problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
5. Find the smallest three-digit number with the property that if a number, which is 1 greater, is appended to it on the right, then the result (a six-digit number) will be a perfect square. Answer: 183 | Solution: Let the required number be \(a\), then \(1000a + a + 1 = n^2\). We can write it as: \(1001a = (n - 1)(n + 1)\). Factorize \(1001 = 7 \times 11 \times 13\), so the product \((n - 1)(n + 1)\) must be divisible by 7, 11, and 13. Moreover, for the square to be a six-digit number, \(n\) must be in the interval \([... | 183 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,151 |
6. Pete came up with all the numbers that can be formed using the digits 2, 0, 1, 8 (each digit can be used no more than once). Find their sum. | Answer: 78331
Solution: First, consider the units place. Each of the digits 1, 2, 8 appears once in this place for single-digit numbers, twice for two-digit numbers, four times for three-digit numbers, and four times for four-digit numbers - a total of 11 times.
In the tens place, each of them appears 3 times for two... | 78331 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,152 |
7. Given a sequence of natural numbers $a_{n}$, the terms of which satisfy
the relations $a_{n+1}=k \cdot \frac{a_{n}}{a_{n-1}}$ (for $n \geq 2$). All terms of the sequence are integers. It is known that $a_{1}=1$, and $a_{2018}=2020$. Find the smallest natural $k$ for which this is possible. | Answer: 2020
Solution: Let $a_2=x$. Then all terms of the sequence will have the form $x^{m} k^{n}$.
The powers of $k$ will repeat with a period of 6: $0,0,1,2,2,1,0,0, \ldots$
The powers of $x$ will also repeat with a period of 6: 0,1,1,0,-1,-1,0,1,...
Since 2018 gives a remainder of 2 when divided by 6, then $a_{... | 2020 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,153 |
1. Find all pairs of natural numbers $x, y \in [1; 8]$, satisfying the equation
$$
\sqrt{x x, x x x \ldots}=y, y y y \ldots
$$
(decimal representation of each of the numbers $x x, x x x \ldots$ and $y, y y y \ldots$ consists of an infinite number of the same digits). | # Problem 1.
Answer: $x=1, y=3$ and $x=4, y=6$.
Solution. Since an infinite decimal of the form $0, x x x \ldots$ is equal to $\frac{x}{9}$, the equation in the condition is equivalent to the relation
$$
11 x+\frac{x}{9}=\left(y+\frac{y}{9}\right)^{2} \Longleftrightarrow 9 x=y^{2}
$$
from which the answer is obtai... | 1,34,6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,154 |
2. Solve the equation
$$
2\left|\log _{1 / 2} x-1\right|-\left|\log _{4}\left(x^{2}\right)+2\right|=-\frac{1}{2} \log _{\sqrt{2}} x
$$ | # Problem 2.
Answer: $\left(0 ; \frac{1}{4}\right] \cup\{1\}$ (variants V-1, V-2, V-4), $\left\{2^{-8 / 5} ; 1\right\}$ (variant V-3).
Solution (variant V-1). Rewrite the equation as
$$
\left|\log _{2} x+2\right|-2\left|\log _{2} x+1\right|=\log _{2} x
$$
and perform the substitution $t=\log _{2} x$. Solving the re... | (0;\frac{1}{4}]\cup{1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,155 |
3. In a circle of radius 3, triangles $A B C$ and $A M N$ are inscribed. The line $A M$ passes through the midpoint $E$ of segment $B C$, and the line $B N$ passes through the midpoint $F$ of segment $A C$. Find the perimeter of triangle $A B C$, if $A M: A E=2: 1$ and $B N: B F=17: 13$. | # Problem 3.
Answer: $\frac{72}{5}$.
Solution. According to the intersecting chords theorem: $A E \cdot E M = B E \cdot E C$. But since by the condition $A E = E M$ and $B E = E C$, then $A E = B E = C E$, i.e., $E$ is the center of the circumscribed circle around $\triangle A B C$. Therefore, angle $A$ is a right an... | \frac{72}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,156 |
5. A square with a side of 12 needs to be completely cut into four squares with an integer side $a$, three squares with an integer side $b$, and ten rectangles with sides $a$ and $b$. Find all values of $a$ and $b$ for which this is possible. | Problem 5.
Answer: $a=2, b=4$.
Solution. A necessary condition for the existence of such a cutting is the equality of areas: $4 a^{2}+3 b^{2}+10 a b=144$. From this, we obtain that $b$ is even and, since $3 b^{2}<144$, the number $b$ can be 2, 4, or 6. For $b=2$, we have: $a(a+5)=33$ and $a$ is not an integer. For $b... | =2,b=4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,158 |
1. Solve the inequality
$$
(2+\sqrt{3})^{x}+2<3(\sqrt{2-\sqrt{3}})^{2 x}
$$
In your answer, specify the sum of all integer values of $x$ that satisfy the given inequality and belong to the interval $(-20 ; 53)$. | Answer: The solution to the inequality is $(-\infty ; 0)$. The desired sum $-19-18-\cdots-2-1=-190$. We write -190 as the answer. | -190 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,159 |
3. The area of triangle $\triangle A B C$ is 10 cm $^{2}$. What is the smallest value in centimeters that the circumference of the circle circumscribed around triangle $\triangle A B C$ can take, given that the midpoints of the heights of this triangle lie on the same line? If the answer is not an integer, round it to ... | Answer: The smallest value is $2 \pi \sqrt{10}$ cm. In the answer, we write 20, since $2 \pi \sqrt{10} \approx 19.8691 \ldots$ | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,161 |
5. Find the set of pairs of real numbers $(x, y)$ satisfying the conditions:
$$
\left\{\begin{array}{l}
3^{-x} y^{4}-2 y^{2}+3^{x} \leq 0 \\
27^{x}+y^{4}-3^{x}-1=0
\end{array}\right.
$$
Calculate the values of the expression $x_{k}^{3}+y_{k}^{3}$ for each solution $\left(x_{k}, y_{k}\right)$ of the system and find th... | Answer: The solution to the system is $(x, y)=(0, \pm 1)$. Answer -1.
## 1 Creative problems (7 items) | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,162 |
1. The function $f(x)=\sin x+\cos x$ is decreasing on the interval $[\pi / 4 ; \pi / 2]$ (monotonicity can be justified by the equality $f(x)=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)$, or using the derivative, or directly using the definition). Therefore, $\sin 1+\cos 1>\sin \frac{\pi}{3}+\cos \frac{\pi}{3}=\frac{1+\... | Answer: the first number is greater.
Answer to variant 172: the first number is greater.
Answer to variant 173: the second number is greater.
Answer to variant 174: the second number is greater. | \frac{1+\sqrt{3}}{2}>\frac{49}{36} | Inequalities | proof | Yes | Yes | olympiads | false | 5,163 |
2. Let $n$ be the time in minutes it takes for the slower boy to complete a lap. Then $n>5, n \in \mathbb{N}$. The speed at which the faster boy catches up to the slower one is
$$
\frac{1}{5}-\frac{1}{n}=\frac{n-5}{5 n}\left(\frac{\text { lap }}{\text { min }}\right)
$$
therefore, the time between meetings is
$$
\fr... | Answer: 6 min.
Answer to option 172: 12 min.
Answer to option $173: 56$ min.
Answer to option $174: 4$ min. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,164 |
4. The equation makes sense when $\left[\log _{3} x\right]>0 \Longleftrightarrow \log _{3} x \geqslant 1 \Longleftrightarrow x \geqslant 3$.
