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LX OM - III - Zadanie 2
Let $ S $ be the set of all points in the plane with both coordinates being integers. Find
the smallest positive integer $ k $ for which there exists a 60-element subset of the set $ S $
with the following property: For any two distinct elements $ A $ and $ B $ of this subset, there exists a po... | Let $ K $ be a subset of the set $ S $ having for a given number $ k $ the property given in the problem statement.
Let us fix any two different points $ (a, b), (c, d) \in K $. Then for some integers
$ x, y $ the area of the triangle with vertices $ (a, b) $, $ (c, d) $, $ (x, y) $ is $ k $, i.e., the equality
$ \frac... | 210 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 929 |
LV OM - III - Task 2
Let $ W $ be a polynomial with integer coefficients, taking relatively prime values for some two different integers. Prove that there exists an infinite set of integers for which the polynomial $ W $ takes pairwise relatively prime values. | A sequence of different integers $ x_0, x_1, x_2, \ldots $ such that the values $ W(x_i) $ are pairwise coprime, can be constructed inductively. The two initial terms $ x_0 = a, x_1 = b $ are numbers whose existence is given in the assumptions.
Assume that different numbers $ x_0, x_1, \ldots, x_n $ have already been d... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 933 |
I OM - B - Task 1
Determine the coefficients $ a $ and $ b $ of the equation $ x^2 + ax + b=0 $ such that the values of $ a $ and $ b $ themselves are roots of this equation. | According to the theorems about the sum and product of the roots of a quadratic equation, the numbers $ a $ and $ b $ are the roots of the equation $ x^2 + ax + b = 0 $ if and only if they satisfy the system of equations
that is,
The second equation of the system (1) is satisfied when $ b = 0 $ or when $ a-1= 0 $... | 1,0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 935 |
I OM - B - Zadanie 1
Wyznaczyć współczynniki $ a $ i $ b $, równania $ x^2 + ax +b=0 $ tak, aby same wartości $ a $ i $ b $ były pierwiastkami tego równania.
|
Liczby $ a $ i $ b $ mają spełniać równanie $ x^2 + ax + b = 0 $. Zatem
czyli
Drugie z równań układu (2) daje $ b = 0 $ lub $ a + b + 1 = 0 $.
Biorąc $ b = 0 $ otrzymujemy rozwiązanie
Biorąc $ a + b + 1 = 0 $ mamy $ b = - a-l $; po podstawieniu do pierwszego z równań układu (2) otrzymujemy równa... | 0,0或1,-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 936 |
IV OM - I - Problem 10
Given are two skew lines $ m $ and $ n $. On line $ m $, a segment $ AB $ of given length $ a $ is measured, and on line $ n $, a segment $ CD $ of given length $ b $ is measured. Prove that the volume of the tetrahedron $ ABCD $ does not depend on the position of segments $ AB $ and $ CD $ on l... | The figure formed by two skew lines $m$ and $n$ is most conveniently represented on a drawing using projections onto two mutually perpendicular planes. We will choose any plane parallel to both lines $m$ and $n$ as the horizontal projection plane (Fig. 33). The vertical projections of the given lines will then be paral... | \frac{1}{6}abd\sin\varphi | Geometry | proof | Yes | Yes | olympiads | false | 939 |
XL OM - III - Task 3
We number the edges of a cube with numbers from 1 to 12.
(a) Prove that for any numbering, there exist at least eight triples of integers $ (i,j,k) $, where $ 1\leq i < j < k\leq 12 $, such that the edges numbered $ i,j,k $ are consecutive sides of a broken line.
(b) Provide an example of a number... | (a) Let $ V $ be any vertex of a cube and let $ p $, $ q $, $ r $ be the numbers of the edges emanating from this vertex, with $ p < q < r $. We will show that
Let $ W $ be the other end of the edge numbered $ q $. Denote the numbers of the other two edges emanating from $ W $ by $ x $ and $ y $; assume that $ x < y $... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 940 |
LVI OM - I - Problem 8
On a circle, there are $ n $ lamps; each can be turned on or off. We perform a series of moves; in each move, we select $ k $ consecutive lamps and change their state: we turn off the ones that are on, and turn on the ones that are off (the number $ k $ does not change during this process). Init... | First, let's assume that $k$ is an even number. We will prove by induction that after each change of state of $k$ consecutive lamps, an even number of lamps are lit.
Let $x_n$ denote the number of lamps that are on after the $n$-th move. Then $x_1 = k$. Assume that $x_{n-1}$ is an even number. If, by performing the $n$... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 941 |
XLVI OM - I - Problem 10
Given a line $ k $ and three distinct points lying on it. Each of them is the origin of a pair of rays; all these rays lie in the same half-plane with edge $ k $. Each of these pairs of rays forms a quadrilateral with each other. Prove that if a circle can be inscribed in two of these quadrila... | Let's denote three given points by $A_1$, $A_2$, $A_3$. For $i = 1,2,3$, let $p_i$, $q_i$ be two given rays with a common starting point $A_i$, and let $l_i$ be the angle bisector of the angle formed by them. These bisectors intersect (pairwise) at points $O_1$, $O_2$, $O_3$:
(figure 5). Additionally, we adopt the fol... | proof | Geometry | proof | Yes | Yes | olympiads | false | 944 |
LII OM - I - Problem 11
A sequence of positive integers $ c_1,c_2,\ldots,c_n $ will be called admissible if using a balance scale and two complete sets of weights with masses $ c_1,c_2,\ldots,c_n $, it is possible to weigh any object with a mass that is a natural number not exceeding $ 2(c_1 + c_2 +\ldots + c_n) $. Fo... | Let $ p_1,p_2,\ldots,p_n $ and $ q_1,q_2,\ldots,q_n $ be two sets of weights with masses $ c_1,c_2,\ldots,c_n $, respectively. First, we will show that with these two sets, we can weigh at most $ \frac{1}{2}(5^n-1) $ objects, regardless of whether the numbers $ c_1,c_2,\ldots,c_n $ form an admissible system or not.
We ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 945 |
XXI OM - II - Problem 6
If $ A $ is a subset of set $ X $, then we define $ A^1 = A $, $ A^{-1} = X - A $. Subsets $ A_1, A_2, \ldots, A_k $ are called mutually independent if the product $ A_1^{\varepsilon_1} \cap A_2^{\varepsilon_2} \ldots A_k^{\varepsilon_k} $ is non-empty for every system of numbers $ \varepsilon_... | Since the number of all $n$-term sequences with values $-1$ and $1$ is $2^n$ (see preparatory problem D of series III), we can assume that the elements of the $2^n$-element set $X$ under consideration are such sequences.
Let $A_i$ for $i = 1, 2, \ldots, n$ denote the set of all sequences in $X$ whose $i$-th term is $1$... | n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 948 |
XXIII OM - III - Problem 2
On a plane, there are $ n > 2 $ points, no three of which are collinear. Prove that the shortest among the closed broken lines passing through these points is a simple broken line. | Let's recall the definition of a simple broken line: A closed broken line with successive vertices $W_1, W_2, \ldots, W_n, W_{n+1}$, where $W_{n+1} = W_1$, is called a simple broken line if the closed segments $\overline{W_iW_{i+1}}$ and $\overline{W_jW_{j+1}}$ are disjoint for $1 \leq i, j \leq n$.
Let no three of the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 950 |
X OM - III - Task 3
Given is a pyramid with a square base $ABCD$ and apex $S$. Find the shortest path that starts and ends at point $S$ and passes through all the vertices of the base. | Let $ a $ denote the length of the side of square $ ABCD $; the lengths of the lateral edges of the pyramid will be denoted by the letters $ k $, $ l $, $ m $, $ n $ in such a way that $ k \leq l \leq m \leq n $.
