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XI OM - III - Task 1
Prove that if $ n $ is an integer greater than $ 4 $, then $ 2^n $ is greater than $ n^2 $. | Since $ 2^5 > 5^2 $, the inequality $ 2^n > n^2 $ is true for $ n = 5 $. Suppose this inequality is true when $ n $ equals some integer $ k > 4 $, i.e., that $ 2^k > k^2 $. Then
the inequality $ 2^n > n^2 $ is also true when $ n = k + 1 $. By the principle of mathematical induction, the inequality is true for every in... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 1,292 |
L OM - I - Problem 9
Points $ D $, $ E $, $ F $ lie on the sides $ BC $, $ CA $, $ AB $ of triangle $ ABC $, respectively. The incircles of triangles $ AEF $, $ BFD $, $ CDE $ are tangent to the incircle of triangle $ DEF $. Prove that the lines $ AD $, $ BE $, $ CF $ intersect at a single point. | Let $ c_1 $ be the incircle of triangle $ DEF $, and $ c_2 $ be the incircle of triangle $ ABC $. Since the incircles of triangles $ AEF $ and $ DEF $ are tangent, we have $ AF + DE = AE + DF $. This means that a circle can be inscribed in quadrilateral $ AEDF $; let this circle be denoted by $ c_A $. Point $ D $ is th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,293 |
LII OM - II - Problem 4
Determine all natural numbers $ n \geq 3 $ for which the following statement is true:
In any $ n $-term arithmetic sequence $ a_1, a_2, \ldots, a_n $, if the number $ 1 \cdot a_1 + 2 \cdot a_2 + \ldots + n \cdot a_n $ is rational, there exists a term that is a rational number. | The considered statement is true only for numbers $ n \geq 3 $ that give a remainder of $ 1 $ when divided by $ 3 $.
Let $ a_i = a + ir $ ($ i = 1,2,\ldots,n $) be an arithmetic sequence. Using the formulas
we determine that the sum given in the problem is equal to
From this, we obtain
First, assume t... | n\geq3thatgiveremainderof1whendivided3 | Number Theory | proof | Yes | Yes | olympiads | false | 1,294 |
VI OM - III - Task 5
In the plane, a line $ m $ and points $ A $ and $ B $ lying on opposite sides of the line $ m $ are given. Find a point $ M $ on the line $ m $ such that the difference in distances from this point to points $ A $ and $ B $ is as large as possible. | Let $ B' $ be the point symmetric to point $ B $ with respect to line $ m $ (Fig. 18). If point $ P $ is any point on line $ m $, then
The difference $ |AP - BP| $ thus reaches its maximum value, equal to the length of segment $ AB $, when $ |AP - BP| $, i.e., when points $ A $, $ B $, and $ P $ lie on the same straig... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,295 |
XXIV OM - II - Problem 5
Prove that if in a tetrahedron $ABCD$ we have $AB = CD$, $AC = BD$, $AD = BC$, then all faces of the tetrahedron are acute triangles. | We will first prove the
Lemma. Let $ \alpha, \beta, \gamma $ be numbers in the interval $ (0; \pi) $. There exists a trihedral angle whose dihedral angles have measures equal to $ \alpha, \beta, \gamma $ if and only if each of the numbers $ \alpha, \beta, \gamma $ is less than the sum of the other two.
Proof. Without l... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,300 |
XVII OM - I - Problem 1
Present the polynomial $ x^5 + x + 1 $ as a product of two polynomials of lower degree with integer coefficients. | Suppose the desired factorization exists. None of the factors of this factorization can be of the first degree. For if for every $x$ the equality held
where the numbers $m, n, a, b, \ldots$ are integers, then it would follow that $ma = 1$, so $m = a = \pm 1$, hence the polynomial $x^5 + x + 1$ would have an integer... | x^5+x+1=(x^2+x+1)(x^3-x^2+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 1,301 |
XLVI OM - III - Zadanie 3
Dana jest liczba pierwsza $ p \geq 3 $. Określamy ciąg $ (a_n) $ wzorami
Wyznaczyć resztę z dzielenia liczby $ a_{p^3} $ przez $ p $.
|
Podany wzór rekurencyjny
możemy zastosować do każdego ze składników jego prawej strony (pod warunkiem, że numer $ n - p $ jest nie mniejszy od $ p $, czyli że $ n \geq 2p $); dostajemy zależność
Stosując ponownie wzór (1) do każdego ze składników ostatniej sumy otrzymujemy równość
(jeśli tylko $ n \g... | p-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 1,306 |
XXVIII - II - Task 3
In a hat, there are 7 slips of paper. On the $ n $-th slip, the number $ 2^n-1 $ is written ($ n = 1, 2, \ldots, 7 $). We draw slips randomly until the sum exceeds 124. What is the most likely value of this sum? | The sum of the numbers $2^0, 2^1, \ldots, 2^6$ is $127$. The sum of any five of these numbers does not exceed $2^2 + 2^3 + 2^4 + 2^5 + 2^6 = 124$. Therefore, we must draw at least six slips from the hat.
Each of the events where we draw six slips from the hat, and the seventh slip with the number $2^{n-1}$ ($n = 1, 2, ... | 127 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 1,307 |
XLIV OM - III - Problem 5
Determine all functions $ f: \mathbb{R} \to \mathbb{R} $ that satisfy the following conditions: | Assume that the function $ f: \mathbb{R} \to \mathrm{R} $ satisfies the given equations. Let $ t $ be a real number different from $ 0 $ and $ -1 $.
In the first equation, we substitute $ x = 1/(t + 1) $; we get the relation
In the second equation, we substitute $ x = 1/t $, $ x = -1/(t + 1) $, $ x = t $; we obta... | f()= | Algebra | proof | Yes | Yes | olympiads | false | 1,309 |
I OM - B - Task 12
How many sides does a polygon have if each of its angles is $ k $ times larger than the adjacent angle? What values can $ k $ take? | If each angle of a polygon is $ k $ times larger than the adjacent angle,
then the sum of all $ n $ angles of the polygon, which - as
we know - equals $ (n - 2) \cdot 180^{\circ} $, is also $ k $ times larger
than the sum of the corresponding adjacent angles (i.e., the exterior angles),
which means it is $ k $ times la... | \frac{1}{2},1,\frac{3}{2},2,\dots | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,310 |
IV OM - III - Task 3
Through each vertex of a tetrahedron of a given volume $ V $, a plane parallel to the opposite face of the tetrahedron has been drawn. Calculate the volume of the tetrahedron formed by these planes. | To make it easier to find a solution, let's first consider an analogous problem on the plane:
Let's draw through each vertex of the triangle $ABC$ a line parallel to the opposite side of the triangle. These lines form a triangle $A_1B_1C_1$ similar to triangle $ABC$ (Fig. 46). The center of similarity is the point $S$,... | 27V | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,311 |
XLII OM - I - Problem 8
Determine the largest natural number $ n $ for which there exist in space $ n+1 $ polyhedra $ W_0, W_1, \ldots, W_n $ with the following properties:
(1) $ W_0 $ is a convex polyhedron with a center of symmetry,
(2) each of the polyhedra $ W_i $ ($ i = 1,\ldots, n $) is obtained from $ W_0 $ by ... | Suppose that polyhedra $W_0, W_1, \ldots, W_n$ satisfy the given conditions. Polyhedron $W_1$ is the image of $W_0$ under a translation by a certain vector $\overrightarrow{\mathbf{v}}$ (condition (2)). Let $O_0$ be the center of symmetry of polyhedron $W_0$ (condition (1)); the point $O_1$, which is the image of $O_0$... | 26 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,312 |
XLVII OM - I - Problem 12
Determine whether there exist two congruent cubes with a common center such that each face of the first cube has a point in common with each face of the second cube. | We will prove that there do not exist two cubes $\mathcal{C}$ and $\mathcal{C}$ for which the following condition is satisfied:
$\quad (*) \quad$ each face of the cube $\mathcal{C}$ has points in common with each face of $\mathcal{C}$.
