problem stringlengths 0 45.3k | solution stringlengths 0 26.8k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 9
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LVI OM - I - Problem 9
Determine all real numbers $ a $ for which the sequence $ (x_n) $ defined by the formulas
satisfies the condition $ x_{n+8} = x_n $ for $ n =0, 1, 2, \ldots $. | For the sequence $ (x_n) $ to be well-defined, it must be $ x_n \neq a $ for every $ n\geq 0 $. Let $ \alpha \in (0;\pi) $ be such a number that $ a = \mathrm{ctg} \alpha $. We will prove by induction that
For $ n = 0 $, equation (1) is satisfied. Assume that it holds for some $ n $. If $ \mathrm{ctg} x \neq \mathrm{c... | \sqrt{2}+1,1,\sqrt{2}-1,0,-\sqrt{2}+1,-1,-\sqrt{2}-1 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
XV OM - II - Task 6
Prove that among any five points in a plane, one can choose three points that are not the vertices of an acute triangle. | If among the given points there are three collinear points, the thesis of the theorem is of course true. If, however, no three of the given points lie on a straight line, then four of these points, for example, $ A $, $ B $, $ C $, $ D $, are the vertices of a convex quadrilateral, as proven in problem 4. The angles of... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XLVI OM - I - Problem 11
Given are natural numbers $ n > m > 1 $. From the set $ \{1,2, \ldots ,n\} $, we draw $ m $ numbers without replacement. Calculate the expected value of the difference between the largest and the smallest drawn number. | An elementary event is the selection of an $m$-element subset from an $n$-element set. These elementary events are equally probable; there are $ {n}\choose{m} $ of them. Let the largest and smallest selected number be denoted by $a$ and $b$, respectively. The random variable $X$ under consideration, which is the differ... | \frac{(n-1)}{n-+1} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
LV OM - I - Task 6
Let $ c $ be a real number such that the polynomial
has five distinct real roots $ x_1 $, $ x_2 $, $ x_3 $, $ x_4 $, $ x_5 $. Determine, depending on $ c $, the sum of the absolute values of the coefficients of the polynomial
| From the equality
we infer that
The expressions on both sides of this equality are polynomials. Therefore, the obtained formula is true for any real number $ x $. Hence, the sum of the absolute values of the coefficients of the polynomial $ Q(x) $ is: | notfound | Algebra | math-word-problem | Incomplete | Incomplete | olympiads | false |
LVII OM - II - Problem 6
Given a prime number $ p $ and an integer $ n $, where $ p \geq n \geq 3 $. The set $ A $ consists of $ n $-term sequences with terms from the set $ \{0,1,2,\dots ,p-1\} $ and has the following property:
For any two sequences $ (x_1,x_2,\dots ,x_n) $ and $ (y_1,y_2,\dots ,y_n) $ from the set $... | Answer: $ p^{n-2} $.
Any two different sequences of set $ A $ must differ on at least one of the first $ n - 2 $ positions. Therefore, the number
of sequences in this set does not exceed the number of all $(n-2)$-term sequences with terms from a $ p $-element set, that is,
the number $ p^{n-2} $.
We will construct $ ... | p^{n-2} | Combinatorics | math-word-problem | Incomplete | Yes | olympiads | false |
XVIII OM - II - Problem 1
Given is a sequence of numbers $ a_1, a_2, \ldots, a_n $ ($ n \geq 3 $), where $ a_1 = a_n = 0 $ and $ a_{k-1}+a_{k+1} \geq 2a_{k+1} $ for $ k = 2, 3, \ldots, (n - 1) $. Prove that this sequence does not contain any positive terms. | In a finite set of numbers $a_1, a_2, \ldots, a_n$, there exists at least one number not less than any of these numbers. Suppose such a number is $a_r$, i.e., $a_i \leq a_r$ for $i = 1, 2, \ldots, n$. Let $s$ be the smallest index with the property that $a_s = a_r$. We will prove that $s=1$. Indeed, if $s > 1$, then th... | proof | Algebra | proof | Yes | Yes | olympiads | false |
XIV OM - II - Task 5
Prove that the polynomial
is divisible by the polynomial $ (x - 1)^3 $. | We need to prove that a given polynomial $P(x)$ can be written in the form $(x - 1)^3 \cdot Q(x)$, where $Q(x)$ is a polynomial in the variable $x$. The appropriate transformation of the polynomial $P(x)$ will be easier if we introduce the variable $y = x - 1$, i.e., if we substitute $x = 1 + y$ in the given polynomial... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
Given an integer $ c \geq 1 $. To each subset $ A $ of the set $ \{1,2, \ldots ,n\} $, we assign a number $ w(A) $ from the set $ \{1,2, \ldots ,c\} $ such that the following condition is satisfied:
Let $ a(n) $ be the number of such assignments. Calculate $ \lim_{n\to \infty}\sqrt[n]{a(n)} $.
Note: $ \min(x,y) $ is t... | Let $ M $ be the set $ \{1,2,\ldots,n\} $, and $ M_i $ be the $(n-1)$-element set obtained from $ M $ by removing the number $ i $:
Given the assignment as mentioned in the problem, let's denote the value $ w(M) $ by $ m $, and the value $ w(M_i) $ by $ f(i) $ (for $ i = 1,2, \ldots ,n $). According to the conditions ... | c | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XIII OM - III - Problem 6
Three lines $ a $, $ b $, $ c $ are pairwise skew. Can one construct a parallelepiped whose edges lie on the lines $ a $, $ b $, $ c $? | The solution to the problem is based on the following theorem:
If $a$ and $b$ are skew lines, then: a) there exists a pair of parallel planes, one of which contains the line $a$ and the other the line $b$; b) such a pair of planes is unique.
Proof: a) Through any point $A$ on the line $a$, we draw a line $b'$ parallel ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false |
LX OM - III - Task 3
Let $ P, Q, R $ be polynomials of degree at least one, with real coefficients,
satisfying for every real number $ x $ the equalities
Prove that $ P = Q = R $. | Method I.
First, we will prove that if polynomials \( W_1, W_2 \) satisfy for every real number \( x \) the equality
\[
W_1(W_1(W_1(x))) = W_2(W_2(W_2(x)))
\]
then \( W_1 = W_2 \).
It is easy to notice that these polynomials have the same degree. For polynomials of degree 0 (constants), there is nothing to prove. L... | proof | Algebra | proof | Incomplete | Incomplete | olympiads | false |
XXXI - I - Task 2
Prove that for any positive numbers $ a, b, x, y, \alpha $ the inequality is true
(Note: The original text did not provide the specific inequality to be proven. If you have the specific inequality, please provide it for a complete translation and proof.) | Given inequality is divided on both sides by the positive number $ \displaystyle \frac{(a+b)^{\alpha+1}}{(x+y)^\alpha} $, we obtain an equivalent inequality
Assuming
we transform this inequality into an equivalent form
To prove the truth of inequality (*), consider the function
defined in the interval $ (0, 1) ... | proof | Inequalities | proof | Incomplete | Yes | olympiads | false |
XXX OM - I - Task 11
Given a positive number $ p $ and three distinct rays $ OA^{\rightarrow} $, $ OB^{\rightarrow} $, $ OC^{\rightarrow} $ contained in a plane. Prove that there exists exactly one such triplet of points $ K, L, M $, that $ K\in OA^{\rightarrow} $, $ L \in OB^{\rightarrow} $, $ M\in OC^{\rightarrow} $... | For the triangles mentioned in the problem to exist, it is of course necessary to assume that no two of the given rays lie on the same line.
