problem
stringlengths 1
7.47k
| solution
stringlengths 0
13.5k
| answer
stringlengths 1
272
| problem_type
stringclasses 8
values | question_type
stringclasses 3
values | problem_is_valid
stringclasses 1
value | solution_is_valid
stringclasses 1
value | source
stringclasses 7
values | synthetic
bool 1
class |
|---|---|---|---|---|---|---|---|---|
4. Let's write the natural number $a$ in its canonical form:
$$
a=p_{1}^{s_{1}} \cdot p_{2}^{s_{2}} \cdot \ldots \cdot p_{n}^{s_{n}}
$$
where $p_{1}, p_{2}, \ldots, p_{n}$ are distinct prime numbers, and $s_{1}, s_{2}, \ldots, s_{n}$ are natural numbers.
It is known that the number of natural divisors of $a$, including 1 and $a$, is equal to $\left(s_{1}+1\right)\left(s_{2}+1\right) \cdot \ldots \cdot\left(s_{n}+1\right)$. According to the problem, this number is 101. Since 101 is a prime number, all the brackets in the product are equal to 1, except for one which is equal to 101. Therefore, the number $a=p^{100}$, and its divisors are $d_{1}=1, d_{2}=p, d_{3}=p^{2}, \ldots$, $d_{101}=p^{100}$.
Let's calculate the sum of the divisors:
$$
\Sigma_{d}=1+p+p^{2}+\ldots+p^{100}=\frac{p^{101}-1}{p-1}=\frac{a \sqrt[100]{a}-1}{\sqrt[100]{a}-1}
$$
Let's calculate the product of the divisors:
$$
\Pi_{d}=p^{1+2+\ldots+100}=p^{50 \cdot 101}=\sqrt{a^{101}}
$$ | Answer: 1) $\Sigma_{d}=\frac{a_{100}^{a}-1}{\sqrt[100]{a}-1}$; 2) $\Pi_{d}=\sqrt{a^{101}}$. | \Sigma_{}=\frac{\sqrt[100]{}-} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. At the intersection of roads $A$ and $B$ (straight lines) is a settlement $C$ (point). Sasha is walking along road $A$ towards point $C$, taking 50 steps per minute, with a step length of 50 cm. At the start of the movement, Sasha was 250 meters away from point $C$. Dan is walking towards $C$ along road $B$ at a speed of 80 steps per minute, with a step length of 40 cm, and at the moment they start moving together, he was 300 meters away from $C$. Each of them, after passing point $C$, continues their movement along their respective roads without stopping. We record the moments of time when both Dan and Sasha have taken an integer number of steps. Find the smallest possible distance between them (along the roads) at such moments of time. How many steps did each of them take by the time this distance was minimized? | Answer: 1) $d_{\text {min }}=15.8$ m; 2) the number of steps Sasha took - 470, the number of steps Dani took - 752. | 15.8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. For what values of $a$ does the equation $\sin 5 a \cdot \cos x - \cos (x + 4 a) = 0$ have two solutions $x_{1}$ and $x_{2}$, such that $x_{1} - x_{2} \neq \pi k, k \in Z$? | Answer: $a=\frac{\pi(4 t+1)}{2}, t \in Z$. | \frac{\pi(4+1)}{2},\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A natural number $a$ has 103 different divisors, including 1 and $a$. Find the sum and product of these divisors. | Answer: 1) $\Sigma_{d}=\frac{a \sqrt[102]{a}-1}{\sqrt[102]{a}-1}$; 2) $\Pi_{d}=\sqrt{a^{103}}$. | \Sigma_{}=\frac{\sqrt[102]{}-1}{\sqrt[102]{}-1};\Pi_{}=\sqrt{} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In a convex quadrilateral $A B C D$, the lengths of sides $B C$ and $A D$ are 2 and $2 \sqrt{2}$ respectively. The distance between the midpoints of diagonals $B D$ and $A C$ is 1. Find the angle between the lines $B C$ and $A D$. | Answer: $\alpha=45^{\circ}$. | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. At the intersection of roads $A$ and $B$ (straight lines) is a settlement $C$ (point). Sasha is walking along road $A$ towards point $C$, taking 40 steps per minute with a step length of 65 cm. At the start of the movement, Sasha was 260 meters away from point $C$. Dan is walking towards $C$ along road $B$ at a speed of 75 steps per minute, with a step length of 50 cm, and at the moment they start moving together, he was 350 meters away from $C$. Each of them, after passing point $C$, continues their movement along their respective roads without stopping. We record the moments of time when both Dan and Sasha have taken an integer number of steps. Find the smallest possible distance between them (along the roads) at such moments of time. How many steps did each of them take by the time this distance was minimized? | Answer: 1) $d_{\text {min }}=18.1$ m; 2) the number of steps Sasha took - 376, the number of steps Dani took - 705. | 18.1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. For which $z$ does the equation $x^{2}+y^{2}+4 z^{2}+2 x y z-9=0$ have a solution for any $y$? | Answer: $1 \leq|z| \leq \frac{3}{2}$. | 1\leq|z|\leq\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A natural number $a$ has 107 different divisors, including 1 and $a$. Find the sum and product of these divisors. | Answer: 1) $\Sigma_{d}=\frac{a \sqrt[106]{a}-1}{\sqrt[106]{a}-1}$; 2) $\Pi_{d}=\sqrt{a^{107}}$. | \Sigma_{}=\frac{\sqrt[106]{}-1}{\sqrt[106]{}-1};\Pi_{}=\sqrt{} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. At the intersection of roads $A$ and $B$ (straight lines) is a settlement $C$ (point). Sasha is walking along road $A$ towards point $C$, taking 45 steps per minute, with a step length of 60 cm. At the start of the movement, Sasha was 290 meters away from point $C$. Dan is walking towards $C$ along road $B$ at a speed of 55 steps per minute, with a step length of 65 cm, and at the moment they start moving together, he was 310 meters away from $C$. Each of them, after passing point $C$, continues their movement along their respective roads without stopping. We record the moments of time when both Dan and Sasha have taken an integer number of steps. Find the smallest possible distance between them (along the roads) at such moments of time. How many steps did each of them take by the time this distance was minimized? | Answer: 1) $d_{\min }=57$ m; 2) the number of steps Sasha took - 396, the number of steps Dani took - 484. | d_{\}=57 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. For which $y$ does the equation $x^{2}+2 y^{2}+8 z^{2}-2 x y z-9=0$ have no solutions for any $z$? | Answer: $\frac{3}{\sqrt{2}}<|y| \leq 2 \sqrt{2}$. | \frac{3}{\sqrt{2}}<|y|\leq2\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A natural number $a$ has 109 different divisors, including 1 and $a$. Find the sum and product of these divisors. | Answer: 1) $\Sigma_{d}=\frac{a \sqrt[108]{a}-1}{\sqrt[108]{a}-1}$; 2) $\Pi_{d}=\sqrt{a^{109}}$. | \sqrt{^{109}} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In a convex quadrilateral $A B C D$, the lengths of sides $B C$ and $A D$ are 4 and 6 respectively. The distance between the midpoints of diagonals $B D$ and $A C$ is 3. Find the angle between the lines $B C$ and $A D$. | Answer: $\alpha=\arccos \frac{1}{3}$. | \alpha=\arccos\frac{1}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Let $A(x ; y)$ be a vertex of a square with its center at the origin, for which $\operatorname{GCD}(x, y)=2$. We will calculate the length of the diagonal of the square $d=2 \sqrt{x^{2}+y^{2}}$ and find the area of the square $S=\frac{d^{2}}{2}=2\left(x^{2}+y^{2}\right)$. According to the problem, $S=10 \operatorname{LCM}(x, y)$. As a result, we have the equation:
$$
2\left(x^{2}+y^{2}\right)=10 \operatorname{LCM}(x, y) .
$$
Using the property $\operatorname{GCD}(x, y) \cdot \operatorname{LCM}(x, y)=x y$, we get
$$
\operatorname{LCM}(x, y)=\frac{x y}{\operatorname{GCD}(x, y)}=\frac{x y}{2} .
$$
Substituting $\operatorname{LCM}(x, y)$ into the equation above, we have
$$
2\left(x^{2}+y^{2}\right)=5 x y .
$$
Dividing the last equation by $y^{2}$, we get a quadratic equation in terms of $\frac{x}{y}$:
$$
2 \frac{x^{2}}{y^{2}}-5 \frac{x}{y}+2=0
$$
This equation has solutions $x=2 y$ and $y=2 x$.
Case 1. $x=2 y$. Then $\operatorname{GCD}(x, y)=\operatorname{GCD}(2 y, y)=y=2$, and $x=4$.
