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10.2. Find all values of the parameter $a$ for which the equation $x^{2}+2 x+2|x+1|=a$ has exactly two roots. | Answer: $a>-1$. Solution. Write the equation in the form $(x+1)^{2}+2|x+1|=a+1$. Let $t=|x+1|, t \geq 0$. Then we get the quadratic equation $t^{2}+2 t=a+1 \Leftrightarrow(t+1)^{2}=a+2$. Considering that $t \geq 0$, the inequality $a+2 \geq 1$ must be satisfied, i.e., $a \geq-1$. Under this condition, $t$ is uniquely d... | >-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,093 |
10.3. Given a trapezoid $A B C D$ and a point $M$ on the lateral side $A B$, such that $D M \perp A B$. It turns out that $M C=C D$. Find the length of the upper base $B C$, if $A D=d$. | Answer: $\frac{d}{2}$. Solution. Let $N$ be the midpoint of the lower base $AD$. Then, by the property of the median of the right triangle $AMD$, we have: $MN=AN=ND=\frac{d}{2}$. Consider $\triangle MCN$ and $\triangle DCN$. They are equal by three sides. Then $\angle MCN=\angle DCN$ and therefore, in the isosceles tri... | \frac{}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,094 |
10.5. Petya and Vasya have equal paper right-angled triangles with legs $a$ and $b$. The boys want to cut out the largest possible square from each triangle, such that one vertex of Petya's square coincides with the right angle of the triangle, and one side of Vasya's square lies on the hypotenuse. a) Find the sizes of... | Answer: a) $\frac{a b}{a+b}$ and $\frac{a b \sqrt{a^{2}+b^{2}}}{a^{2}+a b+b^{2}}$; b) always. Solution. a) The side of Petya's square is calculated in problem 9.5. Let $x$ be the side of Vasya's square. Then the hypotenuse is divided into three segments of lengths, respectively, $\quad x \operatorname{tg} A, \quad x$ a... | \frac{}{+b} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,096 |
11.1. How many roots does the equation $\sqrt{14-x^{2}}(\sin x-\cos 2 x)=0$ have? | Answer: 6 roots. Solution. The domain of the equation: $14-x^{2} \geq 0 \Leftrightarrow|x| \leq \sqrt{14}$. On this domain, we solve the trigonometric equation $\sin x-\cos 2 x=0 \Leftrightarrow \sin x+2 \sin ^{2} x-1=0$; $\sin x=\frac{-1 \pm 3}{4}$, i.e., $\sin x=-1$ or $\sin x=\frac{1}{2}$. From this, we get three se... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,097 |
11.4. Find the smallest natural number that has exactly 55 natural divisors, including one and the number itself. | Answer: $2^{10} \cdot 3^{4}$. Solution. If a natural number $n$ has the form $n=p_{1}^{k_{1}} \cdot p_{2}^{k_{2}} \cdot \ldots \cdot p_{m}^{k_{m}}$, where $p_{1}, p_{2}, \ldots, p_{m}$ are its distinct prime divisors, then the number of natural divisors of $n$ is $\left(k_{1}+1\right)\left(k_{2}+1\right) \ldots\left(k_... | 2^{10}\cdot3^{4} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,100 |
11.5. In the tetrahedron $S A B C$, the edges $S A, S B, S C$ are mutually perpendicular and equal to $a, b, c$ respectively. a) Find the side of the cube with vertex $S$ of the largest volume that is entirely contained within the tetrahedron; b) Determine the dimensions of the rectangular parallelepiped with vertex $S... | Answer: a) $\frac{a b c}{a b+b c+a c} ;$ b) $\frac{a}{3}, \frac{b}{3}, \frac{c}{3}$. Solution. a) Analogous to the solution of problem 9.5, consider a rectangular coordinate system with the origin at point $S$ and coordinate axes along $S A, S B$, $S C$. Then the equation of the plane $A B C$ is $\frac{x}{a}+\frac{y}{b... | )\frac{}{++};b)\frac{}{3},\frac{b}{3},\frac{}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,101 |
10.1. Find all values of the parameter $a$ for which the equation $\left|x^{2}-2 a x+1\right|=a$ has three roots. | Answer: $a=\frac{1+\sqrt{5}}{2}$. Solution. Obviously, $a \geq 0$ (since the left side of the equation is non-negative). If the discriminant of the quadratic trinomial $D=4\left(a^{2}-1\right)$ is negative, then we have the quadratic equation $x^{2}-2 a x+1=a$, and, therefore, it has no more than two roots. Therefore, ... | \frac{1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,102 |
10.2. The number $a$ is a root of the quadratic equation $x^{2}-x-50=0$. Find the value of $a^{4}-101 a$. | Answer: 2550. Solution. We have $a^{2}=a+50$, therefore $a^{4}=(a+50)^{2}=a^{2}+100 a+2500=$ $a+50+100 a+2500=101 a+2550$. Hence $a^{4}-101 a=2550$. | 2550 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,103 |
10.3. Given a right triangle $A B C$. A point $M$ is taken on the hypotenuse $A C$. Let $K, L$ be the centers of the circles inscribed in triangles $A B M$ and $C B M$ respectively. Find the distance from point $M$ to the midpoint of $K L$, if the radius $R$ of the circle circumscribed around triangle $B K L$ is known. | Answer: $\frac{R \sqrt{2}}{2}$. Solution. Note that angle $K M L$ is a right angle, since $\angle K M L=\angle K M B+\angle L M B=$ $=\frac{1}{2}(\angle A M B+\angle C M B)=90^{\circ}$. Similarly, we obtain that $\angle K B L=\frac{1}{2} \cdot 90^{\circ}=45^{\circ}$. Then in triangle $B K L$ we have $K L=2 R \cdot \sin... | \frac{R\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,104 |
10.5. We have 100 sticks of lengths $1, 0.9, (0.9)^{2}, \ldots, (0.9)^{99}$. Can we use some of these sticks, not necessarily all, to form an isosceles triangle? | Answer: No. Solution. Suppose, for the sake of contradiction, that it is possible to form a triangle, and let $0.9^{n_{1}}, 0.9^{n_{2}}, \ldots, 0.9^{n_{k}}$ be the lengths of the sticks forming one side, and $0.9^{m_{1}}$, $0.9^{m_{2}}, \ldots, 0.9^{m_{l}}$ be the lengths of the sticks forming the other side. Then we ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,106 |
9.2. In triangle $A B C$, side $B C$ is equal to segment $A M$, where $M$ is the point of intersection of the medians. Find the angle $\angle B M C$. | Answer: $90^{\circ}$. Solution. Let $A N-$ be the median. By the property of the intersection point of medians, $M N=\frac{1}{2} A M=\frac{1}{2} B C$. Thus, in triangle $B M C$, the median $M N$ is equal to half of side $B C$. Therefore, $\triangle B M N$ is a right triangle with a right angle at $B M N$ (this is a kno... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,107 |
