problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
20.10. (SRP, 78). Non-constant functions $f(n, m)$ defined on the set of all pairs of integers, taking integer values and satisfying the identity
$$
\begin{aligned}
& f(n, m) \equiv \frac{1}{4}(f(n-1, m)+f(n+1, m)+f(n, m-1)+ \\
& +f(n, m+1))
\end{aligned}
$$
$n, m \in \mathbf{Z}$. Prove that: a) such functions exist;... | 20.10. a) For example, the function
$$
f(n, m)=n \quad(n, m \in \mathbb{Z})
$$
satisfies all the conditions of the problem.
b) Suppose the statement is false, i.e., for some number $k \in \mathbf{Z}$, all values of a certain function $f(n, m)$, satisfying the conditions of the problem, for example, do not exceed $k$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,148 |
20.11. (Austria - PRN, 78). For a given subset $S$ of the set of pairs of integers, we call a function $f: S \rightarrow S$ universal if it is invertible and for any pair $(n ; m) \in S$ it satisfies the condition
$f(n, m) \in\{(n-1 ; m) ;(n+1 ; m) ;(n ; m-1) ;(n ; m+1)\}$. Prove that if there exists at least one univ... | 20.11. We will call a point $(n ; m) \in S$ even or odd depending on whether the sum $n+m$ is even or odd, respectively. Suppose there exists a universal function $g(n, m)$, then the function $g^{-1}(n, m)$ is also universal. Consider the function defined as follows:
$$
f(n, m)= \begin{cases}g(n, m), & \text { if the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,149 |
20.12*. (USA, 82). Find all pairs of non-zero integer values \( m \leqslant n \), satisfying the inequality \( m+n \neq 0 \) and the identity
\[
f_{m}(x, y) f_{n}(x, y)=f_{m+n}(x, y), \quad x, y \in \mathbf{R}, \quad x y(x+y) \neq 0,
\]
where it is denoted
\[
f_{k}(x, y)=\left(x^{k}+y^{k}+(-1)^{k}(x+y)^{k}\right) ... | 20.12. We will prove that no other pairs of the required values, except those given in the indication, satisfy the required conditions. Suppose that some integer pair $(m ; n)$, different from $(2 ; 3)$ and $(2 ; 5)$, satisfies the inequalities $m \leqslant n, m n(m+n) \neq 0$ and the identity
$$
f_{m}(x, y) f_{n}(x, ... | (2;3)(2;5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,150 |
20.13*. (Jury, PNR, 77). Prove that if the function $f(x, y)$, defined on the set of all pairs of rational numbers and taking only positive values, satisfies the three identities
$$
\begin{aligned}
f(x y, z) & \equiv f(x, z) f(y, z) \\
f(z, x y) & \equiv f(z, x) f(z, y) \\
f(x, 1-x) & \equiv 1, \quad x, y, z \in \math... | 20.13. Substituting the values $x=y=0$ and $x=y=1$ into the first original identity for the function $f(x, y)$, we obtain the equalities $f(0, z)=1$ and $f(1, z)=1$ respectively. Further, substituting $x=y=-1$, we have
$$
1=f(1, z)=f(-1, z) f(-1, z)=(f(-1, z))^{2}
$$
hence, $f(-1, z)=1$. Similarly, from the second or... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,151 |
20.14*. (Jury, SFRY, 79). Prove that any function $f: \mathbf{R} \rightarrow \mathbf{R}$, satisfying one of the identities
$$
\begin{gathered}
f(x+y) \equiv f(x)+f(y), \quad x, y \in \mathbf{R} \\
f(x y+x+y) \equiv f(x y)+f(x)+f(y), \quad x, y \in \mathbf{R}
\end{gathered}
$$
satisfies the other. | 20.14. If the function $f(x)$ satisfies the first identity, then
$$
f(x y+x+y)=f(x y)+f(x+y) \leftrightharpoons f(x y)+f(x)+f(y)
$$
$(x, y \in \mathrm{R})$, i.e., the second identity is also satisfied for it. Now let the function $f(x)$ satisfy the second identity. By setting $y=u+v+u v$ in it, we get
$f(x+u+v+x u+x... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,152 |
20.15. (Austria, 75). Find all continuous functions $\mathrm{f}:(1 ;+\infty) \rightarrow \mathbf{R}$, satisfying the identity
$$
f(x y) \equiv x f(y)+y f(x), \quad x, y>1
$$ | 20.15. We will prove that for any value of $k>0$ the identity
$$
f\left(x^{k}\right) \equiv k x^{k-1} f(x) \quad(x>1)
$$
is valid. The proof will be conducted in three stages.
1) Let $k \in \mathbb{N}$. If $k=1$, then we have
$$
f\left(x^{1}\right)=1 \cdot x^{0} \cdot f(x)
$$
and if the identity is valid for some ... | f(x)=\lnx | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,153 |
20.16*. (CRR, 82). a) Prove that if a continuous function $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfies the identity
$$
f(f(f(x))) \equiv x, \quad x \in \mathbf{R}
$$
then for any value $x \in \mathbf{R}$, the equality $f(x)=x$ holds.
b) Provide an example of a (discontinuous) function $g: \mathbf{R} \longrightar... | 20.16. a) From the condition of the problem, it follows that the function $f(x)$ does not take any value at more than one point $x \in \mathbb{R}$. Indeed, if $u=f(x)=f(y)$ for some $x, y \in \mathbb{R}$, then
$$
x=f^{3}(x)=f^{2}(u)=f^{3}(y)=y.
$$
From this and the continuity of the function $f(x)$, it follows that i... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,154 |
20.17*. (Jury, France, 79). Find all monotonic invertible functions $f: \mathbf{R} \rightarrow \mathbf{R}$, satisfying the identity
$$
f(x)+f^{-1}(x)=2 x, \quad x \in \mathbb{R}
$$ | 20.17. Consider the function $g(x)=f(x)-x$ and prove that for any value $k \in \mathbf{Z}$ the identity
$$
f(x+k g(x))=x+(k+1) g(x) \quad(x \in \mathbb{R})
$$
holds. For $k=0$, we have the valid identity $f(x)=x+g(x)$. Suppose for some value $k \in \mathbf{N}$ the identities
$$
f(x+(k-1) g(x)) \equiv x+k g(x)
$$
an... | f(x)=x+ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,155 |
20.18*. (New York, 77). Find all differentiable functions $f: \mathbf{R} \longrightarrow \mathbf{R}$, satisfying the identity
$$
f^{\prime}\left(\frac{x+y}{2}\right)=\frac{f(y)-f(x)}{y-x}, \quad x, y \in \mathbf{R}, \quad x \neq y
$$ | 20.18. Since, by the condition of the problem, for any value of $y \neq 0$ the identity
$$
f^{\prime}(x)=\frac{f(x+y)-f(x-y)}{2 y} \quad(x \in \mathbf{R})
$$
holds, and the right-hand side is differentiable with respect to $x$, then
$$
\begin{aligned}
& f^{\prime \prime}(x)=\frac{f^{\prime}(x+y)-f^{\prime}(x-y)}{2 y... | f(x)=^{2}++\text{,where},b,\in{R} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 29,156 |
20.19. (Belgium, 77). Find all infinitely differentiable functions $f: \mathbf{R} \rightarrow \mathbf{R}$, satisfying the identity
$$
f(x+y) \equiv f(x)+f(y)+2xy, \quad x, y \in \mathbf{R}
$$ | 20.19. Substituting $x=y=0$ into the original identity, we get $f(0)=$ $=2 f(0)$, i.e., $f(0)=0$. For any $x \in \mathbb{R}$ from the identity
$$
f(x+y)-f(x)=f(y)+2 x y \quad(y \in \mathbf{R})
$$
## we have
$$
\begin{aligned}
f^{\prime}(x)=\lim _{y \rightarrow 0} \frac{f(x+y)-f(x)}{y}= & \lim _{y \rightarrow 0} \fra... | f(x)=x^2+ax | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,157 |
20.20. (England, 69). Prove that if a function $f: \mathbf{R} \rightarrow \mathbf{R}$, not identically zero, satisfies the identity
$$
f(x) f(y) \equiv f(x+y), \quad x, y \in \mathbf{R}
$$
and is differentiable at the point $x=0$, then it is infinitely differentiable at any point $x \in \mathbf{R}$.