Case 2. $y=2 x$. Then $\operatorname{GCD}(x, y)=\operatorname{GCD}(x, 2 x)=x=2$, and $y=4$. | Answer: 1) $x=2, y=4$; 2) $x=4, y=2$. | 2,4or4,2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Rewrite the original system of equations in the form
$$
\left\{\begin{array}{l}
a=-3 x^{2}+5 x-2 \\
(x+2) a=4\left(x^{2}-1\right)
\end{array}\right.
$$
Notice that $x=-2$ is not a solution to the second equation of the system. Therefore, the system can be rewritten as
$$
\left\{\begin{array}{l}
a=-3 x^{2}+5 x-2 \\
a=\frac{4\left(x^{2}-1\right)}{x+2}
\end{array}\right.
$$
If the system is consistent, then its solution satisfies the equation
$$
\frac{4\left(x^{2}-1\right)}{x+2}=-3 x^{2}+5 x-2
$$
Let's solve this equation. For this, we factorize the left and right sides of the equation:
$$
\frac{4(x-1)(x+1)}{x+2}=(x-1)(-3 x+2)
$$
The last equation can be reduced to the form
$$
x(x-1)(3 x+8)=0
$$ | The solution $x=0$ corresponds to the parameter $a=-2$, the solution $x=1$ corresponds to the parameter $a=0$, and $x=-\frac{8}{3}$ corresponds to the parameter $a=-\frac{110}{3}$.
Answer: 1) The system is consistent for $a=-2,0,-\frac{110}{3},-24$;
2) for $a=0 \quad x=1$; for $a=-2 \quad x=0$; for $a=-\frac{110}{3}$
$$
x=-\frac{8}{3}
$$ | 0,1,-\frac{8}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Let $A$ and $E$ be the centers of circles $\omega_{1}$ and $\omega_{2}$ with radii $r$ and $R$, respectively $(r<R)$, and by the problem's condition, $A \in \omega_{2}$, while $B$ and $C$ are the points of intersection of these circles. Note that their common chord $BC$ is perpendicular to the segment $AE$. We have $AB = AC = r, \quad EB = EC = EA = R$. Let $\square BAE = \square CAE = \alpha, \quad \square BEA = \square CEA = \gamma$.
Write the formula for calculating the area of the union of circles $\Omega_{1}$ and $\Omega_{2}$, bounded by circles $\omega_{1}$ and $\omega_{2}$:
$$
S_{\Omega_{1} \cup \Omega_{2}} = S_{\Omega_{1}} + S_{\Omega_{2}} - S_{\Omega_{1} \cap \Omega_{2}},
$$
where $S_{\Omega_{1}} = \pi r^{2}, S_{\Omega_{2}} = \pi R^{2}$ are the areas of circles $\Omega_{1}$ and $\Omega_{2}$, and $S_{\Omega_{1} \cap \Omega_{2}}$ is the area of their intersection.
Find $S_{\Omega_{1} \cap \Omega_{2}}$. The intersection $\Omega_{1} \cap \Omega_{2}$ consists of two segments $BDC \in \Omega_{1}$ and $BAC \in \Omega_{2}$. Denote the areas of these segments by $S_{1}$ and $S_{2}$, respectively.
The area $S_{1}$ of segment $BDC$ is the difference between the area of sector $BDCA$ and the area of $\square ABC$:
$$
S_{1} = \frac{(2\alpha) r^{2}}{2} - \frac{1}{2} r^{2} \sin 2\alpha
$$
Since $\square ABE$ is isosceles $(AE = BE)$, we have $\cos \alpha = \frac{r}{2R}$. Then
$$
\begin{gathered}
\alpha = \arccos \frac{r}{2R}, \sin \alpha = \sqrt{1 - \left(\frac{r}{2R}\right)^{2}} \\
\sin 2\alpha = 2 \sin \alpha \cos \alpha = \frac{r}{2R} \cdot \sqrt{1 - \left(\frac{r}{2R}\right)^{2}}
\end{gathered}
$$
By the problem's condition, $r = 1, \quad R = 2$. Substituting these values into the above formulas: $\quad \cos \alpha = \frac{1}{4}, \quad \alpha = \arccos \frac{1}{4}$, $\sin \alpha = \frac{\sqrt{15}}{4}, \sin 2\alpha = \frac{\sqrt{15}}{8}$. Then
$$
S_{1} = \arccos \frac{1}{4} - \frac{\sqrt{15}}{16}
$$
The area $S_{2}$ of segment $BAC$ is the difference between the area of sector $BACE$ and the area of $\square CBE$:
$$
S_{2} = \frac{(2\gamma) R^{2}}{2} - \frac{1}{2} R^{2} \sin 2\gamma
$$
Since $\square ABE$ is isosceles $(AE = BE)$, we have $\gamma = \pi - 2\alpha$. Find
$$
\begin{gathered}
\cos \gamma = \cos (\pi - 2\alpha) = -\cos 2\alpha = \sin^{2} \alpha - \cos^{2} \alpha = \frac{15}{16} - \frac{1}{16} = \frac{7}{8} \\
\sin \gamma = \sqrt{1 - \frac{49}{64}} = \frac{\sqrt{15}}{8}, \sin 2\gamma = \frac{7\sqrt{15}}{32}
\end{gathered}
$$
Then
$$
S_{2} = 4\left(\pi - 2 \arccos \frac{1}{4}\right) - \frac{7\sqrt{15}}{16}
$$
Calculate the area of the intersection of circles $\Omega_{1}$ and $\Omega_{2}$:
$$
S_{\Omega_{1} \cap \Omega_{2}} = S_{1} + S_{2} = 4\pi - 7 \arccos \frac{1}{4} - \frac{\sqrt{15}}{2}
$$
Substituting the obtained result into the formula for calculating the area of the union of circles $\Omega_{1}$ and $\Omega_{2}$, we find:
$$
S_{\Omega_{1} \cup \Omega_{2}} = \pi + 4\pi - \left(4\pi - 7 \arccos \frac{1}{4} - \frac{\sqrt{15}}{2}\right) = \pi + 7 \arccos \frac{1}{4} + \frac{\sqrt{15}}{2}.
$$ | Answer: $S=\pi+7 \arccos \frac{1}{4}+\frac{\sqrt{15}}{2}$. | \pi+7\arccos\frac{1}{4}+\frac{\sqrt{15}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. The path from school to home is 140 m. Kolya covered this distance in 200 steps, with the length of each step being no more than $a$ cm, and the sum of the lengths of any two steps being greater than the length of any other step. What values can the number $a$ take under these conditions? | Answer: $a \in\left[70 ; \frac{14000}{199}\right)$. | \in[70;\frac{14000}{199}) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Positive integers $x, y$, for which $\gcd(x, y)=3$, are the coordinates of a vertex of a square with its center at the origin and an area of $20 \cdot \text{lcm}(x, y)$. Find the perimeter of the square. | Answer: $p=24 \sqrt{5}$. | 24\sqrt{5} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. For what values of $x$ does the expression
$$
1+\cos ^{2}\left(\frac{\pi \sin 2 x}{\sqrt{3}}\right)+\sin ^{2}(2 \sqrt{3} \pi \cos x)
$$
take its smallest possible value? | Answer: $x= \pm \frac{\pi}{6}+\pi k, k \in Z$. | \\frac{\pi}{6}+\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. For what values of $a$ is the system $\left\{\begin{array}{l}3 x^{2}-x-a-10=0 \\ (a+4) x+a+12=0\end{array}\right.$ consistent? Solve the system for all permissible $a$. | Answer: 1) the system is consistent for $a=-10,-8,4$;
2) when $a=-8$, $x=1$; when $a=4$, $x=-2$; when $a=-10$, $x=\frac{1}{3}$. | -10,-8,4; | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The center of a circle with radius 2 lies on the circumference of a circle with radius 3. Find the area of the intersection of the circles bounded by these circumferences. | Answer: $S=9 \pi-14 \arccos \frac{1}{3}-4 \sqrt{2}$. | 9\pi-14\arccos\frac{1}{3}-4\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. The path from school to home is 300 m. Vova covered this distance in 400 steps, with the length of each step not exceeding $a$ cm, and the sum of the lengths of any two steps being greater than the length of any other step. What values can the number $a$ take under these conditions? | Answer: $a \in\left[75 ; \frac{10000}{133}\right)$. | \in[75;\frac{10000}{133}) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. For what values of $x$ does the expression
$$
3-|\operatorname{tg}(\sqrt{2} \pi \cos \pi x)|-\left|\operatorname{ctg}\left(\frac{\pi \sqrt{2} \sin 3 \pi x}{2}\right)\right|
$$
take its maximum possible value | Answer: $x=-\frac{1}{4}+\frac{n}{2}, n \in Z$. | -\frac{1}{4}+\frac{n}{2},n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. For what values of $a$ is the system $\left\{\begin{array}{c}10 x^{2}+x-a-11=0 \\ 4 x^{2}+(a+4) x-3 a-8=0\end{array} \mathrm{c}\right.$ consistent? Solve the system for all permissible $a$. | Answer: 1) the system is consistent for $a=0,-2,54$;
2) when $a=0$, $x=1$; when $a=-2$, $x=-1$; when $a=54$, $x=\frac{5}{2}$. | 0,-2,54; | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The center of a circle with a radius of 2 lies on the circumference of a circle with a radius of 5. Find the area of the union of the circles bounded by these circumferences. | Answer: $S=4 \pi+46 \arccos \frac{1}{5}+4 \sqrt{6}$. | 4\pi+46\arccos\frac{1}{5}+4\sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. The path from school to home is 180 meters, and Kostya covered it in 300 steps, with the length of each step being no more than $a$ cm, and the sum of the lengths of any two steps being greater than the length of any other step. What values can the number $a$ take under these conditions? | Answer: $a \in\left[60 ; \frac{18000}{299}\right)$. | \in[60;\frac{18000}{299}) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. For what values of $x$ does the expression
