problem stringlengths 0 45.3k | solution stringlengths 0 26.8k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 9
values | question_type stringclasses 5
values | problem_is_valid stringclasses 5
values | solution_is_valid stringclasses 5
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XXIV OM - I - Problem 11
Prove that if the center of the sphere circumscribed around a tetrahedron coincides with the center of the sphere inscribed in this tetrahedron, then the faces of this tetrahedron are congruent. | Let point $O$ be the center of the inscribed and circumscribed spheres of the tetrahedron $ABCD$. Denote by $r$ and $R$ the radii of these spheres, respectively. If $O$ is the point of tangency of the inscribed sphere with one of the faces, and $P$ is one of the vertices of this face, then applying the Pythagorean theo... | proof | Geometry | proof | Yes | Yes | olympiads | false |
VIII OM - I - Problem 7
Construct triangle $ ABC $ given: side $ AB = c $, circumradius $ R $, and angle $ \delta $ between the angle bisector of $ C $ and the altitude from vertex $ C $. | Let $D$ (Fig. 4) denote the foot of the altitude of triangle $ABC$ drawn from vertex $C$, $E$ - the point of intersection of the angle bisector of angle $C$ with side $AB$, $M$ - the point of intersection of line $CE$ with the circumcircle of triangle $ABC$, and $F$ - the foot of the perpendicular dropped from $M$ to l... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false |
II OM - II - Task 6
Given points $ A $ and $ B $ and a circle $ k $. Construct a circle passing through points $ A $ and $ B $ and intersecting circle $ k $ in a common chord of given length $ d $. | Analyze. The task boils down to determining the center $X$ of the common chord of the given circle and the circle to be found; for, having the point $X$, we can draw in the given circle $k$ a chord $CD$ with the center $X$ and draw a circle through the points $A$, $B$, $C$, $D$.
The geometric locus of the centers $X$ o... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false |
III OM - I - Problem 11
Prove that if none of the angles of a convex quadrilateral $ABCD$ is right, then the following equality holds
| We know that
thus
Hence
Multiplying both sides of this equality by $ (1 - \tan A \tan B) ( 1- \tan C \tan D) $, we get
or, after a simple transformation,
Dividing both sides of this equality by the product $ \tan A \tan B \tan C \tan D $, we get
which is the desired equality (1).
How... | proof | Geometry | proof | Incomplete | Yes | olympiads | false |
XXX OM - I - Task 8
Among the cones inscribed in the sphere $ B $, the cone $ S_1 $ was chosen such that the sphere $ K_1 $ inscribed in the cone $ S_1 $ has the maximum volume. Then, a cone $ S_2 $ of maximum volume was inscribed in the sphere $ B $, and a sphere $ K_2 $ was inscribed in the cone $ S_2 $. Determine w... | Let $ S $ be any cone inscribed in a given sphere $ B $, and $ K $ - the sphere inscribed in the cone $ S $. The axial section of the cone $ S $ is an isosceles triangle $ ACD $, where $ A $ is the vertex of the cone (Fig. 7).
om30_1r_img_7.jpg
Let $ O $ be the center of the sphere $ B $, and $ Q $ - the center of the ... | V(\frac{\sqrt{3}}{3})>V(\frac{1}{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXXI - I - Task 1
Determine for which values of the parameter $ a $ a rhombus with side length $ a $ is a cross-section of a cube with edge length 2 by a plane passing through the center of the cube. | Suppose the rhombus $KLMN$ is a cross-section of the cube $ABCDA_1B_1C_1D_1$ by a plane passing through the center of the cube $O$. Since $O$ is also the center of symmetry of the cross-section, the opposite vertices of the rhombus lie on opposite edges of the cube. Let, for example, as in the figure, $K \in \overline{... | [2,\sqrt{5}] | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XIV OM - II - Task 6
From point $ S $ in space, $ 3 $ rays emerge: $ SA $, $ SB $, and $ SC $, none of which is perpendicular to the other two. Through each of these rays, a plane is drawn perpendicular to the plane containing the other two rays. Prove that the planes drawn intersect along a single line $ d $. | In the case where the half-lines lie in the same plane, the theorem is obvious; the line $d$ is then perpendicular to the given plane. Let us assume, therefore, that the half-lines $SA$, $SB$, $SC$ form a trihedral angle. According to the assumption, at most one of the dihedral angles of this trihedron is a right angle... | proof | Geometry | proof | Yes | Yes | olympiads | false |
LX OM - II - Task 6
For each integer $ n \geqslant 3 $, determine all sequences of real numbers
$ (x_1,x_2, \cdots ,x_n) $, for which
where we assume $ x_0 = x_n $ and $ x_{n+1} = x_1 $. | Suppose the sequence $ (x_1,x_2, \cdots ,x_n) $ has the desired property. Let us assume
In this case, the equality $ y_1 + y_2 + \cdots +y_n = x_1 + x_2 + \cdots + x_n -n $ holds, and thus from the two equalities given in the problem, we have
From this, we obtain
Therefore, $ y_1 = y_2 = \cdots = y_n = 0 ... | (x_1,x_2,\cdots,x_n)=(1,1,\cdots,1) | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
LVII OM - III - Problem 1
Solve in real numbers $ a, b, c, d, e $ the system of equations
(Note: The system of equations is not provided in the original text, so it is not included in the translation.) | Let's assume that the numbers $a, b, c, d, e$ satisfy the given system of equations and that $b$ is the largest among them (without loss of generality, as the system is cyclic). We will show that in this case $b = d$.
