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6. Find the largest ten-digit number of the form $\overline{a_{9} a_{8} a_{7} a_{6} a_{5} a_{4} a_{3} a_{2} a_{1} a_{0}}$, possessing the following property: the digit equal to $\mathrm{a}_{\mathrm{i}}$ appears in its representation exactly $\mathrm{a}_{9-\mathrm{i}}$ times (for example, the digit equal to $\mathrm{a}_... | Solution: We will divide all digits into pairs $a_{0}-a_{9}, \ldots, a_{4}-a_{5}$. We will prove that the digit 9 cannot be present in the given number. Suppose $a_{i}=9$, then the paired digit $a_{9-i}$ appears 9 times. If $a_{9-i} \neq 9$, then 9 appears only once, so we get a number consisting of one nine and nine o... | 8888228888 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,437 |
7. On a circle, 2017 different points $A_{1}, \ldots, A_{2017}$ are marked, and all possible chords connecting these points pairwise are drawn. A line is drawn through the point $A_{1}$, not passing through any of the points $A_{2}, \ldots, A_{2017}$. Find the maximum possible number of chords that can have at least on... | Solution: Let $k$ points be located on one side of the given line, then $2016-k$ points are on the other side. Thus, $k(2016-k)$ chords intersect the line and another 2016 pass through point $A_{1}$. Note that $k(2016-k)=-(k-1008)^{2}+1008^{2}$. The maximum of this expression is achieved when $k=1008$. | 1018080 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,438 |
1.1. (2 points) On two bookshelves, there are mathematics books, with an equal number on each. If 5 books are moved from the first shelf to the second, then the second shelf will have twice as many books as the first. How many books are there in total on both shelves? | Answer: 30.
Solution. If $x$ is the number of books on both shelves, then $\frac{x}{3}+5=\frac{2 x}{3}-5$, from which $x=30$. | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,440 |
2.1. (14 points) A team consisting of juniors and masters from the "Vimpel" sports society went to a shooting tournament. The average number of points scored by the juniors turned out to be 22, by the masters - 47, and the average number of points for the entire team - 41. What is the share (in percent) of masters in t... | Answer: 76.
Solution. Let there be $x$ juniors and $y$ masters in the team. Then the total number of points scored by the team is $22x + 47y = 41(x + y)$, from which we find $19x = 6y$. Therefore, the proportion of masters is $\frac{y}{x+y} = \frac{19y}{19x + 19y} = \frac{19y}{25y} = 0.76$, i.e., $76\%$. | 76 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,441 |
3.1. (14 points) Calculate $x^{3}+\frac{1}{x^{3}}$, given that $x+\frac{1}{x}=3$. | Answer: 18.
Solution. If $x+\frac{1}{x}=a$, then $a^{3}=\left(x+\frac{1}{x}\right)^{3}=x^{3}+3\left(x+\frac{1}{x}\right)+\frac{1}{x^{3}}$, from which $x^{3}+\frac{1}{x^{3}}=a^{3}-3 a=18$. | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,442 |
4.1. (14 points) An eraser, 3 pens, and 2 markers cost 240 rubles, while 2 erasers, 4 markers, and 5 pens cost 440 rubles. What is the total cost (in rubles) of 3 erasers, 4 pens, and 6 markers? | Answer: 520.
Solution. From the condition, it follows that 3 erasers, 8 pens, and 6 markers cost $440+240=680$ rubles. Moreover, 2 erasers, 6 pens, and 4 markers will cost $2 \cdot 240=480$ rubles. Therefore, one pen costs $480-440=40$ rubles. Then, 3 erasers, 4 pens, and 6 markers cost $680-4 \cdot 40=520$ rubles. | 520 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,443 |
5.1. (14 points) In an acute-angled triangle $A B C$, angle $A$ is equal to $35^{\circ}$, segments $B B_{1}$ and $C C_{1}$ are altitudes, points $B_{2}$ and $C_{2}$ are the midpoints of sides $A C$ and $A B$ respectively. Lines $B_{1} C_{2}$ and $C_{1} B_{2}$ intersect at point $K$. Find the measure (in degrees) of ang... | Answer: 75.
Solution. Note that angles $B$ and $C$ of triangle $ABC$ are greater than $\angle A=35^{\circ}$ (otherwise it would be an obtuse triangle), so point $C_{1}$ lies on side $AB$ be... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,444 |
6.1. (14 points) In a hut, several residents of the island gathered, some from the Ah tribe, and the rest from the Ukh tribe. The residents of the Ah tribe always tell the truth, while the residents of the Ukh tribe always lie. One of the residents said: "There are no more than 16 of us in the hut," and then added: "We... | Answer: 15.
Solution. A resident of tribe Ah cannot say "we are all from tribe Ukh," so the first one is from tribe Ukh. Therefore, there are no fewer than 17 people in the hut. So the second one spoke the truth, i.e., he is from tribe Ah. Therefore, there are no more than 17 people in the hut. Thus, there are 17 peop... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,445 |
7.1. (14 points) Find the greatest integer value of $a$ for which the equation
$$
(x-a)(x-7)+3=0
$$
has at least one integer root. | Answer: 11.
Solution. For the desired values of $a$, the numbers $x-a$ and $x-7$ are integers, and their product is -3, so there are 4 possible cases:
$$
\left\{\begin{array} { l }
{ x - a = 1 , } \\
{ x - 7 = - 3 ; }
\end{array} \quad \left\{\begin{array} { l }
{ x - a = 3 , } \\
{ x - 7 = - 1 ; }
\end{array} \qua... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,446 |
1. The second term of a geometric progression is 5, and the third term is 1. Find the first term of this progression. | Answer: 25. Solution. Since $\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=q$, then $a_{1}=\frac{a_{2}^{2}}{a_{3}}=25$. | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,449 |
2. Find the area of a right triangle, one of whose legs is 6, and the hypotenuse is 10. | Answer: 24. Solution. According to the Pythagorean theorem, the second leg is equal to $\sqrt{10^{2}-6^{2}}=8$. Therefore, the area of the triangle is $\frac{6 \cdot 8}{2}=24$.
