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XXIX OM - II - Problem 4
From the vertices of a regular $2n$-gon, 3 different points are chosen randomly. Let $p_n$ be the probability that the triangle with vertices at the chosen points is acute. Calculate $\lim_{n\to \infty} p_n$.
Note. We assume that all choices of three different points are equally probable. | Let $ A_1, A_2, \ldots , A_{2n} $ be the consecutive vertices of a regular $ 2n $-gon, and $ O $ - the center of the circle circumscribed around this polygon. We will investigate how many triangles $ A_1A_iA_j $, where $ i < j $, have the angle $ \measuredangle A_iA_1A_j $ not acute.
Since the inscribed angle in a circ... | \frac{1}{4} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XXXII - III - Task 5
Determine all pairs of integers $ (x,y) $ satisfying the equation
Note: The equation is missing in the provided text. If you have the equation, please provide it for a complete translation. | First, note that if the pair of integers $ (x, y) $ is a solution to the given equation, then the numbers $ x $ and $ y $ are of the same parity, because otherwise the left side of the equation would take an odd value, while the right side would be divisible by $ 8 $. Moreover, we check directly that no pair of equal n... | (2,8)(8,2) | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
XXVIII - III - Task 6
Given is a polynomial
where $ a \neq 0 $, $ Q $ is a non-zero polynomial, $ k $ is a natural number. Prove that $ W $ has at least $ k + 1 $ coefficients different from zero. | \spos{1} We will apply induction on $ k $ for $ k \geq 0 $. If $ k = 0 $, then from (1) we get $ W(x) = Q(x) $. Moreover, the polynomial $ W(x) $ has at least one non-zero coefficient, since the polynomial $ Q(x) $ is non-zero. The theorem given in the problem is therefore true for the number $ k = 0 $.
Next, let $ k $... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
XII OM - II - Problem 6
In the plane of an acute triangle $ABC$, a tape of width $d < AB$ with edges perpendicular to $AB$ is moved. At what position of the tape will it cover the largest part of the triangle? | First, let's note that the part of the triangle covered by the tape cannot be the largest when one of its edges, for example, the left one, is outside the triangle, because in this case, by moving the tape slightly to the right, the covered part increases. The maximum covered area also does not occur when both edges of... | proof | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
XLIII OM - I - Problem 8
In a tetrahedron $ABCD$, the sums of the lengths of opposite edges are $1$. The radii of the circles inscribed in the faces $BCD$, $CDA$, $DAB$, and $ABC$ are $r_A$, $r_B$, $r_C$, $r_D$. Prove that
and equality holds for a regular tetrahedron. | We will use the following fact:
If a triangle with side lengths $a$, $b$, $c$ is circumscribed around a circle with radius $r$, then
\[
a + b + c \geq 6 \sqrt{3} r,
\]
with equality holding only for an equilateral triangle.
(In other words: among all triangles circumscribed around a given circle, the one with the sho... | proof | Geometry | proof | Incomplete | Yes | olympiads | false |
XXXIV - III - Task 3
We consider a single-player game on an infinite chessboard based on the following rule. If two pieces are on adjacent squares and the next square is empty (the three squares being discussed lie on the same horizontal or vertical line), we can remove these pieces and place one piece on the third of... | By introducing a coordinate system on the plane, we can assign a pair of integers to each square of the chessboard, such that two adjacent squares in the same column are assigned the numbers $(m, n)$ and $(m, n+1)$, and two adjacent squares in the same row are assigned the numbers $(m, n)$ and $(m+1, n)$. We can assign... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
XXXV OM - I - Problem 12
In a trihedral angle $ W $, a ray $ L $ passing through its vertex is contained. Prove that the sum of the angles formed by $ L $ with the edges of $ W $ does not exceed the sum of the dihedral angles of $ W $. | Consider a sphere of radius 1 with its center at the vertex of a trihedral angle. The measure of each plane angle, whose sides are rays originating from the center of the sphere, is equal to the length of the corresponding arc of a great circle (the shorter of the two arcs defined by the points of intersection of the g... | proof | Geometry | proof | Yes | Yes | olympiads | false |
V OM - I - Task 3
Prove that if in a quadrilateral $ABCD$ the equality $AB + CD = AD + BC$ holds, then the incircles of triangles $ABC$ and $ACD$ are tangent. | If the incircles of triangles $ABC$ and $ACD$ touch their common side $AC$ at points $M$ and $N$ respectively (Fig. 20), then according to known formulas we have
from which
If $AB + CD = AD + BC$, then from the above equality it follows that $AM = AN$, i.e., that points $M$ and $N$ coincide; both circles are tangent ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
VII OM - III - Task 4
Prove that if natural numbers $ a $, $ b $, $ c $ satisfy the equation
then:
1° at least one of the numbers $ a $ and $ b $ is divisible by $ 3 $,
2° at least one of the numbers $ a $ and $ b $ is divisible by $ 4 $,
3° at least one of the numbers $ a $, $ b $, $ c $ is divisible by $ 5 $... | 1°. Since the square of an integer gives a remainder of $0$ or $1$ when divided by $3$ (see the solution to problem 2), if neither $a$ nor $b$ were divisible by $3$, then the remainder of $a^2 + b^2$ when divided by $3$ would be $2$, which is impossible since $a^2 + b^2$ equals the square of an integer $c$. Therefore, ... | proof | Number Theory | proof | Incomplete | Yes | olympiads | false |
LIII OM - II - Task 1
Given a function $ f: \mathbb{R} \to \mathbb{R} $, such that for every real number $ x $ the following equalities hold
Prove that the function $ f $ is periodic. | Based on the equality data in the problem, we obtain for any real number $ x $:
which proves that the function $ f $ is periodic with a period of $ \frac{1}{2} $. | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
LI OM - I - Task 8
Numbers $ c(n,k) $ are defined for non-negative integers $ n\geq k $ in such a way that the following equalities hold:
Prove that $ c(n,k) = c(n,n-k) $ for $ n\geq k \geq 0 $. | Let $ f $ be a function defined on the set of non-negative integers by the formulas:
Let also
Then $ a(n,k) = a(n,n-k) $. We will show that $ a(n,k) = c(n,k) $.
Since the conditions given in the problem uniquely determine the numbers $ c(n,k) $, it suffices to prove that the numbers $ a(n,k) $ satisfy the s... | proof | Combinatorics | proof | Incomplete | Yes | olympiads | false |
IX OM - II - Task 2
Six equal disks are placed on a plane in such a way that their centers lie at the vertices of a regular hexagon with a side equal to the diameter of the disks. How many rotations will a seventh disk of the same size make while rolling externally on the same plane along the disks until it returns to... | Let circle $K$ with center $O$ and radius $r$ roll without slipping on a circle with center $S$ and radius $R$ (Fig. 16). The rolling without slipping means that different points of one circle successively coincide with different points of the other circle, and in this correspondence, the length of the arc between two ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
X OM - I - Task 7
On a plane, there are four different points $ A $, $ B $, $ C $, $ D $.
a) Prove that among the segments $ AB $, $ AC $, $ AD $, $ BC $, $ BD $, $ CD $, at least two are of different lengths,
b) How are the points $ A $, $ B $, $ C $, $ D $ positioned if the segments mentioned above have only two di... | a) To prove that the segments $ AB $, $ AC $, $ AD $, $ BC $, $ BD $, $ CD $ cannot all be equal to each other, it suffices to show that if five of them have the same length, then the sixth has a different length. Let's assume, for example, that $ AB = AC = AD = BC = BD $. In this case, $ ABC $ and $ ABD $ are equilate... | notfound | Geometry | proof | Yes | Yes | olympiads | false |
XX OM - II - Task 1
Prove that if the real numbers $ a, b, c, d $ satisfy the equations
then
(Note: The equations are missing in the provided text, so they are not translated.) | \spos{1} If we eliminate $ b $ and $ d $ from the given equations, we will obtain an equation of the form $ f(a, c) = 0 $. Then, $ f(b, d) = 0 $ as well, since equations (1) --(3) do not change if we replace $ a, b, c, d $ with $ b, a, d, c $ respectively. Therefore, the numbers $ a, b, c, d $ satisfy the equation $ f(... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
XXI OM - I - Problem 6
Prove that a real number $ a $ is rational if and only if there exist integers $ p > n > m \leq 0 $ such that $ a + m, a + n, a + p $ form a geometric progression. | Let $ a $ be a rational number. We choose a natural number $ m $ such that $ a + m > 0 $. Then
where $ r $ and $ q $ are natural numbers.
