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XXXVI OM - III - Problem 4
Inside the triangle $ ABC $, a point is given that is respectively $ d_a, d_b, d_c $ away from the lines $ BC, CA, AB $. Let $ r $ be the length of the radius of the circle inscribed in triangle $ ABC $. Prove that | The area of triangle $ABC$ can be represented as the sum of the areas of three triangles, whose common vertex is a given point, and the bases are the sides $AB$, $BC$, $CA$ respectively. If the lengths of these sides are $c$, $a$, $b$ respectively, then
On the other hand, the same area is equal to $\frac{1}{2} r(a+b+c... | proof | Geometry | proof | Incomplete | Incomplete | olympiads | false |
XXIII OM - II - Problem 6
Prove that there exists a function $ f $ defined and differentiable on the set of all real numbers, satisfying the conditions
Note: The conditions are not provided in the original text, so they are not translated. | We will prove that the function $ f $ defined by the formula
satisfies the conditions of the problem. Of course, $ f $ is a continuous function, because at the points $ x = 0, \displaystyle \frac{1}{2}, 1 $ the limit of this function is equal to its value, and in some neighborhood of each of the remaining points, the ... | proof | Calculus | proof | Incomplete | Yes | olympiads | false |
VI OM - II - Task 1
Calculate the sum $ x^4 + y^4 + z^4 $ given that $ x + y + z = 0 $ and $ x^2 + y^2 + z^2 = a $, where $ a $ is a given positive number. | Attention. The system of equations $ x+y+z = 0 $ and $ x^2 + y^2 + z^2 = a $, where $ a > 0 $, has infinitely many solutions $ (x, y, z) $. The result of the above problem means that for each of these solutions, the sum $ x^4 + y^4 + z^4 $ has the same value $ \frac{a^2}{2} $. If the condition $ x+y+z =0 $ were replace... | \frac{^2}{2} | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
XIX OM - I - Problem 7
Points $ D, E, F $ are the midpoints of the sides of triangle $ ABC $. Prove that if the circumcircles of triangles $ ABC $ and $ DEF $ are tangent, then the point of tangency is one of the points $ A $, $ B $, $ C $ and triangle $ ABC $ is a right triangle. | Suppose that the circle $c$ circumscribed around triangle $ABC$ and the circle $k$ circumscribed around triangle $DEF$ are tangent at point $T$ (Fig. 5).
Triangle $DEF$ lies inside circle $c$, so the tangency of circles $c$ and $k$ is internal, and the center $S$ of circle $k$ lies on the ray $TO$. Triangle $DEF$ is si... | proof | Geometry | proof | Yes | Yes | olympiads | false |
VIII OM - III - Task 5
Given a line $ m $ and a segment $ AB $ parallel to it. Divide the segment $ AB $ into three equal parts using only a ruler, i.e., by drawing only straight lines. | We first find the midpoint $ S $ of segment $ AB $ as in problem 17 (Fig. 16) (Fig. 23). We draw lines $ AT $ and $ SE $, which intersect at point $ G $, and line $ GB $ intersecting line $ m $ at point $ H $. Since $ AS = SB $ and $ m \parallel AB $, then $ DT = TE $ and $ TE = EH $. Therefore, if $ K $ is the point o... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false |
LIV OM - III - Task 1
In an acute triangle $ABC$, segment $CD$ is an altitude. Through the midpoint $M$ of side $AB$, a line is drawn intersecting rays $CA$ and $CB$ at points $K$ and $L$, respectively, such that $CK = CL$. Point $S$ is the center of the circumcircle of triangle $CKL$. Prove that $SD = SM$. | From Menelaus' theorem applied to triangle $ABC$ we get
Let $E$ be the second intersection point of line $CS$ with the circumcircle of triangle $CKL$ (Fig. 1). From the equality $CK = CL$, it follows that $EK = EL$. Moreover, $\measuredangle AKE = 90^\circ = \measuredangle BLE$.
Therefore, the right triangles $AK... | proof | Geometry | proof | Yes | Yes | olympiads | false |
LII OM - I - Problem 6
Integers $ a $, $ b $, $ x $, $ y $ satisfy the equation
Prove that $ a \geq 44b $. | We will first show that the following equality holds:
Indeed, from the binomial formula of Newton, we obtain
where the integers \( c, d \) are given by:
Using the equality given in the problem statement, we get
From the irrationality of the number \( \sqrt{2001} \), it follows that \( b = d \), ... | proof | Number Theory | proof | Incomplete | Yes | olympiads | false |
XIX OM - I - Problem 2
Let $ p(n) $ denote the number of prime numbers not greater than the natural number $ n $. Prove that if $ n \geq 8 $, then $ p(n) < \frac{n}{2} $ | If $ n $ is an even number, then in the set of natural numbers from $ 1 $ to $ n $, there are $ \frac{n}{2}-1 $ even numbers different from $ 2 $, and thus composite, and at least one odd number, namely $ 1 $, which is not a prime number. Prime numbers belong to the set of the remaining $ \frac{n}{2} $ numbers, so $ p(... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
XXXVII OM - II - Problem 3
Let S be a sphere circumscribed around a regular tetrahedron with an edge length greater than 1. The sphere $ S $ is represented as the union of four sets. Prove that there exists one of these sets that contains points $ P $, $ Q $, such that the length of the segment $ PQ $ exceeds 1. | We will first prove a lemma that is a one-dimensional version of this theorem.
Lemma. The circle $\omega$ circumscribed around an equilateral triangle with a side length greater than $1$ is represented as the union of three sets: $\omega = U \cup V \cup W$. Then, there exist points $P$, $Q$ in one of the sets $U$, $V$,... | proof | Geometry | proof | Yes | Yes | olympiads | false |
X OM - III - Task 6
Given is a triangle in which the sides $ a $, $ b $, $ c $ form an arithmetic progression, and the angles also form an arithmetic progression. Find the ratios of the sides of this triangle. | Suppose that triangle $ABC$ satisfies the conditions of the problem, with $A \leq B \leq C$, and thus $a \leq b \leq c$. In this case
From equation (1) and the equation $A + B + C = \pi$, it follows that $B = \frac{\pi}{3}$; the cosine rule then gives the relationship
Eliminating $b$ from equations (2... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XI OM - I - Problem 9
a) Find a necessary and sufficient condition for the quadratic equations
to have a common root.
b) Prove that if these equations have a common root but are not identical, and the numbers $ p_1 $, $ q_1 $, $ p_2 $, $ q_2 $ are rational, then the roots of the given equations are rationa... | a) Suppose that a number $ x $ satisfies both equations (1). The condition we seek can be found by eliminating $ x $ from these equations, which we can do, for example, in the following way.
If $ p_1 \ne p_2 $, then we can solve equations (1) for $ x $ and $ x^2 $, which gives
and from this, as a result of eliminat... | proof | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
XXI OM - III - Task 5
In how many ways can a set consisting of twelve elements be divided into six disjoint two-element sets? | Let's denote one of the elements of a 12-element set by $a_1$. A two-element subset containing the element $a_1$ can be formed in $11$ ways. Specifically, we need to add one of the remaining eleven elements to $a_1$.
