problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
9.5. Let $M$ - be a finite set of numbers (distinct). It is known that among any three of its elements, there will be two whose sum belongs to $M$. What is the maximum number of elements that can be in $M$? | Answer: 7.
Solution: An example of a set with 7 elements: $\{-3,-2,-1,0,1,2,3\}$.
We will prove that a set $M=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$ with $n>7$ numbers does not have the required property.
We can assume that $a_{1}>a_{2}>a_{3}>\ldots>a_{n}$ and $a_{4}>0$ (changing the signs of all elements does ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.1. In the wagon, several kilograms of apple jam were loaded, of which $20 \%$ was good and $80 \%$ was bad. Every day, half of the existing bad jam rotted, and it was thrown away. After several days, it turned out that $20 \%$ of the jam in the wagon was bad and $80 \%$ was good. How many days have passed since the l... | Answer: 4 days.
Solution 1: Let the initial total amount be $x$, and the final amount be $y$ kilograms of jam. Then, since the amount of good jam did not change, $0.2 x = 0.8 y$, which means $x = 4 y$. Therefore, initially, the amount of bad jam was $0.8 x = 3.2 y$, and it became $0.2 y$, meaning the mass of bad jam d... | 4 | Other | math-word-problem | Yes | Yes | olympiads | false |
8.2. Once Alexei and Daniil were playing such a game. If a number \( x \) is written on the board, it can be erased and replaced with \( 2x \) or \( x - 1000 \). The player who gets a number not greater than 1000 or not less than 4000 loses. Both players aim to win. At some point, the boys stopped playing. Who lost if ... | Answer: no one lost.
Solution: note that if a number is less than 2000 but greater than 1000, then by multiplying by 2, you can get a number that is less than 4000. If a number is less than 4000 but greater than 2000, then by subtracting 1000 (possibly twice), you can get a number between 1000 and 2000. Thus, the only... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.2. Around a round table, 15 boys and 20 girls sat down. It turned out that the number of pairs of boys sitting next to each other is one and a half times less than the number of pairs of girls sitting next to each other. Find the number of boy - girl pairs sitting next to each other. | Answer: 10.
Solution. Let's call a group several children of the same gender sitting in a row, with children of the opposite gender sitting to the left and right of the outermost ones. Let $X$ be the number of groups of boys, which is equal to the number of groups of girls sitting in a row. It is easy to see that the ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.4. What is the maximum number of triangles with vertices at the vertices of a regular 18-gon that can be marked so that no two different sides of these triangles are parallel? The triangles can intersect and have common vertices, coinciding segments are considered parallel. | Answer: 5.
Solution. Estimating the number of triangles. Let's number the vertices of the 18-gon from 1 to 18 clockwise. The sides of the triangles are the sides and diagonals of the regular 18-gon. We will call a diagonal even if an even number of sides lies between its ends, and odd otherwise. The parity of a diagon... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.3. Find the maximum length of a horizontal segment with endpoints on the graph of the function $y=x^{3}-x$ | Answer: 2.
Solution 1. A horizontal segment of length $a>0$ with endpoints on the graph of the function $y=x^{3}-x$ exists if and only if the equation $(x+a)^{3}-(x+a)=x^{3}-x$ has at least one solution for the given value of the parameter $a$. Expanding the brackets, combining like terms, and dividing by $a>0$, we ob... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
8.4. From identical isosceles triangles, where the angle opposite the base is $45^{\circ}$ and the lateral side is 1, a figure was formed as shown in the diagram. Find the distance between points $A$ and $B$. | Answer: 2.
Solution: Let's denote the points $K, L, M$, as shown in the figure. We will construct an isosceles triangle $A K C$ equal to the original one. Connect vertex $C$ to other points as shown in the figure.
In the original triangles, the angle at the vertex is $45^{\circ}$. Therefore, the other two angles are ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.3. On a plane, there are points $A, B, C, D, X$. Some segment lengths are known: $A C=2$, $A X=5$, $A D=11$, $C D=9$, $C B=10$, $D B=1$, $X B=7$. Find the length of the segment $C X$. | Answer: 3.
Solution: Note that $A D=11=2+9=A C+C D$. Therefore, points $A, C, D$ lie on the same line $C D$ (since the triangle inequality becomes an equality). Note that $C B=10=9+1=C D+D B$. Therefore, points $C, D, B$ lie on the same line $C D$ (since the triangle inequality becomes an equality). Therefore, all fou... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.5. In each cell of a 5 by 5 table, a letter is written such that in any row and in any column there are no more than three different letters. What is the maximum number of different letters that can be in such a table | Answer: 11.
Solution: If each row contains no more than two different letters, then the total number of letters does not exceed $10=5 * 2$. Further, we can assume that the first row contains exactly three different letters. If each of the remaining rows has at least one letter in common with the first, then the total ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.1. Let $a^{2}+b^{2}=c^{2}+d^{2}=1$ and $a c+b d=0$ for some real numbers $a, b, c, d$. Find all possible values of the expression $a b+c d$. | Answer: 0.
Solution: Let's first assume $b \neq 0$. From the second equation, express $d=\frac{-a c}{b}$ and substitute it into the equation $c^{2}+d^{2}=1$. Eliminating the denominator, we get $c^{2}\left(a^{2}+b^{2}\right)=b^{2}$, from which, given $a^{2}+b^{2}=1$, we obtain $b= \pm c$. Substituting this into the eq... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.1. Large sandglasses measure an hour, and small ones measure 11 minutes. How can you use these sandglasses to measure a minute? | Solution: We will run the large hourglass twice in a row and the small one eleven times in a row. A minute will be measured between the second time the large hourglass finishes (120 minutes) and the 11th time the small one finishes (121 minutes).
Criteria: Any correct example - 7 points. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.1. Vikentiy walked from the city to the village, and at the same time Afanasiy walked from the village to the city. Find the distance between the village and the city, given that the distance between the pedestrians was 2 km twice: first, when Vikentiy had walked half the way to the village, and then, when Afanasiy ... | Answer: 6 km.
Solution. Let the distance between the village and the city be denoted as $S$ km, the speeds of Vikentiy and Afanasy as $x$ and $y$, and calculate the time spent by the travelers in the first and second cases. In the first case, we get: $\frac{S / 2}{x}=\frac{S / 2-2}{y}$, in the second case $\frac{2 S /... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. From two cities, the distance between which is 105 km, two pedestrians set out simultaneously towards each other at constant speeds and met after 7.5 hours. Determine the speed of each of them, knowing that if the first walked 1.5 times faster, and the second 2 times slower, they would have met after $8 \frac{1}{1... | Answer: 6 and 8 km per hour.
Solution: Let their speeds be $x$ and $y$ km per hour, respectively. From the condition, we get: $\frac{15}{2}(x+y)=105, \frac{105}{13}\left(\frac{3}{2} x+\frac{1}{2} y\right)=105$, from which $x=6, y=8$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.3. Two circles intersect at points A and B. A tangent to the first circle is drawn through point A, intersecting the second circle at point C. A tangent to the second circle is drawn through point B, intersecting the first circle at point D. Find the angle between the lines $\mathrm{AD}$ and $\mathrm{BC}$. | Answer. The lines are parallel, the angle is $0^{\circ}$.
Solution. Mark point $\mathrm{P}$ on the extension of $\mathrm{CA}$ beyond point A and point M on the extension of DB beyond point B. The inscribed angle ABD in the first circle, which subtends the chord AD, is equal to the angle PAD between the chord AD and th... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.1. There are two ingots of different copper and tin alloys weighing 6 and 12 kg respectively. From each of them, a piece of the same weight was cut off and the first piece was alloyed with the remainder of the second ingot, and the second piece - with the remainder of the first ingot, after which the ratio of copper ... | Answer: 4 kilograms.
