problem
stringlengths 28
7.48k
| solution
stringlengths 0
18.5k
| answer
stringclasses 1
value | problem_type
stringclasses 8
values | question_type
stringclasses 1
value | source
stringclasses 6
values | synthetic
bool 1
class | source_dataset
stringclasses 1
value | source_split
stringclasses 1
value | __index_level_0__
int64 1
111k
|
|---|---|---|---|---|---|---|---|---|---|
LV OM - I - Task 1
Given is a polygon with sides of rational length, in which all internal angles are equal to $ 90^{\circ} $ or $ 270^{\circ} $. From a fixed vertex, we emit a light ray into the interior of the polygon in the direction of the angle bisector of the internal angle at that vertex. The ray reflects according to the principle: the angle of incidence is equal to the angle of reflection. Prove that the ray will hit one of the vertices of the polygon.
|
Since every internal angle of the considered polygon $ \mathcal{W} $ is $ 90^{\circ} $ or $ 270^{\circ} $, all sides of this polygon (after an appropriate rotation) are aligned horizontally or vertically (Fig. 1). Let $ p_1/q_1, p_2/q_2, \ldots, p_n/q_n $ be the lengths of the consecutive sides of the polygon $ \mathcal{W} $, where each of the numbers $ p_1, p_2, \ldots, p_n $, $ q_1, q_2, \ldots, q_n $ is a positive integer.
Consider a square grid where each small square has a side length of $ 1/(q_1 q_2 \ldots q_n) $. Then the polygon $ \mathcal{W} $ can be placed on this grid such that each of its sides lies on a line defining the grid (Fig. 1). A light ray emitted from a vertex of the polygon $ \mathcal{W} $ according to the given rules moves along the diagonals of the grid squares and reflects off the sides of the polygon $ \mathcal{W} $ at grid points.
om55_1r_img_1.jpg
Suppose the light ray did not hit any vertex of the polygon $ \mathcal{W} $. It therefore reflected off the sides of the polygon an arbitrary number of times. This implies that the light ray hit a point $ \mathcal{P} $ on the boundary of the polygon $ \mathcal{W} $ at least three times. Thus, the light ray hit the point $ \mathcal{P} $ at least twice moving in the same direction. This means that the trajectory of the light ray is periodic, which is not possible, since the light ray started its journey from a vertex of the polygon $ \mathcal{W} $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 441 |
XLIX OM - I - Problem 7
Determine whether there exists a convex polyhedron with $ k $ edges and a plane not passing through any of its vertices and intersecting $ r $ edges, such that $ 3r > 2k $.
|
Let $ W $ be any convex polyhedron; let $ k $ be the number of its edges, and $ s $ -- the number of faces. The boundary of each face contains at least three edges, and each edge is a common side of exactly two faces. Therefore, $ 2k \geq 3s $.
If a plane not passing through any vertex intersects $ r $ edges, then the section of the polyhedron by this plane is a convex polygon with $ r $ vertices, and thus also $ r $ sides. Each of its sides is contained in some face of the polyhedron $ W $, and each in a different one. Therefore, $ s \geq r $.
Combining the obtained inequalities, we see that $ 2k \geq 3r $. Therefore, there does not exist a polyhedron for which the inequality $ 3r > 2k $ would be satisfied.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 442 |
XXXIV OM - III - Problem 5
In the plane, vectors $ \overrightarrow{a_1}, \overrightarrow{a_2}, \overrightarrow{a_3} $ of length 1 are given. Prove that one can choose numbers $ \varepsilon_1, \varepsilon_2, \varepsilon_3 $ equal to 1 or 2, such that the length of the vector $ \varepsilon_1\overrightarrow{a_1} + \varepsilon_2\overrightarrow{a_2} + \varepsilon_3\overrightarrow{a_3} $ is not less than 2.
|
Let's start from an arbitrary point $0$ with vectors $\overrightarrow{a_1}$, $\overrightarrow{a_2}$, $\overrightarrow{a_3}$. They determine three lines intersecting at point $0$. Among the angles determined by pairs of these lines, there is an angle of measure $\alpha$ not greater than $\frac{\pi}{3}$. Suppose this is the angle between the lines determined by vectors $\overrightarrow{a_1}$ and $\overrightarrow{a_2}$ (otherwise, we can change the numbering of the vectors). Therefore, one of the angles $\measuredangle (\overrightarrow{a_1}, \overrightarrow{a_2})$, $\measuredangle (\overrightarrow{a_1}, -\overrightarrow{a_2})$ has a measure $\alpha \in \left[0, \frac{\pi}{3} \right]$.
We choose $\varepsilon_2$ so that the angle $\measuredangle (\overrightarrow{a_1}, \varepsilon_2 \overrightarrow{a_2})$ has a measure $\alpha$. Let's compute the dot product of the vector $\overrightarrow{b} = \overrightarrow{a_1} + \varepsilon_2 \overrightarrow{a_2}$ with itself:
Therefore, $|\overrightarrow{b}| \geq \sqrt{3}$.
Next, notice that
Therefore, one of the numbers $(\overrightarrow{a_1} + \varepsilon_2 \overrightarrow{a_2} + \overrightarrow{a_3})^2$, $(\overrightarrow{a_1} + \varepsilon_2 \overrightarrow{a_2} - \overrightarrow{a_3})^2$ is not less than $4$. We choose $\varepsilon_3$ so that $(\overrightarrow{a_1} + \varepsilon_2 \overrightarrow{a_2} + \varepsilon_3 \overrightarrow{a_3})^2 \geq 4$. Then, $|\overrightarrow{a_1} + \varepsilon_2 \overrightarrow{a_2} + \varepsilon_3 \overrightarrow{a_3}| \geq 2$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 446 |
XXV OM - II - Problem 3
Prove that the orthogonal projections of vertex $ D $ of the tetrahedron $ ABCD $ onto the bisecting planes of the internal and external dihedral angles at edges $ \overline{AB} $, $ \overline{BC} $, and $ \overline{CA} $ lie on the same plane.
|
The bisecting plane is a plane of symmetry of the dihedral angle. Therefore, the image $D'$ of the vertex $D$ in the symmetry relative to any of the considered bisecting planes lies in the plane $ABC$. It follows that if $P$ is the orthogonal projection of the point $D$ onto the bisecting plane, then $P$ is the midpoint of the segment $\overline{DD'}$. Thus, the point $P$ is the image of the point $D$ in the homothety $\varphi$ with the center at point $D$ and the ratio $\displaystyle \frac{1}{2}$. Therefore, the projections of the point $P$ onto all the considered bisecting planes lie in the plane which is the image of the plane $ABC$ in the homothety $\varphi$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 447 |
XXXV OM - II - Problem 2
On the sides of triangle $ABC$, we construct similar isosceles triangles: triangle $APB$ outside triangle $ABC$ ($AP = PB$), triangle $CQA$ outside triangle $ABC$ ($CQ = QA$), and triangle $CRB$ inside triangle $ABC$ ($CR = RB$). Prove that $APRQ$ is a parallelogram or that points $A, P, R, Q$ lie on a straight line.
|
Consider a similarity with a fixed point $C$ that transforms $B$ into $R$ (this is the composition of a rotation around $C$ by the angle $BCR$ and a homothety with center $C$ and scale equal to the ratio of the base length to the arm length in each of the constructed isosceles triangles). This similarity transforms $A$ into $Q$. It follows that triangles $ABC$ and $QRC$ are similar.
om35_2r_img_8.jpg
Similarly, we observe that a similarity with a fixed point $B$ that transforms $C$ into $R$ transforms $A$ into $P$, so triangles $ABC$ and $PBR$ are similar. Therefore, triangles $QRC$ and $PBR$ are similar. Since $BR = CR$, these triangles are congruent, and thus $QC = PR$ and $QR = PB$. But $QC = AQ$ and $PB = AP$, so $AQ = PR$ and $AP = QR$. Moreover, the angle between vectors $\overrightarrow{BP}$ and $\overrightarrow{RQ}$ has a measure equal to $\measuredangle PBA + \measuredangle BCR = 2 \measuredangle PBA$, and the angle between vectors $\overrightarrow{AP}$ and $\overrightarrow{BP}$ has a measure equal to $180^\circ - 2 \measuredangle PBA$, which implies that vectors $\overrightarrow{RQ}$ and $\overrightarrow{AP}$ are parallel. Therefore, points $A$, $P$, $R$, $Q$ lie on the same line or are consecutive vertices of a parallelogram.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 448 |
XIII OM - I - Problem 3
Prove that the perpendiculars dropped from the centers of the excircles of a triangle to the corresponding sides of the triangle intersect at one point.
|
A simple solution to the problem can be inferred from the observation that on each side of the triangle, the point of tangency of the inscribed circle and the point of tangency of the corresponding excircle are symmetric with respect to the midpoint of that side. For example, if the inscribed circle in triangle $ABC$ touches side $BC$ at point $K$, and the excircle at point $L$, then, using the usual notation for the sides of the triangle, we have $CK = \frac{1}{2} (a + b - c)$, $BL = \frac{1}{2} (a + b - c)$, so $CK = BL$, i.e., points $K$ and $L$ are symmetric with respect to the midpoint $M$ of side $BC$ (Fig. 3).
Let $S$ be the center of the inscribed circle in triangle $ABC$, $O$ - the center of the circumscribed circle of this triangle, and $T$ the point symmetric to point $S$ with respect to point $O$. Since $SO = OT$, $KM = ML$, and lines $SK$ and $OM$ are parallel, line $TL$ is symmetric to line $SK$ with respect to the axis $OM$. Therefore, line $TL$ is perpendicular to side $BC$ at the point of tangency with the excircle, and thus passes through the center of this circle. The same reasoning applies to sides $AB$ and $AC$. Hence, it follows that through point $T$ pass the perpendiculars drawn from the centers of the excircles to the corresponding sides of the triangle.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 450 |
XLV OM - III - Task 4
We have three unmarked vessels: an empty $ m $-liter, an empty $ n $-liter, and a full $(m+n)$-liter vessel of water. The numbers $ m $ and $ n $ are relatively prime natural numbers. Prove that for every number $ k \in \{1,2, \ldots , m+n-1\} $, it is possible to obtain exactly $ k $ liters of water in the third vessel by pouring water.
|
When pouring water from one container to another, we either completely empty the container we are pouring from or completely fill the container we are pouring into. (Since the containers have no markings, this is the only method that allows us to control the volume of water being poured.) Thus, after each pour, we have a non-negative integer number of liters of water in each container (not exceeding, of course, the capacity of the given container). Let's denote:
$ A $ - the $ m $-liter container,
$ B $ - the $ n $-liter container,
$ C $ - the $ (m+n) $-liter container.
Suppose that at some point, container $ C $ contains $ c $ liters of water. We will show that with at most two pours, we can achieve a state where container $ C $ contains $ c $ liters of water, where
Let $ a $ and $ b $ be the volume (in liters) of water in containers $ A $ and $ B $ at the considered moment. Thus, $ a $, $ b $, and $ c $ are integers,
Assume that $ c \leq m $. Then $ a + b \geq n $. We pour water from container $ A $ to container $ B $, completely filling it; that is, we pour $ n - b $ liters from $ A $ to $ B $ (this is possible because $ a \geq n - b $). Then we pour the entire contents of container $ B $, which is $ n $ liters, into $ C $, obtaining $ c + n $ liters in container $ C $.
Consider the second case: $ c > m $. Then $ a + b < n $. We pour the entire contents of container $ A $, which is $ a $ liters, into $ B $. We pour $ m $ liters from $ C $ into the empty container $ A $; in container $ C $, $ c - m $ liters of water remain.
Thus, we have obtained $ c $ liters in container $ C $, where the number $ c $ is given by formula (1). From this formula, it follows that
At the initial moment, container $ C $ contained $ m + n $ liters. By applying the described procedure, we obtain in container $ C $:
\begin{center}
after (at most) two pours - $ c_1 $ liters;\\
after (at most) four pours - $ c_2 $ liters;\\
etc.;\\
after (at most) $ 2i $ pours - $ c_i $ liters,
\end{center}
where
We will prove that every number $ k \in \{1,2,\ldots,m+n-1\} $ is equal to $ c_i $ for some $ i $; it is therefore a possible volume of water in container $ C $.
Let $ c_0 = 0 $. Each of the numbers
belongs to the set $ \{0,1,2,\ldots,m+n-1\} $; it is, according to formula (2), the remainder of the division of the product $ in $ by $ m + n $.
The numbers $ m $ and $ n $ are relatively prime by assumption. Therefore, the numbers $ n $ and $ m + n $ are also relatively prime. It follows that the remainders (3) are all different. Indeed: if for some numbers $ i $, $ j $ such that $ 0 \leq i < j \leq m + n - 1 $ the equality $ c_i = c_j $ held, it would mean that the difference $ jn - in $, i.e., the product $ (j - i)n $, is divisible by $ m + n $. The second factor of this product (i.e., $ n $) has no common prime factors with the number $ m + n $; this number must therefore be a divisor of the difference $ (j - i) $ - a contradiction, since $ 0 < j - i < m + n $.
Thus, indeed, the remainders (3) listed above are different elements of the set $ \{0,1,2,\ldots, m+n-1\} $. There are $ m + n $ of them. Therefore, every number $ k $ belonging to this set is equal to some number $ c_i $. This is precisely what we intended to prove.
Note: The described method allows achieving any given value $ k $ in no more than $ 2(m + n -1) $ pours. This estimate can be improved. If the desired number $ k $ is in the second half of the sequence (3), then using the "reverse" procedure, which with at most two pours will cause the volume of water in container $ C $ to change from value $ c $ to value
We leave it to the Reader as an exercise to refine the details and convince themselves that by choosing one of these procedures, we can achieve the desired value $ k $ in no more than $ 2[\frac{1}{2}(m + n)] $ pours.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 451 |
XLVI OM - II - Problem 3
Given are positive irrational numbers $ a $, $ b $, $ c $, $ d $, such that $ a+b = 1 $. Prove that $ c+d = 1 $ if and only if for every natural number $ n $ the equality $ [na] +[nb] = [nc] + [nd] $ holds.
Note: $ [x] $ is the greatest integer not greater than $ x $.
|
Let's take any natural number $ n \geq 1 $. According to the definition of the symbol $ [x] $,
The products $ na $, $ nb $, $ nc $, $ nd $ cannot be integers (since the numbers $ a $, $ b $, $ c $, $ d $ are irrational by assumption), and therefore in none of the listed relationships can equality hold. Adding the inequalities in the left "column" and the right "column" (and using the assumption that $ a + b = 1 $), we obtain the following relationships
where
The numbers $ n $ and $ k_n $ are integers, so the first of the double inequalities (1) implies the equality $ n = k_n + 1 $.
If now $ c + d = 1 $, then the number $ l_n $, like $ k_n $, must (by the second pair of inequalities (1)) satisfy the equality $ n = l_n + 1 $; thus $ k_n = l_n( = n-1) $ for $ n = 1,2,3,\ldots $.
Conversely, assuming that $ k_n = l_n $ for $ n = 1,2,3,\ldots $, we obtain from the second pair of relationships (1) the double inequality
and since $ k_n = n - 1 $, we have
The only value of the sum $ c + d $ that satisfies this infinite system of double inequalities is the number $ 1 $.
We have thus established the desired equivalence:
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 453 |
LIX OM - III - Task 5
The areas of all sections of the parallelepiped $ \mathcal{R} $ by planes passing through the midpoints of three of its edges,
none of which are parallel and do not have common points, are equal. Prove that the parallelepiped
$ \mathcal{R} $ is a rectangular parallelepiped.
|
Let's assume that the areas of all the sections mentioned in the problem are equal to $S$.
Let $ABCD$ and $EFGH$ be the bases of the parallelepiped $\mathcal{R}$ (Fig. 3), let $O$ be its center of symmetry, and let $I, J, K, L, M, N$ denote the midpoints of the edges $AE, EF, FG, GC, CD, DA$, respectively. Then the following pairs of points: $I$ and $L$, $J$ and $M$, $K$ and $N$, are symmetric with respect to the point $O$.
om59_3r_img_3.jpg
The plane $\pi$ passing through the points $I, K, M$ in the section of the parallelepiped $\mathcal{R}$ defines a figure with area $S$. We will show that this figure is the hexagon $IJKLMN$.
Let $\pi$ be the plane passing through the points $I, N, M$. By the converse of Thales' theorem, we have $NM \parallel AC \parallel IL$. The line $NM$ and the point $I$ lie in the plane $\pi$, so the line passing through the point $I$ and parallel to the line $NM$ (i.e., the line $IL$) also lies in the plane $\pi$. Therefore, the midpoint $O$ of the segment $IL$ lies in the plane $\pi$. This means that the points $J$ and $K$, which are symmetric to the points $M$ and $N$ with respect to the point $O$, also lie in the plane $\pi$. Consequently, the points $I, J, K, L, M, N$ lie in this plane; they thus form a flat hexagon with the center of symmetry $O$.
Moreover, the planes $\pi$ and $\pi$ have three non-collinear points in common: $I, K$, and $M$. This proves that $\pi = \pi$.
In this way, we have shown that the point $O$ is the center of symmetry of the hexagon $IJKLMN$ with area $S$. Therefore, the quadrilateral $ILMN$ has an area of $\frac{1}{2}S$; it is also a trapezoid, as we have shown earlier that the lines $IL$ and $MN$ are parallel.
If we denote by $P$ and $Q$ the midpoints of the edges $AB$ and $BC$, respectively, and conduct similar reasoning, we can justify that the quadrilateral $ILQP$ is a trapezoid with an area of $\frac{1}{2}S$.
Thus, the trapezoids $ILMN$ and $ILQP$ have equal areas; they also have a common base $IL$ and equal second bases ($NM = PQ = \frac{1}{2}AC$). It follows that these trapezoids have equal heights. In other words, the distances from the line $IL$ to the lines $NM$ and $PQ$ are the same.
If $l_1, l_2$ are different parallel lines in space, then any line parallel to them and equally distant from them lies in the plane $\Pi$, with respect to which the lines $l_1$ and $l_2$ are symmetric. The plane $\Pi$ is also perpendicular to the plane defined by the lines $l_1$ and $l_2$.
Therefore, the lines $AC$ and $IL$, which are equidistant from the lines $NM$ and $PQ$, define a plane perpendicular to the plane $MNPQ$. In other words, the plane $\pi_1$ containing the parallelogram $ACGE$ is perpendicular to the base $ABCD$. Similarly, we show that the plane $\pi_2$ containing the parallelogram $BFHD$ is perpendicular to the base $ABCD$. Hence, the edge of the planes $\pi_1$ and $\pi_2$, which is a line parallel to $AE$, is perpendicular to the base $ABCD$.
From this, we obtain the equalities $\measuredangle EAB = \measuredangle EAD = 90^{\circ}$.
Similarly, we prove that $\measuredangle DAB = 90^{\circ}$. Therefore, the parallelepiped $\mathcal{R}$ is a rectangular parallelepiped.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 454 |
LVIII OM - I - Problem 12
The polynomial $ W $ with real coefficients takes only positive values in the interval $ \langle a;b\rangle $ (where $ {a<b} $). Prove that there exist polynomials $ P $ and $ Q_1,Q_2,\ldots,Q_m $ such that
for every real number $ x $.
|
We will conduct the proof by induction on the degree of the polynomial $W$.
If $W(x) \equiv c$ is a constant polynomial, then of course $c > 0$ and in this case we can take $P(x) \equiv \sqrt{c}$, $m=1$, and $Q_1(x) \equiv 0$.
Now suppose that the thesis of the problem is true for all polynomials of degree less than $n$ and let $W$ be a polynomial of degree $n$.
The function
considered in the interval $(a; b)$, is continuous and takes positive values in it, which approach infinity at the ends of the interval. Therefore, the function $f$ attains its minimum value $c$ at some point (not necessarily one) in the interval; this is a positive number.
Thus, the inequality $f(x) \geq c$ holds, i.e., $W(x) \geq c(x-a)(b-x)$ for $x \in [a; b]$, which becomes an equality at some point (points). Speaking figuratively, $c$ is the number for which the parabola $y = c(x-a)(b-x)$ in the interval $[a; b]$ is "tangent from below" to the graph of the polynomial $W$ on this interval, possibly at several points (Fig. 7). Therefore, the polynomial $G(x) = W(x) - c(x-a)(b-x)$ in the considered interval takes non-negative values and has at least one root (Fig. 8).
om58_1r_img_7.jpg
om58_1r_img_8.jpg
Notice that $G(a) = W(a)$ and $G(b) = W(b)$, so the polynomial $G$ takes positive values at the ends of the interval. Therefore, the roots of the polynomial $G$ in the interval $(a, b)$ have even multiplicity. Thus, there exist numbers (not necessarily distinct) $g_1, g_2, \ldots, g_k \in (a, b)$ such that the polynomial $G$ has the form
$$G(x) = (x-g_1)^2(x-g_2)^2 \ldots (x-g_k)^2H(x),$$
where the polynomial $H$ takes only positive values in the interval $[a; b]$. Let $B(x) = (x-g_1)(x-g_2) \ldots (x-g_k)$; thus, we have $G(x) = B(x)^2H(x)$.
If $m$ is the degree of the polynomial $G$, then $m \leq n$, with the exception of the case $n=1$, when $m=2$. Moreover, the polynomial $H$ has a degree no greater than $m-2$. Therefore, $H$ is a polynomial of degree lower than $W$. Hence, by the induction hypothesis, there exists a representation of the form
Thus, we obtain
and this is the desired representation of the polynomial $W$. This completes the solution of the problem.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 455 |
XXVII OM - II - Problem 6
Six points are placed on a plane in such a way that any three of them are vertices of a triangle with sides of different lengths. Prove that the shortest side of one of these triangles is also the longest side of another one of them.
|
Let $P_1, P_2, \ldots, P_6$ be given points. In each of the triangles $P_iP_jP_k$, we paint the shortest side red. In this way, some segments $\overline{P_rP_s}$ are painted red, while others remain unpainted.
It suffices to prove that there exists a triangle with vertices at the given points, all of whose sides are painted red. The longest side of this triangle is also the shortest side of some other triangle, since it has been painted red. From each of the given points, five segments connect it to the other given points. Therefore, either at least 3 of these segments are painted red, or at least 3 are not painted.
If from point $P_1$ at least three segments are painted red (for example, segments $\overline{P_1P_2}$, $\overline{P_1P_3}$, $\overline{P_1P_4}$ are painted red), then in the triangle determined by the other ends of these segments (i.e., in the triangle $P_2P_3P_4$), at least one of the sides (namely, the shortest one) is painted red. Let, for example, segment $\overline{P_2P_3}$ be painted red. Then in the triangle $P_1P_2P_3$, all sides are painted red.
If, however, from point $P_1$ at least three segments are not painted (let these be segments $\overline{P_1P_2}$, $\overline{P_1P_3}$, $\overline{P_1P_4}$), then consider the triangles $P_1P_2P_3$, $P_1P_2P_4$, $P_1P_3P_4$. In each of them, at least one of the sides is painted red, but it is not the side containing vertex $P_1$. Therefore, the segments $\overline{P_2P_3}$, $\overline{P_2P_4}$, $\overline{P_3P_4}$ are painted red, i.e., the triangle $P_2P_3P_4$ has all its sides painted red.
Note 1. In the above solution, we did not use the assumption that the given points lie in a plane.
Note 2. The thesis of the problem can also be obtained by reasoning similarly to the solution of problem 15 (3) from the XXII Mathematical Olympiad.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 456 |
XV OM - I - Problem 9
Prove that the quotient of the sum of all natural divisors of an integer $ n > 1 $ by the number of these divisors is greater than $ \sqrt{n} $.
|
Let $ d_1, d_2, \ldots, d_s $ denote all the natural divisors of a given integer $ n > 1 $. Among these divisors, there are certainly unequal numbers, such as $ 1 $ and $ n $. According to the Cauchy inequality, the arithmetic mean of positive numbers that are not all equal is greater than the geometric mean of these numbers, hence
We will prove that the right-hand side of this inequality is equal to $ \sqrt{n} $. For this purpose, we will distinguish two cases:
1. The number $ n $ is not a square of an integer. In the set of numbers $ d_1, d_2, \ldots, d_s $, for each number $ d_i $ there is a different number $ d_k = \frac{n}{d_i} $, so the number $ s $ is even and the set $ \{d_1,d_2, \ldots,d_s\} $ can be divided into $ \frac{s}{2} $ such pairs, where the product of the numbers in each pair is equal to $ n $. In this case,
2. The number $ n $ is a square of a natural number $ d_r $. For each number $ d_i $ in the set $ \{d_1, d_2, \ldots, d_s\} $ different from $ d_r $, there is a number $ d_k = \frac{n}{d_i} $ different from $ d_i $, so the number $ s - 1 $ is even and the set of numbers $ \{ d_1, \ldots d_{r-1}, d_{r+1} \ldots, d_s\} $ can be divided into $ \frac{s-1}{2} $ pairs with a product equal to $ n $. In this case,
In view of inequality (1), in each of the cases 1 and 2, the inequality holds:
This completes the translation, maintaining the original text's line breaks and formatting.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 457 |
XXXIV OM - I - Problem 7
Let $ S_n $ be the set of sequences of length $ n $ with terms $ -1 $, $ +1 $. We define the function $ f: S_n - \{(-1, 1, 1,\ldots, 1)\} \to S_n $ as follows: if $ (a_1, \ldots, a_n) \in S_n - \{(-1, 1, 1,\ldots, 1)\} $ and $ k = \max_{1\leq j \leq n} \{j \;:\; a_1\cdot a_2\cdot \ldots \cdot a_j = 1\} $, then $ f(a_1,\ldots, a_n) = (a_1, \ldots, a_{k-1}, -a_k, a_{k+1}, \ldots, a_n) $. Prove that
|
To simplify the notation, we introduce the symbol $ f^{(k)} $ to denote the $ k $-fold iteration of the function $ f $:
The thesis of the problem will be proved by induction. The definition of the function $ f $ depends, of course, on the natural number $ n $. Where it helps to avoid misunderstandings, we will write $ f_n $ instead of $ f $ to highlight the value of the natural number $ n $.
First, we will prove two lemmas.
Lemma 1. If $ m $, $ n $ are natural numbers and $ f_n^{(2m)}(1, 1, \ldots, 1)= (a_1, a_2, \ldots, a_n) $, then $ a_1a_2 \ldots a_n=1 $.
Proof. From the definition of the function $ f $, it follows that if $ f(b_1, b_2, \ldots, b_n)= (c_1, c_2, \ldots, c_n) $, then $ c_1c_2\ldots c_n = -b_1 b_2 \ldots b_n $. Therefore, an even number of compositions of the function $ f $ assigns the sequence $ (1, 1,\ldots, 1) $ a sequence whose terms multiply to $ 1 $.
