problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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3-ча 1. On two rays \(l_{1}\) and \(l_{2}\), emanating from point \(O\), segments \(O A_{1}\) and \(O B_{1}\) are laid out on ray \(l_{1}\) and \(O A_{2}\) and \(O B_{2}\) on ray \(l_{2}\); in this case \(\frac{O A_{1}}{O A_{2}} \neq \frac{O B_{1}}{O B_{2}}\).
Determine the geometric locus of points \(S\) of intersect... | Solution 1. Applying Menelaus' theorem to triangle \(S A_{2} B_{2}\) and line \(l_{1}\), we get
\[
\frac{S A_{1}}{A_{2} A_{1}} \cdot \frac{A_{2} O}{B_{2} O} \cdot \frac{B_{2} B_{1}}{S B_{1}}=1
\]
When ray \(l_{2}\) rotates, the ratios \(A_{2} O: B_{2} O\) and \(B_{2} B_{1}: A_{2} A_{1}\) remain constant, so the ratio... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,408 |
3-2. Given four lines \(m_{1}, m_{2}, m_{3}, m_{4}\), intersecting at a single point \(O\). Through an arbitrary point \(A_{1}\) on line \(m_{1}\), we draw a line parallel to line \(m_{4}\), until it intersects line \(m_{2}\) at point \(A_{2}\), through \(A_{2}\) we draw a line parallel to \(m_{1}\), until it intersect... | Solution 2. Let \(\vec{a}=\overrightarrow{O A_{1}}, \vec{b}=\overrightarrow{O A_{2}}, \vec{c}=\overrightarrow{O A_{3}}, \vec{d}=\overrightarrow{O A_{4}}\). We express these vectors through \(e_{1}=\vec{a}\) and \(e_{2}=\vec{d}\). As a result, we get
\[
\begin{aligned}
\vec{b} & =e_{1}+\lambda_{1} e_{2} \\
\vec{c} & =\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,409 |
3-ча 5. Given a segment \(O A\). From the end of the segment \(A\), 5 segments \(A B_{1}, A B_{2}, A B_{3}, A B_{4}, A B_{5}\) extend. From each point \(B_{i}\), five new segments can extend, or no new segments, and so on.
Can the number of free ends of the constructed segments be equal to \(1001\)?
By a free end of ... | Solution 5. Answer: Yes, it can. When five segments are drawn from the end of a segment, 5 new free ends appear and one old end disappears. As a result, the number of free ends increases by 4. Therefore, if the five segments are drawn \(k\) times, the number of free ends is \(4 k+1\). For \(k=250\), we get the required... | 1001 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,410 |
3-4. Does there exist such an \(n\) that \(n^{2}+n+1\) is divisible by 1955? | Solution 4. Answer: No, it does not exist. The number \(n^{2}+n\) when divided by 5 gives a remainder of either 0, 1, or 2. Therefore, \(n^{2}+n+1\) is not divisible by 5, and thus, \(n^{2}+n+1\) is not divisible by 1955. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,417 |
3-ча 1. \(2^{n}=10 a+b\). Prove that if \(n>3\), then \(a b\) is divisible by \(6 . n, a\) and \(b\) are integers, \(b<10\). | Solution 1. Since \(2^{4}=16\), the number \(2^{4 k}\) ends in 6. Accordingly, the numbers \(2^{4 k+1}, 2^{4 k+2}, 2^{4 k+3}\) end in \(2,4,8\). For numbers of the form \(2^{4 k}\), the required statement is obvious, since \(b=6\). Note also that the number \(b\) is always even. Therefore, it is sufficient to check tha... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,419 |
3-5. On a circle, four points \(A, B, C, D\) are given. Through each pair of adjacent points, a circle is drawn. The second points of intersection of the adjacent circles are denoted by \(A_{1}, B_{1}, C_{1}, D_{1}\). (Some of them may coincide with the original points.) Prove that \(A_{1}, B_{1}, C_{1}, D_{1}\) lie on... | Solution 5. Let's use the properties of directed angles between lines:
\[
\begin{aligned}
\angle\left(B_{1} A_{1}, A_{1} D_{1}\right) & =\angle\left(B_{1} A_{1}, A_{1} A\right)+\angle\left(A A_{1}, A_{1} D_{1}\right)= \\
& =\angle\left(B_{1} B, B A\right)+\angle\left(A D, D D_{1}\right) \\
\angle\left(B_{1} C_{1}, C_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,422 |
3-rd 1. The numbers \(1,2, \ldots, k^{2}\) are arranged in a square table
\[
\begin{array}{cccc}
1, & 2, & \ldots, & k \\
k+1, & k+2, & \ldots, & 2 k \\
\ldots & \ldots, & \ldots & \ldots \\
(k-1) k+1, & \ldots, & \ldots, & k^{2}
\end{array}
\]
An arbitrary number is written down, after which the row and column conta... | Solve 1. Answer: \(\frac{k\left(k^{2}+1\right)}{2}\).
Let's write the given table as
\[
\begin{array}{cccc}
k \cdot 0+1, & k \cdot 0+2, & \ldots, & k \cdot 0+k \\
k \cdot 1+1, & k \cdot 1+2, & \ldots, & k \cdot 1+k \\
\ldots & \ldots & \ldots, & \ldots \\
(k-1) k+1, & (k-1) k+2 & \ldots, & (k-1) k+k
\end{array}
\]
E... | \frac{k(k^{2}+1)}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,423 |
3-2. Find the geometric locus of the midpoints of segments with endpoints on two different non-intersecting circles, one lying outside the other. | Solution 2. Let \(S_{1}\) and \(S_{2}\) be the given circles, \(O_{1}\) and \(O_{2}\) their centers. Consider the circle \(S_{2}^{\prime}\), which is obtained from the circle \(S_{2}\) by translation by the vector \(\overrightarrow{O_{2} \overrightarrow{O_{1}}}\); the center of this circle coincides with the center of ... | \frac{R-r}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,424 |
3-cha 4. \(p\) prime numbers \(a_{1}, a_{2}, \ldots, a_{p}\) form an increasing arithmetic progression and \(a_{1}>p\). Prove that if \(p\) is a prime number, then the common difference of the progression is divisible by \(p\). | Solution 4. Consider the remainders of the numbers \(a_{1}, \ldots, a_{p}\) when divided by \(p\). These numbers are prime and all are strictly greater than \(p\), so none of them are divisible by \(p\). Thus, we have \(p\) distinct remainders, all different from \(p\). Therefore, if two numbers \(a_{i}\) and \(a_{j}\)... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,426 |
3-ча 1. A square table with \(n^{2}\) cells is filled with numbers from 1 to \(n\) such that each row and each column contains all these numbers. If \(n\) is odd and the table is symmetric with respect to the diagonal running from the top left corner to the bottom right, then all these numbers \(1,2,3, \ldots, n\) will... | Solve 1. The table is symmetric with respect to the diagonal, so each number located outside the diagonal corresponds to an equal number in the symmetric position. Therefore, outside the diagonal, there is an even number of ones, an even number of twos, and so on. By the condition, in each row, all numbers from 1 to \(... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 26,427 |
3-5. Given a trihedral angle with vertex \(O\). Can we find a plane section \(A B C\) such that the angles \(O A B, O B A, O B C, O C B, O A C, O C A\) are all acute? | Solve 5. Answer: Yes, it is possible. Choose points \(A, B, C\) at the same distance from point \(O\). Then all the specified angles will be the base angles of isosceles triangles, and the base angle of an isosceles triangle is always acute.
