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Problem 6. We will "flip" each of the equations in the system:
$$
\left\{\begin{array}{l}
\frac{a+b}{a b}=1 \\
\frac{b+c}{b c}=\frac{1}{2} \\
\frac{c+a}{c a}=\frac{1}{4}
\end{array}\right.
$$
(The transformation is equivalent since none of the right-hand sides can become zero).
Notice that $\frac{a+b}{a b}=\frac{1}{... | Answer. $a=\frac{8}{3}, b=\frac{8}{5}, c \stackrel{=}{=}-8$. | =\frac{8}{3},b=\frac{8}{5},=-8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,256 |
Problem 4. Let
$$
2^{x}=\left(1+\operatorname{tg} 0.01^{\circ}\right)\left(1+\operatorname{tg} 0.02^{\circ}\right)\left(1+\operatorname{tg} 0.03^{\circ}\right) \ldots\left(1+\operatorname{tg} 44.99^{\circ}\right)
$$
Find $x$. If necessary, round your answer to 0.01. | Problem 4. Let
$$
2^{x}=\left(1+\operatorname{tg} 0.01^{\circ}\right)\left(1+\operatorname{tg} 0.02^{\circ}\right)\left(1+\operatorname{tg} 0.03^{\circ}\right) \ldots\left(1+\operatorname{tg} 44.99^{\circ}\right)
$$
Find $x$. If necessary, round your answer to 0.01.
Answer: 2249.5. | 2249.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,260 |
Problem 6. In quadrilateral $A B C D$, the lengths of sides $B C$ and $C D$ are 2 and 6, respectively. The points of intersection of the medians of triangles $A B C, B C D$, and $A C D$ form an equilateral triangle. What is the maximum value that the area of quadrilateral $A B C D$ can take? If necessary, round your an... | Problem 6. In quadrilateral $A B C D$, the lengths of sides $B C$ and $C D$ are 2 and 6, respectively. The points of intersection of the medians of triangles $A B C, B C D$, and $A C D$ form an equilateral triangle. What is the maximum value that the area of quadrilateral $A B C D$ can take? If necessary, round your an... | 10\sqrt{3}+12\approx29.32 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,262 |
Task 1. Represent in the form of an irreducible fraction:
$$
\frac{12+15}{18}+\frac{21+24}{27}+\ldots+\frac{48+51}{54}
$$ | Answer: $\frac{171}{20}$.
## Solution.
$$
\begin{gathered}
\frac{12+15}{18}+\frac{21+24}{27}+\ldots+\frac{48+51}{54}=\frac{4+5}{6}+\frac{7+8}{9}+\ldots+\frac{16+17}{18}= \\
=\frac{(6-2)+(6-1)}{6}+\frac{(9-2)+(9-1)}{9}+\ldots+\frac{(18-2)+(18-1)}{18}=2+\frac{-3}{6}+2+\frac{-3}{9}+\ldots+2+\frac{-3}{18}= \\
=2 \times 5... | \frac{171}{20} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,263 |
Problem 2. Andrei, Maksim, Igor, and Kolya competed in a cycling race. When asked who took which place, they answered:
Andrei: - I was neither first nor last.
Maksim: - I was not last.
Igor: - I was first.
Kolya: - I was last.
It is known that three boys answered honestly and only one lied. Which of the boys lied? | Answer: Igor lied.
Solution. If Igor and Kolya told the truth, then they took the first and last place, so Andrey and Maksim could not have taken these places, which means they also told the truth and no one lied, which is impossible. Therefore, either Igor or Kolya lied. If Kolya lied, then the last place was not tak... | Igorlied | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,264 |
Problem 3. For natural numbers $x$ and $y$ and an odd integer $z$, it is known that $x! + y! = 24z + 2017$. Find all possible such triples of numbers $(x, y, z)$. (Recall that $1! = 1, 2! = 1 \cdot 2, n! = 1 \cdot 2 \cdot \ldots \cdot n$) | Answer: $(1 ; 4 ;-83),(4 ; 1 ;-83),(1 ; 5 ;-79),(5 ; 1 ;-79)$.
Solution. The expression $x!+y!$ must be an odd number, so $x=1$ or $y=1$.
Let $x=1$, then $1+y!=24z+2017$ or $y!=24z+2016$. Since $2016=24 \cdot 84$, $y!$ is divisible by 24, hence $y \geqslant 4$. If $y \geqslant 6$, then $y!$ is divisible by 48 and $20... | (1;4;-83),(4;1;-83),(1;5;-79),(5;1;-79) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,265 |
Problem 4. Let $L$ be the intersection point of the diagonals $C E$ and $D F$ of a regular hexagon $A B C D E F$ with side length 3. Point $K$ is such that $\overrightarrow{L K}=3 \overrightarrow{A B}-\overrightarrow{A C}$. Determine whether point $K$ lies inside, on the boundary, or outside of $A B C D E F$, and also ... | Answer: $\sqrt{3}$, the point lies outside the hexagon.
Solution. Let $O$ be the center of the hexagon. It is known that then $F E D O$ is a rhombus, from which $F D \perp E O$. Similarly, $E C \perp D O$. Therefore, point $L$ is the center of the equilateral triangle $D E O$.
Next, since $3 \overrightarrow{A B}-\ove... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,266 |
Problem 5. Solve the system of equations in real numbers
$$
\left\{\begin{array}{l}
x+y+2-4xy=0 \\
y+z+2-4yz=0 \\
z+x+2-4zx=0
\end{array}\right.
$$ | Answer: $(1,1,1),\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\right)$.
Solution. Note that $x \neq \frac{1}{4}$ : otherwise, substituting into the first equation, we get $\frac{1}{4}+2=0$, which is false. Therefore, from the first and third equations, we obtain $y=-\frac{x+2}{1-4 x}, z=-\frac{x+2}{1-4 x}$, from which ... | (1,1,1),(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,267 |
Problem 6. Compare the numbers $\frac{\sin 2016^{\circ}}{\sin 2017^{\circ}}$ and $\frac{\sin 2018^{\circ}}{\sin 2019^{\circ}}$. | Answer: The second expression is greater.
Solution. Note that $1800^{\circ}=5 \cdot 360^{\circ}$, so it is sufficient to compare the numbers $\frac{\sin 216^{\circ}}{\sin 217^{\circ}}$ and $\frac{\sin 211^{\circ}}{\sin 219^{\circ}}$. Note that the numbers $\sin 217^{\circ}$ and $\sin 219^{\circ}$ have the same sign, s... | Theexpressionisgreater. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,268 |
Problem 7. In an isosceles trapezoid $A B C D$ with bases $A D$ and $B C (A D > B C)$, the lateral side is 20 cm, and the angle $B A C$ is $45^{\circ}$. Let $O$ be the center of the circle circumscribed around $A B C D$. It turns out that the lines $O D$ and $A B$ are perpendicular. Find the length of the base $A D$ of... | Answer: $10(\sqrt{6}+\sqrt{2})$ cm.
Solution. Since $O$ - the center of the circumscribed circle - lies on the perpendicular bisector of segment $AB$, point $D$ also lies on this perpendicular bisector. Therefore, $AD = BD$, from which $\angle DAB = \angle DBA = \alpha$.
Due to the symmetry of the isosceles trapezoid... | 10(\sqrt{6}+\sqrt{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,269 |
Problem 8. For what values of the parameter $a$ does the equation
$$
4^{|x-a|} \log _{1 / 3}\left(x^{2}-2 x+4\right)+2^{x^{2}-2 x} \log _{\sqrt{3}}(2|x-a|+3)=0
$$
have exactly three solutions? | Answer: For $a=\frac{1}{2} ; 1 ; \frac{3}{2}$.
