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28. Circles $\Omega_{1}$ and $\Omega_{2}$ of equal radii intersect at points $B$ and $C$. A point $A$ is chosen on circle $\Omega_{1}$. Ray $A B$ intersects circle $\Omega_{2}$ at point $D$ (point $B$ lies between points $A$ and $D$). On ray $D C$, a point $E$ is chosen such that $D C = C E$. Find $A E$, if param1.
| ... | 28. Circles $\Omega_{1}$ and $\Omega_{2}$ of equal radii intersect at points $B$ and $C$. A point $A$ is chosen on circle $\Omega_{1}$. Ray $A B$ intersects circle $\Omega_{2}$ at point $D$ (point $B$ lies between points $A$ and $D$). On ray $D C$, a point $E$ is chosen such that $D C = C E$. Find $A E$, if param1.
| ... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
30. On the board, there are param1 natural numbers. It is known that the sum of any five of them is not less than param2. Find the smallest possible value of the sum of all the numbers written on the board.
| param1 | param2 | Answer |
| :---: | :---: | :---: |
| 20 | 117 | |
| 18 | 97 | |
| 19 | 107 | |
| 26 | 153... | 30. On the board, ragat1 natural numbers are written. It is known that the sum of any five of them is not less than param2. Find the smallest possible value of the sum of all numbers written on the board.
| param1 | param2 | Answer |
| :---: | :---: | :---: |
| 20 | 117 | 477 |
| 18 | 97 | 357 |
| 19 | 107 | 415 |
| 2... | 477 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. Given quadratic trinomials $f_{1}(x)=x^{2}-a x-3, f_{2}(x)=x^{2}+2 x-b, f_{3}(x)=3 x^{2}+(2-2 a) x-6-b$ and $f_{4}(x)=3 x^{2}+(4-a) x-3-2 b$. Let the differences of their roots be $A, B, C$ and $D$ respectively. It is known that $|C| \neq|D|$. Find the ratio $\frac{A^{2}-B^{2}}{C^{2}-D^{2}}$. The values of $A, B, C,... | Answer: 3.
Solution. Let $\alpha x^{2}+\beta x+\gamma$ be a quadratic trinomial with a positive discriminant $T$. Then its roots are determined by the formula $x_{1,2}=\frac{-b \pm \sqrt{T}}{2 a}$, so $\left|x_{2}-x_{1}\right|=\left|\frac{-b+\sqrt{T}-(-b-\sqrt{T})}{2 a}\right|=\frac{\sqrt{T}}{|a|}$. Applying this form... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. On the table, there are 100 different cards with numbers $3, 6, 9, \ldots 297, 300$ (each card has exactly one number, and each number appears exactly once). In how many ways can 2 cards be chosen so that the sum of the numbers on the selected cards is divisible by $5?$ | Answer: 990.
Solution. The given numbers, arranged in ascending order, form an arithmetic progression with a common difference of 3. Therefore, the remainders of these numbers when divided by 5 alternate. Indeed, if one of these numbers is divisible by 5, i.e., has the form $5k$, where $k \in \mathbb{N}$, then the nex... | 990 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
|y|+|4+y| \leqslant 4 \\
\frac{x-y^{2}-4 y-3}{2 y-x+3} \geqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area. | Answer: 8.
Solution. Consider the first inequality. To open the absolute values, we consider three possible cases.
1) $y<0$. Then $y+4-y \leqslant 4 \Leftrightarrow 4 \leqslant 4$, i.e., the solution is $y \in \mathbb{R}$, but since $y<0$, we have $y \in (-\infty; 0)$.
2) $y=0$. Then $y+4-y \leqslant 4 \Leftrightarro... | 8 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. Given quadratic trinomials $f_{1}(x)=x^{2}-2 x+a, f_{2}(x)=x^{2}+b x-2, f_{3}(x)=4 x^{2}+(b-6) x+3 a-2$ and $f_{4}(x)=4 x^{2}+(3 b-2) x-6+a$. Let the differences of their roots be $A, B, C$ and $D$ respectively. It is known that $|C| \neq|D|$. Find the ratio $\frac{A^{2}-B^{2}}{C^{2}-D^{2}}$. The values of $A, B, C,... | Answer: 2.
Solution. Let $\alpha x^{2}+\beta x+\gamma$ be a quadratic trinomial with a positive discriminant $T$. Then its roots are determined by the formula $x_{1,2}=\frac{-b \pm \sqrt{T}}{2 a}$, so $\left|x_{2}-x_{1}\right|=\left|\frac{-b+\sqrt{T}-(-b-\sqrt{T})}{2 a}\right|=\frac{\sqrt{T}}{|a|}$. Applying this form... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In a right triangle $ABC (\angle B=90^{\circ})$, a circle $\Gamma$ with center $I$ is inscribed, touching sides $AB$ and $BC$ at points $K$ and $L$ respectively. A line passing through point $I$ intersects sides $AB$ and $BC$ at points $M$ and $N$ respectively. Find the radius of the circle $\Gamma$ if $MK=225$, $NL... | Answer: $R=120, AC=680$.
Solution. Angles $KIM$ and $LNI$ are equal as corresponding angles when lines $BC$ and $KI$ are parallel, so right triangles $KIM$ and $LNI$ are similar. Therefore, $\frac{MK}{KI}=\frac{IL}{LN}$, or (if we denote the radius of the circle by $r$) $\frac{225}{r}=\frac{r}{64}$, from which $r=120$... | 680 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. On the table, there are 150 different cards with numbers $2, 4, 6, \ldots 298, 300$ (each card has exactly one number, and each number appears exactly once). In how many ways can 2 cards be chosen so that the sum of the numbers on the selected cards is divisible by $5?$ | Answer: 2235.
Solution. The given numbers, arranged in ascending order, form an arithmetic progression with a difference of 2. Therefore, the remainders of these numbers when divided by 5 alternate. Indeed, if one of these numbers is divisible by 5, i.e., has the form $5k$, where $k \in \mathbb{N}$, then the next numb... | 2235 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
|x|+|4-x| \leqslant 4 \\
\frac{x^{2}-4 x-2 y+2}{y-x+3} \geqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area. | Answer: 4.
Solution. Consider the first inequality. To open the absolute values, we consider three possible cases.
1) $x<0$. Then $-x-x+4 \leqslant 4 \Leftrightarrow -2x \leqslant 0 \Leftrightarrow x \geqslant 0$, i.e., there are no solutions in this case.
2) $0 \leqslant x \leqslant 4$. Then $x-x+4 \leqslant 4 \Left... | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. There are 306 different cards with numbers $3,19,3^{2}, 19^{2}, \ldots, 3^{153}, 19^{153}$ (each card has exactly one number, and each number appears exactly once). In how many ways can 2 cards be chosen so that the product of the numbers on the chosen cards is a square of an integer? | Answer: 17328.
Solution. To obtain the square of a natural number, it is necessary and sufficient that each factor enters the prime factorization of the number in an even power.
Suppose two cards with powers of three are chosen. We have 76 even exponents $(2,4,6, \ldots, 152)$ and 77 odd exponents $(1,3,5, \ldots, 15... | 17328 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
x-y \geqslant|x+y| \\
\frac{x^{2}-6 x+y^{2}-8 y}{3 y-x+6} \geqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area. | Answer: 3.