Consider a geometric sequence with a common ratio of $ 1 + q $ and the first term $ \frac{r}{q} $. Thus, the subsequent terms of the sequence will be
Assuming then $ n = m + r $ an... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
L OM - III - Task 6
In a convex hexagon $ ABCDEF $, the following equalities hold:
Prove that $ \frac{AB}{BF} \cdot \frac{FD}{DE} \cdot \frac{EC}{CA}=1 $. | Given the assumption $ \measuredangle A + \measuredangle C + \measuredangle E = 360^\circ $, there exists a point $ P $ such that the following angle equalities hold (figure):
om50_3r_img_3.jpg
Triangles $ DEF $ and $ DCP $ are therefore similar, from which we obtain
KUsing the equality given in the problem and t... | proof | Geometry | proof | Incomplete | Yes | olympiads | false |
XLVIII OM - I - Problem 3
Prove that for any real numbers $ a, b \geq 1 $, $ c \geq 0 $, and for every natural number $ n \geq 1 $, the following inequality holds
保留了源文本的换行和格式，但最后一句“保留了源文本的换行和格式”是中文，应翻译为英文：
Preserving the line breaks and format of the original text. | Given the inequality to prove, we rewrite it in the form
Let us denote: $ \alpha = ab + ac $, $ \beta = ab + c $. Since $ a \geq 1 $, $ b \geq 1 $, $ c \geq 0 $, it follows that $ \alpha \geq a $, $ \beta \geq 1 $; moreover, $ \alpha - \beta = (a-1)c $, and therefore
we have obtained inequality (1). | proof | Inequalities | proof | Incomplete | Yes | olympiads | false |
XXIII OM - I - Problem 3
Prove that if $ A, B, C $ are angles of a triangle, and $ t $ is a real number, then
The text above has been translated into English, preserving the original text's line breaks and format. | Since $ A + B + C = \pi $, then $ \cos A = \cos [\pi - (B + C)] = - \cos (B + C) $. Therefore
This implies the inequality given in the problem. | proof | Algebra | proof | Incomplete | Incomplete | olympiads | false |
XXXVI OM - III - Problem 1
Determine the largest number $ k $ such that for every natural number $ n $ there are at least $ k $ natural numbers greater than $ n $, less than $ n+17 $, and coprime with the product $ n(n+17) $. | We will first prove that for every natural number $n$, there exists at least one natural number between $n$ and $n+17$ that is coprime with $n(n+17)$.
In the case where $n$ is an even number, the required property is satisfied by the number $n+1$. Of course, the numbers $n$ and $n+1$ are coprime. If a number $d > 1$ we... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXV OM - III - Task 5
Prove that if natural numbers $ n $, $ r $ satisfy the inequality $ r + 3 \leq n $, then the numbers $ \binom{n}{r} $, $ \binom{n}{r+1} $, $ \binom{n}{r+2} $, $ \binom{n}{r+3} $ are not consecutive terms of any arithmetic sequence. | We will first prove two lemmas.
Lemma 1. For a fixed natural number $n$, there are at most two natural numbers $k \leq n - 2$ such that the numbers $\binom{n}{k}$, $\binom{n}{k+1}$, $\binom{n}{k+2}$ are consecutive terms of an arithmetic sequence.
Proof. If the numbers $\displaystyle \binom{n}{k}, \binom{n}{k+1}, \bino... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
XXXIX OM - II - Problem 4
Prove that for every natural number $ n $ the number $ n^{2n} - n^{n+2} + n^n - 1 $ is divisible by $ (n - 1)^3 $. | Let us denote:
Assume that $ n \geq 2 $, hence $ m \geq 1 $. According to the binomial formula,
where $ A $, $ B $, $ C $ are integers. We transform further:
Thus, $ N = Dm^3 = D(n - 1)^3 $, where $ D $ is an integer (this is also true for $ n = 1 $).
We have thus shown that $ N $ is divisible by $ ... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
XXXVII OM - I - Problem 1
Let the natural numbers $ a_n $ and $ b_n $ satisfy the equality
Calculate $ \lim_{n\to \infty} \frac{a_n}{b_n} $. | We will find a recursive relationship between the terms of the sequences $ (a_n) $ and $ (b_n) $. Notice that
Hence, $ (2a_n+3b_n-a_{n+1}) = (a_{n}+2b_n-b_{n+1})\sqrt{3} $, and from the irrationality of the number $ \sqrt{3} $, we infer that both sides of the last equality must be zeros.
Therefore,
Further... | \sqrt{3} | Algebra | math-word-problem | Incomplete | Incomplete | olympiads | false |
LVIII OM - I - Problem 1
Problem 1.
Solve in real numbers $ x $, $ y $, $ z $ the system of equations | Subtracting the second equation of the system from the first, we get
Similarly, by subtracting the third equation from the first, we obtain the equation $ 0=(x-z)(x+z-2y+5) $. Therefore, the system of equations given in the problem is equivalent to the system
The product of two numbers is equal to zero if and ... | (x,y,z)=(-2,-2,-2),(\frac{1}{3},\frac{1}{3},\frac{1}{3}),(2,2,-3),(-\frac{1}{3},-\frac{1}{3},-\frac{16}{3}),(2,-3,2),(-\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XLIV OM - III - Problem 2
Point $ O $ is the center of the circle $ k $ inscribed in the non-isosceles trapezoid $ ABCD $, whose longer base $ AB $ has midpoint $ M $. The shorter base $ CD $ is tangent to the circle $ k $ at point $ E $, and the line $ OM $ intersects the base $ CD $ at point $ F $. Prove that $ |DE|... | Since the equality $ |DE| = |FC| $ (considered in the thesis of the problem) holds if and only if $ |DF| = |EC| $, the conditions of the assumptions and the thesis will not change when the roles of the legs $ AD $ and $ BC $ of the given trapezoid are swapped. Therefore, we can assume, without loss of generality, that ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XXXII - I - Problem 8
Given a point $ P $ inside a sphere $ S $. The transformation $ f: S\to S $ is defined as follows: for a point $ X \in S $, the point $ f(X) \in S $, $ f(X) \neq X $, and $ P\in \overline{Xf(X)} $. Prove that the image of any circle contained in $ S $ under the transformation $ f $ is a circle. | Let $ k $ be a circle contained in the sphere $ S $, and let $ S_1 $ be a sphere containing $ k $ and the point $ P $, $ P_1 $ be such a point that the segment $ \overline{PP_1} $ is a diameter of $ S_1 $. For every $ X \in k $, the angle $ PXP_1 $ is a right angle. Let $ X $ be the orthogonal projection of the point $... | proof | Geometry | proof | Yes | Yes | olympiads | false |
LVIII OM - II - Problem 3
From $ n^2 $ tiles in the shape of equilateral triangles with a side of $ 1 $, an equilateral triangle with a side of $ n $ was formed. Each tile is white on one side and black on the other. A move consists of the following actions: We select a tile $ P $ that shares sides with at least two o... | Let's call a boundary segment the common side of two tiles whose visible sides have different colors. Of course, the number of boundary segments is non-negative and does not exceed the number $ m $ of all segments that are the common side of two tiles. We will investigate how the number of boundary segments changes as ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
XXXI - II - Task 5
We write down the terms of the sequence $ (n_1, n_2, \ldots, n_k) $, where $ n_1 = 1000 $, and $ n_j $ for $ j > 1 $ is an integer chosen randomly from the interval $ [0, n_{j-1} - 1] $ (each number in this interval is equally likely to be chosen). We stop writing when the chosen number is zero, i.e... | For a given non-negative integer $ n $, consider the random variable $ X_n $ being the length $ k $ of the sequence ($ n_1, n_2, \ldots, n_k $), where $ n_1 = n $, $ n_j $ for $ j > 1 $ is an integer chosen randomly from the interval $ [0, n_{j-1} - 1] $, and $ n_k = 0 $. Let $ E_n $ denote the expected value of this r... | 8.48547 | Combinatorics | proof | Yes | Yes | olympiads | false |
IV OM - II - Task 6
Given a circle and two tangents to this circle. Draw a third tangent to the circle in such a way that the segment of this tangent contained between the given tangents has a given length $ d $. | When the tangent lines $ m $ and $ n $ intersect, the task coincides with the already solved problem No. 11.