Solution. Let's simplify the expression:
$$
4^{|x-a|} \log _{1 / 3}\left(x^{2}-2 x+4\right)+2^{x^{2}-2 x} \log _{\sqrt{3}}(2|x-a|+3)=0
$$
$$
\begin{gathered}
2^{2|x-a|}\left(-\log _{3}\left(x^{2}-2 x+4\right)\right)+2^{x^{2}-2 x} 2 \log _{3}(2|x-a|+3)=0 \\
2^{2|x-a|+3}\... | \frac{1}{2};1;\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,270 |
Problem 9. In a football championship, 20 teams participate, each playing against each other once. What is the minimum number of games that must be played so that among any three teams, there are two that have already played against each other? | Answer: 90 games.
Solution. We will consider the games that have not been played. The condition means that the unplayed games do not form triangles. We will prove by induction on $k$ that for $2k$ teams, the maximum number of unplayed games is no more than $k^2$.
Base case: $k=1$ (the estimate is obvious).
Inductive... | 90 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,271 |
Problem 10. In a triangular pyramid $A B C D$ with base $A B C$, the lateral edges are pairwise perpendicular, $D A=D B=2, D C=5$. A light ray is emitted from a point on the base. After reflecting exactly once from each lateral face (the ray does not reflect from the edges), the ray hits a point on the base of the pyra... | Answer: $\frac{10 \sqrt{6}}{9}$.
Solution. Let $N$ be the point from which the ray is emitted, and $K$ be the point on the base of the pyramid where the ray will finally hit.
We will sequentially "straighten" the path of the ray as follows: if at some moment the ray should reflect off a certain plane, we will assume ... | \frac{10\sqrt{6}}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,272 |
Task 1. Represent in the form of an irreducible fraction:
$$
\frac{8+10}{12}+\frac{14+16}{18}+\ldots+\frac{32+34}{36}
$$ | Answer: $\frac{171}{20}$.
## Solution.
$$
\begin{gathered}
\frac{8+10}{12}+\frac{14+16}{18}+\ldots+\frac{32+34}{36}=\frac{4+5}{6}+\frac{7+8}{9}+\ldots+\frac{16+17}{18}= \\
=\frac{(6-2)+(6-1)}{6}+\frac{(9-2)+(9-1)}{9}+\ldots+\frac{(18-2)+(18-1)}{18}=2+\frac{-3}{6}+2+\frac{-3}{9}+\ldots+2+\frac{-3}{18}= \\
=2 \times 5-... | \frac{171}{20} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,273 |
Problem 2. Misha, Anton, Katya, and Natasha organized a table tennis tournament. When asked who took which place, they answered:
Misha: - I was neither first nor last.
Anton: - I was not last.
Katya: - I was first.
Natasha: - I was last.
It is known that one of the kids lied, while the other three told the truth. ... | Answer: Misha took third place.
Solution. First, let's find out who among the children lied. If Katya and Natasha told the truth, then they took the first and last places, so Misha and Anton cannot have taken these places, which means they also told the truth and no one lied, which is impossible. Therefore, either Kat... | Misha | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,274 |
Problem 3. For natural numbers $x$ and $y$ and an odd integer $z$, it is known that $x! + y! = 48z + 2017$. Find all possible such triples of numbers $(x, y, z)$. (Recall that $1! = 1, 2! = 1 \cdot 2, n! = 1 \cdot 2 \cdot \ldots \cdot n$) Answer: $(1; 6; -27), (6; 1; -27), (1; 7; 63), (7; 1; 63)$. | Solution. The expression $x!+y!$ must be an odd number, so $x=1$ or $y=1$.
Let $x=1$, then $1+y!=48z+2017$ or $y!=48z+2016$. Since $2016=48 \cdot 42$, $y!$ is divisible by 48, hence $y \geqslant 6$. If $y \geqslant 8$, then $y!$ is divisible by 96 and $2016=96 \cdot 21$. In this case, $z$ would be even, which contradi... | (1;6;-27),(6;1;-27),(1;7;63),(7;1;63) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,275 |
Problem 4. Let $L$ be the intersection point of the diagonals $C E$ and $D F$ of a regular hexagon $A B C D E F$ with side length 4. Point $K$ is such that $\overrightarrow{L K}=3 \overrightarrow{F A}-\overrightarrow{F B}$. Determine whether point $K$ lies inside, on the boundary, or outside of $A B C D E F$, and also ... | Answer: $\frac{4 \sqrt{3}}{3}$.
Solution. Let $O$ be the center of the hexagon. It is known that then $F E D O$ is a rhombus, from which $F D \perp E O$. Similarly, $E C \perp D O$. Therefore, point $L$ is the center of the equilateral triangle $D E O$.
Next, since $3 \overrightarrow{F A}-\overrightarrow{F B}=\overri... | \frac{4\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,276 |
Problem 5. Solve the system of equations in real numbers
$$
\left\{\begin{array}{l}
x+y+2+4xy=0 \\
y+z+2+4yz=0 \\
z+x+2+4zx=0
\end{array}\right.
$$ | Answer: No solutions.
Solution. Note that $x \neq-\frac{1}{4}$ : otherwise, substituting into the first equation, we get $-\frac{1}{4}+2=0$, which is false. Therefore, from the first and third equations, we obtain $y=-\frac{x+2}{1+4 x}, z=-\frac{x+2}{1+4 x}$, from which $y=z$. Similarly, $x=y$. Therefore, $x=y=z$.
Su... | Nosolutions | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,277 |
Problem 6. Compare the numbers $\frac{\cos 2014^{\circ}}{\cos 2015^{\circ}}$ and $\frac{\cos 2016^{\circ}}{\cos 2017^{\circ}}$. | Answer: The second expression is greater.
Solution. Note that $1800^{\circ}=5 \cdot 360^{\circ}$, so it is sufficient to compare the numbers $\frac{\cos 214^{\circ}}{\cos 215^{\circ}}$ and $\frac{\cos 216^{\circ}}{\cos 21^{\circ}}$. Note that the numbers $\cos 215^{\circ}$ and $\cos 217^{\circ}$ have the same sign, so... | Theexpressionisgreater. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,278 |
Problem 8. For what values of the parameter $a$ does the equation
$$
8^{|x-a|} \log _{1 / 5}\left(x^{2}+2 x+5\right)+2^{x^{2}+2 x} \log _{\sqrt{5}}(3|x-a|+4)=0
$$
have exactly three solutions? | Answer: For $a=-\frac{7}{4},-1,-\frac{1}{4}$.
Solution. Let's simplify the expression:
$$
\begin{gathered}
8^{|x-a|} \log _{1 / 5}\left(x^{2}+2 x+5\right)+2^{x^{2}+2 x} \log _{\sqrt{5}}(3|x-a|+4)=0 \\
2^{3|x-a|}\left(-\log _{5}\left(x^{2}+2 x+5\right)\right)+2^{x^{2}+2 x} 2 \log _{5}(3|x-a|+4)=0 \\
2^{3|x-a|+4}\left(... | -\frac{7}{4},-1,-\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,279 |
Problem 9. In a football championship, 16 teams participate, each playing against each other once. What is the minimum number of games that must be played so that among any three teams, there are two that have already played against each other?
# | # Answer: 56 games.
Solution. We will consider the games that have not been played. The condition means that there will be no three teams that have not played against each other at all. We will prove by induction on $k$ that for $2k$ teams, the maximum number of unplayed games is no more than $k^2$.
Base case: $k=1$ ... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,280 |
Problem 10. In a triangular pyramid $ABCD$ with base $ABC$, the lateral edges are pairwise perpendicular, $DA=DB=2, DC=5$. A light ray is emitted from a point on the base. After reflecting exactly once from each lateral face (the ray does not reflect from the edges), the ray hits a point on the base of the pyramid. Wha... | Answer: $\frac{10 \sqrt{6}}{9}$.
Solution. Let $N$ be the point from which the ray is emitted, and $K$ be the point on the base of the pyramid where the ray will finally hit.