Each path starting and ending at point $ S $ and passing through all the vertices of the square can be symb... | k++3a | Geometry | math-word-problem | Yes | Yes | olympiads | false | 951 |
XLVII OM - I - Problem 3
In a group of $ kn $ people, each person knows more than $ (k - 1)n $ others ($ k $, $ n $ are natural numbers). Prove that it is possible to select $ k + 1 $ people from this group, such that any two of them know each other.
Note: If person $ A $ knows person $ B $, then person $ B $ knows pe... | A group of people, each of whom knows every other, we will call a {\it clique}. Let $ m $ be the largest natural number for which there exists an $ m $-person clique in the considered group of $ kn $ people. We will show that $ m > k $. This will already imply the thesis to be proven, since any ($ k+1 $)-element subset... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 952 |
XXIV OM - I - Problem 7
Prove that among five segments lying on the same straight line, there are either three segments that have a common point or three segments that are pairwise disjoint. | Let the open intervals be $ I_k = (a_k; b_k) $, where $ k = 1, 2, 3, 4, 5 $. For closed, half-open, etc., intervals, the reasoning proceeds similarly.
Without loss of generality, we can assume that $ a_1 \leq a_2 \leq a_3 \leq a_4 \leq a_5 $. Suppose that there is no point belonging to any three intervals. If $ b_1 > a... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 953 |
VI OM - I - Problem 8
Prove that a circle can be inscribed in a trapezoid if and only if the circles whose diameters are the non-parallel sides of the trapezoid are externally tangent to each other. | A circle can be inscribed in a convex quadrilateral $ABCD$ if and only if the sums of the lengths of opposite sides of the quadrilateral are equal, i.e., when
When the quadrilateral $ABCD$ is a trapezoid with parallel sides $AB$ and $CD$, denoting the midpoints of sides $AD$ and $BC$ by $M$ and $N$ respectively (Fig... | proof | Geometry | proof | Yes | Yes | olympiads | false | 956 |
XIX OM - I - Problem 4
In the plane, given are parallel lines $ a $ and $ b $ and a point $ M $ lying outside the strip bounded by these lines. Determine points $ A $ and $ B $ on lines $ a $ and $ b $, respectively, such that segment $ AB $ is perpendicular to $ a $ and $ b $, and the angle $ AMB $ is the largest. | Let $ AB $ be a segment with endpoints $ A \in a $ and $ B \in b $, perpendicular to the lines $ a $ and $ b $, and $ N $ - the point symmetric to point $ M $ with respect to a line parallel to lines $ a $ and $ b $ and equidistant from them (Fig. 4).
The circle with diameter $ AB $ lies within the strip bounded by the... | ABlieonthecirclewithdiameterMN | Geometry | math-word-problem | Yes | Yes | olympiads | false | 958 |
XXIX OM - I - Problem 6
Prove that on the plane, any closed broken line of length 1 is contained in some circle of radius length $ \frac{1}{4} $ | Let points $A$ and $B$ belonging to a closed broken line of length $1$ divide this broken line into two equal parts, and let $C$ be any point on this broken line. Since a segment is the shortest broken line connecting two points, we have $AC + BC \leq \frac{1}{2}$.
Let $P$ be the midpoint of segment $\overline{AB}$, an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 960 |
XIII OM - III - Task 4
In how many ways can a set of $ n $ items be divided into two sets? | The answer to the posed question will depend on whether - to the considered sets, we include the empty set, i.e., a set that does not have any element (object), or not; in the first case, the sought number is $ 1 $ greater than in the second. We agree to consider only non-empty sets.
Let $ p $ denote the number of all ... | 2^{n-1}-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 961 |
XVII OM - I - Problem 11
Prove that the centers of the excircles of a triangle and the points symmetric to the center of the incircle of the triangle with respect to its vertices lie on a single circle. | Let $A_1$, $B_1$, $C_1$ be the points symmetric to the center $O$ of the incircle of triangle $ABC$ with respect to the vertices $A$, $B$, $C$, and let $S_1$ be the center of the excircle of triangle $ABC$ that is tangent to side $BC$ (Fig. 8). To prove the theorem, it suffices to show that the points $A_1$, $B_1$, $C_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 963 |
XIX OM - III - Problem 5
On a plane, there are $ n $ points ($ n \geq 4 $), any four of which are vertices of a convex quadrilateral. Prove that all these points are vertices of a convex polygon. | We will apply the method of induction. When $ n = 4 $, the thesis of the theorem is true. Assume that the thesis of the theorem is true for some natural number $ n \geq 4 $, and let $ A_1, A_2, \ldots, A_n, A_{n+1} $ be such $ n + 1 $ points in the plane that any four of them are vertices of a convex polygon. According... | proof | Geometry | proof | Yes | Yes | olympiads | false | 964 |
XLIII OM - I - Zadanie 11
Dana jest liczba naturalna $ n\geq 1 $. Rozważamy tabelę zbudowaną z $ n(n+1)/2 $ okienek, ustawionych w $ n $ rzędach: jedno okienko w pierwszym rzędzie, dwa w drugim itd., $ n $ okienek w $ n $-tym rzędzie. W okienka tabeli wpisujemy w sposób losowy liczby od $ 1 $ do $ n(n + l )/2 $. Niech... |
Dla ustalonej liczby naturalnej $ n \geq 2 $ rozważane w zadaniu zdarzenie
jest równoważne koniunkcji zdarzeń
oraz
Warunek określający zdarzenie $ \mathcal{B}_n $ jest równoważny stwierdzeniu, że liczba $ N = n(n+ 1)/2 $ trafia do jednego z $ n $ okienek $ n $-tego rzędu tabeli; w takim razie
... | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 966 | |
LV OM - III - Task 5
Determine the maximum number of lines in space passing through a fixed point and such that any two intersect at the same angle. | Let $ \ell_1,\ldots,\ell_n $ be lines passing through a common point $ O $. A pair of intersecting lines determines four angles on the plane containing them: two vertical angles with measure $ \alpha \leq 90^\circ $ and the other two angles with measure $ 180^\circ - \alpha $. According to the assumption, the value of ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 967 |
XLII OM - II - Problem 6
A parallelepiped contains a sphere of radius $ r $ and is contained in a sphere of radius $ R $. Prove that $ \frac{R}{r} \geq \sqrt{3} $. | om42_2r_img_8.jpg
Let us choose any face of the parallelepiped and denote its vertices by $ A $, $ B $, $ C $, $ D $ such that segment $ AC $ is the longer (i.e., "not shorter") diagonal of the parallelogram $ ABCD $. The remaining vertices of the parallelepiped are denoted by $ E $, $ F $, $ G $, $ H $ such that segme... | R/r\geq\sqrt{3} | Geometry | proof | Yes | Yes | olympiads | false | 968 |
XII OM - III - Task 3
Prove that if the section of a tetrahedron by a plane is a parallelogram, then half of its perimeter is contained between the length of the shortest and the length of the longest edge of the tetrahedron. | Let the parallelogram $MNPQ$ be a planar section of the tetrahedron $ABCD$. The plane $MNPQ$ is, of course, different from the planes of the faces of the tetrahedron, and each side of the parallelogram lies on a different face. Suppose that the sides $MN$ and $QP$ lie on the faces $ABC$ and $BCD$; they are then paralle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 970 |
XXIX OM - I - Problem 11
Let $ f: \mathbb{R} \to \mathbb{R} $ be a continuous function for which there exists a number $ x $ satisfying the conditions: $ f(f(f(x))) = x $, $ f(x) \neq x $.
