface which are so situated that the angles of elevation of the tops of the poles are equal. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
07fh | Let $S$ be an infinite set of positive integers, define:
$$
T = \{x + y \mid x, y \in S, x \neq y\}.
$$
Suppose that there are only finitely many primes $p$ such that:
a) $p \equiv 1 \pmod 4$.
b) There exists a positive integer $s$ such that $p \mid s$, $s \in T$.
Prove that there are infinitely many primes that divide... | [
"Assume the contrary, that there exist only finitely many prime numbers that divide an element of $S$. Let's denote them $p_1, p_2, \\dots, p_t$. Take\n$$\nx_1, x_2, \\dots, x_{2mt+1} \\in S.\n$$\nFor every $x \\in S$ we can write it as\n$$\nx = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_t^{\\alpha_t}.\n$$\nThe pigeo... | Iran | 37th Iranian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
09s9 | Problem:
Bepaal alle polynomen $P(x)$ met reële coëfficiënten waarvoor het polynoom
$$
Q(x) = (x+1) P(x-1) - (x-1) P(x)
$$
constant is. | [
"Solution:\n\nOplossing I. Stel dat $P(x)$ een constant polynoom is, zeg $P(x) = a$ met $a \\in \\mathbb{R}$. Dan is\n$$\nQ(x) = (x+1) a - (x-1) a = a x + a - a x + a = 2 a,\n$$\nen dat is constant. Dus elk constant polynoom $P(x)$ voldoet.\n\nWe nemen nu verder aan dat $P$ niet constant is. We kunnen dan schrijven... | Netherlands | IMO-selectietoets | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | All polynomials of the form P(x) = b x^2 + b x + a for real a, b. | |
0ea3 | Problem:
Dana je funkcija $f$ s predpisom $f(x) = -2x^{2} + 8x - d$. Izračunaj vrednosti parametra $d$ tako,
a) da se bo graf funkcije $f$ dotikal osi $x$,
b) da bo maksimalna vrednost funkcije $f$ enaka $6$,
c) da bo funkcija $f$ povsod pozitivna,
ali pa utemelji, da taka vrednost parametra $d$ ne obstaja. | [] | Slovenia | 14. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a) d = 8; b) d = 2; c) no such value exists | |
0afq | Во четириаголникот **АВСD**, **Е** е средина на страната **ВС**, и плоштината на триаголникот **АЕD** е двпати помала од плоштината на четириаголникот **АВСD**. Докажи дека **АВ** е паралелна со **CD**.
 | [
"Нека **АВСD** е четириаголникот во кој **Е** е средина на **ВС**, при што $P_{АВСD} = 2P_{АЕD}$. Точката **D** ќе ја пресликаме централно симетрично со центар на симетрија **Е**. Нека $D_1$ е нејзината слика. Бидејќи $ECD \\cong EBC$ имаме $P_{ECD} = P_{ЕВС}$, од каде добиваме\n\n\n$$\nP_{... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | Macedonian, English | proof only | null | |
0l5a | Problem:
Right triangle $\triangle DEF$ with $\angle D = 90^{\circ}$ and $\angle F = 30^{\circ}$ is inscribed in equilateral triangle $\triangle ABC$ such that $D$, $E$, and $F$ lie on segments $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Given that $BD = 7$ and $DC = 4$, compute $DE$. | [
"Solution:\n\n\nFrom $\\angle E = 60^{\\circ}$, we get that $\\angle AEF = 120^{\\circ} - \\angle CED = \\angle CDE$. Therefore, $\\triangle AEF \\sim \\triangle CDE$. Since $EF:DE = 2:1$, the ratio of similarity must be $2:1$, so $AE = 2CD = 8$. Recall $ABC$ has side length $7 + 4 = 11$, s... | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | sqrt(13) | |
06hk | How many sets of integers $(a, b, c)$ satisfy $2 \le a \le b \le c$ and $abc = 2013 \times 2014$? (2 marks)
有多少組整數 $(a, b, c)$ 滿足 $2 \le a \le b \le c$ 及 $abc = 2013 \times 2014$? (2分) | [] | Hong Kong | HONG KONG PRELIMINARY SELECTION CONTEST | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English; Chinese | final answer only | 90 | |
0164 | A square $1 \times 1$ is cut into some quadrangles. Prove that the sum of the squares of all sides of all quadrangles isn't less than $4$. | [] | Baltic Way | Baltic Way SHL | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
02jj | Problem:
A figura mostra a marca de uma empresa, formada por dois círculos concêntricos e outros quatro círculos de mesmo raio, cada um deles tangente a dois dos outros e aos dois círculos concêntricos. O raio do círculo menor mede $1~\mathrm{cm}$. Qual é, em centímetros, o raio do círculo maior?
