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XXIX OM - I - Problem 7
For a fixed natural number $ n>2 $, we define: $ x_1 = n $, $ y_1=1 $, $ x_{i+1} = \left[\frac{1}{2}(x_i+y_i)\right] $, $ y_{i+1} = \left[ \frac{n}{x_{i+1}}\right] $. Prove that the smallest of the numbers $ x_1, x_2, \ldots, x_n $ is equal to $ [\sqrt{n}] $.
Note. $ [a] $ denotes the greatest integer not greater than $ a $. | For any integer $ k $, the number $ \left[ \frac{k}{2} \right] $ is equal to $ \frac{k}{2} $ or $ \frac{k-1}{2} $ depending on whether $ k $ is even or odd. Therefore, $ \left[ \frac{k}{2} \right] \geq \frac{k-1}{2} $ for any integer $ k $. We also have $ a \geq [a] > a - 1 $ by the definition of the symbol $ [a] $. Finally, $ \frac{1}{2} \left( x_i + \frac{n}{x_i} \right) \geq \sqrt{n} $, because the geometric mean of the positive numbers $ x_i $ and $ \frac{n}{x_i} $ does not exceed their arithmetic mean.
Taking the above into account, we obtain that
that is, $ x_{i+1} > [\sqrt{n}] - 1 $. It follows that $ x_{i+1} \geq [\sqrt{n}] $ for $ i = 1, 2, \ldots , n -1 $. We also have $ x_1 = n \geq \sqrt{n} \geq [\sqrt{n}] $.
On the other hand, if for some $ i = 1, 2, \ldots , n- 1 $ we have $ x_i > [\sqrt{n}] $, then $ x_i > \sqrt{n} $ and therefore
that is, $ x_{i+1} < x_i $.
If, therefore, all terms of the sequence $ x_1, x_2, \ldots, x_n $ were greater than $ [\sqrt{n}] $, then it would be a decreasing sequence of natural numbers greater than $ [\sqrt{n}] $. This is impossible, since $ x_1 = n $ and the sequence has $ n $ terms. Therefore, some term of the sequence is equal to $ [\sqrt{n}] $ and it is the smallest of the numbers $ x_1, x_2, \ldots, x_n $.
Note. From the above solution, it follows that if for some $ i = 1,2, \ldots ,n- 1 $ the inequality $ x_{i+1} \geq x_i $ holds, then it is the smallest term of the sequence $ x_1, x_2, \ldots, x_n $ and therefore $ x_i= [\sqrt{n}] $.
From this problem, using the above note, we can derive a relatively simple method for calculating approximate values of square roots of natural numbers. We will show this on the example of the number $ n = 1977 $. We calculate the initial terms of the sequences $ (x_i) $ and $ (y_i) $ given in the problem and place the results in a table:
Since $ x_9 = x_8 $, it follows that $ [\sqrt{1977}] = x_8 = 44 $. It was therefore sufficient to calculate only $ 9 $ terms of the sequence $ (x_1, x_2, \ldots , x_{1977}) $ to find $ [\sqrt{1977}] $, instead of all $ 1977 $ of its terms. In this example, we have $ x_i = 44 $ for $ 8 \leq i \leq 1977 $. However, the sequence $ (x_i) $ is not always monotonic. The reader is invited to find an appropriate example. | [\sqrt{n}] | Number Theory | proof | Yes | Yes | olympiads | false | 573 |
XIX OM - III - Problem 6
Given a set of $ n > 3 $ points, no three of which are collinear, and a natural number $ k < n $. Prove the following statements:
1. If $ k \leq \frac{n}{2} $, then each point in the given set can be connected to at least $ k $ other points in the set in such a way that among the drawn segments, there are no three sides of the same triangle.
2. If $ k > \frac{n}{2} $ and each point in the given set is connected to $ k $ other points in the set, then among the drawn segments, there are three sides of the same triangle. | $ 1^\circ $. Suppose $ k \leq \frac{n}{2} $. From the given set $ Z $, select a part $ Z_1 $ consisting of $ \left[ \frac{n}{2} \right] $ points*); the remaining part $ Z_2 $ contains $ \left[ \frac{n}{2} \right] $ points if $ n $ is even, and $ \left[ \frac{n}{2} \right] + 1 $ points if $ n $ is odd. Since $ k $ is an integer, from the condition $ k \leq \frac{n}{2} $ it follows that $ k \leq \left[ \frac{n}{2} \right] $. Therefore, each of the sets $ Z_1 $, $ Z_2 $ contains at least $ k $ points.
Connect each point of set $ Z_1 $ with each point of set $ Z_2 $ by segments. Then each point of set $ Z $ will be connected to at least $ k $ other points of this set. No three of the drawn segments are sides of the same triangle, for if such a triangle existed, then two of its vertices, i.e., both ends of one of the segments, would lie in one of the sets $ Z_1 $, $ Z_2 $, which is impossible, since no such segment was drawn.
$ 2^\circ $ Suppose $ k > \frac{n}{2} $ and that each point of the given set is connected by $ k $ segments to $ k $ other points of the set; let $ AB $ be one of the drawn segments.
From each of the points $ A $ and $ B $, in addition to $ AB $, there are still $ k - 1 $ other segments, i.e., a total of $ 2k - 2 $ segments whose ends belong to the set of the remaining $ n - 2 $ points. But if $ k > \frac{n}{2} $, then $ 2k - 2 > n - 2 $; among those $ 2k - 2 $ segments, there are therefore two segments with a common end $ C $. In the set of drawn segments, there are thus the sides $ AB $, $ AC $, and $ BC $ of triangle $ ABC $.
Note. In a similar way, a more general theorem can be proved. Let there be given: a set $ Z $ consisting of $ n > 3 $ points, no three of which are collinear, a natural number $ p $ satisfying the inequalities $ 3 \leq p < n $, and a natural number $ k < n $. Then:
$ 1 ^\circ $ If $ k \leq \frac{p-2}{p-1}n $, then each of the points of set $ Z $ can be connected by segments to at least $ k $ other points of this set so that in any subset of set $ Z $ consisting of $ p $ points, some two points are not connected by a segment.
$ 2^\circ $ If $ k > \frac{p-2}{p-1}n $ and each point of set $ Z $ is connected by a segment to $ k $ other points of this set, then there exists a subset of set $ Z $ consisting of $ p $ points in which every two points are connected by a segment. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 574 |
LVII OM - II - Problem 3
Positive numbers $ a, b, c $ satisfy the condition $ ab+bc+ca = abc $. Prove that | Dividing the equality $ ab+bc+ca = abc $ by $ abc $ on both sides, we get
Let us substitute: $ x=1/a,\ y =1/b,\ z =1/c $. Then the numbers $ x, y, z $ are positive, and their sum is 1. Moreover,
Thus, the inequality to be proved takes the form
We will show that for any positive numbers $ x, y $, the following relationship holds:
Transforming the above inequality equivalently, we get the following in sequence:
We have obtained a relationship that is true for any positive numbers $ x $ and $ y $, so the inequality (2) is satisfied. Similarly, we prove that
Adding the relations (2), (3), and using the condition $ x + y + z = 1 $, we obtain the inequality (1) to be proved. | proof | Algebra | proof | Yes | Yes | olympiads | false | 575 |
XXIX OM - I - Problem 10
Point $ O $ is an internal point of a convex quadrilateral $ ABCD $, $ A_1 $, $ B_1 $, $ C_1 $, $ D_1 $ are the orthogonal projections of point $ O $ onto the lines $ AB $, $ BC $, $ CD $, and $ DA $, respectively, $ A_{i+1} $, $ B_{i+1} $, $ C_{i+1} $, $ D_{i+1} $ are the orthogonal projections of point $ O $ onto the lines $ A_iB_i $, $ B_iC_i $, $ C_iD_i $, $ D_iA_i $, respectively. Prove that the quadrilaterals $ A_4B_4C_4D_4 $ and $ ABCD $ are similar. | The construction given in the task is not always feasible. For example, three of the points $A_{i+1}$, $B_{i+1}$, $C_{i+1}$, $D_{i+1}$ may lie on the same line. In this case, the task loses its meaning. Therefore, we will provide a solution to the task with the additional assumption that all considered quadrilaterals exist and are convex.
Since $ \measuredangle OD_1A + \measuredangle OA_1A = \frac{\pi}{2} + \frac{\pi}{2} = \pi $, a circle can be circumscribed around quadrilateral $OA_1AD_1$. The measures of angles inscribed in a circle and subtended by the same arc are equal. Therefore, $ \measuredangle OAD = \measuredangle OA_1D_1 $ (Fig. 8). Similarly, we prove that $ \measuredangle OA_2D_2 = \measuredangle OA_2D_2 = \measuredangle OA_3D_3 = \measuredangle OA_4D_4 $.
Reasoning similarly, we obtain $ \measuredangle ODA = \measuredangle OC_1D_1 = \measuredangle OB_2C_2 = \measuredangle OA_3B_3 = \measuredangle OD_4A_4 $. Triangles $OAD$ and $OA_4D_4$ are similar because two corresponding angles in these triangles have equal measures. Therefore, $ \displaystyle \frac{OA}{AD} = \frac{OA_4}{A_4D_4} $. Similarly, we prove that $ \displaystyle \frac{OA}{AB} = \frac{OA_4}{A_4B_4} $. Hence, $ \displaystyle \frac{AB}{A_4B_4} = \frac{AD}{A_4D_4} $, and similarly $ \displaystyle \frac{AB}{A_4B_4} = \frac{BC}{B_4C_4} = \frac{CD}{C_4D_4} $.
We have also proved that $ \measuredangle OAD = \measuredangle OA_4D_4 $. Similarly, we prove that $ \measuredangle OAB = \measuredangle OA_4B_4 $, and thus $ \measuredangle BAD = \measuredangle OAB + \measuredangle OAD = \measuredangle OA_4B_4 + \measuredangle OA_4D_4 = \measuredangle B_4A_4D_4 $. Similarly, we can show that the measures of the remaining angles of quadrilateral $ABCD$ are equal to the measures of the corresponding angles of quadrilateral $A_4B_4C_4D_4$.
Thus, the corresponding sides of these quadrilaterals are proportional, and the measures of the corresponding angles are equal. Therefore, these quadrilaterals are similar. | proof | Geometry | proof | Yes | Yes | olympiads | false | 581 |
LI OM - I - Task 4
Each point of the circle is painted one of three colors. Prove that some three points of the same color are vertices of an isosceles triangle. | Let $A_1, A_2, \ldots, A_{13}$ be the vertices of an arbitrary regular 13-gon inscribed in a given circle. We will show that among any five vertices of this 13-gon, there will be three that form the vertices of an isosceles triangle. Since among any thirteen points on a given circle, there will always be five colored the same color, the problem will be solved.
Suppose, then, that it is possible to choose five vertices from $A_1, A_2, \ldots, A_{13}$ such that no three of the chosen points form an isosceles triangle.
First, assume that among the chosen points, there are no two adjacent vertices of the 13-gon $A_1 A_2 \ldots A_{13}$. Then among the chosen points, we will find two separated by exactly one vertex that was not chosen. Without loss of generality, we can assume these are $A_1$ and $A_3$. Since no three chosen vertices form an isosceles triangle, none of the points $A_{12}$, $A_{13}$, $A_2$, $A_4$, $A_5$ (Fig. 1) were chosen.
om51_1r_img_3.jpg
om51_1r_img_4.jpg
Since triangle $A_1A_6A_{11}$ is isosceles, at least one of the vertices $A_6$, $A_{11}$ was not chosen. Without loss of generality, we can assume that this vertex is $A_6$. Therefore, among the five points $A_7$, $A_8$, $A_9$, $A_{10}$, $A_{11}$, exactly three were chosen, and no two of them are consecutive vertices of the 13-gon $A_1A_2\ldots A_{13}$. Thus, the points $A_7$, $A_9$, $A_{11}$ must have been chosen. However, this is not possible, as these points form an isosceles triangle (Fig. 2). We have reached a contradiction.
The remaining case to consider is when among the chosen vertices, there are two adjacent ones, say $A_1$ and $A_2$. This means that none of the vertices $A_{13}$, $A_3$, $A_8$ (Fig. 3) were chosen.
om51_1r_img_5.jpg
om51_1r_img_6.jpg
Since triangles $A_2A_4A_6$, $A_1A_6A_{11}$, $A_{11}A_1A_4$ are isosceles, at most one of the points $A_4$, $A_6$, $A_{11}$ was chosen. Similarly, we conclude that at most one of the points $A_5$, $A_{10}$, $A_{12}$ was chosen. Therefore, at least one of the points $A_7$, $A_9$ must have been chosen. Without loss of generality, assume that the point $A_7$ was chosen (Fig. 4). The isosceles nature of triangles $A_1A_4A_7$, $A_2A_{10}A_7$, $A_2A_{11}A_7$, and $A_2A_{12}A_7$ proves that the points $A_4$, $A_{10}$, $A_{11}$, and $A_{12}$ were not chosen.
Thus, among the five chosen points, there must be two from $A_5$, $A_6$, $A_9$. Given the isosceles nature of triangle $A_5A_6A_7$, only one of the points $A_5$, $A_6$ could have been chosen. Therefore, the chosen point is $A_9$ (Fig. 5). This, in turn, means that the point $A_5$ was not chosen, as triangle $A_5A_7A_9$ is isosceles.
om51_1r_img_7.jpg
om51_1r_img_8.jpg
Thus, the fifth chosen point must be $A_6$, which leads to a contradiction, as triangle $A_9A_1A_6$ is isosceles (Fig. 6).
Thus, we have shown that it is impossible to choose five vertices of the 13-gon such that no three of them form an isosceles triangle, which completes the solution of the problem. | proof | Geometry | proof | Yes | Yes | olympiads | false | 584 |
IX OM - II - Problem 6
On a plane, there are two circles $ C_1 $ and $ C_2 $ and a line $ m $. Find a point on the line $ m $ from which tangents can be drawn to the circles $ C_1 $ and $ C_2 $ that are equally inclined to the line $ m $. | If the given circles have a common tangent $s$, intersecting the line $m$ at point $S$, then we can consider that point $S$ satisfies the conditions of the problem, as through this point pass two coinciding tangents to the given circles equally inclined to the line $m$.
Ignoring this trivial solution, we will look for other solutions. Suppose that through point $T$ of the line $m$ pass two different lines $t_1$ and $t_2$, tangent respectively to circles $C_1$ and $C_2$ and equally inclined to the line $m$, i.e., symmetric with respect to $m$ (Fig. 26). The circle $C$, symmetric to the circle $C_1$ with respect to $m$, is tangent to the line symmetric to the tangent $t_1$ of the circle $C_1$, i.e., to the line $t_2$. Therefore, the line $t_2$ is a common tangent of the circles $C$ and $C_2$. The solution to the problem is the point of intersection $T$ of each common tangent of the circles $C$ and $C_2$ with the line $m$. Note that point $T$ is also the point of intersection with the line $m$ of the common tangent $t_1$ of the circle $C_1$ and the circle $C$ symmetric to the circle $C_2$ with respect to the line $m$. The problem can have $4$, $3$, $2$, $1$, or $0$ solutions. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 585 |
XXXV OM - I - Problem 8
Let $ n $ be an even natural number. Prove that a quadrilateral can be divided into $ n $ triangles, whose vertices lie at the vertices of the quadrilateral or inside the quadrilateral, and each side of the triangle is either a side of the quadrilateral or a side of another triangle. | At least one diagonal of the quadrilateral is contained within its interior. Let this be, for example, the diagonal $ \overline{AC} $. If $ n = 2 $, then the division of the quadrilateral $ ABCD $ into triangles $ ABC $ and $ ACD $ is the desired division. Suppose that $ n = 2k $ is an even number. Choose any points $ A_1, \ldots, A_{k-1} $ on the diagonal $ AC $ lying on it in the order consistent with the numbering and different from $ A $ and $ C $. Additionally, let $ A_0 = A $, $ A_k = C $. The division of the quadrilateral $ ABCD $ into triangles $ A_{i-1}BA_i $ and $ A_{i-1}DA_i $, $ i = 1, \ldots, k $ satisfies the given conditions.
om35_1r_img_3.jpg | proof | Geometry | proof | Yes | Yes | olympiads | false | 588 |
V OM - I - Problem 7
In the plane, a line $ p $ and points $ A $ and $ B $ are given. Find a point $ M $ on the line $ p $ such that the sum of the squares $ AM^2 + BM^2 $ is minimized. | The geometric locus of points $ X $ in the plane for which $ AX^2 + BX^2 $ has a given value $ 2k^2 $ is a certain circle whose center lies at the midpoint $ S $ of segment $ AB $; the radius of this circle is larger the larger $ k $ is.
The smallest circle centered at $ S $ that intersects a line $ p $ is the circle tangent to line $ p $; the point of tangency is the projection $ M $ of point $ S $ onto line $ p $. Point $ M $ is the sought point; the problem always has a solution and only one.
Note. The theorem about the geometric locus, which we referred to in the above solution, can be proven as follows. Let $ AB = c $ and let $ X $ be any point in the plane (Fig. 22). Then
When $ X $ lies outside the line $ AB $, this equality is a known formula for the square of the median of a triangle (see problem 2). When $ X $ lies on the line $ AB $, verifying the equality is easy and is left to the reader.
From this equality, it follows: 1°) if $ AX^2 + BX^2 $ has a given value $ 2k^2 $, then point $ X $ lies on a circle centered at $ S $ with radius $ \sqrt{k^2 - \frac{1}{4} c^2} $; thus, $ k \geq \frac{1}{2} c $, and when $ k = \frac{1}{2} c $, the circle reduces to point $ S $; 2°) if point $ X $ lies on such a circle, i.e., if $ SX = \sqrt{k^2 - \frac{1}{4} c^2} $, then $ AX^2 + BX^2 = 2k^2 $, Q.E.D. | M | Geometry | math-word-problem | Yes | Yes | olympiads | false | 589 |
XXIX OM - I - Problem 12
Determine the least upper bound of such numbers $ \alpha \leq \frac{\pi}{2} $, that every acute angle $ MON $ of measure $ \alpha $ and every triangle $ T $ on the plane have the following property. There exists a triangle $ ABC $ isometric to $ T $ such that the side $ \overline{AB} $ is parallel to $ OM $ and the lines perpendicular to $ ON $ and passing through the centroids of triangles $ ABC $ and $ ABC' $ respectively intersect the segment $ \overline{AB} $; $ C' $ is the reflection of vertex $ C $ across the perpendicular bisector of side $ \overline{AB} $.
Note. Physically, the problem means: What should be the angle of inclination of an inclined plane so that for any triangle, one can choose a certain side such that the triangle, placed with this side on the inclined plane, will not tip over. | First, note that if $Q$ is the midpoint of side $\widehat{AC}$, then $\measuredangle ABQ \leq \frac{\pi}{6}$. Indeed, denoting by $D$ the projection of point $A$ onto the line $BQ$, we get $AD \leq AQ = \frac{1}{2} AC \leq \frac{1}{2} AB$ (Fig. 9).
Therefore, $\sin \measuredangle ABQ = \frac{AD}{AB} \leq \frac{1}{2}$. Since angle $ABC$ is not the largest angle of triangle $ABC$, it is acute. Hence, angle $ABQ$ is also acute. From the inequality $\sin \measuredangle ABQ \leq \frac{1}{2}$, we obtain that $\measuredangle ABQ \leq \frac{\pi}{6}$.
Let $\overline{AB}$ be the longest side of triangle $ABC$ isometric to triangle $T$, let $P$ be the centroid of triangle $ABC$, and $P$ be the centroid of triangle $ABC$. We can assume that $A = O$, as this does not change the conditions of the problem (Fig. 10).
om29_1r_img_9.jpgom29_1r_img_10.jpg
From the initial observation, it follows that each of the angles $BOP$ and $BOP$ has a measure not greater than $\frac{\pi}{6}$. Therefore, if $\measuredangle MON \leq \frac{\pi}{3}$, then $\measuredangle NOP = \measuredangle MON + \measuredangle BOP \leq \frac{\pi}{2}$ and similarly $\measuredangle NOP \leq \frac{\pi}{2}$. It follows that the lines perpendicular to $ON$ and passing through points $P$ and $P$ respectively intersect the segment $\overline{AB}$. Therefore, the sought upper bound of the numbers $\alpha$ is not less than $\frac{\pi}{3}$.
Conversely, if the line perpendicular to $ON$ passing through point $P$ intersects the segment $\overline{AB}$, then $\measuredangle NOP \leq \frac{\pi}{2}$. Therefore, $\measuredangle MON = \measuredangle NOP - \measuredangle BOP \leq \frac{\pi}{2} - \measuredangle BOP$. In particular, if $T$ is an equilateral triangle, then $\measuredangle BOP = \frac{\pi}{6}$ and hence
$\measuredangle MON \leq \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$. It follows that the sought upper bound of the numbers $\alpha$ is the number $\frac{\pi}{3}$. | \frac{\pi}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 592 |
XII OM - III - Task 4
Prove that if each side of a triangle is less than $ 1 $, then its area is less than $ \frac{\sqrt{3}}{4} $. | In every triangle, at least one angle does not exceed $ 60^\circ $. Suppose that in triangle $ ABC $ with sides less than one unit, $ \measuredangle C \leq 60^\circ $. Then | proof | Geometry | proof | Yes | Yes | olympiads | false | 593 |
XLIII OM - I - Problem 12
On a plane, four lines are drawn such that no two of them are parallel and no three have a common point. These lines form four triangles. Prove that the orthocenters of these triangles lie on a single straight line.
Note: The orthocenter of a triangle is the point where its altitudes intersect. | Let's denote four given lines by $k$, $l$, $m$, $n$, and their points of intersection by $O$, $P$, $Q$, $R$, $S$, $T$, such that
None of these points coincide; this is guaranteed by the conditions of the problem.
Figure 5 shows one possible configuration; however, for the further course of reasoning, it does not matter whether the three points of intersection of any of these lines with the remaining lines lie on the chosen line in the same order as in the figure, or in any other order.
The orthocenters of the four considered triangles
we denote by $E$, $F$, $G$, $H$ respectively. The line $OE$ (see Note 1) contains the altitude of triangle $OPT$, and is therefore perpendicular to the line $PT$, which is identical to the line $QP$.
om43_1r_img_6.jpg
Similarly, we justify each of the following eight perpendicularity relations:
The thesis of the problem can be obtained using vector calculus. From the relations (1), it follows that the following scalar products are equal to zero:
And since $ \overrightarrow{QP} = \overrightarrow{OP} - \overrightarrow{OQ} $, $ \overrightarrow{PE}=\overrightarrow{OE}-\overrightarrow{OP} $, we can rewrite these equalities in the form
The relations (2), (3), (4) are written similarly:
We add the two equalities in (5) side by side:
and transform the obtained relation to the form
Similarly, by adding the two equalities in each of the pairs (6), (7), (8), we get - respectively - the dependencies
Let's denote the vector $ \overrightarrow{OP} - \overrightarrow{OQ} +\overrightarrow{OR} $ by $ \overrightarrow{\mathbf{w}} $. According to the equalities (9) - (12),
Subtracting the first two products, we get the equality , which means that the vectors $ \overrightarrow{EF} $ and $ \overrightarrow{\mathbf{w}} $ are perpendicular. Similarly, by considering the differences of any two products appearing in (13), we conclude that the vector $ \overrightarrow{\mathbf{w}} $ is perpendicular to each of the vectors whose endpoints are any two of the points $ E $, $ F $, $ G $, $ H $.
Notice that $ \overrightarrow{\mathbf{w}} = \overrightarrow{OP} -\overrightarrow{OQ} + \overrightarrow{OR} = \overrightarrow{OP} + \overrightarrow{QR} $ is not a zero vector; the equality $ \overrightarrow{OP} + \overrightarrow{QR} = \overrightarrow{\mathbf{0}} $ would mean that the quadrilateral $ OPQR $ is a parallelogram, contrary to the assumption that the lines $ OP $ and $ QR $ - i.e., $ k $ and $ m $ - are not parallel. Therefore, we can speak of a line perpendicular to $ \overrightarrow{\mathbf{w}} $ passing through a given point; let this point be, for example, $ E $. Since each of the vectors $ \overrightarrow{EF} $, $ \overrightarrow{EG} $, $ \overrightarrow{EH} $ is perpendicular to $ \overrightarrow{\mathbf{w}} $, the points $ F $, $ G $, $ H $ lie on the line defined just now. The proof is complete.
Note 1. When some of the considered lines intersect at a right angle, then the orthocenter of the corresponding triangle coincides with one of the vertices; for example, point $ E $ may be identical to $ O $. One cannot then speak of the {\it line} $ OE $; nevertheless, the equality $ \overrightarrow{OE} \bullet \overrightarrow{QP} = 0 $ is meaningful (and true) in this case as well, because $ \overrightarrow{OE} $ is a zero vector. Similarly, any of the line notations used in the relations (1) - (4) may lose its meaning; but the corresponding relations (5) - (8) are valid in every case. | proof | Geometry | proof | Yes | Yes | olympiads | false | 596 |
XVII OM - I - Problem 12
Prove the theorem: If the sum of the planar angles at the vertex of a regular pyramid is equal to $180^{\circ}$, then in this pyramid, the center of the circumscribed sphere coincides with the center of the inscribed sphere. | We introduce the following notations: $ AB $ - the edge of the base of the given regular pyramid, $ S $ - its apex, $ H $ - the projection of point $ S $ onto the base, $ O $ - the center of the sphere circumscribed around the pyramid, $ M $ - the center of the circle circumscribed around triangle $ ASB $ (Fig. 9).
To prove the theorem, it suffices to state that the distance $ OM $ from point $ O $ to the lateral face $ ASB $ is equal to the distance $ OH $ from point $ O $ to the plane of the base of the pyramid.
According to the assumption, $ \measuredangle ASB = \frac{180^\circ}{n} $ where $ n $ denotes the number of edges of the base, and $ \measuredangle AHB = \frac{360^\circ}{n} $, so $ \measuredangle AHB = 2 \measuredangle ASB $.
