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LVII OM - III - Problem 3
Given a convex hexagon $ABCDEF$, in which $AC = DF$, $CE = FB$ and $EA = BD$. Prove that the lines connecting the midpoints of opposite sides of this hexagon intersect at one point.
|
Let $P, Q, R$ be the midpoints of the diagonals $AD, BE, CF$, respectively.
First, assume that two of the points $P, Q, R$ coincide; let, for example, $P = Q$ (Fig. 1).
Then the quadrilateral $ABDE$ is a parallelogram. Moreover, triangle $ACE$ is congruent to triangle $DFB$, which implies that
$\measuredangle EAC = \measuredangle BDF$. This equality, together with the obtained parallelism $BD || EA$, proves that the segments
$AC$ and $DF$ are parallel. Furthermore, these segments are of equal length, so quadrilateral $ACDF$ is a parallelogram. Point
$R$ coincides, therefore, with points $P$ and $Q$, which means that it is the center of symmetry of the hexagon $ABCDEF$. It remains to note
that in this case, the lines connecting the midpoints of opposite sides of the hexagon $ABCDEF$ pass through its center of symmetry.
om57_3r_img_1.jpg
om57_3r_img_2.jpg
Next, assume that the points $P, Q, R$ are distinct. Let $M$ and $N$ be the midpoints of segments $AB$ and $DE$, respectively (Fig. 2).
From the converse of Thales' theorem, it follows that segments $PN$ and $AE$ are parallel, and moreover,
$PN = \frac{1}{2}AE$. Similarly, we obtain the relations $QM = \frac{1}{2}AE$, $PM = \frac{1}{2}BD$, and $QN = \frac{1}{2}BD$.
From these relations and the equality $AE = BD$, we conclude that $PN = QM = PM = QN$, which means that quadrilateral $MPNQ$ is a rhombus.
Thus, line $MN$ is the perpendicular bisector of segment $PQ$.
Similarly, we prove that the other two lines connecting the midpoints of opposite sides of the given hexagon are the perpendicular bisectors
of segments $QR$ and $RP$. Therefore, the lines connecting the midpoints of opposite sides of the hexagon $ABCDEF$ have a common point, which is
the center of the circle circumscribed around triangle $PQR$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 633 |
XLVI OM - I - Problem 9
Let $ a $ and $ b $ be real numbers whose sum is equal to 1. Prove that if $ a^3 $ and $ b^3 $ are rational numbers, then $ a $ and $ b $ are also rational numbers.
|
We raise the equality $ a + b = 1 $ to the second and third powers on both sides and obtain the relations: $ a^2 +2ab + b^2 = 1 $, that is,
and $ a^3 + 3a^2b + 3ab^2 + b^3 = 1 $, that is,
If, therefore, the numbers $ a^3 $ and $ b^3 $ are rational, then from equation (2) it follows that the product $ ab = (1-a^3-b^3)/3 $ is a rational number, and from equation (1) the sum $ a^2 + b^2 $ is also a rational number. In this case, the number $ a^2 + ab + b^2 $ is also rational - and different from zero, as can be seen, for example, from the transformation
The numbers $ a^3 $ and $ b^3 $ are rational, so their difference $ a^3 - b^3 $ is also rational. This leads to the conclusion that the difference
is a rational number. In combination with the condition $ a + b = 1 $, this gives the desired conclusion: the numbers
are rational.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 636 |
XIV OM - I - Problem 12
In a circle with center $ O $ and radius $ r $, a regular pentagon $ A_1A_2A_3A_4A_5 $ is inscribed, and on the smaller arc with endpoints $ A_1 $, $ A_5 $, a point $ M $ is chosen. Prove that
|
The equality (1) we need to prove is a relationship between the lengths of certain chords of a circle. Another relationship of the same kind is the well-known Ptolemy's theorem: The product of the diagonals of a cyclic quadrilateral equals the sum of the products of its opposite sides1). It suggests using this theorem to prove equality (1). In the considered figure, there are $10$ such cyclic quadrilaterals, one of whose vertices is $M$, and the other three are vertices of the pentagon $A_1A_2A_3A_4A_5$. Given the form of equality (1), we will choose those of these ten quadrilaterals in which segments $MA_1$, $MA_3$, $MA_5$ appear only as sides, and segments $MA_2$ and $MA_4$ only as diagonals, or vice versa. These are the quadrilaterals
Let $a$ denote the length of a side of the given pentagon and $b$ - the length of its diagonal. Applying Ptolemy's theorem to the mentioned $5$ quadrilaterals, we obtain the following equalities, which can be easily written without even looking at the diagram:
Adding these equalities:
and after dividing by $2a + b$, we obtain the desired equality (1).
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 637 |
XXII OM - I - Problem 12
Prove that every convex polyhedron has a triangular face or a trihedral angle.
|
Assume that there exists a convex polyhedron $W$ without any triangular faces or trihedral angles. Let $w$ be the number of vertices, $k$ the number of edges, and $s$ the number of faces of this polyhedron, and let $\varphi$ be the sum of the measures of all dihedral angles at the vertices of the polyhedron $W$. Since at least four edges emanate from each vertex, and each edge connects two vertices, we have $k \geq \frac{1}{2} \cdot 4w = 2w$. Since each face has at least four edges, and for $n \geq 4$ the sum of the interior angles of an $n$-gon is $\geq 2\pi$, we have $\varphi \geq 2\pi s$. Inside each face $S_i$, choose a point $O_i$ and connect it to all vertices of that face. As a result, we obtain a division of the surface of the polyhedron into $2k$ triangles, since each edge is a side of exactly two triangles. Let $\Psi$ be the sum of the measures of the angles in all triangles. Then $4\pi w \leq 2k\pi = \Psi = \varphi + 2\pi s \leq 2\varphi$, and thus $\varphi < 2\pi w$. On the other hand, since the polyhedron $W$ is convex, $\varphi < 2\pi w$. The obtained contradiction proves that there does not exist a polyhedron with the assumed properties.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 639 |
XXXIII OM - II - Problem 6
Given is a finite set $ B $ of points in space, such that any two distances between points of this set are different. Each point of the set $ B $ is connected by a segment to the nearest point of the set $ B $. In this way, we obtain a set of segments, one of which (chosen arbitrarily) is painted red, and all the remaining segments are painted green. Prove that there exist two points of the set $ B $ that cannot be connected by a broken line composed of segments painted green.
|
Suppose the segment $ A_1A_2 $ is red, and the rest are green. If there existed a broken line composed of green segments connecting points $ A_1 $ and $ A_2 $, then there would be points $ A_3, \ldots, A_n $ being the successive endpoints of the segments forming this broken line. The notation $ A_i \to A_j $ is read as "the nearest point to $ A_i $ in set $ B $ is $ A_j $." From the fact that the segment $ \overline{A_iA_j} $ is drawn, it follows that $ A_i \to A_j $ or $ A_j \to A_i $. Given the assumption that any two distances between points in set $ B $ are different, it cannot simultaneously be true for $ k \ne j $ that $ A_i \to A_j $ and $ A_i \to A_k $. Therefore, it must be either $ A_1 \to A_2, A_2 \to A_3, \ldots, A_{n-1} \to A_n, A_n \to A_1 $, or $ A_1 \to A_n, A_n \to A_{n-1}, \ldots, A_3 \to A_2, A_2 \to A_1 $.
In both cases, the distances between successive points would satisfy contradictory inequalities.
or
Given that the considered segments do not have common interior points, the above solution implies a stronger thesis than the one stated in the problem: There exist two points in set $ B $ that cannot be connected by a broken line composed of segments contained in the sum of the green segments.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 642 |
LI OM - II - Task 3
On an $ n \times n $ chessboard, $ n^2 $ different integers are placed, one on each square. In each column, the square with the largest number is painted red. A set of $ n $ squares on the chessboard is called admissible if no two of these squares are in the same row or the same column. Among all admissible sets, the set for which the sum of the numbers placed on its squares is the largest is chosen.
Prove that in such a selected set, there is a red square.
|
We will conduct an indirect proof.
Assume that in the chosen admissible set (denoted by $ A $) there is no red field.
We define a sequence of fields $ D_1, E_1, D_2, E_2, \ldots $ of the given chessboard according to the rule described below. (For $ n = 8 $, the process of selecting fields $ D_1, E_1, D_2, E_2, \ldots $ is shown in Figures 1 and 2. Red fields are shaded, while fields from set $ A $ are marked with a circle.)
om51_2r_img_3.jpg
om51_2r_img_4.jpg
The field $ D_1 $ is any field from set $ A $. Let $ E_1 $ be the red field lying in the same column as $ D_1 $.
Then we consider the field $ D_2 $ from set $ A $, which lies in the same row as $ E_1 $. By $ E_2 $ we denote the red field lying in the same column as $ D_2 $. We continue this process. (In general: the red field $ E_i $ lies in the same column as $ D_i $, while the field $ D_{i+1} $ is a field from set $ A $ lying in the same row as $ E_i $.)
om51_2r_img_5.jpg
In this way, we obtain a sequence of fields $ D_1, E_1, D_2, E_2, D_3, \ldots $, in which each field uniquely determines the next one. Since there are finitely many red fields, we will find such positive integers $ i $, $ t $, for which $ E_i = E_{i+t} $ and $ E_i \neq E_{i+s} $ for $ s = 1, 2, \ldots, t-1 $. (In our example shown in Figures 1 and 2, $ E_2 = E_5 $, so $ i = 2 $, $ t = 3 $.)
We remove from set $ A $ the fields $ D_{i+1}, D_{i+2}, \ldots, D_{i+t} $ and replace them with the fields $ E_{i+1}, E_{i+2}, \ldots, E_{i+t} $ (Fig. 3). The set obtained in this way is denoted by $ B $.
From the definition of the sequence $ D_1, E_1, D_2, E_2, D_3, \ldots $ it follows that the fields
find themselves in the same columns as the fields
and in the same rows as the fields
Therefore, set $ B $ is admissible. Moreover, the numbers written on the fields $ E_{i+1}, E_{i+2}, \ldots, E_{i+t} $ are greater than the numbers written on the fields $ D_{i+1}, D_{i+2}, \ldots, D_{i+t} $, respectively. It follows that the sum of the numbers written on the fields of set $ B $ is greater than the sum of the numbers written on the fields of set $ A $. This, however, contradicts the assumption that $ A $ has the largest sum of numbers among all admissible sets.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 644 |
XXIII OM - III - Problem 1
Polynomials $ u_1(x) = a_ix + b_i $ ($ a_i, b_i $ - real numbers; $ i = 1, 2, 3 $) satisfy for some natural $ n > 2 $ the equation
Udowodnić, że istnieją takie liczby rzeczywiste $ A, B, c_1, c_2, c_3 $, że $ u_i(x)=c_i(Ax+B) $ for $ i = 1, 2, 3 $.
Prove that there exist real numbers $ A, B, c_1, c_2, c_3 $ such that $ u_i(x)=c_i(Ax+B) $ for $ i = 1, 2, 3 $.
|
If $ a_1 = a_2 = 0 $, then the polynomials $ u_1 $ and $ u_2 $ are constant. Therefore, the polynomial $ u_3 $ is also constant, i.e., $ a_3 = 0 $. In this case, it suffices to take $ c_i = b_i $ for $ i = 1, 2, 3 $ and $ A = 0 $ and $ B = 1 $.
Let then at least one of the numbers $ a_1, a_2 $ be different from zero, for example, $ a_1 \ne 0 $. Denoting $ y = a_1x + b_1 $ we get $ u_j(x) = \displaystyle \frac{a_j}{a_1} y + \frac{b_ja_1 - a_jb_1}{a_1} $ for $ j = 2, 3 $, i.e., $ u_j(x) = A_jy + B_j $, where $ A_j = \displaystyle \frac{a_j}{a_1} $, $ \displaystyle \frac{b_ja_1 - a_jb_1}{a_1} $, $ j = 2, 3 $. Equation (1) then takes the form
By comparing the constant terms and the coefficients of $ y $ and $ y^n $ (we use here the assumption $ n > 2 $) on both sides of equation (2), we get
If $ B_2 = 0 $, then from (3) it follows that $ B_3 = 0 $ and therefore $ b_ja_1 - a_jb_1 = 0 $ for $ j = 2, 3 $, i.e., $ b_j = \displaystyle \frac{a_j}{a_1} b_1 $. It suffices then to take $ c_1 = 1 $, $ c_j = \displaystyle \frac{a_j}{a_1} $ for $ j = 2, 3 $, $ A = a_1 $, $ B = b_1 $.
If $ B_2 \ne 0 $, then from (3) it follows that $ B_3 = 0 $ and dividing both sides of (4) by (3) we get $ \displaystyle \frac{A_2}{B_2} = \frac{A_3}{B_3} $. Raising both sides of the last inequality to the $ n $-th power and using (3) we obtain $ A^n_2 = A^n_3 $. This contradicts equation (5). Therefore, this case cannot occur.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 645 |
XXXVI OM - III - Problem 5
Let $ P $ be a polynomial in two variables such that for every real number $ t $ the equality $ P(\cos t, \sin t) = 0 $ holds. Prove that there exists a polynomial $ Q $ such that the identity holds
保留了源文本的换行和格式，但最后一句“保留了源文本的换行和格式”是中文，应该翻译成英文如下：
Preserving the line breaks and format of the source text.
|
Let's arrange the polynomial $ P(x,y) $ in decreasing powers of the variable $ x $. $ P(x,y) = P_n(y) \cdot x^n + P_{n-1}(y) \cdot x^{n-1} + \ldots + P_1(y) \cdot x + P_0(y) $, where $ P_n, P_{n-1}, \ldots, P_1, P_0 $ are polynomials in the variable $ y $. Treating $ y $ as a fixed value, we can compute the quotient and remainder when dividing the polynomial $ P $ by the polynomial $ x^2 + y^2 - 1 $ (treated as a polynomial in the variable $ x $).
We will obtain
where $ Q_i $, $ R_j $ are polynomials in the variable $ y $. Substituting $ x = \cos t $, $ y = \sin t $ in the above equation, we get
since $ \cos^2 t + \sin^2 t - 1 = 0 $ for any $ t $. Given the assumption about the polynomial $ P $, we have
for any real $ t $. Substituting $ \pi - t $ for $ t $, we get
Adding this to the previous equation, we obtain $ R_0(\sin t) = 0 $. The polynomial $ R_0 $ takes the value $ 0 $ for infinitely many arguments, since $ \sin t $ can be any number in the interval $ \langle -1 ; 1 \rangle $. It follows that $ R_0 $ is the zero polynomial. Therefore, we have $ R_1(\sin t) \cdot \cos t = 0 $ for every $ t $, from which we similarly conclude that $ R_1 $ is the zero polynomial. Thus, $ P(x,y) = Q(x,y)(x^2 + y^2 - 1) $, where $ Q(x,y) = Q_{n-2}(y)x^{n-2} + \ldots + Q_n(y) $.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 646 |
XXXIV OM - I - Problem 6
Prove that for any polynomial $ P $, the polynomial $ P\circ P \circ P \circ \ldots \circ P(x)-x $ is divisible by $ P(x)-x $.
|
Putting $ P(x)= a_nx^n + a_{n-1}x^{n-1} + \ldots +a_1x+a_0 $ we get
Since each term of the last sum can be factored:
thus
Substituting in the last relation $ P(x) $ for $ x $ we get $ P \circ P(x)-P(x) | P \circ P \circ P(x) - P \circ P(x) $, from which by the transitivity of the divisibility relation it follows that $ P(x)-x|P \circ P \circ P(x)-P \circ P(x) $. Hence for $ k= 1, 2, \ldots $ we get
The polynomial $ \underbrace{P \circ P \circ \ldots \circ P}_n (x) - x $ can be represented as a sum
By the previous considerations, each term of this sum is divisible by $ P(x) - x $, from which it follows that $ P(x) - x|\underbrace{P \circ P \circ \ldots \circ P}_n(x) - x $.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 647 |
XXXVII OM - II - Problem 2
In a chess tournament, 66 players participate, each playing one game against every other player, and the matches take place in four cities. Prove that there exists a trio of players who play all their games against each other in the same city.
|
Let's choose one of the players, let's call him $Z_1$. He has to play $65$ games, so he plays at least $17$ games in one city. Let's denote this city by $M_1$. Consider the opponents of $Z_1$ in the matches played in $M_1$. There are at least $17$ of them. If there is a pair among them who play a game against each other also in $M_1$, then together with player $Z_1$ they form a trio playing a "triangle of matches" in one city, and the theorem is proved. Therefore, let's assume that the entire group of $17$ (or more) chess players play all their matches against each other in the remaining three cities. Choose one from this group and call him $Z_2$. He has at least $16$ opponents in this group, so he must play with at least $6$ of them in one city, which we will denote by $M_2$. If any pair among them plays their game also in $M_2$, this forms a "triangle of matches" in $M_2$. So let's assume that this is not the case; this means that the considered group of $6$ (or more) chess players play their matches against each other in the remaining two cities. We repeat the same reasoning. Choose one player from this group and call him $Z_3$, and note that having at least $5$ opponents in this group, he must play with at least $3$ of them in one city $M_3$. If there is a pair among them who play their game in $M_3$, then together with player $Z_3$ they form a trio playing a "triangle of matches" in $M_3$. Otherwise, the entire group of $3$ (or more) chess players play their matches against each other in the last remaining city $M_4$. Thus, the theorem is proved.
Note. The given proof is inductive in nature, consisting of several steps, each of which repeats the same reasoning pattern. This observation leads to the following natural generalization; it is convenient to formulate it in the language of so-called graph theory.
Let $(n_k)$ be a sequence of numbers defined by the recurrence relation $n_1 = 2$, $n_k = kn_{k-1} + 1$ for $k \geq 2$. Suppose that $G$ is a set of $n$ points in space, $n > n_k$, and that each of the segments connecting the points of set $G$ has been colored with one of $k$ colors. Then there are three points in set $G$ such that the triangle with vertices at these points has all sides of the same color.
(The initial terms of the sequence $(n_k)$ are the numbers $2$, $5$, $16$, $65$, which are the numbers appearing in the solution of the problem; for $k = 4$ we obtain the theorem given in the problem, where of course the chess players should be interpreted as points of set $G$, the matches as segments, and the cities as colors).
The general theorem formulated just now can be easily proved by induction using the reasoning pattern used in the solution of the problem.
It is worth noting that the sequence $(n_k)$ defined by the given recurrence relation is not optimal; if $k \geq 4$, then there are numbers $n \leq n_k$ such that for any $n$-element set of points $G$ the conclusion of the theorem is true (with $k$ colors); in particular, the theorem from the problem remains true when the number $66$ is replaced by the number $65$. However, no formula is known that gives a non-trivial lower bound for the allowable values of $n$, or the asymptotic behavior of these minimal values as $k \to \infty$.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 650 |
XXIII OM - III - Problem 6
Prove that the sum of the digits of the number $ 1972^n $ tends to infinity as $ n $ tends to infinity.
|
We will prove in general that if $a$ is an even natural number not divisible by $5$ and $s_n$ denotes the sum of the digits of the number $a^n$ for $n = 1, 2, \ldots$, then the sequence $s_n$ tends to infinity.
Let $a_r, a_{r-1}, \ldots, a_2, a_1$, where $a \ne 0$, be the consecutive digits of the number $a^n$, i.e.,
The digit $a_1$ is different from zero, because by assumption the number $a$ is not divisible by $5$, and thus the number $a_n$ is not divisible by $5$. We also have $a^n < 10^r$, so $r > n \log_{10} a \geq n \log_{10} 2$.
We will prove
Lemma.
For every natural number $j$ satisfying the condition
at least one of the digits $a_{j+1}, a_{j+2}, \ldots, a_{4j}$ of the number $a^n$ is different from zero.
Proof. If for some natural number $j$ satisfying (1) it were the case that $a_{j+1} = a_{j+2} = \ldots = a_{4j} = 0$, then denoting
we would have $a^n - c = (a_r a_{r-1} \ldots a_{4j+1} 0 \ldots 0)_{10}$. Thus $10^{4j} \mid a^n - c$ and therefore $2^{4j} \mid a^n - c$. Moreover, from $2 \mid a$ it follows that $2^n \mid a^n$. Hence, given $4^j \leq n$, we obtain $2^{4j} \mid a^n - (a^n - c) = c$. However, $2^{4j} = 16^j > 10^j > c$. Therefore, $c = 0$. This is impossible, however, because the last digit $a_1$ of the number $c$ is different from zero. The obtained contradiction completes the proof of the lemma.
From the assumption and this lemma, it follows in particular that in each of the following sequences of digits there is a term different from zero:
where the number $j = 4^k$ satisfies condition (1), i.e., $4^k \leq \frac{1}{4} \min (r, n)$. Since $\frac{1}{4} \min (r, n) \geq \frac{1}{4} \min (n \log_{10} 2, n) = \frac{1}{4} n \log_{10} 2 > \frac{1}{16} n$, we can take $k$ to be the largest natural number satisfying one of the following equivalent inequalities
The sequences (2) contain different digits of the number $a^n$, there are $k + 2$ of these sequences, and each of them contains a digit different from zero. Therefore, the sum of the digits $s_n$ of the number $a^n$ is not less than $k + 2 = [\log_4 n]$.
Since $s_n \geq [\log_4 n] > \log_4 n - 1$ and $\lim_{n \to \infty} \log_4 n = \infty$, then $\lim_{n \to \infty} s_n = \infty$.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 651 |
XVI OM - I - Problem 3
Through each edge of a trihedral angle, a plane containing the bisector of the opposite planar angle has been drawn. Prove that the three planes intersect along a single line.
|
Consider a trihedral angle with vertex $O$. We measure three equal segments $OA = OB = OC$ on its edges. Consider the plane $\alpha$ passing through the edge $OA$ and the bisector of the angle $BOC$. Since the triangle $BOC$ is isosceles, the bisector of the angle $BOC$ passes through the midpoint $M$ of the side $BC$. The plane $\alpha$, on which points $A$ and $M$ lie, passes through the median $AM$ of the triangle $ABC$ and thus through the centroid $S$ of this triangle. Therefore, the plane $\alpha$ passes through the line $OS$.
In the above reasoning, the plane $\alpha$ can be replaced by a plane passing through another edge of the trihedral angle and the bisector of the opposite face. All three of these planes thus pass through the line $OS$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 652 |
XVIII OM - II - Problem 3
Two circles are internally tangent at point $ A $. A chord $ BC $ of the larger circle is tangent to the smaller circle at point $ D $. Prove that $ AD $ is the angle bisector of $ \angle BAC $.
|
The point of tangency $ A $ of the given circles is their center of homothety. In this homothety, the tangent $ BC $ of the smaller circle corresponds to the tangent to the larger circle at point $ E $, which corresponds to point $ D $ (Fig. 6). As homothetic lines, these tangents are parallel. Therefore, the arcs $ BE $ and $ EC $ of the larger circle, determined by the tangent at point $ E $ and the chord $ BC $ parallel to it, are equal. Hence, the inscribed angles $ BAE $ and $ EAC $ are equal, i.e., $ AE $ is the angle bisector of $ BAC $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 653 |
XLVI OM - II - Problem 2
In a convex hexagon $ABCDEF$, the following equalities hold: $|AB| = |BC|$, $|CD| = |DE|$, $|EF| = |FA|$. Prove that the lines containing the altitudes of triangles $BCD$, $DEF$, and $FAB$ drawn from vertices $C$, $E$, and $A$, respectively, intersect at a single point.
|
Let's adopt the following notations:
$ k_1 $ - the circle with center $ D $ and radius $ |DC| = |DE| $;
$ k_2 $ - the circle with center $ F $ and radius $ |FE| = |FA| $;
$ k_3 $ - the circle with center $ B $ and radius $ |BA| = |BC| $.
Circles $ k_2 $ and $ k_3 $ intersect at point $ A $; let's denote the second point of their intersection by $ A' $ (see Note 2); similarly, denote by $ C' $ the second (other than $ C $) point of intersection of circles $ k_3 $ and $ k_1 $, and by $ E' $ the second (other than $ E $) point of intersection of circles $ k_1 $ and $ k_2 $ (see Figure 7). Points $ A $ and $ A' $ are symmetric with respect to the line passing through the centers $ F $ and $ B $ of circles $ k_2 $ and $ k_3 $; thus, $ AA' $ contains the altitude of triangle $ FAB $ drawn from vertex $ A $; it is therefore one of the three lines mentioned in the problem statement. The other two lines, in question, are - analogously - lines $ CC' $ and $ EE' $.
om46_2r_img_7.jpg
Line $ AA' $ is the power axis of the pair of circles $ k_2 $, $ k_3 $; line $ CC' $ is the power axis of the pair $ k_3 $, $ k_1 $, and line $ EE' $ is the power axis of the pair $ k_1 $, $ k_2 $. It is known (see Note 1) that for any triplet of circles, the three power axes determined by pairs of these circles form a pencil of lines intersecting at one point or are parallel. In the considered situation, lines $ AA' $, $ CC' $, $ EE' $, being perpendicular to the sides $ FB $, $ BD $, $ DF $ of triangle $ DFB $, cannot be parallel - and thus intersect at one point. The proof is complete.