Let's denote by $a_2$ one of the elements of the 10-element set that remains after forming the two-elem... | 11\cdot9\cdot7\cdot5\cdot3 | Combinatorics | math-word-problem | Yes | Incomplete | olympiads | false |
XXII OM - II - Problem 1
In how many ways can $ k $ fields of a chessboard $ n \times n $ ($ k \leq n $) be chosen so that no two of the selected fields lie in the same row or column? | First, one can choose $ k $ rows in which the selected fields will lie. This can be done in $ \displaystyle \binom{n}{k} $ ways. Then, in each of these rows, one must sequentially choose one of $ n $ fields, one of $ n - 1 $ fields lying in the remaining columns, etc. - one of $ n - k + 1 $ fields. Therefore, the total... | \binom{n}{k}n(n-1)\ldots(n-k+1) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
L OM - I - Problem 6
Diagonals $ AC $ and $ BD $ of a convex quadrilateral $ ABCD $ intersect at point $ P $. Point $ M $ is the midpoint of side $ AB $. Line $ MP $ intersects side $ CD $ at point $ Q $. Prove that the ratio of the areas of triangles $ BCP $ and $ ADP $ is equal to the ratio of the lengths of segment... | We will denote by $ [XYZ] $ the area of triangle $ XYZ $. Let $ K $ be the intersection point of segments $ DP $ and $ AQ $, and let $ L $ be the intersection point of segments $ CP $ and $ BQ $. Since point $ M $ is the midpoint of segment $ AD $, therefore (by Ceva's theorem)
We thus obtain
Please not... | proof | Geometry | proof | Yes | Incomplete | olympiads | false |
XXXIII OM - III - Problem 1
Indicate such a way of arranging $ n $ girls and $ n $ boys around a round table so that the number $ d_n - c_n $ is maximized, where $ d_n $ is the number of girls sitting between two boys, and $ c_n $ is the number of boys sitting between two girls. | Consider any arrangement of $ n $ girls and $ n $ boys. Let's call a group of $ k $ girls (for $ k > 1 $) any sequence of $ k $ girls sitting next to each other, if the first and the $ k $-th girl in this sequence are seated next to a boy at the table. Let $ D_n $ be the number of groups of girls in a given arrangement... | [\frac{n}{2}]-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
IX OM - I - Problem 11
Prove that if two quadrilaterals have the same midpoints of their sides, then they have equal areas. Show the validity of an analogous theorem for convex polygons with any even number of sides. | Let $M$, $N$, $P$, $Q$ be the midpoints of the sides $AB$, $BC$, $CD$, $DA$ of quadrilateral $ABCD$ (Fig. 10). By the theorem on the segment joining the midpoints of two sides of a triangle, in triangle $ABD$ we have
and in triangle $BCD$ we have
thus $MQ = NP$ and $MQ \parallel NP$, so quadrilateral $MNPQ$ is a para... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XLVIII OM - II - Problem 1
For each real number $ a $, determine the number of ordered triples of real numbers $ (x, y, z) $ satisfying the system of equations
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”是额外的说明，不是翻译的一部分。正确的翻译结果应不包含这句话。 | Substitution: $ x=u+\frac{1}{2} $, $ y=v+\frac{1}{2} $, $ z=w+\frac{1}{2} $ reduces the given system of equations to the form:
---
from which, by subtracting sides: $ u^2 = v^2 = w^2 $. Therefore, each solution $ (u, v, w) $ of the system (*) has exactly one of the following forms:
---
For $ u = v = w = t $, the sy... | notfound | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
XXIV OM - II - Problem 2
In a given square, there are nine points, no three of which are collinear. Prove that three of them are vertices of a triangle with an area not exceeding $ \frac{1}{8} $ of the area of the square. | Let's first prove the
Lemma. If triangle $ABC$ is contained within a certain rectangle, then the area of the triangle is not greater than half the area of that rectangle.
Proof. Let $A$, $B$, $C$ be the projections of the vertices of triangle $ABC$ onto one of the sides of the rectangle $PQRS$ containing this triangle ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
VIII OM - I - Problem 9
Prove that if $ x $, $ y $, $ z $, and $ \sqrt{x} + \sqrt{y} + \sqrt{z} $ are rational numbers, then $ \sqrt{x} $, $ \sqrt{y} $, and $ \sqrt{z} $ are also rational numbers. | Let
according to the assumption $ w $ is a rational number. Then
Squaring both sides of this equality, we obtain
from which follows
Squaring both sides again, we get
and from this follows
If $ w(w^2 + z - x - y) \ne 0 $, then from equation (2) it follows that
and thus in this c... | proof | Algebra | proof | Yes | Incomplete | olympiads | false |
LI OM - II - Task 2
The bisector of angle $ BAC $ of triangle $ ABC $ intersects the circumcircle of this triangle at point $ D $ different from $ A $. Points $ K $ and $ L $ are the orthogonal projections of points $ B $ and $ C $, respectively, onto the line $ AD $. Prove that | Let $ o $ be the circumcircle of triangle $ ABC $. From the equality of angles $ BAD $ and $ DAC $, it follows that the lengths of arcs $ BD $ and $ DC $ of circle $ o $ are equal (the length of arc $ XY $ is measured from point $ X $ to point $ Y $ in the counterclockwise direction).
om51_2r_img_1.jpg
om51_2r_img_2.jp... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XII OM - II - Task 3
Prove that for any angles $ x $, $ y $, $ z $ the following equality holds | We transform the right side of equation (1) by first applying the formula $ 2 \sin \alpha \sin \beta = \cos (\alpha - \beta) - \cos (\alpha + \beta) $, and then the formulas for $ \cos (\alpha + \beta) $ and $ \cos (\alpha - \beta) $:
After substituting in the last expression $ \sin^2 x \sin^2 y = (1 - \cos^2 x) ... | proof | Algebra | proof | Yes | Yes | olympiads | false |
XXVII OM - I - Problem 11
Transmitting and receiving stations are sequentially connected: $ S_0 $ with $ S_1 $, $ S_1 $ with $ S_2 $, ..., $ S_n $ with $ S_{n+1} $, ... Station $ S_0 $ transmits a signal of 1 or -1, station $ S_1 $ receives the same signal with probability $ 1 - \varepsilon $, and the opposite signal ... | We have of course $ p^n(1 \mid 1) = p^n(-1 \mid -1) $, $ p^n(1 \mid 1) + p^n(1 \mid -1) = 1 $ and $ p^n(-1 \mid 1) + p^n(-1 \mid -1) = 1 $. Therefore, $ p^n(1 \mid -1) = p^n(-1 \mid 1) $.
From the above equalities, it follows that the limits of the sequences $(p^n(1 \mid 1))$ and $(p^n(-1 \mid -1))$ are equal, and simi... | Algebra | math-word-problem | Yes | Yes | olympiads | false | |
XXX OM - II - Task 3
In space, a line $ k $ and a cube with vertex $ M $ and edges $ \overline{MA} $, $ \overline{MB} $, $ \overline{MC} $, each of length 1, are given. Prove that the length of the orthogonal projection of the edge $ MA $ onto the line $ k $ is equal to the area of the orthogonal projection of the squ... | Without loss of generality, we can assume that the line $k$ passes through the point $M$ (Fig. 12) and that the plane $\pi$ perpendicular to the line $k$ also contains the point $M$. Let $A$ be the orthogonal projection of the point $A$ onto the line $k$.
om30_2r_img_12.jpg
As is known, if two planes intersect at an an... | proof | Geometry | proof | Yes | Yes | olympiads | false |
III OM - I - Task 4
a) Given points $ A $, $ B $, $ C $ not lying on a straight line. Determine three mutually parallel lines passing through points $ A $, $ B $, $ C $, respectively, so that the distances between adjacent parallel lines are equal.
b) Given points $ A $, $ B $, $ C $, $ D $ not lying on a plane. Deter... | a) Suppose that the lines $a$, $b$, $c$ passing through points $A$, $B$, $C$ respectively and being mutually parallel satisfy the condition of the problem, that is, the distances between adjacent parallel lines are equal. Then the line among $a$, $b$, $c$ that lies between the other two is equidistant from them. Let th... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
II OM - III - Task 3
Prove that if $ a > 0 $, $ b > 0 $, $ c > 0 $, then the inequality holds
The translation is provided as requested, maintaining the original text's line breaks and format. | We apply the transformation
We obtained the sum of three non-negative numbers, and thus a non-negative number; from this, the inequality given in the problem immediately follows.
