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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 integer $ n > 4 $.
In the above proof, we relied on the fact that when $ k > 4 $, $ k^2 > 2k + 1 $. Indeed, the inequality $ k^2 > 2k + 1 $ is equivalent to the inequality $ k (k - 2) > 1 $, which is true for $ k > 4 $, and even for $ k \geq 3 $.
Note. A more general theorem is true: if $ a $ is an integer greater than $ 1 $, and $ n $ is an integer greater than $ a^2 $, then
For $ a = 2 $, we have already proved the above. Let us assume that $ a > 2 $ and use mathematical induction. The inequality (1) is true when $ n = a^2 $; indeed, $ a^{a^2} $ is greater than $ (a^2)^a = a^{2a} $ when $ a > 2 $, because the inequality $ a > 2 $ implies that $ a^2 > 2a $.
Suppose that for some integer $ k \geq a^2 $,
we need to prove that
According to the inductive hypothesis, $ a^{k+1} = a^k \cdot a > k^a \cdot a $, so it suffices to prove that when $ a > 2 $ and $ k \geq a^2 $, $ k^a \cdot a > (k + 1)^a $, or that
Indeed, when $ a > 2 $ and $ k \geq a^2 $,
which is what we needed to prove. (In the proof, we relied on the theorem that if $ n $ is an integer greater than $ 1 $ and $ d > -1 $, then $ (1 + d)^n > 1 + nd $. The proof of this inequality can be easily obtained by induction. See Mathematical Olympiad Problems, Volume II, Problem 36). | 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 the center of homothety $ j_1 $, with a positive scale, transforming circle $ c_1 $ into $ c_A $, and point $ A $ is the center of homothety $ j_2 $, with a positive scale, transforming circle $ c_A $ into $ c_2 $. Therefore, the homothety $ j $, with a positive scale, transforming circle $ c_1 $ into $ c_2 $ is the composition of homotheties $ j_1 $ and $ j_2 $ — its center thus lies on the line $ AD $. Similarly, we prove that the center of homothety $ j $ lies on the lines $ BE $ and $ CF $.
Conclusion: The lines $ AD $, $ BE $, and $ CF $ have a common point, which is the center of homothety of circles $ c_1 $ and $ c_2 $. | 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 that the number $ n $ gives a remainder of $ 1 $ when divided by $ 3 $. Then the number $ k = \frac{1}{3}(2n +1) $ is an integer. From equation (1) and the fact that $ S $ is a rational number, it follows that the term $ a_k $ is a rational number. Therefore, the considered statement is true in this case.
Next, assume that the number $ n $ does not give a remainder of $ 1 $ when divided by $ 3 $ (in other words, the number $ \frac{1}{3} (2n + 1) $ is not an integer). We will construct an arithmetic sequence $ a_i $ for which the considered statement is not true. Let us take:
From equation (1), we get $ S = 0 $, but for any $ i \in \{1,2,...,n\} $, the number
is irrational. Therefore, the considered statement is not true if $ n $ does not give a remainder of $ 1 $ when divided by $ 3 $. | 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 straight line. If line $ AB $ is not parallel to line $ m $, then the sought point $ M $ is the intersection of lines $ AB $ and $ m $. Line $ m $ is then the angle bisector of $ \angle AMB $. If lines $ AB $ and $ m $ are parallel, i.e., if points $ A $ and $ B $ are equidistant from line $ m $, then the problem has no solution.
The proven theorem allows us to easily justify an important property of the hyperbola. Let $ A_1 $ and $ A_2 $ be the vertices, and $ F_1 $ and $ F_2 $ the foci of the hyperbola (Fig. 19). The hyperbola divides the plane into three regions, which we will denote as $ I $, $ II $, and $ III $, as shown in Fig. 19.
It is known that if point $ P $ lies on the hyperbola, then $ |F_1P - F_2P| = A_1A_2 $; if $ P $ is in region $ I $, then $ |F_1P - F_2P| < A_1A_2 $; and if $ P $ lies in one of the regions $ II $ or $ III $, then $ |F_1P - F_2P| > A_1A_2 $.
Let $ m $ be the tangent line to the hyperbola at point $ M $, and $ P $ be any point on line $ m $. Line $ m $ lies (except at point $ M $) in region $ I $, so the difference $ |F_1P - F_2P| $ is less than $ A_1A_2 $ for every point $ P $ different from $ M $, and in point $ M $ it reaches its maximum value equal to the length of $ A_1A_2 $. From the previously proven theorem, it follows that line $ m $ is the angle bisector of $ \angle F_1MF_2 $, i.e., the following theorem holds:
The tangent to the hyperbola is the angle bisector of the angle formed by the segments connecting the point of tangency with the foci of the hyperbola. | 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 loss of generality, we can assume that $ \alpha \geq \beta \geq \gamma $. Then, of course, $ \beta < \alpha + \gamma $ and $ \gamma < \alpha + \beta $. We need to prove that there exists a trihedral angle whose dihedral angles have measures equal to $ \alpha, \beta, \gamma $ if and only if $ \alpha < \beta + \gamma $.
Let $ t $ be a ray with origin at point $ P $. Rays with origin at point $ P $ that form an angle of a given measure $ \mu $ with $ t $ are the generators of a cone of revolution with vertex at point $ P $, axis $ t $, and vertex angle of measure $ 2\mu $. Conversely, every generator of this cone forms an angle of measure $ \mu $ with ray $ t $ (Fig. 16).
Let rays $ k $ and $ m $ contained in plane $ \pi $ have origin at point $ O $ and form an angle of measure $ \alpha $ (Fig. 17). Consider the cone of revolution $ S_1 $ with vertex at point $ O $, axis $ k $, and vertex angle of measure $ 2\beta $, and the cone of revolution $ S_2 $ with vertex at point $ O $, axis $ m $, and vertex angle of measure $ 2\gamma $. The intersection of the surfaces of these cones is either the point $ O $ (if $ \beta + \gamma < \alpha $), or a certain ray contained in plane $ \pi $ (if $ \beta + \gamma = \alpha $), or a pair of rays $ p_1 $, $ p_2 $, neither of which is contained in plane $ \pi $ (if $ \beta + \gamma > \alpha $). In the last case, rays $ k $, $ m $, $ p_1 $ form a trihedral angle whose dihedral angles at the vertex have measures equal to $ \alpha, \beta, \gamma $. If, however, $ \alpha \geq \beta + \gamma $, then such a trihedral angle does not exist, because the corresponding cones do not have a common generator not contained in plane $ \pi $ in this case.
We now proceed to solve the problem. From the equalities given in the problem, it follows that any two faces of the tetrahedron $ ABCD $ have three edges respectively equal (Fig. 18). They are therefore congruent triangles. Hence, in particular,
Since the sum of the measures of the angles of triangle $ ABC $ is $ \pi $, it follows that the sum of the measures of the dihedral angles at vertex $ D $ is equal to $ \pi $. Therefore, by the lemma,
Hence $ \measuredangle BDC < \displaystyle \frac{\pi}{2} $.
Similarly, we prove that each of the remaining dihedral angles is acute.
Note. The thesis of the problem follows directly from problem 12 of the XII Mathematical Olympiad. | 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 root $-\frac{n}{m}$. This is impossible, however, because for every integer $x$ the number $x^5 + x + 1$ is odd.
Hence, the desired factorization must have the form
where the numbers $m, n, p, a, b, c, d$ are integers.
In this case, $ma = 1$, $pd = 1$, so $m = a = \pm 1$, $p = d = \pm 1$.
Thus, there are four possible cases, but it suffices to consider two of them; the remaining cases reduce to these by changing the sign of both factors in the factorization.
a) When $m = a = 1$, $p = d = 1$, then by comparing the coefficients of like powers of $x$ on both sides of the equality
we obtain
The system of equations (1) - (4) has one solution $n = 1$, $b = -1$, $c = 0$, which gives the factorization
b) When $m = a = 1$, $p = d = -1$, we similarly obtain the system of equations
It is easy to check that this system has no solutions. The solution to the problem is therefore the formula
where $\varepsilon = \pm 1$.
Note. The factorization sought in the problem can be found quickly in the following way:
Such a solution to the problem still requires a proof that there is no other factorization besides the one obtained by changing the signs of both found factors. The easiest way to show this is by using complex numbers. Let $x_1$, $x_2$ denote the roots of the polynomial $x^2 + x + 1$, and $x_3$, $x_4$, $x_5$ the roots of the polynomial $x^3 - x^2 + 1$.
Then
Any quadratic divisor of the polynomial $x^5 + x + 1$ with integer coefficients is of the form $m(x - x_i)(x - x_k)$, where $m$ is an integer and $x_i$, $x_k$ are two roots of the polynomial. It is easy to see that such a divisor can only be $m(x - x_1)(x - x_2)$. The detailed proof of this is left as an exercise.
Hint. One of the numbers $x_3$, $x_4$, $x_5$ is an irrational real number, and the other two are complex conjugates, whose sum and product are irrational. | 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 \geq 3p $). Po $ k $-krotnym zastosowaniu analogicznego przekształcenia uzyskujemy związek
sensowny i słuszny pod warunkiem że $ n \geq kp $.
(Wyprowadzenie wzoru (2) ma charakter indukcyjny; proponujemy Czytelnikowi, jako proste ćwiczenie, przeprowadzenie szczególowego uzasadnienia, przez indukcję względem $ k = 1,2,\ldots,[n/p] $, z wielokrotnym wykorzystaniem wzoru (1).)
Jeśli teraz $ n \geq p^2 $, to możemy w równości (2) podstawić $ k = p $, otrzymując
Dla $ i = 1,2,\ldots,p-1 $ współczynniki dwumianowe $ {{p}\choose{i}} $ są liczbami podzielnymi przez $ p $ (patrz: Uwaga). Jeśli więc z sumy napisanej po prawej stronie równości (3) odrzucimy wszystkie składniki z wyjątkiem dwóch skrajnych (odpowiadających wartościom wskaźnika $ i = 0 $ oraz $ i = p $), to otrzymamy liczbę przystającą do $ a_n $ modulo $ p $:
(dla $ n \geq p^2 $). Przyrównując (modulo $ p $) prawe strony związków (1) i (4) stwierdzamy, że $ a_{n-1}+a_{n-p} \equiv a_{n-p} + a_{n-p^2} $, czyli
Oznaczmy przez $ r_j $ resztę z dzielenia liczby $ a_j $ przez $ p $ (dla $ j = 0,1,2,\ldots $). Uzyskaną zależność przepisujemy w postaci
To znaczy, że ciąg ($ r_j $) jest okresowy, z okresem $ p^2-1 $; zatem
Podstawiamy $ n = p^3 $, $ k = p $, i otrzymujemy wynik:
Pozostaje zauważyć, że (w myśl wzoru (1))
i wobec tego $ r_{p^3} = r_p = p-1 $. Tak więc reszta z dzielenia liczby $ a_{p^3} $ przez $ p $ równa się $ p-1 $.
Uwaga. Fakt, na który powołaliśmy się w rozwiązaniu:
liczba pierwsza $ p $ jest dzielnikiem liczb $ {{p}\choose{1}} $, $ {{p}\choose{2}} $, $ \ldots $, $ {{p}\choose{p-1}} $
jest dobrze znany (oraz banalnie prosty: w równości $ {{p}\choose{i}} \cdot (i!) \cdot ((p-i)!) = p! $ czynniki $ i! $ oraz $ (p-i)! $ są dla $ i \in \{1,2,\ldots,p-1\} $ niepodzielne przez $ p $, więc czynnik $ {{p}\choose{i}} $ musi dzielić się przez $ p $).
| 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 $2^{n-1}$ ($n = 1, 2, \ldots, 7$) remains in the hat, is equally likely. The probability of such an event is thus $\displaystyle \frac{1}{7}$.
The sum of the numbers on the drawn slips is equal to $127 - 2^{n-1}$. If $n = 1$, this sum is $126$; if $n = 2$, it is $125$; if $n = 3, 4, 5, 6$ or $7$, the sum is less than $124$ and we must draw a seventh slip. In this last case, the sum of the numbers on all the drawn slips will be $127$. Therefore, the probability that the sum of the numbers on all the slips drawn according to the conditions of the problem is $125$, $126$, or $127$, is $\displaystyle \frac{1}{7}$, $\displaystyle \frac{1}{7}$, $\displaystyle \frac{5}{7}$, respectively.
Thus, the most probable value of the sum is $127$. | 127 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 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 obtain the dependencies
Finally, in the third equation, we substitute $ x = (t + 1)/t $, $ x = t $, $ x = t + 1 $; we get the equalities
We will transform the value $ f(\frac{t}{t+1}) $ in two ways. From formulas (3), (1), (7), (4) we get
and from formulas (5), (2), (6) we get
Equating the right sides of the obtained expressions:
Multiplying both sides by $ (t+1)^2 $ gives the equality
which means $ f(t) = t $.
We assumed that $ t \neq 0 $, $ t \neq -1 $. But the equality $ f(0) = 0 $ follows directly from the oddness of the function $ f $; hence (using the given equations again) we have $ 0 =f(0) =f((-1) + 1) =f(-1)+1 $, which means $ f(-1) = -1 $. Ultimately, then,
Verifying that this function satisfies the given equations is immediate. | 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 larger than $ 360^{\circ} $. Therefore
from which
Since $ n - 2 $ must be a natural number, then $ 2k $ should
be a natural number, i.e., $ k $ must be half
of a natural number, so $ k = \frac{1}{2}, 1, \frac{3}{2}, 2, \dots $ | \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 $S$, where the lines $AA_1$, $BB_1$, $CC_1$ intersect; point $S$ is the centroid of triangle $ABC$, as well as of triangle $A_1B_1C_1$. The similarity is inverse, because corresponding points, e.g., $A$ and $A_1$, lie on the rays $SA$ and $SA_1$ in opposite directions. The ratio of similarity is
Since the ratio of the areas of similar figures equals the square of the ratio of similarity, the area of triangle $A_1B_1C_1$ is $4$ times larger than the area of triangle $ABC$, which is also directly visible.
Returning to the main topic, let's briefly recall the properties of similar figures in space. Let a figure $F$ be given in space, whose points are $A, B, C, \ldots$ Choose any point $S$ and a positive number $k$. When we measure segments $SA, SB, SC, \ldots$ on the rays $SA, SB, SC, \ldots$, the points $A$ form a figure $F$ directly similar to figure $F$ relative to point $S$ in the ratio $k$. If, however, we measure the same segments $SA_1 = k \cdot SA, SB_1 = k \cdot SB, SC_1 = k \cdot SC, \ldots$ on the extensions of the rays $SA, SB, SC, \ldots$, we obtain a figure $F_1$ inversely similar to figure $F$ relative to point $S$ in the ratio $k$ (Fig. 47). From this definition, it easily follows that the figure similar to the segment $AB$ is the segment $A_1B_1$, with the lines $AB$ and $A_1B_1$ being parallel (or coinciding, which happens when the line $AB$ passes through point $S$). The figure similar to the triangle $ABC$ is, therefore, the triangle $A_1B_1C_1$, with the planes of the triangles being parallel (or coinciding in the case where the plane $ABC$ contains point $S$). The figure similar to the tetrahedron $ABCD$ is the tetrahedron $A_1B_1C_1D_1$. The ratio of similar segments equals the ratio of similarity, the ratio of the areas of similar triangles (or generally: planar figures) equals the square of the ratio of similarity, and the ratio of the volumes of similar tetrahedra (or generally: solids) equals the cube of the ratio of similarity.
To solve the problem, we will also need the following theorem:
The segments connecting the vertices of a tetrahedron with the centroids of the opposite faces intersect at one point, which divides each of these segments in the ratio $3 \colon 1$.
The proof of this theorem is simple; it suffices to prove it for one pair of such segments. Let $M$ be the centroid of the face $BCD$, and $N$ be the centroid of the face $ACD$ of the tetrahedron $ABCD$ (Fig. 47). The lines $BM$ and $AN$ intersect at the midpoint $P$ of the edge $CD$.
The segments $AM$ and $BN$ connecting the vertices $A$ and $B$ of the triangle $ABP$ with the points $M$ and $N$ of the sides $BP$ and $AP$ intersect at a point $S$ inside this triangle. Since $AP = 3 \cdot NP$ and $BP = 3 \cdot MP$, it follows that $MN \parallel AB$ and $AB = 3 MN$; hence, $AS = 3 \cdot SM$ and $BS = 3 \cdot SN$, which is what we had to prove. The point $S$ is the centroid of the tetrahedron.
Applying the above knowledge, we can briefly state the solution to the problem: We construct the figure inversely similar in the ratio $3$ to the tetrahedron $ABCD$ relative to its centroid $S$. This figure is the tetrahedron $A_1B_1C_1D_1$, whose vertices lie on the extensions of the rays $SA$, $SB$, $SC$, $SD$, with $SA_1 = 3 \cdot SA$, $SB_1 = 3 \cdot SB$, $SC_1 = 3 \cdot SC$, $SD_1 = 3 \cdot SD$.
The plane $B_1C_1D_1$ is parallel to the corresponding plane $BCD$, and it passes through the vertex $A$, since from the equality $AS = 3 \cdot SM$ it follows that point $A$ is the point similar to point $M$ of the face $BCD$. Therefore, the tetrahedron $A_1B_1C_1D_1$ is the tetrahedron mentioned in the problem. Since it is similar to the tetrahedron $ABCD$ in the ratio $3$, its volume is $3^3 \cdot V$, or $27 V$. | 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 a translation,
(3) each of the polyhedra $ W_i $ ($ i = 1,\ldots, n $) has a point in common with $ W_0 $,
(4) the polyhedra $ W_0, W_1, \ldots, W_n $ have pairwise disjoint interiors. | Suppose that polyhedra $W_0, W_1, \ldots, W_n$ satisfy the given conditions. Polyhedron $W_1$ is the image of $W_0$ under a translation by a certain vector $\overrightarrow{\mathbf{v}}$ (condition (2)). Let $O_0$ be the center of symmetry of polyhedron $W_0$ (condition (1)); the point $O_1$, which is the image of $O_0$ under the translation by $\overrightarrow{\mathbf{v}}$, is the center of symmetry of $W_1$. Figure 3 illustrates a planar variant of the considered problem (a representation of the spatial configuration would obscure this illustration); polyhedra $W_0$ and $W_1$ are depicted as centrally symmetric polygons.
om42_1r_img_3.jpg
By condition (3), polyhedra $W_0$ and $W_1$ have common points (possibly many). Let $K$ be a common point of $W_0$ and $W_1$ (arbitrarily chosen). Denote by $L$ the image of point $K$ under the central symmetry with respect to $O_1$; thus, $L \in W_1$. Let $N$ be a point such that $\overrightarrow{NL} = \overrightarrow{\mathbf{v}}$ and let $M$ be the midpoint of segment $NK$ (Figure 4). Therefore, $N \in W_0$. According to condition (1), the set $W_0$ is convex; this means that with any two points belonging to $W_0$, the entire segment connecting these points is contained in $W_0$. Since $K \in W_0$ and $N \in W_0$, it follows that $M \in W_0$. Segment $MO_1$ connects the midpoints of segments $KN$ and $KL$, and thus $\overrightarrow{MO_1} = \frac{1}{2} \overrightarrow{NL} = \frac{1}{2} \overrightarrow{\mathbf{v}}$, which means $M$ is the midpoint of segment $O_0O_1$.
Let $U$ be the image of polyhedron $W_0$ under a homothety with center $O_0$ and scale factor 3. We will show that $W_1 \subset U$. Take any point $P \in W_1$: let $Q \in W_0$ be a point such that $\overrightarrow{QP} = \overrightarrow{\mathbf{v}}$ and let $S$ be the center of symmetry of parallelogram $O_0O_1PQ$. The medians $O_0S$ and $QM$ of triangle $O_0O_1Q$ intersect at a point $G$ such that $\overrightarrow{O_0G} = \frac{2}{3}\overrightarrow{O_0S} = \frac{1}{3}\overrightarrow{O_0P}$ (Figure 5). This means that $P$ is the image of point $G$ under the considered homothety. Since $G$ is a point on segment $QM$ with endpoints in the set (convex) $W_0$, it follows that $G \in W_0$. Therefore, $P \in U$ and from the arbitrariness of the choice of point $P \in W_1$ we conclude that $W_1 \subset U$.
om42_1r_img_4.jpg
om42_1r_img_5.jpg
In the same way, we prove that each of the sets $W_i (i=1, \ldots, n)$ is contained in $U$. Of course, also $W_0 \subset U$. Thus, the set $W_0 \cup W_1 \cup \ldots \cup W_n$ is a polyhedron contained in $U$. By conditions (2) and (4), its volume equals the volume of $W_0$ multiplied by $n+1$. On the other hand, the volume of $U$ equals the volume of $W_0$ multiplied by 27. Therefore, $n \leq 26$.
It remains to note that the value $n = 26$ can be achieved (example realization: 27 cubes $W_0, \ldots, W_{26}$ arranged like a Rubik's cube). Thus, the sought number is $26$.
Note. We obtain the same result assuming that $W_0, \ldots, W_n$ are any bounded, closed convex bodies (not necessarily polyhedra), with non-empty interiors, satisfying conditions (1)-(4); the reasoning carries over without any changes. Moreover, condition (4) turns out to be unnecessary. This was proven by Marcin Kasperski in the work 27 convex sets without a center of symmetry, awarded a gold medal at the Student Mathematical Paper Competition in 1991; a summary of the work is presented in Delta, issue 3 (1992). | 26 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 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 needed for anything.
Let then two cubes $\mathcal{C}$ and $\mathcal{C}$ be given. Since in statement (*) the roles of the symbols $\mathcal{C}$ and $\mathcal{C}$ are symmetric, we can assume without loss of generality that the edge length of the cube $\mathcal{C}$ is not less than the edge length of the cube $\mathcal{C}$. We will show in three ways that condition (*) cannot be satisfied.
Let $\mathcal{C}$ be a cube with an edge of unit length, and $\mathcal{C}$ a cube with an edge length of $\geq 1$. Choose two opposite faces $\mathcal{S}$ and $\mathcal{T}$ of the cube $\mathcal{C}$. The plane of the face $\mathcal{S}$ divides space into two half-spaces; denote by $\mathcal{H}$ the half-space that does not contain the square $\mathcal{T}$. Assume that $\mathcal{H}$ is a closed half-space; that is, the plane of the face $\mathcal{S}$ is a subset of the set $\mathcal{H}$.
Suppose that in the set $\mathcal{H}$ there are two vertices $A$, $B$ of the cube $\mathcal{C}$, not being the endpoints of one edge. Denote the midpoint of the segment $AB$ by $M$; it is also a point of the set $\mathcal{H}$.
If $AB$ is one of the four diagonals of the cube $\mathcal{C}$ intersecting its interior, then $M$ is its center. Every point of the cube $\mathcal{C}$ is at a distance from $M$ of at most $\frac{1}{2} \sqrt{3}$. This number is less than $1$. Since the distance from the square $\mathcal{T}$ to the half-space $\mathcal{H}$ is at least $1$, the square $\mathcal{T}$ does not contain any point of the cube $\mathcal{C}$.
If $AB$ is a diagonal of one of the faces of the cube $\mathcal{C}$, then $M$ is the center of that face. All its points are at a distance from $M$ of at most $\frac{1}{2} \sqrt{2}$, so the entire face is disjoint from the square $\mathcal{T}$.
The only remaining case to consider is when in the set $\mathcal{H}$ there is either no vertex of the cube $\mathcal{C}$, or exactly one vertex, or exactly two vertices connected by an edge. The remaining vertices (8, 7, or 6 in number) lie outside the set $\mathcal{H}$: in each case, among them, we can find four points that are the vertices of some face of the cube $\mathcal{C}$. This face has no points in common with the set $\mathcal{H}$, and therefore has no points in common with the face $\mathcal{S}$ of the cube $\mathcal{C}$.
We have shown that the cube $\mathcal{C}$ has a face that is disjoint either from $\mathcal{S}$ or from $\mathcal{T}$. Condition (*) cannot be satisfied.
Note 2. Replacing the word "cubes" everywhere in the problem statement with "parallelepipeds" yields a much more general problem. The author of this solution does not know the answer to the question contained in such a reformulated problem.
Note 3. A four-dimensional cube is a subset of four-dimensional space $\mathbb{R}^4$ similar to the set
(similar in the geometric sense). The subset of the set $\mathcal{C}$ consisting of points with the first coordinate equal to $1$ is isometric to a three-dimensional cube; the same can be said about the other seven subsets of the set $\mathcal{C}$ obtained by fixing any coordinate and assigning it the value $1$ or $-1$. These subsets are called the faces of the cube $\mathcal{C}$. If $\mathcal{C}$ is another cube, obtained from the cube $\mathcal{C}$ by a similarity transformation, then the images of the faces of the cube $\mathcal{C}$ under this similarity transformation form the faces of the cube $\mathcal{C}$.
For four-dimensional cubes, we can pose a problem analogous to the one considered in the problem: do there exist in $\mathbb{R}^4$ two such cubes that each face of one of them has points in common with each face of the other? The answer may seem surprising: in $\mathbb{R}^4$ such pairs do exist. We will give an example - without any justification; verifying the truth of all the statements in the next paragraph may be an ambitious task for the Reader.
One of the cubes will be the set $\mathcal{C}$ defined above. Its vertices are points of the form $(\pm 1, \pm 1, \pm 1, \pm 1)$ (the choice of signs is arbitrary; such a cube has 16 vertices). Now consider another 16 points: eight points of the form $(\pm 1, \pm 1, \pm 1, \pm 1)$, in which the product of the coordinates equals $1$, and eight points having three coordinates equal to $0$ and the remaining one equal to $2$ or $-2$. This 16 points also form the set of vertices of a four-dimensional cube $\mathcal{C}$, and this cube is congruent to $\mathcal{C}$. Moreover, each three-dimensional face of one cube has at least one common vertex with each three-dimensional face of the other cube! | 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
\[
\text{from the rationality of numbers } a, b, s, \text{ and } xy, \text{ we conclude that the difference } x - y \text{ is a rational number.}
\]
Since both the sum and the difference of the numbers $ x $ and $ y $ are rational, both these numbers are rational.
Let's represent $ x $ as an irreducible fraction $ m/n $. Then $ x^3 = m^3/n^3 $ is also an irreducible fraction, and since $ x^3 = a $ is an integer, we have $ n = 1 $. This means that $ x $ is a natural number. Similarly, we prove that $ y $ is a natural number. | 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), it follows that
The number $ |a - x_0| $ is an integer, positive, and is a divisor of unity, hence $ |a - x_0| = 1 $; similarly, we conclude that $ |b - x_0| = 1 $, $ |c - x_0| = 1 $.
It follows that two of the numbers $ a - x_0 $, $ b - x_0 $, $ c - x_0 $, and thus also two of the numbers $ a $, $ b $, $ c $, are equal, which contradicts the assumption. Therefore, the polynomial $ f(x) $ does not have an integer root, Q.E.D.
Note. In the above reasoning, we relied on the fact that the coefficients of the polynomial $ \varphi (x) $ are integers. This is easy to show.
Let
From the equality $ f(x) = (x - x_0) \varphi (x) $, by comparing the coefficients of both sides, we get
thus
From this, it follows that the number $ c_0 $ is an integer, and if $ c_{k-1} $ is an integer ($ k = 1,2, \ldots, n - 1 $), then $ c_k $ is also an integer. Therefore, the numbers $ c_0, c_1, \ldots, c_{n-1} $ are integers. | 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_{333} $ are divisible by 333, or none of them are. | 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 equality (1), it follows that the number $ n_i $ is divisible by 333 if and only if the number $ n_{i+1} $ is divisible by 333.
If the number $ n_1 $ is divisible by 333, then by the above statement, we obtain (by induction) the divisibility by 333 of each of the numbers $ n_2, n_3, \ldots , n_{333} $. If, however, the number $ n_1 $ is not divisible by 333, then similarly as above, we conclude that none of the numbers $ n_2, n_3, \ldots , n_{333} $ is divisible by 333. | 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, such that $ KM = LM $. | 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 measures of these angles are equal or their sum is a straight angle. We have then
If in at least two cases the second part of the alternative is satisfied, for example, $ \measuredangle A +\measuredangle A', $ \measuredangle B +\measuredangle B', $ then, given that the sum of the angles in a triangle is $ 180^\circ, $ we would have
From this, $ \measuredangle C + \measuredangle C', $ which is not possible.
Therefore, in at least two cases the first part of the alternative is satisfied, for example, $ \measuredangle A = \measuredangle A' $ and $ \measuredangle B =\measuredangle B'. $ It follows that triangles $ ABC $ and $ A'B'C' $ are similar.
Corollary. If the corresponding sides of two triangles are perpendicular, then the triangles are similar.
Proof. Let
Rotate triangle $ A_1B_1C_1 $ around any point by a right angle. We will then obtain triangle $ A_2B_2C_2 $ with sides perpendicular to the corresponding sides of triangle $ A_1B_1C_1, $ i.e.,
From (1) and (2), it follows that $ AB \parallel A_2B_2, $ BC \parallel B_2C_2, $ CA \parallel C_2A_2. $ Therefore, by the lemma, triangles $ ABC $ and $ A_2B_2C_2 $ are similar. Since triangles $ A_2B_2C_2 $ and $ A_1B_1C_1 $ are congruent, it follows that triangles $ ABC $ and $ A_1B_1C_1 $ are also similar.
We will now proceed to solve the problem.
If point $ P $ lies on the line containing segment $ CD, $ then $ P = K = L = M $ and therefore $ KM = LM = 0. $
If point $ P $ does not lie on the line containing segment $ CD, $ then from the conditions of the problem it follows that the corresponding sides of triangles $ PKM $ and $ CBE $ are perpendicular, and also the corresponding sides of triangles $ PLM $ and $ CAE $ are perpendicular (Fig. 6). Therefore, by the corollary proved above, triangles $ PKM $ and $ CBE $ and triangles $ PLM $ and $ CAE $ are similar.
Denoting the ratios of these similarities by $ \lambda $ and $ \mu $ respectively, we get that $ PM = \lambda CE, $ $ KM = \lambda BE, $ $ PM = \mu CE, $ $ LM = \mu AE. $ From the first and third equalities, it follows that $ \lambda =\mu. $ Point $ E $ is the midpoint of segment $ AB, $ i.e., $ AE = BE. $ Therefore, from the second and fourth equalities, it follows that $ KM = LM. $ | 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 = MF \colon MB $ holds, so triangles $ AMF $ and $ EMB $ with the common angle $ M $ have proportional sides, and thus are similar, and $ \measuredangle A = \measuredangle E $. This means that the segment $ BF $ is seen from points $ A $ and $ E $ at the same angle, so points $ A $ and $ E $ lie on a circle passing through $ B $ and $ F $. | 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 because, for example, in the trapezoid $1234$, the diagonals $13$ and $24$ divide into parts in the same ratio as the bases $23$ and $14$, i.e., in the ratio $1 \colon 2$, so, for instance, segment $II\ 2$ equals one-third of segment $42$. The hexagon $123456$ consists of $18$ triangles of equal area, and the shaded hexagon $I\ II\ III\ IV\ V\ VI$ consists of $6$ triangles of the same area. Therefore, the area of the shaded hexagon is $\frac{2}{9} S$. | \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 $AOB$ and $AB = OM \sin \alpha$. If $AB = d$, then $OM = \frac{d}{\sin \alpha}$. Therefore, point $M$ lies on the circle $k$ with center $O$ and radius $\frac{d}{\sin \alpha}$.
Conversely, if a point $N$ lies on this circle $k$ (Fig. 28), then its projections $C$ and $D$ onto lines $a$ and $b$ lie on the circle with diameter $ON = \frac{d}{\sin \alpha}$; this circle is the circumcircle of triangle $COD$, so $CD = \frac{d}{\sin \alpha} \cdot \sin \measuredangle COD$. But $\measuredangle COD$ is equal to $\alpha$ or $180^\circ - \alpha$, so $\sin \measuredangle COD = \sin \alpha$ and $CD = d$.
The sought geometric locus is thus the circle drawn with the intersection of the given lines as the center and a radius equal to $\frac{d}{\sin \alpha}$, where $\alpha$ denotes one of the angles between the given lines. | 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 distinct numbers $ a_1, a_2, \ldots, a_{n+1} $ satisfying the condition: for $ i = 1,2,\ldots,n $, the number $ a_i $ divides $ a_{i+1} $. | 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 that for any $ k \in M $ we have $ 1 \leq d(k) \leq n $. Then, by the pigeonhole principle, there exist such different numbers $ k_1,k_2,\ldots,k_{n+1} $ that
From the assumptions of the problem, it follows that for some different numbers $ s $, $ t $, the divisibility $ k_s|k_t $ holds. The condition $ d(k_s) = m $ means that there exist numbers $ a_1 = k_s, a_2, \ldots, a_m $, satisfying: for $ i = 1,2,\ldots,m-1 $, the number $ a_i $ is divisible by $ a_{i+1} $. However, then the numbers $ b_1 = k_t, b_2 =a_1 = k_s , b_3 = a_2, \ldots , b_{m+1} =a_m $ satisfy the condition: for $ i = 1,2,\ldots,m $, the number $ b_i $ is divisible by $ b_{i+1} $. This proves that $ d(k_t) \geq m + 1 $, contradicting the assumption that $ d(k_t) = m $.
We have obtained a contradiction. | 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 homogeneous polynomial of degree two in the variables $ a_1, a_2, \ldots, a_n $, which we can write as
We state that for $ i, k = 1, 2, \ldots, n $, $ i < k $
Hence
| 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 vertices of the cube lie.
Let's first consider case (a). The three vertices of the cube that lie on the face $KLM$ must be vertices of one face of the cube. This means that a certain face of the cube, say $ABCD$, lies inside the triangle $KLM$; the remaining four vertices $A$, $B$, $C$, $D$ are on the faces $KLN$, $LMN$, $MKN$. Without loss of generality, we can assume that the face $KLN$ contains two of the points $A$, $B$, $C$, $D$. They must be two consecutive vertices of the square $A$. Without loss of generality, we can assume that the points $A$, $B$ lie on the face $KLN$, the point $C$ lies on the face $LMN$, and the point $D$ is on the face $MKN$.
Condition (a) thus uniquely determines (up to the adopted notation) the position of the given cube inside the regular tetrahedron with edge $1$. We proceed to determine the length of the edge $a$.
Let $K$, $L$, $M$ be the points of intersection of the edges $KN$, $LN$, $MN$ with the plane $A$. The tetrahedron $K$ is regular. Let its edge be denoted by $b$. Then $K$, from which
The height of a regular tetrahedron with edge $\lambda$ is given by the formula $h = \frac{\sqrt{6}}{3}\lambda$. Comparing the heights of the tetrahedra $KLMN$ and $K$, we obtain the equality $\frac{\sqrt{6}}{3}b + a = \frac{\sqrt{6}}{3}$, from which, using equality (1), we have
The remaining case (b) needs to be considered.
Each of the four segments connecting the vertices of the given cube, lying on the same face of the tetrahedron $KLMN$, is an edge of this cube. These segments have no common endpoints. Suppose the edge $AB$ lies on the face $KLM$. Then one of the edges $A$ or $CD$ lies on one of the faces $KLN$, $LMN$, $MKN$. Without loss of generality, assume that $A$ lies on $KLN$. Then the lines $AB$ and $A$ are parallel to the edge $KL$, and consequently, the lines $CD$ and $C$ are perpendicular to the edge $MN$; they cannot therefore lie on the faces $LMN$ and $KMN$. We can therefore assume that the segment $CC$ lies on the face $LMN$, and the segment $DD$ is on the face $KMN$.
Let $X$, $Y$, $Z$, $T$ be the midpoints of the edges $KN$, $NL$, $LM$, $MK$, respectively. Then the squares $A$ and $D$ have sides parallel to the sides of the square $XYZT$. Hence, there exists (in space) the center of homothety $P$ of the squares $A$ and $XYZT$, lying on the edge $KL$, and the center of homothety $Q$ of the squares $D$ and $XYZT$, lying on the edge $MN$. The scales of homothety in both cases are equal and are $2a$. The distances from the points $P$ and $Q$ to the plane $XYZT$ are equal and are $\sqrt{2}/4$. It follows that the distances from the planes $A$ and $D$ to the plane $XYZT$ are equal — and each of them is $a/2$. These three quantities are related by the equation
It remains to note that the above value can be realized by taking $P$ and $Q$ as the midpoints of the edges $KL$ and $MN$.
In summary: the possible values of $a$ are | 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 triangle $ T $ formed by the medians of triangle $ ABC $, with a scale factor of $ \frac{2}{3} $; the area of triangle $ BSM $ is therefore $ \frac{4}{9} $ of the area of $ T $. However, the area of $ \triangle BSM = \text{area of }\triangle ASB = \frac{1}{3} \cdot \text{area of }\triangle ABC $. It follows that $ \text{area of }T = \frac{3}{4} \cdot \text{area of }\triangle ABC $. | \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 union such that if its members abstain from voting, and members of all other unions vote for one of the two candidates, the candidates for director will not receive an equal number of votes. | 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} $ is an even number, then if the members of the $ i $-th union abstain from voting, the number of remaining votes $ a_1+ \ldots+a_{j-1}+a_{j+1} + \ldots+ a_{100} $ will be odd. Therefore, the number of votes cast for each candidate cannot be equal. If, on the other hand, the sum $ a_1+ \ldots+ a_{100} $ is an odd number, then one of its components must be even (the sum of a hundred odd numbers is an even number), let's say $ a_k = 2s $. If in this case the members of the $ k $-th union abstain from voting, the sum $ a_1 + \ldots a_{k-1} + a_{k+1} + \ldots + a_{100} $ is an odd number. The number of votes cast for each candidate cannot be equal. | 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 consecutive heads.
Let $n = 100k + r$, where $k \geq 0$ and $0 \leq r < 100$. Each $n$-element sequence thus consists of $k$ 100-element sequences and one $r$-element sequence. The total number of 100-element sequences with two values is $2^{100}$; therefore, after excluding the sequence composed of 100 heads, there remain $2^{100} - 1$ 100-element sequences. Each $n$-element sequence not containing 100 consecutive heads thus consists of $k$ such 100-element sequences and some $r$-element sequence. Hence, the number of unfavorable events is not greater than $(2^{100} - 1)^k \cdot 2^r$. Therefore,
If $n$ tends to infinity, then of course $k$ also tends to infinity. Since for $0 < q < 1$ we have $\displaystyle \lim_{k \to \infty} q^k = 0$, then $\displaystyle \lim_{k \to \infty} \left( 1 - \frac{1}{2^{100}} \right)^k = 0$. Therefore, from (1) by the squeeze theorem, we have $\displaystyle \lim_{k \to \infty} q^k = 0$, so $\displaystyle \lim_{n \to \infty} p_n = 1$. | 1 | Combinatorics | proof | Yes | Yes | olympiads | false | 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 + Y_n = 0 $. Calculate $ \lim_{n\to\infty} p_n $. | 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 $ r^2 - 4 \cdot 2^n s \geq 0 $, that is,
We will find the number of such pairs $ (r, s) $ for which (1) holds. For any $ a $, the number of solutions to the inequality $ 0 \leq x \leq a $ in integers $ x $ is $ [a] + 1 $, where $ [a] $ is the greatest integer not exceeding $ a $. Therefore, the number of pairs $ (r, s) $ satisfying
(1) is equal to
Consider the sum
By the formula
we have
Since for any real number $ x $ the inequality
holds, we have
The number of all elementary events, i.e., the considered pairs $ (r, s) $, is $ 2^{2n} $. Therefore, $ \displaystyle p^n = \frac{A^n}{2^{2n}} $ and from (3) it follows that
From (2) we obtain that $ \displaystyle \lim_{n \to \infty} \frac{B_n}{2^{2n}} = \frac{1}{12} $ and therefore from (4) we conclude that $ \displaystyle \lim_{n \to \infty} p_n = \frac{1}{12} $ | \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 (perpendicular to the bisector of angle $ABC$). Since lines $l$ and $m$ do not coincide, we can find a point $P$ on the ellipse lying on the opposite side of line $m$ from points $A$ and $C$ (Figure 3).
om39_1r_img_3.jpg
Let $A'$ and $C'$ be the projections of points $A$ and $C$ onto line $m$, and let $D$ be the point symmetric to $C$ with respect to line $m$. The right triangles $BC'$ and $BD$ are congruent; thus, $|BC| = |BD|$, $|\measuredangle C'$. Since $|\measuredangle C'$ (by the definition of line $m$), we conclude that segment $BD$ is an extension of segment $AB$.
Point $P$ is farther from point $C$ than from point $D$. Therefore,
From this inequality, it follows that the perimeter of triangle $APC$ is greater than the perimeter of triangle $ABC$. We have reached a contradiction with the assumption of the maximality of the perimeter of triangle $ABC$; the proof is complete.
Note: The statement of the problem, as well as the given proof, remains valid if, instead of an ellipse, we consider any smooth closed curve (i.e., having a tangent at every point) bounding a convex region in the plane. | 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 have the same number. | 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 pair of balls in $ \binom{k}{2} $ ways. The total number of ways to choose a pair of balls with the same number (from all $ N $ balls) is
Here is an inductive proof of formula (2): the formula is correct for $ n = 2 $; assuming its validity for some $ n \geq 2 $, we have for $ n+1 $
Thus, formula (2) is proved.
The sought probability is the quotient of the found number (formula (2)) by $ q $ (formula (1)), i.e., it equals $ \dfrac{4}{3(n+2)} $. | \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 if and only if
From the Law of Sines applied to triangle $ ABC $, we have
so the equality (*) is equivalent to the equality
Since $ A = \pi - (B+C) $, then $ \sin A = \sin(B+ C)= \sin B\cos C+\cos B\sin C $, hence the equality (*) is equivalent to the equality
We divide the numerator and the denominator by $ \cos^2B \cos^2 C $:
and transform it further equivalently by applying the relationship $ \frac{1}{\cos^2 x} = \tan^2 x +1 $.
The last equality holds if and only if $ 2= 1-\tan B \tan C $ or $ 2=-(1-\tan B \tan C) $, i.e.,
The first of these cases holds if and only if $ \overrightarrow{BX} = \overrightarrow{CY} $, which is possible only if one of the angles $ B $ or $ C $ is obtuse, the second case can only occur for an acute triangle. | \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 each ray can reflect from the sides of the angle at most $ n $ times. | 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 collinear (Fig. 15). This follows from the equality of angles $\measuredangle CPA = \measuredangle QPO = \measuredangle \varphi(Q)PO$.
om29_3r_img_15.jpgom29_3r_img_16.jpg
Therefore, the problem reduces to the following: From point $O$, we draw half-lines $m_1, m_2, \ldots, m_k, m_{k+1}, \ldots$ such that $\measuredangle (m_i, m_{i+1}) = \theta$ for $i = 1, 2, \ldots$ (Fig. 16). We need to prove that there exists a natural number $n$ with the property that any half-line $m$ passing through a point located inside the angle $\measuredangle (m_1, m_2)$ and intersecting the half-line $m_2$ intersects no more than $n$ of the drawn half-lines.
To prove this, it suffices to choose $k$ such that $(k - 2)\theta < \pi \leq (k - 1)\theta$. Then $\measuredangle (m_2, m_{k+1}) = (k - 1)\theta \geq \pi$ and the half-line $m$ intersecting the half-line $m_2$ will not intersect the half-line $m_{k+1}$, because the half-lines $m_3$ and $m_{k+1}$ do not lie in the same open half-plane determined by the line containing the half-line $m_2$. Thus, the half-line $m$ will intersect at most $k - 1$ of the given half-lines $m_2, m_3, \ldots, m_k$, where $k - 2 < \frac{\pi}{\theta}$, which means $k - 1 < \frac{\pi}{\theta} + 1$. Therefore, the light ray will reflect off the sides of the given angle fewer than $\frac{\pi}{\theta} + 1$ times. It suffices to take $n = \left[ \frac{\pi}{\theta} \right] + 1$. | [\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 points $Y$ and $Z$ analogously.
om56_3r_img_4.jpg
We will show that the area of the hexagon $AZBXCY$ is not less than $3/4$, which will complete the solution of the problem.
Through points $A$, $B$, and $C$, we draw lines parallel to the lines $BC$, $CA$, and $AB$, respectively, forming a triangle with vertices $D$, $E$, $F$ (Fig. 5). From the fact that triangle $ABC$ has the largest area among all triangles with vertices in the vertices of the polygon $\mathcal{W}$, it follows that the polygon $\mathcal{W}$ is contained within the triangle $DEF$.
Through points $X$, $Y$, and $Z$, we draw lines parallel to the lines $BC$, $CA$, and $AB$, respectively, intersecting the sides of the triangle $DEF$ at points $K$, $L$, $M$, $N$, $P$, $Q$, as shown in Fig. 5. From the definition of points $X$, $Y$, and $Z$, it follows that the polygon $\mathcal{W}$ is contained within the hexagon with vertices $K$, $L$, $M$, $N$, $P$, $Q$.
Let $s$ denote the area of triangle $ABC$. Then each of the triangles $BCD$, $CAE$, and $ABF$ has an area equal to $s$. Let $x$, $y$, $z$ denote the areas of triangles $BCX$, $CAY$, and $ABZ$, respectively. We need to show that $x + y + z + s \geq 3/4$.
Let $J$ be a point such that the quadrilateral $CBJL$ is a parallelogram. Then triangles $CBD$ and $JKB$ are similar. Denoting the area of a figure $\mathcal{F}$ by $[\mathcal{F}]$, we have the relationships
thus $[JKB] = x^2/s$. From this, we obtain
Similarly, we express the areas of trapezoids $ACMN$ and $BAPQ$. Therefore, the area of the hexagon with vertices $K$, $L$, $M$, $N$, $P$, $Q$ is
Since this hexagon contains the given polygon $\mathcal{W}$, we have
It suffices to show that
Multiplying both sides of this inequality by $12s$ and reducing similar terms, we obtain its equivalent form:
and this last inequality is clearly satisfied. | 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 quadrilateral $MQNP$ is equal to the sum of the areas of triangles $APD$ and $CQB$. | 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 medians
$AD$, $BE$, and $CF$. Consider the homothety with center $S$ and scale $-1/2$. This homothety maps points $A$, $B$, and $C$
to points $D$, $E$, and $F$ respectively, and the circle of radius $R$ containing triangle $ABC$ to a circle $\omega$ of radius $R/2$
containing triangle $DEF$.
om57_1r_img_3.jpg
om57_1r_img_4.jpg
Draw a tangent $k_A$ to circle $\omega$ that is parallel to side $BC$ and lies on the opposite side of line
$BC$ from point $A$. Such a line exists because circle $\omega$ intersects line $BC$. Similarly, we construct tangents
$k_B$ and $k_C$ parallel to lines $CA$ and $AB$ respectively. Lines $k_A$, $k_B$, and $k_C$ form the sides of triangle
$XYZ$ similar to triangle $ABC$ (Fig. 4). Moreover, circle $\omega$ is inscribed in triangle $XYZ$. Since triangle
$ABC$ is contained within triangle $XYZ$, the radius of the circle inscribed in triangle $ABC$ does not exceed the radius of the circle inscribed
in triangle $XYZ$. Hence $r \leq R/2$, or $R \geq 2r$. This completes the proof of the lemma.
We proceed to solve the problem.
Consider the plane $\pi$ passing through point $S$ and parallel to the plane $ABC$. Plane $\pi$ intersects edges $AD$, $BD$, $CD$ at points $A$, $B$, $C$ respectively, and its intersection with the sphere inscribed in tetrahedron $ABCD$ is a circle $\omega$ of radius 1, contained in triangle $A$. The homothety with center $D$ that maps triangle $A$ to triangle $ABC$ maps circle $\omega$ to a circle of radius greater than 1 and contained in triangle $ABC$. Hence, the radius of the circle inscribed in triangle $ABC$ is greater than 1.
Let $T$ be the point of tangency of the sphere inscribed in tetrahedron $ABCD$ with face $ABC$ and suppose that $SA \leq \sqrt{5}$. Then also $SB \leq \sqrt{5}$ and
$SC \leq \sqrt{5}$. Furthermore
and similarly $TB \leq 2$ and $TC \leq 2$. Therefore, triangle $ABC$ lies inside a circle with center $T$ and radius 2, which means (by the lemma) that the radius of the circle inscribed in triangle $ABC$ is less than or equal to 1. This is a contradiction, since we have already shown that this radius is greater than 1. This contradiction proves that $SA > \sqrt{5}$. | 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 element, so it is contained in any set with which it is not disjoint. On the other hand, for any two of the remaining
$ n - 1 $ sets, the set that is listed earlier in the above sequence is contained in the other set.
In this way, we have proved that
On the other hand, let $ B_1, B_2, \cdots, B_t $ be different
subsets of the set $ \{1, 2, \cdots,m\} $ with the property described in the problem statement. If among the listed subsets the set
$ \{1,2, \cdots,m\} $ does not appear, we can add it, increasing the number of these
subsets by $ 1 $ without violating the validity of the postulated condition.
Let us assume, then, that $ B_t = \{1,2, \cdots,m\} $.
Among the sets $ B_1, B_2, \cdots, B_{t-1} $, let us choose the one that has the most elements. For the sake of argument, let it be $ B_1 $
and let $ d $ denote the number of its elements. Then $ d < n $.
Moreover, the equality $ d = 0 $ means that $ B_1 $ is the empty set, which, given the choice of the set $ B_1 $, is possible only if
$ t = 2 $. In the further part of the reasoning, we assume, then, that $ 0 < d < n $.
By the definition of the set $ B_1 $, each of the sets
$ B_i $ ($ i=1,2,\cdots,t - 1) $ is either a subset of the set $ B_1 $,
or is disjoint from it. By renumbering, we can assume that the sets $ B_1, B_2, \cdots, B_r $ are subsets of the set
$ B_1 $, while the sets $ B_{r+1}, B_{r+2}, \cdots, B_{t-1} $ are
disjoint from $ B_1 $.
The sets $ B_1, B_2, \cdots, B_r $ are different subsets of
the $ d $-element set $ B_1 $, any two of which are either
disjoint or one is contained in the other. It follows that
$ r \leqslant a_d $. Among the $ t-r-1 $ sets
$ B_{r+1}, B_{r+2}, \cdots, B_{t-1} $, any two are either disjoint,
or one is contained in the other, and they are subsets of the
$ (m-d) $-element set $ {1, 2, \cdots,m}\setminus B_1 $,
and moreover, the empty set does not appear among them, as it
is a subset of $ B_1 $. Hence we obtain
$ t-r -1 \leqslant a_{m-d} -1 $ and consequently
As a result, the number of subsets of the set $ \{1,2, \cdots,m\} $
with the desired property does not exceed the number $ a_d +a_{m-d} $.
We have thus shown that $ a_m $ does not exceed the maximum of the numbers
If, then, in the inequality (1) equality holds for
$ n =1, 2, \cdots,m - 1 $, then it also holds for $ n = m $.
Since for $ n = 1 $ the inequality (1) becomes an equality, using
induction, we finally conclude that $ a_n =2n $ for $ n =1,2,3, \cdots $. | 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 squares, so the area of the sought rectangle is
The number $1056 = 2^5 \cdot 3 \cdot 11$ is the product of the lengths of two adjacent sides of the rectangle. To find these lengths, we need to factorize 1056 into two factors, each of which is not less than 18, i.e., the length of the side of the largest square. There are three such factorizations:
The factorizations $22 \cdot 48$ and $24 \cdot 44$ should be rejected. If, in a rectangle with a width of 22 or 24, we place a square with a side of 18, then the remaining part above or below the square will be a strip of length 18 and a width no greater than 6, and such a strip cannot be filled with the given squares, as only squares with sides 1 and 4 can fit, and the sum 1 + 4 = 5 is less than 18.
Therefore, the sought rectangle, if it exists, has sides 32 and 33. We will show that indeed a rectangle ABCD with sides $AB=32$ and $AD=33$ (Fig. 11) can be assembled from the given nine squares in one and only one way.
om1_Br_img_11.jpg
The square with a side of 18 must be located in one of the corners of the rectangle. Otherwise, above and below this square, there would remain strips of length 18 and a combined width of 14. One of these strips would then have a width no greater than 7, and such a strip cannot be filled with the given squares, as only squares with sides 1, 4, and 7 can fit, and the sum 1 + 4 + 7 = 12 is less than 18.
Therefore, we draw the square $AMNP$ with side $AM=18$.
Next, we will decide which squares should be arranged along the segment $MB=14$. The number 14 must be either the side length of one square or the sum of the side lengths of several squares; if the latter case, there are only two possibilities:
om1_Br_img_12.jpg
Both of these possibilities are invalid. If we place a square with a side of 9 or 10 along the segment $MB$, a strip of length 9 or 10 and a width no greater than 5 will remain, which again cannot be filled with the remaining given squares. Therefore, a square $MBKL$ with a side of 14 must be placed along the side $MB$.
Now let's pay attention to point $N$. A dividing line must extend from this point into the interior of the figure $PNLKCD$, either as an extension of segment $MN$ or as an extension of segment $PN$. The first possibility (indicated on Fig. 11 with a dashed line) cannot occur.
In that case, the side $PN=18$ would be the sum of the sides of several squares, and among them, the side of 15, as there would be no other place for a square with a side of 15. But 18 = 15 + 3, and 3 is neither a side nor the sum of the sides of the given squares. Therefore, the dividing line extending from point $N$ must be the extension of side $PN$.
In this case, using similar reasoning as before, we will conclude that the segment $LN=4$ must be the side of one and only one square $LRSN$, then we will conclude that the segment $RK=10$ must also be the side of one and only one square, and then we will easily find the positions of the remaining squares and obtain the solution to the problem as shown in Fig. 12. | 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 sides $BC$, $CA$, $AB$.
om45_1r_img_5.jpg
Furthermore, we assume
Thus, $a = y + z$, $b = z + x$, $c = x + y$. Now, observe that
Therefore, in view of the obvious estimate $a \leq 2R$, we obtain the inequality
which we were to prove.
{\kom Note:} As can be seen, a stronger inequality than the required one holds:
This is indeed a strict inequality, as the estimates $x \leq \sqrt{3}r$ and $a \leq 2R$ used in it cannot simultaneously become equalities: the first one becomes an equality only for an equilateral triangle, and the second one only for a right triangle. | 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^\circ - \measuredangle PCL = \measuredangle BCA $. The obtained relationships prove that triangles $ XLC $ and $ BCA $ are similar. Hence, and from the equality $ BO = CO $, it follows that
which means $ \measuredangle XCL + \measuredangle LCB + \measuredangle BCO = 180^\circ $. The last equality indicates that points $ X $, $ C $, $ O $ lie on the same line.
om53_3r_img_1.jpg | 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 the given number gives a remainder of $1$ when divided by $3$. Since, by assumption, the number is divisible by $3$, at least two of the squares in the considered sum must be squares of numbers divisible by $3$ - they are therefore divisible by $9$. | 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 example, that the numbers $ x_1 $, $ x_2 $, $ x_3 $ are of the same parity.
Reasoning similarly as before, we conclude that among the numbers $ y_1 $, $ y_2 $, $ y_3 $, at least two are of the same parity. Let's assume, for example, that the numbers $ y_1 $, $ y_2 $ are of the same parity. Then the numbers $ x_1 + x_2 $ and $ y_1 + y_2 $ will be even, and thus the midpoint of the segment $ P_1P_2 $ will have integer coordinates $ \left(\frac{1}{2}(x_1+x_2), \frac{1}{2}(y_1+y_2) \right) $, i.e., it will be a lattice point.
Note 1. An analogous theorem holds in $ n $-dimensional space: If there are $ 2^n + 1 $ points with integer coordinates in $ n $-dimensional space, then the midpoint of one of the segments connecting these points has integer coordinates.
Note 2. The problem can also be generalized in another way. If there are $ n $ lattice points on a plane, then the midpoints of at least $ \displaystyle \binom{\left< \frac{n}{4} \right>}{2} $ segments connecting these points are lattice points.
Here $ \left<x\right> $ denotes the smallest integer not less than $ x $.
Proof of Note 2. Divide the given points into classes depending on the parity of their coordinates. There are four classes $ (n, n) $, $ (n, p) $, $ (p, n) $, and $ (p, p) $, where $ n $ indicates that the corresponding coordinate is odd, and $ p $ indicates that it is even. Therefore, in one of the classes, there will be at least $ \frac{n}{4} $ points, and since the number of points in a class is an integer, there will be at least $ \left< \frac{n}{4} \right> $ points in one of the classes.
Since the midpoint of a segment whose endpoints are points in the same class is a lattice point, and for $ r $ points there are $ \binom{r}{2} $ segments connecting them, the number of segments whose midpoints are lattice points is at least $ \displaystyle \binom{\left< \frac{n}{4} \right>}{2} $.
Notice also that | 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 verify this, we can use the known formulas for the length of medians:
\[
m_a = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}, \quad m_b = \frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2}, \quad m_c = \frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2}
\]
From these formulas, it follows that, for example, if $m_a > m_b$, then $m_a^2 > m_b^2$, and according to the above formula, $b^2 > a^2$, hence $b > a$.
Suppose $a \leq b \leq c$, then $m_c \leq m_b \leq m_a$. The triangle with sides $m_a$, $m_b$, $m_c$ is similar to the triangle with sides $a$, $b$, $c$ if and only if
\[
\frac{m_a}{a} = \frac{m_b}{b} = \frac{m_c}{c}
\]
In problem 8, we proved that the area of the triangle equals $\frac{4}{3}$ of the area of the triangle formed by its medians. If these triangles are similar, the ratio of their areas equals the square of the ratio of corresponding sides, so from (2) we have $m_a^2 = \frac{3}{4} c^2$. Therefore, using the first formula from (1),
\[
m_a = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}
\]
Thus,
\[
\frac{3}{4} c^2 = \frac{1}{4} (2b^2 + 2c^2 - a^2)
\]
Simplifying, we get
\[
3c^2 = 2b^2 + 2c^2 - a^2
\]
\[
c^2 = 2b^2 - a^2
\]
Conversely, if condition (3) is satisfied, then the triangle with sides $a$, $b$, $c$ is similar to the triangle formed by its medians. Indeed, considering the relationship (3) in the formulas (1), we obtain
\[
m_a = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2} = \frac{1}{2} \sqrt{2b^2 + 2(2b^2 - a^2) - a^2} = \frac{1}{2} \sqrt{6b^2 - 3a^2} = \frac{\sqrt{3}}{2} \sqrt{2b^2 - a^2} = \frac{\sqrt{3}}{2} c
\]
\[
m_b = \frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2} = \frac{1}{2} \sqrt{2a^2 + 2(2b^2 - a^2) - b^2} = \frac{1}{2} \sqrt{4b^2 - b^2} = \frac{1}{2} \sqrt{3b^2} = \frac{\sqrt{3}}{2} b
\]
\[
m_c = \frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2} = \frac{1}{2} \sqrt{2a^2 + 2b^2 - (2b^2 - a^2)} = \frac{1}{2} \sqrt{3a^2} = \frac{\sqrt{3}}{2} a
\]
From this, it follows that $m_a$, $m_b$, $m_c$ are proportional to $c$, $b$, $a$.
Thus, we have the following result:
A triangle is similar to the triangle formed by its medians if and only if the square of the middle side of the triangle is equal to the arithmetic mean of the squares of the shortest and longest sides.
Note 1. In the solution, we could have avoided referring to problem 8 and derived condition (3) from formulas (1) and (2). This straightforward calculation is proposed as an exercise. | ^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 before. Prove that the number of reflections off the sides of the square is odd. | 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 that side (see Figure 5). Thus, for example, the point $X = (u,v)$ in the square $OABC$ with vertices $O = (0,0)$, $A = (1,0)$, $B = (1,1)$, $C = (0,1)$ is identified with the points $(2-u,v)$ and $(u,2-v)$ in the unit squares adjacent to $OABC$ on the right and above, and also with two points in the two unit squares adjacent to $OABC$ on the left and below.
Through symmetry with respect to further grid lines, this point has its replicas in all squares of the grid - one in each unit square. The center of the square $OABC$, i.e., the point $P=(\frac{1}{2},\frac{1}{2})$, like any other, has infinitely many copies; they are the centers of all the squares in the grid, i.e., points of the form $(\frac{1}{2}+ m, \frac{1}{2} +n)$, where $m, n$ can be any integers.
We can assume that $OABC$ is the square in question, and that a light ray was sent from point $P$ "to the right and up" (along a line with a positive slope). It first reflects off the side $AB$ or $BC$ at some point $Q$. The next time it hits the edge of the square $OABC$ at some point $R$.
The segment $QR$ of its trajectory has a copy $QR$ in the square adjacent to $OABC$ (along the side $AB$ or $BC$), which is a linear extension of the segment $PQ$; the linearity follows from the law of reflection. Indeed: in the situation as shown, the point $R$ is symmetric to $R$ with respect to the line $AB$, so $|\measuredangle BQR| = |\measuredangle BQR|$, and since also $|\measuredangle BQR| = |\measuredangle PQA|$ (law of reflection), the point $R$ lies on the ray $PQ^\to$.
Similarly, the next segment of the ray's path has a copy in the next square of the grid, which is an extension of the segment $QR$. And so on: each subsequent segment of the path can be identified with a further segment of the ray $PQ^\to$. In this way, the entire trajectory of the ray, from its start at point $P$ to its return to point $P$, corresponds to a certain linear path on the plane, starting at point $P$ and ending at some point $Z$, which is one of the replicas of the center of the square $OABC$. Thus, $Z$ is a point of the form $(\frac{1}{2} +m, \frac{1}{2} + n)$ for some pair of natural numbers $m, n$.
The segment $PZ$ contains, in particular, the point $M=(\frac{1}{2}(1+m),\frac{1}{2}(1 + n))$, the midpoint of this segment. If both numbers $m, n$ are even, then $M$ is the center of some square in the grid, i.e., an image of the center $P$ of the square $OABC$. If both numbers $m, n$ are odd, then $M$ is a point with integer coordinates, i.e., an image of a vertex of the square $OABC$. Each of these situations is excluded by the conditions of the problem. Therefore, the numbers $m$ and $n$ are of different parity, and thus their sum is an odd number.
However, note that the segment $PZ$ intersects $m$ vertical lines and $n$ horizontal lines, which are grid lines. Each such intersection point corresponds to a point where the ray reflects off the edge of the square $OABC$. This correspondence is one-to-one, since the segment $PZ$ does not pass through any grid node (the "ray did not hit a vertex of the square"). Hence, the number of reflections is equal to the sum $m + n$; this sum, as we have shown, is an odd number. The theorem is thus proved. | 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 segments $\overline{AA'}$, $\overline{BB'}$, $\overline{CC'}$, and by $A_2$, $B_2$, $C_2$ the midpoints of segments $\overline{AA'}$, $\overline{BB'}$, $\overline{CC'}$ (Fig. 6). We have $\overrightarrow{A_1A_2} = \overrightarrow{A_1A} + \overrightarrow{AA_2} = \frac{1}{2} \overrightarrow{OO'}$. Similarly, we find that $\overrightarrow{B_1B_2} = \frac{1}{2} \overrightarrow{OO'}$ and $\overrightarrow{C_1C_2} = \frac{1}{2} \overrightarrow{OO'}$. Therefore, points $A_2$, $B_2$, $C_2$ are the images of points $A_1$, $B_1$, $C_1$ under the translation by the vector $\frac{1}{2} \overrightarrow{OO'}$. It suffices to prove that points $A_2$, $B_2$, $C_2$ are collinear or are the vertices of an equilateral triangle.
Equilateral triangles $\triangle ABC$ and $\triangle A'B'C'$ are inscribed in the circle $K$ with center $O$. There exists an isometry $\varphi$ that maps points $A$, $B$, $C$ to $A'$, $B'$, $C'$, respectively. This isometry maps the circle $K$ onto itself. Therefore, the isometry $\varphi$ is either a rotation about the point $O$ or a reflection across an axis.
If $\varphi$ is a rotation (Fig. 7), then performing a rotation $S$ by an angle $\frac{2}{3} \pi$ about the point $O$ will map points $A$ and $A'$ to points $B$ and $B'$, respectively. Thus, $S(A_2) = B_2$ and similarly $S(B_2) = C_2$ and $S(C_2) = A_2$. Therefore, $\overline{A_2B_2} = \overline{S(A_2) S(B_2)} = \overline{B_2C_2}$ and similarly $\overline{B_2C_2} = \overline{C_2A_2}$. Hence, triangle $A_2B_2C_2$ is equilateral or $A_2 = B_2 = C_2 = 0$ (Fig. 8).
If the isometry $\varphi$ is a reflection, then points $A_2$, $B_2$, $C_2$ lie on the axis of this reflection and are therefore collinear (Fig. 9). | 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 trapezoid. Each of the triangles $ABM$ and $DCM$ thus has an area smaller than half the area of the trapezoid, and it follows that a line passing through point $M$ and dividing the trapezoid into two parts of equal area must intersect line $BC$ at some point $N$ lying between points $B$ and $C$. Let $MN$ be the sought line, then $\text{area of }ABNM = \text{area of }MNCD$. Drawing segments $AN$ and $ND$, we have $\text{area of }AMN = \text{area of }MDN$, and thus also $\text{area of }ABN = \text{area of }CDN$. Since $\text{area of }ABN = AB \cdot BN \cdot \sin \measuredangle B$, $\text{area of }CDN = CD \cdot CN \cdot \sin \measuredangle C$, and
therefore
the sought point $N$ must divide segment $BC$ in the ratio $\frac{CD}{AB}$. Conversely, if $\frac{BN}{CN} = \frac{CD}{AB}$, then retracing the above reasoning in reverse, we find that indeed $\text{area of }ABNM = \text{area of }MNCD$. The problem always has a solution and only one.
The problem can be solved without using trigonometry, relying instead, for example, on the theorem that the areas of triangles having one pair of equal angles are in the same ratio as the products of the sides containing those equal angles.
For this purpose, on the extension of side $BC$, we measure segment $CE = CN$. Then $\text{area of }DCE = \text{area of }DCN = \text{area of }ABN$, where triangles $ABN$ and $DCE$ have equal angles at vertices $B$ and $C$. Applying the theorem mentioned above, we obtain
i.e., the same equality as before. | 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 the same arc are equal).
Similarly, in the quadrilateral $ SMBP $ we have
But $ \measuredangle SCN = \measuredangle SBP $, since each of these angles equals
$ 90^{\circ} - \measuredangle BAC $; therefore $ \measuredangle SMN = \measuredangle SMP $.
Indeed, the altitude $ AM $ is the angle bisector of $ \angle NMP $.
Note. In an obtuse triangle (Fig. 4) the altitude $ AM $ drawn from the vertex of the obtuse angle $ A $ is the angle bisector of $ \angle NMP $, while the other two altitudes are the bisectors of the exterior angles of triangle $ MNP $; the proof is analogous to the previous one.
om1_Br_img_4.jpg | 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 $X$ and $Y$ on the planes $A$ and $B$.
For this purpose, consider the sheaf of planes with edge $m$, i.e., the set of all planes passing through the line $m$. Each segment $XY$, being parallel to the line $MN$, lies in one of the planes of this sheaf. We will determine the desired geometric locus by finding its points in each of the planes of the sheaf. We distinguish two cases.
1° Let the plane of the sheaf intersect the line $s$ at point $P$, and thus the planes $A$ and $B$ along the lines $PM$ and $PN$ (Fig. 35a). Those of the considered segments $XY$ that lie in the plane $MPN$ have their endpoints on the lines $PM$ and $PN$; the midpoints $Z$ of these segments thus form a line $p$ passing through the points $P$ and $S$.
2° Let the plane of the sheaf be parallel to the line $s$, and thus intersect the planes $A$ and $B$ along the lines $MK$ and $NL$ parallel to the line $s$ (Fig. 35b).
Those of the considered segments $XY$ that lie in the plane $MKS$ have their endpoints on the lines $MK$ and $NL$, and the midpoints $Z$ of these segments form a line $q$ passing through the point $S$ and parallel to the line $s$.
The desired geometric locus thus consists of the points of all lines drawn through the point $S$ to all points $P$ of the line $s$, and the points of the line $q$ drawn through the point $S$ parallel to the line $s$.
These points form a plane determined by the point $S$ and the line $s$.
Note. One might have doubts as to whether the line $s$ belongs to our geometric locus, since a line parallel to the line $m$ and drawn from a point on the line $s$ intersects both planes $A$ and $B$ at the same point and does not have a segment $XY$ on this line. In geometry, however, it is convenient to introduce the convention that if the points $X$ and $Y$ coincide, they form a zero segment, i.e., a segment consisting of only one point, which is then the beginning, end, and midpoint of this segment.
With this convention, the line $s$ belongs to the determined geometric locus. | 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 line $l$, see: Note).
Let the points of intersection of the line $l$ with the boundary of the square be denoted by $A$ and $C$; we can assume (by cyclically reassigning the labels of the vertices of the square if necessary) that point $A$ lies on side $AB$, and point $C$ lies on side $CD$, with $A \neq C$.
Draw a line $m$ through point $O$ perpendicular to $l$. It intersects the boundary of the square at points $B$ and $D$, which lie on sides $BC$ and $DA$ respectively, with $B \neq D$ (see Figure 7).
The lines $l$ and $m$ divide the square into four congruent figures. Triangle $A$ lies on one side of the line $l$, and thus is contained in the union of two of these figures. One of them must contain at least two vertices of triangle $A$, i.e., the endpoints of a segment of length not less than $1$. Let this be the part of the square cut out by the rays $(OA$ and $(OB (such an assumption does not reduce the generality of the considerations).
In the special case where the lines $l$ and $m$ contain the diagonals of the square $ABCD$ (i.e., when $A = A$, $B = B$, $C = C$, $D = D$), the considered figure is the triangle $OAB$. The only segment of length $\geq 1$ contained in this triangle is its hypotenuse $AB$. This would mean that segment $AB$ is one of the sides of triangle $A$ - contrary to the assumption that triangle $A$ lies inside the square $ABCD$.
In the remaining case ($A \neq A$ etc.), the considered figure is the quadrilateral $OA$, where the angles at vertices $O$ and $B$ are right angles - so a circle can be circumscribed around it. Its diameter is the segment $A$ of length less than $1$ (since $A$ is a square inscribed in $ABCD$, not identical to $ABCD$). Therefore, the figure (quadrilateral) $OA$ cannot contain a pair of points at a distance of at least $1$ - contrary to what we stated a moment ago.
The obtained contradiction completes the proof.
Note. In the solution, we used the fact that through any point $O$ lying outside a given triangle $\Delta$, one can draw a line $Z$ such that the triangle $\Delta$ lies entirely on one side of it. It might be worth providing a rigorous justification for this fact.
The triangle is the intersection of three closed half-planes, whose edges are the lines containing its sides. Since point $O$ does not belong to the given triangle, it does not belong to at least one of these half-planes. Let $l$ be its edge (if there are two such half-planes, we choose one of them, and its edge we denote by $l$). The line $Z$, parallel to $V$ and passing through $O$, has the desired property. | 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 points from the set $ R_n $. Let $ p_n $ be the probability that the selected points are the vertices of an obtuse triangle. Calculate $ \lim_{n\to \infty} p_n $.
Note: Three collinear points do not form a triangle. | 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 equality: from the number of all triples, we subtract the number of collinear triples of points - that is, triples of points lying on any of the four edges, each of which contains $ n +1 $ points of the set $ R_n $.)
After expanding the binomial coefficients, we obtain the formulas
Let us divide the set $ R_n $ into four-element "layers" $ H_0, H_1, \ldots, H_n $: the vertices of one of the square faces of the cube $ K_n $ form the layer $ H_0 $, the vertices of the other - the layer $ H_n $; the layer $ H_k $ consists of points of the set $ R_n $ at a distance of $ k $ units from the plane of the layer $ H_0 $.
Since we are interested in the limiting (as $ n \to \infty $) behavior of the quantities under consideration, we can assume that $ n \geq 4 $.
Fix a natural number $ r $, $ 3 \leq r \leq n-1 $, and choose any numbers $ k, m \in \{0,1,\ldots,n\} $ such that $ m - k = r + 1 $. There are thus $ r $ layers between layers $ H_k $ and $ H_m $. By choosing any points $ A \in H_k $, $ C \in H_m $, and $ B \in H_l $, where $ l \in \{k + 1,\ldots,m-1\} $, we obtain either an obtuse triangle or a collinear triple of points (since $ m-k \geq 4 $). We have $ 16 $ ways to position the pair of points $ A $ and $ C $ and $ 4r $ possible positions for point $ B $ - a total of $ 64r $ possibilities; $ 4r $ of them give collinear triples. The remaining $ 60r $ are obtuse triangles.
Since the numbers $ k $, $ m $ (with a given difference $ r+1 $) can be chosen in $ n-r $ ways, the number $ P(n) $ of all possible obtuse triangles is at least the sum
Thus, we have a two-sided estimate
The quantities $ Q(n) $, $ T(n) $, $ S(n) $ are third-degree polynomials (in $ n $). The coefficient of $ n^3 $ in $ Q(n) $ is $ 32/3 $, while in $ T(n) $ and $ S(n) $ the coefficient of $ n^3 $ is $ 10 $. Therefore,
and by the squeeze theorem,
| \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 bounding plane) $ \pi $ bounded by the plane $ \alpha $. Since a segment connecting two points of the half-space $ \pi $ is entirely contained in $ \pi $, all edges of the polyhedron $ W $ lie in $ \pi $.
It follows that all points of the faces of the polyhedron belong to $ \pi $, as each point on a face lies either on an edge or on a segment connecting points of two edges. Similarly, each interior point of the polyhedron belongs to $ \pi $, as each such point lies on a segment connecting points of two faces.
Let polyhedra $ W_1 $ and $ W_2 $ be given. Consider all segments connecting the vertices of the polyhedron $ W_1 $ with the vertices of the polyhedron $ W_2 $. Since the set of these segments is finite, there is at least one segment of maximum length among them. Let such a segment be $ P_1P_2 $ ($ P_1 $ - a vertex of $ W_1 $, $ P_2 $ - a vertex of $ W_2 $) and let $ A_1 $ be any point of the polyhedron $ W_1 $, and $ A_2 $ - any point of the polyhedron $ W_2 $.
We will prove that $ P_1P_2 \geq A_1A_2 $.
Draw through the point $ A_1 $ a plane $ \alpha_1 $ perpendicular to the line $ A_1A_2 $, and through the point $ A_2 $ a plane $ \alpha_2 $ also perpendicular to $ A_1A_2 $. The open half-space $ \pi $, which is bounded by the half-plane $ \alpha_1 $ and contains the point $ A_2 $, does not contain the entire polyhedron $ W_1 $, as it does not contain the point $ A_1 $. Therefore, by the previous observation, there exists a vertex $ B_1 $ of the polyhedron $ W_1 $ not belonging to $ \pi $, i.e., lying on the plane $ \alpha_1 $ or on the opposite side of the plane $ \alpha_1 $ from the point $ A_2 $. Similarly, there exists a vertex $ B_2 $ of the polyhedron $ W_2 $ lying on the plane $ \alpha_2 $ or on the opposite side of the plane $ \alpha_2 $ from the point $ A_1 $. Since $ B_1B_2 \geq A_1A_2 $, and $ P_1P_2 \geq B_1B_2 $, therefore | 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 remaining two. | 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 shown that if the roots of the polynomial $W$ satisfy the relation (1), then for $ \displaystyle m = \frac{1}{2}(x_1 + x_2) $, we have $ W(x+m) = x^4 + px^2 + q $, where $ \displaystyle p = - \frac{1}{4} (x_1 - x_2)^2 - \frac{1}{4} (x_3 - x_4)^2 $, $ \displaystyle q = \left[ \frac{1}{4}(x_1 - x_2)(x_3 - x_4) \right]^2 $.
We will now prove the converse implication. If a real number is a root of the polynomial $x^4 + px^2 + q$, then its opposite number is also a root. Therefore, if for some real number $m$ the polynomial $W(x+m)$ has the form $x^4 + px^2 + q$, then its roots $y_1, y_2, y_3, y_4$ satisfy the equations (2). The roots of the polynomial $W(x)$ are the numbers $x_i = y_i + m$, where $i = 1, 2, 3, 4$. From (2) we thus obtain that $x_1 + x_2 = y_1 + m + (-y_1) + m = 2m$ and similarly $x_3 + x_4 = y_3 + m + (-y_3) + m = 2m$. Therefore, the equation (1) holds.
Note. It is not difficult to notice that there is only one number $m$ that satisfies the conditions of the problem. If $f(x) = x^4 + px^2 + q$, then for $k \ne 0$ the polynomial $f(x+k)$ has the coefficient of $x^3$ equal to $4k$, and thus different from zero. | 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 these segments are all three blue, we have a triangle $ A_2A_3A_4 $ of one color. If, however, one of these segments, say $ A_2A_3 $, is red, then the triangle $ A_1A_2A_3 $ is of one color.
For a smaller number of given points, the thesis of the theorem is not true. For example, in a convex pentagon, where all sides are red and all diagonals are blue, any three consecutive vertices form a triangle with two red sides and one blue side, and any three non-consecutive vertices form a triangle with two blue sides and one red side. | 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 $ASC$ and $DSB$ are similar, from which we particularly obtain $\measuredangle ACS = \measuredangle DBS$. Hence, it follows that the arcs $GA$ and $FD$ of the circle $o$ are equal (by "arc $XY$" we mean the arc running along the circle $o$ from point $X$ to point $Y$ in the counterclockwise direction). Similarly, from the equality $\measuredangle SAD = \measuredangle SCB$, we conclude that the arcs $GB$ and $ED$ are equal. Therefore, the arcs $AB$ and $EF$ are also equal.
om56_3r_img_1.jpg
om56_3r_img_2.jpg
Quadrilateral $ABEF$ is thus an isosceles trapezoid, and point $S$ is the intersection of its diagonals (Fig. 2). Let us assume that point $P$ lies on the arc $AB$ of the circle $o$ that does not contain points $C$ and $D$. From the equality
it follows that line $PQ$ is parallel to lines $BE$ and $AF$. Hence, it follows that points $A$, $P$, $B$ are the images of points $F$, $Q$, $E$ under the reflection about the perpendicular bisector of segments $AF$ and $BE$. Therefore, $PS = QS$. | 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 line.
Determine whether there exists a strategy for the player starting the game
that guarantees him a win. | 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 formation of a sequence of 5 consecutive
checkers of the same color in one line, he makes such a move and wins
the game. Otherwise, the second player places a checker on the field symmetrical
to the center of the board to the field just occupied
by the first player, and in a color opposite to the checker just
placed by the first player.
The dimensions of the board are even numbers, so the field
symmetrical to the center of the board to a given field is a different field.
The described strategy is therefore correctly defined: after each move
of the second player, any pair of fields of the board symmetrical to
its center consists either of two free fields or of two occupied
fields, so after the first player's move, the field symmetrical to
the field he occupied is free.
We will show that it is not possible for the starting player to win
under such a strategy of the second player.
Suppose, on the contrary, that a certain move of the first player
secured his victory. Consider the position on the board before
making this move. As we have shown, in this position any
field is occupied by a checker if and only if the field symmetrical to it is occupied by a checker of the opposite color. Thus,
the arrangement of 4 checkers of one color $ k $, which the first player by
placing a checker on the field $ P $ increased to a winning arrangement
of 5 consecutive checkers of color $ k $ in one line, corresponds in this
position to an arrangement of 4 checkers of color $ k opposite to $ k $, and at the same time
the field $ P symmetrical to the field $ P $ is free. Then, however, in accordance
with the described strategy, the last move of the second player should consist
in placing a checker of color $ k on the field $ P, which leads
to the formation of a sequence of 5 consecutive checkers of color $ k in one line,
and consequently to the victory of the second player — contrary to the assumption
that the first player won by making the next move.
The obtained contradiction proves that under such a strategy of the second
player, the starting player cannot secure a win in the game. | 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 \leq 9 $, $ c > 0 $.
We will first prove that the only possible value of $ c $ is $ 15 $. Indeed, when $ x $, $ y $, $ z $ satisfy the above conditions, and $ 0 < c \leq 14 $, then
while if $ c > 16 $, then
Substituting $ c = 15 $ into equation (1), we obtain the equation
If integers $ x $, $ y $, $ z $ satisfy equation (2), then $ z $ is divisible by the common factor $ 5 $ of the first two terms, and since $ 0 \leq z \leq 9 $, there are only two possibilities:
a) $ z = 0 $; from equation (2) we get
from which it is clear that $ x $ and $ y $ can only have the values $ x = 1 $, $ y = 5 $, which gives the sought number the value $ 150 $. This number indeed satisfies the condition of the problem, as in base $ 15 $ numeration it represents the number $ 15^2 + 5 \cdot 15 = 300 = 2 \cdot 150 $.
b) $ z = 5 $; to determine $ x $ and $ y $ we get from (2) the equation:
from which it is clear that $ x < 3 $, since for $ x \geq 3 $ we have $ -5x + y + 1 \leq -15 + 9 + 1 = -5 $. When $ x = 1 $, then $ y = 4 $, and when $ x = 2 $, we have $ y = 9 $; the corresponding values of the sought number are $ 145 $ and $ 295 $. Both these numbers satisfy the condition of the problem, as
The problem thus has $ 3 $ solutions, which are the numbers $ 145 $, $ 150 $, $ 295 $. The appropriate base of numeration for each of them is $ 15 $. | 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 remainder $ r_j $. Calculate the sum $ a_1 + \ldots + a_n $. | 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
thus | 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 Ceva's theorem, the given equality of products is equivalent to the fact that the lines $AP$, $BQ$, $CR$ have a common point. | 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 angles and segments that do not directly follow from the assumptions, but could only be suggested by a figure drawn in accordance with the proven thesis.
Let $K$, $L$, $M$ be the midpoints of sides $BC$, $CA$, $AB$, and $U$, $V$, $W$ - the feet of the altitudes of triangle $ABC$, dropped to these sides, respectively. Let us assume:
Then
(Figure 7 shows the situation when $x < 0$, $y < 0$, $z > 0$.)
The medians $AK$, $BL$, $CM$ intersect at point $G$ (the centroid of triangle $ABC$), which divides each of them in the ratio $2:1$. Using vector notation, this can be written as follows:
From the first of these equalities and the fact that lines $AU$ and $GP$ are parallel, it follows that triangle $KAU$ is the image of triangle $KGP$ under a homothety with center $K$ and scale $3$ (if point $P$ coincides with $K$, both triangles degenerate into segments). Therefore, $\overrightarrow{KU} = 3 \cdot \overrightarrow{KP}$. Similarly (by invoking the parallelism relations $BV \parallel GQ$, $CW \parallel GR$), we justify the next two of the following equalities:
Hence, in view of (1), we obtain the relations
Since the altitudes of the triangle intersect at one point, by Ceva's theorem we have the equality of products $|BU| \cdot |CV| \cdot |AW| = |UC| \cdot |VA| \cdot |WB|$ (analogous to the equality $|BP| \cdot |CQ| \cdot |AR| = |PC| \cdot |QA| \cdot |RB|$ given in the assumption). Considering the relations (1) and (2), we can rewrite the discussed product equalities as
Using the notation
we give the obtained dependencies a concise form:
Of course, $f(t)$ is a polynomial (of the third degree) in the variable $t$. Let us arrange the expression defining $f(t)$ according to the powers of the variable $t$:
Thus,
where
The relations (3) give the system of equations
whose only solution is $p = q = 0$.
The equality $q = 0$, i.e., $xyz = 0$, means that
In this case, $P = K$ or $Q = L$ or $R = M$: the foot of one of the altitudes coincides with the midpoint of the corresponding side. This means that the triangle is isosceles. | 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. Specifically, if point $K$ is one of the vertices of the square, the only solution is the square $ABCD$ itself. If point $K$ lies on the interior of a side, there are two solutions: the square $ABCD$ and the square whose vertices are obtained by rotating point $K$ around the center of the square $ABCD$ by a multiple of a right angle.
Next, we will consider the cases where point $K$ is inside the square $ABCD$ (Fig. 46a) or outside the square (Fig. 46b).
**Analysis.** Suppose that a side $MN$ or its extension of an inscribed square $MNPQ$ in the given square $ABCD$ passes through the given point $K$ (Figs. 46a and 46b).
The squares $ABCD$ and $MNPQ$ have a common center $O$ (see problem 8 for level B on page 62).
Rotate the square $MNPQ$ by $90^\circ$ around point $O$ in the direction specified by the order of points $A$, $B$, $C$, $D$, and indicated by the arrow on the diagram. Points $M$ and $K$ will be in points $N$ and $L$ after the rotation, respectively. Since $\measuredangle KNL = 90^\circ$, point $N$ lies on the circle with diameter $KL$.
**Construction.** Rotate point $K$ by $90^\circ$ around point $O$ in the direction indicated by the arrow (Fig. 46) to obtain point $L$ lying on the line perpendicular to line $OK$ at point $O$, with $OL = OK$. Draw the circle $\Gamma$ with diameter $KL$.
Suppose the circle $\Gamma$ passes through point $N$ on the perimeter of the square $ABCD$. Construct the inscribed square $MNPQ$ in the square $ABCD$, where $M$ is the vertex of this square that, when rotated by $90^\circ$ around $O$ in the direction of the arrow, passes through point $N$.
We will show that the square $MNPQ$ satisfies the condition of the problem, specifically that line $MN$ passes through point $K$.
Consider two cases:
1° If point $K$ lies inside the square $ABCD$ (Fig. 46a), then $\measuredangle OKM = \measuredangle OLN$, because angle $OLN$ is the result of rotating angle $OKM$, and $\measuredangle OLN + \measuredangle OKN = 180^\circ$, since $\measuredangle OLN$ and $\measuredangle OKN$ are opposite angles of a cyclic quadrilateral.
From this, it follows that
Since the rays $OM$ and $ON$, and thus points $M$ and $N$, lie on opposite sides of line $OK$, it follows from the previous equality that rays $KM$ and $KN$ form a single line.
2° If point $K$ lies outside the square $ABCD$ (Fig. 46b), then $\measuredangle OKM = \measuredangle OLN$, as in case 1°, and $\measuredangle OLN = \measuredangle OKN$, as inscribed angles | 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., $ h_A > h_B $. Then it is not possible to place a segment of length $ h_A $ in this triangle, which would be parallel to the altitude drawn from point $ B $. The longest segment in this direction contained in triangle $ ABC $ is the altitude drawn from point $ B $, which has a length of $ h_B $. Therefore, the triangle $ ABC $ satisfying the conditions of the problem is equilateral. | 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 for such an arrangement of segments so that in each "network node" exactly three segments meet, lying in the same plane and forming angles of $120^\circ$, because the existence of a node of any other form allows for a local modification of the network in the vicinity of such a node, leading to a shortening of the network.)
Let $A_1B_1C_1D_1$ and $A_2B_2C_2D_2$ be two opposite faces of a cube, labeled so that segments $A_1A_2$, $B_1B_2$, $C_1C_2$, $D_1D_2$ are four parallel edges of the cube. Denote the midpoints of edges $A_1B_1$, $C_1D_1$, $A_2B_2$, $C_2D_2$ by $K_1$, $L_1$, $K_2$, $L_2$, respectively, and the midpoints of segments $K_1L_1$ and $K_2L_2$ (i.e., the midpoints of the two mentioned faces) by $O_1$ and $O_2$ (see Figure 4).
om39_1r_img_4.jpg
On segment $O_1O_2$, we find points $P_1$ and $P_2$ such that $|O_1P_1| = |O_2P_2| = \sqrt{3}/6$. Then $|P_1P_2| = 1 - \sqrt{3}/3$. (Segments $K_1P_1$, $L_1P_1$, $P_1P_2$, $P_2K_2$, $P_2L_2$ form a network in the square $K_1L_1L_2K_2$ as described in task $C$; the angles at points $P_1$ and $P_2$ are each $120^\circ$.)
The length of segment $K_1P_1$ is $\sqrt{3}/3$. Let $M_1$ be the midpoint of this segment; thus, $|K_1M_1| = |M_1P_1| = \sqrt{3}/6$. (It is not difficult to verify that now segments $M_1A_1$ and $M_1B_1$ form angles of $120^\circ$ with segment $M_1P_1$). Segments $A_1M_1$ and $B_1M_1$ have lengths equal to $\sqrt{3}/3$. Similarly, let $N_1$, $M_2$, $N_2$ be the midpoints of segments $L_1P_1$, $K_2P_2$, $L_2P_2$, respectively. We consider the set of the following thirteen segments:
Sum of their lengths is $1 + 3\sqrt{3}$. In this way, we have found a set of segments satisfying the given conditions. | 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 $MR = PS = \frac{1}{2} AD$; similarly, $MP \parallel BC \parallel RS$ and $MP = RS = \frac{1}{2} BC$. Quadrilateral $MPSR$ is a parallelogram; according to the theorem on the sum of the squares of the diagonals of a parallelogram, we write
Similarly,
From the above three equations, it is easy to calculate the lengths of $MS$, $PR$, $NQ$. We obtain, for example,
This is the desired formula expressing the distance between the midpoints of segments $AB$ and $CD$ in terms of the mutual distances of points $A$, $B$, $C$, $D$.
The above reasoning applies without change to the case where points $A$, $B$, $C$, $D$ lie on the same plane, but no three of them lie on the same line, and the points $M$, $N$, $P$, $Q$, $R$, $S$ are distinct.
It is not difficult to verify that formula (1) is general, i.e., it is true regardless of the position of points $A$, $B$, $C$, $D$. Some of these points or all of them could even coincide. We propose to the reader to modify the above proof appropriately to demonstrate the validity of formula (1) in the following cases, which together with the cases already considered exhaust all possibilities.
a) Three of the points $A$, $B$, $C$, $D$ lie on the same line, and the fourth is outside this line;
b) points $A$, $B$, $C$, $D$ lie on the same line;
c) no three of the points $A$, $B$, $C$, $D$ lie on the same line, but points $M$ and $S$ or points $N$ and $Q$ or points $P$ and $R$ coincide.
However, the generality of formula (1) could be proven in another way, namely by treating possibilities a), b), c) as limiting cases of the main case, where points $A$, $B$, $C$, $D$ do not lie on the same plane. We will explain this for case a). Suppose that $A$, $B$, $C$ are three distinct points on a line $p$, and point $D$ is outside line $p$. Let $M$ be the midpoint of segment $AB$, and $S$ the midpoint of segment $CD$.
Choose a point $A'$ near point $A$ outside the plane of points $A$, $B$, $C$, $D$. Let $M'$ denote the midpoint of segment $A'B$. Since points $A'$, $B$, $C$, $D$ do not lie in the same plane, as we proved earlier,
Let point $A'$ approach point $A$ indefinitely (or, as it is said, tend to point $A$) along line $A'A$. Then point $M'$ tends to point $M$, and the lengths $M'S$, $A'C$, $A'D$, $A'B$ tend to the lengths $MS$, $AC$, $AD$, $AB$, and from formula (1a) we obtain formula (1) by passing, as it is said in mathematics, to the limit. The above reasoning is based on initial knowledge from the so-called theory of limits, which can be found in the first chapters of any textbook on mathematical analysis (See, for example, F. Leja, Differential and Integral Calculus, 3rd edition, Warsaw 1954, Chapter II.) | 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 $ on diagonal $ NK $, $ a_5, b_5, c_5 $ on diagonal $ PL $. Prove that | 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
Then
Using the sine rule applied to triangles $ B_{i-2}A_iB_{i+2} $, we obtain (Figure 10):
om38_1r_img_10.jpg
From the same rule applied to triangles $ A_{i-1}B_iA_{i+1} $, we have
Multiplying the five equations (1) and the five equations (2) side by side, we get
from which the thesis of the problem immediately follows. | 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 $, $ \measuredangle B = 90^\circ $.
We will determine the geometric locus of the point satisfying condition 2°. We will distinguish two cases:
a) Point $ A $ does not lie on line $ p $. There are then two points $ M $ and $ N $ on line $ p $ of the desired geometric locus (fig. 30); they are the vertices of triangles $ ATM $ and $ ATN $, where $ AT $ is the distance from point $ A $ to line $ p $, and $ \measuredangle AMT = \measuredangle ANT = 30^\circ $.
Let $ C $ be a point satisfying condition 2° and let the angle $ ABC $ be oriented in the same direction as angle $ ATM $.
Point $ M $ then lies on the circumcircle of triangle $ ABC $. Indeed, point $ B $ either coincides with point $ M $, or lies on the ray $ MT $, in which case points $ M $ and $ C $ lie on the same side of line $ AB $ and $ \measuredangle AMB = \measuredangle ACB = 30^\circ $, or finally, point $ B $ lies on the extension of ray $ TM $ beyond point $ M $, in which case points $ M $ and $ C $ are on opposite sides of line $ AB $ and $ \measuredangle AMB + \measuredangle ACB = 180^\circ $, since $ \measuredangle AMB = 180^\circ - \measuredangle AMT = 150^\circ $, and $ \measuredangle ACB = 30^\circ $.
Since the diameter of the circumcircle of triangle $ ABC $ is $ AC $, then $ \measuredangle AMC = 90^\circ $. We have obtained the result that all points of the sought geometric locus, for which $ \measuredangle ABC $ is oriented in the same direction as angle $ ATM $, lie on a line perpendicular to line $ AM $ at point $ M $. Conversely, every point $ C $ of this perpendicular belongs to the geometric locus; if the circumcircle of triangle $ AMC $ passes through point $ B $ on line $ p $, then triangle $ ABC $ satisfies condition 2°, since $ \measuredangle ABC = \measuredangle AMC = 90^\circ $, and $ \measuredangle ACB = \measuredangle AMT = 30^\circ $.
Similarly, all points satisfying condition 2°, and such that angle $ ABC $ is oriented in the same direction as angle $ ATN $, form a line perpendicular to line $ AN $ at point $ N $.
We have proved that the sought geometric locus consists of two lines perpendicular to lines $ AM $ and $ AN $ at points $ M $ and $ N $, respectively.
b) Point $ A $ lies on line $ p $. In this case, we directly state that the sought geometric locus consists of all points of two lines passing through point $ A $ and forming angles of $ 60^\circ $ and $ 120^\circ $ with line $ p $, except for point $ A $ itself (fig. 31).
The solution to the given problem is obtained by finding the points of intersection of the given circle $ k $ with both lines of the determined geometric locus. Depending on the position of the circle relative to these lines, the number of solutions to the problem can be $ 4 $, $ 3 $, $ 2 $, $ 1 $, or $ 0 $. | 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 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
（此处保留了原文的多个空行） | 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 both numbers $ x_k $ and $ y_k $ must be zeros. In this case, inequality (1) is satisfied.
Let us assume, therefore, that the number on the right-hand side of inequality (1) is different from $ 0 $, and thus positive. Then the inequality to be proved can be reduced to the form
Substitute:
The numbers $ a_{1},a_{2}, \ldots, a_{n}, $ $ b_{1},b_{2}, \ldots, b_{n} $ belong to the interval $ \langle 0,1 \rangle $ and $ a_k + b_k = 1 $. From this and the inequality between the geometric mean and the arithmetic mean, we obtain
or
The left-hand side of inequality (2) is not greater than the left-hand side of inequality (3). This means that the inequality (3) just proved implies inequality (2), and thus proves inequality (1). | 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 particular
By adding these inequalities side by side, we get that $a_1 + a_2 + \ldots + a_{100} = 1 + 2 + \ldots + 100 < 10B$, which means $505 < B$.
Thus, the number $A$ defined in the problem satisfies the inequality
On the other hand, consider the following permutation $a_1, a_2, \ldots, a_{100}$ of the set of natural numbers not greater than $100$
This permutation can be defined by the formulas:
We will prove that the sum of any $10$ consecutive terms of this permutation is not greater than $505$.
Indeed, if the first of the considered $10$ terms has an even number $2k$, then
If, however, the first of the considered terms has an odd number $2k + 1$, then
Thus, the sum of any $10$ consecutive terms of this permutation is not greater than $505$. Therefore, the number $A$ defined in the problem satisfies the inequality
From (1) and (2), it follows that $A = 505$.
Note 1. The problem can be generalized as follows: Find the largest integer $A$ such that for any permutation of the set of natural numbers not greater than an even number $n = 2t$, the sum of some $m = 2r$ (where $r$ is a divisor of $t$) consecutive terms is at least $A$.
By making minor changes in the solution provided above, consisting in replacing the number $100$ with $2t$ and the number $10$ with $2r$, it can be proved that $A = \frac{1}{2} m(n + 1)$.
Note 2. In the case where $m \leq n$ are any natural numbers, it is generally not true that every permutation of the set of natural numbers not greater than $n$ contains $m$ consecutive terms with a sum not less than $\frac{1}{2} m(n + 1)$. For example, for $n = 6$, $m = 4$, the permutation $6, 4, 1, 2, 3, 5$ does not contain four consecutive terms with a sum not less than $14$. | 505 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 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 segments $ LC $, $ AK $, $ MB $ are therefore parallel and have a common perpendicular bisector, passing through the point $ O $. This perpendicular bisector also passes through the point $ P $, as the intersection of the diagonals $ MC $ and $ BL $ of the isosceles trapezoid $ MBCL $. Therefore, in particular, $ \measuredangle APO = 90^\circ $. | 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 whose side lengths do not exceed $ k $. Let $ k(n) $ be the largest of such numbers $ k $ for which $ P(k, n) \leq \frac{1}{2} P(n, n) $. Calculate $ \lim_{n\to \infty} \frac{k(n)}{n} $. | 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 inequality
Thus
Multiplying by $ - \frac{1}{n^3} $ the difference of the obtained inequality and $ n(n + 1)(2n + 1) $, we have:
The above inequality takes the form
where $ a_n = 1 - \frac{k(n)}{n} $.
We will prove that the sequence $ \left( 2a_n \left( a_n + \frac{1}{n} \right) \left( a_n + \frac{1}{2n} \right) \right) $ converges to the number $ 1 $. If this were not the case, then in view of inequality (*) and the fact that $ \displaystyle \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right) \left( 1 + \frac{1}{2n} \right) = 1 $, we would have
Let us take $ n_0 $ so large that for every $ n \geq n_0 $ the conditions
are satisfied.
Since
we have the inequality
Based on the last inequality and (***), we obtain
This is contradictory to condition (**). The contradiction arose from the assumption that the number $ 1 $ is not the limit of the sequence $ 2a_n \left( a_n + \frac{1}{n} \right) \left( a_n + \frac{1}{2n} \right) $. Therefore, the limit of this sequence exists and is equal to $ 1 $. Similarly, we conclude that the sequence $ \left( 2a_n \left( a_n - \frac{1}{n} \right) \left( a_n - \frac{1}{2n} \right) \right) $ also converges to $ 1 $. Given the obvious inequality
we obtain, based on the theorem of three sequences, that
Since $ \frac{k(n)}{n} = 1 - a_n $ and $ \displaystyle \lim_{n \to \infty} a_n = \frac{1}{\sqrt[3]{2}} $, we have | \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 lines $ AP $ and $ EC $, we obtain the equality
which proves that points $ A $, $ Q $, $ D $, $ E $ lie on the same circle. Therefore,
From the relationships
it follows that lines $ AP $ and $ FC $ are parallel. Hence, $ \measuredangle AQE = \measuredangle FCE $, which, combined with formula (1), gives the first equality of the thesis. The second is analogous. | 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). In both cases, the reasoning will be the same. Let $k'$ be the line symmetric to the line $k$ with respect to the line $l$. The line $k'$ is, of course, different from the line $k$ and different from the line $l$. We will prove that the line $k'$ is an axis of symmetry of the figure $F$.
If the figure $F$ contains some point $A$, then it also contains the point $B$ symmetric to the point $A$ with respect to the axis $l$, then the point $C$ symmetric to the point $B$ with respect to the axis $k$, and further the point $D$ symmetric to the point $C$ with respect to the axis $l$. Notice that the segment $AD$ is symmetric to the segment $BC$ with respect to the axis $l$, since points $A$ and $D$ are respectively symmetric to points $B$ and $C$ with respect to this axis. In this case, the perpendicular bisector of segment $AD$ is symmetric to the perpendicular bisector of segment $BC$ (i.e., to the line $k$) with respect to the axis $l$, which means that the perpendicular bisector of segment $AD$ is the line $k'$. We have proven that if the figure $F$ contains some point $A$, then it also contains the point $D$, symmetric to the point $A$ with respect to the line $k'$, i.e., that $k'$ is an axis of symmetry of the figure $F$. | 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 parallel to the diagonals of the rhombus. | 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 half the length of the shorter diagonal of the rhombus.
The entire figure is symmetric with respect to each of the lines $ AC $ and $ BD $, as well as with respect to the point $ O $. The drawn circle intersects each side of the rhombus at two points. Let $ M $ be one of the points of intersection of side $ AB $ with the circle, and let $ N $ be the point of intersection of side $ BC $ with the circle that is not symmetric to point $ M $ with respect to the line $ BD $. Let $ P $ and $ Q $ then be the points symmetric to points $ M $ and $ N $ with respect to point $ O $; segments $ MP $ and $ NQ $ are thus diameters of the circle; due to the symmetry of the figure with respect to point $ O $, points $ P $ and $ Q $ are points of intersection of sides $ CD $ and $ DA $ with the circle.
Quadrilateral $ MNPQ $ is a rectangle, as each of its angles is an inscribed angle in the drawn circle and subtends its diameter (e.g., angle $ MNP $ subtends diameter $ MP $). Side $ MN $ of this rectangle is not perpendicular to line $ BD $, as line $ MN $ does not pass through the point symmetric to point $ M $ with respect to line $ BD $; thus, side $ MN $ is not parallel to diagonal $ AC $. Side $ MN $ is also not parallel to diagonal $ BD $, as it connects points $ M $ and $ N $ lying on opposite sides of line $ BD $. The vertices $ M $, $ N $, $ P $, $ Q $ of rectangle $ MNPQ $ lie on the sides $ AB $, $ BC $, $ CD $, $ DA $ of the rhombus, respectively, but the sides of this rectangle are not parallel to the diagonals of the rhombus. Rectangle $ MNPQ $ thus serves as a counterexample refuting statement a).
b) We will prove that the second theorem is true.
Let $ MNPQ $ be a square inscribed in the rhombus $ ABCD $, with vertices $ M $, $ N $, $ P $, $ Q $ of the square lying on sides $ AB $, $ BC $, $ CD $, $ DA $ of the rhombus, respectively.
We will first prove that the center $ O $ of the square is also the center of the rhombus. Rotate the entire figure around point $ O $ by $ 180^\circ $. Vertex $ M $ of the square will end up in the opposite vertex $ P $ after the rotation (Fig. 29), line $ AB $ will become a line passing through point $ P $ and parallel to line $ AB $, i.e., it will coincide with line $ CD $. Since the same reasoning applies to each side of the rhombus, it follows that the rhombus will overlap with itself after the rotation, which means that the center of rotation $ O $ is the center of symmetry of the rhombus.
Now, rotate the entire figure around point $ O $ by $ 90^\circ $ so that ray $ OA $ coincides with ray $ OB $ (the direction of rotation is indicated by an arrow in Fig. 30). After the rotation, square $ MNPQ $ will transform into square $ NPQM $, i.e., it will overlap with itself; rhombus $ ABCD $ will transform into rhombus $ A'BC'D' $, which does not coincide with rhombus $ ABCD $, as the rhombus is not a square by assumption. If, for example, $ OA > OB $, then $ OA' > OB $ and $ OD' > OD $; thus, segments $ D'A' $ and $ AB $ intersect. The point of intersection of these segments is vertex $ M $ of the square, as point $ Q $ of segment $ AD $ will fall on this point after the rotation. Similarly, segments $ BC $ and $ A'D' $ intersect at vertex $ N $ of the square.
The figure composed of both rhombi is symmetric with respect to line $ BD $; in this symmetry, point $ M $ corresponds to point $ N $. Line $ MN $ is therefore perpendicular to the axis of symmetry $ BD $, i.e., it is parallel to line $ AC $, which was to be proved.
Note. If we were to discard the assumption that rhombus $ ABCD $ is not a square, the theorem would not be true, as an infinite number of squares can be inscribed in a square, of which only one has sides parallel to the diagonals of the given square.
From the proven theorem b), it follows that only one square can be inscribed in a rhombus that is not a square, such that each side of the rhombus contains one vertex of the square. A stronger theorem can be proven: there is only one square whose vertices lie on the perimeter of rhombus $ ABCD $ that is not a square. It suffices to show that there is no square whose two vertices lie on the same side of the rhombus, while the other vertices lie on other sides of the rhombus. We leave this as an exercise. | 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 length of segment $MN$. | 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 $DC = AB$, we get $EM = EN$. Therefore, triangle $MEN$ is isosceles, which means $EF \perp MN$. Consequently, the orthogonal projection of segment $EN$ onto line $MN$ is segment $FN$; and since segment $AB$ is parallel to segment $EN$ and has twice the length, the projection of segment $AB$ onto line $MN$ is a segment of length $2 \cdot FN = MN$.
Similarly, we prove that the orthogonal projection of segment $CD$ onto line $MN$ is a segment of length $MN$, which completes the solution of the problem. | 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 does not exceed $ \pi $.
On the other hand, it is known (see the solution to problem 17(5) of the XXIV Mathematical Olympiad), that the sum of the measures of two plane angles at a vertex of a tetrahedron is greater than the measure of the third plane angle at that vertex. If, therefore, one of the plane angles at a vertex of the tetrahedron is not acute, then the sum of the measures of all the plane angles at this vertex is greater than $ \pi $.
From the above, it follows that all the plane angles at vertex $ A $ are acute. | 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 number \( p \) and let \( a = p \). In the difference \( p^{(n-1)!+1} - p \), the subtrahend is divisible by \( p^2 \), but the minuend is not divisible by \( p^2 \), so \( p^{(n-1)!+1} - p \) is not divisible by \( p^2 \), and therefore it is not divisible by the number \( n \), which is a multiple of \( p^2 \).
For the sufficiency of the condition, assume that \( n \) is a square-free number, \( n = p_1 p_2 \ldots p_k \), where \( p_i \) are prime numbers and \( p_i \ne p_j \) for \( i \ne j \). To show that \( n \) divides \( a^{(n-1)!+1} - a \), it suffices to state that \( a^{(n-1)!+1} - a \) is divisible by \( p_i \) for \( i = 1, 2, \ldots, n \). If \( p_i \) divides \( a \), then \( p_i \) obviously divides \( a^{(n-1)!+1} - a \). Let's consider the case when \( p_i \) does not divide \( a \). In this case, \( p_i \) does not divide any of the \( p_i \) numbers \( 1, a, a^2, \ldots, a^{p_i-1} \), so some two of these numbers give the same remainder when divided by \( p_i \); let these be \( a^k \) and \( a^l \) for \( 0 \leq k < l < p_i \): \( a^k = t \cdot p_i + r \), \( a^l = s \cdot p_i + r \). Therefore, the number \( a^l - a^k \) is divisible by \( p_i \), and thus \( a^{l-k} - 1 \) is divisible by \( p_i \). Since \( l - k < p_i \leq n \), \( l - k \) divides \( (n-1)! \). Therefore, \( (n-1)! = (l - k) \cdot w \), and thus \( a^{(n-1)!+1} - a = (a^{l-k} - 1) (a^{(l-k)(w-1)} + a^{(l-k)(w-2)} + \ldots + a^{l-k} + 1) \). Therefore, \( a^{l-k} - 1 \) divides \( a^{(n-1)!} - 1 \), and thus it also divides \( a(a^{(n-1)!} - 1) = a^{(n-1)!+1} - a \). It follows that \( p_i \), being a divisor of \( a^{l-k} - 1 \), also divides \( a^{(n-1)!+1} - a \), which completes the proof.
Note. The proof of the sufficiency of the condition that \( n \) is a square-free number can be slightly shortened by using Fermat's Little Theorem.
For any integer \( a \) and prime number \( p_i \), the number \( a^{p_i} - a \) is divisible by \( p_i \).
In the case where \( p_i \) does not divide \( a \), it follows directly from this theorem that \( p_i \) divides \( a^{p_i-1} - 1 \). | 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 prime number. Hence, the numbers $ p_1 $ and $ p_2 $ are odd, and the number $ r $ equal to the difference $ p_2 - p_1 $ is even and one of the cases holds: $ r = 6k + 2 $ or $ r = 6k + 4 $, where $ k $ is an integer $ \geq 0 $.
We will prove that $ p_1 $ is divisible by $ 3 $. Indeed, if $ p_1 = 3m + 1 $ ($ m $ - an integer) and $ r = 6k + 2 $, it would follow that $ p_2 = 3m + 6k + 3 $ is divisible by $ 3 $, and since $ p_2 > 3 $, $ p_2 $ would not be a prime number. If, on the other hand, $ p_1 = 3m + 1 $ and $ r = 6k + 4 $, then $ p_3 = 3m + 12k + 9 $ would not be a prime number. Similarly, from the assumption that $ p_1 = 3m + 2 $ and $ r = 6k + 2 $, it would follow that $ p_3 = 3m + 12k + 6 $ is not a prime number, and from the assumption that $ p_1 = 3m + 2 $ and $ r = 6k + 4 $, we would get $ p_2 = 3m + 6k + 6 $, and thus $ p_2 $ would not be a prime number.
Therefore, $ p_1 $ is a prime number divisible by $ 3 $, i.e., $ p_1 = 3 $. | 3 | Number Theory | proof | Yes | Yes | olympiads | false | 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 sequence.
Proof of statement (a). Let $ p = 2^km $, $ k,m \geq1 $, $ m $ odd. If $ p = c_n $, then $ c_{n+1} = 2^{k-1}\cdot 3m $; when $ k -1 \geq 1 $, then $ c_{n+2} = 2^{k-2}\cdot 3^2m $; etc., generally, for $ j \in \{1,\ldots, k\} $ we have $ c_{n-j} = 2^{k-j} \cdot 3^j m $. In particular, for $ j = k $ we get $ c_{n+k} = 3^km $. This is a larger than $ p $ odd number.
Proof of statement (b). Now $ q - 1 $ is an even, positive number ($ q - 1 $ cannot be zero, because $ c_n \geq 2 $ for all $ n $; this follows immediately by induction from the definition of the sequence $ (c_n) $). Thus $ q = 2^km+1 $, $ k,m \geq 1 $, $ m $ odd. If $ q = c_n $, then $ c_{n+1} = 2^{k-1}\cdot 3m + 1 $ and generally, as before, for $ j\in \{1, \ldots, k\} $ we have $ c_{n+j} = 2^{k-j}\cdot 3^jm+1 $; in particular $ c_{n+k} = 3^km + 1 $. This is a larger than $ q $ even number. | 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 vertices of the polygon $ W $ and let in this polygon $ \measuredangle A_{i-1}A_iA_{i+1} = \alpha $ (Fig. 13); since $ n \geq 5 $, then $ \alpha \geq 108^\circ $.
According to the inscribed angle theorem, the reflex angle $ A_{i-1}OA_{i+1} = 2 \alpha $, so the convex angle $ A_{i-1}OA_{i+1} = 360^\circ - 2\alpha = \beta $. Rotate the polygon $ W $ around the point $ O $ by the angle $ \beta $, for example, in the direction determined by the traversal $ A_1A_2\ldots A_n $ of the polygon $ W $. The vertex $ A_{i-1} $ will end up at the point $ A_{i+1} $ after this rotation, i.e., each vertex of the polygon $ W $ will cover a vertex with an index 2 greater, with the indices 2 greater than $ n-1 $ and $ n $ being considered as $ 1 $ and $ 2 $. Therefore, the points
will cover the points
Each segment connecting two points of the sequence (1) will cover a segment connecting the corresponding points of the sequence (2). Thus, for odd $ n $
and
all sides of the polygon $ W $ are equal, q.e.d. | 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 $C$ and $D$ - with a sphere twice as heavy with center $N$. The center of gravity of the system of these two new spheres is the same as the center of gravity of the system of the previous four spheres, since the gravitational forces acting at points $A$ and $B$ have been replaced by their resultant acting at point $M$, and similarly, the gravitational forces acting at points $C$ and $D$ have been replaced by their resultant acting at point $N$. Therefore, the center of gravity of the two spheres placed at points $M$ and $N$, i.e., the midpoint of segment $MN$, lies at the center of the considered circle.
Chords $AB$ and $CD$, whose midpoints $M$ and $N$ lie on the diameter of the circle, are perpendicular to this diameter; these chords are therefore parallel. Similarly, the other two sides $BC$ and $AD$ of the quadrilateral $ABCD$ are parallel. This quadrilateral is thus an inscribed parallelogram in a circle, which means it is a rectangle. | 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) $ dzieli się przez $ x_1-x_2 $. [Uzasadnienie: jeżeli $ P(x) = a_0 + a_1x + \ldots + a_nx^n $ ($ a_i $, całkowite), to różnica $ P(x_1)-P(x_2) $ jest sumą składników $ a_i(x_1^i-x_2^i $) dla $ i = 1,\ldots,n $, a każdy z tych składników dzieli się przez $ x_1 -x_2 $.]
Biorąc jako $ x_1 $ liczby $ u $ i $ v $, a jako $ x_2 $ liczby $ a $, $ b $ i $ c $, dostajemy za każdym razem różnicę $ P(x_1) - P(x_2) $ równą $ -5^m $. Skoro ma się ona dzielić przez każdą z rozważanych różnic $ x_1 -x_2 $, muszą zachodzić równości
dla pewnych wykładników całkowitych $ j, k,l,n,p,q\geq 0 $.
Zapiszmy różnicę $ u-v $ w postaci $ u-u = 5^\alpha w $;, gdzie $ \alpha \geq 0 $, $ w > 0 $ są liczbami całkowitymi, przy czym $ w $ nie dzieli się przez $ 5 $. Z dwóch pierwszych równości (1) wynika, że
Liczba w ostatnim nawiasie nie dzieli się przez $ 5 $; jest więc równa $ w $, a wykładnik $ \min\{j,n\} $ jest równy $ \alpha $. A ponieważ $ w>0 $, znak $ \pm $ przed $ 5^{|j-n|} $ można pominąć.
Pisząc następnie liczbę $ u - v $ jako $ (u-b) - (v-b) $ oraz jako $ (u-c) - (v-c) $ i korzystając z pozostałych zależności (1) otrzymujemy analogicznie związki
Zauważmy wszelako, że liczba $ w $ może mieć tylko jedno przedstawienie postaci $ 5^\delta+\varepsilon $ ($ \delta \geq 0 $ całkowite, $ \varepsilon \in \{+1,-1\} $). Z równości (3) wynika więc, że wykładniki $ |j-n| $, $ |k-p| $, $ |l-q| $ są równe. Oznaczmy ich wspólną wartość przez $ \delta $ oraz przyjmijmy $ \beta=\alpha+\delta $.
W myśl związków (2), w każdej z trzech par $ \{j,n\} $, $ \{k,p\} $, $ \{l,q\} $ jedna liczba równa się $ \alpha $; pozostała musi się równać $ \beta $. Zatem pewne dwie spośród liczb $ j $, $ k $, $ l $ są równe; niech na przykład $ j = k $. Wówczas $ n = p $ (jeśli $ j = k = \alpha $, to $ n = p = \beta $, a jeśli $ j = k = \beta $, to $ n = p = \alpha $) i wobec związków (1):
Zgodnie z założeniem, liczby $ a $ i $ b $ nie są równe; otrzymane równości oznaczają w takim razie, że $ (u-a) = (b-u) $ oraz $ (v - a) = (b-v) $. Stąd $ 2u = a + b $ oraz $ 2v = a+b $, wbrew temu, że $ u>v $. Sprzeczność kończy dowód.
| 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 to each side of the $n$-gon. On the other hand, at least two vertices of the $(n + 1)$-gon belong to some side, e.g., $\overline{A_1A_n}$, of the $n$-gon, because $n + 1 > n$.
Let $B_1$ be the vertex of the $(n + 1)$-gon on the side $\overline{A_1A_n}$ that is closer to point $A_1$ than the other vertex of the $(n + 1)$-gon on $\overline{A_1A_n}$. Let $x$ be the length of the segment $\overline{A_1B_1}$. From the above, it follows that $0 \leq x \leq a$. Finally, let $B_2$ be the vertex of the $(n + 1)$-gon different from $B_1$ and lying on the side $A_1A_2$, and let $B_3, \ldots, B_{n+1}$ be the subsequent vertices of the $(n + 1)$-gon.
The difference in the areas of the $n$-gon and the $(n + 1)$-gon, $S_n - S_{n+1}$, is equal to the sum of the areas of the triangles $B_1A_1B_2, B_2A_2B_3, \ldots, B_nA_nB_{n+1}$. In each such triangle, we have:
Therefore, the area $P_i$ of the triangle $B_iA_iB_{i+1}$ is equal to
Since
we obtain from (1) and (3) sequentially
Thus, $A_1B_2 = na - x$. $A_1B_2 + B_2A_2 = (n + 1)a$ and hence
, $B_2A_2 + A_2B_3 = na$ and hence $A_2B_3 = na - (a + x) = (n - 1)a - x$,
$A_2B_2 + B_3A_3 = (n + 1)a$ and hence $B_3A_3 = (n + 1)a - ((n - 1)a - x) = 2a + x$
and by easy induction, generally
Therefore, from (4) and (2) we obtain
Let $Ax^2 + Bx + C$ be the quadratic polynomial written in the last square bracket. Then $A = -n$ and $B = \left( n^2 - 2 \frac{n(n-1)}{2} \right) a = na$. We obtain, therefore, that
which means
The area $S_{n+1}$ of the $(n + 1)$-gon is thus a quadratic function $S_{n+1}(x) = px^2 + qx + r$ of the variable $x$ considered in the interval $0 \leq x \leq a$.
Since $p = \frac{1}{2} \sin \alpha \cdot n > 0$, this function attains its minimum value at $x = \frac{-q}{2p} = \frac{\frac{1}{2} \sin \alpha \cdot na}{\sin \alpha \cdot na} = \frac{a}{2}$. The point $\frac{a}{2}$ is the midpoint of the interval $\langle 0, a \rangle$. Therefore, the numbers $S_{n+1}(0)$ and $S_{n+1}(a)$ are equal and are the largest values taken by the function $S_{n+1}(x)$ in the interval $\langle 0, a \rangle$.
Thus, the area of the $(n + 1)$-gon is greatest when $x = 0$ or $x = a$, i.e., when one of its vertices is a vertex of the regular $n$-gon. The area of the $(n + 1)$-gon is smallest when $x = \frac{a}{2}$, i.e., when one of its sides lies on a side of the regular $n$-gon such that the midpoints of these sides coincide. | 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 belong to the interior of the tetrahedron; hence it does not belong to at least one of these half-spaces. Therefore, there is a face $S$ of the tetrahedron $Q$ such that the opposite vertex of the tetrahedron and the center of the sphere $K$ lie on opposite sides of the plane of this face (or, in the limiting case, the center of the sphere lies on this plane).
Thus, $Q$ is contained in the smaller of the two parts (or in one of the two equal parts) into which the plane of the face $S$ divides the sphere $K$. Therefore, the height of the tetrahedron $Q$ dropped onto the face $S$ has a length not greater than $1$, so the volume of $Q$ does not exceed one third of the area of this face. The area of the face $S$, being a triangle, is not greater than the area of an equilateral triangle inscribed in the circle which is the section of the sphere $K$ by the plane of this face (see Note); and even more so, it is not greater than the area of an equilateral triangle inscribed in the great circle $E$. The latter area equals $3\sqrt{3}/4$. Therefore, the volume of $Q$ is less than or equal to $\sqrt{3}/4$.
In all the estimates mentioned, we can achieve equality by taking $O$ to be a tetrahedron whose base is an equilateral triangle inscribed in $E$, and the fourth vertex is placed at the farthest point from $E$ on the hemisphere $H$ (at the "north pole").
The number $\sqrt{3}/4$ is the sought maximum.
Note: For completeness, we will provide an elementary proof of the fact that among all triangles inscribed in a fixed circle of radius $r$, the one with the largest area is the equilateral triangle.
Let $ABC$ be any such triangle. By placing point $C$ at the midpoint of the larger of the two arcs into which points $A$ and $B$ divide the circle circumscribed around triangle $ABC$, we obtain an isosceles triangle $ABC$, acute or right, with an area not less than that of triangle $ABC$. Therefore, in searching for a triangle with the maximum area (inscribed in a given circle), we can limit our attention to the class of isosceles triangles without obtuse angles (inscribed in this circle).
If now $ABC$ is a triangle from this class ($|AC| = |BC|$), then denoting by $x$ the distance from side $AB$ to the center of the circle, we get the relationships: $|AB| = 2\sqrt{r^2 - x^2}$, $h_C = r + x$ (the height dropped from vertex $C$), and thus
We have used here the inequality between the geometric mean and the arithmetic mean of four numbers: $a_1 = a_2 = a_3 = r + x$, $a_4 = 3r - 3x$. This inequality becomes an equality if and only if these numbers are equal, i.e., for $x = r/2$. It remains to note that the last equality means that triangle $ABC$ is equilateral. | \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 that the number $ N $ admits a representation in the form
Dividing $ N $ by $ n + 1 $, we get a quotient $ q $ and a remainder $ r $:
If $ q = 0 $, then $ N = r $ divides $ (n+1)! $; it is enough to take as (1) a representation with only one term, equal to $ r $.
If $ q > 0 $, then in view of the inequality $ N < (n + 1)! $, it must be that $ q < n! $. By the induction hypothesis, the number $ q $ is a sum of the form
Hence, in view of (2)
The numbers $ d_i(n+1) $ are divisors of $ (n+1)! $. If $ r = 0 $, then we obtain the representation (1) by taking $ m = k $, $ D_i = d_i (n+1) $ for $ i=1, \ldots, k $. If, however, $ r > 0 $, we take $ m = k+1 $, $ D_i=d_i(n + 1) $ for $ i = 1, \ldots, k $, and $ D_{k+1} = r $. The inequalities $ D_1 > \ldots > D_k $ follow from the analogous inequalities between the numbers $ d_i $. The inequality $ D_k > D_{k+1} $ follows from the fact that $ d_k \geq 1 $, and $ r < n+1 $.
By the principle of induction, the theorem is true for every natural number $ n $. | 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 $ w(p) - w(0) = c_1p + c_2p^2 + \ldots + c_rp^r $ is of course divisible by $ p $. Therefore, $ p \mid 3^p - 1 $. Hence $ p \ne 3 $. From Fermat's Little Theorem, it follows that $ p \mid 3^{p-1} - 1 $. Therefore, $ p \mid (3^p - 1) - (3^{p-1} - 1) = 2 \cdot 3^{p-1} $. Since $ p \ne 3 $, it follows that $ p = 2 $. | 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 accordance with the problem statement, that $ a = b = c = 1 $. Fix $ d \in (0;3) $ (for other values of $ d $, quadrilaterals with sides $ 1 $, $ 1 $, $ 1 $, $ d $ do not exist). The maximum of the quantity (1) (for a fixed $ d $) is achieved by taking $ \phi = 90^\circ $, which corresponds to the quadrilateral being inscribed in a circle; in the considered situation ($ a = b = c = 1 $), this is equivalent to $ ABCD $ being an isosceles trapezoid.
For an isosceles trapezoid with sides $ 1 $, $ 1 $, $ 1 $, $ d $, we have $ p = (3+d)/2 $; its area is determined using formula (1), and then estimated using the inequality of means:
In this estimation, equality holds only when the four factors under the square root are equal, i.e., when $ 1+d = 9-3d $, which is for $ d = 2 $.
Hence the answer: in the considered set of quadrilaterals, the largest area is achieved by the isosceles trapezoid with sides $ 1 $, $ 1 $, $ 1 $, $ 2 $ (the result is unique up to isometry). | 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 $0 \leq k \leq 6$, $0 \leq m \leq 6$, $k \ne m$, so $-6 \leq k - m \leq 6$ and $k - m \ne 0$. The number $(k - m) \cdot 30$ is divisible by $7$, and the number $(k - m) \cdot 2$ is not divisible by $7$, because its absolute value is an even natural number not greater than $12$. Therefore, the sum of these numbers, i.e., the number $r$, is not divisible by $7$. Hence, each of the seven numbers gives a different remainder when divided by $7$, and thus one and only one of these seven remainders equals zero, c. n. d. | 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 $, $ \gamma $ has a defined tangent, i.e., that none of these angles has the form , where $ m $ denotes an integer, and that $ \alpha + \beta + \gamma = \frac{1}{2} \pi + k\pi $ ($ k $ - an integer). For the proof, we will distinguish two cases:
1° One of the angles $ \alpha $, $ \beta $, $ \gamma $ is an integer multiple of $ \pi $, e.g., $ \gamma = n\pi $. The formula (1), which needs to be proven, then has the form $ \tan \alpha \tan \beta = 1 $. According to the assumption,
hence,
therefore,
Since according to the assumption $ \beta $ does not have the form $ \frac{1}{2}\pi + m\pi $, then $ \cot \beta \ne 0 $ and the above equality gives
2° None of the angles $ \alpha $, $ \beta $, $ \gamma $ is an integer multiple of $ \pi $. Then,
where the angle $ \frac{1}{2} \pi - \gamma $ does not have the form $ \frac{1}{2}\pi + m\pi $, so there exists $ \tan (\frac{1}{2}\pi - \gamma) = \cot \gamma $ and from the above equality we obtain
and hence, as above, formula (1).
The converse theorem is also true:
If equality (1) holds, then $ \alpha + \beta + \gamma = \frac{1}{2}\pi + k\pi $ ($ k $ - an integer).
We will prove this theorem by distinguishing two cases:
1° $ \tan \gamma = 0 $, so $ \gamma = m\pi $ ($ m $ - an integer). Equality (1) then has the form
hence,
so
and finally,
where $ k = m + n $.
2° $ \tan\gamma \ne 0 $. From equality (1) we have
Indeed, $ 1 - \tan \alpha \tan \beta \ne 0 $, because from the equality $ 1 - \tan \alpha \tan \beta = 0 $ and equality (2) it would follow that $ \tan \alpha + \tan \beta = 0 $, and hence the equality $ \tan \alpha \tan \beta + \tan^2 \beta = 0 $ and finally the equality $ 1 + \tan^2 \beta = 0 $, which is impossible. We can therefore divide both sides of equality (2) by $ 1 - \tan \alpha \tan \beta $; we obtain
hence,
and finally, | 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 to the coordinate axes. These lines intersect at points with integer coordinates, thus forming a rectangle whose area is an integer and on whose sides the vertices of the considered triangle lie. The triangle is formed from this rectangle by cutting off one, two, or three right-angled triangles with legs parallel to the coordinate axes and vertices at points with integer coordinates. The lengths of the legs of these triangles are integers, so the area of each triangle (equal to half the product of the lengths of the legs) is an integer or half an integer. Therefore, the area $ S $ of the considered triangle is the difference between an integer (the area of the rectangle) and one, two, or three numbers, each of which is an integer or half an integer. Thus, $ 2S $ is an integer. | 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 $. Prove that $ \measuredangle ABC = \measuredangle O_1CO_2 $ if and only if $ d^2 = r_1^2 + r_2^2 + R^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 angles at vertex $ C $, we get $ 360^\circ = \measuredangle O_1 CO_2 + \measuredangle ACB + (90^\circ - \measuredangle BAC) + (90^\circ - \measuredangle ABC) = \measuredangle O_1C0_2 + \measuredangle ACB + (180^\circ - (\measuredangle BAC + \measuredangle ABC)) = \measuredangle O_1CO_2 + \measuredangle ACB + \measuredangle ACB $. Therefore, $ \measuredangle ACB = 180^\circ - \frac{1}{2} \measuredangle O_1CO_2 $.