Note 1. For Readers who are not sufficiently familiar with the theorem on the collinearity of power axes, we will sketch its proof, along with the definitions of the used concepts.
Let $ P $ be a point in the plane of circle $ k $ and let $ l $ be any line passing through point $ P $ and intersecting circle $ k $ at two points $ U $ and $ V $. The value of the product $ |PU| \cdot |PV| $ depends only on circle $ k $ and point $ P $, but not on the choice of line $ l $. [Justification: suppose another line $ l' $ intersects circle $ k $ at points $ U' $ and $ V' $; the well-known equality $ |PU| \cdot |PV| = |PU' \cdot |PV'| $ follows from the similarity of triangles $ PUU' $ and $ PUV' $.]
This constant value, taken with a positive sign if point $ P $ lies outside circle $ k $, and with a negative sign if $ P $ lies inside $ k $, is called the power of point $ P $ with respect to circle $ k $; it is equal to the difference $ |OP|^2 - r^2 $, where $ O $ and $ r $ are the center and radius of circle $ k $. [Justification: let $ UV $ be a diameter contained in line $ OP $; one of the segments $ PU $, $ PV $ has length $ |OP| + r $, and the other has length $ \pm (|OP| - r) $; the sign is positive or negative depending on whether point $ P $ lies outside $ k $ or inside $ k $; thus (respectively) $ |PU| \cdot |PV| = \pm (|OP|^2 - r^2) $.]
Let two circles with non-coinciding centers $ O_1 $, $ O_2 $ and radii $ r_1 $, $ r_2 $ be given. If point $ P $ has equal power with respect to both of these circles, then the same property also applies to point $ Q $ defined as the orthogonal projection of point $ P $ onto line $ O_1O_2 $; this follows from the equality $ |O_1Q|^2 - |O_2Q|^2 = |O_1P|^2 - |O_2P|^2 = r_1^2 - r_2^2 $; on line $ O_1O_2 $ there is exactly one point $ Q $ such that $ |O_1Q|^2 - |O_2Q|^2 = r_1^2 - r_2^2 $. Hence, the set of points with equal power with respect to two given circles is a line perpendicular to $ O_1O_2 $; it is called the power axis of this pair of circles. If the circles intersect at two points, then each of these points has zero power with respect to both circles, and thus the power axis is the line passing through the points of intersection. If the circles are tangent, the power axis is the tangent line to both at the point of tangency.
If we consider any triplet of circles $ k_1 $, $ k_2 $, $ k_3 $ (with pairwise distinct centers) and if these centers are not collinear, then the power axis of the pair $ k_1 $, $ k_3 $ intersects the power axis of the pair $ k_2 $, $ k_3 $. The intersection point has the same power with respect to all three circles, and thus lies on the power axis of the pair $ k_1 $, $ k_2 $. If, however, the centers of circles $ k_1 $, $ k_2 $, $ k_3 $ are collinear, then each of these three axes is perpendicular to the line passing through the centers. Hence the thesis of the theorem: the three considered power axes either intersect at one point or are parallel.
Note 2. The problem statement mentions triangles $ BCD $, $ DEF $, $ FAB $; it should be inferred from this that each of the angles $ BCD $, $ DEF $, $ FAB $ is less than a straight angle. However, the reasoning remains valid even if one of these angles is a straight angle (and the corresponding triangle degenerates into a segment; by "line containing the altitude" we then mean the line perpendicular to that segment and passing through the corresponding vertex). Two of the circles $ k_1 $, $ k_2 $, $ k_3 $ are then tangent; but their power axis is still the line containing the altitude (in the sense specified just now), so no modifications to the further reasoning are required.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 654 |
XLV OM - III - Task 5
Points $ A_1, A_2, \ldots , A_8 $ are the vertices of a parallelepiped with center $ O $. Prove that
|
Let's assume that one of the faces of a given parallelepiped is the parallelogram $A_1A_2A_3A_4$, and the opposite face is the parallelogram $A_5A_6A_7A_8$, with segments $A_1A_5$, $A_2A_6$, $A_3A_7$, $A_4A_8$ being four edges of the parallelepiped (Figure 17). Let's denote the distances from the vertices to the point $O$ as follows:
om45_3r_img_17.jpg
The inequality to be proven takes the form
equivalently:
Since
it remains to show that
Segments $OA_1$, $OA_4$, $A_1A_4$ are the three sides of a triangle; similarly, segments $OA_2$, $OA_3$, $A_2A_3$ are the three sides of a triangle. Therefore,
(triangle inequalities). Segments $A_1A_4$ and $A_2A_3$ are opposite sides of the parallelogram $A_1A_2A_3A_4$, so $|A_1A_4| = |A_2A_3|$. Hence, from relation (2), we obtain the inequality
or $a \leq b + c + d$. Multiplying both sides by $a$, we write it as
Similarly,
Adding these four inequalities side by side, we obtain inequality (1), which we wanted to prove.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 655 |
LIX OM - III - Task 4
Each point in the plane with both integer coordinates has been painted either white or black. Prove that from the set of all painted points, an infinite subset can be selected which has a center of symmetry and all of whose points have the same color.
|
Suppose the thesis of the problem is false.
Consider the central symmetry with respect to the point $ (0,0) $. Since there does not exist an infinite set symmetric with respect to this point and composed of points of the same color, only finitely many points with integer coordinates pass to points of the same color under this symmetry. Therefore, there exists an integer $ M $ such that for every point with integer coordinates $ (x,y) $, where $ |y| > M $, the points $ (-x,-y) $ and $ (x,y) $ have different colors.
Similarly, considering the central symmetry with respect to the point $ (\frac{1}{2},0) $, we see that there exists an integer $ N $ such that for every point with integer coordinates $ (x,y) $, where $ |y| > N $, the point $ (x,y) $ has a different color than its image under the considered symmetry, which is the point $ (-x+1,-y) $.
Let $ k = \max\{M,N\} + 1 $ and consider any integer $ s $. Then the point $ (s,k) $ under the symmetry with respect to the point $ (0,0) $ maps to the point $ (-s,-k) $, which is of a different color than the point $ (s,k) $. Furthermore, the point $ (-s,-k) $ under the symmetry with respect to the point $ (\frac{1}{2}, 0) $ maps to the point $ (s +1,k) $, which is of a different color than the point $ (-s,-k) $.
Therefore, the points $ (s,k) $ and $ (s+1,k) $ have the same color. Since $ s $ was an arbitrary integer, it follows that all points with the first coordinate an integer and the second coordinate equal to $ k $ have the same color.
However, these points form an infinite set, whose center of symmetry is the point $ (0,k) $.
The obtained contradiction with the indirect assumption completes the solution.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 656 |
XXVII OM - II - Problem 5
Prove that if $ \cos \pi x =\frac{1}{3} $ then $ x $ is an irrational number.
|
Let $ \cos t = \frac{1}{3} $. We will prove by induction that for every natural number $ n $ the following formula holds:
where $ a_n $ is an integer not divisible by $ 3 $.
For $ n=1 $, it suffices to take $ a_1 = 1 $. For $ n = 2 $ we have
Thus, $ a_2 = -7 $. Suppose next that for some natural number $ k $ the following holds:
where $ a_k $ and $ a_{k+1} $ are integers not divisible by $ 3 $. We have $ \cos (k + 2)t + \cos kt = 2\cos (k + 1)t \cdot \cos t $. Therefore,
The number $ a_{k+2} = 2a_{k+1} - 9a_k $ is clearly an integer and not divisible by $ 3 $. Hence, by the principle of induction, formula (1) holds for every natural number $ n $.
Suppose now that for some rational number $ x = \frac{r}{s} $ the following holds:
$ \cos \pi x = \frac{1}{3} $. Then $ t = \pi x $ and by formula (1) for $ n = 2s $ we have
where $ a_{2s} $ is an integer not divisible by $ 3 $.
On the other hand, we have
The obtained contradiction proves that for no rational number $ x $ does $ \cos \pi x = \frac{1}{3} $ hold.
Note. Reasoning similarly, one can prove that if the number $ \cos \pi x $ is rational and different from $ 0 $, $ \pm \frac{1}{2} $, and $ \pm 1 $, then $ x $ is an irrational number.
See also articles in the journal "Mathematics" No. 5(128) from 1973, pp. 313-317, and No. 3(132) from 1974, pp. 168-172.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 657 |
III OM - III - Task 5
Prove that none of the digits $ 2 $, $ 4 $, $ 7 $, $ 9 $ can be the last digit of the number
where $ n $ is a natural number.
|
If the last digit of a given number is $ x $, then
so
Since $ n $ is an integer, the discriminant of the above equation, i.e., the number
is a square of an integer. The last digit of this discriminant is the last digit of the number $ 8x + 1 $. When $ x $ equals $ 2 $, $ 4 $, $ 7 $, $ 9 $, then $ 8x + 1 $ ends with the digit $ 7 $, $ 3 $, $ 7 $, $ 3 $, respectively.
The square of an integer ending in the digit $ c $ has the form $ (10a + c)^2 = 100a^2 + 20ac + c^2 $, and thus has the same last digit as $ c^2 $. Since the squares of numbers from $ 0 $ to $ 9 $ do not end in the digit $ 7 $ or $ 3 $, $ x $ cannot be any of the digits $ 2 $, $ 4 $, $ 7 $, $ 9 $.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 660 |
LII OM - I - Problem 9
Prove that among any $12$ consecutive positive integers, there exists a number that is not the sum of $10$ fourth powers of integers.
|
A number that is the fourth power of an integer gives a remainder of $0$ or $1$ when divided by $16$. Indeed: for even numbers of the form $2k$ we have $(2k)^4 = 16k^4$, whereas for odd numbers $2k + 1$ we obtain
(The number $\frac{1}{2} k(3k+1)$ is an integer for any integer $k$).
Hence, a number that is the sum of $10$ fourth powers of integers gives one of eleven remainders when divided by $16$: $0,1,2,3,\ldots, 10$. Among $12$ consecutive integers, no two give the same remainder when divided by $16$. Therefore, one of these $12$ integers does not give a remainder from the set $\{0,1,2,3,\ldots,10\}$ when divided by $16$. This number cannot then be the sum of $10$ fourth powers of integers.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 661 |
LIV OM - III - Task 2
The number $ a $ is positive and less than 1. Prove that for every finite, strictly increasing sequence of non-negative integers ($ k_1,\ldots,k_n $) the following inequality holds
|
We apply induction on $ n $. For $ n =1 $, the inequality takes the form
and is obviously satisfied. Fix $ n \geq 2 $ and assume that the inequality holds for any increasing sequence of length $ n-1 $. Consider any increasing sequence of length $ n $: $ 0 \leq k_1 < k_2 < \ldots < k_n $. By dividing both sides of the inequality to be proven by $ a^{2k_l} $, we can assume without loss of generality that $ k_1 = 0 $. Then
From the inequality $ 0 < k_2 < k_3 < \ldots < k_n $ and $ 0 < a < 1 $, it follows that
Therefore, from equation (1) and the induction hypothesis, we obtain
This completes the inductive step and the solution of the problem.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 663 |
XXXVI OM - III - Task 3
Prove that if the function $ f: \mathbb{R} \to \mathbb{R} $ satisfies for every $ x \in \mathbb{R} $ the equation $ f(3x) = 3f(x) - 4(f(x))^3 $ and is continuous at 0, then all its values belong to the interval $ \langle -1;1\rangle $.
|
We will first prove a lemma.
Lemma. If $ |f(x)| \leq 1 $, then $ |f(3x)| \leq 1 $.
Proof.
If $ |f(x)| \leq 1 $, then for some $ y $ we have $ f(x) = \sin y $. Therefore, $ f(3x) = 3 \sin y - (\sin y)^3 = \sin 3y $, which implies that $ |f(3x)| \leq 1 $.
Let's calculate $ f(0) $. Substituting $ x = 0 $ in the given formula, we get $ f(0) = 3f(0) - 4(f(0))^3 $, so $ 4(f(0))^3 = 2f(0) $, hence $ f(0) = 0 $ or $ f(0) = 1/ \sqrt{2} $ or $ f(0) = - 1/\sqrt{2} $. In each case, $ |f(0)| < 1 $. From the continuity of the function $ f $ at the point $ 0 $, it follows that there exists a neighborhood $ (-d;d) $ of the point $ 0 $ such that for $ x $ in this neighborhood, $ |f(x)| < 1 $. For any real number $ t $, there exists a natural number $ n $ such that $ t/3^n \in (-d;d) $. Therefore, $ |f(t/3^n)| \leq 1 $, and by the lemma, $ |f(t/3^{n-1})| = \left| f\left(3 \cot \frac{t}{3^n} \right)\right| \leq 1 $, $ |f(t/3^{n-2})| = \left| f\left(3 \cot \frac{t}{3^{n-1}} \right)\right| \leq 1 $, and so on, finally $ |f(t)| \leq 1 $.
Therefore, all values of the function $ f $ belong to the interval $ [-1; 1] $.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 665 |
XXVIII - I - Problem 4
Inside the tetrahedron $ABCD$, a point $M$ is chosen, and it turns out that for each of the tetrahedra $ABCM$, $ABDM$, $ACDM$, $BCDM$, there exists a sphere tangent to all its edges. Prove that there exists a sphere tangent to all the edges of the tetrahedron $ABCD$.
|
We will use the following theorem, which was given as problem 4 at the VI Mathematical Olympiad.
Theorem. There exists a sphere tangent to all edges of the tetrahedron $ PQRS $ if and only if the sums of the lengths of opposite edges of this tetrahedron are equal, i.e., if
Applying this theorem to the tetrahedra $ ABCM $, $ ACDM $, and $ BCDM $, we obtain in particular
Therefore,
which means
and similarly,
which means
Thus, we have proved that
Therefore, based on the theorem cited above, there exists a sphere tangent to all edges of the tetrahedron $ ABCD $.
Note. In the above solution, we did not use the assumption that there exists a sphere tangent to all edges of the tetrahedron $ ABDM $. By reasoning similarly, we can therefore prove that if for any three of the four tetrahedra $ ABCM $, $ ABDM $, $ ACDM $, $ BCDM $ there exists a sphere tangent to all its edges, then there exists a sphere tangent to all edges of the tetrahedron $ ABCD $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 668 |
XVIII OM - II - Problem 5
On a plane, there are two triangles, one outside the other.
Prove that there exists a line passing through two vertices of one triangle such that the third vertex of this triangle and the other triangle lie on opposite sides of this line.
|
Let's denote the vertices of the given triangles by $A_1, A_2, A_3$ and $B_1, B_2, B_3$. Let $MN$ be a segment such that: $1^\circ$ point $M$ lies on the perimeter of triangle $A_1A_2A_3$, $2^\circ$ point $N$ lies on the perimeter of triangle $B_1B_2B_3$, $3^\circ$ segment $MN$ is not longer than any segment connecting points on the perimeter of triangle $A_1A_2A_3$ with points on the perimeter of triangle $B_1B_2B_3$.
The segment $MN$ can be found as follows. For each of the nine pairs $(A_i, B_jB_k)$, where $i, j, k = 1, 2$ or $3$, $j \ne k$, we choose the shortest segment connecting point $A_i$ with points on segment $B_jB_k$; this will be the altitude of triangle $A_iB_jB_k$ from vertex $A_i$ or one of the segments $A_iB_j$, $A_iB_k$. Similarly, for each of the nine pairs $(B_i, A_jA_k)$, we choose the shortest segment connecting point $B_i$ with points on segment $A_jA_k$. Finally, among the eighteen segments obtained in this way (some of which may coincide), we choose the shortest one, i.e., the one that is not longer than any other. One end of this segment, which we will call $M$, lies on the perimeter of triangle $A_1A_2A_3$, and the other end $N$ lies on the perimeter of triangle $B_1B_2B_3$. Segment $MN$ satisfies the above condition $3^\circ$, since for any segment connecting a point $K$ on the perimeter of the first triangle with a point $L$ on the perimeter of the second triangle (Fig. 7), there exists a parallel and not longer segment connecting a vertex of one triangle with a point on the side of the other triangle, which in turn is not longer than $MN$. Therefore, $MN$ is a segment with the desired properties.
Draw through points $M$ and $N$ the lines $m$ and $n$ perpendicular to $MN$ (Fig. 8). Inside the strip of the plane bounded by lines $m$ and $n$, there are no points of either of the given triangles. For if, for example, a point $P$ of triangle $A_1A_2A_3$ were inside this strip, then the entire segment $PM$ would lie within triangle $A_1A_2A_3$, and since angle $PMN$ is acute, there would be points $Q$ between $P$ and $M$ such that $QN < MN$, which would contradict the definition of segment $MN$.
One of the following cases occurs.
a) Points $M$ and $N$ are vertices of the given triangles. Suppose, for example, that $M$ coincides with $A_1$ and $N$ with $B_1$ (Fig. 9). Consider the angles at vertex $A_1$ formed by the rays $A_1A_2$ and $A_1A_3$ with the rays $m_1$ and $m_2$ of line $m$, and the angles at vertex $B_1$ formed by the rays $B_1B_2$ and $B_1B_3$ with the rays $n_1$ and $n_2$. Let the smallest of these angles be, for example, angle $\alpha$ between ray $A_1A_2$ and ray $m_1$.
Then line $A_1A_2$ is the line we are looking for.
Indeed, angle $\alpha$ is either acute or zero. If $\alpha = 0$, line $A_1A_2$ coincides with line $m$, so point $A_3$ lies on one side of this line, and points $B_1, B_2, B_3$ on the other. The same is true when angle $\alpha$ is acute and line $A_1A_2$ intersects line $n$ at point $C$. In this case, point $A_3$ lies on the opposite side of line $A_1A_2$ from point $B_1$ (since the convex angle between $A_1A_3$ and $m_1$ is $> \alpha$). Points $B_2$ and $B_3$ lie on the same side as $B_1$. If, for example, point $B_2$ were on the opposite side of line $A_1A_2$ from point $B_1$ or on line $A_1A_2$, then there would be some point $D$ on segment $B_1B_2$ on line $A_1A_2$ and $\measuredangle DB_1C = \beta$ would be smaller than $\alpha$, which contradicts the assumption, since $\measuredangle B_1CA_1 = \alpha$ would be an exterior angle of triangle $DB_1C$.
b) Points $M$ and $N$ are not both vertices of the given triangles, for example, point $M$ lies inside side $A_1A_2$. In this case, the entire side $A_1A_2$ lies on line $m$ and line $m$ separates point $A_3$ from points $B_1$, $B_2$, $B_3$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 671 |
LVII OM - III - Problem 5
Given is a tetrahedron $ABCD$ where $AB = CD$. The inscribed sphere of this tetrahedron is tangent to the faces $ABC$ and $ABD$ at points $K$ and $L$, respectively. Prove that if points $K$ and $L$ are the centroids of the faces $ABC$ and $ABD$, then the tetrahedron $ABCD$ is regular.
|
Let $s$ be the sphere inscribed in the tetrahedron $ABCD$. We will start by showing that triangles $ABC$ and $ABD$ are congruent.
Let $E$ be the midpoint of edge $AB$. Since $K$ and $L$ are the points of tangency of the sphere $s$ with the tetrahedron $ABCD$, we have $AK = AL$ and $BK = BL$, which implies that triangles $AKB$ and $ALB$ are congruent. Their medians $KE$ and $LE$ have the same length, so $CE = 3 \cdot KE = 3 \cdot LE = DE$. Therefore, triangles $BEC$ and $BED$ are congruent. Similarly, we prove that triangles $AEC$ and $AED$ are congruent. Hence, triangles $ABC$ and $ABD$ are congruent.
Let $M$ and $N$ be the points of tangency of the sphere $s$ with the faces $BCD$ and $CDA$, respectively. We obtain the following pairs of congruent triangles: $BKC$ and $BMC$; $BLD$ and $BMD$; $AKC$ and $ANC$; $ALD$ and $AND$; $CMD$ and $CND$.
Let $\alpha = \measuredangle AKC$ and $\beta = \measuredangle BKC$. Then $\measuredangle ALD = \alpha$ and $\measuredangle BLD = \beta$, since triangles $ABC$ and $ABD$ are congruent. Therefore,
Thus, we get $\measuredangle DNC = 360^\circ - \measuredangle ANC - \measuredangle AND = 360^\circ - 2\alpha$,
and $\measuredangle DMC = 360^\circ - \measuredangle BMC - \measuredangle BMD = 360^\circ - 2\beta$.
But $\measuredangle DMC = \measuredangle DNC$, so $\alpha = \beta$. Therefore, $\measuredangle AKE = \measuredangle BKE$, so $KE$ is both the angle bisector and the median in triangle $AKB$. Hence, $AK = BK$, and thus $AC = BC$. Ultimately, we get $AC = BC = AD = BD$.
Through each edge of the tetrahedron $ABCD$, draw a plane parallel to the opposite edge of the tetrahedron. These planes form a parallelepiped. In each of its faces, one diagonal is an edge of the tetrahedron $ABCD$, and the other diagonal is equal and parallel to the opposite edge of this tetrahedron. Since the opposite edges of the tetrahedron $ABCD$ are equal, all faces of the resulting parallelepiped are rectangles. Therefore, this parallelepiped is a rectangular parallelepiped. This implies, in particular, that the line connecting the midpoints of edges $AD$ and $BC$ is perpendicular to these edges.
Consider the transformation that is a rotation by $180^\circ$ around the line connecting the midpoints of edges $AD$ and $BC$. As a result of this transformation, the tetrahedron $ABCD$ maps onto itself, so the sphere $s$ must also map onto itself. Moreover, the faces $ABC$ and $ABD$ map to the faces $DCB$ and $DCA$, respectively, and thus the centroids of the faces $DCB$ and $DCA$ coincide with the points $M$ and $N$.
Similarly, we show that $AB = BD = DC = CA$, from which we conclude that all edges of the tetrahedron $ABCD$ are of equal length. This means that the tetrahedron $ABCD$ is regular.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 673 |
III OM - I - Problem 7
Let $ a $, $ b $ denote the legs of a right triangle, $ c $ - its hypotenuse, $ r $ - the radius of the inscribed circle, and $ r_a $, $ r_b $, $ r_c $ - the radii of the excircles of this triangle. Prove that:
1) $ r + r_a + r_b = r_c $
2) the radii $ r $, $ r_a $, $ r_b $, $ r_c $ are simultaneously integers if and only if the sides $ a $, $ b $, $ c $ are expressed by integers.
|
In this task, we are dealing with figure 16, which is a special case of figure 11, as angle \( C \) is a right angle. We see that \( r = CM \), \( r_a = CP \), \( r_b = CR \), \( r_c = CQ \).
By denoting \( a + b + c = 2p \) and applying the formulas given in Note II to problem 16, we obtain
From this, the first part of the theorem immediately follows:
To prove the second part, let us first assume that \( a \), \( b \), and \( c \) are integers. The proof that \( r \), \( r_a \), \( r_b \), and \( r_c \) are also integers, as can be seen from formulas (1), reduces to showing that \( p \) is an integer, i.e., that \( 2p \) is an even number. According to the Pythagorean theorem,
Therefore, either all three numbers \( a^2 \), \( b^2 \), and \( c^2 \) are even, or one of them is even and the other two are odd. Indeed, in any other case, i.e., if two of these numbers were even and the third one odd, or if all three were odd, the left side of equation (2) would not be divisible by 2, and the equation could not hold. Hence, it follows that either all three numbers \( a \), \( b \), and \( c \) are even, or one is even and the other two are odd; in both cases, the sum of the numbers \( a \), \( b \), and \( c \), i.e., \( 2p \), is an even number, which was to be shown.
