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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
04ti | For positive $a$, $b$, $c$ it holds
$$
(a + c)(b^2 + ac) = 4a.
$$
Find the maximal possible value of $b + c$ and find all triples $(a, b, c)$, for which the value is attained. | [
"We use the well known inequality $a^2 + b^2 \\ge 2ab$ to adjust the given one:\n$$\n4a = (a + c)(b^2 + ac) = a(b^2 + c^2) + c(a^2 + b^2) \\ge a(b^2 + c^2) + 2abc = a(b + c)^2.\n$$\nWe can see, that $b + c \\le 2$, and also that the equality holds if and only if $0 < a = b < 2$ and $c = 2 - b > 0$. That's it."
] | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Maximum b + c = 2. All maximizing triples are a = b with 0 < b < 2 and c = 2 − b. | |
0kr1 | Problem:
Suppose point $P$ is inside triangle $ABC$. Let $AP$, $BP$, and $CP$ intersect sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Suppose $\angle APB = \angle BPC = \angle CPA$, $PD = \frac{1}{4}$, $PE = \frac{1}{5}$, and $PF = \frac{1}{7}$. Compute $AP + BP + CP$. | [
"Solution:\nThe key is the following lemma:\nLemma: If $\\angle X = 120^\\circ$ in $\\triangle XYZ$, and the bisector of $X$ intersects $YZ$ at $T$, then\n$$\n\\frac{1}{XY} + \\frac{1}{XZ} = \\frac{1}{XT}\n$$\nProof of the Lemma. Construct point $W$ on $XY$ such that $\\triangle XWT$ is equilateral. We also have $T... | United States | HMMT November 2022 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 19/12 | |
0ffm | Problem:
Decir si es posible elegir $1983$ enteros positivos distintos, todos menores o iguales que $10^{5}$, de forma que tres cualesquiera de ellos no sean términos consecutivos de una progresión aritmética. Justificar la respuesta. | [] | Spain | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Yes | |
0iog | Problem:
Let $A := \mathbb{Q} \setminus \{0, 1\}$ denote the set of all rationals other than $0$ and $1$. A function $f: A \rightarrow \mathbb{R}$ has the property that for all $x \in A$,
$$
f(x) + f\left(1 - \frac{1}{x}\right) = \log |x|
$$
Compute the value of $f(2007)$. | [
"Solution:\nAnswer: $\\log (2007 / 2006)$.\nLet $g: A \\rightarrow A$ be defined by $g(x) := 1 - 1 / x$; the key property is that\n$$\ng(g(g(x))) = 1 - \\frac{1}{1 - \\frac{1}{1 - \\frac{1}{x}}} = x\n$$\nThe given equation rewrites as $f(x) + f(g(x)) = \\log |x|$. Substituting $x = g(y)$ and $x = g(g(z))$ gives the... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | log(2007/2006) | |
0f95 | Problem:
Eight pawns are placed on a chessboard, so that there is one in each row and column. Show that an even number of the pawns are on black squares. | [
"Solution:\n\nLabel the rows and columns of the chessboard from $1$ to $8$. The square in row $i$ and column $j$ is black if and only if $i + j$ is even.\n\nSince there is one pawn in each row and column, the pawns occupy the squares $(1, \\sigma(1)), (2, \\sigma(2)), \\ldots, (8, \\sigma(8))$ for some permutation ... | Soviet Union | 23rd ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof only | null | |
01w3 | Determine all pairs $(n, k)$ of distinct positive integers such that there exists a positive integer $s$ for which the numbers of divisors of $sn$ and of $sk$ are equal. | [] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | All distinct pairs of positive integers n, k such that neither n divides k nor k divides n. | |
0chc | Find all the real numbers $x$ and $y$ so that:
(i) $x \ge 2y^2$;
(ii) $y \ge 2x^2$;
(iii) the number $8(x - y)$ is an integer. | [
"From (i) and (ii) follows $x \\ge 0$ and $y \\ge 0$. Moreover, $x = 0$ if and only if $y = 0$. This points to the solution $(0, 0)$ and the other solutions $(x, y)$ have $x > 0$ and $y > 0$.\nLet $(x, y)$ be a solution with $x > 0$ and $y > 0$.\nFrom (i) and (ii) follows $x \\ge 2y^2 \\ge 8x^4$, therefore $x(8x^3 ... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | All pairs (a, a) with a in [0, 1/2], together with (1/4, 1/8) and (1/8, 1/4). | |
00py | Two circles $\Gamma_1$ and $\Gamma_2$ intersect at points $M$, $N$. A line $l$ is tangent to $\Gamma_1$, $\Gamma_2$ at $A$ and $B$, respectively. The lines passing through $A$ and $B$ and perpendicular to $l$ intersect $MN$ at $C$ and $D$ respectively. Prove that $ABCD$ is a parallelogram. | [
"Let $K$ be the second intersection point of $AD$ with $\\Gamma_1$, $Q$ the second intersection of $BC$ with $\\Gamma_2$, $P$ the intersection of $AC$ and $\\Gamma_1$, and $G$ the intersection of $BD$ and $\\Gamma_2$. We will show that $K$, $G$, $P$, $Q$ all lie on a line perpendicular to both $AD$ and $BC$.\n\nSin... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
058w | A point $M$ is chosen on the side $AC$ of a triangle $ABC$ and a point $K$ is chosen on the line segment $BM$ so that $AM = \frac{1}{3}AC$ and $BK = \frac{1}{4}BM$. Let $N$ be the intersection of the line $AK$ and the side $BC$. What percentage of the area of the triangle $ABC$ is the area of the quadrilateral $MKNC$? | [
"*Answer:* 65%.\n\nLet the area of the triangle $ABC$ be $S$ and the areas of triangles $AKM$, $BKN$, $MKN$ and $CMN$ be $S_1$, $S_2$, $S_3$ and $S_4$, respectively (Fig. 30). Then:\n\n* $S_2 + S_3 + S_4 = \\frac{2}{3}S$ since the l.h.s. is the area of the triangle $MBC$ while triangles $MBC$ and $ABC$ have equal a... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 65% | |
0frn | Problem:
Sea $ABC$ un triángulo con $AB \neq AC$, sea $I$ su incentro, $\gamma$ su circunferencia inscrita y $D$ el punto medio de $BC$. La tangente a $\gamma$ por $D$ diferente de $BC$ toca a $\gamma$ en $E$. Demuestra que $AE$ y $DI$ son paralelas. | [
"Solution:\n\n(Ver la figura 1) Sea $P$ el punto de tangencia de $\\gamma$ con $BC$. Sean $Q$ y $R$ los puntos simétricos de $P$ con respecto a $D$ e $I$, respectivamente. Tenemos que:\n- $ER$ es paralela a $DI$. Por un lado, $DI$ es perpendicular a $PE$ (es de hecho la mediatriz de $PE$). Por otro lado, por ser $R... | Spain | LVII Olimpiada Matemática Española Concurso Final Nacional | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Concurrency and Collinearity > Men... | null | proof only | null | |
09tg | Problem:
Bepaal alle paren priemgetallen $(p, q)$ zodat $p^{2}+5 p q+4 q^{2}$ het kwadraat van een geheel getal is. | [
"Solution:\n\nSchrijf $p^{2}+5 p q+4 q^{2}=a^{2}$, met $a \\geq 0$ een geheel getal. De linkerkant is gelijk aan $(p+2 q)^{2}+p q$, dus we kunnen dit herschrijven tot $p q=a^{2}-(p+2 q)^{2}$, oftewel $p q=(a-p-2 q)(a+p+2 q)$. De tweede factor rechts is groter dan $p$ en groter dan $q$, maar het is wel een deler van... | Netherlands | Selectietoets | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (5, 11), (13, 3), (7, 5) | |
0hzp | Problem:
A cross-section of a river is a trapezoid with bases $10$ and $16$ and slanted sides of length $5$. At this section the water is flowing at $\pi\ \mathrm{mph}$. A little ways downstream is a dam where the water flows through $4$ identical circular holes at $16\ \mathrm{mph}$. What is the radius of the holes? | [
"Solution:\nThe volume of water going through any cross-section of the river in an hour (assuming the cross-sections are parallel) is the area times the velocity. The trapezoid has height $4$, hence area $52$, so the volume of water going through at any hour is $52\\pi$. Let $r$ be the radius of the holes, then the... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Circles"
] | null | final answer only | sqrt(13)/4 | |
06mf | In $\triangle ABC$, $AB < AC$. The internal bisector of $\angle BAC$ meets $BC$ at $D$, while the external bisector of $\angle BAC$ meets $CB$ produced at $E$. If $EB = 2022$ and the lengths of $BD$ and $DC$ are integers, how many possible lengths of $BD$ are there? | [
"Let the lengths of $BD$ and $DC$ be $x$ and $y$ respectively. By the angle bisector theorem, we have $\\frac{x}{y} = \\frac{AB}{AC} = \\frac{EB}{EC}$, i.e.\n$$\n\\frac{x}{y} = \\frac{2022}{2022 + x + y}\n$$\n\n\n\n$$\ny = \\frac{x^2 + 2022x}{2022 - x} = -x - 4044 + \\frac{2022 \\cdot 4044}... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Geometry > Plane Geometry > Triangles",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 16 | |
0cro | Pete and Basil play the following game on a checkered $n \times n$ board. Initially the whole board is white except for one corner square which is black; a rook is put onto this square. The players move alternately, Pete moves first. On each turn, a player moves a rook to another square horizontally or vertically. Imme... | [
"Одна из выигрышных стратегий для Пети состоит в том, чтобы каждым своим ходом делать самый длинный из возможных вертикальных ходов (например, первым ходом он пойдёт по вертикали в другой угол доски). Покажем, что, действуя согласно ей, он выиграет.\n\nНазовём белую клетку *достижимой*, если из текущего положения л... | Russia | XL Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Pete (the first player) has a winning strategy. | |
0gql | Let $I$, $O$, $J_A$ be the centers of the incircle, circumcircle, and excircle corresponding to the side $BC$ of a triangle $ABC$ with $AC > AB$, respectively. Let $r$, $R$, $r_a$ be the radii of these circles, respectively. Let the incircle touch the side $BC$ at $D$ and $E$ be a point on the line segment $BD$ differe... | [
"Let the excircle with center $J_A$ touch $BC$ at $F$, $P$ be the midpoint of the smaller arc $BC$ of the circumcircle of $ABC$, and $M$ be the midpoint of the side $BC$. Then the point $P$ is the midpoint of the line segment $IJ_A$ and hence the area of the triangle $IEP$ is half the area of the triangle $IEB$. Th... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",... | null | proof only | null | |
045r | Given a positive integer $n$, let $D$ be the set of positive divisors of $n$, and let $f: D \to \mathbb{Z}$ be a function. Prove that the following are equivalent:
(A) for any positive divisor $m$ of $n$,
$$
n \mid \sum_{d|m} f(d) \binom{n/d}{m/d};
$$
(B) for any positive divisor $k$ of $n$,
$$
k \mid \sum_{d|k} f(d).
