In $\triangle A B C$, $\angle C=90^{\circ}$, and the sides opposite to $\angle A$, $\angle B$, and $\angle C$ are $a$, $b$, and $c$ respectively. If the sum of the squares of the roots of the equation $c\left(x^{2}+1\right)-2 \sqrt{2} b x-a\left(x^{2}-1\right)=0$ is 10, then the value of $\frac{b}{a}$ is . $\qquad$

Solution:
3. $\sqrt{15}$.

Given the equation transformed as
$$
(c-a) x^{2}-2 \sqrt{2} b x+(c+a)=0 \text {. }
$$

From $\Delta=8 b^{2}-4\left(c^{2}-a^{2}\right)=4 b^{2}>0$, we know the equation has two real roots $x_{1} 、 x_{2}$, and
$$
\begin{array}{l}
x_{1}^{2}+x_{2}^{2}=\left(\frac{2 \sqrt{2} b}{c-a}\right)^{2}-2 \cdot \frac{c+a}{c-a} \\
=8 \cdot \frac{c^{2}-a^{2}}{(c-a)^{2}}-2 \cdot \frac{c+a}{c-a} \\
=6 \cdot \frac{c+a}{c-a}=10 .
\end{array}
$$

Therefore, $c=4 a, b=\sqrt{15} a \Rightarrow \frac{b}{a}=\sqrt{15}$.

Answer: \sqrt{15}


A rectangle is inscribed in a circle of area $32\pi$ and the area of the rectangle is $34$. Find its perimeter.

Solution:
1. **Determine the radius of the circle:**
   The area of the circle is given as \(32\pi\). The formula for the area of a circle is:
   \[
   \pi r^2 = 32\pi
   \]
   Solving for \(r\):
   \[
   r^2 = 32 \implies r = \sqrt{32} = 4\sqrt{2}
   \]

2. **Relate the radius to the rectangle:**
   The rectangle is inscribed in the circle, so the diagonal of the rectangle is the diameter of the circle. Therefore, the diameter \(d\) is:
   \[
   d = 2r = 2 \times 4\sqrt{2} = 8\sqrt{2}
   \]

3. **Express the diagonal in terms of the rectangle's side lengths:**
   Let the side lengths of the rectangle be \(a\) and \(b\). The diagonal of the rectangle can be expressed using the Pythagorean theorem:
   \[
   \sqrt{a^2 + b^2} = 8\sqrt{2}
   \]
   Squaring both sides:
   \[
   a^2 + b^2 = (8\sqrt{2})^2 = 128
   \]

4. **Use the area of the rectangle:**
   The area of the rectangle is given as 34:
   \[
   ab = 34
   \]

5. **Find the sum of the sides:**
   We need to find \(a + b\). We use the identity \((a + b)^2 = a^2 + b^2 + 2ab\):
   \[
   (a + b)^2 = 128 + 2 \times 34 = 128 + 68 = 196
   \]
   Taking the square root of both sides:
   \[
   a + b = \sqrt{196} = 14
   \]

6. **Calculate the perimeter:**
   The perimeter \(P\) of the rectangle is given by:
   \[
   P = 2(a + b) = 2 \times 14 = 28
   \]

The final answer is \(\boxed{28}\).

Answer: 28


A square-based pyramid has all edges 2 units long. We attach a pyramid to one of the side faces of the pyramid, where the side edges of the new pyramid are of equal length. The total length of the edges of the resulting solid is 18 units. What is the volume of the solid?

Solution:
The base of the pyramid $A B C D E$ is a square, and every edge is 2 units long, so the side faces are equilateral triangles. The edges of the added pyramid $B C E F$ are equal, making $B C E F$ a regular pyramid. The side edges of this pyramid are longer than the distance from the center of the equilateral triangle $B C E$ to its vertices, so the total length of the side edges is at least $3 \cdot \frac{2 \sqrt{3}}{2} \cdot \frac{2}{3}=2 \sqrt{3}>2$. Therefore, the total length of the edges of the resulting solid can only be 18 if some edges of the original pyramid coincide with the faces of the new solid. The vertex $F$ is located perpendicularly above the centroid of the triangle $E B C$, so $F$ cannot lie in the plane of $A B C D$. If $F$ lies in the plane of $A B E$, then, since the perpendicular bisector plane of the segment $B C$ is a symmetry plane for both the original pyramid and the pyramid $B C E F$, $F$ must also lie in the plane of $C D E$. Therefore, the edges $B E$ and $C E$ of the original pyramid must coincide with the faces of the new solid. The total length of the edges of the new solid will be exactly 18 units if $B F=C F=E F=2$.

![](https://cdn.mathpix.com/cropped/2024_05_02_17e75dc0d6cc397fdf3cg-1.jpg?height=583&width=775&top_left_y=584&top_left_x=664)

According to this, the resulting solid consists of a square pyramid and a regular tetrahedron. The height of the square pyramid can be easily determined using the Pythagorean theorem in the triangle $A C E$: $\sqrt{2}$, so its volume is $\frac{4 \sqrt{2}}{3}$. The volume of the regular tetrahedron with an edge length of 2 units is $\frac{2 \sqrt{2}}{3}$. Therefore, the volume of the resulting solid is $2 \sqrt{2}$.