Solution. The first inequality is equivalent to the system ${ }^{1}$ $\left\{\begin{array}{l}x+y \leqslant x-y, \\ x+y \geqslant y-x\end{array} \Leftrightarrow\left\{\begin{array}{l}y \leqslant 0, \\ x \geqslant 0 .\end{array}\right.\right.$
Consider the second inequality. It can be written as $\frac{(x-3)... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. There are 294 different cards with numbers $7, 11, 7^{2}, 11^{2}, \ldots, 7^{147}, 11^{147}$ (each card has exactly one number, and each number appears exactly once). In how many ways can 2 cards be chosen so that the product of the numbers on the chosen cards is a square of an integer? | Answer: 15987.
Solution. To obtain the square of a natural number, it is necessary and sufficient that each factor enters the prime factorization of the number in an even power.
Suppose two cards with powers of seven are chosen. We have 73 even exponents $(2,4,6, \ldots, 146)$ and 74 odd exponents $(1,3,5, \ldots, 14... | 15987 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
y+x \geqslant|x-y| \\
\frac{x^{2}-8 x+y^{2}+6 y}{x+2 y-8} \leqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area. | Answer: 8.
Solution. The first inequality is equivalent to the system $\left\{\begin{array}{l}x-y \leqslant x+y, \\ x-y \geqslant-x-y\end{array} \Leftrightarrow\left\{\begin{array}{l}x \geqslant 0, \\ y \geqslant 0 .\end{array}\right.\right.$.
Consider the second inequality. It can be written as $\frac{(x-4)^{2}+(y+3... | 8 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. Find all values of $p$, for each of which the numbers $p-2$, $2 \cdot \sqrt{p}$, and $-3-p$ are respectively the first, second, and third terms of some geometric progression. | Answer: $p=1$.
Solution. For the given numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(2 \sqrt{p})^{2}=(p-2)(-p-3)$, from which
$$
\left\{\begin{array}{l}
p>0 \\
p^{2}+5 p-6=0
\end{array} \Leftrightarrow p=1\right.
$$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In the number $2016 * * * * 02 * *$, each of the 6 asterisks needs to be replaced with any of the digits $0,2,4,5,7,9$ (digits can be repeated) so that the resulting 12-digit number is divisible by 15. In how many ways can this be done? | Answer: 5184.
Solution. For a number to be divisible by 15, it is necessary and sufficient that it is divisible by 5 and by 3. To ensure divisibility by 5, we can choose 0 or 5 as the last digit from the available options (2 ways).
To ensure divisibility by three, we proceed as follows. We will choose four digits arb... | 5184 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find all values of $p$, for each of which the numbers $-p-12, 2 \cdot \sqrt{p}$, and $p-5$ are respectively the first, second, and third terms of some geometric progression. | Answer: $p=4$.
Solution. For the specified numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(2 \sqrt{p})^{2}=(-p-12)(p-5)$, from which
$$
\left\{\begin{array}{l}
p>0, \\
p^{2}+11 p-60=0
\end{array} \Leftrightarrow p=4\right.
$$ | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In the number $2016^{* * * *} 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,6,7,8$ (digits can be repeated) so that the resulting 11-digit number is divisible by 6. In how many ways can this be done | Answer: 2160.
Solution. For a number to be divisible by 6, it is necessary and sufficient that it is divisible by 2 and by 3. To ensure divisibility by 2, we can choose the last digit from the available options as $0, 2, 4, 6, 8$ (5 ways).
To ensure divisibility by three, we proceed as follows. Choose three digits ar... | 2160 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find all values of $p$, for each of which the numbers $p-2$, $3 \cdot \sqrt{p}$, and $-8-p$ are respectively the first, second, and third terms of some geometric progression. | Answer: $p=1$.
Solution. For the given numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(3 \sqrt{p})^{2}=(p-2)(-8-p)$, from which
$$
\left\{\begin{array}{l}
p>0 \\
p^{2}+15 p-16=0
\end{array} \Leftrightarrow p=1\right.
$$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In the number $2016 * * * * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,5,7,9$ (digits can be repeated) so that the resulting 11-digit number is divisible by 15. In how many ways can this be done? | # Answer: 864.
Solution. For a number to be divisible by 15, it is necessary and sufficient that it is divisible by 5 and by 3. To ensure divisibility by 5, we can choose 0 or 5 as the last digit from the available options (2 ways).
To ensure divisibility by three, we proceed as follows. We will choose three digits a... | 864 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find all values of $p$, for each of which the numbers $-p-8$, $3 \cdot \sqrt{p}$, and $p-7$ are the first, second, and third terms, respectively, of some geometric progression. | Answer: $p=4$.
Solution. For the specified numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(3 \sqrt{p})^{2}=(-p-8)(p-7)$, from which
$$
\left\{\begin{array}{l}
p>0, \\
p^{2}+10 p-56=0
\end{array} \Leftrightarrow p=4\right.
$$ | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In the number $2016 * * * * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,7,8,9$ (digits can be repeated) so that the resulting 11-digit number is divisible by 6. In how many ways can this be done? | Answer: 1728.
Solution. For a number to be divisible by 6, it is necessary and sufficient that it is divisible by 2 and by 3. To ensure divisibility by 2, we can choose the last digit from the available options as $0, 2, 4, 8$ (4 ways).
To ensure divisibility by 3, we proceed as follows. Select three digits arbitrari... | 1728 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. On the plane $(x ; y)$, plot the set of points satisfying the equation $|3 x|+|4 y|+|48-3 x-4 y|=48$, and find the area of the resulting figure.
# | # Answer: 96.
Solution. Note that the equality $|a|+|b|+|c|=a+b+c$ holds if and only if the numbers $a, b$, and $c$ are non-negative (since if at least one of them is negative, the left side is greater than the right). Therefore, the first equation is equivalent to the system of inequalities
$$
\left\{\begin{array} {... | 96 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that $\operatorname{tg}(\alpha+2 \gamma)+2 \operatorname{tg} \alpha-4 \operatorname{tg}(2 \gamma)=0, \operatorname{tg} \gamma=\frac{1}{3}$. Find $\operatorname{ctg} \alpha$. | Answer: 2 or $\frac{1}{3}$.
Solution. $\operatorname{tg} 2 \gamma=\frac{2 \operatorname{tg} \gamma}{1-\operatorname{tg}^{2} \gamma}=\frac{2 / 3}{1-1 / 9}=\frac{3}{4}$. Then the given equality can be transformed as follows:
$$
\frac{\operatorname{tg} \alpha+\operatorname{tg} 2 \gamma}{1-\operatorname{tg} \alpha \opera... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. On the plane $(x ; y)$, plot the set of points satisfying the equation $|5 x|+|12 y|+|60-5 x-12 y|=60$, and find the area of the resulting figure.
# | # Answer: 30.
Solution. Note that the equality $|a|+|b|+|c|=a+b+c$ holds if and only if the numbers $a, b$, and $c$ are non-negative (since if at least one of them is negative, the left side is greater than the right). Therefore, the first equation is equivalent to the system of inequalities
$$
\left\{\begin{array} {... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that $\operatorname{tg}(2 \alpha-\beta)-4 \operatorname{tg} 2 \alpha+4 \operatorname{tg} \beta=0, \operatorname{tg} \alpha=-3$. Find $\operatorname{ctg} \beta$. | Answer: -1 or $\frac{4}{3}$.