Suppose, to the contrary, that $b > d$. Subtracting the second equation from the first, we get $a^2 - ... | (0,0,0,0,0)(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}) | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
LV OM - I - Task 9
Given are polynomials $ W_1(x), W_2(x), W_3(x), \ldots, W_n(x) $ of degree at least 1, with integer coefficients. Prove that for some integer $ a $, all the numbers
are composite. | Given polynomials have a degree of at least 1, so there exists a natural number $ x_0 $ such that for $ i = 1,2,\ldots,n $ the numbers $ c_i = |W_i(x_0)| $ are greater than 1 and $ |W_i(x)| > c_i $ for all $ x > x_0 $.
Let $ a = x_0 + c_1 c_2 \ldots c_n $. We will show that the numbers
are composite, which will comple... | proof | Number Theory | proof | Incomplete | Yes | olympiads | false |
XXV - I - Task 2
Prove that for every natural $ m $
The translation is provided as requested, maintaining the original text's line breaks and format. | We need to prove that
From the formula $ \displaystyle \binom{n}{k} = \binom{n}{n-k} $, where $ 0 \leq k \leq n $, it follows in particular that $ \displaystyle \binom{2m+1}{m} = \binom{2m+1}{m+1} $. From the binomial formula we obtain
Therefore, inequality (1) follows. | proof | Number Theory | proof | Incomplete | Yes | olympiads | false |
XXIV OM - I - Problem 9
Real numbers $ x_1, x_2, \ldots, x_n $ satisfy the conditions:
Prove that the number $ n $ is divisible by 4. | From the conditions of the problem, it follows that the numbers $ x_1x_2, x_2x_3, \ldots, x_{n-1}x_n, x_nx_1 $ are equal to $ 1 $ or $ -1 $. Since their sum is equal to zero, $ k $ of these numbers are equal to $ 1 $ and $ k $ are equal to $ -1 $. Therefore, $ n = 2k $. The product of these numbers is equal to $ (-1)^k... | 4s | Algebra | proof | Incomplete | Yes | olympiads | false |
VIII OM - I - Zadanie 5
Jaki warunek powinna spełniać liczbą $ q $, aby istniał trójkąt, którego boki tworzą postęp geometryczny o ilorazie $ q $?
|
Weźmy pod uwagę trzy odcinki, których długości tworzą postęp geometryczny o ilorazie $ q $. Ponieważ stosunek dwóch odcinków jest liczbą niezależną od jednostki, jaką te odcinki mierzymy, więc jeśli za jednostkę obierzemy pierwszy z rozważanych trzech odcinków, to długości tych odcinków wyrażą się liczbami $ 1 $, $ q ... | \frac{-1+\sqrt{5}}{2}<\frac{1+\sqrt{5}}{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
LI OM - III - Task 1
Given an integer $ n \geq 2 $.
Determine the number of solutions $ (x_1,x_2,\ldots,x_n) $ of the system of equations
in non-negative real numbers. | Answer: $ 2^n $.
From the first equation, we obtain
from which it follows that $ x_1 \leq 4 $. Therefore, there exists exactly one number $ \alpha \in \langle 0,\pi/2\rangle $, satisfying the relationship $ x_1 =4\sin^2 \alpha $.
Then $ x_{2}=x_{1}(4-x_{1})=4\sin^{2}\alpha\cdot 4\cos^{2}\alpha=4\sin^{2}2\alpha $.... | 2^n | Combinatorics | math-word-problem | Incomplete | Yes | olympiads | false |
XXXVIII OM - I - Zadanie 2
Warunkiem koniecznym i wystarczającym na to, by liczby dodatnie $ a, b, c $ były długościami boków pewnego trójkąta jest, by spełniały one nierówność $ |a—b| < c < |a+b| $. Sformułować i udowodnić analogiczne warunki konieczne i wystarczające na to, by trzy liczby dodatnie były długośc... |
a) Załóżmy, że liczby $ h_a $, $ h_b $, $ h_c $ są długościami wysokości pewnego trójkąta. Niech $ a $, $ b $, $ c $ będą długościami boków tego trójkąta ($ h_a $ — wysokość opuszczona na bok o długości $ a $ itd.), a $ S $ - jego polem. Z równości $ 2S = ah_a= bh_b = ch_c $ wynika, że liczby $ h_a $, $ h_b $, $ h_c... | proof | Geometry | proof | Yes | Incomplete | olympiads | false |
XL OM - III - Task 5
On a sphere of radius $ r $, there are three circles of radius $ a $, each tangent to the other two and contained within the same hemisphere. Find the radius of the circle lying on the same sphere and tangent to each of these three circles.
