problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
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 point
$ C \in S $ such that the area of triangle $ ABC $ is equal to $ k $. | 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{1}{2}|(a - c)(y - d) - (b - d)(x - c)| = k $ holds. From this, we obtain the condition that the equation
has for any fixed and different $ (a, b), (c, d) \in K $ a solution in integers $ x, y $.
We will prove that if the number $ m $ does not divide $ 2k $, then the set $ K $ has no more than $ m^2 $ elements.
To this end, consider pairs $ (a \mod m, b \mod m) $ of residues of the coordinates of the points of the set $ K $ modulo $ m $.
There are $ m^2 $ of them, so if $ |K| > m^2 $, then by the pigeonhole principle, we will find two different points
$ (a, b) \in K $ and $ (c, d) \in K $ such that $ a \equiv c \mod m $ and $ b \equiv d \mod m $. For such points,
the equation (1) has no solution, since the left side of the equation for any $ x $ and $ y $ is divisible by $ m $,
while the right side is not. Therefore, $ |K| \leqslant m^2 $ for any $ m $ that is not a divisor of $ 2k $.
From the above considerations, it follows that if $ |K| = 60 $, then $ 2k $ must be divisible by all numbers
$ m \leqslant 7 $, since $ 60 > 7^2 $. It is easy to check that the smallest natural number divisible by
2, 3, 4, 5, 6, 7 is $ 2^2 \cdot 3 \cdot 5 \cdot 7 = 420 $, so $ 210|k $.
We will show that for every $ k $ such that $ 210|k $, a 60-element set $ K $ having the property given in the problem statement can be constructed. Let $ K $ be the set of all elements of the set $ S $,
whose both coordinates are in the set $ \{0, 1,..., 7\} $. Fix any two different points
$ A =(a, b) $ and $ B=(c, d) $ from the set $ K $. Then $ a - c, b - d \in \{-7, -6,..., 6, 7\} $, so if
$ a \neq c $, then $ a-c|420 $ and if $ b = d $, then $ b-d|420 $. Without loss of generality, we can assume that $ b = d $.
Then $ b - d|2k $, since $ 420|2k $. The point $ C =(c + \frac{2k}{d-b} ,d) $ is therefore an element of the set $ S $,
and the area of the triangle $ ABC $ is
Moreover, $ |K| = 60 $. As the set $ K $, we can take any 60-element subset of the set $ K $.
Thus, we have shown that a 60-element set having the desired property exists only for positive integers $ k $
that are multiples of 210. The smallest such number is 210. | 210 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 929 |
LV OM - III - Task 2
Let $ W $ be a polynomial with integer coefficients, taking relatively prime values for some two different integers. Prove that there exists an infinite set of integers for which the polynomial $ W $ takes pairwise relatively prime values. | A sequence of different integers $ x_0, x_1, x_2, \ldots $ such that the values $ W(x_i) $ are pairwise coprime, can be constructed inductively. The two initial terms $ x_0 = a, x_1 = b $ are numbers whose existence is given in the assumptions.
Assume that different numbers $ x_0, x_1, \ldots, x_n $ have already been defined for which the values $ y_i = W(x_i) $ are pairwise coprime. The number $ y_0 $ is coprime with the product $ y_1 y_2 \ldots y_n $; there exist, therefore, integers $ k $, $ m $ such that $ ky_0 + my_1y_2 \ldots y_n = 1 $. We define the next term of the sequence:
where $ z $ is a natural number large enough to ensure that the number $ x_{n+1} $ is different from the numbers $ x_0,x_1,\ldots ,x_n $. Of course,
and from the transformation $ x_{n+1} = b + (a - b)(1 - my_1 y_2 \ldots y_n)+zy_0 y_1 \ldots y_n $, it is clear that
Applying the polynomial $ W $ to both sides of each of these two congruences (which is allowed since congruences with a common modulus can be added and multiplied), we obtain the relations
This means that for some integers $ s $, $ t $, the following equalities hold
And since the value $ W(b) = y_1 $ is coprime with $ y_0 $, and the value $ W (a) = y_0 $ is coprime with the product $ y_1y_2 \ldots y_n $, it follows from the obtained equalities that the value $ W(x_{n+1})= y_{n+1} $ is coprime with both $ y_0 $ and $ y_1 y_2 \ldots y_n $.
Continuing this procedure, we obtain an infinite sequence $ (x_i) $ yielding values $ y_i = W(x_i) $ that are pairwise coprime. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 933 |
I OM - B - Task 1
Determine the coefficients $ a $ and $ b $ of the equation $ x^2 + ax + b=0 $ such that the values of $ a $ and $ b $ themselves are roots of this equation. | According to the theorems about the sum and product of the roots of a quadratic equation, the numbers $ a $ and $ b $ are the roots of the equation $ x^2 + ax + b = 0 $ if and only if they satisfy the system of equations
that is,
The second equation of the system (1) is satisfied when $ b = 0 $ or when $ a-1= 0 $, i.e., when $ a = 1 $.
System (1) therefore has two solutions:
The quadratic equations with the desired property are | 1,0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 935 |
I OM - B - Zadanie 1
Wyznaczyć współczynniki $ a $ i $ b $, równania $ x^2 + ax +b=0 $ tak, aby same wartości $ a $ i $ b $ były pierwiastkami tego równania.
|
Liczby $ a $ i $ b $ mają spełniać równanie $ x^2 + ax + b = 0 $. Zatem
czyli
Drugie z równań układu (2) daje $ b = 0 $ lub $ a + b + 1 = 0 $.
Biorąc $ b = 0 $ otrzymujemy rozwiązanie
Biorąc $ a + b + 1 = 0 $ mamy $ b = - a-l $; po podstawieniu do pierwszego z równań układu (2) otrzymujemy równanie
którego pierwiastkami są liczby $ a = 1 $ i $ a = -1 $, przy czym odpowiednimi wartościami $ b $ są liczby $ b = - 2 $ i $ b=-\frac{1}{2} $.
Stąd dalsze rozwiązania układu równań (2):
Ze znalezionych rozwiązań tylko liczby $ a = 0 $, $ b = 0 $ oraz
liczby $ a = 1 $, $ b = - 2 $ spełniają warunki zadania. Liczby $ a = -\frac{1}{2} $,
$ b = -\frac{1}{2} $ warunków tych nie spełniają, gdyż pierwiastkami równania
$ x^2 - \frac{1}{2}x -\frac{1}{2} = 0 $ są liczby $ -\frac{1}{2} $ i 1, a nie liczby $ -\frac{1}{2} $ i $ -\frac{1}{2} $.
Aby wyjaśnić ten paradoks, zauważmy, że układ równań (2)
daje tylko warunki konieczne dla poszukiwanych liczb $ a $ i $ b $,
Nie są to jednak warunki dostateczne: jeśli bowiem liczby
$ a $ i $ b $ spełniają układ (2), to każda z nich jest wprawdzie pierwiastkiem
równania $ x^2 + ax + b = 0 $, ale z tego nie wynika, że $ a $ i $ b $
dają oba pierwiastki tego równania. W przypadku, gdy $ a = b $,
jest to jeden i ten sam pierwiastek, jak to właśnie zachodzi w równaniu
$ x^2 -\frac{1}{2}x -\frac{1}{2} = 0 $.
| 0,0或1,-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 936 |
IV OM - I - Problem 10
Given are two skew lines $ m $ and $ n $. On line $ m $, a segment $ AB $ of given length $ a $ is measured, and on line $ n $, a segment $ CD $ of given length $ b $ is measured. Prove that the volume of the tetrahedron $ ABCD $ does not depend on the position of segments $ AB $ and $ CD $ on lines $ m $ and $ n $. | 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 parallel lines $m$ and $n$, whose distance $l$ is equal to the distance between the skew lines $m$ and $n$. The horizontal projections $m$ and $n$ will be two intersecting lines, forming an angle $\varphi$ equal to the angle between the skew lines $m$ and $n$. The horizontal projections of segments $AB$ and $CD$, which are parallel to the horizontal projection plane, have lengths $A$ and $C$.
To calculate the volume of the tetrahedron $ABCD$, we will first consider a parallelepiped "circumscribed" around this tetrahedron, specifically a parallelepiped for which segments $AB$ and $CD$ are the diagonals of two opposite faces. The other diagonals of these faces of the parallelepiped are: segment $EF = b$, parallel to segment $CD$ and having a common midpoint $M$ with segment $AB$, and segment $GH = a$, parallel to segment $AB$ and having a common midpoint $N$ with segment $CD$.
The volume $V$ of the parallelepiped constructed in this way is equal to the product of the area of the face $AEBF$ by its distance from the opposite face, i.e., by $d$.
But the area of the parallelogram $AEBF$, whose diagonals have lengths $a$ and $b$ and form an angle $\varphi$, is equal to $\frac{1}{2} ab \sin \varphi$; therefore, $V = \frac{1}{2} abd \sin \varphi$.
The tetrahedron $ABCD$ is formed from the parallelepiped by cutting off four corner tetrahedra $EABC$, $FABD$, $GACD$, $HBCD$ (Fig. 34). The volume of each of these tetrahedra, whose bases are half the size of the bases of the parallelepiped and whose heights are equal to $d$, is $\frac{1}{6} V$.
Therefore, the volume of the tetrahedron is $\frac{1}{3} V = \frac{1}{6} abd \sin \varphi$, which depends only on the lengths $a$, $b$, $d$, and the angle $\varphi$, and does not depend on the position of segments $AB$ and $CD$ on lines $m$ and $n$. | \frac{1}{6}abd\sin\varphi | Geometry | proof | Yes | Yes | olympiads | false | 939 |
XL OM - III - Task 3
We number the edges of a cube with numbers from 1 to 12.
(a) Prove that for any numbering, there exist at least eight triples of integers $ (i,j,k) $, where $ 1\leq i < j < k\leq 12 $, such that the edges numbered $ i,j,k $ are consecutive sides of a broken line.
(b) Provide an example of a numbering for which there do not exist nine triples with the properties mentioned in (a). | (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 $.
If $ q < x $, the given conditions are satisfied by the triples $ (p,q,x) $, $ (p,q,y) $.
If $ q > y $, the given conditions are satisfied by the triples $ (x, q, r) $, $ (y, q, r) $.
If $ x < q < y $, the given conditions are satisfied by the triples $ (p, q, y) $, $ (x, q, r) $.
This proves the correctness of statement (1).
Now, let us assign to each of the eight vertices of the cube the average of the three edge numbers emanating from that vertex. One number can be assigned to at most two vertices. Therefore, among the assigned numbers, there are at least four different ones. Each of these is, according to (1), the middle term of two different triples with the desired properties. Thus, there are at least eight such triples.
(b) The four parallel edges are numbered (in any order) with the numbers from $ 1 $ to $ 4 $. The next four parallel edges are numbered with the numbers from $ 5 $ to $ 8 $. Finally, the remaining four parallel edges are numbered with the numbers from $ 9 $ to $ 12 $. If a triple $ (i,j,k) $, $ i < j < k $, satisfies the condition given in (a), then the segments numbered $ i $, $ j $, $ k $ cannot lie in the same plane (otherwise, the extreme segments would be parallel; their numbers would both be greater or both be smaller than the number of the middle edge). Therefore, these segments must represent all three directions, which means that $ i \in \{1,2, 3,4\} $, $ j \in \{5,6, 7, 8\} $, $ k \in \{9,10,11,12\} $. Moreover, each of the segments numbered $ 5 $, $ 6 $, $ 7 $, $ 8 $ is the middle term of exactly two of the triples of interest to us.
Thus, there are exactly eight such triples. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 940 |
LVI OM - I - Problem 8
On a circle, there are $ n $ lamps; each can be turned on or off. We perform a series of moves; in each move, we select $ k $ consecutive lamps and change their state: we turn off the ones that are on, and turn on the ones that are off (the number $ k $ does not change during this process). Initially, all lamps are off. For a given natural number $ n $, determine all natural numbers $ k \leq n $ for which it is possible to achieve a state with exactly one lamp turned on. | 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 $n$-th move, we turn on $a$ lamps and turn off $b$, then $a + b = k$, so $x_n = x_{n-1} + a - b = x_{n-1} + k - 2b$.
Since $x_{n-1}$ and $k$ are even numbers, $x_n$ is also an even number. The inductive proof is complete. Therefore, if $k$ is an even number, it is not possible to achieve a state with exactly one lamp on.
Next, assume that the numbers $n$ and $k$ have a common divisor $d > 1$ and color the lamps cyclically using $d$ colors. (Strictly speaking, we proceed as follows: we number the lamps consecutively from 1 to $n$ moving around the circle in a fixed direction, and then we paint them using $d$ colors, such that the same color is used for lamps whose numbers give the same remainder when divided by $d$.)
Suppose that after a certain move, exactly $s_i$ lamps of the $i$-th color are lit ($i = 1, 2, \ldots, d$). In each move, we change the state of exactly $l = k/d$ lamps of the $i$-th color. If we turn on $a_i$ lamps of the $i$-th color and turn off $b_i$ lamps of the $i$-th color, then the number $s_i$ increases by exactly $a_i - b_i = l - 2b_i$. Since no lamp is initially on, after each move the numbers $s_1, s_2, \ldots, s_d$ give the same remainder when divided by 2. It is therefore not possible to achieve exactly one lamp being on (we would then have $s_j = 1$ for some number $j$ and $s_i = 0$ for all $i \neq j$).
The remaining case to consider is when $k$ is an odd number such that $\gcd(k, n) = 1$. This condition is equivalent to $\gcd(k, 2n) = 1$. We will show that using the considered moves, it is possible to achieve a situation where exactly one lamp is on.
Applying the Euclidean algorithm to the numbers $k$ and $2n$, we conclude that there exists a positive integer $x$ and a non-negative integer $y$ such that
We select a block of $k$ consecutive lamps and change their state to the opposite. Then we select a block of $k$ lamps adjacent to the previous one and change the state of these lamps to the opposite. We continue this procedure moving around the given circle in a fixed direction and perform a total of $x$ moves.
From equation (1), it follows that the state of one lamp will change exactly $2y + 1$ times, and the state of each other lamp exactly $2y$ times. As a result, exactly one lamp will be on, and the rest will be off. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 941 |
XLVI OM - I - Problem 10
Given a line $ k $ and three distinct points lying on it. Each of them is the origin of a pair of rays; all these rays lie in the same half-plane with edge $ k $. Each of these pairs of rays forms a quadrilateral with each other. Prove that if a circle can be inscribed in two of these quadrilaterals, then a circle can also be inscribed in the third. | 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 following notation (for $i, j \in \{1,2,3\}, i \ne j$):
- $h_i$ - the distance from point $O_i$ to the line $k$,
- $O_i'$ - the projection of point $O_i$ onto the line $k$,
- $r_{ij}$ - the distance from point $O_i$ to the ray $p_j$,
- $T_{ij}$ - the projection of point $O_i$ onto the ray $p_j$.
The point $A_3$ is the center of a homothety that maps points $O_1$, $O_1'$, $T_{13}$ to points $O_2$, $O_2'$, $T_{23}$, respectively. This implies the proportion $|O_1T_{13}| : |O_2T_{23}| = |O_1O_1'| : |O_2O_2'|$.
By cyclically shifting the numbers $1, 2, 3$, we obtain (analogously):
\[
\frac{|O_2T_{21}|}{|O_3T_{31}|} = \frac{|O_2O_2'|}{|O_3O_3'|}
\]
Multiplying these three proportions side by side, we get 1 on the right side, and on the left side, we get a fraction whose numerator must be equal to the denominator:
\[
\frac{|O_1T_{12}| \cdot |O_2T_{23}| \cdot |O_3T_{31}|}{|O_2T_{21}| \cdot |O_3T_{32}| \cdot |O_1T_{13}|} = 1
\]
Notice now that if a circle can be inscribed in the quadrilateral determined by the pairs of rays $p_1$, $q_1$ and $p_2$, $q_2$, then its center lies on both bisectors $l_1$ and $l_2$. The center is thus the point $O_3$, and the radius is equal to both distances $r_{31} = |O_3T_{31}|$ and $r_{32} = |O_3T_{32}|$; therefore, $r_{31} = r_{32}$.
Conversely, if the equality $r_{31} = r_{32}$ holds, then the point $O_3$ is the center of a circle tangent to the rays $p_1$, $q_1$, $p_2$, $q_2$, and thus inscribed in the considered quadrilateral. Therefore, the equality $r_{31} = r_{32}$ is a necessary and sufficient condition for the existence of such a circle.
By analogy (through cyclically shifting the numbers), we infer that a circle can be inscribed in the quadrilateral determined by the pairs of rays $p_2$, $q_2$ and $p_3$, $q_3$ if and only if $r_{12} = r_{13}$, and a circle can be inscribed in the quadrilateral determined by the pairs of rays $p_3$, $q_3$ and $p_1$, $q_1$ if and only if $r_{23} = r_{21}$.
The task thus reduces to showing that if two of the equalities $r_{12} = r_{13}$, $r_{23} = r_{21}$, $r_{31} = r_{32}$ hold, then the third also holds. This implication is an immediate consequence of the equality (1) obtained above. | proof | Geometry | proof | Yes | Yes | olympiads | false | 944 |
LII OM - I - Problem 11
A sequence of positive integers $ c_1,c_2,\ldots,c_n $ will be called admissible if using a balance scale and two complete sets of weights with masses $ c_1,c_2,\ldots,c_n $, it is possible to weigh any object with a mass that is a natural number not exceeding $ 2(c_1 + c_2 +\ldots + c_n) $. For each number $ n $, determine the maximum sum of $ n $ numbers forming an admissible sequence.
Note: Weights can be placed on both pans of the balance. | 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 will call an arrangement of weights on the scales optimal if for each $ k \in\{1,2,\ldots,n\} $, the weights $ p_k $ and $ q_k $ do not lie on different scales.
To obtain an optimal arrangement, two weights $ p_k $ and $ q_k $ can be placed on the scales in $ 5 $ ways (see the table below). Therefore, the number of all optimal arrangements of weights is $ 5^n $. This number includes one "empty" arrangement, in which no weights are placed on either scale. The remaining arrangements can be paired symmetrically, where one arrangement is obtained from the other by swapping the contents of both scales.
| Left scale | Right scale |
|------------|-------------|
| 0 | 0 |
| 1 | 0 |
| 2 | 0 |
| 0 | 1 |
| 0 | 2 |
Thus, the number of non-empty, essentially different optimal arrangements of weights is $ \frac{1}{2}(5^n-1) $, which means we can weigh at most $ \frac{1}{2}(5^n-1) $ objects of different masses.
Assume that the numbers $ c_1,c_2,\ldots,c_n $ form an admissible system. In this case, the mass of the heaviest object that can be weighed is $ 2(c_1 + c_2 +\ldots + c_n) $. Since we cannot weigh more than $ \frac{1}{2}(5^n-1) $ objects,
We will now show that the numbers $ c_1 = 1, c_2 = 5, c_3 = 5^2,\ldots ,c_n = 5^{n-1} $ form an admissible system. For these values of $ c_i $, the following equality holds:
Therefore, we need to prove that with two sets of weights of masses $ 1, 5, 5^2,\ldots,5^{n-1} $, we can weigh any object with a mass that is a natural number not exceeding $ \frac{1}{2}(5^n-1) $. We will prove this fact in two ways.
**Induction.** For $ n = 1 $, we have two weights of mass $ 1 $ each, and with them, we can weigh an object of mass $ 1 $ or $ 2 = \frac{1}{2}(5^1 - 1) $.
Assume that we have two sets of weights of masses $ 1, 5, 5^2 ,\ldots , 5^{n-1} $ and with them, we can weigh any object with a mass not exceeding $ \frac{1}{2}(5^n-1) $. Here is how to weigh an object $ p $ with mass $ c $, when:
(a) $ c\in \langle \frac{1}{2}(5^n-1)+1 ,5^n-1 \rangle $.
Using the induction hypothesis, we first weigh an object $ q $ with mass $ 5^n-c $ by placing it on the right scale. Then, we place an additional weight of mass $ 5^n $ on the right scale, remove the object $ q $, and place the object $ p $ on the left scale.
(b) $ c\in \langle 5^n+1,\frac{1}{2}(3\cdot 5^n-1)\rangle $.
Similarly to case (a), we first weigh an object $ q $ with mass $ c - 5^n $ by placing it on the right scale. Then, we place an additional weight of mass $ 5^n $ on the left scale, remove the object $ q $, and place the object $ p $ on the right scale.
(c) $ c\in \langle\frac{1}{2}(3\cdot 5^n-1)+1,2\cdot 5^n-1\rangle $.
We first weigh an object $ q $ with mass $ 2\cdot 5^n - c $ by placing it on the right scale. Then, we place two weights of mass $ 5^n $ on the right scale, remove the object $ q $, and place the object $ p $ on the left scale.
(d) $ c\in \langle 2\cdot 5^n+1,\frac{1}{2}(5^{n+1}-1) \rangle $.
We first weigh an object $ q $ with mass $ c - 2\cdot 5^n $ by placing it on the right scale. Then, we place two weights of mass $ 5^n $ on the left scale, remove the object $ q $, and place the object $ p $ on the right scale.
Objects with masses $ 5^n $ and $ 2\cdot 5^n $ can be weighed using only two weights of mass $ 5^n $. The inductive proof is thus completed. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 945 |
XXI OM - II - Problem 6
If $ A $ is a subset of set $ X $, then we define $ A^1 = A $, $ A^{-1} = X - A $. Subsets $ A_1, A_2, \ldots, A_k $ are called mutually independent if the product $ A_1^{\varepsilon_1} \cap A_2^{\varepsilon_2} \ldots A_k^{\varepsilon_k} $ is non-empty for every system of numbers $ \varepsilon_1, \varepsilon_2, \ldots, \varepsilon_k $, such that $ |\varepsilon_i| = 1 $ for $ i = 1, 2, \ldots, k $.
What is the maximum number of mutually independent subsets of a $ 2^n $-element set? | 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 $1$. We will show that the subsets $A_1, A_2, \ldots, A_n$ are mutually independent.
A sequence $(\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_n)$, where $|\varepsilon_j| = 1$, belongs to each of the sets $A_i^{\varepsilon_i}$ for $i = 1, 2, \ldots, n$. If $\varepsilon_i = 1$, then the considered sequence has its $i$-th term equal to $1$, and thus belongs to $A_i^{\varepsilon_i} = A_i^1$; if, on the other hand, $\varepsilon_i = -1$, then the considered sequence has its $i$-th term equal to $-1$, and thus belongs to $X - A_i = A_i^{-1}$. Therefore, the set $A_1^{\varepsilon_1} \cap A_2^{\varepsilon_2} \cap \ldots \cap A_n^{\varepsilon_n}$ contains the sequence $(\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_n)$, and thus is non-empty.
We will now show that if $r > n$ and $B_1, B_2, \ldots, B_r$ are some subsets of the set $X$, then they are not mutually independent. If the sequences $(\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_r)$ and $(\varepsilon$ with values $-1$ and $1$ are different, then the sets
are disjoint. Specifically, if, for example, $\varepsilon_i = 1$, then we have $B \subset B_i^{\varepsilon_i} = B_i$ and $B \subset X - B_i$, and the sets $B_i$ and $X - B_i$ are obviously disjoint.
Thus, if the sets $B_1, B_2, \ldots, B_r$ were mutually independent, then all sets of the form $B_1^{\varepsilon_1} \cap B_2^{\varepsilon_2} \cap \ldots \cap B_r^{\varepsilon_r}$ would be non-empty and pairwise disjoint. There are $2^r > 2^n$ such sets. This is impossible, since the set $X$ contains only $2^n$ elements.
We have thus proved that $n$ is the maximum number of mutually independent subsets of a $2^n$-element set. | n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 948 |
XXIII OM - III - Problem 2
On a plane, there are $ n > 2 $ points, no three of which are collinear. Prove that the shortest among the closed broken lines passing through these points is a simple broken line. | Let's recall the definition of a simple broken line: A closed broken line with successive vertices $W_1, W_2, \ldots, W_n, W_{n+1}$, where $W_{n+1} = W_1$, is called a simple broken line if the closed segments $\overline{W_iW_{i+1}}$ and $\overline{W_jW_{j+1}}$ are disjoint for $1 \leq i, j \leq n$.
Let no three of the points $A_1, A_2, \ldots, A_n$ on the plane be collinear and let $A_{n+1} = A_1$. Suppose that the closed broken line $L$ passing successively through the points $A_1, A_2, \ldots, A_n, A_{n+1}$ is the shortest and is not a simple broken line. If $i \ne j$, $1 \leq i, j \leq n$, then let $L(A_i, A_j)$ be the part of the broken line $L$ starting at point $A_i$ and containing the points $A_{i+1}, A_{i+2}, \ldots, A_j$ successively.
For $i = 1, 2, \ldots, n$, we have $L(A_i, A_{i+1}) = \overline{A_i A_{i+1}}$, because the shortest broken line connecting two points is a segment.
From the assumption that $L$ is not a simple broken line, it follows that there exist numbers $i$, $j$ such that
and the broken lines $L(A_i, A_{i+1})$ and $L(A_j, A_{j+1})$ have a common point $P$ (Fig. 15). From conditions (1) it follows that the points $A_i, A_{i+1}, A_j, A_{j+1}$ are distinct. Therefore, the point $P$ belongs to the interior of each of the segments $\overline{A_iA_{i+1}}$, $\overline{A_jA_{j+1}}$; for by assumption, no three of the points $A_1, A_2, \ldots, A_n$ lie on the same line.
Since in a triangle the length of any side is less than the sum of the lengths of the other sides, we have
Therefore,
Thus, the closed broken line $L$ composed successively of the broken lines $L(A_{j+1}, A_i)$, $\overline{A_iA_j}$, $L(A_{i+1}, A_j)$, $\overline{A_{i+1}A_{j+1}}$ passes through the points $A_1, A_2, \ldots, A_n$ and is shorter than the closed broken line $L$ which consists successively of the broken lines $L(A_{j+1}, A_i)$, $\overline{A_iA_{i+1}}$, $L(A_{i+1}, A_j)$, $\overline{A_jA_{j+1}}$. The obtained contradiction proves that the shortest closed broken line passing through the points $A_1, A_2, \ldots, A_n$ is a simple broken line.