When the lines $ m $ and $ n $ are parallel, the solution is immediate. From any point $ A $ on the line $ m $ as the center, we draw a circle with radius of length $ d $.
If this circle has a common point $ B $... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XVI OM - III - Task 2
Prove that if the numbers $ x_1 $ and $ x_2 $ are roots of the equation
$ x^2 + px - 1 = 0 $, where $ p $ is an odd number, then for every natural $ n $ the numbers $ x_1^n + x_2^n $ and $ x_1^{n+1} + x_2^{n+1} $ are integers and relatively prime. | We will apply the method of induction. Since $ p $ is an odd number, the numbers
are integers and relatively prime. The theorem is thus true for $ n = 0 $.
Suppose that for some natural number $ n \geq 0 $ the numbers
are integers and relatively prime. We will prove that then $ x_1^{n+2} + x_2^{n+2} $ is an... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
LVI OM - II - Problem 4
Given is the polynomial $ W(x)=x^2+ax+b $, with integer coefficients, satisfying the condition:
For every prime number $ p $ there exists an integer $ k $ such that the numbers $ W(k) $ and $ W(k + 1) $ are divisible by $ p $. Prove that there exists an integer $ m $ for which | The condition $ W(m) = W(m +1) = 0 $ means that $ W(x)=(x-m)(x-m+1) $, i.e., $ W(x)=x^{2}-(2m+1)x+m^{2}+m $. Therefore, it is necessary to show that there exists an integer $ m $ such that
Let us fix a prime number $ p $. Then for some integer $ k $, the numbers
are divisible by $ p $. Therefore, the following nu... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
XL OM - I - Problem 9
Determine all natural numbers $ x $, $ y $, $ z $ satisfying the equation
Note: The original text did not provide the specific equation. If you have the equation, please provide it for a complete translation. | From the given equation, it follows that if $ x $, $ y $, $ z $ is a solution, then $ x, y \geq 2 $, $ z \geq 3 $. We will consider three cases:
{\em Case} I. $ x = 2 $. The equation takes the form $ 3y^{-2} + 4z^{-2} = \frac{1}{2} $, or equivalently: $ 4y^2 + 3z^2 = \frac{1}{2} y^2z^2 $, or yet another way:
Cons... | (3,2,12)(3,3,3) | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
XXVI - I - Problem 8
Given is an acute triangle $ABC$. We append two equilateral triangles $ABC$ and $ACB$ to it externally. Let $K$ and $L$ be the midpoints of segments $\overline{AC}$ and $\overline{B}$, respectively, and let $M$ be a point on segment $\overline{BC}$ such that $BM = 3MC$.
Prove that the angles of tr... | From the conditions of the problem, it follows that the angle $ \measuredangle KAL $ is obtuse. We choose a coordinate system such that the points $ K $, $ L $, and $ A $ have the following coordinates: $ K = (-a, 0) $, $ L = (b, 0) $, $ A = (0, -c) $, where $ a $, $ b $, and $ c $ are certain positive numbers (Fig. 8)... | proof | Geometry | proof | Incomplete | Incomplete | olympiads | false |
XIII OM - I - Problem 6
Factor the quadratic polynomial into real factors
where $ p $ and $ q $ are real numbers satisfying the inequality
Please note that the mathematical expressions and symbols are kept as they are, only the text has been translated. | From the inequality $ p^2 - 4q < 0 $, it follows that the quadratic polynomial $ y^2 + py + q $ has no real roots; therefore, the quartic polynomial $ x^4 + px^2 + q $ also has no real roots. If the number $ x $ were a real root of the second polynomial, then the number $ y = x^2 $ would be a real root of the first pol... | x^4+px^2+(x^2+\alphax+\beta)(x^2-\alphax+\beta) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
LIX OM - I - Task 5
Find all such triples of prime numbers $ (p,q,r) $, that the numbers
are divisible by $ p+ q +r $. | Let the prime numbers $ p, q, r $ satisfy the given conditions. The number
is divisible by $ p+q+r $. Therefore, the numbers $ p^2-qr $ and $ p(p^2-qr)=p^3-pqr $ are also divisible by $ p+q+r $. Similarly, we prove that the numbers
are divisible by $ p + q + r $. Summing these three numbers, we get the divisibility... | (p,q,r)=(3,3,3) | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
XXXIX OM - II - Problem 2
Given are real numbers $ x_i $, $ y_i $ ($ i = 1, 2, \ldots, n $) such that
and
Prove that
(Note: The original text did not provide the specific conditions and the statement to prove. If you have the complete problem statement, please provide it for a more accurate trans... | Let's rewrite assumption (2) in the form
By the inequality between means, we have
Let us introduce the notation:
and additionally let $ s_0 = 0 $. From the inequality (3) obtained a moment ago, it follows that the numbers $ s_k $ are non-negative. Notice that
from which
Therefore,
Gi... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
VIII OM - III - Task 6
Given a cube with base $ABCD$, where $AB = a$ cm. Calculate the distance between the line $BC$ and the line passing through point $A$ and the center $S$ of the opposite face. | Since $ BC\parallel AD $, the line $ BC $ is parallel to the plane $ ADS $. The sought distance between the lines $ BC $ and $ AS $ is equal to the distance from the line $ BC $ to the plane $ ADS $ (Fig. 25).
Consider a parallelepiped whose base is the square $ ABCD $, and one of its side faces is the rectangle $ AMND... | Geometry | math-word-problem | Yes | Yes | olympiads | false | |
XXXVIII OM - III - Zadanie 5
Wyznaczyć najmniejszą liczbę naturalną $ n $, dla której liczba $ n^2-n+11 $ jest iloczynem czterech liczb pierwszych (niekoniecznie różnych).
|
Niech $ f(x) = x^2-x+11 $. Wartości przyjmowane przez funkcję $ f $ dla argumentów całkowitych są liczbami całkowitymi niepodzielnymi przez $ 2 $, $ 3 $, $ 5 $, $ 7 $. Przekonujemy się o tym badając reszty z dzielenia $ n $ i $ f(n) $ przez te cztery początkowe liczby pierwsze:
\begin{tabular}{lllll}
&\multicolum... | 132 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXIII OM - I - Task 1
Draw the graph of the function | If $ x \geq 1 $, then $ x \pm 2 \sqrt{x-1}= (x - 1) \pm 2 \sqrt{x-1} + 1 = (\sqrt{x-1} \pm 1)^2 $ and therefore.