## Chapter 6 Po... | 20.20. Substituting into the original identity for the function $f(x)$ the values $x=y=0$, we get $f(0)(f(0)-1)=0$, i.e., $f(0)=0$ or $f(0)=1$. But if $f(0)=0$, then from the identity
$$
f(0) f(x)=f(x) \quad(x \in \mathbb{R})
$$
it follows that the identity $f(x)=0$ holds, which contradicts the condition of the probl... | proof | Calculus | proof | Yes | Yes | olympiads | false | 29,158 |
21.1. (NBR, 80). Prove that for the roots $x_{1}, x_{2}$ of the polynomial
$$
x^{2}+p x-\frac{1}{2 p^{2}}, \quad \text { where } p \in \mathbf{R}, \quad p \neq 0
$$
the inequality $x_{1}^{4}+x_{2}^{4} \geqslant 2+\sqrt{2}$ holds. | 21.1. Using Vieta's theorem $\left(x_{1}+x_{2}=-p, x_{1} x_{2}=-1 /\left(2 p^{2}\right)\right)$ and the inequality between the arithmetic mean and the geometric mean of two numbers, we get
$$
\begin{aligned}
& x_{1}^{4}+x_{2}^{4}=\left(x_{1}+x_{2}\right)^{4}-2 x_{1} x_{2}\left(2\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}... | 2+\sqrt{2} | Inequalities | proof | Yes | Yes | olympiads | false | 29,159 |
21.2. (PRB, 61). Find all pairs of real numbers $p, q$, for which the polynomial $x^{4}+p x^{2}+q$ has 4 real roots forming an arithmetic progression. | 21.2. The polynomial $x^{4}+p x^{2}+q$ has 4 real roots if and only if the polynomial $y^{2}+p y+q$ (relative to
satisfies the conditions $p^{2} \geqslant 4 q, q \geqslant 0, p \leqslant 0$. ... | p\leqslant0,q=0.09p^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,160 |
21.3. (England, 67). Prove that if the roots of the polynomial $x^{2}+p x+1$ are numbers $\alpha$ and $\beta$, and the roots of the polynomial $x^{2}+q x+1$ are numbers $\gamma$ and $\delta$, then the following equality holds:
$$
(\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta)=q^{2}-p^{2}
$$ | 21.3. By Vieta's theorem, we have the equalities
$$
\alpha+\beta=-p, \alpha \beta=1, \gamma+\delta=-q, \gamma \delta=1
$$
from which we obtain
$$
\begin{aligned}
& (\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta)= \\
& =\left(\alpha \beta-(\alpha+\beta) \gamma+\gamma^{2}\right)\left(\alpha \beta+(\alpha+\b... | q^{2}-p^{2} | Algebra | proof | Yes | Yes | olympiads | false | 29,161 |
21.4. (GDR, 70). Prove that for any non-zero values of $\alpha, \beta$ the roots $x_{1}, x_{2}, x_{3}$ of the polynomial
$$
\alpha x^{3}-\alpha x^{2}+\beta x+\beta
$$
satisfy the equation
$$
\left(x_{1}+x_{2}+x_{3}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}\right)=-1
$$ | 21.4. By Vieta's theorem for the roots $x_{1}, x_{2}, x_{3}$ of the polynomial $\alpha x^{3}-\alpha x^{2}+\beta x+\beta$, we have
$$
x_{1}+x_{2}+x_{3}=1, \quad x_{1} x_{2}+x_{2} x_{3}+x_{1} x_{3}=\beta / \alpha, \quad x_{1} x_{2} x_{3}=-\beta / \alpha .
$$
Therefore,
$$
\begin{aligned}
& \left(x_{1}+x_{2}+x_{3}\righ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,162 |
21.5. (Austria, 83). Find all values of $a$ for which the roots $x_{1}, x_{2}, x_{3}$ of the polynomial $x^{3}-6 x^{2}+a x+a$ satisfy the equation
$$
\left(x_{1}-3\right)^{3}+\left(x_{2}-3\right)^{3}+\left(x_{3}-3\right)^{3}=0
$$ | 21.5. Let's make the substitution $y=x-3$, then the numbers $y_{1}=x_{1}-3, y_{2}=$ $=x_{2}-3$ and $y_{3}=x_{3}-3$ are the roots of the polynomial
$$
(y+3)^{3}-6(y+3)^{2}+a(y+3)+a=y^{3}+3 y^{2}+(a-9) y+4 a-27
$$
By Vieta's theorem, we have the equalities
$$
\begin{aligned}
& y_{1}+y_{2}+y_{3}=-3 \\
& y_{1} y_{2}+y_{... | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,163 |
21.6. (Jury, Canada, 82). Let one of the roots of the polynomial
$$
P(x)=x^{3}+a x^{2}+b x+c, \quad \text { where } a, b, c \in \mathbf{Z}
$$
be the product of the other two. Prove that the number $2 P(-1)$ is divisible by the number $P(1)+P(-1)-2(1+P(0))$. | 21.6. For the roots $u, v$ and $w v$ of the polynomial $P(x)$, by Vieta's theorem we have the equalities
$$
\begin{aligned}
& u+v+u v=-a \\
& u v(1+u+z)=b \\
& u^{2} v^{2}=-c
\end{aligned}
$$
from which, in the case $a \neq 1$, we obtain
$$
b-c=u v(1+u+v+u v)=u v(1-a)
$$
i.e., $u v=(b-c) /(1-a)$ is a rational numbe... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,164 |
21.7. (USA, 77). Let $a$ and $b$ be two of the four roots of the polynomial $x^{4}+x^{3}-1$. Prove that $a b$ is a root of the polynomial $x^{6}+x^{4}+x^{3}-x^{2}-1$. | 21.7. Let \(a, b, c, d\) be the roots of the polynomial
\[
P(x) = x^4 + x^3 - 1 = (x - a)(x - b)(x - c)(x - d)
\]
We will prove the equality
\[
(a b)^3 + (c d)^3 + a b + c d + 1 = 0
\]
(from which the required
\[
\begin{aligned}
(a b)^6 + (a b)^4 + (a b)^3 - (a b)^2 - 1 &= \\
& = (a b)^3 \left((a b)^3 - \left(\fra... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,165 |
21.8. (SFRY, 81). For integers $a, b, c$ it is known that $a>0$, and the polynomial $a x^{2}+b x+c$ has two distinct roots in the interval $(0 ; 1)$. Prove that $a \geqslant 5$. Find at least one pair of numbers $b, c$ for $a=5$. | 21.8. Since the polynomial $P(x)=a x^{2}+b x+c(a>0)$ has two distinct roots $0 < 4ac$ and by Vieta's theorem $0 < c > 0, b < 0$, from which $a+c > -b$. Squaring both (positive) parts of the last inequality, we get $a^{2}+2ac+c^{2}>b^{2}$, from which we have the estimates $(a-c)^{2}>b^{2}-4ac>0$, from which it follows t... | \geqslant5,\quad5x^{2}-5x+1=0 | Algebra | proof | Yes | Yes | olympiads | false | 29,166 |
21.9. (Czechoslovakia, 67). The numbers $a, b, c$ are three of the four roots of the polynomial $x^{4}-a x^{3}-b x+c$. Find all such triples of numbers $a, b, c$. | 21.9. Suppose that $c=0$. Then the number $c$ is a root of the equation $x^{4}-a x^{3}-b x=0$, so it remains to find all polynomials of the form $x^{3}-a x^{2}-b$, for which the numbers $a$ and $b$ are roots. If we denote by $d$ the third root of such a polynomial, then by Vieta's theorem we will have $a+b+d=a$, i.e., ... | \begin{aligned}&(0;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,167 |
21.10. (CSSR, 54). Prove that complex numbers $a$ and $b$ satisfy the condition $a^{2}=2 b \neq 0$ if and only if the roots of the polynomial $x^{2}+a x+b$ form two vertices of an isosceles right triangle on the complex plane, with the right angle vertex located at the origin. | 21.10. Let $x_{1}, x_{2}$ be the roots of the polynomial $x^{2} + a x + b$. The points $x_{1}$, $x_{2}$, and $0$ form the vertices of the required triangle if and only if $x_{1} \neq 0$ and the number $x_{2} / x_{1}$ is either $i$ or $-i$ (since the magnitudes of the numbers $x_{1}$ and $x_{2}$ are the same, and their ... | ^{2}=2b\neq0 | Algebra | proof | Yes | Yes | olympiads | false | 29,168 |
21.11. (VNR, 83). The polynomial
$$
P(x)=x^{n}+a_{1} x^{n-1}+\ldots+a_{n-1} x+1
$$
with non-negative coefficients $a_{1}, \ldots, a_{n-1}$ has d real roots. Prove that
$$
P(2) \geqslant 3^{n}
$$ | 21.11. Since all coefficients of the polynomial $P(x)$ are non-negative, none of its roots $\alpha_{1}, \ldots, \alpha_{n}$ can be positive. Therefore, this polynomial has the form
$$
P(x)=\left(x+\beta_{1}\right) \ldots\left(x+\beta_{n}\right)
$$
where $\beta_{i}=-\alpha_{i}>0, i=1,2, \ldots, n$. Using the mean valu... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,169 |