We will sequentially "straighten" the path of the ray as follows: if at some moment the ray should reflect off a certain plane, we will assume ... | \frac{10\sqrt{6}}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,281 |
Task 1. Represent in the form of an irreducible fraction:
$$
\frac{4+8}{12}+\frac{16+20}{24}+\ldots+\frac{64+68}{72}
$$ | Answer: $\frac{191}{20}$.
## Solution.
$$
\begin{gathered}
\frac{4+8}{12}+\frac{16+20}{24}+\ldots+\frac{64+68}{72}=\frac{1+2}{3}+\frac{4+5}{6}+\ldots+\frac{16+17}{18}= \\
=\frac{(3-2)+(3-1)}{3}+\frac{(6-2)+(6-1)}{6}+\ldots+\frac{(18-2)+(18-1)}{18}= \\
=2+\frac{-3}{3}+2+\frac{-3}{6}+2+\frac{-3}{9}+\ldots+2+\frac{-3}{1... | \frac{191}{20} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,282 |
Problem 2. Anna, Vera, Katya, and Natasha competed in the long jump. When asked who took which place, they answered:
Anna: - I was neither first nor last.
Vera: - I was not last.
Katya: - I was first.
Natasha: - I was last.
It is known that one of the girls lied, while the other three told the truth. Which of the ... | Answer: Vера took the first place.
Solution. If Katya and Natasha told the truth, then they took the first and last places, so Anna and Vера could not have taken these places, which means they also told the truth and no one lied, which is impossible. Therefore, either Katya or Natasha lied. If Natasha lied, then no on... | Vера | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,283 |
Problem 3. For natural numbers $x$ and $y$ and an odd integer $z$, it is known that $x! + y! = 8z + 2017$. Find all possible such triples of numbers $(x, y, z)$. (Recall that $1! = 1, 2! = 1 \cdot 2, n! = 1 \cdot 2 \cdot \ldots \cdot n$) Answer: $(1; 4; -249), (4; 1; -249), (1; 5; -237), (5; 1; -237)$. | Solution. The expression $x!+y!$ must be an odd number, so $x=1$ or $y=1$.
Let $x=1$, then $1+y!=8z+2017$ or $y!=8z+2016$. Since $2016=8 \cdot 252$, $y!$ is divisible by 8, hence $y \geqslant 4$. If $y \geqslant 6$, then $y!$ is divisible by 16 and $2016=16 \cdot 126$. In this case, $z$ would be even, which contradict... | (1;4;-249),(4;1;-249),(1;5;-237),(5;1;-237) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,284 |
Problem 4. Let $L$ be the intersection point of the diagonals $C E$ and $D F$ of a regular hexagon $A B C D E F$ with side length 2. Point $K$ is such that $\overrightarrow{L K}=\overrightarrow{A C}-3 \overrightarrow{B C}$. Determine whether point $K$ lies inside, on the boundary, or outside of $A B C D E F$, and also ... | Answer: $\frac{2 \sqrt{3}}{3}$.
Solution. Let $O$ be the center of the hexagon. It is known that then $F E D O$ is a rhombus, from which $F D \perp E O$. Similarly, $E C \perp D O$. Therefore, point $L$ is the center of the equilateral triangle $D E O$.
Next, since $\overrightarrow{A C}-3 \overrightarrow{B C}=\overri... | \frac{2\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,285 |
Problem 5. Solve the system of equations in real numbers
$$
\left\{\begin{array}{l}
x+y-2-4xy=0 \\
y+z-2-4yz=0 \\
z+x-2-4zx=0
\end{array}\right.
$$ | Answer: No solutions.
Solution. Note that $x \neq \frac{1}{4}$ : otherwise, substituting into the first equation, we get $\frac{1}{4}-2=0$, which is false. Therefore, from the first and third equations, we obtain $y=-\frac{x-2}{1-4 x}, z=-\frac{x-2}{1-4 x}$, from which $y=z$. Similarly, $x=y$. Therefore, $x=y=z$.
Sub... | Nosolutions | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,286 |
Problem 6. Compare the numbers $\frac{\sin 2014^{\circ}}{\sin 2015^{\circ}}$ and $\frac{\sin 2016^{\circ}}{\sin 2017^{\circ}}$. | Answer: The second expression is greater.
Solution. Note that $1800^{\circ}=5 \cdot 360^{\circ}$, so it is sufficient to compare the numbers $\frac{\sin 214^{\circ}}{\sin 215^{\circ}}$ and $\frac{\sin 216^{\circ}}{\sin 217^{\circ}}$. Note that the numbers $\sin 215^{\circ}$ and $\sin 217^{\circ}$ have the same sign, s... | Theexpressionisgreater. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,287 |
Problem 7. In an isosceles trapezoid $ABCD$ with bases $AD$ and $BC (AD > BC)$, the lateral side is 40 cm, and the angle $BAC$ is $45^{\circ}$. Let $O$ be the center of the circle circumscribed around $ABCD$. It turns out that the point $O$ lies on the line connecting point $D$ and the midpoint of side $AB$. Find the l... | Answer: $20(\sqrt{6}+\sqrt{2})$ cm.
Solution. Since $O$ - the center of the circumscribed circle - lies on the perpendicular bisector of segment $AB$, point $D$ also lies on this perpendicular bisector. Therefore, $AD = BD$, from which $\angle DAB = \angle DBA = \alpha$.
Due to the symmetry of the isosceles trapezoid... | 20(\sqrt{6}+\sqrt{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,288 |
Problem 9. In a football championship, 16 teams participate, each playing against each other once. What is the minimum number of games that must be played so that among any three teams, there are two that have already played against each other? | Answer: 56 games.
Solution. We will consider the games that have not been played. The condition means that there will be no three teams that have not played with each other at all. We will prove by induction on $k$ that for $2k$ teams, the maximum number of unplayed games is no more than $k^2$.
Base case: $k=1$ (the ... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,289 |
Problem 10. In a triangular pyramid $A B C D$ with base $A B C$, the lateral edges are pairwise perpendicular, $D A=D B=5, D C=1$. A light ray is emitted from a point on the base. After reflecting exactly once from each lateral face (the ray does not reflect from the edges), the ray hits a point on the base of the pyra... | Answer: $\frac{10 \sqrt{3}}{9}$.
Solution. Let $N$ be the point from which the ray is emitted, and $K$ be the point on the base of the pyramid where the ray will finally hit.
We will sequentially "straighten" the path of the ray as follows: if at some moment the ray should reflect off a certain plane, we will assume ... | \frac{10\sqrt{3}}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,290 |
Task 1. Represent in the form of an irreducible fraction:
$$
\frac{5+10}{15}+\frac{20+25}{30}+\ldots+\frac{80+85}{90}
$$ | Answer: $\frac{191}{20}$.
## Solution.
$$
\begin{gathered}
\quad \frac{5+10}{15}+\frac{20+25}{30}+\ldots+\frac{80+85}{90}=\frac{1+2}{3}+\frac{4+5}{6}+\ldots+\frac{16+17}{18}= \\
=\frac{(3-2)+(3-1)}{3}+\frac{(6-2)+(6-1)}{6}+\ldots+\frac{(18-2)+(18-1)}{18}= \\
=2+\frac{-3}{3}+2+\frac{-3}{6}+2+\frac{-3}{9}+\ldots+2+\fra... | \frac{191}{20} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,291 |
Problem 2. Misha, Anton, Petya, and Fyodor competed in parallel bars exercises. When asked who took which place, they answered:
Misha: - I was neither first nor last.
Anton: - I was not last.
Petya: - I was first.
Fyodor: - I was last.
It is known that one of the boys lied, while the other three told the truth. Wh... | Answer: Fedia took the last place.
Solution. Suppose Fedia lied. Then no one of the boys could take the last place, which is impossible. Therefore, Fedia told the truth and was the last. | Fedia | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,292 |
Problem 3. For natural numbers $x$ and $y$ and an odd integer $z$, it is known that $x! + y! = 16z + 2017$. Find all possible such triples of numbers $(x, y, z)$. (Recall that $1! = 1, 2! = 1 \cdot 2, n! = 1 \cdot 2 \cdot \ldots \cdot n$) | Answer: $(1 ; 6 ;-81),(6 ; 1 ;-81),(1 ; 7 ; 189),(7 ; 1 ; 189)$.