Note 2. Inequality (1) can be generalized. Let $ a_1, a_2, \ldots, a_n $ denote positive numbers. By discarding one of these numbers in... | proof | Inequalities | proof | Incomplete | Incomplete | olympiads | false |
LX OM - I - Task 1
On some fields of a chessboard of size $ m \times n $, rooks are placed.
It is known that any rook is in the attack range of at most
two other rooks.
Determine, depending on $ m, n \geq 2 $, the maximum number of rooks on the chessboard,
for which such a situation is possible. | Answer: $ m+n $ rooks.
First, we will prove that on a rectangular chessboard of size $ m\times n $ $ (m, n \geqslant 1) $, one can place at most $ m+n $ rooks such that each rook is in the attack range of no more than two others. We will conduct the proof in three ways.
Let's call a row connected if it contains a roo... | +n | Combinatorics | math-word-problem | Yes | Incomplete | olympiads | false |
LV OM - I - Task 2
Determine whether there exists a prime number $ p $ and non-negative integers $ x $, $ y $, $ z $ satisfying the equation | Suppose there exist numbers $ p $, $ x $, $ y $, $ z $ satisfying the given equation in the problem. Since $ p $ is a prime number, the divisors of $ p^z $ are only powers of $ p $. Therefore, there exist non-negative integers $ a $, $ b $ such that $ 12x + 5 = p^a $ and $ 12y + 7 = p^b $. From this, it follows that th... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
LVII OM - III - Problem 4
We perform the following operation on a triplet of numbers. We select two of these numbers and replace them with their sum and their product, while the remaining number remains unchanged. Determine whether, starting from the triplet (3,4,5) and performing this operation, we can again obtain a... | If in the first step we choose the numbers 3 and 5, we will get the triplet (4, 8, 15), in which exactly one number is odd. Choosing two even numbers from such a triplet, we replace them with even numbers, whereas deciding on one even and one odd number, we get numbers of different parities. Therefore, from a triplet i... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XLVI OM - I - Problem 8
In a regular $ n $-sided pyramid, the angles of inclination of the lateral face and the lateral edge to the base plane are $ \alpha $ and $ \beta $, respectively. Prove that $ \sin^2 \alpha - \sin^2 \beta \leq \tan ^2 \frac{\pi}{2n} $. | Let's assume that the side of the regular $ n $-gon $ A_1A_2\ldots A_n $, which serves as the base of the pyramid, has a length of $ 2a $. Let $ O $ be the common center of the circle circumscribed around this polygon and the circle inscribed in it; let the radii of these circles be (respectively) $ R $ and $ r $. Deno... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XIV OM - II - Task 1
Prove that if the numbers $ p $, $ q $, $ r $ satisfy the equations
then for any numbers $ a $, $ b $, $ c $ the equality holds | From equation (2) it follows that
Since
then by virtue of (1) and (4)
Therefore | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
LX OM - II - Task 5
Determine all integers $ n \geqslant 4 $ with the following property: Among any $ n $ different 3-element subsets of an $ n $-element set, one can select two subsets that have exactly one element in common. | Let $A_1, A_2, \cdots, A_k$ be distinct 3-element subsets of a fixed $n$-element set.
Suppose, moreover, that any two of these sets are either disjoint or have exactly two elements in common.
Let $\mathcal{A} = \{A_1, A_2, \cdots, A_k\}$ and introduce the following notation:
$X \sim Y$ if and only if the intersection $... | n\geq4 | Combinatorics | math-word-problem | Yes | Incomplete | olympiads | false |
XXV OM - III - Problem 1
In the tetrahedron $ ABCD $, the edge $ \overline{AB} $ is perpendicular to the edge $ \overline{CD} $ and $ \measuredangle ACB = \measuredangle ADB $. Prove that the plane determined by the edge $ \overline{AB} $ and the midpoint of the edge $ \overline{CD} $ is perpendicular to the edge $ \o... | We will first prove the
Lemma. If lines $AB$ and $PQ$ intersect at a right angle at point $P$, then the number of points on the ray $PQ^\to$ from which the segment $\overline{AB}$ is seen at an angle $\alpha$ is $0$, $1$, or $2$.
Proof. As is known, the set of points contained in the half-plane with edge $AB$ and conta... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XXIII OM - I - Problem 6
Determine for which digits $ a $ the decimal representation of a number $ \frac{n(n+1)}{2} $ ($ n\in \mathbb{N} $) consists entirely of the digit $ a $. | Of course, $ a $ cannot be zero. If the decimal representation of the number
$ \displaystyle t_n = \frac{1}{2} n (n + 1) $ consists of $ k $ ones, where $ k \geq 2 $, then $ 9 t_n = 10^k - 1 $. From this, after simple transformations, we obtain $ (3n + 1) \cdot (3n + 2) = 2 \cdot 10^k = 2^{k+1} \cdot 5^k $. Since the c... | 5,6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXXVIII OM - III - Problem 2
A regular $ n $-gon is inscribed in a circle of length 1. From among the arcs of the circle having endpoints at the vertices of this polygon, we randomly draw five arcs $ L_1, \ldots, L_5 $ with replacement, where the probability of drawing each arc is the same. Prove that the expected val... | The vertices of a polygon divide a circle into $n$ arcs of length $1/n$. Let us number these arcs: $J_1, \ldots, J_n$. Denote by $\mathcal{L}$ the set from which the selection is made. The elements of the set $\mathcal{L}$ are arcs of a given circle, each of which has two ends, which are different vertices of the consi... | \frac{1}{32} | Combinatorics | proof | Yes | Yes | olympiads | false |
XXXIX OM - III - Problem 3
Let $ W $ be a polygon (not necessarily convex) with a center of symmetry. Prove that there exists a parallelogram containing $ W $ such that the midpoints of the sides of this parallelogram lie on the boundary of $ W $. | Among all triangles $OAB$, where $A$, $B$ are vertices of the polygon $W$, and $O$ is its center of symmetry, let us choose the triangle with the maximum area (there are at least four such triangles, and there may be more; we choose any one of them). Let $OMN$ be the selected triangle, and let $M$, $N$ be the vertices ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XXI OM - I - Task 1
Solve in integers the equation
| If $ |x| \geq 2 $ and $ |y| \geq 2 $, then
Therefore, if the pair of integers $ x $, $ y $ is a solution to equation (1), then $ |x| = 1 $ or $ |y| = 1 $. If $ |x| = 1 $, the equation becomes $ 1 + \frac{1}{xy} + \frac{1}{y^2} = 1 $, which simplifies to $ \frac{x+y}{xy^2} = 0 $. Therefore, $ x + y = 0 $. It follows th... | (-1,1)(1,-1) | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
XX OM - II - Task 6
Prove that every polyhedron has at least two faces with the same number of sides. | Let $ s $ denote the number of faces of the polyhedron $ W $, and $ P $ be the face with the largest number of sides, which we will denote by $ n $. The face $ P $ is adjacent along its sides to $ n $ other (pairwise distinct) faces of the polyhedron, so $ s \geq n+1 $. On the other hand, the number of sides of each fa... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XLIX OM - I - Problem 12
Let $ g(k) $ be the greatest prime divisor of the integer $ k $, when $ |k|\geq 2 $, and let $ g(-1) = g(0) = g(1) = 1 $. Determine whether there exists a polynomial $ W $ of positive degree with integer coefficients, for which the set of numbers of the form $ g(W(x)) $ ($ x $ - integer) is fi... | We will prove that there does not exist a polynomial with the given property. For a proof by contradiction, suppose that
is a polynomial of degree $ n \geq 1 $ with integer coefficients and that the set of numbers of the form $ g(W(x)) $ (for integer values of $ x $) is finite. This means: there exists a natural numbe... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
XLIX OM - II - Problem 3
(a) Non-negative numbers $ a $, $ b $, $ c $, $ d $, $ e $, $ f $, whose sum is equal to $ 1 $, satisfy the relationship
Prove that
(b) Decide whether there exist six different positive numbers $ a $, $ b $, $ c $, $ d $, $ e $, $ f $ with a sum equal to $ 1 $, for which both writt... | (a) Let $ A = ace + bdf $, $ B = abc + bcd + cde + def + efa + fab $. Using the inequality between the geometric mean and the arithmetic mean, we obtain
Therefore, $ B \leq \frac{1}{27} - A \leq \frac{1}{27} - \frac{1}{108} = \frac{1}{36}. $
(b) Such numbers exist. For the inequalities in the problem to become equali... | notfound | Inequalities | proof | Incomplete | Yes | olympiads | false |
LVII OM - I - Problem 3
An acute triangle $ABC$ is inscribed in a circle with center $O$. Point $D$ is the orthogonal projection of point $C$ onto line $AB$, and points $E$ and $F$ are the orthogonal projections of point $D$ onto lines $AC$ and $BC$, respectively. Prove that the area of quadrilateral $EOFC$ is equal t... | Let $ P $ be the point symmetric to point $ C $ with respect to point $ O $ (Fig. 1). Since triangle $ ABC $ is acute,
points $ A $, $ P $, $ B $, and $ C $ lie on a circle with center $ O $ in this exact order.