Note. It is easy to prove that there exists a shortest closed broken line passing through the given points $A_1, A_2, \ldots, A_n$. Namely, if a broken line passes successively through the points $A_i$ and $A_j$, then the part of the broken line contained between these points is no longer than the segment $\overline{A_iA_j}$ (since the segment is the shortest broken line connecting two points). Therefore, to find the shortest broken line passing through the points $A_1, A_2, \ldots, A_n$, it suffices to consider broken lines composed of segments $\overline{A_iA_j}$. There are a finite number of such segments. Therefore, there are only a finite number of closed broken lines determined by such segments. Hence, there exists among these broken lines the shortest one. | proof | Geometry | proof | Yes | Yes | olympiads | false | 950 |
X OM - III - Task 3
Given is a pyramid with a square base $ABCD$ and apex $S$. Find the shortest path that starts and ends at point $S$ and passes through all the vertices of the base. | Let $ a $ denote the length of the side of square $ ABCD $; the lengths of the lateral edges of the pyramid will be denoted by the letters $ k $, $ l $, $ m $, $ n $ in such a way that $ k \leq l \leq m \leq n $.
Each path starting and ending at point $ S $ and passing through all the vertices of the square can be symbolically written as
where $ W_1, W_2, W_3, W_4 $ represent the vertices $ A $, $ B $, $ C $, $ D $ taken in some order. The part of the path from $ S $ to $ W_1 $ cannot be shorter than the edge $ SW_1 $, and the part of the path from $ W_4 $ to $ S $ cannot be shorter than the edge $ SW_4 $. These two parts of the path together cannot be shorter than the sum of the two shortest lateral edges, i.e., than $ k + l $. The part of the path from $ W_1 $ to $ W_2 $ cannot be shorter than the side of the square, i.e., than $ a $; the same applies to the path from $ W_2 $ to $ W_3 $ and from $ W_3 $ to $ W_4 $. The entire path (1) cannot, therefore, be shorter than
If we can find a path whose length is exactly $ k + l + 3a $, it will be the shortest path. To this end, we will determine the two shortest lateral edges of the pyramid.
Draw 4 axes of symmetry of the square $ ABCD $ (Fig. 31). They divide the plane of the square into 8 angular regions $ \textrm{I}-\textrm{VIII} $. The projection $ O $ of the vertex $ S $ of the pyramid onto the plane of the square lies inside or on the boundary of one of these regions, say region $ \textrm{I} $. Based on the properties of the perpendicular bisector of a segment, the following inequalities then hold
Thus, $ OA $ and $ OB $ are the two shortest of the segments $ OA $, $ OB $, $ OC $, $ OD $. In this case, $ SA $ and $ SB $ are the two shortest lateral edges of the pyramid (the shorter projection corresponds to the shorter slant), namely $ SA = k $, $ SB = l $, from which it follows that the path $ SADCBS $, whose length $ SA + AD + DC + CB + BS = k + l + 3a $ is the sought shortest path.
When the point $ O $ lies on the ray $ MA $ but does not coincide with the center $ M $ of the square, there are two shortest paths $ SADCBS $ and $ SABCDS $. When the point $ O $ coincides with the point $ M $, there are four such paths. Each path can, of course, be traversed in two directions. | k++3a | Geometry | math-word-problem | Yes | Yes | olympiads | false | 951 |
XLVII OM - I - Problem 3
In a group of $ kn $ people, each person knows more than $ (k - 1)n $ others ($ k $, $ n $ are natural numbers). Prove that it is possible to select $ k + 1 $ people from this group, such that any two of them know each other.
Note: If person $ A $ knows person $ B $, then person $ B $ knows person $ A $. | 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 of such a maximal $ m $-person clique is then a ($ k+1 $)-person clique.
Let $ \{p_1,\ldots,p_m\} $ be an $ m $-person clique. Denote by $ N_i $ the set of all people unknown to $ p_i \ (i = 1,\ldots,m) $, and by $ n_i $ the number of people in the set $ N_i $. Each person $ p_i $ knows more than $ kn-n $ others, so the group of her/his unknowns consists of fewer than $ n-1 $ people:
Outside the clique $ \{p_1,\ldots,p_m\} $, there are $ kn-m $ people; let us denote their set by $ Q $.
Take any person $ q $ from the set $ Q $. If she/he were a friend of all people $ p_1,p_2, \ldots, p_m $, we could add her/him to them, and the resulting set $ \{p_1,p_2, \ldots, p_m, q \} $ would be an ($ m+1 $)-person clique. This, however, contradicts the definition of $ m $ as the largest possible number of people in a clique.
Hence, we conclude that every person from the set $ Q $ is unknown to someone from the clique $ \{p_1,\ldots, p_m \} $, and thus belongs to at least one set $ N_i $. Therefore, the number of people in the set $ Q $ (equal to $ kn-m $) does not exceed the total number of people in the sets $ N_1, \ldots , N_m $. The following inequality thus holds:
after reduction: $ k < m $. As we stated at the beginning, this inequality is sufficient to complete the proof. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 952 |
XXIV OM - I - Problem 7
Prove that among five segments lying on the same straight line, there are either three segments that have a common point or three segments that are pairwise disjoint. | Let the open intervals be $ I_k = (a_k; b_k) $, where $ k = 1, 2, 3, 4, 5 $. For closed, half-open, etc., intervals, the reasoning proceeds similarly.
Without loss of generality, we can assume that $ a_1 \leq a_2 \leq a_3 \leq a_4 \leq a_5 $. Suppose that there is no point belonging to any three intervals. If $ b_1 > a_3 $ and $ b_2 > a_3 $, then the intervals $ I_1, I_2, I_3 $ would have a common point. Therefore, $ b_1 \leq a_3 $ or $ b_2 \leq a_3 $. Let $ r $ be the number, $ 1 $ or $ 2 $, for which $ b_r \leq a_3 $ holds.
If $ b_3 > a_5 $ and $ b_4 > a_5 $, then the intervals $ I_3, I_4, I_5 $ would have a common point. Therefore, $ b_3 \leq a_5 $ or $ b_4 \leq a_5 $. Let $ s $ be the number $ 3 $ or $ 4 $, for which $ b_s \leq a_5 $ holds.
We thus have $ a_r < b_r \leq a_3 \leq a_s < b_s \leq a_5 < b_5 $. It follows that the intervals $ I_r, I_s, I_5 $ are pairwise disjoint. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 953 |
VI OM - I - Problem 8
Prove that a circle can be inscribed in a trapezoid if and only if the circles whose diameters are the non-parallel sides of the trapezoid are externally tangent to each other. | A circle can be inscribed in a convex quadrilateral $ABCD$ if and only if the sums of the lengths of opposite sides of the quadrilateral are equal, i.e., when
When the quadrilateral $ABCD$ is a trapezoid with parallel sides $AB$ and $CD$, denoting the midpoints of sides $AD$ and $BC$ by $M$ and $N$ respectively (Fig. 6), we have $AB + CD = 2 MN$, so equality (1) can be replaced by equality
Equality (2), on the other hand, expresses the necessary and sufficient condition for the external tangency of circles with centers at $M$ and $N$ and radii $\frac{1}{2} AD$ and $\frac{1}{2} BC$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 956 |
XIX OM - I - Problem 4
In the plane, given are parallel lines $ a $ and $ b $ and a point $ M $ lying outside the strip bounded by these lines. Determine points $ A $ and $ B $ on lines $ a $ and $ b $, respectively, such that segment $ AB $ is perpendicular to $ a $ and $ b $, and the angle $ AMB $ is the largest. | Let $ AB $ be a segment with endpoints $ A \in a $ and $ B \in b $, perpendicular to the lines $ a $ and $ b $, and $ N $ - the point symmetric to point $ M $ with respect to a line parallel to lines $ a $ and $ b $ and equidistant from them (Fig. 4).
The circle with diameter $ AB $ lies within the strip bounded by the lines $ a $ and $ b $, so point $ M $ lies outside this circle, which means $ \measuredangle AMB = \alpha $ is less than $ 90^\circ $. Therefore, the maximum of angle $ \alpha $ coincides with the maximum of $ \sin \alpha $, since for angles from $ 0^\circ $ to $ 90^\circ $, the sine of the angle increases with the angle.
When segment $ AB $ lies on line $ MN $, then $ \sin \alpha = 0 $.
When points $ A $, $ B $, and $ M $ are vertices of a triangle, and $ r $ denotes the radius of the circumcircle of this triangle, then $ \sin \alpha = \frac{AB}{2r} $. Segment $ MN $ is a chord of this circle, so $ MN \leq 2r $, hence
From this, it follows that the maximum of $ \sin \alpha $, and thus the maximum of $ \alpha $, occurs when $ \sin \alpha = \frac{AB}{MN} $, i.e., when
which means that points $ A $ and $ B $ lie on the circle with diameter $ MN $. | ABlieonthecirclewithdiameterMN | Geometry | math-word-problem | Yes | Yes | olympiads | false | 958 |
XXIX OM - I - Problem 6
Prove that on the plane, any closed broken line of length 1 is contained in some circle of radius length $ \frac{1}{4} $ | Let points $A$ and $B$ belonging to a closed broken line of length $1$ divide this broken line into two equal parts, and let $C$ be any point on this broken line. Since a segment is the shortest broken line connecting two points, we have $AC + BC \leq \frac{1}{2}$.
Let $P$ be the midpoint of segment $\overline{AB}$, and $C'$ be the point symmetric to $C$ with respect to point $P$ (Fig. 7). Then $CP = C'P$ and $BC = AC$. Therefore,
and hence $CP \leq \frac{1}{4}$.
We have thus proved that any point $C$ of the given broken line is at a distance from point $P$ not greater than $\frac{1}{4}$. Therefore, this broken line is contained within a circle with center $P$ and radius of length $\frac{1}{4}$.
om29_1r_img_7.jpg
Note. A similar problem was given in the VII Mathematical Olympiad as problem 23 (5). | proof | Geometry | proof | Yes | Yes | olympiads | false | 960 |
XIII OM - III - Task 4
In how many ways can a set of $ n $ items be divided into two sets? | The answer to the posed question will depend on whether - to the considered sets, we include the empty set, i.e., a set that does not have any element (object), or not; in the first case, the sought number is $ 1 $ greater than in the second. We agree to consider only non-empty sets.
Let $ p $ denote the number of all possible partitions of the set $ Z $ having $ n $ elements, into two sets, i.e., into two parts.
Then $ 2p $ denotes the number of all possible parts, i.e., subsets of the set $ Z $, with the condition that we only consider proper subsets, i.e., different from the set $ Z $ itself.
The number $ 2p $ of all subsets of a set with $ n $ elements can be found by adding the numbers of subsets with $ 1, 2, \ldots, (n - 1) $ elements.
The number of subsets with $ k $ elements is as many as there are combinations of $ n $ elements taken $ k $ at a time, i.e., $ \binom{n}{k} $. Therefore,
According to the binomial theorem of Newton,
Subtracting equations (1) and (2) and considering that $ \binom{n}{0} = \binom{n}{n} = 1 $, we obtain
and from this,
| 2^{n-1}-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 961 |
XVII OM - I - Problem 11
Prove that the centers of the excircles of a triangle and the points symmetric to the center of the incircle of the triangle with respect to its vertices lie on a single circle. | Let $A_1$, $B_1$, $C_1$ be the points symmetric to the center $O$ of the incircle of triangle $ABC$ with respect to the vertices $A$, $B$, $C$, and let $S_1$ be the center of the excircle of triangle $ABC$ that is tangent to side $BC$ (Fig. 8). To prove the theorem, it suffices to show that the points $A_1$, $B_1$, $C_1$, and $S_1$ lie on the same circle. Since $C_1$ and $S_1$ lie on the same side of the line $A_1B_1$, specifically on the same side of the line $A_1B_1$ as triangle $ABC$, it reduces to showing that $\measuredangle A_1S_1B_1 = \measuredangle A_1C_1B_1$. Notice that:
a) Triangles $A_1B_1C_1$ and $ABC$ are similar with respect to point $O$, so
b) Lines $BS_1$ and $CS_1$, as the angle bisectors of the external angles of triangle $ABC$, are perpendicular to the angle bisectors $OB$ and $OC$ of the corresponding angles of this triangle; points $B$, $O$, $C$, $S_1$ as vertices of a quadrilateral with two right angles lie on the same circle, with points $C$ and $S_1$ lying on the same side of the line $OB$, thus
c) Line $BS_1$ is the perpendicular bisector of segment $OB_1$, so
From a), b), and c), it follows that
Note. The theorem we have proved is equivalent to the theorem:
The points symmetric to the orthocenter of triangle $S_1S_2S_3$ with respect to the sides of this triangle lie on the circumcircle of triangle $S_1S_2S_3$ *).
To prove this, it suffices to notice that if $S_1$, $S_2$, $S_3$ are the centers of the excircles of triangle $ABC$, and $O$ is the center of the incircle of triangle $ABC$, then point $O$ is the orthocenter of triangle $S_1S_2S_3$, and points $A_1$, $B_1$, $C_1$ - symmetric to $O$ with respect to points $A$, $B$, $C$ - are also symmetric to $O$ with respect to the sides of triangle $S_1S_2S_3$. Conversely, if $O$ is the orthocenter of a given triangle $S_1S_2S_3$, and $A_1$, $B_1$, $C_1$ are the points symmetric to $O$ with respect to the sides of this triangle, then they are also symmetric to $O$ with respect to the vertices $A$, $B$, $C$ of the "pedal" triangle for triangle $S_1S_2S_3$, and $O$ is the center of the incircle of triangle $ABC$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 963 |
XIX OM - III - Problem 5
On a plane, there are $ n $ points ($ n \geq 4 $), any four of which are vertices of a convex quadrilateral. Prove that all these points are vertices of a convex polygon. | We will apply the method of induction. When $ n = 4 $, the thesis of the theorem is true. Assume that the thesis of the theorem is true for some natural number $ n \geq 4 $, and let $ A_1, A_2, \ldots, A_n, A_{n+1} $ be such $ n + 1 $ points in the plane that any four of them are vertices of a convex polygon. According to the induction hypothesis, the points $ A_1, A_2, \ldots, A_n $ are vertices of a convex polygon $ W $. Let us assume that they are consecutive vertices, which can be achieved by appropriately numbering the points. The point $ A_{n+1} $ does not lie on the boundary of the polygon $ W $, since no three given points are collinear. It also does not lie inside the polygon $ W $, since any point inside the polygon belongs to one of the triangles into which the polygon can be divided by its diagonals, and the point $ A_{n + 1} $ cannot be in such a triangle. Therefore, the point $ A_{n+1} $ lies outside the polygon $ W $. Consider the convex angles with vertex $ A_{n+1} $, whose sides pass through the vertices of the polygon $ W $. The set of these angles is finite, so there is a largest angle among them. Let this be, for example, the angle $ \alpha = A_kA_{n + 1}A_l $. Inside the angle $ \alpha $ lie all the vertices of the polygon $ W $ except $ A_k $ and $ A_l $. In the triangle $ T $ with vertices $ A_k $, $ A_{n+1} $, $ A_l $, none of the vertices $ A_i $ of the polygon $ W $, different from $ A_k $ and $ A_l $, lie, since the points $ A_i $, $ A_k $, $ A_l $, $ A_{n+1} $ are, according to the assumption, vertices of a convex quadrilateral. Therefore, $ A_k $ and $ A_l $ are consecutive vertices of the polygon $ W $, for example, $ l=k+1 $. In this case, the points $ A_1, A_2 \ldots A_k, A_{n+1}, A_{k+1}, \ldots, A_n $ are consecutive vertices of the polygon $ W_1 $, which consists of the polygon $ W $ and the triangle $ T $.
We will prove that the polygon $ W_1 $ is convex, i.e., that any segment whose endpoints lie in the polygon $ W_1 $ is entirely contained in this polygon. Indeed, if the points $ M $ and $ N $ lie in the polygon $ W_1 $, then one of the following cases occurs: a) The points $ M $ and $ N $ both lie in the convex polygon $ W $, or both in the triangle $ T $; then the entire segment $ MN $ is contained in the polygon $ W $ or in the triangle $ T $, and thus is contained in the polygon $ W_1 $; b) One of these points, say the point $ M $, lies in the polygon $ W $, and the point $ N $ in the triangle $ T $. The segment $ MN $ then intersects the boundary of the triangle $ T $ at some point $ P $. The point $ P $ lies on that part of the boundary of the triangle $ T $ which is in the angle $ \alpha $, i.e., on the segment $ A_kA_{k+1} $. The segment $ MP $ therefore belongs to the polygon $ W $, and the segment $ PN $ to the triangle $ T $; each of these segments, and thus their sum $ MN $, belongs to the polygon $ W_1 $. The inductive proof of the theorem has been completed.
Note. A very simple proof of the theorem can be obtained by relying on the following theorem:
For any finite set $ Z $ of points in the plane, consisting of $ n \geq 3 $ points not lying on a single line, there exists a convex polygon $ W $ such that
1. The set $ Z $ is contained in the polygon $ W $.
2. Each vertex of the polygon $ W $ belongs to the set $ Z $.
Such a polygon $ W $ is called the convex hull of the set $ Z $.
Let $ Z $ be a set of $ n \geq 4 $ points in the plane, any four of which are vertices of a convex quadrilateral, and let the polygon $ W $ be the convex hull of the set $ Z $. Then each point of the set $ Z $ is a vertex of the polygon $ W $. Indeed: $ 1^\circ $ no point of the set $ Z $ lies on the boundary of the polygon $ W $ between two of its vertices, since there are no three collinear points in the set $ Z $; $ 2^\circ $ no point of the set $ Z $ lies inside the polygon $ W $, since it would then lie in one of the triangles into which $ W $ can be divided by its diagonals, which would also contradict the assumption. The convex hull $ W $ is thus the polygon whose existence needed to be proved. | proof | Geometry | proof | Yes | Yes | olympiads | false | 964 |
XLIII OM - I - Zadanie 11
Dana jest liczba naturalna $ n\geq 1 $. Rozważamy tabelę zbudowaną z $ n(n+1)/2 $ okienek, ustawionych w $ n $ rzędach: jedno okienko w pierwszym rzędzie, dwa w drugim itd., $ n $ okienek w $ n $-tym rzędzie. W okienka tabeli wpisujemy w sposób losowy liczby od $ 1 $ do $ n(n + l )/2 $. Niech $ m_k $ będzie największą z liczb w $ k $-tym rzędzie. Obliczyć prawdopodobieństwo tego, że $ m_1 < m_2 < \ldots < m_n $.
|
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
Wybierzmy dowolne $ n $ liczb ze zbioru $ \{1,2,\ldots,N\} $ i przyjmijmy, że te liczby zostały wpisane w okienka $ n $-tego rzędu. Pozostaje zbiór $ M $ różnych liczb, gdzie $ M = N - n= n(n - 1 )/2 $; należy je rozmieścić w okienkach od pierwszego do ($ n - 1 $)-go rzędu.
Zbiór ten możemy zastąpić po prostu zbiorem $ \{1,2,\ldots,M\} $; prawdopodobieństwo zajścia nierówności $ m_1 \leq \ldots \leq m_{n} $ nie zmieni się przy tym.
Uwaga ta jest słuszna dla każdego wyboru $ n $ liczb (do $ n $-tego rzędu tabeli); w szczególności dla każdego takiego wyboru, który powoduje zajście zdarzenia $ \mathcal{B}_n $. Wobec tego prawdopodobieństwo warunkowe $ P(\mathcal{C}_n \mid \mathcal{B}_n) $ jest równe prawdopodobieństwu zdarzenia $ \mathcal{A}_{n-1} $ i otrzymujemy wzór rekurencyjny
Ponieważ $ P(\mathcal{A}_1) = 1 $, zatem
| Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 966 | |
LV OM - III - Task 5
Determine the maximum number of lines in space passing through a fixed point and such that any two intersect at the same angle. | Let $ \ell_1,\ldots,\ell_n $ be lines passing through a common point $ O $. A pair of intersecting lines determines four angles on the plane containing them: two vertical angles with measure $ \alpha \leq 90^\circ $ and the other two angles with measure $ 180^\circ - \alpha $. According to the assumption, the value of $ \alpha $ is the same for every pair $ \ell_i, \ell_j $.
Consider a sphere $ S $ centered at $ O $ with an arbitrary radius, and denote by $ A_i $, $ B_i $ the points of intersection of the line $ \ell_i $ with this sphere. Each of the segments $ A_iA_j $, $ A_iB_j $, $ B_iB_j $ (for $ i \neq j $) is a chord of the sphere $ S $, determined by a central angle of measure $ \alpha $ or $ 180^\circ - \alpha $. Therefore, these segments have at most two different lengths $ a $ and $ b $.
Fix the notation so that $ \measuredangle A_iOA_n = \alpha $ for $ i = 1,\ldots,n-1 $. Then the points $ A_1, \ldots, A_{n-1} $ lie on a single circle (lying in a plane perpendicular to the line $ \ell_n $). Let $ A_1C $ be the diameter of this circle. Each of the points $ A_2,\ldots,A_{n-1} $ is at a distance $ a $ or $ b $ from the point $ A_1 $; hence, on each of the two semicircles with endpoints $ A_1 $ and $ C $, there are at most two points from the set $ \{A_2,\ldots ,A_{n-1}\} $. Therefore, this set has at most four elements; which means that $ n \leq 6 $.
On the other hand, if $ n = 6 $, we can place the points $ A_1,A_2,\ldots,A_5 $ at the vertices of a regular pentagon, and the plane of this pentagon at such a distance from the point $ A_6 $ that these six points, together with their antipodal points $ B_1,\ldots,B_6 $, are the vertices of a regular icosahedron. The segments $ A_iB_i $ (diameters of the sphere $ S $) connect opposite vertices of this icosahedron and any two of them form the same angle. Therefore, $ n = 6 $ is the largest possible number of lines $ \ell_i $. | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 967 |
XLII OM - II - Problem 6
A parallelepiped contains a sphere of radius $ r $ and is contained in a sphere of radius $ R $. Prove that $ \frac{R}{r} \geq \sqrt{3} $. | om42_2r_img_8.jpg
Let us choose any face of the parallelepiped and denote its vertices by $ A $, $ B $, $ C $, $ D $ such that segment $ AC $ is the longer (i.e., "not shorter") diagonal of the parallelogram $ ABCD $. The remaining vertices of the parallelepiped are denoted by $ E $, $ F $, $ G $, $ H $ such that segments $ AE $, $ BF $, $ CG $, $ DH $ are its four parallel edges (see Figure 8). Finally, let us assume that $ AG $ is the longer diagonal of the parallelogram $ ACGE $; if the opposite is true, we swap the labels: $ A \longleftrightarrow C $, $ E \longleftrightarrow G $.
From this arrangement of labels, it follows that $ |\measuredangle ABC| \geq 90^\circ $, $ |\measuredangle ACG| \geq 90^\circ $. Applying the cosine rule to triangles $ ACG $ and $ ABC $, we obtain the relationship
Notice that each of the segments $ AB $, $ BC $, $ CG $ has endpoints located on two parallel faces of the parallelepiped $ ABCDEFGH $. Since the parallelepiped contains a sphere of radius $ r $, the distance between each pair of parallel faces is at least $ 2r $. Therefore, we have the inequalities
On the other hand, the length of the diagonal $ AG $ does not exceed the diameter of the sphere in which the parallelepiped is contained: $ |AG| \leq 2R $. Taking into account all these inequalities, we conclude that $ (2R)^2 \geq (2r)^2 + (2r)^2 + (2r)^2 $, and thus $ R^2 \geq 3r^2 $, or $ R/r \geq \sqrt{3} $.
Note. Many participants of the Olympiad began their solution by choosing a vertex of the parallelepiped such that the three faces meeting at this vertex have acute (or right) angles at it. However, such a vertex may not exist; this happens when there is a vertex at which the planar angles of the three faces of the parallelepiped are obtuse. | R/r\geq\sqrt{3} | Geometry | proof | Yes | Yes | olympiads | false | 968 |
XII OM - III - Task 3
Prove that if the section of a tetrahedron by a plane is a parallelogram, then half of its perimeter is contained between the length of the shortest and the length of the longest edge of the tetrahedron. | Let the parallelogram $MNPQ$ be a planar section of the tetrahedron $ABCD$. The plane $MNPQ$ is, of course, different from the planes of the faces of the tetrahedron, and each side of the parallelogram lies on a different face. Suppose that the sides $MN$ and $QP$ lie on the faces $ABC$ and $BCD$; they are then parallel to the edge $BC$, along which these faces meet, and the sides $MQ$ and $NP$ are parallel to the edge $AD$ (Fig. 19).
Applying Thales' theorem to triangles $AMN$ and $ABC$ and to triangles $BMQ$ and $BAD$ we get
Hence
Suppose that $BC \leq AD$; in that case, from the above equality, it follows that
thus
If $a$ denotes the length of the smallest, and $b$ the length of the largest diagonal of the tetrahedron, then $BC \geq a$, and $AD \leq b$, and from the above inequality, it follows that
which was to be proved. | proof | Geometry | proof | Yes | Yes | olympiads | false | 970 |
XXIX OM - I - Problem 11
Let $ f: \mathbb{R} \to \mathbb{R} $ be a continuous function for which there exists a number $ x $ satisfying the conditions: $ f(f(f(x))) = x $, $ f(x) \neq x $.
Prove that there exist numbers $ y_1 $, $ y_2 $, $ y_3 $ such that $ y_i \neq y_j $ for $ i \neq j $ and $ f(f(y_k))=y_k $ for $ k = 1, 2, 3 $.
Note. It is known that every continuous function $ g: \mathbb{R} \to \mathbb{R} $ has the Darboux property, meaning if $ a < b $ and $ d $ is between $ g(a) $ and $ g(b) $, then there exists a number $ c \in [a, b] $ such that $ g(c) = d $. | 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, $ff(x) \ne f(x)$.
Thus, the numbers $x$, $f(x)$, and $f(x)$ are pairwise distinct. Let the smallest of them be denoted by $a$, the largest by $c$, and the remaining one by $b$. Therefore, $a < b < c$. From the conditions of the problem, it follows that the function $f$ defines a permutation of the set $\{a, b, c\}$ and $f(a) \ne a$, $f(b) \ne b$, $f(c) \ne c$.
Consider the case when $f(a) = c$. If $f(a) = b$, the reasoning proceeds similarly. Thus, we have $f(b) = a$ and $f(c) = b$.
