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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0eky | Problem:
Koliko je vrednost izraza $2^{0^{2^{3}}}+0^{2^{3^{2}}}+2^{3^{2^{0}}}+3^{2^{0^{2}}}$?
(A) 3
(B) 4
(C) 7
(D) 12
(E) Večja od 100. | [
"Solution:\nUpoštevamo, da je $0^{m}=0$ in $n^{0}=1$ za katerikoli naravni števili $m$ in $n$, in dobimo\n$$\n2^{0^{2^{3}}}+0^{2^{3^{2}}}+2^{3^{2^{0}}}+3^{2^{0^{2}}}=2^{0}+0+2^{3}+3=1+0+8+3=12\n$$"
] | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | D | |
006j | En un partido de biribol se enfrentan dos equipos de 4 personas cada uno. Se organiza un torneo de biribol en que participan $n$ personas, que forman equipos para cada partido (los equipos no son fijos). Al final del torneo se observó que cada dos personas disputaron exactamente un partido en equipos rivales. ¿Para qué... | [] | Argentina | XXIII Olimpíada Iberoamericana de Matemática | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Spanish | proof and answer | n ≡ 1 (mod 32) | |
0l50 | Problem:
A square of side length $1$ is dissected into two congruent pentagons. Compute the least upper bound of the perimeter of one of these pentagons. | [
"Solution:\n\nLet $P_{1}$ and $P_{2}$ be the two congruent pentagons. Let $p(P)$ denote the perimeter of polygon $P$.\nWe give an upper bound for $p(P_{1}) + p(P_{2})$. Note that since a square has four sides, at least four sides of $P_{1}$ and $P_{2}$ combined lie on the sides of the squar... | United States | HMMT February | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 2 + 3*sqrt(2) | |
044x | Given function $f(x) = |2 - \log_3 x|$, positive real numbers $a, b, c$ satisfy $a < b < c$ and $f(a) = 2f(b) = 2f(c)$. Find the value of $\frac{ac}{b}$. | [
"Note that $f(x) = |\\log_3(\\frac{x}{9})|$ is monotonically decreasing on $(0, 9]$ and monotonically increasing on $[9, +\\infty)$.\nBy the conditions satisfied by $a, b, c$, we know that $0 < a < b < 9 < c$ and\n$$\n\\log_3\\left(\\frac{9}{a}\\right) = 2\\log_3\\left(\\frac{9}{b}\\right) = 2\\log_3\\left(\\frac{c... | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 9 | |
0jei | Problem:
The whole numbers from $1$ to $100$ are each written on an index card, and the $100$ cards shuffled in a hat. Twenty-six cards are drawn out of the hat at random. Prove that two of the numbers drawn have a difference of $1$, $2$, or $3$. | [
"Solution:\n\nDivide the numbers from $1$ to $100$ into groups of four:\n\n\n\nSince $100 / 4 = 25$, there are $25$ groups. If $26$ numbers are selected, then two of them must come from the same group (the famous \"Pigeonhole Principle\"), and so their difference is at most $3$.",
"Soluti... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0c4n | Which three-digit numbers $abc$ have a multiple of the form $ababc$? | [] | Romania | SHORTLISTED PROBLEMS FOR THE 70th NMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | All three-digit numbers ending with zero, and 125. | |
0bo5 | Find all pairs of positive integers $A$ and $B$, having the same number of digits in the decimal representation, such that $2 \cdot A \cdot B = \overline{AB}$.
Here $\overline{AB}$ denotes the number obtained by concatenating $A$ and $B$. | [
"Answer: 36 and 1352.\nLet $n$ be the number of digits of $A$ and $B$. The given relation is the same as $(2A - 1)B = 10^n A$, so $2A - 1 \\mid 10^n A$. Since $(2A - 1, A) = 1$ and $(2, 2A - 1) = 1$, we infer that $2A - 1 \\mid 5^n$, so $2A - 1 \\le 5^n$.\n\nOn the other hand, since $A$ has $n$ digits we have $A \\... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"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 | (A,B) = (3,6) and (13,52) | |
0jj0 | Problem:
Suppose that $m$ and $n$ are integers with $1 \leq m \leq 49$ and $n \geq 0$ such that $m$ divides $n^{n+1}+1$. What is the number of possible values of $m$? | [
"Solution:\n\nAnswer: $\\quad 29$\n\nIf $n$ is even, $n+1 \\mid n^{n+1}+1$, so we can cover all odd $m$.\n\nIf $m$ is even and $m \\mid n^{n+1}+1$, then $n$ must be odd, so $n+1$ is even, and $m$ cannot be divisible by $4$ or any prime congruent to $3 \\pmod{4}$. Conversely, if $m / 2$ has all factors $1 \\pmod{4}$... | United States | HMMT November 2014 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 29 | |
02gd | Prove that
$$
(a+b)(a+c) \ge 2\sqrt{abc(a+b+c)}
$$
for all positive real numbers $a$, $b$ and $c$. | [
"By AM-GM,\n$$\n(a+b)(a+c) = bc + a(a+b+c) \\geq 2\\sqrt{bc \\cdot a(a+b+c)}\n$$"
] | Brazil | XXIII OBM | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0czb | Consider a triangle $ABC$. Let $A_{1}$ be the symmetric point of $A$ with respect to the line $BC$, $B_{1}$ the symmetric point of $B$ with respect to the line $CA$, and $C_{1}$ the symmetric point of $C$ with respect to the line $AB$. Determine the possible set of angles of triangle $ABC$ for which $A_{1}B_{1}C_{1}$ i... | [
"We will use the following relation: For any angle $\\theta$,\n$$\n\\cos 3\\theta = \\cos \\theta - 4 \\cos \\theta \\sin^2 \\theta\n$$\nLet $a, b, c$ be the sides of the triangle and $\\alpha, \\beta, \\gamma$ be the respective angles opposite these sides. Since the triangles $A_{1}BC$, $AB_{1}C$ and $ABC_{1}$ are... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | (60°, 60°, 60°), (30°, 75°, 75°), (150°, 15°, 15°) | |
0e8j | Problem:
Naj bo $m$ realno število in $f(x) = x^{2} - m x + m - 1$ ter $g(x) = x^{2} - 2 x - m$. Izračunaj vrednost parametra $m$ tako, da bosta najmanjši vrednosti funkcij $f$ in $g$ enaki. Pri največjem izmed tako izračunanih vrednostih parametra $m$ reši neenačbo $2 f(x) \geq g(x-1)$. | [] | Slovenia | Državno tekmovanje | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | m = 0 or m = 8; for m = 8, the inequality 2 f(x) ≥ g(x − 1) reduces to x^2 − 12x + 19 ≥ 0, so the solution set is x ≤ 6 − sqrt(17) or x ≥ 6 + sqrt(17). | |
0h89 | Bisector of the angle $\angle ABC$ of the triangle $ABC$ intersects the circumcircle of triangle $ABC$ at the point $K$. Point $N$ belongs to the segment $AB$ and $NK \perp AB$. Let $P$ be a midpoint of the segment $NB$. Consider the line through $P$ that is parallel to $BC$ and intersects line $BK$ at the point $T$. P... | [
"Let $M = NT \\cap AC$ (fig. 23). We note that since $BK$ is a bisector of $ABC$ then $\\angle KBC = \\angle KBA = \\alpha$, and since $PT$ is parallel to $BC$ then $\\angle KBC = \\angle PTB$. Thus, $PT = PB = PN$. It follows that $\\triangle BNT$ is right-angled with hypotenuse $BN$. Since triangle $\\triangle BN... | Ukraine | UkraineMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07ac | Prove that for each $n \in \mathbb{N}$ there exist natural numbers $a_1 < a_2 < \dots < a_n$ such that $\phi(a_1) > \phi(a_2) > \dots > \phi(a_n)$. | [
"First we prove a lemma.\n\n**Lemma.** Let $a > b$ be two positive integers such that $\\frac{a}{b} > 4$. Then there exist some positive integer $l$ such that $b < \\phi(2^l) = 2^{l-1} < 2^l < a$.\n\n**Proof of lemma.** There exists some positive integer $m$ such that $2^{m-1} \\le b < 2^m$. So $a > 4b \\ge 2^{m+1}... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
09em | Find all natural numbers $n$, $m$ that satisfy conditions: $(n + m + 1)|2mn$ and $(n + m - 1)|(n^2 + m^2 - 1)$. | [
