problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
# Problem 4.
Diagonals $A C$ and $B D$ of a convex quadrilateral $A B C D$, equal to 3 and 4 respectively, intersect at an angle of $75^{\circ}$. What is the sum of the squares of the lengths of the segments connecting the midpoints of opposite sides of the quadrilateral? | Answer: 12.5
Solution. Each such segment is found using the cosine theorem in a triangle connecting the midpoints of the sides of the quadrilateral. In this triangle, two sides are equal to half the diagonals, and the angle coincides with the angle between the diagonals or complements it to $180^{\circ}$. In this case... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,202 |
# Problem 5.
The segments of two lines, enclosed between two parallel planes, are in the ratio $5: 9$, and the acute angles between these lines and one of the planes are in the ratio $2: 1$. Find the cosine of the smaller angle. | Answer: 0.9
Solution. If $h$ is the distance between the planes and $\alpha$ is the smaller of the angles, then the lengths of the segments are $h / \sin 2 \alpha$ and $h / \sin \alpha$, respectively. Therefore,
$$
\frac{\sin \alpha}{\sin 2 \alpha}=\frac{5}{9}, \quad \text { from which } \quad \cos \alpha=\frac{\sin ... | 0.9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,203 |
# Problem 7.
Find all values of $a$ for each of which the equation $x^{2}+2 a x=8 a$ has two distinct integer roots. In the answer, write the product of all such $a$, rounding to the hundredths if necessary. | Answer: 506.25
Solution. If $x_{1}$ and $x_{2}$ are the roots of this equation, then $x_{1}+x_{2}=-2a, x_{1}x_{2}=-8a$. Therefore, $x_{1}x_{2}-4(x_{1}+x_{2})=0$, which means $(x_{1}-4)(x_{2}-4)=16$. Let, for definiteness, $x_{1} \leqslant x_{2}$. Then the following cases are possible (see table)
\[
\begin{array}{rlcc... | 506.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,204 |
# Problem 8.
Inside a convex $n$-gon, 100 points are placed such that no three of these $n+100$ points lie on the same line. The polygon is divided into triangles, each of whose vertices are 3 of the given $n+100$ points. For what maximum value of $n$ can there not be more than 300 triangles? | Answer: 102
Solution. If each point is a vertex of some triangle, then the sum of the angles of all the obtained triangles is $180^{\circ} \cdot(n-2)+360^{\circ} \cdot 100=180^{\circ} \cdot(n-2+200)=180^{\circ} \cdot(n+198)$. Therefore, $\frac{180^{\circ} \cdot(n+198)}{180^{\circ}}=n+198$ triangles are obtained. If on... | 102 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,205 |
# Problem 9.
Given a polynomial $P(x)$ of degree 10 with the leading coefficient 1. The graph of $y=P(x)$ lies entirely above the $O x$ axis. The polynomial $-P(x)$ is factored into irreducible factors (i.e., polynomials that cannot be expressed as the product of two non-constant polynomials). It is known that at $x=2... | Answer: 243
Solution. Note that since the graph of $y=P(x)$ is entirely above the $O x$ axis, the polynomial $P(x)$ has no real roots. This means that all irreducible polynomials in the factorization have degree 2, and their number is 5. Since the polynomial $-P(x)$ will have the same number of irreducible polynomials... | 243 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,206 |
# Problem 10.
Set $A$ on the plane $O x y$ is defined by the equation $x^{2}+y^{2}=2 x+2 y+23$. Set $B$ on the same plane is defined by the equation $|x-1|+|y-1|=5$. Set $C$ is the intersection of sets $A$ and $B$. What is the maximum value that the product of the lengths of $n$ segments $X Y_{1} \cdot X Y_{2} \cdot X... | Answer: 1250
Solution. The set $A$ is $(x-1)^{2}+(y-1)^{2}=25$, which is a circle with center $(1 ; 1)$ and radius 5. The set $B$ is a square with vertices $(-4 ; 1),(1 ; 6),(6 ; 1),(1 ;-4)$. The center of the square is the point $(1 ; 1)$, the diagonals of the square are 10, and the sides are $5 \sqrt{2}$. The set $C... | 1250 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,207 |
1. After treating the garden with a caterpillar control agent, the gardener noticed that from 12 blackcurrant bushes he was getting the same harvest as before from 15 bushes. By what percentage did the blackcurrant yield in the garden increase? | Answer: $25 \%$.
Solution. The yield from 12 bushes increased by $15 / 12=1.25$ times, which means the yield of currants increased by $25 \%$. | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,208 |
2. What is the degree measure of angle $\angle A$, if its bisector forms an angle with one of its sides that is three times smaller than the angle adjacent to $\angle A$? | Answer: $72^{\circ}$.
Solution. Let $x$ be the degree measure of the angle formed by the angle bisector of $\angle A$ with one of its sides. Then the degree measure of the angle $\angle A$ is $2 x$, and the degree measure of the adjacent angle is $3 x$. Therefore, $2 x+3 x=180$, from which $x=36$ and $\angle A=72^{\ci... | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,209 |
4. On a vast meadow, a lamb is grazing, tied with two ropes to two stakes (each stake with its own rope).
a) What shape will the area on the meadow be, where the lamb can eat the grass?
b) In the middle between the stakes, a rose grows, and the distance between the stakes is 20 meters. What should the lengths of the ... | Answer: b) Exactly one of the ropes is shorter than 10 m.
