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class | __index_level_0__ int64 0 742k |
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03.3. The point $D$ inside the equilateral triangle $\triangle A B C$ satisfies $\angle A D C=150^{\circ}$. Prove that a triangle with side lengths $|A D|,|B D|,|C D|$ is necessarily a right-angled triangle.
Figure 14 .
|
Solution. (See Figure 14.) We rotate the figure counterclockwise $60^{\circ}$ around $C$. Because $A B C$ is an equilateral triangle, $\angle B A C=60^{\circ}$, so $A$ is mapped on $B$. Assume $D$ maps to $E$. The properties of rotation imply $A D=B E$ and $\angle B E C=150^{\circ}$. Because the triangle $D E C$ is equilateral, $D E=D C$ and $\angle D E C=60^{\circ}$. But then $\angle D E B=150^{\circ}-60^{\circ}=90^{\circ}$. So segments having the lengths as specified in the problem indeed are sides of a right triangle.
| proof | Geometry | proof | Yes | Yes | olympiads | false | 425 |
03.4. Let $\mathbb{R}^{*}=\mathbb{R} \backslash\{0\}$ be the set of non-zero real numbers. Find all functions $f: \mathbb{R}^{*} \rightarrow \mathbb{R}^{*}$ satisfying
$$
f(x)+f(y)=f(x y f(x+y))
$$
for $x, y \in \mathbb{R}^{*}$ and $x+y \neq 0$.
|
Solution. If $x \neq y$, then
$$
f(y)+f(x-y)=f(y(x-y) f(x))
$$
Because $f(y) \neq 0$, we cannot have $f(x-y)=f(y(x-y) f(x))$ or $x-y=y(x-y) f(x)$. So for all $x \neq y, y f(x) \neq 1$. The only remaining possibility is $f(x)=\frac{1}{x}$. - One easily checks that $f, f(x)=\frac{1}{x}$, indeed satisfies the original functional equation.
| f(x)=\frac{1}{x} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 426 |
04.1. 27 balls, labelled by numbers from 1 to 27, are in a red, blue or yellow bowl. Find the possible numbers of balls in the red bowl, if the averages of the labels in the red, blue, and yellow bowl are 15, 3 ja 18, respectively.
|
Solution. Let $R, B$, and $Y$, respectively, be the numbers of balls in the red, blue, and yellow bowl. The mean value condition implies $B \leq 5$ (there are at most two balls with a number $3$ ). $R, B$ and $Y$ satisfy the equations
$$
\begin{aligned}
R+B+Y & =27 \\
15 R+3 S+18 Y & =\sum_{j=1}^{27} j=14 \cdot 27=378
\end{aligned}
$$
We eliminate $S$ to obtain $4 R+5 Y=99$. By checking the possibilities we note that the pairs of positive integers satisfying the last equation are $(R, Y)=(21,3),(16,7),(11,11)$, $(6,15)$, and $(1,19)$. The last two, however, do not satisfy $B=27-(R+Y) \leq 5$. We still have to ascertain that the three first alternatives are possible. In the case $R=21$ we can choose the balls $5,6, \ldots, 25$, in the red bowl, and 2,3 and 4 in the blue bowl; if $P=16$, $7,8, \ldots, 14,16,17, \ldots, 23$, can go to the red bowl and $1,2,4$ and 5 in the blue one, and if $P=11$, the red bowl can have balls $10,11, \ldots 20$, and the blue one $1,2,3,4,5$. The red bowl can contain 21,16 or 11 balls.
| 21,16,11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 427 |
04.2. Let $f_{1}=0, f_{2}=1$, and $f_{n+2}=f_{n+1}+f_{n}$, for $n=1$, 2, ..., be the Fibonacci sequence. Show that there exists a strictly increasing infinite arithmetic sequence none of whose numbers belongs to the Fibonacci sequence. [A sequence is arithmetic, if the difference of any of its consecutive terms is a constant.]
|
Solution. The Fibonacci sequence modulo any integer $n>1$ is periodic. (Pairs of residues are a finite set, so some pair appears twice in the sequence, and the sequence from the second appearance of the pair onwards is a copy of the sequence from the first pair onwards.) There are integers for which the Fibonacci residue sequence does not contain all possible residues. For instance modulo 11 the sequence is $0,1,1,2,3,5,8,2,10,1$, $0,1,1, \ldots$ Wee see that the number 4 is missing. It follows that no integer of the form $4+11 k$ appears in the Fibonacci sequence. But here we have an arithmetic sequence of the kind required.
| proof | Number Theory | proof | Yes | Yes | olympiads | false | 428 |
04.3. Let $x_{11}, x_{21}, \ldots, x_{n 1}, n>2$, be a sequence of integers. We assume that all of the numbers $x_{i 1}$ are not equal. Assuming that the numbers $x_{1 k}, x_{2 k}, \ldots, x_{n k}$ have been defined, we set
$$
\begin{aligned}
x_{i, k+1} & =\frac{1}{2}\left(x_{i k}+x_{i+1, k}\right), i=1,2, \ldots, n-1 \\
x_{n, k+1} & =\frac{1}{2}\left(x_{n k}+x_{1 k}\right)
\end{aligned}
$$
Show that for $n$ odd, $x_{j k}$ is not an integer for some $j, k$. Does the same conclusion hold for $n$ even?
|
Solution. We compute the first index modulo $n$, i.e. $x_{1 k}=x_{n+1, k}$. Let $M_{k}=\max _{j} x_{j k}$ and $m_{k}=\min _{j} x_{j k}$. Evidently $\left(M_{k}\right)$ is a non-increasing and $\left(m_{k}\right)$ a non-decreasing sequence, and $M_{k+1}=M_{k}$ is possible only if $x_{j k}=x_{j+1, k}=M_{k}$ for some $j$. If exactly $p$
consequtive numbers $x_{j k}$ equal $M_{k}$, then exactly $p-1$ consequtive numbers $x_{j, k+1}$ equal $M_{k+1}$ which is equal to $M_{k}$. So after a finite number of steps we arrive at the situation $M_{k+1}m_{k}$ for some $k$ 's. If all the numbers in all the sequences are integers, then all $m_{k}$ 's and $M_{k}$ 's are integers. So after a finite number of steps $m_{k}=M_{k}$, and all numbers $x_{j k}$ are equal. Then $x_{1, k-1}+x_{2, k-1}=x_{2, k-1}+x_{3, k-1}=\cdots=x_{n-1, k-1}+x_{n, k-1}=x_{n, k-1}+x_{1, k-1}$. If $n$ is odd, then $x_{1, k-1}=x_{3, k-1}=\cdots=x_{n, k-1}$ and $x_{1, k-1}=x_{n-1, k-1}=\cdots=x_{2, k-1}$. But then we could show in a similar way that all numbers $x_{j, k-2}$ are equal and finally that all numbers $x_{j, 1}$ are equal, contrary to the assumption. If $n$ is even, then all $x_{i, k}$ 's can be integers. Take, for instance, $x_{1,1}=x_{3,1}=\cdots=x_{n-1,1}=0, x_{2,1}=x_{4,1}=\cdots=x_{n, 1}=2$. Then every $x_{j, k}=1, k \geq 2$.
| proof | Algebra | proof | Yes | Yes | olympiads | false | 429 |
04.4. Let $a, b$, and $c$ be the side lengths of a triangle and let $R$ be its circumradius. Show that
$$
\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a} \geq \frac{1}{R^{2}}
$$
|
Solution 1. By the well-known (Euler) theorem, the inradius $r$ and circumradius $R$ of any triangle satisfy $2 r \leq R$. (In fact, $R(R-2 r)=d^{2}$, where $d$ is the distance between the incenter and circumcenter.) The area $S$ of a triangle can be written as
$$
A=\frac{r}{2}(a+b+c)
$$
and, by the sine theorem, as
$$
A=\frac{1}{2} a b \sin \gamma=\frac{1}{4} \frac{a b c}{R}
$$
Combining these, we obtain
$$
\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}=\frac{a+b+c}{a b c}=\frac{2 A}{r} \cdot \frac{1}{4 R A}=\frac{1}{2 r R} \geq \frac{1}{R^{2}}
$$
| proof | Inequalities | proof | Yes | Yes | olympiads | false | 430 |
05.1. Find all positive integers $k$ such that the product of the digits of $k$, in the decimal system, equals
$$
\frac{25}{8} k-211
$$
|
Solution. Let
$$
a=\sum_{k=0}^{n} a_{k} 10^{k}, \quad 0 \leq a_{k} \leq 9, \text { for } 0 \leq k \leq n-1,1 \leq a_{n} \leq 9
$$
Set
$$
f(a)=\prod_{k=0}^{n} a_{k}
$$
Since
$$
f(a)=\frac{25}{8} a-211 \geq 0
$$
$a \geq \frac{8}{25} \cdot 211=\frac{1688}{25}>66$. Also, $f(a)$ is an integer, and $\operatorname{gcf}(8,25)=1$, so $8 \mid a$. On the other hand,
$$
f(a) \leq 9^{n-1} a_{n} \leq 10^{n} a_{n} \leq a
$$
So
$$
\frac{25}{8} a-211 \leq a
$$
or $a \leq \frac{8}{17} \cdot 211=\frac{1688}{17}<100$. The only multiples of 8 between 66 and 100 are $72,80,88$, and 96. Now $25 \cdot 9-211=17=7 \cdot 2,25 \cdot 10-211=39 \neq 8 \cdot 0,25 \cdot 11-211=64=8 \cdot 8$, and $25 \cdot 12-211=89 \neq 9 \cdot 6$. So 72 and 88 are the numbers asked for.
| 7288 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 431 |
05.2. Let $a, b$, and $c$ be positive real numbers. Prove that
$$
\frac{2 a^{2}}{b+c}+\frac{2 b^{2}}{c+a}+\frac{2 c^{2}}{a+b} \geq a+b+c
$$
|
Solution 1. Use brute force. Removing the denominators and brackets and combining simililar terms yields the equivalent inequality
$$
\begin{gathered}
0 \leq 2 a^{4}+2 b^{4}+2 c^{4}+a^{3} b+a^{3} c+a b^{3}+b^{3} c+a c^{3}+b c^{3} \\
-2 a^{2} b^{2}-2 b^{2} c^{2}-2 a^{2} c^{2}-2 a b c^{2}-2 a b^{2} c-2 a^{2} b c \\
=a^{4}+b^{4}-2 a^{2} b^{2}+b^{4}+c^{4}-2 b^{2} c^{2}+c^{4}+a^{4}-2 a^{2} c^{2} \\
+a b\left(a^{2}+b^{2}-2 c^{2}\right)+b c\left(b^{2}+c^{2}-2 a^{2}\right)+c a\left(c^{2}+a^{2}-2 b^{2}\right) \\
=\left(a^{2}-b^{2}\right)^{2}+\left(b^{2}-c^{2}\right)^{2}+\left(c^{2}-a^{2}\right)^{2} \\
+a b(a-b)^{2}+b c(b-c)^{2}+c a(c-a)^{2} \\
+a b\left(2 a b-2 c^{2}\right)+b c\left(2 b c-2 a^{2}\right)+c a\left(2 c a-2 b^{2}\right)
\end{gathered}
$$
The six first terms on the right hand side are non-negative and the last three can be written as
$$
\begin{gathered}
2 a^{2} b^{2}-2 a b c^{2}+2 b^{2} c^{2}-2 a^{2} b c+2 c^{2} a^{2}-2 a b^{2} c \\
=a^{2}\left(b^{2}+c^{2}-2 b c\right)+b^{2}\left(a^{2}+c^{2}-2 a c\right)+c^{2}\left(a^{2}+b^{2}-2 a b\right) \\
=a^{2}(b-c)^{2}+b^{2}(c-a)^{2}+c^{2}(a-b)^{2} \geq 0
\end{gathered}
$$
So the original inequality is true.
| proof | Inequalities | proof | Yes | Yes | olympiads | false | 432 |
05.4. The circle $\mathcal{C}_{1}$ is inside the circle $\mathcal{C}_{2}$, and the circles touch each other at $A$. A line through $A$ intersects $\mathcal{C}_{1}$ also at $B$ and $\mathcal{C}_{2}$ also at $C$. The tangent to $\mathcal{C}_{1}$ at $B$ intersects $\mathcal{C}_{2}$ at $D$ and $E$. The tangents of $\mathcal{C}_{1}$ passing through $C$ touch $\mathcal{C}_{1}$ at $F$ and $G$. Prove that $D$, $E, F$, and $G$ are concyclic.
Figure 15.
|
Solution. (See Figure 15.) Draw the tangent $\mathrm{CH}$ to $\mathcal{C}_{2}$ at $C$. By the theorem of the angle between a tangent and chord, the angles $A B H$ and $A C H$ both equal the angle at $A$ between $B A$ and the common tangent of the circles at $A$. But this means that the angles $A B H$ and $A C H$ are equal, and $C H \| B E$. So $C$ is the midpoint of the arc $D E$. This again implies the equality of the angles $C E B$ and $B A E$, as well as $C E=C D$. So the triangles $A E C, C E B$, having also a common angle $E C B$, are similar. So
$$
\frac{C B}{C E}=\frac{C E}{A C}
$$
and $C B \cdot A C=C E^{2}=C D^{2}$. But by the power of a point theorem, $C B \cdot C A=C G^{2}=C F^{2}$. We have in fact proved $C D=C E=C F=C G$, so the four points are indeed concyclic.
06.1 Let $B$ and $C$ be points on two fixed rays emanating from a point $A$ such that $A B+A C$ is constant. Prove that there exists a point $D \neq A$ such that the circumcircles of the triangels $A B C$ pass through $D$ for every choice of $B$ and $C$.
Figure 16 .
| proof | Geometry | proof | Yes | Yes | olympiads | false | 433 |
06.2. The real numbers $x, y$ and $z$ are not all equal and satisfy
$$
x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}=k
$$
Determine all possible values of $k$.
|
Solution. Let $(x, y, z)$ be a solution of the system of equations Since
$$
x=k-\frac{1}{y}=\frac{k y-1}{y} \quad \text { and } \quad z=\frac{1}{k-y}
$$
the equation
$$
\frac{1}{k-y}+\frac{y}{k y-1}=k
$$
to be simplified into
$$
\left(1-k^{2}\right)\left(y^{2}-k y+1\right)=0
$$
is true. So either $|k|=1$ or
$$
k=y+\frac{1}{y}
$$
The latter alternative, substituted to the original equations, yields immediately $x=y$ and $z=y$. So $k= \pm 1$ is the only possibility. If $k=1$, for instance $x=2, y=-1$ and $z=\frac{1}{2}$ is a solution; if $k=-1$, a solution is obtained by reversing the signs for a solution with $k=1$. So $k=1$ and $k=-1$ are the only possible values for $k$.
| \1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 434 |
06.3. A sequence of positive integers $\left\{a_{n}\right\}$ is given by
$$
a_{0}=m \quad \text { and } \quad a_{n+1}=a_{n}^{5}+487
$$
for all $n \geq 0$. Determine all values of $m$ for which the sequence contains as many square numbers as possible.
|
Solution. Consider the expression $x^{5}+487$ modulo 4. Clearly $x \equiv 0 \Rightarrow x^{5}+487 \equiv 3$, $x \equiv 1 \Rightarrow x^{5}+487 \equiv 0 ; x \equiv 2 \Rightarrow x^{5}+487 \equiv 3$, and $x \equiv 3 \Rightarrow x^{5}+487 \equiv 2$. Square numbers are always $\equiv 0$ or $\equiv 1 \bmod 4$. If there is an even square in the sequence, then all subsequent numbers of the sequence are either $\equiv 2$ or $\equiv 3 \bmod 4$, and hence not squares. If there is an odd square in the sequence, then the following number in the sequence can be an even square, but then none of the other numbers are squares. So the maximal number of squares in the sequence is two. In this case the first number of the sequence has to be the first square, since no number of the sequence following another one satisfies $x \equiv 1 \bmod 4$. We have to find numbers $k^{2}$ such that $k^{10}+487=n^{2}$. We factorize $n^{2}-k^{10}$. Because 487 is a prime, $n-k^{5}=1$ and $n+k^{5}=487$ or $n=244$ and $k=3$. The only solution of the problem thus is $m=3^{2}=9$.
| 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 435 |
06.4. The squares of a $100 \times 100$ chessboard are painted with 100 different colours. Each square has only one colour and every colour is used exactly 100 times. Show that there exists a row or a column on the chessboard in which at least 10 colours are used.
|
Solution. Denote by $R_{i}$ the number of colours used to colour the squares of the $i$ 'th row and let $C_{j}$ be the number of colours used to colour the squares of the $j$ 'th column. Let $r_{k}$ be the number of rows on which colour $k$ appears and let $c_{k}$ be the number of columns on which colour $k$ appears. By the arithmetic-geometric inequality, $r_{k}+c_{k} \geq 2 \sqrt{r_{k} c_{k}}$. Since colour $k$ appears at most $c_{k}$ times on each of the $r_{k}$ columns on which it can be found, $c_{k} r_{k}$ must be at least the total number of occurences of colour $k$, which equals 100 . So $r_{k}+c_{k} \geq 20$. In the sum $\sum_{i=1}^{100} R_{i}$, each colour $k$ contributes $r_{k}$ times and in the sum $\sum_{j=1}^{100} C_{j}$ each colour $k$ contributes $c_{k}$ times. Hence
$$
\sum_{i=1}^{100} R_{i}+\sum_{j=1}^{100} C_{j}=\sum_{k=1}^{100} r_{k}+\sum_{k=1}^{100} c_{k}=\sum_{k=1}^{100}\left(r_{k}+c_{k}\right) \geq 2000
$$
But if the sum of 200 positive integers is at least 2000, at least one of the summands is at least 10. The claim has been proved.
| proof | Combinatorics | proof | Yes | Yes | olympiads | false | 436 |
07.1. Find one solution in positive integers to the equation
$$
x^{2}-2 x-2007 y^{2}=0
$$
|
Solution. The equation can be written in the form
$$
x(x 2)=223 \cdot(3 y)^{2}
$$
Here the prime number 223 must divide $x$ or $x 2$. In fact, for $x=225$ we get $x(x 2)=$ $15^{2} \cdot 223$, which is equivalent to $223 \cdot(3 y)^{2}$ for $y=5$. Thus, $(x, y)=(225,5)$ is one solution.
| (225,5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 437 |
07.2. A triangle, a line and three rectangles, with one side parallel to the given line, are given in such a way that the rectangles completely cover the sides of the triangle. Prove that the rectangles must completely cover the interior of the triangle.
|
Solution. Take any point $P$ inside the triangle and draw through $P$ the line parallel to the given line as well as the line perpendicular to it. These lines meet the sides of the triangle in four points. Of these four, two must be in one of the three rectangles. Now if the two points are on the same line, then the whole segment between them, $P$ included, is in the same rectangle. If the two points, say $Q$ and $R$, are on perpendicular lines, the perpendicular segments $R P$ and $P Q$ are also in the same rectangle. So in any case, $P$ is in one of the rectangles.
| proof | Geometry | proof | Yes | Yes | olympiads | false | 438 |
07.3. The number $10^{2007}$ is written on a blackboard, Anne and Berit play a game where the player in turn makes one of two operations:
(i) replace a number $x$ on the blackboard by two integer numbers a and $b$ greater than 1 such that $x=a b$;
(ii) erase one or both of two equal numbers on the blackboard.
The player who is not able to make her turn loses the game. Who has a winning strategy?
|
Solution. We describe a winning strategy for Anne. Her first move is
$$
10^{2007} \rightarrow 2^{2007}, 5^{2007}
$$
We want to show that Anne can act in such a way that the numbers on the blackboard after each of her moves are of the form
$$
2^{\alpha_{1}}, \ldots, 2^{\alpha_{k}}, 5^{\alpha_{1}}, \ldots, 5^{\alpha_{k}}
$$
This is the case after Anne's first move. If Berit for example replaces $2^{\alpha_{j}}$ by $2^{\beta_{1}}$ and $2^{\beta_{2}}$, then Anne would replace $5^{\alpha_{j}}$ by $5^{\alpha_{1}}$ and $5^{\alpha_{2}}$. If Berit for example erases $5^{\alpha_{j}}$ or two $5^{\alpha_{j}}$ 's, (which means that there is an $\alpha_{i}=\alpha_{j}$ ) then Anne would erase $2^{\alpha_{j}}$ or $22^{\alpha_{j}}$ 's. Thus for each move Berit makes, Anne can answer with a 'symmetric' move. Since the game is finite, Berit must be the first player failing to make a move. Thus Anne has a winning strategy.
| proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 439 |
07.4. A line through a point $A$ intersects a circle in two points, $B$ and $C$, in such a way that $B$ lies between $A$ and $C$. From the point $A$ draw the two tangents to the circle, meeting the circle at points $S$ and $T$. Let $P$ be the intersection of the lines $S T$ and AC. Show that $A P / P C=2 \cdot A B / B C$.
|
Solution. First we show that if we fix the points $A, B$ and $C$ but vary the circle, then the point $P$ stays fixed. To that end, suppose we have two different circles through $B$ and $C$. Draw the tangents from $A$ to one circle, meeting the circle at points $S_{1}$ and $T_{1}$, and the tangents to the other circle, meeting that circle at points $S_{2}$ and $T_{2}$. Then, according to the power of a point theorem
$$
A S_{1}^{2}=A T_{1}^{2}=A B \cdot A C=A S_{2}^{2}=A T_{2}^{2}
$$
This implies that all the tangent points $S_{1}, T_{1}, S_{2}$ and $T_{2}$ lie on the same circle with center $A$. Let $Q$ be the intersection of $S_{1} T_{1}$ and $S_{2} T_{2}$. Then by applying again the theorem of a power of a point but now with respect to the circle with center A, we have that $Q S_{1} \cdot Q T_{1}=Q S_{2} \cdot Q T_{2}$. But this in turn means that the point $Q$ has the same power with respect to the two circles we started with, and hence lies on the radical axis of those two circles, that is, the line $B C$ (the radical axis is the locus of points of equal power with respect to two given circles). So $Q$ is the intersection of $A C$ and both $S_{1} T_{1}$ and $S_{2} T_{2}$, which proves that the intersection point defined in the problem is the same for both circles.
Since the location of $P$ is independent of the circle through $B$ and $C$ we can, without loss of generality, choose the circle with $B C$ as diameter. Let $O$ be the center of this circle, $R$ its radius, $d=A O$, and $r=P O$. Then the triangles $A S O$ and $S P O$ are similar, so $O S / A O=P O / O S$, that is, $R / d=r / R$, or $R^{2}=d r$. Then finally we have
$$
\frac{A P}{P C}=\frac{d-r}{R+r}=\frac{d^{2}-d r}{d R+d r}=\frac{d^{2}-{ }^{R} 2}{d r+r^{2}}=\frac{d-R}{R}=2 \cdot \frac{d-R}{2 R}=2 \cdot \frac{A B}{B C}
$$
| \frac{AP}{PC}=2\cdot\frac{AB}{BC} | Geometry | proof | Yes | Yes | olympiads | false | 440 |
08.1. Determine all real numbers $A, B$ and $C$ such that there exists a real function $f$ that satisfies
$$
f(x+f(y))=A x+B y+C
$$
for all real $x$ and $y$.
|
Solution. Let $A, B$ and $C$ be real numbers and $f$ a function such that $f(x+f(y))=$ $A x+B y+C$ for all $x$ and $y$. Let $z$ be a real number and set $x=z-f(0)$ and $y=0$. Then
$$
f(z)=f(z-f(0)+f(0))=A(z-f(0))+B \cdot 0+C=A z-A f(0)+C
$$
so there are numbers $a$ and $b$ such that $f(z)=a z+b$ for all $z$. Now $f(x+f(g))=$ $a x+a^{2} y+(a+1) b$, and $(A, B, C)=\left(a, a^{2},(a+1) b\right)$, where $a$ and $b$ are arbitrary real numbers, that is, $(A, B, C)=\left(a, a^{2}, c\right)$, where $a \neq-1$ and $c$ are arbitrary, or $(A, B, C)=(-1,1,0)$
| (A,B,C)=(,^2,)where\neq-1arbitrary,or(A,B,C)=(-1,1,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 441 |
08.2. Assume that $n \geq 3$ people with different names sit around a round table. We call any unordered pair of them, say $M$ and $N$, dominating, if
(i) $M$ and $N$ do not sit on adjacent seats, and
(ii) on one (or both) of the arcs connecting $M$ and $N$ along the table edge, all people have names that come alphabetically after the names of $M$ and $N$.