We will show that for all $x \geqslant 3$, the inequality $\log _{3}[x] \geqslant\left[\log _{3} x\right]$ holds (in fact, this inequality is true for all $x \geqslant 1$, but we... | Answer: $x \in[3 ; 4)$.
Answer to option $\mathbf{1 7 2}: x \in[2 ; 3)$.
Answer to option $173: x \in[3 ; 4)$.
Answer to option $\mathbf{1 7 4}: x \in[2 ; 3)$. | x\in[3;4) | Algebra | proof | Yes | Yes | olympiads | false | 5,166 |
1. The total number of students in the school must be divisible by 7 and 4, which means it must be divisible by 28. Since the number is no more than 40, there were 28 students in total. The number of prize winners is $(1 / 7 + 1 / 4 + 1 / 4) \cdot 28 = 18$ people. Therefore, 10 students did not receive any medals. | Answer: $x=10$.
Answer to the variant: $4-2: x=18$. | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,173 |
2. First method: Since $x_{1}^{3}-2015 x_{1}+2016=0$ and $x_{2}^{3}-2015 x_{2}+2016=0$, then $x_{1}^{3}-x_{2}^{3}=2015 \cdot\left(x_{1}-x_{2}\right)$. Therefore, $x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}=2015$.
Second method: By Vieta's theorem (but then it needs to be justified that there are three distinct roots): $x_{1}+x_{... | Answer: 2015.
Answer to option: 4-2: 2016. | 2015 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,174 |
4. The right side of the inequality is zero when $|x| \leqslant 10$ (for other $x$ it is undefined). Denoting $\alpha=\frac{\pi x}{4}$, we get:
$$
\begin{aligned}
& \frac{\sin 2 \alpha-\cos 2 \alpha+1}{\sin 2 \alpha+\cos 2 \alpha-1} \geqslant 0 \Longleftrightarrow \frac{2 \sin \alpha \cos \alpha+2 \sin ^{2} \alpha}{2 ... | Answer: -12.
Answer to the option: $4-2: 12$. | -12 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,176 |
5. The first equation is reduced to the form $(x+y)^{2}+(y-a)^{2}=0$, from which we obtain $x=-a ; y=a$. Substitution into the inequality gives:
$$
2^{-2-a} \cdot \log _{2}(-a)<1 \Longleftrightarrow \log _{2}(-a)<2^{2+a}
$$
Equality is achieved at $a=-2$, and due to monotonicity we get $a \in$ $(-2 ; 0)$. | Answer: $x=-a ; y=a$ for $a \in(-2 ; 0)$.
Answer to variant: $4-2: x=a ; y=-a$ for $a \in(-3 ; 0)$. | -for\in(-2;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,177 |
2. If there are $x$ black pieces, then there are $32-x$ white pieces. Since each black piece (pentagonal) borders only white pieces, the number of black and white piece borders will be $5 x$. On the other hand, such borders are $3 \cdot(32-x)$. From the equation $5 x=3 \cdot(32-x)$ we get $x=12$. | Answer: 12.
Answer to option: 5-2: 20. | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,179 |
3. The original equation is equivalent to
$$
\begin{aligned}
\left(\frac{1-\cos 2 x}{2}\right)^{5}+\left(\frac{1+\cos 2 x}{2}\right)^{5}=\frac{29}{16} \cos ^{4} 2 x & \Longleftrightarrow \\
& \Longleftrightarrow \frac{2+20 \cos ^{2} 2 x+10 \cos ^{4} 2 x}{32}=\frac{29}{16} \cos ^{4} 2 x
\end{aligned}
$$
From which $24... | Answer: $x=\frac{\pi}{8}+\frac{\pi k}{4}$.
Answer to option: $5-2: x=\frac{\pi}{8}+\frac{\pi k}{4}$. | \frac{\pi}{8}+\frac{\pik}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,180 |
1. Let $N$ - a natural number from the condition of the problem. From the condition, it follows that the number of fours in $N$ is $2m$, and the number of fives is $2m+17$. Using the property of equal remainders of division by 9: the remainder of the division by 9 of a natural number is equal to the remainder of the di... | Answer: 4. Answer to the option: $6-2: 5$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,183 | |
5. After substituting $x=\cos \alpha, y=\sin \alpha$, the second equation is satisfied, and the first equation can be written as $\cos ^{15} 3 \alpha+\sin ^{16} 3 \alpha=1$. From the chain
$$
1=\cos ^{15} 3 \alpha+\sin ^{16} 3 \alpha \leqslant \cos ^{2} 3 \alpha+\sin ^{2} 3 \alpha=1
$$
it follows that $\cos ^{15} 3 \... | Answer: 9.
Answer to option: $6-2$ : 9.
The equation $\cos 3 x=0$ is equivalent to
$$
3 x=\pi / 2+2 \pi k, k \in \mathbb{Z}, \quad 3 x=-\pi / 2+2 \pi n, n \in \mathbb{Z}
$$
Answers and solutions to option $\mathbf{7}-1$ | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,187 |
3. Let $N$ sq.m. need to be painted, $x$ - the number of students, $v-$ the productivity of each student, $t_{1}$ - the time it takes to complete the work. Then it follows that $x v t_{1}=N$ and each student should paint $v t_{1}=\frac{N}{x}$ sq. m.
In the condition of the problem, the master's productivity is greater... | Answer: 9 .
Answer to the option: $7-2: 11$. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,190 |
5. Let $\alpha=\pi x$.
$$
\begin{aligned}
& \cos (8 \alpha)+2 \cos (4 \alpha)-\cos (2 \alpha)+2 \sin (\alpha)+3=0 \Longleftrightarrow \\
& 2 \cos ^{2} 4 \alpha-1+2 \cos 4 \alpha-1+2 \sin ^{2} \alpha+2 \sin \alpha+3=0 \Longleftrightarrow \\
& 2\left(\cos 4 \alpha+\frac{1}{2}\right)^{2}+2\left(\sin \alpha+\frac{1}{2}\ri... | Answer: 5050
Answer to option: 7-2: 4950.
Solutions for option 17-1 and answers to all options | 5050 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,192 |
2. Let $n$ be the time in minutes it takes for the slower boy to complete a lap. Then $n>5, n \in \mathbb{N}$. The speed at which the faster boy catches up to the slower one is
$$
\frac{1}{5}-\frac{1}{n}=\frac{n-5}{5 n}\left(\frac{\text { lap }}{\text { min }}\right)
$$
therefore, the time between meetings is
$$
\le... | Answer: 6 min.
Answer to option 17 - $2: 12$ min.
Answer to option 17 - $\mathbf{\text { : }}$ : 56 min.
Answer to option $17-4: 4$ min. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,193 |
5. Drop perpendiculars $D D_{1}, D D_{2}, D D_{3}$ from point $D$ to the planes $S B C$, $S A C$, and $S A B$ respectively. Let $D D_{1}=x, D D_{2}=y, D D_{3}=z$. According to the condition, we form the system of equations
$$
\left\{\begin{array}{l}
y^{2}+z^{2}=5 \\
x^{2}+z^{2}=13 \\
x^{2}+y^{2}=10
\end{array}\right.
... | Answer: 27.
Answer to option 17-2: 108.
Answer to option $17-3: 27$.
Answer to option $17-4: 108$.