From this, we see that the function $ y $ is defined for $ x \geq 1 $.
We have $ \sqrt{x-1} + 1 \geq 0 $ for every $ x \geq 1 $ and $ \sqrt{x -1} - 1 \geq 0 $ for $ x \geq 2 $ and $ \sqrt{x- 1} \leq 0 $ for ... | Algebra | math-word-problem | Incomplete | Yes | olympiads | false | |
XLVII OM - II - Problem 5
Find all pairs of integers $ (x,y) $ that satisfy the equation
保留了源文本的换行和格式，但请注意，最后一句“保留了源文本的换行和格式”是说明性的，不应包含在翻译文本中。这是翻译结果：
XLVII OM - II - Problem 5
Find all pairs of integers $ (x,y) $ that satisfy the equation
（This note is not part of the translation and is for you... | We substitute $ x = u + 1 $, $ y = v + 1 $ and transform the equation:
One of the factors of the obtained product must be equal to $ 5 $ or $ -5 $, and the other (respectively) $ 1 $ or $ -1 $.
From this, it follows that the numbers $ u + v $ and $ uv $ satisfy one of the following systems of equations:
In ... | (1,2),(2,1),(-5,2),(2,-5) | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
XXV OM - III - Task 2
Salmon swimming upstream must overcome two waterfalls. The probability that a salmon will overcome the first waterfall in a given attempt is $ p > 0 $, and the probability of overcoming the second waterfall in a given attempt is $ q > 0 $. We assume that successive attempts to overcome the waterf... | Let $ A_n $ be the event that the salmon does not overcome the first waterfall in $ n $ attempts, and $ B_n $ - the event that the salmon does not overcome both waterfalls in $ n $ attempts. Since the probability that the salmon does not overcome the first waterfall in one attempt is $ 1 - p $, and the attempts are ind... | \max(0,1-\frac{p}{q}) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XI OM - I - Problem 6
Prove that if the numbers $ a $, $ b $, $ c $, none of which are zero, satisfy the relationship
then they can be arranged in such an order that they form a geometric progression. | From three different non-zero numbers $ a $, $ b $, $ c $, a geometric progression can be formed if and only if one of them is the geometric mean of the other two, i.e., if one of the differences $ ab - c^2 $, $ bc - a^2 $, $ ca - b^2 $ equals zero, which is equivalent to the condition
We need to prove that equality (... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
L OM - III - Task 4
Determine for which natural numbers $ n \geq 2 $ the system of equations
has a solution in integers $ x_1,x_2,x_3,\ldots,x_n $. | The equation $ x_1^2 + x_2^2 + 50 = 16x_1 + 12x_2 $ is equivalent to the equation
Numbers $ x_1, x_2, \ldots, x_n $ satisfy the system of equations given in the problem if and only if the points $ (x_1, x_2), (x_2, x_3), \ldots, (x_n, x_1) $ lie on a circle with center $ (8, 6) $ and radius $ \sqrt{50} $. On this circ... | nisdivisible3 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
XLVIII OM - I - Problem 1
Solve the system of equations
Note: $ [t] $ is the greatest integer less than or equal to $ t $. | Suppose that the numbers $ x $, $ y $ satisfy the given system of equations.
First, consider the case where at least one of these numbers is negative; suppose, for example, that $ y < 0 $. Then $ [y] \leq -1 $, and in this case, by the second equation of the system, the inequality holds
Therefore, $ x \geq 2 $.
S... | (1,0)(0,1) | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
XXXVI OM - I - Problem 8
Prove that if $ (a_n) $ is a sequence of real numbers such that $ a_{n+2}=|a_{n+1}|-a_n $ for $ n = 1,2,\ldots $, then $ a_{k+9} = a_k $ for $ k = 1,2,\ldots $. | Let's consider the transformation $ F $ of the plane into itself given by the formula $ F(x,y) = (y,|y|-x) $. Let $ a $ be a positive number. Let $ I_0 $ be the segment with endpoints $ (0,-a) $ and $ (-a,0) $, let $ I_1 $ be the image of the segment $ I_0 $ under the transformation $ F $, and further, inductively, let... | proof | Algebra | proof | Yes | Yes | olympiads | false |
XX OM - I - Problem 11
In a convex quadrilateral $ABCD$, the sum of the distances from each vertex to the lines $AB$, $BC$, $CD$, $DA$ is the same. Prove that this quadrilateral is a parallelogram. | Let's denote the measures of the angles of a quadrilateral by $A$, $B$, $C$, $D$, and the lengths of the sides $AB$, $BC$, $CD$, $DA$ by $a$, $b$, $c$, $d$ respectively (Fig. 6).
The sum of the distances from vertex $A$ to the lines containing the sides of the quadrilateral is then $a \sin B + d \sin D$. Denoting this ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
LII OM - I - Problem 5
Prove that for any natural number $ n \geq 2 $ and any prime number $ p $, the number $ n^{p^p} + p^p $ is composite. | If $ p $ is an odd prime, then by the identity
in which we set $ x = n^{p^{p-1}} $ and $ y = p $, the number $ n^{p^p} + p^p $ is composite. (For the above values of $ x $, $ y $, the inequalities $ x^p + y^p > x + y > 1 $ hold, so each of the factors on the right-hand side of formula (1) is greater than 1).
On the ot... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
XXI OM - I - Problem 4
In a square $ABCD$ with side length 1, there is a convex quadrilateral with an area greater than $\frac{1}{2}$. Prove that this quadrilateral contains a segment of length $\frac{1}{2}$ parallel to $\overline{AB}$. | We draw through the vertices of a given quadrilateral lines parallel to line $AB$ (Fig. 4). In the case where these lines are different, the quadrilateral is divided into $3$ parts: a triangle with base $p$ and height $h_1$, a trapezoid with bases $p$ and $q$ and height $h_2$, and a triangle with base $q$ and height $h... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XXVI - I - Problem 6
Let $ T_n $ be a sequence of integers defined as follows:
Prove that
and find a formula for $ T_{2k} $. | By writing out the first five terms of the sequence $ T_n $, we can directly verify that formula (3) holds for $ k = 1, 2, 3 $. Let us assume that it holds for every natural number $ k $ not greater than $ n $, where $ n $ is some natural number $ \geq 3 $. We will prove formula (3) for the number $ n+1 $.
From (1) and... | T_{2k}=[\frac{12}{7}\cdot2^{k-3}]-1 | Number Theory | proof | Incomplete | Yes | olympiads | false |
XLIV OM - II - Task 1
Prove that for positive numbers $ x $, $ y $, $ u $, $ v $ the inequality holds
(Note: The original text had a lot of blank lines, which have been preserved in the translation.) | Multiplying both sides by the product of all denominators, we transform the given inequality into an equivalent form
Let's denote the left and right side of (1) by $L$ and $P$ respectively, and assume
Then
from which
Notice now that
and
Therefore, ultimately,
Inequality (... | proof | Inequalities | proof | Incomplete | Yes | olympiads | false |
XIII OM - I - Problem 4
Prove that the line symmetric to the median $CS$ of triangle $ABC$ with respect to the angle bisector of angle $C$ of this triangle divides the side $AB$ into segments proportional to the squares of the sides $AC$ and $BC$. | Let $ CM $ (Fig. 4) be the line symmetric to the median $ CS $ with respect to the angle bisector $ CD $ of angle $ C $ in triangle $ ABC $.
