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 |
|---|---|---|---|---|---|---|---|---|---|
11.2. Solve the equation $\arccos \frac{x+1}{2}=2 \operatorname{arctg} x$. | Answer: $x=\sqrt{2}-1$.
Solution. Taking the cosine of both sides and using the formula $\cos 2 \alpha=\frac{1-\operatorname{tg}^{2} \alpha}{1+\operatorname{tg}^{2} \alpha}$, we get $\frac{x+1}{2}=\frac{1-x^{2}}{1+x^{2}}$. The roots of this equation are $x_{1}=-1 ; x_{2,3}= \pm \sqrt{2}-1$. Checking for $x=-1$ and for... | \sqrt{2}-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,227 |
11.3. On the coordinate plane, a parabola - the graph of a reduced quadratic trinomial with integer coefficients - was drawn. It touches the $O x$ axis. Prove that on this parabola, there is a point with integer coordinates ( $a, b$ ), such that the graph $y=x^{2}+a x+b$ also touches the $O x$ axis. | Solution. Let $x^{2}+p x+q$ be the given quadratic polynomial. Then $p^{2}-4 q=0$ by the condition of the graph touching the $O x$ axis (this condition is equivalent to the discriminant of the quadratic being zero). Note that the point with coordinates $(-p, q)$ lies on the graph of the quadratic, since $(-p)^{2}+p(-p)... | proof | Algebra | proof | Yes | Yes | olympiads | false | 22,228 |
11.4. In a $25 \times 25$ grid, some cells are marked with a plus sign. Prove that there exist two (possibly overlapping) $3 \times 3$ squares with the same arrangement of plus signs (i.e., the plus signs should coincide under a parallel shift). | Solution. First, let's count the number of $3 \times 3$ squares in a $25 \times 25$ grid. Each such square is uniquely determined by the position of its lower left vertex. This vertex can be any node in the part of the grid that remains if we cut off a 3-cell border from the top and right of the $25 \times 25$ square: ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,229 |
11.1. Solve the inequality $\left|x^{2}-x\right| \leq 2 x^{2}-3 x+1$. | Answer: $x \leq 1 / 3 ; x \geq 1$.
Solution. If $x^{2}-x>0$, i.e., when $x>1$ or $x<0$, the original inequality holds. If $x^{2}-x \leq 0$, i.e., when $0 \leq x \leq 1$, the inequality can be written as: $3 x^{2}-4 x+1 \geq 0 \Leftrightarrow(x-1 / 3)(x-1)$ $\geq 0 \Leftrightarrow x \leq 1 / 3$ or $x \geq 1$. Consideri... | x\leq1/3;x\geq1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 22,230 |
11.2. Solve the equation $2 \cos (\pi x)=x+\frac{1}{x}$. | Answer: $x=-1$.
Solution. The left side of the equation is no more than two in absolute value, while the right side is no less than two in absolute value: for the right side, this follows from elementary inequalities, for example (for $x>0$), - from the inequality between the arithmetic mean and the geometric mean, an... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,231 |
11.3. On the hypotenuse $AB$ of a right triangle $ABC$, a square is constructed outward with center $O$. Prove that the ratio of the length $CO$ to the sum of the legs $AC + CB$ is a constant for all right triangles and find this ratio. | Answer: $\sqrt{2} / 2$.
Solution. Let $A B=c, \angle A=\alpha$. Then, by the cosine theorem in triangle $A C O$, we have $\mathrm{CO}^{2}=c^{2} \cos ^{2} \alpha+c^{2} / 2-\sqrt{2} c^{2} \cos \alpha \cos \left(\alpha+45^{\circ}\right)=c^{2} / 2+c^{2} \cos \alpha \sin \alpha=c^{2}(1+2 \sin \alpha \cos \alpha) / 2$.
Fro... | \frac{\sqrt{2}}{2} | Geometry | proof | Yes | Yes | olympiads | false | 22,232 |
11.4. Prove that the number $2^{2022}+1$ a) is divisible by 65; b) can be represented as the product of four natural numbers, different from 1. | Solution. a) Let's denote the given number by N. The prime factorization of 2022 gives $2022 = 2 \cdot 3 \cdot 337$. Therefore, $N = \left(4^{3}\right)^{337} + 1$. As is known, the sum of odd powers of two numbers can be factored: $a^{2n+1} + b^{2n+1} = (a + b)\left(a^{2n} - a^{2n-1}b + \ldots - ab^{2n-1} + b^{2n}\righ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,233 |
7.1. Given a square and a rectangle. The longer side of the rectangle is $11\%$ greater than the side of the square, and the shorter side is $10\%$ less than the side of the square. a) Is the area of the rectangle greater or smaller compared to the area of the square, and by what percentage? b) Answer the same question... | Answer: a) the area of the rectangle is less by $0.1 \%$, b) the perimeter of the rectangle is greater by $0.5 \%$. Solution. Let $a$ be the side of the square, then the sides of the rectangle are $1.11 a$ and $0.9 a$. The area and perimeter of the rectangle are respectively $1.11 \cdot 0.9 a^{2}=0.999 a^{2}$ and $2(1.... | )0.1,b)0.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,234 |
7.2. Agent 007 wants to encrypt his number using two natural numbers $m$ and $n$ such that $0.07=\frac{1}{m}+\frac{1}{n} \cdot$ Can he do this? | Answer: it will. The solution follows from the equality: $0.07=0.05+0.02=\frac{1}{20}+\frac{1}{50}$. | 0.07=\frac{1}{20}+\frac{1}{50} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,235 |
7.3. On an island where knights and liars live, several (more than two) people gathered around a round table. It is known that there were both knights and liars at the table, and each said such a phrase: "Only one of my two neighbors is a knight." Who is more at the table: knights or liars, and by how many times? (Knig... | Answer: There are twice as many knights. Solution. Consider an arbitrary knight (by the condition of the problem, there are knights at the table). From his truthful words, it follows that a knight sits on one side of him, and a liar on the other. Without loss of generality, we can assume that the knight is sitting to h... | There\\twice\\many\knights | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,236 |
7.4. On the board, there were 10 numbers. In one operation, it is allowed to erase any two numbers $a, b$ from the board and write $a+2b$ and $b+2a$ instead. Can it happen that after several operations all the numbers on the board become the same, if initially there were a) the numbers 1, $2, \ldots, 10;$ b) any 10 dif... | Answer. a) It cannot; b) it can. Solution. a) Note that with any operation, the parity of the numbers does not change (since $a+2b$ has the same parity as $a$, and similarly, $b+2a$ has the same parity as $b$). Initially, there were 5 even and 5 odd numbers, so at the end, there should also be 5 even and 5 odd numbers,... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,237 |