10.2. Given three numbers: $x, y, z$. It is known that each of the numbers $2x - y, 3y - 2z$, and $4z - 3x$ is negative. Prove that each of the numbers $x, y, z$ is also negative? | Solution. Let $A=2 x-y, B=3 y-2 z, C=4 z-3 x$. Then $3 A+B+2 C=6 z$, which means $z<0$. Next, from the equalities $3 y=B+2 z$ and $2 x=A+y$, we sequentially obtain that $y<0$ and $x<0$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 22,109 |
10.3. Given a trapezoid $A B C D(B C \| A D)$, in which a circle with center $O$ is inscribed. The line $B O$ intersects the lower base $A D$ at point $M$. Prove the relation for the areas $S_{A O M}+S_{C O D}=\frac{1}{2} S_{A B C D}$. | Solution. Consider two triangles $A B O$ and $A M O$. Triangle $A B O$ is a right triangle, since $\angle A O B=180^{\circ}-\frac{1}{2}(\angle A+\angle B)=90^{\circ}$. (Here, two facts are used: the center of the inscribed circle lies at the intersection of the angle bisectors, and the sum of the angles of a trapezoid ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,110 |
10.4. There are $n$ weights, each weighing an integer number of grams, and their total weight is $2 k$ (g). Is it true that the weights can always be divided into two scales so that they balance, if a) $n=k ;$ b) $n=k+1$? | Answer: a) incorrect; b) correct. Solution. See problem 9.5 (with $n$ and $k$ replaced by 50.) Comment. For odd $n$, in addition to the counterexample ( $n+1,1,1, \ldots, 1$ ), another one can be given $(2,2, \ldots, 2)$ ( $n$ twos), and it can be shown that no other examples exist. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,111 |
10.5. How many a) rectangles; b) right-angled triangles with integer sides exist, for which the area is numerically equal to the perimeter? (Congruent figures are counted as one). | Answer: a) two; b) two. Solution. a) Let the sides of the rectangle be integers $a$ and $b$ ( $a \leq b$ ). Then $a b=2(a+b)$, which is equivalent to the equation $(a-2)(b-2)=4$. The number 4 can be factored into the product of two integers in four ways: $(1 ; 4),(2,2),(-4 ;-1),(-2,-2)$, of which the first two give rec... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,112 |
11.1. Solve the equation $\cos ^{2}(\sqrt{2} x)-\sin ^{2} x=1$. | Answer: $x=0$. Solution. The left side of the equation does not exceed one, and it can equal one only if the following two conditions are simultaneously satisfied: $\cos ^{2}(\sqrt{2} x)=1$ and $\sin x=0$. Hence, $\sqrt{2} x=\pi n$ and $x=\pi k, k, n \in \mathbf{Z}$. Then $\frac{\pi n}{\sqrt{2}}=\pi k$, i.e., $n=\sqrt{... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,113 |
11.2. Find all values of the parameter $a$ for which the equation $\sqrt{x^{2}-2 x-3}\left(x^{2}-3 a x+2 a^{2}\right)=0$ has exactly three roots. | Answer: $-1 \leq a<-\frac{1}{2}$ or $\frac{3}{2}<a \leq 3$. Solution. See problem 10.2. | -1\leq<-\frac{1}{2}or\frac{3}{2}<\leq3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,114 |
11.4. $n$ vectors in space are such that any pair of them forms an obtuse angle. What is the largest value that $n$ can take? | Answer: 4. Solution. Let $\vec{a}_{1}, \vec{a}_{2}, \ldots, \vec{a}_{n}$ be the given vectors. Direct the $z$-axis of the coordinate space along $\vec{a}_{n}$. Then the $z$-coordinate of the other vectors must be negative (this follows from the formula for the cosine of the angle between vectors through the scalar prod... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,116 |
11.1. Solve the equation $(\sin 2 x-\pi \sin x) \sqrt{11 x^{2}-x^{4}-10}=0$. | Answer: $x \in\{-\sqrt{10},-\pi,-1,1, \pi, \sqrt{10}\}$. Solution. Let's find the domain. We have $11 x^{2}-x^{4}-10 \geq 0 \Leftrightarrow\left(x^{2}-10\right)\left(x^{2}-1\right) \leq 0 \Leftrightarrow-\sqrt{10} \leq x \leq-1$ or $1 \leq x \leq \sqrt{10}$. In this case, we obtain four values $x \in\{-\sqrt{10},-1,1, ... | x\in{-\sqrt{10},-\pi,-1,1,\pi,\sqrt{10}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,117 |
11.2. Given a rectangle with integer coordinates of vertices on the coordinate plane. Let $\alpha$ be the angle between its diagonals. Is it necessarily a rational number: a) $\cos \alpha ? ;$ b) $\sin \alpha$ ? | Answer: a) yes; b) yes. Solution. a) Let $\boldsymbol{A} \boldsymbol{B C D}$ be the given rectangle and $O$ its center. Then $\alpha=\angle C O D, \frac{\alpha}{2}=\angle C A D$. Since $\cos ^{2} \frac{\alpha}{2}=(A D)^{2} /(A C)^{2}$ is a rational number, then $\cos \alpha=2 \cos ^{2} \frac{\alpha}{2}-1$ is also a rat... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,118 |
11.1. Solve the inequality $f(f(x))<(f(x))^{2}$, where $f(x)=2 x^{2}-1$. | Answer: $x \in(-1 ; 0) \cup(0 ; 1)$. Solution. Let $y=2 x^{2}-1$. Then
$$
2 y^{2}-1<y^{2} \Leftrightarrow-1<y<1 \Leftrightarrow 0<2 x^{2}<2 \Leftrightarrow 0<|x|<1
$$ | x\in(-1;0)\cup(0;1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 22,121 |
11.2. Find the greatest and the least value of the function $y=(\arcsin x) \cdot(\arccos x)$. | Answer: The maximum value $=\pi^{2} / 16$ (at $x=\sqrt{2} / 2$), the minimum value $=-\pi^{2} / 2$ (at $x=-1$). Solution. The values of $\arcsin x$ and $\arccos x$ for any $x \in[-1 ; 1]$, as is known, are related by the equation $\arcsin x+\arccos x=\pi / 2$. Thus, we need to investigate the function $y(t)=t(\pi / 2-t... | Themaximumvalue=\frac{\pi^2}{16}(atx=\frac{\sqrt{2}}{2}),theminimumvalue=-\frac{\pi^2}{2}(atx=-1). | Calculus | math-word-problem | Yes | Yes | olympiads | false | 22,122 |
11.3. The numbers $x, y$ satisfy the equation $\sqrt{x^{3}+y}+\sqrt{y^{3}+x}=\sqrt{x^{3}+x}+\sqrt{y^{3}+y}$. Can we assert that $x=y$? | Answer: it can be. Solution. See the solution to problem 10.3: everywhere instead of inequalities, equalities should be considered, and then the original equation reduces to the equation $(x-y)^{2}\left(x^{2}+x y+y^{2}\right)=0$. The first factor is zero (only) when $x=y$, and the second when $x=y=0$. | y | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,123 |
8.3. Can five points $A, B, C, D, E$ be marked (and labeled) on a plane so that the five triangles $A B C, B C D, C D E, D E A, E A B$ are acute-angled? | Answer: It is possible. Solution. Consider a regular pentagon, mark its vertices and denote them as shown in the figure (moving clockwise and marking every other vertex). Then all five triangles will be isosceles and equal to each other (which means they are all acute-angled; it is not difficult to calculate their angl... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,128 |