Since the areas of triangles with the same height are proportional to the bases of these triangles, therefore
From the equality $ \measuredangle ACD = \measuredangle DCB $ and... | proof | Geometry | proof | Yes | Yes | olympiads | false |
LIX OM - I - Task 7
In an $ n $-person association, there are $ 2n-1 $ committees (any non-empty set of association members
forms a committee). A chairperson must be selected in each committee. The following condition must be met: If
committee $ C $ is the union $ C = A\cup B $ of two committees $ A $ and $ B $, then ... | Answer: n!.
We will establish a one-to-one correspondence between the possible choices of committee chairs
and the ways of assigning members of the association the numbers $1, 2, \dots, n$ (where different members
are assigned different numbers). There are, of course, as many of the latter as there are permutations of ... | n! | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
LVI OM - II - Task 2
In a convex quadrilateral $ABCD$, point $M$ is the midpoint of diagonal $AC$. Prove that if
then a circle can be circumscribed around quadrilateral $ABCD$.
| Let us denote: $ \alpha=\measuredangle BAD $.
If $ \alpha=90^{\circ} $, then points $ B $, $ M $, $ D $ lie on the same straight line. Then points $ A $ and $ C $ are symmetric with respect to the line $ BD $. Hence $ \measuredangle BCD = \measuredangle BAD = 90^{\circ} $, which means that a circle can be circumscribed... | proof | Geometry | proof | Incomplete | Yes | olympiads | false |
XL OM - I - Problem 7
In space, a finite set of points is given, any four of which are vertices of a tetrahedron with a volume less than or equal to 1. Prove that there exists a tetrahedron with a volume no greater than 27, containing all these points. | Let $ABCD$ be a tetrahedron of maximum volume among all tetrahedra whose vertices are points of the considered set. Denote by $O$ its centroid, that is, the point of intersection of four segments, each connecting a vertex of the tetrahedron with the centroid of the opposite face. It is known that the centroid of a tetr... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XVII OM - I - Problem 4
On a plane, a circle and a point $ M $ are given. Find points $ A $ and $ B $ on the circle such that the segment $ AB $ has a given length $ d $, and the angle $ AMB $ is equal to a given angle $ \alpha $. | Suppose $ A $ and $ B $ are points on a given circle satisfying the conditions $ AB = d $, $ \measuredangle AMB = \alpha $ (Fig. 5).
Let us choose any chord $ CD $ of length $ d $ in the given circle. Rotate the triangle $ AMB $ around the center $ O $ of the given circle by the angle $ HOK $, where $ H $ and $ K $ den... | 4,3,2,1,或0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
IV OM - II - Task 4
Solve the system of equations
| In task No. 2, we considered a special case of the system of equations (1), where $ n = 7 $. When $ n $ is arbitrary, we need to distinguish two cases.
1. The number $ n $ is even, $ n = 2m $. Multiplying the equations: the first, third, ... up to the $(2m - 1)$-th, we get
\[
x_1 x_3 \cdots x_{2m-1} = 1 \cdot 3 \cdots... | notfound | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
XVI OM - II - Task 6
Prove that there does not exist a polyhedron whose every planar section is a triangle. | Every vertex of a polyhedron is the common point of at least three of its edges. Let $ A $ be a vertex of the polyhedron, and segments $ AB $ and $ AC $ as well as $ AB $ and $ AD $ be edges of two faces. Let us choose points $ P $ and $ Q $ on segments $ AC $ and $ AD $, respectively, different from the endpoints of t... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XLIV OM - II - Zadanie 6
Dana jest funkcja ciągła $ f: \mathbb{R} \to \mathbb{R} $ spełniająca warunki:
Obliczyć $ f(500) $.
|
W podanym równaniu
podstawiamy $ x = 1000 $ i otrzymujemy równość $ 999 \cdot f(999) = 1 $, czyli
Z kolei podstawmy w (1) $ x = 999 $: dostaniemy związek $ \frac{1}{999} \cdot f\left(\frac{1}{999}\right) = 1 $, czyli
Widzimy, że liczby $ 1/999 $ oraz $ 999 $ są wartościami funkcji $ f $. Zgodnie z za... | \frac{1}{500} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
XXXII - I - Problem 10
Determine all functions $ f $ mapping the set of all rational numbers $ \mathbb{Q} $ to itself that satisfy the following conditions:
a) $ f(1)=2 $,
b) $ f(xy) = f(x)f(y)-f(x+y)+1 $ for $ x, y \in \mathbb{Q} $. | If the function $ f $ satisfies conditions a) and b), then the function $ g: \mathbb{Q} \to \mathbb{Q} $ defined by the formula $ g(x) = f(x)- 1 $ satisfies the conditions
a') $ g(1) = 1 $
and $ g(xy)+1 = (g(x)+1)(g(y)+1)-g(x+y)-1+1 $, $ g(xy)+1 = g(x)g(y)+g(x)+g(y)+1-g(x+y) $, i.e.
b') $ g(xy)+g(x+y) = g(x)g(y)+g(x)... | f(x)=x+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XXVIII - III - Task 2
For a fixed natural number $ s \geq 3 $, we are given a sequence of circles $ (K_n) $ and a sequence of convex $ s $-gons $ (W_n) $ such that
Prove that the sequence of diameters of the circles $ K_n $ converges to zero. | Let points $P_1, P_2, \ldots, P_{s+1}$ divide the circumference of the circle $K_n$ into $s + 1$ equal arcs (Fig. 13, here we have taken $s = 5$). Consider the segments of the circle $K_n$ determined by the chords $\overline{P_1P_2}, \overline{P_2P_3}, \ldots, \overline{P_sP_{s+1}}, \overline{P_{s+1}P_1}$ and not conta... | proof | Geometry | proof | Incomplete | Yes | olympiads | false |
LVIII OM - II - Problem 1
The polynomial $ P(x) $ has integer coefficients. Prove that if the polynomials $ P(x) $ and $ P(P(P(x))) $ have a common real root, then they also have a common integer root. | Let the real number $ a $ be a common root of the polynomials $ P(x) $ and $ P(P(P(x))) $. From the equalities $ P(a)=0 $ and $ P(P(P(a)))=0 $, we obtain $ P(P(0))=0 $. The integer $ m=P(0) $ thus satisfies the conditions
which means that $ m $ is the desired common integer root. | proof | Algebra | proof | Yes | Yes | olympiads | false |
XXXI - III - Task 1
Calculate the area of an octagon inscribed in a circle, knowing that each of four consecutive sides of this octagon has a length of 1, and each of the remaining four has a length of 2. | The area of the octagon given is the sum of the areas of eight triangles, whose common vertex is the center of the given circle, and the bases are chords, four of which have a length of $1$, and the remaining four - a length of $2$. The same area has an octagon inscribed in a circle, in which the consecutive sides have... | 6+4\sqrt{2}-1 | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
XLII OM - III - Problem 5
Non-intersecting circles $ k_1 $ and $ k_2 $ lie one outside the other. The common tangents to these circles intersect the line determined by their centers at points $ A $ and $ B $. Let $ P $ be any point on circle $ k_1 $. Prove that there exists a diameter of circle $ k_2 $, one end of whi... | Let the centers of circles $k_1$ and $k_2$ be denoted by $O_1$ and $O_2$, and their radii by $r_1$ and $r_2$. Points $A$ and $B$ lie on the line $O_1O_2$, and it does not matter which letter denotes which point. For the sake of clarity, let us agree that point $B$ lies on the segment $O_1O_2$, and point $A$ lies outsid... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XLIII OM - I - Problem 6
The function $ f: \mathbb{R} \to \mathbb{R} $ has the property that the function
is periodic. Prove that the function $ f $ is also periodic. | Notice that $ g(x)=h(f(x)) $, where
We will show that the function $ h $ is strictly increasing. One can of course check that in each interval of the form $ ((2k- 1 )\pi; (2k + 1)\pi) $ (for integer $ k $) its derivative is positive, and therefore the function $ h $ is increasing in each closed interval $ \langle... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
III OM - I - Task 3
Prove that if
then at least two of the numbers $ a $, $ b $, $ c $ are equal. | The construction of the left-hand side components of equation (1) evokes the formula for the tangent of the difference of two angles, leading to a solution using trigonometry. We can write
where $ \alpha $, $ \beta $, $ \gamma $ are certain angles contained within the interval from $ - 90^\circ $ to $ 90^\circ $. Equa... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
XXIX OM - II - Problem 2
In the plane, a set $ M $ of points is given with the following properties:
1. The points of set $ M $ do not lie on a single straight line,
2. If points $ A, B, C, D $ are vertices of a parallelogram and $ A, B, C \in M $, then $ D \in M $,
3. If $ A, B \in M $, then $ AB \geq 1 $.
Prove that... | We will first prove three lemmas.
Lemma 1. From condition $3^\circ$, it follows that every circle contains only a finite number of points of set $M$.
Proof. Let $K$ be a circle with center at point $O$ and radius of length $r$, and let $K'$ be a circle with the same center and radius of length $r + \frac{1}{2}$. From c... | proof | Geometry | proof | Yes | Incomplete | olympiads | false |
XXVIII - III - Task 4
The function $ h: \mathbb{R} \to \mathbb{R} $ is differentiable and satisfies for every $ x $ the condition $ h(ax) = b h(x) $, where $ a $ and $ b $ are certain fixed real numbers and $ 0 \neq |a| \neq 1 $. Additionally, the function $ h $ is continuous at the point $ x = 0 $.
Prove that $ a = b... | From the formula
it follows that for every real number $ x \ne 0 $ the equality
holds, i.e.,
Passing to the limit in (2) as $ x \to 0 $, we obtain $ h $ and thus, since $ h $, that $ a = b $. Therefore, equality (1) takes the form
Hence, $ h(x) = h(a \cdot a^{-1} x) = a \cdot h(a^{-1} x) $, i.e.... | proof | Calculus | proof | Yes | Incomplete | olympiads | false |
XII OM - II - Task 4
Find the last four digits of the number $ 5^{5555} $. | \spos{1} We will calculate a few consecutive powers of the number $ 5 $ starting from $ 5^4 $:
It turned out that $ 5^8 $ has the same last four digits as $ 5^4 $, and therefore the same applies to the numbers $ 5^9 $ and $ 5^5 $, etc., i.e., starting from $ 5^4 $, two powers of the number $ 5 $, whose exponents ... | 8125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XLV OM - I - Problem 6
The function $ f: \mathbb{R} \to \mathbb{R} $ is continuous. Prove that if for every real number $ x $ there exists a natural number $ n $ such that $ (\underbrace{f \circ \ldots \circ f}_{n})(x) = 1 $, then $ f(1)=1 $. | We will denote by $ f^r $ the $ r $-th iterate of the function $ f $, that is, the $ r $-fold composition (superposition) $ f \circ f \circ \ldots \circ f $.
Suppose, contrary to the thesis, that $ f(1) \neq 1 $. Let $ m $ be the smallest positive integer for which the equality
holds (such a number exists by the condi... | proof | Algebra | proof | Yes | Yes | olympiads | false |
I OM - B - Task 13
Prove that if the equality
is true,
then the numbers $ x, y, z $ are proportional to the numbers $ a, b, c $. | When in a given equality we perform multiplication on both sides, and then carry out the reduction of like terms, we obtain the equality
Thus
that is
The sum of the squares of real numbers is only equal to zero when each of these numbers is equal to zero, therefore
Thus
that is
which was to be shown. | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
XIV OM - I - Zadanie 10
Dowieść, że jeżeli liczby $ a $, $ b $, $ c $ są dodatnie, to
|
Dla dowolnych liczb $ a $, $ b $, $ c $ prawdziwe są nierówności
Nierówności (2) i (3) są odpowiednio równoważne nierównościom
Dodając (4) i (5) stronami otrzymujemy
Po podzieleniu obu stron nierówności (6) przez liczbę $ 2abc $, która jest według założenia dodatnia, otrzymujemy nierównoś... | proof | Inequalities | proof | Incomplete | Yes | olympiads | false |
XLVII OM - III - Problem 1
Determine all pairs $ (n,r) $, where $ n $ is a positive integer and $ r $ is a real number, for which the polynomial $ (x + 1)^n - r $ is divisible by the polynomial $ 2x^2 + 2x + 1 $. | Let $ Q_n(x) $ and $ R_n(x) $ denote the quotient and remainder, respectively, when the polynomial $ (x + 1)^n $ is divided by $ 2x^2 + 2x + 1 $. The pair $ (n,r) $ is one of the sought pairs if and only if $ R_n(x) $ is a constant polynomial, identically equal to $ r $.
For $ n = 1,2,3,4 $ we have:
Therefore,
... | (4k,(-\frac{1}{4})^k) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XXIX OM - I - Problem 3
Let $ a, b, c, d $ be the lengths of the sides of any quadrilateral. Prove that there exists a trapezoid whose sides have lengths $ a, b, c, d $. | Without loss of generality, we can assume that $ a \geq b \geq c \geq d $. Then
From the equality $ a + b = c + d $ it follows that $ (a - d) + (b - c) = 0 $, and hence $ a = b = c = d $. From the equality $ a + c = b + d $, it follows that $ (a - b) + (c - d) = 0 $, and hence $ a = b $ and $ c = d $. Therefore, if eq... | proof | Geometry | proof | Yes | Incomplete | olympiads | false |
XVII OM - I - Problem 6
Prove the theorem: If the coefficients $ a, b, c, d $ of the cubic equation $ ax^3 + bx^2 + cx + d = 0 $ are integers, where the number $ ad $ is odd and the number $ bc $ is even, then the equation cannot have three rational roots. | Suppose the given equation has three rational roots $x_1, x_2, x_3$. Let us transform the equation by introducing a new variable $y = ax$, i.e., substituting $\frac{y}{a}$ for $x$ in the equation.
We obtain the equation
whose roots are $y_1 = ax_1$, $y_2 = ax_2$, $y_3 = ax_3$. These numbers, being the products of... | proof | Algebra | proof | Yes | Yes | olympiads | false |
LV OM - I - Task 1
Given is a polygon with sides of rational length, in which all internal angles are equal to $ 90^{\circ} $ or $ 270^{\circ} $. From a fixed vertex, we emit a light ray into the interior of the polygon in the direction of the angle bisector of the internal angle at that vertex. The ray reflects accor... | Since every internal angle of the considered polygon $ \mathcal{W} $ is $ 90^{\circ} $ or $ 270^{\circ} $, all sides of this polygon (after an appropriate rotation) are aligned horizontally or vertically (Fig. 1). Let $ p_1/q_1, p_2/q_2, \ldots, p_n/q_n $ be the lengths of the consecutive sides of the polygon $ \mathca... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XLIX OM - I - Problem 7
Determine whether there exists a convex polyhedron with $ k $ edges and a plane not passing through any of its vertices and intersecting $ r $ edges, such that $ 3r > 2k $. | Let $ W $ be any convex polyhedron; let $ k $ be the number of its edges, and $ s $ -- the number of faces. The boundary of each face contains at least three edges, and each edge is a common side of exactly two faces. Therefore, $ 2k \geq 3s $.
