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1. Let's divide the sequence of natural numbers into groups:
$(1),(2,3),(4,5,6)(7,8,9,10), \ldots$
Denote $S_{n}$ as the sum of the $n$-th group of numbers. Find $S_{16}-S_{4}-S_{1}$. | Solution. Note that in the $n$-th group there are $n$ terms and the first one is $\frac{n(n-1)}{2}+$ 1. The last term of the $n$-th group is $\frac{n(n-1)}{2}+1+(n-1)=\frac{n^{2}+n}{2}$. Therefore, $S_{n}=\left(\frac{n(n-1)}{2}+1+\frac{n(n+1)}{2}\right) \cdot \frac{n}{2}=\frac{n\left(n^{2}+1\right)}{2}$. Hence, $S_{16}... | 2021 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 3.1. Condition:
Danil took a white cube and numbered its faces with numbers from 1 to 6, writing each one exactly once. It turned out that the sum of the numbers on one pair of opposite faces is 11. What can the sum of the numbers on none of the remaining pairs of opposite faces NOT be?
## Answer options:
$\square... | # Solution.
If the sum of the numbers on opposite faces is 11, then the numbers on these faces are 5 and 6. The remaining numbers can be paired in three ways: $(1,2)$ and $(3,4); (1,3)$ and $(2,4); (1,4)$ and $(2,3)$, i.e., the sums that CAN be obtained are: $3,7,4,6$ and 5. Therefore, the sums that CANNOT be obtained... | 9 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
# 5.1. Condition:
Polina came to the cafeteria and saw that 2 puff pastries cost 3 times more than 2 pies. Polina didn't have enough money for 2 puff pastries, but she did have enough for 1 pie and 1 puff pastry. After the purchase, she wondered how many times more money she spent buying 1 puff pastry and 1 pie instea... | Answer: 2
## Solution.
2 puff pastries are 3 times more expensive than 2 pies, so one puff pastry is 3 times more expensive than one pie, which means one puff pastry costs as much as 3 pies. Therefore, 1 puff pastry and 1 pie cost as much as 4 pies. Then they are 2 times more expensive than 2 pies. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 8.1. Condition:
A five-digit number is called a hill if the first three digits are in ascending order and the last three digits are in descending order. For example, 13760 and 28932 are hills, while 78821 and 86521 are not hills. How many hills exist that are greater than the number $77777?$
# | # Answer: 36
## Solution.
Suitable slides must start with 7, so the next digit is either 8 or 9. If the second position is occupied by the digit 9, the third digit will not be greater than the second, so the condition is only satisfied by 8. Therefore, the third position is occupied by 9. The last two positions can b... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 8.2. Condition:
A five-digit number is called a pit if the first three digits are in descending order, and the last three digits are in ascending order. For example, 73016 and 98346 are pits, while 88012 and 56821 are not pits. How many pits are there that are less than the number $22222?$ | Answer: 36
## Solution.
Suitable pits must start with 2, so the next is either 1 or 0. If the second position is occupied by the digit 1, then the third digit will not be greater than the second, so the condition is only satisfied by 0. Therefore, the third position is 0. The last two places can be occupied by any tw... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.1. If the discriminant of the quadratic polynomial $f(x)=a x^{2}+2 b x+c$ is subtracted from the discriminant of the quadratic polynomial $g(x)=(a+1) x^{2}+2(b+2) x+c+4$, the result is 24. Find $f(-2)$. | Answer: 6.
Solution: We have: $D_{1}-D_{2}=4\left(b^{2}-a c-(b+2)^{2}+(a+1)(c+4)\right)=4(-4 b+4 a+c)=$ $4 f(-2)$.
Comment: Correct answer without justification - 0 points. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.2. Six people - liars and knights - sat around a table. Liars always lie, while knights always tell the truth. Each of them was given a coin. Then each of them passed their coin to one of their two neighbors. After that, 3 people said: "I have one coin," while the other 3 said: "I have no coins." What is the maximum ... | Answer: 4.
Solution: After passing the coins, each person sitting at the table can have 0, 1, or 2 coins. The total number of coins will be 6. Note that if a person lies, they will state a number of coins that differs from the actual number by 1 or 2. Since the total number of coins based on the answers differs from t... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.3. On the board, there are $N$ prime numbers (not necessarily distinct). It turns out that the sum of any three numbers on the board is also a prime number. For what largest $N$ is this possible | Answer: $N=4$.
Solution. Consider the remainders when the $N$ written numbers are divided by 3. All three remainders cannot occur, because in this case, the sum of three numbers with different remainders will be divisible by 3 (and will be greater than 3), so it will not be a prime number. Therefore, there can be no m... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.1. The numbers $2^{2019}$ and $5^{2019}$ are written consecutively. How many digits are written in total? | Answer: 2020.
Solution. Let the number $2^{2019}$ contain $\mathrm{m}$ digits, and the number $5^{2019}$ contain $\mathrm{n}$ digits. Then the following inequalities hold: $10^{\mathrm{m}-1}<2^{2019}<10^{\mathrm{m}}, 10^{\mathrm{n}-1}<5^{2019}<10^{\mathrm{n}}$ (the inequalities are strict because the power of two or f... | 2020 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.5. Each cell of a $7 \mathrm{x} 8$ table (7 rows and 8 columns) is painted in one of three colors: red, yellow, or green. In each row, the number of red cells is not less than the number of yellow cells and not less than the number of green cells, and in each column, the number of yellow cells is not less than the n... | Answer: 8.
Solution. 1) In each row of the table, there are no fewer red cells than yellow ones, so in the entire table, there are no fewer red cells than yellow ones. In each column of the table, there are no fewer yellow cells than red ones, so in the entire table, there are no fewer yellow cells than red ones. Thus... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4-1. Katya attached a square with a perimeter of 40 cm to a square with a perimeter of 100 cm as shown in the figure. What is the perimeter of the resulting figure in centimeters?
 | Answer: 120.
