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11.2. Graph the solution set of the system of equations $\left\{\begin{array}{l}x-2 y=1 \\ x^{3}-6 x y-8 y^{3}=1\end{array}\right.$ on the coordinate plane. | Answer. The set of solutions is the line $y=\frac{x-1}{2}$. Solution. The second equation is a consequence of the first, since
$x^{3}-8 y^{3}-6 x y=(x-2 y)\left(x^{2}+2 x y+4 y^{2}\right)-6 x y=$ $=x^{2}+2 x y+4 y^{2}-6 x y=(x-2 y)^{2}=1$. | \frac{x-1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,357 |
10.3. Plot on the coordinate plane the set of points satisfying the equation $2 x^{2}+y^{2}+3 x y+3 x+y=2$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | The sought set of points is the union of two lines $y=-x-2$ and $y=-2x+1$. Solution. Consider the given equation as a quadratic equation in terms of $y: y^{2}+y(3x+1)+2x^{2}+3x-2=0$. Its discriminant is $D=x^{2}-6x+9=(x-3)^{2}$, and the solutions are $y_{1}=-x-2, y_{2}=-2x+1$. These solutions represent two lines on the... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,358 | |
9.1 Given a quadratic trinomial $a x^{2}+b x+c$, which has roots. Is it necessary that the quadratic trinomial a) $a^{2} x^{2}+b^{2} x+c^{2}$ ? b) $a^{3} x^{2}+b^{3} x+c^{3}$ ? | Answer: a) not necessarily; b) necessarily. Solution. a) Consider the example of the quadratic trinomial $x^{2}+x-6$, which has roots (-3 and 2), while the trinomial $x^{2}+x+36$ has no roots. b) We have $b^{2} \geq 4 a c$, since the discriminant of the original trinomial is non-negative. If $a c>0$, then we raise the ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,359 |
9.4. Given 100 positive numbers. Can we assert that: a) the sum of any ten of them is less than 10, if it is known that the sum of any seven of them is less than 7; b) the sum of any seven of them is less than 7, if it is known that the sum of any ten of them is less than 10? | Answer: a) can; b) cannot. Solution. Let $x_{1}, x_{2}, \ldots, x_{10}$ be any 10 numbers from the given 100 numbers. It is required to prove that their sum is less than 10 if the sum of any seven numbers is less than 7. Denote
$$
S_{1}=x_{1}+x_{2}+\ldots+x_{7}, S_{2}=x_{2}+x_{3}+\ldots+x_{8}, \ldots, S_{10}=x_{10}+x_... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 22,362 |
9.1. Given the equation $x^{3}+5 y=y^{3}+5 x$. Do there exist solutions to this equation in the form of a) natural numbers $x \neq y$? ; b) integers $x \neq y$? | Answer: a) do not exist; b) do not exist. Solution. We have $x^{3}-y^{3}=5(x-y)$. If $x \neq y$, then after dividing by ( $x-y$ ) we get $x^{2}+x y+y^{2}=5$. a) For natural $x, y$: the smallest (and different) natural numbers 1 and 2, when substituted into the left side of the last equation, give $7>5$, and therefore, ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,363 |
9.3 Does there exist a convex 27-gon where all angles are different and expressed as whole numbers of degrees? | Answer: does not exist. Solution. The sum of the exterior angles of a convex $n$-gon is $360^{\circ}$. If all the angles of a 27-gon were distinct natural numbers, then the sum of the exterior angles would be no less than $1^{\circ}+2^{\circ}+\ldots+27^{\circ}=378^{\circ}>360^{\circ}$. The obtained contradiction shows ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,365 |
9.4 In the class, each boy is friends with exactly three girls, and each girl is friends with exactly two boys. Can there be a total of: a) 32 people? b) 30 people? | Answer. a) cannot; b) can. Solution. See problem 10.4. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,366 |
10.1. Plot on the coordinate plane the set of points whose coordinates satisfy the equation $4 x^{2} y^{2}=4 x y+3$. | Solution. The given equation can be reduced to an equivalent one: $(2 x y-1)^{2}=4 \Leftrightarrow 2 x y=1 \pm 2$. Therefore, the set of points in question consists of two hyperbolas $y=\frac{3 / 2}{x}$ and $y=\frac{-1 / 2}{x}$ (each quadrant will have one branch of the corresponding hyperbola). | \frac{3/2}{x} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,367 |
10.2. Find all values of the parameter $b$, for each of which the system of equations
$$
\left\{\begin{array}{l}
x^{2}-2 x+y^{2}=0 \\
a x+y=a b
\end{array}\right.
$$
has a solution for any $a$. | Answer. $\quad 0 \leq b \leq 2$.
Solution. On the coordinate plane, the first equation represents a circle $S$ : $(x-1)^{2}+y^{2}=1$ with a radius of 1 centered at the point ( $1 ; 0$ ). The second equation is the equation of a line $l$ passing through the point $M(b ; 0)$ with a slope of $a$. If the point $M$ lies in... | 0\leqb\leq2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,368 |
10.3. Given a triangle $A B C$, point $O$ is the center of the inscribed circle. Find the angle $A$ if it is known that the radii of the circumscribed circles of triangles $A B C$ and $B O C$ are the same. | Answer. $\quad 60^{\circ}$.
Solution. Let $\angle A=\alpha$. Note that $\angle B O C=90^{\circ}+\frac{\alpha}{2}$; indeed, $\angle B O C=180^{\circ}-(\angle O B C+\angle B C O)=180^{\circ}-\frac{1}{2}(\angle B+\angle C)=180^{\circ}-\frac{1}{2}\left(180^{\circ}-\alpha\right)=90^{\circ}+\frac{\alpha}{2}$. Then $B C=2 R_... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,369 |
10.4. Given the function $f(x)=\sqrt{x^{2}-x}$. Find the domain of the function $y=f(f(x))$. | Answer. $\quad\left(-\infty, \frac{1-\sqrt{5}}{2}\right] \cup\{0\} \cup\{1\} \cup\left[\frac{1+\sqrt{5}}{2},+\infty\right)$.
Solution. We have $y=\sqrt{u^{2}-u}$, where $u=\sqrt{x^{2}-x}$. To find the domain, we need to solve the inequality $u^{2}-u \geq 0 \Leftrightarrow u \leq 0$ or $u \geq 1$.
In the first case, w... | (-\infty,\frac{1-\sqrt{5}}{2}]\cup{0}\cup{1}\cup[\frac{1+\sqrt{5}}{2},+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,370 |
10.5. a) Given any ten integers. Prove that it is possible to choose two numbers such that the difference of their cubes is divisible by 27. b) Is the same statement true for any eight integers? | Answer. b) It is true.