Solution. The expression $x!+y!$ must be an odd number, so $x=1$ or $y=1$.
Let $x=1$, then $1+y!=16z+2017$ or $y!=16z+2016$. Since $2016=16 \cdot 126$, $y!$ is divisible by 16, hence $y \geqslant 6$. If $y \geqslant 8$, then $y!$ is divisible by 32 and ... | (1;6;-81),(6;1;-81),(1;7;189),(7;1;189) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,293 |
Problem 4. Let $L$ be the intersection point of the diagonals $C E$ and $D F$ of a regular hexagon $A B C D E F$ with side length 5. Point $K$ is such that $\overrightarrow{L K}=\overrightarrow{F B}-3 \overrightarrow{A B}$. Determine whether point $K$ lies inside, on the boundary, or outside of $A B C D E F$, and also ... | Answer: $\frac{5 \sqrt{3}}{3}$.
Solution. Let $O-$ be the center of the hexagon. It is known that then $F E D O$ is a rhombus, from which $F D \perp E O$. Similarly, $E C \perp D O$. Therefore, point $L$ is the center of the equilateral triangle $D E O$. Furthermore, since $\overrightarrow{F B}-3 \overrightarrow{A B}=... | \frac{5\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,294 |
Problem 5. Solve the system of equations in real numbers
$$
\left\{\begin{array}{l}
x+y-2+4xy=0 \\
y+z-2+4yz=0 \\
z+x-2+4zx=0
\end{array}\right.
$$ | Answer: $(-1,-1,-1),\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$.
Solution. Note that $x \neq-\frac{1}{4}$ : otherwise, substituting into the first equation, we get $-\frac{1}{4}-2=0$, which is false. Therefore, from the first and third equations, we obtain $y=-\frac{x-2}{1+4 x}, z=-\frac{x-2}{1+4 x}$, from whi... | (-1,-1,-1),(\frac{1}{2},\frac{1}{2},\frac{1}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,295 |
Problem 6. Compare the numbers $\frac{\cos 2016^{\circ}}{\cos 2017^{\circ}}$ and $\frac{\cos 2018^{\circ}}{\cos 2019^{\circ}}$. | Answer: The second expression is greater
Solution. Note that $1800^{\circ}=5 \cdot 360^{\circ}$, so it is sufficient to compare the numbers $\frac{\cos 216^{\circ}}{\cos 217^{\circ}}$ and $\frac{\cos 218^{\circ}}{\cos 219^{\circ}}$. Note that the numbers $\cos 217^{\circ}$ and $\cos 219^{\circ}$ have the same sign, so... | \cos3<\cos1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,296 |
Problem 1. Given the polynomial $F(x)=1+2 x+3 x^{2}+4 x^{3}+\ldots+100 x^{99}$. Is it possible, by rearranging the coefficients in it, to obtain a polynomial $G(x)=g_{0}+g_{1} x+g_{2} x^{2}+g_{3} x^{3}+\ldots+g_{99} x^{99}$ such that for all natural numbers $k \geqslant 2$ the difference $F(k)-G(k)$ is not divisible by... | Answer: No, it cannot.
Solution. Suppose the opposite and let such a polynomial $G(x)$ exist. We will use the following well-known lemma:
if $H(x)$ is a polynomial with integer coefficients, then for any integers $a$ and $b$, the number $H(a)-H(b)$ is divisible by $a-b$.
Then the numbers $F(101)-F(1)$ and $G(101)-G(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 9,297 |
Problem 2. For what least $n$ do there exist $n$ numbers from the interval $(-1 ; 1)$ such that their sum is 0 and the sum of their squares is 30? | Answer: 32.
Solution. First, let's provide an example of 32 numbers whose sum is 0 and the sum of their squares is 30. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{16}=\sqrt{\frac{15}{16}}, x_{17}=x_{18}=\ldots=x_{32}=-\sqrt{\frac{15}{16}}$ will work.
Now, we will prove that fewer than 32 numbers will not suffice... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,298 |
Problem 3. Points $A$ and $B$, located on a circular avenue, are connected by a straight highway segment 4 km long, which is the diameter of the circular avenue. A pedestrian left his house and started walking along the avenue from point $A$. After 1 hour, he realized he had forgotten his keys and asked a neighbor-cycl... | Answer: after $\frac{2 \pi-2}{21}$ hours.
Solution. For definiteness, let's assume that the pedestrian went for a walk along the circular alley counterclockwise. At point $A$, the cyclist has three options: | \frac{2\pi-2}{21} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,299 |
Problem 4. In a right triangle $ABC$, a circle is constructed on the leg $AC$ as its diameter, which intersects the hypotenuse $AB$ at point $E$. A tangent to the circle is drawn through point $E$, which intersects the leg $CB$ at point $D$. Find the length of $DB$, if $AE=6$, and $BE=2$. | Answer: 2.
Solution. The solution is based on two simple observations. First, $\angle A E C=90^{\circ}$, since it subtends the diameter. Second, $D E$ and $D C$ are tangents to the circle from the condition, so $D E=D C$. Therefore, in the right triangle $C E B$, a point $D$ is marked on the hypotenuse $B C$ such that... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,301 |
Problem 5. Solve the system of equations
$$
\left\{\begin{array}{l}
12 x^{2}+4 x y+3 y^{2}+16 x=-6 \\
4 x^{2}-12 x y+y^{2}+12 x-10 y=-7
\end{array}\right.
$$ | Answer: $x=-\frac{3}{4}, y=\frac{1}{2}$.
Solution. Add the first equation, multiplied by 3, and the second. We get,
$$
40 x^{2}+10 y^{2}+60 x-10 y=-25
$$
after dividing by 10 and transformations, $4\left(x+\frac{3}{4}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=0$. The sum of two squares can only equal zero if each of... | -\frac{3}{4},\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,302 |
Task 6. Find the greatest negative root of the equation
$$
4 \sin (3 x)+13 \cos (3 x)=8 \sin (x)+11 \cos (x)
$$ | Answer: $\frac{\alpha-\beta}{2} \approx-0.1651$, where $\alpha=\operatorname{arctg} \frac{4}{13}, \beta=\operatorname{arctg} \frac{8}{11}$.
Comment 1. Depending on the choice of the solution method and the inverse trigonometric function, the answer may have a different form. Let's provide several ways to describe $\al... | \frac{\alpha-\beta}{2}\approx-0.1651 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,303 |
Problem 7. For what values of the parameter $a$ does the equation $x^{3}-11 x^{2}+a x-8=0$ have three distinct real roots that form a geometric progression? | Answer: only 22.
Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}-11 x^{2}+a x-8=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a polynomial of the third degree:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=11 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{2... | 22 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,304 |
Problem 8. For the tetrahedron $P Q R S$, it is known that $P Q=4, S R=6, \angle Q R S=\angle P S R=50^{\circ}, \angle Q S R=$ $\angle P R S=40^{\circ}$. A sphere is circumscribed around the tetrahedron. Consider the set of all points on this sphere for which the sum of the spherical distances from these points to $P, ... | Answer: $18 \pi$, i.e., half the area of a sphere with radius 3.
Solution. Since $\angle Q R S+\angle Q S R=\angle P R S+\angle P S R=90^{\circ}$, triangles $Q R S$ and $P R S$ are right triangles with a common hypotenuse $R S$. If $O$ is the midpoint of segment $R S$, then by the property of the median of a right tri... | 18\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,305 |
Problem 9. It is known about the functions $p(x)$ and $q(x)$ that $p(0)=q(0)>0$ and $p^{\prime}(x) \sqrt{q^{\prime}(x)}=\sqrt{2}$ for any $x \in[0 ; 1]$. Prove that if $x \in[0 ; 1]$, then $p(x)+2 q(x)>3 x$. | Solution. Note that $p(0)+2 q(0)>0$, so to prove the inequality, it is sufficient to check that the function $p(x)+2 q(x)-3 x$ is increasing on the interval $[0 ; 1]$. To do this, we will prove that its derivative is non-negative on this interval. This can be done in two ways.