Solution: If we add the perimeters of the two squares, we get $100+40=140$ cm. This is more than the perimeter of the resulting figure by twice the side of the smaller square. The side of the smaller square is $40: 4=10$ cm. Therefore, the answer is $140-20=120$ cm. | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4-5. Given a figure consisting of 33 circles. You need to choose three circles that are consecutive in one of the directions. In how many ways can this be done? The image shows three of the desired ways.
 | Answer: 500.
Solution. The perimeter of the figure consists of 3 segments of length 100 and 4 segments of length 50. Therefore, the length of the perimeter is
$$
3 \cdot 100 + 4 \cdot 50 = 500
$$ | 500 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5-5. Along a straight alley, 400 lamps are placed at equal intervals, numbered in order from 1 to 400. At the same time, from different ends of the alley, Alla and Boris started walking towards each other at different constant speeds (Alla from the first lamp, Boris from the four hundredth). When Alla was at the 55th l... | Answer. At the 163rd lamppost.
Solution. There are a total of 399 intervals between the lampposts. According to the condition, while Allа walks 54 intervals, Boris walks 79 intervals. Note that $54+79=133$, which is exactly three times less than the length of the alley. Therefore, Allа should walk three times more to ... | 163 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5-6. A rectangular table of size $x$ cm $\times 80$ cm is covered with identical sheets of paper of size 5 cm $\times 8$ cm. The first sheet touches the bottom left corner, and each subsequent sheet is placed one centimeter higher and one centimeter to the right of the previous one. The last sheet touches the top right... | Answer: 77.
Solution I. Let's say we have placed another sheet of paper. Let's look at the height and width of the rectangle for which it will be in the upper right corner.
We will call suc... | 77 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7-3. A secret object is a rectangle measuring $200 \times 300$ meters. Outside the object, there is a guard at each of the four corners. An intruder approached the perimeter of the secret object from the outside, and all the guards ran to him along the shortest paths along the external perimeter (the intruder remained ... | Answer: 150.
Solution. Note that no matter where the violator is, two guards in opposite corners will run a distance equal to half the perimeter in total.
Therefore, all four guards will ru... | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7-4. In a giraffe beauty contest, two giraffes, Tall and Spotted, made it to the final. 135 voters are divided into 5 districts, each district is divided into 9 precincts, and each precinct has 3 voters. The voters choose the winner by majority vote in their precinct; in the district, the giraffe that wins the majority... | Answer: 30.
Solution. For High to win the final, he must win in 3 districts. To win in a district, High must win in 5 precincts of that district. In total, he needs to win in at least $3 \cdot 5=15$ precincts. To win in a precinct, at least 2 voters must vote for him. Therefore, at least 30 voters are needed.
Comment... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8-1. Two rectangles $8 \times 10$ and $12 \times 9$ are overlaid as shown in the figure. The area of the black part is 37. What is the area of the gray part? If necessary, round the answer to 0.01 or write the answer as a common fraction.
Since the sum of the angles in triangle $D F E$ is $180^{\circ}$, we have $7 \alpha... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 9.
Solution. Since $O B=O C$, then $\angle B C O=32^{\circ}$. Therefore, to find angle $x$, it is sufficient to find angle $A C O: x=32^{\circ}-\angle A C O$.
Since $O A=O C$, then ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 58.
Solution. Angle $ACD$ is a right angle since it subtends the diameter of the circle.
Therefore, $\angle CAD = 90^{\circ} - \angle CDA = 48^{\circ}$. Also, $AO = BO = CO$ as they... | 58 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.4. What is the minimum number of cells that need to be marked on an $8 \times 9$ board so that among any five consecutive cells in a row, column, or diagonal, there is a marked cell
Fig. 11.4a | Answer: 14 cells.
Solution. Example.
See Fig. 11.4a.
Estimate. Let's highlight 14 rectangles of size $\quad 1 \times 5 \quad$ on the board, touching only two central cells (see Fig. 11.4b). In each of them, there must be at least one marked cell, meaning there are at least 14 marked cells.
 | Solution. Note that $3990=2 \cdot 3 \cdot 5 \cdot 7 \cdot 19=1 \cdot 6 \cdot 5 \cdot 7 \cdot 19$ and $6+5+7+1=19$. Therefore, any four-digit number that contains one 1, one 6, one 5, and one 7 will work, for example, 1567.
Comment. To receive full credit, it is sufficient to provide any correct example. | 1567 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Three motorcyclists start simultaneously from one point on a circular highway in the same direction. The first motorcyclist caught up with the second for the first time after making 4.5 laps from the start, and 30 minutes before that, he caught up with the third motorcyclist for the first time. The second motorcycli... | # Solution
Let $x, y, z$ be the speeds of the first, second, and third motorcyclists in circles per hour, respectively.
Express the time it takes for the first motorcyclist to catch up with the second, the first to catch up with the third, and the second to catch up with the third in different ways.
We obtain the sy... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.1. The hikers had several identical packs of cookies. At the midday break, they opened two packs and divided the cookies equally among all the participants of the hike. One cookie was left over, and the hikers fed it to a squirrel. At the evening break, they opened another three packs and also divided the cookies equ... | Solution: Let there be $n$ tourists in the group, and $m$ cookies in each pack (where $m$ and $n$ are natural numbers, and $n > 13$). According to the problem, the numbers $x = 2m - 1$ and $y = 3m - 13$ are divisible by $n$. Therefore, the number $3x - 2y = 23$ is also divisible by $n$. Since 23 is a prime number and h... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.4. It is known that $a^{2}+2=b^{4}, b^{2}+2=c^{4}, c^{2}+2=a^{4}$. What is the value of the product $\left(a^{2}-1\right)\left(b^{2}-1\right)\left(c^{2}-1\right)$? Find all possible values and prove that there are no others. | Solution: Subtract 1 from both sides of each equation, we get:
$$
\begin{gathered}
a^{2}+1=b^{4}-1=\left(b^{2}-1\right)\left(b^{2}+1\right) \\
b^{2}+1=\left(c^{2}-1\right)\left(c^{2}+1\right), \quad c^{2}+1=\left(a^{2}-1\right)\left(a^{2}+1\right)
\end{gathered}
$$
Multiply the left and right parts of all three equat... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.5. When fully fueled, a motorboat can travel exactly 40 km upstream or exactly 60 km downstream. What is the greatest distance the motorboat can travel along the river if the fuel must be enough for the round trip, back to the starting point? Justify your answer. | # Solution:
Method 1. Let's launch a raft along with the boat, which will move with the current, and observe the boat from this raft. Then, no matter which direction the boat goes, upstream or downstream, it will be at the same distance from the raft by the time the fuel runs out. This means that the raft will be at t... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. We took ten consecutive natural numbers greater than 1, multiplied them, found all the prime divisors of the resulting number, and multiplied these prime divisors (taking each exactly once). What is the smallest number that could have resulted? Fully justify your answer.