Solution: Let the weight of each of the cut pieces be $x$ kg, and the proportions of tin in the first and second ingots be $a \neq b$ respectively. Then, the proportion of tin after remelting in the first ingot will be $\frac{b x + a(6 - x)}{6}$, and in the second ingot $\frac{a x + b(12 - x)}{12}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.4. In the tournament, each of the six teams played against each other exactly once. In the end, the teams scored 12, 10, 9, 8, 7, and 6 points respectively. a) How many points were awarded for a win in a match, if 1 point was awarded for a draw, and 0 points for a loss? The answer, of course, should be a natural numb... | Answer. a) 4 points.
b) The first team had three wins, the second team had two wins and two draws, the third team had two wins and one draw, the fourth team had two wins, the fifth team had one win and three draws, the sixth team had one win and two draws. The rest of the matches were lost by the teams.
c) One exampl... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.2. Three people are playing table tennis, with the player who loses a game giving way to the player who did not participate in it. In the end, it turned out that the first player played 21 games, and the second - 10. How many games did the third player play? | Answer: 11.
Solution: According to the problem, the first player played 21 games, so there were at least 21 games in total. Out of any two consecutive games, the second player must participate in at least one, which means there were no more than \(2 \cdot 10 + 1 = 21\) games. Therefore, a total of 21 games were played... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.5. On the board, 10 natural numbers are written, some of which may be equal, and the square of each of them divides the sum of all the others. What is the maximum number of different numbers that can be among the written ones? | Answer. Four
Solution. An example for four different numbers: 1,1,1,2,2,3,5,5,5,5.
Let among the listed numbers there are exactly $n \geq 2$ different ones, the maximum of which we denote by $x$, and the sum of all numbers by $\mathrm{S}$. Then the sum of all numbers except the maximum does not exceed $(9-n) x + x - ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. The bathtub fills up from the hot water tap in 17 minutes, and from the cold water tap in 11 minutes. After how many minutes from opening the hot water tap should the cold water tap be opened so that by the time the bathtub is full, there is one third more hot water than cold water? | Answer. In 5 minutes.
Solution. Let the volume of the bathtub be $V$, then by the time it is filled, there should be $\frac{3}{7} V$ of cold water and $\frac{4}{7} V$ of hot water in it. The filling rates of cold and hot water are $\frac{V}{11}$ and $\frac{V}{17}$, respectively. Therefore, the desired time is the diff... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.3. Find the smallest natural number $n$ such that in any set of $n$ distinct natural numbers, not exceeding 1000, it is always possible to select two numbers, the larger of which does not divide the smaller one. | Answer. $n=11$.
Solution. Among the first 10 powers of two $1=2^{0}, 2=2^{1}, 4=2^{2}, \ldots, 512=2^{9}$, in each pair of numbers, the larger number is divisible by the smaller one, hence $n \geq 11$.
On the other hand, let there be some set of $n \geq 11$ numbers where the larger number of each pair is divisible by... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.3. Anya wrote down 100 numbers in her notebook. Then Sonya wrote down in her notebook all the pairwise products of the numbers written by Anya. Artem noticed that there were exactly 2000 negative numbers in Sonya's notebook. How many zeros did Anya initially write down in her notebook? | Answer: 10 zeros.
Solution: Let Anya write down $n$ positive numbers, $m$ negative numbers, and $100-n-m$ zeros in her notebook. Then, by the condition, $n m=2000$, since a negative number can only be obtained by multiplying numbers of different signs.
Let's list all the divisors of the number $2000=2^{4} * 5^{3}$:
... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.4. Arseny sat down at the computer between 4 and 5 PM, when the hour and minute hands were pointing in opposite directions, and got up from it on the same day between 10 and 11 PM, when the hands coincided. How long did Arseny sit at the computer? | Solution: Let's see where the hands will be 6 hours after Arseny sat down at the computer. The minute hand will go around the clock 6 times and return to its place. The hour hand will move exactly half a circle. Therefore, the angle between the hands will change by 180 degrees, i.e., the hands will coincide. It is obvi... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.5. Several seventh-graders were solving problems. The teacher does not remember how many children there were and how many problems each of them solved. However, he remembers that, on the one hand, each solved more than a fifth of what the others solved, and on the other hand, each solved less than a third of what the... | Answer: 5 seventh-graders.
Solution: Let one seventh-grader solve $a$ problems, and the rest solve $S - a$. Then
$$
\begin{gathered}
a < (S-a) / 3 \\
3a < S - a \\
4a < S \\
a < S / 4
\end{gathered}
$$
Similarly,
$$
\begin{gathered}
(S-a) / 5 < a \\
S - a < 5a \\
S < 6a \\
S / 6 < a
\end{gathered}
$$
Thus, if ther... | 5 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. Point $M$ lies on the leg $A C$ of the right triangle $A B C$ with a right angle at $C$, such that $A M=2, M C=16$. Segment $M H$ is the altitude of triangle $A M B$. Point $D$ is located on the line $M H$ such that the angle $A D B$ is $90^{\circ}$, and points $C$ and $D$ lie on the same side of the line $A B$. Fin... | Solution. 1. A circle can be circumscribed around quadrilateral $A B C D$ with diameter $A B$ (angles $A D B$ and $A C B$ are right angles). Then $\angle A B D=\angle A C D$,
$$
\angle H A D=90^{\circ}-\angle A B D, \angle A D H=\angle A B D=\angle A C D
$$
Triangles $A C D$ and $A D M$ are similar, and $\frac{A D}{A... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.
Find the roots of the equation $f(x)=8$, if $4 f(3-x)-f(x)=3 x^{2}-4 x-3$ for any real value of $x$. In your answer, specify the product of the found roots. # | # Solution:
Notice that when $x$ is replaced by $3-x$, the expression $3-x$ changes to $x$. That is, the pair $f(x)$ and $f(3-x)$ is invariant under this substitution. Replace $x$ with $3-x$ in the equation given in the problem. We get:
$4 f(x) - f(3-x) = 3(3-x)^2 - 4(3-x) - 3 = 3x^2 - 14x + 12$. Express $f(x)$ from ... | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Find all values of the parameter $a$, for each of which the solution set of the inequality $\frac{x^{2}+(a+1) x+a}{x^{2}+5 x+4} \geq 0$ is the union of three non-overlapping intervals. In your answer, specify the sum of the three smallest integer values of $a$ from the obtained interval. | # Solution
Let's factorize the numerator and the denominator of the left part of the inequality, it will take the form: $\frac{(x+1)(x+a)}{(x+1)(x+4)} \geq 0$. There are five possible cases for the placement of the number ($-a$) relative to the numbers (-4) and (-1). In each case, the inequality is solved using the in... | 9 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
9.
Calculate the number $8^{2021}$, find the sum of the digits in this number, and write down the result. Then, in the newly written number, find the sum of the digits and write down the result again. These actions were repeated until a single-digit number was obtained. Find this number. | # Solution.
Consider the natural powers of 8. Notice that even powers of the number 8 give a remainder of 1 when divided by 9, while odd powers (including the number \(8^{2021}\)) give a remainder of 8. Indeed, let's analyze the powers of 8:
\[
\begin{gathered}
8^{2}=(9-1)^{2}=9 n+1, n \in N \\
8^{3}=(9 n+1) \cdot 8=... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $8 \sin ^{4}(\pi x)-\sin ^{2} x=\cos ^{2} x-\cos (4 \pi x)$. In your answer, specify the sum of the roots that belong to the interval $[-1 ; 2]$.
