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


Question: Let $a, b$ and $c$ be positive real numbers such that $\min \{a b, b c, c a\} \geqslant 1$. Prove that

$$
\sqrt[3]{\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)} \leqslant\left(\frac{a+b+c}{3}\right)^{2}+1\tag{1}
$$

Solution:
We first show the following.

- Claim. For any positive real numbers $x, y$ with $x y \geqslant 1$, we have

$$
\left(x^{2}+1\right)\left(y^{2}+1\right) \leqslant\left(\left(\frac{x+y}{2}\right)^{2}+1\right)^{2}\tag{2}
$$

Proof. Note that $x y \geqslant 1$ implies $\left(\frac{x+y}{2}\right)^{2}-1 \geqslant x y-1 \geqslant 0$. We find that

$$
\left(x^{2}+1\right)\left(y^{2}+1\right)=(x y-1)^{2}+(x+y)^{2} \leqslant\left(\left(\frac{x+y}{2}\right)^{2}-1\right)^{2}+(x+y)^{2}=\left(\left(\frac{x+y}{2}\right)^{2}+1\right)^{2} .
$$

Without loss of generality, assume $a \geqslant b \geqslant c$. This implies $a \geqslant 1$. Let $d=\frac{a+b+c}{3}$. Note that

$$
a d=\frac{a(a+b+c)}{3} \geqslant \frac{1+1+1}{3}=1 .
$$

Then we can apply (2) to the pair $(a, d)$ and the pair $(b, c)$. We get

$$
\left(a^{2}+1\right)\left(d^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \leqslant\left(\left(\frac{a+d}{2}\right)^{2}+1\right)^{2}\left(\left(\frac{b+c}{2}\right)^{2}+1\right)^{2} .\tag{3}
$$

Next, from

$$
\frac{a+d}{2} \cdot \frac{b+c}{2} \geqslant \sqrt{a d} \cdot \sqrt{b c} \geqslant 1
$$

we can apply (2) again to the pair $\left(\frac{a+d}{2}, \frac{b+c}{2}\right)$. Together with (3), we have

$$
\left(a^{2}+1\right)\left(d^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \leqslant\left(\left(\frac{a+b+c+d}{4}\right)^{2}+1\right)^{4}=\left(d^{2}+1\right)^{4} .
$$

Therefore, $\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \leqslant\left(d^{2}+1\right)^{3}$, and (1) follows by taking cube root of both sides.
Subfield: Algebra


Question: Find the smallest real constant $C$ such that for any positive real numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $a_{5}$ (not necessarily distinct), one can always choose distinct subscripts $i, j, k$ and $l$ such that

$$
\left|\frac{a_{i}}{a_{j}}-\frac{a_{k}}{a_{l}}\right| \leqslant C .\tag{1}
$$

Solution:
We first show that $C \leqslant \frac{1}{2}$. For any positive real numbers $a_{1} \leqslant a_{2} \leqslant a_{3} \leqslant a_{4} \leqslant a_{5}$, consider the five fractions

$$
\frac{a_{1}}{a_{2}}, \frac{a_{3}}{a_{4}}, \frac{a_{1}}{a_{5}}, \frac{a_{2}}{a_{3}}, \frac{a_{4}}{a_{5}}\tag{2}
$$

Each of them lies in the interval $(0,1]$. Therefore, by the Pigeonhole Principle, at least three of them must lie in $\left(0, \frac{1}{2}\right]$ or lie in $\left(\frac{1}{2}, 1\right]$ simultaneously. In particular, there must be two consecutive terms in (2) which belong to an interval of length $\frac{1}{2}$ (here, we regard $\frac{a_{1}}{a_{2}}$ and $\frac{a_{4}}{a_{5}}$ as consecutive). In other words, the difference of these two fractions is less than $\frac{1}{2}$. As the indices involved in these two fractions are distinct, we can choose them to be $i, j, k, l$ and conclude that $C \leqslant \frac{1}{2}$.

Next, we show that $C=\frac{1}{2}$ is best possible. Consider the numbers $1,2,2,2, n$ where $n$ is a large real number. The fractions formed by two of these numbers in ascending order are $\frac{1}{n}, \frac{2}{n}, \frac{1}{2}, \frac{2}{2}, \frac{2}{1}, \frac{n}{2}, \frac{n}{1}$. Since the indices $i, j, k, l$ are distinct, $\frac{1}{n}$ and $\frac{2}{n}$ cannot be chosen simultaneously. Therefore the minimum value of the left-hand side of (1) is $\frac{1}{2}-\frac{2}{n}$. When $n$ tends to infinity, this value approaches $\frac{1}{2}$, and so $C$ cannot be less than $\frac{1}{2}$.

These conclude that $C=\frac{1}{2}$ is the smallest possible choice.

Final answer: $\frac{1}{2}$
Subfield: Algebra


Question: 
Prove that for every positive integer $n$, there exists a fraction $\frac{a}{b}$ where $a$ and $b$ are integers satisfying $0<b \leqslant \sqrt{n}+1$ and $\sqrt{n} \leqslant \frac{a}{b} \leqslant \sqrt{n+1}$.

Solution:
Let $r$ be the unique positive integer for which $r^{2} \leqslant n<(r+1)^{2}$. Write $n=r^{2}+s$. Then we have $0 \leqslant s \leqslant 2 r$. We discuss in two cases according to the parity of $s$.

- Case 1. $s$ is even.

Consider the number $\left(r+\frac{s}{2 r}\right)^{2}=r^{2}+s+\left(\frac{s}{2 r}\right)^{2}$. We find that

$$
n=r^{2}+s \leqslant r^{2}+s+\left(\frac{s}{2 r}\right)^{2} \leqslant r^{2}+s+1=n+1 .
$$

It follows that

$$
\sqrt{n} \leqslant r+\frac{s}{2 r} \leqslant \sqrt{n+1} .
$$

Since $s$ is even, we can choose the fraction $r+\frac{s}{2 r}=\frac{r^{2}+(s / 2)}{r}$ since $r \leqslant \sqrt{n}$.

- Case 2. $s$ is odd.

Consider the number $\left(r+1-\frac{2 r+1-s}{2(r+1)}\right)^{2}=(r+1)^{2}-(2 r+1-s)+\left(\frac{2 r+1-s}{2(r+1)}\right)^{2}$. We find that

$$
\begin{aligned}
n=r^{2}+s=(r+1)^{2}-(2 r+1-s) & \leqslant(r+1)^{2}-(2 r+1-s)+\left(\frac{2 r+1-s}{2(r+1)}\right)^{2} \\
& \leqslant(r+1)^{2}-(2 r+1-s)+1=n+1
\end{aligned}
$$

It follows that

$$
\sqrt{n} \leqslant r+1-\frac{2 r+1-s}{2(r+1)} \leqslant \sqrt{n+1} .
$$

Since $s$ is odd, we can choose the fraction $(r+1)-\frac{2 r+1-s}{2(r+1)}=\frac{(r+1)^{2}-r+((s-1) / 2)}{r+1}$ since $r+1 \leqslant \sqrt{n}+1$.
Subfield: Algebra


Question: 
Prove that there are infinitely many positive integers $n$ such that there is no fraction $\frac{a}{b}$ where $a$ and $b$ are integers satisfying $0<b \leqslant \sqrt{n}$ and $\sqrt{n} \leqslant \frac{a}{b} \leqslant \sqrt{n+1}$.

Solution:
(b) We show that for every positive integer $r$, there is no fraction $\frac{a}{b}$ with $b \leqslant \sqrt{r^{2}+1}$ such that $\sqrt{r^{2}+1} \leqslant \frac{a}{b} \leqslant \sqrt{r^{2}+2}$. Suppose on the contrary that such a fraction exists. Since $b \leqslant \sqrt{r^{2}+1}<r+1$ and $b$ is an integer, we have $b \leqslant r$. Hence,

$$
(b r)^{2}<b^{2}\left(r^{2}+1\right) \leqslant a^{2} \leqslant b^{2}\left(r^{2}+2\right) \leqslant b^{2} r^{2}+2 b r<(b r+1)^{2} .
$$

This shows the square number $a^{2}$ is strictly bounded between the two consecutive squares $(b r)^{2}$ and $(b r+1)^{2}$, which is impossible. Hence, we have found infinitely many $n=r^{2}+1$ for which there is no fraction of the desired form.
Subfield: Algebra


Question: The equation

$$
(x-1)(x-2) \cdots(x-2016)=(x-1)(x-2) \cdots(x-2016)
$$

is written on the board. One tries to erase some linear factors from both sides so that each side still has at least one factor, and the resulting equation has no real roots. Find the least number of linear factors one needs to erase to achieve this.

Solution:
Since there are 2016 common linear factors on both sides, we need to erase at least 2016 factors. We claim that the equation has no real roots if we erase all factors $(x-k)$ on the left-hand side with $k \equiv 2,3(\bmod 4)$, and all factors $(x-m)$ on the right-hand side with $m \equiv 0,1(\bmod 4)$. Therefore, it suffices to show that no real number $x$ satisfies

$$
\prod_{j=0}^{503}(x-4 j-1)(x-4 j-4)=\prod_{j=0}^{503}(x-4 j-2)(x-4 j-3) .
\tag{1}
$$

- Case 1. $x=1,2, \ldots, 2016$.

In this case, one side of (1) is zero while the other side is not. This shows $x$ cannot satisfy (1).

- Case 2. $4 k+1<x<4 k+2$ or $4 k+3<x<4 k+4$ for some $k=0,1, \ldots, 503$.

For $j=0,1, \ldots, 503$ with $j \neq k$, the product $(x-4 j-1)(x-4 j-4)$ is positive. For $j=k$, the product $(x-4 k-1)(x-4 k-4)$ is negative. This shows the left-hand side of (1) is negative. On the other hand, each product $(x-4 j-2)(x-4 j-3)$ on the right-hand side of (1) is positive. This yields a contradiction.

- Case 3. $x<1$ or $x>2016$ or $4 k<x<4 k+1$ for some $k=1,2, \ldots, 503$.

The equation (1) can be rewritten as

$$
1=\prod_{j=0}^{503} \frac{(x-4 j-1)(x-4 j-4)}{(x-4 j-2)(x-4 j-3)}=\prod_{j=0}^{503}\left(1-\frac{2}{(x-4 j-2)(x-4 j-3)}\right) .
$$

Note that $(x-4 j-2)(x-4 j-3)>2$ for $0 \leqslant j \leqslant 503$ in this case. So each term in the product lies strictly between 0 and 1 , and the whole product must be less than 1 , which is impossible.

- Case 4. $4 k+2<x<4 k+3$ for some $k=0,1, \ldots, 503$.

This time we rewrite (1) as

$$
\begin{aligned}
1 & =\frac{x-1}{x-2} \cdot \frac{x-2016}{x-2015} \prod_{j=1}^{503} \frac{(x-4 j)(x-4 j-1)}{(x-4 j+1)(x-4 j-2)} \\
& =\frac{x-1}{x-2} \cdot \frac{x-2016}{x-2015} \prod_{j=1}^{503}\left(1+\frac{2}{(x-4 j+1)(x-4 j-2)}\right) .
\end{aligned}
$$

Clearly, $\frac{x-1}{x-2}$ and $\frac{x-2016}{x-2015}$ are both greater than 1 . For the range of $x$ in this case, each term in the product is also greater than 1 . Then the right-hand side must be greater than 1 and hence a contradiction arises.



From the four cases, we conclude that (1) has no real roots. Hence, the minimum number of linear factors to be erased is 2016 .

Final answer: 2016
Subfield: Combinatorics


Question: Find all positive integers $n$ for which all positive divisors of $n$ can be put into the cells of a rectangular table under the following constraints:

- each cell contains a distinct divisor;
- the sums of all rows are equal; and
- the sums of all columns are equal.

Solution:
Suppose all positive divisors of $n$ can be arranged into a rectangular table of size $k \times l$ where the number of rows $k$ does not exceed the number of columns $l$. Let the sum of numbers in each column be $s$. Since $n$ belongs to one of the columns, we have $s \geqslant n$, where equality holds only when $n=1$.

For $j=1,2, \ldots, l$, let $d_{j}$ be the largest number in the $j$-th column. Without loss of generality, assume $d_{1}>d_{2}>\cdots>d_{l}$. Since these are divisors of $n$, we have

$$
d_{l} \leqslant \frac{n}{l}\tag{1}
$$

As $d_{l}$ is the maximum entry of the $l$-th column, we must have

$$
d_{l} \geqslant \frac{s}{k} \geqslant \frac{n}{k}\tag{2}
$$

The relations (1) and (2) combine to give $\frac{n}{l} \geqslant \frac{n}{k}$, that is, $k \geqslant l$. Together with $k \leqslant l$, we conclude that $k=l$. Then all inequalities in (1) and (2) are equalities. In particular, $s=n$ and so $n=1$, in which case the conditions are clearly satisfied.
Clearly $n=1$ works. Then we assume $n>1$ and let its prime factorization be $n=p_{1}^{r_{1}} p_{2}^{r_{2}} \cdots p_{t}^{r_{t}}$. Suppose the table has $k$ rows and $l$ columns with $1<k \leqslant l$. Note that $k l$ is the number of positive divisors of $n$ and the sum of all entries is the sum of positive divisors of $n$, which we denote by $\sigma(n)$. Consider the column containing $n$. Since the column sum is $\frac{\sigma(n)}{l}$, we must have $\frac{\sigma(n)}{l}>n$. Therefore, we have

$$
\begin{aligned}
\left(r_{1}+1\right)\left(r_{2}+1\right) \cdots\left(r_{t}+1\right) & =k l \leqslant l^{2}<\left(\frac{\sigma(n)}{n}\right)^{2} \\
& =\left(1+\frac{1}{p_{1}}+\cdots+\frac{1}{p_{1}^{r_{1}}}\right)^{2} \cdots\left(1+\frac{1}{p_{t}}+\cdots+\frac{1}{p_{t}^{r_{t}}}\right)^{2} .
\end{aligned}
$$

This can be rewritten as

$$
f\left(p_{1}, r_{1}\right) f\left(p_{2}, r_{2}\right) \cdots f\left(p_{t}, r_{t}\right)<1
\tag{3}
$$

where

$$
f(p, r)=\frac{r+1}{\left(1+\frac{1}{p}+\cdots+\frac{1}{p^{r}}\right)^{2}}=\frac{(r+1)\left(1-\frac{1}{p}\right)^{2}}{\left(1-\frac{1}{p^{r+1}}\right)^{2}}
$$



Direct computation yields

$$
f(2,1)=\frac{8}{9}, \quad f(2,2)=\frac{48}{49}, \quad f(3,1)=\frac{9}{8}
$$

Also, we find that

$$
\begin{aligned}
& f(2, r) \geqslant\left(1-\frac{1}{2^{r+1}}\right)^{-2}>1 \quad \text { for } r \geqslant 3, \\
& f(3, r) \geqslant \frac{4}{3}\left(1-\frac{1}{3^{r+1}}\right)^{-2}>\frac{4}{3}>\frac{9}{8} \quad \text { for } r \geqslant 2, \text { and } \\
& f(p, r) \geqslant \frac{32}{25}\left(1-\frac{1}{p^{r+1}}\right)^{-2}>\frac{32}{25}>\frac{9}{8} \quad \text { for } p \geqslant 5 .
\end{aligned}
$$

From these values and bounds, it is clear that (3) holds only when $n=2$ or 4 . In both cases, it is easy to see that the conditions are not satisfied. Hence, the only possible $n$ is 1 .

Final answer: 1
Subfield: Combinatorics


Question: Let $n$ be a positive integer relatively prime to 6 . We paint the vertices of a regular $n$-gon with three colours so that there is an odd number of vertices of each colour. Show that there exists an isosceles triangle whose three vertices are of different colours.

Solution:
For $k=1,2,3$, let $a_{k}$ be the number of isosceles triangles whose vertices contain exactly $k$ colours. Suppose on the contrary that $a_{3}=0$. Let $b, c, d$ be the number of vertices of the three different colours respectively. We now count the number of pairs $(\triangle, E)$ where $\triangle$ is an isosceles triangle and $E$ is a side of $\triangle$ whose endpoints are of different colours.

On the one hand, since we have assumed $a_{3}=0$, each triangle in the pair must contain exactly two colours, and hence each triangle contributes twice. Thus the number of pairs is $2 a_{2}$.

On the other hand, if we pick any two vertices $A, B$ of distinct colours, then there are three isosceles triangles having these as vertices, two when $A B$ is not the base and one when $A B$ is the base since $n$ is odd. Note that the three triangles are all distinct as $(n, 3)=1$. In this way, we count the number of pairs to be $3(b c+c d+d b)$. However, note that $2 a_{2}$ is even while $3(b c+c d+d b)$ is odd, as each of $b, c, d$ is. This yields a contradiction and hence $a_{3} \geqslant 1$.
Subfield: Combinatorics


Question: Find all positive integers $n$ for which we can fill in the entries of an $n \times n$ table with the following properties:

- each entry can be one of $I, M$ and $O$;
- in each row and each column, the letters $I, M$ and $O$ occur the same number of times; and
- in any diagonal whose number of entries is a multiple of three, the letters $I, M$ and $O$ occur the same number of times.

Solution:
We first show that such a table exists when $n$ is a multiple of 9 . Consider the following $9 \times 9$ table.

$$
\left(\begin{array}{ccccccccc}
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M
\end{array}\right)\tag{1}
$$

It is a direct checking that the table (1) satisfies the requirements. For $n=9 k$ where $k$ is a positive integer, we form an $n \times n$ table using $k \times k$ copies of (1). For each row and each column of the table of size $n$, since there are three $I$ 's, three $M$ 's and three $O$ 's for any nine consecutive entries, the numbers of $I, M$ and $O$ are equal. In addition, every diagonal of the large table whose number of entries is divisible by 3 intersects each copy of (1) at a diagonal with number of entries divisible by 3 (possibly zero). Therefore, every such diagonal also contains the same number of $I, M$ and $O$.

Next, consider any $n \times n$ table for which the requirements can be met. As the number of entries of each row should be a multiple of 3 , we let $n=3 k$ where $k$ is a positive integer. We divide the whole table into $k \times k$ copies of $3 \times 3$ blocks. We call the entry at the centre of such a $3 \times 3$ square a vital entry. We also call any row, column or diagonal that contains at least one vital entry a vital line. We compute the number of pairs $(l, c)$ where $l$ is a vital line and $c$ is an entry belonging to $l$ that contains the letter $M$. We let this number be $N$.

On the one hand, since each vital line contains the same number of $I, M$ and $O$, it is obvious that each vital row and each vital column contain $k$ occurrences of $M$. For vital diagonals in either direction, we count there are exactly

$$
1+2+\cdots+(k-1)+k+(k-1)+\cdots+2+1=k^{2}
$$

occurrences of $M$. Therefore, we have $N=4 k^{2}$.



On the other hand, there are $3 k^{2}$ occurrences of $M$ in the whole table. Note that each entry belongs to exactly 1 or 4 vital lines. Therefore, $N$ must be congruent to $3 k^{2} \bmod 3$.

From the double counting, we get $4 k^{2} \equiv 3 k^{2}(\bmod 3)$, which forces $k$ to be a multiple of 3. Therefore, $n$ has to be a multiple of 9 and the proof is complete.

Final answer: $n$ can be any multiple of 9
Subfield: Combinatorics


Question: Let $n \geqslant 2$ be an integer. In the plane, there are $n$ segments given in such a way that any two segments have an intersection point in the interior, and no three segments intersect at a single point. Jeff places a snail at one of the endpoints of each of the segments and claps his hands $n-1$ times. Each time when he claps his hands, all the snails move along their own segments and stay at the next intersection points until the next clap. Since there are $n-1$ intersection points on each segment, all snails will reach the furthest intersection points from their starting points after $n-1$ claps.
(a) Prove that if $n$ is odd then Jeff can always place the snails so that no two of them ever occupy the same intersection point.

Solution:
We consider a big disk which contains all the segments. We extend each segment to a line $l_{i}$ so that each of them cuts the disk at two distinct points $A_{i}, B_{i}$.

For odd $n$, we travel along the circumference of the disk and mark each of the points $A_{i}$ or $B_{i}$ 'in' and 'out' alternately. Since every pair of lines intersect in the disk, there are exactly $n-1$ points between $A_{i}$ and $B_{i}$ for any fixed $1 \leqslant i \leqslant n$. As $n$ is odd, this means one of $A_{i}$ and $B_{i}$ is marked 'in' and the other is marked 'out'. Then Jeff can put a snail on the endpoint of each segment which is closer to the 'in' side of the corresponding line. We claim that the snails on $l_{i}$ and $l_{j}$ do not meet for any pairs $i, j$, hence proving part (a).
<img_3311>

Without loss of generality, we may assume the snails start at $A_{i}$ and $A_{j}$ respectively. Let $l_{i}$ intersect $l_{j}$ at $P$. Note that there is an odd number of points between $\operatorname{arc} A_{i} A_{j}$. Each of these points belongs to a line $l_{k}$. Such a line $l_{k}$ must intersect exactly one of



the segments $A_{i} P$ and $A_{j} P$, making an odd number of intersections. For the other lines, they may intersect both segments $A_{i} P$ and $A_{j} P$, or meet none of them. Therefore, the total number of intersection points on segments $A_{i} P$ and $A_{j} P$ (not counting $P$ ) is odd. However, if the snails arrive at $P$ at the same time, then there should be the same number of intersections on $A_{i} P$ and $A_{j} P$, which gives an even number of intersections. This is a contradiction so the snails do not meet each other.
Subfield: Combinatorics


Question: Let $n \geqslant 2$ be an integer. In the plane, there are $n$ segments given in such a way that any two segments have an intersection point in the interior, and no three segments intersect at a single point. Jeff places a snail at one of the endpoints of each of the segments and claps his hands $n-1$ times. Each time when he claps his hands, all the snails move along their own segments and stay at the next intersection points until the next clap. Since there are $n-1$ intersection points on each segment, all snails will reach the furthest intersection points from their starting points after $n-1$ claps.
Prove that if $n$ is even then there must be a moment when some two snails occupy the same intersection point no matter how Jeff places the snails.

Solution:
We consider a big disk which contains all the segments. We extend each segment to a line $l_{i}$ so that each of them cuts the disk at two distinct points $A_{i}, B_{i}$.

For even $n$, we consider any way that Jeff places the snails and mark each of the points $A_{i}$ or $B_{i}$ 'in' and 'out' according to the directions travelled by the snails. In this case there must be two neighbouring points $A_{i}$ and $A_{j}$ both of which are marked 'in'. Let $P$ be the intersection of the segments $A_{i} B_{i}$ and $A_{j} B_{j}$. Then any other segment meeting one of the segments $A_{i} P$ and $A_{j} P$ must also meet the other one, and so the number of intersections on $A_{i} P$ and $A_{j} P$ are the same. This shows the snails starting from $A_{i}$ and $A_{j}$ will meet at $P$.
Subfield: Geometry


Question: Let $B=(-1,0)$ and $C=(1,0)$ be fixed points on the coordinate plane. A nonempty, bounded subset $S$ of the plane is said to be nice if

(i) there is a point $T$ in $S$ such that for every point $Q$ in $S$, the segment $T Q$ lies entirely in $S$; and

(ii) for any triangle $P_{1} P_{2} P_{3}$, there exists a unique point $A$ in $S$ and a permutation $\sigma$ of the indices $\{1,2,3\}$ for which triangles $A B C$ and $P_{\sigma(1)} P_{\sigma(2)} P_{\sigma(3)}$ are similar.

Prove that there exist two distinct nice subsets $S$ and $S^{\prime}$ of the set $\{(x, y): x \geqslant 0, y \geqslant 0\}$ such that if $A \in S$ and $A^{\prime} \in S^{\prime}$ are the unique choices of points in (ii), then the product $B A \cdot B A^{\prime}$ is a constant independent of the triangle $P_{1} P_{2} P_{3}$.

Solution:
If in the similarity of $\triangle A B C$ and $\triangle P_{\sigma(1)} P_{\sigma(2)} P_{\sigma(3)}, B C$ corresponds to the longest side of $\triangle P_{1} P_{2} P_{3}$, then we have $B C \geqslant A B \geqslant A C$. The condition $B C \geqslant A B$ is equivalent to $(x+1)^{2}+y^{2} \leqslant 4$, while $A B \geqslant A C$ is trivially satisfied for any point in the first quadrant. Then we first define

$$
S=\left\{(x, y):(x+1)^{2}+y^{2} \leqslant 4, x \geqslant 0, y \geqslant 0\right\}
$$

Note that $S$ is the intersection of a disk and the first quadrant, so it is bounded and convex, and we can choose any $T \in S$ to meet condition (i). For any point $A$ in $S$, the relation $B C \geqslant A B \geqslant A C$ always holds. Therefore, the point $A$ in (ii) is uniquely determined, while its existence is guaranteed by the above construction.

<img_3425>

Next, if in the similarity of $\triangle A^{\prime} B C$ and $\triangle P_{\sigma(1)} P_{\sigma(2)} P_{\sigma(3)}, B C$ corresponds to the second longest side of $\triangle P_{1} P_{2} P_{3}$, then we have $A^{\prime} B \geqslant B C \geqslant A^{\prime} C$. The two inequalities are equivalent to $(x+1)^{2}+y^{2} \geqslant 4$ and $(x-1)^{2}+y^{2} \leqslant 4$ respectively. Then we define

$$
S^{\prime}=\left\{(x, y):(x+1)^{2}+y^{2} \geqslant 4,(x-1)^{2}+y^{2} \leqslant 4, x \geqslant 0, y \geqslant 0\right\}
$$



The boundedness condition is satisfied while (ii) can be argued as in the previous case. For (i), note that $S^{\prime}$ contains points inside the disk $(x-1)^{2}+y^{2} \leqslant 4$ and outside the disk $(x+1)^{2}+y^{2} \geqslant 4$. This shows we can take $T^{\prime}=(1,2)$ in (i), which is the topmost point of the circle $(x-1)^{2}+y^{2}=4$.

It remains to check that the product $B A \cdot B A^{\prime}$ is a constant. Suppose we are given a triangle $P_{1} P_{2} P_{3}$ with $P_{1} P_{2} \geqslant P_{2} P_{3} \geqslant P_{3} P_{1}$. By the similarity, we have

$$
B A=B C \cdot \frac{P_{2} P_{3}}{P_{1} P_{2}} \quad \text { and } \quad B A^{\prime}=B C \cdot \frac{P_{1} P_{2}}{P_{2} P_{3}}
$$

Thus $B A \cdot B A^{\prime}=B C^{2}=4$, which is certainly independent of the triangle $P_{1} P_{2} P_{3}$.