Note: Similar to on a plane, we say that two circles lyin... | Let's denote three given circles (of radius $a$) by $k_1$, $k_2$, $k_3$. Suppose $k_0$ is a circle of radius $x$, located on the same sphere and tangent to each of the circles $k_1$, $k_2$, $k_3$. None of the circles $k_1$, $k_2$, $k_3$ are great circles of the given sphere (two different great circles always have two ... | Geometry | math-word-problem | Yes | Yes | olympiads | false | |
VII OM - II - Problem 2
Prove that if $ H $ is the orthocenter of a non-right triangle $ ABC $, then the circumcircles of triangles $ AHB $, $ BHC $, $ CHA $, and $ ABC $ are equal. | Let $ R $ denote the radius of the circumcircle of triangle $ ABC $, and $ R_1 $ - the radius of the circumcircle of triangle $ AHB $ (Fig. 16); then
Indeed, $ \measuredangle AHB = \measuredangle NHM = 180^\circ - \measuredangle NCM = 180^\circ - \measuredangle ACB $, since in quadrilateral $ MHNC $ angles $ M $ ... | proof | Geometry | proof | Yes | Incomplete | olympiads | false |
XXII OM - II - Problem 4
In the plane, a finite set of points $ Z $ is given with the property that no two distances between points in $ Z $ are equal. Points $ A, B $ belonging to $ Z $ are connected if and only if $ A $ is the nearest point to $ B $ or $ B $ is the nearest point to $ A $. Prove that no point in the ... | Suppose that point $ A \in Z $ is connected to some points $ B $ and $ C $ of set $ Z $ and let, for example, $ AB < AC $. Then point $ C $ is not the closest to $ A $, and from the conditions of the problem it follows that point $ A $ is the closest to $ C $. Therefore, $ AC < BC $. Hence, in triangle $ ABC $, side $ ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
V OM - III - Task 2
What algebraic relationship holds between $ A $, $ B $, and $ C $, if
| From equality (1), the following equalities follow:
From equality (2), the alternative follows:
equivalent to the alternative:
where $ k $, $ m $, $ n $ denote integers.
We have obtained the result: If equality (1) holds, then
where $ k $, $ m $, $ n $ are integers.
... | A-B=k\pi,\quadB-C=\pi,\quadA-C=n\pi | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
XXI OM - III - Task 2
Given are three infinite sequences whose elements are natural numbers:
where if $ i \neq j $, then $ a_i \neq a_j $, $ b_i \neq b_j $, $ c_i \neq c_j $.
Prove that there exists a pair of indices $ k, l $ such that $ k < l $ and $ a_k < a_l $, $ b_k < b_l $, $ c_k < c_l $. | Since the terms of the sequence $ a_1, a_2, \ldots $ are distinct natural numbers, and the set of natural numbers less than $ a_1 $ is finite, therefore all sufficiently distant terms of this sequence are greater than $ a_1 $, i.e.,
Similarly, we state that
If $ n $ is a natural number greater than eac... | proof | Combinatorics | proof | Yes | Incomplete | olympiads | false |
LII OM - III - Problem 6
Given positive integers $ n_1 < n_2 < \ldots < n_{2000} < 10^{100} $. Prove that from the set $ \{n_1, n_2, \ldots, n_{2000}\} $, one can select non-empty, disjoint subsets $ A $ and $ B $ having the same number of elements, the same sum of elements, and the same sum of squares of elements. | For a set $ X \subseteq \{n_1,n_2,\ldots,n_{2000}\} $, let $ s_0(X) $, $ s_1(X) $, and $ s_2(X) $ denote the number of elements, the sum of elements, and the sum of squares of elements of the set $ X $, respectively.
It suffices to prove that there exist two different subsets $ C $ and $ D $ of the set $ \{n_1,n_2,\ldo... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
XLVIII OM - II - Problem 6
In a cube with edge length $1$, there are eight points. Prove that some two of them are the endpoints of a segment of length not greater than $1$. | Let the vertices of a given cube $\mathcal{C}$ be denoted by $A_1, \ldots, A_8$. Let $\mathcal{C}_i$ be the cube with edge length $\frac{1}{2}$, having one vertex at point $A_i$ and three faces contained in the faces of cube $\mathcal{C}$. Let $P_1, \ldots, P_8$ be eight given points.
