di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Number Theory


Question: Does there exist a nonnegative integer $a$ for which the equation

$$
\left\lfloor\frac{m}{1}\right\rfloor+\left\lfloor\frac{m}{2}\right\rfloor+\left\lfloor\frac{m}{3}\right\rfloor+\cdots+\left\lfloor\frac{m}{m}\right\rfloor=n^{2}+a
$$

has more than one million different solutions $(m, n)$ where $m$ and $n$ are positive integers? (The expression $\lfloor x\rfloor$ denotes the integer part(or floor) of the real number $x$. Thus $\lfloor\sqrt{2}\rfloor=1,\lfloor\pi\rfloor=\left\lfloor\frac{22}{7}\right\rfloor=$ $3,\lfloor 42\rfloor=42$ and $\lfloor 0\rfloor=0)$

Solution:
Denote the equation from the statement by (1). The left hand side of (1) depends only on $m$, and will throughout be denoted by $L(m)$. Fix an integer $q>10^{7}$ and note that for $m=q^{3}$



$$
L\left(q^{3}\right)=\sum_{k=1}^{q^{3}}\left[\frac{q^{3}}{k}\right] \leq \sum_{k=1}^{q^{3}} \frac{q^{3}}{k} \leq q^{3} \cdot \sum_{k=1}^{q^{3}} \frac{1}{k} \leq q^{3} \cdot q=q^{4} \tag{2}
$$

Indeed, the first inequality results from $[x] \leq x$. The second inequality can be seen (for instance) as follows. We divide the terms in the sum $\sum_{k=1}^{q^{3}} \frac{1}{k}$ into several groups: For $j \geq 0$, the $j$-th group contains the $2^{j}$ consecutive terms $\frac{1}{2^{j}}, \ldots, \frac{1}{2^{j+1}-1}$. Since every term in the $j$-th group is bounded by $\frac{1}{2^{j}}$, the overall contribution of the $j$-th group to the sum is at most 1 . Since the first $q$ groups together would contain $2^{q}-1>q^{3}$ terms, the number of groups does not exceed $q$, and hence the value of the sum under consideration is indeed bounded by $q$.

Call an integer $m$ special, if it satisfies $1 \leq L(m) \leq q^{4}$. Denote by $g(m) \geq 1$ the largest integer whose square is bounded by $L(m)$; in other words $g^{2}(m) \leq L(m)<(g(m)+1)^{2}$. Note that $g(m) \leq$ $q^{2}$ for all special $m$, which implies

$$
0 \leq L(m)-g^{2}(m)<(g(m)+1)^{2}-g^{2}(m)=2 g(m)+1 \leq 2 q^{2}+1 . \tag{3}
$$

Finally, we do some counting. Inequality (2) and the monotonicity of $L(m)$ imply that there exist at least $q^{3}$ special integers. Because of (3), every special integer $m$ has $0 \leq L(m)-g^{2}(m) \leq 2 q^{2}+1$. By averaging, at least $\frac{q^{3}}{2 q^{2}+2}>10^{6}$ special integers must yield the same value $L(m)-g^{2}(m)$. This frequently occurring value is our choice for $\alpha$, which yields more than $10^{6}$ solutions $(m, g(m))$ to equation (1).
Subfield: Geometry


Question: Let $A B C D$ be a convex quadrilateral with $\angle D A B=\angle B C D=90^{\circ}$ and $\angle A B C>\angle C D A$. Let $Q$ and $R$ be points on the segments $B C$ and $C D$, respectively, such that the line $Q R$ intersects lines $A B$ and $A D$ at points $P$ and $S$, respectively. It is given that $P Q=R S$. Let the midpoint of $B D$ be $M$ and the midpoint of $Q R$ be $N$. Prove that $M, N, A$ and $C$ lie on a circle.

Solution:
Note that $N$ is also the midpoint of $P S$. From right-angled triangles $P A S$ and $C Q R$ we obtain $\angle A N P=$ $2 \angle A S P, \angle C N Q=2 \angle C R Q$, hence $\angle A N C=\angle A N P+\angle C N Q=2(\angle A S P+\angle C R Q)=2(\angle R S D+\angle D R S)=$ $2 \angle A D C$.