$$
4+\operatorname{tg}^{2}(2 \pi \sin \pi x)+\operatorname{ctg}^{2}(3 \pi \cos 2 \pi x)
$$
take its smallest possible value | Answer: $x= \pm \frac{1}{6}+2 m, x= \pm \frac{5}{6}+2 m, m \in Z$. | \\frac{1}{6}+2,\\frac{5}{6}+2,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. For what values of $a$ is the system $\left\{\begin{array}{l}12 x^{2}+48 x-a+36=0 \\ (a+60) x-3(a-20)=0\end{array}\right.$ consistent? Solve the system for all permissible $a$. | Answer: 1) the system is consistent for $a=-12,0,180$;
2) when $a=-12$, $x=-2$; when $a=0$, $x=-1$; when $a=180$
$$
x=2 \text { . }
$$ | =-12,0,180;when=-12,-2;when=0,-1;when=180,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. By the condition, $a_{k}=\frac{a_{1}+a_{2}+\ldots a_{k-1}+a_{k+1}+\ldots a_{n}}{k+1}=\frac{120-a_{k}}{k+1}$ for all $k=1,2, \ldots, n-1$. From this, $(k+2) a_{k}=120$ and $a_{k}=\frac{120}{k+2}$. Considering that $a_{k}$ are integers, the maximum value of $n$ is determined by the divisibility of the numerator by $3,4,5, \ldots, n,(n+1)$. Since 120 is divisible by $3,4,5,6$, but not by 7, we get $n+1=6$, hence $\quad n_{\max }=5$. We calculate $\quad a_{1}=\frac{120}{3}=40$, $a_{2}=\frac{120}{4}=30, a_{3}=\frac{120}{5}=24, a_{4}=\frac{120}{6}=20$. Now we find the remaining number $a_{5}=120-(40+30+24+20)=6$. | Answer: $n_{\max }=5, a_{1}=40, a_{2}=30, a_{3}=24, a_{4}=20, a_{5}=6$. | n_{\max}=5,a_{1}=40,a_{2}=30,a_{3}=24,a_{4}=20,a_{5}=6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The race track for car racing consists of three sections: highway, dirt, and mud. The speed of two cars participating in the race is the same on each section of the track, equal to 120, 40, and 10 km/h, respectively. The time started when the red car was on the highway 600 meters ahead of the white car, which at that moment was crossing the starting line at the beginning of the highway section. Find the distance between the cars at the moments when both were on the dirt section of the track. Find the distance between the cars at the moment when they were both on the mud section of the track. | Answer: $s_{1}=200 \mathrm{M}, s_{2}=50 \mathrm{M}$. | s_{1}=200\mathrm{M},s_{2}=50\mathrm{M} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Non-zero integers $a, b, c$ are three consecutive terms of a decreasing arithmetic progression. All six quadratic equations, whose coefficients are the numbers $2a, 2b, c$, taken in any order, have two roots. Find the greatest possible value of the common difference of the progression and the corresponding numbers $a, b, c$. | Answer: $d_{\max }=-5 ; a=4, b=-1, c=-6$. | d_{\max}=-5;=4,b=-1,=-6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In the notebook, $n$ integers are written, ordered in descending order $a_{1}>a_{2}>\ldots>a_{n}$ and having a sum of 2520. It is known that the $k$-th number written, $a_{k}$, except for the last one when $k=n$, is $(k+1)$ times smaller than the sum of all the other written numbers. Find the maximum number $n$ possible under these conditions. Find these numbers for the maximum possible $n$. | Answer: $n_{\max }=9 ; a_{1}=840, a_{2}=630, a_{3}=504, a_{4}=420, a_{5}=360$, $a_{6}=315, a_{7}=280, a_{8}=252, a_{9}=-1081$. | n_{\max}=9;a_{1}=840,a_{2}=630,a_{3}=504,a_{4}=420,a_{5}=360,a_{6}=315,a_{7}=280,a_{8}=252,a_{9}=-1081 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The race track for car racing consists of three sections: highway, dirt, and mud. The speed of two cars participating in the race is the same on each section of the track, equal to 100, 70, and 15 km/h, respectively. The time started when the red car was on the highway 500 m ahead of the white car, which at that moment was crossing the starting line at the beginning of the highway section. Find the distance between the cars at the moments when both were on the dirt section of the track. Find the distance between the cars at the moment when they were both on the mud section of the track. | Answer: $s_{1}=350 \mathrm{m}, s_{2}=75 \mathrm{m}$. | s_{1}=350\mathrm{},s_{2}=75\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Non-zero integers $a, b, c$ are three consecutive terms of an increasing arithmetic progression. All six quadratic equations, whose coefficients are the numbers $a, b, 2c$, taken in any order, have two distinct roots. Find the smallest possible value of the common difference of the progression and the corresponding numbers $a, b, c$. | Answer: $d_{\min }=4 ; a=-5, b=-1, c=3$. | d_{\}=4;=-5,b=-1,=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In the notebook, $n$ integers are written, ordered in descending order $a_{1}>a_{2}>\ldots>a_{n}$ and having a sum of 420. It is known that the $k-$th number written, $a_{k}$, except for the last one when $k=n$, is $(k+1)$ times smaller than the sum of all the other written numbers. Find the maximum number $n$ possible under these conditions. Find these numbers for the maximum possible $n$. | Answer: $n_{\max }=6 ; a_{1}=140, a_{2}=105, a_{3}=84, a_{4}=70, a_{5}=60$, $a_{6}=-39$. | n_{\max}=6;a_{1}=140,a_{2}=105,a_{3}=84,a_{4}=70,a_{5}=60,a_{6}=-39 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The race track for car racing consists of three sections: highway, dirt, and mud. The speed of two cars participating in the race is the same on each section of the track, equal to 150, 60, and 18 km/h, respectively. The time started when the red car was on the highway 300 meters ahead of the white car, which at that moment was crossing the starting line at the beginning of the highway section. Find the distance between the cars at the moments when both were on the dirt section of the track. Find the distance between the cars at the moment when they were both on the mud section of the track. | Answer: $s_{1}=120 \mathrm{~m}, s_{2}=36 \mathrm{~m}$. | s_{1}=120\mathrm{~},s_{2}=36\mathrm{~} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Non-zero integers $a, b, c$ are three consecutive terms of a decreasing arithmetic progression. All six quadratic equations, whose coefficients are the numbers $a, 2b, 4c$, taken in any order, have two roots. Find the greatest possible value of the common difference of the progression and the corresponding numbers $a, b, c$. | Answer: $d_{\max }=-3 ; a=4, b=1, c=-2$. | d_{\max}=-3;=4,b=1,=-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In the notebook, $n$ integers are written, ordered in descending order $a_{1}>a_{2}>\ldots>a_{n}$ and having a sum of 840. It is known that the $k$-th number written in order, $a_{k}$, except for the last one when $k=n$, is $(k+1)$ times smaller than the sum of all the other written numbers. Find the maximum number $n$ possible under these conditions. Find these numbers for the maximum possible $n$. | Answer: $n_{\max }=7 ; a_{1}=280, a_{2}=210, a_{3}=168, a_{4}=140, a_{5}=120$, $a_{6}=105, a_{7}=-183$. | n_{\max}=7;a_{1}=280,a_{2}=210,a_{3}=168,a_{4}=140,a_{5}=120,a_{6}=105,a_{7}=-183 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In triangle $A B C$, the perpendicular bisectors of sides $A B$ and $A C$ intersect lines $A C$ and $A B$ at points $N$ and $M$ respectively. The length of segment $N M$ is equal to the length of side $B C$ of the triangle. The angle at vertex $C$ of the triangle is $40^{\circ}$. Find the angle at vertex $B$ of the triangle. | Answer: $80^{\circ}$ or $20^{\circ}$.