"Note that $(n + m - 1)|(n + m - 1)^2 + 2(n + m - 1) = (n + m - 1)^2 + 2mn$. By combining it with given condition we get $(n + m - 1)|2mn$. Since none of $(n + m - 1)$ and $(n + m + 1)$ equals to $2$, greatest common divisor (GCD) of these numbers is not greater than $2$.\nIf $GCD = 2$ then $\\frac{1}{2}(n + m - 1)... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All pairs of consecutive natural numbers, i.e., |n − m| = 1. | |
0bxq | Let $n$ be a positive integer, let $a_1, \dots, a_n$ be pairwise distinct real numbers, and let $b_1, \dots, b_n$ be arbitrary real numbers. Show that:
a) if the $b_i$ are all positive, then there exists a polynomial $f$ with real coefficients, no root of which is real, such that $f(a_i) = b_i$, $i = 1, \dots, n$;
b) t... | [
"a) We exhibit two examples. Let $f_i = \\prod_{j \\neq i} (X - a_j)$, $i = 1, \\dots, n$. Since the $f_i$ never vanish simultaneously, the Lagrange type interpolation polynomial $f = \\sum_{i=1}^n \\frac{b_i f_i^2}{f_i(a_i)^2}$ clearly satisfies the required conditions.\n\nAnother example may be obtained by adding... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | English | proof only | null | |
04ug | Find the largest integer $d$ for which a $43 \times 47$ table can be filled with numbers $1$ and $2$ such that the sum of the numbers in each column and in each row is a multiple of $d$. (Do not forget to show that no larger $d$ works.) (Tomáš Bárta) | [] | Czech Republic | First Round | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 47 | |
072t | Problem:
In a cyclic quadrilateral $ABCD$, $AB = a$, $BC = b$, $CD = c$, $\angle ABC = 120^\circ$, and $\angle ABD = 30^\circ$. Prove that
i) $c \geq a + b$;
ii) $|\sqrt{c+a} - \sqrt{c+b}| = \sqrt{c-a-b}$. | [
"Solution:\n\nApplying cosine rule to triangle $ABC$, we get\n$$\nAC^2 = a^2 + b^2 - 2ab \\cos 120^\\circ = a^2 + b^2 + ab\n$$\nObserve that $\\angle DAC = \\angle DBC = 120^\\circ - 30^\\circ = 90^\\circ$. Thus we get\n$$\nc^2 = \\frac{AC^2}{\\cos^2 30^\\circ} = \\frac{4}{3}(a^2 + b^2 + ab)\n$$\nSo\n$$\nc^2 - (a+b... | India | INMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0bn2 | Find the smallest positive integer $n$ for which, no matter how we choose to color red $n$ vertices of a cube, there is a vertex of the cube whose three adjacent vertices are all colored red. | [
"Let $ABCD A'B'C'D'$ be a cube. Coloring red the four vertices of a face (e.g. $A, B, C, D$), no vertex of the cube has all three adjacent vertices colored red, so $n \\ge 5$.\n\nNow, let us color 5 vertices of the cube in red. Anyway we do it, one of the faces $ABCD$ and $A'B'C'D'$ has at least three red vertices.... | Romania | 66th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 5 | |
0ake | Let $f(x)$ and $g(x)$ be given by
$$
f(x) = \frac{1}{x} + \frac{1}{x-2} + \dots + \frac{1}{x-2018}
$$
and
$$
g(x) = \frac{1}{x-1} + \frac{1}{x-3} + \dots + \frac{1}{x-2017}.
$$
Prove that
$$
|f(x) - g(x)| > 2
$$
for any non-integer real number $x$ satisfying $0 < x < 2018$. | [] | North Macedonia | Asian-Pacific Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0izy | Problem:
Find two lines of symmetry of the graph of the function $y = x + \frac{1}{x}$. Express your answer as two equations of the form $y = a x + b$. | [
"Solution:\nAnswer: $y = (1 + \\sqrt{2}) x$ and $y = (1 - \\sqrt{2}) x$\n\nThe graph of the function $y = x + \\frac{1}{x}$ is a hyperbola. We can see this more clearly by writing it out in the standard form $x^{2} - x y + 1 = 0$ or $\\left(\\frac{y}{2}\\right)^{2} - \\left(x - \\frac{1}{2} y\\right)^{2} = 1$.\n\nT... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | y = (1 + sqrt(2)) x and y = (1 - sqrt(2)) x | |
0ks5 | Problem:
Suppose $x$ and $y$ are positive real numbers such that
$$
x + \frac{1}{y} = y + \frac{2}{x} = 3.
$$
Compute the maximum possible value of $x y$. | [
"Solution:\nRewrite the equations as $x y + 1 = 3 y$ and $x y + 2 = 3 x$. Let $x y = C$, so $x = \\frac{C+2}{3}$ and $y = \\frac{C+1}{3}$. Then\n$$\n\\left(\\frac{C+2}{3}\\right)\\left(\\frac{C+1}{3}\\right) = C \\Longrightarrow C^{2} - 6C + 2 = 0.\n$$\nThe larger of its two roots is $3 + \\sqrt{7}$.",
"Solution:... | United States | HMMT November 2022 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 3 + sqrt(7) | |
0cc7 | The diagonals of the parallelogram $ABCD$ intersect at $O$ and $M$ is the midpoint of the side $AB$. Let $P$ be a point of the segment $OC$ and $Q$ be the intersection of the lines $MP$ and $BC$. The parallel to $MP$ through $O$ intersects the line $CD$ at the point $N$. Prove that the points $A, N$ are $Q$ collinear i... | [
"If $A, N, Q$ are collinear, then the fundamental theorem for similarity yields $\\frac{CR}{MB} = \\frac{QC}{QB} = \\frac{CN}{AB}$ and from $AB = 2 \\cdot MB$ follows $CN = 2 \\cdot CR$. Now $RP \\parallel ON$ shows that $RP$ is a midline in $\\triangle CON$, therefore $P$ is the midpoint of $OC$.\n\n![](attached_i... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0gjf | $$
\sum_{k=0}^{2^n-1} 2^{r(k)} f(k + (-1)^k x) = (-1)^{x+n}
$$
對於所有整數 $x$ 皆成立。
For each positive integer $k$, define $r(k)$ as the number of runs of $k$ in base-2, where a run is a collection of consecutive 0s or consecutive 1s without a larger one containing it. For example, $(11100100)_2$ has 4 runs, namely $111-00-... | [
"The only solution is $f(x) = (-1)^x$.\n\nTo prove this, let's first consider the case $n = 1$.\n$$\nf(x) + 2f(1-x) = (-1)^{x+1}.\n$$\nBy replacing $x$ with $1-x$, we have\n$$\nf(1-x) + 2f(x) = (-1)^x.\n$$\nHence, $f(x) = (-1)^x$. Also, we can check this is a solution for $n = 1$.\n\nFor general $n$, we can assume ... | Taiwan | IMO 3J, Independent Study 1 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Chinese; English | proof and answer | f(x) = (-1)^x | |
0b1s | Problem:
One of the biggest mathematical breakthroughs in 2019 was progress on an 82-year old problem by the renowned mathematician and Fields medalist Terence Tao.
Consider the function
$$
f(n) = \begin{cases}
\frac{n}{2} & \text{if } n \text{ is even} \\
3n + 1 & \text{if } n \text{ is odd}
\end{cases}
$$
Starting... | [] | Philippines | 22nd Philippine Mathematical Olympiad | [
"Math Word Problems"
] | null | final answer only | Collatz | |
0499 | Mario has written a 30-digit number whose sum of digits is 123. Then he wrote all the digits again in some other order following the original number. Prove that the 60-digit number he obtained is not a perfect square. | [] | Croatia | Hrvatska 2011 | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
09im | Let $ABC$ be a triangle with $\angle A = 90^\circ$ and $AB = 2AC$. A point $M$ is chosen on the side $AC$. Let $P$ be the foot of the perpendicular from $A$ to the line $BM$, and let $Q$ be the foot of the perpendicular from $C$ to the line $BM$. Prove that $4PQ + 2QC = BP$. | [] | Mongolia | Mongolian Mathematical Olympiad Round 1 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0hpz | Problem:
A line of soldiers 1 mile long is jogging. The drill sergeant, in a car, moving at twice their speed, repeatedly drives from the back of the line to the front of the line and back again. When each soldier has marched 15 miles, how much mileage has been added to the car, to the nearest mile? | [
"Solution:\n30 ."