(5 points) | Solution. Considering the basic trigonometric identity, we get
$8 \sin ^{4}(\pi x)-1+\cos (4 \pi x)=0 \quad \Rightarrow \quad 8 \sin ^{4}(\pi x)-2 \sin ^{2}(2 \pi x)=0 \quad \Rightarrow$
$\left(2 \sin ^{2}(\pi x)-2 \sin (\pi x) \cos (\pi x)\right)\left(2 \sin ^{2}(\pi x)+2 \sin (\pi x) \cos (\pi x)\right)=0 \Rightarr... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. What is the smallest area that a right triangle can have, if its hypotenuse lies on the tangent to the graph of the function $y=\sqrt{x-3}$, one of its legs lies on the $y$-axis, and one of its vertices coincides with the point of tangency | Solution. $\quad f(x)=\sqrt{x-3}, \quad f^{\prime}\left(x_{0}\right)=\frac{1}{2 \sqrt{x-3}}$
$$
\begin{aligned}
& S_{A B C}=\frac{1}{2} A B \cdot B C, x_{0}-\text { abscissa of the point of tangency } A, \\
& A\left(x_{0}, f\left(x_{0}\right)\right), \quad B\left(0, f\left(x_{0}\right)\right), \quad C \quad \text { - ... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
7. In triangle $A B C$, altitudes $A D, B E, C F$ are drawn. The length of side $A C$ is $1+\sqrt{3}$. The distances from the center of the inscribed circle in triangle $D E F$ to points $A$ and $C$ are $\sqrt{2}$ and 2, respectively. Find the length of side $A B$. | Solution. $\quad A D, B E, C F$ are the altitudes of triangle $A B C, D A, E B, F C$ are the angle bisectors of angles $D, E, F$ of triangle $D E F, O$ is the point of intersection of the altitudes of triangle $A B C$, which is also the center of the inscribed circle of triangle $D E F$. Thus, $A O=\sqrt{2}, C O=2$. Le... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A vessel with a capacity of 10 liters is filled with air containing $24\%$ oxygen. A certain volume of air was pumped out of the vessel and the same volume of argon was added. Then, the same volume of the mixture as the first time was pumped out and again the same volume of argon was added. In the new mixture, $11.7... | Solution. Let $x$ liters of the mixture be released each time from the vessel. Then, the first time, the amount of oxygen left in the vessel is $2.4 - 0.24x$. The percentage of oxygen in the mixture after adding argon is $(2.4 - 0.24x) \times 10$. The second time, the amount of oxygen left in the vessel is $2.4 - 0.24x... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. In triangle $A B C$, the altitudes $A D, B E, C F$ are drawn. The length of side $A C$ is $\sqrt{6}+\sqrt{2}$. The distances from the center of the inscribed circle of triangle $D E F$ to points $A$ and $C$ are 2 and $2 \sqrt{2}$, respectively. Find the radius of the circumscribed circle around triangle $D E F$. (16... | Solution. $\quad A D, B E, C F$ are the altitudes of triangle $A B C, D A, E B, F C$ are the angle bisectors of angles $D, E, F$ of triangle $D E F, O$ is the point of intersection of the altitudes of triangle $A B C$, which is also the center of the inscribed circle of triangle $D E F$. Thus, $A O=2, C O=2 \sqrt{2}$. ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Given 2019 indistinguishable coins. All coins have the same weight, except for one, which is lighter. What is the minimum number of weighings required to guarantee finding the lighter coin using a balance scale without weights? | Solution. $\quad$ We will prove the following statement by induction on $k$: if there are $N$ visually identical coins, with $3^{k-1}<N \leq 3^{k}$, and one of them is lighter, then it can be found in $k$ weighings. Base case: $k=0, N=1$, no weighing is needed for a single coin. Inductive step: suppose the statement is... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. (15 points) In a convex quadrilateral $A B C D, A B=10, C D=15$. Diagonals $A C$ and $B D$ intersect at point $O, A C=20$, triangles $A O D$ and $B O C$ have equal areas. Find $A O$.
# | # Solution:
From the equality of the areas of triangles $A O D$ and $B O C$ and the equality of angles $\angle A O D=\angle B O C$, it follows that $\frac{A O \cdot O D}{B O \cdot O C}=1$ (by the theorem on the ratio of areas of triangles with one equal angle). From this, we get $\frac{A O}{O C}=\frac{B O}{O D}$. Addi... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Find the largest integer \( a \) such that the expression
\[
a^{2}-15 a-(\tan x-1)(\tan x+2)(\tan x+5)(\tan x+8)
\]
is less than 35 for any value of \( x \in (-\pi / 2, \pi / 2) \).
(6 points) | Solution. Let's make the substitution $t=\operatorname{tg} x$. We need to determine for which values of $a$ the inequality $a^{2}-15 a-(t-1)(t+2)(t+5)(t+8)a^{2}-15 a-35,\left(t^{2}+7 t-8\right)\left(t^{2}+7 t+10\right)>a^{2}-15 a-35$
$z=t^{2}+7 t+1,(z-9)(z+9)>a^{2}-15 a-35, z^{2}>a^{2}-15 a+46$, $0>a^{2}-15 a+46, \sqr... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. On the plane $x O y$, the lines $y=3 x-3$ and $x=-1$ intersect at point $\mathrm{B}$, and the line passing through point $M(1 ; 2)$ intersects the given lines at points A and C respectively. For what positive value of the abscissa of point A will the area of triangle $\mathrm{ABC}$ be the smallest?
(12 points) | # Solution.
$A C: \quad y=k x+d, \quad M \in A C \Rightarrow d=2-k$
$A(a ; 3 a-3) \in A C \Rightarrow 3 a-3=k a+2-k \Rightarrow a=\frac{5-k}{3-k}$,
$C(-1 ; c) \in A C \Rightarrow c=-2 k+2$,
$S_{A B C}=\frac{1}{2}(c+6) \cdot(a+1)=\frac{2(k-4)^{2}}{3-k}$
$S^{\prime}=\frac{2(k-4)(2-k)}{(3-k)^{2}}=0, k_{\min }=2, \qua... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Specify the smallest integer value of \(a\) for which the system has a unique solution
\[
\left\{\begin{array}{l}
\frac{y}{a-\sqrt{x}-1}=4 \\
y=\frac{\sqrt{x}+5}{\sqrt{x}+1}
\end{array}\right.
\] | # Solution.
Solving the system by substitution, we arrive at an equation with constraints on the unknown quantity ${ }^{x}$.
$$
\left\{\begin{array} { l }
{ \frac { y } { a - \sqrt { x } - 1 } = 4 } \\
{ y = \frac { \sqrt { x } + 5 } { \sqrt { x } + 1 } }
\end{array} \Rightarrow \left\{\begin{array} { l }
{ x \geq ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find all pairs of integers $(x, y)$ that satisfy the equation $x^{2}-x y-6 y^{2}-11=0$. For each pair $(x, y)$ found, calculate the product $x y$. In the answer, write the sum of these products. | Solution. $x^{2}-x y-6 y^{2}-11=0,(x-3 y)(x+2 y)=11$. Since $x$ and $y$ are integers, we have four cases:
$\left\{\begin{array}{c}x-3 y=11 \\ x+2 y=1\end{array} \Leftrightarrow\left\{\begin{array}{c}y=-2 \\ x=5\end{array}\right.\right.$
2)
$\left\{\begin{array}{c}x-3 y=-11 \\ x+2 y=-1,\end{array} \Leftrightarrow\left\{... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Find all integer values of the parameter \(a\) for which the system has at least one solution
\[
\left\{\begin{array}{l}
y-2=x(x+2) \\
x^{2}+a^{2}+2 x=y(2 a-y)
\end{array}\right.
\]
In the answer, specify the sum of the found values of the parameter \(a\). | Solution. Transform the system
$$
\left\{\begin{array} { l }
{ y - 1 = ( x + 1 ) ^ { 2 } , } \\
{ ( x + 1 ) ^ { 2 } + ( y - a ) ^ { 2 } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
y-1=(x+1)^{2} \\
y-2+(y-a)^{2}=0
\end{array}\right.\right.
$$
Consider the second equation of the system
$$
y^{2}-y(2 a-1)+a^{... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. During the shooting practice, each soldier fired 10 times. One of them completed the task successfully and scored 90 points. How many times did he score 9 points, if there were 4 tens, and the results of the hits were sevens, eights, and nines. There were no misses at all. | Solution: Since the soldier scored 90 points and 40 of them were scored in 4 attempts, he scored 50 points with the remaining 6 shots. Since the soldier only hit the seven, eight, and nine, let's assume that in three shots (once each in seven, eight, and nine), he scored 24 points. Then, for the remaining three shots, ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Ivan Ivanovich approached a source with two empty cans, one with a capacity of 10 liters, and the other - 8 liters. Water from the source flowed in two streams - one stronger, the other weaker. Ivan Ivanovich simultaneously placed the cans under the streams and, when half of the smaller can was filled, he switched t... | Solution. Let $x$ liters of water fill the larger can while 4 liters fill the smaller can. After the switch, $(10-x)$ liters fill the larger can, and 4 liters fill the smaller can again. Since the flow rates are constant, the ratio of the volumes of water filled in the same time is also constant. We can set up the equa... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Find all values of $n, n \in N$, for which the sum of the first terms of the sequence $a_{k}=3 k^{2}-3 k+1, \quad k \in N, \quad$ is equal to the sum of the first $n$ terms of the sequence $b_{k}=2 k+89, k \in N$ (12 points) | Solution. Note that $a_{k}=3 k^{2}-3 k+1=k^{3}-(k-1)^{3}$, and the sum is $S_{n}=n^{3}$.