... | [
"**Proof:** For map $f: D \\to \\mathbb{Z}$, we define a map $g: D \\to \\mathbb{Z}$ to be\n$$\ng(k) = \\sum_{d|k} f(d), \\quad \\forall k \\in D.\n$$\nBy Möbius transform, the map $f$ is uniquely determined by $g$:\n$$\nf(k) = \\sum_{d|k} \\mu\\left(\\frac{k}{d}\\right) g(d), \\quad \\forall k \\in D,\n$$\nTherefo... | China | 2022 China Team Selection Test for IMO | [
"Number Theory > Number-Theoretic Functions > Möbius inversion",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof only | null | |
0773 | For each positive integer $n \ge 2$, define the polynomial $f_n(x)$ by $f_n(x) = x^n - x^{n-1} - x^{n-2} - \cdots - x - 1$. Prove that
(a) for each positive integer $n \ge 2$, the equation $f_n(x) = 0$ has a unique real positive root, say, $\alpha_n$;
(b) $(\alpha_n)_{n \ge 2}$ is a strictly increasing sequence; and
(c... | [
"(a) Since $f_n(x)$ has one change of sign, it follows that $f_n(x) = 0$ has at most one positive real root by Descartes' Rule. As $f_n(0) = -1 < 0$ and $f_n(2) = 2^n - 2^{n-1} - 2^{n-2} - \\cdots - 2 - 1 = 1 > 0$, we see that $f_n(x) = 0$ has at least one root between $0$ and $2$. Thus $f_n(x) = 0$ has exactly one... | India | India_2017 | [
"Algebra > Algebraic Expressions > Polynomials > Descartes' Rule of Signs",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
02c2 | Problem:
Sete números inteiros positivos estão escritos em ordem crescente numa mesma linha. Coloque entre esses números cinco sinais de "+" e um só de "$=$" para obter uma igualdade. | [
"Solution:\n\nSolução 1 - Os 7 números podem ser escritos como\n\n$$\n\\underbrace{n-3, n-2, n-1}_{3 n-6}, n, \\underbrace{n+1, n+2, n+3}_{3 n+6}\n$$\n\nObservando que $3 n-6+12=3 n+6$, concluímos que $n=12$. Logo, os números são\n\n$$\n9+10+11+12=13+14+15\n$$\n\n\nSolução 2 - Seja $n+1, n+2, \\ldots, n+7$ os sete ... | Brazil | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 9+10+11+12=13+14+15 | |
0atg | Problem:
Rectangle $BRIM$ has $BR = 16$ and $BM = 18$. The points $A$ and $H$ are located on $IM$ and $BM$, respectively, so that $MA = 6$ and $MH = 8$. If $T$ is the intersection of $BA$ and $IH$, find the area of quadrilateral $MATH$. | [] | Philippines | Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 34 | |
0anw | Problem:
Determine the slopes of the lines passing through $P(3,0)$ that intersect the parabola with equation $y=8-x^{2}$ at exactly one point.
(a) $-4,-8$
(b) $-3,-2$
(c) $-5,-7$
(d) $-4,-7$ | [] | Philippines | Qualifying Round | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | (a) | |
07sw | Given are two circles $\Omega_P$ and $\Omega_Q$ which intersect at two distinct points $A$ and $B$. Construct a circle $\Omega$ that contains $\Omega_P$ and $\Omega_Q$ and which is tangent to $\Omega_P$ and $\Omega_Q$ at points $P$ and $Q$, respectively, such that $P$, $A$, $Q$ are collinear. Justify your construction. | [
"**Description of the construction.** Let $O_P$ and $O_Q$ denote the centres of the circles $\\Omega_P$ and $\\Omega_Q$, respectively. Draw the circle centre $A$ that passes through $B$ and let $K$ and $L$ be its intersection points with $\\Omega_P$ and $\\Omega_Q$, respectively. Next construct the angle bisector o... | Ireland | IRL_ABooklet_2020 | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | null | |
0cre | Pete has put several tokens into some squares of a checkered $50 \times 50$ board (at most one token per square). Prove that Bazil can put at most $99$ tokens into empty squares so that each row and each column contains an even number of tokens. | [
"Построим граф с вершинами $r_1, \\dots, r_{50}$, соответствующими строкам доски, и вершинами $c_1, \\dots, c_{50}$, соответствующим её столбцам. Вершины $r_i$ и $c_j$ соединим ребром, если клетка в пересечении соответствующих строки и столбца свободна. Тогда Васина цель переформулируется так: требуется отметить не... | Russia | XL Russian mathematical olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
05hj | Problem:
Deux cercles de centres respectifs $B$ et $C$ et de rayons différents sont tangents extérieurement en un point $A$. Soit $t$ une tangente commune aux deux cercles ne contenant pas le point $A$. La perpendiculaire à la droite $t$ passant par le point $A$ coupe la médiatrice du segment $[BC]$ en un point $F$. M... | [
"Solution:\n\nSoient $X$ et $Y$ les points de tangence de $t$ avec les cercles de centre $B$ et $C$ respectivement. Soit $O$ le point d'intersection de la droite $t$ avec la tangente commune aux deux cercles en $A$. Les points $X$ et $A$ sont symétriques par rapport à la droite $(OB)$ et les points $Y$ et $A$ sont ... | France | ENVOI 1 : GÉOMÉTRIE Corrigé | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0gny | Find all primes $p$ for which there exist an odd integer $n$ and a polynomial $Q(x)$ with integer coefficients such that the polynomial $1 + p n^2 + \prod_{i=1}^{2p-2} Q(x^i)$ has at least one integer root. | [
"Let $P(x) = 1 + p n^2 + \\prod_{i=1}^{2p-2} Q(x^i)$. For $p = 2$, $n = 1$ and $Q(x) = 2x + 1$ work as $P(-1) = 0$.\n\nWe will show that for odd primes no suitable $n$ and $Q(x)$ exist.\n\nSince all $Q(a^i)$ have the same parity for $1 \\le i \\le 2p-2$ for an integer $a$; if $P(a) = 0$, then $p \\equiv 3 \\pmod{4}... | Turkey | Team Selection Test for IMO 2009 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 2 | |
0i5h | Problem:
Find the set consisting of all real values of $x$ such that the three numbers $2^{x}, 2^{x^{2}}, 2^{x^{3}}$ form a non-constant arithmetic progression (in that order). | [
"Solution:\nThe empty set, $\\varnothing$. Trivially, $x=0,1$ yield constant arithmetic progressions; we show that there are no other possibilities. If these numbers do form a progression, then, by the AM-GM (arithmetic mean-geometric mean) inequality,\n$$\n\\begin{aligned}\n& 2 \\cdot 2^{x^{2}}=2^{x}+2^{x^{3}} \\g... | United States | Harvard-MIT Math Tournament | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | ∅ | |
0ed3 | The bisector of the angle at $A$, the altitude from $A$, and the median from $A$ divide the angle at $A$ into 4 equal parts. Determine the sizes of angles of the triangle $ABC$. | [
"Let $D$ be the midpoint of the side $BC$, $E$ be the foot of the altitude from vertex $A$, and $S$ be the intersection of the angle bisector at vertex $A$ with the side $BC$. Since the angle bisector divides the angle into two equal parts it must lie between the altitude and the median. Thus the point $S$ lies bet... | Slovenia | National Math Olympiad 2015 – First Round | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | ∠A = 90°, and ∠B, ∠C are 22.5° and 67.5° (in either order). | |
0aaj | Let $a, b, c, d$ be integers. Prove that for any positive integer $n$, at least $\lfloor \frac{n}{4} \rfloor$ positive integers $m \le n$ are such that $m^5 + d m^4 + c m^3 + b m^2 + 2021 m + a$ is not a perfect square. | [] | North Macedonia | Fourth Memorial Mathematical Contest | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Polynomials"
] | English | proof only | null | |
09tq | Problem:
a) Als $c\left(a^{3}+b^{3}\right)=a\left(b^{3}+c^{3}\right)=b\left(c^{3}+a^{3}\right)$ voor positieve reële getallen $a, b, c$, geldt dan noodzakelijk $a=b=c$ ?