## Main Task | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,450 |
1.1. Businessmen Ivanov, Petrov, and Sidorov decided to create an automobile enterprise. Ivanov bought 70 identical cars for the enterprise, Petrov - 40 such cars, and Sidorov contributed 44 million rubles to the enterprise. It is known that Ivanov and Petrov can divide this money between themselves so that the contrib... | Answer: 40. Solution. Method 1. Each of the businessmen should contribute as much as Sidorov, that is, 44 million rubles. If the car costs $x$, then $\frac{70 x+40 x}{3}=44, x=1.2$. It turns out that Ivanov contributed $70 \cdot 1.2=84$ million rubles, so he should get back $84-44=40$ million rubles. Petrov contributed... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,451 |
2.1. How many 9-digit numbers divisible by 5 can be formed by rearranging the digits of the number 377353752? | Answer: 1120. Solution. Since the number is divisible by 5, the digit in the 9th place can only be five. After this, we need to distribute the remaining 8 digits across the 8 remaining places: 3 sevens, 3 threes, a five, and a two. The total number of permutations will be 8!, but since there are repeating digits, the a... | 1120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,452 |
2.2. How many 9-digit numbers divisible by 2 can be formed by rearranging the digits of the number 131152152? | Answer: 840. Instructions. The number of numbers will be $\frac{8!}{4!2!}=840$. | 840 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,453 |
2.3. How many 9-digit numbers divisible by 5 can be formed by rearranging the digits of the number 137153751? | Answer: 1680. Instructions. The number of numbers will be $\frac{8!}{3!2!2!}=1680$. | 1680 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,454 |
2.4. How many 9-digit numbers divisible by 2 can be formed by rearranging the digits of the number 231157152? | Answer: 3360. Instructions. The number of numbers will be $\frac{8!}{3!2!}=3360$. | 3360 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,455 |
3.1. Determine the total surface area of a cube if the distance between non-intersecting diagonals of two adjacent faces of this cube is 8. If the answer is not an integer, round it to the nearest integer. | Answer: 1152. Solution. Let's take as the diagonals specified in the condition the diagonals $A_{1} C_{1}$ and $A D_{1}$ of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$. Construct two parallel planes $A_{1} C_{1} B$ and $A D_{1} C$, each containing one of these diagonals. The distance between these planes is equal to the... | 1152 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,456 |
4.1. Find the sum of all roots of the equation $x^{2}-31 x+220=2^{x}\left(31-2 x-2^{x}\right)$. | Answer: 7. Solution. The original equation is equivalent to the equation $\left(x+2^{x}\right)^{2}-31\left(x+2^{x}\right)+220=0 \Leftrightarrow\left[\begin{array}{l}x+2^{x}=11, \\ x+2^{x}=20 .\end{array}\right.$
Each of the equations in this system has no more than one root, as the function $f(x)=x+2^{x}$ is monotonic... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,457 |
4.2. Find the sum of all roots of the equation $x^{2}-41 x+330=3^{x}\left(41-2 x-3^{x}\right)$. | Answer: 5. Instructions. Roots of the equation: 2 and 3. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,458 |
5.1. Among all the simple fractions, where the numerator and denominator are two-digit numbers, find the smallest fraction greater than $\frac{3}{4}$. In your answer, specify its numerator. | Answer: 73. Solution. It is required to find a fraction $\frac{a}{b}$ such that $\frac{a}{b}-\frac{3}{4}=\frac{4 a-3 b}{4 b}$ reaches a minimum. Therefore, the maximum two-digit $b$ is sought, for which $4 a-3 b=1$. If in this case $b \geq 50$, then the fraction $\frac{a}{b}-\frac{3}{4}=\frac{1}{4 b}$ will always be le... | 73 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,460 |
5.2. Among all the irreducible fractions, where the numerator and denominator are two-digit numbers, find the smallest fraction greater than $\frac{5}{6}$. In your answer, specify its numerator. | Answer: 81. Instructions. The required fraction: $\frac{81}{97}$. | 81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,461 |
5.3. Among all the irreducible fractions, where the numerator and denominator are two-digit numbers, find the smallest fraction greater than $\frac{4}{5}$. In your answer, specify its numerator. | Answer: 77. Instructions. The required fraction: $\frac{77}{96}$. | 77 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,462 |
5.6. Among all the irreducible fractions, where the numerator and denominator are two-digit numbers, find the smallest fraction greater than $\frac{4}{9}$. In your answer, specify its numerator. | Answer: 41. Instructions. The required fraction: $\frac{41}{92}$. | 41 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,465 |
6.1. In an isosceles triangle \(ABC\), one of the angles is equal to the difference of the other two, and one of the angles is twice another. The angle bisectors of angles \(A\), \(B\), and \(C\) intersect the circumcircle of the triangle at points \(L\), \(O\), and \(M\) respectively. Find the area of triangle \(LOM\)... | Answer: 11. Solution. Let in triangle $ABC$ the angle $\alpha$ be equal to the difference of angles $\beta-\gamma$. Since $\alpha+\beta+\gamma=180^{\circ}$, then $\beta-\gamma+\beta+\gamma=180^{\circ}$ and therefore $\beta=90^{\circ}$. If $\beta=90^{\circ}$ is twice one of the angles $\alpha$ or $\gamma$, then the tria... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,466 |
6.2. In an isosceles triangle \(ABC\), one of the angles is equal to the difference of the other two, and one of the angles is twice another. The angle bisectors of angles \(A\), \(B\), and \(C\) intersect the circumcircle of the triangle at points \(L\), \(O\), and \(M\) respectively. Find the area of triangle \(LOM\)... | Answer: 3. Instructions. Exact answer: $\sqrt{3}+1$. | \sqrt{3}+1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,467 |
6.3. In an isosceles triangle \(ABC\), one of the angles is equal to the difference of the other two, and one of the angles is twice another. The angle bisectors of angles \(A\), \(B\), and \(C\) intersect the circumcircle of the triangle at points \(L\), \(O\), and \(M\) respectively. Find the area of triangle \(LOM\)... | Answer: 44. Instructions. Exact answer: $16 \sqrt{3}+16$. | 16\sqrt{3}+16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,468 |
6.4. In an isosceles triangle \(ABC\), one of the angles is equal to the difference of the other two, and one of the angles is twice another. The angle bisectors of angles \(A\), \(B\), and \(C\) intersect the circumcircle of the triangle at points \(L\), \(O\), and \(M\) respectively. Find the area of triangle \(LOM\)... | Answer: 27. Instructions. Exact answer: $10 \sqrt{3}+10$. | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,469 |
7.1. Calculate: $4\left(\sin ^{3} \frac{49 \pi}{48} \cos \frac{49 \pi}{16}+\cos ^{3} \frac{49 \pi}{48} \sin \frac{49 \pi}{16}\right) \cos \frac{49 \pi}{12}$. | Answer: 0.75. Solution: Let $\alpha=\frac{49 \pi}{48}=\pi+\frac{\pi}{48}$. Then:
$\left(\sin ^{3} \alpha \cos 3 \alpha+\cos ^{3} \alpha \sin 3 \alpha\right) \cos 4 \alpha=\left[\sin ^{3} \alpha\left(4 \cos ^{3} \alpha-3 \cos \alpha\right)+\cos ^{3} \alpha\left(3 \sin \alpha-4 \sin ^{3} \alpha\right)\right] \cos 4 \alp... | 0.75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,470 |
8.2. Solve the inequality $\sqrt{x^{2}-x-56}-\sqrt{x^{2}-25 x+136}<8 \sqrt{\frac{x+7}{x-8}}$, and find the sum of its integer solutions that belong to the interval $[-25 ; 25]$. | Answer: -285. Instructions. Solution of the inequality: $x \in(-\infty ;-7] \cup(18 ; 20)$. | -285 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,472 |
8.3. Solve the inequality $\sqrt{x^{2}+3 x-54}-\sqrt{x^{2}+27 x+162}<8 \sqrt{\frac{x-6}{x+9}}$, and find the sum of its integer solutions that belong to the interval $[-25 ; 25]$. | Answer: 290 . Instructions. Solution of the inequality: $x \in(-21 ;-19) \bigcup[6 ;+\infty)$. | 290 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,473 |
8.4. Solve the inequality $\sqrt{x^{2}+x-56}-\sqrt{x^{2}+25 x+136}<8 \sqrt{\frac{x-7}{x+8}}$, and find the sum of its integer solutions that belong to the interval $[-25 ; 25]$. | Answer: 285. Instructions. Solution of the inequality: $x \in(-20 ;-18) \cup[7 ;+\infty)$. | 285 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,474 |
9.1. Find the maximum value of the expression $(\sqrt{8-4 \sqrt{3}} \sin x-3 \sqrt{2(1+\cos 2 x)}-2) \cdot(3+2 \sqrt{11-\sqrt{3}} \cos y-\cos 2 y)$. If the answer is not an integer, round it to the nearest integer. | Answer: 33. Solution. Let $f(x)=\sqrt{8-4 \sqrt{3}} \sin x-3 \sqrt{2(1+\cos 2 x)}-2$, $g(y)=3+2 \sqrt{11-\sqrt{3}} \cos y-\cos 2 y$ and estimate these two functions.