Solution. a) Among ten numbers, there will be two numbers $a, b$ that give the same remainder when divided by 9. Then $a^{3}-b^{3}=(a-b)\left((a-b)^{2}+3 a b\right)$. Here, the first factor is divisible by 9, and the second by 3, which leads to the result.
b) If among the remainders when the ei... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,371 |
7.1. A natural number $n$ was multiplied by the sum of the digits of the number $3 n$, and the resulting number was then multiplied by 2. As a result, 2022 was obtained. Find $n$. | Answer: 337. Solution. Let $s(N)$ denote the sum of the digits of the number $N$. Then the condition of the problem can be written as $2 n \cdot s(3 n)=2 \cdot 3 \cdot 337$, i.e., $n \cdot s(3 n)=3 \cdot 337=1011$. Therefore, $n \leq 1011$, so $3 n$ is at most a four-digit number, and $s(3 n)<4 \cdot 9=36$. Thus, $s(3 ... | 337 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,372 |
7.2. Let $a$ be the number of six-digit numbers divisible by 13 but not divisible by 17, and $b$ be the number of six-digit numbers divisible by 17 but not divisible by 13.
Find $a-b$. | Answer: 16290. Solution. Let $c$ be the number of six-digit numbers divisible by both 13 and 17. Then $a+c$ is the number of all six-digit numbers divisible by 13, and therefore $a+c=\left[\frac{999999}{13}\right]-\left[\frac{99999}{13}\right]=76923-7692=69231$.
denotes the integer part of the number $x$ ). Similarly,... | 16290 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,373 |
7.3. On side $B C$ of triangle $A B C$, a point $M$ is marked such that $A B=B M$ and $A M=M C$. It is known that angle $B$ is five times angle $C$. Find the angles of the triangle. | Answer. $\angle A=60^{\circ}, \angle B=100^{\circ}, \angle C=20^{\circ}$. Solution. Triangles $A B M$ and $AMC$ are isosceles, so the angles at their bases are equal. Let these angles be $x$ and $y$ respectively. Then, by the property of the exterior angle $A M B$ for triangle $AMC$, we have $x=2 y$. Therefore, the sum... | \angleA=60,\angleB=100,\angleC=20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,374 |
7.4. Find the largest natural number, all digits of which are different, and the product of these digits is a cube of some natural number. | Answer: 984321. Solution. Obviously, among the digits of the desired number $x$, there is no zero. The product of all digits from 1 to 9 is $2^{7} \cdot 3^{4} \cdot 5^{1} \cdot 7^{1}$. Therefore, $x$ cannot contain the digit 5 or the digit 7 (otherwise, the product of the digits of the number $x$ would have to be divis... | 984321 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,375 |
7.5. What is the maximum number of three-cell corners that can be cut out from a grid rectangle of size: a) $5 \times 10$ cells; b) $5 \times 9$ cells? | Answer: a) 16; b) 15. Solution. a) It is clear that no more than 16 corners can be cut out, since
otherwise the rectangle must contain no fewer than \(17 \cdot 3=51>50\) cells. The left image... | )16;b)15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,376 |
8.3. Find the largest natural number, all digits of which are different, and the product of these digits is a cube of some natural number. | Answer: 984321. Solution. See problem 7.4 | 984321 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,379 |
8.4. In an $n \times n$ grid, each cell is colored either white or black. For what smallest $n$ is it always true (i.e., for any coloring) that there exists a rectangle whose vertices coincide with the centers of four cells of the same color? | Answer: $n=5$. Proof. We will prove that for $n=5$ (and thus for $\mathrm{n}>5$) such a rectangle will exist. Consider the bottom row of the table. In it, there are at least 3 cells of the same color. Let's assume for definiteness that these are white cells. Then consider the three columns with these cells at the base,... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,380 |
8.5. Girls stood in a circle, some of them wearing scarves. A circle is called correct if each girl without a scarf has a neighbor wearing a scarf. a) What is the minimum number of scarves in a correct circle of 25 girls? b) Prove that if in this correct circle of 25 girls there are more than 12 scarves, then some girl... | Answer: a) 9. Solution. a) Note that among three consecutive girls, there is at least one handkerchief: otherwise, the girl in the middle would not have a neighbor with a handkerchief. Fix one girl with a handkerchief in the circle, say Tanya, and consider three girls in sequence clockwise after Tanya. There are a tota... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,381 |
9.1. Can the length of one of the medians of a right triangle be a) $51\%$ of the length of the hypotenuse? b) $49\%$ of the length of the hypotenuse? | Answer. a) Can; b) cannot. Solution. It is clear that the median is drawn to the midpoint of one of the legs, otherwise its length would be equal to half the hypotenuse. Let $\triangle A B C$ be a right triangle with a right angle at $C$ and $B C=a, A C=b$. Without loss of generality, we can assume that the median $A M... | )Can;b)cannot | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,382 |
9.2. Two cyclists set off towards each other from points $A$ and $B$ simultaneously. They traveled at constant speeds. After meeting, the first cyclist took 40 minutes to reach point $B$, while the second cyclist took one and a half hours to reach point $A$. Find the time from the start of the journey until they met an... | Answer: Time until the meeting - 1 hour. The speed of the first cyclist is 1.5 times greater than the speed of the second. Solution. See problem 8.3 | Timeuntilthemeeting-10Thespeedofthefirstcyclistis1.5timesgreaterthanthespeedofthe | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,383 |
9.3. The inscribed circle of triangle $A B C$ with center $O$ touches the sides $A B, B C$ and $A C$ at points $M, N$ and $K$ respectively. It turns out that angle $A O C$ is four times the angle $M K N$. Find angle $B$. | Answer: $108^{\circ}$. Solution. Let $\angle A=\alpha, \angle B=\beta, \angle C=\gamma$. Then, in the isosceles triangle $A K M$, the angle at the base is $\angle A K M=\left(180^{\circ}-\alpha\right) / 2$. Similarly, in triangle $C K N$, we find $\angle C K N=\left(180^{\circ}-\gamma\right) / 2$. Therefore, $\angle M ... | 108 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,384 |
9.4 At a round table, 20 shareholders are seated. What is the minimum possible value of the total number of their shares, given that a) any three of them together have more than 1000 shares, b) any three consecutive shareholders together have more than 1000 shares? | Answer: a) 6679; b) 6674. Solution. a) Let $x_{i}$ be the number of shares owned by the $i$-th shareholder (if counted in a circle, starting from some shareholder). Arrange the numbers $x_{i}$ in ascending order, we get: $y_{1} \leq y_{2} \leq y_{3} \leq \ldots \leq$ $y_{20}$. Note that $y_{3} \geq 334$, because otherw... | 6679;6674 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,385 |