7.5. a) Can a $7 \times 7$ checkerboard with the central cell removed be cut into dominoes (rectangles $2 \times 1$) such that the number of horizontal and vertical dominoes is the same? b) The same question for a $7 \times 7$ square with a corner cell removed. | Answer. a) It is possible; b) it is not possible. Solution. a) See the example of cutting in the figure. b) Let's color the square in two colors as shown in the second figure. We will get 27 black cells and 21 white cells. Note that any horizontal domino occupies one black and one white cell, while a vertical domino oc... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,238 |
9.1. Inside the rectangle $A B C D$, a point $M$ is marked, and on the segments $A M, B M, C M$ and $D M$, four circles are constructed with these segments as diameters. Let $S_{A}, S_{B}, S_{C}, S_{D}$ be the areas of these circles, respectively. Prove that $S_{A}+S_{C}=S_{B}+S_{D}$. | Solution. Let $A B=a, A D=b$ and $x, y$ be the distances from point $M$ to sides $A B$ and $A D$ respectively. Then, by the Pythagorean theorem and the formula for the area of a circle, we have $S_{A}=\pi \frac{x^{2}+y^{2}}{4}$, $S_{B}=\pi \frac{x^{2}+(a-y)^{2}}{4}, S_{C}=\pi \frac{(b-x)^{2}+(a-y)^{2}}{4}, S_{D}=\pi \f... | S_{A}+S_{C}=S_{B}+S_{D} | Geometry | proof | Yes | Yes | olympiads | false | 22,239 |
9.4. How many three-digit natural numbers $n$ exist for which the number $n^{3}-n^{2}$ is a perfect square | Answer: 22. Solution. Let $n^{3}-n^{2}=m^{2}$ for some natural number $m$. Then $n^{2}(n-1)=m^{2}$, and therefore $n-1$ must also be a perfect square: $n-1=a^{2}$. Thus, $n=a^{2}+1$ and $m=\left(a^{2}+1\right) a$. Therefore, we need to find all three-digit numbers $n$ that are one more than perfect squares. Such number... | 22 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,242 |
9.1. a) Prove that for any number $d>0$ there exists a right triangle in which the larger leg differs in length from both the smaller leg and the hypotenuse by $d$. b) Prove that the radius of the inscribed circle of such a triangle is $d$. | Solution. a) Let $a<b$ be the lengths of the legs, $c$ be the length of the hypotenuse. Then the equalities $a=b-d, c=b+d$ must hold, and by the Pythagorean theorem $(b+d)^{2}=(b-d)^{2}+b^{2} \Leftrightarrow b^{2}=4 b d \Leftrightarrow b=4 d$ and thus, $a=3 d, b=5 d$. An obvious check (the converse of the Pythagorean t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,243 |
9.3. Given a right triangle, the height dropped to the hypotenuse is 4 times smaller than the hypotenuse. Find the acute angles of this triangle. | Answer: $15^{\circ}$ and $75^{\circ}$. Solution. See problem 8.4. | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,245 |
9.4. Does there exist a natural number $n$ such that the number $n^{2}+6 n+2019$ is divisible by 100? | Answer: does not exist. Solution. Transform the expression $n^{2}+6 n+2019=(n+3)^{2}+2010$. If this sum is divisible by 100 (and thus by 10), then the number $(n+3)^{2}$ must be divisible by 10, and therefore it is divisible by 100 (in the prime factorization of a perfect square, all exponents are even numbers). But 20... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,246 |
8.1. In a four-digit number, Petya erased the first digit and obtained a three-digit number. Then he divided the original number by this three-digit number and got a quotient of 3, with a remainder of 8. What is the original number? (Find all possible numbers). | Answer: 1496 or 2996. Solution. See problem 7.1. | 1496or2996 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,248 |
8.2. In triangle $A B C$, angle $A$ is the largest. Points $M$ and $N$ are symmetric to vertex $A$ with respect to the angle bisectors of angles $B$ and $C$ respectively. Find $\angle A$, if $\angle M A N=50^{\circ}$. | Answer: $80^{\circ}$. Solution. Let $\angle A=\alpha, \angle B=\beta, \angle C=\gamma$. Then $\angle A M B=90^{\circ}-\frac{\beta}{2}, \angle A N C=90^{\circ}-\frac{\gamma}{2}$ (since triangles $A M B$ and $\quad$ are isosceles). Therefore, $\angle M A N=180^{\circ}-\left(90^{\circ}-\frac{\beta}{2}\right)-\left(90^{\ci... | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,249 |
8.5. 25 students in a class, among whom $n$ are boys, are sitting around a large round table. Is it necessarily true that there will be two boys between whom (clockwise) sit exactly 4 people, if a) $n=10$; b) $n=11$? | Answer: a) not necessarily; b) necessarily. Solution. a) Let's construct an example. Number the seats around the table clockwise: $1,2, \ldots, 25$. If 10 boys sit in seats $1,2,3,4,5$ (first group) and $11,12,13,14,15$ (second group), then clockwise between boys of the same group, there are no more than three people, ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,252 |
7.1. Given the number $N=2011 \cdot 2012 \cdot 2013 \cdot 2014+1$. Is this number prime or composite? | Answer. Composite. Solution. The last digit of the number $N$ is equal to the last digit of the number $1 \cdot 2 \cdot 3 \cdot 4+1$, i.e., it is equal to 5. Therefore, $N$ is divisible by 5. (For another, more general solution, see problem 8.2.) | Composite | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,253 |
7.2. The bisector of angle $ABC$ forms an angle with its sides that is three times smaller than the adjacent angle to $ABC$. Find the measure of angle $ABC$. | Answer. $\quad 72^{\circ}$. Solution. Let $x$ be the degree measure of angle $ABC$. From the condition of the problem, we get the equation $\frac{x}{2}=\frac{180-x}{3} \Leftrightarrow 5 x=360 \Leftrightarrow x=72$ (degrees). | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,254 |
7.3. A natural number is called curious if, after subtracting the sum of its digits from it, the result is a number consisting of identical digits. How many three-digit curious numbers exist? | Answer: 30 numbers. Solution: Let $\overline{x y z}$ be a curious three-digit number. Then the number
$$
A=\overline{x y z}-(x+y+z)=100 x+10 y+z-(x+y+z)=9(11 x+y)
$$
is divisible by 9 and consists of identical digits, and $100-27 \leq A \leq 999-1$. Thus, $A$ can be either 99, 333, or 666.