Conversely: if \( r \), \( r_a \), \( r_b \), and \( r_c \) are integers, then \( a \), \( b \), and \( c \) are also integers, as it follows from formulas (1) that
直接输出翻译结果，保留了源文本的换行和格式。
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 675 |
XXXIV OM - III - Problem 1
In the plane, a convex $ n $-gon $ P_1, \ldots, P_n $ and a point $ Q $ inside it are given, which does not lie on any diagonal. Prove that if $ n $ is even, then the number of triangles $ P_iP_jP_k $ ($ i,j,k= 1,2,\ldots,n $) that contain the point $ Q $ is even.
|
om34_3r_img_11.jpg
First, consider a convex quadrilateral $ABCD$ and a point $K$ inside it that does not lie on any diagonal. Point $K$ belongs to one of the two half-planes with edge $AC$. It follows that point $K$ belongs to exactly two of the triangles $ABC$, $ABD$, $ACD$, $BCD$ (in Fig. 11, point $K$ belongs to triangles $ACD$ and $BCD$). In this way, we have proven the theorem for $n = 4$. Suppose that $n$ is an even number greater than $4$. By choosing any four vertices $P_i$, $P_j$, $P_k$, $P_l$ of the given polygon, we obtain a convex quadrilateral. If point $Q$ lies within this quadrilateral, then it belongs to two of the triangles $P_iP_jP_k$, $P_iP_jP_l$, $P_iP_kP_l$, $P_jP_kP_l$; if point $Q$ does not lie within the quadrilateral, then it does not belong to any of these triangles. In each case, the number of triangles determined by the vertices $P_i$, $P_j$, $P_k$, $P_l$ and containing point $Q$ is even. Any arbitrarily chosen triangle $P_iP_jP_k$ is contained in $n-3$ quadrilaterals whose vertices are the vertices of the given polygon. Therefore, when considering all quadrilaterals $P_iP_jP_kP_l$ and the triangles contained in them and containing point $Q$, we count each triangle $(n-3)$ times. By adding the numbers of triangles contained in the quadrilaterals $P_iP_jP_kP_l$ and containing point $Q$, we obtain an even number equal to $(n-3)t$, where $t$ is the number of triangles $P_iP_jP_k$ containing point $Q$. Since $n-3$ is an odd number (because $n$ is an even number), the number of considered triangles is even.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 676 |
XXXIII OM - II - Problem 4
Let $ A $ be a finite set of points in space having the property that for any of its points $ P, Q $ there is an isometry of space mapping the set $ A $ onto the set $ A $ and the point $ P $ onto the point $ Q $. Prove that there exists a sphere passing through all points of the set $ A $.
|
Let $ S $ be the center of gravity of the system of points of set $ A $. Each isometry transforming set $ A $ into set $ A $ also transforms point $ S $ into $ S $. For any $ P, Q \in A $, an isometry transforming $ A $ into $ A $ and $ P $ into $ Q $ also transforms the segment $ \overline{PS} $ into the segment $ \overline{QS} $. Therefore, $ PS = QS $, that is, the distances from any two points of set $ A $ to point $ S $ are equal. Hence, all points of set $ A $ lie on a certain sphere with center $ S $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 679 |
XXXI - II - Problem 4
Prove that if $ a $ and $ b $ are real numbers and the polynomial $ ax^3 - ax^2 + 9bx - b $ has three positive roots, then they are equal.
|
Suppose that the positive numbers $ s $, $ t $, $ u $ are roots of the polynomial $ ax^3 - ax^2 + 9bx - b $. Therefore, $ ax^3 - ax^2 + 9bx - b = a (x - s) (x - t)(x - u) $, and by expanding the right side and equating the coefficients of the successive powers of the variable $ x $, we obtain the so-called Vieta's formulas
Dividing the second of these formulas by the third, we get
Multiplying this equation by the first of Vieta's formulas
Notice that $ \frac{t}{s} + \frac{s}{t} \geq 2 $, with equality holding only when $ s = t $. Indeed, this inequality is equivalent to the inequality $ t^2 + s^2 \geq 2st $ (obtained by multiplying by the positive number $ st $), which in turn is equivalent to the inequality $ t^2 + s^2 - 2st \geq 0 $, or the inequality $ (t - s)^2 \geq 0 $.
Similarly, we observe that
with equality holding only when $ s = t = u $. Therefore, from equation (*), it follows that
hence $ s = t = u $.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 681 |
XXXII - I - Problem 9
In space, there is a set of $3n$ points, no four of which lie on the same plane. Prove that this set can be divided into $n$ three-element sets $\{A_i, B_i, C_i\}$ such that the triangles $A_iB_iC_i$ are pairwise disjoint.
|
Proof by induction. For $ n = 1 $, the thesis of the theorem is obviously satisfied. Assume that the thesis is satisfied for some $ n $ and consider $ 3(n+1) $ points in space, no four of which lie on the same plane. The smallest convex set containing the considered points is a polyhedron, each face of which is a triangle (since no four points lie in the same plane). Let $ A $, $ B $, $ C $ be the vertices of one of these triangles. The set of the remaining $ 3n $ points can be divided into $ n $ sets $ \{A_1, B_1, C_1\} $ with the property that the triangles $ A_iB_iC_i $ are pairwise disjoint. Since all points $ A_1, B_1, C_1 $ lie on one side of the plane $ ABC $, the triangle $ ABC $ is disjoint from each of the triangles $ A_iB_iC_i $. We have thus obtained a partition of the set of $ 3(n+1) $ points into three-element subsets with the required property.
By the principle of induction, the thesis of the theorem is satisfied for every natural number $ n $.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 684 |
XVI OM - III - Problem 3
On a circle, $ n > 2 $ points are chosen and each of them is connected by a segment to every other. Can all these segments be traced in one continuous line, i.e., so that the end of the first segment is the beginning of the second, the end of the second is the beginning of the third, etc., and so that the end of the last segment is the beginning of the first?
|
To facilitate pronunciation, we will introduce a certain convention. Let $ Z $ be a finite set of points on a circle (the condition that the points of set $ Z $ lie on a circle can be replaced by the weaker assumption that no $ 3 $ points of this set lie on the same straight line. The same remark applies to the further course of the solution). If a closed broken line, whose vertices are the points of set $ Z $, has the property that in the sequence of consecutive sides of the broken line, each segment connecting two points of set $ Z $ occurs once and only once, then we will call such a broken line $ L(Z) $.
Suppose that for some set $ Z $ consisting of $ n \geq 3 $ points of a circle, there exists a broken line $ L(Z) $. Then at each point of set $ Z $, $ n - 1 $ sides of the broken line meet. Drawing the broken line in one continuous stroke, we reach each of its vertices as many times as we leave it; the number $ n - 1 $ is therefore even, i.e., $ n $ is an odd number. The broken line $ L(Z) $ cannot, therefore, exist when $ n $ is an even number. We will prove that when $ n $ is odd, then the broken line $ L(Z) $ does exist (there may, of course, be different broken lines $ L(Z) $ for the same set $ Z $).
\spos{1} (induction method). Let $ n = 2m + 1 $ ($ m $ - a natural number). For $ m = 1 $, i.e., for a set of three points $ A_1 $, $ A_2 $, $ A_3 $, the theorem is true; the desired broken line is $ A_1A_2A_3A_1 $.
Suppose that the theorem is true when $ m = k - 1 $, i.e., when the set has $ 2k - 1 $ points ($ k $ - a natural number $ \geq 2 $); we will prove that it is then true when $ m = k $, i.e., when the set has $ n = 2k + 1 $ points.
Consider the set $ Z $ consisting of points $ A_1A_2, \ldots, A_{2k-1}, A_{2k}, A_{2k+1} $ of the circle and let $ U $ denote the set containing only the points $ A_1, A_2, \ldots, A_{2k-1} $. According to the induction hypothesis, for the set $ U $ there exists some broken line $ L(U) $.
It is easy to see that there exists a closed broken line $ K $ whose sides are segments connecting each of the points $ A_1, A_2, \ldots, A_{2k-1} $ with the points $ A_{2k} $ and $ A_{2k+1} $ and the segment $ A_{2k}A_{2k+1} $, and in the sequence of consecutive sides of the broken line $ K $, each of these segments occurs only once. This condition is satisfied, for example, by the broken line $ K $:
Notice that the sides of the broken line $ K $ are all and only those among the segments connecting the points of set $ Z $ that are not sides of the broken line $ L(U) $.
We now form a single closed broken line from the broken lines $ L(U) $ and $ K $ in the following way. We choose point $ A_1 $ as the beginning of the closed broken line $ L(U) $ and traverse this broken line, eventually reaching point $ A_1 $; we then traverse the broken line $ K $ starting from point $ A_1 $ and ending again at point $ A_1 $. We obtain a broken line whose beginning and end is point $ A_1 $, and in the sequence of consecutive sides of this broken line, each of the segments connecting the points of set $ Z $ appears only once. This is, therefore, the sought broken line $ L(Z) $. The theorem has been proved.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 685 |
LVI OM - II - Problem 5
Given is a rhombus $ABCD$, where $ \measuredangle BAD > 60^{\circ} $. Points $E$ and $F$ lie on sides $AB$ and $AD$ respectively, such that $ \measuredangle ECF = \measuredangle ABD $. Lines $CE$ and $CF$ intersect diagonal $BD$ at points $P$ and $Q$ respectively. Prove that
|
From the equality $ \measuredangle FDQ = \measuredangle PCQ $, it follows that triangles $ FDQ $ and $ PCQ $ are similar. Since points $ A $ and $ C $ are symmetric with respect to the line $ BD $, triangles $ PCQ $ and $ PAQ $ are congruent, and
This equality means that points $ A $, $ P $, $ Q $, and $ F $ lie on the same circle. Similarly, we prove that points $ A $, $ P $, $ Q $, and $ E $ lie on the same circle. Therefore, there exists a circle passing through points $ A $, $ P $, $ Q $, $ E $, and $ F $.
om56_2r_img_3.jpg
From the relationship $ \measuredangle CPQ = \measuredangle CFE $, we conclude that triangles $ CPQ $ and $ CFE $ are similar. Similarly, we obtain the similarity of triangles $ DQF $ and $ DAP $. Moreover, triangles $ PAQ $ and $ ABQ $ are similar. Hence
This completes the proof.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 686 |
XXVI - I - Problem 7
Let $ Z $ be the set of all finite sequences with terms $ a, b, c $ and let $ Z_1 $ be a subset of $ Z $ containing for each natural number $ k $ exactly one $ k $-term sequence.
We form the smallest set $ Z_2 $ with the property that $ Z_2 \supset Z_1 $ and if a certain $ k $-term sequence belongs to $ Z_1 $, then every $ (k+r) $-term sequence ($ r = 1, 2, \ldots $) obtained from this $ k $-term sequence by appending an $ r $-term sequence at the beginning or the end, belongs to $ Z_2 $.
Prove that for every $ k $ there exists in $ Z $ a $ k $-term sequence that does not belong to $ Z_2 $.
|
Every $ n $-term sequence belonging to $ Z_2 $ arises from an $ (n - k) $-term sequence (where $ 1 \leq k \leq n $) belonging to $ Z_1 $ by appending a certain $ k $-term sequence at the beginning or end, or it is an $ n $-term sequence belonging to $ Z_1 $. The number of such $ k $-term sequences is $ 3^k $, because each term of the sequence can take one of three values. Therefore, the number of $ n $-term sequences belonging to $ Z_2 $ that arise from a certain $ (n-k) $-term sequence belonging to $ Z_1 $ is no greater than $ 2 \cdot 3^k $. Hence, the number of all $ k $-term sequences belonging to $ Z_2 $ is no greater than
The number of all $ n $-term sequences with terms belonging to a three-element set is $ 3^n $. Therefore, for every natural number $ n $, there are at least two $ n $-term sequences that do not belong to $ Z_2 $.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 688 |
XLVIII OM - I - Problem 4
Prove that a natural number $ n \geq 2 $ is composite if and only if there exist natural numbers $ a,b,x,y \geq 1 $ satisfying the conditions: $ a+b=n $, $ \frac{x}{a}+\frac{y}{b}=1 $.
|
Let $ n $ be a composite number: $ n = qr $, $ q \geq 2 $, $ r \geq 2 $. Assume:
These are positive integers with the properties:
We have thus shown that if $ n \geq 2 $ is a composite number, then there exist integers $ a, b, x, y \geq 1 $ satisfying conditions (1).
The remaining part to prove is the converse implication. Assume that for some integer $ n \geq 2 $ there exists a quadruple of integers $ a, b, x, y \geq 1 $ satisfying equations (1). From the second of these equations, it follows that $ x < a $, $ y < b $, and
Let $ d $ be the greatest common divisor of $ a $ and $ b $, and let us assume: $ a = \alpha d $, $ b = \beta d $. The numbers $ \alpha $ and $ \beta $ are coprime. The first equation (1) gives the relationship: $ n = (\alpha + \beta)d $. The factor $ \alpha + \beta $ is greater than $ 1 $. To obtain the required conclusion that $ n $ is a composite number, it suffices to show that $ d > 1 $.
Equation (2), when divided by $ d $, takes the form $ \alpha y = \beta (a - x) $. Therefore, the number $ \beta $, being coprime with $ \alpha $, must be a divisor of the number $ y $. Hence $ \beta \leq y $. Since $ y < b $, we get the inequality $ \beta < b $, and consequently $ d = b / \beta > 1 $.
Conclusion: The existence of numbers $ a, b, x, y \geq 1 $ satisfying conditions (1) implies that $ n $ is a composite number. Combined with the previously proven converse implication, this gives the thesis of the problem.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 690 |
Let $ a_1,a_2, \ldots, a_n $, $ b_1,b_2, \ldots, b_n $ be integers. Prove that
|
Sums appearing on both sides of the given inequality in the task will not change if we arbitrarily change the order of numbers $ a_i $ and numbers $ b_i $. Without loss of generality, we can therefore assume that $ a_1 \leq a_2 \leq \ldots \leq a_n $ and $ b_1 \leq b_2 \leq \ldots \leq b_n $. In this case
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 691 |
XXV OM - III - Problem 6
A convex $ n $-gon was divided into triangles by diagonals in such a way that
1° an even number of diagonals emanate from each vertex,
2° no two diagonals have common interior points.
Prove that $ n $ is divisible by 3.
|
We will first prove the
Lemma. If $F$ is a figure on a plane and it is divided into parts by $r$ lines, then these parts can be painted with two colors in such a way that any two parts sharing a segment have different colors.
Proof. We will use induction with respect to $r$. In the case of $r = 1$, the thesis is obvious, because a line divides the plane into two half-planes. Parts of the figure $F$ contained in one half-plane are painted one color, and parts contained in the other are painted another color.
Let $r$ be a natural number. We assume that in the case of $r$ lines, the parts of the figure $F$ can be painted with two colors in such a way that the condition of the lemma is satisfied. By drawing the $(r + 1)$-th line, we divide the plane into two half-planes. Parts of the figure $F$ contained in one of these half-planes are left as they are, and parts contained in the other are painted with the opposite color (we change the color of each part). Then, of course, the condition of the lemma will be satisfied.
We proceed to solve the problem. Since the given $n$-gon has been divided using a certain number of lines (diagonals), the parts obtained from the division (triangles) can be painted with two colors, according to the lemma, so that triangles sharing a side have different colors.
Since the number of diagonals considered from any vertex $A$ of the given $n$-gon is even, the number of triangles for which point $A$ is a vertex is odd. Therefore, since these triangles have different colors in sequence, the extreme triangles have the same color. It follows that the triangles whose one side is a side of the given polygon are all painted the same color.
Thus, the sum of the number of sides of the $n$-gon and the number of all considered diagonals is equal to the number of sides of the triangles painted with this color. This number is therefore divisible by $3$. On the other hand, the number of considered diagonals is equal to the number of sides of the triangles painted with the other color. This number is also divisible by $3$. Therefore, the difference between these numbers, i.e., the number $n$, is also divisible by $3$.
Note. If in the problem statement triangles are replaced by $k$-gons, where $k > 3$, then by replacing triangles with $k$-gons in the above solution, we obtain that the number $n$ is divisible by $k$.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 692 |
XXXVI OM - II - Problem 6
In space, there are distinct points $ A, B, C_0, C_1, C_2 $, such that $ |AC_i| = 2 |BC_i| $ for $ i = 0,1,2 $ and $ |C_1C_2|=\frac{4}{3}|AB| $. Prove that the angle $ C_1C_0C_2 $ is a right angle and that the points $ A, B, C_1, C_2 $ lie in the same plane.
|
The proof will be based on the following theorem of Apollonius:
On the plane $\pi$, there are two points $A$ and $B$; in addition, there is a positive number $\lambda \ne 1$. The set of points $X$ on the plane $\pi$ that satisfy the condition $|AX| = \lambda |BX|$ forms a circle, and the center of this circle lies on the line $AB$.
The proof of this theorem using analytic geometry is immediate: we can assume that $A = (-1,0)$, $B = (1,0)$; the condition imposed on the point $X = (x,y)$ takes the form of the equation
after simple transformations
where $a = (\lambda^2+1)/(\lambda^2-1)$, $r^2 = a^2-1$; hence the thesis.
The circle mentioned in the above theorem is called the Apollonian circle (for points $A$, $B$ and ratio $\lambda$). Readers interested in other proofs of this theorem and information about topics related to the Apollonian circle are referred to the article by Jerzy Bednarczuk in the monthly Delta, issue 7/1985.
Transferring the considerations to space, note that by rotating the plane $\pi$ around the line $AB$, we obtain the spatial version of the discussed theorem:
The set of points $X$ satisfying the condition $|AX| = \lambda |BX|$ forms a sphere, and its center lies on the line $AB$ ($A$, $B$ - given points, $\lambda$ - given number, $\lambda > 0$, $\lambda \ne 1$).
We will call this sphere the Apollonian sphere (for points $A$, $B$ and ratio $\lambda$).
om36_2r_img_7.jpg
The solution to the problem is now very simple. Denoting by $S$ the Apollonian sphere for the given points $A$, $B$ and ratio $\lambda = 2$, we see that $C_i \in S (i = 0,1,2)$. The sphere $S$ contains, in particular, two points on the line $AB$: the point $P \in \overline{AB}$ such that $|BP| = \frac{1}{3} |AB|$ and the point $Q \ne A$ such that $|BQ| = |AB|$ (see figure 7); their distance is $\frac{4}{3} |AB|$, and since the center of $S$ lies on the line $AB$, $\overline{PQ}$ is a diameter of the sphere $S$. Therefore, the diameter of $S$ equals $\frac{4}{3} |AB|$ (this could also be inferred from the analytical formula obtained in the proof of Apollonius' theorem). It follows that the segment $\overline{C_1C_2}$ must also be a diameter of the sphere $S$. The angle $C_1C_0C_2$ is a right angle, because $C_0 \in S$. The lines $AB$ and $C_1C_2$ intersect (at the center of the sphere $S$), so the points $A$, $B$, $C_1$, $C_2$ are coplanar.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 693 |
XXXIX OM - II - Problem 5
Determine whether every rectangle that can be covered by 25 circles of radius 2 can also be covered by 100 circles of radius 1.
|
The answer is affirmative. Let $ R $ be a rectangle that can be covered by 25 circles of radius 2. The axes of symmetry of the rectangle $ R $ divide it into four rectangles similar to it on a scale of $ 1 /2 $. Each of them can be covered by a family of 25 circles of radius 1; it is enough to transform the given 25 circles of radius 2 by a homothety of scale $ 1/2 $ with respect to the appropriate vertex. In this way, we obtain a total of 100 circles of radius 1 covering the entire rectangle $ R $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 694 |
VII OM - III - Problem 5
Prove that any polygon with a perimeter equal to $2a$ can be covered by a disk with a diameter equal to $a$.
|
Let $W$ be a polygon whose perimeter has length $2a$. Choose two points $A$ and $B$ on its perimeter that divide the perimeter into halves, i.e., into two parts of length $a$; then $AB < a$ (Fig. 20).
We will prove that the disk of radius $\frac{a}{2}$, whose center lies at the midpoint $O$ of segment $AB$, completely covers the polygon $W$.
We will use the method of proof by contradiction. Suppose that the disk does not cover the polygon. Then part of the polygon protrudes beyond the disk, so there exists a point $C$ on the perimeter of the polygon such that $OC > \frac{a}{2}$. Point $C$ is different from $A$ and $B$, since $OA = OB < \frac{a}{2}$, and it divides one of the halves of the perimeter defined by $A$ and $B$ into two parts. Let $p$ be the length of the part whose endpoints are $A$ and $C$, and $q$ the length of the part whose endpoints are $C$ and $B$; obviously $p + q = a$. Since $p \geq AC$ and $q \geq CB$, therefore
But
From (1) and (2) we obtain the contradiction $a > a$; the assumption that the disk does not cover the polygon is therefore false.
In writing inequality (2), we applied the theorem that the sum of the sides $AC$ and $CB$ of triangle $ABC$ is greater than twice the median $OC$. This theorem is easy to prove: Let $C'$ be the point symmetric to point $C$ with respect to point $O$, then
Note 1. If the disk has a diameter smaller than $a$, for example, equal to $a - d$ ($d < a$), then not every polygon with perimeter $2a$ can be covered by it. For example, it cannot cover a triangle with sides $a - \frac{d}{2}$, $\frac{a}{2} + \frac{d}{4}$, $\frac{a}{2} + \frac{d}{4}$.
Note 2. A more general theorem is true: a disk of diameter $a$ can cover any piece of the plane bounded by a curve of length $2a$. The proof remains the same as before.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 696 |
IX OM - II - Task 3
Prove that if the polynomial $ f(x) = ax^3 + bx^2 + cx + d $ with integer coefficients takes odd values for $ x = 0 $ and $ x = 1 $, then the equation $ f(x) = 0 $ has no integer roots.
|
According to the assumption, the numbers $ f(0) = d $ and $ f(1) = a + b + c + d $ are odd. We will show that $ f(x) $ is then an odd integer for every integer $ x $. Indeed, if $ x $ is an even number, then $ f(x) = ax^3 + bx^2 + cx + d $ is the sum of even numbers $ ax^3 $, $ bx^2 $, $ cx $, and the odd number $ d $. If, on the other hand, $ x $ is an odd number, then $ f(x) = a (x^3 - 1) + b(x^2 - 1) + c (x - 1) + (a + b + c + d) $ is the sum of even numbers $ a(x^3 - 1) $, $ b(x^2 - 1) $, $ c(x - 1) $, and the odd number $ a + b + c + d $. Therefore, $ f(x) \ne 0 $, Q.E.D.
Note. The above reasoning can be applied to any polynomial
with integer coefficients $ a_0, a_1, \ldots, a_n $.
When $ x $ is an even integer, then
is also an even number, so the number $ f(x) $ has the same parity as $ a_n $, i.e., $ f(0) $.
When $ x $ is an odd integer, then
is an even number, so the number $ f(x) $ has the same parity as the number $ a_0 + a_1 + \ldots + a_n $ equal to $ f(1) $.
If, therefore, $ f(0) $ and $ f(1) $ are odd numbers, then for every integer $ x $ the number $ f(x) $ is odd, hence $ f(x) \ne 0 $.
We can also say that if there exist integers $ k $ and $ l $ such that $ f(2k) $ and $ f(2l + 1) $ are odd numbers, then $ f(x) $ is an odd number for every integer $ x $, hence $ f(x) \ne 0 $.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 699 |
XLVII OM - III - Problem 2
Inside a given triangle $ABC$, a point $P$ is chosen satisfying the conditions: $|\measuredangle PBC|=|\measuredangle PCA| < |\measuredangle PAB|$. The line $BP$ intersects the circumcircle of triangle $ABC$ at points $B$ and $E$. The circumcircle of triangle $APE$ intersects the line $CE$ at points $E$ and $F$. Prove that points $A$, $P$, $E$, $F$ are consecutive vertices of a quadrilateral and that the ratio of the area of quadrilateral $APEF$ to the area of triangle $ABP$ does not depend on the choice of point $P$.
|
Let's denote the circumcircle of triangle $ABC$ by $\Omega$, and the circumcircle of triangle $APE$ by $\omega$ (Figure 10). Through point $E$, we draw a tangent line to circle $\omega$ and choose an arbitrary point $Q$ on this line, lying on the same side of line $AE$ as points $B$, $C$, and $P$ (it could be, for example, the intersection point of this line with circle $\Omega$, different from $E$). It lies on the opposite side of line $EP$ from point $A$ - that is, on the same side of line $EP$ as point $C$.
The following angle equalities hold:
equalities (1) connect pairs of inscribed angles in circle $\Omega$, and equality (2) follows from the fact that line $EQ$ is tangent to circle $\omega$ (and from the choice of point $Q$).
The conditions given in the problem regarding the angles determined by point $P$, combined with the first equality (1), lead to the following conclusions:
and
Using successively the relations (3), (4), and the middle equality (1), we have:
In view of equality (2), we can rewrite the obtained relationship as
Earlier, we stated that points $Q$ and $C$ lie on the same side of line $EP$. From inequality (5), it follows that points $P$ and $C$ lie on opposite sides of line $EQ$.
Furthermore, note that $P$ and $F$, as points of the circle tangent to line $EQ$, lie on the same side of this line. Therefore, points $C$ and $F$ lie on opposite sides of this line. Hence, we conclude that points $C$, $E$, and $F$ lie on line $CE$ in that exact order.
Equality (3) shows that line $AE$ is parallel to $PC$. Since point $E$ lies between points $C$ and $F$, this means that point $F$ lies on the opposite side of line $AE$ from any point on line $PC$; in particular, it lies on the opposite side of this line from point $P$. Thus, points $A$, $P$, $E$, and $F$ are consecutive vertices of a quadrilateral; this is the first part of the thesis of the problem.
From the proven parallelism $AE \parallel PC$, it also follows that triangles $APE$ and $ACE$ have equal areas; they have a common side $AE$ and equal heights dropped from vertices $P$ and $C$ to this side. Therefore, the area of quadrilateral $APEF$ is equal to the area of triangle $ACF$.