1) $f(x)=\sqrt{8-4 \sqrt{3}} \sin x-6|\cos x|-2 \leq \sqrt{8-4 \sqrt{3}} \cdot 1-6 \cdot 0-2=\sqrt{8-4 \sqrt{3}}-2=\sqrt{(\sqrt{6}-\sqrt{2})^{2}}-2$ $=\s... | 33 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,475 |
9.2. Find the minimum value of the expression
$(3 \sqrt{2(1+\cos 2 x)}-\sqrt{8-4 \sqrt{3}} \sin x+2) \cdot(3+2 \sqrt{11-\sqrt{3}} \cos y-\cos 2 y) \cdot$ If the answer is not an integer, round it to the nearest integer. | Answer: -33. Instructions. Exact answer: $4 \sqrt{3}-40$. | 4\sqrt{3}-40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,476 |
9.3. Find the maximum value of the expression $(\sqrt{3-\sqrt{2}} \sin x-\sqrt{2(1+\cos 2 x)}-1) \cdot(3+2 \sqrt{7-\sqrt{2}} \cos y-\cos 2 y) \cdot$ If the answer is not an integer, round it to the nearest integer. | Answer: 9. Instructions. Exact answer: $12-2 \sqrt{2}$. | 12-2\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,477 |
9.4. Find the minimum value of the expression $(\sqrt{2(1+\cos 2 x)}-\sqrt{3-\sqrt{2}} \sin x+1) \cdot(3+2 \sqrt{7-\sqrt{2}} \cos y-\cos 2 y)$. If the answer is not an integer, round it to the nearest integer. | Answer: -9 . Instructions. Exact answer: $2 \sqrt{2}-12$. | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,478 |
9.5. Find the maximum value of the expression $(\sqrt{36-4 \sqrt{5}} \sin x-\sqrt{2(1+\cos 2 x)}-2) \cdot(3+2 \sqrt{10-\sqrt{5}} \cos y-\cos 2 y) \cdot$ If the answer is not an integer, round it to the nearest integer. | Answer: 27. Instructions. Exact answer: $36-4 \sqrt{5}$. | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,479 |
9.6. Find the minimum value of the expression $(\sqrt{2(1+\cos 2 x)}-\sqrt{36-4 \sqrt{5}} \sin x+2) \cdot(3+2 \sqrt{10-\sqrt{5}} \cos y-\cos 2 y) \cdot$ If the answer is not an integer, round it to the nearest integer. | Answer: -27 . Instructions. Exact answer: $4 \sqrt{5}-36$. | -27 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,480 |
9.7. Find the maximum value of the expression $(\sqrt{9-\sqrt{7}} \sin x-\sqrt{2(1+\cos 2 x)}-1) \cdot(3+2 \sqrt{13-\sqrt{7}} \cos y-\cos 2 y)$. If the answer is not an integer, round it to the nearest integer. | Answer: 19. Instructions. Exact answer: $24-2 \sqrt{7}$. | 24-2\sqrt{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,481 |
9.8. Find the minimum value of the expression
$(\sqrt{2(1+\cos 2 x)}-\sqrt{9-\sqrt{7}} \sin x+1) \cdot(3+2 \sqrt{13-\sqrt{7}} \cos y-\cos 2 y) \cdot$ If the answer is not an integer, round it to the nearest integer. | Answer: -19. Instructions. Exact answer: $2 \sqrt{7}-24$. | 2\sqrt{7}-24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,482 |
10.1. Find the largest integer $k$ such that for at least one natural number $n>1000$, the number $n!=1 \cdot 2 \cdot \ldots \cdot n$ is divisible by $2^{n+k+2}$. | Answer: -3. Solution. The highest power of two that divides $n!$ is the finite sum $\left[\frac{n}{2^{1}}\right]+\left[\frac{n}{2^{2}}\right]+\ldots \leq \frac{n}{2^{1}}+\frac{n}{2^{2}}+\ldots<n$, so it does not exceed $n-1$. At the same time, the equality of this power to the value $n-1$ is achieved at powers of two (... | -3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,483 |
2. Several boys and girls gathered around a round table. It is known that exactly for 7 girls, girls are sitting to their left, and for 12 - boys. It is also known that for $75\%$ of the boys, girls are sitting to their right. How many people are sitting at the table? | Answer: 35 people.
Solution: From the condition, it is clear that there are exactly 19 girls.
Note that the number of girls who have boys sitting to their left is equal to the number of boys who have girls sitting to their right. Thus, $75\%$ of the boys equals 12, i.e., there are 16 boys in total sitting at the tabl... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,485 |
5. Find the sum of the digits of the number $\underbrace{44 \ldots 4}_{2012 \text { times }} \cdot \underbrace{99 \ldots 9}_{2012 \text { times }}$ | Answer: 18108.
Solution: Note that $\underbrace{4 \ldots 4}_{2012} \cdot \underbrace{9 \ldots 9}_{2012}=\underbrace{4 \ldots 4}_{2012} \underbrace{0 \ldots 0}_{2012}-\underbrace{4 \ldots 4}_{2012}=\underbrace{4 \ldots 4}_{2011} 3 \underbrace{5 \ldots 5}_{2011} 6$. The sum of the digits is $4 \cdot 2011+3+5 \cdot 2011+... | 18108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,486 |
6. A mad painter runs through the cells of a $2012 \times 2013$ board, initially painted in black and white. At the very beginning, he runs into a corner cell. After the painter leaves a cell, the color of that cell changes. Can the painter always run across the board and jump off from one of the border cells so that a... | Answer: Yes, always.