First, note that if points $P$ and $Q$ belong to different sides of an angle with vertex $O$ and point $Q$ moves away from $O$, then the perimeter of triangle $OPQ$ increases with... | proof | Geometry | proof | Yes | Yes | olympiads | false |
III OM - I - Problem 8
In triangle $ ABC $, point $ M $ lies on side $ BC $, point $ N $ lies on side $ AC $, and segments $ AM $ and $ BN $ intersect at point $ P $. Given the ratios
calculate the ratios $ AP \colon PM $ and $ BP \colon PN $. | The task can be easily solved by relying on Thales' theorem. Let's draw $MK \parallel BN$ (Fig. 17); then in triangles $BCN$ and $MAK$ we have respectively: $\frac{NK}{KC} = \frac{BM}{MC} = m$ and $\frac{AP}{PM} = \frac{AN}{NK}$. Therefore,
Similarly, we obtain
Note. The solution to problem 23 immediately f... | Geometry | math-word-problem | Incomplete | Yes | olympiads | false | |
II OM - II - Task 5
Prove that if between the sides and opposite angles $ A $ and $ B $ of triangle $ ABC $ there is a relationship
then such a triangle is right-angled or isosceles. | We transform equation (1) by substituting the known expansions of $ \sin (A - B) $ and $ \sin (A + B) $ and writing this equality in the form of a proportion
from which, after simple calculations, the proportion follows
We now replace the ratio of the sides with the ratio of the sines of the opposite angles:
From... | proof | Geometry | proof | Incomplete | Yes | olympiads | false |
LV OM - III - Task 3
In a certain tournament, $ n $ players participated $ (n \geq 3) $. Each played against each other exactly once, and there were no draws. A three-element set of players is called a draw triplet if the players can be numbered in such a way that the first won against the second, the second against t... | Instead of determining the maximum number of draw triples, we will determine the minimum number of non-draw triples.
Let's number the players from 1 to $ n $ and assume that player number $ i $ won $ x_i $ times. Then the total number of matches is
Each non-draw triple is uniquely determined by the player who won... | \frac{1}{24}n(n^2-1) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XXIV OM - I - Problem 10
Find the smallest natural number $ n > 1 $ with the following property: there exists a set $ Z $ consisting of $ n $ points in the plane such that every line $ AB $ ($ A, B \in Z $) is parallel to some other line $ CD $ ($ C, D \in Z $). | We will first prove that the set $ Z $ of vertices of a regular pentagon has the property given in the problem, that is, $ n \leq 5 $. We will show that each side of the regular pentagon is parallel to a certain diagonal and vice versa, each diagonal is parallel to a corresponding side.
It suffices to prove that $ AB \... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
VIII OM - I - Problem 4
Draw a line through point $ A $ lying outside a given circle $ K $, which intersects this circle at points $ B $ and $ C $ in such a way that $ AB = BC $. | Let $ AB $ (Fig. 1) be the sought line, and $ D $ - the point symmetric to point $ O $ with respect to the midpoint $ B $. Since points $ A $ and $ C $ are also symmetric with respect to point $ B $, it follows that $ AD = OC $. Point $ D $ is thus the common point of circles with centers at $ A $ and $ O $ and radii $... | notfound | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
XXV - I - Task 1
During World War I, a battle took place near a certain castle. One of the shells destroyed a statue of a knight with a spear standing at the entrance to the castle. This happened on the last day of the month. The product of the day of the month, the month number, the length of the spear expressed in f... | The last day of the month can only be $28$, $29$, $30$, or $31$. Of these numbers, only $29$ is a divisor of the number $451,066 = 2 \cdot 7 \cdot 11 \cdot 29 \cdot 101$. Therefore, the battle took place on February $29$ in a leap year. During World War I, only the year $1916$ was a leap year. From the problem statemen... | 1714 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
XL OM - III - Task 4
Let $ n, k $ be natural numbers. We choose a sequence of sets $ A_0, \ldots, A_k $ such that $ A_0 = \{1, \ldots, n\} $, and for $ i = 1, \ldots, k $, the set $ A_i $ is a randomly chosen subset of $ A_{i-1} $, with each subset being equally likely. We consider the random variable equal to the num... | We introduce random variables $ X_1, \ldots , X_n $ defined as follows:
\
($ i = 1,\ldots ,n $). The random variable $ X $ (the number of elements in the set $ A_k $) considered in the problem is the sum of the variables introduced just now:
Let us fix a number $ i \in A_0 = \{1, \ldots, n\} $. Since $ A_1 ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
VII OM - II - Task 3
A homogeneous horizontal plate weighing $ Q $ kG in the shape of a circle is supported at points $ A $, $ B $, $ C $ lying on the circumference of the plate, with $ AC = BC $ and $ ACB = 2\alpha $. What weight $ x $ kG should be placed on the plate at the other end $ D $ of the diameter drawn from... | The weight $ Q $ of a plate supported at points $ A $, $ B $, $ C $ (Fig. 17) is balanced by the reactions at these points, acting vertically upwards. The center of gravity $ O $ of the plate, as the point of application of the weight $ Q $ acting downwards, must lie within the triangle $ ABC $, i.e., the angle $ 2\alp... | Geometry | math-word-problem | Yes | Yes | olympiads | false | |
LI OM - II - Task 6
A polynomial $ w(x) $ of degree two with integer coefficients takes values that are squares of integers for integer $ x $. Prove that the polynomial $ w(x) $ is the square of some polynomial. | Let $ w(x) = ax^2 + bx + c $. Introduce the notation: $ k_n = \sqrt{w(n)} $ for any positive integer $ n $. Then $ k_n $ can be zero for at most two values of $ n $. For the remaining $ n $ we have
Dividing the numerator and the denominator of the obtained fraction by $ n $ and passing to the limit as $ n $ goes to in... | proof | Algebra | proof | Yes | Yes | olympiads | false |
XXXVIII OM - I - Problem 3
Solve the equation in natural numbers
Note: The original text is quite sparse and doesn't provide the actual equation to solve. If you have the specific equation, please provide it, and I can help translate and solve it. | Assume that natural numbers $ n $, $ k $ satisfy the given equation. For $ n = 1 $, the corresponding $ k $ does not exist; hence $ n > 1 $.
The left side of the equation is an odd number, so the number $ k $ must also be odd.
Let $ m = [n/2] $ (the integer part of the number $ m/2 $). Thus, $ n = 2m + r $, where $ m >... | n=4,k=13 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
V OM - III - Task 3
A uniform circular disk was suspended in a horizontal position from a string attached to its center $ O $. Three weights $ p_1 $, $ p_2 $, $ p_3 $ were placed at three different points $ A $, $ B $, $ C $ on the edge of the disk, after which the disk remained in equilibrium. Calculate the angles $ ... | Since the shield remained in equilibrium after the weights were placed, the resultant of the parallel forces $ p_1 $, $ p_2 $, $ p_3 $ passes through the point $ O $ (Fig. 41). Of the three angles $ AOB $, $ BOC $, $ COA $ considered in the same direction of rotation, at least two are less than $ 180^\circ $, since the... | notfound | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
XXI OM - III - Task 1
Diameter $ \overline{AB} $ divides the circle into two semicircles. On one semicircle, n points $ P_1 P_2, \ldots, P_n $ are chosen such that $ P_1 $ lies between $ A $ and $ P_2 $, $ P_2 $ lies between $ P_1 $ and $ P_3 $, $ \ldots $, $ P_n $ lies between $ P_{n-1} $ and $ B $. How should point ... | Notice that the sum of the areas of triangles $ CP_1P_2, CP_2P_3, CP_3P_4, \ldots, CP_{n-1}P_n $ is equal to the sum of the areas of the polygon $ P_1P_2 \ldots P_n $ and triangle $ CP_1P_n $ (Fig. 13). The first area does not depend on the choice of point $ C $. The second area will be maximal when the distance from p... | C | Geometry | math-word-problem | Yes | Yes | olympiads | false |
LI OM - I - Task 1
Given a natural number $ n \geq 3 $. Prove that the sum of the cubes of all natural numbers less than $ n $ and coprime with $ n $ is divisible by $ n $. | Every natural number less than $ n $ and coprime with $ n $ is equal to $ k $ or $ n-k $, where $ k $ is some natural number less than $ n/2 $ and coprime with $ n $. Therefore, the given sum of cubes is equal to
and thus is divisible by $ n $. | proof | Number Theory | proof | Yes | Incomplete | olympiads | false |
XXII OM - II - Problem 5
Given is the set of numbers $ \{1, 2, 3, \ldots, 100\} $. From this set, form 10 pairwise disjoint subsets $ N_i = \{a_{i,1}, a_{i,2}, \ldots a_{i,10}} $ ($ i = 1, 2, \ldots, 10 $) such that the sum of the products
is maximized. | We will first prove the following
Lemma. If $ \{a_1, \ldots, a_r, b_1, \ldots, b_r \} $ is a set of $ 2r $ distinct positive numbers and $ a_i < b_j $ for $ i,j = 1,2, \ldots,r $, then for every $ k $ $ (1 \leq k < r) $
In other words, the lemma states that if a given set is divided into two $ r $-element subsets, ... | notfound | Combinatorics | math-word-problem | Incomplete | Yes | olympiads | false |
L OM - I - Task 2
To prove that for any real numbers $ a $, $ b $, $ c $, $ d $ the inequality holds
| I way:
We get the thesis immediately from the following equality:
II way:
By the power of the inequality between the quadratic mean and the arithmetic mean, we obtain $ 3((a + b)^2+ c^2+ d^2) \geq ((a+b) + c+d)^2 $, which is the inequality we needed to prove. | proof | Inequalities | proof | Incomplete | Yes | olympiads | false |
XXXIII OM - II - Task 3
Prove that for every natural number $ n \geq 2 $ the following inequality holds | For non-negative numbers $ a $, $ b $, the inequality
holds.