9.5. Given a scalene triangle with sides $a, b, c$. If there exists a triangle with sides $a+b-c, b+c-a, a+c-b$, then the same procedure is applied to this new triangle (and so on), otherwise the process stops. a) Can a triangle similar to the original one appear in this process? b) Can this process continue indefinite... | Answer: a) no; b) no. Solution. a) Let, for definiteness, $a \leq b \leq c$, then the three new numbers will obviously satisfy the inequalities $a+b-c \leq a+c-b \leq b+c-a$. Therefore, the quantity $\Delta=c-a$, equal to the difference between the largest and smallest side, after one procedure will be $\Delta_{1}=2(c-... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,386 |
10.3. Given a triangle $A B C$ inscribed in a circle $\omega$. Point $M$ is the foot of the perpendicular from point $B$ to the line $A C$, and point $N$ is the foot of the perpendicular from point $A$ to the tangent to $\omega$ drawn through point $B$. Prove that $M N \| B C$. | Solution. Consider the quadrilateral ANBM. A circle can be circumscribed around it (with diameter $AB$, since angles $ANB$ and $AMB$ are right angles). Therefore, $\angle BNM = \angle BAC$ (by the property of inscribed angles). Furthermore, the angle between the tangent at point $B$ and the chord $BC$ is also equal to ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,388 |
10.4. a) Investigate the function $y=\frac{\sqrt{x^{2}+1}+x-1}{\sqrt{x^{2}+1}+x+1}$ for evenness (oddness). b) Find the domain and range of this function. | Answer: a) the function is odd; b) domain of definition ( $-\infty, \infty$ ), range of values $(-1 ; 1)$. Solution. Since $\sqrt{x^{2}+1}>|x|$, then $\sqrt{x^{2}+1}+x+1>0$ for all $x$. Therefore, the domain of definition is ( $-\infty, \infty$ ). We transform the expression for $y$ by multiplying the numerator and den... | )thefunctionisodd;b)domainofdefinition(-\infty,\infty),rangeofvalues(-1;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,389 |
11.4. Petya has accumulated many pieces of plasticine of three colors, and he tightly filled a hollow cube with a side of 5 cm with plasticine, so that there was no free space left in the cube. Prove that there are two points of the same color inside the cube at a distance of exactly 7 cm from each other. | Solution. In the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, consider 4 vertices $A, C, B_{1}, D_{1}$. They are the vertices of a regular tetrahedron with side $a \sqrt{2}$, where $a=5$ is the edge of the cube. Since $5 \sqrt{2}>7$, consider a similar tetrahedron with a similarity coefficient $\mathrm{k}=\frac{7}{5 \sqrt{2... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,395 |
11.5. Does there exist a tenth-degree polynomial that takes integer values for all integer arguments, with the leading coefficient not exceeding $10^{-6}$ in absolute value? | Answer. There exists. The solution. An example of such a polynomial is $P(x)=\frac{1}{10!} x(x-1)(x-2) \cdot \ldots \cdot(x-9)$. Let's show that this polynomial satisfies the condition of the problem. Its leading coefficient is equal to 1/(10!). We will check that $1 /(10!)10^{5}$. Indeed, $9!=2^{7} \cdot 3^{4} \cdot 5... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,396 |
10.1. Append a digit to the left and right of the eight-digit number 20222023 so that the resulting 10-digit number is divisible by 72. (List all possible solutions.) | Answer: 3202220232.
Solution. Since $72=8 \cdot 9$, it is required to append digits so that the resulting number is divisible by both 8 and 9. Divisibility by 8 is determined by the last three digits: thus, to the two-digit number 23, we need to append a digit on the right to form a three-digit number that is a multip... | 3202220232 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,397 |
10.2. Solve the equation $\frac{(2 x-1)^{2}}{x^{3}-2 x^{2}+1}+\frac{4 x-4 x^{2}-1}{x^{3}-3 x^{2}+2}=0$. | Answer: $x_{1}=\frac{1}{2}, x_{2}=-1$.
Solution. Rewrite the equation as $\frac{(2 x-1)^{2}}{x^{3}-2 x^{2}+1}=\frac{(2 x-1)^{2}}{x^{3}-3 x^{2}+2}$. Therefore, either $2 x-1=0$, or $\frac{1}{x^{3}-2 x^{2}+1}=\frac{1}{x^{3}-3 x^{2}+2}$. From the first equation, $x=1 / 2$, and for this value, the denominators of the seco... | x_{1}=\frac{1}{2},x_{2}=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,398 |
10.3. Given a quadrilateral $A B C D$ circumscribed by a circle of radius $R$. Can we assert that if $A B^{2}+B C^{2}+C D^{2}+A D^{2}=8 R^{2}$, then at least one of the diagonals of the quadrilateral is a diameter of the circumscribed circle | Answer: No, it cannot.
Solution. For example, consider an inscribed quadrilateral with mutually perpendicular diagonals that are not diameters. Such an example can be derived by introducing the angular measures of the arcs $\alpha, \beta, \gamma, \delta$, into which the circle is divided. Then $A B^{2}+B C^{2}+C D^{2}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,399 |
10.4. In a $25 \times 25$ grid, some cells are marked with a plus sign. Prove that there exist two (possibly overlapping) $3 \times 3$ squares with the same arrangement of plus signs (i.e., the plus signs should coincide under a parallel shift). | Solution. First, let's count the number of $3 \times 3$ squares in a $25 \times 25$ grid. Each such square is uniquely determined by the position of its lower left vertex. This vertex can be any node in the part of the grid that remains if we cut off a 3-cell border from the top and right of the $25 \times 25$ square: ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,400 |
10.1. Solve the inequality $\left|x^{2}-x\right| \leq 2 x^{2}-3 x+1$. | Answer. $x \leq 1 / 3 ; x \geq 1$.
Solution. If $x^{2}-x>0$, i.e., when $x>1$ or $x<0$, the original inequality holds. If $x^{2}-x \leq 0$, i.e., when $0 \leq x \leq 1$, the inequality can be written as: $3 x^{2}-4 x+1 \geq 0 \Leftrightarrow(x-1 / 3)(x-1) \geq 0 \Leftrightarrow x \leq 1 / 3$ or $x \geq 1$. Considering... | x\leq1/3;x\geq1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 22,401 |
10.2. Let $s(n)$ denote the sum of the digits of a natural number $n$ (in decimal notation). Does there exist an $n$ such that $n \cdot s(n)=20222022?$ | Answer. Does not exist.
Solution. By the divisibility rule for 3 and 9, the numbers $n$ and $s(n)$ either both are divisible by 3 (by 9), or both are not divisible by 3 (respectively, by 9). First, consider the case where $n$ and $s(n)$ are divisible by 3. Then the product $n \cdot s(n)$ is divisible by 9. In the seco... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,402 |
10.3. On the hypotenuse $AB$ of a right triangle $ABC$, a square is constructed outward with center $O$. Prove that the ratio of the length $CO$ to the sum of the legs $AC + CB$ is a constant for all right triangles and find this ratio. | Answer: $\sqrt{2} / 2$.