Determine the minimal number of dominating pairs.
|
Solution. We will show by induction that the number of dominating pairs (hence also the minimal number of dominating pairs) is $n-3$ for $n \geq 3$. If $n=3$, all pairs of people sit on adjacent seats, so there are no dominating pairs. Assume that the number of dominating pairs is $n-3$ for some $n>3$. If there are $n+1$ people around the table, let the person whose name is alphabetically last leave the table. The two people sitting next to that person, who formed a dominating pair, no longer do. On the other hand, any other dominating pair remains a dominating pair in the new configuration of $n$ people, and any dominating pair in the new configuration was also a dominating pair in the old. The number of dominating pairs in the new configuration is $n-3$, so the number in the old was $(n+1)-3$.
| n-3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 442 |
08.3. Let $A B C$ be a triangle and let $D$ and $E$ be points on $B C$ and $C A$, respectively, such that $A D$ and $B E$ are angle bisectors of $A B C$. Let $F$ and $G$ be points on the circumcircle of $A B C$ such that $A F$ and $D E$ are parallel and $F G$ and $B C$ are parallel. Show that
$$
\frac{A G}{B G}=\frac{A C+B C}{A B+C B}
$$
|
Solution. Let $A B=c, B C=a$ and $C A=b$. Then it follows from the angle bisector theorem that $C D=$ $a b /(b+c)$. Similarly, $C E=a b /(a+c)$, so $C D / C E=(a+c) /(b+c)$. The angles $\angle A B G, \angle A F G$ and $\angle E D C$ are equal, and so are $\angle A G B$ and $\angle A C B$, and consequently, the triangles $C E D$ and $G A B$ are similar. The conclusion follows.
| \frac{AG}{BG}=\frac{AC+BC}{AB+CB} | Geometry | proof | Yes | Yes | olympiads | false | 443 |
08.4. The difference between the cubes of two consecutive positive integers is a square $n^{2}$, where $n$ is a positive integer. Show that $n$ is the sum of two squares.
|
Solution. Assume that $(m+1)^{3}-m^{3}=n^{2}$. Rearranging, we get $3(2 m+1)^{2}=(2 n+$ $1)(2 n-1)$. Since $2 n+1$ and $2 n-1$ are relatively prime (if they had a common divisor, it would have divided the difference, which is 2 , but they are both odd), one of them is a square (of an odd integer, since it is odd) and the other divided by 3 is a square. An odd number squared minus 1 is divisible by 4 since $(2 t+1)^{2}-1=4\left(t^{2}+t\right)$. From the first equation we see that $n$ is odd, say $n=2 k+1$. Then $2 n+1=4 k+3$, so the square must be $2 n-1$, say $2 n-1=(2 t+1)^{2}$. Rearrangement yields $n=t^{2}+(t+1)^{2}$. (An example: $8^{3}-7^{3}=\left(2^{2}+3^{2}\right)^{2}$. $)$
| proof | Number Theory | proof | Yes | Yes | olympiads | false | 444 |
09.1. A point $P$ is chosen in an arbitrary triangle. Three lines are drawn through $P$ which are parallel to the sides of the triangle. The lines divide the triangle into three smaller
triangles and three parallelograms. Let $f$ be the ratio between the total area of the three smaller triangles and the area of the given triangle. Show that $f \geq \frac{1}{3}$ and determine those points $P$ for which $f=\frac{1}{3}$.
|
Solution. Let $A B C$ be the triangle and let the lines through $P$ parallel to its sides intersect the sides in the points $D, E ; F, G$ and $H, I$. The triangles $A B C$, $D E P, P F G$ and $I P H$ are similar and $B D=I P$, $E C=P F$. If $B C=a, I P=a_{1}, D E=a_{2}$ ja $P F=a_{3}$, then $a_{1}+a_{2}+a_{3}=a$. There is a posi-
tive $k$ such that the areas of the triangles are $k a^{2}, k a_{1}^{2}$, $k a_{2}^{2}$ and $k a_{3}^{2}$. But then
$$
f=\frac{k a_{1}^{2}+k a_{2}^{2}+k a_{3}^{2}}{k a^{2}}=\frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}{\left(a_{1}+a_{2}+a_{3}\right)^{2}}
$$
By the arithmetic-quadratic inequality,
$$
\frac{\left(a_{1}+a_{2}+a_{3}\right)^{2}}{9} \leq \frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}{3}
$$
where equality holds if and only if $a_{1}=a_{2}=a_{3}$. It is easy to see that $a_{1}=a_{2}=a_{3}$ implies that $P$ is the centroid of $A B C$. So $f \geq \frac{1}{3}$, and $f=\frac{1}{3}$ if and only if $P$ is the centroid of $A B C$.
| f\geq\frac{1}{3},f=\frac{1}{3}ifonlyifPisthecentroidofABC | Geometry | proof | Yes | Yes | olympiads | false | 445 |
09.2. On a faded piece of paper it is possible, with some effort, to discern the following:
$$
\left(x^{2}+x+a\right)\left(x^{15}-\ldots\right)=x^{17}+x^{13}+x^{5}-90 x^{4}+x-90
$$
Some parts have got lost, partly the constant term of the first factor of the left side, partly the main part of the other factor. It would be possible to restore the polynomial forming the other factor, but we restrict ourselves to asking the question: What is the value of the constant term a? We assume that all polynomials in the statement above have only integer coefficients.
|
Solution. We denote the polynomial $x^{2}+x+a$ by $P_{a}(x)$, the polynomial forming the other factor of the left side by $Q(x)$ and the polynomial on the right side by $R(x)$. The polynomials are integer valued for every integer $x$. For $x=0$ we get $P_{a}(0)=a$ and $R(0)=-90$, so $a$ is a divisor of $90=2 \cdot 3 \cdot 3 \cdot 5$. For $x=-1$ we get $P_{a}(-1)=-184$, so $a$ is also a divisor of $184=2 \cdot 2 \cdot 2 \cdot 23$. But the only prime factor in common is 2 . So the only possibilities for $a$ are $\pm 2$ and $\pm 1$. If $a=1$, we get for $x=1$ that $P_{1}(1)=3$, while $R(1)=4-180=-176$, which cannot be divided by 3 . If $a=-2$ we get for $x=1$ that $P_{2}(1)=0$, i.e. the left side is equal to 0 , while the right side is equal to $R(1)=-176$, which is different from 0 . Neither $a=1$ nor $a=-2$ will thus work. It remains to check $a=2$ and $a=-1$. Before we use the procedure above again, we need a factorization of $R(x)$. We observe that $x^{4}+1$ is a divisor of $R(x)$, since the right side may be written as $\left(x^{4}+1\right)\left(x^{13}+x-90\right)$. If $a=-1$ we get for $x=2$ that $P_{1}(2)=5$, while $x^{4}+1=17$ and $x^{13}+x-90=8104$. So the right hand side is not divisible by 5 . Now, the only remaining possibility is $a=2$, i.e. $x^{2}+x+2$ is a divisor of $R(x)$.
| 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 446 |
09.3. The integers 1, 2, 3, 4 and 5 are written on a blackboard. It is allowed to wipe out two integers $a$ and $b$ and replace them with $a+b$ and $a b$. Is it possible, by repeating this procedure, to reach a situation where three of the five integers on the blackboard are 2009?
|
Solution. The answer is no. First notice that in each move two integers will be replaced with two greater integers (except in the case where the number 1 is wiped out). Notice also that from the start there are three odd integers. If one chooses to replace two odd integers on the blackboard, the number of odd integers on the blackboard decreases. If one chooses to replace two integers, which are not both odd, the number of odd integers on the blackboard is unchanged. To end up in a situation, where three of the integers on the blackboard are 2009, then it is not allowed in any move to replace two odd integers. Hence the number 2009 can only be obtained as a sum $a+b$. In the first move that gives the integer 2009 on the blackboard, one has to choose $a$ and $b$ such that $a+b=2009$. In this case either $a b>2009$ or $a b=2008$. In the case $a b=2008$, one of the factors is equal to 1 , and hence 1 does no longer appear on the blackboard. The two integers $a+b=2009$ and $a b$ that appears in the creation of the first 2009 cannot be used any more in creation of the remaining two integers of 2009. Also the next 2009 can only be obtained if one chooses c and d such that $c+d=2009$ and $c d>2009$ or $c d=2008$, and in the last case 1 does not appear any longer on the blackboard. The numbers $c+d=2009$ and $c d$ cannot be used in obtaining the last integer 2009. Hence the last integer 2009 cannot be obtained.
| proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 447 |
09.4. There are 32 competitors in a tournament. No two of them are equal in playing strength, and in a one against one match the better one always wins. Show that the gold, silver, and bronze medal winners can be found in 39 matches.
|
Solution. To determine the gold medalist, we organize 16 pairs and matches, then 8 matches of the winners, 4 matches of the winners, 2 and finally one match, 31 matches altogether. Now the silver medal winner has at some point lost to number 1 ; as there were 5 rounds, there are 5 candidates. Let $C_{i}$ be the candidate who lost to the gold medalist in round $i$. Now let $C_{l}$ and $C_{2}$ play, the winner then play with $C_{3}$ etc. After 4 matches we know the silver medalist; assume she was $C_{k}$. Now the bronze medalist must have lost against the gold medalist or against $C_{k}$ or both. (If she lost to someone else, this someone else was below the second place). Now the silver medalist $C_{k}$ won $k-1$ times in the first rounds and the $5-k$ players $C_{k+1}, \ldots, C_{5}$, and if $\mathrm{k} i 1$ one player $C_{j}$ with $j<k$. So there are either $k-1+5-k=4$ or 5 candidates for the third place. At most 4 matches are again needed to determine the bronze winner.
| proof | Combinatorics | proof | Yes | Yes | olympiads | false | 448 |
10.1. A function $f: \mathbb{Z} \rightarrow \mathbb{Z}_{+}$, where $\mathbb{Z}_{+}$is the set of positive integers, is non-decreasing and satisfies $f(m n)=f(m) f(n)$ for all relatively prime positive integers $m$ and $n$. Prove that $f(8) f(13) \geq(f(10))^{2}$.
|
Solution. Since $\mathrm{f}$ is non-decreasing, $f(91) \geq f(90)$, which (by factorization into relatively prime factors) implies $f(13) f(7) \geq f(9) f(10)$. Also $f(72) \geq f(70)$, and therefore $f(8) f(9) \geq f(7) f(10)$. Since all values of $\mathrm{f}$ are positive, we get $f(8) f(9) \cdot f(13) f(7) \geq$ $f(7) f(10) \cdot f(9) f(10)$, and dividing both sides by $f(7) f(9)>0, f(8) f(13) \geq f(l 0) f(10)=$ $(f(10))^{2}$.
| proof | Number Theory | proof | Yes | Yes | olympiads | false | 449 |
10.2. Three circles $\Gamma_{A}, \Gamma_{B}$ and $\Gamma_{C}$ share a common point of intersection $O$. The other common of $\Gamma_{A}$ and $\Gamma_{B}$ is $C$, that of $\Gamma_{A}$ and $\Gamma_{C}$ is $B$ and that of $\Gamma_{C}$ and $\Gamma_{B}$ is $A$. The
line $A O$ intersects the circle $\Gamma_{C}$ in the poin $X \neq O$. Similarly, the line $B O$ intersects the circle $\Gamma_{B}$ in the point $Y \neq O$, and the line $C O$ intersects the circle $\Gamma_{C}$ in the point $Z \neq O$. Show that
$$
\frac{|A Y||B Z||C X|}{|A Z||B X||C Y|}=1
$$
|
Solution 1. Let $\angle A O Y=\alpha, \angle A O Z=\beta$ and $\angle Z O B=\gamma$. So $\alpha+\beta+\gamma=180^{\circ}$. Also $\angle B O X=\alpha$ (vertical angles) and $\angle A C Y=\alpha=\angle B C X$ (angles subtending equal arcs); similarly $\angle C O X=\beta$, $\angle A B Z=\beta=\angle C B X ; \angle C O Y=\gamma ; \angle B A Z=\gamma=$ $\angle C A Y$. Each of the triangles $C Y A, C B X$ and $Z B A$ have two angles from the set $\{\alpha, \beta, \gamma\}$. All triangles are then similar.
Similarity implies
$$
\frac{A Y}{C Y}=\frac{A B}{B Z}, \quad \frac{C X}{B X}=\frac{A Z}{A B}
$$
Consequently
$$
\frac{A Y}{A Z} \cdot \frac{B Z}{B X} \cdot \frac{C X}{C Y}=\frac{A B}{B Z} \cdot \frac{A Z}{A B} \cdot \frac{B Z}{A Z}=1
$$
| proof | Geometry | proof | Yes | Yes | olympiads | false | 450 |
10.3. Laura has 2010 lamps connected with 2010 buttons in front of her. For each button, she wants to know the corresponding lamp. In order to do this, she observes which lamps are lit when Richard presses a selection of buttons. (Not pressing anything is also a possible selection.) Richard always presses the buttons simultaneously, so the lamps are lit simultaneously, too.
a) If Richard chooses the buttons to be pressed, what is the maximum number of different combinations of buttons he can press until Laura can assign the buttons to the lamps correctly?
b) Supposing that Laura will choose the combinations of buttons to be pressed, what is the minimum number of attempts she has to do until she is able to associate the buttons with the lamps in a correct way?
|
Solution. a) Let us say that two lamps are separated, if one of the lamps is turned on while the other lamp remains off. Laura can find out which lamps belong to the buttons if every two lamps are separated. Let Richard choose two arbitrary lamps. To begin with, he turns both lamps on and then varies all the other lamps in all possible ways. There are $2^{2008}$ different combinations for the remaining $2010-2=2008$ lamps. Then Richard turns
the two chosen lamps off. Also, at this time there are $2^{2008}$ combinations for the remaining lamps. Consequently, for the $2^{2009}$ combinations in all, it is not possible to separate the two lamps of the first pair. However, we cannot avoid the separation if we add one more combination. Indeed, for every pair of lamps, we see that if we turn on a combination of lamps $2^{2009}+1$ times, there must be at least one setup where exactly one of the lamps is turned on and the other is turned off. Thus, the answer is $2^{2009}+1$.
b) For every new step with a combination of lamps turned on, we get a partition of the set of lamps into smaller and smaller subsets where elements belonging to the same subset cannot be separated. In each step every subset is either unchanged or divided into two smaller parts, i.e. the total number of subsets after $\mathrm{k}$ steps will be at most $2^{k}$. We are finished when the number of subsets is equal to 2010 , so the answer is at least $\left\lceil\log _{2} 2010\right\rceil=11$. But it is easy to see that Laura certainly can choose buttons in every step in such a way that there are at most $2^{11-k}$ lamps in every part of the partition after $k$ steps. Thus, the answer is 11 .
| 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 451 |
10.4. A positive integer is called simple if its ordinary decimal representation consists entirely of zeroes and ones. Find the least positive integer $k$ such that each positive integer $n$ can be written as $n=a_{1} \pm a_{2} \pm a_{3} \pm \cdots \pm a_{k}$, where $a_{1}, \ldots, a_{k}$ are simple.
|
Solution. We can always write $n=a_{l}+a_{2}+\cdots+a_{9}$ where $a_{j}$ has 1 's in the places where $n$ has digits greater or equal to $j$ and 0 's in the other places. So $k \leq 9$. To show that $k \geq 9$, consider $n=10203040506070809$. Suppose $n=a_{l}+a_{2}+\cdots+a_{j}-a_{j+l}-a_{j+2}-\cdots-a_{k}$, where $a_{l}, \ldots, a_{k}$ are simple, and $k<9$. Then all digits of $b_{l}=a_{l}+\cdots+a_{j}$ are not greater than $j$ and all digits of $b_{2}=a_{j+l}+\cdots+a_{k}$ are not greater than $k-j$. We have $n+b_{2}=b_{l}$. We perform column addition of $n$ and $b_{2}$ and consider digit $j+1$ in the number $n$. There will be no carry digit coming from lower decimal places, since the sum there is less that $10 \ldots 0+88 \ldots 8=98 \ldots 8$. So in the column of $j+1$ we get the sum of $j+1$ and the corresponding digit in $b_{2}$. The resulting digit should be less than $j+1$. Thus in the corresponding place in $b_{2}$ we have at least $9-j$. But $9-j \leq k-j$, implying $k \geq 9$. Hence, we have proved that the maximal $k$ is 9 .
| 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 452 |
11.1. When $a_{0}, a_{1}, \ldots, a_{1000}$ denote digits, can the sum of the 1001-digit numbers $a_{0} a_{1} \ldots a_{1000}$ and $a_{1000} a_{999} \ldots a_{0}$ have odd digits only?
|
Solution. The answer is no. The following diagram illustrates the calculation of the sum digit by digit.
| $a_{0}$ | $a_{1}$ | $\ldots$ | $a_{i}$ | $\ldots$ | $a_{500}$ | $\ldots$ | $a_{1000-i}$ | $\ldots$ | $a_{999}$ | $a_{1000}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $a_{1000}$ | $a_{999}$ | $\ldots$ | $a_{1000-i}$ | $\ldots$ | $a_{500}$ | $\ldots$ | $a_{i}$ | $\ldots$ | $a_{1}$ | $a_{0}$ |
| $s_{1000}$ | $s_{999}$ | $\ldots$ | $s_{1000-i}$ | $\ldots$ | $s_{500}$ | $\ldots$ | $s_{i}$ | $\ldots$ | $s_{1}$ | $s_{0}$ |
Thus $s_{i}$ are the digits of the sum. The digit $s_{\text {lool }}$ may be absent. We call column $i$ the column in the diagram with the digit $s_{i}$. Assume that $s_{i}$ is odd for $i=0,1, \ldots, 1000$. By induction on $i$ we prove that $a_{2 i}+a_{\text {looo }-2 i}$ is odd for $i=0,1, \ldots, 250$. This implies that $a_{2.250}+a_{l 000-2.250}=2 a_{500}$ is odd, which is a contradiction. Here is the proof: Since $s_{0}$ is odd, $a_{0}+a_{1000}$ is odd, so the statement is true for $\mathrm{i}=0$. Assume that $a_{2 i}+a_{1000-2 i}$ is odd for some $i \in\{0,1, \ldots, 224\}$. Since $s_{1000-2 i}$ is odd and $a_{2 i}+a_{1000-2 i}$ is odd, there is no
carry in column $1000-2 i$, so $a_{2 i+1}+a_{1000-(2 i+1)} \leq 9$. But then because $s_{2 i+1}$ is odd and $a_{2 i+1}+a_{1000-(2 i+1)} \leq 9$, there is no carry in column $2 i+2$. Hence $a_{2 i+2}+a_{1000-(2 i+2)}$ is odd because $s_{2 i+2}$ is odd. This completes the induction step.
| proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 453 |
11.2. In a triangle $A B C$ assume $A B=A C$, and let $D$ and $E$ be points on the extension of segment $B A$ beyond $A$ and on the segment $B C$, respectively, such that the lines $C D$ and $A E$ are parallel. Prove that $C D \geq \frac{4 h}{B C} C E$, where $h$ is the height from $A$ in triangle ABC. When does equality hold?
|
Solution. Because $A E \| D C$, the triangles $A B E$ and $D B C$ are similar. So
$$
C D=\frac{B C}{B E} \cdot A E
$$
$\mathrm{ja}$
$$
C D=\frac{A E \cdot B C}{B E \cdot C E} \cdot C E
$$
Let $A F$ be an altitude of $A B C$. Then $A E \geq A F=h$, and equality holds if and only if $E=F$. Because $A B C$ is isosceles, $F$ is the midpoint of $B C$. The arithmetic-geometric mean inequality yields
$$
B E \cdot C E \leq\left(\frac{B E+E C}{2}\right)^{2}=\left(\frac{B C}{2}\right)^{2}
$$
and equality holds if and only if $E$ is the midpoint of $B C$ i.e. $E=F$. The conclusion folows when these estimates are inserted in (1); furthermore, equality is equivalent to $E=F$
| proof | Geometry | proof | Yes | Yes | olympiads | false | 454 |
11.3. Find all functions $f$ such that
$$
f(f(x)+y)=f\left(x^{2}-y\right)+4 y f(x)
$$
for all real numbers $x$ and $y$.
|
Solution. Substituting $y=x^{2}$ yields $f\left(f(x)+x^{2}\right)=f(0)+4 x^{2} f(x)$ for all real $x$. And $y=-f(x)$ gives $f(0)=f\left(x^{2}+f(x)\right)-4 f(x)^{2}$ for all $x$. Combining these two equations gives $4 f(x)^{2}=4 x^{4} f(x)$, so for each $x$ either $f(x)=0$ or $f(x)=x^{2}$. In particular $f(0)=0$. Now suppose there exists a real number $a \neq 0$ such that $f(a) \neq 0$. Then $f(a)=a^{2}$ and $f\left(a^{2}+y\right)=f\left(a^{2}-y\right)+4 y a^{2}$ for all $y$. If $f\left(a^{2}-y\right)=0$ for some $y \neq 0$, then $f\left(a^{2}+y\right)=4 y a^{2} \neq 0$, so $f\left(a^{2}+y\right)=\left(a^{2}+y\right)^{2}$ must hold and then $\left(a^{2}+y\right)^{2}=4 y a^{2}$, which yields $\left(a^{2}-y\right)^{2}=0$, or $a^{2}-y=0$. This shows that if $f(a) \neq 0$ for some $a$ then $f(x)=0$ only when $x=0$. So either $f(x)=0$ for all $x$ or $f(x)=x^{2}$ for all $x$. It can easily be verified that these two functions are indeed solutions to the given equation.
| proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 455 |
## Problem 1
Determine all sequences of non-negative integers $a_{1}, \ldots, a_{2016}$ all less than or equal to 2016 satisfying $i+j \mid i a_{i}+j a_{j}$ for all $i, j \in\{1,2, \ldots, 2016\}$.
|
Solution Answer: All constant sequences of non-negative integers.
The condition rewrites to $i+j \mid i\left(a_{i}-a_{j}\right)$. Since $2 k-1$ and $k$ are coprime, we see that $2 k-1 \mid a_{k}-a_{k-1}$. Thus if $2 k-1>2016$, then $a_{k}=a_{k-1}$ since $a_{k}$ and $a_{k-1}$ are non-negative and at most 2016. All together $a_{1009}=a_{1010}=\cdots=a_{2016}$.
If $i2016$ we conclude as before that $a_{i}=a_{j}=a_{2016}$. So any such sequence is constant.
| proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 457 |
## Problem 2
Let $A B C D$ be a cyclic quadrilateral satisfying $A B=A D$ and $A B+B C=C D$.
Determine $\angle C D A$.
|
Solution 2 Answer: $\angle C D A=60^{\circ}$.
Choose the point $E$ on the segment $C D$ such that $D E=A D$. Then $C E=C D-A D=$ $C D-A B=B C$, and hence the triangle $C E B$ is isosceles.
Now, since $A B=A D$ then $\angle B C A=\angle A C D$. This shows that $C A$ is the bisector of $\angle B C D=\angle B C E$. In an isosceles triangle, the bisector of the apex angle is also the perpendicular bisector of the base. Hence $A$ is on the perpendicular bisector of $B E$, and $A E=A B=A D=D E$. This shows that triangle $A E D$ is equilateral, and thus $\angle C D A=60^{\circ}$.
| 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 458 |
## Problem 3
Find all $a \in \mathbb{R}$ for which there exists a function $f: \mathbb{R} \rightarrow \mathbb{R}$, such that
(i) $f(f(x))=f(x)+x$, for all $x \in \mathbb{R}$,
(ii) $f(f(x)-x)=f(x)+$ ax, for all $x \in \mathbb{R}$.
|
Solution 3 Answer: $a=\frac{1 \pm \sqrt{5}}{2}$.
From (i) we get $f(f(f(x))-f(x))=f(x)$. On the other hand (ii) gives
$$
f(f(f(x))-f(x))=f(f(x))+a f(x)
$$
Thus we have $(1-a) f(x)=f(f(x))$. Now it follows by (i) that $(1-a) f(x)=f(x)+x$, and hence $f(x)=-\frac{1}{a} x$, since $a=0$ obviously does not give a solution.