Prove that there exist numbers $ y_1 $, $ y_2 $, $ y_3 $ such that $ y_i \neq y_j $ for $ i \neq j $ and $ f(f(y_k))=y_k $ for $ k... | If $ff(x) = x$, then we would have $fff(x) = f(x)$, which means $x = f(x)$, contradicting the assumption. Therefore, $ff(x) \ne x$. Similarly, from the equation $ff(x) = f(x)$, it follows that $fff(x) = ff(x)$, which means $x = ff(x)$. However, we have already proven that this last equality does not hold. Therefore, $f... | proof | Algebra | proof | Yes | Yes | olympiads | false | 971 |
XXXIX OM - I - Problem 12
Given are two concentric spheres $ S_R $ and $ S_r $ with radii $ R $ and $ r $, respectively, where $ R > r $. Let $ A_0 $ be a point on $ S_R $. Let $ T_{A_0} $ be the cone with vertex at $ A_0 $ and tangent to $ S_r $. Denote $ K_1 = T_{A_0} \cap S_R $. Then, for each point $ A \in K_1 $, ... | In this task, by a cone, we mean the surface of an unbounded circular cone, and by the tangency of a cone to a sphere - tangency along a certain circle. (This lack of precision in the formulation was noticed and pointed out by many participants of the olympiad.)
Let $O$ be the common center of spheres $S_r$ and $S_R$. ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 973 |
V OM - I - Problem 4
Inside a given triangle $ ABC $, find a point $ O $ such that the areas of triangles $ AOB $, $ BOC $, $ COA $ are in the ratio $ 1 \colon 2 \colon 3 $. | If the areas of triangles $AOB$, $BOC$, $COA$ are in the ratio $1 \colon 2 \colon 3$, then these areas constitute respectively $\frac{1}{6}$, $\frac{2}{6}$, $\frac{3}{6}$ of the area of triangle $ABC$. The required point $O$ thus lies at the intersection of the line parallel to line $AC$ drawn through the midpoint of s... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 974 |
L OM - III - Zadanie 2
Given are non-negative integers $ a_1 <a_2 <a_3 < \ldots < a_{101} $ less than $ 5050 $. Prove that among them, one can choose four different $ a_k $, $ a_l $, $ a_m $, $ a_n $, such that the number $ a_k + a_l -a_m -a_n $ is divisible by $ 5050 $. | We consider all expressions of the form $ a_k + a_l $, where $ 1 \leq k < l \leq 101 $. There are $ {101 \choose 2} = 5050 $ such expressions. If we find two of them, say $ a_k + a_l $ and $ a_m + a_n $, whose values give the same remainder when divided by 5050, then the numbers $ a_k $, $ a_l $, $ a_m $, $ a_n $ satis... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 977 |
LI OM - I - Task 3
The sum of positive numbers $ a $, $ b $, $ c $ is equal to $ 1 $. Prove that | It is enough to prove that for any positive numbers $ a $, $ b $, $ c $ the inequality holds
Transforming the above relationship equivalently, we get
By making the substitution $ x = bc $, $ y = ca $, $ z = ab $, we reduce the inequality to be proved to the form
Squaring both sides of the last inequal... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 979 |
V OM - III - Task 1
Prove that in an isosceles trapezoid circumscribed around a circle, the segments connecting the points of tangency of opposite sides with the circle pass through the point of intersection of the diagonals. | Let $E$, $F$, $G$, $H$ denote the points of tangency of the sides $AB$, $BC$, $CD$, $DA$ with the incircle of trapezoid $ABCD$.
Let $M$ be the point of intersection of segments $EG$ and $HF$ (Fig. 37a). Since trapezoid $ABCD$ is isosceles, the line $EG$ is the axis of symmetry of the figure, and point $F$ is symmetric ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 981 |
X OM - II - Task 6
From point $ M $ on the surface of a sphere, three mutually perpendicular chords of the sphere $ MA $, $ MB $, $ MC $ were drawn. Prove that the segment connecting point $ M $ with the center of the sphere intersects the plane of triangle $ ABC $ at the centroid of this triangle. | The plane $ AMB $ intersects the surface of the sphere along a circle passing through points $ A $, $ M $, and $ B $, which is the circumcircle of the right triangle $ AMB $; the center of this circle is therefore at the midpoint $ K $ of segment $ AB $ (Fig. 25).
The center $ O $ of the sphere lies on the perpendicula... | proof | Geometry | proof | Yes | Yes | olympiads | false | 983 |
XLIX OM - I - Problem 10
Medians $ AD $, $ BE $, $ CF $ of triangle $ ABC $ intersect at point $ G $. Circles can be circumscribed around quadrilaterals $ AFGE $ and $ BDGF $. Prove that triangle $ ABC $ is equilateral. | Let's denote the measures of angles $CAB$, $ABC$, $BCA$ by $\alpha$, $\beta$, $\gamma$ respectively. Points $F$ and $D$ are the midpoints of sides $AB$ and $BC$, so $DF \parallel CA$, and therefore $|\measuredangle BDF| = \gamma$ (Figure 5).
The opposite angles of the cyclic quadrilateral $AFGE$ sum up to $180^\circ$. ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 984 |
XXII OM - I - Problem 11
We toss a coin $2n$ times. Let $p_n$ denote the probability that we will get a series of length $r > n$. Prove that $\lim p_n = 0$. | Elementary events are all $2n$-term sequences with terms being heads or tails; all of them are equally probable and there are $2^{2n}$ of them. Let $A_n$ be the event that we get a series of length at least $n$.
Consider the following three-stage procedure:
1) Selecting some $n$ consecutive positions out of $2n$;
2) Pl... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 985 |
XIX OM - III - Problem 3
In the tetrahedron $ABCD$, the edges $AD$, $BD$, $CD$ are equal. On the plane $ABC$, points $A_1, B_1, C_1$ are chosen such that they are not collinear. The lines $DA_1, DB_1, DC_1$ intersect the surface of the sphere circumscribed around the tetrahedron at points $A_2, B_2, C_2$, respectively... | The sphere (i.e., the surface of a sphere) $ S $ circumscribed around the tetrahedron $ ABCD $ intersects the plane $ ABC $ along the circle $ k $ circumscribed around the triangle $ ABC $. From the equality $ DA = DB = DC $, it follows that the sphere $ t $ with center $ D $ and radius $ r = DA $ passes through the po... | proof | Geometry | proof | Yes | Yes | olympiads | false | 989 |
XXV OM - II - Problem 5
Given are real numbers $ q,t \in \langle \frac{1}{2}; 1) $, $ t \in (0; 1 \rangle $. Prove that there exists an increasing sequence of natural numbers $ {n_k} $ ($ k = 1,2, \ldots $), such that | The sequence $ \{n_k\} $ $ (k = 1,2, \ldots) $ is defined inductively as follows. Since $ 0 < q < 1 $, we have $ q^0 = 1 $ and $ \displaystyle \lim_{n \to \infty} q^n = 0 $. For each number $ t $ in the interval $ (0; 1 \rangle $, there exists a natural number $ n_1 $ such that
Wthen
since $ 1 - q \leq q $ for $ \dis... | proof | Algebra | proof | Yes | Yes | olympiads | false | 993 |
XXIII OM - I - Problem 4
Points $ A $ and $ B $ do not belong to the plane $ \pi $. Find the set of all points $ M \in \pi $ with the property that the lines $ AM $ and $ BM $ form equal angles with the plane $ \pi $. | Let $A'$ and $B'$ denote the projections of points $A$ and $B$ onto the plane $\pi$ (Fig. 4). If $A' = A$ and $AA' = 0$, then the sought set of points is the entire plane. If $A' \neq A$ and $AA' \neq 0$, then only point $A$ satisfies the conditions of the problem. For if $M \neq A$ and $M \in \pi$, then the angles $\m... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 994 |
XVI OM - III - Task 5
Points $ A_1 $, $ B_1 $, $ C_1 $ divide the sides $ BC $, $ CA $, $ AB $ of triangle $ ABC $ in the ratios $ k_1 $, $ k_2 $, $ k_3 $. Calculate the ratio of the areas of triangles $ A_1B_1C_1 $ and $ ABC $. | According to the assumption
Hence
Let the symbol $ p (ABC) $ denote the area of triangle $ ABC $ (Fig. 21). Since
then
By applying the theorem that the ratio of the areas of two triangles having a common angle equals the ratio of the products of the sides of each triangle forming that angle, we ... | \frac{k_1k_2k_3}{(1+k_1)(1+k_2)(1+k_3)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 995 |
XLII OM - II - Problem 4
Find all monotonic functions $ f: \mathbb{R} \to \mathbb{R} $ that satisfy the equation | For any $ x > 0 $, we have the inequality $ f(4x) - f(3x) > 0 $, which implies that the (monotonic) function $ f $ is increasing in the interval $ (0;\infty) $. Similarly, for $ x < 0 $, the inequality $ f(3x) - f(4x) > 0 $ holds, so $ f $ is also increasing in $ (-\infty;0) $.
