| # | id | problem_markdown | solutions_markdown | images | country | competition | topics_flat | language | problem_type | final_answer |
|---|
1
|
0caz
|
The continuous function $f : \mathbb{R} \to \mathbb{R}$ has the property $(f \circ f)(x) = x^{2023}$, for every $x \in \mathbb{R}$. Prove that:
a) $2023\sqrt[2023]{f(x)} = f(2023\sqrt[2023]{x})$, for every $x \in \mathbb{R}$;
b) for every $n \ge 3$ there exists an arithmetic progression $x_1, x_2, \dots, x_n$ so that $\sum_{k=1}^{n} f(x_k) = 0$.
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[]
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[]
|
Romania
|
SHORTLISTED PROBLEMS FOR THE 73rd NMO
|
[
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
]
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proof only
|
|
2
|
0awa
|
Problem:
If $\tan x + \tan y = 5$ and $\tan (x + y) = 10$, find $\cot^2 x + \cot^2 y$.
|
[
"Solution:\nWe know that\n$$\n\\tan(x + y) = \\frac{\\tan x + \\tan y}{1 - \\tan x \\tan y} = 10\n$$\nThis leads to\n$$\n\\tan x \\tan y = \\frac{1}{2}\n$$\nHence,\n$$\n\\begin{aligned}\n\\cot^2 x + \\cot^2 y &= \\frac{1}{\\tan^2 x} + \\frac{1}{\\tan^2 y} \\\\\n&= \\frac{\\tan^2 y + \\tan^2 x}{\\tan^2 x \\tan^2 y} \\\\\n&= \\frac{(\\tan x + \\tan y)^2 - 2 \\tan x \\tan y}{(\\tan x \\tan y)^2} \\\\\n&= \\frac{5^2 - 1}{\\frac{1}{4}} \\\\\n&= 96\n\\end{aligned}\n$$"
]
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[]
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Philippines
|
18th PMO National Stage Oral Phase
|
[
"Precalculus > Trigonometric functions"
]
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final answer only
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96
|
3
|
0609
|
Problem:
Soit $n$ un entier vérifiant $n \geqslant 2$. On note $d$ le plus grand diviseur de $n$ différent de $n$. On suppose que $d > 1$. Démontrer que $n + d$ n'est pas une puissance de 2.
|
[
"Solution:\n\nSupposons par l'absurde que $n + d$ est une puissance de $2$. Notons que $d$ divise $n$, donc $d$ divise $n + d$, donc $d$ divise une puissance de $2$. Ainsi $d$ est une puissance de $2$ : on pose $d = 2^{k}$ pour un certain $k \\in \\mathbb{N}$. On a $k \\geqslant 1$ car $d \\neq 1$. Posons $n = d a = 2^{k} a$ avec $a \\in \\mathbb{N}^{*}$. Comme $d$ est diff\u00e9rent de $n$, on a $a \\geqslant 2$. Ainsi $2^{k-1} a$ est un diviseur de $n$, sup\u00e9rieur ou \u00e9gal \u00e0 $2^{k} = d$, et diff\u00e9rent de $n$. Par hypoth\u00e8se $d = 2^{k-1} a$, donc $a = 2$. Ainsi $n = 2^{k+1}$.\n\nOn trouve alors $n + d = 2^{k}(2 + 1) = 3 \\times 2^{k}$ qui n'est pas une puissance de $2$, ce qui est une contradiction. Ainsi $n + d$ n'est pas une puissance de $2$."