\section*{Second Round}
\section*{7th Grade}
3 - part 1. Solve the equation... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,429 |
3-2. The quadratic trinomial \(a x^{2}+b x+c\) is a perfect fourth power for all integer \(x\). Prove that then \(a=b=0\). | Solution 2. It is clear that \(a \geqslant 0\) and \(c \geqslant 0\). Consider the values of \(x\) equal to \(1,2, \ldots, n\). If one of the numbers \(a\) and \(b\) is different from zero, then the quadratic polynomial \(a x^{2}+b x+c\) for such \(x\) takes at least \(n / 2\) different values. These values are between... | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,430 |
3-1. The quadratic trinomial \(a x^{2}+b x+c\) is a perfect square for all integer \(x\). Prove that then \(a x^{2}+\) \(b x+c=(d x+e)^{2}\). | Solution 1. Let \(f(x)=a x^{2}+b x+c\). Then
\[
f(x+1)-f(x)=\frac{(f(x+1))^{2}-(f(x))^{2}}{f(x+1)+f(x)}=\frac{a(2 x+1)+b}{f(x+1)+f(x)}
\]
therefore \(\lim _{x \rightarrow \infty}(f(x+1)-f(x))=\sqrt{a}\). For integer \(x\), the number \(f(x+1)-f(x)\) is an integer, so \(\sqrt{a}=d\), where \(d\) is an integer. Moreove... | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,433 |
3-ча 3. Point \(O\) lies inside a convex \(n\)-gon \(A_{1} \ldots A_{n}\) and is connected to the vertices by segments. The sides of the \(n\)-gon are numbered from 1 to \(n\), with different sides having different numbers. The same is done with the segments \(O A_{1}, \ldots, O A_{n}\).
a) For \(n=9\), find a numberin... | Solve 3. a) See fig.???
b) The sum of all the side numbers of the triangles \(A_{1} O A_{2}, \ldots, A_{n} O A_{1}\) is \(3(1+2+\cdots+n)=\frac{3 n(n+1)}{2}\). If for each of the \(n\) triangles the sum of the side numbers is the same, then it is equal to \(\frac{3(n+1)}{2}\). But for \(n=10\) the number \(\frac{3(n+1... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,434 |
3-4. Inequality
\[
A a(B b+C c)+B b(C c+A a)+C c(A a+B b)>\frac{1}{2}\left(A B c^{2}+B C a^{2}+C A b^{2}\right)
\]
where \(a>0, b>0, c>0\) are given numbers, holds for all \(A>0, B>0, C>0\). Can a triangle be formed from the segments \(a, b, c\)? | Solution 4. Answer: Yes, it can. Let \(A=B=1, C=\varepsilon\). Then
\[
a(b+\varepsilon c)+b(\varepsilon c+a)+\varepsilon c(a+b)>\frac{1}{2}\left(c^{2}+\varepsilon a^{2}+\varepsilon b^{2}\right)
\]
Therefore, the inequality \(2 a b \geqslant \frac{1}{2} c^{2}\) must hold, since otherwise inequality (1) would not be sa... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 26,435 |
3-ча 1. Given a triangle \(A B C\). On the sides \(A B, B C, C A\), points \(C_{1}, A_{1}, B_{1}\) are taken respectively such that
\[
\frac{A C_{1}}{C_{1} B}=\frac{B A_{1}}{A_{1} C}=\frac{C B_{1}}{B_{1} A}=\frac{1}{n}
\]
On the sides \(A_{1} B_{1}, B_{1} C_{1}, C_{1} A_{1}\) of triangle \(A_{1} B_{1} C_{1}\), points... | Let \((n+1) \vec{a}=\overrightarrow{B C}, (n+1) \vec{b}=\overrightarrow{C A}, (n+1) \vec{c}=\overrightarrow{A B}\). Then
\[
\overrightarrow{A_{2} C_{2}}=\frac{1}{n+1}(n \vec{b}+\vec{c})+n \vec{c}+\vec{a}+\frac{n}{n+1}(n \vec{a}+\vec{b})
\]
In the obtained expression, the coefficients of \(\vec{a}\) and \(\vec{c}\) ar... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,436 |
3-ча 3. Given the equation
\[
x^{n}-a_{1} x^{n-1}-a_{2} x^{n-2}-\cdots-a_{n-1} x-a_{n}=0
\]
where \(a_{1} \geqslant 0, a_{2} \geqslant 0, a_{n} \geqslant 0\). Prove that this equation cannot have two positive roots. | Solution 3. Rewrite the given equation as
\[
1=\frac{a_{1}}{x}+\frac{a_{2}}{x^{2}}+\cdots+\frac{a_{n}}{x^{n}}
\]
For \(x>0\), the function \(f(x)=\frac{a_{1}}{x}+\frac{a_{2}}{x^{2}}+\cdots+\frac{a_{n}}{x^{n}}\) is monotonically decreasing, so it cannot take the value 1 at two different positive values of \(x\). | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,437 |
3-5. Five people play several games of dominoes (two on two) so that each player has each other as a partner once and as an opponent twice. Find the number of games played and all possible ways of distributing the players. | Solution 5. Answer: 5 games; the distribution of players is unique (up to their numbering). First, let's see who the 1st player played with. We can assume that the 1st and 2nd played against the 3rd and 4th, and the 1st and 5th played against the 2nd and 3rd (this can be achieved by swapping the 3rd and 4th). The 3rd p... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,438 |
3-ча 1. Prove that if \(\frac{p}{q}\) is an irreducible rational fraction that is a root of the polynomial
\[
f(x)=a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n}
\]
with integer coefficients, then \(p-k q\) is a divisor of the number \(f(k)\) for any integer \(k\). | Solution 1. The polynomial \(f(x)\) is divisible by \(x-\frac{p}{q}\), so \(f(x)=g(x)\left(x-\frac{p}{q}\right)\). Let \(g(x)=b_{0} x^{n-1}+b_{2} x^{n-2}+\) \(\cdots+b_{n-1}\). Then \(a_{0}=b_{0}, a_{1}=b_{1}-b_{0} \frac{p}{q}, a_{2}=b_{2}-b_{2} \frac{p}{q}, \ldots, a_{n-1}=b_{n-1}-b_{n-2} \frac{p}{q}, a_{n}=-b_{n-1} \... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,439 |
3-5. Given a triangle \(A_{0} B_{0} C_{0}\). On its sides \(A_{0} B_{0}, B_{0} C_{0}, C_{0} A_{0}\), points \(C_{1}, A_{1}, B_{1}\) are taken, respectively. On the sides \(A_{1} B_{1}, B_{1} C_{1}, C_{1} A_{1}\) of triangle \(A_{1} B_{1} C_{1}\), points \(C_{2}, A_{2}\), \(B_{2}\) are taken, respectively, and generally... | Solve 5. That the triangle \(A B C\) is contained within the triangle \(A_{1} B_{1} C_{1}\) is obvious. We will show that the points \(A_{2}, B_{2}, C_{2}\) are the points of intersection of the sides of the triangle \(A_{1} B_{1} C_{1}\) with the lines \(A_{0} A_{1}, B_{0} B_{1}, C_{0} C_{1}\). Place masses \(1+k^{3},... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,442 |
3-ча 1. Prove that there do not exist four points \(A, B, C\) and \(D\) on a plane such that all triangles \(A B C, B C D, C D A, D A B\) are acute-angled. | Solve 1. We assume that no three of the four given points lie on the same line. There are two different arrangements of four points on a plane.
1) Points \(A, B, C\) and \(D\) are the vertices of a convex quadrilateral. The sum of the angles of a quadrilateral is \(360^{\circ}\), so not all its angles are acute. Take ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,443 |
3-4. Find all numbers by which the fraction \(\frac{5 l+6}{8 l+7}\) can be reduced when \(l\) is an integer. | Solution 4. We will compute \(\text{GCD}(5l+6, 8l+7)\), using the fact that \(\text{GCD}(a, b) = \text{GCD}(a-b, b)\). As a result, we get \(\text{GCD}(5l+6, 8l+7) = \text{GCD}(5l+6, 3l+1) = \text{GCD}(2l+5, 3l+1) = \text{GCD}(2l+5, l-4) = \text{GCD}(l+9, l-4) = \text{GCD}(13, l-4)\). The number 13 is prime, so the fra... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,445 |
3-5. What is the smallest number of points that can be chosen on a circle of length 1956 so that for each of these points there is exactly one chosen point at a distance of 1 and exactly one at a distance of 2 (distances are measured along the circumference)? | Solve 5. Answer: 1304.
