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3-5. The city's bus network is organized as follows: 1) from any stop to any other stop, you can get without transferring;
2) for any pair of routes, there is, and only one, stop where you can transfer from one of these routes to the other;
3) on each route, there are exactly three stops.
How many bus routes are ther... | Solve 5. Answer: 7. We will prove that if the given conditions are satisfied, then the number of stops $n$ and the number of routes $N$ are related by the formula $N=n(n-1)+1$. Let $a$ be one of the routes, and $B$ be a stop that route $a$ does not pass through. Each route passing through $B$ intersects route $a$. Ther... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,249 |
3-ча 1. In a chess tournament, students from grades IX and X participated. There were 10 times more students from grade X than from grade IX, and they scored 4.5 times more points in total than all the students from grade IX. How many points did the students from grade IX score? Find all solutions. | Solution 1. Let $x$ be the number of ninth graders in the tournament. Then there were a total of $11x$ participants, and they scored $\frac{11x(11x-1)}{2}$ points. According to the problem, the ratio of the number of points scored by the ninth graders to the number of points scored by the tenth graders is $1:4.5$. Ther... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,250 |
3-3. On the sides $P Q, Q R, R P$ of triangle $P Q R$, segments $A B, C D, E F$ are laid out. Inside the triangle, a point $S_{0}$ is given. Find the geometric locus of points $S$ lying inside triangle $P Q R$ for which the sum of the areas of triangles $S A B, S C D, S E F$ is equal to the sum of the areas of triangle... | Solve 3. Let's assume that $\frac{A B}{P Q} \geqslant \frac{C D}{Q R} \geqslant \frac{E F}{R P}$. We will lay off segments $Q D^{\prime}=C D \cdot \frac{P Q}{A B}$ and $P E^{\prime}=F E \cdot \frac{P Q}{A B}$ on the sides $Q R$ and $P R$. Then
$$
\begin{aligned}
S_{\triangle S A B}+S_{\triangle S C D}+S_{\triangle S E... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,251 |
3-4. In the city, there are 57 bus routes. It is known that:
1) from any stop to any other stop, one can travel without transferring;
2) for any pair of routes, there is one, and only one, stop where one can transfer from one of these routes to the other;
3) each route has no fewer than three stops.
How many stops do... | Solution 4. Answer: 8. We will prove that if the given conditions are satisfied, then the number of stops $n$ and the number of routes $N$ are related by the formula $N=n(n-1)+1$. First, we will show that if a route has $n$ stops, then any other route also has $n$ stops, and furthermore, each stop is served by $n$ rout... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,252 |
3-ча 1. Determine the coefficients that will stand by $x^{17}$ and $x^{18}$ after expanding the brackets and combining like terms in the expression
$$
\left(1+x^{5}+x^{7}\right)^{20}
$$ | Solution 1. The number 18 cannot be represented as the sum of the numbers 5 and 7, so the coefficient of $x^{18}$ will be zero.
The number 17 can be represented as the sum of the numbers 5 and 7 as follows: $17=7+5+5$; this representation is unique up to the order of the terms. In one of the 20 expressions $1+x^{5}+x^... | 3420 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,253 |
3-4. Given a convex pentagon $A B C D E$. The sides opposite to the vertices $A, B, C, D$, $E$ are the segments $C D, D E, E A, A B, B C$, respectively. Prove that if an arbitrary point $M$, lying inside the pentagon, is connected by lines to all its vertices, then of these lines, either exactly one, or exactly three, ... | Solution 4. Let's draw the diagonals of the given pentagon. They divide it into 11 regions: one pentagon, 5 internal triangles (whose sides are the diagonals), and 5 external triangles (one of the sides of each being a side of the pentagon). If point $M$ belongs to an external triangle, the number of required lines is ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,256 |
3-5. Point $O$ is the orthocenter of an acute triangle $ABC$. Prove that the 3 circles passing through: the first through points $O, A, B$, the second through points $O, B, C$, and the third through points $O, C, A$ are equal to each other. | Solution 5. It is easy to check that $\angle A O B=90^{\circ}+\angle C$. Therefore, the radius of the circumcircle of triangle $A O B$ is $\frac{1}{2} A B \sin \angle A O B=\frac{1}{2} A B \sin \angle C=R$, where $R$ is the radius of the circumcircle of triangle $A B C$. The radii of the other considered circles are al... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,257 |
3-ча 1. In which of the expressions:
$$
\left(1-x^{2}+x^{3}\right)^{1000}, \quad\left(1+x^{2}-x^{3}\right)^{1000}
$$
after expanding the brackets and combining like terms, is the coefficient of $x^{20}$ greater? | Solve 1. Answer: In the expression $\left(1+x^{2}-x^{3}\right)^{1000}$. Let $P(x)=\left(1-x^{2}+x^{3}\right)^{1000}$ and $Q(x)=\left(1+x^{2}-x^{3}\right)^{1000}$. The coefficient of $x^{20}$ in the polynomial $P(x)$ is the same as in the polynomial $P(-x)=\left(1-x^{2}-x^{3}\right)^{1000}$, and the coefficient of $x^{2... | 0.89001 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,258 |
3-5. Find all lines in space passing through a given point $M$ at a given distance $d$ from a given line $AB$. | Solve 5. Answer: lines lying in two planes tangent to a cylinder of radius $d$ with axis $A B$ and passing through point $M$, except for the line parallel to $A B$ (if point $M$ is located at a distance $d$ from line $A B$, then there will be only one tangent plane and there is no need to exclude the line). Lines at a ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,260 |
3-3. From two hundred numbers: $1,2,3,4,5,6,7, \ldots, 199,200$ one hundred and one numbers are chosen arbitrarily. Prove that among the chosen numbers, there will be two such that one divides the other. | Solution 3. Consider the largest odd divisors of the selected numbers. Among the numbers from 1 to 200, there are exactly 100 different largest odd divisors (the numbers $1,3, \ldots, 199$). Therefore, two of the selected numbers have the same largest odd divisor. This means that the two selected numbers differ only in... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 26,263 |
3-cha 2. $k$ wire triangles are arranged in space such that:
1) any 2 of them have exactly one common vertex,
2) in each vertex, the same number $p$ of triangles meet.
Find all values of $k$ and $p$ for which the specified arrangement is possible. | Solution 2. Answer: $(k, p)=(1,1),(4,2)$ or $(7,3)$.
First, we prove that $p \leqslant 3$. Suppose that $p \geqslant 4$. Take triangle $\Delta_{1}$. To its vertex $A$ adjoins triangle $\Delta_{2}$. To the vertex $B \neq A$ of triangle $\Delta_{2}$ adjoin triangles $\Delta_{3}, \Delta_{4}, \Delta_{5}$; all of them are ... | (k,p)=(1,1),(4,2)or(7,3) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,266 |
3-3. In a numerical triangle, each number is equal to the sum of three numbers from the previous row:
| | | | | | | | | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | | | | 1 | | | | |
| | | | 1 | 1 | 1 | | | |
| | | 1 | 2 | 3 | 2 | 1 | | |
| | 1 | 3 | 6 | 7... | Sure, here is the translated text:
```
```
Starting from the third line, write the first four numbers in each line, replacing each even number with 0 and each odd number with 1:
```
,(2,3,6)\) or \((3,3,3)\).