Therefore, for $ k = 1,2,\ldots,n $ we have
From inequality (*) it also follows that
We obtain the inequalities:
and
Suppose $ n $ is an odd number. By combining the first and last terms, the second and second-to-last terms, etc., in t... | proof | Inequalities | proof | Yes | Yes | olympiads | false |
XLVIII OM - II - Problem 5
We throw $ k $ white and $ m $ black six-sided dice. Calculate the probability that the remainder of the total number of eyes thrown on the white dice divided by $ 7 $ is equal to the remainder of the total number of eyes thrown on the black dice divided by $ 7 $. | The set of elementary events in this task is
where all elementary events are equally probable. Let $ A $ denote the event whose probability we are looking for:
Let $ B $ be the event that the total number of eyes thrown on all $ k + m $ dice is divisible by $ 7 $:
The transformation given by the formu... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
IV OM - I - Problem 12
Given a line $ p $ and points $ A $ and $ B $ lying on opposite sides of this line. Draw a circle through points $ A $ and $ B $ such that the chord lying on the line $ p $ is as short as possible. | Let $ M $ denote the point of intersection of the lines $ AB $ and $ p $, and let the circle passing through points $ A $ and $ B $ intersect the line $ p $ at points $ C $ and $ D $ (Fig. 35).
The chord $ CD $ is the sum of segments $ CM $ and $ MD $, whose product, by the theorem of intersecting chords of a circle, e... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XLII OM - II - Problem 5
$ P_1, P_2, \ldots, P_n $ are distinct two-element subsets of the set $ \{1,2,\ldots,n\} $. Sets $ P_i $, $ P_j $ for $ i\neq j $ have a common element if and only if the set $ \{i,j\} $ is one of the sets $ P_1, P_2, \ldots, P_n $. Prove that each of the numbers $ 1,2,\ldots,n $ is a common e... | For each $ k \in \{1,2,\ldots,n\} $, let $ m_k $ be the number of sets $ P_i $ that contain the element $ k $. The sum of these numbers (as $ k $ ranges from $ 1 $ to $ n $) is $ 2n $, because each of the sets $ P_1,\ldots, P_n $ has two elements and is thus counted twice. We therefore have the equality
\[
\sum_{k=1}^... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
II OM - I - Task 1
Prove that the product of two factors, each of which is the sum of the squares of two integers, is also the sum of the squares of two integers. | We will apply the transformation
If $ a $, $ b $, $ c $, $ d $ are integers, then $ ac + bd $ and $ ad - bc $ are also integers; the theorem has thus been proven.
Notice that a more general theorem holds:
The product of $ n $ factors, each of which is the sum of the squares of two integers, is also the sum of the... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
XVI OM - III - Task 4
Prove that if integers $ a $ and $ b $ satisfy the equation
then the numbers $ a - b $ and $ 2a + 2b + 1 $ are squares of integers. | Suppose that the integers $ a $ and $ b $ satisfy equation (1). When $ a = 0 $, then $ 3b^2 + b = b (3b + 1) = 0 $, and since for an integer $ b $, $ 3b + 1 \neq 0 $, it follows that $ b = 0 $. The thesis of the theorem is true in this case, as $ a - b = 0 $, $ 2a+2b+1 = 1 $.
It remains to consider the case when $ a \n... | proof | Number Theory | proof | Incomplete | Yes | olympiads | false |
XLIV OM - II - Problem 2
Given a circle with center $O$ and a point $P$ lying outside this circle. Through point $P$, we draw a line $l$ intersecting the given circle at points $A$ and $B$. Let $C$ be the point symmetric to $B$ with respect to the line $OP$, and let $m$ be the line passing through points $A$ and $C$. ... | The task does not specify how points $A$ and $B$ are situated on line $l$, meaning which one is closer to point $P$; Figure 5 illustrates these two situations. Let $Q$ be the point of intersection of lines $OP$ and $AC$; the thesis of the task will be proven if we show that the position of point $Q$ on the ray $OP^\to$... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XXVI - I - Task 3
Prove that for every natural $ n $ | The terms corresponding to $ j = 1 $ and $ j = n-1 $ are equal. Similarly, the terms corresponding to $ j = 2 $ and $ j = n-2 $ are equal, and so on. Moreover, when $ n $ is an even number, the term corresponding to $ \displaystyle j - \frac{n}{2} $ is equal to zero. Therefore, the given sum is equal to
where $ \dis... | \frac{n}{2} | Number Theory | proof | Incomplete | Yes | olympiads | false |
XXXII - I - Problem 6
Given numbers $ a_1\geq a_2 \geq \ldots \geq a_n \geq 0 $ satisfying the condition $ \sum_{i=1}^n a_i = 1 $. Prove that there exist integers $ k_1\geq k_2 \geq \ldots \geq k_n \geq 0 $ such that | Let $ [x] $ be the greatest integer not greater than $ x $.
Assume $ \widetilde{k_j} = [2na_j] $. The numbers $ \widetilde{k_j} $ satisfy the condition $ \widetilde{k_1} \geq \widetilde{k_2} \geq \ldots \geq \widetilde{k_n} \geq 0 $, and moreover $ 2na_j-1 < \widetilde{k_j} \leq 2na_j $, so $ 2a_j - \frac{1}{n} < \frac... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
LVIII OM - II - Problem 5
A convex quadrilateral $ABCD$, where $AB \ne CD$, is inscribed in a circle. Quadrilaterals $AKDL$ and $CMBN$ are rhombuses with side lengths of $a$. Prove that points $K$, $L$, $M$, $N$ lie on the same circle. | Since the chords $ AB $ and $ CD $ are of different lengths, the lines $ AD $ and $ BC $ are not parallel. Let's denote their intersection point by $ P $ (Fig. 10). We will show that the points $ K $, $ L $, $ M $, and $ N $ lie on a circle with center $ P $.