We now need to check whether (i) and (ii) hold for this function for some values of $a$ and all real $x$. We have
$$
f(f(x))=-\frac{1}{a} f(x)=\frac{1}{a^{2}} x, \text { and } f(x)+x=-\frac{1}{a} x+x=\frac{a-1}{a} x
$$
Thus (i) will hold for all real $x$ iff $\frac{1}{a^{2}}=\frac{a-1}{a}$, i.e. iff $a=\frac{1 \pm \sqrt{5}}{2}$. For these values of $a$ we have
$$
f(f(x)-x)=-\frac{1}{a}(f(x)-x)=-\frac{1}{a}\left(-\frac{1}{a} x-x\right)=\left(\frac{1}{a^{2}}+\frac{1}{a}\right) x=\frac{a+1}{a^{2}} x=x
$$
and
$$
f(x)+a x=-\frac{1}{a} x+a x=\frac{a^{2}-1}{a} x=x
$$
so that for these two values of $a$ both (i) and (ii) hold for all real $x$. Thus the values of $a$ such that there exists a function $f$ with the desired properties are $a=\frac{1 \pm \sqrt{5}}{2}$.
| \frac{1\\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 459 |
## Problem 4
King George has decided to connect the 1680 islands in his kingdom by bridges. Unfortunately the rebel movement will destroy two bridges after all the bridges have been built, but not two bridges from the same island.
What is the minimal number of bridges the King has to build in order to make sure that it is still possible to travel by bridges between any two of the 1680 islands after the rebel movement has destroyed two bridges?
|
Solution 4 Answer: 2016
An island cannot be connected with just one bridge, since this bridge could be destroyed. Consider the case of two islands, each with only two bridges, connected by a bridge. (It is not possible that they are connected with two bridges, since then they would be isolated from the other islands no matter what.) If they are also connected to two separate islands, then they would be isolated if the rebel movement destroys the two bridges from these islands not connecting the two. So the two bridges not connecting them must go to the same island. That third island must have at least two other bridges, otherwise the rebel movement could cut off these three islands.
Suppose there is a pair of islands with exactly two bridges that are connected to each other. From the above it is easy to see that removing the pair (and the three bridges connected to them) must leave a set of islands with the same properties. Continue removing such pairs, until there are none left. (Note that the reduced set of islands could have a new such pair and that also needs to be removed.) Suppose we are left with $n$ islands and since two islands are removed at a time, $n$ must be an even number. And from the argument above it is clear that $n \geq 4$.
Consider the remaining set of islands and let $x$ be the number of islands with exactly two bridges (which now are not connected to each other). Then $n-x$ islands have at least three bridges each. Let $B^{\prime}$ be the number of bridges in the reduced set. Now $B^{\prime} \geq 2 x$ and $2 B^{\prime} \geq 2 x+3(n-x)=3 n-x$. Hence $2 B^{\prime} \geq \max (4 x, 3 n-x) \geq 4 \cdot \frac{3 n}{5}$, and thus $B^{\prime} \geq \frac{6 n}{5}$. Now let $B$ be the number of bridges in the original set. Then
$$
B=B^{\prime}+3 \cdot \frac{1680-n}{2} \geq \frac{6 n}{5}+\frac{6(1680-n)}{4} \geq \frac{6 \cdot 1680}{5}=2016
$$
It is possible to construct an example with exactly 2016 bridges: Take 672 of the islands and number them $0,1,2, \ldots 671$. Connect island number $i$ with the islands numbered $i-1$, $i+1$ and $i+336$ (modulo 672). This gives 1008 bridges. We now have a circular path of 672 bridges: $0-1-2-\cdots-671-0$. If one of these 672 bridges are destroyed, the 672 islands are still connected. If two of these bridges are destroyed, the path is broken into two parts. Let $i$ be an island on the shortest path (if they have the same length, just pick a random one). Then island $i+336$ (modulo 672) must be on the other part of the path, and the bridge connecting these two islands will connect the two paths. Hence no matter which two bridges the rebel movement destroys, it is possible to travel between any of the 672 islands.
Now for every of the 1008 bridges above, replace it with two bridges with a new island between the two. This increases the number of bridges to 2016 and the number of islands to $672+1008=1680$ completing the construction. Since the rebel movement does not destroy two bridges from the same island, the same argument as above shows that with this construction it is possible to travel between any of the 1680 islands after the destruction of the two bridges.
| 2016 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 460 |
## Problem 1
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(f(x) f(1-x))=f(x) \quad \text { and } \quad f(f(x))=1-f(x)
$$
for all real $x$.
|
Solution 1. Notice that $f(f(f(x)))=^{2} 1-f(f(x))={ }^{2} f(x)$. This is equation 3 . By substituting $f(x)$ for $x$ in the first equation we get:
$$
f(\underline{f(x)})={ }^{1} f(f(\underline{f(x)}) f(1-\underline{f(x)}))={ }^{2} f(f(f(x)) f(f(f(x))))={ }^{3} f(f(f(x)) f(x))
$$
Again we substitute $f(x)$ for $x$ above:
$$
f(f(\underline{f(x)}))=f(f(f(\underline{f(x)})) f(\underline{f(x)}))
$$
Equation 3 applied on both sides gives us:
$$
f(x)=f(f(x) f(f(x)))
$$
But this is the same as what we started with so
$$
f(x)=f(f(x) f(f(x)))=f(f(x))={ }^{2} 1-f(x)
$$
Therefore $f(x)=\frac{1}{2}$, which is a solution.
| f(x)=\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 461 |
## Problem 3
Anton and Britta play a game with the set $M=\{1,2,3, \ldots, n-1\}$ where $n \geq 5$ is an odd integer. In each step Anton removes a number from $M$ and puts it in his set $A$, and Britta removes a number from $M$ and puts it in her set $B$ (both $A$ and $B$ are empty to begin with). When $M$ is empty, Anton picks two distinct numbers $x_{1}, x_{2}$ from $A$ and shows them to Britta. Britta then picks two distinct numbers $y_{1}, y_{2}$ from $B$. Britta wins if
$$
\left(x_{1} x_{2}\left(x_{1}-y_{1}\right)\left(x_{2}-y_{2}\right)\right)^{\frac{n-1}{2}} \equiv 1 \quad \bmod n
$$
otherwise Anton wins. Find all $n$ for which Britta has a winning strategy.
|
Solution. Britta wins if and only if $n$ is prime.
If $n$ is not prime, then Anton can add any prime divisor $p<n$ of $n$ to his set $A$ in the first round and choose $x_{1}=p$ which means that the product $\left(x_{1} x_{2}\left(x_{1}-y_{1}\right)\left(x_{2}-y_{2}\right)\right)^{\frac{n-1}{2}}$ is divisible by $p$ and is not $1 \bmod n$ no matter what Britta chooses. And so Britta loses.
If $n$ is prime, then $x_{1} x_{2} \not \equiv 0 \bmod n$, and there exists a number $\alpha$ such that $x^{2} \equiv \alpha$ $\bmod n$ has no solution. Then Britta can always add the number $Y \in M, Y \equiv \alpha X^{-1}$ to $B$, if Anton adds the number $X$ to $A$ in each round. Notice that Anton can never have chosen the number $Y$ beforehand, since $Y \equiv \alpha X^{-1} \Longleftrightarrow X \equiv \alpha Y^{-1}$ and $X \neq Y$ (as $X^{2} \equiv \alpha \bmod n$ is not possible). This means that Britta can always choose the
numbers $y_{1}=\alpha x_{2}^{-1}, y_{2}=\alpha x_{1}^{-1}$ from $B$. This will result in
$$
\begin{aligned}
\left(x_{1} x_{2}\left(x_{1}-y_{1}\right)\left(x_{2}-y_{2}\right)\right)^{\frac{n-1}{2}} & \equiv\left(\left(x_{1} x_{2}-x_{2} y_{1}\right)\left(x_{1} x_{2}-x_{1} y_{2}\right)\right)^{\frac{n-1}{2}} \\
& \equiv\left(\left(x_{1} x_{2}-\alpha\right)\left(x_{1} x_{2}-\alpha\right)\right)^{\frac{n-1}{2}} \\
& \equiv\left(x_{1} x_{2}-\alpha\right)^{n-1} \\
& \equiv 1 \quad \bmod n
\end{aligned}
$$
The last equation is true by Fermat's little theorem, because $n$ is prime and $x_{1} x_{2}-\alpha \not \equiv$ $0 \bmod n\left(\right.$ since $x_{1} \not \equiv y_{1}=\alpha x_{2}^{-1}$ ).
Alternative strategy for Britta is to choose $n-a$ when Anton pick up $a$, which is always possible because $n$ is an odd number and if one of the numbers $a, n-a$ was already chosen before the same is true for the second one. At the end Britta chooses $y_{i}=-x_{i}$ and gets
$$
\left(x_{1} x_{2}\left(x_{1}-y_{1}\right)\left(x_{2}-y_{2}\right)\right)^{\frac{n-1}{2}}=\left(\left(2 x_{1}\right)^{2}\left(2 x_{2}\right)^{2}\right)^{\frac{n-1}{2}} \equiv 1 \quad \bmod n
$$
| proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 463 |
## Problem 4
Let $A B C$ be an acute-angled triangle with circumscribed circle $k$ and centre of the circumscribed circle $O$. A line through $O$ intersects the sides $A B$ and $A C$ at $D$ and $E$. Denote by $B^{\prime}$ and $C^{\prime}$ the reflections of $B$ and $C$ over $O$, respectively. Prove that the circumscribed circles of $O D C^{\prime}$ and $O E B^{\prime}$ concur on $k$.
|
Solution. Let $P$ be the intersection of the circles $k$ and the circumscribed circle of triangle $A D E^{1}$. Let $C_{1}$ be the second intersection of the circumscribed circle of $\triangle D O P$ with $k$. We will prove that $C_{1}=C^{\prime}$, i.e. the reflection of $C$ over $O$. We know that $\left|O C_{1}\right|=|O P|$, and hence $\measuredangle C_{1} P O=\measuredangle O C_{1} P$, furthermore $\measuredangle O C_{1} P=\measuredangle O D P=$ $\measuredangle E D P$, since the quadrilateral $C_{1} P O D$ by assumption is inscribed and the points $O, D$ and $E$ are collinear. Now, since $P$ is the centre of spiral similarity sending $D E$ to $B C$ the triangles $P D E$ and $P B C$ are similar, and we have $\measuredangle E D P=\measuredangle C B P$, and finally, from the inscribed angle theorem we have
$$
\measuredangle O P C=90^{\circ}-\frac{\measuredangle C O P}{2}=90^{\circ}-\measuredangle C B P=90^{\circ}-\measuredangle C_{1} P O
$$
The conclusion follows, since $90^{\circ}=\measuredangle C_{1} P O+\measuredangle O P C$, and since $C_{1}$ is by assumption on $k$, it must be the antipodal point of $C$ with respect to $k$.
[^0]
[^0]: ${ }^{1}$ That is, the Miquel point of quadrilateral BCED.
| proof | Geometry | proof | Yes | Yes | olympiads | false | 464 |
Problem 1 Let $n$ be a positive integer. Show that there exist positive integers $a$ and $b$ such that:
$$
\frac{a^{2}+a+1}{b^{2}+b+1}=n^{2}+n+1
$$
|
Solution 1 Let $P(x)=x^{2}+x+1$. We have $P(n) P(n+1)=\left(n^{2}+n+1\right)\left(n^{2}+3 n+3\right)=$ $n^{4}+4 n^{3}+7 n^{2}+6 n+3$. Also, $P\left((n+1)^{2}\right)=n^{4}+4 n^{3}+7 n^{2}+6 n+3$. By choosing $a=(n+1)^{2}$ and $b=n+1$ we get $P(a) / P(b)=P(n)$ as desired.
| proof | Number Theory | proof | Yes | Yes | olympiads | false | 465 |
Problem 2 Let $a, b, \alpha, \beta$ be real numbers such that $0 \leq a, b \leq 1$, and $0 \leq \alpha, \beta \leq \frac{\pi}{2}$. Show that if
$$
a b \cos (\alpha-\beta) \leq \sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}
$$
then
$$
a \cos \alpha+b \sin \beta \leq 1+a b \sin (\beta-\alpha)
$$
|
Solution 2 The condition can be rewritten as
$$
a b \cos (\alpha-\beta)=a b \cos \alpha \cos \beta+a b \sin \alpha \sin \beta \leq \sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}
$$
Set $x=a \cos \alpha, y=b \sin \beta, z=b \cos \beta, t=a \sin \alpha$. We can now rewrite the condition as
$$
x z+y t \leq \sqrt{\left(1-x^{2}-t^{2}\right)\left(1-y^{2}-z^{2}\right)}
$$
whereas the inequality we need to prove now looks like
$$
x+y \leq 1+x y-z t
$$
Since $x, y, z, t \geq 0$, and $1+x y-z t=1+a b \sin (\beta-\alpha) \geq 0$, we can square both sides of both inequalities, and get equivalent ones. After a couple of cancelations the condition yields
$$
2 x y z t \leq 1-x^{2}-y^{2}-z^{2}-t^{2}+x^{2} y^{2}+z^{2} t^{2}
$$
so that
$$
x^{2}+y^{2}+z^{2}+t^{2} \leq(x y-z t)^{2}+1
$$
which is equivalent to
$$
x^{2}+y^{2}+z^{2}+t^{2}+2 x y-2 z t \leq(1+x y-z t)^{2}
$$
or
$$
(x+y)^{2}+(z-t)^{2} \leq(1+x y-z t)^{2}
$$
Since $(x+y)^{2} \leq(x+y)^{2}+(z-t)^{2}$, the desired inequality follows.
| proof | Inequalities | proof | Yes | Yes | olympiads | false | 466 |
Problem 3 Let $M$ and $N$ be the midpoints of the sides $A C$ and $A B$, respectively, of an acute triangle $A B C, A B \neq A C$. Let $\omega_{B}$ be the circle centered at $M$ passing through $B$, and let $\omega_{C}$ be the circle centered at $N$ passing through $C$. Let the point $D$ be such that $A B C D$ is an isosceles trapezoid with $A D$ parallel to $B C$. Assume that $\omega_{B}$ and $\omega_{C}$ intersect in two distinct points $P$ and $Q$. Show that $D$ lies on the line $P Q$.
|
Solution 3 Let $E$ be such that $A B E C$ is a parallelogram with $A B \| C E$ and $A C \| B E$, and let $\omega$ be the circumscribed circle of $\triangle A B C$ with centre $O$.
It is known that the radical axis of two circles is perpendicular to the line connecting the two centres. Since $B E \perp M O$ and $C E \perp N O$, this means that $B E$ and $C E$ are the radical axes of $\omega$ and $\omega_{B}$, and of $\omega$ and $\omega_{C}$, respectively, so $E$ is the radical centre of $\omega$, $\omega_{B}$, and $\omega_{C}$.
Now as $B E=A C=B D$ and $C E=A B=C D$ we find that $B C$ is the perpendicular bisector of $D E$. Most importantly we have $D E \perp B C$. Denote by $t$ the radical axis of $\omega_{B}$ and $\omega_{C}$, i.e. $t=P Q$. Then since $t \perp M N$ we find that $t$ and $D E$ are parallel. Therefore since $E$ lies on $t$ we get that $D$ also lies on $t$.
Alternative solution Reflect $B$ across $M$ to a point $B^{\prime}$ forming a parallelogram $A B C B^{\prime}$. Then $B^{\prime}$ lies on $\omega_{B}$ diagonally opposite $B$, and since $A B^{\prime} \| B C$ it lies on $A D$. Similarly reflect $C$ across $N$ to a point $C^{\prime}$, which satisfies analogous properties. Note that $C B^{\prime}=A B=C D$, so we find that triangle $C D B^{\prime}$ and similarly triangle $B D C^{\prime}$ are isosceles.
Let $B^{\prime \prime}$ and $C^{\prime \prime}$ be the orthogonal projections of $B$ and $C$ onto $A D$. Since $B B^{\prime}$ is a diameter of $\omega_{B}$ we get that $B^{\prime \prime}$ lies on $\omega_{B}$, and similarly $C^{\prime \prime}$ lies on $\omega_{C}$. Moreover $B B^{\prime \prime}$ is an altitude of the isosceles triangle $B D C^{\prime}$ with $B D=B C^{\prime}$, hence it coincides with the median from $B$, so $B^{\prime \prime}$ is in fact the midpoint of $D C^{\prime}$. Similarly $C^{\prime \prime}$ is the midpoint of $D B^{\prime}$. From this we get
$$
2=\frac{D C^{\prime}}{D B^{\prime \prime}}=\frac{D B^{\prime}}{D C^{\prime \prime}}
$$
which rearranges as $D C^{\prime} \cdot D C^{\prime \prime}=D B^{\prime} \cdot D B^{\prime \prime}$. This means that $D$ has same the power with respect to $\omega_{B}$ and $\omega_{C}$, hence it lies on their radical axis $P Q$.
| proof | Geometry | proof | Yes | Yes | olympiads | false | 467 |
PROBLEM 1. The real numbers $a, b, c$ are such that $a^{2}+b^{2}=2 c^{2}$, and also such that $a \neq b, c \neq-a, c \neq-b$. Show that
$$
\frac{(a+b+2 c)\left(2 a^{2}-b^{2}-c^{2}\right)}{(a-b)(a+c)(b+c)}
$$
is an integer.
|
SolUTiON. Let us first note that
$$
\frac{a+b+2 c}{(a+c)(b+c)}=\frac{(a+c)+(b+c)}{(a+c)(b+c)}=\frac{1}{a+c}+\frac{1}{b+c}
$$
Further we have
$$
2 a^{2}-b^{2}-c^{2}=2 a^{2}-\left(2 c^{2}-a^{2}\right)-c^{2}=3 a^{2}-3 c^{2}=3(a+c)(a-c)
$$
and
$$
2 a^{2}-b^{2}-c^{2}=2\left(2 c^{2}-b^{2}\right)-b^{2}-c^{2}=3 c^{2}-3 b^{2}=3(b+c)(c-b)
$$
so that
$$
\frac{(a+b+2 c)\left(2 a^{2}-b^{2}-c^{2}\right)}{(a-b)(a+c)(b+c)}=\frac{3(a-c)+3(c-b)}{a-b}=\frac{3(a-b)}{a-b}=3
$$
an integer.
| 3 | Algebra | proof | Yes | Yes | olympiads | false | 470 |
Problem 2. Given a triangle $A B C$, let $P$ lie on the circumcircle of the triangle and be the midpoint of the arc $B C$ which does not contain $A$. Draw a straight line $l$ through $P$ so that $l$ is parallel to $A B$. Denote by $k$ the circle which passes through $B$, and is tangent to $l$ at the point $P$. Let $Q$ be the second point of intersection of $k$ and the line $A B$ (if there is no second point of intersection, choose $Q=B)$. Prove that $A Q=A C$.
|
Solution I. There are three possibilities: $Q$ between $A$ and $B, Q=B$, and $B$ between $A$ and $Q$. If $Q=B$ we have that $\angle A B P$ is right, and $A P$ is a diameter
of the circumcircle. The triangles $A B P$ and $A C P$ are then congruent (they have $A P$ in common, $P B=P C$, and both have a right angle opposite to $A P$ ). Hence ir follows that $A B=A C$.
The solutions in the other two cases are very similar. We present the one in the case when $Q$ lies between $A$ and $B$.
The segment $A P$ is the angle bisector of the angle at $A$, since $P$ is the midpoint of the arc $B C$ of the circumcircle which does not contain $A$. Also, $P C=P B$. Since the segment $Q B$ is parallel to the tangent to $k$ at $P$, it is orthogonal to the diameter of $k$ through $P$. Thus this diameter cuts $Q B$ in halves, to form two congruent right triangles, and it follows that $P Q=P B$. We have (in the usual notation) $\angle P C B=\angle P B C=\frac{\alpha}{2}$, and
$$
\angle A Q P=180^{\circ}-\angle B Q P=180^{\circ}-\angle Q B P=180^{\circ}-\beta-\frac{\alpha}{2}=\frac{\alpha}{2}+\gamma=\angle A C P
$$
Hence the triangles $A Q P$ and $A C P$ are congruent (two pairs of equal angles and one pair of equal corresponding sides), and it follows that $A C=A Q$.
| proof | Geometry | proof | Yes | Yes | olympiads | false | 471 |
Problem 3. Find the smallest positive integer $n$, such that there exist $n$ integers $x_{1}, x_{2}, \ldots, x_{n}$ (not necessarily different), with $1 \leq x_{k} \leq n, 1 \leq k \leq n$, and such that
$$
x_{1}+x_{2}+\cdots+x_{n}=\frac{n(n+1)}{2}, \quad \text { and } \quad x_{1} x_{2} \cdots x_{n}=n!
$$
but $\left\{x_{1}, x_{2}, \ldots, x_{n}\right\} \neq\{1,2, \ldots, n\}$.
|
Solution. If it is possible to find a set of numbers as required for some $n=k$, then it will also be possible for $n=k+1$ (choose $x_{1}, \ldots, x_{k}$ as for $n=k$, and
let $x_{k+1}=k+1$ ). Thus we have to find a positive integer $n$ such that a set as required exists, and prove that such a set does not exist for $n-1$.
For $n=9$ we have $8+6+3=9+4+4$, and $8 \cdot 6 \cdot 3=9 \cdot 4 \cdot 4$, so that a set of numbers as required will exist for all $n \geq 9$. It remains to eliminate $n=8$.
Assume $x_{1}, \ldots, x_{8}$ are numbers that satisfy the conditions of the problem. Since 5 and 7 are primes, and since $2 \cdot 5>8$ and $2 \cdot 7>8$, two of the $x$-numbers have to be equal to 5 and 7 ; without loss of generality we can assume that $x_{1}=5, x_{2}=7$. For the remaining numbers we have $x_{3} x_{4} \cdots x_{8}=2^{7} \cdot 3^{2}$, and $x_{3}+x_{4}+\cdots+x_{8}=36-12=24$. Since $3^{2}=9>8$, it follows that exactly two of the numbers $x_{3}, \ldots, x_{8}$ are divisible by 3 , and the rest of the numbers are powers of 2. There are three possible cases to consider: two of the numbers are equal to 3 ; two of the numbers are equal to 6 ; one number is equal to 3 and another one is equal to 6 .
Case 1. $x_{3}=x_{4}=3$
We then have $x_{5}+x_{6}+x_{7}+x_{8}=18$, and $x_{5} x_{6} x_{7} x_{8}=2^{7}$. The possible powers of 2 with sum 18 are $(1,1,8,8)$ and $(2,4,4,8)$, none of them gives the product $2^{7}$.
Case 2. $x_{3}=3, x_{4}=6$
We have $x_{5}+x_{6}+x_{7}+x_{8}=15$, and $x_{5} x_{6} x_{7} x_{8}=2^{6}$. It is immediate to check that the only possibility for the remaining numbers is $(1,2,4,8)$, which is not allowed, since it gives $\left\{x_{1}, x_{2}, \ldots, x_{8}\right\}=\{1,2, \ldots, 8\}$.
Case 3. $x_{3}=x_{4}=6$
Now we have $x_{5}+x_{6}+x_{7}+x_{8}=12$, and $x_{5} x_{6} x_{7} x_{8}=2^{5}$. The possible powers of 2 which give the correct sum are $(1,1,2,8)$ and $(2,2,4,4)$, but again, they do not give the desired product.
Thus the smallest positive integer with the required property is 9 .
| 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 472 |
Problem 4. The number 1 is written on the blackboard. After that a sequence of numbers is created as follows: at each step each number $a$ on the blackboard is replaced by the numbers $a-1$ and $a+1$; if the number 0 occurs, it is erased immediately; if a number occurs more than once, all its occurrences are left on the blackboard. Thus the blackboard will show 1 after 0 steps; 2 after 1 step; 1,3 after 2 steps; 2, 2, 4 after 3 steps, and so on. How many numbers will there be on the blackboard after $n$ steps?
|
Solution I. Let $S$ be a set of different numbers, all of them less than $2^{n-1}$, and create two new sets as follows: $S_{1}$, consisting of all the numbers in $S$ except
the smallest one, and $S_{2}$, with elements the smallest element of $S$ and all the numbers we get by adding $2^{n-1}$ to each number in $S$. Note that if the number of elements in $S$ is $a$, then $S_{1}$ has $a-1$ elements, and $S_{2}$ has $a+1$ elements. This corresponds to the operations we are allowed to perform on the blackboard, if we throw away all empty sets. If we now operate simultaneously on the sets and on the numbers, then after $n$ steps the number of sets will be exactly equal to the number of numbers on the blackboard.