Similarly, using right-angled triangles $B A D$ and $B C D$, we obtain $\angle A M C=2 \angle A D C$.

Thus $\angle A M C=\angle A N C$, and the required statement follows.

<img_3169>
In this proof we show that we have $\angle N C M=\angle N A M$ instead. From right-angled triangles $B C D$ and $Q C R$ we get $\angle D R S=\angle C R Q=\angle R C N$ and $\angle B D C=\angle D C M$. Hence $\angle N C M=\angle D C M-\angle R C N$. From rightangled triangle $A P S$ we get $\angle P S A=\angle S A N$. From right-angled triangle $B A D$ we have $\angle M A D=\angle B D A$. Moreover, $\angle B D A=\angle D R S+\angle R S D-\angle R D B$.

Therefore $\angle N A M=\angle N A S-\angle M A D=\angle C D B-\angle D R S=\angle N C M$, and the required statement follows.
As $N$ is also the midpoint of $P S$, we can shrink triangle $A P S$ to a triangle $A_{0} Q R$ (where $P$ is sent to $Q$ and $S$ is sent to $R$ ). Then $A_{0}, Q, R$ and $C$ lie on a circle with center $N$. According to the shrinking the line $A_{0} R$ is parallel to the line $A D$. Therefore $\angle C N A=\angle C N A_{0}=2 \angle C R A_{0}=2 \angle C D A=\angle C M A$. The required statement follows.
Subfield: Algebra


Question: Let $n$ be an odd positive integer, and let $x_{1}, x_{2}, \ldots, x_{n}$ be non-negative real numbers. Show that

$$
\min _{i=1, \ldots, n}\left(x_{i}^{2}+x_{i+1}^{2}\right) \leq \max _{j=1, \ldots, n}\left(2 x_{j} x_{j+1}\right)
$$

where $x_{n+1}=x_{1}$.

Solution:
In what follows, indices are reduced modulo $n$. Consider the $n$ differences $x_{k+1}-x_{k}, k=1, \ldots, n$. Since $n$ is odd, there exists an index $j$ such that $\left(x_{j+1}-x_{j}\right)\left(x_{j+2}-x_{j+1}\right) \geq 0$. Without loss of generality, we may and will assume both factors non-negative, so $x_{j} \leq x_{j+1} \leq x_{j+2}$. Consequently,

$$
\min _{k=1, \ldots, n}\left(x_{k}^{2}+x_{k+1}^{2}\right) \leq x_{j}^{2}+x_{j+1}^{2} \leq 2 x_{j+1}^{2} \leq 2 x_{j+1} x_{j+2} \leq \max _{k=1, \ldots, n}\left(2 x_{k} x_{k+1}\right)
$$
Subfield: Geometry


Question: Let $A B C D$ be a cyclic quadrilateral, and let diagonals $A C$ and $B D$ intersect at $X$. Let $C_{1}, D_{1}$ and $M$ be the midpoints of segments $C X$, $D X$ and $C D$, respectively. Lines $A D_{1}$ and $B C_{1}$ intersect at $Y$, and line $M Y$ intersects diagonals $A C$ and $B D$ at different points $E$ and $F$, respectively. Prove that line $X Y$ is tangent to the circle through $E, F$ and $X$.

<img_3388>

Solution:
We are to prove that $\angle E X Y=\angle E F X$; alternatively, but equivalently, $\angle A Y X+\angle X A Y=\angle B Y F+\angle X B Y$.

Since the quadrangle $A B C D$ is cyclic, the triangles $X A D$ and $X B C$ are similar, and since $A D_{1}$ and $B C_{1}$ are corresponding medians in these triangles, it follows that $\angle X A Y=\angle X A D_{1}=\angle X B C_{1}=\angle X B Y$.

Finally, $\angle A Y X=\angle B Y F$, since $X$ and $M$ are corresponding points in the similar triangles $A B Y$ and $C_{1} D_{1} Y$ : indeed, $\angle X A B=\angle X D C=\angle M C_{1} D_{1}$, and $\angle X B A=\angle X C D=\angle M D_{1} C_{1}$.
Subfield: Geometry


Question: Let $A B C D$ be a cyclic quadrilateral, and let diagonals $A C$ and $B D$ intersect at $X$. Let $C_{1}, D_{1}$ and $M$ be the midpoints of segments $C X$, $D X$ and $C D$, respectively. Lines $A D_{1}$ and $B C_{1}$ intersect at $Y$, and line $M Y$ intersects diagonals $A C$ and $B D$ at different points $E$ and $F$, respectively. Prove that line $X Y$ is tangent to the circle through $E, F$ and $X$.