## Final round of the "Rosatom" Olympiad, 9th grade, CIS, February 2020
# | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. The chord $A B$ of the parabola $y=x^{2}$ intersects the y-axis at point $C$ and is divided by it in the ratio $A C: C B=5: 3$. Find the abscissas of points $A$ and $B$, if the ordinate of point $C$ is 15. | Answer: $\left(x_{A}=-5, x_{B}=3\right),\left(x_{A}=5, x_{B}=-3\right)$. | (x_{A}=-5,x_{B}=3),(x_{A}=5,x_{B}=-3) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The sum $b_{7}+b_{6}+\ldots+b_{2019}$ of the terms of the geometric progression $\left\{b_{n}\right\}, b_{n}>0$ is 27, and the sum of their reciprocals $\frac{1}{b_{7}}+\frac{1}{b_{6}}+\ldots+\frac{1}{b_{2019}}$ is 3. Find the product $b_{7} \cdot b_{6} \cdot \ldots \cdot b_{2019}$. | Answer: $b_{7} \cdot b_{6} \cdot \ldots \cdot b_{2019}=3^{2013}$. | 3^{2013} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. It is known that a circle can be inscribed in a trapezoid with an angle of $60^{\circ}$ at the base, and a circle can also be circumscribed around it. Find the ratio of the perimeter of the trapezoid to the length of the inscribed circle. Find the ratio of the perimeter of the trapezoid to the length of the circumscribed circle. | Answer: $\quad \frac{P}{L_{1}}=\frac{8 \sqrt{3}}{3 \pi}, \frac{P}{L_{2}}=\frac{4 \sqrt{21}}{7 \pi}$. | \frac{P}{L_{1}}=\frac{8\sqrt{3}}{3\pi},\frac{P}{L_{2}}=\frac{4\sqrt{21}}{7\pi} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. The sum $b_{6}+b_{7}+\ldots+b_{2018}$ of the terms of the geometric progression $\left\{b_{n}\right\}, b_{n}>0$ is 6. The sum of the same terms taken with alternating signs $b_{6}-b_{7}+b_{8}-\ldots-b_{2017}+b_{2018}$ is 3. Find the sum of the squares of the same terms $b_{6}^{2}+b_{7}^{2}+\ldots+b_{2018}^{2}$. | Answer: $b_{6}^{2}+b_{7}^{2}+\ldots+b_{2018}^{2}=18$. | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. It is known that in trapezoid $A B C D$, where diagonal $B D$ forms an angle of $45^{0}$ with the base, a circle can be inscribed and a circle can be circumscribed around it. Find the ratio of the area of the trapezoid to the area of the inscribed circle. Find the ratio of the area of the trapezoid to the area of the circumscribed circle. | Answer: $\frac{S}{S_{1}}=\frac{4}{\pi}, \quad \frac{S}{S_{2}}=\frac{2}{\pi}$. | \frac{S}{S_{1}}=\frac{4}{\pi},\quad\frac{S}{S_{2}}=\frac{2}{\pi} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. The chord $A B$ of the parabola $y=x^{2}$ intersects the y-axis at point $C$ and is divided by it in the ratio $A C: C B=5: 2$. Find the abscissas of points $A$ and $B$, if the ordinate of point $C$ is 20. | Answer: $\left(x_{A}=-5 \sqrt{2}, x_{B}=2 \sqrt{2}\right),\left(x_{A}=5 \sqrt{2}, x_{B}=-2 \sqrt{2}\right)$. | (x_{A}=-5\sqrt{2},x_{B}=2\sqrt{2}),(x_{A}=5\sqrt{2},x_{B}=-2\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The sum $b_{8}^{2}+b_{9}^{2}+\ldots+b_{2020}^{2}$ of the squares of the terms of the geometric progression $\left\{b_{n}\right\}, b_{n}>0$ is 4. The sum of their reciprocals $\frac{1}{b_{8}^{2}}+\frac{1}{b_{9}^{2}}+\ldots+\frac{1}{b_{2020}^{2}}$ is 1. Find the product $b_{8}^{2} \cdot b_{9}^{2} \cdot \ldots \cdot b_{2020}^{2}$. | Answer: $b_{8}^{2} \cdot b_{9}^{2} \cdot \ldots \cdot b_{2020}^{2}=2^{2013}$. | 2^{2013} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. It is known that in trapezoid $A B C D$, where diagonal $B D$ forms an angle of $30^{\circ}$ with the base, a circle can be inscribed and a circle can be circumscribed around it. Find the ratio of the perimeter of the trapezoid to the length of the inscribed circle. Find the ratio of the perimeter of the trapezoid to the length of the circumscribed circle. | Answer: $\frac{P}{L_{1}}=\frac{4 \sqrt{3}}{\pi}, \frac{P}{L_{2}}=\frac{2}{\pi}$. | \frac{P}{L_{1}}=\frac{4\sqrt{3}}{\pi},\frac{P}{L_{2}}=\frac{2}{\pi} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. On the number line, there is a set of positive, three-digit integers $a$, for which the sum of the digits in their notation equals 16. Find the greatest and the smallest distance between two numbers from this set. | 2. Solution. Let $x, y, z$ be the hundreds, tens, and units of the number $a$. According to the condition, $x+y+z=16$. The smallest number satisfying these conditions is $a_{\min }=169$, and the largest number is $a_{\max }=970$. The greatest distance is their difference $d_{\max }=a_{\max }-a_{\min }=801$. The smallest distance between numbers from this set is the smallest distance between adjacent numbers. Let some number from this set be of the form $a=N+10 x+y(N \geq 100)$, then the next number from this set is $b=N+10(x+1)+(y-1)$. The difference between them is $b-a=9$, for example, one can choose $a=169$, and $b=178$. Thus, $d_{\max }=801$, and $d_{\min }=9$. | d_{\max}=801,d_{\}=9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Calculate the value of the expression $\sin \frac{b \pi}{36}$, where $b$ is the sum of all distinct numbers obtained from the number $a=987654321$ by cyclic permutations of its digits (in a cyclic permutation, all digits of the number, except the last one, are shifted one place to the right, and the last one is moved to the first place). | 3. Solution. When summing all different numbers obtained from the number $a=987654321$, including the number $a$ itself, in each digit place we will get the sum of all its digits $9+8+7+\ldots+3+2+1=45$. Therefore, $b=45\left(10^{8}+10^{7}+10^{6}+\ldots+10^{2}+10+1\right)=45 \cdot 111111111=4999999995$. Calculate $\frac{\pi b}{36}=\frac{\pi \cdot 45 \cdot 111111111}{36}=\frac{\pi \cdot 555555555}{4}=138888888 \pi+\frac{3}{4} \pi$. From this, it follows that
$\sin \frac{\pi b}{36}=\sin \left(138888888 \pi+\frac{3}{4} \pi\right)=\sin \frac{3 \pi}{4}=\frac{\sqrt{2}}{2}$. | \frac{\sqrt{2}}{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Let $r_{n}$ denote the remainder of the division of $n$ by 3. Find the sequence of numbers $X_{n}$, which are the roots of the equation $(n+1) x^{r_{n}}-n(n+2) x^{r_{n+1}}+n^{2} x^{r_{n+2}}=0$ for any natural number $n$, and its limit is equal to 1. | 4. Solution. Note that, since $r_{n}$ is the remainder of the division of $n$ by 3, it can take three different values: 0, 1, and 2.
Case 1. $n=3 k, k \in \square$, then $r_{n}=0, r_{n+1}=1, r_{n+2}=2$. The equation is: $n^{2} x^{2}-n(n+2) x+n+1=0$.
Its roots are
$x_{1, n}=\frac{1}{n}, \quad x_{2, n}=\frac{n+1}{n}$.
The condition of the problem ($x_{n} \rightarrow 1$ as $n \rightarrow \infty$) is satisfied only by $x_{2, n}$.
Case 2. $n=3 k+1, k \in \square \cup\{0\}$, then $r_{n}=1, r_{n+1}=2, r_{n+2}=0$. The equation is: $-n(n+2) x^{2}+(n+1) x+n^{2}=0$.
Its roots are
$x_{1, n}=\frac{n+1-\sqrt{4 n^{4}+8 n^{3}+n^{2}+2 n+1}}{2 n(n+2)}, \quad x_{2, n}=\frac{n+1+\sqrt{4 n^{4}+8 n^{3}+n^{2}+2 n+1}}{2 n(n+2)}$.
The condition of the problem ($x_{n} \rightarrow 1$ as $n \rightarrow \infty$) is satisfied only by $x_{2, n}$.
Case 3. $n=3 k+2, k \in \square \cup\{0\}$, then $r_{n}=2, r_{n+1}=0, r_{n+2}=1$. The equation is: $(n+1) x^{2}+n^{2} x-n(n+2)=0$.
Its roots are
$x_{1, n}=\frac{-n^{2}-\sqrt{n^{4}+4 n^{3}+12 n^{2}+8 n}}{2(n+1)}$
$x_{2, n}=\frac{-n^{2}+\sqrt{n^{4}+4 n^{3}+12 n^{2}+8 n}}{2(n+1)}=\frac{4 n^{3}+12 n^{2}+8 n}{2(n+1)\left(\sqrt{n^{4}+4 n^{3}+12 n^{2}+8 n}+n^{2}\right)}$.
The condition of the problem ($x_{n} \rightarrow 1$ as $n \rightarrow \infty$) is satisfied only by $x_{2, n}$.
Thus, the desired sequence is
$x_{n}=x_{2, n}=\left\{\begin{array}{cc}\frac{n+1}{n}, & n=3 k, k \in \square, \\ \frac{n+1+\sqrt{4 n^{4}+8 n^{3}+n^{2}+2 n+1}}{2 n(n+2)}, & n=3 k+1, k \in \square \cup\{0\}, \\ \frac{-n^{2}+\sqrt{n^{4}+4 n^{3}+12 n^{2}+8 n}}{2(n+1)}, & n=3 k+2, k \in \square \cup\{0\} .\end{array}\right.$ | x_{n}=x_{2,n}={\begin{pmatrix}\frac{n+1}{n},&n=3k,k\in\,\\\frac{n+1+\sqrt{4n^{4}+8n^{3}+n^{2}+2n+1}}{2n(n+2)},&n=3k+1,k\in\\cup{0},\} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Circles $K_{1}$ and $K_{2}$ with centers at points $O_{1}$ and $O_{2}$ have the same radius $r$ and touch the circle $K$ of radius $R$ with center at point $O$ internally. The angle $O_{1} O O_{2}$ is $120^{\circ}$. Find the radius $q$ of the circle that touches $K_{1}$ and $K_{2}$ externally and the circle $K$ internally. | 5. Solution. Let $Q$ be the center of the desired circle, and $q$ be its radius.