Solution. $\operatorname{tg} 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha}=\frac{-6}{1-9}=\frac{3}{4}$. Then the given equality can be transformed as follows:
$$
\frac{\operatorname{tg} 2 \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} 2 \alpha \operatorna... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. [4 points] Find the number of eight-digit numbers, the product of whose digits equals 1400. The answer should be presented as an integer. | Answer: 5880.
Solution. Since $1400=7 \cdot 2^{3} \cdot 5^{2}$, the sought numbers can consist of the following digits: (a) three twos, two fives, one seven, and two ones, (b) one four, one two, two fives, one seven, and three ones, or (c) one eight, two fives, one seven, and four ones. We will calculate the number of... | 5880 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. [5 points] Two circles of the same radius 9 intersect at points $A$ and $B$. On the first circle, a point $C$ is chosen, and on the second circle, a point $D$ is chosen. It turns out that point $B$ lies on the segment $C D$, and $\angle C A D=90^{\circ}$. On the perpendicular to $C D$ passing through point $B$, a po... | Answer: 18.
Solution. Let $R=9$ be the radii of the circles given in the condition, $\angle B A D=\alpha, \angle B C F=\beta$. Then $\angle B A C=\frac{\pi}{2}-\alpha$, and by the sine rule $B D=2 R \sin \alpha, B C=2 R \sin \left(\frac{\pi}{2}-\alpha\right)=2 R \cos \alpha$. Therefore, $C F^{2}=B C^{2}+B D^{2}=4 R^{2... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. [4 points] Find the number of eight-digit numbers, the product of whose digits equals 7000. The answer should be presented as an integer. | Answer: 5600.
Solution. Since $7000=7 \cdot 2^{3} \cdot 5^{3}$, the sought numbers can consist of the following digits: (a) three twos, three fives, one seven, and one one, (b) four, two, three fives, one seven, and two ones, or (c) eight, three fives, one seven, and three ones. We will calculate the number of variant... | 5600 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. [5 points] Two circles of the same radius 7 intersect at points $A$ and $B$. On the first circle, a point $C$ is chosen, and on the second circle, a point $D$ is chosen. It turns out that point $B$ lies on the segment $C D$, and $\angle C A D=90^{\circ}$. On the perpendicular to $C D$ passing through point $B$, a po... | Answer: 14.
Solution. Let $R=7$ be the radii of the circles given in the condition, $\angle B A D=\alpha, \angle B C F=\beta$. Then $\angle B A C=\frac{\pi}{2}-\alpha$, and by the sine rule $B D=2 R \sin \alpha, B C=2 R \sin \left(\frac{\pi}{2}-\alpha\right)=2 R \cos \alpha$. Therefore, $C F^{2}=B C^{2}+B D^{2}=4 R^{2... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] On a plane with a given rectangular Cartesian coordinate system, a square is drawn with vertices at points $(0 ; 0),(0 ; 59),(59 ; 59)$, and $(59 ; 0)$. Find the number of ways to choose two grid nodes inside this square (not including its boundary) such that at least one of these nodes lies on one of the... | Answer: 370330
Solution. There are two possible cases.
1) Both selected nodes lie on the lines specified in the condition. Each of them contains 58 points inside the square, and there are no repetitions among them (the intersection point of the lines has non-integer coordinates). There are 116 ways to choose the firs... | 370330 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. [5 points] Point $D$ lies on side $A C$ of triangle $A B C$. The circle with diameter $B D$ intersects sides $A B$ and $B C$ at points $P$ and $T$ respectively. Points $M$ and $N$ are the midpoints of segments $A D$ and $C D$ respectively. It is known that $P M \| T N$.
a) Find the angle $A B C$.
b) Suppose additi... | Answer: (a) $90^{\circ} ;$ ( $\mathbf{\text { ( ) }} 5$.
Solution. (a) Points $P$ and $T$ lie on the circle with diameter $B D$, therefore $\angle B P D=\angle B T D=90^{\circ}$. Consequently, triangles $A D P$ and $D C T$ are right-angled; $P M$ and $T N$ are their medians. Since the median of a right-angled triangle... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] On a plane with a given rectangular Cartesian coordinate system, a square is drawn with vertices at points $(0 ; 0),(0 ; 69),(69 ; 69)$, and ( $69 ; 0)$. Find the number of ways to choose two grid nodes inside this square (not including its boundary) such that at least one of these nodes lies on one of th... | Answer: 601460.
Solution. There are two possible cases.
1) Both selected nodes lie on the specified lines. There are 68 points on each of them inside the square, and there are no duplicates among them (the intersection point of the lines has non-integer coordinates). There are 136 ways to choose the first point, and ... | 601460 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] On a plane with a given rectangular Cartesian coordinate system, a square is drawn with vertices at points $(0 ; 0),(0 ; 65),(65 ; 65)$ and ( $65 ; 0)$. Find the number of ways to choose two grid nodes inside this square (not including its boundary) such that at least one of these nodes lies on one of the... | # Answer: 500032.
Solution. There are two possible cases.
1) Both selected nodes lie on the lines specified in the condition. There are 64 points on each of them inside the square, and there are no repetitions among them (the intersection point of the lines has non-integer coordinates). There are 128 ways to choose t... | 500032 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. [6 points] Solve the equation $\sqrt{x+1}-\sqrt{4-x}+3=2 \sqrt{4+3 x-x^{2}}$. | Answer: $3, \frac{3-2 \sqrt{6}}{2}$.
Solution. Let $\sqrt{x+1}-\sqrt{4-x}=t$. Squaring both sides of this equation, we get $(x+1)-2 \sqrt{(x+1)(4-x)}+(4-x)=t^{2}$, from which $2 \sqrt{4+3 x-x^{2}}=5-t^{2}$. The equation becomes $t+3=5-t^{2}$; hence $t^{2}+t-2=0$, i.e., $t=1$ or $t=-2$. We consider each case separately... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] On a plane with a given rectangular Cartesian coordinate system, a square is drawn with vertices at points $(0 ; 0),(0 ; 63),(63 ; 63)$, and $(63 ; 0)$. Find the number of ways to choose two grid nodes inside this square (not including its boundary) such that at least one of these nodes lies on one of the... | Answer: 453902.
Solution. There are two possible cases.
1) Both selected nodes lie on the specified lines. There are 62 points on each of them inside the square, and there are no repetitions among them (the intersection point of the lines has non-integer coordinates). There are 124 ways to choose the first point, and... | 453902 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On the sides of triangle $A B C$, points were marked: 10 - on side $A B$, 11 - on side $B C$, 12 - on side $A C$. At the same time, none of the vertices of the triangle were marked. How many triangles exist with vertices at the marked points? | Answer: 4951.
Solution. Three points out of the 33 given can be chosen in $C_{33}^{3}=5456$ ways. In this case, a triangle is formed in all cases except when all three points lie on one side of the triangle. Thus, $C_{12}^{3}+C_{11}^{3}+C_{10}^{3}=220+165+120=505$ ways do not work. Therefore, there are $5456-505=4951$... | 4951 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. On the coordinate plane, squares are considered, all vertices of which have integer non-negative coordinates, and the center is located at the point $(50 ; 30)$. Find the number of such squares. | Answer: 930.
Solution. Draw through the given point $(50 ; 30)$ vertical and horizontal lines $(x=50$ and $y=30)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "lower" vertex of the square can be located in 30 ways: $(... | 930 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. On the coordinate plane, squares are considered, all vertices of which have natural coordinates, and the center is located at the point $(55 ; 25)$. Find the number of such squares. | Answer: 600.