b) Als $a\left(a^{3}+b^{3}\right)=b\left(b^{3}+c^{3}\right)=c\left(c^{3}+a^{3}\right)$ voor positieve reële getallen $a, b, c$, geldt dan noodzakelijk... | [
"Solution:\na) We beweren dat $(a, b, c)=(2,2,-1+\\sqrt{5})$ aan de gegeven gelijkheden voldoet. In dit drietal zijn alle getallen positief reëel en ze zijn niet allemaal gelijk, dus het antwoord op de vraag is dan nee.\nWe berekenen $c^{3}=(-1+\\sqrt{5})^{3}=-1+3 \\cdot \\sqrt{5}-3 \\cdot 5+5 \\sqrt{5}=-16+8 \\sqr... | Netherlands | IMO-selectietoets II | [
"Algebra > Equations and Inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a) No; for example, a equals b equals 2 and c equals minus one plus the square root of five. b) Yes; it implies all three are equal. | |
0i9v | Problem:
You are given a $10 \times 2$ grid of unit squares. Two different squares are adjacent if they share a side. How many ways can one mark exactly nine of the squares so that no two marked squares are adjacent? | [
"Solution:\nSince each row has only two squares, it is impossible for two marked squares to be in the same row. Therefore, exactly nine of the ten rows contain marked squares.\n\nConsider two cases:\n\nCase 1: The first or last row is empty. These two cases are symmetrical, so assume without loss of generality that... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 36 | |
0l0x | Problem:
Let $x < y$ be positive real numbers such that
$$
\sqrt{x} + \sqrt{y} = 4 \quad \text{and} \quad \sqrt{x+2} + \sqrt{y+2} = 5
$$
Compute $x$. | [
"Solution:\nAdding and subtracting both equations gives\n$$\n\\begin{aligned}\n& \\sqrt{x+2} + \\sqrt{x} + \\sqrt{y+2} + \\sqrt{y} = 9 \\\\\n& \\sqrt{x+2} - \\sqrt{x} + \\sqrt{y+2} - \\sqrt{y} = 1\n\\end{aligned}\n$$\nSubstitute $a = \\sqrt{x} + \\sqrt{x+2}$ and $b = \\sqrt{y} + \\sqrt{y+2}$. Then since $(\\sqrt{x+... | United States | HMMT February 2024 Guts Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 49/36 | |
07um | For which positive integers $n$ can $n$ positive integers $a_1, a_2, \dots, a_n$ be found such that:
$$
a_1 + a_2 + a_3 + \dots + a_n = 2n \quad \text{and}
$$
$$
a_1 + 2a_2 + 3a_3 + \dots + n a_n = n^2 ?
$$ | [
"There is no positive solution for $n = 1$ or $n = 2$ but positive solutions exist for any $n \\ge 3$. Eliminate the small cases first. If $n = 1$ then $1 = a_1 = 2$, a contradiction. With $n = 2$, we must solve $a_1 + a_2 = 4$ and $a_1 + 2a_2 = 4$. The unique solution is $a_1 = 4$ and $a_2 = 0$, and as $a_2$ is no... | Ireland | IRL_ABooklet | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | n ≥ 3 | |
0h5y | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$
4f(x+f(y)) = f(x)+f(y)+f(xy)+1
$$
for all real numbers $x, y$. | [
"**Answer:** $f(x) \\equiv 1$.\n\nFirst suppose that there exists $a \\neq 0$ such that $f(a) = f(0) = b$. By (0) we denote the main equation.\nSubstituting $y=0$ in (0), we get:\n$$\n4f(x+b) = f(x)+b+b+1.\n$$\n(1)\n\nSubstituting $y=a$ in (0), we get:\n$$\n4f(x+b) = f(x)+b+f(xa)+1.\n$$\n(2)\n\nCombining (1) and (2... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) ≡ 1 | |
076y | Define an $n$-magic square to mean an $n \times n$ square matrix of non-negative integers such that the sum of all the entries in each row and each column is $m$ for some $m \in \mathbb{N}$. Also define an $n$-permutation matrix to mean an $n \times n$ square matrix of $n(n-1)$ zeroes and $n$ ones such that every row a... | [
"Fix an $n \\in \\mathbb{N}$. We will prove the following statement by Induction:\n$P(m)$ : Every $n$-magic square with common sum $m$ can be written as a sum of $m$ permutation matrices.\nFor the base case, we verify that $P(1)$ is true. Since all the entries are non-negative integers, the only way possible $n$-ma... | India | India_2017 | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
008i | On a square $2012 \times 2012$ checkerboard, L-triominoes like those shown below are placed without overlapping (each L-triomino covers exactly $3$ squares of the board). Determine how many L-triominoes can be placed as a maximum on the board if for any two rows and any two columns, at least one of the four intersectio... | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1012036 | |
0dxu | Problem:
Dan je ostrokotni trikotnik $ABC$ in središče njemu očrtane krožnice $O$. Naj bo $O_1$ točka na simetrali daljice $AB$, ki leži na nasprotnem bregu premice $AB$ kot točka $O$. Označimo krožnico s središčem $O_1$ in polmerom $AO_1$ s $\mathcal{K}$. Naj premici $CA$ in $CB$ sekata krožnico $\mathcal{K}$ še v to... | [
"Solution:\n\nNaj bo $D$ presečišče premic $A_1B$ in $AB_1$ ter označimo $\\angle A_1AB_1 = \\alpha$. Tedaj je $\\angle CAD = \\pi - \\angle DAA_1 = \\pi - \\alpha$. Ker so točke $A, C, B$ in $D$ konciklične, je $\\angle CBD = \\pi - \\angle CAD = \\pi - (\\pi - \\alpha) = \\alpha$. Zato je $\\angle A_1BB_1 = \\pi ... | Slovenia | 51. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0fkx | Problem:
Sean $ABC$ un triángulo acutángulo, $I$ el centro del círculo inscrito en el triángulo $ABC$, $r$ su radio y $R$ el radio del círculo circunscrito al triángulo $ABC$. Se traza la altura $AD = h_{a}$, con $D$ perteneciente al lado $BC$. Demuestra que
$$
DI^{2} = (2R - h_{a})(h_{a} - 2r)
$$ | [
"Solution:\n\nSean $E$ y $M$ las proyecciones ortogonales de $I$ sobre $BC$ y $AD$, respectivamente.\n\nSe tiene: $AI = \\frac{r}{\\operatorname{sen} \\frac{A}{2}} ; \\quad r = \\frac{S}{p} \\Rightarrow AI = \\frac{S}{p \\cdot \\operatorname{sen} \\frac{A}{2}}$ (1) donde, evidentemente,\n$S$ es el área del triángul... | Spain | Fase Nacional de la XLV Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Method... | null | proof only | null | |
04fs | Let $ABCD$ be a parallelogram with acute angle at $A$. Let $E$ be the foot of the perpendicular from $C$ to the line $AB$, and let $F$ be the foot of the perpendicular from $C$ to the line $AD$.
