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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0axv | Problem:
Three of the roots of $x^{4} + a x^{2} + b x + c = 0$ are $3$, $-2$, and $5$. Find the value of $a + b + c$. | [
"Solution:\n\nThe coefficient of $x^{3}$ is $0$, so by Vieta's formula, the sum of the roots must be $0$, and the fourth root must be $-6$. Therefore, the polynomial factors as $(x-3)(x+2)(x-5)(x+6)$. Setting $x=1$ gives $1 + a + b + c = (1-3)(1+2)(1-5)(1+6) = 168$, so $a + b + c = 167$."
] | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 167 | |
04bo | A triangle with an area of $1.5 \text{ cm}^2$ is inscribed in a circle of radius $1.25$ cm, with one side of the triangle as the diameter of the circle. What are the lengths of the sides of the triangle? | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | final answer only | 1.5 cm, 2 cm, 2.5 cm | |
0784 | In triangle $ABC$, let $D$ be the foot of the perpendicular from $A$ to line $BC$. Point $K$ lies inside triangle $ABC$ such that $\angle KAB = \angle KCA$ and $\angle KAC = \angle KBA$. The line through $K$ perpendicular to line $DK$ meets the circle with diameter $BC$ at points $X$ and $Y$. Prove that $AX \cdot DY = ... | [
"Assume $AB \\neq AC$ for all solutions as in the case that $AB = AC$, $X, Y$ are symmetric about $AD$ and the result is immediate.\nLet $M$ be the midpoint of $BC$. Let $\\omega = (BC)$ and $\\Omega = (ABC)$.\nWe prove some preliminary results here which will be freely used across different proofs.\nLet $T$ be the... | India | IMO TST Day 2 | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Coaxal circl... | null | proof only | null | |
0i5e | Problem:
The expression $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$. Find the value of
$$
\left\lfloor\frac{2002!}{2001!+2000!+1999!+\cdots+1!}\right\rfloor .
$$ | [
"Solution:\n\n2000\n\nWe break up $2002! = 2002 \\cdot 2001!$ as\n$$\n\\begin{gathered}\n2000(2001!)+2 \\cdot 2001(2000!)=2000(2001!)+2000(2000!)+2002 \\cdot 2000(1999!) \\\\\n>2000(2001!+2000!+1999!+\\cdots+1!)\n\\end{gathered}\n$$\nOn the other hand,\n$$\n2001(2001!+2000!+\\cdots+1!)>2001(2001!+2000!)=2001(2001!)... | United States | Harvard-MIT Math Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 2000 | |
0hog | Problem:
Two points $A$ and $C$ are marked on a circle; assume the tangents to the circle at $A$ and $C$ meet at $P$. Let $B$ be another point on the circle, and suppose $P B$ meets the circle again at $D$. Show that $A B \cdot C D = B C \cdot D A$. | [
"Solution:\n\nSince $P A$ is tangent to the circle and $\\angle A B D$ is the inscribed angle opposite $A D$, $\\angle P A D = \\angle A B P$. Since $\\angle A P B$ is shared, it follows from AA similarity that $\\triangle P A D \\sim \\triangle P B A$. Thus we get\n$$\n\\frac{A B}{A D} = \\frac{P B}{P A}\n$$\nSimi... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fv2 | Problem:
Finde alle Funktionen $f: \mathbb{R} \rightarrow \mathbb{R}$, sodass für alle $x, y \in \mathbb{R}$ gilt
$$
y f(2 x)-x f(2 y)=8 x y\left(x^{2}-y^{2}\right)
$$ | [
"Solution:\n\nSetze $y=1/2$ und $x=z/2$, dann folgt\n$$\n\\frac{f(z)}{2}-\\frac{f(1) z}{2}=\\frac{z^{3}-z}{2}\n$$\nMit $c=f(1)-1$ gilt also $f(z)=z^{3}+c z$ für alle $z \\in \\mathbb{R}$. Einsetzen zeigt, dass diese Funktionen für alle $c \\in \\mathbb{R}$ Lösungen der Gleichung sind.\n\n\nSolution 2:\n\nMit $y=0$ ... | Switzerland | SMO Finalrunde | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = x^3 + c x for all real constants c | |
09xc | Determine all triples $(x, y, z)$ of real numbers satisfying:
$$
\begin{aligned}
x^2 - yz &= |y - z| + 1, \\
y^2 - zx &= |z - x| + 1, \\
z^2 - xy &= |x - y| + 1.
\end{aligned}
$$ | [
"The system of equations is symmetric: if you swap $x$ and $y$, for example, then the third equation stays the same and the first two equations are swapped. Hence, we can assume without loss of generality that $x \\ge y \\ge z$. Then the system of equations becomes:\n$$\n\\begin{aligned}\nx^2 - yz &= y - z + 1, \\\... | Netherlands | BxMO Team Selection Test | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | All permutations of (4/3, 4/3, -5/3) and (5/3, -4/3, -4/3). | |
0adi | Solve the equation $\log_{3x+4}(2x+1)^2 + \log_{2x+1}(6x^2 + 11x + 4) = 4$. | [
"By definition $2x+1>0$, $3x+4>0$, $6x^2+11x+4>0$ and $2x+1, 3x+4 \\neq 1$. Therefore $x \\in (-\\frac{1}{2}, \\infty)$. Because $6x^2+11x+4 = (2x+1)(3x+4)$ the equation is equivalent to\n\n$$\n2 \\log_{3x+4}(2x+1) + \\log_{2x+1}((2x+1)(3x+4)) = 4 \\Leftrightarrow\n$$\n$$\n2 \\log_{3x+4}(2x+1) + \\log_{2x+1}(2x+1) ... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 3/4 | |
0fyx | Problem:
Finde natürliche Zahlen $a, b, c$, so dass die Quersumme von $a+b, b+c$ und $c+a$ jeweils kleiner als 5 ist, die Quersumme von $a+b+c$ aber grösser als 50. | [
"Solution:\nDie Idee ist, zuerst die paarweisen Summen $u = a + b$, $v = b + c$ und $w = c + a$ festzulegen. Damit diese Summen wirklich zu natürlichen Zahlen $a, b, c$ gehören, müssen sie die Seitenlängen eines nichtdegenerierten Dreiecks mit geradem Umfang sein. Wir wählen diese Summen so, dass deren Dezimaldarst... | Switzerland | IMO Selektion | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a = 5554445555, b = 5555554445, c = 4445555555 | |
0akk | Find all pairs $(p,q)$, $p,q \in \mathbb{N}$ such that
$$
(p+1)^{p-1} + (p-1)^{p+1} = q^q.
$$ | [
"First, we have\n$$\n(p+1)^{p-1} + (p-1)^{p+1} \\ge (p+1)^{p-1} \\ge (p-1)^{p-1} \\quad (1)\n$$\n$$\n(p+1)^{p-1} + (p-1)^{p+1} < (p+1)^{p+1} + (p+1)^{p-1} = 2(p+1)^{p+1} < (p+2)^{p+2} \\quad (2)\n$$\nFrom (1) and (2) follows\n$$\n(p-1)^{p-1} \\le q^q < (p+2)^{p+2}.\n$$\n1) Let\n$$\nq = p-1, \\quad (p+1)^{p-1} + (p-... | North Macedonia | Junior Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | [(1,1), (2,2)] | |
0est | In triangle $ABC$, $AB = AC$, and $D$ is on $BC$. A point $E$ is chosen on $AC$, and a point $F$ is chosen on $AB$, such that $DE = DC$ and $DF = DB$. It is given that $\frac{DC}{BD} = 2$ and $\frac{AF}{AE} = 5$. Determine the value of $\frac{AB}{BC}$. | [
"Note that triangles $ABC$ and $DFB$ are both isosceles, and they share the angle at $B$. Therefore, they must be similar. Likewise, triangle $DCE$ is similar to these two. Let $BF = x$ and $BD = FD = y$. Then we have $DC = 2BD = 2y$, and since $DFB$ and $DCE$ are similar, it also follows that $CE = 2x$. Next, we l... | South Africa | The South African Mathematical Olympiad Third Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(3)/2 | |
0jsa | Problem:
Define $\phi^{!}(n)$ as the product of all positive integers less than or equal to $n$ and relatively prime to $n$. Compute the remainder when
$$
\sum_{\substack{2 \leq n \leq 50 \\ \operatorname{gcd}(n, 50)=1}} \phi^{!}(n)
$$
is divided by $50$. | [
"Solution:\nFirst, $\\phi^{!}(n)$ is even for all odd $n$, so it vanishes modulo $2$.\n\nTo compute the remainder modulo $25$, we first evaluate $\\phi^{!}(3) + \\phi^{!}(7) + \\phi^{!}(9) \\equiv 2 + 5 \\cdot 4 + 5 \\cdot 3 \\equiv 12$ $ (\\bmod 25) $. Now, for $n \\geq 11$ the contribution modulo $25$ vanishes as... | United States | HMMT February | [
"Number Theory > Modular Arithmetic",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | final answer only | 12 | |
09r1 | Problem:
Bepaal alle positieve gehele getallen die niet geschreven kunnen worden als $\frac{a}{b}+\frac{a+1}{b+1}$ met $a, b$ positief en geheel. | [
"Solution:\nEr geldt\n$$\n\\frac{a}{b}+\\frac{a+1}{b+1}=\\frac{2 a b+a+b}{b(b+1)}\n$$\nNeem aan dat dit gelijk is aan een geheel getal $n$. Dan geldt $b \\mid 2 a b+a+b$ en $b+1 \\mid 2 a b+a+b$. Uit het eerste volgt $b \\mid a$ en dus ook $b \\mid a-b$. Uit het tweede volgt $b+1 \\mid (2 a b+a+b)-(b+1) \\cdot 2 a=... | Netherlands | Toets 6 juni 2012 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | The non-representable positive integers are 1 and all numbers of the form 2^m + 2 with m ≥ 0. | |
0au7 | Problem:
In how many ways can Alex, Billy, and Charles split $7$ identical marbles among themselves so that no two have the same number of marbles? It is possible for someone not to get any marbles. | [
"Solution:\n\nLet the numbers of marbles that Alex, Billy, and Charles get be $a$, $b$, and $c$, respectively, with $a$, $b$, $c \\geq 0$ and $a + b + c = 7$. We require that $a$, $b$, and $c$ are all distinct (no two are equal).\n\nSince the marbles are identical and the people are distinguishable, we are counting... | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 24 | |
0egg | Problem:
Za realni števili $x$ in $y$, kjer $x \neq 0$, $y \notin \{-2, 0, 2\}$ in $x + y \neq 0$, poenostavi izraz:
$$
\frac{x y^{2018} + 2 x y^{2017}}{y^{2016} - 4 y^{2014}} \cdot \left(\left(\frac{x^{2}}{y^{3}} + x^{-1}\right) : \left(x y^{-2} - \frac{1}{y} + x^{-1}\right)\right) : \frac{(x-y)^{2} + 4 x y}{1 + \fra... | [
"Solution:\n\nŠtevec prvega ulomka $x y^{2018} + 2 x y^{2017}$ preoblikujemo v $x y^{2017}(y + 2)$, imenovalec $y^{2016} - 4 y^{2014}$ pa preoblikujemo v $y^{2014}(y - 2)(y + 2)$. Prvi ulomek okrajšamo in dobimo $\\frac{x y^{3}}{y - 2}$.\n\nIzraz $\\frac{x^{2}}{y^{3}} + x^{-1}$ preoblikujemo v $\\frac{x^{3} + y^{3}... | Slovenia | 18. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Odbirno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 2y/(y - 2) | |
049z | Let $a$, $b$, $c$ be real numbers and let $a \neq 0$. Prove that at least one of the equations
$$
a \sin x + b \cos x + c = 0,
$$
$$
a \tgg y + b \ctg y + 2c = 0
$$
has real solutions. | [
"Let us consider the first equation:\n$$\na \\sin x + b \\cos x + c = 0.\n$$\nThis can be rewritten as:\n$$\na \\sin x + b \\cos x = -c.\n$$\nLet $R = \\sqrt{a^2 + b^2}$ and let $\\varphi$ be such that $a = R \\sin \\varphi$, $b = R \\cos \\varphi$.\nThen:\n$$\na \\sin x + b \\cos x = R \\sin x \\sin \\varphi + R \... | Croatia | CroatianCompetitions2011 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0ads | Дали постои неконстантна низа природни броеви $a_1, a_2, ..., a_n, ...$, таква што за секој природен број $k \ge 2$ е исполнето равенството
$$
a_k = \frac{2a_{k-1}a_{k+1}}{a_{k-1} + a_{k+1}}?
$$ | [
"Нека претпоставиме дека таква низа природни броеви постои. Тогаш за низата од реципрочни вредности $b_n = \\frac{1}{a_n}, n \\in N$, имаме\n$$\nb_n = \\frac{1}{a_n} = \\frac{1}{\\frac{2a_{n-1}a_{n+1}}{a_{n-1} + a_{n+1}}} = \\frac{1}{2\\left(\\frac{1}{a_{n-1}} + \\frac{1}{a_{n+1}}\\right)} = \\frac{b_{n-1} + b_{n+1... | North Macedonia | Републички натпревар по математика за средно образование | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | Macedonian, English | proof and answer | No; no such non-constant sequence of natural numbers exists. | |
07w7 | Suppose $ABCD$ is a cyclic quadrilateral, with side lengths $a = AB$, $b = BC$, $c = CD$, $d = DA$. Prove that its area ($ABCD$) doesn't exceed the following expression
$$
\frac{ab + ac + ad + bc + bd + cd}{6},
$$
with equality iff the quadrilateral is a square. | [
"$$\n(ABCD) = \\sqrt{(\\sigma - a)(\\sigma - b)(\\sigma - c)(\\sigma - d)},\n$$\nwhence pairing the factors in the expression $\\sqrt{(\\sigma - a)(\\sigma - b)(\\sigma - c)(\\sigma - d)}$ in three different ways, and noting that by AM-GM, e.g.,\n$$\n\\sqrt{(\\sigma - a)(\\sigma - b)} \\le \\frac{2\\sigma - a - b}{... | Ireland | IRL_ABooklet_2023 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
00by | Given is a convex quadrilateral $ABCD$ with $AB = BD = a$ and $CD = DA = b$. Let $P$ in side $AB$ such that $DP$ is bisector of $\vec{ADB}$ and $Q$ in side $BC$ such that $DQ$ is bisector of $\vec{CDB}$. Find the circumradius of triangle $DPQ$. | [
"Let the parallel to $AD$ through $P$ intersect $BD$ at $O$. We show that $O$ is the circumcenter of $DPQ$.\n\nOne has $\\vec{ADP} = \\vec{BDP}$ ($DP$ bisects $\\vec{ADB}$) and $\\vec{ADP} = \\vec{OPD}$ (as $PO \\parallel AD$). Hence $\\vec{ODP} = \\vec{OPD}$ and so $OP = OD$. Also $DO = AP$, $BO = BP$ as triangle ... | Argentina | Argentina_2018 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | ab/(a+b) | |
03w7 | As shown in the Fig. 1, $M$, $N$ are the midpoints of arcs $\widehat{BC}$, $\widehat{AC}$ respectively, which are on the circumscribed circle of an acute triangle $\triangle ABC$ ($\angle A < \angle B$). Through point $C$ draw $PC \parallel MN$, intercepting the circle $\Gamma$ at point $P$. $I$ is the inner center of ... | [
"(1) As shown in Fig. 2, join $NI$, $MI$. Since $PC \\parallel MN$ and $P, C, M, N$ are concyclic, $PCMN$ is an isosceles trapezoid. Therefore, $NP = MC$, $PM = NC$.\nJoin $AM$, $CI$. Then $AM$ intercepts $CI$ at $I$. We have\n$$\n\\begin{align*}\n\\angle MIC &= \\angle MAC + \\angle ACI \\\\\n&= \\angle MCB + \\an... | China | China Mathematical Competition (Complementary Test) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry... | English | proof only | null | |
0e2b | Let $K_1$ and $K_2$ be the circles centered at $O_1$ and $O_2$, respectively, meeting at the points $A$ and $B$. Let $p$ be the line through the point $A$ meeting the circles $K_1$ and $K_2$ again at $C_1$ and $C_2$. Assume that $A$ lies between $C_1$ and $C_2$. Denote the intersection of the lines $C_1O_1$ and $C_2O_2... | [
"Write $\\angle ABC_1 = \\alpha$ and $\\angle C_2BA = \\beta$. Then $\\angle C_2BC_1 = \\alpha + \\beta$. The points $C_1, C_2, B$ and $D$ are concyclic if and only if $\\angle C_2BC_1 = \\angle C_2DC_1$. Let us show that $\\angle C_2DC_1 = \\alpha + \\beta$.\n\nThe central angle equals twice the inscribed angle, s... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0aho | An acute-angled triangle $ABC$ is given. The points $A_1$, $A_2$, $B_1$, $B_2$, $C_1$ and $C_2$ lie on its sides in such a way that $\overline{AA_1} = \overline{A_1A_2} = \overline{A_2B} = \frac{1}{3}\overline{AB}$, $\overline{BB_1} = \overline{B_1B_2} = \overline{B_2C} = \frac{1}{3}\overline{BC}$ and $\overline{CC_1} ... | [
"Let us denote the intersections of $a_B$, $b_C$ and $c_A$ with $b_A$, $c_B$ and $a_C$ respectively by $A'$, $B'$ and $C'$. The triangle $AA_1C_2$ is similar to $ABC$, since they have a common angle and their sides are at a ratio $1:3$. Let $O_A$ be the center of the circumscribed circle around the triangle $AA_1C_... | North Macedonia | Team Selection Test for IMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0f9f | Problem:
A graph has $n$ points and $n(n-1)/2$ edges. Each edge is colored with one of $k$ colors so that there are no closed monochrome paths. What is the largest possible value of $n$ (given $k$)? | [] | Soviet Union | 24th ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof and answer | 2k | |
0efx | Problem:
Poišči vsa realna števila $x$, ki rešijo enačbo
$$
\log_{2}\left(\log_{2}(7x-6)\right)+1=\log_{2}\left(\log_{2}(3x-2)\right)+\log_{2} 3
$$ | [
"Solution:\nZ upoštevanjem zveze $1=\\log_{2} 2$ in formule za vsoto logaritmov $\\log_{2} a+\\log_{2} b=\\log_{2}(a \\cdot b)$ enačbo najprej preoblikujemo v\n$$\n\\log_{2}\\left(2 \\log_{2}(7x-6)\\right)=\\log_{2}\\left(3 \\log_{2}(3x-2)\\right)\n$$\nNato uporabimo še zvezo $a \\log_{2} b=\\log_{2} b^{a}$, da dob... | Slovenia | Slovenian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 2 | |
072s | Let $p$ be a prime and $X$ be a finite set containing at least $p$ elements. A collection of mutually disjoint $p$-element subsets of $X$ is called a $p$-family. The empty collection itself is regarded as a $p$-family. Let $A$ (respectively, $B$) denote the number of $p$-families having an even (respectively, odd) numb... | [
"We observe that\n$$\nA = \\sum_{r \\ge 0,\\ r\\ \\text{even}} \\binom{n}{rp} \\frac{(rp)!}{(p!)^r r!}\n$$\nand\n$$\nB = \\sum_{r \\ge 0,\\ r\\ \\text{odd}} \\binom{n}{rp} \\frac{(rp)!}{(p!)^r r!}\n$$\nThe empty collection corresponds to $r=0$ in the summation for $A$. If $u_r = \\frac{(rp)!}{(p!)^r r!}$, $r \\ge 0... | India | Indija TS 2006 | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
05aq | All integers from $1$ to $12$ are written on the edges of a cube so that every edge has exactly one integer. Two robot ants stand in the same vertex of the cube and wish to arrive at the vertex that is furthest away from them. Each ant picks a path consisting of exactly three edges of the cube. They multiply the intege... | [
"Since $100$ is divisible by $5^2$, but none of the numbers on the edges of the cube are divisible by $5^2$, the path of the first ant has to contain two numbers that are divisible by $5$. The only two such numbers are $5$ and $10$. Their product is divisible by $2$, but not by $2^2$. Since $100 = 2^2 \\cdot 5^2$, ... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | First ant: 300; Second ant: 110 | |
0e8i | Problem:
Reši enačbo
$$
3 + \frac{11}{x-1} + \frac{10}{1-2x} - \frac{2}{4x^2 - 4x + 1} = 0
$$
Rešitve zapiši v obliki okrajšanih ulomkov z racionaliziranimi imenovalci. | [] | Slovenia | Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | x = 0, x = sqrt(3)/6, x = -sqrt(3)/6 | |
06vc | On a flat plane in Camelot, King Arthur builds a labyrinth $\mathfrak{L}$ consisting of $n$ walls, each of which is an infinite straight line. No two walls are parallel, and no three walls have a common point. Merlin then paints one side of each wall entirely red and the other side entirely blue.