Solution. a) The desired figure is the intersection of two circles with centers at the stakes and radii equal to the lengths $l_{1}$ and $l_{2}$ of the first and second ropes, respectively. Therefore, it can be either the figure shaded in Fig. 3, or a complete ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,211 |
6. Can the figure shown in the diagram be cut into four equal parts along the grid lines so that they can be used to form a square? Answer: Yes. | Solution. A way to cut the figure and assemble it from the resulting pieces
 \leftarrow(50,25)$. Then the complete sequence of actions will look like this:
$(25,25) \leftarrow(50,25) \leftarrow(50,... | 850g | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,217 |
3. Find the number of numbers from 1 to 3400 that are divisible by 34 and have exactly 2 odd natural divisors. For example, the number 34 has divisors $1,2,17$ and 34, exactly two of which are odd. | Answer: 7. Solution. It is obvious that if a number is divisible by 34, then its divisors will always include 1 and 17. According to the condition, there should be no other odd divisors, so these must be numbers of the form $17 \cdot 2^{k}$, $k \geqslant 1$ (and only they). These numbers fall within the specified range... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,218 |
4. It is known that the fraction $\frac{a}{b}$ is less than the fraction $\frac{c}{d}$ and $b>d>0$. Determine which is smaller: the arithmetic mean of these two fractions or the fraction $\frac{a+c}{b+d}$. | Answer: The fraction $\frac{a+c}{b+d}$ is smaller. Solution. $\frac{a}{b}<\frac{c}{d} \Leftrightarrow a d<b c \Leftrightarrow a d(b-d)<$ $b c(b-d) \Leftrightarrow a b d+b c d<b^{2} c+a d^{2} \Leftrightarrow 2 a b d+2 b c d<a b d+b c d+b^{2} c+a d^{2}$ $\Leftrightarrow 2(a+c) b d<(a d+b c)(b+d) \Leftrightarrow \frac{a+c... | \frac{+}{b+}<\frac{1}{2}(\frac{}{b}+\frac{}{}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,219 |
5. An isosceles triangle with sides $A B=B C=3, A C=4$ has an inscribed circle that touches the sides of the triangle at points $K, L$, and $M$. Find the ratio of the areas $S(\triangle A B C): S(\triangle K L M)$. | Answer: $\frac{9}{2}$. Solution. Since the triangle is isosceles, the inscribed circle touches the base at its midpoint, so $A M=M C=$ $A K=C L=2, B K=B L=1$. We will find the areas of the triangles (in relation to the area of $\triangle A B C): S(\triangle A K M)=S(\triangle C L M)=\frac{1}{3} S(\triangle A B C)$,
$S(... | \frac{9}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,220 |
6. Find the sum
$$
\begin{aligned}
& \frac{1}{(\sqrt[4]{1}+\sqrt[4]{2})(\sqrt{1}+\sqrt{2})}+\frac{1}{(\sqrt[4]{2}+\sqrt[4]{3})(\sqrt{2}+\sqrt{3})}+ \\
& +\ldots+\frac{1}{(\sqrt[4]{9999}+\sqrt[4]{10000})(\sqrt{9999}+\sqrt{10000})}
\end{aligned}
$$ | Answer: 9. Solution. If we multiply the numerator and denominator of the fraction $\frac{1}{(\sqrt[4]{n}+\sqrt[4]{n+1})(\sqrt{n}+\sqrt{n+1})}$ by $\sqrt[4]{n}-\sqrt[4]{n+1}$, we get $\sqrt[4]{n+1}-\sqrt[4]{n}$. The specified sum transforms to $(\sqrt[4]{2}-\sqrt[4]{1})+(\sqrt[4]{3}-\sqrt[4]{2})+\ldots+(\sqrt[4]{10000}-... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,221 |
7. In a $5 \times 5$ table, numbers (not necessarily integers) are arranged such that each number is three times smaller than the number in the adjacent cell to the right and twice as large as the number in the adjacent cell below. Find the number in the central cell if it is known that the sum of all numbers in the ta... | Answer: $\frac{36}{11}$. Solution. Let the number in the lower left corner be $x$. We get the equation $x(1+2+4+8+16)(1+3+9+27+81)=341$, from which $x=\frac{1}{11}$. The number in the center is $36 x=\frac{36}{11}$. | \frac{36}{11} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,222 |
8. In trapezoid $A B C D$, base $A D$ is four times larger than base $B C$, and angle $\angle B C D$ is twice the angle $\angle B A D$. Find the ratio $C D$ : $P Q$, where $P Q-$ is the midline of the trapezoid. | Answer: 6:5. Solution. Let $B C=a, A D=4 a$, then $P Q=(a+4 a) / 2=$ $2.5 a$. Draw the bisector $C L$ of angle $\angle B C D$ (see figure). It will divide the trapezoid into a parallelogram and an isosceles triangle $\triangle C L D$ with sides $C D=L D=3 a$. Therefore, $C D: P Q=3 a: 2.5 a=6: 5$.
![](https://cdn.math... | 6:5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,223 |
7. In "Dragon Poker," the deck has four suits. An Ace brings 1 point, a Jack -2 points, a Two $-2^{2}$, a Three $-2^{3}, \ldots$, a Ten $-2^{10}=1024$ points. Kings and Queens are absent. You can choose any number of cards from the deck. In how many ways can you score 2018 points? | Answer: $C_{2021}^{3}=1373734330$.
Solution. In any suit, you can score any number of points from 0 to 2047, and this can be done in a unique way. This can be done as follows: write down this number in binary and select the cards corresponding to the binary positions containing 1 (i.e., if there is a 1 in the $k$-th p... | 1373734330 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,224 |
5.1. $[5-6.4$ (а. - 10 points, б. - 20 points)] On a grid paper, a figure is drawn (see the picture). It is required to cut it into several parts and assemble a square from them (parts can be rotated, but not flipped). Is it possible to do this under the condition that a) there are no more than four parts; b) there are... | Answer: a) yes; b) yes.
Solution. Possible cutting options for parts a) and b) are shown in Fig. 5, a) and b).
Fig. 5, a)
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,226 |
1. Find $15 \%$ of 9 . | Answer: 1.35. Solution. $9 \cdot \frac{15}{100}=1.35$. | 1.35 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,227 |
1.1. Masha thought of a 10-digit number and told Vasya that the remainder of dividing this number by 9 is 3. Then Masha crossed out one digit and told Vasya that the remainder of dividing the resulting 9-digit number by 9 is 7. Help Vasya guess the digit that Masha crossed out. Write this digit in the answer. | Answer: 5. Solution. According to the divisibility rule for 9, the remainder of the division of the sum of the digits of the number by 9 is 3 (so the sum of the digits of this number is $9n+3$), and the remainder of the division of the sum of the digits of the resulting number by 9 is 7 (so the sum of the digits of the... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,229 |
2.1. Trapezoid $A B C D$ with base $A D=6$ is inscribed in a circle. The tangent to the circle at point $A$ intersects lines $B D$ and $C D$ at points $M$ and $N$ respectively. Find $A N$, if $A B \perp M D$ and $A M=3$. | Answer: 12. Solution. The condition $A B \perp M D$ means that $\angle A B D=90^{\circ}$, that is, the base $A D$ is the diameter of the circle. Since the trapezoid is inscribed, it is isosceles. $\triangle D N A$ and $\triangle D A C$ are similar as right triangles with a common angle. From the equality of angles $\an... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,230 |
3.1. The parabola $y=x^{2}$ intersects with the line $y=25$. A circle is constructed on the segment between the intersection points of the parabola and the line as its diameter. Find the area of the convex polygon whose vertices are the intersection points of the given circle and the parabola. Provide the nearest integ... | Answer: 10. Solution. The parabola $y=x^{2}$ intersects the line $y=a$ at points with coordinates $( \pm \sqrt{a} ; a)$. Each of the intersection points of the circle and the parabola mentioned in the problem has coordinates $\left(x ; x^{2}\right)$. For such a point, the distance to the line $y=a$ is $\left|a-x^{2}\ri... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,231 |
4.1. Solve the equation $\left(x^{2}-2 x+4\right)^{x^{2}-2 x+3}=625$. In the answer, specify the sum of the squares of all its roots. If there are no roots, put 0. | Answer: 6. Solution. Since $x^{2}-2 x+3>0$, then $x^{2}-2 x+4>1$. The function $f(z)=z^{z-1}$ is increasing for $z>1$ (if $1<z_{1}<z_{2}$, then $f\left(z_{1}\right)=z_{1}^{z_{1}-1}<z_{1}^{z_{2}-1}<z_{2}^{z_{2}-1}=f\left(z_{2}\right)$. Therefore, the original equation, which is of the form $f\left(x^{2}-2 x+4\right)=f(5... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,232 |