[^0]: ${ }^{1}$ This equality can be proven by expressing $B C^{2}$ from two triangles $B A C$ and $B D C$ using the planimetric cosine theorem. | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,195 |
7. A natural number $N$ ends in ...70, and it has exactly 72 natural divisors (including 1 and itself). How many natural divisors does the number 80N have? ANSWER: 324. | Solution: Each divisor of the number $\mathrm{N}$ can be represented in the form $2^{a} \cdot 5^{b} \cdot q$, where q is not divisible by 2 and 5, and the numbers $a, b$ are 0 or 1. Therefore, there are only 4 different combinations for the pair a and b, which means there are $72 / 4=18$ possible values for q. These co... | 324 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,202 |
1. Solve the inequality
$$
\frac{4+\sqrt{-3-x}}{3-|x+4|} \leqslant 1
$$
In your answer, specify the sum of all integer values of $x$ that satisfy the given inequality and do not exceed 12 in absolute value. | Solution. Transferring 1 to the left side and bringing to a common denominator:
$$
\frac{1+\sqrt{-3-x}+|x+4|}{3-|x+4|} \leqslant 0
$$
For $x>-3$, the expression under the square root is negative, meaning there are no solutions. For $x \leqslant-3$, the numerator is positive, and thus,
$$
3-|x+4|3 \Longleftrightarrow... | -50 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,204 |
2. Solve the equation $\cos 8 x=\frac{14}{3}(\cos 2 x-\sin 2 x)^{2}-1$. In your answer, specify the number equal to the sum of the roots of the equation that belong to the interval $\left[\frac{11 \pi}{2} ; \frac{13 \pi}{2}\right]$, rounding this number to two decimal places if necessary. | Solution. The original equation is equivalent to the equation
$$
\begin{aligned}
1-2 \sin ^{2} 4 x=\frac{14}{3}(1-\sin 4 x)-1 & \Longleftrightarrow 2 \sin ^{2} 4 x-\frac{14}{3} \sin 4 x+\frac{8}{3}=0 \Longleftrightarrow \\
& \Longleftrightarrow \sin 4 t=1 \Longleftrightarrow x=\frac{\pi}{8}+\frac{\pi k}{2}, k \in \mat... | 36.91 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,205 |
3. Find the side $B C$ of the quadrilateral $A B C D$, if $\angle B A C=\alpha, \angle A C D=\beta, \angle B C A+\angle C A D=\frac{\pi}{2}$ and $A D=a$. In the answer, write the result rounded to two decimal places.
$$
\alpha=\arcsin \frac{5}{13}, \beta=\arcsin \frac{12}{13}, a=24
$$ | Solution. Since the sum of angles $\angle B A C+\angle A C D=\arcsin \frac{5}{13}+\arcsin \frac{12}{13}=\frac{\pi}{2}$, then $\angle B A D+\angle B C D=$ $\pi$. Therefore, a circle can be circumscribed around quadrilateral $A B C D$. Next, by the Law of Sines $\frac{A D}{\sin \beta}=\frac{B C}{\sin \alpha} \Rightarrow ... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,206 |
4. Solve the system
$$
\left\{\begin{array}{c}
\log _{x+a}\left(y^{2}-2 c y+c^{2}\right)+\log _{y-c}\left(x^{2}+2 x a+a^{2}\right)=4 \\
\log _{x+a}(-2 y+3+2 c)+\log _{y-c}(2 x-1+2 a)=2
\end{array}\right.
$$
if $a=8, c=20$. Calculate the values of the expression $x_{k} \cdot y_{k}$ for each solution $\left(x_{k}, y_{k... | Solution. The first equation of the system for $x>-a, x \neq 1-a, y>c, y \neq 1+c$ is equivalent to the equation
$$
\log _{x+a}(y-c)+\log _{y-c}(x+a)=2 \Longleftrightarrow \log _{x+a}(y-c)=1 \Longleftrightarrow y=x+a+c
$$
or for these values, $y=x+28$. Substituting into the second equation leads to the equation
$$
\... | -152.44 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,207 |
5. When drying, apricots lose $10 \%$ of their mass, and grapes lose $30 \%$ of their mass. Determine the proportion in which apricots and grapes must be taken so that after drying, the mixture contains twice as much dried apricot as raisins. In your answer, indicate the ratio of the initial mass of apricots to the mas... | Solution. If we take $x$ (kg) of apricots and $y$ (kg) of grapes, then after drying, the apricots will weigh $0.9 x$, and the grapes will weigh $0.7 y$. Since the first mass should be twice the second, then $0.9 x = 1.4 y$, which means $\frac{x}{y} = \frac{14}{9} \approx 1.56$.
Answer: 1.56.
## Set of Creative Proble... | 1.56 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,208 |
1.24. $l=224, m=2001, n=2014$. Answer: $839\left(x=\frac{180}{4015}=\frac{36}{803}\right)$.
II. Compute the value of the expression
$$
((\ldots((2018 \oplus 2017) \oplus 2016) \oplus \ldots \oplus 2) \oplus 1)
$$
if $x \oplus y=\frac{3(x+y)}{x y+9}$. | Solution. Since $x \oplus 3=\frac{3(x+3)}{3 x+9}=1$, regardless of the value of $x$, the desired expression is $(((x \oplus 3) \oplus 2) \oplus 1)=((1 \oplus 2) \oplus 1)=\left(\left(\frac{3(1+2)}{1 \cdot 2+9}\right) \oplus 1\right)=\left(\frac{9}{11} \oplus 1\right)=\frac{3\left(\frac{9}{11}+1\right)}{\frac{9}{11} \cd... | 0.56 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,209 |
2.16. If $x \oplus y=\frac{4(x+10 y)}{x y+160}$. Answer: 0.26.
III. For each $a$, for which the equation $x^{3}-x^{2}-4 x-a=0$ has three distinct real roots, denote these roots, ordered in descending order $\left(x_{1}>x_{2}>x_{3}\right)$, by $x_{1}=x_{1}(a), x_{2}=x_{2}(a), x_{3}=x_{3}(a)$. Determine for which of the... | Solution. Let's find for which $a$ there will be exactly three distinct solutions. For this, consider the expression $\tilde{f}(x)=x^{3}-x^{2}-4 x$. Three solutions $\tilde{f}(x)=a$ will exist if and only if the line $y=a$ intersects the graph of the function $y=\tilde{f}(x)$ at three points. Let's find the derivative ... | -3.14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,210 |
3.12. $f(x)=\frac{1}{4} x^{3}-\frac{1}{4} x^{2}-x-a$. Answer: $-0.79$.