In the case of $A=99$, we ... | 30 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,255 |
7.4. We have a set consisting of $n$ weights weighing $1,2, \ldots, n$ (grams). Can all the weights be divided into two piles of equal weight if: a) $n=30$, b) $n=31$. | Answer. a) No; b) Yes. Solution. a) The total weight of all weights is odd, since the number of weights with an odd weight is an odd number (these are 15 weights with weights $1,3,5, \ldots, 29$). Therefore, it is impossible to divide the weights into two piles of equal weight.
b) First, place the weight 31 in the lef... | )No;b)Yes | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,256 |
7.5. In the company, 11 people gathered. It turned out that each person is friends with at least six of those present. Prove that in this company, there will be three friends (each is friends with the other two). | Solution. Let's take any two friends $A$ and $B$. From the remaining nine people, $A$ has at least 5 friends, and $B$ has at least 5 friends. Therefore, among the friends of $A$ and $B$, there is at least one common friend (otherwise, it would be $5+5 \leq 9$). Together with $A$ and $B$, this common friend forms the re... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,257 |
11.1. Given the sequence $a_{n}=(-1)^{1+2+\ldots+n}$. Find $a_{1}+a_{2}+\ldots+a_{2017}$. | Answer: -1. Solution. We have $a_{n}=(-1)^{\text {sn }}$, where $S_{n}=\frac{n(n+1)}{2}$. It is easy to notice and prove that the parity of the number $S_{n}$ repeats with a period of 4: indeed, $S_{n+4}-S_{n}=\frac{(n+4)(n+5)-n(n+1)}{2}=\frac{8 n+20}{2}=4 n+10$, i.e., an even number. Therefore, $a_{1}+a_{2}+\ldots+a_{... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,258 |
11.2. Plot on the coordinate plane the set of points satisfying the inequality $\arcsin x+\arcsin y>\frac{\pi}{2}$.
 | Solution. Note that the given set is in the first quadrant, since if $x\frac{\pi}{2}-\arcsin x>\frac{\pi}{2}$. Therefore, we consider $x, y \in[0,1]$. We have $\arcsin y>\frac{\pi}{2}-\arcsin x \Leftrightarrow \sin (\arcsin y) \geq \sin \left(\frac{\pi}{2}-\arcsin x\right) \quad$ (the inequalities are equivalent, since... | y^{2}>1-x^{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 22,259 |
11.3. Given a trapezoid $A B C D(B C \| A D)$, in which a circle with center $O$ is inscribed. The lines $B O$ and $C O$ intersect the lower base $A D$ at points $M$ and $N$ respectively. Prove the relationship for the areas $S_{\text {AON }}+S_{\text {DOM }}+2 S_{\text {NOM }}=\frac{1}{2} S_{\text {ABCD }}$. | Solution. Note that triangles $B O C$ and $M O N$ are equal ($B O=O M, C O=O N$, since $O$ lies on the midline of the trapezoid, and $\angle B O C=\angle M O N$ - vertical angles). Therefore, $S_{A O N}+2 S_{N O M}+S_{M O D}=S_{A O D}+S_{\text {BOC }}=\frac{1}{2} S_{A B C D}$ (just as in problem 10.3). Another way foll... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,260 |
11.5. For a certain quadratic trinomial, the following information is known: its leading coefficient is one, it has integer roots, and its graph (parabola) intersects the line $y=2017$ at two points with integer coordinates. Can the ordinate of the vertex of the parabola be uniquely determined from this information? | Answer: It is possible. Solution. Let $n_{1}, n_{2}\left(n_{1}<n_{2}\right)$ be the roots of the given quadratic trinomial $f(x)$. Then $f(x)=\left(x-n_{1}\right)\left(x-n_{2}\right)$. From the condition about the integer roots of the equation $\left(x-n_{1}\right)\left(x-n_{2}\right)=2017$, we obtain (due to the simpl... | -1008^2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,262 |
11.6 Find $\sin \left(\frac{\pi}{2}\left(2 x^{2}+1\right)\right)+\cos \left(\pi\left(x^{2}+4 x\right)\right)=0$. | Answer: $-1+\sqrt{\frac{3}{2}}$.
Solution. Writing the first term as $\cos \pi x^{2}$ and transforming the sum of cosines into a product, we obtain two series of roots $4 x=2 n+1$ and $2 x^{2}+4 x=2 m+1$ (where $m, n$ are integers). The positive roots are of the form $x=\frac{n}{2}+\frac{1}{4}$ and $x=-1+\sqrt{m+\frac... | -1+\sqrt{\frac{3}{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,263 |
11.7 Find all values of the parameter \(a\), for each of which the system of equations
\[
\left\{
\begin{array}{l}
x^{2}+y^{2}=2 a, \\
x+\log _{2}\left(y^{2}+1\right)=a
\end{array}
\right.
\]
has a unique solution. | Answer: $a=0$.
Solution. Note that if $(x, y)$ is a solution to the system, then $(x, -y)$ is also a solution to this system. Therefore, if for a parameter $a$ the system has a unique soluti... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,264 |
11.8 Find the area of the figure defined on the coordinate plane by the inequality $x^{2}+y^{2} \leq 2(|x|-|y|)$. | Answer: $2 \pi-4$.
Solution. It is obvious that the figure
is symmetric with respect to the coordinate axes and the origin (since the inequality does not change when the signs of $x, y$ are ... | 2\pi-4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,265 |
11.9 Does there exist a 2010-gon with side lengths 1, 2, .., 2010 (in some order) into which a circle can be inscribed? | Answer: Does not exist.
Solution. If a circle can be inscribed in a polygon with an even number of sides, then the sum of the lengths of the sides with odd numbers (counting counterclockwise from a fixed vertex) is equal to the sum of the lengths of the sides with even numbers. This fact is proven in the same way as t... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,266 |
11.10 a) Find the reduced quadratic trinomial $P(x)$, for which the graph $\mathrm{y}=P(x)$ is symmetric with respect to the Oy axis and is tangent to the line $y=x$.
b) Let $P(x)$ be a quadratic trinomial with a positive leading coefficient. Prove that if the equation $P(x)=x$ has a unique solution, then the equation... | Answer: a) $P(x)=x^{2}+1 / 4$.
Solution. a) The condition of symmetry with respect to the Oy axis means that $P(x)$ has the form $P(x)=$ $x^{2}+a$. The condition of tangency to the line $y=x$ means that the quadratic equation $x^{2}+a=x$ has a unique solution, i.e., its discriminant $1-4 a$ is equal to zero.
b) Since... | P(x)=x^{2}+1/4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,267 |
7.1. Do there exist such natural numbers $m, n$ that $m n(m+n)=2020$? | Answer: They do not exist. Solution: Factorize 2020 on the right side of the equation into prime factors: $2020=2 \cdot 2 \cdot 5 \cdot 101$. Since the number 101 is prime, one of the three factors on the left side ( $m, n$ or $m+n$ ) must be divisible by 101. If 101 divides $m$ or $n$, then $m \geq 101$ or, respective... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,268 |
7.2. Kolya wants to represent the fraction $\frac{3}{7}$ in decimal form, writing down 0 whole numbers and 1000 digits after the decimal point. Then he plans to erase the 500th digit after the decimal point. Will the resulting number be greater or less than $\frac{3}{7}$? | Answer. More. Solution. To answer the question of the problem, it is necessary to find out which is greater: the 500th or the 501st digit after the decimal point? The fraction $\frac{3}{7}$ in decimal form is periodic with a period of 428571 of length 6, that is, $\frac{3}{7}=0.428571428571 \ldots$. Since 500 gives a r... | More | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,269 |
7.3. In the 7th grade, there are 25 people, and each attends a dance club or a drama club (some attend both clubs). Everyone took a math test, and the teacher calculated the percentage of students who received a failing grade (2) separately among the "dancers" and among the "actors." It turned out that the percentage i... | Answer. It can. Solution. Consider the following example. Suppose there are 10 "dancers" and 20 "actors" in the class, then the number of "dancing actors" will be $10+20-25=5$ people, and the number of "pure dancers" will be $10-5=5$ and the number of "pure actors" will be $20-5=15$. Suppose that three "dancers" and si... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,270 |
7.4. Given two equal-legged acute triangles. It is known that the first triangle has an angle equal to some angle of the second triangle, and a side equal to some side of the second triangle. Can we assert that the triangles are equal? | Answer. No. Solution. See the example of unequal triangles $ABC$ and $ACD$ in the figure, which satisfy the conditions of the problem.
 | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,271 |
7.5. Petya has 4 Soviet copper coins - one each of 1, 2, 3, and 5 kopecks. He learned from the internet the following fact: these coins should weigh exactly as many grams as their denomination. Petya wants to check this fact using a balance scale. Will he be able to do this if he has only one 9-gram weight? | Answer: It will. Solution: Let $x$ be the weight of a 1-kopeck coin, $y$ be the weight of a 2-kopeck coin. If the fact being checked is true, then $(x+y)$ is the weight of a 3-kopeck coin, and ($x+2y$) is the weight of a 5-kopeck coin. To verify this, we will perform two weighings: 1) 1 kopeck + 2 kopecks = 3 kopecks (... | 1,2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,272 |
8.2. Find all prime numbers $p$ for which $8 p+1$ is: a) a perfect square? b) a perfect cube | Answer a) $p=3$; b) such $p$ does not exist. Solution. a) The equation $8 p+1=n^{2}$ is not satisfied by $p=2$, and thus $p$ is odd. Write the equation in the form $(n-1)(n+1)=8 p$. The number on the right side has the following natural divisors: $1,2,4,8, p, 2 p, 4 p, 8 p$. The factors on the left side differ by 2 and... | )p=3;b)suchpdoesnotexist | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,274 |
8.3. $n$ chips numbered $1,2, \ldots, n$ are arranged in a row in ascending order. In one move, it is allowed to swap any two chips that have exactly two chips between them. Does there exist an $n$ for which it is possible to rearrange all the chips in reverse order after several moves? | Answer. Does not exist. Solution. Suppose, for the sake of contradiction, that for some $n$ it was possible to rearrange the chips in reverse order. Assume that the chips are located on a coordinate line at points with coordinates $1, 2, ..., n$. With any move, the coordinates of the two chips that swap places change b... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,275 |
8.4. On the sides $C D$ and $A D$ of the square $A B C D$, points $M$ and $N$ are marked, respectively. It turns out that $C M + A N = B N. \quad$ Prove that $\angle C B M = \angle M B N$ | Solution. Rotate triangle $B C M$ around point $B$ by $90^{\circ}$ clockwise (see fig.). Then point $C$ will move to point $A$, and point $M$ will move to point $M^{\prime}$ on line $A D$. From the condition $C M + A N = B N$, it follows that $M^{\prime} N = B N$, i.e., triangle $M^{\prime} N B$ is isosceles and the an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,276 |
8.2. Replace the two asterisks with two different numbers so that the identity equality is obtained:
$$
(3 x-*)(2 x+5)-x=6 x^{2}+2(5 x-*)
$$ | Answer: 2 and 5. Solution. Let $A$ and $B$ be the numbers corresponding to the first and second asterisks. Then, multiplying the brackets according to algebraic rules and equating the coefficients of $x$ and the constant terms, we get: $14-2 A=10, -5 A=-2 B$. From this, $A=2, B=5$. | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,279 |
8.3. Given a convex quadrilateral $A B C D$ and a point $M$ inside it, not lying on the diagonals. Prove that at least one of the angles $\angle A M C$ or $\angle B M D$ is obtuse. | Solution. Let $O$ be the point of intersection of the diagonals, and let, for definiteness, point $M$ lie inside triangle $B O C$. Then
$$
\begin{aligned}
& \angle B M D + \angle A M C = \angle B M A + \angle A M D + \angle A M D + \angle D M C > \\
& > \angle B M A + \angle A M D + \angle D M C = 360^{\circ} - \angle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,280 |
8.2. Can a cube be divided into 2020 smaller cubes? | Answer: It is possible. Solution. Let's say we have a unit cube. We can construct its partition, for example, based on the equality $2020=999+999+6+8+8$, specifically as follows. First, we divide the cube into 1000 identical smaller cubes with edge length $\frac{1}{10}$, then we divide one of these smaller cubes into 1... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,283 |
8.2. Prove that for all natural $\mathrm{n}>1$ the number $n^{2020}+4$ is composite. | The solution follows from the equalities $n^{2020}+4=n^{2020}+4+4 n^{1010}-4 n^{1010}=\left(n^{1010}+2\right)^{2}-\left(2 n^{505}\right)^{2}=$
$$
=\left(n^{1010}+2+2 n^{505}\right)\left(n^{1010}+2-2 n^{505}\right) . \text { Clearly, for } n>1 \text { both brackets are }>1 .
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,286 |
8.3. Find a six-digit number that, when multiplied by 9, is written with the same digits as the original number but in reverse order? How many such six-digit numbers are there? | Answer. 109989 is the only number. Solution See problem 7.3. | 109989 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,287 |
8.4. a) Prove that in any convex quadrilateral, there are two sides that are shorter in length than the longest diagonal. b) Can there be exactly two such sides? | Answer: b) Yes, it can. Solution a) Of the four angles of the quadrilateral $ABCD$, at least one is not acute (since the angles sum up to $360^{\circ}$). Let this be angle $A$. Then in $\triangle ABD$ we have $BD > AB$ and $BD > AD$ (i.e., $AB$ and $AD$ are smaller than the diagonal $BD$, and thus smaller than the larg... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,288 |
3. Find the set of values and the smallest period of the function $y=\cos ^{2} 3 x+\sin 6 x$. | Answer. The set of values $\left[\frac{1}{2}-\frac{\sqrt{5}}{2}, \frac{1}{2}+\frac{\sqrt{5}}{2}\right]$, the smallest period $\pi / 3$. Hint. Transform the function $y$ to the form $y=\frac{1+\cos 6 x}{2}+\sin 6 x=\frac{1}{2}+\frac{\sqrt{5}}{2} \sin (6 x+\alpha)$, where $\alpha-$ is an auxiliary angle ( $\sin \alpha=\s... | [\frac{1}{2}-\frac{\sqrt{5}}{2},\frac{1}{2}+\frac{\sqrt{5}}{2}],\frac{\pi}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,290 |
4. Given a triangle $A B C$, where side $B C$ is twice the length of side $A B$, and a bisector $B M$ is drawn. a) Find the ratio $R_{1} / R_{2}$, where $R_{1}, R_{2}$ are the radii of the circumcircles of triangles $A B M$ and $C B M$ respectively. b) Prove that $3 / 4 < r_{1} / r_{2} < 1$, where $r_{1}, r_{2}$ are th... | Answer. a) $R_{1} / R_{2}=1 / 2$. Hint. a) Let $\angle B M A=\alpha$, then $\angle B M C=\pi-\alpha$. Then in triangles $A B M$ and $C B M$ we have the relations $A B=2 R_{1} \sin \alpha, B C=2 R_{2} \sin (\pi-\alpha)=2 R_{2} \sin \alpha$. Hence $R_{1} / R_{2}=A B / B C=1 / 2$. (Note that in this solution, the fact tha... | R_{1}/R_{2}=1/2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,291 |
11.1. Solve the equation $\arccos \frac{x+1}{2}=2 \operatorname{arctg} x$. | Answer: $x=\sqrt{2}-1$. Hint. Taking the cosine of both sides and using the formula $\cos 2 \alpha=\frac{1-\operatorname{tg}^{2} \alpha}{1+\operatorname{tg}^{2} \alpha}$. As a result, we get $\frac{x+1}{2}=\frac{1-x^{2}}{1+x^{2}}$. The roots of this equation are: $x_{1}=-1 ; x_{2,3}= \pm \sqrt{2}-1$. By determining the... | \sqrt{2}-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,293 |
8.1. The older brother's journey to school takes 12 minutes, while the younger brother's journey (along the same route) takes 20 minutes. How many minutes will pass from the moment the younger brother leaves home until he is caught up by the older brother, if the older brother left 5 minutes after the younger brother? | Answer: In 12.5 minutes.