Quadrilateral $APEF$ is inscribed in a circle, so
We rewrite the third equality (1) as $|\measuredangle ACF| = |\measuredangle ABP|$. From these equalities, we infer that triangle $ACF$ is similar to triangle $ABP$. Therefore,
The obtained number is a constant for a given triangle $ABC$, independent of the choice of point $P$ satisfying the given conditions.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 700 |
XLV OM - I - Problem 9
In a conference, $2n$ people are participating. Each participant has at least $n$ acquaintances among the other participants. Prove that all participants of the conference can be accommodated in double rooms so that each participant shares a room with an acquaintance.
|
In the problem, it is tacitly assumed that if one person knows another, then the latter also knows the former, and that no one is their own acquaintance.
Let $ m $ be the maximum number of disjoint pairs of acquaintances that can be formed among the conference participants. The task reduces to proving that $ m = n $. Suppose, then, that $ m < n $ and let $ P_1, \ldots, P_m $ be a system of $ m $ disjoint pairs of acquaintances. Denote the set of remaining conference participants (those who are not in any of the pairs $ P_1, \ldots, P_m $) by $ Q $.
Consider any two participants $ A $ and $ B $ from the set $ Q $. Neither of them has an acquaintance in the set $ Q $ (if they did, they would form a ($ m + 1 $)-th pair, contradicting the maximality of $ m $). Therefore, according to the assumption, both participant $ A $ and $ B $ have at least $ n $ acquaintances in the set $ P_1 \cup \ldots \cup P_m $.
Assign to each conference participant the number $ 2 $, $ 1 $, or $ 0 $, depending on whether they know both $ A $ and $ B $, exactly one of them, or neither. For $ i = 1, \ldots, m $, let $ s_i $ be the sum of the numbers assigned to the two conference participants forming the pair $ P_i $. From the conclusion of the last sentence in the previous paragraph, the inequality $ s_1 + \ldots + s_m \geq 2n $ follows, since the sum $ s_1 + \ldots + s_m $ represents the number of all "connections" (in the sense of the "being acquainted" relation) between the elements of the set $ P_1 \cup \ldots \cup P_m $ and the elements of the set $ \{A, B\} $.
If all the terms in the sum $ s_1 + \ldots + s_m $ were less than or equal to $ 2 $, the value of this sum would not exceed $ 2m $, and thus would be less than $ 2n $. Therefore, at least one term is greater than $ 2 $. Without loss of generality, assume that $ s_m \geq 3 $ (the order of numbering can be changed if necessary). Let the participants forming the pair $ P_m $ be denoted by $ C $ and $ D $. The sum of the numbers assigned to them (i.e., $ s_m $) is $ 3 $ or $ 4 $, which means that at least one of these participants knows both $ A $ and $ B $, and the other knows at least one of these people. We can assume (again, a matter of notation) that $ C $ is acquainted with both $ A $ and $ B $, while $ D $ is acquainted with $ B $.
We split the pair $ \{C, D\} $ and form two new pairs: $ \{A, C\} $ and $ \{B, D\} $. Together with the remaining pairs $ P_1, \ldots, P_{m-1} $, we thus have $ m + 1 $ disjoint pairs of acquaintances. The obtained contradiction with the maximality of $ m $ proves the thesis of the problem.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 701 |
I OM - B - Task 14
Prove that the sum of natural numbers is divisible by 9 if and only if the sum of all digits of these numbers is divisible by 9.
|
We will first prove that the difference between a natural number $ a $ and the sum $ s $ of its digits is divisible by 9. Let $ C_O, C_1, C_2, C_3, \dots $ denote the units, tens, hundreds, thousands, and further place values of this number. Then
Subtracting these equations side by side, we get
Since each term on the right side is divisible by 9, the number $ a - s $ is also divisible by 9.
Let $ a_1, a_2, \dots, a_n $ denote natural numbers, and $ s_1, s_2, \dots, s_n $ denote the sums of the digits of these numbers, respectively.
In the identity
the term on the right side within the broken brackets is divisible by 9, as it is the sum of numbers divisible by 9. Therefore, the number $ a_1 + a_2 + \dots + a_n $ is divisible by 9 if and only if the term $ s_1 + s_2 + \dots + s_n $, i.e., the sum of all the digits of the numbers $ a_1, a_2, \dots, a_n $, is divisible by 9, which was to be proved.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 702 |
XXXII - III - Task 4
On the table lie $ n $ tokens marked with integers. If among these tokens there are two marked with the same number, for example, the number $ k $, then we replace them with one token marked with the number $ k+1 $ and one with the number $ k-1 $. Prove that after a finite (non-negative) number of such changes, all tokens will be marked with different numbers.
|
We will first prove the following lemma:
Lemma. Let $m_0$ be the minimum number among the numbers written on the tokens on the table before any changes are made. If after $k$ changes, the numbers $m_1^{(k)}, m_2^{(k)}, \ldots, m_{j_k}^{(k)}$ are all the numbers not greater than $m_0$, with $m_{j_k}^{(k)} \leq m_{j_k-1}^{(k)} \leq \ldots \leq m_2^{(k)} \leq m_1^{(k)} \leq m_0^{(k)} = m_0$, then $m_i^{(k)} - m_{i+1}^{(k)} \leq 2$ for $i = 0, 1, \ldots, j_k-1$.
We will prove the lemma by induction on $k$.
For $k = 1$, there can only be $m_1^{(1)} = m_0 - 1$, and this situation occurs when there were at least two tokens with the number $m_0$ at the beginning. Of course, then $m_0^{(1)} - m_1^{(1)} \leq 2$.
Assume that for some $k$, the numbers $m_1^{(k)}, \ldots, m_{j_k}^{(k)}$ are all the numbers not greater than $m_0$ written on the tokens after $k$ changes, and
If in the $(k+1)$-th step we exchange two tokens with numbers greater than $m_0$, then each term $m_i^{(k+1)}$ is equal to the corresponding $m_i^{(k)}$, so the thesis is obviously satisfied. Let $m_i^{(k)} = m_{i+1}^{(k)} = m$ and the $(k+1)$-th step of our procedure involves replacing tokens with the number $m$ with tokens with the numbers $m+1$ and $m-1$. The sequence $(m_i^{(k+1)})$ thus arose from the sequence $(m_i^{(k)})$ by replacing two terms equal to $m$ with terms equal to $m+1$ and $m-1$. The difference between these terms is $2$, the difference between each of these terms and the neighboring term does not exceed $2$, and the difference between each pair of other neighboring terms has not changed, so it also does not exceed $2$. This completes the inductive proof of the lemma.
From the lemma, it follows that after each change, the number written on any token will not be less than $m_0 - 2n$.
Notice now that if before starting the procedure described in the problem we add to each of the numbers written on the tokens a fixed integer, this will not change our problem in any way. By adding the number $2n - m_0 + 1$ if necessary, we can assume that both at the beginning and after each change, all the numbers written on the tokens are positive. Consider now the product $P_0$ of the numbers initially written on the tokens, the product $P_1$ of the numbers written on the tokens after the first change, the product $P_2$ of the numbers on the tokens after the second change, etc. Each term of the sequence $P_i$ is the product of natural numbers, and $P_{i+1}$ arises from $P_i$ by replacing two factors equal to $k$ with the product $(k+1)(k-1)$, i.e., $P_i = k^2c$, $P_{i+1} = (k+1)(k-1)c$. Since $c > 0$ and $k^2 > k^2 - 1 = (k+1)(k-1)$, it follows that $P_i > P_{i+1}$. The sequence $(P_i)$ is therefore a decreasing sequence of natural numbers, and such a sequence must be finite. This proves that after a finite number of steps, all tokens will be marked with different numbers.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 704 |
XII OM - III - Task 1
Prove that every natural number which is not an integer power of the number $ 2 $, is the sum of two or more consecutive natural numbers.
|
Let $ N $ be a natural number which is not an integer power of two, hence
where $ m $ and $ n $ are integers, with $ m \geq 1 $, $ n \geq 0 $.
We need to prove that there exist integers $ a $ and $ k $ such that
and
i.e.,
Condition (2) is satisfied if $ 2a + k - 1 = 2^{n+1} $, $ k = 2m + 1 $, i.e., if
The solution $ (a, k) $ of equation (2), defined by formulas (3), satisfies condition (1) only if $ a > 0 $, i.e., if
Condition (2) is also satisfied when $ 2a + k - 1 = 2m + 1 $, $ k = 2^{n+1} $, i.e., if
The numbers $ a $ and $ k $ defined by formulas (5) are integers; they satisfy condition (1) only if
Since one of conditions (4) and (6) is always satisfied, equation (2) always has an integer solution $ (a, k) $ that satisfies inequalities (1).
For example, if $ N = 30 $, we get the solution $ a = 6 $, $ k = 4 $, and $ N = 6 + 7 + 8 + 9 $; if $ N = 225 $, then $ a = 112 $, $ k = 2 $, and $ N = 112 + 113 $.
Note. Let's consider whether the above-found decomposition of the natural number $ N $ into a sum of consecutive natural numbers is the only such decomposition.
Suppose, as before, that the integers $ a $ and $ k $ satisfy conditions (1) and (2). Only the following two cases are possible:
A) The number $ k $ is even, so the number $ 2a + k - 1 $ is odd.
From equation (2), it follows that $ k $ is divisible by $ 2^{n+1} $, hence
and
where $ p $ and $ q $ are non-negative integers. Since $ a \geq 1 $, it follows from equation (8) that
Conversely, if the number $ 2m + 1 $ has the form (9), where $ p $ and $ q $ are integers, with $ p \geq 0 $, and $ q $ satisfies condition (10), then the numbers $ a $ and $ k $ calculated according to formulas (7) and (8) define the decomposition of the number $ N $ into a sum of consecutive natural numbers. If $ p = 0 $, this is the same decomposition we found earlier [formulas (5)].
B) The number $ k $ is odd, so $ 2a + k - 1 $ is divisible by $ 2^{n+1} $. In this case,
from which, as before,
where the integers $ p $ and $ q $ satisfy the inequalities
since $ k > 1 $, $ a > 0 $.
Conversely, if the number $ 2m + 1 $ has the form (9), where $ p $ and $ q $ are integers satisfying conditions (13), then the numbers $ a $ and $ k $ calculated according to formulas (11) and (12) define the desired decomposition of the number $ N $. If $ p = 0 $, this is the decomposition found earlier [formulas (3)].
We have obtained the following result. If the number $ 2m + 1 $ is a prime number, then there is only one decomposition of the number $ N $ into a sum of consecutive natural numbers. This is the decomposition found earlier, defined by formulas (3) or formulas (5), depending on whether inequality (4) or inequality (6) holds. If, however, $ 2m + 1 $ is a composite number, then there are other decompositions. Specifically, each decomposition of the number $ 2m + 1 $ into two factors unequal and greater than 1 corresponds to two decompositions of the number $ N $. It is easy to show (which we propose to the reader) that either one of these decompositions corresponds to case A) and the other to case B), or both belong to case B). Finally, if there is a decomposition of the number $ 2m + 1 $ into two equal factors, i.e., if it is a square of an integer, then in addition to the decompositions already mentioned, there is one more decomposition belonging to case B).
For example:
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 705 |
XLIII OM - I - Problem 9
Prove that among any $ n+2 $ integers, there exist two such that their sum or difference is divisible by $ 2n $.
|
Let the considered numbers be denoted by $ k_1, \ldots, k_{n+2} $ and let $ r_i $ be the remainder of the division of $ k_i $ by $ 2n $.
Suppose that among these remainders there are two equal ones:
In this case, the difference $ k_i - k_j $ is divisible by $ 2n $; the thesis holds.
Suppose, on the other hand, that among the remainders $ r_1, \ldots, r_{n+2} $ there are no two equal. These remainders are then different elements of the set $ \{0,1, \ldots, 2n-1\} $. At least $ n $ of them do not equal $ 0 $ or $ n $; by changing the numbering if necessary, we can assume that the remainders $ r_1, \ldots, r_n $ are different from $ 0 $ and $ n $, i.e., they belong to the set
This set is the union of the following two-element sets:
We thus have $ n $ different numbers $ r_1, \ldots, r_n $ distributed among the union of $ n-1 $ sets $ A_1, \ldots, A_{n-1} $. At least two of them belong to the same set $ A_l $, by the pigeonhole principle. Their sum equals $ r_i + r_j = 2n $. This means that the sum $ k_i + k_j $ is divisible by $ 2n $; the thesis holds in this case as well.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 707 |
XLII OM - III - Problem 6
Prove that for every triple of real numbers $ x,y,z $ such that $ x^2+y^2+z^2 = 2 $, the inequality $ x+y+z\leq 2+xyz $ holds, and determine when equality occurs.
|
Let's adopt the notations
From the assumption that $ x^2 + y^2 + z^2 = 2 $, the following equalities follow:
The left sides are non-negative numbers. Hence, $ u, v, w \leq 1 $. Therefore, we have the inequality
From the adopted notations, it follows that
which means
Furthermore,
and
We need to prove that
We transform the left side of inequality (1), using the relationships obtained above:
According to (1), this is a non-negative number. Therefore,
Since,
we conclude that $ (s-q)^2 \leq 4 $, which means
The inequality to be proven (2) follows from this immediately.
For it to become an equality, equality signs must hold in the estimates (1) and (3). Equality in (3) means that one of the numbers $ x $, $ y $, $ z $ is zero. Let's assume, for example, that $ z = 0 $. Then $ u = v = 0 $ and equality in (1) means that $ w = 1 $, i.e., $ xy = 1 $. Since $ q = 0 $, the proposed equality in (2) takes the form $ s = 2 $, i.e., $ x + y = 2 $.
The only solution to the system of equations: $ x + y = 2 $, $ xy = 1 $ is $ x = y = 1 $.
We assumed, for example, that $ z = 0 $. Of course, if $ x = 0 $ or $ y = 0 $, then, respectively, we obtain the relationship $ y = z = 1 $ or $ z = x = 1 $ as a necessary condition for the equality in (2) to hold.
And conversely: if two of the numbers $ x $, $ y $, $ z $ are ones and one is zero, then the proven inequality becomes an equality. This is therefore a necessary and sufficient condition.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 708 |
XXXVIII OM - II - Problem 3
On a chessboard with dimensions 1000 by 1000 and fields colored in the usual way with white and black, there is a set A consisting of 1000 fields. Any two fields of set A can be connected by a sequence of fields from set A such that consecutive fields share a common side. Prove that set A contains at least 250 white fields.
|
A finite sequence of chessboard squares (of arbitrary dimensions) having the property that any two consecutive squares share a side, we will call a path, and the number of squares in this sequence decreased by $1$ - the length of the path (a single square constitutes a path of length $0$).
A non-empty set $A$ composed of any number of chessboard squares and such that any two of its squares are connected by a path contained in it, we will call connected.
We will prove by induction on $n$ that the number of white squares in any connected set $A$ composed of $n$ squares is not less than $(n-1)/4$. For $n = 1000$, the thesis of the problem will follow from this.
When $n = 1$, the given value is $0$, so there is nothing to prove.
Let us fix a natural number $n > 1$ and assume that every $m$-square connected set, where $m$ is any number less than $n$, has at least $(m-1)/4$ white squares. Let a connected set $A$ composed of $n$ squares be given, and let $b$ be the number of white squares in the set $A$. We need to prove that
If we prove this, the proof of our statement will be completed by the principle of induction.
We will prove inequality (1) in two ways.
Choose one white square from the set $A$ (its existence follows from the fact that $n > 1$),
call it $P$ and remove it from the set $A$. The resulting set of $n-1$ squares we will denote by $A'$. If $Q$ is any square in $A'$, then the set of all squares in $A'$ that can be connected to $Q$ by a path contained in $A'$ is connected. Since every square in $A'$ is connected to the square $P$ by a path contained in $A$, which must pass through one of the four squares adjacent to $P$, it follows that every square in $A'$ is connected to one of these four squares by a path contained in $A'$.
It follows that $A'$ is the union of at most four non-empty disjoint connected sets. Let us call them $A_1, \ldots, A_k$ ($1 \leq k \leq 4$) and assume that the set $A_j$ consists of $m_j$ squares, of which $b_j$ are white ($j = 1, \ldots, k$).
Obviously
\[ b = b_1 + b_2 + \cdots + b_k + 1 \]
(because the square $P$ is white). From the induction hypothesis
\[ b_j \geq \frac{m_j - 1}{4} \]
for $j = 1, \ldots, k$.
It suffices to add these inequalities to obtain
\[ b_1 + b_2 + \cdots + b_k \geq \frac{m_1 - 1}{4} + \frac{m_2 - 1}{4} + \cdots + \frac{m_k - 1}{4} \]
The inequality (1) immediately follows from this, because $k \leq 4$.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 709 |
XXIV OM - III - Problem 4
On a line, a system of segments with a total length $ < 1 $ is given. Prove that any system of $ n $ points on the line can be moved along it by a vector of length $ \leq \frac{n}{2} $ so that after the move, none of the points lies on any of the given segments.
|
Let $ P_i $, where $ i = 1, 2, \ldots, n $, be given points and let $ A $ be the union of given segments with a total length less than $ 1 $. For $ i = 1, 2, \ldots, n $, let $ I_i $ be a segment of length $ n $, whose center is the point $ P_i $.
Of course, $ A \cap I_i $ is the union of segments with a total length less than $ 1 $. Let $ \varphi_i $ be the translation by the vector $ \overrightarrow{P_iP_1} $ for $ i = 1, 2, \ldots, n $. Since translation preserves the length of a segment and $ \varphi_i(I_i) = I_1 $, the set
\[
\varphi_1(A) \cup \varphi_2(A) \cup \ldots \cup \varphi_n(A)
\]
is the union of segments with a total length less than $ n $ contained in $ I_i $. As we know, $ I_1 $ has a length of $ n $. Therefore, there exists a point $ Q $ belonging to $ I_1 - I $.
We will show that the translation $ \varphi $ by the vector $ \overrightarrow{P_1Q} $ satisfies the conditions of the problem. Since $ Q \in I_1 $, $ P_1 $ is the center of the segment $ I_1 $, and the length $ l_1 $ is equal to $ n $, we have $ P_1Q \leq \displaystyle \frac{n}{2} $. Therefore, for each $ i = 1, 2, \ldots, n $, we have $ \varphi (P_i) \in I_i $.
If for some $ i $ it were $ \varphi (P_i) \in A $, then in view of $ \varphi (P_i) \in I_i $ we would have $ \varphi (P_i) \in A \cap I_i $. Hence
\[
\varphi_i \varphi (P_i) = Q.
\]
The transformation $ \varphi_i \varphi $ is a translation by the vector $ \overrightarrow{P_1Q} + \overrightarrow{P_iP_1} = \overrightarrow{P_iQ} $. Therefore, $ \varphi_i \varphi(P_i) = Q $. From (1) it follows that $ Q \in I $. This contradicts the definition of the point $ Q $.
The obtained contradiction proves that for each $ i = 1, 2, \ldots, n $ we have $ \varphi(P_i) \not \in A $.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 710 |
IX OM - III - Problem 6
Prove that among all quadrilaterals circumscribed around a given circle, the one with the smallest perimeter is a square.
|
Since the area of a polygon circumscribed about a circle is proportional to its perimeter, the theorem we need to prove is equivalent to the theorem that among all quadrilaterals circumscribed about a given circle, the square has the smallest area.
Let $Q$ be a square circumscribed about the circle $K$, $ABCD$ a quadrilateral circumscribed about the same circle, but not a square, and $K$ the circle circumscribed about the square. For greater clarity, in Fig. 34, the quadrilaterals are described on two equal circles instead of one.
Notice that at least one vertex of the quadrilateral $ABCD$ must lie inside the circle $K$, since at least one of its angles is greater than a right angle.
The sides of the square cut off four equal circular segments from the circle $K$; denoting the areas of these segments by $s$, we have
The lines $AB$, $BC$, $CD$, $DA$ cut off from the circle $K$ the same segments of area $s$; since at least one of the points $A$, $B$, $C$, $D$ lies inside the circle $K$, at least two of these segments partially overlap. Therefore, the difference $K$ is smaller than the area of that part of the quadrilateral $ABCD$ which lies within the circle $K$, and thus even more so
Considering (1), we obtain
Note. The theorem we have proved is a special case of the following theorem.
Among all polygons with the same number $n$ of sides circumscribed about a given circle, the regular polygon has the smallest perimeter.
To obtain a proof of this theorem, it suffices to replace the words "square" and "quadrilateral" in the reasoning given above in method 1 with the words "regular polygon with $n$ sides" and "polygon with $n$ sides" and the number $4$ with $n$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 711 |
XXV OM - III - Problem 3
Let $ r $ be a natural number. Prove that the quadratic polynomial $ x^2 - rx - 1 $ is not a divisor of any non-zero polynomial $ p(x) $ with integer coefficients, all of which are less in absolute value than $ r $.
|
\spos{1} Suppose the polynomial
where $ a_n \ne 0 $, has integer coefficients, each in absolute value less than $ r $, and is divisible by the quadratic polynomial $ f(x) = x^2 - rx - 1 $. We then have
where the polynomial $ h(x) = b_0 + b_1x + \ldots + b_{n-2}x^{n-2} $ has integer coefficients, which follows from the polynomial division algorithm.
By comparing the corresponding coefficients of the polynomials on both sides of equation (1), we obtain
[ (2.0) \qquad a_0 = -b_0, \]
[ (2.1) \qquad a_1 = -b_1 - rb_0, \]
[ (2.2) \qquad a_2 = -b_2 - rb_1 + b_0, \]
[ \vdots \]
[ (2.k) \qquad a_k = -b_k - rb_{k-1} + b_{k-2} \ \text{for} \ k = 2, 3, \ldots, n-2,\]
[ \vdots \]
[ (2.n-2) \qquad a_{n-2} = - b_{n-2} - rb_{n-3} + b_{n-4}, \]
[ (2.n-1) \qquad a_{n-1} = - rb_{n-2} + b_{n-3}, \]
[ (2.n) \qquad a_n = b_{n-2} \]
From equation (2.n), it follows that $ b_{n-2} \ne 0 $. Then from (2.n - 1), it follows that $ b_{n-3} \ne 0 $ and the numbers $ b_{n-2} $ and $ b_{n-3} $ have the same sign. Otherwise, we would have $ |a_{n-1}| = |-rb_{n-2} + b_{n-3}| = |-rb_{n-2}| + |b_{n-3}| \geq r |b_{n-2}| \geq r $, which contradicts the assumption that $ |a_{n-1}| < r $. Similarly, from (2.n - 2), we obtain that $ b_{n-4} \ne 0 $ and the numbers $ b_{n-3} $ and $ b_{n-4} $ have the same sign. Otherwise, we would have $ |a_{n-2}| = |-b_{n-2} - rb_{n-3} + b_{n-4}| = |-b_{n-2}| + |-rb_{n-3}| + |b_{n-4}| \geq r |b_{n-3}| \geq r $, which contradicts the assumption that $ |a_{n-2}| < r $.
In general, if for some number $ k \in \{2, 3, \ldots, n - 2\} $ the numbers $ b_k $ and $ b_{k-1} $ have the same sign, then $ b_{k-2} \ne 0 $ and the numbers $ b_{k-1} $ and $ b_{k-2} $ have the same sign. Otherwise, from equation (2.k), we would obtain that $ |a_k| = |-b_k| + |-rb_{k-1}| + |b_{k-2}| \geq r $, which contradicts the assumption.
By the principle of induction, it follows that the numbers $ b_{n-2}, b_{n-3}, \ldots, b_1, b_0 $ are non-zero and have the same sign. Therefore, from equation (2.1), we obtain $ |a_1| = |-b_1 - rb_0| = |b_1| + |rb_0| \geq r $, which contradicts the assumption.
The obtained contradiction proves that such a polynomial $ p $ does not exist.
Note 1. In the above solution, we did not use the assumption that the polynomial $ p $ has integer coefficients in an essential way. It would have been sufficient to assume that $ |p_n| \geq 1 $ and $ |p_i| \leq r - 1 $ for $ i = 0, 1, \ldots, n-1 $.