Solution. Draw through the given point $(55 ; 25)$ vertical and horizontal lines $(x=55$ and $y=25)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "lower" vertex of the square can be located in 24 ways: $(... | 600 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of pairs of integers $(a ; b)$ such that $1 \leqslant a \leqslant 70, 1 \leqslant b \leqslant 50$, and the area $S$ of the figure defined by the system of inequalities
$$
\left\{\begin{array}{l}
\frac{x}{a}+\frac{y}{b} \geqslant 1 \\
x \leqslant a \\
y \leqslant b
\end{array}\right.
$$
is such that... | Answer: 1260.
Solution. The given system of inequalities defines a triangle on the plane with vertices $(a ; 0),(0 ; b)$, and $(a ; b)$. This triangle is right-angled, and its doubled area is equal to the product of the legs, i.e., $a b$. According to the condition, $a b: 5$, so one of the numbers $a$ or $b$ must be d... | 1260 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. Plot the figure $\Phi$ on the plane, consisting of points $(x ; y)$ of the coordinate plane such that the system of inequalities is satisfied
$$
\left\{\begin{array}{l}
\sqrt{x^{2}-3 y^{2}+4 x+4} \leqslant 2 x+1 \\
x^{2}+y^{2} \leqslant 4
\end{array}\right.
$$
Determine how many parts the figure $\Phi$ consists of... | Solution. The first inequality is equivalent to the system of inequalities
$$
\left\{\begin{array} { l }
{ x ^ { 2 } - 3 y ^ { 2 } + 4 x + 4 \leqslant 4 x ^ { 2 } + 4 x + 1 , } \\
{ ( x + 2 ) ^ { 2 } - 3 y ^ { 2 } \geqslant 0 , } \\
{ 2 x + 1 \geqslant 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}+y^{... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of pairs of integers $(a ; b)$ such that $1 \leqslant a \leqslant 80,1 \leqslant b \leqslant 30$, and the area $S$ of the figure defined by the system of inequalities
$$
\left\{\begin{array}{l}
\frac{x}{a}+\frac{y}{b} \geqslant 1 \\
x \leqslant a \\
y \leqslant b
\end{array}\right.
$$
is such that ... | Answer: 864.
Solution. The given system of inequalities defines a triangle on the plane with vertices $(a ; 0),(0 ; b)$, and $(a ; b)$. This triangle is right-angled, and its doubled area is equal to the product of the legs, i.e., $a b$. According to the condition, $a b: 5$, so one of the numbers $a$ or $b$ must be di... | 864 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. Plot the figure $\Phi$ on the plane, consisting of points $(x ; y)$ of the coordinate plane such that the system of inequalities is satisfied
$$
\left\{\begin{array}{l}
\sqrt{y^{2}-8 x^{2}-6 y+9} \leqslant 3 y-1 \\
x^{2}+y^{2} \leqslant 9
\end{array}\right.
$$
Determine how many parts the figure $\Phi$ consists of... | Solution. The first inequality is equivalent to the system of inequalities
$$
\left\{\begin{array} { l }
{ y ^ { 2 } - 8 x ^ { 2 } - 6 y + 9 \leqslant 9 y ^ { 2 } - 6 y + 1 , } \\
{ ( y - 3 ) ^ { 2 } - 8 x ^ { 2 } \geqslant 0 , } \\
{ 3 y - 1 \geqslant 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}+y^{... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
10.2. There are 90 cards - 10 with the digit 1, 10 with the digit $2, \ldots, 10$ with the digit 9. From all these cards, two numbers were formed, one of which is three times the other. Prove that one of these numbers can be factored into four not necessarily distinct natural factors, greater than one. | Solution. Let these numbers be $A$ and $B=3A$. Then the sum of the digits of the number $B$ is divisible by 3. But the sum of the digits on all cards is divisible by 9 (and therefore by 3), so the sum of the digits of the number $A$ is divisible by 3. This means that the number $A$ is divisible by 3. But then the numbe... | 40 | Number Theory | proof | Yes | Yes | olympiads | false |
10.4. Quadrilateral $A B C D$ is inscribed in a circle. The perpendicular to side $B C$, drawn through its midpoint - point $M$, intersects side $A B$ at point $K$. The circle with diameter $K C$ intersects segment $C D$ at point $P(P \neq C)$. Find the angle between the lines $M P$ and $A D$. | Answer: $90^{\circ}$.
Solution. We will prove that the lines $M P$ and $A D$ are perpendicular. Let $\omega$ be the circle constructed with $K C$ as its diameter, then point $M$ lies on $\omega$, since angle $C M K$ is a right angle. Therefore, $\angle C P M = \angle C K M = \alpha$ (they subtend the arc $C M$ of circ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
30. It is known that for pairwise distinct numbers $a, b, c$, the equality param1 holds. What is the smallest value that the expression $a+b+c$ can take?
| param1 | Answer |
| :---: | :---: |
| $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)+2\left(a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)\right)=0$ | -2 |
| $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b... | 30. It is known that for pairwise distinct numbers $a, b, c$, the equality param1 holds. What is the smallest value that the expression $a+b+c$ can take?
| param1 | Answer |
| :---: | :---: |
| $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)+2\left(a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)\right)=0$ | -2 |
| $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b... | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. For each natural $n$, which is not a perfect square, all values of the variable $x$ are calculated for which both numbers $x+\sqrt{n}$ and $x^{3}+$ param1 $\sqrt{n}$ are integers. Find the total number of such values of $x$.
| param1 | answer |
| :---: | :---: |
| 1524 | |
| 1372 | |
| 1228 | |
| 1092 | |
| 964... | 8. For each natural $n$, which is not a perfect square, all values of the variable $x$ are calculated for which both numbers $x+\sqrt{n}$ and $x^{3}+$ param1 $\sqrt{n}$ are integers. Find the total number of such values of $x$.
| param1 | answer |
| :---: | :---: |
| 1524 | 39 |
| 1372 | 33 |
| 1228 | 35 |
| 1092 | 27... | 33 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. A regular param 1 -sided polygon is inscribed in a circle, with different natural numbers written at its vertices. A pair of non-adjacent vertices of the polygon $A$ and $B$ is called interesting if, on at least one of the two arcs $A B$, all the numbers written at the vertices of the arc are greater than the number... | 9. A regular param 1-gon is inscribed in a circle, with different natural numbers written at its vertices. A pair of non-adjacent vertices of the polygon $A$ and $B$ is called interesting if, on at least one of the two arcs $A B$, all the numbers written at the vertices of the arc are greater than the numbers written a... | 92 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. (12 points) The sequence of functions is defined by the formulas:
$$
f_{0}(x)=2 \sqrt{x}, f_{n+1}(x)=\frac{4}{2-f_{n}(x)}, n=0,1,2 \ldots, x \in[4 ; 9]
$$
Find $f_{2023}(4)$. | Answer: -2.
Solution. It is easy to calculate that $f_{3}(x)=f_{0}(x)$, therefore
$$
f_{2023}(x)=f_{1}(x)=\frac{2}{1-\sqrt{x}}
$$
Then $f_{2023}(4)=-2$.
Remark. One can immediately compute the values of the functions at the given point. The sequence will be $f_{0}(4)=4, f_{1}(4)=-2, f_{2}(4)=1, f_{3}(4)=4 \ldots$
... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (15 points) The vertices of the broken line $A B C D E F G$ have coordinates $A(-1 ; -7), B(2 ; 5), C(3 ; -8), D(-3 ; 4), E(5 ; -1), F(-4 ; -2), G(6 ; 4)$.