![](https://cdn.mathpix.com/cropped/2023_12_21_2fe10796898d797e3ea2g-1.jpg?height=557&width=556&top_left_y=1943&top_left_x=750)

Solution:
We are to prove that $\angle E X Y=\angle E F X$; alternatively, but equivalently, $\angle A Y X+\angle X A Y=\angle B Y F+\angle X B Y$.

Since the quadrangle $A B C D$ is cyclic, the triangles $X A D$ and $X B C$ are similar, and since $A D_{1}$ and $B C_{1}$ are corresponding medians in these triangles, it follows that $\angle X A Y=\angle X A D_{1}=\angle X B C_{1}=\angle X B Y$.

Finally, $\angle A Y X=\angle B Y F$, since $X$ and $M$ are corresponding points in the similar triangles $A B Y$ and $C_{1} D_{1} Y$ : indeed, $\angle X A B=\angle X D C=\angle M C_{1} D_{1}$, and $\angle X B A=\angle X C D=\angle M D_{1} C_{1}$.

Final answer: 证明题，略
Subfield: Combinatorics


Question: Let $m$ be a positive integer. Consider a $4 m \times 4 m$ array of square unit cells. Two different cells are related to each other if they are in either the same row or in the same column. No cell is related to itself. Some cells are coloured blue, such that every cell is related to at least two blue cells. Determine the minimum number of blue cells.

Solution:
The required minimum is $6 m$ and is achieved by a diagonal string of $m$ $4 \times 4$ blocks of the form below (bullets mark centres of blue cells):

<img_3402>

In particular, this configuration shows that the required minimum does not exceed $6 m$.

We now show that any configuration of blue cells satisfying the condition in the statement has cardinality at least $6 \mathrm{~m}$.

Fix such a configuration and let $m_{1}^{r}$ be the number of blue cells in rows containing exactly one such, let $m_{2}^{r}$ be the number of blue cells in rows containing exactly two such, and let $m_{3}^{r}$ be the number of blue cells in rows containing at least three such; the numbers $m_{1}^{c}, m_{2}^{c}$ and $m_{3}^{c}$ are defined similarly.

Begin by noticing that $m_{3}^{c} \geq m_{1}^{r}$ and, similarly, $m_{3}^{r} \geq m_{1}^{c}$. Indeed, if a blue cell is alone in its row, respectively column, then there are at least two other blue cells in its column, respectively row, and the claim follows.

Suppose now, if possible, the total number of blue cells is less than $6 \mathrm{~m}$. We will show that $m_{1}^{r}>m_{3}^{r}$ and $m_{1}^{c}>m_{3}^{c}$, and reach a contradiction by the preceding: $m_{1}^{r}>m_{3}^{r} \geq m_{1}^{c}>m_{3}^{c} \geq m_{1}^{r}$.

We prove the first inequality; the other one is dealt with similarly. To this end, notice that there are no empty rows - otherwise, each column would contain at least two blue cells, whence a total of at least $8 m>6 m$ blue cells, which is a contradiction. Next, count rows to get $m_{1}^{r}+m_{2}^{r} / 2+m_{3}^{r} / 3 \geq 4 m$, and count blue cells to get $m_{1}^{r}+m_{2}^{r}+m_{3}^{r}<6 m$. Subtraction of the latter from the former multiplied by $3 / 2$ yields $m_{1}^{r}-m_{3}^{r}>m_{2}^{r} / 2 \geq 0$, and the conclusion follows.
To prove that a minimal configuration of blue cells satisfying the condition in the statement has cardinality at least $6 m$, consider a bipartite graph whose vertex parts are the rows and the columns of the array, respectively, a row and a column being joined by an edge if and only if the two cross at a blue cell. Clearly, the number of blue cells is equal to the number of edges of this graph, and the relationship condition in the statement reads: for every row $r$ and every column $c, \operatorname{deg} r+\operatorname{deg} c-\epsilon(r, c) \geq 2$, where $\epsilon(r, c)=2$ if $r$ and $c$ are joined by an edge, and $\epsilon(r, c)=0$ otherwise.

