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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
014a | Problem:
For a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of real numbers it is known that
$$
a_{n} = a_{n-1} + a_{n+2} \quad \text{for } n = 2, 3, 4, \ldots
$$
What is the largest number of its consecutive elements that can all be positive? | [
"Solution:\n\nThe initial segment of the sequence could be $1; 2; 3; 1; 1; -2; 0$. Clearly it is enough to consider only initial segments. For each sequence the first 6 elements are $a_{1}; a_{2}; a_{3}; a_{2} - a_{1}; a_{3} - a_{2}; a_{2} - a_{1} - a_{3}$. As we see, $a_{1} + a_{5} + a_{6} = a_{1} + (a_{3} - a_{2}... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 5 | |
05z4 | Problem:
Dans un tournoi organisé entre 6 équipes, chaque équipe joue contre chaque autre équipe exactement une fois. Lorsqu'une équipe gagne, elle obtient 3 points, et l'équipe perdante reçoit 0 point. Si la partie est nulle, les deux équipes reçoivent un point. Déterminez les $a$ pour lesquels il est possible que le... | [
"Solution:\n\nLors de chaque match, entre deux et trois points sont attribués au total. Étant donné qu'il y a 15 matchs au total, il y a donc eu entre 30 et 45 points de répartis entre les équipes. On doit donc avoir:\n$$\n\\begin{aligned}\n& 30 \\leqslant a+(a+1)+(a+2)+\\cdots+(a+5) \\leqslant 45 \\\\\n& 30 \\leqs... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 4 | |
0ji9 | Problem:
Let $z$ be a non-real complex number with $z^{23}=1$. Compute
$$
\sum_{k=0}^{22} \frac{1}{1+z^{k}+z^{2 k}}
$$ | [
"Solution:\nAnswer: $\\frac{46}{3}$\n\nFirst solution: Note that\n$$\n\\sum_{k=0}^{22} \\frac{1}{1+z^{k}+z^{2 k}}=\\frac{1}{3}+\\sum_{k=1}^{22} \\frac{1-z^{k}}{1-z^{3 k}}=\\frac{1}{3}+\\sum_{k=1}^{22} \\frac{1-\\left(z^{24}\\right)^{k}}{1-z^{3 k}}=\\frac{1}{3}+\\sum_{k=1}^{22} \\sum_{\\ell=0}^{7} z^{3 k \\ell}\n$$\... | United States | HMMT | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | 46/3 | |
01i0 | Let $O$ be the circumcenter of triangle $ABC$. Let $D$, $E$ and $F$ be the reflection of $O$ over lines $BC$, $CA$ and $AB$, respectively. Show that lines $AD$, $BE$ and $CF$ are concurrent. | [
"Refer to figure 12.\n\nAs $O$ is the circumcenter of $\\triangle ABC$ it follows that line segments $OB$ and $OC$ are concurrent. As $D$ is the reflection of $O$ across $BC$ it follows that segments $OB$ and $DB$ are concurrent and segments $OC$ and $DC$ are concurrent. It follows that $OBDC$ is a rhombus and ther... | Baltic Way | Baltic Way 2021 Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Transformations > Translation"
] | null | proof only | null | |
0bny | Let $ABC$ be a triangle, let $A_1$, $B_1$, $C_1$ be the antipodes of the vertices $A$, $B$, $C$, respectively, in the circle $ABC$, and let $X$ be a point in the plane $ABC$, collinear with no two vertices of the triangle $ABC$. The line through $B$, perpendicular to the line $XB$, and the line through $C$, perpendicul... | [
"Let the lines $B_2C_2$, $C_2A_2$, $A_2B_2$ meet the circle $ABC$ again at $A_3$, $B_3$, $C_3$, respectively. The angle $AA_3A_1$ is a right angle, so $A_1A_3$ is perpendicular to $B_2C_2$ and hence parallel to $XA$, and the parallel lines $XA$ and $A_1A_3$ are equidistant from the circumcentre $O$ of the triangle ... | Romania | 2015 Thirteenth IMAR Mathematical Competition | [
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof only | null | |
0507 | Consider a parallelogram $ABCD$.
a) Prove that if the incenter of the triangle $ABC$ is located on the diagonal $BD$, then the parallelogram $ABCD$ is a rhombus.
b) Is the parallelogram $ABCD$ a rhombus whenever the circumcenter of the triangle $ABC$ is located on the diagonal $BD$? | [
"a) As the incenter of the triangle $ABC$ is located on diagonal $BD$ (Fig. 1), we can conclude that $BD$ is the bisector of $\\angle ABC$. Therefore $\\angle ABD = \\angle CBD$. However, since $ABCD$ is a parallelogram, $\\angle ABD = \\angle CDB$. Hence the triangle $BCD$ is isosceles, i.e. $|BC| = |CD|$. Thus, $... | Estonia | Selected Problems from Open Contests | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | a) Yes, ABCD is a rhombus. b) No; for example, a non-square rectangle. | |
0bbk | Triangle $ABC$ has $\angle ABC = 60^\circ$. Points $M$ and $D$ are placed on the sides $(AC)$ and $(AB)$ respectively, so that $\angle BCA = 2\angle MBC$ and $BD = MC$. Find the measure of the angle $\angle DMB$. | [
"Take $G$ so that $CG = CM$, $C \\in (BG)$ and denote $x = \\angle MBC$. Since $\\angle MCB$ is exterior to the isosceles triangle $MCG$, $\\angle MGC = x$. This means that triangle $MBG$ is isosceles and similar to triangle $MCG$. Thus, $\\frac{BG}{MG} = \\frac{MB}{MC} = \\frac{MG}{BD}$.\n\nDenote $H$ the common p... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 30° | |
0ki4 | For how many ordered pairs $(b, c)$ of positive integers does neither $x^2 + bx + c = 0$ nor $x^2 + cx + b = 0$ have two distinct real solutions?
(A) 4 (B) 6 (C) 8 (D) 12 (E) 16 | [] | United States | AMC 12 A | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | B | |
03zp | Determine, with proof, all pairs $(a,b)$ of integers, such that for any positive integer $n$, one has $n \mid (a^n + b^{n-1})$. | [
"The solution pairs consist of $(0,0)$ and $(-1,-1)$.\n\nIf one of $a$ and $b$ is $0$, it is obvious that the other is also $0$.\n\nNow we assume that $ab \\neq 0$, select a large prime $p$ such that $p > |a + b^2|$, it follows from Fermat's little theorem that\n$$\na^p + b^{p+1} \\equiv a + b^2 \\pmod p,\n$$\nAs $... | China | 2011 China Western Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | [(0, 0), (-1, -1)] | |
01oj | Points $X$ and $Y$ are marked on the sides $AB$ and $AD$ of the convex quadrilateral $ABCD$ respectively.
Find $AX : BX$ if $CX \parallel DA$, $DX \parallel CB$, $BY \parallel CD$, $CY \parallel BA$. | [
"Answer: $(\\sqrt{5} + 1)/2$.\nLet $\\lambda = AX : BX$ be the required ratio. On one hand, by Thales' theorem\n$$\n\\begin{aligned}\n\\lambda &= \\frac{AX}{BX} = \\frac{YR}{RB} = \\\\\n&= [YR = CD, \\text{ since } CDYR \\text{ is a parallelogram}] = \\\\\n&= \\frac{CD}{RB}.\n\\end{aligned} \n\\quad (1)\n$$\n![](at... | Belarus | BelarusMO 2013_s | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | (1+sqrt(5))/2 | |
0j4w | Problem:
Tac is dressing his cat to go outside. He has four indistinguishable socks, four indistinguishable shoes, and four indistinguishable snow-shoes. In a hurry, Tac randomly pulls pieces of clothing out of a drawer and tries to put them on a random one of his cat's legs; however, Tac never tries to put more than ... | [
"Solution:\n\nOn each leg, Tac's cat will get a shoe, a sock, and a snow-shoe in a random order. Thus, the probability that they will be put on in order for any given leg is $\\frac{1}{3!} = \\frac{1}{6}$. Thus, the probability that this will occur for all 4 legs is $\\left(\\frac{1}{6}\\right)^4 = \\frac{1}{1296}$... | United States | Harvard-MIT November Tournament | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | final answer only | 1/1296 | |
0d0t | In an $n \times n$ board, the numbers $0$ through $n^2 - 1$ are written so that the number in row $i$ and column $j$ is equal to
$$
(i-1) + n(j-1) \text{ where } 1 \le i, j \le n.
$$
Suppose we select $n$ different cells of the board, where no two cells are in the same row or column. Find the maximum possible product o... | [
"The answer is $n! \\cdot (n-1)^n$. This is achievable by choosing the numbers $n-1, 2(n-1), 3(n-1), \\dots, n(n-1)$, which are at positions $(n, 1), (n-1, 2), (n-2, 3), \\dots, (1, n)$. To show this is the best possible, we begin with a lemma.\n\n**Lemma.** If $a < b < c < d$ are real numbers such that $a+d = b+c$... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | n!·(n−1)^n | |
0bu1 | Problem:
Fie $f: \mathbb{R} \rightarrow (0, \infty)$ o funcție continuă și periodică. Dacă $2$ este perioadă a lui $f$, arătați că:
a. $\int_{0}^{2} \frac{f(x+1)}{f(x)} \, \mathrm{d}x \geq 2$.