Solution: Let $S$ be the distance from home to school. Since the younger brother covers this distance in 20 minutes, in 5 minutes he will cover the distance $\frac{5 S}{20}=\frac{S}{4}$. After the older brother leaves the house, the rate at which the distance between them decreases will be $\f... | 12.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,296 |
8.2. In the decimal representation of a natural number $N$, a 0 was inserted between the second and third digits from the right, resulting in $9N$. What can $N$ be? (list all possible values). | Answer: 225; 450; 675.
Solution. Let $x$ be the third digit from the right of the number $N$. The two-digit number formed by the last two digits of the number $N$ will be denoted by $B$, and the number obtained from $N$ by erasing its last three digits will be denoted by $A$. Thus, $N=1000 A+100 x+B$. The new number, ... | 225;450;675 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,297 |
8.3. Given a triangle $A B C$. On the side $A C$, the largest in the triangle, points $M$ and $N$ are marked such that $A M=A B$ and $C N=C B$. It turns out that angle $N B M$ is three times smaller than angle $A B C$. Find $\angle A B C$. | Answer: $108^{\circ}$.
Solution. Point $N$ lies between $A$ and $M$, since $A M+C V=A B+B C > AC$. Let $\angle N B M = x$, then $\angle A B M + \angle N B C = \angle A B C + \angle N B M = 3... | 108 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,298 |
8.4. In the gym, 10 boys and 10 girls stand in a circle in a random order. Will the coach be able to draw 10 non-intersecting segments on the floor with chalk so that each of them connects a boy and a girl (regardless of the arrangement of the children)? | Answer: It is possible.
Solution: Let's consider the following model of the problem: on a circle, there are 10 points marked with the letter m (boys) and 10 points marked with the letter d (girls). We need to draw 10 non-intersecting chords, each of which has different letters at its ends (m - d). First, we draw a cho... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,299 |
8.1. Two cars are driving on a highway at a speed of 80 km/h and with an interval of 10 m. At the speed limit sign, the cars instantly reduce their speed to 60 km/h. What will be the interval between them after the speed limit sign? | Answer: $7.5 \mathrm{M}$.
Solution. Let $v$ (m/hour) be the speed of the cars before the sign, and $u$ (m/hour) be the speed of the cars after the sign. The second car will pass the sign later than the first by $10 / v$ (hours). During this time, the first car will travel $10 u / v$ (meters) $= 10 \cdot 6 / 8 = 7.5$ m... | 7.5\mathrm{M} | Other | math-word-problem | Yes | Yes | olympiads | false | 22,300 |
8.2. Given a convex quadrilateral $A B C D$. Can we assert that $A B<C D$, if a) angles $A$ and $B$ are obtuse; b) angle $A$ is greater than angle $D$, and angle $B$ is greater than angle $C$? | Answer. a) yes; b) yes.
Solution. a) The solution to part b) also implies part a), but for part a) a simpler solution can be provided. Draw perpendiculars from points $A$ and $B$ to line $DC$... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,301 |
8.3. There are 10 weights, and it is known that if any one of them is removed, the remaining 9 weights can be divided into three groups of equal weight. Must all 10 weights necessarily be of the same weight? | Answer: Not necessarily.
Solution. Here is an example: suppose there are seven weights of 1 g and three weights of 7 g. Then, if we remove a 1 g weight, we can arrange them as follows: $1+1+7=1+1+7=1+1+7$, and if we remove a 7 g weight, then $1+1+1+1+1+1+1=7=7$. | Notnecessarily | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,302 |
8.4. Given three distinct numbers $a, b$, and $c$, among which there is a number greater than 2022. Can it be that the numbers $a^{2}-b^{2}, b^{2}-c^{2}$, and $c^{2}-a^{2}$ are three consecutive integers? | Answer: it can.