Answer: 2\sqrt{2}


Given $P(2,1)$, draw a line $l$ through point $P$ that intersects the positive $x$-axis and $y$-axis at points $A$ and $B$ respectively. Then the equation of the line $l$ that minimizes the perimeter of $\triangle AOB$ (where $O$ is the origin) is $\qquad$

Solution:
$$
4.3 x+4 y-10=0 \text {. }
$$

As shown in Figure 2, let $\angle B A O = \theta, t = \tan \frac{\theta}{2}$. Then the perimeter of $\triangle A O B$ is
$$
\begin{aligned}
s= & (2+\cot \theta)+ \\
& (1+2 \tan \theta)+ \\
& \left(\frac{1}{\sin \theta}+\frac{2}{\cos \theta}\right) \\
= & 3+\frac{1+\cos \theta}{\sin \theta}+\frac{2(1+\sin \theta)}{\cos \theta} \\
= & 1+\frac{1}{t}+\frac{4}{1-t} \\
= & 6+\frac{1-t}{t}+\frac{4 t}{1-t} \geqslant 10 .
\end{aligned}
$$

Equality holds if and only if $\frac{1-t}{t}=\frac{4 t}{1-t}$, i.e., $\tan \frac{\theta}{2}=\frac{1}{3}$.
Then the slope of $l$ is $k=-\tan \theta=-\frac{2 \tan \frac{\theta}{2}}{1-\tan ^{2} \frac{\theta}{2}}=-\frac{3}{4}$.
Therefore, the equation of $l$ is $3 x+4 y-10=0$.

Answer: 3 x+4 y-10=0


Find the sum of all real roots of the equation $3 \tan ^{2} x+8 \tan x+3=0$ in the range $0<x<2 \pi$.

Solution:
18. $5 \pi$
18. Being a quadratic equation in $\tan x$ with discriminant $8^{2}-4(3)(3)>0$, we see that there are two possible values of $\tan x$ (say $\tan x_{1}$ and $\tan x_{2}$ ) and hence four possible values of $x$ in the range $0<x<2 \pi$. Since $\tan x_{1}+\tan x_{2}=-\frac{8}{3}$ and $\tan x_{1} \tan x_{2}=\frac{3}{3}=1$, both $\tan x_{1}$ and $\tan x_{2}$ are negative, so we have two possible values of $x$ in the second quadrant and two in the fourth quadrant.
Consider $x_{1}, x_{2}$ in the second quadrant satisfying the equation. From $\tan x_{1} \tan x_{2}=1$, we get $\tan x_{2}=\cot x_{1}=\tan \left(\frac{3 \pi}{2}-x_{1}\right)$. As $\frac{3 \pi}{2}-x_{1}$ is also in the second quadrant, we must have $x_{2}=\frac{3 \pi}{2}-x_{1}$. On the other hand, it is easy to see that the two possible values of $x$ in the fourth quadrant are simply $x_{1}+\pi$ and $x_{2}+\pi$, as $\tan (x+\pi)=\tan x$ for all $x$ (except at points where $\tan x$ is undefined). It follows that the answer is
$$
x_{1}+\left(\frac{3 \pi}{2}-x_{1}\right)+\left(x_{1}+\pi\right)+\left(\frac{3 \pi}{2}-x_{1}+\pi\right)=5 \pi \text {. }
$$

Answer: 5\pi


Let \(\left(a_{n}\right)_{n=1}^{\infty}\) be an infinite sequence such that for all positive integers \(n\) we have

\[
a_{n+1}=\frac{a_{n}^{2}}{a_{n}^{2}-4 a_{n}+6}
\]

a) Find all values \(a_{1}\) for which the sequence is constant.

b) Let \(a_{1}=5\). Find \(\left\lfloor a_{2018}\right\rfloor\).

Solution:
Solution. a) Assume $\left(a_{n}\right)_{n=1}^{\infty}$ is constant. Then $a_{2}=a_{1}$, hence

$$
a_{1}=\frac{a_{1}^{2}}{a_{1}^{2}-4 a_{1}+6}
$$

which rewrites as $a_{1}\left(a_{1}-2\right)\left(a_{1}-3\right)=0$. It follows that $a_{1} \in\{0,2,3\}$. On the other hand, once $a_{2}=a_{1}$, the sequence is clearly constant (formally e.g. by mathematical induction), thus any such $a_{1}$ indeed works.

b) Let $a_{1}=5$. Using the recurrence, we obtain $a_{2} \approx 2.27, a_{3} \approx 2.49, a_{4} \approx 2.77$, and so on. This leads to the following conjecture: For any $n \geqslant 2$ we have $20, \\
& a_{n+1}-2=\frac{a_{n}^{2}}{a_{n}^{2}-4 a_{n}+6}-2=\frac{\left(6-a_{n}\right)\left(a_{n}-2\right)}{\left(a_{n}-2\right)^{2}+2}>0 .
\end{aligned}
$$

This proves $2<a_{n+1}<3$, hence the proof of the conjecture is complete. In particular, $2<a_{2018}<3$ and the answer is $\left\lfloor a_{2018}\right\rfloor=2$.