Let us see what the set operations look like. We must start with a set, consisting only of the number 0 . Next we get an empty set (thrown away), and the set $\{0,1\}$; next the sets $\{1\}$ and $\{0,2,3\}$; next again (an empty set and) $\{1,5\},\{2,3\},\{0,4,6,7\}$, etc.
It is now fairly easy to prove by induction that after $n$ steps
(1) each number less than $2^{n}$ appears in exactly one set;
(2) the number of elements in the sets corresponds exactly to the numbers on the blackboard;
(3) if the numbers in each set are written in increasing order, then the difference between two neighbours is a power of 2 ; thus the binary representations of two neighbours differ in exactly one position (in the binary system the example above looks like this: $\{0\} ;\{0,1\} ;\{01\},\{00,10,11\} ;\{001,101\},\{010,011\}$, $\{000,100,110,111\}) ;$
(4) if $k$ is the number of ones in the binary code of the smallest number of a set, and $l$ the number of ones in the largest number of the same set, then $k+l=n$; (5) each set contains exactly one number with $\left\lfloor\frac{n}{2}\right\rfloor$ ones.
The last property tells us that the number of sets after $n$ steps is equal to the number of numbers such that their binary representation contains exactly $\left\lfloor\frac{n}{2}\right\rfloor$ ones out of $n$ digits, i.e. the number of numbers on the blackboard after $n$ steps will be equal to $\binom{n}{\left\lfloor\frac{n}{2}\right\rfloor}$.
| \binom{n}{\lfloor\frac{n}{2}\rfloor} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 473 |
## Problem 1.
Let $A B C$ be a triangle and $\Gamma$ the circle with diameter $A B$. The bisectors of $\angle B A C$ and $\angle A B C$ intersect $\Gamma$ (also) at $D$ and $E$, respectively. The incircle of $A B C$ meets $B C$ and $A C$ at $F$ and $G$, respectively. Prove that $D, E, F$ and $G$ are collinear.
|
Solution 1. Let the line $E D$ meet $A C$ at $G^{\prime}$ and $B C$ at $F^{\prime} . A D$ and $B E$ intersect at $I$, the incenter of $A B C$. As angles subtending the same arc $\widehat{B D}$, $\angle D A B=\angle D E B=\angle G^{\prime} E I$. But $\angle D A B=\angle C A D=$ $\angle G^{\prime} A I$. This means that $E, A, I$ and $G^{\prime}$ are concyclic, and $\angle A E I=\angle A G^{\prime} I$ as angles subtending the same chord $A I$. But $A B$ is a diameter of $\Gamma$, and so $\angle A E B=$ $\angle A E I$ is a right angle. So $I G^{\prime} \perp A C$, or $G^{\prime}$ is the foot of the perpendicular from $I$ to $A C$. This implies $G^{\prime}=G$. In a similar manner we prove that $F^{\prime}=F$, and the proof is complete.
| proof | Geometry | proof | Yes | Yes | olympiads | false | 474 |
## Problem 2.
Find the primes $p, q, r$, given that one of the numbers $p q r$ and $p+q+r$ is 101 times the other.
|
Solution. We may assume $r=\max \{p, q, r\}$. Then $p+q+r \leq 3 r$ and $p q r \geq 4 r$. So the sum of the three primes is always less than their product. The only relevant requirement thus is $p q r=101(p+q+r)$. We observe that 101 is a prime. So one of $p, q, r$ must be 101. Assume $r=101$. Then $p q=p+q+101$. This can be written as $(p-1)(q-1)=102$. Since $102=1 \cdot 102=2 \cdot 51=3 \cdot 34=6 \cdot 17$, the possibilities for $\{p, q\}$ are $\{2,103\},\{3,52\},\{4,35\},\{7,18\}$ The only case, where both the numbers are primes, is $\{2,103\}$. So the only solution to the problem is $\{p, q, r\}=\{2,101,103\}$.
| {2,101,103} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 475 |
## Problem 3.
Let $n>1$ and $p(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}$ be a polynomial with $n$ real roots (counted with multiplicity). Let the polynomial $q$ be defined by
$$
q(x)=\prod_{j=1}^{2015} p(x+j)
$$
We know that $p(2015)=2015$. Prove that $q$ has at least 1970 different roots $r_{1}, \ldots, r_{1970}$ such that $\left|r_{j}\right|<2015$ for all $j=1, \ldots, 1970$.
|
Solution. Let $h_{j}(x)=p(x+j)$. Consider $h_{2015}$. Like $p$, it has $n$ real roots $s_{1}, s_{2}, \ldots, s_{n}$, and $h_{2015}(0)=p(2015)=2015$. By Viète, the product $\left|s_{1} s_{2} \cdots s_{n}\right|$ equals 2015. Since $n \geq 2$, there is at least one $s_{j}$ such that $\left|s_{j}\right| \leq \sqrt{2015}<\sqrt{2025}=45$. Denote this $s_{j}$ by $m$. Now for all $j=0,1, \ldots, 2014, h_{2015-j}(m+j)=p(m+j+2015-j)=p(m+2015)=$ $h_{2015}(m)=0$. So $m, m+1, \ldots, m+2014$ are all roots of $q$. Since $0 \leq|m|<45$, the condition $|m+j|<2015$ is satisfied by at least 1970 different $j, 0 \leq j \leq 2014$, and we are done.
| proof | Algebra | proof | Yes | Yes | olympiads | false | 476 |
## Problem 4.
An encyclopedia consists of 2000 numbered volumes. The volumes are stacked in order with number 1 on top and 2000 in the bottom. One may perform two operations with the stack:
(i) For $n$ even, one may take the top $n$ volumes and put them in the bottom of the stack without changing the order.
(ii) For $n$ odd, one may take the top $n$ volumes, turn the order around and put them on top of the stack again.
How many different permutations of the volumes can be obtained by using these two operations repeatedly?
|
Solution 1. (By the proposer.) Let the positions of the books in the stack be $1,2,3, \ldots, 2000$ from the top (and consider them modulo 2000). Notice that both operations fix the parity of the number of the book at a any given position. Operation (i) subtracts an even integer from the number of the book at each position. If $A$ is an operation of type (i), and $B$ is an operation of type (ii), then the operation $A^{-1} B A$ changes the order of the books in the positions $n+1$ to $n+m$ where $n$ is even and $m$ is odd. This is called turning the interval.
Now we prove that all the volumes in odd positions can be placed in the odd positions in every way we like: If the volume we want in position 1 is in position $m_{1}$, we turn the interval 1 to $m_{1}$. Now if the volume we want in position 3 is in position $m_{3}$, we turn the interval 3 to $m_{3}$, and so on. In this way we can permute the volumes in odd positions exactly as we want to.
Now we prove that we can permute the volumes in even positions exactly as we want without changing the positions of the volumes in the odd positions: We can make a transposition of the two volumes in position $2 n$ and $2 n+2 m$ by turning the interval $2 n+1$ to $2 n+2 m-1$, then turning the interval $2 n+2 m+1$ to $2 n-1$, then turning the interval $2 n+1$ to $2 n-1$, and finally adding $2 m$ to the number of the volume in each position.
Since there are 1000! permutations of the volumes in the odd positions, and 1000! permutations of the volumes in the even positions, altogether we have (1000!) ${ }^{2}$ different permutations.
| (1000!)^2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 477 |
Problem 2. In a football tournament there are $n$ teams, with $n \geq 4$, and each pair of teams meets exactly once. Suppose that, at the end of the tournament, the final scores form an arithmetic sequence where each team scores 1 more point than the following team on the scoreboard. Determine the maximum possible score of the lowest scoring team, assuming usual scoring for football games (where the winner of a game gets 3 points, the loser 0 points, and if there is a tie both teams get 1 point).
|
Solution. Note that the total number of games equals the number of different pairings, that is, $n(n-1) / 2$. Suppose the lowest scoring team ends with $k$ points. Then the total score for all teams is
$$
k+(k+1)+\cdots+(k+n-1)=n k+\frac{(n-1) n}{2}
$$
Some games must end in a tie, for otherwise, all team scores would be a multiple of 3 and cannot be 1 point apart. Since the total score of a tie is only 2 points compared to 3 points if one of the teams wins, we therefore know that
$$
n k+\frac{(n-1) n}{2}<3 \cdot \frac{n(n-1)}{2}
$$
so $n k<n(n-1)$, and hence $k<n-1$. It follows that the lowest scoring team can score no more than $n-2$ points.
We now show by induction that it is indeed possible for the lowest scoring team to score $n-2$ points.
The following scoreboard shows this is possible for $n=4$ :
| - | 3 | 1 | 1 | 5 |
| :---: | :---: | :---: | :---: | :---: |
| 0 | - | 1 | 3 | 4 |
| 1 | 1 | - | 1 | 3 |
| 1 | 0 | 1 | - | 2 |
Now suppose we have a scoreboard for $n$ teams labelled $T_{n-2}, \ldots, T_{2 n-3}$, where team $T_{i}$ scores $i$ points. Keep the results among these teams unchanged while adding one more team.
Write $n=3 q+r$ with $r \in\{1,-1,0\}$, and let the new team tie with just one of the original teams, lose against $q$ teams, and win against the rest of them. The new team thus wins $n-1-q$ games, and gets $1+3(n-1-q)=3 n-2-3 q=2 n-2+r$ points.
Moreover, we arrange for the $q$ teams which win against the new team to form an arithmetic sequence $T_{j}, T_{j+3}, \ldots, T_{j+3(q-1)}=T_{j+n-r-3}$, so that each of them, itself having gained three points, fills the slot vacated by the next one.
(i) If $r=1$, then let the new team tie with team $T_{n-2}$ and lose to each of the teams $T_{n-1}, T_{n+2}, \ldots, T_{n-1+n-r-3}=T_{2 n-5}$.
Team $T_{n-2}$ now has $n-1$ points and takes the place vacated by $T_{n-1}$. At the other end, $T_{2 n-5}$ now has $2 n-2$ points, just one more than the previous top team $T_{2 n-3}$. And the new team has $2 n-2+r=2 n-1$ points, becoming the new top team. The teams now have all scores from $n-1$ up to $2 n-1$.
(ii) If $r=-1$, then let the new team tie with team $T_{2 n-3}$ and lose to each of the teams $T_{n-2}, T_{n+1}, \ldots, T_{n-2+n-r-3}=T_{2 n-4}$.
The old top team $T_{2 n-3}$ now has $2 n-2$ points, and its former place is filled by the new team, which gets $2 n-2+r=2 n-3$ points. $T_{2 n-4}$ now has $2 n-1$ points and is the new top team. So again we have all scores ranging from $n-1$ up to $2 n-1$.
(iii) If $r=0$, then let the new team tie with team $T_{n-2}$ and lose to teams $T_{n-1}, T_{n+2}, \ldots, T_{n-1+n-r-3}=T_{2 n-4}$.
Team $T_{n-2}$ now has $n-1$ points and fills the slot vacated by $T_{n-1}$. At the top end, $T_{2 n-4}$ now has $2 n-1$ points, while the new team has $2 n-2+r=2 n-2$ points, and yet again we have all scores from $n-1$ to $2 n-1$.
This concludes the proof.
| n-2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 478 |
Problem 3. Define a sequence $\left(n_{k}\right)_{k \geq 0}$ by $n_{0}=n_{1}=1$, and $n_{2 k}=n_{k}+n_{k-1}$ and $n_{2 k+1}=n_{k}$ for $k \geq 1$. Let further $q_{k}=n_{k} / n_{k-1}$ for each $k \geq 1$. Show that every positive rational number is present exactly once in the sequence $\left(q_{k}\right)_{k \geq 1}$.
|
Solution. Clearly, all the numbers $n_{k}$ are positive integers. Moreover,
$$
q_{2 k}=\frac{n_{2 k}}{n_{2 k-1}}=\frac{n_{k}+n_{k-1}}{n_{k-1}}=q_{k}+1
$$
and similarly,
$$
\frac{1}{q_{2 k+1}}=\frac{n_{2 k}}{n_{2 k+1}}=\frac{n_{k}+n_{k-1}}{n_{k}}=\frac{1}{q_{k}}+1
$$
In particular, $q_{k}>1$ when $k$ is even, and $q_{k}s$. We know that $r / s=q_{m}$ is only possible if $m$ is even. But
$$
\frac{r}{s}=q_{2 k} \Leftrightarrow \frac{r-s}{s}=q_{k}
$$
by (1), and moreover, the latter equality holds for precisely one $k$ according to the induction hypothesis, since $\operatorname{gcd}(r-s, s)=1$ and $(r-s)+s=r \leq t$.
Next, assume that $r<s$. We know that $r / s=q_{m}$ is only possible if $m$ is odd. But
$$
\frac{r}{s}=q_{2 k+1} \Leftrightarrow \frac{s}{r}=\frac{1}{q_{2 k+1}} \Leftrightarrow \frac{s-r}{r}=\frac{1}{q_{k}}
$$
by (2), and moreover, the latter equality holds for precisely one $k$ according to the induction hypothesis, since $\operatorname{gcd}(s-r, r)=1$ and $(s-r)+r=s \leq t$.
| proof | Number Theory | proof | Yes | Yes | olympiads | false | 479 |
Problem 4. Let $A B C$ be an acute angled triangle, and $H$ a point in its interior. Let the reflections of $H$ through the sides $A B$ and $A C$ be called $H_{c}$ and $H_{b}$, respectively, and let the reflections of $H$ through the midpoints of these same sides be called $H_{c}^{\prime}$ and $H_{b}^{\prime}$, respectively. Show that the four points $H_{b}, H_{b}^{\prime}, H_{c}$, and $H_{c}^{\prime}$ are concyclic if and only if at least two of them coincide or $H$ lies on the altitude from $A$ in triangle $A B C$.
|
Solution. If at least two of the four points $H_{b}, H_{b}^{\prime}, H_{c}$, and $H_{c}^{\prime}$ coincide, all four are obviously concyclic. Therefore we may assume that these four points are distinct.
Let $P_{b}$ denote the midpoint of segment $H H_{b}, P_{b}^{\prime}$ the midpoint of segment $H H_{b}^{\prime}, P_{c}$ the midpoint of segment $H H_{c}$, and $P_{c}^{\prime}$ the midpoint of segment $H H_{c}^{\prime}$.
The triangle $H H_{b} H_{b}^{\prime}$ being right-angled in $H_{b}$, it follows that the perpendicular bisector $\ell_{b}$ of the side $H_{b} H_{b}^{\prime}$ goes through the point $P_{b}^{\prime}$. Since the segments $P_{b} P_{b}^{\prime}$ and $H_{b} H_{b}^{\prime}$ are parallel and $P_{b}^{\prime}$ is the midpoint of the side $A C$, we then conclude that $\ell_{b}$ also goes through the circumcentre $O$ of triangle $A B C$.
Similarly the perpendicular bisector $\ell_{c}$ of the segment $H_{c} H_{c}^{\prime}$ also goes through $O$. Hence the four points $H_{b}, H_{b}^{\prime}, H_{c}$, and $H_{c}^{\prime}$ are concyclic if and only if also the perpendicular bisector $\ell$ of the segment $H_{b}^{\prime} H_{c}^{\prime}$ goes through the point $O$. Since $H_{b}^{\prime} H_{c}^{\prime}\left\|P_{b}^{\prime} P_{c}^{\prime}\right\| B C$, this is the case if and only if $\ell$ is the perpendicular bisector $m$ of the segment $B C$.
Let $k$ denote the perpendicular bisector of the segment $P_{b}^{\prime} P_{c}^{\prime}$. Since the lines $\ell$ and $m$ are obtained from $k$ by similarities of ratio 2 and centres $H$ and $A$, respectively, they coincide if and only if $H A$ is parallel to $m$. Thus $H_{b}, H_{b}^{\prime}, H_{c}$, and $H_{c}^{\prime}$ are concyclic if and only if $H$ lies on the altitude from $A$ in triangle $A B C$.
Click here to experiment with the figure in GeoGebra.
| proof | Geometry | proof | Yes | Yes | olympiads | false | 480 |
LIV OM - II - Task 3
Given is the polynomial $ W(x) = x^4 - 3x^3 + 5x^2 - 9x $. Determine all pairs of different integers $ a $, $ b $ satisfying the equation | Notice that $ W(x) = (x - 1)(x - 2)(x^2 + 3) - 6 $. For $ n > 3 $, the following inequalities therefore hold
and moreover
From this, we obtain
Thus, the values of the polynomial $ W $ at the points $ -2, \pm 3, \pm 4, \ldots $ are pairwise distinct. We directly calculate that
and from inequality (1), we conclude that the values of the polynomial $ W $ at the remaining integer points are greater than $ W(3) = 18 $. Therefore, we obtain a total of four pairs of numbers satisfying the conditions of the problem:
(Note: The specific inequalities and calculations are not provided in the original text, so they are left as blank spaces in the translation.) | (-1,0),(0,-1),(1,2),(2,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 481 |
LII OM - I - Problem 10
Prove that inside any triangle $ABC$ there exists a point $P$ with the following property:
Every line passing through point $P$ divides the perimeter of triangle $ABC$ in the same ratio as it divides its area. | We will show that the property described in the problem statement is possessed by the center of the inscribed circle.
om52_1r_img_17.jpg
Let $ \ell $ be any line passing through point $ P $, which is the center of the circle inscribed in triangle $ ABC $ (Fig. 1). Without loss of generality, assume that line $ \ell $ intersects sides $ AC $ and $ BC $ at points $ D $ and $ E $, respectively. We need to show that
where $ [\mathcal{F}] $ is the area of figure $ \mathcal{F} $. Denoting by $ r $ the radius of the circle inscribed in triangle $ ABC $, we get
and
Dividing the sides of equations (2) and (3), we obtain equation (1). | proof | Geometry | proof | Yes | Yes | olympiads | false | 483 |
XXXVII OM - III - Problem 1
A square with a side length of 1 is covered by $ m^2 $ rectangles. Prove that the perimeter of one of these rectangles is greater than or equal to $ 4/m $. | Pole $ P = ab $ i obwód $ p = 2(a+b) $ prostokąta o bokach długości $ a $, $ b $ związane są nierównością
The area $ P = ab $ and the perimeter $ p = 2(a+b) $ of a rectangle with side lengths $ a $, $ b $ are related by the inequality
Gdyby więc każdy z rozważanych $ m^2 $ prostokątów miał obwód mniejszy od
$ 4/m $, to pole każdego z nich byłoby mniejsze od $ \frac{1}{16} \cdot \left( \frac{4}{m} \right)^2 $ (czyli od $ 1/m^2 $), a więc suma ich pól byłaby mniejsza od $ 1 $ - wbrew założeniu, że pokrywają one kwadrat jednostkowy.
If, therefore, each of the $ m^2 $ rectangles considered had a perimeter less than
$ 4/m $, then the area of each of them would be less than $ \frac{1}{16} \cdot \left( \frac{4}{m} \right)^2 $ (i.e., less than $ 1/m^2 $), and thus the sum of their areas would be less than $ 1 $ - contrary to the assumption that they cover the unit square. | proof | Geometry | proof | Yes | Yes | olympiads | false | 484 |
XI OM - II - Task 1
Prove that if real numbers $ a $ and $ b $ are not both equal to zero, then for every natural $ n $ | When one of the numbers $ a $ and $ b $ is equal to zero or when both are of the same sign, inequality (1) is obvious, since in that case no term on the left side $ L $ of the inequality is negative. It remains to prove the case when $ a $ and $ b $ are of different signs. Due to the symmetry of $ L $ with respect to $ a $ and $ b $, it suffices to consider the case $ a > 0 $, $ b < 0 $.
If $ b = aq $, then $ q < 0 $ and
therefore, according to the known formula for the sum of a geometric progression
Since $ q < 0 $ and $ q^{2n+1} < 0 $, then $ 1 - q > 0 $ and $ 1 - q^{2n+1} > 0 $, and since $ a^{2n} > 0 $, indeed $ L > 0 $. | proof | Algebra | proof | Yes | Yes | olympiads | false | 487 |
XXIV OM - I - Problem 8
Find a polynomial with integer coefficients of the lowest possible degree, for which $ \sqrt{2} + \sqrt{3} $ is a root. | We will find a polynomial with integer coefficients, of which the root is the number $ a = \sqrt{2} + \sqrt{3} $. It is, of course, a root of the polynomial
This polynomial, however, does not have integer coefficients. Let's multiply the polynomial (1) by $ x - \sqrt{2} + \sqrt{3} $. We get $ (x - \sqrt{2})^2 - 3 $, that is,
Of course, the number $ \alpha $ is a root of the polynomial (2). However, not all of its coefficients are integers. Let's then multiply the polynomial (2) by $ x^2 + 2\sqrt{2}x - 1 $. We get,
The number $ \alpha $ is a root of the polynomial (3), which has integer coefficients.
We will prove that (3) is the polynomial of the lowest degree among such polynomials with integer coefficients, of which the root is the number $ \alpha $.
First, we will show that the polynomial (3) is not the product of two polynomials of positive degree with rational coefficients. If the polynomial (3) were divisible by a polynomial of the first degree with rational coefficients, then it would have a root in the set of rational numbers. Any such root is an integer that divides the constant term of the polynomial (3). It is easy to check that neither $ 1 $ nor $ - 1 $ is a root of this polynomial.
Next, we will prove that the polynomial (3) is not divisible by any quadratic polynomial with rational coefficients. If,
where $ a, b, c, d \in \mathbb{Q} $, then by comparing the coefficients of the same powers of $ x $ on both sides of equation (7), we would get the system of equations,
Solving for $ c $ from the first equation and for $ d $ from the last equation, and substituting into the remaining equations, we get,
Hence, $ a = 0 $, or $ b = \pm 1 $.
If $ a = 0 $, then $ b + \frac{1}{b} = - 10 $, that is, $ b^2 - 10 b + 1 = 0 $. This last equation has no roots in the set of rational numbers.
If $ b = 1 $, then $ a^2 = 12 $, and if $ b = - 1 $, then $ a^2 = 8 $. In neither of these cases is $ a $ a rational number.
If the number $ \sqrt{2} + \sqrt{3} $ were a root of some non-zero polynomial $ f $ of degree at most three with rational coefficients and this degree were minimal, then denoting by $ g $ and $ r $ the quotient and remainder of the division of the polynomial (3) by $ f $, we would get,
where $ \mathrm{deg} \, r < \mathrm{deg} \, f $. Here, $ r $ is not the zero polynomial, since, as we have shown, the polynomial (3) is not the product of two polynomials of positive degree with rational coefficients.
Substituting $ x = \sqrt{2} + \sqrt{3} $ in the last equation, we get $ r( \sqrt{2} + \sqrt{3}) = 0 $. However, the polynomial $ f $ had the smallest degree among non-zero polynomials with rational coefficients, of which the root is the number $ \sqrt{2} + \sqrt{3} $. Since $ \mathrm{deg} \, r < \mathrm{deg} \, f $, we have reached a contradiction. | x^4-10x^2+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 489 |
XLVII OM - I - Problem 2
A palindromic number is defined as a natural number whose decimal representation read from left to right is the same as when read from right to left. Let $ (x_n) $ be the increasing sequence of all palindromic numbers. Determine all prime numbers that are divisors of at least one of the differences $ x_{n+1}-x_n $. | The decimal representation of the palindromic number $ x_n $ (respectively: ($ 2m $)-digit or ($ 2m-1 $)-digit) looks like this:
or
($ c_i \in \{ 0,1,2,3,4,5,6,7,8,9\} $ for $ i = 1, \ldots ,m;\ c_1 \ne 0 $). There are three possible situations:
Case I. $ c_m \ne 9 $. If $ x_n $ has the form (1), then $ x_{n+1} $ (the smallest palindromic number greater than $ x_n $) has the decimal representation
where $ d_m = c_m + 1 $.
Justification: Example (3) shows that there are palindromic numbers greater than $ x_n $ and having the same initial digits $ c_1,c_2,\ldots,c_{m-1} $ as $ x_n $; thus, $ x_{n+1} $ must be one of these numbers. Its next ($ m $-th) digit cannot be $ c_m $, because then the number would be identical to $ x_n $ (formula (1)). By taking $ d_m= c_m +1 $ as the next digit, we get the only palindromic number with the initial digits $ c_1,c_2,\ldots,c_{m-1},d_m $ (its further digits are already uniquely determined; see formula (3)); and by writing any digit greater than $ d_m $ on the $ m $-th place, we get even larger numbers; therefore, formula (3) indeed represents the number $ x_{n+1} $.
