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VI OM - II - Task 3
What should be the angle at the vertex of an isosceles triangle so that a triangle can be constructed with sides equal to the height, base, and one of the remaining sides of this isosceles triangle? | We will adopt the notations indicated in Fig. 9. A triangle with sides equal to $a$, $c$, $h$ can be constructed if and only if the following inequalities are satisfied:
Since in triangle $ADC$ we have $a > h$, $\frac{c}{2} + h > a$, the first two of the above inequalities always hold, so the necessary and sufficient condition for the existence of a triangle with sides $a$, $c$, $h$ is the inequality
From triangle $ADC$ we have $h = a \cos \frac{x}{2}$, $\frac{c}{2} = a \sin \frac{x}{2}$; substituting into inequality (1) gives
or
and since $\frac{x}{4} < 90^\circ$, the required condition takes the form
or
Approximately, $4 \arctan -\frac{1}{2} \approx 106^\circ$ (with a slight deficit). | 106 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
IX OM - II - Task 2
Six equal disks are placed on a plane in such a way that their centers lie at the vertices of a regular hexagon with a side equal to the diameter of the disks. How many rotations will a seventh disk of the same size make while rolling externally on the same plane along the disks until it returns to its initial position? | Let circle $K$ with center $O$ and radius $r$ roll without slipping on a circle with center $S$ and radius $R$ (Fig. 16). The rolling without slipping means that different points of one circle successively coincide with different points of the other circle, and in this correspondence, the length of the arc between two points of one circle equals the length of the arc between the corresponding points of the other circle.
Consider two positions $K$ and $K_1$ of the moving circle, corresponding to centers $O$ and $O_1$ and points of tangency with the fixed circle $A$ and $B_1$; let $B$ be the point on the circumference of circle $K$ that, in the new position, passes to the point of tangency $B_1$, and let $\alpha$ and $\beta$ denote the radian measures of angles $ASB_1$ and $AOB$. Since the length of arc $AB_1$ of the fixed circle equals the length of arc $AB$ of the moving circle, we have $R \alpha = r \beta$. The radius $OA$ of the moving circle will take the new position $O_1A_1$, with $\measuredangle A_1O_1B_1 = \measuredangle AOB = \beta$. Draw the radius $O_1A_0$ parallel to $OA$; we get $\measuredangle A_0O_1B_1 = \measuredangle ASB_1 = \alpha$. The angle $A_0O_1A_1 = \alpha + \beta$ is equal to the angle through which circle $K$ has rotated when moving from position $K$ to position $K_1$. If $R = r$, then $\beta = \alpha$, and the angle of rotation is then $2 \alpha$.
Once the above is established, it is easy to answer the posed question.
The moving disk $K$ rolls successively on six given disks; the transition from one to another occurs at positions where disk $K$ is simultaneously tangent to two adjacent fixed disks. Let $K_1$ and $K_2$ be the positions where disk $K$ is tangent to fixed disk $N$ and to one of the adjacent disks (Fig. 17). Then $\measuredangle O_1SO_2 = 120^\circ$, so when moving from position $K_1$ to position $K_2$, the disk rotates by $240^\circ$. The disk will return to its initial position after $6$ such rotations, having rotated a total of $6 \cdot 240^\circ = 4 \cdot 360^\circ$, i.e., it will complete $4$ full rotations.
Note. We can more generally consider the rolling of a disk with radius $r$ on any curve $C$ (Fig. 18). When the point of tangency $A$ traverses an arc $AB_1$ of length $l$ on the curve, the angle through which the disk rotates is $\alpha + \beta$, where $\alpha$ is the angle between the lines $OA$ and $O_1B_1$, i.e., the angle between the normals to the curve $C$ at points $A$ and $B$, and $\beta = \frac{l}{r}$. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXXVI OM - III - Problem 1
Determine the largest number $ k $ such that for every natural number $ n $ there are at least $ k $ natural numbers greater than $ n $, less than $ n+17 $, and coprime with the product $ n(n+17) $. | We will first prove that for every natural number $n$, there exists at least one natural number between $n$ and $n+17$ that is coprime with $n(n+17)$.
In the case where $n$ is an even number, the required property is satisfied by the number $n+1$. Of course, the numbers $n$ and $n+1$ are coprime. If a number $d > 1$ were a common divisor of $n+1$ and $n+17$, then it would divide the difference $(n+17)-(n+1) = 16$, and thus would be an even number. However, since $n+1$ is an odd number, such a $d$ does not exist. Therefore, the numbers $n+1$ and $n(n+17)$ are coprime.
In the case where $n$ is an odd number, the required property is satisfied by the number $n + 16$. Indeed, the numbers $n+16$ and $n+17$ are coprime. Similarly, as above, we conclude that $n$ and $n+16$ are also coprime. Therefore, $n+16$ and $n(n+17)$ are coprime.
In this way, we have shown that $k \geq 1$.
Consider the number $n = 16!$. The numbers $n+2, n+3, \ldots, n+16$ are not coprime with $n(n+17)$, because for $j = 2, 3, \ldots, 16$, the numbers $n+j$ and $n$ are divisible by $j$. Only the number $n+1$ is coprime with $n(n+17)$. From this example, it follows that $k \leq 1$. Therefore, $k = 1$. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXXVIII OM - III - Zadanie 5
Wyznaczyć najmniejszą liczbę naturalną $ n $, dla której liczba $ n^2-n+11 $ jest iloczynem czterech liczb pierwszych (niekoniecznie różnych).
|
Niech $ f(x) = x^2-x+11 $. Wartości przyjmowane przez funkcję $ f $ dla argumentów całkowitych są liczbami całkowitymi niepodzielnymi przez $ 2 $, $ 3 $, $ 5 $, $ 7 $. Przekonujemy się o tym badając reszty z dzielenia $ n $ i $ f(n) $ przez te cztery początkowe liczby pierwsze:
\begin{tabular}{lllll}
&\multicolumn{4}{l}{Reszty z dzielenia:}\\
&przez 2&przez 3&przez 5&przez 7\\
$ n $&0 1&0 1 2 &0 1 2 3 4&0 1 2 3 4 5 6\\
$ f(n) $&1 1&2 2 1&1 1 3 2 3& 4 4 6 3 2 3 6
\end{tabular}
Zatem dowolna liczba $ N $ będąca wartością $ f $ dla argumentu naturalnego i spełniająca podany w zadaniu warunek musi mieć postać $ N = p_1p_2p_3p_4 $, gdzie czynniki $ p_i $ są liczbami pierwszymi $ \geq 11 $.
Najmniejsza z takich liczb $ N= 11^4 $ prowadzi do równania kwadratowego $ x^2-x+11=11^4 $ o pierwiastkach niewymiernych. Ale już druga z kolei $ N = 11^3 \cdot 13 $ jest równa wartości $ f(132) $. Funkcja $ f $ jest ściśle rosnąca w przedziale $ \langle 1/2; \infty) $ wobec czego znaleziona minimalna możliwa wartość $ N $ wyznacza minimalną możliwą wartość $ n $. Stąd odpowiedź: $ n = 132 $.
| 132 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XII OM - II - Task 4
Find the last four digits of the number $ 5^{5555} $. | \spos{1} We will calculate a few consecutive powers of the number $ 5 $ starting from $ 5^4 $:
It turned out that $ 5^8 $ has the same last four digits as $ 5^4 $, and therefore the same applies to the numbers $ 5^9 $ and $ 5^5 $, etc., i.e., starting from $ 5^4 $, two powers of the number $ 5 $, whose exponents differ by a multiple of $ 4 $, have the same last four digits. The number $ 5^{5555} = 5^{4\cdot 1388+3} $ therefore has the same last $ 4 $ digits as the number $ 5^7 $, i.e., the digits $ 8125 $. | 8125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XVIII OM - II - Problem 4
Solve the equation in natural numbers
Note: The equation itself is not provided in the original text, so it is not included in the translation. | Suppose that the triple $(x, y, z)$ of natural numbers satisfies equation (1). After dividing both sides of equation (1) by $xyz$ we get
One of the numbers $x$, $y$, $z$ is less than $3$; for if $x \geq 3$, $y \geq 3$, $z \geq 3$, then the left side of equation (2) is $\leq 1$, while the right side is $> 1$.
If $x < 3$, then one of the following cases occurs:
a) $x = 1$; from equation (1) it follows that
b) $x = 2$; from equation (1) we get $2y + yz + 2z = 2 yz + 2$, hence
Therefore, $y-2 = 1$, $z-2 = 2$ or $y-2 = 2$, $z-2 = 1$, i.e., $y = 3$, $z = 4$ or $y = 4$, $z = 3$.
Analogous solutions are obtained assuming that $y = 2$ or that $z = 2$.
Equation (1) thus has $7$ solutions in natural numbers:
Note: Since
equation (1) is equivalent to the equation
which, after substituting $x-1 = X$, $y -1 = Y$, $z-1 = Z$, takes the form
The solution of equation (1b) in natural numbers was the subject of No. 9 in the XVII Mathematical Olympiad. | 7 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
III OM - I - Task 4
a) Given points $ A $, $ B $, $ C $ not lying on a straight line. Determine three mutually parallel lines passing through points $ A $, $ B $, $ C $, respectively, so that the distances between adjacent parallel lines are equal.
b) Given points $ A $, $ B $, $ C $, $ D $ not lying on a plane. Determine four mutually parallel planes passing through points $ A $, $ B $, $ C $, $ D $, respectively, so that the distances between adjacent parallel planes are equal. | a) Suppose that the lines $a$, $b$, $c$ passing through points $A$, $B$, $C$ respectively and being mutually parallel satisfy the condition of the problem, that is, the distances between adjacent parallel lines are equal. Then the line among $a$, $b$, $c$ that lies between the other two is equidistant from them. Let this line be, for example, line $b$. In this case, points $A$ and $C$ are equidistant from line $b$ and lie on opposite sides of it; therefore, line $b$ intersects segment $AC$ at its midpoint $M$. From the fact that points $A$, $B$, $C$ do not lie on a straight line, it follows that point $M$ is different from point $B$.
Hence the construction: we draw line $b$ through point $B$ and through the midpoint $M$ of segment $AC$, and then we draw through points $A$ and $C$ lines $a$ and $c$ parallel to line $b$ (Fig. 12). The parallel lines $a$, $b$, $c$ determined in this way solve the problem, since points $A$ and $C$, and thus lines $a$ and $c$, are equidistant from line $b$ and lie on opposite sides of this line.
We found the above solution assuming that among the sought lines $a$, $b$, $c$, the line $b$ lies between lines $a$ and $c$; since the "inner" line can equally well be $a$ or $c$, the problem has three solutions.
b) Suppose that the planes $\alpha$, $\beta$, $\gamma$, $\delta$, passing through points $A$, $B$, $C$, $D$ respectively and being mutually parallel, satisfy the condition of the problem, that is, the distances between adjacent planes are equal. Let these planes be in the order $\alpha$, $\beta$, $\gamma$, $\delta$. We mean by this that plane $\beta$ is equidistant from planes $\alpha$ and $\gamma$, and plane $\gamma$ is equidistant from planes $\beta$ and $\delta$.
In this case, points $A$ and $C$ are equidistant from plane $\beta$ and lie on opposite sides of it, so plane $\beta$ passes through the midpoint $M$ of segment $AC$. Similarly, plane $\gamma$ passes through the midpoint $N$ of segment $BD$. From the fact that points $A$, $B$, $C$, $D$ do not lie in a plane, it follows that point $M$ is different from point $B$, and point $N$ is different from point $C$.
From this, we derive the following construction. We connect point $B$ with the midpoint $M$ of segment $AC$, and point $C$ with the midpoint $N$ of segment $BD$ (Figure 13 shows a parallel projection of the figure). Lines $BM$ and $CN$ are skew; if they lay in the same plane, then points $A$, $B$, $C$, $D$ would lie in the same plane, contrary to the assumption. We know from stereometry that through two skew lines $BM$ and $CN$ one can draw two and only two parallel planes $\beta$ and $\gamma$.
To do this, we draw through point $M$ a line $m$ parallel to line $CN$, and through point $N$ a line $n$ parallel to line $BM$; plane $\beta$ is then determined by lines $m$ and $BM$, and plane $\gamma$ by lines $n$ and $CN$.
Finally, we draw through points $A$ and $D$ planes $\alpha$ and $\delta$ parallel to planes $\beta$ and $\gamma$; we can determine them, as indicated in Figure 13, by drawing through each of points $A$ and $D$ lines parallel to lines $BM$ and $CN$.
The planes $\alpha$, $\beta$, $\gamma$, $\delta$ determined in this way solve the problem, since points $A$ and $C$, and thus planes $\alpha$ and $\gamma$, are equidistant from plane $\beta$ and lie on opposite sides of this plane - and similarly, planes $\beta$ and $\delta$ are equidistant from plane $\gamma$ and lie on opposite sides of this plane.
We found the above solution assuming that the sought planes are in the order $\alpha$, $\beta$, $\gamma$, $\delta$. For other orders of the sought planes, we will find other solutions in the same way. The number of all possible orders, or permutations of the letters $\alpha$, $\beta$, $\gamma$, $\delta$, is $4!$, i.e., $24$. However, note that two "reverse" permutations, such as $\alpha$, $\beta$, $\gamma$, $\delta$ and $\delta$, $\gamma$, $\beta$, $\alpha$, give the same solution. Therefore, the problem has $\frac{24}{2} = 12$ solutions corresponding to the permutations. | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
L OM - I - Task 3
In an isosceles triangle $ ABC $, angle $ BAC $ is a right angle. Point $ D $ lies on side $ BC $, such that $ BD = 2 \cdot CD $. Point $ E $ is the orthogonal projection of point $ B $ onto line $ AD $. Determine the measure of angle $ CED $. | Let's complete the triangle $ABC$ to a square $ABFC$. Assume that line $AD$ intersects side $CF$ at point $P$, and line $BE$ intersects side $AC$ at point $Q$. Since
$ CP= \frac{1}{2} CF $. Using the perpendicularity of lines $AP$ and $BQ$ and the above equality, we get $ CQ= \frac{1}{2} AC $, and consequently $ CP=CQ $. Points $C$, $Q$, $E$, $P$ lie on the same circle, from which $ \measuredangle CED =\measuredangle CEP =\measuredangle CQP = 45^\circ $. | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
LX OM - III - Zadanie 2
Let $ S $ be the set of all points in the plane with both coordinates being integers. Find
the smallest positive integer $ k $ for which there exists a 60-element subset of the set $ S $
with the following property: For any two distinct elements $ A $ and $ B $ of this subset, there exists a 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 |
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 |
XXVI - I - Problem 9
Calculate the limit
| For any natural numbers $ k $ and $ n $, where $ n \leq k $, we have
In particular, $ \displaystyle \binom{2^n}{n} \leq (2^n)^n = 2^{n^2} $. By the binomial formula, we have
and hence $ \displaystyle 2^n > \frac{n^2}{24} $. Therefore,
Since $ \displaystyle \lim_{n \to \infty} \frac{1}{n}= 0 $, it follows that $ \displaystyle \lim_{n \to \infty} \frac{1}{2^n} \log \binom{2^n}{n}= 0 $. | 0 | Calculus | math-word-problem | Incomplete | Yes | olympiads | false |
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 |
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 |
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 |
XXI OM - III - Task 6
Find the smallest real number $ A $ such that for every quadratic trinomial $ f(x) $ satisfying the condition
the inequality $ f.
holds. | Let the quadratic trinomial $ f(x) = ax^2 + bx + c $ satisfy condition (1). Then, in particular, $ |f(0)| \leq 1 $, $ \left| f \left( \frac{1}{2} \right) \right| \leq 1 $, and $ |f(1)| \leq 1 $.
Since
and
thus
Therefore, $ A \leq 8 $.
On the other hand, the quadratic trinomial $ f(x) = -8x^2 + 8x - 1 $ satisfies condition (1), which can be easily read from its graph (Fig. 15). Moreover, $ f. Thus, $ A \geq 8 $.
The sought number $ A $ is therefore equal to $ 8 $. | 8 | Inequalities | math-word-problem | Incomplete | Yes | olympiads | false |
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 |
VI OM - II - Task 2
Find the natural number $ n $ knowing that the sum
is a three-digit number with identical digits. | A three-digit number with identical digits has the form $ 111 \cdot c = 3 \cdot 37 \cdot c $, where $ c $ is one of the numbers $ 1, 2, \ldots, 9 $, and the sum of the first $ n $ natural numbers is $ \frac{1}{2} n (n + 1) $, so the number $ n $ must satisfy the condition
Since $ 37 $ is a prime number, one of the numbers $ n $ and $ n+1 $ must be divisible by $ 37 $. There are two possible cases:
a) $ n = 37 k $, where $ k $ is a natural number; equation (1) then gives
The right side here is at most $ 54 $, so the number $ k $ can be at most $ 1 $, i.e., $ n = 37 $. This value, however, is not a solution to the problem, as $ \frac{1}{2} n (n + 1) $ then equals $ 37 \cdot 19 = 703 $.
b) $ n + 1 = 37 k $ ($ k $ = natural number); equation (1) gives $ (37 k - 1) k = 2 \cdot 3 \cdot c $.