Solution. The sum of the numbers $a^{2}-b^{2}, b^{2}-c^{2}$, and $c^{2}-a^{2}$ is zero, and therefore, if these are consecutive integers, they must be $(-1), 0$, and 1. From the equation $b^{2}-c^{2}=0$, it follows that either $b=c$ or $b=-c$, but since the numbers $a, b$, and $c$ must be distinct, onl... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,303 |
7.1. The older brother's journey to school takes 12 minutes, while the younger brother's journey (along the same route) takes 20 minutes. How many minutes will pass from the moment the younger brother leaves home until he is caught up by the older brother, if the older brother left 5 minutes after the younger brother? | Answer: In 12.5 minutes. Solution: Let $S$ be the distance from home to school. Since the younger brother covers this distance in 20 minutes, in 5 minutes he will cover the distance $\frac{5 S}{20}=\frac{S}{4}$. After the older brother leaves the house, the rate at which the brothers are closing the distance will be $\... | 12.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,304 |
7.2. The length of the rectangle was increased by $12 \%$, and the width - by $15 \%$. a) By what percentage is the area of the new rectangle larger than the area of the original? b) Find the ratio of the length to the width of the original rectangle, if it is known that the perimeter of the new rectangle is $13 \%$ la... | Answer. a) by $28.8 \%$. b) 2:1. Solution. a) Let $a$ and $b$ be the length and width of the rectangle, respectively. Then the sides of the new rectangle are $1.12 a$ and $1.15 b$, and its area $S=1.12 a \cdot 1.15 b=1.288 a b$, which is $128.8 \%$ of the area of the original rectangle. Thus, the area has increased by ... | )28.8\,b)2:1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,305 |
7.3. Given natural numbers $a$ and $b$. Can we assert that they end with the same digit in decimal notation if it is known that the numbers $2a + b$ and $2b + a$ end with the same digit? | Answer: Yes, it is possible. Solution: Let's calculate the difference between the numbers $2a + b$ and $2b + a$, then we get that $a - b$ is divisible by 10, i.e., $a$ and $b$ end with the same digit. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,306 |
7.4. In the gym, 10 boys and 10 girls stand in a circle in a random order. Will the coach be able to draw 10 non-intersecting segments on the floor with chalk so that each of them connects a boy and a girl (regardless of the arrangement of the children)? | Answer: It is possible. Solution: Let's consider the following model of the problem: on a circle, there are 10 points marked with the letter m (boys) and 10 points marked with the letter d (girls). We need to draw 10 non-intersecting chords, each of which has different letters at its ends (m - d). First, we draw a chor... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,307 |
7.1. Two cars are driving on a highway at a speed of 80 km/h and with an interval of 10 m. At the speed limit sign, the cars instantly reduce their speed to 60 km/h. What will be the interval between them after the speed limit sign? | Answer: 7.5 m. Solution: Let $v$ (m/hour) be the speed of the cars before the sign, and $u$ (m/hour) be the speed of the cars after the sign. The second car will pass the sign later than the first by $10 / v$ (hours). In this time, the first car will travel $10 u / v$ (meters) $= 10 \cdot 6 / 8 = 7.5$ meters. This inte... | 7.5 | Other | math-word-problem | Yes | Yes | olympiads | false | 22,308 |
7.2. Two isosceles triangles are given, with the same perimeters. The base of the second triangle is $15\%$ larger than the base of the first, and the lateral side of the second triangle is $5\%$ shorter than the lateral side of the first. Find the ratio of the sides of the first triangle. | Answer. The base relates to the lateral side as 2:3. Solution. Let $a$ and $b$ be the length of the base and the lateral side, respectively, of the first triangle. Then its perimeter is $a+2 b$. The base and the lateral side of the second triangle are, respectively, $1.15 a$ and $0.95 b$, and its perimeter is $1.15 a+1... | :b=2:3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,309 |
7.3. There are seven weights, and it is known that if any one of them is removed, the remaining six weights can be divided into three groups of equal weight. Is it necessarily true that all seven weights are of the same weight? | Answer. Not necessarily. Solution. An example can be given as follows: let the weights of the weights be $\{1,1,1,1,4,4,4\}$ (in grams). Then, if a 1 g weight is removed, the weights can be divided into three groups as follows: $1+4=1+4=1+4$, and if a 4 g weight is removed, then $1+1+1+1=4=4$. | Notnecessarily | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 22,310 |
7.4. Given three distinct numbers $a, b$, and $c$, among which there is a number greater than 2022. Can it be that the numbers $a^{2}-b^{2}, b^{2}-c^{2}$, and $c^{2}-a^{2}$ are three consecutive integers? | Answer. Yes. Solution. The sum of the numbers $a^{2}-b^{2}, b^{2}-c^{2}$, and $c^{2}-a^{2}$ is zero, and therefore, if these are consecutive integers, they must be equal to (-1), 0, and 1. From the equation $b^{2}-c^{2}=0$ it follows that either $b=c$ or $b=-c$, but since the numbers $a, b$, and $c$ must be different, ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,311 |
7.1. Find the smallest natural number divisible by 5, with the sum of its digits being 100. Justify your answer. | Answer: 599999999995 (between two fives there are 10 nines). Solution. Due to divisibility by 5, the last digit of the number $N$ can be either 5 or 0. If the last digit is 0, without changing the sum of the digits, we can replace 0 with 5 and subtract one from each of the five non-zero digits in other positions. Then ... | 599999999995 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,312 |
7.2. In triangle $ABC$, the point of intersection of the medians is equidistant from all three vertices. Prove that triangle $ABC$ is equilateral. | Solution. Let $B M$ be the median and $K$ be the point of intersection of the medians. Then, in the isosceles triangle $A K C$, the median $K M$ is also an altitude. From this, we obtain that in triangle $A B C$, the median $B M$ is also an altitude. Therefore, the right triangles $A B M$ and $C B M$ are congruent (by ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,313 |
7.3. Does there exist a set of 25 different integers with the following property: the sum of all 25 numbers is zero, and for any 24 of them, the absolute value of the sum is greater than $25?$ | Answer: It exists. Solution. If we consider a set of 25 numbers with a zero sum, then for any 24 numbers of this set, the sum will be equal to ( $-x$ ), where $x$ is the remaining (discarded) number. Therefore, we need to find a set of 25 different integers with a zero sum, where all numbers in absolute value are great... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,314 |
7.4. In the astronaut squad, there are 20 people, and each has at least 14 friends in the squad. A crew is needed for space, in which everyone is friends with each other. Is it necessarily possible to form a crew of four people? | Answer: definitely. Solution. For an arbitrary cosmonaut K, consider the group G of his 14 friends, so that in the additional group G* there will be 6 people together with K. Within G, each cosmonaut, say C, has at least 8 friends ($8=14-6$: from all friends of C, who are no less than 14, the cosmonauts of the addition... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,315 |
7.5. In each cell of a $9 \times 9$ grid, zeros are written. In one move, it is allowed to choose a row and add an arbitrary positive number to any two adjacent cells in the chosen row (the number added can be changed from move to move). Is it possible to form a square in which the sums in all nine columns are the same... | Answer: No. Solution. Let $A$ be the sum of 45 numbers in the five columns with odd numbers, and $B$ be the sum of 36 numbers in the remaining four columns (with even numbers). Note that with each move, the same number (which generally depends on the move) is added to both $A$ and $B$. Therefore, the difference $A-B$ r... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,316 |
8.3. There are five sticks, each longer than 2 cm but shorter than $8 \mathrm{~cm}$. Prove that it is possible to take three of these five sticks and form a triangle from them. | Solution. Let's order the lengths of the sticks in ascending order: $a_{1} \leq a_{2} \leq a_{3} \leq a_{4} \leq a_{5}$. Assuming the opposite, we will have: $a_{3} \geq a_{1}+a_{2}>2+2=4$. Next, $a_{4} \geq a_{2}+a_{3}>2+4=6 \quad$ and $a_{5} \geq a_{3}+a_{4}>4+6=10 /$ Thus, we have obtained a contradiction with the c... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,319 |