The first method, substitution.
$$
p^{\p... | proof | Calculus | proof | Yes | Yes | olympiads | false | 9,306 |
Task 10. Pete needs to solder an electrical circuit consisting of 10 chips connected to each other by wires (one wire connects two different chips; two chips can be connected by no more than one wire), with one chip having 9 wires coming out of it, one chip having 8 wires, one chip having 7 wires, two chips having 5 wi... | Answer: No, it cannot.
Solution. Let's introduce a graph where the vertices are the chips, and an edge between vertices is drawn if and only if the corresponding chips are connected by wires.
We will describe the graph as a sequence of integers $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$, where $a_{i}$ represents the ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,307 |
Problem 1. Given the polynomial $P(x)=1+2 x+3 x^{2}+4 x^{3}+\ldots+101 x^{100}$. Is it possible, by rearranging the coefficients in it, to obtain a polynomial $Q(x)=q_{0}+q_{1} x+q_{2} x^{2}+q_{3} x^{3}+\ldots+q_{100} x^{100}$ such that for all natural numbers $k \geqslant 2$ the difference $P(k)-Q(k)$ is not divisible... | Answer: No, it cannot.
Solution. Suppose the opposite and let such a polynomial $Q(x)$ exist. We will use the following well-known lemma:
if $H(x)$ is a polynomial with integer coefficients, then for any integers $a$ and $b$, the number $H(a)-H(b)$ is divisible by $a-b$.
Then the numbers $P(2021)-P(1)$ and $Q(2021)-... | proof | Algebra | proof | Yes | Yes | olympiads | false | 9,308 |
Problem 2. For what least $n$ do there exist $n$ numbers from the interval $(-1 ; 1)$ such that their sum is 0 and the sum of their squares is 42? | Answer: 44.
Solution. First, let's provide an example of 44 numbers whose sum is 0 and the sum of their squares is 42. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{22}=\sqrt{\frac{21}{22}}, x_{23}=x_{24}=\ldots=x_{44}=-\sqrt{\frac{21}{22}}$ will work.
Now, we will prove that it is impossible to achieve this with ... | 44 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,309 |
Problem 3. Points $A$ and $B$, located on a circular avenue, are connected by a straight highway segment 4 km long, which is the diameter of the circular avenue. A pedestrian left his house and started walking along the avenue from point $A$. After 1 hour, he realized he had forgotten his keys and asked a neighbor-cycl... | Answer: after $\frac{4 \pi-7}{27}$ hours.
Solution. For definiteness, let's assume that the pedestrian went for a walk along the circular alley counterclockwise. At point $A$, the cyclist has three options: | \frac{4\pi-7}{27} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,310 |
Problem 4. In a right triangle $P Q R$, a circle is constructed on the leg $P R$ as its diameter, which intersects the hypotenuse $P Q$ at point $T$. A tangent to the circle is drawn through point $T$, which intersects the leg $R Q$ at point $S$. Find the length of $S Q$, if $P T=15$, and $Q T=5$. | Answer: 5.
Solution. The solution is based on two simple observations. First, $\angle P T R=90^{\circ}$, since it subtends the diameter. Second, $S T$ and $S R$ are tangents to the circle from the given conditions, so $S T=S R$. Therefore, in the right triangle $R T Q$, a point $S$ is marked on the hypotenuse $Q R$ su... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,312 |
Problem 5. Solve the system of equations
$$
\left\{\begin{array}{l}
3 x^{2}+4 x y+12 y^{2}+16 y=-6 \\
x^{2}-12 x y+4 y^{2}-10 x+12 y=-7
\end{array}\right.
$$ | Answer: $x=\frac{1}{2}, y=-\frac{3}{4}$.
Solution. Add the first equation, multiplied by 3, and the second. We get,
$$
10 x^{2}+40 y^{2}-10 x+60 y=-25
$$
after dividing by 10 and transformations, $\left(x-\frac{1}{2}\right)^{2}+4\left(y+\frac{3}{4}\right)^{2}=0$. The sum of two squares can only equal zero if each of... | \frac{1}{2},-\frac{3}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,313 |
Problem 6. Find the greatest negative root of the equation
$$
\sin (x)+8 \cos (x)=4 \sin (8 x)+7 \cos (8 x)
$$ | Answer: $\frac{\alpha+\beta-2 \pi}{9} \approx-0.6266$, where $\alpha=\operatorname{arctg} \frac{1}{8}, \beta=\operatorname{arctg} \frac{4}{7}$.
Comment 1. Depending on the method of solution chosen and the inverse trigonometric function used, the answer can take different forms. Let's present a few more ways to descri... | \frac{\alpha+\beta-2\pi}{9}\approx-0.6266 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,314 |
Problem 7. For what values of the parameter $a$ does the equation $x^{3}-14 x^{2}+a x-27=0$ have three distinct real roots that form a geometric progression? | Answer: only 42.
Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}-14 x^{2}+a x-27=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a cubic polynomial:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=14 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{2} x_{3}=27
\e... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,315 |
Problem 8. For the tetrahedron $X Y Z T$, it is known that $X Y=6, T Z=8, \angle Y Z T=\angle X T Z=25^{\circ}, \angle Y T Z=$ $\angle X Z T=65^{\circ}$. A sphere is circumscribed around the tetrahedron. Consider the set of all points on this sphere for which the sum of the spherical distances from these points to $X, ... | Answer: $32 \pi$, i.e., half the area of a sphere with radius 4.
Solution. Since $\angle Y Z T+\angle Y T Z=\angle X Z T+\angle X T Z=90^{\circ}$, triangles $Y Z T$ and $X Z T$ are right triangles with a common hypotenuse $Z T$. If $O$ is the midpoint of segment $Z T$, then by the property of the median of a right tri... | 32\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,316 |
Problem 9. It is known about the functions $s(x)$ and $t(x)$ that $s(0)=t(0)>0$ and $s^{\prime}(x) \sqrt{t^{\prime}(x)}=5$ for any $x \in[0 ; 1]$. Prove that if $x \in[0 ; 1]$, then $2 s(x)+5 t(x)>15 x$. | Solution. Note that $2 s(0)+5 t(0)>0$, so to prove the inequality, it is sufficient to check that the function $2 s(x)+5 t(x)-15 x$ is increasing on the interval $[0 ; 1]$. To do this, we will prove that its derivative is non-negative on this interval. This can be done in two ways.
The first method, substitution.
$$
... | proof | Calculus | proof | Yes | Yes | olympiads | false | 9,317 |
Problem 10. A club has 10 members, some of whom are friends with each other. Is it possible that in this club one member is friends with 9 others, one member - with 7 others, one member - with 6 others, two members - with 5 others each, two members - with 3 others each, one member - with 2 others, two members - with 1 ... | Answer: No, it cannot.
Solution. Let's introduce a graph where the vertices are the participants of the club, and an edge between vertices is drawn if and only if the corresponding participants of the club are friends.
We will describe the graph as a sequence of integers $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$, wh... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,318 |
Problem 1. Given the polynomial $S(x)=1+3 x+5 x^{2}+7 x^{3}+\ldots+201 x^{100}$. Is it possible, by rearranging the coefficients in it, to obtain a polynomial $T(x)=t_{0}+t_{1} x+t_{2} x^{2}+t_{3} x^{3}+\ldots+t_{100} x^{100}$ such that for all natural numbers $k \geqslant 2$ the difference $S(k)-T(k)$ is not divisible... | Answer: No, it cannot.
Solution. Suppose the opposite and let such a polynomial $T(x)$ exist. We will use the following well-known lemma:
if $H(x)$ is a polynomial with integer coefficients, then for any integers $a$ and $b$, the number $H(a)-H(b)$ is divisible by $a-b$.