21.12. (NBR, 84). The polynomial
$$
a x^{n}-a x^{n-1}+c_{2} x^{n-2}+\ldots+c_{n-2} x^{2}-n^{2} b x+b
$$
has exactly $n$ positive roots. Prove that all these roots are equal to each other. | 21.12. Since the original polynomial has $n$ positive roots $x_{1}, \ldots, x_{n}$, its degree is at least $n$. Therefore, $a \neq 0$, and by Vieta's theorem we have
$$
\begin{aligned}
(-1)^{n} \sum_{i=1}^{n} x_{1} x_{2} \ldots x_{i-1} x_{i+1} \ldots x_{n} & =n^{2} \frac{b}{a} \\
(-1)^{n} x_{1} x_{2} \ldots x_{n} & =\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,170 |
21.13. (NBR, 83). Can the polynomials
$$
x^{5}-x-1 \text{ and } x^{2}+a x+b, \text{ where } a, b \in \mathbf{Q},
$$
have common complex roots? | 21.13. Suppose that some number $\alpha$ is a solution to each of the two given equations, i.e.,
$$
\alpha^{3}=\alpha+1 \text { and } \alpha^{2}=-a \alpha-b
$$
## Then
$\alpha+1=\alpha^{3}=\alpha\left(\alpha^{2}\right)^{2}=\alpha(-a \alpha-b)^{2}=\alpha\left(a^{2}(-a \alpha-b)+2 a b \alpha+b^{2}\right)=$ $=\left(2 a... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,171 |
21.15. (GDR, 70). Prove that for any value of $n \in \mathbf{Z}^{+}$, the polynomial
$$
f_{n}(x)=1+x+\frac{x^{2}}{2!}+\ldots+\frac{x^{n}}{n!}
$$
cannot have more than one real root. | 21.15. We will prove by induction that if $n$ is even, then the polynomial $P_{n}(x)$ takes positive values for all $x \in \mathbb{R}$ (hence, it has no real roots), and if $n$ is odd, then the polynomial $P_{n}(x)$ has exactly one real root. For $n=0$, we have $P_{n}(x)=1>0$ for all $x$. Suppose the statement is true ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,173 |
21.16*. (GDR, 69; GDR, 71). Prove that if a polynomial $P(x)$ of degree $n$ with real coefficients has no real roots, then the polynomial
$$
Q(x)=P(x)+\alpha P^{\prime}(x)+\ldots+\alpha^{n} P^{(n)}(x)
$$
for any value of $\alpha \in \mathbf{R}$ also has no real roots. | 21.16. We have
$$
Q^{\prime}(x)=P^{\prime}(x)+\alpha P^{n}(x)+\ldots+\alpha^{n-1} P^{(n)}(x)
$$
(since $\operatorname{Pr}^{(n+1)}(x)=0$). Therefore,
$$
Q(x)-\alpha Q^{\prime}(x)=P(x)
$$
Without loss of generality, we assume that the leading coefficient of the polynomial $P(x)$ is positive. Since this polynomial has... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,174 |
21.17*. (PDR, 79). Prove that for any polynomial $P(x)$ of degree $n>1$, having $n$ distinct real roots $x_{1}, x_{2}, \ldots, x_{n}$, the following equality holds
$$
\frac{1}{P^{\prime}\left(x_{1}\right)}+\frac{1}{P^{\prime}\left(x_{2}\right)}+\ldots+\frac{1}{P^{\prime}\left(x_{n}\right)}=0
$$ | 21.17. Let's factor the original polynomial:
$$
P(x)=a\left(x-x_{1}\right) \ldots\left(x-x_{n}\right) \text {, where } a \neq 0 .
$$
Then
$$
P^{\prime}(x)=P_{1}(x)+\ldots+P_{n}(x)
$$
where $P_{k}(x) (k=1,2, \ldots, n)$ denotes a polynomial of degree $n-1 \geqslant 1$ that satisfies the identity
$$
\left(x-x_{k}\ri... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,175 |
21.18*. (New York, 75). Let $P(x)$ be a polynomial with real coefficients, all roots of which are purely imaginary numbers. Prove that all roots of the polynomial $P^{\prime}(x)$, except one, are also purely imaginary. | 21.18. According to Theorem 57, a polynomial with real coefficients can only have an even number of purely imaginary roots, which are divided into pairs of mutually conjugate roots. Therefore, the polynomial \( P(x) \) can be represented as \( P(x) = a \left(x - i a_1\right) \ldots \left(x - i a_{2n}\right) = a \left(x... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,176 |
21.19*. (SRP, 78). Prove that non-zero polynomials $P$ and $Q$ with complex coefficients have the same roots (of the same multiplicity) if and only if the function $f(z)=|P(z)|-|Q(z)|$ has a constant sign at all points $z \in$ C, where it is non-zero.
## § 22. Divisibility and Equality of Polynomials
(see Appendix G:... | 21.19. Let the roots of polynomials $P$ and $Q$ coincide (along with their multiplicities), then we have
$$
\begin{aligned}
& P(z)=a\left(z-z_{1}\right)^{n_{1}}\left(z-z_{2}\right)^{n_{2}} \ldots\left(z-z_{k}\right)^{n_{k}} \\
& Q(z)=b\left(z-z_{1}\right)^{n_{1}}\left(z-z_{2}\right)^{n_{2}} \ldots\left(z-z_{k}\right)^... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,177 |
22.1. (New York, 73; Belgium, 81). Prove that for any value of $n \in \mathbf{Z}^{+}$, the polynomial $(x+1)^{2 n+1} + x^{n+2}$ is divisible by the polynomial $x^{2} + x + 1$. | 22.1. We will prove this by induction on $n \in \mathbf{Z}^{+}$. For $n=0$, the statement is true because in this case $(x+1)^{2 n+1}+x^{n+2} = x^{2}+x+1$. Assume that for some value $n-1$ the statement holds, i.e., the polynomial $(x+1)^{2 n-1}+x^{n+1}$ is divisible by the polynomial $x^{2}+x+1$. Then the polynomial
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,178 |
22.2. (SRP, 62). Prove that for any values of $n \in \mathbf{N}$ and $\alpha \in \mathbf{R}$, satisfying the conditions $n \neq 1$ and $\sin \alpha \neq$ $\neq 0$, the polynomial
$$
P(x)=x^{n} \sin \alpha-x \sin n \alpha+\sin (n-1) \alpha
$$
is divisible by the polynomial
$$
Q(x)=x^{2}-2 x \cos \alpha+1
$$ | 22.2. Let $x_{\ell}=\cos \alpha+i \varepsilon \sin \alpha$, where $\varepsilon \in\{-1 ; 1\}$. Then the polynomial $Q(x)$ is represented as
$$
Q(x)=(x-\cos \alpha-i \sin \alpha)(x-\cos \alpha+i \sin \alpha)=\left(x-x_{1}\right)\left(x-x_{-1}\right)
$$
By De Moivre's formula, we have $x_{\mathrm{e}}^{n}=(\cos 8 \alpha... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,179 |
22.3. (SRP, 66). Find all polynomials $R(x)$ of degree less than 4, for each of which there exists a polynomial $P(x)$ satisfying the identity $7 \sin ^{31} t+8 \sin ^{18} t-5 \sin ^{5} t \cos ^{4} t-10 \sin ^{2} t+5 \sin ^{5} t-2=$ $=P(\sin t)\left(\sin ^{4} t-(1+\sin t)\left(\cos ^{2} t-2\right)\right)+R(\sin t), \qu... | 22.3. Let's make the substitutions $\cos ^{2} t=1-\sin ^{2} t$ and $\sin t=x$. We will obtain that the desired polynomial $R(x)$ of degree less than 4 is the remainder from the division of the polynomial $S(x)=7 x^{31}+8 x^{13}-5 x^{9}-2$ by the polynomial $Q(x)=x^{4}+x^{3}+x^{2}+x+1$, since the identity $S(x)=P(x) Q(x... | 13x^{3}+5x^{2}+12x+3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,180 |
22.4. (USA, 77). Find all pairs of numbers $m, n \in \mathbf{N}$, for which the polynomial $1+x^{n}+x^{2 n}+\ldots+x^{m n}$ is divisible by the polynomial $1+x+x^{2}+\ldots+x^{m}$. | 22.4. The polynomials $P(x)=1+x+\ldots+x^{m}$ and $Q(x)=1+x^{n}+\ldots+x^{m n}$ do not have multiple roots, since the polynomials $x^{m+1}-1=(x-1) P(x)$ and $x^{n(n+1)}-1=\left(x^{n}-1\right) Q(x)$ do not have multiple roots. Therefore, according to Theorem 54, the polynomial $Q(x)$ is divisible by $P(x)$ if and only i... | (+1,n)=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,181 |
22.5. (USA, 76). Let polynomials $P(x), Q(x), R(x)$, and $S(x)$ satisfy the identity
$$
P\left(x^{5}\right)+x Q\left(x^{5}\right)+x^{2} R\left(x^{6}\right) \equiv\left(x^{4}+x^{3}+x^{2}+x+1\right) S(x) .