Solution: If the painter runs from a corner cell to an arbitrary white cell, then returns to the corner cell along the same route, the specified cell will change color, while all other cells will remain the same color. This way, the color of all cells except the corner cell can be changed. If the ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,487 |
8. Point $P$ lies inside triangle $A B C$. It is connected to all vertices of the triangle, and perpendiculars are dropped from it to the sides, forming 6 triangles. It turns out that 4 of them are equal. Does this always mean that the triangle is isosceles? | Answer: Not necessarily.
Solution: An example of such a triangle that is not isosceles. Here, $\triangle A B C$ is a right triangle, and point $P$ is obtained by the intersection of the angle bisector of $\angle B C A$ and the perpendicular bisector of side $A C$.
 Find the sum of the squares of two numbers if it is known that their arithmetic mean is 8, and the geometric mean is $2 \sqrt{5}$. | Answer: 216.
Solution. If $a$ and $b$ are the numbers in question, then $a+b=16, a b=(2 \sqrt{5})^{2}=20$, therefore
$$
a^{2}+b^{2}=(a+b)^{2}-2 a b=256-40=216 .
$$ | 216 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,490 |
1.2.1. (2 points) Let $x_{1}$ and $x_{2}$ be the roots of the equation $\sqrt{14} x^{2}-\sqrt{116} x+\sqrt{56}=0$. Calculate $\left|\frac{1}{x_{1}^{2}}-\frac{1}{x_{2}^{2}}\right|$. Answer: $\frac{\sqrt{29}}{14} \approx 0.38$. | Solution. For the equation $a x^{2}+b x+c=0$ we have
$$
\left|\frac{1}{x_{1}^{2}}-\frac{1}{x_{2}^{2}}\right|=\frac{\left|x_{2}-x_{1}\right| \cdot\left|x_{2}+x_{1}\right|}{x_{1}^{2} \cdot x_{2}^{2}}=\frac{\left|\frac{\sqrt{D}}{a}\right| \cdot\left|\frac{-b}{a}\right|}{\frac{c^{2}}{a^{2}}}=\frac{|b| \sqrt{D}}{c^{2}}
$$ | \frac{\sqrt{29}}{14} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,491 |
2.1.1. (2 points) Find the greatest value of $x$ that satisfies the inequality
$$
\left(6+5 x+x^{2}\right) \sqrt{2 x^{2}-x^{3}-x} \leqslant 0
$$ | Answer: 1.
Solution.
$$
\left[\begin{array} { c }
{ 2 x ^ { 2 } - x ^ { 3 } - x = 0 ; } \\
{ \{ \begin{array} { c }
{ 2 x ^ { 2 } - x ^ { 3 } - x > 0 ; } \\
{ 6 + 5 x + x ^ { 2 } \leqslant 0 ; }
\end{array} }
\end{array} \Longleftrightarrow \left[\begin{array} { l }
{ - x ( x - 1 ) ^ { 2 } = 0 ; } \\
{ \{ \begin{a... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,493 |
3.1.1. (12 points) Calculate $\sin (\alpha+\beta)$, if $\sin \alpha+\cos \beta=\frac{1}{4}$ and $\cos \alpha+\sin \beta=-\frac{8}{5}$. Answer: $\frac{249}{800} \approx 0.31$. | Solution. Squaring the given equalities and adding them, we get
$$
\sin ^{2} \alpha+2 \sin \alpha \cos \beta+\cos ^{2} \beta+\cos ^{2} \alpha+2 \cos \alpha \sin \beta+\sin ^{2} \beta=\frac{1}{16}+\frac{64}{25}=\frac{1049}{400}
$$
from which we find $2 \sin (\alpha+\beta)+2=\frac{1049}{400}$, i.e., $\sin (\alpha+\beta... | \frac{249}{800} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,494 |
3.2.1. (12 points) The equation $x^{2}+5 x+1=0$ has roots $x_{1}$ and $x_{2}$. Find the value of the expression
$$
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}
$$ | Answer: 220.
Solution. Since $x_{1}^{2}=-5 x_{1}-1$, then $\left(1+x_{1}\right)^{2}=1+2 x_{1}-5 x_{1}-1=-3 x_{1}$. Therefore,
$$
\begin{gathered}
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}=6\left(\frac{-5 x_{1}-1}{-3 x_{2}}+\frac{-5 x_{2}-1}{-3 x_{1}}\right)=\frac{... | 220 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,498 |
3.3.1. (12 points) Simplify the expression
$$
\left(\frac{2}{\sqrt[3]{3}}+3\right)-\left(\frac{\sqrt[3]{3}+1}{2}-\frac{1}{\sqrt[3]{9}+\sqrt[3]{3}+1}-\frac{2}{1-\sqrt[3]{3}}\right): \frac{3+\sqrt[3]{9}+2 \sqrt[3]{3}}{2}
$$
Answer: 3. | Solution. Let's introduce the notation $a=\sqrt[3]{3}$. Then the expression transforms to
$$
\left(\frac{2}{a}+3\right)-\left(\frac{a+1}{a^{3}-1}-\frac{1}{a^{2}+a+1}-\frac{2}{1-a}\right): \frac{a^{3}+a^{2}+2 a}{a^{3}-1}
$$
and after simplifications and performing the division, it transforms to $\left(\frac{2}{a}+3\ri... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,499 |
4.1.1. (12 points) From point $A$ to point $B$, a bus and a cyclist departed simultaneously at 13:00. After arriving at point $B$, the bus, without stopping, headed back and met the cyclist at point $C$ at 13:10. Upon returning to point $A$, the bus again, without stopping, headed to point $B$ and caught up with the cy... | Answer: 40.
Solution. Let $v_{1}$ and $v_{2}$ be the speeds (in km/h) of the cyclist and the bus, respectively. By the time of the first meeting, they have collectively traveled $\frac{1}{6} v_{1}+\frac{1}{6} v_{2}=8$ km. Until the next meeting, the time that has passed is equal to $\frac{s_{0}}{v_{1}}=\frac{2 \cdot \... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,500 |
4.2.1. (12 points) Every morning, each member of the Ivanov family drinks an 180-gram cup of coffee with milk. The amount of milk and coffee in their cups varies. Masha Ivanova found out that she drank $\frac{2}{9}$ of all the milk consumed that morning and $\frac{1}{6}$ of all the coffee consumed that morning. How man... | Answer: 5.
Solution. Let there be $x$ (for example, grams) of milk and $y$ grams of coffee, and $n$ people in the family. Since each family member drank the same amount of coffee with milk, then $\left(\frac{2 x}{9}+\frac{y}{6}\right) n=$ $=x+y \Leftrightarrow(4 x+3 y) n=18 x+18 y \Leftrightarrow 2 x(2 n-9)=3 y(6-n)$.... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,501 |
4.3.1. (12 points) On the table, there are 13 weights arranged in a row by mass (the lightest on the left, the heaviest on the right). It is known that the mass of each weight is an integer number of grams, the masses of any two adjacent weights differ by no more than 5 grams, and the total mass of the weights does not... | Answer: 185.