Replacing $ 4x $ with the letter $ z $, w... | proof | Algebra | proof | Yes | Yes | olympiads | false | 997 |
XLIV OM - II - Problem 3
On the edge $ |OA_1| $ of the tetrahedron $ |OA_1B_1C_1| $, points $ A_2 $, $ A_3 $ are chosen such that $ |OA_1| > |OA_2| > |OA_3| > O $. Let $ B_2 $, $ B_3 $ be points on the edge $ |OB_1| $, and $ C_2 $, $ C_3 $ be points on the edge $ |OC_1| $, such that the planes $ |A_1B_1C_1| $, $ |A_2B... | Let $ B_1, $ B_2 be the orthogonal projections of points $ B_1, $ B_2 onto the plane $ OA_1C_1 $, and $ C_1, $ C_3 be the projections of points $ C_1, $ C_3 onto the plane $ OA_1B_1 $ (Figure 7; it is convenient to imagine the tetrahedron each time so that the face we are projecting onto is its "horizontal" base). From... | V_1+V_2+V_3\geq3V | Geometry | proof | Yes | Yes | olympiads | false | 1,000 |
XXXII - I - Problem 7
Let $ (x_n) $ be a sequence of natural numbers satisfying the conditions
a) $ x_1 < x_2 < x_3 < \ldots $,
b) there exists $ j \in \mathbb{N} $, $ j > 1 $, such that $ x_j = j $,
c) $ x_{kl} = x_kx_l $, for coprime $ k,l \in \mathbb{N} $.
Prove that $ x_n = n $ for every $ n $. | Since the number of natural numbers less than a given natural number $n$ is $n-1$, if for some $j$ we have $x_j = j$, then by condition a) it must be that $x_k = k$ for $k \leq j$. If the number $j$ mentioned in condition b) is greater than $2$, then the number $j-1$ is greater than $1$ and is relatively prime to the n... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,001 |
XXVI - I - Problem 4
Given are pairwise disjoint spheres $ K_1, K_2, K_3 $ with pairwise distinct radii. Let $ A_{ij} $ be the vertex of the cone circumscribed around the spheres $ K_i $ and $ K_j $. Prove that the points $ A_{12} $, $ A_{23} $, $ A_{31} $ are collinear. | If $ A $ is the vertex of a cone circumscribed around spheres $ K $ and $ L $ of different radii, then there exists exactly one such homothety $ j $ such that $ j(K) = L $. Its center is the point $ A $, and the coefficient is the ratio of the radii of the spheres $ L $ and $ K $.
Without loss of generality, we can ass... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,002 |
LVII OM - I - Problem 1
Determine all non-negative integers $ n $ for which the number
$ 2^n +105 $ is a perfect square of an integer. | If $ n $ is an odd number, then the number $ 2^n +105 $ gives a remainder of 2 when divided by 3. Since a number that is the square of an integer cannot give a remainder of 2 when divided by 3, the number $ n $ satisfying the conditions of the problem must be even.
Let's assume that $ n =2k $ for some non-negative int... | 4,6,8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 1,008 |
LVI OM - I - Problem 11
A circle with center $ I $ is inscribed in a convex quadrilateral $ ABCD $, with the point $ I $ not lying on the line $ AC $. The diagonals $ AC $ and $ BD $ intersect at point $ E $. The line passing through point $ E $ and perpendicular to the line $ BD $ intersects the lines $ AI $, $ CI $ ... | Let $ o_1 $ be a circle with center $ P $ and tangent to the lines $ AD $ and $ AB $ at points $ S $ and $ T $, respectively. Similarly, let $ o_2 $ be a circle with center $ Q $ and tangent to the lines $ CB $ and $ CD $ at points $ U $ and $ W $, respectively (Fig. 5). Let $ o $ be the incircle of quadrilateral $ ABC... | PE=EQ | Geometry | proof | Yes | Yes | olympiads | false | 1,009 |
XVI OM - II - Task 4
Find all prime numbers $ p $ such that $ 4p^2 +1 $ and $ 6p^2 + 1 $ are also prime numbers. | To solve the problem, we will investigate the divisibility of the numbers \( u = 4p^2 + 1 \) and \( v = 6p^2 + 1 \) by \( 5 \). It is known that the remainder of the division of the product of two integers by a natural number is equal to the remainder of the division of the product of their remainders by that number. B... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 1,010 |
X OM - I - Task 4
Prove that if point $ P $ moves in the plane of triangle $ ABC $, then the triangle $ S_1S_2S_3 $, whose vertices are the centroids $ S_1 $, $ S_2 $ and $ S_3 $ of triangles $ PBC $, $ PCA $ and $ PAB $ does not change its shape or size. | If the vertex $ P $ of the triangle $ PAB $ is moved to the point $ P' $ (Fig. 6a), then the centroid $ S_3 $ of this triangle will move to the point $ S $.
The vector $ S_3S $ has the same direction as the vector $ PP' $ and is three times shorter, since these vectors are similar (directly) relative to the point $ N $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,012 |
LI OM - I - Task 7
Prove that for any positive integer $ n $ and any number $ t\in(\frac{1}{2},1) $ there exist such numbers $ a,b \in (1999,2000) $, that | Let $ c = 3999/2 $. We seek numbers $ a $, $ b $ in the form $ a = c+x $, $ b = c-x $, where $ x \in (0, \frac{1}{2}) $. Then
where $ s = 2t - 1 > 0 $. After expanding using the binomial theorem, the last expression takes the form
where $ p(x) $ is some polynomial.
We want to show that for some number $ y \in (0,\fra... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,015 |
VII OM - III - Task 3
On a straight line, there are three different points $ M $, $ D $, $ H $. Construct a right-angled triangle for which $ M $ is the midpoint of the hypotenuse, $ D $ - the intersection point of the angle bisector of the right angle with the hypotenuse, and $ H $ - the foot of the altitude on the h... | Let $ABC$ be the sought right-angled triangle. Since points $M$, $D$, $H$ are distinct, the legs $AC$ and $BC$ are not equal. Let, for example, $AC > BC$. Then point $D$ lies between points $M$ and $B$, and point $H$ lies between points $D$ and $B$, so point $D$ lies between points $M$ and $H$ (Fig. 19).
Since
an... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,016 |
XXXVIII OM - II - Problem 5
Determine all prime numbers $ p $ and natural numbers $ x, y $, for which $ p^x - y^3 = 1 $. | Suppose that the numbers $ p $, $ x $, $ y $ satisfy the given conditions. Then
Let's consider two cases:
1. $ y = 1 $. Then (1) reduces to the equation $ p^x = 2 $, from which $ p = 2 $, $ x = 1 $.