Each cube $\mathcal{C}_i$ has a di... | proof | Geometry | proof | Yes | Yes | olympiads | false |
L OM - I - Problem 1
Prove that among numbers of the form $ 50^n + (50n + 1)^{50} $, where $ n $ is a natural number, there are infinitely many composite numbers. | I way:
If $ n $ is an odd number, then the number $ 50^n $ when divided by $ 3 $ gives a remainder of $ 2 $. If, moreover, $ n $ is divisible by $ 3 $, then $ (50n+1)^{50} $ when divided by $ 3 $ gives a remainder of $ 1 $. Hence, the number $ 50^n + (50n+1)^{50} $ is divisible by $ 3 $ for numbers $ n $ of the form $... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
XXXIII OM - II - Problem 2
The plane is covered with circles in such a way that the center of each of these circles does not belong to any other circle. Prove that each point of the plane belongs to at most five circles. | Suppose that point $ A $ belongs to six circles from the considered family and that $ O_1 $, $ O_2 $, $ O_3 $, $ O_4 $, $ O_5 $, $ O_6 $ are the centers of these circles. It follows that one of the angles $ O_iAO_j $ $ (i,j = 1, 2, \ldots, 6) $ has a measure not greater than $ 60^\circ $ (six of the angles $ O_iAO_j $ ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
LVII OM - II - Problem 5
Point $ C $ is the midpoint of segment $ AB $. Circle $ o_1 $ passing through points $ A $ and $ C $ intersects circle $ o_2 $ passing
through points $ B $ and $ C $ at different points $ C $ and $ D $. Point $ P $ is the midpoint of the arc $ AD $ of circle $ o_1 $, which does not
contain poi... | If $ AC = CD $, then also $ BC = CD $. Then segments $ PC $ and $ QC $ are diameters of circles $ o_1 $ and $ o_2 $, respectively. Therefore, $ \measuredangle CDP = \measuredangle CDQ = 90^{\circ} $, which implies that point $ D $ lies on segment $ PQ $, and lines $ PQ $ and $ CD $ are perpendicular.
Let us assume in t... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XVIII OM - II - Task 2
In a room, there are 100 people, each of whom knows at least 66 of the remaining 99 people. Prove that it is possible that in every quartet of these people, there are two who do not know each other. We assume that if person $ A $ knows person $ B $, then person $ B $ also knows person $ A $. | Let's denote the people in the room by the letters $A_1, A_2, \ldots, A_{100}$. Let $M$ be the set of people $\{A_1, A_2, \ldots, A_{33}\}$, $N$ be the set of people $\{A_{34}, A_{35}, \ldots, A_{66}\}$, and $P$ be the set of people $\{A_{67}, A_{68}, \ldots, A_{100}\}$. The case described in the problem occurs, for ex... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
LV OM - I - Task 7
Find all solutions to the equation $ a^2+b^2=c^2 $ in positive integers such that the numbers $ a $ and $ c $ are prime, and the number $ b $ is the product of at most four prime numbers. | There are three solutions $(a,b,c): (3,4,5), (5,12,13), (11,60,61)$. Let $a$, $b$, $c$ be numbers satisfying the conditions of the problem. Then
from the assumption that the number $a$ is prime, we obtain $c = b + 1$. Therefore, $a^2 = 2b + 1$, which implies that the number $a$ is odd.
Let $a = 2n + 1$. Then
if the n... | (3,4,5),(5,12,13),(11,60,61) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XLV OM - III - Task 2
In the plane, there are two parallel lines $ k $ and $ l $ and a circle disjoint from line $ k $. From a point $ A $ lying on line $ k $, we draw two tangents to this circle intersecting line $ l $ at points $ B $, $ C $. Let $ m $ be the line passing through point $ A $ and the midpoint of segme... | Note 1. The conditions of the problem do not change if the given line $ l $ is replaced by another line $ l' $ parallel to $ k $ and different from $ k $. Indeed: if two tangents to a given circle emanating from point $ A $ intersect the line $ l $ at points $ B $ and $ C $, then the midpoints of segments $ BC $ and $ ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XIV OM - II - Task 4
In triangle $ABC$, the angle bisectors of the internal and external angles at vertices $A$ and $B$ have been drawn. Prove that the perpendicular projections of point $C$ onto these bisectors lie on the same line. | Let $M$ and $N$ denote the projections of point $C$ onto the angle bisector $AM$ of angle $A$ of triangle $ABC$ and onto the bisector of the adjacent angle (Fig. 22).
Since the bisectors $AM$ and $AN$ are perpendicular, the quadrilateral $AMCN$ is a rectangle. The point $S$, where the diagonals $MN$ and $AC$ intersect,... | proof | Geometry | proof | Yes | Yes | olympiads | false |
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 |
XXX OM - I - Task 10
Given are sequences of polynomials $ (P_n) $, $ (Q_n) $ defined as follows:
Let $ a_n $ be the number of coefficients of the polynomial $ P_n $ equal to 1. Prove that the limit $ \lim_{n\to \infty} \frac{a_n}{2^n} $ exists and calculate it. | From the definition of polynomials $P_n$ and $Q_n$, it follows that for $n \geq 1$, each of them has degree $2^n - 1$ and each of the coefficients of these polynomials is equal to $1$ or $-1$. If $a_n$ is the number of coefficients of the polynomial $P_n$ that are equal to $1$, and $b_n$ is the number of coefficients o... | \frac{1}{2} | Algebra | proof | Incomplete | Yes | olympiads | false |
XXXI - I - Problem 3
A triangle $ABC$ is inscribed in a circle $k$. On the circle $k$, points $A$ are chosen such that the lines $AA$ contain the medians of the triangle $ABC$. Prove that the area of triangle $A$ is not less than the area of triangle $ABC$.
(Note: There seems to be a typo in the original text. The ph... | Let $ AB = c $, $ BC = a $, $ CA = b $, $ \measuredangle BAC = \alpha $, $ \measuredangle ABC = \beta $, $ \measuredangle BCA = \gamma $. Let $ C_1 $ be the midpoint of side $ \overline{AB} $.
Applying the lemma for the median $ \overline{CC_1} $ proved on p. 32, we get
Since $ \measuredangle B and $ \measuredang... | proof | Geometry | proof | Incomplete | Incomplete | olympiads | false |
II OM - II - Zadanie 4
Dowieść, że jeżeli równania
mają wspólny pierwiastek, to między współczynnikami tych równań zachodzi związek
|
Jeżeli któryś z pierwiastków $ x_1 $, $ x_2 $ pierwszego z danych równań (1) równa się któremuś z pierwiastków drugiego równania, to któraś z różnic $ x_1 - x_3 $, $ x_2 - x_3 $, $ x_1 - x_4 $, $ x_2 - x_4 $ równa się zeru. Zachodzi to wtedy i tylko wtedy, gdy
Jeżeli wykonamy mnożenie po lewej stronie tej równoś... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
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 |
III OM - I - Task 9
Solve the equation
where $ a $ denotes a given positive number. | The definition of the absolute value $ |l| $ of a real number $ l $ is as follows:
To solve equation (1), it is best to replace it with an equation that does not contain the absolute value symbol. We will look for solutions to the equation in the intervals $ (-\infty, 0) $, $ (0,1) $, $ (1,2) $, $ (2,\infty) $.
1. If ... | notfound | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
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 |
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 |
XLIII OM - I - Task 4
Calculate
| Denoting the $ n $-th term of the considered sum by $ a_n $, we obtain the equality
Thus, $ a_n $, as a positive number, equals $ (\sqrt{n+1}- \sqrt{n-1})/\sqrt{2} $. Therefore,
We add all these equalities side by side, reduce what can be reduced, divide both sides by $ \sqrt{2} $, and obtain the result: | notfound | Other | math-word-problem | Incomplete | Yes | olympiads | false |
XXVIII - I - Zadanie 6
Obliczyć
|
Wyraz o numerze $ n $ ciągu, którego granicę mamy obliczyć, jest iloczynem ułamków postaci
Mamy rozkłady $ k^3 - 1 = (k - 1) (k^2 + k + 1) $ oraz $ k^3 + 1 = (k+ 1)(k^2 - k + 1) $. Ponieważ $ (k + 1)^2 - (k + 1) + 1 = k^2+2k+1 $, więc czynnik $ k^2 + k + 1 $ występujący w liczniku ułamka $ \displaystyle \frac{k^... | \frac{2}{3} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
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 |
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 |
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 |
XVIII OM - III - Problem 1
Find the highest power of the number 2 that is a divisor of the number
where $ n $ is a natural number. | \spos{1} Notice that $ n!L_n = (2n)! $. Therefore,
From this, it follows that $ L_n $ is divisible by $ 2^n $, but not by $ 2^{n+1} $. | 2^n | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
XV OM - III - Task 1
Prove that the inequality
is not true for any value of $ \alpha $. | We will apply the method of reduction to absurdity. Suppose that a number $ \alpha $ satisfies inequality (1). In such a case, $ \alpha $ satisfies the inequalities
where $ k $ denotes any integer. Indeed, when $ \alpha = k \cdot \frac{\pi}{2} $ ($ k $ = integer), then the number $ \frac{1}{\tan \alpha} $ d... | proof | Inequalities | proof | Incomplete | Yes | olympiads | false |
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 |
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 |
XXXVII OM - II - Problem 6
In triangle $ABC$, point $A$ is chosen on side $BC$, point $B$ on side $AC$, and point $C$ on side $AB$ such that the lines $AA$, $BB$, $CC$ intersect at a single point, or equivalently $|BA|. Prove that the area of triangle $A$ is not greater than $1/4$ of the area of triangle $ABC$. | Attention. The equivalence between the conditions given in the assumptions is the content of Ceva's Theorem. We will show two methods of solving; in the first, we will use the equality of the given products (without directly using the fact that the three considered lines have a common point), and in the second method, ... | proof | Geometry | proof | Incomplete | Yes | olympiads | false |
LIV OM - II - Task 2
Quadrilateral $ABCD$ is inscribed in circle $o$. The angle bisectors of $\angle DAB$ and $\angle ABC$ intersect at point $P$, and the angle bisectors of $\angle BCD$ and $\angle CDA$ intersect at point $Q$. Point $M$ is the midpoint of the arc $BC$ of circle $o$ that does not contain points $D$ an... | First, assume that lines $AD$ and $BC$ are parallel. Then quadrilateral $ABCD$ is an isosceles trapezoid. Its bases $BC$ and $AD$ are perpendicular to line $MN$. Moreover, points $P$ and $Q$ lie on a line parallel to the bases of the trapezoid and equidistant from them. Hence the thesis.
om54_2r_img_1.jpg
om54_2r_img_2... | proof | Geometry | proof | Yes | Incomplete | olympiads | false |
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 |
XI OM - I - Problem 12
Let for the tetrahedron $ ABCD $ the letters $ S_A $, $ S_B $, $ S_C $, $ S_D $ denote the areas of the faces of the tetrahedron opposite to the vertices $ A $, $ B $, $ C $, $ D $, and $ \alpha $, $ \beta $, $ \gamma $ the dihedral angles of the tetrahedron at the edges $ AD $, $ BD $, and $ CD... | The theorem we are to prove is analogous to the law of cosines for a triangle; it could be called the law of cosines for a tetrahedron. The idea suggests that the proofs of both theorems can also be conducted analogously. To this end, we choose the simplest geometric proof of the law of cosines for a triangle, which re... | proof | Geometry | proof | Yes | Incomplete | olympiads | false |
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 |
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 |
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 |
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 |
XXXIV OM - I - Problem 2
Prove that if $ A, B, C $ are angles of a triangle, then
(Note: The original text did not provide the complete statement of the problem to be proved. If you have the full problem statement, please provide it for a complete translation.) | Since $ A+B+C = 180^\circ $, then $ A+ \frac{B}{2} = 90^\circ + \frac{A-C}{2} $ and using reduction formulas we obtain $ \sin \left( A + \frac{B}{2} \right) = \cos \frac{A-C}{2} $. Similarly, we conclude that $ \sin \left( B+ \frac{C}{2} \right) = \cos \frac{B-A}{2} $, $ \sin \left( C +\frac{A}{2} \right) = \cos \frac{... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
VII OM - I - Task 1
Solve the system of equations
| Since $ x^3 + y^3 = (x + y) (x^2 - xy + y^2) $, taking into account the second equation of the system (1), we can replace the first equation of this system with the equation
1) When $ a = 0 $, equation ($ \alpha $) is identically satisfied; the given system of equations (1) reduces to the second equation; the soluti... | 4solutionsfor(u,v) | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
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 |
XXXVII OM - II - Zadanie 1
Wyznaczyć wszystkie funkcje $ f : \mathbb{R} \to \mathbb{R} $ ciągłe w zerze i takie, że dla każdej liczby rzeczywistej $ x $ zachodzi równość
|
Załóżmy, że funkcja $ f : \mathbb{R} \to \mathbb{R} $, ciągła w punkcie $ 0 $, spełnia dane równanie funkcyjne.
W celu pozbycia się drugiego składnika po prawej stronie równania wprowadzamy funkcję pomocniczą $ h(x) =f(x)-\frac{1}{3}x $; jest ona również ciągła w punkcie $ 0 $. Wstawiając $ f(x) = h(x) + \frac{1}{3} x... | f(x)=\frac{1}{3}x | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
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 |
XLVII OM - III - Problem 4
In the tetrahedron $ABCD$, the following equalities hold:
Prove that the edges $AB$ and $CD$ have the same length. | In the proof, we will use the following (known) fact:
(This simple theorem appeared, for example, as a lemma in the solution of one of the problems of the 24th Mathematical Olympiad; a proof can therefore be found in the brochure: 24th Mathematical Olympiad. Report of the Main Committee, Warsaw 1974, p. 66 - but ... | proof | Geometry | proof | Incomplete | Yes | olympiads | false |
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 |
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 |
XLV OM - I - Problem 3
Prove that if $ a $, $ b $, $ c $ are the lengths of the sides of a triangle, then
Note: The original text had a blank space after "then," which I have preserved in the translation. | If the numbers $ a $, $ b $, $ c $ are the lengths of the sides of a triangle, then the numbers $ x $, $ y $, $ z $ given by the formulas
are positive. Treating this system of three equations as a system of equations with unknowns $ a $, $ b $, $ c $, we obtain: $ a = y + z $, $ b = z + x $, $ c = x + y $. The inequa... | proof | Inequalities | proof | Incomplete | Yes | olympiads | false |
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 |
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 |
LVII OM - II - Problem 4
Let $ c $ be a fixed positive integer. The sequence $ (a_n) $ is defined by the conditions
where $ d(m) $ denotes the number of positive divisors of the number $ m $. Prove that there exists a positive integer $ k $ such that the sequence $ a_k,a_{k+1},a_{k+2},\ldots $ is periodic. | We will show that for $ n =1,2,3,\ldots $ the inequalities $ 1 \leq a_n \leq 2c+2 $ hold. The first one is obviously satisfied.
The second one we will prove by induction. For $ n = 1 $ we have $ a_1 = 1 \leq 2c + 2 $, which is true. Let us then fix $ n>1 $ and assume that
$ a_n \leq 2c+2 $.
If $ k \geq 3 $ is an integ... | proof | Number Theory | proof | Incomplete | Yes | olympiads | false |
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 | |
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 |
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 |
II OM - II - Zadanie 3
Dowieść, że równanie
gdzie $ m \ne 0 $, $ n \ne 0 $, $ a \ne b $, ma pierwiastki rzeczywiste ($ m $, $ n $, $ a $, $ b $ oznaczają liczby rzeczywiste).
|
Przepiszmy dane równanie w postaci
Zakładając, że $ x \ne a $ i $ x \ne b $, pomnóżmy dane równanie przez $ (a - x) (b - x) $; otrzymamy wówczas równanie
Równania (1) i (2) są równoważne, tzn. są spełnione przez te same wartości $ x $. Dla wartości $ x $ różnych od $ a $ i od $ b $ wynika to z faktu, że le... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
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 |
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 |
VII OM - II - Problem 5
Prove that the numbers $ A $, $ B $, $ C $ defined by the formulas
where $ \alpha>0 $, $ \beta > 0 $, $ \gamma > 0 $ and $ \alpha + \beta + \gamma = 90^\circ $, satisfy the inequality
(Note: The formulas for $A$, $B$, and $C$ are missing in the original text, so they are not in... | When $ \alpha $, $ \beta $, $ \gamma $ are acute angles that sum to a right angle, then $ \tg \beta \cdot \tg \gamma + \tg \gamma \cdot \tg \alpha + \tg \alpha \cdot \tg \beta = 1 $ (see problem 10), therefore
The problem reduces to showing that non-negative numbers $ A $, $ B $, $ C $ satisfying condition (2) al... | proof | Inequalities | proof | Incomplete | Yes | olympiads | false |
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 |
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 |
VIII OM - I - Task 2
Find the formula for the sum of the fourth powers of consecutive integers from $ 1 $ to $ n $. | Let $ S_p (n) $ denote the sum of the $ p $-th powers of consecutive integers from $ 1 $ to $ n $, i.e.,
To find the formula for the sum $ S_4(n) $, we assume that the formulas for
are already known.
Substituting in the formula
the values $ k = 1, 2, \ldots, n $, we obtain the equations
Adding these equations side... | notfound | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
XXXV OM - I - Problem 11
Provide explicit formulas for the $ n $-th term of sequences defined by the following recurrence relations:
Note: The translation preserves the original text's line breaks and formatting. | Consider the sequence $ (c_n) $ defined by the formula
Since $ c_{n+1} = a_{n+1}+b_{n+1} = 3a_n + b_n - a_n +b_n = 2a_n + 2b_n = 2c_n $, the sequence $ (c_n) $ is defined by the recursive formula
By induction, we establish that $ c_n = 2^{n + 2} $.
Since $ b_n = c_n-a_n = 2^{n+2} -a_n $, we have
From the last formul... | a_n=2^n(2n+5),\,b_n=-2^n(2n+1) | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
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 |
IX OM - II - Problem 5
Outside the triangle $ ABC $, equilateral triangles $ BMC $, $ CNA $, and $ APB $ are constructed. Prove that the centers $ S $, $ T $, $ U $ of these triangles form an equilateral triangle. | Notice that $ \measuredangle TAC = 30^\circ $ and $ \measuredangle BAU = 30^\circ $ (Fig. 22). If we rotate triangle $ TAU $ around point $ A $ by $ 30^\circ $ in the direction of the arrow, then ray $ AT $ will coincide with ray $ AC $, and ray $ AU $ will coincide with ray $ AP $, so $ \measuredangle TAU = \measureda... | proof | Geometry | proof | Yes | Incomplete | olympiads | false |
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 |
XX OM - I - Task 9
Prove that for every positive integer $ n $ | Let's transform the left side of inequality (1):
According to the known formula
From (2) and (3), it follows that
We know that the arithmetic mean of $ n $ positive numbers is not less than the geometric mean of these numbers; thus,
From (4) and (5), we obtain | proof | Number Theory | proof | Incomplete | Yes | olympiads | false |
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 |
LIII OM - II - Problem 2
In a convex quadrilateral $ABCD$, the following equalities hold:
Prove that $AD = CD$. | From the equality of angles and the inequality $ \measuredangle ADC < 180^\circ $, it follows that
Angles $ ACB $ and $ BAC $ are therefore acute, which means that the center $ O $ of the circumcircle of triangle $ ABC $ lies inside angle $ ABC $ (Fig. 1). Hence, in particular, points $ O $ and $ D $ lie on the same s... | AD=CD | Geometry | proof | Incomplete | Yes | olympiads | false |
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 |
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 |
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 |
XLIX OM - II - Problem 1
Let $ A_n = \{1, 2,\ldots, n\} $. Determine whether for every natural number $ n \geq 2 $ there exist functions $ f: A_n \to A_n $, $ g: A_n \to A_n $, satisfying the conditions: | For every natural number $ n \geq 2 $, there exist functions with the required properties. Here is an example: | proof | Combinatorics | proof | Yes | Problem not solved | olympiads | false |
LIV OM - II - Task 1
Prove that there exists an integer $ n > 2003 $, such that in the sequence
every term is a divisor of all subsequent terms. | The number $ {n \choose {k+1}} $ is divisible by $ {n \choose k}, $ if and only if the number
is an integer. Therefore, the number $ n $ satisfies the conditions of the problem if and only if the number $ n +1 $ is divisible by each of the numbers $ 1, 2, 3, 4, \ldots, 2003 $. Such a number is, for example, $ n = 20... | 2003!-1 | Number Theory | proof | Incomplete | Yes | olympiads | false |
XX OM - II - Task 4
Prove that for any natural numbers min, the inequality holds
Note: The original text seems to be incomplete, as the inequality or the specific statement to prove is missing. If you have the full statement, please provide it for a more accurate translation. | We know (see problem 5) that for any positive numbers $ a $ and $ b $ and any natural number $ m $, the following inequality holds:
We will apply this inequality to the sum appearing on the left side of inequality (1). We combine the terms of this sum in pairs: the first with the last, the second with the second-to-la... | proof | Inequalities | proof | Incomplete | Incomplete | olympiads | false |
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 |
XXXVIII OM - II - Problem 6
To any quadrilateral $ABCD$, we assign the centers of the circumcircles of triangles $BCD$, $CDA$, $DAB$, $ABC$. Prove that if the vertices of a convex quadrilateral $Q$ do not lie on a circle, then
a) the four points assigned to quadrilateral $Q$ in the above manner are the vertices of a c... | a) Let the vertices of the quadrilateral $Q$ be denoted sequentially by $A$, $B$, $C$, $D$, and the centers of the circumcircles of triangles $BCD$, $CDA$, $DAB$, $ABC$ by $A'$, $B'$, $C'$, $D'$, respectively. The quadrilateral $Q$ is not cyclic by assumption; hence $|\measuredangle A| + |\measuredangle B| \neq |\measu... | proof | Geometry | proof | Yes | Incomplete | olympiads | false |
VI OM - I - Problem 7
On the median $ CM $ of triangle $ ABC $, a point $ N $ is chosen, and a line $ AN $ is drawn intersecting side $ BC $ at point $ Q $ and a line $ BN $ is drawn intersecting side $ AC $ at point $ P $. Prove that line $ PQ $ is parallel to line $ AB $. | We will apply the method of reduction to absurdity. Suppose that a line parallel to line $AB$ drawn through point $P$ intersects line $BC$ at a point $Q$ different from point $Q$ (Fig. 5). Line $AQ$ then intersects line $BP$ at a point $N$ different from point $N$; line $CN$ intersects line $AB$ at a point $M$ differen... | proof | Geometry | proof | Yes | Incomplete | olympiads | false |
XXXIX OM - III - Task 1
Numbers $ x_1, x_2, \ldots, x_n $ belonging to the interval $ \langle 0; 1\rangle $ satisfy the condition
where $ m $ is an integer, $ r \in \langle 0; 1) $. Prove that
(Note: The original text had some missing or incomplete mathematical conditions. If you have the complete pro... | Let's first consider the special case when all numbers $x_1, \ldots, x_n$, except for at most one, are equal to $0$ or $1$. Since $x_1 + \ldots + x_n = m + r$, where $m$ is an integer and $0 \leq r < 1$, then (in this case) exactly $m$ of the numbers $x_i$ are ones, one equals $r$, and the rest are zeros. Thus, $x_i^2 ... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
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 |
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 |
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 |
IV OM - II - Task 1
A table has been arranged
Write a formula for the $ n $-th row of the table, which when substituting $ n = 1, 2, 3, 4 $ would give the above four rows of the table and be true for any natural number $ n $. | The left sides of the four written rows of the table can be created in the following way: In the first row, we have the number $1$, equal to $1^2$; in the second row, we write the sum of consecutive integers from $1^2 + 1$ to $2^2$; in the third row, the sum of consecutive integers from $2^2 + 1$ to $3^2$; and finally,... | proof | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
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 |
XXXI - III - Task 4
Prove that for every polynomial $ W $ of three variables there exist such polynomials $ U $ and $ V $ that the following equalities hold identically
| Suppose that polynomials $ U $ and $ V $ satisfy the conditions of the problem. Then
From the first two equations we calculate
From the next two equations of the system (*) we obtain
thus
Finally, from the last two equations of the system (*) it follows
thus
Therefore
... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
LVIII OM - II - Problem 6
Positive numbers $ a $, $ b $, $ c $, $ d $ satisfy the condition
Prove that
(Note: The original text had blank spaces or missing conditions in the mathematical statements. Please ensure the problem statement is complete for an accurate translation.) | We will show that for any positive numbers $ x $, $ y $ the following inequality holds
Indeed, transforming the above inequality equivalently, we have
The last inequality is satisfied, since the numbers $ x-y $ and $ x^3-y^3 $ have the same sign.
Using the relation (1), we see that the problem will be solved if we pr... | proof | Inequalities | proof | Incomplete | Yes | olympiads | false |
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