Solution. Let $A B=c, \angle A=\alpha$. Then, by the cosine theorem in triangle $A C O$, we have $\mathrm{CO}^{2}=c^{2} \cos ^{2} \alpha+c^{2} / 2-\sqrt{2} c^{2} \cos \alpha \cos \left(\alpha+45^{\circ}\right)=c^{2} / 2+c^{2} \cos \alpha \sin \alpha=c^{2}(1+2 \sin \alpha \cos \alpha) / 2$. Ther... | \frac{\sqrt{2}}{2} | Geometry | proof | Yes | Yes | olympiads | false | 22,403 |
10.4. In some cells of an $8 \times 8$ square table, a plus sign has been placed. Prove that there exist two (possibly intersecting) $4 \times 4$ squares, each containing the same number of plus signs. | Solution. First, let's count the number of $4 \times 4$ squares in an $8 \times 8$ table. Each such square is uniquely determined by the position of its lower left vertex. This vertex can be any node point of the part of the table that remains if a border of 4 cells is cut off from the top and right of the $8 \times 8$... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,404 |
11.1. Solve the equation $\left|x^{2}-100\right|=10 x+90$. | Answer. $x_{1}=5+\sqrt{215}, x_{2}=-5+\sqrt{35}$. Solution. See problem 10.1. | x_{1}=5+\sqrt{215},x_{2}=-5+\sqrt{35} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,405 |
11.3. Find all values of the parameter $a$ for which the equation $\left|x^{2}-4\right|=a x+6$ has four distinct roots. | Answer: $-30$. Let's construct the graph of $y=\left|x^{2}-4\right|$ on the coordinate plane and draw a tangent from the point $M_{1}(0 ; 6)$ to the branch of the graph located between points $M_{1}$ and $M_{2}(-2 ; 0)$. We need to find the point of tangency $M_{3}\left(x_{0} ; y_{0}\right)$. This point satisfies the e... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,407 |
11.4. Does there exist a function $f$, defined for all real numbers and satisfying the identities: $f\left(2 \cos ^{2} x\right)=1+\cos 4 x$ and $f\left(x+\frac{1}{x}\right)=x^{3}+\frac{1}{x^{3}}$? | Answer: exists. Solution. The value $x+\frac{1}{x} \geq 2$ for all positive $x$ (which follows from the inequality between the arithmetic and geometric means). Therefore, $x+\frac{1}{x} \leq-2$ for all negative $x$. Let $t=x+\frac{1}{x}$. Then $t^{3}=x^{3}+\frac{1}{x^{3}}+3\left(x+\frac{1}{x}\right)$, i.e., $x^{3}+\fra... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,408 |
11.1. Find the largest term of the sequence a) $a_{n}=\frac{n}{n^{2}+2020}$, б) $a_{n}=\frac{2020^{n}}{n!}$ (where $n!=1 \cdot 2 \cdot 3 \cdots n)$ | Answer. a) $a_{45}=\frac{45}{4045}$, b) $a_{2019}=a_{2020}=\frac{2020^{2019}}{2019!}$. Solution. a) Let's investigate the inequality $a_{n}<a_{n+1}$, i.e., $a_{n}=\frac{n}{n^{2}+2020}<\frac{n+1}{(n+1)^{2}+2020}=a_{n+1}$. Transforming it, we get $n^{2}+n<2020$. This inequality holds, as it is easy to verify, for $n=44$ ... | )\frac{45}{4045},b)\frac{2020^{2019}}{2019!} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,409 |
11.1. Find the set of values of the function $y=|2 \sin x+3 \cos x+4|$. | Answer: $[4-\sqrt{13}, 4+\sqrt{13}]$. Solution. By transforming $2 \sin x+3 \cos x$ using an additional angle $\alpha$, we get $2 \sin x+3 \cos x+4=\sqrt{13} \sin (x+\alpha)+4$. The maximum value of this expression is $4+\sqrt{13}$, and the minimum value is $4-\sqrt{13}>0$. Therefore, the result follows. | [4-\sqrt{13},4+\sqrt{13}] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,412 |
11.4. For a polynomial $P(x)$ (not a constant) all coefficients are non-negative. Can $P(x)$ be divisible by a polynomial with a positive leading coefficient and a negative constant term? | Answer: No, it cannot. Solution. Let $P(x)$ be divisible by $Q(x)=a x^{k}+\ldots+b$, where $a>0, b<0$. Then $Q(x)$ has a positive root. Indeed, $Q(0)=b<0$ and $Q(x) \rightarrow+\infty$ as $x \rightarrow+\infty$, so the graph of the continuous function $\mathrm{y}=Q(x)$ must intersect the positive half-axis Ox. Therefor... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,415 |
8.2. Given an isosceles triangle $A B C$ with base $A C$. Points $K$ and $N$ are marked on the side $B C$ ( $K$ lies between $B$ and $N$ ). It turns out that $K N=A N$ and $\angle B A K=\angle N A C$. Find $\angle B A N$. | Answer: $60^{\circ}$. Solution. Let $\angle B A K=\angle N A C=x, \angle K A N=y$. In the isosceles triangle $A K N$, the angle $A K N$ (at the base) is also equal to $y$. The angles at the base of triangle $A B C$ are equal to $2 x+y$, and the angle at the vertex $B$ is therefore equal to $180^{\circ}-2(2 x+y)$. The e... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,417 |
8.3. What is the smallest natural number $n$ for which there exist natural numbers $x$ and $y$ satisfying the equation a) $x \cdot(x+n)=y^{2}$; b) $x \cdot(x+n)=y^{3}$? | Answer: a) $n=3$; b) $n=2$. Solution. a) For $n=3$ such $x$ and $y$ exist: $x=1, y=2$. The impossibility of the values $n=1$ and $n=2$ follows from the inequalities $x^{2}<x(x+1)<(x+1)^{2}$ and $x^{2}<x(x+2)<(x+1)^{2}$. b) For $n=2$ such $x$ and $y$ exist: $x=2, y=2$. For $n=1$, from the equation $x(x+1)=y^{3}$ it foll... | )n=3;b)n=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,418 |
8.1. $\quad$ A cyclist planned to travel from point $A$ to point $B$ in 5 hours, moving at a constant speed. He traveled at the planned speed until the midpoint of the journey, then increased his speed by $25 \%$. He completed the journey to point $B$ at the new speed. How long did the entire trip take? | Answer: 4 hours 30 minutes. Solution. See problem 7.1. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,420 |
8.2. Several integers are written on the board. Petya replaced each number (by erasing it) as follows: he wrote down half of it if it was even, and doubled it if it was odd. Could the sum of the new numbers be equal to the sum of the original numbers if the sum of the original numbers was a) 2021; b) 2022? | Answer: a) could not. b) could. Let $A$ be the initial sum of even numbers on the board, and $B$ be the sum of odd numbers, and let $n=A+B$. Then we have the equation $\frac{A}{2}+2 B=A+B \Leftrightarrow A=2 B$. Therefore, the sum on the board should be $A+B=3 B=n$. In case a) $n=2021$, and this leads to a contradictio... | )couldnot.b)could | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,421 |
8.4. In the class, each boy is friends with exactly three girls, and each girl is friends with exactly two boys. Can there be a total of: a) 32 people? b) 30 people? | Answer. a) cannot; b) can. Solution. a) Let $m$ be the number of boys, $d$-- the number of girls. On one hand, the number of friendly connections is $3 m$ (if we sum their number for all boys), on the other hand, it is $2 d$. Thus, $3 m = 2 d$. Therefore, $m$ is an even number: $m = 2 k$ for some natural number $k$, an... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,422 |
7.1. Given a rectangle where the length is three times the width. It is known that the perimeter of the rectangle is numerically equal to its area. Find the sides of the rectangle. | Answer: $\frac{8}{3}$ and 8. Solution. Let $x$ be the width, then $3 x$ is the length of the rectangle. From the conditions of the problem $2(3 x+x)=3 x \cdot x \Leftrightarrow 8 x=3 x^{2}$. From this, dividing by $x(\neq 0)$, we get $x=\frac{8}{3} ; 3 x=8$. | \frac{8}{3}8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,423 |
7.2. Find the sum of all three-digit natural numbers that do not contain the digits 0 or 9 in their representation. | Answer: 255 744. Solution. We will add the numbers in a column. Each last digit (for example, two) appears in the units place as many times as there are three-digit numbers of the specified form with this last digit (in our case -- these are numbers of the form $\overline{x y 2}$, where $x, y$ are any digits from 1 to ... | 255744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,424 |
7.3. Can the 200 consecutive numbers: $1,2, \ldots, 200$ be rearranged so that adjacent numbers differ by either 3 or $5?$ | Answer: It is possible. Solution. Rearrange the numbers as follows:
$1,4,7,2,5,8,3,6,9 \ldots$ this way we can transition to the next set of eight numbers. And for the next set of eight numbers, a similar rearrangement leads to the start of the next set:
$8 n+1,8 n+4,8 n+7,8 n+2,8 n+5,8 n+8,8 n+3,8 n+6,8(n+1)+1$.