2. $ y > 1 $. Then both factors of the product on the right side of (1) are numbers greater than $ 1 $ and as divis... | (2,1,1)(3,2,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 1,017 |
XLII OM - III - Problem 4
In the plane with a Cartesian coordinate system, we consider the set $ V $ of all free vectors, both of whose coordinates are integers. Determine all functions $ f $, defined on the set $ V $, with real values, satisfying the conditions:
(a) $ f(v) = 1 $ for each of the four vectors $ v\in V ... | Suppose that the function $ f $ satisfies the given conditions. Using these conditions, let's list the values of $ f(\overrightarrow{\mathbf{v}}) $ for a few vectors with small integer coordinates:
(When calculating the last of the found values, we used the orthogonality of the vectors $ [3,0] $ and $ [0,1] $ and the ... | f([x,y])=x^2+y^2 | Algebra | proof | Yes | Yes | olympiads | false | 1,023 |
XII OM - II - Problem 5
Prove that if the real numbers $ a $, $ b $, $ c $ satisfy the inequalities
[ (1) \qquad a + b + c > 0, \]
[ (2) \qquad ab + bc + ca > 0, \]
[ (3) \qquad abc > 0, \]
then $ a > 0 $, $ b > 0 $, $ c > 0 $. | From inequalities (1) and (2), it can be inferred that at least two of the numbers $a$, $b$, $c$ are positive. For from inequality (1) it follows first that at least one of these numbers is positive, let's say $c > 0$. Indeed, for any $a$ and $b$
and from inequality (2) it follows that
therefore
transferring t... | proof | Algebra | proof | Yes | Yes | olympiads | false | 1,025 |
XXIX OM - II - Problem 3
Given a sequence of natural numbers $ (a_i) $, such that for every natural number $ n $, the sum of the terms of the sequence that are not greater than $ n $ is not less than $ n $. Prove that for every natural number $ k $, one can choose a finite subsequence from $ (a_i) $ whose sum of terms... | We will apply induction with respect to $k$. Consider the case $k = 1$. By assumption, the sum of those terms of the given sequence that are not greater than $1$ (and thus are equal to $1$) is not less than $1$. Therefore, there exists a term in the given sequence that is equal to $1$. Hence, the sum of the terms of a ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,026 |
XLVII OM - II - Problem 6
Inside a parallelepiped, whose edges have lengths $ a $, $ b $, $ c $, there is a point $ P $. Prove that there exists a vertex of the parallelepiped whose distance from point $ P $ does not exceed $ \frac{1}{2}\sqrt{a^2 + b^2 + c^2} $. | The walls of the parallelepiped define six planes. Let $ \pi $ be the plane whose distance from point $ P $ is the smallest; if there are two such planes (or more), we choose any one of them and denote it by $ \pi $. Let $ ABCD $ be the face of the parallelepiped contained in the plane $ \pi $ and let $ N $ be the orth... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,028 |
XXIX OM - I - Zadanie 2
Dla danej liczby naturalnej $ k>7 $ skonstruować ciąg $ S = (a_0, a_1, \ldots, a_{k-1}) $ o tej własności, że dla każdego $ i $ ($ 0 < i < k - 1 $) $ a_i $ równe liczbie wyrazów ciągu $ S $ równych $ i $.
|
Dla dowolnej liczby naturalnej $ k $ znajdziemy wszystkie ciągi $ S $ spełniające warunki zadania.
Z warunków zadania wynika, że w ciągu $ S $ jest $ a_0 $ wyrazów równych $ 0 $, $ a_1 $ wyrazów równych $ 1 $, $ \ldots $ , $ a_{k-1} $ wyrazów równych $ k - 1 $ i innych wyrazów nie ma. Zatem liczba wszystkich wyrazów c... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1,029 |
XXVI - I - Problem 5
Determine all integers $ m $ for which the polynomial $ x^3-mx^2+mx-(m^2+1) $ has an integer root. | Suppose that the integers $ x $ and $ m $ satisfy the equation $ x^3 - mx^2+mx-(m^2+1) = 0 $, which is $ (x^2+m)(x-m) = 1 $. Then $ x^2+m = 1 $ and $ x- m = 1 $ or $ x^2+m = -1 $ and $ x-m = -1 $. Adding the equations in each of these systems, we get $ x^2+x = 2 $ or $ x^2+x = - 2 $. The roots of the first equation are... | =0=-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 1,032 |
XLIV OM - I - Problem 12
Prove that the polynomial $ x^n + 4 $ is a product of two polynomials of lower degree with integer coefficients if and only if $ n $ is divisible by $ 4 $. | \spos{I} Suppose that the polynomial $ x^4 + 4 $ is the product of two polynomials,
with the properties under consideration:
In the product $ F(x)G(x) $, the coefficient of $ x^n $ is $ a_kb_m $, and the constant term equals $ a_0b_0 $. Therefore, the following equations hold:
The numbers ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 1,033 |
XIII OM - I - Problem 5
Prove that all powers of a number whose last eight digits are $12890625$ also end with the digits $12890625$. | We need to prove that if $ n $ and $ l $ are integers, with $ n > 0 $, $ l \geq 0 $, then
where $ m $ is a non-negative integer.
First, observe that in the expansion of the left side of the above equality according to the binomial theorem for the power of a binomial, all terms except for $ (12890625)^n $ have a factor... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,034 |
XV OM - III - Task 6
Given is a pyramid SABCD, whose base is a convex quadrilateral $ABCD$ with perpendicular diagonals $AC$ and $BD$, and the orthogonal projection of vertex S onto the base is point O, the intersection of the diagonals of the base. Prove that the orthogonal projections of point O onto the lateral fac... | Let $M$, $N$, $P$, $Q$ be the orthogonal projections of point $O$ onto the planes $ASB$, $BSC$, $CSD$, $DSA$ (Fig. 23).
The plane $SOM$ is perpendicular to the plane of quadrilateral $ABCD$ and to the plane $ASB$, as it contains the perpendiculars $SO$ and $OM$ to these planes. Therefore, the plane $SOM$ is perpendicul... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,035 |
LIX OM - I - Task 6
Determine all polynomials $ W(x) $ with real coefficients such that for every real number
$ x $ the equality is satisfied | First, note that the only constant polynomials satisfying the conditions of the problem are $ W(x) \equiv 0 $ and $ W(x) \equiv 1 $. Let us assume from now on that the polynomial $ W(x) $ is not constant.