Notice that there are no empty rows/columns, so the graph has no isolated vertices. By the preceding, the cardinality of every connected component of the graph is at least 4 , so there are at most $2 \cdot 4 m / 4=2 m$ such and, consequently, the graph has at least $8 m-2 m=6 m$ edges. This completes the proof.

Final answer: $6m$
Subfield: Geometry


Question: Two circles, $\omega_{1}$ and $\omega_{2}$, of equal radius intersect at different points $X_{1}$ and $X_{2}$. Consider a circle $\omega$ externally tangent to $\omega_{1}$ at a point $T_{1}$, and internally tangent to $\omega_{2}$ at a point $T_{2}$. Prove that lines $X_{1} T_{1}$ and $X_{2} T_{2}$ intersect at a point lying on $\omega$.

Solution:
Let the line $X_{k} T_{k}$ and $\omega$ meet again at $X_{k}^{\prime}, k=1,2$, and notice that the tangent $t_{k}$ to $\omega_{k}$ at $X_{k}$ and the tangent $t_{k}^{\prime}$ to $\omega$ at $X_{k}^{\prime}$ are parallel. Since the $\omega_{k}$ have equal radii, the $t_{k}$ are parallel, so the $t_{k}^{\prime}$ are parallel, and consequently the points $X_{1}^{\prime}$ and $X_{2}^{\prime}$ coincide (they are not antipodal, since they both lie on the same side of the line $T_{1} T_{2}$ ). The conclusion follows.
The circle $\omega$ is the image of $\omega_{k}$ under a homothety $h_{k}$ centred at $T_{k}, k=1,2$. The tangent to $\omega$ at $X_{k}^{\prime}=h_{k}\left(X_{k}\right)$ is therefore parallel to the tangent $t_{k}$ to $\omega_{k}$ at $X_{k}$. Since the $\omega_{k}$ have equal radii, the $t_{k}$ are parallel, so $X_{1}^{\prime}=X_{2}^{\prime}$; and since the points $X_{k}, T_{k}$ and $X_{k}^{\prime}$ are collinear, the conclusion follows.
<img_3212>
Invert from $X_{1}$ and use an asterisk to denote images under this inversion. Notice that $\omega_{k}^{*}$ is the tangent from $X_{2}^{*}$ to $\omega^{*}$ at $T_{k}^{*}$, and the pole $X_{1}$ lies on the bisectrix of the angle formed by the $\omega_{k}^{*}$, not containing $\omega^{*}$. Letting $X_{1} T_{1}^{*}$ and $\omega^{*}$ meet again at $Y$, standard angle chase shows that $Y$ lies on the circle $X_{1} X_{2}^{*} T_{2}^{*}$, and the conclusion follows.
Subfield: Combinatorics


Question: Let $k$ and $n$ be integers such that $k \geq 2$ and $k \leq n \leq 2 k-1$. Place rectangular tiles, each of size $1 \times k$ or $k \times 1$, on an $n \times n$ chessboard so that each tile covers exactly $k$ cells, and no two tiles overlap. Do this until no further tile can be placed in this way. For each such $k$ and $n$, determine the minimum number of tiles that such an arrangement may contain.

Solution:
The required minimum is $n$ if $n=k$, and it is $\min (n, 2 n-2 k+2)$ if $k<n<2 k$.

The case $n=k$ being clear, assume henceforth $k<n<2 k$. Begin by describing maximal arrangements on the board $[0, n] \times[0, n]$, having the above mentioned cardinalities.

If $k<n<2 k-1$, then $\min (n, 2 n-2 k+2)=2 n-2 k+2$. To obtain a maximal arrangement of this cardinality, place four tiles, $[0, k] \times[0,1]$, $[0,1] \times[1, k+1],[1, k+1] \times[k, k+1]$ and $[k, k+1] \times[0, k]$ in the square $[0, k] \times[0, k]$, stack $n-k-1$ horizontal tiles in the rectangle $[1, k+1] \times[k+1, n]$, and erect $n-k-1$ vertical tiles in the rectangle $[k+1, n] \times[1, k+1]$.