5.1. Calculate: $\frac{1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots+2013 \cdot 2014}{(1+2+3+\ldots+2014) \cdot \frac{1}{5}}$. If necessary, round the answer to the nearest hundredths. | Answer: 6710. Solution. Let's calculate the sum in the numerator: $\sum_{k=1}^{n} k(k+1)=\sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} k$ $=\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{3}$. Here, the well-known formula for the sum of squares of natural numbers is used, which can be proven in various ways (these pr... | 6710 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,233 |
6.1. Find the greatest root of the equation $|\sin (2 \pi x)-\cos (\pi x)|=|| \sin (2 \pi x)|-| \cos (\pi x) \|$, belonging to the interval $\left(\frac{1}{4} ; 2\right)$. | Answer: 1.5. Solution. Since $|A-B|=|| A|-| B|| \Leftrightarrow A^{2}-2 A B+B^{2}=A^{2}-2|A B|+B^{2} \Leftrightarrow$ $A B=|A B| \Leftrightarrow A B \geq 0$, then $\sin 2 \pi x \cdot \cos \pi x \geq 0 \Leftrightarrow 2 \sin \pi x \cdot \cos ^{2} \pi x \geq 0 \Leftrightarrow\left[\begin{array}{c}\cos \pi x=0, \\ \sin \p... | 1.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,234 |
7.1. A team of athletes, one third of whom are snowboarders, descended from the mountain. Some of them got into a cable car carriage, which can accommodate no more than 10 people, while all the others descended on their own, and their number turned out to be more than $45 \%$, but less than $50 \%$ of the total number.... | Answer: 5. Solution. If there were $x$ snowboarders and $y$ people descended on the cable car, then there were a total of $3 x$ athletes and $\left\{\begin{array}{l}\frac{9}{20}=\frac{45}{100}<\frac{3 x-y}{3 x}<\frac{50}{100}=\frac{10}{20}, \\ y \leq 10\end{array} \Rightarrow\right.$ $\left\{\begin{array}{l}x<\frac{20}... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,235 |
8.1. On the lateral edge $S B$ of a regular hexagonal pyramid $S A B C D E F$, a point $P$ is chosen such that $S P$ is 10 times greater than $P B$. The pyramid is divided into two parts by a plane passing through point $P$ and parallel to the edges $S E$ and $F E$. Find the ratio of the volume of the larger of these p... | Answer: 116.44. Solution. Let $S P=c x, P B=b x$ (for the original problem $c=10$, $b=1$). Draw a line parallel to $E F$ from point $P$, and denote the intersection of this line with edge $S C$ as $Q$. Through points $P$ and $Q$, draw planes perpendicular to $P Q$ (Fig. 1). These planes are obviously perpendicular to t... | 116.44 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,236 |
9.1. Find the number of roots of the equation
$$
\operatorname{arctg}\left(\operatorname{tg}\left(\sqrt{13 \pi^{2}+12 \pi x-12 x^{2}}\right)\right)=\arcsin \left(\sin \sqrt{\frac{13 \pi^{2}}{4}+3 \pi x-3 x^{2}}\right)
$$ | Answer: 9. Solution. Let's make the substitution $t=\sqrt{\frac{13 \pi^{2}}{4}+3 \pi x-3 x^{2}}$. Since $\frac{13 \pi^{2}}{4}+3 \pi x-3 x^{2}=4 \pi^{2}-3\left(x-\frac{\pi}{2}\right)^{2}$, then $0 \leq t \leq 2 \pi$. The original equation, after the specified substitution, transforms into the equation $\operatorname{arc... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,237 |
10.1. Find the sum of all integer values of $a$ belonging to the interval $[-10 ; 10]$, for each of which the double inequality $5 \leq x \leq 10$ implies the inequality $a x+3 a^{2}-12 a+12>a^{2} \sqrt{x-1}$. | Answer: -47. Solution. Let $t=\sqrt{x-1}$, then the original problem reduces to finding such values of $a$ for which the inequality $f(t) \equiv a t^{2}-a^{2} t+3 a^{2}-11 a+12>0$ holds for all $t \in[2 ; 3]$. This means that the minimum of the function $f(t)$ on the interval $2 \leq t \leq 3$ is positive.
If $a=0$, t... | -47 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,238 |
# Task 2.
While a lion cub, who was 6 minutes away from the water hole, set off to drink, a second one, having already quenched his thirst, started heading back along the same path at 1.5 times the speed of the first. At the same time, a turtle, who was 32 minutes away from the water hole, set off along the same path ... | Answer: 28.8
Solution. Let $x$ be the speed of the turtle, and $y$ be the speed of the first lion cub. Then the speed of the second lion cub is $1.5 y$. The entire path to the watering hole for the first lion cub is $6 y$, and for the turtle, it is $32 x$. Therefore, the initial distance between them was $6 y - 32 x$,... | 28.8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,239 |
1.1. (2 points) In a large family, one of the children has 3 brothers and 6 sisters, while another has 4 brothers and 5 sisters. How many boys are there in this family | Answer: 4.
Solution. In the family, boys have fewer brothers than girls have, and girls have fewer sisters than boys have. Therefore, there are 4 boys and 6 girls in the family. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,240 |
2.1. (14 points) When going from the first to the third floor, Petya walks 36 steps. When going from the first floor to his own floor in the same entrance, Vasya walks 72 steps. On which floor does Vasya live? | Answer: 5.
Solution. From the first to the third floor, there are 36 steps - the same number as from the third to the fifth. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,241 |
3.1. (14 points) Svetlana, Katya, Olya, Masha, and Tanya attend a math club, in which more than $60 \%$ of the students are boys. What is the smallest number of schoolchildren that can be in this club? | Answer: 13.
Solution. Let $M$ be the number of boys, $D$ be the number of girls in the club. Then $\frac{M}{M+D}>\frac{3}{5}$, hence $M>\frac{3}{2} D \geqslant \frac{15}{2}$. The minimum possible value of $M$ is 8, and the minimum possible value of $D$ is 5, making a total of 13 children. | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,242 |
4.1. (14 points) Find all integer values that the fraction $\frac{8 n+157}{4 n+7}$ can take for natural $n$. In your answer, write the sum of the found values. | Answer: 18.
Solution. We have $\frac{8 n+157}{4 n+7}=2+\frac{143}{4 n+7}$. Since the divisors of the number 143 are only $1, 11, 13$, and 143, integer values of the fraction are obtained only when $n=1$ and $n=34$, which are 15 and 3, respectively, and their sum is 18. | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,243 |
5.1. (14 points) A guard has detained a stranger and wants to drive him away. But the person caught said that he had made a bet with his friends for 100 coins that the guard would not drive him away (if the guard does, he has to pay his friends 100 coins, otherwise they pay him), and, deciding to bribe the guard, offer... | Answer: 199.
Solution. If the guard asks for 199 coins, then by agreeing to give him this amount, the outsider will win the dispute and receive 100 coins. In total, he will lose 99 coins. If the outsider refuses, he will lose the dispute and lose 100 coins, which is less favorable (by 1 coin) for the one caught. If th... | 199 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,244 |
7.1. (14 points) We will call a natural number a snail if its representation consists of the representations of three consecutive natural numbers, concatenated in some order: for example, 312 or 121413. Snail numbers can sometimes be squares of natural numbers: for example, $324=18^{2}$ or $576=24^{2}$. Find a four-dig... | Answer: 1089.
Solution. Note that a snail number can only be a four-digit number if the three numbers from which its record is formed are $8, 9, 10$ (numbers $7, 8, 9$ and smaller still form a three-digit number, while $9, 10, 11$ and larger - already form a number of no less than five digits). It remains to figure ou... | 1089 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,247 |
1.1. The sequence $\left\{x_{n}\right\}$ is defined by the conditions $x_{1}=20, x_{2}=17, x_{n+1}=x_{n}-x_{n-1}(n \geqslant 2)$. Find $x_{2018}$. | Answer: 17.
