exam: BalticWay

problem: Let $a \circ b=a+b-a b$. Find all triples $(x, y, z)$ of integers such that $(x \circ y) \circ z+(y \circ z) \circ x+(z \circ x) \circ y=0$.

solution:
Note that

$$
(x \circ y) \circ z=x+y+z-x y-y z-x z+x y z=(x-1)(y-1)(z-1)+1 .
$$

Hence

$$
(x \circ y) \circ z+(y \circ z) \circ x+(z \circ x) \circ y=3((x-1)(y-1)(z-1)+1) \text {. }
$$

Now, if the required equality holds we have $(x-1)(y-1)(z-1)=-1$. There are only four possible decompositions of -1 into a product of three integers. Thus we have four such triples, namely $(0,0,0)$, $(0,2,2),(2,0,2)$ and $(2,2,0)$.

solution_match: Solution.


problem: Let $a_{1}, a_{2}, \ldots, a_{9}$ be any non-negative numbers such that $a_{1}=a_{9}=0$ and at least one of the numbers is non-zero. Prove that for some $i, 2 \leq i \leq 8$, the inequality $a_{i-1}+a_{i+1}<2 a_{i}$ holds. Will the statement remain true if we change the number 2 in the last inequality to $1.9 ?$

solution:
Suppose we have the opposite inequality $a_{i-1}+a_{i+1} \geq 2 a_{i}$ for all $i=2, \ldots, 8$. Let $a_{k}=\max _{1 \leq i \leq 9} a_{i}$. Then we have $a_{k-1}=a_{k+1}=a_{k}, a_{k-2}=a_{k-1}=a_{k}$, etc. Finally we get $a_{1}=a_{k}$, a contradiction.

Suppose now $a_{i-1}+a_{i+1} \geq 1.9 a_{i}$, i.e., $a_{i+1} \geq 1.9 a_{i}-a_{i-1}$ for all $i=2, \ldots, 8$, and let $a_{k}=\max _{1 \leq i \leq 9} a_{i}$. We can multiply all numbers $a_{1}, \ldots, a_{9}$ by the same positive constant without changing the situation in any way, so we assume $a_{k}=1$. Then we have $a_{k-1}+a_{k+1} \geq 1.9$ and hence $0.9 \leq a_{k-1}, a_{k+1} \leq 1$. Moreover, at least one of the numbers $a_{k-1}, a_{k+1}$ must be greater than or equal to 0.95 - let us assume $a_{k+1} \geq 0.95$. Now, we consider two sub-cases:

(a) $k \geq 5$. Then we have

$$
\begin{aligned}
1 \geq a_{k+1} & \geq 0.95>0 \\
1 \geq a_{k+2} & \geq 1.9 a_{k+1}-a_{k} \geq 1.9 \cdot 0.95-1=0.805>0 \\
a_{k+3} & \geq 1.9 a_{k+2}-a_{k+1} \geq 1.9 \cdot 0.805-1=0.5295>0 \\
a_{k+4} & \geq 1.9 a_{k+3}-a_{k+2} \geq 1.9 \cdot 0.5295-1=0.00605>0
\end{aligned}
$$

So in any case we have $a_{9}>0$, a contradiction.

(b) $k \leq 4$. In this case we obtain

$$
\begin{aligned}
1 \geq a_{k-1} & \geq 0.9>0 \\
a_{k-2} & \geq 1.9 a_{k-1}-a_{k} \geq 1.9 \cdot 0.9-1=0.71>0 \\
a_{k-3} & \geq 1.9 a_{k-2}-a_{k-1} \geq 1.9 \cdot 0.71-1=0.349>0
\end{aligned}
$$

and hence $a_{1}>0$, contrary to the condition of the problem.


problem:
Find the largest value of the expression

$$
x y+x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}-\sqrt{\left(1-x^{2}\right)\left(1-y^{2}\right)}
$$

solution:
The expression is well-defined only for $|x|,|y| \leq 1$ and we can assume that $x, y \geq 0$. Let $x=\cos \alpha$ and $y=\cos \beta$ for some $0 \leq \alpha, \beta \leq \frac{\pi}{2}$. This reduces the expression to

$$
\cos \alpha \cos \beta+\cos \alpha \sin \beta+\cos \beta \sin \alpha-\sin \alpha \sin \beta=\cos (\alpha+\beta)+\sin (\alpha+\beta)=\sqrt{2} \cdot \sin \left(\alpha+\beta+\frac{\pi}{4}\right)
$$

which does not exceed $\sqrt{2}$. The equality holds when $\alpha+\beta+\frac{\pi}{4}=\frac{\pi}{2}$, for example when $\alpha=\frac{\pi}{4}$ and $\beta=0$, i.e., $x=\frac{\sqrt{2}}{2}$ and $y=1$.


problem: Is there an integer $n$ such that $\sqrt{n-1}+\sqrt{n+1}$ is a rational number?

solution:
Inverting the relation gives

$$
\frac{q}{p}=\frac{1}{\sqrt{n+1}+\sqrt{n-1}}=\frac{\sqrt{n+1}-\sqrt{n-1}}{(\sqrt{n+1}+\sqrt{n-1})(\sqrt{n+1}-\sqrt{n-1})}=\frac{\sqrt{n+1}-\sqrt{n-1}}{2} .
$$

Hence we get the system of equations

$$
\left\{\begin{array}{l}
\sqrt{n+1}+\sqrt{n-1}=\frac{p}{q} \\
\sqrt{n+1}-\sqrt{n-1}=\frac{2 q}{p}
\end{array}\right.
$$

Adding these equations and dividing by 2 gives $\sqrt{n+1}=\frac{2 q^{2}+p^{2}}{2 p q}$. This implies $4 n p^{2} q^{2}=4 q^{4}+p^{4}$.

Suppose now that $n, p$ and $q$ are all positive integers with $p$ and $q$ relatively prime. The relation $4 n p^{2} q^{2}=$ $4 q^{4}+p^{4}$ shows that $p^{4}$, and hence $p$, is divisible by 2 . Letting $p=2 P$ we obtain $4 n P^{2} q^{2}=q^{4}+4 P^{4}$ which shows that $q$ must also be divisible by 2 . This contradicts the assumption that $p$ and $q$ are relatively prime.


problem: Let $p(x)$ be a polynomial with integer coefficients such that both equations $p(x)=1$ and $p(x)=3$ have integer solutions. Can the equation $p(x)=2$ have two different integer solutions?

solution: Observe first that if $a$ and $b$ are two different integers then $p(a)-p(b)$ is divisible by $a-b$. Suppose now that $p(a)=1$ and $p(b)=3$ for some integers $a$ and $b$. If we have $p(c)=2$ for some integer $c$, then $c-b= \pm 1$ and $c-a= \pm 1$, hence there can be at most one such integer $c$.


problem: Prove that any irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are positive integers and $q$ is odd, is equal to a fraction $\frac{n}{2^{k}-1}$ for some positive integers $n$ and $k$.

solution: Since the number of congruence classes modulo $q$ is finite, there exist two non-negative integers $i$ and $j$ with $i>j$ which satisfy $2^{i} \equiv 2^{j}(\bmod q)$. Hence, $q$ divides the number $2^{i}-2^{j}=2^{j}\left(2^{i-j}-1\right)$. Since $q$ is odd, $q$ has to divide $2^{i-j}-1$. Now it suffices to multiply the numerator and denominator of the fraction $\frac{p}{q}$ by $\frac{2^{i-j}-1}{q}$.


problem: Let $p>2$ be a prime number and $1+\frac{1}{2^{3}}+\frac{1}{3^{3}}+\cdots+\frac{1}{(p-1)^{3}}=\frac{m}{n}$ where $m$ and $n$ are relatively prime. Show that $m$ is a multiple of $p$.

solution:
The sum has an even number of terms; they can be joined in pairs in such a way that the sum is the sum of the terms

$$
\frac{1}{k^{3}}+\frac{1}{(p-k)^{3}}=\frac{p^{3}-3 p^{2} k+3 p k^{2}}{k^{3}(p-k)^{3}} .
$$

The sum of all terms of this type has a denominator in which every prime factor is less than $p$ while the numerator has $p$ as a factor.


problem: Show that for any integer $a \geq 5$ there exist integers $b$ and $c, c \geq b \geq a$, such that $a, b, c$ are the lengths of the sides of a right-angled triangle.

solution:
We first show this for odd numbers $a=2 i+1 \geq 3$. Put $c=2 k+1$ and $b=2 k$. Then $c^{2}-b^{2}=(2 k+1)^{2}-(2 k)^{2}=4 k+1=a^{2}$. Now $a=2 i+1$ and thus $a^{2}=4 i^{2}+4 i+1$ and $k=i^{2}+i$. Furthermore, $c>b=2 i^{2}+2 i>2 i+1=a$.

Since any multiple of a Pythagorean triple (i.e., a triple of integers $(x, y, z)$ such that $x^{2}+y^{2}=z^{2}$ ) is also a Pythagorean triple we see that the statement is also true for all even numbers which have an odd factor. Hence only the powers of 2 remain. But for 8 we have the triple $(8,15,17)$ and hence all higher powers of 2 are also minimum values of such a triple.


problem: Find all pairs of positive integers $(a, b)$ such that $2^{a}+3^{b}$ is the square of an integer.

solution:
Considering the equality $2^{a}+3^{b}=n^{2}$ modulo 3 it is easy to see that $a$ must be even. Obviously $n$ is odd so we may take $a=2 x, n=2 y+1$ and write the equality as $4^{x}+3^{b}=(2 y+1)^{2}=4 y^{2}+4 y+1$. Hence $3^{b} \equiv 1(\bmod 4)$ which implies $b=2 z$ for some positive integer $z$. So we get $4^{x}+9^{z}=(2 y+1)^{2}$ and $4^{x}=\left(2 y+1-3^{z}\right)\left(2 y+1+3^{z}\right)$. Both factors on the right-hand side are even numbers but at most one of them is divisible by 4 (since their sum is not divisible by 4 ). Hence $2 y+1-3^{z}=2$ and $2 y+1+3^{z}=2^{2 x-1}$. These two equalities yield $2 \cdot 3^{z}=2^{2 x-1}-2$ and $3^{z}=4^{x-1}-1$. Clearly $x>1$ and a simple argument
modulo 10 gives $z=4 d+1, x-1=2 e+1$ for some non-negative integers $d$ and $e$. Substituting, we get $3^{4 d+1}=4^{2 e+1}-1$ and $3 \cdot(80+1)^{d}=4^{2 e+1}-1$. If $d \geq 1$ then $e \geq 1$, a contradiction (expanding the left-hand expression and moving everything to the left we find that all summands but one are divisible by $4^{2}$ ). Hence $e=d=0, z=1, b=2, x=2$ and $a=4$, and we obtain the classical $2^{4}+3^{2}=4^{2}+3^{2}=5^{2}$.


problem:
How many positive integers satisfy the following three conditions:

(i) All digits of the number are from the set $\{1,2,3,4,5\}$;

(ii) The absolute value of the difference between any two consecutive digits is 1 ;

(iii) The integer has 1994 digits?

solution:
Consider all positive integers with $2 n$ digits satisfying conditions $(i)$ and (ii) of the problem. Let the number of such integers beginning with $1,2,3,4$ and 5 be $a_{n}, b_{n}, c_{n}, d_{n}$ and $e_{n}$, respectively. Then, for $n=1$ we have $a_{1}=1$ (integer 12), $b_{1}=2$ (integers 21 and 23), $c_{1}=2$ (integers 32 and 34$), d_{1}=2$ (integers 43 and 45) and $e_{1}=1$ (integer 54). Observe that $c_{1}=a_{1}+e_{1}$.

Suppose now that $n>1$, i.e., the integers have at least four digits. If an integer begins with the digit 1 then the second digit is 2 while the third can be 1 or 3 . This gives the relation

$$
a_{n}=a_{n-1}+c_{n-1} \text {. }
$$

Similarly, if the first digit is 5 , then the second is 4 while the third can be 3 or 5 . This implies

$$
e_{n}=c_{n-1}+e_{n-1}
$$

If the integer begins with 23 then the third digit is 2 or 4 . If the integer begins with 21 then the third digit is 2 . From this we can conclude that

$$
b_{n}=2 b_{n-1}+d_{n-1} \text {. }
$$

In the same manner we can show that

$$
d_{n}=b_{n-1}+2 d_{n-1} .
$$

If the integer begins with 32 then the third digit must be 1 or 3 , and if it begins with 34 the third digit is 3 or 5 . Hence

$$
c_{n}=a_{n-1}+2 c_{n-1}+e_{n-1} \text {. }
$$

From (1), (2) and (5) it follows that $c_{n}=a_{n}+e_{n}$, which is true for all $n \geq 1$. On the other hand, adding the relations (1)-(5) results in

$$
a_{n}+b_{n}+c_{n}+d_{n}+e_{n}=2 a_{n-1}+3 b_{n-1}+4 c_{n-1}+3 d_{n-1}+2 e_{n-1}
$$

and, since $c_{n-1}=a_{n-1}+e_{n-1}$,

$$
a_{n}+b_{n}+c_{n}+d_{n}+e_{n}=3\left(a_{n-1}+b_{n-1}+c_{n-1}+d_{n-1}+e_{n-1}\right)
$$

Thus the number of integers satisfying conditions $(i)$ and $(i i)$ increases three times when we increase the number of digits by 2 . Since the number of such integers with two digits is 8 , and $1994=2+2 \cdot 996$, the number of integers satisfying all three conditions is $8 \cdot 3^{996}$.


problem: Let $N S$ and $E W$ be two perpendicular diameters of a circle $\mathcal{C}$. A line $l$ touches $\mathcal{C}$ at point $S$. Let $A$ and $B$ be two points on $\mathcal{C}$, symmetric with respect to the diameter $E W$. Denote the intersection points of $l$ with the lines $N A$ and $N B$ by $A^{\prime}$ and $B^{\prime}$, respectively. Show that $\left|S A^{\prime}\right| \cdot\left|S B^{\prime}\right|=|S N|^{2}$.

solution:
We have $\angle N A S=\angle N B S=90^{\circ}$ (see Figure 1). Thus, the triangles $N A^{\prime} S$ and $N S A$ are similar. Also, the triangles $B^{\prime} N S$ and $S N B$ are similar and the triangles $N S A$ and $S N B$ are congruent. Hence, the triangles $N A^{\prime} S$ and $B^{\prime} N S$ are similar which implies $\frac{S A^{\prime}}{S N}=\frac{S N}{S B^{\prime}}$ and $S A^{\prime} \cdot S B^{\prime}=S N^{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_bbd51418f58e9fd65d6eg-4.jpg?height=457&width=1079&top_left_y=228&top_left_x=500)

Figure 1


problem: The inscribed circle of the triangle $A_{1} A_{2} A_{3}$ touches the sides $A_{2} A_{3}, A_{3} A_{1}$ and $A_{1} A_{2}$ at points $S_{1}, S_{2}, S_{3}$, respectively. Let $O_{1}, O_{2}, O_{3}$ be the centres of the inscribed circles of triangles $A_{1} S_{2} S_{3}, A_{2} S_{3} S_{1}$ and $A_{3} S_{1} S_{2}$, respectively. Prove that the straight lines $O_{1} S_{1}, O_{2} S_{2}$ and $O_{3} S_{3}$ intersect at one point.

solution:
We shall prove that the lines $S_{1} O_{1}, S_{2} O_{2}, S_{3} O_{3}$ are the bisectors of the angles of the triangle $S_{1} S_{2} S_{3}$. Let $O$ and $r$ be the centre and radius of the inscribed circle $C$ of the triangle $A_{1} A_{2} A_{3}$. Further, let $P_{1}$ and $H_{1}$ be the points where the inscribed circle of the triangle $A_{1} S_{2} S_{3}$ (with the centre $O_{1}$ and radius $r_{1}$ ) touches its sides $A_{1} S_{2}$ and $S_{2} S_{3}$, respectively (see Figure 2). To show that $S_{1} O_{1}$ is the bisector of the angle $\angle S_{3} S_{1} S_{2}$ it is sufficient to prove that $O_{1}$ lies on the circumference of circle $C$, for in this case the arcs $O_{1} S_{2}$ and $O_{1} S_{3}$ will obviously be equal. To prove this, first note that as $A_{1} S_{2} S_{3}$ is an isosceles triangle the point $H_{1}$, as well as $O_{1}$, lies on the straight line $A_{1} O$. Now, it suffices to show that $\left|O H_{1}\right|=r-r_{1}$. Indeed, we have

$$
\begin{aligned}
& \frac{r-r_{1}}{r}=1-\frac{r_{1}}{r}=1-\frac{\left|O_{1} P_{1}\right|}{\left|O S_{2}\right|}=1-\frac{\left|P_{1} A_{1}\right|}{\left|S_{2} A_{1}\right|}=\frac{\left|S_{2} A_{1}\right|-\left|P_{1} A_{1}\right|}{\left|S_{2} A_{1}\right|} \\
& =\frac{\left|S_{2} P_{1}\right|}{\left|S_{2} A_{1}\right|}=\frac{\left|S_{2} H_{1}\right|}{\left|S_{2} A_{1}\right|}=\frac{\left|O H_{1}\right|}{\left|O S_{2}\right|}=\frac{\left|O H_{1}\right|}{r} .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_bbd51418f58e9fd65d6eg-4.jpg?height=391&width=671&top_left_y=1489&top_left_x=744)

Figure 2


problem: Find the smallest number $a$ such that a square of side $a$ can contain five disks of radius 1 so that no two of the disks have a common interior point.

solution: Let $P Q R S$ be a square which has the property described in the problem. Clearly, $a>2$. Let $P^{\prime} Q^{\prime} R^{\prime} S^{\prime}$ be the square inside $P Q R S$ whose sides are at distance 1 from the sides of $P Q R S$, and, consequently, are of length $a-2$. Since all the five disks are inside $P Q R S$, their centres are inside $P^{\prime} Q^{\prime} R^{\prime} S^{\prime}$. Divide $P^{\prime} Q^{\prime} R^{\prime} S^{\prime}$ into four congruent squares of side length $\frac{a}{2}-1$. By the pigeonhole principle, at least two of the five centres are in the same small square. Their distance, then, is at most $\sqrt{2}\left(\frac{a}{2}-1\right)$. Since the distance has to be at least 2 , we have $a \geq 2+2 \sqrt{2}$. On the other hand, if $a=2+2 \sqrt{2}$, we can place the five disks in such a way that one is centred at the centre of $P Q R S$ and the other four have centres at $P^{\prime}, Q^{\prime}$, $R^{\prime}$ and $S^{\prime}$.


problem:
Let $\alpha, \beta, \gamma$ be the angles of a triangle opposite to its sides with lengths $a, b$ and $c$, respectively. Prove the inequality

$$
a \cdot\left(\frac{1}{\beta}+\frac{1}{\gamma}\right)+b \cdot\left(\frac{1}{\gamma}+\frac{1}{\alpha}\right)+c \cdot\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \geq 2 \cdot\left(\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\right) \cdot
$$

solution:
Clearly, the inequality $a>b$ implies $\alpha>\beta$ and similarly $a<b$ implies $\alpha<\beta$, hence $(a-b)(\alpha-\beta) \geq$ 0 and $a \alpha+b \beta \geq a \beta+b \alpha$. Dividing the last equality by $\alpha \beta$ we get

$$
\frac{a}{\beta}+\frac{b}{\alpha} \geq \frac{a}{\alpha}+\frac{b}{\beta}
$$

Similarly we get

$$
\frac{a}{\gamma}+\frac{c}{\alpha} \geq \frac{a}{\alpha}+\frac{c}{\gamma}
$$

and

$$
\frac{b}{\gamma}+\frac{c}{\beta} \geq \frac{b}{\beta}+\frac{c}{\gamma}
$$

To finish the proof it suffices to add the inequalities (6)-(8).


problem: Does there exist a triangle such that the lengths of all its sides and altitudes are integers and its perimeter is equal to 1995 ?

solution:
Consider a triangle $A B C$ with all its sides and heights having integer lengths. From the cosine theorem we conclude that $\cos \angle A, \cos \angle B$ and $\cos \angle C$ are rational numbers. Let $A H$ be one of the heights of the triangle $A B C$, with the point $H$ lying on the straight line determined by the side $B C$. Then $|B H|$ and $|C H|$ must be rational and hence integer (consider the Pythagorean theorem for the triangles $A B H$ and $A C H$ ). Now, if $|B H|$ and $|C H|$ have different parity then $|A B|$ and $|A C|$ also have different parity and $|B C|$ is odd. If $|B H|$ and $|C H|$ have the same parity then $|A B|$ and $|A C|$ also have the same parity and $|B C|$ is even. In both cases the perimeter of triangle $A B C$ is an even number and hence cannot be equal to 1995

Remark. In the solution we only used the fact that all three sides and one height of the triangle $A B C$ are integers.

![](https://cdn.mathpix.com/cropped/2024_04_17_bbd51418f58e9fd65d6eg-5.jpg?height=249&width=380&top_left_y=1446&top_left_x=884)

Figure 3


problem: The Wonder Island is inhabited by Hedgehogs. Each Hedgehog consists of three segments of unit length having a common endpoint, with all three angles between them equal to $120^{\circ}$ (see Figure 3). Given that all Hedgehogs are lying flat on the island and no two of them touch each other, prove that there is a finite number of Hedgehogs on Wonder Island.

solution:
It suffices to prove that if the distance between the centres of two Hedgehogs is less than 0.2, then these Hedgehogs intersect. To show this, consider two Hedgehogs with their centres at points $O$ and $M$, respectively, such that $|O M|<0.2$. Let $A, B$ and $C$ be the endpoints of the needles of the first Hedgehog (see Figure 4) and draw a straight line $l$ parallel to $A C$ through the point $M$. As $|A C|=\sqrt{3}$ implies $|K L| \leq$ $\frac{0.2}{0.5}|A C|<1$ and the second Hedgehog has at least one of its needles pointing inside the triangle $O K L$, this needle intersects the first Hedgehog.

Remark. If the Hedgehogs can move their needles so that the angles between them can take any positive value then there can be an infinite number of Hedgehogs on the Wonder Island.

![](https://cdn.mathpix.com/cropped/2024_04_17_bbd51418f58e9fd65d6eg-6.jpg?height=405&width=437&top_left_y=206&top_left_x=867)

Figure 4


problem: In a certain kingdom, the king has decided to build 25 new towns on 13 uninhabited islands so that on each island there will be at least one town. Direct ferry connections will be established between any pair of new towns which are on different islands. Determine the least possible number of these connections.

solution: Let $a_{1}, \ldots, a_{13}$ be the numbers of towns on each island. Suppose there exist numbers $i$ and $j$ such that $a_{i} \geq a_{j}>1$ and consider an arbitrary town $A$ on the $j$-th island. The number of ferry connections from town $A$ is equal to $25-a_{j}$. On the other hand, if we "move" town $A$ to the $i$-th island then there will be $25-\left(a_{i}+1\right)$ connections from town $A$ while no other connections will be affected by this move. Hence, the smallest number of connections will be achieved if there are 13 towns on one island and one town on each of the other 12 islands. In this case there will be $13 \cdot 12+\frac{12 \cdot 11}{2}=222$ connections.


problem: There are $n$ lines $(n>2)$ given in the plane. No two of the lines are parallel and no three of them intersect at one point. Every point of intersection of these lines is labelled with a natural number between 1 and $n-1$. Prove that, if and only if $n$ is even, it is possible to assign the labels in such a way that every line has all the numbers from 1 to $n-1$ at its points of intersection with the other $n-1$ lines.

solution:
Suppose we have assigned the labels in the required manner. When a point has label 1 then there can be no more occurrences of label 1 on the two lines that intersect at that point. Therefore the number of intersection points labelled with 1 has to be exactly $\frac{n}{2}$, and so $n$ must be even. Now, let $n$ be an even number and denote the $n$ lines by $l_{1}, l_{2}, \ldots, l_{n}$. First write the lines $l_{i}$ in the following table:

$$
\begin{array}{llllll}
& & l_{3} & l_{4} & \ldots & l_{n / 2+1} \\
l_{1} & l_{2} & & & & \\
& & l_{n} & l_{n-1} & \ldots & l_{n / 2+2}
\end{array}
$$

and then rotate the picture $n-1$ times:

$$
\begin{array}{llllll}
& & l_{2} & l_{3} & \ldots & l_{n / 2} \\
l_{1} & l_{n} & l_{n-1} & l_{n-2} & \ldots & l_{n / 2+1} \\
& & & & & \\
& & l_{n} & l_{2} & \ldots & l_{n / 2-1} \\
l_{1} & l_{n-1} & & & & l_{n / 2}
\end{array}
$$

etc.

According to these tables, we can join the lines in pairs in $n-1$ different ways $-l_{1}$ with the line next to it and every other line with the line directly above or under it. Now we can assign the label $i$ to all the intersection points of the pairs of lines shown in the $i$ th table.


problem: The Wonder Island Intelligence Service has 16 spies in Tartu. Each of them watches on some of his colleagues. It is known that if spy $A$ watches on spy $B$ then $B$ does not watch on $A$. Moreover, any 10 spies can be numbered in such a way that the first spy watches on the second, the second watches on the third, .., the tenth watches on the first. Prove that any 11 spies can also be numbered in a similar manner.

solution:
We call two spies $A$ and $B$ neutral to each other if neither $A$ watches on $B$ nor $B$ watches on $A$.

Denote the spies $A_{1}, A_{2}, \ldots, A_{16}$. Let $a_{i}, b_{i}$ and $c_{i}$ denote the number of spies that watch on $A_{i}$, the number of that are watched by $A_{i}$ and the number of spies neutral to $A_{i}$, respectively. Clearly, we have

$$
\begin{aligned}
a_{i}+b_{i}+c_{i} & =15, \\
a_{i}+c_{i} & \leq 8, \\
b_{i}+c_{i} & \leq 8
\end{aligned}
$$

for any $i=1, \ldots, 16$ (if any of the last two inequalities does not hold then there exist 10 spies who cannot be numbered in the required manner). Combining the relations above we find $c_{i} \leq 1$. Hence, for any spy, the number of his neutral colleagues is 0 or 1 .

Now suppose there is a group of 11 spies that cannot be numbered as required. Let $B$ be an arbitrary spy in this group. Number the other 10 spies as $C_{1}, C_{2}, \ldots, C_{10}$ so that $C_{1}$ watches on $C_{2}, \ldots, C_{10}$ watches on $C_{1}$. Suppose there is no spy neutral to $B$ among $C_{1}, \ldots, C_{10}$. Then, if $C_{1}$ watches on $B$ then $B$ cannot watch on $C_{2}$, as otherwise $C_{1}, B, C_{2}, \ldots, C_{10}$ would form an 11-cycle. So $C_{2}$ watches on $B$, etc. As some of the spies $C_{1}, C_{2}, \ldots, C_{10}$ must watch on $B$ we get all of them watching on $B$, a contradiction. Therefore, each of the 11 spies must have exactly one spy neutral to him among the other 10 - but this is impossible.


problem: An equilateral triangle is divided into 9000000 congruent equilateral triangles by lines parallel to its sides. Each vertex of the small triangles is coloured in one of three colours. Prove that there exist three points of the same colour being the vertices of a triangle with its sides parallel to the sides of the original triangle.

solution:
Consider the side $A B$ of the big triangle $A B C$ as "horizontal" and suppose the statement of the problem does not hold. The side $A B$ contains 3001 vertices $A=A_{0}, A_{1}, \ldots, A_{3000}=B$ of 3 colours. Hence, there are at least 1001 vertices of one colour, e.g., red. For any two red vertices $A_{k}$ and $A_{n}$ there exists a unique vertex $B_{k n}$ such that the triangle $B_{k n} A_{k} A_{n}$ is equilateral. That vertex $B_{k n}$ cannot be red. For different pairs $(k, n)$ the corresponding vertices $B_{k n}$ are different, so we have at least $\left(\begin{array}{c}1001 \\ 2\end{array}\right)>500000$ vertices of type $B_{k n}$ that cannot be red. As all these vertices are situated on 3000 horizontal lines, there exists a line $L$ which contains more than 160 vertices of type $B_{k n}$, each of them coloured in one of the two remaining colours. Hence there exist at least 81 vertices of the same colour, e.g., blue, on line $L$. For every two blue vertices $B_{k n}$ and $B_{m l}$ on line $L$ there exists a unique vertex $C_{k n m l}$ such that:

(i) $C_{k n m l}$ lies above the line $L$;

(ii) The triangle $C_{k n m l} B_{k n} B_{m l}$ is equilateral;

(iii) $C_{k n m l}=B_{p q}$ where $p=\min (k, m)$ and $q=\max (n, l)$.

Different pairs of vertices $B_{k n}$ belonging to line $L$ define different vertices $C_{k n m l}$. So we have at least $\left(\begin{array}{c}81 \\ 2\end{array}\right)>3200$ vertices of type $C_{k n m l}$ that can be neither blue nor red. As the number of these vertices exceeds the number of horizontal lines, there must be two vertices $C_{k n m l}$ and $C_{p q r s}$ on one horizontal line. Now, these two vertices define a new vertex $D_{\text {knmlpqrs }}$ that cannot have any of the three colours, a contradiction.

Remark. The minimal size of the big triangle that can be handled by this proof is 2557 .