Lemma 2. If $ f_n^{(m)} (1,1,\ldots,1) = (a_1, a_2, \ldots, a_n) $, then $ f_{n+1}^{(2m)}(1, 1, \ldots,1)= (a_1, a_2, \ldots, a_n, a) $, where $ a = 1 $ or $ a = - 1 $.
Proof. We will use induction on $ m $.
If $ m= 1 $, then $ f_n(1, 1,\ldots, 1)= (1, 1,\ldots, 1, - 1) $, and $ f_{n+1}^{(2)} (1,1,\ldots,1) = f_{n+1}(1,1,\ldots, 1,-1)= (1,1,\ldots,-1,-1) $.
Assume that for some $ m $, if $ f_n^{(m)}(1,1,\ldots, 1) =(a_1,a_2,\ldots,a_n) $, then $ f_{n+1}^{(2m)}(1,1,\ldots, 1) =(a_1, a_2,\ldots,a_n, a) $. From the definition of the function $ f $, it follows that $ f_n^{(m+1)}(1,1,\ldots, 1) =f_n(a_1, a_2, \ldots, a_n)= (a_1, a_2, \ldots, -a_k, \ldots, a_n) $. By Lemma 1, $ a_1a_2\ldots a_na= 1 $, so $ f_{n+1}(a_1, a_2, \ldots, a_n, a)= (a_1, a_2, \ldots, a_n, -a) $, and by setting $ a_{n + 1} = -a $ we have $ f_{n+1}(a_1,a_2,\ldots, a_n,a)= (a_1, a_2, \ldots, a_n, a_{n+1}) $. Since $ a_1a_2\ldots a_{n + 1} = - 1 $, it follows that
$ f_{n+1}(a_1, a_2, \ldots, a_{n+1})= (a_1,a_2,\ldots,-a_k, \ldots, a_{n+1}) $. Therefore, $ f_{n+1}^{(2(k+1))}(1, 1, \ldots, 1)= f_{n+1}^{(2)} (a_1, a_2, \ldots, a_n, -a_{n+1}) = (a_1, a_2, \ldots, -a_k, \ldots, a_n, a_{n+1}) $. This completes the proof of the inductive step. By the principle of induction, the thesis of the lemma is satisfied for every natural $ k $.
We will now proceed to the proof of the theorem stated in the problem.
For $ n =2 $, the theorem is obviously satisfied.
Assume for some $ n $ that
Based on Lemma 2,
where $ a = 1 $ or $ a = -1 $.
However, from Lemma 1, $ -1 \cdot 1 \cdot \ldots \cdot 1 \cdot a = 1 $, so $ a = - 1 $. Therefore,
By the principle of induction, for every $ n $, $ f^{2^{n-1}} (1,1,\ldots, 1)= (-1,1,\ldots, 1) $.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 459 |
XIX OM - I - Problem 7
Points $ D, E, F $ are the midpoints of the sides of triangle $ ABC $. Prove that if the circumcircles of triangles $ ABC $ and $ DEF $ are tangent, then the point of tangency is one of the points $ A $, $ B $, $ C $ and triangle $ ABC $ is a right triangle.
|
Suppose that the circle $c$ circumscribed around triangle $ABC$ and the circle $k$ circumscribed around triangle $DEF$ are tangent at point $T$ (Fig. 5).
Triangle $DEF$ lies inside circle $c$, so the tangency of circles $c$ and $k$ is internal, and the center $S$ of circle $k$ lies on the ray $TO$. Triangle $DEF$ is similar to triangle $ABC$ in the ratio $1:2$, so the radius $TS$ of circle $k$ is half the radius $TO$ of circle $c$; hence, point $S$ is the midpoint of segment $TO$, with $TO$ being the diameter of circle $k$.
Points $D$, $E$, $F$ of circle $k$ are distinct, so at most one of them can coincide with point $O$. Suppose, for example, that points $D$ and $E$ are different from point $O$.
The chord $BC$ of circle $c$, whose midpoint is point $D$, is perpendicular to the line $OD$, so one of the angles $ODB$ and $ODC$ is inscribed in circle $k$, which means that point $B$ or point $C$ is at the end $T$ of the diameter $OT$ of circle $k$. Similarly, one of the endpoints of the chord $AC$ lies at point $T$. Therefore, vertex $C$ of triangle $ABC$ is at point $T$. In this case, the midpoint $F$ of side $AB$ lies at point $O$. If point $F$ were different from point $O$, then the chord $AB$ of circle $c$ would be perpendicular to the line $OF$, so one of the right angles $OFA$ and $OFB$ would be inscribed in circle $k$, which means that point $A$ or point $B$ would be at point $T$, i.e., would coincide with point $C$, which is impossible.
Angle $C$ of triangle $ABC$ is an inscribed angle in circle $c$ whose sides pass through the endpoints of the diameter $AB$ of this circle, i.e., it is a right angle.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 462 |
LIV OM - III - Task 1
In an acute triangle $ABC$, segment $CD$ is an altitude. Through the midpoint $M$ of side $AB$, a line is drawn intersecting rays $CA$ and $CB$ at points $K$ and $L$, respectively, such that $CK = CL$. Point $S$ is the center of the circumcircle of triangle $CKL$. Prove that $SD = SM$.
|
From Menelaus' theorem applied to triangle $ABC$ we get
Let $E$ be the second intersection point of line $CS$ with the circumcircle of triangle $CKL$ (Fig. 1). From the equality $CK = CL$, it follows that $EK = EL$. Moreover, $\measuredangle AKE = 90^\circ = \measuredangle BLE$.
Therefore, the right triangles $AKE$ and $BLE$ are congruent. Hence, we get $AE = BE$, which means $EM \perp AB$, and this gives $CD \parallel EM$.
om54_3r_img_1.jpg
Since $S$ is the midpoint of segment $CE$, its orthogonal projection onto line $AB$ coincides with the midpoint of segment $DM$. Therefore, $SD = SM$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 463 |
XIX OM - I - Problem 2
Let $ p(n) $ denote the number of prime numbers not greater than the natural number $ n $. Prove that if $ n \geq 8 $, then $ p(n) < \frac{n}{2} $
|
If $ n $ is an even number, then in the set of natural numbers from $ 1 $ to $ n $, there are $ \frac{n}{2}-1 $ even numbers different from $ 2 $, and thus composite, and at least one odd number, namely $ 1 $, which is not a prime number. Prime numbers belong to the set of the remaining $ \frac{n}{2} $ numbers, so $ p(n) \leq \frac{n}{2} $.
If $ n $ is an odd number at least equal to $ 9 $, then in the set of natural numbers from $ 1 $ to $ n $, there are $ \frac{n-1}{2}-1 $ even numbers different from $ 2 $, and there are at least two odd numbers, namely $ 1 $ and $ 9 $, which are not prime numbers. Prime numbers belong to the set of the remaining $ n - \left(\frac{n-1}{2}-1 \right) - 2 = \frac{n-1}{2} $ numbers, so $ p(n) < \frac{n}{2} $.
Therefore, for every natural number $ n $:
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 465 |
XXXVII OM - II - Problem 3
Let S be a sphere circumscribed around a regular tetrahedron with an edge length greater than 1. The sphere $ S $ is represented as the union of four sets. Prove that there exists one of these sets that contains points $ P $, $ Q $, such that the length of the segment $ PQ $ exceeds 1.
|
We will first prove a lemma that is a one-dimensional version of this theorem.
Lemma. The circle $\omega$ circumscribed around an equilateral triangle with a side length greater than $1$ is represented as the union of three sets: $\omega = U \cup V \cup W$. Then, there exist points $P$, $Q$ in one of the sets $U$, $V$, $W$ such that $|PQ| > 1$.
om37_2r_img_7.jpg
Proof of the lemma. For any two non-antipodal points $X, Y \in \omega$, we will denote by $\breve{XY}$ the closed arc of the circle $\omega$ with endpoints $X$ and $Y$, contained in a semicircle. Suppose that the thesis of the lemma is not true and take any equilateral triangle $ABC$ inscribed in $\omega$. Its vertices must belong to different sets (among $U$, $V$, $W$); we can assume that $A \in U$, $B \in V$, $C \in W$. Then $W \cap \breve{AB} = \emptyset$ (because the distance from any point of the arc $\breve{AB}$ to the point $C$ exceeds $1$). Let
and let $E \in \breve{CA}$, $F \in \breve{CB}$ be points such that $|CE| = a$, $|CF| = b$ (figure 7). The segment $EF$ has a length not greater than $1$; if $|EF| > 1$, then since, according to (1), arbitrarily close to the points $E$ and $F$ there are points of the set $W$, we would find points $P, Q \in W$ such that $|PQ| > 1$ - contrary to our assumption. Therefore, $|EF| \leq 1$, and thus the arc $EF$ is smaller than $1/3$ of the circle. (This is, of course, the arc with endpoints $E$ and $F$ that contains the point $C$, because the other of these arcs contains the points $A$ and $B$). There are therefore points $P, Q, R \in \omega$ lying outside the arc $\breve{EF}$ and being the vertices of an equilateral triangle.
From the definition of the numbers $a$ and $b$ and the points $E$ and $F$, it follows that $W \cap \breve{CA} \subset \breve{CE}$, $W \cap \breve{CB} \subset \breve{CF}$, and since $W \cap \breve{AB} = \emptyset$, we conclude from this that $W \subset \breve{EF}$. Therefore, the points $P$, $Q$, $R$ belong to the union of the sets $U$ and $V$, and thus two of them must belong to one of these sets. This completes the proof of the lemma, because $|PQ| = |QR| = |RP| > 1$.
The thesis of the problem immediately follows from the lemma: we take a regular tetrahedron $ABCD$ inscribed in the sphere $S$ and the circle $\omega$ circumscribed around the triangle $ABC$. The point $D$ belongs to one of the four sets under consideration. If this set contains any point $X \in \omega$, then $|DX| > 1$. Otherwise, the circle $\omega$ is contained in the union of the remaining three sets, and by the lemma, one of these sets contains points $P$, $Q$ such that $|PQ| > 1$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 466 |
IX OM - I - Problem 11
Prove that if two quadrilaterals have the same midpoints of their sides, then they have equal areas. Show the validity of an analogous theorem for convex polygons with any even number of sides.
|
Let $M$, $N$, $P$, $Q$ be the midpoints of the sides $AB$, $BC$, $CD$, $DA$ of quadrilateral $ABCD$ (Fig. 10). By the theorem on the segment joining the midpoints of two sides of a triangle, in triangle $ABD$ we have
and in triangle $BCD$ we have
thus $MQ = NP$ and $MQ \parallel NP$, so quadrilateral $MNPQ$ is a parallelogram; line $BD$ divides it into parallelograms $QMKL$ and $PNKL$. The area of parallelogram $QMKL$ is half the area of triangle $ABD$, since side $MQ$ and the height of the parallelogram relative to this side are equal to half of side $BD$ and half of the height relative to $BD$ in triangle $ABD$. Similarly, the area of parallelogram $PNKL$ is half the area of triangle $BCD$. Therefore, the area of quadrilateral $ABCD$ is twice the area of parallelogram $MNPQ$.
If, therefore, two quadrilaterals have the same midpoints of sides $M$, $N$, $P$, $Q$, then their areas are equal, and specifically equal to twice the area of parallelogram $MNPQ$.
We will prove a more general theorem, that if two convex polygons with $2n$ sides have the same midpoints of sides, then they have equal areas. We will use complete induction. For $n = 2$ the theorem is true; assume it is true for $n = k - 1$ ($k \geq 3$). Let $A_1A_2 \ldots A_{2k}$ and $B_1B_2 \ldots B_{2k}$ be two convex polygons with $2k$ sides having the same midpoints of sides $M_1, M_2, \ldots M_{2k}$, where $M_i$ is the common midpoint of sides $A_iA_{i+1}$ and $B_iB_{i+1}$. In Fig. 11, polygon $B_1B_2 \ldots B_{2k}$ is not drawn; the reader is asked to imagine it.
Draw diagonals $A_1A_4$ and $B_1B_4$ in these polygons, dividing them into quadrilaterals $A_1A_2A_3A_4$ and $B_1B_2B_3B_4$ and convex polygons $A_1A_4 \ldots A_{2k}$ and $B_1 B_4 \ldots B_{2k}$ with $2k - 2$ sides. The midpoint $N$ of segment $A_1A_4$ is uniquely determined by points $M_1$, $M_2$, $M_3$, as it is the fourth vertex of the parallelogram whose other three consecutive vertices are points $M_1$, $M_2$, $M_3$. This means that the same point $N$ is also the midpoint of diagonal $B_1B_4$ of polygon $B_1B_2\ldots B_{2k}$. This implies that both quadrilaterals $A_1A_2A_3A_4$ and $B_1B_2B_3B_4$, as well as both polygons $A_1A_4\ldots A_{2k}$ and $B_1B_4 \ldots B_{2k}$, have the same midpoints of sides. Therefore, by the previous theorem
and by the induction hypothesis
Adding these equalities side by side, we get
Note 1. The proof of the theorem for quadrilaterals applies to both convex and concave quadrilaterals (see Fig. 10). However, in the proof of the theorem for polygons with an even number of sides greater than 4, we relied on the assumption that the polygons are convex, specifically on the fact that drawing a diagonal through the first and fourth vertices (counted in a certain direction along the perimeter) in a convex polygon with $2k$ sides divides the polygon into a quadrilateral and a polygon with $2k - 2$ sides. The question arises whether the theorem is true for concave polygons. The answer is affirmative, but the proof must be conducted differently, as there are concave polygons in which there is no diagonal that cuts off a quadrilateral, such as the 12-pointed star in Fig. 12. We do not provide this proof, leaving it as an interesting but challenging task for the reader.
Note 2. The theorem is also true for polygons with an odd number of sides. In this case, it is obvious, as two polygons with an odd number of sides having the same midpoints of sides must coincide. See Problems from Mathematical Olympiads, Warsaw 1956, PZWS, problem no. 123.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 468 |
XXIV OM - II - Problem 2
In a given square, there are nine points, no three of which are collinear. Prove that three of them are vertices of a triangle with an area not exceeding $ \frac{1}{8} $ of the area of the square.
|
Let's first prove the
Lemma. If triangle $ABC$ is contained within a certain rectangle, then the area of the triangle is not greater than half the area of that rectangle.
Proof. Let $A$, $B$, $C$ be the projections of the vertices of triangle $ABC$ onto one of the sides of the rectangle $PQRS$ containing this triangle (Fig. 14). Suppose, for example, that $B$ and let $D$ be the point of intersection of line $AC$ with the line perpendicular to $PQ$ passing through point $B$.
Triangle $ABC$ is the sum of triangles $ABD$ and $CBD$ (one of them may be degenerate to a segment if $A$ or $B$). Therefore, $S_{ABC} = S_{ABD} + S_{BCD} = \frac{1}{2} A$.
Since $A$ and $BD \leq QR$, it follows that
We now proceed to solve the problem. We divide the given square by lines parallel to its sides into four congruent squares (or four congruent rectangles) - Fig. 15. The area of such a square (rectangle) is equal to $\frac{1}{4}$ of the area of the given square. Among the nine given points, at least three belong to one of the obtained squares (rectangles). The area of the triangle with vertices at these points is, by the lemma, not greater than $\frac{1}{8}$ of the area of the given square.
Note. In a similar way, it can be shown that the statement of the problem remains true if the numbers $9$ and $\frac{1}{8}$ are replaced by the numbers $2n+1$ and $\frac{1}{2n}$, respectively. It is sufficient to consider the division of the square by lines parallel to its sides into $n$ congruent rectangles.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 469 |
LI OM - II - Task 2
The bisector of angle $ BAC $ of triangle $ ABC $ intersects the circumcircle of this triangle at point $ D $ different from $ A $. Points $ K $ and $ L $ are the orthogonal projections of points $ B $ and $ C $, respectively, onto the line $ AD $. Prove that
|
Let $ o $ be the circumcircle of triangle $ ABC $. From the equality of angles $ BAD $ and $ DAC $, it follows that the lengths of arcs $ BD $ and $ DC $ of circle $ o $ are equal (the length of arc $ XY $ is measured from point $ X $ to point $ Y $ in the counterclockwise direction).
om51_2r_img_1.jpg
om51_2r_img_2.jpg
Let $ E $ be the point symmetric to point $ D $ with respect to the perpendicular bisector of side $ AB $, and let $ N $ be the orthogonal projection of point $ E $ onto line $ AD $ (Fig. 1 and 2). Then $ AD = BE $. Moreover, the lengths of arcs $ DC $ and $ EA $ are equal, which means that $ EN = CL $.
Points $ B $ and $ E $ lie on opposite sides of line $ AD $. Therefore, the length of segment $ BE $ is not less than the sum of the distances from points $ B $ and $ E $ to line $ AD $. In other words, $ BE \geq BK + EN $, which means $ AD \geq BK + CL $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 471 |
XII OM - II - Task 3
Prove that for any angles $ x $, $ y $, $ z $ the following equality holds
|
We transform the right side of equation (1) by first applying the formula $ 2 \sin \alpha \sin \beta = \cos (\alpha - \beta) - \cos (\alpha + \beta) $, and then the formulas for $ \cos (\alpha + \beta) $ and $ \cos (\alpha - \beta) $:
After substituting in the last expression $ \sin^2 x \sin^2 y = (1 - \cos^2 x) (1 - \cos^2 y) $, and expanding and reducing the brackets, we obtain an expression equal to the left side of equation (1).
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 472 |
XXX OM - II - Task 3
In space, a line $ k $ and a cube with vertex $ M $ and edges $ \overline{MA} $, $ \overline{MB} $, $ \overline{MC} $, each of length 1, are given. Prove that the length of the orthogonal projection of the edge $ MA $ onto the line $ k $ is equal to the area of the orthogonal projection of the square with sides $ MB $ and $ MC $ onto a plane perpendicular to the line $ k $.
|
Without loss of generality, we can assume that the line $k$ passes through the point $M$ (Fig. 12) and that the plane $\pi$ perpendicular to the line $k$ also contains the point $M$. Let $A$ be the orthogonal projection of the point $A$ onto the line $k$.
om30_2r_img_12.jpg
As is known, if two planes intersect at an angle $\varphi$ and one of them contains a figure with area $S$, then the orthogonal projection of this figure onto the other plane has an area of $S \cdot \cos \varphi$.
In our problem, let the angle between the planes $\pi$ and $MBC$ be $\varphi$. Then the angle between the lines perpendicular to these planes will also be $\varphi$, i.e., $\measuredangle AMA$.
The area of the square with sides $\overline{MB}$ and $\overline{MC}$ is $1$. Therefore, the area of the projection of this square onto the plane $\pi$ will be $\cos \varphi$. On the other hand, we have $MA$. It follows from this the thesis of the problem.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 473 |
XLVI OM - I - Problem 8
In a regular $ n $-sided pyramid, the angles of inclination of the lateral face and the lateral edge to the base plane are $ \alpha $ and $ \beta $, respectively. Prove that $ \sin^2 \alpha - \sin^2 \beta \leq \tan ^2 \frac{\pi}{2n} $.
|
Let's assume that the side of the regular $ n $-gon $ A_1A_2\ldots A_n $, which serves as the base of the pyramid, has a length of $ 2a $. Let $ O $ be the common center of the circle circumscribed around this polygon and the circle inscribed in it; let the radii of these circles be (respectively) $ R $ and $ r $. Denote the vertex of the pyramid by $ S $, and the length of the segment $ OS $ (the height of the pyramid) by $ h $.
om46_1r_img_4.jpg
Take one of the edges of the base - for example, $ A_1A_2 $ - and let $ M $ be its midpoint. Let $ \varphi = | \measuredangle A_1OM| = | \measuredangle A_2OM| = \pi/n $. The dihedral angle between the plane of the base (i.e., the plane $ A_1A_2O $) and the plane of the lateral face $ A_1 A_2S $ has a measure equal to the planar angle $ OMS $. From the adopted notation, the following relationships (figure 4) follow:
It is necessary to show that
The useful trigonometric relationships are:
Using formulas (4) and (2), we obtain the equality
Based on the relationships (5) and (1), we have:
Therefore,
And since
the numerator of the fraction obtained on the right side of equation (7) is not greater than the denominator. Therefore, the value of the quotient (7) does not exceed $ 1 $; thus, inequality (3) has been proven.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 475 |
XXV OM - III - Problem 1
In the tetrahedron $ ABCD $, the edge $ \overline{AB} $ is perpendicular to the edge $ \overline{CD} $ and $ \measuredangle ACB = \measuredangle ADB $. Prove that the plane determined by the edge $ \overline{AB} $ and the midpoint of the edge $ \overline{CD} $ is perpendicular to the edge $ \overline{CD} $.
|
We will first prove the
Lemma. If lines $AB$ and $PQ$ intersect at a right angle at point $P$, then the number of points on the ray $PQ^\to$ from which the segment $\overline{AB}$ is seen at an angle $\alpha$ is $0$, $1$, or $2$.
Proof. As is known, the set of points contained in the half-plane with edge $AB$ and containing point $Q$, from which the segment $\overline{AB}$ is seen at an angle $\alpha$, is an arc of a certain circle. This arc has $0$, $1$, or $2$ points in common with the ray $PQ^\to$ depending on the position of the segment $AB$ relative to point $P$ and the size of the angle $\alpha$ (Fig. 19).
Corollary. Let line $AB$ be perpendicular to the plane $\pi$. The set of points in the plane $\pi$ from which the segment $\overline{AB}$ is seen at an angle $\alpha (0 < \alpha < \pi)$ is empty, is a circle, or is the union of two concentric circles.
Proof. Let $P$ be the point of intersection of line $AB$ with the plane $\pi$, and let $Q$ be any point in the plane $\pi$ different from $P$. By the lemma, the ray $PQ^\to$ contains at most two points from which the segment $\overline{AB}$ is seen at an angle $\alpha$. When rotated around line $AB$, these points form at most two concentric circles, and from each point of these circles, the segment $\overline{AB}$ is seen at an angle $\alpha$. By the lemma, from each point in the plane $\pi$ not belonging to these circles, the segment $\overline{AB}$ is seen at an angle different from $\alpha$.
We proceed to the solution of the problem. Let $\pi$ be a plane perpendicular to $AB$ and containing the segment $\overline{CD}$, and let $P$ be the point of intersection of line $AB$ with the plane $\pi$. By the conditions of the problem, the segment $\overline{AB}$ is seen from points $C$ and $D$ at the same angle. From the corollary to the lemma, it follows that either
1) points $C$ and $D$ belong to one circle centered at point $P$, or,
2) points $C$ and $D$ belong to different circles centered at point $P$.
In the first case, the perpendicular bisector of the segment $\overline{CD}$ contains point $P$, and therefore the plane determined by the edge $\overline{AB}$ and the midpoint of the edge $\overline{CD}$ is perpendicular to the edge $\overline{CD}$. In the second case, the perpendicular bisector of the segment $\overline{CD}$ does not contain point $P$. If it did, then point $P$ would be equidistant from points $C$ and $D$. These points would then belong to one circle centered at point $P$, which is not the case. Therefore, the plane determined by the edge $\overline{AB}$ and the midpoint of the edge $\overline{CD}$ does not contain the perpendicular bisector of the segment $\overline{CD}$, and thus is not perpendicular to it.
The thesis of the problem is therefore not true in the second case. The first case occurs, for example, when point $P$ lies on the segment $\overline{AB}$, and thus, for example, when the angle $\measuredangle ACB$ is obtuse.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 477 |
XXXIX OM - III - Problem 3
Let $ W $ be a polygon (not necessarily convex) with a center of symmetry. Prove that there exists a parallelogram containing $ W $ such that the midpoints of the sides of this parallelogram lie on the boundary of $ W $.
|
Among all triangles $OAB$, where $A$, $B$ are vertices of the polygon $W$, and $O$ is its center of symmetry, let us choose the triangle with the maximum area (there are at least four such triangles, and there may be more; we choose any one of them). Let $OMN$ be the selected triangle, and let $M$, $N$ be the vertices of the polygon $W$ that are symmetric (respectively) to $M$, $N$ with respect to $O$. Through points $M$ and $M'$, we draw lines $m$ and $m'$ parallel to $NN'$; through points $N$ and $N'$, we draw lines $n$ and $n'$ parallel to $MM'$ (see Figure 10).
om39_3r_img_10.jpg
From the maximality of the area of triangle $OMN$, it follows that no vertex of the polygon $W$ can lie farther from the line $MM'$ than point $N$, or farther from the line $NN'$ than point $M$. Therefore, all vertices lie within the parallelogram $R$ formed by the lines $m$, $m'$, $n$, $n'$, and thus $W \subset R$. The midpoints of the sides of $R$ are the points $M$, $N$, $M'$, $N'$, which are vertices of the polygon. Therefore, $R$ is the desired parallelogram.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 479 |
XX OM - II - Task 6
Prove that every polyhedron has at least two faces with the same number of sides.
|
Let $ s $ denote the number of faces of the polyhedron $ W $, and $ P $ be the face with the largest number of sides, which we will denote by $ n $. The face $ P $ is adjacent along its sides to $ n $ other (pairwise distinct) faces of the polyhedron, so $ s \geq n+1 $. On the other hand, the number of sides of each face of the polyhedron $ W $ is one of the numbers $ 3, 4, \ldots, n $; there are $ n-2 $ such numbers. If each face of the polyhedron had a different number of sides, then the inequality $ s \leq n-2 $ would hold, which contradicts the previous one. Therefore, such a polyhedron does not exist.
Note. The reasoning above demonstrates a stronger theorem: In every polyhedron, there are at least three faces such that each of them has the same number of sides as some other face of the polyhedron.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 480 |
XLIX OM - I - Problem 12
Let $ g(k) $ be the greatest prime divisor of the integer $ k $, when $ |k|\geq 2 $, and let $ g(-1) = g(0) = g(1) = 1 $. Determine whether there exists a polynomial $ W $ of positive degree with integer coefficients, for which the set of numbers of the form $ g(W(x)) $ ($ x $ - integer) is finite.
|
We will prove that there does not exist a polynomial with the given property. For a proof by contradiction, suppose that
is a polynomial of degree $ n \geq 1 $ with integer coefficients and that the set of numbers of the form $ g(W(x)) $ (for integer values of $ x $) is finite. This means: there exists a natural number $ m $ and there are prime numbers $ p_1, \ldots, p_m $ with the following property:
if $ x $ is an integer and $ W(x) \neq Q $, then the value $ W(x) $ is not divisible by any prime number different from $ p_1, \ldots, p_m $.
We will consider two cases.
If $ a_0 = 0 $, then for every integer $ x \neq 0 $, the value $ W(x) $ is divisible by $ x $. Let $ b = 1 + p_1 p_2 \ldots p_m $. We find a natural number $ k $ for which $ W(kb) \neq 0 $; such a number exists because the polynomial $ W $ has only finitely many roots.
The value $ W(kb) $ is divisible by $ kb $, and therefore by $ b $; hence it has at least one prime divisor different from $ p_1, \ldots, p_m $. This is a contradiction to condition (2).