Solution. We will prove that among ten conse... | Answer: 2310.
## Municipal Stage of the All-Russian Mathematics Olympiad 2018-2019 Academic Year
10th Grade
Grading Criteria
| Score | Points for |
| :---: | :---: | :--- |
| 7 | Complete solution, a correct algorithm of actions for Zhenya is provided, and all conditions are verified. |
| 6 | Complete solution, a c... | 2310 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. How many 4-digit numbers exist where the digit in the thousands place is greater than the digit in the hundreds place? | Solution. The digit in the thousands place can take one of 9 possible values: $1,2,3, \ldots, 9$ (we cannot take 0, since the number is four-digit). For each of these options, we can specify the corresponding number of options for the hundreds digit: 1, $2,3, \ldots, 9$. That is, there are a total of 45 options. The ot... | 4500 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.5. What is the smallest value that the GCD $(x, y)$ can take, where $x$ and $y$ are natural numbers, if the LCM $(x, y)=(x-y)^{2}$? | Solution:
When $x=4$ and $y=2$, LCM $(x, y)=(x-y)^{2}=4$, and GCD $(x, y)=2$. Thus, 2 is achievable.
Assume that GCD $(x, y)=1$ (the numbers are coprime). Then LCM $(x, y)=x y$. From the condition, we get that $x y=x^{2}-2 x y+y^{2}$. We obtain $x^{2}-3 x y+y^{2}=0$. Solving the obtained quadratic equation with respe... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2.1. Sasha solved the quadratic equation $3 x^{2}+b x+c=0$ (where $b$ and $c$ are some real numbers). In his answer, he got exactly one root: $x=-4$. Find $b$. | Answer: 24
Solution. By Vieta's theorem $x_{1}+x_{2}=-b / 3$. In our case $x_{1}=x_{2}=-4$, hence $-b / 3=-8$ and $b=24$.
Remark. Another solution can be obtained by noticing that the equation should reduce to a perfect square $3(x+4)^{2}=0$. | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1. Find the sum
$$
\sqrt[7]{(-7)^{7}}+\sqrt[8]{(-8)^{8}}+\sqrt[9]{(-9)^{9}}+\ldots+\sqrt[100]{(-100)^{100}}
$$
(Each term is of the form $\left.\sqrt[k]{(-k)^{k}}\right)$ | Answer: 47
Solution. Note that $\sqrt[k]{(-k)^{k}}$ equals $k$ when $k$ is even, and equals $-k$ when $k$ is odd. Therefore, our sum is $-7+8-9+10-\ldots-99+100=(-7+8)+(-9+10)+\ldots+(-99+100)$. In the last expression, there are 47 parentheses, each of which equals 1. | 47 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. For triangle $A B C$, the following is known: $A B=12, B C=10, \angle A B C=120^{\circ}$. Find $R^{2}$, where $R-$ is the radius of the smallest circle in which this triangle can be placed.
Solution. Since segment $AC$ fits inside the circle, $2R \geqslant AC$. On the other hand, the circle constructed with $AC$ as its diameter covers the triangle $ABC$, since $\angl... | 91 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.1. Let's consider the number 616. The sum of its digits is 13, and the product of its digits is 36. What is the largest natural number whose sum of digits is 13 and the product of its digits is 36? | Answer: 3322111
Solution. In the decimal representation of this number, there cannot be any zeros. Let's list all the digits of this number that are greater than 1; their product is still 36. Since 36 can be factored into prime factors as $36=2 \cdot 2 \cdot 3 \cdot 3$, the possible sets of digits greater than 1 are: ... | 3322111 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.1. On a plane, 55 points are marked - the vertices of a certain regular 54-gon and its center. Petya wants to paint a triplet of the marked points in red so that the painted points are the vertices of some equilateral triangle. In how many ways can Petya do this? | Answer: 72
Solution. Assume that our 54-gon is inscribed in a circle, and its vertices divide this circle into 54 arcs of length 1.
If one of the painted points is the center, then the other two painted points must be the ends of an arc of size $54 / 6=9$. There are 54 such possibilities.
Otherwise, all three painte... | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.1. In a row, the numbers are written: $100^{100}, 101^{101}, 102^{102}, \ldots, 234^{234}$ (i.e., the numbers of the form $n^{n}$ for natural numbers n from 100 to 234). How many of the listed numbers are perfect squares? (A perfect square is the square of an integer.) | # Answer: 71
Solution. Consider a number of the form $m^{k}$, where $m$ and $k$ are natural numbers. If $k$ is even, then $m^{k}$ is a perfect square. If $k$ is odd, then $m^{k}$ is a perfect square if and only if $m$ is a perfect square. Thus, the answer to our problem is the total number of even numbers and odd perf... | 71 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.1. On the coordinate plane, a parallelogram $O A B C$ is drawn, with its center located at the point $\left(\frac{19}{2}, \frac{15}{2}\right)$, and points $A, B$, and $C$ have natural coordinates. Find the number of such parallelograms. (Here, $O$ denotes the origin - the point $(0,0)$; two parallelograms with the sa... | Answer: 126.
Solution. Note that $O A B C$ is a parallelogram if and only if the point $E\left(\frac{19}{2}, \frac{15}{2}\right)$ is the midpoint of segments $O B$ and $A C$, and moreover, points $O, A, B, C$ do not lie on the same line.