IV. The decagon $A B C D P Q R S T U V W$ has two parallel bases: square $A B C D$ and octagon $P Q R S T U V W$, all angles of which are equal, and eight lateral faces: triangles $A P Q, B R S, C T U, D V W$ and rectangles $D A P W, A B R Q, B C T... | # Solution:
The procedure for constructing the section of a decagon by the plane $A S U$ consists of two steps:
1) Mark the point $K$ of intersection of the lines $S U$ and $Q R$, the point $L$ of intersection of the lines $S U$ and $W V$, and the point $M$ of intersection of the lines $S U$ and $P W$;
2) Draw the li... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,211 |
1. The decagon $A B C D P Q R S T U V W$ has two parallel bases: square $A B C D$ and octagon $P Q R S T U V W$, all angles of which are equal, as well as eight lateral faces: triangles $A P Q, B R S, C T U, D V W$ and rectangles $D A P W, A B R Q, B C T S$ and $C D V U$. It is known that the area of the section of thi... | Answer: 3.61 . (Exact value $\sqrt{13}(m=1, k=1)$.) | 3.61 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,212 |
3. The decagon $A B C D P Q R S T U V W$ has two parallel bases: square $A B C D$ and octagon $P Q R S T U V W$, all angles of which are equal, as well as eight lateral faces: triangles $A P Q, B R S, C T U, D V W$ and rectangles $D A P W, A B R Q, B C T S$ and $C D V U$. It is known that the area of the section of the... | Answer: 8.49 . (Exact value $6 \sqrt{2}(m=\sqrt{2}, k=1)$.) | 8.49 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,214 |
4. The decagon $A B C D P Q R S T U V W$ has two parallel bases: square $A B C D$ and octagon $P Q R S T U V W$, all angles of which are equal, and eight lateral faces: triangles $A P Q, B R S, C T U, D V W$ and rectangles $D A P W, A B R Q, B C T S$ and $C D V U$. It is known that the area of the section of the decago... | Answer: 4.58. (Exact value $\sqrt{21}(m=2, k=1)$ ). | 4.58 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,215 |
9. The decagon $A B C D P Q R S T U V W$ has two parallel bases: square $A B C D$ and octagon $P Q R S T U V W$, all angles of which are equal, and eight lateral faces: triangles $A P Q, B R S, C T U, D V W$ and rectangles $D A P W, A B R Q, B C T S$ and $C D V U$. It is known that the area of the section of the decago... | Answer: 2.65 . (Exact value $\sqrt{7}(m=1 / 3, k=2)$.) | 2.65 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,220 |
10. The decagon $A B C D P Q R S T U V W$ has two parallel bases: square $A B C D$ and octagon $P Q R S T U V W$, all angles of which are equal, and eight lateral faces: triangles $A P Q, B R S, C T U, D V W$ and rectangles $D A P W, A B R Q, B C T S$ and $C D V U$. It is known that the area of the section of the decag... | Answer: 3.32 . (The exact value $\sqrt{11}(m=1 / 9, k=2)$.) | 3.32 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,221 |
11. The decagon $A B C D P Q R S T U V W$ has two parallel bases: square $A B C D$ and octagon $P Q R S T U V W$, all angles of which are equal, and eight lateral faces: triangles $A P Q, B R S, C T U, D V W$ and rectangles $D A P W, A B R Q, B C T S$ and $C D V U$. It is known that the area of the section of the decag... | Answer: 1.73 . (Exact value $\sqrt{3}(m=5 / 9, k=2)$.) | 1.73 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,222 |
12. The decagon $A B C D P Q R S T U V W$ has two parallel bases: square $A B C D$ and octagon $P Q R S T U V W$, all angles of which are equal, and eight lateral faces: triangles $A P Q, B R S, C T U, D V W$ and rectangles $D A P W, A B R Q, B C T S$ and $C D V U$. It is known that the area of the section of the decag... | Answer: 2.83 . (Exact value $\sqrt{8}(m=11 / 9, k=2)$.) | 2.83 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,223 |
2. In quadrilateral $A B C D$, it is known that $A B=B C, \angle A B C=$ $\angle A D C=90^{\circ}$. A perpendicular $B H$ is dropped from vertex $B$ to side $A D$. Find the area of quadrilateral $A B C D$, if it is known that $B H=h$. | Answer: $h^{2}$.
Solution:
Cut off the triangle $A B H$, attach it to the top - we get a square with side $h$. | ^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,228 |
3. We will denote $\max (A, B, C)$ as the greatest of the numbers $A, B, C$. Find the smallest value of the quantity $\max \left(x^{2}+|y|,(x+2)^{2}+|y|, x^{2}+|y-1|\right)$. | Answer: 1.5.
Solution: Note that when $x^{2}>(x+2)^{2}$ for $x-1$. If $x=-1$, then both of these values are equal to 1, otherwise one of them is greater than 1. Similarly, for $|y|$ and $|y-1|$, when $y=1 / 2$ they are both equal to $1 / 2$, and when $y \neq 1 / 2$ one of them is greater. Therefore, the minimum is ach... | 1.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,229 |
4. The number 2015 can be represented as the sum of consecutive integers in various ways, for example, $2015=1007+1008$ or $2015=$ $401+402+403+404+405$. In how many ways can this be done? | Answer: 16.
Solution: The sum of $k$ numbers starting from $n$ is $S(k, n)=\frac{1}{2}(2 n+k-1) \cdot k$. That is, we need to solve the equation $(2 n+k-1) \cdot k=4030$ in integers. Obviously, $k$ can be any divisor of $4030=2 \times 5 \times 13 \times 31$. Note that each of the prime factors $(2,5,13$ and 31$)$ can ... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,230 |
7. Find the minimum value of the function $f(x, y)=\frac{2015(x+y)}{\sqrt{2015 x^{2}+2015 y^{2}}}$ and specify all pairs $(x, y)$ at which it is achieved. | Answer: a) $-\sqrt{4030}$. b) $x=y<0$.
Solution: $f(x, y)= \pm \sqrt{2015} \sqrt{\frac{(x+y)^{2}}{x^{2}+y^{2}}}$. The minimum value is achieved when $\frac{(x+y)^{2}}{x^{2}+y^{2}}$ reaches its maximum, and $x+y<0$. Note that $\frac{(x+y)^{2}}{x^{2}+y^{2}}=1+\frac{2 x y}{x^{2}+y^{2}} \leqslant 2$ and equality is achiev... | -\sqrt{4030} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,231 |
8. Given 2015 pairwise coprime natural numbers, not exceeding $10^{7}$. Can they all be composite? | Answer: No.
 | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,232 |
9. Given triangles $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$, the areas of which are 1 and 2025, respectively. It is known that rays $A B$ and $A^{\prime} B^{\prime}$ are parallel and point in opposite directions (see figure). The same is true for pairs $B C$ and $B^{\prime} C^{\prime}, C A$ and $C^{\prime} A^{\pr... | Answer: 484.
Solution: Obviously, triangles $A B C A^{\prime} B^{\prime} C^{\prime}$ are similar with a coefficient of 45.
Three trapezoids are formed, in which $A^{\prime \prime} B^{\prime \prime}, B^{\prime \prime} C^{\prime \prime}$ and $C^{\prime \prime} A^{\prime \prime}$ are segments connecting the midpoints of... | 484 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,233 |
10. Find the function $f(x)$, which is known to satisfy
$$
f(x)= \begin{cases}x \cdot f\left(\frac{2 x+3}{x-2}\right)+3, & \text { for } x \neq 2 \\ 0, & \text { for } x=2\end{cases}
$$ | Answer: $f(x)=\frac{3(x+1)(2-x)}{2\left(x^{2}+x+1\right)}$. | f(x)=\frac{3(x+1)(2-x)}{2(x^{2}+x+1)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,234 |
4. A school coach decided to reward 12 students who ran the distance in the best time. Each of them must be awarded a "gold", "silver", or "bronze" medal. All three types of medals must be used, and the one who finished earlier cannot be awarded a less valuable medal than the one who finished later.
How many ways can ... | Answer: $C_{11}^{2}=55$.
Solution: see 9th grade. | 55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,238 |
5. On a line, there are 16 points $A_{1}, \ldots A_{16}$, spaced 1 cm apart. Misha constructs circles according to the following rules:
a) The circles do not intersect or touch.
b) Inside each circle, there is at least one of the specified points $\mathrm{A}_{1}, \ldots \mathrm{A}_{16}$.
c) None of these points lie ... | Answer: 31.
Solution: see 9th grade.
## Tasks for 7-8 grades
## Option 3b (Moscow) | 31 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,239 |
1. Each worker on the construction site has at least one construction specialty. 10 people have the specialty of a mason, 9 - the specialty of a painter, 8 - the specialty of a plasterer. It is known that at least four people have the specialties of a mason and a plasterer simultaneously, at least five people - the spe... | # Answer 18.