At the intersection of... | [
"First we show by induction that the $n$ walls divide the plane into $\\binom{n+1}{2}+1$ regions. The claim is true for $n=0$ as, when there are no walls, the plane forms a single region. When placing the $n^{\\text {th }}$ wall, it intersects each of the $n-1$ other walls exactly once and hence splits each of $n$ ... | IMO | IMO 2019 Shortlisted Problems | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | n+1 | |
0hvc | Problem:
If triangle $ABC$ has perimeter $2$, prove that not all its altitudes can exceed $1 / \sqrt{3}$ in length. | [
"Solution:\nLet $a, b, c$ be the side lengths of the triangle, where $a$ is the longest side, so $a \\geq 2 / 3$, and let $h_{a}$ be the length of the altitude to that side. The semiperimeter is $1$, so by Heron's formula we get\n$$\n\\frac{1}{2} a h_{a} = \\sqrt{(1-a)(1-b)(1-c)} \\Longrightarrow h_{a} \\leq 3 \\sq... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
037b | Problem:
The tangent lines to the circumcircle $k$ of an isosceles $\triangle ABC$, $AC = BC$, at the points $B$ and $C$ meet at point $X$. If $AX$ meets $k$ at point $Y$, find the ratio $\frac{AY}{BY}$. | [
"Solution:\nIf $\\angle BAY = \\alpha$ and $\\angle ABY = \\beta$ then $\\angle BYX = \\beta + \\alpha$. Furthermore $\\angle ACB = \\angle AYB = 180^\\circ - \\alpha - \\beta$, implying $\\angle BAC = \\angle ABC = \\frac{\\beta + \\alpha}{2}$. Thus $\\angle AYC = \\frac{\\beta + \\alpha}{2}$ and $\\angle YCX = \\... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 2 | |
0ck8 | Alexia has several marbles and her friend, Cristina, has none. Each day of one week, starting on Monday, Alexia gave Cristina some of her marbles, so that each day Alexia gave more marbles than the day before. On Monday Alexia gave five times less marbles than on Friday, on Tuesday she gave six times less marbles than ... | [
"Denote $a_1, a_2, \\dots, a_7$ the number of marbles given by Alexia from Monday until Sunday, where $1 \\le a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < a_7$ and $a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 = 72$. Then $a_5 = 5a_1$, $a_6 = 6a_2$, $a_7 = 7a_3$, hence $6a_1 + 7a_2 + 8a_3 + a_4 = 72$ (1).\n\nIf $a_1 \\ge 3$, the... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 7 | |
0br0 | Find all finite groups which have proper subgroups and all of these have order $2$ or $3$. | [] | Romania | 67th NMO Shortlisted Problems | [
"Algebra > Abstract Algebra > Group Theory"
] | English | proof and answer | C4, V4, C6, S3, C9, C3×C3 | |
0h9m | For positive integers, prove the inequality:
$$
\frac{y}{2x + y} + \frac{z}{2y + z} + \frac{x}{2z + x} \geq 1.
$$ | [
"We prove it by using Cauchy–Schwarz inequality:\n$$\n\\left(\\frac{y}{2x+y} + \\frac{z}{2y+z} + \\frac{x}{2z+x}\\right) (x+y+z)^2 = \\left(\\frac{y}{2x+y} + \\frac{z}{2y+z} + \\frac{x}{2z+x}\\right) \\left(y(2x+y) + z(2y+z) + x(2z+x)\\right) \n\\geq \\left(\\sqrt{\\frac{y}{2x+y}} \\cdot \\sqrt{y(2x+y)} + \\sqrt{\\... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0iwd | Problem:
Regular hexagon $A B C D E F$ has side length $2$. A laser beam is fired inside the hexagon from point $A$ and hits $\overline{B C}$ at point $G$. The laser then reflects off $\overline{B C}$ and hits the midpoint of $\overline{D E}$. Find $B G$. | [
"Solution:\n\nAnswer: $\\frac{2}{5}$ Look at the diagram below, in which points $J, K, M, T$, and $X$ have been defined. $M$ is the midpoint of $\\overline{D E}$, $B C J K$ is a rhombus with $J$ lying on the extension of $\\overline{C D}$, $T$ is the intersection of lines $\\overline{C D}$ and $\\overline{G M}$ whe... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | 2/5 | |
00cj | Sean $\Gamma$ una circunferencia de centro $S$ y radio $r$ y $A$ un punto exterior a la circunferencia. Sea $BC$ un diámetro de $\Gamma$ tal que $B$ no pertenece a la recta $AS$, y consideramos el punto $O$ en el que se cortan las mediatrices del triángulo $ABC$, o sea, el circuncentro del $ABC$.
Determinar todas las p... | [] | Argentina | Nacional OMA 2019 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | Spanish | proof and answer | The locus of O is the fixed line perpendicular to AS whose signed distance from S equals (SA^2 − r^2)/(2·SA). Equivalently, it is the set of points O satisfying SA · SO = (SA^2 − r^2)/2. | |
0jzx | Problem:
Determine the largest real number $c$ such that for any $2017$ real numbers $x_{1}, x_{2}, \ldots, x_{2017}$, the inequality
$$
\sum_{i=1}^{2016} x_{i}\left(x_{i}+x_{i+1}\right) \geq c \cdot x_{2017}^{2}
$$
holds. | [
"Solution:\nLet $n=2016$. Define a sequence of real numbers $\\{p_{k}\\}$ by $p_{1}=0$, and for all $k \\geq 1$,\n$$\np_{k+1}=\\frac{1}{4\\left(1-p_{k}\\right)}\n$$\nNote that, for every $i \\geq 1$,\n$$\n\\left(1-p_{i}\\right) \\cdot x_{i}^{2}+x_{i} x_{i+1}+p_{i+1} x_{i+1}^{2}=\\left(\\frac{x_{i}}{2 \\sqrt{p_{i+1}... | United States | February 2017 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | -1008/2017 | |
0g61 | 設 $O$ 為銳角三角形 $ABC$ 之外心, 且 $AO$ 與 $BC$ 交於 $D$ 點。由 $D$ 點做兩射線分別垂直 $AB$ 與 $AC$, 並分別交 $AB$, $AC$ 於 $E$, $F$ 與交 $ABC$ 的外接圓於 $K$, $L$。試證: 若 $K$, $E$, $F$, $L$ 四點共圓, 則 $AB = AC$. | [
"因 $\\angle EKL = \\angle DFE = \\angle DAE$, $KL \\perp AD$。故 $AO$ 平分 $KL$, 從而 $K$, $L$ 對 $AO$ 對稱。故 $\\angle ADK = \\angle ADL$, $\\angle BAD = \\angle CAD$。故 $AB = AC$.\n\n"
] | Taiwan | 二〇一一數學奧林匹亞競賽第一階段選訓營,模擬競賽(二) | [
"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 | |
01qj | Find all sequences $(a_n)$ of positive integers satisfying the equality $a_n = a_{n-1} + a_{n+1}$
a) for all $n \ge 2$;
b) for all $n \ge 3$. | [
"a) there are no such sequences;\n\nb) there exists a unique sequence: $a_1 = a_2 = 1$, $a_i = 2$ for all $i \\ge 3$."