The continuous function $f(t)$ at the endpoints of the interval $\langle a, b \rangle$ takes the values $c$ and $a$. Therefore, by the Darboux property, there is a point in this interval where the function $f(t)$ takes an intermediate value $b$, that is, $a < z < b$ and $f(z) = b$. To solve the problem, it is sufficient to prove that the function $g(t) = ff(t) - t$ has at least three zeros. We have
Thus, the function $g(t)$ at the endpoints of each of the intervals $\langle a, z \rangle$, $\langle z, b \rangle$, $\langle b, c \rangle$ takes values of opposite signs. Therefore, by the Darboux property, it has at least one zero in each of these intervals.
Note. For any natural number $n$, let $f_n$ be the $n$-fold iteration of the function $f$, that is, $f_n(x) = f(f(\ldots (f(x))\ldots))$ ($n$ times). We say that a real number $x$ has period $n$ if $f_n(x) = x$ and $f_m(x) \ne x$ for $m < n$. Our problem can thus be formulated as: *Prove that if for a continuous function $f \colon \mathbb{R} \to \mathbb{R}$ there exists a number $x$ of period $3$, then there exist three numbers $y_1, y_2, y_3$ of period $2$ or $1$.*
The following general theorem of Sharkovsky holds: Arrange all natural numbers as follows
At the beginning is an increasing sequence of all prime numbers, then a sequence of prime numbers multiplied by $2$, by $4$, etc., and finally a decreasing sequence of all powers of $2$.
Sharkovsky's theorem states that if for a continuous function $f \colon \mathbb{R} \to \mathbb{R}$ there exists a number of period $n$, then there also exists a number of period $m$, provided that in the sequence (1) the number $m$ appears after the number $n$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 971 |
XXXIX OM - I - Problem 12
Given are two concentric spheres $ S_R $ and $ S_r $ with radii $ R $ and $ r $, respectively, where $ R > r $. Let $ A_0 $ be a point on $ S_R $. Let $ T_{A_0} $ be the cone with vertex at $ A_0 $ and tangent to $ S_r $. Denote $ K_1 = T_{A_0} \cap S_R $. Then, for each point $ A \in K_1 $, we draw the cone $ T_A $ and define $ K_2 = \bigcup_{A\in K_1} T_A \cap S_R $, and further by induction $ K_{n+1}= \bigcup_{A\in K_n} T_A \cap S_R $, where $ T_A $ is the cone with vertex at $ A $ and tangent to $ S_r $. Prove that for some $ n $, $ K_n = S_R $. | 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 $S_R$. Let $u$ be the angle at the vertex of the cone $T_{A_0}$ (and thus of any cone $T_A$ considered in the problem); $a = 2 \arcsin (r/R)$. Let us take $N = [\pi/2\alpha]+1$. We will prove by induction on $n = 1, \ldots, N$ the following implication:
From this, it will follow that $K_{2N} = S_R$.
For $n = 1$, the antecedent of the implication (1) takes the form: . The cone $T_P$ then has points in common with the set $K_1 = T_{A_0} \cap S_R$. Let $Q \in T_P \cap K_1$ (see figure 5). Note that the statements: $Q \in T_P$ and $P \in T_Q$ are equivalent (each of them means exactly that the segment $PQ$ is tangent to the sphere $S_r$). Thus, in this case,
which means $P \in K_2$.
Now assume that the implication (1) to be proved holds for some natural number $n \leq N-1 = [\pi/2\alpha]$. We will show the validity of the analogous implication for $n+1$.
om39_1r_img_5-6.jpg
Let $P \in S_R$ be a point such that $|\measuredangle A_0OP| \leq 2(n+1) \alpha$ (figure 6 illustrates the situation when $n = 2$, $n+1 = 3$; it shows the section of the considered spheres by the plane $A_0PP$). Consider the great circle of the sphere $S_R$ passing through the points $A_0$ and $P$. Let $\textit{ł}$ be the shorter of the two arcs into which the points $A_0$ and $P$ divide this circle (in the case where $P$ is the antipodal point to $A_0$, we take any great circle with diameter $A_0P$ and any semicircle $A_0P$). On the arc $\textit{ł}$, we find points $L$ and $M$ such that $|\measuredangle LOM| = |\measuredangle MOP| - \alpha$. Then $|\measuredangle LOP| = 2 \alpha$, $|\measuredangle A_0OL|$, $|\measuredangle A_0OP|????$, so by the induction hypothesis $L \in K_{2n}$. Let $M$ be the antipodal point to $M$. In the considered great circle, the inscribed angle $LM$ is half the central angle $LOP$. Therefore, $|\measuredangle LM| = \alpha$, which means that points $L$ and $P$ belong to the cone $T_M$. Segments $LM$ and $PM$ are thus tangent to the sphere $S_r$; hence $M \in K_{2n+1}$, in accordance with the definition of the set $K_{2n+1}$ as the union of sets of the form $T_A \cap S_R$ for $A \in K_{2n}$. Since, in turn, the set $K_{2n+2}$ is the union of sets $T_A \cap S_R$ for $A \in K_{2n+1}$, we conclude that $P \in T_{M}$. This completes the inductive step.
By the principle of induction, the implication (1) is valid for all considered values of $n$ (i.e., for $n = 1, \ldots, N$). In particular, it holds for $n = N$. When $n = N$, the antecedent of this implication is true for any point of the sphere $S_R$: the inequality $|\measuredangle A_0OP| \leq 2N \alpha$ is automatically satisfied because $2N\alpha > \pi$ (in accordance with the definition of the number $N$). Therefore, the consequent is also true; which means that every point $P$ of the sphere $S_R$ belongs to the set $K_{2N}$. Thus, ultimately, $S_R = K_{2N}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 973 |
V OM - I - Problem 4
Inside a given triangle $ ABC $, find a point $ O $ such that the areas of triangles $ AOB $, $ BOC $, $ COA $ are in the ratio $ 1 \colon 2 \colon 3 $. | If the areas of triangles $AOB$, $BOC$, $COA$ are in the ratio $1 \colon 2 \colon 3$, then these areas constitute respectively $\frac{1}{6}$, $\frac{2}{6}$, $\frac{3}{6}$ of the area of triangle $ABC$. The required point $O$ thus lies at the intersection of the line parallel to line $AC$ drawn through the midpoint of side $AB$ and the line parallel to line $BC$ cutting off one-third of side $AB$, measured from point $B$. A line parallel to side $AB$ cutting off on side $AC$, measured from vertex $A$, a segment equal to $\frac{1}{6}AC$ also passes through this point (Fig. 21). The problem always has a solution and only one.
The problem is solved in the same way if instead of the ratio $1 \colon 2 \colon 3$, the ratio $m \colon n \colon p$ of any natural numbers is taken. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 974 |
L OM - III - Zadanie 2
Given are non-negative integers $ a_1 <a_2 <a_3 < \ldots < a_{101} $ less than $ 5050 $. Prove that among them, one can choose four different $ a_k $, $ a_l $, $ a_m $, $ a_n $, such that the number $ a_k + a_l -a_m -a_n $ is divisible by $ 5050 $. | We consider all expressions of the form $ a_k + a_l $, where $ 1 \leq k < l \leq 101 $. There are $ {101 \choose 2} = 5050 $ such expressions. If we find two of them, say $ a_k + a_l $ and $ a_m + a_n $, whose values give the same remainder when divided by 5050, then the numbers $ a_k $, $ a_l $, $ a_m $, $ a_n $ satisfy the conditions of the problem — these numbers are pairwise distinct (if, for example, $ a_k = a_m $, then also $ a_l = a_n $, since $ 0 \leq a_l, a_n < 5050 $).
It remains to consider the case where all the above sums give different remainders when divided by 5050. We will show that this case cannot occur. If it were so, then the considered sums would give all possible remainders when divided by 5050 — each one exactly once. Hence
which proves that $ S $ is an odd number.
On the other hand,
which means that $ S $ is an even number. We have reached a contradiction. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 977 |
LI OM - I - Task 3
The sum of positive numbers $ a $, $ b $, $ c $ is equal to $ 1 $. Prove that | It is enough to prove that for any positive numbers $ a $, $ b $, $ c $ the inequality holds
Transforming the above relationship equivalently, we get
By making the substitution $ x = bc $, $ y = ca $, $ z = ab $, we reduce the inequality to be proved to the form
Squaring both sides of the last inequality and transforming it equivalently, we get
which is true. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 979 |
V OM - III - Task 1
Prove that in an isosceles trapezoid circumscribed around a circle, the segments connecting the points of tangency of opposite sides with the circle pass through the point of intersection of the diagonals. | Let $E$, $F$, $G$, $H$ denote the points of tangency of the sides $AB$, $BC$, $CD$, $DA$ with the incircle of trapezoid $ABCD$.
Let $M$ be the point of intersection of segments $EG$ and $HF$ (Fig. 37a). Since trapezoid $ABCD$ is isosceles, the line $EG$ is the axis of symmetry of the figure, and point $F$ is symmetric to point $H$ with respect to $EG$, and $HF \bot EG$. The parallel lines $AB$, $HF$, and $DC$ determine proportional segments on the lines $EG$ and $BC$, hence
Let $N$ be the point of intersection of the diagonals $AC$ and $BD$ of the trapezoid. Since these diagonals are symmetric with respect to the line $EG$, the point $N$ lies on the segment $EG$ (Fig. 37b). Triangles $AEN$ and $CGN$ are similar with respect to point $N$, hence
But $AE = EB = BF$, and $FC = CG$, so $\frac{AE}{CG} = \frac{BF}{FC}$; from the above proportions, it follows that
Points $M$ and $N$ divide the segment $EG$ in the same ratio, hence these points coincide, which was to be proved. | proof | Geometry | proof | Yes | Yes | olympiads | false | 981 |
X OM - II - Task 6
From point $ M $ on the surface of a sphere, three mutually perpendicular chords of the sphere $ MA $, $ MB $, $ MC $ were drawn. Prove that the segment connecting point $ M $ with the center of the sphere intersects the plane of triangle $ ABC $ at the centroid of this triangle. | The plane $ AMB $ intersects the surface of the sphere along a circle passing through points $ A $, $ M $, and $ B $, which is the circumcircle of the right triangle $ AMB $; the center of this circle is therefore at the midpoint $ K $ of segment $ AB $ (Fig. 25).
The center $ O $ of the sphere lies on the perpendicular line erected at point $ K $ to the plane $ AMB $ on the same side of the plane $ AMB $ as point $ C $. Since the line $ MC $ is perpendicular to the plane $ AMB $, the lines $ KO $ and $ MC $ lie - as parallel lines - in the same plane. Segments $ MO $ and $ KC $ are the diagonals of the trapezoid $ MKOC $, and therefore intersect at some point $ S $. It turns out, then, that segment $ MO $ intersects the median $ KC $ of triangle $ ABC $. Similarly, it intersects each of the other medians of triangle $ ABC $, and since segment $ MO $ does not lie in the plane $ ABC $, it follows that it passes through the centroid of triangle $ ABC $. | proof | Geometry | proof | Yes | Yes | olympiads | false | 983 |
XLIX OM - I - Problem 10
Medians $ AD $, $ BE $, $ CF $ of triangle $ ABC $ intersect at point $ G $. Circles can be circumscribed around quadrilaterals $ AFGE $ and $ BDGF $. Prove that triangle $ ABC $ is equilateral. | Let's denote the measures of angles $CAB$, $ABC$, $BCA$ by $\alpha$, $\beta$, $\gamma$ respectively. Points $F$ and $D$ are the midpoints of sides $AB$ and $BC$, so $DF \parallel CA$, and therefore $|\measuredangle BDF| = \gamma$ (Figure 5).
The opposite angles of the cyclic quadrilateral $AFGE$ sum up to $180^\circ$. Meanwhile, the existence of a circumcircle around quadrilateral $BDGF$ implies the equality of angles $BDF$ and $BGF$ (both inscribed in the same circle). Therefore,
Thus,
Analogously, we prove that $\gamma = \beta$. Triangle $ABC$ is therefore equilateral.
Note: There are many variants of this solution, utilizing the equality of different angles (resulting from the problem's assumptions); we have chosen the shortest one. | proof | Geometry | proof | Yes | Yes | olympiads | false | 984 |
XXII OM - I - Problem 11
We toss a coin $2n$ times. Let $p_n$ denote the probability that we will get a series of length $r > n$. Prove that $\lim p_n = 0$. | Elementary events are all $2n$-term sequences with terms being heads or tails; all of them are equally probable and there are $2^{2n}$ of them. Let $A_n$ be the event that we get a series of length at least $n$.
Consider the following three-stage procedure:
1) Selecting some $n$ consecutive positions out of $2n$;
2) Placing only heads or only tails on the selected positions;
3) Filling the remaining positions with heads and tails arbitrarily. As a result of each such procedure, we obtain a sequence that is an elementary event favorable to the event $A_n$, because such a sequence contains a series of length $\geq n$. Conversely, any sequence containing a series of length $\geq n$ can be obtained as a result of such a procedure. Of course, the same sequence can be obtained several times. Therefore, the number of the described ways of proceeding is greater than the number of elementary events favorable to the event $A_n$.
We will now determine the number of the described ways of proceeding. Action 1) can be performed in $n + 1$ ways, because the first of the $n$ consecutive positions chosen can only be position number $1, 2, \ldots, n$ or $n + 1$. Action 2) can be performed in two ways: either by placing heads or by placing tails. Finally, action 3) can be performed in $2^n$ ways, because we have $n$ remaining positions to fill with heads or tails.
Thus, the number of all the described ways of proceeding is equal to $2(n + 1) 2^n$.
From the binomial theorem, we obtain that for $n \geq 2$
Therefore
Since $ {\displaystyle \lim_{n \to \infty}} \frac{4}{n} = 0 $ and $ {\displaystyle \lim_{n \to \infty}} \frac{2}{n-1} = 0 $, it follows that
$ \displaystyle \lim_{n \to \infty} p(A_n) = 0 $. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 985 |
XIX OM - III - Problem 3
In the tetrahedron $ABCD$, the edges $AD$, $BD$, $CD$ are equal. On the plane $ABC$, points $A_1, B_1, C_1$ are chosen such that they are not collinear. The lines $DA_1, DB_1, DC_1$ intersect the surface of the sphere circumscribed around the tetrahedron at points $A_2, B_2, C_2$, respectively, different from point $D$. Prove that the points $A_1, B_1, C_1, A_2, B_2, C_2$ lie on the surface of some sphere. | 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 points $ B $ and $ C $, so it intersects the sphere $ s $ and the plane $ ABC $ along the same circle $ k $. Therefore, the orthogonal projection $ H $ of the point $ D $ onto the plane $ ABC $ is the center of the circle $ k $. The plane $ A_1HD $ then contains a certain diameter $ MN $ of the circle $ k $ and intersects the sphere $ s $ along the circle $ l $ passing through the points $ M $, $ N $, and $ D $ (Fig. 16).
According to the power of a point theorem with respect to a circle,
Hence,
so,
This equality can be written in the form,
Similarly,
From (1) and (2) it follows that,
From equality (3) and the fact that the pairs of points $ A_1 $, $ A_2 $, and $ B_1 $, $ B_2 $ lie on two different rays with origin $ D $ that do not form a single line, it is easy to deduce that the points $ A_1 $, $ A_2 $, $ B_1 $, $ B_2 $ lie on a circle.
Indeed, let $ m $ be a circle passing through the non-collinear points $ A_1 $, $ A_2 $, $ B_1 $ (Fig. 17).
The point $ D $ lies outside the circle $ m $, so,
where $ X $ is the point of intersection of the circle $ m $ with the ray $ DB_1 $.
From (3) and (4) we obtain,
and since the points $ X $ and $ B_2 $ lie on the same ray with origin $ D $, the point $ X $ coincides with the point $ B_2 $.
Similarly, we conclude that the points $ B_1 $, $ B_2 $, $ C_1 $, $ C_2 $ lie on some circle $ n $.
The circles $ m $ and $ n $ have the common points $ B_1 $ and $ B_2 $, and they lie in different planes, since according to the assumption, the common points of these circles with the plane $ ABC $, i.e., the points $ A_1 $, $ B_1 $, $ C_1 $, are not collinear.
Hence, the circles $ m $ and $ n $, and thus the points $ A_1 $, $ B_1 $, $ C_1 $, $ A_2 $, $ B_2 $, $ C_2 $, lie on one sphere, q.e.d.
Note. The beginning of the above proof can be significantly shortened by referring to the properties of inversion. Specifically, in the inversion with respect to the sphere with center $ D $ and radius $ r = DA $ intersecting the plane $ ABC $ along the circle $ k $, the image of the plane $ ABC $ is a sphere passing through the center of inversion $ D $ and the circle $ k $, i.e., the sphere $ s $. The images of the points $ A_1 $, $ B_1 $, $ C_1 $ are then the points $ A_2 $, $ B_2 $, $ C_2 $, respectively, so,
The rest of the proof remains the same as before. | proof | Geometry | proof | Yes | Yes | olympiads | false | 989 |
XXV OM - II - Problem 5
Given are real numbers $ q,t \in \langle \frac{1}{2}; 1) $, $ t \in (0; 1 \rangle $. Prove that there exists an increasing sequence of natural numbers $ {n_k} $ ($ k = 1,2, \ldots $), such that | The sequence $ \{n_k\} $ $ (k = 1,2, \ldots) $ is defined inductively as follows. Since $ 0 < q < 1 $, we have $ q^0 = 1 $ and $ \displaystyle \lim_{n \to \infty} q^n = 0 $. For each number $ t $ in the interval $ (0; 1 \rangle $, there exists a natural number $ n_1 $ such that
Wthen
since $ 1 - q \leq q $ for $ \displaystyle q \in \langle \frac{1}{2}; 1) $.
Next, assume that for some natural number $ k $, the increasing sequence of natural numbers $ n_1, n_2, \ldots, n_k $ has already been defined such that
We take $ n_{k+1} $ to be the natural number $ m $ such that
From (1) and (2), it follows that $ q^m < q^{n_k} $, and thus $ m > n_k $, i.e., $ n_{k+1} > n_k $. Moreover, from (2) we obtain
since $ 1 - q \leq q $ for $ q \in \langle \frac{1}{2}; 1) $.
Therefore, by the principle of induction, there exists an infinite increasing sequence of natural numbers $ n_1, n_2, \ldots $ such that condition (1) is satisfied for $ k = 1, 2, \ldots $.
Since $ \displaystyle \lim_{n \to \infty} q^{n_k} = 0 $, it follows from (1) that
Note. The sequence $ \{n_k \} $ satisfying the conditions of the problem is generally not uniquely determined. Consider the polynomial $ f(x) = x^3 + x^2 - 1 $. It has a root in the interval $ \displaystyle \left( \frac{1}{2}; 1 \right) $, because $ f(x) $ takes values of opposite signs at the endpoints of this interval. Let us denote such a root by $ q $. Then, by calculating the sums of geometric series, we obtain
From the equality $ q^3 + q^2 - 1 = 0 $, it follows that
Thus | proof | Algebra | proof | Yes | Yes | olympiads | false | 993 |
XXIII OM - I - Problem 4
Points $ A $ and $ B $ do not belong to the plane $ \pi $. Find the set of all points $ M \in \pi $ with the property that the lines $ AM $ and $ BM $ form equal angles with the plane $ \pi $. | Let $A'$ and $B'$ denote the projections of points $A$ and $B$ onto the plane $\pi$ (Fig. 4). If $A' = A$ and $AA' = 0$, then the sought set of points is the entire plane. If $A' \neq A$ and $AA' \neq 0$, then only point $A$ satisfies the conditions of the problem. For if $M \neq A$ and $M \in \pi$, then the angles $\measuredangle AMA'$ and $\measuredangle BMB'$ are acute. Therefore, $\tan \measuredangle BMB' \neq \tan \measuredangle AMA'$, and thus the considered angles are different.
Finally, if $A' \neq A$, the condition stated in the problem takes the form $\measuredangle (AM, \pi) = \measuredangle (BM, \pi)$, which is equivalent to $\measuredangle AMA' = \measuredangle BMB'$. Since both these angles are acute, this condition is equivalent to $\tan \measuredangle AMA' = \tan \measuredangle BMB'$, or $\frac{AA'}{A'M} = \frac{BB'}{B'M}$. Therefore, the sought point $M$ must satisfy the condition $\frac{AA'}{A'M} = \frac{BB'}{B'M}$.
In other words, the ratio of the distances of point $M$ from points $A$ and $B$ must be constant. As is known, if this ratio is $1$, then the geometric locus of points $M$ is the perpendicular bisector of segment $A$, and if it is different from $1$, then this geometric locus is a circle with a diameter determined by the points that internally and externally divide segment $\overline{A}$ in this ratio. This is the so-called Apollonian circle. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 994 |
XVI OM - III - Task 5
Points $ A_1 $, $ B_1 $, $ C_1 $ divide the sides $ BC $, $ CA $, $ AB $ of triangle $ ABC $ in the ratios $ k_1 $, $ k_2 $, $ k_3 $. Calculate the ratio of the areas of triangles $ A_1B_1C_1 $ and $ ABC $. | According to the assumption
Hence
Let the symbol $ p (ABC) $ denote the area of triangle $ ABC $ (Fig. 21). Since
then
By applying the theorem that the ratio of the areas of two triangles having a common angle equals the ratio of the products of the sides of each triangle forming that angle, we obtain
Substituting into equation (2) the values of the ratios (1) gives the equality
after transforming the right side, we obtain | \frac{k_1k_2k_3}{(1+k_1)(1+k_2)(1+k_3)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 995 |
XLII OM - II - Problem 4
Find all monotonic functions $ f: \mathbb{R} \to \mathbb{R} $ that satisfy the equation | For any $ x > 0 $, we have the inequality $ f(4x) - f(3x) > 0 $, which implies that the (monotonic) function $ f $ is increasing in the interval $ (0;\infty) $. Similarly, for $ x < 0 $, the inequality $ f(3x) - f(4x) > 0 $ holds, so $ f $ is also increasing in $ (-\infty;0) $.
Replacing $ 4x $ with the letter $ z $, we rewrite the given equation as
Take any number $ x \ne 0 $ and any natural number $ n $. Applying the formula (1) $ n $ times, we obtain the equality
which is
We now want to pass from $ n $ to infinity. For $ x > 0 $, the sequence $ ((\frac{3}{4})^nx) $ is decreasing, so the sequence of values $ f((\frac{3}{4})^nx) $ is also decreasing, and moreover, bounded from below by the value $ f(0) $ (here we use the assumption of the monotonicity of the function $ f $ over the entire set $ \mathbf{R} $). Similarly, for $ x < 0 $, the sequence $ ((\frac{3}{4})^nx) $ is increasing, so the sequence $ (f(\frac{3}{4})^n x) $ is increasing, bounded from above by $ f(0) $. In either case, the sequence is convergent. Its limit may {\it a priori} depend on $ x $; let us denote its value by $ g(x) $:
Suppose that for some two positive numbers $ u $, $ v $, the inequality
holds.
Take a number $ \varepsilon $ such that
By the definition of the limit of a sequence,
and
We now find a natural number $ k $ large enough so that for $ n = k $, inequality (4) holds. Then we find a natural number $ m $ large enough so that for $ n = m $, inequality (5) holds, and moreover,
We then have (by (3))
This, however, contradicts inequality (6); we remember that the function $ f $ is increasing in $ (0;\infty) $.
The obtained contradiction proves that the inequality $ g(u) < g(v) $ is not possible for any pair of numbers $ u, v > 0 $. Hence, the value $ g(x) $ is the same for all $ x > 0 $. Let us denote this value by $ c $. Thus,
Similarly, we prove that the value of the limit $ g(x) $ is the same for all $ x < 0 $. Denoting this value by $ a $, we have
We can now take the limit in formula (2) (as $ n \to \infty $). We obtain
Given the requirement of the monotonicity of the function $ f $ on the set $ \mathbf{R} $, the inequality $ a \leq c $ must hold, and the value $ f(0) $ must lie between $ a $ and $ c $. Ultimately, then,
where $ a $, $ b $, $ c $ are any constants such that $ a \leq b \leq c $.
Verifying that indeed any function of this form satisfies the given functional equation is immediate. | proof | Algebra | proof | Yes | Yes | olympiads | false | 997 |
XLIV OM - II - Problem 3
On the edge $ |OA_1| $ of the tetrahedron $ |OA_1B_1C_1| $, points $ A_2 $, $ A_3 $ are chosen such that $ |OA_1| > |OA_2| > |OA_3| > O $. Let $ B_2 $, $ B_3 $ be points on the edge $ |OB_1| $, and $ C_2 $, $ C_3 $ be points on the edge $ |OC_1| $, such that the planes $ |A_1B_1C_1| $, $ |A_2B_2C_2| $, $ |A_3B_3C_3| $ are parallel. Denote by $ V_i $ ($ i=1,2,3 $) the volume of the tetrahedron $ |OA_iB_iC_i| $, and by $ V $ the volume of the tetrahedron $ |OA_1B_2C_3| $. Prove that $ V_1 + V_2 + V_3 \geq 3V $. | Let $ B_1, $ B_2 be the orthogonal projections of points $ B_1, $ B_2 onto the plane $ OA_1C_1 $, and $ C_1, $ C_3 be the projections of points $ C_1, $ C_3 onto the plane $ OA_1B_1 $ (Figure 7; it is convenient to imagine the tetrahedron each time so that the face we are projecting onto is its "horizontal" base). From the similarity of pairs of triangles $ OB_1B_1 $ and $ OB_2B_2 $ as well as $ OC_1C_1 $ and $ OC_3C_3 $, we obtain the proportions
from the similarity of pairs of triangles $ OA_1B_1 $ and $ OA_2B_2 $ as well as $ OA_1C_1 $ and $ OA_3C_3 $, we obtain the proportions
Let the lengths of segments $ OA_1 $, $ OA_2 $, $ OA_3 $ be denoted briefly by $ a_1 $, $ a_2 $, $ a_3 $. From the written proportions, we obtain the relationships
Consider the tetrahedra $ OA_1B_2C_3 $ and $ OA_1B_2C_1 $. They have a common face (the "base") $ OA_1B_2 $, so the ratio of their volumes equals the ratio of the heights dropped onto this face:
In turn, the tetrahedra $ OA_1B_2C_1 $ and $ OA_1B_1C_1 $ have a common face $ OA_1C_1 $, so (analogously) the ratio of their volumes equals the ratio of the corresponding heights:
Multiplying the relationships (1) and (2) (and using the notations $ V $, $ V_1 $ introduced in the problem statement), we get the equality
In deriving this formula, we did not use the assumption that $ a_1 > a_2 > a_3 $ anywhere. We can repeat the same reasoning, arbitrarily changing the roles of the indices $ 1 $, $ 2 $, $ 3 $. We obtain analogous proportions
Multiplying these three proportions side by side gives the equality $ V^3/(V_1V_2 V_3) = 1 $, or $ V^3 = V_1V_2V_3 $. From this, by the inequality between the geometric mean and the arithmetic mean,
| V_1+V_2+V_3\geq3V | Geometry | proof | Yes | Yes | olympiads | false | 1,000 |
XXXII - I - Problem 7
Let $ (x_n) $ be a sequence of natural numbers satisfying the conditions
a) $ x_1 < x_2 < x_3 < \ldots $,
b) there exists $ j \in \mathbb{N} $, $ j > 1 $, such that $ x_j = j $,
c) $ x_{kl} = x_kx_l $, for coprime $ k,l \in \mathbb{N} $.