For the second sequence $\quad b_{k}=2 k+89=(k+45)^{2}-(k+44)^{2}, \quad$ the sum is $S_{n}=(n+45)^{2}-45^{2}=n(n+90)$.
We get the equation $n^{3}=n(n+90) \Rightarrow n^{2}-n-90=0 \Rightarrow n=10$.
Answer: 10. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 6. (Option 2).
Translate the text above into English, preserving the original text's line breaks and formatting, and output the translation result directly. | Solution: A total of $68+59+46=173$ "yes" answers were given to the three questions. Note that since each resident of Kashino lives in exactly one district, if every resident were a knight (i.e., told the truth only), the number of "yes" answers would be equal to the number of residents in the city (a knight says "yes"... | 12 | Number Theory | proof | Yes | Yes | olympiads | false |
1. $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}$, but according to Vieta's theorem $\left\{\begin{array}{l}D \geq 0 \\ x_{1}+x_{2}=-(m+1) . \text { Then, } \mathrm{c} \\ x_{1} x_{2}=2 m-2\end{array}\right.$
considering that $D=(m+3)^{2} \geq 0$, we have $x_{1}^{2}+x_{2}^{2}=$
$(-(m+1))^{2}-2(2 m-2)... | Answer: For the equation $x^{2}+(m+1) x+2 m-2=0$, the smallest sum of the squares of its roots is 4 when $m=1$.
Grading criteria.
| 15 points | Correct and justified solution. |
| :--- | :--- |
| 10 points | Using Vieta's theorem, the expression for the sum of the squares of the roots is correctly written, but there ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (10 points) Solve the equation
$$
\sqrt{7-x^{2}+6 x}+\sqrt{6 x-x^{2}}=7+\sqrt{x(3-x)}
$$ | # Solution:
The domain of the variable x in our problem is the interval [0;3]. By completing the square in the expressions under the square roots on the left side of the equation or by plotting the graphs, we notice that the values of the first root do not exceed 4, and the second does not exceed 3, with the minimum v... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Find the value of the expression $2 a-\left(\frac{2 a-3}{a+1}-\frac{a+1}{2-2 a}-\frac{a^{2}+3}{2 a^{2-2}}\right) \cdot \frac{a^{3}+1}{a^{2}-a}+\frac{2}{a}$ when $a=1580$. | Solution:
1) $2 a-\frac{2(a-1)(2 a-3)+(a+1)(a+1)-\left(a^{2}+3\right)}{2(a-1)(a+1)} \cdot \frac{(a+1)\left(a^{2}-a+1\right)}{a^{2}-a}+\frac{2}{a}$
2) $2 a-\frac{2(a-1)(2 a-3)+(a+1)(a+1)-\left(a^{2}+3\right)}{2(a-1)} \cdot \frac{\left(a^{2}-a+1\right)}{a^{2}-a}+\frac{2}{a}$
3) $2 a-\frac{\left(-4 a+2+2 a^{2}\right)}{(a... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In a convex quadrilateral $A B C D$, the angles at vertices $B, C$, and $D$ are $30^{\circ}, 90^{\circ}$, and $120^{\circ}$ respectively. Find the length of segment $A B$, if $A D=C D=2$. | Solution:
Extend lines $A B$ and $C D$ to intersect at point $E$, the resulting triangle $A D E$ is equilateral, so $E D=E A=2$, the leg $E C=E D+D C=2+2=4$, since it lies in the right triangle $B C E$ opposite the angle $30^{\circ}$, then the hypotenuse $B E=8$, and the segment $A B=B E-E A=8-2=6$.
Answer: 6. | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. In an acute-angled triangle $ABC$ with sides $AB=4, AC=3$, a point $N$ is marked on the median $AM$ such that $\angle BNM = \angle MAC$. Find the length of the segment $BN$. | Solution:
Let's make an additional construction, doubling the median $A M$ beyond point $M$, thereby obtaining point $K$ such that $K \in A M, K M=A M$. Triangles $K M B$
Preliminary (correspondence) online stage of the "Step into the Future" School Students' Olympiad in the subject of Mathematics
and $A M C$ are eq... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. Solve the equation $a^{2}+2=b$ !, given that a, b belong to N. In the answer, indicate the sum of the product of all possible a and the product of all possible b (if the equation has no solutions, indicate 0; if there are infinitely many solutions, indicate 1000). | Solution:
$b!-2=a^{2} ; x, y \in N$
$a \geq 1$, i.e. $a^{2} \geq 1 \Rightarrow b!\geq 3$, i.e. $b \geq 3$
If $x \geq 5$, then $x!$ ends in 0, then $y^{2}$ ends in 8, but there is no number whose square ends in 8, i.e. $x<5$.
This gives us:
$\left[\begin{array}{l}b=3 \\ b=4\end{array} \Rightarrow\left[\begin{array}... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) For what values of the parameter $a$ does the equation
$$
(a+1)(|x-2.3|-1)^{2}-2(a-3)(|x-2.3|-1)+a-1=0
$$
have exactly two distinct solutions? | Solution. Let $|x-2.3|-1=t$ (1), then the original equation will take the form: $(a+1) t^{2}-2(a-3) t+a-1=0$ (2). Let's analyze equation (1): when $t>-1$, it corresponds to two different values of x. Thus, the original equation can have from zero to four solutions. It has two distinct roots in the following three cases... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) Given a cyclic quadrilateral $A B C D$. The rays $A B$ and $D C$ intersect at point $E$, and the rays $D A$ and $C B$ intersect at point $F$. The ray $B A$ intersects the circumcircle of triangle $D E F$ at point $L$, and the ray $B C$ intersects the same circle at point $K$. The length of segment $L K$ ... | # Solution.
$\angle F L E = \angle F D E = \angle F K E = \alpha$, since these angles subtend the arc $F E$.
$\angle E B K = \angle F D E = \alpha$, since quadrilateral $A B C D$ is cyclic... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) A circle passes through the vertices $A$ and $C$ of an isosceles triangle $ABC (AB = BC)$ and intersects the sides $AB$ and $BC$ at points $M$ and $N$, respectively. $MK$, a chord of this circle, equal in length to $2 \sqrt{5}$, contains point $H$, which lies on $AC$ and is the foot of the altitude of tr... | # Solution.
Quadrilateral $A M N C$ is an isosceles trapezoid. $\triangle A M H = \Delta H N C$ - by two sides and the angle between them.
$$
\begin{gathered}
\angle A H M = \angle H M N =... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Find the minimum value of the expression $\frac{3 f(1)+6 f(0)-f(-1)}{f(0)-f(-2)}$, if $f(x)=a x^{2}+b x+c$ is an arbitrary quadratic function satisfying the condition $b>2 a$ and taking non-negative values for all real $x$.
(12 points) | Solution. We have $f(1)=a+b+c, \quad f(0)=c, \quad f(-1)=a-b+c, f(-2)=4 a-2 b+c$, $\frac{3 f(1)+6 f(0)-f(-1)}{f(0)-f(-2)}=\frac{3(a+b+c)+6 c-a+b-c}{c-4 a+2 b-c}=\frac{2 a+4 b+8 c}{2 b-4 a}=\frac{a+2 b+4 c}{b-2 a}$.