Let \(A\) be one of the selected points, \(B\) and \(C\) be the selected points, located at distances 1 and 2 from it, respectively. The arrangement \(A B C\) is impossible, since in this case, for point \(B\), there are two selected points at a distance of 1. Therefore, the points are arranged ... | 1304 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,446 |
3-ча 1. On the sides \(A B\) and \(C B\) of triangle \(A B C\), equal segments of arbitrary length \(A D\) and \(C E\) are laid off. Find the geometric locus of the midpoints of segments \(D E\). | Solution 1. Let \(X\) be the midpoint of segment \(DE\), and \(M\) be the midpoint of segment \(AC\). Extend triangles \(ADX\) and \(CEX\) to parallelograms \(ADX A'\) and \(CEX C'\). Since point \(X\) is the midpoint of segment \(DE\), segments \(AA'\) and \(C'C\) are equal. It is also clear that these segments are pa... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,447 |
3-2. In the decimal representation of a positive number \(\alpha\), all decimal places starting from the third digit after the decimal point are discarded (i.e., an approximation of \(\alpha\) with a deficiency is taken to an accuracy of 0.01). The resulting number is divided by \(\alpha\), and the quotient is again ro... | Solve 2. Answer: \(0, \frac{50}{100}, \frac{51}{100}, \ldots, \frac{99}{100}, 1\).
Let \(\alpha=\frac{n}{100}+\alpha_{1}\), where \(n\) is an integer and \(0 \leqslant \alpha_{1}0\), then \(\frac{n}{n+100 \alpha_{1}}>\frac{1}{2}\), since \(100 \alpha_{1}<1\). The fraction \(\frac{n}{n+100 \alpha_{1}}\) can take all va... | 0,\frac{50}{100},\frac{51}{100},\ldots,\frac{99}{100},1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,448 |
3-ча 4. Let \(a, b, c, d, l\) be integers. Prove that if the fraction \(\frac{a l+b}{c l+d}\) is reducible by a number \(k\), then \(a d - b c\) is divisible by \(k\). | Solution 4. Suppose that \(a l+b=k m\) and \(c l+d=k n\). Multiply the first equation by \(-c\), and the second by \(a\). Adding these equations, we get \(a d-b c=k(n a-m c)\). Therefore, \(a d-b c\) is divisible by \(k\). | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,450 |
3-5. On a grid paper, a table is written, and in each cell, there is a number equal to the arithmetic mean of the four numbers in the adjacent cells. All numbers in the table are distinct. Prove that the largest number is on the edge (i.e., at least one of the adjacent cells is missing). | Solve 5. Suppose the largest number \(a\) is not at the edge. Then it has all four neighboring numbers \(a_{1}, a_{2}, a_{3}, a_{4}\) in the table, and \(a=\left(a_{1}+a_{2}+a_{3}+a_{4}\right) / 4\). But \(a>a_{1}, a>a_{2}, a>a_{3}, a>a_{4}\). Therefore, \(a=\left(a_{1}+a_{2}+a_{3}+a_{4}\right) / 4\). This leads to a c... | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,451 |
3-ча 1. In a convex quadrilateral \(A B C D\), a quadrilateral \(K L M N\) is formed by the centroids of triangles \(A B C, B C D, D B A\) and \(C D A\). Prove that the lines connecting the midpoints of opposite sides of quadrilateral \(A B C D\) intersect at the same point as the lines connecting the midpoints of oppo... | Solve 1. The specified lines intersect at the center of mass of the quadrilateral \(A B C D\), i.e., at the center of mass of the system of points \(A, B, C, D\) with equal masses. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,452 |
3-2. In the decimal representation of a positive number \(\alpha\), all decimal places starting from the fourth digit after the decimal point are discarded (i.e., an approximation of \(\alpha\) with a deficiency is taken with an accuracy of 0.001). The resulting number is divided by \(\alpha\), and the quotient is agai... | Solve 2. Answer: \(0, \frac{500}{1000}, \frac{501}{1000}, \ldots, \frac{999}{1000}, 1\). See the solution to problem 2 for 8th grade. | 0,\frac{500}{1000},\frac{501}{1000},\ldots,\frac{999}{1000},1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,453 |
3-4. Given positive numbers \(h, s_{1}, s_{2}\) and a triangle \(A B C\) located in space. In how many ways can a point \(D\) be chosen so that in the tetrahedron \(A B C D\), the height dropped from vertex \(D\) is equal to \(h\), and the areas of the faces \(A C D\) and \(B C D\) are \(s_{1}\) and \(s_{2}\) respectiv... | Solution 4. Answer: \(0, 2, 4\) or 8.
To ensure that the height dropped from vertex \(D\) is equal to \(h\), point \(D\) must lie in one of two planes \(\Pi_{1}\) and \(\Pi_{2}\), parallel to the plane \(A B C\). To ensure that the area of face \(A C D\) is equal to \(s_{1}\), point \(D\) must lie on a cylinder with a... | 0,2,4,8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,455 |
3-ча 1. A square is inscribed in a triangle such that two of its vertices lie on the base, and the other two vertices lie on the sides of the triangle. Prove that the side of the square is less than \(2 r\) but greater than \(\sqrt{2} r\), where \(r\) is the radius of the circle inscribed in the triangle. | Solution 1. Let two vertices of the considered square lie on side \(A B\). The circle \(S_{1}\), inscribed in this square, touches side \(A B\) and is strictly inside the given triangle \(A B C\) (it definitely does not touch sides \(A C\) and \(B C\)). Therefore, there exists a triangle \(A_{1} B_{1} C_{1}\), whose si... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,457 |
3-2. In the decimal representation of a positive number \(\alpha\), all decimal places starting from the fifth digit after the decimal point are discarded (i.e., an approximation of \(\alpha\) with a deficiency is taken with an accuracy of 0.0001). The resulting number is divided by \(\alpha\), and the quotient is agai... | Solve 2. Answer: \(0, \frac{5000}{10000}, \frac{5001}{10000}, \ldots, \frac{9999}{10000}, 1\). See the solution to problem 2 for 8th grade. | 0,\frac{5000}{10000},\frac{5001}{10000},\ldots,\frac{9999}{10000},1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,458 |
3-4. Given a closed spatial broken line. A certain plane intersects all its segments: \(A_{1} A_{2}\) at point \(B_{1}, A_{2} A_{3}\) - at point \(B_{2}, \ldots, A_{n} A_{1}\) - at point \(B_{n}\). Prove that
\[
\frac{A_{1} B_{1}}{B_{1} A_{2}} \frac{A_{2} B_{2}}{B_{2} A_{3}} \cdots \frac{A_{n} B_{n}}{B_{n} A_{1}}=1
\] | Solution 4. Consider the projection onto a line orthogonal to the plane of the section. All points \(B_{1}, \ldots, B_{n}\) are projected onto the same point \(B\). Let \(A_{1}^{\prime}, \ldots, A_{n}^{\prime}\) be the projections of points \(A_{1}, \ldots, A_{n}\). Then
\[
\frac{A_{1} B_{1}}{B_{1} A_{2}} \frac{A_{2} ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,459 |
3-5. Prove that the system of equations
\[
\begin{aligned}
x_{1}-x_{2} & =a, \\
x_{3}-x_{4} & =b, \\
x_{1}+x_{2}+x_{3}+x_{4} & =1
\end{aligned}
\]
has at least one positive solution
\[
x_{1}>0, x_{2}>0, x_{3}>0, x_{4}>0
\]
if and only if \(|a|+|b|<1\). | Solve 5. If \(a \geqslant 0\), then write the first equation in the form \(x_{1}=x_{2}+a\), and if \(a<0\), then write it in the form \(x_{1}=x_{2}-a\). Similarly, if \(b \geqslant 0\), then write the second equation in the form \(x_{3}=x_{4}+b\), and if \(b<0\), then write it in the form \(x_{3}=x_{4}-b\). In both cas... | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,460 |
3-ча 1. Point \(O\) is the center of the circle circumscribed around triangle \(A B C\). Points \(A_{1}, B_{1}\), and \(C_{1}\) are symmetric to point \(O\) with respect to the sides of triangle \(A B C\). Prove that all altitudes of triangle \(A_{1} B_{1} C_{1}\) pass through point \(O\), and all altitudes of triangle... | Solution 1. Let \(A_{2}, B_{2}, C_{2}\) be the midpoints of sides \(CB, BA, AC\). It is clear that \(OA_{2} \perp BC\) and \(BC \parallel B_{2}C_{2} \parallel B_{1}C_{1}\). Therefore, \(OA_{1} \perp B_{1}C_{1}\), i.e., \(OA_{1}\) is the altitude of triangle \(A_{1}B_{1}C_{1}\). Similarly, it can be shown that \(OB_{1}\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,461 |
3-2. Points \(A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}\) divide a circle of radius 1 into six equal parts. From \(A_{1}\), a ray \(l_{1}\) is drawn in the direction of \(A_{2}\), from \(A_{2}\) - a ray \(l_{2}\) in the direction of \(A_{3}, \ldots\), from \(A_{6}\) - a ray \(l_{6}\) in the direction of \(A_{1}\). From ... | Solve 2. Answer: \(B_{1} A_{1}=2\).