10.1. On the coordinate plane, plot the set of points whose coordinates satisfy the equation $|x|+|y|=x^{2}$. | Solution. Note that if the point ( $x_{0} ; y_{0}$ ) belongs to our set, then the points $\left(-x_{0} ; y_{0}\right),\left(x_{0} ;-y_{0}\right)$ and $\left(-x_{0} ;-y_{0}\right)$ also belong to it. Therefore, it is sufficient to depict our set in the first quadrant and then reflect it symmetrically relative to the axe... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,130 |
10.3. Find all values of the parameter $a$ for which the equation $a x^{2}+\sin ^{2} x=a^{2}-a$ has a unique solution. | Answer: $a=1$. Solution. Note that only $x=0$ can be the unique root of the equation, since due to the evenness of the functions involved, for any solution $x_{0} \neq 0$, $(-x_{0})$ will also be a solution. Therefore, we necessarily get $a^{2}-a=0 \Leftrightarrow a=0$ or $a=1$. Let's check these values. When $a=0$, we... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,132 |
7.1. In 7a class, $52\%$ are girls. All students in the class can line up in such a way that boys and girls alternate. How many students are in the class? | Answer: 25 students. Solution: Considering that there are more girls than boys in the class, from the condition of alternation, we get that there are exactly one more girl than boys. Therefore, one person constitutes $52-48=4 \%$ of the class size. Thus, the number of students in the class (i.e., $100 \%$) is $100 / 4=... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,134 |
7.2. Do there exist three integers (which may be the same) such that if the product of any two of them minus the third one equals $2018?$ | Answer: they exist. Solution. We can directly check that the triplet of numbers -1, -1, -2017 satisfies the conditions of the problem. Let's show how to obtain such a triplet. Denote the required integers as $a, b, c$. Consider the two equations $a \cdot b - c = 2018$ and $b \cdot c - a = 2018$. Subtracting the second ... | -1,-1,-2017 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,135 |
7.3. Given a rectangular grid of $7 \times 14$ (cells). What is the maximum number of three-cell corners that can be cut out from this rectangle? | Answer: 32 corners. Solution: It is obvious that no more than 32 corners can be cut out, as otherwise, the rectangle must contain no fewer than $33 \cdot 3=99>98$ cells. The image below shows an example of cutting one
 Prove that there exists a pair of two-digit numbers such that if 15 is added to the first number and 20 is subtracted from the second, the resulting numbers will remain two-digit, and their product will be equal to the product of the original numbers. b) How many such pairs are there? | Answer: b) 16 pairs. Solution. The sought numbers $x$ and $y$ must satisfy the condition $(x+15)(y-20)=xy$. Expanding the brackets and transforming, we get the equation $3y-4x=60$. From this equation, it follows that the number $x$ must be divisible by 3, and $y$ by 4, i.e., $x=3x_{1}, y=4y_{1}$ for some natural $x_{1}... | 16 | Algebra | proof | Yes | Yes | olympiads | false | 22,137 |
7.5. In an isosceles triangle $ABC$, the lateral sides $AB$ and $BC$ are divided into $n$ and $n+1$ equal parts, respectively ( $n>1$ ). From vertex $A$, $n$ segments are drawn to the division points on side $BC$, and from vertex $C$, $(n-1)$ segments are drawn to the division points on side $AB$. Then, the median from... | Answer. They cannot. Solution. Suppose, for the sake of contradiction, that segments $A D, C E$, and $B M$ intersect at a single point $O$. Let segment $A E$ contain $x$ out of $n$ parts of the division on side $A B$, and segment $C D$ contain $y$ out of $(n+1)$ parts of the division on side $B C$. Under symmetry relat... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,138 |
11.1. On the coordinate plane, plot the set of points satisfying the inequality $\log _{x} y+\log _{y} x>2$. | Answer. See fig.Solution. The domain of the inequality is $x>0, y>0, x \neq 1, y \neq 1$. Using the property of logarithms $\log _{x} y=\frac{1}{\log _{y} x}$, we obtain the equivalent inequality $\log _{x} y+\frac{1}{\log _{x} y}>2$. Let $t=\log _{x} y$. Solving the inequality $t+\frac{1}{t}>2$, we get $\frac{(t-1)^{2... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 22,139 |
11.2. There are $n$ weights weighing $1,2 \ldots, n$ (g) and a two-pan balance. Can all the weights be arranged on the balance so that one pan has twice as many weights as the other, and the balance is in equilibrium: a) for $n=90$; b) for $n=99$? | Answer. a) cannot; b) can. Solution. a) The sum of the weights of all weights $\frac{90 \cdot 91}{2}=4095$ is an odd number, therefore, 90 weights cannot be divided into two piles with equal sums. b) Let's try to find 33 weights with consecutive weights, the total weight of which is equal to half the sum of the weights... | )cannot;b)can | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,140 |
11.4. On the coordinate plane, the parabola $y=x^{2}$ is drawn. On the positive $O y$-axis, a point $A$ is taken, and through it, two lines with positive slopes are drawn. Let $M_{1}, N_{1}$ and $M_{2}, N_{2}$ be the points of intersection with the parabola of the first and second line, respectively. Find the ordinate ... | Answer. 1. Solution. Let $a$ be the ordinate of point A. The line passing through point $A$ has the equation $\mathrm{y}=k \cdot x+a$, and the abscissas $x_{1}, x_{2}$ of points M and $N$ of intersection of the line with the parabola are the roots of the equation $x^{2}=k \cdot x+a$. From Vieta's theorem, we have $x_{1... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,142 |
11.5. Solve the system of equations $\left\{\begin{array}{l}3^{x}=\sqrt{y} \\ 2^{-y}=x^{3}\end{array}\right.$. | Answer. $x=1 / 2, y=3$. Solution. It is directly verified that the numbers $x=1 / 2, y=3$ are a solution to the system. We will prove that the solution is unique. For this, we will show that the function $y(x)$, defined by the first equation, is strictly decreasing, and the function defined by the second equation is st... | \frac{1}{2},3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,143 |