Let \(x \leqslant y \leqslant z\) be natural numbers, and \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\). Then \(x1000^{10}\), so the number \(2^{100}\) has at least 31 digits. On the other hand,
\[
\frac{1024^{10}}{1000^{10}}<\left(\frac{1025}{1000}\right)^{10}=\left(\... | (2,4,4),(2,3,6),(3,3,3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,269 |
3-ча 1. If the number \(\frac{2^{n}-2}{n}\) is an integer, then the number \(\frac{2^{2^{n}-1}-2}{2^{n}-1}\) is also an integer. Prove. | Let \(2^{n}-2=n m\). Then
\[
\begin{aligned}
\frac{2^{2^{n}-1}-2}{2^{n}-1} & =2 \frac{2^{2^{n}-2}-1}{2^{n}-1}=\frac{2^{n m}-1}{2^{n}-1}= \\
& =2\left(2^{n(m-1)}+2^{n(m-2)}+\cdots+2^{n}+1\right)
\end{aligned}
\] | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,271 |
3-ча 2. Prove without using tables that
\[
\frac{1}{\log _{2} \pi}+\frac{1}{\log _{5} \pi}>2
\] | Solution 2. Let \(\log _{2} \pi=a\) and \(\log _{5} \pi=b\). Then \(2^{a}=\pi\) and \(5^{b}=\pi\), i.e. \(\pi^{1 / a}=2\) and \(\pi^{1 / b}=5\). Therefore, \(\pi^{1 / a+1 / b}=2 \cdot 5=10\). Considering that \(\pi^{2}<10\), we obtain the required. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 26,272 |
3-ча 3. Given two triangular pyramids \(A B C D\) and \(A^{\prime} B C D\) with a common base \(B C D\), and the point \(A^{\prime}\) lies inside the pyramid \(A B C D\). Prove that the sum of the plane angles at vertex \(A^{\prime}\) of the pyramid \(A^{\prime} B C D\) is greater than the sum of the plane angles at ve... | Solution 3. First, we will prove the required statement in the case when point \(A^{\prime}\) lies on edge \(A B\). It is clear that \(\angle B A^{\prime} C = \angle B A C + \angle A C A^{\prime}\) and \(\angle B A^{\prime} D = \angle B A D + \angle A D A^{\prime}\). Therefore, the required inequality can be transforme... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,273 |
3-ча 1. Solve the equation \(x^{y}=y^{x}(x \neq y)\) in natural numbers. Find all solutions. | Solve 1. Answer: \(x=2, y=4\) or \(x=4, y=2\). From the equality \(x^{y}=y^{x}\) it follows that the prime divisors of the numbers \(x\) and \(y\) are the same, i.e., \(x=p_{1}^{a_{1}} p_{2}^{a_{2}} \ldots p_{n}^{a_{n}}\) and \(y=p_{1}^{b_{1}} p_{2}^{b_{2}} \ldots p_{n}^{n_{n}}\). The same equality shows that \(a_{1} y... | 2,4or4,2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,274 |
3-ча 2. Prove that in any triangle the inequality holds: \(R \geqslant 2 r\) (where \(R\) and \(r\) are the radii of the circumscribed and inscribed circles, respectively), and the equality \(R=2 r\) holds only for an equilateral triangle. | Solution 2. Let \(A_{1}, B_{1}\) and \(C_{1}\) be the midpoints of sides \(B C, A C\) and \(A B\) respectively. Under a homothety with the center at the intersection point of the medians of triangle \(A B C\) and a homothety coefficient of \(-1 / 2\), the circumcircle \(S\) of triangle \(A B C\) transforms into the cir... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 26,275 |
3-ча 1. Find all positive rational solutions of the equation \(x^{y}=y^{x}(x \neq y)\). | Solution 1. Let \(y=k x\). Then \(x^{k x}=(k x)^{x}\). Extracting the root of degree \(x\), and then dividing by \(x\), we get \(x^{k-1}=k\). Thus, \(x=k^{\frac{1}{k-1}}\) and \(y=k k^{\frac{1}{k-1}}=k^{\frac{k}{k-1}}\). The number \(k\) is rational; let \(\frac{1}{k-1}=\frac{p}{q}\) be an irreducible fraction. Substit... | (\frac{p+1}{p})^{p},(\frac{p+1}{p})^{p+1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,277 |
3-ча 1. Show that
\[
27195^{8}-10887^{8}+10152^{8}
\]
is divisible by 26460. | Solution 1. Note that \(26460=2^{2} \cdot 3^{3} \cdot 5 \cdot 7^{2}\). Represent the number \(A=27195^{8}-10887^{8}+10152^{8}\) as \(27195^{8}-\left(10887^{8}-10152^{8}\right)\). This number is divisible by \(5 \cdot 7^{2}\). Indeed, \(27195=3 \cdot 5 \cdot 7^{2} \cdot 37\), and the difference \(10887^{8}-10152^{8}\) i... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,281 |
3-ча 3. Prove that the equality:
\[
x^{2}+y^{2}+z^{2}=2 x y z
\]
for integers \(x, y\) and \(z\) is possible only when \(x=y=z=0\). | Solution 3. Suppose \(x=2^{m} x_{1}, y=2^{n} y_{1}, z=2^{k} z_{1}\), where the numbers \(x_{1}, y_{1}, z_{1}\) are odd. We can assume that \(m \leqslant n \leqslant k\). Then both sides of the equation can be divided by \(\left(2^{m}\right)^{2}\). As a result, we get
\[
x_{1}^{2}+2^{(n-m)} y_{1}^{2}+2^{(k-m)} z_{1}^{2... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,283 |
3-4. Given a closed broken line in the plane with a perimeter of 1. Prove that a circle with radius \(\frac{1}{4}\) can be drawn to cover the entire broken line. | Solution 4. Take two points \(A\) and \(B\) on the broken line, dividing its perimeter in half. Then \(A B \leqslant 1 / 2\). We will prove that all points of the broken line lie inside a circle of radius \(1 / 4\) with the center at the midpoint \(O\) of segment \(A B\). Let \(M\) be an arbitrary point on the broken l... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,284 |
3-2. How are the planes of symmetry of a bounded body arranged if it has two axes of rotation? (An axis of rotation of a body is a line about which the body can be rotated by any angle and still coincide with itself.)
| Solution 2. The axes of rotation of a bounded body intersect at its center of mass. If the body has two axes of rotation intersecting at point \(O\), and point \(A\) belongs to the body, then the body contains the entire sphere of radius \(O A\) centered at \(O\). Therefore, any plane passing through point \(O\) is a p... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,286 |
3-ча 3. Find the real roots of the equation:
\[
x^{2}+2 a x+\frac{1}{16}=-a+\sqrt{a^{2}+x-\frac{1}{16}} \quad\left(0<a<\frac{1}{4}\right)
\] | Solution 3. Let \(f(x)\) be the left-hand side of the given equation. The graph of the function \(y=f(x)\) is a parabola. If \(00\). Therefore, the original equation is equivalent to the equation
\[
x^{2}+2 a x+\frac{1}{16}=-a \pm \sqrt{a^{2}+x-\frac{1}{16}}
\]
The right-hand side of equation (1) represents the inver... | x_{1,2}=\frac{1-2}{2}\\sqrt{(\frac{1-2}{2})^{2}-\frac{1}{1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,287 |
3-ча 5. Prove that if a hexagon has opposite sides parallel and diagonals connecting opposite vertices are equal, then a circle can be circumscribed around it. | Solution 5. Let \(A B C D E F\) be a hexagon satisfying the condition of the problem. Quadrilateral \(F B C E\) is an isosceles trapezoid, so the line \(M M_{1}\), connecting the midpoints of its bases, is perpendicular to them and serves as the bisector of the angle between the diagonals \(B E\) and \(F C\). Similarly... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,289 |
3-ча 1. 12 fields are arranged in a circle: on four adjacent fields, there are four differently colored chips: red, yellow, green, and blue.
One move allows you to move any chip from the field it occupies, over four fields to the fifth (if it is free) in either of the two possible directions. After several moves, the ... | Solve 1. Answer: KZHS, HSZK, HSKZ, ZKHJ.