Find the sum of the angles with vertices at points $B, E, C, F, D$. | Answer: $135^{\circ}$.
Solution. The closed broken line $B C D E F B$ forms a five-pointed "star". The sum of the angles at the rays of this star is $180^{\circ}$.
We will prove that the sum of the angles at the rays of any five-pointed star $B C D E F B$ is $180^{\circ}$. Let $O$ be the point of intersection of the ... | 135 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (12 points) The sequence of functions is defined by the formulas:
$$
f_{0}(x)=2 \sqrt{x}, f_{n+1}(x)=\frac{4}{2-f_{n}(x)}, n=0,1,2 \ldots, x \in[4 ; 9]
$$
Find $f_{2023}(9)$. | Answer: -1.
Solution. It is easy to calculate that $f_{3}(x)=f_{0}(x)$, therefore,
$$
f_{2023}(x)=f_{1}(x)=\frac{2}{1-\sqrt{x}}
$$
Then $f_{2023}(9)=-1$.
Remark. One can immediately compute the values of the functions at the given point. The sequence will be $f_{0}(9)=6, f_{1}(9)=-1, f_{2}(9)=\frac{4}{3}, f_{3}(9)=... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (15 points) The vertices of the broken line $A B C D E F G$ have coordinates $A(0 ; -5), B(3 ; 7), C(4 ; -6), D(-2 ; 6), E(6 ; 1), F(-3 ; 0), G(7 ; 6)$.
Find the sum of the angles with vertices at points $B, E, C, F, D$. | Answer: $135^{\circ}$.
Solution. The closed broken line $B C D E F B$ forms a five-pointed "star". The sum of the angles at the rays of this star is $180^{\circ}$.
We will prove that the sum of the angles at the rays of any five-pointed star $B C D E F B$ is $180^{\circ}$. Let $O$ be the point of intersection of the ... | 135 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Nils has a goose farm. Nils calculated that if he sells 75 geese, the feed will last 20 days longer than if he doesn't sell any. If he buys an additional 100 geese, the feed will run out 15 days earlier than if he doesn't make such a purchase. How many geese does Nils have? | Answer: 300.
Solution. Let $A$ be the total amount of feed (in kg), $x$ be the amount of feed per goose per day (in kg), $n$ be the number of geese, and $k$ be the number of days the feed will last. Then
$$
\begin{gathered}
A=k x n=(k+20) x(n-75)=(k-15) x(n+100) \\
k n=(k+20)(n-75)=(k-15)(n+100)
\end{gathered}
$$
So... | 300 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Nils has a goose farm. Nils calculated that if he sells 50 geese, the feed will last 20 days longer than if he doesn't sell any. If he buys an additional 100 geese, the feed will run out 10 days earlier than if he doesn't make such a purchase. How many geese does Nils have? | Answer: 300.
Solution. Let $A$ be the total amount of feed (in kg), $x$ be the amount of feed per goose per day (in kg), $n$ be the number of geese, and $k$ be the number of days the feed will last. Then
$$
\begin{gathered}
A=k x n=(k+20) x(n-50)=(k-10) x(n+100) \\
k n=(k+20)(n-50)=(k-10)(n+100)
\end{gathered}
$$
So... | 300 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Solve the equation
$$
2 x+2+\operatorname{arctg} x \cdot \sqrt{x^{2}+1}+\operatorname{arctg}(x+2) \cdot \sqrt{x^{2}+4 x+5}=0
$$ | Answer: -1.
Solution. Let $f(x)=x+\operatorname{arctg} x \cdot \sqrt{x^{2}+1}$. The original equation can be rewritten as $f(x)+f(x+2)=0$. Note that the function $f(x)$ is odd. It is increasing on the positive half-axis (as the sum of increasing functions). Due to its oddness, it is increasing on the entire real line.... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem No. 6 (10 points)
The density of a body is defined as the ratio of its mass to the volume it occupies. A homogeneous cube with a volume of \( V = 8 \, \text{m}^3 \) is given. As a result of heating, each of its edges increased by 4 mm. By what percentage did the density of this cube change?
Answer: decrease... | # Solution and evaluation criteria:
Volume of the cube: $v=a^{3}$, where $a$ is the length of the edge, therefore:
$a=2 \partial m=200 mm$.
Final edge length: $a_{\kappa}=204$ mm.
Thus, the final volume: $V_{\kappa}=a_{K}^{3}=2.04^{3}=8.489664 \partial \mu^{3} \approx 1.06 V$.
Therefore, the density:
$\rho_{K}=\f... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem № 7 (10 points)
A person and his faithful dog started moving along the perimeter of a block from point A at the same time $t_{0}=0$ min. The person moved with a constant speed clockwise, while the dog ran with a constant speed counterclockwise (see fig.). It is known that they met for the first time after $t... | # Answer: in 9 min
## Solution and evaluation criteria:
In $t_{1}=1$ min, the person and the dog together covered a distance equal to the perimeter of the block, with the person moving 100 meters from the starting point of the journey.
That is, during each subsequent meeting, the person will be 100 meters away from ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem No. 6 (10 points)
The density of a body is defined as the ratio of its mass to the volume it occupies. A homogeneous cube with a volume of \( V = 27 \partial \mu^{3} \) is given. As a result of heating, each of its edges increased by 9 mm. By what percentage did the density of this cube change? | # Answer: decreased by $8 \%$
## Solution and evaluation criteria:
Volume of a cube: $v=a^{3}$, where $a$ is the length of the edge, therefore:
$a=3 dm=300 mm$.
Final edge length: $a_{K}=309 mm$.
Thus, the final volume: $V_{K}=a_{K}^{3}=3.09^{3}=29.503629 dm^{3} \approx 1.09 V$.
Therefore, the density:
$\rho_{K}... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem № 7 (10 points)
A person and his faithful dog started moving along the perimeter of a block from point $A$ simultaneously at time $t_{0}=0$ min. The person moved with a constant speed clockwise, while the dog ran with a constant speed counterclockwise (see figure). It is known that they met for the first tim... | # Answer: in 14 min
## Solution and evaluation criteria:
In $t_{1}=2$ min, the person and the dog together covered a distance equal to the perimeter of the block, with the person moving 100 meters from the starting point of the journey.
That is, during each subsequent meeting, the person will be 100 meters away from... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem № 8 (15 points)
In some English-speaking countries, temperature is measured in degrees Fahrenheit. An English schoolboy, observing a thermometer in a glass of cooling water, noticed that it cooled by $30^{\circ} \mathrm{F}$. He became curious about how much heat was released. In books, he found the following... | # Answer: 140 kJ
## Solution and grading criteria:
The change in temperature in degrees Fahrenheit is related to the change in temperature in degrees Celsius:
That is, in degrees Celsius,... | 140 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. 20 balls of the same mass are moving along a chute towards a metal wall with the same speed. Coming towards them at the same speed are 16 balls of the same mass. When two balls collide, they fly apart with the same speed. After colliding with the wall, a ball bounces off it with the same speed. (The balls move only ... | # Answer: 510.