The only possible value for $ k $ is similarly to a) the number $ 1 $, in which case $ n = 36 $ and $ \frac{1}{2} n(n + 1) = 18 \cdot 37 = 666 $. Therefore, the only solution to the problem is $ n = 36 $. | 36 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
XXXIX OM - I - Problem 1
For each positive number $ a $, determine the number of roots of the polynomial $ x^3+(a+2)x^2-x-3a $. | Let's denote the considered polynomial by $ F(x) $. A polynomial of the third degree has at most three real roots. We will show that the polynomial $ F $ has at least three real roots - and thus has exactly three real roots (for any value of the parameter $ a > 0 $).
It is enough to notice that
If a continuous function with real values takes values of different signs at two points, then at some point in the interval bounded by these points, it takes the value zero (Darboux property).
Hence, in each of the intervals $ (-a-3; -2) $, $ (-2; 0) $, $ (0; 2) $, there is a root of the polynomial $ F $. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XXIV OM - II - Problem 3
Let $ f:\mathbb{R} \to \mathbb{R} $ be an increasing function satisfying the conditions:
1. $ f(x+1) = f(x) + 1 $ for every $ x \in \mathbb{R} $,
2. there exists an integer $ p $ such that $ f(f(f(0))) = p $. Prove that for every real number $ x $
where $ x_1 = x $ and $ x_n = f(x_{n-1}) $ for $ n = 2, 3, \ldots $. | We will first provide several properties of the function $f$ satisfying condition $1^\circ$ of the problem. Let $f_n(x)$ be the $n$-fold composition of the function $f$, i.e., let $f_1(x) = f(x)$ and $f_{n+1}(x) = f_n(f_1(x))$ for $n = 1, 2, \ldots$. It follows that if $n = k + m$, where $k$ and $m$ are natural numbers, then
We will prove that for any integer $r$, the formula
Consider first the case when $r$ is a natural number. We will use induction. For $r = 1$, formula (2) follows from condition $1^\circ$ of the problem. Assume that for some natural number $r$, formula (2) holds. We will prove the analogous formula for the number $r + 1$. From condition $1^\circ$ and the induction hypothesis, we obtain
Thus, by the principle of induction, formula (2) holds for every natural number $r$.
If $r = 0$, then formula (2) is obviously true. If finally $r = -s$ is a negative integer, then $s$ is a natural number, and therefore
Substituting $y = x - s$ here, we get $f(x) = f(x - s) + s$, i.e., $f(x - s) = f(x) - s$. This is formula (2) for $r = -s$.
By induction with respect to $n$, from formula (2), it follows that for any integer $r$ and natural number $n$,
We now proceed to solve the problem. Notice first that $x_{n+1} = f_n(x)$ for $n = 1, 2, \ldots$, so we need to find the limit of the sequence
$\displaystyle \frac{f_n(x)}{n+1}$. We will first prove that
Let $r$ be the remainder of the division of the natural number $n$ by $3$, i.e., let $n = 3k + r$, where $r = 0, 1$, or $2$ and $k \geq 0$. By induction with respect to $k$, we will prove that
If $k = 0$, then $n = r$ and formula (5) is true. If formula (5) holds for some $k \geq 0$, then the formula for the number $k + 1$ has the form
Formula (6) follows from (1), (3), and (5), since $f_{n+3}(0) = f_n(f_3(0)) = f_n(p) = f_n(0 + p) = f_n(0) + p = kp + f_r(0) + p = (k + 1)p + f_r(0)$.
Thus, by the principle of induction, formula (5) holds for any $k \geq 0$.
Since $k = \displaystyle \frac{n-r}{3} = \frac{n+1}{3} - \frac{1-r}{3}$, it follows from (5) that
Since the expression $(1-r)p + 3f_r(0)$ takes only three values (for $r = 0, 1, 2$), it is bounded. Therefore,
and from (7) it follows that (4).
Now, for any $x \in \mathbb{R}$, let $a = [x]$. Then $a \leq x < a + 1$. Since the function $f$, and therefore $f_n$, is increasing, it follows that
The number $a$ is an integer. Therefore, from (3) and (8), we obtain
Since $\displaystyle \lim_{n \to \infty} \frac{a}{n+1} = 0$, it follows from (4) and (9) by the squeeze theorem that $\displaystyle \lim_{n \to \infty} \frac{f_n(x)}{n+1} = 3$. | 3 | Algebra | proof | Incomplete | Yes | olympiads | false |
XLIV OM - III - Problem 3
Let $ g(k) $ denote the greatest odd divisor of the positive integer $ k $, and let us assume
The sequence $ (x_n) $ is defined by the relations $ x_1 = 1 $, $ x_{n+1} = f(x_n) $. Prove that the number 800 appears exactly once among the terms of this sequence. Determine $ n $ for which $ x_n = 800 $. | Let's list the first fifteen terms of the sequence $ (x_n) $, grouping them into blocks consisting of one, two, three, four, and five terms respectively:
We have obtained five rows of the infinite system (U), which can be continued according to the following rules: the $ j $-th row consists of $ j $ numbers, the last of which is $ 2j - 1 $; each number that is not the last term of any row is twice the number directly above it.
Let's adopt these rules as the definition of the system (U).
Every integer $ k \geq 1 $ can be uniquely represented in the form
and thus appears exactly once in the system (U), specifically at the $ m $-th position in the $ j $-th column (positions in the column are numbered from zero).
If $ k $ is an even number, and thus has the form (1) with $ m \geq 1 $, then $ g(k) = 2j - 1 $, and therefore
This means that $ f(k) $ is in the $ (j + 1) $-th column at the $ (m-1) $-th position, i.e., directly to the right of $ k $.
If $ k $ is an odd number, $ k = 2j - 1 $ (and thus forms the right end of the $ j $-th row), then
In this case, the number $ f(k) $ is the first term of the $ (j + 1) $-th (i.e., the next) row.
Therefore, reading the terms of the table (U) lexicographically, i.e., row by row (and in each row from left to right), we will traverse the consecutive terms of the sequence $ (x_n) $, in accordance with the recursive relationship defining this sequence $ x_{n+1} = f(x_n) $.
Every natural number $ k \geq 1 $ appears in the scheme (U), and thus in the sequence $ x_1, x_2, x_3, \ldots $ exactly once. This applies in particular to the number $ k = 800 $. It remains to locate this number. According to the general rule, the number $ 800 $, equal to $ 2^5(2\cdot 13 -1) $, is in the third column, at the fifth position. Therefore (see formulas (1) and (2)), the number $ f(800) $ is in the fourth column at the fourth position, the number $ f \circ f(800) $ is in the fifth column at the third position, and so on, the number $ f \circ f \circ f \circ f \circ f (800) $ is in the eighth column at the zeroth position, i.e., it forms the right end of the eighteenth row.
Since the $ j $-th row contains $ j $ terms, the number ending the eighteenth row appears in the sequence $ (x_n) $ with the number $ 1 + 2 + 3 + \ldots + 18 = 171 $. If $ 800 = x_N $, then we have the equality $ x_{N+5} = f \circ f \circ f \circ f \circ f (800) = x_{171} $. Therefore (due to the distinctness of the sequence $ (x_n) $), we conclude that $ N + 5 = 171 $, i.e., $ N = 166 $. | 166 | Number Theory | proof | Incomplete | Yes | olympiads | false |
XX OM - II - Task 2
Find all four-digit numbers in which the thousands digit is equal to the hundreds digit, and the tens digit is equal to the units digit, and which are squares of integers. | Suppose the number $ x $ satisfies the conditions of the problem and denote its consecutive digits by the letters $ a, a, b, b $. Then
The number $ x $ is divisible by $ 11 $, so as a square of an integer, it is divisible by $ 11^2 $, i.e., $ x = 11^2 \cdot k^2 $ ($ k $ - an integer), hence
Therefore,
The number $ a+b $ is thus divisible by $ 11 $. Since $ 0 < a \leq 9 $, $ 0 \leq b \leq 9 $, then $ 0 < a+b \leq 18 $, hence
Therefore, we conclude that $ b \ne 0 $, $ b \ne 1 $; since $ b $ is the last digit of the square of an integer, it cannot be any of the digits $ 2, 3, 7, 8 $. Thus, $ b $ is one of the digits $ 4, 5, 6, 9 $. The corresponding values of $ a $ are $ 7, 6, 5, 2 $, so the possible values of $ x $ are only the numbers $ 7744 $, $ 6655 $, $ 5566 $, $ 2299 $. Only the first one is a square of an integer.
The problem has one solution, which is the number $ 7744 $. | 7744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
V OM - I - Task 2
Investigate when the sum of the cubes of three consecutive natural numbers is divisible by $18$. | Let $ a - 1 $, $ a $, $ a + 1 $, be three consecutive natural numbers; the sum of their cubes
can be transformed in the following way:
Since one of the numbers $ a - 1 $, $ a $, $ a + 1 $ is divisible by $ 3 $, then one of the numbers $ a $ and $ (a + 1) (a - 1) + 3 $ is also divisible by $ 3 $. Therefore, the sum $ S $ of the cubes of three consecutive natural numbers is always divisible by $ 9 $.
The sum $ S $ is thus divisible by $ 18 = 9 \cdot 2 $ if and only if it is divisible by $ 2 $. Since $ S = 3a^3 + 6a $, $ S $ is even if and only if $ a $ is even, i.e., when $ a - 1 $ is odd. Hence, the conclusion:
The sum of the cubes of three consecutive natural numbers is divisible by $ 18 $ if and only if the first of these numbers is odd. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXXV OM - II - Task 5
Calculate the lower bound of the areas of convex hexagons, all of whose vertices have integer coordinates. | We will use the following lemma, which was the content of a competition problem in the previous Mathematical Olympiad.
Lemma. Twice the area of a triangle, whose all vertices have integer coordinates, is an integer.
Proof. The area of triangle $ABC$ is equal to half the absolute value of the determinant formed from the coordinates of vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$. If the coordinates of vertices $A$, $B$, $C$ are integers, then the coordinates of vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are also integers, the determinant formed from these coordinates is also an integer, and consequently, twice the area of triangle $ABC$ is an integer.
Every point on the plane with integer coordinates belongs to exactly one of the following four sets:
$A$ - the set of points whose both coordinates are even,
$B$ - the set of points whose both coordinates are odd,
$C$ - the set of points whose first coordinate is even and the second is odd,
$D$ - the set of points whose first coordinate is odd and the second is even.
When two points on the plane belong to one of these sets, we say they have the same parity type. If points $P = (p_1, p_2)$ and $Q = (q_1, q_2)$ have the same parity type, then the midpoint of segment $\overline{PQ}$ has coordinates $\left( \frac{1}{2} (p_1 + q_1), \frac{1}{2} (p_2 + q_2) \right)$, which are integers (the sum of numbers of the same parity is an even number).
Let's return to convex hexagons whose each vertex has integer coordinates. Among the six vertices of the hexagon, we can find two having the same parity type.
If two consecutive vertices (e.g., $K$ and $L$) of hexagon $KLMNOP$ have the same parity type, then connecting the midpoint $S$ of segment $\overline{KL}$ with point $M$ results in hexagon $KSMNOP$ having a smaller area than hexagon $KLMNOP$. When determining the lower bound of the areas of hexagons, we can therefore limit ourselves to hexagons in which consecutive vertices do not have the same parity type. Therefore, there exist some two non-consecutive vertices having the same parity type. The midpoint of the segment connecting these vertices has integer coordinates and lies inside the hexagon. Connecting this midpoint with all vertices of the hexagon results in six triangles, whose sum of areas equals the area of the hexagon. Therefore, by the lemma, the area of the hexagon is not less than $6 \cdot \frac{1}{2} = 3$. An example of a convex hexagon with vertices having integer coordinates and an area of $3$ is the hexagon with vertices $K = (-1, -1)$, $L = (0, -1)$, $M = (1, 0)$, $N = (1, 1)$, $O = (0, 1)$, $P = (-1, 0)$.
The sought lower bound is therefore $3$. | 3 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
LIX OM - II - Task 1
Determine the maximum possible length of a sequence of consecutive integers, each of which can be expressed in the form $ x^3 + 2y^2 $ for some integers $ x, y $. | A sequence of five consecutive integers -1, 0, 1, 2, 3 satisfies the conditions of the problem: indeed, we have
On the other hand, among any six consecutive integers, there exists a number, say
$ m $, which gives a remainder of 4 or 6 when divided by 8. The number $ m $ is even; if there were a representation
in the form $ m = x^3 +2y^2 $ for some integers $ x, y $, then the number $ x $ would be even. In this case, however,
we would obtain the divisibility $ 8|x^3 $ and as a result, the numbers $ m $ and $ 2y^2 $ would give the same remainder (4 or 6) when divided by 8.
Therefore, the number $ y^2 $ would give a remainder of 2 or 3 when divided by 4. This is impossible: the equalities
prove that the square of an integer can give a remainder of only 0 or 1 when divided by 4.
We have thus shown that among any six consecutive integers, there exists a number that cannot be represented
in the form $ x^3 +2y^2 $.
Answer: The maximum possible length of such a sequence is 5. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XV OM - I - Problem 11
In triangle $ ABC $, angle $ A $ is $ 20^\circ $, $ AB = AC $. On sides $ AB $ and $ AC $, points $ D $ and $ E $ are chosen such that $ \measuredangle DCB = 60^\circ $ and $ \measuredangle EBC = 50^\circ $. Calculate the angle $ EDC $. | Let $ \measuredangle EDC = x $ (Fig. 9). Notice that $ \measuredangle ACB = \measuredangle $ABC$ = 80^\circ $, $ \measuredangle CDB = 180^\circ-80^\circ-60^\circ = 40^\circ $, $ \measuredangle CEB = 180^\circ - 80^\circ-50^\circ = \measuredangle EBC $, hence $ EC = CB $. The ratio $ \frac{DC}{CE} $ of the sides of triangle $ EDC $ equals the ratio of the sides $ \frac{DC}{CB} $ of triangle $ BDC $, so the ratios of the sines of the angles opposite the corresponding sides in these triangles are equal:
The right side of the obtained equation can be transformed:
We need to find the convex angle $ x $ that satisfies the equation
or the equation
By transforming the products of sines into differences of cosines, we obtain an equivalent equation
Considering the condition $ 0 < x < 180^\circ $, we get $ x = 30^\circ $.
Note. The last part of the solution can be slightly shortened. Specifically, from the form of equation (1), it is immediately clear that it has a root $ x = 30^\circ $. No other convex angle satisfies this equation; if
then
so
thus, if $ 0 < x < 180^\circ $ and $ 0 < y < 180^\circ $, then $ x = y $. | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Given an integer $ n \geq 5 $. Determine the number of solutions in real numbers $ x_1, x_2, x_3, \ldots, x_n $ of the system of equations
where $ x_{-1}=x_{n-1}, $ $ x_{0}=x_{n}, $ $ x_{1}=x_{n+1}, $ $ x_{2}=x_{n+2} $. | Adding equations side by side, we obtain
Thus, the numbers $ x_1, x_2, \ldots, x_n $ take values 0 or 1. The task reduces to finding the number of solutions to the system of equations
in numbers belonging to the set $ \{0,1\} $.
Notice that the system of equations (1) is satisfied when $ x_i = x_2 = \dots = x_n = 1 $ or when $ x_1 = x_2 = \ldots = x_n = 0 $.
Assume that the number $ n $ is not divisible by 3. Then we obtain
from which $ x_n = x_2 $. Similarly, we prove that $ x_i = x_{i+2} $ for $ i = 1,2,\ldots,n-1 $. Therefore, if $ n $ is an odd number not divisible by 3, the only solutions $ (x_1, x_2, \ldots, x_n) $ to the system (1) are $ (0,0,0,\ldots,0) $ and $ (1,1,1,\ldots, 1) $. For even numbers $ n $ not divisible by 3, we obtain two additional solutions $ (x_1, x_2, \ldots, x_n) $, namely
Now assume that the number $ n $ is divisible by 3. Then the system of equations (1) has at least 8 solutions: the unknowns can repeat periodically with a period of 3.
Let's examine the existence of solutions $ (x_1, x_2, x_3, \ldots, x_n) $ that are not periodic with a period of 3. Without loss of generality, assume that $ x_1 = x_4 $. Then the equality $ x_1 + x_2 = x_4 + x_5 $ is possible only if $ x_1 + x_2 = 1 $. Hence, $ x_1 = x_2 $ and $ x_2 = x_5 $.