$$
Prove that the polynomial $P(x)$ is divisible by the polynomial $x-1$. | 22.5. Let $s_{0}, s_{1}, \ldots, s_{n}$ be the coefficients of the polynomial $S(x)=$ $=s_{0}+s_{1} x+\ldots+s_{n} x^{n}$. Multiplying both sides of the original identity by the polynomial $x-1$, we obtain the identity
$$
(x-1)\left(P\left(x^{5}\right)+x Q\left(x^{5}\right)+x^{2} R\left(x^{5}\right)\right) \equiv\left... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,182 |
22.6. (New-York, 75). Find all polynomials $P(x)$ satisfying the condition $P(0)=0$ and the identity
$$
P(x)=\frac{1}{2}(P(x+1)+P(x-1)), \quad x \in \mathbf{R}
$$ | 22.6. Any polynomial of the form $P(x)=a x$, where $a$ is a constant, satisfies the conditions of the problem. We will prove by induction on $n \in \mathbf{Z}^{+}$ that for each desired polynomial $P(x)$, the equalities $P(n)=n P(1)$ hold. For $n=0$ and $n=1$, these equalities are true. Suppose they have already been p... | P(x)= | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,183 |
22.7. (GDR, 77). Find all polynomials $P(x)$ that satisfy the identity
$$
x P(x-1) \equiv (x-2) P(x), \quad x \in \mathbf{R}
$$ | 22.7. Substituting into the original identity the values $x=0$; 2, we get that the polynomial $P(x)$ has roots 0 and 1, and therefore, is divisible by the polynomial $x^{2}-x$. Further, substituting into the identity the expression $P(x)=$ $=\left(x^{2}-x\right) Q(x)$, we obtain for the polynomial $Q(x)$ the identity $... | P(x)=(x^2-x) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,184 |
22.8. (New York, 76). Find all polynomials $P(x)$ satisfying the identity
$$
(x-1) P(x+1)-(x+2) P(x) \equiv 0, \quad x \in \mathbf{R}
$$
3 Edited by I. N. Sergeev | 22.8. Substituting into the original identity the values $x=1; -2; 0$, we get that the desired polynomial $P(x)$ has roots 0, $\pm 1$, which means it is divisible by the polynomial $x^{3}-x$. Next, substituting into the identity the expression
$$
P(x)=\left(x^{3}-x\right) Q(x),
$$
we obtain for the polynomial $Q(x)$ ... | P(x)=(x^3-x) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,185 |
22.9. (SRP, 80). Find all non-zero polynomials $P(x)$ that satisfy the identity
$$
P\left(x^{2}\right) \equiv (P(x))^{2}, \quad x \in \mathbf{R}
$$ | 22.9. Let the desired polynomial have the form
$$
P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0} \text {, where } a_{n} \neq 0 \text {. }
$$
Assume that at least one of the coefficients $a_{n-1}, \ldots, a_{1}, a_{0}$ is different from zero. Choose the largest value of $k<n$ for which $a_{k} \neq 0$. Then we h... | P(x)=x^{n}, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,186 |
22.10. (NBR, 76). Find all non-zero polynomials $P(x)$ that satisfy the identity
$$
P\left(x^{2}-2 x\right) \equiv(P(x-2))^{2}, \quad x \in \mathbf{R}
$$ | 22.10. Let $y=x-1, Q(y)=P(y-1)$. Then we have
$$
\begin{aligned}
& (P(x-2))^{2}=(P(y-1))^{2}=(Q(y))^{2}, \\
& P\left(x^{2}-2 x\right)=P\left(y^{2}-1\right)=Q\left(y^{2}\right),
\end{aligned}
$$
and the original identity can be written as
$$
Q\left(y^{2}\right)=(Q(y))^{2}, \quad y \in \mathbb{R},
$$
which coincides,... | P(y)=(y+1)^{n}, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,187 |
22.11. (Jury, NRB, 79). Find all non-zero polynomials $P(x)$ with real coefficients, satisfying the identity
$$
P(x) P\left(2 x^{2}\right) \equiv P\left(2 x^{3}+x\right), \quad x \in \mathbf{R}
$$ | 22.11. Let a non-zero polynomial $P(x)=a_{n} x^{n}+\ldots+a_{1} x+a_{0}$, where $a_{n} \neq 0$, satisfy the conditions of the problem. Then, by comparing the coefficients of $x^{3 n}$ and $x^{0}$ in the original identity, we obtain the equalities $a_{n}^{2}=a_{n}, a_{0}^{2}=a_{0}$, from which $a_{n}=1$ and $a_{0}=1$ (i... | P(x)=(x^2+1)^k,wherek\in{Z}^+ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,188 |
22.12. (SRP, 78). Prove that for any polynomial $P(x) \neq x$ and any number $n \in \mathbf{N}$, the polynomial
$$
Q_{n}(x)=P \underbrace{(P(\ldots P}_{n}(x) \ldots))-x
$$
is divisible by the polynomial $Q_{1}(x)=P(x)-x$. | 22.12. We will prove the statement by induction on $n \in N$. Since the polynomial $Q_{i}(x)$ divides itself, the statement is true for $n=1$. Now suppose that for some value of $n \in N$, the identity $Q_{n}(x)=R_{n}(x) Q_{1}(x)$ has already been proven, where $R_{n}(x)$ is a polynomial. Then we have
$$
\begin{aligne... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,189 |
22.13. (CPR, 78). Prove that if polynomials $P(x), Q(x), R(x)$ of the third degree with real coefficients satisfy the inequalities $P(x) \leqslant Q(x) \leqslant$ $\leqslant R(x)$ for all values of $x \in \mathbf{R}$ and at least at one point $x_{0} \in \mathbf{R}$ the equality $P\left(x_{0}\right)=R\left(x_{0}\right)$... | 22.13. Note that if a polynomial $S(x)$ of degree not higher than three satisfies the condition $S(x) \geqslant 0$ for all $x \in \mathbb{R}$ and the equality $S\left(x_{0}\right)=0$,
then it can be represented in the form $S(x)=a\left(x-x_{0}\right)^{2}$, where $a \geqslant 0$. Therefore,
$$
R(x)=P(x)+a\left(x-x_{0}\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,190 |
22.14. (Jury, VNR, 79). Given a polynomial $P(x) = a x^{2} + b x + c$, where $a \neq 0$. Prove that for any number $n \in \mathrm{N}$, there cannot be more than one polynomial $Q(x)$ of degree $n$ satisfying the identity $Q(P(x)) \equiv P(Q(x)), \quad x \in \mathbf{R}$. | 22.14. If a polynomial $Q(x)=q_{n} x^{n}+\ldots+q_{0}$ satisfies the conditions $q_{n} \neq 0$ and $Q(P(x))=P(Q(x))$, then by equating the leading coefficients of the polynomials $Q(P(x))$ and $P(Q(x))$ of degree $2 n$, we get $q_{n} a^{n}=a q_{n}^{2}$, i.e., $q_{n}=a^{n-1}$. Therefore, if, contrary to the statement of... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,191 |
22.15. (SRP, 79). Prove that the polynomial $P(z)$ is an even function of $z \in \mathbf{C}$ if and only if there exists a polynomial $Q(z)$ satisfying the identity
$$
P(z) \equiv Q(z) Q(-z), \quad z \in \mathbf{C}
$$ | 22.15. If $P(z) \equiv Q(z) Q(-z)$, then $P(-z) \equiv Q(-z) Q(z) \equiv P(z)$, i.e., $P(z)$ is an even function. Let us now assume that a non-zero polynomial $P(z)$ is an even function (if $P(z) \equiv 0$, then let $Q(z)=0$). We will prove by induction on the number $m$ of non-zero roots of the polynomial $P(z)$ that ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,192 |
22.16. (Jury, VNR, 79). Prove that if a polynomial $P(x)$ with real coefficients takes only non-negative values for all $x \in \mathbf{R}$, then it can be represented in the form
$$
P(x)=Q_{1}^{2}(x)+\ldots+Q_{n}^{2}(x)
$$
where $Q_{1}(x), \ldots, Q_{n}(x)$ are some polynomials with real coefficients. | 22.16. We will prove the statement of the problem by induction on the number $m$ of the roots of the non-zero polynomial $P(x)$ that are not real numbers (if $P(x)=0$, then let $n=1$ and $Q_{1}(x)=0$). Let $m=0$ and the polynomial $P(x)$ satisfies the condition of the problem. Then each of its roots is real and has eve... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,193 |