b. $\int_{0}^{2} \frac{f(x+1)}{f(x)} \, \mathrm{d}x = 2$ dacă și numai dacă $1$ este perioadă a lui $f$. | [
"Solution:\n\na. Inegalitatea rezultă din relațiile de mai jos:\n$$\n\\begin{aligned}\n\\int_{0}^{2} \\frac{f(x+1)}{f(x)} \\, \\mathrm{d}x &= \\int_{0}^{1} \\frac{f(x+1)}{f(x)} \\, \\mathrm{d}x + \\int_{1}^{2} \\frac{f(x+1)}{f(x)} \\, \\mathrm{d}x \\\\\n&= \\int_{0}^{1} \\frac{f(x+1)}{f(x)} \\, \\mathrm{d}x + \\int... | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0lec | Let $ABC$ be an acute, non-isosceles triangle with altitudes $AD$, $BE$ and $CF$. The circle ($AD$) intersects $DE$, $DF$ at $M$, $N$. Let $P$, $Q$ be the points on $AB$, $AC$ respectively such that $NP$ is perpendicular to $AB$ and $MQ$ is parallel to $AC$. Let ($I$) be the circumcircle of triangle $APQ$.
a) Prove th... | [
"a) Let $T$ be the foot of $A$ on $EF$. Note that $FC$ is the internal bisector of $\\angle DFE$ so $FM$, $FT$ are symmetric with respect to $AB$. On the other hand, $\\angle FNA = \\angle FTA = 90^\\circ$ then $N$, $T$ are symmetric with respect to $AB$. Therefore, $N$, $P$, $T$ are collinear and $TP \\perp AB$. S... | Vietnam | VMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
017d | Given a circle such that it is possible to fit inside it six circles with radius $r$ so that they do not overlap. Prove that it is also possible to fit inside it seven circles with radius $r$ so that they do not overlap. | [
"We prove that the radius of a big circle is at least $3r$. If so then seven circles can be fitted inside it in a standard way as in figure 2.\n\nAssume the contrary – that the radius of it is less than $3r$. Denote the center of the big circle $O$ and the centers of small circles $A_1, A_2, \\dots, A_6$ respective... | Baltic Way | BALTIC WAY | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
08b7 | Problem:
Due polinomi monici (cioè con coefficiente di grado massimo uguale a 1) a coefficienti interi $p(x)$ e $q(x)$ sono tali che il loro massimo comun divisore sia $(x-1)(x-2)$, il loro minimo comune multiplo sia $(x-1)^2(x-2)^3(x-3)(x+1)$ e il grado di $p(x)$ sia minore o uguale al grado di $q(x)$. In quanti modi... | [
"Solution:\n\nLa risposta è $\\mathbf{(C)}$. Consideriamo i quattro fattori del minimo comune multiplo $A(x) = (x-1)^2$, $B(x) = (x-2)^3$, $C(x) = x-3$ e $D(x) = x+1$. Perché il minimo comune multiplo di $p(x)$ e $q(x)$ sia $A(x) B(x) C(x) D(x)$, è necessario che ciascuno dei quattro fattori divida almeno uno dei d... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | C | |
03yu | Let $n > 1$ be an integer, $k$ be the number of distinct prime factors of $n$. Prove that there exists an integer $a$, $1 < a < \frac{n}{k} + 1$, such that $n \mid a^2 - a$. (posed by Yu Hongbing) | [
"Let $n = p_1^{a_1} \\cdots p_k^{a_k}$ be the standard factorization of $n$. Since $p_1^{a_1}, \\dots, p_k^{a_k}$ are pairwise coprime, by the Chinese Remainder Theorem, for each $i$, $1 \\le i \\le k$, congruence equations\n$$\n\\begin{cases} x \\equiv 1 \\pmod{p_i^{a_i}} \\\\ x \\equiv 0 \\pmod{p_j^{a_j}}, \\quad... | China | China National Team Selection Test | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0dna | Problem:
Нека су $A' $ и $B'$ подножја висина из темена $A$ и $B$, редом, оштроуглог троугла $ABC$ ($AC \neq BC$). Кружница $k$ садржи тачке $A'$ и $B'$ и додирује страницу $AB$ у тачки $D$. Ако троуглови $ADA'$ и $BDB'$ имају једнаке површине, доказати да је
$$
\varangle A' D B' = \varangle ACB.
$$ | [
"Solution:\n\nНека је без смањења општости $BC > AC$. Тада се праве $A'B'$ и $AB$ секу у тачки $P$, при чему је $A$ између $P$ и $B$. Из једнакости површина $ADA'$ и $BDB'$ следи да је $\\frac{AD}{DB} = \\frac{PB'}{PA'}$. С друге стране, важи и $PD^2 = PA' \\cdot PB' = PA \\cdot PB$, одакле је $\\frac{PD}{PB} = \\f... | Serbia | Serbian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > ... | null | proof only | null | |
0ddw | Prove that it is impossible to fill the cells of an $8 \times 8$ table with the numbers from $1$ to $64$ (each number must be used once) so that for each $2 \times 2$ square, the difference between products of the numbers on its diagonals will be equal to $1$. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | null | proof only | null | |
00px | Determine all quadruplets $(x, y, z, t)$ of positive integers, such that
$$
12^x + 13^y - 14^z = 2013^t.
$$ | [
"Consider equation modulo $13$. We have $12^x + 13^y - 14^z \\equiv (-1)^x + 0 - 1 \\pmod{13}$, and $13 \\nmid 2013$, so $x$ must be odd. Also, $2013^t \\equiv -2 \\pmod{13}$, which implies $t = 12t_1 + 1$, $t_1 \\in \\mathbb{N}_0$.\n\nConsider equation modulo $3$. We have $12^x + 13^y - 14^z \\equiv 1 - (-1)^z \\p... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof and answer | (1, 3, 2, 1) | |
0e9n | Problem:
V podjetju izdelujejo vijake s šestimi enakimi stroji. V devetih urah bi s temi stroji izdelali 900 vijakov. V koliko urah bodo izdelali 900 vijakov, če se po petih urah pokvarita dva stroja? | [] | Slovenia | 14. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 11 hours | |
0imo | Problem:
In $\triangle ABC$, $D$ and $E$ are two points on segment $\overline{BC}$ such that $BD = CE$ and $\angle BAD = \angle CAE$. Prove that $\triangle ABC$ is isosceles.
 | [
"Solution:\n\nTranslate $\\triangle BDA$ horizontally until its side $BD$ coincides with side $EC$, and label the image of point $A$ by $A'$. We now have two triangles $\\triangle ECA$ and $\\triangle ECA'$ which share the same base $EC$, have the same height (equal to the height $AH$ of the original $\\triangle AB... | United States | 9th Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hp6 | Problem:
a. Prove that for all real numbers $x$ and $y$,
$$
x^{2}-2 y^{2}=-\left[(x+2 y)^{2}-2(x+y)^{2}\right] .
$$
b. How many positive integer solutions does the equation $x^{2}-2 y^{2}=1$ have?
c. How many positive integer solutions does the equation $x^{2}-2 y^{2}=5$ have? | [
"Solution:\n\na. This is simply an algebra calculation:\n$$\n\\begin{aligned}\n-\\left[(x+2 y)^{2}-2(x+y)^{2}\\right] & =-\\left[x^{2}+4 x y+4 y^{2}-2\\left(x^{2}+2 x y+y^{2}\\right)\\right] \\\\\n& =-\\left[x^{2}+4 x y+4 y^{2}-2 x^{2}-4 x y-2 y^{2}\\right] \\\\\n& =-\\left[-x^{2}+2 y^{2}\\right] \\\\\n& =x^{2}-2 y... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence r... | null | proof and answer | a) The identity holds. b) Infinitely many positive integer solutions. c) No positive integer solutions. | |
020j | Problem:
Find all functions $f: \mathbb{R} \rightarrow \mathbb{Z}$ such that
$$
(f(f(y)-x))^{2}+f(x)^{2}+f(y)^{2}=f(y) \cdot(1+2 f(f(y)))
$$
for all $x, y \in \mathbb{R}$. | [
"Solution:\nTake $x=y=0$ and write $c=f(0)$, then we find $f(c)^{2}+c^{2}+c^{2}=c+2 c f(c)$, so $(f(c)-c)^{2}=c-c^{2}$. The left-hand side is non-negative, so the right-hand side must be non-negative as well, hence $c-c^{2} \\geqslant 0$, so $c(1-c) \\geqslant 0$. This implies $0 \\leqslant c \\leqslant 1$, and sin... | Benelux Mathematical Olympiad | 8th Benelux Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | The only solutions are the constant functions f(x) = 0 for all real x and f(x) = 1 for all real x. | |
06ga | Suppose $a_1 = 4$ and $a_{n+1} \ge a_n + n + 2$ for all natural numbers $n$. Show that
$$
\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_{2010}} < \frac{3}{2}.