Solution: From the condition, it follows that $x_{n+3}=x_{n+2}-x_{n+1}=x_{n+1}-x_{n}-x_{n+1}=-x_{n}$, therefore $x_{n+6}=$ $-x_{n+3}=x_{n}$, i.e., the sequence is periodic with a period of 6. Since $2018=6 \cdot 336+2$, we get $x_{2018}=x_{2}=17$. | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,250 |
2.1. Find all values of $x$ for which the greatest of the numbers $x^{2}$ and $\cos 2 x$ is less than $\frac{1}{2}$. In your answer, record the total length of the intervals found on the number line, rounding it to the nearest hundredths if necessary. | Answer: 0.37 (exact value: $\sqrt{2}-\frac{\pi}{3}$).
Solution. The inequality $\max \left\{x^{2}, \cos 2 x\right\}<\frac{1}{2}$ is equivalent to the system of inequalities
$$
\left\{\begin{array}{l}
x^{2}<\frac{1}{2}, \\
\cos 2 x<\frac{1}{2}
\end{array} \Leftrightarrow x \in\left(-\frac{\sqrt{2}}{2} ;-\frac{\pi}{6}\... | 0.37 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,251 |
2.3. Find all values of $x$ for which the smallest of the numbers $\frac{1}{x}$ and $\sin x$ is greater than $\frac{1}{2}$. In your answer, write the total length of the found intervals on the number line, rounding it to the nearest hundredths if necessary. | Answer: 1.48 (exact value: $2-\frac{\pi}{6}$ ). | 1.48 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,253 |
2.4. Find all values of $x$ for which the smallest of the numbers $8-x^{2}$ and $\operatorname{ctg} x$ is not less than -1. In the answer, record the total length of the found intervals on the number line, rounding it to hundredths if necessary. | Answer: 4.57 (exact value: $3+\frac{\pi}{2}$ ). | 4.57 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,254 |
3.1. Philatelist Andrey decided to distribute all his stamps equally into 2 envelopes, but it turned out that one stamp was left over. When he distributed them equally into 3 envelopes, one stamp was again left over; when he distributed them equally into 5 envelopes, 3 stamps were left over; finally, when he tried to d... | Answer: 223.
Solution. If the desired number is $x$, then from the first sentence it follows that $x$ is odd, and from the rest it follows that the number $x+2$ must be divisible by 3, 5, and 9, i.e., has the form $5 \cdot 9 \cdot p$. Therefore, $x=45(2 k-1)-2=90 k-47$. According to the condition $150<x \leqslant 300$... | 223 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,255 |
4.1. Through the vertex $A$ of the parallelogram $A B C D$, a line is drawn intersecting the diagonal $B D$, the side $C D$, and the line $B C$ at points $E, F$, and $G$ respectively. Find the ratio $B E: E D$, if $F G: F E=4$. Round your answer to the nearest hundredth if necessary. | Answer: 2.24 (exact value: $\sqrt{5}$).
Solution. Draw the line $E H$ parallel to $A B$ (point $H$ lies on side $B C$), and denote $B H=a, H C=b, B E=x, E D=y$ (see Fig. 1). By Thales' theorem, the desired ratio is $t=\frac{x}{y}=\frac{a}{b}$. From the similarity of triangles $B E G$ and $A E D$, we get $\frac{x}{a+5 ... | 2.24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,256 |
4.3. Through the vertex $A$ of the parallelogram $A B C D$, a line is drawn intersecting the diagonal $B D$, the side $C D$, and the line $B C$ at points $E, F$, and $G$ respectively. Find $E D$, if $F G$ : $F E=7, B E=8$. Round your answer to the nearest hundredth if necessary. | Answer: 2.83 (exact value: $2 \sqrt{2}$ ). | 2.83 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,257 |
4.4. Through the vertex $A$ of the parallelogram $A B C D$, a line is drawn intersecting the diagonal $B D$, the side $C D$, and the line $B C$ at points $E, F$, and $G$ respectively. Find $B E$, if $F G$ : $F E=9, E D=1$. Round your answer to the nearest hundredth if necessary. | Answer: 3.16 (exact value: $\sqrt{10}$ ). | 3.16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,258 |
5.1. Find the sum of the digits in the decimal representation of the integer part of the number $\sqrt{\underbrace{11 \ldots 11}_{2017} \underbrace{22 \ldots .22}_{2018} 5}$. | Answer: 6056 (the number from the condition of the problem is $\underbrace{33 \ldots 33}_{2017} 5$).
Solution. Since
$\underbrace{11 \ldots 11}_{2017} \underbrace{22 \ldots 22}_{2018} 5=\frac{10^{2017}-1}{9} \cdot 10^{2019}+\frac{10^{2018}-1}{9} \cdot 20+5=\frac{10^{4036}+10^{2019}+25}{9}=\left(\frac{10^{2018}+5}{3}\... | 6056 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,259 |
5.2. Find the sum of the digits in the decimal representation of the integer part of the number $\sqrt{\underbrace{11 \ldots 11}_{2018} \underbrace{55 \ldots 55}_{2017} 6}$. | Answer: 6055 (the number from the condition of the problem is $\underbrace{33 \ldots 33}_{2017} 4$ ). | 6055 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,260 |
5.3. Find the sum of the digits in the decimal representation of the integer part of the number $\sqrt{\underbrace{44.44}_{2017} \underbrace{22 \ldots 22}_{2018}} 5$. | Answer: 12107 (the number from the condition of the problem is $\underbrace{66 \ldots 66}_{2017} 5$ ). | 12107 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,261 |
5.4. Find the sum of the digits in the decimal representation of the integer part of the number $\sqrt{\underbrace{44.44}_{2018} \underbrace{88 \ldots 88}_{2017} 9}$. | Answer: 12109 (the number from the condition of the problem is equal to $\underbrace{66 \ldots 66}_{2017} 7$ ). | 12109 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,262 |
6.1. Find all integer solutions of the equation $x \ln 27 \log _{13} e=27 \log _{13} y$. In your answer, specify the sum $x+y$ for the solution $(x, y)$, where $y$ is the smallest, exceeding 70. | Answer: 117.
Solution: The original equation is equivalent to the equation $\frac{x \ln 27}{\ln 13}=\frac{27 \ln y}{\ln 13} \Leftrightarrow x \ln 27=27 \ln y \Leftrightarrow$ $\ln 27^{x}=\ln y^{27} \Leftrightarrow 27^{x}=y^{27} \Leftrightarrow 3^{x}=y^{9}$, with $y \geqslant 1$, and thus $x \geqslant 0$. Since 3 is a ... | 117 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,263 |
7.1. Find the minimum value of the expression $\frac{5 x^{2}+8 x y+5 y^{2}-14 x-10 y+30}{\left(4-x^{2}-10 x y-25 y^{2}\right)^{7 / 2}}$. Round the answer to hundredths if necessary. | Answer: 0.16 (exact value: $\frac{5}{32}$).
Solution. The expression from the problem condition is equal to
$$
\frac{(2 x+y)^{2}+(x+2 y)^{2}-14 x-10 y+30}{\left(4-(x+5 y)^{2}\right)^{7 / 2}}=\frac{(2 x+y-3)^{2}+(x+2 y-1)^{2}+20}{\left(4-(x+5 y)^{2}\right)^{7 / 2}}
$$
The numerator of this fraction is no less than 20... | \frac{5}{32} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,264 |
7.2. Find the minimum value of the expression $\frac{5 x^{2}-8 x y+5 y^{2}-10 x+14 y+55}{\left(9-25 x^{2}+10 x y-y^{2}\right)^{5 / 2}}$. Round the answer to the nearest hundredths if necessary. | Answer: 0.19 (exact value: $\frac{5}{27}$ ). | 0.19 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,265 |
7.3. Find the minimum value of the expression $\frac{13 x^{2}+24 x y+13 y^{2}-14 x-16 y+61}{\left(4-16 x^{2}-8 x y-y^{2}\right)^{7 / 2}}$. Round the answer to the nearest hundredths if necessary. | Answer: 0.44 (exact value: $\frac{7}{16}$ ). | 0.44 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,266 |
7.4. Find the minimum value of the expression $\frac{13 x^{2}+24 x y+13 y^{2}+16 x+14 y+68}{\left(9-x^{2}-8 x y-16 y^{2}\right)^{5 / 2}}$. Round the answer to hundredths if necessary. | Answer: 0.26 (exact value: $\frac{7}{27}$).