Let's make copies of our 12 fields and arrange them in a circle in the following order: \(1,6,11,4,9,2,7\), \(12,5,10,3,8\). On the new circle, the chips move simply to the adjacent field (to the right or left). Therefore, any move in which the chips swap places represents a mo... | HSZK,HSKZ,ZKHJ | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,290 |
3-2. Given two triangles: \(\triangle A B C\) and \(\triangle D E F\) and a point \(O\). Any point \(X\) is taken in \(\triangle A B C\) and any point \(Y\) in \(\triangle D E F\); the triangle \(O X Y\) is completed to form a parallelogram \(O X Z Y\).
a) Prove that all points obtained in this way form a polygon.
b)... | Solution 2. We will solve the problem in a more general form, when instead of two triangles, we have a convex \(n_{1}\)-gon and an \(n_{2}\)-gon. First, we will prove that the resulting figure is convex. Let \(A_{1}\) and \(A_{2}\) be points of this figure. Then there exist parallelograms \(O B_{1} A_{1} C_{1}\) and \(... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,291 |
3-3. There are 13 weights, each weighing a whole number of grams. It is known that any 12 of them can be divided into 2 groups of 6 weights each, such that the scales will balance. Prove that all the weights have the same weight. | Solution 3. First, let's prove that all weights are either simultaneously even or odd. Suppose the 12 weights are distributed on two scales, 6 weights on each, such that equilibrium is achieved. Assume a weight \(a\) g is set aside, and one of the weights on the scales weighs \(b\) g, where \(a\) and \(b\) have differe... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 26,292 |
3-4. In an arbitrary (convex - ed. note) hexagon, the midpoints of the sides are connected every other one. Prove that the points of intersection of the medians of the two resulting triangles coincide. | Solution 4. The point of intersection of the medians of each of the considered triangles serves as the center of mass of the hexagon (i.e., the center of mass of the six vertices of the hexagon, into which identical masses are placed).
Part 5. If there are 100 arbitrary (integer - ed. note) numbers, then among them, o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,293 |
3-6. Given a circle and a point outside it; from this point, we travel along a closed broken line consisting of straight line segments tangent to the circle, and end the path at the starting point. The segments of the path along which we approached the center of the circle are taken with a “plus” sign, and the segments... | Problem 6. Let \(A B C D \ldots Y Z\) be the given closed broken line, \(t_{A}, t_{B}, \ldots, t_{Z}\) be the lengths of the tangents to the circle drawn from the vertices of the broken line. According to the sign convention, the algebraic length of the path segment from \(A\) to \(B\) is \(t_{A}-t_{B}\). Therefore, th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,294 |
3-5. Prove that numbers of the form \(2^{n}\) for different positive integers \(n\) can start with any predetermined combination of digits. | Solution 5. Let \(A\) be a given natural number. We will show that a natural number \(n\) can be chosen such that \(10^{m} A<2^{n}<10^{m}(A+1)\), i.e., \(m+\lg A<n \lg 2<m+\lg (A+1)\). An equivalent condition is that there exist natural numbers \(m\) and \(n\) such that \(\lg A<n \lg 2-m<\lg (A+1)\). The number \(\lg 2... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,297 |
3-2. There are 555 weights weighing: 1 g, 2 g, 3 g, 4 g, .. 555 g. Divide them into 3 equal weight piles. | Solution 2. Nine weights with weights \(n, n+1, \ldots, n+8\) can be divided into three equal-weight piles: 1) \( n\), \(n+4, n+8 ; 2) n+1, n+5, n+6 ; 3) n+2, n+3, n+7\). This allows the weights \(1,2, \ldots, 549=61 \cdot 9\) to be divided into three equal-weight piles. The remaining six weights \(550,551, \ldots, 555... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,300 |
3-ча 3. Given 3 circles \(O_{1}, O_{2}, O_{3}\), passing through one point \(O\). The second points of intersection of \(O_{1}\) with \(O_{2}\), \(O_{2}\) with \(O_{3}\), and \(O_{3}\) with \(O_{1}\) are denoted by \(A_{1}, A_{2}\), and \(A_{3}\) respectively. On \(O_{1}\), we take an arbitrary point \(B_{1}\). If \(B_... | Solution 3. Let \(\angle(A B, C D)\) be the directed angle between lines \(A B\) and \(C D\). Then \(\angle\left(A_{3} O, A_{3} B_{1}\right)+\) \(\angle\left(A_{3} B_{3}, A_{3} O\right)=\angle\left(A_{1} O, A_{1} B_{1}\right)+\angle\left(A_{2} B_{3}, A_{2} O\right)=\angle\left(A_{1} O, A_{1} B_{2}\right)+\angle\left(A_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,301 |
3-rd 5. From point \(A\) to other points, one can get there in two ways:
1. Exit immediately and walk.
2. Call a car and, after waiting for a certain amount of time, ride in it.
In each case, the mode of transportation that requires the least time is used. It turns out that
\begin{tabular}{|c|c|}
\hline if the fina... | Solution 5. Answer: 25 min.
Let \(v\) be the pedestrian's speed, \(V\) be the car's speed, and \(T\) be the waiting time for the car. (We will measure speed in km/h and time in hours.) If walking, the time required for 1.2 km and 3 km would be \(1 / v\), \(2 / v\), and \(3 / v\) hours, respectively, and if traveling b... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,303 |
3-ча 1. Let \(A\) be an arbitrary angle, \(B\) and \(C\) be acute angles. Does there always exist an angle \(X\) such that
\[
\sin X=\frac{\sin B \sin C}{1-\cos A \cos B \cos C} ?
\]
(From "Imaginary Geometry" by N. I. Lobachevsky). | Solution 1. Answer: yes, always. By the condition \(\cos B \cos C>0\). Moreover, \(\sin B \sin C+\cos B \cos C=\cos (B-\) \(C) \leqslant 1\) and \(\cos A \leqslant 1\). Therefore, \(\sin B \sin C \leqslant 1-\cos B \cos C \leqslant 1-\cos A \cos B \cos C\) and
\[
0<\frac{\sin B \sin C}{1-\cos A \cos B \cos C} \leqslan... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,304 |
3-2. From two triangular pyramids with a common base, one lies inside the other. Can the sum of the edges of the inner pyramid be greater than the sum of the edges of the outer pyramid? | Solution 2. Answer: Yes, it can. Consider a regular triangular pyramid with base \(B C D\) and vertex \(A\). Let the length of the side of the base be \(\varepsilon\), and the length of the lateral edge be 1. Take a point \(D^{\prime}\) on the side \(A D\) such that \(A D^{\prime}=\varepsilon\). If \(\varepsilon\) is s... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,305 |
3-3. There are 81 weights weighing \(1^{2} g, 2^{2} g, 3^{2} g, \ldots, 81^{2}\) g. Divide them into 3 equal-weight piles. | Solution 3. Let's distribute nine weights with masses \(n^{2},(n+1)^{2}, \ldots,(n+8)^{2}\) into three piles as follows: 1) \(n^{2}\), \((n+5)^{2},(n+7)^{2} ; 2) (n+1)^{2},(n+3)^{2},(n+8)^{2} ; 3) (n+2)^{2},(n+4)^{2},(n+6)^{2}\). The first two piles weigh the same, while the third pile weighs 18 units less. Therefore, ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,306 |
3-4. Solve the equation:
\[
\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1
\] | Solution 4. Answer: \(5 \leqslant x \leqslant 10\). Notice that
\[
\begin{aligned}
& x+3-4 \sqrt{x-1}=(\sqrt{x-1}-2)^{2} \\
& x+8-6 \sqrt{x-1}=(\sqrt{x-1}-3)^{2}
\end{aligned}
\]
Therefore, the original equation can be written as
\[
|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1
\]
(we consider all roots to be positive). Let's co... | 5\leqslantx\leqslant10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,307 |
3-5. Given \(n\) circles: \(O_{1}, O_{2}, \ldots O_{n}\), passing through a common point \(O\). The second points of intersection of \(O_{1}\) with \(O_{2}, O_{2}\) with \(O_{3}, \ldots, O_{3}\) with \(O_{1}\) are denoted by \(A_{1}, A_{2}, \ldots, A_{n}\), respectively. On \(O_{1}\), take an arbitrary point \(B_{1}\).... | Solution 5. Let \(\angle(A B, C D)\) be the directed angle between lines \(A B\) and \(C D\) (measured up to \(\left.180^{\circ}\right)\). Then \(\left.\angle\left(B_{n} A_{n}, A_{n} O\right)=\angle\left(B_{n} A_{n-1}, A_{n-1} O\right)=\angle B_{n-1} A_{n-1}, A_{n-1} O\right)\). Similarly, we obtain \(\angle\left(B_{n-... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,308 |
3-ча 1. In a convex 13-sided polygon, all diagonals are drawn. They divide it into polygons. Let's take among them a polygon with the largest number of sides. What is the maximum number of sides it can have | Solve 1. Answer: 13. From each vertex of the original 13-gon, no more than two diagonals emerge that are sides of the considered polygon. Each diagonal corresponds to two vertices, so the number of sides of the considered polygon does not exceed 13. An example of a regular 13-gon shows that the number of sides of the p... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,309 |
3-ча 2. Prove that
\[
\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \cdots \frac{99}{100}<\frac{1}{10}
\] | Solution 2. Let \(A=\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \ldots \frac{99}{100}\) and \(B=\frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \ldots \frac{98}{99}\). It is clear that \(B>A\) and \(A B=\frac{1}{100}\). Therefore, \(A^{2}<A B=\frac{1}{100}\).