$$ | [
"Adding the inequalities\n$$\n\\begin{align*}\na_n &\\ge a_{n-1} + n + 1, \\\\\na_{n-1} &\\ge a_{n-2} + n, \\\\\n&\\vdots, \\\\\na_2 &\\ge a_1 + 3,\n\\end{align*}\n$$\nwe obtain $a_n \\ge a_1 + (3+4+\\cdots+(n+1)) = \\frac{n^2+3n+4}{2} > \\frac{(n+1)(n+2)}{2}$. It follows\nthat\n$$\n\\sum_{n=1}^{2010} \\frac{1}{a_n... | Hong Kong | IMO HK TST | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
06dr | Let $ABCD$ be a cyclic quadrilateral. $K, L, M, N$ are midpoints of sides $AB, BC, CD$ and $DA$ respectively. Prove that the orthocentres of triangles $AKN, BKL, CLM, DMN$ are vertices of a parallelogram. | [
"\n\nLet $H_1, H_2, H_3, H_4$ be the orthocentres of $\\triangle AKN, \\triangle BKL, \\triangle CLM, \\triangle DMN$\nrespectively, and let $O$ be the centre of $(ABCD)$.\nSince $K$ is the midpoint of $AB$, we have $OK \\perp AB$. Thus, $OK // H_1N$. Similarly, $ON // H_1K$. This shows $H_... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0j6y | Problem:
Find the sum of the coefficients of the polynomial $P(x) = x^{4} - 29 x^{3} + a x^{2} + b x + c$, given that $P(5) = 11$, $P(11) = 17$, and $P(17) = 23$. | [
"Solution:\nAnswer: $-3193$\n\nDefine $Q(x) = P(x) - x - 6 = x^{4} - 29 x^{3} + a x^{2} + (b-1) x + (c-6)$ and notice that $Q(5) = Q(11) = Q(17) = 0$. $Q(x)$ has degree 4 and by Vieta's Formulas the sum of its roots is $29$, so its last root is $29 - 17 - 11 - 5 = -4$, giving us $Q(x) = (x-5)(x-11)(x-17)(x+4)$. Thi... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | -3193 | |
0h5k | Solve the system of equations:
$$
\begin{cases} x^2 + 3xy = 3y + x, \\ y^2 - yx = 3x + y. \end{cases}
$$ | [
"Add up the equations and get\n$$\nx^2 + 2xy + y^2 = 4y + 4x \\text{ or } (x+y)^2 = 4(x+y).\n$$\n\nIf $x = -y$, the second equation implies\n$$\n2y^2 = -2y, \\text{ hence } y = 0 \\text{ or } y = -1.\n$$\nWe get two solutions, (0; 0) and (1; -1).\n\nIf $x = 4 - y$, the first equation implies\n$$\ny^2 - y(4 - y) = 3... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Other"
] | English | proof and answer | [(0, 0), (1, -1), (1, 3), (6, -2)] | |
0knt | Adults made up $\frac{5}{12}$ of the crowd of people at a concert. After a bus carrying 50 more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived. | [] | United States | AIME II | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 154 | |
00ku | In each move, one player cuts the string between two pearls and the other player chooses one of the resulting parts of the string while the other part is discarded.
In the first move, Alice cuts the string, thereafter, the players take turns.
A player loses if he or she obtains a string with a single pearl such that no... | [
"We claim that the winning situations are exactly the strings of an even number of pearls. We prove this claim by induction.\n\nA string with one pearl is a losing situation by definition.\n\nA string with an even number $n$ of pearls can easily be cut into two odd parts. These parts are a losing situation for the ... | Austria | Austrian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Bob has a winning strategy. | |
09ww | Problem:
Bepaal alle paren $(a, b)$ van positieve gehele getallen waarvoor
$$
a+b=\varphi(a)+\varphi(b)+\operatorname{ggd}(a, b) .
$$
Hier is $\varphi(n)$ het aantal getallen $k$ uit $\{1,2, \ldots, n\}$ met $\operatorname{ggd}(n, k)=1$. | [
"Solution:\nOplossing I. Stel eerst dat $a=1$. Er geldt $\\varphi(1)=1$. Voor alle positieve gehele $b$ is nu $\\operatorname{ggd}(a, b)=1$, dus de vergelijking wordt $1+b=1+\\varphi(b)+1$, dus $\\varphi(b)=b-1$. Er is dus precies één getal uit $\\{1,2, \\ldots, b\\}$ dat niet copriem met $b$ is; dat moet $b$ zelf ... | Netherlands | IMO-selectietoets II | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequ... | null | proof and answer | (1, p) and (p, 1) for any prime p; and (2^k, 2^k) for any positive integer k | |
00b7 | Given are 16 balls with weights $13$, $14$, $15$, $\ldots$, $28$ grams. Determine the balls with weights $13$, $14$, $27$, $28$ grams, by using a two-pan balance at most $26$ times. | [
"One can find the lightest ball $13$ via direct elimination by pairs: we make $8$ pairs and determine the $8$ lightest on each pair, now we make $4$ pairs and determine the lightest on each pair, then we select the two lightest and finally, the ball with the minimum weight. This takes $8+4+2+1=15$ attempts. Ball $1... | Argentina | Argentina_2018 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms"
] | English | proof only | null | |
0i63 | Problem:
Two $4 \times 4$ squares are randomly placed on an $8 \times 8$ chessboard so that their sides lie along the grid lines of the board. What is the probability that the two squares overlap? | [
"Solution:\n\n$529/625$. Each square has $5$ horizontal $\\cdot$ $5$ vertical $=25$ possible positions, so there are $625$ possible placements of the squares. If they do not overlap, then either one square lies in the top four rows and the other square lies in the bottom four rows, or one square lies in the left fo... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 529/625 | |
02zj | Problem:
Um fotógrafo deve tirar fotos de uma festa com 10 membros de uma mesma família. Cada um dos 45 possíveis pares de pessoas dessa família devem aparecer juntos em exatamente uma foto. Além disso, existem apenas dois tipo de fotos: as que possuem 2 ou 3 pessoas.
a) Verifique que cada pessoa da família deverá ap... | [
"Solution:\n\na) Cada pessoa deve aparecer com outros 9 familiares exatamente uma vez em alguma foto. Nas fotos com 3 pessoas, uma pessoa aparece em exatamente 2 pares e, como 9 é ímpar, é necessário que cada pessoa apareça em pelo menos uma foto com exatamente 2 pessoas.\n\nb) Sejam $x$ o número de fotos com 3 pes... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 19 | |
0eg2 | Problem:
Določi parametra $a$ in $b$ tako, da bosta premici podani z enačbama
$$
\begin{aligned}
(a+b) x - a y + a - 2 &= 0 \\
(2b - a) x + (a - 4b) y - a &= 0
\end{aligned}
$$
identični (sovpadali, se prekrivali). | [
"Solution:\n\nEnačbi premic zapišemo v eksplicitni obliki:\n$$\ny = \\frac{a+b}{a} x + \\frac{a-2}{a}\n$$\nin\n$$\ny = \\frac{a-2b}{a-4b} x + \\frac{a}{a-4b}.\n$$\nČe sta premici identični, imata zapisani v eksplicitni obliki enaka smerna koeficienta in enaka odseka na ordinatni osi. Dobimo enačbi\n$$\n\\frac{a+b}{... | Slovenia | 17. tekmovanje dijakov srednjih tehniških in strokovnih šol v znanju matematike | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a = 4, b = -1 | |
0jnb | Problem:
Let $N$ be the number of functions $f$ from $\{1,2, \ldots, 101\} \rightarrow \{1,2, \ldots, 101\}$ such that $f^{101}(1)=2$. Find the remainder when $N$ is divided by 103. | [
"Solution:\nFor convenience, let $n=101$. Compute the number of functions such that $f^{n}(1)=1$. Since $n$ is a prime, there are 2 cases: the order of $1$ is either $1$ or $n$. The first case gives $n^{n-1}$ functions, and the second case gives $(n-1)!$ functions. By symmetry, the number of ways for $f^{n}(1)=2$ i... | United States | HMMT November 2015 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | 43 | |
0hn3 | Problem:
Two regular polygons are said to be matching if the double of the interior angle of one of them equals the triple of the exterior angle of the other. Find all pairs of matching polygons. | [
"Solution:\n\nThe answers, expressed in terms of numbers of sides, are $(3,9)$, $(4,6)$, $(5,5)$, and $(8,4)$. They are easily verified to be matching.\n\nTo prove that there are no others, first note that if the first polygon has at most five sides, the second one necessarily has one of the numbers of sides necess... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (3,9), (4,6), (5,5), (8,4) | |
0e0t | Let $k$ be a positive integer greater than $1$. Prove: for every non-negative integer $m$ there exist $k$ positive integers $n_1, n_2, \dots, n_k$, such that
$$
n_1^2 + n_2^2 + \dots + n_k^2 = 5^{m+k}.