Repeating the reasoning, we find that $ x_2 = x_3 $ and $ x_3 = x_6 $, so for any $ i $ we have $ x_i = x_{i+1} $. Therefore, the unknowns must alternate between 0 and 1, which is possible only for even $ n $.
In summary: the number of solutions to the given system of equations is:
2 for $ n $ relatively prime to 6;
4 for even $ n $ not divisible by 3;
8 for odd $ n $ divisible by 3;
10 for $ n $ divisible by 6. | 2 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
XXVIII - II - Task 3
In a hat, there are 7 slips of paper. On the $ n $-th slip, the number $ 2^n-1 $ is written ($ n = 1, 2, \ldots, 7 $). We draw slips randomly until the sum exceeds 124. What is the most likely value of this sum? | The sum of the numbers $2^0, 2^1, \ldots, 2^6$ is $127$. The sum of any five of these numbers does not exceed $2^2 + 2^3 + 2^4 + 2^5 + 2^6 = 124$. Therefore, we must draw at least six slips from the hat.
Each of the events where we draw six slips from the hat, and the seventh slip with the number $2^{n-1}$ ($n = 1, 2, \ldots, 7$) remains in the hat, is equally likely. The probability of such an event is thus $\displaystyle \frac{1}{7}$.
The sum of the numbers on the drawn slips is equal to $127 - 2^{n-1}$. If $n = 1$, this sum is $126$; if $n = 2$, it is $125$; if $n = 3, 4, 5, 6$ or $7$, the sum is less than $124$ and we must draw a seventh slip. In this last case, the sum of the numbers on all the drawn slips will be $127$. Therefore, the probability that the sum of the numbers on all the slips drawn according to the conditions of the problem is $125$, $126$, or $127$, is $\displaystyle \frac{1}{7}$, $\displaystyle \frac{1}{7}$, $\displaystyle \frac{5}{7}$, respectively.
Thus, the most probable value of the sum is $127$. | 127 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XLII OM - I - Problem 8
Determine the largest natural number $ n $ for which there exist in space $ n+1 $ polyhedra $ W_0, W_1, \ldots, W_n $ with the following properties:
(1) $ W_0 $ is a convex polyhedron with a center of symmetry,
(2) each of the polyhedra $ W_i $ ($ i = 1,\ldots, n $) is obtained from $ W_0 $ by a translation,
(3) each of the polyhedra $ W_i $ ($ i = 1,\ldots, n $) has a point in common with $ W_0 $,
(4) the polyhedra $ W_0, W_1, \ldots, W_n $ have pairwise disjoint interiors. | Suppose that polyhedra $W_0, W_1, \ldots, W_n$ satisfy the given conditions. Polyhedron $W_1$ is the image of $W_0$ under a translation by a certain vector $\overrightarrow{\mathbf{v}}$ (condition (2)). Let $O_0$ be the center of symmetry of polyhedron $W_0$ (condition (1)); the point $O_1$, which is the image of $O_0$ under the translation by $\overrightarrow{\mathbf{v}}$, is the center of symmetry of $W_1$. Figure 3 illustrates a planar variant of the considered problem (a representation of the spatial configuration would obscure this illustration); polyhedra $W_0$ and $W_1$ are depicted as centrally symmetric polygons.
om42_1r_img_3.jpg
By condition (3), polyhedra $W_0$ and $W_1$ have common points (possibly many). Let $K$ be a common point of $W_0$ and $W_1$ (arbitrarily chosen). Denote by $L$ the image of point $K$ under the central symmetry with respect to $O_1$; thus, $L \in W_1$. Let $N$ be a point such that $\overrightarrow{NL} = \overrightarrow{\mathbf{v}}$ and let $M$ be the midpoint of segment $NK$ (Figure 4). Therefore, $N \in W_0$. According to condition (1), the set $W_0$ is convex; this means that with any two points belonging to $W_0$, the entire segment connecting these points is contained in $W_0$. Since $K \in W_0$ and $N \in W_0$, it follows that $M \in W_0$. Segment $MO_1$ connects the midpoints of segments $KN$ and $KL$, and thus $\overrightarrow{MO_1} = \frac{1}{2} \overrightarrow{NL} = \frac{1}{2} \overrightarrow{\mathbf{v}}$, which means $M$ is the midpoint of segment $O_0O_1$.
Let $U$ be the image of polyhedron $W_0$ under a homothety with center $O_0$ and scale factor 3. We will show that $W_1 \subset U$. Take any point $P \in W_1$: let $Q \in W_0$ be a point such that $\overrightarrow{QP} = \overrightarrow{\mathbf{v}}$ and let $S$ be the center of symmetry of parallelogram $O_0O_1PQ$. The medians $O_0S$ and $QM$ of triangle $O_0O_1Q$ intersect at a point $G$ such that $\overrightarrow{O_0G} = \frac{2}{3}\overrightarrow{O_0S} = \frac{1}{3}\overrightarrow{O_0P}$ (Figure 5). This means that $P$ is the image of point $G$ under the considered homothety. Since $G$ is a point on segment $QM$ with endpoints in the set (convex) $W_0$, it follows that $G \in W_0$. Therefore, $P \in U$ and from the arbitrariness of the choice of point $P \in W_1$ we conclude that $W_1 \subset U$.
om42_1r_img_4.jpg
om42_1r_img_5.jpg
In the same way, we prove that each of the sets $W_i (i=1, \ldots, n)$ is contained in $U$. Of course, also $W_0 \subset U$. Thus, the set $W_0 \cup W_1 \cup \ldots \cup W_n$ is a polyhedron contained in $U$. By conditions (2) and (4), its volume equals the volume of $W_0$ multiplied by $n+1$. On the other hand, the volume of $U$ equals the volume of $W_0$ multiplied by 27. Therefore, $n \leq 26$.
It remains to note that the value $n = 26$ can be achieved (example realization: 27 cubes $W_0, \ldots, W_{26}$ arranged like a Rubik's cube). Thus, the sought number is $26$.
Note. We obtain the same result assuming that $W_0, \ldots, W_n$ are any bounded, closed convex bodies (not necessarily polyhedra), with non-empty interiors, satisfying conditions (1)-(4); the reasoning carries over without any changes. Moreover, condition (4) turns out to be unnecessary. This was proven by Marcin Kasperski in the work 27 convex sets without a center of symmetry, awarded a gold medal at the Student Mathematical Paper Competition in 1991; a summary of the work is presented in Delta, issue 3 (1992). | 26 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXIV OM - III - Task 2
Let $ p_n $ be the probability that a series of 100 consecutive heads will appear in $ n $ coin tosses. Prove that the sequence of numbers $ p_n $ is convergent and calculate its limit. | The number of elementary events is equal to the number of $n$-element sequences with two values: heads and tails, i.e., the number $2^n$. A favorable event is a sequence containing 100 consecutive heads. We estimate the number of unfavorable events from above, i.e., the number of sequences not containing 100 consecutive heads.
Let $n = 100k + r$, where $k \geq 0$ and $0 \leq r < 100$. Each $n$-element sequence thus consists of $k$ 100-element sequences and one $r$-element sequence. The total number of 100-element sequences with two values is $2^{100}$; therefore, after excluding the sequence composed of 100 heads, there remain $2^{100} - 1$ 100-element sequences. Each $n$-element sequence not containing 100 consecutive heads thus consists of $k$ such 100-element sequences and some $r$-element sequence. Hence, the number of unfavorable events is not greater than $(2^{100} - 1)^k \cdot 2^r$. Therefore,
If $n$ tends to infinity, then of course $k$ also tends to infinity. Since for $0 < q < 1$ we have $\displaystyle \lim_{k \to \infty} q^k = 0$, then $\displaystyle \lim_{k \to \infty} \left( 1 - \frac{1}{2^{100}} \right)^k = 0$. Therefore, from (1) by the squeeze theorem, we have $\displaystyle \lim_{k \to \infty} q^k = 0$, so $\displaystyle \lim_{n \to \infty} p_n = 1$. | 1 | Combinatorics | proof | Yes | Yes | olympiads | false |
XXVIII - I - Problem 11
From the numbers $ 1, 2, \ldots, n $, we choose one, with each of them being equally likely. Let $ p_n $ be the probability of the event that in the decimal representation of the chosen number, all digits: $ 0, 1, \ldots, 9 $ appear. Calculate $ \lim_{n\to \infty} p_n $. | Let the number $ n $ have $ k $ digits in its decimal representation, i.e., let $ 10^{k-1} \leq n < 10^k $, where $ k $ is some natural number. Then each of the numbers $ 1, 2, \ldots, n $ has no more than $ k $ digits. We estimate from above the number $ A_0 $ of such numbers with at most $ k $ digits that do not contain the digit $ 0 $ in their decimal representation.
The number of $ r $-digit numbers whose decimal representation does not contain the digit $ 0 $ is equal to $ 9^r $, because each of the $ r $ digits of such a number belongs to the nine-element set $ \{1, 2, \ldots, 9\} $. Therefore,
Similarly, we observe that for $ i = 1, 2, \ldots, 9 $ the number $ A_i $ of numbers with at most $ k $ digits that do not contain the digit $ i $ in their decimal representation satisfies the inequality $ A_i < 9^{k+1} $.
It follows that the number $ A $ of natural numbers among $ 1, 2, \ldots, n $ that do not contain at least one of the digits $ 0, 1, 2, \ldots, 9 $ in their decimal representation satisfies the inequality
Therefore, the number $ B_n $ of numbers among $ 1, 2, \ldots, n $ whose decimal representation contains all the digits $ 0, 1, 2, \ldots, 9 $ satisfies
Thus,
If $ n $ tends to infinity, then $ k $ also tends to infinity. Since for any number $ q $ in the interval $ (-1; 1) $ we have $ \displaystyle \lim_{k \to \infty} q^k = 0 $, in particular $ \displaystyle \lim_{k \to \infty} \left( \frac{9}{10} \right)^k = 0 $. Therefore, from (1) by the squeeze theorem, it follows that $ \lim_{n \to \infty} p_n = 1 $. | 0 | Combinatorics | math-word-problem | Yes | Incomplete | olympiads | false |
XXII OM - III - Problem 5
Find the largest integer $ A $ such that for every permutation of the set of natural numbers not greater than 100, the sum of some 10 consecutive terms is at least $ A $. | The sum of all natural numbers not greater than $100$ is equal to $1 + 2 + \ldots + 100 = \frac{1 + 100}{2} \cdot 100 = 5050$. If $a_1, a_2, \ldots, a_{100}$ is some permutation of the set of natural numbers not greater than $100$ and the sum of any $10$ terms of this permutation is less than some number $B$, then in particular
By adding these inequalities side by side, we get that $a_1 + a_2 + \ldots + a_{100} = 1 + 2 + \ldots + 100 < 10B$, which means $505 < B$.
Thus, the number $A$ defined in the problem satisfies the inequality
On the other hand, consider the following permutation $a_1, a_2, \ldots, a_{100}$ of the set of natural numbers not greater than $100$
This permutation can be defined by the formulas:
We will prove that the sum of any $10$ consecutive terms of this permutation is not greater than $505$.
Indeed, if the first of the considered $10$ terms has an even number $2k$, then
If, however, the first of the considered terms has an odd number $2k + 1$, then
Thus, the sum of any $10$ consecutive terms of this permutation is not greater than $505$. Therefore, the number $A$ defined in the problem satisfies the inequality
From (1) and (2), it follows that $A = 505$.
Note 1. The problem can be generalized as follows: Find the largest integer $A$ such that for any permutation of the set of natural numbers not greater than an even number $n = 2t$, the sum of some $m = 2r$ (where $r$ is a divisor of $t$) consecutive terms is at least $A$.
By making minor changes in the solution provided above, consisting in replacing the number $100$ with $2t$ and the number $10$ with $2r$, it can be proved that $A = \frac{1}{2} m(n + 1)$.
Note 2. In the case where $m \leq n$ are any natural numbers, it is generally not true that every permutation of the set of natural numbers not greater than $n$ contains $m$ consecutive terms with a sum not less than $\frac{1}{2} m(n + 1)$. For example, for $n = 6$, $m = 4$, the permutation $6, 4, 1, 2, 3, 5$ does not contain four consecutive terms with a sum not less than $14$. | 505 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XV OM - II - Task 3
Prove that if three prime numbers form an arithmetic progression with a difference not divisible by 6, then the smallest of these numbers is $3$. | Suppose that the prime numbers $ p_1 $, $ p_2 $, $ p_3 $ form an arithmetic progression with a difference $ r > 0 $ not divisible by $ 6 $, and the smallest of them is $ p_1 $. Then
Therefore, $ p_1 \geq 3 $, for if $ p_1 = 2 $, the number $ p_3 $ would be an even number greater than $ 2 $, and thus would not be a prime number. Hence, the numbers $ p_1 $ and $ p_2 $ are odd, and the number $ r $ equal to the difference $ p_2 - p_1 $ is even and one of the cases holds: $ r = 6k + 2 $ or $ r = 6k + 4 $, where $ k $ is an integer $ \geq 0 $.
We will prove that $ p_1 $ is divisible by $ 3 $. Indeed, if $ p_1 = 3m + 1 $ ($ m $ - an integer) and $ r = 6k + 2 $, it would follow that $ p_2 = 3m + 6k + 3 $ is divisible by $ 3 $, and since $ p_2 > 3 $, $ p_2 $ would not be a prime number. If, on the other hand, $ p_1 = 3m + 1 $ and $ r = 6k + 4 $, then $ p_3 = 3m + 12k + 9 $ would not be a prime number. Similarly, from the assumption that $ p_1 = 3m + 2 $ and $ r = 6k + 2 $, it would follow that $ p_3 = 3m + 12k + 6 $ is not a prime number, and from the assumption that $ p_1 = 3m + 2 $ and $ r = 6k + 4 $, we would get $ p_2 = 3m + 6k + 6 $, and thus $ p_2 $ would not be a prime number.
Therefore, $ p_1 $ is a prime number divisible by $ 3 $, i.e., $ p_1 = 3 $. | 3 | Number Theory | proof | Yes | Yes | olympiads | false |
LII OM - III - Task 4
Given such integers $ a $ and $ b $ that for every non-negative integer $ n $ the number $ 2^na + b $ is a square of an integer. Prove that $ a = 0 $.
| If $ b = 0 $, then $ a = 0 $, because for $ a \ne 0 $, the numbers $ a $ and $ 2a $ cannot both be squares of integers.
If the number $ a $ were negative, then for some large natural number $ n $, the number $ 2^n a + b $ would also be negative, and thus could not be a square of an integer.
The only case left to consider is when $ a \geq 0 $ and $ b \neq 0 $.
For every positive integer $ k $, the numbers
are squares of different non-negative integers, say
Then $ x_{k}+y_{k}\leq(x_{k}+y_{k})|x_{k}-y_{k}|=|x_{k}^{2}-y_{k}^{2}|=|3b| $, hence
Thus the sequence $ (x_k) $ is bounded, which is only possible if $ a = 0 $. | 0 | Number Theory | proof | Yes | Incomplete | olympiads | false |
XLIII OM - I - Problem 2
In square $ABCD$ with side length $1$, point $E$ lies on side $BC$, point $F$ lies on side $CD$, the measures of angles $EAB$ and $EAF$ are $20^{\circ}$ and $45^{\circ}$, respectively. Calculate the height of triangle $AEF$ drawn from vertex $A$. | The measure of angle $ FAD $ is $ 90^\circ - (20^\circ + 45^\circ) = 25^\circ $. From point $ A $, we draw a ray forming angles of $ 20^\circ $ and $ 25^\circ $ with rays $ AE $ and $ AF $, respectively, and we place a segment $ AG $ of length $ 1 $ on it (figure 2).
From the equality $ |AG| =|AB| = 1 $, $ | \measuredangle EAG| = | \measuredangle EAB| = 20^\circ $, it follows that triangle $ EAG $ is congruent to $ EAB $.
Similarly, from the equality $ |AG| = |AD| = 1 $, $ | \measuredangle FAG| = | \measuredangle FAD| = 25^\circ $, it follows that triangle $ FAG $ is congruent to $ FAD $.