Part 3. A circle is inscribed in a triangle. A square i... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 26,310 |
3-4. On a circle, there are 20 points. These 20 points are connected by 10 (non-intersecting - ed. note) chords. In how many ways can this be done? | Solution 4. Answer: 16796. Let \(a_{n}\) be the number of ways to connect \(2 n\) points on a circle with \(n\) non-intersecting chords. It is clear that \(a_{1}=1\) and \(a_{2}=2\). We will show that
\[
a_{n}=a_{n-1}+a_{n-2} a_{1}+a_{n-3} a_{2}+\cdots+a_{1} a_{n-2}+a_{n-1}.
\]
Fix one of the \(2 n\) points. The chor... | 16796 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,311 |
3-ча 1. In a convex 1950-gon, all diagonals are drawn. They divide it into polygons. Among them, we take the polygon with the largest number of sides. What is the maximum number of sides it can have? | Solution 1. Answer: 1949. The same reasoning as in the solution to problem 1 for grades 7-8 shows that the resulting polygon has no more than 1950 sides, and if the number of its sides is 1950, then exactly two diagonals emanate from each vertex of the original polygon, bounding the resulting polygon. Suppose two diago... | 1949 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,312 |
3-2. The numbers \(1,2,3, \ldots 101\) are written in a row in some order. Prove that 90 of them can be erased so that the remaining 11 are arranged in order of their magnitude (either increasing or decreasing).
(This problem was not solved by any of the participants in the olympiad.) | Solution 2. We will prove that any sequence of \(m n+1\) pairwise distinct numbers contains either an increasing subsequence of \(m+1\) numbers or a decreasing subsequence of \(n+1\) numbers. We associate with each term \(a_{k}\) of the given sequence two numbers \(x_{k}\) and \(y_{k}\), where \(x_{k}\) is the greatest... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 26,313 |
3-4. Can 10 bus routes be laid out in a city and stops be set up on them in such a way that for any 8 routes chosen, there is a stop not lying on any of them, while any 9 routes pass through all stops. | Solution 4. Answer: Yes, it is possible. Let's draw 10 pairwise intersecting lines. Suppose the routes follow these lines, and the stops are the points of intersection of the lines. Any 9 routes pass through all the stops, since each stop lying on the remaining line is intersected by one of the 9 lines corresponding to... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,315 |
3-ча 1. Prove that the polynomial
\[
x^{12}-x^{9}+x^{4}-x+1
\]
is positive for all values of \(x\). | Solve 1. If \(x \leqslant 0\), then \(x^{12},-x^{9}, x^{4},-x \geqslant 0\). If \(0 < x < 1\), then \(x^{12}+x^{9}(1-x^{3})+x^{4}(1-x^{3})+x(1-x^{3})+1>0\). If \(x \geqslant 1\), then \(x^{9}\left(x^{3}-1\right)+x\left(x^{3}-1\right)+1>0\). | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,316 |
3-4. A point inside an isosceles trapezoid is connected to all vertices. Prove that from the four resulting segments, a quadrilateral can be formed that is inscribed in this trapezoid. | Solution 4. Let \(A B C D\) be the given trapezoid (\(A B\) and \(C D\) are its bases), and \(P\) be the given point. Attach to trapezoid \(A B C D\) an equal trapezoid \(A^{\prime} B^{\prime} C^{\prime} D^{\prime}\) such that vertex \(C^{\prime}\) coincides with vertex \(A\), and vertex \(B^{\prime}\) coincides with v... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,318 |
3-5. There is a piece of chain consisting of 60 links, each weighing 1 g. What is the smallest number of links that need to be unbuckled so that from the resulting parts, all weights of 1 g, 2 g, 3 g, \(\ldots, 60\) g can be formed (an unbuckled link also weighs 1 g) | Solution 5. Answer: 3 links. Let's determine the greatest \(n\) for which it is sufficient to break \(k\) links of an \(n\)-link chain so that all weights from 1 to \(n\) can be formed from the resulting parts. If \(k\) links are broken, then any number of links from 1 to \(k\) can be formed from them. But \(k+1\) link... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 26,319 |
3-2. Prove that the first three digits of the quotient
\[
\frac{0.1234567891011 \ldots 4748495051}{0.51504948 \ldots 4321}
\]
are \(0.239\). | Problem 2. Let \(a=0.1234 \ldots 5051\) and \(b=0.5150 \ldots 321\). It is required to prove that \(0.239 b \leqslant a\), \(0.24 \cdot 0.515=0.1236>a\). | 0.239 | Number Theory | proof | Yes | Yes | olympiads | false | 26,321 |
3-4. There is a piece of chain consisting of 150 links, each weighing 1 g. What is the smallest number of links that need to be unbuckled so that from the resulting parts, all weights of 1 g, 2 g, 3 g, \(\ldots, 150\) g can be formed (an unbuckled link also weighs 1 g)? | Solution 4. Answer: 4 links. According to the solution of problem 5 for grades \(7-8\) for a chain consisting of \(n\) links, where \(64 \leqslant n \leqslant 159\), it is sufficient to unfasten 4 links. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 26,323 |
3-ча 3. At the consultation, there were 20 schoolchildren and 20 problems were discussed. It turned out that each of the schoolchildren solved two problems and each problem was solved by two schoolchildren. Prove that it is possible to organize the discussion of the problems in such a way that each schoolchild presents... | Solution 3. Let's start analyzing the problem with an arbitrary student telling one of the problems they have solved. Suppose this student solved problems \(a_{1}\) and \(a_{2}\), and he told problem \(a_{1}\). Then there is exactly one student who also solved problem \(a_{2}\) (and another problem \(a_{3}\)). This stu... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 26,326 |
3-4. The projection of a point \(A\) from a point \(O\) onto a plane \(P\) is the point \(A^{\prime}\) where the line \(O A\) intersects the plane \(P\). The projection of a triangle is the figure consisting of the projections of all its points. What figures can the projection of a triangle be if the point \(O\) does n... | Solution 4. Let \(A B C\) be the given triangle. Consider the complete trihedral angle \(O A B C\) with vertex \(O\), consisting of two trihedral angles (the edges of one angle are the rays \(O A, O B, O C\), and the edges of the other are their extensions). The projection of the triangle \(A B C\) onto the plane \(P\)... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,327 |
3-rd 5. When dividing the polynomial \(x^{1951}-1\) by \(x^{4}+x^{3}+2 x^{2}+x+1\), a quotient and a remainder are obtained. Find the coefficient of \(x^{14}\) in the quotient. | Solve 5. Answer: -1. The equalities \(x^{4}+x^{3}+2 x^{2}+x+1=\left(x^{2}+1\right)\left(x^{2}+x+1\right)\) and \(x^{12}-1=(x-1)\left(x^{2}+x+\right.\)
1) \(\left(x^{3}+1\right)\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)\) show that
\[
\begin{aligned}
x^{4}+x^{3}+2 x^{2}+x+1 & =\frac{x^{12}-1}{(x-1)\left(x^{3}+1\ri... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,328 |
3-ча 1. All edges of a triangular pyramid are equal to \(a\). Find the maximum area that the orthogonal projection of this pyramid onto a plane can have. | Solution 1. Answer: \(a^{2} / 2\). The projection of the tetrahedron can be a triangle or a quadrilateral. In the first case, it is the projection of one of the faces, so its area does not exceed \(\sqrt{3} a^{2} / 4\). In the second case, the diagonals of the quadrilateral are projections of the edges of the tetrahedr... | ^{2}/2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,329 |