Solution. If the mass of the heaviest weight is $m$, then the masses of the other weights will be no less than $m-5, m-10, \ldots, m-60$ grams, and their total mass will be no less than $13 m-390$ grams. Then $13 m-390 \leq 2019$, from which $m \leq 185$. It remains to verify that the set of weights with ... | 185 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,502 |
4.4.1. (12 points) A goat eats 1 hay wagon in 6 weeks, a sheep in 8 weeks, and a cow in 3 weeks. How many weeks will it take for 5 goats, 3 sheep, and 2 cows to eat 30 such hay wagons together? | Answer: 16.
Solution. The goat eats hay at a rate of $1 / 6$ cart per week, the sheep - at a rate of $1 / 8$ cart per week, the cow - at a rate of $1 / 3$ cart per week. Then 5 goats, 3 sheep, and 2 cows together will eat hay at a rate of $\frac{5}{6}+\frac{3}{8}+\frac{2}{3}=\frac{20+9+16}{24}=\frac{45}{24}=\frac{15}{... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,503 |
5.1.1. (12 points) In an acute-angled triangle $A B C$, angle $A$ is equal to $35^{\circ}$, segments $B B_{1}$ and $C C_{1}$ are altitudes, points $B_{2}$ and $C_{2}$ are the midpoints of sides $A C$ and $A B$ respectively. Lines $B_{1} C_{2}$ and $C_{1} B_{2}$ intersect at point $K$. Find the measure (in degrees) of a... | Answer: 75.
Solution. Note that angles $B$ and $C$ of triangle $ABC$ are greater than $\angle A=35^{\circ}$ (otherwise it would be an obtuse triangle), so point $C_{1}$ lies on side $AB$ between points $B$ and $C_{2}$, and point $B_{1}$ lies on side $AC$ between points $C$ and $B_{2}$. Therefore, the point $K$ of inte... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,504 |
5.2.1. (12 points) $B L$ is the bisector of triangle $A B C$. Find its area, given that $|A L|=2,|B L|=3 \sqrt{10},|C L|=3$. | Answer: $\frac{15 \sqrt{15}}{4} \approx 14.52$.
Solution. By the property of the angle bisector of a triangle, $A B: B C=A L: L C=2: 3$, so if $A B=2 x$, then $B C=3 x$. Then, since $B L^{2}=A B \cdot B C-A L \cdot L C$, we get $90=6 x^{2}-6$, from which $x=4$. Thus, triangle $A B C$ has sides 8, 12, and 5. Therefore,... | \frac{15\sqrt{15}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,505 |
5.2.2. $B L$ is the bisector of triangle $A B C$. Find its area, given that $|A L|=3$, $|B L|=6 \sqrt{5},|C L|=4$. | Answer: $\frac{21 \sqrt{55}}{4} \approx 38.94$. | \frac{21\sqrt{55}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,506 |
5.2.3. $B L$ is the bisector of triangle $A B C$. Find its area, given that $|A L|=2$, $|B L|=\sqrt{30},|C L|=5$. | Answer: $\frac{7 \sqrt{39}}{4} \approx 10.93$. | \frac{7\sqrt{39}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,507 |
5.2.4. $B L$ is the bisector of triangle $A B C$. Find its area, given that $|A L|=4$, $|B L|=2 \sqrt{15},|C L|=5$. | Answer: $\frac{9 \sqrt{231}}{4} \approx 34.2$. | \frac{9\sqrt{231}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,508 |
5.3.1. (12 points) Among all possible triangles $ABC$ such that $BC=2 \sqrt[4]{3}, \angle BAC=\frac{\pi}{3}$, find the one with the maximum area. What is this area? | Answer: 3.
Solution. The locus of points from which the segment $B C$ is "seen" at an angle $\alpha$ consists of arcs of two circles, from the centers of which the segment $B C$ is "seen" at an angle $2 \pi-2 \alpha$. The points on these arcs that are farthest from the segment are their midpoints. Therefore, the desir... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,509 |
6.1.1. (12 points) Solve the equation
$$
5^{\sqrt{x^{3}+3 x^{2}+3 x+1}}=\sqrt{\left(5 \sqrt[4]{(x+1)^{5}}\right)^{3}}
$$
In the answer, write the root if there is only one, or the sum of the roots if there are several. | Answer: $\frac{49}{16} \approx 3.06$.
Solution. The equation is equivalent to the following: $(x+1)^{3 / 2}=\frac{3}{2}(x+1)^{5 / 4}$, or $(x+1)\left((x+1)^{1 / 4}-\frac{3}{2}\right)=$ 0 , from which we find the roots: $x=-1, x=\frac{81}{16}-1=\frac{65}{16}$. Their sum is $\frac{49}{16} \approx 3.06$. | \frac{49}{16} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,510 |
6.2.1. (12 points) Solve the system of equations $\left\{\begin{array}{l}9 y^{2}-4 x^{2}=144-48 x, \\ 9 y^{2}+4 x^{2}=144+18 x y .\end{array}\right.$
Given the solutions $\left(x_{1} ; y_{1}\right),\left(x_{2} ; y_{2}\right), \ldots,\left(x_{n} ; y_{n}\right)$, write the sum of the squares
$$
x_{1}^{2}+x_{2}^{2}+\ldo... | Answer: 68.
Solution. From the first equation, we get $(3 y)^{2}=(2 x-12)^{2}$ and sequentially substitute into the second equation $y=\frac{2 x-12}{3}$ and $y=\frac{12-2 x}{3}$. The solutions obtained are: $(x ; y)=(0 ; 4),(0 ;-4),(6 ; 0)$. Answer: $6^{2}+4^{2}+(-4)^{2}=68$. | 68 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,512 |
6.3.1. (12 points) Solve the equation $2 x^{3}+24 x=3-12 x^{2}$.
Answer: $\sqrt[3]{\frac{19}{2}}-2 \approx 0.12$. | Solution. $2 x^{3}+24 x=3-12 x^{2} \Leftrightarrow 2\left(x^{3}+6 x^{2}+12 x+8\right)-16-3=0 \Leftrightarrow(x+2)^{3}=\frac{19}{2} \Leftrightarrow x=\sqrt[3]{\frac{19}{2}}-2$. | \sqrt[3]{\frac{19}{2}}-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,513 |
6.4.1. (12 points) Solve the inequality $\sqrt{x-4}+\sqrt{x+1}+\sqrt{2 x}-\sqrt{33-x}>4$. In your answer, write the sum of all its integer solutions. | Answer: 525.
Solution. Rewrite the inequality as
$$
\sqrt{x-4}+\sqrt{x+1}+\sqrt{2 x}>\sqrt{33-x}+4
$$
Find the domain of the variable $x$:
$$
\left\{\begin{array}{l}
x-4 \geqslant 0 \\
x+1 \geqslant 0 \\
2 x \geqslant 0 \\
33-x \geqslant 0
\end{array}\right.
$$
from which $x \in[4,33]$. We are interested in the in... | 525 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,516 |
7.1.1. (12 points) Find the greatest negative root of the equation
$$
\frac{\sin \pi x - \cos 2 \pi x}{(\sin \pi x - 1)^2 + \cos^2 \pi x - 1} = 0
$$ | Answer: -0.5.