Now we will consider the case when $ a_0 \neq 0 $. Consider the number $ c = a_0 p_1 p_2 \ldots p_m $. We find a natural number $ k \geq 2 $ for which $ W(kc) \neq a_0 $; such a number exists because the value $ a_0 $ is taken by the polynomial $ W $ at only finitely many points.
From formula (1) we obtain the equality
where $ q $ denotes the number in the square brackets. This is an integer different from zero (since $ W(kc) \neq a_0 $). Substituting into (3) the expression defining the number $ c $, we get the equality
Since $ k \geq 2 $ and $ q \neq 0 $, the number $ w $ is not equal to $ 1 $, $ -1 $, or $ 0 $; it therefore has a prime divisor different from $ p_1, \ldots, p_m $. Thus, in this case as well, we have obtained a contradiction to condition (2); the proof is complete.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 481 |
LVII OM - I - Problem 3
An acute triangle $ABC$ is inscribed in a circle with center $O$. Point $D$ is the orthogonal projection of point $C$ onto line $AB$, and points $E$ and $F$ are the orthogonal projections of point $D$ onto lines $AC$ and $BC$, respectively. Prove that the area of quadrilateral $EOFC$ is equal to half the area of triangle $ABC$.
|
Let $ P $ be the point symmetric to point $ C $ with respect to point $ O $ (Fig. 1). Since triangle $ ABC $ is acute,
points $ A $, $ P $, $ B $, and $ C $ lie on a circle with center $ O $ in this exact order.
om57_1r_img_1.jpg
The areas of triangles $ COE $ and $ POE $ are equal, as these triangles have a common height dropped from vertex $ E $, and segments $ CO $ and $ OP $ are of equal length. Similarly, the areas of triangles $ COF $ and $ POF $ are equal. Therefore, we obtain $ [EOFC] = \frac{1}{2} \cdot [EPFC] $, where the symbol $ [\mathcal{F}] $ denotes the area of figure $ \mathcal{F} $.
To complete the solution, we need to prove that $ [EPFC] = [ABC] $.
Segment $ CP $ is the diameter of the circumcircle of triangle $ ABC $, so lines $ AP $ and $ AC $ are perpendicular. Line $ DE $ is
perpendicular to line $ AC $, so it is parallel to line $ AP $. Therefore, the areas of triangles $ DEA $ and $ DEP $ are equal. Similarly,
the areas of triangles $ DFB $ and $ DFP $ are equal. Thus,
which completes the solution of the problem.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 483 |
XL OM - I - Task 4
Prove that it is impossible to cut a square along a finite number of segments and arcs of circles in such a way that the resulting pieces can be assembled into a circle (pieces can be flipped).
|
Suppose the division mentioned in the task is feasible. The resulting parts of the square are denoted by $K_1, \ldots, K_n$. The boundary of each figure $K_i$ consists of a finite number of line segments and arcs of circles. Among the arcs of circles that are parts of the boundary of $K_i$, some are convex outward from $K_i$, and others are convex inward toward $K_i$. Let the sum of the lengths of the first type of arcs be denoted by $a_i$, and the lengths of the second type by $b_i$. We adopt $d_i = a_i - b_i$.
Consider the sum $s = d_1 + \ldots + d_n$.
The figures $K_i$ result from the division of the square. The boundary of the square consists of straight line segments. Therefore, each arc of a circle that is a division line of the square is simultaneously a subset of the boundary of two figures $K_i$, with its length contributing to the sum $s$ with a positive sign in one of these figures and with a negative sign in the other. Thus, $s = 0$.
We assumed that the figures $K_i$ can be assembled into a circle. Some of the considered arcs must then be used to form the boundary of this circle. When calculating the sum $s$, the lengths of these arcs will be counted once, always with a positive sign. On the other hand, the lengths of the remaining arcs of circles, forming the division lines of the circle, will be reduced, as in the case of the square. The sum $s$ will therefore be equal to the circumference of the circle.
The obtained contradiction ($s = 0$ and simultaneously $s > 0$) completes the proof.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 484 |
XIII OM - I - Problem 11
Given is a quadrilateral $ABCD$ whose diagonals intersect at a right angle at point $M$. Prove that the $8$ points where perpendiculars drawn from point $M$ to the lines $AB$, $BC$, $CD$, and $DA$ intersect the sides of the quadrilateral lie on a circle.
|
Let $P$, $Q$, $R$, $S$ denote the feet of the perpendiculars dropped from point $M$ to the sides $AB$, $BC$, $CD$, $DA$ respectively (Fig. 16).
First, we will prove that the points $P$, $Q$, $R$, $S$ lie on a circle by showing that the sum of two opposite angles of quadrilateral $PQRS$ equals $180^\circ$.
Each of the quadrilaterals $MPAS$, $MQBP$, $MRCQ$, and $MSDR$ has two right angles, so the vertices of each of them lie on a circle. According to the theorem of inscribed angles,
Thus,
since $AM \bot DM$.
Similarly,
so,
Adding equations (1) and (2) and considering that $\measuredangle MPS + \measuredangle MPQ = \measuredangle SPQ$ and $\measuredangle MRS + \measuredangle MRQ = \measuredangle SRQ$, we get:
Therefore, the points $P$, $Q$, $R$, $S$ indeed lie on a circle. It remains to prove that the points $P_1$, $Q_1$, $R_1$, $S_1$, where the lines $MP$, $MQ$, $MR$, and $MS$ intersect the sides $CD$, $DA$, $AB$, and $BC$ respectively, also lie on the same circle. It is sufficient to prove this for one of these points, for example, $Q_1$, since the reasoning for each of the others is the same (Fig. 17).
Indeed,
From the equality of angles $PSQ_1$ and $PQQ_1$, it follows that the points $P$, $Q$, $S$, $Q_1$ lie on a circle, c.n.d.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 485 |
XV OM - III - Task 4
Prove that if the roots of the equation $ x^3 + ax^2 + bx + c = 0 $, with real coefficients, are real, then the roots of the equation $ 3x^2 + 2ax + b = 0 $ are also real.
|
The task boils down to showing that the given assumptions imply the inequality
Let $ x_1 $, $ x_2 $, $ x_3 $ denote the roots of equation (2); according to the assumption, they are real numbers.
We know that
Hence
Note. Using elementary knowledge of derivatives, the problem can be solved much more simply. It is enough to notice that the function $ 3x^2 + 2ax + b $ is the derivative of the function $ x^3 + ax^2 + bx + c $ and to apply the theorem that between any two roots of a differentiable function, there lies some root of the derivative of that function, and that a multiple root of a function is also a root of its derivative.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 486 |
IV OM - II - Task 3
A triangular piece of sheet metal weighs $ 900 $ g. Prove that by cutting this sheet along a straight line passing through the center of gravity of the triangle, it is impossible to cut off a piece weighing less than $ 400 $ g.
|
We assume that the sheet is of uniform thickness everywhere; the weight of a piece of sheet is then proportional to the area of the plane figure represented by that piece of sheet. The task is to show that after cutting a triangle along a straight line passing through its center of gravity, i.e., through the point of intersection of its medians, each part of the triangle has an area of at least $ \frac{4}{9} $ of the area of the entire triangle.
Let $ P $ be the center of gravity of triangle $ ABC $ with area $ S $ (Fig. 36). When the triangle is cut along one of its medians, for example, along $ AD $, it is divided into two triangles with areas equal to $ \frac{1}{2}S $. By cutting the triangle $ ABC $ along a segment passing through point $ P $ and parallel to one of the sides of the triangle, for example, along the segment $ EF $ parallel to side $ AB $, we divide this triangle into triangle $ EFC $ and trapezoid $ ABFE $. Triangle $ EFC $ is similar to triangle $ ABC $ in the ratio $ \frac{EC}{AC} = \frac{2}{3} $. Since the ratio of the areas of similar figures equals the square of the similarity ratio, the area of triangle $ EFG $ equals $ \frac{4}{9} S $, and the area of trapezoid $ ABFE $ equals $ \frac{5}{9}S $.
It turns out that in the two cases considered, the thesis of the theorem holds. Let us draw through point $ P $ any line not passing through any of the vertices of the triangle and not parallel to any of its sides. This line will intersect two sides of the triangle, for example, side $ AC $ at point $ M $ and side $ BC $ at point $ N $. Points $ M $ and $ N $ lie on opposite sides of line $ EF $: for example, let point $ M $ lie on segment $ AE $, and point $ N $ on segment $ FD $. Our theorem will be proved when we show that
To this end, let us note that
Let $ K $ be the point symmetric to point $ A $ with respect to point $ P $. Point $ K $ lies on the extension of segment $ PD $ beyond point $ D $, since $ PK = AP = 2PD $. Triangle $ PFK $ is symmetric to triangle $ PEA $ with respect to point $ P $. Let $ L $ be the point on segment $ FK $ symmetric to point $ M $. Then
From the equality (1) and the inequality (2), it follows that
which was to be proved.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 490 |
XII OM - III - Problem 5
Four lines intersecting at six points form four triangles. Prove that the circumcircles of these triangles have a common point.
|
Four lines intersecting at $6$ points form a figure known as a complete quadrilateral (see Problems from Mathematical Olympiads, vol. I, Warsaw 1960. Problem 79). Let us denote these points by $A$, $B$, $C$, $D$, $E$, $F$ (Fig. 21) in such a way that on the given lines lie the respective triplets of points $(A, B, C)$, $(A, D, E)$, $(C, F, D)$, $(B, F, E)$, with point $B$ lying inside segment $AC$, point $D$ - inside segment $AE$, and point $F$ being a common internal point of segments $CD$ and $BE$.
On this figure, there are triangles: $ABE$, $BCF$, $ACD$, and $DEF$; let $K_1$, $K_2$, $K_3$, $K_4$ denote the circumcircles of these triangles, respectively. Circle $K_2$ passes through point $F$ lying inside circle $K_1$ and through point $C$ lying outside this circle. Therefore, circles $K_1$ and $K_2$ intersect at $2$ points; one of them is point $B$, so the second intersection point $M$ lies on the opposite side of chord $FC$. According to the inscribed angle theorem,
therefore,
Since points $D$ and $M$ lie on the same side of line $AC$, it follows from the equality $\measuredangle AMC = \measuredangle ADC$ that point $M$ lies on circle $K_3$. Circles $K_1$, $K_2$, $K_3$ thus have a common point $M$.
In the same way, we can prove that circles $K_1$, $K_3$, $K_4$ have a common point $N$. Points $M$ and $N$ are common points of circles $K_1$ and $K_3$; neither of them can coincide with the common point $A$ of these circles, since point $A$ lies outside circles $K_2$ and $K_4$, so $M$ and $N$ coincide, i.e., circles $K_1$, $K_2$, $K_3$, $K_4$ have a common point.
Note. The above proof requires supplementation. We have assumed without justification that the intersection points of the $4$ lines can always be labeled in such a way that on the given lines lie the respective triplets $(A, B, C)$, $(A, D, E)$, $(C, F, D)$, $(B, F, E)$, with points $B$, $D$, $F$, and $F$ being internal points of these $4$ triplets. We inferred this from observing the diagram, which is not a rigorous mathematical proof. We will now provide a proof without using the diagram.
Among the given $6$ points, there are $4$ triplets of collinear points, i.e., lying on the same line, and in each triplet, one of the points lies between the other two. Therefore, one of the given $6$ points is not an internal point of any triplet; let us call it $P_1$. Through $P_1$ pass $2$ of the given lines; on one of them lie $2$ more of the given points, which we can label as $P_2$ and $P_3$ in such a way that point $P_2$ lies between points $P_1$ and $P_3$; similarly, on the other line lie $2$ other points, which we will label as $P_4$ and $P_5$ such that $P_4$ lies between $P_1$ and $P_5$. The sixth point we will label as $P_6$, and there can be $2$ cases, which we will consider in turn: 1) $P_6$ is the intersection point of lines $P_3P_4$ and $P_2P_5$; in this case, $P_6$ lies between points $P_3$ and $P_4$, since line $P_2P_5$ intersects side $P_1P_3$ of triangle $P_1P_3P_4$ at point $P_2$, so it must intersect side $P_3P_4$ (Theorem: If a line has a point in common with a side of a triangle and does not pass through any vertex of the triangle, then it has a point in common with one of the other sides of the triangle, is a fundamental fact in geometry usually taken as an axiom known as Pasch's axiom. It was formulated in 1882 by Moritz Pasch (1843-1930), a German mathematician and professor at the University of Giessen.); similarly, point $P_6$ lies between points $P_2$ and $P_5$. If we label points $P_1$, $P_2$, $P_3$, $P_4$, $P_5$, $P_6$ as $A$, $B$, $C$, $D$, $E$, $F$, respectively, we obtain the desired labeling of the points. 2) $P_6$ is the intersection point of lines $P_2P_4$ and $P_3P_5$. In this case, one of two scenarios may occur: a) Point $P_2$ lies between points $P_4$ and $P_6$; in this case, point $P_3$ lies between points $P_5$ and $P_6$, since line $P_1P_3$ passing through point $P_2$ of side $P_4P_6$ of triangle $P_4P_5P_6$ must intersect side $P_5P_6$. We will then obtain the desired labeling by calling points $P_1$, $P_2$, $P_3$, $P_4$, $P_5$, $P_6$ $C$, $F$, $D$, $B$, $A$, $E$, respectively; b) point $P_4$ lies between points $P_2$ and $P_6$; in this case, point $P_5$ lies between points $P_3$ and $P_6$, since line $P_1P_5$ passing through point $P_4$ of side $P_2P_6$ of triangle $P_2P_3P_6$ must intersect some point of side $P_3P_6$. In this case, we label points $P_1$, $P_2$, $P_3$, $P_4$, $P_5$, $P_6$ as $C$, $B$, $A$, $F$, $D$, $E$, respectively, and obtain the desired labeling.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 491 |
XLIII OM - II - Problem 1
Each vertex of a certain polygon has both coordinates as integers; the length of each side of this polygon is a natural number. Prove that the perimeter of the polygon is an even number.
|
Let $ A_1,\ldots,A_n $ be the consecutive vertices of a polygon; it will be convenient for us to denote the vertex $ A_n $ also by $ A_0 $. The boundary of the polygon forms a closed broken line, hence
Let us denote the coordinates of the $ i $-th vector of the above sum by $ u_i $, $ v_i $, and its length by $ w_i $:
The numbers $ u_i $, $ v_i $, $ w_i $ are assumed to be integers. Of course, .
In the solution, we will use the simple observation that the square of any integer $ m $ is a number of the same parity as $ m $ (since $ m^2 = m + m(m- 1) $, and the product $ m(m- 1) $ is an even number). Hence, for any integers $ m_1 ,m_2,\ldots, m_n $, we have the equivalence:
(Indeed: by the previous observation, both of these sums have the same number of odd terms.)
Equality (1) is equivalent to the statement that
By taking $ m_i= u_i $ and then $ m_i = v_i $ in (2), we conclude that the sums
$ \displaystyle \sum_{i=1}^n u_i^2 $ and $ \displaystyle \sum_{i=1}^n v_i^2 $ are even numbers. Therefore, the sum
is also an even number. Finally, applying equivalence (2) to the numbers $ m_i = w_i $, we infer that the sum $ \displaystyle \sum_{i=1}^n v_i^2 $ is also an even number. And this sum — is precisely the perimeter of the polygon.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 493 |
XXIX OM - I - Problem 4
Let $ Y $ be a figure consisting of closed segments $ \overline{OA} $, $ \overline{OB} $, $ \overline{OC} $, where point O lies inside the triangle $ ABC $. Prove that in no square can one place infinitely many mutually disjoint isometric images of the figure $ Y $.
|
Let $ d $ be the smallest of the numbers $ OA $, $ OB $, $ OC $. Choose points $ P \in \overline{OA} $, $ Q \in \overline{OB} $, $ R \in \overline{OC} $ such that $ OP= OQ = OR= \frac{d}{2} $ (Fig. 6).
Then $ PQ < OP + OQ = d $ and similarly $ PR < d $ and $ QR < d $. If $ Y $ is a figure consisting of closed segments $ \overline{OA} $, $ \overline{OB} $, and $ \overline{OC} $, and is isometric to the figure $ Y $, and point $ O $ belongs to the triangle $ OPQ $, then each of the segments $ \overline{OA} $, $ \overline{OB} $, and $ \overline{OC} $ intersects the boundary of the triangle $ OPQ $, because in the triangle no segment longer than the longest side of the triangle is contained. From the conditions of the problem, it follows that each of the angles $ \measuredangle A $, $ \measuredangle B $, and $ \measuredangle C $ has a measure less than $ \pi $ and together they form a full angle. Therefore, at least one of the segments $ \overline{OA} $, $ \overline{OB} $, and $ \overline{OC} $ intersects the side $ \overline{OP} $ or $ \overline{OQ} $. The figures $ Y $ and $ Y $ are therefore not disjoint.
om29_1r_img_6.jpg
Similarly, we prove that if point $ O $ belongs to the triangle $ OPR $ or $ OQR $, then the figures $ Y $ and $ Y $ are not disjoint. If, therefore, the figures $ Y $ and $ Y $ are disjoint, then point $ O $ does not belong to the triangle $ PQR $. Let $ r $ be the length of the radius of a circle centered at point $ O $ contained in the triangle $ PQR $. From the above, it follows that if the figures $ Y $ and $ Y $ are disjoint, then the points $ O $ and $ O $ are more than $ r $ apart.
Now consider a square $ K $ with side length $ a $. Choose a natural number $ n $ greater than $ \displaystyle \frac{a}{r} \sqrt{2} $ and divide the square $ K $ by lines parallel to the sides into $ n^2 $ small squares. The length of the side of such a small square is therefore $ \displaystyle \frac{a}{n} $, and the length of its diagonal is $ \displaystyle \frac{a}{n} \sqrt{2} $. Since $ n > \frac{a}{r} \sqrt{2} $, then $ \frac{a}{n} \sqrt{2} < r $. The distance between any two points belonging to a square does not exceed the length of its diagonal. Therefore, the distance between any two points belonging to a small square is less than $ r $. From the initial part of the solution, it follows that if $ f_1 $ and $ f_2 $ are isometries of the figure $ Y $ into the square $ K $ and the figures $ f_1(Y) $ and $ f_2(Y) $ are disjoint, then the points $ f_1(0) $ and $ f_2(0) $ are more than $ r $ apart. These points do not therefore belong to the same small square. Thus, in the square $ K $, one can place at most $ n^2 $ mutually disjoint isometric images of the figure $ Y $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 495 |
XLV OM - I - Problem 5
Prove that if the polynomial $ x^3 + ax^2 + bx + c $ has three distinct real roots, then the polynomial $ x^3 + ax^2 + \frac{1}{4}(a^2 + b)x +\frac{1}{8}(ab - c) $ also has three distinct real roots.
|
Let's denote the polynomials in the problem by $P(x)$ and $Q(x)$:
Replacing the variable $x$ in the polynomial $P(x)$ with the difference $2x - a$ and transforming the obtained expression:
From the obtained identity:
it immediately follows that if a number $\xi$ is a root of the polynomial $P(x)$, then the number $-\frac{1}{2}(\xi + a)$ is a root of the polynomial $Q(x)$.
Therefore, if the numbers $x_1$, $x_2$, $x_3$ are three distinct roots of the polynomial $P(x)$, then the numbers $-\frac{1}{2}(x_1 + a)$, $-\frac{1}{2}(x_2 + a)$, $-\frac{1}{2}(x_3 + a)$ are three distinct roots of the polynomial $Q(x)$.
{\kom Note.} The idea to transform the expression $P(2x - a)$ does not require any kind of "revelation." The form of the polynomials $P(x)$ and $Q(x)$ (and the thesis of the problem \ldots) suggests that there might be constants $\alpha$, $\beta$, $\gamma$ for which the equality
is satisfied identically (with $\alpha \neq 0$, $\gamma \neq 0$). By comparing the coefficients of $x^3$, we see that if such constants exist, then $\gamma$ must be equal to $\alpha^3$. The proposed identity (2) takes the form
Substituting the expressions defining the polynomials $P(x)$ and $Q(x)$, and equating the coefficients of the polynomials obtained on the left and right sides of formula (2), we obtain a system of three equations with two unknowns ($\alpha$ and $\beta$). It turns out that, regardless of the values of the given constants (parameters) $a$, $b$, $c$, this system always has a solution $\alpha = -2$, $\beta = -a$ (so $\gamma = -8$); in general, this is its only solution, but for some special values of the constants $a$, $b$, $c$, there may be other solutions. However, this does not matter: for the found values of the numbers $\alpha$, $\beta$, $\gamma$, the identity (2) is always satisfied; substituting $x$ with $-x$ it takes the form (1).
Of course, when writing the solution to the problem, no one is required to "explain" where the idea to examine the expression $P(2x-a)$ comes from.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 496 |
IX OM - III - Task 1
Prove that the product of three consecutive natural numbers, of which the middle one is a cube of a natural number, is divisible by $ 504 $.
|
We need to prove that if $a$ is a natural number greater than $1$, then the number
is divisible by $504 = 7 \cdot 8 \cdot 9$. Since the numbers $7$, $8$, and $9$ are pairwise coprime, the task reduces to proving the divisibility of $N$ by each of these numbers.
a) The number $a$ can be represented in the form $a = 7k + r$, where $k$ is a natural number, and $r$ is one of the numbers $0$, $1$, $2$, $3$, $4$, $5$, $6$. Then $a^3 = 7^3k^3 + 3 \cdot 7^2k^2r + 3 \cdot 7kr^2 + r^3$, so when divided by $7$, the number $a^3$ gives the same remainder as $r^3$. But $r^3$ is one of the numbers $0$, $1$, $8$, $27$, $64$, $125$, $216$, so the remainder of $a^3$ when divided by $7$ is one of the numbers $0$, $1$, $6$, which means that one of the numbers $a^3$, $a^3 - 1$, $a^3 + 1$ is divisible by $7$.
b) To establish the divisibility of the number $N$ by $8$, it suffices to note that if $a$ is an even number, then $a^3$ is divisible by $8$, and if $a$ is odd, then $a^3 - 1$ and $a^3 + 1$ are two consecutive even numbers, one of which is divisible by $4$, and their product by $8$.
c) The number $a$ can be represented in the form $a = 3l + s$, where $l$ is a natural number, and $s$ is one of the numbers $0$, $1$, $2$. Then $a^3 = 3^3l^3 + 3 \cdot 3l^2s + 3 \cdot 3l \cdot s^2 + s^3$, from which we see that $a^3$ gives a remainder of $s^3$ when divided by $9$, i.e., $0$, $1$, or $8$. Therefore, one of the numbers $a^3$, $a^3 - 1$, or $a^3 + 1$ is divisible by $9$.
Note 1. It is easy to observe that in the statement of the theorem, we can speak more generally about integers instead of natural numbers.
Note 2. Using simple facts from number theory, the arguments presented above in points a) and c) can be replaced with simpler ones. According to Fermat's Little Theorem (see Seventh Mathematical Olympiad, Warsaw 1957, problem no. 2), if an integer $a$ is not divisible by $7$, then the congruence
holds, from which it follows that
Thus, if $a$ is any integer, one of the numbers $a^3$, $a^3 - 1$, $a^3 + 1$ is divisible by $7$.
When it comes to divisibility by $9$, which is not a prime number, a more general theorem than Fermat's must be applied, namely Euler's Theorem (the famous Swiss mathematician, 1707-1783). Let $m$ be a natural number, and let $\varphi(m)$ denote the number of natural numbers not greater than $m$ and coprime with $m$. For example, $\varphi(1) = \varphi(2) = 1$, $\varphi(3) = \varphi(4) = 2$, $\varphi(5) = 4$, $\varphi(6) = 2$, $\varphi(7) = 6$, $\varphi(8) = 4$, $\varphi(9) = 6$, etc.
Euler's Theorem states: If the integers $a$ and $m > 1$ are coprime, then
For example, if $m = 9$, we get the theorem: if the integer $a$ is not divisible by $3$, then
from which - as above - it follows that for any integer $a$, one of the numbers $a^3$, $a^3 - 1$, $a^3 + 1$ is divisible by $9$.
A proof of Euler's Theorem can be conducted as follows. Let $r_1, r_2, \ldots, r_\varphi$, where $\varphi = \varphi(m)$, be the increasing sequence of all natural numbers not greater than $m$ and coprime with $m$. Then each of the numbers $ar_1, ar_2, \ldots, ar_\varphi$ is also coprime with $m$; let $\rho_1, \rho_2, \ldots, \rho_\varphi$ denote the remainders of these numbers when divided by $m$, so
Each of the numbers $\rho_i$ is coprime with $m$, because according to (1) $\alpha r_i - \varphi_i$ is divisible by $m$, and $\alpha r_i$ is coprime with $m$. Moreover, the numbers $\rho_i$ are all different, because from the equality $\rho_i = \rho_k$ it would follow that $\alpha r_i \equiv \alpha r_k \pmod m$, and thus $r_i \equiv r_k \pmod m$, i.e., $r_i \equiv r_k$, contrary to the definition of the numbers $r_i$. Therefore, the numbers $\rho_1, \rho_2, \ldots, \rho_\varphi$ differ from the numbers $r_1, r_2, \ldots, r_\varphi$ at most in order, and the equality
holds.
Multiplying the congruences (1) side by side, we get
Since each of the numbers $r_i$ is coprime with $m$, the same applies to the product of these numbers, so from (2) and (3) it follows that
which was to be proved.
Note that if $m$ is a prime number, then $\varphi(m) = m - 1$; in this case, Euler's Theorem gives Fermat's Little Theorem.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 500 |
XXXIV OM - III - Problem 6
Prove that if all dihedral angles of a tetrahedron are acute, then all its faces are acute triangles.
|
We assign a unit vector perpendicular to each face and directed outward from the tetrahedron. Consider two faces forming an angle $ \alpha $. The vectors assigned to these faces form an angle $ \pi - \alpha $ (Fig. 12). Therefore, for all dihedral angles of the tetrahedron to be acute, it is necessary and sufficient that the angles between each pair of the vectors $ \overrightarrow{a} $, $ \overrightarrow{b} $, $ \overrightarrow{c} $, $ \overrightarrow{d} $ assigned to the respective faces be obtuse, and thus that
om34_3r_img_12.jpg
om34_3r_img_13.jpg
Assume that the above inequalities are satisfied. We will prove that every planar angle of the tetrahedron is acute. Consider the angle $ ABC $. The vector $ \overrightarrow{a} $ assigned to the face $ BCD $ is perpendicular to the line $ BC $. Therefore, the projection $ \overrightarrow{a}_{ABC} $ of the vector $ \overrightarrow{a} $ onto the plane $ ABC $ is also perpendicular to the line $ BC $. Similarly, the vector $ \overrightarrow{c} $ assigned to the face $ ABD $ is perpendicular to the line $ AB $, so its projection $ \overrightarrow{c}_{ABC} $ onto the plane $ ABC $ is perpendicular to $ AB $. It follows that
Therefore, the angle $ ABC $ is acute if and only if $ \overrightarrow{a}_{ABC} \cdot \overrightarrow{c}_{ABC} < 0 $.
om34_3r_img_14.jpg
The vectors $ \overrightarrow{a} $, $ \overrightarrow{b} $, $ \overrightarrow{c} $, $ \overrightarrow{d} $ have length $ 1 $, so (Fig. 14)
Therefore,
since each of the scalar products appearing here is negative. We have thus shown that the angle $ ABC $ is acute. The proof for the remaining planar angles is analogous.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 501 |
LIX OM - III - Task 1
In the fields of an $ n \times n $ table, the numbers $ 1, 2, \dots , n^2 $ are written, with the numbers $ 1, 2, \dots , n $ being in the first row (from left to right), the numbers $ n +1, n +2, \dots ,2n $ in the second row, and so on.