Solution. We will assume that initially, each ball moving towards the wall has a red flag, and the rest of the balls have blue flags. Imagine that when the balls collide, they exchange flags. Then each blue flag moves at a constant speed in one direction (away from the wall), and each red flag reaches t... | 510 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Andrei was driving to the airport of a neighboring city. After an hour of driving at a speed of 60 km/h, he realized that if he did not change his speed, he would be 20 minutes late. Then he sharply increased his speed, as a result of which he covered the remaining part of the journey at an average speed of 90 km/h ... | Answer: 180 km.
Solution. Let the distance from Andrey's house to the airport be $s$ km, and the time he intended to spend on the road be $1+t$ hours. Then
$$
s=60+60\left(t+\frac{1}{3}\right)=60+90\left(t-\frac{1}{3}\right)
$$
From this,
$$
60 t+20=90 t-30, \quad t=\frac{5}{3}, \quad s=180
$$
Evaluation. Full sco... | 180 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A natural number $n$ is such that the number $36 n^{2}$ has exactly 51 different natural divisors. How many natural divisors does the number $5 n$ have? | Answer: 16.
Solution. Let $m=p_{1}^{k_{1}} \cdot p_{2}^{k_{2}} \cdot \ldots \cdot p_{l}^{k_{l}}$, where $p_{1}, p_{2}, \ldots, p_{l}$ are pairwise distinct prime numbers. Then the number of natural divisors of the number $m$ is
$$
\tau(m)=\left(k_{1}+1\right)\left(k_{2}+1\right) \ldots\left(k_{l}+1\right)
$$
Indeed,... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Viktor was driving to the airport of a neighboring city. After half an hour of driving at a speed of 60 km/h, he realized that if he did not change his speed, he would be 15 minutes late. Then he increased his speed, as a result of which he covered the remaining part of the journey at an average speed of 80 km/h and... | Answer: 150 km.
Solution. Let the distance from Viktor's house to the airport be $s$ km, and the time he intended to spend on the road $\frac{1}{2}+t$ hours. Then
$$
s=30+60\left(t+\frac{1}{4}\right)=30+80\left(t-\frac{1}{4}\right)
$$
From this,
$$
60 t+15=80 t-20, \quad t=\frac{7}{4}, \quad s=150
$$
Evaluation. F... | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A natural number $n$ is such that the number $100 n^{2}$ has exactly 55 different natural divisors. How many natural divisors does the number 10n have? | Answer: 18.
Solution. Let $m=p_{1}^{k_{1}} \cdot p_{2}^{k_{2}} \cdot \ldots \cdot p_{l}^{k_{l}}$, where $p_{1}, p_{2}, \ldots, p_{l}$ are pairwise distinct prime numbers. Then the number of natural divisors of the number $m$ is
$$
\tau(m)=\left(k_{1}+1\right)\left(k_{2}+1\right) \ldots\left(k_{l}+1\right)
$$
Indeed,... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. (17 points) The medians drawn from vertices $A$ and $B$ of triangle $ABC$ are perpendicular to each other. Find the area of the square with side $AB$, if $BC=28, AC=44$.
# | # Answer: 544
Solution. Let $D$ be the midpoint of $B C$, $E$ be the midpoint of $A C$, and $M$ be the point of intersection of the medians. Let $M D=a$, $M E=b$. Then $A M=2 a$, $B M=2 b$. From the right triangles $B M D$ and $A M E$, we have $a^{2}+4 b^{2}=B D^{2}=14^{2}$ and $4 a^{2}+b^{2}=A E^{2}=22^{2}$. Adding t... | 544 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (17 points) In a $4 \times 5$ grid, 5 crosses need to be placed such that each row and each column contains at least one cross. How many ways can this be done? | # Answer: 240
Solution. From the condition, it follows that in some row, two cells are marked (while in the others, only one each). This row can be chosen in 4 ways, and the two crosses in it can be chosen in $5 \cdot 4 / 2=10$ ways. The remaining three crosses can be chosen in $3 \cdot 2 \cdot 1=6$ ways. By the rule ... | 240 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (16 points) Fresh mushrooms contain $80 \%$ water by mass, while dried mushrooms contain $20 \%$ water. How many kg of dried mushrooms can be obtained from 20 kg of fresh mushrooms? | # Answer: 5 kg
Solution. The dry matter in fresh mushrooms is 4 kg, which constitutes $80 \%$ of the weight in dried mushrooms. By setting up the corresponding proportion, we will find the weight of the dried mushrooms. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (17 points) The medians drawn from vertices $A$ and $B$ of triangle $ABC$ are perpendicular to each other. Find the area of the square with side $AB$, if $BC=36, AC=48$.
# | # Answer: 720
Solution. Let $D$ be the midpoint of $B C$, $E$ be the midpoint of $A C$, and $M$ be the point of intersection of the medians. Let $M D=a$, $M E=b$. Then $A M=2 a$, $B M=2 b$. From the right triangles $B M D$ and $A M E$, we have $a^{2}+4 b^{2}=B D^{2}=18^{2}$ and $4 a^{2}+b^{2}=A E^{2}=24^{2}$. Adding t... | 720 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (17 points) In a $3 \times 4$ grid, 4 crosses need to be placed such that each row and each column contains at least one cross. How many ways can this be done? | # Answer: 36
Solution. From the condition, it follows that in some row, two cells are marked (while in the others, only one each). This row can be chosen in 3 ways, and the two crosses in it can be placed in $4 \cdot 3 / 2=6$ ways. The remaining two crosses can be chosen in 2 ways. By the multiplication rule, the tota... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) A ball was thrown from the surface of the Earth at an angle of $30^{\circ}$ with a speed of $v_{0}=20 \mathrm{M} / \mathrm{c}$. How long will it take for the velocity vector of the ball to turn by an angle of $60^{\circ}$? Neglect air resistance. The acceleration due to gravity is $g=10 \mathrm{M} / \mat... | Answer: $t=2 s$
Solution. In fact, the problem requires finding the total flight time of the ball. The $y$-coordinate of the ball changes according to the law: $y=v_{0 y} t-\frac{g t^{2}}{2}=v_{0} \sin 30^{\circ} t-\frac{g t^{2}}{2}=10 \cdot t-5 t^{2}=0$
From this, we obtain that: $t=2 s$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. In a row, the numbers $1,2,3, \ldots, 2014,2015$ are written. We will call a number from this row good if, after its removal, the sum of all the remaining 2014 numbers is divisible by 2016. Find all the good numbers. | Answer: The only good number is 1008.
Solution. The remainder of the division of the sum of all 2015 numbers by 2016 is 1008:
$(1+2015)+(2+2014)+\ldots+(1007+1009)+1008=2016 \cdot 1007+1008$.
Therefore, only 1008 can be crossed out.
Scoring. 12 points for a correct solution. If it is shown that 1008 is a good numbe... | 1008 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Nils has a goose farm. Nils calculated that if he sells 75 geese, the feed will last 20 days longer than if he doesn't sell any. If he buys an additional 100 geese, the feed will run out 15 days earlier than if he doesn't make such a purchase. How many geese does Nils have? | Answer: 300.
Solution. Let $A$ be the total amount of feed (in kg), $x$ be the amount of feed per goose per day (in kg), $n$ be the number of geese, and $k$ be the number of days the feed will last. Then
$$
\begin{gathered}
A=k x n=(k+20) x(n-75)=(k-15) x(n+100) \\
k n=(k+20)(n-75)=(k-15)(n+100)
\end{gathered}
$$
So... | 300 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Nils has a goose farm. Nils calculated that if he sells 50 geese, the feed will last 20 days longer than if he doesn't sell any. If he buys an additional 100 geese, the feed will run out 10 days earlier than if he doesn't make such a purchase. How many geese does Nils have? | Answer: 300.