Answer: 2


Suppose the real numbers $x$ and $y$ satisfy the equations
$$
x^{3}-3 x^{2}+5 x=1 \text { and } y^{3}-3 y^{2}+5 y=5
$$

Find $x+y$.

Solution:
12. Answer: 2
Solution. From $x^{3}-3 x^{2}+5 x=1$, we have
$$
(x-1)^{3}+2(x-1)=-2,
$$
and from $y^{3}-3 y^{2}+5 y=5$, we have
$$
(y-1)^{3}+2(y-1)=2 .
$$

Thus
$$
\begin{array}{c}
0=(x-1)^{3}+2(x-1)+(y-1)^{3}+2(y-1) \\
=(x+y-2)\left((x-1)^{2}-(x-1)(y-1)+(y-1)^{2}\right)+2(x+y-2) \\
=(x+y-2)\left(2+(x-1)^{2}-(x-1)(y-1)+(y-1)^{2}\right) .
\end{array}
$$

For any real numbers $x$ and $y$, we always have
$$
(x-1)^{2}-(x-1)(y-1)+(y-1)^{2} \geq 0
$$
and thus $x+y-2=0$, implying that $x+y=2$.

Answer: 2


A row of soldiers is called incorrect if no three consecutive soldiers stand in order of height (neither in ascending nor in descending order). How many incorrect rows can be formed from $n$ soldiers of different heights if
a) $n=4$;

b) $n=5$?

Solution:
Let's place a "+" between neighboring soldiers if the right one is taller than the left one, and a "-" otherwise. According to the condition, the plus and minus signs in the incorrect rows should alternate. It is clear that the number of rows starting with a plus is equal to the number of rows starting with a minus.

a) Let's find the number of rows of the type *+*_*+*. Denote the soldiers by the letters $A, B, C$ and $D$ in decreasing order of height. In the row of the specified type, $A$ can stand in the second or fourth position (after a plus).

In the row ${ }^{*} A^{*+*}$, soldier $B$ can stand in the first or fourth position. In the first case, we have one row - $B A D C$, in the second case - two: $C$ and $D$ can be placed in the remaining positions arbitrarily.

In the row *+*_*A, soldier $B$ can stand only in the second position. There are again two such rows.

In total, we have 5 rows of the type *+*_*+*, and all incorrect rows are twice as many.

b) Let's find the number of rows of the type *+*_*+*_*. Denote the soldiers by the letters $A, B, C, D$ and $E$ in decreasing order of height. Here, $A$ can stand in the second or fourth position. But these cases are symmetric, so it is enough to consider only the first: * $A^{*+*}-*$. In this case, $B$ can stand in the first or fourth position.

In the first case ( $B A^{*+*}{ }^{*}$ ), $C$ can stand only in the fourth position, and $D$ and $E$ can be placed in the remaining positions arbitrarily (2 rows).

In the second case $\left({ }^{*} A^{*} B^{*}\right)$, we have 6 rows: $C, D$ and $E$ can be placed arbitrarily.

In total, we have 8 rows of the type * $A^{*+*-*}$, and all incorrect rows are four times as many.

## Answer

a) 10 rows;
b) 32 rows.

Answer: 32


Three customers are at the hairdresser, each paying their bill at the cash register.

- The first customer pays an amount equal to the amount in the cash register and receives 10 reais in change.
- The second customer performs the same operation as the first.
- The third customer performs the same operation as the first.

Find the initial amount in the cash register, knowing that, at the end of the three operations, the cash register was empty.

Solution:
Let $x$ be the initial amount in the cash register. This amount plus what the three customers paid will result in an empty cash register.

- The first customer pays $x-10$ and, after him, there are $x+x-10=2 x-10$ reais in the cash register.
- The second customer pays $(2 x-10)-10=2 x-20$ and, after him, there are $(2 x-10)+(2 x-20)=4 x-30$ in the cash register.
- The third customer pays $(4 x-30)-10=4 x-40$ and, after him, there are $(4 x-30)+(4 x-40)=8 x-70$ in the cash register.

Since the cash register is empty after the third customer, $8 x-70=0$, that is,

$$
x=70 / 8=8.75 \text { reais. }
$$

Answer: 8.75


Solve the equation $\arccos \left(3 x^{2}\right)+2 \arcsin x=0$.