If $n=2 k-1$, then $\min (n, 2 n-2 k+2)=n=2 k-1$. A maximal arrangement of $2 k-1$ tiles is obtained by stacking $k-1$ horizontal tiles in the rectangle $[0, k] \times[0, k-1]$, another $k-1$ horizontal tiles in the rectangle $[0, k] \times[k, 2 k-1]$, and adding the horizontal tile $[k-1,2 k-1] \times[k-1, k]$.

The above examples show that the required minimum does not exceed the mentioned values.

To prove the reverse inequality, consider a maximal arrangement and let $r$, respectively $c$, be the number of rows, respectively columns, not containing a tile.

If $r=0$ or $c=0$, the arrangement clearly contains at least $n$ tiles.

If $r$ and $c$ are both positive, we show that the arrangement contains at least $2 n-2 k+2$ tiles. To this end, we will prove that the rows, respectively columns, not containing a tile are consecutive. Assume this for the moment, to notice that these $r$ rows and $c$ columns cross to form an $r \times c$ rectangular array containing no tile at all, so $r<k$ and $c<k$ by maximality. Consequently, there are $n-r \geq n-k+1$ rows containing at least one horizontal tile each, and $n-c \geq n-k+1$ columns containing at least one vertical tile each, whence a total of at least $2 n-2 k+2$ tiles.

We now show that the rows not containing a tile are consecutive; columns are dealt with similarly. Consider a horizontal tile $T$. Since $n<2 k$, the nearest horizontal side of the board is at most $k-1$ rows away from the row containing $T$. These rows, if any, cross the $k$ columns $T$ crosses to form a rectangular array no vertical tile fits in. Maximality forces each of these rows to contain a horizontal tile and the claim follows.

Consequently, the cardinality of every maximal arrangement is at least $\min (n, 2 n-2 k+2)$, and the conclusion follows.

Final answer: $n$ if $n=k$, and it is $\min (n, 2 n-2 k+2)$ if $k<n<2 k$
Subfield: Number Theory


Question: Let $m>1$ be an integer. A sequence $a_{1}, a_{2}, a_{3}, \ldots$ is defined by $a_{1}=a_{2}=1$, $a_{3}=4$, and for all $n \geq 4$,

$$
a_{n}=m\left(a_{n-1}+a_{n-2}\right)-a_{n-3} .
$$

Determine all integers $m$ such that every term of the sequence is a square.

Solution:
Consider an integer $m>1$ for which the sequence defined in the problem statement contains only perfect squares. We shall first show that $m-1$ is a power of 3 .

Suppose that $m-1$ is even. Then $a_{4}=5 m-1$ should be divisible by 4 and hence $m \equiv 1(\bmod 4)$. But then $a_{5}=5 m^{2}+3 m-1 \equiv 3(\bmod 4)$ cannot be a square, a contradiction. Therefore $m-1$ is odd.

Suppose that an odd prime $p \neq 3$ divides $m-1$. Note that $a_{n}-a_{n-1} \equiv a_{n-2}-a_{n-3}(\bmod p)$. It follows that modulo $p$ the sequence takes the form $1,1,4,4,7,7,10,10, \ldots$; indeed, a simple induction shows that $a_{2 k} \equiv a_{2 k-1} \equiv 3 k-2(\bmod p)$ for $k \geq 1$. Since $\operatorname{gcd}(p, 3)=1$ we get that the sequence $a_{n}(\bmod p)$ contains all the residues modulo $p$, a contradiction since only $(p+1) / 2$ residues modulo $p$ are squares. This shows that $m-1$ is a power of 3 .

Let $h, k$ be integers such that $m=3^{k}+1$ and $a_{4}=h^{2}$. We then have $5 \cdot 3^{k}=(h-2)(h+2)$. Since $\operatorname{gcd}(h-2, h+2)=1$, it follows that $h-2$ equals either $1,3^{k}$ or 5 , and $h+2$ equals either $5 \cdot 3^{k}, 5$ or $3^{k}$, respectively. In the first two cases we get $k=0$ and in the last case we get $k=2$. This implies that either $m=2$ or $m=10$.