Solution. Transforming the equation, we get
$$
\frac{\sin \pi x-1+2 \sin ^{2} \pi x}{2 \sin \pi x-1}=0, \quad \frac{(\sin \pi x+1)(2 \sin \pi x-1)}{2 \sin \pi x-1}=0
$$
from which $\sin \pi x=-1, x=-\frac{1}{2}+2 n, n \in \mathbb{Z}$. The greatest negative number among these is $-\frac{1}{2}$. | -0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,517 |
7.1.3. Find the greatest negative root of the equation
$$
\frac{\sin \pi x - \cos 2 \pi x}{(\sin \pi x + 1)^2 + \cos^2 \pi x} = 0.
$$ | Answer: $-\frac{7}{6} \approx-1.17$ | -\frac{7}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,518 |
7.2.1. (12 points) Calculate the value of the expression $\arccos \frac{\sqrt{6}+1}{2 \sqrt{3}}-\arccos \sqrt{\frac{2}{3}}$. Write the obtained expression in the form $\frac{a \pi}{b}$, where $a$ and $b$ are integers, and are coprime, and specify the value of $|a-b|$.
# | # Answer: 7.
Solution. Since $\alpha=\arccos \frac{\sqrt{6}+1}{2 \sqrt{3}} \in\left(0 ; \frac{\pi}{2}\right)$ and $\beta=\arccos \sqrt{\frac{2}{3}} \in\left(0 ; \frac{\pi}{2}\right)$, then $A=\alpha-\beta \in\left(-\frac{\pi}{2} ; \frac{\pi}{2}\right)$. Therefore, $\sin A=\sin \alpha \cos \beta-\cos \alpha \sin \beta$... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,519 |
7.3.1. (12 points) Find the sum of all real roots of the equation
$$
\sin \left(\pi\left(x^{2}-x+1\right)\right)=\sin (\pi(x-1))
$$
belonging to the interval $[0 ; 2]$. | Answer: 4.73.
Solution. The equation is equivalent to the system
$$
\left[\begin{array}{l}
\pi\left(x^{2}-x+1\right)=\pi(x-1)+2 \pi k \\
\pi\left(x^{2}-x+1\right)=\pi-(x-1) \pi+2 \pi n
\end{array}(k, n \in \mathbb{Z})\right.
$$
By transforming the system, we get that
$$
\left[\begin{array} { l }
{ x ^ { 2 } - x + ... | 4.73 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,520 |
7.4.1. (12 points) How many integer roots of the equation
$$
\cos 2 \pi x + \cos \pi x = \sin 3 \pi x + \sin \pi x
$$
lie between the roots of the equation $x^{2} + 10 x - 17 = 0$? | # Answer: 7.
Solution. If $x=2 k, k \in \mathbb{Z}$, then $\cos 2 \pi x+\cos \pi x=\cos 4 \pi k+\cos 2 \pi k=2, \sin 3 \pi x+\sin \pi x=$ $\sin 6 \pi k+\sin 2 \pi k=0$, so there are no even numbers among the roots of the first equation.
If, however, $x=2 k+1, k \in \mathbb{Z}$, then $\cos 2 \pi x+\cos \pi x=\cos (4 \... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,521 |
8.1.1. (12 points) Among the first hundred elements of the arithmetic progression $3,7,11, \ldots$ find those that are also elements of the arithmetic progression $2,9,16, \ldots$ In your answer, indicate the sum of the found numbers. | Answer: 2870.
Solution.
$$
\begin{gathered}
a_{n}=a_{1}+(n-1) d_{1}=3+(n-1) \cdot 4 \\
b_{m}=b_{1}+(m-1) d_{2}=2+(m-1) \cdot 7 \\
3+(n-1) \cdot 4=2+(m-1) \cdot 7 \\
4(n+1)=7 m, \quad m=4 k, \quad n=7 k-1
\end{gathered}
$$
Consider the sequence of coinciding terms of the progressions $A_{k}$. For $k=1$, we find $n=6$... | 2870 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,522 |
8.2.1. (12 points) The first, second, and third terms of a geometric progression are pairwise distinct and are equal to the second, fourth, and seventh terms of a certain arithmetic progression, respectively, and the product of these three numbers is 64. Find the first term of the geometric progression. | Answer: $\frac{8}{3} \approx 2.67$.
Solution. Let the first term of the geometric progression be denoted by $b$. Then the second term of the geometric progression is $b q=b+2 d$, and the third term is $b q^{2}=b+5 d$. The product of the three numbers is $(b q)^{3}=64$, from which we obtain $b q=4$. Therefore, $b=4-2 d... | \frac{8}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,523 |
8.3.1. (12 points) Five numbers form an increasing arithmetic progression. The sum of their cubes is zero, and the sum of their squares is 70. Find the smallest of these numbers. | Answer: $-2 \sqrt{7} \approx-5.29$.
Solution. Let's represent the terms of the progression as
$$
a-2 d, a-d, a, a+d, a+2 d
$$
Then the sum of the squares is
$$
(a-2 d)^{2}+(a-d)^{2}+a^{2}+(a+d)^{2}+(a+2 d)^{2}=5 a^{2}+10 d^{2}=70
$$
and the sum of the cubes is
$$
(a-2 d)^{3}+(a-d)^{3}+a^{3}+(a+d)^{3}+(a+2 d)^{3}=... | -2\sqrt{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,524 |
8.3.3. Five numbers form a decreasing arithmetic progression. The sum of their cubes is zero, and the sum of their fourth powers is 136. Find the smallest of these numbers. | Answer: $-2 \sqrt{2} \approx-2.83$. | -2\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,526 |
8.3.5. Five numbers form a decreasing arithmetic progression. The sum of their cubes is zero, and the sum of their fourth powers is 306. Find the smallest of these numbers. | Answer: $-2 \sqrt{3} \approx-3.46$. | -2\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,528 |
9.1.1. (12 points) Find the minimum value of the function $f(x)=\operatorname{tg}^{2} x+3 \operatorname{tg} x+6 \operatorname{ctg} x+4 \operatorname{ctg}^{2} x-1$ on the interval $\left(0 ; \frac{\pi}{2}\right)$. | Answer: $3+6 \sqrt{2} \approx 11.49$.