11.1. Solve the inequality $2 \cos (\cos x)>1$. | Answer: $x$ is any real number.
Solution. Since $\cos x \in[-1,1]$ for all $x$, from the inequality $1\cos 1>\cos \pi / 3=1 / 2$. Hence the result. | x | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 22,144 |
11.2. Solve the equation $(\sqrt{2023}+\sqrt{2022})^{x}-(\sqrt{2023}-\sqrt{2022})^{x}=\sqrt{8088}$. | Answer: $x=1$.
Solution. Notice that $\sqrt{8088}=2 \sqrt{2022}$. This observation suggests that $x=1$ is a root of the equation. We will show that there are no other roots. Indeed, the left side represents the difference of two exponential functions: the base of the first is greater than one, and the base of the seco... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,145 |
11.3. Given a triangle with side lengths being rational numbers. Prove that the following are rational numbers: a) the ratio $R / r$, where $R$ and $r$ are the radii of the circumscribed and inscribed circles; b) the value $\sin \frac{\alpha}{2} \sin \frac{\beta}{2} \sin \frac{\chi}{2}$, where $\alpha, \beta, \chi$ are... | Solution. a) Let $a, b, c$ be the lengths of the sides, $p$ be the semiperimeter, and $S$ be the area of the triangle. From the known relations $S=p r$, $S=\frac{a b c}{4 R}$, and Heron's formula $S=\sqrt{p(p-a)(p-b)(p-c)}$, it follows that $\frac{R}{r}=\frac{a b c p}{4 S^{2}}=\frac{a b c p}{4 p(p-a)(p-b)(p-c)}=\frac{a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,146 |
11.4. Given several rectangular parallelepipeds in space. It is known that each pair of parallelepipeds has at least one common point, and their edges are respectively parallel. Is it necessarily true that all parallelepipeds have a common point? | Answer. Mandatory.
Solution. Since the edges of the parallelepipeds are respectively parallel, we can introduce a Cartesian coordinate system, directing the axes along three edges adjacent to one vertex (which will become the origin of coordinates) of the chosen parallelepiped. In this coordinate system, the edges of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,147 |
11.5. a) Prove that the first 11 natural numbers $1, 2, \ldots, 11$ cannot be rearranged so that adjacent numbers differ by either 3 or 5. b) Can this be done for the numbers 1, 2, .., 12? | Answer: b) is possible.
Solution. a) Let's draw a graph of possible neighbors (see the figure). Consider three pairs of vertices in the quadrilaterals on the graph (they are marked with bold dots), namely: $(2,10),(1,9)$ and $(3,11)$ - these are vertices with the smallest degree (number of neighbors), equal to 2. Supp... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,148 |
8.2. Given a natural number n. Let $N=n^{4}-90 n^{2}-91 n-90$. Prove that for $n>10$ a) $N$ is a composite natural number; b) $N$ can be represented as the product of three natural factors, each greater than one. | Solution. a) Of course, part a) follows from the factorization in part b), but another solution can be provided. The fact that $N$ is a composite number follows from its evenness (provided that $N>2$), and the evenness follows from the fact that the numbers $n^{4}$ and $91 n$ have the same parity. Further, $N=n^{2}\lef... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,150 |
8.3. In triangle $A B C$, the bisector $A M$ and the median $B N$ intersect at point $O$. It turns out that the areas of triangles $A B M$ and $M N C$ are equal. Find $\angle M O N$. | Answer: $90^{\circ}$. Solution. Since $S_{A M N}=S_{M N C}$ (because $A N=N C$), from the condition of the problem we have $S_{A B M}=S_{A N M}$. Therefore, in triangles $A B M$ and $A M N$, the heights drawn from vertices $B$ and $N$ are equal. Let $B_{1}$ and $N_{1}$ be the bases of these heights. Then the right tria... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,151 |
8.4. Natural numbers $m$ and $n$ are such that $m \cdot n$ is divisible by $m+n$. Can we assert that $m$ is divisible by $n$ if it is known that a) $n$ is a prime number? b) $n$ is the product of two different prime numbers? | Answer: a) can; b) cannot. Solution. a) Let $n=p$ - a prime number. We have $m p=k(m+p)$ for some natural number $k$. If we assume, to the contrary, that $m$ does not divide $p$, then from the equality $k m=p(m-k)$ it would follow that $k$ divides $p$, i.e., $k=k_{1} \cdot p$ for some natural number $k_{1}$. Then, by c... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,152 |
8.5. 10 girls and 10 boys stood in a row such that girls and boys alternate, specifically from left to right: girl-boy-girl-boy and so on. Every minute, in one (any) pair of neighbors "girl-boy," the children can swap places, provided that the girl is to the left of the boy. Can such an "exchange process" continue for ... | Answer: It cannot. Solution. Consider the numbers in order (from left to right) of all ten boys. Initially, these were all even numbers $2,4,6, \ldots, 20$. Every minute, the number of one of the boys decreases by one, and the process will continue until the numbers of the boys become $1,2,3, \ldots, 10$ (i.e., until a... | 55 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,153 |
9.1. Prove that the equation $x^{99}=2013 y^{100}$ has solutions in natural numbers $x, y$. | Solution. See the solution to problem 8.2. In this equation, we can take $x=2013^{99}$, $y=2013^{98}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,154 |
9.2. The number $a$ is a root of the quadratic equation $x^{2}-x-50=0$. Find the value of $a^{3}-51 a$. | Answer: 50. Solution. We have $a^{2}=a+50$, therefore $a^{3}=a^{2}+50 a=a+50+50 a=51 a+50$. Hence $a^{3}-51 a=50$. | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,155 |
9.3. From 60 numbers $1,2, \ldots, 60$, 25 numbers were chosen. It is known that the sum of any two of the chosen numbers is not equal to 60. Prove that among the chosen numbers, there are multiples of five. | Solution. Suppose the opposite, then the chosen numbers are among the first 48 natural numbers not divisible by five. Note that among the chosen numbers, there cannot be a pair of the form ( $a, 60-a)$, i.e., in each such pair, there is no more than one number among the chosen ones. Since there are a total of $48: 2=24... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,156 |
9.4. In triangle $ABC$ with sides $AB=c, BC=a, AC=b$, the median $BM$ is drawn. Incircles are inscribed in triangles $ABM$ and $BCM$. Find the distance between the points of tangency of these incircles with the median $BM$. | Answer: $\quad \frac{|a-c|}{2}$. Solution. Let $P_{1}, Q_{1}, L_{1}$ be the points of tangency of the incircle of triangle $A B M$ with sides $A B, A M$ and $B M$ respectively. Similarly, the points of tangency of the second circle with sides $B C, M C$ and $B M$ are denoted as $P_{2}, Q_{2}, L_{2}$. We need to find $L... | \frac{|-|}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,157 |
9.5. How many solutions in natural numbers $x, y$ does the system of equations have
$$
\left\{\begin{array}{l}
\text { GCD }(x, y)=20! \\
\text { LCM }(x, y)=30!