Solution. Let $A$ be the total amount of feed (in kg), $x$ be the amount of feed per goose per day (in kg), $n$ be the number of geese, and $k$ be the number of days the feed will last. Then
$$
\begin{gathered}
A=k x n=(k+20) x(n-50)=(k-10) x(n+100) \\
k n=(k+20)(n-50)=(k-10)(n+100)
\end{gathered}
$$
So... | 300 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Let
$$
\sqrt{49-a^{2}}-\sqrt{25-a^{2}}=3
$$
Calculate the value of the expression
$$
\sqrt{49-a^{2}}+\sqrt{25-a^{2}} .
$$ | Answer: 8.
Solution. Let
$$
\sqrt{49-a^{2}}+\sqrt{25-a^{2}}=x
$$
Multiplying this equality by the original one, we get $24=3x$.
Evaluation. Full solution: 11 points. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (10 points) Five identical non-ideal ammeters are connected as shown in the figure. Ideal power supply is connected to points $A$ and $B$. Determine the sum of the readings of all ammeters, given that the reading of the first ammeter $I_{1}=2 \mathrm{MA}$.
$ is defined by the following relations: $a_{1}=1, a_{2}=2, a_{n}=a_{n-1}-a_{n-2}+n$ (for $n \geqslant 3$). Find $a_{2019}$. | Answer: 2020.
Solution. Let's list the first terms of the sequence:
$$
1,2,4,6,7,7,7,8,10,12,13,13,13,14,16,18,19,19,19,20,21,23, \ldots
$$
We can observe a pattern: $a_{n+6}=a_{n}+6$. Let's prove it. We have
$a_{n+1}=a_{n}-a_{n-1}+n+1=\left(a_{n-1}-a_{n-2}+n\right)-a_{n-1}+n+1=-a_{n-2}+2 n+1$.
By substituting $n$... | 2020 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The infantry column stretched out over 1 km. Sergeant Kim, riding out on a gyro-scooter from the end of the column, reached its beginning and returned to the end. The infantrymen walked 2 km 400 m during this time. How far did the sergeant travel during this time? | Answer: 3 km $600 \mathrm{~m}$
Solution. Let the speed of the column be $x$ km/h, and the sergeant travels $k$ times faster, i.e., at a speed of $k x$ km/h. To reach the end of the column, Kim traveled $t_{1}=\frac{1}{k x-x}$ hours (catching up), and in the opposite direction, $t_{2}=\frac{1}{k x+x}$ hours (meeting he... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. The sequence $\left(a_{n}\right)$ is defined by the following relations: $a_{1}=1, a_{2}=3, a_{n}=a_{n-1}-a_{n-2}+n$ (for $n \geqslant 3$). Find $a_{1000}$. | Answer: 1002.
Solution. Let's list the first terms of the sequence:
$$
1,3,5,6,6,6,7,9,11,12,12,12,13,15,17,18,18,18,19,21, \ldots
$$
We can see a pattern: $a_{n+6}=a_{n}+6$. Let's prove it. We have
$a_{n+1}=a_{n}-a_{n-1}+n+1=\left(a_{n-1}-a_{n-2}+n\right)-a_{n-1}+n+1=-a_{n-2}+2 n+1$.
By substituting $n$ with $n+2... | 1002 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (17 points) The price of an entry ticket to the stadium is 400 p. After reducing the entry fee, the number of spectators increased by $25 \%$, and the revenue increased by $12.5 \%$. What is the new price of the entry ticket after the price reduction? | Answer: 360
Solution. Let the number of viewers before the ticket price reduction be 1 person. Then the revenue was 400 rubles. Let $x$ rubles be the new ticket price. We get the equation $x \cdot 1.25 = 400 - 1.125$. From which $x = 360$. | 360 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (17 points) A tourist travels from point $A$ to point $B$ in 1 hour 56 minutes. The route from $A$ to $B$ goes uphill first, then on flat ground, and finally downhill. What is the length of the road on flat ground if the tourist's speed downhill is 6 km/h, uphill is 4 km/h, and on flat ground is 5 km/h, and the tota... | # Answer: 3
Solution. Let $x$ km be the distance the tourist walks uphill, $y$ km - on flat ground, then $9-x-y$ km - downhill. We get $\frac{x}{4}+\frac{y}{5}+\frac{9-x-y}{6}=\frac{29}{15}$. After transformations, $5 x+2 y=26$. It is obvious that $x$ must be even and $x+y \leq 9$. The only solution is $x=4, y=3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (16 points) A chess player played 40 chess games and scored 25 points (1 point for each win, -0.5 points for a draw, 0 points for a loss). Find the difference between the number of his wins and the number of his losses.
# | # Answer: 10
Solution. Let the chess player have $n$ wins and $m$ losses. Then we get $n+0.5 \cdot(40-n-m)=25$. In the end, $n-m=10$. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (17 points) The price of an entry ticket to the stadium is 400 p. After increasing the entry fee, the number of spectators decreased by $20 \%$, but the revenue increased by $5 \%$. What is the new price of the entry ticket after the price increase? | Answer: 525
Solution. Let the number of viewers before the ticket price reduction be 1 person. Then the revenue was 400 rubles. Let $x$ rubles be the new ticket price. We get the equation $x \cdot 0.8 = 400 \cdot 1.05$. From which $x=525$. | 525 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (17 points) A tourist travels from point $A$ to point $B$ in 2 hours and 14 minutes. The route from $A$ to $B$ goes uphill first, then on flat terrain, and finally downhill. What is the length of the uphill road if the tourist's speed downhill is 6 km/h, uphill is 4 km/h, and on flat terrain is 5 km/h, and the total... | Answer: 6
Solution. Let $x$ km be the distance the tourist walks uphill, $y$ km - on flat ground, then $10-x-y$ km - downhill. We get $\frac{x}{4}+\frac{y}{5}+\frac{10-x-y}{6}=\frac{67}{30}$. After transformations, $5 x+2 y=34$. It is obvious that $x$ must be even and $x+y \leq 10$. The only solution is $x=6, y=2$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (16 points) A chess player played 42 games and scored 30 points (1 point for each win, -0.5 points for each draw, 0 points for each loss). Find the difference between the number of his wins and the number of his losses. | Answer: 18
Solution. Let the chess player have $n$ wins and $m$ losses. Then we get $n+0.5 \cdot(42-n-m)=30$. In the end, $n-m=18$. | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) An aluminum and a copper part have the same volume. The density of aluminum $\rho_{A}=2700 \mathrm{kg} / \mathrm{m}^{3}$, the density of copper $\rho_{M}=8900 \mathrm{kg} / \mathrm{m}^{3}$. Find the mass of copper, if it is known that the masses of the parts differ by $\Delta m=60 \mathrm{g}$. | Answer: 862
Solution. Volume of aluminum: $V=\frac{m_{M}-\Delta m}{\rho_{A}}$, volume of copper: $V=\frac{m_{M}}{\rho_{M}}$. We get: $\frac{m_{M}}{\rho_{M}}=\frac{m_{M}-\Delta m}{\rho_{A}} . \quad$ From this, the mass of aluminum: $m_{A}=\frac{\Delta m \cdot \rho_{M}}{\rho_{M}-\rho_{A}}=\frac{0.06 \cdot 8900}{8900-270... | 862 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A mowing crew mowed the entire meadow in two days. On the first day, half of the meadow and another 3 hectares were mowed, and on the second day, a third of the remaining area and another 6 hectares were mowed. What is the area of the meadow? | Answer: 24 hectares.