]
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[]
|
France
|
PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES
|
[
"Number Theory > Divisibility / Factorization > Factorization techniques"
]
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proof only
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4
|
05gq
|
Problem:
Soit $ABC$ un triangle, $H$ le pied de la hauteur issue de $A$, $L$ le pied de la bissectrice issue de $B$ et $M$ le milieu de $[AB]$. On suppose que dans le triangle $HLM$ on a les deux propriétés : $(AH)$ est une hauteur et $(BL)$ est une bissectrice. Montrer que $(CM)$ est une médiane.
|
[
"Solution:\n\nOn prend le triangle $ABC$, on va d\u00e9montrer qu'il est \u00e9quilat\u00e9ral, en utilisant les deux propri\u00e9t\u00e9s, la troisi\u00e8me d\u00e9coulera.\nOn a $(AH) \\perp (ML)$ et $(AH) \\perp (BC)$, donc $(BC) // (ML)$. On trouve ainsi que $L$ est le milieu de $[AC]$. Cependant, $\\frac{AL}{LC} = \\frac{AB}{BC}$ ce qui montre que $AB = BC$, et $(BL)$ est une bissectrice, une hauteur et une m\u00e9diane.\n\nIl faut maintenant que $(BL)$ bissecte $\\widehat{MLH}$, ou encore $(AC)$ comme bissectrice ext\u00e9rieure, donc par chasse aux angles $\\beta = \\widehat{HLC} = \\widehat{MLA} = \\alpha$. Donc $\\alpha = \\beta$ et $ABC$ est \u00e9quilat\u00e9ral, donc en particulier $(CM)$ est la m\u00e9diane dans $MHL$."
]
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[]
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France
|
Préparation Olympique Française de Mathématiques
|
[
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
]
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proof only
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|
5
|
0kq5
|
Problem:
In convex quadrilateral $ABCD$ with $AB = 11$ and $CD = 13$, there is a point $P$ for which $\triangle ADP$ and $\triangle BCP$ are congruent equilateral triangles. Compute the side length of these triangles.
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[
"Solution:\n\nEvidently $ABCD$ is an isosceles trapezoid with $P$ as its circumcenter. Now, construct isosceles trapezoid $AB'BC$ (that is, $BB'$ is parallel to $AC$.) Then $AB'PD$ is a rhombus, so $\\angle B'CD = \\frac{1}{2} \\angle B'PD = 60^{\\circ}$ by the inscribed angle theorem. Also, $B'C = 11$ because the quadrilateral $B'APC$ is a $60^{\\circ}$ rotation of $ADPB$ about $P$. Since $CD = 13$, we use the law of cosines to get that $B'D = 7\\sqrt{3}$. Hence $AP = 7$."
]
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[]
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United States
|
HMMT November 2022
|
[
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals"
]
|
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final answer only
|
7
|
6
|
074g
|
Let $f(x)$ and $g(y)$ be two monic polynomials with complex coefficients, and of the same degree $n$, such that
$$
f(x) - g(y) = \prod_{j=1}^{n} (a_j x + b_j y + c_j),
$$
where $a_j, b_j, c_j$ are complex numbers, $1 \le j \le n$. Prove that there exist complex numbers $a, b, c$ such that
$$
f(x) = (x+a)^n + c, \quad g(y) = (y+b)^n + c.