3-2. There are several numbers, each of which is less than 1951. The least common multiple of any two of them is greater than 1951. Prove that the sum of the reciprocals of these numbers is less than 2. | Solution 2. Let \(a_{1}, \ldots, a_{n}\) be the given numbers. The number of terms in the sequence \(1,2,3, \ldots, 1951\) that are divisible by \(a_{k}\) is \(\left[\frac{1951}{a_{k}}\right]\). By the condition, the least common multiple of any two of the numbers \(a_{1}, \ldots, a_{n}\) is greater than 1951, so there... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,330 |
3-3. A bus route contains 14 stops (including the two terminal stops). The bus can carry no more than 25 passengers at any time. Prove that during the bus trip from one end to the other
a) there will be eight different stops \(A_{1}, B_{1}, A_{2}, B_{2}, A_{3}, B_{3}, A_{4}, B_{4}\), such that no passenger travels fro... | 3. a) We will only consider the passengers who are in the bus when it travels from the 7th stop to the 8th. These will be exactly the passengers traveling from stop number \(i \leqslant 7\) to stop number \(j \geqslant 8\). Let's take a \(7 \times 7\) square, the rows of which are numbered by the numbers \(1, \ldots, 7... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 26,331 |
3-4. A circle has the property that a regular triangle can be moved inside it so that each vertex of the triangle describes this circle. Find a closed non-intersecting curve, different from a circle, inside which a regular triangle can also be moved so that each of its vertices describes this curve. | Solution 4. Take a circle whose radius is equal to the side of an equilateral triangle, and cut off two segments from it, the chords of which subtend an angle of \(120^{\circ}\). Combine these two segments into a figure by placing their chords against each other. The boundary of this figure has the required property. I... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,332 |
3-ча 1. In \(\triangle A B C\), an inscribed circle touches its sides at points \(L, M\) and \(N\). Prove that \(\Delta L M N\) is always acute (regardless of the type of \(\triangle A B C\)). | Let \(L, M\) and \(N\) lie on sides \(B C, C A\) and \(A B\) respectively. Triangles \(B N L\) and \(C L M\) are isosceles; the angles at their bases are \(\frac{1}{2}\left(180^{\circ}-\angle B\right)\) and \(\frac{1}{2}\left(180^{\circ}-\angle C\right)\). Therefore, \(\angle N L M=\) \(\frac{1}{2}\left(180^{\circ}-\an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,333 |
3-ча 2. Prove the identity
\[
\begin{aligned}
& (a x+b y+c z)^{2}+(b x+c y+a z)^{2}+(c x+a y+b z)^{2}= \\
& =(c x+b y+a z)^{2}+(b x+a y+c z)^{2}+(a x+c y+b z)^{2}
\end{aligned}
\] | Solve 2. It is easy to check that both expressions are equal
\[
\left(x^{2}+y^{2}+z^{2}\right)\left(a^{2}+b^{2}+c^{2}\right)+2(x y+y z+z x)(a b+b c+c a)
\] | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,334 |
3-3. If all 6 faces of a parallelepiped are equal parallelograms, then they are rhombuses. Prove it. | Solution 3. Let the edges of length \(a, b, c\) extend from a vertex of the parallelepiped. Then one face is a parallelogram with sides \(a\) and \(b\), and another face is a parallelogram with sides \(b\) and \(c\). These parallelograms are equal, so \(a=c\). Similarly, it can be proven that \(a=b=c\). | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,335 |
3-4. Two people \(A\) and \(B\) need to travel from point \(M\) to point \(N\), which is 15 km away from \(M\). On foot, they can travel at a speed of 6 km/h. In addition, they have a bicycle that can be ridden at a speed of 15 km/h. \(A\) sets out on foot, while \(B\) rides the bicycle until meeting pedestrian \(C\), ... | To solve 4. For \(A\) and \(B\) to arrive at point \(N\), they must walk the same distance \(x\) and cycle the same distance \(15-x\). Then \(C\) must also walk the distance \(x\) before meeting \(B\). Therefore, \(C\) will cycle \(15-2x\) before meeting \(A\), and after that, \(A\) will cycle \(15-x\). In total, \(A\)... | \frac{3}{11} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 26,336 |
3-ча 1. Prove that if the orthocenter divides the altitudes of a triangle in the same ratio, then the triangle is equilateral. | Solution 1. Let \(H\) be the intersection point of the altitudes \(A A_{1}, B B_{1}\), and \(C C_{1}\) of triangle \(A B C\). By the given condition, \(A_{1} H \cdot B H = B_{1} H \cdot A H\). On the other hand, since points \(A_{1}\) and \(B_{1}\) lie on the circle with diameter \(A B\), we have \(A_{1} H \cdot A H = ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,337 |
3-2. Prove the identity
\[
\begin{aligned}
& (a x+b y+c z+d u)^{2}+(b x+c y+d z+a u)^{2}+(c x+d y+a z+b u)^{2}+ \\
& +(d x+a y+b z+c u)^{2}= \\
& =(d x+c y+b z+a u)^{2}+(c x+b y+a z+d u)^{2}+(b x+a y+d z+c u)^{2}+ \\
& +(a x+d y+c z+b u)^{2}
\end{aligned}
\] | Solve 2. It is easy to check that both expressions are equal
\[
\left(x^{2}+y^{2}+z^{2}+u^{2}\right)\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+2(x y+y z+z u+u x)(a b+b c+c d+d a)+4(x z+y u)(a c+b d)
\] | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,338 |
3-4. Two people \(A\) and \(B\) need to get from point \(M\) to point \(N\) as quickly as possible, with \(N\) located 15 km from \(M\). On foot, they can travel at a speed of 6 km/h. Additionally, they have a bicycle that can be ridden at a speed of 15 km/h. \(A\) sets out on foot, while \(B\) rides the bicycle until ... | Solution 4. For \(A\) and \(B\) to spend the least amount of time on the road, they must arrive at \(N\) simultaneously, i.e., they must walk the same distances. Indeed, let \(A\) walk a distance of \(x\) and ride a bicycle for \(15-x\), and let \(B\) walk \(y\) and ride \(15-y\). Then the time spent by \(A\) is \(\fra... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 26,339 | |
3-ча 1. Given a geometric progression with an integer denominator (not equal to 0 and -1). Prove that the sum of any number of arbitrarily chosen terms of this progression cannot equal any term of this progression. | Solution 1. Each term of a geometric progression is represented in the form \(a q^{n}, n \geqslant 0\). The case when \(q=1\) is obvious, so we will assume that \(q \neq 1\). Suppose there exist distinct non-negative integers \(k_{1}, k_{2}, \ldots, k_{m+1}(m \geqslant 2)\), for which
\[
a q^{k_{1}}+a q^{k_{2}}+\cdots... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,340 |
3-5. Given a sequence of integers constructed as follows:
\(a_{1}\) - an arbitrary three-digit number,
\(a_{2}\) - the sum of the squares of its digits,
\(a_{3}\) - the sum of the squares of the digits of \(a_{2}\), and so on. Prove that in the sequence \(a_{1}, a_{2}, a_{3}, \ldots\) there will inevitably be either... | Solution 5. First, note that \(a_{2} \leqslant 9^{2} \cdot 3=243\), and therefore, \(a_{3} \leqslant 2^{2}+9^{2} \cdot 2=166\). If \(100 \leqslant a_{3} \leqslant 166\), then \(a_{4} \leqslant 1+6^{2}+9^{2}=118\), and if \(100 \leqslant a_{4} \leqslant 166\), then \(a_{5} \leqslant 2+64<100\). Therefore, it is sufficie... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,344 |
3-ча 2. Prove that for an integer \(n \geqslant 2\) and \(|x| < 1\),
\[
(1-x)^{n}+(1+x)^{n} < 2^{n}
\] | Solution 2. We apply induction on \(n\). For \(n=2\), we get \((1-x)^{2}+(1+x)^{2}=2\left(1+x^{2}\right)<4\). Now suppose that \((1-x)^{n}+(1+x)^{n}<2^{n}\). Then
\[
(1-x)^{n+1}+(1+x)^{n+1}<\left((1-x)^{n}+(1+x)^{n}\right)((1-x)+(1+x))<2^{n} \cdot 2=2^{n+1}
\] | proof | Inequalities | proof | Yes | Yes | olympiads | false | 26,346 |
3-4. If for any positive \(p\) all roots of the equation
\[
a x^{2}+b x+c+p=0
\]
are real and positive, then the coefficient \(a\) is zero. Prove. | Solution 4. Suppose that \(a>0\). Then for large positive \(p\) the discriminant \(D=b^{2}-4 a c-4 a p\) is negative, so the given equation has no real roots at all. Suppose that \(a<0\). Then for large positive \(p\) the product of the roots, which is \(\frac{c+p}{a}\), is negative.