8.4. Given a triangle $A B C$ with median $B M$. On the median, an arbitrary point $P$ is marked, and through $P$ a line parallel to $A B$ is drawn, and through point $C$ a line parallel to $B M$ is drawn. These lines intersect at point $Q$. Prove that the segment $B P$ is bisected at the point of intersection with the... | Solution. Let $D$ be the intersection point of lines $A B$ and $C Q$, and $E$ be the intersection point of lines $A P$ and $C Q$. Consider triangle $A D E$ and show that $\triangle P B Q$ is the medial triangle in $\triangle A D E$. Indeed, point $P$ is the midpoint of segment $A E$ (since $M$ is the midpoint of $A C$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,320 |
8.3. On the edges of a cube, numbers $1,2, \ldots, 12$ were placed in some order, and for each face, the sum of the four numbers on its edges was calculated. Prove that there is a face for which this sum is greater than 25. | Solution. Let's calculate the corresponding sum on each face and then add these sums for all six faces. We will get $(1+2+\ldots+12) \cdot 2$ as a result, since in this calculation any edge will be counted twice. Thus, the total sum is 156, and then the sum for at least one face is not less than $\frac{156}{6}=26$. (In... | 26 | Combinatorics | proof | Yes | Yes | olympiads | false | 22,323 |
8.4. Given a right triangle, the height dropped to the hypotenuse is 4 times smaller than the hypotenuse. Find the acute angles of this triangle. | Answer: $15^{\circ}$ and $75^{\circ}$. Solution. Let $A B C$ be the given triangle, $C M$ the height from vertex $C$ of the right angle, and $C O$ the median. Consider the right triangle $C M O$. By the property of the median from the vertex of the right angle, $C O=\frac{A B}{2}$, and therefore (by the condition) $C O... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,324 |
7.1. At the final test among the seventh graders of the school, 80 people were present. As a result, everyone received positive grades (threes, fours, or fives), and the average score was 3.45. Prove that the number of fours was even. | Solution. Let $x, y, z$ be the number of threes, fours, and fives, respectively. Then we will have two equations: $x+y+z=80$ and $3 x+4 y+5 z=3.45 \cdot 80=276$. If we subtract the first equation, multiplied by 3, from the second equation, we get $y+2 z=36$. Therefore, $y=36-2 z$ - an even number. | 36-2z | Number Theory | proof | Yes | Yes | olympiads | false | 22,325 |
7.2. Petya says to Kolya: «I placed some numbers at the vertices of a cube, and then on each face I wrote the sum of the four numbers at its vertices. Then I added all six numbers on the faces and got 2019. Can you figure out what the sum of the eight numbers at the vertices of the cube is?» How would you answer this q... | Answer: 673. Solution. Each vertex of the cube belongs to three faces (which meet at this vertex) and therefore the number placed at this vertex participates three times in the calculation of the sums on the faces. Thus, the sum of the numbers at the vertices is $2019: 3=673$. Comment: another way to solve (direct) is ... | 673 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,326 |
7.3. Prove that $(n+20)(n+201)(n+2020)$ is divisible by 6 for any natural $n$. | Solution. The neighboring factors $n+20, n+201$ have different parity, and therefore their product is divisible by 2. Next, we note that the three factors give different remainders when divided by 3, since the difference between the numbers of any pair of these factors is not divisible by 3. This means they give all th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,327 |
7.4. What is the minimum number of kings that need to be placed on a chessboard so that they attack all unoccupied squares? (A king attacks the squares that are adjacent to its square by side or corner).
 will be $6 \cdot 6=36$. | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,333 |
10.3 Given a convex quadrilateral $A B C D$ and a point $M$ inside it. It turns out that all triangles $A B M, B C M, C D M$ and $D A M$ are isosceles. Prove that among the segments $A M, B M, C M$ and $D M$ there are at least two of the same length. | Solution. Since $\angle A M B+\angle B M C+\angle C M D+\angle D M A=360^{\circ}$, at least one of these four angles is not acute. Let, for definiteness, $\angle A M B \geq 90^{\circ}$. Then $A B>A M$ and $A B>B M$. Therefore, the equal sides in $\triangle A M B$ are $A M$ and $B M$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,334 |
10.4 a) Prove that there exists an increasing geometric progression from which three terms (not necessarily consecutive) can be selected to form an arithmetic progression. b) Can the common ratio of such a geometric progression be greater than $1.9 ?$ | Answer. b) It can. Solution. a) As an example, we can take the geometric progression 1; $q; q^{2}; q^{3}$, where $q=(1+\sqrt{5}) / 2$; for this progression, the equality $1+q^{3}=2 q^{2}$ holds, which, taking into account $q \neq 1$, is equivalent to the quadratic equation $q^{2}-q-1=0$. b) Let $1, q, q^{2}, \ldots, q^... | proof | Algebra | proof | Yes | Yes | olympiads | false | 22,335 |
11.2. Given the quadratic equation $a^{3} x^{2}+b^{3} x+c^{3}=0$, which has two roots. Prove that the equation $a^{5} x^{2}+b^{5} x+c^{5}=0$ also has two roots. | Solution. We have $b^{6}>4 a^{3} c^{3}$, since the quadratic equation has a positive discriminant. It is required to prove that $b^{10}>4 a^{5} c^{5}$. If $a c<0$, the inequality is obvious. If $a c>0$, then $b^{6}>4(a c)^{3}$. Due to the monotonically increasing function $y=x^{5 / 3}$, we get $\left(b^{2}\right)^{5}>4... | proof | Algebra | proof | Yes | Yes | olympiads | false | 22,336 |
11.3. Given a circle of unit radius, $AB$ is its diameter. Point $M$ moves along the circle, $M_{1}$ is its projection on $AB$. Denote $f(M)=AB+MM_{1}-AM-BM$. Find the maximum and minimum values of the function $f(M)$. | Answer: the minimum value is 0; the maximum value is $3-2 \sqrt{2}$.
Solution. Let $\angle M A B=\alpha, \alpha \in\left[0, \frac{\pi}{2}\right]$. Then $f(M)=2+2 \cos \alpha \sin \alpha-2 \cos \alpha-2 \sin \alpha$. Denote this expression by $g(\alpha)$. We have $g^{\prime}(\alpha)=2\left(\cos ^{2} \alpha-\sin ^{2} \a... | 3-2\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,337 |
11.4. How many Pythagorean triangles exist with one of the legs equal to 2013? (A Pythagorean triangle is a right triangle with integer sides. Equal triangles are counted as one.). | Answer: 13. Solution. From the Pythagorean theorem, we obtain the equation in integers $2013^{2}+x^{2}=y^{2} \Leftrightarrow(y-x)(y+x)=2013^{2}=3^{2} \cdot 11^{2} \cdot 61^{2}$. This equation is equivalent to the system $\left\{\begin{array}{l}y-x=d_{1} \\ y+x=d_{2}\end{array}\right.$, where $d_{1}, d_{2}=\frac{2013^{2... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,338 |
7.1 The length of a rectangle is $25 \%$ greater than its width. By a straight cut, parallel to the shorter side, this rectangle is cut into a square and a rectangular strip. By what percentage is the perimeter of the square greater than the perimeter of the strip? | Answer: By $60 \%$. Solution: Let $a$ be the width of the rectangle, then its length is $1.25 a$. The perimeter of the resulting square is $4 a$, and the perimeter of the strip is $2 a+0.5 a=2.5 a$. The ratio of the perimeters is $4 a: 2.5 a=1.6$. Therefore, the perimeter of the square is greater than the perimeter of ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,340 |
7.2 Can 8 numbers be chosen from the first hundred natural numbers so that their sum is divisible by each of them | Answer. Yes. Solution. An example of the required numbers can be $1 ; 2 ; 3 ; 6 ; 12 ; 24 ; 48 ; 96.3$ here the sum 192 is divisible by each of the numbers. This example is not unique; others can be provided, such as $1 ; 2 ; 3 ; 4 ; 5 ; 15 ; 30 ; 60$ with a sum of 120. | 120 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,341 |
7.3 In a six-digit number, one digit was crossed out to obtain a five-digit number. The five-digit number was subtracted from the original number, and the result was 654321. Find the original number | Answer: 727 023. Solution: Note that the last digit was crossed out, as otherwise the last digit of the number after subtraction would have been zero. Let $y$ be the last digit of the original number, and $x$ be the five-digit number after crossing out. Then the obtained number is $10 x + y$ $x = 9 x + y = 654321$. Div... | 727023 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,342 |
7.4 In a box, there are 25 colored pencils. It is known that among any five pencils, there will be at least two pencils of the same color. Prove that there are 7 pencils of the same color in the box. | Solution. From the condition of the problem, it follows that the pencils have no more than 4 colors. Then, reasoning by contradiction, we get that there will be 7 pencils of the same color: indeed, otherwise, the total number of pencils would not exceed $6 \cdot 4=24$.