$$
|
[
"Observe that $\\prod_{j=1}^{n} a_j = 1$. Hence by dividing by this product, we may write\n$$\nf(x) - g(y) = \\prod_{j=1}^{n} x - \\alpha_j y + \\beta_j.\n$$\nWe observe that\n$$\n\\prod_{j=1}^{n} (x - \\alpha_j y) = x^n - y^n = \\prod_{j=1}^{n} (x - w^j y),\n$$\nwhere $w$ is the primitive $d$-th root of unity. Hence, after renumbering if necessary, we may write $\\alpha_j = w^j$, $1 \\le j \\le n$. Thus we have\n$$\nf(x) - g(y) = \\prod_{j=1}^{n} (x - w^j y + \\beta_j).\n$$\nDefine\n$$\na = \\frac{\\beta_1 - w\\beta_n}{w - 1}, \\quad b = \\frac{\\beta_1 - \\beta_n}{w - 1}.\n$$\nConsider $F(x) = f(x + a)$, $G(y) = g(y + b)$. It is easy to check that\n$$\nF(x) - G(y) = (x - y)(x - wy) \\prod_{j=2}^{n-1} (x - w^j y + \\gamma_j),\n$$\nwhere $\\gamma_j = \\beta_j + a - w^j b$, $2 \\le j \\le n-1$. Putting $y = x$, we see that $F(x) = G(x)$. Putting $x = wy$, we also see that $F(wy) = G(y) = F(y)$. Thus whenever $\\alpha$ is a root of $F(x) = 0$, so is $w\\alpha$. Thus $F(y) = 0$ has roots $\\alpha, w\\alpha, \\dots, w^{n-1}\\alpha$. We thus obtain\n$$\nF(y) = G(y) = (y - \\alpha)(y - w\\alpha) \\cdots (y - w^{n-1}\\alpha) = y^n - \\alpha^n.\n$$\nThis gives\n$$\nf(x) = F(x - a) = (x - a)^n - \\alpha^n, \\quad g(y) = (y - b)^n - \\alpha^n.\n$$"
]
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[]
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India
|
Indija TS 2009
|
[
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
]
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proof only
|
|
7
|
0dgx
|
Let $ABCD$ be a convex quadrilateral such that the line $CD$ is a tangent to the circle on $AB$ as diameter. Prove that the line $AB$ is a tangent to the circle on $CD$ as diameter if and only if the lines $BC$ and $AD$ are parallel.
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[]
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[]
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Saudi Arabia
|
Saudi Arabian IMO Booklet
|
[
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals"
]
|
English
|
proof only
|
|
8
|
0hpb
|
Problem:
A natural number $n$ is chosen between two consecutive square numbers. The smaller square is obtained by subtracting $k$ from $n$, and the larger one is obtained by adding $\ell$ to $n$. Prove that the number $n-k \ell$ is the square of an integer.
|
[
"Solution:\n\nLet $n-k = x^{2}$ and $n+\\ell = (x+1)^{2}$. Then $k = n - x^{2}$ and $\\ell = (x+1)^{2} - n$. We express everything in terms of $n$ and $x$:\n$$\n\\begin{aligned}\nn - k \\ell & = n - \\left(n - x^{2}\\right)\\left((x+1)^{2} - n\\right) \\\\\n& = n + \\left(x^{2} - n\\right)\\left(x^{2} + 2x + 1 - n\\right) \\\\\n& = n + \\left(x^{2} - n\\right)\\left(\\left(x^{2} - n\\right) + 2x + 1\\right) \\\\\n& = n + \\left(x^{2} - n\\right)^{2} + 2x\\left(x^{2} - n\\right) + x^{2} - n \\\\\n& = \\left(x^{2} - n\\right)^{2} + 2x\\left(x^{2} - n\\right) + x^{2} \\\\\n& = \\left(x^{2} - n + x\\right)^{2}.\n\\end{aligned}\n$$\nThis is the square of an integer."
]
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[]
|
United States
|
Berkeley Math Circle Monthly Contest 4
|
[
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
]
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proof only
|
|
9
|
0beo
|
Problem:
Adottak az $a, b \in \mathbb{R}^{*}$ számok és az $f: \mathbb{R} \rightarrow \mathbb{R}$,
$$
f(x)= \begin{cases}a x, & x \in \mathbb{Q} \\ b x, & x \in \mathbb{R} \backslash \mathbb{Q}\end{cases}
$$
függvény.
Igazold, hogy $f$ akkor és csak akkor injektív, ha $f$ szürjektív.
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[]
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[]
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Romania
|
Matematika tantárgyverseny Megyei szakasz
|
[
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
]
|
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proof only
|
|
10
|
0cih
|
Show that there exist infinitely many natural numbers $n$ for which $2025 \cdot n$ is a perfect cube, and $5202 \cdot n$ is a perfect square.
|
[]
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[]
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Romania
|
75th NMO
|
[
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
]
|
English
|
proof only
|
|