] | Belarus | Selection and Training Session | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a) No such sequences exist.
b) The unique sequence is a1 = 1, a2 = 1, and ai = 2 for all i at least 3. | |
0fdi | Problem:
Dado el polinomio $P(X) = X^{4} + \square X^{3} + \square X^{2} + \square X + \square$, en el que cada cuadrado representa un hueco donde se colocará un coeficiente, se plantea el siguiente juego entre dos jugadores: Alternativamente, el primer y el segundo jugador eligen un hueco vacío y colocan en él un ent... | [
"Solution:\n\nNota: El enunciado presenta cierta ambigüedad: hay que distinguir si \"dos raíces enteras\" significa que éstas son distintas, o por el contrario pueden ser raíces dobles.\nEl caso de dos raíces distintas es sencillo; basta con la primera jugada para el primer jugador. En el caso de raíces dobles tien... | Spain | null | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
045v | (1) Prove that on the complex plane, the convex hull of all complex roots of the equation
$$
z^{20} + 63z + 22 = 0
$$
is larger than $\pi$.
(2) Let $n$ be a positive integer, and $1 \le k_1 < k_2 < \dots < k_n$ be $n$ odd integers. Prove that for any $n$ complex numbers $a_1, a_2, \dots, a_n$ with sum 1 and any comple... | [
"(1) We first prove the following.\n\n**Lemma:** [Gauss-Lucas Theorem] All zeros of the polynomial $f'(z)$ lie in the closure formed by zeros of $f(z)$.\n\n**Proof of Lemma:** By the fundamental theorem of algebra, we know that $f(z)$ admits a decomposition:\n$$\nf(z) = A(z - z_1)^{\\alpha_1} (z - z_2)^{\\alpha_2} ... | China | 2022 China Team Selection Test for IMO | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | English | proof only | null | |
042z | Suppose all the edges of regular triangular pyramid $P$-$ABC$ have length $1$ and $L$, $M$, $N$ are the midpoints of edges $PA$, $PB$ and $PC$, respectively. The area of the cross section of the circumscribed sphere of this regular triangular pyramid intercepted by plane $LMN$ is ______. | [
"The given conditions show that plane $LMN$ is parallel to plane $ABC$ and the ratio of the distances from point $P$ to planes $LMN$ and $ABC$ is $1 : 2$. Let $H$ be the centroid of face $ABC$ in regular triangular pyramid $P$-$ABC$ and $PH$ intersects plane $LMN$ at point $K$. Then $PH \\perp ABC$ and $PK \\perp L... | China | China Mathematical Competition | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | π/3 | |
0h6j | Andriiko has unlimited amount of chips painted in 6 colours. He wants to put some of the chips in one row so that for any two different colours in that row there are two adjacent chips of those colours. What is the minimum amount of chips he can put in a row?
**Answer:** 18. | [
"The chip of each colour has to be adjacent at least 5 times, to each of other colour. But any chip has maximum 2 neighbours, thus there is at least 3 chips of each colour in a row. Here is an example that 18 is a correct answer: 123456246325164135."
] | Ukraine | UkraineMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 18 | |
00ps | Let $p$ be a prime number. Determine all triples $(a, b, c)$ of positive integers such that $a + b + c < 2p\sqrt{p}$ and
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{p}.
$$ | [
"Given equation is equivalent to $abc = p(ab + bc + ca)$, and therefore $p \\mid abc$. So, at least one of $a, b, c$, w.l.o.g. $a$, is divisible by $p$ and let $a = pa_1$.\nFirst, let us suppose that $p \\nmid bc$. Then\n$$\n\\frac{b + c}{bc} = \\frac{1}{b} + \\frac{1}{c} = \\frac{1}{p} - \\frac{1}{a} = \\frac{a_1 ... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | All solutions are permutations of the following triples, depending on the prime:
- For p ≥ 31: (3p, 3p, 3p), (2p, 4p, 4p), (2p, 3p, 6p)
- For p = 29: (3p, 3p, 3p), (2p, 4p, 4p)
- For p = 23: (3p, 3p, 3p)
There are no solutions for smaller primes. | |
0gg9 | 設點 $A,B,C,D$ 為圓 $\omega$ 上的四點，$ABCD$ 形成一個凸四邊形。設 $AB$ 與 $CD$ 交於點 $E$，且 $A$ 位於 $B,E$ 之間；又設 $BD$ 與 $AC$ 交於點 $F$。令點 $X$ 為圓 $\omega$ 上異於 $D$ 的一點使得 $DX$ 平行於 $EF$。設點 $Y$ 為點 $D$ 關於直線 $EF$ 的對稱點，且知 $Y$ 位於圓 $\omega$ 內。證明 $A,X,Y$ 三點共線。 | [] | Taiwan | 2022 數學奧林匹亞競賽第五階段培訓營, 國際競賽實作(二) | [
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Chinese; English | proof only | null | |
0l5r | Problem:
For any integer $x$, let
$$
f(x) = 100! \left(1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots + \frac{x^{100}}{100!}\right).
$$
A positive integer $a$ is chosen such that $f(a) - 20$ is divisible by $101^{2}$. Compute the remainder when $f(a + 101)$ is divided by $101^{2}$. | [
"Solution:\n\nBy the binomial theorem,\n$$\n(a + 101)^{n} \\equiv a^{n} + \\binom{n}{1} a^{n - 1}101 = a^{n} + 101n a^{n - 1} \\pmod {101^{2}}.\n$$\nUsing this gives (all congruences are modulo $101^{2}$)\n$$\n\\begin{align*}\nf(a + 101) &= 100!\\sum_{n = 0}^{100}\\frac{(a + 101)^{n}}{n!} \\\\\n&\\equiv 100!\\sum_{... | United States | HMMT February | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1939 | |
04kx | A city has $M$ streets and $N$ squares, where $M$ and $N$ are positive integers such that $M > N$. Each street connects two squares and does not go through any other squares.
The citizens wish to change the appearance of the city. This year, each street will be coloured for the first time, in either red or blue. It is ... | [
"Let us call one arrangement of colours on the streets a *colouring*, and one selection of a square and the change of colours of all streets leading to that square a *transformation*.\nNote that if we can get from one colouring to another by a sequence of transformations, then we can also get from the second colour... | Croatia | Mathematical competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0220 | Problem:
No jogo - Aldo, Bernardo e Carlos jogam baralho. No início, a quantia em dinheiro que eles tinham estava na proporção $7:6:5$. No final do jogo, a proporção era $6:5:4$. Um dos jogadores ganhou 1200 reais. Qual a quantidade de dinheiro com que ficou cada jogador, no final da partida? | [
"Solution:\n\nSeja $T$ a quantidade total de dinheiro no jogo. Assim, no início, os jogadores possuíam:\n$$\n\\begin{array}{ll}\n\\text{Aldo:} & \\frac{7}{18} T \\\\\n\\text{Bernardo:} & \\frac{6}{18} T \\\\\n\\text{Carlos:} & \\frac{5}{18} T\n\\end{array}\n$$\nNo final eles possuíam:\n$$\n\\begin{array}{ll}\n\\tex... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | Aldo: 43200; Bernardo: 36000; Carlos: 28800 | |
09rl | Problem:
Vind alle functies $f: \mathbb{R} \rightarrow \mathbb{R}$ die voldoen aan
$$
f(x+y f(x))=f(x f(y))-x+f(y+f(x))
$$
voor alle $x, y \in \mathbb{R}$. | [
"Solution:\n\nVul in $x=y=0$ : dat geeft\n$$\nf(0)=f(0)-0+f(f(0))\n$$\ndus $f(f(0))=0$.\n\nVul in $x=y=1$ : dat geeft\n$$\nf(1+f(1))=f(f(1))-1+f(1+f(1))\n$$\ndus $f(f(1))=1$.\n\nVul nu in $x=1, y=0$ :\n$$\nf(1)=f(f(0))-1+f(f(1))\n$$\nWe weten dat $f(f(0))=0$ en $f(f(1))=1$, dus we vinden nu $f(1)=0$.\nOmdat $f(f(1)... | Netherlands | Selectietoets | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 1 - x | |
04vh | Given an acute-angled scalene triangle $ABC$. The angle bisector of the angle $BAC$ and the perpendicular bisectors of the sides $AB$, $AC$ define a triangle. Prove that its orthocenter lies on the median from the vertex $A$.