This means that
point $ G $ lies on segment $ EF $ and is the foot of the altitude of triangle $ AEF $ drawn from vertex $ A $. Its length $ |AG| = 1 $. | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XIV OM - I - Task 5
How many digits does the number
have when written in the decimal system? | The task can be easily solved using 5-digit decimal logarithm tables. According to the tables, $\log 2$ is approximately $0.30103$, with the error of this approximation being less than $0.00001$, hence
therefore
after multiplying by $2^{16} = 65536$, we obtain the inequality
From this, it follows that the number $2^{2^{16}}$ has $19728$ or $19729$ digits, which does not yet provide an answer to the question. To obtain it, we need to estimate $\log 2$ more accurately. For this purpose, we will use the approximate value of the logarithm of the number $2^{10} = 1024$ given in the tables. According to the tables, $\log 1024 \approx 3.01030$ with an error less than $0.00001$, hence
therefore
so
Thus, the number $2^{2^{16}}$ has $19729$ digits. Since the number $2^{2^{16}}$ is even, its last digit is not $9$, hence the number $2^{2^{16}} + 1$ also has $19729$ digits. | 19729 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
XLIV OM - I - Problem 6
The sequence $ (x_n) $ is defined as follows:
Calculate the sum $ \sum_{n=0}^{1992} 2^n x_n $. | Let's replace the parameter $1992$ with any arbitrarily chosen natural number $N \geq 1$ and consider the sequence $(x_n)$ defined by the formulas
when $N = 1992$, this is the sequence given in the problem.
Let's experiment a bit. If $N = 3$, the initial terms of the sequence $(x_n)$ are the numbers $3$, $-9$, $9$, $-3$. If $N = 4$, the initial terms of the sequence $(x_n)$ are the numbers $4$, $-16$, $24$, $-16$, $4$. If $N = 5$, the initial terms of the sequence $(x_n)$ are the numbers $5$, $-25$, $50$, $-50$, $25$, $-5$. These observations lead to the conjecture that for every natural number $N \geq 1$, the equality
holds.
To prove this, we will use the auxiliary equality
which holds for any natural numbers $N \geq n \geq 1$. Here is its justification:
We treat $N$ as fixed. For $n = 1$, formula (2) holds. Assume that formula (2) is true for some $n$ ($1 \leq n \leq N-1$). We will use the basic property of binomial coefficients:
Multiplying both sides of this equality by $(-1)^n$ and adding to the corresponding sides of equality (2):
more concisely:
This is equality (2) with $n$ replaced by $n+1$. By the principle of induction, formula (2) holds for $n = 1, 2, \ldots, N$.
We proceed to the proof of formula (1). For $n = 0$, the formula holds. Take any natural number $n$ (satisfying the conditions $1 \leq n \leq N$) and assume inductively that the equality $x_k = (-1)^k N \binom{N}{k}$ (i.e., (1) with $n$ replaced by $k$) holds for $k = 0, 1, \ldots, n-1$. We transform the expression defining $x_n$, using formula (2):
We have shown the validity of formula (1) for the chosen number $n$. Therefore, by induction, we conclude that equality (1) holds for all $n \in \{1, 2, \ldots, N\}$.
Thus,
The sum appearing in the last expression is the expansion (by Newton) of the $N$-th power of the binomial $(-2 + 1)$. Therefore,
This is the desired value. For $N = 1992$, it is $1992$. | 1992 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
XLVIII OM - I - Problem 8
Let $ a_n $ be the number of all non-empty subsets of the set $ \{1,2,\ldots,6n\} $, the sum of whose elements gives a remainder of 5 when divided by 6, and let $ b_n $ be the number of all non-empty subsets of the set $ \{1,2,\ldots,7n\} $, the product of whose elements gives a remainder of 5 when divided by 7. Calculate the quotient $ a_n/b_n $. | For a finite set of numbers $A$, the symbols $s(A)$ and $p(A)$ will denote the sum and the product of all numbers in this set, respectively. Let $n$ be a fixed natural number. According to the problem statement,
Notice that in the definition of the number $b_n$, we can remove all multiples of the number $7$ from the set $\{1,2,\ldots,7n\}$; if a set $B$ contains any number divisible by $7$, then the condition $p(B) \equiv 5 (\mathrm{mod} 7)$ is certainly not satisfied. By adopting the notation
and
we can thus rewrite the definitions of the numbers $a_n$ and $b_n$ as follows:
Consider any number $x \in X$. It is an element of exactly one of the sets $X_0, X_1, \ldots, X_{n-1}$. The number $3^x$ is not divisible by $7$, so in each of the sets $Y_0, Y_1, \ldots, Y_{n-1}$ there is exactly one element congruent to $3^x$ modulo $7$. Among them, we choose the element $y$ belonging to the set $Y_k$ with the same index $k$ as the index of the set $X_k$ containing $x$. Denote this element $y$ by $h(x)$. In this way, a function $h: X \to Y$ is defined with the property:
We will show that this function is injective.
Suppose that $h(u) = h(v)$ for some numbers $u, v \in X$. Assume that $u \leq v$ and $u \in X_k$, $v \in X_l$. Then $h(u) \in Y_k$, $h(v) \in Y_l$, so from the equality $h(u) = h(v)$ it follows that $k = l$, and moreover,
The key to further considerations is the following, easy to verify, property:
Since $u \leq v$ and $u, v \in X_k$, we have $0 \leq v - u \leq 5$. Multiplying the relation (3) by the integer $3^{6(k+1)-u}$ on both sides, we get
From this, using formulas (4), we first infer that $1 \equiv 3^{v-u} (\mathrm{mod} 7)$, and then that $v - u = 0$, i.e., $u = v$. The injectivity of the function $h$ has been established.
The function $h$ maps the set $X$, which has $6n$ elements, to the set $Y$, which also has $6n$ elements. Hence, it follows that it is a bijective mapping from the set $X$ onto the entire set $Y$.
Now let $A$ be any non-empty subset of the set $X$ and let $h(A)$ be the image of the set $A$ under the mapping $h$. We will prove that the implication holds:
Assume that $A = \{x_1, \ldots, x_m\}$. Then $h(A) = \{y_1, \ldots, y_m\}$, where
(see (2)). If $s(A) \equiv r (\mathrm{mod} 6)$, $0 \leq r < 6$, then there exists an integer $q \geq 0$ such that $x_1 + \ldots + x_m = 6q + r$. Using the first formula (4) again, we calculate:
This proves the validity of the implication (5).
Using property (4), it is not difficult to verify that $3^5 \equiv 5 (\mathrm{mod} 7)$ and $3^r \not\equiv 5 (\mathrm{mod} 7)$ for $r \neq 5 \ (0 \leq r < 6)$. From statement (5), it follows that
And since $h$ is a bijective mapping from the set $X$ to $Y$, we conclude from this that there are as many subsets $A$ of the set $X$ that satisfy the condition $s(A) \equiv 5 (\mathrm{mod} 6)$ as there are subsets $B$ of the set $Y$ that satisfy the condition $p(B) \equiv 5 (\mathrm{mod} 7)$. This means that $a_n = b_n$.
We obtain the answer: for every natural number $n$, the quotient $a_n / b_n$ is equal to $1$.
Note: Property (5) can be expressed in words—suggestively, though not strictly: the mapping $h$ "translates" addition modulo $6$ into multiplication modulo $7$. Readers familiar with the simplest facts and language of group theory know that the set $\{0, \ldots, 5\}$ is a group under the operation of addition modulo $6$, and the set $\{1, \ldots, 6\}$ is a group under the operation of multiplication modulo $7$.
The mentioned property, already in precise terminology, states that the operation which assigns to a number $x \in \{0, \ldots, 5\}$ the remainder of the division of $3^x$ by $7$ is an *isomorphism* of the first of these groups onto the second one—and this observation underlies the entire solution above.
A crucial role is played by the fact that the powers of three $3^0, 3^1, \ldots, 3^5$ give different remainders when divided by $7$—see formulas (4). The same property holds for five; therefore, one can consistently replace exponentiation with base $3$ with exponentiation with base $5$ everywhere, and the modified solution will also be correct. | 1 | Combinatorics | math-word-problem | Yes | Incomplete | olympiads | false |
LI OM - I - Task 10
In space, there are three mutually perpendicular unit vectors $ \overrightarrow{OA} $, $ \overrightarrow{OB} $, $ \overrightarrow{OC} $. Let $ \omega $ be a plane passing through the point $ O $, and let $ A', $ B', $ C' $ be the projections of points $ A $, $ B $, $ C $ onto the plane $ \omega $, respectively. Determine the set of values of the expression
for all planes $ \omega $.
| We will show that the value of expression (1) is equal to $ 2 $, regardless of the choice of plane $ \omega $.
Let $ \omega $ be any plane passing through point $ O $, and $ \pi $ be a plane containing points $ O $, $ A $, $ B $. Denote by $ \ell $ the common line of planes $ \pi $ and $ \omega $. Let $ X $, $ Y $ be the orthogonal projections of points $ A $, $ B $ onto the line $ \ell $. Then, regardless of the position of points $ A $, $ B $ relative to the line $ \ell $ (Fig. 1 and 2),
om51_1r_img_10.jpg
om51_1r_img_11.jpg
Hence
Moreover, $ \measuredangle AXA = (\text{angle between planes } \pi \text{ and } \omega) = \beta $. Therefore, by equality (2),
By the Pythagorean theorem and the above equality, we obtain
The vector $ \overrightarrow{OC} $ is perpendicular to the plane $ \pi $, so $ \measuredangle COC. Therefore,
Adding the equations (3) and (4) side by side, we get
regardless of the choice of plane $ \omega $. | 2 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
L OM - I - Problem 11
In an urn, there are two balls: a white one and a black one. Additionally, we have 50 white balls and 50 black balls at our disposal. We perform the following action 50 times: we draw a ball from the urn, and then return it to the urn along with one more ball of the same color as the drawn ball. After completing these actions, we have 52 balls in the urn. What is the most probable number of white balls in the urn? | Let $ P(k,n) $, where $ 1 \leq k\leq n-1 $, denote the probability of the event that when there are $ n $ balls in the urn, exactly $ k $ of them are white. Then
Using the above relationships, we prove by induction (with respect to $ n $) that $ P(k,n) = 1/(n-1) $ for $ k = 1,2,\ldots,n-1 $. In particular
Therefore, each possible number of white balls after $ 50 $ draws (from $ 1 $ to $ 51 $) is equally likely. | 51 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
LII OM - I - Task 1
Solve in integers the equation
| We reduce the given equation to the form $ f (x) = f (2000) $, where
Since $ f $ is an increasing function on the interval $ \langle 1,\infty ) $, the given equation has only one solution in this interval, which is $ x = 2000 $. On the set $ (-\infty,0\rangle $, the function $ f $ is decreasing, so there is at most one negative integer $ a $ for which $ f (a) = f (2000) $. However,
and
Therefore, $ a \in (-2000,-1999) $, which contradicts the fact that $ a $ is an integer.
Answer: The given equation has one integer solution: $ x = 2000 $. | 2000 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
XLVI OM - III - Problem 2
The diagonals of a convex pentagon divide this pentagon into a pentagon and ten triangles. What is the maximum possible number of triangles with equal areas? | om46_3r_img_12.jpg
Let's denote the considered pentagon by $ABCDE$, and the pentagon formed by the intersection points of the diagonals by $KLMNP$ so that the following triangles are those mentioned in the problem:
$\Delta_0$: triangle $LEM$; $\quad \Delta_1$: triangle $EMA$;
$\Delta_2$: triangle $MAN$; $\quad \Delta_3$: triangle $ANB$;
$\Delta_4$: triangle $NBP$; $\quad \Delta_5$: triangle $BPC$;
$\Delta_6$: triangle $PCK$; $\quad \Delta_7$: triangle $CKD$;
$\Delta_8$: triangle $KDL$; $\quad \Delta_9$: triangle $DLE$
(figure 12). We will start by proving the implication:
We assume that the numbering of the triangles is cyclic (that is, we assume: $\Delta_{-1} = \Delta_9$; $\Delta_{10} = \Delta_0$).
Suppose, for example, that triangles $\Delta_4$, $\Delta_5$, and $\Delta_6$ have equal areas. From the equality of the areas of triangles $NBP$ and $BPC$, it follows that segment $BP$ is a median in triangle $BCN$; from the equality of the areas of triangles $BPC$ and $PCK$, it follows that segment $CP$ is a median in triangle $BCK$. Point $P$ would then be the common midpoint of diagonals $BK$ and $CN$ of quadrilateral $BCKN$, which should therefore be a parallelogram - but the lines $BN$ and $CK$ intersect at point $E$. This is a contradiction; implication (1) is proven.
We proceed to the main solution. The example of a regular pentagon shows that it is possible to obtain five triangles $\Delta_i$ with equal areas. We will prove that it is not possible to obtain seven such triangles.
Suppose, therefore, that there exists a seven-element set $Z$, contained in the set $\{0,1,\ldots,9\}$, such that triangles $\Delta_i$ with indices $i \in Z$ have equal areas. Let $k$, $l$, $m$ be three numbers from the set $\{0,1,\ldots,9\}$, not belonging to $Z$. Consider all triples of consecutive indices:
where, as before, the addition $i \pm 1$ should be understood modulo $10$. Each number from the set $\{0,1,\ldots,9\}$ belongs to exactly three triples $T_i$. This applies, in particular, to each of the numbers $k$, $l$, $m$. Therefore, the three-element set $\{k,l,m\}$ has a non-empty intersection with at most nine triples $T_i$. At least one triple $T_j$ remains that does not contain any of the numbers $k$, $l$, $m$, and is therefore contained in the set $Z$. The number $j$ must be even, according to observation (1).
Without loss of generality, assume that $j = 2$. This means that the numbers $1$, $2$, $3$ belong to the set $Z$. Therefore, the numbers $0$ and $4$ cannot belong to it (since, again by observation (1), the triples $\{0,1,2\}$ and $\{2,3,4\}$ are excluded); thus, one of the numbers $k$, $l$, $m$ equals $0$, and another equals $4$. The third of these numbers must be part of the triple $\{6,7,8\}$ (which otherwise would be contained in the set $Z$, contrary to implication (1)). The roles of the numbers $6$ and $8$ are symmetric in this context. We therefore have to consider two essentially different cases:
The set $Z$, respectively, has one of the following forms:
Let $S_i$ be the area of triangle $\Delta_i$ (for $i = 0,1,\ldots,9$). We will show that the implications hold:
This will be a proof that none of the seven-element sets $Z$ listed above can be the set of indices of triangles with equal areas. According to earlier statements, this will also justify the conclusion that among the ten triangles $\Delta_i$, there are never seven triangles with the same area.
Proof of implication (2). From the given equalities of areas:
and
it follows (respectively) that points $N$ and $P$ are the midpoints of segments $BM$ and $BK$, and points $M$ and $L$ are the midpoints of segments $EN$ and $EK$ (figure 13).
Thus, segment $NP$ connects the midpoints of two sides of triangle $BMK$, and segment $ML$ connects the midpoints of two sides of triangle $ENK$, and therefore
- and
The obtained parallelism relations show that quadrilateral $AMKN$ is a parallelogram; hence $|MK| = |AN|$. Line $MK$ is parallel to $NC$, so triangles $ENC$ and $EMK$ are similar in the ratio $|EN| : |EM| = 2$. Therefore, $|NC| = 2 \cdot |MK|$, and consequently
The segments $PC$ and $AN$, lying on the same line, are the bases of triangles $ANB$ and $BPC$ with a common vertex $B$. The ratio of the lengths of these bases is therefore also the ratio of the areas of the triangles: area($BPC$) : area($ANB$) = $3 : 2$, i.e., $S_5 : S_3 = 3 : 2$. The conclusion of implication (2) is thus proven.
om46_3r_img_13.jpg
om46_3r_img_14.jpg
Proof of implication (3). The given equalities of the areas of triangles:
and
show that points $M$ and $N$ divide segment $EB$ into three equal parts, and similarly, points $L$ and $K$ divide segment $EC$ into three equal parts. Therefore, line $BC$ is parallel to $ML$ (i.e., to $AM$). Since $N$ is the midpoint of segment $BM$, triangles $AMN$ and $CBN$ are similar in the ratio $-1$ (figure 14). Hence, area($AMN$) = area($CBN$) $> $ area($CBP$), i.e., $S_2 > S_5$. This completes the proof of implication (3), and thus also the proof of the general theorem: it is not possible to have seven triangles $\Delta_i$ with equal areas.
The reasoning conducted in the last case (proof of (3)) also provides a hint on how to obtain six triangles $\Delta_i$ with equal areas. We take any isosceles triangle $BCE$ where $|EB| = |EC|$. On the sides $EB$ and $EC$, we find points $M$ and $N$ and $L$ and $K$, dividing these sides into three equal parts:
The intersection points of line $ML$ with lines $CN$ and $BK$ are denoted by $A$ and $D$, respectively (figure 14 can still serve as an illustration; one only needs to imagine moving point $E$ to the perpendicular bisector of side $BC$ and appropriately repositioning the other points). The equalities (4) and (5) then hold; and thanks to the assumption that triangle $BCE$ is isosceles (and the resulting symmetry of the entire configuration), all areas appearing in relations (4) and (5) are equal.