Let \(B_{2}, B_{3}, \ldots, B_{7}\) be the feet of the perpendiculars dropped from \(l_{6}, l_{5}, \ldots, l_{1} ; x_{1}=A_{2} B_{1}, x_{2}=\) \(A_{1} B_{2}, x_{3}=A_{6} B_{3}, \ldots, x_{7}=A_{2} B_{7}\). Then \(x_{k+1}=\frac{1}{2}\left(1+x_{k}\right)\). By the condition \(x_{1}=x_... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,462 |
3-rd 3. 100 numbers, including positive and negative, are written in a row. Underlined, first, each positive number, and second, each number whose sum with the next one is positive. Can the sum of all underlined numbers be negative? Equal to zero? | Solution 3. Answer: No, it cannot.
We will underline positive numbers with one line, and negative numbers, the sum of which with the next number is positive, with two lines. It is clear that after each number \(a\) underlined with two lines, there is a number \(b\) underlined with one line, and \(a+b>0\). The underlin... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,463 |
3-4. 64 non-negative numbers, the sum of which is 1956, are arranged in the form of a square table: eight numbers in each row and in each column. The sum of the numbers on one of the diagonals is 112. The numbers symmetrically located relative to this diagonal are equal. Prove that the sum of the numbers in any column ... | Solution 4. Suppose the sum of the numbers in some column is \(S \geqslant 1035\). Consider the row symmetric to this column with respect to the highlighted diagonal. The sum of the numbers in this row is also \(S\), and the sum of all numbers in this column and this row is \(2 S-s\), where \(s\) is the number at their... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 26,464 |
3-5. There are 15 magazines on the table, covering it completely. Prove that you can remove seven magazines so that the remaining magazines cover no less than \(8 / 15\) of the table's area.
(This problem was not solved by any of the participants in the competition.) | Solution 5. First, we will prove that if \(n\) magazines cover an area \(S\), then one magazine can be removed such that the remaining magazines cover an area of at least \((n-1) S / n\). If after removing a certain magazine, the remaining magazines cover an area less than \((n-1) S / n\), then the area covered only by... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 26,465 |
3-ча 1. A cargo weighing 13.5 tons is packed into boxes such that the weight of each box does not exceed 350 kg. Prove that this cargo can be transported using 11 one-and-a-half-ton trucks. (The weight of the empty boxes can be neglected.) | Solution 1. We will allocate 8 trucks and load them sequentially, each time loading the box that cannot be loaded anymore next to the truck. The loaded boxes together with the boxes standing next to the trucks weigh more than \(8 \cdot 1.5=12\) tons, so the remaining boxes weigh less than 1.5 tons; they can be transpor... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 26,466 |
3-cha 2. 64 non-negative numbers, the sum of which is 1956, are arranged in the form of a square table with eight numbers in each row and each column. The sum of the numbers on the two diagonals is 112. Numbers that are symmetrically located relative to any diagonal are equal. Prove that the sum of the numbers in any r... | Solution 2. Suppose the sum of the numbers in some row is \(S \geqslant 518\). Consider two columns symmetric to this row with respect to the two diagonals, and another row symmetric to these columns. The number 8 is even, so we will get two different columns and two different rows. At the intersections of these rows a... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 26,467 |
3-4. 100 numbers, including positive and negative, are written in a row. Underlined, first, every positive number, second, every number whose sum with the next one is positive, and third, every number whose sum with the next two is positive. Can the sum of all underlined numbers be negative? Equal to zero? | Solution 4. Answer: No, it cannot.
We will underline numbers as follows: 1) positive numbers with one line; 2) negative numbers, the sum of which with the next number is positive, with two lines; 3) negative numbers, for which the sum with the next number is non-positive, but the sum with the next two numbers is posit... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 26,469 |
3-5. In a rectangle with an area of 5 square units, nine rectangles, each with an area of one square unit, are placed. Prove that the area of the common part of some two rectangles is greater than or equal to \(1 / 9\). | Solution 5. Suppose the area of the common part of any two rectangles is less than \(1/9\). We will show that then they occupy an area greater than 5. Number the rectangles arbitrarily. The first rectangle occupies an area of 1. Add the second rectangle. The area of the common part of the first and second rectangles is... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 26,470 |
3-2. In a cube with an edge length of 13, 1956 points are selected. Is it possible to place a cube with an edge length of 1 inside the larger cube so that it does not contain any of the selected points? | Solution 2. Answer: Yes, it is possible. Let's cut the given cube into \(13^{3}=2197\) unit cubes. If there were a chosen point inside each of these unit cubes, the number of chosen points would be no less than 2197, which contradicts the condition. Therefore, there is at least one of these unit cubes that does not con... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,471 |
3-3. We took three numbers \(x, y, z\). We calculated the absolute values of the pairwise differences \(x_{1}=|x-y|\), \(y_{1}=|y-z|, z_{1}=|z-x|\). In the same way, from the numbers \(x_{1}, y_{1}, z_{1}\) we constructed the numbers \(x_{2}, y_{2}, z_{2}\) and so on. It turned out that for some \(n\), \(x_{n}=x\), \(y... | Solve 3. Answer: \(y=z=0\).
The numbers \(x_{n}, y_{n}, z_{n}\) are non-negative, so the numbers \(x, y, z\) are also non-negative. If all the numbers \(x\), \(y, z\) were positive, then the largest of the numbers \(x_{1}, y_{1}, z_{1}\) would be strictly less than the largest of the numbers \(x, y, z\), and then the ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,472 |
3-4. A quadrilateral is circumscribed around a circle. Prove that the lines connecting adjacent points of tangency and not intersecting at one of these points, intersect on the extension of a diagonal or are parallel to it. | Solution 4. When solving this problem, it is convenient to consider that the quadrilateral is composed of two triangles, so we will introduce the following notations: \(A C B C^{\prime}\) - the given quadrilateral, \(A_{1}\) and \(B_{1}\) - the points of tangency with sides \(B C\) and \(A C, A_{1}^{\prime}\) and \(B_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,473 |
3-ча 1. Consecutively, \(n\) numbers are written, among which there are positive and negative numbers. Each positive number is underlined, as well as each number whose sum with several immediately following numbers is positive. Prove that the sum of all underlined numbers is positive. | Solution 1. Answer: No, it cannot.