Note. In variants 7.1 and 7.3, the exact values 0.15625 and 0.4375, respectively, are also counted as correct. | \frac{7}{27} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,267 |
8.1. Points $K, L, M, N$ are the centers of the circles inscribed in the faces $S A B, S A C$, $S B C$ and $A B C$ of the tetrahedron $S A B C$. It is known that $A B=S C=5, A C=S B=7, B C=S A=8$. Find the volume of the tetrahedron $K L M N$. Round your answer to the nearest hundredth if necessary. | Answer: 0.66 (exact value: $\frac{\sqrt{11}}{5}$).
Solution. We will solve the problem in a general form. Let $BC = SA = a$, $AC = SB = b$, $AB = SC = c$. The faces of the tetrahedron $SABC$ are equal triangles (such a tetrahedron is called equifacial). Draw a plane through each edge of the tetrahedron, parallel to th... | 0.66 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,268 |
8.2. Points $A, B, C, D$ are the centers of the circles inscribed in the faces $P Q S, P R S$, $Q R S$ and $P Q R$ of the tetrahedron $P Q R S$. It is known that $P Q=R S=7, P R=Q S=8, P S=Q R=9$. Find the volume of the tetrahedron $A B C D$. Round your answer to the nearest hundredth if necessary. | Answer: 1.84 (exact value: $\frac{5 \sqrt{11}}{9}$ ). | 1.84 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,269 |
8.3. In the tetrahedron $K L M N$, it is known that $K L = M N = 4$, $K M = L N = 5$, $K N = M L = 6$. Points $P, Q, R, S$ are the centers of the inscribed circles of triangles $K L M, K L N$, $K M N$ and $L M N$. Find the volume of the tetrahedron $P Q R S$. Round the answer to the nearest hundredth if necessary. | Answer: 0.29 (exact value: $\frac{7 \sqrt{6}}{60}$ ). | 0.29 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,270 |
8.4. In the tetrahedron $E F G H$, it is known that $E F=G H=7, E G=F H=10, E H=F G=11$. Points $K, L, M, N$ are the centers of the circles inscribed in triangles $E F G, E F H$, $E G H$ and $F G H$. Find the volume of the tetrahedron $K L M N$. Round the answer to the nearest hundredth if necessary. | Answer: 2.09 (exact value: $\frac{\sqrt{215}}{7}$ ). | 2.09 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,271 |
8.5. In the tetrahedron $K L M N$, the lengths of the edges are known: $K L = M N = 9$, $K M = L N = 15$, $K N = L M = 16$. Points $P, Q, R, S$ are the centers of the circles inscribed in triangles $K L M$, $K L N$, $K M N$, and $L M N$. Find the volume of the tetrahedron $P Q R S$. Round the answer to the nearest hund... | Answer: 4.85 (exact value: $\frac{11 \sqrt{7}}{6}$ ). | 4.85 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,272 |
# Problem 1.
V-1 Which of the numbers is larger:
$$
A=\frac{3}{(1 \cdot 2)^{2}}+\frac{5}{(2 \cdot 3)^{2}}+\ldots+\frac{87}{(43 \cdot 44)^{2}}+\frac{89}{(44 \cdot 45)^{2}} \text { or } B=\frac{\sqrt[6]{4-2 \sqrt{3}} \cdot \sqrt[3]{\sqrt{3}+1}}{\sqrt[3]{2}} ?
$$ | Answer: $B$
Solution. Since $\frac{2 n+1}{(n(n+1))^{2}}=\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}$, then $A=\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)+\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)+\ldots+\left(\frac{1}{43^{2}}-\frac{1}{44^{2}}\right)+\left(\frac{1}{44^{2}}-\frac{1}{45^{2}}\right)=\frac{1}{1^{2}}-\frac{1}{4... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,273 |
# Problem 2.
In-1 A 2022-digit natural number is guessed, any two adjacent digits of which (arranged in the same order) form a two-digit number that is divisible either by 19 or by 23. The guessed number starts with the digit 4. What digit does it end with? | Answer: 6 or 8.
Solution. Two-digit numbers divisible by 19 are 19, 38, 57, 76, 95. Two-digit numbers divisible by 23 are $23, 46, 69, 92$. Since the first digit is 4, the second digit is 6, the third is 9, and the fourth is 2 or 5. If the fourth digit is 2, the continuation is: $2-3-8$ - there is no further continuat... | 6or8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,274 |
# Problem 3.
In-1 There is a function
$$
f(x)=\frac{1}{\sqrt[5]{1-x^{5}}}
$$
Calculate
$$
f(f(f(f(f(\ldots f(2022)))) \ldots))
$$
where the function $f$ is applied 1303 times. | Answer: $1 / \sqrt[5]{1-2022^{5}}$
Solution. For odd $n$
$$
\begin{gathered}
f(x)=\frac{1}{\sqrt[n]{1-x^{n}}} \\
f(f(x))=\frac{1}{\sqrt[n]{1-\frac{1}{1-x^{n}}}}=\sqrt[n]{1-\frac{1}{x^{n}}} \\
f(f(f(x)))=\sqrt[n]{1-\left(1-x^{n}\right)}=x \\
f(f(f(f(x))))=f(x)
\end{gathered}
$$
We see periodicity, period $=3$. The re... | \frac{1}{\sqrt[5]{1-2022^{5}}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,275 |
# Problem 4.
In-1 The angle at the vertex in the axial section of a cone is $60^{\circ}$. Outside this cone, there are 11 spheres of radius 3, each touching two adjacent spheres, the lateral surface of the cone, and the plane of its base. Find the radius of the base of the cone. | Answer: $\frac{3}{\sin \frac{\pi}{11}}-\sqrt{3}$.
Solution. The general condition: Around (or inside) a cone with an angle at the vertex in the axial section equal to $2 \alpha$, there are $n$ balls of radius $r$, each of which touches two adjacent balls, the lateral surface of the cone, and the plane of its base. Fin... | \frac{3}{\sin\frac{\pi}{11}}-\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,276 |
# Problem 5.
In-1 If the real numbers $a, b, c$ are ordered in non-decreasing order, resulting in the triplet $x_{1} \leqslant x_{2} \leqslant x_{3}$, then the number $x_{2}$ will be called the middle number of $a, b, c$. Find all values of $t$ for which the middle number of the three numbers
$$
a=t^{3}-100 t ; \quad... | Answer: $\left(-10 ;-4 \pi+\frac{5 \pi}{6}\right) \cup\left(-2 \pi+\frac{\pi}{6} ;-2 \pi+\frac{5 \pi}{6}\right) \cup\left(2 \pi+\frac{\pi}{6} ; 2 \pi+\frac{5 \pi}{6}\right) \cup(10 ;+\infty)$
Solution. Directly comparing the values of $a, b, c$ is difficult. However, for the average of three numbers to be positive, it... | (-10;-4\pi+\frac{5\pi}{6})\cup(-2\pi+\frac{\pi}{6};-2\pi+\frac{5\pi}{6})\cup(2\pi+\frac{\pi}{6};2\pi+\frac{5\pi}{6})\cup(10;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,277 |
# Problem 6.
B-1 For which values of the parameter $a \in \mathbb{R}$ does the maximum distance between the roots of the equation
$$
a \operatorname{tg}^{3} x+\left(2-a-a^{2}\right) \operatorname{tg}^{2} x+\left(a^{2}-2 a-2\right) \operatorname{tg} x+2 a=0
$$
belonging to the interval $\left(-\frac{\pi}{2} ; \frac{\... | Answer: $\frac{\pi}{4}$ when $a=0$.