$n$ fields of the table have been selected, such that no two lie in the same row or column. Let $ a_i $ be the number found in the selected field that lies in the row numbered $ i $. Prove that
|
We will first prove that
Let $ b_i $ denote the column number in which the number $ a_i $ is located. Then,
From the conditions of the problem, it follows that the sequence $ (b_1,b_2,\dots ,b_n) $ is a permutation of the sequence $ (1, 2,\dots ,n) $. Therefore,
and in view of formula (2), the proof of relation (1) reduces to a calculation:
Using now equality (1) and the inequality between the arithmetic mean and the harmonic mean, we obtain
which is the thesis of the problem.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 503 |
XVIII OM - III - Problem 3
In a room, there are 100 people, each of whom knows at least 67 others. Prove that there is a quartet of people in the room where every two people know each other. We assume that if person $A$ knows person $B$, then person $B$ also knows person $A$.
|
In the set $M$ of people in the room, there are at most $32$ people who do not know a certain person $A$ (assuming that person $A$ knows themselves).
If $A$, $B$, and $C$ are any three people in the set $M$, then there is a person $D$ in the set $M$, different from $A$, $B$, and $C$, who knows these three people. Indeed, if all the people who do not know $A$, all the people who do not know $B$, and all the people who do not know $C$ leave the room, then at most $3 \cdot 32 = 96$ people will leave. Each of the remaining people then knows $A$, $B$, and $C$, and since there are at least $100-96 = 4$ of them, at least one of them is different from $A$, $B$, and $C$.
We can thus easily find a quartet of people in the set $M$ who know each other. Specifically, $A$ can be any person in this set, $B$ a person who knows $A$ and is different from $A$, $C$ a person who knows $A$ and $B$ and is different from $A$ and $B$, and finally $D$ a person who knows $A$, $B$, and $C$ and is different from $A$, $B$, and $C$.
Note. In a similar way, we can prove a more general theorem: If there are $n$ people in the room, each of whom knows at least $\left[ \frac{2n}{3} \right] +1$ of the remaining people, then there is a quartet of people in the room, in which every two people know each other.
To prove this, we first note that the number of people who do not know a certain person is at most $n - \left[ \frac{2n}{3} \right] -2$. The number of people who do not know at least one of the three people $A$, $B$, and $C$ is therefore at most $m = 3n - 3 \left[ \frac{2n}{3} \right] - 6$. Since $\left[ \frac{2n}{3} \right] > \frac{2n}{3} -1$, it follows that $3 \left[ \frac{2n}{3} \right] > 2n - 3$, so $3 \left[ \frac{2n}{3} \right] \geq 2n-2$, and thus $m \leq 3n - (2n-2) - 6 = n-4$. The number of people who know $A$, $B$, and $C$ is therefore at least $n-m \geq 4$, so at least one of these people is different from $A$, $B$, and $C$. The rest of the proof remains the same as before.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 505 |
XLII OM - I - Problem 1
On the shore of a lake in the shape of a circle, there are four piers $K$, $L$, $P$, $Q$. A kayak departs from pier $K$ heading towards pier $Q$, and a boat departs from pier $L$ heading towards pier $P$. It is known that if, maintaining their speeds, the kayak headed towards pier $P$ and the boat towards pier $Q$, a collision would occur. Prove that the kayak and the boat will reach their destinations at the same time.
|
om42_1r_img_1.jpg
Suppose the kayak is moving at a speed of $ v_k $, and the boat is moving at a speed of $ v_l $. Let $ X $ be the point where a "collision" would occur. The lengths of the segments into which point $ X $ divides the chords $ KP $ and $ LQ $ are related by $ |KX| \cdot |PX| = |LX| \cdot |QX| $ (Figure 1).
From the conditions of the problem, it follows that the ratio of speeds $ v_k \colon v_l = |KX| \colon |LX| $. From these two equalities, we obtain the proportion $ v_k \colon v_l = |QX| \colon |PX| $.
The vertical angles $ KXQ $ and $ LXP $ are equal. Therefore, triangles $ KXQ $ and $ LXP $ are similar in the ratio $ v_k \colon v_l $, and thus $ |KQ| \colon |LP| = v_k \colon v_l $, that is,
The left side of the last proportion expresses the time used by the kayak to travel from $ K $ to $ Q $; the right side - the time for the boat to travel from $ L $ to $ P $. Hence the thesis.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 506 |
LIX OM - III - Task 6
Let $ S $ be the set of all positive integers that can be represented in the form $ a^2 + 5b^2 $
for some relatively prime integers $ a $ and $ b $. Furthermore, let $ p $ be a prime number that gives a remainder
of 3 when divided by 4. Prove that if some positive multiple of the number $ p $ belongs to the set $ S $, then the number
$ 2p $ also belongs to the set $ S $.
|
Given the conditions of the problem, the number $a^2 + 5b^2$ is divisible by $p$ for some relatively prime integers $a$ and $b$.
The number $b$ is not divisible by $p$; otherwise, from the divisibility $p \mid b$ and $p \mid a^2 + 5b^2$, we would obtain $p \mid a$, contradicting the relative primality of $a$ and $b$.
Therefore, none of the $p - 1$ numbers: $b, 2b, \dots, (p - 1)b$ is divisible by $p$. Moreover, these numbers give different remainders when divided by $p$. Indeed, if we have $p \mid ib - jb = (i - j)b$ for some indices $1 \leqslant i, j \leqslant p - 1$, then the prime number $p$ is a divisor of the difference $i - j$, implying that $i = j$. Consequently, one of the considered numbers gives a remainder of 1 when divided by $p$. Let us assume that this number is $kb$ (for some value $k \in \{1, 2, \dots, p - 1\}$). Then we have
Let $z$ denote the largest integer not exceeding $p$. Consider the set of remainders when the numbers $0, m, 2m, \dots, zm$ are divided by $p$. Arrange these $z + 1$ remainders in non-decreasing order to obtain the sequence $r_1 \leqslant r_2 \leqslant \dots \leqslant r_{z+1}$. Then each of the $z + 1$ numbers:
is non-negative, and their sum is $p$. It follows that at least one of these numbers is no greater than
Thus, among the numbers $r_2 - r_1, r_3 - r_2, \dots, r_{z+1} - r_z, p - (r_{z+1} - r_1)$, we can find a number not exceeding $z$. This means that for some two different numbers, say $cm$ and $dm$ ($0 \leqslant c, d \leqslant z$), their remainders when divided by $p$ differ by no more than $z$, or differ by at least $p - z$. Let $y = |c - d|$; then one of the numbers $ym, -ym$ gives a remainder $x \leqslant z$ when divided by $p$. Since $p \mid m^2 + 5$, the numbers
give the same remainder when divided by $p$. One of the factors $x - my, x + my$ is divisible by $p$. Therefore, $p \mid x^2 + 5y^2$. Moreover, $0 \leqslant x \leqslant z$ and $0 \leqslant y \leqslant z$, which, combined with the inequality $z < p$, gives
Hence, we conclude that $x^2 + 5y^2$ is one of the numbers $p, 2p, 3p, 4p$, or $5p$.
The equality $x^2 + 5y^2 = p$ would mean that the number $x^2 + 5y^2$ gives a remainder of 3 when divided by 4. This is not possible, since the numbers $x$ and $5y^2$ give remainders of 0 or 1 when divided by 4. Therefore, $x^2 + 5y^2 = p$. This also proves that the equalities $x^2 + 5y^2 = 4p$ and $x^2 + 5y^2 = 5p$ cannot hold. The first of these equalities would mean that the numbers $x$ and $y$ are even (in no other case is $x^2 + 5y^2$ divisible by 4) and
\[
\left(\frac{1}{2}x\right)^2 + 5\left(\frac{1}{2}y\right)^2 = p;
\]
the second equality leads to the conclusion that $5 \mid x$ and
\[
y^2 + 5\left(\frac{1}{2}x\right)^2 = p.
\]
In both cases, we arrive at a contradiction with the previously proven fact that a number of the form $e^2 + 5f^2$ is different from $p$ for any integer values of $e$ and $f$.
Thus, we have established that $x^2 + 5y^2 = 2p$ or $x^2 + 5y^2 = 3p$.
Suppose that $x^2 + 5y^2 = 3p$. Then the numbers $x$ and $y$ are not divisible by 3 (otherwise, we would have $3 \mid x$ and $3 \mid y$, but then $3^2 \mid x^2 + 5y^2 = 3p$; this can only happen if $p = 3$, but then we obtain the following contradiction: $9 = 3p = x^2 + 5y^2 \geqslant 3^2 + 5 \cdot 3^2 = 54$). By changing the signs of the numbers $x$ and $y$ if necessary, we can assume that $x$ and $y$ are integers giving a remainder of 1 when divided by 3. Moreover,
The numbers $x + 5y$ and $x - y$ are divisible by 3; if $x + 5y = 3g$ and $x - y = 3h$, then from the above equality, we conclude that $2p = g^2 + 5h^2$.
Ultimately, we have proven that the number $2p$ can be expressed in the form $s^2 + 5t^2$ for some integers $s$ and $t$. It remains to note that in this case, the numbers $s$ and $t$ are relatively prime; if they had a common prime divisor $q$, we would obtain $q^2 \mid s^2 + 5t^2 = 2p$, leading to a contradiction, since $p$ is an odd prime. Thus, the proof of the relation $2p \in S$ is complete.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 507 |
XXVII OM - II - Problem 3
We consider a hemispherical cap that does not contain any great circle. The distance between points $A$ and $B$ on such a cap is defined as the length of the arc of the great circle of the sphere with endpoints at points $A$ and $B$, which is contained in the cap. Prove that there is no isometry mapping this cap onto a subset of the plane.
Note. A hemispherical cap is any of the two parts into which a plane intersecting a sphere divides its surface.
|
Let a given spherical cap correspond to the central angle $ \alpha $. From the assumption, we have $ 0 < \alpha < \frac{\pi}{2} $. For any angle $ \beta $ satisfying $ 0 < \beta < \alpha $, consider a regular pyramid $ OABCD $, whose base is a square $ ABCD $ inscribed in the given spherical cap, and whose vertex $ O $ is the center of the sphere containing this cap, such that $ \measuredangle AOC = \beta $ (Fig. 13). Let $ \measuredangle AOD = \gamma $ and let points $ P $ and $ Q $ be the midpoints of segments $ \overline{AD} $ and $ \overline{AC} $, respectively.
We have the following relationships
Therefore
If $ R $ is the radius of the sphere containing the given spherical cap, then the distances between points $ A $ and $ C $ and points $ B $ and $ D $ on the cap are equal to $ R\beta $, and the distances between points $ A $ and $ B $, $ B $ and $ C $, $ C $ and $ D $, and $ D $ and $ A $ on the cap are equal to $ R\gamma $.
If there exists an isometry $ \varphi $ of the given spherical cap onto a subset of the plane, then $ \varphi(A) \varphi (B)\varphi (C) \varphi (D) $ is a quadrilateral on the plane such that $ \varphi(A)\varphi(B) = \varphi(B)\varphi(C) = \varphi(C)\varphi(D) = \varphi(D)\varphi(A) = R \gamma $. This is therefore a rhombus. Its diagonals have equal lengths: $ \varphi(A)\varphi(C) = \varphi(B) \varphi(D) = R\beta $. The considered quadrilateral is thus a square. Therefore, $ \varphi(A)\varphi(C) = \sqrt{2}\varphi(A)\varphi(B) $, i.e.,
From (1) and (2), it follows that $ \sin \frac{\beta}{2} = \sqrt{2} \sin \frac{\beta}{2\sqrt{2}} $ for every angle $ \beta $ satisfying $ 0 < \beta < \alpha $. Substituting in particular the angle $ \frac{\beta}{\sqrt{2}} $ for $ \beta $ here, we get $ \sin \frac{\beta}{2\sqrt{2}}= \sqrt{2} \sin \frac{\beta}{4} $. From the last two equalities, it follows that $ \sin \frac{\beta}{2} =2 \sin \frac{\beta}{4} $. On the other hand, we have $ \sin \frac{\beta}{2} = 2 \sin \frac{\beta}{4} \cos \frac{\beta}{4} $. Comparing the obtained results, we get that $ \cos \frac{\beta}{4} = 1 $. This is impossible, since $ 0 < \frac{\beta}{4} < \frac{\alpha}{4} < \frac{\pi}{8} $.
The obtained contradiction proves that there is no isometry $ \varphi $ of the given spherical cap onto a subset of the plane.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 508 |
XXVIII - I - Problem 8
Prove that the set $ \{1, 2, \ldots, 2^{s+1}\} $ can be partitioned into two $ 2s $-element sets $ \{x_1, x_2, \ldots, x_{2^s}\} $, $ \{y_1, y_2, \ldots, y_{2^s}\} $, such that for every natural number $ j \leq s $ the following equality holds
|
We will apply induction with respect to $s$. For $s = 1$, we have the set $\{1, 2, 3, 4\}$. By splitting it into subsets $\{1, 4\}$ and $\{2, 3\}$, we will have the conditions of the problem satisfied, since $1 + 4 = 2 + 3$.
Next, assume that for some natural number $s$, there exists such a partition of the set $\{1, 2, \ldots, 2^{s+1}\}$ into subsets $\{x_1, x_2, \ldots, x_{2^s}\}$ and $\{y_1, y_2, \ldots, y_{2^s}\}$, that
\[
\sum_{i=1}^{2^s} x_i^r = \sum_{i=1}^{2^s} y_i^r
\]
for $r = 1, 2, \ldots, s + 1$.
We will prove that the partition of the set $\{1, 2, \ldots, 2^{s+2}\}$ into subsets
\[
\{x_1, x_2, \ldots, x_{2^s}, x_1 + 2^{s+1}, x_2 + 2^{s+1}, \ldots, x_{2^s} + 2^{s+1}\}
\]
and
\[
\{y_1, y_2, \ldots, y_{2^s}, y_1 + 2^{s+1}, y_2 + 2^{s+1}, \ldots, y_{2^s} + 2^{s+1}\}
\]
satisfies the conditions of the problem.
For $r = 1, 2, \ldots, s + 1$, by the binomial formula, the sum of the $r$-th powers of the elements of the first subset is equal to
\[
\sum_{i=1}^{2^s} (x_i + 2^{s+1})^r = \sum_{i=1}^{2^s} \sum_{k=0}^r \binom{r}{k} x_i^k (2^{s+1})^{r-k}
\]
Similarly, we calculate the sum of the $r$-th powers of the elements of the second subset:
\[
\sum_{i=1}^{2^s} (y_i + 2^{s+1})^r = \sum_{i=1}^{2^s} \sum_{k=0}^r \binom{r}{k} y_i^k (2^{s+1})^{r-k}
\]
Comparing the obtained results, we see that they differ only in that one formula contains the sum
\[
\sum_{i=1}^{2^s} x_i^k
\]
and the other contains the sum
\[
\sum_{i=1}^{2^s} y_i^k
\]
From formula (1), it follows that these sums are equal.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 510 |
VI OM - I - Problem 10
Prove that the number $ 53^{53} - 33^{33} $ is divisible by $ 10 $.
|
It is enough to prove that the last digits of the numbers $53^{53}$ and $33^{33}$ are the same. The last digit of the product $(10k + a)(10m + b)(10n + c) \ldots$, where $k, m, n, \ldots a, b, c, \ldots$ are natural numbers, is equal to the last digit of the product $abc\ldots$. The number $53^{53}$ therefore has the same last digit as the number $3^{53}$. But $3^{53} = 3^{52} \cdot 3 = (81)^{13} \cdot 3$, so the number $3^{53}$ has the same last digit as the number $1^{13} \cdot 3$, i.e., the digit $3$. Similarly, the number $33^{33}$ has the same last digit as the number $3^{33} = 3^{32} \cdot 3 = (3^4)^8 \cdot 3 = (81)^8 \cdot 3$, i.e., the digit $3$, q.e.d.
The above reasoning can be given a concise and clear form using the concepts and simplest properties of congruences (see Problems from Mathematical Olympiads, problem No. 8).
Since $53 \equiv 3 \pmod{10}$, $33 \equiv 3 \pmod{10}$, $3^4 \equiv 1 \pmod{10}$, therefore
\[53^{53} \equiv 3^{53} \equiv (3^4)^{13} \cdot 3 \equiv 3 \pmod{10},\]
\[33^{33} \equiv 3^{33} \equiv (3^4)^8 \cdot 3 \equiv 3 \pmod{10},\]
therefore
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 511 |
XIX OM - I - Problem 6
Prove that if the polynomial $ F(x) $ with integer coefficients has a root $ \frac{p}{q} $, where $ p $ and $ q $ are relatively prime integers, then $ p - q $ is a divisor of the number $ F(1) $, and $ p + q $ is a divisor of the number $ F(-1) $.
|
Let us substitute $ x $ with the variable $ y = qx $. Equation (1) then takes the form
Consider the polynomial
whose coefficients are integers. It has an integer root $ y=p $, hence
where $ Q(y) $ is a polynomial with integer coefficients*).
Substituting $ y = q $ into this equality, we get
The numbers $ q^n $ and $ q - p $ are coprime; if some prime number were a common divisor of $ q^n $ and $ q - p $, it would also be a common divisor of $ q $ and $ q - p $, and thus also a common divisor of $ q $ and $ q-(q-p)=p $, contradicting the assumption.
Therefore, the number $ q - p $ is a divisor of the number $ F(1) $, Q.E.D.
Note. A shorter solution to the problem can be obtained by relying on the following theorem:
If a polynomial $ F(x) $ with integer coefficients has a rational root $ \frac{q}{p} $ ($ p $, $ q $ - integers that are coprime), then the quotient of the polynomial $ F(x) $ divided by $ x - \frac{p}{q} $ is a polynomial with integer coefficients.
We will prove this theorem. Let
where the numbers $ a_0, a_1, \ldots,a_n $ are integers. We need to prove that the numbers $ c_0, c_1, \ldots, c_{n-1} $ are integers.
The coefficients of $ x^n $ on both sides of the equality (*) are equal, so $ c_0 = a_0 $.
Notice that
which implies that $ a_0 $, and thus $ c_0 $, is divisible by $ q $. We will prove the theorem by induction.
When $ n = 1 $, the thesis of the theorem is obvious; the quotient of $ F(x) $ divided by $ x - \frac{p}{q} $ is then equal to $ c_0 $.
Suppose the thesis of the theorem is true for some $ n $ and let
Form the polynomial of degree $ n $
Since the number $ c_0 = a_0 $ is divisible by $ q $, it is a polynomial with integer coefficients. According to the induction hypothesis, the numbers $ c_1, c_2, \ldots, c_n $ are integers, i.e., the thesis of the theorem is true for polynomials of degree $ n + 1 $.
The theorem is proved.
The solution to the previous problem is now immediate. If
and the polynomial $ F(x) $ has integer coefficients, then the polynomial $ Q(x) $ also has integer coefficients and
thus
which implies that the number $ p - q $ is a divisor of the number $ F(1) $.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 512 |
XLVII OM - I - Problem 5
In the plane, a triangle $ ABC $ is given, where $ |\measuredangle CAB| = \alpha > 90^{\circ} $, and a segment $ PQ $, whose midpoint is point $ A $. Prove that
|
Let $ M $ be the midpoint of side $ BC $ of triangle $ ABC $, and $ O $ - the center of the circle $ \omega $ circumscribed around this triangle. Angle $ CAB $ is obtuse by assumption. It follows that points $ A $ and $ O $ lie on opposite sides of line $ BC $; in particular, point $ O $ does not coincide with $ M $.
Let $ ON $ be the radius of circle $ \omega $ passing through point $ M $ (see figure 4). The circle with center $ M $ and radius $ MN $ is tangent to circle $ \omega $ at point $ N $, and point $ A $ lies outside circle $ \omega $ (or on it - when it coincides with $ N $). Therefore, $ |MA| \geq |MN| $.
om47_1r_img_4.jpg
Points $ B $ and $ C $ are symmetric with respect to line $ NO $, which, therefore, bisects angle $ BNC $. And since $ | \measuredangle BNC | = |\measuredangle BAC| = \alpha $ (angles inscribed in the same arc $ BC $), we get the equality
On the extension of side $ CA $, we lay off a segment $ AD $ equal in length to it. Triangle $ BCD $ is formed; points $ A $ and $ M $ are the midpoints of its sides $ CD $ and $ CB $. Therefore, triangles $ BCD $ and $ MCA $ are similar. We obtain the relationship
Point $ A $ is the common midpoint of segments $ PQ $ and $ CD $. Therefore, quadrilateral $ CPDQ $ is a parallelogram (which can degenerate into a segment; note that the reasoning in this part is completely independent of the positions of points $ P $ and $ Q $ relative to the previously considered segments and circles). Thus, $ |CQ|=|PD| $, from which we get
We multiply the inequalities (1) and (2) side by side and obtain the inequality
which was to be proven.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 513 |
XXIII OM - III - Problem 4
On a straight line without common points with the sphere $K$, points $A$ and $B$ are given. The orthogonal projection $P$ of the center of the sphere $K$ onto the line $AB$ lies between points $A$ and $B$, and $AP$ and $BP$ are greater than the radius of the sphere. We consider the set $Z$ of triangles $ABC$ whose sides $\overline{AC}$ and $\overline{BC}$ are tangent to the sphere $K$. Prove that $T$ is a triangle with the largest perimeter among the triangles in the set $Z$ if and only if $T$ is a triangle with the largest area among the triangles in the set $Z$.
|
We will first prove the
Lemma. If $ S $ is a vertex, $ O $ is the center of the base of a right circular cone, and point $ T $ lies in the plane of the base, then the line $ ST $ forms the largest angle with such a generatrix $ SR $ of the cone that $ O \in \overline{TR} $ (Fig. 16).
Proof. For any point $ R $ on the circumference of the base of the cone, by the Law of Cosines, we have
Since the lengths $ SR $ and $ TS $ are constant (do not depend on the position of point $ R $), the angle $ \measuredangle TSR $ will be the largest when its cosine is the smallest, i.e., when $ TR $ is as large as possible. This occurs when $ O \in \overline{TR} $.
We now proceed to solve the problem. Let $ ABC_O $ be the triangle in the family $ Z $ that contains the center $ O $ of the sphere $ K $ (this triangle certainly belongs to the family $ Z $).
Let $ \alpha_0 = \measuredangle BAC_0 $, $ \beta_0 = \measuredangle ABC_0 $, and for any triangle $ ABC $ in the family $ Z $, let $ \alpha = \measuredangle BAC $, $ \beta = \measuredangle ABC $. Since the line $ AC $ is tangent to the sphere $ K $, it is a generatrix of a right circular cone with axis $ AO $. By the lemma, we have $ \alpha_0 \geq \alpha $ and similarly $ \beta_0 \geq \beta $. Drawing the triangles $ ABC_0 $ and $ ABC $ in the same plane (Fig. 17), we observe that $ \triangle ABC \subset \triangle ABC_0 $ and the triangles share a common base. Therefore, the area and perimeter of triangle $ ABC_0 $ are not less than the area and perimeter of triangle $ ABC $, respectively.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 515 |
LIV OM - II - Task 4
Prove that for every prime number $ p > 3 $ there exist integers $ x $, $ y $, $ k $ satisfying the conditions: $ 0 < 2k < p $ and
|
Let $ A $ be the set of numbers of the form $ x^2 $, where $ 0 \leq x < p/2 $, and $ B $ the set of numbers of the form $ 3 - y^2 $, where $ 0 \leq y < p/2 $; the numbers $ x $, $ y $ are integers. We will show that the numbers in the set $ A $ give different remainders when divided by $ p $.
Suppose that the numbers $ x_1^2 $, $ x_2^2 $ from the set $ A $ give the same remainder when divided by $ p $. Then
from which we have the divisibility
Since $ 0 \leq x_1 $, $ x_2 < p/2 $, we have $ -p/2 < x_1 - x_2 < p/2 $ and $ 0 \leq x_1 + x_2 < p $. Therefore, and from the above divisibility, we conclude that $ x_1 = x_2 $.
Thus, the numbers in the set $ A $ give different remainders when divided by $ p $. Similarly, we prove that the numbers in the set $ B $ give different remainders when divided by $ p $.
Since the sets $ A $ and $ B $ together have $ p + 1 $ elements, there exist two numbers - one from the set $ A $, the other from the set $ B $ - that give the same remainder when divided by $ p $. Let $ x^2 $ and $ 3 - y^2 $ be these numbers. Then the number $ x^2 + y^2 - 3 $ is divisible by $ p $, so we have
for some integer $ k $. Since $ p > 3 $, $ k $ is a positive number and
The conditions given in the problem statement are thus satisfied.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 517 |
XXI OM - III - Task 4
In the plane, there are $ n $ rectangles with sides respectively parallel to two given perpendicular lines. Prove that if any two of these rectangles have at least one point in common, then there exists a point belonging to all the rectangles.
|
Let us choose a simple set of axes for the coordinate system (Fig. 14). Then the vertices of the $i$-th rectangle $P_i$ will have coordinates $(x_i, y_i)$, $(x_i, y)$, $(x, y_i)$, $(x, y)$, where $x_i < x$, $y_i < y$. The point $(x, y)$ will therefore belong to the rectangle $P_i$ if and only if $x_i \leq x \leq x$ and $y_i \leq y \leq y$.
By assumption, any two rectangles $P_i$ and $P_j$ have a common point. Therefore, there exist numbers $x$, $y$ such that
and
Thus,
Let $a$ be the largest of the numbers $x_i$, and $b$ be the largest of the numbers $y_i$. Then, by definition,
The inequalities (2) and (3) prove that the point $(a, b)$ belongs to each of the rectangles $P_j$.
Note 1. In the given solution, we did not significantly use the fact that the number of given rectangles is finite. If a certain infinite family of rectangles with similar properties were given, the solution would proceed similarly, except that the numbers $a$ and $b$ would be defined as the least upper bounds of the sets of numbers $x_j$ and $y_j$, respectively.