![](attached_image_1... | [
"Solution:\n\nSeja $r$ o raio das quatro circunferências iguais. Ligando os centros $A$ e $B$ de duas destas circunferências ao centro $O$ das circunferências concêntricas, obtemos o triângulo $OAB$ como na figura ao lado. Lembrando que a reta que une os centros de duas circunferências tangentes passa pelo ponto de... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 3 + 2√2 | |
0241 | Problem:
Sejam $ABCD$ um losango, com $\angle BAD > \angle ABC$, e $P$ e $Q$ pontos nos lados $AB$ e $AD$, respectivamente, tais que o triângulo $PCQ$ é equilátero, com lado igual ao lado do losango. Encontre as medidas dos ângulos do losango. | [
"Solution:\n\nComo $\\angle ABC = \\angle QDC = \\alpha$ e $BC = PC = QC = CD$, segue que $\\angle BCP = \\angle DCQ = 180^\\circ - 2\\alpha$. Assim,\n$$\n\\begin{aligned}\n180^\\circ - \\alpha & = \\angle BCD \\\\\n& = \\left(180^\\circ - 2\\alpha\\right) + 60^\\circ + \\left(180^\\circ - 2\\alpha\\right) \\\\\n& ... | Brazil | null | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 80°, 80°, 100°, 100° | |
09uf | Points $A$, $B$, and $C$ lie on a circle with centre $M$. The reflection of point $M$ in the line $AB$ lies inside triangle $ABC$ and is the intersection of the angular bisectors of angles $A$ and $B$. (The angular bisector of an angle is the line that divides the angle into two equal angles.) Line $AM$ intersects the ... | [
"Let $I$ be the reflection of point $M$ in the line $AB$. We define $\\alpha = \\angle CAI$ and $\\beta = \\angle CBI$. Since $AI$ is the angular bisector of $\\angle CAB$, we find that $\\angle IAB = \\alpha$. Since $I$ is the reflection of $M$ in the line $AB$, we find that $\\angle BAM = \\alpha$. Triangle $AMC$... | Netherlands | Final Round, September 2019 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles"
] | English | proof only | null | |
0kb7 | Let $S$ be a set of 16 points in the plane, no three collinear. Let $\chi(S)$ denote the number of ways to draw 8 line segments with endpoints in $S$, such that no two drawn segments intersect, even at endpoints. Find the smallest possible value of $\chi(S)$ across all such $S$. | [
"The answer is 1430. In general, we prove that with $2n$ points the answer is the $n^{\\text{th}}$ Catalan number $C_n = \\frac{1}{n+1}\\binom{2n}{n}$.\nFirst of all, it is well-known that if $S$ is a convex $2n$-gon, then $\\chi(S) = C_n$.\n\nIt remains to prove the lower bound. We proceed by (strong) induction on... | United States | USA TSTST | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | null | proof and answer | 1430 | |
04ya | Find all functions $f: (0, +\infty) \to \mathbb{R}$ satisfying
$$
f(x) - f(x + y) = f\left(\frac{x}{y}\right) f(x + y) \quad \text{for all } x, y > 0.
$$ | [
"Suppose $f(t) = 0$ for some $t > 0$. For $0 < x < t$ we choose $y = t - x > 0$ and find $f(x) = 0$. From setting $x = y = 1$ we conclude that $f(1) \\ne -1$. Hence by setting $x = y$ we get $f(x) - f(2x) = f(1)f(2x)$ for $x > 0$. Inductively we find\n$$\nf(2^n x) = f(x)(1 + f(1))^{-n}. \\quad (2)\n$$\nHence for an... | Czech-Polish-Slovak Mathematical Match | null | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = 0 for all x > 0, or f(x) = 1/x for all x > 0 |
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