Solution. For $x>0$, we have $t=\operatorname{tg} x+2 \operatorname{ctg} x \geqslant 2 \sqrt{2}$ (equality is achieved at $x=\operatorname{arctg} \sqrt{2}$), and $f(x)=(\operatorname{tg} x+2 \operatorname{ctg} x)^{2}-3(\operatorname{tg} x+2 \operatorname{ctg} x)-5$. Therefore, we ... | 3+6\sqrt{2}\approx11.49 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,529 |
9.1.2. Find the minimum value of the function $f(x)=\operatorname{tg}^{2} x-4 \operatorname{tg} x-12 \operatorname{ctg} x+9 \operatorname{ctg}^{2} x-3$ on the interval $\left(-\frac{\pi}{2} ; 0\right)$. | Answer: $3+8 \sqrt{3} \approx 16.86$. | 3+8\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,530 |
9.1.3. Find the minimum value of the function $f(x)=\operatorname{tg}^{2} x-4 \operatorname{tg} x-8 \operatorname{ctg} x+4 \operatorname{ctg}^{2} x+5$ on the interval $\left(\frac{\pi}{2} ; \pi\right)$. | Answer: $9+8 \sqrt{2} \approx 20.31$. | 9+8\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,531 |
9.1.4. Find the minimum value of the function $f(x)=\operatorname{tg}^{2} x+2 \operatorname{tg} x+6 \operatorname{ctg} x+9 \operatorname{ctg}^{2} x+4$ on the interval $\left(0 ; \frac{\pi}{2}\right)$. | Answer: $10+4 \sqrt{3} \approx 16.93$ | 10+4\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,532 |
9.2.1. (12 points) Find the minimum value of the expression $\cos (x+y)$, given that $\cos x+\cos y=\frac{1}{3}$. | Answer: $-\frac{17}{18} \approx-0.94$.
Solution. We have $2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}=\frac{1}{3}$, from which $\left|\cos \frac{x+y}{2}\right| \geqslant \frac{1}{6}$, therefore by the double-angle cosine formula $\cos (x+y)=2\left|\cos \frac{x+y}{2}\right|^{2}-1 \geqslant-\frac{17}{18}$. The minimum value... | -\frac{17}{18} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,533 |
9.2.2. Find the maximum value of the expression $\cos (x+y)$, given that $\cos x - \cos y = \frac{1}{4}$. | Answer: $\frac{31}{32} \approx 0.97$. | \frac{31}{32} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,534 |
9.2.4. Find the maximum value of the expression $\cos (x-y)$, given that $\sin x-\sin y=$ $\frac{3}{4}$. | Answer: $\frac{23}{32} \approx 0.72$. | \frac{23}{32} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,535 |
10.1.1. (12 points) The numbers $a$ and $b$ are such that the polynomial $x^{4}+x^{3}+2 x^{2}+a x+b$ is the square of some other polynomial. Find $b$. | Answer: $\frac{49}{64} \approx 0.77$.
Solution. Let $\left(x^{2}+A x+B\right)^{2}=x^{4}+x^{3}+2 x^{2}+a x+b$. Expanding the brackets, we get
$$
x^{4}+2 A x^{3}+\left(A^{2}+2 B\right) x^{2}+2 A B x+B^{2}=x^{4}+x^{3}+2 x^{2}+a x+b,
$$
therefore, by equating the coefficients of the powers, we find $2 A=1, A^{2}+2 B=2,2... | \frac{49}{64} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,536 |
10.2.1. (12 points) Find the greatest integer value of $a$ for which the equation
$$
\sqrt[3]{x^{2}-(a+7) x+7 a}+\sqrt[3]{3}=0
$$
has at least one integer root. | Answer: 11.
Solution. For the desired values of $a$, the numbers $x-a$ and $x-7$ are integers, and their product is -3, so there are 4 possible cases:
$$
\left\{\begin{array} { l }
{ x - a = 1 , } \\
{ x - 7 = - 3 ; }
\end{array} \quad \left\{\begin{array} { l }
{ x - a = 3 , } \\
{ x - 7 = - 1 ; }
\end{array} \qua... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,540 |
10.3.1. (12 points) The curve given by the equation $y=2^{p} x^{2}+5 p x-2^{p^{2}}$ intersects the $O x$ axis at points $A$ and $B$, and the $O y$ axis at point $C$. Find the sum of all values of the parameter $p$ for which the center of the circle circumscribed around triangle $A B C$ lies on the $O x$ axis. | Answer: -1.
Solution. Let $x_{1}$ and $x_{2}$ be the roots of the quadratic trinomial $a x^{2}+b x+c (c \neq 0)$, the graph of which intersects the $O x$ axis at points $A$ and $B$, and the $O y$ axis at point $C$. Then $A\left(x_{1} ; 0\right), B\left(x_{2} ; 0\right), C(0 ; c)$. The fact that the center of the circl... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,541 |
10.4.1. (12 points) Find all pairs of integers $(x, y)$ that are solutions to the equation
$$
7 x y-13 x+15 y-37=0 \text {. }
$$
In your answer, indicate the sum of the found values of $x$. | Answer: 4.
Solution. Multiplying both sides of the equation by 7, we get
$$
7 \cdot 7 x y - 13 \cdot 7 x + 15 \cdot 7 y - 37 \cdot 7 = 0
$$
Rewriting this equation, we have
$$
(7 x + 15)(7 y - 13) = 64
$$
If $x$ and $y$ are integers, then the numbers $7 x + 15$ and $7 y - 13$ are also integers and leave a remainde... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,542 |
1. Know-it-all informed the little people that the average consumption of watermelon syrup in Green City was 10 barrels per day in December and 5 barrels per day in January, respectively. From this, Don't-know-it-all concluded that the number of days in December when syrup consumption was at least 10 barrels was certai... | Answer: No.
Solution. We will provide an example to show that Nезнайка (Nезнайка is a character name, often translated as "Dontknow") is incorrect. Suppose from December 1 to December 30, the dwarfs drank 0 barrels, and on December 31, they drank 310 barrels. Then, on average, they drank $\frac{310}{31}=10$ barrels of... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,543 |
2. A kitten bites off a quarter of a sausage from one end, after which a puppy bites off a third of the remaining piece from the opposite end, then the kitten again - a quarter from its end, and the puppy - a third from its end, and so on. It is required to tie a thread around the sausage in advance so that no one eats... | Answer: $1: 1$.
Solution. Let $\ell$ be the length of the sausage remaining after the same number of "bites" by the kitten and the puppy. Then on the next step, the kitten will bite off $\frac{\ell}{4}$, and the puppy will bite off $-\frac{1}{3} \frac{3 \ell}{4}=\frac{\ell}{4}$. This means both will eat the same part ... | 1:1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,544 |
3. The sequence $a_{1}, a_{2}, \ldots$ is defined by the equalities
$$
a_{1}=100, \quad a_{n+1}=a_{n}+\frac{1}{a_{n}}, \quad n \in \mathbb{N}
$$
Find the integer closest to $a_{2013}$. | # Answer: 118.
Solution.
\[
\begin{aligned}
a_{2013}^{2}=\left(a_{2012}+\frac{1}{a_{2012}}\right)^{2}=a_{2012}^{2}+2+\frac{1}{a_{2012}^{2}}=a_{2011}^{2} & +2 \cdot 2+\frac{1}{a_{2011}^{2}}+\frac{1}{a_{2012}^{2}}=\ldots \\
& =a_{1}^{2}+2 \cdot 2012+\frac{1}{a_{1}^{2}}+\ldots+\frac{1}{a_{2011}^{2}}+\frac{1}{a_{2012}^{2... | 118 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,545 |
4. The quiz participants were asked four questions: 90 participants answered the first question correctly, 50 answered the second, 40 answered the third, and 20 answered the fourth, and no one was able to answer more than two questions correctly. What is the minimum number of participants in the quiz under these condit... | Answer: 100.