Si... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,425 |
7.4. In a row, 23 identical-looking coins are laid out: genuine and counterfeit. It is known that there are six counterfeit coins in total, and they lie in a row. They differ in weight from the genuine ones (but may also differ in weight from each other, while genuine coins weigh the same). Can at least one counterfeit... | Answer: Yes. Solution. Let's highlight the 6th, 12th, and 18th coins in the row. Since there are exactly 5 other coins between adjacent highlighted coins, and also 5 coins from the extreme highlighted ones to the end of the row, one of the highlighted coins must be counterfeit. We will take two of these three highlight... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,426 |
7.1. $\quad$ A cyclist planned to travel from point $A$ to point $F$ in 5 hours, moving at a constant speed. With the planned speed, he traveled halfway, then increased his speed by $25 \%$. He continued to point $B$ with the new speed. How long did the entire journey take? | Answer: 4 hours 30 minutes. Solution. Let $a$ be the distance between points $A$ and $B$, and $v$ be the planned speed. Then $\frac{a}{v}=5$. The increased speed is $1.25v$. The cyclist will complete the entire journey in $\frac{a}{2v} + \frac{a}{2 \cdot 1.25v} = \frac{4.5}{5} \frac{a}{v} = 4.5$ (hours). | 4.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,427 |
7.2. Find the sum of all four-digit natural numbers composed of the digits 3, 6, and 9. | Answer: 539946. Solution. We will add the numbers in a column. Each last digit (for example, six) appears in the units place as many times as there are three-digit numbers of the specified type with this last digit (in our case -- these are numbers of the form $\overline{x y z 6}$, where $x, y, z$ are arbitrary sets of... | 539946 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,428 |
7.3. Prove that from any six integers, one can select four numbers and denote them as $a, b, c, d$ such that the two numbers $a b$ and $c d$ give the same remainder when divided by 3. | Solution. From the given six integers, choose $a, c$ that have the same remainders when divided by 3, and then from the remaining four numbers, choose $b, d$ that have the same remainders when divided by 3. Then $a, b, c, d$ are the desired numbers. Indeed, since $a-c=3 k$ and $b-d=3 n$ for some integers $k, n$, the di... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,429 |
7.4. A square was cut (with a straight cut) into two rectangles. It turned out that the perimeters of both rectangles are integers. Is it true that the perimeter of the original square is also an integer? | Answer: Incorrect. Solution. Here is an example: a square with side $\frac{2}{3}$ (perimeter $\frac{8}{3}$) can be cut into two equal rectangles: each has a length of $\frac{2}{3}$ and a width of $\frac{1}{3}$, so their perimeters are whole numbers, equal to 2. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,430 |
10.3. Find all values of the parameter $a$ for which the equation $x^{2}+y^{2}+2 x=|x-a|-1$ has two solutions. | Answer: such $a$ does not exist. Solution. We have one equation with two unknowns $x$, y. In the general (non-exceptional) case, it represents some curve on the coordinate plane, i.e., the equation has an infinite set of solutions. Let's check that the given equation is not an exception. Write the equation in the form ... | such\\does\not\exist | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,433 |
10.4. The sequence $a_{n}$ is defined as follows: $a_{1}=2^{20}, a_{n+1}=s\left(a_{n}\right)$ for all $n$, where $s(a)$ denotes the sum of the digits of the natural number $a$. Find $a_{100}$. | Answer: 5. Solution. The main fact used is that the sum of the digits of any number has the same remainder when divided by 9 as the number itself. This fact is proven in the same way as the well-known divisibility rule for 9. We will show that the sequence $a_{n}$ decreases rapidly until it becomes less than 10, and af... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,434 |
9.3 Can the digits in the 30-digit number 11...122...233...3 (which contains 10 ones, 10 twos, and 10 threes) be rearranged so that the result is a square of a natural number? | Answer: No. Solution: The sum of the digits of this number is 60 and it will not change with any permutation of the digits. Since 60 is divisible by 3 but not by 9, assuming the opposite, that is, that there exists a natural number $n$, the square of which has a sum of digits equal to 60, and using the divisibility rul... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,437 |
9.4 Prove that if the area of a convex quadrilateral is equal to the product of its midlines, then its diagonals are equal to each other (a midline is a segment connecting the midpoints of opposite sides). | Solution. Let $A B C D$ be the given quadrilateral and points $A_{1}, B_{1}, C_{1}, D_{1}$ be the midpoints of sides $A B, B C, C D$ and $D A$ respectively. Then $A_{1} B_{1} C_{1} D_{1}$ is a parallelogram, the sides of which are parallel to the diagonals $A C$ and $B D$: this well-known fact follows from the equaliti... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,438 |
9.1 Prove that for all natural $n$ the number $n^{3}+6 n^{2}+12 n+7$ is composite. | Solution. Let's represent the expression as $(n+2)^{3}-1=(n+1)\left(n^{2}+5 n+7\right)$, where, obviously, each factor is greater than 1. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,439 |
9.2 Prove the inequality $|a+1| \leq a^{2}-a+2$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution. When $a \geq-1$ the inequality is reduced to $(a-1)^{2} \geq 0$, and when $a<-1-$ it is reduced to $a^{2}+3 \geq 0$. | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 22,440 |
10.2. Find all values of the parameter $a$, for each of which the system of equations
$$
\left\{\begin{array}{l}
|x|+|y-1|=1 \\
y=a \cdot x+2012
\end{array}\right. \text { has a unique solution. }
$$ | Answer. $a= \pm 2011$. Solution. Let's draw on the coordinate plane the set of solutions of the first equation. This is the boundary of a square (see figure). The line representing the second equation passes through the point $M(0 ; 2012)$. Therefore, the condition for the uniqueness of the solution will be satisfied o... | \2011 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,444 |
10.5. Can the angles of a cyclic quadrilateral represent: a) an arithmetic progression with a non-zero difference; b) a geometric progression with a ratio different from one. | Answer. a) They can; b) they cannot. Solution. a) We can take the angles of a quadrilateral equal to $\alpha, 2 \alpha, 3 \alpha, 4 \alpha$ (in some order), where $\alpha=36^{\circ}$. If we take a quadrilateral $A B C D$ with angles $\angle A=\alpha$, $\angle B=2 \alpha, \angle C=4 \alpha, \angle D=3 \alpha$, then the ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,447 |