Let \(a_{1}, a_{2}, \ldots, a_{n}\). Underline the number \(a_{i} k+1\) with lines if \(k\) is the smallest number for which \(a_{i}+a_{i+1}+a_{i+2}+\cdots+a_{i+k}>0\) (a positive number \(a_{1}\) is underlined with one line). Clearly, if the number \(a_{i}\) is underlined with \(k+1... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 26,474 |
3-5. On the extensions of the sides \(A_{1} A_{2}, A_{2} A_{3}, \ldots, A_{n} A_{1}\) of a regular \(n\)-gon ( \(n \geqslant 5\) - note by the editor) \(A_{1} A_{2} \ldots A_{n}\), construct points \(B_{1}, B_{2}, \ldots, B_{n}\) such that \(B_{1} B_{2}\) is perpendicular to \(A_{1} A_{2}, B_{2} B_{3}\) is perpendicula... | Solution 5. Let \(x_{k}=A_{k+1} B_{k}, \alpha\) be the exterior angle of a regular \(n\)-gon, and \(a\) be the length of its side. Then \(x_{1}=\left(a+x_{2}\right) \cos \alpha, x_{2}=\left(a+x_{3}\right) \cos \alpha, \ldots, x_{n}=\left(a+x_{1}\right) \cos \alpha\). Then \(x_{1}=a_{1}+b_{1} x_{2}=a_{2}+b_{2} x_{3}=\) ... | x:=\cos\alpha:(1-\cos\alpha) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,477 |
3-2. It is known that \(a x^{3}+b x^{2}+c x+d\), where \(a, b, c, d\) are given integers, is divisible by 5 for any integer \(x\). Prove that all numbers \(a, b, c, d\) are divisible by 5. | Solve 2. Substituting \(x=0\), we get that \(d\) is divisible by 5. Considering this and substituting \(x= \pm 1\), we get that \(a+b+c\) and \(-a+b-c\) are divisible by 5. Therefore, \(2 b\) and \(2 a+2 c\) are divisible by 5, which means \(b\) and \(a+c\) are divisible by 5. Substituting \(x=2\), we get that \(4(2 a+... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,479 |
3-ча 3. A snail is crawling on a table at a constant speed. Every 15 minutes, it turns \(90^{\circ}\) to the left or right, and in the intervals between turns, it crawls in a straight line. Prove that it can return to the starting point only after an integer number of hours. | Solution 3. Let the snail crawl forward \(a_{1}\) 15-minute intervals (compared to the direction of movement from the starting point), \(a_{2}\) backward, \(b_{1}\) to the right, and \(b_{2}\) to the left before returning to the starting point. Then \(a_{1}=a_{2}\) and \(b_{1}=b_{2}\) (since the snail returned to the s... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 26,480 |
3-4. In a rectangular table composed of positive numbers, the product of the sum of the numbers in any column and the sum of the numbers in any row equals the number at their intersection. Prove that the sum of all numbers in the table is one. | Solution 4. First solution. Let \(x_{1}, \ldots, x_{n}\) be the sums of the numbers in the rows, and \(y_{1}, \ldots, y_{m}\) be the sums of the numbers in the columns. At the intersection of the \(i\)-th row and the \(j\)-th column, the number \(x_{i} y_{j}\) is placed. Therefore, the sum of the numbers in the \(i\)-t... | 1 | Algebra | proof | Yes | Yes | olympiads | false | 26,481 |
3-5. The distance from \(A\) to \(B\) is 999 km. Along the road, there are kilometer markers indicating the distances to \(A\) and to \(B: 0\) ।999, 1 ।998, \(\ldots, 999\) ।0. How many of these markers have only two different digits? | Solve 5. Answer: 40.
Assume that on the kilometer post, the number \(\overline{a b d a_{1} b_{1} c_{1}}\) is written. Then \(\overline{a b c} + \overline{a_{1} b_{1} c_{1}} = 999\), so \(a_{1} = 9 - a\), \(b_{1} = 9 - b\), and \(c_{1} = 9 - c\). If \(a = b = c\), then the required condition is satisfied. There will be... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,482 |
3-4. In a rectangular table, the product of the sum of the numbers in any column and the sum of the numbers in any row equals the number at their intersection. Prove that the sum of all numbers in the table is one, or all numbers are zero. | Solution 4. Let \(x_{1}, \ldots, x_{n}\) be the sums of the numbers in the rows, and \(y_{1}, \ldots, y_{m}\) be the sums of the numbers in the columns. At the intersection of the \(i\)-th row and the \(j\)-th column, the number \(x_{i} y_{j}\) is placed. Therefore, the sum of the numbers in the \(i\)-th row is \(x_{i}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,485 |
3-5. It is known that \(a x^{4}+b x^{3}+c x^{2}+d x+e\), where \(a, b, c, d, e\) are given integers, is divisible by 7 for any integer \(x\). Prove that all integers \(a, b, c, d, e\) are divisible by 7. | Solution 5. Substituting \(x=0\), we get that \(e\) is divisible by 7. Considering this and substituting \(x= \pm 1\), we get that the numbers \(a \pm b+c \pm d\) are divisible by 7. Therefore, \(2(a+c)\) and \(2(b+d)\) are divisible by 7, which means \(a+c\) and \(b+d\) are divisible by 7. Substituting \(x= \pm 2\) an... | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,486 |
3-ча 3. In a convex quadrilateral \(A B C D\), point \(M\) is the midpoint of diagonal \(A C\), and point \(N\) is the midpoint of diagonal \(B D\). Line \(M N\) intersects sides \(A B\) and \(C D\) at points \(M^{\prime}\) and \(N^{\prime}\). Prove that if \(M M^{\prime}=N N^{\prime}\), then \(B C \| A D\). | If points \(M\) and \(N\) coincide, then \(A B C D\) is a parallelogram. Therefore, we will assume that points \(M\) and \(N\) are distinct. Suppose that lines \(A D\) and \(B C\) are not parallel. Let \(M^{\prime \prime}, K, N^{\prime \prime}\) be the midpoints of sides \(A B, B C, C D\) respectively. If \(M N \| B C\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,487 |
3-5. A flat polygon \(A_{1} A_{2} \ldots A_{n}\) is composed of \(n\) rigid rods connected by hinges. Prove that if \(n>4\), it can be deformed into a triangle. | Solution 5. Let \(a\) be the largest side of the given polygon (if there are several largest sides, we choose any one of them). Consider the part of the polygon that remains after removing side \(a\), and take the point that bisects the perimeter of this part. If this point is a vertex of the polygon, then we can obvio... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,488 |
3-ча 1. For which integers \(n\) is the number \(20^{n}+16^{n}-3^{n}-1\) divisible by 323? | Solution 1. Answer: for even \(n\). First, note that \(323=17 \cdot 19\), so a number is divisible by 323 if and only if it is divisible by 17 and 19. The number \(20^{n}-3^{n}\) is divisible by \(20-3=17\). Furthermore, \(16^{n} \equiv(-1)^{n}(\bmod 17)\), so the number \(16^{n}-1\) is divisible by 17 if and only if \... | forevenn | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,489 |
3-2. In space, a closed broken line is constructed such that all segments have the same length and every three consecutive segments are pairwise perpendicular. Prove that the number of segments is divisible by 6. | Solution 2. Let \(e_{1}, e_{2}, e_{3}\) be the vectors of three consecutive segments of the given broken line. After the vectors \(e_{2}\), \(e_{3}\), there must be a vector perpendicular to them, i.e., the vector \(\pm e_{1}\). Continuing this reasoning, we get that the sequence of segment vectors has the form \(e_{1}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,490 |
3-5. A flat polygon \(A_{1} A_{2} \ldots A_{n}\) is made of \(n\) rigid rods connected by hinges. Can it be deformed into a triangle? | Solve 5. If \(n>4\), then it is possible: see the solution to problem 5 for 9th grade. It is also proven there that when \(n=4\), any quadrilateral can be deformed except for a parallelogram.
\section*{Second Round}
\section*{7th Grade} | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,491 |
3-5. Kolya and Petya are dividing \(2 n+1\) nuts, \(n \geqslant 2\), and each wants to get as many as possible. There are three stages of division (each stage is repeated three times).
1st stage: Petya divides all the nuts into two parts, each containing no fewer than two nuts.
2nd stage: Kolya divides each part agai... | Solve 5.
\section*{10th Grade} | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 26,515 |
3-5. From 28 domino tiles, all tiles with sixes were removed. Can the remaining ones be laid out in a chain? | Solve 5.
\section*{9-11 grades} | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 26,542 |
3-ча 1. Prove that there exists a number \(q\) such that in the decimal representation of the number \(q \cdot 2^{1967}\) there is not a single zero. | Solve 1.
Part 2. Let \(d(N)\) denote the number of divisors of \(N\). Find all \(N\) such that \(\frac{N}{d(N)}=P\) is an integer and a prime number (the numbers 1 and \(N\) are also considered divisors).