\section*{Second Round}
\section*... | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,348 |
3-ча 1. Solve the system of fifteen equations with fifteen unknowns:
\[
\left\{\begin{aligned}
1-x_{1} x_{2} & =0 \\
1-x_{2} x_{3} & =0 \\
1-x_{3} x_{4} & =0 \\
\cdots & \cdots \\
1-x_{14} x_{15} & =0 \\
1-x_{15} x_{1} & =0
\end{aligned}\right.
\] | Solve 1. Answer: \(x_{1}=x_{2}=\cdots=x_{15}= \pm 1\). Clearly, \(x_{2} \neq 0\), so from the first and second equations we get \(x_{1}=x_{3}\). From the second and third equations we get \(x_{2}=x_{4}\) and so on. Moreover, from the first and last equations we get \(x_{15}=x_{2}\). In the end, we get \(x_{1}=x_{3}=\cd... | x_{1}=x_{2}=\cdots=x_{15}=\1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,349 |
3-2. From point \(C\), tangents \(C A\) and \(C B\) are drawn to circle \(O\). From an arbitrary point \(N\) on the circle, perpendiculars \(N D, N E, N F\) are dropped to the lines \(A B, C A\), and \(C B\) respectively. Prove that \(N D\) is the mean proportional between \(N E\) and \(N F\). | Solution 2. Right triangles \(N B D\) and \(N A E\), \(N B F\) and \(N A D\) are similar, since \(\angle N B D=\) \(\angle N A E\) and \(\angle N B F=\angle N A D\). Therefore, \(N B: N A=N D: N E\) and \(N B: N A=N F: N D\), which means \(N D: N E=N F: N D\), i.e., \(N D^{2}=N E \cdot N F\). | ND^{2}=NE\cdotNF | Geometry | proof | Yes | Yes | olympiads | false | 26,354 |
3-ча 1. Solve the system of equations:
\[
\left\{\begin{aligned}
1-x_{1} x_{2} & =0 \\
1-x_{2} x_{3} & =0 \\
1-x_{3} x_{4} & =0 \\
\cdots & \cdots \\
1-x_{n-1} x_{n} & =0 \\
1-x_{n} x_{1} & =0
\end{aligned}\right.
\] | Solve 1. Answer: \(x_{1}=x_{2}=\cdots=x_{n}= \pm 1\) for odd \(n\), \(x_{1}=x_{3}=\cdots=x_{n-1}=a\) and \(x_{2}=x_{4}=\) \(\cdots=x_{n}=\frac{1}{a}(a \neq 0)\) for even \(n\). For odd \(n\), the solution is essentially provided in the solution to problem 1 for 7th grade. For even \(n\), we similarly obtain \(x_{1}=x_{... | x_{1}=x_{2}=\cdots=x_{n}=\1foroddn,\,x_{1}=x_{3}=\cdots=x_{n-1}=x_{2}=x_{4}=\cdots=x_{n}=\frac{1}{}(\neq0)forevenn | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,357 |
3-4. In isosceles \(\triangle A B C \angle A B C=20^{\circ}\). On the equal sides \(C B\) and \(A B\) are taken points \(P\) and \(Q\) respectively, such that \(\angle P A C=50^{\circ}\) and \(\angle Q C A=60^{\circ}\). Prove that \(\angle P Q C=30^{\circ}\). | Solution 4. Choose a point \(R\) on side \(B C\) such that \(\angle C A R=60^{\circ}\). Let \(S\) be the intersection point of lines \(C Q\) and \(A R\) (Fig.???). It is sufficient to prove that \(Q P\) is the bisector of angle \(S Q R\), which is equal to \(60^{\circ}\). Simple angle calculations show that \(\angle A ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,359 |
3-ча 5. 200 students are lined up in a rectangle with 10 people in each transverse row and 20 people in each longitudinal row. In each longitudinal row, the tallest student is selected, and then the shortest of the 10 selected people is chosen. On the other hand, in each transverse row, the shortest student is selected... | Problem 5. Let \(A\) be the shortest among the tall, and \(B\) be the tallest among the short. Let \(C\) stand at the intersection of the longitudinal row in which \(A\) stands and the transverse row in which \(B\) stands. Then \(A\) is not shorter than \(C\) and \(C\) is not shorter than \(B\), so \(A\) is not shorter... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 26,360 |
3-rd 1. Prove that the sum
\[
\cos 32 x+a_{31} \cos 31 x+a_{30} \cos 30 x+\cdots+a_{1} \cos x
\]
takes both positive and negative values. | Solution 1. Suppose that the sum
\[
\cos 32 x+a_{31} \cos 31 x+a_{30} \cos 30 x+\cdots+a_{1} \cos x
\]
takes only positive values for all \(x\). By replacing \(x\) with \(x+\pi\), we get that the expression
\[
\cos 32 x-a_{31} \cos 31 x+a_{30} \cos 30 x-\cdots+a_{2} \cos 2 x-a_{1} \cos x
\]
takes positive values fo... | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,361 |
3-2. Find the smallest number, written with only ones, that is divisible by the number composed of a hundred threes ( \(333 \ldots 33\) ). | Solve 2. Answer: \(\underbrace{11 \ldots 1}_{300}\). The number \(a_{n}=\underbrace{11 \ldots 1}_{n}\) is divisible by \(\underbrace{33 \ldots 3}_{100}\) if and only if \(n\) is divisible by 3 and \(a_{n}\) is divisible by \(a_{100}\). We will show that \(a_{n}\) is divisible by \(a_{m}\) if and only if \(n\) is divisi... | \underbrace{11\ldots1}_{300} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,364 |
3-4. Prove that for any positive integer \(n\) the number \(n^{2}+8 n+15\) is not divisible by \(n+4\). | Solution 4. It is enough to notice that \(n^{2}+8 n+15=(n+4)^{2}-1\).