Analogous reasoning shows that if $ x_n $ has the form (2), then $ x_{n+1} $ has the decimal representation
where $ d_m = c_m + 1 $.
We calculate the difference $ x_{n+1} -x_n $ using the usual subtraction algorithm (that is, by writing the digits of the number $ x_n $ under the corresponding digits of the number $ x_{n+1} $ and performing "written subtraction"). The result of the subtraction:
Case II. Not all digits of the number $ x_n $ (given by formula (1) or (2)) are nines, but the digit $ c_m $ is a nine: $ c_m = 9 $. Let $ c_k $ be the last digit different from $ 9 $ in the sequence $ c_1,c_2,\ldots,c_{m-1},c_m $. Then
where $ l = 2m-2k $ or $ l = 2m-1-2k $, depending on whether the number $ x_n $ is expressed by formula (1) or (2). In each of these cases, the smallest palindromic number greater than $ x_n $ has the representation
where $ d_k = c_k + 1 $. (Detailed justification - very similar to the one conducted in Case I.) The subtraction algorithm now gives the result
Case III. All digits of the number $ x_n $ are nines:
Then $ x_{n+1} = \overline{1 \underbrace{00\ldots 0}_{p-1}1} $, so
Cases I, II, III exhaust all possibilities. The results obtained in formulas (5), (6), (7) show that the only prime divisors of the differences $ x_{n+1}-x_n $ are the numbers $ 2 $, $ 5 $, and $ 11 $. | 2,5,11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 490 |
VI OM - I - Problem 9
Present the polynomial $ x^4 + x^3 + x^2 + x + 1 $ as the difference of squares of two polynomials of different degrees with real coefficients. | If the polynomial $ W (x) = x^4 + x^3 + x^2 + x + 1 $ is equal to the difference $ U(x)^2 - V(x)^2 $, where $ U(x) $ and $ V(x) $ are polynomials of different degrees, then the polynomial $ U(x) $ must be of the second degree and the polynomial $ V(x) $ must be of the first or zero degree. In this case, the polynomial $ V(x)^2 $ does not contain terms of degree higher than two, so the first two terms of the polynomial $ U(x)^2 $ must be the same as in the given polynomial $ W(x) $, i.e., $ U(x)^2 $ has the form $ x^4 + x^3 + \ldots $. From this, we conclude that the polynomial $ U(x) $ has the form $ U(x) = x^2 + \frac{1}{2}x + a $. Therefore,
From the expression obtained for $ V(x)^2 $, we see that $ V(x) $ cannot be of zero degree, as it cannot be simultaneously $ 2a - \frac{3}{4} = 0 $ and $ a - 1 = 0 $; $ V(x) $ must therefore be of the first degree. It follows further that the square of the function $ V(x) $ must be a quadratic trinomial, in which the coefficient of $ x^2 $ is positive, and the discriminant equals $ 0 $, thus
The second condition gives, after simplification, the equation
This equation has three roots $ a = 1 $ and $ a = \frac{1}{4} $ ($ -1 \pm \sqrt{5} $), of which only the root $ a = 1 $ satisfies the condition $ 2a - \frac{3}{4} > 0 $. Ultimately, we obtain the factorization
as the only solution to the problem. | x^4+x^3+x^2+x+1=(x^2+\frac{1}{2}x+1)^2-(\frac{1}{2}x)^2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 491 |
XXXVII OM - I - Problem 12
Prove that if the line connecting the midpoints of two opposite edges of a tetrahedron passes through the center of the inscribed sphere of this tetrahedron, then it also passes through the center of the circumscribed sphere of this tetrahedron. | We will first prove a lemma.
Lemma. Given a convex dihedral angle formed by half-planes $\alpha$ and $\beta$ with a common edge $l$. Let $\pi$ be the bisecting half-plane of this dihedral angle. Suppose that point $P \in \pi$ is the midpoint of segment $AB$ with endpoints $A \in \alpha$, $B \in \beta$ (where $A$, $B$, $P \in l$). Let $O$ denote the orthogonal projection of point $P$ onto line $l$. Then the lines $OP$ and $AB$ are perpendicular.
Proof of the lemma. Lines $l$ and $AB$ are skew. Therefore, there exists exactly one plane $\sigma$ containing line $AB$ and parallel to $l$. This plane intersects the half-planes $\alpha$, $\beta$, and $\pi$ along lines $\alpha \cap \sigma$, $\beta \cap \sigma$, $\pi \cap \sigma$ parallel to $l$ (Figure 5). Lines $\pi \cap \sigma$ and $OP$ are thus perpendicular.
om37_1r_img_5.jpg
Consider triangle $OST$ located in a plane perpendicular to $l$, with vertices $S \in \alpha \cap \sigma$, $T \in \beta \cap \sigma$. Right triangles $ASP$ and $BTP$ have equal angles ($|\measuredangle SPA| = |\measuredangle TPB|$ as vertical angles) and equal hypotenuses ($|PA| = |PB|$) - they are therefore congruent and thus $|PS| = |PT|$. Therefore, segment $OP$ is the median in triangle $OST$. It is also the angle bisector of $\angle SOT$ (since the half-plane $\pi$ bisects the dihedral angle between $\alpha$ and $\beta$). It follows that this segment is also the altitude in triangle $OST$, so $OP \bot ST$. Line $OP$, being perpendicular to lines $ST$ and $\pi \cap \sigma$ lying in plane $\sigma$, is perpendicular to this plane - and thus to line $AB$.
Proof of the theorem. Let $P$ and $Q$ be the midpoints of edges $AB$ and $CD$ of tetrahedron $ABCD$ (Figure 6) and let, according to the problem's assumption, line $PQ$ pass through the center of the inscribed sphere of the tetrahedron. Then the half-plane $CDP^\rightarrow$ bisects the dihedral angle formed by faces $CDA$ and $CDB$. Denoting by $O$ the orthogonal projection of point $P$ onto line $l = CD$, we have by the lemma that $OP \bot AB$. Similarly, the half-plane $ABQ^\rightarrow$ bisects the dihedral angle between faces $ABC$ and $ABD$, and by the lemma $QR \bot CD$, where $R$ is the projection of point $Q$ onto line $AB$.
om37_1r_img_6.jpg
Thus, each of the lines $OP$ and $QR$ is perpendicular to both lines $AB$ and $CD$. But for any pair of skew lines, there is only one line that intersects both and is perpendicular to them. Consequently, lines $OP$ and $QR$ coincide. This means that line $PQ$ is perpendicular to lines $AB$ and $CD$. Therefore, planes $CDP$ and $ABQ$ are the perpendicular bisector planes of segments $AB$ and $CD$ (respectively). The center of the circumscribed sphere must lie in both of these planes, and thus on their line of intersection, which is line $PQ$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 492 |
I OM - B - Task 2
Prove that if the number $ a $ is the sum of the squares of two different natural numbers, then $ a^2 $ is the sum of the squares of two natural numbers. | If $ a=x^2+y^2 $, then
保留了源文本的换行和格式。请注意，最后一行的“保留了源文本的换行和格式”是中文，翻译成英文应为：“The line breaks and formatting of the source text have been preserved.” 但为了保持格式一致，这里没有将其翻译出来。 | proof | Number Theory | proof | Yes | Yes | olympiads | false | 494 |
XXVIII - I - Problem 5
Prove that if $ P(x, y) $ is a polynomial in two variables such that $ P(x, y) = P(y, x) $ for all real $ x, y $ and the polynomial $ (x-y) $ divides $ P(x, y) $, then the polynomial $ (x - y)^2 $ also divides $ P(x, y) $. | By assumption, there exists a polynomial $ Q(x, y) $ such that $ P(x,y) = (x - y) \cdot Q(x, y) $. From the equality $ P(x, y) = P(y, x) $, we obtain that $ (x - y) \cdot Q(x, y) = (y - x) \cdot Q(y, x) $, which means $ (x - y) \cdot (Q(x, y) + Q(y, x)) = 0 $. Since the product of polynomials is a zero polynomial only if one of the factors is a zero polynomial, and the polynomial $ x - y $ is non-zero, it follows from the last equality that $ Q(x, y) + Q(y, x) = 0 $. By substituting $ x = y $ here, we particularly obtain that $ Q(y, y) = 0 $, i.e., $ Q(y, y) $ is a zero polynomial.
Treating the polynomials $ Q(x, y) $ and $ x - y $ as polynomials in the variable $ x $ with coefficients being polynomials in the variable $ y $, we divide $ Q(x, y) $ by $ x - y $. We obtain some quotient $ U(x, y) $ and a remainder $ R(y) $ independent of $ x $. From the division algorithm, it follows that $ U(x, y) $ and $ R(y) $ are polynomials in the variables $ x $, $ y $, and $ y $ respectively. Thus, we have $ Q(x, y) = (x - y) \cdot U(x, y) + R(y) $. Substituting $ x=y $ here, we get $ Q(y, y) = R(y) $. On the other hand, we have proven that $ Q(y, y) = 0 $. Therefore, $ R(y) = 0 $, which means $ Q(x, y) = (x - y) \cdot U(x, y) $ and hence $ P(x, y) = (x - y)^2 \cdot U(x, y) $. | proof | Algebra | proof | Yes | Yes | olympiads | false | 495 |
IV OM - I - Problem 11
Prove that if $ A + B + C $ or $ A + B - C $ or $ A - B + C $ or $ A - B - C $ equals an odd number of straight angles, then $ \cos^2A + \cos^2B + \cos^2C + 2 \cos A \cos B \cos C = 1 $ and that the converse theorem is also true. | \spos{1} We need to prove that
The task can be solved in a very simple way when it is noticed that the equation $ x = (2k + 1) \cdot 180^\circ $ is equivalent to the equation $ \cos \frac{x}{2} = 0 $. The necessary and sufficient condition for one of the angles $ A + B + C $, $ A + B - C $, $ A - B + C $, $ A - B - C $ to equal an odd multiple of a straight angle can thus be expressed by the following equation (2)
The task, therefore, reduces to showing that equality (1) is equivalent to equality (2). We will achieve this by transforming the left side of equality (2) using known formulas for sums and products of trigonometric functions:
It turns out that the left side of equality (2) is identically equal to $ \frac{1}{4} $ of the left side of equality (1). The equalities are therefore equivalent, and it follows that equality (1) holds if and only if one of the angles $ A + B + C $, $ A + B - C $, $ A - B + C $, $ A - B - C $ is an odd multiple of a straight angle. | proof | Algebra | proof | Yes | Yes | olympiads | false | 497 |
XL OM - III - Task 2
In the plane, there are three circles $ k_1 $, $ k_2 $, $ k_3 $. Circles $ k_2 $ and $ k_3 $ are externally tangent at point $ P $, circles $ k_3 $ and $ k_1 $ — at point $ Q $, and circles $ k_1 $ and $ k_2 $ — at point $ R $. The line $ PQ $ intersects circle $ k_1 $ again at point $ S $, and the line $ PR $ — at point $ T $. The line $ SR $ intersects circle $ k_2 $ again at point $ U $, and the line $ TQ $ intersects $ k_3 $ again at point $ V $. Prove that point $ P $ lies on the line $ UV $. | Let's denote the incircle of triangle $O_1O_2O_3$ by $k$, its center by $I$, and the centers of circles $k_1$, $k_2$, $k_3$ by $O_1$, $O_2$, $O_3$. Circle $k$ is tangent to the sides of triangle $O_1O_2O_3$ at points $P$, $Q$, $R$; this follows from the equalities $|O_1Q| = |O_1R|$, $|O_2R| = |O_2P|$, $|O_3P| = |O_3Q|$.
The following angle equalities hold (see Figure 6):
From these equalities, we obtain the relationship
which means that triangles $PRS$ and $PQT$ are right triangles: $PT \perp SU$, $PS \perp TY$.
Thus, angles $PRU$ and $PQV$ are right angles. These are inscribed angles in circles $k_2$ and $k_3$ and must subtend semicircles. Therefore, segments $PU$ and $PV$ are diameters of these circles; they are thus perpendicular to the common tangent (line $PI$). Hence, points $U$, $P$, $V$ are collinear. | proof | Geometry | proof | Yes | Yes | olympiads | false | 499 |
XV OM - II - Task 6
Prove that among any five points in a plane, one can choose three points that are not the vertices of an acute triangle. | If among the given points there are three collinear points, the thesis of the theorem is of course true. If, however, no three of the given points lie on a straight line, then four of these points, for example, $ A $, $ B $, $ C $, $ D $, are the vertices of a convex quadrilateral, as proven in problem 4. The angles of the quadrilateral $ ABCD $ are convex, and their sum is $ 360^\circ $, so at least one of these angles, for example, angle $ BAC $, is a convex angle not less than $ 90^\circ $. The points $ A $, $ B $, $ C $ are therefore not the vertices of an acute triangle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 501 |
XLVI OM - I - Problem 11
Given are natural numbers $ n > m > 1 $. From the set $ \{1,2, \ldots ,n\} $, we draw $ m $ numbers without replacement. Calculate the expected value of the difference between the largest and the smallest drawn number. | An elementary event is the selection of an $m$-element subset from an $n$-element set. These elementary events are equally probable; there are $ {n}\choose{m} $ of them. Let the largest and smallest selected number be denoted by $a$ and $b$, respectively. The random variable $X$ under consideration, which is the difference $a - b$, can take any integer value from $m - 1$ to $n - 1$.
Let us fix a natural number $r$ satisfying the inequalities $m - 1 \leq r \leq n - 1$. In the set $\{1, 2, \ldots, n\}$, there are $n - r$ pairs of numbers $a$, $b$ with the difference $a - b = r$. For a fixed position of the numbers $a$ and $b$, there are $ {r-1}\choose{m-2} $ ways to choose the remaining $m-2$ numbers from the set $\{b+1, \ldots, a-1\}$. Therefore, the product $(n - r) {{r-1}\choose{m-2}}$ represents the number of $m$-element subsets of the set $\{1, 2, \ldots, n\}$ in which the difference between the largest and smallest element is $r$—that is, the number of elementary events in which the random variable $X$ takes the value $r$.
Dividing this product by the number of all elementary events gives the probability that $X = r$; let us denote this probability by $p_r$. Thus,
\[
p_r = \frac{(n - r) {{r-1}\choose{m-2}}}{{n}\choose{m}}
\]
Using the easily justified relationships
\[
{n \choose m} = \frac{n!}{m!(n-m)!} \quad \text{and} \quad {r-1 \choose m-2} = \frac{(r-1)!}{(m-2)!(r-m+1)!}
\]
we transform the obtained expression as follows:
\[
p_r = \frac{(n - r) \frac{(r-1)!}{(m-2)!(r-m+1)!}}{\frac{n!}{m!(n-m)!}} = \frac{(n - r) (r-1)! m! (n-m)!}{(m-2)! (r-m+1)! n!}
\]
Since $r$ takes values from $m-1$ to $n-1$, the expected value of the random variable $X$, which is the number
\[
E(X) = \sum_{r=m-1}^{n-1} r p_r
\]
is equal to
\[
E(X) = \sum_{r=m-1}^{n-1} r \frac{(n - r) (r-1)! m! (n-m)!}{(m-2)! (r-m+1)! n!}
\]
To simplify the sums in the square brackets, we use the formula
\[
\sum_{r=j}^{N} {r \choose j} = {N+1 \choose j+1}
\]
(see: Note). By setting $N = n - 1$ and $j = m - 1$ in this formula, and then $N = n$ and $j = m$, we obtain, respectively:
\[
\sum_{r=m-1}^{n-1} {r \choose m-1} = {n \choose m}
\]
and
\[
\sum_{r=m}^{n} {r \choose m} = {n+1 \choose m+1}
\]
(applying a change of the summation index: $s = r + 1$). We can now complete the calculation (1):
\[
E(X) = \sum_{r=m-1}^{n-1} r \frac{(n - r) (r-1)! m! (n-m)!}{(m-2)! (r-m+1)! n!} = \frac{m(n-1)}{n-m+1}
\]
This is the sought value.
Note. Formula (2) is well-known. It can be easily proven by induction on $N$ for a fixed $j$. It can also be "seen" by carefully examining Pascal's triangle; we recommend this as a pleasant exercise in observation for the readers. | \frac{(n-1)}{n-+1} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 502 |
XVIII OM - II - Problem 1
Given is a sequence of numbers $ a_1, a_2, \ldots, a_n $ ($ n \geq 3 $), where $ a_1 = a_n = 0 $ and $ a_{k-1}+a_{k+1} \geq 2a_{k+1} $ for $ k = 2, 3, \ldots, (n - 1) $. Prove that this sequence does not contain any positive terms. | In a finite set of numbers $a_1, a_2, \ldots, a_n$, there exists at least one number not less than any of these numbers. Suppose such a number is $a_r$, i.e., $a_i \leq a_r$ for $i = 1, 2, \ldots, n$. Let $s$ be the smallest index with the property that $a_s = a_r$. We will prove that $s=1$. Indeed, if $s > 1$, then the inequalities would hold:
Adding these inequalities, we get
which contradicts the assumption of the theorem.
If, however, $s = 1$, this means that $a_r = a_1 = 0$, so $a_i \leq 0$ for $i = 1, 2, \ldots, n$.
Note. In the above proof, we relied on the theorem that in any finite set of numbers $a_1, a_2, \ldots, a_n$, there exists a number $a_r$ such that $a_i \leq a_r$ for $i = 1, 2, \ldots, n$.
This theorem, known as the maximum principle, is an easy consequence of the principle of induction. Conversely, the principle of induction follows from the maximum principle. We propose proving this equivalence as an exercise. | proof | Algebra | proof | Yes | Yes | olympiads | false | 505 |
Given an integer $ c \geq 1 $. To each subset $ A $ of the set $ \{1,2, \ldots ,n\} $, we assign a number $ w(A) $ from the set $ \{1,2, \ldots ,c\} $ such that the following condition is satisfied:
Let $ a(n) $ be the number of such assignments. Calculate $ \lim_{n\to \infty}\sqrt[n]{a(n)} $.
Note: $ \min(x,y) $ is the smaller of the numbers $ x $, $ y $. | Let $ M $ be the set $ \{1,2,\ldots,n\} $, and $ M_i $ be the $(n-1)$-element set obtained from $ M $ by removing the number $ i $:
Given the assignment as mentioned in the problem, let's denote the value $ w(M) $ by $ m $, and the value $ w(M_i) $ by $ f(i) $ (for $ i = 1,2, \ldots ,n $). According to the conditions of the problem, the values $ f(1), \ldots ,f(n) $ and $ m $ are numbers from the set $ \{1,2, \ldots ,c\} $, and for each $ i $, the following equality holds:
This is simply a fancy way of writing the inequality:
Thus, the assignment $ w(\cdot) $ is associated with the function:
From the condition imposed on the assignment $ w(\cdot) $:
it follows by induction that the equality:
holds for any sets $ A_1,\ldots,A_k $ contained in $ M $. (The symbol $ \min(x_1,\ldots,x_k) $ denotes the smallest of the numbers $ x_1,\ldots,x_k $; it can, of course, be the common value of several of these numbers.)
Let $ A $ be any subset of the set $ M $, not identical to the entire set $ M $, and let $ i_1,\ldots,i_k $ be all the numbers in $ M $ that do {\it not} belong to $ A $:
The set $ A $ is then exactly the intersection of the sets $ M_{i_1},\ldots,M_{i_k} $:
Indeed: a number $ i $ belongs simultaneously to all sets $ M_{i_1},\ldots,M_{i_k} $ if and only if it belongs to $ M $ and is different from the numbers $ i_1,\ldots,i_k $ - that is, (in accordance with (3)) if and only if it is an element of the set $ A $.
From the connections (4), (2), (3), the equality follows:
(The symbol $ \min_{i \not \in A}f(i) $ denotes the minimum value taken by the function $ f $ on those elements of the set $ M $ that do not belong to the set $ A $.)
Conclusion: the assignment $ w(\cdot) $ is fully determined by specifying the value $ m = w(M) $ and the values $ f(i) = w(M_i) $ for $ i = 1,2,\ldots ,n $. In other words, it is determined by choosing the number $ m $ and the function (1); it is then given by the formula:
And conversely: if $ m $ is any positive integer not exceeding $ c $, and if $ f $ is any mapping of the set $ M $ into the set $ \{1,2,\ldots,m\} $, then the formula (5) defines an assignment $ A \mapsto w(A) $ satisfying the condition stated in the problem.
For a fixed $ m $, there are exactly $ m^n $ functions (1). Therefore, the number of admissible assignments (5) is:
We obtain the two-sided estimate $ c^n \leq a(n) \leq c^n \cdot c $; and hence:
As $ n $ approaches infinity, the $ n $-th root of any fixed positive number approaches $ 1 $. Therefore, the expression on the right side of the inequality (6) approaches $ c $, and by the squeeze theorem, $ \displaystyle \lim_{n \to \infty} \sqrt[n]{a(n)} = c $.
The answer is: The sought limit value is the number $ c $. | c | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 507 |
XIII OM - III - Problem 6
Three lines $ a $, $ b $, $ c $ are pairwise skew. Can one construct a parallelepiped whose edges lie on the lines $ a $, $ b $, $ c $? | The solution to the problem is based on the following theorem:
If $a$ and $b$ are skew lines, then: a) there exists a pair of parallel planes, one of which contains the line $a$ and the other the line $b$; b) such a pair of planes is unique.
Proof: a) Through any point $A$ on the line $a$, we draw a line $b'$ parallel to $b$, and similarly, through any point $B$ on the line $b$, we draw a line $a'$ parallel to $a$. The plane $\alpha$ of lines $a$ and $b'$ and the plane $\beta$ of lines $a'$ and $b$ (different from $\alpha$) are parallel, since two intersecting lines of one of these planes are respectively parallel to two intersecting lines of the other.
b) If planes $\alpha$ and $\alpha'$ pass through the line $a$, and planes $\beta$ and $\beta'$ pass through the line $b$, with $\alpha \parallel \beta$ and $\alpha \neq \alpha'$, then planes $\alpha$ and $\alpha'$ must coincide. Indeed, if $\alpha$ and $\alpha'$ did not coincide, they would intersect along the line $a$; since $\alpha$ and $\alpha'$ are parallel to the line $b$ lying on planes $\beta$ and $\beta'$, the line $a$ would have to be parallel to the line $b$, contradicting the assumption that lines $a$ and $b$ are skew. Therefore, planes $\alpha$ and $\alpha'$ and similarly planes $\beta$ and $\beta'$ coincide.
Notice now that through each of the two skew edges of a parallelepiped, there pass two of its faces, and that among these four faces, which are all different, two are parallel. Indeed, a parallelepiped has 6 faces in pairs of parallel faces, so among the four different faces, some two must belong to one of these pairs.
Thus, if two skew lines $a$ and $b$ are given, on which lie two edges of a parallelepiped, then according to the previous theorem, lines $a$ and $b$ uniquely determine the planes of two opposite faces of the parallelepiped.
Three lines $a$, $b$, $c$ pairwise skew uniquely determine 3 pairs of planes: $(\alpha_1, \alpha_2)$, $(\beta_1, \beta_2)$, $(\gamma_1, \gamma_2)$ passing through lines $a$, $b$, $c$ respectively, such that $\alpha_1 \parallel \beta_1$, $\alpha_2 \parallel \gamma_1$, $\beta_2 \parallel \gamma_2$. There are two possible cases:
1) The planes of one of the pairs of parallel planes $\alpha_1 \parallel \beta_1$, $\alpha_2 \parallel \gamma_1$, $\beta_2 \parallel \gamma_2$ are parallel to the planes of another pair (considering the coincidence of planes as a special case of parallelism), e.g., $\alpha_1 \parallel \beta_1 \parallel \alpha_2 \parallel \gamma_1$, then $\alpha_1$ and $\alpha_2$, having a common line, coincide; similarly, $\beta_1$ and $\beta_2$ as well as $\gamma_1$ and $\gamma_2$ must coincide, as they are two pairs of planes passing through lines $b$ and $c$ respectively, and the planes of one pair are parallel to the planes of the other pair.
In the considered case, lines $a$, $b$, $c$ lie on 3 mutually parallel planes $\alpha_1$, $\beta_1$, $\gamma_1$. A parallelepiped whose 3 edges lie on lines $a$, $b$, $c$ does not exist, as all its faces would have to lie on the three planes $\alpha_1$, $\beta_1$, $\gamma_1$, which is impossible.