3. In triangle $A B C$, angle $A$ is equal to $60^{\circ}$, the distances from vertices $B$ and $C$ to the center of the inscribed circle of triangle $A B C$ are 3 and 4, respectively. Find the radius of the circle circumscribed around triangle $A B C$. | Answer. $\quad \sqrt{\frac{37}{3}}$. Hint. See problem 4 for 9th grade. | \sqrt{\frac{37}{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,449 |
10.2. What is the smallest positive number that the leading coefficient of a quadratic trinomial $P(x)$, which takes integer values for all integer $x$, can equal? | Answer: $\frac{1}{2}$. Hint. Let $P(x)=a x^{2}+b x+c$. Then $c=P(0)$ is an integer. Substituting $x= \pm 1$, we get that $P(-1)+P(1)=2 a+2 c$. Therefore, $2 a-$ is an integer, and thus $|a| \geq \frac{1}{2}$. The value $a=\frac{1}{2}$ is achieved, for example, for the polynomial $P(x)=\frac{x(x+1)}{2}$. | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,451 |
10.3. On the hypotenuse $AB$ of a right triangle $ABC$, a square is constructed outward (i.e., $AB$ is a side of this square). Let $O$ be the center of the square. Prove that the ratio of the length $AO$ to the sum of the legs $AC + CB$ is a constant for all right triangles and find this ratio. | Answer: $\frac{\sqrt{2}}{2}$. Hint. The result is obtained by direct calculation using the cosine theorem (and the formula $\cos \left(\alpha+45^{\circ}\right)$) in triangle $A C O$. | \frac{\sqrt{2}}{2} | Geometry | proof | Yes | Yes | olympiads | false | 22,452 |
9.1. For what values of the number $a$ do the three graphs $y=a x+a, y=x$ and $y=2-2 a x$ intersect at one point | Answer: when $a=1 / 2$ and $a=-2$. Solution. The problem reduces to solving the given three equations with three unknowns $x, y$ and $a$. Substitute the expression for $y$ from the second equation $y=x$ into the first and third equations: $\left\{\begin{array}{l}x=a x+a \\ x=2-2 a x\end{array}\right.$. Express from the... | =1/2=-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,454 |
9.2. In triangle $A B C$, the median $B M$ is twice as small as the bisector $A N$. It is known that angle $C B M$ is three times larger than angle $C A N$. Find the angles of triangle $A B C$. | # Answer: $\angle A=\angle C=36^{\circ}, \angle B=108^{\circ}$. Solution.
Let $\alpha$ be the angle between the angle bisector $A N$ and the sides $A B$ and $A C$. Draw a line through point $M$ parallel to the angle bisector $A N$, and let $K$ be the point of intersection of this line with side $B C$. Since $M$ is the... | \angleA=\angleC=36,\angleB=108 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,455 |
9.3. For a natural number $n$, denote by $T(n)$ the product of all its natural divisors (including $n$). a) Calculate $T(2022)$. b) Does there exist a prime number $p$ and a natural number $n$ such that $T(n)=p^{2022}$? | Answer: a) $2022^{4}$; b) does not exist. Solution. a) Factorize 2022 into prime factors: $2022=2 \cdot 3 \cdot 337$. Each even divisor of this number has the form $2 \cdot 3^{a} \cdot 337^{b}$, where the exponents $a, b$ take two values 0 or 1. The number of such ordered pairs $(a, b)$ is $4=2 \cdot 2$. Therefore, the... | )2022^{4};b)doesnotexist | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,456 |
11.1. Solve the equation $1-\cos 2 x+2 \cos ^{2} 2 x=4 \sin x \cdot \cos 2 x$. | Answer: $x=-\frac{\pi}{2}+2 \pi k, x=(-1)^{n} \frac{\pi}{6}+\pi n, n, k \in \mathbf{Z}$. Solution. The original equation is equivalent to $2\left(\sin ^{2} x+\cos ^{2} 2 x-2 \sin x \cdot \cos 2 x\right)=0 \Leftrightarrow(\sin x-\cos 2 x)^{2}=0 \Leftrightarrow \sin x-1+2 \sin ^{2} x=0 \Leftrightarrow$ $(\sin x)_{1,2}=\f... | -\frac{\pi}{2}+2\pik,(-1)^{n}\frac{\pi}{6}+\pin,n,k\in{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,459 |
11.4. Does there exist an $n$-sided pyramid such that directions (arrows) can be chosen on its edges so that the sum of all $2 n$ edge vectors equals the zero vector a) for $n=9$; b) for $n=10$? | Answer: a) does not exist; b) exists. Solution. a) Consider an axis perpendicular to the base of the pyramid with the origin of coordinates on the plane of the base. If we project the lateral edge vectors onto this axis, then all projections will be equal in magnitude, and since their number is odd, the sum of the proj... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,461 |
8.1. Find the smallest natural number divisible by 5, with the sum of its digits being 100. Justify your answer. | Answer: 599999999995 (between two fives there are 10 nines).
Solution. Due to divisibility by 5, the last digit of the number $N$ can be either 5 or 0. If the last digit is 0, without changing the sum of the digits, we can replace 0 with 5 and subtract one from each of the five non-zero digits in other positions. Then... | 599999999995 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,463 |
8.2. Given two coprime natural numbers \( p \) and \( q \), differing by more than one. Prove that there exists a natural number \( n \) such that the numbers \( p+n \) and \( q+n \) will not be coprime. | Solution. Let, for definiteness, $q>p$. Denote $m=q-p>1$. Notice that $p+n$ and $q+n$ for any fixed $n$ give the same remainders when divided by $m$, since their difference is constant and equal to $q-p=m$. As $n$ changes from 1 to $m$, these remainders run through all possible $m$ values (starting from $p(\bmod m)$). ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,464 |
8.3. For a given triangle and a given quadrilateral, the following is known: for any two angles of the triangle, there is an angle of the quadrilateral that is equal in magnitude to the sum of these two angles of the triangle. Prove that the triangle is isosceles. | Solution. Let $\alpha, \beta, \gamma$ be the angles of a triangle. If we assume, for the sake of contradiction, that all these angles are different, then the numbers $(\alpha+\beta)$, $(\beta+\gamma)$, and $(\alpha+\gamma)$ will be different, and in the quadrilateral there will be three different vertices with such ang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,465 |
8.4. In a group of astronauts, there are 20 people, and each has at least 15 friends in the group. A crew is needed for space, in which everyone is friends with each other. Is it necessarily possible to form a crew a) of four people? b) of five people? | Answer: a) necessarily; b) not necessarily.
Solution. a) See the solution to problem 7.4 (the condition in 7.4 is even more general than in part a) of problem 8.4). b) Consider the following counterexample. Suppose the astronauts are divided into 4 groups of 5 people each, and let each astronaut not be friends with an... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,466 |
8.5. In each cell of a $9 \times 9$ grid, zeros are written. It is required to form a magic square (the sum in all rows and columns should be equal to the same number) in several moves. Is it possible to do this if in one move it is allowed to: a) select a row and add a positive number to any two adjacent cells in the ... | Answer: a) no; b) yes.