Solution: According to the principle of inclusion-exclusion, the total number of workers is K+M+S - KM-KS-MS + KMS = 10+9+8-4-5-3 + KMS = 15 + KMS. Note that the number of workers proficient in all three specialties cannot exceed 3. | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,240 |
1. Natural numbers, for which the sum of the digits equals 5, were arranged in ascending order. What number is in the $122-$nd position? | Answer: 40001.
Solution: Let's calculate the number of such numbers for different digit counts.
Let $n$ be the number of digits. Subtract 1 from the most significant digit, we get a number (which can now start with zero), the sum of whose digits is 4. Represent this as follows - there are 4 balls, between which $n-1$... | 40001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,245 |
3. On a line, there are 16 points $A_{1}, \ldots A_{16}$, spaced $1 \mathrm{~cm}$ apart. Misha constructs circles according to the following rules:
a) The circles do not intersect or touch.
b) Inside each circle, there is at least one of the specified points $\mathrm{A}_{1}, \ldots \mathrm{A}_{16}$.
c) None of these... | Solution: We can represent such a system of circles as a tree with 16 leaves.
In such a tree, there cannot be more than 31 nodes. It is easy to construct a binary tree in which there are exactly 31 nodes. | 31 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,246 |
5. The sequence $a_{n}$ is defined as follows:
$a_{1}=2, a_{n+1}=a_{n}+\frac{2 a_{n}}{n}$, for $n \geq 1$. Find $a_{999}$. | Answer: 999000.
Solution: Consider the beginning of the sequence: $a_{1}=2, a_{2}=6, a_{3}=12, a_{4}=20, \ldots$
We can notice a pattern - the difference between consecutive terms forms an arithmetic progression: $4,6,8, \ldots$.
From this, we get the formula for the $\mathrm{n}$-th term $a_{n}=n \cdot(n+1)$, which ... | 999000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,248 |
2. The robotics club accepts only those who know mathematics, physics, or programming. It is known that 8 members of the club know physics, 7 know mathematics, and 11 know programming. Additionally, it is known that at least two know both physics and mathematics, at least three know both mathematics and programming, an... | Answer 19.
3 Natural numbers, the sum of whose digits is 5, were arranged in ascending order. What number is in the $111-$th place?
Answer: 23000. | 23000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,253 |
5. On a line, there are 15 points $A_{1}, \ldots A_{15}$, spaced 1 cm apart. Petya constructs circles according to the following rules:
a) The circles do not intersect or touch.
b) Inside each circle, there is at least one of the specified points $\mathrm{A}_{1}, \ldots \mathrm{A}_{15}$.
c) None of these points lie o... | Answer: 29.
## Option 2b (Saratov) | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,255 |
1. The digits of numbers from 20 to 99 occupy the first $80 \cdot 2=160$ places in this sequence. There are $2021-160=1861$ places left. The digits of numbers from 100 to 719 occupy the next $(719-99) \cdot 3=1860$ places. Therefore, the 2021st place is the first digit of the number 720, which is 7. | Answer: 7. Answer to option: $1-2: 0$. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,257 |
2. When $b=-1$, the equation is reduced to $|x|-\arcsin x-\arccos x-|x|+1+a=0$, which for $x \in[-1 ; 1]$ is equivalent to $1+a-\frac{\pi}{2}=0$. Thus, when $b=-1$, a solution exists only when $a=\frac{\pi}{2}-1$.
On the other hand, when $a=\frac{\pi}{2}-1$, the equation
$$
|x|-\arcsin x+b \cdot(\arccos x+|x|-1)+\fra... | Answer: $\frac{\pi}{2}-1$.
Answer to option: $1-2: \frac{\pi}{2}-2$. | \frac{\pi}{2}-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,258 |
3. The equation is transformed into the form
$$
t^{\alpha}=\alpha(t+1), \quad \alpha=\lg 2 \in(0,1), \quad t=x^{2}-3>0
$$
where:
1) the left side of the equation is a concave up function (since $\alpha \in(0 ; 1)$), defined for $t \geqslant 0$
2) the right side of the equation is a linear function with a positive sl... | Answer: 4. Answer to option: $1-2: 4$.
保留源文本的换行和格式，直接输出翻译结果。 | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,259 |
4.
$$
\frac{S_{M N C}}{S_{A B C}}=\frac{(p-c)^{2}}{a b}=\frac{(a+b-c)^{2}}{4 a b}=\frac{4(a+b)^{2}}{81 a b}
$$
Since $a+b \geqslant 2 \sqrt{a b}$ (equality is achieved only when $a=b$), then $\frac{16(a+b)^{2}}{81 a b} \geqslant \frac{4 \cdot 4 a b}{81 a b}=\frac{16}{81}$.
Rewrite the ratio of the areas in the foll... | Answer: a) $16 / 81$. b) $[16 / 81 ; 2 / 7)$. Answer to variant: 2-2: a) $64 / 225$. b) $[64 / 225 ; 4 / 11)$. | [16/81;2/7) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,265 |
5. Substitution $2^{x}=t>0$. The equation $t^{4}-6 t^{3}+8 t^{2}+2(1-a) t-(a-1)^{2}=0$ is quadratic relative to $(a-1):(a-1)^{2}+2 t(a-1)-t^{4}+6 t^{3}-8 t^{2}=0$. Since $D / 4=t^{2}+t^{4}-6 t^{3}+8 t^{2}=t^{2}(t-3)^{2}$, then
$$
\left[\begin{array}{l}
a-1=-t+t^{2}-3 t=t^{2}-4 t \\
a-1=-t-t^{2}+3 t=-t^{2}+2 t
\end{arr... | Answer: $(-3 ;-2) \cup(-2 ; 1) \cup(1 ; 2)$.
Answer to variant: 2-2: $(-2 ;-1) \cup(-1 ; 2) \cup(2 ; 3)$.
## Lomonosov Moscow State University
Olympiad "Conquer Sparrow Hills"
Variant 2-1 (Kemerovo) | (-3;-2)\cup(-2;1)\cup(1;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,266 |
1. The inequality is equivalent to
$$
\begin{gathered}
\frac{\sqrt{x^{2}+14 x+40}-2 \sqrt{x^{2}-x-2}}{\left(\sqrt{x^{2}-x-2}-2\right)\left(\sqrt{x^{2}+14 x+40}-4\right)} \leqslant 0 \Leftrightarrow \\
\left\{\begin{array} { l }
{ \frac { x ^ { 2 } + 1 4 x + 4 0 - 4 ( x ^ { 2 } - x - 2 ) } { ( x ^ { 2 } - x - 6 ) ( x ... | Answer: $(-\infty ;-12) \cup(-2 ;-1] \cup[2 ; 3) \cup[8 ;+\infty)$.
Answer to option $6-2: (-\infty ;-13) \cup(-3 ;-2] \cup[1 ; 2) \cup[7 ;+\infty)$. | (-\infty;-12)\cup(-2;-1]\cup[2;3)\cup[8;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,267 |
2. If in the septenary system a number is written as $\overline{a b c}=49 a+7 b+c$, then in the system of base 11 it equals $121 c+11 b+a$. Therefore, $49 a+7 b+c=121 c+11 b+a \Longleftrightarrow b=12 a-30 c=6(2 a-5 c)$. Since $b \in[0 ; 6]$, either $b=6,2 a-5 c=1$ (then $c=1, a=3$), or $b=0,2 a-5 c=0$ (then $c=2, a=5$... | Answer: 190 and 247.