Solution. The given equation can be rewritten as $(\operatorname{tg} x-1)(\operatorname{tg} x-a)(a \operatorname{tg} x+2)=0$, from which, for $x \in\left(-\frac{\pi}{2} ; \frac{\pi}{2}\right)$, either $\operatorname{tg} x=1$ and $x=\frac{\pi}{4}$, or $\operatorname{tg} x=a$ and $x=\... | \frac{\pi}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,278 |
# Problem 7.
B-1 The height $B D$ of an acute-angled triangle $A B C$ intersects with its other heights at point $H$. Point $K$ lies on segment $A C$ such that the measure of angle $B K H$ is maximized. Find $D K$, if $A D=2, D C=3$. | Answer: $\sqrt{6}$
Solution. The larger the acute angle, the larger its tangent. Therefore, the condition of the maximum angle $B K H$ can be replaced by the condition of the maximum of its tangent. By the tangent of the difference formula, we have
$$
\begin{aligned}
\operatorname{tg} \angle B K H & =\operatorname{tg... | \sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,279 |
1.1. Find the smallest 12-digit natural number that is divisible by 36 and contains each of the 10 digits at least once. | Answer: 100023457896.
Solution. The number must be divisible by 4 and by 9. Since the sum of ten different digits is 45, the sum of the two remaining digits must be 0, 9, or 18. We need the smallest number, so we add two digits 0 to the digits $0,1, \ldots, 9$ and place the digits 10002345 at the beginning of the desi... | 100023457896 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,281 |
2.1. The sine of the dihedral angle at the lateral edge of a regular quadrilateral pyramid is $\frac{15}{17}$. Find the area of the lateral surface of the pyramid if the area of its diagonal section is $3 \sqrt{34}$. | Answer: 68.
Solution. Let the linear angle of the dihedral angle given in the problem be denoted as $\alpha$. This angle is always obtuse, so $\cos \alpha=-\frac{8}{17}$.
The projection of the lateral face onto the diagonal section is a triangle, the area of which is half the area of this section. Since the dihedral ... | 68 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,282 |
3.1. For what greatest $a$ is the inequality $\frac{\sqrt[3]{\operatorname{tg} x}-\sqrt[3]{\operatorname{ctg} x}}{\sqrt[3]{\sin x}+\sqrt[3]{\cos x}}>\frac{a}{2}$ satisfied for all permissible $x \in\left(\frac{3 \pi}{2} ; 2 \pi\right)$? Round the answer to the nearest hundredth if necessary. | Answer: 4.49 (exact value: $4 \sqrt[6]{2}$).
Solution. Transform the expression on the left side of the inequality as follows:
$$
\frac{\sqrt[3]{\operatorname{tg} x}-\sqrt[3]{\operatorname{ctg} x}}{\sqrt[3]{\sin x}+\sqrt[3]{\cos x}}=\frac{\sqrt[3]{\sin ^{2} x}-\sqrt[3]{\cos ^{2} x}}{\sqrt[3]{\sin x \cos x}(\sqrt[3]{\... | 4.49 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,283 |
3.3. For what least $a$ is the inequality $\frac{\sqrt[3]{\operatorname{ctg}^{2} x}-\sqrt[3]{\operatorname{tg}^{2} x}}{\sqrt[3]{\sin ^{2} x}-\sqrt[3]{\cos ^{2} x}}<a$ satisfied for all permissible $x \in\left(-\frac{3 \pi}{2} ;-\pi\right)$? Round the answer to the nearest hundredth if necessary. | Answer: $-2.52$ (exact value: $-2 \sqrt[3]{2}$ ). | -2.52 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,285 |
3.4. For what least positive $a$ is the inequality $\frac{\sqrt[3]{\sin ^{2} x}-\sqrt[3]{\cos ^{2} x}}{\sqrt[3]{\operatorname{tg}^{2} x}-\sqrt[3]{\operatorname{ctg}^{2} x}}<\frac{a}{2}$ satisfied for all permissible $x \in\left(\frac{3 \pi}{2} ; 2 \pi\right)$? Round the answer to the nearest hundredths if necessary. | Answer: 0.79 (exact value: $\frac{1}{\sqrt[3]{2}}$ ). | 0.79 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,286 |
4.1. A batch of tablets from four different brands was delivered to a computer store. Among them, Lenovo, Samsung, and Huawei tablets made up less than a third, with Samsung tablets being 6 more than Lenovo. All other tablets were Apple iPads, and there were three times as many of them as Huawei. If the number of Lenov... | Answer: 94.
Solution. Let $n$ be the total number of tablets we are looking for, with $x$ being the number of Lenovo brand tablets, and $y$ being the number of Huawei brand tablets. Then the number of Samsung tablets is $x+6$, and the number of Apple iPad tablets is $n-2x-y-6=3y$. From the problem statement, we also h... | 94 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,287 |
5.1. Let $S(n)$ be the sum of the digits in the decimal representation of the number $n$. Find $S\left(S\left(S\left(S\left(2017^{2017}\right)\right)\right)\right)$. Answer. 1. | Solution. Since $2017^{2017}<10000^{2017}$, the number of digits in the representation of $2017^{2017}$ does not exceed $4 \cdot 2017=$ 8068, and their sum $S\left(2017^{2017}\right)$ does not exceed $9 \cdot 8068=72612$. Then we sequentially obtain $S\left(S\left(2017^{2017}\right)\right) \leqslant 6+9 \cdot 4=42, S\l... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,288 |
6.1. Let $f(x)=x^{2}+p x+q$. It is known that the inequality $|f(x)|>\frac{1}{2}$ has no solutions on the interval $[1 ; 3]$. Find $\underbrace{f(f(\ldots f}_{2017}\left(\frac{3+\sqrt{7}}{2}\right)) \ldots)$. Round your answer to the nearest hundredth if necessary. | Answer: 0.18 (exact value: $\frac{3-\sqrt{7}}{2}$).