] | United States | null | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 30 | |
00ft | Let $\Gamma_{1}$ and $\Gamma_{2}$ be two circles intersecting at $P$ and $Q$. The common tangent, closer to $P$, of $\Gamma_{1}$ and $\Gamma_{2}$ touches $\Gamma_{1}$ at $A$ and $\Gamma_{2}$ at $B$. The tangent of $\Gamma_{1}$ at $P$ meets $\Gamma_{2}$ at $C$, which is different from $P$ and the extension of $A P$ meet... | [
"Let $\\alpha=\\angle P A B$, $\\beta=\\angle A B P$ and $\\gamma=\\angle Q A P$. Then, since $P C$ is tangent to $\\Gamma_{1}$, we have $\\angle Q P C= \\angle Q B C=\\gamma$. Thus $A, B, R, Q$ are concyclic.\n\nSince $A B$ is a common tangent to $\\Gamma_{1}$ and $\\Gamma_{2}$ then $\\angle A Q P=\\alpha$ and $\\... | Asia Pacific Mathematics Olympiad (APMO) | XI APMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06uz | Let $n > 1$ be a positive integer. Each cell of an $n \times n$ table contains an integer. Suppose that the following conditions are satisfied:
(i) Each number in the table is congruent to $1$ modulo $n$;
(ii) The sum of numbers in any row, as well as the sum of numbers in any column, is congruent to $n$ modulo $n^{2}$... | [
"Let $A_{i, j}$ be the entry in the $i^{\\text{th}}$ row and the $j^{\\text{th}}$ column; let $P$ be the product of all $n^{2}$ entries. For convenience, denote $a_{i, j} = A_{i, j} - 1$ and $r_{i} = R_{i} - 1$. We show that\n$$\n\\sum_{i=1}^{n} R_{i} \\equiv (n-1) + P \\quad (\\bmod\\ n^{4}) \\tag{1}\n$$\nDue to s... | IMO | IMO Shortlisted Problems | [
"Number Theory > Other"
] | null | proof only | null | |
06mw | Let $n$ and $k$ be two integers with $n > k \ge 1$. There are $2n+1$ students standing in a circle. Each student $S$ has $2k$ neighbours—namely, the $k$ students closest to $S$ on the right, and the $k$ students closest to $S$ on the left.
Suppose that $n+1$ of the students are girls, and the other $n$ are boys. Prove ... | [
"4. (IMO Shortlist 2021 C5) See the official solution."
] | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
03wb | Let $n \ge 3$ be any given integer. Determine the smallest positive integer $k$, for which there exists a set $A$ of $k$ real numbers and $n$ real numbers $x_1, x_2, \dots, x_n$, which are distinct from each other such that
$$
x_1 + x_2,\ x_2 + x_3,\ \dots,\ x_{n-1} + x_n,\ x_n + x_1
$$
are all in the set $A$. | [
"Let $m_1 = x_1 + x_2$, $m_2 = x_2 + x_3$, $\\dots$, $m_{n-1} = x_{n-1} + x_n$, $m_n = x_n + x_1$.\nFirst, note that $m_1 \\neq m_2$, otherwise $x_1 = x_3$, which contradicts the fact that $x_i$ are distinct. Similarly, $m_i \\neq m_{i+1}$, for $i=1, 2, \\dots, n$, where $m_{n+1} = m_1$, as usual. It follows that $... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 3 | |
01d2 | A graph has $2016$ vertices. All the edges of the graph are coloured blue or red. It is known that the graph contains no blue path on $1062$ vertices. Prove that it is possible to select two disjoint sets of $477$ vertices each, such that all the edges between these sets are red. | [
"Denote the graph by $G$, let $n = 1060$. Let $P_n$ denote a path on $n$ vertices. We perform the following algorithm on $G$ and construct a blue path $P$.\n\nLet $v_1$ be an arbitrary vertex of $G$, let $P = (v_1)$, $U = V \\setminus \\{v_1\\}$, and $W = \\emptyset$. We investigate all edges from $v_1$ to $U$ sear... | Baltic Way | Baltic Way 2016 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
05pb | Problem:
Soit $ABC$ un triangle dont les trois angles sont aigus et $\Gamma$ son cercle circonscrit. La tangente à $\Gamma$ en $A$ recoupe $(BC)$ en $P$. On note $M$ le milieu de $[AP]$. La droite $(BM)$ recoupe $\Gamma$ en $R$ et la droite $(PR)$ recoupe $\Gamma$ en $S$.
Montrer que $(AP)$ et $(CS)$ sont parallèles.... | [
"Solution:\n\nEn écrivant la puissance de $M$ par rapport à $\\Gamma$ puis le fait que $M$ est le milieu de $[AP]$, on obtient\n$$\nMR \\times MB = MA^{2} = MP^{2}\n$$\nLes triangles $MRP$ et $MPB$ sont donc indirectement semblables, donc $\\widehat{RPM} = \\widehat{PBM}$. On a donc\n$$\n\\begin{aligned}\n\\widehat... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03rx | Let $x$ and $y$ be positive real numbers with $x^3 + y^3 = x - y$. Prove that $x^2 + 4y^2 < 1$. (posed by Xiong Bin) | [
"In view of $x^3 + y^3 > 0$, it suffices to show that $(x^2 + 4y^2)(x - y) < x^3 + y^3$. Expanding the left-hand side of the last inequality and canceling the like terms we obtain $4xy^2 < x^2y + 5y^3$. By the AM-GM inequality, we have $x^2y + 5y^3 \\ge 2\\sqrt{5xy^2} > 4xy^2$."
] | China | China Girls' Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0c8v | Determine the numbers $\overline{abc}$, with $a < b < c$, knowing that the remainders of divisions of the numbers $\overline{abc}, \overline{bca}, \overline{cab}$ by 27 belong to the set $\{1, 2, 3, 4, 5\}$. | [
"The numbers $\\overline{abc}, \\overline{bca}, \\overline{cab}$ have the same sum of digits therefore they will have the same remainder $r$ when divided by 9.\nSince the remainders modulo 27 are small, they are preserved modulo 9. Indeed if $n = 27k + r$, then $n = 9 \\cdot 3k + r$, so $r$ will be a common remaind... | Romania | Romanian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 138, 246, 489, 678 | |
09ny | Show that the equation $x^4 + x^3 + x^2 + x = y^6 + 61$ has no integer solutions. | [
"Suppose there exist integers $x$ and $y$ such that $x^4 + x^3 + x^2 + x = y^6 + 61$.\n\nLet us consider the equation modulo $7$.\n\nFirst, note that $y^6 \\equiv 0$ or $1 \\pmod{7}$ for any integer $y$, since by Fermat's Little Theorem, $y^6 \\equiv 1$ if $y$ is not divisible by $7$, and $0$ otherwise.\n\nSo $y^6 ... | Mongolia | MMO2025 Round 2 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof only | null | |
0hdu | Let $M$ be the middle point of the side $AC$ of triangle $ABC$. Inside $\triangle BMC$, there is such point $P$ that $\angle BMP = 90^\circ$, $\angle ABC + \angle APC = 180^\circ$. Prove that $\angle PBM + \angle CBM = \angle PCA$.
(Anton Tryhub) | [
"We construct point $B'$ such that $ABCB'$ is a parallelogram. Then (Fig. 26)\n$$\n\\angle ABC + \\angle APC = 180^\\circ = \\angle AB'C + \\angle APC = 180^\\circ,\n$$\nhence, quadrilateral $APCB'$ is inscribed. Obviously, $\\triangle BPB'$ is isosceles, which yields\n$$\n\\angle PCA = \\angle PB'A = \\angle PB'M ... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06kv | Find all integers $n \ge 3$ with the following property: there exist $n$ distinct points on the plane such that each point is the circumcentre of a triangle formed by 3 of the points. | [
"The answer is any integer $n \\ge 6$.\nWhen $n = 6$, consider two equilateral triangles of side lengths 1 having parallel sides and the same orientation such that each pair of corresponding vertices is at a distance 1 apart. It is clear that each point has a distance 1 to three other points, so that it is the circ... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous... | null | proof and answer | all integers n ≥ 6 | |
0cl5 | Let $n \in \mathbb{N}$, with $n \ge 2$, and let $x_1, x_2, \dots, x_n$ be positive real numbers such that $x_1 + x_2 + \dots + x_n = 1$. Define
$$
m = \min \left\{ \frac{x_1}{1+x_1}, \frac{x_2}{1+x_1+x_2}, \dots, \frac{x_n}{1+x_1+x_2+\dots+x_n} \right\}
$$
and
$$
M = \max \left\{ \frac{x_1}{1+x_1}, \frac{x_2}{1+x_1+x_2... | [
"a.\nLet us denote\n$$\na_k = \\frac{x_k}{1+x_1+\\cdots+x_k} \\quad \\text{and} \\quad b_k = 1-a_k = \\frac{1+x_1+\\cdots+x_{k-1}}{1+x_1+\\cdots+x_k}, \\quad 1 \\le k \\le n.\n$$\nBy the inequality of arithmetic and geometric means, we have\n$$\nb_1 + b_2 + \\cdots + b_n \\ge n \\sqrt[n]{b_1 b_2 \\cdots b_n} = \\fr... | Romania | 75th NMO Selection Tests | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | a) 1 - 2^{-1/n}; b) 1 - 2^{-1/n} | |
0bpb | Problem:
Határozd meg azokat az $x, y$ egész számokat, amelyekre
$$
5^{x}-\log_{2}(y+3)=3^{y} \text{ és } 5^{y}-\log_{2}(x+3)=3^{x}.