(Josef Tkadlec) | [
"Let us denote $M$ the midpoint of $AB$, $N$ the midpoint of $AC$, $K$ and $L$ the intersections of the angle $CAB$ bisector with perpendicular bisectors of $AB$ and $AC$, respectively. The intersection of perpendicular bisectors of $AB$ and $AC$ is denoted by $O$. The triangle $KLO$ is therefore the triangle from ... | Czech Republic | 72nd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distanc... | English | proof only | null | |
007a | Consider a semicircle with diameter $AB$, center $O$ and radius $r$. Let $C$ be the point on segment $AB$ such that $AC = \frac{2r}{3}$. Line $l$ is perpendicular to $AB$ at $C$ and $D$ is the common point of $l$ and the semicircle. Let $H$ be the foot of the perpendicular from $O$ to $AD$ and $E$ the intersection of l... | [
"a) Since $CO = \\frac{1}{3}r$ and $OD = r$, Pythagoras' theorem in triangles $COD$ and $ACD$ gives $CD = \\sqrt{OD^2 - OC^2} = \\frac{2\\sqrt{2}}{3}r$, $AD = \\sqrt{AC^2 + CD^2} = \\frac{2\\sqrt{3}}{3}r$.\n\nb) As $AD$ is a chord in the semicircle and $O$ its center, $OH \\perp AD$ implies that $H$ is the midpoint... | Argentina | Mathematical Olympiad Rioplatense | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | AD = (2√3/3)·r; ∠MHN = 90° | |
07l8 | In the triangle $ABC$ we have $|AB| < |AC|$. The bisectors of the angles at $B$ and $C$ meet $AC$ and $AB$ at $D$ and $E$ respectively. The lines $BD$ and $CE$ intersect at the incentre $I$ of $\triangle ABC$.
Prove that $\angle BAC = 60^\circ$ if and only if $|IE| = |ID|$. | [
"Let $\\angle BAC = 2\\alpha$, $\\angle CBA = 2\\beta$ and $\\angle ACB = 2\\gamma$. Assume first that $2\\alpha = \\angle BAC = 60^\\circ$. This implies $2\\beta + 2\\gamma = 120^\\circ$, i.e. $\\beta + \\gamma = 60^\\circ$. Hence, $\\angle DIE = \\angle BIC = 120^\\circ$. Therefore, $\\angle BAC + \\angle DIE = 1... | Ireland | Irska | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > ... | English | proof only | null | |
0b48 | Problem:
Find all ordered pairs $(a, b)$ of positive integers such that $a^{2}+b^{2}+25=15 a b$ and $a^{2}+a b+b^{2}$ is prime. | [
"Solution:\nThe only solutions are $(1,2)$ and $(2,1)$.\n\nFirst, note that $a \\neq b$; otherwise we have $13 a^{2}=25$ which is impossible. We can assume without loss of generality that $1 \\leq a < b$; we then take $(b, a)$ as well.\n\nIf $a=1$, we get $b^{2}+26=15 b$ which yields $b=2$ or $b=13$. For $b=2$ we g... | Philippines | 25th Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | [(1,2), (2,1)] | |
0jht | Problem:
Find the number of positive divisors $d$ of $15! = 15 \cdot 14 \cdots 2 \cdot 1$ such that $\operatorname{gcd}(d, 60) = 5$. | [
"Solution:\nSince $\\operatorname{gcd}(d, 60) = 5$, we know that $d = 5^{i} d^{\\prime}$ for some integer $i > 0$ and some integer $d^{\\prime}$ which is relatively prime to $60$. Consequently, $d^{\\prime}$ is a divisor of $(15!) / 5$; eliminating common factors with $60$ gives that $d^{\\prime}$ is a factor of $\... | United States | HMMT 2013 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | final answer only | 36 | |
0iic | Problem:
Find, with proof, all positive integer palindromes whose square is also a palindrome. | [
"Solution:\nLet $n := \\sum_{i=0}^{d} a_{i} \\cdot 10^{i}$ be a palindrome, where the $a_{i}$ are digits with $a_{i} = a_{d-i}$ and $a_{d} \\neq 0$. Then, if we let\n$$\nb_{k} := \\sum_{i+j=k} a_{i} a_{j}\n$$\nfor all $0 \\leq k \\leq 2d$, then\n$$\nn^{2} = \\sum_{k=0}^{2d} b_{k} \\cdot 10^{k}\n$$\n(this is not nec... | United States | Harvard-MIT Mathematics Tournament, Team Round A | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | Exactly the palindromes whose digits satisfy that the sum of their squares is less than ten; equivalently, a palindrome has a palindromic square if and only if the sum of squares of its decimal digits is less than ten. | |
0kfw | Let $P_1P_2 \cdots P_{100}$ be a cyclic 100-gon, and let $P_i = P_{i+100}$ for all $i$. Define $Q_i$ as the intersection of diagonals $\overline{P_{i-2}P_{i+1}}$ and $\overline{P_{i-1}P_{i+2}}$ for all integers $i$.
Suppose there exists a point $P$ satisfying $\overline{PP_i} \perp \overline{P_{i-1}P_{i+1}}$ for all in... | [
"We show two solutions.\n\n**Solution to proposed problem** We let $\\overline{PP_2}$ and $\\overline{P_1P_3}$ intersect (perpendicularly) at point $K_2$, and define $K_\\bullet$ cyclically.\n\n\n\n**Claim** — The points $K_\\bullet$ are concyclic say with circumcircle $\\gamma$.\n\n*Proof*... | United States | USA IMO TST | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > ... | null | proof only | null | |
08pi | Problem:
Consider a regular $2n$-gon $P, A_{1}A_{2}\ldots A_{2n}$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $SE$ contains no other points that lie on the sides of $P$ except $S$. We color t... | [
"Solution:\n\nFor $n=2$, the answer is $36$; for $n=3$, the answer is $30$ and for $n \\geq 4$, the answer is $6n$.\n\nLemma 1. Given a regular $2n$-gon in the plane and a sequence of $n$ consecutive sides $s_{1}, s_{2}, \\ldots, s_{n}$ there is an external point $Q$ in the plane, such that the color of each $s_{i}... | JBMO | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 36 if n = 2; 30 if n = 3; 6n if n ≥ 4 | |
08y2 | Let $a$ be a positive integer. Show the following assertion is valid for sufficiently large integer $n$:
From a grid of squares extending to $\pm\infty$ in both $x$- and $y$-directions, choose $n$ unit (i.e., $1 \times 1$) squares and color them black. Let $K$ be the number of $a \times a$ square blocks which contain e... | [
"For a pair of integers $u, v$ and a positive integer $m$ we write $u \\equiv v \\pmod m$ to mean that $u - v$ is divisible by $m$.\n\nLet us call an $a \\times a$ square box of the given grid a region. When $n \\ge a$, if we color black all of $n \\times 1$ square boxes stacked consecutively on a vertical strip, t... | Japan | Japan 2015 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | null | proof only | null | |
07r9 | $$
x^2(2 - x)^2 = 1 + 2(1 - x)^2.
$$ | [
"Let $y = 1 - x$. Then $x(2 - x) = (1 - y)(1 + y) = 1 - y^2$, and so an equivalent version of the equation is $(1 - y^2)^2 = 1 + 2y^2$, i.e., $y^4 = 4y^2$. Thus $y = 0, \\pm 2$ and $x = -1, 1, 3$.\n\nAlternatively, the given equation could be written as $x^2(2 - x)^2 - 1 = 2(1 - x)^2$.\nUsing the difference of two ... | Ireland | Ireland_2017 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | -1, 1, 3 | |
0l5j | Problem:
In a two-dimensional cave with a parallel floor and ceiling, two stalactites of lengths $16$ and $36$ hang perpendicularly from the ceiling, while two stalagmites of heights $25$ and $49$ grow perpendicularly from the ground. If the tips of these four structures form the vertices of a square in some order, co... | [
"Solution:\n\nNote that the difference in heights between the two stalactites does not equal the difference in heights between the stalagmites. This tells us that the two stalactites form a pair of opposite vertices of the square, and likewise for the stalagmites. As the midpoint of the pairs of structures must the... | United States | HMMT February 2025 | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 63 | |
008t | In the triangle $ABC$, the incircle is tangent to the sides $AB$ and $AC$ in $D$ and $E$, respectively. The line $DE$ meets the circumcircle in $P$ and $Q$, with $P$ on the minor arc $AB$ and $Q$ on the minor arc $AC$. It is known that $P$ is the midpoint of $AB$. Find $A$ and the ratio $\frac{PQ}{BC}$. | [] | Argentina | XXIX Olimpíada Matemática Argentina National Round | [
"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"
] | English | proof and answer | A = 60° and PQ/BC = 1 | |
0le9 | a) Let $a$, $b$ and $c$ be real numbers that satisfy $a^2 + b^2 + c^2 = 1$. Prove that
$$
|a - b| + |b - c| + |c - a| \le 2\sqrt{2}.