![](https://cdn.mathpix.com/cropped/2024_05_06_6bdc1b7f380f62058621g-09.jpg?heig... | 126 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In the class, two students sit at each desk. The number of desks where two boys sit is twice the number of desks where two girls sit. And the number of desks where two girls sit is twice the number of desks where a boy and a girl sit. How many boys are there in the class if there are 10 girls?
Answer: there... | Solution. Let the number of desks where a boy and a girl sit be $x$. Then the number of desks with two girls is $2x$. Therefore, the total number of girls is $2 \cdot 2x + x = 5x = 10$, from which $x = 2$. Then the number of desks with two boys is $4x = 8$. Therefore, the total number of boys is $2 \cdot 8 + 2 = 18$.
... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. A natural number is written on the board. Nikolai noticed that he can append a digit to the right of it in two ways so that the resulting number is divisible by 9. In how many ways can he append a digit to the right of the given number so that the resulting number is divisible by 3? | Answer: 4 ways.
Solution. Note that the difference between the two numbers "noticed" by Nikolai is less than 10, but is divisible by 9. Therefore, this difference is 9. This is only possible if the appended digits are 0 and 9. Then it is easy to see that for divisibility by 3, in addition to these two digits, the digi... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. An archipelago consists of several small islands and one large island. It was decided to build bridges between the islands so that the large island would be connected to each small island by two bridges, and any two small islands would be connected by one bridge.
By November 1, all the bridges between the s... | Solution. Let's number the small islands of the archipelago. If a bridge connects islands with numbers $a$ and $b$, we write the smaller of these two numbers on this bridge.
Suppose the number of small islands in the archipelago is no more than six. Then there are no more than 5 bridges with the number 1, no more than... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.7. Given a cyclic quadrilateral $A B C D$. The rays $A B$ and $D C$ intersect at point $K$. It turns out that points $B, D$, as well as the midpoints of segments $A C$ and $K C$, lie on the same circle. What values can the angle $A D C$ take?
(G. Zhukov $)$ | Answer: $90^{\circ}$.
Solution. Let $N$ and $M$ be the midpoints of segments $K C$ and $A C$, respectively. Then $M N$ is the midline of triangle $A K C$, so $\angle B A C = \angle N M C$. Additionally, $\angle B A C = \angle B D C$, since quadrilateral $A B C D$ is cyclic.
Suppose points $M$ and $N$ lie on the same ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.6. Given a cyclic quadrilateral $A B C D$. The rays $A B$ and $D C$ intersect at point $K$. It turns out that points $B, D$, as well as the midpoints of segments $A C$ and $K C$, lie on the same circle. What values can the angle $A D C$ take?
(G. Zhukov) | Answer: $90^{\circ}$.
Solution. Let $N$ and $M$ be the midpoints of segments $K C$ and $A C$, respectively. Then $M N$ is the midline of triangle $A K C$, so $\angle B A C = \angle N M C$. Additionally, $\angle B A C = \angle B D C$, since quadrilateral $A B C D$ is cyclic.
Suppose points $M$ and $N$ lie on the same ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.7. Given a polynomial
$$
P(x)=a_{2 n} x^{2 n}+a_{2 n-1} x^{2 n-1}+\ldots+a_{1} x+a_{0}
$$
where each coefficient $a_{i}$ belongs to the interval $[100,101]$. For what minimal $n$ can such a polynomial have a real root? (I. Bogdanov, K. Sukhov) | Answer. $n=100$.
Solution. Let's call a polynomial that satisfies the condition of the problem beautiful. The polynomial $P(x)=100\left(x^{200}+x^{198}+\ldots+x^{2}+\right.$ $+1)+101\left(x^{199}+x^{197}+\ldots+x\right)$ is beautiful and has a root -1. Therefore, when $n=100$, the required is possible.
It remains to ... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. For what values of $a$ do the equations
$$
x^{2}+a x+1=0 \quad \text { and } \quad x^{2}+x+a=0
$$
have at least one common root? | Solution. Let $x_{0}$ be the common root of the equations. Substitute into the equations, equate the expressions for $x_{0}^{2}$.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x_{0}^{2} + a x_{0} + 1 = 0, \\
x_{0}^{2} + x_{0} + a = 0
\end{array}\right. \\
& (a-1) x_{0} = a-1 \quad a x_{0} + 1 = x_{0} + a \\
&
\end{alig... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let $F(x)$ and $G(x)$ be polynomials of degree 2021. It is known that for all real $x$, $F(F(x)) = G(G(x))$ and there exists a real number $k, k \neq 0$, such that for all real $x$, $F(k F(F(x))) = G(k G(G(x)))$. Find the degree of the polynomial $F(x) - G(x)$. | Solution. Since $F(F(x))=G(G(x))$, then $F(k G(G(x)))=G(k G(G(x)))$. This equality holds for all $x$, i.e., for more than 2022 values of the variable (Since the polynomial $G(x)$ takes an infinite number of values). Therefore, the polynomials coincide, their difference has degree 0.
Answer. The degree of $F(x)-G(x)$ i... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. "That" and "this", plus half of "that" and "this" - what percentage of three quarters of "that" and "this" would this be? | 3. "That" and "this", plus half of "that" and "this" - this is one and a half "that" and "this", which is twice as much as three quarters of "that" and "this", i.e., it constitutes $200 \%$ of them.
Answer: $200 \%$. | 200 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In two groups, there is an equal number of students. Each student studies at least one language: English or French. It is known that 5 people in the first group and 5 in the second group study both languages. The number of students studying French in the first group is three times less than in the second group. The ... | 4. Let the number of people studying French in the first group be $x$, then in the second group it is $3x$. Let the number of people studying English in the second group be $y$, then in the first group it is $4y$. The total number of people in the first group is $x+4y-5$, and in the second group it is $3x+y-5$. By equa... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. In 8 "Y" grade, there are quite a few underachievers, but Vovochka performs the worst of all. The pedagogical council decided that either Vovochka must correct his failing grades by the end of the quarter, or he will be expelled. If Vovochka corrects his grades, then 24% of the class will be underachievers, but if h... | # Solution
Let there be $n$ students in the class now. According to the condition, $0.24 n = 0.25(n-1)$, which means $0.01 n = 0.25$. Therefore, $n = 25$. One person makes up $4 \%$ of 25, so there are now $24 + 4 = 28 \%$ of underachievers.
