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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0go3 | In a country $2010$ cities are connected directly to the capital by a highway. The number of cities connected directly to any other given city is less than $2010$, and if this number is the same for two cities, then it is even. $k$ of the highways connecting the capital directly to various cities will be closed to traf... | [
"The answer is $503$.\nWe want the connected components of the graph $G$ representing the cities and the highways to remain the same when we remove $k$ of the edges incident with the vertex $v_0$ representing the capital. Without loss of generality we may assume that $G$ is connected.\n\nLet $G'$ be the graph obtai... | Turkey | 18th Turkish Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 503 | |
0gmc | In an acute triangle $ABC$, let $H$ be the intersection of its heights and $D$ the midpoint of $[AC]$. Show that the line $DH$ passes through an intersection point of the circumcircle of $ABC$ with the circle for which $[BH]$ is a diameter. | [] | Turkey | TEAM SELECTION EXAMINATION FOR THE 42nd INTERNATIONAL MATH- EMATICAL OLYMPIAD. TURKEY. | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic ... | English | proof only | null | |
0454 | Let $\{a_n\}$ and $\{b_n\}$ be two sequences of positive real numbers such that, for any positive integer $n$,
$$
a_{n+1} = a_n - \frac{1}{1 + \sum_{i=1}^{n} \frac{1}{a_i}}, \quad \text{and} \quad b_{n+1} = b_n + \frac{1}{1 + \sum_{i=1}^{n} \frac{1}{b_i}}.
$$
(1)
If $a_{100}b_{100} = a_{101}b_{101}$, find the value of ... | [
"(1) Set $a_1 = a$ and $b_1 = b$ ($a, b > 0$). The recurrence formula of $a_n$ implies that, for any positive integer $n \\ge 2$, we have\n$$\n\\frac{1}{a_n - a_{n+1}} = 1 + \\sum_{i=1}^{n} \\frac{1}{a_i} \\implies \\frac{1}{a_n - a_{n+1}} = \\frac{1}{a_{n-1} - a_n} + \\frac{1}{a_n}.\n$$\nThat is, $\\frac{a_n}{a_n ... | China | Chinese Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | Part (1): 199; Part (2): a_100 + b_100 is larger than a_101 + b_101. | |
0b9d | Let $n$ be an integer number greater than or equal to $2$, and let $K$ be a closed convex set of area greater than or equal to $n$, contained in the open square $(0, n) \times (0, n)$. Prove that $K$ contains some point of the integral lattice $\mathbb{Z} \times \mathbb{Z}$. | [
"Transform $K$ by a suitable two-step Steiner-Edler symmetrization. First, perform a horizontal translation of each slice $K \\cap (\\mathbb{R} \\times y)$ to place it symmetrically about the vertical line $x = \\frac{1}{2}$. It is readily checked that the image $K'$ of $K$ under this transformation is a closed con... | Romania | NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Geometry > Plane Geometry > Combinatorial Geometry > Minkowski's theorem"
] | English | proof only | null | |
044d | Suppose $a$, $b$, $c > 1$ and $(a^2b)^{\log_a c} = a \cdot (ac)^{\log_a b}$ is satisfied. Then the value of $\log_c(ab)$ is ______. | [
"Taking the logarithm of the original equation with respect to an arbitrary base $a$ on both sides, we get\n$$\n\\log_a c \\cdot (2 + \\log_a b) = 1 + \\log_a b \\cdot (1 + \\log_a c).\n$$\nSimplifying the above equation gives $2\\log_a c = 1 + \\log_a b$. Therefore, $c^2 = ab$, and then $\\log_c(ab) = \\log_c c^2 ... | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | 2 | |
02ev | The polynomial $x^3 + px + q$ has three distinct real roots. Show that $p < 0$. | [
"The sum of the squares of the roots $\\alpha, \\beta, \\gamma$ is $s_2 = \\alpha^2 + \\beta^2 + \\gamma^2 = (\\alpha + \\beta + \\gamma)^2 - 2(\\alpha\\beta + \\beta\\gamma + \\gamma\\alpha) = 0^2 - 2p$. Since the roots are distinct, $s_2 > 0 \\iff p < 0$."
] | Brazil | XIV OBM | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof only | null | |
09px | Problem:
Zij $ABC$ een driehoek, punt $P$ het midden van $BC$ en punt $Q$ op lijnstuk $CA$ zodat $|CQ|=2|QA|$. Zij $S$ het snijpunt van $BQ$ en $AP$. Bewijs dat $|AS|=|SP|$. | [
"Solution:\nOplossing I. Trek een lijn door $P$ evenwijdig aan $AC$ en zij $T$ het snijpunt van deze lijn met $BQ$. Dan is $PT$ een middenparallel in driehoek $BCQ$, dus geldt $|PT|=\\frac{1}{2}|CQ|=|QA|$. Nu is $ATPQ$ een vierhoek met een paar even lange, evenwijdige zijden, dus $ATPQ$ is een parallellogram. Van e... | Netherlands | Dutch TST | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
088y | Problem:
Una successione $\{x_{n} \mid n=0,1,2, \ldots\}$ di numeri reali è definita, al variare del parametro reale $a$, come segue:
$$
\left\{
\begin{array}{l}
x_{0}=a \\
x_{n+1}=2-x_{n}^{2} \quad \text{ per } n \geq 1
\end{array}
\right.
$$
a. Trovare tutti i valori di $a$ per cui $x_{n}$ è costante (cioè vale $x_... | [
"Solution:\n\na. Se $x_{n}$ deve essere costante, in particolare deve valere che\n$$\nx_{1}=x_{0}=a.\n$$\nMa allora\n$$\na=x_{1}=2-x_{0}^{2}=2-a^{2},\n$$\nquindi $a$ deve soddisfare l'equazione\n$$\na^{2}+a-2=0\n$$\nche ha come soluzioni $a=-2$ e $a=1$. Nessun altro valore può rendere la successione costante (ma a ... | Italy | Olimpiadi di Matematica | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | Constant sequences occur only for a equal to 1 or −2. Taking y = −2, if the initial absolute value is less than 2 then all terms have absolute value less than 2; if the initial absolute value is greater than 2 then the sequence is strictly decreasing. | |
0h4l | Find all values of the parameter $a$ such that the parabolas $y = x^2 + 2013x + a$ and $y = -x^2 + ax + 2013$ are tangent to each other. | [
"Two parabolas with different leading coefficients are tangent if and only if they have exactly one common point, so the equation $x^2 + 2013x + a = -x^2 + ax + 2013$, that is equivalent to the equation $2x^2 + (2013-a)x + (a-2013) = 0$, must have exactly one root. Therefore,\n$$\nD = (a - 2013)^2 - 8(a - 2013) = (... | Ukraine | Ukrainian National Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | 2013 or 2021 | |
0jui | Problem:
Meghal is playing a game with 2016 rounds $1,2, \cdots, 2016$. In round $n$, two rectangular double-sided mirrors are arranged such that they share a common edge and the angle between the faces is $\frac{2 \pi}{n+2}$. Meghal shoots a laser at these mirrors and her score for the round is the number of points o... | [
"Solution:\n\nLet points $O, A_{1}, A_{2}$ lie in a plane such that $\\angle A_{1} O A_{2}=\\frac{2 \\pi}{n+2}$. We represent the mirrors as line segments extending between $O$ and $A_{1}$, and $O$ and $A_{2}$. Also let points $A_{3}, A_{4}, \\cdots, A_{n+2}$ lie in the plane such that $A_{i+1}$ is the reflection o... | United States | HMMT November 2016 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 1019088 | |
07n3 | Suppose $a$, $b$, $c$, $d$ are positive numbers such that
$$
1 = 3abcd + 2(abc + bcd + dca + dab) + (ab + bc + cd + da + ac + bd).
$$
Prove that
$$
abcd \le \frac{1}{81},
$$
and that the inequality is strict unless $a = b = c = d = 1/3$. | [
"By the AM-GM inequality,\n$$\n\\begin{aligned}\n&abc + bcd + dca + dab \\ge 4\\sqrt[4]{(abcd)^3} \\quad \\text{and} \\\\\n&ab + bc + cd + da + ac + bd \\ge 6\\sqrt[6]{(abcd)^3},\n\\end{aligned}\n$$\nwith equality in both inequalities iff $a = b = c = d$. Hence, letting $x = \\sqrt[4]{abcd}$, we have that\n$$\n1 \\... | Ireland | Ireland | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | abcd ≤ 1/81, with equality only when a = b = c = d = 1/3 | |
06f8 | Determine if there exists a positive integer pair $(m, n)$, such that
(i) the greatest common divisor of $m$ and $n$ is $1$, and $m \le 2007$,
(ii) for any $k = 1, 2, \dots, 2007$, $\lfloor \frac{nk}{m} \rfloor = \lceil \sqrt{2k} \rceil$.
(Here $\lfloor x \rfloor$ stands for the greatest integer less than or equal to ... | [
"Yes, it exists.\nThere are finitely many fractions $\\frac{a}{b}$ of lowest term such that $a \\ge 1$, $1 \\le b \\le 2007$ and $\\frac{a}{b} < \\sqrt{2}$. Let $\\frac{n}{m}$ be the largest such fraction. If $\\lfloor \\frac{nk}{m} \\rfloor \\ne \\lfloor \\sqrt{2}k \\rfloor$ for some $k = 1, 2, \\dots, 2007$, then... | Hong Kong | CHKMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Yes, such a pair exists. | |
0i1c | Problem:
Let $ABCD$ be a quadrilateral, and let $O$ be the intersection of $AC$ and $BD$. Quadrilateral $A'B'C'D'$ is obtained by rotating $ABCD$ about $O$ by some angle. Let $A_1, B_1, C_1, D_1$ be the intersection points of the lines $A'B'$ and $AB$, $B'C'$ and $BC$, $C'D'$ and $CD$, $D'A'$ and $DA$, respectively. P... | [
"Solution:\n\nFirst we prove a lemma: Suppose some line $ST$ is rotated by angle $\\theta$ about $O$ to obtain line $S'T'$. Let $ST$ and $S'T'$ intersect at $S_1$, and let $S_2$ be the foot of the perpendicular from $O$ to $ST$. Then a rotation about $O$ of angle $\\theta/2$, together with a homothety about $O$ of ... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > ... | null | proof only | null | |
06hs | Let $A$, $B$, $C$ be points on the same plane with $\angle ACB = 120^\circ$. There is a sequence of circles $\omega_0, \omega_1, \omega_2, \dots$ on the same plane (with corresponding radii $r_0, r_1, r_2, \dots$, where $r_0 > r_1 > r_2 > \dots$) such that each circle is tangent to both segments $CA$ and $CB$. Furtherm... | [
"Let $S$ be the sum and $O$ the centre of $\\omega_0$. Then $2S = CO + r_0 = \\frac{3}{\\sin 60^\\circ} + 3 = 2\\sqrt{3} + 3$, and hence $S = \\sqrt{3} + \\frac{3}{2}$."
] | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | √3 + 3/2 | |
0b8b | Let $n$ be a positive integer and consider the integers $x_1, x_2, \dots, x_n$, $y_1, y_2, \dots, y_n$ such that
a) $x_1 + x_2 + \dots + x_n = y_1 + y_2 + \dots + y_n = 0$;
b) $x_1^2 + y_1^2 = x_2^2 + y_2^2 = \dots = x_n^2 + y_n^2 = 0$.
Prove that $n$ is an even number. | [
"Set $a = x_1^2 + y_1^2$. If $a$ is odd, the numbers $x_i$ and $y_i$ do not have the same parity, so $x_i + y_i$ is odd. Since $\\sum (x_i + y_i) = 0$, it follows that $n$ is even.\n\nSuppose $a = 4k+2$. Then $x_i$ and $y_i$ are both odd. The equality $x_1+x_2+\\dots+x_n = 0$ implies $n$ is even.\n\nFinally, if $a ... | Romania | NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
07gl | We call a polynomial $x^{n_1} + x^{n_2} + \dots + x^{n_{1398}} + 1$ *special* if $n_1, n_2, \dots, n_{1398}$ are distinct positive integers. Do there exist an infinite set of polynomials with real coefficients such that the product of each two of them is special? | [
"We first prove the following lemma.\n\n**Lemma.** Let $f(x)$ be a polynomial with complex coefficients such that its leading coefficient is rational. If for some positive integer $k$ we have $f(x)^k \\in \\mathbb{Z}[X]$, then the polynomial $f(x)$ would also be in $\\mathbb{Z}[X]$.\n\n*Proof.* Assume $f(x) = a_n x... | Iran | 38th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | null | proof and answer | No | |
0fq2 | Problem:
Sean $p$ un primo impar y $S_{q} = \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{5 \cdot 6 \cdot 7} + \ldots + \frac{1}{q(q+1)(q+2)}$, donde $q = \frac{3p-5}{2}$. Escribimos $\frac{1}{p} - 2 S_{q}$ en la forma $\frac{m}{n}$, donde $m$ y $n$ son enteros. Demuestra que $m \equiv n \pmod{p}$; es decir, que $m$ y $n$ da... | [
"Solution:\n\nSe tiene que\n$$\n\\begin{aligned}\n& \\frac{2}{k(k+1)(k+2)} = \\frac{(k+2)-k}{k(k+1)(k+2)} = \\frac{1}{k(k+1)} - \\frac{1}{(k+1)(k+2)} \\\\\n= & \\left(\\frac{1}{k} - \\frac{1}{k+1}\\right) - \\left(\\frac{1}{k+1} - \\frac{1}{k+2}\\right) = \\frac{1}{k} + \\frac{1}{k+1} + \\frac{1}{k+2} - \\frac{3}{k... | Spain | LIII Olimpiada matemática Española | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
04kk | Let $p$ be a prime number. Determine all pairs $(a, b)$ of integers such that
$$
p(a - 2) = a(b - 1).
$$ | [] | Croatia | Mathematical competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All integer pairs (a, b) with a ≠ 0 and a | 2p, given by b = p + 1 − 2p/a. Equivalently, for a ∈ {±1, ±2, ±p, ±2p}, b = p + 1 − 2p/a. | |
01eb | Olga and Sasha play a game on an infinite hexagonal grid. They take alternating turns in placing a counter on a free hexagon of their choice, with Olga opening the game. Beginning from the 2018th move, a new rule will come into play. A counter may now be placed only on those free hexagons having at least two occupied n... | [
"Answer: Olga has a winning strategy.\nThe game cannot go on forever. Draw a large hexagon enclosing all 2017 counters in play after the 2017th move, as in Figure 1. While it will be possible to place future counters in the hexagonal frame at distance 1 from the shaded part (i.e. immediately surrounding it), where ... | Baltic Way | Baltic Way shortlist | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Olga has a winning strategy. | |
0ayb | Problem:
A line passes through $(k, -9)$ and $(7, 3k)$ and has slope $2k$. Find the possible values of $k$. | [
"Solution:\nWe have $\\frac{3k - (-9)}{7 - k} = 2k$, so $3k + 9 = 2k(7 - k)$ or $2k^2 - 11k + 9 = (2k - 9)(k - 1) = 0$. Hence $k = \\frac{9}{2}$ or $k = 1$."
] | Philippines | 20th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | k = 1 or k = 9/2 | |
0hbu | For a quadrilateral $ABCD$, $\angle ABD = \angle DBC$ and $AD = CD$. Let $DH$ be the height of $\triangle ABD$. Prove that $|BC - BH| = HA$.
(Danylo Khilko)
 | [
"On the ray $BA$, we put down a segment $BE = BC$. If point $E$ belongs to $AB$ (fig. 19), then $\\triangle BCD = \\triangle BED$ due to two pairs of equal sides and the angle between them. Then, $AD = CD = ED$, which yields that $\\triangle ADE$ is isosceles. There, $HD$ is the height and, hence, the median. There... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
00s6 | Construct outside the acute-angled triangle $ABC$ the isosceles triangles $ABA_B$, $ABB_A$, $ACA_C$, $ACC_A$, $BCB_C$ and $BCC_B$, so that
$$
AB = AB_A = BA_B, \quad AC = AC_A = CA_C, \quad BC = BC_B = CB_C
$$
and
$$
\angle BAB_A = \angle ABA_B = \angle CAC_A = \angle ACA_C = \angle BCB_C = \angle CBC_B = \alpha < 90^\... | [
"**Lemma.** If $BCD$ is the isosceles triangle which is outside the triangle $ABC$ and has\n$$\n\\angle CBD = \\angle BCD = 90^\\circ - \\alpha := \\beta,\n$$\nthen $AD \\perp BA_C$.\n\n*Proof of the lemma.* Construct an isosceles triangle $ABE$ outside the triangle $ABC$, so that $\\angle ABE = \\angle AEB = \\bet... | Balkan Mathematical Olympiad | BMO 2017 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > An... | English | proof only | null | |
01i1 | Point $P$ lies inside triangle $ABC$, $M$ is the midpoint of side $BC$ and $P'$ is symmetric to $P$ with respect to $M$. $K$ and $H$ are projections of $P$ on $AB$ and $AC$ respectively and $KM = HM$. Prove that $\angle PAB = \angle CAP'$. | [
"Denote by $A'$, $K'$, $H'$ the points symmetrical to $A$, $K$, $H$ with respect to $M$, as in figure 13.\nIt is clear that $KH'H'K$ is a rectangle and the quadrilateral $AKPH$ is inscribed, denote its circumcircle by $\\omega$. Then\n$$\n\\angle CHK' = 90^\\circ - \\angle AHK = \\angle PHK = \\angle PAK.\n$$\nLet ... | Baltic Way | Baltic Way 2021 Shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00js | By $\lfloor x \rfloor$ we denote the largest integer that is smaller or equal to $x$ and by $\lceil x \rceil$ we denote the smallest integer that is greater or equal to $x$.
For every given pair $(a, b)$ of positive natural numbers find all natural numbers $n$ with
$$
b + \lfloor \frac{n}{a} \rfloor = \lfloor \frac{n+... | [
"We set $k := \\lfloor \\frac{n}{a} \\rfloor$ and $l := \\lfloor \\frac{n+b}{a} \\rfloor$. We thus have to find all nonnegative integers $n$ such that there exist integers $k$ and $l$ satisfying\n$$\nb+k=l \\text{ and } k \\le \\frac{n}{a} < k+1 \\text{ and } l-1 < \\frac{n+b}{a} \\le l.\n$$\nBy substituting $l = b... | Austria | AustriaMO2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All solutions are classified by cases:
- If a = 1, all nonnegative integers n are solutions.
- If b = 1, all nonnegative integers n are solutions.
- If b = 2, the solutions are exactly those n with n congruent to −1 modulo a (equivalently, n ≡ a − 1 mod a). For a = 1 this again gives all n.
- If a ≥ 2 and b ≥ 3, there ... | |
0218 | Problem:
A positive integer $n$ is friendly if every pair of neighbouring digits of $n$, written in base 10, differs by exactly 1. For example, $6787$ is friendly, but $211$ and $901$ are not.