Then the numbers $S(2021)-S(1)$ and $T(2021)-... | proof | Algebra | proof | Yes | Yes | olympiads | false | 9,319 |
Problem 2. For what least $n$ do there exist $n$ numbers from the interval $(-1 ; 1)$ such that their sum is 0 and the sum of their squares is 36?
# | # Answer: 38.
Solution. First, let's provide an example of 38 numbers whose sum is 0 and the sum of their squares is 36. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{19}=\sqrt{\frac{18}{19}}, x_{20}=x_{21}=\ldots=x_{38}=-\sqrt{\frac{18}{19}}$ will work.
Now, we will prove that fewer than 38 numbers will not suffi... | 38 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,320 |
Problem 3. Points $A$ and $B$, located on a circular avenue, are connected by a straight highway segment 4 km long, which is the diameter of the circular avenue. A pedestrian left his house from point $A$ and started walking along the avenue. After 1 hour, he realized he had forgotten his keys and asked a neighbor-cycl... | Answer: after $\frac{21-4 \pi}{43}$ hours.
Solution. For definiteness, let's assume that the pedestrian went for a walk along the circular alley counterclockwise. At point $A$, the cyclist has three options: | \frac{21-4\pi}{43} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,321 |
Problem 4. In a right triangle $X Y Z$, a circle is constructed on the leg $X Z$ as its diameter, which intersects the hypotenuse $X Y$ at point $W$. A tangent to the circle is drawn through point $W$, which intersects the leg $Z Y$ at point $V$. Find the length of $V Y$, if $X W=12$, and $Y W=4$. | Answer: 4.
Solution. The solution is based on two simple observations. First, $\angle X W Z=90^{\circ}$, since it subtends the diameter. Second, $V W$ and $V Z$ are tangents to the circle from the condition, so $V W=V Z$. Therefore, in the right triangle $Z W Y$, a point $V$ is marked on the hypotenuse $Y Z$ such that... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,323 |
Problem 5. Solve the system of equations
$$
\left\{\begin{array}{l}
16 x^{2}+8 x y+4 y^{2}+20 x+2 y=-7 \\
8 x^{2}-16 x y+2 y^{2}+20 x-14 y=-11
\end{array}\right.
$$ | Answer: $x=-\frac{3}{4}, y=\frac{1}{2}$.
Solution. Add the first equation, multiplied by 2, and the second. We get,
$$
40 x^{2}+10 y^{2}+60 x-10 y=-25
$$
after dividing by 10 and transformations, $4\left(x+\frac{3}{4}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=0$. The sum of two squares can equal zero only if each of... | -\frac{3}{4},\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,324 |
Problem 6. Find the smallest positive root of the equation
$$
2 \sin (6 x)+9 \cos (6 x)=6 \sin (2 x)+7 \cos (2 x)
$$ | Answer: $\frac{\alpha+\beta}{8} \approx 0.1159$, where $\alpha=\operatorname{arctg} \frac{2}{9}, \beta=\operatorname{arctg} \frac{6}{7}$.
Comment 1. Depending on the choice of the solution method and the inverse trigonometric function, the answer can have a different form. Let's provide several ways to describe $\alph... | \frac{\alpha+\beta}{8}\approx0.1159 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,325 |
Problem 7. For what values of the parameter $a$ does the equation $x^{3}-15 x^{2}+a x-64=0$ have three distinct real roots that form a geometric progression? | Answer: only 60.
Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}-15 x^{2}+a x-64=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a polynomial of the third degree:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=15 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,326 |
Problem 8. For the tetrahedron $K L M N$, it is known that $K L=5, N M=6, \angle L M N=\angle K N M=35^{\circ}, \angle L N M=$ $\angle K M N=55^{\circ}$. A sphere is circumscribed around the tetrahedron. Consider the set of all points on this sphere for which the sum of the spherical distances from these points to $K, ... | Answer: $18 \pi$, i.e., half the area of a sphere with radius 3.
Solution. Since $\angle L M N+\angle L N M=\angle K M N+\angle K N M=90^{\circ}$, triangles $L M N$ and $K M N$ are right triangles with a common hypotenuse $M N$. If $O$ is the midpoint of segment $M N$, then by the property of the median of a right tri... | 18\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,327 |
Problem 9. It is known about the functions $a(x)$ and $b(x)$ that $a(0)=b(0)>0$ and $a^{\prime}(x) \sqrt{b^{\prime}(x)}=2$ for any $x \in[0 ; 1]$. Prove that if $x \in[0 ; 1]$, then $a(x)+8 b(x)>6 x$. | Solution. Note that $a(0)+8 b(0)>0$, so to prove the inequality, it is sufficient to check that the function $a(x)+8 b(x)-6 x$ is increasing on the interval $[0 ; 1]$. To do this, we will prove that its derivative is non-negative on this interval. This can be done in two ways.
The first method, substitution.
$$
a^{\p... | proof | Calculus | proof | Yes | Yes | olympiads | false | 9,328 |
Task 10. In the squad, there are 10 people. Each day, two of them need to be chosen as duty officers (with the same pair not being able to serve twice). Could it be that after several days, one person has been on duty 9 times, two people - 8 times each, two people - 5 times each, four people - 3 times each, and one per... | Answer: No, it could not.
Solution. Let's introduce a graph where the vertices are the members of the squad, and an edge between vertices is drawn if and only if the corresponding members of the squad went on duty together.
We will describe the graph as a sequence of integers $\left(a_{1}, a_{2}, \ldots, a_{n}\right)... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,329 |
Problem 1. Given the polynomial $A(x)=1+3 x+5 x^{2}+7 x^{3}+\ldots+199 x^{99}$. Is it possible, by rearranging the coefficients in it, to obtain a polynomial $B(x)=b_{0}+b_{1} x+b_{2} x^{2}+b_{3} x^{3}+\ldots+b_{99} x^{99}$ such that for all natural numbers $k \geqslant 2$ the difference $A(k)-B(k)$ is not divisible by... | Answer: No, it cannot.
Solution. Suppose the opposite and let such a polynomial $B(x)$ exist. We will use the following well-known lemma:
if $H(x)$ is a polynomial with integer coefficients, then for any integers $a$ and $b$, the number $H(a)-H(b)$ is divisible by $a-b$.
Then the numbers $A(200)-A(1)$ and $B(200)-B(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 9,330 |
Problem 2. For what least $n$ do there exist $n$ numbers from the interval $(-1 ; 1)$ such that their sum is 0 and the sum of their squares is 40? | Answer: 42.
Solution. First, let's provide an example of 42 numbers whose sum is 0 and the sum of their squares is 40. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{21}=\sqrt{\frac{20}{21}}, x_{22}=x_{23}=\ldots=x_{42}=-\sqrt{\frac{20}{21}}$ will work.
Now, we will prove that fewer than 42 numbers will not suffice... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,331 |
Problem 3. Points $A$ and $B$, located on a circular avenue, are connected by a straight highway segment 4 km long, which is the diameter of the circular avenue. A pedestrian left his house from point $A$ and started walking along the avenue. After 1 hour, he realized he had forgotten his keys and asked a neighbor-cycl... | Answer: after $\frac{11}{51}$ hours.
Solution. For definiteness, let's assume that the pedestrian went for a walk along the circular alley counterclockwise. At point $A$, the cyclist has three options: | \frac{11}{51} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,332 |
Problem 4. In a right triangle $KLM$, a circle is constructed on the leg $KM$ as its diameter, which intersects the hypotenuse $KL$ at point $G$. A tangent to the circle is drawn through point $G$, intersecting the leg $ML$ at point $F$. Find the length of $FL$, if $KG=5$ and $LG=4$. | Answer: 3.