Answer: $28 \%$. | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The heights of an acute-angled triangle $ABC$, drawn from points $B$ and $C$, were extended to intersect the circumscribed circle at points $B_{1}$ and $C_{1}$. It turned out that the segment $B_{1} C_{1}$ passes through the center of the circumscribed circle. Find the angle $BAC$. | # Solution
Since $B_{1} C_{1}$ is the diameter of the circle, $\angle B_{1} B C_{1}=\angle B_{1} C C_{1}=90^{\circ}$, therefore, $B C_{1} \| A C$ and $C B_{1} \| A B$ (see figure). Since $B C_{1} \| A C$, then $\angle C_{1} B A=\angle A=\alpha$. Similarly, $\angle B_{1} C A=\angle A=\alpha$. The degree measure of the ... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 6. CONDITION
On an 8x8 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 "Queen... | Solution. Obviously, each checker is flipped as many times as its number has divisors. 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. Thus, the numbers of the "queens" remaining on the board will b... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task 8.5
For a triangle, one of whose angles is equal to $120^{\circ}$, it is known that it can be cut into two isosceles triangles. What can the other two angles of the original triangle be?
## Number of points 7 | Answer:
$40^{\circ}$ and $20^{\circ}$ or $45^{\circ}$ and $15^{\circ}$.
## Solution
Let in triangle $ABC$, $\angle B=120^{\circ}$. The cut specified in the problem must pass through one of the vertices of the triangle (otherwise, two triangles will not be formed upon division). It can pass through vertex $B$ or anot... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.1. What is the largest number of different natural numbers that can be chosen so that the sum of any three of them is a prime number? | Solution. One example of four numbers that satisfy the condition of the problem is $1,3,7,9$. Indeed, the numbers $1+3+7=11, 1+3+9=13, 1+7+9=17, 3+7+9=19$ are prime.
Suppose it was possible to choose five numbers. Consider the remainders of these numbers when divided by 3. If there are three identical remainders among... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.3. Solve the equation:
$1+\frac{3}{x+3}\left(1+\frac{2}{x+2}\left(1+\frac{1}{x+1}\right)\right)=x$.
# | # Solution.
$1+\frac{1}{x+1}=\frac{x+2}{x+1}$, therefore the given equation is equivalent to the equation $1+\frac{3}{x+3}\left(1+\frac{2}{x+1}\right)=x$ under the condition that $\mathrm{x} \neq-2$. Proceeding similarly, we get that $1+\frac{3}{x+3}=x$, where $\mathrm{x} \neq-2$ and $\mathrm{x} \neq-3$. The roots of ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.5. Three families of parallel lines have been drawn, with 10 lines in each. What is the maximum number of triangles they can cut out of the plane? | Solution. Consider 100 nodes - the intersection points of lines from the first and second directions. Divide them into 10 sectors: the first sector - nodes lying on the first lines of the first and second directions. The second sector - nodes lying on the second lines (excluding points lying in the first sector) and so... | 150 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. For what smallest natural $k$ does the number 2016 $\cdot$ 20162016 $\cdot$ 201620162016 $\cdot$... 20162016...2016( $k$ factors) divide without remainder by $3^{67}$? | Solution.
We will show that in the case of 27 factors, the product is divisible by $3^{67}$, but not by $3^{68}$.
$2016 \cdot 20162016 \cdot \ldots \cdot \underbrace{20162016 . .2016}_{27 \text { times }}=$
$=2016^{27} \cdot(1 \cdot 1001 \cdot 1001001 \cdot \ldots \cdot \underbrace{1001001 \ldots 1001}_{27 \text { t... | 27 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Each of the equations $a x^{2}-b x+c=0$ and $c x^{2}-a x+b=0$ has two distinct real roots. The sum of the roots of the first equation is non-negative, and the product of the roots of the first equation is 9 times the sum of the roots of the second equation. Find the ratio of the sum of the roots of the first equatio... | # Solution.
From the condition, it follows that the coefficients $a, c \neq 0$.
By Vieta's theorem, from the condition it follows that $\frac{c}{a}=9 \frac{a}{c}$. Hence, $c^{2}=9 a^{2}$, which means $\left[\begin{array}{l}c=3 a, \\ c=-3 a .\end{array}\right.$
1 case.
$c=3 a$.
We get the equations $a x^{2}-b x+3 a... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A circle is inscribed in triangle $A B C$. Two points $E$ and $F$ are marked on the largest side of the triangle $A C$ such that $A E=A B$, and $C F=C B$. Segment $B E$ intersects the inscribed circle at points $P$ and $Q$, with $B P=1, P Q=8$. What is the length of segment $E F$? | Solution.
Let the sides of the triangle be $AB=c, AC=b, BC=a$. Then $EF=a+c-b$.
Let $G$ be the point of tangency of the inscribed circle with side $BC$. Then $BG^2 = BQ \cdot BP = 9$. There... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. In the parliament of a certain state, there are 2016 deputies, who are divided into 3 factions: "blues," "reds," and "greens." Each deputy either always tells the truth or always lies. Each deputy was asked the following three questions:
1) Are you a member of the "blues" faction?
2) Are you a member of the "reds" f... | # Solution.
Let the number of deputies telling the truth in the "blue," "red," and "green" factions be $r_{1}, r_{2},$ and $r_{3}$ respectively, and the number of deputies lying in the "blue," "red," and "green" factions be $l_{1}, l_{2},$ and $l_{3}$ respectively.
According to the problem:
$\left\{\begin{array}{l}r... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Calculate:
$$
\left(\frac{1+2}{3}+\frac{4+5}{6}+\frac{7+8}{9}+\ldots+\frac{2017+2018}{2019}\right)+\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{673}\right)
$$ | Answer: 1346.