Note 2. Let $ H(f) $ be the maximum absolute value of the coefficients of the polynomial $ f $. For polynomials with integer coefficients, it is generally not true that if $ f \mid g $, then $ H(f) \leq H(g) $. For example,
The translation is complete, and the original formatting and line breaks have been preserved.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 712 |
L OM - II - Task 2
A cube $ S $ with an edge of $ 2 $ is built from eight unit cubes. A block will be called a solid obtained by removing one of the eight unit cubes from the cube $ S $. A cube $ T $ with an edge of $ 2^n $ is built from $ (2^n)^3 $ unit cubes. Prove that after removing any one of the $ (2^n)^3 $ unit cubes from the cube $ T $, the resulting solid can be tightly filled with blocks.
|
We will conduct an inductive proof. For $ n = 1 $, the thesis of the problem is obviously true. Assume, therefore, that the thesis is true for some $ n $ and consider a cube $ C $ with edge $ 2^{n+1} $ and a distinguished unit cube $ D $. The cube $ C $ is divided into eight congruent cubes $ C_1, C_2, \ldots, C_8 $, each with edge $ 2^n $. One of them (let's assume it is $ C_8 $) contains the unit cube $ D $. Place a block in the very center of the cube $ C $ such that its intersection with each of the cubes $ C_1, C_2, \ldots, C_7 $ is a unit cube. Then we are left to fill with blocks eight cubes $ C_1, C_2, \ldots, C_8 $, each with a removed unit cube. This, by the inductive hypothesis, is feasible. The inductive proof is thus completed.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 713 |
LX OM - I - Task 3
The incircle of triangle $ABC$ is tangent to sides $BC$, $CA$, $AB$ at points $D$, $E$, $F$ respectively. Points $M$, $N$, $J$ are the centers of the incircles of triangles $AEF$, $BDF$, $DEF$ respectively. Prove that points $F$ and $J$ are symmetric with respect to the line $MN$.
|
om60_1r_img_2.jpg
First, we will prove that points $M$ and $N$ are the midpoints of the shorter arcs $FE$ and $FD$ of the incircle of triangle $ABC$.
Indeed, let $M$ be the midpoint of the shorter arc $FE$ of this
circle. The line $AC$ is tangent to it at point $E$, so (Fig. 2)
Since point $M$ is the midpoint of arc $EF$, triangle
$EM$ is isosceles. Therefore, $\measuredangle EFM$, which together
with equality (1) proves that point $M$ lies on the angle bisector of $\measuredangle AEF$. Similarly, we prove
that this point lies on the angle bisector of $\measuredangle AFE$, so it
coincides with the center $M$ of the incircle of triangle $AEF$. We reason similarly
for point $N$.
The incircle of triangle $ABC$ is also the circumcircle
of triangle $DEF$. Since $M$ is the midpoint of the shorter arc $EF$ of
this circle, line $DM$ contains the angle bisector of $\measuredangle EDF$. Therefore,
point $J$, which is the center of the incircle of this angle, lies on this line.
Similarly, we conclude that point $J$ lies on line $EN$.
To justify that points $J$ and $F$ are symmetric with respect to line
$MN$, it suffices to show that triangles $MFN$ and $MJN$ are congruent,
since this would mean that they are symmetric with respect to line $MN$.
However, we have
where the second equality follows from the fact that $N$ is the midpoint of arc $FD$.
Similarly, we get $\measuredangle JNM = \measuredangle FNM$. Triangles $MFN$ and $MJN$ therefore have
equal corresponding angles and a common side $MN$, so they are congruent
(by the angle-side-angle criterion). This completes the solution of the problem.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 715 |
X OM - I - Problem 12
Prove that if in a triangle $ ABC $ the numbers $ \tan A $, $ \tan B $, and $ \tan C $ form an arithmetic progression, then the numbers $ \sin 2A $, $ \sin 2B $, and $ \sin 2C $ also form an arithmetic progression.
|
From the assumptions of the problem, it follows that
thus
therefore
so
and finally
which was to be proved.
Note. The converse statement is also true: if the angles of triangle $ABC$ satisfy equation (2), then they also satisfy equation (1). Indeed, based on the transformations performed above in method 1, we can state that if (2) holds, then (2a) also holds, and therefore (1a), and thus (1). We can also use the transformations from method 2, performing them in reverse order, but we then need to justify the last transformation, i.e., dividing the equality by $ \cos A \cdot \cos B \cdot \cos C $, by showing that under assumption (2), none of the angles $A$, $B$, $C$ can be right angles. We leave this as an easy exercise for the reader.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 726 |
XIX OM - II - Task 3
Prove that if at least 5 people are sitting around a round table, then they can be rearranged so that each person has two different neighbors than before.
|
\spos{1} Let us denote the given people by natural numbers from $1$ to $n$ ($n \geq 5$) and assume that person $1$ is seated next to persons $n$ and $2$, person $2$ is seated next to persons $1$ and $3$, and so on, finally person $n$ is seated next to persons $n-1$ and $1$. We will simply say that the numbers from $1$ to $n$ are arranged in a cycle.
Let $n$ be an odd number. Arrange these same numbers in a cycle
in which the consecutive odd numbers and the consecutive even numbers appear separately. It is easy to observe that in cycle (2) each number has different neighbors than in cycle (1). Specifically: $1^\circ$ Each odd number, except for $1$ and $n$, is adjacent to even numbers in cycle (1), and to odd numbers in cycle (2); $2^\circ$ Each even number, except for $2$ and $n-1$, is adjacent to odd numbers in cycle (1), and to even numbers in cycle (2); $3^\circ$ The number $1$ is adjacent to numbers $n$ and $2$ in cycle (1), and to numbers $n-1$ and $3$ in cycle (2), which are different from $n$ and $2$ since $n \geq 5$; the same applies to the numbers $n$, $2$, and $n-1$.
Now, let $n$ be an even number and arrange the numbers from $1$ to $n$ in a cycle
in which, similarly to (2), the consecutive odd numbers and the consecutive even numbers appear separately, and then swap the numbers $n-2$ and $n$:
In cycle (3), each number has different neighbors than in cycle (1), which we observe in exactly the same way as in the case of $n$ being odd.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 729 |
VIII OM - III - Task 2
Prove that between the sides $ a $, $ b $, $ c $ and the opposite angles $ A $, $ B $, $ C $ of a triangle, the following relationship holds
|
Substituting in equation (1)
($ R $ - the radius of the circle circumscribed around the triangle), after dividing both sides by $ R^2 $, we get:
It suffices to justify formula (3), since formula (1) immediately follows from formulas (3) and (2). We will transform the right side of formula (3)
Thus, formula (3) is correct.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 730 |
XLVII OM - II - Problem 2
A circle with center $ O $ inscribed in a convex quadrilateral $ ABCD $ is tangent to the sides $ AB $, $ BC $, $ CD $, $ DA $ at points $ K $, $ L $, $ M $, $ N $, respectively, and the lines $ KL $ and $ MN $ intersect at point $ S $. Prove that the lines $ BD $ and $ OS $ are perpendicular.
|
Let the orthogonal projections of points $ B $ and $ D $ onto the line $ OS $ be denoted by $ U $ and $ V $, respectively. The angles $ OKB $, $ OLB $, and $ OUB $ are right angles, and thus the points $ K $, $ L $, and $ U $ lie on a circle whose diameter is the segment $ OB $ (Figure 7). (The point $ U $ may coincide with $ O $, and in this case, we cannot speak of the angle $ OUB $; but in such a case, the statement that $ U $ is a point on the mentioned circle does not require any justification.) Similarly, the points $ M $, $ N $, and $ V $ lie on a circle whose diameter is the segment $ OD $. Applying the theorem of intersecting chords to these circles, we obtain the equalities
The same theorem, applied to the circle $ w $ inscribed in the quadrilateral $ ABCD $, shows that the left sides of the above relationships are equal. Therefore, the right sides must also be equal, i.e., the equality $ |SU|=|SV| $ holds; and the segments $ SU $ and $ SV $ are contained in the ray $ SO^\to $. Hence, the conclusion is that the points $ U $ and $ V $ coincide, and thus the line $ BD $ is perpendicular to $ OS $.
Note: The reasoning conducted does not depend on whether $ S $ is the intersection point of the rays $ KL^\to $ and $ NM^\to $, or the rays $ LK^\to $ and $ MN^\to $. The same note applies to the other methods of solution.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 731 |
XXXVI OM - II - Problem 2
Prove that for a natural number $ n > 2 $, the number $ n! $ is the sum of $ n $ distinct divisors of itself.
|
We prove by induction a slightly stronger thesis: For every natural number $ n > 2 $, the number $ n! $ is the sum of $ n $ distinct divisors of itself, the smallest of which is $ 1 $. For $ n = 3 $, this is true: $ 3! = 6 = 1+2+3 $. Assume the thesis is true for some $ n $:
where the natural numbers $ k $ are divisors of $ n! $, and multiply the written equality by $ n+1 $ on both sides. We get
where $ l_1 = 1 $, $ l_2 = n $, $ l_i = (n+1)k_{i-1} $ for $ i > 2 $. The numbers $ l_i $ are divisors of $ (n+1)! $ and satisfy the relations $ 1 = l_1 < l_2 < \ldots < l_{n+1} $. This completes the inductive proof.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 732 |
XVIII OM - I - Problem 8
Given a sphere $ k $, a circle $ c $ lying on its surface, and a point $ S $ not lying on this surface, such that every line passing through $ S $ and through some point $ M $ of the circle $ c $ intersects the surface of the sphere $ k $ at another point $ N $ different from $ M $. Prove that the set $ Z $ of all points $ N $ forms a circle.
|
When point $ S $ lies in the plane of circle $ c $, the proof of the theorem's thesis is immediate, for $ S $ is then inside circle $ c $ (since points lying outside $ c $ do not satisfy the adopted assumption) and the set of all points $ N $ is circle $ c $.
Assume that point $ S $ does not lie in the plane of circle $ c $ and let $ H $ be the orthogonal projection of point $ S $ onto this plane; points $ S $ and $ H $ are different (Fig. 3).
First, suppose that point $ H $ does not lie on circle $ c $. Let $ M $ be a point on circle $ c $, and $ N $ the second point of intersection of line $ SM $ with the surface of sphere $ k $. Draw through point $ N $ in the plane $ SHM $ a perpendicular to $ SM $; it intersects line $ SH $ at some point $ L $, since lines $ SH $ and $ SM $ are not perpendicular.
Right triangles $ SHM $ and $ SNL $ are similar, so $ \frac{SH}{SM}= \frac{SN}{SL} $, and thus $ SL = \frac{SM \cdot SN}{SH} $.
The value of the product $ SM \cdot SN $ does not depend on the position of point $ M $ on the surface of the sphere; this number is known as the power of point $ S $ relative to the sphere. The length $ SH $ also does not depend on the position of point $ M $. It follows from the above equality that the length $ SL $ is constant, i.e., all points $ M $ on circle $ c $ correspond to the same point $ L $. Since $ \measuredangle SNL $ is a right angle, point $ N $ lies on the surface of sphere $ k_1 $ with diameter $ SL $, and thus belongs to circle $ c_1 $, which is the intersection of the surface of sphere $ k_1 $ with the surface of sphere $ k $. The set $ Z $ of all points $ N $ is therefore contained in circle $ c_1 $.
Conversely, if $ N $ is a point on circle $ c_1 $, and $ M $ is a point on ray $ SN $ such that $ SM = \frac{SH \cdot SL}{SN} $, then point $ M $ lies on the surface of sphere $ k $, since then $ SM \cdot SN=SH \cdot SL $ is equal to the power of point $ S $ relative to $ k $. Since triangles $ SHM $ and $ SNL $ are similar (because $ SM \colon SH = SL \colon SN $, and $ \measuredangle HSM = \measuredangle NSL $), $ \measuredangle SHM = \measuredangle SNL = 90^\circ $, which implies that point $ M $ lies in the plane of circle $ c $. Therefore, point $ M $ lies on circle $ c $, and point $ N $ belongs to set $ Z $.
We conclude from this that set $ Z $ is circle $ c_1 $.
In the case where point $ H $ lies on circle $ c_1 $, the above proof requires a minor modification. From the equality $ SH \cdot SL = SM \cdot SN $, it follows that point $ L $ lies on the surface of sphere $ k $, and is thus one of the points in set $ Z $. The further reasoning, which proceeds as before, consists in showing that the remaining points of set $ Z $ form the set of points of circle $ c_1 $ different from $ L $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 733 |
XII OM - I - Problem 12
Prove that if all the faces of a tetrahedron are congruent triangles, then these triangles are acute.
|
Suppose each face of the tetrahedron $ABCD$ (Fig. 11) is a triangle with sides $a$, $b$, $c$; let, for example, $BC = a$, $CA = b$, $AB = c$. There are 3 cases, which we will consider in turn. 1° - None of the lengths $a$, $b$, $c$ are equal. Two edges emanating from the same vertex cannot then be equal, since they are sides of one of the triangles; therefore, opposite edges are equal, i.e., $AD = a$, $BD = b$, $CD = c$. Let $M$ be the midpoint of one of the edges, for example, $AB$. The medians $MD$ and $MC$ of triangles $ABD$ and $ABC$ are equal and are sides of triangle $DMC$, and since $MD + MC > DC$, it follows that $MD > \frac{1}{2} DC = \frac{1}{2} AB$. If, in triangle $ADB$, the median $MD$ is greater than half the base $AB$, then angle $ADB$ is acute. Indeed, from the inequalities $AM < MD$ and $BM < MD$, the inequalities $\measuredangle ADM < \measuredangle DAM$ and $\measuredangle MDB < \measuredangle DBM$ follow, hence $\measuredangle ADB < \measuredangle DAM + \measuredangle DBM$, and thus $2\measuredangle ADB < \measuredangle DAM + \measuredangle DBM + \measuredangle ADB = 180^\circ$.
2°. The faces of the tetrahedron are isosceles triangles, let, for example, $a = b \neq c$. Then, in triangle $ADB$, where $AB = c$, it must be that $AD = BD = a$, and since $BC = a$, it follows that $CD = c$, so in this case, too, the opposite edges of the tetrahedron are equal; the rest of the proof remains the same as before.
3°. If $a = b = c$, the faces of the tetrahedron are equilateral triangles, and thus they are acute.
Note 1. The theorem can be proved in a way that does not require distinguishing several cases; such a proof is considered more elegant in mathematics.
We will show that if the faces of the tetrahedron are congruent triangles with sides $a$, $b$, $c$, then the opposite edges of the tetrahedron are equal. Suppose this is not the case. Let, for example, $AB = c$, $CD = a$, with $a \neq c$.
One of the sides $BC$ and $AC$ of triangle $ABC$ equals $a$, let, for example, $BC = a$. One of the sides of triangle $BCD$ equals $c$, and since $BC = a$ and $CD = a$, it follows that $BD = c$. In this case, triangles $ABC$ and $BCD$ cannot be congruent, as the first has 2 sides equal to $c$, and the second has 2 sides not equal to $c$. The assumption that $AB \neq CD$ leads to a contradiction, so $AB = CD$ and similarly $BC = AD$, $AC = BD$. The rest of the proof proceeds as in the previous method.
Note 2. The question arises whether there exists a tetrahedron whose faces are congruent to any given acute triangle $ABC$. We will prove that this is indeed the case. Draw through points $A$, $B$, $C$ lines parallel to lines $BC$, $CA$, and $AB$, respectively. They will form a triangle $D_1D_2D_3$ (Fig. 12) consisting of triangle $ABC$ and 3 triangles congruent to it, and it is easy to see that the tetrahedron with base $ABC$ and height $h = HD$, where $H$ is the orthocenter (the orthocenter of a triangle is the point of intersection of its altitudes) of triangle $D_1D_2D_3$, is the desired tetrahedron. To prove this, we need to show that $DB = AC$, $DA = BC$, and $DC = AB$. Indeed, $DB^2 = h^2 + HB^2$, so by formula (1) $DB = DXB = AC$. Let the altitudes $D_1M$ and $D_2N$ intersect $BC$ and $AC$ at points $K$ and $L$, respectively. Note that triangles $D_1HN$ and $D_2HM$ are similar, so $\frac{D_1H}{HN} = \frac{D_2H}{HM}$, hence $\frac{D_1K+KH}{LN-LH}=\frac{D_2L+LH}{KN-KH}$, and since $LN=D_2L$ and $KM = D_1K$, from the previous equality we get $D_1K^2- KH^2 = D_2L^2 - LH^2$ and thus (based on the theorem that in a triangle, the difference of the squares of two sides equals the difference of the squares of the projections of these sides on the third side), $D_1B^2 - HB^2 = D_2A^2 - HA^2 = h^2$. Since also $DA^2 - HA^2 = h^2$, it follows that $DA = D_2A = BC$. In a similar way, we prove that $DC = AB$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 741 |
XXXIV OM - II - Problem 3
Point $ P $ lies inside triangle $ ABC $, such that $ \measuredangle PAC = \measuredangle PBC $. Points $ L $ and $ M $ are the projections of $ P $ onto lines $ BC $ and $ CA $, respectively, and $ D $ is the midpoint of segment $ AB $. Prove that $ DL = DM $.
|
Let $ Q $ and $ R $ be the midpoints of segments $ AP $ and $ BP $, respectively. These points are the centers of the circles circumscribed around triangles $ APM $ and $ BPL $, which implies that
as central angles subtending the same arcs as the corresponding inscribed angles. Angles $ PAM $ and $ PBL $ are equal by assumption. Therefore,
om34_2r_img_6.jpg
Segment $ \overline{DQ} $ connects the midpoints of sides $ \overline{AB} $ and $ \overline{AP} $ of triangle $ ABP $. It follows that $ \overline{DQ} || \overline{BP} $. Similarly, $ \overline{DR} || \overline{AP} $. Therefore, quadrilateral $ DRPQ $ is a parallelogram, which implies that
From the equality (1) and (2), we obtain that $ \measuredangle MQD = \measuredangle DRL $. Moreover, $ MQ= QP= DR $ and $ LR= RP= DQ $. It follows that triangles $ MQD $ and $ DRL $ are congruent, and thus $ DM = DL $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 743 |
XXXIII OM - III - Problem 2
In a cyclic quadrilateral $ABCD$, a line passing through the midpoint of side $\overline{AB}$ and the intersection of the diagonals is perpendicular to side $\overline{CD}$. Prove that sides $\overline{AB}$ and $\overline{CD}$ are parallel or the diagonals of the quadrilateral are perpendicular.
|
Let $O$ be the point of intersection of the diagonals, $E$ - the midpoint of side $\overline{AB}$, $F$ - the point of intersection of side $\overline{CD}$ and line $EC$. Since quadrilateral $ABCD$ is inscribed in a circle, therefore
Thus
By applying the Law of Sines to triangles $AOE$ and $BOF$ in sequence, we get
Since $EA = BE$, dividing the last equations side by side, we obtain
From this, using the appropriate reduction formula
Since $0 < \alpha < 90^\circ$, $0 < \beta < 90^\circ$, either a) $2\alpha = 2\beta$, or b) $2\gamma = 180^\circ - 2\beta$. In case a), $\alpha = \beta$, and thus lines $AB$ and $CD$ intersected by line $AC$ form congruent alternate interior angles;
This implies that $AB \parallel CD$.
om33_3r_img_7.jpg
In case b), we get $\alpha = 90^\circ - \beta$, i.e., $\alpha + \beta = 90^\circ$, and in triangle $ABO$, $|\measuredangle BAC| + |\measuredangle ABO| = 90^\circ$. This implies that $|\measuredangle AOB| = 90^\circ$, that is, diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 744 |
XI OM - III - Task 3
On a circle, 6 different points $ A $, $ B $, $ C $, $ D $, $ E $, $ F $ are chosen in such a way that $ AB $ is parallel to $ DE $, and $ DC $ is parallel to $ AF $. Prove that $ BC $ is parallel to $ EF $.
|
The given points can lie on the circle in various ways. To solve the problem correctly, we need to provide a proof that is independent of the specific arrangement of the points. Let's reason without any diagram.
First, note that if two chords of a circle are parallel, for example, \( MN \parallel PQ \), then the points \( M \), \( N \), \( P \), \( Q \) are the vertices of an isosceles trapezoid with bases \( MN \) and \( PQ \). Therefore, the remaining sides and diagonals are equal, i.e., \( MP = NQ \) and \( MQ = NP \).
Given that \( AB \parallel DE \), it follows that \( AD = BE \). Since \( DC \parallel AF \), it follows that \( AD = CF \). From both of these equalities, we conclude that \( BE = CF \). This implies that the points \( B \), \( C \), \( E \), \( F \) are the vertices of an isosceles trapezoid, where \( BE \) and \( CF \) are either the equal sides or the diagonals. Therefore, the parallel sides of the trapezoid are either \( BC \) and \( EF \) or \( BF \) and \( CE \). If \( BC \parallel EF \), the thesis of our theorem is true; we will prove that if \( BF \parallel CE \), then \( BC \parallel EF \) as well. Indeed, from the fact that \( AB \parallel DE \), \( AF \parallel DC \), and \( BF \parallel CE \), it follows that triangles \( ABF \) and \( DEC \) are similar. The center of similarity is the center \( O \) of the given circle, because the circle is circumscribed around both triangles, and in a similarity, the center of the circumscribed circle of one triangle corresponds to the center of the circumscribed circle of the other triangle, so point \( O \) corresponds to itself. Therefore, the similarity of triangles \( ABF \) and \( DEC \) is a central symmetry. Since in this symmetry, the points corresponding to \( B \) and \( C \) are \( E \) and \( F \), respectively, it follows that \( BC \parallel EF \), Q.E.D.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 745 |
XXIII OM - II - Problem 5
Prove that in a convex quadrilateral inscribed in a circle, the lines passing through the midpoints of the sides and perpendicular to the opposite sides intersect at one point.
|
Let the convex quadrilateral $ABCD$ be inscribed in a circle. Denote by $P$, $Q$, $R$, $S$ the midpoints of the sides of this quadrilateral, and by $O$ - the midpoint of the segment $PR$ (Fig. 14).
Thus *)
Since $\displaystyle \frac{1}{2} (Q + S) = \frac{1}{4} (A + B + C + D) = O$, the point $O$ is also the midpoint of the segment $\overline{QS}$.
Let $A'$ be the image of the quadrilateral $ABCD$ under the symmetry with respect to the point $O$. Since in central symmetry, the midpoint of a segment maps to the midpoint of a segment, and in the considered symmetry, the points $P$ and $R$ map to $R$ and $P$, and the points $Q$ and $S$ map to $S$ and $Q$, the points $P$, $Q$, $R$, $S$ are also the midpoints of the sides of the quadrilateral $A'$. The quadrilateral $A'$ can be inscribed in a circle because the quadrilateral $ABCD$ can be inscribed in a circle and this property is preserved under central symmetry. The sides of the quadrilateral $A'$ are parallel to the corresponding sides of the quadrilateral $ABCD$. Therefore, the lines passing through the midpoints of the sides of the quadrilateral $ABCD$ and perpendicular to its opposite sides are the same as the lines passing through the midpoints of the corresponding sides of the quadrilateral $A'$ and perpendicular to those sides. These lines pass through the center of the circle circumscribed around the quadrilateral $A'$, since the sides of this quadrilateral are chords of the circle, and a line passing through the midpoint of a chord and perpendicular to it passes through the center of the circle. The sought point is thus the center of the circle circumscribed around the quadrilateral $A'$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 747 |
XXIV OM - I - Problem 1
Prove that for every triangle, there exists a line dividing the triangle into two figures with equal perimeters and equal areas.
|
In an isosceles triangle, the line containing the bisector of the angle formed by its legs satisfies the conditions of the problem. Let the side lengths of triangle $ABC$ be different, for example, $a < b < c$, where $a = BC$, $b = AC$, $c = AB$ (Fig. 6). Let $p = \frac{1}{2} (a + b + c)$.
If $p - a \leq x \leq c$, then $p - c \leq p - x \leq a$. Therefore, for every $x \in \langle p-a; c \rangle$, there exist points $P \in \overline{AB}$ and $Q \in \overline{BC}$ such that $BP = x$ and $BQ = p - x$. Then the line $PQ$ divides the perimeter of triangle $ABC$ into two equal parts. We will show that there exists an $x \in \langle p-a; c \rangle$ such that the area $S_1$ of triangle $BPQ$ is equal to half the area $S$ of triangle $ABC$. We have $S - 2S_1 = \frac{1}{2} AB \cdot BC \cdot \sin \measuredangle B - BP \cdot BQ \cdot \sin \measuredangle B = \frac{1}{2} \sin \measuredangle B (ac - 2x (p - x))$. Let $f(x) = ac - 2x (p - x)$. We calculate that $f(p - a) = a (a - b) < 0$ and $f(c) = c (c - b) > 0$. Since the polynomial $f$ takes values of opposite signs at the endpoints of the interval $\langle p - a; c \rangle$, it has a root $x_0$ in this interval. Therefore, for $x = x_0$, we have $S = 2S_1$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 748 |
IV OM - I - Task 4
Prove that the number $ 2^{55} + 1 $ is divisible by $ 11 $.
|
It is known that if $ n $ is a natural number, and $ a $ and $ b $ are any numbers, then
if we substitute in this equality $ a = 2^5 $, $ b = - 1 $, $ n = 11 $, we get
where $ C $ is an integer. The number $ 2^{55} + 1 $ is therefore divisible by $ 33 $, and thus also by $ 11 $.
Note 1. In the same way, we can generally prove that the number $ 2^{5n} + 1 $, where $ n $ represents any odd number, is divisible by $ 11 $.