Since $f(x)=a x^{2}+b x+c \quad-$ is an arbitrary quadratic function that takes non-negative values for ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In triangle $A B C$, the bisector $A D$ is drawn. It is known that the centers of the inscribed circle of triangle $A B D$ and the circumscribed circle of triangle $A B C$ coincide. Find $C D$, if $A C=\sqrt{5}+1$. The answer should not include trigonometric function notations or their inverses.
(20 points) | Solution: Let $\angle A=\alpha, \quad B=\beta$. Point $O$ is the center of the inscribed circle of triangle $ABD$. $\angle BAO=\alpha / 4, \angle ABO=\beta / 2$. Since $O$ is the center of the circumscribed circle around triangle $ABC$, triangle $AOB$ is isosceles, and $\angle BAO=\angle ABO, \beta=\alpha / 2$. Triangl... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A workshop produces transformers of types $A$ and $B$. For one transformer of type $A$, 5 kg of transformer iron and 3 kg of wire are used, and for a transformer of type $B$, 3 kg of iron and 2 kg of wire are used. The profit from selling a transformer of type $A$ is 12 thousand rubles, and for type $B$ it is 10 tho... | Solution. Let $x$ be the number of transformers of type $A$, and $y$ be the number of transformers of type $B$. Then the profit per shift is calculated by the formula $D=12 x+10 y$, with the conditions
 \Longrightarrow 1000^{n} \equiv(-1)^{n}(\bmod 7)$. Therefore, the given number is congruent modulo 7 to the number
$$
999-998+997-996+\ldots+101-100=450 \equiv 2(\bmod 7)
$$
Answer: 2 | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Two balls, the sizes of which can be neglected in this problem, move along a circle. When moving in the same direction, they meet every 20 seconds, and when moving in opposite directions - every 4 seconds. It is known that when moving towards each other along the circle, the distance between the approaching balls de... | # Solution:
Let the speed of the faster ball be $v$, and the slower one be $u$. When moving in the same direction, the faster ball catches up with the slower one when the difference in the distances they have traveled equals the length of the circle. According to the problem, we set up a system of two linear equations... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. For what values of the parameter $\boldsymbol{a}$ does the equation $|f(x)-4|=p(x)$, where $f(x)=\left|\frac{x^{2}+3 x}{x+3}-\frac{x^{2}-4 x+4}{2-x}\right|$, $p(x)=a$ have three solutions? If there is more than one value of the parameter, indicate their product in the answer. | # Solution:
Simplify $f(x)=\left|\frac{x^{2}+3 x}{x+3}-\frac{x^{2}-4 x+4}{2-x}\right|$, we get $f(x)=|2 x-2|$, where $x \neq-3, x \neq 2$.
Solve the equation || $2 x-2|-4|=a$, where $x \neq-3, x \neq 2$ graphically in the system $x O a$.
The equation has three solutions when $a=2$.
The product is 2.
 be Masha's labor productivity, $y$ be Dasha's, and $z$ be Sasha's. Since the labor productivity adds up when working together, we can set up a system of equations based on the problem's conditions: $\left\{\begin{array}{l}(x+y) \cdot 7.5=1 ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. On the sides $AB$ and $BC$ of triangle $ABC$, points $M$ and $N$ are marked respectively such that $\angle CMA = \angle ANC$. Segments $MC$ and $AN$ intersect at point $O$, and $ON = OM$. Find $BC$, if $AM = 5 \, \text{cm}, BM = 3 \, \text{cm}$. | # Solution:
Triangles $A M O$ and $C N O$ are congruent ($\angle A M O=\angle C N O, O N=O M, \angle M O A=\angle N O C$, as vertical angles). From the congruence of the triangles, it follow... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. During the shooting practice, each soldier fired 10 times. One of them completed the task successfully and scored 90 points. How many times did he score 7 points, if he hit the bullseye 4 times, and the results of the other hits were sevens, eights, and nines? There were no misses at all. | # Solution:
Since the soldier scored 90 points and 40 of them were scored in 4 shots, he scored 50 points with the remaining 6 shots. As the soldier only hit the 7, 8, and 9, let's assume that in three shots (once each in 7, 8, and 9), he scored 24 points. Then, for the remaining three shots, he scored 26 points, whic... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Boys are dividing the catch. The first one took $r$ fish and a seventh part of the remainder; the second - $2 r$ fish and a seventh part of the new remainder; the third - $3 r$ fish and a seventh part of the new remainder, and so on. It turned out that in this way all the caught fish were divided equally. How many b... | # Solution:
Let $x$ be the number of boys; $y$ be the number of fish each received. Then the last boy took $x r$ fish (there could be no remainder, otherwise there would not have been an even distribution), so $y=x r$. The second-to-last boy took $(x-1) r+\frac{x r}{6}=y$; i.e., $x r$ is $\frac{6}{7}$ of the second-to... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. In rectangle $A B C D$, point $E$ is located on diagonal $A C$ such that $B C=E C$, point $M$ is on side $B C$ such that $E M=M C$. Find the length of segment $M C$, if $B M=5, A E=2$. | # Solution:
Draw $A F$ parallel to $B E$ (point $F$ lies on line $B C$), then $\angle C B E=\angle C F A$, $\angle C E B=\angle C A F$. Considering that $B C=C E$, we get that triangle $FCA$ ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Task: On an $8 \times 8$ chessboard, 64 checkers numbered from 1 to 64 are placed. 64 students take turns approaching the board and flipping only those checkers whose numbers are divisible by the ordinal number of the current student. A "Queen" is a checker that has been flipped an odd number of times. How many "Que... | Solution: It is clear that each checker is flipped as many times as the number of divisors of its number. Therefore, the number of "queens" will be the number of numbers from 1 to 64 that have an odd number of divisors, and this property is only possessed by perfect squares. That is, the numbers of the "queens" remaini... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. One worker in two hours makes 5 more parts than the other, and accordingly spends 2 hours less to manufacture 100 parts. How much time does each worker spend on manufacturing 100 parts?
# | # Solution:
Let the second worker make 100 parts in $x$ hours, and the first worker in $x-2$ hours.
$\frac{100}{x-2}-\frac{100}{x}=\frac{5}{2} ; \frac{40}{x-2}-\frac{40}{x}=1 ; x^{2}-2 x-80=0, x=1+9=10$. Answer: 8 and 10 hours. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. In triangle $A B C$, angles $A$ and $B$ are $45^{\circ}$ and $30^{\circ}$ respectively, and $C M$ is the median. The incircles of triangles $A C M$ and $B C M$ touch segment $C M$ at points $D$ and $E$. Find the radius of the circumcircle of triangle $A B C$ if the length of segment $D E$ is $4(\sqrt{2}-1)$. | # Solution:
By the property of tangents to a circle, we have:
$$
A G=A K=x, C G=C D=y, C E=C F=z, B F=B H=u, D M=\frac{A B}{2}-A K=\frac{A B}{2}-x
$$
$M E=\frac{A B}{2}-B H=\frac{A B}{2}-u$,
Then $D E=z-y, D E=D M-M E=u-x$. Therefore, $2 D E=z-y+u-x=C B-$ AC. Let $C B=a, A C=b$. Then $a-b=8(\sqrt{2}-1)$. By the Law... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. One tourist covers a distance of 20 km 2.5 hours faster than the other. If the first tourist reduced their speed by 2 km/h, and the second increased their speed by 50%, they would spend the same amount of time on the same distance. Find the speeds of the tourists. | Solution: Let $\mathrm{x}$ be the speed of the first tourist, and $y$ be the speed of the second tourist.
\[
\frac{20}{x}+\frac{5}{2}=\frac{20}{y}, \quad x-2=1.5 y, \quad \frac{4}{x}+\frac{1}{2}=\frac{6}{x-2}, \quad x^{2}-6 x-16=0, \quad x=8, \quad y=4
\]
Answer: $8 \mathrm{km} / \mathrm{u}, 4 \mathrm{km} / \mathrm{q... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Since $\mathrm{x} \leq \mathrm{P}$ and Р:х, let $2 \mathrm{x} \leq \mathrm{P} => \mathrm{x} \leq \mathrm{P} / 2$, and also $\mathrm{y} \leq \mathrm{P} ; \kappa \leq \mathrm{P} ; \mathrm{e} \leq \mathrm{P} ; =>$
$$
\mathrm{x}+\mathrm{y}+\mathrm{K}+\mathrm{e} \leq 3.5 \mathrm{P} ; => \mathrm{P} \geq \frac{2}{7}(\math... | Answer: 2 l
## №6: Planimetry. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. (10 points) In a row, 2018 digits are written consecutively. It is known that in this row, every two-digit number formed by two adjacent digits (in the order they are written) is divisible by 17 or 23. The last digit in this row is 5. What is the first digit in the row? Provide a justified answer.