Note 2. A similar theorem is true if, instead of rectangles, we consider any convex figures. Specifically, the following holds:
Helly's Theorem. If on a plane there is a non-empty family of convex sets such that any three sets of this family have a common point, then there exists a point belonging to all sets of this family.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 520 |
XXIV OM - I - Problem 11
Prove that if the center of the sphere circumscribed around a tetrahedron coincides with the center of the sphere inscribed in this tetrahedron, then the faces of this tetrahedron are congruent.
|
Let point $O$ be the center of the inscribed and circumscribed spheres of the tetrahedron $ABCD$. Denote by $r$ and $R$ the radii of these spheres, respectively. If $O$ is the point of tangency of the inscribed sphere with one of the faces, and $P$ is one of the vertices of this face, then applying the Pythagorean theorem to triangle $OO'$ (where $O'$ is the point of tangency), we get $O$. Therefore, the distance from point $O$ to each vertex of this face is the same, i.e., $O$ is the center of the circle circumscribed around this face. It also follows from this that the radii of the circles circumscribed around each face of the tetrahedron have the same length $\sqrt{R^2 - r^2}$.
Point $O$ lies inside the face of the tetrahedron, as it is the point of tangency of the inscribed sphere in the tetrahedron. Since $O$ is also the center of the circle circumscribed around this face, it follows that all angles of this face are acute. These are inscribed angles subtended by arcs smaller than a semicircle (Fig. 12).
Applying the Law of Sines to triangles $ABC$ and $CBD$ (Fig. 13), we get $\sin \measuredangle BAC = \frac{BC}{2 \sqrt{R^2 - r^2}}$ and $\sin \measuredangle BDC = \frac{BC}{2 \sqrt{R^2 - r^2}}$. Therefore, $\sin \measuredangle BAC = \sin \measuredangle BDC$, and since both these angles are acute, $\measuredangle BAC = \measuredangle BDC$. Similarly, we prove that any two angles of the faces of the tetrahedron opposite the same edge are equal, i.e., $\measuredangle ABC = \measuredangle ADC$, $\measuredangle ACB = \measuredangle ADB$, $\measuredangle ABD = \measuredangle ACD$, $\measuredangle BAD = \measuredangle BCD$, $\measuredangle CAD = \measuredangle CBD$. Denoting the common measures of the angles in the last six equalities by $\alpha, \beta, \gamma, \delta, \varepsilon, \mu$ respectively, we have
since the sum of the angles of any face is $\pi$.
Adding the equations (1) and (2) side by side, we get
Subtracting the equations (3) side by side, we obtain $\beta = \varepsilon$. Since $\beta = \measuredangle ABC$ and $\varepsilon = \measuredangle BCD$, it follows that triangles $ABC$ and $DCB$ are congruent (Fig. 13).
Similarly, we prove that any two faces of the tetrahedron $ABCD$ are congruent.
Note. If the center of the circle circumscribed around a triangle is the center of the circle inscribed in this triangle, then the triangle is equilateral. An analogous theorem in space does not hold: There are tetrahedra that are not regular, for which the center of the inscribed sphere is the center of the circumscribed sphere.
Consider, for example, the tetrahedron $ABCD$ where $A = (1, a, 0)$, $B = (-1, a, 0)$, $C = (0, -a, 1)$, $D = (0, -a, -1)$, and $a$ is a positive number different from $\frac{\sqrt{2}}{2}$. Then $AB = 2$ and $AC = \sqrt{2 + 4a^2} \ne 2$. Therefore, the tetrahedron $ABCD$ is not regular. Each of the following isometries $f_1(x, y, z) = (x, y, -z)$, $f_2(x, y, z) = (-x, y, z)$, $f_3(x, y, z) = (z, -y, x)$ maps the set $\{A, B, C, D\}$ onto itself. Therefore, under each of these isometries, the center $P = (r, s, t)$ of the sphere circumscribed around the tetrahedron $ABCD$ maps onto itself. From $f_1(P) = P$ it follows that $t = 0$, from $f_2(P) = P$ that $r = 0$, and from $f_3(P) = P$ that $s = 0$. Therefore, $P = (0, 0, 0)$. Similarly, we conclude that $P$ is the center of the sphere inscribed in the tetrahedron $ABCD$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 523 |
XIV OM - II - Task 6
From point $ S $ in space, $ 3 $ rays emerge: $ SA $, $ SB $, and $ SC $, none of which is perpendicular to the other two. Through each of these rays, a plane is drawn perpendicular to the plane containing the other two rays. Prove that the planes drawn intersect along a single line $ d $.
|
In the case where the half-lines lie in the same plane, the theorem is obvious; the line $d$ is then perpendicular to the given plane. Let us assume, therefore, that the half-lines $SA$, $SB$, $SC$ form a trihedral angle. According to the assumption, at most one of the dihedral angles of this trihedron is a right angle; let us assume, for example, that $\measuredangle ASB \ne 90^\circ$ and $\measuredangle ASC \ne 90^\circ$.
Choose an arbitrary point $A_1$ on the edge $SA$ different from $S$ and draw through $A_1$ a plane $\pi$ perpendicular to the line $SA_1$; it intersects the lines $SB$ and $SC$ at some points $B_1$ and $C_1$ (Fig. 25).
Let $\alpha$, $\beta$, $\gamma$ denote the planes drawn through the edges $SA$, $SB$, $SC$ of the trihedron $SABC$ perpendicular to the faces $BSC$, $CSA$, $ASB$, and let $m$, $n$, $p$ be the lines of intersection of the planes $\alpha$, $\beta$, $\gamma$ with the plane $\pi$.
Since $\alpha \bot \pi$ and $\alpha \bot \textrm{plane } BSC$, then $\alpha \bot B_1C_1$, so the line $m$ intersects the line $B_1C_1$ at some point $M$ and is perpendicular to it, $A_1M \bot B_1C_1$. Since $\beta \bot \textrm{plane } ASC$ and $\pi \bot \textrm{plane } ASC$, then $\pi \bot \textrm{plane } ASC$, so $\pi \bot A_1C_1$, and since $n$ and $A_1C_1$ lie in the plane $\pi$, they intersect at some point $N$ and $B_1N \bot A_1C_1$. Similarly, we conclude that the line $p$ intersects the line $A_1B_1$ at some point $P$ and is perpendicular to it, i.e., $C_1P \bot A_1B_1$.
The lines $A_1M$, $B_1N$, and $C_1P$ thus contain the altitudes of the triangle $A_1B_1C_1$, so they intersect at a single point $H$. We have proved that the planes $\alpha$, $\beta$, $\gamma$ have a common line $d = SH$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 525 |
XXII OM - II - Problem 4
In the plane, a finite set of points $ Z $ is given with the property that no two distances between points in $ Z $ are equal. Points $ A, B $ belonging to $ Z $ are connected if and only if $ A $ is the nearest point to $ B $ or $ B $ is the nearest point to $ A $. Prove that no point in the set $ Z $ will be connected to more than five other points.
|
Suppose that point $ A \in Z $ is connected to some points $ B $ and $ C $ of set $ Z $ and let, for example, $ AB < AC $. Then point $ C $ is not the closest to $ A $, and from the conditions of the problem it follows that point $ A $ is the closest to $ C $. Therefore, $ AC < BC $. Hence, in triangle $ ABC $, side $ \overline{BC} $ is the longest, and thus angle $ \measuredangle BAC $ is the largest. It follows that $ \measuredangle BAC > \frac{\pi}{3} $.
We have thus proved that if point $ A \in Z $ is connected to some points $ B $ and $ C $ of set $ Z $, then $ \measuredangle BAC > \frac{\pi}{3} $.
If, therefore, point $ A \in Z $ is connected (Fig. 13) to points $ B_1, B_2, \ldots, B_n $ of set $ Z $, then
Adding these inequalities side by side, we obtain that $ 2\pi > n \frac{\pi}{3} $, i.e., $ n < 6 $. Therefore, the number $ n $ of points of set $ Z $ to which point $ A $ is connected is not greater than $ 5 $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 531 |
LII OM - III - Problem 6
Given positive integers $ n_1 < n_2 < \ldots < n_{2000} < 10^{100} $. Prove that from the set $ \{n_1, n_2, \ldots, n_{2000}\} $, one can select non-empty, disjoint subsets $ A $ and $ B $ having the same number of elements, the same sum of elements, and the same sum of squares of elements.
|
For a set $ X \subseteq \{n_1,n_2,\ldots,n_{2000}\} $, let $ s_0(X) $, $ s_1(X) $, and $ s_2(X) $ denote the number of elements, the sum of elements, and the sum of squares of elements of the set $ X $, respectively.
It suffices to prove that there exist two different subsets $ C $ and $ D $ of the set $ \{n_1,n_2,\ldots,n_{2000}\} $ such that $ s_i(C) = s_i(D) $ for $ i = 0,1,2 $. Then the sets $ A = C \setminus D $ and $ B = D\setminus C $ are non-empty and satisfy the conditions of the problem.
For any subset $ X $ of the set $ \{n_1 , n_2, \ldots ,n_{2000}\} $ we have
Denoting $ S(X) = s_0(X) + 10^4s_1(X) + 10^{108}s_2(X) $, we obtain
Hence, there exist two different subsets $ C $, $ D $ of the set $ \{n_1 , n_2, \ldots ,n_{2000}\} $ such that $ S(C) = S(D) $. From inequality (1), it follows that $ s_i(C) = s_i(D) $ for $ i = 0, 1, 2 $, which completes the solution of the problem.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 533 |
XLVIII OM - II - Problem 6
In a cube with edge length $1$, there are eight points. Prove that some two of them are the endpoints of a segment of length not greater than $1$.
|
Let the vertices of a given cube $\mathcal{C}$ be denoted by $A_1, \ldots, A_8$. Let $\mathcal{C}_i$ be the cube with edge length $\frac{1}{2}$, having one vertex at point $A_i$ and three faces contained in the faces of cube $\mathcal{C}$. Let $P_1, \ldots, P_8$ be eight given points.
Each cube $\mathcal{C}_i$ has a diameter of $\frac{1}{2}\sqrt{3} < 1$. Therefore, if two different points $P_k$ and $P_l$ lie in any of the cubes $\mathcal{C}_i$, then $P_kP_l < 1$; the required condition is satisfied.
Assume, then, that each of these cubes contains exactly one point $P_i$. Fix the numbering so that $P_i \in \mathcal{C}_i$ for $i = 1, \ldots, 8$. Each cube $\mathcal{C}_i$ has exactly three faces in common with the surface of cube $\mathcal{C}$; the point $P_i$ is at a distance of no more than $\frac{1}{2}$ from each of these three faces. Denote these three distances by $a_i$, $b_i$, $c_i$. Let $d$ be the largest of the 24 numbers: $a_1, b_1, c_1, \ldots, a_8, b_8, c_8$. Without loss of generality, we can assume that $d = P_1Q$, where $Q$ is the orthogonal projection of point $P_1$ onto a certain face of cube $\mathcal{C}$, and that $A_1A_2$ is an edge of cube $\mathcal{C}$ parallel to the line $P_1Q$.
Let $R$ be the orthogonal projection of point $P$ onto the line $A_1A_2$. Consider the rectangular prism whose one edge is the segment $A_2R$, and one of the faces perpendicular to $A_2R$ is a square with vertex $A_2$ and side length $d$, contained in a face of cube $\mathcal{C}$. From the definition of the number $d$, it follows that this rectangular prism contains points $P_1$ and $P_2$. Its diameter is equal to
since $0 \leq d \leq \frac{1}{2} < \frac{2}{3}$. Therefore, $P_1P_2 \leq 1$, which completes the proof.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 534 |
L OM - I - Problem 1
Prove that among numbers of the form $ 50^n + (50n + 1)^{50} $, where $ n $ is a natural number, there are infinitely many composite numbers.
|
I way:
If $ n $ is an odd number, then the number $ 50^n $ when divided by $ 3 $ gives a remainder of $ 2 $. If, moreover, $ n $ is divisible by $ 3 $, then $ (50n+1)^{50} $ when divided by $ 3 $ gives a remainder of $ 1 $. Hence, the number $ 50^n + (50n+1)^{50} $ is divisible by $ 3 $ for numbers $ n $ of the form $ 6k + 3 $.
II way:
For $ n $ divisible by $ 5 $, the number $ 50^n +(50n+1)^{50} $ is the sum of fifth powers. Therefore, this number, by virtue of the identity
is composite.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 535 |
XXXIII OM - II - Problem 2
The plane is covered with circles in such a way that the center of each of these circles does not belong to any other circle. Prove that each point of the plane belongs to at most five circles.
|
Suppose that point $ A $ belongs to six circles from the considered family and that $ O_1 $, $ O_2 $, $ O_3 $, $ O_4 $, $ O_5 $, $ O_6 $ are the centers of these circles. It follows that one of the angles $ O_iAO_j $ $ (i,j = 1, 2, \ldots, 6) $ has a measure not greater than $ 60^\circ $ (six of the angles $ O_iAO_j $ sum up to a full angle), hence in the triangle $ O_iAO_j $ the side $ O_iO_j $ opposite this angle is not the longest side, since $ O_iO_j \leq \max (AO_i,AO_j) $. Assume that $ O_iO_j \leq AO_i $. It follows that: $ O_j $ belongs to the circle with center $ O_i $. This contradicts the assumption that the center of each of the given circles is not a point of any other circle. Therefore, there are no points in the plane belonging to more than five circles of the given family.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 536 |
LVII OM - II - Problem 5
Point $ C $ is the midpoint of segment $ AB $. Circle $ o_1 $ passing through points $ A $ and $ C $ intersects circle $ o_2 $ passing
through points $ B $ and $ C $ at different points $ C $ and $ D $. Point $ P $ is the midpoint of the arc $ AD $ of circle $ o_1 $, which does not
contain point $ C $. Point $ Q $ is the midpoint of the arc $ BD $ of circle $ o_2 $, which does not contain point $ C $. Prove that the lines $ PQ $ and $ CD $
are perpendicular.
|
If $ AC = CD $, then also $ BC = CD $. Then segments $ PC $ and $ QC $ are diameters of circles $ o_1 $ and $ o_2 $, respectively. Therefore, $ \measuredangle CDP = \measuredangle CDQ = 90^{\circ} $, which implies that point $ D $ lies on segment $ PQ $, and lines $ PQ $ and $ CD $ are perpendicular.
Let us assume in the further part of the solution that $ AC \neq CD $.
Let $ E $ be a point lying on ray $ CD $ such that $ CE = AC $ (Fig. 3). Then also $ CE = BC $.
From the relationships $ CE = AC $ and $ \measuredangle ACP = \measuredangle ECP $, it follows that triangles $ ACP $ and $ ECP $ are congruent
(side-angle-side criterion). Therefore, $ EP = AP = DP $.
Similarly, we prove that $ EQ = DQ $.
om57_2r_img_3.jpg
From the obtained equalities, it follows that points $ D $ and $ E $ are symmetric with respect to line $ PQ $. Line $ DE $ is therefore perpendicular
to line $ PQ $, which completes the solution of the problem.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 537 |
XVIII OM - II - Task 2
In a room, there are 100 people, each of whom knows at least 66 of the remaining 99 people. Prove that it is possible that in every quartet of these people, there are two who do not know each other. We assume that if person $ A $ knows person $ B $, then person $ B $ also knows person $ A $.
|
Let's denote the people in the room by the letters $A_1, A_2, \ldots, A_{100}$. Let $M$ be the set of people $\{A_1, A_2, \ldots, A_{33}\}$, $N$ be the set of people $\{A_{34}, A_{35}, \ldots, A_{66}\}$, and $P$ be the set of people $\{A_{67}, A_{68}, \ldots, A_{100}\}$. The case described in the problem occurs, for example, when each person in set $M$ knows only and all the people in sets $N$ and $P$ (a total of 67 people), and similarly, each person in set $N$ knows all the people in sets $P$ and $M$ and only those people (67 people), and each person in set $P$ knows all the people in sets $M$ and $N$ and only those people (66 people). If ($A_i, A_j, A_k, A_l$) is any quartet of these people, then two of them are in the same set $M$, $N$, or $P$, and therefore, these two people do not know each other.
Note. The theorem above can be easily generalized. Suppose there are $N$ people in the room, each of whom knows at least $\left[ \frac{2n}{3} \right]$ of the other people. It can then happen that in every quartet of these people, there are two who do not know each other. To prove this, divide the set $M$ of all people into 3 sets $M_1$, $M_2$, and $M_3$ containing $m_1 = \left[ \frac{n+1}{3} \right]$, $m_2 = \left[ \frac{n+1}{3} \right]$, and $m_3 = n - 2 \left[ \frac{n+1}{3} \right]$ people, respectively. If each person in set $M_i$ knows the people in sets $M_j$ and $M_k$ ($i, j, k = 1, 2, 3; i \ne j \ne k \ne i$) and only those people, then in every quartet of people in set $M$, there are two people who belong to the same set $M_i$, so they do not know each other. Each person in set $M$ then knows at least $\left[ \frac{2n}{3} \right]$ people. This is easiest to verify separately for each of the three possible cases: $n = 3k$, $n = 3k+1$, $n = 3k + 2$.
Namely
We see from this that the given inequalities are satisfied for any natural number $n$.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 538 |
XLV OM - III - Task 2
In the plane, there are two parallel lines $ k $ and $ l $ and a circle disjoint from line $ k $. From a point $ A $ lying on line $ k $, we draw two tangents to this circle intersecting line $ l $ at points $ B $, $ C $. Let $ m $ be the line passing through point $ A $ and the midpoint of segment $ BC $. Prove that all such lines $ m $ (corresponding to different choices of point $ A $ on line $ k $) have a common point.
|
Note 1. The conditions of the problem do not change if the given line $ l $ is replaced by another line $ l' $ parallel to $ k $ and different from $ k $. Indeed: if two tangents to a given circle emanating from point $ A $ intersect the line $ l $ at points $ B $ and $ C $, then the midpoints of segments $ BC $ and $ B $ are collinear with point $ A $, and thus the line $ m $ defined in the problem coincides with the line $ m' $ obtained in the same way after replacing the line $ l $ with the line $ l' $. We can therefore assume that $ l $ is a line parallel to $ k $, chosen in a convenient position for the considerations being made.
Let $ I $ and $ r $ be the center and radius of the given circle $ \Omega $, and let $ N $ and $ S $ be the points of this circle that are, respectively, closest and farthest from the line $ k $. In light of the observation just made, we can assume that the line $ l $ is tangent to the circle $ \Omega $ at point $ S $. Let the distance between the lines $ k $ and $ l $ be denoted by $ h $.
Fix a point $ A $ on the line $ k $. Let $ B $ and $ C $ be the points on the line $ l $ determined according to the problem statement. The circle $ \Omega $ is inscribed in the triangle $ ABC $. Denote the midpoint of side $ BC $ by $ M $, and the foot of the perpendicular from vertex $ A $ by $ D $ (so $ |AD| = h $). Without loss of generality, we can assume that $ |AB| \leq |AC| $. Then the points $ B $, $ S $, $ M $, $ C $ lie on the line $ l $ in that order (in a special case, points $ S $ and $ M $ may coincide). Figure 13 shows the situation when angle $ ABC $ is acute; but the considerations (and calculations) that follow also cover the case when this angle is right or obtuse.
om45_3r_img_13.jpg
Since $ S $ is the point of tangency of the inscribed circle in triangle $ ABC $ with side $ BC $, the length of segment $ BS $ is given by the known formula
\[
BS = \frac{a + c - b}{2}
\]
where, as usual, $ a = |BC| $, $ b = |CA| $, $ c = |AB| $. Therefore,
\[
BS = \frac{a + c - b}{2}
\]
Further, denoting by $ d $ the distance between points $ D $ and $ M $, we have the equalities
\[
d = \frac{1}{2}a - \frac{a + c - b}{2} = \frac{b - c}{2}
\]
(The sign of the difference $ \frac{1}{2}a - d $ depends on whether angle $ ABC $ is acute or not). From this (and the relationships in the right triangles $ ABD $ and $ ACD $), we get the relationship
\[
h^2 = AD^2 = AM^2 - DM^2 = \left(\frac{a}{2}\right)^2 - \left(\frac{b - c}{2}\right)^2 = \frac{a^2 - (b - c)^2}{4}
\]
Thus,
\[
h^2 = \frac{a^2 - (b - c)^2}{4}
\]
A useful equality is $ ah = (a + b + c)r $, both sides of which express twice the area of triangle $ ABC $. It follows that $ r(b + c) = a(h - r) $, or
\[
r(b + c) = a(h - r)
\]
Let $ X $ be the point of intersection of the line $ m $ (i.e., $ AM $) with the line $ NS $. From the similarity of triangles $ MDA $ and $ MSX $, we have the proportion $ |SX|:|SM| = |AD| : |MD| $. Considering the relationships (2), (3), (4), we obtain the equality
\[
|SX| = \frac{h \cdot |SM|}{|MD|} = \frac{h \cdot \frac{a}{2}}{\frac{b - c}{2}} = \frac{ha}{b - c}
\]
The last quantity depends only on $ h $ and $ r $; it is therefore determined solely by the line $ k $ and the circle $ \Omega $, and does not depend on the choice of point $ A $. This means that the point $ X $, located on the segment $ SN $ at a distance $ \frac{hr}{h - r} $ from point $ S $, is the common point of all the lines $ m $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 539 |
XIV OM - II - Task 4
In triangle $ABC$, the angle bisectors of the internal and external angles at vertices $A$ and $B$ have been drawn. Prove that the perpendicular projections of point $C$ onto these bisectors lie on the same line.
|
Let $M$ and $N$ denote the projections of point $C$ onto the angle bisector $AM$ of angle $A$ of triangle $ABC$ and onto the bisector of the adjacent angle (Fig. 22).
Since the bisectors $AM$ and $AN$ are perpendicular, the quadrilateral $AMCN$ is a rectangle. The point $S$, where the diagonals $MN$ and $AC$ intersect, is the midpoint of $AC$. Notice that the line $NM$ is symmetric to the line $AC$ with respect to the line $ST$ parallel to $AM$, and the line $AC$ is symmetric to the line $AB$ with respect to the line $AM$. Successive applications of mirror reflections (symmetries) with respect to the parallel lines $ST$ and $AM$ transform the line $NM$ into the line $AB$.
Since the composition of two mirror reflections of a figure with respect to two parallel axes is a translation of that figure, the line $NM$ is parallel to the line $AB$. The line $NM$ is thus the line passing through the midpoints of sides $AC$ and $BC$. The same applies to the line $PQ$ drawn through the projections $P$ and $Q$ of point $C$ onto the angle bisector of angle $B$ of triangle $ABC$ and onto the bisector of the adjacent angle. Therefore, the lines $MN$ and $PQ$ coincide, q.e.d.
Note. In the above solution, we have relied on an important theorem in plane geometry, that the transformation of a figure consisting of the successive application of two mirror reflections (axial symmetries) with respect to two parallel lines is a translation of that figure. The proof of this theorem is very simple.
Let $a$ and $b$ be two parallel lines, and $\overline{W}$ the vector of this translation parallel to $a$, which transforms $a$ into $b$. Let $P$ be any point in the plane, $Q$ the point symmetric to $P$ with respect to line $a$, $R$ the point symmetric to $Q$ with respect to line $b$, and finally let $A$ and $B$ be the midpoints of segments $PQ$ and $QR$ (Fig. 23).
Then, the point $R$ is obtained by translating the point $P$ in a direction perpendicular to the lines $a$ and $b$ by a distance equal to twice the distance between the lines $a$ and $b$.
Such a translation will be experienced by any figure when it is first transformed by symmetry with respect to line $a$ and then with respect to line $b$. Conversely, any translation can be replaced by the composition of two axial symmetries with respect to axes perpendicular to the direction of the translation, whose distance is half the length of the translation. One of these axes can be drawn through any point in the plane.
Equally important is the theorem:
The transformation of the plane consisting of the composition of two axial symmetries with respect to axes $a$ and $b$ intersecting at point $O$ is a rotation around point $O$ by an angle twice the angle between $a$ and $b$.
The proof, illustrated by Fig. 24, is analogous to the previous one: ,
Conversely, any rotation in the plane can be replaced by the composition of two mirror reflections with respect to two axes passing through the center of rotation, whose angle is half the angle of rotation. One of these axes can be drawn through any point in the plane.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 540 |
LV OM - III - Task 2
Let $ W $ be a polynomial with integer coefficients, taking relatively prime values for some two different integers. Prove that there exists an infinite set of integers for which the polynomial $ W $ takes pairwise relatively prime values.
|
A sequence of different integers $ x_0, x_1, x_2, \ldots $ such that the values $ W(x_i) $ are pairwise coprime, can be constructed inductively. The two initial terms $ x_0 = a, x_1 = b $ are numbers whose existence is given in the assumptions.
Assume that different numbers $ x_0, x_1, \ldots, x_n $ have already been defined for which the values $ y_i = W(x_i) $ are pairwise coprime. The number $ y_0 $ is coprime with the product $ y_1 y_2 \ldots y_n $; there exist, therefore, integers $ k $, $ m $ such that $ ky_0 + my_1y_2 \ldots y_n = 1 $. We define the next term of the sequence:
where $ z $ is a natural number large enough to ensure that the number $ x_{n+1} $ is different from the numbers $ x_0,x_1,\ldots ,x_n $. Of course,
and from the transformation $ x_{n+1} = b + (a - b)(1 - my_1 y_2 \ldots y_n)+zy_0 y_1 \ldots y_n $, it is clear that
Applying the polynomial $ W $ to both sides of each of these two congruences (which is allowed since congruences with a common modulus can be added and multiplied), we obtain the relations
This means that for some integers $ s $, $ t $, the following equalities hold
And since the value $ W(b) = y_1 $ is coprime with $ y_0 $, and the value $ W (a) = y_0 $ is coprime with the product $ y_1y_2 \ldots y_n $, it follows from the obtained equalities that the value $ W(x_{n+1})= y_{n+1} $ is coprime with both $ y_0 $ and $ y_1 y_2 \ldots y_n $.
Continuing this procedure, we obtain an infinite sequence $ (x_i) $ yielding values $ y_i = W(x_i) $ that are pairwise coprime.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 544 |
XL OM - III - Task 3
We number the edges of a cube with numbers from 1 to 12.
(a) Prove that for any numbering, there exist at least eight triples of integers $ (i,j,k) $, where $ 1\leq i < j < k\leq 12 $, such that the edges numbered $ i,j,k $ are consecutive sides of a broken line.
(b) Provide an example of a numbering for which there do not exist nine triples with the properties mentioned in (a).
|
(a) Let $ V $ be any vertex of a cube and let $ p $, $ q $, $ r $ be the numbers of the edges emanating from this vertex, with $ p < q < r $. We will show that
Let $ W $ be the other end of the edge numbered $ q $. Denote the numbers of the other two edges emanating from $ W $ by $ x $ and $ y $; assume that $ x < y $.