Solution. We have
$$
\begin{aligned}
& \left(\frac{1+2}{3}+\frac{4+5}{6}+\frac{7+8}{9}+\ldots+\frac{2017+2018}{2019}\right)+\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{673}\right)= \\
= & \left(\frac{(3-2)+(3-1)}{3}+\frac{(6-2)+(6-1)}{6}+\frac{(9-2)+(9-1)}{9}+\ldots\right. \\
& \left.\ldots+\frac{(2... | 1346 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Square $A B C D$ is inscribed in circle $\omega$. On the smaller arc $C D$ of circle $\omega$, an arbitrary point $M$ is chosen. Inside the square, points $K$ and $L$ are marked such that $K L M D$ is a square. Find $\angle A K D$. | Answer: $135^{\circ}$.
Solution.
We will prove the equality of triangles $D A K$ and $D C M$. Let's check that the conditions of the first criterion for the equality of triangles are met. Se... | 135 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. Given a positive number $a$. It is known that the equation $x^{3}+1=a x$ has exactly two positive roots, and the ratio of the larger to the smaller one is 2018. The equation $x^{3}+1=a x^{2}$ also has exactly two positive roots. Prove that the ratio of the larger to the smaller one is also 2018. | Solution. Let the positive roots of the equation $x^{3}+1=a x$ be denoted by $x_{1}$ and $x_{2}$ $\left(0<x_{1}<x_{2}, x_{2}: x_{1}=2018\right)$. Substitute them into the equation and divide the two resulting equations by $x_{1}^{3}$ and $x_{2}^{3}$:
\[
\begin{aligned}
& x_{1}^{3}+1=a x_{1} \Longleftrightarrow 1+\left... | 2018 | Algebra | proof | Yes | Yes | olympiads | false |
Problem 4.2. Petya took half of the candies from the box and put them in two pockets. Deciding that he took too many, Petya took out 6 candies from each pocket and put them back into the box. By how many more candies did the box have than Petya's pockets? | Answer: 24.
Solution. Let $x$ be the number of candies in one of Petya's pockets before he returned some of the candies to the box. At that moment, there were $2 x$ candies in the box.
After Petya took out 6 candies from each pocket, he had
$$
(x-6)+(x-6)=2 x-12 \text{ candies }
$$
and there were $2 x+12$ candies i... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.5. On the meadow, there were 12 cows. The shepherds brought a flock of sheep. There were more sheep than the cows' ears, but fewer than the cows' legs. How many sheep were there if there were 12 times more sheep than shepherds? | Answer: 36 sheep.
Solution. Since there are 12 times more sheep than shepherds, the number of sheep is divisible by 12. In addition, the number of sheep is more than $12 \cdot 2=24$ (the number of ears of cows) and less than $12 \cdot 4=$ 48 (the number of legs of cows). The only number divisible by 12 and between 24 ... | 36 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.7. Vanya goes to the swimming pool every Wednesday and Friday. After one of his visits, he noticed that he had already gone 10 times this month. What will be the date of the third visit in the next month if he continues to go on Wednesdays and Fridays? | Answer: 12.
Solution. Vanya visited the swimming pool 10 times in a month, which means he visited it 5 times on Wednesdays and 5 times on Fridays.
If the first visit to the swimming pool in this month fell on a Friday, then from the first Friday to the last Friday he visited (including the first and last Friday), 29 ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1.4. From sticks of the same length, a row of 700 octagons was laid out as shown in the figure. How many sticks were used in total?
Answer, option 1. 7001.
Answer, option 2. 3501.
Answer,... | Solution option 1. In the first octagon, there are 8 sticks. Constructing each subsequent octagon adds 7 sticks. In total, it will be $8+7 \cdot 999=7001$ sticks. | 7001 | Geometry | MCQ | Yes | Yes | olympiads | false |
2.4. The standard 12-hour clock face is arranged as shown in the figure. At noon, both the hour and minute hands were vertical, and now the hour hand is exactly pointing to the "18 minutes" mark.
 divisions. Subtracting 60 divisions from 204 as many times as possible (considering that the min... | 24 | Geometry | MCQ | Yes | Yes | olympiads | false |
4.4. In parallelogram $A B C D$, the angle at vertex $A$ is $60^{\circ}$, $A B=73$, and $B C=88$. The bisector of angle $A B C$ intersects segment $A D$ at point $E$, and ray $C D$ at point $F$. Find the length of segment $E F$.
Answer, option 1. 9.
Answer, option 2. 13.
Answer, option 3. 12.
Answer, option 4. 15.
... | Solution 1. Lines $A D$ and $B C$ are parallel, therefore
$$
\angle A B C=180^{\circ}-\angle B A D=120^{\circ}
$$
(angles $A B C$ and $B A D$ are consecutive interior angles). Then $\angle A B E=60^{\circ}$ and $\angle C B E=60^{\circ}$, since $B E$ is the bisector of angle $A B C$. In triangle $A B E$, the angles at... | 9 | Geometry | MCQ | Yes | Yes | olympiads | false |
6.4. To be freed from Ivan the Fisherman's net, the Golden Fish gives him a reward: either 4 pearls and 1 diamond; or 2 emeralds and 2 diamonds; or 2 pearls, 1 emerald, and 1 diamond. After several releases of the Golden Fish, Ivan the Fisherman received 1000 pearls, 800 emeralds, and some number of diamonds. How many ... | Solution 1. Let there be $x$ actions of the first type, $y$ actions of the second type, and $z$ actions of the third type. From the problem statement, we get two equations $4 x+2 z=2000, 2 y+z=800$, and we need to find $x+y+z$. By adding the first equation to twice the second, we get $4 x+4 y+4 z=3600$, from which $x+y... | 900 | Number Theory | MCQ | Yes | Yes | olympiads | false |
8.4. In the Go section, there are 55 boys of different ratings. The boys decided to play a tournament, each with each other in one game. To make it more interesting, some of the boys were allowed to use a computer's help exactly once during the tournament. If in a game, one of the boys uses a computer's help and the ot... | Solution Option 1. Let's number the players in descending order of their strength from first to fiftieth.