$$
b) Given 2019 real numbers $a_1, a_2, \dots, a_{2019}$ such that $a_1^2 + a_2^2 + \dots + a_{2019}^2 = 1$. Find the maximum value of
$$
S = |a_1 - a_2| + |a_2 - a_3| + \dots + |a_{201... | [
"Using Cauchy-Schwarz inequality, we get\n$$\nx_1 + x_2 + \\dots + x_n \\le |x_1| + |x_2| + \\dots + |x_n| \\\\\n\\le \\sqrt{n(x_1^2 + x_2^2 + \\dots + x_n^2)}.\n$$\n\na) Without loss of generality, we can assume $a \\le b \\le c$ and the inequality turns out to\n$$\n|a - b| + |b - c| + |c - a| = (a - b) + (b - c) ... | Vietnam | VMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | a) 2√2; b) 2√2018 | |
0jye | Problem:
There are 2017 jars in a row on a table, initially empty. Each day, a nice man picks ten consecutive jars and deposits one coin in each of the ten jars. Later, Kelvin the Frog comes back to see that $N$ of the jars all contain the same positive integer number of coins (i.e. there is an integer $d>0$ such that... | [
"Solution:\n\nAnswer: 2014\n\nLabel the jars $1,2, \\ldots, 2017$. I claim that the answer is 2014. To show this, we need both a construction and an upper bound.\n\nFor the construction, for $1 \\leq i \\leq 201$, put a coin in the jars $10i+1, 10i+2, \\ldots, 10i+10$. After this, each of the jars $1,2, \\ldots, 20... | United States | February 2017 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 2014 | |
0g28 | Problem:
Seien $A$, $B$, $C$ und $D$ vier Punkte, die in dieser Reihenfolge auf einem Kreis liegen. Nehme an, es gibt einen Punkt $K$ auf der Strecke $A B$, sodass $B D$ die Strecke $K C$ und $A C$ die Strecke $K D$ halbiert. Bestimme den kleinstmöglichen Wert, den $\left|\frac{A B}{C D}\right|$ annehmen kann. | [
"Solution:\n\nOn n'utilise que des distances non-orientées (i.e. positives).\nOn obtient une construction si $A B C D$ est un trapèze isocèle avec base $A B=2 C D$ et $K$ le milieu de $A B$.\n\nLa condition sur les milieux implique que $[A D C]=[A K C]$ où la notation crochets signifie l'aire du triangle. Si l'on p... | Switzerland | SMO-Selektion | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequa... | null | proof and answer | 2 | |
0cwp | Let $N$ be a positive integer. A cube $(2N+1) \times (2N+1) \times (2N+1)$ is made of $(2N+1)^3$ unit cubes; each unit cube is either black or white. It turned out that among any 8 unit cubes which form a cube $2 \times 2 \times 2$ the number of black unit cubes is not greater than 4. Find the greatest possible total n... | [
"**Ответ.** $(N+1)^2(4N+1)$.\n\nLet $k = (N+1)^2(4N+1)$. Introduce a coordinate system where all vertices of unit cubes have integer coordinates from $0$ to $2N+1$.\n\nFirst, we present an example showing that the number of black cubes can indeed be $k$. For each cube, consider its vertex closest to the origin (its... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Solid Geometry > 3D Shapes"
] | Russian | proof and answer | (N+1)^2(4N+1) | |
0jw5 | Problem:
A bird thinks the number $2 n^{2}+29$ is prime for every positive integer $n$. Find a counterexample to the bird's conjecture. | [
"Solution:\nSimply taking $n=29$ works, since $2 \\cdot 29^{2}+29=29(2 \\cdot 29+1)=29 \\cdot 59$."
] | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 29 | |
0b87 | An $8 \times 8$ array is divided into 64 unit squares. In some of the unit squares a diagonal is drawn, such that no two diagonals share a common point – not even an endpoint. Find the maximum number of diagonals that can be drawn under these circumstances. | [
"Look at two contiguous rows of a $2n \\times 2n$ array, containing $k$ diagonals in the cells of row $A$ and $\\ell$ diagonals in the cells of row $B$. The ends of these diagonals on the separating line of the two rows have to be distinct points, and they can only occupy at most the $2n + 1$ available positions, t... | Romania | NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 36 | |
0jld | Problem:
Let $D$ be the set of divisors of $100$. Let $Z$ be the set of integers between $1$ and $100$, inclusive. Mark chooses an element $d$ of $D$ and an element $z$ of $Z$ uniformly at random. What is the probability that $d$ divides $z$? | [
"Solution:\n\nThe answer is $\\frac{217}{900}$.\n\nAs $100 = 2^{2} \\cdot 5^{2}$, there are $3 \\cdot 3 = 9$ divisors of $100$, so there are $900$ possible pairs of $d$ and $z$ that can be chosen.\n\nIf $d$ is chosen, then there are $\\frac{100}{d}$ possible values of $z$ such that $d$ divides $z$, so the total num... | United States | HMMT 2014 | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 217/900 | |
0c66 | Alice and Bob are given an integer $n \ge 2$ and a token to play the following game. Initially, Alice chooses an order of the numbers $1, 2, \dots, n$, and writes them down in a row, in that order, on a sheet of paper. Next, Bob chooses one of these numbers and places the token on that number. Continuing, Alice moves t... | [
"Alice has a winning strategy if and only if $n$ is congruent to $0$ or $-1$ modulo $4$.\n\nLooking upon placing the token on a number as a one unit drop of that number, a move is possible if and only if the current position of the token is not flanked by two zeroes. Notice that each player places the token on posi... | Romania | SELECTION TESTS FOR THE 2019 BMO AND IMO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | n ≡ 0 or 3 (mod 4) | |
09pt | Problem:
Laat $m, n$ positieve gehele getallen zijn. Bekijk een rijtje positieve gehele getallen $a_{1}$, $a_{2}, \ldots, a_{n}$ dat voldoet aan $m=a_{1} \geq a_{2} \geq \cdots \geq a_{n} \geq 1$. Voor $1 \leq i \leq m$ definiëren we
$$
b_{i}=\#\left\{j \in\{1,2, \ldots, n\}: a_{j} \geq i\right\}
$$
dus $b_{i}$ is het... | [] | Netherlands | TOETS TRAININGSKAMP | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0f3n | Problem:
A square is divided into $n$ parallel strips (parallel to the bottom side of the square). The width of each strip is integral. The total width of the strips with odd width equals the total width of the strips with even width. A diagonal of the square is drawn which divides each strip into a left part and a ri... | [
"Solution:\n\nLet $LO$ be the total area of the left parts of the odd strips, $LE$ the total area of the left parts of the even strips, and $RE$ the total area of the right parts of the even strips. Since the diagonal bisects the square, $LO + LE = A / 2$, where $A$ is the area of the square. Also $RE + LE = A / 2$... | Soviet Union | ASU | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0299 | Problem:
(a) Um quadrilátero com lados opostos iguais é um paralelogramo. As figuras a seguir mostram dois paralelogramos com os mesmos lados, sendo o segundo um retângulo.
Determine a maior área possível de um paralelogramo de lados $b$ e $h$.
(b) Considere as duas figuras a seguir, onde $... | [
"Solution:\n\n(a) Considerando todos os paralelogramos com lados $b$ e $h$, observe que o retângulo é o que possui maior altura em relação à base fixa $b$. Na figura a seguir, pela Desigualdade Triangular, temos $h' < h$.\n\n\n\nComo cada triângulo representa metade do paralelogramo, então ... | Brazil | NÍVEL 3 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | b*h | |
03zn | Suppose that the sequence $\{a_n\}$ defined by $a_1 = a_2 = 1$, $a_n = 7a_{n-1} - a_{n-2}$, $n \ge 3$. Prove that $a_n + a_{n+1} + 2$ is a perfect square for any positive integer $n$. (posed by Tao Pingsheng) | [
"It is well known that the solution of the sequence can be obtained by solving two geometric sequences. We get $a_n = C_1\\lambda_1^n + C_2\\lambda_2^n$, where\n$$\n\\lambda_1 = \\frac{7 + \\sqrt{45}}{2} = \\left( \\frac{3 + \\sqrt{5}}{2} \\right)^2, \\quad \\lambda_2 = \\frac{7 - \\sqrt{45}}{2} = \\left( \\frac{3 ... | China | China Southeastern Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0gtj | Find the smallest value of
$$
xy + yz + zx + \frac{1}{x} + \frac{2}{y} + \frac{5}{z},
$$
where $x, y, z$ are positive real numbers. | [
"Answer: $3 \\cdot \\sqrt[3]{36}$.\nUsing AM-GM inequality, we get\n$$\n\\begin{aligned}\nyx + \\frac{1}{3x} + \\frac{1}{2y} &\\ge 3\\sqrt[3]{\\frac{1}{6}}, \\\\\nyz + \\frac{3}{2y} + \\frac{3}{z} &\\ge 3\\sqrt[3]{\\frac{9}{2}}, \\\\\nzx + \\frac{2}{z} + \\frac{2}{3x} &\\ge 3\\sqrt[3]{\\frac{4}{3}}.\n\\end{aligned}... | Turkey | Team Selection Test | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 3*36^(1/3) | |
0bdr | Let $f$ and $g$ be two quadratic functions. Prove that, if the functions $\lfloor f \rfloor$ and $\lfloor g \rfloor$ are equal, then the functions $f$ and $g$ are equal. | [] | Romania | Shortlisted Problems for the 64th NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
02gm | A finite collection of squares has total area $4$. Show that they can be arranged to cover a square of side $1$. | [
"Let the sides of the squares be equal to $a_i$ for $i = 1, 2, \\ldots, N$ ($N$ is the number of squares).\n\nIf some $a_k > 1$ then the $k$th square will cover the unit square. Now let's assume the case when $a_i < 1$ for all $i$.\n\nEvery number $a_i$ must satisfy $2^{-k_i} \\le a_i < 2^{-k_i+1}$ for some integer... | Brazil | XXIV OBM | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