om57_1r_img_1.jpg
The areas of triangles $ COE $ and $ POE $ are equal, as these triangles have a common he... | proof | Geometry | proof | Yes | Yes | olympiads | false |
LVI OM - II - Task 1
Determine all positive integers $ n $ for which
are prime numbers. | Let $a \geq 2$, $m \geq 1$ be integers. We will first show that if the number $m$ has an odd divisor greater than 1, then the number $a^m + 1$ is composite. Indeed: if $l$ is an odd divisor of $m$, then $m = ld$ for some positive integer $d$, and
From the above formula, it follows that the number $a^d + 1$ is a diviso... | n=1orn=2 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
XL OM - I - Task 4
Prove that it is impossible to cut a square along a finite number of segments and arcs of circles in such a way that the resulting pieces can be assembled into a circle (pieces can be flipped). | Suppose the division mentioned in the task is feasible. The resulting parts of the square are denoted by $K_1, \ldots, K_n$. The boundary of each figure $K_i$ consists of a finite number of line segments and arcs of circles. Among the arcs of circles that are parts of the boundary of $K_i$, some are convex outward from... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XXVI - I - Problem 11
A ship is moving at a constant speed along a straight line with an east-west direction. Every $ T $ minutes, the direction of movement is randomly chosen: with probability $ p $, the ship moves in the eastern direction for the next $ T $ minutes, and with probability $ q= 1-p $, it moves in the w... | Let's assume that a submarine fires a torpedo at the moment when a lottery is taking place on the ship. In the case of firing the torpedo at a different time, the reasoning proceeds similarly. We can also assume, without loss of generality, that $ p \geq q $, i.e., $ \displaystyle p \geq \frac{1}{2} $. After $ T $ minu... | \frac{4}{9} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XIII OM - I - Problem 11
Given is a quadrilateral $ABCD$ whose diagonals intersect at a right angle at point $M$. Prove that the $8$ points where perpendiculars drawn from point $M$ to the lines $AB$, $BC$, $CD$, and $DA$ intersect the sides of the quadrilateral lie on a circle. | Let $P$, $Q$, $R$, $S$ denote the feet of the perpendiculars dropped from point $M$ to the sides $AB$, $BC$, $CD$, $DA$ respectively (Fig. 16).
First, we will prove that the points $P$, $Q$, $R$, $S$ lie on a circle by showing that the sum of two opposite angles of quadrilateral $PQRS$ equals $180^\circ$.
Each of the q... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XV OM - III - Task 4
Prove that if the roots of the equation $ x^3 + ax^2 + bx + c = 0 $, with real coefficients, are real, then the roots of the equation $ 3x^2 + 2ax + b = 0 $ are also real. | The task boils down to showing that the given assumptions imply the inequality
Let $ x_1 $, $ x_2 $, $ x_3 $ denote the roots of equation (2); according to the assumption, they are real numbers.
We know that
Hence
Note. Using elementary knowledge of derivatives, the problem can be solved much more simply. It i... | proof | Algebra | proof | Yes | Yes | olympiads | false |
XXXI - III - Task 2
Prove that for any number $ n $ there exists a solution to the equation
in natural numbers $ a, b, c $ greater than $ n $. | First, let's note that there are solutions to the given equation in natural numbers, for example, $ (1, 1, 1) $, $ (2, 1, 1) $, $ (5, 2, 1) $, etc.
We will show that if a triplet of natural numbers $ (a, b, c) $ satisfying the condition $ a > b > c $ is a solution to the given equation, then there exists a natural numb... | proof | Number Theory | proof | Incomplete | Yes | olympiads | false |
LVIII OM - III - Task 2
A positive integer will be called white if it is equal to 1
or is the product of an even number of prime numbers (not necessarily distinct).
The remaining positive integers will be called black.
Determine whether there exists a positive integer such that the sum of its white
divisors is equal ... | For an integer $ k>1 $, let us denote
We will prove that for any relatively prime positive integers $ l $, $ m $, the equality
is true.