We now show the converse. Suppose that $m=2$ or $m=10$. Let $t=1$ or $t=3$ so that $m=t^{2}+1$. Let $b_{1}, b_{2}, b_{3}, \ldots$ be a sequence of integers defined by $b_{1}=1, b_{2}=1, b_{3}=2$, and

$$
b_{n}=t b_{n-1}+b_{n-2}, \quad \text { for all } n \geq 4
$$

Clearly, $a_{n}=b_{n}^{2}$ for $n=1,2,3$. Note that if $m=2$ then $a_{4}=9$ and $b_{4}=3$, and if $m=10$ then $a_{4}=49$ and $b_{4}=7$. In both the cases we have $a_{4}=b_{4}^{2}$.

If $n \geq 5$ then we have

$$
b_{n}^{2}+b_{n-3}^{2}=\left(t b_{n-1}+b_{n-2}\right)^{2}+\left(b_{n-1}-t b_{n-2}\right)^{2}=\left(t^{2}+1\right)\left(b_{n-1}^{2}+b_{n-2}^{2}\right)=m\left(b_{n-1}^{2}+b_{n-2}^{2}\right) .
$$

Therefore, it follows by induction that $a_{n}=b_{n}^{2}$ for all $n \geq 1$. This completes the solution.
We present an alternate proof that $m=2$ and $m=10$ are the only possible values of $m$ with the required property.

Note that

$$
\begin{aligned}
& a_{4}=5 m-1, \\
& a_{5}=5 m^{2}+3 m-1, \\
& a_{6}=5 m^{3}+8 m^{2}-2 m-4 .
\end{aligned}
$$

Since $a_{4}$ and $a_{6}$ are squares, so is $a_{4} a_{6}$. We have

$$
4 a_{4} a_{6}=100 m^{4}+140 m^{3}-72 m^{2}-72 m+16 .
$$

Notice that

$$
\begin{aligned}
& \left(10 m^{2}+7 m-7\right)^{2}=100 m^{4}+140 m^{3}-91 m^{2}-98 m+49<4 a_{4} a_{6}, \\
& \left(10 m^{2}+7 m-5\right)^{2}=100 m^{4}+140 m^{3}-51 m^{2}-70 m+25>4 a_{4} a_{6},
\end{aligned}
$$

so we must have

$$
4 a_{4} a_{6}=\left(10 m^{2}+7 m-6\right)^{2}=100 m^{4}+140 m^{3}-71 m^{2}-84 m+36 .
$$

This implies that $m^{2}-12 m+20=0$, so $m=2$ or $m=10$.

Final answer: 1,2
Subfield: Geometry


Question: Let $A B C$ be a triangle with $C A=C B$ and $\angle A C B=120^{\circ}$, and let $M$ be the midpoint of $A B$. Let $P$ be a variable point on the circumcircle of $A B C$, and let $Q$ be the point on the segment $C P$ such that $Q P=2 Q C$. It is given that the line through $P$ and perpendicular to $A B$ intersects the line $M Q$ at a unique point $N$.

Prove that there exists a fixed circle such that $N$ lies on this circle for all possible positions of $P$.

Solution:
Let $O$ be the circumcenter of $A B C$. From the assumption that $\angle A C B=120^{\circ}$ it follows that $M$ is the midpoint of $C O$.

Let $\omega$ denote the circle with center in $C$ and radius $C O$. This circle in the image of the circumcircle of $A B C$ through the translation that sends $O$ to $C$. We claim that $N$ lies on $\omega$.
<img_3895>

Let us consider the triangles $Q N P$ and $Q M C$. The angles in $Q$ are equal. Since $N P$ is parallel to $M C$ (both lines are perpendicular to $A B$ ), it turns out that $\angle Q N P=\angle Q M C$, and hence the two triangles are similar. Since $Q P=2 Q C$, it follows that

$$
N P=2 M C=C O
$$

which proves that $N$ lies on $\omega$.
Let $M^{\prime}$ denote the symmetric of $M$ with respect to $O$.