The line $ AD $ is the perpendicular bisector of segment $ ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XXXI - III - Problem 5
In a tetrahedron, the areas of the six triangles, whose sides are the edges and whose vertices are the midpoints of the opposite edges of the tetrahedron, are equal. Prove that the tetrahedron is regular. | The assumption of the task guarantees the equality of the areas of six triangles, each of which has a base being an edge of the tetrahedron, and the opposite vertex is the midpoint of the opposite edge. On the tetrahedron $ABCD$, we describe a parallelepiped $A_1CB_1DAC_1BD_1$, whose each face contains a certain edge o... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XV OM - III - Task 3
Given is a tetrahedron $ABCD$, whose edges $AB, BC, CD, DA$ are tangent to a certain sphere: Prove that the points of tangency lie in the same plane. | We distinguish two cases:
a) $ AM = CN $, thus also $ AQ = CP $. Triangles $ MBN $ and $ ABC $ are then similar with respect to point $ B $, so $ MN \parallel AC $ and similarly $ QP \parallel AC $. Therefore, $ QP \parallel MN $, which means points $ MNPQ $ lie in the same plane.
b) $ AM \ne CN $, let's say $ AM > CN ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XXVI - I - Problem 10
Let $ \alpha $ be an irrational number, $ A_1 $ - a point on the circle $ S $ with center $ O $. Consider the infinite sequence $ A_n $ of points on the circle $ S $, where the point $ A_{k+1} $ is the image of the point $ A_k $ under a rotation about the point $ O $ by an angle $ \alpha m $. Pro... | If for certain natural numbers $k$ and $n$, where $k \ne n$, it was $A_k = A_n$, then the rotations by angles $(k-1)\alpha\pi$ and $(n-1)\alpha\pi$ would be equal. This means that the numbers $(k-1)\alpha\pi$ and $(n-1)\alpha\pi$ would differ by an integer multiple of $2\pi$, i.e., $(n-k)\alpha\pi = 2m\pi$, where $m$ i... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
XIII OM - I - Problem 8
Given a skew quadrilateral $ABCD$ and a plane intersecting the lines $AB$, $BC$, $CD$, $DA$ at points $M$, $N$, $P$, $Q$ respectively, different from $A$, $B$, $C$, $D$. Prove that | Let $ A, $ B, $ C, $ D denote the orthogonal projections of points $ A, $ B, $ C, $ D onto a given plane (Fig. 14). The right triangles $ AA i $ BB have equal acute angles at vertex $ M $, so they are similar. Therefore,
Hence,
Note. The theorem can be generalized by replacing the quadrilateral with any closed ... | proof | Geometry | proof | Yes | Incomplete | olympiads | false |
LIX OM - I -Zadanie 1
Rozwiązać w liczbach rzeczywistych $ x, y, z $ układ równań
|
Jeżeli trójka liczb $ (x,y,z) $ spełnia dany układ równań, to spełniają go również trójki liczb
$ (y,z,x) $ i $ (z,x,y) $. Zatem bez ograniczenia ogólności rozumowania możemy przyjąć, że $ x $ jest
najmniejszą wśród liczb $ x, y, z $ (pozostałe rozwiązania otrzymamy przez cykliczne przestawienie zmiennych).
Rozp... | (x,y,z)\in{(-2,-2,-2),(-1,-1,-1),(0,0,0),(1,1,1),(2,2,2)} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
LVI OM - I - Problem 10
Among all subsets of a fixed $ n $-element set $ X $, we sequentially draw with replacement three sets $ A $, $ B $, $ C $. Each time, the probability of drawing any of the $ 2^n $ subsets of set $ X $ is equally likely. Determine the most probable number of elements in the set $ A\cap B\cap C ... | The set of elementary events $\Omega$ consists of all triples $(A,B,C)$, where $A$, $B$, and $C$ are subsets of a given $n$-element set $S$. The selection of each of the $(2^n)^3 = 8^n$ triples $(A,B,C)$ is equally probable. Let $X_k$ $(k = 0,1,2,\ldots,n)$ denote the event that the triple of sets $(A,B,C)$ satisfies t... | [\frac{1}{8}(n+1)] | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XVI OM - II - Task 5
Prove that a square can be divided into any number of squares greater than 5, but it cannot be divided into 5 squares. | a) First, let us note that having a square divided into $ m $ squares, we can divide one of these squares into four squares by connecting the midpoints of its opposite sides. The entire square will then be divided into $ m + 3 $ squares.
Let $ n $ be a natural number greater than $ 1 $. Divide each side of the square $... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XL OM - I - Task 3
Triangles $ABC$ and $ABC$ have equal perimeters, and the bisectors of the external angles at vertices $C$ and $C$ intersect at point $O$. Prove that angles $AOC$ and $BOC$ are equal. | Let $ |AC| + |BC| $ and $ |AC| $ be equal; denote their common value by $ s $. On the extensions of segments $ AC $ and $ BC $, we lay off segments $ CD $ and $ CE $ such that...
Triangles $ BCD $ and $ ACE $ are isosceles, and point $ O $ lies on the axis of symmetry of each of them (Figure 1). Therefore,
an... | proof | Geometry | proof | Incomplete | Yes | olympiads | false |
LVII OM - I - Problem 2
Solve in non-negative real numbers $ x $ the equation
Note: $ [t] $ denotes the greatest integer less than or equal to $ t $. | Let $ x $ be a number satisfying the given equation. Then the number $ y = \sqrt[5]{x} $ is an integer. Moreover, $ y \geq 0 $, since $ x \geq 0 $. Substituting $ x = y^5 $ into the considered equation, we get
This means that
The first of these inequalities is true for any non-negative number $ y $. The second, h... | 0,1,2^5,3^5,4^5 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
XLIX OM - II - Problem 2
In triangle $ABC$, angle $BCA$ is obtuse and $\measuredangle BAC = 2\measuredangle ABC$. The line passing through point $B$ and perpendicular to $BC$ intersects line $AC$ at point $D$. Point $M$ is the midpoint of side $AB$. Prove that $\measuredangle AMC = \measuredangle BMD$. | Let a line parallel to $ AB $ and passing through point $ C $ intersect segment $ BD $ at point $ E $. Denote by $ N $ the midpoint of segment $ CE $. Then points $ M $, $ N $, $ D $ are collinear. Triangle $ BCE $ is a right triangle, so $ N $ is the center of the circle circumscribed around it. This implies that tria... | proof | Geometry | proof | Yes | Yes | olympiads | false |
X OM - III - Task 1
Prove that for any numbers $ a $ and $ b $ the inequality holds
| We will first prove that
Inequality (2) is equivalent to the inequality
Indeed,
We see that the left side of inequality (3) is the product of two non-negative factors; therefore, inequality (3), and thus (2), is true. We will now prove the inequality
It is equivalent to the inequality
We transform the left side of... | proof | Inequalities | proof | Incomplete | Yes | olympiads | false |
XLIX OM - I - Problem 9
Let $ a_0 = 0.91 $ and
Calculate $ \lim_{n\to\infty}(a_0a_1\ldots a_n) $. | According to the given definition,
For better readability of the following expressions, let's denote the number $ \frac{1}{10} $ by $ \beta $ and assume $ \alpha = 1 + \beta + \beta^2 = 1.11 $. The following equalities hold:
And further, by obvious induction,
Dividing both sides by $ \alpha = \frac{111}{10... | \frac{100}{111} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
XXII OM - I - Problem 10
Given is a table with $ n $ rows and $ n $ columns. The number located in the $ m $-th column and $ k $-th row is equal to $ n(k - 1) + m $. How should $ n $ numbers be chosen, one from each row and each column, so that the product of these numbers is the largest? | Let's denote $ a_{k,m} = n (k - 1) + m $, where $ 1 \leq k, m \leq n $. Suppose we select the number in the column with number $ m_i $ from the $ i $-th row, where $ i = 1, 2, \ldots, n $. From the conditions of the problem, it follows that the sequence $ m_1, m_2, \ldots, m_n $ has distinct terms. Each such sequence c... | a_{1n}a_{2n-1}\ldotsa_{n1}=n(2n-1)(3n-2)(4n-3)\ldots(n^2-(n-1)) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XX OM - II - Task 3
Given is a quadrilateral $ABCD$ inscribed in a circle. The images of points $A$ and $C$ under reflection across the line $BD$ are points $A$ and $C$, respectively, and the images of points $B$ and $D$ under reflection across the line $AC$ are points $B$ and $D$, respectively. Prove that points $A$,... | Line $ A $ is symmetric to line $ AC $ with respect to line $ BD $, so it passes through the intersection point $ S $ of line $ AC $ with the axis of symmetry $ BD $. Similarly, line $ B $ intersects line $ BD $ at the axis of symmetry of $ AC $, meaning it also passes through point $ S $. The equality of symmetric seg... | proof | Geometry | proof | Yes | Incomplete | olympiads | false |
LIV OM - I - Task 12
For positive numbers $ a $, $ b $, $ c $, $ d $ we define
Prove the inequality | Let $ S = a + b + c + d $. Then $ S^3 = A + 6B + 3Q $, where
Using the inequality
valid for any real number $ x $, we get
Thus
which gives $ S^3 < 4A + 24B $, which is the inequality to be proven. | S^3<4A+24B | Inequalities | proof | Incomplete | Yes | olympiads | false |
XXXIX OM - II - Problem 6
A convex polyhedron is given with $ k $ faces $ S_1, \ldots, S_k $. Denote the unit vector perpendicular to the face $ S_i $ ($ i = 1, \ldots, k $) directed outward from the polyhedron by $ \overrightarrow{n_i} $, and the area of this face by $ P_i $. Prove that | We start with the observation that if in space a convex planar polygon $S$ and a plane $\pi$ are given, and if $\overrightarrow{n}$ is a unit vector perpendicular to the plane of the polygon $S$, and $\overrightarrow{w}$ is a unit vector perpendicular to the plane $\pi$, then denoting by $W$ the orthogonal projection o... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XIX OM - III - Task 4
Given a natural number $ n>2 $. Provide a set of $ n $ natural numbers $ a_1, a_2, \ldots, a_n $, so that in the set of sums
there are as few different numbers as possible, and provide a set of $ n $ natural numbers $ b_1, b_2, \ldots, b_n $, so that in the set of sums
there are as many d... | When $Z$ is a finite set of numbers, we agree to denote by $l(Z)$ the number of different values that the sums of two numbers belonging to $Z$ can take.