Indeed, every positive divisor $ d $ of the product $ lm $ has a unique representation in the form $ d=ab $, where $ a $ is a divisor of the number $ l $, and $ b $ is a ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XLIV OM - II - Problem 5
On the sides $ BC $, $ CA $, $ AB $ of triangle $ ABC $, points $ D $, $ E $, and $ F $ are chosen respectively such that the incircles of triangles $ AEF $, $ BFD $, and $ CDE $ have radii equal to $ r_1 $. The incircles of triangles $ DEF $ and $ ABC $ have radii $ r_2 $ and $ r $, respectiv... | Let $ P $, $ Q $, $ R $ be the points of tangency of the incircle of triangle $ ABC $ with the sides $ BC $, $ CA $, $ AB $, respectively, and let $ K $ be the point of tangency of the incircle of triangle $ EAF $ with the side $ AF $. The centers of these circles are denoted by $ I $ and $ J $, respectively (see Figur... | r_1+r_2 | Geometry | proof | Yes | Yes | olympiads | false |
XXXII - I - Problem 5
Prove that if $ O $ is the center, $ P $ - any point on the inscribed circle of an equilateral triangle $ A_lA_2A_3 $ different from the point of tangency, then | Since $ \overrightarrow{A_iP} = \overrightarrow{A_iO} + \overrightarrow{OP} $, therefore
If the radius of the inscribed circle is $ r $, then $ OP= r $, $ A_iO = 2r $ $ (i = 1, 2, 3) $. Let $ \varphi $ be the angle of minimum measure among the angles formed by the vector $ \overrightarrow{OP} $ with the vectors $... | proof | Geometry | proof | Incomplete | Yes | olympiads | false |
II OM - I - Task 2
Find the condition that the coefficients of the quadratic polynomials
should satisfy so that between the roots of each of these polynomials lies a root of the other polynomial. The letters $ m $, $ n $, $ p $, $ q $ denote real numbers here. | The task is to deduce the relationship that must hold between the coefficients $ m $, $ n $, $ p $, and $ q $ under the assumption that the quadratic polynomials (1) have real roots such that the pair of roots of one equation and the pair of roots of the other equation interlace each other. Let the roots of the first q... | notfound | Algebra | math-word-problem | Incomplete | Incomplete | olympiads | false |
IV OM - II - Task 3
A triangular piece of sheet metal weighs $ 900 $ g. Prove that by cutting this sheet along a straight line passing through the center of gravity of the triangle, it is impossible to cut off a piece weighing less than $ 400 $ g. | We assume that the sheet is of uniform thickness everywhere; the weight of a piece of sheet is then proportional to the area of the plane figure represented by that piece of sheet. The task is to show that after cutting a triangle along a straight line passing through its center of gravity, i.e., through the point of i... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XII OM - III - Problem 5
Four lines intersecting at six points form four triangles. Prove that the circumcircles of these triangles have a common point. | Four lines intersecting at $6$ points form a figure known as a complete quadrilateral (see Problems from Mathematical Olympiads, vol. I, Warsaw 1960. Problem 79). Let us denote these points by $A$, $B$, $C$, $D$, $E$, $F$ (Fig. 21) in such a way that on the given lines lie the respective triplets of points $(A, B, C)$,... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XXIX OM - II - Task 1
Prove that if $ 0 < x \leq \frac{1}{2} $, $ 0 < y \leq \frac{1}{2} $ | \spos{1} Since by the conditions of the problem the numbers $ x $, $ y $, $ 1-x $, $ 1-y $ are positive, multiplying both sides of the given inequality by $ xy (1 -x) (1 -y) $ we obtain an equivalent inequality
Multiplying out the brackets and moving all terms to the left side we get
that is, $ (x - y)^2(1 - x ... | proof | Inequalities | proof | Incomplete | Yes | olympiads | false |
XXX OM - I - Task 7
For a given natural number $ n $, calculate the number of integers $ x $ in the interval $ [1, n] $ for which $ x^3 - x $ is divisible by $ n $. | If $ n = p^k $, where $ k \geq 1 $ and $ p $ is an odd prime, then for any integer $ x $, at most one of the numbers $ x - 1, x, x + 1 $ is divisible by $ p $. Therefore, the number $ x^3 - x = (x - 1) x(x + 1) $ is divisible by $ p^k $ if and only if one of the numbers $ x - 1, x, x + 1 $ is divisible by $ p^k $, whic... | Number Theory | math-word-problem | Yes | Yes | olympiads | false | |
XLIII OM - II - Problem 1
Each vertex of a certain polygon has both coordinates as integers; the length of each side of this polygon is a natural number. Prove that the perimeter of the polygon is an even number. | Let $ A_1,\ldots,A_n $ be the consecutive vertices of a polygon; it will be convenient for us to denote the vertex $ A_n $ also by $ A_0 $. The boundary of the polygon forms a closed broken line, hence
Let us denote the coordinates of the $ i $-th vector of the above sum by $ u_i $, $ v_i $, and its length by $ w... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XLII OM - III - Problem 2
Let $ X $ be the set of points in the plane $ (x.y) $ with both coordinates being integers. A path of length $ n $ is any sequence $ (P_0, P_1,\ldots, P_n) $ of points in the set $ X $ such that $ |P_{i-1}P_{i}|=1 $ for $ i \in \{1,2,\ldots,n\} $. Let $ F(n) $ be the number of different paths... | For any path of length $ n $, with the starting point $ P_0 = (0,0) $, we associate a sequence of $ 2n $ symbols, each of which is a zero or a one. We do this as follows. If the path has the form $ (P_0,P_1,\ldots,P_n) $, then the corresponding zero-one sequence
called the code of the path, is determined by the set of... | \binom{2n}{n} | Combinatorics | proof | Yes | Yes | olympiads | false |
XVII OM - II - Problem 1
Solve the equation in natural numbers
Note: The equation to be solved is not provided in the original text. If you have the specific equation, please provide it for a complete translation. | Suppose that the quadruple of natural numbers $(x, y, z, t)$ is a solution to equation (1) and that
From (2) it follows that $x + y + z + t \leq 4t$ and $xyzt \geq x^3t$, so by (1) $x^3t \leq 4t$, hence $x^3 \leq 4$, and since $x$ is a natural number, $x = 1$. Substituting this value into equation (1) we get
Since ac... | (1,1,2,4)itspermutations,(- | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
XLVII OM - III - Problem 6
Among all permutations $ f $ of the set $ \{1,2, \ldots , n\} $ satisfying the condition
we choose one (each choice is equally probable). Let $ p_n $ be the probability that the chosen permutation satisfies the condition
Determine all natural numbers $ n $ for which $ p_n > \frac... | Let $ V_n $ be the set of those permutations of the set $ \{1,2,\ldots,n\} $ (i.e., distinct-value functions $ f: \{1,2,\ldots,n\} \to \{1,2,\ldots,n\} $) that satisfy the first of the given conditions:
Let $ U_n $ be the set of those permutations $ f $ that satisfy both given conditions: (1) and
(Obviously, $ U_n $ ... | 1,2,3,4,5,6 | Combinatorics | math-word-problem | Incomplete | Yes | olympiads | false |
XXIX OM - I - Problem 4
Let $ Y $ be a figure consisting of closed segments $ \overline{OA} $, $ \overline{OB} $, $ \overline{OC} $, where point O lies inside the triangle $ ABC $. Prove that in no square can one place infinitely many mutually disjoint isometric images of the figure $ Y $. | Let $ d $ be the smallest of the numbers $ OA $, $ OB $, $ OC $. Choose points $ P \in \overline{OA} $, $ Q \in \overline{OB} $, $ R \in \overline{OC} $ such that $ OP= OQ = OR= \frac{d}{2} $ (Fig. 6).
Then $ PQ < OP + OQ = d $ and similarly $ PR < d $ and $ QR < d $. If $ Y $ is a figure consisting of closed segments ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XLV OM - I - Problem 5
Prove that if the polynomial $ x^3 + ax^2 + bx + c $ has three distinct real roots, then the polynomial $ x^3 + ax^2 + \frac{1}{4}(a^2 + b)x +\frac{1}{8}(ab - c) $ also has three distinct real roots. | Let's denote the polynomials in the problem by $P(x)$ and $Q(x)$:
Replacing the variable $x$ in the polynomial $P(x)$ with the difference $2x - a$ and transforming the obtained expression:
From the obtained identity:
it immediately follows that if a number $\xi$ is a root of the polynomial $P(x)$, then the number $-... | proof | Algebra | proof | Yes | Yes | olympiads | false |
LX OM - I - Task 11
Prove that for any integers
$ k>m \geqslant 1 $ the inequality is satisfied | The given inequality to prove can be rewritten in the form
Thus, the thesis of the problem reduces to showing that the sequence $ (a_n) $ given by the formula
is strictly decreasing. In other words, it suffices to prove the inequality
We will proceed with further reasoning in two ways.
Fix a number $ t = 1, 2, \cdo... | proof | Inequalities | proof | Incomplete | Yes | olympiads | false |
LI OM - I - Task 5
Determine all pairs $ (a,b) $ of natural numbers for which the numbers $ a^3 + 6ab + 1 $ and $ b^3 + 6ab + 1 $ are cubes of natural numbers. | Let $ a $ and $ b $ be numbers satisfying the conditions of the problem. Without loss of generality, we can assume that $ a \leq b $. Then
Since the number $ b^3+6ab + 1 $ is a cube of an integer, the following equality must hold
Transforming the above relationship equivalently, we obtain step by step
... | (1,1) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
LIII OM - II - Problem 4
Determine all such triples of prime numbers $ p \leq q \leq r $, that the numbers
are prime. | If all three prime numbers $ p $, $ q $, $ r $ were greater than $ 2 $, then the number $ pq + r $ as an even number greater than $ 2 $ would be composite. Therefore, at least one of the numbers $ p $, $ q $, $ r $ is equal to $ 2 $, which, given the inequality $ p \leq q \leq r $, gives $ p = 2 $.