\end{array} \quad(\text { where } n!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot n) ?\right.
$$ | Answer: $2^{8}$. Solution. If for the given two numbers $x, y$ the set of their prime divisors is denoted as $p_{1}, p_{2}, \cdots, p_{k}$ and we write $x=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}, y=p_{1}^{\beta_{1}} p_{2}^{\beta_{2}} \cdots p_{k}^{\beta_{k}}$ (where $\alpha_{i}, \beta_{i}$ are n... | 256 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,158 |
9.1. Do there exist such non-integer numbers $x, y$ that both numbers $5 x+7 y$ and $7 x+10 y$ are integers? | Answer: They do not exist. Solution. Let $5 x+7 y=m, 7 x+10 y=n$, where $m$ and $n$ are integers. Solve this system of equations by multiplying the first equation by 10, and the second by 7. Subtracting the obtained equations, we will have $x=10 m-7 n$, i.e., $x$ is an integer. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,159 |
9.2. In triangle $A B C$, angle $A$ is the largest. Points $M$ and $N$ are symmetric to vertex $A$ with respect to the angle bisectors of angles $B$ and $C$ respectively. Find $\angle A$, if $\angle M A N=50^{\circ}$. | Answer: $80^{\circ}$. Solution. See problem 8.2 | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,160 |
9.3 How many right-angled triangles with integer sides exist, where one of the legs is equal to 2021. | Answer: 4. Solution. Let the hypotenuse of a right triangle be $x$, one of the legs be $-y$, and the other be 2021. Then, by the Pythagorean theorem, $x^{2}-y^{2}=2021^{2}$, i.e., $(x-y) \cdot(x+y)=2021^{2}$. Considering that $x>y$, we have: $x+y>x-y>0$. Since the prime factorization of 2021 is $2021=43 \cdot 47$, the ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,161 |
9.5. Given 10 numbers: $10, 20, 30, \ldots, 100$. With them, you can perform the following operation: choose any three and add one to each of the chosen numbers. The same operation is then performed on the resulting 10 numbers, and so on. Can you, after several operations, obtain: a) all the same numbers? b) all number... | Answer: a) yes, b) no. Solution. a) We will show how to perform three operations on the numbers $a_{1}, a_{2}, \ldots, a_{10}$ so that after several sets of three operations, all numbers become equal. Let $m$ be the smallest and $M$ be the largest of the numbers $a_{1}, a_{2}, \ldots, a_{10}$. We calculate the sum
$$
... | )yes,b)no | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,163 |
8.3. In triangle $A B C$, the median $B M$ is drawn. Sides $A B$ and $B C$ form angles of $100^{\circ}$ and $40^{\circ}$ with the median, respectively, and side $A B$ is equal to 1. Find the length of $B M$. | Answer: $B M=1 / 2$. Solution. Mark point $D$, symmetric to vertex $A$ with respect to point $M$. Quadrilateral $A B C D$ is a parallelogram because its diagonals are bisected at the point of intersection. Therefore, $\quad \angle B D C=\angle \mathrm{ABD}=100^{\circ} \quad$ and $\quad \angle$ $B C D=180^{\circ}-\angle... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,167 |
8.4. In grade 8, there are 30 people, among them 22 attend the French language club, 21 - the German language club, and 18 - the Chinese language club. Prove that there is a student in the class who attends all three clubs. | Solution. Suppose, for the sake of contradiction, that there is no student attending three clubs. Let every student attending a club receive a badge for each club they attend. Then, on one hand, a total of $22+21+18=61$ badges were distributed. On the other hand, no more than $30 \cdot 2=60$ badges were distributed, si... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,168 |
11.1 Given a triangle, where two angles $\alpha, \beta$ satisfy the relation $\cos \alpha+\cos \beta=\sin \alpha+\sin \beta$. Is this triangle necessarily a right triangle? | Answer: necessarily. Solution. Rewrite the given relation as $2 \cos (\alpha+\beta) / 2) \cos ((\alpha-\beta) / 2)=2 \sin ((\alpha+\beta) / 2) \cos ((\alpha-\beta) / 2)$.
Notice that $\cos ((\alpha-\beta) / 2) \neq 0$, otherwise $\alpha-\beta=180^{\circ}$. Then $\cos (\alpha+\beta) / 2)=\sin ((\alpha+\beta) / 2)$, and... | \alpha+\beta=90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,170 |
11.4. In a class of 30 people, for New Year's, each person sent greeting cards to no fewer than 16 classmates. Prove that there were no fewer than 45 pairs of mutual greetings. | Solution. A total of no less than $30 \cdot 16=480$ letters were sent, while the number of pairs of classmates is $(30 \cdot 29) / 2=435$. For each pair of classmates, there can be one of three situations: a) neither of them wrote to the other; b) only one wrote to the other; c) they exchanged letters. Let the number o... | 45 | Combinatorics | proof | Yes | Yes | olympiads | false | 22,173 |
11.1. $\quad$ Find the smallest period of the function $y=\cos ^{10} x+\sin ^{10} x$. | Answer: $\frac{\pi}{2}$. Solution. First, we check that $\frac{\pi}{2}$ is a period of the function. Indeed,
$$
y\left(x+\frac{\pi}{2}\right)=\left(\cos ^{2}\left(x+\frac{\pi}{2}\right)\right)^{5}+\left(\sin ^{2}\left(x+\frac{\pi}{2}\right)\right)^{5}=\sin ^{10} x+\cos ^{10} x
$$
To prove that this is the smallest pe... | \frac{\pi}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,174 |
11.2. In the piggy bank, there are 1000 coins of 1 ruble, 2 rubles, and 5 rubles, with a total value of 2000 rubles. How many coins of each denomination are in the piggy bank, given that the number of 1-ruble coins is a prime number. | Answer: 1-ruble coins - 3, 2-ruble coins - 996, 5-ruble coins - 1. Solution. Let $x, y, z$ be the number of coins worth 1 ruble, 2 rubles, and 5 rubles, respectively. Then $x+y+z=1000$ and $x+2 y+5 z=2000$. Subtract the first equation, multiplied by 2, from the second. We get $x=3 z$. Thus, from the condition that $x$ ... | 1-ruble-3,2-ruble-996,5-ruble-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,175 |
11.1. Solve the equation $x^{10}-3 x^{4}+x^{2}+1=0$. | Answer: $x= \pm 1$. Solution. Making the substitution $t=x^{2}, t \geq 0$, we get the equation $t^{5}-3 t^{2}+t+1=0$. Noticing that $t=1$ is a root and dividing the left side by $(t-1)$, we will have $(t-1)\left(t^{4}+t^{3}+t^{2}-2 t-1\right)=0$. The polynomial $t^{4}+t^{3}+t^{2}-2 t-1$ also has a root $t=1$. After div... | \1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,178 |
11.2. Prove the inequality $\sin \alpha \cos \frac{\alpha}{2} \leq \sin \left(\frac{\pi}{4}+\alpha\right)$ for all $\alpha \in\left[0, \frac{\pi}{2}\right]$. | Solution. In the right-hand side, by the sine sum formula, we have $\sin \left(\frac{\pi}{4}+\alpha\right)=\frac{\sqrt{2}}{2}(\cos \alpha+\sin \alpha)$. To the left-hand side, we apply the double-angle cosine formula $\cos \frac{\alpha}{2}=\frac{\sqrt{1+\cos \alpha}}{\sqrt{2}}$ (here we consider that $\cos \frac{\alpha... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 22,179 |
11.5. On the coordinate plane, the graph of $y=\frac{2020}{x}$ is constructed. How many points on the graph have a tangent that intersects both coordinate axes at points with integer coordinates? | Answer: 40 points. Solution. The equation of the tangent at the point $\left(x_{0}, y_{0}\right)$ to the hyperbola $y=k / x$ is $y-y_{0}=-\left(k / x_{0}^{2}\right)\left(x-x_{0}\right)$, where $y_{0}=k / x_{0}$. From this equation, the coordinates $x_{1}$ and $y_{1}$ of the points of intersection with the axes O $x$ an... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,182 |
9.1. For what values of the parameter $a$ do the equations $a x+a=7$ and $3 x-a=17$ have a common integer root? | Answer: $a=1$.