If $ q < x $, the given conditions are satisfied by the triples $ (p,q,x) $, $ (p,q,y) $.
If $ q > y $, the given conditions are satisfied by the triples $ (x, q, r) $, $ (y, q, r) $.
If $ x < q < y $, the given conditions are satisfied by the triples $ (p, q, y) $, $ (x, q, r) $.
This proves the correctness of statement (1).
Now, let us assign to each of the eight vertices of the cube the average of the three edge numbers emanating from that vertex. One number can be assigned to at most two vertices. Therefore, among the assigned numbers, there are at least four different ones. Each of these is, according to (1), the middle term of two different triples with the desired properties. Thus, there are at least eight such triples.
(b) The four parallel edges are numbered (in any order) with the numbers from $ 1 $ to $ 4 $. The next four parallel edges are numbered with the numbers from $ 5 $ to $ 8 $. Finally, the remaining four parallel edges are numbered with the numbers from $ 9 $ to $ 12 $. If a triple $ (i,j,k) $, $ i < j < k $, satisfies the condition given in (a), then the segments numbered $ i $, $ j $, $ k $ cannot lie in the same plane (otherwise, the extreme segments would be parallel; their numbers would both be greater or both be smaller than the number of the middle edge). Therefore, these segments must represent all three directions, which means that $ i \in \{1,2, 3,4\} $, $ j \in \{5,6, 7, 8\} $, $ k \in \{9,10,11,12\} $. Moreover, each of the segments numbered $ 5 $, $ 6 $, $ 7 $, $ 8 $ is the middle term of exactly two of the triples of interest to us.
Thus, there are exactly eight such triples.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 546 |
XLVI OM - I - Problem 10
Given a line $ k $ and three distinct points lying on it. Each of them is the origin of a pair of rays; all these rays lie in the same half-plane with edge $ k $. Each of these pairs of rays forms a quadrilateral with each other. Prove that if a circle can be inscribed in two of these quadrilaterals, then a circle can also be inscribed in the third.
|
Let's denote three given points by $A_1$, $A_2$, $A_3$. For $i = 1,2,3$, let $p_i$, $q_i$ be two given rays with a common starting point $A_i$, and let $l_i$ be the angle bisector of the angle formed by them. These bisectors intersect (pairwise) at points $O_1$, $O_2$, $O_3$:
(figure 5). Additionally, we adopt the following notation (for $i, j \in \{1,2,3\}, i \ne j$):
- $h_i$ - the distance from point $O_i$ to the line $k$,
- $O_i'$ - the projection of point $O_i$ onto the line $k$,
- $r_{ij}$ - the distance from point $O_i$ to the ray $p_j$,
- $T_{ij}$ - the projection of point $O_i$ onto the ray $p_j$.
The point $A_3$ is the center of a homothety that maps points $O_1$, $O_1'$, $T_{13}$ to points $O_2$, $O_2'$, $T_{23}$, respectively. This implies the proportion $|O_1T_{13}| : |O_2T_{23}| = |O_1O_1'| : |O_2O_2'|$.
By cyclically shifting the numbers $1, 2, 3$, we obtain (analogously):
\[
\frac{|O_2T_{21}|}{|O_3T_{31}|} = \frac{|O_2O_2'|}{|O_3O_3'|}
\]
Multiplying these three proportions side by side, we get 1 on the right side, and on the left side, we get a fraction whose numerator must be equal to the denominator:
\[
\frac{|O_1T_{12}| \cdot |O_2T_{23}| \cdot |O_3T_{31}|}{|O_2T_{21}| \cdot |O_3T_{32}| \cdot |O_1T_{13}|} = 1
\]
Notice now that if a circle can be inscribed in the quadrilateral determined by the pairs of rays $p_1$, $q_1$ and $p_2$, $q_2$, then its center lies on both bisectors $l_1$ and $l_2$. The center is thus the point $O_3$, and the radius is equal to both distances $r_{31} = |O_3T_{31}|$ and $r_{32} = |O_3T_{32}|$; therefore, $r_{31} = r_{32}$.
Conversely, if the equality $r_{31} = r_{32}$ holds, then the point $O_3$ is the center of a circle tangent to the rays $p_1$, $q_1$, $p_2$, $q_2$, and thus inscribed in the considered quadrilateral. Therefore, the equality $r_{31} = r_{32}$ is a necessary and sufficient condition for the existence of such a circle.
By analogy (through cyclically shifting the numbers), we infer that a circle can be inscribed in the quadrilateral determined by the pairs of rays $p_2$, $q_2$ and $p_3$, $q_3$ if and only if $r_{12} = r_{13}$, and a circle can be inscribed in the quadrilateral determined by the pairs of rays $p_3$, $q_3$ and $p_1$, $q_1$ if and only if $r_{23} = r_{21}$.
The task thus reduces to showing that if two of the equalities $r_{12} = r_{13}$, $r_{23} = r_{21}$, $r_{31} = r_{32}$ hold, then the third also holds. This implication is an immediate consequence of the equality (1) obtained above.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 548 |
XXIII OM - III - Problem 2
On a plane, there are $ n > 2 $ points, no three of which are collinear. Prove that the shortest among the closed broken lines passing through these points is a simple broken line.
|
Let's recall the definition of a simple broken line: A closed broken line with successive vertices $W_1, W_2, \ldots, W_n, W_{n+1}$, where $W_{n+1} = W_1$, is called a simple broken line if the closed segments $\overline{W_iW_{i+1}}$ and $\overline{W_jW_{j+1}}$ are disjoint for $1 \leq i, j \leq n$.
Let no three of the points $A_1, A_2, \ldots, A_n$ on the plane be collinear and let $A_{n+1} = A_1$. Suppose that the closed broken line $L$ passing successively through the points $A_1, A_2, \ldots, A_n, A_{n+1}$ is the shortest and is not a simple broken line. If $i \ne j$, $1 \leq i, j \leq n$, then let $L(A_i, A_j)$ be the part of the broken line $L$ starting at point $A_i$ and containing the points $A_{i+1}, A_{i+2}, \ldots, A_j$ successively.
For $i = 1, 2, \ldots, n$, we have $L(A_i, A_{i+1}) = \overline{A_i A_{i+1}}$, because the shortest broken line connecting two points is a segment.
From the assumption that $L$ is not a simple broken line, it follows that there exist numbers $i$, $j$ such that
and the broken lines $L(A_i, A_{i+1})$ and $L(A_j, A_{j+1})$ have a common point $P$ (Fig. 15). From conditions (1) it follows that the points $A_i, A_{i+1}, A_j, A_{j+1}$ are distinct. Therefore, the point $P$ belongs to the interior of each of the segments $\overline{A_iA_{i+1}}$, $\overline{A_jA_{j+1}}$; for by assumption, no three of the points $A_1, A_2, \ldots, A_n$ lie on the same line.
Since in a triangle the length of any side is less than the sum of the lengths of the other sides, we have
Therefore,
Thus, the closed broken line $L$ composed successively of the broken lines $L(A_{j+1}, A_i)$, $\overline{A_iA_j}$, $L(A_{i+1}, A_j)$, $\overline{A_{i+1}A_{j+1}}$ passes through the points $A_1, A_2, \ldots, A_n$ and is shorter than the closed broken line $L$ which consists successively of the broken lines $L(A_{j+1}, A_i)$, $\overline{A_iA_{i+1}}$, $L(A_{i+1}, A_j)$, $\overline{A_jA_{j+1}}$. The obtained contradiction proves that the shortest closed broken line passing through the points $A_1, A_2, \ldots, A_n$ is a simple broken line.
Note. It is easy to prove that there exists a shortest closed broken line passing through the given points $A_1, A_2, \ldots, A_n$. Namely, if a broken line passes successively through the points $A_i$ and $A_j$, then the part of the broken line contained between these points is no longer than the segment $\overline{A_iA_j}$ (since the segment is the shortest broken line connecting two points). Therefore, to find the shortest broken line passing through the points $A_1, A_2, \ldots, A_n$, it suffices to consider broken lines composed of segments $\overline{A_iA_j}$. There are a finite number of such segments. Therefore, there are only a finite number of closed broken lines determined by such segments. Hence, there exists among these broken lines the shortest one.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 552 |
XLVII OM - I - Problem 3
In a group of $ kn $ people, each person knows more than $ (k - 1)n $ others ($ k $, $ n $ are natural numbers). Prove that it is possible to select $ k + 1 $ people from this group, such that any two of them know each other.
Note: If person $ A $ knows person $ B $, then person $ B $ knows person $ A $.
|
A group of people, each of whom knows every other, we will call a {\it clique}. Let $ m $ be the largest natural number for which there exists an $ m $-person clique in the considered group of $ kn $ people. We will show that $ m > k $. This will already imply the thesis to be proven, since any ($ k+1 $)-element subset of such a maximal $ m $-person clique is then a ($ k+1 $)-person clique.
Let $ \{p_1,\ldots,p_m\} $ be an $ m $-person clique. Denote by $ N_i $ the set of all people unknown to $ p_i \ (i = 1,\ldots,m) $, and by $ n_i $ the number of people in the set $ N_i $. Each person $ p_i $ knows more than $ kn-n $ others, so the group of her/his unknowns consists of fewer than $ n-1 $ people:
Outside the clique $ \{p_1,\ldots,p_m\} $, there are $ kn-m $ people; let us denote their set by $ Q $.
Take any person $ q $ from the set $ Q $. If she/he were a friend of all people $ p_1,p_2, \ldots, p_m $, we could add her/him to them, and the resulting set $ \{p_1,p_2, \ldots, p_m, q \} $ would be an ($ m+1 $)-person clique. This, however, contradicts the definition of $ m $ as the largest possible number of people in a clique.
Hence, we conclude that every person from the set $ Q $ is unknown to someone from the clique $ \{p_1,\ldots, p_m \} $, and thus belongs to at least one set $ N_i $. Therefore, the number of people in the set $ Q $ (equal to $ kn-m $) does not exceed the total number of people in the sets $ N_1, \ldots , N_m $. The following inequality thus holds:
after reduction: $ k < m $. As we stated at the beginning, this inequality is sufficient to complete the proof.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 553 |
XXIV OM - I - Problem 7
Prove that among five segments lying on the same straight line, there are either three segments that have a common point or three segments that are pairwise disjoint.
|
Let the open intervals be $ I_k = (a_k; b_k) $, where $ k = 1, 2, 3, 4, 5 $. For closed, half-open, etc., intervals, the reasoning proceeds similarly.
Without loss of generality, we can assume that $ a_1 \leq a_2 \leq a_3 \leq a_4 \leq a_5 $. Suppose that there is no point belonging to any three intervals. If $ b_1 > a_3 $ and $ b_2 > a_3 $, then the intervals $ I_1, I_2, I_3 $ would have a common point. Therefore, $ b_1 \leq a_3 $ or $ b_2 \leq a_3 $. Let $ r $ be the number, $ 1 $ or $ 2 $, for which $ b_r \leq a_3 $ holds.
If $ b_3 > a_5 $ and $ b_4 > a_5 $, then the intervals $ I_3, I_4, I_5 $ would have a common point. Therefore, $ b_3 \leq a_5 $ or $ b_4 \leq a_5 $. Let $ s $ be the number $ 3 $ or $ 4 $, for which $ b_s \leq a_5 $ holds.
We thus have $ a_r < b_r \leq a_3 \leq a_s < b_s \leq a_5 < b_5 $. It follows that the intervals $ I_r, I_s, I_5 $ are pairwise disjoint.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 554 |
VI OM - I - Problem 8
Prove that a circle can be inscribed in a trapezoid if and only if the circles whose diameters are the non-parallel sides of the trapezoid are externally tangent to each other.
|
A circle can be inscribed in a convex quadrilateral $ABCD$ if and only if the sums of the lengths of opposite sides of the quadrilateral are equal, i.e., when
When the quadrilateral $ABCD$ is a trapezoid with parallel sides $AB$ and $CD$, denoting the midpoints of sides $AD$ and $BC$ by $M$ and $N$ respectively (Fig. 6), we have $AB + CD = 2 MN$, so equality (1) can be replaced by equality
Equality (2), on the other hand, expresses the necessary and sufficient condition for the external tangency of circles with centers at $M$ and $N$ and radii $\frac{1}{2} AD$ and $\frac{1}{2} BC$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 556 |
XXIX OM - I - Problem 6
Prove that on the plane, any closed broken line of length 1 is contained in some circle of radius length $ \frac{1}{4} $
|
Let points $A$ and $B$ belonging to a closed broken line of length $1$ divide this broken line into two equal parts, and let $C$ be any point on this broken line. Since a segment is the shortest broken line connecting two points, we have $AC + BC \leq \frac{1}{2}$.
Let $P$ be the midpoint of segment $\overline{AB}$, and $C'$ be the point symmetric to $C$ with respect to point $P$ (Fig. 7). Then $CP = C'P$ and $BC = AC$. Therefore,
and hence $CP \leq \frac{1}{4}$.
We have thus proved that any point $C$ of the given broken line is at a distance from point $P$ not greater than $\frac{1}{4}$. Therefore, this broken line is contained within a circle with center $P$ and radius of length $\frac{1}{4}$.
om29_1r_img_7.jpg
Note. A similar problem was given in the VII Mathematical Olympiad as problem 23 (5).
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 558 |
XVII OM - I - Problem 11
Prove that the centers of the excircles of a triangle and the points symmetric to the center of the incircle of the triangle with respect to its vertices lie on a single circle.
|
Let $A_1$, $B_1$, $C_1$ be the points symmetric to the center $O$ of the incircle of triangle $ABC$ with respect to the vertices $A$, $B$, $C$, and let $S_1$ be the center of the excircle of triangle $ABC$ that is tangent to side $BC$ (Fig. 8). To prove the theorem, it suffices to show that the points $A_1$, $B_1$, $C_1$, and $S_1$ lie on the same circle. Since $C_1$ and $S_1$ lie on the same side of the line $A_1B_1$, specifically on the same side of the line $A_1B_1$ as triangle $ABC$, it reduces to showing that $\measuredangle A_1S_1B_1 = \measuredangle A_1C_1B_1$. Notice that:
a) Triangles $A_1B_1C_1$ and $ABC$ are similar with respect to point $O$, so
b) Lines $BS_1$ and $CS_1$, as the angle bisectors of the external angles of triangle $ABC$, are perpendicular to the angle bisectors $OB$ and $OC$ of the corresponding angles of this triangle; points $B$, $O$, $C$, $S_1$ as vertices of a quadrilateral with two right angles lie on the same circle, with points $C$ and $S_1$ lying on the same side of the line $OB$, thus
c) Line $BS_1$ is the perpendicular bisector of segment $OB_1$, so
From a), b), and c), it follows that
Note. The theorem we have proved is equivalent to the theorem:
The points symmetric to the orthocenter of triangle $S_1S_2S_3$ with respect to the sides of this triangle lie on the circumcircle of triangle $S_1S_2S_3$ *).
To prove this, it suffices to notice that if $S_1$, $S_2$, $S_3$ are the centers of the excircles of triangle $ABC$, and $O$ is the center of the incircle of triangle $ABC$, then point $O$ is the orthocenter of triangle $S_1S_2S_3$, and points $A_1$, $B_1$, $C_1$ - symmetric to $O$ with respect to points $A$, $B$, $C$ - are also symmetric to $O$ with respect to the sides of triangle $S_1S_2S_3$. Conversely, if $O$ is the orthocenter of a given triangle $S_1S_2S_3$, and $A_1$, $B_1$, $C_1$ are the points symmetric to $O$ with respect to the sides of this triangle, then they are also symmetric to $O$ with respect to the vertices $A$, $B$, $C$ of the "pedal" triangle for triangle $S_1S_2S_3$, and $O$ is the center of the incircle of triangle $ABC$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 560 |
XIX OM - III - Problem 5
On a plane, there are $ n $ points ($ n \geq 4 $), any four of which are vertices of a convex quadrilateral. Prove that all these points are vertices of a convex polygon.
|
We will apply the method of induction. When $ n = 4 $, the thesis of the theorem is true. Assume that the thesis of the theorem is true for some natural number $ n \geq 4 $, and let $ A_1, A_2, \ldots, A_n, A_{n+1} $ be such $ n + 1 $ points in the plane that any four of them are vertices of a convex polygon. According to the induction hypothesis, the points $ A_1, A_2, \ldots, A_n $ are vertices of a convex polygon $ W $. Let us assume that they are consecutive vertices, which can be achieved by appropriately numbering the points. The point $ A_{n+1} $ does not lie on the boundary of the polygon $ W $, since no three given points are collinear. It also does not lie inside the polygon $ W $, since any point inside the polygon belongs to one of the triangles into which the polygon can be divided by its diagonals, and the point $ A_{n + 1} $ cannot be in such a triangle. Therefore, the point $ A_{n+1} $ lies outside the polygon $ W $. Consider the convex angles with vertex $ A_{n+1} $, whose sides pass through the vertices of the polygon $ W $. The set of these angles is finite, so there is a largest angle among them. Let this be, for example, the angle $ \alpha = A_kA_{n + 1}A_l $. Inside the angle $ \alpha $ lie all the vertices of the polygon $ W $ except $ A_k $ and $ A_l $. In the triangle $ T $ with vertices $ A_k $, $ A_{n+1} $, $ A_l $, none of the vertices $ A_i $ of the polygon $ W $, different from $ A_k $ and $ A_l $, lie, since the points $ A_i $, $ A_k $, $ A_l $, $ A_{n+1} $ are, according to the assumption, vertices of a convex quadrilateral. Therefore, $ A_k $ and $ A_l $ are consecutive vertices of the polygon $ W $, for example, $ l=k+1 $. In this case, the points $ A_1, A_2 \ldots A_k, A_{n+1}, A_{k+1}, \ldots, A_n $ are consecutive vertices of the polygon $ W_1 $, which consists of the polygon $ W $ and the triangle $ T $.
We will prove that the polygon $ W_1 $ is convex, i.e., that any segment whose endpoints lie in the polygon $ W_1 $ is entirely contained in this polygon. Indeed, if the points $ M $ and $ N $ lie in the polygon $ W_1 $, then one of the following cases occurs: a) The points $ M $ and $ N $ both lie in the convex polygon $ W $, or both in the triangle $ T $; then the entire segment $ MN $ is contained in the polygon $ W $ or in the triangle $ T $, and thus is contained in the polygon $ W_1 $; b) One of these points, say the point $ M $, lies in the polygon $ W $, and the point $ N $ in the triangle $ T $. The segment $ MN $ then intersects the boundary of the triangle $ T $ at some point $ P $. The point $ P $ lies on that part of the boundary of the triangle $ T $ which is in the angle $ \alpha $, i.e., on the segment $ A_kA_{k+1} $. The segment $ MP $ therefore belongs to the polygon $ W $, and the segment $ PN $ to the triangle $ T $; each of these segments, and thus their sum $ MN $, belongs to the polygon $ W_1 $. The inductive proof of the theorem has been completed.
Note. A very simple proof of the theorem can be obtained by relying on the following theorem:
For any finite set $ Z $ of points in the plane, consisting of $ n \geq 3 $ points not lying on a single line, there exists a convex polygon $ W $ such that
1. The set $ Z $ is contained in the polygon $ W $.
2. Each vertex of the polygon $ W $ belongs to the set $ Z $.
Such a polygon $ W $ is called the convex hull of the set $ Z $.
Let $ Z $ be a set of $ n \geq 4 $ points in the plane, any four of which are vertices of a convex quadrilateral, and let the polygon $ W $ be the convex hull of the set $ Z $. Then each point of the set $ Z $ is a vertex of the polygon $ W $. Indeed: $ 1^\circ $ no point of the set $ Z $ lies on the boundary of the polygon $ W $ between two of its vertices, since there are no three collinear points in the set $ Z $; $ 2^\circ $ no point of the set $ Z $ lies inside the polygon $ W $, since it would then lie in one of the triangles into which $ W $ can be divided by its diagonals, which would also contradict the assumption. The convex hull $ W $ is thus the polygon whose existence needed to be proved.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 561 |
XII OM - III - Task 3
Prove that if the section of a tetrahedron by a plane is a parallelogram, then half of its perimeter is contained between the length of the shortest and the length of the longest edge of the tetrahedron.
|
Let the parallelogram $MNPQ$ be a planar section of the tetrahedron $ABCD$. The plane $MNPQ$ is, of course, different from the planes of the faces of the tetrahedron, and each side of the parallelogram lies on a different face. Suppose that the sides $MN$ and $QP$ lie on the faces $ABC$ and $BCD$; they are then parallel to the edge $BC$, along which these faces meet, and the sides $MQ$ and $NP$ are parallel to the edge $AD$ (Fig. 19).
Applying Thales' theorem to triangles $AMN$ and $ABC$ and to triangles $BMQ$ and $BAD$ we get
Hence
Suppose that $BC \leq AD$; in that case, from the above equality, it follows that
thus
If $a$ denotes the length of the smallest, and $b$ the length of the largest diagonal of the tetrahedron, then $BC \geq a$, and $AD \leq b$, and from the above inequality, it follows that
which was to be proved.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 565 |
XXIX OM - I - Problem 11
Let $ f: \mathbb{R} \to \mathbb{R} $ be a continuous function for which there exists a number $ x $ satisfying the conditions: $ f(f(f(x))) = x $, $ f(x) \neq x $.
Prove that there exist numbers $ y_1 $, $ y_2 $, $ y_3 $ such that $ y_i \neq y_j $ for $ i \neq j $ and $ f(f(y_k))=y_k $ for $ k = 1, 2, 3 $.
Note. It is known that every continuous function $ g: \mathbb{R} \to \mathbb{R} $ has the Darboux property, meaning if $ a < b $ and $ d $ is between $ g(a) $ and $ g(b) $, then there exists a number $ c \in [a, b] $ such that $ g(c) = d $.
|
If $ff(x) = x$, then we would have $fff(x) = f(x)$, which means $x = f(x)$, contradicting the assumption. Therefore, $ff(x) \ne x$. Similarly, from the equation $ff(x) = f(x)$, it follows that $fff(x) = ff(x)$, which means $x = ff(x)$. However, we have already proven that this last equality does not hold. Therefore, $ff(x) \ne f(x)$.
Thus, the numbers $x$, $f(x)$, and $f(x)$ are pairwise distinct. Let the smallest of them be denoted by $a$, the largest by $c$, and the remaining one by $b$. Therefore, $a < b < c$. From the conditions of the problem, it follows that the function $f$ defines a permutation of the set $\{a, b, c\}$ and $f(a) \ne a$, $f(b) \ne b$, $f(c) \ne c$.
Consider the case when $f(a) = c$. If $f(a) = b$, the reasoning proceeds similarly. Thus, we have $f(b) = a$ and $f(c) = b$.
The continuous function $f(t)$ at the endpoints of the interval $\langle a, b \rangle$ takes the values $c$ and $a$. Therefore, by the Darboux property, there is a point in this interval where the function $f(t)$ takes an intermediate value $b$, that is, $a < z < b$ and $f(z) = b$. To solve the problem, it is sufficient to prove that the function $g(t) = ff(t) - t$ has at least three zeros. We have
Thus, the function $g(t)$ at the endpoints of each of the intervals $\langle a, z \rangle$, $\langle z, b \rangle$, $\langle b, c \rangle$ takes values of opposite signs. Therefore, by the Darboux property, it has at least one zero in each of these intervals.
Note. For any natural number $n$, let $f_n$ be the $n$-fold iteration of the function $f$, that is, $f_n(x) = f(f(\ldots (f(x))\ldots))$ ($n$ times). We say that a real number $x$ has period $n$ if $f_n(x) = x$ and $f_m(x) \ne x$ for $m < n$. Our problem can thus be formulated as: *Prove that if for a continuous function $f \colon \mathbb{R} \to \mathbb{R}$ there exists a number $x$ of period $3$, then there exist three numbers $y_1, y_2, y_3$ of period $2$ or $1$.*
The following general theorem of Sharkovsky holds: Arrange all natural numbers as follows
At the beginning is an increasing sequence of all prime numbers, then a sequence of prime numbers multiplied by $2$, by $4$, etc., and finally a decreasing sequence of all powers of $2$.
Sharkovsky's theorem states that if for a continuous function $f \colon \mathbb{R} \to \mathbb{R}$ there exists a number of period $n$, then there also exists a number of period $m$, provided that in the sequence (1) the number $m$ appears after the number $n$.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 566 |
XXXIX OM - I - Problem 12
Given are two concentric spheres $ S_R $ and $ S_r $ with radii $ R $ and $ r $, respectively, where $ R > r $. Let $ A_0 $ be a point on $ S_R $. Let $ T_{A_0} $ be the cone with vertex at $ A_0 $ and tangent to $ S_r $. Denote $ K_1 = T_{A_0} \cap S_R $. Then, for each point $ A \in K_1 $, we draw the cone $ T_A $ and define $ K_2 = \bigcup_{A\in K_1} T_A \cap S_R $, and further by induction $ K_{n+1}= \bigcup_{A\in K_n} T_A \cap S_R $, where $ T_A $ is the cone with vertex at $ A $ and tangent to $ S_r $. Prove that for some $ n $, $ K_n = S_R $.
|
In this task, by a cone, we mean the surface of an unbounded circular cone, and by the tangency of a cone to a sphere - tangency along a certain circle. (This lack of precision in the formulation was noticed and pointed out by many participants of the olympiad.)
Let $O$ be the common center of spheres $S_r$ and $S_R$. Let $u$ be the angle at the vertex of the cone $T_{A_0}$ (and thus of any cone $T_A$ considered in the problem); $a = 2 \arcsin (r/R)$. Let us take $N = [\pi/2\alpha]+1$. We will prove by induction on $n = 1, \ldots, N$ the following implication:
From this, it will follow that $K_{2N} = S_R$.
For $n = 1$, the antecedent of the implication (1) takes the form: . The cone $T_P$ then has points in common with the set $K_1 = T_{A_0} \cap S_R$. Let $Q \in T_P \cap K_1$ (see figure 5). Note that the statements: $Q \in T_P$ and $P \in T_Q$ are equivalent (each of them means exactly that the segment $PQ$ is tangent to the sphere $S_r$). Thus, in this case,
which means $P \in K_2$.