Note 2. The above task, as well as many other tasks concerning the divisibility of numbers, can be quickly solved using the concept and properties of *congruences*.
Two integers, $ a $ and $ b $, are said to be *in congruence* modulo $ k $ or that they *congrue* modulo $ k $, if the difference $ a - b $ is divisible by $ k $; we write this in the form of the formula
which we call a congruence.
We can also say that two numbers are in congruence modulo $ k $ if they give the same remainder when divided by $ k $.
From the definition of congruence, we can immediately draw the following conclusions:
1° $ a \equiv a \pmod k $ for any integer $ a $ and natural $ k $.
2° If $ a \equiv b \pmod k $, then $ b \equiv a \pmod k $.
3° If $ a \equiv b \pmod k $ and $ b \equiv c \pmod k $, then $ a \equiv c \pmod k $.
4° If $ a \equiv b \pmod k $ and $ c \equiv d \pmod k $, then $ a+c \equiv b+d \pmod k $.
The proofs of theorems 1 - 4 present no difficulties; let the reader conduct them.
5° If $ a \equiv b \pmod k $ and $ c \equiv d \pmod k $, then $ ac \equiv bd \pmod k $.
For the proof, it suffices to note that $ ac - bd = ac - bc + bc - bd = (a - b) c + (c - d) b $.
From theorem 5°, the following theorem follows:
6° If $ a \equiv b \pmod k $, then $ a^2 \equiv b^2 \pmod k $, and generally: $ a^n \equiv b^n \pmod k $ for any natural $ n $. The last conclusion we draw by complete induction.
Using congruences, we will solve the above task No. 18 in the following simple way:
Since $ 2^5 \equiv - 1 \pmod {11} $, then $ (2^5)^{11} = (- 1)^{11} \pmod {11} $, hence $ (2^5)^{11} = - 1 \pmod {11} $, which means that $ 2^{55} + 1 $ is divisible by $ 11 $.
As an example of the application of congruences, we will solve another task:
Find the last digit of the number $ 2^{1000} $.
Since $ 2^5 \equiv 2 \pmod {10} $, then
The last digit of the number $ 2^{1000} $ is therefore $ 6 $.
A lecture on the theory of congruences and many tasks can be found in the book: W. Sierpiński *Theory of Numbers*, Warsaw-Wrocław 1950.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 751 |
XXVI - III - Task 4
In the decimal expansion of a certain natural number, the digits 1, 3, 7, and 9 appear. Prove that by permuting the digits of this expansion, one can obtain the decimal expansion of a number divisible by 7.
|
By performing an appropriate permutation of the digits of a given natural number, we can assume that its last four digits are $1$, $3$, $7$, and $9$. Thus, the considered natural number $n$ is the sum of the number $1379$ and some non-negative integer $a$, whose last four digits are zeros. We will prove that by performing an appropriate permutation of the last four digits of the number $n$, we will obtain a number $n$ divisible by $7$.
Numbers
give the remainders $0$, $1$, $2$, $3$, $4$, $5$, $6$ when divided by $7$, respectively. Therefore, if the number $a$ gives a remainder $r$ when divided by $7$, then by adding to it the number (1) that gives a remainder $7 - r$ when divided by $7$, we will obtain the desired number $n$ divisible by $7$ and formed by permuting the digits of the number $n$.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 754 |
XXIII OM - II - Problem 3
The coordinates of the vertices of a triangle in the Cartesian coordinate system $ XOY $ are integers. Prove that the diameter of the circumcircle of this triangle is not greater than the product of the lengths of the sides of the triangle.
|
If the lengths of the sides of a triangle are $a$, $b$, $c$, and the length of the radius of the circle circumscribed around the triangle is $R$, then the area of the triangle is given by the formula
If the coordinates of the vertices of the triangle are $A = (a_1, a_2)$, $B = (b_1, b_2)$, $C = (c_1, c_2)$, then its area is given by the formula
It follows that if the coordinates of the vertices of the triangle are integers, then $2S$ is a positive integer, i.e., $2S \geq 1$. Therefore, from formula (1) we obtain $\displaystyle \frac{abc}{2R} = 2S \geq 1$, which means $abc \geq 2R$.
Note. We can prove in another way that the number $2S$ is a natural number. Specifically, inscribe the given triangle $ABC$ in a rectangle whose sides are parallel to the coordinate axes (Fig. 13). Let $x_1 = \min (a_1, b_1, c_1)$, $x_2 = \max(a_1, b_1, c_1)$, $y_1 = \min(a_2, b_2, c_2)$, $y_2 = \max(a_2, b_2, c_2)$. Then the vertices of this rectangle will be the points $(x_i, y_j)$, where $i, j = 1, 2$. Therefore, the area of the rectangle $N = (x_2 - x_1)(y_2 - y_1)$ is a natural number. The difference between this rectangle and triangle $ABC$ is the sum of at most three right triangles, whose legs have lengths expressed by natural numbers. Therefore, the areas of these triangles are $\displaystyle \frac{a}{2}, \frac{b}{2}, \frac{c}{2}$, where $a, b, c$ are natural numbers. Therefore, the area $S$ of triangle $ABC$ is expressed by a number of the form $\displaystyle \frac{n}{2}$, where $n \in \mathbb{N}$, specifically $S = N - \displaystyle \frac{a+b+c}{2}$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 756 |
IX OM - I - Problem 1
Prove that if each of the functions
[ (1) \qquad (a_1x + b_1)^2 + (a_2x + b_2)^2, \]
[ (2) \qquad (b_1x + c_1)^2 + (b_2x + c_2)^2 \]
is the square of a linear function that is not constant, then the function
[ (3) \qquad (c_1x + a_1)^2 + (c_2x + a_2)^2 \]
is the square of a linear function.
|
First, let us note that if the function $ Ax^2 + 2Bx + C $ is the square of a linear function $ mx + n $, then its discriminant vanishes, i.e., $ B^2 - AC = 0 $. Indeed, from the identity $ Ax^2 + 2Bx + C = m^2x^2 + 2mnx + n^2 $, it follows that $ B^2 - AC = (mn)^2 - m^2n^2 = 0 $.
Conversely, if $ B^2 - AC = 0 $ and $ C > 0 $, then the function $ Ax^2 + 2Bx + C $ is the square of a linear function. For if $ A \ne 0 $, then from the conditions $ B^2 = AC $ and $ C > 0 $ it follows that $ A > 0 $, hence
while if $ A = 0 $, then $ B = 0 $ and
The function (1) can be written in the form
The discriminant of this function,
is, according to the above remark, equal to zero, so
Similarly for function (2)
Eliminating $ b_2 $ and $ b_1 $ from (4) and (5) respectively, we obtain
Since function (2) is not constant, at least one of the numbers $ b_1 $ and $ b_2 $ is different from zero, hence from equations (6) and (7) it follows that
From equality (8), it follows that the function (3), i.e., the function
has a discriminant equal to zero, since this discriminant equals $ - 4 (a_1c_2 - a_2c_1)^2 $. The constant term $ a_1^2 + a_2^2 $ of this function cannot be equal to zero, for otherwise it would be $ a_1 = 0 $, $ a_2 = 0 $, contrary to the assumption that function (1) is not constant. In such a case, $ a_1^2 + a_2^2 > 0 $, from which, according to the remark mentioned at the beginning, we conclude that function (3) is the square of a linear function.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 757 |
XV OM - I - Problem 4
In the plane, there are 5 points, no three of which are collinear. Prove that among them, there are four points that are the vertices of a convex quadrilateral.
|
Let $A$, $B$, $C$, $D$, $E$ denote given points. Since no three of them are collinear, points $A$, $B$, $C$ are the vertices of a triangle, and each of the points $D$ and $E$ lies either inside or outside this triangle. Therefore, one of the following cases occurs:
a) Points $D$ and $E$ lie inside the triangle $ABC$. The line $DE$ does not pass through any of the vertices $A$, $B$, $C$, so it intersects two sides of the triangle, for example, $AC$ at point $M$ and $BC$ at point $N$ (Fig. 3).
We can arrange the labels such that point $D$ lies between points $M$ and $E$. In this case, the quadrilateral $ADEB$ is convex, as we can easily verify that for each of the sides $AD$, $DE$, $EB$, $BA$, the other two vertices of the quadrilateral lie on the same side of that side.
b) One of the points $D$ and $E$, say point $D$, lies inside the triangle $ABC$, and the other, i.e., $E$, lies outside this triangle. The rays $DA$, $DB$, $DC$ divide the plane into 3 convex angular regions $ADB$, $BDC$, $CDA$; point $E$ lies inside one of them, for example, inside the angle $BDC$. The quadrilateral $DBEC$ is convex, as it is the intersection of two convex angular regions $BDC$ and $BEC$.
c) Points $D$ and $E$ lie outside the triangle $ABC$. If point $D$ lies inside one of the convex angles $BAC$, $CBA$, $ACB$, for example, inside the angle $BAC$ (Fig. 5), then the quadrilateral $ABDC$ is convex, as the intersection of the convex angular regions $BAC$ and $BDC$.
If, however, point $D$ lies inside one of the vertical angles to the angles of the triangle, for example, inside the vertical angle to angle $A$, then point $A$ lies inside the triangle $BDC$ (Fig. 6), in which case we have a situation analogous to case a) or case b), for which we have already provided a proof.
Note. A figure $F$ is called convex if every segment whose endpoints belong to the figure $F$ is entirely contained within the figure $F$. From this definition, it follows directly that the points common to two or more convex figures also form a convex figure. We leave the easy proof of this theorem to the Reader.
If $AB$ is a side of a convex polygon $W$, then all points of this polygon, except for the points of side $AB$, lie on one side of the line $AB$. Indeed, if points $C$ and $D$ of the polygon $W$ lay on opposite sides of the line $AB$, then connecting points $C$ and $D$ with segments to points $A$ and $B$, we would obtain a quadrilateral $ACBD$, all of whose sides would belong to $W$; thus, the entire quadrilateral $ACBD$ would be contained in $W$, which means that its diagonal $AB$ could not be a side of the polygon $W$.
Conversely, if the entire polygon $W$ lies on one side of each line containing one of its sides, except for the points of that side, then the polygon $W$ is convex. If there were a point $M$ on the segment $PQ$ connecting points $P$ and $Q$ of the polygon $W$ that does not belong to the polygon $W$, then the segment $PM$ would have a point in common with the boundary of the polygon, say with side $AB$, and in that case, points $P$ and $Q$ would not lie on the same side of the line $AB$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 760 |
LVIII OM - I - Problem 7
Given is a tetrahedron $ABCD$. The angle bisector of $\angle ABC$ intersects the edge $AC$ at point $Q$. Point $P$ is symmetric to $D$ with respect to point $Q$. Point $R$ lies on the edge $AB$, such that $BR = \frac{1}{2}BC$. Prove that a triangle can be formed with segments of lengths $BP$, $CD$, and $2 \cdot QR$.
|
Let $ S $ be the midpoint of edge $ BC $, and $ T $ - the point symmetric to point $ C $ with respect to point $ Q $ (Fig. 3).
om58_1r_img_3.jpg
Since $ BS={1\over 2}BC=BR $, points $ R $ and $ S $ are symmetric with respect to the angle bisector $ BQ $ of angle $ ABC $, and thus $ QR=QS $. Furthermore, triangle $ CTB $ is the image of triangle $ CQS $ under a homothety with center at point $ C $ and scale factor $ 2 $, which gives $ BT=2\cdot QS=2\cdot QR $. Finally, the endpoints of segments $ TP $ and $ CD $ are symmetric to each other with respect to point $ Q $, hence $ PT=CD $.
From the derived relationships, we obtain that triangle $ BTP $ is constructed from segments of lengths $ BP $, $ BT=2\cdot QR $, and $ PT=CD $, from which the thesis follows (points $ B $, $ T $, $ P $ do not lie on the same line, as point $ P $ is outside the plane $ BCT $).
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 763 |
XXXI - II - Problem 1
Students $ A $ and $ B $ play according to the following rules: student $ A $ chooses a vector $ \overrightarrow{a_1} $ of length 1 on the plane, then student $ B $ gives a number $ s_1 $, equal to $ 1 $ or $ -1 $; then student $ A $ chooses a vector $ \overrightarrow{a_1} $ of length $ 1 $, and in turn, student $ B $ gives a number $ s_2 $ equal to $ 1 $ or $ -1 $, and so on. Student $ B $ wins if for some $ n $ the vector $ \sum_{j=1}^n \varepsilon_j \overrightarrow{a_j} $ has a length greater than a predetermined number $ R $ before the game starts. Prove that student $ B $ can achieve a win in no more than $ R^2 + 1 $ steps, regardless of the actions of partner $ A $.
|
We will prove by induction that student $ B $ can choose the numbers $ \varepsilon_1, \varepsilon_2, \ldots, \varepsilon_n $ such that the length of the vector $ \overrightarrow{\omega_n} = \sum_{k=1}^n \varepsilon_k \overrightarrow{a_k} $ is not less than $ \sqrt{n} $.
For $ n = 1 $, the vector has a length of $ 1 $ regardless of whether $ \varepsilon_1 = 1 $ or $ \varepsilon_1 = - 1 $.
Assume that for some $ n \geq 1 $, the length of the vector $ \overrightarrow{\omega_n} = \sum_{k=1}^n \varepsilon_k \overrightarrow{a_k} $ is $ \geq \sqrt{n} $. If $ \overrightarrow{a_{n+1}} $ is the next vector of length $ 1 $ given by player $ A $, then player $ B $ should take $ \varepsilon_{n+1} = 1 $ if the vectors $ \overrightarrow{a_{n+1}} $ and $ \overrightarrow{w_n} $ form an angle of measure $ \leq \frac{\pi}{2} $, and $ \varepsilon_{n+1} = - 1 $ if this angle has a measure $ > \frac{\pi}{2} $ (in which case the vectors $ \overrightarrow{w_n} $ and $ - \overrightarrow{a_{n+1}} $ form an angle of measure $ < \frac{\pi}{2} $). Thanks to this choice, $ \cos \measuredangle (\overrightarrow{\omega_n}, \varepsilon_{n+1} \overrightarrow{a_{n+1}}) \geq 0 $. Therefore,
and thus
Hence, by the principle of induction, it follows that player $ B $ can choose the numbers $ \varepsilon_k $ such that for every natural number $ n $, $ \omega_n \geq \sqrt{n} $.
For any non-negative real number $ R $, the number $ [R^2 + 1] $ is the smallest natural number greater than $ R^2 $, so after $ n = [R^2 + 1] $ steps,
player $ B $ can ensure that the length of the vector $ \omega_n $ is greater than or equal to $ \sqrt{[R^2 + 1]} > \sqrt{R^2} = R $. If the predetermined number $ R $ is negative, student $ B $ wins already in the first step, $ R^2 + 1 > 1 $.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 764 |
XXIX OM - III - Task 2
On a plane, there are points with integer coordinates, at least one of which is not divisible by A. Prove that these points cannot be paired in such a way that the distance between the points of each pair is equal to 1; this means that an infinite chessboard with fields cut out at coordinates divisible by 4 cannot be covered with domino tiles.
|
Consider the set $ A $ of points on the plane with integer coordinates $ (i, j) $, where $ 0 \leq i, j \leq 4k $. Notice that if we pair the points according to the conditions of the problem, the sum of the coordinates of one point in each pair will be odd, and the other will be even.
The set $ A $ has $ (4k + 1)^2 $ points, of which $ (k + 1)^2 $ points have both coordinates divisible by $ 4 $ and, according to the conditions of the problem, such points are not considered. The number of points in set $ A $ whose sum of coordinates is even is one more than the number of points in set $ A $ whose sum of coordinates is odd. Therefore, the number of the former points is $ 8k^2 + 4k + 1 $, and the latter is $ 8k^2 + 4k $.
On the boundary of the square defined by the points in set $ A $, there are $ 8k $ points with an odd sum of coordinates (with $ 2k $ on each side of the square).
According to the conditions of the problem, $ (8k^2 + 4k) - 8k = 8k^2 - 4k $ points in set $ A $ with an odd sum of coordinates, which lie inside the considered square, need to be paired with some of the $ (8k^2 + 4k + 1) - (k + 1)^2 = 7k^2 + 2k $ points in set $ A $ with an even sum of coordinates, but not both coordinates divisible by $ 4 $.
If it were possible to pair the points on the plane according to the conditions of the problem, then for every natural number $ k $, the inequality
would hold,
i.e., $ k^2 \leq 6k $. However, for $ k = 7 $, this inequality does not hold.
Therefore, it is not possible to pair the given points according to the conditions of the problem.
Note. If the squares of the chessboard are alternately colored white and black and one of the cut-out squares is black, then all the cut-out squares are black. Since a domino tile covers one white and one black square, the number of white and black squares covered by domino tiles is the same. If we consider a sufficiently large finite part of an infinite chessboard, there will be too many white squares to cover them all with domino tiles. This idea is the basis for the solution given above.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 765 |
XLV OM - I - Problem 12
Prove that the sums of opposite dihedral angles of a tetrahedron are equal if and only if the sums of opposite edges of this tetrahedron are equal.
|
For a quadruple of non-coplanar points $ U $, $ V $, $ X $, $ Y $, we will denote by $ |\measuredangle X(UV)Y| $ the measure of the convex dihedral angle formed by the half-planes with common edge $ UV $, passing through points $ X $ and $ Y $.
Let us consider a tetrahedron $ ABCD $. We will prove the equivalence of the following statements:
From this equivalence, the thesis of the problem will follow.
Proof of the implication (1) $ \Longrightarrow $(2). Let us assume that the incircle of triangle $ ABC $ has center $ I $ and is tangent to sides $ AB $, $ AC $, $ BC $ at points $ E $, $ F $, $ K $, respectively, and the incircle of triangle $ BCD $ has center $ J $ and is tangent to sides $ BD $, $ CD $, $ BC $ at points $ G $, $ H $, $ L $, respectively (Figure 6). We rewrite equality (1) as
The following segments tangent to the considered circles have equal lengths:
Thus, the first and last terms on the left side of equality (3) cancel out with the first and last terms on the right side, and the equality of the remaining (middle) terms can be rewritten as
Since also $ |BK| + |CK| = |BL| + |CL|\ (= |BC|) $, it follows that $ |BK| = |BL| $ and $ |CK| = |CL| $, which means that point $ L $ coincides with $ K $.
The plane $ IJK $ is perpendicular to the line $ BC $. The lines lying in this plane, passing through points $ I $ and $ J $ and perpendicular to the planes $ ABC $ and $ BCD $, respectively, intersect at a point we will denote by $ S $. Segments $ IE $, $ IF $, $ IK $ are of equal length (as radii of the same circle). Therefore, the right triangles $ SIE $, $ SIF $, $ SIK $ are congruent, and we obtain the equalities
Similarly, from the congruence of triangles $ SJG $, $ SJH $, $ SJK $, we infer that
(It is worth noting that point $ S $ is the center of a sphere tangent to the edges $ AB $, $ AC $, $ BC $, $ BD $, $ CD $ at points $ E $, $ F $, $ K $, $ G $, $ H $, respectively.)
Let $ P $ and $ Q $ be the orthogonal projections of point $ S $ onto the planes $ ABD $ and $ ACD $ (Figure 7). The right triangles $ SPE $ and $ SPG $ have a common side $ SP $ and equal hypotenuses $ SE $ and $ SG $ (segments of the same length as $ SK $; see (4), (5)); thus, they are congruent, and the equality of angles holds
Analogous reasoning shows that
Points $ E $, $ F $, $ G $, $ H $ are the orthogonal projections of point $ S $ onto the edges $ AB $, $ AC $, $ BD $, $ CD $ (Figures 6 and 7). Thus:
- Points $ S $, $ E $, $ P $, $ I $ lie in a plane perpendicular to $ AB $;
- Points $ S $, $ F $, $ Q $, $ I $ lie in a plane perpendicular to $ AC $;
- Points $ S $, $ G $, $ P $, $ J $ lie in a plane perpendicular to $ BD $;
- Points $ S $, $ H $, $ Q $, $ J $ lie in a plane perpendicular to $ CD $.
Hence (Figure 8) the first of the following equalities; the others follow by analogy:
The equality (2) to be proved is now a direct consequence of relations (4), (5), (6), (7). The implication (1) $ \Longrightarrow $ (2) is thus established.
Note 1: In the proof of the converse implication, we will use the following (known) fact:
(8) In any tetrahedron, the sum of two dihedral angles at any vertex is greater than the third dihedral angle at that vertex.
In other words, the "distance" between two rays with a common origin, measured by the angle between them, satisfies the "triangle inequality": three rays with a common origin determine, pairwise, three such angles, each of which is less than the sum of the other two.
(This simple theorem appeared, for example, as a lemma in the solution to problem 5 of the second round of the 24th Mathematical Olympiad; of course, the reader will find a proof in the publication: 24th Mathematical Olympiad. Report of the Main Committee, Warsaw 1974, p. 66 - but we encourage the reader to independently conduct this proof; it is a simple exercise.)
Proof of the implication (2) $ \Longrightarrow $ (1). Suppose that equality (2) holds and equality (1) does not hold; let, for example,
(assuming the opposite inequality, it is of course sufficient to change the notation accordingly).
Figure 9
Consider a moving point $ X $ sliding along the edge $ CD $. When $ X $ coincides with $ C $, the difference $ |BX|-|CX| = |BC| $ is greater than $ |AB|-|AC| $ (the inequality for the sides of triangle $ ABC $). When $ X $ coincides with $ D $, the difference $ |BX|-|CX| = |BD|-|CD| $ is less than $ |AB|- |AC| $ (which follows from (9)). Therefore, there exists a position of point $ X $ such that the difference $ |BX|-|CX| $ equals $ |AB| - |AC| $, i.e.,
In the following, $ X $ denotes a fixed point on the segment $ CD $ satisfying this equality. It is analogous to equality (1) for the tetrahedron $ ABCX $. By the first part of the proof, an analogous equality to (2) holds for the sums of dihedral angles:
Since point $ X $ lies on the segment $ CD $, we have
and
Substituting relations (11) into equality (2) and relation (12) into (10), we obtain the equalities
and
which, when subtracted side by side, yield the relation
Let $ Z $ be any point inside the tetrahedron $ ABDX $. Denote by $ U $, $ V $, $ W $ the orthogonal projections of point $ Z $ onto the planes $ BDX $, $ BAX $, $ BAD $ (Figure 9). Projecting points $ V $ and $ W $ onto the line $ AB $, we obtain the same point $ Y $ (the projection of point $ Z $ onto $ AB $). In the quadrilateral $ ZVYW $, the angles at vertices $ V $ and $ W $ are right angles. Thus,
Similarly, we show that
Taking into account these equalities, relation (13) takes the form
This is already a contradiction: since $ ABDX $ is a (non-degenerate) tetrahedron, points $ Z $, $ U $, $ V $, $ W $ are non-coplanar and $ ZUVW $ is a (non-degenerate) tetrahedron. According to theorem (8), the sum of angles $ UZW $ and $ UZV $ should be strictly greater than angle $ VZW $. The contradiction completes the proof of the implication (2) $ \Longrightarrow $ (1).
Proof of the thesis of the problem. Let us denote the lengths of the edges of the given tetrahedron $ ABCD $ as follows:
and assume that the dihedral angle formed by the faces with common edge $ a_i $ has measure $ \alpha_i $, and the dihedral angle formed by the faces with common edge $ b_i $ has measure $ \beta_i $. We have proved the equivalence between conditions (1) and (2), i.e., the equivalence
By analogy, we infer that the following equivalences are also valid:
The set of statements (14), (15), (16) implies the equivalence
($ i,j \in \{1,2,3\} $), which is the thesis of the problem.
Note 2: In the solution, we have shown more than the problem required: we have shown the validity of the equivalence (14) (involving only two pairs of opposite edges and opposite dihedral angles), without using the logical dependencies between the conditions appearing in (15) and (16). As we noted earlier, statements (15) and (16), and consequently (17), are already simple consequences of this.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 768 |
XLVI OM - I - Problem 5
Given are positive numbers $ a $, $ b $. Prove the equivalence of the statements:
保留了源文本的换行和格式，但最后一句是中文，翻译成英文应为: "The statements are to be proven equivalent while retaining the original text's line breaks and format." 但为了保持格式的一致性，这里不添加额外的句子。
|
Assume the truth of statement (2). Substituting $ x = 1+\frac{1}{\sqrt{a}} $, we obtain the inequality
thus $ a + \sqrt{a} + \sqrt{a} + 1 > b $, or equivalently,
Since $ b $ is a positive number, it follows that condition (1) holds.
Conversely, from condition (1), we derive (2) as follows. Let $ x $ be any number greater than $ 1 $. Denote the difference $ x - 1 $ by $ t $; this is a positive number. We transform the left side of the inequality in condition (2):
If condition (1) is satisfied, then the number $ (\sqrt{a}+1)^2 $ is greater than $ b $, so by the above estimate, the value of the expression $ ax + \frac{x}{x-1} $ is greater than $ b $. The number $ x > 1 $ was chosen arbitrarily; thus, we have statement (2).