If the polynomial $ W(x) $ is of the form $ W(x)=cx^n $ for some real number $ c \neq 0 $ and integer $ n \geqslan... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 1,036 |
XVII OM - I - Problem 5
Given positive numbers $ p $ and $ q $. Prove that a rectangular prism, in which the sum of the edges equals $ 4p $, and the surface area equals $ 2q $, exists if and only if $ p^2 \geq 3q $. | a) Suppose that the desired rectangular parallelepiped exists. Then there are numbers $ x $, $ y $, $ z $ satisfying the equations
In such a case, the numbers $ p $ and $ q $ satisfy the inequality $ p^2 \geq 3q $, since from (1) and (2) it follows that
b) Suppose that given positive numbers $ p $ and... | proof | Algebra | proof | Yes | Yes | olympiads | false | 1,037 |
LVI OM - II - Task 3
In the data space, there are $ n $ points ($ n\geq 2 $) of which no four lie in the same plane. Some of these points have been connected by segments. Let $ K $ be the number of segments drawn ($ K\geq 1 $), and $ T $ the number of triangles formed. Prove that | Let us number from 1 to $ m $ those points from which at least one segment starts, and assume that from the point numbered $ i $ ($ i=1,2,\ldots,m $) exactly $ k_i $ segments start. Then $ k_i>0 $. Furthermore, let $ t_i $ ($ i=1,2,\ldots,m $) denote the number of those triangles, one of whose vertices is the point num... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 1,038 |
XL OM - I - Task 6
Calculate the sum of the series $ \sum_{n\in A}\frac{1}{2^n} $, where the summation runs over the set $ A $ of all natural numbers not divisible by 2, 3, 5. | A natural number $ n $ is not divisible by $ 2 $, $ 3 $, and $ 5 $ if and only if, when divided by 30, it gives one of the following eight remainders:
Therefore, the sum we need to calculate is equal to
where $ s_i $ denotes the sum of an analogous series formed from terms corresponding to those values of $... | \frac{1}{1-2^{-30}}(\frac{1}{2^1}+\frac{1}{2^7}+\frac{1}{2^{11}}+\frac{1}{2^{13}}+\frac{1}{2^{17}}+\frac{1}{2^{19}}+\frac{1}{2^{23}}+\frac{1}{2^{} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 1,039 |
XL OM - I - Task 1
Prove that if the numbers $ k $, $ n $ ($ k < n $) are coprime, then the number $ \binom{n-1}{k-1} $ is divisible by $ k $. | Let's denote:
These are integers. The equality
holds, that is, $ M \cdot n = N \cdot k $. Since $ k $ is relatively prime to $ n $, it must be a divisor of $ M $; which is exactly what we needed to prove. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,041 |
XXXVIII OM - III - Problem 4
Among the set of tetrahedrons with a base area of 1, a total surface area of 4, and having equal angles of inclination of the lateral faces to the base, find the tetrahedron with the maximum volume. | Let $ABCD$ be any tetrahedron satisfying the given conditions:
From condition (2), it follows that the point $D$, which is the projection of point $D$ onto the plane $ABC$, is at the same distance from the lines $BC$, $CA$, $AB$ (equal to $r = h \ctg \phi$, where $h = |DD'|$ and $\phi$ is the dihedral angle... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,049 |
XXXIX OM - II - Problem 3
Inside an acute triangle $ ABC $, consider a point $ P $ and its projections $ L, M, N $ onto the sides $ BC, CA, AB $, respectively. Determine the point $ P $ for which the sum $ |BL|^2 + |CM|^2 + |AN|^2 $ is minimized. | Let $ P $ be any point inside the triangle $ ABC $. Considering the right triangles $ ANP $, $ BLP $, $ CMP $ (Figure 7), we observe that
Similarly, from the triangles $ AMP $, $ BNP $, $ CLP $ we get
Thus, by equating the right sides of the obtained equalities:
The left side of equation (1) is precis... | PisthecircumcenteroftriangleABC | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,050 |
XLVIII OM - I - Problem 2
Point $ P $ lies inside the parallelogram $ ABCD $, and the equality $ |\measuredangle ABP| = |\measuredangle ADP| $ holds. Prove that $ |\measuredangle PAB| = |\measuredangle PCB| $. | We translate the triangle $ ADP $ in parallel such that the image of side $ AD $ is segment $ BC $. Let the image of vertex $ P $ be denoted by $ Q $. Therefore, we have the equality $ |\measuredangle ADP| = |\measuredangle BCQ| $ (Figure 1). Segments $ AB $ and $ PQ $ are parallel, so $ |\measuredangle ABP| = |\measur... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,054 |
XI OM - I - Problem 7
A triangle $ABC$ is inscribed in a given circle. On the sides $AC$ and $BC$, segments $AM$ and $BN$ of a given length $d$ are measured. Find the geometric locus of the midpoint $S$ of segment $MN$, when points $A$ and $B$ are fixed, and point $C$ moves along the given circle. | We draw from point $ S $ parallel to the lines $ AC $ and $ BC $ and from points $ A $ and $ B $ parallel to the line $ MN $; we obtain parallelograms $ AMSP $ and $ BNSR $ (Fig. 5).
If triangle $ ABC $ is not isosceles, for example if (as in Fig. 5) $ AC > BC $, then $ MC > NC $, so $ \measuredangle CMN < \measuredang... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,055 |
XXXVI OM - I - Zadanie 9
W urnie jest 1985 kartek z napisanymi liczbami 1,2,3,..., 1985, każda lczba na innej kartce. Losujemy bez zwracania 100 kartek. Znaleźć wartość oczekiwaną sumy liczb napisanych na wylosowanych kartkach.
|
Losowanie $ 100 $ kartek z urny zawierającej $ 1985 $ kartek można interpretować jako wybieranie $ 100 $-elementowego podzbioru zbioru $ 1985 $-elementowego. Zamiast danych liczb $ 1985 $ i $ 100 $ weźmy dowolne liczby naturalne $ n $ i $ k $, $ n \geq k $. Dla dowolnego $ k $-elementowego zbioru $ X $ będącego podzbi... | 99300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1,058 |
LII OM - I - Task 4
Determine whether 65 balls with a diameter of 1 can fit into a cubic box with an edge of 4. | Answer: It is possible.
The way to place the balls is as follows.
At the bottom of the box, we place a layer consisting of 16 balls. Then we place a layer consisting of 9 balls, each of which is tangent to four balls of the first layer (Fig. 1 and 2). The third layer consists of 16 balls that are tangent to the balls o... | 66 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,060 |
XII OM - II - Task 2
Prove that all altitudes of a tetrahedron intersect at one point if and only if the sums of the squares of opposite edges are equal. | The solution to the problem will help us observe that the segment connecting the midpoints of two edges of a tetrahedron belonging to the same face is equal to half of the third edge of that face (and is parallel to it). Let's denote the midpoints of the edges of the tetrahedron $ABC$ as shown in Fig. 13 with the lette... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,062 |
XXVII OM - I - Zadanie 12
Ciąg $ (x_n) $ określony jest wzorami $ x_0 = 25 $, $ x_n = x_{n-1} + \frac{1}{x_{n-1}} $ ($ n = 1, 2, \ldots $) Dowieść, że $ x_n > 1975 $ dla $ n > 1950000 $.
|
Udowodnimy ogólniejsze
Twierdzenie. Jeżeli $ \displaystyle x_0 > \frac{1}{2} $ i $ x_{n} = x_{n-1} + \frac{1}{x_{n-1}} $ dla $ n = 1, 2, \ldots $, to dla każdej liczby naturalnej $ N $ zachodzi nierówność
Dowód. Wyrazy ciągu $ (x_{n}) $ są oczywiście liczbami dodatnimi. Wobec tego $ x_n = x_{n-1} + \frac{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 1,063 |
II OM - III - Task 2
What digits should be placed instead of zeros in the third and fifth positions in the number $ 3000003 $ to obtain a number divisible by $ 13 $? | Let the sought digits be denoted by $ x $ and $ y $, and write the number $ 30x0y03 $ in the form
We need to find such integers $ x $ and $ y $ that the number $ N $ is divisible by $ 13 $ and that the inequalities
are satisfied.
The numbers $ 10^6 $, $ 10^4 $, $ 10^2 $ give remainders $ 1 $, $ 3 $, $ 9 $ respective... | 3080103,3040203,3020303,3090503,3060603,3030703,3000803 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 1,064 |
XLIII OM - III - Problem 6
Prove that for every natural number $ k $, the number $ (k!)^{k^2+k+1} $ is a divisor of the number $ (k^3)! $. | For every pair of natural numbers $ n,l \geq 1 $, the following equality holds:
Let us fix natural numbers $ n,m \geq 1 $. Substitute $ l = 1,2,\ldots,m $ into (1) and multiply the resulting equations side by side:
We transform the left side of equation (2):
and after reducing the repeated factors in the numerators ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,066 |
LVII OM - III - Problem 3
Given a convex hexagon $ABCDEF$, in which $AC = DF$, $CE = FB$ and $EA = BD$. Prove that the lines connecting the midpoints of opposite sides of this hexagon intersect at one point. | Let $P, Q, R$ be the midpoints of the diagonals $AD, BE, CF$, respectively.
First, assume that two of the points $P, Q, R$ coincide; let, for example, $P = Q$ (Fig. 1).