1) Let $A$ be a set of $n > 2$ natural numbers $a_i$ ($i = 1,2,\ldots, n$) so labeled that
Since
thus in the set of sums $a_i + a_j$ there are at least $2n - 3$ diff... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXXVIII OM - III - Problem 1
In a square with side 1, there are $ n $ points ($ n > 2 $). Prove that they can be numbered $ P_1, P_2, ..., P_n $ such that \sum_{i=1}^n |P_{i-1}P_i|^2 \leq 4 (we assume $ P_0=P_n $). | The proof will be based on the following lemma:
Lemma. In a right-angled isosceles triangle $ABC$, where $|AB| = |BC| = a$, $|\measuredangle ABC| = 90^\circ$, there are $k$ points ($k \geq 1$). These points can then be numbered $P_1, \ldots, P_k$ such that
(we assume $P_0 = A$, $P_{k+1} = C$).
Proof. We use induction.... | proof | Geometry | proof | Yes | Yes | olympiads | false |
VI OM - I - Task 6
What necessary and sufficient conditions should integers $ a $ and $ b $, where $ b $ is not a square of an integer, satisfy so that there exist integers $ x $ and $ y $ satisfying the equation
| Suppose the equality holds
where $ a $, $ b $, $ x $, $ y $ are integers, and $ b $ is not a perfect square; in this case, the numbers $ a $, $ b $, $ x $, $ y $ are positive, i.e., they are natural numbers and $ x > y $. Squaring the equality (1) we obtain the equality
and from this, by squaring again, we get the eq... | proof | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
XXIII OM - I - Problem 5
The sequence of polynomials $ P_0(x), P_1(x), P_2(x), \ldots $ is defined by the formulas $ P_0(x) = 2 $, $ P_1(x) = x $,
Prove the existence of real numbers $ a, b, c $ such that for every natural number $ n $ the following equality holds: | Suppose that for certain numbers $ a $, $ b $, $ c $, equation (2) holds. Substituting $ n = 1 $ in it, we get $ (x^2 - 4)^2 = [ax^2 + bx + 2 (c - a)]^2 $. By comparing the coefficients of $ x^4 $, $ x^3 $, $ x^2 $ in the polynomials on both sides of the last equality, we obtain $ a^2 = 1 $, $ 2ab = 0 $, $ b^2 + 4a (c ... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
XXV OM - II - Problem 4
In a convex quadrilateral $ABCD$ with area $S$, each side is divided into 3 equal parts, and segments connecting corresponding division points of opposite sides are drawn in such a way that the quadrilateral is divided into 9 smaller quadrilaterals. Prove that the sum of the areas of the follow... | We will first prove the
Lemma. The segment connecting the corresponding division points of opposite sides of the given quadrilateral intersects with analogous segments connecting the corresponding division points of the remaining sides of the quadrilateral at points that divide this segment into three equal parts.
Proo... | proof | Geometry | proof | Yes | Yes | olympiads | false |
V OM - I - Zadanie 5
Rozwiązać układ równań
|
Dodajemy drugie i trzecie równanie układu (1); otrzymujemy $ xy + zt + xz + yt = 25 $ lub $ (x + t) (y + z) = 25 $; biorąc to równanie zamiast trzeciego równania układu (1) otrzymujemy równoważny układ równań
W pierwszych dwóch równaniach układu (2) możemy potraktować jako niewiadome $ x + t $ i $ y + z $; mając... | Algebra | math-word-problem | Incomplete | Yes | olympiads | false | |
VII OM - II - Problem 4
Prove that the equation $ 2x^2 - 215y^2 = 1 $ has no solutions in integers. | Let $ x $ and $ y $ denote integers. The number $ 215y^2 $ is divisible by $ 5 $, so the number $ 215y^2 + 1 $ gives a remainder of $ 1 $ when divided by $ 5 $. The number $ x^2 $ has one of the forms $ 5k $, $ 5k + 1 $, $ 5k + 4 $ ($ k $ - an integer, see problem 2), so the number $ 2x^2 $ has one of the forms $ 10k $... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
XXXIX OM - II - Problem 1
Let $ f(x) $ be a polynomial, $ n $ - a natural number. Prove that if $ f(x^{n}) $ is divisible by $ x-1 $, then it is also divisible by $ x^{n-1} + x^{n-2} + \ldots + x + 1 $. | From the condition of the problem, it follows that the polynomial $ F(x) = f(x) $ takes the value $ 0 $ at the point $ x = 1 $. Therefore, the polynomial $ f $ also takes the value $ 0 $ at the point $ 1 $. By the virtue of the Remainder Theorem, the polynomial $ f $ is thus divisible by the binomial $ x-1 $. In other ... | proof | Algebra | proof | Yes | Incomplete | olympiads | false |
VIII OM - I - Problem 3
Prove that for every triangle the following inequalities hold
where $ r $ denotes the radius of the inscribed circle of the triangle, and $ h_1 $ and $ h_2 $ - two altitudes of the triangle. | Let $ S $ denote the area of a triangle, $ 2p=a+b+c $ - its perimeter, where $ a $ and $ b $ are the sides of the triangle to which the corresponding altitudes $ h_1 $ and $ h_2 $ are dropped. The inequalities (1) are equivalent to the inequalities
Since $ \frac{S}{r} = p = \frac{a+b+c}{2} $, $ \frac{2S}{h_1} = a $, $... | proof | Inequalities | proof | Incomplete | Yes | olympiads | false |
XXXV OM - I - Problem 10
In the plane, there are $3n$ points, among which no three points are collinear. Prove that there exist $n$ disjoint triangles with vertices at the given points. | Consider all lines, each of which passes through two points of a given set of $3n$ points. There are finitely many such lines, so there exists a line $l$ that is not perpendicular to any of them. Project all the given points perpendicularly onto the line $l$, ensuring that the projections of any two points do not coinc... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
IX OM - II - Problem 4
Prove that if
where $ a $, $ b $, $ c $, $ x $, $ y $, $ z $ represent real numbers, then $ a = b = c $ and $ x = y = z = 0 $. | Transforming equation (1), we obtain successively
Since the left side of equation (2) is a non-negative number, and the right side is a non-positive number, both are equal to zero, from which
| proof | Algebra | proof | Incomplete | Yes | olympiads | false |
L OM - III - Task 1
Point $ D $ lies on side $ BC $ of triangle $ ABC $, such that $ AD > BC $. Point $ E $ lies on side $ AC $ and satisfies the condition
Prove that $ AD > BE $. | We complete the triangle $ BDA $ to the parallelogram $ BDAF $. On the ray $ BC^\to $, we lay off the segment $ BK $ of length $ AD $. The given relationship now takes the form
From this equality and the parallelism of segments $ CK $ and $ AF $, it follows that points $ F $, $ E $, $ K $ are collinear.