Solution. a) Let $A$ be the sum of 45 numbers in the five columns with odd numbers, and $B$ be the sum of 36 numbers in the remaining four columns (with even numbers). Note that with each move, the same number (which generally depends on the move) is added to both $A$ and $B$. Therefore, the dif... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,467 |
11.2. The tangent to the graph of $y=x^{2}$ intersects the $O x$ and $O y$ axes at points $A$ and $B$. Find the lengths of the legs $O A$ and $O B$, if the area of triangle $A O B$ is 16. | Answer: $O A=2, O B=16$. Solution. Let $\left(x_{0}, x_{0}^{2}\right)$ be the point of tangency, for definiteness $x_{0}>0$ (if $x_{0}<0$, the tangent line intercepts segments of the same length on the axes as at ( $-x_{0}$ )). The equation of the tangent line is $y-x_{0}^{2}=2 x_{0}\left(x-x_{0}\right)$, (since $\left... | OA=2,OB=16 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 22,470 |
11.4. Prove the inequality $\cos ^{2} \alpha+\cos ^{2} \beta \leq 1+|\cos \alpha \cdot \cos \beta|$. | Solution. Since $\cos ^{2} \alpha \leq|\cos \alpha|, \quad \cos ^{2} \beta \leq|\cos \beta|$ it is sufficient to prove that $|\cos \alpha|+|\cos \beta| \leq 1+|\cos \alpha| \cdot|\cos \beta| \Leftrightarrow(1-|\cos \alpha|)(1-|\cos \beta|) \geq 0$. The last inequality is obvious (since $|\cos x| \leq 1$ for all $x$ ). | proof | Inequalities | proof | Yes | Yes | olympiads | false | 22,472 |
11.1. For all possible values of the parameter $a$, solve the equation
$\log _{2} \sqrt{\cos x}=\log _{4}(a \cdot \sin x)$ | Answer. When $a=0$ the equation makes no sense;
when $a>0 \quad x=\operatorname{arcctg} a+2 \pi n, n \in \mathbb{Z}$;
when $a<0 \quad x=\pi+\operatorname{arcctg} a+2 \pi m, m \in \mathbb{Z}$. On the domain of definition, we have the trigonometric equation $\cos x=a \sin x \Leftrightarrow$ $\operatorname{ctg} x=a \Lef... | When\=0\the\equation\makes\no\sense;\when\>0\\operatorname{arcctg}+2\pin,\n\in\mathbb{Z};\when\<0\\pi+\operatorname{arcctg}+2\pi,\\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,473 |
11.2. Solve the system of equations
$$
\left\{\begin{array}{l}
x y+x=15 \\
x^{2}+y^{2}+x y+y=41
\end{array}\right.
$$ | Answer: (3;4), (5;2). Solution. By adding the equations, we get $(x+y)^{2}+(x+y)=56$. Let $x+y=t$, then $t^{2}+t-56=0 \Leftrightarrow t_{1}=-8, t_{2}=7$. Thus, we have a combination of two systems equivalent to the original system: 1) $\left\{\begin{array}{l}x+y=-8 \\ x y+x=15\end{array}\right.$ and 2) $\left\{\begin{a... | (3;4),(5;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,474 |
11.3. Find the domain and range of the function
$$
y=\sqrt{x^{2}+3 x+2}+\frac{1}{\sqrt{2 x^{2}+6 x+4}}
$$ | Answer. Domain of definition $x-1$; range of values $y \geq \sqrt[4]{8}$.
Solution. The domain of definition is given by the inequality $x^{2}+3 x+2>0 \Leftrightarrow(x+2)(x+1)>0 \Leftrightarrow$ $x-1$. To find the range of values, let $t=\sqrt{x^{2}+3 x+2}$. Then $t>0$ and $y=t+\frac{1}{\sqrt{2}}$. We can investigate... | y\geq\sqrt[4]{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,475 |
11.6. Edges $S A$, $S B$, and $S C$ of the tetrahedron $S A B C$ are mutually perpendicular, and their lengths are 2, 3, and 6, respectively. a) Find the radius $R$ of the sphere circumscribed around the tetrahedron. b) Does there exist a sphere with a radius smaller than $R$ that contains the given tetrahedron? | Answer. a) $R=7 / 2$; b) exists. Solution. Consider a rectangular coordinate system in space with the origin at point $S$ and coordinate axes along the edges $O A, O B$ and $O C$. The center of the circumscribed sphere can be found in different ways. One way is direct: if we write the equality of the squares of the dis... | \frac{7}{2}; | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,477 |
7.4. In triangle $A B C$, the angles $A$ and $C$ at the base are $20^{\circ}$ and $40^{\circ}$, respectively. It is known that $A C - A B = 5$ (cm). Find the length of the angle bisector of angle $B$. | Answer: 5 cm. Solution. Let $B M$ be the bisector of angle $B$. Mark a point $N$ on the base $A C$ such that $A N=A B$. Then triangle $A B N$ is isosceles and $\angle A B N=\angle A N B=80^{\circ}$. Since $\angle A B M=\frac{180^{\circ}-20^{\circ}-40^{\circ}}{2}=60^{\circ}$, then $\angle B M N=\angle A+\angle A B M=20^... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,478 |
10.2. In triangle $A B C$, the median $B M$ is twice as small as the bisector $A N$. It is known that angle $C B M$ is three times larger than angle $C A N$. Find the angles of triangle $A B C$. | Answer: $\angle A=\angle C=36^{\circ}, \angle B=108^{\circ}$. Solution. See problem 9.2. | \angleA=\angleC=36,\angleB=108 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,480 |
10.3. Prove the inequality $\sqrt{x^{3}+y}+\sqrt{y^{3}+x} \geq \sqrt{x^{3}+x}+\sqrt{y^{3}+y}$. | Solution. First, we investigate the domain of the variables. Since $x^{3}+x \geq 0 \Leftrightarrow x\left(x^{2}+1\right) \geq 0$, then $x \geq 0$. Similarly, $y \geq 0$. Thus, for non-negative $x, y$, both sides of the inequality are meaningful and non-negative. Therefore, squaring both sides leads to an equivalent ine... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 22,481 |
10.4. Prove that there exists an infinite set of triples of natural numbers $x, y, z$, satisfying the relation $x^{2}+y^{2}=z^{2022}$. | Solution. Let's take a Pythagorean triple, for example, ( $3 ; 4 ; 5$ ), and consider the relations $(3 t)^{2}+(4 t)^{2}=(5 t)^{2}$ for various natural $t$. If we set $x^{2}=9 t^{2}, y^{2}=16 t^{2}, z^{2022}=25 t^{2}$, then taking the number $z$, divisible by 5, i.e., $z=5 n$ for a natural $n$, we get $25 t^{2}=5^{2022... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,482 |
10.5. On the coordinate plane, there is a rectangle with integer coordinates of vertices, different from a square. Prove that several lines can be drawn, parallel to the sides of the rectangle, so that the rectangle is divided into squares with integer coordinates of vertices. | Solution. Let $ABCD$ be the given rectangle. Without loss of generality, we can assume that $A=O$ is the origin: otherwise, we can shift the origin to point $A$, and at the end, make a shift by the integer vector $\overrightarrow{AO}$. Let the vectors be $\vec{u}=\overrightarrow{OD}=(p; q)$, $\vec{v}=\overrightarrow{OB... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,483 |
10.3. The bisector $B K$ of triangle $A B C$ at the point $I$ of intersection of the bisectors is divided in the ratio $B I: I K=10: 7$. Prove that angle $B$ is acute. | Solution. Let $A B=c, B C=a, A C=b$. By the property of the angle bisector in triangles $A B K$ and $B C K$, we have: $\frac{A K}{c}=\frac{7}{10}=\frac{K C}{a}$. Let $t=\frac{7}{10}$. Then $A K=c t, K C=a t$. Therefore, $b=(a+c) t$. To determine whether angle $B$ is acute, we need to find the sign of the expression $a^... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,486 |