Conclusion: six of the triangles $\Delta_i$ can have equal areas, and this number cannot be increased. | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXII OM - I - Problem 4
Determine the angles that a plane passing through the midpoints of three skew edges of a cube makes with the faces of the cube. | Let $P$, $Q$, $R$ be the midpoints of three skew edges of a cube with edge length $a$ (Fig. 7). Let $n$ be the plane of triangle $PQR$, and $\pi$ be the plane of the base $ABCD$.
Triangle $PQR$ is equilateral because by rotating the cube $90^\circ$ around a vertical axis (passing through the centers of faces $ABCD$ and $A$), and then around a horizontal axis (passing through the centers of faces $ADD$ and $BCC$), points $P$, $O$, $R$ will respectively move to $Q$, $R$, $P$. Therefore, $PQ = QR = RP$.
Let $E$ be the midpoint of segment $RQ$. Then point $E$ is equidistant from faces $ABCD$ and $A$, and thus line $PE$ is parallel to plane $\pi$, and therefore also to the common edge of planes $\pi$ and $\pi$.
Line $PE$ is perpendicular to line $QR$ because the median in an equilateral triangle is also an altitude. Therefore, line $QR$ is perpendicular to the common edge of planes $\pi$ and $\pi$. Hence, the angle $\alpha$ between planes $\pi$ and $\pi$ is equal to the angle between line $QR$ and its projection $QR$ on plane $\pi$. Thus, $\alpha = \measuredangle RQR$.
Using the Pythagorean theorem, we calculate that $QR$, i.e., $QR$. Therefore, $\tan \alpha = \frac{RR$. From the tables, we read that $\alpha = 54^\circ45$.
Similarly, we prove that the angle between plane $\pi$ and any face of the cube is also equal to $\alpha$. | 5445 | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
LII OM - I - Problem 3
Find all natural numbers $ n \geq 2 $ such that the inequality
holds for any positive real numbers $ x_1,x_2,\ldots,x_n $. | The only number satisfying the conditions of the problem is $ n = 2 $.
For $ n = 2 $, the given inequality takes the form
Thus, the number $ n = 2 $ satisfies the conditions of the problem.
If $ n \geq 3 $, then inequality (1) is not true for any positive real numbers. The numbers
serve as a counterexample. Indeed, substituting the above values into inequality (1) yields
By equivalently transforming the above inequality, we obtain $ 3n \leq 7 $, which is not true when $ n \geq 3 $. | 2 | Inequalities | math-word-problem | Incomplete | Yes | olympiads | false |
XLIV OM - I - Problem 11
In six different cells of an $ n \times n $ table, we place a cross; all arrangements of crosses are equally probable. Let $ p_n $ be the probability that in some row or column there will be at least two crosses. Calculate the limit of the sequence $ (np_n) $ as $ n \to \infty $. | Elementary events are determined by six-element subsets of the set of $n^2$ cells of the table; there are $\binom{n^2}{6}$ of them. Let $\mathcal{Z}$ be the complementary event to the event considered in the problem. The configurations favorable to event $\mathcal{Z}$ are obtained as follows: we place the first cross in any arbitrarily chosen cell: here we have $n^2$ possibilities. We then "cross out" the entire horizontal row and the entire vertical row intersecting at the cell where we placed the first cross, and place the second cross in any of the remaining cells: thus, we now have $(n-1)^2$ possibilities. We repeat this scheme four more times and obtain the number of possibilities equal to $n^2 (n -1 )^2 (n - 2)^2 (n - 3)^2 (n - 4 )^2 (n - 5)^2$. This number must still be divided by $6!$ (the number of permutations of a six-element set) to make the reasoning independent of the order of placing the crosses.
Thus, the probability of event $\mathcal{Z}$ (equal to $1 - p_n$) is
One should not multiply all these factors! Let us denote the numerator of the obtained fraction by $L_n$, and the denominator by $M_n$. It is enough to notice that
where $\phi(n)$ and $f(n)$ are polynomials (in the variable $n$), of degree (at most) 4 and 10, respectively; similarly,
where $\psi(n)$ and $g(n)$ are polynomials (in the variable $n$), of degree (at most) 8 and 10, respectively. Therefore,
Since the polynomials $f(n)$ and $g(n)$ are of degree (at most) tenth, then
and consequently, $\displaystyle \lim_{n\to \infty} np_n = 30$. | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XX OM - I - Task 4
Given points $ A, B, C, D $ that do not lie on the same plane. Determine the plane such that the distances from these points to the plane are equal. | Suppose the plane $ \pi $ is equidistant from points $ A $, $ B $, $ C $, $ D $. None of the given points lie on the plane $ \pi $, for otherwise all four would have to lie on it, contrary to the assumption. The points $ A $, $ B $, $ C $, $ D $ do not lie on the same side of the plane $ \pi $, for they would then have to belong to the same plane parallel to $ \pi $, which again contradicts the assumption. Therefore, one of the following cases applies:
1. Three of the given points, for example, $ A $, $ B $, $ C $ lie on one side, and the fourth point $ D $ lies on the other side of the plane $ \pi $. In this case, the plane must pass through the midpoints $ M $, $ N $, $ P $ of the edges $ AD $, $ BD $, $ CD $ of the tetrahedron $ ABCD $ (Fig. 2). Conversely, the plane passing through the points $ M $, $ N $, $ P $ is equidistant from the points $ A $, $ B $, $ C $, $ D $. The points $ M $, $ N $, $ P $ are non-collinear, so they determine exactly one plane.
Since instead of the edges emanating from vertex $ D $, we can take the edges emanating from another vertex of the tetrahedron, there are four planes equidistant from the points $ A $, $ B $, $ C $, $ D $ and satisfying condition 1.
2. Two of the given points, for example, $ A $ and $ B $ lie on one side, and the other two $ C $ and $ D $ lie on the other side of the plane $ \pi $. In this case, the plane passes through the midpoints $ M $, $ N $, $ R $, $ Q $ of the edges $ AD $, $ BD $, $ AC $, $ BC $ of the tetrahedron $ ABCD $ (Fig. 3). Conversely, the plane passing through the points $ M $, $ N $, $ R $, $ Q $ is equidistant from the points $ A $, $ B $, $ C $, $ D $. The points $ M $, $ N $, $ R $, $ Q $ determine exactly one plane, as they are the vertices of a parallelogram. Indeed, $ MN \parallel AB $ and $ RQ \parallel AB $, so $ MN \parallel RQ $ and similarly $ MR \parallel NQ $.
Since the set of points $ \{A, B, C, D\} $ can be divided into two pairs in three ways, there are three planes equidistant from $ A $, $ B $, $ C $, $ D $ and satisfying condition 2.
The problem thus has 7 solutions. | 7 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
XV OM - I - Problem 7
Given a circle and points $ A $ and $ B $ inside it. Find a point $ P $ on this circle such that the angle $ APB $ is subtended by a chord $ MN $ equal to $ AB $. Does the problem have a solution if the given points, or only one of them, lie outside the circle? | Suppose that point $ P $ of a given circle $ C $ with radius $ r $ is a solution to the problem (Fig. 7).
Since points $ A $ and $ B $ lie inside the circle $ C $, points $ M $ and $ N $ lie on the rays $ PA $ and $ PB $ respectively, and angle $ APB $ coincides with angle $ MPN $. Triangles $ APB $ and $ MPN $ have equal bases $ AB = MN = a $ and a common angle at the vertex $ \measuredangle APB = \measuredangle MPN = \alpha $; hence, the circles circumscribed around these triangles have the same radius $ r = \frac{a}{2 \sin \alpha} $, and point $ P $ lies on the circle with radius $ r $ passing through points $ A $ and $ B $. Therefore, the construction of the desired point is as follows (Fig. 7a). We construct circles with radius $ r $ passing through points $ A $ and $ B $. Since $ AB < 2r $, there are 2 such circles: $ K_1 $ and $ K_2 $. If $ P $ is a common point of circles $ K_1 $ and $ C $, or circles $ K_2 $ and $ C $, then $ P $ is a solution to the problem. Indeed, the rays $ PA $ and $ PB $ intersect the given circle $ C $ at points $ M $ and $ N $; from triangle $ MPN $ we have $ MN = 2r \sin \measuredangle MPN = 2r \sin \measuredangle APB $; but in triangle $ APB $ we have $ \sin \measuredangle APB = AB \colon 2r $, so $ MN = AB $.
Each of the circles $ K_1 $ and $ K_2 $ intersects the circle $ C $ at two points; if $ O_1 $ and $ O_2 $ are the centers of circles $ K_1 $ and $ K_2 $, then
since $ OA < r $, $ AO_1 = r $; similarly $ 0 < OO_2 < 2r $.
The problem thus has 4 solutions.
Let us examine whether the above reasoning remains valid for any position of points $ A $ and $ B $. If point $ P $ is a solution to the problem, then points $ M $ and $ N $ lie on the lines $ PA $ and $ PB $, but not necessarily on the rays $ PA $ and $ PB $, so angle $ MPN $ coincides either with angle $ APB $ or with the adjacent angle to angle $ APB $. The conclusion that triangles $ APB $ and $ MPN $ have the same circumradius remains valid, hence point $ P $ lies, as before, on the circle with radius $ r $ passing through $ A $ and $ B $. Conversely, if point $ P $ is a common point of the given circle and a circle $ K $ with radius $ r $ passing through $ A $ and $ B $, then this point, provided it is different from $ A $ and $ B $, is a solution to the problem, as we conclude in the same way as before.
For such a point $ P $ to exist, the following conditions must be met:
1. There must exist a circle with radius $ r $ passing through points $ A $ and $ B $; this is the case when $ AB \leq 2r $. If $ AB < 2r $, there are two such circles $ K_1 $ and $ K_2 $ with centers $ O_1 $ and $ O_2 $. If $ AB = 2r $, there is one such circle: $ K_1 = K_2 $, $ O_1 = O_2 $.
2. There must exist a common point of circle $ K_1 $ or $ K_2 $ with the given circle $ C $, different from $ A $ and $ B $.
The table below gives the number of solutions for different cases of the position of points $ A $ and $ B $.
We propose to the Reader to provide a detailed justification of the above data for each of the cases 2-6 and to make the appropriate drawings. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
LV OM - II - Task 3
Determine the number of infinite sequences $ a_1,a_2,a_3,\dots $ with terms equal to $ +1 $ and $ -1 $, satisfying the equation
and the condition: in every triplet of consecutive terms $ (a_n, a_{n+1}, a_{n+2}) $, both $ +1 $ and $ -1 $ appear. | Let $ (a_n) $ be such a sequence. We will start by showing that
Suppose $ a_{3k+1} = a_{3k+2} = a $. Then it must be that $ a_{3k} = a_{3k+3} = b $ and $ b \neq a $. Multiplying the above equalities by $ a_2 $, we get $ a_{6k+2} = a_{6k+4} = c $ and $ a_{6k} = a_{6k+6} = d $, where $ c \neq d $. In the triplet $ (a_{6k+2},a_{6k+3},a_{6k+4}) $, +1 and -1 appear, which implies that $ a_{6k+3} = d $. Therefore, $ a_{6k} = a_{6k+3} = a_{6k+6} $. Dividing these equalities by $ a_3 $, we get $ a_{2k} = a_{2k+1} = a_{2k+2} $. Contradiction.
Next, we will prove by induction that
Of course, $ a_1 = a_1^2 = 1 $; and from property (1), it follows that $ a_{50} \neq a_{49} = a_7^2 = 1 $, so $ a_{50} = -1 $; but $ a_{50} = a_2a_5^2 = a_2 $, so $ a_2 = -1 $; the equalities (2) hold for $ k = 0 $.
Fix $ k \geq 1 $ and assume that $ a_{3i+1} = 1 $, $ a_{3i+2} = -1 $ for $ i < k $. Write the number $ k $ in the form $ 2j $ or $ 2j+1 $. If $ k = 2j $, then $ 3k + 2 = 2(3j + 1) $, and consequently $ a_{3k+2} = a_2a_{3j+1} = -1 $; if $ k = 2j + 1 $, then $ 3k + 1 = 2(3j + 2) $, so $ a_{3k+1} = a_2a_{3j+2} = 1 $. In both cases, one of the equalities (2) has been established; the other follows from relation (1). This completes the inductive proof of theorem (2).
Let $ a_3 = c $. Every natural number $ n $ has a unique representation in the form $ n = 3^rm $, where $ r > 0 $ and $ m $ is not divisible by 3. From the multiplicative condition, it follows that $ a_n = a_3^r a_m = c^r a_m $, and according to relations (2):
Thus, the sequence $ (a_n) $ is determined by the value $ c = a_3 $, which can be 1 or -1. Conversely, for both $ c = 1 $ and $ c = -1 $, formula (3) defines a sequence $ (a_n) $ that satisfies the given conditions. Therefore, there are two such sequences. | 2 | Combinatorics | math-word-problem | Incomplete | Yes | olympiads | false |
XXVI - I - Task 1
At the ball, there were 42 people. Lady $ A_1 $ danced with 7 gentlemen, Lady $ A_2 $ danced with 8 gentlemen, ..., Lady $ A_n $ danced with all the gentlemen. How many gentlemen were at the ball? | The number of ladies at the ball is $ n $, so the number of gentlemen is $ 42-n $. The lady with number $ k $, where $ 1 \leq k \leq n $, danced with $ k+6 $ gentlemen. Therefore, the lady with number $ n $ danced with $ n+ 6 $ gentlemen. These were all the gentlemen present at the ball. Thus, $ 42-n = n + 6 $. Solving this equation, we get $ n = 18 $. The number of gentlemen at the ball is therefore $ 42-18 = 24 $. | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XXXVI OM - I - Problem 11
Provide an example of a convex polyhedron having 1985 faces, among which there are 993 faces such that no two of them share a common edge. | It is easy to obtain a number of faces significantly greater than $993$, and not only without common edges, but completely disjoint. For this purpose, consider any convex polyhedron $W$ having $5$ faces and vertices; we assume that no two faces lie in the same plane, i.e., that all dihedral angles are less than a straight angle. We will now perform the operation of "cutting off vertices": near each vertex, we draw a plane intersecting all edges emanating from that vertex at points less than $1/2$ the length of the edge from the vertex; this plane cuts off the given vertex along with a small pyramid. As a result of this operation, we obtain a convex polyhedron $W$ with $s + w$ faces, because each face of the polyhedron $W$, after trimming the corners, becomes a face of the polyhedron $W$, and in addition, in place of each cut-off vertex of the polyhedron $W$, an additional face of the polyhedron $W$ (the base of the cut-off pyramid) appears. These "new" faces (and there are $w$ of them) are pairwise disjoint.
Now, we need to choose values of $s$ and $w$ such that $s + w = 1985$, $w \geq 993$; the polyhedron $W$ will then be one of the solutions to the problem. Let us take as the polyhedron $W$ a regular prism with an $n$-sided base; it has $w = 2n$ vertices and $s = n + 2$ faces. For $n = 661$, we get $s + w = 1985$, $w = 1322$. Therefore, the polyhedron $W$ resulting from $W$ through the above-described operation of "cutting off vertices" has $1985$ faces, and among them, $1322$ faces are pairwise disjoint. Of course, among these $1322$ faces, we can arbitrarily choose $993$ faces that are pairwise disjoint to satisfy the condition of the problem. | 993 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
L OM - II - Task 5
Let $ S = \{1, 2,3,4, 5\} $. Determine the number of functions $ f: S \to S $ satisfying the equation $ f^{50} (x) = x $ for all $ x \in S $.
Note: $ f^{50}(x) = \underbrace{f \circ f \circ \ldots \circ f}_{50} (x) $. | Let $ f $ be a function satisfying the conditions of the problem. For numbers $ x \neq y $, we get $ f^{49}(f(x)) = x \neq y = f^{49}(f(y)) $, hence $ f(x) \neq f(y) $. Therefore, $ f $ is a permutation of the set $ S $. Denote by $ r(x) $ ($ x \in S $) the smallest positive integer such that $ f^{r(x)}(x) = x $. Then $ r(x) \leq 5 $ and $ r(x) \mid 50 $, so $ r(x) \in \{1, 2, 5\} $.
If there exists a number $ a \in S $ such that $ r(a) = 5 $, then the numbers $ a $, $ f(a) $, $ f^2(a) $, $ f^3(a) $, $ f^4(a) $ are distinct; they thus exhaust the set $ S $. Then for any number $ x \in S $, $ r(x) = 5 $. The function $ f $ is thus uniquely determined by the permutation $ (f(1), f^2(1), f^3(1), f^4(1)) $ of the set $ \{2, 3, 4, 5\} $; hence it can be defined in $ 4! = 24 $ ways.
If for all $ x \in S $, $ r(x) = 1 $, then $ f $ is the identity function. Such a function is unique.
The remaining case to consider is when the maximum value attained by the function $ r $ is $ 2 $. Let $ a $ be an element of the set $ S $ such that $ r(a) = 2 $. Then also $ r(b) = 2 $, where $ b = f(a) $.