Point $ E $... | proof | Geometry | proof | Incomplete | Yes | olympiads | false |
X OM - I - Task 1
Prove that when $ n $ is a natural number greater than $ 1 $, then
The text has been translated while preserving the original formatting and line breaks. | We transform the left side $ L $ of equation (1):
after simplifying the fraction by $ 2 \cdot 3^2 \cdot 4 \cdot \ldots \cdot (n - 1)^2 \cdot n $ we obtain
| proof | Number Theory | proof | Incomplete | Yes | olympiads | false |
LIII OM - II - Problem 3
In an $ n $-person association, there are six committees. Each of them includes at least $ n/4 $ people. Prove that there exist two committees and a group of at least $ n/30 $ people, who belong to both of these committees. | Let's number the committees by $1, 2, \ldots, 6$ and denote by $K_i$ the number of members of the $i$-th committee who are not members of any committee with a number less than $i$.
$K_1$ is the number of all members of the first committee, which gives $K_1 \geq n/4$.
Assume that the intersection of any two committees h... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
LI OM - I - Task 2
In an acute triangle $ ABC $, the condition
is satisfied.
Point $ D $ lies on side $ BC $, such that $ \measuredangle BAD =\frac{1}{2} \measuredangle ABC $. Prove that
Note: The original text had blank lines and symbols that were not translated as they do not contain text. If there ... | We will show that equality (1) is true for any triangle ABC, not necessarily acute.
Let $ E $ be a point on the ray $ BC^\to $ such that $ AD $ is the bisector of angle $ BAE $ (Fig. 1 and 2).
om51_1r_img_1.jpg
om51_1r_img_2.jpg
Then $ \measuredangle EBA = \measuredangle EAB $. If point $ E $ lies on side $ BC $ (Fig. ... | proof | Geometry | proof | Incomplete | Yes | olympiads | false |
XX OM - I - Task 10
Prove that if $ a_1, a_2, b_1, b_2 $ are rational numbers satisfying the conditions
then the numbers
are squares of rational numbers. | Let $ a_1-a_2 = k $, $ b_1- b_2 = l $; then
the conditions (1) take the form
It follows from them that the rational number $ \frac{k}{l} $ satisfies the quadratic equation
thus the discriminant of this equation, i.e.,
is the square of a rational number, so the number $ 1-a_1b_1 $ is also the square of a rational nu... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
V OM - II - Task 6
Prove that if $ x_1, x_2, \ldots, x_n $ are angles contained between $ 0^\circ $ and $ 180^\circ $, and $ n $ is any natural number greater than $ 1 $, then | We will prove a "stronger" theorem, namely, we will show that if $0^\circ < x_i < 180^\circ$ for $i = 1, 2, \ldots, n$, where $n \geq 2$, then
(If $|a| < b$, then $a < b$ (but not vice versa), so inequality (2) implies inequality (1)).
In the proof, we will use known properties of the absolute value:
as well as the f... | proof | Algebra | proof | Yes | Yes | olympiads | false |
XIII OM - III - Task 5
Prove that if $ n $ is a natural number greater than $ 2 $, then | Raising both sides of the inequality (1) to the power of $ n (n +1) $, we obtain an equivalent inequality
which we will in turn replace with the equivalent inequality $ \left( \frac{n+1}{n} \right)^n < n $, i.e., the inequality
The proof of inequality (2) for every natural number $ n $ greater than $ 2 $ can be... | proof | Number Theory | proof | Incomplete | Yes | olympiads | false |
XLV OM - I - Problem 2
The sequence of functions $ f_0, f_1, f_2, \ldots $ is defined as follows:
For each natural number $ n $, solve the equation $ f_n(x) = 1 $. | Let the symbol $ M_k $ denote the set of all odd integers in the interval $ \langle 1; 2k+1 \rangle $. Take any integers $ m \geq 1 $, $ k \geq 0 $. According to the definition of the sequence of functions under consideration, the following statements (1), (2), (3) are successively equivalent:
All fun... | Foranyfixednaturaln,thesetofsolutionstotheequationf_n(x)=1consistsofalloddintegersintheinterval\langle-2n-1;2n+1\rangle | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
For a natural number $ k \geq 1 $, let $ p(k) $ denote the smallest prime number that is not a divisor of $ k $. If $ p(k) > 2 $, then we define $ q(k) $ as the product of all prime numbers less than $ p(k) $; if $ p(k) = 2 $, then we set $ q(k) = 1 $. We define the sequence $ (x_n) $ by the formulas:
Determine a... | Let $ p_0 = 2 $, $ p_1 = 3 $, $ p_2 = 5 $, $ p_3 = 7 $, $ \ldots $ be the increasing sequence of all prime numbers. The terms of the sequence $(x_n)$ under consideration can be characterized by the digits of the binary representation of the index $ n $. Specifically, for $ n \geq 1 $, the implication holds:
For $ n = ... | 2106 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
XXIX OM - I - Problem 7
For a fixed natural number $ n>2 $, we define: $ x_1 = n $, $ y_1=1 $, $ x_{i+1} = \left[\frac{1}{2}(x_i+y_i)\right] $, $ y_{i+1} = \left[ \frac{n}{x_{i+1}}\right] $. Prove that the smallest of the numbers $ x_1, x_2, \ldots, x_n $ is equal to $ [\sqrt{n}] $.
Note. $ [a] $ denotes the greatest ... | For any integer $ k $, the number $ \left[ \frac{k}{2} \right] $ is equal to $ \frac{k}{2} $ or $ \frac{k-1}{2} $ depending on whether $ k $ is even or odd. Therefore, $ \left[ \frac{k}{2} \right] \geq \frac{k-1}{2} $ for any integer $ k $. We also have $ a \geq [a] > a - 1 $ by the definition of the symbol $ [a] $. Fi... | [\sqrt{n}] | Number Theory | proof | Yes | Yes | olympiads | false |
XIX OM - III - Problem 6
Given a set of $ n > 3 $ points, no three of which are collinear, and a natural number $ k < n $. Prove the following statements:
1. If $ k \leq \frac{n}{2} $, then each point in the given set can be connected to at least $ k $ other points in the set in such a way that among the drawn segment... | $ 1^\circ $. Suppose $ k \leq \frac{n}{2} $. From the given set $ Z $, select a part $ Z_1 $ consisting of $ \left[ \frac{n}{2} \right] $ points*); the remaining part $ Z_2 $ contains $ \left[ \frac{n}{2} \right] $ points if $ n $ is even, and $ \left[ \frac{n}{2} \right] + 1 $ points if $ n $ is odd. Since $ k $ is an... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
LVII OM - II - Problem 3
Positive numbers $ a, b, c $ satisfy the condition $ ab+bc+ca = abc $. Prove that | Dividing the equality $ ab+bc+ca = abc $ by $ abc $ on both sides, we get
Let us substitute: $ x=1/a,\ y =1/b,\ z =1/c $. Then the numbers $ x, y, z $ are positive, and their sum is 1. Moreover,
Thus, the inequality to be proved takes the form
We will show that for any positive numbers $ x, y $, th... | proof | Algebra | proof | Yes | Yes | olympiads | false |
XVII OM - I - Problem 10
Problem 10.