The implications (2) $ \Rightarrow $ (1) and (1) $ \Rightarrow $ (2) together constitute the proof of the desired equivalence.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 769 |
XI OM - III - Task 1
Prove that if $ n $ is an integer greater than $ 4 $, then $ 2^n $ is greater than $ n^2 $.
|
Since $ 2^5 > 5^2 $, the inequality $ 2^n > n^2 $ is true for $ n = 5 $. Suppose this inequality is true when $ n $ equals some integer $ k > 4 $, i.e., that $ 2^k > k^2 $. Then
the inequality $ 2^n > n^2 $ is also true when $ n = k + 1 $. By the principle of mathematical induction, the inequality is true for every integer $ n > 4 $.
In the above proof, we relied on the fact that when $ k > 4 $, $ k^2 > 2k + 1 $. Indeed, the inequality $ k^2 > 2k + 1 $ is equivalent to the inequality $ k (k - 2) > 1 $, which is true for $ k > 4 $, and even for $ k \geq 3 $.
Note. A more general theorem is true: if $ a $ is an integer greater than $ 1 $, and $ n $ is an integer greater than $ a^2 $, then
For $ a = 2 $, we have already proved the above. Let us assume that $ a > 2 $ and use mathematical induction. The inequality (1) is true when $ n = a^2 $; indeed, $ a^{a^2} $ is greater than $ (a^2)^a = a^{2a} $ when $ a > 2 $, because the inequality $ a > 2 $ implies that $ a^2 > 2a $.
Suppose that for some integer $ k \geq a^2 $,
we need to prove that
According to the inductive hypothesis, $ a^{k+1} = a^k \cdot a > k^a \cdot a $, so it suffices to prove that when $ a > 2 $ and $ k \geq a^2 $, $ k^a \cdot a > (k + 1)^a $, or that
Indeed, when $ a > 2 $ and $ k \geq a^2 $,
which is what we needed to prove. (In the proof, we relied on the theorem that if $ n $ is an integer greater than $ 1 $ and $ d > -1 $, then $ (1 + d)^n > 1 + nd $. The proof of this inequality can be easily obtained by induction. See Mathematical Olympiad Problems, Volume II, Problem 36).
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 773 |
L OM - I - Problem 9
Points $ D $, $ E $, $ F $ lie on the sides $ BC $, $ CA $, $ AB $ of triangle $ ABC $, respectively. The incircles of triangles $ AEF $, $ BFD $, $ CDE $ are tangent to the incircle of triangle $ DEF $. Prove that the lines $ AD $, $ BE $, $ CF $ intersect at a single point.
|
Let $ c_1 $ be the incircle of triangle $ DEF $, and $ c_2 $ be the incircle of triangle $ ABC $. Since the incircles of triangles $ AEF $ and $ DEF $ are tangent, we have $ AF + DE = AE + DF $. This means that a circle can be inscribed in quadrilateral $ AEDF $; let this circle be denoted by $ c_A $. Point $ D $ is the center of homothety $ j_1 $, with a positive scale, transforming circle $ c_1 $ into $ c_A $, and point $ A $ is the center of homothety $ j_2 $, with a positive scale, transforming circle $ c_A $ into $ c_2 $. Therefore, the homothety $ j $, with a positive scale, transforming circle $ c_1 $ into $ c_2 $ is the composition of homotheties $ j_1 $ and $ j_2 $ — its center thus lies on the line $ AD $. Similarly, we prove that the center of homothety $ j $ lies on the lines $ BE $ and $ CF $.
Conclusion: The lines $ AD $, $ BE $, and $ CF $ have a common point, which is the center of homothety of circles $ c_1 $ and $ c_2 $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 774 |
XXIV OM - II - Problem 5
Prove that if in a tetrahedron $ABCD$ we have $AB = CD$, $AC = BD$, $AD = BC$, then all faces of the tetrahedron are acute triangles.
|
We will first prove the
Lemma. Let $ \alpha, \beta, \gamma $ be numbers in the interval $ (0; \pi) $. There exists a trihedral angle whose dihedral angles have measures equal to $ \alpha, \beta, \gamma $ if and only if each of the numbers $ \alpha, \beta, \gamma $ is less than the sum of the other two.
Proof. Without loss of generality, we can assume that $ \alpha \geq \beta \geq \gamma $. Then, of course, $ \beta < \alpha + \gamma $ and $ \gamma < \alpha + \beta $. We need to prove that there exists a trihedral angle whose dihedral angles have measures equal to $ \alpha, \beta, \gamma $ if and only if $ \alpha < \beta + \gamma $.
Let $ t $ be a ray with origin at point $ P $. Rays with origin at point $ P $ that form an angle of a given measure $ \mu $ with $ t $ are the generators of a cone of revolution with vertex at point $ P $, axis $ t $, and vertex angle of measure $ 2\mu $. Conversely, every generator of this cone forms an angle of measure $ \mu $ with ray $ t $ (Fig. 16).
Let rays $ k $ and $ m $ contained in plane $ \pi $ have origin at point $ O $ and form an angle of measure $ \alpha $ (Fig. 17). Consider the cone of revolution $ S_1 $ with vertex at point $ O $, axis $ k $, and vertex angle of measure $ 2\beta $, and the cone of revolution $ S_2 $ with vertex at point $ O $, axis $ m $, and vertex angle of measure $ 2\gamma $. The intersection of the surfaces of these cones is either the point $ O $ (if $ \beta + \gamma < \alpha $), or a certain ray contained in plane $ \pi $ (if $ \beta + \gamma = \alpha $), or a pair of rays $ p_1 $, $ p_2 $, neither of which is contained in plane $ \pi $ (if $ \beta + \gamma > \alpha $). In the last case, rays $ k $, $ m $, $ p_1 $ form a trihedral angle whose dihedral angles at the vertex have measures equal to $ \alpha, \beta, \gamma $. If, however, $ \alpha \geq \beta + \gamma $, then such a trihedral angle does not exist, because the corresponding cones do not have a common generator not contained in plane $ \pi $ in this case.
We now proceed to solve the problem. From the equalities given in the problem, it follows that any two faces of the tetrahedron $ ABCD $ have three edges respectively equal (Fig. 18). They are therefore congruent triangles. Hence, in particular,
Since the sum of the measures of the angles of triangle $ ABC $ is $ \pi $, it follows that the sum of the measures of the dihedral angles at vertex $ D $ is equal to $ \pi $. Therefore, by the lemma,
Hence $ \measuredangle BDC < \displaystyle \frac{\pi}{2} $.
Similarly, we prove that each of the remaining dihedral angles is acute.
Note. The thesis of the problem follows directly from problem 12 of the XII Mathematical Olympiad.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 778 |
XLVII OM - I - Problem 12
Determine whether there exist two congruent cubes with a common center such that each face of the first cube has a point in common with each face of the second cube.
|
We will prove that there do not exist two cubes $\mathcal{C}$ and $\mathcal{C}$ for which the following condition is satisfied:
$\quad (*) \quad$ each face of the cube $\mathcal{C}$ has points in common with each face of $\mathcal{C}$.
The additional assumptions given in the problem (congruence, common center) are not needed for anything.
Let then two cubes $\mathcal{C}$ and $\mathcal{C}$ be given. Since in statement (*) the roles of the symbols $\mathcal{C}$ and $\mathcal{C}$ are symmetric, we can assume without loss of generality that the edge length of the cube $\mathcal{C}$ is not less than the edge length of the cube $\mathcal{C}$. We will show in three ways that condition (*) cannot be satisfied.
Let $\mathcal{C}$ be a cube with an edge of unit length, and $\mathcal{C}$ a cube with an edge length of $\geq 1$. Choose two opposite faces $\mathcal{S}$ and $\mathcal{T}$ of the cube $\mathcal{C}$. The plane of the face $\mathcal{S}$ divides space into two half-spaces; denote by $\mathcal{H}$ the half-space that does not contain the square $\mathcal{T}$. Assume that $\mathcal{H}$ is a closed half-space; that is, the plane of the face $\mathcal{S}$ is a subset of the set $\mathcal{H}$.
Suppose that in the set $\mathcal{H}$ there are two vertices $A$, $B$ of the cube $\mathcal{C}$, not being the endpoints of one edge. Denote the midpoint of the segment $AB$ by $M$; it is also a point of the set $\mathcal{H}$.
If $AB$ is one of the four diagonals of the cube $\mathcal{C}$ intersecting its interior, then $M$ is its center. Every point of the cube $\mathcal{C}$ is at a distance from $M$ of at most $\frac{1}{2} \sqrt{3}$. This number is less than $1$. Since the distance from the square $\mathcal{T}$ to the half-space $\mathcal{H}$ is at least $1$, the square $\mathcal{T}$ does not contain any point of the cube $\mathcal{C}$.
If $AB$ is a diagonal of one of the faces of the cube $\mathcal{C}$, then $M$ is the center of that face. All its points are at a distance from $M$ of at most $\frac{1}{2} \sqrt{2}$, so the entire face is disjoint from the square $\mathcal{T}$.
The only remaining case to consider is when in the set $\mathcal{H}$ there is either no vertex of the cube $\mathcal{C}$, or exactly one vertex, or exactly two vertices connected by an edge. The remaining vertices (8, 7, or 6 in number) lie outside the set $\mathcal{H}$: in each case, among them, we can find four points that are the vertices of some face of the cube $\mathcal{C}$. This face has no points in common with the set $\mathcal{H}$, and therefore has no points in common with the face $\mathcal{S}$ of the cube $\mathcal{C}$.
We have shown that the cube $\mathcal{C}$ has a face that is disjoint either from $\mathcal{S}$ or from $\mathcal{T}$. Condition (*) cannot be satisfied.
Note 2. Replacing the word "cubes" everywhere in the problem statement with "parallelepipeds" yields a much more general problem. The author of this solution does not know the answer to the question contained in such a reformulated problem.
Note 3. A four-dimensional cube is a subset of four-dimensional space $\mathbb{R}^4$ similar to the set
(similar in the geometric sense). The subset of the set $\mathcal{C}$ consisting of points with the first coordinate equal to $1$ is isometric to a three-dimensional cube; the same can be said about the other seven subsets of the set $\mathcal{C}$ obtained by fixing any coordinate and assigning it the value $1$ or $-1$. These subsets are called the faces of the cube $\mathcal{C}$. If $\mathcal{C}$ is another cube, obtained from the cube $\mathcal{C}$ by a similarity transformation, then the images of the faces of the cube $\mathcal{C}$ under this similarity transformation form the faces of the cube $\mathcal{C}$.
For four-dimensional cubes, we can pose a problem analogous to the one considered in the problem: do there exist in $\mathbb{R}^4$ two such cubes that each face of one of them has points in common with each face of the other? The answer may seem surprising: in $\mathbb{R}^4$ such pairs do exist. We will give an example - without any justification; verifying the truth of all the statements in the next paragraph may be an ambitious task for the Reader.
One of the cubes will be the set $\mathcal{C}$ defined above. Its vertices are points of the form $(\pm 1, \pm 1, \pm 1, \pm 1)$ (the choice of signs is arbitrary; such a cube has 16 vertices). Now consider another 16 points: eight points of the form $(\pm 1, \pm 1, \pm 1, \pm 1)$, in which the product of the coordinates equals $1$, and eight points having three coordinates equal to $0$ and the remaining one equal to $2$ or $-2$. This 16 points also form the set of vertices of a four-dimensional cube $\mathcal{C}$, and this cube is congruent to $\mathcal{C}$. Moreover, each three-dimensional face of one cube has at least one common vertex with each three-dimensional face of the other cube!
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 784 |
XXXVI OM - II - Problem 4
Prove that if for natural numbers $ a, b $ the number $ \sqrt[3]{a} + \sqrt[3]{b} $ is rational, then $ a, b $ are cubes of natural numbers.
|
Let's denote the numbers $ \sqrt[3]{a} $ and $ \sqrt[3]{b} $ by $ x $ and $ y $, respectively, and their sum by $ s $. We need to prove that $ x $ and $ y $ are natural numbers. By assumption, the number $ s = x + y $ is rational. Since
\[
xy \text{ is a rational number.}
\]
Further, we have
\[
\text{from the rationality of numbers } a, b, s, \text{ and } xy, \text{ we conclude that the difference } x - y \text{ is a rational number.}
\]
Since both the sum and the difference of the numbers $ x $ and $ y $ are rational, both these numbers are rational.
Let's represent $ x $ as an irreducible fraction $ m/n $. Then $ x^3 = m^3/n^3 $ is also an irreducible fraction, and since $ x^3 = a $ is an integer, we have $ n = 1 $. This means that $ x $ is a natural number. Similarly, we prove that $ y $ is a natural number.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 785 |
XIX OM - II - Problem 1
Prove that a polynomial in the variable $ x $ with integer coefficients, whose absolute value for three different integer values of $ x $ equals 1, does not have integer roots.
|
Suppose that $ f(x) $ is a polynomial in the variable $ x $ with integer coefficients and that
where $ a $, $ b $, $ c $ are three distinct integers.
Assume that $ x_0 $ is an integer root of the polynomial $ f(x) $, so for every $ x $
where $ \varphi (x) $ is a polynomial with integer coefficients.
From (1) and (2), it follows that
The number $ |a - x_0| $ is an integer, positive, and is a divisor of unity, hence $ |a - x_0| = 1 $; similarly, we conclude that $ |b - x_0| = 1 $, $ |c - x_0| = 1 $.
It follows that two of the numbers $ a - x_0 $, $ b - x_0 $, $ c - x_0 $, and thus also two of the numbers $ a $, $ b $, $ c $, are equal, which contradicts the assumption. Therefore, the polynomial $ f(x) $ does not have an integer root, Q.E.D.
Note. In the above reasoning, we relied on the fact that the coefficients of the polynomial $ \varphi (x) $ are integers. This is easy to show.
Let
From the equality $ f(x) = (x - x_0) \varphi (x) $, by comparing the coefficients of both sides, we get
thus
From this, it follows that the number $ c_0 $ is an integer, and if $ c_{k-1} $ is an integer ($ k = 1,2, \ldots, n - 1 $), then $ c_k $ is also an integer. Therefore, the numbers $ c_0, c_1, \ldots, c_{n-1} $ are integers.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 787 |
LIV OM - I - Task 5
A natural number $ n_1 $ is written in the decimal system using 333 digits, none of which are zero. For $ i = 1 , 2, 3, \ldots , 332 $, the number $ n_{i+1} $ is formed from the number $ n_i $ by moving the units digit to the beginning. Prove that either all the numbers $ n_1, n_2, n_3, \ldots , n_{333} $ are divisible by 333, or none of them are.
|
Let $ j_i $ $ (i = 1,2,3,\ldots ,333) $ denote the units digit of the number $ n_i $. Then for $ i =1,2,\ldots,332 $ we have
The number $ 10^{333} - 1 = (10^3 - 1) \cdot (10^{330} + 10^{327} + 10^{324} +\ldots + 10^3 + 1) $ is divisible by 333, and the numbers 10 and 333 are relatively prime. From the second equality (1), it follows that the number $ n_i $ is divisible by 333 if and only if the number $ n_{i+1} $ is divisible by 333.
If the number $ n_1 $ is divisible by 333, then by the above statement, we obtain (by induction) the divisibility by 333 of each of the numbers $ n_2, n_3, \ldots , n_{333} $. If, however, the number $ n_1 $ is not divisible by 333, then similarly as above, we conclude that none of the numbers $ n_2, n_3, \ldots , n_{333} $ is divisible by 333.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 790 |
XXII OM - I - Problem 3
Through a point $ P $, belonging to the plane of triangle $ ABC $, three lines perpendicular to the lines $ BC $, $ AC $, and the line containing the median $ \overline{CE} $ have been drawn. Prove that they intersect the line containing the altitude $ CD $ at points $ K, L, M $, respectively, such that $ KM = LM $.
|
We will first prove a few auxiliary facts.
Lemma. If the corresponding sides of two triangles are parallel, then the triangles are similar.
Proof. Let $ AB\parallel A', $ BC \parallel B', $ CA \parallel C'. $ Then the corresponding angles of triangles $ ABC $ and $ A'B'C' $ have sides that are parallel. Therefore, the measures of these angles are equal or their sum is a straight angle. We have then
If in at least two cases the second part of the alternative is satisfied, for example, $ \measuredangle A +\measuredangle A', $ \measuredangle B +\measuredangle B', $ then, given that the sum of the angles in a triangle is $ 180^\circ, $ we would have
From this, $ \measuredangle C + \measuredangle C', $ which is not possible.
Therefore, in at least two cases the first part of the alternative is satisfied, for example, $ \measuredangle A = \measuredangle A' $ and $ \measuredangle B =\measuredangle B'. $ It follows that triangles $ ABC $ and $ A'B'C' $ are similar.
Corollary. If the corresponding sides of two triangles are perpendicular, then the triangles are similar.
Proof. Let
Rotate triangle $ A_1B_1C_1 $ around any point by a right angle. We will then obtain triangle $ A_2B_2C_2 $ with sides perpendicular to the corresponding sides of triangle $ A_1B_1C_1, $ i.e.,
From (1) and (2), it follows that $ AB \parallel A_2B_2, $ BC \parallel B_2C_2, $ CA \parallel C_2A_2. $ Therefore, by the lemma, triangles $ ABC $ and $ A_2B_2C_2 $ are similar. Since triangles $ A_2B_2C_2 $ and $ A_1B_1C_1 $ are congruent, it follows that triangles $ ABC $ and $ A_1B_1C_1 $ are also similar.
We will now proceed to solve the problem.
If point $ P $ lies on the line containing segment $ CD, $ then $ P = K = L = M $ and therefore $ KM = LM = 0. $
If point $ P $ does not lie on the line containing segment $ CD, $ then from the conditions of the problem it follows that the corresponding sides of triangles $ PKM $ and $ CBE $ are perpendicular, and also the corresponding sides of triangles $ PLM $ and $ CAE $ are perpendicular (Fig. 6). Therefore, by the corollary proved above, triangles $ PKM $ and $ CBE $ and triangles $ PLM $ and $ CAE $ are similar.
Denoting the ratios of these similarities by $ \lambda $ and $ \mu $ respectively, we get that $ PM = \lambda CE, $ $ KM = \lambda BE, $ $ PM = \mu CE, $ $ LM = \mu AE. $ From the first and third equalities, it follows that $ \lambda =\mu. $ Point $ E $ is the midpoint of segment $ AB, $ i.e., $ AE = BE. $ Therefore, from the second and fourth equalities, it follows that $ KM = LM. $
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 793 |
X OM - I - Problem 10
Prove that if a circle can be circumscribed around each of the quadrilaterals $ABCD$ and $CDEF$, and the lines $AB$, $CD$, $EF$ intersect at a single point $M$, then a circle can be circumscribed around the quadrilateral $ABEF$.
|
We apply the theorem about secants of a circle passing through one point: 1° to lines $ AB $ and $ CD $, 2° to lines $ CD $ and $ EF $ (Fig. 16).
We obtain
From this last equality, it follows that points $ A $, $ B $, $ E $, $ F $ lie on a circle. Indeed, due to this equality, the proportion $ MA \colon ME = MF \colon MB $ holds, so triangles $ AMF $ and $ EMB $ with the common angle $ M $ have proportional sides, and thus are similar, and $ \measuredangle A = \measuredangle E $. This means that the segment $ BF $ is seen from points $ A $ and $ E $ at the same angle, so points $ A $ and $ E $ lie on a circle passing through $ B $ and $ F $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 794 |
LI OM - I - Problem 11
Given a positive integer $ n $ and a set $ M $, consisting of $ n^2 + 1 $ different positive integers and having the following property: among any $ n+1 $ numbers chosen from the set $ M $, there is a pair of numbers, one of which divides the other. Prove that in the set $ M $ there exist distinct numbers $ a_1, a_2, \ldots, a_{n+1} $ satisfying the condition: for $ i = 1,2,\ldots,n $, the number $ a_i $ divides $ a_{i+1} $.
|
Let $ k \in M $. Denote by $ d(k) $ the largest natural number $ \ell $ for which there exist different numbers $ a_1 = k, a_2, a_3, \ldots, a_\ell $ satisfying the condition: for $ i = 1,2,\ldots,\ell-1 $, the number $ a_i $ is divisible by $ a_{i+1} $.
Suppose the thesis of the problem is not satisfied. This means that for any $ k \in M $ we have $ 1 \leq d(k) \leq n $. Then, by the pigeonhole principle, there exist such different numbers $ k_1,k_2,\ldots,k_{n+1} $ that
From the assumptions of the problem, it follows that for some different numbers $ s $, $ t $, the divisibility $ k_s|k_t $ holds. The condition $ d(k_s) = m $ means that there exist numbers $ a_1 = k_s, a_2, \ldots, a_m $, satisfying: for $ i = 1,2,\ldots,m-1 $, the number $ a_i $ is divisible by $ a_{i+1} $. However, then the numbers $ b_1 = k_t, b_2 =a_1 = k_s , b_3 = a_2, \ldots , b_{m+1} =a_m $ satisfy the condition: for $ i = 1,2,\ldots,m $, the number $ b_i $ is divisible by $ b_{i+1} $. This proves that $ d(k_t) \geq m + 1 $, contradicting the assumption that $ d(k_t) = m $.
We have obtained a contradiction.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 795 |
XX OM - I - Task 1
Prove that if the numbers $ a_1, a_2, \ldots, a_n $ and $ b_1, b_2, \ldots, b_n $ are positive, then
|
The left side $ L $ of inequality (1) is transformed as follows.
Since for any positive numbers $ x $, $ y $, the inequality holds
thus
which means
Note. The above solution can be presented in the following form. Both the left side $ L $ and the right side $ P $ of inequality (1) are equal to a homogeneous polynomial of degree two in the variables $ a_1, a_2, \ldots, a_n $, which we can write as
We state that for $ i, k = 1, 2, \ldots, n $, $ i < k $
Hence
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 796 |
XXXIII OM - I - Problem 5
In a certain workplace, each employee is a member of exactly one of 100 trade unions. The employees are to elect a director from among two candidates. Members of each union agree on whether to abstain from voting or which of the two candidates they will vote for. Prove that there exists a union such that if its members abstain from voting, and members of all other unions vote for one of the two candidates, the candidates for director will not receive an equal number of votes.
|
Let $ a_1, a_2, \ldots, a_{100} $ be the number of members of each of the respective unions. We can assume that one of the numbers $ a_1, \ldots, a_{100} $ (for example, $ a_j $) is odd, because otherwise we could divide all $ a_i $ by the highest power of two that divides them all. If now the sum $ a_1+ \ldots+ a_{100} $ is an even number, then if the members of the $ i $-th union abstain from voting, the number of remaining votes $ a_1+ \ldots+a_{j-1}+a_{j+1} + \ldots+ a_{100} $ will be odd. Therefore, the number of votes cast for each candidate cannot be equal. If, on the other hand, the sum $ a_1+ \ldots+ a_{100} $ is an odd number, then one of its components must be even (the sum of a hundred odd numbers is an even number), let's say $ a_k = 2s $. If in this case the members of the $ k $-th union abstain from voting, the sum $ a_1 + \ldots a_{k-1} + a_{k+1} + \ldots + a_{100} $ is an odd number. The number of votes cast for each candidate cannot be equal.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 797 |
XXXIX OM - I - Problem 4
A triangle of maximum perimeter is inscribed in an ellipse. Prove that at each vertex, the angles between the tangent to the ellipse and the sides emanating from that vertex are equal.
|
Let $ABC$ be a triangle of maximum perimeter inscribed in a given ellipse. Suppose the statement of the problem is not true and, for example, the line $l$ tangent to the ellipse at point $B$ forms different angles with the sides $AB$ and $BC$. Draw a line $m$ through point $B$ that forms equal angles with these sides (perpendicular to the bisector of angle $ABC$). Since lines $l$ and $m$ do not coincide, we can find a point $P$ on the ellipse lying on the opposite side of line $m$ from points $A$ and $C$ (Figure 3).
om39_1r_img_3.jpg
Let $A'$ and $C'$ be the projections of points $A$ and $C$ onto line $m$, and let $D$ be the point symmetric to $C$ with respect to line $m$. The right triangles $BC'$ and $BD$ are congruent; thus, $|BC| = |BD|$, $|\measuredangle C'$. Since $|\measuredangle C'$ (by the definition of line $m$), we conclude that segment $BD$ is an extension of segment $AB$.
Point $P$ is farther from point $C$ than from point $D$. Therefore,
From this inequality, it follows that the perimeter of triangle $APC$ is greater than the perimeter of triangle $ABC$. We have reached a contradiction with the assumption of the maximality of the perimeter of triangle $ABC$; the proof is complete.