Prove that $ x_n = n $ for every $ n $. | Since the number of natural numbers less than a given natural number $n$ is $n-1$, if for some $j$ we have $x_j = j$, then by condition a) it must be that $x_k = k$ for $k \leq j$. If the number $j$ mentioned in condition b) is greater than $2$, then the number $j-1$ is greater than $1$ and is relatively prime to the number $j$. At the same time, $x_{j-1} = j-1$, so by condition c) it is also $x_{(j-1)j} = (j-1)j$. Since $(j-1)^2 < (j-1)j$, we have $x_{(j-1)^2} = (j-1)^2$, and since the numbers $(j-1)^2$ and $j$ are relatively prime, we obtain by condition c) that $x_{(j-1)^2j} = (j-1)^2j$. Reasoning analogously, we get $x_{(j-1)^mj} = (j-1)^mj$ for $m = 1, 2, \ldots$.
Since for every natural number $n$ there exists an $m$ such that $n < (j-1)^mj$, by what we have established at the beginning of the solution, it is $x_n = n$ for every $n$.
The case where the $j$ mentioned in condition b) is equal to $2$ remains to be considered. Then $x_2 = 2$. Let $x_3 = 3 + q$. Therefore, $x_6 = x_2 \cdot x_3 = 2 \cdot (3 + q) = 6 + 2q$. Therefore, $x_5 < 6 + 2q$, so $x_5 \leq 5 + 2q$, and thus $x_{10} = x_2x_5 \leq 10 + 4q$, from which $x_9 \leq 9 + 4q$, and $x_{18} = x_2x_9 \leq 18 + 8q$. From the last inequality, it follows that $x_{15} \leq 15 + 8q$. On the other hand, $x_5 \geq x_3 + 2 = 5 + q$, so $x_{15} = x_3x_5 \geq (3 + q)(5 + q) = 15 + 8q + q^2$. From this, we get
from which it follows that $q = 0$. Therefore, $x_3 = 3$ and the problem has been reduced to the case already considered above. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,001 |
XXVI - I - Problem 4
Given are pairwise disjoint spheres $ K_1, K_2, K_3 $ with pairwise distinct radii. Let $ A_{ij} $ be the vertex of the cone circumscribed around the spheres $ K_i $ and $ K_j $. Prove that the points $ A_{12} $, $ A_{23} $, $ A_{31} $ are collinear. | If $ A $ is the vertex of a cone circumscribed around spheres $ K $ and $ L $ of different radii, then there exists exactly one such homothety $ j $ such that $ j(K) = L $. Its center is the point $ A $, and the coefficient is the ratio of the radii of the spheres $ L $ and $ K $.
Without loss of generality, we can assume that the radii of the spheres $ K_1 $, $ K_2 $, $ K_3 $ form an increasing sequence. Let $ s $ and $ t $ be such homotheties that $ s(K_1) = K_2 $ and $ t(K_2) = K_3 $. Then $ ts(K_1) = K_3 $. To solve the problem, it is sufficient to prove that the composition $ ts $ of homotheties $ t $ and $ s $ with coefficients greater than $ 1 $ is a homothety and the centers of homotheties $ t $, $ s $, and $ ts $ are collinear.
Let us choose a coordinate system such that the center of homothety $ s $ is the point $ (0, 0, 0) $, and the center of homothety $ t $ is the point $ (a, 0, 0) $.
Then $ s(x, y, z) = (kx, ky, kz) $, $ t(x, y, z) = (mx - am + a, my, mz) $, where $ k, m > 1 $. We calculate that $ ts(x,y, z) = t(kx, ky, kz) = (mkx- am+a, kmy, kmz) $.
Therefore, $ ts $ is a homothety with the center $ \displaystyle\left(\frac{a(m-1)}{mk-1}, 0, 0\right) $. The centers of homotheties $ s $, $ t $, and $ ts $ are thus collinear. | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,002 |
LVII OM - I - Problem 1
Determine all non-negative integers $ n $ for which the number
$ 2^n +105 $ is a perfect square of an integer. | If $ n $ is an odd number, then the number $ 2^n +105 $ gives a remainder of 2 when divided by 3. Since a number that is the square of an integer cannot give a remainder of 2 when divided by 3, the number $ n $ satisfying the conditions of the problem must be even.
Let's assume that $ n =2k $ for some non-negative integer $ k $. We want to solve the equation $ 2^{2k} +105 = m^2 $ in non-negative integers $ k $ and $ m $. We rewrite this equation as $ (m-2^k)(m+2^k) = 105 $.
From the obtained equation, it follows that $ m-2^k >0 $. Of course, the inequality $ m+2^k > m-2^k $ is also satisfied. Since $ 105=3\cdot 5\cdot 7 $, finding non-negative integers $ m $ and $ k $ that satisfy the relation $ (m-2^k)(m+2^k)=105 $ reduces to solving four systems of equations:
By subtracting the first equation from the second, we conclude that the number $ 2 \cdot 2^k =2^{k+1} $ equals 8, 16, 32, or 104. From this, we obtain three possible values for $ k $; they are: 2, 3, or 4. Thus, the possible values of the number $ n $ are 4, 6, or 8.
We directly check that for the three obtained values of $ n $, the number $ 2^n +105 $ is the square of an integer:
直接检查结果如下：
- 当 $ n = 4 $ 时, $ 2^4 + 105 = 16 + 105 = 121 = 11^2 $.
- 当 $ n = 6 $ 时, $ 2^6 + 105 = 64 + 105 = 169 = 13^2 $.
- 当 $ n = 8 $ 时, $ 2^8 + 105 = 256 + 105 = 361 = 19^2 $.
Directly checking the results:
- For $ n = 4 $, $ 2^4 + 105 = 16 + 105 = 121 = 11^2 $.
- For $ n = 6 $, $ 2^6 + 105 = 64 + 105 = 169 = 13^2 $.
- For $ n = 8 $, $ 2^8 + 105 = 256 + 105 = 361 = 19^2 $. | 4,6,8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 1,008 |
LVI OM - I - Problem 11
A circle with center $ I $ is inscribed in a convex quadrilateral $ ABCD $, with the point $ I $ not lying on the line $ AC $. The diagonals $ AC $ and $ BD $ intersect at point $ E $. The line passing through point $ E $ and perpendicular to the line $ BD $ intersects the lines $ AI $, $ CI $ at points $ P $, $ Q $, respectively. Prove that $ PE = EQ $. | Let $ o_1 $ be a circle with center $ P $ and tangent to the lines $ AD $ and $ AB $ at points $ S $ and $ T $, respectively. Similarly, let $ o_2 $ be a circle with center $ Q $ and tangent to the lines $ CB $ and $ CD $ at points $ U $ and $ W $, respectively (Fig. 5). Let $ o $ be the incircle of quadrilateral $ ABCD $. Without loss of generality, assume that point $ Q $ lies on segment $ CI $.
om56_1r_img_5.jpg
Using Menelaus' theorem for triangle $ PIQ $, we obtain
where $ r $, $ r_1 $, and $ r_2 $ are the radii of circles $ o $, $ o_1 $, and $ o_2 $, respectively. Hence,
Let us introduce the following notation: $ a=BT $, $ b=DS $, $ c=BU $, $ d=DW $. Then we obtain the relationships:
If $ a=b $, then $ AB =AD $ and $ CB =CD $; in this case, point $ I $ lies on diagonal $ AC $, which contradicts the assumptions. Therefore, $ a \neq b $, and thus $ c \neq d $. From equation (2), we then have $ a + b = c + d $, which, combined with the relationship $ a-b = c-d $, gives $ a = c $ and $ b = d $. Therefore,
Equations (1) and (3) imply the relationship $ (\lambda^{2}-1)(r_{1}^{2}-r_{2}^{2})=0 $.
Suppose that $ \lambda = 1 $. Then $ EQ = r_2 $, so circle $ o_2 $ is inscribed in triangle $ BCD $. Since a circle can be inscribed in quadrilateral $ ABCD $, the circle $ \omega $ inscribed in triangle $ ABD $ is tangent to circle $ o_2 $. This implies that points $ Q $, $ E $, and the center of circle $ \omega $ are collinear. However, this is not possible, as these points lie on the sides $ IC $, $ CA $, and $ AI $ of triangle $ AIC $, respectively, and are different from its vertices. We have reached a contradiction, which proves that $ \lambda\neq 1 $. From the equation obtained above, it follows that $ r_1 = r_2 $, i.e., $ PE = EQ $. | PE=EQ | Geometry | proof | Yes | Yes | olympiads | false | 1,009 |
XVI OM - II - Task 4
Find all prime numbers $ p $ such that $ 4p^2 +1 $ and $ 6p^2 + 1 $ are also prime numbers. | To solve the problem, we will investigate the divisibility of the numbers \( u = 4p^2 + 1 \) and \( v = 6p^2 + 1 \) by \( 5 \). It is known that the remainder of the division of the product of two integers by a natural number is equal to the remainder of the division of the product of their remainders by that number. Based on this, we can easily find the following:
\begin{center}
When \( p \equiv 0 \pmod{5} \), then \( u \equiv 1 \pmod{5} \), \( v \equiv 1 \pmod{5} \);\\
When \( p \equiv 1 \pmod{5} \), then \( u \equiv 0 \pmod{5} \), \( v \equiv 2 \pmod{5} \);\\
When \( p \equiv 2 \pmod{5} \), then \( u \equiv 2 \pmod{5} \), \( v \equiv 0 \pmod{5} \);\\
When \( p \equiv 3 \pmod{5} \), then \( u \equiv 2 \pmod{5} \), \( v \equiv 0 \pmod{5} \);\\
When \( p \equiv 4 \pmod{5} \), then \( u \equiv 0 \pmod{5} \), \( v \equiv 2 \pmod{5} \).\\
\end{center}
From the above, it follows that the numbers \( u \) and \( v \) can both be prime numbers only when \( p \equiv 0 \pmod{5} \), i.e., when \( p = 5 \), since \( p \) must be a prime number. In this case, \( u = 4 \cdot 5^2 + 1 = 101 \), \( v = 6 \cdot 5^2 + 1 = 151 \), so they are indeed prime numbers.
Therefore, the only solution to the problem is \( p = 5 \). | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 1,010 |
X OM - I - Task 4
Prove that if point $ P $ moves in the plane of triangle $ ABC $, then the triangle $ S_1S_2S_3 $, whose vertices are the centroids $ S_1 $, $ S_2 $ and $ S_3 $ of triangles $ PBC $, $ PCA $ and $ PAB $ does not change its shape or size. | If the vertex $ P $ of the triangle $ PAB $ is moved to the point $ P' $ (Fig. 6a), then the centroid $ S_3 $ of this triangle will move to the point $ S $.
The vector $ S_3S $ has the same direction as the vector $ PP' $ and is three times shorter, since these vectors are similar (directly) relative to the point $ N $ in a scale of $ \frac{1}{3} $. Each of the points $ S_1 $ and $ S_2 $ will also move by the same vector, and thus the entire triangle $ S_1S_2S_3 $ will move. It turns out that when the position of the point $ P $ changes, the triangle $ S_1S_2S_3 $ undergoes a parallel shift, and thus does not change in size or shape.
Note. The problem can be generalized by replacing the triangle $ ABC $ with any polygon. | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,012 |
LI OM - I - Task 7
Prove that for any positive integer $ n $ and any number $ t\in(\frac{1}{2},1) $ there exist such numbers $ a,b \in (1999,2000) $, that | Let $ c = 3999/2 $. We seek numbers $ a $, $ b $ in the form $ a = c+x $, $ b = c-x $, where $ x \in (0, \frac{1}{2}) $. Then
where $ s = 2t - 1 > 0 $. After expanding using the binomial theorem, the last expression takes the form
where $ p(x) $ is some polynomial.
We want to show that for some number $ y \in (0,\frac{1}{2}) $, the number $ w(y) $ is positive. From equation (1), it follows that
Therefore, there exists a number $ y \in (0,\frac{1}{2}) $ such that the expression $ w(y)/y $, and consequently the expression $ w(y) $, has a positive value. This completes the proof. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,015 |
VII OM - III - Task 3
On a straight line, there are three different points $ M $, $ D $, $ H $. Construct a right-angled triangle for which $ M $ is the midpoint of the hypotenuse, $ D $ - the intersection point of the angle bisector of the right angle with the hypotenuse, and $ H $ - the foot of the altitude on the hypotenuse. | Let $ABC$ be the sought right-angled triangle. Since points $M$, $D$, $H$ are distinct, the legs $AC$ and $BC$ are not equal. Let, for example, $AC > BC$. Then point $D$ lies between points $M$ and $B$, and point $H$ lies between points $D$ and $B$, so point $D$ lies between points $M$ and $H$ (Fig. 19).
Since
and
it follows that $CD$ is the angle bisector of the angle at vertex $C$ of triangle $MCH$. Therefore, $MD > DH$, since $MD \colon DH = MC \colon CH$, and $MC > CH$.
The problem has a solution only if the given points $M$, $D$, $H$ are positioned such that $D$ lies between $M$ and $H$ and $MD > CH$. These conditions are also sufficient. The vertex $C$ can be found as the intersection of the Apollonian circle for segment $MH$ and the ratio $MD \colon DH$ with the perpendicular to the line $MH$ at point $H$, which lies inside that circle. The other vertices of the triangle can be found by measuring segments $MA = MC$ and $MB = MC$ on the line $MH$ on both sides of point $M$. The resulting triangle $ABC$ satisfies the conditions of the problem. It is a right-angled triangle because point $C$ lies on the circle with diameter $AB$, $CM$ is its median, $CH$ is its altitude; finally, $CD$ is the angle bisector of angle $C$, since $\measuredangle ACD = \measuredangle ACM + \measuredangle MCD = \measuredangle A + \measuredangle MCD = \measuredangle BCH + \measuredangle HCD = \measuredangle BCD$.
The problem has two solutions that are symmetric with respect to the given line $MH$. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,016 |
XXXVIII OM - II - Problem 5
Determine all prime numbers $ p $ and natural numbers $ x, y $, for which $ p^x - y^3 = 1 $. | Suppose that the numbers $ p $, $ x $, $ y $ satisfy the given conditions. Then
Let's consider two cases:
1. $ y = 1 $. Then (1) reduces to the equation $ p^x = 2 $, from which $ p = 2 $, $ x = 1 $.
2. $ y > 1 $. Then both factors of the product on the right side of (1) are numbers greater than $ 1 $ and as divisors of the power of the prime number $ p $, they must themselves be positive powers of this prime number:
Now notice that
Thus, $ 3 $ is divisible by $ p $, which means that $ p = 3 $. Equation (3) takes the form
which is
The right side of the last equation is not divisible by $ 3 $. Hence $ l - 1 = 0 $ and $ 2k-1=k $, so $ k = l = 1 $ and by the relation (2) $ x = 2 $, $ y = 2 $.
In summary: the solutions $ (p, x, y) $ of the given equation can only be the triples $ (2, 1, 1) $ and $ (3, 2, 2) $. Verifying that these are indeed solutions is immediate. | (2,1,1)(3,2,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 1,017 |
XLII OM - III - Problem 4
In the plane with a Cartesian coordinate system, we consider the set $ V $ of all free vectors, both of whose coordinates are integers. Determine all functions $ f $, defined on the set $ V $, with real values, satisfying the conditions:
(a) $ f(v) = 1 $ for each of the four vectors $ v\in V $ of length $ 1 $;
(b) $ f(v+w) = f(v) + f(w) $ for every pair of perpendicular vectors $ v, w \in V $.
Note: The zero vector is perpendicular to every vector. | Suppose that the function $ f $ satisfies the given conditions. Using these conditions, let's list the values of $ f(\overrightarrow{\mathbf{v}}) $ for a few vectors with small integer coordinates:
(When calculating the last of the found values, we used the orthogonality of the vectors $ [3,0] $ and $ [0,1] $ and the resulting equality $ f([3,0]) = f([3,1])-f([0,1]) $ from condition (b).)
From this table, we see that $ f([k,0])=k^2 $ for $ k = 1,2,3 $. This leads to the hypothesis that for all natural numbers $ k $, the equality $ f([k,0]) = k^2 $ should hold, and also - by symmetry - $ f([0,k]) = k^2 $, $ f([-k,0]) = k^2 $, $ f([0,-k]) = k^2 $, and if so -
for the vector $ [x,y] $ is the sum of the orthogonal vectors $ [x,0] $ and $ [0,y] $.
We will now prove the correctness of hypothesis (1).
The zero vector $ \overrightarrow{\mathbf{0}} = [0,0] $ is orthogonal to every vector, including itself. Therefore, $ f(\overrightarrow{\mathbf{0}}) = f(\overrightarrow{\mathbf{0}}+\overrightarrow{\mathbf{0}}) = f(\overrightarrow{\mathbf{0}}) + f(\overrightarrow{\mathbf{0}}) $, which means $ f(\overrightarrow{\mathbf{0}}) = 0 $. Thus, equality (1) holds for $ x = y = 0 $.
We will prove by induction on $ n $ that
For $ n = 1 $, this is true; this can be seen from the table.
Assume the statement (2) is true for some natural number $ n \geq 1 $. In particular, we then have
The four vectors appearing in these equalities are pairwise orthogonal. Therefore, according to condition (b),
At the same time, we have the equality
Thus, using the first of the formulas (3) and the induction hypothesis, we get:
Similarly, using the remaining three formulas (3), we get the equalities
We have thus shown that
for any integer $ k $ whose absolute value does not exceed $ n + 1 $. (For $ |k| = n+1 $, these are the formulas obtained just now; and for $ |k| \leq n $, this is a consequence of the induction hypothesis.) If $ [x,y] $ is any vector with integer coordinates whose absolute values do not exceed $ n+1 $, then
This proves statement (2) with $ n $ replaced by $ n+1 $.
By the principle of induction, (2) holds for every natural number $ n \geq 1 $.
Therefore, the only function satisfying the given conditions is the function $ f $ defined by formula (1). | f([x,y])=x^2+y^2 | Algebra | proof | Yes | Yes | olympiads | false | 1,023 |
XII OM - II - Problem 5
Prove that if the real numbers $ a $, $ b $, $ c $ satisfy the inequalities
[ (1) \qquad a + b + c > 0, \]
[ (2) \qquad ab + bc + ca > 0, \]
[ (3) \qquad abc > 0, \]
then $ a > 0 $, $ b > 0 $, $ c > 0 $. | From inequalities (1) and (2), it can be inferred that at least two of the numbers $a$, $b$, $c$ are positive. For from inequality (1) it follows first that at least one of these numbers is positive, let's say $c > 0$. Indeed, for any $a$ and $b$
and from inequality (2) it follows that
therefore
transferring terms to one side, we obtain from this
since $c > 0$, then $a + b + 2c = (a + b + c) + c > 0$, hence from the above inequality it follows that
In this case, at least one of the numbers $a$ and $b$, for example $b$, is positive.
If, however, two of the numbers $a$, $b$, $c$ are positive, then by inequality (3) the third one must also be positive.
Note. The theorem proved above is a special case of the theorem: If for certain values of $a_1, a_2, \ldots, a_n$ all the elementary symmetric functions (see problem 10) have positive values, then all the values $a_1, a_2, \ldots, a_n$ are positive. | proof | Algebra | proof | Yes | Yes | olympiads | false | 1,025 |
XXIX OM - II - Problem 3
Given a sequence of natural numbers $ (a_i) $, such that for every natural number $ n $, the sum of the terms of the sequence that are not greater than $ n $ is not less than $ n $. Prove that for every natural number $ k $, one can choose a finite subsequence from $ (a_i) $ whose sum of terms equals $ k $. | We will apply induction with respect to $k$. Consider the case $k = 1$. By assumption, the sum of those terms of the given sequence that are not greater than $1$ (and thus are equal to $1$) is not less than $1$. Therefore, there exists a term in the given sequence that is equal to $1$. Hence, the sum of the terms of a one-term sequence composed of this very term is equal to $1$.
Next, let $k$ be a fixed natural number and assume that for every natural number not greater than $k$, one can select a finite sequence from the given sequence whose sum of terms is equal to $i$. We will prove that from the given sequence, one can select a finite sequence whose sum of terms is equal to $k + 1$.
By the inductive hypothesis, from the given sequence, one can select a finite sequence whose sum of terms is equal to $k$. Let $a_m$ be the smallest term of the given sequence that does not appear in sequence (1).
Since, by the conditions of the problem, the sum of all terms of the given sequence that do not exceed $k + 1$ is not less than $k + 1$, and the sum of the terms of sequence (1) is equal to $k$, some term of the given sequence not greater than $k + 1$ does not appear in sequence (1). Therefore, $a_m \leq k+1$.
If $a_m = 1$, then the sum of the terms of the sequence $(a_{i_1}, a_{i_2}, \ldots, a_{i_r}, a_m)$ is equal to $k + 1$.
If $a_m > 1$, then $1 \leq a_m - 1 \leq k$. Therefore, by the inductive hypothesis, the number $a_m - 1$ is the sum of some terms of the given sequence. All these terms appear in sequence (1), because the smallest term of the given sequence not appearing in (1) is $a_m$. Let
where $s \leq r$. Then the sum of the terms of the sequence
is equal to
By the principle of induction, for every natural number $k$, one can select from the given sequence a finite sequence whose sum of terms is equal to $k$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,026 |
XLVII OM - II - Problem 6
Inside a parallelepiped, whose edges have lengths $ a $, $ b $, $ c $, there is a point $ P $. Prove that there exists a vertex of the parallelepiped whose distance from point $ P $ does not exceed $ \frac{1}{2}\sqrt{a^2 + b^2 + c^2} $. | The walls of the parallelepiped define six planes. Let $ \pi $ be the plane whose distance from point $ P $ is the smallest; if there are two such planes (or more), we choose any one of them and denote it by $ \pi $. Let $ ABCD $ be the face of the parallelepiped contained in the plane $ \pi $ and let $ N $ be the orthogonal projection of point $ P $ onto this plane.
Assume that the four edges of the parallelepiped, connecting the vertices $ A $, $ B $, $ C $, $ D $ with the vertices of the face $ A $ parallel to $ ABCD $, have length $ c $. (We do not lose generality by this, as we can change the notation of the edge lengths if necessary.)
Point $ P $ is no less distant from the face $ A $ than from the face $ ABCD $. Therefore, $ |NP| \leq \frac{1}{2} c $.
We will show that point $ N $ lies within the parallelogram $ ABCD $. Suppose otherwise. Then the segment $ PN $ intersects the edge of the parallelepiped at some point $ X $, belonging to some other face. The distance from point $ P $ to the plane of this face does not exceed the length of segment $ PX $, and thus is less than the length of segment $ PN $, i.e., the distance from point $ P $ to the plane $ \pi $. This, however, contradicts the choice of the plane $ \pi $; the contradiction proves the falsity of the made assumption.
Thus, $ N $ is a point of the parallelogram $ ABCD $, whose sides have lengths $ a $ and $ b $. We repeat the reasoning: among the lines $ AB $, $ BC $, $ CD $, $ DA $, we choose the one whose distance from point $ N $ is the smallest; if there is more than one such line, we choose any one of them.
Assume that this is the line $ AB $ and that $ |AB| = a $ (we change the notation if necessary). Let $ K $ be the orthogonal projection of point $ N $ onto the line $ AB $. Reasoning similarly as before, we notice that $ |KN| \leq \frac{1}{2}b $ and that $ K $ is a point on the segment $ AB $; if the point $ K $ were on the line $ AB $ outside the segment $ AB $, then the segment $ NK $ would intersect another side of the parallelogram $ ABCD $, and thus the distance from point $ N $ to this side would be less than the distance from $ N $ to the line $ AB $ - contrary to the choice of this line.
We can finally assume, without loss of generality, that point $ K $ lies on the segment $ AB $ at a distance no greater from the end $ A $ than from $ B $. Therefore, $ |AK| \leq \frac{1}{2} a $.
The directions of the lines $ AK $, $ KN $, $ NP $ are pairwise perpendicular. The segments $ AK $, $ KN $, $ NP $ are thus three perpendicular edges of a rectangular parallelepiped, whose internal diagonal is the segment $ AP $. The lengths of these segments do not exceed (respectively) the numbers $ \frac{1}{2}a $, $ \frac{1}{2}b $, $ \frac{1}{2}c $. Therefore, ultimately,
point $ A $ is the vertex whose existence needed to be demonstrated. | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,028 |
XXIX OM - I - Zadanie 2
Dla danej liczby naturalnej $ k>7 $ skonstruować ciąg $ S = (a_0, a_1, \ldots, a_{k-1}) $ o tej własności, że dla każdego $ i $ ($ 0 < i < k - 1 $) $ a_i $ równe liczbie wyrazów ciągu $ S $ równych $ i $.
|
Dla dowolnej liczby naturalnej $ k $ znajdziemy wszystkie ciągi $ S $ spełniające warunki zadania.