Solve 2. | Number Theory | proof | Yes | Yes | olympiads | false | 26,546 | |
3-ча 1. In a chess tournament, 12 people participated. After the tournament, each participant made 12 lists. The first list includes only himself, the second list includes him and those he defeated, the third list includes all people from the second list and those they defeated, and so on. The 12th list includes all pe... | Solution 1. Answer: 54. If the \((k+1)\)-th list is the same as the \(k\)-th, then the lists numbered \(k+2, \ldots, 11, 12\) will also be exactly the same. However, according to the condition, the 11th list and the 12th list are different. Therefore, each participant's \(k\)-th list contains exactly \(k\) people. In p... | 54 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,553 |
3-ча 3. Can a convex heptagon \(A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{7}\) with angles \(A_{1}=\) \(140^{\circ}, A_{2}=120^{\circ}, A_{3}=130^{\circ}, A_{4}=120^{\circ}, A_{5}=130^{\circ}, A_{6}=110^{\circ}, A_{7}=150^{\circ} \) be inscribed in a circle? | Solve 3. Answer: No, it is not possible. Suppose the convex heptagon \(A_{1} A_{2} \ldots A_{7}\) is inscribed in a circle with center \(O\). Then \(\angle A_{1} O A_{3}=\angle A_{3} O A_{5}=2\left(180^{\circ}-120^{\circ}\right)=120^{\circ}\) and \(\angle A_{5} O A_{7}=2\left(180^{\circ}-110^{\circ}\right)=140^{\circ}\... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,554 |
3-ча 1. On an infinite chessboard, two black checkers are placed on two diagonally adjacent black squares. Is it possible to additionally place some number of black checkers and one white checker on this board in such a way that the white checker can capture all the black checkers, including the two initially placed, i... | Solution 1.
Part 2. On 99 cards, the numbers \(1,2,3, \ldots, 99\) are written. Then the cards are shuffled, laid out with their clean sides up, and on the clean sides, the numbers \(1,2,3,4, \ldots, 99\) are written again. For each card, the numbers on it are added, and the 99 resulting sums are multiplied. Prove tha... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 26,567 |
3-ча 1. Find all natural numbers \(n\), for which the number \(n \cdot 2^{n}+1\) is divisible by three. | Solve 1.
Part 2. Find a point on the plane such that the sum of the distances from it to four given points is minimal.
Solve 2. | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,618 | |
3-6. All values of the quadratic trinomial \(a x^{2}+b x+c\) on the interval \([0,1]\) do not exceed 1 in absolute value. What is the maximum value that the quantity \(|a|+|b|+|c|\) can have in this case? | Solve 6.
\section*{9th Grade} | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,643 |
1.1. The bases $A D$ and $B C$ of trapezoid $A B C D$ are equal to $a$ and $b(a>b)$.
a) Find the length of the segment cut off by the diagonals on the midline.
b) Find the length of the segment $M N$, the ends of which divide the sides $A B$ and $C D$ in the ratio $A M: M B=D N: N C=p: q$. | 1.1. a) Let $P$ and $Q$ be the midpoints of sides $AB$ and $CD$, and $K$ and $L$ be the points of intersection of line $PQ$ with diagonals $AC$ and $BD$. Then $PL = a / 2$ and $PK = b / 2$, so $KL = PL - PK = (a - b) / 2$.
b) Take a point $F$ on side $AD$ such that $BF \| CD$. Let $E$ be the point of intersection of s... | \frac{b}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,644 |
1.2. Prove that the midpoints of the sides of any quadrilateral are the vertices of a parallelogram. For which quadrilaterals is this parallelogram a rectangle, a rhombus, or a square? | 1.2. Let $K, L, M$ and $N$ be the midpoints of the sides $A B, B C, C D$ and $D A$ of quadrilateral $A B C D$, respectively. Then $K L = M N = A C / 2$ and segment $K L$ is parallel to $M N$, i.e., $K L M N$ is a parallelogram. It is now clear that $K L M N$ is a rectangle if the diagonals $A C$ and $B D$ are perpendic... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,645 |
1.3. a) Points $A_{1}$ and $B_{1}$ divide the sides $B C$ and $A C$ of triangle $A B C$ in the ratios $B A_{1}: A_{1} C=1: p$ and $A B_{1}: B_{1} C=1: q$. In what ratio does the segment $A A_{1}$ divide the segment $B B_{1}$?
b) Points $A_{1}$ and $B_{1}$ are taken on the sides $B C$ and $A C$ of triangle $A B C$. Seg... | 1.3. a) Let the intersection point of segments $A A_{1}$ and $B B_{1}$ be denoted as $O$. In triangle $B_{1} B C$, draw segment $A_{1} A_{2} \| B B_{1}$. Then $B_{1} C / B_{1} A_{2}=1+p$, so $A O: O A_{1}=A B_{1}: B_{1} A_{2}=B_{1} C: q B_{1} A_{2}=(1+p): q$.
b) Let $B A_{1}: A_{1} C=1: p$ and $A B_{1}: B_{1} C=1: q$.... | \frac{1+p}{q} | Geometry | proof | Yes | Yes | olympiads | false | 26,646 |
1.4. Through point $P$ of the median $C C_{1}$ of triangle $A B C$, lines $A A_{1}$ and $B B_{1}$ are drawn (points $A_{1}$ and $B_{1}$ lie on sides $B C$ and $C A$). Prove that $A_{1} B_{1} \| A B$. | 1.4. Let $A_{2}$ be the midpoint of segment $A_{1} B$. Then $C A_{1}: A_{1} A_{2}=C P: P C_{1}$ and $A_{1} A_{2}: A_{1} B=1: 2$, so $C A_{1}: A_{1} B=C P: 2 P C_{1}$. Similarly, $C B_{1}: B_{1} A=$ $=C P: 2 P C_{1}=C A_{1}: A_{1} B$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,647 |
1.5. The line connecting point $P$ of intersection of the diagonals of quadrilateral $A B C D$ with point $Q$ of intersection of lines $A B$ and $C D$ bisects side $A D$. Prove that it also bisects side $B C$. | 1.5. Point $P$ lies on the median $Q M$ of triangle $A Q D$ (or on its extension). It is easy to verify that the solution to problem 1.4 remains valid even when point $P$ lies on the extension of the median. Consequently, $B C \| A D$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,648 |
1.6. On the side $A D$ of the parallelogram $A B C D$, a point $P$ is taken such that $A P: A D=1: n ; Q$ is the point of intersection of the lines $A C$ and $B P$. Prove that $A Q: A C=1:(n+1)$. | 1.6. Since $\triangle A Q P \sim \triangle C Q B$, then $A Q: Q C=$ $=A P: B C=1: n$. Therefore, $A C=A Q+Q C=(n+$ $+1) A Q$. | AC=(n+1)AQ | Geometry | proof | Yes | Yes | olympiads | false | 26,649 |
1.7. The vertices of the parallelogram $A_{1} B_{1} C_{1} D_{1}$ lie on the sides of the parallelogram $A B C D$ (point $A_{1}$ lies on side $A B$, point $B_{1}$ - on side $B C$, and so on). Prove that the centers of both parallelograms coincide. | 1.7. The center of parallelogram $A_{1} B_{1} C_{1} D_{1}$ as the midpoint of segment $B_{1} D_{1}$ lies on the segment connecting the midpoints of sides $A B$ and $C D$. Similarly, it lies on the segment connecting the midpoints of sides $B C$ and $A D$. The point of intersection of these segments is the center of par... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,650 |
1.8. On the diagonal $B D$ of parallelogram $A B C D$, a point $K$ is taken. Line $A K$ intersects lines $B C$ and $C D$ at points $L$ and $M$. Prove that $A K^{2}=L K \cdot K M$. | 1.8. It is clear that $A K: K M = B K: K D =$ $= L K: A K$, i.e., $A K^{2} = L K \cdot K M$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,651 |
1.9. One of the diagonals of a cyclic quadrilateral is a diameter. Prove that the projections of opposite sides onto the other diagonal are equal. | 1.9. Let $AC$ be the diameter of the circle circumscribed around quadrilateral $ABCD$.