22.17*. (PRC, 76; Jury, Sweden, 76). Let the polynomial $P(x)$ with real coefficients satisfy the inequality $P(x)>0$ for all values of $x>0$. Prove that there exist polynomials $Q(x)$ and $R(x)$ with non-negative coefficients, for which the identity
$$
P(x) \equiv Q(x) / R(x)
$$
holds. | 22.17. The original polynomial is represented in the form (see Theorem 58)
$$
P(x)=a F_{1}(x) \ldots F_{n}(x) G_{1}(x) \ldots G_{k}(x)
$$
where the polynomials $F_{i}(x)$ and $G_{j}(x)$ have the form
$$
\begin{gathered}
F_{i}(x)=x-\alpha_{i}, \quad \alpha_{i} \leq 0, \quad i=1, \ldots, m ; \\
G_{j}(x)=\left(x-\beta_... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,194 |
22.18*. (Jury, GDR, 83). Let $A(n)$ be the set of polynomials of the form
$$
P(x)=a_{0}+a_{\mathrm{i}} x+\ldots+a_{n} x^{n}
$$
where $0 \leqslant a_{0}=a_{n} \leqslant a_{i}=a_{n-1} \leqslant \ldots \leqslant a_{[n / 2]}=a_{[(n+1) / 2]}$. Prove that if $P(x) \in A(n)$ and $Q(x) \in A(m)$, then the polynomial $P(x) Q(... | 22.18. Let us introduce the notation $R_{n, i}(x)=x^{i}+x^{i+1}+\ldots+x^{n-i}$, where $i=0,1, \ldots,[n / 2]$. Then any polynomial $P(x) \in A(n)$ can be represented as
$$
\begin{aligned}
P(x)=a_{0} R_{n, 0}(x) & +\left(a_{1}-a_{0}\right) R_{n, 1}(x)+\ldots \\
& \ldots+\left(a_{[n / 2]}-a_{[n / 2]-1}\right) R_{n,[n /... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,195 |
22.19*. (Jury, VNR, 77). For which values of $n \in \mathbf{N}$ do there exist non-zero polynomials $P$ and $Q$ in $n$ variables with integer coefficients, satisfying the identity
$$
\begin{aligned}
& \left(x_{1}+x_{2}+\ldots+x_{n}\right) P\left(x_{1}, x_{2}, \ldots, x_{n}\right) \equiv Q\left(x_{1}^{2}, x_{2}^{2}, \l... | 22.19. Let $A_{n}$ denote the set of $2^{n}$ all possible sequences $\varepsilon=\left(\varepsilon_{i} ; \ldots ; \varepsilon_{n}\right)$, consisting of numbers $\varepsilon_{i}$, each of which is either 1 or -1. Also, let $x_{\varepsilon}=\mathrm{e}_{1} x_{1}+\ldots+\mathrm{e}_{n} x_{n}$, and we will prove that the pr... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,196 |
22.20*. (Jury, 81). Let for polynomials $P(x)$, $Q(x)$ of degree greater than 0 be denoted
$$
P_{c}=\{z \in \mathbf{C} \mid P(z)=c\}, \quad Q_{c}=\{z \in \mathbf{C} \mid Q(z)=c\}
$$
Prove that if $P_{0}=Q_{0}$ and $P_{1}=Q_{1}$, then $P(x)=Q(x)$, $x \in \mathbf{R}$. | 22.20. If the polynomial $P(x)$ has roots $\alpha_{\mathrm{I}}, \ldots, \alpha_{s}$ of multiplicities $k_{1}, \ldots, k_{s}$ respectively, then the polynomial $Q(x)$ has the same roots (but possibly of different multiplicities), since $P_{0}=Q_{0}$. Similarly, if the polynomial $P(x)-1$ has roots $\beta_{1}, \ldots, \b... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,197 |
22.21*. (PDR, 78). Prove that if polynomials $P(x, y), Q(x, y), R(x, y)$ of degree less than $m \in \mathbf{N}$ satisfy the identity
$$
x^{2 m} P(x, y)+y^{2 m} Q(x, y) \equiv(x+y)^{2 m} R(x, y), \quad x, y \in \mathbb{R}
$$
then
$$
P(x, y) \equiv Q(x, y) \leftrightharpoons R(x, y) \equiv 0
$$
## § 23. Various Prope... | 22.21. Suppose that at least one of the polynomials $P, Q, R$ satisfying the condition of the problem is not identically zero. Then there exist numbers $x_{0}$ and $y_{0}^{*} \neq 0$ such that at least one of the numbers $P\left(x_{0}, y_{0}\right), Q\left(x_{0}, y_{0}\right), R\left(x_{0}, y_{0}\right)$ is not zero. C... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,198 |
23.1. (SRP, 62). Under what restrictions on the integers $p$ and $q$:
a) the polynomial $P(x)=x^{2}+p x+q$ takes even (odd) values for all $x \in \mathbf{Z}$;
b) the polynomial $Q(x)=x^{3}+p x+q$ takes values divisible by 3 for all $x \in \mathbf{Z}$? | 23.1. a) The values of $P(x)$ for all $x \in \mathbf{Z}$ have the same parity if and only if each of the numbers
$$
P(x+1)-P(x)=\left((x+1)^{2}+p(x+1)+q\right)-\left(x^{2}+p x+q\right)=2 x+1+p
$$
is divisible by 2, i.e., when $p$ is odd. In this case, the parity of all values of $P(x)$ is uniquely determined by the p... | q\equiv0(\bmod3),\quadp\equiv2(\bmod3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,199 |
23.2. (GDR, 83). Prove that the polynomial
$$
P(x)=\frac{1}{630} x^{0}-\frac{1}{21} x^{7}+\frac{13}{30} x^{5}-\frac{82}{63} x^{3}+\frac{32}{35} x
$$
takes integer values for all $x \in \mathbf{Z}$. | 23.2. Note that the original polynomial can be represented as $P(x)=\frac{1}{2 \cdot 5 \cdot 7 \cdot 9}(x-4)(x-3)(x-2)(x-1) x(x+1)(x+2)(x+3)(x+4)$.
Since among nine consecutive integers there are always numbers divisible by $2,5,7,9$, then for any $x \in \mathbf{Z}$ the product $\prod_{i=-4}^{4}(x+i)$ is divisible by ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,200 |
23.3. (Czechoslovakia, 62). Find all values of $x \in \mathbf{Z}$ for which the polynomial $2 x^{2}-x-36$ takes values equal to the squares of prime numbers. | 23.3. Let $2 x^{2}-x-36=p^{2}$, where $p$ is a prime number. Then $p^{2}=$ $=(x+4)(2 x-9)=a b$, where it is denoted $a=x+4, b=2 x-9$, and $a, b \in \mathbb{Z}$ with $2 a-b=17$. Since $a$ is an integer that divides $p^{2}$, there are only the following 6 cases:
1) $a=p^{2}, b=1$, then $2 p^{2}-1=17$ and $p=3$, so $x=a-... | 513 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,201 |
23.4. (SRP, 75). For given numbers $p, q \in \mathbf{R}$, find all values that the polynomial $P(x)=x^{2}+p x+q$ takes for $x \in[-1 ; 1]$. | 23.4. The function $P(x)$ has one minimum point $x_{0}=-p / 2$ on the line. For $xx_{0}$ - it increases. Therefore, for the set $A$ of values of the function $P(x)$ on the interval $[-1 ; 1]$, we have the following: if $p1$ and
$$
A=[P(1) ; P(-1)]=[1+p+q ; 1-p+q]
$$
if $-2 \leqslant p \leqslant 2$, then $-1 \leqslant... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,202 | |
23.5. (Beijing, 63). The polynomial $P(x)$ with integer coefficients takes the value 2 at four different values of $x \in \mathbf{Z}$. Prove that for no $x \in \mathbf{Z}$ does this polynomial take the values $1,3,5,7$ and 9. | 23.5. Consider the polynomial $Q(x)=P(x)-2$ and prove the following statement: if the polynomial $Q(x)$ with integer coefficients has four distinct integer roots, then for any value of $x \in \mathbf{Z}$, the integer $|Q(x)|$ is either zero or composite (in particular, it cannot be equal to 1). Let $a, b, c, d$ be dist... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,203 |
23.6. (England, 80). Find at least one set $M$ consisting of 7 consecutive natural numbers, for which there exists a polynomial $P(x)$ of the fifth degree with the following properties:
a) all coefficients of the polynomial $P(x)$ are integers;
b) for five numbers $k \in M$, including the smallest and largest numbers... | 23.6. The set $M=\{25 ; 26 ; 27 ; 28 ; 29 ; 30 ; 31\}$ and the polynomial $P(x)=x+(x-25)(x-27)(x-28)(x-29)(x-31)$ satisfy all the conditions of the problem (in this case, condition b) is satisfied for $k=25,27,28$, 29,31, and condition c) - for $k=30$ ). | M={25;26;27;28;29;30;31},P(x)=x+(x-25)(x-27)(x-28)(x-29)(x-31) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,204 |
23.7. (GDR, 74). a) Prove that there does not exist a polynomial $P(x)$ for which for any $x \in \mathbf{R}$ the inequalities are satisfied: 1) $P^{\prime}(x)>P^{\prime \prime}(x)$ and 2) $P(x)>P^{\prime \prime}(x)$.