- Compute the cosine of both sides of the equation $\arccos \left(3 x^{2}\right)=-2 \arcsin x$

$3 x^{2}=\cos (-2 \arcsin x)=1-2 \sin ^{2}(\arcsin x)=1-2 x^{2}$

$$
\begin{gathered}
3 x^{2}=1-2 x^{2} \\
x^{2}=\frac{1}{5} ; \quad x= \pm \frac{1}{\sqrt{5}}
\end{gathered}
$$

Instead of checking, note that $3 x^{2} \geq 0$;

$$
\begin{gathered}
0 \leq \arccos \left(3 x^{2}\right) \leq \frac{\pi}{2} \\
-\frac{\pi}{2} \leq 2 \arcsin x \leq 0 \\
-\frac{\pi}{4} \leq \arcsin x \leq 0 ;-\frac{\sqrt{2}}{2} \leq x \leq 0
\end{gathered}
$$

Solution: Answer: $-\frac{1}{\sqrt{5}}$.

Answer: -\frac{1}{\sqrt{5}}


Find all odd prime $p$ such that $1+k(p-1)$ is prime for all integer $k$ where $1 \le k \le \dfrac{p-1}{2}$.

Solution:
1. **Restate the problem**: We need to find all odd primes \( p \) such that \( 1 + k(p-1) \) is prime for all integers \( k \) where \( 1 \le k \le \frac{p-1}{2} \).

2. **Consider the case \( p \ge 13 \)**:
   - Let \( q \) be an odd prime such that \( q \le \frac{p-1}{2} \).
   - We need \( 1 + k(p-1) \) to be prime for \( k = 1, 2, \ldots, q \).
   - By the Pigeonhole Principle, there exist \( a \) and \( b \) such that \( 1 + a(p-1) \equiv 1 + b(p-1) \pmod{q} \).
   - This implies \( q \mid (a-b)(p-1) \) and since \( |a-b| < q \), we have \( q \mid p-1 \).
   - Therefore, \( q \mid \frac{p-1}{2} \).

3. **Apply Bertrand's Postulate**:
   - Bertrand's Postulate states that for any integer \( n > 1 \), there is always at least one prime \( r \) such that \( n < r < 2n \).
   - Applying this to \( \frac{p-1}{4} \), we get a prime \( r \) such that \( \frac{p-1}{4} < r < \frac{p-1}{2} \).
   - Since \( r \) is a prime and \( r \neq 3 \), we have \( r \mid \frac{p-1}{2} \) and \( 3 \mid \frac{p-1}{2} \).

4. **Derive a contradiction**:
   - If \( 3r \mid \frac{p-1}{2} \), then \( \frac{3(p-1)}{2} < 3r \le \frac{p-1}{2} \), which is a contradiction because \( \frac{3(p-1)}{2} \) is greater than \( \frac{p-1}{2} \).

5. **Consider the case \( p \le 12 \)**:
   - For \( p = 11 \): \( 1 + 2(11-1) = 21 \) is not prime.
   - For \( p = 7 \): \( 1 + (7-1) = 7 \), \( 1 + 2(7-1) = 13 \), \( 1 + 3(7-1) = 19 \) are all primes.
   - For \( p = 5 \): \( 1 + 2(5-1) = 9 \) is not prime.
   - For \( p = 3 \): \( 1 + (3-1) = 3 \) is prime.

6. **Conclusion**:
   - The only odd primes \( p \) that satisfy the condition are \( p = 3 \) and \( p = 7 \).

The final answer is \(\boxed{3, 7}\)

Answer: 3, 7


In a certain populated place, all telephone numbers consist of six digits arranged in strictly ascending or strictly descending order, with the first digit in the number not being 0. What is the maximum number of telephone numbers that can exist in this place?

Solution:
Solution. Let's set $n=n_{1}+n_{2}$, where $n_{1}$ is the number of telephone numbers with digits in strictly increasing order, and $n_{2}$ is the number of telephone numbers with digits in strictly decreasing order. Each number whose digits are in strictly decreasing order is obtained by selecting six digits out of a total of 10, which are then arranged from the largest to the smallest. Therefore, such numbers can total $n_{2}=\binom{10}{6}=210$. Similarly, each number whose digits are in strictly increasing order is obtained by selecting the digits and then arranging them from the smallest to the largest. Here, we have 9 digits available, as numbers cannot start with 0, which would be the smallest digit in any possible combination that includes it. Thus, $n_{1}=\left({ }_{6}^{9}\right)=84$. Therefore, the maximum number of telephone numbers that can exist in this place is

$$
n=84+210=294
$$

NOTE. Prove that

$$
\frac{1}{2021}<\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \ldots \cdot \frac{2019}{2020}<\sqrt{\frac{1}{2021}}
$$

Answer: 294


Does there exist a fraction equal to $\frac{7}{13}$, the difference between the denominator and the numerator of which is 24?

Solution:
Answer: Yes, it exists, $\frac{28}{52}$.

Solution. Since $\frac{7}{13}$ is an irreducible fraction, any fraction equal to it has the form $\frac{7 x}{13 x}$, where $x$ is some natural number. In this case, the difference between the denominator and the numerator of such a fraction will be equal to $6 x$. We have $6 x=24$, hence $x=4$. This is the only fraction that fits the condition of the problem.