$$
Problem:
Să se determine numerele întregi $x, y$, pentru care
$$
5^{x}-\log_{2}(y+3)=3^{y} \text{ şi } 5^{y}-\log_{2}(x+3)=3^{x}
$$ | [
"Solution:\nScăzând egalităţile, se obţine\n$$\n5^{x}+3^{x}+\\log_{2}(x+3)=5^{y}+3^{y}+\\log_{2}(y+3)\n$$\nCum funcţia $f(t)=5^{t}+3^{t}+\\log_{2}(t+3)$ este strict crescătoare, rezultă $x=y$.\nPentru rezolvarea în $\\mathbb{Z}$ a ecuaţiei\n$$\n5^{x}=3^{x}+\\log_{2}(x+3)\n$$\nse observă că $x \\in\\{-2,-1,0\\}$ nu ... | Romania | Olimpiada Naţională de Matematică, Etapa Judeţeană şi a Municipiului Bucureşti | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | x = y = 1 | |
0hn8 | Problem:
Let $d$ be a fixed positive integer. Prove that there exists a unique polynomial $S(n)$ such that for every integer $n \geq 0$, $S(n) = \sum_{k=0}^{n} k^{d} = 0^{d} + 1^{d} + \cdots + n^{d}$. Also prove that $S(n)$ can be expressed in the form
$$
c_{0} + c_{1}(2n+1) + c_{2}(2n+2)^{2} + \cdots + c_{d+1}(2n+1)^{... | [
"Solution:\nWe shall prove by strong induction on $d \\geq 0$ the existence of a polynomial $P_{d}(x)$ of degree $d+1$ with rational coefficients such that\n- the exponents of $x$ which occur with nonzero coefficients are all of opposite parity from $d$;\n- $P_{d}(2n+1) - P_{d}(2n-1) = n^{d}$ (defined to be $1$ whe... | United States | Berkeley Math Circle Take-Home Contest #6 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0cym | Let $ABC$ be a triangle with $\angle A = 90^\circ$ and let $P$ be a point on the hypotenuse $BC$. Prove that
$$
\frac{AB^2}{PC} + \frac{AC^2}{PB} \geq \frac{BC^3}{PA^2 + PB \cdot PC}
$$ | [
"Applying Cauchy-Schwarz inequality we have\n$$\n\\begin{gathered}\n\\frac{AB^2}{PC} + \\frac{AC^2}{PB} = \\frac{AB^4}{AB^2 \\cdot PC} + \\frac{AC^4}{AC^2 \\cdot PB} \\geq \\frac{(AB^2 + AC^2)^2}{AB^2 \\cdot PC + AC^2 \\cdot PB} \\\\\n= \\frac{BC^4}{AB^2 \\cdot PC + AC^2 \\cdot PB}\n\\end{gathered}\n$$\n![](attache... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
02wx | Problem:
Na figura a seguir, $ABC$ é um triângulo isósceles com $BA = BC$. O ponto $D$ está em seu interior de modo que $\angle ABD = 13^{\circ}$, $\angle ADB = 150^{\circ}$ e $\angle ACD = 30^{\circ}$. Além disso, $ADE$ é um triângulo equilátero. Determine o valor do ângulo $\angle DBC$.
 | [
"Solution:\n\nTrace os segmentos indicados na figura a seguir.\n\n\n\nComo $\\triangle ADE$ é equilátero, segue que $\\angle ADE = 60^{\\circ}$. Daí, $\\angle BDE = 360^{\\circ} - 150^{\\circ} - 60^{\\circ} = 150^{\\circ}$. Além disso, de $AD = DE$, $BD = BD$ e $\\angle ADB = \\angle BDE$, ... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 39° | |
0j4p | Problem:
Let $ABC$ be a triangle, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$, respectively. If $A$ is not a right angle, prove that the circumcenter of triangle $AEF$ lies on the incircle of triangle $ABC$ if and only if the incenter of triangle $ABC$ lies on the circumcircle of triangle $AEF$. | [
"Solution:\nLet $D$ be the foot of the altitude from $A$. Let $H$ be the orthocenter of triangle $ABC$. Let $M$ be the midpoint of $AH$. Let $I$ be the incenter of triangle $ABC$. Let $\\omega$ be the incircle of triangle $ABC$. Let $\\gamma$ be the circumcircle of $AEF$. Let $\\eta$ be the nine-point circle of tri... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Polar tr... | null | proof only | null | |
0ij9 | Problem:
At a nursery, $2006$ babies sit in a circle. Suddenly each baby pokes the baby immediately to either its left or its right, with equal probability. What is the expected number of unpoked babies? | [
"Solution:\n\nThe probability that any given baby goes unpoked is $1/4$. So the answer is $2006/4 = 1003/2$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | 1003/2 | |
07ot | Suppose $t$ is a real number and $\sin(2t) > 0$. Prove that
$$
1 + 6 \sin t \cos t \geq (\sin^3 t + \cos^3 t)(\sin t + \cos t)^3 + 16 \sin^3 t \cos^3 t,
$$
with equality iff $\cos(2t) = 0$. | [
"Let $s$ stand for $\\sin t$, $c$ for $\\cos t$, and let $x = 2sc = \\sin(2t)$. Then\n$$\n\\begin{aligned}\n& (\\sin^3 t + \\cos^3 t)(\\sin t + \\cos t)^3 + 16 \\sin^3 t \\cos^3 t \\\\\n&= (s^3 + c^3)(s + c)^3 + 2x^3 \\\\\n&= (s^2 - sc + c^2)(s + c)^4 + 2x^3 \\\\\n&= (1 - sc)(s^2 + 2sc + c^2)^2 + 2x^3 \\\\\n&= \\le... | Ireland | Irska 2014 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0hph | Problem:
Given the hypotenuse and the difference of the two legs of a right triangle, show how to reconstruct the triangle with ruler and compass. | [
"Solution:\nHere is one construction. Let $AB = d$ be the segment representing the difference of the two legs. Extend $AB$ to $C$ and raise a perpendicular $BD$ to $AC$ at $B$. Bisect angle $DBC$ to make ray $BE$ with $\\angle EBC = 45^{\\circ}$. Now set the compass to the length $c$ of the hypotenuse and draw a ci... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0azd | Problem:
In a right triangle $ABC$ with $\angle C = 90^\circ$, $BC = 10$, and $AC = 15$. Two squares are inscribed in $ABC$ as shown in the figure. Find the minimum sum of the areas of the squares.
 | [
"Solution:\n\nLet $NM = s$ and $SR = t$ be the side lengths of the two squares. By the Pythagorean theorem, we have $AB = 5\\sqrt{13}$. From the above figure, triangles $NMB$, $ACB$ and $ARS$ are similar. Thus, $MB = \\frac{BC \\cdot NM}{AC} = \\frac{2s}{3}$ and $AR = \\frac{AC \\cdot RS}{BC} = \\frac{3t}{2}$. We s... | Philippines | 20th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | 36 | |
0g1m | Problem:
Finde alle monoton steigenden Folgen $a_{1}, a_{2}, a_{3}, \ldots$ natürlicher Zahlen, sodass $i+j$ und $a_{i}+a_{j}$ für alle $i, j \in \mathbb{N}$ die gleiche Anzahl Teiler haben. | [
"Solution:\n\nErst zeigen wir, dass die Folge streng monoton steigend ist, dann, dass sie unendlich viele Fixpunkte enthält. Für eine natürliche Zahl $n$, sei $d(n)$ die Anzahl positiver Teiler von $n$.\n\n- Streng monoton steigend: Nehme an, es gibt natürliche Zahlen $i<j$ sodass $a_{i}=a_{j}$; wegen Montonie führ... | Switzerland | IMO-Selektion | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a_i = i for all natural i | |
06uk | There are $2017$ mutually external circles drawn on a blackboard, such that no two are tangent and no three share a common tangent. A tangent segment is a line segment that is a common tangent to two circles, starting at one tangent point and ending at the other one. Luciano is drawing tangent segments on the blackboar... | [
"First, consider a particular arrangement of circles $C_{1}, C_{2}, \\ldots, C_{n}$ where all the centers are aligned and each $C_{i}$ is eclipsed from the other circles by its neighbors - for example, taking $C_{i}$ with center $(i^{2}, 0)$ and radius $i / 2$ works. Then the only tangent segments that can be drawn... | IMO | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 6048 | |
0cl8 | Let $m \in \mathbb{N}$, $m \ge 2$ be a fixed natural number, and let $(a_n)_{n \ge 1}$ be a sequence of nonnegative real numbers such that $a_{n+1} \le a_n - a_{mn}$, $\forall n \ge 1$.
a) Prove that the sequence $(b_n)_{n \ge 1}$, $b_n = \sum_{k=1}^{n} a_k$ is bounded above.