Solution. The total number of answers is 200. Since no one answered more than two questions, the minimum possible number of participants in the quiz is 100, and in this case, each participant in the quiz must answer exactly 2 questions correctly. We will provide an example of a quiz where the described si... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,546 |
5. A fixed light ray falls on a mirror, forming an acute angle $\alpha$ with its projection on the mirror's plane. The mirror is rotated around the indicated projection by an acute angle $\beta$. Find the angle between the two reflected rays obtained before and after the rotation. | Answer: $\arccos \left(1-2 \sin ^{2} \alpha \sin ^{2} \beta\right)$.
Solution. Let $A O$ be the incident ray, $O N_{1}$ be the normal to the initial position of the mirror, and $O N_{2}$ be the normal to the second position of the mirror. Consider the trihedral angle $O A N_{1} N_{2}$. Its dihedral angles are equal to... | \arccos(1-2\sin^{2}\alpha\sin^{2}\beta) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,547 |
6. The figure on the coordinate plane consists of points $(x, y)$, satisfying for any $t \in \mathbb{R}$ the two inequalities
$$
x^{2}+y^{2}<\pi^{2}+t^{2}, \quad \cos y<2+\cos 2 x+\cos x(4 \sin t-1)-\cos 2 t
$$
Find the area of this figure. | Answer: $\pi^{3}-2 \pi^{2}$.
Solution. The figure, the coordinates of the points of which satisfy the inequality $x^{2}+y^{2}\cos y+\cos x$
From the last inequality, it follows that $(x, y)$ satisfying this inequality for all $t$ are precisely $(x, y)$ satisfying the inequality
$$
\cos y+\cos x 0 } \\
{ \operatornam... | \pi^{3}-2\pi^{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,548 |
7. Vovochka wrote on the board the equality $101=11011$. The computer science teacher said that this equality would be true if it is understood as the same number written in different number systems. Find the bases of these number systems. | # Answer: 18 and 4.
Solution. Let the base of the first number be $n$, and the second $-k$. Then the equality written by Vovochka is equivalent to the equation
$$
n^{2}+1=k^{4}+k^{3}+k+1 \Leftrightarrow n^{2}=k^{4}+k^{3}+k, \quad n, k \in \mathbb{N}
$$
Multiply the last equation by 4 and factor out perfect squares:
... | 184 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,549 |
8. Find the minimum value of the discriminant of a quadratic trinomial, the graph of which has no common points with the regions located below the x-axis and above the graph of the function $y=\frac{1}{\sqrt{1-x^{2}}}$. | Answer: -4.
Solution. Since the discriminant of a quadratic trinomial
$$
D=b^{2}-4 a c
$$
we will find the maximum value of the product $a c$. Since for all $x$ the inequality $a x^{2}+b x+c \geqslant 0$ holds, then $a \geqslant 0, c \geqslant 0$. Let
$$
f(x)=\frac{1}{\sqrt{1-x^{2}}}-\left(a x^{2}+b x+c\right)
$$
... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,550 |
9. In triangle $ABC$, the bisectors $AL, BM$, and $CN$ are drawn, and $\angle ANM = \angle ALC$. Find the radius of the circumcircle of triangle $LMN$, two sides of which are equal to 3 and 4. | Answer: 2
Solution.
The following statement is true:
Let $A L, B M$ and $C N$ be the bisectors of the internal angles of a triangle. If $\angle A N M = \angle A L C$, then $\angle B C A = 120^{\circ}$.
Proof. Draw $M D \| A L$, then $\angle C M D = \frac{\alpha}{2}$. The angle $\angle A L C$ is an external angle of... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,551 |
10. For which natural numbers $n$ and $k$ do the inequalities $\left|x_{1}\right|+\ldots+\left|x_{k}\right| \leqslant n$ and $\left|y_{1}\right|+\ldots+\left|y_{n}\right| \leqslant k$ have the same number of integer solutions $\left(x_{1}, \ldots, x_{k}\right)$ and $\left(y_{1}, \ldots, y_{n}\right)$? | Answer: for any natural $n$ and $k$.
## Solution.
I. First, we will prove that for any natural $n$ and $k$ and any $l$, the inequalities:
$$
x_{1}+\ldots+x_{k} \leqslant n \quad \text { (1) } \quad \text { and } \quad y_{1}+\ldots+y_{n} \leqslant k
$$
have the same number of solutions in which all variable values a... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,552 |
Problem 1. Vasya and Petya ran out from the starting point of a circular running track at the same time and ran in opposite directions. They met at some point on the track. Vasya ran a full lap and, continuing to run in the same direction, reached the place of their previous meeting at the moment when Petya had run a f... | Answer: $\frac{\sqrt{5}+1}{2}$.
Solution. Let $v-$ be Petya's speed, $x v$ - Vasya's speed, $t$ - the time it took for them to reach the meeting point. Then from the condition we have the equation $\frac{(1+x) v t}{x v}=\frac{x v t}{v}$, from which $x^{2}=1+x, x=\frac{\sqrt{5}+1}{2}$.
Answer to variant 2: $\frac{\sqr... | \frac{\sqrt{5}+1}{2} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,553 |
Problem 2. The distance between the roots of the quadratic trinomial $x^{2}+p x+q$ is 1. Find the coefficients $p$ and $q$, given that they are prime numbers.
Answer: $p=3, q=2$. | Solution. The square of the distance between the roots of the quadratic polynomial is
$$
\left|x_{1}-x_{2}\right|^{2}=x_{1}^{2}+x_{2}^{2}-2 x_{1} x_{2}=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}=p^{2}-4 q=1
$$
from which we get $(p-1)(p+1)=4 q$. Both factors on the left side are even, and one of them is divisible by ... | p=3,q=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,554 |
Problem 3. A clock shows exactly one o'clock. A mosquito and a fly are sitting at the same distance from the center on the hour and minute hands, respectively. When the hands coincide, the insects swap places. How many times greater is the distance that the mosquito has traveled in half a day compared to the distance t... | Answer: $83 / 73$.
Solution. The mosquito and the fly move in a circle. In the first hour, the mosquito will cover $11 / 12$ of this circle (at the beginning, it was on the hour hand, pointing to 1, at the end - on the minute hand, pointing to 12). In the second hour, the mosquito will cover $3 / 12$ of the circle (it... | \frac{83}{73} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,555 |
Problem 4. Each cell of a $3 \times 3$ table is painted in one of three colors such that cells sharing a side have different colors. Among all possible such colorings, find the proportion of those in which exactly two colors are used. | Answer: $1 / 41$.
Solution. The central cell can be painted in any of the three colors, let's call this color $a$. Each of the four cells that share a side with the central one can be painted in any of the two remaining colors. Let the cell above the central one be painted in color $b$. The third color we will call $c... | \frac{1}{41} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,556 |
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