Z warunków zadania wynika, że w ciągu $ S $ jest $ a_0 $ wyrazów równych $ 0 $, $ a_1 $ wyrazów równych $ 1 $, $ \ldots $ , $ a_{k-1} $ wyrazów równych $ k - 1 $ i innych wyrazów nie ma. Zatem liczba wszystkich wyrazów ciągu $ S $ jest równa $ a_0 + a_1 + \ldots + a_{k-1} $ i stąd
Z określenia ciągu $ S $ wynika, że $ a_0 \ne 0 $. Niech $ a_{i_1}, a_{i_2}, \ldots, a_{i_s} $, gdzie $ 0 = i_1 < i_2 < \ldots < i_s $ będą wszystkimi różnymi od zera wyrazami ciągu $ S $. Wtedy $ k - s $ wyrazów tego ciągu jest równych zero, a więc $ a_{i_0} = a_0 = k - s $ i $ a_{k-1} \geq 1 $.
Z (1) otrzymujemy, że $ a_{i_1} + a_{i_2} + \ldots + a_{i_s} = k $ i wobec tego
Ponieważ liczby $ a_{i_2}, \ldots, a_{i_s} $ są naturalne i jest ich $ s - 1 $, więc z (2) wynika, ze $ s - 2 $ z tych liczb jest równych $ 1 $, a jedna jest równa $ 2 $.
1). Jeżeli $ a_0 = 1 $, to w ciągu $ S $ jeden wyraz jest równy $ 0 $, $ s - 2 $
wyrazy są równe $ 1 $ i jeden wyraz równy jest $ 2 $. Wobec tego $ a_2 = 1 $ i $ a_j = 0 $ dla $ j > 2 $. Wynika stąd, że $ a_1 = 2 $ (ponieważ tylko ten wyraz może być różny od $ 0 $ i $ 1 $) oraz $ a_3 = 0 $ i $ k - 1 = 3 $ (ponieważ tylko jeden wyraz ciągu $ S $ jest równy $ 0 $), Zatem $ S = (1,2,1,0) $.
Sprawdzamy bez trudu, że ten ciąg $ S $ spełnia warunki zadania.
2). Jeżeli $ a_0 = 2 $, to w ciągu $ S $ dwa wyrazy są równe $ 0 $, $ s - 2 $ wyrazy są równe $ 1 $ i dwa wyrazy są równe $ 2 $, wliczając $ a_0 $. Wobec tego $ a_2 = 2 $ i $ a_j = 0 $ dla $ j > 2 $, a także $ a_1 \leq 1 $.
Jeżeli $ a_1 = 0 $, to również $ a_3 = 0 $ i $ k - 1 = 3 $ (ponieważ nie ma wyrazów równych $ 1 $ i są dwa wyrazy równe $ 0 $). Otrzymujemy ciąg $ S = (2, 0, 2, 0) $, który oczywiście spełnia warunki zadania.
Jeżeli zaś $ a_1 = 1 $, to $ a_3 = a_4 = 0 $ i $ k - 1 =4 $ (ponieważ są dwa wyrazy równe $ 0 $). Otrzymujemy w tym przypadku ciąg $ S = (2, 1, 2, 0, 0) $ który również spełnia warunki zadania.
3). Jeżeli $ a_0 > 2 $, to w ciągu $ S $ jest $ a_0 = k - s $ wyrazów równych $ 0 $, $ s- 2 $ wyrazy równe $ 1 $ i jeden wyraz jest równy $ 2 $. Wobec tego $ a_2 = 1 $, $ a_{k-s} = 1 $ (i przy tym $ k - s > 2 $) oraz $ a_j = 0 $ dla $ j > 2 $, $ j \ne k - s $. Wynika stąd, że $ a_1 \geq 2 $. Zatem w ciągu $ S $ dokładnie dwa wyrazy są równe $ 1 $, a więc $ a_1 = 2 $ i $ s - 2 = 2 $, czyli $ s = 4 $.
Otrzymujemy więc ciąg $ S $, w którym $ a_0 = k - 4 $, $ a_1 = 2 $, $ a_2 = 1 $, $ a_{k-4} = 1 $, a pozostałe wyrazy są równe zeru. Przy tym $ k - 4 > 2 $, tzn. $ k \geq 7 $. Sprawdzamy bez trudu, że na odwrót, dla każdej liczby naturalnej $ k \geq 7 $ określony wyżej ciąg $ S $ spełnia warunki zadania.
Uwaga. Rozwiązaliśmy zadanie ogólniejsze. Udowodniliśmy, że dla $ k \geq 7 $ istnieje dokładnie jeden ciąg $ S $ spełniający warunki zadania (i skonstruowaliśmy taki ciąg), ponadto dla $ k = 4 $ istnieją dwa takie ciągi $ (2, 0, 2, 0) $ i $ (1, 2, 1, 0) $, a dla $ k = 5 $ jeden ciąg $ (2, 1, 2, 0, 0) $. Dla $ k = 2 $, $ 3 $ i $ 6 $ takich ciągów $ S $ nie ma.
| proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1,029 |
XXVI - I - Problem 5
Determine all integers $ m $ for which the polynomial $ x^3-mx^2+mx-(m^2+1) $ has an integer root. | Suppose that the integers $ x $ and $ m $ satisfy the equation $ x^3 - mx^2+mx-(m^2+1) = 0 $, which is $ (x^2+m)(x-m) = 1 $. Then $ x^2+m = 1 $ and $ x- m = 1 $ or $ x^2+m = -1 $ and $ x-m = -1 $. Adding the equations in each of these systems, we get $ x^2+x = 2 $ or $ x^2+x = - 2 $. The roots of the first equation are $ x = 1 $ and $ x = -2 $. They correspond to the numbers $ m = 0 $ and $ m = - 3 $. The second equation has no solutions in the set of integers. | =0=-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 1,032 |
XLIV OM - I - Problem 12
Prove that the polynomial $ x^n + 4 $ is a product of two polynomials of lower degree with integer coefficients if and only if $ n $ is divisible by $ 4 $. | \spos{I} Suppose that the polynomial $ x^4 + 4 $ is the product of two polynomials,
with the properties under consideration:
In the product $ F(x)G(x) $, the coefficient of $ x^n $ is $ a_kb_m $, and the constant term equals $ a_0b_0 $. Therefore, the following equations hold:
The numbers $ a_k $, $ b_m $ are integers; thus $ a_k = b_m = 1 $ or $ a_k = b_m = -1 $. Additionally, let us assume:
Let $ \alpha $ be the smallest index such that $ a_\alpha $ is an odd number, and let $ \beta $ be the smallest index such that $ b_\beta $ is an odd number. Since $ a_k = b_m = \pm 1 $, it follows that $ \alpha \leq k $, $ \beta \leq m $. The coefficient of $ x^{\alpha+\beta} $ in the product $ F(x)G(x) $ is equal to
(due to the assumption (2), all symbols make sense even when $ \alpha+ \beta $ exceeds $ k $ or $ m $). In the sum (3), the term $ a_\alpha b_\beta $ is an odd number; each of the remaining terms is an even number (since it is a product of the form $ a_ib_j $, where $ i < \alpha $ or $ j < \beta $, so at least one of the factors $ a_i $, $ b_j $ is even). Therefore, the entire sum (3) is odd.
In the polynomial (1), only $ x^n $ has an odd coefficient. Hence, $ \alpha + \beta = n $, which means the equations $ \alpha = k $ and $ \beta = m $ must hold. According to the definitions of $ \alpha $ and $ \beta $, we obtain the conclusion:
Since $ a_0b_0 = 4 $, it follows that $ a_0 = b_0 = 2 $ or $ a_0 = b_0 = -2 $.
We will show that $ k = m $. Suppose that $ k \neq m $; let, for example, $ k < m $ (such a restriction does not reduce the generality of the considerations due to the symmetry of the roles of $ k $ and $ m $). We write the product of the polynomials $ F(x) $ and $ G(x) $ in the form
where
Given the property (4), all coefficients of the polynomial $ P(x) $ are divisible by $ 4 $. The coefficient of $ x^k $ in the polynomial $ Q(x) $ is $ a_kb_0 $. We previously stated that $ a_k = \pm 1 $ and $ b_0 = \pm 2 $. Therefore, the coefficient of $ x^k $ in the polynomial $ P(x) + Q(x) $ is an even number not divisible by $ 4 $. This is a contradiction, because according to the equalities (1) and (5), the sum $ P(x) + Q(x) $ is a polynomial with a constant value of $ 4 $.
This contradiction proves that $ k = m $; hence, $ n = k + m = 2k $, and consequently
From this, it follows that neither of the polynomials $ F(x) $, $ G(x) $ has real roots, so they must be of even degree. This means that $ k $ is an even number, and thus $ n $ is divisible by $ 4 $.
Conversely, if $ n = 4l $, $ l \in \mathbb{N} $, then the desired factorization exists and has the form
直接输出翻译结果。 | proof | Algebra | proof | Yes | Yes | olympiads | false | 1,033 |
XIII OM - I - Problem 5
Prove that all powers of a number whose last eight digits are $12890625$ also end with the digits $12890625$. | We need to prove that if $ n $ and $ l $ are integers, with $ n > 0 $, $ l \geq 0 $, then
where $ m $ is a non-negative integer.
First, observe that in the expansion of the left side of the above equality according to the binomial theorem for the power of a binomial, all terms except for $ (12890625)^n $ have a factor of $ 108 $. We can move these terms to the right side and reduce the problem to the following form:
Prove that for every natural number $ n $
where $ k_n $ is an integer, of course dependent on $ n $.
When $ n = 1 $, formula (1) is true, $ k_1 = 0 $.
We will prove that formula (1) is true for $ n = 2 $; to do this, we need to calculate the square of the number $ 12890625 $; however, we do not need to determine all the digits of this square, but only the last $ 8 $ digits. We can, for example, use the fact that the square of the sum of several numbers equals the sum of the squares of these numbers and twice the products of all pairs of these numbers and calculate in this way:
If formula (1) is true for some natural number $ n $, then it is also true for $ n + 1 $, since
where $ k_{n+1} = k_2 + 12890625 k_n $.
Based on the principle of mathematical induction, we conclude from this that formula (1) is true for every natural number $ n $.
Note. The interesting property of the number $ 12890625 $, which we have learned, suggests the question of whether there are more numbers with the same or fewer digits that have this property. Thus, the problem is to find a $ k $-digit number $ A $ such that the last $ k $ digits of each power of this number form the number $ A $.
Just as in the previous solution, we state that if the square of the number $ A $ satisfies the condition of the problem, then each power of this number satisfies it. Therefore, the problem ultimately reduces to finding a $ k $-digit number $ A $ such that
where $ m $ is an integer.
When $ k=1 $, the solution is immediate; the numbers sought are $ 0 $, $ 1 $, $ 5 $, and $ 6 $. Therefore, we will assume in the following that $ k>1 $.
Suppose that equality (2) holds for some natural $ k $-digit number $ A $. We can write this equality in the form
The numbers $ A $ and $ A-1 $ are relatively prime, since any common divisor of these numbers is also a divisor of their difference, which is $ 1 $, so the only such divisor is $ 1 $ or $ -1 $. Since $ A < 10^k $, one of these numbers is divisible by $ 2^k $, and the other by $ 5^k $, i.e., there are only two cases:
a) $ A = 2^k \cdot x $ and $ A-1 = 5^k \cdot y $
b) $ A = 5^k \cdot z $ and $ A-1 = 2^k \cdot u $
In case a)
and in case b)
Conversely, if a pair of natural numbers $ (x, y) $ satisfies equation (4), then the number $ A = 2^k \cdot x $ satisfies condition a), and therefore also condition (2). To be a solution to the problem, this number must also have $ k $ digits, i.e., it must satisfy the condition
Similarly, if a pair of natural numbers $ (z, u) $ satisfies equation (5), then the number $ A = 5^k \cdot z $ is a solution to the problem if it satisfies the condition
To investigate whether equations (4) and (5) have such solutions, we will consider the general equation
where $ a $ and $ b $ are integers greater than $ 1 $ and relatively prime.
We will prove that there exists one and only one pair $ (x_0, y_0) $ of positive integers satisfying this equation and the condition $ x_0 < b $. Indeed, consider the remainders of the division of the positive numbers $ ax_k - 1 $ by $ b $, where $ x_k $ takes the values $ 1, 2, \ldots, b $. Among these remainders, there are no equal ones; if the equalities
were to hold, where $ x_1 $ and $ x_2 $ are two different numbers among the numbers $ 1, 2, \ldots, b $ and $ q_1 $ and $ q_2 $ are integers, then the equality
would hold, so the number $ a (x_1 - x_2) $ would be divisible by $ b $, which is impossible since $ a $ and $ b $ are relatively prime and the difference $ x_1 - x_2 $ is less in absolute value than $ b $. One of these $ b $ remainders must therefore be equal to $ 0 $; let the corresponding value of $ x_k $ be $ x_0 $. The number $ x_0 $ cannot be equal to $ b $, since $ ab - 1 $ is not divisible by $ b $, so $ x_0 < b $.
The positive integer $ \frac{ax_0 - 1}{b} $ will be denoted by $ y_0 $, so
If $ x $ and $ y $ denote any integers satisfying equation (8), then by subtracting equations (8) and (9) we obtain
Since $ a $ and $ b $ are relatively prime, $ x - x_0 $ is divisible by $ b $, so $ x - x_0 = b \cdot t $, where $ t $ is an integer, and $ y - y_0 = at $, from which
All solutions of equation (8) in integers $ x $, $ y $ are thus obtained from formulas (10) by substituting integers for $ t $. These numbers are positive and satisfy the condition $ x < b $ if and only if $ t = 0 $, i.e., if $ x = x_0 $, $ y = y_0 $.
From the above, it follows that there exists one and only one pair of integers $ (x_0, y_0) $ satisfying equation (4) and the conditions $ x_0 > 0 $, $ y_0 > 0 $, $ x_0 < 5^k $, so $ A = 2^k \cdot x_0 < 10^k $; the number $ A $ is thus at most $ k $-digit. However, it may happen that $ A $ has fewer than $ k $ digits. We agree that in such a case we will prepend the appropriate number of zeros to the left of $ A $ instead of the missing digits and consider that we have obtained a $ k $-digit solution to the problem. With this agreement, the problem has for every $ k > 1 $ one solution of the form $ A = 2^k \cdot x $ ($ x - 1 $ natural). Similarly, equation (5) leads to a second solution of the form $ A = 5^k \cdot z $ ($ z - 1 $ natural).
For example, for $ k = 3 $ we have solutions $ A_1 = 625 $ and $ A_2 = 376 $, for $ k = 4 $ the solutions are $ A_1 = 0625 $ and $ A_2 = 9376 $, for $ k = 9 $ one solution is $ A_1 = 212890625 $; let the reader find the second solution themselves. To find it without lengthy calculations, one needs to familiarize themselves with the theory of linear Diophantine equations in two variables; the reader will find it, for example, in the books of Prof. W. Sierpiński, Theory of Numbers and Theoretical Arithmetic. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,034 |
XV OM - III - Task 6
Given is a pyramid SABCD, whose base is a convex quadrilateral $ABCD$ with perpendicular diagonals $AC$ and $BD$, and the orthogonal projection of vertex S onto the base is point O, the intersection of the diagonals of the base. Prove that the orthogonal projections of point O onto the lateral faces of the pyramid lie on a circle. | Let $M$, $N$, $P$, $Q$ be the orthogonal projections of point $O$ onto the planes $ASB$, $BSC$, $CSD$, $DSA$ (Fig. 23).
The plane $SOM$ is perpendicular to the plane of quadrilateral $ABCD$ and to the plane $ASB$, as it contains the perpendiculars $SO$ and $OM$ to these planes. Therefore, the plane $SOM$ is perpendicular to the line of intersection $AB$ of the planes $ABCD$ and $ASB$ and intersects it at point $M$ on the line $SM$, with $AB \perp OM$, i.e., $M$ is the orthogonal projection of point $O$ onto the line $AB$. Similarly, the lines $SN$, $SP$, and $SQ$ intersect the lines $BC$, $CD$, and $DA$ at points $N$, $P$, and $Q$, which are the orthogonal projections of point $O$ onto these lines. The points $M$, $N$, $P$, $Q$ lie on a certain circle $\alpha$. The points $M$, $N$, $M$, $N$ also lie on a circle. Indeed, the segment $OB$ is seen from each of these points at a right angle, so they lie on a sphere (i.e., on the surface of a sphere) with diameter $OB$; they also lie in the plane $SMN$, so they lie on the line of intersection of the sphere with the plane, i.e., on a certain circle $\beta$. The circles $\alpha$ and $\beta$ lie in different planes and have two common points $M$ and $N$; therefore, there exists a sphere $\gamma$ passing through both these circles; the points $M$, $N$, $P$, $Q$, $M$, $N$ lie on the sphere $\gamma$. Similarly, the points $M$, $N$, $P$, $Q$, $N$, $P$ lie on a certain sphere $\gamma$, and the points $M$, $N$, $P$, $Q$, $P$, $Q$ lie on a certain sphere $\gamma$. The surfaces $\gamma$, $\gamma$, $\gamma$ are identical, since $\gamma$ and $\gamma$ pass through the vertices of the tetrahedron $M$, and $\gamma$ and $\gamma$ through the vertices of the tetrahedron $M$. Therefore, the points $M$ lie on the sphere $\gamma$. On the other hand, the points $M$, $N$, $P$, $Q$ lie on a sphere with diameter $OS$, since from each of these points the segment $OS$ is seen at a right angle; this sphere is different from the sphere $\gamma$, as it does not pass through the points $M$, $N$, $P$, $Q$, being tangent to the plane $ABCD$ at point $O$. The points $M$, $N$, $P$, $Q$ therefore lie on the line of intersection of two different spheres, i.e., on a circle.
Note. The proof above can be significantly shortened by considering the stereographic projection of the sphere with diameter $OS$ from point $S$ onto the plane $ABCD$. It is known that in such a projection, a circle lying in the plane corresponds to a circle on the sphere. The circle passing through the points $M$, $N$, $P$, $Q$ corresponds to a circle on the sphere, which passes through the corresponding points of the sphere, i.e., through the points $M$, $N$, $P$, $Q$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,035 |
LIX OM - I - Task 6
Determine all polynomials $ W(x) $ with real coefficients such that for every real number
$ x $ the equality is satisfied | First, note that the only constant polynomials satisfying the conditions of the problem are $ W(x) \equiv 0 $ and $ W(x) \equiv 1 $. Let us assume from now on that the polynomial $ W(x) $ is not constant.
If the polynomial $ W(x) $ is of the form $ W(x)=cx^n $ for some real number $ c \neq 0 $ and integer $ n \geqslant 1 $, then in condition (1) we have
which implies $ c^2=c^5 $ and $ c=1 $. Therefore, among polynomials of the considered form, only polynomials $ W(x)= x $ for $ n \geqslant 1 $ satisfy the conditions of the problem.
The case remains to be considered where the polynomial $ W(x) $ is the sum of at least two non-zero monomials. In such a case, we can write
where $ n>l \geqslant 0 $, $ a_n \neq 0 $, $ a_l \neq 0 $, and the polynomial $ G(x) $ is either the zero polynomial or has a degree of at most $ l -1 $. Thus,
where $ P(x) $ is a polynomial of degree at most $ 2n+3l -1 $, and
where $ Q(x) $ is a polynomial of degree at most $ 3n +2l $. The numbers $ 2n +3l $ and $ 3n+2l $ are less than $ 4n+l $. Therefore, the coefficient of the power $ x^{4n+l} $ in the polynomial $ W(x^2)\cdot W(x^3) $ is zero, while in the polynomial $ (W(x))^5 $ it is $ 5a_na_l \neq 0 $.
We have thus arrived at a contradiction. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 1,036 |
XVII OM - I - Problem 5
Given positive numbers $ p $ and $ q $. Prove that a rectangular prism, in which the sum of the edges equals $ 4p $, and the surface area equals $ 2q $, exists if and only if $ p^2 \geq 3q $. | a) Suppose that the desired rectangular parallelepiped exists. Then there are numbers $ x $, $ y $, $ z $ satisfying the equations
In such a case, the numbers $ p $ and $ q $ satisfy the inequality $ p^2 \geq 3q $, since from (1) and (2) it follows that
b) Suppose that given positive numbers $ p $ and $ q $ satisfy the inequality $ p^2 \geq 3q. $
We will show that there are then positive numbers $ x $, $ y $, $ z $ satisfying equations (1) and (2). If $ \sigma $ denotes a positive number less than $ \frac{p}{3} $, then the numbers defined by the formulas
are positive and satisfy equation (1). From equality (3) it follows that
Taking $ \sigma = \frac{1}{3} \sqrt{p^2-3q} $ we state that $ 0 < \sigma < \frac{p}{3} $ and that
The numbers $ x $, $ y $, $ z $ determined in this way also satisfy equation (2), so the rectangular parallelepiped with edges $ x $, $ y $, $ z $ has the desired properties. | proof | Algebra | proof | Yes | Yes | olympiads | false | 1,037 |
LVI OM - II - Task 3
In the data space, there are $ n $ points ($ n\geq 2 $) of which no four lie in the same plane. Some of these points have been connected by segments. Let $ K $ be the number of segments drawn ($ K\geq 1 $), and $ T $ the number of triangles formed. Prove that | Let us number from 1 to $ m $ those points from which at least one segment starts, and assume that from the point numbered $ i $ ($ i=1,2,\ldots,m $) exactly $ k_i $ segments start. Then $ k_i>0 $. Furthermore, let $ t_i $ ($ i=1,2,\ldots,m $) denote the number of those triangles, one of whose vertices is the point numbered $ i $. Then
The set $ A_i $ of those points that are connected by a segment to the point numbered $ i $ contains exactly $ k_i $ elements. Moreover, the number of triangles having a vertex at the point numbered $ i $ is the same as the number of segments with endpoints in the set $ A_i $. Therefore,
On the other hand, the number of these segments is less than the number of all drawn segments. Therefore,
Multiplying the inequalities (2) and (3) side by side, we get $ t_i^2 < \frac{1}{2}k_i^2K $. From this and equation (1), we obtain
Raising the last inequality to the power of two on both sides, we get the thesis. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 1,038 |
XL OM - I - Task 6
Calculate the sum of the series $ \sum_{n\in A}\frac{1}{2^n} $, where the summation runs over the set $ A $ of all natural numbers not divisible by 2, 3, 5. | A natural number $ n $ is not divisible by $ 2 $, $ 3 $, and $ 5 $ if and only if, when divided by 30, it gives one of the following eight remainders:
Therefore, the sum we need to calculate is equal to
where $ s_i $ denotes the sum of an analogous series formed from terms corresponding to those values of $ n $ that have the form
Thus,
(The correctness of the applied transformation and the validity of equation (1) follow from the fact that the terms of the considered series are positive; they can therefore be grouped and rearranged in any way without affecting the convergence and value of the series sum.)
The right-hand side of equation (2) represents a geometric series with a common ratio of $ 2^{-30} $. Its sum is equal to
Substituting these numbers into equation (1), we get
This is the desired value. | \frac{1}{1-2^{-30}}(\frac{1}{2^1}+\frac{1}{2^7}+\frac{1}{2^{11}}+\frac{1}{2^{13}}+\frac{1}{2^{17}}+\frac{1}{2^{19}}+\frac{1}{2^{23}}+\frac{1}{2^{} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 1,039 |
XL OM - I - Task 1
Prove that if the numbers $ k $, $ n $ ($ k < n $) are coprime, then the number $ \binom{n-1}{k-1} $ is divisible by $ k $. | Let's denote:
These are integers. The equality
holds, that is, $ M \cdot n = N \cdot k $. Since $ k $ is relatively prime to $ n $, it must be a divisor of $ M $; which is exactly what we needed to prove. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,041 |
XXXVIII OM - III - Problem 4
Among the set of tetrahedrons with a base area of 1, a total surface area of 4, and having equal angles of inclination of the lateral faces to the base, find the tetrahedron with the maximum volume. | Let $ABCD$ be any tetrahedron satisfying the given conditions:
From condition (2), it follows that the point $D$, which is the projection of point $D$ onto the plane $ABC$, is at the same distance from the lines $BC$, $CA$, $AB$ (equal to $r = h \ctg \phi$, where $h = |DD'|$ and $\phi$ is the dihedral angle). Moreover, point $D$ lies within the triangle $ABC$; if it were outside this triangle, then among the dihedral angles mentioned in (2), there would be at least one acute angle and at least one obtuse angle, contradicting their assumed equality. Therefore, point $D$ is the incenter of triangle $ABC$. The radius of this circle is $r$.
The height of each lateral face dropped from vertex $D$ has length $w = \sqrt{h^2 + r^2}$. Denoting by $p$ half the perimeter of triangle $ABC$, we obtain from equality (1):
and
Thus, $p^2 h^2 = 8$, or $h = \frac{2\sqrt{2}}{p}$. The volume of the tetrahedron $ABCD$ is therefore
Among all triangles with area $1$, the one with the smallest perimeter is the equilateral triangle (see Note 2). By (3), the minimum value of $p$ gives the maximum value of $V$. Therefore, among the considered class of tetrahedra, the one with the largest volume is the tetrahedron $ABCD$ with the following properties: the base $ABC$ is an equilateral triangle with area $1$; the foot of the height from $D$ coincides with the incenter of triangle $ABC$; the lateral surface has area $3$.
The only (up to isometry) tetrahedron with these properties is the regular tetrahedron with faces of area $1$. It constitutes the solution to the problem.
Note 1. If the condition given in the problem is interpreted as the equality of the inclination of the lateral faces to the base plane (not necessarily to the face $ABC$ itself), the reasoning would require the following modification: point $D$, being at the same distance $r$ from the lines $BC$, $CA$, $AB$, must be the incenter of triangle $ABC$ or one of the excenters. The equalities $3 = p \sqrt{h^2 + r^2}$ and $V = \frac{h}{3}$ remain valid. For a fixed triangle $ABC$, the maximum value of $V$ will be obtained by choosing the minimum (of the four possible) value of $r$—that is, the radius of the incircle. This means that other possibilities can be neglected.
Note 2. In the final part of the solution, we referred to a known theorem stating that among all triangles with area $1$, the one with the smallest perimeter is the equilateral triangle. For completeness, we will provide a proof of this fact.