Answer to option 7-2: 248. (This number is $305_{9}$.) | 190247 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,268 |
3. The equation is equivalent to
$$
\begin{gathered}
5(\operatorname{tg} x+\operatorname{ctg} 3 x)=\operatorname{tg} 2 x-\operatorname{tg} x \Leftrightarrow 5 \frac{\cos 2 x}{\cos x \sin 3 x}=\frac{\sin x}{\cos x \cos 2 x} \Leftrightarrow \\
\left\{\begin{array} { l }
{ 1 0 \operatorname { c o s } ^ { 2 } 2 x = \oper... | Answer: $\pm \frac{1}{2} \arccos \left(-\frac{1}{4}\right)+\pi n, \pm \frac{1}{2} \arccos \left(\frac{1}{3}\right)+\pi n, n \in Z$.
Answer to variant $6-2: \pm \frac{1}{2} \arccos \left(-\frac{1}{2}\right)+\pi n, \pm \frac{1}{2} \arccos \left(\frac{1}{3}\right)+\pi n, n \in Z$. | \\frac{1}{2}\arccos(-\frac{1}{4})+\pin,\\frac{1}{2}\arccos(\frac{1}{3})+\pin,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,269 |
4. Since
$$
f(n)+f\left(\frac{1}{n}\right)=-\frac{n^{2}}{1+n^{2}}-\frac{1 / n^{2}}{1+1 / n^{2}}=-\frac{n^{2}+1}{1+n^{2}}=-1
$$
then the original sum is equal to
$$
f(1)+\sum_{k=2}^{2018}\left(f(k)+f\left(\frac{1}{k}\right)\right)=-\frac{1}{1+1}-2017=-2017.5
$$ | Answer: $-2017.5$.
Answer to option: 3-2: 2017.5 . | -2017.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,274 |
2. Let there be $n$ voters, and $k$ votes were cast for a given candidate. Then $0.50332 \leqslant$ $\frac{k}{n} \leqslant 0.50333$, and $1.00664 \leqslant \frac{2 k}{n} \leqslant 1.00666$. If we denote $m=2 k-n$, then $0.00664 \leqslant \frac{m}{n} \leqslant$ 0.00666.
- If $m=1$, then $150.1<\frac{1}{0.00666} \leqsla... | Answer: 451. Answer to the option: $5-2: 341$.
# | 451 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,281 |
# 3. We have
$$
\begin{aligned}
& 42=\frac{1}{\sqrt{x+2}+\sqrt{x+3}}+\frac{1}{\sqrt{x+3}+\sqrt{x+4}}+\ldots+\frac{1}{\sqrt{x+2017}+\sqrt{x+2018}} \equiv \\
& \equiv(\sqrt{x+3}-\sqrt{x+2})+(\sqrt{x+4}-\sqrt{x+3})+\ldots(\sqrt{x+2018}-\sqrt{x+2017}) \equiv \\
& \equiv \sqrt{x+2018}-\sqrt{x+2} \Leftrightarrow \\
& \Leftr... | Answer: $x=7$. Answer to option $5-2: x=6$. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,282 |
5. Let $g(x)=\left(x+\log _{4}|a|\right)\left(x+\log _{|a|} 4\right), h(x)=x^{2}+10 \cdot 2^{a} x+a^{2}-3$ and $d_{h}(a)$ - the quarter of the discriminant of the quadratic trinomial $h(x)$. The inequality holds for all $x$ if and only if:
a) the roots of $g(x)$ coincide, and the roots of $h(x)$ either do not exist or... | Answer: $a=-4 ; a=-2$. Answer to option $5-2: a=4 ; a=2$. | =-4;=-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,284 |
1. The villages "Upper Vasyuki" and "Lower Vasyuki" are located on the bank of a river. A steamboat takes one hour to travel from Upper Vasyuki to Lower Vasyuki, while a motorboat takes 45 minutes. It is known that the speed of the motorboat in still water is twice the speed of the steamboat (also in still water). Dete... | Answer: 90 minutes.
Solution: Let's take the distance between the villages as a unit of length (unit). Then the speed of the steamboat downstream is 1 unit/h, and the speed of the motorboat is $4 / 3$ units/h. Therefore, the own speed of the steamboat is $1 / 3$ units/h, from which the speed of the current is $2 / 3$ ... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,285 |
3. Let $\Sigma(n)$ denote the sum of the digits of the number $n$. Find the smallest three-digit $n$ such that $\Sigma(n)=\Sigma(2 n)=\Sigma(3 n)=\ldots=\Sigma\left(n^{2}\right)$ | Answer: 999
Solution: Let the desired number be $\overline{a b c}$. Note that this number is not less than 101 (since 100 does not work). Therefore, $101 \cdot \overline{a b c}=\overline{a b c 00}+\overline{a b c}$ also has the same sum of digits. But the last digits of this number are obviously $b$ and $c$, so the su... | 999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,287 |
4. Compare the numbers
$$
\left(1+\frac{1}{1755}\right)\left(1+\frac{1}{1756}\right) \ldots\left(1+\frac{1}{2015}\right) \text { and } \sqrt{\frac{8}{7}} .
$$
Indicate in the answer "1" if the first number is greater; "2", if the second number is greater; "0", if the numbers are equal. | Answer: The first is greater
Solution: (Solution: on the left, bring each parenthesis to a common denominator, use the inequality $\frac{n \cdot n}{(n-1)(n+1)}>1$ and we get that the square of the first number is greater than $\frac{2015}{1755}=\frac{31}{27}>\frac{8}{7$.) | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,288 |
5. Two perpendicular chords $A C$ and $B D$ are drawn in a circle. Find the radius of the circle, given that $A B=3, C D=4$. Answer: $\frac{5}{2}$. | Solution: If we reflect $C$ across the perpendicular bisector of $B D$, we obtain a point $C^{\prime}$ on the circle such that $\angle A B C^{\prime}=90^{\circ}$, so $B C^{\prime}=C D=4,2 R=\sqrt{A B^{2}+B C^{\prime 2}}=5$.) | \frac{5}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,289 |
7. Several buses (more than three) at the beginning of the workday sequentially leave from one point to another at constant and equal speeds. Upon arrival at the final point, each of them, without delay, turns around and heads back in the opposite direction. All buses make the same number of trips back and forth, and t... | Answer: 52 or 40.
Solution: If we denote $n-$ as the number of buses and $k-$ as the number of trips, then we can calculate the number of encounters (for example, by a schedule illustrating the movement of buses). It turns out that every two buses meet $2 k-1$ times. In total, we get $n(n-1)(2 k-1)=$ 300, i.e., we nee... | 52or40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,291 |
8. Find all values of the parameter $a$ for which the quadratic trinomial $\frac{1}{3} x^{2}+\left(a+\frac{1}{2}\right) x+\left(a^{2}+a\right)$ has two roots, the sum of the cubes of which is exactly three times their product. In your answer, specify the largest such $a$. | Answer: $a=-1 / 4$.
Solution: By Vieta's theorem, we express the equality $x_{1}^{3}+x_{2}^{3}=3 x_{1} x_{2}$ in terms of $a$, from which a quadratic equation in $a$ is obtained, and two options for $a$ are found, but only one is given in the answer, since the original equation does not have roots for the other value. | -\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,292 |
9. Calculate (without using a calculator) $\sqrt[3]{9+4 \sqrt{5}}+\sqrt[3]{9-4 \sqrt{5}}$, given that this number is an integer. | Answer: 3.