Furthermore, $ q > 2... | (2,3,5) | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
XXXIII OM - I - Problem 11
Real numbers $ u, v, w $ are roots of the equation $ x^3 + qx + r = 0 $, where $ r \neq 0 $. Express the roots of the equation $ r^2x^3 + q^3x + q^3 = 0 $ in terms of $ u, v, w $ and show that it has no roots in the interval $ (-1, 3) $. | Based on Vieta's formulas is
Suppose the roots of the equation $ r^2x^3 + q^3x + q^3 = 0 $ are
$ x_1 = \frac{q}{r}a $, $ x_2 = \frac{q}{r}b $, $ x_3 = \frac{q}{r}c $. Therefore,
it follows that (after possibly reordering) $ a = u $, $ b = v $, $ c = w $.
Therefore, the roots of the e... | proof | Algebra | proof | Yes | Incomplete | olympiads | false |
X OM - I - Task 2
To prove that
| Given
therefore
from where
The equation above has a root $ x = 4 $; by factoring the left side, it can be written in the form:
from which it is clear that it has no other real roots. Therefore, equality (1) holds.
Note 1. Since
taking into account equality (1), we see that $ \sqrt[3]{20 +... | proof | Other | proof | Incomplete | Yes | olympiads | false |
IX OM - III - Task 1
Prove that the product of three consecutive natural numbers, of which the middle one is a cube of a natural number, is divisible by $ 504 $. | We need to prove that if $a$ is a natural number greater than $1$, then the number
is divisible by $504 = 7 \cdot 8 \cdot 9$. Since the numbers $7$, $8$, and $9$ are pairwise coprime, the task reduces to proving the divisibility of $N$ by each of these numbers.
a) The number $a$ can be represented in the form $a = 7k ... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
XXXV OM - I - Problem 9
Three events satisfy the conditions:
a) their probabilities are equal,
b) any two of them are independent,
c) they do not occur simultaneously.
Determine the maximum value of the probability of each of these events. | om35_1r_img_4.jpg
Suppose that $ A $, $ B $, $ C $ are these events, each with a probability of $ p $. Given the assumption of independence between any two of the considered events, the probability of each intersection $ A \cap B $, $ B \cap C $, $ C \cap A $ is $ p^2 $. Furthermore, $ A \cap B \cap C \ne 0 $. Therefor... | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XXXIV OM - III - Problem 6
Prove that if all dihedral angles of a tetrahedron are acute, then all its faces are acute triangles. | We assign a unit vector perpendicular to each face and directed outward from the tetrahedron. Consider two faces forming an angle $ \alpha $. The vectors assigned to these faces form an angle $ \pi - \alpha $ (Fig. 12). Therefore, for all dihedral angles of the tetrahedron to be acute, it is necessary and sufficient th... | proof | Geometry | proof | Yes | Yes | olympiads | false |
VII OM - III - Zadanie 1
Rozwiązać układ równań
|
1) Wyznaczymy najpierw rozwiązania rzeczywiste układu (1) spełniające warunek $ x \ne 0 $, $ y \ne 0 $, $ z \ne 0 $. Przy tym założeniu układ (1) jest równoważny układowi równań
który otrzymujemy dzieląc każde z równań (1) obustronnie przez $ xyz $.
Łatwy rachunek pokazuje, że układ (2) jest równoważny układowi ... | Algebra | math-word-problem | Incomplete | Incomplete | olympiads | false | |
V OM - I - Zadanie 6
Rozłożyć na czynniki wyrażenie
|
Zauważmy najpierw, że wyrażenie $ W $ dopuszcza podstawienie cykliczne, tzn. nie ulega zmianie, gdy jednocześnie zastąpimy w nim $ x $ przez $ y $, $ y $ przez $ z $ i $ z $ przez $ x $.
Jeżeli w wyrażeniu $ W $, które jest wielomianem stopnia czwartego względem $ x $, podstawimy $ x = y $, otrzymamy $ W = 0 $; wielom... | -1,-1 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
XLVI OM - I - Problem 12
The sequence $ (x_n) $ is defined as follows:
Prove that for every natural number $ n \geq 1 $ the inequality holds: | Let's consider the sequence of positive numbers $ (y_n) $ defined as follows:
The given recursive formula for the sequence $ (x_n) $ implies an analogous formula for the sequence $ (y_n) $:
that is,
Assume additionally that $ y_0 = 1 $. The obtained recursive formula is then valid also for $ n= 1 $. N... | proof | Inequalities | proof | Incomplete | Yes | olympiads | false |
XLIV OM - I - Zadanie 1
Rozwiązać równanie
Uwaga: $ [t] $ jest największą liczbą całkowitą nie większą od $ t $,
|
Oznaczmy lewą i prawą stronę równania odpowiednio przez $ L(x) $ i $ P(x) $; z postaci równania wynika, że $ x\neq 0 $. Dla wszystkich liczb $ x \neq 0 $ mianownik ułamka definiującego wyrażenie $ L(x) $ jest różny od zera, a więc wyrażenie to jest dobrze określone i mamy równość
stąd zaś
Wartość bezwzględ... | -\sqrt{2},-1,\sqrt{3},2 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
LIX OM - III - Task 1
In the fields of an $ n \times n $ table, the numbers $ 1, 2, \dots , n^2 $ are written, with the numbers $ 1, 2, \dots , n $ being in the first row (from left to right), the numbers $ n +1, n +2, \dots ,2n $ in the second row, and so on.
$n$ fields of the table have been selected, such that no ... | We will first prove that
Let $ b_i $ denote the column number in which the number $ a_i $ is located. Then,
From the conditions of the problem, it follows that the sequence $ (b_1,b_2,\dots ,b_n) $ is a permutation of the sequence $ (1, 2,\dots ,n) $. Therefore,
and in view of formula (2), the proof ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
XVI OM - I - Problem 1
Prove that for any real numbers $ a, b, c $ | Let's introduce the notations: $ bc = u $, $ ca = v $, $ ab = w $. The theorem we need to prove takes a simpler form:
or
The above inequality is true because
Note. The proven theorem has a simple geometric interpretation. Let $ A $ and $ B $ be vectors in three-dimensional space. From the inequa... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
XVIII OM - III - Problem 3
In a room, there are 100 people, each of whom knows at least 67 others. Prove that there is a quartet of people in the room where every two people know each other. We assume that if person $A$ knows person $B$, then person $B$ also knows person $A$. | In the set $M$ of people in the room, there are at most $32$ people who do not know a certain person $A$ (assuming that person $A$ knows themselves).
If $A$, $B$, and $C$ are any three people in the set $M$, then there is a person $D$ in the set $M$, different from $A$, $B$, and $C$, who knows these three people. Indee... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
XLII OM - I - Problem 1
On the shore of a lake in the shape of a circle, there are four piers $K$, $L$, $P$, $Q$. A kayak departs from pier $K$ heading towards pier $Q$, and a boat departs from pier $L$ heading towards pier $P$. It is known that if, maintaining their speeds, the kayak headed towards pier $P$ and the b... | om42_1r_img_1.jpg
Suppose the kayak is moving at a speed of $ v_k $, and the boat is moving at a speed of $ v_l $. Let $ X $ be the point where a "collision" would occur. The lengths of the segments into which point $ X $ divides the chords $ KP $ and $ LQ $ are related by $ |KX| \cdot |PX| = |LX| \cdot |QX| $ (Figure ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
LIX OM - III - Task 6
Let $ S $ be the set of all positive integers that can be represented in the form $ a^2 + 5b^2 $
for some relatively prime integers $ a $ and $ b $. Furthermore, let $ p $ be a prime number that gives a remainder
of 3 when divided by 4. Prove that if some positive multiple of the number $ p $ bel... | Given the conditions of the problem, the number $a^2 + 5b^2$ is divisible by $p$ for some relatively prime integers $a$ and $b$.