2) The planes of each of the pairs of parallel planes $(\alpha_1, \beta_1)$, $(\alpha_2, \gamma_1)$, $(\beta_2, \gamma_2)$ intersect the planes of the other two pairs. Each plane intersects 4 of the remaining ones, so we have 12 lines of intersection, among which are lines $a$, $b$, $c$. The 6 planes thus determine the desired parallelepiped in space. Specifically, the parallel planes $\alpha_1$ and $\beta_1$ define a layer of space between them, the planes $\alpha_2$ and $\gamma_1$ cut a 4-sided column from this layer, and finally, the planes $\beta_2$ and $\gamma_2$ cut a parallelepiped from this column.
The answer to the posed question is therefore: If three lines $a$, $b$, $c$ are pairwise skew, then a parallelepiped whose 3 edges lie on lines $a$, $b$, $c$ can be constructed if and only if the lines $a$, $b$, $c$ do not lie on 3 mutually parallel planes. Such a parallelepiped is unique. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 508 |
XXX OM - I - Task 11
Given a positive number $ p $ and three distinct rays $ OA^{\rightarrow} $, $ OB^{\rightarrow} $, $ OC^{\rightarrow} $ contained in a plane. Prove that there exists exactly one such triplet of points $ K, L, M $, that $ K\in OA^{\rightarrow} $, $ L \in OB^{\rightarrow} $, $ M\in OC^{\rightarrow} $ and the perimeter of each of the triangles $ OKL $, $ OLM $, $ OMK $ is equal to $ 2p $. | For the triangles mentioned in the problem to exist, it is of course necessary to assume that no two of the given rays lie on the same line.
First, note that if points $P$ and $Q$ belong to different sides of an angle with vertex $O$ and point $Q$ moves away from $O$, then the perimeter of triangle $OPQ$ increases without bound. Indeed, let point $Q$ lie between points $O$ and $Q$ (Fig. 9).
om30_1r_img_9.jpg
Then $PQ < QQ$ and therefore
the perimeter of triangle $OPQ$ is less than the perimeter of triangle $OPQ$. Moreover, the perimeter of triangle $OPQ$ can be arbitrarily large if we choose point $Q$ appropriately on the ray $OQ^\to$.
Let the number $x$ satisfy $0 < x < p$. Let $K(x)$ be the point on the ray $OA^\to$ such that $OK(x) = x$. Such a point is unique. Next, let $L(x) \in OB^\to$ and $M(x) \in OC^\to$ be such points that the perimeter of each of the triangles $OK(x)L(x)$ and $OK(x)M(x)$ is equal to $2p$. By the initial observation, these conditions uniquely determine the points $L(x)$ and $M(x)$.
Since
when $x \to 0$, then $OK(x) \to 0$, $OL(x) \to p$, $K(x)L(x) \to p$, $OM(x) \to p$, $K(x)M(x) \to p$.
Similarly, when $x \to p$, then $OK(x) \to p$, $OL(x) \to 0$, $K(x)L(x) \to p$, $OM(x) \to 0$, $K(x)M(x) \to p$.
Therefore, when $x \to 0$, the perimeter of triangle $OL(x)M(x)$ approaches a limit greater than $2p$, and when $x \to p$, this perimeter approaches zero. Since this perimeter is a continuous function of $x$, there exists some $x = x_0$ in the interval $(0, p)$ such that the perimeter of triangle $OL(x_0)M(x_0)$ is equal to $2p$.
Thus, the points $K = K(x_0)$, $L = L(x_0)$, $M = M(x_0)$ satisfy the conditions of the problem.
We will prove that such a triplet of points is unique. If points $K$, $L$, $M$ also satisfy the conditions of the problem and, for example, $OK < OK$, then from the initial observation it follows that $OL > OL$ and $OM > OM$. In this case, however, the perimeter of triangle $OLM$ would be less than the perimeter of triangle $OL$. The obtained contradiction proves that there is only one triplet of points satisfying the conditions of the problem. | proof | Geometry | proof | Yes | Yes | olympiads | false | 511 |
LV OM - III - Task 3
In a certain tournament, $ n $ players participated $ (n \geq 3) $. Each played against each other exactly once, and there were no draws. A three-element set of players is called a draw triplet if the players can be numbered in such a way that the first won against the second, the second against the third, and the third against the first. Determine the maximum number of draw triplets that could have appeared in the tournament. | Instead of determining the maximum number of draw triples, we will determine the minimum number of non-draw triples.
Let's number the players from 1 to $ n $ and assume that player number $ i $ won $ x_i $ times. Then the total number of matches is
Each non-draw triple is uniquely determined by the player who won against the other two players in that triple. Therefore, player number $ i $ determines $ x_i \choose 2 $ non-draw triples, and the total number of non-draw triples in the tournament is
In the further part of the solution, we will consider two cases:
The number $ n $ is odd, i.e., $ n = 2k + 1 $. Then, by the inequality between the quadratic mean and the arithmetic mean, we obtain
Thus, the number of draw triples does not exceed
The above number of draw triples will occur only when each player wins exactly $ k $ times (equality holds in inequality (1) at that time). Such a result of the matches can be achieved in the following way. We seat the players around a round table and assume that each of them won only against the players sitting to their right at a distance of no more than $ k $ places. In such a setup of matches, each player won exactly $ k $ times.
The number $ n $ is even, i.e., $ n = 2k $. Then $ (x_i - k)(x_i - k + 1) \geq 0 $ for $ i = 1,2,\ldots,n $, from which we obtain $ x_i^2 \geq (2k - 1)x_i - k(k - 1) $. Therefore,
From this, it follows that the number of draw triples does not exceed
The above number of draw triples will occur only when each player wins $ k $ or $ k-1 $ times (equality holds in inequality (2) at that time). Such a result of the matches can be achieved in the following way. As above, we seat the players around a round table and assume that each of them won against the players sitting to their right at a distance of no more than $ k-1 $ places. The result of the match between players sitting opposite each other is determined arbitrarily. In this way, each player won $ k $ or $ k-1 $ times.
Summarizing: the maximum number of draw triples that can appear in the tournament is $ \frac{1}{24}n(n^2 - 1) $ if $ n $ is an odd number, and $ \frac{1}{24}n(n^2 - 4) $ if $ n $ is an even number. | \frac{1}{24}n(n^2-1) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 514 |
XXIV OM - I - Problem 10
Find the smallest natural number $ n > 1 $ with the following property: there exists a set $ Z $ consisting of $ n $ points in the plane such that every line $ AB $ ($ A, B \in Z $) is parallel to some other line $ CD $ ($ C, D \in Z $). | We will first prove that the set $ Z $ of vertices of a regular pentagon has the property given in the problem, that is, $ n \leq 5 $. We will show that each side of the regular pentagon is parallel to a certain diagonal and vice versa, each diagonal is parallel to a corresponding side.
It suffices to prove that $ AB \parallel CE $ (Fig. 11). Since a circle can be circumscribed around quadrilateral $ ABCE $ (namely, it is the circle circumscribed around the given regular pentagon), we have $ \measuredangle A + \measuredangle BCE = \pi $. Since $ \measuredangle A = \measuredangle B $, it follows that $ \measuredangle BCE = \pi - \measuredangle B $, which proves that $ AB \parallel CE $.
On the other hand, from the conditions of the problem, it follows that $ n \geq 4 $, since there are at least two different parallel lines, each containing at least two points of the set $ Z $. If $ n = 4 $ and points $ A, B, C, D $ satisfied the conditions of the problem, they would be the vertices of a trapezoid. However, none of the diagonals of a trapezoid is parallel to another line determined by its vertices. Therefore, $ n > 4 $, and from the previously proven inequality $ n \leq 5 $, it follows that $ n = 5 $. | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 515 |
XXV - I - Task 1
During World War I, a battle took place near a certain castle. One of the shells destroyed a statue of a knight with a spear standing at the entrance to the castle. This happened on the last day of the month. The product of the day of the month, the month number, the length of the spear expressed in feet, half the age of the battery commander firing at the castle expressed in years, and half the time the statue stood expressed in years equals 451,066. In which year was the statue erected? | The last day of the month can only be $28$, $29$, $30$, or $31$. Of these numbers, only $29$ is a divisor of the number $451,066 = 2 \cdot 7 \cdot 11 \cdot 29 \cdot 101$. Therefore, the battle took place on February $29$ in a leap year. During World War I, only the year $1916$ was a leap year. From the problem statement, it follows that a divisor of the number $7 \cdot 11 \cdot 101$ is half the age of the battery commander. Only the number $11$ satisfies this condition, so the battery commander was $22$ years old. The length of the pike is a divisor of the number $7 \cdot 101$. Therefore, the pike was $7$ feet long. Thus, half the time the statue stood is $101$ years. It was erected $202$ years before the year $1916$, i.e., in the year $1714$. | 1714 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 517 |
XL OM - III - Task 4
Let $ n, k $ be natural numbers. We choose a sequence of sets $ A_0, \ldots, A_k $ such that $ A_0 = \{1, \ldots, n\} $, and for $ i = 1, \ldots, k $, the set $ A_i $ is a randomly chosen subset of $ A_{i-1} $, with each subset being equally likely. We consider the random variable equal to the number of elements in $ A_k $. Prove that its expected value is $ n2^{-k} $. | We introduce random variables $ X_1, \ldots , X_n $ defined as follows:
\
($ i = 1,\ldots ,n $). The random variable $ X $ (the number of elements in the set $ A_k $) considered in the problem is the sum of the variables introduced just now:
Let us fix a number $ i \in A_0 = \{1, \ldots, n\} $. Since $ A_1 $ is a randomly chosen subset of $ A_0 $, the number $ i $ will be in the set $ A_1 $ with probability
This follows from the fact that the set $ A_0 $ has as many subsets containing a fixed element $ i $ as subsets not containing this element.
With the same probability, the number $ i $ will end up in the set $ A_2 $ - provided it previously ended up in $ A_1 $:
Thus,
(To put it less formally and more vividly: the number $ i $ will be in the set $ A_2 $ if two consecutive drawings are "successful" for it.) Continuing this reasoning, we conclude inductively that
For $ j = k $, the obtained equality $ P(i \in A_k) = 2^{-k} $ means that the random variable $ X_i $ takes the value $ 1 $ with probability $ 2^{-k} $ (and the value $ 0 $ with probability $ 1-2^{-k} $). Therefore, its expected value is $ 2^{-k} $.
This conclusion is valid for any choice of $ i \in \{1,\ldots, n\} $. This means that the number $ 2^{-k} $ is the expected value of each of the random variables $ X_1, \ldots , X_n $. Therefore,
| proof | Combinatorics | proof | Yes | Yes | olympiads | false | 518 |
VII OM - II - Task 3
A homogeneous horizontal plate weighing $ Q $ kG in the shape of a circle is supported at points $ A $, $ B $, $ C $ lying on the circumference of the plate, with $ AC = BC $ and $ ACB = 2\alpha $. What weight $ x $ kG should be placed on the plate at the other end $ D $ of the diameter drawn from point $ C $, so that the pressure of the plate on the support at $ C $ is zero? | The weight $ Q $ of a plate supported at points $ A $, $ B $, $ C $ (Fig. 17) is balanced by the reactions at these points, acting vertically upwards. The center of gravity $ O $ of the plate, as the point of application of the weight $ Q $ acting downwards, must lie within the triangle $ ABC $, i.e., the angle $ 2\alpha $ must be acute. Due to the symmetry of the plate relative to the line $ CD $, the reactions at $ A $ and $ B $ are equal, their resultant $ R $ passes through the midpoint $ E $ of the segment $ AB $; we could remove the supports at $ A $ and $ B $ and support the plate at $ E $. If we place a weight $ x $ on the plate at point $ D $, the reactions at the supports will change. We need to determine the weight $ x $ in such a way that the reaction at $ C $ (equal to the pressure of the plate on the support at $ C $) becomes zero. This will happen when the weights $ Q $ and $ x $ and the reaction $ R $ are in equilibrium.
According to the known law of the lever, the following equality holds
and since $ OE = OA \cos 2\alpha $, $ DE = OD - OE = OD (1 - \cos 2\alpha) $, therefore
| Geometry | math-word-problem | Yes | Yes | olympiads | false | 519 | |
LI OM - II - Task 6
A polynomial $ w(x) $ of degree two with integer coefficients takes values that are squares of integers for integer $ x $. Prove that the polynomial $ w(x) $ is the square of some polynomial. | Let $ w(x) = ax^2 + bx + c $. Introduce the notation: $ k_n = \sqrt{w(n)} $ for any positive integer $ n $. Then $ k_n $ can be zero for at most two values of $ n $. For the remaining $ n $ we have
Dividing the numerator and the denominator of the obtained fraction by $ n $ and passing to the limit as $ n $ goes to infinity, we see that the sequence $ (k_{n+1} - k_n) $ is convergent and its limit is $ \sqrt{a} $. Since the terms of the sequence $ (k_{n+1} - k_n) $ are integers, the limit of this sequence, the number $ \sqrt{a} $, is also an integer. Moreover, there exists a natural number $ m $ such that
Let $ d = k_m - m\sqrt{a} $. From the relation (1) it follows that
Consider $ p(x) = (x\sqrt{a} + d)^2 $. By the definition of the numbers $ k_n $ and the equality (2), we obtain the equality $ p(n) = w(n) $ for $ n \geq m $. This means that $ p(x) = w(x) $, hence | proof | Algebra | proof | Yes | Yes | olympiads | false | 520 |
XXI OM - III - Task 1
Diameter $ \overline{AB} $ divides the circle into two semicircles. On one semicircle, n points $ P_1 P_2, \ldots, P_n $ are chosen such that $ P_1 $ lies between $ A $ and $ P_2 $, $ P_2 $ lies between $ P_1 $ and $ P_3 $, $ \ldots $, $ P_n $ lies between $ P_{n-1} $ and $ B $. How should point $ C $ be chosen on the other semicircle so that the sum of the areas of triangles $ CP_1P_2, CP_2P_3, CP_3P_4, \ldots, CP_{n-1}P_n $ is maximized? | Notice that the sum of the areas of triangles $ CP_1P_2, CP_2P_3, CP_3P_4, \ldots, CP_{n-1}P_n $ is equal to the sum of the areas of the polygon $ P_1P_2 \ldots P_n $ and triangle $ CP_1P_n $ (Fig. 13). The first area does not depend on the choice of point $ C $. The second area will be maximal when the distance from point $ C $ to the line $ P_1P_n $ is as large as possible. The point on the circle farthest from a given chord lies on the perpendicular bisector of that chord. Therefore, $ C $ should be the point of intersection of the perpendicular bisector of segment $ \overline{P_1P_n} $ with the other semicircle. | C | Geometry | math-word-problem | Yes | Yes | olympiads | false | 523 |
XXXIII OM - II - Task 3
Prove that for every natural number $ n \geq 2 $ the following inequality holds | For non-negative numbers $ a $, $ b $, the inequality
holds.
Therefore, for $ k = 1,2,\ldots,n $ we have
From inequality (*) it also follows that
We obtain the inequalities:
and
Suppose $ n $ is an odd number. By combining the first and last terms, the second and second-to-last terms, etc., in the product $ \log_n 2 \cdot \log_n 4 \cdot \ldots \cdot \log_n (2n - 2) $, we get a product of expressions of the form $ \log_n k \log_n (2n - k) $, each of which, by the previous considerations, is a positive number not greater than $ 1 $. This leads to the desired inequality.
If $ n $ is an even number, then a similar procedure as for odd numbers leads to a product of expressions of the form $ \log_n k \cdot \log_n (2n - k) $ and the factor $ \log_n (2n - n) = \log_n n = 1 $. Therefore, in this case as well, we obtain the inequality | proof | Inequalities | proof | Yes | Yes | olympiads | false | 527 |
XLVIII OM - II - Problem 5
We throw $ k $ white and $ m $ black six-sided dice. Calculate the probability that the remainder of the total number of eyes thrown on the white dice divided by $ 7 $ is equal to the remainder of the total number of eyes thrown on the black dice divided by $ 7 $. | The set of elementary events in this task is
where all elementary events are equally probable. Let $ A $ denote the event whose probability we are looking for:
Let $ B $ be the event that the total number of eyes thrown on all $ k + m $ dice is divisible by $ 7 $:
The transformation given by the formula
is a one-to-one mapping of the set $ A $ onto the set $ B $. Therefore, sets $ A $ and $ B $ are equinumerous, and thus the probabilities of events $ A $ and $ B $ are the same.
Let $ p_{n,r} $ ($ n \geq 1 $, $ r = 0,1,\ldots, 6 $) denote the probability that the remainder when the total number of eyes thrown on $ n $ dice is divided by $ 7 $ is equal to $ r $. Then $ p_{1,0} = 0 $, $ p_{1,1} = p_{1,2} = \ldots = p_{1,6} = 1/6 $.
From the formula for total probability, we get for $ n = 1,2,3,\ldots $ the equality:
Introducing the notation $ q_n = p_{n,0} - \frac{1}{7} $, from formula (*) we obtain the equality
From this, by easy induction, we get
Finally, therefore, | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 528 |
IV OM - I - Problem 12
Given a line $ p $ and points $ A $ and $ B $ lying on opposite sides of this line. Draw a circle through points $ A $ and $ B $ such that the chord lying on the line $ p $ is as short as possible. | Let $ M $ denote the point of intersection of the lines $ AB $ and $ p $, and let the circle passing through points $ A $ and $ B $ intersect the line $ p $ at points $ C $ and $ D $ (Fig. 35).
The chord $ CD $ is the sum of segments $ CM $ and $ MD $, whose product, by the theorem of intersecting chords of a circle, equals the product of the given segments $ AM $ and $ MB $. From arithmetic, we know that the sum of positive numbers with a given product is the smallest when these numbers are equal. The chord $ CD $ is thus the shortest when $ CM = MD $, i.e., when point $ M $ is the midpoint of this chord. The center of the sought circle, therefore, lies on the perpendicular line drawn to the line $ p $ through point $ M $. Since the center of this circle must also lie on the perpendicular bisector of the chord $ AB $, we determine it as the point of intersection of the two mentioned lines. The problem always has a solution, and it is unique. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 529 |
XLII OM - II - Problem 5
$ P_1, P_2, \ldots, P_n $ are distinct two-element subsets of the set $ \{1,2,\ldots,n\} $. Sets $ P_i $, $ P_j $ for $ i\neq j $ have a common element if and only if the set $ \{i,j\} $ is one of the sets $ P_1, P_2, \ldots, P_n $. Prove that each of the numbers $ 1,2,\ldots,n $ is a common element of exactly two of the sets $ P_1, P_2, \ldots, P_n $. | For each $ k \in \{1,2,\ldots,n\} $, let $ m_k $ be the number of sets $ P_i $ that contain the element $ k $. The sum of these numbers (as $ k $ ranges from $ 1 $ to $ n $) is $ 2n $, because each of the sets $ P_1,\ldots, P_n $ has two elements and is thus counted twice. We therefore have the equality
\[
\sum_{k=1}^n m_k = 2n.
\]
We need to prove that $ m_k = 2 $ for $ k = 1, \ldots, n $.
According to the problem statement, the family of $ n $ sets $ P_1,\ldots, P_n $ consists exactly of those pairs $ \{i,j\} $ of elements of the set $ \{1,2,\ldots,n\} $ for which the sets $ P_i $, $ P_j $ have a common element. This common element can be any number $ k \in \{1,2,\ldots,n\} $. For a fixed $ k $, there are $ m_k $ sets $ P_i $ that contain $ k $; thus, there are
\[
\binom{m_k}{2}
\]
pairs $ \{ i,j \} $ such that $ P_i \cap P_j = \{k\} $. The number of all such pairs is the sum of expressions (2) as $ k $ ranges from $ 1 $ to $ n $. Since each such pair $ \{i,j\} $ is one of the sets $ P_1,\ldots, P_n $, the number of these pairs is $ n $. We obtain the equality
\[
\sum_{k=1}^n \binom{m_k}{2} = n.
\]
Thus, in view of (1),
\[
\sum_{k=1}^n \frac{m_k(m_k-1)}{2} = n.
\]
Hence, from (1),
\[
\sum_{k=1}^n m_k^2 - \sum_{k=1}^n m_k = 2n.
\]
Therefore,
\[
\sum_{k=1}^n m_k^2 - 2n = 2n,
\]
which simplifies to
\[
\sum_{k=1}^n m_k^2 = 4n.
\]
We want to prove that all numbers $ m_k $ equal $ 2 $. Consider the differences $ m_k - 2 $. Let's compute the sum of the squares of these differences, using equations (1) and (3):
\[
\sum_{k=1}^n (m_k - 2)^2 = \sum_{k=1}^n (m_k^2 - 4m_k + 4) = \sum_{k=1}^n m_k^2 - 4 \sum_{k=1}^n m_k + 4n.
\]
Substituting the values from (1) and (3):
\[
\sum_{k=1}^n (m_k - 2)^2 = 4n - 4 \cdot 2n + 4n = 0.
\]
The sum of non-negative numbers equals zero only if all these numbers are zero. Therefore, $ m_1 = \ldots = m_n = 2 $.
This means that each number $ k \in \{1,2,\ldots,n\} $ belongs to exactly two sets $ P_i $. The proof is complete.
**Note 1.** By modifying the transformation used at the end of the solution, we can easily obtain a proof of the inequality between the *arithmetic mean*
\[
A(x_1,\ldots,x_n) = \frac{x_1 + \ldots + x_n}{n}
\]
of a system of non-negative numbers $ x_1,\ldots,x_n $ and the *quadratic mean* of these numbers, defined by the formula
\[
Q(x_1,\ldots,x_n) = \sqrt{\frac{x_1^2 + \ldots + x_n^2}{n}}.
\]
Writing $ A $ and $ Q $ instead of $ A(x_1,\ldots,x_n) $ and $ Q(x_1,\ldots,x_n) $, we have:
\[
\sum_{k=1}^n (x_k - A)^2 = \sum_{k=1}^n x_k^2 - 2A \sum_{k=1}^n x_k + nA^2 = nQ^2 - 2nA^2 + nA^2 = n(Q^2 - A^2).
\]
Thus,
\[
Q^2 \geq A^2,
\]
with equality holding only if $ x_1 = \ldots = x_n = A $.
Conversely, assuming the inequality (4) (along with the information when it becomes an equality) as a known fact, we can immediately deduce the proven theorem ($ m_1 = \ldots = m_n = 2 $) from equations (1) and (3).
**Note 2.** In this problem, we are dealing with a pair $ (V,E) $, where $ V $ is the set $ \{1,2,\ldots,n\} $, and $ E $ is a family of two-element sets $ P_1,\ldots,P_n $, each of which is contained in $ V $. This type of configuration forms a graph. Generally, a *graph* is a pair $ (V,E) $, where $ V $ is any set, and $ E $ is a certain family of distinguished two-element subsets of $ V $. (In applications, $ V $ is most often a finite set.) It is convenient to interpret $ V $ as a set of points (on a plane or in space), and each two-element set $ \{u,v\} $ belonging to the family $ E $ as a segment connecting points $ u $ and $ v $. This interpretation justifies the terminology used in graph theory: the elements of the set $ V $ are called the vertices of the graph, and the elements of the set $ E $ are called its edges (see problem 4 from the 32nd International Mathematical Olympiad).
The number of edges that have a given vertex as an endpoint is called the *degree* of that vertex.
Two edges of a graph are called *adjacent* if they have a common endpoint. Two vertices of a graph are called *adjacent* if they are connected by an edge.
It should be emphasized that this interpretation is intended only to facilitate the visualization of the given sets and their mutual relations (of a set-theoretic and combinatorial nature); there is no "geometry" involved here. The lengths of the edges and the angles between them, as well as their possible intersections outside the vertices of the graph, are abstracted away. For example, the graph $ (V_1,E_1) $, where $ V_1 $ is the set of vertices of a square, and $ E_1 $ consists of its three sides and both diagonals, is indistinguishable from the graph $ (V_2, E_2) $, where $ V_2 $ is the set of vertices of a tetrahedron, and $ E_2 $ is the set of any five of its edges.