We will assume that each won game brings the player one point. Consider two players who ultimately scored more points than the two strongest participants. It is clear that the number of one of them is not less tha... | 45 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
3. Given a triangle $A B C$, where $\angle B A C=60^{\circ}$. Point $S$ is the midpoint of the angle bisector $A D$. It is known that $\angle S B A=30^{\circ}$. Find DC/BS. | 3. Answer: 2. Solution: We have $\mathrm{BS}=\mathrm{AS}=\mathrm{SD}$, therefore, triangle $\mathrm{ABD}$ is a right triangle (in it, the median is equal to half the side), then $\angle A C D=30^{\circ}$, hence $A D=D C$, from which we get that the required ratio is 2.
Criteria: correct solution - 7 points, proved tha... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. On 52 cards, the numbers from 1 to 52 are written. It is considered that 1 is older than 52, and in all other pairs, the card with the larger number is older. The cards are lying on the table in a random order face down. In one question, Petya can find out which card is older in any pair. Can Petya guarantee to find... | 6. Answer: it can. We will divide the cards into 13 quartets. In each quartet, we will make the following comparisons: by dividing the quartet into pairs, we will determine the higher cards in the pairs (let $\mathrm{a}>\mathrm{b}$ and $c>d$), then we will determine the higher of the two higher cards (let $a>c$). Card ... | 53 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. Inside the square $A B C D$, points $K$ and $M$ are marked (point $M$ is inside triangle $A B D$, point $K$ is inside $B M C$) such that triangles $B A M$ and $D K M$ are equal $(A M=K M, B M=M D, A B=K D)$. Find $\angle K C M$, if $\angle A M B=100^{\circ}$ | Answer: $35^{\circ}$.
Solution. Note that triangles $A B M$ and $A M D$ are also equal by three sides. Thus, point $M$ lies on the diagonal $A C$ of the square, meaning $\angle M C D=\angle M A D=\angle M A B=45^{\circ}($ Fig. 1$)$.
In addition, from the equality of triangles $B A M$ and $D K M$, it follows that $\an... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1.1. For what least natural value of \( b \) does the equation
$$
x^{2}+b x+25=0
$$
have at least one root? | Answer: 10
Solution. The equation has at least one root if and only if the discriminant $D=b^{2}-4 \cdot 25=$ $=b^{2}-100$ is greater than or equal to 0. For positive $b$, this condition is equivalent to the condition $b \geqslant 10$. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.1. Every month Ivan pays a fixed amount from his salary for a mortgage, and the remaining part of the salary is spent on current expenses. In December, Ivan paid $40 \%$ of his salary for the mortgage. In January, Ivan's salary increased by $9 \%$. By what percentage did the amount spent on current expenses increase ... | Answer: 15
Solution. Let Ivan's December salary be $100 r$. Then Ivan paid $40 r$ for the mortgage, and in December, he spent $60 r$ on current expenses. In January, Ivan's salary was $109 r$, so he spent $109 r - 40 r = 69 r$ on current expenses. Thus, the amount spent on current expenses increased by $9 r$, which is... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1. It is known that the area of the shaded region of the figure is $\frac{32}{\pi}$, and the radius of the smaller circle is 3 times smaller than the radius of the larger circle. What is the length of the smaller circle?
^{2}=9\pi R^{2}$. Therefore, the area of the shaded part is $S_{2}-S_{1}=8\pi R^{2}$. We have $8\pi R^... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4.1. In the product
$$
24^{a} \cdot 25^{b} \cdot 26^{c} \cdot 27^{d} \cdot 28^{e} \cdot 29^{f} \cdot 30^{g}
$$
the seven exponents $a, b, c, d, e, f, g$ were replaced by the numbers $1, 2, 3, 5, 8, 10, 11$ in some order. Find the maximum number of zeros that the decimal representation of this product can end with. | Answer: 32
Solution. The prime factor 5 enters the factorization of this product with multiplicity $2 b+g$, so there are no more than $2 b+g$ zeros in this product. Since $b \leqslant 11$ (the largest of the given exponents) and $b+g \leqslant 11+10$ (the sum of the two largest of the given exponents), then $2 b+g \le... | 32 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.1. The geometric progression $b_{1}, b_{2}, \ldots$ is such that $b_{25}=2 \operatorname{tg} \alpha, b_{31}=2 \sin \alpha$ for some acute angle $\alpha$. Find the number $n$ for which $b_{n}=\sin 2 \alpha$.
# | # Answer: 37
Solution. Let $q$ be the common ratio of our progression. Using the fact that $b_{31}=b_{25} q^{6}$ and $\sin 2 \alpha=2 \sin \alpha \cos \alpha$. We have $q^{6}=\frac{b_{31}}{b_{25}}=\frac{2 \sin \alpha}{2 \tan \alpha}=\cos \alpha$.
On the other hand, $\frac{\sin 2 \alpha}{2 \sin \alpha}=\cos \alpha$, f... | 37 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.1. Given a rectangular parallelepiped $2 \times 3 \times 2 \sqrt{3}$. What is the smallest value that the sum of the distances from an arbitrary point in space to all eight of its vertices can take?
 | Answer: 24.
Solution. In a horizontal strip $1 \times 5$, there are 1 five-cell, 2 four-cell, 3 three-cell, 4 two-cell, and 5 one-cell rectangles. In total, $1+2+3+4+5=15$ rectangles.
In a vertical strip $1 \times 4$, there are 1 four-cell, 2 three-cell, 3 two-cell, and 4 one-cell rectangles. In total, $1+2+3+4=10$ r... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.3. A cuckoo clock is hanging on the wall. When a new hour begins, the cuckoo says "cuckoo" a number of times equal to the number the hour hand points to (for example, at 19:00, "cuckoo" sounds 7 times). One morning, Maxim approached the clock when it was 9:05. He started turning the minute hand until he advan... | Answer: 43.