b) Will the statement in a) be true if inequality 1) is replaced by the inequality $\left.1^{\prime}\r... | 23.7. a) If $P(x)$ is a constant, then $P^{\prime}(x) \Rightarrow P^{\prime \prime}(x) \Rightarrow 0$, and inequality 1) is not satisfied. Let $\operatorname{deg} P(x)=n \geqslant 1$, then if $n$ is odd, $\operatorname{deg}\left(P(x)-P^{\prime \prime}(x)\right)=n$ is an odd number, from which $P(x)-P^{\prime \prime}(x)... | proof | Calculus | proof | Yes | Yes | olympiads | false | 29,205 |
23.8. (SRP, 82). Let polynomials $P_{0}(x)$, $P_{i}(x), \ldots, P_{n}(x)$ with real coefficients and numbers $a_{i}, \ldots, a_{n} \in \mathbf{R}$ be given. Prove that if the function
$$
f(x)=P_{0}(x)+\sum_{k=1}^{n} a_{k}\left|P_{k}(x)\right|
$$
does not take any value more than once for $x \in \mathbf{R}$, then the ... | 23.8. We can assume that all polynomials $P_{1}(x), \ldots, P_{n}(x)$ are non-zero. Let us choose some number $x^{+}$, greater than all the real roots of any of these polynomials. Then for all $x \geqslant x^{+}$, the function $f(x)$ coincides with the polynomial
$$
P^{+}(x)=P_{0}(x)+\sum_{k=1}^{n} a_{k}\left(\operato... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,206 |
23.11. (Finland, 80). Let on the coordinate plane be given a curve which is the graph of some polynomial
$$
P(x)=x^{4}+p x^{3}+q x^{2}+r x+s(p, q, r, s \in \mathbf{R})
$$
A line on this plane is called horizontal if it is parallel to the x-axis and intersects the curve at four points $A, B, C, D$ (listed from left to... | 23.11. Let an arbitrary horizontal line $y=y_{0}$ intersect the curve $y=P(x)$ at four points $A, B, C, D$ with abscissas $x_{1}, x_{2}, x_{3}, x_{4}$, respectively. We need to prove that $|A B|+|B C|>|C D|$, i.e., when
$$
\left(x_{2}-x_{1}\right)+\left(x_{3}-x_{1}\right)>\left(x_{4}-x_{1}\right)
$$
which is equivale... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,208 |
23.12. (NBR, 77). Let $Q(x)$ be a non-zero polynomial. Prove that for each $n \in \mathbf{Z}^{+}$ the polynomial $P(x)=$ $=(x-1)^{n} Q(x)$ has at least $n+1$ non-zero coefficients. | 23.12. We will prove the required statement by induction on the number $n \in \mathbf{Z}^{+}$. For $n=0$, the polynomial $P(x)=Q(x)$ has at least one non-zero coefficient, since $Q(x) \neq 0$. Suppose that for some $n \geq 1$ it has already been proven that if the polynomial $R(x)$ is non-zero, then the polynomial $(x-... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,209 |
23.13*. (CSSR, 74). Let $M$ be the set of all polynomials of the form
$$
P(x)=a x^{3}+b x^{2}+c x+d \quad(a, b, c, d \in \mathbf{R})
$$
satisfying the inequality $|P(x)| \leqslant 1$ for $x \in[-1 ; 1]$. Prove that some number $k$ provides the estimate $|a| \leqslant k$ for all polynomials $P(x) \in M$. Find the smal... | 23.13. The polynomial $P_{0}(x)=4 x^{3}-3 x$ belongs to the set $M$, since $P_{0}(-1)=-1, P_{0}(1)=1$, and at its extremum points we have $P_{0}(-1 / 2)=1, P_{0}(1 / 2)=-1$. Let's prove that for any polynomial $P(x) \in M$ the estimate $|a| \leqslant 4$ holds. Suppose, on the contrary, that there exists a polynomial $P... | 4 | Algebra | proof | Yes | Yes | olympiads | false | 29,210 |
23.14*. (Jury, Finland, 83). Let $p$ and $q$ be arbitrary natural numbers. Prove that there exists a polynomial $P(x)$ with integer coefficients such that for all values from some interval $I \subset \mathbf{R}$ of length $1 / q$ the inequality $|P(x)-p / q|<1 / q^{2}$ is satisfied. | 23.14. If $q=1$, then let $P(x)=p$. Let $q>1$, then consider the interval $I=(1 /(2 q), 3 /(2 q))$ of length $1 / q$. Since $3 /(2 q)<1$, there exists a number $m \in \mathbb{N}$ such that $(3 /(2 q))^{m}<1 / q$. Let $a=1 - (1 /(2 q))^{m}$, then we have the inequalities $0<1-q x^{m}<a<1$ for all $x \in I$. Choose a num... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,211 |
23.15. (GDR, 80). Find all pairs of polynomials $P(x)$, $Q(x)$ of the third degree with real coefficients, satisfying the following four conditions:
a) both polynomials at each of the points $x=1,2,3,4$ take the value 0 or 1;
b) if $P(1)=0$ or $P(2)=1$, then $Q(1)=Q(3)=1$;
c) if $P(2)=0$ or $P(4)=0$, then $Q(2)=Q(4)... | 23.15. Let $\alpha_{k}=P(k), \beta_{k}=Q(k)$, where $k=1,2,3,4$, and the polynomials $P(x)$ and $Q(x)$ satisfy the conditions of the problem. Then the "four-digit" numbers $\overline{\alpha_{1} \alpha_{2} \alpha_{3} \alpha_{1}}$ and $\overline{\beta_{1} \beta_{2} \beta_{3} \beta_{4}}$ are not equal to any of the number... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,212 |
23.16. (USA, 75). The polynomial $P(x)$ of degree $n$ satisfies the equalities $P(k)=k /(k+1)$ for $k=0,1, \ldots, n$. Find $P(n+1)$. | 23.16. Note that there exists only one polynomial $P(x)$ satisfying the conditions of the problem, since if there were another polynomial $Q(x) \not \equiv P(x)$ with the same properties, then the polynomial $P(x)-Q(x)$ of degree not greater than $n$ would have at least $n+1$ roots. Since for the polynomial $R(x)=x+\fr... | \frac{n+1+(-1)^{n+1}}{n+2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,213 |
23.17. (Jury, 81). The polynomial $P(x)$ of degree $n$ satisfies the equalities $P(k)=1 / C_{n+1}^{k}$ for $k=0,1, \ldots, n$. Find $P(n+1)$. | 23.17. Using the Lagrange interpolation formula, we obtain (below it is implied that $1 \leqslant i \leqslant n$)
$$
\begin{aligned}
& P(x)=\sum_{k=0}^{n} \frac{1}{C_{n+1}^{k}} \prod_{i \neq k} \frac{(x-i)}{(k-i)}=\sum_{k=0}^{n} \frac{\prod_{i \neq k}(x-i)}{\left.C_{n+1}^{k}(-1)^{n-k}(n-k)!k\right)}= \\
& =\sum_{k=0}^... | P(n+1)=\begin{cases}0& | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,214 |
24.1. (Czechoslovakia, 73). How many different pairs of non-intersecting subsets does a set consisting of $n$ elements have? | 24.1. Let's find the number of different pairs of non-intersecting subsets under the condition that the first and second subsets are distinguished in the pair. For each of the $n$ elements, there are 3 possibilities: it can either be included in the first subset, or included in the second subset, or not included in eit... | (3^{n}+1)/2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,217 |
24.2. (PRN, 78). The set $X$ consists of $n$ elements. For two subsets $A_{1}, A_{2} \subset X$, the number of elements in the set $A_{1} \cap A_{2}$ is counted. Prove that the sum of all obtained numbers is $n \cdot 4^{n-1}$. | 24.2. Since there are only $2^{n}$ different subsets of the set $X$, there exist $\left(2^{n}\right)^{2}=4^{n}$ different ordered pairs of subsets. We will divide all these pairs into $4^{n-1}$ quartets, including each pair ( $A_{1} ; A_{2}$ ) in one quartet with the pairs ( $\bar{A}_{1} ; A_{2}$ ), $\left(A_{1} ; \ove... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,218 |
24.3. (New York, 78). For numbers $n>3, k=$ $=\left[\frac{1}{6} n(n+1)\right]$ and the set $X_{n}$, consisting of $n(n+1) / 2$ elements, it is known that $k$ elements of this set are blue, $k$ elements are red, and all the other elements are white. Prove that the set $X_{n}$ can be partitioned into $n$ pairwise non-int... | 24.3. We will prove the statement by induction on $n>3$. For $n=4,5$, $6,7,8,9$, the possible partitions of the set $X_{n}$ into subsets $A_{1}, \ldots, A_{n}$ are given in the table, where for each value of $n$ and each of the three colors, the numbers $m$ of those sets $A_{m}$ are indicated which should consist of el... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,219 |
24.4. (SFRY, 72). For each value of $n \in \mathbf{N}$, find the largest number $k \in \mathbf{N}$ with the following property: in a set consisting of $n$ elements, one can choose $k$ different subsets, any two of which have a non-empty intersection. | 24.4. Let us fix the element $a_{1}$ of the set $X=\left\{a_{1} ; a_{2} ; \ldots ; a_{n}\right\}$ and consider only subsets containing $a_{1}$. The number of such subsets is equal to the number of subsets of the set $\left\{a_{2} ; \ldots ; a_{n}\right\}$, i.e., the number $2^{n-1}$. Therefore, $k \geqslant 2^{n-1}$. O... | 2^{n-1} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,220 |
24.5. (England, 76). Let 50 subsets $A_{1}, \ldots, A_{50}$ be chosen in a finite set $X$, each of which contains more than half of the elements of the set $X$. Prove that one can find a subset $B \subset X$, containing no more than 5 elements and having at least one element in common with each of the sets $A_{1}, \ldo... | 24.5. Let the number of elements in the set $X$ be $n$. Each of the selected subsets $A_{1}, \ldots, A_{50}$ contains more than $n / 2$ elements, so the sum of the number of elements in all these sets exceeds $50 \cdot(n / 2)=25 n$. By the pigeonhole principle, there exists an element in the set $X$ that belongs to at ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,221 |
24.6. (Austria - PR China, 78). Given 1978 sets, each of which has 40 elements. It is known that any two of these sets have exactly one common element. Prove that there exists an element that belongs to all 1978 sets. | 24.6. Consider any set $A$ from the given 1978 sets. It intersects with each of the other 1977 sets, so there exists an element $a \in A$ that belongs to at least 50 of these sets. (Indeed, if each of the 40 elements of set $A$ belongs to no more than 49 sets, then there are no more than $40 \times 49 = 1960$ sets diff... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,222 |