b) Prove that the sequence $(c_n)_{n \ge ... | [
"a) We notice that $(b_n)_{n \\ge 1}$ is non-decreasing. Also, the sequence $(a_n)_{n \\ge 1}$ is non-increasing, since $0 \\le a_{mn} \\le a_n - a_{n+1}$. Moreover,\n$$\n\\sum_{k=1}^{n} a_{mk} \\le a_1 - a_{n+1} \\le a_1.\n$$\n\nUsing the monotonicity of $(a_n)$ and $(b_n)$, we have:\n$$\nb_n \\le b_{mn} = \\sum_{... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0cg9 | Determine all positive integers $n$, with $n \ge 2$, such that the equation
$$
x^2 - 3 \cdot x + 5 = 0
$$
has a unique solution in the ring $(\mathbb{Z}_n, +, \cdot)$. | [
"We shall denote by $M$ the set of all positive integers $n$, with $n \\ge 2$, such that the equation (1) has a unique solution in the ring $(\\mathbb{Z}_n, +, \\cdot)$.\nWe will show that $M = \\{11\\}$.\n\nIn the ring $(\\mathbb{Z}_{11}, +, \\cdot)$, the equation (1) can be equivalently written as\n$$\nx^2 - \\ha... | Romania | 74th Romanian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 11 | |
0haj | In triangle $ABC$, $H$ is the orthocenter and $AK$ is an altitude. Circle $w$ goes through points $A$ and $K$ and intersects sides $AB$ and $AC$ at points $M$ and $N$ respectively. The line through point $A$ parallel to $BC$ intersects the circumscribed circles of triangles $AHM$ and $AHN$ a second time at points $X$ a... | [
"Let $Z$ be the point of intersection of circle $w$ with line $BC$, then $AZ$ is the diameter of $w$ (see Fig. 35).\n\nReally, if $K = Z$, $w$ is tangent to $BC$, so, as $AK \\perp BC$, the center of $w$ is on $AK$. If $K \\neq Z$, then $\\angle AKZ = 90^\\circ$ and $AZ$ is the diameter of $w$. Then $\\angle AMZ = ... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Circles > Tangents... | English | proof only | null | |
09jv | Consider a positive integer $n$ that can be expressed as the sum of positive integers $a_1 > a_2 > \cdots > a_m$ such that
$$
\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_m} = 1,
$$
where $m \ge 2$. Prove that $n + 7 \le a_1 a_2$ and determine the cases when equality holds. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | Equality holds if and only if m = 3 with (a1, a2, a3) = (6, 3, 2) (in decreasing order), in which case n = 11 and a1 a2 = 18. | |
0axt | Problem:
A semicircle $\Gamma$ has diameter $AB = 25$. Point $P$ lies on $AB$ with $AP = 16$ and $C$ is on the semicircle such that $PC \perp AB$. A circle $\omega$ is drawn so that it is tangent to segment $PC$, segment $PB$, and $\Gamma$. What is the radius of $\omega$? | [] | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 4 | |
0508 | The numbers $0$, $1$, and $2$ are written in the vertices of a triangle. One step involves increasing two of the three numbers by $m$ or decreasing one of the three numbers by $n$. Is it possible that after some steps there are numbers $1$, $2$, and $3$ (in an arbitrary order) written in the vertices if
a) $m = 3$, $n... | [
"a) Both the step that involves increasing two of the numbers by $3$ and the step that involves decreasing one of the numbers by $6$ result in the sum of all three numbers being changed by $6$. Thus the remainder when the sum of the three numbers is divided by $6$ will always be the same regardless of the number of... | Estonia | Selected Problems from Open Contests | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | a) No; b) Yes | |
00tz | Let $n \ge 3$ be an odd positive integer, and consider an $n \times n$ grid containing $n^2$ cells. Dionysus colours each cell either red or blue. A frog can hop directly between two cells if they have the same colour and share at least one vertex. Xanthias views the colouring, and wants to place frogs on $k$ of the ce... | [
"Let $G$ be the graph whose vertices are all $(n+1)^2$ vertices of the grid and where two vertices are adjacent if and only if they are adjacent in the grid and moreover the two cells in either side of the corresponding edge have different colours.\nThe connected components of $G$, excluding the isolated vertices, ... | Balkan Mathematical Olympiad | BMO 2022 shortlist | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | ((n+1)^2)/4 + 1 | |
0cul | A number $x$ is chosen so that each of the sums $S = \sin 64x + \sin 65x$ and $C = \cos 64x + \cos 65x$ is a rational number. Prove that in one of these sums, both summands are rational. | [
"Since $S^2 + C^2 = 2 + 2 \\cos x \\in \\mathbb{Q}$, we have $\\cos x \\in \\mathbb{Q}$ and hence $\\cos 64x \\in \\mathbb{Q}$.",
"Заметим, что число\n$$\n\\begin{aligned}\nS^2 + C^2 &= (\\sin^2 64x + \\cos^2 64x) + (\\sin^2 65x + \\cos^2 65x) + \\\\\n&\\quad + 2(\\sin 64x \\sin 65x + \\cos 64x \\cos 65x) = \\\\\... | Russia | XLIII Russian mathematical olympiad | [
"Precalculus > Trigonometric functions"
] | English; Russian | proof only | null | |
053t | Manni and Miku play the following game with rooks on an $8 \times 8$ chessboard. At the beginning of the game, Miku places 8 rooks to the squares of the board according to his will. Then both players make moves alternately, Manni starts. On any move, each player shifts exactly one rook along a rank or file (i.e. row or... | [
"We show at first that Miku can play in such a way that Manni can never remove a rook. Let there be one rook in each rank and file in the initial configuration. Suppose that Manni moves a rook from square $(x, y)$ to square $(x, z)$. As a consequence, each file contains one rook but there are no rooks in rank $y$ a... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0iox | Problem:
A triangle has sides of length $9$, $40$, and $41$. What is its area? | [
"Solution:\nObserve that $9^{2} + 40^{2} = 41^{2}$, so this triangle is right and therefore has area $\\frac{1}{2} \\cdot 9 \\cdot 40 = 180$."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Triangles"
] | null | final answer only | 180 | |
03bk | Let $n \ge 2$ be a positive integer and $a_1 < a_2 < \dots < a_{2n}$ be real numbers. If $S = \sum_{i=1}^{2n} a_i$, $A_1 = \sum_{i,j,i<j} a_{2i}a_{2j}$ and $A_2 = \sum_{i,j,i<j} a_{2i-1}a_{2j-1}$, prove the inequality
$$
(n-1)S^2 > 4n(A_1 + A_2).