The number $b$ is not divisible by $p$; otherwise, from the divisibility $p \mid b$ and $p \mid a^2 + 5b^2$, we would obtain $p \mid a$, contradicting the relative primality of $a$ and $b$. ... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
XXVII OM - II - Problem 3
We consider a hemispherical cap that does not contain any great circle. The distance between points $A$ and $B$ on such a cap is defined as the length of the arc of the great circle of the sphere with endpoints at points $A$ and $B$, which is contained in the cap. Prove that there is no isome... | Let a given spherical cap correspond to the central angle $ \alpha $. From the assumption, we have $ 0 < \alpha < \frac{\pi}{2} $. For any angle $ \beta $ satisfying $ 0 < \beta < \alpha $, consider a regular pyramid $ OABCD $, whose base is a square $ ABCD $ inscribed in the given spherical cap, and whose vertex $ O $... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XXXII - I - Task 2
Given a non-negative integer $ n $. Prove that the number of systems of integers $ n_1 \geq n_2 \geq \ldots \geq n_k \geq 0 $, such that
is not less than $ 2^n $. | We will conduct an inductive proof with respect to $ n $.
For $ n = 0 $, we have $ 2^0 = 1 $, and of course, a system containing the single number $ 1 $ satisfies the requirements of the problem.
Assume that for every non-negative integer $ m $ less than the natural number $ n $, the number of systems $ n_1 \geq n_2 \g... | proof | Combinatorics | proof | Incomplete | Yes | olympiads | false |
XXVIII - I - Problem 8
Prove that the set $ \{1, 2, \ldots, 2^{s+1}\} $ can be partitioned into two $ 2s $-element sets $ \{x_1, x_2, \ldots, x_{2^s}\} $, $ \{y_1, y_2, \ldots, y_{2^s}\} $, such that for every natural number $ j \leq s $ the following equality holds | We will apply induction with respect to $s$. For $s = 1$, we have the set $\{1, 2, 3, 4\}$. By splitting it into subsets $\{1, 4\}$ and $\{2, 3\}$, we will have the conditions of the problem satisfied, since $1 + 4 = 2 + 3$.
Next, assume that for some natural number $s$, there exists such a partition of the set $\{1, 2... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
VII OM - I - Task 3
In a square $ABCD$ with area $S$, vertex $A$ is connected to the midpoint of side $BG$, vertex $B$ is connected to the midpoint of side $CD$, vertex $C$ is connected to the midpoint of side $DA$, and vertex $D$ is connected to the midpoint of side $AB$. Calculate the area of the part of the square ... | The solution to the problem is illustrated in Fig. 1, where $ M $, $ N $, $ P $, $ Q $ denote the midpoints of the sides, and $ O $ the center of the square; through the vertices $ A $, $ B $, $ C $, $ D $ lines parallel to the lines $ DQ $, $ AM $, $ BN $, $ CP $ have been drawn. If the figure is rotated by $ 90^\circ... | \frac | Geometry | math-word-problem | Yes | Yes | olympiads | false |
VI OM - I - Problem 10
Prove that the number $ 53^{53} - 33^{33} $ is divisible by $ 10 $. | It is enough to prove that the last digits of the numbers $53^{53}$ and $33^{33}$ are the same. The last digit of the product $(10k + a)(10m + b)(10n + c) \ldots$, where $k, m, n, \ldots a, b, c, \ldots$ are natural numbers, is equal to the last digit of the product $abc\ldots$. The number $53^{53}$ therefore has the s... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
XIX OM - I - Problem 6
Prove that if the polynomial $ F(x) $ with integer coefficients has a root $ \frac{p}{q} $, where $ p $ and $ q $ are relatively prime integers, then $ p - q $ is a divisor of the number $ F(1) $, and $ p + q $ is a divisor of the number $ F(-1) $. | Let us substitute $ x $ with the variable $ y = qx $. Equation (1) then takes the form
Consider the polynomial
whose coefficients are integers. It has an integer root $ y=p $, hence
where $ Q(y) $ is a polynomial with integer coefficients*).
Substituting $ y = q $ into this equality, we get
The numbers $ q^n $ and... | proof | Algebra | proof | Yes | Yes | olympiads | false |
XLVII OM - I - Problem 5
In the plane, a triangle $ ABC $ is given, where $ |\measuredangle CAB| = \alpha > 90^{\circ} $, and a segment $ PQ $, whose midpoint is point $ A $. Prove that | Let $ M $ be the midpoint of side $ BC $ of triangle $ ABC $, and $ O $ - the center of the circle $ \omega $ circumscribed around this triangle. Angle $ CAB $ is obtuse by assumption. It follows that points $ A $ and $ O $ lie on opposite sides of line $ BC $; in particular, point $ O $ does not coincide with $ M $.
L... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XI OM - II - Task 2
Given are equations
each pair of which has a common root, but all three do not have a common root. Prove that
| Let $ \alpha $ be the common root of the first two equations, and $ \beta $ and $ \gamma $ the remaining roots of the first and second equations; then the roots of the third equation are the numbers $ \beta $ and $ \gamma $.
According to the formulas for the sum and product of the roots of a quadratic equation
Fr... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
XXIII OM - III - Problem 4
On a straight line without common points with the sphere $K$, points $A$ and $B$ are given. The orthogonal projection $P$ of the center of the sphere $K$ onto the line $AB$ lies between points $A$ and $B$, and $AP$ and $BP$ are greater than the radius of the sphere. We consider the set $Z$ o... | We will first prove the
Lemma. If $ S $ is a vertex, $ O $ is the center of the base of a right circular cone, and point $ T $ lies in the plane of the base, then the line $ ST $ forms the largest angle with such a generatrix $ SR $ of the cone that $ O \in \overline{TR} $ (Fig. 16).
Proof. For any point $ R $ on the c... | proof | Geometry | proof | Yes | Yes | olympiads | false |
X OM - III - Task 5
In the plane of triangle $ ABC $, a line moves, intersecting sides $ AC $ and $ BC $ at points $ D $ and $ E $ such that $ AD = BE $. Find the geometric locus of the midpoint $ M $ of segment $ DE $. | When $ AC = BC $, the sought geometric locus is the angle bisector of $ C $. It remains to consider only the case where $ AC \ne BC $. We will further assume that $ AC < BC $.
First, note that when $ AC < BC $, the line $ DE $ intersects the line $ AB $ at a point lying on the extension of $ AB $ beyond point $ B $, si... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false |
LII OM - II - Problem 1
Given natural numbers $ k $, $ n $ greater than $ 1 $, where the number $ p = 2k-1 $ is prime. Prove that if the number
is divisible by $ p $, then it is divisible by $ p^2 $. | The number $ p $ is a divisor of the number
Thus, $ p $, being an odd number, is a divisor of the number $ (2n - 1)^2 - p^2 $. Since $ p $ is prime, it follows that $ p | (2n-1) $. From this divisibility, it follows that $ p^2 | (2n - 1)^2 $, which means $ p^2 | a $. | proof | Number Theory | proof | Incomplete | Yes | olympiads | false |
XXXVII OM - III - Problem 2
Determine the maximum volume of tetrahedra having three faces with an area of 1. | Let $ABCD$ be a tetrahedron where $S_A = S_B = S_D = 1$; by $S_X$ we denote here the area of the face not containing vertex $X$ ($X = A, B, C, D$). Further, let us denote
(figure 10 shows the net of the surface of tetrahedron $ABCD$ unfolded onto the plane $ABC$).
om37_3r_img_10.jpg
om37_3r_img_11.jpg
The following re... | \frac{\sqrt{2}}{3} | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
LIV OM - II - Task 4
Prove that for every prime number $ p > 3 $ there exist integers $ x $, $ y $, $ k $ satisfying the conditions: $ 0 < 2k < p $ and | Let $ A $ be the set of numbers of the form $ x^2 $, where $ 0 \leq x < p/2 $, and $ B $ the set of numbers of the form $ 3 - y^2 $, where $ 0 \leq y < p/2 $; the numbers $ x $, $ y $ are integers. We will show that the numbers in the set $ A $ give different remainders when divided by $ p $.