# | # Solution:
All two-digit numbers divisible by 17 or 23:
$$
\begin{aligned}
& 17,34,51,68,85 \\
& 23,46,69,92
\end{aligned}
$$
The following diagram shows with arrows which digit can follow which in the row:
$$
\begin{aligned}
& 1 \rightarrow 7 \quad 9 \rightarrow 2 \rightarrow 3 \\
& \uparrow \quad \uparrow \\
& 5... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. Rectangle $ABCD$ with sides $AB=1$ and $AD=10$ serves as the base of pyramid $SABCD$, and the edge $SA=4$ is perpendicular to the base. Find a point $M$ on edge $AD$ such that triangle $SMC$ has the smallest perimeter. Find the area of this triangle.
# | # Solution:
Let's lay off $A T=A S$ on the extension of edge $A B$. For any position of point $M$ on side $A D$, $T M=S M$, so the minimum value of the sum $S M+M C$ will be when $M=T C \cap A D$.
Let's introduce the notation
$A B=a, A D=b, A S=c, A M=x$. From
$\triangle T A M \sim \triangle C D M$ it follows that
... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. (Option 1)
Calculate $x^{3}+3 x$, where $x=\sqrt[3]{2+\sqrt{5}}-\frac{1}{\sqrt[3]{2+\sqrt{5}}}$. | Solution: Let $\sqrt[3]{2+\sqrt{5}}=a$, then $x=a-\frac{1}{a}$,
$$
\begin{aligned}
& x^{3}+3 x=\left(a-\frac{1}{a}\right)^{3}+3\left(a-\frac{1}{a}\right)=a^{3}-\frac{1}{a^{3}}=2+\sqrt{5}-\frac{1}{2+\sqrt{5}}=\frac{(2+\sqrt{5})^{2}-1}{2+\sqrt{5}}= \\
& =\frac{8+4 \sqrt{5}}{2+\sqrt{5}}=4
\end{aligned}
$$
Answer: 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. There are two lead-tin alloys. In the first alloy, the mass of lead is to the mass of tin as $1: 2$; in the second - as $2: 3$. How many grams of the first alloy are needed to obtain 22 g of a new alloy with the mass ratio of lead to tin being 4:7? | Solution. Let the first alloy contain x g of lead and 2x g of tin. In the second alloy, there are 2y g of lead and 3y g of tin. Then $k \cdot 3x + n \cdot 5y = 22 ; \frac{kx + n \cdot 2y}{k \cdot 2x + n \cdot 3y} = \frac{4}{7} ; \quad$ we need to find $k \cdot 3x$ and $5ny$. Let $ny = b ; kx = a \cdot \frac{a + 2b}{2a ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The number $\overline{6 x 62 y 4}$ is divisible by 11, and when divided by 9, it gives a remainder of 6. Find the remainder when this number is divided by $13 .(15$ points) | # Solution
By the divisibility rule for 11, we get
$((x+2+4)-(6+6+y)) \vdots 11$ or $(x-6-y) \vdots 11$
Let's find suitable options: $(0 ; 5)(1 ; 6)(2 ; 7)(3 ; 8)(4 ; 9)(6 ; 0)(7 ; 1)(8 ; 2)$ $(9 ; 3)$.
If the number $\overline{6 x 62 y 4}$ gives a remainder of 6 when divided by 9, then by the divisibility rule for... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Find the sum of the integers that belong to the set of values of the function $f(x)=\log _{3}(40 \cos 2 x+41)$ for $x \in[(5 / 3)(\operatorname{arctg}(1 / 5)) \cos (\pi-\arcsin (-0.8)) ; \operatorname{arctg} 3]$ (10 points) | Solution: Since $\cos (\pi-\arcsin (-0.8))=\cos (\pi+\arcsin 0.8)=-\cos (\arcsin 0.8)=-0.6$, then $x \in[(5 / 3)(\operatorname{arctg}(1 / 5)) \cos (\pi-\arcsin (-0.8)) ; \operatorname{arctg} 3]=[-\operatorname{arctg}(1 / 5) ; \operatorname{arctg} 3]$. $2 x \in[-2 \operatorname{arctg}(1 / 5) ; 2 \operatorname{arctg} 3]$... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. On the side $AC$ of triangle $ABC$, a circle is constructed with $AC$ as its diameter, which intersects sides $AB$ and $BC$ at points $D$ and $E$ respectively. The angle $EDC$ is $30^{\circ}$, $AE = \sqrt{3}$, and the area of triangle $DBE$ is to the area of triangle $ABC$ as $1:2$. Find the length of segment $BO$, ... | # Solution:
1) $\angle E D C=\angle E A C=30^{\circ}$ (inscribed angles subtending the same arc);
2) $A C$ - diameter of the circle $\Rightarrow \triangle A E C$ - right triangle, $\angle A E C=90^{\circ}, \angle E C A=60^{\circ}$, $A C=\frac{A E}{\cos 30^{\circ}}=2, E C=A C \sin 30^{\circ}=1$
=\log _{3}(10 \cos 2 x+17)$ for $x \in[1,25(\operatorname{arctg} 0.25) \cos (\pi-\arcsin (-0.6)) ; \operatorname{arctg} 3] . \quad(10$ points $)$ | # Solution:
Since $\quad$ $\cos (\pi-\arcsin (-0.6))=\cos (\pi+\arcsin 0.6)=-\cos (\arcsin 0.6)=-0.8, \quad$ then $x \in[1.25(\operatorname{arctg} 0.25) \cos (\pi-\arcsin (-0.6)) ; \operatorname{arctg} 3]=[-\operatorname{arctg} 0.25 ; \operatorname{arctg} 3] \quad$ Therefore, $2 x \in[-2 \operatorname{arctg} 0.25 ; 2 ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. On the side $AC$ of triangle $ABC$, a circle is constructed with $AC$ as its diameter, which intersects sides $AB$ and $BC$ at points $D$ and $E$ respectively. The angle $EDC$ is $30^{\circ}$, $EC=1$, and the area of triangle $DBE$ is to the area of triangle $ABC$ as $1:2$. Find the length of segment $BO$, where $O$... | # Solution:
1) $\angle E D C=\angle E A C=30^{\circ}$ (inscribed angles subtending the same arc);
2) $A C$ - diameter of the circle $\Rightarrow \triangle A E C$ - right triangle, $\angle A E C=90^{\circ}, \angle E C A=60^{\circ}$,
$$
A C=\frac{E C}{\sin 30^{\circ}}=2, \quad A E=E C \operatorname{tg} 60^{\circ}=\sqrt... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Find the sum of the integers that belong to the set of values of the function $f(x)=\log _{2}(5 \cos 2 x+11)$ for $x \in[1,25(\operatorname{arctg}(1 / 3)) \cos (\pi+\arcsin (-0.6)) ; \operatorname{arctg} 2] \quad$ (10 points) | # Solution:
Since $\quad$ $\cos (\pi+\arcsin (-0.6))=\cos (\pi-\arcsin 0.6)=-\cos (\arcsin 0.6)=-0.8, \quad$ then $x \in[1.25(\operatorname{arctg}(1 / 3)) \cos (\pi+\arcsin (-0.6)) ; \operatorname{arctg} 2]=[-\operatorname{arctg}(1 / 3) ; \operatorname{arctg} 2]$ Therefore, $2 x \in[-2 \operatorname{arctg}(1 / 3) ; 2 ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. On the side $AC$ of triangle $ABC$, a circle is constructed with $AC$ as its diameter, which intersects sides $AB$ and $BC$ at points $D$ and $E$ respectively. The angle $EDC$ is $30^{\circ}$, the area of triangle $AEC$ is $\sqrt{3} / 2$, and the area of triangle $DBE$ is to the area of triangle $ABC$ as $1: 2$. Fin... | Solution:
1) $\angle E D C=\angle E A C=30^{\circ}$ (inscribed angles subtending the same arc);
2) $A C$ - diameter of the circle $\Rightarrow \triangle A E C$ - right-angled, $\angle A E C=90^{\circ}, \angle E C A=60^{\circ}$, $A C=\frac{E C}{\sin 30^{\circ}}=2 E C, A E=E C \operatorname{tg} 60^{\circ}=\sqrt{3} E C ;... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation $\quad \sqrt{x+\sqrt{x}-\frac{71}{16}}-\sqrt{x+\sqrt{x}-\frac{87}{16}}=\frac{1}{2}$. | Solution. Taking into account that $x$ is non-negative, we make the substitution $u=\sqrt{x+\sqrt{x}-\frac{71}{16}}, v=\sqrt{x+\sqrt{x}-\frac{87}{16}}, u \geq 0, v \geq 0$.