Let $a$, $b$, $c$ be the lengths of the sides of a triangle with area $1$ and let $p = \frac{a+b+c}{2}$. We need to prove that $p \geq \sqrt[4]{27}$ (half the perimeter of an equilateral triangle with area $1$). From Heron's formula and the inequality of means, we have
Thus, $p^4 \geq 27$, as required. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,049 |
XXXIX OM - II - Problem 3
Inside an acute triangle $ ABC $, consider a point $ P $ and its projections $ L, M, N $ onto the sides $ BC, CA, AB $, respectively. Determine the point $ P $ for which the sum $ |BL|^2 + |CM|^2 + |AN|^2 $ is minimized. | Let $ P $ be any point inside the triangle $ ABC $. Considering the right triangles $ ANP $, $ BLP $, $ CMP $ (Figure 7), we observe that
Similarly, from the triangles $ AMP $, $ BNP $, $ CLP $ we get
Thus, by equating the right sides of the obtained equalities:
The left side of equation (1) is precisely the quantity we are considering in the problem; let's denote it by $ f(P) $. We need to find the point $ P $ for which $ f(P) $ reaches its minimum.
The sum of the left and right sides of equation (1) is $ 2f(P) $. Therefore,
Now notice that
The value of this expression is the smallest when the difference $ |BL| - |CL| $ is zero (since $ |BC| $ is constant) - that is, when $ L $ is the midpoint of segment $ BC $. Similarly, the other two sums in parentheses on the right side of (2) have the smallest value when points $ M $ and $ N $ are the midpoints of sides $ CA $ and $ AB $.
These conditions can be satisfied simultaneously: the projections of point $ P $ onto the sides of the triangle coincide with the midpoints of these sides if and only if $ P $ lies at the intersection of the perpendicular bisectors of the sides. Hence the answer: the quantity $ f(P) $ reaches its minimum when $ P $ is the circumcenter of triangle $ ABC $ (it lies in the interior of triangle $ ABC $, since the triangle is acute). | PisthecircumcenteroftriangleABC | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,050 |
XLVIII OM - I - Problem 2
Point $ P $ lies inside the parallelogram $ ABCD $, and the equality $ |\measuredangle ABP| = |\measuredangle ADP| $ holds. Prove that $ |\measuredangle PAB| = |\measuredangle PCB| $. | We translate the triangle $ ADP $ in parallel such that the image of side $ AD $ is segment $ BC $. Let the image of vertex $ P $ be denoted by $ Q $. Therefore, we have the equality $ |\measuredangle ADP| = |\measuredangle BCQ| $ (Figure 1). Segments $ AB $ and $ PQ $ are parallel, so $ |\measuredangle ABP| = |\measuredangle QPB| $.
At the same time, $ |\measuredangle ABP| = |\measuredangle ADP| $ (according to the condition of the problem). Therefore, $ |\measuredangle BCQ| = |\measuredangle QPB| $.
Segment $ BQ $ is thus visible at the same angle from points $ P $ and $ C $, which lie on the same side of the line $ BQ $. Conclusion: points $ B $, $ P $, $ C $, $ Q $ lie on the same circle. Hence, given that points $ C $ and $ Q $ lie on the same side of the line $ PB $, it follows that $ |\measuredangle PCB| = |\measuredangle PQB| $.
It remains to note that quadrilateral $ PABQ $ is a parallelogram, so $ |\measuredangle PQB| = |\measuredangle PAB| $. We obtain the equality $ |\measuredangle PAB| = |\measuredangle PCB| $, which was to be proven. | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,054 |
XI OM - I - Problem 7
A triangle $ABC$ is inscribed in a given circle. On the sides $AC$ and $BC$, segments $AM$ and $BN$ of a given length $d$ are measured. Find the geometric locus of the midpoint $S$ of segment $MN$, when points $A$ and $B$ are fixed, and point $C$ moves along the given circle. | We draw from point $ S $ parallel to the lines $ AC $ and $ BC $ and from points $ A $ and $ B $ parallel to the line $ MN $; we obtain parallelograms $ AMSP $ and $ BNSR $ (Fig. 5).
If triangle $ ABC $ is not isosceles, for example if (as in Fig. 5) $ AC > BC $, then $ MC > NC $, so $ \measuredangle CMN < \measuredangle CNM $, from which it is easy to deduce that $ \measuredangle CMN < \measuredangle A $, and $ \measuredangle CNM > \measuredangle B $ (Indeed, drawing a segment $ MK $ parallel to $ AB $ in the triangle, we have $ BK < AM $, because $ AM $ and $ BK $ are perpendicular to $ AC $ and $ BC $. Hence $ BK < BN $, so point $ N $ lies on segment $ KC $). Therefore, point $ P $ lies inside triangle $ ABC $, and point $ R $ outside this triangle, and segment $ PR $ intersects side $ AB $ at some point $ T $. Triangles $ ATP $ and $ BTR $ are similar with respect to point $ T $, since $ AP $ is parallel to $ RB $; because $ AP = MS = SN = RB $, this similarity is a central symmetry, and $ T $ is the common midpoint of segments $ AB $ and $ PR $.
In the case where triangle $ ABC $ is isosceles, the midpoint $ T $ of segment $ AB $ also coincides with the midpoint of segment $ PR $; this follows from the symmetry of the entire figure with respect to the perpendicular bisector of segment $ AB $.
Triangle $ PSR $ is isosceles, since $ PS = AM = d $ and $ RS = NB = d $; angle $ PSR $ equals angle $ ACB $ (angles with parallel and similarly directed sides), which we denote by $ \gamma $. Therefore,
When point $ C $ moves along one of the arcs of a given circle with endpoints $ A $ and $ B $, for example, along the upper side of $ AB $ in Fig. 5, the measure of angle $ \gamma $ does not change, so the length of $ TS $ is constant. This means that point $ S $ moves along a circle $ \Gamma $ with center $ T $ and radius equal to $ d \cos \frac{\gamma}{2} $.
It remains to determine which part of circle $ \Gamma $ is the geometric locus of point $ S $ as point $ C $ traverses the considered arc $ AB $. Due to the symmetry of the figure with respect to the perpendicular bisector of segment $ AB $, it suffices to consider points on one side of this bisector, for example, the side containing $ TB $. Notice that point $ N $ moves along a circle with center $ B $ and radius $ d $ (Fig. 6). Point $ N_1 $, where this circle intersects the given circle, is the extreme position of point $ N $ and simultaneously the corresponding extreme position of point $ C $. Point $ S $ is then at the midpoint $ S_1 $ of segment $ M_1N_1 $, where $ AM_1 = d $. Point $ S_1 $ is the endpoint of the arc of circle $ \Gamma $ that constitutes the desired geometric locus.
The problem can be stated more broadly than the text of the problem suggests, namely, not to limit ourselves to measuring segments of length $ d $ on the sides $ AC $ and $ BC $ of triangle $ ABC $, but to measure them on the rays $ AC $ and $ BC $. In this case, point $ C $ can traverse the entire arc $ AB $. As it approaches point $ B $, the ray $ BC $ rotating around point $ B $ approaches the tangent at point $ B $ to the given circle, point $ N $ approaches point $ N_0 $ of this tangent, where $ BN_0 = d $, point $ M $ approaches point $ M_0 $ ($ AM_0 = d $), and point $ S $ approaches the midpoint $ S_0 $ of segment $ M_0N_0 $. Point $ S_0 $ is then the endpoint of the arc of circle $ \Gamma $ that is the desired geometric locus. The other endpoint is point $ S $ symmetric to $ S_0 $ with respect to the perpendicular bisector of segment $ AB $.
The second part of the geometric locus is found analogously on the other side of line $ AB $. It is an arc of a circle with center $ T $ and radius $ d \sin \frac{\gamma}{2} $. When $ AB $ is a diameter of the given circle, both parts of the geometric locus are arcs of the same circle.
Note 1. The position of point $ S_0 $ can also be specified by giving the angle $ S_0TB $. We can easily calculate it by applying the Law of Sines to triangles $ S_0M_0T $ and $ N_0M_0B $; specifically,
Dividing these equations side by side and considering that $ TS_0 = d \cos \frac{\gamma}{2} $, $ N_0B = d $, $ M_0S_0 = \frac{1}{2} M_0N_0 $, and $ \measuredangle M_0BN_0 = 180^\circ - \gamma $, we obtain
It turns out that $ \measuredangle S_0TB $ does not depend on the length $ d $. Changing this length thus yields as the corresponding geometric loci arcs that are similar with respect to the center $ T $ of segment $ AB $.
Note 2. In the above solution, we showed that every point of the considered geometric locus lies on the arc $ S_0S $. We did not, however, prove the converse, that every point of the arc $ S_0S $ belongs to this geometric locus. Although it seems obvious that when point $ C $ traces the arc $ AB $ continuously, point $ S $ also traces - continuously - the arc $ S_0S $, this needs to be justified more rigorously. Using concepts from mathematical analysis and analytic geometry, it would suffice to state that the coordinates of point $ S $ (in a conveniently chosen coordinate system) are continuous functions of the coordinates of point $ C $. A proof not requiring these concepts can be obtained by reversing the previous reasoning.
Let $ S $ be an internal point of the arc $ S_0S $ \left( ST = d \cos \frac{\gamma}{2} \right). We construct an isosceles triangle $ PSR $ with height $ ST $ and vertex angle $ S $ equal to $ \gamma $. Triangles $ ATP $ and $ BTR $ are symmetric with respect to $ T $, so segments $ AP $ and $ BR $ are equal and parallel. Drawing lines through $ A $ and $ B $ parallel to $ PS $ and $ RS $; since they form an angle equal to angle $ PSR $, i.e., angle $ \gamma $, they intersect at point $ C $ of the given circle. Drawing a line through $ S $ parallel to $ AP $ and $ RB $ we obtain points $ M $ and $ N $ at the intersections with $ AC $ and $ BC $, respectively, such that $ AM = PS = d $, $ BN = RS = d $, $ MS = AP = RB = SN $. Point $ S $ therefore belongs to the geometric locus, c.n.d. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,055 |
XXXVI OM - I - Zadanie 9
W urnie jest 1985 kartek z napisanymi liczbami 1,2,3,..., 1985, każda lczba na innej kartce. Losujemy bez zwracania 100 kartek. Znaleźć wartość oczekiwaną sumy liczb napisanych na wylosowanych kartkach.
|
Losowanie $ 100 $ kartek z urny zawierającej $ 1985 $ kartek można interpretować jako wybieranie $ 100 $-elementowego podzbioru zbioru $ 1985 $-elementowego. Zamiast danych liczb $ 1985 $ i $ 100 $ weźmy dowolne liczby naturalne $ n $ i $ k $, $ n \geq k $. Dla dowolnego $ k $-elementowego zbioru $ X $ będącego podzbiorem zbioru $ Z = \{1,2,\ldots,n\} $ oznaczmy przez $ s(X) $ sumę liczb w zbiorze $ X $. Ponumerujmy wszystkie $ k $-elementowe podzbiory $ Z $ liczbami od $ 1 $ do $ ???????????????? = \binom{n}{k}: X_1, \ldots, X_N $. Wybieramy losowo jeden z tych zbiorów. Prawdopodobieństwo każdego wyboru jest takie samo, a więc równa się $ p = 1 /N $. Wartość oczekiwana sumy liczb w tym zbiorze równa się
Policzymy, w ilu zbiorach $ X_i $ występuje dowolnie ustalona liczba $ x \in Z $. Liczbie $ x $ towarzyszy w każdym z tych zbiorów $ k-1 $ liczb dowolnie wybranych ze zbioru $ Z-\{x\} $. Możliwości jest $ M = \binom{n-1}{k-1} $. Wobec tego każda
liczba $ x \in Z $ występuje w $ M $ zbiorach $ X_i $; tak więc w sumie każdy składnik $ x \in Z $ pojawia się $ M $ razy. Stąd $ s = M(1 + 2+ \ldots+n) = Mn(n+1)/2 $ i szukana wartość oczekiwana wynosi
W naszym zadaniu $ n = 1985 $, $ k = 100 $, zatem $ E = 99 300 $.
| 99300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1,058 |
LII OM - I - Task 4
Determine whether 65 balls with a diameter of 1 can fit into a cubic box with an edge of 4. | Answer: It is possible.
The way to place the balls is as follows.
At the bottom of the box, we place a layer consisting of 16 balls. Then we place a layer consisting of 9 balls, each of which is tangent to four balls of the first layer (Fig. 1 and 2). The third layer consists of 16 balls that are tangent to the balls of the second layer (Fig. 4 and 5). Similarly, we place two more layers (Fig. 6).
om52_1r_img_2.jpg
om52_1r_img_3.jpg
om52_1r_img_4.jpg
In total, we have placed $ 16 + 9 + 16 + 9 + 16 = 66 $ balls. It remains to calculate how high the fifth layer reaches.
om52_1r_img_5.jpg
om52_1r_img_6.jpg
om52_1r_img_7.jpg
Let's choose any ball from the second layer; this ball is tangent to four balls of the first layer. The centers of these five balls are the vertices of a regular square pyramid, each edge of which has a length of 1 (Fig. 3). By the Pythagorean theorem, the height of this pyramid is $ \frac{\sqrt{2}}{2} $. Therefore, the highest point that the fifth layer reaches is at a distance of $ \frac{1}{2} + 4 \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} = 1 + 2\sqrt{2} < 4 $ from the base plane. The 66 balls placed in this way fit into a cubic box with an edge of 4. | 66 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,060 |
XII OM - II - Task 2
Prove that all altitudes of a tetrahedron intersect at one point if and only if the sums of the squares of opposite edges are equal. | The solution to the problem will help us observe that the segment connecting the midpoints of two edges of a tetrahedron belonging to the same face is equal to half of the third edge of that face (and is parallel to it). Let's denote the midpoints of the edges of the tetrahedron $ABC$ as shown in Fig. 13 with the letters $M$, $N$, $P$, $Q$, $R$, $S$.
We will first prove that the sums of the squares of opposite edges of the tetrahedron $ABCD$ are equal if and only if the segments $MP$, $NQ$, and $RS$, i.e., the segments connecting the midpoints of opposite edges, are equal. Indeed, the equality
is equivalent to the equality
which in turn is equivalent to the equality
since the quadrilaterals $MNPQ$ and $MRPS$ are parallelograms, so in each of them the sum of the squares of the diagonals is equal to twice the sum of the squares of two adjacent sides. Finally, this last equality can be replaced by the equality
Similarly, the equality $AC^2 + BD^2 = AB^2 + CD^2$ is equivalent to the equality $MP = RS$.
The equality of the segments $MP$, $NQ$, and $RS$ holds if and only if the parallelograms $MNPQ$, $MRPS$, $NRQS$ are rectangles, and thus if and only if each pair of opposite edges of the tetrahedron are perpendicular.
We will then prove that if two opposite edges of a tetrahedron are perpendicular, then the altitudes of the tetrahedron drawn from the endpoints of one of these edges intersect.
Let, for example, $AB \bot DC$ (Fig. 14). Draw a perpendicular $CK$ to $AB$; since $CK \bot AB$ and $CD \bot AB$, the line $AB$ is perpendicular to the plane $CKD$. It follows that the altitudes $DH$ and $CG$ of the triangle $CKD$ are altitudes of the tetrahedron, since the line $DH$ is perpendicular to $CK$ and to $AB$, and thus is perpendicular to the plane $ABC$, and similarly the line $CG$ is perpendicular to the plane $ABD$.
Conversely, if the altitudes $DH$ and $CG$ of the tetrahedron intersect, then the line $AB$ is perpendicular to the plane containing $DH$ and $CG$, since $AB \bot DH$ and $AB \bot CG$, and thus the line $AB$ is also perpendicular to the line $CD$.
Therefore, the perpendicularity of each pair of opposite edges of the tetrahedron holds if and only if any two altitudes of the tetrahedron intersect.
Finally, we will prove that if any two altitudes of the tetrahedron intersect, then all four altitudes pass through one point.
Suppose that the altitudes $h_B$ and $h_C$ of the tetrahedron drawn from the vertices $B$ and $C$ lie in the plane $\alpha$, and the altitudes $h_C$ and $h_A$ drawn from the vertices $C$ and $A$ - in the plane $\beta$. Then the point of intersection of the altitudes $h_A$ and $h_B$ lies on the line of intersection of the planes $\alpha$ and $\beta$, i.e., on the line $h_C$; similarly, we conclude that it lies on the line $h_D$, so all the altitudes intersect at one point.
The theorem has been proved.
Note. A tetrahedron in which all four altitudes pass through one point (orthocenter) is called an orthocentric tetrahedron. In the above proof, we have stated that it can also be defined by any of the following properties:
1) perpendicularity of opposite edges,
2) equality of segments connecting the midpoints of opposite edges,
3) intersection of any two altitudes.
We will add one more property to this. A parallelepiped can be described on any tetrahedron, i.e., a parallelepiped can be constructed such that the edges of the tetrahedron are the diagonals of the faces of the parallelepiped (Fig. 15). A tetrahedron is orthocentric if and only if it has the property:
4) the parallelepiped described on the tetrahedron has all edges equal (i.e., its faces are rhombuses). | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,062 |
XXVII OM - I - Zadanie 12
Ciąg $ (x_n) $ określony jest wzorami $ x_0 = 25 $, $ x_n = x_{n-1} + \frac{1}{x_{n-1}} $ ($ n = 1, 2, \ldots $) Dowieść, że $ x_n > 1975 $ dla $ n > 1950000 $.
|
Udowodnimy ogólniejsze
Twierdzenie. Jeżeli $ \displaystyle x_0 > \frac{1}{2} $ i $ x_{n} = x_{n-1} + \frac{1}{x_{n-1}} $ dla $ n = 1, 2, \ldots $, to dla każdej liczby naturalnej $ N $ zachodzi nierówność
Dowód. Wyrazy ciągu $ (x_{n}) $ są oczywiście liczbami dodatnimi. Wobec tego $ x_n = x_{n-1} + \frac{1}{x_{n-1}} > x_{n-1} $ dla $ n=1, 2, \ldots $ tzn. ciąg
$ (x_{n}) $ jest rosnący.
Podnosząc obustronnie równość $ x_{n} = x_{n-1}+ \frac{1}{x_{n-1}} $ do kwadratu otrzymujemy
i stąd $ x_n^2 - x_{n-1}^2 > 2 $. Podstawiając tu kolejno $ n = 1, 2, \ldots, N $ i dodając stronami otrzymane nierówności uzyskujemy $ x^2_N - x^2_0 > 2N $, czyli
Z drugiej strony, ponieważ ciąg $ (x_{n}) $ jest rosnący i jego wyrazy są liczbami dodatnimi, więc
Wobec tego z (2) otrzymujemy
Podstawiając tu kolejno $ n = 1,2, \ldots, N $ i dodając stronami otrzymane nierówności uzyskujemy
Wobec tego
ponieważ $ x_0^2 > \frac{1}{2} $.
Stąd otrzymujemy $ x_N - \frac{1}{2x_0} < \sqrt{x_0^2 + 2N} $ i wobec tego na mocy (3) mamy tezę twierdzenia.
Jeżeli $ x_0 = 25 $ i $ N > 1950000 $, to z (1) wynika, że $ x_N > \sqrt{x_0^2}+ 2N > \sqrt{25^2 + 2 \cdot 1950000} = \sqrt{1975^2} = 1975 $, co daje rozwiązanie zadania.
Zauważmy jeszcze, że dla $ x_0 = 25 $ i $ N= 1950000 $ z (1) wynika, że
$ x_N < \frac{1}{2 \cdot 25} + \sqrt{25^2 + 2 \cdot 1950000} = \frac{1}{50} + 1975 = 1975,02 $.
Wzór (1) pozwala więc wyznaczyć liczbę $ x_{N} $ z błędem mniejszym od $ 0,01 $. Mianowicie, z powyższego wynika, że
| proof | Algebra | proof | Yes | Yes | olympiads | false | 1,063 |
II OM - III - Task 2
What digits should be placed instead of zeros in the third and fifth positions in the number $ 3000003 $ to obtain a number divisible by $ 13 $? | Let the sought digits be denoted by $ x $ and $ y $, and write the number $ 30x0y03 $ in the form
We need to find such integers $ x $ and $ y $ that the number $ N $ is divisible by $ 13 $ and that the inequalities
are satisfied.
The numbers $ 10^6 $, $ 10^4 $, $ 10^2 $ give remainders $ 1 $, $ 3 $, $ 9 $ respectively when divided by $ 13 $, so
\[ 3 \cdot 10^6 = 13k_1 + 3, \]
\[ x \cdot 10^4 = 13k_2 + 3x, \]
\[ y \cdot 10^2 = 13k_3 + 9y, \]
where $ k_1 $, $ k_2 $, $ k_3 $ are natural numbers. Therefore,
\[ 30x0y03 = 3 \cdot 10^6 + x \cdot 10^4 + y \cdot 10^2 + 3 = 13k_1 + 3 + 13k_2 + 3x + 13k_3 + 9y + 3 \]
\[ = 13(k_1 + k_2 + k_3) + 3 + 3x + 9y + 3 \]
\[ = 13(k_1 + k_2 + k_3) + 3x + 9y + 6 \]
where $ k $ is a natural number.
For the number $ N $ to be divisible by $ 13 $, it is necessary and sufficient that the number $ x + 3y + 2 $ is divisible by $ 13 $, i.e., that $ x $ and $ y $ satisfy the equation
\[ x + 3y + 2 = 13m \]
where $ m $ is a natural number.
From the inequalities $ x \leq 9 $, $ y \leq 9 $, it follows that
\[ 0 \leq x + 3y + 2 \leq 38 \]
so the number $ m $ must satisfy the inequality $ 13m \leq 38 $, i.e., $ m $ can only be $ 1 $ or $ 2 $.
1° Taking $ m = 1 $, we get the equation for $ x $ and $ y $
\[ x + 3y + 2 = 13 \]
Under the constraints $ 0 \leq x \leq 9 $, $ 0 \leq y \leq 9 $, this equation has three integer solutions:
\[ (x, y) = (8, 1), (4, 2), (2, 3) \]
2° Taking $ m = 2 $, we get the equation
\[ x + 3y + 2 = 26 \]
and we obtain the solutions:
\[ (x, y) = (9, 5), (6, 6), (3, 7), (0, 8) \]
Thus, the problem has $ 7 $ solutions, corresponding to the numbers: $ 3080103 $, $ 3040203 $, $ 3020303 $, $ 3090503 $, $ 3060603 $, $ 3030703 $, $ 3000803 $.
Note. In problem 35, it was about changing the digits of a certain specified number. Let's consider two more general examples of such problems:
I. Given a natural number $ N $ not divisible by a prime number $ p $ and having a digit $ 0 $ in the $ k $-th decimal place. Can the zero be replaced by a digit $ x $ so that the new number is divisible by $ p $?
First, note that if $ p $ is $ 2 $ or $ 5 $, the problem has no solution. Therefore, we assume in the following that $ p \ne 2 $ and $ p \ne 5 $.
When we place the digit $ x $ in the $ k $-th decimal place of the number $ N $, the resulting number $ N(x) $ is given by the formula
\[ N(x) = N + x \cdot 10^{k-1} \]
Consider the values of the function $ N(x) $ when $ x $ takes the values $ 0, 1, 2, \ldots, (p - 1) $:
\[ N(0), N(1), N(2), \ldots, N(p-1) \]
Each of the $ p $ numbers in the sequence (2) gives a different remainder when divided by $ p $. Indeed, according to formula (1), the difference between two numbers in the sequence (2) is given by
\[ N(i) - N(j) = (i - j) \cdot 10^{k-1} \]
When $ i \ne j $, this number is not divisible by $ p $, because the factor $ (i - j) $ is non-zero and less in absolute value than $ p $, and the factor $ 10^{k-1} $ has only the prime divisors $ 2 $ and $ 5 $. Therefore, the numbers $ N(i) $ and $ N(j) $ give different remainders when divided by $ p $.
Since the remainders of the numbers in the sequence (2) are different and there are $ p $ of them, one and only one of these remainders is zero, i.e., one and only one of the numbers in the sequence (2) is divisible by $ p $; let this number be $ N(x_0) $.
If $ x_0 \leq 9 $, then $ x_0 $ gives a solution to the problem.
If $ x_0 > 9 $, the problem has no solution.
Hence the conclusion:
a) When $ p = 3 $, the problem always has three solutions; the sought digits are either $ 1 $, $ 4 $, $ 7 $ or $ 2 $, $ 5 $, $ 8 $.
b) When $ p = 7 $, the problem has either two solutions - the digits $ 1 $, $ 8 $ or $ 2 $, $ 9 $, or one solution - one of the digits $ 3 $, $ 4 $, $ 5 $, $ 6 $.
c) When $ p \geq 11 $, the problem may have one solution or no solution. If $ p = 11 $, the first of these cases occurs, for example, for the number $ 10 $, and the second case is given by the number $ 109 $.
II. Given a natural number $ N $ not divisible by a prime number $ p \geq 11 $ and having a digit $ 0 $ in the $ k $-th and $ l $-th decimal places. Can the zeros be replaced by digits $ x $ and $ y $ so that the new number is divisible by $ p $?
When we place the digit $ x $ in the $ k $-th decimal place and the digit $ y $ in the $ l $-th decimal place of the number $ N $, we obtain the number $ N(x, y) $ given by the formula
\[ N(x, y) = N + x \cdot 10^{k-1} + y \cdot 10^{l-1} \]
Consider the values of the function $ N(x, y) $ obtained by substituting $ 0, 1, 2, \ldots, (p - 1) $ for $ x $ and $ y $. We get $ p^2 $ numbers:
\[ N(0, 0), N(0, 1), \ldots, N(0, p-1), N(1, 0), N(1, 1), \ldots, N(1, p-1), \ldots, N(p-1, 0), N(p-1, 1), \ldots, N(p-1, p-1) \]
We state, as in problem I, that in each row and each column of this table there is one and only one number divisible by $ p $; let the number in the $ i $-th row be denoted by $ N(i, y_i) $.
The table thus contains $ p $ numbers divisible by $ p $; they are the numbers
\[ N(0, y_0), N(1, y_1), N(2, y_2), \ldots, N(p-1, y_{p-1}) \]
where the sequence of numbers $ y_0, y_1, y_2, \ldots, y_{p-1} $ contains the same numbers as the sequence $ 0, 1, 2, \ldots, (p - 1) $, but in a different order. Note also that $ y_0 $ is certainly not equal to $ 0 $, since the number $ N(0, 0) = N $, by assumption, is not divisible by $ p $.