\section*{8th grade} | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,366 |
3-ча 1. Three circles touch each other pairwise. A circle is drawn through the three points of tangency. Prove that this circle is perpendicular to each of the three original circles. (The angle between two circles at their point of intersection is defined as the angle formed by their tangents at that point.) | Let \(A, B, C\) be the points of tangency, \(A_{1}, B_{1}\) and \(C_{1}\) be the centers of the given circles, and let points \(A, B\) and \(C\) lie on segments \(B_{1} C_{1}, C_{1} A_{1}\) and \(A_{1} B_{1}\) respectively. Then \(A_{1} B=A_{1} C, B_{1} A=B_{1} C\) and \(C_{1} A=C_{1} B\). From this it follows that \(A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,367 |
3-4. Prove the inequality
\[
\frac{2-\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}{2-\underbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}_{n-1 \text { times }}}>\frac{1}{4}
\] | Solution 4. Let \(a=\underbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}_{n-1}\). Then \(\underbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}_{n}=\sqrt{2+a}\).
Thus, we need to prove that
\[
\frac{2-\sqrt{2+a}}{2-a}>\frac{1}{4}
\]
By induction on \(n\), it is easy to prove that \(a2-a \\
6+a & >4 \sqrt{2+a}
\end{array}
\]
Afte... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 26,368 |
3-ча 2. \(A B\) and \(A_{1} B_{1}\) are two skew segments. \(O\) and \(O_{1}\) are their respective midpoints. Prove that the segment \(O O_{1}\) is less than half the sum of the segments \(A A_{1}\) and \(B B_{1}\). | Solution 2. Add the equations \(\overrightarrow{A A_{1}}=\overrightarrow{A O}+\overrightarrow{O O_{1}}+\overrightarrow{O_{1} A_{1}}\) and \(\overrightarrow{B B_{1}}=\overrightarrow{B O}+\overrightarrow{O O_{1}}+\overrightarrow{O_{1} B_{1}}\). Considering that \(\overrightarrow{A O}+\overrightarrow{B O}=\overrightarrow{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,370 |
3-ча 1. Prove that the greatest common divisor of the sum of two numbers and their least common multiple is equal to the greatest common divisor of the numbers themselves. | Solve 1. The least common multiple of numbers \(a\) and \(b\) includes only those prime divisors that are present in \(a\) and \(b\). Only they can be included in the greatest common divisor of the sum and the least common multiple. Therefore, it is sufficient to track the degree of each prime factor separately. Let \(... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,373 |
3-ча 4. A thousand points are vertices of a convex thousand-sided polygon, inside which there are another five hundred points such that no three of the five hundred lie on the same line. The thousand-sided polygon is cut into triangles in such a way that all the 1500 points are vertices of the triangles and these trian... | Solve 4. Answer: 1998. The sum of all angles of the obtained triangles is equal to the sum of the angles of a 1000-gon and 500 angles of \(360^{\circ}\), corresponding to 500 internal points. Therefore, the sum of the angles of the triangles is \(998 \cdot 180^{\circ} + 500 \cdot 360^{\circ} = (998 + 2 \cdot 500) 180^{... | x_{1}=x_{3}=x_{5}=1,x_{2}=x_{4}=-1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,376 |
3-ча 1. \(a, b, c\) and \(d\) are the lengths of the consecutive sides of a quadrilateral. Let \(S\) be its area. Prove that
\[
S \leqslant \frac{1}{4}(a+b)(c+d)
\] | Solve 1. Cut the quadrilateral into two triangles with sides \(a\) and \(d, b\) and \(c\). As a result, we get \(2 S \leqslant a d + b c\). It remains to prove the inequality \(2 S \leqslant a c + b d\). To do this, proceed as follows. Cut off the triangle with sides \(c\) and \(d\), flip it, and attach it to the trian... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 26,377 |
3-ча 2. 1953 digits are written in a circle. It is known that if you read these digits clockwise, starting from a certain specific point, the resulting 1953-digit number is divisible by 27. Prove that if you start reading clockwise from any other point, the resulting number will also be divisible by 27. | Solution 2. We will prove the required statement by replacing 1953 with an arbitrary natural number \(n\), divisible by 3. Let the number \(a_{0}+a_{1} \cdot 10+a_{2} \cdot 10^{2}+\cdots+a_{n} \cdot 10^{n-1}=a_{0}+10 a\) be divisible by 27. It is sufficient to prove that then the number \(a_{1}+a_{2} \cdot 10+\cdots+a_... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,378 |
3-3. On a circle, points \(A_{1}, A_{2}, \ldots, A_{16}\) are given. We will construct all possible convex polygons whose vertices are among the points \(A_{1}, A_{2}, \ldots, A_{16}\). We will divide these polygons into two groups. The first group will include all polygons where \(A_{1}\) is a vertex. The second group... | Solve 3. Answer: in the second. Each polygon not containing point \(A_{1}\) can be associated with a polygon containing point \(A_{1}\) by simply adding \(A_{1}\) to its vertices. The reverse operation (removing vertex \(A_{1}\)) is only possible for polygons with no fewer than 4 vertices. | inthe | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,379 |
3-4. In a plane, there are \(n\) cogwheels arranged such that the first wheel is engaged with the second by its teeth, the second with the third, and so on. Finally, the last wheel is engaged with the first. Can the wheels in such a system rotate? | Solution 4. Answer: they can when \(n\) is even, they cannot when \(n\) is odd. See the solution to problem 3 for 7th grade. | they\can\when\n\is\even,\they\cannot\when\n\is\odd | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 26,380 |
3-5. Solve the system
\[
\left\{\begin{aligned}
x_{1}+2 x_{2}+2 x_{3}+\cdots+2 x_{100} & =1 \\
x_{1}+3 x_{2}+4 x_{3}+\cdots+4 x_{100} & =2 \\
x_{1}+3 x_{2}+5 x_{3}+\cdots+6 x_{100} & =3 \\
\cdots & \cdots \\
x_{1}+3 x_{2}+5 x_{3}+\cdots+199 x_{100} & =100
\end{aligned}\right.
\] | Solve 5. Answer: \(x_{1}=x_{3}=\cdots=x_{99}=-1, x_{2}=x_{4}=\cdots=x_{100}=1\). The solution is similar to the solution of problem 5 for 7th grade.
\section*{9th grade} | x_{1}=x_{3}=\cdots=x_{99}=-1,x_{2}=x_{4}=\cdots=x_{100}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,381 |
3-2. In the plane, given \(\Delta A_{1} A_{2} A_{3}\) and a line \(l\) outside it, forming angles \(\alpha_{3}, \alpha_{1}, \alpha_{2}\) with the extensions of the sides \(A_{1} A_{2}, A_{2} A_{3}, A_{3} A_{1}\) respectively. Through the points \(A_{1}, A_{2}, A_{3}\), lines are drawn forming angles \(2 d-\alpha_{1}, 2... | Solution 2. Introduce a coordinate system on the plane, choosing the line \(l\) as the \(x\)-axis. Let \(\left(a_{1}, b_{1}\right)\), \(\left(a_{2}, b_{2}\right),\left(a_{3}, b_{3}\right)\) be the coordinates of the vertices \(A_{1}, A_{2}, A_{3}\). The line \(A_{2} A_{3}\) is given by the equation
\[
\frac{x-a_{2}}{a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,382 |
3-ча 3. Given the equations \(a x^{2}+b x+c=0\) (1) and \(-a x^{2}+b x+c\) (2). Prove that if \(x_{1}\) and \(x_{2}\) are any roots of equations (1) and (2), respectively, then there exists a root \(x_{3}\) of the equation \(\frac{a}{2} x^{2}+b x+c=0\) such that either \(x_{1} \leqslant x_{3} \leqslant x_{2}\) or \(x_{... | Solve 3. When \(x=x_{1}\) and when \(x=x_{2}\), the quadratic polynomial \(\frac{a}{2} x^{2}+b x+c\) takes the values \(-a x_{1}^{2} / 2\) and \(3 a x_{2}^{2} / 2\). These values have different signs, so one of the roots of the quadratic polynomial is located between \(x_{1}\) and \(x_{2}\). | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,383 |
3-5. Cut a cube into three equal pyramids. | Solution 5. One of the vertices of the cube can be taken as the apex of the pyramids, and the three faces of the cube that do not contain this vertex can serve as their bases.