Solution. The condition of the problem means that for all $x \in [1; 3]$, the inequality $|f(x)| \leqslant \frac{1}{2}$ holds. By making the substitution $t = x + 2$, we obtain the inequality $\left|t^{2} + (p+4)t + (2p+q+4)\right| \leqslant \frac{1}{2}$, which holds... | 0.18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,289 |
6.2. Let $f(x)=x^{2}+p x+q$. It is known that the inequality $|f(x)|>\frac{1}{2}$ has no solutions on the interval $[2 ; 4]$. Find $\underbrace{f(f(\ldots f}_{2017}\left(\frac{5-\sqrt{11}}{2}\right)) \ldots)$. Round your answer to the nearest hundredth if necessary. | Answer: 4.16 (exact value: $\frac{5+\sqrt{11}}{2}$ ). | 4.16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,290 |
6.4. Let $f(x)=x^{2}+p x+q$. It is known that the inequality $|f(x)|>\frac{1}{2}$ has no solutions on the interval $[4 ; 6]$. Find $\underbrace{f(f(\ldots f}_{2017}\left(\frac{9-\sqrt{19}}{2}\right)) \ldots)$. Round your answer to the nearest hundredth if necessary. | Answer: 6.68 (exact value: $\frac{9+\sqrt{19}}{2}$ ). | 6.68 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,292 |
7.1. In triangle $A B C$, the median $A M$ is drawn, point $O$ is the center of the circumscribed circle around it, and point $Q$ is the center of the inscribed circle in it. Segments $A M$ and $O Q$ intersect at point $S$, and $2 \frac{O S}{M S}=3 \sqrt{3} \frac{Q S}{A S}$. Find the sum of the sines of the measures of... | Answer: 1.13 (exact value: 9/8).
Solution. Point $Q$ is the intersection point of the angle bisectors of triangle $ABC$. Draw the angle bisector of $\angle BAC$, and denote the intersection point of this bisector with the circumcircle of triangle $ABC$ by $L$. Connect points $C$ and $Q$.
Since $\angle BAL = \angle CA... | \frac{9}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,293 |
7.2. In triangle $K L M$, the median $K P$ is drawn, point $O$ is the center of the circumscribed circle around it, and point $Q$ is the center of the inscribed circle in it. Segments $K P$ and $O Q$ intersect at point $S$, and $\frac{O S}{P S}=\sqrt{6} \frac{Q S}{K S}$. Find the product of the cosines of the angles $K... | Answer: $-0.38$ (exact value: $-3 / 8$ ). | -0.38 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,294 |
7.3. In triangle $A B C$, the median $A M$ is drawn, point $O$ is the center of the circumscribed circle around it, and point $Q$ is the center of the inscribed circle in it. Segments $A M$ and $O Q$ intersect at point $R$, and $\frac{O R}{M R}=\sqrt{7} \frac{Q R}{A R}$. Find the cosine of the difference in the measure... | Answer: $-0.13$ (exact value: $-1 / 8$ ). | -0.13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,295 |
7.4. In triangle $K L M$, the median $K P$ is drawn, point $O$ is the center of the circumscribed circle around it, and point $Q$ is the center of the inscribed circle in it. Segments $K P$ and $O Q$ intersect at point $R$, and $\frac{O R}{P R}=\sqrt{14} \frac{Q R}{K R}$. Find the product of the sines of the angles $K ... | Answer: 0.63 (exact value: $5 / 8$).
Note. The exact values $0.125, -0.375, -0.125$ and 0.625 are also accepted as correct. | 0.63 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,296 |
8.1. Solve the equation $x^{x+y}=y^{y-x}$ in natural numbers. In your answer, specify the sum $x+y$ for the solution $(x, y)$, where $y$ is the smallest, exceeding 1500. | Answer: 2744.
Solution. Let $y = t x$, where $t \in \mathbb{Q}$ for $x, y \in \mathbb{N}$. Then, substituting this into the equation, we find
$$
x = t^{\frac{t-1}{2}}, \quad y = t^{\frac{t+1}{2}}
$$
We will show that the number $t$ can only be an integer. Suppose the contrary: let $t$ be a rational number different ... | 2744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,297 |
# Problem 1.
In-1 Find $f(1)+f(2)+f(3)+\ldots+f(13)$, if $f(n)=4 n^{3}-6 n^{2}+4 n+13$. | Answer: 28743.
Solution. Since $f(n)=n^{4}-(n-1)^{4}+14$, then
$f(1)+f(2)+f(3)+\ldots+f(13)=1^{4}+2^{4}+\ldots+13^{4}-\left(0^{4}+1^{4}+\ldots+12^{4}\right)+14 \cdot 13=13^{4}+14 \cdot 13=28743$.
V-2 Find $f(1)+f(2)+f(3)+\ldots+f(11)$, if $f(n)=4 n^{3}+6 n^{2}+4 n+9$.
Answer: 20823.
V-3 Find $f(1)+f(2)+f(3)+\ldots... | 28743 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,298 |
# Problem 2.
In-1 Solve the system of equations:
$$
\left\{\begin{array}{l}
\sqrt{x^{2}+y}+|y+8|=1 \\
\sqrt{x^{2}+y-1}+|x+8|=5
\end{array}\right.
$$ | Answer: $x=-3, y=-8$.
Solution. From the second equation of the system, it follows that $x^{2}+y \geqslant 1$, which together with the first equation gives $y=-8, x= \pm 3$. Considering the second equation, we get $x=-3$.
V-2 Solve the system of equations:
$$
\left\{\begin{array}{l}
\sqrt{x^{2}-y}+|y-5|=2 \\
\sqrt{x... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,299 |
# Problem 3.
In-1 The numbers $a, b, c$ are such that each of the two equations $x^{2}+b x+a=0$ and $x^{2}+c x+a=1$ has two integer roots, and all these roots are less than (-1). Find the smallest value of $a$. | Answer: 15.
Solution. By Vieta's theorem, the product of the roots of the first equation is $a$, and the product of the roots of the second equation is $a-1$. Since the roots are integers and less than -1, their product is greater than 1, so each of the two consecutive numbers $a-1$ and $a$ is the product of two diffe... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,300 |
# Problem 4.
B-1 Two circular tracks $\alpha$ and $\beta$ of the same radius touch each other. Car $A$ is driving clockwise on track $\alpha$, and car $B$ is driving counterclockwise on track $\beta$. At the start, cars $A$ and $B$ are on the same straight line with the center of track $\alpha$, and this line is tange... | Answer: $\frac{1}{2}$ hour $=30$ minutes.
Solution. Let the radius of the circular track be 1. Introduce a rectangular coordinate system with the origin at the starting point of car $B$, the $O x$ axis passing through the center of the track $\beta$, and the $O y$ axis passing through the center of the track $\alpha$.... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,301 |
# Problem 5.
B-1 From an isosceles triangle with an angle $\alpha$ at the vertex and an area of 1, a circle of maximum area is cut out, and from it, a triangle similar to the original and of maximum area is cut out. What are the greatest and least values that the area $S(\alpha)$ of the resulting triangle can take for... | Answer: $\frac{1}{4} ; 7-4 \sqrt{3}$.
Solution. Let $\beta=\frac{\alpha}{2}$ and $r$ be the radius of the circle. Then the halves of the bases of the original and cut-out triangles are respectively
$$
a=\frac{r(1+\sin \beta)}{\cos \beta}, \quad b= \begin{cases}r \sin 2 \beta, & \beta \leqslant 45^{\circ} \\ r, & \bet... | \frac{1}{4};7-4\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,302 |
# Problem 6.
In-1 In an irregular pyramid $A B C D$, the sum of the planar angles at vertex $A$ is $180^{\circ}$. Find the surface area of this pyramid if the area of face $B C D$ is $s$ and $A B=$ $C D, A D=B C$. | Answer: $4 s$.
Solution. We will prove that the faces of the pyramid are equal triangles. For this, consider the net $A D^{\prime \prime \prime} B D^{\prime} C D^{\prime \prime}$ of the pyramid $A B C D$, where $A D^{\prime \prime}=A D^{\prime \prime \prime}=A D, B D^{\prime \prime \prime}=B D^{\prime}=B D, C D^{\prim... | 4s | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,303 |
Problem 2. Baba Yaga must arrive at Bald Mountain exactly at midnight. She calculated that if she flies on a broom at a speed of 50 km/h, she will be 2 hours late, and if she flies on an electric broom at a speed of 150 km/h, she will arrive 2 hours early. To arrive at Bald Mountain exactly on time, Baba Yaga used a wa... | Answer: At 20:00 at a speed of 75 km/h.