Suppose that the numbers $... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
XLVIII OM - III - Problem 2
Find all triples of real numbers $ x $, $ y $, $ z $ satisfying the system of equations | Suppose that the numbers $ x $, $ y $, $ z $ satisfy the given system; from the second equation it follows that
Further, we have
which, by the first equation of the system, gives:
There is also the inequality
that is,
By property (1), we can multiply inequalities (3) and (4) side by side:
... | (x,y,z)=(\1/3,\1/3,\1/3)(x,y,z)=(\1/\sqrt{3},0,0),(0,\1/\sqrt{3},0),(0,0,\1/\sqrt{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
L OM - I - Task 3
In an isosceles triangle $ ABC $, angle $ BAC $ is a right angle. Point $ D $ lies on side $ BC $, such that $ BD = 2 \cdot CD $. Point $ E $ is the orthogonal projection of point $ B $ onto line $ AD $. Determine the measure of angle $ CED $. | Let's complete the triangle $ABC$ to a square $ABFC$. Assume that line $AD$ intersects side $CF$ at point $P$, and line $BE$ intersects side $AC$ at point $Q$. Since
$ CP= \frac{1}{2} CF $. Using the perpendicularity of lines $AP$ and $BQ$ and the above equality, we get $ CQ= \frac{1}{2} AC $, and consequently $... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXXIII OM - I - Problem 12
Let $ S $ be a set of $ n $ ($ n \geq 3 $) points in space. For each plane $ \pi $, consider the number $ R(\pi) $ equal to the sum of the distances from the points of set $ S $ to $ \pi $. Let $ \pi_0 $ be a plane such that the number $ R(\pi_0) $ is minimal. Prove that some three points of... | We will first prove a lemma.
Lemma. Let $\pi$ be a plane and $l$ a line contained in it. No point of the set $S$ belongs to $\pi - l$. On the other hand, $\pi_1$ is a plane obtained from $\pi$ by rotating around $l$ until it encounters a point of the set $S$. Then the following inequality holds: $R(\pi) > R(\pi_1)$.
Pr... | proof | Geometry | proof | Yes | Incomplete | olympiads | false |
XXII OM - II - Zadanie 6
Dany jest ciąg nieskończony $ \{a_n\} $. Dowieść, że jeżeli
to
dla $ n = 1, 2, \ldots $.
|
Nierówność $ a_n + a_{n+2} \geq 2a_{n+1} $ daną w założeniu przekształcamy do postaci równoważnej
Oznaczając $ b_{n+1} = a_{n+1} - a_n $ dla $ n = 1, 2, \ldots $ nierówność (1) można zapisać w postaci $ b_{n+2} > b_{n+1} $. Zatem ciąg $ \{b_n\} $ jest niemalejący.
Nierówność daną w tezie przekształcamy do pos... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
V OM - III - Task 4
Find the values of $ x $ satisfying the inequality
where $ a $ is a given positive number. | Suppose that the number $ x $ satisfies inequality (1), then $ x \geq a $, from inequality (1) follow the inequalities
\[ (2) \qquad \sqrt{x} > \sqrt{x - a} + 2, \]
\[ (3) \qquad x > (\sqrt{x - a} + 2)^2 = x - a + 4 \sqrt{x - a} + 4, \]
\[ (4) \qquad \sqrt{x - a} < \frac{a-4}{4}. \]
From inequality (4) it follows t... | x>(\frac{4}{4})^2 | Inequalities | math-word-problem | Incomplete | Yes | olympiads | false |
XXI OM - III - Task 4
In the plane, there are $ n $ rectangles with sides respectively parallel to two given perpendicular lines. Prove that if any two of these rectangles have at least one point in common, then there exists a point belonging to all the rectangles. | Let us choose a simple set of axes for the coordinate system (Fig. 14). Then the vertices of the $i$-th rectangle $P_i$ will have coordinates $(x_i, y_i)$, $(x_i, y)$, $(x, y_i)$, $(x, y)$, where $x_i < x$, $y_i < y$. The point $(x, y)$ will therefore belong to the rectangle $P_i$ if and only if $x_i \leq x \leq x$ and... | proof | Geometry | proof | Yes | Yes | olympiads | false |
LI OM - III - Task 4
In a regular pyramid with vertex $ S $ and base $ A_1A_2\ldots A_n $, each lateral edge forms an angle of $ 60^\circ $ with the base plane. For each natural number $ n \geq 3 $, determine whether it is possible to choose points $ B_2,B_3,\ldots,B_n $ lying on the edges $ A_2S, A_3S,\ldots,A_nS $, ... | Such points exist for any $ n \geq 3 $.
Let $ O $ be the foot of the height of the pyramid, and $ \alpha $ the measure of the angle of the lateral face at vertex $ S $. From the equality $ A_1S = 2A_1O $ we get
from which
om51_3r_img_3.jpg
Let's cut the lateral surface of the pyramid along the edge $ A_1S $... | proof | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
XXXVIII OM - I - Problem 4
For a given natural number $ n $, we consider the family $ W $ of all sets $ A $ of pairs of natural numbers not greater than $ n $, such that if $ (a, b) \in A $, then $ (x, y) \in A $ for $ 1 \leq x \leq a $, $ 1 \leq y \leq b $.
Prove that the family $ W $ consists of $ \binom{2n}{n} $ di... | For any pair of natural numbers $ (x,y) $, we denote by $ Q_{xy} $ the unit square with vertices $ (x-1,y-1) $, $ (x,y-1) $, $ (x,y) $, $ (x-1,y) $, and by $ R_{xy} $ - the rectangle with vertices $ (0,0) $, $ (x,0) $, $ (x,y) $, $ (0,y) $. If $ A $ is any set of pairs of natural numbers, then the symbol $ Z(A) $ will ... | \binom{2n}{n} | Combinatorics | proof | Yes | Yes | olympiads | false |
XXV - I - Problem 4
Prove that if $ x_1, \ldots, x_n $ are real numbers such that $ \sum_{i=1}^n x_i = 0 $, then for any real numbers $ t_1, t_2, \ldots, t_n $ the inequality is true | Since $ \displaystyle \sum_{k=1}^n x_k = 0 $, therefore | proof | Inequalities | proof | Yes | Problem not solved | olympiads | false |
XXXV OM - I - Problem 6
Solve the system of equations
Note: The original text was very minimal and did not contain the actual system of equations. If you have the specific equations, please provide them, and I can help solve them. | If the pair of numbers $ (x,y) $ is a solution to this system, then $ xy \ne 1 $, $ xy \ne -1 $. Let's consider a few of the simplest cases.
If $ x = 1 $, the equations take the form
they are thus, in the case when $ a = b = 1 $, satisfied by any number $ y $ different from $ 1 $, $ - 1 $, and for other $ a $, $ ... | notfound | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
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