In the circle circumscribed around triangle $ ASB $, angle $ AMB $ is a central angle, thus $ \measuredangle AMB = 2 \measuredangle ASB $, hence $ \measuredangle AHB = \measuredangle AMB $. The isosceles triangles $ AHB $ and $ AMB $ are therefore congruent, $ AH = AM $. From the equality $ OS = AO $ and $ MS = AM = AH $, the congruence of the right triangles $ OSM $ and $ AOH $ follows, so $ OM = OH $, Q.E.D. | proof | Geometry | proof | Yes | Yes | olympiads | false | 597 |
XLV OM - I - Problem 4
Given a circle with center $ O $, a point $ A $ inside this circle, and a chord $ PQ $, which is not a diameter, passing through $ A $. Lines $ p $ and $ q $ are tangent to the considered circle at points $ P $ and $ Q $, respectively. Line $ l $ passing through point $ A $ and perpendicular to $ OA $ intersects lines $ p $ and $ q $ at points $ K $ and $ L $, respectively. Prove that $ |AK|=|AL| $. | om45_1r_img_1.jpg
We start by observing that points $P$ and $K$ lie on one side of the line $OA$, while points $Q$ and $L$ lie on the other side (regardless of whether point $A$ is closer to the endpoint $P$ or $Q$ on the chord $PQ$). The chord $PQ$ of the given circle is not a diameter, so the tangents $p$ and $q$ intersect. Let's denote the point of intersection by $S$.
We circumscribe a circle $\omega_1$ around triangle $OAK$. We circumscribe a circle $\omega_2$ around triangle $OAL$. Angles $OAK$ and $OAL$ are right angles by assumption, so the diameters of these circles are (respectively) segments $OK$ and $OL$ (see Figure 1).
The lines $p$ and $q$, tangent to the given circle, are perpendicular to its radii $OP$ and $OQ$; thus, angles $OPK$ and $OQL$ are right angles. Therefore, circle $\omega_1$ passes through point $P$, and circle $\omega_2$ passes through point $Q$. (In the special case where point $A$ is the midpoint of the chord $PQ$ and consequently points $K$ and $L$ coincide with points $P$ and $Q$, respectively, we cannot strictly speak of angles $OPK$ and $OQL$, but the conclusion that $P \in \omega_1$ and $Q \in \omega_2$ is still valid.)
Angles $OPA$ and $OKA$ are inscribed angles in circle $\omega_1$, subtending the same arc $OA$ (we use the earlier observation that points $P$ and $K$ lie on the same side of the line $OA$). Therefore, these angles have the same measure: $|\measuredangle OPA| = |\measuredangle OKA|$; similarly, $|\measuredangle OQA| = |\measuredangle OLA|$ (inscribed angles in circle $\omega_2$, subtending the same arc $OA$). Finally, note that
since triangle $POQ$ is isosceles.
From the obtained angle equalities, it follows that
Therefore, the right triangles $OAK$ and $OAL$ are congruent, and we obtain the desired equality of segments $|AK| = |AL|$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 598 |
XL OM - I - Task 5
For a given natural number $ n $, determine the number of sequences $ (a_1, a_2, \ldots , a_n) $ whose terms $ a_i $ belong to the set $ \{0,1,2,3,4\} $ and satisfy the condition $ |a_i - a_{i+1}| = 1 $ for $ i = 1,2,\ldots,n-1 $. | We will divide all sequences satisfying the conditions given in the problem into three types:
sequences of type I — ending with the symbol $ 0 $ or $ 4 $;\\
sequences of type II — ending with the symbol $ 1 $ or $ 3 $;\\
sequences of type III — ending with the symbol $ 2 $.
Let $ x_n $, $ y_n $, $ z_n $ denote the number of $ n $-term sequences of types I, II, III, respectively. Each sequence of type I (length $ n $) can be extended in only one way to a sequence of length $ n + 1 $; it will be a sequence of type II (since after a zero there can only be a one, and after a four there can only be a three). From each sequence of type II (length $ n $), two sequences of length $ n+1 $ can be formed, one of type I and one of type III (after a one, a zero, and after a three, a four, or after a one or three - a two). Finally, a sequence of type III (length $ n $) will give rise to two sequences of length $ n+1 $, both of type II (after a two - a three or a one).
Thus, we obtain the relationship
From this, we get:
From equations (1), we see that
Let the sum $ x_n + y_n + z_n $ (i.e., the number of all admissible sequences of length $ n $) be denoted by $ s_n $. Adding the equations (2) and considering the equality (3), we obtain the relation:
We need to know the values of $ s_1 $, $ s_2 $, and $ s_3 $. A sequence of length $ 1 $ is a single symbol. We have five symbols available, so $ s_1 = 5 $. Now let's list all admissible sequences of length 2 and 3:
and
Therefore,
From this and the recursive formula (4), we obtain by obvious induction the answer: | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 601 |
LVII OM - I - Problem 4
Participants in a mathematics competition solved six problems, each graded with one of the scores 6, 5, 2, 0. It turned out that
for every pair of participants $ A, B $, there are two problems such that in each of them $ A $ received a different score than $ B $.
Determine the maximum number of participants for which such a situation is possible. | We will show that the largest number of participants for which such a situation is possible is 1024. We will continue to assume that the permissible ratings are the numbers 0, 1, 2, 3 (instead of 5 points, we give 4, and then divide each rating by 2).
Let $ P = \{0,1,2,3\} $ and consider the set
Set $ X $ obviously has 4096 elements. We will consider subsets $ A $ of set $ X $ with the following property (*):
(*) If $ (a_1,a_2,\dots,a_6) $, $ (b_1,b_2,\dots,b_6) \in A $, then there exist $i, j$ such that
It suffices to show that the largest number of elements in set $ A $ with property (*) is 1024.
First, we show that if set $ A $ has property (*), then it has at most 1024 elements. Assume, therefore, that we have a subset $ A $ of set $ X $ with property (*) and suppose that it has at least 1025 elements. Since there are exactly 1024 sequences of length 5 with terms from the four-element set $ P $, it follows from the pigeonhole principle that in set $ A $ there are at least two sequences that have the same terms from the first to the fifth. These sequences differ, therefore, only in one term—the sixth, which contradicts property (*). Therefore, set $ A $ has at most 1024 elements.
Now we show that there exists a set $ A $ with at least 1024 elements and having property (*). It suffices to take the following set:
First, we show that set $ A $ has at least 1024 elements. Take any numbers $ a_1,a_2,\dots,a_5 \in P $. We can make such a choice in 1024 ways. Let $ r $ be the remainder of the division of the sum $ a_1+a_2+\dots+a_5 $ by 4, and let $ a_6 = 4 - r $. Then, of course, $ (a_1,a_2,\dots,a_6) \in A $, so we have indicated at least 1024 different sequences in set $ A $.
Finally, we show that set $ A $ has property (*). Suppose that
and sequences $ (a_1,a_2,\dots,a_6) $ and $ (b_1,b_2,\dots,b_6) $ differ in only one term, say the term with index $ k: \; a_k \neq b_k $, where $ 1 \leq k \leq 6 $ and $ a_i = b_i $ for $ i \neq k $. Since the numbers $ a_1 +a_2 +\dots +a_6 $ and $ b_1 +b_2 +\dots+b_6 $ are divisible by 4, their difference is also divisible by 4. But
Thus, the number $ a_k - b_k $ is divisible by 4. Since $ a_k, b_k \in P $, then
In the set $ \{-3,-2, -1,0,1,2,3\} $, there is only one number divisible by 4, namely 0. Therefore, $ a_k = b_k $, contrary to the assumption that sequences $ (a_1,a_2,\dots,a_6) $ and $ (b_1,b_2,\dots,b_6) $ differ in the term with index $ k $. This contradiction proves that set $ A $ has property (*), which completes the proof. | 1024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 602 |
XXVI - I - Problem 12
In space, there is a cube with side $ a $, and spheres $ B_1, B_2, \ldots, B_n $ of arbitrary radii such that every point of the cube belongs to at least one of the spheres. Prove that among these spheres, one can select pairwise disjoint spheres such that the sum of their volumes is not less than $ \left(\frac{a}{5}\right)^3 $. | By induction on $ N $ we will prove
Theorem. If in a space there is a set $ F $ of volume $ V $ contained in the sum of $ N $ open balls $ B_1, B_2, \ldots, B_N $, then there exists a subset $ B_{i_1}, B_{i_2}, \ldots, B_{i_r} $ of these balls such that the balls belonging to this subset are pairwise disjoint and the sum of the volumes of the balls $ B_{i_1}, B_{i_2}, \ldots, B_{i_r} $ is greater than $ \displaystyle \frac{1}{27} V $.
Proof. If $ N = 1 $, the theorem is obvious. The ball $ B_1 $ contains the figure $ F $ of volume $ V $. Then the volume of the ball $ B_1 $ is not less than $ V $, and therefore is greater than $ \displaystyle \frac{1}{27} V $.
Next, assume that the above theorem is true for some natural number $ N $. We will prove it for the number $ N+1 $. Let the set $ F $ of volume $ V $ be contained in the sum of balls $ B_1, B_2, \ldots, B_N, B_{N+1} $. We can assume that the volume $ V_{N+1} $ of the ball $ B_{N+1} $ is not less than the volume of each of the other balls. Let $ B $ be the ball with the same center as the ball $ B_{N+1} $ and with radius of length $ 3r $, where $ r $ is the length of the radius of the ball $ B_{N+1} $. Then the volume $ V $ of the ball $ B $ is equal to $ 27 V_{N+1} $. Let $ V_0 $ be the volume of the set $ F_0 = F-B $. Since the sets $ F_0 $ and $ B $ are disjoint and $ F $ is contained in the sum of the sets $ F_0 $ and $ B $, then $ V \leq V_0 + 27 V_{N+1} $.
Every point of the set $ F_0 $ belongs to at least one of the balls $ B_1, B_2, \ldots, B_N $. Without loss of generality, we can assume that for some $ k, 0 \leq k \leq N $ each of the balls $ B_1, B_2, \ldots, B_k $ has a point in common with the set $ F_0 $, and each of the balls $ B_{k+1}, B_{k+2}, \ldots, B_N $ is disjoint from $ F_0 $. Then the set $ F_0 $ is contained in the sum of the balls $ B_1, B_2, \ldots, B_k $.
The distance from the center of the ball $ B_{N+1} $ to any point of the set $ F_0 $ is not less than $ 3r $. Therefore, the distance from any point of the ball $ B_{N+1} $ to any point of the set $ F_0 $ is not less than $ 2r $. The diameter of each of the balls $ B_1, B_2, \ldots, B_k $ is not greater than the diameter of the ball $ B_{N+1} $, i.e., not greater than $ 2r $. Therefore, each of the balls $ B_1, B_2, \ldots, B_k $ is disjoint from the ball $ B_{N+1} $. The number of balls $ B_1, B_2, \ldots, B_k $ is not greater than $ N $. By the induction hypothesis, there exists a subset $ B_{i_1}, B_{i_2}, \ldots, B_{i_r} $ of the set of balls $ B_1, B_2, \ldots, B_k $ such that the balls belonging to this subset are pairwise disjoint, and the sum of their volumes is greater than $ \displaystyle \frac{1}{27} V_0 $, and therefore - greater than $ \displaystyle \frac{1}{27} V - V_{N+1} $. Since the ball $ B_{N+1} $ is disjoint from each of the balls $ B_1, B_2, \ldots, B_k $, the balls $ B_{i_1}, B_{i_2}, \ldots, B_{i_r} $ are pairwise disjoint. The sum of their volumes is greater than $ \displaystyle \left( \frac{1}{27} V - V_{N+1} \right) + V_{N+1} = \frac{1}{27} V $. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 604 |
XXXII - I - Problem 4
On the sides of an acute triangle $ ABC $, squares $ ABED $, $ BCGF $, $ ACHI $ are constructed outside the triangle. Prove that the medians of the triangle formed by the lines $ EF $, $ GH $, $ DI $ are perpendicular to the sides of triangle $ ABC $. | Let $K$, $L$, $M$ be the vertices of the formed triangle lying at the squares $ABED$, $BCGF$, $ACHI$, respectively. Let $KO$ and $LO$ be lines perpendicular to segments $\overline{AB}$ and $\overline{BC}$, respectively (Fig. 8).
om32_1r_img_8.jpg
Triangles $KLO$ and $BEF$ have corresponding sides that are parallel, so they are similar. From this, we obtain
Suppose the perpendicular to $AC$ passing through point $M$ does not pass through $O$, then it intersects line $KO$ at some point $P$, and line $LO$ at some point $R$. If at the same time $KP > KO$ (the opposite case $KP < KO$ is considered analogously), then $MR > MP$ and $LO > LR$. Similarly, as stated above, we have
Dividing these equalities side by side, we get
Thus, based on the previously written inequalities, we have
The obtained contradiction is a consequence of the assumption that the perpendiculars to the sides of triangle $ABC$ drawn from the corresponding vertices of triangle $KLM$ do not intersect at one point. Therefore, $O = P = R$. We will show that $O$ is the centroid of triangle $KLM$. From equalities (1) and (2), it follows that
Considering angles $ABC$, $CBF$, $FBE$, $EBA$, we note that $\measuredangle EBF = 360^\circ - 2 \cdot 90^\circ - \measuredangle ABC = 180^\circ - \measuredangle ABC$. Therefore, $\measuredangle KOL = 180^\circ - \measuredangle ABC$, and from this, it follows that the area of triangle $KLO$ is
Similarly, we obtain that
The rays $OK^\to$, $OL^\to$, $OM^\to$ thus divide triangle $KLM$ into three triangles of equal area, which implies that $O$ is the centroid of triangle $KLM$, i.e., the lines $KO$, $LO$, $MO$ contain the medians of the corresponding sides of triangle $KLM$. Thus, the medians of this triangle are perpendicular to the corresponding sides of triangle $ABC$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 608 |
LVIII OM - III - Problem 1
In an acute triangle $ABC$, point $O$ is the center of the circumscribed circle, segment $CD$ is the altitude, point $E$ lies on side $AB$, and point $M$ is the midpoint of segment $CE$. The line perpendicular to line $OM$ and passing through point $M$ intersects lines $AC$, $BC$ at points $K$, $L$ respectively. Prove that | Let's draw a line through point $ M $ parallel to side $ AB $, which intersects segments $ AC $ and $ BC $ at points $ P $ and $ Q $ respectively (Fig. 12). Then points $ P $ and $ Q $ are the midpoints of sides $ AC $ and $ BC $ respectively. Since point $ O $ is the center of the circumcircle of triangle $ ABC $, we have $ \measuredangle APO=90^\circ $. Moreover, $ \measuredangle KMO=90^\circ $. Therefore, points $ K $, $ P $, $ M $, $ O $ lie on the same circle, whose diameter is segment $ KO $. Hence, $ \measuredangle OKM=\measuredangle OPM $.
Similarly, we prove that $ \measuredangle OLM=\measuredangle OQM $. From the equality of these angles, it follows that triangles $ OKL $ and $ OPQ $ are similar. Denote by $ S $ the orthogonal projection of point $ O $ onto line $ PQ $, and we obtain
om58_3r_img_12.jpg
om58_3r_img_13.jpg
Let $ X $ be the midpoint of segment $ OC $, and let $ Y $, $ R $ be the orthogonal projections of points $ X $ and $ C $ onto line $ PQ $ (Fig. 13). Since $ \measuredangle OPC=\measuredangle OQC=90^\circ $, points $ C $, $ P $, $ O $, $ Q $ lie on the same circle, whose center is point $ X $. Therefore, $ PX=XQ $, and thus $ PY=YQ $. Moreover, point $ X $ is the midpoint of segment $ OC $, so $ RY=YS $. From this, we obtain $ PR=QS $.
Therefore,
Combining equalities (1) and (2), we obtain the thesis.
Note: After obtaining relation (1), we can complete the solution in a slightly different way.
Let $ T $ be the midpoint of segment $ AB $, and let $ G $ denote the centroid of triangle $ ABC $ (Fig. 14).
om58_3r_img_14.jpg
Then point $ G $ divides each of the medians $ AQ $, $ BP $, $ CT $ of triangle $ ABC $ in the ratio $ 2:1 $. Therefore, the homothety $ j $ with center at point $ G $ and scale $ -1/2 $ maps points $ A $, $ B $, $ C $ to points $ Q $, $ P $, $ T $ respectively.
On the other hand, the line passing through point $ O $ and perpendicular to lines $ AB $ and $ QP $ contains both point $ S $ (since $ S $ is the orthogonal projection of point $ O $ onto line $ QP $) and point $ T $ (the projection of the center of the circumcircle of triangle $ ABC $ onto a side of the triangle is the midpoint of that side). In other words, segment $ TS $ is the altitude of triangle $ QPT $.
As a result, the homothety $ j $ maps the foot $ D $ of the altitude $ CD $ in triangle $ ABC $ to the foot $ S $ of the altitude $ TS $ in triangle $ QPT $. This means that $ j $ maps segments $ AD $ and $ DB $ to segments $ QS $ and $ SP $ respectively. It follows that the ratios are equal,
which, together with equality (1), implies the thesis of the problem. | proof | Geometry | proof | Yes | Yes | olympiads | false | 609 |
XXIII OM - I - Problem 7
A broken line contained in a square with a side length of 50 has the property that the distance from any point of this square to it is less than 1. Prove that the length of this broken line is greater than 1248. | Let the broken line $ A_1A_2 \ldots A_n $ have the property given in the problem. Denote by $ K_i $ ($ i= 1, 2, \ldots, n $) the circle with center at point $ A_i $ and radius of length $ 1 $, and by $ F_i $ ($ i= 1, 2, \ldots, n-1 $) the figure bounded by segments parallel to segment $ \overline{A_iA_{i+1}} $ and at a distance of $ 1 $ from it, as well as by arcs of circles $ K_i $ and $ K_{i+1} $ (in Fig. 5, the figure $ F_i $ is shaded).
The set of points in the plane at a distance of less than $ 1 $ from some point of segment $ \overline{A_i A_{i+1}} $ is contained in the union of circles $ K_i $ and $ K_{i+1} $ and figure $ F_i $. From the conditions of the problem, it follows that the given square is contained in the set
The area of the figure $ K_i \cup F_i $ is not less than $ 2A_iA_{i+1} $ (Fig. 5), and the area of the circle $ K_n $ is equal to $ \pi $. Therefore, the area of the given square does not exceed the sum of the areas of these figures, i.e.,
Hence the length of the broken line $ \displaystyle = \sum_{i=1}^{n-1} A_iA_{i+1} \geq 1250 - \frac{\pi}{2} > 1248 $. | 1248 | Geometry | proof | Yes | Yes | olympiads | false | 613 |
XXV OM - II - Problem 6
Given is a sequence of integers $ a_1, a_2, \ldots, a_{2n+1} $ with the following property: after discarding any term, the remaining terms can be divided into two groups of $ n $ terms each, such that the sum of the terms in the first group is equal to the sum of the terms in the second. Prove that all terms of the sequence are equal. | \spos{1} First, note that if a sequence of real numbers satisfies the conditions of the problem, then so does any sequence of the form
where $ k $ is any real number.
Suppose that a sequence (1) of integers satisfies the conditions of the problem and $ a_1 \ne a_2 $. Without loss of generality, we can assume that $ a_1 = 0 $. Otherwise, we would add $ -a_1 $ to each term of the sequence. Among the sequences (1) of integers that satisfy the conditions of the problem, with $ a_1 \ne a_2 $ and $ a_1 = 0 $, let us choose one in which the number $ |a_2| $ is minimal.
By discarding the term $ a_1 = 0 $, the sum of the remaining terms is even by the conditions of the problem. Therefore, the sum of all terms of the sequence (1) is even. Similarly, by discarding any term $ a_j $, the sum of the remaining terms is even. Hence, any term $ a_j $ of the sequence (1) is even.
It follows that the terms of the sequence $ b_1, b_2, \ldots, b_{2n+1} $, where $ b_i = \displaystyle \frac{1}{2} a_i $ for $ i= 1,2,\ldots, 2n+1 $, are integers. We have $ |b_2| = \displaystyle \frac{1}{2} |a_2| < |a_2| $, since $ a_2 \ne 0 $. This contradicts the minimality of the number $ |a_2| $.
The obtained contradiction proves that in every sequence of integers (1) that satisfies the conditions of the problem, all terms are equal. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 614 |
XXXVIII OM - III - Problem 3
Given is a polynomial $ W $ with non-negative integer coefficients. We define a sequence of numbers $ (p_n) $, where $ p_n $ is the sum of the digits of the number $ W(n) $. Prove that some number appears in the sequence $ (p_n) $ infinitely many times. | Let $ W(x) = a_mx^m + \ldots +a_1x+a_0 $ ($ a_i $ - non-negative integers) and let $ q $ be the number of digits in the decimal representation of the largest of the numbers $ a_0, \ldots, a_m $. Let $ \overline{a_i} $ denote the $ q $-digit decimal representation of $ a_i $; if $ a_i $ has fewer than $ q $ digits, we prepend zeros. Then for $ r > q $, the decimal representation of the number $ W(10^r) $ has the form
where $ \bar{b} $ is a block consisting of $ r-q $ zeros. The sum of the digits of each of these numbers equals the sum of the digits of all the coefficients of the polynomial $ W $. This is therefore the common value of all terms of the sequence $ (p_n) $ with indices $ n = 10^r $, $ r \leq q $. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 615 |
XXI OM - I - Problem 2
Given is the sequence $ \{c_n\} $ defined by the formulas $ c_1=\frac{a}{2} $, $ c_{n+1}=\frac{a+c_n^2}{2} $ where $ a $ is a given number satisfying the inequality $ 0 < a < 1 $. Prove that for every $ n $ the inequality $ c_n < 1 - \sqrt{1 - a} $ holds. Prove the convergence of the sequence $ \{c_n\} $ and calculate its limit. | From the definition of the sequence $ \{c_n\} $, it follows that all its terms are positive. By applying the method of induction, we will show that for every natural number $ n $, the inequality
holds.
We first state that $ c_1 < 1 - \sqrt{1 - a} $, i.e.,
Specifically, we have
This implies (2).
If for some natural number $ n $ the inequality
holds,
By induction, inequality (1) is therefore true for every natural number $ n $.
We will prove that the sequence $ \{c_n\} $ is increasing, i.e., that for every natural number $ n $, the inequality
holds.
We calculate that
We have proven that $ c_n < 1 - \sqrt{1 - a} $, and thus $ c_n < 1 + \sqrt{1 - a} $. Therefore, $ c_n - 1 + \sqrt{1 - a} < 0 $ and $ c_n - 1 - \sqrt{1 - a} < 0 $. Hence, $ c_{n+1} - c_n > 0 $. We have thus shown that the sequence $ \{c_n\} $ is increasing and bounded above.
As is known, every increasing and bounded above sequence is convergent. Let $ \displaystyle \lim_{n \to \infty} c_n = g $.
From the equality $ c_{n+1} = \frac{a + c_n^2}{2} $, it follows that $ g = \frac{a + g^2}{2} $, i.e., $ g^2 - 2g + a = 0 $. From this, we obtain that
Since, however, all terms of the sequence $ \{c_n\} $ are less than the number $ 1 - \sqrt{1 - a} $, and $ 1 - \sqrt{1 - a} < 1 + \sqrt{1 - a} $, the number $ 1 + \sqrt{1 - a} $ is not the limit of the sequence $ \{c_n\} $. Therefore, $ \displaystyle \lim_{n \to \infty} c_n = 1 - \sqrt{1 - a} $. | 1-\sqrt{1-} | Algebra | proof | Yes | Yes | olympiads | false | 617 |
XLIX OM - I - Problem 11
In a tennis tournament, $ n $ players participated. Each played one match against each other; there were no draws. Prove that there exists a player $ A $ who has either directly or indirectly defeated every other player $ B $, i.e., player $ A $ won against $ B $ or player $ A $ defeated some player $ C $ who won against player $ B $. | Let $ A $ be the participant in the tournament who defeated the largest number of opponents. (If several players have the same maximum number of victories, we choose any one of them.) We claim that the player $ A $ selected in this way has defeated, directly or indirectly, all other players.
Let $ B $ be any player who has won against $ A $. We need to show that among all the players $ C_1, \ldots, C_m $ defeated by $ A $, there exists one who has won against $ B $.
Suppose this is not the case. This means that player $ B $ has defeated players $ C_1, \ldots, C_m $ as well as player $ A $ — he would then have more victories on his record than player $ A $.
However, this is not possible, since $ A $ is the player who has achieved the most victories. The obtained contradiction completes the proof. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 619 |
II OM - III - Task 1
A beam of length $ a $ has been suspended horizontally by its ends on two parallel ropes of equal length $ b $. We rotate the beam by an angle $ \varphi $ around a vertical axis passing through the center of the beam. By how much will the beam be raised? | When solving geometric problems, a properly executed drawing is an important aid to our imagination. We represent spatial figures through mappings onto the drawing plane. There are various ways of such mapping. In elementary geometry, we most often draw an oblique parallel projection of the figure; in many cases, it is convenient to use the method of orthogonal projections onto two perpendicular planes, i.e., the so-called Monge's method (compare, for example, problem no. 23 on page 113).
We will present the solution to our problem in two variants, using once one and once the other of the mentioned mapping methods.
When using the method of oblique parallel projection, we will take as the projection plane - the plane passing through the beam $ AB $ and through the suspension points $ M $ and $ N $. We therefore draw the quadrilateral $ ABNM $ "in natural size" (fig. 59 a and b). Let $ S $ denote the midpoint of the beam. After twisting the beam, it will take the position $ CD $. The midpoint of the segment $ CD $ lies on the projection plane; let us assume that this is the point $ T $.
The position of the projection of point $ C $ depends on the direction of projection; for the oblique projection of point $ C $, we can take any point $ C $, for example, as in figure 59 a or 59 b. The projection of point $ D $ will be the point symmetric to point $ C $ with respect to point $ T $.