## Criteria

4 p. A correct example is provided, or it is proven that such a fraction exists.

Answer: \frac{28}{52}


Given an integer $n \geqslant 2$, let integers $a_{0}$, $a_{1}, \cdots, a_{n}$ satisfy
$$
0=a_{0}<a_{1}<\cdots<a_{n}=2 n-1 .
$$

Find the minimum possible number of elements in the set $\left\{a_{i}+a_{j} \mid 0 \leqslant i \leqslant j \leqslant n\right\}$.

Solution:
Let $S=\left\{a_{i}+a_{j} \mid 0 \leqslant i \leqslant j \leqslant n\right\}$.
If we take $a_{k}=k(k=1,2, \cdots, n-1)$, then
$$
S=\{0,1, \cdots, 3 n-2,4 n-2\} .
$$

Therefore, $|S|=3 n$.
We now prove: $|S| \geqslant 3 n$ always holds.
It is easy to see that the following $2 n+1$ numbers
$$
a_{0}|X|$, then $Y \cap Z \neq \varnothing$.
Thus, there exists
$a_{j}=b_{i}+a_{n}-a_{k}$ or $a_{j}+a_{n}=b_{i}+a_{n}-a_{k}$.
Hence $b_{i}+a_{n}=a_{j}+a_{k} \in S$ or $b_{i}=a_{j}+a_{k} \in S$.
In conclusion, the minimum possible value of $|S|$ is $3 n$.

Answer: 3n


There are two numbers 365 and 24. Now subtract 19 from the first number and add 12 to the second number, which counts as one operation. How many operations will it take for the two numbers to be equal.

Solution:
【Answer】11 times.
【Analysis】Key point: Difference Analysis
Xueersi corresponding lecture: Age problems involving catching up differences in the third-grade autumn class,
Profit and loss problems in the third-grade summer class, are actually all difference analysis.
The original difference between the two numbers: $365-24=341$, each operation reduces the first number by 19 and increases the second number by 12, so the difference between the two numbers decreases by $19+12=31$. It requires $341 \div 31=11$ (times) of operations to make the two numbers equal.

Answer: 11


$\mathrm{ABCD}$ is a trapezoid with bases $\mathrm{AD}=6$ and $\mathrm{BC}=10$. It turns out that the midpoints of all four sides of the trapezoid lie on the same circle. Find its radius.

Solution:
Answer: 4.

## Examples of answer notation:

45

$4 ; 5$

#

Answer: 4


In an $8 \times 8$ frame that is 2 cells wide (see figure), there are a total of 48 cells.

How many cells are in a $254 \times 254$ frame that is 2 cells wide?

Solution:
Answer: 2016.

## Solution

First method. Cut the frame into four identical rectangles as shown in the figure. The width of the rectangles is equal to the width of the frame, i.e., 2 cells. The length of each

![](https://cdn.mathpix.com/cropped/2024_05_06_833ed34de7b72168187fg-2.jpg?height=414&width=374&top_left_y=410&top_left_x=1572)
rectangle is 2 less than the side of the frame: $254-2=252$ cells. Then the area of one rectangle is $2 \cdot 252=504$. Therefore, the total number of cells in the frame is $4 \cdot 504=2016$.

![](https://cdn.mathpix.com/cropped/2024_05_06_833ed34de7b72168187fg-2.jpg?height=416&width=366&top_left_y=937&top_left_x=408)

Second method. The area of the frame can be obtained by subtracting the area of the inner square from the area of the large square $254 \times 254$. The side of the inner square is 4 cells less than the side of the large square. Therefore, the area of the frame is $254^{2}-250^{2}=(254-250)(254+250)=2016$.

Note. If the side of the frame is denoted by $n$, it can be proven (for example, by the methods described above) that its area will be $(8n - 16)$ cells.

## Grading Criteria.

- Any complete correct solution - 7 points.
- Correct approach, but an arithmetic error was made - 3 points.
- Correct reasoning, but an error was made in estimating the dimensions (for example, in the second method, it is incorrectly assumed that the side of the inner square is 2 cells less than the large one) - 2 points.
- Only the correct answer - 1 point.

Answer: 2016


In the plane a point $O$ is and a sequence of points $P_1, P_2, P_3, \ldots$ are given. The distances $OP_1, OP_2, OP_3, \ldots$ are $r_1, r_2, r_3, \ldots$ Let $\alpha$ satisfies $0 < \alpha < 1.$ Suppose that for every $n$ the distance from the point $P_n$ to any other point of the sequence is $\geq r^{\alpha}_n.$ Determine the exponent $\beta$, as large as possible such that for some $C$ independent of $n$
\[r_n \geq Cn^{\beta}, n = 1,2, \ldots\]

Solution:
1. **Understanding the Problem:**
   We are given a point \( O \) and a sequence of points \( P_1, P_2, P_3, \ldots \) in the plane. The distances from \( O \) to \( P_n \) are denoted by \( r_n \). We are also given that for every \( n \), the distance from \( P_n \) to any other point in the sequence is at least \( r_n^\alpha \), where \( 0 < \alpha < 1 \). We need to determine the largest possible exponent \( \beta \) such that \( r_n \geq Cn^\beta \) for some constant \( C \) independent of \( n \).