Case 1. The center $O$ does not lie inside the desired circle (Fig. 1).
Fig 1
From the equality of triangles $O_{1} O Q$ and $Q O O_{2}$ and the condition of the problem, it follows that $\angle O_{1} O Q = \angle O_{2} O Q = 60^{\circ}$.
Consider triangle $O O_{1} Q$. Write the cosine theorem:
$(r+q)^{2} = (R-r)^{2} + (R-q)^{2} - 2(R-r)(R-q) \cos 60^{\circ}$.
From this, we find: $q = \frac{R(R-r)}{R+3r}$.
Case 2. The center $O$ lies inside the desired circle (Fig. 2).
From the equality of triangles $O_{1} O Q$ and $Q_{2}$ and the condition of the problem, it follows that $\angle O_{1} O Q = \angle O_{2} O Q = 120^{\circ}$.
Consider triangle $O O_{1} Q$. Write the cosine theorem:
$(r+q)^{2} = (R-r)^{2} + (R-q)^{2} - 2(R-r)(R-q) \cos 120^{\circ}$.
From this, we find: $q = \frac{3R(R-r)}{3R+r}$.
Fig 2 | \frac{R(R-r)}{R+3r} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Let $A=\overline{a b c b a}$ be a five-digit symmetric number, $a \neq 0$. If $1 \leq a \leq 8$, then the last digit of the number $A+11$ will be $a+1$, and therefore the first digit in the representation of $A+11$ should also be $a+1$. This is possible only with a carry-over from the digit, i.e., when $b=c=9$. Then $A+11=(a+1) 999(a+1)$ is a symmetric number for any $a=1,2, \ldots, 8$. The case $a=9$ is impossible, since $A+11$ ends in zero, and thus, due to symmetry, it should start with zero. But a number cannot start with zero.
The total number of solutions is equal to the number of possible choices for the number $a$, which is eight. | Answer: eight numbers of the form $\overline{a 999 a}$, where $a=1,2, \ldots, 8$. | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Integers, the decimal representation of which reads the same from left to right and from right to left, we will call symmetric. For example, the number 513315 is symmetric, while 513325 is not. How many six-digit symmetric numbers exist such that adding 110 to them leaves them symmetric? | Answer: 81 numbers of the form $\overline{a b 99 b a}$, where $a=1,2, \ldots, 9, b=0,1,2, \ldots, 8$. | 81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In city "N", there are 12 horizontal and 16 vertical streets, of which a pair of horizontal and a pair of vertical streets form the rectangular boundary of the city, while the rest divide it into blocks that are squares with a side length of 100m. Each block has an address consisting of two integers $(i ; j), i=1,2, . ., 11, j=1,2, \ldots, 15-$ the numbers of the streets that bound it from below and from the left. Taxis transport passengers from one block to another, adhering to the following rules: 1) pick-up and drop-off can be made at any point on the boundary of the block at the passenger's request; 2) it is forbidden to enter inside the block; 3) transportation is carried out along the shortest path; 4) a fee of 1 coin is charged for every 100m traveled (rounding the distance to the nearest 100m in favor of the driver). How many blocks are there in the city? What is the maximum and minimum fare that a driver can charge a passenger for a ride from block $(7,2)$ to block $(2 ; 1)$ without violating the rules? | Answer: 165 blocks; $c_{\min }=4$ coins, $c_{\max }=8$ coins. | 165 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In the confectionery, there were either people with excess weight or sweet lovers (possibly both). It turned out that $80 \%$ of people with excess weight love sweets, and $70 \%$ of sweet lovers have excess weight. What fraction of those present love sweets but do not suffer from excess weight? | Answer: $\frac{12}{47}$. | \frac{12}{47} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Integers, the decimal representation of which reads the same from left to right and from right to left, we will call symmetric. For example, the number 5134315 is symmetric, while 5134415 is not. How many seven-digit symmetric numbers exist such that adding 1100 to them leaves them symmetric? | Answer: 810 numbers of the form $\overline{a b c 9 c b a}$, where $a=1,2, \ldots, 9$, $b=0,1,2, \ldots, 9, c=0,1,2, \ldots, 8$. | 810 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In city "N", there are 7 horizontal and 13 vertical streets, of which a pair of horizontal and a pair of vertical streets form the rectangular boundary of the city, while the rest divide it into blocks that are squares with a side length of 100 m. Each block has an address consisting of two integers $(i ; j), i=1,2, . ., 6, j=1,2, \ldots, 12$ - the numbers of the streets that bound it from below and from the left. Taxis transport passengers from one block to another, adhering to the following rules: 1) pick-up and drop-off can be made at any point on the boundary of the block at the passenger's request; 2) it is forbidden to enter inside the block; 3) transportation is carried out along the shortest path; 4) a fee of 1 coin is charged for every 100 m traveled (rounding the distance to the nearest 100 m in favor of the driver). How many blocks are there in the city? What is the maximum and minimum fare that the driver can charge the passenger for a ride from block $(4,2)$ to block $(1 ; 9)$ without violating the rules. | # Answer: 72 blocks; $c_{\min }=8$ coins, $c_{\max }=12$ coins. | 72 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Integers, whose decimal notation reads the same from left to right and from right to left, we will call symmetric. For example, the number 513151315 is symmetric, while 513152315 is not. How many nine-digit symmetric numbers exist such that adding 11000 to them leaves them symmetric? | Answer: 8100 numbers of the form $\overline{a b c d 9 d c b a}$, where $a=1,2, \ldots, 9$, $b=0,1,2, \ldots, 9, c=0,1,2, \ldots, 9, d=0,1,2, \ldots, 8$. | 8100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In city "N", there are 10 horizontal and 12 vertical streets, of which a pair of horizontal and a pair of vertical streets form the rectangular boundary of the city, while the rest divide it into blocks that are squares with a side length of 100 meters. Each block has an address consisting of two integers $(i ; j), i=1,2, . ., 9, j=1,2, \ldots, 11-$ the numbers of the streets that bound it from below and from the left. Taxis transport passengers from one block to another, adhering to the following rules: 1) pick-up and drop-off can be made at any point on the boundary of the block at the passenger's request; 2) it is forbidden to enter inside the block; 3) transportation is carried out along the shortest path; 4) a fee of 1 coin is charged for every 100 meters traveled (rounding the distance to the nearest 100 meters in favor of the driver). How many blocks are there in the city? What is the maximum and minimum fare that the driver can charge the passenger for a ride from block $(7,1)$ to block $(2 ; 10)$ without violating the rules? | Answer: 99 blocks; $c_{\min }=10$ coins, $c_{\max }=14$ coins. | 99 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Kuzya the flea can make a jump in any direction on a plane for exactly 19 mm. Her task is to get from point $A$ to point $B$ on the plane, the distance between which is 1812 cm. What is the minimum number of jumps she must make to do this? | Answer: $n_{\min }=954$ jumps. | 954 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. For which $x$ does the inequality $\frac{(x-y)^{2}(x+2 y)}{x+y+1} \geq 0$ hold for all solutions $y$ of the equation $4 \sqrt{y}=y+3$? | Task 1. Answer: $x \in(-\infty ;-18] \cup(-10 ;-2) \cup(-2 ;+\infty)$ | x\in(-\infty;-18]\cup(-10;-2)\cup(-2;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. Consider a rectangular parallelepiped, the volume of which is numerically equal to the sum of the lengths of its edges. Find all possible values of its volume given that the lengths of its edges are integers. | Task 4. Answer: $V=120 ; 84 ; 72 ; 60 ; 48$
Possible ordered values of triples: (1;;;24), (1;6;14), (1;8;9), (2;3;10), (2;4;6) | 120,84,72,60,48 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. The function $f(x)$ at each point $x$ takes a value equal to the maximum of the expression $(x+2) y^{2}+(3-2 x) y+3 x+a$ with respect to the variable $y$ on the interval $[1 ; 4]$. For what values of $a$ does the inequality $f(x) \geq 0$ hold for all $x \geq -3$? | Problem 5. Answer: $a \in[-11 ;+\infty)$ | \in[-11;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. A plot of land in the form of a right-angled triangle with legs of 4 and 3 needs to be divided by a line $L$ into two plots such that 1) the plots have equal area; 2) the length of the common boundary (fence) of the plots is minimized. Indicate the points on the sides of the triangle through which the desired line $L$ passes and find the length of the fence. | Problem 6. Answer: 1) The line intersects the larger leg $B C$ (angle $\measuredangle C=90^{\circ}$) at point $M: B M=\sqrt{10}$ and the hypotenuse $B A$ at point $N: B N=\sqrt{10}$.