The additional assumptions given in the problem (congruence, common center) are not ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,314 |
XXXVI OM - II - Problem 4
Prove that if for natural numbers $ a, b $ the number $ \sqrt[3]{a} + \sqrt[3]{b} $ is rational, then $ a, b $ are cubes of natural numbers. | Let's denote the numbers $ \sqrt[3]{a} $ and $ \sqrt[3]{b} $ by $ x $ and $ y $, respectively, and their sum by $ s $. We need to prove that $ x $ and $ y $ are natural numbers. By assumption, the number $ s = x + y $ is rational. Since
\[
xy \text{ is a rational number.}
\]
Further, we have
\[
\text{from the ration... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,315 |
XIX OM - II - Problem 1
Prove that a polynomial in the variable $ x $ with integer coefficients, whose absolute value for three different integer values of $ x $ equals 1, does not have integer roots. | Suppose that $ f(x) $ is a polynomial in the variable $ x $ with integer coefficients and that
where $ a $, $ b $, $ c $ are three distinct integers.
Assume that $ x_0 $ is an integer root of the polynomial $ f(x) $, so for every $ x $
where $ \varphi (x) $ is a polynomial with integer coefficients.
From (1) and (2),... | proof | Algebra | proof | Yes | Yes | olympiads | false | 1,317 |
LIV OM - I - Task 5
A natural number $ n_1 $ is written in the decimal system using 333 digits, none of which are zero. For $ i = 1 , 2, 3, \ldots , 332 $, the number $ n_{i+1} $ is formed from the number $ n_i $ by moving the units digit to the beginning. Prove that either all the numbers $ n_1, n_2, n_3, \ldots , n_... | Let $ j_i $ $ (i = 1,2,3,\ldots ,333) $ denote the units digit of the number $ n_i $. Then for $ i =1,2,\ldots,332 $ we have
The number $ 10^{333} - 1 = (10^3 - 1) \cdot (10^{330} + 10^{327} + 10^{324} +\ldots + 10^3 + 1) $ is divisible by 333, and the numbers 10 and 333 are relatively prime. From the second equ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,320 |
XXII OM - I - Problem 3
Through a point $ P $, belonging to the plane of triangle $ ABC $, three lines perpendicular to the lines $ BC $, $ AC $, and the line containing the median $ \overline{CE} $ have been drawn. Prove that they intersect the line containing the altitude $ CD $ at points $ K, L, M $, respectively, ... | We will first prove a few auxiliary facts.
Lemma. If the corresponding sides of two triangles are parallel, then the triangles are similar.
Proof. Let $ AB\parallel A', $ BC \parallel B', $ CA \parallel C'. $ Then the corresponding angles of triangles $ ABC $ and $ A'B'C' $ have sides that are parallel. Therefore, the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,323 |
X OM - I - Problem 10
Prove that if a circle can be circumscribed around each of the quadrilaterals $ABCD$ and $CDEF$, and the lines $AB$, $CD$, $EF$ intersect at a single point $M$, then a circle can be circumscribed around the quadrilateral $ABEF$.
| We apply the theorem about secants of a circle passing through one point: 1° to lines $ AB $ and $ CD $, 2° to lines $ CD $ and $ EF $ (Fig. 16).
We obtain
From this last equality, it follows that points $ A $, $ B $, $ E $, $ F $ lie on a circle. Indeed, due to this equality, the proportion $ MA \colon ME ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,324 |
XI OM - I - Task 3
Each side of a triangle with a given area $ S $ is divided into three equal parts, and the points of division are connected by segments, skipping one point to form two triangles. Calculate the area of the hexagon that is the common part of these triangles. | The solution to the problem can be easily read from Fig. 1. The area of the hexagon $123456$ equals $\frac{2}{3}S$, as this hexagon is obtained from the triangle $ABC$ by cutting off three equal corner triangles with an area of $\frac{1}{9}S$. The sides of triangles $135$ and $246$ bisect each other into equal segments... | \frac{2}{9}S | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,326 |
V OM - I - Task 11
Given are two intersecting lines $ a $ and $ b $. Find the geometric locus of point $ M $ having the property that the distance between the orthogonal projections of point $ M $ on lines $ a $ and $ b $ is constant, equal to a given segment $ d $. | Let $A$ and $B$ denote the projections of point $M$ onto lines $a$ and $b$ intersecting at point $O$, and let $\alpha$ denote the convex angle $AOB$ (Fig. 27). The circle with diameter $OM$ passes through points $A$ and $B$, since angles $OAM$ and $OBM$ are right angles; this circle is thus the circumcircle of triangle... | The\geometric\locus\is\the\circle\with\center\O\\radius\\frac{}{\sin\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,327 |
LI OM - I - Problem 11
Given a positive integer $ n $ and a set $ M $, consisting of $ n^2 + 1 $ different positive integers and having the following property: among any $ n+1 $ numbers chosen from the set $ M $, there is a pair of numbers, one of which divides the other. Prove that in the set $ M $ there exist distin... | Let $ k \in M $. Denote by $ d(k) $ the largest natural number $ \ell $ for which there exist different numbers $ a_1 = k, a_2, a_3, \ldots, a_\ell $ satisfying the condition: for $ i = 1,2,\ldots,\ell-1 $, the number $ a_i $ is divisible by $ a_{i+1} $.
Suppose the thesis of the problem is not satisfied. This means th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 1,328 |
XX OM - I - Task 1
Prove that if the numbers $ a_1, a_2, \ldots, a_n $ and $ b_1, b_2, \ldots, b_n $ are positive, then | The left side $ L $ of inequality (1) is transformed as follows.