Fig. 1.3 Drop perpendiculars $AA_1$ and $CC_1$ from $A$ and $C$ to $BD$ (Fig. 1.3). We need to prove... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,652 |
1.10. On the base $AD$ of trapezoid $ABCD$, a point $E$ is taken such that $AE = BC$. Segments $CA$ and $CE$ intersect diagonal $BD$ at points $O$ and $P$ respectively. Prove that if $BO = PD$, then $AD^2 = BC^2 + AD \cdot BC$. | 1.10. Since $B O=P D$, then $B O: O D=D P: P B=k$. Let $B C=1$. Then $A D=k$ and $E D=1 / k$. Therefore, $k=A D=A E+E D=1+(1 / k)$, i.e., $k^{2}=1+k$. It remains to note that $k^{2}=A D^{2}$ and $1+k=B C^{2}+B C \cdot A D$. | AD^2=BC^2+AD\cdotBC | Geometry | proof | Yes | Yes | olympiads | false | 26,653 |
1.11. Points $A$ and $B$ intercept an arc of $60^{\circ}$ on a circle with center $O$. A point $M$ is taken on this arc. Prove that the line passing through the midpoints of segments $M A$ and $O B$ is perpendicular to the line passing through the midpoints of segments $M B$ and $O A$. | 1.11. Let $C, D, E, F$ be the midpoints of the sides $A O, O B, B M, M A$ respectively of the quadrilateral $A O B M$. Since $A B = M O = R$, where $R$ is the radius of the given circle, then according to problem $1.2 C D E F$ is a rhombus. Therefore, $C E \perp D F$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,654 |
1.12. a) Points $A, B$ and $C$ lie on one line, and points $A_{1}, B_{1}$, and $C_{1}$ lie on another. Prove that if $A B_{1} \| B A_{1}$ and $A C_{1} \| C A_{1}$, then $B C_{1} \| C B_{1}$.
b) Points $A, B$ and $C$ lie on one line, and points $A_{1}, B_{1}$ and $C_{1}$ are such that $A B_{1}\left\|B A_{1}, A C_{1}\ri... | 1.12. a) If the lines on which the given points lie are parallel, the statement of the problem is obvious. Let us assume that these lines intersect at point $O$. Then $O A: O B = O B_{1}: O A_{1}$, and $O C: O A = O A_{1}: O C_{1}$, so $O C: O B = O B_{1}: O C_{1}$, which means $B C_{1} \| C B_{1}$ (the ratios of segme... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,655 |
1.13. In triangle $A B C$, the angle bisectors $A A_{1}$ and $B B_{1}$ are drawn. Prove that the distance from any point $M$ on the segment $A_{1} B_{1}$ to the line $A B$ is equal to the sum of the distances from $M$ to the lines $A C$ and $B C$. | 1.13. A point lying on the bisector of an angle is equidistant from its sides.
Fig. 1.4 is removed from its sides. Let \( a \) be the distance from point \( A_1 \) to the lines \( AC \) an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,656 |
1.14. Let $M$ and $N$ be the midpoints of sides $A D$ and $B C$ of rectangle $A B C D$. On the extension of segment $D C$ beyond point $D$, take point $P$; $Q$ is the point of intersection of lines $P M$ and $A C$. Prove that $\angle Q N M = \angle M N P$. | 1.14. Let the line passing through the center $O$ of the given rectangle parallel to $B C$ intersect the segment $Q N$ at point $K$ (Fig. 1.4). Since $M O \| P C$, then $Q M: M P=Q O: O C$, and since $K O \| B C$, then $Q O: O C=Q K: K N$. Therefore, $Q M: M P=$ $=Q K: K N$, i.e., $K M \| N P$. Hence, $\angle M N P=$ $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,657 |
1.15. On the extensions of the bases $A D$ and $B C$ of trapezoid $A B C D$ beyond points $A$ and $C$, points $K$ and $L$ are taken. Segment $K L$ intersects sides $A B$ and $C D$ at points $M$ and $N$, and diagonals $A C$ and $B D$ at points $O$ and $P$. Prove that if $K M=N L$, then $K O=P L$. | 1.15. Draw a line $E F$ through point $M$ parallel to $C D$ (points $E$ and $F$ lie on lines $B C$ and $A D$). Then $P L: P K=B L: K D$ and $O K: O L=K A: C L=K A: K F=B L: E L$. Since $K D=E L$, then $P L: P K=O K: O L$, and therefore, $P L=O K$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,658 |
1.16*. On the sides $AB, BC, CD$, and $DA$ of a convex quadrilateral $ABCD$, points $P, Q, R$, and $S$ are taken such that $BP: AB = CR: CD = \alpha$ and $AS: AD = BQ: BC = \beta$. Prove that the segments $PR$ and $QS$ are divided by their point of intersection in the ratios $\beta:(1-\beta)$ and $\alpha:(1-\alpha)$.
... | 1.16. Consider the parallelogram $A B C D_{1}$. We can assume that points $D$ and $D_{1}$ do not coincide (otherwise, the statement of the problem is obvious). Take points $S_{1}$ and $R_{1}$ on sides $A D_{1}$ and $C D_{1}$ such that $S S_{1} \| D D_{1}$ and $R R_{1} \| D D_{1}$. Let $N$ be the intersection point of s... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,659 |
1.17. a) In triangle $A B C$, the bisector $B D$ of an internal or external angle is drawn. Prove that $A D: D C=A B: B C$.
b) Prove that the center $O$ of the inscribed circle of triangle $A B C$ divides the bisector $A A_{1}$ in the ratio $A O: O A_{1}=(b+c): a$, where $a, b, c$ are the lengths of the sides of the t... | 1.17. a) Drop perpendiculars $A K$ and $C L$ from vertices $A$ and $C$ to the line $B D$. Since $\angle C B L = \angle A B K$ and $\angle C D L = \angle K D A$, we have $\triangle B L C \sim \triangle B K A$ and $\triangle C L D \sim \triangle A K D$. Therefore, $A D: D C = A K: C L = A B: B C$.
b) Considering that $B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,660 |
1.18. The lengths of two sides of a triangle are $a$, and the length of the third side is $b$. Calculate the radius of its circumscribed circle. | 1.18. Let $O$ be the center of the circumscribed circle of isosceles triangle $ABC$, $B_{1}$ the midpoint of the base $AC$, and $A_{1}$ the midpoint of the lateral side $BC$. Since $\triangle B O A_{1} \sim \triangle B C B_{1}$, then $B O: B A_{1}=B C: B B_{1}$, which means $R=B O=$ $=a^{2} / \sqrt{4 a^{2}-b^{2}}$. | \frac{^2}{\sqrt{4a^2-b^2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,661 |
1.19. A line passing through vertex $A$ of square $A B C D$ intersects side $C D$ at point $E$ and line $B C$ at point $F$. Prove that
$$
\frac{1}{A E^{2}}+\frac{1}{A F^{2}}=\frac{1}{A B^{2}}
$$ | 1.19. If $\angle E A D=\varphi$, then $A E=A D / \cos \varphi=A B / \cos \varphi$ and $A F=A B / \sin \varphi$. Therefore $\frac{1}{A E^{2}}+\frac{1}{A F^{2}}=\frac{\left(\cos ^{2} \varphi+\sin ^{2} \varphi\right)}{A B^{2}}=\frac{1}{A B^{2}}$. | \frac{1}{AB^{2}} | Geometry | proof | Yes | Yes | olympiads | false | 26,662 |
1.20. On the heights $B B_{1}$ and $C C_{1}$ of triangle $A B C$, points $B_{2}$ and $C_{2}$ are taken such that $\angle A B_{2} C=\angle A C_{2} B=90^{\circ}$. Prove that $A B_{2}=A C_{2}$. | 1.20. It is easy to verify that $A B_{2}^{2}=A B_{1} \cdot A C=A C_{1} \cdot A B=A C_{2}^{2}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,663 |
1.21. In trapezoid $ABCD (BC \| AD)$, a circle is inscribed, touching the lateral sides $AB$ and $CD$ at points $K$ and $L$ respectively, and the bases $AD$ and $BC$ at points $M$ and $N$.
a) Let $Q$ be the point of intersection of segments $BM$ and $AN$. Prove that $KQ \| AD$.
b) Prove that $AK \cdot KB = CL \cdot L... | 1.21. a) Since $B Q: Q M=B N: A M=B K: A K$, then $K Q \| A M$.