## 1st round. 11.11 .2018
7th grade | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,343 |
7.1 In a bookstore, Vasya and Tolya were interested in the same book. Vasya was short of 150 rubles to buy it, and Tolya was short of 200 rubles. When Vasya asked Tolya to lend him half of his cash, Vasya was able to buy the book and still had 100 rubles left for transportation. How much did the book cost? | Answer: 700 rubles. Solution: If $x$ is the price of the book, then Vasya had $(x-150)$ rubles, and Petya had $(x-200)$ rubles. By setting up the equation $x-150+\frac{x-200}{2}=x+100$ and solving it, we find $x=700$. | 700 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,344 |
7.2 In the glass, there was a solution in which water made up $99 \%$. The glass with the solution was weighed, and the weight turned out to be 500 gr. After that, part of the water evaporated, so that in the end, the proportion of water was $98 \%$. What will be the weight of the glass with the resulting solution, if ... | Answer: 400 g. Indication. Initially, the weight of the solution was $500-300=200$ (g)., and the amount of water was $0.99 \cdot 200=198$ (g.), and thus, the substance was $200-198=2$ (g.). After the evaporation of water, 2 g of the substance make up $100 \% -98 \% =2 \%$ of the weight of the solution, so the entire so... | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,345 |
7.3 Petya wrote down all natural numbers from 1 to $n$ in a row and counted the total number of digits written. Then he called Kolya and asked: "What is $n$ if a total of 2018 digits are written?" Kolya said: "Count again, you made a mistake." Who is right? | Answer. Kolya is right. Note. If 2018 digits are written, then the number $n$ must be a three-digit number: indeed, in the case of a two-digit $n$, no more than $9+2 \cdot 90=189$ digits would be written, and in the case of a four-digit (or more) number, more than $9+2 \cdot 90+3 \cdot 900=2889$ digits would be written... | Kolya | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,346 |
7.4 There are $n$ sticks of lengths $1,2, \ldots, n$. Can a square be formed from these sticks, and if not, what is the minimum number of sticks that can be broken in half to form a square: a) for $n=12$; b) for $n=15$? (All sticks must be used). | Answer. a) 2 sticks; b) Yes. Solution. a) Since the sum $1+2+\ldots+12=78$ is not divisible by 4, it is impossible to form a square. The side of the square would have to be $\frac{78}{4}=19.5$. If only one stick is broken, its parts might end up on two different sides of the square, but the other two sides cannot have ... | )2sticks;b)Yes | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,347 |
9.1. Find all prime numbers $p$ for which $8 p+1$ is: a) a perfect square? b) a perfect cube | Answer. a) $p=3 ;$ b) such $p$ does not exist. Solution. See problem 8.2. | )p=3;b)suchpdoesnotexist | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,348 |
9.2. Given two quadratic functions $f(x)=a x^{2}+b x+c$ and $g(x)=c x^{2}+b x+a$. It turns out that the function $f(x)+g(x)$ has a unique root. Prove that the functions $f(x)$ and $g(x)$ have a common root. | Solution. The quadratic trinomial $f(x)+g(x)=(a+c) x^{2}+2 b x+(a+c)$ has a unique root under the condition that the discriminant is zero: $b^{2}-(a+c)^{2}=0$. From this, it follows that either $a+c=b$ or $a+c=-b$. In the case when $a+c=b$, we have $f(-1)=g(-1)=0$, and in the case when $a+c=-b$, we get $f(1)=g(1)=0$. T... | proof | Algebra | proof | Yes | Yes | olympiads | false | 22,349 |
9.3. $n$ chips numbered $1,2, \ldots, n$ are arranged in a row in ascending order. In one move, it is allowed to swap any two chips that have either two or five chips between them. Does there exist an $n$ for which it is possible to rearrange all the chips in reverse order after several moves? | Answer. Does not exist. Solution. See the solution to problem 8.3. Compared to the conditions of problem 8.3, here there is an additional possibility to swap two chips that have five chips between them. This allows changing the coordinates by 6 units, but this additional possibility does not change the remainder when t... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,350 |
11.2. Prove the inequality $\sqrt{a^{2}+1}+\sqrt{2}>|a+1|$. | Solution. The inequality is not difficult to prove algebraically, getting rid of the roots twice by squaring. However, it is simpler to prove it geometrically by considering points $A(-a ; 0)$, $B(0 ; 1)$, and $C(1 ; 0)$ on the coordinate plane. Then the inequality can be written as $A B+B C>A C$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 22,352 |
11.3. Given a triangle $A B C$, point $O$ is the center of the inscribed circle. Prove that $R_{B O C}$ $<2 R_{A B C}$, where $R_{B O C}$ and $R_{A B C}$ are the radii of the circumcircles of triangles $B O C$ and $A B C$. | Solution. When solving problem 10.3, we obtained $R_{A B C} \sin \alpha=R_{B O C} \cos \frac{\alpha}{2}$. From this, $R_{B O C}=2 R_{A B C} \sin \frac{\alpha}{2} \leq 2 R_{A B C}$, and the inequality is actually strict, since otherwise we would have $\frac{\alpha}{2}=90^{\circ}$, i.e., $\alpha=180^{\circ}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,353 |
9.4. In a school mathematics olympiad for 9th graders, 20 people participated. As a result, all participants scored different points, and each participant's score was less than the sum of the scores of any two other participants. Prove that each participant scored more than 18 points. | Solution. Let $x, y$ be the scores of the participants who took the last and second-to-last place, respectively. If we assume the opposite of the statement of the problem, then $x \leq 18$. Since $x < y$, the next scores in ascending order should be at least $y+1, y+2, \ldots, y+18$, respectively. Thus, the winner has ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 22,356 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.