\section*{10th Grade} | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,385 |
3-ча 1. Find the roots of the equation
\[
1-\frac{x}{1}+\frac{x(x-1)}{2!}-\cdots+(-1)^{n} \frac{x(x-1) \ldots(x-n+1)}{n!}=0
\] | Solve 1. Answer: \(1,2, \ldots, n\). The equality
\[
1-\frac{k}{1}+\frac{k(k-1)}{2!}-\cdots+(-1)^{k} \frac{k(k-1) \ldots 2 \cdot 1}{k!}=(1-1)^{k}=0
\]
shows that the numbers \(k=1,2, \ldots, n\) are roots of the given equation. The equation cannot have more than \(n\) roots, since its degree is \(n\). | 1,2,\ldots,n | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,386 |
3-ча 3. Let \(x_{0}=10^{9}, x_{n}=\frac{x_{n-1}^{2}+2}{2 x_{n-1}}\). Prove that \(0<x_{36}-\sqrt{2}<10^{-9}\). | Solution 3. From the identity
\[
x_{n}-\sqrt{2}=\frac{x_{n-1}^{2}+2}{2 x_{n-1}}-\sqrt{2}=\frac{1}{2 x_{n-1}}\left(x_{n-1}-\sqrt{2}\right)^{2}
\]
we immediately get that \(x_{n}-\sqrt{2}>0\). Moreover, considering that \(\frac{x_{n-1}-\sqrt{2}}{x_{n-1}}1\), we obtain
\[
x_{n}-\sqrt{2}<\frac{\left(x_{n-1}-\sqrt{2}\rig... | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,387 |
3-4. Do there exist integers \(m\) and \(n\) satisfying the equation \(m^{2}+1954=n^{2}\) ? | Solution 4. Answer: no, they do not exist. Suppose that \(n^{2}-m^{2}=1954\). If one of the numbers \(m\) and \(n\) is even and the other is odd, then the number \(n^{2}-m^{2}\) is odd. Therefore, both numbers \(m\) and \(n\) are either even or odd. It is easy to check that in this case the number \(n^{2}-m^{2}\) is di... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,391 |
3-5. Determine the maximum value of the ratio of a three-digit number to the number equal to the sum of the digits of this number. | Solve 5. Answer: 100. The sum of the digits of a three-digit number \(100a + 10b + c\) is \(a + b + c\). It is clear that \(\frac{100a + 10b + c}{a + b + c} \leqslant 100\). Moreover, for the number 100, the given ratio is 100.
\section*{8th Grade} | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,392 |
3-ча 1. Prove that if
\[
\begin{aligned}
x_{0}^{4}+a_{1} x_{0}^{3}+a_{2} x_{0}^{2}+a_{3} x_{0}+a_{4} & =0 \\
4 x_{0}^{3}+3 a_{1} x_{0}^{2}+2 a_{2} x_{0}+a_{3} & =0
\end{aligned}
\]
then \(x^{4}+a_{1} x^{3}+a_{2} x^{2}+a_{3} x+a_{4}\) is divisible by \(\left(x-x_{0}\right)^{2}\). | Let \(f(x)=x^{4}+a_{1} x^{3}+a_{2} x^{2}+a_{3} x+a_{4}\). By the condition \(f\left(x_{0}\right)=f^{\prime}\left(x_{0}\right)=0\). Therefore, \(x_{0}\) is a double root of the polynomial \(f(x)\), i.e., the polynomial \(f(x)\) is divisible by \(\left(x-x_{0}\right)^{2}\). | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,395 |
3-4. Given a triangle \(A B C\). Let \(A_{1}, B_{1}, C_{1}\) be the points of intersection of the lines \(A S, B S, C S\) with the sides \(B C, C A, A B\) of the triangle, respectively, where \(S\) is an arbitrary internal point of the triangle \(A B C\). Prove that, in at least one of the quadrilaterals \(A B_{1} S C_... | Solution 4. Suppose that in each of the obtained quadrilaterals \(A B_{1} S C_{1}, C_{1} S A_{1} B, A_{1} S B_{1} C\) at least one of the angles at each pair of vertices \(C_{1}\) and \(B_{1}, C_{1}\) and \(A_{1}, A_{1}\) and \(B_{1}\) is acute. For example, in the quadrilateral \(A B_{1} S C_{1}\), the angle at vertex... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,397 |
3-ча 1. Find all real solutions of the equation
\[
x^{2}+2 x \sin (x y)+1=0
\] | Solve 1. Answer: \(x= \pm 1, y=-\frac{\pi}{2}+k \pi\).
If we consider the given equation as a quadratic equation in terms of \(x\), then its discriminant will be equal to \(4\left(\sin ^{2}(x y)-1\right)\). The discriminant must be non-negative, so \(\sin ^{2}(x y) \geqslant 1\), i.e., \(\sin (x y)= \pm 1\). The solut... | \1,-\frac{\pi}{2}+k\pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,399 |
3-ча 2. See problem 2 for 9th grade.
3-ча 3. Given 100 numbers \(a_{1}, a_{2}, a_{3}, \ldots, a_{100}\), satisfying the conditions:
\[
\begin{array}{r}
a_{1}-4 a_{2}+3 a_{3} \geqslant 0, \\
a_{2}-4 a_{3}+3 a_{4} \geqslant 0, \\
a_{3}-4 a_{4}+3 a_{5} \geqslant 0, \\
\ldots \\
\ldots \\
\ldots \\
a_{99}-4 a_{100}+3 a_{... | Solve 3. Answer: \(a_{1}=a_{2}=\cdots=a_{100}=1\). See the solution to problem 3 for 9th grade. | a_{1}=a_{2}=\cdots=a_{100}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,400 |
3-ча 1. A square \(17 \times 17\) is cut out from a grid paper. The cells of the square are arbitrarily filled with the numbers \(1,2,3, \ldots, 70\), with one and only one number in each cell.
Prove that there exist four different cells with centers at points \(A, B, C, D\) such that \(A B=C D, A D=B C\) and the sum ... | Solution 1. Consider all possible pairs of cells that are symmetric with respect to the center of the square. The number of such pairs is \(\left(17^{2}-1\right) / 2=144\). The sum of the numbers written in two cells can be equal to 2, \(3, \ldots, 140\). Therefore, there will be two pairs of cells symmetric with respe... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 26,404 |
3-3. Given the number
\[
H=2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 \cdot 37
\]
(the product of prime numbers). Let \(1, 2, 3, 5, 6, 7, 10, 11, 13, 14, \ldots, H\) be all its divisors, listed in ascending order. Below the sequence of divisors, write a sequence of +1 and... | Solution 3. We will call a divisor "even" if it can be decomposed into an even number of prime factors (1 is considered an "even" divisor), and "odd" if it can be decomposed into an odd number of prime factors. We will prove by induction on \(k\) that the number \(N_{k}=2 \cdot 3 \cdot 5 \cdot 7 \cdots \cdots p_{k}\) (... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,406 |
3-5. How many axes of symmetry can a heptagon have? | Solution 5. Answer: 0.1 or 7.
The axis of symmetry of a heptagon must pass through one of its vertices (the other vertices are divided into pairs of symmetric vertices). Suppose the heptagon has an axis of symmetry. Then it has three pairs of equal angles and three pairs of equal sides. The second axis of symmetry can... | 0or7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,407 |
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