039y | There are 1000 cities $A_1, A_2, \dots, A_{1000}$ in a country and some of them are connected by airlines. It is known that the $i$-th city is connected to $d_i$ other cities, where $d_1 \le d_2 \le \dots \le d_{1000}$ and $d_j \ge j + 1$ for every $j = 1, 2, \dots, 999 - d_{999}$. Prove that if the airport of a city i... | [
"Denote the non-closed airports with $B_1, \\dots, B_{999}$ where $B_i = A_i$ for $i = 1, \\dots, k-1$ and $B_i = A_{i+1}$ for $i = k+1, \\dots, 1000$. Denote the number of the airlines from $B_i$ by $d'_i$. It is clear that $d'_i \\ge d_i - 1$ for every $i = 1, \\dots, 999$. We can assume without loss of generalit... | Bulgaria | Fall Mathematical Competition | [
"Discrete Mathematics > Graph Theory"
] | English | proof only | null | |
06xj | Let $N$ be a positive integer. Prove that there exist three permutations $a_{1}, a_{2}, \ldots, a_{N}$; $b_{1}, b_{2}, \ldots, b_{N}$; and $c_{1}, c_{2}, \ldots, c_{N}$ of $1,2, \ldots, N$ such that
$$
\left|\sqrt{a_{k}}+\sqrt{b_{k}}+\sqrt{c_{k}}-2 \sqrt{N}\right|<2023
$$
for every $k=1,2, \ldots, N$. | [
"The idea is to approximate the numbers $\\sqrt{1}, \\sqrt{2}, \\ldots, \\sqrt{N}$ by the nearest integer with errors $<0.5$. This gives the following sequence\n$$\n1,1,2,2,2,2,3,3,3,3,3,3,4, \\ldots .\n$$\nMore precisely, for each $k \\geqslant 1$, we round $\\sqrt{k^{2}-k+1}, \\ldots, \\sqrt{k^{2}+k}$ to $k$, so ... | IMO | International Mathematical Olympiad Shortlist | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Algebra > Equations and Inequalities > Li... | null | proof only | null | |
0457 | Let $n$ be a positive integer. There are $3n$ women's volleyball teams attending a tournament. Each pair of teams play at most once (there are no ties in volleyball games). Assume that a total number of $3n^2$ games have been played. Prove that there exists a team whose number of winning games and number of losing game... | [
"*Proof.* We prove by contradiction: assume the conclusion is false. Suppose that the teams $P_1, \\cdots, P_k$ all have won less than $\\frac{n}{4}$ games, while the remaining $3n-k$ teams, $Q_1, \\cdots, Q_{3n-k}$ all have won at least $\\frac{n}{4}$ games. Based on the assumption of proof by contradiction, the n... | China | 2022 CGMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
04zy | Find all integral solutions of the equation $x^3 - y^3 = 3xy + 1$. | [
"First assume $x > y$. Then $3xy + 1 = x^3 - y^3 = (x-y)(x^2 + xy + y^2) \\ge (x-y) \\cdot 3xy$. Thus $3xy = x^3 - y^3 - 1 \\ge 1 - 1 = 0$ because $x > y$. If $3xy > 0$, then $3xy \\ge 3$, hence the inequality $3xy + 1 \\ge (x-y) \\cdot 3xy$ derived above implies $x-y = 1$. If $3xy = 0$, then either $x = 0$ or $y =... | Estonia | Selected Problems from the Final Round of National Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | All integer pairs (x, y) of the form (n+1, n) for any integer n, together with the pair (-1, 1). | |
085n | Problem:
Un intero positivo si dice triangolare se si può scrivere nella forma $\frac{n(n+1)}{2}$ per qualche intero positivo $n$. Quante sono le coppie $(a, b)$ di numeri triangolari tali che $b-a=2007$? (Si ricorda che 223 è un numero primo). | [
"Solution:\n\nCi sono 6 coppie di numeri triangolari che soddisfano la condizione richiesta. Il problema è equivalente a trovare le coppie di interi positivi $(n, m)$ tali che\n$$\n\\frac{n(n+1)}{2}-\\frac{m(m+1)}{2}=2007.\n$$\nDifatti, poiché valori diversi di $n$ determinano valori diversi di $\\frac{n(n+1)}{2}$,... | Italy | Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 6 | |
085e | Problem:
In un triangolo isoscele $ABC$ con $AC = BC \neq AB$, si fissi un punto $P$ sulla base $AB$. Quante posizioni può assumere nel piano un punto $Q$ se vogliamo che i punti $A, P$ e $Q$, presi in ordine qualsiasi, siano i vertici di un triangolo simile ad $ABC$?
(A) 0
(B) 2
(C) 3
(D) 4
(E) 6. | [
"Solution:\n\nLa risposta è $\\mathbf{(E)}$. Il triangolo $APQ$ può essere costruito in modo tale che $AQ = QP$, oppure in modo tale che $AP = QP$, oppure in modo tale che $AQ = AP$. Per ciascuna di queste 3 possibilità, ci sono due scelte per la collocazione del punto $P$ in posizioni simmetriche rispetto alla ret... | Italy | Progetto Olimpiadi di Matematica 2007 GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | MCQ | E | |
07o3 | Prove that the cubic $x^3 - 3a x^2 + 2x - 4a$ has only one real root for every real number $a$. | [
"Reduce the given polynomial to normal form, which can be done by noting that\n$$\n\\begin{aligned}\nx^3 - 3a x^2 + 2x - 4a &= (x-a)^3 - 3a^2 x + a^3 + 2(x-a) - 2a \\\\\n&= (x-a)^3 + (2 - 3a^2)(x-a) - 2a^3 - 2a \\\\\n&= X^3 + pX + q,\n\\end{aligned}\n$$\nwhere $X = x - a$, and $p = (2 - 3a^2)$, $q = -2a(a^2 + 1)$. ... | Ireland | Ireland | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
062q | Problem:
Man bestimme alle Paare $(m, n)$ nicht-negativer ganzer Zahlen, die der Gleichung
$$
3^{m}-7^{n}=2
$$
genügen. | [
"Solution:\nDie Paare $(1,0)$ und $(2,1)$ sind Lösungen. Es sei $(m, n)$ eine Lösung mit $m, n \\geq 2$. Die Lösung besteht aus drei Schritten:\n\n1. Es ist $7^{n} \\equiv -2 \\bmod 9 \\Longleftrightarrow n \\equiv 1 \\bmod 3$.\n2. Es ist $3^{m} \\equiv 2 \\bmod 49 \\Longleftrightarrow m \\equiv 26 \\bmod 42$.\n3. ... | Germany | 1. IMO-Auswahlklausur | [
"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"
] | null | proof and answer | (m, n) = (1, 0) and (2, 1) | |
06nq | Given an acute $\triangle ABC$ with $AB > AC$, let $M$ be the midpoint of the side $BC$. Let $I$ be the incentre of $\triangle ABC$, and let $D$ be the foot of perpendicular from $I$ to $BC$. The line passing through $A$ and perpendicular to $BC$ meets the circumcircle of $\triangle ABC$ again at $E \neq A$. Suppose $\... | [
"Let $O$ and $H$ be the circumcentre and the orthocentre of $\\triangle ABC$ respectively. Reflect $A$ about $O$ to a point $A'$. Reflect $D$ about $M$ and $I$ to points $D'$ and $P$ respectively. Also, let $Q$ be the midpoint of $AH$. Notice the following well-known results.\n\n* $E$ and $H$ are symmetric about th... | Hong Kong | The 26th Hong Kong (China) Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | null | proof only | null | |
0iat | Let $a_1, a_2, \dots, a_n, b_1, b_2, \dots, b_n$ be real numbers such that
$$
(a_1^2 + a_2^2 + \cdots + a_n^2 - 1)(b_1^2 + b_2^2 + \cdots + b_n^2 - 1)
> (a_1b_1 + a_2b_2 + \cdots + a_nb_n - 1)^2.
$$
Show that $a_1^2 + a_2^2 + \cdots + a_n^2 > 1$ and $b_1^2 + b_2^2 + \cdots + b_n^2 > 1$. | [
"If exactly one of $a_1^2 + a_2^2 + \\dots + a_n^2$ and $b_1^2 + b_2^2 + \\dots + b_n^2$ is no greater than $1$, then the left-hand side of the inequality is no greater than $0$, and inequality $(*)$ is trivial. Now we assume that both are no greater than $1$. Set $a = 1 - a_1^2 - a_2^2 - \\dots - a_n^2$ and $b = 1... | United States | USA IMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Linear Algebra > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometr... | null | proof only | null | |
03wz | Let $D$ be a point on the side $BC$ of an acute triangle $ABC$. The circle with diameter $BD$ meets the lines $AB$ and $AD$ respectively at the points $X$ and $P$, which are different from the points $B$ and $D$. The circle with diameter $CD$ meets the lines $AC$ and $AD$ respectively at the points $Y$ and $Q$, which a... | [
"**Proof** Join the segments $XY$ and $DX$. It follows from the given conditions that $B$, $P$, $D$, $X$ are concyclic, and $C$, $Y$, $Q$, $D$ are concyclic. Then\n$$\n\\angle AXM = \\angle BXP = \\angle BDP \\\\\n= \\angle QDC = \\angle AYN.\n$$\n\nAnd it follows from $\\angle AMX = \\angl... | China | China Western Mathematical Olympiad | [
"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",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
0icq | Problem:
Andrea flips a fair coin repeatedly, continuing until she either flips two heads in a row (the sequence $H H$) or flips tails followed by heads (the sequence $T H$). What is the probability that she will stop after flipping $H H$? | [
"Solution:\n\nThe only way that Andrea can ever flip $H H$ is if she never flips $T$, in which case she must flip two heads immediately at the beginning. This happens with probability $\\frac{1}{4}$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 1/4 | |
0idy | Problem:
A classroom consists of a $5 \times 5$ array of desks, to be filled by anywhere from $0$ to $25$ students, inclusive. No student will sit at a desk unless either all other desks in its row or all others in its column are filled (or both). Considering only the set of desks that are occupied (and not which stude... | [
"Solution:\nThe set of empty desks must be of the form (non-full rows) $\\times$ (non-full columns): each empty desk is in a non-full column and a non-full row, and the given condition implies that each desk in such a position is empty. So if there are fewer than $25$ students, then both of these sets are nonempty;... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 962 | |
01tp | Three positive integers are written on a blackboard. Per move one replaces the set of these numbers by the new set in accordance with the following rule: each number of the set is replaced by the quotient of the sum of the squares of two other numbers and this number.