Note: The statement of the problem, as well as the given proof, remains valid if, instead of an ellipse, we consider any smooth closed curve (i.e., having a tangent at every point) bounding a convex region in the plane.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 801 |
LVI OM - III - Problem 6
Prove that every convex polygon with an area of 1 contains a convex hexagon with an area of at least 3/4.
|
From the vertices of a given polygon $\mathcal{W}$, we select three $A$, $B$, and $C$ which are the vertices of a triangle with the largest area, equal to $s$. Let $X$ be a point of the polygon $\mathcal{W}$, lying on the opposite side of the line $BC$ from point $A$ and farthest from the line $BC$ (Fig. 4). We define points $Y$ and $Z$ analogously.
om56_3r_img_4.jpg
We will show that the area of the hexagon $AZBXCY$ is not less than $3/4$, which will complete the solution of the problem.
Through points $A$, $B$, and $C$, we draw lines parallel to the lines $BC$, $CA$, and $AB$, respectively, forming a triangle with vertices $D$, $E$, $F$ (Fig. 5). From the fact that triangle $ABC$ has the largest area among all triangles with vertices in the vertices of the polygon $\mathcal{W}$, it follows that the polygon $\mathcal{W}$ is contained within the triangle $DEF$.
Through points $X$, $Y$, and $Z$, we draw lines parallel to the lines $BC$, $CA$, and $AB$, respectively, intersecting the sides of the triangle $DEF$ at points $K$, $L$, $M$, $N$, $P$, $Q$, as shown in Fig. 5. From the definition of points $X$, $Y$, and $Z$, it follows that the polygon $\mathcal{W}$ is contained within the hexagon with vertices $K$, $L$, $M$, $N$, $P$, $Q$.
Let $s$ denote the area of triangle $ABC$. Then each of the triangles $BCD$, $CAE$, and $ABF$ has an area equal to $s$. Let $x$, $y$, $z$ denote the areas of triangles $BCX$, $CAY$, and $ABZ$, respectively. We need to show that $x + y + z + s \geq 3/4$.
Let $J$ be a point such that the quadrilateral $CBJL$ is a parallelogram. Then triangles $CBD$ and $JKB$ are similar. Denoting the area of a figure $\mathcal{F}$ by $[\mathcal{F}]$, we have the relationships
thus $[JKB] = x^2/s$. From this, we obtain
Similarly, we express the areas of trapezoids $ACMN$ and $BAPQ$. Therefore, the area of the hexagon with vertices $K$, $L$, $M$, $N$, $P$, $Q$ is
Since this hexagon contains the given polygon $\mathcal{W}$, we have
It suffices to show that
Multiplying both sides of this inequality by $12s$ and reducing similar terms, we obtain its equivalent form:
and this last inequality is clearly satisfied.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 808 |
XVIII OM - I - Problem 3
In a convex quadrilateral $ABCD$, the midpoint $M$ of side $AB$ is connected to vertices $C$ and $D$, and the midpoint $N$ of side $CD$ is connected to vertices $A$ and $B$. Segments $AN$ and $DM$ intersect at point $P$, and segments $BN$ and $CM$ intersect at point $Q$. Prove that the area of quadrilateral $MQNP$ is equal to the sum of the areas of triangles $APD$ and $CQB$.
|
The area of a polygon $ABC\ldots$ will be denoted by the symbol $ (ABC\ldots) $. We state that (Fig. 1)
Thus,
therefore,
But,
and from (1) and (2) it follows that,
Similarly,
Indeed,
From (3), (4), and (5) we obtain,
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 809 |
XLV OM - I - Problem 11
A triangle with perimeter $ 2p $ is inscribed in a circle with radius $ R $ and circumscribed around a circle with radius $ r $. Prove that $ p < 2(R + r) $.
|
Let $ABC$ be the considered triangle; we adopt the usual notations
assuming (without loss of generality) that $a$ is the largest angle of this triangle. Thus, $a \geq 60^\circ$. Let $I$ be the center of the inscribed circle of triangle $ABC$, and let $A'$, $B'$, $C'$ be the points of tangency of this circle with the sides $BC$, $CA$, $AB$.
om45_1r_img_5.jpg
Furthermore, we assume
Thus, $a = y + z$, $b = z + x$, $c = x + y$. Now, observe that
Therefore, in view of the obvious estimate $a \leq 2R$, we obtain the inequality
which we were to prove.
{\kom Note:} As can be seen, a stronger inequality than the required one holds:
This is indeed a strict inequality, as the estimates $x \leq \sqrt{3}r$ and $a \leq 2R$ used in it cannot simultaneously become equalities: the first one becomes an equality only for an equilateral triangle, and the second one only for a right triangle.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 811 |
LIII OM - III - Problem 2
On the sides $ AC $ and $ BC $ of the acute-angled triangle $ ABC $, rectangles $ ACPQ $ and $ BKLC $ of equal areas are constructed on its external side. Prove that the midpoint of segment $ PL $, point $ C $, and the center of the circumcircle of triangle $ ABC $ lie on one straight line.
|
Let $ O $ be the center of the circle circumscribed around triangle $ ABC $ (Fig. 1). Complete triangle $ PCL $ to parallelogram $ PCLX $. It suffices to prove that points $ X $, $ C $, $ O $ are collinear.
From the equality of the areas of the given rectangles, we obtain
Moreover, $ \measuredangle XLC = 180^\circ - \measuredangle PCL = \measuredangle BCA $. The obtained relationships prove that triangles $ XLC $ and $ BCA $ are similar. Hence, and from the equality $ BO = CO $, it follows that
which means $ \measuredangle XCL + \measuredangle LCB + \measuredangle BCO = 180^\circ $. The last equality indicates that points $ X $, $ C $, $ O $ lie on the same line.
om53_3r_img_1.jpg
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 812 |
XXXVI OM - I - Problem 1
Prove that if an integer divisible by 3 is the sum of five squares of integers, then at least two of these squares are divisible by 9.
|
The square of an integer not divisible by $3$ gives a remainder of $1$ when divided by $3$. Therefore, if a certain number is the sum of five squares and none of these squares is divisible by $3$, then the given number gives a remainder of $2$ when divided by $3$; if only one of these squares is divisible by $3$, then the given number gives a remainder of $1$ when divided by $3$. Since, by assumption, the number is divisible by $3$, at least two of the squares in the considered sum must be squares of numbers divisible by $3$ - they are therefore divisible by $9$.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 813 |
XXII OM - I - Problem 7
On a plane, there are five lattice points (i.e., points with integer coordinates). Prove that the midpoint of one of the segments connecting these points is also a lattice point.
|
Let the lattice points be $ P_i = (x_i, y_i) $, where $ i = 1, 2, 3, 4, 5 $. Among the numbers $ x_1 $, $ x_2 $, $ x_3 $, $ x_4 $, $ x_5 $, there are at least three of the same parity (i.e., three even or three odd), because if, for example, at most two are even, then the remaining ones are odd. Let's assume, for example, that the numbers $ x_1 $, $ x_2 $, $ x_3 $ are of the same parity.
Reasoning similarly as before, we conclude that among the numbers $ y_1 $, $ y_2 $, $ y_3 $, at least two are of the same parity. Let's assume, for example, that the numbers $ y_1 $, $ y_2 $ are of the same parity. Then the numbers $ x_1 + x_2 $ and $ y_1 + y_2 $ will be even, and thus the midpoint of the segment $ P_1P_2 $ will have integer coordinates $ \left(\frac{1}{2}(x_1+x_2), \frac{1}{2}(y_1+y_2) \right) $, i.e., it will be a lattice point.
Note 1. An analogous theorem holds in $ n $-dimensional space: If there are $ 2^n + 1 $ points with integer coordinates in $ n $-dimensional space, then the midpoint of one of the segments connecting these points has integer coordinates.
Note 2. The problem can also be generalized in another way. If there are $ n $ lattice points on a plane, then the midpoints of at least $ \displaystyle \binom{\left< \frac{n}{4} \right>}{2} $ segments connecting these points are lattice points.
Here $ \left<x\right> $ denotes the smallest integer not less than $ x $.
Proof of Note 2. Divide the given points into classes depending on the parity of their coordinates. There are four classes $ (n, n) $, $ (n, p) $, $ (p, n) $, and $ (p, p) $, where $ n $ indicates that the corresponding coordinate is odd, and $ p $ indicates that it is even. Therefore, in one of the classes, there will be at least $ \frac{n}{4} $ points, and since the number of points in a class is an integer, there will be at least $ \left< \frac{n}{4} \right> $ points in one of the classes.
Since the midpoint of a segment whose endpoints are points in the same class is a lattice point, and for $ r $ points there are $ \binom{r}{2} $ segments connecting them, the number of segments whose midpoints are lattice points is at least $ \displaystyle \binom{\left< \frac{n}{4} \right>}{2} $.
Notice also that
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 814 |
XLVII OM - I - Problem 8
A light ray emanates from the center of a square, reflecting off the sides of the square according to the principle that the angle of incidence equals the angle of reflection. After some time, the ray returns to the center of the square. The ray never hit a vertex or passed through the center before. Prove that the number of reflections off the sides of the square is odd.
|
On the plane with a rectangular coordinate system $Oxy$, we draw all lines with equations $x = c$ ($c$ integers) and $y = c$ ($c$ integers); in this way, the entire plane is covered by an infinite grid of unit squares. In each pair of squares sharing a common side, we identify points that are symmetric with respect to that side (see Figure 5). Thus, for example, the point $X = (u,v)$ in the square $OABC$ with vertices $O = (0,0)$, $A = (1,0)$, $B = (1,1)$, $C = (0,1)$ is identified with the points $(2-u,v)$ and $(u,2-v)$ in the unit squares adjacent to $OABC$ on the right and above, and also with two points in the two unit squares adjacent to $OABC$ on the left and below.
Through symmetry with respect to further grid lines, this point has its replicas in all squares of the grid - one in each unit square. The center of the square $OABC$, i.e., the point $P=(\frac{1}{2},\frac{1}{2})$, like any other, has infinitely many copies; they are the centers of all the squares in the grid, i.e., points of the form $(\frac{1}{2}+ m, \frac{1}{2} +n)$, where $m, n$ can be any integers.
We can assume that $OABC$ is the square in question, and that a light ray was sent from point $P$ "to the right and up" (along a line with a positive slope). It first reflects off the side $AB$ or $BC$ at some point $Q$. The next time it hits the edge of the square $OABC$ at some point $R$.
The segment $QR$ of its trajectory has a copy $QR$ in the square adjacent to $OABC$ (along the side $AB$ or $BC$), which is a linear extension of the segment $PQ$; the linearity follows from the law of reflection. Indeed: in the situation as shown, the point $R$ is symmetric to $R$ with respect to the line $AB$, so $|\measuredangle BQR| = |\measuredangle BQR|$, and since also $|\measuredangle BQR| = |\measuredangle PQA|$ (law of reflection), the point $R$ lies on the ray $PQ^\to$.
Similarly, the next segment of the ray's path has a copy in the next square of the grid, which is an extension of the segment $QR$. And so on: each subsequent segment of the path can be identified with a further segment of the ray $PQ^\to$. In this way, the entire trajectory of the ray, from its start at point $P$ to its return to point $P$, corresponds to a certain linear path on the plane, starting at point $P$ and ending at some point $Z$, which is one of the replicas of the center of the square $OABC$. Thus, $Z$ is a point of the form $(\frac{1}{2} +m, \frac{1}{2} + n)$ for some pair of natural numbers $m, n$.
The segment $PZ$ contains, in particular, the point $M=(\frac{1}{2}(1+m),\frac{1}{2}(1 + n))$, the midpoint of this segment. If both numbers $m, n$ are even, then $M$ is the center of some square in the grid, i.e., an image of the center $P$ of the square $OABC$. If both numbers $m, n$ are odd, then $M$ is a point with integer coordinates, i.e., an image of a vertex of the square $OABC$. Each of these situations is excluded by the conditions of the problem. Therefore, the numbers $m$ and $n$ are of different parity, and thus their sum is an odd number.
However, note that the segment $PZ$ intersects $m$ vertical lines and $n$ horizontal lines, which are grid lines. Each such intersection point corresponds to a point where the ray reflects off the edge of the square $OABC$. This correspondence is one-to-one, since the segment $PZ$ does not pass through any grid node (the "ray did not hit a vertex of the square"). Hence, the number of reflections is equal to the sum $m + n$; this sum, as we have shown, is an odd number. The theorem is thus proved.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 816 |
XXIII OM - I - Problem 8
On a plane, there are two congruent equilateral triangles $ABC$ and $A'B'C'$. Prove that if the midpoints of segments $\overline{AA'},$ $\overline{BB'},$ $\overline{CC'}$ are not collinear, then they are the vertices of an equilateral triangle.
|
Let points $O$ and $O'$ be the centers of the circumcircles of triangles $ABC$ and $A'B'C'$, respectively. Denote by $T$ the translation of the plane by the vector $\overrightarrow{OO'}$, and let $A'$, $B'$, $C'$ be the images of $A$, $B$, $C$ under this translation. Denote by $A_1$, $B_1$, $C_1$ the midpoints of segments $\overline{AA'}$, $\overline{BB'}$, $\overline{CC'}$, and by $A_2$, $B_2$, $C_2$ the midpoints of segments $\overline{AA'}$, $\overline{BB'}$, $\overline{CC'}$ (Fig. 6). We have $\overrightarrow{A_1A_2} = \overrightarrow{A_1A} + \overrightarrow{AA_2} = \frac{1}{2} \overrightarrow{OO'}$. Similarly, we find that $\overrightarrow{B_1B_2} = \frac{1}{2} \overrightarrow{OO'}$ and $\overrightarrow{C_1C_2} = \frac{1}{2} \overrightarrow{OO'}$. Therefore, points $A_2$, $B_2$, $C_2$ are the images of points $A_1$, $B_1$, $C_1$ under the translation by the vector $\frac{1}{2} \overrightarrow{OO'}$. It suffices to prove that points $A_2$, $B_2$, $C_2$ are collinear or are the vertices of an equilateral triangle.
Equilateral triangles $\triangle ABC$ and $\triangle A'B'C'$ are inscribed in the circle $K$ with center $O$. There exists an isometry $\varphi$ that maps points $A$, $B$, $C$ to $A'$, $B'$, $C'$, respectively. This isometry maps the circle $K$ onto itself. Therefore, the isometry $\varphi$ is either a rotation about the point $O$ or a reflection across an axis.
If $\varphi$ is a rotation (Fig. 7), then performing a rotation $S$ by an angle $\frac{2}{3} \pi$ about the point $O$ will map points $A$ and $A'$ to points $B$ and $B'$, respectively. Thus, $S(A_2) = B_2$ and similarly $S(B_2) = C_2$ and $S(C_2) = A_2$. Therefore, $\overline{A_2B_2} = \overline{S(A_2) S(B_2)} = \overline{B_2C_2}$ and similarly $\overline{B_2C_2} = \overline{C_2A_2}$. Hence, triangle $A_2B_2C_2$ is equilateral or $A_2 = B_2 = C_2 = 0$ (Fig. 8).
If the isometry $\varphi$ is a reflection, then points $A_2$, $B_2$, $C_2$ lie on the axis of this reflection and are therefore collinear (Fig. 9).
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 818 |
I OM - B - Task 8
Prove that the altitudes of an acute triangle are the angle bisectors of the triangle whose vertices are the feet of these altitudes.
|
Let $ S $ be the point of intersection of the altitudes $ AM $, $ BN $, and $ CP $ of an acute triangle $ ABC $ (Fig. 3).
om1_Br_img_3.jpg
The quadrilateral $ SMCN $, in which the angles at vertices $ M $ and $ N $ are right angles, is a cyclic quadrilateral with diameter $ SC $; therefore,
(angles subtended by the same arc are equal).
Similarly, in the quadrilateral $ SMBP $ we have
But $ \measuredangle SCN = \measuredangle SBP $, since each of these angles equals
$ 90^{\circ} - \measuredangle BAC $; therefore $ \measuredangle SMN = \measuredangle SMP $.
Indeed, the altitude $ AM $ is the angle bisector of $ \angle NMP $.
Note. In an obtuse triangle (Fig. 4) the altitude $ AM $ drawn from the vertex of the obtuse angle $ A $ is the angle bisector of $ \angle NMP $, while the other two altitudes are the bisectors of the exterior angles of triangle $ MNP $; the proof is analogous to the previous one.
om1_Br_img_4.jpg
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 821 |
XLI OM - III - Task 4
Inside a square with a side length of 1, a triangle was drawn, each side of which has a length of at least 1. Prove that the center of the square belongs to the triangle.
|
We apply proof by contradiction.
Suppose that the center $O$ of the given square $ABCD$ does not belong to the drawn triangle $A$. There then exists a line $l$ passing through $O$ such that triangle $A$ lies entirely on one side of this line (i.e., it is contained in one of the two open half-planes determined by the line $l$, see: Note).
Let the points of intersection of the line $l$ with the boundary of the square be denoted by $A$ and $C$; we can assume (by cyclically reassigning the labels of the vertices of the square if necessary) that point $A$ lies on side $AB$, and point $C$ lies on side $CD$, with $A \neq C$.
Draw a line $m$ through point $O$ perpendicular to $l$. It intersects the boundary of the square at points $B$ and $D$, which lie on sides $BC$ and $DA$ respectively, with $B \neq D$ (see Figure 7).
The lines $l$ and $m$ divide the square into four congruent figures. Triangle $A$ lies on one side of the line $l$, and thus is contained in the union of two of these figures. One of them must contain at least two vertices of triangle $A$, i.e., the endpoints of a segment of length not less than $1$. Let this be the part of the square cut out by the rays $(OA$ and $(OB (such an assumption does not reduce the generality of the considerations).
In the special case where the lines $l$ and $m$ contain the diagonals of the square $ABCD$ (i.e., when $A = A$, $B = B$, $C = C$, $D = D$), the considered figure is the triangle $OAB$. The only segment of length $\geq 1$ contained in this triangle is its hypotenuse $AB$. This would mean that segment $AB$ is one of the sides of triangle $A$ - contrary to the assumption that triangle $A$ lies inside the square $ABCD$.
In the remaining case ($A \neq A$ etc.), the considered figure is the quadrilateral $OA$, where the angles at vertices $O$ and $B$ are right angles - so a circle can be circumscribed around it. Its diameter is the segment $A$ of length less than $1$ (since $A$ is a square inscribed in $ABCD$, not identical to $ABCD$). Therefore, the figure (quadrilateral) $OA$ cannot contain a pair of points at a distance of at least $1$ - contrary to what we stated a moment ago.
The obtained contradiction completes the proof.
Note. In the solution, we used the fact that through any point $O$ lying outside a given triangle $\Delta$, one can draw a line $Z$ such that the triangle $\Delta$ lies entirely on one side of it. It might be worth providing a rigorous justification for this fact.
The triangle is the intersection of three closed half-planes, whose edges are the lines containing its sides. Since point $O$ does not belong to the given triangle, it does not belong to at least one of these half-planes. Let $l$ be its edge (if there are two such half-planes, we choose one of them, and its edge we denote by $l$). The line $Z$, parallel to $V$ and passing through $O$, has the desired property.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 822 |
XVI OM - I - Problem 12
In space, there are two polyhedra. Prove that the longest of the segments connecting points of one polyhedron with points of the other has its ends at the vertices of the polyhedra.
|
First, let us note that if all vertices of a polyhedron lie on one side of a certain plane, then all other points of the polyhedron lie on that same side of the plane.
Suppose, for instance, that all vertices of the polyhedron $ W $ are located in an open half-space (i.e., a half-space without the points of the bounding plane) $ \pi $ bounded by the plane $ \alpha $. Since a segment connecting two points of the half-space $ \pi $ is entirely contained in $ \pi $, all edges of the polyhedron $ W $ lie in $ \pi $.
It follows that all points of the faces of the polyhedron belong to $ \pi $, as each point on a face lies either on an edge or on a segment connecting points of two edges. Similarly, each interior point of the polyhedron belongs to $ \pi $, as each such point lies on a segment connecting points of two faces.
Let polyhedra $ W_1 $ and $ W_2 $ be given. Consider all segments connecting the vertices of the polyhedron $ W_1 $ with the vertices of the polyhedron $ W_2 $. Since the set of these segments is finite, there is at least one segment of maximum length among them. Let such a segment be $ P_1P_2 $ ($ P_1 $ - a vertex of $ W_1 $, $ P_2 $ - a vertex of $ W_2 $) and let $ A_1 $ be any point of the polyhedron $ W_1 $, and $ A_2 $ - any point of the polyhedron $ W_2 $.
We will prove that $ P_1P_2 \geq A_1A_2 $.
Draw through the point $ A_1 $ a plane $ \alpha_1 $ perpendicular to the line $ A_1A_2 $, and through the point $ A_2 $ a plane $ \alpha_2 $ also perpendicular to $ A_1A_2 $. The open half-space $ \pi $, which is bounded by the half-plane $ \alpha_1 $ and contains the point $ A_2 $, does not contain the entire polyhedron $ W_1 $, as it does not contain the point $ A_1 $. Therefore, by the previous observation, there exists a vertex $ B_1 $ of the polyhedron $ W_1 $ not belonging to $ \pi $, i.e., lying on the plane $ \alpha_1 $ or on the opposite side of the plane $ \alpha_1 $ from the point $ A_2 $. Similarly, there exists a vertex $ B_2 $ of the polyhedron $ W_2 $ lying on the plane $ \alpha_2 $ or on the opposite side of the plane $ \alpha_2 $ from the point $ A_1 $. Since $ B_1B_2 \geq A_1A_2 $, and $ P_1P_2 \geq B_1B_2 $, therefore
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 824 |
XXVI - II - Task 1
Given the polynomial $ W(x) = x^4 + ax^3 + bx + cx + d $. Prove that for the equation $ W(x) = 0 $ to have four real roots, it is necessary and sufficient for there to exist an $ m $ such that $ W(x+m) = x^4+px^2+q $, that the sum of some two roots of the equation $ W(x) = 0 $ equals the sum of the remaining two.
|
If the roots of the polynomial $W(x)$ are the numbers $x_1, x_2, x_3, x_4$ and
then by taking $ \displaystyle m = \frac{1}{2} (x_1 + x_2) $, we obtain that the roots of the polynomial $W(x+m)$ are the numbers $ y_i = x_i - m $ for $ i = 1, 2, 3, 4 $, i.e., the numbers
Thus,
It follows from this that
We have thus shown that if the roots of the polynomial $W$ satisfy the relation (1), then for $ \displaystyle m = \frac{1}{2}(x_1 + x_2) $, we have $ W(x+m) = x^4 + px^2 + q $, where $ \displaystyle p = - \frac{1}{4} (x_1 - x_2)^2 - \frac{1}{4} (x_3 - x_4)^2 $, $ \displaystyle q = \left[ \frac{1}{4}(x_1 - x_2)(x_3 - x_4) \right]^2 $.
We will now prove the converse implication. If a real number is a root of the polynomial $x^4 + px^2 + q$, then its opposite number is also a root. Therefore, if for some real number $m$ the polynomial $W(x+m)$ has the form $x^4 + px^2 + q$, then its roots $y_1, y_2, y_3, y_4$ satisfy the equations (2). The roots of the polynomial $W(x)$ are the numbers $x_i = y_i + m$, where $i = 1, 2, 3, 4$. From (2) we thus obtain that $x_1 + x_2 = y_1 + m + (-y_1) + m = 2m$ and similarly $x_3 + x_4 = y_3 + m + (-y_3) + m = 2m$. Therefore, the equation (1) holds.
Note. It is not difficult to notice that there is only one number $m$ that satisfies the conditions of the problem. If $f(x) = x^4 + px^2 + q$, then for $k \ne 0$ the polynomial $f(x+k)$ has the coefficient of $x^3$ equal to $4k$, and thus different from zero.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 826 |
XVII OM - II - Problem 3
In the plane, 6 points are chosen, no three of which lie on the same line, and all segments connecting these points in pairs are drawn. Some of these segments are drawn in red, and others in blue. Prove that some three of the given points are vertices of a triangle with sides of the same color.
|
Let $ A_1 $, $ A_2 $, $ A_3 $, $ A_4 $, $ A_5 $, $ A_6 $ be given points. Among the five segments $ A_1A_i $ ($ i = 2, 3, 4, 5, 6 $) at least three are of the same color. Suppose, for example, that the segments $ A_1A_2 $, $ A_1A_3 $, $ A_1A_4 $ are red and consider the segments $ A_2A_3 $, $ A_3A_4 $, $ A_4A_2 $. If these segments are all three blue, we have a triangle $ A_2A_3A_4 $ of one color. If, however, one of these segments, say $ A_2A_3 $, is red, then the triangle $ A_1A_2A_3 $ is of one color.
For a smaller number of given points, the thesis of the theorem is not true. For example, in a convex pentagon, where all sides are red and all diagonals are blue, any three consecutive vertices form a triangle with two red sides and one blue side, and any three non-consecutive vertices form a triangle with two blue sides and one red side.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 827 |
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