2) $L_{\text {min }}=2$ | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Note that the coefficients of the equation 3 and 453 are divisible by 3, then the term $5 y^{2}$ is also divisible by 3, so $y=3 t$, $y^{2}=9 t^{2}$, where $t \in \mathbb{Z}$. The equation takes the form: $3 x^{2}+5 \cdot 9 t^{2}=453, x^{2}+5 \cdot 3 t^{2}=151, x^{2}=151-15 t^{2}, \quad$ where $\quad x \in \mathbb{Z}$. Substitute possible values of $t \geq 0$.
| Values of $t$ | $15 t^{2}$ | $x^{2}=151-15 t^{2}$ | Integer value of $x$. |
| :--- | :--- | :--- | :--- |
| $t=0$ | 0 | $x^{2}=151$ | no |
| $t=1$ | 15 | $x^{2}=151-15, x^{2}=136$ | no |
| $t=2$ | 60 | $x^{2}=151-60, x^{2}=91$ | no |
| $t=3$ | 135 | $x^{2}=151-135, x^{2}=16$ | $x= \pm 4$ |
| $t=4$ | 240 | $x^{2}=151-240<0$ | no |
Thus, $x= \pm 4, t= \pm 3, y= \pm 9$. | Answer: 4 pairs, $( \pm 4 ; \pm 9)$. | (\4;\9) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. When purchasing goods for an amount of no less than 1500 rubles, the store provides a discount on subsequent purchases of $26 \%$. Having 1800 rubles in his pocket, Sasha wanted to buy 5 kg of shashlik and 1 can of tomato sauce. In the store, shashlik was sold at a price of 350 rubles per kg, and the sauce was priced at 70 rubles per can. Realizing that he didn't have enough money for the purchase, Sasha still managed to buy what he intended. How did he do it? | First purchase: 4 kg of shashlik, 1 can of sauce. Its cost is $350 \times 4 + 2 \times 70 = 1540$ p. Second purchase: 1 kg of shashlik. With a $26\%$ discount, its price is $350 \cdot 0.74 = 259$. The total amount of both purchases - $1799 \mathrm{p}$. | 1799\mathrm{p} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Twenty-seven identical cubes, on the faces of which the digits from 1 to 6 are drawn such that the sum of the digits on opposite faces of the cube is constant and equal to 7 (dice), are assembled into a cube. What values can the sum of the numbers drawn on all faces of this cube take? | Answer: all integers in the interval $[90 ; 288]$. | [90;288] | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. When purchasing goods for an amount of no less than 1000 rubles, the store provides a discount on subsequent purchases of $50 \%$. Having 1200 rubles in her pocket, Dasha wanted to buy 4 kg of strawberries and 6 kg of sugar. In the store, strawberries were sold at a price of 300 rubles per kg, and sugar - at a price of 30 rubles per kg. Realizing that she didn't have enough money for the purchase, Dasha still managed to buy what she intended. How did she do it? | First purchase: 3 kg of strawberries, 4 kg of sugar. Its cost is $300 \times 3 + 4 \times 30 = 1020$ rubles. Second purchase: 1 kg of strawberries, 2 kg of sugar. With a $50\%$ discount, its price is $(300 + 60) \cdot 0.5 = 180$ rubles. The total amount of both purchases is $1200$ rubles. | 1200 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. How many different pairs of integers $(x ; y)$ exist for which $4 x^{2}+7 y^{2}=1600$? Find all such pairs. | Answer: 6 pairs, $( \pm 20 ; 0),( \pm 15 ; \pm 10)$. | 6pairs,(\20;0),(\15;\10) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Sixty-four identical cubes, on the faces of which the digits from 1 to 6 are drawn such that the sum of the digits on opposite faces of the cube is constant and equal to 7 (dice), are assembled into a cube. What values can the sum of the numbers drawn on all the faces of this cube take? | Answer: all integers in the interval $[144 ; 528]$. | allintegersintheinterval[144;528] | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. When purchasing goods for an amount of no less than 900 rubles, the store provides a discount on subsequent purchases of $25 \%$. Having 1200 rubles in his pocket, Petya wanted to buy 3 kg of meat and 1 kg of onions. In the store, meat was sold at 400 rubles per kg, and onions at 50 rubles per kg. Realizing that he did not have enough money for the purchase, Petya still managed to buy what he intended. How did he do it? | First purchase: 2 kg of meat and 2 kg of onions. Its cost is 900 p. Second purchase: 1 kg of meat. With a $25 \%$ discount, its price is $400 \cdot 0.75=300$. The total amount of both purchases $-1200 \mathrm{p}$. | 1200\mathrm{p} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. How many different pairs of integers $(x ; y)$ exist for which $12 x^{2}+7 y^{2}=4620$? Find all such pairs. | Answer: 8 pairs, $( \pm 7 ; \pm 24),( \pm 14 ; \pm 18)$. | 8pairs,(\7;\24),(\14;\18) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. One hundred and twenty-five identical cubes, on the faces of which the digits from 1 to 6 are drawn such that the sum of the digits on opposite faces of the cube is constant and equal to 7 (dice), are assembled into a cube. What values can the sum of the numbers drawn on all the faces of this cube take? | Answer: all integers in the interval $[210 ; 840]$. | [210;840] | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Express $x$ from the 3rd equation
$$
x=-3 z+5
$$
Substitute it into the 1st equation
$$
\left\{\begin{array}{r}
-3 z+2 y=a-5 \\
2 y+3 z=a+1
\end{array}\right.
$$
Adding the 1st and 2nd equations, we get $4 y=2 a-2$. From this, we find $y=\frac{a-2}{2}$. Then $z=\frac{a+1-2 y}{3}=1$, and $x=2$.
Substitute the solution we found into the equation of the sphere
$$
1+\frac{(a+2)^{2}}{4}+9=19
$$
Rewrite this equation as $(a+2)^{2}=36$. From this, we find $a_{1}=4, a_{2}=-8$. For $a_{1}=4$, we get $x=2, y=1, z=1$, and for $a_{2}=-8 \quad x=2, y=-5, z=1$. | Answer: 1) when $a=4, x=2, y=1, z=1$;
2) when $a=-8 \quad x=2, y=-5, z=1$. | 1)when=4,2,1,1;\quad2)when=-8,2,-5,1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. For which $a$
1) the equation $f(f(x))=x$ has an infinite number of solutions;
2) the equation $f(f(f(f(x))))=x$ has an infinite number of solutions for functions of the form
$$
f(x)=\frac{(2 a+8) x-5}{2 x-a} ?
$$ | Answer: 1) $a=-8, f(x)=\frac{-8 x-5}{2 x+8}$
$$
\text { 2) } \begin{aligned}
a & =-8, f(x)=\frac{-8 x-5}{2 x+8} ; a=-2, f(x)=\frac{4 x-5}{2 x+2} \\
a & =-\frac{22}{5}, f(x)=\frac{-4 x-25}{10 x+22}
\end{aligned}
$$ | =-8,=-2,=-\frac{22}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Each of the four numbers $-\sqrt{2} ; -1 ; -\frac{1}{2} ; 0$ is the result of one of the four arithmetic operations $\cos x + \cos y$, $\cos x - \cos y$, $\cos x \cdot \cos y$, $\cos x : \cos y$ on the numbers $\cos x, \cos y$ for some $x$ and $y$. Find all permissible pairs $(x ; y)$ for this. | Answer: $x= \pm \frac{3 \pi}{4}+2 \pi k, k \in Z, y= \pm \frac{\pi}{4}+2 \pi n, n \in Z$. | \\frac{3\pi}{4}+2\pik,k\inZ,\\frac{\pi}{4}+2\pin,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. For which $a$ does the solution $(x, y, z)$ of the system $\left\{\begin{array}{l}2 x+3 y=a-1 \\ 3 y+4 z=a+1 \\ x+2 z=a-2\end{array}\right.$ satisfy the equation $(x+1)^{2}+y^{2}+(z+2)^{2}=\frac{49}{9}$? Find these solutions. | Answer: 1) when $a=-3, x=-3, y=\frac{2}{3}, z=-1$;
2) when $a=\frac{9}{5}, x=-\frac{3}{5}, y=\frac{2}{3}, z=\frac{1}{5}$. | 1) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Points $M, N, P$ are the feet of the altitudes dropped from the vertices of triangle $A B C$ with angles $30^{\circ}, 75^{\circ}, 75^{\circ}$ to its sides. Find the ratio of the areas of triangles $M N P$ and $A B C$. | Answer: $(2 \sqrt{3}-3): 4$ | (2\sqrt{3}-3):4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. For which $a$
1) the equation $f(f(x))=x$ has an infinite number of solutions;
2) the equation $f(f(f(f(x))))=x$ has an infinite number of solutions for the function of the form
$$
f(x)=\frac{(5-3 a) x-2}{5 x+a-1} ?
$$ | Answer: 1) $a=2, f(x)=\frac{-x-2}{5 x+1}$
2) $a=2, f(x)=\frac{-x-2}{5 x+1} ; a=3, f(x)=\frac{-4 x-2}{5 x+2}$
$$
a=\frac{1}{5}, f(x)=\frac{22 x-10}{25 x-4}
$$ | =2,=3,=\frac{1}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. For which $a$ does the solution $(x, y, z)$ of the system $\left\{\begin{array}{l}x-2 y=a-1 \\ 2 z-y=a+2 \\ x+4 z=a+3\end{array}\right.$ satisfy the equation $(x+1)^{2}+y^{2}+(z-1)^{2}=20$ ? Find these solutions. | Answer: 1) when $a=8, x=-1, y=-4, z=3$;
2) when $a=-8, x=-1, y=4, z=-1$. | 1)when=8,-1,-4,3;2)when=-8,-1,4,-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Points $M, N, P$ are the feet of the altitudes dropped from the vertices of triangle $A B C$ with angles $37.5^{\circ}, 60^{\circ}, 82.5^{\circ}$ to its sides. Find the ratio of the areas of triangles $M N P$ and $A B C$. | Answer: $(\sqrt{2}-1): 4$.