$$ | [
"First, we shall prove the following\n**Lemma.** If $P(x) = b_0x^n + b_1x^{n-1} + b_2x^{n-2} + \\dots + b_{n-1}x + b_n$ has $n$ real distinct roots then $(n-1)b_1^2 - 2nb_0b_2 > 0$.\n\n*Proof.* First differentiate $n-2$ times the function $f(x)$. As a result we have quadratic function having two real distinct roots... | Bulgaria | Team selection test for the 54th IMO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof only | null | |
0iad | Problem:
A cell of a 2-configuration of a set $A$ is a nonempty subset $C$ of $A$ such that
i. for any two distinct elements $a, b$ of $C$, there exists a sequence $c_{0}, c_{1}, \ldots, c_{n}$ of elements of $A$ with $c_{0}=a, c_{n}=b$, and such that $\left\{c_{0}, c_{1}\right\},\left\{c_{1}, c_{2}\right\}, \ldots,\le... | [
"Solution:\nFirst, given $a$, let $C_{a}$ be the set of all $b \\in A$ for which there exists a sequence $a=c_{0}, c_{1}, \\ldots, c_{n}=b$ as in the definition of a cell. Certainly $a \\in C_{a}$ (take $n=0$); we claim that $C_{a}$ is a cell. If $b, b^{\\prime} \\in C_{a}$, then there exist sequences $a=c_{0}, c_{... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
0f5d | Problem:
Three positive integers are written on a blackboard. A move consists of replacing one of the numbers by the sum of the other two less one. For example, if the numbers are $3$, $4$, $5$, then one move could lead to $4$, $5$, $8$ or $3$, $5$, $7$ or $3$, $4$, $6$. After a series of moves the three numbers are $... | [] | Soviet Union | 17th ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | Initial 2,2,2: No. Initial 3,3,3: No. | |
0d1q | Adel draws an $m \times n$ grid of dots on the coordinate plane, at the points of integer coordinates $(a, b)$ where $1 \leq a \leq m$ and $1 \leq b \leq n$. He proceeds to draw a closed path along $k$ of these dots, $(a_{1}, b_{1}),(a_{2}, b_{2}), \ldots,(a_{k}, b_{k})$, such that $(a_{i}, b_{i})$ and $(a_{i+1}, b_{i+... | [
"If $m$ is even, Adel can draw the following closed path which passes through all the dots of his grid. Therefore, the maximum possible value of $k$ is $mn$.\n\n\n\nIf $n$ is even, Adel can draw a similar closed path obtained by symmetry with respect to the first diagonal. Therefore, the ma... | Saudi Arabia | Selection tests for the International Mathematical Olympiad 2013 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | If both m and n are odd, the maximum k is mn − 1; otherwise, the maximum k is mn. | |
0kha | Problem:
Alice draws three cards from a standard 52-card deck with replacement. Ace through 10 are worth 1 to 10 points respectively, and the face cards King, Queen, and Jack are each worth 10 points. The probability that the sum of the point values of the cards drawn is a multiple of 10 can be written as $\frac{m}{n}... | [
"Solution:\n\nThe probability that all three cards drawn are face cards is $\\left(\\frac{3}{13}\\right)^{3}=\\frac{27}{2197}$. In that case, the sum is 30 and therefore a multiple of 10. Otherwise, one of the cards is not a face card, so its point value $p$ is drawn uniformly from values from 1 to 10. The sum of t... | United States | HMMT November 2021 | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 26597 | |
0l7x | Let $N$ denote the number of ordered triples of positive integers $(a, b, c)$ such that $a, b, c \le 3^6$ and $a^3 + b^3 + c^3$ is a multiple of $3^7$. Find the remainder when $N$ is divided by 1000. | [] | United States | 2025 AIME I | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | final answer only | 441 | |
09ev | Prove that if $a_1, a_2, \dots, a_{2014}$ are positive real numbers and there is no integer among them then there exists infinitely many $n$ such that $(n; [a_1n] + [a_2n] + \dots + [a_{2014n}]) = 1$. | [
"Proceeding by contradiction, suppose that there exists $M$ such that $(n; [a_1n] + [a_2n] + \\dots + [a_{2014n}]) \\neq 1$ for all $n \\ge M$. Consequently, for any prime $p_n$, (for all $p_n \\ge M$) there exists $x_n \\in \\mathbb{N}$ such that: $[a_1p_n] + [a_2p_n] + \\dots + [a_{2014p_n}] = p_nx_n$. It is obvi... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
06no | Let $m$ and $n$ be positive integers such that $\sqrt{m} + \sqrt{n} = \sqrt{2023}$. Find the greatest possible value of $m+n$. | [
"Note that $\\sqrt{2023} = 17\\sqrt{7}$. Hence we must have $\\sqrt{m} = p\\sqrt{7}$ and $\\sqrt{n} = q\\sqrt{7}$, where $p$ and $q$ are positive integers with sum $17$. We have\n$$\nm + n = 7p^2 + 7q^2 = 7p^2 + 7(17-p)^2 = 14 \\left[ \\left( p - \\frac{17}{2} \\right)^2 + \\frac{289}{4} \\right],\n$$\nwhich is max... | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 1799 | |
0efp | Problem:
Dani sta funkciji $f, g: \mathbb{R} \rightarrow \mathbb{R}$, podani s predpisoma $f(x)=6-4x-2x^{2}$ in $g(x)=-x+4$.
a) V istem koordinatnem sistemu nariši grafa obeh funkcij.
b) Zapiši enačbo tangente na graf funkcije $f$, ki je vzporedna grafu funkcije $g$.
 | [
"Solution:\n\na) Ničli funkcije $f$ sta $-3$ in $1$, teme grafa funkcije $f$ je v točki $T(-1,8)$. Graf funkcije $g$ ima smerni koeficient $-1$, ordinatno os pa seka v točki $N(0,4)$. Narišemo oba grafa.\n\n\n\nIzračun ničel in začetne vrednosti funkcije $f$ ..... 1 točka\nIzračun temena gr... | Slovenia | 17. tekmovanje dijakov srednjih tehniških in strokovnih šol v znanju matematike | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | y = -x + 57/8 | |
08hi | Problem:
Let $a, b, c, d \geq 1$ be arbitrary positive numbers. Prove that the equations system $a x - y z = c$, $b x - y t = -d$ has at least a solution $(x, y, z, t)$ in positive integers. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0kxo | Problem:
Let $S$ be the set of ordered pairs $(a, b)$ of positive integers such that $\operatorname{gcd}(a, b)=1$. Compute
$$
\sum_{(a, b) \in S}\left\lfloor\frac{300}{2 a+3 b}\right\rfloor
$$ | [
"Solution: The key claim is the following.\nClaim: The sum in the problem is equal to the number of solutions of $2 x+3 y \\leq 300$ where $x, y$ are positive integers.\n\nProof. The sum in the problem is the same as counting the number of triples $(a, b, d)$ of positive integers such that $\\operatorname{gcd}(a, b... | United States | HMMT February | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Diophan... | null | proof and answer | 7400 | |
0dwn | Problem:
Izračunaj vsoto $[\log_{2} 1]+[\log_{2} 2]+[\log_{2} 3]+\cdots+[\log_{2} 256]$. (Izraz $[x]$ pomeni največje celo število, ki ni večje od $x$.) | [
"Solution:\nVemo, da je\n$$\n[\\log_{2} n]= \\begin{cases}0, & n=1 \\\\ 1, & 2 \\leq n<2^{2} \\\\ 2, & 2^{2} \\leq n<2^{3} \\\\ \\vdots \\\\ 7, & 2^{7} \\leq n<2^{8} \\\\ 8, & n=2^{8}=256\\end{cases}\n$$\nZato je iskana vsota enaka\n$$\n\\begin{aligned}\n& 1 \\cdot (2^{2}-2)+2 \\cdot (2^{3}-2^{2})+\\cdots+7 \\cdot ... | Slovenia | 49. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | 1546 | |
020x | Problem:
a. Prove that for all $a, b, c, d \in \mathbb{R}$ with $a+b+c+d=0$,
$$
\max (a, b)+\max (a, c)+\max (a, d)+\max (b, c)+\max (b, d)+\max (c, d) \geqslant 0
$$
b. Find the largest non-negative integer $k$ such that it is possible to replace $k$ of the six maxima in this inequality by minima in such a way that ... | [
"Solution:\n\nThe left-hand-side of the inequality is invariant under permutations of $a, b, c, d$. We may therefore suppose that $a \\geqslant b \\geqslant c \\geqslant d$, so that the inequality reduces to\n$$\n0 \\leqslant 3 a+2 b+c=a+(a+b)+(a+b+c)\n$$\nWe claim that each of the terms on the right-hand side is n... | Benelux Mathematical Olympiad | 13th Benelux Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 2 | |
03q6 | Suppose points $F_1$, $F_2$ are the foci of the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$, $P$ is a point on the ellipse, and $|PF_1| : |PF_2| = 2 : 1$. Then the area of $\triangle PF_1F_2$ is equal to ________. | [
"$|PF_1| + |PF_2| = 2a = 6$ by definition of an ellipse. Since $|PF_1| : |PF_2| = 2 : 1$, then $|PF_1| = 4$ and $|PF_2| = 2$. Notice that $|F_1F_2| = 2c = 2\\sqrt{5}$, and\n$$\n|PF_1|^2 + |PF_2|^2 = 4^2 + 2^2 = 20 = |F_1F_2|^2.\n$$\nThen $\\triangle PF_1F_2$ is a right triangle. So $S_{\\triangle PF_1F_2} = \\frac{... | China | China Mathematical Competition (Shaanxi) | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | final answer only | 4 | |
04q2 | The point $A(0, 2)$ is given on the parabola $y^2 = x + 4$. Find all points $B$ on the given parabola, different from $A$, for which there exists a point $C$, also on the parabola, such that the angle $\angle ACB$ is right. (China) | [] | Croatia | Croatian Mathematical Society Competitions | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | All points B on y^2 = x + 4 with coordinates B = (t^2 − 4, t) where t ≤ 0 or t ≥ 4 (in particular, y ≤ 0 or y ≥ 4). | |
08ra | Find every integer $k$ that satisfies the following condition.