Calculating the sought length $ ST = x $ is simple. We draw a segment $ TK $ parallel and equal to segment $ SA $; then
So $ AM = b $, and segment $ KM $ is the leg of the right triangle $ KMC $ with the hypotenuse $ MC = b $ and the leg $ KC $. Segment $ KC $ is the base of the isosceles triangle $ KTC $, in which $ TK = TC = \frac{1}{2} AB = \frac{1}{2} a $, $ \measuredangle KTC = \varphi $.
Therefore,
Finally, we get
If $ b < a $, the angle of twist $ \varphi $ cannot be greater than the angle $ \varphi_0 $ given by the formula
For the value $ \varphi = \varphi_0 $, we have $ x = b $. Further increase of the angle is not possible without stretching the lines.
If $ b \geq a $, the maximum value of $ \varphi $ is $ 180^\circ $. For the value $ \varphi = 180^\circ $, the lines cross if $ b > a $, and overlap each other if $ b = a $.
In the above solution, the goal was to calculate the elevation of the beam after its twisting. The figure in parallel projection was only a possibly simple illustration necessary for this calculation. If we want the drawing to be a graphical solution to the problem, i.e., to give the correct length of segment $ ST $ given the lengths $ a $, $ b $, and the angle $ \varphi $, it should be done a bit differently. Specifically, the point $ T $, which we chose arbitrarily in figures 59 a and b, must be constructed from the given quantities $ a $, $ b $, $ \varphi $.
For this purpose, note that in the right triangle $ KMC $, we know the hypotenuse $ MC = MA = b $ and the leg $ KC $, which is the base of the isosceles triangle $ KCT $, where $ TK = TC = \frac{1}{2} a $, and $ \measuredangle KTC = \varphi $. From these data, we can construct the triangle to find the length of $ KM $ and the length of $ ST = AM - KM $.
The construction is shown in figure 60.
We construct triangle $ ASP $, where $ AS = SP = \frac{1}{2}a $, $ \measuredangle ASP = \varphi $. We draw a semicircle with diameter $ AM $ and a chord $ AL = AP $ in it. We measure on the line $ MA $ a segment $ MK $ equal to segment $ ML $. Point $ K $ determines the level of point $ T $; the sought elevation will be determined by segment $ TS = KA $.
The projection of the twisted beam $ C $ will be drawn as before, choosing point $ C $ arbitrarily. | \sqrt{b^2-(\frac{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 621 |
XIV OM - III - Task 1
Prove that two natural numbers, whose digits are all ones, are relatively prime if and only if the numbers of their digits are relatively prime. | Let $ J_m $ denote the $ m $-digit number whose digits are all ones:
If $ m $ is divisible by $ d $ ($ m $, $ d $ - natural numbers), then $ J_m $ is divisible by $ J_d $. Indeed, if $ m = k \cdot d $ ($ k $ - a natural number), then
where the number $ M $ is an integer.
$ \beta $) If $ m > n $, then
Let the numbers $ J_m $ and $ J_n $ be given, where $ m > n > 1 $.
a) The proof of the theorem that if $ J_m $ and $ J_n $ are coprime, then $ m $ and $ n $ are also coprime, is immediate, because according to $ \alpha $) if $ d $ is a common divisor of the natural numbers $ m $ and $ n $, then $ J_d $ is a common divisor of the numbers $ J_m $ and $ J_n $.
b) We will prove the converse theorem. Suppose the natural numbers $ m $ and $ n $ are coprime. Applying the Euclidean algorithm, i.e., dividing $ m $ by $ m $, $ n $ by the obtained remainder $ r $, the remainder $ r $ by the new remainder $ r_1 $, etc., we get smaller and smaller remainders, so we eventually reach a remainder equal to zero:
From the sequence of equalities (1), it follows that the natural number $ r_k $ is a common divisor of the numbers $ r_{k-1}, r_{k-2}, \ldots, r, n, m $, so $ r_k = 1 $.
From the first equality (1), we infer, based on $ \beta $), that
Let $ D $ be a common divisor of the natural numbers $ J_m $ and $ J_n $. Since $ J_{nq} $ is divisible by $ J_n $ according to $ \alpha $), $ D $ is also a divisor of the number $ J_r \cdot 10^{nq} $ according to (2). The number $ 10^{nq} $ has no other prime divisors except for the numbers $ 2 $ and $ 5 $, which are not divisors of the number $ J_m $, so $ D $ is coprime with $ 10^{nq} $. Therefore, $ D $ is a divisor of $ J_r $.
In the same way, based on the next equalities (1), we will conclude successively that $ D $ is a divisor of the numbers $ J_{r_1}, J_{r_2}, \ldots, J_{r_k} $. But $ r_k = 1 $, so
This means that $ J_m $ and $ J_n $ are coprime, which is what we had to prove.
Note 1. The part b) of the above proof can be replaced by a much shorter argument, using the following well-known theorem of number theory:
If the natural numbers $ m $ and $ n $ are coprime, then there exist natural numbers $ x $ and $ y $ such that
The converse theorem also holds: if the natural numbers $ m $, $ n $, $ x $, $ y $ satisfy equation (3), then $ m $ and $ n $ are coprime. This follows from equation (3) because any common divisor of the numbers $ m $ and $ n $ is a divisor of the number $ 1 $.
Assume the natural numbers $ m $, $ n $ are coprime, so for some natural numbers $ x $ and $ y $, (3) holds. Then
and since according to ($ \alpha $): $ J_{mx} = J_m \cdot M $ and $ J_{my} = J_m \cdot N $, where the numbers $ M $ and $ N $ are natural, then:
from which it follows that $ J_m $ and $ J_n $ are coprime.
Note 2. The proof of the theorem cited in Note 1 can be derived from the sequence of equalities (1). Substituting the value of $ r $ from the first equality into the second, and then after this substitution the value of $ r_1 $ from the second equality into the third, etc., we eventually arrive, considering that $ r_k = 1 $, at an equality of the form (3).
Note 3. In the above arguments, we assumed that the base of numeration is $ 10 $. The theorem is, however, true for any base of numeration $ g $. The proof remains the same, except that everywhere $ 10 $ must be replaced by $ g $. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 623 |
LX OM - III - Task 1
Each vertex of a convex hexagon is the center of a circle with a radius equal to the length of the longer of the two sides of the hexagon containing that vertex. Prove that if the intersection of all six circles (considered with their boundaries) is non-empty, then the hexagon is regular. | Let's denote the consecutive vertices of the given hexagon by the letters $A, B, C, D, E$, and $F$.
Assume that the intersection of the six circles mentioned in the problem is non-empty and that point
$P$ belongs to this intersection. Consider any side of the hexagon, for the sake of argument, let it be
side $AB$. From the conditions of the problem, it follows that $AP \leqslant AB$ and $BP \leqslant AB$, so
if $P$ does not lie on the line $AB$, then segment $AB$ is the longest side of triangle $ABP$.
Angle $\measuredangle APB$ is therefore its largest angle, i.e., $\measuredangle APB \geqslant 60^{\circ}$.
If, however, point $P$ lies on the line $AB$, then it lies on segment $AB$ (otherwise $AP > AB$ or $BP > AB$).
If it is not at the end of this segment, then $\measuredangle APB = 180^{\circ}$. In summary,
if $P \neq A$ and $P \neq B$, then $\measuredangle APB \geqslant 60^{\circ}$. Similar reasoning can be
applied to the other sides of the hexagon.
We will show that point $P$ does not lie outside the hexagon $ABCDEF$. If it did, then for some
vertices $X, Y$, the convex angle $\measuredangle XPY < 180^{\circ}$ would cover the entire hexagon
(Fig. 1). Given the convexity of hexagon $ABCDEF$,
this would imply
which is contradictory to $\measuredangle XPY < 180^{\circ}$.
Since point $P$ does not lie outside the hexagon, it must lie in its interior or on its boundary.
First, consider the case where $P$ is one of the vertices of the hexagon. Without loss of
generality, we can assume that $P = A$. Then, due to the convexity of the hexagon, we have
which is impossible. If $P$ lies on the boundary of hexagon $ABCDEF$ but is not a vertex,
then it lies between some consecutive vertices. Without loss of generality, we can
assume that $P$ lies on side $AB$ of the hexagon. Then we have
which is also impossible.
Therefore, $P$ lies in the interior of the hexagon. Angles $\measuredangle APB$, $\measuredangle BPC$, $\measuredangle CPD$,
$\measuredangle DPE$, $\measuredangle EPF$, and $\measuredangle FPA$ then sum up to a full angle, and the measure
of each of them is at least $60^{\circ}$. It follows that the measure of each of them is
$60^{\circ}$ (Fig. 2). Since $\measuredangle APB = 60^{\circ}$ is the largest angle of triangle
$ABP$, the other angles also measure $60^{\circ}$. Triangle $ABP$ is therefore equilateral, and similarly,
the other triangles with vertex $P$ and base being one of the sides are also equilateral. Thus, the hexagon
is regular. | proof | Geometry | proof | Yes | Yes | olympiads | false | 624 |
LV OM - I - Task 8
Point $ P $ lies inside the tetrahedron $ ABCD $. Prove that | Let $ K $ be the point of intersection of the plane $ CDP $ with the edge $ AB $, and let $ L $ be the point of intersection of the plane $ ABP $ with the edge $ CD $ (Fig. 4).
Similarly, let $ M $ be the point of intersection of the plane $ ADP $ and the edge $ BC $, and let $ N $ denote the common point of the plane $ BCP $ and the edge $ AD $.
om55_1r_img_4.jpg
The points $ K $, $ P $, and $ L $ belong to the common part of the planes $ ABP $ and $ CDP $, so they lie on the same line. Similarly, the points $ M $, $ P $, and $ N $ lie on the same line. Therefore, the points $ K $, $ L $, $ M $, $ N $, and $ P $ lie in the same plane. This concludes the solution of the problem. | proof | Geometry | proof | Yes | Yes | olympiads | false | 625 |
LVII OM - III - Problem 2
Determine all positive integers $ k $ for which the number $ 3^k+5^k $ is a power of an integer with an exponent greater than 1. | If $ k $ is an even number, then the numbers $ 3^k $ and $ 5^k $ are squares of odd numbers, giving a remainder of 1 when divided by 4. Hence, the number $ 3^k + 5^k $ gives a remainder of 2 when divided by 4, and thus is divisible by 2 but not by $ 2^2 $. Such a number cannot be a power of an integer with an exponent greater than 1.
If $ k $ is an odd number, then
The second factor on the right side of the above relationship contains an odd (equal to $ k $) number of odd summands.
Hence, the number $ 3^k + 5^k $ is divisible by 8 and not by 16. If this number is a power of an integer with an exponent greater than 1, then it must be a cube of an integer.
If $ k = 1 $, then the considered number is a cube of an integer: $ 3^1 + 5^1 = 2^3 $. Let us assume in the further part of the reasoning that $ k \geq 3 $. From the relationship
it follows that cubes of integers give remainders of 0, 1, 8 when divided by 9. For $ k \geq 3 $ we have $ 9 | 3^k $, so $ 3^k + 5^k \equiv 5^k (\mod 9) $.
The remainders of the numbers $ 5, 5^2, 5^3, 5^4, 5^5, 5^6 $ when divided by 9 are 5, 7, 8, 4, 2, 1, respectively. Therefore, if $ 3^k + 5^k $ is a cube of an integer for $ k \geq 3 $, then $ 3 | k $. We have previously shown that $ k $ cannot be an even number.
Thus, the number $ k $ is of the form $ 6l + 3 $, where $ l $ is a non-negative integer.
From the relationship $ 3^3 \equiv 5^3 \equiv 6 (\mod 7) $ and $ 3^6 \equiv 5^6 \equiv 1 (\mod 7) $, it follows that
However, from direct verification, we obtain that the cube of an integer gives a remainder of 0, 1, or 6 when divided by 7:
Therefore, the considered number for $ k \geq 3 $ cannot be a cube of an integer, which concludes the solution of the problem.
Answer: $ k = 1 $. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 626 |
L OM - I - Task 4
Given real numbers $ x $, $ y $, such that the numbers $ x + y $, $ x^2 + y^2 $, $ x^3 + y^3 $, and $ x^4 + y^4 $ are integers. Prove that for every positive integer $ n $, the number $ x^n + y^n $ is an integer. | Since $ 2xy= (x + y)^2 - (x^2 + y^2) $ and $ 2x^2y^2 = (x^2 + y^2)^2 - (x^4 + y^4) $, the numbers $ 2xy $ and $ 2x^2y^2 $ are integers. If the number $ xy $ were not an integer, then $ 2xy $ would be odd. But then the number $ 2x^2y^2 = (2xy)^2/2 $ would not be an integer. The conclusion is that the number $ xy $ is an integer.
The integrality of the number $ x^n + y^n $ is proved by induction using the identity $ x^n + y^n = (x^{n-1} +y^{n-1})(x + y)-xy(x^{n-2} +y^{n-2}) $. | proof | Algebra | proof | Yes | Yes | olympiads | false | 627 |
V OM - III - Task 5
Prove that if in a tetrahedron $ABCD$ the opposite edges are equal, i.e., $AB = CD$, $AC = BD$, $AD = BC$, then the lines passing through the midpoints of opposite edges are mutually perpendicular and are axes of symmetry of the tetrahedron. | Let $ K $, $ L $, $ M $, $ N $, $ P $, $ Q $ denote the midpoints of the edges of the tetrahedron $ ABCD $, as indicated in Fig. 42. It is sufficient to prove that any of the lines $ KL $, $ MN $, $ PQ $, for example, $ KL $, is an axis of symmetry of the tetrahedron and that it is perpendicular to one of the two remaining lines, for example, to $ PQ $.
\spos{1} Since, by assumption, $ AD = BC $ and $ BD = AC $, the triangles $ ABD $ and $ ABC $ are congruent, as they have corresponding equal sides, so the medians $ DK $ and $ CK $ of these triangles are equal, which means that triangle $ DKC $ is isosceles and the median $ KL $ of this triangle is its altitude, i.e., $ KL \bot DC $; similarly, $ KL \bot AB $. It follows that the line $ KL $ is an axis of symmetry of the tetrahedron, as point $ S $ is symmetric to point $ A $, and point $ C $ is symmetric to point $ D $ with respect to the line $ KL $. Segment $ BC $ is symmetric to segment $ AD $, so the midpoint $ Q $ of segment $ BC $ is symmetric to the midpoint $ P $ of segment $ AD $. The line $ PQ $ passing through points $ P $ and $ Q $ symmetric with respect to the line $ KL $ intersects this line and is perpendicular to it. | proof | Geometry | proof | Yes | Yes | olympiads | false | 630 |
XXIV OM - I - Problem 12
In a class of n students, a Secret Santa event was organized. Each student draws the name of the person for whom they are to buy a gift, so student $ A_1 $ buys a gift for student $ A_2 $, $ A_2 $ buys a gift for $ A_3 $, ..., $ A_k $ buys a gift for $ A_1 $, where $ 1 \leq k \leq n $. Assuming that all drawing outcomes are equally probable, calculate the probability that $ k = n $. | In the result of any drawing, each student draws a certain student from the same class (possibly themselves), and different students draw different students. Therefore, the result of each drawing defines a certain one-to-one mapping (permutation) of the set of all students in the class onto itself. Conversely, each permutation can be the result of some drawing. Therefore, the number of elementary events is equal to the number of permutations of an $n$-element set, i.e., $n!$.
The favorable event will occur when a certain student $A_1$ draws one of the remaining $n-1$ students, denoted by $A_2$; student $A_2$ draws one of the $n-2$ students different from $A_1$ and $A_2$, denoted by $A_3$; $\ldots$; finally, student $A_{n-1}$ draws student $A_n$ different from $A_1, A_2, \ldots, A_{n-1}$. Since student $A_1$ can draw a student different from themselves in $n-1$ ways, student $A_2$ can draw a student different from $A_1$ and $A_2$ in $n-2$ ways, etc., the number of favorable events is equal to $(n-1)(n-2) \ldots 2 \cdot 1 = (n-1)!$.
Therefore, the probability that $k = n$ is $\displaystyle \frac{(n-1)!}{n!} = \frac{1}{n}$. | \frac{1}{n} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 636 |
VIII OM - I - Task 6
Find a four-digit number, whose first two digits are the same, the last two digits are the same, and which is a square of an integer. | If $ x $ is the number sought, then
where $ a $ and $ b $ are integers satisfying the inequalities $ 0 < a \leq 9 $, $ 0 \leq b \leq 9 $. The number $ x $ is divisible by $ 11 $, since
Since $ x $ is a perfect square, being divisible by $ 11 $ it must be divisible by $ 11^2 $, so the number
is divisible by $ 11 $. It follows that $ a + b $ is divisible by $ 11 $, and since $ 0 < a + b \leq 18 $, then $ a + b = 11 $. Therefore,
from which we infer that $ 9a + 1 $ is the square of some natural number $ m $:
Since $ 9a + 1 \leq 82 $, then $ m \leq 9 $.
From the above,
It follows from this equality that the product $ (m + 1) (m - 1) $ is divisible by $ 9 $, and since at most one of the numbers $ m + 1 $ and $ m - 1 $ is divisible by $ 3 $, then one of them is divisible by $ 9 $. Considering that the natural number $ m $ is less than $ 10 $, we conclude from this that $ m + 1 = 9 $, so $ m = 8 $. In this case, $ a = 7 $, $ b = 4 $, and the sought number is $ 7744 = (88)^2 $. | 7744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 638 |
XXV - I - Problem 3
Prove that the bisectors of the angles formed by the lines containing the opposite sides of a convex quadrilateral inscribed in a circle are respectively parallel to the bisectors of the angles formed by the lines containing the diagonals of this quadrilateral. | Let $ A $, $ B $, $ C $, $ D $ be consecutive vertices of a quadrilateral inscribed in a circle. Let $ O $ be the point of intersection of the lines $ AB $ and $ CD $, and let $ k $ be the angle bisector of $ \measuredangle AOD $ (Fig. 12). If $ A' $ and $ C' $ are the images of points $ A $ and $ C $, respectively, under reflection in the line $ k $, then the lines $ A'C' $ and $ BD $ are parallel. We have $ \measuredangle OBD = \measuredangle A'CO $ (as inscribed angles subtended by the same arc) and $ \measuredangle A'CO = \measuredangle ACO $ (symmetry preserves angle measures). Therefore, the angle bisectors of the angles formed by the lines $ AC $ and $ BD $, and $ AC $ and $ A'C' $ are parallel. The latter angle bisector is the line $ k $. | proof | Geometry | proof | Yes | Yes | olympiads | false | 639 |
XIX OM - I - Problem 5
Given a set of $ 2n $ ($ n \geq 2 $) different numbers. How to divide this set into pairs so that the sum of the products of the numbers in each pair is a) the smallest, b) the largest? | By a partition of a set of numbers $ Z $ into pairs, we mean a set of pairs of numbers from $ Z $ such that each number in $ Z $ belongs to one and only one pair.
Let us denote the given numbers by the letters $ a_1, a_2, \ldots, a_{2n} $ in such a way that
The number of all partitions of the set of numbers (1) into pairs is finite, so there exists a partition $ P $ for which the sum of the products of the numbers in each pair has the smallest value. Let us denote this sum by the letter $ S_P $.
One of the terms in the sum $ S_P $ must be the product $ a_1a_{2n} $. There is a term $ a_1a_k $ in the sum $ S_P $, one of whose factors is $ a_1 $. If $ a_k \ne a_{2n} $, then the sum $ S_P $ would have another term with the factor $ a_{2n} $, for example, $ a_la_{2n} $, where $ l \ne k $ and $ l \ne 1 $. Then
thus, by (1), the inequality
would hold.
In that case, by replacing the pairs $ (a_1, a_k) $ and $ (a_l, a_{2n}) $ in the partition $ P $ with the pairs $ (a_1, a_{2n}) $ and $ (a_k, a_l) $, we would obtain a partition $ Q $ of the given numbers into pairs, for which the sum $ S_Q $ of the products of the numbers in each pair would satisfy, due to (2), the inequality
contrary to the assumption that the smallest such sum is $ S_P $.
Similarly, we conclude that the sum $ S_P - a_1a_{2n} $ must have the term $ a_2a_{2n-1} $, the sum $ S_P -a_1a_{2n} - a_2a_{2n-1} $ - the term $ a_3a_{2n-2} $, etc.
Thus
i.e., in the partition $ P $, the pairs $ (a_1 a_{2n}), (a_2, a_{2n-1}), \ldots, (a_n, a_{n-1}) $ appear.
In a similar way, it can be shown that the largest sum $ S_T $ of the products of the numbers in each pair occurs in the partition $ T $ of the given set into pairs $ (a_1 a_2), (a_3, a_4),\ldots ,(a_{2n-1}, a_{2n}) $, i.e., that | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 641 |
XXXIV OM - I - Problem 1
$ A $ tosses a coin $ n $ times, $ B $ tosses it $ n+1 $ times. What is the probability that $ B $ will get more heads than $ A $? | Consider the following events. $ Z_0 $ - the event that player $ B $ has flipped more heads than player $ A $, $ Z_r $ - the event that $ B $ has flipped more tails than player $ A $. These events are equally probable: if the coin is fair, it does not matter which side is called heads and which is called tails. Thus, $ P(Z_0) = P(Z_r) $. Now let $ a_0 $, $ a_r $, $ b_0 $, $ b_r $ be the number of heads and tails flipped by $ A $ and the number of heads and tails flipped by $ B $, respectively. According to the conditions of the problem, $ a_0+ a_r = n $, $ b_0+b_r=n+1 $. Therefore, by subtracting the equations, $ (b_0-a_0)+ (b_r-a_r) = 1 $. From this equation, it follows that one and only one of the numbers $ b_0-a_0 $ and $ b_r-a_r $ is positive, meaning that exactly one of the events $ Z_0 $ and $ Z_r $ must occur. In other words, these events are mutually exclusive and cover all possibilities. Therefore, $ P(Z_0)=P(Z_r)= 1/2 $. | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 643 |
XXXVI OM - III - Problem 6
Prove that if in a convex polyhedron with $ k $ faces there are more than $ k/2 $ faces, no two of which have a common edge, then it is impossible to inscribe a sphere in this polyhedron. | Suppose that a sphere can be inscribed in a given polyhedron. Let $S_1, \ldots, S_k$ be the faces of the polyhedron, and $P_1, \ldots, P_k$ the points of tangency of the respective faces with the sphere. If $A$ and $B$ are common vertices of the faces $S_i$ and $S_j$, then the triangles $ABP_i$ and $ABP_j$ are congruent (the segments $AP_i$ and $AP_j$ tangent to the sphere from point $A$ have equal lengths, similarly $|BP_i| = |BP_j|$). In this way, the full angle at vertex $P_i$ located on face $S_i$ can be represented as the sum of the corresponding angles located on faces sharing an edge with $S_i$. The sum of the measures of the angles determined in this way on all faces is $k \cdot 2\pi$. According to the assumption, there are more than $k/2$ faces, no two of which share an edge. Of course, the sum of the measures of the angles centered at the points of tangency of these faces with the sphere is more than $k\pi$. Each of these angles is equal to a corresponding angle on one of the remaining faces. But there are fewer than $k/2$ of these remaining faces, so the sum of the measures of the angles considered on them is less than $k\pi$. We have obtained a contradiction. Therefore, a sphere cannot be inscribed in such a polyhedron.
Note. The existence of polyhedra satisfying such a condition was demonstrated in problem 11 of the first stage of the competition. | proof | Geometry | proof | Yes | Yes | olympiads | false | 645 |
XXIII OM - III - Problem 5
Prove that all subsets of a finite set can be arranged in a sequence, where consecutive terms differ by one element. | Assume that a given finite set $A$ has $n$ elements. By applying induction on $n$, we will prove that there exists such a sequence of all subsets of set $A$ that
(1)
all subsets of set $A$ are included, and
(2)
for any two consecutive subsets $A_i$ and $A_{i+1}$ in the sequence, one of the sets $A_{i+1} - A_i$ and $A_i - A_{i+1}$ is empty, and the other is a singleton.
For $n = 1$, the sequence $\emptyset$, $A$ obviously satisfies (2). Assume that for some natural number $n$, there exists such a sequence (1) of all subsets of an $n$-element set $A$ that satisfies (2). We will prove that there exists a sequence of all subsets of an $(n+1)$-element set $A \cup \{b\}$ that satisfies (2).
We apply induction on $n$. If the sequence (1) of all subsets of an $n$-element set $A$ satisfies condition (2), then the following sequence of all subsets of an $(n+1)$-element set $A \cup \{b\}$ also satisfies this condition:
(5)
This sequence contains all sets $A_i$ and all sets $A_i \cup \{b\}$ for $i = 1, 2, \ldots, 2^n$, i.e., all subsets of set $A \cup \{b\}$. Consider the consecutive terms of sequence (5). From the induction hypothesis, we know that one of the sets $A_{i+1} - A_i$ and $A_i - A_{i+1}$ is empty, and the other is a singleton; it follows that one of the sets $(A_i \cup \{b\}) - (A_{i+1} \cup \{b\}) = A_i - A_{i+1}$ and $(A_{i+1} \cup \{b\}) - (A_i \cup \{b\}) = A_{i+1} - A_i$ is empty, and the other is a singleton.
Finally, $(A_{2^n} \cup \{b\}) - A_{2^n} = \{b\}$ and $A_{2^n} - (A_{2^n} \cup \{b\}) = \emptyset$. Therefore, sequence (5) satisfies condition (2).
Note. This problem has an interesting geometric interpretation. Consider the case $n = 3$. If $A = \{a_1, a_2, a_3\}$, then each subset of set $A$ can be identified with a point $(e_1, e_2, e_3)$, whose coordinates are either $0$ or $1$. Specifically, we define
(6)
The points thus defined are the vertices of a cube determined by the unit vectors of the coordinate axes.
Let $A$ and $A'$ be subsets of set $A$. One of the sets $A' - A$ and $A - A'$ is empty, and the other is a singleton if and only if the points corresponding to sets $A$ and $A'$ differ in one coordinate. Such points belong to some edge of the cube. Therefore, our problem can be formulated as follows: There exists a broken line formed by some edges of the cube that passes exactly once through each of its vertices.
In the case $n \ne 3$, an analogous interpretation of the problem can be given. Instead of a cube, one should consider an $n$-dimensional hypercube. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 646 |
XI OM - II - Task 3
Given are two circles with a common center $ O $ and a point $ A $. Construct a circle with center $ A $ intersecting the given circles at points $ M $ and $ N $, so that the line $ MN $ passes through the point $ O $. | Suppose a circle with center $A$ intersects the given circles at points $M$ and $N$ lying on a straight line with point $O$. Let $S$ be the midpoint of segment $MN$. The problem will be solved when we can find point $S$. If points $M$ and $N$ lie on the same side of point $O$, then
if, however, $M$ and $N$ lie on opposite sides of point $O$, then
where $r_1$ and $r_2$ denote the radii of the given circles, with $r_1 > r_2$ (Fig. 20).
Angle $OSA$ is a right angle, since in the isosceles triangle $MAN$ ($MA = NA$) the median $AS$ is also an altitude.
Therefore, point $S$ is the intersection of one of the circles $k_1$ and $k_2$ drawn from center $O$ with radii $\frac{r_1 + r_2}{2}$ and $\frac{r_1 - r_2}{2}$ and the circle $k_3$ with diameter $OA$.
From this, the method of determining point $S$, and thus the construction of the desired circle, immediately follows. The existence and number of solutions depend on the existence of common points of circles $k_1$ and $k_2$ with circle $k_3$. The condition for the intersection of circles $k_1$ and $k_3$ is the inequality $\frac{r_1 + r_2}{2} < OA$; when $\frac{r_1 + r_2}{2} = OA$, the circles are tangent.
Similarly, the condition for the intersection of circles $k_2$ and $k_3$ is the inequality $\frac{r_1 - r_2}{2} < OA$; when $\frac{r_1 - r_2}{2} = OA$, the circles are tangent.
Hence the conclusion:
a) when $OA \geq \frac{r_1 + r_2}{2}$, the problem has 2 solutions, and when $OA = \frac{r_1 + r_2}{2}$, one of the obtained circles is tangent to both given circles;
b) when $\frac{r_1 - r_2}{2} \leq OA < \frac{r_1 + r_2}{2}$, the problem has one solution, which in the case of equality $OA = \frac{r_1 - r_2}{2}$ is a circle tangent to both given circles.
c) When $OA < \frac{r_1 - r_2}{2}$, the problem has no solutions. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 648 |
LIV OM - II - Task 5
Point $ A $ lies outside the circle $ o $ with center $ O $. From point $ A $, two tangent lines to circle $ o $ are drawn at points $ B $ and $ C $. A certain tangent to circle $ o $ intersects segments $ AB $ and $ AC $ at points $ E $ and $ F $, respectively. Lines $ OE $ and $ OF $ intersect segment $ BC $ at points $ P $ and $ Q $, respectively. Prove that from segments $ BP $, $ PQ $, and $ QC $, a triangle similar to triangle $ AEF $ can be constructed. | Let $ X $ be the point of tangency of circle $ o $ with line $ EF $ (Fig. 3). From the equality
of
it follows that triangles $ BEP $ and $ XEP $ are congruent. Therefore, $ BP = XP $.
Similarly, we prove the equality $ CQ = XQ $. This means that triangle $ PQX $ is constructed from segments $ BP $, $ PQ $, and $ QC $.
om54_2r_img_3.jpg
It remains to show that triangle $ PQX $ is similar to triangle $ FEA $. Let $ \measuredangle FEA = \alpha $, $ \measuredangle EFA = \beta $. Triangle $ ABC $ is isosceles, from which we calculate that $ \measuredangle ABC = \frac{1}{2} (\alpha+\beta) $. Furthermore, $ \measuredangle BEP = 90^{\circ} - \frac{1}{2} \alpha $. The last two equalities imply
that
which results in
Similarly, we prove that $ \measuredangle XQP = \alpha $, which completes the solution of the problem. | proof | Geometry | proof | Yes | Yes | olympiads | false | 649 |
XXXVII OM - III - Problem 3
Prove that if $ p $ is a prime number, and the integer $ m $ satisfies the inequality $ 0 \leq m < p—1 $, then $ p $ divides the number $ \sum_{j=1}^p j^m $. | Let's denote the sum $1^m + 2^m + \ldots + p^m$ by $A(p, m)$. We need to prove that under the given conditions
Proof is conducted by induction on $m$ (for a fixed $p$). When $m = 0$, we get the number $A(p, 0) = p$, which is clearly divisible by $p$.
Fix an integer $m$ satisfying $0 < m < p - 1$ and assume that
We want to show that $p \mid A(p, m)$.
According to the binomial theorem, for each natural number $j$ the following equality holds
Substituting $j = 1, \ldots, p$ we get a sequence of equalities
Adding these equalities side by side, we obtain
The left-hand side of equation (2) can be rewritten as . This is a number divisible by $p$. In the last sum on the right-hand side of equation (2), all terms are divisible by $p$ by the inductive hypothesis (1). Therefore, the remaining term in (2) is also divisible by $p$:
Since $p$ is a prime number, it must divide one of the factors of the obtained product. The number $m + 1$, being smaller than $p$, does not divide $p$. Therefore, ultimately $p \mid A(p, m)$, which is what we needed to prove.
By the principle of induction, the theorem given in the problem is proved.
Note. Equation (2) is a recursive formula, from which one can compute the sums of powers (with increasingly larger natural exponents) of finite sequences of consecutive natural numbers. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 651 |
XLVI OM - II - Zadanie 1
Wielomian $ P(x) $ ma współczynniki całkowite. Udowodnić, że jeżeli liczba $ P(5) $ dzieli się przez 2, a liczba $ P(2) $ dzieli się przez 5, to liczba $ P(7) $ dzieli się przez 10.
|
Dowód opiera się na spostrzeżeniu, że jeśli $ P(x) = a_0 + a_1x + \ldots + a_nx^n $ jest wielomianem o współczynnikach całkowitych, to dla każdej pary różnych liczb całkowitych $ u $, $ v $ różnica $ P(u) - P(v) $ jest podzielna przez różnicę $ u-v $. Istotnie:
a każda z różnic $ u^k - v^k $ rozkłada się na czynniki:
Przyjmując w szczególności $ u = 7 $, $ v = 5 $ widzimy, że różnica $ P(7)-P(5) $ dzieli się przez $ 2 $; przyjmując następnie $ u = 7 $, $ v = 2 $ stwierdzamy, że różnica $ P(7)-P(2) $ dzieli się przez $ 5 $. Stąd, wobec warunków danych w założeniu zadania, wnosimy, że liczba $ P(7) $ dzieli się zarówno przez $ 2 $, jak i przez $ 5 $. A ponieważ liczby $ 2 $ i $ 5 $ są względnie pierwsze, zatem liczba $ P(7) $ dzieli się przez iloczyn $ 2 \cdot 5 $, czyli przez $ 10 $.
| proof | Algebra | proof | Yes | Yes | olympiads | false | 652 |
LIX OM - II - Task 6
Given a positive integer $ n $ not divisible by 3. Prove that there exists a number $ m $ with the following property:
Every integer not less than $ m $ is the sum of the digits of some multiple of the number $ n $. | A good number will be understood as a positive integer that is the sum of the digits of some multiple of the number $ n $.
First of all, note that the sum of two good numbers is also a good number. Indeed, let $ k_1, k_2 $ be multiples of the number $ n $ with the sums of their digits being $ s_1, s_2 $, respectively. Let $ t $ be an integer such that $ 10^t > k_2 $. Then the number $ k = 10^t \cdot k_1 + k_2 $ is a multiple of the number $ n $, and the sum of the digits of the number $ k $ is equal to $ s_1 + s_2 $, since when adding the numbers $ 10^t \cdot k_1 $ and $ k_2 $, non-zero digits appear in different positions.
The number $ n $ is not divisible by 3, so its sum of digits is also not divisible by 3.
Now let $ b $ be an integer such that $ 10^{3^b} > n $. Assume that the digits of the decimal representation of the number $ n $ are $ c_1, c_2, \dots, c_l $. We can of course assume that $ c_l \neq 0 $, since dividing the number $ n $ by 10 does not change the conditions of the problem. Then the digits of the decimal representation of the number
$ (10^{3^b} -1)n = \underbrace{999\dots 99}_{3^b} \cdot n $, divisible by $ n $, are the digits:
Hence, the sum of the digits of the number $ (10^{3^b} -1)n $ is $ 9 \cdot 3^b = 3^{b+2} $.
We have thus shown that the good numbers are: the number $ a $, which is not divisible by 3, and the number $ 3^d $, where $ d = b+2 $.
We will show that every integer not less than $ a \cdot 3^d $ is good. This will imply that the number
$ m = a \cdot 3^d $ satisfies the thesis of the problem.
Let $ s \geqslant a \cdot 3^d $ be any integer. The numbers
are then positive. Moreover, no two of them give the same remainder when divided by $ 3^d $ (since the divisibility
$ 3^d | (s - ia) - (s - ja) = (j - i)a $ for some indices $ 0 \leqslant i, j \leqslant 3^d - 1 $ due to the non-divisibility of the number $ a $ by 3 is only possible if $ 3^d | j - i $, which implies $ i = j $). There are $ 3d $ numbers in the sequence (*), so there exists a number among them that is divisible by $ 3^d $. Let, for example, $ s - va = w \cdot 3^d $. Then
$ v, w \geqslant 0 $ and
The numbers $ a $ and $ 3^d $ are good. Therefore, from the above representation, we conclude that the number $ s $, as the sum of good numbers, is also good, which is what we were proving. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 653 |
XXVIII - I - Problem 7
Three circles with radii equal to 1 intersect such that their common part is bounded by three arcs, each belonging to a different circle. The endpoints of these arcs are denoted by $ K, L, M $. The first circle intersects the second at points $ C $ and $ M $, the second intersects the third at points $ A $ and $ K $, and the third intersects the first at points $ B $ and $ L $.
Prove that the sum of the lengths of the arcs $ AM, BK, CL $ equals $ \pi $. | If two circles of equal radii intersect at points $ P $ and $ Q $, then the lengths of the arcs determined by these points on one circle are equal to the lengths of the corresponding arcs determined by these points on the other circle. By performing a symmetry with respect to the line $ PQ $, the arcs with endpoints $ P $ and $ Q $ on one circle are transformed into the corresponding arcs with endpoints $ P $ and $ Q $ on the other circle.
From this, the following equalities of the arc lengths of the circles given in the problem (Fig. 5) follow:
By adding these equalities side by side, we conclude that
The measure of an inscribed angle in a circle is half the measure of the central angle subtended by the same arc. In a circle with a radius of length $ 1 $, the measure of the central angle expressed in radians is equal to the length of the arc on which this angle is subtended. Therefore,
and similarly,
Since the sum of the measures of the angles of triangle $ KLM $ is $ \pi $, by adding these equalities side by side, we obtain, by virtue of (1),
| proof | Geometry | proof | Yes | Yes | olympiads | false | 656 |
XXXII - I - Problem 3
A cone $ S_1 $ is circumscribed around a sphere $ K $. Let $ A $ be the center of the circle of tangency of the cone with the sphere. A plane $ \Pi $ is drawn through the vertex of the cone $ S_1 $ perpendicular to the axis of the cone. A second cone $ S_2 $ is circumscribed around the sphere $ K $, with its vertex lying on the plane $ \Pi $. Prove that the plane of the circle of tangency of the cone $ S_2 $ with the sphere $ K $ passes through the point $ A $. | Let $ O $ be the center of the sphere, $ r $ - the length of its radius. $ W_1 $, $ W_2 $ are the vertices of the cones $ S_1 $ and $ S_2 $ (Fig. 7).
om32_1r_img_7.jpg
The plane determined by the points $ O $, $ W_1 $, $ W_2 $ contains the point $ A $, which is the center of the circle of tangency of the cone $ S_1 $ with the sphere $ K $ (point $ A $ lies on the segment $ \overline{OW_1} $), and also contains a certain diameter $ \overline{BC} $ of this circle, the center $ R $ of the circle of tangency of the cone $ S_2 $ with the sphere $ K $, and the diameter $ \overline{MN} $ of this circle. Let $ D $ be the point of intersection of the segment $ \overline{MN} $ and the line $ W_1O $. We will show that $ D = A $. Suppose that the point $ D $ lies between $ M $ and $ R $. Then $ MD = MR - RD $, $ ND = MR + RD $, and $ MD \cdot ND = (MR - RD) (MR + RD) = MR^2 - RD^2 $. Applying the Pythagorean theorem to triangles $ MRW_2 $ and $ DRW_2 $ we get from this
however, from the Pythagorean theorem applied to triangles $ MW_2O $ and $ W_1DW_2 $ it follows that
From the theorem of the secant and the tangent, it follows that
thus
If $ DW_1 < AW_1 $, then $ MD \cdot DN > W_1B^2 - AW_1^2 = AB^2 $, since $ AB^2 = BA \cdot AC $, so $ MD \cdot DN > BA \cdot AC $, which is not possible, because the segment $ \overline{MN} $ intersects the segment $ \overline{AB} $ at some point $ T $, such that $ MT \cdot TN = BT \cdot TC \leq AB^2 $. If $ DW_1 > AW_1 $, then $ W_1O^2 - DW_1 < W_1O^2 - AW_1^2 = AB^2 $, which is also not possible. Therefore, it must be $ DW_1 = AW_1 $, from which it follows that $ A = D $. Therefore, the point $ A $ lies on the line $ MN $, and thus in the plane of the circle of tangency of the cone $ S_2 $ with the sphere $ K $. | proof | Geometry | proof | Yes | Yes | olympiads | false | 657 |
XXVII OM - II - Problem 4
Inside the circle $ S $, place the circle $ T $ and the circles $ K_1, K_2, \ldots, K_n $ that are externally tangent to $ T $ and internally tangent to $ S $, with the circle $ K_1 $ tangent to $ K_2 $, $ K_2 $ tangent to $ K_3 $, and so on. Prove that the points of tangency between the circles $ K_1 $ and $ K_2 $, $ K_2 $ and $ K_3 $, and so on, lie on a circle. | Let circles $S$ and $T$ be concentric. We will show that the points of tangency $K_1$ with $K_2$, $K_2$ with $K_3$, etc., are equidistant from the common center $O$ of circles $S$ and $T$.
Let for $i = 1, 2, \ldots, A_i$ be the point of tangency of circles $K_i$ and $T$, $B_i$ - the point of tangency of circles $K_i$ and $K_{i+1}$, $C_i$ - the point of tangency of circles $K_i$ and $S$, $O_i$ - the center of circle $K_i$ (Fig. 14). Let $r$ and $R$ be the lengths of the radii of circles $T$ and $S$ respectively. Then the length of the radius of circle $K_i$ is equal to $\frac{1}{2} A_iC_i = \frac{1}{2}(R-r)$, and thus circles $K_1, K_2, \ldots$ have radii of equal length. Point $B_i$ is then the midpoint of segment $\overline{O_iO_{i+1}}$, and angle $OB_iO_i$ is a right angle. By the Pythagorean theorem, we then have
From this, it follows that $OB_1 = OB_2 = \ldots = \sqrt{Rr}$, and thus points $B_1, B_2, \ldots$ are equidistant from point $O$.
Suppose now that circles $S$ and $T$ are not concentric. Then, by the remark to the solution of problem 7, there exists a circle $K$ such that the inversion $\varphi$ with respect to $K$ maps circles $S$ and $T$ to concentric circles and the center $P$ of circle $K$ lies outside circle $S$.
Inversion $\varphi$ is a one-to-one transformation of the set of points of the plane without point $P$ and maps each circle not containing point $P$ to some circle not containing point $P$. The inverse transformation to inversion $\varphi$ is the same inversion, i.e., $\varphi(\varphi(A)) = A$ for any point $A$ different from $P$.
Circles $S$ and $T$ are then concentric, and circles $K_1, K_2, \ldots$ tangent to circles $S$ and $T$ are mapped by inversion $\varphi$ to circles $K_1$ tangent to circles $S$ and $T$. In this case, circle $T$ contains circle $S$, and point $P$ lies outside circle $T$. Moreover, circle $K_1$ is tangent to circle $K_2$, circle $K_2$ is tangent to $K_3$, etc. Since circles $S$ and $T$ are concentric, it follows from the case considered at the beginning that the points of tangency of $K_1$ with $K_2$, $K_2$ with $K_3$, etc., belong to some circle $U$ lying inside $T$. Therefore, point $P$ does not belong to $L$, and thus figure $\varphi(L)$ is a circle.
By performing inversion $\varphi$ again, we conclude that the points of tangency of circles $\varphi(K_1)$ with $\varphi(K_2)$, $\varphi(K_2)$ with $\varphi(K_3)$, etc., belong to circle $\varphi(L)$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 658 |
LI OM - II - Problem 4
Point $ I $ is the center of the circle inscribed in triangle $ ABC $, where $ AB \neq AC $. Lines $ BI $ and $ CI $ intersect sides $ AC $ and $ AB $ at points $ D $ and $ E $, respectively. Determine all possible measures of angle $ BAC $ for which the equality $ DI = EI $ can hold. | We will show that the only value taken by angle $ BAC $ is $ 60^\circ $.
By the Law of Sines applied to triangles $ ADI $ and $ AEI $, we obtain $ \sin \measuredangle AEI = \sin \measuredangle ADI $. Hence,
om51_2r_img_6.jpg
First, suppose that the equality $ \measuredangle AEI = \measuredangle ADI $ holds (Fig. 1). Then also $ \measuredangle AIE = \measuredangle AID $, which means that triangles $ AEI $ and $ ADI $ are congruent (angle-side-angle criterion). Therefore, $ AD = AE $. This proves that triangles $ ADB $ and $ AEC $ are also congruent (angle-side-angle criterion). Hence, we obtain $ AB = AC $, which contradicts the assumptions made in the problem statement.
om51_2r_img_7.jpg
The remaining case to consider is when $ \measuredangle AEI + \measuredangle ADI = 180^\circ $ (Fig. 2). Then points $ A $, $ E $, $ I $, $ D $ lie on a single circle. Therefore,
From this, we obtain
which means $ \measuredangle BAC = 60^\circ $.
To complete the solution, it remains to show that there exists a triangle $ ABC $ in which $ AB \neq AC $, $ \measuredangle BAC = 60^\circ $, and $ DI = EI $. We will show more: in any triangle $ ABC $ with $ \measuredangle BAC = 60^\circ $, the equality $ DI = EI $ holds.
If $ \measuredangle BAC = 60^\circ $, then
Therefore, a circle can be circumscribed around quadrilateral $ AEID $. Since $ AI $ is the angle bisector of $ \angle EAD $, then $ DI = EI $. | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 659 |
XXXIV OM - II - Problem 6
For a given number $ n $, let $ p_n $ denote the probability that when a pair of integers $ k, m $ satisfying the conditions $ 0 \leq k \leq m \leq 2^n $ is chosen at random (each pair is equally likely), the number $ \binom{m}{k} $ is even. Calculate $ \lim_{n\to \infty} p_n $. | om34_2r_img_10.jpg
The diagram in Figure 10 shows Pascal's triangle written modulo $2$, meaning it has zeros and ones in the places where the usual Pascal's triangle has even and odd numbers, respectively. Just like in the usual Pascal's triangle, each element here is the sum of the elements directly above it, according to the addition table modulo 2: $0+0=0$, $0+1=1$, $1+0=1$, $1+1 = 0$. The drawn horizontal lines divide this triangle into layers:
- layer $W_0$ consisting of row number $0$,
- layer $W_1$ consisting of row number $1$,
- layer $W_2$ consisting of rows number $2$ and $3$,
- layer $W_3$ consisting of rows number $4$, $5$, $6$, $7$,
- $\ldots$
- layer $W_j$ consisting of rows with numbers from $2^{j-1}$ to $2^j-1$
- $\ldots$
We will prove by induction that in any layer $W_j$ there are two disjoint triangles, each identical to the triangle formed by the layers $W_0, W_1, \ldots, W_{j-1}$, and that these two triangles are separated by an inverted triangle composed entirely of zeros; the top row of the layer consists of extreme ones and zeros, and the bottom row of the layer consists of all ones.
For $j = 1, 2, 3$, this is visible from the diagrams (the diagonal lines define the division into the triangles mentioned). Assume the above relationships are true for some $j$ and consider the layer $W_{j+1}$. By the induction hypothesis, the last row of layer $W_j$ has the form $111\ldots 111$, and thus, according to the addition rule given above, the first row of layer $W_{j+1}$ has the form $100\ldots 001$. The extreme ones of this row give rise to two new copies of Pascal's triangle, and between them are zeros. Layer $W_{j+1}$ has as many rows as the layers $W_0, W_1, \ldots, W_j$ taken together, so at the base of each of these two triangles at the bottom of layer $W_{j+1}$ will be the same pattern of digits as in the last row of layer $W_j$, i.e., a row of ones. They will fill the last row of layer $W_{j+1}$. This completes the proof of the inductive thesis.
The number $p_n$ defined in the problem is the ratio of the number of zeros in the rows numbered from $0$ to $2^n$ to the number of all elements in these rows. These rows cover the layers $W_0, W_1, \ldots, W_n$ and the first row of layer $W_{n+1}$. From the relationship proved above, it follows that in the sum of layers $W_0, \ldots, W_j$ there are three times as many ones as in the sum of layers $W_0, \ldots, W_{j-1}\ (j = 1, 2, 3, \ldots)$, and since there is one one in layer $W_0$, there are $3^n$ ones in the sum of layers $W_0, \ldots, W_n$. Adding to this the two extreme ones of the next row, we see that there are $a_n = 3^n + 2$ ones in the considered fragment of Pascal's triangle. The total number of elements in this fragment is $b_n = 1 + 2 + 3 + \ldots + (2^n + 1) = (2^n + 1)(2^n + 2) / 2$, and thus there are $b_n - a_n$ zeros. The probability in question is
\[
\frac{b_n - a_n}{b_n} = \frac{(2^n + 1)(2^n + 2) / 2 - (3^n + 2)}{(2^n + 1)(2^n + 2) / 2}
\]
Hence,
\[
\lim_{n \to \infty} p_n = 1
\] | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 663 |
L OM - I - Problem 8
Given a natural number $ n \geq 2 $ and an $ n $-element set $ S $. Determine the smallest natural number $ k $ for which there exist subsets $ A_1, A_2, \ldots, A_k $ of the set $ S $ with the following property:
for any two distinct elements $ a, b \in S $, there exists a number $ j \in \{1, 2, \ldots, k\} $ such that the set $ A_j \cap \{a, b\} $ is a singleton. | We will show that $ k = [\log_2(n-1) +1] $, i.e., $ 2^{k-1} < n \leq 2^k $.
Given the number $ k $ defined as above, we construct sets $ A_j $ as follows: we number the elements of set $ S $ with $ k $-digit binary numbers (leading zeros are allowed). We thus have $ 2^k \geq n $ numbers available. Then, we take $ A_i $ to be the set of elements of $ S $ whose number has a 1 in the $ i $-th position. Any two different elements $ a $ and $ b $ of set $ S $ are then assigned different numbers, which differ, say, in the $ j $-th position. In this case, exactly one of the elements $ a $, $ b $ belongs to the set $ A_j $.
The number $ k $ found in this way is the smallest. Suppose, for instance, that there exist sets $ A_1, A_2, \ldots, A_k $ satisfying the conditions of the problem for $ 2^k < n $. We assign to each element $ x $ of set $ S $ a sequence of $ k $ numbers $ (c_1, c_2, \ldots, c_k) $ according to the following rule:
Since $ 2^k < n $, there exist two different elements $ a, b \in S $ that are assigned the same sequence of numbers. This means that element $ a $ belongs to exactly the same sets among $ A_1, A_2, \ldots, A_k $ as element $ b $. | [\log_2(n-1)+1] | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 664 |
XLVIII OM - II - Problem 3
Given a set of $ n $ points ($ n \geq 2 $), no three of which are collinear. We color all segments with endpoints in this set such that any two segments sharing a common endpoint have different colors. Determine the smallest number of colors for which such a coloring exists. | Let's assume that the given points are vertices of some $n$-gon (if $n = 2$, of course, one color is enough). All segments of the same color determine disjoint pairs of vertices of the polygon. The maximum number of disjoint two-element subsets of an $n$-element set is $[n/2]$. Therefore, the number of colors is not less than
We will show that this number of colors is sufficient to meet the desired condition.
When $n$ is an odd number, we take the set of vertices of an $n$-sided regular polygon and paint the side and all diagonals parallel to it with the same color; segments not parallel are painted with different colors. We used $n$ colors.
When $n$ is an even number, we take the set of vertices of an $(n-1)$-sided regular polygon $V$ and one more point $P_0$. We color the sides and diagonals of the polygon $V$ in the manner described above, using $n - 1$ colors. For each family of parallel segments of the same color $k_i$, there remains one vertex $P_i$ of the polygon $V$ that is not an endpoint of any of these segments. We paint the segment $P_0P_i$ with color $k_i$. We used $n - 1$ colors, meeting the required condition. | [n/2] | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 666 |
XXII OM - II - Problem 3
Given 6 lines in space, of which no 3 are parallel, no 3 pass through the same point, and no 3 are contained in the same plane. Prove that among these 6 lines, there are 3 lines that are mutually skew. | From the conditions of the problem, it follows that in every triple of the considered lines, there is a pair of skew lines. Let us associate the given lines with the vertices of a convex hexagon marked with the numbers $1, 2, \ldots, 6$ (Fig. 11). If two lines are skew, let us connect the corresponding vertices of the hexagon with a solid line, and if they are not skew, with a dashed line.
Thus, we have reduced the problem to the following: Six points are given. Each pair of them is connected by a solid or a dashed line. Moreover, every triangle with vertices belonging to the given set of points has at least one side that is a solid line. Prove that there exists a triangle whose all sides are solid lines.
Suppose there is no triangle whose all sides are solid lines. Therefore, there is a pair of points connected by a dashed line. Let this be, for example, the pair $1, 2$. Since in each of the triangles $1, 2, 3$; $1, 2, 4$; $1, 2, 5$; $1, 2, 6$ there is a side drawn with a solid line and the side $1, 2$ is not one of them, each of the points $3, 4, 5, 6$ is connected by a solid line to at least one of the points $1, 2$.
If one of the points $1, 2$ were connected by a solid line to at least three of the points $3, 4, 5, 6$, for example, point $1$ with points $3, 4, 5$, then considering the triangles $1, 3, 4$; $1, 4, 5$; $1, 3, 5$, in which two sides are drawn with solid lines, we conclude that the sides $3, 4$; $4, 5$ and $3, 5$ are drawn with dashed lines (Fig. 11). In the triangle $3, 4, 5$, none of the sides would be drawn with a solid line. This contradicts the conditions of the problem.
Therefore, each of the points $1$ and $2$ is connected by a solid line to exactly two of the points $3, 4, 5, 6$. Without loss of generality, we can assume that point $1$ is connected by a solid line to points $3$ and $4$, and point $2$ is connected by a solid line to points $5$ and $6$. Meanwhile, point $1$ is connected by a dashed line to points $5$ and $6$, and point $2$ is connected by a dashed line to points $3$ and $4$ (Fig. 12). Considering the triangle $2, 5, 6$, we conclude that points $5$ and $6$ are connected by a dashed line. Then, however, the triangle $1, 5, 6$ has all sides drawn with dashed lines. This contradicts the conditions of the problem. The obtained contradiction proves that there exists a triangle whose all sides are drawn with solid lines. | proof | Geometry | proof | Yes | Yes | olympiads | false | 667 |
XII OM - I - Problem 2
Prove that if $ a_1 \leq a_2 \leq a_3 $ and $ b_1 \leq b_2 \leq b_3 $, then | Inequality (1) will be replaced by an equivalent inequality
This, in turn, by the inequality
According to the assumption $ a_1 - a_2 \leq 0 $, $ b_1 -b_2 \leq 0 $, therefore
Hence
Similarly
And
Adding the above 3 inequalities side by side, we obtain inequality (2) equivalent to the desired inequality (1).
Note 1. A more general theorem can be proven. If
then
Indeed, inequality (4) is equivalent to the inequality
which can be easily proven by induction. This inequality is obviously true when $ n = 1 $, since $ a_1b_1 \leq a_1b_1 $; it is therefore sufficient to show that if this inequality holds for $ n \leq k $, where $ k $ is any chosen natural number, then it also holds for $ n = k + 1 $. We assert this as follows:
Note 2. Let the numbers $ a_i $ and $ b_i $ ($ i = 1, 2, \ldots,n $) satisfy condition (3) and let $ i_1,i_2, \ldots , i_n $ be any permutation (i.e., arrangement) of the numbers $ 1, 2, \ldots, n $. Then
Inequality (6) seems almost obvious; we suggest that the reader justify it precisely using induction. From inequality (6), inequality (5) can be deduced without using induction:
Note 3. The principle of mathematical induction is usually formulated in the following form.
$ Z_1 $: Let $ T (n) $ denote a statement expressing some property of the natural number $ n $. If
[a)] $ T (1) $ is true,
[b)] the truth of $ T (n) $ implies the truth of $ T (n + 1) $,
then the statement $ T (n) $ is true for every natural number $ n $.
In Note 1, we used the following form of this principle:
$ Z_2 $: Let $ S (n) $ denote a statement expressing some property of the natural number $ n $. If
[$ \alpha $)] $ S (1) $ is true,
[$ \beta $)] the truth of $ S(1), S(2), \ldots, S (n) $ implies the truth of $ S (n + 1) $,
then the statement $ S (n) $ is true for every natural number $ n $.
Both forms $ Z_1 $ and $ Z_2 $ are equivalent. Suppose, for instance, that the principle $ Z_2 $ is valid and that the statement $ T (n) $ satisfies conditions a) and b). If the truth of $ T (n) $ implies the truth of $ T (n + 1) $, then the truth of all statements $ T (1), T (2), \ldots, T (n) $ even more so implies the truth of $ T (n + 1) $, the statement $ T (n) $ thus satisfies conditions $ \alpha $) and $ \beta $), and by the principle $ Z_2 $, the statement $ T (n) $ is true for every natural $ n $, so the principle $ Z_1 $ is valid.
Conversely, suppose that the principle $ Z_1 $ is valid and that some statement $ S (n) $ satisfies conditions $ \alpha $) and $ \beta $). Let $ T (n) $ denote the statement asserting that each of the statements $ S (1), S (2), \ldots, S (n) $ holds. If $ T (n) $ is true for some natural $ n $, then by $ \beta $), $ S (n + 1) $ is true, and thus $ T (n + 1) $ is also true. Therefore, conditions a) and b) are satisfied, and by the principle $ Z_1 $, the statement $ T (n) $ is true for every natural $ n $; this means that $ S (n) $ is true for every natural $ n $, which means that the principle $ Z_2 $ is valid. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 669 |
LX OM - II - Zadanie 3
Rozłączne okręgi $ o_1 $ i $ o_2 $ o środkach odpowiednio $ I_1 $ i $ I_2 $ są styczne do prostej
$ k $ odpowiednio w punktach $ A_1 $ i $ A_2 $ oraz leżą po tej samej jej stronie. Punkt $ C $ leży na odcinku
$ I_1I_2 $, przy czym $ \meqsuredangle A_1CA_2 =90^{\circ} $. Dla $ i=1,2 $ niech $ B_i $ będzie punktem różnym od
$ A_i $, w którym prosta $ A_iC $ przecina okrąg $ o_i $. Dowieść, że prosta $ B_1B_2 $ jest styczna do okręgów
$ o_1 $ i $ o_2 $.
|
Niech prosta styczna do okręgu $ o_1 $ w punkcie $ B_1 $ przecina prostą $ A_1A_2 $ w punkcie $ D $.
(Punkt przecięcia istnieje, gdyż punkt $ C $ nie leży na prostej $ A_1I_1 $, a więc odcinek $ A_1B_1 $
nie jest średnicą okręgu $ o_1 $.) Trójkąt $ A_1B_1D $ jest wówczas równoramienny, gdyż dwa jego boki
są odcinkami stycznych do okręgu $ o_1 $ poprowadzonymi z punktu $ D $. Wobec tego prosta
$ DI_1 $ jako dwusieczna kąta $ \measuredangle A_1DB_1 $ jest prostopadła do cięciwy $ A_1B_1 $,
a więc równoległa do prostej $ CA_2 $.
Udowodnimy, że proste $ A_1B_1 $ i $ DI_2 $ są równoległe.
Jeżeli $ A_1A_2 || I_1I_2 $, to $ A_1A_2 = I_1I_2 $ i czworokąt $ I_1CA_2D $ jest równoległobokiem.
W takim razie $ CI_2 = A_1D $ i czworokąt $ CI_2DA_1 $ także jest równoległobokiem. W szczególności
więc $ A_1B_1 || DI_2 $.
Jeżeli natomiast proste $ A_1A_2 $ i $ I_1I_2 $ przecinają się w punkcie $ E $, to stosując dwukrotnie
twierdzenie Talesa do kąta $ \measuredangle A_1EI_1 $ przeciętego prostymi równoległymi $ I_1D $, $ CA_2 $
oraz $ I_1A_1 $, $ I_2A_2 $ otrzymujemy
co w połączeniu z twierdzeniem odwrotnym do twierdzenia Talesa znowu daje zależność $ A_1B_1 || DI_2 $.
Wykazana właśnie równoległość wraz z danym w treści zadania warunkiem $ \measuredangle A_1CA_2 = 90^{\circ} $
pozwala wnioskować, że proste $ DI_2 $ i $ B_2A_2 $ są prostopadłe. Zauważmy teraz, że prosta $ DI_2 $ przechodzi
przez środek okręgu $ o_2 $. W efekcie punkty $ B_2 $ i $ A_2 $ są symetryczne względem prostej $ DI_2 $
(gdyż są końcami cięciwy prostopadłej do tej prostej) oraz proste $ DB_2 $ i $ DA_2 $ są symetryczne
względem tej prostej. Skoro zaś prosta $ DA_2 $ jest styczna do okręgu $ o_2 $, to również prosta $ DB_2 $
jest doń styczna, co kończy rozwiązanie zadania.
| proof | Geometry | proof | Yes | Yes | olympiads | false | 670 |
LVII OM - III - Problem 6
Determine all pairs of integers $ a $, $ b $, for which there exists a polynomial $ P(x) $ with integer coefficients, such that the product $ (x^2 + ax + b)\cdot P(x) $ is a polynomial of the form
where each of the numbers $ c_0,c_1,\dots ,c_{n-1} $ is equal to 1 or -1. | If the quadratic polynomial $x^2 + ax + b$ is a divisor of some polynomial $Q(x)$ with integer coefficients, then of course the constant term $b$ is a divisor of the constant term of the polynomial $Q(x)$. Therefore, in the considered situation, $b=\pm 1$.
No polynomial $Q(x)$ of the form $x^n \pm x^{n-1} \pm x^{n-2} \pm \ldots \pm x \pm 1$ has real roots with an absolute value greater than or equal to 2. Suppose, for instance, that a number $r$ with $|r| \geq 2$ satisfies the equation $Q(r) = 0$. Then
skipping some steps, we conclude that $|r| < 2$. This is a contradiction.
On the other hand, every root of the quadratic polynomial $T(x) = x^2 + ax + b$ is also a root of the polynomial $T(x) \cdot P(x)$.
Therefore, the quadratic polynomial $T(x)$ has no roots in the set $(-\infty, -2] \cup [2, +\infty)$.
Thus, $T(-2) > 0$ and $T(2) > 0$, which means $4 - 2a + b > 0$ and $4 + 2a + b > 0$.
If $b = 1$, then we have $5 > 2a$ and $5 > -2a$, which implies that $a$ is one of the numbers $-2, -1, 0, 1, 2$. If, however, $b = -1$, then $3 > 2a$ and $3 > -2a$, and the cases to consider are when $a$ is one of the numbers $-1, 0, 1$.
Direct verification shows that all the obtained pairs $(a, b)$ satisfy the conditions of the problem:
Answer: The conditions of the problem are satisfied by the following eight pairs $(a, b)$: (-2,1), (-1,-1), (-1,1), (0,-1), (0,1), (1,-1), (1,1), (2,1). | (-2,1),(-1,-1),(-1,1),(0,-1),(0,1),(1,-1),(1,1),(2,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 672 |
II OM - III - Problem 5
In a circle, a quadrilateral $ABCD$ is inscribed. Lines $AB$ and $CD$ intersect at point $E$, and lines $AD$ and $BC$ intersect at point $F$. The angle bisector of $\angle AEC$ intersects side $BC$ at point $M$ and side $AD$ at point $N$; and the angle bisector of $\angle BFD$ intersects side $AB$ at point $P$ and side $CD$ at point $Q$. Prove that quadrilateral $MPNQ$ is a rhombus. | Ignoring for now the condition that points $A$, $B$, $C$, $D$ lie on a circle, consider any convex quadrilateral $ABCD$ where the extensions of sides $AB$ and $CD$ intersect at point $E$, and the extensions of sides $AD$ and $BC$ intersect at point $F$.
Draw the angle bisectors $EO$ and $FO$ of angles $E$ and $F$, and the segment $EF$ as indicated in Figure 66, and consider triangles $EAF$, $ECF$, $EOF$ with the common base $EF$.
Each of the angles at the base $EF$ in triangle $EOF$ is the arithmetic mean of the angles of triangles $EAF$ and $ECF$ at the same vertex; it follows that the third angle $x$ of triangle $EOF$ is also the arithmetic mean of the remaining angles $\alpha$ and $\gamma$ of triangles $EAF$ and $ECF$:
We can reach the same conclusion in the following way:
From any point $M$ (Figure 67), draw rays $a_1$, $b_1$, $c_1$, $d_1$, $e_1$, $f_1$ having the directions of rays $a$, $b$, $c$, $d$, $e$, $f$ according to the labels in Figure 66.
Since $e_1$ is the bisector of the angle between $a_1$ and $c_1$, and $f_1$ is the bisector of the angle between $d_1$ and $b_1$, the angle between rays $e_1$ and $f_1$ is the arithmetic mean of the angle between rays $a_1$ and $d_1$ and the angle between rays $c_1$ and $b_1$, which we can write as:
However, $\measuredangle (e_1, f_1) = \measuredangle (e, f)$, $\measuredangle (a_1, d_1) = \measuredangle (a, d)$, $\measuredangle (c_1, b_1) = \measuredangle (c, b)$, thus
We have obtained, as is easily verified, the same equality as before.
Now assume that the quadrilateral $ABCD$ is inscribed in a circle (Figure 68).
In this case, $\alpha + \gamma = 180^\circ$ and the previous equality gives
This means that the diagonals of quadrilateral $MPNQ$ are perpendicular.
Notice that in this case, in triangle $PEQ$, the angle bisector $EO$ of angle $E$ is perpendicular to side $PQ$, so triangle $PEQ$ is isosceles, and point $O$ is the midpoint of segment $PQ$. Similarly, point $O$ is the midpoint of segment $MN$.
Quadrilateral $MPNQ$, whose diagonals bisect each other and are perpendicular, is therefore a rhombus, which is what we needed to prove. | proof | Geometry | proof | Yes | Yes | olympiads | false | 674 |
XXX OM - II - Task 1
Given are points $ A $ and $ B $ on the edge of a circular pool. An athlete needs to get from point $ A $ to point $ B $ by walking along the edge of the pool or swimming in the pool; he can change his mode of movement multiple times. How should the athlete move to get from point $ A $ to $ B $ in the shortest time, given that he moves in the water twice as slowly as on land? | Let the radius of the pool be $r$, and the arc $\widehat{AB}$ corresponds to the central angle $\alpha$, where $0 < \alpha \leq \pi$. Then the length of the arc $\widehat{AB}$ is $r\alpha$, and the length of the segment $\overline{AB}$ is (Fig. 10)
om30_2r_img_10.jpg
We will prove that $r\alpha < 2AB$. It will follow that a sportsman will reach point $B$ from point $A$ faster by moving along the arc $\widehat{AB}$ than by moving along the segment $AB$. Therefore, he should walk along the edge of the pool, not swim in the pool.
By (1), the inequality $r\alpha < 2AB$ that we need to prove is equivalent to the following
From the known formula $\beta < \tan \beta$ for $0 < \beta < \frac{\pi}{2}$, it follows in particular that
Since $0 < \tan \frac{\alpha}{4} \leq 1$, using the half-angle formula, we get
by (3). It follows that inequality (2) holds.
om30_2r_img_11.jpg
Note. Let us divide the given angle $\alpha$ into four equal parts by the lines $OC$, $OD$, $OE$, and let $AC \bot OA$, $CE \bot OD$, $BE \bot OB$ (Fig. 11). Then $AC = CD = DE = EB = r \tan \frac{\alpha}{4}$. The length of the broken line $ACDEB$ is $4r \tan \frac{\alpha}{4}$. It does not exceed, by (4), twice the length of the segment $\overline{AB}$, which is $4r \sin \frac{\alpha}{2}$. Therefore, even by walking along this broken line, the sportsman will reach point $B$ from point $A$ no later than if he swam in the pool. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 676 |
XXVIII - I - Task 3
Let $ a $ and $ b $ be natural numbers. A rectangle with sides of length $ a $ and $ b $ has been divided by lines parallel to the sides into unit squares. Through the interiors of how many squares does the diagonal of the rectangle pass? | Consider a rectangle $ABCD$ with base $\overline{AB}$ of length $a$ and height $\overline{BC}$ of length $b$. Let's consider its diagonal $\overline{AC}$ connecting the lower left vertex $A$ with the upper right vertex $C$ (Fig. 4).
First, assume that the numbers $a$ and $b$ are relatively prime. If a vertex $E$ of a unit square contained within the rectangle $ABCD$ lies on $\overline{AC}$, then we have $\displaystyle \frac{a}{b} = \frac{AB}{BC} = \frac{AF}{EF}$, which means $a \cdot EF = b \cdot AF$, where $F$ is the projection of point $E$ onto line $AB$. Since $a$ and $b$ are relatively prime and $EF$ and $AF$ are natural numbers, it follows from the last equation that $a$ is a divisor of $AF$. This is impossible because $AF < AB = a$.
The obtained contradiction proves that the diagonal $\overline{AC}$ does not pass through any vertex of a unit square contained within the rectangle $ABCD$. Therefore, it defines a sequence of unit squares whose interiors it consecutively intersects. Each subsequent square in the sequence shares a side with the previous one and is located to the right or above it. By following the path defined by the squares of this sequence, we perform $a - 1$ moves to the right and $b - 1$ moves upwards. In total, we have $a + b - 2$ moves, meaning the path consists of $a + b - 1$ squares, i.e., the diagonal $\overline{AC}$ passes through the interiors of $a + b - 1$ unit squares.
If, however, the numbers $a$ and $b$ are not relatively prime and their greatest common divisor is $d$, then let $a = da'$ and $b = db'$. In this case, the numbers $a'$ and $b'$ are relatively prime. The diagonal of a rectangle with sides of length $a'$ and $b'$ consists of $d$ equal segments, each of which is the diagonal of a rectangle with sides of length $a'$ and $b'$. Based on the already considered case, each such segment of the diagonal passes through the interiors of $a' + b' - 1$ unit squares, and therefore the entire diagonal passes through the interiors of $d(a' + b' - 1)$ unit squares. | \gcd(,b) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 679 |
XLVI OM - II - Problem 5
The circles inscribed in the faces $ABC$ and $ABD$ of the tetrahedron $ABCD$ are tangent to the edge $AB$ at the same point. Prove that the points of tangency of these circles with the edges $AC$, $BC$ and $AD$, $BD$ lie on a single circle. | Let's assume that the circle inscribed in the face $ABC$ is tangent to the edges $AC$, $BC$, and $AB$ at points $P$, $Q$, and $T$, respectively, and the circle inscribed in the face $ABD$ is tangent to the edges $AD$, $BD$, and $AB$ at points $R$, $S$, and $T$ (the same point $T$, according to the assumption). We have the following equalities of tangent segments:
If point $T$ is the midpoint of edge $AB$, then triangles $ABC$ and $ABD$ are isosceles, and lines $PQ$ and $RS$ are parallel to line $AB$; points $P$ and $Q$ are symmetric with respect to the plane passing through $T$ and perpendicular to $AB$, and points $R$ and $S$ are also symmetric with respect to this plane, so segments $PQ$ and $RS$ are the bases of an isosceles trapezoid on which a circle can be circumscribed.
We further assume that point $T$ is not the midpoint of edge $AB$; without loss of generality, let's assume that $|AT| > |BT|$. Neither of the lines $PQ$ and $RS$ is parallel to $AB$. Points $A$, $B$, $P$, and $Q$ lie in the plane of face $ABC$, so line $PQ$ intersects line $AB$; let's denote the intersection point by $U$. Similarly, points $A$, $B$, $R$, and $S$ lie in the plane of face $ABD$, so line $RS$ intersects line $AB$; let's denote the intersection point by $V$.
Points $U$ and $V$ lie on the ray $AB^\to$, beyond segment $AB$; therefore,
(Figure 10 shows the arrangement of the considered points and lines separately on the plane $ABC$ and on the plane $ABD$.)
By the theorem of tangent and secant segments (applied to the circles inscribed in triangles $ABC$ and $ABD$), we have the equalities:
Now let's apply Menelaus' theorem to triangle $ABC$ intersected by line $PQ$ and to triangle $ABD$ intersected by line $RS$:
From relations (2) and (1), we get the proportions
Replacing $|AU|$ and $|AV|$ with the right-hand sides of equations (3), we obtain equations from which we calculate:
Thus, $|BU| = |BV|$. Therefore, point $U$ coincides with $V$; it is the intersection point of lines $PQ$ and $RS$.
A pair of intersecting lines defines a plane. Thus, points $P$, $Q$, $R$, and $S$ lie in the same plane. Since points $U$ and $V$ coincide, from relations (4) we obtain the equality $|UP| \cdot |UQ| = |UR| \cdot |US|$. It follows that points $P$, $Q$, $R$, and $S$ lie on the same circle.
Note 1. Menelaus' theorem was used only to derive equality (5); however, it is not necessary for this. We encourage readers to find other justifications for formulas (5).
Note 2. Once we conclude that points $P$, $Q$, $R$, and $S$ are coplanar, we can reason as follows: through the center of circle $PQT$, we draw a line perpendicular to the plane $ABC$, and through the center of circle $RST$, we draw a line perpendicular to the plane $ABD$. These lines lie in the same plane (perpendicular to $AB$ and passing through point $T$), and they are not parallel, so they intersect. From equality (1), it easily follows that the intersection point is equidistant from points $P$, $Q$, $T$, $R$, and $S$ - it is, therefore, the center of a sphere on which these points lie. The intersection of this sphere with the plane $PQRS$ gives the desired circle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 682 |
LX OM - II - Task 1
Real numbers $ a_1, a_2, \cdots , a_n $ $ (n \geqslant 2) $ satisfy the condition
$ a_1 \geqslant a_2 \geqslant \cdots \geqslant a_n > 0 $. Prove the inequality | We will apply mathematical induction with respect to the value of the number $ n $.
For $ n = 2 $, the thesis of the problem is true, as both sides of the inequality are equal to $ 2a_2 $.
Moving on to the inductive step, consider the real numbers
$ a_1, a_2, \cdots, a_{n+1} $, for which $ a_1 \geqslant a_2 \geqslant \cdots \geqslant a_{n+1} > 0 $.
We need to prove the inequality
Let us introduce the notation
Then the left side of the inequality (1) can be written in the form
Applying the inductive hypothesis to $ n $ numbers $ a_2 \geqslant a_3 \geqslant \cdots \geqslant a_{n+1} \geqslant 0 $,
we obtain
Furthermore, for any real numbers $ x \geqslant y > 0 $, the double inequality
$ -x \leqslant -x + 2(x - y) = x - 2y < x $ holds, which implies $ |x - 2y| \leqslant x $, and consequently,
This means that the number $ K - L $ is non-negative. The number $ t = a_1 - a_2 $ is also non-negative, so
continuing the inequality (2) and using (3), we get
We have obtained the right side of the inequality (1), which completes the inductive proof. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 683 |
XVIII OM - II - Task 6
Prove that points $ A_1, A_2, \ldots, A_n $ ($ n \geq 7 $) lying on the surface of a sphere lie on a circle if and only if the tangent planes to the sphere at these points have a common point or are parallel to a single line. | a) Suppose that points $ A_1, A_2, \ldots, A_n $ on the surface of a sphere with center $ O $ lie on a circle $ k $. Let $ \alpha_1 $ denote the tangent plane to the sphere at point $ A_i $ ($ i= 1,2, \ldots, n $), and $ \pi $ - the plane of the circle $ k $.
If the center of the circle $ k $ is point $ O $, then the plane $ \alpha_i $ is perpendicular to the radius $ OA_i $, so it is perpendicular to the plane $ \pi $ containing the line $ OA_i $. Therefore, the plane $ \alpha_i $ is parallel to any line perpendicular to $ \pi $.
If, however, the center of the circle $ k $ is a point $ S $ different from $ O $ (Fig. 10), then the plane $ \pi $ is perpendicular to the line $ OS $, so in the triangle $ OSA_i $ the angle $ S $ is a right angle, and the angle $ O $ is acute, which means that the line $ OS $ intersects the plane $ \alpha_i $ at some point $ T_i $. In the triangle $ OA_iT_i $ the angle $ A_i $ is a right angle, hence the equality $ OA_i^2 = OT_i \cdot OS $ holds, therefore
The segment $ OA_i $ equals the radius of the sphere, so the segment $ OT_i $ has the same length for every $ i $, and since all points $ T_i $ lie on the same ray $ OS $, they coincide.
b) Suppose that all planes $ \alpha_i $ are parallel to a line $ l $ or all have a common point $ T $. In the first case, all segments $ OA_i $ are perpendicular to the line $ l $, so they lie in the same plane $ \pi $, which means that all points $ A_i $ lie on a circle $ k $, which is the intersection of the plane $ \pi $ with the surface of the given sphere. In the second case, the line $ TA_i $ is for each $ i $ tangent to the surface of the sphere, so it is perpendicular to the radius $ OA_i $. Let $ A_iS_i $ be the altitude of the right triangle $ OA_iT $ to the hypotenuse $ OT $; then $ OA_i^2 = OT \cdot OS_i $, hence
it follows that all segments $ OS_i $ are equal. Since all points $ S_i $ lie on the segment $ OT $, they coincide. Therefore, all segments $ A_iS_i $ lie in the same plane $ \pi $, which means that all points $ A_i $ lie on the intersection of the plane $ \pi $ with the surface of the sphere, i.e., on some circle $ k $. | proof | Geometry | proof | Yes | Yes | olympiads | false | 684 |
XVI OM - I - Problem 4
The school organized three trips for its 300 students. The same number of students participated in each trip. Each student went on at least one trip, but half of the participants in the first trip, one-third of the participants in the second trip, and one-fourth of the participants in the third trip only went on one trip.
How many students went on each trip? How many participants in the first trip also participated in the second, and how many of them also participated in the third trip? | Let $ x $ denote the number of participants in each trip, and $ y $, $ z $, $ u $, $ w $ denote, respectively, the numbers of students who went: a) only on the first and second trip, b) only on the first and third trip, c) only on the second and third trip, d) on all three trips.
The total number of students equals the sum of: 1) the number of participants in only one trip, i.e., the number $ \frac{x}{2} + \frac{x}{3} + \frac{x}{4} = \frac{13}{12}x $, 2) the number of participants in only two trips, and 3) the number of participants in all three trips, thus
The number of participants in the first trip is the sum of the numbers: 1) students who went only on the first trip, 2) students who went on the first and second or first and third trip, and 3) students who went on all three trips, thus
Similarly, for the number of participants in the second and third trips, we obtain the equations
We need to find non-negative integer solutions to the system of equations (1) - (4). Suppose such a solution is the set of numbers $ x $, $ y $, $ z $, $ u $, $ w $. From equation (1), it follows that $ \frac{13}{12} x $ is an integer, so $ x $ is divisible by $ 12 $. Let $ x = 12t $; equations (1) - (4) take the form
Adding equations (5) and (8) side by side, we get the equation
Since $ y \geq 0 $, it follows from (9) that $ t \leq \frac{300}{22}=13.6\ldots $, thus
Adding (6) and (7) side by side and subtracting (8), we get
Since $ w \geq 0 $, it follows from equation (11) that $ y \leq \frac{5}{2} t $, from which, using equation (9), we conclude that $ 22t + \frac{5}{2}t \geq 300 $, thus
$ t \geq \frac{600}{49} = 12.2\ldots, $ hence
From (10) and (11), it follows that
From equations (9), (11), (6), and (7), we obtain sequentially $ y = 14 $, $ w = 37 $, $ z = 27 $, $ u = 53 $. It is easy to verify that the found numbers satisfy the system of equations (1) - (4), thus they constitute the only solution to the problem. | x=120,y=14,z=27,u=53,w=37 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 685 |
XI OM - I - Problem 11
Prove that if a quadrilateral is circumscribed around a circle, then a circle can be circumscribed around it if and only if the chords connecting the points of tangency of the opposite sides of the quadrilateral with the circle are perpendicular. | Let $ABCD$ be a quadrilateral inscribed in a certain circle, and let $M$, $N$, $P$, $Q$ denote the points of tangency of the sides $AB$, $BC$, $CD$, $DA$ with the inscribed circle, and let $S$ be the point of intersection of segments $MP$ and $NQ$ (Fig. 16).
Since tangents to a circle drawn from the ends of a chord form equal angles with that chord, if we denote $\measuredangle AMS = \delta$, $\measuredangle AQS = \varepsilon$, then $\measuredangle DPS = \delta$, $\measuredangle BNS = \varepsilon$, hence $\measuredangle CPS = 180^\circ - \delta$, $\measuredangle CNS = 180^\circ - \varepsilon$. Let $\measuredangle MSQ = \varphi$.
In quadrilateral $AMSQ$
and in quadrilateral $CNSP$
Adding these equalities, we get
A circle can be circumscribed around quadrilateral $ABCD$ if and only if the sums of the opposite angles are each $180^\circ$. From the last equality, it follows that $\measuredangle A + \measuredangle C = 180^\circ$ if and only if $\varphi = 90^\circ$, i.e., when the chords $MP$ and $NQ$ are perpendicular. Q.E.D. | proof | Geometry | proof | Yes | Yes | olympiads | false | 686 |
LIV OM - III - Task 3
Determine all polynomials $ W $ with integer coefficients that satisfy the following condition: for every natural number $ n $, the number $ 2^n-1 $ is divisible by $ W(n) $. | Let $ W(x) = a_0 + a_1x +\ldots + a_rx^r $ be a polynomial with integer coefficients, satisfying the given condition. Suppose that for some natural number $ k $, the value $ W(k) $ is different from $ 1 $ and $ -1 $; it therefore has a prime divisor $ p $. Let $ m = k + p $. The difference $ W(m) - W(k) $ is the sum of differences of the form $ a_j(m^j - k^j) $, which are numbers divisible by $ m - k $, i.e., by $ p $. Therefore, $ W(m) $ is divisible by $ p $.
From the condition of the problem, it follows that $ p $ is a divisor of the numbers $ 2^k - 1 $ and $ 2^m - 1 $; it is therefore an odd prime divisor of the number $ 2^m - 2^k = 2^k (2^p - 1) $, i.e., it is a divisor of the number $ 2^p - 1 $. This, however, is not possible, because
the values of the binomial coefficient $ {p \choose i} $ are for $ i = 1,\ldots,p- 1 $ numbers divisible by $ p $.
The contradiction proves that $ W(n) = \pm 1 $ for every natural number $ n $. Hence, it follows that $ W $ is a polynomial identically equal to $ 1 $ or identically equal to $ -1 $. Of course, each of these two polynomials satisfies the postulated condition. | W(x)=1orW(x)=-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 687 |
XXXVII OM - II - Problem 5
Prove that if a polynomial $ f $ not identically zero satisfies for every real $ x $ the equation $ f(x)f(x + 3) = f(x^2 + x + 3) $, then it has no real roots. | Suppose a polynomial $ f $ satisfying the given conditions has real roots. Let $ x_0 $ be the largest of them. The number $ y_0 = x_0^2 + x_0 + 3 > x_0 $ is also a root of the polynomial $ f $, because
- contrary to the fact that $ x_0 $ is the largest root. The obtained contradiction proves that $ f $ cannot have real roots. | proof | Algebra | proof | Yes | Yes | olympiads | false | 689 |
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 | 690 |
XXXIV OM - I - Problem 5
Prove the existence of a number $ C_0 $ with the property that for any sequence $ x_1, x_2, \ldots, x_N $ of positive numbers and for any positive number $ K $, if the number of terms $ x_j $ not less than $ K $ is not greater than $ \frac{N}{K} $, then | Suppose the sequence $ x_1, x_2, \ldots, x_N $ satisfies the given condition. By rearranging the terms if necessary, we can assume that $ x_1 \geq x_2 \geq \ldots \geq x_N $. We will show that $ x_j \leq N/j $ for $ j = 1, \ldots, N $. Suppose this is not the case, i.e., for some index $ m $, the inequality $ x_m > N/m $ holds. Take any number $ K $ such that $ N/m < K < x_m $. According to the assumption, the number of terms $ x_j $ greater than $ K $ is no more than $ N/K $, and thus is less than $ m $. This leads to a contradiction because $ x_1, \ldots, x_m > K $. Therefore, we have $ x_j \leq N/j $ for $ j = 1, \ldots, N $. We can now estimate the left-hand side of the inequality given in the problem:
We will prove by induction that for any natural number $ N $, the inequality $ N^N/N! < e^N $ holds, where $ e $ is the limit of the increasing sequence $ \left( \left(1+\frac{1}{n} \right) \right) $. For $ N = 1 $, the last inequality is satisfied. Assume the inequality holds for some $ N $. Then
\[
\frac{(N+1)^{N+1}}{(N+1)!} = \frac{(N+1)^N}{N!} = \frac{(N+1)^N}{N^N} \cdot \frac{N^N}{N!} < \left(1 + \frac{1}{N}\right)^N \cdot e^N < e \cdot e^N = e^{N+1},
\]
which completes the inductive proof. Continuing the estimation of the quantity $ L(N) $, we get
\[
L(N) = \sum_{j=1}^N \log x_j \leq \sum_{j=1}^N \log \left( \frac{N}{j} \right) = \sum_{j=1}^N \left( \log N - \log j \right) = N \log N - \sum_{j=1}^N \log j.
\]
We can use the inequality $ \sum_{j=1}^N \log j \geq \int_1^N \log x \, dx = N \log N - N + 1 $, so
\[
L(N) \leq N \log N - (N \log N - N + 1) = N - 1.
\]
Thus, we can take the constant $ C_0 $ to be $ \log e = 1 $.
Note. The problem statement did not specify whether the symbol $ \log $ denotes the common logarithm (base 10) or the natural logarithm (base $ e $). This does not matter. The theorem is valid for logarithms of any base greater than 1. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 692 |
XXXVI OM - III - Problem 2
Given a square with side length 1 and positive numbers $ a_1, b_1, a_2, b_2, \ldots, a_n, b_n $ not greater than 1 such that $ a_1b_1 + a_2b_2 + \ldots + a_nb_n > 100 $. Prove that the square can be covered by rectangles $ P_i $ ($ i = 1,2,\ldots,n $) with side lengths $ a_i, b_i $ parallel to the sides of the square. | Let $ AB $ be the base of a square. We can assume that $ 1 \geq a_x \geq a_2 \geq \ldots \geq a_n $. On the ray $ AB^\to $, we sequentially lay down segments of lengths $ b_1, b_2, \ldots, b_k $ until we exceed point $ B $, and over each subsequent segment of length $ b_i $, we construct a rectangle whose other side has length $ a_i $. In this way, we obtain a layer of rectangles, the sum of whose areas $ S_1 $ satisfies the inequality
Next, we draw a line parallel to $ AB $ at a distance of $ a_k $ from it, intersecting the sides of the square at points $ A_1 $ and $ B_1 $, and on this line, we build the next layer of rectangles: we sequentially lay down segments of lengths $ b_{k+1}, b_{k+2}, \ldots, b_r $ until we exceed point $ B_1 $, and over each subsequent segment of length $ b_i $, we construct a rectangle whose other side has length $ a_i $. The sum $ S_2 $ of the areas of the rectangles in this layer satisfies the inequality
We continue this process until we exhaust all the numbers $ a_1b_1, \ldots, a_nb_n $. Suppose that as a result of this process, we do not cover the entire square. In that case,
On the other hand,
but
Thus,
contradicting inequality (1). Therefore, the method of arranging the rectangles described here leads to covering the given square.
Note. The number $ 100 $ given in the problem statement can be replaced by a smaller number. The solution presented here works when $ a_1b_1 + a_2b_2 + \ldots + a_nb_n \geq 6 $. | proof | Geometry | proof | Yes | Yes | olympiads | false | 693 |
XVII OM - I - Problem 8
In space, three lines are given. Find a line that intersects them at points $ A, B, C $ such that $ \frac{AB}{BC} $ equals a given positive number $ k $. | Let us first consider the case where the three given lines are pairwise skew, and at the same time do not lie in three mutually parallel planes.
Let the given lines be denoted by the letters $a$, $b$, $c$ and let $l$ be the line intersecting them at points $A$, $B$, $C$ respectively, such that $\frac{AB}{BC}=k$.
Let $\alpha$ and $\gamma$ be parallel planes passing through the lines $a$ and $c$ respectively; we know that such a pair of planes exists and is unique. Note that point $B$ is such a point on line $b$ that its distances from planes $\alpha$ and $\gamma$ are in the ratio $\frac{AB}{BC}=k$ and that $l$ is the line of intersection of planes passing through point $B$ and containing lines $a$ and $c$.
From the above analysis, we derive the following construction (Fig. 7):
Having marked the given lines in any way with the letters $a$, $b$, $c$, we draw parallel planes $\alpha$ and $\gamma$ passing through lines $a$ and $c$ respectively*). We then draw a plane $\beta$ parallel to $\alpha$ and $\gamma$, whose distances from planes $\alpha$ and $\gamma$ are in the ratio $k$.
Plane $\beta$ intersects line $b$ at some point $B$, since according to the assumption, line $b$ does not lie in a plane parallel to $\alpha$ and $\gamma$. Through point $B$ we draw planes $\alpha_1$ and $\gamma_1$ containing lines $a$ and $c$ respectively; they intersect according to a line, which we will denote by $l$. Line $l$ intersects lines $a$ and $c$ at some points $A$ and $C$ respectively; if it were parallel to one of them, it would lie in plane $\beta$, so it would also be parallel to the other, contrary to the assumption that $a$ and $c$ are skew. Since segments $AB$ and $BC$ are proportional to the distances of point $B$ from planes $\alpha$ and $\gamma$, $\frac{AB}{BC} = k$, i.e., the triplet of points $A$, $B$, $C$ is a solution to the problem.
The construction is always feasible. When $k \ne 1$, plane $\beta$ can have one of two positions: in the layer between planes $\alpha$ and $\gamma$ or outside this layer. For the chosen labeling of the given lines, there are thus two solutions. Since we have 6 ways to label the given lines with the letters $a$, $b$, $c$, the problem has 12 solutions. However, it should be noted that there are such values of the ratio $k$ for which some of these solutions are the same in the sense that they consist of the same points only differently named. It is easy to verify that this occurs in the case of the so-called golden section of a segment. Namely, when $k= \frac{\sqrt{5}-1}{2}$ and the triplet of points $A$, $B$, $C$ is a solution to the problem, then $\frac{AB}{BC}= \frac{BC}{AC}$ or $\frac{AB}{BC}= \frac{AC}{AB}$ depending on whether point $B$ or point $A$ lies between the other two. In the first case, from the solution $A$, $B$, $C$ we obtain another solution by renaming these points as $C$, $A$, $B$, in the second case by renaming them as $B$, $C$, $A$. Similarly, when $k=\frac{\sqrt{5}+1}{2}$, then from the solution $A$, $B$, $C$ we obtain another solution by changing the names of the points to $B$, $C$, $A$, when $B$ lies between $A$ and $C$, and to $C$, $A$, $B$, when $C$ lies between $A$ and $B$. In the first case, $\frac{AB}{BC}= \frac{AC}{AB}$, in the second $\frac{AB}{BC}= \frac{BC}{AC}$. In each of these two cases, among the 12 solutions, there are only 6 different triplets of points.
When $k=1$, there are 6 solutions, but they represent only 3 different triplets of points. Namely, from each solution $A$, $B$, $C$ another solution is obtained by renaming the points as $C$, $B$, $A$.
For other cases of the mutual position of the given lines, the solution to the problem is very simple; we will limit ourselves to brief explanations, proposing to the reader a detailed justification of the given theorems.
When the given lines are pairwise skew and lie in three parallel planes, the problem has a solution only if the distances of one of these planes from the other two are in the ratio $k$; in this case, there are infinitely many solutions.
When two given lines - let us call them $m$ and $n$ - lie in one plane, and the third line $p$ is skew to each of them, a solution can exist only if $p$ intersects the plane of lines $m$ and $n$ at some point $P$. If lines $m$ and $n$ intersect and $k \ne 1$, there are 12 solutions, of which for $k = \frac{\sqrt{5}+1}{2}$ and for $k = \frac{\sqrt{5}-1}{2}$, these solutions form only 6 different triplets of points; if $k=1$, there are 6 solutions, of which only three are different triplets of points. If lines $m$ and $n$ are parallel, solutions exist only if the ratio of two of the numbers $\lambda$, $\mu$, $\nu$, where $\lambda$ is the distance between lines $m$ and $n$, and $\mu$ and $\nu$ are the distances of point $P$ from $m$ and $n$, equals the number $k$; if such a case occurs, there are infinitely many solutions.
When a pair of given lines lies in one plane and another pair in another plane, there are no solutions.
When all three lines lie in one plane but are not pairwise parallel, there are infinitely many solutions.
Finally, when the given lines lie in one plane and are pairwise parallel, a solution exists only if the distances of one of these lines from the other two are in the ratio $k$; if this case occurs, there are infinitely many solutions. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 695 |
XLV OM - I - Problem 8
Given natural numbers $ a $, $ b $, $ c $, such that $ a^3 $ is divisible by $ b $, $ b^3 $ is divisible by $ c $, and $ c^3 $ is divisible by $ a $. Prove that the number $ (a+b+c)^{13} $ is divisible by $ abc $. | The number $ (a + b + c)^{13} $ is the product of thirteen factors, each equal to $ a + b + c $. Multiplying these thirteen identical expressions, we obtain a sum of many (specifically $ 3^{13} $) terms of the form
It suffices to show that each such term is divisible by the product $ abc $. This is obvious when none of the exponents $ k $, $ m $, $ n $ is zero, because then $ a^kb^mc^n = abc \cdot a^{k-1}b^{m-1}c^{n-1} $. It remains to consider the case where at least one of these exponents is zero.
Since the system of assumptions is invariant under cyclic permutation of the variables ( $ a \mapsto b \mapsto c \mapsto a $ ), we do not lose generality by assuming that $ n = 0 $. The product (1) then has the form
The problem has been reduced to showing that for each such pair of exponents $ k $, $ m $, the number $ a^kb^m $ is divisible by $ abc $.
From the assumptions given in the problem, it follows that
A number of the form (2) can be written as the product of four factors (separated by dots):
In each of the above representations, the first factor is divisible by $ a $, the second by $ b $, and the third by $ c $; the fourth is a certain natural number. Thus, in each case, the product $ a^kb^m $ is a number divisible by the product $ abc $. In light of the previous remarks, this completes the proof. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 696 |
XLVII OM - III - Problem 3
Given a natural number $ n \geq 2 $ and positive numbers $ a_1, a_2, \ldots , a_n $, whose sum equals 1.
(a) Prove that for any positive numbers $ x_1, x_2, \ldots , x_n $ whose sum equals 1, the following inequality holds:
(b) Determine all systems of positive numbers $ x_1, x_2, \ldots , x_n $ whose sum equals 1 for which the above inequality becomes an equality.
Note: The symbol $ \sum_{i<j} x_i x_j $ denotes the sum of $ \binom{n}{2} $ terms corresponding to all pairs of indices $ i, j $ from the set $ \{1, 2, \ldots , n\} $ satisfying the condition $ i < j $. | From the given conditions ($ n \geq 2 $; $ a_i>0 $; $ \sum a_i =1 $), it follows that the numbers $ a_1, a_2, \ldots, a_n $ belong to the interval $ (0;\ 1) $. Let us assume
these are also numbers in the interval $ (0;\ 1) $. For them, the equality
holds.
Let $ x_1, x_2,\ldots, x_n $ be any real numbers whose sum is equal to $ 1 $. Then
The inequality to be proven takes the form
or, equivalently,
Using the notation (1), we get another equivalent form of the inequality to be proven:
For each index $ i \in \{1,2,\ldots,n\} $, we have the obvious inequality
We add these $ n $ inequalities side by side (expanding the squares of the differences):
We use formula (2) and the fact that $ \sum x_i=1 $, and we obtain
A simple transformation yields the inequality (3) from this.
We have thus proven the thesis (a).
From the reasoning conducted, it also immediately follows the resolution of issue (b): the inequality (3) (equivalent to the inequality given in the problem) becomes an equality if and only if each of the inequalities (4) (for $ i = 1,2, \ldots,n $) turns into an equality; that is, when $ x_i/c_i= c_i/(n - 1) $, or, according to notation (1), when
| proof | Inequalities | proof | Yes | Yes | olympiads | false | 698 |
XXIX OM - I - Problem 1
The sequence of numbers $ (p_n) $ is defined as follows: $ p_1 = 2 $, $ p_n $ is the largest prime divisor of the sum $ p_1p_2\ldots p_{n-1} + 1 $. Prove that the number 5 does not appear in the sequence $ (p_n) $. | We have $ p_1 = 2 $ and $ p_2 = 3 $, because $ p_2 $ is the largest prime divisor of the number $ p_1 + 1 = 3 $. Therefore, for $ n > 2 $, the number $ p_1p_2 \ldots p_{n-1} + 1 = 6p_3 \ldots p_n + 1 $ is not divisible by either $ 2 $ or $ 3 $. If, for some natural number $ n $, the largest prime divisor of the number $ p_1p_2 \ldots p_{n-1} + 1 $ were equal to $ 5 $, then it would be its only prime divisor, i.e., it would be equal to $ 5^r $ for some $ r \geq 1 $. Therefore,
The number $ 5^r - 1 $ is divisible by the number $ 5 - 1 = 4 $. Hence, from (1) it follows that $ p_k = 2 $ for some $ k \geq 3 $. This is impossible, because the number $ p_k $, by definition, is a divisor of the number $ p_1p_2 \ldots p_{k-1} + 1 = 2p_2 \ldots p_{k-1} + 1 $, and this number is odd.
The obtained contradiction proves that the number 5 does not appear in the sequence $ (p_n) $. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 699 |
XXIX OM - II - Problem 4
From the vertices of a regular $2n$-gon, 3 different points are chosen randomly. Let $p_n$ be the probability that the triangle with vertices at the chosen points is acute. Calculate $\lim_{n\to \infty} p_n$.
Note. We assume that all choices of three different points are equally probable. | Let $ A_1, A_2, \ldots , A_{2n} $ be the consecutive vertices of a regular $ 2n $-gon, and $ O $ - the center of the circle circumscribed around this polygon. We will investigate how many triangles $ A_1A_iA_j $, where $ i < j $, have the angle $ \measuredangle A_iA_1A_j $ not acute.
Since the inscribed angle in a circle has a measure that is half the measure of the central angle subtended by the same arc, therefore
Thus, $ \measuredangle A_iA_1A_j $ is not acute if and only if $ j - i \geq n $. We will calculate the number of such pairs $ (i, j) $, such that
From these inequalities, it follows that $ i \leq j - n \leq 2n - n = n $. For each $ i $ satisfying $ 2 \leq i \leq n $, there are $ n - i + 1 $ values for $ j $ satisfying (1), namely $ j = i + n, i + n + 1, \ldots, 2n $. Therefore, the number of such pairs is
We have thus proved that the number of non-acute triangles $ A_1A_iA_j $, in which the angle $ \measuredangle A_iA_1A_j $ is not acute, is equal to $ \frac{n(n-1)}{2} $. Since in a triangle at most one angle is not acute, the number of all non-acute triangles $ A_iA_jA_k $ is equal to
The number of all triangles $ A_iA_jA_k $ is equal to the number of three-element subsets of the $ 2n $-element set $ \{A_1, A_2, \ldots , A_{2n} \} $, that is, the number
Therefore,
Thus, $ \displaystyle \lim_{n \to \infty} p_n = 1 - \frac{3}{4} = \frac{1}{4} $.
om29_2r_img_12.jpg
Note. The triangle $ A_1A_iA_j $, where $ 1 < i < j \leq 2n $, is acute if and only if $ i \leq n $, $ j > n $, and $ j - i < n $. By associating the triangle $ A_1A_iA_j $ with the point $ (i,j) $, we see that the acute triangles correspond to lattice points located in the shaded triangle (Fig. 12), and any triangles correspond to lattice points belonging to the triangle $ AOB $. The ratio of the number of lattice points in the shaded triangle to the number of all lattice points in the triangle $ AOB $ approaches the ratio of the areas of these triangles as $ n \to \infty $. This ratio of areas is equal to $ \frac{1}{4} $. | \frac{1}{4} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 700 |
XXXIV - III - Task 3
We consider a single-player game on an infinite chessboard based on the following rule. If two pieces are on adjacent squares and the next square is empty (the three squares being discussed lie on the same horizontal or vertical line), we can remove these pieces and place one piece on the third of these squares (which was empty).
Prove that if in the initial position the pieces fill a rectangle with a number of squares divisible by 3, then we cannot achieve a position where there is only one piece on the chessboard. | By introducing a coordinate system on the plane, we can assign a pair of integers to each square of the chessboard, such that two adjacent squares in the same column are assigned the numbers $(m, n)$ and $(m, n+1)$, and two adjacent squares in the same row are assigned the numbers $(m, n)$ and $(m+1, n)$. We can assign to each square of the chessboard the sum $m+n$ of its coordinates, and then the remainder of dividing $m+n$ by $3$. In this way, we divide all the squares into three sets:
Pawns placed in the initial position occupy a rectangle of dimensions $3k \times l$ or $l \times 3k$. In each row parallel to the side of length $3k$, every third pawn stands on a square belonging to $A_0$, every third on a square belonging to $A_1$, and every third on a square belonging to $A_2$ (the sums of the coordinates of consecutive squares are consecutive integers). Therefore, in the initial position, the number of pawns occupying squares belonging to $A_0$, to $A_1$, and to $A_2$ are equal. If in a certain position exactly $a_0$ pawns occupy squares from the set $A_0$, $a_1$ pawns - squares from the set $A_1$, and $a_2$ pawns - from the set $A_2$, and by making a move allowed in the game, we take away two pawns and place a third, and we do this on three squares that are adjacent in one row, then each of these squares belongs to a different set $A_0$, $A_1$, $A_2$. Thus, two of the numbers $a_0$, $a_1$, $a_2$ decrease by $1$, and the remaining number increases by $1$. Since in the initial position the numbers of pawns occupying squares from the sets $A_0$, $A_1$, $A_2$ were equal, after each series of allowed moves, the numbers of pawns occupying squares from the sets $A_0$, $A_1$, $A_2$ are numbers of the same parity. If there is only one pawn left on the chessboard, it belongs to one of the sets $A_0$, $A_1$, or $A_2$, for example, to $A_0$, and in this case the numbers $a_0 = 1$, $a_1 = 0$, $a_2 = 0$ do not have the same parity. It follows from this that it is not possible to reach a position in which only one pawn remains on the chessboard. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 705 |
XXXV OM - I - Problem 12
In a trihedral angle $ W $, a ray $ L $ passing through its vertex is contained. Prove that the sum of the angles formed by $ L $ with the edges of $ W $ does not exceed the sum of the dihedral angles of $ W $. | Consider a sphere of radius 1 with its center at the vertex of a trihedral angle. The measure of each plane angle, whose sides are rays originating from the center of the sphere, is equal to the length of the corresponding arc of a great circle (the shorter of the two arcs defined by the points of intersection of the given rays with the surface of the sphere). All arcs on the surface of the sphere that we will discuss in the course of this solution will be arcs of the corresponding great circles.
om35_1r_img_6.jpg
om35_1r_img_7.jpg
The sum of the measures of the plane angles of the trihedral angle is equal to the sum of the lengths of the arcs $ AB $, $ BC $, and $ CA $, whose endpoints are the points of intersection of the edges of the trihedral angle and the surface of the sphere. The sum of the measures of the angles formed by $ L $ with the edges of the trihedral angle is equal to the sum of the lengths of the arcs $ AD $, $ BD $, and $ CD $, where $ D $ is the point of intersection of the ray $ L $ and the surface of the sphere.
The length of the arc (in accordance with the adopted convention - the arc of a great circle) connecting two points on the surface of the sphere satisfies the properties of distance (see, e.g., R. Courant, H. Robbins, What is Mathematics, Chapter VII), in particular the so-called triangle inequality.
Based on the triangle inequality, the lengths of the corresponding arcs satisfy
\[
\text{length}(AB) + \text{length}(BC) \geq \text{length}(AC)
\]
\[
\text{length}(BC) + \text{length}(CA) \geq \text{length}(AB)
\]
\[
\text{length}(CA) + \text{length}(AB) \geq \text{length}(BC)
\]
Thus, we have
\[
\text{length}(AD) + \text{length}(BD) \geq \text{length}(AB)
\]
By analogous reasoning, we can state that
\[
\text{length}(BD) + \text{length}(CD) \geq \text{length}(BC)
\]
and
\[
\text{length}(CD) + \text{length}(AD) \geq \text{length}(CA)
\]
Adding the last three inequalities side by side, we get
\[
2(\text{length}(AD) + \text{length}(BD) + \text{length}(CD)) \geq \text{length}(AB) + \text{length}(BC) + \text{length}(CA)
\]
Thus,
\[
\text{length}(AD) + \text{length}(BD) + \text{length}(CD) \geq \frac{1}{2} (\text{length}(AB) + \text{length}(BC) + \text{length}(CA))
\]
Therefore, the sum of the measures of the angles formed by $ L $ with the edges of the trihedral angle $ W $ does not exceed the sum of the measures of the plane angles of $ W $. | proof | Geometry | proof | Yes | Yes | olympiads | false | 706 |
V OM - I - Task 3
Prove that if in a quadrilateral $ABCD$ the equality $AB + CD = AD + BC$ holds, then the incircles of triangles $ABC$ and $ACD$ are tangent. | If the incircles of triangles $ABC$ and $ACD$ touch their common side $AC$ at points $M$ and $N$ respectively (Fig. 20), then according to known formulas we have
from which
If $AB + CD = AD + BC$, then from the above equality it follows that $AM = AN$, i.e., that points $M$ and $N$ coincide; both circles are tangent to the line $AC$ at the same point, so the circles are tangent to each other.
Note. The equality (1) also implies the converse theorem: if the incircles of triangles $ABC$ and $ACD$ are tangent, then $AM = AN$, hence $AB + CD = AD + BC$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 707 |
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 | 711 |
X OM - I - Task 7
On a plane, there are four different points $ A $, $ B $, $ C $, $ D $.
a) Prove that among the segments $ AB $, $ AC $, $ AD $, $ BC $, $ BD $, $ CD $, at least two are of different lengths,
b) How are the points $ A $, $ B $, $ C $, $ D $ positioned if the segments mentioned above have only two different lengths? | a) To prove that the segments $ AB $, $ AC $, $ AD $, $ BC $, $ BD $, $ CD $ cannot all be equal to each other, it suffices to show that if five of them have the same length, then the sixth has a different length. Let's assume, for example, that $ AB = AC = AD = BC = BD $. In this case, $ ABC $ and $ ABD $ are equilateral triangles sharing the side $ AB $ and forming a rhombus $ ACBD $, in which $ \angle C = 60^\circ $. The diagonals of such a rhombus are not equal; specifically, $ CD > AB $, Q.E.D.
b) Suppose the segments $ AB $, $ AC $, $ AD $, $ BC $, $ BD $, $ CD $ are of two different lengths $ a $ and $ b $, with $ a < b $. Then the points $ A $, $ B $, $ C $, $ D $ cannot lie on the same line; if they did, for example in the order $ A $, $ B $, $ C $, $ D $, then it would be $ AB < AC < AD $, contradicting the assumption that only two of the mutual distances between the points $ A $, $ B $, $ C $, $ D $ are different. We will prove that no three of the given points can lie on a line. Indeed, if point $ B $ were on the line $ AC $ between points $ A $ and $ C $, then it would be $ AB = BC = a $, $ AC = b $. Point $ D $ would be outside the line $ AC $, and the segment $ DB $ being the median of triangle $ ACD $ would be smaller than one of the sides $ DA $ and $ DC $ of this triangle, so it would be $ DB = a $, and one of the sides $ DA $ and $ DC $ would be equal to $ b $. In this case, point $ D $ would lie on the circle with diameter $ AC $, implying that one of the chords $ AD $ and $ DC $ of the circle is equal to the diameter $ AC $ of the circle, which is impossible.
Notice that if no three of the points $ A $, $ B $, $ C $, $ D $ lie on a line, then there are two possible cases. If point $ D $ lies in one of the regions bounded by one side of triangle $ ABC $ and the extensions of the other two sides, like point $ D_1 $ in Fig. 7, then the given points form a convex quadrilateral, in this case quadrilateral $ ABCD_1 $, composed of triangles $ ABC $ and $ ACD_1 $ sharing the side $ AC $. If, however, point $ D $ lies in one of the corner regions or inside triangle $ ABC $, then the given points form a concave quadrilateral like $ ABD_2C $ or $ ABCD_3 $ in Fig. 7; one of the given points then lies inside the triangle formed by the other three points. We will consider these two cases separately.
1) Suppose the given points form a convex quadrilateral $ ABCD $. Several cases can arise, which we will consider using the theorem that in a convex quadrilateral, the sum of the diagonals is greater than half the perimeter (This theorem can be easily proven by applying to each of the four triangles into which the diagonals divide the quadrilateral the theorem that the sum of two sides of a triangle is greater than the third side).
1a) Quadrilateral $ ABCD $ has four sides equal. The sum of the diagonals is then greater than twice the side, so the sides have length $ a $, and the diagonals are either both equal to $ b $ or one is equal to $ b $ and the other to $ a $. The quadrilateral is either a square with side $ a $, in which case $ b = a\sqrt{2} $, or a rhombus with side $ a $ and an acute angle of $ 60^\circ $ (Figs. 8 and 9), in which case $ b = a\sqrt{3} $.
1b) Quadrilateral $ ABCD $ has three sides equal, say $ AB = BC = CD $. These sides cannot have length $ b $, because then half the perimeter would be greater than $ a + b $, so each of the diagonals would have length $ b $, i.e., five of the six mutual distances between the given points would be equal to $ b $, and the sixth would be equal to $ a $, thus smaller than the others, which would contradict the result obtained in a). Therefore, $ AB = BC = CD = a $, $ AD = b $. The sum of the diagonals must be greater than $ 2a $, so the diagonals have either lengths $ a $, $ b $ or $ b $, $ b $. The first case is, however, excluded, because if, for example, $ AC = a $ and $ BD = b $, then triangle $ ABC $ would be equilateral, so $ \angle ABC $ would be $ 60^\circ $, and in the isosceles triangle $ ABD $, angle $ ABD $ would be greater than $ 60^\circ $, which is impossible, since $ \angle ABD $ is part of angle $ ABC $. Therefore, $ AC = BD = b $; triangles $ ABC $ and $ DCB $ are congruent, from which it easily follows that quadrilateral $ ABCD $ is an isosceles trapezoid (Fig. 10).
According to Ptolemy's theorem, $ AC \cdot BD = AB \cdot DC + AD \cdot BC $, i.e., $ b^2 = ab + a^2 $, from which we see that $ b $ is the length of the diagonal of a regular pentagon with side equal to $ a $.
1c) Two sides of quadrilateral $ ABCD $ have length $ a $, and the other two have length $ b $.
In this case, the sides of equal length cannot be opposite sides, because the quadrilateral would then be a parallelogram and would have a diagonal longer than $ b $, which contradicts the assumption. The quadrilateral is therefore a kite, with each of the diagonals having length $ b $, since the sum of the diagonals is longer than $ a + b $.
Denoting the vertices of the kite as in Fig. 11 and applying the law of cosines to triangle $ ACD $, where $ \angle DAC = 30^\circ $, we obtain the relation
\[
b^2 = a^2 + a^2 - 2a^2 \cos(30^\circ) = 2a^2 - 2a^2 \cdot \frac{\sqrt{3}}{2} = 2a^2 - a^2\sqrt{3}
\]
\[
b^2 = a^2(2 - \sqrt{3})
\]
\[
b = a\sqrt{2 - \sqrt{3}}
\]
We observe that $ b $ is the length of the radius of the circle circumscribed around a regular dodecagon with side equal to $ a $.
2) Suppose the points $ A $, $ B $, $ C $, $ D $ form a concave quadrilateral: let, for example, point $ D $ lie inside triangle $ ABC $. Notice that any segment (open) lying inside a triangle is shorter than the longest side of the triangle. It follows that 1° at least one side of triangle $ ABC $ is equal to $ b $, 2° each of the segments $ DA $, $ DB $, $ DC $ is equal to $ a $, i.e., point $ D $ is the center of the circle circumscribed around triangle $ ABC $. The following cases can arise.
2a) $ AB = BC = CA = b $. The given points form the figure in Fig. 12, where $ b = a\sqrt{3} $.
2b) $ AB = BC = b $, $ CA = a $. In this case, points $ A $, $ B $, $ C $, $ D $ form the figure in Fig. 13. Angle $ ABC $ is $ 30^\circ $, as it is half of angle $ ADC $, which is $ 60^\circ $. Thus, $ a $ is the length of the side of a regular dodecagon inscribed in a circle with radius $ b $, and we have the relation
\[
b = a\sqrt{3}
\]
The case where two sides of triangle $ ABC $, say $ AB $ and $ BC $, are both equal to $ a $ cannot occur, because $ AB + BC > AD + DC $, i.e., $ AB + BC > 2a $.
Summarizing the obtained results, we can say that there are only 6 different figures formed by four points $ A $, $ B $, $ C $, $ D $ such that the segments connecting them have two different lengths; the ratio of the larger to the smaller of these lengths can only have the values $ \sqrt{2} $ (Fig. 8), $ \sqrt{3} $ (Figs. 9 and 12), $ \frac{\sqrt{5} + 1}{2} $ (Fig. 10), and $ \frac{\sqrt{6} + \sqrt{2}}{2} $ (Figs. 11 and 13).
We propose to the reader to solve a similar problem for figures formed by five points in space. | notfound | Geometry | proof | Yes | Yes | olympiads | false | 712 |
XXI OM - I - Problem 6
Prove that a real number $ a $ is rational if and only if there exist integers $ p > n > m \leq 0 $ such that $ a + m, a + n, a + p $ form a geometric progression. | Let $ a $ be a rational number. We choose a natural number $ m $ such that $ a + m > 0 $. Then
where $ r $ and $ q $ are natural numbers.
Consider a geometric sequence with a common ratio of $ 1 + q $ and the first term $ \frac{r}{q} $. Thus, the subsequent terms of the sequence will be
Assuming then $ n = m + r $ and $ p = m + 2r + rq $, we conclude that $ m $, $ n $, $ p $ are integers satisfying the condition $ 0 \leq m < n < p $. Moreover, the numbers $ a + m $, $ a + n $, $ a + p $ form a geometric sequence.
Conversely, suppose that the integers $ m $, $ n $, $ p $ satisfy the condition $ 0 \leq m < n < p $ and that for some real number $ a $ the numbers $ a + m $, $ a + n $, $ a + p $ form a geometric sequence.
Let $ 1 + q $ be the common ratio of this sequence. Then
Therefore
Since $ n > m $, it follows from (1) that $ a + m \ne 0 $ and $ q \ne 0 $. Dividing (2) by (1) we get
Therefore, from (1) it follows that $ a = \frac{n-m}{q} - m $ is also a rational number. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 714 |
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 | 718 |
XXV OM - III - Task 5
Prove that if natural numbers $ n $, $ r $ satisfy the inequality $ r + 3 \leq n $, then the numbers $ \binom{n}{r} $, $ \binom{n}{r+1} $, $ \binom{n}{r+2} $, $ \binom{n}{r+3} $ are not consecutive terms of any arithmetic sequence. | We will first prove two lemmas.
Lemma 1. For a fixed natural number $n$, there are at most two natural numbers $k \leq n - 2$ such that the numbers $\binom{n}{k}$, $\binom{n}{k+1}$, $\binom{n}{k+2}$ are consecutive terms of an arithmetic sequence.
Proof. If the numbers $\displaystyle \binom{n}{k}, \binom{n}{k+1}, \binom{n}{k+2}$ are consecutive terms of an arithmetic sequence, then
Multiplying this equality by $\displaystyle \frac{(k+2)!(n-k)!}{n!}$, we get
This is a quadratic equation in terms of $k$. Therefore, it has at most two solutions.
Lemma 2. For a fixed natural number $n$, there is at most one natural number $k \leq n - 1$ such that $\displaystyle \binom{n}{k} = \binom{n}{k+1}$.
Proof. Multiplying the above equality on both sides by $\displaystyle \frac{(k+1)!(n-k)!}{n!}$, we get $k + 1 = n - k$. This is a first-degree equation in terms of $k$. Therefore, it has at most one solution in the set of natural numbers less than $n$.
We now proceed to solve the problem. For simplicity of notation, let us assume that $a_j = \displaystyle \binom{n}{j}$ for $j = 1, 2, \ldots, n$. Suppose that the numbers
are consecutive terms of an arithmetic sequence. By the formula $\displaystyle \binom{n}{k} = \binom{n}{n-k}$, i.e., $a_k = a_{n-k}$, the numbers $a_{n-r-3}, a_{n-r-2}, a_{n-r-1}, a_{n-r}$ are also consecutive terms of an arithmetic sequence. Therefore, we have the following three-term arithmetic sequences
From Lemma 1, the set $\{r, r + 1, n - r - 3, n - r - 2\}$ contains at most two different numbers. Since $r$ and $r + 1$ as well as $n - r - 3$ and $n - r - 2$ are consecutive natural numbers, we have $r = n - r - 3$ and $r + 1 = n - r - 2$.
Thus, $a_{r+1} = a_{n-r-2} = a_{r+2}$, meaning the arithmetic sequence (1) is constant. From Lemma 2, this is impossible.
The obtained contradiction proves that the numbers (1) are not consecutive terms of an arithmetic sequence.
Note. It can be proven that if for some natural numbers $n$ and $r$, where $r \leq n - 2$, the numbers $\displaystyle \binom{n}{r}, \binom{n}{r+1}, \binom{n}{r+2}$ are consecutive terms of an arithmetic sequence, then there exists a natural number $m \geq 3$ such that $n = m^2 - 2$ and $r = \displaystyle \frac{1}{2} (m^2 - m) - 2$ or $r = \displaystyle \frac{1}{2} (m^2 + m) - 2$.
Conversely, if the numbers $n$ and $r$ are defined by the above formulas, then the numbers $\displaystyle \binom{n}{k}, \binom{n}{k+1}, \binom{n}{k+2}$ are consecutive terms of an arithmetic sequence.
For example, taking $m = 3$ gives that the numbers $\displaystyle \binom{7}{1}, \binom{7}{2}, \binom{7}{3}$ form an arithmetic sequence. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 719 |
XXXIX OM - II - Problem 4
Prove that for every natural number $ n $ the number $ n^{2n} - n^{n+2} + n^n - 1 $ is divisible by $ (n - 1)^3 $. | Let us denote:
Assume that $ n \geq 2 $, hence $ m \geq 1 $. According to the binomial formula,
where $ A $, $ B $, $ C $ are integers. We transform further:
Thus, $ N = Dm^3 = D(n - 1)^3 $, where $ D $ is an integer (this is also true for $ n = 1 $).
We have thus shown that $ N $ is divisible by $ (n - 1)^3 $. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 720 |
LVIII OM - I - Problem 1
Problem 1.
Solve in real numbers $ x $, $ y $, $ z $ the system of equations | Subtracting the second equation of the system from the first, we get
Similarly, by subtracting the third equation from the first, we obtain the equation $ 0=(x-z)(x+z-2y+5) $. Therefore, the system of equations given in the problem is equivalent to the system
The product of two numbers is equal to zero if and only if at least one of the factors is equal to zero. Therefore, system (1) is equivalent to the following alternative of four systems:
In the first of the systems (2), the second and third equations lead to the conclusion that $ x=y=z $. Using this relationship in the first equation of this system, we obtain the quadratic equation $ 3x^2+5x=2 $, which has two solutions: $ x=-2 $ and $ x={1\over 3} $. Thus, the considered system of equations has two solutions: $ (-2,-2,-2) $ and $ ({1\over 3},{1\over 3},{1\over 3}) $.
Now let's examine the second system (2). The second equation of this system means that $ x=y $; combining this with the third equation, we see that $ z=-x+2y-5=x-5 $. Therefore, the first equation of the system takes the form
$$2=x^2+2yz+5x=x^2+2x(x-5)+5x=3x^2-5x.$$
The quadratic equation $ 3x^2-5x=2 $ has solutions $ x=2 $ and $ x=-{1\over 3} $. This gives two solutions of the system: $ (2,2,-3) $ and $ (-{1\over 3},-{1\over 3},-{16\over 3}) $.
Similarly, we solve the third system (2), obtaining the solutions $ (2,-3,2) $ and $ (-{1\over 3},-{16\over 3},-{1\over 3}) $.
Finally, let's move on to the last system (2). Combining the second and third equations, we see that $ y-2z=z-2y $, which means $ y=z $. Therefore, the second equation takes the form $ y=x+5 $, and the first can be written as
$$2=x^2+2yz+5x=(y-5)^2+2y^2+5(y-5)=3y^2-5y,$$
which, after solving the quadratic equation, gives $ y=2 $ and $ y=-{1\over 3} $. As a result, we get two solutions: $ (-3,2,2) $ and $ (-{16\over 3},-{1\over 3},-{1\over 3}) $.
Finally, it remains to verify that each of the 8 triples indeed represents a solution to the system of equations given in the problem.
Answer: The solutions $ (x, y, z) $ of the given system of equations are: | (x,y,z)=(-2,-2,-2),(\frac{1}{3},\frac{1}{3},\frac{1}{3}),(2,2,-3),(-\frac{1}{3},-\frac{1}{3},-\frac{16}{3}),(2,-3,2),(-\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 722 |
XLIV OM - III - Problem 2
Point $ O $ is the center of the circle $ k $ inscribed in the non-isosceles trapezoid $ ABCD $, whose longer base $ AB $ has midpoint $ M $. The shorter base $ CD $ is tangent to the circle $ k $ at point $ E $, and the line $ OM $ intersects the base $ CD $ at point $ F $. Prove that $ |DE| = |FC| $ if and only if $ |AB| = 2\cdot |CD| $. | Since the equality $ |DE| = |FC| $ (considered in the thesis of the problem) holds if and only if $ |DF| = |EC| $, the conditions of the assumptions and the thesis will not change when the roles of the legs $ AD $ and $ BC $ of the given trapezoid are swapped. Therefore, we can assume, without loss of generality, that $ |AD| > |BC| $; this is the situation depicted in Figure 9.
om44_3r_img_9.jpg
om44_3r_img_10.jpg
We extend the legs $ AD $ and $ BC $ to intersect at point $ S $. Let $ k $ be the incircle of triangle $ CDS $, $ K $ the point of tangency of circle $ k $ with side $ CD $, $ N $ the midpoint of side $ CD $, and $ L $ the point of tangency of circle $ k $ with base $ AB $. We have the equalities
the first of these equalities is a well-known formula expressing the distance from the point of tangency of the incircle (of a triangle) with a side to the vertex; the second is an analogous formula for the distance from the point of tangency of the excircle with the corresponding side to the vertex (note that $ k $ is the excircle of triangle $ CDS $); this formula may not be as widely known as the first equality in (1); we leave its proof to the Reader as a not difficult exercise.
Since $ N $ is the midpoint of segment $ CD $, it follows from the equalities (1) that $ |NK| = |NE| $, i.e.,
The homothety with center $ S $ and scale $ |AB|: |CD| $ maps triangle $ SCD $ to triangle $ SBA $, the incircle of the first of these triangles to the incircle of the second (i.e., circle $ k $ to circle $ k $), the point of tangency $ K $ to the point of tangency $ L $, and the midpoint $ N $ of side $ CD $ to the midpoint $ M $ of side $ AB $. Therefore,
The right triangles $ OLM $ and $ OEF $ are congruent (segments $ OL $ and $ OE $ are radii of circle $ k $, and angles $ OLM $ and $ OEF $ are equal as vertical angles). Therefore, the following equality holds:
By virtue of the relationships (3), (2), (4), and (1), we have
We obtain the equivalence
which is the thesis of the problem. | proof | Geometry | proof | Yes | Yes | olympiads | false | 723 |
XXXII - I - Problem 8
Given a point $ P $ inside a sphere $ S $. The transformation $ f: S\to S $ is defined as follows: for a point $ X \in S $, the point $ f(X) \in S $, $ f(X) \neq X $, and $ P\in \overline{Xf(X)} $. Prove that the image of any circle contained in $ S $ under the transformation $ f $ is a circle. | Let $ k $ be a circle contained in the sphere $ S $, and let $ S_1 $ be a sphere containing $ k $ and the point $ P $, $ P_1 $ be such a point that the segment $ \overline{PP_1} $ is a diameter of $ S_1 $. For every $ X \in k $, the angle $ PXP_1 $ is a right angle. Let $ X $ be the orthogonal projection of the point $ f(X) $ onto the line $ PP_1 $. The triangles $ f(X)PX $ and $ P_1PX $ are similar, so
The value of the last expression does not depend on the position of the point $ X $ on the circle $ k $ by a theorem stating that the product of the segments into which a fixed point divides the chords of a circle is constant. Therefore, for every point $ X \in k $, the projection of the point $ f(X) $ onto $ PP_1 $ is the same point. It follows that all images $ f(X) $ belong to a plane perpendicular to the line $ PP_1 $. Since these images belong to the sphere $ S $, they belong to the intersection of the sphere and the plane, i.e., to some circle $ o $. Therefore, $ f(k) \subset o $. Since, however, the transformation $ f $ is invertible, with $ f^{-1} = f $, and by analogous reasoning we determine that $ f(o) \subset k $, the circle $ o $ is the image of the circle $ k $ (and simultaneously $ k $ is the image of the circle $ o $). | proof | Geometry | proof | Yes | Yes | olympiads | false | 724 |
LVIII OM - II - Problem 3
From $ n^2 $ tiles in the shape of equilateral triangles with a side of $ 1 $, an equilateral triangle with a side of $ n $ was formed. Each tile is white on one side and black on the other. A move consists of the following actions: We select a tile $ P $ that shares sides with at least two other tiles, whose visible sides have a color different from the visible side of tile $ P $. Then we flip tile $ P $ to the other side.
For each $ n\ge 2 $, determine whether there exists an initial arrangement of tiles that allows for an infinite sequence of moves. | Let's call a boundary segment the common side of two tiles whose visible sides have different colors. Of course, the number of boundary segments is non-negative and does not exceed the number $ m $ of all segments that are the common side of two tiles. We will investigate how the number of boundary segments changes as a result of performing an allowed move.
Each of the $ n^2 $ tiles shares common sides with at most three other tiles. Flipping a tile is allowed if at least two of these sides are boundary segments. Suppose we flip tile $ P $. Then boundary segments can only appear or disappear on the sides of tile $ P $. Moreover, a side of tile $ P $ is a boundary segment after flipping if and only if it was not a boundary segment before flipping. It follows that as a result of performing an allowed move, the number of boundary segments decreases.
In this way, we have proved that from any initial arrangement of tiles, no more than $ m $ moves can be made.
Answer: For no $ n\ge 2 $ does there exist an arrangement allowing an infinite sequence of moves. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 725 |
XXXI - II - Task 5
We write down the terms of the sequence $ (n_1, n_2, \ldots, n_k) $, where $ n_1 = 1000 $, and $ n_j $ for $ j > 1 $ is an integer chosen randomly from the interval $ [0, n_{j-1} - 1] $ (each number in this interval is equally likely to be chosen). We stop writing when the chosen number is zero, i.e., $ n_{k-1} $, $ n_k = 0 $. The length $ k $ of the sequence $ (n_1, n_2, \ldots, n_k) $ is a random variable. Prove that the expected value of this random variable is greater than 7. | For a given non-negative integer $ n $, consider the random variable $ X_n $ being the length $ k $ of the sequence ($ n_1, n_2, \ldots, n_k $), where $ n_1 = n $, $ n_j $ for $ j > 1 $ is an integer chosen randomly from the interval $ [0, n_{j-1} - 1] $, and $ n_k = 0 $. Let $ E_n $ denote the expected value of this random variable. Of course, $ E_0 = 1 $, since for $ n = 0 $ there is only one sequence of length $ 1 $, whose only term is $ 0 $. Let $ n $ be a natural number. Suppose we have determined the expected values $ E_m $ for all $ m < n $. We consider sequences $ (n_1, n_2, \ldots, n_k) $ where $ n_1 = n $. There are two possible events:
1) Either $ n_2 = n - 1 $ (the probability of this event is $ \frac{1}{n} $)
and in this case the expected length of the sequence ($ n_2, \ldots, n_k $) is $ E_{n-1} $, so the expected length of the sequence ($ n_1, n_2, \ldots, n_k $) is $ E_{n-1} + 1 $,
2) or $ n_2 < n - 1 $ (the probability of this event is $ \frac{n-1}{n} $) and in this case the expected length of the sequence $ (n_1, n_2, \ldots, n_k) $ is exactly the same as the expected length of the sequence $ (m_1, m_2, \ldots, m_k) $ where $ m_1 = n - 1 $; this value is $ E_{n-1} $.
Therefore
Since $ E_0 = 1 $, then $ E_1 = 1 + \frac{1}{1} = 2, E_2 = 2 + \frac{1}{2}, \ldots, E_n = 2 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} $.
The expected value we are interested in is
The number in the last parenthesis is less than $ \frac{24}{1000} $, while each
of the numbers in the previous parentheses (of which there are eight) is greater than $ \frac{1}{2} $ since in the $ k $-th parenthesis is the number
being the sum of $ 2^k $ terms, all of which are smaller (only the last one is equal). Therefore
Note. The approximate value of $ E_{1000} $ is $ 8.48547 $. | 8.48547 | Combinatorics | proof | Yes | Yes | olympiads | false | 726 |
IV OM - II - Task 6
Given a circle and two tangents to this circle. Draw a third tangent to the circle in such a way that the segment of this tangent contained between the given tangents has a given length $ d $. | When the tangent lines $ m $ and $ n $ intersect, the task coincides with the already solved problem No. 11.
When the lines $ m $ and $ n $ are parallel, the solution is immediate. From any point $ A $ on the line $ m $ as the center, we draw a circle with radius of length $ d $.
If this circle has a common point $ B $ with the line $ n $, we draw tangents to the given circle parallel to the line $ AB $. The task has $ 4 $ solutions, $ 2 $ solutions, or no solution depending on whether $ d $ is greater than the distance between the lines $ m $ and $ n $, equal to that distance, or less than it. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 727 |
XVI OM - III - Task 2
Prove that if the numbers $ x_1 $ and $ x_2 $ are roots of the equation
$ x^2 + px - 1 = 0 $, where $ p $ is an odd number, then for every natural $ n $ the numbers $ x_1^n + x_2^n $ and $ x_1^{n+1} + x_2^{n+1} $ are integers and relatively prime. | We will apply the method of induction. Since $ p $ is an odd number, the numbers
are integers and relatively prime. The theorem is thus true for $ n = 0 $.
Suppose that for some natural number $ n \geq 0 $ the numbers
are integers and relatively prime. We will prove that then $ x_1^{n+2} + x_2^{n+2} $ is an integer relatively prime to $ x_1^{n+1} + x_2^{n+1} $. Indeed, considering the equalities $ x_1 + x_2 = -p $, $ x_1x_2 = -1 $, we obtain from this
From this equality, it follows that 1) $ x_1^{n+2} + x_2^{n+2} $ is the sum of two integers, so it is an integer, 2) any common divisor of the numbers $ x_1^{n+2} + x_2^{n+2} $ and $ x_1^{n+1} + x_2^{n+1} $ is a divisor of the number $ x_1^{n} + x_2^{n} $, and thus is a common divisor of the relatively prime numbers $ x_1^{n+1} + x_2^{n+1} $ and $ x_1^{n} + x_2^{n} $, the numbers $ x_1^{n+2} + x_2^{n+2} $ and $ x_1^{n+1} + x_2^{n+1} $ are therefore relatively prime. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 728 |
LVI OM - II - Problem 4
Given is the polynomial $ W(x)=x^2+ax+b $, with integer coefficients, satisfying the condition:
For every prime number $ p $ there exists an integer $ k $ such that the numbers $ W(k) $ and $ W(k + 1) $ are divisible by $ p $. Prove that there exists an integer $ m $ for which | The condition $ W(m) = W(m +1) = 0 $ means that $ W(x)=(x-m)(x-m+1) $, i.e., $ W(x)=x^{2}-(2m+1)x+m^{2}+m $. Therefore, it is necessary to show that there exists an integer $ m $ such that
Let us fix a prime number $ p $. Then for some integer $ k $, the numbers
are divisible by $ p $. Therefore, the following numbers are divisible by $ p $:
The number $ E $ is determined by the coefficients of the polynomial $ W $ and does not depend on $ k $. By conducting this reasoning for each prime number $ p $, we conclude that the number $ E $ is divisible by every prime number $ p $, and consequently $ E = 0 $. This implies that the number $ a $ is odd, i.e., $ a = -2m-1 $ for some integer $ m $, and $ b= \frac{1}{4}(a^{2}-1)=m^{2}+m $. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 729 |
XIII OM - I - Problem 6
Factor the quadratic polynomial into real factors
where $ p $ and $ q $ are real numbers satisfying the inequality
Please note that the mathematical expressions and symbols are kept as they are, only the text has been translated. | From the inequality $ p^2 - 4q < 0 $, it follows that the quadratic polynomial $ y^2 + py + q $ has no real roots; therefore, the quartic polynomial $ x^4 + px^2 + q $ also has no real roots. If the number $ x $ were a real root of the second polynomial, then the number $ y = x^2 $ would be a real root of the first polynomial. Therefore, the desired factorization can only have the form
Comparing the coefficients of the same powers of $ x $ on both sides of the above equality, we obtain the equations
\[ (1) \qquad \alpha + \gamma = 0, \]
\[ (2) \qquad \beta + \delta + \alpha \gamma = p, \]
\[ (3) \qquad \alpha \delta + \beta \gamma = 0, \]
\[ (4) \qquad \beta \delta = q. \]
From equation (1) we have $ \gamma = - \alpha $, substituting into equation (3) gives
a) If $ \alpha = 0 $, then from equations (2) and (4) it follows that
This system of equations has no real solutions, because $ p^2 - 4q < 0 $.
b) $ \delta = \beta $; from equation (4) we then obtain
The above values of $ \beta $ and $ \delta $ are real, since from the inequality $ p^2 - 4q < 0 $ it follows that $ 4q > p^2 $, hence $ q > 0 $.
According to equation (2) $ \alpha \gamma = p - \beta - \delta $, so to determine $ \alpha $ and $ \gamma $ we have the system of equations
Therefore
From the inequality $ 4q > p^2 $ it follows that $ 2 \sqrt{q} > |p| \geq p $, hence $ 2 \sqrt{q} - p > 0 $; Since $ 4q - p^2 = (2 \sqrt{q} - p) (2 \sqrt{q} + p) > 0 $, then also $ 2 \sqrt{q} + p > 0 $.
To obtain a real value for $ \alpha $, we need to take the upper sign in the above formula for $ \alpha^2 $. We obtain the following real values of the coefficients $ \alpha $, $ \beta $, $ \gamma $, $ \delta $:
The desired factorization is expressed by the formula: | x^4+px^2+(x^2+\alphax+\beta)(x^2-\alphax+\beta) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 732 |
VIII OM - III - Task 6
Given a cube with base $ABCD$, where $AB = a$ cm. Calculate the distance between the line $BC$ and the line passing through point $A$ and the center $S$ of the opposite face. | Since $ BC\parallel AD $, the line $ BC $ is parallel to the plane $ ADS $. The sought distance between the lines $ BC $ and $ AS $ is equal to the distance from the line $ BC $ to the plane $ ADS $ (Fig. 25).
Consider a parallelepiped whose base is the square $ ABCD $, and one of its side faces is the rectangle $ AMND $, which is a section of the cube by the plane $ ADS $. The volume of this parallelepiped is equal to the volume $ a^3 $ of the given cube. On the other hand, the volume of the parallelepiped is equal to the product of the area $ AM \cdot MN $ of the rectangle $ AMND $ and the distance $ x $ from the edge $ BC $ to the plane of the rectangle. We have $ AM = \sqrt{AA^2 + A_1M^2} = \sqrt{a^2 + \left( \frac{a}{2} \right)^2} = \frac{a\sqrt{5}}{2} $, $ MN = a $, thus we obtain the equation | Geometry | math-word-problem | Yes | Yes | olympiads | false | 735 | |
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 | 736 |
XXV OM - III - Task 2
Salmon swimming upstream must overcome two waterfalls. The probability that a salmon will overcome the first waterfall in a given attempt is $ p > 0 $, and the probability of overcoming the second waterfall in a given attempt is $ q > 0 $. We assume that successive attempts to overcome the waterfalls are independent. Calculate the probability of the event that the salmon will not overcome the first waterfall in $ n $ attempts, given that in $ n $ attempts it will not overcome both waterfalls. | Let $ A_n $ be the event that the salmon does not overcome the first waterfall in $ n $ attempts, and $ B_n $ - the event that the salmon does not overcome both waterfalls in $ n $ attempts. Since the probability that the salmon does not overcome the first waterfall in one attempt is $ 1 - p $, and the attempts are independent, therefore
Event $ B_n $ will occur when either the salmon does not overcome the first waterfall in $ n $ attempts, or it overcomes the first waterfall in the $ k $-th attempt $ (1 \leq k \leq n) $ and does not overcome the second waterfall in the remaining $ n-k $ attempts. Therefore,
If $ p = q $, then formula (2) takes the form
If $ q = 1 $, then
In this case, the salmon would overcome the second waterfall in the first attempt. Therefore, event $ B_n $ will occur when the salmon does not overcome the first waterfall in $ n $ attempts or overcomes it only in the $ n $-th attempt.
Formula (4) follows from formula (2) if we assume that $ 0^0 = 1 $.
Finally, if $ p \ne q $ and $ q < 1 $, then using the formula for the sum of the terms of a geometric sequence, we transform (2) as follows:
We thus have in this case
If event $ A_n $ occurs, then event $ B_n $ will certainly occur. Therefore, $ A_n \cap B_n = A_n $. From the formula for conditional probability, we obtain
If $ n = 1 $, then from the conditions of the problem, $ p(A_n) = 1 - p $ and $ p(B_n) = 1 $. Therefore, from (6) we obtain that $ p(A_n \mid B_n) = 1 - p $. Henceforth, we will assume that $ n \geq 2 $.
Notice that if $ p \ne q $, then $ p(B_n) \ne 0 $. Indeed, if it were, then from formula (5) we would obtain $ p(1 - q)^n = q(1 - p)^n $. From this,
If, for example, $ p < q $, then $ 1 - p > 1 - q $ and equality (7) cannot hold. Similarly, we prove that (7) does not hold if $ p > q $.
If $ p = q $ and $ p(B_n) = 0 $, then from formula (3) it follows that $ p = 1 $. Therefore, in the case where $ p = q = 1 $ (and only in this case) the conditional probability $ p(A_n \mid B_n) $ does not exist. We calculate it in the remaining cases.
If $ p = q < 1 $, then from (1), (3), and (6) we obtain
If $ p < q = 1 $, then from (1), (4), and (6) we obtain
If $ p \ne q < 1 $, then from (1), (5), and (6) we obtain
Note.
We will examine in each case the number $ g = \lim_{n \to \infty} p(A_n \mid B_n) $.
If $ p = q < 1 $, then from formula (8) it follows that $ g = 0 $.
If $ p < q = 1 $, then from formula (9) it follows that $ g = 1 - p = 1 - \frac{p}{1} $.
If $ p < q < 1 $, then $ 1 - p > 1 - q $ and $ \lim_{n \to \infty} \left( \frac{1-p}{1-q} \right)^n = \infty $. Therefore, from (10) it follows that $ g = \frac{p-q}{-q} = 1 - \frac{p}{q} $.
If $ q < p \leq 1 $, then $ 1 - p < 1 - q $ and $ \lim_{n \to \infty} \left( \frac{1-p}{1-q} \right)^n = 0 $. Therefore, from (10) we obtain that $ g = 0 $.
Thus, in all cases, $ g = \max \left(0, 1 - \frac{p}{q} \right) $. | \max(0,1-\frac{p}{q}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 739 |
XXXVI OM - I - Problem 8
Prove that if $ (a_n) $ is a sequence of real numbers such that $ a_{n+2}=|a_{n+1}|-a_n $ for $ n = 1,2,\ldots $, then $ a_{k+9} = a_k $ for $ k = 1,2,\ldots $. | Let's consider the transformation $ F $ of the plane into itself given by the formula $ F(x,y) = (y,|y|-x) $. Let $ a $ be a positive number. Let $ I_0 $ be the segment with endpoints $ (0,-a) $ and $ (-a,0) $, let $ I_1 $ be the image of the segment $ I_0 $ under the transformation $ F $, and further, inductively, let $ I_n = F(I_{n-1}) $. The segment $ I_0 $ has the parametric representation
Acting on the point $ (x,y) = (-t, -a+t) $ with the transformation $ F $, we obtain
and therefore $ I_1 $ is a segment with the parametric representation
In the same way, we check that the subsequent sets $ I_n $ are segments with the following parametric representations (the parameter $ t $ always runs through the interval $ \langle 0;a \rangle $):
Thus, $ I_9 $ coincides with $ I_0 $. The sum of the segments $ I_0, \ldots, I_8 $ forms a closed broken line, which we will denote by $ L_a $. Interpreting the parameter $ t $ as time, we see that when the point $ P = (x,y) $ moves uniformly along the segment $ I_n $ in the direction indicated on the figure 4 by an arrow, its image $ F(P) $ moves uniformly along the segment $ I_{n+1} $. This means that the point $ F(P) $ divides the segment $ I_{n+1} $ in the same ratio in which $ P $ divides $ I_n $. From this and the observed cyclicity $ (I_9 = I_0) $, it follows that every point of the broken line $ L_a $, after applying the transformation $ F^9 $ (ninefold composition of the transformation $ F $), returns to its original position.
om36_1r_img_4.jpg
All broken lines $ L_a $ (for different values of $ a $) are similar sets on the plane, and their sum fills the entire plane except for the point $ (0,0) $, which is, of course, a fixed point of the transformation $ F $. Therefore, $ F^9 $ is the identity transformation.
From this, the thesis of the problem immediately follows: taking $ x $ and $ y $ as two consecutive terms of the sequence $ (a_n) $, we see that
and therefore $ F^9(a_n,a_{n+1}) = (a_{n+9}, a_{n+10}) $, and since $ F^9 $ is the identity, then $ a_{n+9} = a_n $. | proof | Algebra | proof | Yes | Yes | olympiads | false | 743 |
XX OM - I - Problem 11
In a convex quadrilateral $ABCD$, the sum of the distances from each vertex to the lines $AB$, $BC$, $CD$, $DA$ is the same. Prove that this quadrilateral is a parallelogram. | Let's denote the measures of the angles of a quadrilateral by $A$, $B$, $C$, $D$, and the lengths of the sides $AB$, $BC$, $CD$, $DA$ by $a$, $b$, $c$, $d$ respectively (Fig. 6).
The sum of the distances from vertex $A$ to the lines containing the sides of the quadrilateral is then $a \sin B + d \sin D$. Denoting this sum by $s$, we have
\[ a \sin B + d \sin D = s \]
and similarly
\[ b \sin C + a \sin A = s \]
\[ c \sin D + b \sin B = s \]
\[ d \sin A + c \sin C = s. \]
Solving the above system of equations for the sines of the angles of the quadrilateral, we obtain
\[ \sin A = \frac{s - c \sin D}{d} \]
\[ \sin B = \frac{s - d \sin A}{a} \]
\[ \sin C = \frac{s - a \sin B}{b} \]
\[ \sin D = \frac{s - b \sin C}{c}. \]
From the equalities (1)--(4), we get the equality
\[ \sin A + \sin C = \sin B + \sin D, \]
which we can write as
\[ \sin \frac{A+C}{2} \cos \frac{A-C}{2} = \sin \frac{B+D}{2} \cos \frac{B-D}{2}. \]
Since $A + B + C + D = 2\pi$, we have $\frac{B+D}{2} = \pi - \frac{A+C}{2}$, so $\sin \frac{A+C}{2} = \sin \frac{B+D}{2}$, and from equality (5) we get
\[ \cos \frac{A-C}{2} = \cos \frac{B-D}{2}. \]
Therefore,
\[ \frac{A-C}{2} = \pm \frac{B-D}{2} + 2k\pi, \quad k \in \mathbb{Z}. \]
Since $ \left| \frac{A-C}{2} \right| < \pi $ and $ \left| \frac{B-D}{2} \right| < \pi $, in equality (6) $k = 0$ holds, and one of the following cases occurs:
a) $\frac{A-C}{2} = \frac{B-D}{2}$; then $A + D = B + C$, so $A + D = \pi$, which implies that sides $AB$ and $DC$ are parallel and $\sin D = \sin A$. Therefore, from equalities (1) and (4) we get $a = c$.
b) $\frac{A-C}{2} = -\frac{B-D}{2}$; then $A + B = C + D$, so $A + B = \pi$, which implies that lines $AD$ and $BC$ are parallel and $\sin A = \sin B$. Therefore, from equalities (1) and (2) we get $b = d$.
We have thus shown that in the given quadrilateral, two opposite sides are parallel and equal, so the quadrilateral is a parallelogram. | proof | Geometry | proof | Yes | Yes | olympiads | false | 744 |
LII OM - I - Problem 5
Prove that for any natural number $ n \geq 2 $ and any prime number $ p $, the number $ n^{p^p} + p^p $ is composite. | If $ p $ is an odd prime, then by the identity
in which we set $ x = n^{p^{p-1}} $ and $ y = p $, the number $ n^{p^p} + p^p $ is composite. (For the above values of $ x $, $ y $, the inequalities $ x^p + y^p > x + y > 1 $ hold, so each of the factors on the right-hand side of formula (1) is greater than 1).
On the other hand, for $ p = 2 $ we get $ n^4 + 4 = (n^2 + 2n + 2)(n^2 - 2n + 2) $. Since $ n \geq 2 $, both factors are greater than $ 1 $. Hence, for $ n \geq 2 $ the number $ n^4 + 4 $ is composite. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 745 |
XXI OM - I - Problem 4
In a square $ABCD$ with side length 1, there is a convex quadrilateral with an area greater than $\frac{1}{2}$. Prove that this quadrilateral contains a segment of length $\frac{1}{2}$ parallel to $\overline{AB}$. | We draw through the vertices of a given quadrilateral lines parallel to line $AB$ (Fig. 4). In the case where these lines are different, the quadrilateral is divided into $3$ parts: a triangle with base $p$ and height $h_1$, a trapezoid with bases $p$ and $q$ and height $h_2$, and a triangle with base $q$ and height $h_3$. By assumption, the area of the quadrilateral $P$ is greater than $\frac{1}{2}$. On the other hand,
Therefore,
Hence,
Since $h_1 + h_2 \leq 1$ and $h_2 + h_3 \leq 1$, we obtain from this that $p + q \geq p(h_1 + h_2) + q(h_2 + h_3) > 1$.
Thus, at least one of the numbers $p$ and $q$ is greater than $\frac{1}{2}$.
In special positions of the given quadrilateral, the considered triangles or trapezoid may degenerate into a segment, i.e., it may be $h_1 = 0$ or $h_2 = 0$ or $h_3 = 0$. However, even in such a case, the above reasoning remains valid.
Remarks
1. This problem can be generalized to the case of any convex figures: If the area of a certain convex figure contained in square $ABCD$ with side $1$ is greater than $\frac{1}{2}$, then this figure contains a segment of length $\frac{1}{2}$ parallel to $\overline{AB}$.
2. Similarly, the theorem is also true for a non-convex quadrilateral; however, a similar theorem does not hold for a non-convex hexagon. | proof | Geometry | proof | Yes | Yes | olympiads | false | 746 |
XIII OM - I - Problem 4
Prove that the line symmetric to the median $CS$ of triangle $ABC$ with respect to the angle bisector of angle $C$ of this triangle divides the side $AB$ into segments proportional to the squares of the sides $AC$ and $BC$. | Let $ CM $ (Fig. 4) be the line symmetric to the median $ CS $ with respect to the angle bisector $ CD $ of angle $ C $ in triangle $ ABC $.
Since the areas of triangles with the same height are proportional to the bases of these triangles, therefore
From the equality $ \measuredangle ACD = \measuredangle DCB $ and $ \measuredangle MCD = \measuredangle DCS $, it follows that $ \measuredangle ACM = \measuredangle SCB $ and $ \measuredangle ACS = \measuredangle MCB $. The areas of triangles having one pair of equal angles are proportional to the products of the sides forming these angles, thus
From the equality (1) and (2), it follows that
Multiplying these equalities side by side and considering that $ SB = AS $, we obtain
which was to be proved. | proof | Geometry | proof | Yes | Yes | olympiads | false | 749 |
LIX OM - I - Task 7
In an $ n $-person association, there are $ 2n-1 $ committees (any non-empty set of association members
forms a committee). A chairperson must be selected in each committee. The following condition must be met: If
committee $ C $ is the union $ C = A\cup B $ of two committees $ A $ and $ B $, then the chairperson of committee $ C $ is also
the chairperson of at least one of the committees $ A $, $ B $. Determine the number of possible selections of chairpersons. | Answer: n!.
We will establish a one-to-one correspondence between the possible choices of committee chairs
and the ways of assigning members of the association the numbers $1, 2, \dots, n$ (where different members
are assigned different numbers). There are, of course, as many of the latter as there are permutations of an $n$-element set,
which is $n!$.
Suppose, then, that in each committee a chair has been chosen in accordance with the assumptions of the problem.
We assign the members of the association numbers inductively as follows:
The number 1 is assigned to the chair of the committee that includes all members of the association.
The number 2 is assigned to the chair of the committee that includes all members of the association except the member with number 1.
The number 3 is assigned to the chair of the committee that includes all members of the association except those with numbers 1 and 2, and so on.
We will show that with this method of assigning numbers, the following condition is satisfied:
(*) The chair of any committee is the one among its members who has been assigned the smallest number.
Indeed, consider any committee $A$ and let $k$ be the smallest of the numbers assigned to the members
of this committee. Let $B$ be the committee consisting of all members of the association who have been assigned numbers
not less than $k$. From the method of assigning numbers, it follows that the chair of committee $B$ is the member
with number $k$. Moreover, committee $B$ is the union of two committees: $A$ and $B \setminus A$. Since the member with number
$k$ does not belong to committee $B \setminus A$, according to the conditions of the problem, he must be the chair of committee $A$.
We have thus defined a mapping that assigns a way of assigning the members of the association the numbers $1, 2, \dots, n$ to a choice of committee chairs. To complete the solution of the problem, we will now define the inverse mapping.
Namely, if the members of the association have been assigned the numbers $1, 2, \dots, n$, then in each committee
we choose as chair the member who has the smallest number (condition (*) will then be satisfied).
If committee $C$ is the union $C = A \cup B$ of two committees $A, B$ and the member with number $k$ is the chair
of committee $C$, then he belongs to one of the committees $A, B$, in which he is the member with the lowest number — and thus is
its chair. We see, then, that the condition given in the problem statement is satisfied. Therefore, if
we start with the choice of committee chairs, assign numbers to the members of the association according to
the previously defined rule, and then reconstruct the committee chairs from these numbers, we
will obtain the original choice of chairs. Similarly, if we assign numbers to the members of the association, choose the committee chairs
based on these numbers, and then assign the members numbers according to the described inductive procedure, the
numbers will be identical to the original numbers. Thus, the defined mappings are mutually
inverse, which completes the solution. | n! | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 750 |
XL OM - I - Problem 7
In space, a finite set of points is given, any four of which are vertices of a tetrahedron with a volume less than or equal to 1. Prove that there exists a tetrahedron with a volume no greater than 27, containing all these points. | Let $ABCD$ be a tetrahedron of maximum volume among all tetrahedra whose vertices are points of the considered set. Denote by $O$ its centroid, that is, the point of intersection of four segments, each connecting a vertex of the tetrahedron with the centroid of the opposite face. It is known that the centroid of a tetrahedron divides each of these four segments in the ratio $3:1$.
Let $A'$ be the image of the tetrahedron $ABCD$ under a homothety with center $O$ and scale $-3$ ($A'$ - the image of point $A$ and so on). According to the remark made a moment ago, the centroid of each face of the tetrahedron is mapped to the opposite vertex in this homothety.
By the assumption of the problem, the volume of the tetrahedron $ABCD$ does not exceed $1$. Therefore, the volume of the tetrahedron $A'$ does not exceed $27$ (the ratio of the volumes of similar bodies equals the cube of the similarity ratio). We will show that the tetrahedron $A'$ contains all the considered points - it is thus the sought tetrahedron.
Suppose that one of these points - let's call it $P$ - lies outside the tetrahedron $A'$. Each tetrahedron is the intersection of four half-spaces obtained by dividing the entire space by the planes of the faces. If, therefore, the point $P$ does not belong to $A'$, it means that $P$ does not belong to one of these half-spaces; say, to the half-space determined by the plane $A'$. In other words, $P$ lies on the opposite side of the plane $A'$ from the point $O$. But the plane $A'$, parallel to the face $ABC$ of the tetrahedron $ABCD$, passes through its vertex $D$ (since, as noted above, the vertex $D$ is the image of the centroid of the face $ABC$ in the homothety defined at the beginning).
Hence, the distance from the point $P$ to the plane $ABC$ is greater than the distance from the point $D$ to this plane. Consequently, the volume of the tetrahedron $ABCP$ is greater than the volume of the tetrahedron $ABCD$. Yet, the tetrahedron $ABCD$ was defined as the one among all tetrahedra with vertices in the given points, whose volume is maximum.
The obtained contradiction proves that all points of the given set lie within the tetrahedron $A'$. The proof is complete. | proof | Geometry | proof | Yes | Yes | olympiads | false | 752 |
XVII OM - I - Problem 4
On a plane, a circle and a point $ M $ are given. Find points $ A $ and $ B $ on the circle such that the segment $ AB $ has a given length $ d $, and the angle $ AMB $ is equal to a given angle $ \alpha $. | Suppose $ A $ and $ B $ are points on a given circle satisfying the conditions $ AB = d $, $ \measuredangle AMB = \alpha $ (Fig. 5).
Let us choose any chord $ CD $ of length $ d $ in the given circle. Rotate the triangle $ AMB $ around the center $ O $ of the given circle by the angle $ HOK $, where $ H $ and $ K $ denote the midpoints of the chords $ AB $ and $ CD $. After the rotation, point $ A $ will be at point $ C $, and point $ B $ at point $ D $ or vice versa; point $ M $ will move to some point $ N $. Point $ N $ a) lies on the circle with center $ O $ and radius $ OM $, b) lies on such an arc of the circle whose endpoints are points $ C $ and $ D $ and which contains the inscribed angle $ \alpha $. Conversely, if we determine a point $ N $ satisfying the above conditions a) and b), then by rotating the triangle $ CND $ around $ O $ by the angle $ NOM $, we obtain the triangle $ AMB $, whose base is the desired segment. Since there are two arcs with endpoints $ C $, $ D $ containing the inscribed angle $ \alpha $, the number of solutions will be $ 4 $, $ 3 $, $ 2 $, $ 1 $, or $ 0 $, depending on the length of the segment $ OM $. | 4,3,2,1,或0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 753 |
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