Solution. 6 hectares made up two-thirds of the remainder. Therefore, 9 hectares were mowed on the second day. Together with the 3 hectares from the first day, this totals 12 hectares, which constitute half of the meadow's area. Therefore, the total area of the meadow is 24 hectares.
Evaluation. 1... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (16 points) After walking one-fifth of the way from home to school, Olya realized she had forgotten her notebook. If she does not go back for it, she will arrive at school 6 minutes before the bell, but if she returns, she will be 2 minutes late. How much time (in minutes) does the journey to school take? | # Answer: 20 min
Solution. The extra $2 / 5$ of the journey takes 8 min. Therefore, the entire journey to school will take 20 min. | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (17 points) Find the smallest root of the equation
$$
\sqrt{x+2}+2 \sqrt{x-1}+3 \sqrt{3 x-2}=10
$$ | Answer: 2
Solution. It is clear that 2 is a root of the equation. The function on the left side of the equation is increasing (as the sum of increasing functions). Therefore, there are no other roots. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (17 points) In a $5 \times 5$ grid, 6 crosses need to be placed such that each row and each column contains at least one cross. How many ways can this be done? | # Answer: 4200
Solution. From the condition, it follows that in some row $a$ and some column $b$ there are two crosses (and in all other rows and columns - one each). Both row $a$ and column $b$ can be chosen in 5 ways. There are two possible cases.
First. At the intersection of $a$ and $b$, there is a cross. We choo... | 4200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) A bullet with a mass of $m=10$ g, flying horizontally at a speed of $v_{1}=500 \mathrm{M} / \mathrm{c}$, penetrates a massive board and exits with a speed of $v_{2}=200 \mathrm{M} / \mathrm{c}$. Find the magnitude of the work done on the bullet by the resistance force of the board. | Answer: 1050 J
Solution. From the law of conservation of energy, it follows that
$$
A=\frac{m v_{1}^{2}}{2}-\frac{m v_{2}^{2}}{2}=\frac{0.01 \cdot 500^{2}}{2}-\frac{0.01 \cdot 200^{2}}{2}=1050 \text { J. }
$$ | 1050 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (15 points) A light ray falls at an angle $\alpha=30^{\circ}$ on the front surface of a parallel glass plate. The refractive index of the glass is $n=1.5$. By what angle is the ray, reflected from the back surface of the plate and exiting back through the front surface, deflected from the direction of the incident r... | Answer: $120^{\circ}$
Solution. The described situation is illustrated in the figure.
We need to find the angle between rays 1 and 2. It is clear that the required angle is $180^{\circ}-2 \... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (16 points) Kolya, after walking a quarter of the way from home to school, realized he had forgotten his workbook. If he does not go back for it, he will arrive at school 5 minutes before the bell, but if he does go back, he will be one minute late. How much time (in minutes) does the journey to school take? | # Answer: 12 min
Solution. The extra $2 / 4$ of the journey takes 6 min. Therefore, the entire journey to school will take 12 min. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (17 points) Find the largest root of the equation
$$
3 \sqrt{x-2}+2 \sqrt{2 x+3}+\sqrt{x+1}=11
$$ | Answer: 3
Solution. It is clear that 3 is a root of the equation. The function on the left side of the equation is increasing (as the sum of increasing functions). Therefore, there are no other roots. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (17 points) In a $4 \times 4$ grid, 5 crosses need to be placed such that each row and each column contains at least one cross. How many ways can this be done? | # Answer: 432
Solution. From the condition, it follows that in some row $a$ and some column $b$ there are two crosses (and in all other rows and columns - one each). Both row $a$ and column $b$ can be chosen in 4 ways. There are two possible cases.
First. At the intersection of $a$ and $b$, there is a cross. We choos... | 432 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) A bullet with a mass of $m=10$ g, flying horizontally at a speed of $v_{1}=400 \mathrm{m} / \mathrm{s}$, penetrates a massive board and exits with a speed of $v_{2}=100 \mathrm{m} / \mathrm{s}$. Find the magnitude of the work done on the bullet by the resistance force of the board. | Answer: 750 J
Solution. From the law of conservation of energy, it follows that
$$
A=\frac{m v_{1}^{2}}{2}-\frac{m v_{2}^{2}}{2}=\frac{0.01 \cdot 400^{2}}{2}-\frac{0.01 \cdot 100^{2}}{2}=750 \text { J. }
$$ | 750 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (15 points) A light ray falls at an angle $\alpha=60^{\circ}$ on the front surface of a parallel glass plate. The refractive index of the glass is $n=1.6$. By what angle is the ray, reflected from the back surface of the plate and exiting back through the front surface, deflected from the direction of the incident r... | Answer: $60^{\circ}$
Solution. The described situation is illustrated in the figure.
We need to find the angle between rays 1 and 2. It is clear that the required angle is $180^{\circ}-2 \a... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem №4 (10 points)
A parallel beam of light falls on the base of a glass cone (refractive index $n=1.5$) along its axis (see fig.). The cross-section of the beam coincides with the base of the cone, the radius
$\operatorname{tg} \alpha=\frac{h}{R}=1.73$, i.e., $\alpha=60^{\circ}$
The law of refraction for rays passing through the cone:
$\frac{\sin \alpha}{\sin \beta}=\frac{1}{1.5}$
A... | 34 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (16 points) A truck left the village of Mirny at a speed of 40 km/h. At the same time, a car left the city of Tikhaya in the same direction as the truck. In the first hour of the journey, the car traveled 50 km, and in each subsequent hour, it traveled 5 km more than in the previous hour. How many hours will it take... | # Answer: 6.
Solution. In the first hour of travel, the pursuit speed of the car relative to the truck was $50-40=10$ km/h. In each subsequent hour, the pursuit speed increases by 5 km/h. Thus, the pursuit speeds form an arithmetic progression $10,15,20 \ldots$ km/h. Let's find the number of hours $n$ until the cars m... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (16 points) A truck left the village of Mirny at a speed of 40 km/h. At the same time, a car left the city of Tikhaya in the same direction as the truck. In the first hour of the journey, the car traveled 50 km, and in each subsequent hour, it traveled 5 km more than in the previous hour. How many hours will it take... | # Answer: 7.
Solution. In the first hour of travel, the pursuit speed of the car relative to the truck was $50-40=10$ km/h. In each subsequent hour, the pursuit speed increases by 5 km/h. Thus, the pursuit speeds form an arithmetic progression $10,15,20 \ldots$ km/h. Let's find the number of hours $n$ until the cars m... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (15 points) Two heaters are connected sequentially to the same DC power source. The water in the pot boiled after $t_{1}=3$ minutes from the first heater. The same water, taken at the same initial temperature, boiled after $t_{2}=6$ minutes from the second heater. How long would it take for the water to boil if the ... | Answer: 2 min.
Solution. The amount of heat required for heating in the first case $Q=I^{2} R_{1} t_{1}$. The amount of heat required for heating in the second case $Q=$ $I^{2} R_{2} t_{2}$. The amount of heat required for heating in the case of parallel connection $Q=I^{2} \frac{R_{1} \cdot R_{2}}{R_{1}+R_{2}} t$. As... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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