Solution. The solution is based on two simple observations. First, $\angle K G M=90^{\circ}$, since it subtends the diameter. Second, $F G$ and $F M$ are tangents to the circle from the condition, so $F G=F M$. Therefore, in the right triangle $M G L$, a point $F$ is marked on the hypotenuse $L M$ such that... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,334 |
Problem 5. Solve the system of equations
$$
\left\{\begin{array}{l}
4 x^{2}+8 x y+16 y^{2}+2 x+20 y=-7 \\
2 x^{2}-16 x y+8 y^{2}-14 x+20 y=-11
\end{array}\right.
$$ | Answer: $x=\frac{1}{2}, y=-\frac{3}{4}$.
Solution. Add the first equation, multiplied by 2, and the second. We get,
$$
10 x^{2}+40 y^{2}-10 x+60 y=-25
$$
after dividing by 10 and transformations, $\left(x-\frac{1}{2}\right)^{2}+4\left(y+\frac{3}{4}\right)^{2}=0$. The sum of two squares can only equal zero if each of... | \frac{1}{2},-\frac{3}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,335 |
Problem 6. Find the smallest positive root of the equation
$$
14 \sin (3 x)-3 \cos (3 x)=13 \sin (2 x)-6 \cos (2 x) .
$$ | Answer: $\frac{2 \pi-\alpha-\beta}{5} \approx 0.7570$, where $\alpha=\operatorname{arctg} \frac{14}{3}, \beta=\operatorname{arctg} \frac{13}{6}$.
Comment 1. Depending on the choice of the solution method and the inverse trigonometric function, the answer can have a different form. Let's provide several ways to describ... | \frac{2\pi-\alpha-\beta}{5}\approx0.7570 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,336 |
Problem 7. For what values of the parameter $a$ does the equation $x^{3}+16 x^{2}+a x+64=0$ have three distinct real roots that form a geometric progression? | Answer: only 64.
Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}+16 x^{2}+a x+64=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a cubic polynomial:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=-16 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{2} x_{3}=-64
... | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,337 |
Problem 8. For the tetrahedron $ABCD$, it is known that $AB=7, DC=8, \angle BCD=\angle ADC=75^{\circ}, \angle BDC=$ $\angle ACD=15^{\circ}$. A sphere is circumscribed around the tetrahedron. Consider the set of all points on this sphere for which the sum of the spherical distances to the points $A, B, C, D$ is no more ... | Answer: $32 \pi$, i.e., half the area of a sphere with radius 4.
Solution. Since $\angle B C D+\angle B D C=\angle A C D+\angle A D C=90^{\circ}$, triangles $B C D$ and $A C D$ are right triangles with a common hypotenuse $C D$. If $O$ is the midpoint of segment $C D$, then by the property of the median of a right tri... | 32\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,338 |
Problem 9. For functions $f(x)$ and $g(x)$, it is known that $f(0)=g(0)>0$ and $f^{\prime}(x) \sqrt{g^{\prime}(x)}=3$ for any $x \in[0 ; 1]$. Prove that if $x \in[0 ; 1]$, then $2 f(x)+3 g(x)>9 x$. | Solution. Note that $2 f(0)+3 g(0)>0$, so to prove the inequality, it is sufficient to check that the function $2 f(x)+3 g(x)-9 x$ is increasing on the interval $[0 ; 1]$. To do this, we will prove that its derivative is non-negative on this interval. This can be done in two ways.
The first method, substitution.
$$
2... | proof | Calculus | proof | Yes | Yes | olympiads | false | 9,339 |
Task 10. In a country, there are 10 cities, some of which are connected by roads (each road connects exactly two cities, there is no more than one road between any two cities, and you can only switch from one road to another in a city). Is it possible that from one city 9 roads lead out, from one - 8, from two - 7 each... | Answer: No, it cannot.
Solution. Let's introduce a graph where the vertices are cities, and an edge between vertices is drawn if and only if the corresponding cities are connected by a road.
We will describe the graph as a sequence of integers $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$, where $a_{i}$ represents the d... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,340 |
Problem 4. Using vectors.
$\overline{H K}=\overline{A K}-\overline{A H}=\frac{1}{2}(\overline{A Q}+\overline{A N})-\frac{1}{2}(\overline{A M}+\overline{A P})=$
$=\frac{1}{2}\left(\frac{1}{2}(\overline{A E}+\overline{A D})+\frac{1}{2}(\overline{A B}+\overline{A C})\right)-$
$-\frac{1}{2}\left(\frac{1}{2} \overline{A ... | Answer: 1.75.
 | 1.75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,341 |
Problem 6. Suppose $T$ is the period of the function $y=f(x)$. From the condition, we get:
$$
\sin x=f(x)-0.5 f(x-\pi)=f(x-T)-0.5 f(x-T-\pi)=\sin (x-T)
$$
Since this equality holds for all $x$, the number $T$ is the period of the function $y=\sin x$, i.e., $T=2 \pi n$, where $n$ is some natural number. From the given... | Answer: $f(x)=\frac{2}{3} \sin x$. | f(x)=\frac{2}{3}\sinx | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,342 |
Problem 9. Each knight gives one affirmative answer to four questions, while a liar gives three. In total, there were $105+45+85+65=300$ affirmative answers. If all the residents of the city were knights, the total number of affirmative answers would be 200. The 100 extra "yes" answers come from the lies of the liars. ... | Answer: in Block B, on 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,344 |
Problem 10. The sum of the surface areas of the polyhedra into which a parallelepiped is divided by sections is equal to the sum of the surface area of the parallelepiped and the areas of the internal surfaces. The sum of the areas of the internal surfaces is equal to twice the sum of the areas of the sections.
Let's ... | Answer: 194.
[^0]: ${ }^{1}$ This statement is a particular case of Cauchy's inequality. | 194 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,345 |
Problem 3. The faces of a die are numbered from 1 to 6. However, the weight of the die is distributed unevenly, and the probability of the number $k$ appearing is directly proportional to $k$. The die is rolled twice in a row. What is the probability that the sum of the numbers rolled will be 7? If necessary, round you... | Answer: $8 / 63 \approx 0.13$. | 8/63\approx0.13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,347 |
Problem 6. Suppose $T$ is the period of the function $y=f(x)$. From the condition, we get:
$$
\sin x=f(x)-0.4 f(x-\pi)=f(x-T)-0.4 f(x-T-\pi)=\sin (x-T)
$$
Since this equality holds for all $x$, the number $T$ is the period of the function $y=\sin x$, i.e., $T=2 \pi n$, where $n$ is some natural number. From the given... | Answer: $f(x)=\frac{5}{7} \sin x$. | f(x)=\frac{5}{7}\sinx | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,348 |
Task 1. Solve the equation $2|x-1| \sin x=x-1$.
---
The text has been translated while preserving the original line breaks and format. | Answer. $1 ; \frac{\pi}{6}+2 \pi n, n=1,2, \ldots ;-\frac{\pi}{6}+2 \pi n, n=0,-1,-2, \ldots ; \pm\left(\frac{5 \pi}{6}+2 \pi n\right), n=0,1,2, \ldots$. | 1;\frac{\pi}{6}+2\pin,n=1,2,\ldots;-\frac{\pi}{6}+2\pin,n=0,-1,-2,\ldots;\(\frac{5\pi}{6}+2\pin),n=0,1,2,\ldots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,351 |
Problem 2. Vanya took three Unified State Exams (USE). He scored 5 points less in Russian than in Physics, and in Physics, he scored 9 points less than in Mathematics. The golden fish that appeared in Vanya's dream promised to fulfill any number of wishes of the following types:
- add one point to each exam;
- decreas... | Answer: No.
Solution: For Vanya to score more than 100 points on more than one exam, the difference in points between these two exams must become zero. However, the difference in points between any two exams either remains unchanged or changes by 4. And initially, the difference in points does not divide by 4 for any ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,352 |
Problem 3. One notebook, 3 notepads, and 2 pens cost 98 rubles, while 3 notebooks and a notepad are 36 rubles cheaper than 5 pens. How much does each item cost, if the notebook costs an even number of rubles? (Each of these items costs a whole number of rubles.) | Answer: $4,22,14$.