2. **Assumption and Condition:**
   We assume the additional condition \( r_1 \leq r_2 \leq r_3 \leq \cdots \) as mentioned in the IMO Compendium. This condition ensures that the sequence \( r_n \) is non-decreasing.

3. **Analyzing the Distance Condition:**
   The condition that the distance from \( P_n \) to any other point in the sequence is at least \( r_n^\alpha \) implies that the points \( P_n \) are spread out in the plane. Specifically, for any \( m \neq n \), the distance \( P_nP_m \geq r_n^\alpha \).

4. **Bounding \( r_n \):**
   To find the largest possible \( \beta \), we need to analyze how \( r_n \) grows with \( n \). We use the fact that the points \( P_n \) are spread out and the distances \( r_n \) are non-decreasing.

5. **Using the Distance Condition:**
   Consider the circle of radius \( r_n^\alpha \) centered at \( P_n \). No other point \( P_m \) (for \( m \neq n \)) can lie within this circle. This implies that the area around each point \( P_n \) is essentially "forbidden" for other points.

6. **Estimating the Growth of \( r_n \):**
   The area of the circle with radius \( r_n^\alpha \) is \( \pi (r_n^\alpha)^2 = \pi r_n^{2\alpha} \). Since the points are spread out, the total area covered by \( n \) such circles should be proportional to the area of a larger circle with radius \( r_n \). Thus, we have:
   \[
   n \cdot \pi r_n^{2\alpha} \leq \pi r_n^2
   \]
   Simplifying, we get:
   \[
   n \leq r_n^{2 - 2\alpha}
   \]
   Taking the logarithm on both sides, we get:
   \[
   \log n \leq (2 - 2\alpha) \log r_n
   \]
   Solving for \( r_n \), we get:
   \[
   r_n \geq n^{\frac{1}{2 - 2\alpha}}
   \]

7. **Determining \( \beta \):**
   Comparing this with \( r_n \geq Cn^\beta \), we see that:
   \[
   \beta = \frac{1}{2 - 2\alpha}
   \]

8. **Conclusion:**
   The largest possible exponent \( \beta \) such that \( r_n \geq Cn^\beta \) for some constant \( C \) independent of \( n \) is:
   \[
   \beta = \frac{1}{2(1 - \alpha)}
   \]

The final answer is \( \boxed{ \beta = \frac{1}{2(1 - \alpha)} } \)

Answer: \beta = \frac{1}{2(1 - \alpha)}


Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=a_{2}=1, a_{n}=\sqrt{3} a_{n-1}-a_{n-2}(n \geqslant 3) \text {. }
$$

Then $a_{2013}=$ $\qquad$

Solution:
4. $1-\sqrt{3}$.

From the given, we know
$$
\begin{array}{l}
a_{3}=\sqrt{3}-1, a_{4}=2-\sqrt{3}, a_{5}=\sqrt{3}-2, \\
a_{6}=1-\sqrt{3}, a_{7}=-1, a_{8}=-1 .
\end{array}
$$

Thus, $a_{n+6}=-a_{n}$.
Therefore, $\left\{a_{n}\right\}$ is a periodic sequence with a period of 12.
Hence, $a_{2013}=a_{9}=1-\sqrt{3}$.

Answer: 1-\sqrt{3}


Given that 5 students are proficient in Chinese, Mathematics, Physics, Chemistry, and History respectively, there are 5 test papers (one each for Chinese, Mathematics, Physics, Chemistry, and History). The teacher randomly distributes one test paper to each student. The probability that at least 4 students receive a test paper that does not match their proficient subject is $\qquad$ .

Solution:
In a permutation of $n$ elements, if all elements are not in their original positions, such a permutation is called a derangement of the original permutation, and the number of derangements of $n$ elements is denoted as $D(n)$.
Consider element $n$, placing it at position $k$, there are $n-1$ choices: if the element numbered $k$ is placed at position $n$, then the placement of the remaining $n-2$ elements has $D(n-2)$ methods; if the element numbered $k$ is not placed at position $n$, then the placement of the remaining $n-1$ elements has $D(n-1)$ methods, i.e., $D(n)=(n-1)[D(n-1)+D(n-2)]$. Let $D(n)=n!N(n)$, thus $D(1)=0, D(2)=1 \Rightarrow N(1)=0, N(2)=\frac{1}{2}$.
$$
\begin{array}{l}
n!N(n)=(n-1) \cdot(n-1)!N(n-1)+(n-1)!N(n-2) \\
\Rightarrow n N(n)=(n-1) N(n-1)+N(n-2) \\
\Rightarrow n[N(n)-N(n-1)]=-[N(n-1)-N(n-2)] \\
\Rightarrow N(n)-N(n-1)=-\frac{1}{n}[N(n-1)-N(n-2)]=\frac{(-1)^{2}}{n(n-1)}[N(n-2)-N(n-3)] \\
=\cdots=\frac{(-1)^{n-2}}{n(n-1) \cdots 3}[N(2)-N(1)]=\frac{(-1)^{n}}{n!},
\end{array}
$$

Therefore, $D(n)=n!\left[\frac{(-1)^{2}}{2!}+\frac{(-1)^{3}}{3!}+\cdots+\frac{(-1)^{n}}{n!}\right]$.
For example, $D(3)=6\left(\frac{1}{2}-\frac{1}{6}\right)=2, D(4)=24\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{24}\right)=9, D(5)=120\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{24}-\frac{1}{120}\right)=44$.
Returning to the original problem, let $A=$ “at least 4 students receive a test paper that does not match their strong subject”,
so $P(A)=\frac{5 \cdot 9+44}{120}=\frac{89}{120}$.

Answer: \frac{89}{120}


On the table lie 54 piles of stones with $1,2,3, \ldots, 54$ stones. In each step, we select any pile, say with $k$ stones, and remove it from the table along with $k$ stones from each pile that has at least $k$ stones. For example, after the first step, if we select the pile with 52 stones, the piles remaining on the table will have $1,2,3, \ldots, 51,1$ and 2 stones. Suppose that after a certain number of steps, only one pile remains on the table. Justify how many stones can be in it.

Solution:
1. If in each step we choose the pile with the most stones, we will gradually remove piles with $54, 53, 52, \ldots$ stones, and after 53 steps, only one pile with one stone will remain on the table.

We can prove that regardless of the procedure, the last pile will always contain a single stone. We will show that after each step, as long as there is at least one pile left on the table, the number of stones in each pile will always form the entire set $\{1, 2, \ldots, n\}$ for some natural number $n$ (we do not exclude the possibility that some numbers may correspond to multiple piles with the same number of stones). This means that there will always be at least one pile with exactly one stone on the table.

At the beginning, the number of stones in the piles forms the set $\{1, 2, \ldots, 54\}$. Suppose that after a certain number of steps, the number of stones in the piles forms the set $\{1, 2, \ldots, n\} (n \geq 2)$. If we now choose a pile with $n$ stones or a pile with one stone, the number of stones in the piles in the next step will form the set $\{1, 2, \ldots, n-1\}$. If we choose a pile with $m$ stones, where $m \notin \{1, n\}$, the number of stones in the piles in the next step will form the set $\{1, 2, \ldots, m-1\} \cup \{1, 2, \ldots, n-m\} = \{1, 2, \ldots, p\}$, where $p = \max \{m-1, n-m\}$. This proves the statement about the number of stones in the piles.

Answer: The last pile will always contain a single stone, regardless of the chosen procedure.

For a complete solution, award 6 points, of which 4 points for formulating the hypothesis that after each step, the number of stones in the piles forms the entire set $\{1, 2, \ldots, n\}$.

Answer: 1


Calculate $2 \operatorname{arctg} 4+\arcsin \frac{8}{17}$.

Solution:
Answer: $\pi$.

Solution. First solution. Let $\operatorname{arctg} 4$ be denoted by $\alpha$, and $\arcsin 8 / 17$ by $\beta$. Note that $\beta \in(0, \pi / 2)$, and $\operatorname{tg}^{2} \beta=\sin ^{2} \beta /\left(1-\sin ^{2} \beta\right)=8^{2} / 15^{2}$, from which $\operatorname{tg} \beta=8 / 15$; also $\operatorname{tg} \alpha=4, \alpha \in(0, \pi / 2)$.

We find:

$$
\begin{gathered}
\operatorname{tg}(2 \alpha)=\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha}=\frac{2 \cdot 4}{1-4^{2}}=-\frac{8}{15} \\
\operatorname{tg}(2 \alpha+\beta)=\frac{\operatorname{tg}(2 \alpha)+\operatorname{tg} \beta}{1-\operatorname{tg}(2 \alpha) \operatorname{tg} \beta}=\frac{-8 / 15+8 / 15}{1-(-8 / 15) \cdot(8 / 15)}=0
\end{gathered}
$$

Finally, since $0<\alpha<\pi / 2,0<\beta<\pi / 2$, then $0<2 \alpha+\beta<3 \pi / 2$. Therefore, $2 \alpha+\beta=\pi$.

Answer: \pi


In an equilateral triangle, a point $T$ is given whose distances from the sides of the triangle are 1, 2, and 3. What is the length of the side of this triangle?

Solution:
Solution.

We see that triangle $ABC$ can be divided into three triangles $CAT, ABT$, and $BCT$. The area of triangle $ABC$ is equal to the sum of the areas of these three triangles:

$$
P(ABC) = P(CAT) + P(ABT) + P(BCT)
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_460b0238d86ae9459a11g-4.jpg?height=534&width=571&top_left_y=650&top_left_x=761)

The area of triangle $CAT$ is equal to $\frac{1}{2}|CA| \cdot v_{b}$. Here, $v_{b}$, the length of the height of this triangle from vertex $T$, is the distance from point $T$ to side $\overline{CA}$.