Solution. Solving these two equations as a system with unknowns $x$ and $a$, express $a$ from the second equation and substitute it into the first. We get the quadratic equation $3 x^{2}-14 x-24=0$. It has two roots: 6 and (-4/3). For the integer root $x=6$, the corresponding value of $a=3 x-17=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,183 |
9.2. Given two coprime natural numbers $p$ and $q$, differing by more than one. a) Prove that there exists a natural number $n$ such that the numbers $p+n$ and $q+n$ will not be coprime. b) Find the smallest such $n$ for $p=2, q=2023$. | Answer: b) 41.
Solution. See the solution to problem 8.2 (including the notation and comments). If both numbers $p+n$ and $q+n$ are divisible by some $k>1$, then $m=q-p$ is also divisible by $k$, and therefore $k$ is not less than the smallest prime divisor of the number $m$. For $p=2, q=2023$, we have $m=2021=43 \cdo... | 41 | Number Theory | proof | Yes | Yes | olympiads | false | 22,184 |
9.3. Given a convex quadrilateral $A B C D$, which is not a kite (a kite is a quadrilateral symmetric with respect to one of its diagonals). It is known that the bisectors of angles $A$ and $C$ intersect at a point on diagonal $B D$. Prove that the bisectors of angles $B$ and $D$ intersect on diagonal $A C$. | Solution. Let $M$ be the point of intersection of the bisectors of angles $A$ and $C$ (since $ABCD$ is not a kite, these bisectors do not coincide and have a unique point of intersection). By the condition, $M$ lies on the diagonal $BD$ and therefore, by the property of angle bisectors, $\frac{DA}{AB}=\frac{DM}{MB}=\fr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,185 |
9.4. Given the equation $x^{3}+2^{n} \cdot y=y^{3}+2^{n} \cdot x$. Prove that a) if natural numbers $x, y, n$ satisfy this equation, then $x=y ;$ b) if non-zero integers $x, y$ and non-negative integers $n$ satisfy this equation, then $|x|=|y|$. | Solution. a) We have $x^{3}-y^{3}=2^{n}(x-y)$. If $x \neq y$, then after dividing by ($x-y$) we get $x^{2}+xy+y^{2}=2^{n}$. If at least one of the numbers $x$ or $y$ is odd, then the left side will be an odd number, while the right side is even. Therefore, $x=2x_{1}, y=2y_{1}$ for some natural numbers $x_{1}$, $y_{1}$.... | proof | Algebra | proof | Yes | Yes | olympiads | false | 22,186 |
9.5. In a financial company, there are 20 shareholders, and their total package is 2000 shares. The shareholders need to be divided into two groups of 10 people each, with packages of 1000 shares in each group. Prove that there will be two such shareholders that if one of them sells a part of their shares to the other,... | Solution. Let's assign shareholders numbers in ascending order of their shares: let $x_{1} \leq x_{2} \leq \ldots \leq x_{20}$, where $x_{i}$ is the number of shares of the shareholder with number i. Next, we divide the shareholders into two groups of 10 people with odd and even numbers. Let $S_{1}$ and $S_{2}$ be the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,187 |
11.1 Prove the inequality $\cos ^{6} \alpha+\sin ^{6} \alpha \geq 1 / 4$ for any $\alpha$. | Solution. Let's write the left part of the inequality as a sum of cubes and use the fundamental trigonometric identity and the double-angle sine formula: $\cos ^{6} \alpha+\sin ^{6} \alpha=\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)$. $\left(\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)^{2}-3 \cos ^{2} \alpha \cdot... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 22,188 |
11.2 What is the maximum value that the ratio of the radius of the inscribed circle to the radius of the circumscribed circle of a right triangle can reach? | Answer $\sqrt{2}-1$. Solution. Let $c$ be the hypotenuse and $\alpha$ be the acute angle. Then $R=c / 2$ and $r=(c \cos \alpha+c \sin \alpha-c) / 2=(c \sqrt{2} \sin (\alpha+\pi / 4)-c) / 2 \leq c(\sqrt{2}-1) / 2$. Therefore, the maximum value of the ratio $r / R$ is $\sqrt{2}-1$ and is achieved when $\alpha=\pi / 4$, i... | \sqrt{2}-1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,189 |
11.3 From 25 natural numbers $1,2, \ldots, 25$, several different numbers need to be selected and arranged in a circle such that the sum of the squares of any three consecutive numbers is divisible by 10. Is it possible to choose a) 8 numbers?; b) 9 numbers? | Answer. a) No, b) Yes. Solution. a) Suppose, to the contrary, that such an arrangement is possible. Take adjacent triples of numbers and subtract the sum of the squares of the second triple from the sum of the squares of the first. We get that the square of each number on the circle gives the same remainder when divide... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,190 |
11.4 a) A rectangle with an area of 2018 is placed on the coordinate plane such that its sides are parallel to the coordinate axes, and all four vertices lie within different quadrants and have integer coordinates. Find the length of its diagonal. b) Does there exist a rectangle with an area of 2018 and integer vertice... | Answer. a) $\sqrt{1018085} ;$ b) exists. Solution. See problem 10.4.
## 1st round. 11.11.2018
## 11th grade | \sqrt{1018085} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,191 |
11.1 For what values of the parameter $a$ does the equation $x^{2}-6 a x-2+2 a+9 a^{2}=0$ have at least one negative root?