Find all odd natural numbers $m$ for which there exists a friendly integer divisible by $64 m$. | [
"Solution:\nAny friendly number divisible by $64$ is divisible by $4$, and hence the number formed by its last two digits is a multiple of $4$, so ends in $00, 04, 08, \\ldots$, or $96$. A friendly number divisible by $4$ must therefore end in $12, 32, 56$, or $76$, so cannot be divisible by $5$. In particular, if ... | Benelux Mathematical Olympiad | 15th Benelux Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All odd positive integers not divisible by five. | |
05cy | Determine all pairs $(m, n)$ of natural numbers that satisfy $m - n = 96$ and $\text{lcm}(m, n) = 2024$. | [
"As $\\text{lcm}(m,n) = 2024 = 8 \\cdot 253$ and $8 = 2^3$, at least one of the numbers $m$ and $n$ is divisible by $8$. Since $8 \\mid 96 = m-n$, the other one must also be divisible by $8$. Both $m$ and $n$ are divisors of $2024$. All divisors of $2024$ that are divisible by $8$ are $8$, $88$, $184$ and $2024$. T... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (184, 88) | |
0ghc | 設 $ABCD$ 為圓內接四邊形。已知 $Q$, $A$, $B$, $P$ 依序排在一條直線上, 且直線 $AC$ 與圓 $ADQ$ 相切、直線 $BD$ 與圓 $BCP$ 相切。令點 $M$, $N$ 分別為邊 $BC$ 與 $AD$ 的中點。證明以下三條直線共點：直線 $CD$；圓 $ANQ$ 在點 $A$ 的切線；圓 $BMP$ 在點 $B$ 的切線。
(註：圓 $ADQ$ 指的是過 $A$, $D$, $Q$ 三點的圓；其餘類推。)
Let $ABCD$ be a cyclic quadrilateral. Assume that the points $Q$, $A$, $B$, $P$ are collinear ... | [
"解法一. 由於 $ABCD$ 有外接圓, 有 $\\angle DAQ = \\angle DCB$。又因於直線 $AC$ 與圓 $AQD$ 相切, 得 $\\angle CBD = \\angle CAD = \\angle AQD$。所以三角形 $ADQ$ 與 $CDB$ 相似 (AA)。\n令 $R$ 為線段 $CD$ 的中點。利用上面的三角形相似, $N$ 與 $R$ 為對應點。所以有\n$\\angle QNA = \\angle BRC$。\n\n\n\n設點 $K$ 為直線 $CD$ 與圓 $ABR$ 的第二個交點 (如果 $CD$ 與圓 $ABR$ 交於兩點... | Taiwan | 2023 數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Chinese (Traditional) | proof only | null | |
0elm | Problem:
Dan je karirast list papirja velikosti $7 \times 7$ kvadratkov in enako velika plastična karirasta šablona, na kateri so nekateri kvadratki zeleni, ostali pa prosojni. Če šablono postavimo na list papirja tako, da se stranice šablone ujemajo s stranicami papirja, se karirast vzorec na šabloni ujema s karirast... | [
"Solution:\n\nŠablono lahko na papir postavimo na 8 različnih načinov; na 4 načine, če je obrnjena na prednjo stran, in na 4 načine, če je obrnjena na zadnjo stran. Med različnimi postavitvami prehajamo z rotacijo šablone za $90^\\circ$ okrog njenega središča in z obračanjem šablone na drugo stran preko ene od njen... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | 10 | |
0fj0 | Problem:
Los números enteros desde $1$ hasta $9$ se distribuyen en las casillas de una tabla $3 \times 3$. Después se suman seis números de tres cifras: los tres que se leen en filas de izquierda a derecha y los tres que se leen en columnas de arriba abajo. ¿Hay alguna disposición para la cual el valor de esa suma sea... | [
"Solution:\n\nConsideremos la distribución\n\n| $a$ | $b$ | $c$ |\n| :--- | :--- | :--- |\n| $d$ | $e$ | $f$ |\n| $g$ | $h$ | $i$ |\n\nResulta\n$$\n\\begin{aligned}\n& S = abc + def + ghi + adg + beh + cfi = \\\\\n& = 100(a + c + b + a + d + g) + 10(d + e + f + b + e + h) + (g + h + i + c + f + i) = \\\\\n& = 200a ... | Spain | Olimpiada Matemática Española | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0acy | Calculate the angles of the triangle if it is known that the sum of two of its angles is $\frac{5}{6}$ from the right angle and one of these angles is 20° bigger than the other. | [
"We have that $\\alpha + \\beta = \\frac{5}{6} \\cdot 90^\\circ = 75^\\circ$ and $\\alpha - \\beta = 20^\\circ$.\n\nThen $\\beta = (75^\\circ - 20^\\circ) : 2 = 27^\\circ 30'$ and $\\alpha = 75^\\circ - 27^\\circ 30' = 47^\\circ 30'$, so we have that $\\gamma = 180^\\circ - (\\alpha + \\beta) = 180^\\circ - 75^\\ci... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 47° 30', 27° 30', 105° | |
0bky | Let $k \in \mathbb{N}$ and define the sets
$$A_k = \bigcup_{i=0}^{k+1} \{(m, n) \in \mathbb{N} \times \mathbb{N} \mid m \ge 2i,\ n \ge 2(k-i) + 1\},$$
$$B_k = \bigcup_{i=-1}^{k} \{(m, n) \in \mathbb{N} \times \mathbb{N} \mid m \ge 2i + 1,\ n \ge 2(k-i)\}.$$
Prove that
$$A_k \cup B_k = \{(m, n) \in \mathbb{N} \times \m... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0hr6 | Problem:
Eight friends, Aerith, Bob, Chebyshev, Descartes, Euler, Fermat, Gauss, and Hilbert, bought tickets for adjacent seats at the opera. However when they arrived they mixed up their seats:
- Bob sat in his assigned seat,
- Chebyshev sat two seats to the right of Gauss' assigned seat,
- Descartes sat one seat to ... | [
"Solution:\n\nNumber the seats $1$ through $8$ and let $a, \\ldots, h$ be the seat assignments. Let $A$ be the seat occupied by Aerith. As each seat is assigned to exactly one person we must have $a+\\cdots+h=1+\\cdots+8$. As each seat is occupied by exactly one person we must have\n$$\n\\begin{aligned}\n1+\\cdots+... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | Chebyshev | |
0d7w | There are totally 16 teams participating in a football tournament; each team playing with every other exactly 1 time. In each match, the winner gains 3 points, the loser gains 0 point and each team gains 1 point for the tie match. Suppose that at the end of the tournament, each team gains the same number of points. Pro... | [
"First, we can see that if a team got $x$ win matches, $y$ lose matches and $z$ tie matches then they got $3x + z$ points and $x + y + z = 15$. We call two teams having the same number of win matches, lose matches and tie matches as \"relate\".\n\nDenote the number of win matches, lose matches and tie matches of tw... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
05fs | Problem:
Résoudre $x^{4}-6 x^{2}+1=7 \times 2^{y}$ pour $x$ et $y$ entiers. | [
"Solution:\n\nL'équation se réécrit $\\left(x^{2}-3\\right)^{2}=7 \\times 2^{y}+8$.\n\nSi $y \\geq 3$, alors le côté droit s'écrit $8\\left(7 \\times 2^{y-3}+1\\right)$. Le côté gauche étant un carré, sa valuation 2-adique est paire. Ainsi, $7 \\times 2^{y-3}+1$ doit être pair, donc $y=3$. Dans ce cas $x^{2}-3=8$ d... | France | Envoi 1 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (3, 2), (-3, 2) | |
0kxm | Let $ABC$ be an equilateral triangle with side length $1$. Points $A_1$ and $A_2$ are chosen on side $BC$, points $B_1$ and $B_2$ are chosen on side $CA$, and points $C_1$ and $C_2$ are chosen on side $AB$ such that $BA_1 < BA_2$, $CB_1 < CB_2$, and $AC_1 < AC_2$.
Suppose that the three line segments $B_1C_2$, $C_1A_2$... | [
"**Claim** ($p = 1$ implies concurrence) — Suppose the six points are chosen so that triangles $AB_2C_1$, $BC_2A_1$, $CA_2B_1$ all have perimeter $1$. Then lines $\\overline{B_1C_2}$, $\\overline{C_1A_2}$, and $\\overline{A_1B_2}$ are concurrent.\n\n*Proof*. The perimeter conditions mean that $\\overline{B_2C_1}$, ... | United States | USA TSTST | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry >... | null | proof and answer | 1 | |
02w4 | Problem:
Determine se o número $\underbrace{11 \ldots 1}_{2016} 2 \underbrace{11 \ldots 1}_{2016}$ é um número primo ou um número composto. | [
"Solution:\nSeja $x=\\underbrace{11 \\ldots 1}_{2017}$. Daí,\n$$\n\\begin{aligned}\n\\underbrace{11 \\ldots 1}_{2016}2\\underbrace{11 \\ldots 1}_{2016} & =10^{2016} \\cdot x + x \\\\\n& = x\\left(10^{2016}+1\\right)\n\\end{aligned}\n$$\nComo $x$ e $10^{2016}+1$ são divisores maiores que 1 do número dado, podemos co... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | composite | |
0142 | Problem:
Let $\alpha$, $\beta$ and $\gamma$ be three angles with $0 \leq \alpha, \beta, \gamma < 90^\circ$ and $\sin \alpha + \sin \beta + \sin \gamma = 1$. Show that
$$
\tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma \geq \frac{3}{8}
$$ | [
"Solution:\nSince $\\tan^2 x = \\frac{1}{\\cos^2 x} - 1$, the inequality to be proved is equivalent to\n$$\n\\frac{1}{\\cos^2 \\alpha} + \\frac{1}{\\cos^2 \\beta} + \\frac{1}{\\cos^2 \\gamma} \\geq \\frac{27}{8}\n$$\nThe AM-HM inequality implies\n$$\n\\begin{aligned}\n\\frac{3}{\\frac{1}{\\cos^2 \\alpha} + \\frac{1... | Baltic Way | Baltic Way 2005 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
01wr | The medians $AA_1$ and $BB_1$ of the triangle $ABC$ intersect at the point $G$. Let $M$ and $N$ be the midpoints of the segments $GA$ and $GB$, and let $K$ and $L$ be the midpoints of the segments $CB_1$ and $CA_1$, respectively. The segments $KN$ and $LM$ intersect at the point $S$.
Find the ratio $CS : SG$. | [
"Answer: $CS : SG = 2 : 1$."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | proof and answer | 2:1 | |
0209 | Problem:
Find the smallest possible value of the expression
$$
\left\lfloor\frac{a+b+c}{d}\right\rfloor+\left\lfloor\frac{b+c+d}{a}\right\rfloor+\left\lfloor\frac{c+d+a}{b}\right\rfloor+\left\lfloor\frac{d+a+b}{c}\right\rfloor,
$$
in which $a$, $b$, $c$ and $d$ vary over the set of positive integers.
(Here $\lfloor x\r... | [
"Solution:\nThe answer is $9$.\n\nNotice that $\\lfloor x\\rfloor > x-1$ for all $x \\in \\mathbb{R}$. Therefore the given expression is strictly greater than\n$$\n\\frac{a+b+c}{d}+\\frac{b+c+d}{a}+\\frac{c+d+a}{b}+\\frac{d+a+b}{c}-4,\n$$\nwhich can be rewritten as\n$$\n\\left(\\frac{a}{b}+\\frac{b}{a}\\right)+\\le... | Benelux Mathematical Olympiad | Benelux Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 9 | |
006b | Se tiene una bolsa con 99 bolitas de diferentes colores (cada bolita tiene un solo color y se desconoce la cantidad de colores). Si se sacan de la bolsa 21 bolitas al azar, siempre hay cuatro o más de un mismo color. Decidir si es necesariamente cierto que la bolsa contiene 18 o más bolitas de un mismo color. ¿Y 17 o m... | [] | Argentina | Argentina 2008 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | Eighteen or more: no. Seventeen or more: yes. | |
04vq | Find all the pairs of natural numbers $(k, n)$ such that there exist natural numbers $a, b$ satisfying:
$$
\text{gcd}(a + k, b) = n \cdot \text{gcd}(a, b).
$$ | [
"We shall prove that every pair $(k, n)$ works.\n\nFirst, if $n = 1$, we can just take $(a, b) = (k, k)$, then\n$$\n\\text{gcd}(a + k, b) = \\text{gcd}(2k, k) = k = \\text{gcd}(k, k) = \\text{gcd}(a, b).\n$$\n\nNow, assume that $n > 1$, then $nk - k > 0$ is a natural number and we can take $(a, b) = ((n-1)k, nk)$. ... | Czech Republic | Second Round of the 73rd Czech and Slovak Mathematical Olympiad (January 16th, 2024) | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | All pairs of natural numbers. | |
08ig | Problem:
Find all the functions $f: N^{*} \rightarrow N^{*}$ which verify the relation $f(2x+3y) = 2f(x) + 3f(y) + 4$ for every positive integers $x, y \geq 1$. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | All solutions are f(n) = k·n − 1 with integer k ≥ 2. | |
0en7 | A school has $n$ students and there are some extra classes provided for them so that each student can participate in any number of them. We know that there are at least two participants in any class. We also know that if two different classes have two common students, then the numbers of their participants are differen... | [
"Let $A_i$, $2 \\le i \\le n$, be the set of all classes with $i$ participants. We will show that $|A_i| \\le \\frac{n(n-1)}{i(i-1)}$. Since every pair of students is together in at most one class of each $A_i$, we have that $\\binom{i}{2} |A_i| \\le \\binom{n}{2}$, giving\n$$\n|A_i| \\le \\frac{\\binom{n}{2}}{\\bi... | South Africa | South-Afrika 2011-2013 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
0fet | Problem:
Se considera la inecuación
$$
|x-1|<a x
$$
donde $a$ es un parámetro real.
a) Discutir la inecuación según los valores de $a$.
b) Caracterizar los valores de $a$ para los cuales la inecuación tiene exactamente DOS soluciones enteras. | [
"Solution:\n\na) En principio distinguiremos dos casos, según que $x \\geq 1$ ó $x<1$.\n\nCaso I: $x \\geq 1$. La desigualdad es equivalente a la siguiente:\n$$\nx-1<a x \\Leftrightarrow (1-a)x<1\n$$\nSubcaso I.1: Supongamos $1-a>0$, es decir, $a<1$. Entonces $x<\\frac{1}{1-a}$, $\\frac{1}{1-a}>1 \\Leftrightarrow a... | Spain | TANDA I | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1/2 < a ≤ 2/3 | |
0ck2 | Solve in $\mathbb{R}$ the equation:
$$
\frac{1}{\{x\}} + \frac{1}{[x]} + \frac{1}{x} = 0,
$$
where $[x]$ and $\{x\}$ denote the integer part and the fractional part of the real number $x$, respectively. | [
"First, we have that $x \\in \\mathbb{R} \\setminus (\\mathbb{Z} \\cup [0, 1))$. The given equation is equivalent to:\n$$\n\\frac{[x] + \\{x\\}}{[x]\\{x\\}} = -\\frac{1}{x} \\Leftrightarrow -x^2 = [x]\\{x\\}. \\quad (*)\n$$\nSince $-x^2 \\le 0$ and $x \\ne 0$, we have $[x]\\{x\\} < 0$, hence $[x] \\le -1$. Because ... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (1 - sqrt(5))/2, 1 - sqrt(5) | |
03rc | When the unit squares at the four corners are removed from a three by three square, the resulting shape is called a cross. What is the maximum number of non-overlapping crosses placed within the boundary of a $10 \times 11$ chessboard? (Each cross covers exactly five unit squares on the board.) (posed by Feng Zuming) | [
"The centers of the crosses (denoted by $*$) must lie in the $8 \\times 9$ subboard in the middle. We tile this central board by three $8 \\times 3$ boards, and label these three boards (a), (b) and (c), from left to right. We consider the number of centers placed in the three boards.\n\n\n... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 15 | |
0ie8 | Problem:
Let $AB$ be the diameter of a semicircle $\Gamma$. Two circles, $\omega_1$ and $\omega_2$, externally tangent to each other and internally tangent to $\Gamma$, are tangent to the line $AB$ at $P$ and $Q$, respectively, and to semicircular arc $AB$ at $C$ and $D$, respectively, with $AP < AQ$. Suppose $F$ lies... | [
"Solution:\n\nExtend the semicircle centered at $O$ to an entire circle $\\omega$, and let the reflection of $F$ over $AB$ be $F'$. Then $CQF'$ is a straight line. Also, the homothety centered at $C$ taking $\\omega_1$ into $\\omega$ takes $P$ to a point $X$ on $\\omega$ and $AB$ to the parallel line tangent to $\\... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 35 | |
07v2 | Let $n \ge 5$ be an odd number and $r$ an integer such that $1 \le r \le (n-1)/2$. In a sports tournament, $n$ players take part in a series of contests. Each contest involves $2r + 1$ players, and the scores obtained by the players are the numbers
$$
-r, -(r-1), \dots, -1, 0, 1, \dots, r-1, r
$$
in some order. Each po... | [
"First, note that each player will participate in $\\binom{n-1}{2r}$ contests. Therefore, the minimum possible score of any player is $-r\\binom{n-1}{2r} = -\\frac{n-1}{2}\\binom{n-2}{2r-1}$ and the maximum possible score of any player is $r\\binom{n-1}{2r} = \\frac{n-1}{2}\\binom{n-2}{2r-1}$. Thus the score of eac... | Ireland | IRL_ABooklet | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof and answer | binom(n-2, 2r-1) | |
0hyo | Problem:
Let $a, b, c, d, e, f$ be positive integers, each at least $2$, whose sum is $S$. Prove that
$$
a(a-1)+b(b-1)+c(c-1)+d(d-1)+e(e-1)+f(f-1) \leq (S-10)(S-11)+10.
$$
When is equality achieved? | [
"Solution:\nSolution I. Adding $-3S+24$ to both sides makes the inequality equivalent to\n$$\n(a-2)^2 + (b-2)^2 + (c-2)^2 + (d-2)^2 + (e-2)^2 + (f-2)^2 \\leq (S-12)^2.\n$$\nSubstituting $A = a-2$, $B = b-2$, etc., this is the same as\n$$\nA^2 + B^2 + C^2 + D^2 + E^2 + F^2 \\leq (A+B+C+D+E+F)^2.\n$$\nOn the left sid... | United States | BAMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | Equality holds if and only if five of the numbers are 2 and the sixth is S − 10. | |
06ys | Problem:
The function $f(n)$ is defined on the positive integers and takes non-negative integer values. It satisfies
(1) $f(m n) = f(m) + f(n)$,
(2) $f(n) = 0$ if the last digit of $n$ is $3$,
(3) $f(10) = 0$.
Find $f(1985)$. | [
"Solution:\n\nIf $f(m n) = 0$, then $f(m) + f(n) = 0$ (by (1)). But $f(m)$ and $f(n)$ are non-negative, so $f(m) = f(n) = 0$. Thus $f(10) = 0$ implies $f(5) = 0$. Similarly $f(3573) = 0$ by (2), so $f(397) = 0$. Hence $f(1985) = f(5) + f(397) = 0$."
] | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 0 | |
09cj | $a_n = a + a^{2n+1}$, $n \ge 1$ дараалал өгөгдөв.
$a_1, a_2, \dots, a_{2012}$ нь бүгд $a$-тай харилцан анхны хоёр бүхэл тооны квадратуудын нийлбэрт задач байх $a$ натурал тоо төгсгөлгүй олон олдохыг харуил. | [
"$a = p$ ба $p$ нь $4k + 1$ хэлбэрийн анхны тоо байхаар авья.\n\n1. $4k+1$ хэлбэрийн анхны тоо төгсгөлгүй олон (MMK-1.8(в))\n2. Аливаа $4k + 1$ хэлбэрийн анхны тоог 2 бүтэн квадратын нийлбэрт задалж болно. (MMK-II), тул\n\n$$\np = u^2 + v^2 \\text{ байх } \\exists u, v \\in \\mathbb{Z} \\text{ ба}\n$$\n\n$$\na_k = ... | Mongolia | ОУМО-53 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | Mongolian | proof only | null | |
0koh | Problem:
Suppose $\omega$ is a circle centered at $O$ with radius $8$. Let $AC$ and $BD$ be perpendicular chords of $\omega$. Let $P$ be a point inside quadrilateral $ABCD$ such that the circumcircles of triangles $ABP$ and $CDP$ are tangent, and the circumcircles of triangles $ADP$ and $BCP$ are tangent. If $AC=2\sqr... | [
"Solution:\n\nLet $X = AC \\cap BD$, $Q = AB \\cap CD$ and $R = BC \\cap AD$. Since $QA \\cdot QB = QC \\cdot QD$, $Q$ is on the radical axis of $(ABP)$ and $(CDP)$, so $Q$ lies on the common tangent at $P$. Thus, $QP^{2} = QA \\cdot QB$. Similarly, $RA \\cdot RC = RP^{2}$. Let $M$ be the Miquel point of quadrilate... | United States | HMMT February 2022 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals... | null | final answer only | 103360 | |
02kz | Problem:
Quais números naturais $m$ e $n$ satisfazem a $2^{n}+1=m^{2}$? | [
"Solution:\n\n$$\n2^{n}=m^{2}-1=(m+1)(m-1)\n$$\n\nTemos que $m-1$ e $m+1$ são potências de $2$ cuja diferença é $2$. Logo, a única solução possível é $m-1=2$ e $m+1=2^{2}$, donde $m=3$. Segue que $2^{n}+1=3^{2}$, e obtemos $n=3$."
] | Brazil | Nível 3 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | m=3, n=3 | |
062o | Problem:
Das Viereck $ABCD$ sei eine Raute mit spitzem Winkel bei $A$. Die Punkte $M$ und $N$ mögen so auf den Strecken $AC$ und $BC$ gelegen sein, dass $|DM| = |MN|$. Ferner sei $P$ der Schnittpunkt von $AC$ und $DN$ sowie $R$ der Schnittpunkt von $AB$ und $DM$. Man beweise, dass $|RP| = |PD|$. | [
"Solution:\n\nDer Fall $N = B$ sei im Folgenden ausgeschlossen. Dann ist $DN$ nicht orthogonal zu $AC$, und $M$ ist eindeutig charakterisiert als Schnittpunkt der Mittelsenkrechte von $DN$ mit $AC$. Im Dreieck $DNC$ ist $AC$ die Winkelhalbierende bei $C$, und in jedem Dreieck schneiden sich die Winkelhalbierende un... | Germany | 1. IMO-Auswahlklausur | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jg8 | Problem:
Determine all real values of $A$ for which there exist distinct complex numbers $x_{1}, x_{2}$ such that the following three equations hold:
$$
\begin{aligned}
x_{1}(x_{1}+1) & =A \\
x_{2}(x_{2}+1) & =A \\
x_{1}^{4}+3 x_{1}^{3}+5 x_{1} & =x_{2}^{4}+3 x_{2}^{3}+5 x_{2} .
\end{aligned}
$$ | [
"Solution:\n\nApplying polynomial division,\n$$\n\\begin{aligned}\nx_{1}^{4}+3 x_{1}^{3}+5 x_{1} & =\\left(x_{1}^{2}+x_{1}-A\\right)\\left(x_{1}^{2}+2 x_{1}+(A-2)\\right)+(A+7) x_{1}+A(A-2) \\\\\n& =(A+7) x_{1}+A(A-2)\n\\end{aligned}\n$$\nThus, in order for the last equation to hold, we need $(A+7) x_{1}=(A+7) x_{2... | United States | HMMT | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | -7 | |
02e9 | Given a point $p$ inside a convex polyhedron $P$. Show that there is a face $F$ of $P$ such that the foot of the perpendicular from $p$ to $F$ lies in the interior of $F$. | [
"Let $F$ be the face of $P$ closer to $p$ and $p'$ be the projection of $p$ in the plane of $F$. Suppose $p'$ lies outside of $F$. The line through $p$ and $p'$ cuts $P$ in two points $A$ and $B$. Let $A$ be the point between $p$ and $p'$. $A$ belongs to a face different from $F$ and the distance from $p$ to $A$ is... | Brazil | IX OBM | [
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof only | null | |
08ym | In a triangle $ABC$, we suppose that points $D$ and $E$ lie on the segments $AB$ and $AC$, respectively. Let $D$, $B$, $C$, $E$ lie on the same circumference and let point $P$ lie inside quadrilateral $DBCE$ with $\angle BDP = \angle BPC = \angle PEC$. Calculate $\frac{BP}{CP}$, given that $AB = 9$, $AC = 11$, $DP = 1$... | [
"$$\n\\frac{\\sqrt{33}}{11}\n$$\nLet $Q$ be the intersection of line $EP$ and $AB$, and $R$ be the intersection of line $DP$ and $AC$. Since $\\angle BPC + \\angle CPE = \\angle PEC + \\angle CPE = 180^\\circ - \\angle ECP$, we have $\\angle QPB = \\angle ECP$. Similarly, we have $\\angle RPC = \\angle DBP$. Hence ... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | sqrt(33)/11 | |
0d7h | In a school, there are totally $n$ students, with $n \geq 2$. The students take part in $m$ clubs and in each club, there are at least $2$ members (a student may take part in more than $1$ club). Eventually, the Principal notices that: If $2$ clubs share at least $2$ common members then they have different numbers of m... | [
"Let $a_i$ be the number of clubs that have $i$ members. Here, $2 \\leq i \\leq n$ and\n$$\nm = a_2 + a_3 + \\cdots + a_n.\n$$\nWe will count the tuples $(A, B, C)$ in which the students $A, B$ take part in the same club $C$ that has $i$ members.\n\n1. The first way of counting:\n- Choosing $1$ club among the clubs... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof only | null | |
0dvq | Problem:
Naj bo $D$ razpolovišče hipotenuze $AB$ pravokotnega trikotnika $ABC$. Označimo z $O_1$ in $O_2$ središči trikotnikoma $ADC$ in $DBC$ očrtanih krožnic. Dokaži, da je $AB$ tangenta na krožnico s premerom $O_1O_2$. | [
"Solution:\n\nOznačimo s $k$ krožnico, ki se dotika stranice $AB$ v $D$ in na kateri leži točka $C$. Naj bo $O$ središče krožnice $k$. Potem so točke $O$, $O_1$ in $O_2$ kolinearne, saj ležijo na simetrali daljice $CD$.\n\nOznačimo $\\angle BAC = \\alpha$. Potem je središčni kot $\\angle DO_1C$ krožnice $k$ enak $2... | Slovenia | 47. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0279 | Problem:
Cortando papéis - No início de uma brincadeira, André tinha sete pedaços de papel. Na primeira rodada da brincadeira, ele pegou alguns destes pedaços e cortou cada um deles em sete pedaços, que foram misturados aos pedaços de papel que não foram cortados nessa rodada. Na segunda rodada, ele novamente pegou al... | [
"Solution:\n\nSe na primeira rodada André pega $n_{1}$ pedaços de papel para cortar cada um deles em sete pedaços, ao final dessa rodada ele ficará com $7-n_{1}$ pedaços sem cortar, mais $7 n_{1}$ pedaços cortados, totalizando $(7-n_{1})+7 n_{1}=7+6 n_{1}$ pedaços de papel. Analogamente, se na segunda rodada André ... | Brazil | Nível 2 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | null | proof and answer | No | |
0fkk | Problem:
Sea $p \geq 3$ un número primo. Se divide cada lado de un triángulo en $p$ partes iguales y se une cada uno de los puntos de división con el vértice opuesto. Calcula el número máximo de regiones, disjuntas dos a dos, en que queda dividido el triángulo. | [
"Solution:\n\nEn primer lugar veremos que tres de estos segmentos (cevianas) no pueden ser concurrentes. Sea el triángulo $ABC$ y $X, Y, Z$ puntos de las divisiones interiores de los lados $BC, AC, AB$ respectivamente. Si $AX, BY$ y $CZ$ fueran concurrentes aplicando el teorema de Ceva tendríamos\n$$\n\\frac{AZ}{ZB... | Spain | XLIV Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 3p^2 - 3p + 1 | |
02t5 | Problem:
No desenho abaixo, $ABCD$ é um retângulo e os pontos $P$ e $Q$ pertencem à diagonal $AC$ de modo que $AQ = PQ = PC = 1$ e $\angle AQD = \angle BPC = 90^\circ$. Encontre a área do retângulo $ABCD$.
 | [
"Solution:\n\nPelas relações métricas no triângulo retângulo, temos $DQ^{2} = AQ \\cdot QC = 2$. Pelo Teorema de Pitágoras nos triângulos $\\triangle DAQ$ e $\\triangle DQC$, temos:\n$$\n\\begin{aligned}\nAD^{2} & = DQ^{2} + AQ^{2} \\\\\n& = 2 + 1 \\\\\n& = 3 \\\\\nDC^{2} & = DQ^{2} + QC^{2} \\\\\n& = 2 + 4 \\\\\n&... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 3*sqrt(2) | |
0awr | Problem:
In how many ways can nine chips be selected from a bag that contains three red chips, three blue chips, three white chips, and three yellow chips? (Assume that the order of selection is irrelevant and that the chips are identical except for their color.) | [
"Solution:\n\nLet $r$, $b$, $w$, and $y$ denote the number of red, blue, white, and yellow chips selected, respectively. We want the number of integer solutions to\n$$\nr + b + w + y = 9\n$$\nsubject to $0 \\leq r \\leq 3$, $0 \\leq b \\leq 3$, $0 \\leq w \\leq 3$, $0 \\leq y \\leq 3$.\n\nSince there are only 3 of ... | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof and answer | 20 | |
059l | Mari chooses five distinct positive integers not greater than $2021$. From these five numbers, it must be possible to choose two numbers with sum $1919$ in two different ways. Likewise, from these five numbers, it must be possible to choose two numbers with sum $2929$ in two different ways. Find all possibilities of wh... | [
"Answer: $1, 908, 1011, 1918, 2021$ is the only possibility.\n\nLet $(a, 1919 - a)$ and $(b, 1919 - b)$ be the two pairs of numbers with sum $1919$. If these sums had a common addend then both addends would be the same whence the choices of two numbers would not be different. Thus $a, b, 1919 - a$ and $1919 - b$ ar... | Estonia | Estonian Math Competitions | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 1, 908, 1011, 1918, 2021 | |
0gip | 令 $n$ 為正整數。我們稱一個嚴格遞增的等差數列 $x_0, x_1, \dots, x_n$ 為 $n$-數列，若且唯若存在正整數 $a_1, a_2, \dots, a_n, b_1, b_2, \dots, b_n$，滿足
$$
x_0 = a_1 \times a_2 \times a_3 \times \dots \times a_n,
$$
$$
x_1 = b_1 \times a_2 \times a_3 \times \dots \times a_n,
$$
$$
x_2 = b_1 \times b_2 \times a_3 \times \dots \times a_n,
$$
$\vdots$
$$
x_n... | [
"最小可能公差為 $n!$。注意到公差為\n$$\nD = (b_1 - a_1)a_2a_3 \\cdots a_n = b_1(b_2 - a_1)a_3a_4 \\cdots = \\cdots = b_1b_2 \\cdots b_{n-1}(b_n - a_n).\n$$\n又由於數列嚴格遞增，$D > 0$，從而 $b_i > a_i$ 對所有 $i$ 都成立。故上式等價於\n$$\n(b_i - a_i)a_{i+1} = b_i(b_{i+1} - a_{i+1}) \\text{ 對所有 } i \\text{ 皆成立。}\n$$\n此外，由於若 $k_i = \\text{gcd}(a_i, b_i)$，... | Taiwan | IMO 2J, Mock Exam 2 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Chinese; English | proof and answer | n! | |
033u | Problem:
In every cell of an $n \times n$ table one of the numbers $-1, 0$ and $1$ is written. Is it possible the sums of the numbers in every row and every column to be $2n$ mutually different numbers, if:
a) $n=4$;
b) $n=5$? | [
"Solution:\n\na) Yes. Here is an example:\n$$\n\\left(\\begin{array}{rrrr}\n1 & 0 & 1 & 1 \\\\\n1 & -1 & -1 & -1 \\\\\n1 & -1 & 1 & 0 \\\\\n1 & -1 & 1 & -1\n\\end{array}\\right)\n$$\n\nb) No. We have 11 possibilities for these sums: $0, \\pm 1, \\pm 2, \\pm 3, \\pm 4$ and $\\pm 5$. Denote by $a_{i}$ the sum of the ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) Yes; b) No | |
0lag | A sequence of real numbers $(x_n)$ is given by
$$
x_1 = \frac{1}{2} \quad \text{and} \quad x_n = \frac{\sqrt{x_{n-1}^2 + 4x_{n-1}} + x_{n-1}}{2} \quad \text{for all } n \ge 2.
$$
For each non-negative integer $n$, let
$$
y_n = \sum_{i=1}^{n} \frac{1}{x_i^2}.
$$
Show that the sequence $(y_n)$ has a finite limit when $n ... | [
"From its definition, it is easy to see that $x_n > 0$ for all $n \\ge 1$.\n\nReformulate the defining relation for the sequence $(x_n)$ in the following form:\n$$\n2x_n - x_{n-1} = \\sqrt{x_{n-1}^2 + 4x_{n-1}} \\quad \\forall n \\ge 2.\n$$\nIt follows that:\n$$\nx_{n-1} = x_n^2 - x_n x_{n-1} \\quad \\forall n \\ge... | Vietnam | Vijetnam 2009 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 6 | |
0bn4 | Find all integers $x, y$ such that
$$
5^x - \log_2 (y+3) = 3^y \quad \text{and} \quad 5^y - \log_2 (x+3) = 3^x.
$$ | [
"Answer: $x = y = 1$.\n\nSubtracting the equalities yields\n$$\n5^x + 3^x + \\log_2 (x+3) = 5^y + 3^y + \\log_2 (y+3).\n$$\nFunction $t \\mapsto 5^t + 3^t + \\log_2 (t+3)$ is increasing, so the hypothesis leads to $x = y$ and becomes now $5^x = 3^x + \\log_2 (x+3)$, with integer $x$. We notice that $x > -3, -2, -1,... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | x = y = 1 | |
0cvw | Let $n$ be a positive integer. Compose a $3 \times 3 \times 3$ cube of 26 white unit cubes and one black unit cube by putting the black one into the center. Compose a $3n \times 3n \times 3n$ cube of $n^3$ such $3 \times 3 \times 3$ cubes. Determine the smallest number $k$ such that it is possible to paint $k$ white un... | [
"Введём систему координат так, чтобы центры кубиков имеют координаты от $1$ до $3n$ по каждой оси. Каждому кубику приписываем координаты его центра. Таким образом, кубик чёрный тогда и только тогда, когда все его координаты дают остаток $2$ при делении на $3$.\n\nОкрасим красным все белые кубики с координатами $(a,... | Russia | Final round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English; Russian | proof and answer | (n+1)n^2 | |
04rf | Given a parallelogram $ABCD$ with center $S$, denote by $O$ the incenter of triangle $ABD$ and by $T$ the point of contact of the incircle of triangle $ABD$ with the diagonal $BD$. Prove that lines $OS$ and $CT$ are parallel. (Jaromír Šimša) | [
"Denote the lengths of $AB$, $AD$, and $BD$ by $a$, $b$, and $c$, respectively. If $a = b$ then both $OS$ and $CT$ coincide with $AC$ and the conclusion is trivial. Suppose $a > b$ (the case $b > a$ being completely analogous).\nLet $T'$ be the reflection of $T$ in $S$ (Fig. 1). As $CT \\parallel AT'$, it suffices ... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plan... | null | proof only | null | |
00m4 | Es sei $ABCD$ ein konvexes Sehnenviereck mit dem Umkreismittelpunkt $U$, in dem die Diagonalen aufeinander normal stehen. Es sei $g$ die Gerade, die man erhält, wenn man die Diagonale $AC$ an der Winkelsymmetrale von $\prec BAD$ spiegelt.
Man zeige, dass der Punkt $U$ auf der Geraden $g$ liegt. | [
"Es sei $X$ der Diagonalenschnittpunkt des Sehnenvierecks $ABCD$ und $E$ der zweite Schnittpunkt der Geraden $g$ mit dem Umkreis $k$ von $ABCD$.\nDie Gerade $g$ erhält man durch Spiegelung der Diagonale $AC$ an der Winkelsymmetrale $w_\\alpha$ von $\\prec BAD$. Daraus folgt\n$$\n\\prec EAD = \\prec BAC = \\prec BAX... | Austria | 48. Österreichische Mathematik-Olympiade | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triang... | German | proof only | null | |
0jws | Problem:
Trapezoid $ABCD$, with bases $AB$ and $CD$, has side lengths $AB = 28$, $BC = 13$, $CD = 14$, and $DA = 15$. Let diagonals $AC$ and $BD$ intersect at $P$, and let $E$ and $F$ be the midpoints of $AP$ and $BP$, respectively. Find the area of quadrilateral $CDEF$. | [
"Solution:\n\nNote that $EF$ is a midline of triangle $APB$, so $EF$ is parallel to $AB$ and $EF = \\frac{1}{2} AB = 14 = CD$. We also have that $EF$ is parallel to $CD$, and so $CDEF$ is a parallelogram. From this, we have $EP = PC$ as well, so $\\frac{CE}{CA} = \\frac{2}{3}$. It follows that the height from $C$ t... | United States | HMMT November | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 112 | |
0e8b | Problem:
Naj bo $D$ razpolovišče stranice $BC$, $E$ razpolovišče stranice $CA$ in $T$ težišče trikotnika $ABC$. Premice $AT$, $BT$ in $CT$ naj sekajo trikotniku $ABC$ očrtano krožnico še v točkah $P$, $Q$ in $R$. Denimo, da je $\angle ACB = \angle RQP$. Dokaži, da je štirikotnik $DCE T$ tetiven. | [
"Solution:\n\n\n\nZaradi tetivnosti štirikotnika $RBCQ$ je $\\angle RQB = \\angle RCB$, torej iz pogoja naloge sledi $\\angle ECT = \\angle ACR = \\angle BQP$. Ker je štirikotnik $QABP$ tetiven, pa je $\\angle BQP = \\angle BAP$. Ker sta $E$ in $D$ razpolovišči stranic, sta premici $AB$ in ... | Slovenia | 57. matematično tekmovanje srednješolcev Slovenije | [
"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 | |
0ipx | Problem:
Let $ABC$ be a right triangle with $\angle A = 90^{\circ}$. Let $D$ be the midpoint of $AB$ and let $E$ be a point on segment $AC$ such that $AD = AE$. Let $BE$ meet $CD$ at $F$. If $\angle BFC = 135^{\circ}$, determine $BC / AB$. | [
"Solution:\nAnswer: $\\frac{\\sqrt{13}}{2}$\n\nLet $\\alpha = \\angle ADC$ and $\\beta = \\angle ABE$. By the exterior angle theorem, $\\alpha = \\angle BFD + \\beta = 45^{\\circ} + \\beta$. Also, note that $\\tan \\beta = AE / AB = AD / AB = 1/2$. Thus,\n$$\n1 = \\tan 45^{\\circ} = \\tan (\\alpha - \\beta) = \\fra... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | sqrt(13)/2 | |
0a3c | Problem:
Vind alle functies $f: \mathbb{R}_{\geq 0} \rightarrow \mathbb{R}$ die voldoen aan
$$
2 x^{3} z f(z)+y f(y) \geq 3 y z^{2} f(x)
$$
voor alle $x, y, z \in \mathbb{R}_{\geq 0}$. | [
"Solution:\nAntwoord: alle functies van de vorm $f_{c, d}(x)=\\left\\{\\begin{array}{ll}c x^{2} & \\text{als } x>0 \\\\ d & \\text{als } x=0\\end{array}\\right.$ met $c \\geq 0$ en $d \\leq 0$.\n\nInvullen van $x=0$ en $y=1$ levert $f(1) \\geq 3 f(0) z^{2}$ voor alle $z \\geq 0$. Als $f(0)>0$, dan is de rechterkant... | Netherlands | IMO-selectietoets | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | All functions of the form f(x) = c x^2 for x > 0 and f(0) = d, with c ≥ 0 and d ≤ 0. | |
0bma | If $k$ and $n$ are positive integers, and $k \le n$, let $M(n, k)$ denote the least common multiple of the numbers $n, n-1, \dots, n-k+1$. Let $f(n)$ be the largest positive integer $k \le n$ such that $M(n, 1) < M(n, 2) < \dots < M(n, k)$. Prove that:
a) $f(n) < 3\sqrt{n}$, for all positive integers $n$;
b) if $N$ i... | [
"a) Clearly, $f(1) = 1$. Notice that\n$$\nM(n, k + 1) = \\operatorname{lcm}(M(n, k), n - k), \\quad 1 \\le k < n. \\quad (*)\n$$\nThus, $M(n, k) \\le M(n, k+1)$, and equality holds if and only if $n-k$ divides $M(n, k)$. If $m > 1$, then $M(m^2, 2) = m^2(m^2 - 1) = (m^2 - m)(m^2 + m)$, so $M(m^2, 2)$ is divisible b... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0k3z | Problem:
Compute the smallest positive integer $n$ for which
$$
\sqrt{100+\sqrt{n}}+\sqrt{100-\sqrt{n}}
$$
is an integer. | [
"Solution:\nThe number $\\sqrt{100+\\sqrt{n}}+\\sqrt{100-\\sqrt{n}}$ is a positive integer if and only if its square is a perfect square. We have\n$$\n\\begin{aligned}\n(\\sqrt{100+\\sqrt{n}}+\\sqrt{100-\\sqrt{n}})^2 & =(100+\\sqrt{n})+(100-\\sqrt{n})+2 \\sqrt{(100+\\sqrt{n})(100-\\sqrt{n})} \\\\\n& =200+2 \\sqrt{1... | United States | HMMT November 2018 | [
"Algebra > Intermediate Algebra > Other",
"Number Theory > Other"
] | null | final answer only | 6156 | |
0fmr | Prueba que las sumas de las primeras, segundas y terceras potencias de las raíces del polinomio $p(x) = x^3 + 2x^2 + 3x + 4$ valen lo mismo. | [
"Sean $r, s$ y $t$ las raíces, reales o complejas, del polinomio $p(x)$ y sea $S_n$ la suma de sus $n$-ésimas potencias, esto es, $S_n = r^n + s^n + t^n$. Por un lado, teniendo en cuenta las fórmulas de Cardano-Vièta resulta que $S_1 = r + s + t = -2$, $rs + st + tr = 3$ y $rst = -4$ lo que nos permite calcular $S_... | Spain | Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | Spanish | proof only | null | |
0dv4 | Problem:
Dani so štirje pravokotniki z dolžino $a=18$. Njihove širine tvorijo geometrijsko zaporedje. Obseg drugega pravokotnika je 60, tretji pravokotnik je kvadrat. Določi širine pravokotnikov. | [
"Solution:\n\nOznačimo širine geometrijskega zaporedja z $b$, $bq$, $bq^2$, $bq^3$.\n\nObseg drugega pravokotnika je $2a + 2bq = 60$.\n\n$a + bq = 30 \\Rightarrow bq = 12$\n\nTretji pravokotnik je kvadrat: $a = bq^2 \\Rightarrow bq^2 = 18$\n\nRešimo sistem:\n\n$12 \\cdot q = 18 \\Rightarrow q = \\frac{3}{2}$\n\n$b ... | Slovenia | 2. matematično tekmovanje dijakov srednjih tehniških in strokovnih šol | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 8, 12, 18, 27 | |
0gt6 | In an acute triangle $ABC$ points $D$ and $E$ are on the sides $[BC]$ and $[AC]$, respectively, such that $BD$ and $CE$ are angle bisectors. Projections of $D$ onto $BC$ and $BA$ are $P$ and $Q$, respectively, projections of $E$ onto $CA$ and $CB$ are $R$ and $S$, respectively. Let $X$ be intersection of $AP$ and $CQ$,... | [
"Let $H$ be the projection of $B$ onto $AC$. Since $B, P, Q, H, D$ lie on the semicircle of diameter $[BD]$, we obtain\n$$\n\\angle PHB = \\angle PDB = 90^\\circ - \\frac{\\hat{B}}{2} = \\angle QDB = \\angle QHB.\n$$\n\n\n\nWe will show that if $BH \\perp AC$ and $\\angle PHB = \\angle QHB$... | Turkey | Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadr... | null | proof only | null | |
01ds | Let $n$ be a positive integer and let $a, b, c, d$ be integers such that $n|a + b + c + d$ and $n|a^2 + b^2 + c^2 + d^2$. Show that
$$
n|a^4 + b^4 + c^4 + d^4 + 4abcd.
$$ | [
"Let\n$$\nw(x) = (x-a)(x-b)(x-c)(x-d) = x^4 + A x^3 + B x^2 + C x + D.\n$$\nIt is clear that $w(a) = w(b) = w(c) = w(d) = 0$. By adding these values we get\n$$\n\\begin{aligned}\nw(a) + w(b) + w(c) + w(d) = a^4 + b^4 + c^4 + d^4 \\\\\n\\quad + A(a^3 + b^3 + c^3 + d^3) + B(a^2 + b^2 + c^2 + d^2) \\\\\n\\quad + C(a +... | Baltic Way | Baltic Way 2016 | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0ldq | A sequence $(x_n)$ is defined as follows
$$
x_1 = 2, \quad x_{n+1} = \sqrt{x_n + 8} - \sqrt{x_n + 3}
$$
for all positive integers $n$.
a) Prove that $(x_n)$ has a finite limit and find that limit.
b) For every positive integer $n$, prove that
$$
n \le x_1 + x_2 + \dots + x_n \le n + 1.
$$ | [
"a) It is easy to see that $x_n > 0$ for all $n \\in \\mathbb{N}^*$. For every positive integer $n$, we have\n$$\n\\begin{aligned}\n|x_{n+1} - 1| &= |\\sqrt{x_n + 8} - 3 + 2 - \\sqrt{x_n + 3}| \\\\\n&= |(x_n - 1)\\left(\\frac{1}{\\sqrt{x_n + 8} + 3} - \\frac{1}{\\sqrt{x_n + 3} + 2}\\right)| \\\\\n&\\le |x_n - 1|\\l... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | Limit = 1; and for all positive integers n, n ≤ x_1 + x_2 + ⋯ + x_n ≤ n + 1. | |
0kl8 | Problem:
Let $ABCD$ be a trapezoid with $AB \parallel CD$, $AB=5$, $BC=9$, $CD=10$, and $DA=7$. Lines $BC$ and $DA$ intersect at point $E$. Let $M$ be the midpoint of $CD$, and let $N$ be the intersection of the circumcircles of $\triangle BMC$ and $\triangle DMA$ (other than $M$). If $EN^{2}=\frac{a}{b}$ for relative... | [
"Solution:\n\nFrom $\\triangle EAB \\sim \\triangle EDC$ with length ratio $1:2$, we have $EA=7$ and $EB=9$. This means that $A$, $B$, $M$ are the midpoints of the sides of $\\triangle ECD$. Let $N'$ be the circumcenter of $\\triangle ECD$. Since $N'$ is on the perpendicular bisectors of $EC$ and $CD$, we have $\\a... | United States | HMMT Spring 2021 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Ang... | null | proof and answer | 90011 | |
0f3z | Problem:
Find all solutions $(x, y)$ in positive integers to $x^{3} - y^{3} = xy + 61$. | [
"Solution:\nPut $x = y + a$. Then $(3a - 1)y^{2} + a(3a - 1)y + (a^{3} - 61) = 0$. The first two terms are positive, so the last term must be negative, so $a = 1, 2, 3$. Trying each case in turn, we get $(y + 6)(y - 5) = 0$, $5y^{2} + 10y - 53 = 0$, $4y^{2} + 12y - 17 = 0$. The last two equations have no integer so... | Soviet Union | 15th ASU | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (6, 5) | |
0hun | Problem:
Prove that there exist pairwise distinct positive integers $a_{0}, a_{1}, a_{2}, \ldots, a_{1000}$ such that
$$
a_{0}! = a_{1}!\, a_{2}!\, \ldots\, a_{1000}!
$$
Here $n! = 1 \times 2 \times \cdots \times n$ as usual. | [
"Solution:\nWe proceed by induction on $n \\geq 2$. First, we can have $a_{1}=3$, $a_{2}=5$ and $a_{0}=6$. Now, given a working tuple $\\left(a_{0}, a_{1}, \\ldots, a_{n}\\right)$, note that the tuple\n$$\n\\left(a_{0}!, a_{1}, \\ldots, a_{n},\\left(a_{0}-1\\right)!\\right)\n$$\nis a working tuple of length $n+1$. ... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0ido | Problem:
Yet another trapezoid $ABCD$ has $AD$ parallel to $BC$. $AC$ and $BD$ intersect at $P$. If $[ADP]/[BCP] = 1/2$, find $[ADP]/[ABCD]$. (Here the notation $[P_1 \cdots P_n]$ denotes the area of the polygon $P_1 \cdots P_n$.) | [
"Solution:\n\nA homothety (scaling) about $P$ takes triangle $ADP$ into $BCP$, since $AD$, $BC$ are parallel and $A, P, C; B, P, D$ are collinear. The ratio of homothety is thus $\\sqrt{2}$. It follows that, if we rescale to put $[ADP]=1$, then $[ABP]=[CDP]=\\sqrt{2}$, just by the ratios of lengths of bases. So $[A... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 3 - 2√2 | |
0e9w | Problem:
Za realno število $x$ označimo $[x]$ največje celo število, ki ni večje od $x$.
a. Dokaži, da za vsa naravna števila $a$, $b$ in $c$ velja
$$
\left[\frac{\left[\frac{c}{a}\right]}{b}\right]=\left[\frac{c}{ab}\right]
$$
b. S primerom pokaži, da gornja enakost ne velja za vsa pozitivna realna števila $a$, $b$... | [
"Solution:\n\na.\nŠtevilo $c$ lahko zapišemo v obliki $c = k a b + r$, kjer je $k$ neko nenegativno celo število, $r < a b$ pa ostanek števila $c$ pri deljenju z $a b$. Število $r$ lahko nadalje zapišemo v obliki $r = m a + n$, kjer je $m$ neko nenegativno celo število, $n < a$ pa ostanek števila $r$ pri deljenju z... | Slovenia | 58. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | The identity holds for all natural numbers. A counterexample for positive reals is a equals 2, b equals one half, c equals 1. | |
0crs | Положительные рациональные числа $a$ и $b$ записаны в виде десятичных дробей, у каждой из которых минимальный период состоит из 30 цифр. У десятичной записи числа $a-b$ длина минимального периода равна 15. При каком наименьшем натуральном $k$ длина минимального периода десятичной записи числа $a+kb$ может также оказать... | [
"Ответ. $k=6$.\n\nДомножив, если нужно, числа $a$ и $b$ на подходящую степень десятки, мы можем считать, что десятичные записи чисел $a, b, a-b$ и $a+kb$ — чисто периодические (то есть периоды начинаются сразу после запятой).\n\nВоспользуемся следующим известным фактом: десятичная запись рационального числа $r$ — ч... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 6 | |
026l | Problem:
Equação de duas variáveis - Determine todos os pares de inteiros $(x, y)$ tais que $9 x y - x^{2} - 8 y^{2} = 2005$. | [
"Solution:\n\nTemos:\n$$\n\\begin{aligned}\n9 x y - x^{2} - 8 y^{2} = 2005 &\\Leftrightarrow x y - x^{2} + 8 x y - 8 y^{2} = 2005 \\\\\n&\\Leftrightarrow x(y - x) + 8 y(x - y) = 2005 \\\\\n&\\Leftrightarrow (x - y)(8 y - x) = 2005\\ (*)\n\\end{aligned}\n$$\nObservemos que a fatoração em primos de $2005$ é $5 \\cdot... | Brazil | Nível 3 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (63, 58), (459, 58), (-63, -58), (-459, -58) | |
0gh2 | 死靈法師和聖騎士在一個 $666 \times 666$ 的方陣上大戰。方陣上原本空無一物。每一回合，死靈法師先在方陣上指定一格，並在以該格為中心的 $3 \times 3$ 範圍內的所有格子（至多九格）各增加一隻骷髏，然後聖騎士指定方陣上任意四格，並從這四格中各消滅一隻骷髏（格子中的骷髏數是非負整數）。
如果在某回合，一個格子內有 $10^6$ 或更多骷髏，則稱這個格子在該回合處於完蛋狀態。試求最大的正整數 $K$，使得死靈法師有策略能保證，不論聖騎士如何行動，都能在某個有限回合內，讓方陣上同時存在至少 $K$ 個完蛋的格子。
A necromancer and a paladin combat on a $666 \times... | [
"最大的 $K = 5 \\times 222^2$；一般性的，對於 $3N \\times 3N$ 的方陣，$K = 5N^2$。\n\n\n\n首先證明 $K \\le 5N^2$。如上圖將方陣著色，並注意到每回合死靈法師只能在至多四個白色格子上增加骷髏，因此聖騎士永遠可以讓白色區域的骷髏數為 $0$，從而死靈法師有可能保證製造出完蛋的格子，至多只能是灰色格子數，也就是 $5N^2$。\n\n接著證明 $K \\ge 5N^2$。事實上，我們可以考慮以下的加強版遊戲：當輪到聖騎士時，他改為選擇死靈法師剛剛增加骷髏的 $3 \\times 3$ 區域中的五個格子，並將全方陣... | Taiwan | 2023 數學奧林匹亞競賽第二階段選訓營 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressio... | Chinese (Traditional) | proof and answer | 246420 | |
0e4o | How many pairs $(m, n)$ of positive integers satisfy the condition $\frac{3}{m} + \frac{2}{n} = 1$?
(A) 2
(B) 3
(C) 4
(D) 5
(E) More than 5. | [
"Clearly $m > 3$ and $n > 2$. Since $n$ decreases as $m$ increases $n$, we see that the only possible solutions are $(m, n) \\in \\{(4, 8), (5, 5), (6, 4), (9, 3)\\}$."
] | Slovenia | National Math Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | MCQ | C | |
00ra | Two positive integers $m$ and $n$ will be called *anagrams*, if each decimal digit $a$ appears as many times in the decimal representation of $m$ as in that of $n$. Is it possible to find four different positive integers such that each of them is an anagram of the sum of the other three? | [
"Let $p$ be a prime number such that its index modulo $10$ be equal to $p-1$ (i.e. the numbers $0$, $1$, $10$, $\\ldots$, $10^{p-2}$ form a complete residue system modulo $p$.) Let $N(p)$ be the number $\\frac{10^{p-1}-1}{p}$ with added leading zeroes in order to be a $(p-1)$-digit number. Then the numbers $iN(p)$ ... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n"
] | null | proof and answer | Yes. For example, let s = 142857 and t = (10^16 − 1)/17. Then the four integers 10^16·s + i·t for i = 1, 2, 3, 4 satisfy the condition. | |
01u5 | A positive integer is called *nice* if it is equal to the sum of the fourth powers of certain five distinct its divisors. (A divisor may be equal to $1$ or to the number itself.)
a) Prove that any nice number is divisible by $5$.
b) Are there infinitely many nice numbers? | [
"Answer: b) there are an infinite number of nice numbers.\n\na) Let $N$ be a nice number, i.e. $N = d_1^4 + d_2^4 + d_3^4 + d_4^4 + d_5^4$, where $d_i$, $i = 1, 2, 3, 4, 5$, are the distinct divisors of $N$. If some divisor of $N$ is divisible by $5$, then $N$ is divisible by $5$.\nSo we suppose that $d_1, d_2, d_3... | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | a) Every nice number is divisible by five. b) Yes, there are infinitely many nice numbers. | |
0022 | El entero positivo $n$ tiene exactamente 18 divisores positivos, contando $1$ y $n$. Se numeran los divisores de $n$ de menor a mayor (el primero es $1$ y el décimo octavo es $n$) y se denota $x$ al sexto de estos divisores. Se sabe que el decimotercer divisor, multiplicado por la suma del primero más el segundo más el... | [] | Argentina | XX OLIMPIADA MATEMÁTICA ARGENTINA | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | español | proof and answer | 3332 | |
09pb | Prove that the inequality
$$
\frac{\cos \alpha}{\sin \alpha} + \frac{\cos \beta}{\sin \beta} + \frac{\cos \gamma}{\sin \gamma} \ge 2
$$
holds for all non-obtuse triangles with angles $\alpha$, $\beta$, $\gamma$ satisfying $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. | [] | Mongolia | MMO2025 Round 2 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0cyg | Let $I$ be the incenter of a triangle $ABC$ and let $A'$, $B'$, $C'$ be midpoints of sides $BC$, $CA$, $AB$, respectively. If $IA' = IB' = IC'$, then prove that triangle $ABC$ is equilateral. | [
"Let $A_1$, $B_1$, $C_1$ be the tangency points of the incircle of triangle $ABC$ with the sides $BC$, $CA$, $AB$, respectively.\n\n\n\nSince $IA' = IB' = IC'$ it follows that triangles $IA_1A'$, $IB_1B'$, and $IC_1C'$ are congruent. We get $A_1A' = B_1B' = C_1C'$, hence\n$$\n\\begin{equati... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents... | English | proof only | null | |
0kcq | Problem:
Let $f(n)$ be the number of distinct prime divisors of $n$ less than $6$. Compute
$$
\sum_{n=1}^{2020} f(n)^2
$$ | [
"Solution:\nDefine\n$$\n\\mathbf{1}_{a \\mid n}= \\begin{cases}1 & a \\mid n \\\\ 0 & \\text{otherwise}\\end{cases}\n$$\nThen\n$$\n\\begin{aligned}\nf(n)^2 & =\\left(\\mathbf{1}_{2 \\mid n}+\\mathbf{1}_{3 \\mid n}+\\mathbf{1}_{5 \\mid n}\\right)^2 \\\\\n& =\\mathbf{1}_{2 \\mid n}+\\mathbf{1}_{3 \\mid n}+\\mathbf{1}... | United States | HMMO 2020 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | final answer only | 3431 | |
087x | Problem:
È dato un trapezio con le basi lunghe $1$ e $4$, rispettivamente. Lo suddividiamo in due trapezi mediante un taglio parallelo alle basi, lungo $3$. Vogliamo ora suddividere i due nuovi trapezi, sempre mediante tagli paralleli alle basi, in $m$ ed $n$ trapezi, rispettivamente, in modo che tutti gli $m+n$ trape... | [
"Solution:\n\nSia $ABCD$ il trapezio, con $AB$ base maggiore, e siano $P$ e $Q$ gli estremi del taglio già effettuato, posti rispettivamente su $AD$ e $BC$. Si prolunghino i lati obliqui fino a farli incontrare in un punto che chiamiamo $E$; i triangoli $DCE$, $PQE$, $ABE$ sono simili (hanno tutti gli angoli congru... | Italy | Cesenatico | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Minimum m+n = 15, with cuts of lengths sqrt(2), sqrt(3), ..., sqrt(15) (including the initial cut of length 3 = sqrt(9)). | |
0d0k | Prove that for every real number $x$ the following inequality holds:
$$
x^6 + x^4 - x^3 - x + \frac{3}{4} > 0.
$$ | [
"The inequality is equivalent to\n$$\nx^6 - x^3 + \\frac{1}{4} + x^4 - x^2 + \\frac{1}{4} + x^2 - x + \\frac{1}{4} > 0,\n$$\nwhich in turn is equivalent to\n$$\n\\left(x^3 - \\frac{1}{2}\\right)^2 + \\left(x^2 - \\frac{1}{2}\\right)^2 + \\left(x - \\frac{1}{2}\\right)^2 > 0.\n$$\n\nSince $\\frac{1}{\\sqrt{2}} \\neq... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities"
] | English | proof only | null | |
0fnh | Sea $ABCD$ un cuadrilátero convexo tal que:
$$
|AB| + |CD| = \sqrt{2} |AC|
$$
y
$$
|BC| + |DA| = \sqrt{2} |BD|.
$$
¿Qué forma tiene el cuadrilátero $ABCD$? | [] | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | Spanish | proof and answer | square |
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