Solution. Let the required costs be $-x, y$ and $z$. The condition gives the system
$$
\left\{\begin{array}{l}
x+3 y+2 z=98 \\
3 x+y-5 z=-36
\end{array}\right.
$$
Multiply the first equation by 3 and subtract the second from it. We get that $8 y+11 z=330$. It is clear that $y$ must be divisible by... | 4,22,14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,353 |
Problem 4. Each of the two workers was assigned to process the same number of parts. The first completed the work in 8 hours. The second spent more than 2 hours on setting up the equipment and with its help finished the work 3 hours earlier than the first. It is known that the second worker processed as many parts in 1... | Answer: 4 times.
Solution: Let $x$ be the time spent on equipment setup. Then the second worker worked (on the equipment) $8-3-x=5-x$ hours, producing as much per hour as the first worker in $x+1$ hours. Therefore, $\frac{8}{5-x}=\frac{x+1}{1}$. We get $x^{2}-4 x+3=0$. But by the condition $x>2$, so $x=3$, and the req... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,354 |
Problem 5. Three regular pentagons have a common center, their sides are respectively parallel. The sides of two pentagons are 4 cm and 12 cm. The third pentagon divides the area of the figure enclosed between the first two in the ratio $1: 3$, counting from the smaller pentagon. Find the side of the third pentagon. | Answer: $4 \sqrt{3}$.
Solution: Let the side of the third pentagon be $x$. The areas of similar figures are in the ratio of the squares of corresponding sides. Therefore, the areas enclosed between the pentagons are in the ratio $\left(x^{2}-4^{2}\right):\left(12^{2}-x^{2}\right)=1: 3$, from which the answer follows. | 4\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,355 |
Problem 6. The function $f$ is such that $f(2 x-3 y)-f(x+y)=-2 x+8 y$ for all $x, y$. Find all possible values of the expression $\frac{f(5 t)-f(t)}{f(4 t)-f(3 t)}$. | Answer: 4.
Solution. By substituting $y=-x$, we get that $f(5 x)=-10 x+f(0)$, i.e., $f(t)=-2 t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,356 |
Problem 7. In an isosceles triangle with a perimeter of 60 cm, the point of intersection of the medians lies on the inscribed circle. Find the sides of the triangle. | Answer: 25, 25, 10.
Solution. Let in the original triangle $ABC$, $AB=BC$, $BM$ be the median, and $O$ be the point of intersection of the medians. Note that $OM$ is the diameter of the circle, so $BM=3OM=6r$ (where $r$ is the radius of the inscribed circle).
The area of the triangle, on one hand, is $\frac{1}{2} AC ... | 25,25,10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,357 |
Problem 8. Solve the system
$$
\left\{\begin{aligned}
x+y+z & =13 \\
x^{2}+y^{2}+z^{2} & =61 \\
x y+x z & =2 y z
\end{aligned}\right.
$$ | Answer. $(4 ; 3 ; 6),(4 ; 6 ; 3)$.
Solution. Let $y+z$ be denoted by $p$, and $yz$ by $q$. Then the system becomes
$$
\left\{\begin{aligned}
x+p & =13 \\
x^{2}+p^{2}-2 q & =61 \\
x p & =2 q
\end{aligned}\right.
$$
From the last two equations, $(x+p)^{2}-3 x p=61$, and considering the first equation, $x p=36$. Solvin... | (4;3;6),(4;6;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,358 |
Problem 9. The polyhedron shown in the figure has all dihedral angles that are right angles. Sasha claims that the shortest path along the surface of this polyhedron from vertex $X$ to vertex $Y$ has a length of 4. Is he correct?
. He scored 3 points less in Russian than in Physics, and in Physics, he scored 7 points less than in Mathematics. The golden fish that appeared in Vanya's dream promised to fulfill any number of wishes of the following types:
- add one point to each exam;
- decreas... | Answer: No.
Solution: For Vanya to score more than 100 points on more than one exam, the difference in points between these two exams must become zero. However, the difference in points between any two exams either remains unchanged or changes by 4. And initially, the difference in points does not divide by 4 for any ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,362 |
Problem 5. Three regular octagons have a common center, their sides are respectively parallel. The sides of two octagons are 7 cm and 42 cm. The third octagon divides the area of the figure enclosed between the first two in the ratio $1: 6$, counting from the smaller octagon. Find the side of the third octagon. | Answer: $7 \sqrt{6}$.
Solution: Let the side of the third octagon be $x$. The areas of similar figures are in the ratio of the squares of corresponding sides. Therefore, the areas enclosed between the octagons are in the ratio $\left(x^{2}-6^{2}\right):\left(42^{2}-x^{2}\right)=1: 6$, from which the answer follows. | 7\sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,363 |
Problem 6. The function $f$ is such that $f(2 x-3 y)-f(x+y)=-2 x+8 y$ for all $x, y$. Find all possible values of the expression $\frac{f(4 t)-f(t)}{f(3 t)-f(2 t)}$. | Answer: 3.
Solution. By substituting $y=-x$, we get that $f(5 x)=-10 x+f(0)$, i.e., $f(t)=-2 t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,364 |
Task 1. Solve the equation $\sqrt{2}|x+1| \sin x=x+1$.
The text above has been translated into English, maintaining the original text's line breaks and format. | Answer. $-1 ; \frac{\pi}{4}+2 \pi n, n=0,1,2, \ldots ;-\frac{\pi}{4}+2 \pi n, n=-1,-2, \ldots ; \pm\left(\frac{3 \pi}{4}+2 \pi n\right), n=0,1,2, \ldots$ | -1;\frac{\pi}{4}+2\pin,n=0,1,2,\ldots;-\frac{\pi}{4}+2\pin,n=-1,-2,\ldots;\(\frac{3\pi}{4}+2\pin),n=0,1,2,\ldots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,367 |
Problem 3. A native from the Tanga-Tanga tribe could, in natural exchange, receive 2 drums, 3 wives, and one leopard skin for 111 arrows. Two leopard skins were valued 8 arrows less than 3 drums and 4 wives. How many arrows did a drum, a wife, and a leopard skin cost separately, if a leopard skin required an even numbe... | Solution. Let the required costs be $-x, y$ and $z$. The condition gives the system
$$
\left\{\begin{aligned}
2 x+3 y+z & =111 \\
3 x+4 y-2 z & =8
\end{aligned}\right.
$$
By adding twice the first equation to the second, we get that $7 x+10 y=230$. It is clear that $x$ must be divisible by 10. Therefore, $x$ can be 3... | 20,9,44 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,368 |
Problem 4. If a passenger travels from Moscow to St. Petersburg by a regular train, they will arrive in 10 hours. If they take an express train, which they have to wait more than 2.5 hours for, they will arrive 3 hours earlier than the regular train. Find the ratio of the speeds of the express train and the regular tra... | Answer. The express is 2.5 times faster.
Solution. Let $x$ be the time spent waiting for the express. Then the express traveled for $10-3-x=7-x$ hours, covering in 2 hours the same distance that the train covers in $x+2$ hours. Therefore, $\frac{10}{7-x}=\frac{x+2}{2}$. We get $x^{2}-5 x+6=0$. But by the condition $x>... | 2.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,369 |
Problem 5. Three regular heptagons have a common center, their sides are respectively parallel. The sides of two heptagons are 6 cm and 30 cm. The third heptagon divides the area of the figure enclosed between the first two in the ratio $1: 5$, counting from the smaller heptagon. Find the side of the third heptagon. | Answer: $6 \sqrt{5}$.
Solution: Let the side of the third heptagon be $x$. The areas of similar figures are in the ratio of the squares of corresponding sides. Therefore, the areas enclosed between the heptagons are in the ratio $\left(x^{2}-6^{2}\right):\left(30^{2}-x^{2}\right)=1: 5$, from which the answer follows. | 6\sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,370 |
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