The number $ N(i, y_i) $ in the sequence (4) gives a solution to the problem only if $ i \leq 9 $ and $ y_i \leq 9 $. It is easy to count how many such numbers the sequence (4) can contain. They should be sought among the numbers
\[ N(0, y_0), N(1, y_1), N(2, y_2), \ldots, N(9, y_9) \]
In the case where none of the numbers $ y_0, y_1, \ldots, y_9 $ is greater than $ 9 $, the problem has $ 10 $ solutions given by the numbers in the sequence (5). This is the maximum possible number of solutions.
If among the numbers $ y_0, y_1, \ldots, y_9 $ there are numbers greater than $ 9 $ (there can be at most $ p - 10 $ such numbers), the corresponding terms should be crossed out from the sequence (5). The problem thus has at least $ 10 - (p - 10) $, i.e., $ 20 - p $ solutions.
Hence the conclusion:
a) When $ p = 11 $, the problem has at least $ 9 $ solutions. An example of the case where there are the maximum number of $ 10 $ solutions is the number $ 101015 $.
b) When $ p = 13 $, the problem has at least $ 7 $ solutions; the case where there are only $ 7 $ solutions was in problem 35.
c) When $ p = 17 $, we have at least three solutions.
d) When $ p = 19 $, the problem has at least one solution.
e) When $ p \geq 23 $, the problem may have no solutions. An appropriate example for the case $ p = 23 $ can be found easily if one considers that the number $ 10^{22} $ gives a remainder of $ 1 $ when divided by $ 23 $. We propose this as an exercise. | 3080103,3040203,3020303,3090503,3060603,3030703,3000803 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 1,064 |
XLIII OM - III - Problem 6
Prove that for every natural number $ k $, the number $ (k!)^{k^2+k+1} $ is a divisor of the number $ (k^3)! $. | For every pair of natural numbers $ n,l \geq 1 $, the following equality holds:
Let us fix natural numbers $ n,m \geq 1 $. Substitute $ l = 1,2,\ldots,m $ into (1) and multiply the resulting equations side by side:
We transform the left side of equation (2):
and after reducing the repeated factors in the numerators and denominators:
Equating the right sides of equations (2) and (3) gives the equality:
Thus, for every pair of natural numbers $ n,m \geq 1 $, the quotient $ \frac{(mn)!}{(n!)^m m!} $ is a natural number.
Let $ k $ be a given natural number. Substituting into (4) first $ n=m=k $, and then $ n=k $, $ m = k^2 $, we get natural numbers on the right side, which we will denote by $ A_1(k) $ and $ A_2(k) $, respectively:
Hence, by multiplying the sides:
Thus, indeed, the number $ (k^3)! $ is divisible by $ (k!)^{k^2+k+1} $.
Note: Let us assume in (4): $ n = k $, $ m=k^j $, and denote the value of the expression (4) obtained by $ A_j(k) $:
Take any natural number $ r \geq 1 $ and multiply the sides of equations (5) applied successively for $ j = 1,\ldots,r $. The result of this operation is:
which, after reducing the common factors in the numerator and denominator, is:
We have thus obtained a strengthened version of the theorem: The number $ (k!)^{1+k+k^2 + \ldots+k^r} $ is a divisor of the number $ (k^{r+1})! $. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,066 |
LVII OM - III - Problem 3
Given a convex hexagon $ABCDEF$, in which $AC = DF$, $CE = FB$ and $EA = BD$. Prove that the lines connecting the midpoints of opposite sides of this hexagon intersect at one point. | Let $P, Q, R$ be the midpoints of the diagonals $AD, BE, CF$, respectively.
First, assume that two of the points $P, Q, R$ coincide; let, for example, $P = Q$ (Fig. 1).
Then the quadrilateral $ABDE$ is a parallelogram. Moreover, triangle $ACE$ is congruent to triangle $DFB$, which implies that
$\measuredangle EAC = \measuredangle BDF$. This equality, together with the obtained parallelism $BD || EA$, proves that the segments
$AC$ and $DF$ are parallel. Furthermore, these segments are of equal length, so quadrilateral $ACDF$ is a parallelogram. Point
$R$ coincides, therefore, with points $P$ and $Q$, which means that it is the center of symmetry of the hexagon $ABCDEF$. It remains to note
that in this case, the lines connecting the midpoints of opposite sides of the hexagon $ABCDEF$ pass through its center of symmetry.
om57_3r_img_1.jpg
om57_3r_img_2.jpg
Next, assume that the points $P, Q, R$ are distinct. Let $M$ and $N$ be the midpoints of segments $AB$ and $DE$, respectively (Fig. 2).
From the converse of Thales' theorem, it follows that segments $PN$ and $AE$ are parallel, and moreover,
$PN = \frac{1}{2}AE$. Similarly, we obtain the relations $QM = \frac{1}{2}AE$, $PM = \frac{1}{2}BD$, and $QN = \frac{1}{2}BD$.
From these relations and the equality $AE = BD$, we conclude that $PN = QM = PM = QN$, which means that quadrilateral $MPNQ$ is a rhombus.
Thus, line $MN$ is the perpendicular bisector of segment $PQ$.
Similarly, we prove that the other two lines connecting the midpoints of opposite sides of the given hexagon are the perpendicular bisectors
of segments $QR$ and $RP$. Therefore, the lines connecting the midpoints of opposite sides of the hexagon $ABCDEF$ have a common point, which is
the center of the circle circumscribed around triangle $PQR$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,068 |
XII OM - I - Problem 1
$k$ parallel lines were drawn and intersected by $n$ parallel lines. How many parallelograms were formed? | Every parallelogram defines (by extending its sides) one pair $ (a, b) $ among the given $ k $ parallel lines and one pair $ (c, d) $ among the $ n $ parallel lines; conversely, any two such pairs $ (a, b) $ and $ (c, d) $ of parallel lines define one parallelogram of the figure. Therefore, the figure has as many parallelograms as there are different quadruples $ [(a, b), (c, d)] $.
Indeed, from $ k $ lines, one can form $ \frac{1}{2} k (k - 1) $ different pairs $ (a, b) $. Indeed, each of the given $ k $ lines can be paired with each of the $ k - 1 $ remaining lines, giving $ k (k - 1) $ pairs, among which each desired pair appears twice, once as $ (a, b) $ and once as $ (b, a) $. Similarly, from $ n $ given lines, one can form $ \frac{1}{2}n(n-1) $ different pairs $ (c, d) $. The number of different quadruples $ [(a, b), (c, d)] $ is therefore
For example, when $ k = 3 $, $ n = 4 $, there are $ 18 $ parallelograms in the figure, as illustrated in Fig. 1. | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1,071 | |
XLVI OM - I - Problem 9
Let $ a $ and $ b $ be real numbers whose sum is equal to 1. Prove that if $ a^3 $ and $ b^3 $ are rational numbers, then $ a $ and $ b $ are also rational numbers. | We raise the equality $ a + b = 1 $ to the second and third powers on both sides and obtain the relations: $ a^2 +2ab + b^2 = 1 $, that is,
and $ a^3 + 3a^2b + 3ab^2 + b^3 = 1 $, that is,
If, therefore, the numbers $ a^3 $ and $ b^3 $ are rational, then from equation (2) it follows that the product $ ab = (1-a^3-b^3)/3 $ is a rational number, and from equation (1) the sum $ a^2 + b^2 $ is also a rational number. In this case, the number $ a^2 + ab + b^2 $ is also rational - and different from zero, as can be seen, for example, from the transformation
The numbers $ a^3 $ and $ b^3 $ are rational, so their difference $ a^3 - b^3 $ is also rational. This leads to the conclusion that the difference
is a rational number. In combination with the condition $ a + b = 1 $, this gives the desired conclusion: the numbers
are rational. | proof | Algebra | proof | Yes | Yes | olympiads | false | 1,072 |
XIV OM - I - Problem 12
In a circle with center $ O $ and radius $ r $, a regular pentagon $ A_1A_2A_3A_4A_5 $ is inscribed, and on the smaller arc with endpoints $ A_1 $, $ A_5 $, a point $ M $ is chosen. Prove that | The equality (1) we need to prove is a relationship between the lengths of certain chords of a circle. Another relationship of the same kind is the well-known Ptolemy's theorem: The product of the diagonals of a cyclic quadrilateral equals the sum of the products of its opposite sides1). It suggests using this theorem to prove equality (1). In the considered figure, there are $10$ such cyclic quadrilaterals, one of whose vertices is $M$, and the other three are vertices of the pentagon $A_1A_2A_3A_4A_5$. Given the form of equality (1), we will choose those of these ten quadrilaterals in which segments $MA_1$, $MA_3$, $MA_5$ appear only as sides, and segments $MA_2$ and $MA_4$ only as diagonals, or vice versa. These are the quadrilaterals
Let $a$ denote the length of a side of the given pentagon and $b$ - the length of its diagonal. Applying Ptolemy's theorem to the mentioned $5$ quadrilaterals, we obtain the following equalities, which can be easily written without even looking at the diagram:
Adding these equalities:
and after dividing by $2a + b$, we obtain the desired equality (1). | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,073 |
XXII OM - I - Problem 12
Prove that every convex polyhedron has a triangular face or a trihedral angle. | Assume that there exists a convex polyhedron $W$ without any triangular faces or trihedral angles. Let $w$ be the number of vertices, $k$ the number of edges, and $s$ the number of faces of this polyhedron, and let $\varphi$ be the sum of the measures of all dihedral angles at the vertices of the polyhedron $W$. Since at least four edges emanate from each vertex, and each edge connects two vertices, we have $k \geq \frac{1}{2} \cdot 4w = 2w$. Since each face has at least four edges, and for $n \geq 4$ the sum of the interior angles of an $n$-gon is $\geq 2\pi$, we have $\varphi \geq 2\pi s$. Inside each face $S_i$, choose a point $O_i$ and connect it to all vertices of that face. As a result, we obtain a division of the surface of the polyhedron into $2k$ triangles, since each edge is a side of exactly two triangles. Let $\Psi$ be the sum of the measures of the angles in all triangles. Then $4\pi w \leq 2k\pi = \Psi = \varphi + 2\pi s \leq 2\varphi$, and thus $\varphi < 2\pi w$. On the other hand, since the polyhedron $W$ is convex, $\varphi < 2\pi w$. The obtained contradiction proves that there does not exist a polyhedron with the assumed properties. | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,075 |
V OM - III - Problem 6
Inside a hoop of radius $2r$, a disk of radius $r$ rolls without slipping along the hoop. What line does a point chosen arbitrarily on the edge of the disk trace? | When a disk rolls along the circumference, the points on the edge of the disk become points of contact with the circumference one after another. The condition of rolling without slipping means that the length of the arc $PQ$ between two points $P$ and $Q$ on the edge of the disk is equal to the length of the arc on the circumference that the points of the arc $PQ$ will sequentially touch during rolling.
Since the radius of the disk is half the radius of the circumference, the edge of the disk continuously passes through the center $O$ of the circumference.
Let us choose a specific point $P$ on the edge of the disk, and let this point $P$ be at point $A$ of the circumference at the initial position of the disk (Fig. 45).
When the disk rolls in such a way that the point of contact of the disk with the circumference traverses a quarter arc $AB$ of the circumference, the corresponding points on the edge of the disk, i.e., those that sequentially touch the circumference at the points of the arc $AB$, form a semicircle; when the point of contact of the disk with the circumference reaches point $B$, point $P$ will be at point $O$. We will show that during this motion, point $P$ traces the radius $AO$. For this purpose, suppose that at some moment the disk touches the circumference at point $Q$ of the arc $AB$. The center $S$ of the disk is then the midpoint of segment $OQ$. Point $P$ is then in such a position that the lengths of the arcs $AQ$ and $QP$ (both less than semicircles) are equal. From this it follows that
On the other hand, by the external angle theorem of triangle $\measuredangle PSQ = \measuredangle POS + \measuredangle OPS$, and since $\measuredangle OPS = \measuredangle POS$, we have
From the equality (1) and (2) we get
Thus, point $P$ lies on the radius $AO$.
Conversely, if $P$ is any internal point of segment $AO$, then there exists a position of the disk such that point $P$ will be at point $P$; the center of the disk is then at the intersection of the perpendicular bisector of segment $OP$ with that arc of the circle with center $O$ and radius $r$ that lies in the angle $AOB$.
We have shown that when the disk rolls along the quarter arc $AB$ of the circumference, point $P$ traces the radius $AO$. When rolling along the quarter arc $BC$, the disk assumes positions symmetric to the previous ones with respect to the line $OB$; point $P$ thus traces the radius $OC$ symmetric to $OA$, i.e., forming a diameter $AC$ of the circumference with $OA$. When rolling along the remaining quarter arcs of the circumference, point $P$ traces the diameter $CA$, as it then assumes positions symmetric to the previous ones with respect to the line $AC$.
We have obtained the result: when the disk rolls inside the circumference with twice the radius, each point of the disk traces a certain diameter of that circumference. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,076 |
L OM - I - Task 5
Find all pairs of positive integers $ x $, $ y $ satisfying the equation $ y^x = x^{50} $. | We write the given equation in the form $ y = x^{50/x} $. Since for every $ x $ being a divisor of $ 50 $, the number on the right side is an integer, we obtain solutions of the equation for $ x \in \{1,2,5,10,25,50\} $. Other solutions of this equation will only be obtained when $ x \geq 2 $ and for some $ k \geq 2 $, the number $ x $ is simultaneously the $ k $-th power of some natural number and a divisor of the number $ 50k $. If $ p $ is a prime divisor of such a number $ x $, then $ p^k|50^k $. Since $ p^k > k $, it cannot be that $ p^k|k $, so $ p \in \{2,5\} $. If $ p = 2 $, then $ 2^k|2k $, from which $ k = 2 $. If, however, $ p = 5 $, then $ 5^k|25k $, from which again $ k = 2 $. Therefore, $ x $ must simultaneously be the square of some natural number and a divisor of the number $ 100 $.
We obtain two new values of $ x $ in this case: $ x = 4 $ and $ x = 100 $. Thus, the given equation has 8 solutions: | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 1,080 |
XXXIII OM - II - Problem 6
Given is a finite set $ B $ of points in space, such that any two distances between points of this set are different. Each point of the set $ B $ is connected by a segment to the nearest point of the set $ B $. In this way, we obtain a set of segments, one of which (chosen arbitrarily) is painted red, and all the remaining segments are painted green. Prove that there exist two points of the set $ B $ that cannot be connected by a broken line composed of segments painted green. | Suppose the segment $ A_1A_2 $ is red, and the rest are green. If there existed a broken line composed of green segments connecting points $ A_1 $ and $ A_2 $, then there would be points $ A_3, \ldots, A_n $ being the successive endpoints of the segments forming this broken line. The notation $ A_i \to A_j $ is read as "the nearest point to $ A_i $ in set $ B $ is $ A_j $." From the fact that the segment $ \overline{A_iA_j} $ is drawn, it follows that $ A_i \to A_j $ or $ A_j \to A_i $. Given the assumption that any two distances between points in set $ B $ are different, it cannot simultaneously be true for $ k \ne j $ that $ A_i \to A_j $ and $ A_i \to A_k $. Therefore, it must be either $ A_1 \to A_2, A_2 \to A_3, \ldots, A_{n-1} \to A_n, A_n \to A_1 $, or $ A_1 \to A_n, A_n \to A_{n-1}, \ldots, A_3 \to A_2, A_2 \to A_1 $.
In both cases, the distances between successive points would satisfy contradictory inequalities.
or
Given that the considered segments do not have common interior points, the above solution implies a stronger thesis than the one stated in the problem: There exist two points in set $ B $ that cannot be connected by a broken line composed of segments contained in the sum of the green segments. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 1,081 |
VIII OM - I - Problem 8
In the rectangular prism $ ABCDA_1B_1C_1D_1 $, the lengths of the edges are given as $ AA_1 = a $, $ AB = b $, $ AD = c $. On the face $ A_1B_1C_1D_1 $, a point $ M $ is chosen at a distance $ p $ from the side $ A_1B_1 $, and at a distance $ q $ from the side $ A_1D_1 $, and a parallelepiped is constructed with the base $ ABCD $ and the lateral edge $ AM $. Calculate the area of the lateral faces of this parallelepiped. | Accepting the notation indicated in figure $5$, we calculate the area of the parallelogram $AMQD$ which is a face of the parallelepiped $ABCDMNPQ$.
The parallelogram $AMQD$ and the rectangle $AKLD$ have a common base $AD$ and height $AK$, thus
But $AK = \sqrt{AA_1^2 + A_1K^2} = \sqrt{a^2 + q^2},\ KL = AD = c$; hence
Similarly, we calculate that the area of $AMNB = b \sqrt{a^2 + p^2}$. | \sqrt{^2+} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,083 |
LI OM - II - Task 3
On an $ n \times n $ chessboard, $ n^2 $ different integers are placed, one on each square. In each column, the square with the largest number is painted red. A set of $ n $ squares on the chessboard is called admissible if no two of these squares are in the same row or the same column. Among all admissible sets, the set for which the sum of the numbers placed on its squares is the largest is chosen.
Prove that in such a selected set, there is a red square. | We will conduct an indirect proof.
Assume that in the chosen admissible set (denoted by $ A $) there is no red field.
We define a sequence of fields $ D_1, E_1, D_2, E_2, \ldots $ of the given chessboard according to the rule described below. (For $ n = 8 $, the process of selecting fields $ D_1, E_1, D_2, E_2, \ldots $ is shown in Figures 1 and 2. Red fields are shaded, while fields from set $ A $ are marked with a circle.)
om51_2r_img_3.jpg
om51_2r_img_4.jpg
The field $ D_1 $ is any field from set $ A $. Let $ E_1 $ be the red field lying in the same column as $ D_1 $.
Then we consider the field $ D_2 $ from set $ A $, which lies in the same row as $ E_1 $. By $ E_2 $ we denote the red field lying in the same column as $ D_2 $. We continue this process. (In general: the red field $ E_i $ lies in the same column as $ D_i $, while the field $ D_{i+1} $ is a field from set $ A $ lying in the same row as $ E_i $.)
om51_2r_img_5.jpg
In this way, we obtain a sequence of fields $ D_1, E_1, D_2, E_2, D_3, \ldots $, in which each field uniquely determines the next one. Since there are finitely many red fields, we will find such positive integers $ i $, $ t $, for which $ E_i = E_{i+t} $ and $ E_i \neq E_{i+s} $ for $ s = 1, 2, \ldots, t-1 $. (In our example shown in Figures 1 and 2, $ E_2 = E_5 $, so $ i = 2 $, $ t = 3 $.)
We remove from set $ A $ the fields $ D_{i+1}, D_{i+2}, \ldots, D_{i+t} $ and replace them with the fields $ E_{i+1}, E_{i+2}, \ldots, E_{i+t} $ (Fig. 3). The set obtained in this way is denoted by $ B $.
From the definition of the sequence $ D_1, E_1, D_2, E_2, D_3, \ldots $ it follows that the fields
find themselves in the same columns as the fields
and in the same rows as the fields
Therefore, set $ B $ is admissible. Moreover, the numbers written on the fields $ E_{i+1}, E_{i+2}, \ldots, E_{i+t} $ are greater than the numbers written on the fields $ D_{i+1}, D_{i+2}, \ldots, D_{i+t} $, respectively. It follows that the sum of the numbers written on the fields of set $ B $ is greater than the sum of the numbers written on the fields of set $ A $. This, however, contradicts the assumption that $ A $ has the largest sum of numbers among all admissible sets. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 1,084 |
XXIII OM - III - Problem 1
Polynomials $ u_1(x) = a_ix + b_i $ ($ a_i, b_i $ - real numbers; $ i = 1, 2, 3 $) satisfy for some natural $ n > 2 $ the equation
Udowodnić, że istnieją takie liczby rzeczywiste $ A, B, c_1, c_2, c_3 $, że $ u_i(x)=c_i(Ax+B) $ for $ i = 1, 2, 3 $.
Prove that there exist real numbers $ A, B, c_1, c_2, c_3 $ such that $ u_i(x)=c_i(Ax+B) $ for $ i = 1, 2, 3 $. | If $ a_1 = a_2 = 0 $, then the polynomials $ u_1 $ and $ u_2 $ are constant. Therefore, the polynomial $ u_3 $ is also constant, i.e., $ a_3 = 0 $. In this case, it suffices to take $ c_i = b_i $ for $ i = 1, 2, 3 $ and $ A = 0 $ and $ B = 1 $.
Let then at least one of the numbers $ a_1, a_2 $ be different from zero, for example, $ a_1 \ne 0 $. Denoting $ y = a_1x + b_1 $ we get $ u_j(x) = \displaystyle \frac{a_j}{a_1} y + \frac{b_ja_1 - a_jb_1}{a_1} $ for $ j = 2, 3 $, i.e., $ u_j(x) = A_jy + B_j $, where $ A_j = \displaystyle \frac{a_j}{a_1} $, $ \displaystyle \frac{b_ja_1 - a_jb_1}{a_1} $, $ j = 2, 3 $. Equation (1) then takes the form
By comparing the constant terms and the coefficients of $ y $ and $ y^n $ (we use here the assumption $ n > 2 $) on both sides of equation (2), we get
If $ B_2 = 0 $, then from (3) it follows that $ B_3 = 0 $ and therefore $ b_ja_1 - a_jb_1 = 0 $ for $ j = 2, 3 $, i.e., $ b_j = \displaystyle \frac{a_j}{a_1} b_1 $. It suffices then to take $ c_1 = 1 $, $ c_j = \displaystyle \frac{a_j}{a_1} $ for $ j = 2, 3 $, $ A = a_1 $, $ B = b_1 $.
If $ B_2 \ne 0 $, then from (3) it follows that $ B_3 = 0 $ and dividing both sides of (4) by (3) we get $ \displaystyle \frac{A_2}{B_2} = \frac{A_3}{B_3} $. Raising both sides of the last inequality to the $ n $-th power and using (3) we obtain $ A^n_2 = A^n_3 $. This contradicts equation (5). Therefore, this case cannot occur. | proof | Algebra | proof | Yes | Yes | olympiads | false | 1,085 |
XXXVI OM - III - Problem 5
Let $ P $ be a polynomial in two variables such that for every real number $ t $ the equality $ P(\cos t, \sin t) = 0 $ holds. Prove that there exists a polynomial $ Q $ such that the identity holds
保留了源文本的换行和格式，但最后一句“保留了源文本的换行和格式”是中文，应该翻译成英文如下：
Preserving the line breaks and format of the source text. | Let's arrange the polynomial $ P(x,y) $ in decreasing powers of the variable $ x $. $ P(x,y) = P_n(y) \cdot x^n + P_{n-1}(y) \cdot x^{n-1} + \ldots + P_1(y) \cdot x + P_0(y) $, where $ P_n, P_{n-1}, \ldots, P_1, P_0 $ are polynomials in the variable $ y $. Treating $ y $ as a fixed value, we can compute the quotient and remainder when dividing the polynomial $ P $ by the polynomial $ x^2 + y^2 - 1 $ (treated as a polynomial in the variable $ x $).
We will obtain
where $ Q_i $, $ R_j $ are polynomials in the variable $ y $. Substituting $ x = \cos t $, $ y = \sin t $ in the above equation, we get
since $ \cos^2 t + \sin^2 t - 1 = 0 $ for any $ t $. Given the assumption about the polynomial $ P $, we have
for any real $ t $. Substituting $ \pi - t $ for $ t $, we get
Adding this to the previous equation, we obtain $ R_0(\sin t) = 0 $. The polynomial $ R_0 $ takes the value $ 0 $ for infinitely many arguments, since $ \sin t $ can be any number in the interval $ \langle -1 ; 1 \rangle $. It follows that $ R_0 $ is the zero polynomial. Therefore, we have $ R_1(\sin t) \cdot \cos t = 0 $ for every $ t $, from which we similarly conclude that $ R_1 $ is the zero polynomial. Thus, $ P(x,y) = Q(x,y)(x^2 + y^2 - 1) $, where $ Q(x,y) = Q_{n-2}(y)x^{n-2} + \ldots + Q_n(y) $. | proof | Algebra | proof | Yes | Yes | olympiads | false | 1,086 |
XXXIV OM - I - Problem 6
Prove that for any polynomial $ P $, the polynomial $ P\circ P \circ P \circ \ldots \circ P(x)-x $ is divisible by $ P(x)-x $. | Putting $ P(x)= a_nx^n + a_{n-1}x^{n-1} + \ldots +a_1x+a_0 $ we get
Since each term of the last sum can be factored:
thus
Substituting in the last relation $ P(x) $ for $ x $ we get $ P \circ P(x)-P(x) | P \circ P \circ P(x) - P \circ P(x) $, from which by the transitivity of the divisibility relation it follows that $ P(x)-x|P \circ P \circ P(x)-P \circ P(x) $. Hence for $ k= 1, 2, \ldots $ we get
The polynomial $ \underbrace{P \circ P \circ \ldots \circ P}_n (x) - x $ can be represented as a sum
By the previous considerations, each term of this sum is divisible by $ P(x) - x $, from which it follows that $ P(x) - x|\underbrace{P \circ P \circ \ldots \circ P}_n(x) - x $. | proof | Algebra | proof | Yes | Yes | olympiads | false | 1,087 |
VI OM - I - Problem 12
In what part of an equilateral triangle should point $ P $ lie so that from segments equal to the distances of this point from the sides of the triangle, a triangle can be constructed? | Let $ k_1 $, $ k_2 $, $ k_3 $ denote the distances from a point $ P $ inside an equilateral triangle $ ABC $ to the sides $ BC $, $ CA $, $ AB $ of the triangle, respectively. We can form a triangle with segments of lengths $ k_1 $, $ k_2 $, $ k_3 $ if and only if they satisfy the following inequalities:
Draw a line through point $ P $ parallel to side $ AB $, and let $ D $ and $ E $ be the points where this line intersects sides $ AC $ and $ BC $ (Fig. 8).