If $ r(x) = 1 $ for all $ x \in S \setminus \{a, b\} $, then $ f $ is determined by the choice of a two-element subset $ \{a, b\} $ of the set $ S $, which can be done in $ {5 \choose 2} = 10 $ ways.
If, however, there exists a number $ c \in S \setminus \{a, b\} $ such that $ r(c) = 2 $, then by setting $ d = f(c) $ and denoting by $ e $ the unique element of the set $ S \setminus \{a, b, c, d\} $, we have
Such a function $ f $ is determined by the choice of the number $ e $ (which can be done in $ 5 $ ways) and the partition of the set $ S \setminus \{e\} $ into two two-element subsets $ \{a, b\} $ and $ \{c, d\} $ (there are $ 3 $ such partitions). We thus get $ 15 $ functions of the form (1).
In total, there are $ 50 $ functions satisfying the conditions of the problem. | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
V OM - I - Problem 9
Points $ A, B, C, D, \ldots $ are consecutive vertices of a certain regular polygon, and the following relationship holds
How many sides does this polygon have? | \spos{1} Let $ r $ denote the radius of the circle circumscribed around a regular polygon $ ABCD\ldots $, and $ 2x $ - the central (convex) angle of this circle corresponding to the chord $ AB $. Then (Fig. 25)
Substituting these expressions into equation (1), we get
To solve equation (2), we multiply it by the product $ \sin x \cdot \sin 2x \cdot \sin 3x $ and move all terms to one side.
We will transform equation (3) aiming to represent the left side as a product. We obtain successively:
[ \sin 2x (\sin 3x - \sin x) - \sin x \sin 3x = 0, \]
[ \sin 2x \cdot 2 \sin x \cos 2x - \sin x \sin 3x = 0,\]
[ \sin x \sin 4x - \sin x \sin 3x = 0,\]
[ \sin x (\sin 4x - \sin 3x) = 0 \]
and finally
Equations (3) and (4) are equivalent, but they are not equivalent to equation (2), as only those solutions of equations (3) and (4) satisfy equation (2) for which $ \sin x \sin 2x \sin 3x \ne 0 $, i.e., $ x \ne \frac{k\pi}{2} $ and $ x \ne \frac{k\pi}{3} $, where $ k $ is any integer. Hence, the roots of equation (2) are those among the numbers (5) that satisfy the condition $ x \ne \frac{k\pi}{2} $ and $ x \ne \frac{k\pi}{3} $. The reader will verify that this condition means that in formula (5), the integer $ n $ is not arbitrary but must satisfy the inequalities $ n \ne \frac{7k-2}{4} $ and $ n \ne \frac{7k-3}{6} $, where $ k $ is any integer.
Since from the problem statement it follows that $ 0 < x \leq \frac{\pi}{4} $, in formula (5) we must take $ n = 0 $, and the problem has a unique solution
which means that the regular polygon $ ABCD \ldots $ is a heptagon. | 7 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
XII OM - III - Task 6
Someone wrote six letters to six people and addressed six envelopes to them. In how many ways can the letters be placed into the envelopes so that no letter ends up in the correct envelope? | Let $ F (n) $ denote the number of all ways of placing $ n $ letters $ L_1, L_2, \ldots, L_n $ into $ n $ envelopes $ K_1, K_2, \ldots, K_n $ so that no letter $ L_i $ ends up in the correct envelope $ K_i $, or more simply, so that there is no "hit". We need to calculate $ F (6) $.
Suppose we incorrectly place all letters by putting letter $ L_1 $ in envelope $ K_i $ ($ i \ne 1 $). There are $ 2 $ cases:
a) Letter $ L_i $ ends up in envelope $ K_1 $. The remaining $ 4 $ letters must then be incorrectly placed in the remaining $ 4 $ envelopes, which can be done in $ F (4) $ ways. Since $ K_i $ can be any of the $ 5 $ envelopes $ K_2 $, $ K_3 $, $ K_4 $, $ K_5 $, $ K_6 $, case a) can occur $ 5 \cdot F (4) $ times.
b) Letter $ L_i $ does not end up in envelope $ K_1 $. By assigning envelope $ K_1 $ to letter $ L_i $ as its "pseudo-correct" envelope, we can say that none of the letters $ L_2 $, $ L_3 $, $ L_4 $, $ L_5 $, $ L_6 $ end up in their correct envelopes, which can be done in $ F (5) $ ways. Since $ K_i $ can be any of the $ 5 $ envelopes, case b) can occur $ 5 \cdot F (5) $ times.
Since cases a) and b) cover all possibilities, therefore
Analogously
\[ F (5) = 4F (4) + 4F (3) \]
\[ F (4) = 3F (3) + 3F (2) \]
\[ F (3) = 2F (2) + 2F (1). \]
But of course
\[ F (2) = 1 \]
\[ F (1) = 0 \]
Therefore, from the above equations, we obtain sequentially,
\[ F (3) = 2 \cdot 1 + 2 \cdot 0 = 2 \]
\[ F (4) = 3 \cdot 2 + 3 \cdot 1 = 9 \]
\[ F (5) = 4 \cdot 9 + 4 \cdot 2 = 44 \]
Note. In the above solutions, the number $ 6 $ does not play any significant role. The same methods can be applied to $ n $ letters and $ n $ envelopes. Method I leads to the formula
\[ F (n) = (n - 1) [F (n - 1) + F (n - 2)] \]
This is the so-called recursive or reduction formula, which allows us to calculate $ F (n) $ if the values of $ F (n - 1) $ and $ F (n - 2) $ are known. Similarly, Method II leads to the recursive formula
\[ F (n) = nF (n - 1) + (-1)^n \]
Formula (1) allows us to calculate $ F (n) $ in terms of $ n $. We can write this formula as
\[ F (n) = nF (n - 1) + (-1)^n \]
From formula (3), we see that the function $ F (k) - kF (k - 1) $ has the same absolute value for every natural number $ k > 1 $, but changes sign when transitioning from $ k $ to $ k + 1 $. And since
\[ F (2) - 2F (1) = 1 \]
we have
\[ F (3) - 3F (2) = -1 \]
\[ F (4) - 4F (3) = 1 \]
\[ \ldots \]
From equality (4), we get
\[ F (n) - nF (n - 1) = (-1)^n \]
We introduce the notation
\[ \varphi (k) = \frac{F (k)}{k!} \]
Formula (5) becomes
\[ \varphi (k) - \varphi (k - 1) = \frac{(-1)^k}{k!} \]
Analogously
\[ \varphi (n) - \varphi (n - 1) = \frac{(-1)^n}{n!} \]
\[ \varphi (n - 1) - \varphi (n - 2) = \frac{(-1)^{n-1}}{(n-1)!} \]
\[ \ldots \]
\[ \varphi (2) - \varphi (1) = \frac{1}{2}. \]
Adding these formulas and considering that $ \varphi (1) = \frac{F (1)}{1} = 0 $, we get
\[ \varphi (n) = \sum_{k=1}^{n} \frac{(-1)^k}{k!} \]
Therefore,
\[ F (n) = n! \sum_{k=1}^{n} \frac{(-1)^k}{k!} \]
The problem we solved is known as the Bernoulli-Euler problem of the mislaid letters. | 265 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XXII OM - I - Problem 6
Let $ f(x,y,z) = \max(x^2 - yz, y^2 - xz, z^2 - xy) $. Find the set of values of the function $ f(x,y,z) $ considered for numbers $ x,y,z $ satisfying the conditions:
Note: $ \max (a, b, c) $ is the greatest of the numbers $ a, b, c $. | From (1) it follows that
thus $ x + y + z = 3 $. It follows from this that the conditions (1) added in the problem are equivalent to the conditions:
If we perform any permutation of the letters $ x $, $ y $, $ z $, the conditions of the problem will not change. Therefore, without loss of generality, we can limit ourselves to considering only such numbers $ x $, $ y $, $ z $, that $ 0 < x < y < z $. Then, of course, $ x^2 - yz \leq y^2 - xz < z^2 - xy $, and thus $ f(x, y, z) = z^2 - xy $. Using (1) and (2), we calculate that $ z^2 - xy = z^2 + yz + zx - 2 = z^2 + z (3 - z) - 2 = 3z - 2 $.
We have thus reduced our problem to the following: Find the set of values of the function $ g(z) = 3z - 2 $ considered for numbers $ z $ satisfying the condition: there exist such numbers $ x $, $ y $ that
The equation $ x + y + z = 3 $ defines a plane $ P $ at a distance of $ \sqrt{3} $ from the origin of the coordinate system, and the equation $ x^2 + y^2 + z^2 = 5 $ defines a sphere $ S $ with center at the point $ (0, 0, 0) $ and radius $ \sqrt{5} $. Since $ \sqrt{3} < \sqrt{5} $, the system of equations (3) defines a circle $ K $ which is the intersection of the plane $ P $ and the sphere $ S $. We are interested in such points $ (x, y, z) $ of the circle $ K $ that satisfy the condition $ 0 \leq x \leq y \leq z $. To determine them, we intersect the circle $ K $ with the planes $ x = 0 $, $ x = y $, $ y = z $. To determine the points of intersection (Fig. 8), we solve the following systems of equations:
Solutions: $ A = (0, 1, 2) $ and $ B = (0, 2, 1) $.
Solutions:
Solutions:
Thus, the condition $ x \geq 0 $ is satisfied by points belonging to the arc $ \widehat{AFB} $, the condition $ x \leq y $ - points belonging to the arc $ \widehat{CED} $, and the condition $ y \leq z $ - points belonging to the arc $ \widehat{EDF} $ (Fig. 8). Therefore, the condition $ 0 \leq x \leq y \leq z $ is satisfied by points belonging to the common part of these arcs, i.e., points belonging to the arc $ \widehat{AD} $.
The projection of the arc $ \widehat{AD} $ onto the z-axis is a segment whose endpoints are the projections of points $ A $ and $ D $, i.e., points $ (0, 0, 2) $ and $ \left( 0, 1 + \frac{2}{3} \sqrt{3} \right) $. Since the function $ g(z) = 3z - 2 $ is increasing, the set of values of this function considered for points of the arc $ \widehat{AD} $ is a segment with endpoints $ g(2) = 4 $ and $ g \left(1 + \frac{2}{3} \sqrt{3} \right) = 1 + 2 \sqrt{3} $. | 4 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
XXII OM - III - Task 3
How many locks at least need to be placed on the treasury so that with a certain distribution of keys among the 11-member committee authorized to open the treasury, any 6 members can open it, but no 5 can? Determine the distribution of keys among the committee members with the minimum number of locks. | Suppose that for some natural number $ n $ there exists a key distribution to $ n $ locks among an 11-member committee such that the conditions of the problem are satisfied. Let $ A_i $ denote the set of locks that the $ i $-th member of the committee can open, where $ i = 1, 2, \ldots, 11 $, and let $ A $ denote the set of all locks. Then from the conditions of the problem, we have
for any five-element subset $ \{ i_1, \ldots, i_5\} $ of the set $ \{1, 2, \ldots, 11\} $ and
for any six-element subset $ \{j_1, \ldots, j_6\} $ of the set $ \{1,2,\ldots, 11\} $.
From (1), it follows that the set $ A - (A_{i_1} \cup \ldots \cup A_{i_5}) $ is non-empty. Let $ x_{i_1}, \ldots, x_{i_5} $ be one of its elements, i.e., a lock that the group of committee members numbered $ i_1, \ldots, i_5 $ cannot open. From (2), it follows that for every $ j \not \in \{i_1, \ldots, i_5 \} $ we have $ x_{i_1 , \ldots, i_5} \in A_j $.
Suppose that $ x_{i_1,\ldots,i_5} = x_{k_1, \ldots, k_5} $ for some subsets $ \{i_1, \ldots, i_5\} $ and $ \{k_1, \ldots, k_5\} $. If these subsets were different, then, for example, $ i_t \not \in \{ k_1, \ldots, k_5 \} $. Therefore, $ x_{k_1, \ldots, k_5} \in A_{i_t} $; but on the other hand, this leads to a contradiction. The obtained contradiction proves that $ \{i_1, \ldots,i_5\} = \{ k_1, \ldots, k_5 \} $.
In other words, different five-element subsets $ \{i_1, \ldots, i_5\} $ correspond to different locks. Therefore, the number of locks is not less than the number of five-element subsets of an 11-element set, i.e., $ n \geq \binom{11}{5} = 462 $.
We will now prove that if we install $ \binom{11}{5} $ locks on the treasury, then we can distribute the keys to them among the members of the 11-member committee in such a way that the conditions of the problem are satisfied.
Let us associate each of the $ \binom{11}{5} $ locks with a five-element subset of the set $ \{1, 2, \ldots, 11\} $ in a one-to-one manner. If a lock corresponds to the subset $ \{i_1, \ldots, i_5\} $, then the keys to it are given to all members of the committee whose numbers are different from $ i_1, \ldots, i_5 $.
We will show that no five members of the committee can open a certain lock, and therefore the treasury. Indeed, the members of the committee numbered $ i_1, \ldots, i_5 $ do not have the key to the lock corresponding to the subset $ \{i_1, \ldots, i_5\} $.
We will show that any six members of the committee can open any lock, and therefore the treasury. If the members of the committee have numbers $ j_1, \ldots, j_6 $ and want to open a lock corresponding to the subset $ \{i_1, \ldots, i_5\} $, then one of the six numbers $ j_1, \ldots, j_6 $ does not belong to this five-element subset, say $ j_t \not \in \{i_1, \ldots, i_5 \} $. Therefore, the member of the committee numbered $ j_t $ has the key to the lock corresponding to the subset $ \{i_1, \ldots, i_5\} $.
Thus, the smallest number satisfying the conditions of the problem is $ \binom{11}{5} = 462 $. | 462 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Consider the set $M$ of integers $n \in[-100 ; 500]$, for which the expression $A=n^{3}+2 n^{2}-5 n-6$ is divisible by 11. How many integers are contained in $M$? Find the largest and smallest of them? | Answer: 1) 164 numbers; 2) $n_{\text {min }}=-100, n_{\text {max }}=497$. | 164 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. A crew of four pirates docked at an island to divide a treasure of gold coins left there. It was late, so they decided to postpone the division until morning. The first pirate woke up in the middle of the night and decided to take his share. He couldn't divide the coins into four equal parts, so he took two coins first, and then a quarter of the remaining coins, and went back to sleep. Unaware of this, the second pirate woke up during the night and did the same as the first. The third and fourth pirates repeated what the first and second had done. In the morning, without saying a word to each other, they divided the remaining coins equally among themselves. How many coins did each pirate get if the original treasure contained no fewer than 3000 and no more than 4000 coins? | Answer: the first pirate 1178 coins, the second pirate 954 coins, the third pirate 786 coins, the fourth - 660 coins. | 1178 | Logic and Puzzles | math-word-problem | Yes | Problem not solved | olympiads | false |
1. Solution. According to the problem, the sum of the original numbers is represented by the expression:
$$
\begin{aligned}
& \left(a_{1}+2\right)^{2}+\left(a_{2}+2\right)^{2}+\ldots+\left(a_{50}+2\right)^{2}=a_{1}^{2}+a_{2}^{2}+\ldots+a_{50}^{2} \rightarrow \\
& {\left[\left(a_{1}+2\right)^{2}-a_{1}^{2}\right]+\left[\left(a_{2}+2\right)^{2}-a_{2}^{2}\right]+\ldots\left[\left(a_{50}+2\right)^{2}-a_{50}^{2}\right]=0 \rightarrow} \\
& \rightarrow 4\left(a_{1}+1\right)+4\left(a_{2}+1\right)+\ldots+4\left(a_{50}+1\right)=0 \rightarrow a_{1}+a_{2}+\ldots+a_{50}=-50
\end{aligned}
$$
Then, if we add 3, we get:
$$
\begin{aligned}
& \left(a_{1}+3\right)^{2}+\left(a_{2}+3\right)^{2}+\ldots+\left(a_{50}+3\right)^{2}-\left(a_{1}^{2}+a_{2}^{2}+\ldots+a_{50}^{2}\right)= \\
& =\left[\left(a_{1}+3\right)^{2}-a_{1}^{2}\right]+\left[\left(a_{2}+3\right)^{2}-a_{2}^{2}\right]+\ldots\left[\left(a_{50}+3\right)^{2}-a_{50}^{2}\right]= \\
& =3\left(2 a_{1}+3\right)+3\left(2 a_{2}+3\right)+\ldots+3\left(2 a_{50}+3\right)=6\left(a_{1}+a_{2}+\ldots+a_{50}\right)+9 \cdot 50= \\
& =-300+450=150
\end{aligned}
$$ | Answer: will increase by 150. | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Solution. In the figure, equal angles are marked with the same numbers. Triangle $A P M$ is similar to triangle $C P B$ with a similarity coefficient $k_{1}=\lambda$
$$
\begin{aligned}
& A P=\lambda x, P C=x, P Q+\lambda x=x \rightarrow P Q=x(1-\lambda) \\
& \lambda x+x=a \rightarrow x=\frac{a}{1+\lambda} \rightarrow P Q=\frac{a(1-\lambda)}{1+\lambda}
\end{aligned}
$$
Triangle $B P Q$ is similar to triangle $B M N$ with a similarity coefficient
$$
k_{2}=\frac{B P}{B M}=\frac{B P}{B P+P M}=\frac{1}{1+\frac{P M}{B P}}=\frac{1}{1+\lambda}
$$
Then $\frac{P Q}{M N}=k_{2} \rightarrow M N=P Q \cdot \frac{1}{k_{2}}=P Q \cdot(1+\lambda)=a(1-\lambda)$. | Answer: $M N=a(1-\lambda)=2$. | 2 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
1. A set of 60 numbers is such that adding 3 to each of them does not change the value of the sum of their squares. By how much will the sum of the squares of these numbers change if 4 is added to each number? | Answer: will increase by 240. | 240 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The length of the diagonal $AC$ of the rhombus $ABCD$ with an acute angle at vertex $A$ is 4. Points $M$ and $N$ on sides $DA$ and $DC$ are the feet of the altitudes of the rhombus dropped from vertex $B$.