Prove that
$$
|AB| \cdot |AE| + |AD| \cdot |AF| = |AC|^2. \quad (\text{USSR})
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
0iot | Prove that for each positive integer $n$, there are pairwise relatively prime integers $k_0, k_1, \dots, k_n$, all strictly greater than 1, such that $k_0 k_1 \cdots k_n - 1$ is the product of two consecutive integers. | [
"We proceed by induction. The case $n=1$ is clear, since we may pick $k_0 = 3$ and $k_1 = 7$. Let us assume now that for a certain $n$ there are pairwise relatively prime integers $1 < k_0 < k_1 < \\cdots < k_n$ such that $k_0 k_1 \\cdots k_n - 1 = a_n(a_n - 1)$, for some positive integer $a_n$. Then choosing $k_{n... | United States | USAMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Resid... | null | proof only | null | |
07f8 | $\triangle ABC$ is an isosceles triangle with $AB = AC$. Point $X$ is an arbitrary point on side $BC$. Points $Y, Z$ are on the sides $AB, AC$, respectively, such that $\angle BXY = \angle ZXC$. A line parallel to $YZ$ and passing through $B$ cuts $XZ$ at $T$. Prove that $AT$ bisects $\angle A$. | [
"Let us denote by $K$ the intersection point of lines $BT$, $XY$. Note that $\\triangle XYB \\sim \\triangle XZC$. Using the fact that $BT \\parallel YZ$, we get\n$$\n\\frac{XK}{YK} = \\frac{XT}{ZT}.\n$$\nTherefore, $K$ and $T$ are corresponding points in triangles $\\triangle XYB$ and $\\triangle XZC$.\nWhich give... | Iran | 37th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0g2a | Problem:
Sei $n$ eine natürliche Zahl. Sei $k$ die Anzahl Möglichkeiten, $n$ als Summe von einer oder mehreren aufeinanderfolgenden natürlichen Zahlen darzustellen. Zeige, dass $k$ der Anzahl ungerader positiver Teiler von $n$ entspricht. | [
"Solution:\n\nWir konstruieren zuerst für jeden ungeraden Teiler von $n$ eine Darstellung als Summe aufeinanderfolgender natürlicher Zahlen:\nSei $a \\in \\mathbb{N}_{0}$ und $n=(2a+1) \\cdot b$ für ein $b \\in \\mathbb{N}$. Betrachte die Nullsumme\n$$\n(-a)+(-a+1)+\\cdots+(a-1)+a=0\n$$\nAddieren wir zu jedem der $... | Switzerland | SMO - Finalrunde | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0eui | Let $\triangle ABC$ be a triangle with $\angle B < \angle C$. Let $O$ be the circle tangent to the line $AC$ at the point $C$ and passing through the point $B$. The lines $AB$ and $CO$ meet the circle $O$ at the points $D (\neq B)$ and $P (\neq C)$, respectively. Let $E$ be the intersection of $AC$ and the line passing... | [
"Let $K'$ be the intersection of the line $CD$ and the line passing through $E$ and parallel to the line $CL$. Since $\\angle EBD = \\angle DCL = \\angle EK'D$, we have that the four points $B$, $D$, $E$ and $K'$ are on one circle, say $O_1$. Since $AE^2 = AC^2 = AD \\cdot AB$, we have that the line $AE$ is tangent... | South Korea | 22nd Korean Mathematical Olympiad Final Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
092h | Problem:
Let $I$ be the incentre of triangle $A B C$ with $A B > A C$ and let the line $A I$ intersect the side $B C$ at $D$. Suppose that point $P$ lies on the segment $B C$ and satisfies $P I = P D$. Further, let $J$ be the point obtained by reflecting $I$ over the perpendicular bisector of $B C$, and let $Q$ be the ... | [
"Solution:\nLet $A I$ intersect the circumcircle of triangle $A B C$ for the second time at $T$. It is known that $T$ is the centre of the circumcircle of triangle $B I C$ and due to symmetry the point $J$ lies on this circle as well.\nSince\n$$\n\\angle B Q P = \\angle A Q P - \\angle A Q B = \\pi - \\angle A D P ... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"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"
] | null | proof only | null | |
0jzk | Problem:
In convex quadrilateral $A B C D$ we have $A B=15$, $B C=16$, $C D=12$, $D A=25$, and $B D=20$. Let $M$ and $\gamma$ denote the circumcenter and circumcircle of $\triangle A B D$. Line $C B$ meets $\gamma$ again at $F$, line $A F$ meets $M C$ at $G$, and line $G D$ meets $\gamma$ again at $E$. Determine the a... | [
"Solution:\n\nNote that $\\angle A D B = \\angle D C B = 90^{\\circ}$ and $B C \\parallel A D$. Now by Pascal's theorem on $D D E B F A$ implies that $B, M, E$ are collinear. So $[A D E] = [A B D] = 150$ and $[B C D] = 96$, so the total area is $396$."
] | United States | February 2017 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 396 | |
0e3d | Problem:
Izračunaj $\cos (\pi+2 x)$, če je $\cos x=\frac{1}{4}$. | [
"Solution:\n\nUporabimo zvezo za prehod na oster kot $\\cos (\\pi+2 x)=-\\cos 2 x$, nato še zvezo za dvojne kote $-\\cos 2 x=-\\cos ^{2} x+\\sin ^{2} x$. Uporabimo zvezo za $\\sin ^{2} x=1-\\cos ^{2} x$. Izraz uredimo, vstavimo $\\cos x=\\frac{1}{4}$. Dobimo $1-2 \\cos ^{2} x=1-2 \\cdot\\left(\\frac{1}{4}\\right)^{... | Slovenia | 10. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Precalculus > Trigonometric functions"
] | null | final answer only | 7/8 | |
07nx | Let $ABC$ be an isosceles triangle with $|AB| = |AC|$. The points $B'$ and $C'$ are on the line $BC$ (extended in both directions) such that $B$ is between $B'$ and $C$ and $C$ is between $B$ and $C'$. Moreover, $B'BA$ and $ACC'$ are isosceles triangles. Let $X$ be a point on the line $BC$ such that $B'$ is not between... | [
"Because $|B'B| = |BA| = |AC| = |CC'|$, we have $|B'X| = |AB| + |BX|$ and $|BC'| = |BC| + |AB|$. For the first equation we used that $B$ is between $B'$ and $X$. To see this, we have to exclude that $X$ is between $B$ and $B'$ (by assumption, $B'$ is not between $X$ and $B$). If $|AX| = |BX|$, $X$ cannot be between... | Ireland | Ireland | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hv3 | Problem:
A triangle has sides of length $888$, $925$, and $x > 0$. Find the value of $x$ that minimizes the area of the circle circumscribed about the triangle. | [
"Solution: 259 ."
] | United States | null | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 259 | |
0f1r | Problem:
$n$ is a positive integer. A universal sequence of length $m$ is a sequence of $m$ integers each between $1$ and $n$ such that one can obtain any permutation of $1, 2, \ldots, n$ by deleting suitable members of the sequence. For example, $1, 2, 3, 1, 2, 1, 3$ is a universal sequence of length $7$ for $n = 3$.... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | There exists a universal sequence of length n^2 − n + 1; any universal sequence must have length at least n(n + 1) / 2; the shortest sequence for n = 4 has length 12. | |
0fch | Problem:
Se dispone de pequeñas piezas de madera de tamaño $4 \times 5 \times 10$. Decidir si es posible o no apilarlas, sin dejar huecos y apoyándolas siempre sobre cualquiera de sus caras, para formar un ortoedro de dimensiones $2^{2003} \times 3^{2003} \times 5^{2003}$. | [
"Solution:\n\nLa superficie de cada una de las caras del ortoedro es:\n$$\nC_{1}=2^{2003} \\times 3^{2003}=6^{2003}, \\quad C_{2}=2^{2003} \\times 5^{2003}=10^{2003} \\text{ y } C_{3}=3^{2003} \\times 5^{2003}=15^{2003}\n$$\nY, de ser posible el apilamiento, debería ser combinación lineal (con coeficientes naturale... | Spain | Fase Local | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > Other 3D problems",
"Number Theory > Other"
] | null | proof and answer | Impossible | |
03vh | Suppose that the convex quadrilateral $ABCD$ satisfies $AB = BC$, $AD = DC$. $E$ is a point on $AB$, and $F$ on $AD$, such that $B$, $E$, $F$, $D$ are concyclic. Draw $\triangle DPE$ directly similar to $\triangle ADC$, and $\triangle BQF$ directly similar to $\triangle ABC$. Prove that $A$, $P$, $Q$ are collinear. (Po... | [
"In $\\triangle BDF$, $O$ is the circumcenter, so $\\angle BOF = 2\\angle BDA$; And $\\triangle ABD \\sim \\triangle CBD$, so $\\angle CDA = 2\\angle BDA$. Hence, $\\angle BOF = \\angle CDA = \\angle EPD$, which implies that the isosceles triangles\n$$\n\\triangle BOF \\sim \\triangle EPD. \\qquad \\textcircled{1}\... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ht2 | Problem:
Let $n > 1$ be an odd integer. Prove that every integer $l$ satisfying $1 \leq l \leq n$ can be represented as a sum or difference of two integers each of which is less than $n$ and relatively prime to $n$. | [
"Solution:\nWe will use the following lemma (it is known as the Chinese Remainder Theorem).\n\nLemma. Let $m_{1}, m_{2}, \\ldots, m_{k}$ be different relatively prime numbers. If $q_{1}, q_{2}, \\ldots, q_{k}$ are arbitrary non-negative integers then there exists a natural number $x$ less than $m_{1} m_{2} \\ldots ... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
03c4 | One hundred and one of the squares of an $n \times n$ table are colored blue. It is known that there exists a unique way to cut the table to rectangles along boundaries of its squares with the following property: every rectangle contains exactly one blue square. Find the smallest possible $n$. | [
"Answer: 101. We shall prove the following more general assertion.\n\n**Lemma.** Some of the squares of a table $P$ are colored blue. A partition of $P$ into rectangles of integer sides is called *good* if every rectangle contains exactly one blue square. Then $P$ possesses a unique good partition if and only if th... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 101 | |
050b | Consider a positive integer $N$ with exactly 6 positive divisors $d_1, \ldots, d_6$ such that $1 = d_1 < d_2 < d_3 < d_4 < d_5 < d_6 = N$. Call such an integer $N$ good if the sum $d_4 + d_5$ is divisible by the sum $d_2 + d_3$.
a) Find the smallest positive integer $N$ which has exactly 6 positive divisors and which ... | [
"a) Considering the numbers from 1 to 20 we see that exactly three of them have 6 divisors: 12 (the divisors are 1, 2, 3, 4, 6, 12), 18 (1, 2, 3, 6, 9, 18), and 20 (1, 2, 4, 5, 10, 20). For 12 the sum $d_4 + d_5 = 4 + 6$ is divisible by the sum $d_2 + d_3 = 2 + 3$ and similarly, for 18 the sum $d_4 + d_5 = 6 + 9$ i... | Estonia | Selected Problems from Open Contests | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | a) 20. b) Infinitely many, for example all numbers of the form N = 4p with p a prime greater than 4 (more generally, all N = p^2 q with distinct primes p, q and q > p^2). | |
07io | In triangle $ABC$ the points $M$ and $N$ are the midpoints of the sides $AB$ and $AC$, respectively. The lines $BN$ and $CM$ intersect the circumcircle of $ABC$, for the second time, at $N'$ and $M'$, respectively. Points $X$ and $Y$ are on the extension of $BC$ such that, point $B$ is between $X$ and $C$ and point $C$... | [
"Since $\\angle BXM' = \\angle ACM$, $\\angle M'BX = \\angle M'AC$ it follows that $\\triangle M'AC \\sim \\triangle M'BX$. Hence, $M'B/M'A = BX/AC$. Yielding, $BX = AC \\cdot \\frac{M'B}{M'A} = b \\cdot \\frac{\\sin \\angle MCB}{\\sin \\angle MCA}$. On the other hand, $\\frac{\\sin \\angle MCB}{\\sin \\angle MCA} ... | Iran | 41th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
07o8 | Let $a$, $b$, $c$ be real numbers and let
$$
x = a + b + c, \quad y = a^2 + b^2 + c^2, \quad z = a^3 + b^3 + c^3 \quad \text{and} \quad S = 2x^3 - 9xy + 9z.
$$
a. Prove that $S$ is unchanged when $a$, $b$, $c$ are replaced by $a + t$, $b + t$, $c + t$, respectively, for any real number $t$.
b. Prove that $(3y - x^2)^... | [
"a.\nLet $x'$, $y'$, $z'$ and $S'$ be obtained from $x$, $y$, $z$ and $S$ by replacing $a$, $b$, $c$ by $a + t$, $b + t$, $c + t$, respectively. So\n$$\nx' = (a + t) + (b + t) + (c + t) = x + 3t,\n$$\n$$\n\\begin{aligned}\ny' &= (a + t)^2 + (b + t)^2 + (c + t)^2 = (a^2 + b^2 + c^2) + 2(a + b + c)t + 3t^2 \\\\\n&= y... | Ireland | Ireland | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
07da | Find the maximum possible value of integer $k$ for which there exist distinct real numbers $x_1, x_2, \dots, x_k$ greater than $1$ such that
$$
\forall\ 1 \le i,j \le k : x_i^{[x_j]} = x_j^{[x_i]}
$$
(By $[x]$ we mean the greatest integer less than or equal to $x$.) | [
"We start with some lemmas.\n\n**Lemma.** For positive integers $m > 3$ and $k > 2$ we have $m^{m+k} > (m+k+1)^m$.\n*Proof.* We must show that $m^k > \\left(\\frac{m+k+1}{m}\\right)^m = \\left(1 + \\frac{k+1}{m}\\right)^m$. But we have $\\left(1 + \\frac{k+1}{m}\\right)^m < e^{k+1}$ and so\n$$\n\\left(1 + \\frac{k+... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 4 | |
01hs | Anne, Ellie and Milo play a game on a hexagonal board tiled with hexagons, $n > 2$ hexagons on each side. The game begins with a token on a tile in one of corner of the board. Ellie and Milo are on the same team, playing against Anne, and they win if the token lands on center of the board. Anne, Ellie and Milo take tur... | [
"We colour the board in three colours in such a way that no neighbouring tiles are of the same colour. We can give each hexagon a coordinate using $\\vec{e}_1 = (1, 0)$ and $\\vec{e}_2 = (\\cos(120^\\circ), \\sin(120^\\circ)) = (\\frac{-1}{2}, \\frac{\\sqrt{3}}{2})$ as basis. Let the center square be the origin. Th... | Baltic Way | Baltic Way 2021 Shortlist | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | n ≡ 1 (mod 3) | |
0hn4 | Problem:
Oscar draws a triangle $ABC$ on a sheet of paper. He finds that the side lengths of $ABC$ are all powers of $2$ (i.e. among $1, 2, 4, 8, \ldots$). Prove that Oscar's triangle is isosceles. | [
"Solution:\n\nConsider a longest side of the triangle, $2^{a}$. We claim that another side must have this length too. Otherwise, suppose for contradiction they are $2^{b}$ and $2^{c}$ where $b, c < a$. Then\n$$\n2^{b} + 2^{c} < 2^{a-1} + 2^{a-1} = 2^{a}\n$$\nwhich contradicts the triangle inequality.\nHence there m... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof only | null | |
0h63 | Determine all pairs $(a, b)$ for which the following condition holds: there exists index $n_0$ such that the sequence $(a_n + a_{n+1}), n \ge n_0$ is non-decreasing.
Consider a sequence $(a_n)$: $a_1 = a, a_2 = b$, where $a$ and $b$ are positive integers, and for all $n \ge 2$ $a_{n+1}$ equals the number of indexes $i... | [
"If $a = b = 1$ then $(a_n)$ has terms $(1; 1; 2; 1; 3; 1; 4; 1; 5; 1; ...)$, and $(a_n + a_{n+1})$ has terms $(2; 3; 3; 4; 4; 5; 5; 6; 6; 7; ...)$.\nTherefore $a = b = 1$ satisfies the condition of the question.\n\nLet $a_1 = a$ and $a_2 = b$. Suppose that $a_1 \\ne 1$ or $a_2 \\ne 1$. Without loss of generality i... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | (1, 1) | |
0g1p | Problem:
Sei $ABC$ ein rechtwinkliges Dreieck mit Hypotenuse $AB$. Ein Kreis um $C$ schneide die Strecke $AB$ zweimal in den Punkten $P$ und $Q$, wobei $P$ zwischen $A$ und $Q$ liegt. Sei $R$ der Punkt auf der Strecke $BC$ mit $\angle RAC = \frac{1}{2} \angle PCQ$ und sei $S$ der Punkt auf der Strecke $AC$ mit $\angle... | [
"Solution:\n\nEs gilt:\n$$\n\\angle RAS = \\angle RAC = \\frac{1}{2} \\angle PCQ = \\angle CBS = \\angle RBS\n$$\nMit der Umkehrung des Peripheriewinkelsatzes folgt, dass $RSAB$ ein Sehnenviereck ist.\nDa $P$ und $Q$ auf einem Kreis mit Mittelpunkt $C$ liegen, ist das Dreieck $CPQ$ gleichschenklig. Man erhält:\n$$\... | Switzerland | SMO - Vorrunde | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0hw5 | Problem:
For all positive integers $n$, prove that
$$
\sum_{k=1}^{n} \phi(k)\left\lfloor\frac{n}{k}\right\rfloor=\frac{n(n+1)}{2}
$$
(For a positive integer $n$, $\phi(n)$ denotes the number of positive integers less than or equal to $n$ and relatively prime to $n$. For a real number $x$, $\lfloor x\rfloor$ denotes th... | [
"Solution:\n\nConsider the fractions $a / b$, where $a$ and $b$ range over integers such that $1 \\leq a \\leq b \\leq n$. We will count these fractions in two ways:\n\na. By unreduced form. For each denominator $b$, there are $b$ possible numerators $a=1,2, \\ldots, b$, so the total number of fractions is\n$$\n1+2... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0cqh | $2N$ points are marked on a circle ($N$ is a positive integer) so that no point in the interior of the circle belongs to three chords with the endpoints at marked points. A collection of $N$ chords with the endpoints at marked points is called a *matching* if every marked point is the endpoint of exactly one of these c... | [
"**Answer.** $1$.\n\n**First solution.** By induction on $N$, we prove that the number of even matchings is one more than the number of odd matchings. For $N=1$ the statement is obvious: there is only one matching, and it is even. Now, prove the statement for $2N$ points, assuming it is true for $2(N-1)$ points. De... | Russia | Russian mathematical olympiad | [
"Geometry > Plane Geometry > Circles",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's ... | English; Russian | proof and answer | 1 | |
0fze | Problem:
Ein $7 \times 7$ Quadrat ist in 49 kleine $1 \times 1$ Quadrate unterteilt. Zwei Ameisen laufen den Seiten der kleinen Quadrate entlang, wobei jede Ameise ihren eigenen geschlossenen Weg läuft und alle 64 Eckpunkte der kleinen Quadrate genau einmal besucht.
Welches ist die minimale Anzahl Seiten der kleinen Q... | [
"Solution:\n\nWir werden zeigen, dass es minimal 16 Seiten gibt, über die beide Ameisen laufen. Dafür zeigen wir zuerst die untere Schranke und dann geben wir eine Konstruktion an.\n\na. 1. Schranke: Jede Ameise muss genau 64 Kanten entlang laufen, um alle Punkte einmal zu besuchen und dann noch zum Anfangspunkt zu... | Switzerland | IMO-Selektionsprüfung | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 16 | |
0duz | Problem:
Naj bo $\mathcal{K}$ krožnica v ravnini, $\mathcal{K}_1$ in $\mathcal{K}_2$ pa disjunktni krožnici, ki se od znotraj dotikata krožnice $\mathcal{K}$ v točkah $A$ in $B$. Naj bo $t$ skupna tangenta krožnic $\mathcal{K}_1$ in $\mathcal{K}_2$, ki se ju dotika v točkah $C$ in $D$ tako, da sta $\mathcal{K}_1$ in $... | [
"Solution:\n\nOznačimo z $O$, $O_1$ in $O_2$ središča krožnic, presečišči premice $t$ z daljicama $O A$ in $O B$ pa označimo z $A'$ in $B'$. Naj bo $\\alpha = \\Varangle C A O_1$. Ker je trikotnik $A O_1 C$ enakokrak, je $\\Varangle C O_1 A' = 2 \\alpha$. Ker je $C D \\perp O_1 C$, je $\\Varangle O_1 A' C = \\pi / ... | Slovenia | 46. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07y4 | Find all quadruples $(a, b, c, d)$ of positive integers for which
$$
a^2 + b^2 + c^2 + d^2 = 2025 \quad \text{and} \quad ab = cd.
$$ | [
"We will use the fact that $n^2 \\equiv 0 \\pmod{3}$ if $3 \\mid n$ and $n^2 \\equiv 1 \\pmod{3}$ otherwise. A consequence is that if the sum of the squares of $k$ integers is divisible by $3$ and $k$ is not divisible by $3$, then at least one of the $k$ integers is divisible by $3$.\n\nBecause $2025 = 3^4 \\cdot 5... | Ireland | IRL_ABooklet_2025 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | [[18, 18, 9, 36], [18, 18, 36, 9], [9, 36, 18, 18], [36, 9, 18, 18]] | |
07ji | Let $n$ be a positive integer such that $n \ge 4$. For every permutation $(s_1, s_2, \dots, s_n)$ of the numbers $1, 2, \dots, n$, we calculate the following expression and write the result on the board:
$$
1s_1 + 2s_2 + 3s_3 + \dots + ns_n
$$
In the end, how many distinct values will be written on the board? | [
"According to the permutation inequality, we always have:\n$$\n1 \\times n + 2 \\times (n-1) + \\cdots + n \\times 1 \\le 1 \\times s_1 + 2 \\times s_2 + \\cdots + n \\times s_n \\le 1 \\times 1 + 2 \\times 2 + \\cdots + n \\times n\n$$\nWe want to prove that all numbers in this range are covered. First, note that ... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | C(n+1, 3) + 1 | |
0gcb | 求滿足以下條件的最小正整數 $n$, 或證明滿足以下條件的正整數 $n$ 不存在:
存在無窮多組由 $n$ 個相異正有理數所構成的數組 $(a_1, a_2, \cdots, a_n)$, 使得
$$
a_1 + a_2 + \cdots + a_n \quad \text{和} \quad \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}
$$
皆為整數。 | [
"$n = 3$\n\nIt's clear that the only solution when $n = 1$ is $a_1 = 1$. Now we show that:\n\n(1) There are only finitely many $(x, y) \\in \\mathbb{Q}_{>0}^2$ such that $x + y$ and $\\frac{1}{x} + \\frac{1}{y}$ are both integers.\n\nWrite $x = a/b$ and $y = c/d$ in the standard form. Then, $x + y \\in \\mathbb{Z}$... | Taiwan | 2018 數學奧林匹亞競賽第二階段選訓營, 模擬競賽(一) | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order ana... | null | proof and answer | 3 | |
06y0 | Let $n$ be a positive integer. Find the minimum possible value of
$$
S = 2^{0} x_{0}^{2} + 2^{1} x_{1}^{2} + \cdots + 2^{n} x_{n}^{2}
$$
where $x_{0}, x_{1}, \ldots, x_{n}$ are nonnegative integers such that $x_{0} + x_{1} + \cdots + x_{n} = n$. | [
"Answer: The minimum value is $\\frac{n(n+1)}{2}$.\n\nFor a fixed $n$, let $f(n)$ denote the minimum possible value of $S$. Consider the following variant: among all infinite sequences of nonnegative integers $x_{0}, x_{1}, \\ldots$, only finitely many of which are nonzero, satisfying $x_{0} + x_{1} + \\cdots = n$,... | IMO | IMO2024 Shortlisted Problems | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | n(n+1)/2 | |
094a | Problem:
We call a positive integer cheesy if we can obtain the average of the digits in its decimal representation by putting a decimal separator after the leftmost digit. Prove that there are only finitely many cheesy numbers.
Example. For instance, $2250$ is cheesy, as the average of the digits is $2.250$. | [
"Solution:\n\nLet $n$ be a positive integer and $k$ be the number of digits of $n$. Let $m$ be the sum of digits of $n$. We prove that if $k \\geq 2^{6}$ then $n$ is not reflexive.\n\nLet $a = v_{2}(k)$ be the nonnegative integer such that $2^{a} \\leq k < 2^{a+1}$. It is easy to see that $n$ is reflexive exactly i... | Middle European Mathematical Olympiad (MEMO) | MEMO Team Competition | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0ktr | Problem:
The formula to convert Celsius to Fahrenheit is
$$
F^{\circ} = 1.8 \cdot C^{\circ} + 32
$$
In Celsius, it is $10^{\circ}$ warmer in New York right now than in Boston. In Fahrenheit, how much warmer is it in New York than in Boston? | [
"Solution:\n\nLet $x$ and $y$ be the temperatures in New York and Boston, respectively, in Celsius. Then $x - y = 10$, so we compute\n$$\n(1.8 \\cdot x + 32) - (1.8 \\cdot y + 32) = 1.8 \\cdot (x - y) = 18\n$$"
] | United States | HMMT November 2023 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 18 | |
059k | Does there exist a positive integer whose
a. digit sum is 100 more than the product of digits?
b. product of digits is 100 more than the digit sum?
c. product of digits is 100 times greater than the digit sum? | [
"**Answer:** (a) Yes; (b) Yes; (c) Yes.\n\n(a) The number $111\\ldots1$, consisting of $101$ ones, works. Its product of digits is $1$ and digit sum $101$.\n\n(b) The number $1111112345$ works. Its product of digits is $120$ and digit sum $20$.\n\n(c) The number $1225555$ has product of digits $2500$ and digit sum ... | Estonia | Estonian Math Competitions | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | (a) Yes; (b) Yes; (c) Yes | |
0bmo | Problem:
a. Számítsd ki:
$$
\int_{0}^{1} x \sin \left(\pi x^{2}\right) \mathrm{d} x
$$
b. Számítsd ki:
$$
\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1} k \int_{\frac{k}{n}}^{\frac{k+1}{n}} \sin \left(\pi x^{2}\right) \mathrm{d} x
$$
Problem:
a. Calculaţi
$$
\int_{0}^{1} x \sin \left(\pi x^{2}\right) \ma... | [
"Solution:\n\na. Făcând substituţia $t=\\pi x^{2}$, integrala devine\n$$\n\\frac{1}{2 \\pi} \\int_{0}^{\\pi} \\sin t \\, \\mathrm{d} t = \\left.\\frac{1}{2 \\pi}(-\\cos t)\\right|_{0}^{\\pi} = \\frac{1}{\\pi}\n$$\n\nb. Fie $f:[0,1] \\rightarrow \\mathbb{R}$, $f(x)=\\sin \\left(\\pi x^{2}\\right)$, şi $F:[0,1] \\rig... | Romania | Olimpiada Naţională de Matematică | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Integral Calculus > Applications",
"Precalculus > Limits"
] | null | proof and answer | a: 1/π; b: 1/π | |
01if | Let $n$ be a positive integer and let $a_1, a_2, \dots, a_n$ and $b_1, b_2, \dots, b_n$ be positive real numbers. Prove that
$$
a_1 \sqrt{\frac{a_1}{b_1}} + a_2 \sqrt{\frac{a_2}{b_2}} + \dots + a_n \sqrt{\frac{a_n}{b_n}} \ge (a_1 + a_2 + \dots + a_n) \sqrt{\frac{a_1 + a_2 + \dots + a_n}{b_1 + b_2 + \dots + b_n}}
$$ | [
"For $a, b \\in \\mathbb{R}_{>0}$ we consider the function $f: \\mathbb{R}_{\\ge 0} \\to \\mathbb{R}$ with $f(x) = b x^3 - a x$ for all $x \\in \\mathbb{R}_{\\ge 0}$.\nFor $x \\ge 0$ we have $b(x - \\sqrt{\\frac{a}{3b}})^2 (x + 2\\sqrt{\\frac{a}{3b}}) \\ge 0$ which is equivalent to\n$$\nf(x) = b x^3 - a x \\geq -\\... | Baltic Way | Baltic Way 2023 Shortlist | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
00jx | We order the positive integers in two rows in the following manner:
1 3 6 11 19 32 53 ...
2 4 5 7 8 9 10 12 13 14 15 16 17 18 20 to 31 33 to 52 54 ...
We first write $1$ in the first row, $2$ in the second and $3$ in the first. After this, the following integers are written in such a way that an individual integer is a... | [
"We first note that $a_1 = 1$, $a_2 = 3$ and $a_3 = 6$ hold. It is quite straight-forward to note that a block of length $a_{n-1} + 1$ starts with the number $a_n + 1$, and that this block therefore ends on the number $a_n + (a_{n-1} + 1)$, which yields $a_{n+1} = a_n + a_{n-1} + 2$.\n\nThis recursion has the const... | Austria | AustriaMO2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | a_n = F_{n+3} - 2, where F_0 = 0 and F_1 = 1 | |
0iq0 | Problem:
Let $ABCD$ be a regular tetrahedron, and let $O$ be the centroid of triangle $BCD$. Consider the point $P$ on $AO$ such that $P$ minimizes $PA + 2(PB + PC + PD)$. Find $\sin \angle PBO$. | [
"Solution:\n\n$\\boxed{\\dfrac{1}{6}}$\n\nWe translate the problem into one about 2-D geometry. Consider the right triangle $ABO$, and $P$ is some point on $AO$. Then, the choice of $P$ minimizes $PA + 6PB$. Construct the line $\\ell$ through $A$ but outside the triangle $ABO$ so that $\\sin \\angle(AO, \\ell) = \\... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 1/6 | |
0gnh | Some unit squares of $2007 \times 2007$ square board are colored. Let $(i, j)$ be a unit square belonging to the $i$-th line and $j$-th column and $S_{i,j}$ be the set of all colored unit squares $(x, y)$ satisfying $x \leq i$ and $y \leq j$. At the first step in each colored unit square $(i, j)$ we write the number of... | [
"Let $f_{(i,j)}$ be the number written on $(i, j)$ in mod $2$. We can suppose that at the $0$th step $f_{(i,j)} = 1$ for all colored unit squares $(i, j)$. We prove the statement by induction with respect to $n$, the total number of colored unit squares.\n\nIf $n=1$, then after the first step $f_{(i,j)} = 1$ for th... | Turkey | Turkish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0awu | Problem:
Find the smallest positive integer whose cube ends in $2017$. | [] | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | final answer only | 9073 | |
07ow | Suppose $n \ge 2$, and $a_1, a_2, \dots, a_n$ are positive numbers that sum to $1$. Prove that
$$
\frac{n}{n-1} \le \sum_{i=1}^{n} \frac{a_i}{1-a_i},
$$
with equality iff $a_i = 1/n$, $i = 1, 2, \dots, n$. | [
"The function $f(x) = \\frac{1}{1-x}$, $0 \\le x < 1$, is strictly convex on $[0, 1)$, and so, if $\\alpha_i$, $i = 1, 2, \\dots, n$ are non-negative numbers that sum to $1$, then\n$$\nf\\left(\\sum_{i=1}^{n} \\alpha_i a_i\\right) \\le \\sum_{i=1}^{n} \\alpha_i f(a_i).\n$$\n\n**Solution 1.** In particular, with $\\... | Ireland | Irska 2014 | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0dw9 | Problem:
Naj bo $ABCD$ tetivni štirikotnik, pri katerem si nobeni 2 nasprotni stranici nista vzporedni. Presečišče premic $AB$ in $CD$ označimo z $E$, presečišče premic $AD$ in $BC$ pa s $F$. Simetrala kota $\angle AFB$ seka daljico $AB$ v točki $P$, daljico $CD$ pa v točki $R$. Simetrala kota $\angle BEC$ seka daljic... | [
"Solution:\n\nZaradi simetrije lahko predpostavimo, da leži točka $A$ med $E$ in $B$, točka $C$ pa med $F$ in $B$. Presečišče obeh simetral označimo s $T$, notranje kote štirikotnika $ABCD$ pa z $\\alpha, \\beta, \\gamma$ in $\\delta$. Potem je\n$$\n\\begin{aligned}\n\\angle FTE &= \\angle TFQ + \\angle FQT = \\\\\... | Slovenia | 48. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03vo | As shown in the figure, $BC$ and $AC$ are tangent to the inscribed circle $I$ of $\triangle ABC$ at $M$ and $N$. $E$ and $F$ are the midpoints of $AB$, $AC$ respectively. $EF$ intersects $BI$ at $D$. Prove that $M$, $N$, $D$ are collinear.
 | [
"Join $AD$, and it is obvious that $\\angle ADB = 90^\\circ$. Then join $AI$ and $DM$. Suppose $DM$ intersects $AC$ at $G$. Since $\\angle ABI = \\angle DBM$, we obtain $\\frac{AB}{BD} = \\frac{BI}{BM}$. Hence, $\\triangle ABI \\sim \\triangle DBM$, and\n$$\n\\angle DMB = \\angle AIB = 90^\\circ + \\frac{1}{2} \\an... | China | China Southeastern 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 > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0dtc | In an acute triangle $ABC$, $AC > AB$, $D$ is the point on $BC$ such that $AD = AB$. Let $\omega_1$ be the circle through $C$ tangent to $AD$ at $D$, and $\omega_2$ the circle through $C$ tangent to $AB$ at $B$. Let $F (\neq C)$ be the second intersection of $\omega_1$ and $\omega_2$. Prove that $F$ lies on $AC$. | [
"\n\nLet $\\omega_2$ intersect $AC$ at $F'$. We shall prove $F'$ lies on $\\omega_1$. Thus $F = F'$ lies on $AC$.\n\nReferring to the figure on the right. First $\\angle ABF' = \\angle BCA = \\alpha$ as $AB$ is tangent to $\\omega_2$ at $B$. Let $\\angle DBF' = \\be... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07dp | Let $X$ be an arbitrary point on the circumcircle of the triangle $ABC$. The perpendicular lines from $X$ to $AB$ and $AC$ cut $BC$ at $P$ and $Q$ respectively. The point $Y$ is the circumcenter of the circle passing through $X, P, Q$. (If $X, P, Q$ are coincident, then consider $Y$ the same point as them.)
a. Prove t... | [
"a. We know that triangle $ABC$ is an equilateral. Thus we have $\\widehat{PRQ} = \\widehat{PQR} = 90^\\circ - 60^\\circ = 30^\\circ$ and by this we have $PQ = PR$. Suppose that $S$ is the circumcenter of triangle $PQR$.\n\n\n\nWe have $\\widehat{QPR} = 180^\\circ - 2 \\times 30^\\circ = 12... | Iran | Iranian 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 > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0km0 | Problem:
The taxicab distance between points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $|x_{2}-x_{1}|+|y_{2}-y_{1}|$. A regular octagon is positioned in the $xy$ plane so that one of its sides has endpoints $(0,0)$ and $(1,0)$. Let $S$ be the set of all points inside the octagon whose taxicab distance from some octagon ... | [
"Solution:\n\nIn the taxicab metric, the set of points that lie at most $d$ units away from some fixed point $P$ form a square centered at $P$ with vertices at a distance of $d$ from $P$ in directions parallel to the axes. The diagram above depicts the intersection of an octagon with eight ... | United States | HMMT November 2021 Team Round | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 2309 | |
0fts | Problem:
Finde alle natürlichen Zahlen, die sich in der Form
$$
\frac{(a+b+c)^{2}}{a b c}
$$
darstellen lassen, wobei $a, b$ und $c$ natürliche Zahlen sind. | [
"Solution:\nNehme an, die natürliche Zahl $n$ lasse sich in dieser Form darstellen, also\n$$\n\\frac{(a+b+c)^{2}}{a b c}=n\n$$\nHalte $n$ fest und wähle $a, b, c$ mit $a \\geq b \\geq c$ und $a+b+c$ minimal. Die obige Gleichung lässt sich umschreiben zu\n$$\na^{2}+a(2 b+2 c-n b c)+(b+c)^{2}=0\n$$\nDies ist eine qua... | Switzerland | IMO Selektion | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1, 2, 3, 4, 5, 6, 8, 9 | |
0db5 | Let non-integer real numbers $a$, $b$, $c$, $d$ are given, such that the sum of each 3 of them is integer. May it happen that $a b + c d$ is an integer. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Other"
] | English | proof and answer | No | |
0cx6 | Let $a \geq b \geq c > 0$. Prove that
$$
(a - b + c)\left(\frac{1}{a} - \frac{1}{b} + \frac{1}{c}\right) \geq 1
$$ | [] | Saudi Arabia | SAMC | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
07r4 | For which prime numbers $p$ do there exist positive rational numbers $x, y$ and a positive integer $n$ such that
$$
x + y + \frac{p}{x} + \frac{p}{y} = 3n?
$$ | [
"If $p \\equiv 2 \\pmod{3}$ and $x = y = 1$, the number\n$$\nx + y + \\frac{p}{x} + \\frac{p}{y} = 2(1 + p)\n$$\nis an integer that is divisible by 3, hence for all such primes the required solutions $x, y, n$ exist.\n\nWe now show that solutions can only exist if $p \\equiv 2 \\pmod{3}$.\n\nSuppose\n$$\nx = \\frac... | Ireland | Ireland_2017 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof and answer | All primes p with p ≡ 2 (mod 3), and no others. | |
04gy | Let $n \ge 2$ be an integer and let $a_0, a_1, \ldots, a_n$ be consecutive terms of an arithmetic progression. Prove that
$$
a_0 - \binom{n}{1}a_1 + \dots + (-1)^k \binom{n}{k}a_k + \dots + (-1)^n \binom{n}{n}a_n = 0.
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
00zd | Problem:
Using each of the eight digits $1, 3, 4, 5, 6, 7, 8$ and $9$ exactly once, a three-digit number $A$, two two-digit numbers $B$ and $C$, $B < C$, and a one-digit number $D$ are formed. The numbers are such that $A + D = B + C = 143$. In how many ways can this be done? | [
"Solution:\n\nFrom $A = 143 - D$ and $1 \\leq D \\leq 9$, it follows that $134 \\leq A \\leq 142$. The hundreds digit of $A$ is therefore $1$, and the tens digit is either $3$ or $4$. If the tens digit of $A$ is $4$, then the sum of the units digits of $A$ and $D$ must be $3$, which is impossible, as the digits $0$... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 24 | |
0iti | Problem:
Let $T = \int_{0}^{\ln 2} \frac{2 e^{3x} + e^{2x} - 1}{e^{3x} + e^{2x} - e^{x} + 1} \, dx$. Evaluate $e^{T}$. | [
"Solution:\nDivide the top and bottom by $e^{x}$ to obtain that\n$$\nT = \\int_{0}^{\\ln 2} \\frac{2 e^{2x} + e^{x} - e^{-x}}{e^{2x} + e^{x} - 1 + e^{-x}} \\, dx\n$$\nNotice that $2 e^{2x} + e^{x} - e^{-x}$ is the derivative of $e^{2x} + e^{x} - 1 + e^{-x}$, and so\n$$\nT = \\left[ \\ln \\left| e^{2x} + e^{x} - 1 +... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Calculus > Derivatives"
] | null | proof and answer | 11/4 | |
097c | Problem:
Fie matricea $A=\left(\begin{array}{cc}-a & \frac{1}{1-a} \\ a^{3}-1 & 1+a\end{array}\right)$, unde $a \in \mathbb{R} \backslash\{1\}$. Determinați $A^{2021}-A^{2020}$. | [
"Solution:\nObservăm că\n$$\nA^{2}=\\left(\\begin{array}{cc}\n-a-1 & \\frac{1}{1-a} \\\\\na^{3}-1 & a\n\\end{array}\\right), \\quad A^{3}=\\left(\\begin{array}{cc}\n-1 & 0 \\\\\n0 & -1\n\\end{array}\\right)=-I_{2}\n$$\nAtunci $A^{2020}=\\left(A^{3}\\right)^{673} \\cdot A=-A$, $A^{2021}=\\left(A^{3}\\right)^{673} \\... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Linear Algebra > Matrices"
] | null | proof and answer | I_2 | |
0fc3 | Problem:
Determina el triángulo de menor perímetro entre todos los que tienen la circunferencia inscrita con el mismo radio y el mismo valor de un ángulo. | [
"Solution:\n\nSean $ABC$ todos los triángulos con esta propiedad; es decir, que el ángulo común con el mismo valor es $A$ y el radio inscrito también común es $r$.\n\nEntonces el perímetro\n$$\n\\begin{aligned}\n& 2p = a + b + c = \\\\\n& r\\left(\\cot \\frac{B}{2} + \\cot \\frac{C}{2}\\right) + r\\left(\\cot \\fra... | Spain | null | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0356 | Problem:
The excircle to the side $AB$ of a triangle $ABC$ is tangent to the circle with diameter $BC$. Find $\Varangle ACB$ if the lengths of the sides $BC$, $CA$ and $AB$ form (in this order) an arithmetic progression. | [
"$$\n\\left(\\frac{a}{2}+r_{c}\\right)^{2}=r_{c}^{2}+\\left(p-\\frac{a}{2}\\right)^{2}\n$$\n\nThen $a r_{c}=p(p-a)$. Since $r_{c}=\\frac{S}{p-c}$, we obtain by using Heron's formula\n$$\na S=p(p-a)(p-c)=\\frac{S^{2}}{p-b}\n$$\n$$\na(p-b)=S\n$$\nSince $a$, $b$ and $c$ form (in this order) an... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | 90° | |
05v0 | Problem:
Soit $f: \mathbb{N}^{\star} \rightarrow \mathbb{N}^{\star}$ une fonction telle que pour tout entier $n \geqslant 1$, $f(f(n))$ soit égal au nombre de diviseurs positifs de $n$. Montrer que si $p$ est un nombre premier, alors $f(p)$ est aussi un nombre premier. | [
"Solution:\nDans la suite, on note $d(n)$ le nombre de diviseurs positifs de $n$. Si $p$ est premier, on a $f(f(p))=2$ par définition. En appliquant $f$ des deux côtés de l'égalité, on voit que $f(2)=d(f(p))$. On veut donc montrer que $f(2)=2$, ce qui prouvera que $f(p)$ a exactement deux diviseurs et est donc prem... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
03g8 | Problem:
Suppose that $n$ people each know exactly one piece of information, and all $n$ pieces are different. Every time person $A$ phones person $B$, $A$ tells $B$ everything that $A$ knows, while $B$ tells $A$ nothing. What is the minimum number of phone calls between pairs of people needed for everyone to know eve... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2n - 2 | |
0hel | Find all four-digit positive integers $\overline{abcd}$, which satisfy the conditions:
1) $a \le b \le c \le d$;
2) $a^2 + b^2 + c^2 + d^2$ is divisible by 4;
3) $\overline{abcd}$ divided by $c$ gives the remainder 7. | [
"From the second condition, it becomes clear that all the digits are either even or odd, since the square of the integer gives the remainder 0 or 1 modulo 4. From the third condition, digit $c$ equals 8 or 9.\nIf $c=8$, then $abcd$ is even, but then it cannot give an odd remainder modulo 8, because then the equalit... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 7999 | |
073z | Let $P$ be an interior point of a triangle $ABC$. Show, with usual notations, that
$$
\frac{PA}{a} + \frac{PB}{b} + \frac{PC}{c} \ge \sqrt{3}.
$$ | [
"Let $G$ denote the centroid of $ABC$. We have\n$$\n\\sum \\frac{PA}{a} = \\sum \\frac{PAGA}{aGA} \\ge \\frac{\\sum PAGA}{\\max\\{aGA, bGB, cGC\\}}.\n$$\nWe observe that $\\vec{PA} \\cdot \\vec{GA} \\le PAGA$. Thus\n$$\n\\begin{align*}\n\\sum PAGA & \\ge \\sum \\vec{PA} \\cdot \\vec{GA} \\\\\n&= \\sum (\\vec{PG} + ... | India | Indija TS 2009 | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0bji | Let $P$ and $Q$ be the midpoints of the diagonals $BD$ and $AC$ of the quadrilateral $ABCD$. Consider the points $M \in (BC)$, $N \in (CD)$, $R \in (PQ)$ and $S \in (AC)$ such that $\frac{BM}{MC} = \frac{DN}{NC} = \frac{PR}{RQ} = \frac{AS}{SC} = k$. Prove that the centroid of the triangle $AMN$ lies on the segment $[RS... | [
"Let $G$ be the centroid of the triangle $AMN$. Then we have\n$$\n\\overrightarrow{GR} = \\frac{\\overrightarrow{GP} + k\\overrightarrow{GQ}}{1+k} = \\frac{\\overrightarrow{GB} + \\overrightarrow{GD} + k(\\overrightarrow{GA} + \\overrightarrow{GC})}{2(1+k)}\n$$\nand\n$$\n\\overrightarrow{GS} = \\frac{\\overrightarr... | Romania | 65th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circ... | null | proof only | null | |
0iag | Problem:
How many solutions in nonnegative integers $(a, b, c)$ are there to the equation
$$
2^{a} + 2^{b} = c! ?
$$ | [
"Solution:\nWe can check that $2^{a} + 2^{b}$ is never divisible by $7$, so we must have $c < 7$. The binary representation of $2^{a} + 2^{b}$ has at most two $1$'s. Writing $0!$, $1!$, $2!$, $\\ldots$, $6!$ in binary, we can check that the only possibilities are $c = 2, 3, 4$, giving solutions $(0, 0, 2)$, $(1, 2,... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 5 | |
0clt | Let $n \ge 3$ be a natural number, and let $\mathcal{F}$ be a family consisting of at most $n$ distinct subsets of the set $\{1, 2, \dots, n\}$ with the property that one can consider $n$ distinct points in the plane, labeled with the numbers $1, 2, \dots, n$, then draw segments between some of these points such that, ... | [
"For $k \\in \\{1, 2, \\dots, n\\}$, denote by $a_k = \\text{card}\\{F \\in \\mathcal{F} \\mid k \\in F\\}$.\n\nThen $a_k \\in \\{0, 1, \\dots, n\\}$, and the sum whose maximum we want to estimate is\n$$\ns = \\sum_{F \\in \\mathcal{F}} \\text{card}(F) = \\sum_{k=1}^{n} a_k.\n$$\nThe connection condition implies th... | Romania | 75th NMO Selection Tests | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Functional equations"
] | English | proof and answer | n(n+1)/2 | |
0cgv | Fix an integer $n \ge 2$. Consider $n$ real numbers $a_1, a_2, \dots, a_n$, not all equal, and let
$$
d = d(a_1, a_2, \dots, a_n) = \max_{1 \le i < j \le n} |a_i - a_j|
$$
and
$$
s = s(a_1, a_2, \dots, a_n) = \sum_{1 \le i < j \le n} |a_i - a_j|.
$$
Determine, in terms of $n$, the smallest and the largest values the q... | [
"The required minimum is $n-1$ and is achieved, for instance, by $a_1 < a_2 = \\dots = a_n$. The maximum is $\\lfloor \\frac{1}{2}n \\rfloor \\lfloor \\frac{1}{2}(n+1) \\rfloor$ and is achieved, for instance, by\n$$\na_1 = \\dots = a_{\\lfloor n/2 \\rfloor} < a_{\\lfloor n/2 \\rfloor+1} = \\dots = a_n.\n$$\nIn each... | Romania | 74th NMO Selection Tests for BMO and IMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | minimum: n−1; maximum: floor(n/2) * floor((n+1)/2) |
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