# | (\sqrt{2}-1):4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. For which $a$ does the solution $(x, y, z)$ of the system $\left\{\begin{array}{c}3 x-y=a-2 \\ y+2 z=-a-1 \\ 3 x-2 z=2 a+3\end{array}\right.$ satisfy the equation $x^{2}+(y-2)^{2}+(z-1)^{2}=2$? Find these solutions. | Answer: 1) when $a=-3, x=-1, y=2, z=0$;
2) when $a=-\frac{51}{13}, x=-\frac{17}{13}, y=2, z=\frac{6}{13}$. | 1) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Points $M, N, P$ are the feet of the altitudes dropped from the vertices of triangle $A B C$ with angles $37.5^{0}, 67.5^{0}, 75^{0}$ to its sides. Find the ratio of the areas of triangles $M N P$ and $A B C$. | Answer: $(2 \sqrt{3}+3 \sqrt{2}-\sqrt{6}-4): 8$.
## Final round of the "Rosatom" Olympiad, 10th grade, CIS, February 2020
# | (2\sqrt{3}+3\sqrt{2}-\sqrt{6}-4):8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Transform the right side of the equation
$2 \sin x(\sin x-\cos x)=2 \sin ^{2} x-\sin 2 x=1-\cos 2 x-\sin 2 x$.
Let $t=\sin 2 x+\cos 2 x-1$, then the original inequality can be rewritten as
$$
\sin t \geq -t
$$
The solutions to this inequality are $t: t \geq 0$. As a result, we arrive at the inequality
$$
\sin 2 x+\cos 2 x \geq 1
$$
Divide both sides of the obtained inequality by $\sqrt{2}$
$$
\frac{1}{\sqrt{2}} \cdot \sin 2 x+\frac{1}{\sqrt{2}} \cdot \cos 2 x \geq \frac{1}{\sqrt{2}}
$$
and rewrite it as
$$
\cos \left(2 x-\frac{\pi}{4}\right) \geq \frac{1}{\sqrt{2}}
$$
Its solutions are $-\frac{\pi}{4}+2 \pi k \leq 2 x-\frac{\pi}{4} \leq \frac{\pi}{4}+2 \pi k, k \in Z$, from which we find $\pi k \leq x \leq \frac{\pi}{4}+\pi k, k \in Z$ | Answer: $\pi k \leq x \leq \frac{\pi}{4}+\pi k, k \in Z$. | \pik\leqx\leq\frac{\pi}{4}+\pik,k\inZ | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5. Let's introduce the notation: $\measuredangle A B C=2 \theta, \quad \measuredangle B C A=2 \beta$, $\measuredangle C A B=2 \varphi$. By the theorem of the sum of angles in triangle $B O A$ we get
$$
\measuredangle B O A=180^{\circ}-\theta-\varphi=180^{\circ}-\frac{180^{\circ}-2 \beta}{2}=90^{\circ}+\beta .
$$
By the condition, quadrilateral $E O D C$ is inscribed, therefore,
$$
\angle D O E=180^{\circ}-2 \beta .
$$
From the condition $\measuredangle B O A=\measuredangle D O E$ (vertical angles) we get the equation
$$
90^{\circ}+\beta=180^{\circ}-2 \beta
$$
From this, we find $\beta=30^{\circ}$. Therefore, $\measuredangle B C A=2 \beta=60^{\circ}$. In triangle $O D C \quad \angle D O C=58^{\circ}$ by the condition, $\measuredangle D C O=\beta=30^{\circ}$, since $O C$ is the bisector of angle $\measuredangle B C A$. Then
$$
\measuredangle O D C=180^{\circ}-\angle D O C-\angle D C O=180^{\circ}-58^{\circ}-30^{\circ}=92^{\circ} .
$$
Let's find $\angle D A C$ of triangle $D A C$
$$
\angle D A C=\varphi=180^{\circ}-\angle O D C-\angle D C A=180^{\circ}-92^{\circ}-60^{\circ}=28^{\circ} .
$$
Therefore, $\measuredangle C A B=2 \varphi=56^{\circ}$. It remains to find $\measuredangle A B C$ :
$$
\measuredangle A B C=180^{\circ}-\measuredangle C A B-\measuredangle B C A=180^{\circ}-56^{\circ}-60^{\circ}=64^{\circ} .
$$ | Answer: $56^{\circ}, 64^{\circ}, 60^{\circ}$. | 56,64,60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the inequality:
$$
\sin (\sin 2 x-\sqrt{3} \cos 2 x-\sqrt{3}) \leq 2 \cos x(\sqrt{3} \cos x-\sin x)
$$ | Answer: $\frac{\pi}{2}+\pi k \leq x \leq \frac{4 \pi}{3}+\pi k, k \in Z$. | \frac{\pi}{2}+\pik\leqx\leq\frac{4\pi}{3}+\pik,k\inZ | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the inequality:
$$
\sin (\sin 4 x+\cos 4 x+1)+2 \cos 2 x(\sin 2 x+\cos 2 x)>0
$$ | Answer: $-\frac{\pi}{8}+\frac{\pi k}{2}<x<\frac{\pi}{4}+\frac{\pi k}{2}, k \in Z$. | -\frac{\pi}{8}+\frac{\pik}{2}<x<\frac{\pi}{4}+\frac{\pik}{2},k\inZ | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. Pete and Vova want to cross a river 160 (m) wide. They have a small rubber boat, in which only one person can paddle at a speed of 32 (m/min). They started the crossing simultaneously: Pete in the boat, Vova - swimming. At some point, Pete left the boat and continued swimming on his own, while Vova, catching up with the boat, completed the crossing in it. Pete swims at a speed of 16, Vova - 24 (m/min). What distance should Pete paddle in the boat so that the total crossing time is minimized? (time for transferring into the boat and river current are not considered). | Answer: $120 \mathrm{m}$. | 120\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. According to the properties of logarithms, after transformations we get
$$
\begin{aligned}
& \log _{2}\left(b_{2} b_{3} \ldots b_{n}\right)=\log _{2} b_{1}^{2} \\
& b_{2} b_{3} \ldots b_{n}=b_{1}^{2}
\end{aligned}
$$
Using the formula for the general term of a geometric progression, we get
$$
b_{2}=b_{1} q, b_{3}=b_{1} q^{2}, \ldots, b_{n}=b_{1} q^{n-1}
$$
and using these relations and the formula for the sum of an arithmetic progression
$$
1+2+\ldots+(n-1)=\frac{n(n-1)}{2}
$$
we transform the obtained equation to
$$
b_{1}^{n-1} q^{1+2+\ldots+(n-1)}=b_{1}^{2} \Rightarrow b_{1}^{n-3} q^{n(n-1) / 2}=1
$$
Taking the logarithm with base $q$, we find
$$
\begin{aligned}
& (n-3) \log _{q} b_{1}+\frac{n(n-1)}{2}=0 \\
& \frac{n-3}{2} \log _{q} b_{1}^{2}+\frac{n(n-1)}{2}=0 \\
& \log _{q} b_{1}^{2}=-\frac{n(n-1)}{n-3}
\end{aligned}
$$
Since
$$
\frac{n(n-1)}{n-3}=\frac{n(n-3)+2(n-3)+6}{n-3}=n+2+\frac{6}{n-3}
$$
then
$$
\log _{q} b_{1}^{2}=-n-2-\frac{6}{n-3} .
$$
According to the problem's condition, the quantity
$$
\frac{6}{n-3}
$$
must be an integer. This is possible only for
$$
n \in\{2 ; 4 ; 5 ; 6 ; 9\}
$$
Substituting the specified values into the formula obtained above, we find all possible values of the desired quantity
$$
\begin{aligned}
& n=2 \Rightarrow \log _{q} b_{1}^{2}=2 \\
& n=4 \Rightarrow \log _{q} b_{1}^{2}=-12 \\
& n=5 \Rightarrow \log _{q} b_{1}^{2}=-10 \\
& n=6 \Rightarrow \log _{q} b_{1}^{2}=-10 \\
& n=9 \Rightarrow \log _{q} b_{1}^{2}=-12
\end{aligned}
$$
From this, it is clear that the smallest value of the desired quantity is -12 and it is achieved for two values: $n=4$ and $n=9$. | Answer: -12 when $n=4$ and $n=9$. | -12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Let $x$ and $y$ be the lengths of two segments on the left side. The length of the third segment is $L-(x+y)$. The sample space consists of all pairs $(x, y)$ for which
$$
\left\{\begin{array} { l }
{ 0 \leq x \leq L } \\
{ 0 \leq y \leq L } \\
{ 0 \leq L - ( x + y ) \leq L }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
0 \leq x \leq L \\
0 \leq y \leq L \\
0 \leq x+y \leq L
\end{array}\right.\right.