The theorem to be proved in this problem can be formulated in the language of graphs as follows:
Given a graph $ (V,E) $, where the sets $ V $ and $ E $ have the same finite number of elements, numbered from $ 1 $ to $ n $. For each pair of different natural numbers $ i,j \in \{1,2,\ldots,n\} $, the vertices numbered $ i $ and $ j $ are adjacent if and only if the edges numbered $ i $ and $ j $ are adjacent. Under these assumptions, the degree of each vertex is $ 2 $. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 530 |
II OM - I - Task 1
Prove that the product of two factors, each of which is the sum of the squares of two integers, is also the sum of the squares of two integers. | We will apply the transformation
If $ a $, $ b $, $ c $, $ d $ are integers, then $ ac + bd $ and $ ad - bc $ are also integers; the theorem has thus been proven.
Notice that a more general theorem holds:
The product of $ n $ factors, each of which is the sum of the squares of two integers, is also the sum of the squares of two integers.
We will prove this theorem using the principle of mathematical induction. Let's briefly recall how this principle can be formulated:
Let $ T_n $ denote a statement asserting some property of the natural number $ n $, and suppose it is known that:
1° The statement $ T_1 $ is true, i.e., the number $ 1 $ has this property.
2° If for some number $ k $ the statement $ T_k $ is true, then the statement $ T_{k+1} $ is also true, i.e., if some natural number $ k $ has this property, then the next natural number $ k + 1 $ also has this property.
Then the statement $ T_n $ is true for every natural number $ n $, i.e., every natural number has this property. (This conclusion is the essence of the principle of induction).
Proceeding to the proof of the theorem mentioned above, we will give it a more convenient formulation:
Theorem ($ T_n $). If the numbers $ a_1, b_1, a_2, b_2, \ldots, a_n, b_n $ are integers, then the product
equals the sum of the squares of two integers.
(We wrote the number 1 at the beginning so that the above expression also has the form of a product in the case $ n = 1 $).
The proof will consist in showing that conditions 1°-2°, appearing in the principle of induction, are satisfied. Indeed:
1° The statement $ T_1 $ is true, since
2° Suppose that for some $ k $ the statement $ T_k $ is true, i.e., that
where $ A $ and $ B $ are integers. We will prove that under this assumption, the statement $ T_{k+1} $ is also true.
Indeed,
Based on the principle of induction, we infer from premises 1° and 2° that the statement $ T_n $ is true for every natural $ n $.
Note. It can be proven that the principle of mathematical induction is equivalent to the following theorem:
In any set of natural numbers (containing at least one number), there exists a smallest number. The stipulation that the set of natural numbers under consideration contains at least one number is necessary because in mathematics, sets of numbers or other objects (called elements of the set) include the so-called empty set, i.e., a set containing no elements. This is very convenient in certain reasoning.
We will conduct the proof of the previous theorem $ T_n $ based on the theorem mentioned above.
Suppose there exist natural numbers for which the statement $ T_n $ is not true. In the set of these numbers, there exists a smallest number $ k $. There exist, therefore, integers $ a_1, b_1, a_2, b_2, \ldots, a_k, b_k $ such that the product
does not equal the sum of the squares of two integers, while at the same time
where $ A $ and $ B $ are integers. But then we can write:
According to this equality, the product (1) equals the sum of the squares of two integers, which contradicts the previous conclusion. The assumption that there exist values of $ n $ for which the statement $ T_n $ is false - led to a contradiction; therefore, the statement $ T_n $ is true for every natural $ n $. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 531 |
XLIV OM - II - Problem 2
Given a circle with center $O$ and a point $P$ lying outside this circle. Through point $P$, we draw a line $l$ intersecting the given circle at points $A$ and $B$. Let $C$ be the point symmetric to $B$ with respect to the line $OP$, and let $m$ be the line passing through points $A$ and $C$. Prove that all such lines $m$ (corresponding to different choices of line $l$) have a common point. | The task does not specify how points $A$ and $B$ are situated on line $l$, meaning which one is closer to point $P$; Figure 5 illustrates these two situations. Let $Q$ be the point of intersection of lines $OP$ and $AC$; the thesis of the task will be proven if we show that the position of point $Q$ on the ray $OP^\to$ does not depend on the choice of line $l$ (which would mean that $Q$ is the sought common point of all considered lines $m$).
Triangle $AOC$ is isosceles; let us adopt the notation
Thus, the convex angle $AOC$ has a measure of
om44_2r_img_5.jpg
Angle $ABC$ is an inscribed angle subtended by the same arc as the convex or concave central angle $AOC$ (depending on whether point $A$ lies between $P$ and $B$, or point $B$ lies between $P$ and $A$). In the first case,
and in the second case,
In the first case, angle $PBC$ is identical to angle $ABC$, and in the second case, angle $PBC$ complements angle $ABC$ to a straight angle. Thus, in both cases, the equality $|\measuredangle PBC| = 90^\circ - \alpha$ holds. Further reasoning does not depend on the case.
Points $B$ and $C$ are symmetric with respect to line $OP$, so triangle $BPC$ is isosceles; hence,
and therefore,
Now, let's look at triangles $OPA$ and $OAQ$. They share a common angle at vertex $O$; angles $OPA$ and $OAQ$ have the same measure $\alpha$. Therefore, these triangles are similar, and the proportion $|OP|: |OA| = |OA|: |OQ|$ holds, which means
This indicates that the position of point $Q$ on the ray $OP^\to$ is determined by the radius of the circle and the position of point $P$; more precisely: point $Q$ is the image of point $P$ under inversion with respect to the given circle. The proof is complete. | proof | Geometry | proof | Yes | Yes | olympiads | false | 533 |
XXXII - I - Problem 6
Given numbers $ a_1\geq a_2 \geq \ldots \geq a_n \geq 0 $ satisfying the condition $ \sum_{i=1}^n a_i = 1 $. Prove that there exist integers $ k_1\geq k_2 \geq \ldots \geq k_n \geq 0 $ such that | Let $ [x] $ be the greatest integer not greater than $ x $.
Assume $ \widetilde{k_j} = [2na_j] $. The numbers $ \widetilde{k_j} $ satisfy the condition $ \widetilde{k_1} \geq \widetilde{k_2} \geq \ldots \geq \widetilde{k_n} \geq 0 $, and moreover $ 2na_j-1 < \widetilde{k_j} \leq 2na_j $, so $ 2a_j - \frac{1}{n} < \frac{\widetilde{k_j}}{n} \leq 2a_j $ and therefore
It follows that $ \sum_{j=1}^n \widetilde{k_j} > n $. We will now define the numbers $ k_j $ by subtracting appropriately chosen non-negative integers from $ \widetilde{k_j} $. Let $ d_n $ be the maximum non-negative integer for which
We set $ k_n = \widetilde{k_n}-d_n $. Of course, $ \frac{k_n}{n} \leq 2a_n $. Next, we define $ d_{n-1} $ as the maximum non-negative integer for which
We set $ k_{n-1} = \widetilde{k_{n-1}}-d_{n-1} $ and so on we define $ d_j $ as the maximum non-negative integer for which
and we set $ k_j = \widetilde{k_j} -d_j $.
Of course, $ \frac{k_j}{n} \leq 2a_j $ for $ j = 1,2,\ldots,n $ and $ \sum_{j=1}^n k_j = n $, since otherwise it would be $ \sum_{j=1}^n k_j > n $, which would mean that at least one term of this sum could be decreased by $ 1 $ while maintaining the conditions of the problem, contradicting the fact that at each step we subtracted the maximum possible $ d_j $. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 535 |
LVIII OM - II - Problem 5
A convex quadrilateral $ABCD$, where $AB \ne CD$, is inscribed in a circle. Quadrilaterals $AKDL$ and $CMBN$ are rhombuses with side lengths of $a$. Prove that points $K$, $L$, $M$, $N$ lie on the same circle. | Since the chords $ AB $ and $ CD $ are of different lengths, the lines $ AD $ and $ BC $ are not parallel. Let's denote their intersection point by $ P $ (Fig. 10). We will show that the points $ K $, $ L $, $ M $, and $ N $ lie on a circle with center $ P $.
The line $ AD $ is the perpendicular bisector of segment $ KL $, from which we get $ PK=PL $. Similarly, $ PM=PN $. It is therefore sufficient to prove that $ PK=PN $.
Let $ Q $ be the intersection point of the diagonals of the rhombus $ AKDL $. Then, by the Pythagorean theorem, we obtain
om58_2r_img_10.jpg
Similarly, we prove that
The points $ A $, $ B $, $ C $, $ D $ lie on the same circle, so the equality
is satisfied.
Combining dependencies (1), (2), and (3), we conclude that
since $ AK=BN=a $, we obtain from this the equality $ PK=PN $, which completes the solution of the problem. | proof | Geometry | proof | Yes | Yes | olympiads | false | 536 |
XXXI - III - Problem 5
In a tetrahedron, the areas of the six triangles, whose sides are the edges and whose vertices are the midpoints of the opposite edges of the tetrahedron, are equal. Prove that the tetrahedron is regular. | The assumption of the task guarantees the equality of the areas of six triangles, each of which has a base being an edge of the tetrahedron, and the opposite vertex is the midpoint of the opposite edge. On the tetrahedron $ABCD$, we describe a parallelepiped $A_1CB_1DAC_1BD_1$, whose each face contains a certain edge of the tetrahedron and is parallel to the opposite edge of the tetrahedron. From the assumption of the equality of the areas of the six triangles, whose sides are edges and whose vertices are the midpoints of the opposite edges of the tetrahedron, it follows that the sections of the parallelepiped by planes containing opposite edges have equal areas, because, for example, the area of $A_1B_1BA$ is equal to $2 \cdot$ the area of $\triangle ABK_1$, the area of $CDD_1C_1$ is equal to $2 \cdot$ the area of $\triangle CDK$, etc., for the remaining four sections.
om31_3r_img_20.jpgom31_3r_img_21.jpg
Now, let's draw a plane perpendicular to a certain edge of the considered parallelepiped, for example, to $\overline{AA_1}$. It intersects the lines $AA_1$, $BB_1$, $CC_1$, $DD_1$ at points $A_2$, $B_2$, $C_2$, $D_2$, respectively, and $A_2C_2B_2D_2$ is a parallelogram. Since the area of $ABB_1A_1$ is equal to $AA_1 \cdot A_2B_2$ and the area of $CDD_1C_1$ is equal to $CC_1 \cdot C_2D_2$, and these areas are equal and $AA_1 = CC_1$, then $A_2B_2 = C_2D_2$. Therefore, the parallelogram $A_2C_2B_2D_2$ has diagonals of equal length, so it is a rectangle. It follows that the faces $AC_1CA_1$ and $CB_1BC_1$; $CB_1BC_1$ and $B_1DD_1B$; $B_1DD_1B$ and $A_1DD_1A$ are perpendicular to each other.
By drawing a plane perpendicular to the edge $AC_1$ and then a plane perpendicular to the edge $AD_1$, we will similarly conclude that any two faces of the parallelepiped having a common edge are perpendicular to each other, and thus the parallelepiped is a rectangular parallelepiped. If $x$, $y$, $z$ are the lengths of the edges of this rectangular parallelepiped, then since (as we determined above) the sections drawn through opposite edges have equal areas, we have
Therefore,
The first of these equalities implies $x^2z^2 = y^2z^2$, so $x = y$, and similarly from the second equality, we get $y = z$.
Therefore, the parallelepiped described on the given tetrahedron is a cube, and hence the given tetrahedron is regular. | proof | Geometry | proof | Yes | Yes | olympiads | false | 537 |
XV OM - III - Task 3
Given is a tetrahedron $ABCD$, whose edges $AB, BC, CD, DA$ are tangent to a certain sphere: Prove that the points of tangency lie in the same plane. | We distinguish two cases:
a) $ AM = CN $, thus also $ AQ = CP $. Triangles $ MBN $ and $ ABC $ are then similar with respect to point $ B $, so $ MN \parallel AC $ and similarly $ QP \parallel AC $. Therefore, $ QP \parallel MN $, which means points $ MNPQ $ lie in the same plane.
b) $ AM \ne CN $, let's say $ AM > CN $, thus also $ AQ > CP $. (Fig. 17).
A line parallel to $ MN $ drawn through point $ C $ then intersects segment $ AB $ at some point $ E $, and a line parallel to $ PQ $ drawn through $ C $ intersects segment $ AD $ at some point $ F $. Since $ BM = BN $ and $ DP = DQ $, we have $ EM = CN = CP = QF $, hence $ MQ \parallel EF $.
The lines $ PQ $, $ QM $, $ MN $ are respectively parallel to the lines $ CF $, $ FE $, $ EC $, so the planes $ PQM $ and $ QMN $ are parallel to the plane $ CFE $; since these planes have a common line $ QM $, they coincide, meaning points $ P $, $ Q $, $ M $, $ N $ lie in the same plane. | proof | Geometry | proof | Yes | Yes | olympiads | false | 538 |
XXVI - I - Problem 10
Let $ \alpha $ be an irrational number, $ A_1 $ - a point on the circle $ S $ with center $ O $. Consider the infinite sequence $ A_n $ of points on the circle $ S $, where the point $ A_{k+1} $ is the image of the point $ A_k $ under a rotation about the point $ O $ by an angle $ \alpha m $. Prove that every arc of the circle $ S $ contains some points of the sequence $ A_n $. | If for certain natural numbers $k$ and $n$, where $k \ne n$, it was $A_k = A_n$, then the rotations by angles $(k-1)\alpha\pi$ and $(n-1)\alpha\pi$ would be equal. This means that the numbers $(k-1)\alpha\pi$ and $(n-1)\alpha\pi$ would differ by an integer multiple of $2\pi$, i.e., $(n-k)\alpha\pi = 2m\pi$, where $m$ is some integer. Hence, $\alpha = \frac{2m}{n-k}$, which contradicts the irrationality of $\alpha$. Therefore, all terms of the sequence $(A_n)$ are distinct.
Let $L$ be an arc of length $d$ contained in the circle $S$ of radius $r$. Let $t$ be a natural number greater than $\frac{2\pi r}{d}$. By dividing the circle $S$ into $t$ equal arcs of length $\frac{2\pi r}{t}$, we observe that there exists among them an arc containing at least two of the points $A_1, A_2, \ldots, A_{t+1}$. Let these points be $A_i$ and $A_j$, where $i < j$. Then the arc $\widehat{A_iA_j}$ has a length no greater than $\frac{2\pi r}{t}$. Since $t > \frac{2\pi r}{d}$, it follows that $\frac{2\pi r}{t} < d$, and thus the length of the arc $\widehat{A_iA_j}$ is less than $d$. This implies that the rotation by the angle $(j-i)\alpha\pi$ maps the point $A_i$ to $A_j$ and generally maps the point $A_{i+n}$ to $A_{j+n}$ for $n = 0, 1, 2, \ldots$. Therefore, the terms of the subsequence $A_1, A_{1 + (j-i)}, A_{1+2(j-i)}, \ldots$ determine arcs of length $\leq \frac{2\pi r}{t}$, which is less than $d$. Hence, some term of this subsequence must belong to the arc $L$ of length $d$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 539 |
LVI OM - I - Problem 10
Among all subsets of a fixed $ n $-element set $ X $, we sequentially draw with replacement three sets $ A $, $ B $, $ C $. Each time, the probability of drawing any of the $ 2^n $ subsets of set $ X $ is equally likely. Determine the most probable number of elements in the set $ A\cap B\cap C $. | The set of elementary events $\Omega$ consists of all triples $(A,B,C)$, where $A$, $B$, and $C$ are subsets of a given $n$-element set $S$. The selection of each of the $(2^n)^3 = 8^n$ triples $(A,B,C)$ is equally probable. Let $X_k$ $(k = 0,1,2,\ldots,n)$ denote the event that the triple of sets $(A,B,C)$ satisfies the condition $|A \cap B \cap C| = k$, where $|F|$ denotes the number of elements in the set $F$. We seek the natural number $k$ for which the probability of the event $X_k$ is the greatest.
Let us fix a number $k \in \{0,1,2, \ldots,n\}$. We will calculate the number of triples $(A, B, C)$ for which $|A \cap B \cap C| = k$.
If $(A,B,C)$ is a triple of subsets of the set $S$, then each element of the set $S$ is in exactly one of the following eight sets
In order to obtain a triple of sets $(A, B, C)$ for which $|A \cap B \cap C| = k$, we first choose $k$ elements of the set $S$ to place in the set $A \cap B \cap C$ — this can be done in $n \choose k$ ways. Each of the remaining $n-k$ elements of $k$ is assigned one of the remaining seven possibilities listed in (1) — this can be done in $7^{n-k}$ ways. The described procedure uniquely determines the triple $(A,B,C)$. Therefore, the number of triples $(A,B,C)$ for which $|A \cap B \cap C| = k$ is ${n \choose k} \cdot 7^{n-k}$. Hence, we obtain
To determine the number $k$ for which the probability of the event $X_k$ is the greatest, we calculate the ratio
From this, it follows that
Denoting $\lambda = [(n-7)/8]$, where $[x]$ is the greatest integer not greater than $x$, we obtain
Thus, the probability $P(X_k)$ is the greatest for $k = \lambda + 1 = [\frac{1}{8} (n+1)]$.
If $\frac{1}{8}(n-7)$ is an integer, then there are two values of $k$ for which the number $P(X_k)$ is the greatest, namely $k = \frac{1}{8} (n-7)$ and $k = \frac{1}{8} (n+1)$. | [\frac{1}{8}(n+1)] | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 542 |
XVI OM - II - Task 5
Prove that a square can be divided into any number of squares greater than 5, but it cannot be divided into 5 squares. | a) First, let us note that having a square divided into $ m $ squares, we can divide one of these squares into four squares by connecting the midpoints of its opposite sides. The entire square will then be divided into $ m + 3 $ squares.
Let $ n $ be a natural number greater than $ 1 $. Divide each side of the square $ Q $ into $ n $ equal parts and connect the corresponding points of division of opposite sides (two points of opposite sides of the square are called corresponding here if they lie on a line perpendicular to those sides). The square $ Q $ will then be divided into $ n^2 $ smaller squares $ Q_i $, with $ n $ squares $ Q_i $ adjacent to each side of the square $ Q $.
Consider two adjacent sides of the square $ Q $; $ 2n - 1 $ squares $ Q_i $ are adjacent to them, since one of them is adjacent to both of these sides; the remaining squares $ Q $ exactly fill a square $ R $ with a side equal to $ \frac{n-1}{n} $ of the side of the square $ Q $. When we erase all the lines of the division that are inside the square $ R $, we obtain a division of the square $ Q $ into $ 2n - 1 $ squares $ Q_i $ and the square $ R $, i.e., into $ (2n - 1) + 1 = 2n $ squares.
Thus, a square can be divided into any even number of squares greater than $ 2 $.
In that case, by the previous remark, a square can also be divided into $ 2n + 3 = 2 (n + 1) + 1 $ ($ n > 1 $) squares, i.e., into any odd number of squares greater than $ 5 $.
We have proved that a square can be divided into any number of squares greater than $ 5 $.
b) When a square is divided into squares only, in such a figure only right angles or straight angles appear, and the dividing squares have sides parallel to the sides of the entire square.
Suppose the square $ Q $ with side length $ l $ and vertices $ A $, $ B $, $ C $, $ D $ is divided into $ 5 $ squares $ Q_1 $, $ Q_2 $, $ Q_3 $, $ Q_4 $, $ Q_5 $. Each vertex of the square $ Q $ is a vertex of one of the squares $ Q_i $ ($ i = 1, 2, \ldots, 5 $), and two different vertices of the square $ Q $ cannot belong to the same square $ Q_i $, since their distance is $ \geq l $, which is greater than the side length of $ Q_i $. Let $ A $, $ B $, $ C $, $ D $ be the vertices of the squares $ Q_1 $, $ Q_2 $, $ Q_3 $, $ Q_4 $ with side lengths $ a $, $ b $, $ c $, $ d $, respectively.
If all the vertices of the square $ Q_5 $ were inside the square $ Q $, then the sides of the squares $ Q_i $ would completely cover the sides of the square $ Q $, i.e., the equality
would hold, which would imply that $ c = a $, $ d = b $,
and thus the area of $ Q $ would be expressed by the formula
as well as by the formula
from which it would follow that
which is impossible.
If one of the vertices of the square $ Q_5 $ were on one of the sides of the square $ Q $, for example, on $ AB $, then one of the sides of the square $ Q_5 $, for example, $ MN $, would lie on $ AB $. The remaining $ 2 $ vertices would lie on lines perpendicular to $ AB $ at points $ M $ and $ N $ at a distance less than $ l $ from $ AB $, i.e., inside the square $ Q_5 $. This would imply that
This system of conditions is, however, contradictory, since from the first three conditions it follows that $ (b + c) - (c + d) + (d + a) = l - l + l = l $, i.e., that $ a + b = l $, which contradicts the last one.
The assumption that the square $ Q $ was divided into $ 5 $ squares led to a contradiction. Such a division is therefore impossible. | proof | Geometry | proof | Yes | Yes | olympiads | false | 543 |
XLIX OM - II - Problem 2
In triangle $ABC$, angle $BCA$ is obtuse and $\measuredangle BAC = 2\measuredangle ABC$. The line passing through point $B$ and perpendicular to $BC$ intersects line $AC$ at point $D$. Point $M$ is the midpoint of side $AB$. Prove that $\measuredangle AMC = \measuredangle BMD$. | Let a line parallel to $ AB $ and passing through point $ C $ intersect segment $ BD $ at point $ E $. Denote by $ N $ the midpoint of segment $ CE $. Then points $ M $, $ N $, $ D $ are collinear. Triangle $ BCE $ is a right triangle, so $ N $ is the center of the circle circumscribed around it. This implies that triangle $ BCN $ is isosceles, which gives the following equalities:
Cyclic quadrilateral $ ABNC $ is thus an isosceles trapezoid, not a parallelogram. Therefore, point $ N $ is the image of point $ C $ under the reflection about the perpendicular bisector of segment $ AB $. This means that $ \measuredangle AMC =\measuredangle BMN = \measuredangle BMD $. | proof | Geometry | proof | Yes | Yes | olympiads | false | 546 |
XXII OM - I - Problem 10
Given is a table with $ n $ rows and $ n $ columns. The number located in the $ m $-th column and $ k $-th row is equal to $ n(k - 1) + m $. How should $ n $ numbers be chosen, one from each row and each column, so that the product of these numbers is the largest? | Let's denote $ a_{k,m} = n (k - 1) + m $, where $ 1 \leq k, m \leq n $. Suppose we select the number in the column with number $ m_i $ from the $ i $-th row, where $ i = 1, 2, \ldots, n $. From the conditions of the problem, it follows that the sequence $ m_1, m_2, \ldots, m_n $ has distinct terms. Each such sequence corresponds to the product $ a_{1m_1} \cdot a_{2m_2} \cdot \ldots \cdot a_{nm_n} $.
We will prove that if the sequence $ m_1, m_2, \ldots, m_n $ is not decreasing, then the corresponding product is not the largest.
For example, let $ m_i < m_{i+1} $ for some $ i $. Consider the products $ A $ and $ B $ corresponding to the sequences:
and
(these sequences differ only in the terms with numbers $ i $ and $ i + 1 $). Therefore,
and
Let $ \displaystyle C= \prod_{k \ne i, i+1} a_{km_k} $ and using the definition of the numbers $ a_{km} $, we get:
Thus, $ B > A $, which means the product corresponding to the second sequence is greater than the product corresponding to the first. Therefore, we have proven that if the sequence $ m_1, m_2, \ldots, m_n $ is not decreasing, then the corresponding product is not the largest. Hence, the largest product corresponds to a decreasing sequence. Since from the numbers $ 1, 2, \ldots, n $, we can form only one $ n $-term decreasing sequence, namely the sequence $ n, n - 1, \ldots, 2, 1 $, the largest product is equal to $ a_{1n}a_{2n-1} \ldots a_{n1} = n(2n - 1) (3n - 2) (4n - 3) \ldots (n^2 - (n -1)) $. | a_{1n}a_{2n-1}\ldotsa_{n1}=n(2n-1)(3n-2)(4n-3)\ldots(n^2-(n-1)) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 549 |
XXXIX OM - II - Problem 6
A convex polyhedron is given with $ k $ faces $ S_1, \ldots, S_k $. Denote the unit vector perpendicular to the face $ S_i $ ($ i = 1, \ldots, k $) directed outward from the polyhedron by $ \overrightarrow{n_i} $, and the area of this face by $ P_i $. Prove that | We start with the observation that if in space a convex planar polygon $S$ and a plane $\pi$ are given, and if $\overrightarrow{n}$ is a unit vector perpendicular to the plane of the polygon $S$, and $\overrightarrow{w}$ is a unit vector perpendicular to the plane $\pi$, then denoting by $W$ the orthogonal projection of the polygon $S$ onto the plane $\pi$, we have the following relationship between the areas of the polygons $S$ and $W$:
\[
\text{Area}(W) = |\overrightarrow{n} \cdot \overrightarrow{w}| \cdot \text{Area}(S)
\]
(The dot between the vectors denotes the dot product.)