24.7. (Beijing, 63). A set of $2^{n}$ elements is partitioned into pairwise non-intersecting subsets. An operation is considered, consisting of transferring some elements from one subset to another, with the number of transferred elements being equal to the number of elements in the second set (which should contain no ... | 24.7. Consider subsets (if any) that contain an odd number of elements. Since the total number of elements is even, the number of sets with an odd number of elements is also even. We will divide these sets into pairs (arbitrarily). With the sets from each pair, we will perform the operation specified in the problem, i.... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,223 |
24.8. (Jury, CSSR, 79). Let $M$ be a subset of the set of all pairs of natural numbers $i<k$, not exceeding a given natural number $n \geqslant 2$. If a pair $i<k$ belongs to the set $M$, then no pair $k<m$ belongs to it. What is the maximum number of pairs that can be in the set $M$? | 24.8. Let $A$ and $B$ denote the sets of all smaller and, respectively, larger numbers that form the pairs of the set $M$. Then, by the condition of the problem, no element from $B$ is contained in the set $A$, and therefore, $A \cap B=\varnothing$. Let the number of elements in the sets $A$ and $B$ be $a$ and $b$ resp... | [n^{2}/4] | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,224 |
24.10*. (USA, 79). In a set consisting of $n \geqslant 5$ elements, $n+1$ different three-element subsets are chosen. Prove that there will be two chosen subsets having exactly one common element. | 24.10. Suppose the opposite. Then any two of the selected subsets either do not intersect or have exactly two common elements. If subsets $A$ and $B$ have exactly two common elements, we will write $A \sim B$. Let $A, B, C$ be three subsets. We will prove that if $A \sim B$ and $B \sim C$, then $A \sim C$. Indeed, let ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,226 |
24.11*. (CRR, 78). The set $X$ is partitioned into pairwise disjoint subsets $A_{1}, \ldots, A_{n}$, and also partitioned into pairwise disjoint subsets $\boldsymbol{B}_{i}, \ldots, B_{n}$. It is known that the union $A_{i} \cup B_{j}$ of any two disjoint subsets $A_{i}, B_{j}(1 \leqslant i \leqslant n, 1 \leqslant$ $\... | 24.11. Let $k$ be the number of elements in the set among $A_{\mathrm{i}}, \ldots, A_{n}$, $B_{1}, \ldots, B_{n}$, which has the least number of elements. For definiteness, let's assume that the set $A_{1}$ contains $k$ elements. The sets $B_{1}, \ldots, B_{n}$ do not intersect with each other. Therefore, the condition... | \frac{n^2}{2} | Combinatorics | proof | Yes | Yes | olympiads | false | 29,227 |
25.1. (England, 72). On the set $S$, a relation $\rightarrow$ is defined, which is satisfied for pairs of elements from the set $S$ and has the following properties: 1) for any distinct elements $a, b \in S$, exactly one of the relations $a \rightarrow b$ or $b \rightarrow a$ holds; 2) for any three distinct elements $... | 25.1. Suppose that in the set $S$ there are three elements $a, b, c$ such that $a \longrightarrow b$ and $a \longrightarrow c$. If $b \longrightarrow c$, then from $a \longrightarrow b$ and $b \longrightarrow c$ it follows that $c \longrightarrow a$, which contradicts the condition $a \longrightarrow c$. If $c \longrig... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,228 |
25.2. (USA, 82). In a society consisting of 1982 people, among any four people, at least one can be chosen who is acquainted with the other three. What is the minimum possible number of people who are acquainted with everyone? | 25.2. If there are no strangers, then the number of people who know everyone is 1982. Let $A$ and $B$ not know each other. Then all other people know each other (if $C$ does not know $D$, then in the group $A, B, C, D$ no one knows the other three). If $A$ and $B$ know all the others, then 1980 people know everyone. If... | 1979 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,229 |
25.3. (NBR, 78). In a company of five people, among any three people, there will be two who know each other, and two who are not acquainted. Prove that the company can be seated around a round table so that on both sides of each person sit their acquaintances. | 25.3. Note that each guest knows exactly two of the other four guests. Indeed, suppose guest $A$ knows three guests: $B, C$, and $D$. Then if $B$ and $C$ know each other, there are no strangers among $A, B$, and $C$. Therefore, $B$ and $C$ (similarly $B$ and $D, C$ and $D$) are not acquainted, and there are no acquaint... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,230 |
25.4. (USA, 78). Nine mathematicians met at an international conference and found that among any three of them, at least two speak the same language. In addition, each mathematician can speak no more than three languages. Prove that at least three of them speak the same language. | 25.4. Suppose that no three mathematicians speak the same language. Consider an arbitrary mathematician $A$. He speaks no more than three languages, and on each of these languages, no more than one other mathematician speaks (otherwise we would get a contradiction with the assumption). Therefore, there are five mathema... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,231 |
25.5. a) (England, 80). In a room, there are 10 people, and among any three of them, there are two who are acquainted with each other. Prove that there will be four people, any two of whom are acquainted with each other.
6) (Curie, PDR, 77). Will the statement in part a) remain true if the number 10 is replaced with 9... | 25.5. a) Suppose that among any four people there are two who are not acquainted. Then, for example, person $A$ cannot have more than three strangers: if $A$ has four strangers, then according to the assumption, among them there will be two who are not acquainted with each other, and they, together with $A$, would form... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,232 |
25.6. (USSR, 81; USA, 81). In a certain country, any two cities are directly connected by one of the following means of transportation: bus, train, or airplane. It is known that there is no city provided with all three types of transportation, and at the same time, there do not exist three cities such that any two of t... | 25.6. Suppose there are five cities connected in the manner indicated in the problem. First, let's prove that no city has three lines of the same type of transport leading out of it. Let city \( A \) be connected to cities \( B, C \), and \( D \), for example, by airplane. Then, according to the condition, no pair of c... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,233 |
25.7. (Yugoslavia, 75). In a certain society, any two acquaintances do not have common acquaintances, and any two non-acquaintances have exactly two common acquaintances. Prove that in this society everyone has the same number of acquaintances. | 25.7. First, we will prove that any two acquaintances have the same number of acquaintances. Indeed, let $A$ and $B$ be acquaintances, and let $A_{1}, A_{2}, \ldots, A_{n}$ be all the acquaintances of $A$, except for $B$. According to the problem's condition, no two among $B, A_{1}, A_{2}, \ldots, A_{n}$ are acquainted... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,234 |
25.8. (VNR, 77). In each of three schools, there are $n$ students. Any student has a total of $n+1$ acquaintances among students from the other two schools. Prove that it is possible to choose one student from each school such that all three chosen students are acquainted with each other. | 25.8. Among all $3 n$ students, we will choose a student (more precisely, one of such students) who has the largest number $k$ of acquaintances in one of the two other schools. Let's assume for definiteness that this student is $A$ from the first school, who knows $k$ students, for example, from the second school. Then... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,235 |
25.9*. (Austria, 73). In space, there are $2 n$ distinct points $A_{1}, A_{2}, \ldots, A_{2 n}(n>1)$. No three of them lie on the same line. Let $M$ be the set of $\left(n^{2}+1\right)$ segments whose endpoints are the given points. Prove that there exists at least one triangle with vertices at some points $A_{r}, A_{s... | 25.9. We will prove by induction on $n \in N$ that if in a graph with $2 n$ vertices no three edges form a triangle, then the number of edges does not exceed $n^{2}$. For $n=1$, the number of edges is always no more than $1=n^{2}$. Suppose the statement is proven for the number $n$. We will prove it for the number $n+1... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,236 |
25.10*. (CRR, 78). At an evening gathering, several young men and women were present. It turned out that if any group of young men is chosen, the number of women who are acquainted with at least one of the young men in this group is no less than the number of young men in the group. Prove that all the young men can sim... | 25.10. We will solve the problem by induction on \( n \) - the number of young men. For \( n=1 \), the statement follows from the fact that for one young man, there will definitely be an acquainted girl. Suppose the statement is proven for all numbers less than \( n \). Let's prove it for the number \( n \). Consider t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,237 |
26.1. (SFRY, 75). Four ones and five zeros are written in a circle in some arbitrary order. The following operation is performed on them: a zero is written between identical digits, and a one between different digits, after which the original digits are erased. Then the same operation is performed on the resulting digi... | 26.1. Suppose that after several operations specified in the problem, we ended up with a set of all ones. Let this be the first time it happened after the $k$-th operation. Then after the ( $k-1$ )-th operation, all the digits on the circle were the same and not equal to zero, and therefore equal to one. This means tha... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,239 |