If a plane not passing through any vertex intersects $ r $ edges, then the ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
IX OM - I - Problem 12
Given a circle with radius $ r $ and rays $ AB $ and $ AC $ tangent to this circle at points $ B $ and $ C $. Determine the tangent to the given circle such that the segment contained within the angle $ BAC $ is the shortest. | The desired straight line should of course be sought among the tangents to the given circle that intersect the segments $AB$ and $AC$, because for other tangents, the segments contained in the angle $BAC$ are larger than the segments of the tangents parallel to them. Let $MN$ (Fig. 13) be the segment of the tangent lin... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false |
X OM - I - Task 5
Solve the equation
| First, note that the roots of the given equation can only be numbers not less than $1$. When $x \geq 1$, we have $x - 2 \sqrt{x - 1} = x - 1 - 2\sqrt{x-1} + 1 = (\sqrt{x - 1} - 1)^2$, and $x + 3 - 4 \sqrt{x - 1} = x - 1 - 4 \sqrt{x - 1} + 4 = (\sqrt{x - 1} - 2)^2$, so the given equation can be written in the form
o... | x\geq5 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
LI OM - I - Problem 6
Point $ X $ lies inside or on the boundary of triangle $ ABC $, where angle $ C $ is a right angle. Points $ P $, $ Q $, and $ R $ are the projections of point $ X $ onto sides $ BC $, $ CA $, and $ AB $, respectively. Prove that the equality
holds if and only if point $ X $ lies on side $ ... | Let's introduce the following notation for brevity:
It is necessary to prove that $ w = 0 $ if and only if $ XR = 0 $.
By the Pythagorean theorem, we obtain the following relationship (Fig. 1):
Transforming the above equality equivalently, we obtain in sequence:
om51_1r_img_9.jpg
From the last equalit... | proof | Geometry | proof | Incomplete | Incomplete | olympiads | false |
V OM - I - Task 1
Find such numbers $ a $, $ b $, $ c $, $ d $, so that the equality
is an identity. | First, note that $ 4x^2 + 16x + 15 = 4 \left( x^2 + 4x + \frac{15}{4} \right) = 4 \left( x + \frac{5}{2} \right) + \left( x + \frac{3}{2} \right) $; the left side of equation (1) therefore has a numerical value (or: numerical sense) only for values of $ x $ different from $ - \frac{5}{2} $ and from $ -\frac{3}{2} $. Th... | 3,\,-\frac{5}{2} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
XXXVIII OM - II - Zadanie 4
Wyznaczyć wszystkie pary liczb rzeczywistych $ a, b $, dla których wielomiany $ x^4 + 2ax^2 + 4bx + a^2 $ i $ x^3 + ax - b $ mają dwa różne wspólne pierwiastki rzeczywiste.
|
Oznaczmy rozważane wielomiany przez $ P $ i $ Q $:
Załóżmy, że wielomiany $ P $ i $ Q $ mają wspólne pierwiastki rzeczywiste $ x_1 $, $ x_2 $ ($ x_1\neq x_2 $). Liczby $ x_1 $ $ x_2 $ są wówczas także pierwiastkami wielomianu
Stąd $ a \neq 0 $ i $ \Delta = 9b^2--4a^3 > 0 $.
Zgodnie z wzorami Viete'a... | (,0),where0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
LI OM - II - Task 1
Determine whether every positive rational number can be represented in the form
where $ a $, $ b $, $ c $, $ d $ are positive integers. | For any positive integers $ p $, $ q $, the following equalities hold:
Therefore, every positive rational number can be represented in the desired form. | proof | Number Theory | proof | Incomplete | Yes | olympiads | false |
VII OM - III - Task 2
Prove that if
and $ n $ is any odd natural number, then
Note: The mathematical expressions are not provided in the original text, so they are left as blank spaces in the translation. | We transform equality (1); we move the expression $ \frac{1}{a + b + c} $ to the left side and multiply both sides by $ abc (a + b + c) $:
We open the parentheses and arrange the left side according to the powers of $ a $; we will get
and finally
From equality (3), it follows that two of the num... | proof | Number Theory | proof | Incomplete | Yes | olympiads | false |
XXXIV OM - III - Problem 5
In the plane, vectors $ \overrightarrow{a_1}, \overrightarrow{a_2}, \overrightarrow{a_3} $ of length 1 are given. Prove that one can choose numbers $ \varepsilon_1, \varepsilon_2, \varepsilon_3 $ equal to 1 or 2, such that the length of the vector $ \varepsilon_1\overrightarrow{a_1} + \varep... | Let's start from an arbitrary point $0$ with vectors $\overrightarrow{a_1}$, $\overrightarrow{a_2}$, $\overrightarrow{a_3}$. They determine three lines intersecting at point $0$. Among the angles determined by pairs of these lines, there is an angle of measure $\alpha$ not greater than $\frac{\pi}{3}$. Suppose this is ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XXV OM - II - Problem 3
Prove that the orthogonal projections of vertex $ D $ of the tetrahedron $ ABCD $ onto the bisecting planes of the internal and external dihedral angles at edges $ \overline{AB} $, $ \overline{BC} $, and $ \overline{CA} $ lie on the same plane. | The bisecting plane is a plane of symmetry of the dihedral angle. Therefore, the image $D'$ of the vertex $D$ in the symmetry relative to any of the considered bisecting planes lies in the plane $ABC$. It follows that if $P$ is the orthogonal projection of the point $D$ onto the bisecting plane, then $P$ is the midpoin... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XXXV OM - II - Problem 2
On the sides of triangle $ABC$, we construct similar isosceles triangles: triangle $APB$ outside triangle $ABC$ ($AP = PB$), triangle $CQA$ outside triangle $ABC$ ($CQ = QA$), and triangle $CRB$ inside triangle $ABC$ ($CR = RB$). Prove that $APRQ$ is a parallelogram or that points $A, P, R, Q$... | Consider a similarity with a fixed point $C$ that transforms $B$ into $R$ (this is the composition of a rotation around $C$ by the angle $BCR$ and a homothety with center $C$ and scale equal to the ratio of the base length to the arm length in each of the constructed isosceles triangles). This similarity transforms $A$... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XVIII OM - II - Problem 4
Solve the equation in natural numbers
Note: The equation itself is not provided in the original text, so it is not included in the translation. | Suppose that the triple $(x, y, z)$ of natural numbers satisfies equation (1). After dividing both sides of equation (1) by $xyz$ we get
One of the numbers $x$, $y$, $z$ is less than $3$; for if $x \geq 3$, $y \geq 3$, $z \geq 3$, then the left side of equation (2) is $\leq 1$, while the right side is $> 1$.