Prove that if real numbers $ a, b, c $ satisfy the inequalities
then for every natural number $ n $ the inequality holds
(Note: The original text had blank lines and symbols that were not filled in, so the translation retains these gaps and the overall format.) | From assumption (2), it follows that the numbers $a$, $b$, $c$ are either all positive or one is positive and two are negative. In the first case, the thesis (3) is obvious, so it suffices to prove the theorem assuming, for example, that $a>0$, $b<0$, $c<0$, and limiting ourselves to odd values of $n$, since for even $... | proof | Inequalities | proof | Incomplete | Yes | olympiads | false |
LV OM - III - Task 4
Prove that if $ a $, $ b $, $ c $ are real numbers, then | By squaring both sides of the given inequality, we obtain an equivalent inequality to prove, where the left side is equal to
while the right side is $ 6(a^2 + b^2 + c^2) + 6(ab+bc+ca) $. After reducing similar terms and dividing both sides by 2, we are left to prove that
By the inequality $ \sqrt{2x^2 + 2y^2} \ge... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
XLII OM - III - Problem 1
Determine whether there exist tetrahedra \( T_1 \) and \( T_2 \) with the following two properties:
(a) the volume of tetrahedron \( T_1 \) is greater than the volume of tetrahedron \( T_2 \);
(b) the area of each face of tetrahedron \( T_1 \) does not exceed the area of any face of tetrahedr... | There are such pairs of tetrahedra. More specifically: for any tetrahedron $T_1$, one can construct a tetrahedron $T_2$ such that conditions (a) and (b) are satisfied. Here is an example method:
Let $T_1$ be any tetrahedron: let $V_1$ be its volume, and $S_1$ be the area of its largest face. Take a positive number $a$ ... | proof | Geometry | proof | Yes | Incomplete | olympiads | false |
II OM - II - Problem 2
In triangle $ ABC $, points $ D $, $ E $, $ F $ are chosen on sides $ BC $, $ CA $, $ AB $ respectively, in such a way that
where $ k $ is a given positive number. Given the area $ S $ of triangle $ ABC $, calculate the area of triangle $ DEF $.
| Using the symbol $ \triangle $ to denote the area of a triangle, we have (Fig. 54):
where $ S = \triangle ABC $.
Indeed,
[ \triangle AFE = \frac{1}{2} AF \cdot AE \cdot \sin A, \]
[ AF = \frac{k}{k+1} \cdot AB, \]
[ AE = \frac{1}{k+1} \cdot AC, \]
thus
and since $ \frac{1}{2} AB \cdot AC \cd... | \frac{k^2-k+1}{(k+1)^2} | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
XLIII OM - I - Problem 10
Given are $ n $ numbers $ a_1\leq a_2 \leq \ldots \leq a_n $. We assume
Prove that | In the definition of the expression $ Q $, the difference $ M_1^2 - M_2 $ is under the square root sign. To ensure that the definition of $ Q $ is correct, it would be necessary to justify that this difference is non-negative. (One could also assume that since the problem statement mentions "the square root of a certai... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
XXIX OM - I - Problem 10
Point $ O $ is an internal point of a convex quadrilateral $ ABCD $, $ A_1 $, $ B_1 $, $ C_1 $, $ D_1 $ are the orthogonal projections of point $ O $ onto the lines $ AB $, $ BC $, $ CD $, and $ DA $, respectively, $ A_{i+1} $, $ B_{i+1} $, $ C_{i+1} $, $ D_{i+1} $ are the orthogonal projectio... | The construction given in the task is not always feasible. For example, three of the points $A_{i+1}$, $B_{i+1}$, $C_{i+1}$, $D_{i+1}$ may lie on the same line. In this case, the task loses its meaning. Therefore, we will provide a solution to the task with the additional assumption that all considered quadrilaterals e... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XII OM - III - Task 2
Prove that if $ a + b = 1 $, then | \spos{I} If $ a = \frac{1}{2} + d $, then $ b = \frac{1}{2} - d $
therefore
Since $ d^2 \geq 0 $ and $ d^4 \geq 0 $, we will obtain from this
Note. The theorem proved above is a special case of the theorem: If $ a + b = 1 $, then
The proof of this theorem can easily be conducted in the man... | proof | Algebra | proof | Incomplete | Incomplete | olympiads | false |
LIII OM - II - Problem 5
Triangle $ ABC $, where $ \measuredangle A = 90^\circ $, is the base of the pyramid $ ABCD $. Additionally, the following equalities hold
Prove that $ \measuredangle ACD \geq 30^\circ $ . | Triangle $ BAC $ is completed to form rectangle $ BACE $. From the equality $ AD = BD $, it follows that $ CD = ED $. This, combined with $ AB = CD $, proves that triangle $ CDE $ is equilateral. Therefore | proof | Geometry | proof | Incomplete | Yes | olympiads | false |
LI OM - I - Task 4
Each point of the circle is painted one of three colors. Prove that some three points of the same color are vertices of an isosceles triangle. | Let $A_1, A_2, \ldots, A_{13}$ be the vertices of an arbitrary regular 13-gon inscribed in a given circle. We will show that among any five vertices of this 13-gon, there will be three that form the vertices of an isosceles triangle. Since among any thirteen points on a given circle, there will always be five colored t... | proof | Geometry | proof | Yes | Yes | olympiads | false |
IX OM - II - Problem 6
On a plane, there are two circles $ C_1 $ and $ C_2 $ and a line $ m $. Find a point on the line $ m $ from which tangents can be drawn to the circles $ C_1 $ and $ C_2 $ that are equally inclined to the line $ m $. | If the given circles have a common tangent $s$, intersecting the line $m$ at point $S$, then we can consider that point $S$ satisfies the conditions of the problem, as through this point pass two coinciding tangents to the given circles equally inclined to the line $m$.
Ignoring this trivial solution, we will look for ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXXIII OM - I - Problem 10
In triangle $ABC$, points $A$ are the midpoints of sides $\overline{BC}, \overline{CA}, \overline{AB}$, respectively, and $H$ is the orthocenter of the triangle. The circle with center $H$ intersects the line $B$ at points $D_1$ and $D_2$, $E_1$ and $E_2$, $F_1$ and $F_2$. Prove that the len... | Line $ B $ is parallel to $ BC $, and is therefore perpendicular to the line $ AH $ containing the height $ \overline{h_a} = \overline{AA_1} $ dropped onto side $ \overline{BC} $. Line $ AH $ is thus the perpendicular bisector of the chord $ \overline{D_1D_2} $. Segments $ \overline{AD_1} $ and $ \overline{AD_2} $ have... | proof | Geometry | proof | Incomplete | Incomplete | olympiads | false |
X OM - I - Task 3
On one arm of the angle, points $ A_1 $, $ A_2 $, $ A_3 $ are chosen, and on the other arm, points $ B_1 $, $ B_2 $, $ B_3 $ are chosen in such a way that the line $ A_1B_2 $ is parallel to the line $ A_2B_1 $, and the line $ A_2B_3 $ is parallel to the line $ A_3B_2 $. Prove that the lines $ A_1B_3 ... | We apply Thales' theorem to the sides of a given angle $O$ (Fig. 1) intersected by two pairs of parallel lines: $A_1B_2 \parallel A_2B_1$ and $A_2B_3 \parallel A_3B_2$; we obtain the proportions
Multiplying these equalities, we get the proportion
from which, by the converse of Thales' theorem, it follows that $A_1B_3... | proof | Geometry | proof | Yes | Incomplete | olympiads | false |
XXXV OM - I - Problem 8
Let $ n $ be an even natural number. Prove that a quadrilateral can be divided into $ n $ triangles, whose vertices lie at the vertices of the quadrilateral or inside the quadrilateral, and each side of the triangle is either a side of the quadrilateral or a side of another triangle. | At least one diagonal of the quadrilateral is contained within its interior. Let this be, for example, the diagonal $ \overline{AC} $. If $ n = 2 $, then the division of the quadrilateral $ ABCD $ into triangles $ ABC $ and $ ACD $ is the desired division. Suppose that $ n = 2k $ is an even number. Choose any points $ ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
V OM - I - Problem 7
In the plane, a line $ p $ and points $ A $ and $ B $ are given. Find a point $ M $ on the line $ p $ such that the sum of the squares $ AM^2 + BM^2 $ is minimized. | The geometric locus of points $ X $ in the plane for which $ AX^2 + BX^2 $ has a given value $ 2k^2 $ is a certain circle whose center lies at the midpoint $ S $ of segment $ AB $; the radius of this circle is larger the larger $ k $ is.