$$ | [
"Let $k = 2$. If $n_1 = 5^l$ and $n_2 = 2 \\cdot 5^l$, then $n_1^2 + n_2^2 = 5^{2l+1}$. If $n_1 = 3 \\cdot 5^l$ and $n_2 = 4 \\cdot 5^l$, then $n_1^2 + n_2^2 = 5^{2l+2}$. In both cases, $l$ is an arbitrary non-negative integer. Hence, the equation $n_1^2 + n_2^2 = 5^l$ has solutions for all positive integers $t$.\n... | Slovenia | Selection Examinations for the IMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Algebraic Number Theory > Quadratic forms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
07b8 | Suppose that $A_1A_2A_3A_4A_5$ is a convex pentagon in the cartesian plane with all its vertices having rational coordinates. For each $1 \le i \le 5$, define $B_i$ as the intersection point of the extensions of the sides $A_{i+1}A_{i+2}$ and $A_{i+3}A_{i+4}$ (the vertices of the pentagon are numbered in a cyclic manne... | [
"First, note that if $(x_1, y_1)$ and $(x_2, y_2)$ are two points with rational coordinates, then the equation of the line passing through them in the cartesian coordinates is\n$$\n\\frac{y - y_1}{x - x_1} = \\frac{y_2 - y_1}{x_2 - x_1}.\n$$\nTherefore, the equation of this line can be written using only rational n... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | English | proof only | null | |
0ek8 | Problem:
Kaj je rešitev enačbe $\sqrt[5]{9^{x-3}}-\frac{1}{\sqrt{3}}=0$?
(A) $x=-\frac{7}{4}$
(B) $x=-\frac{4}{7}$
(C) $x=\frac{4}{7}$
(D) $x=\frac{7}{4}$
(E) $x=-\frac{9}{4}$ | [
"Solution:\nEnačbo preoblikujemo v obliko $9^{\\frac{x-3}{5}}=\\frac{1}{\\sqrt{3}}$ in nadalje v obliko $3^{\\frac{2x-6}{5}}=3^{-\\frac{1}{2}}$. Ko rešimo to eksponentno enačbo, dobimo rešitev $\\frac{7}{4}$. Pravilen je odgovor $D$."
] | Slovenia | 22. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | MCQ | D | |
01oy | Point $X$ is marked inside an acute-angled triangle $ABC$ such that $\angle AXC = \angle A + \angle B$, $\angle BXA = \angle B + \angle C$, $\angle BXC = \angle C + \angle A$.
Prove that $\frac{BC \cdot AX}{AB} = \frac{AB \cdot CX}{AC}$. | [
"Extend the cevians $AX$, $BX$, $CX$; let their intersection points with the sides $BC$, $AC$, $BA$ be $A_1$, $B_1$, $C_1$ respectively. We have\n$$\n\\angle AXB_1 = 180^\\circ - \\angle BXA = 180^\\circ - (\\angle B + \\angle C) = \\angle A = \\alpha.\n$$\nSimilarly, $\\angle BXC_1 = \\angle B = \\beta$, $\\angle ... | Belarus | BelarusMO 2013_s | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
048o | Spouses Ana and Tomislav attended a party with four other couples. Upon arrival there was a certain number of handshaking, but no one shook hands with his or her spouse nor with himself/herself. When later Tomislav asked everyone at the party how many people they shook hands with, he got nine different answers. How man... | [
"No one shook hands with more than eight people so the responses that Tomislav got are: \"0\", \"1\", \"2\", \"3\", \"4\", \"5\", \"6\", \"7\", \"8\".\n\nAna certainly did not shake hands with eight people. Namely, if we assume that she did shake hands with eight people, all the others (except Tomislav) had to shak... | Croatia | Hrvatska 2011 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 4 | |
0kn3 | Problem:
For positive integers $a$ and $b$, let $M(a, b)=\frac{\operatorname{lcm}(a, b)}{\operatorname{gcd}(a, b)}$, and for each positive integer $n \geq 2$, define
$$
x_{n}=M(1, M(2, M(3, \ldots, M(n-2, M(n-1, n)) \ldots)))
$$
Compute the number of positive integers $n$ such that $2 \leq n \leq 2021$ and $5 x_{n}^{2... | [
"Solution:\n\nThe desired condition is that $x_{n}=5 x_{n+1}$ or $x_{n+1}=5 x_{n}$.\n\nNote that for any prime $p$, we have $\\nu_{p}(M(a, b))=\\left|\\nu_{p}(a)-\\nu_{p}(b)\\right|$. Furthermore, $\\nu_{p}(M(a, b)) \\equiv \\nu_{p}(a)+\\nu_{p}(b) \\bmod 2$. So, we have that\n$$\n\\nu_{p}\\left(x_{n}\\right) \\equi... | United States | HMMT Spring 2021 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 20 | |
0h3k | For arbitrary positive numbers $a$, $b$, $c$ prove the inequality
$$
\frac{ab}{a+3b+2c} + \frac{bc}{b+3c+2a} + \frac{ca}{c+3a+2b} \le \frac{1}{6}(a+b+c).
$$ | [
"Використаємо відому нерівність $(x + y + z) \\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right) \\geq 9$, $x, y, z > 0$, яку неважко довести за допомогою нерівності Коші. Перепишемо її у вигляді $\\frac{1}{x+y+z} \\leq \\frac{1}{9}\\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right)$. Оскільки $a + 3b +... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0acw | Let $k$ is the circumscribed circle for the quadrangle $ABCD$. The angle at the vertex $B$ is twice bigger than the angle at the vertex $A$ and for $40^\circ$ smaller than the angle at the vertex $D$. Calculate the angles of $ABCD$. | [
"From the condition in the problem we have that $\\beta = 2\\alpha$ and $\\beta = \\delta - 40^\\circ$. From this we have $\\delta = 2\\alpha + 40^\\circ$. Because $ABCD$ can be inscribed in a circle we have that $\\alpha + \\gamma = \\beta + \\delta$ and because $\\alpha + \\gamma + \\beta + \\delta = 360^\\circ$ ... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | A = 35°, B = 70°, C = 145°, D = 110° | |
0fk0 | Problem:
Encontrar todas las soluciones enteras posibles, $x$ e $y$, de la ecuación:
$$
p(x+y)=x y
$$
siendo $p$ un cierto número primo. | [
"Solution:\nDe $p(x+y)=x y$ y del hecho que $p$ es un número primo se deduce que $p$ divide a $x$ o a $y$. Puesto que, en el enunciado, los papeles de $x$ e $y$ son completamente simétricos, se puede, sin pérdida de generalidad, suponer que $p$ divide a $x$ y que, en consecuencia, hay un número $k$ tal que\n$$\nx=k... | Spain | Spanish Mathematical Olympiad - Local Stage | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All integer solutions are the ordered pairs (x, y) with (x − p)(y − p) = p^2, i.e., (x, y) ∈ {(0, 0), (2p, 2p), (p(p+1), p+1), (p+1, p(p+1)), (p(1−p), p−1), (p−1, p(1−p))}. | |
08zy | Let $N$ be the set of positive integers. A function $f: N \to N$ satisfies $f(34) = 2024$, and for any positive integers $a, b, c$, there exists a triangle with side lengths $a+f(b)$, $b+f(c)$, $c+f(a)$. Find the smallest possible value of $f(100) + f(101) + \dots + f(199)$ under these conditions. Note that three point... | [
"The existence of a triangle with side lengths $a + f(b)$, $b + f(c)$, $c + f(a)$ is equivalent to the following conditions:\n$$\n\\begin{cases} a + f(b) < b + f(c) + c + f(a), \\\\ b + f(c) < c + f(a) + a + f(b), \\\\ c + f(a) < a + f(b) + b + f(c). \\end{cases}\n$$\nBy symmetry of $a, b, c$, the existence of a tr... | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 102050 | |
0i3p | Problem:
Find the number of positive integer solutions to $n^{x} + n^{y} = n^{z}$ with $n^{z} < 2001$. | [
"Solution:\nIf $n = 1$, the relation can not hold, so assume otherwise. If $x > y$, the left hand side factors as $n^{y}(n^{x-y} + 1)$ so $n^{x-y} + 1$ is a power of $n$. But it leaves a remainder of 1 when divided by $n$ and is greater than 1, a contradiction. We reach a similar contradiction if $y > x$. So $y = x... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 10 | |
05w9 | Problem:
Soit $ABCD$ un parallélogramme. Une droite $d$ passant par $C$ coupe respectivement les droites $(AB)$ et $(AD)$ en $X$ et en $Y$. Les tangentes en $X$ et en $Y$ au cercle circonscrit à $AXY$ se coupent en un point $T$, puis la droite $(CT)$ recoupe le cercle circonscrit à $CDY$ en un point $P$.