10.2. Given numbers $a, b$ (not necessarily distinct) and the corresponding quadratic trinomials $x^{2}+a x+b^{2}$ and $x^{2}+b x+a^{2}$. It turns out that these trinomials have a common root. Can $a$ and $b$ be non-zero numbers? | Answer: they cannot. Solution. Let $x^{2}+a x+b^{2}=0$ and $x^{2}+b x+a^{2}=0$, where $x$. is a common root. Subtracting these equations, we get $(a-b)(x-a-b)=0$. If $a=b$, then $x^{2}+a x+a^{2}=0$, but the discriminant of this equation $D=-3 a^{2}$ is negative for $a \neq 0$. If $x=a+b$, then $(a+b)^{2}+a(a+b)+b^{2}=0... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,489 |
10.4. Given a triangle and a quadrilateral, it is known that for any two angles of the triangle, there is an angle in the quadrilateral that is equal in magnitude to the sum of these two angles of the triangle. Prove that the triangle is isosceles. | Solution. Let $\alpha, \beta, \gamma$ be the angles of a triangle. If we assume, for the sake of contradiction, that all these angles are different, then the numbers $(\alpha+\beta), (\beta+\gamma)$, and $(\alpha+\gamma)$ will all be different, and in the quadrilateral, there will be three different vertices with such ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,491 |
7.1. During the test in $7^{\mathrm{a}}$, there were three more boys than girls. According to the test results, the number of fours was two more than the number of fives, and the number of threes was twice the number of fours. Prove that someone received a two or a one. | Solution. Let $x$ be the number of fives, then the total number of positive grades is $x+(x+2)+2(x+2)=4 x+6$. On the other hand, there were $2 m+3$ people in the test, where $m$ is the number of girls. Since the numbers $4 x+6$ and $2 m+3$ have different parities, they do not coincide, and thus there were also bad grad... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 22,492 |
7.2. What is the smallest number of digits that can be appended to the right of the number 2013 so that the resulting number is divisible by all natural numbers less than 10? | Answer: three digits. Solution. The least common multiple of the numbers $(1,2, \ldots, 9)$ is 2520. If two digits are appended to 2013, the resulting number will not exceed 201399. Dividing 201399 by 2520 with a remainder, we get a remainder of 2319. Since 2319 > 99, there is no integer multiple of 2520 between the nu... | 2013480 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,493 |
7.3. We have 10 sticks with lengths 1 (cm), 2 (cm), $2^{2}$ (cm),.., $2^{9}$ (cm). Can we form an isosceles triangle using these sticks, not necessarily all of them? | Answer: No. Solution. Suppose, for the sake of contradiction, that the triangle can be formed, and let $2^{n_{1}}, 2^{n_{2}}, \ldots, 2^{n_{k}}$ be the lengths of the sticks forming one side, and $2^{m_{1}}, 2^{m_{2}}, \ldots, 2^{m_{l}}$ be the lengths of the sticks forming the other side. Then we have the equation $2^... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,494 |
7.4. A row contains 100 numbers. Is it necessarily true that the sum of all these numbers is positive if it is known that a) the sum of any seven numbers is positive? b) the sum of any seven consecutive numbers is positive? | Answer: a) necessarily; b) no. Solution. a) Let $a_{1}, a_{2} \ldots a_{100}$ be the given numbers. We compute 100 positive sums $S_{1}=a_{1}+a_{2}+\ldots+a_{7}, S_{2}=a_{2}+a_{3}+\ldots+a_{8}, \ldots, S_{100}=a_{100}+a_{1}+\ldots+a_{6}$ (seven consecutive numbers are added in a cyclic arrangement). Then $S_{1}+S_{2}+\... | )necessarily;b)no | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 22,495 |
7.5. Kolya placed 17 kings on a chessboard. Petya must remove 12 kings from the board so that the remaining 5 kings do not attack each other. Can he always do this? (A king attacks all squares adjacent to its square, either by side or corner.) | Answer: Yes. Solution. Let's color the board in 4 colors as follows: divide the board into 16 $2 \times 2$ squares and color each such square as follows:
| 1 | 2 |
| :--- | :--- |
| 3 | 4 | (you can color the corner square and then parallel transfer the coloring to other squares):
| 1 | 2 | 1 | 2 | $\ldots$ |
| :---:... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,496 |
7.1. Find two natural numbers $m$ and $n$, given that $m<n<2m$ and $mn=2013$. | Answer: $\quad m=33, n=61$.
The solution follows from the prime factorization $2013=3 \cdot 11 \cdot 61$ and a small enumeration of the divisors of 2013; there are 8 in total, and the "candidates" for the role of the number $m$ are the smaller divisors, namely the four numbers $1,3,11,33$, and among them, only $m=33$ ... | =33,n=61 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,497 |
7.2. Find the sum of all three-digit natural numbers that do not contain the digit 0 or the digit 5. | Answer: 284160. Solution. We will add the numbers in a column. Each last digit appears in the units place as many times as there are three-digit numbers ending with this digit. Therefore, it will appear $8 \cdot 8=64$ times (since a total of 8 digits are used for the hundreds and tens places). Thus, the sum of the digi... | 284160 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,498 |
7.4. On an island, there live knights and liars. Knights always tell the truth, liars always lie (the islanders know who is who). A tourist who arrived on the island met a group of 10 islanders and started asking each one in turn: "Who is more in your group: knights, liars, or are they equal in number?" Five of them sa... | Answer: "Equally". Solution. Suppose first that there are more than five liars in the company. Then the first five told the truth, and thus they are knights. So, there are at least five knights, which means our assumption is incorrect. Now suppose that there are more than five knights in the company. Then among the fir... | Equally | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,500 |
8.1. A hundred-digit natural number $N$ is composed of ones and twos, and between any two twos, there is an even number of digits. It is known that $N$ is divisible by 3. How many ones and how many twos are in the number $N$? | Answer: two twos and 98 ones. Solution. If the number $N$ contains more than two twos, then considering any three twos, we get a contradiction: indeed, between the first and second twos there is an even number of digits, between the second and third twos there is an even number, but together with the second (i.e., the ... | two\twos\\98\ones | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,501 |
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