Let us consider the quadrilateral $M M^{\prime} P N$. The lines $M M^{\prime}$ and $N P$ are parallel by construction. Also the lines $P M^{\prime}$ and $N M$ are parallel (homothety from $C$ with coefficient 3). It follows that $M M^{\prime} P N$ is a parallelogram, and hence $P N=M M^{\prime}=O C$.
There are many computation approaches to this problem. For example, we can set Cartesian coordinates so that

$$
A=\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right), \quad B=\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right), \quad C=(0,1), \quad M=\left(0, \frac{1}{2}\right)
$$

Setting $P=(a, b)$, we obtain that $Q=(a / 3,(2+b) / 3)$. The equation of the line through $P$ and perpendicular to $A B$ is $x=a$. The equation of the line $M Q$ (if $a \neq 0$ ) is

$$
y-\frac{1}{2}=\frac{x}{a}\left(\frac{1}{2}+b\right)
$$

The intersection of the two lines is therefore

$$
N=(a, 1+b)=P+(0,1)
$$

This shows that the map $P \rightarrow N$ in the translation by the vector $(0,1)$. This result is independent of the position of $P$ (provided that $a \neq 0$, because otherwise $N$ is not well-defined).

When $P$ lies on the circumcircle of $A B C$, with the exception of the two points with $a=0$, then necessarily $N$ lies on the translated circle (which is the circle with center in $C$ and radius 1).
Subfield: Algebra


Question: Consider the set

$$
A=\left\{1+\frac{1}{k}: k=1,2,3, \ldots\right\}
$$
Prove that every integer $x \geq 2$ can be written as the product of one or more elements of $A$, which are not necessarily different.

Solution:
Every integer $x \geq 2$ can be written as the telescopic product of $x-1$ elements of $A$ as

$$
x=\left(1+\frac{1}{x-1}\right) \cdot\left(1+\frac{1}{x-2}\right) \cdot \ldots \cdot\left(1+\frac{1}{2}\right) \cdot\left(1+\frac{1}{1}\right),
$$

which is enough to establish the solution.
Subfield: Combinatorics


Question: A domino is a $1 \times 2$ or $2 \times 1$ tile.

Let $n \geq 3$ be an integer. Dominoes are placed on an $n \times n$ board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap.

The value of a row or column is the number of dominoes that cover at least one cell of this row or column. The configuration is called balanced if there exists some $k \geq 1$ such that each row and each column has a value of $k$.

Prove that a balanced configuration exists for every $n \geq 3$, and find the minimum number of dominoes needed in such a configuration.

Solution:
The minimal number of dominoes required in a balanced configuration is $2 n / 3$ if $n$ is a multiple of 3 , and $2 n$ otherwise.

In order to show that this number is necessary, we count in two different ways the number of elements of the set $S$ of all pairs $(\ell, d)$, where $\ell$ is a row or a column of the board, and $d$ is a domino that covers at least one cell of that row or column. On the one hand, since each row or column intersects the same number $k$ of dominoes, the set $S$ has $2 n k$ elements. On the other hand, since each domino intersects 3 rows/columns, the set $S$ has $3 D$ elements, where $D$ is the total number of dominoes on the board. This leads to the equality

$$
2 n k=3 D \text {. }
$$

If $n$ is a multiple of 3 , from the trivial inequality $k \geq 1$ we obtain that $D \geq 2 n / 3$. If $n$ is not a multiple of 3 , then $k$ is a multiple of 3 , which means that $k \geq 3$ and hence $D \geq 2 n$.

Now we need to exhibit a balanced configuration with this number of dominoes. The following diagram shows a balanced configuration with $n=3$ and $k=1$.

<img_3670>

If $n$ is any multiple of 3 , we can obtain a balanced configuration with $k=1$ by using $n / 3$ of these $3 \times 3$ blocks along the principal diagonal of the board.

The following diagrams show balanced configurations with $k=3$ and $n \in\{4,5,6,7\}$.
<img_3264>

Any $n \geq 8$ can be written in the form $4 A+r$ where $A$ is a positive integer and $r \in\{4,5,6,7\}$. Therefore, we can obtain a balanced configuration with $n \geq 8$ and $k=3$ by using one block with size $r \times r$, and $A$ blocks with size $4 \times 4$ along the principal diagonal of the board. In particular, this construction covers all the cases where $n$ is not a multiple of 3 .