Since for any positive numbers $ x $, $ y $, the inequality holds
thus
which means
Note. The above solution can be presented in the following form. Both the left side $ L $ and the right side $ P $ of inequality (1) are equal to a ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 1,329 |
L OM - I - Problem 12
All vertices of a cube with edge $ a $ lie on the surface of a regular tetrahedron with edge 1. Determine the possible values of $ a $. | From the conditions of the problem, it follows that a given cube $ABCDA$ lies inside a given tetrahedron $KLMN$. There are two possible cases:
- There exists a face of the tetrahedron $KLMN$ (for example, $KLM$) on which at least three vertices of the cube lie;
- On each face of the tetrahedron $KLMN$, exactly two ver... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,330 |
X OM - I - Task 8
The sides of a triangle are the medians of another triangle. Calculate the ratio of the areas of both triangles. Can a triangle be constructed from the medians of any triangle? | Let $ S $ be the centroid of triangle $ ABC $. Extend the median $ AD $ by a segment $ DM = SD $ (Fig. 15). We obtain triangle $ BSM $, in which $ BS = \frac{2}{3} BE $, $ SM = \frac{2}{3} AD $, and $ BM = SC = \frac{2}{3} CF $ (since quadrilateral $ SBMC $ is a parallelogram). Triangle $ BSM $ is thus similar to trian... | \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,331 |
XXXIII OM - I - Problem 5
In a certain workplace, each employee is a member of exactly one of 100 trade unions. The employees are to elect a director from among two candidates. Members of each union agree on whether to abstain from voting or which of the two candidates they will vote for. Prove that there exists a uni... | Let $ a_1, a_2, \ldots, a_{100} $ be the number of members of each of the respective unions. We can assume that one of the numbers $ a_1, \ldots, a_{100} $ (for example, $ a_j $) is odd, because otherwise we could divide all $ a_i $ by the highest power of two that divides them all. If now the sum $ a_1+ \ldots+ a_{100... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 1,332 |
XXIV OM - III - Task 2
Let $ p_n $ be the probability that a series of 100 consecutive heads will appear in $ n $ coin tosses. Prove that the sequence of numbers $ p_n $ is convergent and calculate its limit. | The number of elementary events is equal to the number of $n$-element sequences with two values: heads and tails, i.e., the number $2^n$. A favorable event is a sequence containing 100 consecutive heads. We estimate the number of unfavorable events from above, i.e., the number of sequences not containing 100 consecutiv... | 1 | Combinatorics | proof | Yes | Yes | olympiads | false | 1,334 |
XXV - I - Problem 11
Let $ X_n $ and $ Y_n $ be independent random variables with the same distribution $ \left{ \left(\frac{k}{2^n}, \frac{1}{2^n}\right) : k = 0, 1, \ldots, 2^n-1\right} $. Denote by $ p_n $ the probability of the event that there exists a real number $ t $ satisfying the equation $ t^2 + X_n \cdot t... | A favorable event consists in the discriminant of the quadratic polynomial $ t^2 + X_n t + Y_n $ being a non-negative number. Let $ \displaystyle X_n = \frac{r}{2^n} $, $ \displaystyle Y_n = \frac{s}{2^n} $, where $ r, s = 0, 1, \ldots, 2^n - 1 $. The inequality $ X_n^2 - 4 Y_n \geq 0 $ is equivalent to the inequality ... | \frac{1}{12} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 1,336 |
XXXIX OM - I - Problem 4
A triangle of maximum perimeter is inscribed in an ellipse. Prove that at each vertex, the angles between the tangent to the ellipse and the sides emanating from that vertex are equal. | Let $ABC$ be a triangle of maximum perimeter inscribed in a given ellipse. Suppose the statement of the problem is not true and, for example, the line $l$ tangent to the ellipse at point $B$ forms different angles with the sides $AB$ and $BC$. Draw a line $m$ through point $B$ that forms equal angles with these sides (... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,337 |
XXXVIII OM - II - Task 1
From an urn containing one ball marked with the number 1, two balls marked with the number 2, ..., $ n $ balls marked with the number $ n $, we draw two balls without replacement. We assume that drawing each ball from the urn is equally probable. Calculate the probability that both drawn balls... | To make the content of the task meaningful, we need to assume that $ n \geq 2 $.
In total, there are $ N = 1+2+ ... +n = n(n+1)/2 $ balls in the urn. Two balls can be chosen from $ N $ balls in $ q $ ways, where
Let us fix $ k \in \{2, \ldots,n\} $. From the $ k $ balls marked with the number $ k $, we can choose a pa... | \frac{4}{3(n+2)} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1,338 |
XXXII - III - Task 2
The perpendicular bisectors of sides $ \overline{AB} $ and $ \overline{AC} $ of triangle $ ABC $ intersect the line containing side $ \overline{BC} $ at points $ X $ and $ Y $. Prove that $ BC = XY $ if and only if $ \tan B \cdot \tan C= 3 $ or $ \tan B\cdot \tan C=-1 $. | Let $ M $, $ N $, be the midpoints of sides $ \overline{AB} $ and $ \overline{AC} $ (Fig. 14).
(if $ B = \frac{\pi}{2} $ or $ C = \frac{\pi}{2} $, then one of the perpendicular bisectors does not intersect the line $ BC $, so this case can be disregarded). Therefore,
and the equality $ BC = XY $ holds... | \tanB\cdot\tanC=3or\tanB\cdot\tanC=-1 | Geometry | proof | Yes | Yes | olympiads | false | 1,339 |
XXIX OM - III - Problem 1
In a given convex angle on a plane, a light ray runs, reflecting from the sides of the angle according to the principle that the angle of incidence equals the angle of reflection. A ray that hits the vertex of the angle is absorbed. Prove that there exists a natural number $ n $ such that eac... | If a light ray emanating from point $C$ reflects off one of the sides of a given angle at point $P$, and then off the other side at point $Q$, then by performing a symmetry $\varphi$ with respect to the line $OP$, where $O$ is the vertex of the given angle, we observe that points $C$, $P$, and $\varphi(Q)$ are collinea... | [\frac{\pi}{\theta}]+1 | Geometry | proof | Yes | Yes | olympiads | false | 1,341 |
LVI OM - III - Problem 6
Prove that every convex polygon with an area of 1 contains a convex hexagon with an area of at least 3/4. | From the vertices of a given polygon $\mathcal{W}$, we select three $A$, $B$, and $C$ which are the vertices of a triangle with the largest area, equal to $s$. Let $X$ be a point of the polygon $\mathcal{W}$, lying on the opposite side of the line $BC$ from point $A$ and farthest from the line $BC$ (Fig. 4). We define ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,345 |
XVIII OM - I - Problem 3
In a convex quadrilateral $ABCD$, the midpoint $M$ of side $AB$ is connected to vertices $C$ and $D$, and the midpoint $N$ of side $CD$ is connected to vertices $A$ and $B$. Segments $AN$ and $DM$ intersect at point $P$, and segments $BN$ and $CM$ intersect at point $Q$. Prove that the area of... | The area of a polygon $ABC\ldots$ will be denoted by the symbol $ (ABC\ldots) $. We state that (Fig. 1)
Thus,
therefore,
But,
and from (1) and (2) it follows that,
Similarly,
Indeed,
From (3), (4), and (5) we obtain,
| proof | Geometry | proof | Yes | Yes | olympiads | false | 1,346 |
LVII OM - I - Problem 8
Tetrahedron $ABCD$ is circumscribed around a sphere with center $S$ and radius 1, such that $SA \geq SB \geq SC$. Prove that $SA > \sqrt{5}$. | We will first prove the following lemma.
Lemma
Triangle $ABC$ is contained in a circle of radius $R$. A circle of radius $r$ is inscribed in triangle $ABC$. Then $R \geq 2r$.
Proof
Let $D, E, F$ be the midpoints of sides $BC, CA, AB$ respectively (Fig. 3). Furthermore, let $S$ be the point of intersection of the median... | SA>\sqrt{5} | Geometry | proof | Yes | Yes | olympiads | false | 1,347 |
LX OM - I - Task 5
For each integer $ n \geqslant 1 $, determine the largest possible number of different subsets of the set $ \{1,2,3, \cdots,n\} $ with the following property: Any two of these subsets are either disjoint or one is contained in the other. | Answer: $ 2n $ subsets.
Let the number of sought subsets be denoted by $ a_n $.
It is not difficult to observe that for $ n =1,2,3,\cdots $ the following $ 2n $
subsets of the set $ \{1,2,3, \cdots,n\} $ have the property described in the problem statement:
Indeed, any of the first $ n+1 $ sets has at most one el... | 2n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1,348 |
I OM - B - Task 16
Form a rectangle from nine squares with sides 1, 4, 7, 8, 9, 10, 14, 15, and 18. | From the given squares, a rectangle can be formed in only one way. We will demonstrate this by conducting the following reasoning: First of all, it is clear that the squares filling the rectangle must have sides parallel to the sides of the rectangle. The area of the rectangle must equal the sum of the areas of all the... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 1,351 |
XLV OM - I - Problem 11
A triangle with perimeter $ 2p $ is inscribed in a circle with radius $ R $ and circumscribed around a circle with radius $ r $. Prove that $ p < 2(R + r) $. | Let $ABC$ be the considered triangle; we adopt the usual notations
assuming (without loss of generality) that $a$ is the largest angle of this triangle. Thus, $a \geq 60^\circ$. Let $I$ be the center of the inscribed circle of triangle $ABC$, and let $A'$, $B'$, $C'$ be the points of tangency of this circle with the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,353 |
LIII OM - III - Problem 2
On the sides $ AC $ and $ BC $ of the acute-angled triangle $ ABC $, rectangles $ ACPQ $ and $ BKLC $ of equal areas are constructed on its external side. Prove that the midpoint of segment $ PL $, point $ C $, and the center of the circumcircle of triangle $ ABC $ lie on one straight line. | Let $ O $ be the center of the circle circumscribed around triangle $ ABC $ (Fig. 1). Complete triangle $ PCL $ to parallelogram $ PCLX $. It suffices to prove that points $ X $, $ C $, $ O $ are collinear.