The smallest circle centered at $ S $ that intersects a line $ p $ is the circle t... | M | Geometry | math-word-problem | Yes | Yes | olympiads | false |
VII OM - II - Problem 6
Prove that if in a tetrahedron $ABCD$ the segments connecting the vertices of the tetrahedron with the centers of the inscribed circles of the opposite faces intersect at one point, then
and that the converse theorem also holds. | 1°. Let $ S $ be the center of the circle inscribed in triangle $ ABC $, and $ T $ - the center of the circle inscribed in triangle $ BCD $ (Fig. 18). According to the assumption, segments $ AT $ and $ DS $ intersect at some point $ P $. The plane passing through edge $ AD $ and point $ P $ intersects edge $ BC $ at po... | proof | Geometry | proof | Incomplete | Yes | olympiads | false |
XXX OM - I - Task 6
Prove that if in a triangle the center of the inscribed circle, the orthocenter, and the center of the circumscribed circle lie on the same line, then the triangle is isosceles. | In triangle $ABC$, let $O$, $P$, $Q$ be the orthocenter, the incenter, and the circumcenter, respectively. By the conditions of the problem, points $O$, $P$, $Q$ lie on the same line.
If $O = Q$, then each altitude of the triangle is the perpendicular bisector of the corresponding side. In this case, triangle $ABC$ is ... | proof | Geometry | proof | Yes | Incomplete | olympiads | false |
XXIX OM - I - Problem 12
Determine the least upper bound of such numbers $ \alpha \leq \frac{\pi}{2} $, that every acute angle $ MON $ of measure $ \alpha $ and every triangle $ T $ on the plane have the following property. There exists a triangle $ ABC $ isometric to $ T $ such that the side $ \overline{AB} $ is para... | First, note that if $Q$ is the midpoint of side $\widehat{AC}$, then $\measuredangle ABQ \leq \frac{\pi}{6}$. Indeed, denoting by $D$ the projection of point $A$ onto the line $BQ$, we get $AD \leq AQ = \frac{1}{2} AC \leq \frac{1}{2} AB$ (Fig. 9).
Therefore, $\sin \measuredangle ABQ = \frac{AD}{AB} \leq \frac{1}{2}$. ... | \frac{\pi}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XII OM - III - Task 4
Prove that if each side of a triangle is less than $ 1 $, then its area is less than $ \frac{\sqrt{3}}{4} $. | In every triangle, at least one angle does not exceed $ 60^\circ $. Suppose that in triangle $ ABC $ with sides less than one unit, $ \measuredangle C \leq 60^\circ $. Then | proof | Geometry | proof | Yes | Yes | olympiads | false |
LV OM - III - Task 6
Given an integer $ m > 1 $. An infinite sequence of integers $ x_0, x_1, x_2, \ldots $ is defined by the conditions
Determine the largest natural number $ k $ for which there exist $ k $ consecutive terms of this sequence that are divisible by $ m $. | The recursive formula
works both forward and backward; that is, any segment of $m$ consecutive terms of the sequence $(x_i)$ determines both all subsequent terms and all preceding terms. The same remark applies to the sequence $r_0, r_1, r_2, \ldots$, where $r_i$ is the remainder of $x_i$ divided by $m$; formula ... | -1 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
XLII OM - III - Problem 3
Let $ N $ be any number of the form
where each of the coefficients $ \varepsilon_1,\ldots, \varepsilon_{60} $ equals $ 1 $ or $ -1 $. Prove that $ N $ is not the fifth power of any natural number.
Note: $ k^{k^k} = k^{(k^k)} $. | Notice that the number $ 60^{60^{60}} $ is the fifth power of some natural number $ M $; namely
Among the terms of the sum defining the number $ N $, the largest (in absolute value) is, of course, the last one ($ 60^{60^{60}} $). All the other terms, taken together, with any signs, constitute such a negligible fractio... | proof | Number Theory | proof | Incomplete | Yes | olympiads | false |
XLIII OM - I - Problem 12
On a plane, four lines are drawn such that no two of them are parallel and no three have a common point. These lines form four triangles. Prove that the orthocenters of these triangles lie on a single straight line.
Note: The orthocenter of a triangle is the point where its altitudes intersec... | Let's denote four given lines by $k$, $l$, $m$, $n$, and their points of intersection by $O$, $P$, $Q$, $R$, $S$, $T$, such that
None of these points coincide; this is guaranteed by the conditions of the problem.
Figure 5 shows one possible configuration; however, for the further course of reasoning, it does not matte... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XVII OM - I - Problem 12
Prove the theorem: If the sum of the planar angles at the vertex of a regular pyramid is equal to $180^{\circ}$, then in this pyramid, the center of the circumscribed sphere coincides with the center of the inscribed sphere. | We introduce the following notations: $ AB $ - the edge of the base of the given regular pyramid, $ S $ - its apex, $ H $ - the projection of point $ S $ onto the base, $ O $ - the center of the sphere circumscribed around the pyramid, $ M $ - the center of the circle circumscribed around triangle $ ASB $ (Fig. 9).
To ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XLV OM - I - Problem 4
Given a circle with center $ O $, a point $ A $ inside this circle, and a chord $ PQ $, which is not a diameter, passing through $ A $. Lines $ p $ and $ q $ are tangent to the considered circle at points $ P $ and $ Q $, respectively. Line $ l $ passing through point $ A $ and perpendicular to ... | om45_1r_img_1.jpg
We start by observing that points $P$ and $K$ lie on one side of the line $OA$, while points $Q$ and $L$ lie on the other side (regardless of whether point $A$ is closer to the endpoint $P$ or $Q$ on the chord $PQ$). The chord $PQ$ of the given circle is not a diameter, so the tangents $p$ and $q$ int... | proof | Geometry | proof | Yes | Yes | olympiads | false |
XXIX OM - I - Problem 5
For which natural numbers $ n $ do there exist non-zero polynomials
$ f(x_1,x_2, \ldots ,x_n) $, $ g(x_1,x_2, \ldots ,x_n) $ with integer coefficients satisfying the equality
Please note that the equality part of the problem statement is missing. If you have the complete problem s... | We will first prove a simple
Lemma. If the polynomial $W(x_1, x_2, \ldots, x_n)$ does not change when we substitute $(-x_i)$ for the variable $x_i$, then this variable appears in it only with even exponents; that is, there exists a polynomial $V(x_1, x_2, \ldots, x_n)$ such that $W(x_1, x_2, \ldots, x_i, \ldots, x_n) =... | proof | Algebra | math-word-problem | Incomplete | Incomplete | olympiads | false |
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