There are infinitely many triplet $(a, b, c)$ of integers such that $(a^2 - k)(b^2 - k) = c^2 - k$. | [
"Consider an arbitrary integer $k$. Take a complex number $\\alpha$ that meets $\\alpha^2 = k$. (For example, let $\\alpha = \\sqrt{k}$ if $k$ is nonnegative, and $\\alpha = i\\sqrt{-k}$ if $k$ negative.)\nIt is easy verify the following equalities:\n$$\n(n + \\alpha)(n + 1 - \\alpha) = ((n(n + 1) - k) + \\alpha), ... | Japan | The 16th Japanese Mathematical Olympiad - The Final Round | [
"Number Theory > Diophantine Equations",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | all integers | |
0ags | Let $S(n)$ denote the sum of the digits of the natural number $n$. For example: $n = 2456$, $S(n) = S(2456) = 2+4+5+6=16$, $S(S(n)) = S(S(2456)) = S(16)=1+6=7$. Is there a natural number $n$ for which $n + S(n) + S(S(n)) = 2011$. | [
"The numbers $n$ and $S(n)$ have the same remainder when divided by $3$ and $9$. Namely, if $n = \\overline{a_m a_{m-1} \\dots a_1 a_0}$ then\n$$\n\\begin{aligned}\nn &= \\overline{a_m a_{m-1} \\dots a_1 a_0} = 10^m a_m + 10^{m-1} a_{m-1} + \\dots + 10 a_1 + a_0 \\\\\n&= [(10^m - 1)a_m + (10^{m-1} - 1)a_{m-1} + \\d... | North Macedonia | XV Junior Macedonian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0fse | Problem:
Ist es möglich, die Menge $\{1,2, \ldots, 33\}$ derart in 11 disjunkte Teilmengen zu zerlegen, dass jede Teilmenge 3 Elemente enthält, von denen eines die Summe der beiden anderen ist? | [
"Solution:\n\nNein, dies ist nicht möglich. Nehme an, doch. Jede der 11 Teilmengen ist dann von der Form $\\{a, b, a+b\\}$, insbesondere ist die Summe $2(a+b)$ dieser drei Elemente gerade. Da die Teilmengen disjunkt sind, ist daher auch die Summe aller 33 Elemente gerade. Im Widerspruch dazu ist aber\n$$\n1+2+\\ldo... | Switzerland | IMO - Selektion | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | No | |
0c7e | Let $G$ be a finite group, and let $x_1, \dots, x_n$ be a labeling of its elements. Consider the $n \times n$ matrix $(a_{ij})$, where $a_{ij} = 1$ if $x_i x_j^{-1} \neq x_j x_i^{-1}$, and $a_{ij} = 0$ otherwise. Establish the parity of the integer $\det(a_{ij})$.
Amer. Math. Monthly | [
"The determinant under consideration is an even integer. To prove this, we show the determinant divisible by the cardinality of the set $S = \\{x: x \\in G, x \\neq x^{-1}\\}$. Since a member of $G$ is one of $S$ if and only if its inverse is, $|S|$ is even (possibly zero), and the conclusion follows.\n\nTo establi... | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Abstract Algebra > Group Theory",
"Algebra > Linear Algebra > Determinants"
] | English | proof only | null | |
0h3s | Let a point $A$ lay outside a given circle $\omega$. Through $A$ two lines are drawn, the first intersect $\omega$ at $B$ and $C$, the second, at $D$ and $E$ ($D$ is between $A$ and $E$). The line going through $D$ and parallel to $BC$ intersects $\omega$ at $F \neq D$, and the line $AF$ intersects $\omega$ at $T \neq ... | [
"Оскільки $DF \\parallel BC$, то $\\angle DFA = \\angle CAF$. До того ж, $\\angle DFA = \\angle DFT = \\angle DET$ як вписані, що спираються на одну дугу. Отже, $\\triangle AMT \\sim \\triangle EMA$ за двома кутами, звідки $\\frac{AM}{MT} = \\frac{EM}{AM}$, тобто $AM^2 = EM \\cdot MT$. За властивістю січних, $ME \\... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | English | proof only | null | |
04jy | Let $I$ be the incentre of the triangle $ABC$ and let $D$ be the point on side $AC$ such that $|AB| = |DB|$. Incircle of the triangle $BCD$ touches the lines $AC$ and $BD$ in points $E$ and $F$, respectively. Prove that the line $EF$ passes through the midpoint of the segment $DI$. | [
"We denote by $a$, $b$ and $c$ the lengths of the sides $BC$, $CA$ and $AB$ respectively.\nLet $K$ be the point at which the incircle of the triangle $ABC$ touches the line $AC$. Let $N$ be the foot of the altitude from $B$ in triangle $ABC$ and let $J$ be the intersection of line $EF$ and the altitude $BN$.\nNote ... | Croatia | Croatian 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 | |
0578 | Circle $\omega_2$ is tangent to circle $\omega_1$ at point $A$ and passes through its center $O$. Point $C$ is chosen on $\omega_2$ in such a way that the ray $AC$ intersects $\omega_1$ the second time at point $D$, the ray $OC$ intersects $\omega_1$ at point $E$ and the line $DE$ is parallel to the line $AO$. Find the... | [
"Denote $\\angle DAE = \\alpha$; then $\\angle DOE = 2\\alpha$ (Fig. 5). From the isosceles triangle $DOE$, we obtain $\\angle OED = \\frac{180^\\circ - 2\\alpha}{2} = 90^\\circ - \\alpha$. Since $DE$ and $AO$ are parallel, $\\angle AOE = 90^\\circ - \\alpha$.\n\nAs the common tangent to $\\omega_1$ and $\\omega_2$... | Estonia | Open Contests | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 30° | |
0iql | Problem:
Sandra the Maverick has 5 pairs of shoes in a drawer, each pair a different color. Every day for 5 days, Sandra takes two shoes out and throws them out the window. If they are the same color, she treats herself to a practice problem from a past HMMT. What is the expected value (average number) of practice pro... | [
"Solution:\n\nAnswer: $\\frac{5}{9}$\n\nOn any given day, there is a $\\frac{1}{9}$ chance that the second shoe that Sandra chooses makes a pair with the first shoe she chose. Thus the average number of problems she does in a day is $\\frac{1}{9}$, so, by the linearity of expectation, she does $\\frac{5}{9}$ proble... | United States | 1st Annual Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 5/9 | |
0lc1 | Several natural numbers are given on a line. We perform a transformation as follows: for every pair of consecutive integers on the line, write the sum of those two numbers in the middle of them. After 2013 such steps, how many times number 2013 are there on the line if
a) The given numbers are $1$ and $1000$?
b) The ... | [
"a) We first observe that one cannot write the number $2013$ between a pair $(a, b)$ with $a + b > 2013$. Using this simple observation, it is easy to check that the number $2013$ is written only twice after $2013$ steps in the $8$th and the $1013$th steps.\n\nb) We add an extra number $1$ after $1000$ on the line.... | Vietnam | VMO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | English | proof and answer | a) 2; b) 1198 | |
097u | Problem:
a) Arătați că mulțimea $M$ nu este vidă.
b) Găsiți toate numerele $a \in \mathbb{R}$ astfel încât $f(a)=0$ are loc pentru cel puțin o funcție $f \in M$.
c) Găsiți toate numerele $a \in \mathbb{R}$ astfel încât $f(a)=0$ are loc pentru toate funcțiile $f \in M$.
unde $M=\{f: \mathbb{R} \rightarrow \mathbb{R} \m... | [
"Solution:\na) Să arătăm că funcția $f: \\mathbb{R} \\rightarrow \\mathbb{R}$, $f(x)=x-1$ aparține mulțimii $M$. Într-adevăr, $f(2 f(x)+x)+f(x)=f(3 x-2)+x-1=4 x-4$. Deci $M \\neq \\varnothing$.\n\nb) Vom arăta că mulțimea acestor valori $a$ nu este vidă, adică există un zerou pentru cel puțin o funcție din $M$. Pen... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | a) M is nonempty, for example with f(x) = x minus 1. b) a = 1. c) a = 1. | |
03x7 | There are $n$ ($n > 12$) students participating in a mathematics contest. The examination paper consists of 15 fill-in-the-blank questions. For each question, the score of a correct answer is 1 point, and no point will be awarded if the answer is wrong or left blank. After analyzing all the possible cases of score dist... | [
"The smallest $n$ is 911. We divide the proof into two parts:\n\n(1) We first prove that $n = 911$ satisfies the conditions. If each student answers at least 3 questions correctly, then for any student there are $\\binom{15}{3} = 455$ ways for him to have exactly 3 correct answers. If there are 911 students partici... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 911 | |
0k5t | Problem:
Let $ABC$ be an acute scalene triangle with incenter $I$. Show that the circumcircle of $BIC$ intersects the Euler line of $ABC$ in two distinct points.