From the equality of the areas of the given rectangles, we obtain
Moreover, $ \measuredangle XLC = 180^\cir... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,356 |
XXXVI OM - I - Problem 1
Prove that if an integer divisible by 3 is the sum of five squares of integers, then at least two of these squares are divisible by 9. | The square of an integer not divisible by $3$ gives a remainder of $1$ when divided by $3$. Therefore, if a certain number is the sum of five squares and none of these squares is divisible by $3$, then the given number gives a remainder of $2$ when divided by $3$; if only one of these squares is divisible by $3$, then ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,357 |
XXII OM - I - Problem 7
On a plane, there are five lattice points (i.e., points with integer coordinates). Prove that the midpoint of one of the segments connecting these points is also a lattice point. | Let the lattice points be $ P_i = (x_i, y_i) $, where $ i = 1, 2, 3, 4, 5 $. Among the numbers $ x_1 $, $ x_2 $, $ x_3 $, $ x_4 $, $ x_5 $, there are at least three of the same parity (i.e., three even or three odd), because if, for example, at most two are even, then the remaining ones are odd. Let's assume, for examp... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,358 |
X OM - II - Task 2
Under what relationship between the sides of a triangle is it similar to the triangle formed by its medians? | Let's denote the medians drawn to the sides $a$, $b$, $c$ of triangle $ABC$ by $m_a$, $m_b$, $m_c$ respectively. We need to investigate in which triangles the medians are proportional to the sides of the triangle.
First, it is important to note that the larger of two unequal medians is drawn to the shorter side. To v... | ^2=2b^2-^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,359 |
XLVII OM - I - Problem 8
A light ray emanates from the center of a square, reflecting off the sides of the square according to the principle that the angle of incidence equals the angle of reflection. After some time, the ray returns to the center of the square. The ray never hit a vertex or passed through the center ... | On the plane with a rectangular coordinate system $Oxy$, we draw all lines with equations $x = c$ ($c$ integers) and $y = c$ ($c$ integers); in this way, the entire plane is covered by an infinite grid of unit squares. In each pair of squares sharing a common side, we identify points that are symmetric with respect to ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,361 |
XXIII OM - I - Problem 8
On a plane, there are two congruent equilateral triangles $ABC$ and $A'B'C'$. Prove that if the midpoints of segments $\overline{AA'},$ $\overline{BB'},$ $\overline{CC'}$ are not collinear, then they are the vertices of an equilateral triangle. | Let points $O$ and $O'$ be the centers of the circumcircles of triangles $ABC$ and $A'B'C'$, respectively. Denote by $T$ the translation of the plane by the vector $\overrightarrow{OO'}$, and let $A'$, $B'$, $C'$ be the images of $A$, $B$, $C$ under this translation. Denote by $A_1$, $B_1$, $C_1$ the midpoints of segme... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,363 |
VI OM - I - Task 3
Draw a line through the midpoint of one of the non-parallel sides of a trapezoid, which divides the trapezoid into two parts of equal area. | Let $ABCD$ be a trapezoid, in which the parallel sides are $AB$ and $DC$, and let $M$ be the midpoint of side $AD$ (Fig. 1). Notice that $\text{area of }ABM = \frac{1}{4} AB \cdot h$, $\text{area of }DCM = \frac{1}{4} DC \cdot h$, $\text{area of }ABCD = \frac{1}{2}(AB + CD) \cdot h$, where $h$ denotes the height of the... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,364 |
I OM - B - Task 8
Prove that the altitudes of an acute triangle are the angle bisectors of the triangle whose vertices are the feet of these altitudes. | Let $ S $ be the point of intersection of the altitudes $ AM $, $ BN $, and $ CP $ of an acute triangle $ ABC $ (Fig. 3).
om1_Br_img_3.jpg
The quadrilateral $ SMCN $, in which the angles at vertices $ M $ and $ N $ are right angles, is a cyclic quadrilateral with diameter $ SC $; therefore,
(angles subtended by t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,370 |
II OM - I - Task 4
Given are two intersecting planes $ A $ and $ B $ and a line $ m $ intersecting planes $ A $ and $ B $. Find the geometric locus of the midpoints of segments parallel to line $ m $, whose ends lie on planes $ A $ and $ B $. | Let $M$ and $N$ be the points of intersection of the line $m$ with the planes $A$ and $B$, $S$ the midpoint of the segment $MN$, and $s$ the line of intersection of the planes $A$ and $B$. We are to find the figure formed by the midpoints of all segments $XY$ parallel to the segment $MN$ and having their endpoints $X$ ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,371 |
XLI OM - III - Task 4
Inside a square with a side length of 1, a triangle was drawn, each side of which has a length of at least 1. Prove that the center of the square belongs to the triangle. | We apply proof by contradiction.
Suppose that the center $O$ of the given square $ABCD$ does not belong to the drawn triangle $A$. There then exists a line $l$ passing through $O$ such that triangle $A$ lies entirely on one side of this line (i.e., it is contained in one of the two open half-planes determined by the li... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,372 |
XLII OM - I - Problem 12
For any natural number $ n $, consider the rectangular prism $ K_n $ with edge lengths $ 1,1,n $ and the set $ R_n $ consisting of $ 4n+1 $ points: the vertices of the rectangular prism $ K_n $ and the points dividing its longer edges into unit segments. We randomly select three different poin... | Let $ Q(n) $ denote the number of all three-element subsets of the set $ R_n $, $ T(n) $ - the number of non-degenerate triangles with vertices in the set $ R_n $, and $ P(n) $ - the number of obtuse triangles. The probability of interest is the fraction $ p_n = P(n)/Q(n) $. Notice that
(Justification of the second ... | \frac{15}{16} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1,373 |
XVI OM - I - Problem 12
In space, there are two polyhedra. Prove that the longest of the segments connecting points of one polyhedron with points of the other has its ends at the vertices of the polyhedra. | First, let us note that if all vertices of a polyhedron lie on one side of a certain plane, then all other points of the polyhedron lie on that same side of the plane.
Suppose, for instance, that all vertices of the polyhedron $ W $ are located in an open half-space (i.e., a half-space without the points of the boundin... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,375 |
XXVI - II - Task 1
Given the polynomial $ W(x) = x^4 + ax^3 + bx + cx + d $. Prove that for the equation $ W(x) = 0 $ to have four real roots, it is necessary and sufficient for there to exist an $ m $ such that $ W(x+m) = x^4+px^2+q $, that the sum of some two roots of the equation $ W(x) = 0 $ equals the sum of the ... | If the roots of the polynomial $W(x)$ are the numbers $x_1, x_2, x_3, x_4$ and
then by taking $ \displaystyle m = \frac{1}{2} (x_1 + x_2) $, we obtain that the roots of the polynomial $W(x+m)$ are the numbers $ y_i = x_i - m $ for $ i = 1, 2, 3, 4 $, i.e., the numbers
Thus,
It follows from this that
We have thus sh... | proof | Algebra | proof | Yes | Yes | olympiads | false | 1,378 |
XVII OM - II - Problem 3
In the plane, 6 points are chosen, no three of which lie on the same line, and all segments connecting these points in pairs are drawn. Some of these segments are drawn in red, and others in blue. Prove that some three of the given points are vertices of a triangle with sides of the same color... | Let $ A_1 $, $ A_2 $, $ A_3 $, $ A_4 $, $ A_5 $, $ A_6 $ be given points. Among the five segments $ A_1A_i $ ($ i = 2, 3, 4, 5, 6 $) at least three are of the same color. Suppose, for example, that the segments $ A_1A_2 $, $ A_1A_3 $, $ A_1A_4 $ are red and consider the segments $ A_2A_3 $, $ A_3A_4 $, $ A_4A_2 $. If t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 1,379 |
Points $ A $, $ B $, $ C $, $ D $ lie, in that exact order, on circle $ o $. Point $ S $ lies inside circle $ o $ and satisfies the conditions
The line containing the bisector of angle $ ASB $ intersects circle $ o $ at points $ P $ and $ Q $. Prove that $ PS = QS $. | Let us assume that the lines $AS$, $BS$, $CS$ intersect the circle $o$ for the second time at points $E$, $F$, $G$ (Fig. 1). From the given equalities in the problem, it follows that triangles $ASD$ and $CSB$ are similar, and thus $\measuredangle ASC = \measuredangle DSB$. These equalities prove that triangles $ASC$ an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,380 |
LX OM - I - Task 9
Given is a $2008 \times 2008$ board. Two players take turns
making moves, each of which consists of choosing a white or
black pawn and placing it on a selected free field. The player wins
whose move results in a sequence of 5 consecutive
pawns of the same color in a vertical, horizontal, or diagonal... | Answer: It does not exist.
We will indicate a strategy for the player who does not start
(called the second player in the further part of the solution), which
will prevent the starting player (the first player) from winning.
This strategy can be described as follows: If the second player can
make a move leading to the ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1,382 |
XIII OM - II - Task 6
Find a three-digit number with the property that the number represented by these digits and in the same order, but in a different base of numeration than $ 10 $, is twice as large as the given number. | If $ x $, $ y $, $ z $ denote the consecutive digits of the sought number, and $ c $ - the new base of numeration, then
the equation can be replaced by the equation
The task is to solve equation (1) in integers $ x $, $ y $, $ z $, $ c $, satisfying the conditions $ 1 \leq x \leq 9 $, $ 0 \leq y \leq 9 $, $ 0 \leq z ... | 145,150,295 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 1,384 |
XXXII - II - Task 4
Given are natural numbers $ k, n $. We define inductively two sequences of numbers $ (a_j) $ and $ (r_j) $ as follows:
First step: divide $ k $ by $ n $ and obtain the quotient $ a_1 $ and the remainder $ r_1 $,
j-th step: divide $ k + r_{j-1} $ by $ n $ and obtain the quotient $ a_j $ and the rema... | According to the definition
Adding these equalities we get
thus
The number $ r_n $ as the remainder of the division of $ k+r_{n-1} $ by $ n $ satisfies the condition $ 0 \leq r_n < n $, but from the last equality, it follows that $ r_n $ is a number divisible by $ n $. Therefore, $ r_n = 0 $ and
... | k | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 1,386 |
XLIII OM - II - Problem 3
Through the centroid of an acute triangle $ABC$, lines perpendicular to the sides $BC$, $CA$, $AB$ are drawn, intersecting them at points $P$, $Q$, $R$ respectively. Prove that if $|BP| \cdot |CQ| \cdot |AR| = |PC| \cdot |QA| \cdot |RB|$, then triangle $ABC$ is isosceles.
Note: According to C... | om43_2r_img_7.jpg
Figure 7 shows an "arbitrary" acute triangle $ABC$; it is not isosceles, and thus (if the thesis of the problem is to be valid) - this figure is not an illustration of the problem. Nevertheless, or rather precisely because of this, such a figure is "safe"; it does not suggest using the equality of ang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,388 |
II OM - I - Task 11
In a given square, inscribe a square such that one of its sides or its extension passes through a given point $ K $. | A square inscribed in a square $ABCD$ is defined as any square whose vertices lie on the perimeter of the square $ABCD$. According to this definition, the square $ABCD$ itself is also considered an inscribed square.
When a given point $K$ lies on the perimeter of the square $ABCD$, the problem is solved immediately. Sp... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,390 |
XXV - I - Problem 7
Prove that if in an acute triangle each altitude can be moved inside the triangle by a rigid motion so that its foot takes the place of the vertex, and the vertex takes the place of the foot, then the triangle is equilateral. | If $ h $ is the length of a certain altitude of triangle $ ABC $, then from the conditions of the problem, it follows that in triangle $ ABC $, a segment of length $ h $ can be placed having any predetermined direction. If triangle $ ABC $ is not equilateral, then the lengths of some of its altitudes are different, e.g... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,392 |
XXXIX OM - I - Problem 8
For a given cube with edge length 1, find the set of segments with a total length not exceeding $1 + 3\sqrt{3}$, having the property that any two vertices of the cube are the endpoints of some broken line composed of segments from this set. | We will provide an example of a set of segments with the required property. This example can be considered the spatial counterpart of the "optimal road network connecting the vertices of a rectangle," constructed in preparatory task $C$. (A hint for the construction is the observation that it is purposeful to search fo... | 1+3\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,393 |
V OM - I - Problem 8
Given are the mutual distances of four points $ A $, $ B $, $ C $, $ D $ in space. Calculate the length of the segment connecting the midpoint of segment $ AB $ with the midpoint of segment $ CD $. | Assume that points $A$, $B$, $C$, $D$ do not lie on the same plane. The midpoints $M$, $N$, $P$, $Q$, $R$, $S$ of segments $AB$, $BC$, $AC$, $AD$, $BD$, $CD$ (Fig. 24) are then distinct points. By the theorem on the segment connecting the midpoints of two sides of a triangle, we have $MR \parallel AD \parallel PS$ and ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,395 |
XXXVIII OM - I - Problem 9
In a convex pentagon $ KLMNP $, we draw diagonals which, intersecting, determine disjoint segments: on diagonal $ KM $ in the order from $ K $ to $ M $ segments of lengths $ a_1, b_1, c_1 $; similarly $ a_2, b_2, c_2 $ on diagonal $ LN $, $ a_3, b_3, c_3 $ on diagonal $ MP $, $ a_4, b_4, c_4... | We change the notation of vertices $ L $, $ M $, $ N $, $ P $, $ K $ to $ A_1 $, $ A_2 $, $ A_3 $, $ A_4 $, $ A_5 $, respectively. We introduce further notations (Figure 9):
for $ i = 1, 2, 3, 4, 5 $; the addition $ i\pm 1 $, $ i\pm 2 $ in the indices should be understood modulo $ 5 $.
om38_1r_img_9.jpg... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,396 |
V OM - II - Task 3
Given: point $ A $, line $ p $, and circle $ k $. Construct triangle $ ABC $ with angles $ A = 60^\circ $, $ B = 90^\circ $, where vertex $ B $ lies on line $ p $, and vertex $ C $ lies on circle $ k $. | (method of geometric loci).
The task boils down to finding a point $ C $ that satisfies two conditions:
1° Point $ C $ lies on a given circle $ k $;
2° point $ C $ is a vertex of triangle $ ABC $ with a given vertex $ A $, whose vertex $ B $ lies on a given line $ p $ and in which $ \measuredangle A = 60^\circ $, $ \me... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,397 |
LIII OM - II - Problem 6
Determine all natural numbers \( n \) such that for any real numbers \( x_{1}, x_{2}, \ldots, x_{n}, \) \( y_{1}, y_{2}, \ldots, y_{n} \) the following inequality holds
保留了源文本的换行和格式，但请注意，最后一句“保留了源文本的换行和格式”是额外的说明，不是翻译的一部分。正确的翻译结果如下：
LIII OM - II - Problem 6
Determine all natural n... | For $ n = 1 $, the given inequality takes the form
which is not true for arbitrary real numbers $ x $, $ y $ - it suffices to take $ x_1 = y_1 = 1 $.
We will show that inequality (1) is true if $ n \geq 2 $.
If one of the factors on the right-hand side of inequality (1), for example,
is equal to $ 0 $, then bot... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 1,399 |
XXII OM - III - Problem 5
Find the largest integer $ A $ such that for every permutation of the set of natural numbers not greater than 100, the sum of some 10 consecutive terms is at least $ A $. | The sum of all natural numbers not greater than $100$ is equal to $1 + 2 + \ldots + 100 = \frac{1 + 100}{2} \cdot 100 = 5050$. If $a_1, a_2, \ldots, a_{100}$ is some permutation of the set of natural numbers not greater than $100$ and the sum of any $10$ terms of this permutation is less than some number $B$, then in p... | 505 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1,400 |
L OM - II - Task 4
Point $ P $ lies inside triangle $ ABC $ and satisfies the conditions: $ \measuredangle PAB = \measuredangle PCA $ and $ \measuredangle PAC = \measuredangle PBA $. Point $ O $ is the center of the circumcircle of triangle $ ABC $. Prove that if $ O \ne P $, then angle $ APO $ is a right angle. | Let $ K $, $ L $, $ M $ be the points of intersection of the lines $ AP $, $ BP $, $ CP $ with the circumcircle of triangle $ ABC $. By the equality $ \measuredangle BAK = \measuredangle ACM $, the lengths of the arcs $ BK $ and $ AM $ are equal. Similarly, the lengths of the arcs $ KC $ and $ LA $ are equal. The segme... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,401 |
XXXIII OM - I - Problem 9
In a chessboard created by dividing a square of side length $ n $ into unit squares using lines parallel to the sides of the square, we consider all squares whose sides are contained in the lines forming the chessboard. Let $ 1 \leq k \leq n $ and $ P(k,n) $ denote the number of these squares... | On the side of the chessboard, there are $ n - k + 1 $ segments of length $ k $ with endpoints at the division points. Hence, the total number of squares of side length $ k $ contained in the lines forming the chessboard is $ (n - k + 1)^2 $. Therefore, we have the following formulas:
The number $ k(n) $ satisfies the... | \frac{\sqrt[3]{4}}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1,402 |
XLVIII OM - III - Problem 5
Given is a convex pentagon $ABCDE$, where $DC = DE$ and $\measuredangle DCB = \measuredangle DEA = 90^\circ$. Let $F$ be the point on side $AB$ determined by the condition $AF: BF = AE: BC$. Prove that $\measuredangle FCE = \measuredangle ADE$ and $\measuredangle FEC = \measuredangle BDC$. | On the extension of segment $ BC $, we lay down segment $ CP = AE $. From the conditions of the problem, it follows that the right triangles $ PCD $ and $ AED $ are congruent. Therefore, $ AD = DP $ and
Since also $ ED = DC $, triangles $ ADP $ and $ EDC $ are similar. Denoting by $ Q $ the intersection point of lin... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,403 |
IV OM - I - Problem 6
Prove that if a plane figure has two and only two axes of symmetry, then these axes are perpendicular. | The theorem will be proven when we show that if a plane figure has two axes of symmetry that are not perpendicular, then it has at least one more axis of symmetry.
Suppose that the plane figure $F$ has two axes of symmetry $k$ and $l$, which are not perpendicular. They can intersect (Fig. 26) or be parallel (Fig. 27). ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,405 |
III OM - II - Task 3
Are the statements true:
a) if four vertices of a rectangle lie on the four sides of a rhombus, then the sides of the rectangle are parallel to the diagonals of the rhombus;
b) if four vertices of a square lie on the four sides of a rhombus, which is not a square, then the sides of the square are ... | a) The first theorem is not true; to demonstrate this, it suffices to provide a counterexample, i.e., to show a figure contradicting the theorem.
From the center $ O $ of the rhombus $ ABCD $ (Fig. 28), draw a circle with a radius greater than the distance from the center of the rhombus to its side, but smaller than ha... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,406 |
LVIII OM - I - Problem 3
In a convex quadrilateral $ABCD$, which is not a parallelogram, the equality $AB=CD$ holds. Points $M$ and $N$ are the midpoints of diagonals $AC$ and $BD$, respectively. Prove that the orthogonal projections of segments $AB$ and $CD$ onto line $MN$ are segments of equal length, equal to the l... | Let point $E$ be the midpoint of side $AD$, and point $F$ be the midpoint of segment $MN$ (Fig. 1).
om58_1r_img_1.jpg
From the converse of Thales' theorem, it follows that $EM \parallel DC$ and $EM = \frac{1}{2}DC$. Similarly, we obtain the relationships $EN \parallel AB$ and $EN = \frac{1}{2}AB$. From the equality $DC... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,407 |
XXVIII - I - Problem 10
Prove that in every tetrahedron there exists a vertex at which all three dihedral angles are acute. | The sum of the measures of the plane angles of any face of a tetrahedron is equal to $ \pi $. Therefore, the sum of the measures of all the plane angles of the tetrahedron is $ 4\pi $. It follows that there exists a vertex $ A $ of the tetrahedron such that the sum of the measures of the plane angles at this vertex doe... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,408 |
XXXI - I - Problem 9
Prove that a natural number $ n > 1 $ divides the number $ a^{(n-1)! + 1} - a $ for every integer $ a $ if and only if $ n $ is a product of distinct prime numbers. | Integers that are not divisible by the square of any prime number are called square-free numbers. First, we will show that the condition for \( n \) to be a square-free number is necessary for \( n \) to divide \( a^{(n-1)!+1} - a \) for every integer \( a \). Suppose \( n \) is divisible by the square of a prime numbe... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,409 |
XV OM - II - Task 3
Prove that if three prime numbers form an arithmetic progression with a difference not divisible by 6, then the smallest of these numbers is $3$. | Suppose that the prime numbers $ p_1 $, $ p_2 $, $ p_3 $ form an arithmetic progression with a difference $ r > 0 $ not divisible by $ 6 $, and the smallest of them is $ p_1 $. Then
Therefore, $ p_1 \geq 3 $, for if $ p_1 = 2 $, the number $ p_3 $ would be an even number greater than $ 2 $, and thus would not be a pri... | 3 | Number Theory | proof | Yes | Yes | olympiads | false | 1,411 |
XL OM - II - Task 5
Given is a sequence $ (c_n) $ of natural numbers defined recursively: $ c_1 = 2 $, $ c_{n+1} = \left[ \frac{3}{2}c_n\right] $. Prove that there are infinitely many even numbers and infinitely many odd numbers among the terms of this sequence. | It is enough to prove the following two statements:
(a) For any even number $ p $ being a term of the sequence $ (c_n) $, there exists a larger odd number that is a term of this sequence.
(b) For any odd number $ q $ being a term of the sequence $ (c_n) $, there exists a larger even number that is a term of this sequen... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,416 |
XVIII OM - III - Problem 5
Prove that if a polygon with an odd number of sides inscribed in a circle has all angles equal, then the polygon is regular. | Assume that a polygon $ W $ with consecutive vertices $ A_1, A_2, \ldots, A_n $, where $ n $ is an odd number, has all angles equal and is inscribed in a circle $ O(r) $. When $ n = 3 $, the thesis of the theorem is obvious, so we will assume that $ n \geq 5 $.
Let $ A_{i-1}, A_i, A_{i+1} $ be three consecutive vertice... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,417 |
I OM - B - Task 19
The centers of four identical spheres lie on the circumference of a circle, and the center of mass of these spheres is at the center of the circle. Show that the centers of the spheres are the vertices of a rectangle. | Let $A, B, C, D$ denote the consecutive vertices of a quadrilateral formed by the centers of the spheres, and $M$ and $N$ - the midpoints of two opposite sides $AB$ and $CD$ of this quadrilateral.
Replace the spheres with centers $A$ and $B$ with a sphere twice as heavy with center $M$, and the spheres with centers $C$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,418 |
XLVIII OM - I - Zadanie 11
Dana jest liczba naturalna $ m\geq 1 $ oraz wielomian $ P(x) $ stopnia dodatniego o współczynnikach całkowitych mający co najmniej trzy różne pierwiastki całkowite. Dowieść, że wielomian $ P(x) + 5^m $ ma co najwyżej jeden pierwiastek całkowity.
|
Załóżmy, że liczby całkowite $ a $, $ b $, $ c $ są różnymi pierwiastkami wielomianu $ P(x) $. Przypuśćmy, wbrew dowodzonej tezie, że wielomian $ P(x) + 5^m $ ma dwa różne pierwiastki całkowite $ u $ i $ v $. Można przyjąć, że $ u>v $.
Dla każdej pary różnych liczb całkowitych $ x_1 $, $ x_2 $ liczba $ P(x_1)-P(x_2... | proof | Algebra | proof | Yes | Yes | olympiads | false | 1,421 |
XXI OM - II - Task 2
On the sides of a regular $ n $-gon, $ n + 1 $ points have been chosen, dividing the perimeter into equal parts. For which arrangement of the chosen points is the area of the convex polygon with these $ n + 1 $ vertices a) the largest, b) the smallest? | Let's denote by $a$ the $(n + 1)$-th part of the side length of the regular $n$-gon $A_1A_2 \ldots A_n$ (Fig. 10), i.e., the side length of this $n$-gon is $(n + 1)a$. Therefore, the perimeter of the $n$-gon is $n(n + 1)a$. Since $na < (n + 1)a < 2na$, at least one and at most two vertices of the $(n + 1)$-gon belong t... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,425 |
XXXIX OM - III - Problem 6
Calculate the maximum volume of a tetrahedron contained within a hemisphere of radius 1. | Let us denote the given hemisphere by $H$, the whole sphere by $K$, and the circle forming the flat part of the boundary of $H$ by $E$. Let $O$ be any tetrahedron contained in $H$. Its interior is the intersection of four open half-spaces determined by the planes of the faces. The center of the sphere $K$ does not belo... | \frac{\sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,427 |
XXXVII OM - I - Problem 9
Prove that every natural number less than $ n! $ is the sum of at most $ n $ different natural numbers dividing $ n! $. | We prove by induction on $ n $. For $ n = 1 $, the theorem is satisfied "in a vacuum" (there are no natural numbers less than $ 1! $). Fix a natural number $ n $ and assume that the thesis of the theorem is true for it. We will prove the thesis for $ n + 1 $.
Take any natural number $ N < (n + 1)! $. We need to show th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,428 |
XXIV OM - I - Problem 3
Let $ w(x) $ be a polynomial with integer coefficients. Prove that if a natural number $ d $ is a divisor of each of the numbers $ a_n = 3^n + w(n) $, where $ n = 0, 1, 2, \ldots $, then $ d $ is a power of 2 with an integer exponent. | Let the prime number $ p $ be a divisor of each of the numbers $ a_n $, where $ n = 0, 1, 2, \ldots $. It suffices to prove that $ p = 2 $.
In particular, we have $ p \mid a_0 $ and $ p \mid a_p $. Hence . Let $ w(x) = c_0 + c_1x + \ldots + c_rx^r $, where the numbers $ c_0, c_1, \ldots, c_r $ are integers. The number ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,431 |
XXXVIII OM - I - Problem 6
Which of the quadrilaterals having three sides of length 1 has the largest area? | We will use formula (1)
where
Formula (1) is valid for any quadrilateral, not necessarily convex; a proof can be found, for example, in S. Straszewicz's {\em Collection of Problems from Mathematical Olympiads} (PZWS 1956) in the commentary to the solution of problem 101 (p. 227).
Let us assume, in acc... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,432 |
VI OM - III - Task 2
Prove that among seven natural numbers forming an arithmetic progression with a difference of $30$, one and only one is divisible by $7$. | Seven numbers forming an arithmetic progression with a difference of $30$ can be written in the form: $a, a + 30, a + 2 \cdot 30, \ldots, a + 6 \cdot 30$. The difference between any two of these numbers has the form
\[
(k - m) \cdot 30
\]
where $k$ and $m$ are integers, with $0 \leq k \leq 6$, $0 \leq m \leq 6$, $k \... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,433 |
VII OM - I - Problem 10
Prove that if $ \alpha > 0 $, $ \beta > 0 $, $ \gamma > 0 $ and $ \alpha + \beta + \gamma = \frac{1}{2} \pi $, then | From the assumption, it follows that $ \alpha $, $ \beta $, $ \gamma $, and $ \alpha + \beta = \frac{1}{2} \pi - \gamma $ are acute angles. Therefore,
hence,
and finally,
Note. The theorem can be generalized. For formula (1) to be true, it suffices to assume that each of the angles $ \alpha $, $ \beta... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,435 |
XXXIV OM - II - Problem 1
In a plane with a given coordinate system, there is a convex polygon whose all vertices have integer coordinates. Prove that twice the area of this polygon is an integer. | We choose any vertex of the polygon and draw all diagonals from it. In this way, we divide the given polygon into triangles, whose vertices have integer coordinates. It is therefore sufficient to prove the considered theorem for a triangle.
om34_2r_img_5.jpg
We draw lines through the vertices of the triangle parallel t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,436 |
XXXV OM - I - Problem 4
Given are two intersecting circles $ o(O_1, r_1) $ and $ o(O_2, r_2) $ and a line tangent to them at points $ A $ and $ B $, respectively. Point $ C $ is the common point of these circles closer to the line $ AB $, $ R $ is the radius of the circumcircle of triangle $ ABC $, and $ d = O_1O_2 $.... | From the isosceles triangle $ AC0_1 $ we determine
Since $ \measuredangle O_1AB = 90^\circ $, then $ \cos O_1AC = \sin (90^\circ - O_1AC) = \sin BAC $. Applying the Law of Sines to triangle $ ABC $, we get
thus
Similarly,
From the last two equalities, it follows that
Calculating the sum of... | ^2=r_1^2+r_2^2+R^2 | Geometry | proof | Yes | Yes | olympiads | false | 1,438 |
XI OM - III - Task 5
From the digits $ 1 $, $ 2 $, $ 3 $, $ 4 $, $ 5 $, $ 6 $, $ 7 $, $ 8 $, $ 9 $, all possible four-digit numbers with different digits were formed. Find the sum of these numbers. | We will calculate how many four-digit numbers with different digits can be formed from the digits $1$ to $9$. The first digit $c_1$ can be chosen in $9$ ways; when $c_1$ is already set, the second digit $c_2$ can be chosen in $8$ ways; when $c_1$ and $c_2$ are set, there are $7$ possibilities for $c_3$; and when $c_1$,... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1,441 |
XXII OM - III - Problem 2
On a billiard table in the shape of a triangle, whose angles are proportional, a ball was struck from a certain internal point. The ball bounces off the walls according to the law "angle of incidence equals angle of reflection." Prove that the number of directions in which the ball can move i... | We will first prove the following
Lemma. After a ball is reflected twice in succession from the sides $\overline{CA}$ and $\overline{AB}$ of triangle $ABC$ (Fig. 14), the direction of the ball's movement changes by an angle of $2 \measuredangle BAC$.
Proof. Consider the image $AB'$ of triangle $ABC$ in the symmetry wit... | 2n | Geometry | proof | Yes | Yes | olympiads | false | 1,442 |
XLV OM - II - Task 4
Each vertex of a cube is assigned the number $ 1 $ or $ -1 $, and each face is assigned the product of the numbers assigned to the vertices of that face. Determine the set of values that the sum of the $ 14 $ numbers assigned to the faces and vertices can take. | Let's consider an arbitrary assignment of the numbers $1$ and $-1$ to the vertices of a cube; denote the sum of these numbers by $S_1$. Assign numbers to the faces according to a given rule; denote the sum of these numbers by $S_2$. The sum of all $14$ considered numbers is $S_1 + S_2$.