Démontrer qu... | [
"Solution:\n\nDans la suite, on note $\\mathcal{C}_{\\mathcal{P}}$ le cercle circonscrit à un triangle $\\mathcal{P}$.\n\nPuisque $(TX)$ est tangente à $\\mathcal{C}_{AXY}$, une chasse aux angles indique que\n$$\n(XT, XY) = (AX, AY) = (DC, DY) = (PC, PY) = (PT, PY)\n$$\nce qui signifie que $P$ appartient à $\\mathc... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
090u | A cada punto del conjunto $A = \{(x, y, z) \in \mathbb{Z}^3\}$, formado por los puntos del espacio tridimensional cuyas coordenadas son enteras, le asignamos un color de entre $p$ colores posibles. Demuestra que forzosamente existe algún paralelepípedo recto (poliedro de seis caras en el que cada cara es un rectángulo)... | [
"Ponemos $n = p \\binom{p+1}{2}$. Para cada $j \\in \\{0, 1, \\dots, n\\}$, pensamos en el conjunto $A_j^0 = \\{(i, j, 0) \\in A; 0 \\le i \\le p\\}$. Por el Principio del Palomar, podemos asegurar que, para cada $j$ hay dos puntos de $A_j^0$ del mismo color. Puede que haya más de dos o que ocurra para más de un co... | Mexico | LVI Olimpiada Matemática Española (Concurso Final) | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof only | null | |
05lj | Problem:
Soient $n$ et $p$ des entiers $\geqslant 1$. Dans une assemblée de $n$ personnes, deux personnes quelconques ont au plus $p$ connaissances communes ; bien sûr, si $A$ connaît $B$, alors $B$ connaît $A$. Montrer que le nombre de paires non ordonnées $\{A, B\}$ de personnes qui se connaissent est inférieur ou é... | [
"Solution:\n\nNotons $A_{1}, \\cdots, A_{n}$ les personnes et considérons le graphe simple et non orienté dont les sommets sont les $A_{i}$, et où deux personnes $A_{i}$ et $A_{j}$ sont reliées par une arête si et seulement si elles se connaissent.\n\nPour tout $i$, notons $d_{i}$ le degré de $A_{i}$ (c'est-à-dire ... | France | Olympiades Françaises de Mathématiques - Test de Février | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
0d6h | On the Cartesian coordinate system $Oxy$, consider a sequence of points $A_{n}(x_{n}, y_{n})$ in which $(x_{n})_{n=1}^{\infty}, (y_{n})_{n=1}^{\infty}$ are two sequences of positive numbers satisfying the following conditions:
$$
x_{n+1} = \sqrt{\frac{x_{n}^{2} + x_{n+2}^{2}}{2}}, \quad y_{n+1} = \left(\frac{\sqrt{y_{n... | [
"From $O, A_{1}, A_{2016}$ are collinear, there exists some positive number $k$ such that\n$$\n\\frac{x_{1}}{y_{1}} = \\frac{x_{2016}}{y_{2016}} = k > 0.\n$$\nWe shall prove that for all $i = \\overline{2,2015}$, $\\frac{x_{i}}{y_{i}} > k$. Indeed,\nNotice that\n$$\nx_{n+1} = \\sqrt{\\frac{x_{n}^{2} + x_{n+2}^{2}}{... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0jp6 | Problem:
A line in the Cartesian plane is called stable if it passes through at least two points $(x, y)$ such that $x$ and $y$ are rational numbers. Prove or disprove: every point lies on some stable line. | [
"Solution:\nThe assertion is false: we will show that the point $(\\sqrt{2}, \\sqrt{3})$ does not lie on a stable line.\n\nNote that the slope of any stable line must be a rational number. Now assume for contradiction that $(a, b)$ lies on a stable line through $(\\sqrt{2}, \\sqrt{3})$, where $a$ and $b$ are both r... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
070c | Problem:
$n > 2$ is an integer. Consider the pairs $(a, b)$ of relatively prime positive integers, such that $a < b \leq n$ and $a + b > n$. Show that the sum of $1 / ab$ taken over all such pairs is $1 / 2$. | [
"Solution:\n\nInduction on $n$. It is obvious for $n = 3$, because the only pairs are $(1, 3)$ and $(2, 3)$, and $1 / 3 + 1 / 6 = 1 / 2$.\n\nNow suppose it is true for $n$. As we move to $n + 1$, we introduce the new pairs $(a, n + 1)$ with $a$ relatively prime to $n + 1$ and we lose the pairs $(a, n + 1 - a)$ with... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 1/2 | |
04of | Determine all pairs $(m, n)$ of positive integers such that
$2^m = 7n^2 + 1.$ | [
"Note that $2^m \\equiv 1 \\pmod{7}$, which means that $m = 3k$ for some positive integer $k$. Now we have\n$$\n2^{3k} - 1 = (2^k - 1)(2^{2k} + 2^k + 1) = 7n^2.\n$$\nDenote $A = 2^k - 1$ and $B = 2^{2k} + 2^k + 1$. Let $d$ be the greatest common divisor of $A$ and $B$, and note that $d$ is odd. Furthermore, $d$ div... | Croatia | Croatian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots >... | English | proof and answer | (3, 1) and (6, 3) | |
017m | There are a few cities in a country; one of them is the capital. There is a direct flight between any two of the cities, and any flight connecting two given cities has the same price. Suppose that all trips that begin and end in the same city and go exactly once through every other city have the same total cost. Prove ... | [
"Let $C$ be the capital and $C_1, C_2, \\dots, C_n$ be the remaining cities. Denote by $d(x, y)$ the price of the connection between the cities $x$ and $y$, and let $\\sigma$ be the total price of a round trip going exactly once through each city.\n\nNow consider a round trip missing the capital and visiting every ... | Baltic Way | BALTIC WAY | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
03mb | Let $d$ be a positive integer. Show that for every integer $S$, there exists an integer $n > 0$ and a sequence $\epsilon_1, \epsilon_2, \dots, \epsilon_n$, where for any $k$, $\epsilon_k = 1$ or $\epsilon_k = -1$, such that
$$
S = \epsilon_1(1+d)^2 + \epsilon_2(1+2d)^2 + \epsilon_3(1+3d)^2 + \dots + \epsilon_n(1+nd)^2.... | [
"Let $U_k = (1 + k d)^2$. We calculate $U_{k+3} - U_{k+2} - U_{k+1} + U_k$. This turns out to be $4d^2$, a constant. Changing signs, we obtain the sum $-4d^2$.\nThus if we have found an expression for a certain number $S_0$ as a sum of the desired type, we can obtain an expression of the desired type for $S_0 + (4d... | Canada | Kanada 2011 | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0k12 | Problem:
What is the largest factor of $130000$ that does not contain the digit $0$ or $5$? | [
"Solution:\n\nIf the number is a multiple of $5$, then its units digit will be either $0$ or $5$. Hence, the largest such number must have no factors of $5$.\n\nWe have $130000 = 2^{4} \\cdot 5^{4} \\cdot 13$. Removing every factor of $5$, we get that our number must be a factor of $2^{4} \\cdot 13 = 208$.\n\nIf ou... | United States | HMMT November 2018 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 26 | |
0g3q | Problem:
Anaëlle hat $2 n$ Steine, welche mit $1,2,3, \ldots, 2 n$ beschriftet sind, sowie eine rote und eine blaue Schachtel. Sie will nun alle $2 n$ Steine in die beiden Schachteln verteilen, sodass die Steine $k$ und $2 k$ für jedes $k=1,2, \ldots, n$ in unterschiedlichen Schachteln landen. Wie viele Möglichkeiten ... | [
"Solution:\n\nAntwort: Anaëlle hat $2^{n}$ Möglichkeiten.\n\nLösung 1 (bijektiv): Für jede ungerade ganze Zahl $1 \\leq t < 2 n$, nenne die Menge aller Steine, deren Beschriftung die Form $t \\cdot 2^{k}$ hat, die Kette ab $t$. Sobald wir einen Stein aus einer Kette in eine Schachtel platzieren, dann ist wegen der ... | Switzerland | Zweite Runde 2021 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2^n | |
0l3m | Problem:
Let $A B C D E F$ be a regular hexagon with $P$ as a point in its interior. Prove that of the three values $\tan \angle A P D$, $\tan \angle B P E$, and $\tan \angle C P F$, two of them sum to the third one. | [
"Solution:\n\nWLOG let the side length of the hexagon be 1. Let $O$ be the center of the hexagon. Consider drawing in the circles $(A P D)$, $(B P E)$, and $(C P F)$. Note that $O$ lies on the radical axis of all three circles, since $A O \\cdot O D = B O \\cdot O E = C O \\cdot O F$. Since... | United States | HMMT February 2024 | [
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Meth... | null | proof only | null | |
0dkq | Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ satisfying
$$
f(f(xy) + 1) = x f(x + f(y)), \forall x, y > 0.
$$ | [
"First, replace $x \\to \\frac{z}{y}$ then\n$$\nf(f(z) + 1) = \\frac{z}{y} f\\left(\\frac{z}{y} + f(y)\\right), \\forall y, z > 0. \\quad (1)\n$$\nIn (1), continue to set $z = 1$ and set $f(f(1) + 1) = c$, we have $f\\left(\\frac{1}{y} + f(y)\\right) = c y$ for all $y > 0$. The right-hand side takes on any values o... | Saudi Arabia | Saudi Booklet | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | All solutions are f(x) = c/x for some constant c > 0. | |
0gwl | We know that $|x+y|+|x-y|=1$. Find the least and the greatest value of expression $x^2 - 6x + y^2 - 6y$. | [
"Expression $f = x^2 - 6x + y^2 - 6y = (x-3)^2 + (y-3)^2 - 18 = R^2 - 18$ (fig.6) is at maximum (minimum) if expression $R^2 = (x-3)^2 + (y-3)^2$ being an equation of the circle of radius $R$ with the center at point $(3,3)$ is at maximum (minimum). The graph of equation $|x+y|+|x-y|=1$ is a square formed by lines ... | Ukraine | Ukrajina 2008 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | Least value is -11/2; greatest value is 13/2. | |
0e0e | At a contest students were presented with 24 multiple choice questions. Contestants who gave no answer or gave more than one answer were awarded 0 points for that question. For the correct answer they received 1 point and for a wrong answer $\frac{1}{4}$ of a point was deducted. If a student received 13 points, at most... | [
"Let $P$ be the number of questions for which the contestant was awarded 1 point and let $N$ denote the number of those for which $\\frac{1}{4}$ of a point was deducted. Then $P - \\frac{1}{4}N = 13$ or $N = 4P - 52$. Since there were 24 questions, we get $P + N \\le 24$ or $5P - 52 \\le 24$. So, $5P \\le 76$ and w... | Slovenia | National Math Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 15 | |
0ih2 | Problem:
What is the maximum number of bishops that can be placed on an $8 \times 8$ chessboard such that at most three bishops lie on any diagonal? | [
"Solution:\n\nIf the chessboard is colored black and white as usual, then any diagonal is a solid color, so we may consider bishops on black and white squares separately. In one direction, the lengths of the black diagonals are $2, 4, 6, 8, 6, 4$, and $2$. Each of these can have at most three bishops, except the fi... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 38 | |
0ie0 | Problem:
A sequence of positive integers is defined by $a_{0}=1$ and $a_{n+1}=a_{n}^{2}+1$ for each $n \geq 0$. Find $\operatorname{gcd}\left(a_{999}, a_{2004}\right)$. | [
"Solution:\nIf $d$ is the relevant greatest common divisor, then $a_{1000}=a_{999}^{2}+1 \\equiv 1=a_{0} \\pmod{d}$, which implies (by induction) that the sequence is periodic modulo $d$, with period $1000$. In particular, $a_{4} \\equiv a_{2004} \\equiv 0$. So $d$ must divide $a_{4}$. Conversely, we can see that $... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 677 | |
0hxm | Let $f(n)$ be the minimal number of colors needed to properly color the directed edges of a tournament on $n$ vertices so that no two edges of the same color form a directed path of length $2$. Determine $f(n)$ for all $n \ge 1$. | [
"Let $f(n)$ be the answer to the problem. We claim that $f(n) = \\lceil \\log n \\rceil$ for all $n \\ge 1$. The proof is divided into showing that this quantity is both an upper and a lower bound.\n\n**Lower Bound.** We show that every tournament on $n$ vertices has directed-edge-chromatic number at least $\\lceil... | United States | TST | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | ⌈log n⌉ | |
09v4 | We compute the product of two numbers, $99\ldots99 \times 99\ldots99,$ where the first number consists of 20 nines, and the second of 21 nines. Which number do you get if you add up the digits of the outcome of this multiplication? | [
"189"
] | Netherlands | Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 189 | |
02va | Problem:
Um tabuleiro de xadrez é um quadrado $8 \times 8$ em que as casinhas estão distribuídas em 8 linhas e 8 colunas.
Uma torre em um tabuleiro de xadrez ataca todas as peças que estejam na sua linha ou na sua coluna. Sabendo disso, determine:
a) Qual o número máximo de torres que podemo... | [
"Solution:\n\na) A seguir, temos um exemplo com 8 torres sem que duas estejam se atacando.\n\nSe colocarmos 9 ou mais torres, como existem apenas 8 linhas, haverá duas numa mesma linha e uma poderá atacar a outra. Portanto, o número máximo é 8.\n\n\nb) A seguir, temos um exemplo com 10 torr... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) 8; b) 10 | |
0lbf | Trên một cánh đồng hình chữ nhật kích thước $m \times n$ ô vuông (gồm $m$ hàng và $n$ cột) người ta đặt máy tưới nước vào bên trong một số các ô vuông. Mỗi máy trong một ô vuông không chỉ tưới được ô vuông này và các ô vuông có cạnh chung với nó, mà còn tưới đến được các ô vuông cùng cột với nó và cách nó đúng một ô. H... | [] | Vietnam | Kì thi chọn học sinh vào Đội tuyển Quốc gia Dự thi IMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Vietnamese | proof and answer | a) n; b) n | |
0aui | Problem:
Find the numerical value of $\left(1-\cot 37^\circ\right)\left(1-\cot 8^\circ\right)$. | [] | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | 2 | |
0g7p | 令 $f: \mathbb{N} \to \mathbb{N}$, 且令 $f^m$ 為 $f$ 作用 $m$ 次。假設對於所有的 $n \in \mathbb{N}$, 存在一個 $k$ 使得 $f^{2k}(n) = n + k$, 且令 $k_n$ 為滿足前式裡的 $k$ 中最小的。試證數列 $k_1, k_2, \dots$ 無界。 | [
"我們關心集合 $S = \\{1, f(1), f^2(1), \\dots\\}$。觀察到 $S$ 是無界的，因為對於所有 $S$ 中的 $n$，存在一個 $k > 0$ 使得在 $S$ 中，$f^{2k}(n) = n + k$。顯然 $f$ 把 $S$ 映射到 $S$ 本身，且在 $S$ 上 $f$ 是一對一。確實若 $f^i(1) = f^j(1)$ ($i$ 與 $j$ 不相等)，則從某一個 $m$ 開始 $f^m(1)$ 會週期性的出現，且 $S$ 會有限。\n\n定義 $g : S \\to S$ 為 $g(n) = f^{2k_n}(n) = n + k_n$。我們證明 $g$ 也是一對一。\n\n假設當 ... | Taiwan | 二〇一三數學奧林匹亞競賽第三階段選訓營,模擬競賽(一) | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
08xo | Find the value of
$$
\sum 17C_a \cdot 17C_b \cdot 17C_c,
$$
where the sum is taken over all triples $(a, b, c)$ of non-negative integers satisfying $a+b+c=5$. | [
"$$\n\\boxed{2349060}\n$$\nConsider a particle moving on the $xy$-plane according to the following rule: if the particle is at the point $(n, m)$ where $n$ and $m$ are integers, it moves in one step either to the point $(n+1, m)$ or to the point $(n, m+1)$. Consider the set $\\Omega$ of all paths that a particle ca... | Japan | Japan Mathematical Olympiad Initial Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 2349060 | |
06q5 | Let $ABCD$ be a convex quadrilateral and let $P$ and $Q$ be points in $ABCD$ such that $PQDA$ and $QPBC$ are cyclic quadrilaterals. Suppose that there exists a point $E$ on the line segment $PQ$ such that $\angle PAE = \angle QDE$ and $\angle PBE = \angle QCE$. Show that the quadrilateral $ABCD$ is cyclic. | [
"Let $F$ be the point on the line $AD$ such that $EF \\parallel PA$. By hypothesis, the quadrilateral $PQDA$ is cyclic. So if $F$ lies between $A$ and $D$ then $\\angle EFD = \\angle PAD = 180^{\\circ} - \\angle EQD$; the points $F$ and $Q$ are on distinct sides of the line $DE$ and we infer that $EFDQ$ is a cyclic... | IMO | 49th International Mathematical Olympiad Spain | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
066n | We consider on a circle $n$ different points such that there are no three chords with ends from these points passing through the same point lying in the interior of the circle.
a. Find the value of $n$, if you know that the number of triangles with vertices three from the $n$ points is equal to $2n$.
β. Find the valu... | [
"(a) Any three from the $n$ points lying on the circle define a triangle. Therefore the number of triangles defined by the $n$ points is equal to the number of combinations of $n$ elements by $3$, that is\n$$\n\\binom{n}{3} = \\frac{n(n-1)(n-2)}{6}\n$$\nTherefore we get the equation: $\\frac{n(n-1)(n-2)}{6} = 2n \\... | Greece | Selection Examination A | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals"
] | English | proof and answer | a: 5; b: 7 | |
02w6 | Problem:
Os pontos $X$, $Y$ e $Z$ estão marcados nos lados $AD$, $AB$ e $BC$ do retângulo $ABCD$, respectivamente. Dado que $AX = CZ$, mostre que $XY + YZ \geq AC$.
 | [
"Solution:\n\n\nConstrua o ponto $E$ sobre a reta $AD$ de modo que $AE = AX$, como indicado na figura acima. Como $AY$ é altura e mediana do triângulo $EYX$, podemos concluir que $YE = YX$. Além disso, como $AE = AX = CZ$ e $AE \\parallel CZ$, segue que $EZCA$ é um paralelogramo e, conseque... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0drt | For any integer $n \ge 1$, show that
$$
\sum_{k=1}^{n} \frac{2^k}{\sqrt{k+0.5}} \le 2^{n+1}\sqrt{n+1} - \frac{4n^{3/2}}{3}.
$$ | [
"First, note that\n$$\n\\frac{1}{\\sqrt{k+0.5}} \\le 2(\\sqrt{k+1}-\\sqrt{k})\n$$\nwhich follows from the fact that\n$$\n\\sqrt{k+1}-\\sqrt{k} = \\frac{1}{\\sqrt{k+1}+\\sqrt{k}} \\ge \\frac{1}{2\\sqrt{k}+0.5}\n$$\nsince $\\sqrt{x}$ is concave. Hence\n$$\n\\frac{2^k}{\\sqrt{k+0.5}} \\le 2^{k+1}(\\sqrt{k+1}-\\sqrt{k}... | Singapore | Singapur | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
02vm | Problem:
Quantos divisores de $88^{10}$ deixam resto 4 quando divididos por 6? | [
"Solution:\nComo $88=2^{3} \\cdot 11$, temos $88^{10}=\\left(2^{3} \\cdot 11\\right)^{10}=2^{30} \\cdot 11^{10}$. Se um inteiro $x$ deixa resto 4 por 6, então $x+2$ é múltiplo de 6. Consequentemente $x$ é par e deixa resto 1 na divisão por 3. Perceba agora que todos os divisores primos de $88^{10}$ deixam resto 2 n... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 165 | |
04ba | If $a + b = 4$ and $a^2 + b^2 = 14$, determine $a^3 + b^3$. | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | final answer only | 52 | |
0jsy | Problem:
For which integers $n \in \{1,2, \ldots, 15\}$ is $n^{n}+1$ a prime number? | [
"Solution:\n\n$n=1$ works. If $n$ has an odd prime factor, you can factor, and this is simulated also by $n=8$:\n$$\na^{2k+1}+1 = (a+1)\\left(\\sum_{i=0}^{2k}(-a)^i\\right)\n$$\nwith both parts larger than one when $a>1$ and $k>0$. So it remains to check $2$ and $4$, which work. Thus the answers are $1,2,4$."
] | United States | HMMT February 2016 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1, 2, 4 | |
058z | Points $A$, $D$, $E$ and $C$ lie on a line in this order. Point $B$ is chosen such that triangles $ADB$ and $BEC$ are similar (in this specific order of vertices), moreover $EC = 2AD$ and $\angle ABC = 120^\circ$. Find $\frac{AC}{AD}$. | [
"**Answer:** $3 + \\sqrt{2}$.\n\nDenote $\\angle EDB = \\alpha$ (Fig. 32). The similarity of $ADB$ and $BEC$ yields $\\angle DAB = \\angle EBC$ and $\\angle BDA = \\angle CEB$; the second of which gives $\\angle DEB = 180^\\circ - \\angle CEB = 180^\\circ - \\angle BDA = \\angle EDB = \\alpha$. Therefore $DB = BE$.... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 3 + sqrt(2) | |
0hsu | Problem:
Let $p$ and $q$ be positive real numbers with $p+q<1$. Teams $A$ and $B$ play a series of games. For each game, $A$ wins with probability $p$, $B$ wins with probability $q$, and they tie with probability $1-p-q$. The series ends when one team has won two more games than the other, that team being declared the... | [
"Solution:\n\nWe can break the series into rounds, where each round is a series of ties followed by one team winning a game. Thus, a round that $A$ wins has the form $A$, $tA$, $ttA$, $tttA$, $\\ldots$, where $t$ represents a tied game and $A$ a game $A$ wins. Thus, the probability $A$ wins a round is the geometric... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | p^2/(p^2+q^2) | |
01ja | Does there exist a scalene triangle whose incenter, circumcenter and orthocenter are collinear? | [
"See Problem G 3."
] | Baltic Way | Baltic Way 2023 Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | English | proof and answer | No | |
06gs | Among $45$ persons in a party, it is found that every person knows at least one other person in the party (i.e. they are friends), and any two persons with the same number of friends in the party do not know each other. What is the maximum number of pairs of persons who are friends? | [] | Hong Kong | Year 2012 | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 870 | |
0iif | Problem:
Cyclic quadrilateral $ABCD$ has side lengths $AB = 1$, $BC = 2$, $CD = 3$ and $DA = 4$. Points $P$ and $Q$ are the midpoints of $\overline{BC}$ and $\overline{DA}$. Compute $PQ^{2}$. | [
"Solution:\n\nConstruct $\\overline{AC}$, $\\overline{AQ}$, $\\overline{BQ}$, $\\overline{BD}$, and let $R$ denote the intersection of $\\overline{AC}$ and $\\overline{BD}$. Because $ABCD$ is cyclic, we have that $\\triangle ABR \\sim \\triangle DCR$ and $\\triangle ADR \\sim \\triangle BCR$. Thus, we may write $AR... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 116/35 | |
0bez | Let $n \ge 2$ be an integer and let $(K, +, \cdot)$ be a commutative field with the property:
$$
\underbrace{1 + \cdots + 1}_{m \text{ times}} \neq 0, \quad m = 2, \dots, n.
$$
Consider a polynomial $f \in K[X]$ of degree $n$ and $G$ a subgroup of the additive group $(K, +)$, $G \neq K$. Prove that there exists $a \in... | [
"Let $g \\in K[X]$ be a polynomial of degree $m \\in \\{2, \\dots, n\\}$. The polynomial\n$$\nh(X) = g(X + 1) - g(X)\n$$\nhas degree $m-1$ and if $\\mathrm{Im}\\, g \\subseteq G$, then $\\mathrm{Im}\\, h \\subseteq G$.\n\nSuppose $\\mathrm{Im}\\, f \\subseteq G$. Consider the polynomials $f_0, f_1, \\dots, f_{n-1}$... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Abstract Algebra > Field Theory"
] | null | proof only | null | |
0k8r | Problem:
Call a positive integer $n$ weird if $n$ does not divide $(n-2)!$. Determine the number of weird numbers between 2 and 100 inclusive. | [
"Solution:\n\nWe claim that all the weird numbers are all the prime numbers and $4$.\n\nSince no numbers between $1$ and $p-2$ divide prime $p$, $(p-2)!$ will not be divisible by $p$. We also have $2! = 2$ not being a multiple of $4$.\n\nNow we show that all other numbers are not weird. If $n = pq$ where $p \\neq q... | United States | HMMT February 2019 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 26 | |
05ri | Problem:
Soit $ABC$ un triangle dont tous les angles sont aigus, avec $AB \neq AC$. Soit $\Gamma$ le cercle circonscrit à $ABC$, et $D$ le milieu de l'arc $\widehat{BC}$ ne contenant pas $A$. Soit $E$ et $F$ des points appartenant respectivement aux segments $[AB]$ et $[AC]$, de sorte que $AE = AF$. Soit $P$ le point ... | [
"Solution:\n\nCommençons par tracer une figure, et notons $X$ le milieu de $[BC]$, que l'on espère être également le point d'intersection des droites $(BC)$ et $(JK)$. Sans perte de généralité, on suppose également que $AB > AC$.\n\n\n\nUne première remarque que l'on peut formuler est que, ... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Analytic / Coordinate Methods... | null | proof only | null | |
0gj7 | 令 $n \ge 5$ 為一正整數。有價值為 1 到 $n$ 的星星各一枚。安妮亞和貝琪玩一個遊戲。
在遊戲開始時，安妮亞先將這 $n$ 枚星星依照她想要的順序，在桌面上排成一橫排。
接著，從貝琪開始，兩個人輪流取走桌面上最左邊或最右邊的一枚星星。當所有星星都被取走時，取得的星星價值總和較高的人獲勝；若總和相同，則兩人平手。試求所有讓貝琪有必勝法的 $n$。
Let $n \ge 5$ be a positive integer. There are $n$ stars with values 1 to $n$, respectively. Anya and Becky play a game. Before the game s... | [
"答案是所有型如 $4k+2$ 的 $n$。為方便討論，將星星從左至右視為在 1 號位置至 $n$ 號位置。以下分別討論。\n\n1. 偶數 $n$:\n注意到在偶數 $n$ 的情況下，貝琪必然可以選擇取得所有奇數位置的星星（先拿走 1 號位置，之後每次都跟安妮亞拿同一側），或是取得所有偶數位置的星星（先拿走 $n$ 號位置，之後每次都跟安妮亞拿同一側）。令 $O$ 為奇數位置星星價值和，$E$ 為偶數位置星星價值和，則 $O+E = n(n+1)/2$。\n\n- 當 $n = 4k + 2$ 時，$n(n-1)/2$ 為奇數，故貝琪只要從 $O$ 和 $E$ 中選擇總和較大的一方取即可。\n- 當 $n = 4k$ 時，... | Taiwan | IMO 1J, Independent Study 1 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Chinese; English | proof and answer | All n of the form 4k+2 | |
0cbd | Let $SABCD$ be a pyramid with the apex $S$ and whose base $ABCD$ is a parallelogram. We consider the points $M$, $N$, $P$ and $Q$ on the edges $SA$, $SB$, $SC$ and $SD$, respectively, such that $MNPQ$ is also a parallelogram.
a) If $ABCD$ is a rhombus, prove that $MNPQ$ is also a rhombus.
b) If $ABCD$ is a rectangle,... | [
"The planes $(SAB)$ and $(SCD)$ meet along the line $d_1$, and let $d_2$ be the intersection line of the planes $(SBC)$ and $(SDA)$.\nAs $AB \\parallel CD$, $AB \\subset (SAB)$ and $CD \\subset (SCD)$, from the 'roof theorem' it follows that $AB \\parallel CD \\parallel d_1$. Because $MN \\parallel PQ$, $MN \\subse... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0de7 | Given $n \geq 2$ distinct positive integers $a_1, a_2, \ldots, a_n$ none of which is a perfect cube. Find the maximal possible number of perfect cubes among their pairwise products. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | floor(n^2/4) | |
06du | Find, with reasons, all integers $a$, $b$, and $c$ such that
$$
\frac{1}{2}(a+b)(b+c)(c+a) + (a+b+c)^3 = 1 - abc.
$$ | [
"The solutions are $(a, b, c) = (1, 0, 0), (2, -1, -1)$ up to permutation.\nLet $s = a + b + c$. Consider\n$$\nP(x) = (x - a)(x - b)(x - c) = x^3 - sx^2 + (ab + bc + ca)x - abc.\n$$\nPutting $x = s$, we obtain $(b+c)(c+a)(a+b) = (ab+bc+ca)s - abc$. Therefore, the given equation becomes\n$$\n(ab + bc + ca)s - abc + ... | Hong Kong | CHKMO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All permutations of (1, 0, 0) and (2, -1, -1). | |
01ne | Let $\Gamma$ be the incircle of a none-isosceles triangle $ABC$, $I$ be its center. Let $A_1$, $B_1$, $C_1$ be the tangency points of $\Gamma$ with the sides $BC$, $AC$, $AB$, respectively. Let $A_2 = \Gamma \cap AA_1$, $M = C_1B_1 \cap AI$, $P$ and $Q$ be the other (different from $A_1$, $A_2$) intersection points of ... | [
"We first show that the quadrilateral $A_1IMA_2$ is cyclic. Indeed, $C_1B_1 \\perp AI$ and $IC_1 \\perp AB$. So, $C_1M$ is an altitude in the right-angled triangle $AIC_1$, whence $AM \\cdot AI = AC_1^2$. Further, by the power of a point theorem, $AC_1^2 = A_2A \\cdot AA_1$. So,\n$$\nAM \\cdot AI = AC_1^2 = A_2A \\... | Belarus | Belorusija 2012 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis ... | English | proof only | null | |
0dre | In a convex hexagon $ABCDEF$, $AB$ is parallel to $DE$, $BC$ is parallel to $EF$ and $CD$ is parallel to $FA$. Prove that the triangles $ACE$ and $BDF$ have the same area. | [
"Let the coordinates of $A$ be $(0,0)$, $B$ be $(1,0)$, $D$ be $(c,b)$, $E$ be $(a,b)$, $F$ be $(e,f)$ and $C$ be $(x,y)$. Then $AB \\parallel ED$. For $AF \\parallel CD$, we need $(b-y)/(c-x) = f/e$. For $EF \\parallel CB$ we need $(x-1)/y = (a-e)/(b-f)$. From these equations, we get\n$$\nxb - ay = b - f + cf - be... | Singapore | Singapur 2015 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Linear Algebra > Determinants"
] | null | proof only | null | |
0gjc | 設 $ABCDE$ 是一個凸五邊形，使得 $\angle ABC = \angle AED = 90^\circ$。假設 $CD$ 的中點是三角形 $ABE$ 的外接圓心。令 $O$ 是三角形 $ACD$ 的外接圓心。證明直線 $AO$ 通過線段 $BE$ 的中點。
Let $ABCDE$ be a convex pentagon such that $\angle ABC = \angle AED = 90^\circ$. Suppose that the midpoint of $CD$ is the circumcentre of triangle $ABE$. Let $O$ be the circumcentre of ... | [
"Let $M$ be the midpoint of $CD$ and $X = BC \\cap ED$. Since $\\angle ABX = \\angle AEX = 90^\\circ$,\n$AX$ is the diameter of the circumcircle of $\\triangle ABX$, thus $ACXD$ is a parallelogram.\n\nNow, it is sufficient to show that $[OAB] = [OAE]$ where $[OAB]$ denotes the area of $\\triangle OAB$.\nLet $C', D'... | Taiwan | IMO 1J, Independent Study 1 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spira... | Chinese; English | proof only | null | |
0ilh | Problem:
Compute
$$
\left\lfloor\frac{2007!+2004!}{2006!+2005!}\right\rfloor
$$
(Note that $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$.) | [
"Solution:\n\nAnswer: $2006$. We have\n$$\n\\left\\lfloor\\frac{2007!+2004!}{2006!+2005!}\\right\\rfloor = \\left\\lfloor\\frac{\\left(2007 \\cdot 2006 + \\frac{1}{2005}\\right) \\cdot 2005!}{(2006+1) \\cdot 2005!}\\right\\rfloor = \\left\\lfloor\\frac{2007 \\cdot 2006 + \\frac{1}{2005}}{2007}\\right\\rfloor = \\le... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 2006 | |
0ju2 | Problem:
Define a number to be an anti-palindrome if, when written in base 3 as $a_{n} a_{n-1} \ldots a_{0}$, then $a_{i}+a_{n-i}=2$ for any $0 \leq i \leq n$. Find the number of anti-palindromes less than $3^{12}$ such that no two consecutive digits in base 3 are equal. | [
"Solution:\nAnswer: 126\nNote once the middle digit/pair of digits is determined, it suffices to choose the digits in the left half of the number and ensure no pair of consecutive digits are equal. For a number with an even number of digits, the middle pair is 02 or 20 while for a number with an odd number of digit... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 126 | |
02jf | Problem:
Geni é cliente de uma companhia telefônica que oferece o seguinte plano:
- tarifa mensal fixa de $R\$ 18,00$
- gratuidade em 10 horas de ligações por mês
- $R\$ 0,03$ por cada minuto que exceder às 10 horas.
Em janeiro, Geni usou seu telefone por 15 horas e 17 minutos, e em fevereiro por 9 horas e 55 minutos. ... | [
"Solution:\nVejamos a despesa em janeiro. Como 10 horas são gratuitas e Geni usou seu telefone por 15 horas e 17 minutos, ela deve pagar o custo de apenas 5 horas e 17 minutos mais a tarifa fixa mensal de 18 reais. Como o preço é dado em minutos, vamos reduzir a minutos o tempo a pagar. Sabemos que 1 hora $=60$ min... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | A | |
0blt | Find all differentiable functions $f : \mathbb{R} \to \mathbb{R}$ for which:
i) $f'(x) = 0$, for all $x \in \mathbb{Z}$;
ii) if $f'(x) = 0$ for some $x \in \mathbb{R}$, then $f(x) = 0$. | [
"Obviously, one solution is the zero function. Suppose now that $f : \\mathbb{R} \\to \\mathbb{R}$ is another solution. Then $f$ is continuous and $f(x) = f'(x) = 0$, for all $x \\in \\mathbb{Z}$. Consider $x_0 \\in \\mathbb{R} \\setminus \\mathbb{Z}$ such that $f(x_0) \\neq 0$, and let $k = \\lfloor x_0 \\rfloor \... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | null | proof and answer | The zero function: f(x) = 0 for all real x. | |
01fg | On a board the numbers $1, 2, 3, \ldots, 500$ are written. In a game two players $A$ and $B$ take turns alternately and $A$ starts. In each turn they have to erase two numbers $n$ and $2n$ on the board. If a player can not take a turn, the player loses. Determine who has a winning strategy. | [
"First we notice that for each odd number $u$, $1 \\le u < 500$, the numbers $u, 2u, 2^2u, \\dots, 2^{k_u}u$, where $k_u$ is the largest integer such that $u2^{k_u} \\le 500$, form a string of length $k_u + 1$ such that it is only possible to erase each number in the string if you also erase one of its neighbours. ... | Baltic Way | Baltic Way 2019 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | A | |
0k16 | Problem:
David and Evan are playing a game. Evan thinks of a positive integer $N$ between 1 and 59, inclusive, and David tries to guess it. Each time David makes a guess, Evan will tell him whether the guess is greater than, equal to, or less than $N$. David wants to devise a strategy that will guarantee that he knows... | [
"Solution:\n\nWe can represent each strategy as a binary tree labeled with the integers from 1 to 59, where David starts at the root and moves to the right child if he is too low and to the left child if he is too high. Our tree must have at most 6 layers as David must guess at most 5 times. Once David has been tol... | United States | HMMT November 2018 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 36440 |
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