Formula (I) is obvious in the case where the considered planes are parallel or perpendicular (because in these cases, respectively, $|\overrightarrow{n} \cdot \overrightarrow{w}|$ equals $1$ or $0$). In the remaining case, when the planes intersect along a line $l$ forming an angle $\varphi (0 < \varphi < \pi/2)$, formula (1) follows from the following observations:
If $S$ is a rectangle with one side parallel and the other perpendicular to the line $l$, its projection $W$ onto the plane $\pi$ is also such a rectangle. The length of the side parallel to $l$ does not change during projection, while the length of the side perpendicular to $l$ is shortened by a factor of $|\cos \varphi|$. The area of the rectangle changes by the same factor; thus, for a rectangle $S$, we obtain formula (1), because $|\overrightarrow{n} \cdot \overrightarrow{w}| = |\cos \measuredangle (\overrightarrow{n}, \overrightarrow{w})| = |\cos \varphi|$.
From this, it immediately follows that (1) is true for any right triangle with one leg parallel and the other perpendicular to $l$, because such a triangle can be completed to a rectangle (whose two sides are the legs of the given triangle). The area of the rectangle changes during projection by a factor of $|\cos \varphi| = |\overrightarrow{n} \cdot \overrightarrow{w}|$, so the same happens to the area of the triangle, which is half the area of the rectangle (Figure 8).
It is enough to notice that any convex polygon $S$ is the sum of a finite number of right triangles with legs parallel and perpendicular to the line $l$, with disjoint interiors. The area of $S$ is the sum of the areas of these triangles, and the area of $W$ (the projection of $S$ onto $\pi$) is the sum of the areas of the projections of the individual triangles (Figure 8). The area of the projection of each of these triangles equals the area of the triangle multiplied by the same factor equal to $|\overrightarrow{n} \cdot \overrightarrow{w}|$, and by summing up, we obtain the formula (1) to be proved.
We proceed to the proof of the theorem. Let
\[
\overrightarrow{v} = \sum_{i=1}^{k} \text{Area}(S_i) \overrightarrow{n_i}
\]
We need to prove that $\overrightarrow{v}$ is the zero vector.
Choose any unit vector $\overrightarrow{w}$ in space. Let $\pi$ be any plane perpendicular to the vector $\overrightarrow{w}$. The orthogonal projection of the considered polyhedron onto the plane $\pi$ is a convex polygon $W$. The projection of any face $S_i$ is a convex polygon $W_i$ (degenerate to a segment if $\overrightarrow{n_i} \bot \overrightarrow{w}$).
We partition the set of indices $\{1, \ldots, k\}$ into three subsets:
\[
I_+ = \{i : \overrightarrow{n_i} \cdot \overrightarrow{w} > 0\}
\]
\[
I_- = \{i : \overrightarrow{n_i} \cdot \overrightarrow{w} < 0\}
\]
\[
I_0 = \{i : \overrightarrow{n_i} \cdot \overrightarrow{w} = 0\}
\]
(Looking at the polyhedron from the outside, in the direction of the vector $\overrightarrow{w}$, we see the faces $S_i$ with indices $i \in I_-$, and we do not see the faces with indices $i \in I_+$) (Figure 9). The projections of the faces with indices $i \in I_+ \cup I_0$ have no common interior points and fill the entire polygon $W$. The same can be said about the projections of the faces with indices $i \in I_- \cup I_0$. Therefore,
\[
\sum_{i \in I_+ \cup I_0} \text{Area}(W_i) = \text{Area}(W)
\]
and simultaneously,
\[
\sum_{i \in I_- \cup I_0} \text{Area}(W_i) = \text{Area}(W)
\]
Notice now that according to formula (1),
\[
\text{Area}(W_i) = |\overrightarrow{n_i} \cdot \overrightarrow{w}| \cdot \text{Area}(S_i)
\]
Therefore,
\[
\sum_{i \in I_+ \cup I_0} |\overrightarrow{n_i} \cdot \overrightarrow{w}| \cdot \text{Area}(S_i) = \text{Area}(W)
\]
and
\[
\sum_{i \in I_- \cup I_0} |\overrightarrow{n_i} \cdot \overrightarrow{w}| \cdot \text{Area}(S_i) = \text{Area}(W)
\]
Since the vector $\overrightarrow{w}$ was chosen arbitrarily, we have thus shown that the vector $\overrightarrow{v}$ is perpendicular to any unit vector. It is therefore the zero vector.
(Figure 9) | proof | Geometry | proof | Yes | Yes | olympiads | false | 552 |
XIX OM - III - Task 4
Given a natural number $ n>2 $. Provide a set of $ n $ natural numbers $ a_1, a_2, \ldots, a_n $, so that in the set of sums
there are as few different numbers as possible, and provide a set of $ n $ natural numbers $ b_1, b_2, \ldots, b_n $, so that in the set of sums
there are as many different numbers as possible. | When $Z$ is a finite set of numbers, we agree to denote by $l(Z)$ the number of different values that the sums of two numbers belonging to $Z$ can take.
1) Let $A$ be a set of $n > 2$ natural numbers $a_i$ ($i = 1,2,\ldots, n$) so labeled that
Since
thus in the set of sums $a_i + a_j$ there are at least $2n - 3$ different numbers, i.e., $l(A) \geq 2n - 3$.
If we then form a set $A$ such that $l(A) = 2n - 3$, it will be the set we are looking for. To achieve this, we need to choose the numbers $a_1, a_2, \ldots, a_n$ in such a way that each sum $a_i + a_j$ is equal to one of the sums appearing in (2), i.e., one of the sums $a_1 + a_j$ ($j = 2,3, \ldots, n$), or one of the sums $a_i + a_n$ ($i = 2,3, \ldots, (n-1)$).
We will show that this case occurs when the numbers of the set $A$ form an arithmetic progression.
Assume that
According to known formulas
Consider the sum $a_i + a_j$, where $1 \leq i < j \leq n$.
When $i + j \leq n$, then by (3)
and when $i + j > n$, then by (3)
From (4) and (5) it follows that each sum $a_i + a_j$ is equal to one of the sums (2), so $l(A) = 2n - 3$, q.e.d.
b) The set of natural numbers $B = \{b_1, b_2, \ldots, b_n\}$ with the largest possible number $l(B)$ of different values of sums $b_i + b_j$ ($i, j = 1,2, \ldots, n$, $i \ne j$), will be obtained when all sums $b_i + b_j$ are different numbers; in this case $l(B) = \frac{1}{2} n (n - 1)$.
Such a set is, for example, the set of terms of a geometric progression
whose ratio $q$ is a natural number greater than $1$.
Suppose that for some terms of the sequence (6) the equality
holds, then
Since the numbers $1 + q^{j-i}$ and $1 + q^{l-k}$ are relatively prime to $q$, it follows from (8) that $q^k$ is divisible by $q^i$, and $q^i$ by $q^k$. Therefore, $k = i$, and by (7) also $l = j$, i.e., the equality (7) cannot hold if $q^i$, $q^j$ are not the same terms of the sequence (6) as $q^k$, $q^l$.
Note 1. Let $A$ be a set of numbers $a_1 < a_2 < \ldots < a_n$ ($n > 2$), for which the number $l(A)$ of different values of sums $a_i + a_j$ is $2n - 3$, i.e., it is as small as possible. We ask whether the sequence $a_1, a_2, \ldots, a_n$ must be an arithmetic progression?
It is easy to see that this is not the case when $n = 3$, nor when $n = 4$. From the inequality $a_1 < a_2 < a_3$ it follows that $a_1 + a_2 < a_1 + a_3 < a_2 + a_3$, so for any set $A$ of three different numbers, $l(A) = 3$. And for a set $A$ of numbers $a_1 < a_2 < a_3 < a_4$ to have $l(A) = 5$, it is only necessary that the equality $a_4 - a_3 = a_2 - a_1$ holds.
We will prove that when $n \geq 5$, then any set $A$ composed of $n$ numbers, for which $l(A) = 2n - 3$, is a set of terms of an arithmetic progression.
For $n = 5$ the thesis of this theorem is true. Indeed, if a set $A$ of five numbers
is given, then
and
so if $l(A) = 2 \cdot 5 - 3 = 7$, then each of the sums appearing in (3) must be equal to one of the seven sums appearing in (2). Since each of the sums (3) is greater than $a_1 + a_3$ and less than $a_3 + a_5$, the equalities
follow, from which it follows that
which means that the sequence (1) is an arithmetic progression.
Assume that the thesis of the theorem is true for some $n \geq 5$. Let $A$ be a set of numbers
for which $l(A) = 2(n + 1) - 3$.
Consider the sets
We will show that
Indeed, if the numbers $a_i$, $a_j$ ($i < j$) of the set $A$ belong to the set $A$, then
so the number of different values of sums $a_i + a_j$ is at least 2 less for the set $A$ than for the set $A$, i.e.,
We know that
Thus, $l(A) \leq 2n - 3$.
Similarly, if the numbers $a_i$, $a_j$ ($i < j$) of the set $A$ belong to the set $A$, then
from which it follows that $l(A) \leq 2n - 3$, so $l(A) \leq 2n - 3$.
By the induction hypothesis, the sequences $a_1, a_2, \ldots, a_n$ and $a_2, a_3, \ldots, a_{n+1}$ are arithmetic progressions, and therefore the sequence $a_1, a_2, \ldots, a_n, a_{n+1}$ is also an arithmetic progression, i.e., the thesis of the theorem is true for sets composed of $n + 1$ numbers. By the principle of induction, it is true for any natural number $n \geq 5$.
Note 2. Examples of such a set $B = \{b_1, b_2, \ldots, b_n\}$ that $l(B) = \frac{1}{2} n(n-1)$ can be given many. This property has, for example, the set of $n$ consecutive terms of the Fibonacci sequence *), i.e., the sequence of numbers $b_n$ defined by the formulas $b_1 = 0$, $b_2 = 1$, and for each $n > 2$ $b_n = b_{n-1} + b_{n-2}$.
The first terms of this sequence are therefore the numbers $0, 1, 1, 2, 3, 5, 8, 13, \ldots$.
The proof that each of the sums $b_i + b_j$ is a different number can be easily carried out by induction. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 553 |
XXXVIII OM - III - Problem 1
In a square with side 1, there are $ n $ points ($ n > 2 $). Prove that they can be numbered $ P_1, P_2, ..., P_n $ such that \sum_{i=1}^n |P_{i-1}P_i|^2 \leq 4 (we assume $ P_0=P_n $). | The proof will be based on the following lemma:
Lemma. In a right-angled isosceles triangle $ABC$, where $|AB| = |BC| = a$, $|\measuredangle ABC| = 90^\circ$, there are $k$ points ($k \geq 1$). These points can then be numbered $P_1, \ldots, P_k$ such that
(we assume $P_0 = A$, $P_{k+1} = C$).
Proof. We use induction. When $k = 1$, we have one point $P$ and need to prove the inequality $|AP|^2 + |PC|^2 \leq 2a^2$, which follows from the fact that $2a^2 = |AC|^2$, and the angle $APC$ is not obtuse.
Fix a natural number $m \geq 2$ and assume that the lemma holds for any right-angled isosceles triangle and any set of $k$ points contained in it, where $k$ is any natural number less than $m$.
Consider a triangle $ABC$ ($|AB| = |BC| = a$, $|\measuredangle B| = 90^\circ$), and a set $Z$ of $m$ points. Let $A_0B_0C_0$ be the smallest triangle with sides parallel to the corresponding sides of triangle $ABC$ that contains the set $Z$. (Construction: each of the three lines containing the sides of triangle $ABC$ is translated parallel until it meets the first point of set $Z$; the translated lines form triangle $A_0B_0C_0$ as described.)
Let $E_0$ be the foot of the perpendicular from vertex $B_0$ to the hypotenuse $A_0C_0$. Of course,
Denote the triangles (closed) $A_0E_0B_0$ and $B_0E_0C_0$ by $\Delta$ and $\Delta'$, and their intersection, the segment $B_0E_0$, by $I$. Each of the triangles $\Delta$ and $\Delta'$ contains points of set $Z$; otherwise, triangle $A_0B_0C_0$ would not be minimal. We now divide the set $Z$ into two non-empty subsets $Z_1$ and $Z_2$ contained in $\Delta$ and $\Delta'$, respectively: points of the set $Z \cap \Delta$ are assigned to $Z_1$, points of the set $Z \cap \Delta'$ are assigned to $Z_2$, and points of the set $Z \cap I$ are distributed between $Z_1$ and $Z_2$ in any way, with the only condition that the resulting sets $Z_1$ and $Z_2$ are non-empty; this is possible because, by the previous conclusion, $Z \cap \Delta \neq \emptyset$, $Z \cap \Delta' \neq \emptyset$, and $m \geq 2$.
Let $k$ be the number of elements in set $Z_1$, and $m-k$ the number of elements in set $Z_2$ ($0 < k < m$). By the induction hypothesis applied to triangles $\Delta$ and $\Delta'$ and the sets $Z_1$ and $Z_2$ contained in them, we can number the points of these sets as $P_1, \ldots, P_k$ and $P_{k+1}, \ldots, P_m$ respectively, such that
Notice now that
(because $|\measuredangle P_kB_0P_{k+1}| \leq 90^\circ$), and moreover
(see Figure 17). Therefore,
This means that the numbering $P_1, \ldots, P_m$ of the points in set $Z$ satisfies the given condition. The inductive proof of the lemma is thus completed.
om38_2r_img_17.jpg
To deduce the theorem from the lemma, we divide the given square $ABCD$ (with side length 1) into two right-angled triangles by the diagonal $AC$. The given set of $n$ points is divided into two subsets, contained in one and the other triangle, respectively, and counting $k$ and $n-k$ points ($0 \leq k \leq n$); points lying on the diagonal are assigned to either of these sets.
We consider two cases:
1. $0 < k < n$. From the lemma, it follows that the points in each of these subsets can be numbered as $P_1, \ldots, P_k$ and $P_{k+1}, \ldots, P_n$ such that
Similarly, as in the proof of the lemma, we use the inequalities
which follow from the fact that $|\measuredangle P_kCP_{k+1}| \leq 90^\circ$, $|\measuredangle P_nAP_1| \leq 90^\circ$ (see Figure 18). We get the inequality
Thus, the numbering $P_1, \ldots, P_n$ satisfies the condition of the problem.
om38_3r_img_18.jpg
2. $k = 0$ or $k = n$. In this case, all given points lie in one of the triangles $ABC$ or $CDA$, and according to the lemma, we can number them as $P_1, \ldots, P_n$ such that
Therefore,
which, combined with the obvious inequality $|P_nP_1|^2 \leq 2$, gives the desired condition | proof | Geometry | proof | Yes | Yes | olympiads | false | 554 |
XXV OM - II - Problem 4
In a convex quadrilateral $ABCD$ with area $S$, each side is divided into 3 equal parts, and segments connecting corresponding division points of opposite sides are drawn in such a way that the quadrilateral is divided into 9 smaller quadrilaterals. Prove that the sum of the areas of the following three quadrilaterals resulting from the division: the one containing vertex $A$, the central one, and the one containing vertex $C$ is equal to $\frac{S}{3}$. | We will first prove the
Lemma. The segment connecting the corresponding division points of opposite sides of the given quadrilateral intersects with analogous segments connecting the corresponding division points of the remaining sides of the quadrilateral at points that divide this segment into three equal parts.
Proof. Let points $S$, $Z$, $W$, $R$ divide the sides $\overline{AB}$, $\overline{BC}$, $\overline{DC}$, $\overline{AD}$ of the given quadrilateral $ABCD$ in the ratio $1:2$ (Fig. 16). Let $E$ be the intersection point of segments $\overline{RZ}$ and $\overline{SW}$. It suffices to prove that point $E$ divides each of these segments in the ratio $1:2$.
Since
\[
\frac{AS}{AB} = \frac{AR}{AD} = \frac{1}{3},
\]
by the converse of Thales' theorem, we have $RS \parallel DB$. It follows that triangles $ARS$ and $ADB$ are similar in the ratio $1:3$. Hence
\[
\frac{RS}{DB} = \frac{1}{3}.
\]
Similarly, from the equality $\frac{CW}{CD} = \frac{CZ}{CB} = \frac{2}{3}$, it follows that $WZ \parallel DB$. Therefore, triangles $CWZ$ and $CDB$ are similar in the ratio $2:3$. Hence
\[
\frac{WZ}{DB} = \frac{2}{3}.
\]
From (1) and (2), it follows that $\frac{RS}{WZ} = \frac{1}{2}$. Moreover, we have $RS \parallel WZ$. Therefore, triangles $RSE$ and $ZWE$ are similar in the ratio $1:2$. In particular, it follows that $\frac{RE}{EZ} = \frac{SE}{EW} = \frac{1}{2}$.
We now proceed to solve the problem. Let points $R, G \in \overline{AD}$, $W, P \in \overline{DC}$, $Q, Z \in \overline{BC}$, $S, H \in \overline{AB}$ divide each of the segments $\overline{AD}$, $\overline{DC}$, $\overline{BC}$, $\overline{AB}$ into three equal parts (Fig. 17). Then, by the lemma, points $E, U \in \overline{RZ}$, $T, F \in \overline{GQ}$, $E, T \in \overline{WS}$, $F, U \in \overline{HP}$ divide each of the segments $\overline{RZ}$, $\overline{GQ}$, $\overline{WS}$, $\overline{HP}$ into three equal parts.
\spos{1} Since $RE = EU$, $TE = ES$, it follows from the converse of Thales' theorem that $RS \parallel TU$. Therefore, triangles $ERS$ and $ETU$ are congruent. Hence $RS = TU$.
In proving the lemma, we showed that $RS \parallel DB$ and $RS = \frac{1}{3} DB$. Similarly, $PQ \parallel DB$ and $PQ = \frac{1}{3} DB$. We have then $TU \parallel RS \parallel DB \parallel PQ$ and
\[
RS = TU = PQ = \frac{1}{3} DB.
\]
By drawing the heights in triangles
\[
\triangle ARS, \triangle ETF, \triangle FPC
\]
to the bases $\overline{RS}$, $\overline{TU}$, $\overline{PQ}$, respectively, we see that the sum of the lengths of these heights is equal to the sum of the lengths of the heights $h$ and $h$ of triangles $ABD$ and $CBD$ drawn to the base $\overline{DB}$. Since the lengths of the bases of triangles (4) are equal (by (3) they are equal to $\frac{1}{3} DB$), the sum of their areas, i.e., the sum of the areas of quadrilaterals $ARSE$, $ETFU$, $FPCQ$ is equal to half the product of the number $\frac{1}{3} DB$ and the sum of their heights, i.e., the number
\[
\frac{1}{3} DB \cdot \frac{1}{2} (h + h) = \frac{1}{6} DB \cdot (h + h) = \frac{1}{6} S_{ABCD}.
\]
Note. An analogous theorem is true in the case of dividing the sides of the quadrilateral $ABCD$ into $n$ equal parts, where $n > 3$. Then the sum of the areas of the quadrilaterals located on the diagonal is equal to $\frac{1}{n} S_{ABCD}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 557 |
VII OM - II - Problem 4
Prove that the equation $ 2x^2 - 215y^2 = 1 $ has no solutions in integers. | Let $ x $ and $ y $ denote integers. The number $ 215y^2 $ is divisible by $ 5 $, so the number $ 215y^2 + 1 $ gives a remainder of $ 1 $ when divided by $ 5 $. The number $ x^2 $ has one of the forms $ 5k $, $ 5k + 1 $, $ 5k + 4 $ ($ k $ - an integer, see problem 2), so the number $ 2x^2 $ has one of the forms $ 10k $, $ 10k + 2 $, $ 10k + 8 = 10k + 5 + 3 $, and thus gives one of the remainders $ 0 $, $ 2 $, or $ 3 $ when divided by $ 5 $. Therefore, the equation $ 2x^2 = 215y^2 + 1 $ cannot hold, i.e., the given equation has no integer solutions.
Note. The solution to the problem can be stated more concisely if we replace the given equation with an equivalent equation $ (2x)^2 = 430y^2 + 2 $ and consider the remainders when divided by $ 10 $, i.e., the last digits of the terms of this equation when $ x $ and $ y $ are integers. The last digit of the square of an even number $ 2x $ can only be $ 0 $, $ 4 $, or $ 6 $, and the last digit of the number $ 430y^2 $ is $ 0 $, so the last digit of the number $ 430y^2 + 2 $ is $ 2 $. Therefore, the given equation cannot have integer solutions $ (x, y) $. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 559 |
XXXV OM - I - Problem 10
In the plane, there are $3n$ points, among which no three points are collinear. Prove that there exist $n$ disjoint triangles with vertices at the given points. | Consider all lines, each of which passes through two points of a given set of $3n$ points. There are finitely many such lines, so there exists a line $l$ that is not perpendicular to any of them. Project all the given points perpendicularly onto the line $l$, ensuring that the projections of any two points do not coincide. Number the given points $A_1, \ldots, A_{3n}$ according to the order in which their projections are located on the line $l$. The triangles $A_{3i-2} A_{3i-1} A_{3i}$ for $i = 1, \ldots, n$ are disjoint. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 562 |
LIII OM - II - Problem 3
In an $ n $-person association, there are six committees. Each of them includes at least $ n/4 $ people. Prove that there exist two committees and a group of at least $ n/30 $ people, who belong to both of these committees. | Let's number the committees by $1, 2, \ldots, 6$ and denote by $K_i$ the number of members of the $i$-th committee who are not members of any committee with a number less than $i$.
$K_1$ is the number of all members of the first committee, which gives $K_1 \geq n/4$.
Assume that the intersection of any two committees has fewer than $n/30$ members.
Since the second committee includes at least $n/4$ members, the number of members of the second committee who are not in the first committee must be greater than $n/4 - n/30$. Therefore,
\[
K_2 > \frac{n}{4} - \frac{n}{30} = \frac{7n}{30}
\]
Similarly, we conclude that
\[
K_3 > \frac{n}{4} - \frac{n}{30} = \frac{7n}{30}
\]
\[
K_4 > \frac{n}{4} - \frac{n}{30} = \frac{7n}{30}
\]
\[
K_5 > \frac{n}{4} - \frac{n}{30} = \frac{7n}{30}
\]
\[
K_6 > \frac{n}{4} - \frac{n}{30} = \frac{7n}{30}
\]
Adding these inequalities, we get
\[
K_1 + K_2 + K_3 + K_4 + K_5 + K_6 > \frac{n}{4} + 5 \cdot \frac{7n}{30} = \frac{n}{4} + \frac{35n}{30} = \frac{n}{4} + \frac{7n}{6} = \frac{3n + 14n}{12} = \frac{17n}{12}
\]
We have obtained a contradiction, since the quantity $K_1 + K_2 + \ldots + K_6$ is not greater than the total number of members of the association, which is $n$.
Thus, the intersection of some two committees must contain at least $n/30$ members. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 566 |
V OM - II - Task 6
Prove that if $ x_1, x_2, \ldots, x_n $ are angles contained between $ 0^\circ $ and $ 180^\circ $, and $ n $ is any natural number greater than $ 1 $, then | We will prove a "stronger" theorem, namely, we will show that if $0^\circ < x_i < 180^\circ$ for $i = 1, 2, \ldots, n$, where $n \geq 2$, then
(If $|a| < b$, then $a < b$ (but not vice versa), so inequality (2) implies inequality (1)).
In the proof, we will use known properties of the absolute value:
as well as the fact that if $0^\circ < x < 180^\circ$, then $|\sin x| = \sin x > 0$, $|\cos x| < 1$, and for any $x$, $|\cos x| \leq 1$.
We will prove this by using the principle of mathematical induction. When $n = 2$, the theorem is true, since
so
Suppose that for some $k \geq 2$,
and let $0 < x_{k+1} < 180^\circ$. Then
By the principle of mathematical induction, we conclude that theorem (2) is true for all $n \geq 2$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 569 |
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