What is the maximum value of the sum of the initial... | [
"Let $a_0, b_0, c_0$ be the initial numbers on the blackboard, and $a_i, b_i, c_i$ be the numbers written on the blackboard after the $i$-th move; let $S(a_0, b_0, c_0)$ and $S(a_i, b_i, c_i)$ be the sums of these numbers, respectively. Then after the $i+1$-th move we obtain the following numbers\n$$\na_{i+1} = \\f... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 63 | |
07rd | Let $a$ and $b$ be positive integers that are co-prime and let $p$ be a prime number. Prove that
$$
\gcd(ab, a^2 + pb^2) = \begin{cases} 1 & \text{if } p \nmid a \\ p & \text{if } p \mid a. \end{cases}
$$ | [
"Let $d = \\gcd(ab, a^2 + pb^2)$ and suppose $d > 1$. Let $q$ be a prime factor of $d$. Because $a^3 = a(a^2 + pb^2) - pb(ab)$ we see that $q \\mid a^3$ and so $q \\mid a$ as well. As $q \\mid a^2 + pb^2$, this implies that $q \\mid pb^2$. Because $\\gcd(a, b) = 1$ this can only happen if $q = p$.\n\nSo far we have... | Ireland | Ireland_2017 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | gcd(ab, a^2 + p b^2) = 1 if p does not divide a, and gcd(ab, a^2 + p b^2) = p if p divides a. | |
0g0d | Problem:
Sei $n \geq 2$ eine natürliche Zahl. In der Mitte eines kreisförmigen Gartens steht ein Wachturm. Am Rand des Gartens stehen $n$ gleichmässig verteilte Gartenzwerge. Auf dem Wachturm wohnen aufmerksame Wächter. Jeder Wächter überwacht einen Bereich des Gartens, der von zwei verschiedenen Gartenzwergen begrenz... | [
"Solution:\n\nWir nennen den Bereich zwischen zwei Gartenzwergen ein Intervall. Definiere eine Kette als ein $n-1$-Tupel von Intervallen $\\left(I_{1}, \\ldots, I_{n-1}\\right)$, wobei $I_{1} \\subset I_{2} \\subset \\cdots \\subset I_{n-1}$ und alle Inklusionen strikt sind.\n\nLemma: Jedes Intervall ist in genau $... | Switzerland | IMO-Selektion | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof only | null | |
014r | Problem:
The director has found out that six conspiracies have been set up in his department, each of them involving exactly three persons. Prove that the director can split the department in two laboratories so that none of the conspirative groups is entirely in the same laboratory. | [
"Solution:\n\nLet the department consist of $n$ persons. Clearly $n > 4$ (because $\\binom{4}{3} < 6$). If $n = 5$, take three persons who do not make a conspiracy and put them in one laboratory, the other two in another. If $n = 6$, note that $\\binom{6}{3} = 20$, so we can find a three-person set such that neithe... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
08cg | Problem:
Quante sono le coppie di numeri reali $(x, y)$ che soddisfano entrambe le equazioni $x + y^{2} = y^{3}$ e $y + x^{2} = x^{3}$?
(A) 1
(B) 3
(C) 5
(D) 9
(E) Infinite | [
"Solution:\n\nLa risposta è (B). Sottraendo le equazioni membro a membro otteniamo\n$$\n(y - x)(-1 + x + y) = (y - x)\\left(y^{2} + x y + x^{2}\\right),\n$$\nda cui vediamo che deve valere almeno una delle due uguaglianze $x = y$ e $x^{2} + y^{2} + x y - x - y + 1 = 0$. Se $x = y$, sostituendo nelle equazioni date ... | Italy | Gara di Febbraio | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | B | |
0l66 | Problem:
Jacob rolls two fair six-sided dice. If the outcomes of these dice rolls are the same, he rolls a third fair six-sided die. Compute the probability that the sum of outcomes of all the dice he rolls is even. | [
"Solution:\n\nThere's a $\\frac{1}{2} - \\frac{1}{6} = \\frac{1}{3}$ probability that he rolls an even number without getting doubles: whatever the first roll is, there is a $\\frac{1}{2}$ chance that the second roll is of opposite parity, and we subtract the $\\frac{1}{6}$ chance that the second roll is the same.\... | United States | HMMT February | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 5/12 | |
0l80 | Problem:
Let $a_1$, $a_2$, $r$, and $s$ be positive integers with $r$ and $s$ odd. The sequence $a_1$, $a_2$, $a_3$, $\dots$ is defined by
$$
a_{n+2} = r a_{n+1} + s a_n
$$
for all $n \ge 1$. Determine the maximum possible number of integers $1 \le l \le 2025$ such that $a_l$ divides $a_{l+1}$, over all possible choice... | [] | United States | USA TST Selection Test for 67th IMO and 15th EGMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | null | proof and answer | 2 | |
06i3 | Find all pairs $(a, b)$ of integers $a$ and $b$ satisfying
$$
(b^2 + 11(a-b))^2 = a^3 b.
$$ | [
"(Belarus MO 2014 Category B Problem 2 modified) The only solutions are $(a, b) = (t, t), (0, 11)$ where $t$ is any integer.\nWe have\n$$\n\\begin{aligned}\n(b^2 + 11(a-b))^2 &= a^3b \\\\\n\\Leftrightarrow \\quad b^4 + 22b^2(a-b) + 121(a-b)^2 &= a^3b \\\\\n\\Leftrightarrow \\quad (a-b)(121(a-b) + 22b^2 - b(a^2 + ab... | Hong Kong | CHKMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (a, b) = (t, t) for any integer t, and (a, b) = (0, 11) | |
01rg | Let $\Omega$ be the circumcircle of a triangle $ABC$. A circle passing through the vertex $A$ and touching $BC$ at point $X$ meets $\Omega$ at point $Y$ (different from $A$). Let point $Z$ (different from $Y$) be the intersection point of the ray $YX$ and $\Omega$.
$$
\angle CAX = \angle ZAB.
$$ | [
"Let $\\omega$ be the circle passing through $A$ and touching the side $BC$ at $X$. Since the angle $YAX$ subtends the arc $YX$, we have $\\angle YAX = \\frac{1}{2} \\angle YX$.\n\nFurther, $\\angle YXC = \\frac{1}{2} \\angle YX$ (as the angle between the tangent $BC$ and the chord $XY$). So, $\\angle YAX = \\angle... | Belarus | Final Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08im | Problem:
Let $a, b, c$ be real numbers such that $a^{2} + b^{2} + c^{2} = 1$. Prove that $P = ab + bc + ca - 2(a + b + c) \geq -\frac{5}{2}$. Are there values of $a, b, c$ such that $P = -\frac{5}{2}$?
Problem:
Fie $a, b, c$ numere reale astfel încât $a^{2} + b^{2} + c^{2} = 1$. Demonstrați că $P = ab + bc + ca - 2(a ... | [
"Solution:\nWe have $ab + bc + ca = \\frac{(a + b + c)^{2} - c^{2} - b^{2} - a^{2}}{2} = \\frac{(a + b + c)^{2} - 1}{2}$.\nIf we put $t = a + b + c$ we obtain\n$$\nP = \\frac{t^{2} - 1}{2} - 2t = \\frac{t^{2} - 4t - 1}{2} = \\frac{(t - 2)^{2} - 5}{2} \\geq -\\frac{5}{2}\n$$\nObviously $P = -\\frac{5}{2}$ when $t = ... | JBMO | 7th JBMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | -5/2; equality is not attainable | |
0jh7 | Problem:
Let $R$ be the region in the Cartesian plane of points $(x, y)$ satisfying $x \geq 0$, $y \geq 0$, and $x+y+\lfloor x\rfloor+\lfloor y\rfloor \leq 5$. Determine the area of $R$. | [
"Solution:\nWe claim that a point in the first quadrant satisfies the desired property if the point is below the line $x+y=3$ and does not satisfy the desired property if it is above the line.\n\nTo see this, for a point inside the region, $x+y<3$ and $\\lfloor x\\rfloor+\\lfloor y\\rfloor \\leq x+y<3$. However, $\... | United States | HMMT 2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 9/2 | |
0hhz | 11 points are given on a circle. Petrik numbered them by the numbers $1, 2, \ldots, 11$. After that, pairs of points were connected by segments: $1$ and $2$, $2$ and $3$, $\ldots$, $10$ and $11$, $11$ and $1$. What is the largest possible number of intersection points of these segments? The given 11 points are not coun... | [
"Consider one of the drawn segments. It cannot intersect itself, nor the adjacent segments on either side, e.g., the segment $3$–$4$ does not intersect the segments $2$–$3$, $3$–$4$, and $4$–$5$. Thus the maximum number of possible intersection points occurs when each of the segments intersects all the other $8$ no... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 44 |
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