The height $BM$ intersects the diagonal $AC$ at point $P$ such that $AP: PC=1: 4$. Find the length of the segment $MN$. | Answer: $M N=a(1-\lambda)=3$.
# | 3 | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
1. A set of 70 numbers is such that adding 4 to each of them does not change the magnitude of the sum of their squares. By how much will the sum of the squares of these numbers change if 5 is added to each number? | Answer: will increase by 350. | 350 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The length of the diagonal $AC$ of the rhombus $ABCD$ with an acute angle at vertex $A$ is 12. Points $M$ and $N$ on sides $DA$ and $DC$ are the bases of the heights of the rhombus dropped from vertex $B$. The height $BM$ intersects the diagonal $AC$ at point $P$ such that $AP: PC = 2: 3$. Find the length of the segment $MN$. | Answer: $M N=a(1-\lambda)=4$.
# | 4 | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
1. A set of 80 numbers is such that adding 5 to each of them does not change the magnitude of the sum of their squares. By how much will the sum of the squares of these numbers change if 6 is added to each number? | Answer: will increase by 480. | 480 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The length of the diagonal $AC$ of the rhombus $ABCD$ with an acute angle at vertex $A$ is 20. Points $M$ and $N$ on sides $DA$ and $DC$ are the bases of the heights of the rhombus dropped from vertex $B$. The height $BM$ intersects the diagonal $AC$ at point $P$ such that $AP: PC=3: 4$. Find the length of the segment $MN$. | Answer: $M N=a(1-\lambda)=5$.
Grading criteria, 8th grade
Preliminary round of the ROSATOM industry physics and mathematics school olympiad, mathematics
# | 5 | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
3. The sum of two natural numbers is 2013. If you erase the last two digits of one of them, add one to the resulting number, and then multiply the result by five, you get the other number. Find these numbers. Enter the largest of them in the provided field. | 3. The sum of two natural numbers is 2013. If you erase the last two digits of one of them, add one to the resulting number, and then multiply the result by five, you get the other number. Find these numbers. Enter the largest of them in the provided field.
Answer: 1913 | 1913 | Number Theory | math-word-problem | Yes | Problem not solved | olympiads | false |
4. The sum of two natural numbers is 2014. If you strike out the last two digits of one of them, multiply the resulting number by three, you get a number that is six more than the other number. Find these numbers. Enter the smallest of them in the provided field. | 4. The sum of two natural numbers is 2014. If you strike out the last two digits of one of them, multiply the resulting number by three, you get a number that is six more than the other number. Find these numbers. Enter the smallest of them in the provided field.
Answer: 51 | 51 | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false |
5. Find the fraction $\frac{p}{q}$ with the smallest possible natural denominator, for which $\frac{1}{2014}<\frac{p}{q}<\frac{1}{2013}$. Enter the denominator of this fraction in the provided field | 5. Find the fraction $\frac{p}{q}$ with the smallest possible natural denominator, for which
$\frac{1}{2014}<\frac{p}{q}<\frac{1}{2013}$. Enter the denominator of this fraction in the provided field
Answer: 4027 | 4027 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. How many pairs of natural numbers $(a ; b)$ exist such that the number $5 a-3$ is divisible by $b$, and the number $5 b-1$ is divisible by $a$? Enter the number of such pairs of numbers in the provided field. | 6. How many pairs of natural numbers $(a ; b)$ exist such that the number $5 a-3$ is divisible by $b$, and the number $5 b-1$ is divisible by $a$? Enter the number of pairs of the specified numbers in the provided field. Answer: 18 | 18 | Number Theory | math-word-problem | Yes | Incomplete | olympiads | false |
7. The coordinates $(x ; y ; z)$ of point $M$ are consecutive terms of a geometric progression, and the numbers $x y, y z, x z$ in the given order are terms of an arithmetic progression, with $z \geq 1$ and $x \neq y \neq z$. Find the smallest possible value of the square of the distance from point $M$ to point $N(1 ; 1 ; 1)$. Enter your answer in the provided field. | 7. The coordinates $(x ; y ; z)$ of point $M$ are consecutive terms of a geometric progression, and the numbers $x y, y z, x z$ in the given order are terms of an arithmetic progression, with $z \geq 1$ and $x \neq y \neq z$. Find the smallest possible value of the square of the distance from point $M$ to point $N(1 ; 1 ; 1)$. Enter your answer in the provided field.
Answer: 18 | 18 | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
9. Find the last two digits of the number $14^{14^{14}}$. Enter your answer in the provided field. | 9. Find the last two digits of the number $14^{14^{14}}$. Enter your answer in the provided field.
Answer: 36 | 36 | Number Theory | math-word-problem | Yes | Problem not solved | olympiads | false |
10. Find the number of twos in the prime factorization of the number $2011 \cdot 2012 \cdot 2013 \cdot \ldots .4020$. Enter your answer in the provided field. | 10. Find the number of twos in the prime factorization of the number $2011 \cdot 2012 \cdot 2013 \cdot \ldots .4020$. Enter your answer in the provided field.
Answer: 2010 | 2010 | Number Theory | math-word-problem | Yes | Problem not solved | olympiads | false |
11. For what values of $a$ does the equation $|x|=a x-2$ have no solutions? Enter the length of the interval of values of the parameter $a$ in the provided field. | 11. For what values of $a$ does the equation $|x|=a x-2$ have no solutions? Enter the length of the interval of parameter $a$ values in the provided field.
Answer: 2 | 2 | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false |
13. For what value of $a$ does the equation $|x-2|=a x-2$ have an infinite number of solutions? Enter the answer in the provided field | 13. For what value of $a$ does the equation $|x-2|=a x-2$ have an infinite number of solutions? Enter the answer in the provided field
Answer: 1 | 1 | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false |
# 2.4. Final round of the "Rosatom" Olympiad, 11th grade, set 4
## Answers and solutions
Problem 1 Answer: $x=2$ | Solution.
Transform the equation $f\left(\log _{2}\left(x\left(x^{2}-1\right)\right)-0.5 \log _{2}(x-1)^{2}\right)=f\left(\log _{2} 6\right)$. From the monotonicity of the function $f$, it follows that the equality is possible if $\log _{2}\left(x\left(x^{2}-1\right)\right)-0.5 \log _{2}(x-1)^{2}=\log _{2} 6\left(^{*}\right)$. The domain of valid values is $x(x-1)(x+1)>0 \rightarrow x \in(-1 ; 0) \cup(1 ;+\infty)$.
Transforming the equation:
1) In the domain $x \in(-1 ; 0): \log _{2} \frac{x(x-1)(x+1)}{1-x}=\log _{2} 6 \rightarrow x^{2}+x+6=0$ - no solutions.
2) In the domain $x \in(1 ;+\infty): \log _{2} \frac{x(x-1)(x+1)}{x-1}=\log _{2} 6 \rightarrow x^{2}+x-6=0 \rightarrow x=2$. | 2 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
4. If $\quad a=\overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}}, \quad$ then $\quad P(a)=\overline{a_{6} a_{1} a_{2} a_{3} a_{4} a_{5}}$, $P(P(a))=\overline{a_{5} a_{6} a_{1} a_{2} a_{3} a_{4}} \quad$ with $\quad a_{5} \neq 0, a_{6} \neq 0, a_{1} \neq 0 . \quad$ From the equality $P(P(a))=a$ it follows that $a_{1}=a_{5}, a_{2}=a_{6}, a_{3}=a_{1}$, $a_{4}=a_{2}, a_{5}=a_{3}, a_{6}=a_{4}$, that is, $a_{1}=a_{3}=a_{5}=t, t=1,2, \ldots, 9$ and $a_{2}=a_{4}=a_{6}=u, u=1,2, \ldots, 9$. Thus, the sought $a=\overline{\text { tututu }}$ and there are 81 such different numbers ( $t$ and $u$ can take any values of the decimal system digits from 1 to 9).
Let $n>2-$ be a prime number, $a=\overline{a_{1} a_{2} a_{3} a_{4} \ldots a_{n-3} a_{n-2} a_{n-1} a_{n}}$. Then
$$
\begin{gathered}
P(a)=\overline{a_{n} a_{1} a_{2} a_{3} a_{4} \ldots a_{n-3} a_{n-2} a_{n-1}} \\
P(P(a))=\overline{a_{n-1} a_{n} a_{1} a_{2} a_{3} a_{4} \ldots a_{n-3} a_{n-2}}
\end{gathered}
$$
The property $P(P(a))=a$ gives the relations $a_{1}=a_{n-1}=a_{n-3}=\ldots=a_{1}$. For a prime $n>2$, all the digits of the number $a$ are involved in the chain, so they are all equal to each other. | Answer: 1) 81 is the number; 2) $a=\overline{t u t u t u}, t, u$, where $t, u$ - are any digits, not equal to zero. | 81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Square the numbers $a=101, b=10101$. Extract the square root of the number $c=102030405060504030201$. | Answer: 1) $a^{2}=10201 ; 2$) $b^{2}=102030201 ;$ 3) $\sqrt{c}=10101010101$. | 10101010101 | Number Theory | math-word-problem | More than one problem | Yes | olympiads | false |
3. Square the numbers $a=1001, b=1001001$. Extract the square root of the number $c=1002003004005004003002001$. | 1) $a^{2}=1002001$; 2) $b^{2}=1002003002001$; 3) $\sqrt{c}=1001001001001$. | 1001001001001 | Algebra | math-word-problem | More than one problem | Yes | olympiads | false |
3. Square the numbers $a=10001, b=100010001$. Extract the square root of the number $c=1000200030004000300020001$. | 1) $a^{2}=100020001$; 2) $b^{2}=10002000300020001$; 3) $\sqrt{c}=1000100010001$. | 1000100010001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Answer: 10 boys, 13 girls. | Solution. Notation: $n$ - number of boys, $m$ - number of girls, $k$ - number of laps run by each boy, $r$ - number of candies received by each boy.
Problem statement:
$$
\left\{\begin{array}{l}
n k+m(k-1)=286 \\
n r+m(r+1)=243
\end{array}, k \geq 2, r \geq 1\right.
$$
Add the equations of the system:
$$
n(k+r)+m(k+r)=529=23^{2} \rightarrow(n+m)(k+r)=23^{2}
$$
Considering the integrality of the variables and the simplicity of the number 23, we have
$$
\left\{\begin{array}{l}
n+m=23 \\
k+r=23
\end{array}\right.
$$
From the first equation, we have:
$$
(m+n) k=286+m \rightarrow m=23 k-286
$$
$1 \leq m \leq 22 \rightarrow 1 \leq 23 k-286 \leq 22 \rightarrow \frac{287}{23} \leq k \leq \frac{308}{23} \rightarrow 12.4 \leq k \leq 13.4 \rightarrow k=13$ $m=13 \rightarrow n=10 \rightarrow r=10$ | 10 | Other | math-word-problem | Incomplete | Yes | olympiads | false |
2. By what natural number can the numerator and denominator of the ordinary fraction of the form $\frac{5 n+3}{7 n+8}$ be reduced? For which integers $n$ can this occur? | Answer: it can be reduced by 19 when $n=19k+7, k \in Z$. | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. By what natural number can the numerator and denominator of the ordinary fraction of the form $\frac{4 n+3}{5 n+2}$ be reduced? For which integers $n$ can this occur? | Answer: can be reduced by 7 when $n=7 k+1, k \in Z$. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. The angle at vertex $B$ of triangle $A B C$ is $130^{\circ}$. Through points $A$ and $C$, lines perpendicular to line $A C$ are drawn and intersect the circumcircle of triangle $A B C$ at points $E$ and $D$. Find the acute angle between the diagonals of the quadrilateral with vertices at points $A, C, D$ and $E$.
Problem 1 Answer: 12 students. | Solution. Let $a$ be the number of students in the first category, $c$ be the number of students in the third category, and $b$ be the part of students from the second category who will definitely lie in response to the first question (and say "YES" to all three questions), while the rest of the students from this category will answer "NO" to all three questions. Then, "YES" to the first question will be given by $a+b+c=25$ students. "YES" to the second question will be given by $b+c=21$ students. "YES" to the third question will be given by $b=6$ students. Solving the system, we get: $a=4, b=6, c=15$. Then, to the second category, we should assign $31-a-c=12$ students.
Problem 2 Answer: $a= \pm 5-2 \sqrt{6}$.
Solution. By Vieta's theorem, the roots of the equation are $m=(a+2 \sqrt{6})$ and $n=\left(\frac{1}{a}-2 \sqrt{6}\right)$.
Then we have:
$$
\left[\begin{array}{l}
a=m-2 \sqrt{6} \\
\frac{1}{a}=n+2 \sqrt{6}
\end{array} \rightarrow 1=m n+2 \sqrt{6}(m-n)-24 \rightarrow 2 \sqrt{6}(m-n)=25-m n\right.
$$
If $m, n \in Z, m \neq n$, then $2 \sqrt{6}=\frac{25-m n}{m-n}$ is a rational number, which is incorrect. Therefore, $m=n$ and then $m^{2}=25 \rightarrow m= \pm 5 \rightarrow a= \pm 5-2 \sqrt{6}$.
Problem 3 Answer: 5400 numbers.
Solution. Write $496125=3^{4} \cdot 5^{3} \cdot 7^{2}$. There are a total of $3^{4} \cdot 5^{3}=10125$ multiples of 49. Among them, $3^{3} \cdot 5^{3}=3375$ numbers are divisible by 3, $3^{4} \cdot 5^{2}=2025$ numbers are divisible by 5, and $3^{3} \cdot 5^{2}=675$ numbers are divisible by 15. Then, among the numbers that are multiples of 49, there are $3375+2025-675=4725$ numbers that are divisible by either 3 or 5. Therefore, the numbers that are not divisible by either 3 or 5 will be $10125-4775=$ 5400 desired numbers.
Problem 4 Answer: $n=210$.
Solution. According to the problem, $n=7 k$ and the expression
$$
n^{2}+25 n+100=(n+5)(n+20)=(7 k+5)(7 k+20)
$$
must be divisible by 5 and 23. Note that if one of the factors is divisible by 5, then the other is also divisible by 5 and vice versa, so:
$$
7 k+5=5 m \rightarrow 7 k=5(m-1) \rightarrow\left\{\begin{array}{c}
k=5 t \\
m=7 t+1
\end{array} \rightarrow n=35 t, t \in Z\right.
$$
Case 1. $7 k+5$ is divisible by 23:
$$
35 t+5=23 u \rightarrow\left\{\begin{array}{l}
t=23 v-10 \\
u=35 v-15
\end{array} \rightarrow n=35 t=35(23 v-10) \rightarrow n_{\min }=455\right.
$$
Case 2. $7 k+20$ is divisible by 23:
$$
35 t+20=23 u \rightarrow\left\{\begin{array}{c}
t=23 v+6 \\
u=35 v+10
\end{array} \rightarrow n=35 t=35(23 v+6) \rightarrow n_{\min }=210\right.
$$
Choosing the smallest of the found $n_{\min }$, we get $n_{\min }=210$.
Problem 5 Answer: $80^{\circ}$.
Solution. Figures 1 and 2 show the possible geometric configurations.
Fig 1
Fig 2
In the figures, point $O$ is the center of the circle $K$ circumscribed around triangle $A B C$.