Then we obtain the system $\left\{\begin{array}{l}u-v=1 / 2, \\ u^{2}-v^{2}=1,\end{array} \Rightarrow\left\{\begin{array}{l}u-v=1 / 2, \\ u+v=2,\e... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. What is the maximum possible number of consecutive terms of an increasing geometric progression that can be three-digit natural numbers? Provide an example of such a sequence. (16 points) | Solution. Let the required members of the progression be $a_{0}, a_{1}, \ldots, a_{n}, a_{k}=a_{0} q^{k}$, the common ratio - an irreducible fraction $q=r / s, r>s$. Then $a_{0}=b s^{n}, a_{n}=b r^{n}, b \in \mathbb{N}$, since $r^{n}$ and $s^{n}$ are coprime. We obtain the restriction
$$
r^{n}<1000 / b, \quad s^{n} \g... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. $f(-x)=3(-x)^{3}-(-x)=-3 x^{3}+x=-\left(3 x^{3}-x\right)=-f(x)$ $g(-x)=f^{3}(-x)+f\left(\frac{1}{-x}\right)-8(-x)^{3}-\frac{2}{-x}=-f^{3}(x)-f\left(\frac{1}{x}\right)+8 x^{3}+\frac{2}{x}=-g(x)$
Therefore, $g$ is an odd function $\Rightarrow$ if $x_{0}$ is a root of the original equation, then $-x_{0}$ is also a roo... | Answer: 0.
Problem 8 (2nd version).
Find the sum of the roots of the equation $g^{3}(x)-g\left(\frac{1}{x}\right)=5 x^{3}+\frac{1}{x}$, where $g(x)=x^{3}+x$.
## Solution.
Consider the function $f(x)=g^{3}(x)-g\left(\frac{1}{x}\right)-5 x^{3}-\frac{1}{x}$, then the roots of the original equation are the roots of the... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. $g(-x)=(-x)^{3}+(-x)=-x^{3}-x=-\left(x^{3}+x\right)=-g(x)$
$$
f(-x)=g^{3}(-x)-g\left(\frac{1}{-x}\right)-5(-x)^{3}-\frac{1}{-x}=-g^{3}(x)+g\left(\frac{1}{x}\right)+5 x^{3}+\frac{1}{x}=-f(x)
$$
Therefore, $f$ is an odd function $\Rightarrow$ if $x_{0}$ is a root of the original equation, then $-x_{0}$ is also a roo... | # Answer: 0.
Task 9 (1st variant). In each vertex of an equilateral triangle with side $\sqrt{10}$, circles of radius $\sqrt{5}$ were constructed, the inner areas of which were painted gray, brown, and raspberry. Find the area of the gray-brown-raspberry region.
Solution. By symmetry, the areas of the figures formed ... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
1. Two trucks were transporting fertilizers, making the same number of trips. It turned out that 4 tons less could be loaded onto the first truck and 3 tons less onto the second truck than planned, so each truck had to make 10 extra trips. As a result, the first truck transported 60 tons more than the second, as planne... | # Solution:
Let $x, y$ - capacity, $t-$ number of trips as planned.
$$
\left\{\begin{array}{l}
x t=(x-4)(t+10), \\
y t=(y-3)(t+10), \Leftrightarrow\left\{\begin{array} { l }
{ 1 0 x - 4 t = 4 0 } \\
{ \quad x t - y t = 6 0 }
\end{array} \quad \left\{\begin{array}{l}
10 y-3 t=30, \\
(x-y) t=60
\end{array} \Rightarrow... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The numerical sequence $\left\{a_{n}\right\}_{n=1}^{\infty}$ is defined such that $a_{1}=\log _{2}\left(\log _{2} f(2)\right), \quad a_{2}=$ $\log _{2}\left(\log _{2} f(f(2))\right), \ldots, a_{n}=\log _{2}(\log _{2} \underbrace{f(f(\ldots f}_{n}(2)))), \ldots$, where $f(x)=x^{x}$. Determine the index $n$ for which ... | # Solution.
If $\log _{2}\left(\log _{2} u\right)=t$, then $u=2^{2^{t}}, f(u)=\left(2^{2^{t}}\right)^{2^{2^{t}}}=2^{2^{t+2^{t}}}, \log _{2}\left(\log _{2} f(u)\right)=t+2^{t} . \quad$ If $u=2,2=2^{2^{t}}, t=0, a_{1}=\log _{2}\left(\log _{2} f(2)\right)=0+2^{0}=1$. If $u=f(2)$, then $t=\log _{2}\left(\log _{2} u\right)... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. More and more countries are exploring space. The number of states that launch their satellites using their own launch vehicles is already 12. There are also countries that use the services of the main space powers to launch their satellites for economic purposes. Due to the increasing number of participants in space... | Solution. According to the condition, at the current moment, all satellites should be located on a sphere with radius $R+H$. Let $O$ be the center of the sphere, and its radius $R_{H}=R+H$. Denote the satellites by points $C_{i}, i=1, \ldots, n$. We need to determine the maximum value of $n$.
Let's find the distance $... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1. Option I.
The decimal representation of a natural number $N$ contains 1580 digits. Among these digits are threes, fives, and sevens, and no other digits. It is known that the number of sevens is 20 less than the number of threes. Find the remainder when the number $N$ is divided by 3. | Solution. Let $x$ be the number of threes in the number $N$. The sum of the digits of the number $N$ is $S=3 x+7(x-20)+5(1580-(2 x-20))=7860$.
The remainder of $S$ divided by 3 is equal to the remainder of $N$ divided by 3 and is 0.
Answer: 0. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 1. II variant.
The decimal representation of a 2015-digit natural number $N$ contains the digits 5, 6, 7 and no other digits. Find the remainder of the division of the number $N$ by 9, given that the number of fives in the representation of the number is 15 more than the number of sevens. | Solution. Let the number $N$ contain $x$ sevens. Then the sum of the digits of the number $N$ is $S=7x+5(x+15)+6(2015-(2x+15))=12075$.
$N \equiv S \equiv 6(\bmod 9)$.
Answer: 6. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4. Option I.
Find the set of values of the parameter $a$, for which the sum of the cubes of the roots of the equation $x^{2}-a x+a+2=0$ is equal to -8.
# | # Solution.
1) $x_{1}^{3}+x_{2}^{3}=\left(x_{1}+x_{2}\right)\left(\left(x_{1}+x_{2}\right)^{2}-3 x_{1} x_{2}\right)=a\left(a^{2}-3(a+2)\right)=a^{3}-3 a(a+2)$.
2) $a^{3}-3 a(a+2)=-8 \Leftrightarrow\left[\begin{array}{l}a=-2, \\ a=1, \\ a=4 .\end{array}\right.$
3) $D=a^{2}-4 a-8$.
For $a=-2 \quad D=4+8-8=4>0$, therefo... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. II variant.
Find the set of values of the parameter $a$, for which the sum of the cubes of the roots of the equation $x^{2}+a x+a+1=0$ is equal to 1. | # Solution.