---
The text has been translated while preserving the original formatting and line breaks. | Answer. $a3 a$. For $a0$. Solving it, we get $(-1-\sqrt{19}) / 9<a<(-1+\sqrt{19}) / 9$. Combining this with the solution for $a<0$, we obtain the answer. Another solution method (more visual) is based on considering two cases depending on the sign of the abscissa $x=3 a$ of the vertex of the parabola $y=x^{2}-6 a x-2+2... | \frac{-1+\sqrt{19}}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,192 |
11.2 Solve the equation $3 \sin x + 4 \cos x = 2^{x+3} + 2^{-x}$. | Answer: No roots. Solution. The left side of the equation, by introducing an auxiliary angle, is reduced to the form $5 \sin (x+\alpha)$. This means it does not exceed 5. For the right side, by the inequality between the arithmetic mean and the geometric mean (for positive numbers), we will have: $2^{x+3}+2^{-x} \geq 2... | Noroots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,193 |
11.4 a) Prove that there exists an increasing geometric progression from which three terms (not necessarily consecutive) can be selected to form an arithmetic progression. b) Can the common ratio of such a geometric progression be a rational number? | Answer. b) It cannot. Solution. a) See problem 10.4a. b) Suppose, for the sake of contradiction, that such a geometric progression exists, and let its first term be $b$, and the common ratio be $q$, where $q>1$ is a rational number. Suppose three numbers $b q^{m}, b q^{n}, b q^{k}$ form an arithmetic progression, where... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,195 |
9.1. Do there exist numbers $a, b$ that satisfy the relation $a^{2}+3 b^{2}+2=3 a b$? | Answer: do not exist. Solution. The result follows from the relation
$$
a^{2}-3 a b+3 b^{2}+2=\left(a-\frac{3}{2} b\right)^{2}+\frac{3}{4} b^{2}+2 \geq 2
$$ | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,196 |
9.2. Prove that there exist two prime numbers $p < q$ such that all natural numbers between $p$ and $q$ are composite. | Solution. Let $n=2015!=1 \cdot 2 \cdot \ldots \cdot 2015$. Then all 2015 numbers: $n+2, n+3, \ldots, n+2015, n+2016$ are composite. Let $p-$ be the largest prime number not exceeding $n+1$, and $q$ - the smallest prime number greater than $n+2016$. Then $q-p>2015$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,197 |
9.1. Given three positive numbers, not necessarily distinct. It is known that if the product of any two of them is subtracted by the third, the result is the same number $a$. Prove that $a \geq-\frac{1}{4}$. | Solution. When solving problem 7.2, we found that for positive numbers, only the first case is realized, which means that all three numbers must coincide, i.e., they must satisfy the equation $x^{2}-x=a$. This equation has real roots under the condition $D=1+4 a \geq 0$, and thus, $a \geq-\frac{1}{4}$ | \geq-\frac{1}{4} | Algebra | proof | Yes | Yes | olympiads | false | 22,200 |
9.2. Is there a point with integer coordinates on the coordinate plane, the distance from which to the origin is equal to $\sqrt{2 \cdot 2017^{2}+2 \cdot 2018^{2}}$? | Answer. There is. Solution. We will prove that for any natural number $n$, there exists a point with integer coordinates, the distance from which to the origin is $\sqrt{2 n^{2}+2(n+1)^{2}}$. To do this, we transform the last expression $2 n^{2}+2(n+1)^{2}=2 n^{2}+2 n^{2}+4 n+2=\left(4 n^{2}+4 n+1\right)+$ $1=(2 n+1)^{... | (2n+1;1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,201 |
9.4. Petya says to Vasya: «I constructed a non-isosceles triangle $A B C$ and drew the angle bisectors $A M$ and $C N$. It turned out that $O M=O N$, where $O$ is the point of intersection of the bisectors. Can you determine what the angle $B$ is?» Vasya replies: «No, that can't be, for the segments $O M$ and $O N$ to ... | Answer: Petya is right. Solution: We will prove that for a non-isosceles triangle with the property \( OM = ON \), the angle \( B \) is uniquely determined, and specifically, \( \angle B = 60^\circ \). Moreover, we will show that any non-isosceles triangle with an angle of \( 60^\circ \) at the vertex has this property... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,203 |
9.5. Find all pairs of natural numbers $m, n$ for which $n!+4!=m^{2}$ (where $n!=1 \cdot 2 \cdot \ldots \cdot n$ ). | Answer: Two pairs $n=1, m=5$ and $n=5, m=12$. Solution. We will show that $n<6$. Indeed, for $n \geq 6$ the number $n!$ is divisible by 16. Therefore, the sum $n!+4$ can be represented as $8 \cdot(2 m+3)$. This number cannot be a square of an integer, since in its prime factorization, the factor 2 appears in an odd pow... | n=1,=5n=5,=12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,204 |
9.1. 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 | 22,205 |
9.2. The number $a$ is a root of the quadratic equation $x^{2}-x-100=0$. Find the value of $a^{4}-201 a$ | Answer: 10100.
Solution. Squaring the expression $a^{2}=a+100$, we get $a^{4}=a^{2}+200a+10000=a+100+200a+10000=201a+10100$. From this, we obtain the answer. | 10100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,206 |
9.3. 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 | 22,207 |
9.4. a) Given natural numbers $a$ and $b$. Can we assert that they have the same remainders when divided by 10, if it is known that the numbers $3a + b$ and $3b + a$ have the same remainders when divided by 10? b) Given natural numbers $a, b$, and $c$. It is known that the numbers $2a + b, 2b + c$, and $2c + a$ have th... | Answer. a) No, it is not possible.
Solution. a) For example, we can take $a=1$ and $b=6$, then both numbers $3a+b$ and $3b+a$ have a remainder of 9. b) Let $s=a+b+c$. By subtracting $s$ from each of the numbers $2a+b, 2b+c$, and $2c+a$, we get the numbers $a-c, b-a, c-b$ (some of which may be negative), and these numb... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,208 |
9.1. Prove the inequality $|a-1| \leq a^{2}-a+1$. | Solution. When $a \geq 1$, the inequality takes the form $a-1 \leq a^{2}-a+1 \Leftrightarrow (a-1)^{2}+1 \geq 0$. When $a<1$, the inequality can be written as: $-a+1 \leq a^{2}-a+1 \Leftrightarrow a^{2} \geq 0$. In both cases, we obtain true (obvious) inequalities, and thus, for all $a$, the original inequality holds. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 22,209 |
9.2. Given a convex quadrilateral $A B C D$. Can we assert that $A B<C D$, if a) angles $A$ and $B$ are obtuse; b) angle $A$ is greater than angle $D$, and angle $B$ is greater than angle $C$? | Answer. a) it is possible; b) it is possible.