Now assume that the implication (1) to be proved holds for some natural number $n \leq N-1 = [\pi/2\alpha]$. We will show the validity of the analogous implication for $n+1$.
om39_1r_img_5-6.jpg
Let $P \in S_R$ be a point such that $|\measuredangle A_0OP| \leq 2(n+1) \alpha$ (figure 6 illustrates the situation when $n = 2$, $n+1 = 3$; it shows the section of the considered spheres by the plane $A_0PP$). Consider the great circle of the sphere $S_R$ passing through the points $A_0$ and $P$. Let $\textit{ł}$ be the shorter of the two arcs into which the points $A_0$ and $P$ divide this circle (in the case where $P$ is the antipodal point to $A_0$, we take any great circle with diameter $A_0P$ and any semicircle $A_0P$). On the arc $\textit{ł}$, we find points $L$ and $M$ such that $|\measuredangle LOM| = |\measuredangle MOP| - \alpha$. Then $|\measuredangle LOP| = 2 \alpha$, $|\measuredangle A_0OL|$, $|\measuredangle A_0OP|????$, so by the induction hypothesis $L \in K_{2n}$. Let $M$ be the antipodal point to $M$. In the considered great circle, the inscribed angle $LM$ is half the central angle $LOP$. Therefore, $|\measuredangle LM| = \alpha$, which means that points $L$ and $P$ belong to the cone $T_M$. Segments $LM$ and $PM$ are thus tangent to the sphere $S_r$; hence $M \in K_{2n+1}$, in accordance with the definition of the set $K_{2n+1}$ as the union of sets of the form $T_A \cap S_R$ for $A \in K_{2n}$. Since, in turn, the set $K_{2n+2}$ is the union of sets $T_A \cap S_R$ for $A \in K_{2n+1}$, we conclude that $P \in T_{M}$. This completes the inductive step.
By the principle of induction, the implication (1) is valid for all considered values of $n$ (i.e., for $n = 1, \ldots, N$). In particular, it holds for $n = N$. When $n = N$, the antecedent of this implication is true for any point of the sphere $S_R$: the inequality $|\measuredangle A_0OP| \leq 2N \alpha$ is automatically satisfied because $2N\alpha > \pi$ (in accordance with the definition of the number $N$). Therefore, the consequent is also true; which means that every point $P$ of the sphere $S_R$ belongs to the set $K_{2N}$. Thus, ultimately, $S_R = K_{2N}$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 568 |
L OM - III - Zadanie 2
Given are non-negative integers $ a_1 <a_2 <a_3 < \ldots < a_{101} $ less than $ 5050 $. Prove that among them, one can choose four different $ a_k $, $ a_l $, $ a_m $, $ a_n $, such that the number $ a_k + a_l -a_m -a_n $ is divisible by $ 5050 $.
|
We consider all expressions of the form $ a_k + a_l $, where $ 1 \leq k < l \leq 101 $. There are $ {101 \choose 2} = 5050 $ such expressions. If we find two of them, say $ a_k + a_l $ and $ a_m + a_n $, whose values give the same remainder when divided by 5050, then the numbers $ a_k $, $ a_l $, $ a_m $, $ a_n $ satisfy the conditions of the problem — these numbers are pairwise distinct (if, for example, $ a_k = a_m $, then also $ a_l = a_n $, since $ 0 \leq a_l, a_n < 5050 $).
It remains to consider the case where all the above sums give different remainders when divided by 5050. We will show that this case cannot occur. If it were so, then the considered sums would give all possible remainders when divided by 5050 — each one exactly once. Hence
which proves that $ S $ is an odd number.
On the other hand,
which means that $ S $ is an even number. We have reached a contradiction.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 569 |
LI OM - I - Task 3
The sum of positive numbers $ a $, $ b $, $ c $ is equal to $ 1 $. Prove that
|
It is enough to prove that for any positive numbers $ a $, $ b $, $ c $ the inequality holds
Transforming the above relationship equivalently, we get
By making the substitution $ x = bc $, $ y = ca $, $ z = ab $, we reduce the inequality to be proved to the form
Squaring both sides of the last inequality and transforming it equivalently, we get
which is true.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 571 |
V OM - III - Task 1
Prove that in an isosceles trapezoid circumscribed around a circle, the segments connecting the points of tangency of opposite sides with the circle pass through the point of intersection of the diagonals.
|
Let $E$, $F$, $G$, $H$ denote the points of tangency of the sides $AB$, $BC$, $CD$, $DA$ with the incircle of trapezoid $ABCD$.
Let $M$ be the point of intersection of segments $EG$ and $HF$ (Fig. 37a). Since trapezoid $ABCD$ is isosceles, the line $EG$ is the axis of symmetry of the figure, and point $F$ is symmetric to point $H$ with respect to $EG$, and $HF \bot EG$. The parallel lines $AB$, $HF$, and $DC$ determine proportional segments on the lines $EG$ and $BC$, hence
Let $N$ be the point of intersection of the diagonals $AC$ and $BD$ of the trapezoid. Since these diagonals are symmetric with respect to the line $EG$, the point $N$ lies on the segment $EG$ (Fig. 37b). Triangles $AEN$ and $CGN$ are similar with respect to point $N$, hence
But $AE = EB = BF$, and $FC = CG$, so $\frac{AE}{CG} = \frac{BF}{FC}$; from the above proportions, it follows that
Points $M$ and $N$ divide the segment $EG$ in the same ratio, hence these points coincide, which was to be proved.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 573 |
X OM - II - Task 6
From point $ M $ on the surface of a sphere, three mutually perpendicular chords of the sphere $ MA $, $ MB $, $ MC $ were drawn. Prove that the segment connecting point $ M $ with the center of the sphere intersects the plane of triangle $ ABC $ at the centroid of this triangle.
|
The plane $ AMB $ intersects the surface of the sphere along a circle passing through points $ A $, $ M $, and $ B $, which is the circumcircle of the right triangle $ AMB $; the center of this circle is therefore at the midpoint $ K $ of segment $ AB $ (Fig. 25).
The center $ O $ of the sphere lies on the perpendicular line erected at point $ K $ to the plane $ AMB $ on the same side of the plane $ AMB $ as point $ C $. Since the line $ MC $ is perpendicular to the plane $ AMB $, the lines $ KO $ and $ MC $ lie - as parallel lines - in the same plane. Segments $ MO $ and $ KC $ are the diagonals of the trapezoid $ MKOC $, and therefore intersect at some point $ S $. It turns out, then, that segment $ MO $ intersects the median $ KC $ of triangle $ ABC $. Similarly, it intersects each of the other medians of triangle $ ABC $, and since segment $ MO $ does not lie in the plane $ ABC $, it follows that it passes through the centroid of triangle $ ABC $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 575 |
XLIX OM - I - Problem 10
Medians $ AD $, $ BE $, $ CF $ of triangle $ ABC $ intersect at point $ G $. Circles can be circumscribed around quadrilaterals $ AFGE $ and $ BDGF $. Prove that triangle $ ABC $ is equilateral.
|
Let's denote the measures of angles $CAB$, $ABC$, $BCA$ by $\alpha$, $\beta$, $\gamma$ respectively. Points $F$ and $D$ are the midpoints of sides $AB$ and $BC$, so $DF \parallel CA$, and therefore $|\measuredangle BDF| = \gamma$ (Figure 5).
The opposite angles of the cyclic quadrilateral $AFGE$ sum up to $180^\circ$. Meanwhile, the existence of a circumcircle around quadrilateral $BDGF$ implies the equality of angles $BDF$ and $BGF$ (both inscribed in the same circle). Therefore,
Thus,
Analogously, we prove that $\gamma = \beta$. Triangle $ABC$ is therefore equilateral.
Note: There are many variants of this solution, utilizing the equality of different angles (resulting from the problem's assumptions); we have chosen the shortest one.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 576 |
XXII OM - I - Problem 11
We toss a coin $2n$ times. Let $p_n$ denote the probability that we will get a series of length $r > n$. Prove that $\lim p_n = 0$.
|
Elementary events are all $2n$-term sequences with terms being heads or tails; all of them are equally probable and there are $2^{2n}$ of them. Let $A_n$ be the event that we get a series of length at least $n$.
Consider the following three-stage procedure:
1) Selecting some $n$ consecutive positions out of $2n$;
2) Placing only heads or only tails on the selected positions;
3) Filling the remaining positions with heads and tails arbitrarily. As a result of each such procedure, we obtain a sequence that is an elementary event favorable to the event $A_n$, because such a sequence contains a series of length $\geq n$. Conversely, any sequence containing a series of length $\geq n$ can be obtained as a result of such a procedure. Of course, the same sequence can be obtained several times. Therefore, the number of the described ways of proceeding is greater than the number of elementary events favorable to the event $A_n$.
We will now determine the number of the described ways of proceeding. Action 1) can be performed in $n + 1$ ways, because the first of the $n$ consecutive positions chosen can only be position number $1, 2, \ldots, n$ or $n + 1$. Action 2) can be performed in two ways: either by placing heads or by placing tails. Finally, action 3) can be performed in $2^n$ ways, because we have $n$ remaining positions to fill with heads or tails.
Thus, the number of all the described ways of proceeding is equal to $2(n + 1) 2^n$.
From the binomial theorem, we obtain that for $n \geq 2$
Therefore
Since $ {\displaystyle \lim_{n \to \infty}} \frac{4}{n} = 0 $ and $ {\displaystyle \lim_{n \to \infty}} \frac{2}{n-1} = 0 $, it follows that
$ \displaystyle \lim_{n \to \infty} p(A_n) = 0 $.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 577 |
XIX OM - III - Problem 3
In the tetrahedron $ABCD$, the edges $AD$, $BD$, $CD$ are equal. On the plane $ABC$, points $A_1, B_1, C_1$ are chosen such that they are not collinear. The lines $DA_1, DB_1, DC_1$ intersect the surface of the sphere circumscribed around the tetrahedron at points $A_2, B_2, C_2$, respectively, different from point $D$. Prove that the points $A_1, B_1, C_1, A_2, B_2, C_2$ lie on the surface of some sphere.
|
The sphere (i.e., the surface of a sphere) $ S $ circumscribed around the tetrahedron $ ABCD $ intersects the plane $ ABC $ along the circle $ k $ circumscribed around the triangle $ ABC $. From the equality $ DA = DB = DC $, it follows that the sphere $ t $ with center $ D $ and radius $ r = DA $ passes through the points $ B $ and $ C $, so it intersects the sphere $ s $ and the plane $ ABC $ along the same circle $ k $. Therefore, the orthogonal projection $ H $ of the point $ D $ onto the plane $ ABC $ is the center of the circle $ k $. The plane $ A_1HD $ then contains a certain diameter $ MN $ of the circle $ k $ and intersects the sphere $ s $ along the circle $ l $ passing through the points $ M $, $ N $, and $ D $ (Fig. 16).
According to the power of a point theorem with respect to a circle,
Hence,
so,
This equality can be written in the form,
Similarly,
From (1) and (2) it follows that,
From equality (3) and the fact that the pairs of points $ A_1 $, $ A_2 $, and $ B_1 $, $ B_2 $ lie on two different rays with origin $ D $ that do not form a single line, it is easy to deduce that the points $ A_1 $, $ A_2 $, $ B_1 $, $ B_2 $ lie on a circle.
Indeed, let $ m $ be a circle passing through the non-collinear points $ A_1 $, $ A_2 $, $ B_1 $ (Fig. 17).
The point $ D $ lies outside the circle $ m $, so,
where $ X $ is the point of intersection of the circle $ m $ with the ray $ DB_1 $.
From (3) and (4) we obtain,
and since the points $ X $ and $ B_2 $ lie on the same ray with origin $ D $, the point $ X $ coincides with the point $ B_2 $.
Similarly, we conclude that the points $ B_1 $, $ B_2 $, $ C_1 $, $ C_2 $ lie on some circle $ n $.
The circles $ m $ and $ n $ have the common points $ B_1 $ and $ B_2 $, and they lie in different planes, since according to the assumption, the common points of these circles with the plane $ ABC $, i.e., the points $ A_1 $, $ B_1 $, $ C_1 $, are not collinear.
Hence, the circles $ m $ and $ n $, and thus the points $ A_1 $, $ B_1 $, $ C_1 $, $ A_2 $, $ B_2 $, $ C_2 $, lie on one sphere, q.e.d.
Note. The beginning of the above proof can be significantly shortened by referring to the properties of inversion. Specifically, in the inversion with respect to the sphere with center $ D $ and radius $ r = DA $ intersecting the plane $ ABC $ along the circle $ k $, the image of the plane $ ABC $ is a sphere passing through the center of inversion $ D $ and the circle $ k $, i.e., the sphere $ s $. The images of the points $ A_1 $, $ B_1 $, $ C_1 $ are then the points $ A_2 $, $ B_2 $, $ C_2 $, respectively, so,
The rest of the proof remains the same as before.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 581 |
XXV OM - II - Problem 5
Given are real numbers $ q,t \in \langle \frac{1}{2}; 1) $, $ t \in (0; 1 \rangle $. Prove that there exists an increasing sequence of natural numbers $ {n_k} $ ($ k = 1,2, \ldots $), such that
|
The sequence $ \{n_k\} $ $ (k = 1,2, \ldots) $ is defined inductively as follows. Since $ 0 < q < 1 $, we have $ q^0 = 1 $ and $ \displaystyle \lim_{n \to \infty} q^n = 0 $. For each number $ t $ in the interval $ (0; 1 \rangle $, there exists a natural number $ n_1 $ such that
Wthen
since $ 1 - q \leq q $ for $ \displaystyle q \in \langle \frac{1}{2}; 1) $.
Next, assume that for some natural number $ k $, the increasing sequence of natural numbers $ n_1, n_2, \ldots, n_k $ has already been defined such that
We take $ n_{k+1} $ to be the natural number $ m $ such that
From (1) and (2), it follows that $ q^m < q^{n_k} $, and thus $ m > n_k $, i.e., $ n_{k+1} > n_k $. Moreover, from (2) we obtain
since $ 1 - q \leq q $ for $ q \in \langle \frac{1}{2}; 1) $.
Therefore, by the principle of induction, there exists an infinite increasing sequence of natural numbers $ n_1, n_2, \ldots $ such that condition (1) is satisfied for $ k = 1, 2, \ldots $.
Since $ \displaystyle \lim_{n \to \infty} q^{n_k} = 0 $, it follows from (1) that
Note. The sequence $ \{n_k \} $ satisfying the conditions of the problem is generally not uniquely determined. Consider the polynomial $ f(x) = x^3 + x^2 - 1 $. It has a root in the interval $ \displaystyle \left( \frac{1}{2}; 1 \right) $, because $ f(x) $ takes values of opposite signs at the endpoints of this interval. Let us denote such a root by $ q $. Then, by calculating the sums of geometric series, we obtain
From the equality $ q^3 + q^2 - 1 = 0 $, it follows that
Thus
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 585 |
XLII OM - II - Problem 4
Find all monotonic functions $ f: \mathbb{R} \to \mathbb{R} $ that satisfy the equation
|
For any $ x > 0 $, we have the inequality $ f(4x) - f(3x) > 0 $, which implies that the (monotonic) function $ f $ is increasing in the interval $ (0;\infty) $. Similarly, for $ x < 0 $, the inequality $ f(3x) - f(4x) > 0 $ holds, so $ f $ is also increasing in $ (-\infty;0) $.
Replacing $ 4x $ with the letter $ z $, we rewrite the given equation as
Take any number $ x \ne 0 $ and any natural number $ n $. Applying the formula (1) $ n $ times, we obtain the equality
which is
We now want to pass from $ n $ to infinity. For $ x > 0 $, the sequence $ ((\frac{3}{4})^nx) $ is decreasing, so the sequence of values $ f((\frac{3}{4})^nx) $ is also decreasing, and moreover, bounded from below by the value $ f(0) $ (here we use the assumption of the monotonicity of the function $ f $ over the entire set $ \mathbf{R} $). Similarly, for $ x < 0 $, the sequence $ ((\frac{3}{4})^nx) $ is increasing, so the sequence $ (f(\frac{3}{4})^n x) $ is increasing, bounded from above by $ f(0) $. In either case, the sequence is convergent. Its limit may {\it a priori} depend on $ x $; let us denote its value by $ g(x) $:
Suppose that for some two positive numbers $ u $, $ v $, the inequality
holds.
Take a number $ \varepsilon $ such that
By the definition of the limit of a sequence,
and
We now find a natural number $ k $ large enough so that for $ n = k $, inequality (4) holds. Then we find a natural number $ m $ large enough so that for $ n = m $, inequality (5) holds, and moreover,
We then have (by (3))
This, however, contradicts inequality (6); we remember that the function $ f $ is increasing in $ (0;\infty) $.
The obtained contradiction proves that the inequality $ g(u) < g(v) $ is not possible for any pair of numbers $ u, v > 0 $. Hence, the value $ g(x) $ is the same for all $ x > 0 $. Let us denote this value by $ c $. Thus,
Similarly, we prove that the value of the limit $ g(x) $ is the same for all $ x < 0 $. Denoting this value by $ a $, we have
We can now take the limit in formula (2) (as $ n \to \infty $). We obtain
Given the requirement of the monotonicity of the function $ f $ on the set $ \mathbf{R} $, the inequality $ a \leq c $ must hold, and the value $ f(0) $ must lie between $ a $ and $ c $. Ultimately, then,
where $ a $, $ b $, $ c $ are any constants such that $ a \leq b \leq c $.
Verifying that indeed any function of this form satisfies the given functional equation is immediate.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 586 |
XXXII - I - Problem 7
Let $ (x_n) $ be a sequence of natural numbers satisfying the conditions
a) $ x_1 < x_2 < x_3 < \ldots $,
b) there exists $ j \in \mathbb{N} $, $ j > 1 $, such that $ x_j = j $,
c) $ x_{kl} = x_kx_l $, for coprime $ k,l \in \mathbb{N} $.
Prove that $ x_n = n $ for every $ n $.
|
Since the number of natural numbers less than a given natural number $n$ is $n-1$, if for some $j$ we have $x_j = j$, then by condition a) it must be that $x_k = k$ for $k \leq j$. If the number $j$ mentioned in condition b) is greater than $2$, then the number $j-1$ is greater than $1$ and is relatively prime to the number $j$. At the same time, $x_{j-1} = j-1$, so by condition c) it is also $x_{(j-1)j} = (j-1)j$. Since $(j-1)^2 < (j-1)j$, we have $x_{(j-1)^2} = (j-1)^2$, and since the numbers $(j-1)^2$ and $j$ are relatively prime, we obtain by condition c) that $x_{(j-1)^2j} = (j-1)^2j$. Reasoning analogously, we get $x_{(j-1)^mj} = (j-1)^mj$ for $m = 1, 2, \ldots$.
Since for every natural number $n$ there exists an $m$ such that $n < (j-1)^mj$, by what we have established at the beginning of the solution, it is $x_n = n$ for every $n$.
The case where the $j$ mentioned in condition b) is equal to $2$ remains to be considered. Then $x_2 = 2$. Let $x_3 = 3 + q$. Therefore, $x_6 = x_2 \cdot x_3 = 2 \cdot (3 + q) = 6 + 2q$. Therefore, $x_5 < 6 + 2q$, so $x_5 \leq 5 + 2q$, and thus $x_{10} = x_2x_5 \leq 10 + 4q$, from which $x_9 \leq 9 + 4q$, and $x_{18} = x_2x_9 \leq 18 + 8q$. From the last inequality, it follows that $x_{15} \leq 15 + 8q$. On the other hand, $x_5 \geq x_3 + 2 = 5 + q$, so $x_{15} = x_3x_5 \geq (3 + q)(5 + q) = 15 + 8q + q^2$. From this, we get
from which it follows that $q = 0$. Therefore, $x_3 = 3$ and the problem has been reduced to the case already considered above.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 590 |
XXVI - I - Problem 4
Given are pairwise disjoint spheres $ K_1, K_2, K_3 $ with pairwise distinct radii. Let $ A_{ij} $ be the vertex of the cone circumscribed around the spheres $ K_i $ and $ K_j $. Prove that the points $ A_{12} $, $ A_{23} $, $ A_{31} $ are collinear.
|
If $ A $ is the vertex of a cone circumscribed around spheres $ K $ and $ L $ of different radii, then there exists exactly one such homothety $ j $ such that $ j(K) = L $. Its center is the point $ A $, and the coefficient is the ratio of the radii of the spheres $ L $ and $ K $.
Without loss of generality, we can assume that the radii of the spheres $ K_1 $, $ K_2 $, $ K_3 $ form an increasing sequence. Let $ s $ and $ t $ be such homotheties that $ s(K_1) = K_2 $ and $ t(K_2) = K_3 $. Then $ ts(K_1) = K_3 $. To solve the problem, it is sufficient to prove that the composition $ ts $ of homotheties $ t $ and $ s $ with coefficients greater than $ 1 $ is a homothety and the centers of homotheties $ t $, $ s $, and $ ts $ are collinear.
Let us choose a coordinate system such that the center of homothety $ s $ is the point $ (0, 0, 0) $, and the center of homothety $ t $ is the point $ (a, 0, 0) $.
Then $ s(x, y, z) = (kx, ky, kz) $, $ t(x, y, z) = (mx - am + a, my, mz) $, where $ k, m > 1 $. We calculate that $ ts(x,y, z) = t(kx, ky, kz) = (mkx- am+a, kmy, kmz) $.
Therefore, $ ts $ is a homothety with the center $ \displaystyle\left(\frac{a(m-1)}{mk-1}, 0, 0\right) $. The centers of homotheties $ s $, $ t $, and $ ts $ are thus collinear.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 591 |
X OM - I - Task 4
Prove that if point $ P $ moves in the plane of triangle $ ABC $, then the triangle $ S_1S_2S_3 $, whose vertices are the centroids $ S_1 $, $ S_2 $ and $ S_3 $ of triangles $ PBC $, $ PCA $ and $ PAB $ does not change its shape or size.
|
If the vertex $ P $ of the triangle $ PAB $ is moved to the point $ P' $ (Fig. 6a), then the centroid $ S_3 $ of this triangle will move to the point $ S $.
The vector $ S_3S $ has the same direction as the vector $ PP' $ and is three times shorter, since these vectors are similar (directly) relative to the point $ N $ in a scale of $ \frac{1}{3} $. Each of the points $ S_1 $ and $ S_2 $ will also move by the same vector, and thus the entire triangle $ S_1S_2S_3 $ will move. It turns out that when the position of the point $ P $ changes, the triangle $ S_1S_2S_3 $ undergoes a parallel shift, and thus does not change in size or shape.
Note. The problem can be generalized by replacing the triangle $ ABC $ with any polygon.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 595 |
LI OM - I - Task 7
Prove that for any positive integer $ n $ and any number $ t\in(\frac{1}{2},1) $ there exist such numbers $ a,b \in (1999,2000) $, that
|
Let $ c = 3999/2 $. We seek numbers $ a $, $ b $ in the form $ a = c+x $, $ b = c-x $, where $ x \in (0, \frac{1}{2}) $. Then
where $ s = 2t - 1 > 0 $. After expanding using the binomial theorem, the last expression takes the form
where $ p(x) $ is some polynomial.
We want to show that for some number $ y \in (0,\frac{1}{2}) $, the number $ w(y) $ is positive. From equation (1), it follows that
Therefore, there exists a number $ y \in (0,\frac{1}{2}) $ such that the expression $ w(y)/y $, and consequently the expression $ w(y) $, has a positive value. This completes the proof.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 597 |
XII OM - II - Problem 5
Prove that if the real numbers $ a $, $ b $, $ c $ satisfy the inequalities
[ (1) \qquad a + b + c > 0, \]
[ (2) \qquad ab + bc + ca > 0, \]
[ (3) \qquad abc > 0, \]
then $ a > 0 $, $ b > 0 $, $ c > 0 $.
|
From inequalities (1) and (2), it can be inferred that at least two of the numbers $a$, $b$, $c$ are positive. For from inequality (1) it follows first that at least one of these numbers is positive, let's say $c > 0$. Indeed, for any $a$ and $b$
and from inequality (2) it follows that
therefore
transferring terms to one side, we obtain from this
since $c > 0$, then $a + b + 2c = (a + b + c) + c > 0$, hence from the above inequality it follows that
In this case, at least one of the numbers $a$ and $b$, for example $b$, is positive.
If, however, two of the numbers $a$, $b$, $c$ are positive, then by inequality (3) the third one must also be positive.
Note. The theorem proved above is a special case of the theorem: If for certain values of $a_1, a_2, \ldots, a_n$ all the elementary symmetric functions (see problem 10) have positive values, then all the values $a_1, a_2, \ldots, a_n$ are positive.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 604 |
XXIX OM - II - Problem 3
Given a sequence of natural numbers $ (a_i) $, such that for every natural number $ n $, the sum of the terms of the sequence that are not greater than $ n $ is not less than $ n $. Prove that for every natural number $ k $, one can choose a finite subsequence from $ (a_i) $ whose sum of terms equals $ k $.
|
We will apply induction with respect to $k$. Consider the case $k = 1$. By assumption, the sum of those terms of the given sequence that are not greater than $1$ (and thus are equal to $1$) is not less than $1$. Therefore, there exists a term in the given sequence that is equal to $1$. Hence, the sum of the terms of a one-term sequence composed of this very term is equal to $1$.
Next, let $k$ be a fixed natural number and assume that for every natural number not greater than $k$, one can select a finite sequence from the given sequence whose sum of terms is equal to $i$. We will prove that from the given sequence, one can select a finite sequence whose sum of terms is equal to $k + 1$.
By the inductive hypothesis, from the given sequence, one can select a finite sequence whose sum of terms is equal to $k$. Let $a_m$ be the smallest term of the given sequence that does not appear in sequence (1).
Since, by the conditions of the problem, the sum of all terms of the given sequence that do not exceed $k + 1$ is not less than $k + 1$, and the sum of the terms of sequence (1) is equal to $k$, some term of the given sequence not greater than $k + 1$ does not appear in sequence (1). Therefore, $a_m \leq k+1$.
If $a_m = 1$, then the sum of the terms of the sequence $(a_{i_1}, a_{i_2}, \ldots, a_{i_r}, a_m)$ is equal to $k + 1$.
If $a_m > 1$, then $1 \leq a_m - 1 \leq k$. Therefore, by the inductive hypothesis, the number $a_m - 1$ is the sum of some terms of the given sequence. All these terms appear in sequence (1), because the smallest term of the given sequence not appearing in (1) is $a_m$. Let
where $s \leq r$. Then the sum of the terms of the sequence
is equal to
By the principle of induction, for every natural number $k$, one can select from the given sequence a finite sequence whose sum of terms is equal to $k$.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 605 |
XLVII OM - II - Problem 6
Inside a parallelepiped, whose edges have lengths $ a $, $ b $, $ c $, there is a point $ P $. Prove that there exists a vertex of the parallelepiped whose distance from point $ P $ does not exceed $ \frac{1}{2}\sqrt{a^2 + b^2 + c^2} $.
|
The walls of the parallelepiped define six planes. Let $ \pi $ be the plane whose distance from point $ P $ is the smallest; if there are two such planes (or more), we choose any one of them and denote it by $ \pi $. Let $ ABCD $ be the face of the parallelepiped contained in the plane $ \pi $ and let $ N $ be the orthogonal projection of point $ P $ onto this plane.
Assume that the four edges of the parallelepiped, connecting the vertices $ A $, $ B $, $ C $, $ D $ with the vertices of the face $ A $ parallel to $ ABCD $, have length $ c $. (We do not lose generality by this, as we can change the notation of the edge lengths if necessary.)