b) Let $O$ be the center of the inscribed circle. Since $\angle C B A+\angle B A D=180^{\circ}$, then $\angle A B O+\angle B A O=90^{\circ}$. Therefore, $\triangle A K O \sim \triangle O K B$, i.e., $A K: K O=O K: K B$. Hence, $A K \cdot K B=K O^{2}=R^{2}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,664 |
1.22. Perpendiculars $A M$ and $A N$ are dropped from the sides $B C$ and $C D$ of parallelogram $A B C D$ (or their extensions). Prove that $\triangle M A N \sim \triangle A B C$. | 1.22. If angle $ABC$ is obtuse (respectively acute), then angle $MAN$ is also obtuse (respectively acute). Moreover, the sides of these angles are mutually perpendicular. Therefore, $\angle ABC = \angle MAN$. Right triangles $ABM$ and $ADN$ have equal angles $ABM$ and $ADN$, so $AM: AN = AB: AD = AB: CB$, i.e., $\trian... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,665 |
1.23. A line $l$ intersects sides $A B$ and $A D$ of parallelogram $A B C D$ at points $E$ and $F$ respectively. Let $G$ be the point of intersection of line $l$ with diagonal $A C$. Prove that $\frac{A B}{A E}+\frac{A D}{A F}=\frac{A C}{A G}$. | 1.23. Let's take points \(D'\) and \(B'\) on the diagonal \(AC\) such that \(BB' \parallel l\) and \(DD' \parallel l\). Then \(AB: AE = AB': AG\) and \(AD: AF = AD': AG\). Since the sides of triangles \(ABB'\) and \(CDD'\) are pairwise parallel and \(AB = CD\), these triangles are equal and \(AB' = CD'\). Therefore, \(... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,666 |
1.24. Let $AC$ be the longer diagonal of parallelogram $ABCD$. Perpendiculars $CE$ and $CF$ are dropped from point $C$ to the extensions of sides $AB$ and $AD$ respectively. Prove that $AB \cdot AE + AD \cdot AF = AC^2$. | 1.24. Drop a perpendicular $B G$ from vertex $B$ to $A C$ (Fig. 1.5). From the similarity of triangles $A B G$ and $A C E$, we get $A C \cdot A G=A E \cdot A B$. Lines $A F$ and $C B$ are parallel, so angles $G C B$ and $C A F$ are equal, and right triangles $C B G$ and $A C F$ are similar. From the similarity of these... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,667 |
1.25. The angles of triangle $ABC$ are related by the equation $3 \alpha + 2 \beta = 180^{\circ}$. Prove that $a^2 + bc = c^2$. | 1.25. Since $\alpha+\beta=90^{\circ}-(\alpha / 2)$, then $\gamma=$ $=180^{\circ}-\alpha-\beta=90^{\circ}+(\alpha / 2)$. Therefore, on the side
Fig. 1.5 not $AB$ one can choose a point $D$... | ^2+=^2 | Geometry | proof | Yes | Yes | olympiads | false | 26,668 |
1.26. The ends of segments $A B$ and $C D$ move along the sides of a given angle, with lines $A B$ and $C D$ moving parallel to each other; $M$ is the point of intersection of segments $A B$ and $C D$. Prove that the value $\frac{A M \cdot B M}{C M \cdot D M}$ remains constant. | 1.26. When segments $A B$ and $C D$ are moved, triangle $A M C$ will be replaced by another triangle similar to the original one. Therefore, the ratio $A M / C M$ remains constant. Similarly, the ratio $B M / D M$ remains constant. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,669 |
1.27. Through an arbitrary point $P$ on the side $A C$ of triangle $A B C$, lines are drawn parallel to its medians $A K$ and $C L$, intersecting sides $B C$ and $A B$ at points $E$ and $F$ respectively. Prove that the medians $A K$ and $C L$ divide the segment $E F$ into three equal parts. | 1.27. Let the point of intersection of the medians be denoted by $O$, the points of intersection of the median $A K$ with the lines $F P$ and $F E$ - by $Q$ and $M$, the points of intersection of the median $C L$ with the lines $E P$ and $F E$ - by $R$ and $N$ respectively (Fig. 1.6). It is clear that $F M: F E = F Q: ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,670 |
1.28. On the bisector of the angle with vertex $C$, a point $P$ is taken. A line passing through point $P$ intersects the sides of the angle at segments of lengths $a$ and $b$. Prove that the value $\frac{1}{a}+\frac{1}{b}$ does not depend on the choice of this line. | 1.28. Let $A$ and $B$ be the points of intersection of a given
Fig. 1.6 straight line with the sides of the angle. Take points $K$ and $L$ on segments $A C$ and $B C$ such that $P K \| B C$... | \frac{1}{}+\frac{1}{b}=\frac{1}{p} | Geometry | proof | Yes | Yes | olympiads | false | 26,671 |
1.29. On the side $B C$ of an equilateral triangle $A B C$, a semicircle is constructed externally with $B C$ as its diameter. Points $K$ and $L$ are taken on the semicircle, dividing it into three equal arcs. Prove that the lines $A K$ and $A L$ divide the segment $B C$ into equal parts. | 1.29. Let the midpoint of side $B C$ be denoted by $O$, and the points of intersection of $A K$ and $A L$ with side $B C$ be denoted by $P$ and $Q$. We can assume that $B P < B Q$. Triangle $L C O$ is equilateral and $L C \| A B$. Therefore, $\triangle A B Q \sim \triangle L C Q$, i.e., $B Q: Q C = A B: L C = 2: 1$. Co... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,672 |
1.31*. Prove that if $a_{1}=a_{2}$ and $b_{1}=b_{2}$ (Fig. 1.1), then $x=y$. | 1.31. Let's draw additional lines, parallel to the lines on which the segments of lengths $a_{1}=a_{2}=a$ and $b_{1}=$ $=b_{2}=b$ lie (Fig. 1.7). From the similarity of triangles, we get the equalities $\frac{x}{b}=\frac{x+z_{1}}{\beta}$, $\frac{a+z_{3}}{a}=\frac{\beta}{b+z_{2}}$, $\frac{y+z_{1}}{\alpha}=\frac{y}{a}$, ... | y | Geometry | proof | Yes | Yes | olympiads | false | 26,674 |
1.32*. On the segment $M N$, similar and identically oriented triangles $A M N, N B M$, and $M N C$ are constructed (Fig. 1.2). Prove that triangle $A B C$ is similar to all these triangles, and the center of its circumscribed circle is equidistant from points $M$ and $N$. | 1.32. Since $\angle A M N = \angle M N C$ and $\angle B M N = \angle M N A$, then $\angle A M B = \angle A N C$. Moreover, $A M : A N = N B : N M = B M : C N$. Therefore, $\triangle A M B \sim \triangle A N C$, which means $\angle M A B = \angle N A C$. Consequently, $\angle B A C = \angle M A N$. The proof for other a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,675 |
1.33*. Segment $B E$ divides triangle $A B C$ into two similar triangles, with the similarity coefficient being $\sqrt{3}$. Find the angles of triangle $A B C$.
## §3. Ratio of Areas of Similar Triangles | 1.33. Since $\angle A E B + \angle B E C = 180^{\circ}$, these angles cannot be different angles of similar triangles $A B E$ and $B E C$, i.e., they are equal and $B E$ is a perpendicular.
There are two possible cases: $\angle A B E = \angle C B E$ or $\angle A B E = \angle B C E$. The first case is ruled out, as in ... | 90,60,30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,676 |
1.34. On side $AC$ of triangle $ABC$, a point $E$ is taken. Through point $E$, a line $DE$ parallel to side $BC$ and a line $EF$ parallel to side $AB$ are drawn ($D$ and $F$ are points on these sides, respectively). Prove that $S_{BDEF}=2 \sqrt{S_{ADE} \cdot S_{EFC}}$. | 1.34. $S_{B D E F} / 2 S_{A D E}=S_{B D E} / S_{A D E}=D B / A D=E F / A D=\sqrt{S_{E F C} / S_{A D E}}$. Therefore $S_{B D E F}=2 \sqrt{S_{A D E} \cdot S_{E F C}}$. | S_{BDEF}=2\sqrt{S_{ADE}\cdotS_{EFC}} | Geometry | proof | Yes | Yes | olympiads | false | 26,677 |
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