Solution. The electric broom flies three times faster than the cauldron, and at the same time, it saves 4 hours of time. This means that two-thirds of the flight time in the cauldron is 4 hours, the entire flight in the cauldron would have taken 6 hours, and the entire flight on... | At20:00atspeedof75/ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,305 |
1. At Andrei's birthday, Yana was the last to arrive, giving him a ball, and Eduard was the second to last, giving him a calculator. Testing the calculator, Andrei noticed that the product of the total number of his gifts and the number of gifts he had before Eduard arrived is exactly 16 more than the product of his ag... | Answer: 18. Solution. If $n$ is the number of gifts, and $a$ is Andrei's age, then $n(n-2)=a(n-1)$. From this, $a=\frac{n^{2}-2 n-16}{n-1}=n-1-\frac{17}{n-1}$. Therefore, $n-1=17, a=16$. | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,306 |
2. In an equilateral triangle $ABC$, points $A_{1}$ and $A_{2}$ are chosen on side $BC$ such that $B A_{1}=A_{1} A_{2}=A_{2} C$. On side $AC$, a point $B_{1}$ is chosen such that $A B_{1}: B_{1}C=1: 2$. Find the sum of the angles $\angle A A_{1} B_{1}+\angle A A_{2} B_{1}$. | Answer: $30^{\circ}$. Solution. Since $A_{1} B_{1} \| A B$, then $\angle B A A_{1}=\angle A A_{1} B_{1}$. From symmetry, $\angle B A A_{1}=\angle C A A_{2}$. It remains to note that $\angle C B_{1} A_{2}=$ $\angle B_{1} A A_{2}+\angle A A_{2} B_{1}$ as an exterior angle in $\triangle A A_{2} B_{1}$. | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,307 |
3. All natural numbers are divided into "good" and "bad" according to the following rules:
a) From any bad number, you can subtract some natural number not exceeding its half so that the resulting difference becomes "good".
b) From a "good" number, you cannot subtract no more than half of it so that it remains "good"... | Answer: 2047. Solution. Note that good numbers have the form $2^{n}-1$. Indeed, let a number have the form $M=2^{n}+k$, where $k=0,1, \ldots, 2^{n}-2$. Then from such a number, you can subtract $k+1 \leqslant \frac{1}{2}\left(2^{n}+k\right)=\frac{M}{2}$. On the other hand, from a number of the form $N=2^{n}-1$, you nee... | 2047 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,308 |
4. In triangle $\triangle A B C$, the base $A B$ of which lies on the x-axis, altitudes $A M, B N$ and $C K$ are drawn. Find the length of the base $A B$, if the coordinates of points $M(2,2)$ and $N(4,4)$ are known. | Answer: $A B=4 \sqrt{5}$. Solution. The circle that has $A B$ as its diameter contains points $M$ and $N$. Its center $D$ is equidistant from $M$ and $N$.
Since the line $M N$ is given by $y=x$, the perpendicular line to it, passing through the midpoint of $M N$ - the point $(3,3)$ - has the form $y=6-x$. Therefore, t... | 4\sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,309 |
1. After getting rid of the Colorado potato beetle, the farmer started to harvest as much potato from 24 hectares as he used to from 27 hectares. By what percentage did the potato yield increase? | Answer: $12.5 \%$.
Solution. From 24 hectares, the farmer began to collect $27 / 24=1.125$ times more potatoes than before. This means that the potato yield increased by $12.5 \%$. | 12.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,310 |
2. Can two angle bisectors of the interior angles of a triangle intersect at a right angle? | Answer: Hem.
Solution. Suppose the opposite: let in some triangle $ABC$ the bisectors of angles $\angle A$ and $\angle C$ intersect at a right angle at point $K$ (see Fig. 1). Since the sum of the angles in triangle $AKC$ is $180^{\circ}$, and $\angle AKC=90^{\circ}$, the sum of the remaining two angles is $\angle KAC... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,311 |
3. Solve the equation
$$
\frac{x^{7}-1}{x^{5}-1}=\frac{x^{5}-1}{x^{3}-1}
$$ | Answer: $0, -1$.
Solution. For $x \neq 1$, both denominators are non-zero, and by the rule of proportion, the given equation takes the form
$$
\left(x^{7}-1\right)\left(x^{3}-1\right)=\left(x^{5}-1\right)^{2} \Leftrightarrow x^{7}-2 x^{5}+x^{3}=0 \Leftrightarrow x^{3}\left(x^{2}-1\right)^{2}=0 \Leftrightarrow x^{3}(x... | 0,-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,312 |
5. Two trains departed simultaneously from points $A$ and $B$ towards each other. It is known that they met at 14:00 and, without changing their speeds, continued their journey. One train arrived at point $B$ at 18:00, and the other train arrived at point $A$ at 23:00. At what time did the trains depart on their journe... | Answer: 8:00.
Solution. Let the trains departed $t$ hours before the moment of meeting, and let $v_{1}$ be the speed of the first train, $v_{2}$ be the speed of the second. Then the first train traveled a distance of $t v_{1}$ from point $A$ to the meeting point with the second train, and the second train traveled thi... | 8:00 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,314 |
6. Calculate
$$
\frac{2 a b\left(a^{3}-b^{3}\right)}{a^{2}+a b+b^{2}}-\frac{(a-b)\left(a^{4}-b^{4}\right)}{a^{2}-b^{2}} \quad \text { for } \quad a=-1, \underbrace{5 \ldots 5}_{2010} 6, \quad b=5, \underbrace{4 \ldots 44}_{2011}
$$
Answer: 343. | Solution. Given $a$ and $b$, we get
$$
\begin{gathered}
\frac{2 a b\left(a^{3}-b^{3}\right)}{a^{2}+a b+b^{2}}-\frac{(a-b)\left(a^{4}-b^{4}\right)}{a^{2}-b^{2}}=2 a b(a-b)-(a-b)\left(a^{2}+b^{2}\right)= \\
\quad=-(a-b)\left(a^{2}-2 a b+b^{2}\right)=-(a-b)^{3}=-(-7)^{3}=343
\end{gathered}
$$ | 343 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,315 |
7. Petya sequentially writes down integers, starting from 21, such that each subsequent number is 4 less than the previous one, and Vasya, looking at the current number, calculates the sum of all the numbers written down by this point. Which of the sums found by Vasya will be closest to $55?$
Answer: 56. | Solution. Let's create a table, in the first row of which we will record the numbers following Petya, and in the second row we will place the results of Vasya's calculations.
| 21 | 17 | 13 | 9 | 5 | 1 | -3 | -7 | -11 | $\cdots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :--- |
| 21 | ... | 56 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,316 |
9. Find the smallest natural number that is greater than the sum of its digits by 1755 (the year of the founding of Moscow University). | Answer: 1770.
Solution. From the condition, it follows that the desired number cannot consist of three or fewer digits. We will look for the smallest such number in the form $\overline{a b c d}$, where $a, b, c, d$ are digits, and $a \neq 0$. We will form the equation:
$$
\begin{gathered}
\overline{a b c d}=a+b+c+d+1... | 1770 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,318 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.