Point $ P $ is no less distant from the face $ A $ than from the face $ ABCD $. Therefore, $ |NP| \leq \frac{1}{2} c $.
We will show that point $ N $ lies within the parallelogram $ ABCD $. Suppose otherwise. Then the segment $ PN $ intersects the edge of the parallelepiped at some point $ X $, belonging to some other face. The distance from point $ P $ to the plane of this face does not exceed the length of segment $ PX $, and thus is less than the length of segment $ PN $, i.e., the distance from point $ P $ to the plane $ \pi $. This, however, contradicts the choice of the plane $ \pi $; the contradiction proves the falsity of the made assumption.
Thus, $ N $ is a point of the parallelogram $ ABCD $, whose sides have lengths $ a $ and $ b $. We repeat the reasoning: among the lines $ AB $, $ BC $, $ CD $, $ DA $, we choose the one whose distance from point $ N $ is the smallest; if there is more than one such line, we choose any one of them.
Assume that this is the line $ AB $ and that $ |AB| = a $ (we change the notation if necessary). Let $ K $ be the orthogonal projection of point $ N $ onto the line $ AB $. Reasoning similarly as before, we notice that $ |KN| \leq \frac{1}{2}b $ and that $ K $ is a point on the segment $ AB $; if the point $ K $ were on the line $ AB $ outside the segment $ AB $, then the segment $ NK $ would intersect another side of the parallelogram $ ABCD $, and thus the distance from point $ N $ to this side would be less than the distance from $ N $ to the line $ AB $ - contrary to the choice of this line.
We can finally assume, without loss of generality, that point $ K $ lies on the segment $ AB $ at a distance no greater from the end $ A $ than from $ B $. Therefore, $ |AK| \leq \frac{1}{2} a $.
The directions of the lines $ AK $, $ KN $, $ NP $ are pairwise perpendicular. The segments $ AK $, $ KN $, $ NP $ are thus three perpendicular edges of a rectangular parallelepiped, whose internal diagonal is the segment $ AP $. The lengths of these segments do not exceed (respectively) the numbers $ \frac{1}{2}a $, $ \frac{1}{2}b $, $ \frac{1}{2}c $. Therefore, ultimately,
point $ A $ is the vertex whose existence needed to be demonstrated.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 607 |
XLIV OM - I - Problem 12
Prove that the polynomial $ x^n + 4 $ is a product of two polynomials of lower degree with integer coefficients if and only if $ n $ is divisible by $ 4 $.
|
\spos{I} Suppose that the polynomial $ x^4 + 4 $ is the product of two polynomials,
with the properties under consideration:
In the product $ F(x)G(x) $, the coefficient of $ x^n $ is $ a_kb_m $, and the constant term equals $ a_0b_0 $. Therefore, the following equations hold:
The numbers $ a_k $, $ b_m $ are integers; thus $ a_k = b_m = 1 $ or $ a_k = b_m = -1 $. Additionally, let us assume:
Let $ \alpha $ be the smallest index such that $ a_\alpha $ is an odd number, and let $ \beta $ be the smallest index such that $ b_\beta $ is an odd number. Since $ a_k = b_m = \pm 1 $, it follows that $ \alpha \leq k $, $ \beta \leq m $. The coefficient of $ x^{\alpha+\beta} $ in the product $ F(x)G(x) $ is equal to
(due to the assumption (2), all symbols make sense even when $ \alpha+ \beta $ exceeds $ k $ or $ m $). In the sum (3), the term $ a_\alpha b_\beta $ is an odd number; each of the remaining terms is an even number (since it is a product of the form $ a_ib_j $, where $ i < \alpha $ or $ j < \beta $, so at least one of the factors $ a_i $, $ b_j $ is even). Therefore, the entire sum (3) is odd.
In the polynomial (1), only $ x^n $ has an odd coefficient. Hence, $ \alpha + \beta = n $, which means the equations $ \alpha = k $ and $ \beta = m $ must hold. According to the definitions of $ \alpha $ and $ \beta $, we obtain the conclusion:
Since $ a_0b_0 = 4 $, it follows that $ a_0 = b_0 = 2 $ or $ a_0 = b_0 = -2 $.
We will show that $ k = m $. Suppose that $ k \neq m $; let, for example, $ k < m $ (such a restriction does not reduce the generality of the considerations due to the symmetry of the roles of $ k $ and $ m $). We write the product of the polynomials $ F(x) $ and $ G(x) $ in the form
where
Given the property (4), all coefficients of the polynomial $ P(x) $ are divisible by $ 4 $. The coefficient of $ x^k $ in the polynomial $ Q(x) $ is $ a_kb_0 $. We previously stated that $ a_k = \pm 1 $ and $ b_0 = \pm 2 $. Therefore, the coefficient of $ x^k $ in the polynomial $ P(x) + Q(x) $ is an even number not divisible by $ 4 $. This is a contradiction, because according to the equalities (1) and (5), the sum $ P(x) + Q(x) $ is a polynomial with a constant value of $ 4 $.
This contradiction proves that $ k = m $; hence, $ n = k + m = 2k $, and consequently
From this, it follows that neither of the polynomials $ F(x) $, $ G(x) $ has real roots, so they must be of even degree. This means that $ k $ is an even number, and thus $ n $ is divisible by $ 4 $.
Conversely, if $ n = 4l $, $ l \in \mathbb{N} $, then the desired factorization exists and has the form
直接输出翻译结果。
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 610 |
XIII OM - I - Problem 5
Prove that all powers of a number whose last eight digits are $12890625$ also end with the digits $12890625$.
|
We need to prove that if $ n $ and $ l $ are integers, with $ n > 0 $, $ l \geq 0 $, then
where $ m $ is a non-negative integer.
First, observe that in the expansion of the left side of the above equality according to the binomial theorem for the power of a binomial, all terms except for $ (12890625)^n $ have a factor of $ 108 $. We can move these terms to the right side and reduce the problem to the following form:
Prove that for every natural number $ n $
where $ k_n $ is an integer, of course dependent on $ n $.
When $ n = 1 $, formula (1) is true, $ k_1 = 0 $.
We will prove that formula (1) is true for $ n = 2 $; to do this, we need to calculate the square of the number $ 12890625 $; however, we do not need to determine all the digits of this square, but only the last $ 8 $ digits. We can, for example, use the fact that the square of the sum of several numbers equals the sum of the squares of these numbers and twice the products of all pairs of these numbers and calculate in this way:
If formula (1) is true for some natural number $ n $, then it is also true for $ n + 1 $, since
where $ k_{n+1} = k_2 + 12890625 k_n $.
Based on the principle of mathematical induction, we conclude from this that formula (1) is true for every natural number $ n $.
Note. The interesting property of the number $ 12890625 $, which we have learned, suggests the question of whether there are more numbers with the same or fewer digits that have this property. Thus, the problem is to find a $ k $-digit number $ A $ such that the last $ k $ digits of each power of this number form the number $ A $.
Just as in the previous solution, we state that if the square of the number $ A $ satisfies the condition of the problem, then each power of this number satisfies it. Therefore, the problem ultimately reduces to finding a $ k $-digit number $ A $ such that
where $ m $ is an integer.
When $ k=1 $, the solution is immediate; the numbers sought are $ 0 $, $ 1 $, $ 5 $, and $ 6 $. Therefore, we will assume in the following that $ k>1 $.
Suppose that equality (2) holds for some natural $ k $-digit number $ A $. We can write this equality in the form
The numbers $ A $ and $ A-1 $ are relatively prime, since any common divisor of these numbers is also a divisor of their difference, which is $ 1 $, so the only such divisor is $ 1 $ or $ -1 $. Since $ A < 10^k $, one of these numbers is divisible by $ 2^k $, and the other by $ 5^k $, i.e., there are only two cases:
a) $ A = 2^k \cdot x $ and $ A-1 = 5^k \cdot y $
b) $ A = 5^k \cdot z $ and $ A-1 = 2^k \cdot u $
In case a)
and in case b)
Conversely, if a pair of natural numbers $ (x, y) $ satisfies equation (4), then the number $ A = 2^k \cdot x $ satisfies condition a), and therefore also condition (2). To be a solution to the problem, this number must also have $ k $ digits, i.e., it must satisfy the condition
Similarly, if a pair of natural numbers $ (z, u) $ satisfies equation (5), then the number $ A = 5^k \cdot z $ is a solution to the problem if it satisfies the condition
To investigate whether equations (4) and (5) have such solutions, we will consider the general equation
where $ a $ and $ b $ are integers greater than $ 1 $ and relatively prime.
We will prove that there exists one and only one pair $ (x_0, y_0) $ of positive integers satisfying this equation and the condition $ x_0 < b $. Indeed, consider the remainders of the division of the positive numbers $ ax_k - 1 $ by $ b $, where $ x_k $ takes the values $ 1, 2, \ldots, b $. Among these remainders, there are no equal ones; if the equalities
were to hold, where $ x_1 $ and $ x_2 $ are two different numbers among the numbers $ 1, 2, \ldots, b $ and $ q_1 $ and $ q_2 $ are integers, then the equality
would hold, so the number $ a (x_1 - x_2) $ would be divisible by $ b $, which is impossible since $ a $ and $ b $ are relatively prime and the difference $ x_1 - x_2 $ is less in absolute value than $ b $. One of these $ b $ remainders must therefore be equal to $ 0 $; let the corresponding value of $ x_k $ be $ x_0 $. The number $ x_0 $ cannot be equal to $ b $, since $ ab - 1 $ is not divisible by $ b $, so $ x_0 < b $.
The positive integer $ \frac{ax_0 - 1}{b} $ will be denoted by $ y_0 $, so
If $ x $ and $ y $ denote any integers satisfying equation (8), then by subtracting equations (8) and (9) we obtain
Since $ a $ and $ b $ are relatively prime, $ x - x_0 $ is divisible by $ b $, so $ x - x_0 = b \cdot t $, where $ t $ is an integer, and $ y - y_0 = at $, from which
All solutions of equation (8) in integers $ x $, $ y $ are thus obtained from formulas (10) by substituting integers for $ t $. These numbers are positive and satisfy the condition $ x < b $ if and only if $ t = 0 $, i.e., if $ x = x_0 $, $ y = y_0 $.
From the above, it follows that there exists one and only one pair of integers $ (x_0, y_0) $ satisfying equation (4) and the conditions $ x_0 > 0 $, $ y_0 > 0 $, $ x_0 < 5^k $, so $ A = 2^k \cdot x_0 < 10^k $; the number $ A $ is thus at most $ k $-digit. However, it may happen that $ A $ has fewer than $ k $ digits. We agree that in such a case we will prepend the appropriate number of zeros to the left of $ A $ instead of the missing digits and consider that we have obtained a $ k $-digit solution to the problem. With this agreement, the problem has for every $ k > 1 $ one solution of the form $ A = 2^k \cdot x $ ($ x - 1 $ natural). Similarly, equation (5) leads to a second solution of the form $ A = 5^k \cdot z $ ($ z - 1 $ natural).
For example, for $ k = 3 $ we have solutions $ A_1 = 625 $ and $ A_2 = 376 $, for $ k = 4 $ the solutions are $ A_1 = 0625 $ and $ A_2 = 9376 $, for $ k = 9 $ one solution is $ A_1 = 212890625 $; let the reader find the second solution themselves. To find it without lengthy calculations, one needs to familiarize themselves with the theory of linear Diophantine equations in two variables; the reader will find it, for example, in the books of Prof. W. Sierpiński, Theory of Numbers and Theoretical Arithmetic.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 611 |
XV OM - III - Task 6
Given is a pyramid SABCD, whose base is a convex quadrilateral $ABCD$ with perpendicular diagonals $AC$ and $BD$, and the orthogonal projection of vertex S onto the base is point O, the intersection of the diagonals of the base. Prove that the orthogonal projections of point O onto the lateral faces of the pyramid lie on a circle.
|
Let $M$, $N$, $P$, $Q$ be the orthogonal projections of point $O$ onto the planes $ASB$, $BSC$, $CSD$, $DSA$ (Fig. 23).
The plane $SOM$ is perpendicular to the plane of quadrilateral $ABCD$ and to the plane $ASB$, as it contains the perpendiculars $SO$ and $OM$ to these planes. Therefore, the plane $SOM$ is perpendicular to the line of intersection $AB$ of the planes $ABCD$ and $ASB$ and intersects it at point $M$ on the line $SM$, with $AB \perp OM$, i.e., $M$ is the orthogonal projection of point $O$ onto the line $AB$. Similarly, the lines $SN$, $SP$, and $SQ$ intersect the lines $BC$, $CD$, and $DA$ at points $N$, $P$, and $Q$, which are the orthogonal projections of point $O$ onto these lines. The points $M$, $N$, $P$, $Q$ lie on a certain circle $\alpha$. The points $M$, $N$, $M$, $N$ also lie on a circle. Indeed, the segment $OB$ is seen from each of these points at a right angle, so they lie on a sphere (i.e., on the surface of a sphere) with diameter $OB$; they also lie in the plane $SMN$, so they lie on the line of intersection of the sphere with the plane, i.e., on a certain circle $\beta$. The circles $\alpha$ and $\beta$ lie in different planes and have two common points $M$ and $N$; therefore, there exists a sphere $\gamma$ passing through both these circles; the points $M$, $N$, $P$, $Q$, $M$, $N$ lie on the sphere $\gamma$. Similarly, the points $M$, $N$, $P$, $Q$, $N$, $P$ lie on a certain sphere $\gamma$, and the points $M$, $N$, $P$, $Q$, $P$, $Q$ lie on a certain sphere $\gamma$. The surfaces $\gamma$, $\gamma$, $\gamma$ are identical, since $\gamma$ and $\gamma$ pass through the vertices of the tetrahedron $M$, and $\gamma$ and $\gamma$ through the vertices of the tetrahedron $M$. Therefore, the points $M$ lie on the sphere $\gamma$. On the other hand, the points $M$, $N$, $P$, $Q$ lie on a sphere with diameter $OS$, since from each of these points the segment $OS$ is seen at a right angle; this sphere is different from the sphere $\gamma$, as it does not pass through the points $M$, $N$, $P$, $Q$, being tangent to the plane $ABCD$ at point $O$. The points $M$, $N$, $P$, $Q$ therefore lie on the line of intersection of two different spheres, i.e., on a circle.
Note. The proof above can be significantly shortened by considering the stereographic projection of the sphere with diameter $OS$ from point $S$ onto the plane $ABCD$. It is known that in such a projection, a circle lying in the plane corresponds to a circle on the sphere. The circle passing through the points $M$, $N$, $P$, $Q$ corresponds to a circle on the sphere, which passes through the corresponding points of the sphere, i.e., through the points $M$, $N$, $P$, $Q$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 612 |
XVII OM - I - Problem 5
Given positive numbers $ p $ and $ q $. Prove that a rectangular prism, in which the sum of the edges equals $ 4p $, and the surface area equals $ 2q $, exists if and only if $ p^2 \geq 3q $.
|
a) Suppose that the desired rectangular parallelepiped exists. Then there are numbers $ x $, $ y $, $ z $ satisfying the equations
In such a case, the numbers $ p $ and $ q $ satisfy the inequality $ p^2 \geq 3q $, since from (1) and (2) it follows that
b) Suppose that given positive numbers $ p $ and $ q $ satisfy the inequality $ p^2 \geq 3q. $
We will show that there are then positive numbers $ x $, $ y $, $ z $ satisfying equations (1) and (2). If $ \sigma $ denotes a positive number less than $ \frac{p}{3} $, then the numbers defined by the formulas
are positive and satisfy equation (1). From equality (3) it follows that
Taking $ \sigma = \frac{1}{3} \sqrt{p^2-3q} $ we state that $ 0 < \sigma < \frac{p}{3} $ and that
The numbers $ x $, $ y $, $ z $ determined in this way also satisfy equation (2), so the rectangular parallelepiped with edges $ x $, $ y $, $ z $ has the desired properties.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 613 |
LVI OM - II - Task 3
In the data space, there are $ n $ points ($ n\geq 2 $) of which no four lie in the same plane. Some of these points have been connected by segments. Let $ K $ be the number of segments drawn ($ K\geq 1 $), and $ T $ the number of triangles formed. Prove that
|
Let us number from 1 to $ m $ those points from which at least one segment starts, and assume that from the point numbered $ i $ ($ i=1,2,\ldots,m $) exactly $ k_i $ segments start. Then $ k_i>0 $. Furthermore, let $ t_i $ ($ i=1,2,\ldots,m $) denote the number of those triangles, one of whose vertices is the point numbered $ i $. Then
The set $ A_i $ of those points that are connected by a segment to the point numbered $ i $ contains exactly $ k_i $ elements. Moreover, the number of triangles having a vertex at the point numbered $ i $ is the same as the number of segments with endpoints in the set $ A_i $. Therefore,
On the other hand, the number of these segments is less than the number of all drawn segments. Therefore,
Multiplying the inequalities (2) and (3) side by side, we get $ t_i^2 < \frac{1}{2}k_i^2K $. From this and equation (1), we obtain
Raising the last inequality to the power of two on both sides, we get the thesis.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 614 |
XL OM - I - Task 1
Prove that if the numbers $ k $, $ n $ ($ k < n $) are coprime, then the number $ \binom{n-1}{k-1} $ is divisible by $ k $.
|
Let's denote:
These are integers. The equality
holds, that is, $ M \cdot n = N \cdot k $. Since $ k $ is relatively prime to $ n $, it must be a divisor of $ M $; which is exactly what we needed to prove.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 615 |
XLVIII OM - I - Problem 2
Point $ P $ lies inside the parallelogram $ ABCD $, and the equality $ |\measuredangle ABP| = |\measuredangle ADP| $ holds. Prove that $ |\measuredangle PAB| = |\measuredangle PCB| $.
|
We translate the triangle $ ADP $ in parallel such that the image of side $ AD $ is segment $ BC $. Let the image of vertex $ P $ be denoted by $ Q $. Therefore, we have the equality $ |\measuredangle ADP| = |\measuredangle BCQ| $ (Figure 1). Segments $ AB $ and $ PQ $ are parallel, so $ |\measuredangle ABP| = |\measuredangle QPB| $.
At the same time, $ |\measuredangle ABP| = |\measuredangle ADP| $ (according to the condition of the problem). Therefore, $ |\measuredangle BCQ| = |\measuredangle QPB| $.
Segment $ BQ $ is thus visible at the same angle from points $ P $ and $ C $, which lie on the same side of the line $ BQ $. Conclusion: points $ B $, $ P $, $ C $, $ Q $ lie on the same circle. Hence, given that points $ C $ and $ Q $ lie on the same side of the line $ PB $, it follows that $ |\measuredangle PCB| = |\measuredangle PQB| $.
It remains to note that quadrilateral $ PABQ $ is a parallelogram, so $ |\measuredangle PQB| = |\measuredangle PAB| $. We obtain the equality $ |\measuredangle PAB| = |\measuredangle PCB| $, which was to be proven.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 624 |
XII OM - II - Task 2
Prove that all altitudes of a tetrahedron intersect at one point if and only if the sums of the squares of opposite edges are equal.
|
The solution to the problem will help us observe that the segment connecting the midpoints of two edges of a tetrahedron belonging to the same face is equal to half of the third edge of that face (and is parallel to it). Let's denote the midpoints of the edges of the tetrahedron $ABC$ as shown in Fig. 13 with the letters $M$, $N$, $P$, $Q$, $R$, $S$.
We will first prove that the sums of the squares of opposite edges of the tetrahedron $ABCD$ are equal if and only if the segments $MP$, $NQ$, and $RS$, i.e., the segments connecting the midpoints of opposite edges, are equal. Indeed, the equality
is equivalent to the equality
which in turn is equivalent to the equality
since the quadrilaterals $MNPQ$ and $MRPS$ are parallelograms, so in each of them the sum of the squares of the diagonals is equal to twice the sum of the squares of two adjacent sides. Finally, this last equality can be replaced by the equality
Similarly, the equality $AC^2 + BD^2 = AB^2 + CD^2$ is equivalent to the equality $MP = RS$.
The equality of the segments $MP$, $NQ$, and $RS$ holds if and only if the parallelograms $MNPQ$, $MRPS$, $NRQS$ are rectangles, and thus if and only if each pair of opposite edges of the tetrahedron are perpendicular.
We will then prove that if two opposite edges of a tetrahedron are perpendicular, then the altitudes of the tetrahedron drawn from the endpoints of one of these edges intersect.
Let, for example, $AB \bot DC$ (Fig. 14). Draw a perpendicular $CK$ to $AB$; since $CK \bot AB$ and $CD \bot AB$, the line $AB$ is perpendicular to the plane $CKD$. It follows that the altitudes $DH$ and $CG$ of the triangle $CKD$ are altitudes of the tetrahedron, since the line $DH$ is perpendicular to $CK$ and to $AB$, and thus is perpendicular to the plane $ABC$, and similarly the line $CG$ is perpendicular to the plane $ABD$.
Conversely, if the altitudes $DH$ and $CG$ of the tetrahedron intersect, then the line $AB$ is perpendicular to the plane containing $DH$ and $CG$, since $AB \bot DH$ and $AB \bot CG$, and thus the line $AB$ is also perpendicular to the line $CD$.
Therefore, the perpendicularity of each pair of opposite edges of the tetrahedron holds if and only if any two altitudes of the tetrahedron intersect.
Finally, we will prove that if any two altitudes of the tetrahedron intersect, then all four altitudes pass through one point.
Suppose that the altitudes $h_B$ and $h_C$ of the tetrahedron drawn from the vertices $B$ and $C$ lie in the plane $\alpha$, and the altitudes $h_C$ and $h_A$ drawn from the vertices $C$ and $A$ - in the plane $\beta$. Then the point of intersection of the altitudes $h_A$ and $h_B$ lies on the line of intersection of the planes $\alpha$ and $\beta$, i.e., on the line $h_C$; similarly, we conclude that it lies on the line $h_D$, so all the altitudes intersect at one point.
The theorem has been proved.
Note. A tetrahedron in which all four altitudes pass through one point (orthocenter) is called an orthocentric tetrahedron. In the above proof, we have stated that it can also be defined by any of the following properties:
1) perpendicularity of opposite edges,
2) equality of segments connecting the midpoints of opposite edges,
3) intersection of any two altitudes.
We will add one more property to this. A parallelepiped can be described on any tetrahedron, i.e., a parallelepiped can be constructed such that the edges of the tetrahedron are the diagonals of the faces of the parallelepiped (Fig. 15). A tetrahedron is orthocentric if and only if it has the property:
4) the parallelepiped described on the tetrahedron has all edges equal (i.e., its faces are rhombuses).
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 629 |
XXVII OM - I - Zadanie 12
Ciąg $ (x_n) $ określony jest wzorami $ x_0 = 25 $, $ x_n = x_{n-1} + \frac{1}{x_{n-1}} $ ($ n = 1, 2, \ldots $) Dowieść, że $ x_n > 1975 $ dla $ n > 1950000 $.
|
Udowodnimy ogólniejsze
Twierdzenie. Jeżeli $ \displaystyle x_0 > \frac{1}{2} $ i $ x_{n} = x_{n-1} + \frac{1}{x_{n-1}} $ dla $ n = 1, 2, \ldots $, to dla każdej liczby naturalnej $ N $ zachodzi nierówność
Dowód. Wyrazy ciągu $ (x_{n}) $ są oczywiście liczbami dodatnimi. Wobec tego $ x_n = x_{n-1} + \frac{1}{x_{n-1}} > x_{n-1} $ dla $ n=1, 2, \ldots $ tzn. ciąg
$ (x_{n}) $ jest rosnący.
Podnosząc obustronnie równość $ x_{n} = x_{n-1}+ \frac{1}{x_{n-1}} $ do kwadratu otrzymujemy
i stąd $ x_n^2 - x_{n-1}^2 > 2 $. Podstawiając tu kolejno $ n = 1, 2, \ldots, N $ i dodając stronami otrzymane nierówności uzyskujemy $ x^2_N - x^2_0 > 2N $, czyli
Z drugiej strony, ponieważ ciąg $ (x_{n}) $ jest rosnący i jego wyrazy są liczbami dodatnimi, więc
Wobec tego z (2) otrzymujemy
Podstawiając tu kolejno $ n = 1,2, \ldots, N $ i dodając stronami otrzymane nierówności uzyskujemy
Wobec tego
ponieważ $ x_0^2 > \frac{1}{2} $.
Stąd otrzymujemy $ x_N - \frac{1}{2x_0} < \sqrt{x_0^2 + 2N} $ i wobec tego na mocy (3) mamy tezę twierdzenia.
Jeżeli $ x_0 = 25 $ i $ N > 1950000 $, to z (1) wynika, że $ x_N > \sqrt{x_0^2}+ 2N > \sqrt{25^2 + 2 \cdot 1950000} = \sqrt{1975^2} = 1975 $, co daje rozwiązanie zadania.
Zauważmy jeszcze, że dla $ x_0 = 25 $ i $ N= 1950000 $ z (1) wynika, że
$ x_N < \frac{1}{2 \cdot 25} + \sqrt{25^2 + 2 \cdot 1950000} = \frac{1}{50} + 1975 = 1975,02 $.
Wzór (1) pozwala więc wyznaczyć liczbę $ x_{N} $ z błędem mniejszym od $ 0,01 $. Mianowicie, z powyższego wynika, że
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 630 |
XLIII OM - III - Problem 6
Prove that for every natural number $ k $, the number $ (k!)^{k^2+k+1} $ is a divisor of the number $ (k^3)! $.
|
For every pair of natural numbers $ n,l \geq 1 $, the following equality holds:
Let us fix natural numbers $ n,m \geq 1 $. Substitute $ l = 1,2,\ldots,m $ into (1) and multiply the resulting equations side by side:
We transform the left side of equation (2):
and after reducing the repeated factors in the numerators and denominators:
Equating the right sides of equations (2) and (3) gives the equality:
Thus, for every pair of natural numbers $ n,m \geq 1 $, the quotient $ \frac{(mn)!}{(n!)^m m!} $ is a natural number.
Let $ k $ be a given natural number. Substituting into (4) first $ n=m=k $, and then $ n=k $, $ m = k^2 $, we get natural numbers on the right side, which we will denote by $ A_1(k) $ and $ A_2(k) $, respectively:
Hence, by multiplying the sides:
Thus, indeed, the number $ (k^3)! $ is divisible by $ (k!)^{k^2+k+1} $.
Note: Let us assume in (4): $ n = k $, $ m=k^j $, and denote the value of the expression (4) obtained by $ A_j(k) $:
Take any natural number $ r \geq 1 $ and multiply the sides of equations (5) applied successively for $ j = 1,\ldots,r $. The result of this operation is:
which, after reducing the common factors in the numerator and denominator, is:
We have thus obtained a strengthened version of the theorem: The number $ (k!)^{1+k+k^2 + \ldots+k^r} $ is a divisor of the number $ (k^{r+1})! $.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 632 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.