Solution. The cuckoo will say "cuckoo" from 9:05 to 16:05. At 10:00 it will say "cuckoo" 10 times, at 11:00 - 11 times, at 12:00 - 12 times. At 13:00 (when the hand points to the number 1) "cuckoo" will sound 1 time. Similarly, at 14:00 - 2 times, at 15:00 - 3 times, at 16:00 - 4 times. In total
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. From a $6 \times 6$ grid square, gray triangles have been cut out. What is the area of the remaining figure? The side length of each cell is 1 cm. Give your answer in square centimeters.
.
For convenience, let's introduce some notations. Let the top-left corner cell of the $5 \times 5$ table be called $A$, and the bottom-right corner ce... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common s... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
Then
$$
\angle BAC = \angle BAK + \angle CAK = \angle BCA ... | 21 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. Given an isosceles triangle $ABC$, where $AB = AC$ and $\angle ABC = 53^{\circ}$. Point $K$ is such that $C$ is the midpoint of segment $AK$. Point $M$ is chosen such that:
- $B$ and $M$ are on the same side of line $AC$;
- $KM = AB$
- angle $MAK$ is the maximum possible.
How many degrees does angle $BAM... | Answer: 44.
Solution. Let the length of segment $AB$ be $R$. Draw a circle with center $K$ and radius $R$ (on which point $M$ lies), as well as the tangent $AP$ to it such that the point of tangency $P$ lies on the same side of $AC$ as $B$. Since $M$ lies inside the angle $PAK$ or on its boundary, the angle $MAK$ does... | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. An equilateral triangle with a side of 10 is divided into 100 small equilateral triangles with a side of 1. Find the number of rhombi consisting of 8 small triangles (such rhombi can be rotated).
It is clear that the number of rhombuses of each orientation will be the same, so let's consider ... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.5. A circle $\omega$ is inscribed in trapezoid $A B C D$, and $L$ is the point of tangency of $\omega$ and side $C D$. It is known that $C L: L D=1: 4$. Find the area of trapezoid $A B C D$, if $B C=9$, $C D=30$.
. Petya wrote down the abscissas of all six points of intersection and multiplied them. Prove that the result does not depend on the choice of the number $k$. | 10.1. A line parallel to the line $y=k x$ has the equation $y=k x+b$ The abscissas of its intersection points with the hyperbola are both roots of the equation $\frac{k}{x}=k x+b, \quad$ equivalent to the equation $k x^{2}+b x-k=0$. The product of the roots of this equation is -1. Multiplying three such products, we ge... | -1 | Algebra | proof | Yes | Yes | olympiads | false |
2. The plane departed from Perm on September 28 at noon and arrived in Kirov at 11:00 AM (all departure and arrival times mentioned in the problem are local). At 7:00 PM the same day, the plane departed from Kirov to Yakutsk and arrived there at 7:00 AM. Three hours later, it departed from Yakutsk to Perm and returned ... | # Solution.
The plane was absent in Perm for 23 hours. Out of these, it was stationed in Kirov for 8 hours (from 11 to 19) and for 3 hours in Yakutsk. In total, out of these 23 hours, it was stationed $8+3=11$ (hours), i.e., the plane was in the air for $23-11=12$ (hours).
## Grading Criteria.
- Correct solution -7 ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. On a glade, 25 gnomes gathered. It is known that 1) every gnome who put on a hat also put on shoes; 2) 12 gnomes came without a hat; 3) 5 gnomes came barefoot. Which gnomes are more and by how many: those who came in shoes but without a hat, or those who put on a hat? | Answer. There are 6 more gnomes who put on a cap.
## Solution.
From condition 2, it follows that $25-12=13$ gnomes came in a cap.
From condition 1, we get that exactly 13 gnomes came both in a cap and in shoes.
From condition 3, it follows that a total of $25-5=20$ gnomes came in shoes.
Thus, $20-13=7$ gnomes came... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. The difference of the squares of two numbers is 6, and if each of these numbers is decreased by 2, then the difference of their squares becomes equal to 18. What is the sum of these numbers? | Answer: -2.
## Solution.
Given:
\[
\begin{aligned}
& a^{2}-b^{2}=6 \\
& (a-2)^{2}-(b-2)^{2}=18
\end{aligned}
\]
There are different ways to proceed.
## Method 1.
\((a-2)^{2}-(b-2)^{2}=a^{2}-4a+4-b^{2}+4b-4=a^{2}-b^{2}-4(a-b)\). Since from the first condition \(a^{2}-b^{2}=6\), we get \(6-4(a-b)=18\). Hence, \(a-b... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.1. At the stadium, through the checkpoints numbered 1, 2, 3, and 4, an equal number of fans entered. They then passed through several more checkpoints, moving along the arrows, and at the fork, the fans were evenly distributed. At checkpoint 7, 15 people passed the control.
![](https://cdn.mathpix.com/cropped/2024_0... | Answer: 24
Solution. Let $2x$ people pass through each of the first four checkpoints. Then, through checkpoint 7, $2x$ people passed from the first two checkpoints and $x$ from the third, totaling $5x=15$ people, from which $x=3$. Therefore, each of the first four checkpoints had 6 people, and checkpoint 8 was passed ... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3.1. A gardener grows white and red flowers: peonies and roses. In his garden, there are 301 stems, among which 135 are roses and 182 are red flowers. What is the smallest number of red peonies that can grow in the gardener's garden? | Answer: 47
Solution. Estimation: The number of red peonies is the difference between the number of red flowers and the number of red roses. The number of red roses is no more than the total number of roses, i.e., no more than 135. Therefore, the number of red peonies is no less than $182-135=47$.
Let's provide an exa... | 47 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.1. In February of a non-leap year, Kirill and Vova decided to eat ice cream according to certain rules. If the date was even and the day of the week was Wednesday or Thursday, they each ate seven portions of ice cream. If the day of the week was Monday or Tuesday and the date was odd, they each ate three portions of ... | # Answer: 110
Solution. February has 28 days, which is exactly 4 Mondays, 4 Tuesdays, and so on. Moreover, if Monday falls on an even date in one week, the following Monday will fall on an odd date, and vice versa. This means that the same day of the week falls on an even date twice and an odd date twice, regardless o... | 110 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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