Then the quadrilateral $ABDE$ is a parallelogram. Moreover, triangle $ACE$ is congruent to triangle $DFB$, which implies that
$\measuredangle EAC = \me... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,068 |
XII OM - I - Problem 1
$k$ parallel lines were drawn and intersected by $n$ parallel lines. How many parallelograms were formed? | Every parallelogram defines (by extending its sides) one pair $ (a, b) $ among the given $ k $ parallel lines and one pair $ (c, d) $ among the $ n $ parallel lines; conversely, any two such pairs $ (a, b) $ and $ (c, d) $ of parallel lines define one parallelogram of the figure. Therefore, the figure has as many paral... | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1,071 | |
XLVI OM - I - Problem 9
Let $ a $ and $ b $ be real numbers whose sum is equal to 1. Prove that if $ a^3 $ and $ b^3 $ are rational numbers, then $ a $ and $ b $ are also rational numbers. | We raise the equality $ a + b = 1 $ to the second and third powers on both sides and obtain the relations: $ a^2 +2ab + b^2 = 1 $, that is,
and $ a^3 + 3a^2b + 3ab^2 + b^3 = 1 $, that is,
If, therefore, the numbers $ a^3 $ and $ b^3 $ are rational, then from equation (2) it follows that the product $ ab = (... | proof | Algebra | proof | Yes | Yes | olympiads | false | 1,072 |
XIV OM - I - Problem 12
In a circle with center $ O $ and radius $ r $, a regular pentagon $ A_1A_2A_3A_4A_5 $ is inscribed, and on the smaller arc with endpoints $ A_1 $, $ A_5 $, a point $ M $ is chosen. Prove that | The equality (1) we need to prove is a relationship between the lengths of certain chords of a circle. Another relationship of the same kind is the well-known Ptolemy's theorem: The product of the diagonals of a cyclic quadrilateral equals the sum of the products of its opposite sides1). It suggests using this theorem ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,073 |
XXII OM - I - Problem 12
Prove that every convex polyhedron has a triangular face or a trihedral angle. | Assume that there exists a convex polyhedron $W$ without any triangular faces or trihedral angles. Let $w$ be the number of vertices, $k$ the number of edges, and $s$ the number of faces of this polyhedron, and let $\varphi$ be the sum of the measures of all dihedral angles at the vertices of the polyhedron $W$. Since ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,075 |
V OM - III - Problem 6
Inside a hoop of radius $2r$, a disk of radius $r$ rolls without slipping along the hoop. What line does a point chosen arbitrarily on the edge of the disk trace? | When a disk rolls along the circumference, the points on the edge of the disk become points of contact with the circumference one after another. The condition of rolling without slipping means that the length of the arc $PQ$ between two points $P$ and $Q$ on the edge of the disk is equal to the length of the arc on the... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,076 |
L OM - I - Task 5
Find all pairs of positive integers $ x $, $ y $ satisfying the equation $ y^x = x^{50} $. | We write the given equation in the form $ y = x^{50/x} $. Since for every $ x $ being a divisor of $ 50 $, the number on the right side is an integer, we obtain solutions of the equation for $ x \in \{1,2,5,10,25,50\} $. Other solutions of this equation will only be obtained when $ x \geq 2 $ and for some $ k \geq 2 $,... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 1,080 |
XXXIII OM - II - Problem 6
Given is a finite set $ B $ of points in space, such that any two distances between points of this set are different. Each point of the set $ B $ is connected by a segment to the nearest point of the set $ B $. In this way, we obtain a set of segments, one of which (chosen arbitrarily) is pa... | Suppose the segment $ A_1A_2 $ is red, and the rest are green. If there existed a broken line composed of green segments connecting points $ A_1 $ and $ A_2 $, then there would be points $ A_3, \ldots, A_n $ being the successive endpoints of the segments forming this broken line. The notation $ A_i \to A_j $ is read as... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 1,081 |
VIII OM - I - Problem 8
In the rectangular prism $ ABCDA_1B_1C_1D_1 $, the lengths of the edges are given as $ AA_1 = a $, $ AB = b $, $ AD = c $. On the face $ A_1B_1C_1D_1 $, a point $ M $ is chosen at a distance $ p $ from the side $ A_1B_1 $, and at a distance $ q $ from the side $ A_1D_1 $, and a parallelepiped i... | Accepting the notation indicated in figure $5$, we calculate the area of the parallelogram $AMQD$ which is a face of the parallelepiped $ABCDMNPQ$.
The parallelogram $AMQD$ and the rectangle $AKLD$ have a common base $AD$ and height $AK$, thus
But $AK = \sqrt{AA_1^2 + A_1K^2} = \sqrt{a^2 + q^2},\ KL = AD = c$; he... | \sqrt{^2+} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,083 |
LI OM - II - Task 3
On an $ n \times n $ chessboard, $ n^2 $ different integers are placed, one on each square. In each column, the square with the largest number is painted red. A set of $ n $ squares on the chessboard is called admissible if no two of these squares are in the same row or the same column. Among all a... | We will conduct an indirect proof.
Assume that in the chosen admissible set (denoted by $ A $) there is no red field.
We define a sequence of fields $ D_1, E_1, D_2, E_2, \ldots $ of the given chessboard according to the rule described below. (For $ n = 8 $, the process of selecting fields $ D_1, E_1, D_2, E_2, \ldots ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 1,084 |
XXIII OM - III - Problem 1
Polynomials $ u_1(x) = a_ix + b_i $ ($ a_i, b_i $ - real numbers; $ i = 1, 2, 3 $) satisfy for some natural $ n > 2 $ the equation
Udowodnić, że istnieją takie liczby rzeczywiste $ A, B, c_1, c_2, c_3 $, że $ u_i(x)=c_i(Ax+B) $ for $ i = 1, 2, 3 $.
Prove that there exist real numbers $ A, ... | If $ a_1 = a_2 = 0 $, then the polynomials $ u_1 $ and $ u_2 $ are constant. Therefore, the polynomial $ u_3 $ is also constant, i.e., $ a_3 = 0 $. In this case, it suffices to take $ c_i = b_i $ for $ i = 1, 2, 3 $ and $ A = 0 $ and $ B = 1 $.
Let then at least one of the numbers $ a_1, a_2 $ be different from zero, f... | proof | Algebra | proof | Yes | Yes | olympiads | false | 1,085 |
XXXVI OM - III - Problem 5
Let $ P $ be a polynomial in two variables such that for every real number $ t $ the equality $ P(\cos t, \sin t) = 0 $ holds. Prove that there exists a polynomial $ Q $ such that the identity 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，应该翻译成英文如下：
Preserving the line breaks and f... | Let's arrange the polynomial $ P(x,y) $ in decreasing powers of the variable $ x $. $ P(x,y) = P_n(y) \cdot x^n + P_{n-1}(y) \cdot x^{n-1} + \ldots + P_1(y) \cdot x + P_0(y) $, where $ P_n, P_{n-1}, \ldots, P_1, P_0 $ are polynomials in the variable $ y $. Treating $ y $ as a fixed value, we can compute the quotient an... | proof | Algebra | proof | Yes | Yes | olympiads | false | 1,086 |
XXXIV OM - I - Problem 6
Prove that for any polynomial $ P $, the polynomial $ P\circ P \circ P \circ \ldots \circ P(x)-x $ is divisible by $ P(x)-x $. | Putting $ P(x)= a_nx^n + a_{n-1}x^{n-1} + \ldots +a_1x+a_0 $ we get
Since each term of the last sum can be factored:
thus
Substituting in the last relation $ P(x) $ for $ x $ we get $ P \circ P(x)-P(x) | P \circ P \circ P(x) - P \circ P(x) $, from which by the transitivity of the divisibility relation it foll... | proof | Algebra | proof | Yes | Yes | olympiads | false | 1,087 |
VI OM - I - Problem 12
In what part of an equilateral triangle should point $ P $ lie so that from segments equal to the distances of this point from the sides of the triangle, a triangle can be constructed? | Let $ k_1 $, $ k_2 $, $ k_3 $ denote the distances from a point $ P $ inside an equilateral triangle $ ABC $ to the sides $ BC $, $ CA $, $ AB $ of the triangle, respectively. We can form a triangle with segments of lengths $ k_1 $, $ k_2 $, $ k_3 $ if and only if they satisfy the following inequalities:
Draw a line t... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,088 |
XXXVII OM - II - Problem 2
In a chess tournament, 66 players participate, each playing one game against every other player, and the matches take place in four cities. Prove that there exists a trio of players who play all their games against each other in the same city. | Let's choose one of the players, let's call him $Z_1$. He has to play $65$ games, so he plays at least $17$ games in one city. Let's denote this city by $M_1$. Consider the opponents of $Z_1$ in the matches played in $M_1$. There are at least $17$ of them. If there is a pair among them who play a game against each othe... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 1,092 |
XI OM - I - Problem 8
In the plane, there are three rays $ OX $, $ OY $, $ OZ $, dividing the plane into three convex angles, and points $ A $, $ B $, $ C $ lying inside the angles $ YOZ $, $ ZOX $, $ XOY $, respectively. Construct a triangle whose vertices lie on $ OX $, $ OY $, $ OZ $, and whose sides pass through $... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,094 | |
XXIII OM - III - Problem 6
Prove that the sum of the digits of the number $ 1972^n $ tends to infinity as $ n $ tends to infinity. | We will prove in general that if $a$ is an even natural number not divisible by $5$ and $s_n$ denotes the sum of the digits of the number $a^n$ for $n = 1, 2, \ldots$, then the sequence $s_n$ tends to infinity.