Solution: Note that $2<\sqrt[3]{9+4 \sqrt{5}}<3$ and $0<\sqrt[3]{9-4 \sqrt{5}}<1$. Therefore, $2<\sqrt[3]{9+4 \sqrt{5}}+\sqrt[3]{9-4 \sqrt{5}}<4$, hence this number is 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,293 |
10. Point $C$ divides the diameter $A B$ in the ratio $A C: B C=2: 1$. A point $P$ is chosen on the circle. Determine the values that the ratio $\operatorname{tg} \angle P A C: \operatorname{tg} \angle A P C$ can take. In your answer, specify the smallest such value. | Answer: $\frac{1}{2}$.
Solution: The angle $\angle A P B=90^{\circ}$, as it subtends the diameter. Drop a perpendicular $C H$ to $A P$, it will be parallel to $P B$. Therefore, $\triangle A P B \propto \triangle A H C$ with a ratio of $\frac{2}{3}$. Hence, $\operatorname{tg} \angle A P C=$ $\frac{H C}{P C}=2 \frac{P B... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,294 |
6. Let's call "sawing" the following operation on a polygon (see fig.):
a) Each side of the polygon is divided into three equal parts.
b) The middle part is chosen as the base of an equilateral triangle located outside the polygon.
c) The base is removed and the other two sides are added.
Let $K_{0}$ be a square wi... | # Answer: 28.33.
Solution: On each step, the number of sides increases by a factor of 4. Therefore, on the $(n-1)$-th step, a $4^n$-gon is obtained. At the same time, the length of the side decreases by a factor of 3, so on the $(n-1)$-th step, the side length is $\frac{5}{2} \cdot 3^{1-n}$. Thus, on the $n$-th step, ... | 28.33 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,296 |
1. Determine which of the numbers is greater:
$$
\frac{\lg 2017}{2 \lg 2} \quad \text { or } \quad 2 \lg \frac{2017}{2}
$$ | # Problem 1.
Answer: the first number is smaller (in all cases).
Solution. Let $A$ denote the numbers 2011, 2013, 2015, or 2017, participating in the different variants, and compare the proposed numbers: $\frac{\lg A}{2 \lg 2} \equiv 0.5 \log _{2} A$ and $2 \lg (A / 2)$.
Since $2000<A<2^{11}$, then $0.5 \log _{2} A... | thefirstissmaller | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,297 |
2. Find all pairs of natural numbers $x, y$ that satisfy the equation
$$
6 x^{2} y+2 x^{2}+3 x y+x-9 y=2016
$$ | # Problem 2.
Answer: $x=4, y=20$.
Solution. Factoring the left side of the equation, we get
$$
(3 y+1)(2 x+3)(x-1)=2013
$$
Since $2013=33 \cdot 11 \cdot 61$ and the number $3 y+1$ has a remainder of 1 when divided by 3, then $3 y+1=61 \Longleftrightarrow y=20$. Since $2 x+3>x-1$, then $x-1=3 ; 2 x+3=11$, i.e., $x=4... | 4,20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,298 |
3. The altitudes of an acute-angled triangle $A B C$ intersect at point $O$. Find the angle $A C B$, if $A B: O C=5$. | # Problem 3.
Answer: $\operatorname{arctg} 2$ (in variant II-1); $\operatorname{arctg} 3$ (in variant II-2); $\operatorname{arctg} 4$ (in variant II-3); $\operatorname{arctg} 5$ (in variant II-4).
Solution. Let $k=A B: O C$. If $B B_{1}$ is the height of $\triangle A B C$, then $\triangle A B B_{1} \sim \triangle C O... | \operatorname{arctg}5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,299 |
4. Determine how many roots the equation has on the interval $[-\pi ; \pi]$
$$
\frac{2 \cos 4 x+1}{2 \cos x-\sqrt{3}}=\frac{2 \sin 4 x-\sqrt{3}}{2 \sin x-1}
$$
and specify these roots | # Problem 4.
Answer: 4 roots: $-5 \pi / 6, -\pi / 3, \pi / 2, 2 \pi / 3$.
Solution. Admissible values are determined by the conditions: $\cos x \neq \frac{\sqrt{3}}{2}, \sin x \neq \frac{1}{2}$, and therefore on the interval $[-\pi, \pi]: x \neq \pm \frac{\pi}{6}, x \neq \frac{5 \pi}{6}$.
Rewrite the equation as
$$... | -\frac{5\pi}{6},-\frac{\pi}{3},\frac{\pi}{2},\frac{2\pi}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,300 |
5. Five edges of a tetrahedron have lengths $2, 4, 5, 9$ and 13. Determine whether the length of the sixth edge can:
a) equal 11;
b) equal 11.1. | # Problem 5.
Answer: a) no; b) no.
Solution. Consider any face of the tetrahedron bounded by three known edges. If the length of one of the edges of this face is 2, then by the triangle inequality, the other two edges can only be 4 and 5. Similarly, if the length of one of the edges of this face is 13, then the lengt... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,301 |
4. Igor Gorshkov has all seven books about Harry Potter. In how many ways can Igor arrange these seven volumes on three different bookshelves, so that each shelf has at least one book? (Arrangements that differ by the order of books on the shelf are considered different). | Answer: $C_{6}^{2} \times 7!=75600$.
Solution: First, we can arrange the books in any order, which gives 7! options. Place two dividers into 6 gaps between the books - this can be done in $C_{6}^{2}$ ways. The dividers will divide the books into three parts, which need to be placed on the 1st, 2nd, and 3rd shelves. In... | 75600 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,304 |
5. Find the sum of all natural numbers that have exactly four natural divisors, three of which (of the divisors) are less than 15, and the fourth is not less than 15. | Answer: $\left((2+3+5+7+11+13)^{2}-\left(2^{2}+3^{2}+5^{2}+7^{2}+11^{2}+13^{2}\right)\right) / 2-$ $6-10-14+27=649$.
Solution: The numbers $N$ specified have exactly 4 divisors either if $N=$ $p^{3}$, or if $N=p q$, where $p$ and $q$ are primes. In the first case, only 27 fits. In the second case, we need to consider ... | 649 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,305 |
6. Famous skater Tony Hawk is riding a skateboard (segment $A B$) in a ramp, which is a semicircle with diameter $P Q$. Point $M$ is the midpoint of the skateboard, $C$ is the foot of the perpendicular dropped from point $A$ to the diameter $P Q$. What values can the angle $\angle A C M$ take if it is known that the an... | Answer: $12^{\circ}$.
Solution: Extend the line $A C$ to intersect the circle at point $D$ (see figure). The chord $A D$ is perpendicular to the diameter $P Q$, therefore, it is bisected by it. Thus, $C M$ is the midline of triangle $A B D$, so $C M \| B D$ and, therefore, $\angle A C M=\angle A D B$. The angle $\angl... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,306 |
7. In the equation $x^{2}+p x+q=0$, in one step, it is allowed to increase or decrease one of the coefficients ($p$ or $q$) by 1. Can the equation $x^{2}-2013 x-13=0$ be transformed into the equation $x^{2}+13 x+2013=0$ in some number of steps, such that no intermediate equation has integer roots? | Answer: No.
Solution: The sum of the coefficients of the equation $x^{2}-2013 x-13=0$ is -2025, and the sum of the coefficients of the equation $x^{2}+13 x+2013=0$ is 2027. In one step, the sum of the coefficients changes by 1. Therefore, during the transition from the first equation to the second, this sum will at so... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,307 |
8. In the school Spartakiad, teams from classes $8^{\mathrm{A}}$, $8^{\mathrm{E}}$, and $8^{\mathrm{B}}$ participated. In each of the competitions, one of these teams took 1st place, another took 2nd place, and another took 3rd place. After the Spartakiad, points were tallied: $x$ points were awarded for 1st place, $y$... | Answer: 5 competitions, $8^{\text{B}}$.