$$
On the coordinate plane, they form a triangle with vertices $(0,0),(L, 0),(0, L):$
The area of this triangle is $L^{2} / 2$. To solve the problem, it is more convenient to calculate the probability of unfavorable events, which are determined by the system of inequalities
$$
\left\{\begin{array} { l }
{ x \leq \frac { 5 } { 1 2 } L } \\
{ y \leq \frac { 5 } { 1 2 } L } \\
{ L - ( x + y ) \leq \frac { 5 } { 1 2 } L }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x \leq \frac{5}{12} L \\
y \leq \frac{5}{12} L \\
x+y \geq \frac{7}{12} L
\end{array}\right.\right.
$$
On the coordinate plane, they also form a right triangle with legs $L / 4$:
The area of this triangle is $\frac{1}{32} L^{2}$. The probability of unfavorable events is the ratio of the areas
$$
q=\frac{\frac{1}{32} L^{2}}{\frac{1}{2} L^{2}}=\frac{1}{16}
$$
And the desired probability is
$$
p=1-q=1-\frac{1}{16}=\frac{15}{16}
$$ | Answer: $\frac{15}{16}$. | \frac{15}{16} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $[\sin 2 x]-2[\cos x]=3[\sin 3 x]$, where $[a]$ is the integer part of $a$, i.e., the greatest integer not exceeding $a$. | $$
\begin{gathered}
x \in\left(2 \pi n ; \frac{\pi}{6}+2 \pi n\right) \cup\left(\frac{\pi}{6}+2 \pi n ; \frac{\pi}{4}+2 \pi n\right) \cup \\
\cup\left(\frac{\pi}{4}+2 \pi n ; \frac{\pi}{3}+2 \pi n\right), n \in \mathbb{Z}
\end{gathered}
$$ | x\in(2\pin;\frac{\pi}{6}+2\pin)\cup(\frac{\pi}{6}+2\pin;\frac{\pi}{4}+2\pin)\cup(\frac{\pi}{4}+2\pin;\frac{\pi}{3}+2\pin),n\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. For which $a$ does the system
$$
\left\{\begin{array}{c}
x \sin a-(y-6) \cos a=0 \\
\left((x-3)^{2}+(y-3)^{2}-1\right)\left((x-3)^{2}+(y-3)^{2}-9\right)=0
\end{array}\right.
$$
have two solutions? | $$
\begin{aligned}
& a \in\left(\frac{\pi}{2}+\pi n ; \frac{3 \pi}{4}-\arcsin \frac{\sqrt{2}}{6}+\pi n\right) \cup \\
& \cup\left(\frac{3 \pi}{4}+\arcsin \frac{\sqrt{2}}{6}+\pi n ; \pi+\pi n\right), n \in Z
\end{aligned}
$$ | \in(\frac{\pi}{2}+\pin;\frac{3\pi}{4}-\arcsin\frac{\sqrt{2}}{6}+\pin)\cup(\frac{3\pi}{4}+\arcsin\frac{\sqrt{2}}{6}+\pin;\pi+\pin),n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. For which $a$ does the system
$$
\left\{\begin{array}{c}
(x-5) \sin a-(y-5) \cos a=0 \\
\left((x+1)^{2}+(y+1)^{2}-4\right)\left((x+1)^{2}+(y+1)^{2}-16\right)=0
\end{array}\right.
$$
have three solutions? | Answer: $a=\frac{\pi}{4} \pm \arcsin \frac{\sqrt{2}}{6}+\pi n, n \in Z$ | \frac{\pi}{4}\\arcsin\frac{\sqrt{2}}{6}+\pin,n\inZ | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $2[\cos 2 x]-[\sin x]=3[\sin 4 x]$, where $[a]$ is the integer part of $a$, i.e., the greatest integer not exceeding $a$. | $$
\left[\begin{array}{l}
x \in\left(2 \pi n ; \frac{\pi}{8}+2 \pi n\right), n \in \mathbb{Z} \\
x=\left(\frac{\pi}{8}+2 \pi n ; \frac{\pi}{4}+2 \pi n\right], n \in \mathbb{Z} \\
x=\frac{3 \pi}{4}+2 \pi n, n \in \mathbb{Z}
\end{array}\right.
$$ | \ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. For which $a$ does the system
$$
\left\{\begin{array}{c}
(x-8) \sin a-(y+2) \cos a=0 \\
\left((x-3)^{2}+(y-3)^{2}-1\right)\left((x-3)^{2}+(y-3)^{2}-9\right)=0
\end{array}\right.
$$
have four solutions | Answer:
$$
a \in\left(\frac{3 \pi}{4}-\arcsin \frac{\sqrt{2}}{10}+\pi n ; \frac{3 \pi}{4}+\arcsin \frac{\sqrt{2}}{10}+\pi n\right), n \in Z
$$ | \in(\frac{3\pi}{4}-\arcsin\frac{\sqrt{2}}{10}+\pin;\frac{3\pi}{4}+\arcsin\frac{\sqrt{2}}{10}+\pin),n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Note that $(n+1)^{2}-(n-1)^{2}=4 n$, i.e., numbers $a=4 n$ that are multiples of four can be represented as the difference of squares of two integers, and therefore, with the condition of zero, can be represented as the sum or difference of four squares. If $a=4 n \pm 1$, then the representation is
$$
a=(n+1)^{2}-(n-1)^{2} \pm 1^{2}
$$
If $a=4 n+2$, then it can be represented as the sum or difference of four squares:
$$
a=(n+1)^{2}-(n-1)^{2}+1^{2}+1^{2}
$$
The number 2021 can be written using representation (1), so, for example, $2021=4 \cdot 505+1, n=505,2021=506^{2}-504^{2}+1^{2}$. | Answer: $2021=506^{2}-504^{2}+1^{2}$. | 2021=506^{2}-504^{2}+1^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Let's highlight the complete squares: $\left\{\begin{array}{l}(x-2)^{2}+(y+1)^{2}=5, \\ (x-2)^{2}+(z-3)^{2}=13, \\ (y+1)^{2}+(z-3)^{2}=10 .\end{array}\right.$
Add all the equations $\left\{\begin{array}{c}(x-2)^{2}+(y+1)^{2}=5, \\ (x-2)^{2}+(z-3)^{2}=13, \\ (x-2)^{2}+(y+1)^{2}+(z-3)^{2}=14 .\end{array}\right.$
And, by subtracting the first and second from the third, we get:
$$
\left\{\begin{array}{l}
(x-2)^{2}=4 \\
(z-3)^{2}=9 \\
(y+1)^{2}=1
\end{array}\right.
$$
Solve the quadratic equations and write down all possible solutions. The system has eight solutions. | Answer: $(0 ; 0 ; 0),(0 ;-2 ; 0),(0 ; 0 ; 6),(0 ;-2 ; 6),(4 ; 0 ; 0),(4 ;-2 ; 0)$,
$(4 ; 0 ; 6),(4 ;-2 ; 6)$. | (0;0;0),(0;-2;0),(0;0;6),(0;-2;6),(4;0;0),(4;-2;0),(4;0;6),(4;-2;6) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Represent the number 1917 as the sum or difference of the squares of three integers. Prove that any integer can be represented as the sum or difference of the squares of four integers. | Answer: $1917=480^{2}-478^{2}+1^{2}$. | 1917=480^{2}-478^{2}+1^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the system $\left\{\begin{array}{c}x^{2}+y^{2}+2 x+6 y=-5 \\ x^{2}+z^{2}+2 x-4 z=8 \\ y^{2}+z^{2}+6 y-4 z=-3\end{array}\right.$. | Answer: $(1 ;-2 ;-1),(1 ;-2 ; 5),(1 ;-4 ;-1),(1 ;-4 ; 5)$,
$$
(-3 ;-2 ;-1),(-3 ;-2 ; 5),(-3 ;-4 ;-1),(-3 ;-4 ; 5)
$$ | (1;-2;-1),(1;-2;5),(1;-4;-1),(1;-4;5),(-3;-2;-1),(-3;-2;5),(-3;-4;-1),(-3;-4;5) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Represent the number 1947 as the sum or difference of the squares of three integers. Prove that any integer can be represented as the sum or difference of the squares of four integers. | Answer: $1947=488^{2}-486^{2}-1^{2}$. | 1947=488^{2}-486^{2}-1^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the system $\left\{\begin{array}{c}x^{2}+y^{2}+8 x-6 y=-20 \\ x^{2}+z^{2}+8 x+4 z=-10 \\ y^{2}+z^{2}-6 y+4 z=0\end{array}\right.$. | Answer: $(-3 ; 1 ; 1),(-3 ; 1 ;-5),(-3 ; 5 ; 1),(-3 ; 5 ;-5)$,
$$
(-5 ; 1 ; 1),(-5 ; 1 ;-5),(-5 ; 5 ; 1),(-5 ; 5 ;-5)
$$ | (-3;1;1),(-3;1;-5),(-3;5;1),(-3;5;-5),(-5;1;1),(-5;1;-5),(-5;5;1),(-5;5;-5) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.