If we change the number at one v... | {14,10,6,2,-2,-6,-10,-14}\setminus{10,-14}={14,6,2,-2,-6,-10} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1,444 |
XLVI OM - I - Zadanie 1
Wyznaczyć wszystkie pary $ (x,y) $ liczb naturalnych, dla których liczby $ \frac{x+1}{y} $ oraz $ \frac{y+1}{x} $ są naturalne.
|
Niech $ (x,y) $ będzie jedną z szukanych par. Liczby naturalne $ y \geq 1 $ oraz $ x \geq 1 $ są odpowiednio dzielnikami liczb $ x + 1 $ oraz $ y + 1 $. Zatem iloczyn $ xy $ jest dzielnikiem iloczynu $ (x + 1)(y+ 1) $, równego sumie $ xy + x + y + 1 $. Jest więc też dzielnikiem sumy $ x + y + 1 $, a zatem zachodzi nie... | (3,2),(2,3),(1,1),(2,1),(1,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 1,445 |
XXX OM - III - Task 3
The experiment consists of performing $ n $ independent trials. The probability of a positive outcome in the $ i $-th trial is $ p_i $. Let $ r_k $ be the probability that exactly $ k $ trials yield a positive result. Prove that | Let $ \{1,2, \ldots,n\} = A_k \cup B_k $, where $ A_k = \{i_1, i_2, \ldots, i_k \} $ and $ B_k = \{i_{k+1}, i_{k+2}, \ldots, i_n \} $, be a partition of the set $ \{1, 2, \ldots, n \} $ into a $ k $-element set and a $ (n-k) $-element set.
The probability of the event that the trials numbered $ i_1, i_2, \ldots,i_k $ y... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 1,447 |
XVIII OM - I - Zadanie 10
Trójka liczb całkowitych tworzy postęp geometryczny o ilorazie całkowitym. Gdy najmniejszą z nich zwiększymy o 9, powstanie postęp arytmetyczny. Jakie to liczby?
|
Poszukiwana trójka liczb ma postać $ (a, aq, aq^2) $, gdzie $ a $ i $ q $ są liczbami całkowitymi różnymi od zera. Rozróżnimy dwa przypadki.
a) Najmniejszą z liczb tej trójki jest $ a $ lub $ aq^2 $. Nową trójką liczb jest wtedy $ (a+9, aq, aq^2) $ lub $ (a, aq, aq^2+9) $; warunek, aby to był postęp arytmetyczny, wyra... | (-1,-4,-16),(-1,2,-4),(-9,-18,-36),(2,-4,8) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 1,448 |
LV OM - I - Task 11
Point $ O $ is the center of the circle circumscribed around the isosceles trapezoid $ ABCD $ with bases $ AB $ and $ CD $. Points $ K $, $ L $, $ M $, $ N $ lie on sides $ AB $, $ BC $, $ CD $, $ DA $, respectively, and quadrilateral KLMN is a rhombus. Prove that point $ O $ lies on the line $ KM ... | Let $ P $, $ Q $, $ R $ be the midpoints of segments $ AD $, $ BC $, $ KM $ (Fig. 5). Then points $ P $, $ Q $, and $ R $ lie on a single straight line - parallel to the bases of trapezoid $ ABCD $. Draw a line through point $ L $ parallel to the bases of trapezoid $ ABCD $, intersecting segment $ AD $ at point $ S $.
... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,449 |
XVII OM - II - Problem 4
Prove that if natural numbers $ a $ and $ b $ satisfy the equation $ a^2 + a = 3b^2 $, then the number $ a + 1 $ is a square of an integer. | Let the equality $ b = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_s^{\alpha_s} $ represent the prime factorization of the number $ b $. According to the assumption, the following equality holds:
Numbers $ a $ and $ a+1 $ are coprime, since any common divisor of these numbers is a divisor of the number $ (a+1)-a=1 $. Ther... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 1,450 |
LV OM - II - Task 4
Determine all positive integers $ n $ that have exactly $ \sqrt{n} $ positive divisors. | The number $ n = 1 $ satisfies the conditions of the problem. Let's assume, for the rest of the solution, that $ n > 1 $.
From the conditions of the problem, it follows that $ n = k^2 $ for some positive integer $ k $. The number of positive divisors of $ n $ that are less than $ k $ is equal to the number of its divis... | n=1n=9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 1,451 |
XLI OM - I - Problem 5
Prove that the edges of a cube cannot be numbered with the numbers from 1 to 12 in such a way that the sum of the numbers of the edges emanating from each vertex is the same.
Can this condition be met by numbering the edges with twelve different numbers from the set $ \{1,2,\ldots,13\} $? | Suppose there exists a way of numbering as mentioned in the first sentence. Each edge is adjacent to two vertices at its two ends. Therefore, twice the sum of the numbers of all edges should be equal to $8s$, where $s$ represents the sum of the numbers of edges emanating from one vertex (a common value for all edges of... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 1,452 |
Circles $ k_1 $, $ k_2 $, $ k_3 $, and $ k_4 $ are externally tangent: $ k_1 $ to $ k_2 $ at point $ A $, $ k_2 $ to $ k_3 $ at point $ B $, $ k_3 $ to $ k_4 $ at point $ C $, and $ k_4 $ to $ k_1 $ at point $ D $. Lines $ AB $ and $ CD $ intersect at point $ S $. A line $ p $ through point $ S $ is tangent to $ k_4 $ ... | Let $O_i$ be the center of the circle $k_i$, and $r_i$ be the length of its radius ($i = 1,2,3,4$). Thus,
om43_2r_img_8.jpg
We will show that
\begin{center}
(2) \qquad the closed broken line $O_1O_2O_3O_4O_1$ has no self-intersections.
\end{center}
Suppose, contrary to statement (2), that for example the segments... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,454 |
XXVII OM - I - Zadanie 4
Samolot leci bez zatrzymywania się po najkrótszej drodze z Oslo do miasta $ X $ leżącego na równiku w Ameryce Południowej. Z Oslo startuje dokładnie w kierunku zachodnim. Wiedząc, że współrzędne geograficzne Oslo są : $ 59^{\circ}55 szerokości północnej i $ 10^{\circ}43 długości wschodniej, ob... |
Najkrótsza droga na powierzchni kuli łącząca dwa punkty tej kuli jest łukiem koła wielkiego, tzn. łukiem okręgu, którego środkiem jest środek kuli. Zatem samolot leci po łuku koła wielkiego przechodzącego przez Oslo. Przy tym Oslo jest najbardziej na północ wysuniętym punktem tego koła wielkiego, ponieważ samolot star... | Quito,10000 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,456 |
IX OM - I - Problem 4
Given a segment $ AH $ and a point $ M $ on it; construct a triangle $ ABC $ in such a way that the segment $ AH $ is one of the altitudes of the triangle $ ABC $, and the point $ M $ is the midpoint of one of the other altitudes. | Let $AH$ and $CL$ be the altitudes of triangle $ABC$, with $CM = ML$ (Fig. 4). Measure on $MA$ a segment $MN = HM$. Triangle $LMN$ is congruent to triangle $CMH$ ($ML = CM$, $MN = HM$, $\measuredangle LMN = \measuredangle CMH$), so $\measuredangle MNL = \measuredangle MHC = 90^\circ$. In the right triangle $ALM$, we kn... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 1,459 |
VI OM - III - Task 3
A regular triangle $ABC$ is inscribed in a circle; prove that if $M$ is any point on the circle, then one of the distances $MA$, $MB$, $MC$ is equal to the sum of the other two. | When point $ M $ is at one of the vertices of triangle $ ABC $, the theorem is obvious, since in this case one of the distances $ MA $, $ MB $, $ MC $ equals zero. It remains to consider the case where $ M $ lies inside one of the arcs into which points $ A $, $ B $, $ C $ divide the circle, for example, inside the arc... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,462 |
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