(Recall that the Euler line of a scalene triangle is the line that passes through its circumcenter, centroid, orthocenter, and the nine-point center.) | [
"Solution:\n\nLet $O$ and $H$ be the circumcenter and orthocenter of $ABC$. Recall that\n$$\n\\begin{aligned}\n\\angle BOC & = 2\\angle A, \\\\\n\\angle BHC & = 180^\\circ - \\angle A, \\\\\n\\angle BIC & = 90^\\circ + \\frac{1}{2} \\angle A.\n\\end{aligned}\n$$\nAs $ABC$ is acute, $A$, $I$, $O$, $H$ all lie on the... | United States | HMIC | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0d20 | In triangle $ABC$, $AB = AC = 3$ and $\angle A = 90^{\circ}$. Let $M$ be the midpoint of side $BC$. Points $D$ and $E$ lie on sides $AC$ and $AB$ respectively such that $AD > AE$ and $ADME$ is a cyclic quadrilateral. Given that triangle $EMD$ has area $2$, find the length of segment $CD$. | [
"Because $AEMD$ is cyclic, we have $\\angle BEM = \\angle ADM$. But $\\angle MBE = \\angle MAD = 45^{\\circ}$ and $BM = AM$. We deduce that triangles $BME$ and $AMD$ are congruent and therefore $ME = MD$.\n\nBecause $AEMD$ is cyclic, $\\angle DME = 90^{\\circ}$. Therefore, the area of triangle $EMD$ is\n$$\n2 = \\f... | Saudi Arabia | Selection tests for the Balkan Mathematical Olympiad 2013 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscella... | English | proof and answer | (3 - sqrt(7))/2 | |
0hqf | Problem:
Find all pairs $(a, b)$ of positive integers such that
$$
1 + 5^{a} = 6^{b}.
$$ | [
"Solution:\nThe only solution is $(1, 1)$.\n\nIt is clear that if $b = 1$ then $a = 1$, and that $(1, 1)$ is a solution. Consequently, assume $b > 1$. Then $6^{b}$ is divisible by $4$. On the other hand, since $5^{a} \\equiv 1^{a} = 1 \\pmod{4}$ for all $a$, the left side is $2 \\pmod{4}$. Thus there are no solutio... | United States | Berkeley Math Circle | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (1, 1) | |
05c8 | In an acute triangle $ABC$ with $AB < AC$, points $D$, $E$ and $F$ are the feet of the altitudes drawn from vertices $A$, $B$ and $C$, respectively. Let the orthocenter of $ABC$ be $H$ and the midpoint of the side $BC$ be $M$. Point $K$ on the prolongation of the line segment $EM$ beyond $M$ and point $L$ on the line s... | [
"Firstly, we prove that $BDLF$ is an isosceles trapezium (Fig. 44). As $\\angle CEB = 90^\\circ = \\angle CFB$, points $B$, $C$, $E$ and $F$ lie on a circle with diameter $BC$. Hence $M$ is the center of this circle. Consequently, $MB = MF$, implying\n\n\nFig. 44\n\n$DB = MB - MD = MF - ML ... | Estonia | Estonian Mathematical Olympiad | [
"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 > Miscellaneou... | English | proof only | null | |
0c54 | A regular triangular pyramid $SABC$ has vertex $S$ and the planes $\alpha, \beta$ are such that $BC \subset \alpha, \alpha \perp (SAB), AB \subset \beta$ and $\beta \perp (SBC)$.
a) Show that if $\angle(\alpha, \beta) = 120^\circ$, then $SABC$ is a regular tetrahedron.
b) If $H$ is the orthocentre of the tetrahedron ... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof only | null | |
0h0o | Olesya wrote down a natural number $N$. After this, Andriy wrote down one sixth, one fifth, one fourth, one third and a half of $N$. It turns out that the sum of all numbers that are written is integer. What is the least possible number that Olesya could write? | [
"The sum is given by: $\\left(\\frac{1}{6} + \\frac{1}{5} + \\frac{1}{4} + \\frac{1}{3} + \\frac{1}{2}\\right)N = \\frac{29N}{20}$, and therefore $N$ is divisible by $20$ and the result follows."
] | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | 20 | |
0l3x | Problem:
Let $P$ be a point in the interior of quadrilateral $ABCD$ such that the circumcircles of triangles $PDA$, $PAB$, and $PBC$ are pairwise distinct but congruent. Let the lines $AD$ and $BC$ meet at $X$. If $O$ is the circumcenter of triangle $XCD$, prove that $OP \perp AB$. | [
"Solution:\n\nBecause the circles have equal radii, $\\angle PDA=\\angle ABP$, so if $(PDA)$ intersects line $AB$ again at a point $B'$, then we have $\\angle PB'B=\\angle PBB'$, which means $PB=PB'$, similarly for the second intersection of $(PCB)$ with $AB$, $A'$; thus, $(PDA)$ and $(PCB)... | United States | HMMT February 2024 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Pla... | null | proof only | null | |
05io | Problem:
Soit $ABC$ un triangle et $O$ un point intérieur à ce triangle. $D$ et $E$ sont respectivement les pieds des perpendiculaires abaissées de $O$ sur les droites $(BC)$ et $(AC)$ et $F$ est le milieu du segment $AB$. Montrer que si $DF = EF$ alors les angles $\widehat{OBD}$ et $\widehat{OAE}$ sont égaux. | [
"Solution:\n\nDans cet exercice il y a une hypothèse superflue, à savoir que le point $O$ est à l'intérieur du triangle $ABC$. Nous allons résoudre l'exercice sans cette hypothèse.\n\nCommençons par tracer un angle droit $AEO$ et un segment $AB$ et plaçons $F$ au milieu du segment $AB$. Notre but maintenant est de ... | France | Olympiades Françaises de Mathématiques - Envoi Numéro 1 - Corrigé | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
036g | Problem:
Solve the equation
$$
\log_{a}\left(a^{2\left(x^{2}+x\right)}+a^{2}\right)=x^{2}+x+\log_{a}\left(a^{2}+1\right)
$$
where $a$ is a real number. | [] | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | For all real a > 0 with a ≠ 1, the solutions are x ∈ {−2, −1, 0, 1}. | |
03nd | Problem:
1. Amy has drawn three points in a plane, $A$, $B$, and $C$, such that $AB = BC = CA = 6$. Amy is allowed to draw a new point if it is the circumcenter of a triangle whose vertices she has already drawn. For example, she can draw the circumcenter $O$ of triangle $ABC$, and then afterwards she can draw the cir... | [
"Solution:\n\n(a) Given triangle $\\triangle ABC$, Amy can draw the following points:\n- $O$ is the circumcenter of $\\triangle ABC$\n- $A_1$ is the circumcenter of $\\triangle BOC$\n- $A_2$ is the circumcenter of $\\triangle OBA_1$\n- $A_3$ is the circumcenter of $\\triangle BA_2A_1$\n\nWe claim that $AA_3 > 7$. W... | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Co... | null | proof only | null | |
0ja7 | Problem:
Let $Q(x) = x^{2} + 2x + 3$, and suppose that $P(x)$ is a polynomial such that
$$
P(Q(x)) = x^{6} + 6x^{5} + 18x^{4} + 32x^{3} + 35x^{2} + 22x + 8
$$
Compute $P(2)$. | [
"Solution:\n\nNote that $Q(-1) = 2$. Therefore, $P(2) = P(Q(-1)) = (-1)^{6} + 6(-1)^{5} + 18(-1)^{4} + 32(-1)^{3} + 35(-1)^{2} + 22(-1) + 8 = 1 - 6 + 18 - 32 + 35 - 22 + 8 = 2$."
] | United States | HMMT November 2012 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 2 | |
0bih | In the acute triangle $ABC$, with $AB \ne BC$, let $T$ denote the midpoint of the side $[AC]$, $A_1$ and $C_1$ denote the feet of the altitudes drawn from $A$ and $C$, respectively. Let $Z$ be the point of intersection of the tangents in $A$ and $C$ to the circumcircle of triangle $ABC$, $X$ be the point of intersectio... | [
"a. From $AZ = ZC$, it follows immediately that $[ZT]$ is the angle bisector of $\\angle AZC$. Notice that $\\angle XAB = \\angle ACB = \\angle BC_1A_1 = \\angle AC_1X$ and\n\n\n\nb. We may assume that $AB < BC$; in this case, $D$ is on the minor arc $AB$. Let $O$ denote the circumcenter of... | Romania | 65th NMO Selection Tests for JBMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadril... | null | proof only | null | |
0a40 | Find all functions $f: \mathbb{Z}_{>0} \to \mathbb{Z}_{>0}$ such that for all positive integers $m$ and $n$ it holds that
$$
(f(m))^2 + 2mf(n) + f(n^2)
$$
is the square of an integer. | [
"First, note that the function $\\iota(m) = m$ is such a function, as\n$$\n(\\iota(m))^2 + 2m\\iota(n) + \\iota(n^2) = (m+n)^2.\n$$\n\nWe show that $\\iota$ is the only such function.\nLet $f$ be any such function. Substituting $m = n = 1$, we see that $f(1)^2 + 3f(1)$ must be a square. As $(f(1)+1)^2 \\le f(1)^2 +... | Netherlands | IMO Team Selection Test 2 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | f(n) = n for all positive integers n | |
068b | Let $a$, $b$, $c$, $d$ be positive real numbers such that
$$
a^2 + b^2 + c^2 + d^2 = 4.
$$
Prove that there are two of $a$, $b$, $c$, $d$ with sum greater or equal to $2$. | [
"Without loss of generality, let $a \\ge b \\ge c \\ge d$ and we will prove that $a+b \\ge 2$.\nWe have that $ab \\ge c^2$ and $ab \\ge d^2$, so by adding them we have: $2ab \\ge c^2 + d^2$.\nTherefore,\n$$\n(a+b)^2 = a^2 + b^2 + 2ab \\ge a^2 + b^2 + c^2 + d^2 = 4,\n$$\nso $a+b \\ge 2$."
] | Greece | Selection Examination | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null |
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