Solution: See the solution to problem 7 for 8th grade. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,308 |
9. Plot on the coordinate plane the set of points $(p, q)$ such that the equation $x^{2}+2 p x+q=0$ has two roots, one of which is greater than 2, and the other is less than 0. | Answer: The interior of the angle formed by the lines $q=0$ and $q=-4 p-4$, which lies below the first and to the left of the second line (see figure).
Solution: The given condition is equiva... | q<0q<-4-4p | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,309 |
10. Find the area of the triangle if it is known that the radius of the inscribed circle is 1, and the lengths of all three altitudes are expressed as integers. | Answer: $3 \sqrt{3}$.
Solution: Let $S-$ be the area, $a, b, c-$ be the sides of the triangle, and $h_{a}, h_{b}, h_{c}$ be the corresponding heights. Then $S=\frac{1}{2} a h_{a}=\frac{1}{2} b h_{b}=\frac{1}{2} c h_{c}$, from which we get $a=\frac{2 S}{h_{a}}, b=\frac{2 S}{h_{b}}$ and $c=\frac{2 S}{h_{c}}$. Substituti... | 3\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,310 |
1. Find all values of $x$, for each of which the expressions
$$
\log _{2011}\left(\sqrt{1+\operatorname{ctg}^{2} x}+\operatorname{ctg} x\right) \quad \text { and } \quad \log _{2013}\left(\sqrt{1+\operatorname{ctg}^{2} x}-\operatorname{ctg} x\right)
$$
are equal to each other. | # Problem 1.
Answer: $\pi n, n \in \mathbb{Z}$ (in variants IV-1 and IV-3) and $\frac{\pi}{2}+\pi n, n \in \mathbb{Z}$ (in variants IV-2 and IV-4).
Solution. The product of the arguments of the logarithms is 1, so one logarithm is non-negative while the other is non-positive. Therefore, the logarithms coincide only ... | \pin,n\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,311 |
2. Points $M$ and $N$ divide side $A C$ of triangle $A B C$ into three parts, each of which is equal to 5, such that $A B \perp B M, B C \perp B N$. Find the area of triangle $A B C$. | # Problem 2.
Answer: $3 \sqrt{3}$ (Option IV-1), $27 \sqrt{3} / 4$ (Option IV-2), $12 \sqrt{3}$ (Option IV-3), $75 \sqrt{3} / 4$ (Option IV-4).
Solution. Points $M$ and $N$ are located on side $A C$ in the following order: $A, N$, $M, C$ (otherwise, angle $B$ would be greater than $\left.180^{\circ}\right)$.
Let $A ... | \frac{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,312 |
3. Solve the inequality
$$
4 x+2+\sqrt{4-x}>x^{2}+\sqrt{x^{2}-5 x+2}
$$ | # Problem 3.
Answer: $\left(2-\sqrt{6}, \frac{5-\sqrt{17}}{2}\right]$.
Solution. Rewrite the inequality as
$$
\sqrt{4-x}-\sqrt{x^{2}-5 x+2}>x^{2}-4 x-2
$$
then multiply and divide the left part by its conjugate expression:
$$
\begin{aligned}
& \frac{-\left(x^{2}-4 x-2\right)}{\sqrt{4-x}+\sqrt{x^{2}-5 x+2}}>x^{2}-4... | (2-\sqrt{6},\frac{5-\sqrt{17}}{2}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,313 |
4. Find the set of values of the expression $x-y+1$ under the condition
$$
(x-y)^{2}=2|2 y-x|+x+15
$$ | # Problem 4.
Answer: $(-\infty ;-2] \cup[6 ;+\infty)$.
Solution. Let's perform a change of variables: $x-y+1=a, 2 y-x=t$. Then the condition of the problem can be reformulated as follows: "Find the set of values of $a$ under the condition $2|t|+t=a^{2}-4 a-12$."
On the plane of variables $(a, t)$, this condition def... | (-\infty;-2]\cup[6;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,314 |
5. A right-angled triangle is called Pythagorean if the lengths of all its sides are natural numbers. Find the greatest integer that divides the product of the lengths of the sides of any Pythagorean triangle. | # Problem 5.
Answer: 60.
Solution. Since a triangle with sides $3,4,5$ is a right triangle, the desired number cannot be greater than 60. We will show that the product of the lengths of the sides of any Pythagorean triangle is divisible by 60, i.e., by 3, by 4, and by 5.
The remainders when the square of an integer ... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,315 |
7. Given three numbers $a, b, c$. It is known that the arithmetic mean of the numbers $a$ and $b$ is 5 more than the arithmetic mean of all three numbers. And the arithmetic mean of the numbers $a$ and $c$ is 8 less than the arithmetic mean of all three numbers. By how much does the arithmetic mean of the numbers $b$ a... | Answer: 3 more.
Solution: The arithmetic mean of the numbers $\frac{a+b}{2}, \frac{b+c}{2}, \frac{a+c}{2}$, obviously, coincides with the arithmetic mean of the numbers $a, b, c$. Therefore, if one $\frac{a+b}{2}$ is 5 more, and $\frac{a+c}{2}$ is 8 less, then, to get the same result, $\frac{b+c}{2}$ must be 3 more. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,316 |
8. Let's call a number "remarkable" if it has exactly 4 distinct natural divisors, and among them, there are two such that neither is a multiple of the other. How many "remarkable" two-digit numbers exist?
# | # Answer 36.
Solution: Such numbers must have the form $p_{1} \cdot p_{2}$, where $p_{1}, p_{2}$ are prime numbers. Note that the smaller of these prime numbers cannot be greater than 7, because otherwise the product will be at least 121. It is sufficient to check $p_{1}=2,3,5,7$. For 2, we get the second factor: 3, 5... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,317 |
9. Assemble a square of the smallest area from squares of size $1 \times 1$, $2 \times 2$, and $3 \times 3$, such that the number of squares of each size is the same. | Answer:
Solution. Let $n$ be the number of squares of each type. Then $n + 4n + 9n = 14n$ must be a perfect square. The smallest $n$ for which this is possible is 14. The figure shows an ex... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,318 |
10. The company conducted a survey among its employees - which social networks they use: VKontakte or Odnoklassniki. Some employees said they use VKontakte, some - Odnoklassniki, some said they use both social networks, and 40 employees said they do not use social networks. Among all those who use social networks, 75% ... | Answer: 540
Solution: Since 75% of social media users use VKontakte, it follows that only 25% use Odnoklassniki. Additionally, 65% use both networks, so the total percentage of users of Odnoklassniki is $65 + 25 = 90\%$ of social media users. These $90\%$ constitute $5 / 6$ of the company's employees, so $100\%$ const... | 540 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,319 |
11. In a regular 2017-gon, all diagonals are drawn. Petya randomly selects some $\mathrm{N}$ diagonals. What is the smallest $N$ such that among the selected diagonals, there are guaranteed to be two of the same length? | Answer: 1008.
Solution: Let's choose an arbitrary vertex and consider all the diagonals emanating from it. There are 2014 of them, and by length, they are divided into 1007 pairs. Clearly, by rotating the polygon, any of its diagonals can be aligned with one of these. Therefore, there are only 1007 different sizes of ... | 1008 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,320 |
12. In how many ways can the number 10000 be factored into three natural factors, none of which is divisible by 10? Factorizations that differ only in the order of the factors are considered the same.
# | # Answer: 6.
Solution: Each of the factors should only include powers of 2 and 5 (they cannot be included simultaneously, as it would be a multiple of 10). There can be two factors that are powers of two, in which case the third factor is 625. Or conversely, two factors are powers of five, and the third factor is 16. ... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,321 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.