26.2. (Jury, NRB, 79). On a grid paper, $n$ cells are marked arbitrarily. Prove that it is always possible to select at least $n / 4$ cells from them, such that no two selected cells touch each other (cells are considered touching if they share at least one vertex). | 26.2. Let's divide the entire plane into identical squares, each consisting of four cells. Further, in each of these squares, the "upper left" cell will be called black, the "upper right" - white, the "lower left" - red, and the "lower right" - blue. On the one hand, any two cells of the same color do not touch. On the... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,240 |
26.4. (CSSR, 77). On a line, $n$ distinct points $A_{1}, A_{2}, \ldots, A_{n}(n \geqslant 4)$ are marked. Each of these points is painted in one of four colors, and all four colors are present. Prove that there exists a segment of the line containing exactly one point of two colors and at least one point of the two rem... | 26.4. Without loss of generality, assume that $A_{1}<A_{2}<\ldots<A_{n}$. We will choose the smallest index $i$ for which among the points $A_{\mathbf{i}}, \ldots, A_{i}$ there are points of all four colors. Then the color of $A_{i}$ is different from the colors of $A_{1}, \ldots, A_{i-1}$. Now we will choose the large... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,242 |
26.5. (SRP, 78). On a plane, there is a set $M$ consisting of $n$ points, no three of which lie on the same line. Each segment with endpoints in $M$ is assigned either the number +1 or the number -1, and the number of segments assigned the number -1 is $m$. A triangle with vertices in $M$ is called negative if the prod... | 26.5. Let $k$ be the number of negative triangles. For each triangle, we multiply the numbers corresponding to its sides, and then multiply all the obtained products. In the end, we get the number $(-1)^{\text {R }}$. Note that the number corresponding to any segment appears in this product $n-2$ times, since any segme... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,243 |
26.6. (Beijing, 64). On a circular road, there are $n$ gas stations, which together contain enough gasoline for one car to make a complete trip around the circle exactly once. Prove that a car with an empty tank (of unlimited capacity) can start from one of the gas stations (after refueling there) and complete a full t... | 26.6. Suppose a car has enough fuel to travel the entire road without refueling. Let it start moving from any station (picking up all the fuel at each station along the way) and end its journey at the same point with the same amount of fuel. Let $A$ be the station, approaching which the car had the least amount of fuel... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 29,244 |
26.7. (Yugoslavia, 74). On an $8 \times 8$ chessboard, 8 white chips are placed on the first row and 8 black chips on the eighth row. Players take turns (whites start) making moves, which consist of moving one of their chips vertically forward or backward by one or several squares. It is forbidden to remove chips from ... | 26.7. Let's divide the board into four parts, with two verticals in each. Suppose that in response to the white's move in any part, the blacks make a move in the same part. Then, if the blacks deprive the whites of the ability to move in each of the parts, the whites will not be able to move at all. Therefore, it is su... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 29,245 |
26.8. (SFRY, 83). A rectangular strip of size $1 \times n(n \geqslant 4)$ is composed of unit squares numbered $1,2, \ldots, n$. There is a chip on each of the squares numbered $n-2$, $n-1, n$. Two players play the following game: each player, on their turn, can move any chip to any free square with a smaller number. T... | 26.8. Let's divide all integers starting from two into non-overlapping pairs of the form ( $2 k ; 2 k+1$ ), $k \in \mathbb{N}$. Then among the three numbers $n$, $n-1, n-2$, two must necessarily form a pair. The beginner should, on the first move, move the chip standing on the field with the number that did not fall in... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,246 |
26.9. (SFRY, 83). "Dolphin" - a figure that moves one field up, to the right, or diagonally to the left
| | | | | | | | |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| | | | | | | | |
| | | | | | | | |
| | | | | 1 | | | |
| | | | | | $\rightarrow$ | | |
| | | | |... | 26.9. We will number the horizontal rows of the board from bottom to top with the numbers \(0, 1, \ldots, 7\) and the vertical columns of the board from left to right with the same numbers. We will assign to each cell of the board the sum of the numbers of the vertical and horizontal lines at the intersection of which ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 29,247 |
26.10. (Jury, PDR, 82). On an infinite chessboard in all directions, where initially tokens are placed exactly covering a rectangle of size $3 k \times n$, a game proceeds according to the following rules: any token can jump over any adjacent (vertically or horizontally) token, provided that the next cell is unoccupied... | 26.10. Let's divide all the cells of an infinite board into 3 sets as shown in Fig. 129 (different numbers denote cells of different sets). Then, with each move, the number of chips in two sets increases by one, and in one set, it increases by one. In this case, the parity of the number of chips in each set changes to ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,248 |
26.11. (SRP, 78). Given a convex polyhedron with $n \geqslant 5$ faces, from each vertex of which exactly three edges emerge. Two players play the following game: each in turn writes their name on one of the free faces. To win, a player must write their name on one of the free faces. faces, having a common vertex. Prov... | 26.11. We will prove that at least one of the faces of the given polyhedron is not a triangle. Suppose, on the contrary, that all $n$ of its faces are triangles. Then this polyhedron has $3n / 2$ edges (since each of the three edges of any face belongs simultaneously to two faces) and $n$ vertices (since each of the th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,249 |
26.12. (Jury, SRP, 77). Let $A=\left(a_{1} ; \ldots ; a_{m}\right)$ be a set of $m=2^{n}(n \in \mathbf{N})$ numbers $a_{i} \in\{1 ;-1\}, i=1,2, \ldots, m$. The operation $S$ is defined by the formula
$$
S(A)=\left(a_{1} a_{2} ; a_{2} a_{3} ; \ldots ; a_{m} a_{1}\right)
$$
Prove that for any set $A$ in the sequence $A... | 26.12. We will prove by induction on $n \in \mathbf{Z}^{+}$ that the result of applying $2^n$ operations $S$ is a set of $m=2^n$ ones. For $n=0$, we have $S(A)=(a_1 a_1)=1$. Suppose the statement is already proven for the number $n-1$. We will prove it for the number $n$. Notice that in the set
\[
\begin{aligned}
T(A)... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,250 |
26.13. (New York, 76). Prove that a square board of size $2 n \times 2 n$, where $n$ is not divisible by 3, from which one arbitrary cell is removed, can be covered by a shape of size $2 \times 2$, from which one cell is removed.) | 26.13. Note that a rectangle $3 \times 2$ can be covered by a triomino (Fig. 130). Therefore, if the removed cell does not lie in a figure that is the union of strips six cells wide, adjacent to two adjacent sides of the square, then we can cover
. a) Let each cell of a rectangular board of size $4 \times 7$ be colored either white or black. Prove that there must be a rectangle on the board, formed by horizontal and vertical lines of the board, all four corner cells of which are colored the same color.
b) Provide an example of a coloring of a r... | 26.14. a) Consider all possible pairs of cells of the same color located in the same column (assuming that the columns have a height of 4 and the rows have a length of 7). In each column, there are at least two such pairs, so there are at least 14 such pairs on the board. Therefore, there exists a color that is present... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,252 |
26.15. (Sweden, 82). On a plane with a rectangular coordinate system, consider the set $M$ of points $(x, y)$, where $x, y \in \mathbf{N}$, and $x \leqslant 12, y \leqslant 12$. Each of these 144 points is colored either red, white, or blue. Prove that there exists a rectangle (with sides parallel to the axes) all of w... | 26.15. Naturally, we will divide all points from $M$ into 12 rows (by the first coordinate) and 12 columns (by the second coordinate). There exists a color in which at least $144 / 3 = 48$ points are painted. We will select exactly 48 points from them. The number of selected points in the column with the second coordin... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,253 |
26.16. a) (Jury, SRV, 77). On the coordinate plane, $n \geqslant 3$ points with integer coordinates are marked such that any three of them form a triangle whose medians do not intersect at a point with integer coordinates. Find the largest number $n$ for which this is possible.
b) (Jury, SRR, 77). In space, 37 differe... | 26.16. a) If $\left(x_{1} ; y_{1}\right),\left(x_{2} ; y_{2}\right),\left(x_{3} ; y_{3}\right)$ are the coordinates of the vertices of a triangle, then the point of intersection of its medians has coordinates
$$
\left(\frac{x_{1}+x_{2}+x_{3}}{3} ; \frac{y_{1}+y_{2}+y_{3}}{3}\right)
$$
We will call a point $(x ; y)$ a... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,254 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.