Solution. Answer. a) it is possible; b) it is possible. Solution. a) The solution to part b) also implies part a), but for
part a) a simpler solu... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,210 |
9.3. 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 | 22,211 |
9.4. 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 | 22,212 |
11.2. Plot on the coordinate plane the set of points satisfying the inequality $\frac{\sqrt{y-x}}{x} \leq 1$ | Solution. For $x>0$, the original inequality can be written as $\sqrt{y-x} \leq x \quad \Leftrightarrow$ $0 \leq y-x \leq x^{2} \quad \Leftrightarrow x \leq y \leq x+x^{2}, \quad$ i.e., the set in the right half-plane lies between the graphs of $y=x$ and $y=x^{2}+x$. It is easy to verify that the parabola $y=x^{2}+x$ i... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 22,213 |
11.3. How many points are there on the hyperbola $y=\frac{2013}{x}$ such that the tangent at these points intersects both coordinate axes at points with integer coordinates | Answer: 48 points. Solution. Let $k=2013$. The equation of the tangent to the hyperbola $y=\frac{k}{x}$ at the point $\left(x_{0}, y_{0}\right)$ is $y-y_{0}=-\frac{k}{x_{0}^{2}}\left(x-x_{0}\right)$, where $y_{0}=\frac{k}{x_{0}}$. From this, we find the coordinates of the points of intersection of the tangent with the ... | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,214 |
10.1. Does there exist a rectangle with irrational sides, for which a) the area and perimeter are integers? b) the area, perimeter, and diagonal are integers? | Answer: a) exists; b) exists. Solution. If the sides of the rectangle are of the form $a=m+\sqrt{n}, b=m-\sqrt{n}$, where $m$ and $n$ are natural numbers, with $n<m^{2}$ and $n$ not being a perfect square, then the area of the rectangle is $(m+\sqrt{n})(m-\sqrt{n})=m^{2}-n$, and the perimeter is $4 m$, thus satisfying ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,216 |
10.2. Given the coefficients $a, b, c$ of the quadratic trinomial $a x^{2}+b x+c$. Its graph intersects the coordinate axes at three points, and a circle passing through these points intersects the y-axis at one more point. Find the ordinate of this fourth point. | Answer: $1 / a$. Solution. First, consider the case when the parabola intersects the $O x$ axis at points $x_{1}, x_{2}$ on the same side of the origin $O$. If $a>0$, then $c>0$ and applying the theorem of intersecting chords for the circle, we get that the desired ordinate $y_{0}$ satisfies the equation $y_{0} \cdot c... | \frac{1}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,217 |
10.3. On the lateral sides $AB$ and $CD$ of trapezoid $ABCD$, points $M$ and $N$ are taken such that $AN=BN$ and $\angle ABN = \angle CDM$. Prove that $CM=MD$. | Solution. From the equality $A N=B N$ it follows that $\angle A B N=\angle B A N$ in the isosceles triangle $A B N$. Then, by the condition of the problem, $\angle C D N=\angle B A N$, and therefore, a circle can be circumscribed around quadrilateral $A M N D$. Thus, $\angle M A D=180^{\circ}-\angle M N D=\angle C N M$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,218 |
10.4. How many solutions in integers $x, y$ does the equation $|3 x+2 y|+|2 x+y|=100$ have? | Answer: 400. Solution. Note that for any integers $a, b$ the system of equations $\left\lvert\,\left\{\begin{array}{l}3 x+2 y=a \\ 2 x+y=b\end{array}\right.$ has an integer solution $\left\lvert\,\left\{\begin{array}{l}x=2 b-a \\ y=2 a-3 b\end{array}\right.$, and different ordered pairs $(a, b)$ correspond to different... | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,219 |
10.1. Plot on the coordinate plane the set of points whose coordinates satisfy the equation $\left(x^{2}+3 x y+2 y^{2}\right)\left(x^{2} y^{2}-1\right)=0$. | Solution. Factoring the first bracket into $(x+2 y)(x+y)$, and the second into $(x y+1)(x y-1)$, we obtain that the desired set is the union of two lines $y=-x, y=-x / 2$ and two hyperbolas $y=1 / x, y=-1 / x$. | -x,-\frac{x}{2},\frac{1}{x},-\frac{1}{x} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,221 |
10.2. How many solutions in natural numbers $x, y$ does the equation $x+y+2 x y=2023$ have? | Answer: 6.
Solution. Multiply the equation by 2, add one to both sides, and factor the left side, while expressing the right side as a product of prime factors:
$$
2 x+4 x y+2 y+1=4047 \Leftrightarrow(2 x+1)(2 y+1)=4047 \Leftrightarrow(2 x+1)(2 y+1)=3 \cdot 19 \cdot 71
$$
Since \(x\) and \(y\) are natural numbers, e... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,222 |
10.3. Given a triangle with side lengths being rational numbers. Prove that the ratio $R / r$ is a rational number, where $R$ and $r$ are the radii of the circumscribed and inscribed circles of the triangle. | Solution. Let $a, b, c$ be the lengths of the sides, $p$ be the semiperimeter, and $S$ be the area of the triangle. From the known relations $S=p r$, $S=\frac{a b c}{4 R}$, and Heron's formula $S=\sqrt{p(p-a)(p-b)(p-c)}$, it follows that $\frac{R}{r}=\frac{a b c p}{4 S^{2}}=\frac{a b c p}{4 p(p-a)(p-b)(p-c)}=\frac{a b ... | \frac{}{4(p-)(p-b)(p-)} | Geometry | proof | Yes | Yes | olympiads | false | 22,223 |
10.4. In a financial company, there are 20 shareholders, and their total package is 2000 shares. The shareholders need to be divided into two groups of 10 people each, with packages of 1000 shares in each group. Prove that there will be two such shareholders that if one of them sells a part of their shares to the other... | Solution. Let's assign shareholders numbers in ascending order of their shares: let $x_{1} \leq x_{2} \leq \ldots \leq x_{20}$, where $x_{i}$ is the number of shares of the shareholder with number $i$. Next, we divide the shareholders into two groups of 10 people with odd and even numbers, respectively. Let $S_{1}$ and... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,224 |
10.5. a) Prove that the first 11 natural numbers $1,2, \ldots, 11$ cannot be rearranged so that adjacent numbers differ by either 3 or 5. b) Can this be done for the numbers $1,2, \ldots, 12 ?$ | Answer: b) it is possible.
Solution. a) Let's draw a graph of possible neighbors (see the figure). Consider three pairs of vertices in the quadrilaterals on the graph (they are marked with bold dots), namely: $(2,10),(1,9)$ and $(3,11)$ - these are vertices with the smallest degree (number of neighbors), equal to 2. S... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,225 |
11.1. On the board, there is a 30-digit number $22 \ldots 211 \ldots 100 \ldots 0$, containing 10 twos, 10 ones, and 10 zeros. Can the digits in this number be rearranged to form a square of a natural number? | Answer: No.
Solution. Let the given number be $N$ and assume, for the sake of contradiction, that $N=n^{2}$ for some natural number $n$. The sum of the digits of $N$ is $2 \cdot 10 + 1 \cdot 10 = 30$. From the divisibility rules for 3 and 9, it follows that $N$ is divisible by 3 but not by 9. Since the sum of the digi... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,226 |
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