Let $a_r, a_{r-1}, \ldots, a_2, a_1$, where $a \ne 0$, be the consecutive digits of the number $a^n$, i.e.,
... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,095 |
XVI OM - I - Problem 3
Through each edge of a trihedral angle, a plane containing the bisector of the opposite planar angle has been drawn. Prove that the three planes intersect along a single line. | Consider a trihedral angle with vertex $O$. We measure three equal segments $OA = OB = OC$ on its edges. Consider the plane $\alpha$ passing through the edge $OA$ and the bisector of the angle $BOC$. Since the triangle $BOC$ is isosceles, the bisector of the angle $BOC$ passes through the midpoint $M$ of the side $BC$.... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,096 |
XVIII OM - II - Problem 3
Two circles are internally tangent at point $ A $. A chord $ BC $ of the larger circle is tangent to the smaller circle at point $ D $. Prove that $ AD $ is the angle bisector of $ \angle BAC $. | The point of tangency $ A $ of the given circles is their center of homothety. In this homothety, the tangent $ BC $ of the smaller circle corresponds to the tangent to the larger circle at point $ E $, which corresponds to point $ D $ (Fig. 6). As homothetic lines, these tangents are parallel. Therefore, the arcs $ BE... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,097 |
XLVI OM - II - Problem 2
In a convex hexagon $ABCDEF$, the following equalities hold: $|AB| = |BC|$, $|CD| = |DE|$, $|EF| = |FA|$. Prove that the lines containing the altitudes of triangles $BCD$, $DEF$, and $FAB$ drawn from vertices $C$, $E$, and $A$, respectively, intersect at a single point. | Let's adopt the following notations:
$ k_1 $ - the circle with center $ D $ and radius $ |DC| = |DE| $;
$ k_2 $ - the circle with center $ F $ and radius $ |FE| = |FA| $;
$ k_3 $ - the circle with center $ B $ and radius $ |BA| = |BC| $.
Circles $ k_2 $ and $ k_3 $ intersect at point $ A $; let's denote the second ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,098 |
XIII OM - I - Problem 2
How many digits do all natural numbers with at most $ m $ digits have in total? | The number of one-digit natural numbers is $ c_1 = 9 $; the number of two-digit numbers is $ c_2= 10^2 -10= 10 \cdot 9 $, they have $ 2 \cdot 10 \cdot 9 $ digits; the number of three-digit numbers is $ c_3=10^3-10^2= 10^2 \cdot 9 $, the number of their digits is $ 3 \cdot 10^2 \cdot 9 $, etc. Finally, the number of $ m... | 9\cdot\frac{10^-1}{9}\cdot\frac{(+1)}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1,099 |
XLV OM - III - Task 5
Points $ A_1, A_2, \ldots , A_8 $ are the vertices of a parallelepiped with center $ O $. Prove that | Let's assume that one of the faces of a given parallelepiped is the parallelogram $A_1A_2A_3A_4$, and the opposite face is the parallelogram $A_5A_6A_7A_8$, with segments $A_1A_5$, $A_2A_6$, $A_3A_7$, $A_4A_8$ being four edges of the parallelepiped (Figure 17). Let's denote the distances from the vertices to the point ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,100 |
LIX OM - III - Task 4
Each point in the plane with both integer coordinates has been painted either white or black. Prove that from the set of all painted points, an infinite subset can be selected which has a center of symmetry and all of whose points have the same color. | Suppose the thesis of the problem is false.
Consider the central symmetry with respect to the point $ (0,0) $. Since there does not exist an infinite set symmetric with respect to this point and composed of points of the same color, only finitely many points with integer coordinates pass to points of the same color und... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 1,102 |
XXVII OM - II - Problem 5
Prove that if $ \cos \pi x =\frac{1}{3} $ then $ x $ is an irrational number. | Let $ \cos t = \frac{1}{3} $. We will prove by induction that for every natural number $ n $ the following formula holds:
where $ a_n $ is an integer not divisible by $ 3 $.
For $ n=1 $, it suffices to take $ a_1 = 1 $. For $ n = 2 $ we have
Thus, $ a_2 = -7 $. Suppose next that for some natural number $ k $ the foll... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,103 |
LVIII OM - I - Problem 11
For each positive integer $ n $, determine the number of permutations $ (x_1,x_2,\ldots,x_{6n-1}) $ of the set $ \{1,2,\ldots,6n{-}1\} $, satisfying the conditions: | The sought number is
For any permutation satisfying the conditions of the problem, the following sequences of inequalities hold
Notice that each of the elements $ x_1 $, $ x_2 $, $ \ldots $, $ x_{6n-1} $ has been listed exactly once in the considered rows. Indeed, in the inequalities (1), all odd indice... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1,104 |
XV OM - I - Problem 8
On three pairwise skew edges of a cube, choose one point on each in such a way that the sum of the squares of the sides of the triangle formed by them is minimized. | Let $ABCD$ be the base of a cube and $AA_1$, $BB_1$, $CC_1$, $DD_1$ its edges perpendicular to $ABCD$. Let $X$, $Y$, $Z$ be points chosen on three pairwise skew edges of the cube. It is known that by rotating the cube around certain axes passing through its center, the cube can be superimposed onto itself in such a way... | \frac{9}{2}^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,107 |
III OM - III - Task 5
Prove that none of the digits $ 2 $, $ 4 $, $ 7 $, $ 9 $ can be the last digit of the number
where $ n $ is a natural number. | If the last digit of a given number is $ x $, then
so
Since $ n $ is an integer, the discriminant of the above equation, i.e., the number
is a square of an integer. The last digit of this discriminant is the last digit of the number $ 8x + 1 $. When $ x $ equals $ 2 $, $ 4 $, $ 7 $, $ 9 $, then $ 8x +... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,108 |
LII OM - I - Problem 9
Prove that among any $12$ consecutive positive integers, there exists a number that is not the sum of $10$ fourth powers of integers. | A number that is the fourth power of an integer gives a remainder of $0$ or $1$ when divided by $16$. Indeed: for even numbers of the form $2k$ we have $(2k)^4 = 16k^4$, whereas for odd numbers $2k + 1$ we obtain
(The number $\frac{1}{2} k(3k+1)$ is an integer for any integer $k$).
Hence, a number that is the sum of $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,110 |
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