In the equilateral triangle $ CDE $, the sum $ k_1 + k_2 $ of the distances from point $ P $ on side $ DE $ to the other sides equals the height $ CF $ of this triangle. Indeed, $\text{area of } CEP + \text{area of } CDP = \text{area of } CDE $, so $ CE \cdot k_1 + CD \cdot k_2 = DE \cdot CF $, from which $ k_1 + k_2 = CF $. The condition $ k_3 < k_1 + k_2 $ can thus be written as $ k_3 < CF $, and since $ k_3 + CF $ equals the height $ h $ of triangle $ ABC $, this condition is equivalent to
Similarly, the inequalities $ k_1 < k_2 + k_3 $ and $ k_2 < k_3 + k_1 $ are equivalent to the inequalities
A point $ P $ satisfies these three conditions if and only if it lies inside the triangle formed by the midpoints of the sides of triangle $ ABC $. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,088 |
XXXVII OM - II - Problem 2
In a chess tournament, 66 players participate, each playing one game against every other player, and the matches take place in four cities. Prove that there exists a trio of players who play all their games against each other in the same city. | Let's choose one of the players, let's call him $Z_1$. He has to play $65$ games, so he plays at least $17$ games in one city. Let's denote this city by $M_1$. Consider the opponents of $Z_1$ in the matches played in $M_1$. There are at least $17$ of them. If there is a pair among them who play a game against each other also in $M_1$, then together with player $Z_1$ they form a trio playing a "triangle of matches" in one city, and the theorem is proved. Therefore, let's assume that the entire group of $17$ (or more) chess players play all their matches against each other in the remaining three cities. Choose one from this group and call him $Z_2$. He has at least $16$ opponents in this group, so he must play with at least $6$ of them in one city, which we will denote by $M_2$. If any pair among them plays their game also in $M_2$, this forms a "triangle of matches" in $M_2$. So let's assume that this is not the case; this means that the considered group of $6$ (or more) chess players play their matches against each other in the remaining two cities. We repeat the same reasoning. Choose one player from this group and call him $Z_3$, and note that having at least $5$ opponents in this group, he must play with at least $3$ of them in one city $M_3$. If there is a pair among them who play their game in $M_3$, then together with player $Z_3$ they form a trio playing a "triangle of matches" in $M_3$. Otherwise, the entire group of $3$ (or more) chess players play their matches against each other in the last remaining city $M_4$. Thus, the theorem is proved.
Note. The given proof is inductive in nature, consisting of several steps, each of which repeats the same reasoning pattern. This observation leads to the following natural generalization; it is convenient to formulate it in the language of so-called graph theory.
Let $(n_k)$ be a sequence of numbers defined by the recurrence relation $n_1 = 2$, $n_k = kn_{k-1} + 1$ for $k \geq 2$. Suppose that $G$ is a set of $n$ points in space, $n > n_k$, and that each of the segments connecting the points of set $G$ has been colored with one of $k$ colors. Then there are three points in set $G$ such that the triangle with vertices at these points has all sides of the same color.
(The initial terms of the sequence $(n_k)$ are the numbers $2$, $5$, $16$, $65$, which are the numbers appearing in the solution of the problem; for $k = 4$ we obtain the theorem given in the problem, where of course the chess players should be interpreted as points of set $G$, the matches as segments, and the cities as colors).
The general theorem formulated just now can be easily proved by induction using the reasoning pattern used in the solution of the problem.
It is worth noting that the sequence $(n_k)$ defined by the given recurrence relation is not optimal; if $k \geq 4$, then there are numbers $n \leq n_k$ such that for any $n$-element set of points $G$ the conclusion of the theorem is true (with $k$ colors); in particular, the theorem from the problem remains true when the number $66$ is replaced by the number $65$. However, no formula is known that gives a non-trivial lower bound for the allowable values of $n$, or the asymptotic behavior of these minimal values as $k \to \infty$. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 1,092 |
XI OM - I - Problem 8
In the plane, there are three rays $ OX $, $ OY $, $ OZ $, dividing the plane into three convex angles, and points $ A $, $ B $, $ C $ lying inside the angles $ YOZ $, $ ZOX $, $ XOY $, respectively. Construct a triangle whose vertices lie on $ OX $, $ OY $, $ OZ $, and whose sides pass through $ A $, $ B $, $ C $. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,094 | |
XXIII OM - III - Problem 6
Prove that the sum of the digits of the number $ 1972^n $ tends to infinity as $ n $ tends to infinity. | We will prove in general that if $a$ is an even natural number not divisible by $5$ and $s_n$ denotes the sum of the digits of the number $a^n$ for $n = 1, 2, \ldots$, then the sequence $s_n$ tends to infinity.
Let $a_r, a_{r-1}, \ldots, a_2, a_1$, where $a \ne 0$, be the consecutive digits of the number $a^n$, i.e.,
The digit $a_1$ is different from zero, because by assumption the number $a$ is not divisible by $5$, and thus the number $a_n$ is not divisible by $5$. We also have $a^n < 10^r$, so $r > n \log_{10} a \geq n \log_{10} 2$.
We will prove
Lemma.
For every natural number $j$ satisfying the condition
at least one of the digits $a_{j+1}, a_{j+2}, \ldots, a_{4j}$ of the number $a^n$ is different from zero.
Proof. If for some natural number $j$ satisfying (1) it were the case that $a_{j+1} = a_{j+2} = \ldots = a_{4j} = 0$, then denoting
we would have $a^n - c = (a_r a_{r-1} \ldots a_{4j+1} 0 \ldots 0)_{10}$. Thus $10^{4j} \mid a^n - c$ and therefore $2^{4j} \mid a^n - c$. Moreover, from $2 \mid a$ it follows that $2^n \mid a^n$. Hence, given $4^j \leq n$, we obtain $2^{4j} \mid a^n - (a^n - c) = c$. However, $2^{4j} = 16^j > 10^j > c$. Therefore, $c = 0$. This is impossible, however, because the last digit $a_1$ of the number $c$ is different from zero. The obtained contradiction completes the proof of the lemma.
From the assumption and this lemma, it follows in particular that in each of the following sequences of digits there is a term different from zero:
where the number $j = 4^k$ satisfies condition (1), i.e., $4^k \leq \frac{1}{4} \min (r, n)$. Since $\frac{1}{4} \min (r, n) \geq \frac{1}{4} \min (n \log_{10} 2, n) = \frac{1}{4} n \log_{10} 2 > \frac{1}{16} n$, we can take $k$ to be the largest natural number satisfying one of the following equivalent inequalities
The sequences (2) contain different digits of the number $a^n$, there are $k + 2$ of these sequences, and each of them contains a digit different from zero. Therefore, the sum of the digits $s_n$ of the number $a^n$ is not less than $k + 2 = [\log_4 n]$.
Since $s_n \geq [\log_4 n] > \log_4 n - 1$ and $\lim_{n \to \infty} \log_4 n = \infty$, then $\lim_{n \to \infty} s_n = \infty$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,095 |
XVI OM - I - Problem 3
Through each edge of a trihedral angle, a plane containing the bisector of the opposite planar angle has been drawn. Prove that the three planes intersect along a single line. | Consider a trihedral angle with vertex $O$. We measure three equal segments $OA = OB = OC$ on its edges. Consider the plane $\alpha$ passing through the edge $OA$ and the bisector of the angle $BOC$. Since the triangle $BOC$ is isosceles, the bisector of the angle $BOC$ passes through the midpoint $M$ of the side $BC$. The plane $\alpha$, on which points $A$ and $M$ lie, passes through the median $AM$ of the triangle $ABC$ and thus through the centroid $S$ of this triangle. Therefore, the plane $\alpha$ passes through the line $OS$.
In the above reasoning, the plane $\alpha$ can be replaced by a plane passing through another edge of the trihedral angle and the bisector of the opposite face. All three of these planes thus pass through the line $OS$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,096 |
XVIII OM - II - Problem 3
Two circles are internally tangent at point $ A $. A chord $ BC $ of the larger circle is tangent to the smaller circle at point $ D $. Prove that $ AD $ is the angle bisector of $ \angle BAC $. | The point of tangency $ A $ of the given circles is their center of homothety. In this homothety, the tangent $ BC $ of the smaller circle corresponds to the tangent to the larger circle at point $ E $, which corresponds to point $ D $ (Fig. 6). As homothetic lines, these tangents are parallel. Therefore, the arcs $ BE $ and $ EC $ of the larger circle, determined by the tangent at point $ E $ and the chord $ BC $ parallel to it, are equal. Hence, the inscribed angles $ BAE $ and $ EAC $ are equal, i.e., $ AE $ is the angle bisector of $ BAC $. | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,097 |
XLVI OM - II - Problem 2
In a convex hexagon $ABCDEF$, the following equalities hold: $|AB| = |BC|$, $|CD| = |DE|$, $|EF| = |FA|$. Prove that the lines containing the altitudes of triangles $BCD$, $DEF$, and $FAB$ drawn from vertices $C$, $E$, and $A$, respectively, intersect at a single point. | Let's adopt the following notations:
$ k_1 $ - the circle with center $ D $ and radius $ |DC| = |DE| $;
$ k_2 $ - the circle with center $ F $ and radius $ |FE| = |FA| $;
$ k_3 $ - the circle with center $ B $ and radius $ |BA| = |BC| $.
Circles $ k_2 $ and $ k_3 $ intersect at point $ A $; let's denote the second point of their intersection by $ A' $ (see Note 2); similarly, denote by $ C' $ the second (other than $ C $) point of intersection of circles $ k_3 $ and $ k_1 $, and by $ E' $ the second (other than $ E $) point of intersection of circles $ k_1 $ and $ k_2 $ (see Figure 7). Points $ A $ and $ A' $ are symmetric with respect to the line passing through the centers $ F $ and $ B $ of circles $ k_2 $ and $ k_3 $; thus, $ AA' $ contains the altitude of triangle $ FAB $ drawn from vertex $ A $; it is therefore one of the three lines mentioned in the problem statement. The other two lines, in question, are - analogously - lines $ CC' $ and $ EE' $.
om46_2r_img_7.jpg
Line $ AA' $ is the power axis of the pair of circles $ k_2 $, $ k_3 $; line $ CC' $ is the power axis of the pair $ k_3 $, $ k_1 $, and line $ EE' $ is the power axis of the pair $ k_1 $, $ k_2 $. It is known (see Note 1) that for any triplet of circles, the three power axes determined by pairs of these circles form a pencil of lines intersecting at one point or are parallel. In the considered situation, lines $ AA' $, $ CC' $, $ EE' $, being perpendicular to the sides $ FB $, $ BD $, $ DF $ of triangle $ DFB $, cannot be parallel - and thus intersect at one point. The proof is complete.
Note 1. For Readers who are not sufficiently familiar with the theorem on the collinearity of power axes, we will sketch its proof, along with the definitions of the used concepts.
Let $ P $ be a point in the plane of circle $ k $ and let $ l $ be any line passing through point $ P $ and intersecting circle $ k $ at two points $ U $ and $ V $. The value of the product $ |PU| \cdot |PV| $ depends only on circle $ k $ and point $ P $, but not on the choice of line $ l $. [Justification: suppose another line $ l' $ intersects circle $ k $ at points $ U' $ and $ V' $; the well-known equality $ |PU| \cdot |PV| = |PU' \cdot |PV'| $ follows from the similarity of triangles $ PUU' $ and $ PUV' $.]
This constant value, taken with a positive sign if point $ P $ lies outside circle $ k $, and with a negative sign if $ P $ lies inside $ k $, is called the power of point $ P $ with respect to circle $ k $; it is equal to the difference $ |OP|^2 - r^2 $, where $ O $ and $ r $ are the center and radius of circle $ k $. [Justification: let $ UV $ be a diameter contained in line $ OP $; one of the segments $ PU $, $ PV $ has length $ |OP| + r $, and the other has length $ \pm (|OP| - r) $; the sign is positive or negative depending on whether point $ P $ lies outside $ k $ or inside $ k $; thus (respectively) $ |PU| \cdot |PV| = \pm (|OP|^2 - r^2) $.]
Let two circles with non-coinciding centers $ O_1 $, $ O_2 $ and radii $ r_1 $, $ r_2 $ be given. If point $ P $ has equal power with respect to both of these circles, then the same property also applies to point $ Q $ defined as the orthogonal projection of point $ P $ onto line $ O_1O_2 $; this follows from the equality $ |O_1Q|^2 - |O_2Q|^2 = |O_1P|^2 - |O_2P|^2 = r_1^2 - r_2^2 $; on line $ O_1O_2 $ there is exactly one point $ Q $ such that $ |O_1Q|^2 - |O_2Q|^2 = r_1^2 - r_2^2 $. Hence, the set of points with equal power with respect to two given circles is a line perpendicular to $ O_1O_2 $; it is called the power axis of this pair of circles. If the circles intersect at two points, then each of these points has zero power with respect to both circles, and thus the power axis is the line passing through the points of intersection. If the circles are tangent, the power axis is the tangent line to both at the point of tangency.
If we consider any triplet of circles $ k_1 $, $ k_2 $, $ k_3 $ (with pairwise distinct centers) and if these centers are not collinear, then the power axis of the pair $ k_1 $, $ k_3 $ intersects the power axis of the pair $ k_2 $, $ k_3 $. The intersection point has the same power with respect to all three circles, and thus lies on the power axis of the pair $ k_1 $, $ k_2 $. If, however, the centers of circles $ k_1 $, $ k_2 $, $ k_3 $ are collinear, then each of these three axes is perpendicular to the line passing through the centers. Hence the thesis of the theorem: the three considered power axes either intersect at one point or are parallel.
Note 2. The problem statement mentions triangles $ BCD $, $ DEF $, $ FAB $; it should be inferred from this that each of the angles $ BCD $, $ DEF $, $ FAB $ is less than a straight angle. However, the reasoning remains valid even if one of these angles is a straight angle (and the corresponding triangle degenerates into a segment; by "line containing the altitude" we then mean the line perpendicular to that segment and passing through the corresponding vertex). Two of the circles $ k_1 $, $ k_2 $, $ k_3 $ are then tangent; but their power axis is still the line containing the altitude (in the sense specified just now), so no modifications to the further reasoning are required. | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,098 |
XIII OM - I - Problem 2
How many digits do all natural numbers with at most $ m $ digits have in total? | The number of one-digit natural numbers is $ c_1 = 9 $; the number of two-digit numbers is $ c_2= 10^2 -10= 10 \cdot 9 $, they have $ 2 \cdot 10 \cdot 9 $ digits; the number of three-digit numbers is $ c_3=10^3-10^2= 10^2 \cdot 9 $, the number of their digits is $ 3 \cdot 10^2 \cdot 9 $, etc. Finally, the number of $ m $-digit numbers is $ c_m = 10^m -10^{m-1} = 10^{m-1} \cdot 9 $; they have $ c_m =m \cdot 10^{m-1} \cdot 9 $ digits.
All numbers with at most $ m $ digits have in total
digits. To calculate the value of the expression in parentheses in formula (1), we will first solve a more general problem, namely find the formula for the sum
where $ a \ne 1 $, depending on $ a $. From equality (2) we get
Subtracting equality (3) from (2) we have
from which
After arranging the terms, we get the formula
Taking $ a = 10 $ in (2) and (4) and substituting the obtained value of the sum (2) into equality (1), we get the desired number of
Note. We derived formula (4) under the assumption that $ a \ne 1 $. When $ a = 1 $, formula (4) makes no sense; in this case, the value of $ S $ can be obtained by substituting $ a = 1 $ in formula (2):
thus, according to the known formula for an arithmetic progression,
We can, however, arrive at formula (5) based on formula (4) by the following reasoning.
Let $ a -1= x $; substituting $ a = 1 + x $ in formula (4) we get
Expanding $ (1 + x)^{m+1} $ and $ (1 + x)^m $ according to the binomial theorem of Newton. We will only calculate the terms containing $ x^0 $, $ x $, and $ x^2 $; the remaining terms form a polynomial in $ x $, in which all terms are of at least the third degree and we can write
where $ f(x) $ denotes some polynomial in $ x $.
Hence,
or
Substituting $ a = 1 $ in formula (6) we get formula (5). It turns out, therefore, that formula (6) gives the value of the sum (2) for any value of $ a $. However, it is useful for calculating this sum only when $ a = 1 $; when $ a \ne 1 $, it must be restored to the form (4). | 9\cdot\frac{10^-1}{9}\cdot\frac{(+1)}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1,099 |
XLV OM - III - Task 5
Points $ A_1, A_2, \ldots , A_8 $ are the vertices of a parallelepiped with center $ O $. Prove that | Let's assume that one of the faces of a given parallelepiped is the parallelogram $A_1A_2A_3A_4$, and the opposite face is the parallelogram $A_5A_6A_7A_8$, with segments $A_1A_5$, $A_2A_6$, $A_3A_7$, $A_4A_8$ being four edges of the parallelepiped (Figure 17). Let's denote the distances from the vertices to the point $O$ as follows:
om45_3r_img_17.jpg
The inequality to be proven takes the form
equivalently:
Since
it remains to show that
Segments $OA_1$, $OA_4$, $A_1A_4$ are the three sides of a triangle; similarly, segments $OA_2$, $OA_3$, $A_2A_3$ are the three sides of a triangle. Therefore,
(triangle inequalities). Segments $A_1A_4$ and $A_2A_3$ are opposite sides of the parallelogram $A_1A_2A_3A_4$, so $|A_1A_4| = |A_2A_3|$. Hence, from relation (2), we obtain the inequality
or $a \leq b + c + d$. Multiplying both sides by $a$, we write it as
Similarly,
Adding these four inequalities side by side, we obtain inequality (1), which we wanted to prove. | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,100 |
LIX OM - III - Task 4
Each point in the plane with both integer coordinates has been painted either white or black. Prove that from the set of all painted points, an infinite subset can be selected which has a center of symmetry and all of whose points have the same color. | Suppose the thesis of the problem is false.
Consider the central symmetry with respect to the point $ (0,0) $. Since there does not exist an infinite set symmetric with respect to this point and composed of points of the same color, only finitely many points with integer coordinates pass to points of the same color under this symmetry. Therefore, there exists an integer $ M $ such that for every point with integer coordinates $ (x,y) $, where $ |y| > M $, the points $ (-x,-y) $ and $ (x,y) $ have different colors.
Similarly, considering the central symmetry with respect to the point $ (\frac{1}{2},0) $, we see that there exists an integer $ N $ such that for every point with integer coordinates $ (x,y) $, where $ |y| > N $, the point $ (x,y) $ has a different color than its image under the considered symmetry, which is the point $ (-x+1,-y) $.
Let $ k = \max\{M,N\} + 1 $ and consider any integer $ s $. Then the point $ (s,k) $ under the symmetry with respect to the point $ (0,0) $ maps to the point $ (-s,-k) $, which is of a different color than the point $ (s,k) $. Furthermore, the point $ (-s,-k) $ under the symmetry with respect to the point $ (\frac{1}{2}, 0) $ maps to the point $ (s +1,k) $, which is of a different color than the point $ (-s,-k) $.
Therefore, the points $ (s,k) $ and $ (s+1,k) $ have the same color. Since $ s $ was an arbitrary integer, it follows that all points with the first coordinate an integer and the second coordinate equal to $ k $ have the same color.
However, these points form an infinite set, whose center of symmetry is the point $ (0,k) $.
The obtained contradiction with the indirect assumption completes the solution. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 1,102 |
XXVII OM - II - Problem 5
Prove that if $ \cos \pi x =\frac{1}{3} $ then $ x $ is an irrational number. | Let $ \cos t = \frac{1}{3} $. We will prove by induction that for every natural number $ n $ the following formula holds:
where $ a_n $ is an integer not divisible by $ 3 $.
For $ n=1 $, it suffices to take $ a_1 = 1 $. For $ n = 2 $ we have
Thus, $ a_2 = -7 $. Suppose next that for some natural number $ k $ the following holds:
where $ a_k $ and $ a_{k+1} $ are integers not divisible by $ 3 $. We have $ \cos (k + 2)t + \cos kt = 2\cos (k + 1)t \cdot \cos t $. Therefore,
The number $ a_{k+2} = 2a_{k+1} - 9a_k $ is clearly an integer and not divisible by $ 3 $. Hence, by the principle of induction, formula (1) holds for every natural number $ n $.
Suppose now that for some rational number $ x = \frac{r}{s} $ the following holds:
$ \cos \pi x = \frac{1}{3} $. Then $ t = \pi x $ and by formula (1) for $ n = 2s $ we have
where $ a_{2s} $ is an integer not divisible by $ 3 $.
On the other hand, we have
The obtained contradiction proves that for no rational number $ x $ does $ \cos \pi x = \frac{1}{3} $ hold.
Note. Reasoning similarly, one can prove that if the number $ \cos \pi x $ is rational and different from $ 0 $, $ \pm \frac{1}{2} $, and $ \pm 1 $, then $ x $ is an irrational number.
See also articles in the journal "Mathematics" No. 5(128) from 1973, pp. 313-317, and No. 3(132) from 1974, pp. 168-172. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,103 |
LVIII OM - I - Problem 11
For each positive integer $ n $, determine the number of permutations $ (x_1,x_2,\ldots,x_{6n-1}) $ of the set $ \{1,2,\ldots,6n{-}1\} $, satisfying the conditions: | The sought number is
For any permutation satisfying the conditions of the problem, the following sequences of inequalities hold
Notice that each of the elements $ x_1 $, $ x_2 $, $ \ldots $, $ x_{6n-1} $ has been listed exactly once in the considered rows. Indeed, in the inequalities (1), all odd indices from the interval $ \langle 1;2n-1\rangle $, even indices from the interval $ \langle 2n;4n\rangle $, and odd indices from the interval $ \langle 4n+1;6n-1\rangle $ appear, while in the inequalities (2) - all even indices from the interval $ \langle 2;2n-2\rangle $, odd indices from the interval $ \langle 2n+1; 4n-1\rangle $, and even indices from the interval $ \langle 4n+2;6n-2\rangle $. By examining the number of even and odd numbers in the respective intervals, we calculate that in relation (1) there are $ n+(n+1)+n=3n+1 $ numbers, and in relation (2) there are $ (n-1)+n+(n-1)=3n-2 $ numbers.
It is not difficult to observe that every inequality of the form $ x_i < x_j $ or $ x_i > x_j $ that can be inferred from the conditions of the problem is found in relation (1) or (2). This means that if for a certain permutation both of these sequences of inequalities are satisfied, then the permutation has the postulated properties. Therefore, each permutation satisfying the conditions of the problem is uniquely determined by specifying the $ 3n+1 $ numbers listed in the inequalities (1) - it is necessary to arrange them in ascending order, take the smallest of these numbers as $ x_{2n} $, the next number as $ x_{4n+1} $, etc., which allows us to uniquely determine the terms of the permutation appearing in relations (1); the remaining $ 3n-2 $ numbers, when arranged in ascending order, similarly determine the terms of the permutation appearing in relations (2).
Thus, the number of permutations having the desired property is equal to the number of $ (3n+1) $-element subsets of a $ (6n-1) $-element set, i.e., the number | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1,104 |
XV OM - I - Problem 8
On three pairwise skew edges of a cube, choose one point on each in such a way that the sum of the squares of the sides of the triangle formed by them is minimized. | Let $ABCD$ be the base of a cube and $AA_1$, $BB_1$, $CC_1$, $DD_1$ its edges perpendicular to $ABCD$. Let $X$, $Y$, $Z$ be points chosen on three pairwise skew edges of the cube. It is known that by rotating the cube around certain axes passing through its center, the cube can be superimposed onto itself in such a way that any chosen edge can cover any other edge. Therefore, without loss of generality, we can assume that point $X$ lies on edge $A_1$. The skew edges to $AA_1$ are $BC$, $CD$, $B_1C_1$, $C_1D_1$. (Fig. 8); among them, the mutually skew ones are $BC$ and $C_1D_1$ as well as $B_1C_1$ and $CD$. Thus, we have two sets of three pairwise skew edges: $AA_1$, $BC$, $C_1D_1$ and $AA_1$, $B_1C_1$, $CD$, which transform into each other when the cube is reflected through the plane $AA_1C_1C$, which is a plane of symmetry of the cube. Therefore, we can assume that the considered set of three pairwise skew edges is $AA_1$, $BC$, $C_1D_1$ and that, for example, point $Y$ lies on $BC$ and point $Z$ on $C_1D_1$. Let $AX = x$, $BY = y$, $C_1Z = z$, $AA_1 = a$. Then
from the obtained formula, it is clear that the sum of the squares of the sides of triangle $XYZ$ reaches a minimum when $x = y = z = \frac{a}{2}$, i.e., when points $X$, $Y$, $Z$ are chosen at the midpoints of the respective edges. The minimum value is $\frac{9}{2} a^2$. | \frac{9}{2}^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,107 |
III OM - III - Task 5
Prove that none of the digits $ 2 $, $ 4 $, $ 7 $, $ 9 $ can be the last digit of the number
where $ n $ is a natural number. | If the last digit of a given number is $ x $, then
so
Since $ n $ is an integer, the discriminant of the above equation, i.e., the number
is a square of an integer. The last digit of this discriminant is the last digit of the number $ 8x + 1 $. When $ x $ equals $ 2 $, $ 4 $, $ 7 $, $ 9 $, then $ 8x + 1 $ ends with the digit $ 7 $, $ 3 $, $ 7 $, $ 3 $, respectively.
The square of an integer ending in the digit $ c $ has the form $ (10a + c)^2 = 100a^2 + 20ac + c^2 $, and thus has the same last digit as $ c^2 $. Since the squares of numbers from $ 0 $ to $ 9 $ do not end in the digit $ 7 $ or $ 3 $, $ x $ cannot be any of the digits $ 2 $, $ 4 $, $ 7 $, $ 9 $. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,108 |
LII OM - I - Problem 9
Prove that among any $12$ consecutive positive integers, there exists a number that is not the sum of $10$ fourth powers of integers. | A number that is the fourth power of an integer gives a remainder of $0$ or $1$ when divided by $16$. Indeed: for even numbers of the form $2k$ we have $(2k)^4 = 16k^4$, whereas for odd numbers $2k + 1$ we obtain
(The number $\frac{1}{2} k(3k+1)$ is an integer for any integer $k$).
Hence, a number that is the sum of $10$ fourth powers of integers gives one of eleven remainders when divided by $16$: $0,1,2,3,\ldots, 10$. Among $12$ consecutive integers, no two give the same remainder when divided by $16$. Therefore, one of these $12$ integers does not give a remainder from the set $\{0,1,2,3,\ldots,10\}$ when divided by $16$. This number cannot then be the sum of $10$ fourth powers of integers. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,110 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.