1) $x_{1}^{3}+x_{2}^{3}=\left(x_{1}+x_{2}\right)\left(\left(x_{1}+x_{2}\right)^{2}-3 x_{1} x_{2}\right)=-a\left(a^{2}-3(a+1)\right)=-a^{3}+3 a(a+1)$.
2) $-a^{3}+3 a(a+1)=1 \Leftrightarrow\left[\begin{array}{l}a=-1, \\ a=2 \pm \sqrt{3} \text {. }\end{array}\right.$
3) $D=a^{2}-4 a-4$.
For $a=-1 \quad D=1+4... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 7. Option 1.
Given a triangle with sides 6, 8, and 10. Find the length of the shortest segment connecting points on the sides of the triangle and dividing it into two equal areas. | Solution. 1) Note that the given triangle is a right triangle: $6^{2}+8^{2}=10^{2}$.
2) Suppose first (and we will justify this later) that the ends of the desired segment $D E=t$ lie on the larger leg $A C=8$ and the hypotenuse $A B=10$ (see figure). Let $A D=x$, $A E=y, \angle B A C=\alpha$. Then
$S_{A D E}=\frac{1... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3.
One side of the parallelogram is $\sqrt{3}$ times larger than the other side. One diagonal of the parallelogram is $\sqrt{7}$ times larger than the other diagonal. How many times larger is one angle of the parallelogram than the other angle? | Solution. Let $x$ be the smaller side, then $\sqrt{3} x$ is the larger side. Let $y$ be the smaller diagonal, then $\sqrt{7} y$ is the larger diagonal. We have:
$2 x^{2}+2(\sqrt{3} x)^{2}=y^{2}+(\sqrt{7} y)^{2}$,
from which $x=y$.
We get: the acute angle of the parallelogram is $30^{\circ}$, the obtuse angle is $150^... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. From point $A$ to point $B$, which are 8 km apart, a tourist and a cyclist set out simultaneously. The cyclist, who took no less than half an hour to travel from $A$ to $B$, without stopping, turned back and started moving towards point $A$, increasing his speed by $25 \%$. After 10 minutes from his departure from p... | Solution: Let $x$ (km/h) be the speed of the tourist, $y$ (km/h) be the initial speed of the cyclist, and $t$ (h) be the time spent by the cyclist traveling from $A$ to $B$. Then
$$
\left\{\begin{array}{c}
x(t+1 / 6)+5 y / 24=8, \\
y t=8, \\
t \geq 0.5,
\end{array} \Rightarrow x(8 / y+1 / 6)+5 y / 24=8, \Rightarrow 5 ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Two numbers x and y satisfy the equation $280 x^{2}-61 x y+3 y^{2}-13=0$ and are the fourth and ninth terms, respectively, of a decreasing arithmetic progression consisting of integers. Find the common difference of this progression. | Solution: We factorize the expression $280 x^{2}-61 x y+3 y^{2}$. For $y \neq 0$, we have $y^{2}\left(280\left(\frac{x}{y}\right)^{2}-61\left(\frac{x}{y}\right)+3\right)=280 y^{2}\left(\frac{x}{y}-\frac{3}{40}\right)\left(\frac{x}{y}-\frac{1}{7}\right)=(40 x-3 y)(7 x-y)$. This formula is also valid for all real numbers... | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. From point $A$ to point $B$, which are 24 km apart, a pedestrian and a cyclist set out simultaneously. The cyclist, who spent no less than two hours on the journey from $A$ to $B$, without stopping, turned back and started moving towards point $A$ at a speed twice the initial speed. After 24 minutes from his departu... | Solution: Let $x$ (km/h) be the speed of the cyclist, $y$ (km/h) be the initial speed of the truck, and $t$ (h) be the time spent by the truck traveling from $A$ to $B$. Then
$$
\left\{\begin{array}{rl}
x(t+0.4)+0.8 y=24, \\
y t & =24, \\
t & \geq 2,
\end{array} \Rightarrow x(24 / y+0.4)+0.8 y=24, \Rightarrow 2 y^{2}+... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Two numbers \(x\) and \(y\) satisfy the equation \(26 x^{2} + 23 x y - 3 y^{2} - 19 = 0\) and are the sixth and eleventh terms, respectively, of a decreasing arithmetic progression consisting of integers. Find the common difference of this progression. | Solution: Factorize the expression $26 x^{2}+23 x y-3 y^{2}$. For $y \neq 0$, we have $y^{2}\left(26\left(\frac{x}{y}\right)^{2}+23\left(\frac{x}{y}\right)-3\right)=26 y^{2}\left(\frac{x}{y}+1\right)\left(\frac{x}{y}-\frac{3}{26}\right)=(x+y)(26 x-3 y)$. This formula is valid for all real numbers $y$. According to the ... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Indicate the smallest value of the parameter $a$ for which the equation has at least one solution
$2 \sin \left(\pi-\frac{\pi x^{2}}{12}\right) \cos \left(\frac{\pi}{6} \sqrt{9-x^{2}}\right)+1=a+2 \sin \left(\frac{\pi}{6} \sqrt{9-x^{2}}\right) \cos \left(\frac{\pi x^{2}}{12}\right)$. | Solution.
Rewrite the equation as
$2 \sin \left(\pi-\frac{\pi x^{2}}{12}\right) \cos \left(\frac{\pi}{6} \sqrt{9-x^{2}}\right)-2 \sin \left(\frac{\pi}{6} \sqrt{9-x^{2}}\right) \cos \left(\pi-\frac{\pi x^{2}}{12}\right)=a-1$, or
$\sin \left(\pi-\frac{\pi x^{2}}{12}-\frac{\pi}{6} \sqrt{9-x^{2}}\right)=\frac{a-1}{2}$.
... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Rhombus $ABCD$ is the base of a pyramid with vertex $S$. All its lateral faces are inclined to the base plane at the same angle of $60^{\circ}$. Points $M, N, K$, and $L$ are the midpoints of the sides of rhombus $ABCD$. A rectangular parallelepiped is constructed on rectangle $MNKL$ as its base. The edges of the up... | Solution.
The height of the pyramid $SO = h$. The diagonals of the rhombus $AC = d_1, BD = d_2$.
The height of the parallelepiped is $h / 2$, and the sides of the base of the parallelepiped are $d_1 / 2$ and $d_2 / 2$.
The volume of the parallelepiped is
$$
V_{\Pi} = \frac{d_1}{2} \cdot \frac{d_2}{2} \cdot \frac{h}... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Specify the greatest value of the parameter $p$ for which the equation has at least one solution
$2 \cos \left(2 \pi-\frac{\pi x^{2}}{6}\right) \cos \left(\frac{\pi}{3} \sqrt{9-x^{2}}\right)-3=p-2 \sin \left(-\frac{\pi x^{2}}{6}\right) \cos \left(\frac{\pi}{3} \sqrt{9-x^{2}}\right)$. (16 points) | # Solution.
Let's rewrite the equation as
$$
\cos \left(2 \pi-\frac{\pi x^{2}}{6}\right) \cos \left(\frac{\pi}{3} \sqrt{9-x^{2}}\right)+\sin \left(2 \pi-\frac{\pi x^{2}}{6}\right) \sin \left(\frac{\pi}{3} \sqrt{9-x^{2}}\right)=\frac{p+3}{2} \text {, or }
$$
$\cos \left(2 \pi-\frac{\pi x^{2}}{6}-\frac{\pi}{3} \sqrt{9... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (1b, 8-11) The probability that a purchased light bulb will work is 0.95. How many light bulbs need to be bought to ensure that with a probability of 0.99, there are at least 5 working ones among them? | # Solution.
Let's take 6 light bulbs. The probability that at least 5 of them will work is the sum of the probabilities that 5 will work and 1 will not, and that all 6 will work, i.e.,
$$
C_{6}^{5} \cdot 0.95^{5} \cdot 0.05 + C_{6}^{6} \cdot 0.95^{6} = 6 \cdot 0.95^{5} \cdot 0.05 + 0.95^{6} = 0.9672
$$
Let's take 7 ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.