Case 1. $\alpha>90^{\circ}$ (Fig. 1). Quadrilateral $A B C D$ is inscribed in circle $K$ by construction, so point $O$ is equidistant from points $C$ and $D$ at a distance equal to the radius of circle $K$. Similarly, point $O$ is equidistant from points $A$ and $E$ at the same distance. Therefore, point $O$ is the intersection point of the diagonals of rectangle $A C D E$. The angle $C D O$, as the opposite angle to $\alpha$ in the inscribed quadrilateral $A B C D$, is equal to $180^{\circ}-\alpha$. The desired angle between the diagonals of rectangle $A C D E$ is: $180^{\circ}-2\left(180^{\circ}-\alpha\right)=2\left(\alpha-90^{\circ}\right)$.
Case 2. $\alpha<90^{\circ}$ (Fig. 2). Point $O$ is equidistant from points $A, C, D$ and $E$, so it is the intersection point of the diagonals of rectangle $A C D E$ with these vertices. The angle $A D C$ is equal to $\alpha$, as it is inscribed and subtends the same arc of circle $K$ as angle $A B C$. Then the desired angle $D O C$ is: $180^{\circ}-2 \alpha=2\left(90^{\circ}-\alpha\right)$.
In case 1, $\alpha=130^{\circ}$, so case 1 applies, and $2\left(\alpha-90^{\circ}\right)=80^{\circ}$. | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Answer: 1) 44 integer solutions 2 ) $a=k \in Z$ | Solution.
Rewrite the system as $\left\{\begin{array}{c}3 \leq x \leq 12 \\ -12-3 a \leq x+3 y \leq-3 a\end{array}\right.$. The set of integer solutions of this system is the union over the parameter $b$ of the integer solutions of the systems $\left\{\begin{array}{c}3 \leq x \leq 12, \\ x+3 y=b, \quad b \in[-12-3 a ;-3 a]\end{array}\right.$. The equation $x+3 y=b$ has integer solutions only for integer $b$, and $\left\{\begin{array}{c}x=b-3 t \\ y=t, t \in Z\end{array}\right.$.
The restriction on $x$ leads to the inequality $3 \leq b-3 t \leq 12 \rightarrow b / 3-4 \leq t \leq b / 3-1$. If $b$ is divisible by 3, then it corresponds to 4 (maximum number) of integers $t$ on the interval $[b / 3-4 ; b / 3-1]$ of length 3, meaning 4 integer solutions of the system. If $b$ is not divisible by 3, then the number of such solutions is 3.
Case 1. $a$ is an integer.
Then on the interval $[-12-3 a ;-3 a]$ there are five (maximum number) of integers $b$ that are divisible by 3, each giving 4 integer solutions of the system, and 8 integers $b$ that are not divisible by 3, each giving 3 solutions. The total number of solutions of the system in this case is $N_{1}=5 \cdot 4+8 \cdot 3=44$.
Case 2. $a=\frac{k}{3}, k \in Z$.
Then on the interval $[-12-3 a ;-3 a]$ there are also 13 (maximum number) of integer values of $b$, of which only 3 are divisible by 3, and the remaining 10 are not divisible by 3 and give 3 solutions each.
The total number of integer solutions is $N_{2}=3 \cdot 4+10 \cdot 3=42$.
Case 3. $3 a$ is not an integer.
Then on the interval $[-12-3 a ;-3 a]$ there are only 11 integers $b$, among which no more than 3 are divisible by 3.
The total sum $N_{3}$ of integer solutions does not exceed $N_{2}$. | 44 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
3. Answer: $\operatorname{GCD}\left(x_{1}, x_{2}, x_{3}\right)=2, \operatorname{LCM}\left(x_{1}, x_{2}, x_{3}\right)=210$ | Solution.
Let's factorize the free term of the equation into prime factors: $4200=2^{3} \cdot 3 \cdot 5^{2} \cdot 7$.
If $p$ is a prime common divisor of the roots, $4200=x_{1} \cdot x_{2} \cdot x_{3}$ is divisible by $p^{3}$. The factorization of the number 4200 into factors shows that such a common divisor can only be $p=2$.
We will prove that each of the roots is divisible by 2. Suppose the opposite, for example, $x_{1}$ is not divisible by 2. By the condition, $x_{1}^{3}-86 x_{1}^{2}+1180 x_{1}-4200=0 \rightarrow x_{1}^{3}=86 x_{1}^{2}-1180 x_{1}+4200$. Such an equality cannot exist, since the left side is an odd number, while the right side is even, i.e., $N O D\left(x_{1}, x_{2}, x_{3}\right)=2$. We will show that none of the roots of the equation is divisible by $5^{2}$. Indeed,
if one of the roots, for example, $x_{1}$, is divisible by $5^{2}$, then from the equality $x_{1}^{3}-86 x_{1}^{2}+1180 x_{1}-4200=0$ it follows that $x_{1}^{3}-86 x_{1}^{2}+1180 x_{1}=4200$. In the left part of the equality, $x_{1}^{3}$ and $x_{1}^{2}$ are divisible by $5^{3}$. Since $1180=5 \cdot 2^{2} \cdot 59, 1180 x_{1}$ is also divisible by $5^{3}$, i.e., the left part of the equality is divisible by $5^{3}$, while the right part is not. Thus, in the LCM of the roots, the factor 5 enters in the first power. The same can be said about the factor 2. Obviously, the other divisors of the number 4200, which entered it in the first powers, also enter the LCM in the first powers. LCM $\left(x_{1}, x_{2}, x_{3}\right)=2 \cdot 5 \cdot 3 \cdot 7=210$. | 210 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
1. Answer: 15 weights, 23 weights | Solution.
$x$ - the number of used weights of 3 kg, $y$ - the number of used weights of 5 kg.
Condition: $3 x+5 y=71, x \geq 0, y \geq 0$. The general solution of the equation:
$\left\{\begin{array}{l}x=7+5 t \\ y=10-3 t\end{array}, t \in Z \rightarrow 7+5 t \geq 0,10-3 t \geq 0 \rightarrow t \in[-1 ; 3]\right.$
The total number of weights: $x+y=17+2 t$ takes the smallest value at $t=-1$ equal to 15 and the largest value at $t=3$ equal to 23. | 15 | Logic and Puzzles | math-word-problem | Incomplete | Yes | olympiads | false |
2. Masha chose five digits: $2,3,5,8$ and 9 and used only them to write down all possible four-digit numbers. For example, 2358, 8888, 9235, etc. Then, for each number, she multiplied the digits in its decimal representation, and then added up all the results. What number did Masha get? | Answer: $27^{4}=531441$. | 531441 | Combinatorics | math-word-problem | Yes | Problem not solved | olympiads | false |
3. In the line, integers are recorded one after another, starting with 3, and each subsequent number, except the first, is the sum of the two numbers adjacent to it. The number 7 appeared at the 1947th position. | Answer: $\Sigma_{851}=7$. | 7 | Number Theory | math-word-problem | Incomplete | Problem not solved | olympiads | false |
5. In a convex quadrilateral $A B C D$, the lengths of sides $B C$ and $A D$ are 2 and $2 \sqrt{2}$ respectively. The distance between the midpoints of diagonals $B D$ and $A C$ is 1. Find the angle between the lines $B C$ and $A D$. | Answer: $\alpha=45^{\circ}$. | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In a convex quadrilateral $A B C D$, the lengths of sides $B C$ and $A D$ are 6 and 8, respectively. The distance between the midpoints of diagonals $B D$ and $A C$ is 5. Find the angle between the lines $B C$ and $A D$. | Answer: $\alpha=90^{\circ}$. | 90 | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
5. Let's introduce the notation: $A B=2 c, A C=2 b, \measuredangle B A C=\alpha$. The feet of the perpendicular bisectors are denoted by points $P$ and $Q$. Then, in the right triangle $\triangle A M Q$, the hypotenuse $A M=\frac{b}{\cos \alpha}$. And in the right triangle $\triangle A N P$, the hypotenuse $A N=\frac{c}{\cos \alpha}$. By the cosine theorem for triangles $A M N$ and $A B C$ respectively, we have
$$
\begin{aligned}
& N M^{2}=A M^{2}+A N^{2}-2 A M \cdot A N \cdot \cos \alpha=\frac{c^{2}+b^{2}-2 b c \cos \alpha}{\cos ^{2} \alpha} \\
& B C^{2}=A B^{2}+A C^{2}-2 A B \cdot A C \cdot \cos \alpha=4\left(c^{2}+b^{2}-2 b c \cos \alpha\right)
\end{aligned}
$$
By the condition, $M N=B C$, therefore $\cos ^{2} \alpha=\frac{1}{4} \Leftrightarrow \cos \alpha= \pm \frac{1}{2}$, from which $\alpha=60^{\circ}$ or $\alpha=120^{\circ}$. We will show that both cases are possible, that is, if $\alpha=60^{\circ}$ or $\alpha=120^{\circ}$, then $M N=B C$.
Case 1. If $\alpha=60^{\circ}$, then $\measuredangle P N A=30^{\circ}$, so $A N=2 c=A B$, and $\measuredangle A M Q=30^{\circ}$, so $A M=2 b=A C$. Therefore, $\triangle A N M=\triangle A B C$ by two sides and the angle $\alpha$ between them. Consequently, $M N=B C$.
Case 2. If $\alpha=120^{\circ}$, then $\measuredangle B A N=60^{\circ}$. Further, $A N=N B$, so $\triangle N A B$ is isosceles, hence $\measuredangle A B N=60^{\circ}$, thus $\triangle N A B$ is equilateral. Therefore, $A N=A B$. Similarly, $\triangle M A C$ is equilateral. Therefore, $A M=A C$. Hence,
$\triangle A N M=\triangle A B C$ by two sides and the angle $\alpha$ between them. Consequently, $M N=B C$. | Answer: $60^{\circ}$ or $120^{\circ}$. | 60 | Geometry | proof | Yes | Yes | olympiads | false |
5. In triangle $A B C$, the perpendicular bisectors of sides $A B$ and $A C$ intersect lines $A C$ and $A B$ at points $N$ and $M$ respectively. The length of segment $N M$ is equal to the length of side $B C$ of the triangle. The angle at vertex $C$ of the triangle is $40^{\circ}$. Find the angle at vertex $B$ of the triangle. | Answer: $80^{\circ}$ or $20^{\circ}$.
## Final round of the "Rosatom" Olympiad, 9th grade, CIS, February 2020
# | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. The sum $b_{6}+b_{7}+\ldots+b_{2018}$ of the terms of the geometric progression $\left\{b_{n}\right\}, b_{n}>0$ is 6. The sum of the same terms taken with alternating signs $b_{6}-b_{7}+b_{8}-\ldots-b_{2017}+b_{2018}$ is 3. Find the sum of the squares of the same terms $b_{6}^{2}+b_{7}^{2}+\ldots+b_{2018}^{2}$. | Answer: $b_{6}^{2}+b_{7}^{2}+\ldots+b_{2018}^{2}=18$. | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Let $A=\overline{a b c b a}$ be a five-digit symmetric number, $a \neq 0$. If $1 \leq a \leq 8$, then the last digit of the number $A+11$ will be $a+1$, and therefore the first digit in the representation of $A+11$ should also be $a+1$. This is possible only with a carry-over from the digit, i.e., when $b=c=9$. Then $A+11=(a+1) 999(a+1)$ is a symmetric number for any $a=1,2, \ldots, 8$. The case $a=9$ is impossible, since $A+11$ ends in zero, and thus, due to symmetry, it should start with zero. But a number cannot start with zero.
The total number of solutions is equal to the number of possible choices for the number $a$, which is eight. | Answer: eight numbers of the form $\overline{a 999 a}$, where $a=1,2, \ldots, 8$. | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Nine horizontal and fifteen vertical streets form a total of $\quad(9-1)(15-1)=112 \quad$ square blocks in the city "N". Let $A$ and $C$ be the nearest points of blocks $(2 ; 3)$ and $(5 ; 12)$, then the path between them (according to the rules) has
a length of $100(|5-2|+|12-3|-2)=1000$. One possible route is the broken line $A E C$. Now let $B$ and $D$ be the farthest points of the blocks, then the path between them (according to the rules) has a length of $100(|5-2|+|12-3|+2)=1400$. One possible route is the broken line $B F D$. Therefore, the minimum number of coins is $c_{\text {min }}=1000: 100=10$, and the maximum is $-c_{\max }=1400: 100=14$. | Answer: 112 blocks; $c_{\min }=10$ coins, $c_{\max }=14$ coins. | 112 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
5. Let the distance between points $A$ and $B$ be $L$ mm, and the length of the flea's jump be $d$ mm. If $L$ were divisible by $d$ exactly, the flea's strategy of jumping in a straight line connecting points $A$ and $B$ would be optimal. It would take $\frac{L}{d}=n, n \in \mathbb{N}$ jumps to reach the goal. Given $L=20190, d=13$, we have $\frac{L}{d}=\frac{20190}{13}=1553.07 \ldots$, and moving in a straight line $AB$, the flea would not be able to reach point $B$ from point $A$ in an integer number of steps $n=1553$. We will show that there exists a strategy to reach the goal in $n+1$ steps.
Since the double inequality $n d < L < (n+1) d$ holds, there exists a triangle with sides $BC = d = 13$, $AC = n d = 20189$, and $AB = L = 20190$. Using the cosine rule, we have
$$
\alpha = \arccos \frac{L^2 + n^2 d^2 - d^2}{2 \cdot n d \cdot L} = \arccos \frac{20190^2 + 20189^2 - 13^2}{2 \cdot 20189 \cdot 20190} \approx 0.04^\circ
$$
Then, the flea, making 1553 jumps in a direction that forms an angle $\alpha$ with the direction of ray $AB$, reaches point $C$. From there, it makes the last, 1554th, jump to point $B$.
Remark: There are other strategies that allow reaching point $B$ in 1554 jumps. | Answer: $\quad n_{\min }=1554$ jumps. | 1554 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
2. Integers, the decimal representation of which reads the same from left to right and from right to left, we will call symmetric. For example, the number 513315 is symmetric, while 513325 is not. How many six-digit symmetric numbers exist such that adding 110 to them leaves them symmetric? | Answer: 81 numbers of the form $\overline{a b 99 b a}$, where $a=1,2, \ldots, 9, b=0,1,2, \ldots, 8$. | 81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In city "N", there are 12 horizontal and 16 vertical streets, of which a pair of horizontal and a pair of vertical streets form the rectangular boundary of the city, while the rest divide it into blocks that are squares with a side length of 100m. Each block has an address consisting of two integers $(i ; j), i=1,2, . ., 11, j=1,2, \ldots, 15-$ the numbers of the streets that bound it from below and from the left. Taxis transport passengers from one block to another, adhering to the following rules: 1) pick-up and drop-off can be made at any point on the boundary of the block at the passenger's request; 2) it is forbidden to enter inside the block; 3) transportation is carried out along the shortest path; 4) a fee of 1 coin is charged for every 100m traveled (rounding the distance to the nearest 100m in favor of the driver). How many blocks are there in the city? What is the maximum and minimum fare that a driver can charge a passenger for a ride from block $(7,2)$ to block $(2 ; 1)$ without violating the rules? | Answer: 165 blocks; $c_{\min }=4$ coins, $c_{\max }=8$ coins. | 165 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Kuzya the flea can make a jump in any direction on a plane for exactly 15 mm. Her task is to get from point $A$ to point $B$ on the plane, the distance between which is 2020 cm. What is the minimum number of jumps she must make to do this? | Answer: $n_{\min }=1347$ jumps.
# | 1347 | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
2. Integers, the decimal representation of which reads the same from left to right and from right to left, we will call symmetric. For example, the number 5134315 is symmetric, while 5134415 is not. How many seven-digit symmetric numbers exist such that adding 1100 to them leaves them symmetric? | Answer: 810 numbers of the form $\overline{a b c 9 c b a}$, where $a=1,2, \ldots, 9$, $b=0,1,2, \ldots, 9, c=0,1,2, \ldots, 8$. | 810 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In city "N", there are 7 horizontal and 13 vertical streets, of which a pair of horizontal and a pair of vertical streets form the rectangular boundary of the city, while the rest divide it into blocks that are squares with a side length of 100 m. Each block has an address consisting of two integers $(i ; j), i=1,2, . ., 6, j=1,2, \ldots, 12$ - the numbers of the streets that bound it from below and from the left. Taxis transport passengers from one block to another, adhering to the following rules: 1) pick-up and drop-off can be made at any point on the boundary of the block at the passenger's request; 2) it is forbidden to enter inside the block; 3) transportation is carried out along the shortest path; 4) a fee of 1 coin is charged for every 100 m traveled (rounding the distance to the nearest 100 m in favor of the driver). How many blocks are there in the city? What is the maximum and minimum fare that the driver can charge the passenger for a ride from block $(4,2)$ to block $(1 ; 9)$ without violating the rules. | # Answer: 72 blocks; $c_{\min }=8$ coins, $c_{\max }=12$ coins. | 72 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits