di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Combinatorics


Question: Queenie and Horst play a game on a $20 \times 20$ chessboard. In the beginning the board is empty. In every turn, Horst places a black knight on an empty square in such a way that his new knight does not attack any previous knights. Then Queenie places a white queen on an empty square. The game gets finished when somebody cannot move.

Find the maximal positive $K$ such that, regardless of the strategy of Queenie, Horst can put at least $K$ knights on the board.

Solution:
We show two strategies, one for Horst to place at least 100 knights, and another strategy for Queenie that prevents Horst from putting more than 100 knights on the board.

A strategy for Horst: Put knights only on black squares, until all black squares get occupied.

Colour the squares of the board black and white in the usual way, such that the white and black squares alternate, and let Horst put his knights on black squares as long as it is possible. Two knights on squares of the same colour never attack each other. The number of black squares is $20^{2} / 2=200$. The two players occupy the squares in turn, so Horst will surely find empty black squares in his first 100 steps.

A strategy for Queenie: Group the squares into cycles of length 4, and after each step of Horst, occupy the opposite square in the same cycle.

Consider the squares of the board as vertices of a graph; let two squares be connected if two knights on those squares would attack each other. Notice that in a $4 \times 4$ board the squares can be grouped into 4 cycles of length 4 , as shown in Figure 1. Divide the board into parts of size $4 \times 4$, and perform the same grouping in every part; this way we arrange the 400 squares of the board into 100 cycles (Figure 2).

<img_3696>

Figure 1

<img_4002>

Figure 2

<img_3138>

Figure 3

The strategy of Queenie can be as follows: Whenever Horst puts a new knight to a certain square $A$, which is part of some cycle $A-B-C-D-A$, let Queenie put her queen on the opposite square $C$ in that cycle (Figure 3). From this point, Horst cannot put any knight on $A$ or $C$ because those squares are already occupied, neither on $B$ or $D$ because those squares are attacked by the knight standing on $A$. Hence, Horst can put at most one knight on each cycle, that is at most 100 knights in total.

Final answer: 100
Subfield: Combinatorics


Question: Let $n$ be a given positive integer. Sisyphus performs a sequence of turns on a board consisting of $n+1$ squares in a row, numbered 0 to $n$ from left to right. Initially, $n$ stones are put into square 0, and the other squares are empty. At every turn, Sisyphus chooses any nonempty square, say with $k$ stones, takes one of those stones and moves it to the right by at most $k$ squares (the stone should stay within the board). Sisyphus' aim is to move all $n$ stones to square $n$.

Prove that Sisyphus cannot reach the aim in less than

$$
\left\lceil\frac{n}{1}\right\rceil+\left\lceil\frac{n}{2}\right\rceil+\left\lceil\frac{n}{3}\right\rceil+\cdots+\left\lceil\frac{n}{n}\right\rceil
$$

turns. (As usual, $\lceil x\rceil$ stands for the least integer not smaller than $x$.)

Solution:
The stones are indistinguishable, and all have the same origin and the same final position. So, at any turn we can prescribe which stone from the chosen square to move. We do it in the following manner. Number the stones from 1 to $n$. At any turn, after choosing a square, Sisyphus moves the stone with the largest number from this square.

This way, when stone $k$ is moved from some square, that square contains not more than $k$ stones (since all their numbers are at most $k$ ). Therefore, stone $k$ is moved by at most $k$ squares at each turn. Since the total shift of the stone is exactly $n$, at least $\lceil n / k\rceil$ moves of stone $k$ should have been made, for every $k=1,2, \ldots, n$.

By summing up over all $k=1,2, \ldots, n$, we get the required estimate.
Subfield: Combinatorics


Question: An anti-Pascal pyramid is a finite set of numbers, placed in a triangle-shaped array so that the first row of the array contains one number, the second row contains two numbers, the third row contains three numbers and so on; and, except for the numbers in the bottom row, each number equals the absolute value of the difference of the two numbers below it. For instance, the triangle below is an anti-Pascal pyramid with four rows, in which every integer from 1 to $1+2+3+4=10$ occurs exactly once:

<img_3201>

Prove that it is impossible to form an anti-Pascal pyramid with 2018 rows, using every integer from 1 to $1+2+\cdots+2018$ exactly once.

Solution:
Let $T$ be an anti-Pascal pyramid with $n$ rows, containing every integer from 1 to $1+2+\cdots+n$, and let $a_{1}$ be the topmost number in $T$ (Figure 1). The two numbers below $a_{1}$ are some $a_{2}$ and $b_{2}=a_{1}+a_{2}$, the two numbers below $b_{2}$ are some $a_{3}$ and $b_{3}=a_{1}+a_{2}+a_{3}$, and so on and so forth all the way down to the bottom row, where some $a_{n}$ and $b_{n}=a_{1}+a_{2}+\cdots+a_{n}$ are the two neighbours below $b_{n-1}=a_{1}+a_{2}+\cdots+a_{n-1}$. Since the $a_{k}$ are $n$ pairwise distinct positive integers whose sum does not exceed the largest number in $T$, which is $1+2+\cdots+n$, it follows that they form a permutation of $1,2, \ldots, n$.

<img_3838>

Figure 1

<img_4049>

Figure 2

Consider now (Figure 2) the two 'equilateral' subtriangles of $T$ whose bottom rows contain the numbers to the left, respectively right, of the pair $a_{n}, b_{n}$. (One of these subtriangles may very well be empty.) At least one of these subtriangles, say $T^{\prime}$, has side length $\ell \geqslant\lceil(n-2) / 2\rceil$. Since $T^{\prime}$ obeys the anti-Pascal rule, it contains $\ell$ pairwise distinct positive integers $a_{1}^{\prime}, a_{2}^{\prime}, \ldots, a_{\ell}^{\prime}$, where $a_{1}^{\prime}$ is at the apex, and $a_{k}^{\prime}$ and $b_{k}^{\prime}=a_{1}^{\prime}+a_{2}^{\prime}+\cdots+a_{k}^{\prime}$ are the two neighbours below $b_{k-1}^{\prime}$ for each $k=2,3 \ldots, \ell$. Since the $a_{k}$ all lie outside $T^{\prime}$, and they form a permutation of $1,2, \ldots, n$, the $a_{k}^{\prime}$ are all greater than $n$. Consequently,

$$
\begin{array}{r}
b_{\ell}^{\prime} \geqslant(n+1)+(n+2)+\cdots+(n+\ell)=\frac{\ell(2 n+\ell+1)}{2} \\
\geqslant \frac{1}{2} \cdot \frac{n-2}{2}\left(2 n+\frac{n-2}{2}+1\right)=\frac{5 n(n-2)}{8},
\end{array}
$$

which is greater than $1+2+\cdots+n=n(n+1) / 2$ for $n=2018$. A contradiction.

So it is not possible.
Subfield: Combinatorics


Question: An anti-Pascal pyramid is a finite set of numbers, placed in a triangle-shaped array so that the first row of the array contains one number, the second row contains two numbers, the third row contains three numbers and so on; and, except for the numbers in the bottom row, each number equals the absolute value of the difference of the two numbers below it. For instance, the triangle below is an anti-Pascal pyramid with four rows, in which every integer from 1 to $1+2+3+4=10$ occurs exactly once:

![](https://cdn.mathpix.com/cropped/2023_12_21_9a7a2c8e401502a85653g-1.jpg?height=208&width=274&top_left_y=553&top_left_x=911)

Prove that it is impossible to form an anti-Pascal pyramid with 2018 rows, using every integer from 1 to $1+2+\cdots+2018$ exactly once.

Solution:
Let $T$ be an anti-Pascal pyramid with $n$ rows, containing every integer from 1 to $1+2+\cdots+n$, and let $a_{1}$ be the topmost number in $T$ (Figure 1). The two numbers below $a_{1}$ are some $a_{2}$ and $b_{2}=a_{1}+a_{2}$, the two numbers below $b_{2}$ are some $a_{3}$ and $b_{3}=a_{1}+a_{2}+a_{3}$, and so on and so forth all the way down to the bottom row, where some $a_{n}$ and $b_{n}=a_{1}+a_{2}+\cdots+a_{n}$ are the two neighbours below $b_{n-1}=a_{1}+a_{2}+\cdots+a_{n-1}$. Since the $a_{k}$ are $n$ pairwise distinct positive integers whose sum does not exceed the largest number in $T$, which is $1+2+\cdots+n$, it follows that they form a permutation of $1,2, \ldots, n$.

![](https://cdn.mathpix.com/cropped/2023_12_21_9a7a2c8e401502a85653g-1.jpg?height=389&width=442&top_left_y=1413&top_left_x=450)

Figure 1

![](https://cdn.mathpix.com/cropped/2023_12_21_9a7a2c8e401502a85653g-1.jpg?height=389&width=459&top_left_y=1413&top_left_x=1164)

Figure 2

Consider now (Figure 2) the two 'equilateral' subtriangles of $T$ whose bottom rows contain the numbers to the left, respectively right, of the pair $a_{n}, b_{n}$. (One of these subtriangles may very well be empty.) At least one of these subtriangles, say $T^{\prime}$, has side length $\ell \geqslant\lceil(n-2) / 2\rceil$. Since $T^{\prime}$ obeys the anti-Pascal rule, it contains $\ell$ pairwise distinct positive integers $a_{1}^{\prime}, a_{2}^{\prime}, \ldots, a_{\ell}^{\prime}$, where $a_{1}^{\prime}$ is at the apex, and $a_{k}^{\prime}$ and $b_{k}^{\prime}=a_{1}^{\prime}+a_{2}^{\prime}+\cdots+a_{k}^{\prime}$ are the two neighbours below $b_{k-1}^{\prime}$ for each $k=2,3 \ldots, \ell$. Since the $a_{k}$ all lie outside $T^{\prime}$, and they form a permutation of $1,2, \ldots, n$, the $a_{k}^{\prime}$ are all greater than $n$. Consequently,

$$
\begin{array}{r}
b_{\ell}^{\prime} \geqslant(n+1)+(n+2)+\cdots+(n+\ell)=\frac{\ell(2 n+\ell+1)}{2} \\
\geqslant \frac{1}{2} \cdot \frac{n-2}{2}\left(2 n+\frac{n-2}{2}+1\right)=\frac{5 n(n-2)}{8},
\end{array}
$$

which is greater than $1+2+\cdots+n=n(n+1) / 2$ for $n=2018$. A contradiction.

So it is not possible.

Final answer: 证明题，略
Subfield: Geometry


Question: Let $A B C$ be a triangle with $A B=A C$, and let $M$ be the midpoint of $B C$. Let $P$ be a point such that $P B<P C$ and $P A$ is parallel to $B C$. Let $X$ and $Y$ be points on the lines $P B$ and $P C$, respectively, so that $B$ lies on the segment $P X, C$ lies on the segment $P Y$, and $\angle P X M=\angle P Y M$. Prove that the quadrilateral $A P X Y$ is cyclic.

Solution:
Since $A B=A C, A M$ is the perpendicular bisector of $B C$, hence $\angle P A M=$ $\angle A M C=90^{\circ}$.

<img_3626>

Now let $Z$ be the common point of $A M$ and the perpendicular through $Y$ to $P C$ (notice that $Z$ lies on to the ray $A M$ beyond $M$ ). We have $\angle P A Z=\angle P Y Z=90^{\circ}$. Thus the points $P, A, Y$, and $Z$ are concyclic.

Since $\angle C M Z=\angle C Y Z=90^{\circ}$, the quadrilateral $C Y Z M$ is cyclic, hence $\angle C Z M=$ $\angle C Y M$. By the condition in the statement, $\angle C Y M=\angle B X M$, and, by symmetry in $Z M$, $\angle C Z M=\angle B Z M$. Therefore, $\angle B X M=\angle B Z M$. It follows that the points $B, X, Z$, and $M$ are concyclic, hence $\angle B X Z=180^{\circ}-\angle B M Z=90^{\circ}$.

Finally, we have $\angle P X Z=\angle P Y Z=\angle P A Z=90^{\circ}$, hence the five points $P, A, X, Y, Z$ are concyclic. In particular, the quadrilateral $A P X Y$ is cyclic, as required.
Subfield: Geometry


Question: A circle $\omega$ of radius 1 is given. A collection $T$ of triangles is called good, if the following conditions hold:

(i) each triangle from $T$ is inscribed in $\omega$;

(ii) no two triangles from $T$ have a common interior point.

Determine all positive real numbers $t$ such that, for each positive integer $n$, there exists a good collection of $n$ triangles, each of perimeter greater than $t$.

Solution:
First, we show how to construct a good collection of $n$ triangles, each of perimeter greater than 4 . This will show that all $t \leqslant 4$ satisfy the required conditions.

Construct inductively an $(n+2)$-gon $B A_{1} A_{2} \ldots A_{n} C$ inscribed in $\omega$ such that $B C$ is a diameter, and $B A_{1} A_{2}, B A_{2} A_{3}, \ldots, B A_{n-1} A_{n}, B A_{n} C$ is a good collection of $n$ triangles. For $n=1$, take any triangle $B A_{1} C$ inscribed in $\omega$ such that $B C$ is a diameter; its perimeter is greater than $2 B C=4$. To perform the inductive step, assume that the $(n+2)$-gon $B A_{1} A_{2} \ldots A_{n} C$ is already constructed. Since $A_{n} B+A_{n} C+B C>4$, one can choose a point $A_{n+1}$ on the small arc $\widehat{C A_{n}}$, close enough to $C$, so that $A_{n} B+A_{n} A_{n+1}+B A_{n+1}$ is still greater than 4 . Thus each of these new triangles $B A_{n} A_{n+1}$ and $B A_{n+1} C$ has perimeter greater than 4 , which completes the induction step.

<img_3904>

We proceed by showing that no $t>4$ satisfies the conditions of the problem. To this end, we assume that there exists a good collection $T$ of $n$ triangles, each of perimeter greater than $t$, and then bound $n$ from above.

Take $\varepsilon>0$ such that $t=4+2 \varepsilon$.

Claim. There exists a positive constant $\sigma=\sigma(\varepsilon)$ such that any triangle $\Delta$ with perimeter $2 s \geqslant 4+2 \varepsilon$, inscribed in $\omega$, has area $S(\Delta)$ at least $\sigma$.

Proof. Let $a, b, c$ be the side lengths of $\Delta$. Since $\Delta$ is inscribed in $\omega$, each side has length at most 2. Therefore, $s-a \geqslant(2+\varepsilon)-2=\varepsilon$. Similarly, $s-b \geqslant \varepsilon$ and $s-c \geqslant \varepsilon$. By Heron's formula, $S(\Delta)=\sqrt{s(s-a)(s-b)(s-c)} \geqslant \sqrt{(2+\varepsilon) \varepsilon^{3}}$. Thus we can set $\sigma(\varepsilon)=\sqrt{(2+\varepsilon) \varepsilon^{3}}$.

Now we see that the total area $S$ of all triangles from $T$ is at least $n \sigma(\varepsilon)$. On the other hand, $S$ does not exceed the area of the disk bounded by $\omega$. Thus $n \sigma(\varepsilon) \leqslant \pi$, which means that $n$ is bounded from above.

Final answer: $t(0,4]$
Subfield: Geometry


Question: A point $T$ is chosen inside a triangle $A B C$. Let $A_{1}, B_{1}$, and $C_{1}$ be the reflections of $T$ in $B C, C A$, and $A B$, respectively. Let $\Omega$ be the circumcircle of the triangle $A_{1} B_{1} C_{1}$. The lines $A_{1} T, B_{1} T$, and $C_{1} T$ meet $\Omega$ again at $A_{2}, B_{2}$, and $C_{2}$, respectively. Prove that the lines $A A_{2}, B B_{2}$, and $C C_{2}$ are concurrent on $\Omega$.

Solution:
By $\sphericalangle(\ell, n)$ we always mean the directed angle of the lines $\ell$ and $n$, taken modulo $180^{\circ}$.

Let $C C_{2}$ meet $\Omega$ again at $K$ (as usual, if $C C_{2}$ is tangent to $\Omega$, we set $T=C_{2}$ ). We show that the line $B B_{2}$ contains $K$; similarly, $A A_{2}$ will also pass through $K$. For this purpose, it suffices to prove that

$$
\sphericalangle\left(C_{2} C, C_{2} A_{1}\right)=\sphericalangle\left(B_{2} B, B_{2} A_{1}\right) .
\tag{1}
$$

By the problem condition, $C B$ and $C A$ are the perpendicular bisectors of $T A_{1}$ and $T B_{1}$, respectively. Hence, $C$ is the circumcentre of the triangle $A_{1} T B_{1}$. Therefore,

$$
\sphericalangle\left(C A_{1}, C B\right)=\sphericalangle(C B, C T)=\sphericalangle\left(B_{1} A_{1}, B_{1} T\right)=\sphericalangle\left(B_{1} A_{1}, B_{1} B_{2}\right) .
$$

In circle $\Omega$ we have $\sphericalangle\left(B_{1} A_{1}, B_{1} B_{2}\right)=\sphericalangle\left(C_{2} A_{1}, C_{2} B_{2}\right)$. Thus,

$$
\sphericalangle\left(C A_{1}, C B\right)=\sphericalangle\left(B_{1} A_{1}, B_{1} B_{2}\right)=\sphericalangle\left(C_{2} A_{1}, C_{2} B_{2}\right) .
\tag{2}
$$

Similarly, we get

$$
\sphericalangle\left(B A_{1}, B C\right)=\sphericalangle\left(C_{1} A_{1}, C_{1} C_{2}\right)=\sphericalangle\left(B_{2} A_{1}, B_{2} C_{2}\right) .
\tag{3}
$$

The two obtained relations yield that the triangles $A_{1} B C$ and $A_{1} B_{2} C_{2}$ are similar and equioriented, hence

$$
\frac{A_{1} B_{2}}{A_{1} B}=\frac{A_{1} C_{2}}{A_{1} C} \quad \text { and } \quad \sphericalangle\left(A_{1} B, A_{1} C\right)=\sphericalangle\left(A_{1} B_{2}, A_{1} C_{2}\right) \text {. }
$$

The second equality may be rewritten as $\sphericalangle\left(A_{1} B, A_{1} B_{2}\right)=\sphericalangle\left(A_{1} C, A_{1} C_{2}\right)$, so the triangles $A_{1} B B_{2}$ and $A_{1} C C_{2}$ are also similar and equioriented. This establishes (1).

<img_3351>
Subfield: Number Theory


Question: Determine all pairs $(n, k)$ of distinct positive integers such that there exists a positive integer $s$ for which the numbers of divisors of $s n$ and of $s k$ are equal.

Solution:
As usual, the number of divisors of a positive integer $n$ is denoted by $d(n)$. If $n=\prod_{i} p_{i}^{\alpha_{i}}$ is the prime factorisation of $n$, then $d(n)=\prod_{i}\left(\alpha_{i}+1\right)$.

We start by showing that one cannot find any suitable number $s$ if $k \mid n$ or $n \mid k$ (and $k \neq n$ ). Suppose that $n \mid k$, and choose any positive integer $s$. Then the set of divisors of $s n$ is a proper subset of that of $s k$, hence $d(s n)<d(s k)$. Therefore, the pair $(n, k)$ does not satisfy the problem requirements. The case $k \mid n$ is similar.

Now assume that $n \nmid k$ and $k \nmid n$. Let $p_{1}, \ldots, p_{t}$ be all primes dividing $n k$, and consider the prime factorisations

$$
n=\prod_{i=1}^{t} p_{i}^{\alpha_{i}} \quad \text { and } \quad k=\prod_{i=1}^{t} p_{i}^{\beta_{i}}
$$

It is reasonable to search for the number $s$ having the form

$$
s=\prod_{i=1}^{t} p_{i}^{\gamma_{i}}
$$

The (nonnegative integer) exponents $\gamma_{i}$ should be chosen so as to satisfy

$$
\frac{d(s n)}{d(s k)}=\prod_{i=1}^{t} \frac{\alpha_{i}+\gamma_{i}+1}{\beta_{i}+\gamma_{i}+1}=1
\tag{1}
$$

First of all, if $\alpha_{i}=\beta_{i}$ for some $i$, then, regardless of the value of $\gamma_{i}$, the corresponding factor in (1) equals 1 and does not affect the product. So we may assume that there is no such index $i$. For the other factors in (1), the following lemma is useful.

Lemma. Let $\alpha>\beta$ be nonnegative integers. Then, for every integer $M \geqslant \beta+1$, there exists a nonnegative integer $\gamma$ such that

$$
\frac{\alpha+\gamma+1}{\beta+\gamma+1}=1+\frac{1}{M}=\frac{M+1}{M}
$$

Proof.

$$
\frac{\alpha+\gamma+1}{\beta+\gamma+1}=1+\frac{1}{M} \Longleftrightarrow \frac{\alpha-\beta}{\beta+\gamma+1}=\frac{1}{M} \Longleftrightarrow \gamma=M(\alpha-\beta)-(\beta+1) \geqslant 0 .
$$

Now we can finish the solution. Without loss of generality, there exists an index $u$ such that $\alpha_{i}>\beta_{i}$ for $i=1,2, \ldots, u$, and $\alpha_{i}<\beta_{i}$ for $i=u+1, \ldots, t$. The conditions $n \nmid k$ and $k \nmid n$ mean that $1 \leqslant u \leqslant t-1$.

Choose an integer $X$ greater than all the $\alpha_{i}$ and $\beta_{i}$. By the lemma, we can define the numbers $\gamma_{i}$ so as to satisfy

$$
\begin{array}{lr}
\frac{\alpha_{i}+\gamma_{i}+1}{\beta_{i}+\gamma_{i}+1}=\frac{u X+i}{u X+i-1} & \text { for } i=1,2, \ldots, u, \text { and } \\
\frac{\beta_{u+i}+\gamma_{u+i}+1}{\alpha_{u+i}+\gamma_{u+i}+1}=\frac{(t-u) X+i}{(t-u) X+i-1} & \text { for } i=1,2, \ldots, t-u .
\end{array}
$$

Then we will have

$$
\frac{d(s n)}{d(s k)}=\prod_{i=1}^{u} \frac{u X+i}{u X+i-1} \cdot \prod_{i=1}^{t-u} \frac{(t-u) X+i-1}{(t-u) X+i}=\frac{u(X+1)}{u X} \cdot \frac{(t-u) X}{(t-u)(X+1)}=1,
$$

as required.

Final answer: All pairs $(n, k)$ such that $n \nmid k$ and $k \nmid n$
Subfield: Algebra


Question: Prove that in any set of 2000 distinct real numbers there exist two pairs $a>b$ and $c>d$ with $a \neq c$ or $b \neq d$, such that

$$
\left|\frac{a-b}{c-d}-1\right|<\frac{1}{100000}
$$

Solution:
For any set $S$ of $n=2000$ distinct real numbers, let $D_{1} \leqslant D_{2} \leqslant \cdots \leqslant D_{m}$ be the distances between them, displayed with their multiplicities. Here $m=n(n-1) / 2$. By rescaling the numbers, we may assume that the smallest distance $D_{1}$ between two elements of $S$ is $D_{1}=1$. Let $D_{1}=1=y-x$ for $x, y \in S$. Evidently $D_{m}=v-u$ is the difference between the largest element $v$ and the smallest element $u$ of $S$.

If $D_{i+1} / D_{i}<1+10^{-5}$ for some $i=1,2, \ldots, m-1$ then the required inequality holds, because $0 \leqslant D_{i+1} / D_{i}-1<10^{-5}$. Otherwise, the reverse inequality

$$
\frac{D_{i+1}}{D_{i}} \geqslant 1+\frac{1}{10^{5}}
$$

holds for each $i=1,2, \ldots, m-1$, and therefore

$$
v-u=D_{m}=\frac{D_{m}}{D_{1}}=\frac{D_{m}}{D_{m-1}} \cdots \frac{D_{3}}{D_{2}} \cdot \frac{D_{2}}{D_{1}} \geqslant\left(1+\frac{1}{10^{5}}\right)^{m-1} .
$$

From $m-1=n(n-1) / 2-1=1000 \cdot 1999-1>19 \cdot 10^{5}$, together with the fact that for all $n \geqslant 1$, $\left(1+\frac{1}{n}\right)^{n} \geqslant 1+\left(\begin{array}{l}n \\ 1\end{array}\right) \cdot \frac{1}{n}=2$, we get

$$
\left(1+\frac{1}{10^{5}}\right)^{19 \cdot 10^{5}}=\left(\left(1+\frac{1}{10^{5}}\right)^{10^{5}}\right)^{19} \geqslant 2^{19}=2^{9} \cdot 2^{10}>500 \cdot 1000>2 \cdot 10^{5},
$$

and so $v-u=D_{m}>2 \cdot 10^{5}$.

Since the distance of $x$ to at least one of the numbers $u, v$ is at least $(u-v) / 2>10^{5}$, we have

$$
|x-z|>10^{5} .
$$

for some $z \in\{u, v\}$. Since $y-x=1$, we have either $z>y>x$ (if $z=v$ ) or $y>x>z$ (if $z=u$ ). If $z>y>x$, selecting $a=z, b=y, c=z$ and $d=x$ (so that $b \neq d$ ), we obtain

$$
\left|\frac{a-b}{c-d}-1\right|=\left|\frac{z-y}{z-x}-1\right|=\left|\frac{x-y}{z-x}\right|=\frac{1}{z-x}<10^{-5}
$$

Otherwise, if $y>x>z$, we may choose $a=y, b=z, c=x$ and $d=z$ (so that $a \neq c$ ), and obtain

$$
\left|\frac{a-b}{c-d}-1\right|=\left|\frac{y-z}{x-z}-1\right|=\left|\frac{y-x}{x-z}\right|=\frac{1}{x-z}<10^{-5}
$$

The desired result follows.
Subfield: Algebra


Question: Let $\mathbb{Q}_{>0}$ be the set of positive rational numbers. Let $f: \mathbb{Q}_{>0} \rightarrow \mathbb{R}$ be a function satisfying the conditions

$$
f(x) f(y) \geqslant f(x y)
\tag{1}
$$
$$
f(x+y) \geqslant f(x)+f(y)
\tag{2}
$$

for all $x, y \in \mathbb{Q}_{>0}$. Given that $f(a)=a$ for some rational $a>1$, prove that $f(x)=x$ for all $x \in \mathbb{Q}_{>0}$.

Solution:
Denote by $\mathbb{Z}_{>0}$ the set of positive integers.

Plugging $x=1, y=a$ into (1) we get $f(1) \geqslant 1$. Next, by an easy induction on $n$ we get from (2) that

$$
f(n x) \geqslant n f(x) \text { for all } n \in \mathbb{Z}_{>0} \text { and } x \in \mathbb{Q}_{>0} \text {. }
\tag{3}
$$

In particular, we have

$$
f(n) \geqslant n f(1) \geqslant n \text { for all } n \in \mathbb{Z}_{>0}
\tag{4}
$$

From (1) again we have $f(m / n) f(n) \geqslant f(m)$, so $f(q)>0$ for all $q \in \mathbb{Q}_{>0}$.

Now, (2) implies that $f$ is strictly increasing; this fact together with (4) yields

$$
f(x) \geqslant f(\lfloor x\rfloor) \geqslant\lfloor x\rfloor>x-1 \quad \text { for all } x \geqslant 1
$$

By an easy induction we get from (1) that $f(x)^{n} \geqslant f\left(x^{n}\right)$, so

$$
f(x)^{n} \geqslant f\left(x^{n}\right)>x^{n}-1 \quad \Longrightarrow \quad f(x) \geqslant \sqrt[n]{x^{n}-1} \quad \text { for all } x>1 \text { and } n \in \mathbb{Z}_{>0}
$$

This yields

$$
f(x) \geqslant x \text { for every } x>1 \text {. }\tag{5}
$$

(Indeed, if $x>y>1$ then $x^{n}-y^{n}=(x-y)\left(x^{n-1}+x^{n-2} y+\cdots+y^{n}\right)>n(x-y)$, so for a large $n$ we have $x^{n}-1>y^{n}$ and thus $f(x)>y$.)

Now, (1) and (5) give $a^{n}=f(a)^{n} \geqslant f\left(a^{n}\right) \geqslant a^{n}$, so $f\left(a^{n}\right)=a^{n}$. Now, for $x>1$ let us choose $n \in \mathbb{Z}_{>0}$ such that $a^{n}-x>1$. Then by (2) and (5) we get

$$
a^{n}=f\left(a^{n}\right) \geqslant f(x)+f\left(a^{n}-x\right) \geqslant x+\left(a^{n}-x\right)=a^{n}
$$

and therefore $f(x)=x$ for $x>1$. Finally, for every $x \in \mathbb{Q}_{>0}$ and every $n \in \mathbb{Z}_{>0}$, from (1) and (3) we get

$$
n f(x)=f(n) f(x) \geqslant f(n x) \geqslant n f(x)
$$

which gives $f(n x)=n f(x)$. Therefore $f(m / n)=f(m) / n=m / n$ for all $m, n \in \mathbb{Z}_{>0}$.