If $x < 3... | 7 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
I OM - I - Task 4
In triangle $ ABC $, angle $ A $ is given. Can angles $ B $ and $ C $ be calculated if it is known that a certain line passing through point $ A $ divides the triangle into two isosceles triangles? | Let the sought angles of triangle $ ABC $ be $ x $ and $ y $, and let segment $ AD $ divide the triangle into two isosceles triangles $ T_1 $ and $ T_2 $. | Geometry | math-word-problem | Yes | Incomplete | olympiads | false | |
XVI OM - I - Problem 2
A segment whose endpoints $ M $ and $ N $ lie on the perimeter of a triangle is drawn through the centroid $ S $ of the triangle. Prove that
(Note: The original text had a series of blank lines at the end, which have been preserved in the translation.) | Suppose that point $ M $ lies on side $ AB $, and point $ N $ lies on side $ BC $ of triangle $ ABC $ (Fig. 1). Let $ C', $ S', $ N' $ be the points symmetric to points $ C $, $ S $, $ N $ with respect to line $ AB $. The distance from the centroid $ S $ to the base $ AB $ of the triangle is equal to $ \frac{1}{3} $ of... | proof | Geometry | proof | Incomplete | Yes | olympiads | false |
XIII OM - I - Problem 3
Prove that the perpendiculars dropped from the centers of the excircles of a triangle to the corresponding sides of the triangle intersect at one point. | A simple solution to the problem can be inferred from the observation that on each side of the triangle, the point of tangency of the inscribed circle and the point of tangency of the corresponding excircle are symmetric with respect to the midpoint of that side. For example, if the inscribed circle in triangle $ABC$ t... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XLV OM - III - Task 4
We have three unmarked vessels: an empty $ m $-liter, an empty $ n $-liter, and a full $(m+n)$-liter vessel of water. The numbers $ m $ and $ n $ are relatively prime natural numbers. Prove that for every number $ k \in \{1,2, \ldots , m+n-1\} $, it is possible to obtain exactly $ k $ liters of w... | When pouring water from one container to another, we either completely empty the container we are pouring from or completely fill the container we are pouring into. (Since the containers have no markings, this is the only method that allows us to control the volume of water being poured.) Thus, after each pour, we have... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
XXVI - III - Task 5
Prove that a triangle with angle $ \alpha $ can be inscribed in a circle of radius $ R $ and circumscribed about a circle of radius $ r $ if and only if | Let circle $K$ with center at point $O$ and radius of length $r$ be tangent at points $P$ and $Q$ to the sides of angle $\alpha$ with vertex $A$. Let $B_0$ be the point of intersection of line $AP$ and line $m$ parallel to $AQ$ and different from $AQ$, and tangent to circle $K$. If point $B$ lies on the ray $t$ of line... | proof | Geometry | proof | Incomplete | Yes | olympiads | false |
XLVI OM - II - Problem 3
Given are positive irrational numbers $ a $, $ b $, $ c $, $ d $, such that $ a+b = 1 $. Prove that $ c+d = 1 $ if and only if for every natural number $ n $ the equality $ [na] +[nb] = [nc] + [nd] $ holds.
Note: $ [x] $ is the greatest integer not greater than $ x $. | Let's take any natural number $ n \geq 1 $. According to the definition of the symbol $ [x] $,
The products $ na $, $ nb $, $ nc $, $ nd $ cannot be integers (since the numbers $ a $, $ b $, $ c $, $ d $ are irrational by assumption), and therefore in none of the listed relationships can equality hold. Addi... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
LIX OM - III - Task 5
The areas of all sections of the parallelepiped $ \mathcal{R} $ by planes passing through the midpoints of three of its edges,
none of which are parallel and do not have common points, are equal. Prove that the parallelepiped
$ \mathcal{R} $ is a rectangular parallelepiped. | Let's assume that the areas of all the sections mentioned in the problem are equal to $S$.
Let $ABCD$ and $EFGH$ be the bases of the parallelepiped $\mathcal{R}$ (Fig. 3), let $O$ be its center of symmetry, and let $I, J, K, L, M, N$ denote the midpoints of the edges $AE, EF, FG, GC, CD, DA$, respectively. Then the fol... | proof | Geometry | proof | Yes | Yes | olympiads | false |
LVIII OM - I - Problem 12
The polynomial $ W $ with real coefficients takes only positive values in the interval $ \langle a;b\rangle $ (where $ {a<b} $). Prove that there exist polynomials $ P $ and $ Q_1,Q_2,\ldots,Q_m $ such that
for every real number $ x $. | We will conduct the proof by induction on the degree of the polynomial $W$.
If $W(x) \equiv c$ is a constant polynomial, then of course $c > 0$ and in this case we can take $P(x) \equiv \sqrt{c}$, $m=1$, and $Q_1(x) \equiv 0$.
Now suppose that the thesis of the problem is true for all polynomials of degree less than $n... | proof | Algebra | proof | Yes | Yes | olympiads | false |
XXVIII - I - Task 1
Find all solutions of the system of equations
Note: The original text did not contain any equations, so the translated text also does not include any equations. If there were supposed to be equations in the original text, they would need to be provided for an accurate translation. | Transferring all terms to the left side of the equations and factoring the resulting polynomials, we obtain
From the first of these equations, it follows that $ y - 2 = 0 $ or $ x - y - 2 = 0 $, which means $ y = 2 $ or $ y = x - 2 $. Substituting $ y = 2 $ into the second equation, we get $ (x - 1)(x + 3) ... | (1,2),(-3,2),(1,-1),(3,1),(\frac{1}{2},-\frac{3}{2}) | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
XXVII OM - II - Problem 6
Six points are placed on a plane in such a way that any three of them are vertices of a triangle with sides of different lengths. Prove that the shortest side of one of these triangles is also the longest side of another one of them. | Let $P_1, P_2, \ldots, P_6$ be given points. In each of the triangles $P_iP_jP_k$, we paint the shortest side red. In this way, some segments $\overline{P_rP_s}$ are painted red, while others remain unpainted.
It suffices to prove that there exists a triangle with vertices at the given points, all of whose sides are pa... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XV OM - I - Problem 9
Prove that the quotient of the sum of all natural divisors of an integer $ n > 1 $ by the number of these divisors is greater than $ \sqrt{n} $. | Let $ d_1, d_2, \ldots, d_s $ denote all the natural divisors of a given integer $ n > 1 $. Among these divisors, there are certainly unequal numbers, such as $ 1 $ and $ n $. According to the Cauchy inequality, the arithmetic mean of positive numbers that are not all equal is greater than the geometric mean of these n... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
III OM - III - Task 4
Prove that if the angles $ A $, $ B $, $ C $ of a triangle satisfy the equation
then one of these angles equals $ 120^\circ $. | We will obtain the proof of the theorem by transforming equation (1). To perform the transformation in a proper way, one must consider what it is aiming to achieve.
It suffices to prove that one of the numbers $ \sin \frac{3}{2} A $, $ \sin \frac{3}{2} B $, $ \sin \frac{3}{2} C $ equals zero, because then one of the a... | proof | Geometry | proof | Incomplete | Incomplete | olympiads | false |
XLII OM - I - Problem 2
Determine all real numbers $ x $ satisfying the inequality
Note: The translation is provided as requested, preserving the original text's line breaks and format. However, the mathematical problem statement itself is incomplete in the provided text, as the inequality is not specified... | Let's denote the left and right sides of the inequality by $ L(x) $ and $ P(x) $. Thus,
where
Notice that
the function $ f $ takes only positive values, and the function $ g $ - only negative values. Therefore,
Substituting the expressions defining the functions $ f $ and $ g $, we obtain the fo... | (-\infty,\frac{1}{3})\cup(3,\infty) | Inequalities | math-word-problem | Incomplete | Yes | olympiads | false |
XXXIV OM - I - Problem 7
Let $ S_n $ be the set of sequences of length $ n $ with terms $ -1 $, $ +1 $. We define the function $ f: S_n - \{(-1, 1, 1,\ldots, 1)\} \to S_n $ as follows: if $ (a_1, \ldots, a_n) \in S_n - \{(-1, 1, 1,\ldots, 1)\} $ and $ k = \max_{1\leq j \leq n} \{j \;:\; a_1\cdot a_2\cdot \ldots \cdot ... | To simplify the notation, we introduce the symbol $ f^{(k)} $ to denote the $ k $-fold iteration of the function $ f $:
The thesis of the problem will be proved by induction. The definition of the function $ f $ depends, of course, on the natural number $ n $. Where it helps to avoid misunderstandings, we will write $... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
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