Thus, the equality $ \measuredangle ACB = \measuredangle O_1 CO_2 $ is satisfied if and only if $ \measuredangle ACB = 120^\circ $, and therefore if and only if $ \cos \measuredangle ACB = -\frac{1}{2} $ and
$ 0 \leq \measuredangle ACB \leq 180^\circ $. From the Law of Cosines, we get
The equality $ \measuredangle ACB = \measuredangle O_1CO_2 $ is thus equivalent to the equality | ^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 $c_1$, $c_2$, and $c_3$ are set, there are $6$ possibilities left for $c_4$. Therefore, the set of all considered numbers consists of $9 \cdot 8 \cdot 7 \cdot 6$ numbers.
For each number in this set, there is a specific other number in the set whose digits complement the corresponding digits of the given number to ten. All numbers in the set can thus be paired, with each pair consisting of two numbers whose corresponding digits sum to $10$, such as $3562$ and $7548$.
There are $\frac{1}{2} \cdot 9 \cdot 8 \cdot 7 \cdot 6$ pairs; the sum of the numbers in one pair is $1000 \cdot 10 + 100 \cdot 10 + 10 \cdot 10 + 10 = 11110$, so the sum of all numbers in the considered set is equal to | 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 is finite. (Assume that the ball does not hit the vertex of the triangle). | 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 with respect to line $AC$ and the image $AB''$ of triangle $AB'$ in the symmetry with respect to line $AB$. Denote by $k$, $l$, $m$ the vectors representing the ball's movement, and by $l'$, $m'$ the images of vectors $l$, $m$ in the first symmetry, and by $m''$ the image of vector $m$ in the second symmetry (Fig. 14).
From the law "the angle of incidence is equal to the angle of reflection," it follows that vectors $k$, $l'$, $m'$ have the same direction. Since the composition of the considered symmetries is a rotation around point $A$ by an angle of $2\measuredangle BAC$, triangle $AB''$ is the image of triangle $ABC$ under this rotation, and vector $m''$ is the image of vector $m$. It follows that the directions of vectors $k$ and $m''$ differ by an angle of $2 \measuredangle BAC$.
We will now proceed to solve the problem. Since the measures of the angles of triangle $ABC$ are proportional, there exists a number $\lambda$ and natural numbers $r$, $s$, $t$ such that $\measuredangle BAC = r\lambda$, $\measuredangle ABC = s\lambda$, $\measuredangle ACB = t\lambda$. Therefore, $\lambda (r + s + t) = \pi$. Denoting $n = r + s + t$, we get that $\lambda = \frac{\pi}{n}$.
Thus, by the lemma, after an even number of reflections, the direction of the ball's movement changes by an angle whose measure is an even multiple of the number $\lambda = \frac{\pi}{n}$, i.e., one of the numbers $2\lambda, 4\lambda, 6\lambda, \ldots, 2n\lambda = 2\pi$. Therefore, after an even number of reflections, the ball can move in one of $n$ directions. Similarly, after an odd number of reflections, the ball can move in one of $n$ directions. The total number of directions in which the ball can move is therefore no greater than $2n$.
Note. Similarly, it can be proven that if in the problem the triangle is replaced by any polygon whose angle measures are proportional, then the number of directions in which the ball can move in such a polygon is also finite. | 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 vertex (it doesn't matter which one), this will affect the numbers assigned to three faces. The value of the sum $S_1$ will increase or decrease by $2$. The value of the sum $S_2$ will increase or decrease by $2$ or $6$. The total sum $S = S_1 + S_2$ will increase or decrease by $4$ or $8$, or remain unchanged.
If all vertices are assigned the number $1$, then all faces are also assigned the number $1$, and in this case $S = 8 + 6 = 14$. (This is, of course, the maximum possible value of the sum $S$.) From this configuration, any other can be reached by changing the numbers at the vertices one by one. According to the previous observation, at each step of this procedure, the sum $S$ will change by $\pm 4$, $\pm 8$, or $0$. Therefore, the values taken by the sum $S$ will differ from the initial value $14$ by multiples of the number $4$. The set of possible values is thus a subset of the set $\{14, 10, 6, 2, -2, -6, -10, -14\}$.
The number $-14$ can be immediately eliminated: we would need to have $-1$ at all vertices and all faces, but this is contradictory to the conditions of the problem.
It is also not possible to achieve the value $S = 10$. This would require $12$ ones to appear with a positive sign and two with a negative sign. If all vertices were assigned a positive one, then all faces would also be assigned a positive one. If exactly one vertex were assigned a negative one, then three faces would receive a negative sign. If, finally, exactly two vertices were assigned a negative sign, then some faces would also receive a negative sign. Therefore, the configuration "10 positives, 2 negatives" is not achievable in any way.
The range of possible values for $S$ is thus limited to the set
Of these six values, each can be achieved in various ways. Here are examples of configurations that realize these values:
If all vertices of the cube are assigned the number $+1$, then $S_1 = 8$, $S_2 = 6$, and $S = 14$.
If all vertices of the cube are assigned the number $-1$, then $S_1 = -8$, $S_2 = 6$, and $S = -2$.
If one vertex is assigned the number $+1$ and the rest $-1$, then $S_1 = -6$, $S_2 = 0$, and $S = -6$.
If one vertex is assigned the number $-1$ and the rest $+1$, then $S_1 = 6$, $S_2 = 0$, and $S = 6$.
If two opposite vertices of the cube are assigned the number $+1$ and the rest $-1$, then $S_1 = -4$, $S_2 = -6$, and $S = -10$.
If two opposite vertices of one face are assigned the number $-1$ and the rest of the vertices of the cube are assigned the number $+1$, then $S_1 = 4$, $S_2 = -2$, and $S = 2$. | {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 nierówność $ xy \leq x + y + 1 $, czyli $ (x-1)(y -1) \leq 2 $. Wartość iloczynu $ (x-1)(y-1) $ może być w takim razie równa tylko $ 2 $, $ 1 $ lub $ 0 $.
W pierwszym przypadku jeden z czynników $ (x - 1) $ i $ (y- 1) $ jest równy $ 2 $, a drugi $ 1 $, czyli para $ (x,y) $ jest jedną z par $ (3,2) $ i $ (2,3) $.
W drugim przypadku mamy $ x -1 = y -1 = 1 $, czyli $ x = y = 2 $. Jednak para $ (x,y) = (2,2) $ nie spełnia warunku zadania (iloraz $ 3/2 $ nie jest liczbą naturalną).
W trzecim przypadku co najmniej jeden z czynników $ (x - 1) $ i $ (y- 1) $ jest zerem. Jeśli na przykład $ y - 1 = 0 $, czyli $ y = 1 $, to liczba $ x $, jako dzielnik sumy $ y + 1 = 2 $, musi być równa $ 1 $ lub $ 2 $; tak więc $ x = y = 1 $ lub $ y = 1 $, $ x = 2 $ lub - przez symetrię - $ x = 1 $, $ y = 2 $.
Rekapitulując, widzimy, że tylko następujące pary $ (x,y) $ mogą spełniać postawione warunki: $ (3,2) $, $ (2,3) $, $ (1,1) $, $ (2,1) $, $ (1,2) $ - i istotnie je spełniają; sprawdzenie jest natychmiastowe.
| (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 $ yield a positive result, while the remaining trials yield a negative result, is equal to
Therefore,
where the summation is over all $ k $-element subsets $ A_k = \{ i_1, i_2, \ldots, i_k\} $ of the set $ \{1, 2, \ldots, n\} $.
Consider the polynomial
Multiplying the outer parentheses by virtue of (1) we get
We calculate the derivative of the polynomial at $ w $ using formula (2):
Therefore, $ w $, since $ w(1) = 1 $ by virtue of (2).
On the other hand, we calculate the derivative of the polynomial at $ w $ using formula (3):
and hence
Comparing the obtained values of $ w(1) $, we get the thesis of the problem. | 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, wyraża równość $ 2aq = aq^2+a+9 $, którą możemy napisać w postaci
Stąd wynika, że
bądź też
Otrzymujemy trójki liczb: $ ( - 1, -4, -16) $, $ ( - 1, 2, -4) $, $ ( - 9, -18, -36) $; stwierdzamy, że są one istotnie rozwiązaniami zadania.
b) Najmniejszą z liczb $ a, aq, aq^2 $ jest $ aq $. Nową trójkę tworzą wtedy liczby $ a, aq+ 9, aq^2 $; warunek, aby to był postęp arytmetyczny wyraża równość $ 2(aq+9) = aq^2+a $, której nadamy postać
Z tej równości wynika, że
bądź też
Otrzymujemy trójki liczb: $ (2, 8, 32) $, $ (2, -4, 8) $, $ (18, 36, 72) $, z których tylko druga spełnia warunki zadania.
Zadanie ma więc cztery rozwiązania: $ ( - 1, -4, -16) $, $ ( -1,2, -4) $, $ (-9, -18, -36) $, $ (2, -4, 8) $.
| (-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 $.
om55_1r_img_5.jpg
Since point $ R $ is the midpoint of diagonal $ KM $ of rhombus $ KLMN $, it is also the midpoint of the other diagonal of this rhombus, i.e., segment $ NL $. Therefore, from the parallelism of lines $ PR $ and $ SL $, we conclude that $ NP = PS $. Point $ P $ is the midpoint of segment $ AD $, so $ OP \perp AD $. This property, the equality $ NP = PS $, and the fact that point $ O $ lies on the perpendicular bisector of segments $ AB $, $ CD $, and $ LS $ imply
Therefore, point $ O $ lies on the perpendicular bisector of segment $ NL $, i.e., on line $ KM $. | 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 $. Therefore, each of the factors $ 3, p_1^{\alpha_1}, p_2^{\alpha_2}, \ldots, p_s^{\alpha_s} $ on the right side of equation (1) is a divisor of one and only one of the numbers $ a $ and $ a+1 $. Therefore, one of the following cases occurs:
where $ p $ and $ q $ denote natural numbers satisfying the equation $ p^2q^2=b^2 $.
In case $ \alpha $), it would be $ 3q^2-p^2=1 $, so $ p^2 \equiv 2 (\bmod 3) $, which is impossible, since the square of a natural number gives a remainder of $ 0 $ or $ 1 $ when divided by $ 3 $.
Therefore, case $ \beta $) holds, i.e., $ a+1 $ is the square of an integer, q.e.d.
Note. From the theorem proved above, it follows that if natural numbers $ a $ and $ b $ satisfy the equation $ a^2-a=3b^2 $, then $ a $ is the square of an integer. For $ a^2-a = a_1^2 + a_1 $ where $ a_1=a- 1 $. | 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 divisors that are greater than $ k $. Indeed: to each divisor $ d < k $ of $ n $, there corresponds a divisor $ n/d $ of $ n $, which is greater than $ k $. Additionally, the number $ k $ itself is a divisor of $ n $, which implies that $ n $ has an odd number of divisors. Therefore, $ k $, and thus $ n $, is an odd number.
The problem thus reduces to finding such odd numbers $ n = k^2 $ $ (n > 1) $, for which the number of divisors less than $ k $ is equal to $ \frac{1}{2} (k-1) $.
Since $ k^2 $ is an odd number, no even number is a divisor of $ k^2 $. Therefore, for the number of divisors less than $ k $ of $ k^2 $ to be equal to $ \frac{1}{2} (k-1) $, every odd number less than $ k $ must be a divisor of $ k^2 $. In particular, $ k^2 $ must be divisible by $ k-2 $. Since $ k^2 = (k-2)(k+2)+4 $, the number $ k-2 $ is an odd divisor of 4. This means that $ k - 2 = 1 $, or $ k = 3 $.
We directly verify that the number $ n = 3^2 = 9 $ satisfies the conditions of the problem. Therefore, the sought numbers are $ n = 1 $ and $ n = 9 $. | 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 the cube). This would imply the equation $8s = 2(1 + \ldots + 12) = 12 \cdot 13$; a contradiction, because $s$ should be an integer.
However, if it is allowed to use any twelve different numbers from the set $\{1,2,\ldots,13\}$ for numbering, then achieving equal sums at all vertices is possible, and in many ways. An example of such a realization is shown in Figure 3.
om41_1r_img_3.jpg | 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 $ at point $ F $. Prove that $ |SE|=|SF| $. | 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 $O_1O_2$ and $O_3O_4$ intersect at a point $P$ (different from $O_i$, $i = 1,2,3,4$). Then,
This is not possible, however, because from equality (1) it follows that both the sum $|O_2O_3| + |O_4O_1|$ and the sum $|O_1O_2| + |O_3O_4|$ have the value $r_1 + r_2 + r_3 + r_4$. The obtained contradiction proves the validity of statement (2).
Therefore, the broken line $O_1O_2O_3O_4O_1$ is the boundary of a quadrilateral. It can be convex or concave; figures 8 and 9 illustrate these two situations.
In the following, we adopt the following convention: for any three different points $X$, $Y$, $Z$, the symbol $|\measuredangle XYZ|$ will always denote the measure of the convex angle with sides $YX^\to$, $YZ^\to$ (i.e., a value in the interval $[0^\circ; 180^\circ]$).
Let us denote the measures of the internal angles of the quadrilateral $O_1O_2O_3O_4$ by
When the quadrilateral is convex, then of course,
(This case also includes the situation where some three vertices are collinear and the quadrilateral degenerates into a triangle.)
om43_2r_img_9.jpg
If, however, $O_1O_2O_3O_4$ is a concave quadrilateral, then one of its internal angles has a measure greater than $180^\circ$. Let, for example, $|\measuredangle O_4| > 180^\circ$. The system of equalities (3) should then be replaced by,
The triangles $DO_1A$ and $A0_2B$ are isosceles. Therefore,
When the quadrilateral $O_1O_2O_3O_4$ is convex, we can rewrite the above relationships as,
and thus,
Similarly, using the fact that the triangles $BO_3C$ and $CO_4D$ are also isosceles, we conclude that,
Considering now the situation where the quadrilateral $O_1O_2O_3O_4$ is concave and, for example, the internal angle at vertex $O_4$ is concave, such that the relationships (4) hold. The derivation of formula (5) remains valid; the same applies to formula (6). Further,
while,
so,
Thus, equality (7) also holds in this case. Similarly, we prove formula (8).
From equalities (5), (6), (7), (8) (valid in any case), it follows that,
Therefore, a circle can be circumscribed around the quadrilateral $ABCD$. Let us denote this circle by $k_0$.
Draw a tangent from point $S$ to $k_0$ at point $G$ (figure 10). By the theorem on the segments of a secant and a tangent, applied to the circles $k_2$, $k_0$, $k_4$, we have the equalities:
Thus, ultimately, $|SE| = |SF|$.
om43_2r_img_10.jpg | 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, obliczyć współrzędne geograficzne miasta $ X $. Jakie to miasto? Obliczyć długość drogi samolotu z dokładnością do 100 km. Zakładamy, że Ziemia jest idealną kulą o długości równika 40 000 km oraz, że samolot leci na wysokości nie większej niż 10 km.
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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 startuje w kierunku zachodnim. Wynika stąd, że łuk o końcach w Oslo i mieście $ X $ jest czwartą częścią koła wielkiego. Długość tego łuku jest więc równa $ 10 000 $ km, a długości geograficzne miasta $ X $ i Osio różnią się o $ 90^\circ $. Zatem miasto $ X $ leży na równiku w punkcie mającym $ 79^\circ 17 długości zachodniej. Przekonujemy się za pomocą mapy, że jedynym większym miastem posiadającym lotnisko, w pobliżu tego punktu jest stolica Ekwadoru Quito.
Jeżeli uwzględnimy, że droga samolotu znajduje się nie na powierzchni ziemi, lecz np. $ 10 $ km nad ziemią, to otrzymamy, że długość tej drogi jest równa $ \frac{1}{4} [2\pi (R + 10)] = \frac{1}{4} 2\pi R + 5\pi \approx 10 000 + 15 $ km, gdzie $ R $ jest długością promienia Ziemi.
Uwaga: Szerokość geograficzna Oslo nie ma wpływu na rozwiązanie. Gdyby więc samolot startował z dowolnego punktu Ziemi, o długości geograficznej wschodniej $ 10^\circ 43 i spełnione były pozostałe warunki zadania, to wylądowałby w Quito po przebyciu drogi długości około $ 10 000 $ km.
| 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 know the hypotenuse $AM$ and the projection $NM$ of the leg $LM$ onto the hypotenuse, so we can construct triangle $ALM$. The construction is as follows.
Draw a circle with diameter $AM$. At point $N$ on segment $AM$ ($NM = MH$), draw a perpendicular to line $AH$. If this perpendicular intersects the circle at point $L$, draw lines $AL$, $LM$, and a line perpendicular to line $AH$ at point $H$, which will intersect lines $AL$ and $LM$ at points $B$ and $C$ respectively. Triangle $ABC$ satisfies the conditions of the problem, as according to the construction, $AH \bot BC$, $CL \bot AB$, $\measuredangle CMH = \measuredangle LMN$, hence $CM = ML$.
The construction is possible if point $N$ lies inside the circle with diameter $AM$, i.e., if $MH < AM$, and two symmetric solutions are obtained relative to $AH$. | 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 $ BC $.
An immediate proof of the theorem is obtained by applying Ptolemy's theorem to the quadrilateral $ ABMC $:
since $ BC = AC = AB $, it follows from the above equality that | proof | Geometry | proof | Yes | Yes | olympiads | false | 1,462 |
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