(5 points)

Therefore, (see the figure) $P(CAT) = \frac{a \cdot 1}{2}$,

(2 points)

and similarly $P(ABT) = \frac{a \cdot 2}{2}, P(BCT) = \frac{a \cdot 3}{2}$,

(3 points)

so we get $\frac{a \cdot 1}{2} + \frac{a \cdot 2}{2} + \frac{a \cdot 3}{2} = \frac{a^2 \sqrt{3}}{4}$, and from this, finally, $3a = \frac{a^2 \sqrt{3}}{4}, a = 4 \sqrt{3}$.

(5 points)

Answer: 4\sqrt{3}


Vasya bought $n$ pairs of identical socks. For $n$ days, Vasya had no problems: every morning he took a new pair from the closet and wore it all day. After $n$ days, Vasya's mother washed all the socks in the washing machine and paired them up as they came out, since, to reiterate, the socks are identical. We will call a pair of socks successful if both socks in this pair were worn by Vasya on the same day.

a) Find the probability that all the resulting pairs are successful.

b) Prove that the expected number of successful pairs is greater than 0.5.

Solution:
a) The probability that the first pair is a match is $1 / 2 n-1$, since when the mother picked up the first sock, $2 n-1$ socks remained, and only one of them forms a matching pair with the first. Similarly, the probability that the second pair is a match (given that the first pair is a match) is $1 / 2 n-3$, and so on. The probability that all pairs are matches is obtained as the product $\frac{1}{2 n-1} \cdot \frac{1}{2 n-3} \cdot \ldots \cdot \frac{1}{3}=\frac{1}{(2 n-1)!!}=\frac{(2 n)!!}{(2 n)!}=\frac{2^{n} n!}{(2 n)!}$.

To find an approximate expression, we apply Stirling's formula: $m!\approx \sqrt{2 \pi m} \cdot m^{m} e^{-m}:$

$$
\frac{2^{n} n!}{(2 n)!} \approx \frac{2^{n} \sqrt{2 \pi n} \cdot n^{n} e^{-n}}{\sqrt{4 \pi n} \cdot(2 n)^{2 n} e^{-2 n}}=\frac{1}{\sqrt{2}}\left(\frac{e}{2 n}\right)^{n}
$$

b) The probability that a specific pair is a match, found in part a), is $1 / 2 n-1$. Let's number the pairs from 1 to n. We introduce indicators:

$\xi_{k}=1$, if the $k$-th pair is a match, $\xi_{k}=0$, if the $k$-th pair is not a match. Then $E \xi_{k}=1 / 2 n-1$.

The random variable $X$ "number of matching pairs" is equal to the sum of the indicators: $X=\xi_{1}+\xi_{2}+\ldots+\xi_{n}$.

Therefore, $\mathrm{E} X=n \mathrm{E} \xi_{1}=n / 2 n-1>0.5$.

## Answer

a) $\frac{2^{n} n!}{(2 n)!} \approx \frac{1}{\sqrt{2}}\left(\frac{e}{2 n}\right)^{n}$

Answer: \frac{2^{n}n!}{(2n)!}\approx\frac{1}{\sqrt{2}}(\frac{e}{2n})^{n}


Arrange $1,2,3,4,5,6$ randomly in a row, denoted as $a, b, c, d, e, f$. Then the probability that $a b c + d e f$ is an even number is $\qquad$

Solution:
3. $\frac{9}{10}$.

Consider the case where $a b c+d e f$ is odd. In this case, $a b c$ and $d e f$ are one odd and one even. If $a b c$ is odd, then $a, b, c$ are permutations of 1, 3, 5, and thus, $d, e, f$ are permutations of 2, 4, 6, which gives $3! \times 3! = 36$ cases.

By symmetry, the number of cases where $a b c+d e f$ is odd is $36 \times 2 = 72$.
Therefore, the probability that $a b c+d e f$ is even is
$$
1-\frac{72}{6!}=1-\frac{72}{720}=\frac{9}{10} .
$$

Answer: \frac{9}{10}


Let $k$, $m$, $n$ be integers. From a point $P\left(m^{3}-m, n^{3}-n\right)$ outside the circle $x^{2}+y^{2}=(3 k+1)^{2}$, two tangents are drawn to the circle, touching it at points $A$ and $B$. The number of integer points (points with both coordinates as integers) on the line $AB$ is ( ).
(A) 2
(B) 1
(C) 0
(D) infinitely many

Solution:
10. C.

It is known that the equation of the line $AB$ on which the chord of contact lies is
$$
\left(m^{3}-m\right) x+\left(n^{3}-n\right) y=(3 k+1)^{2} \text {. }
$$

If the line $AB$ contains an integer point $\left(x_{0}, y_{0}\right)$, then
$$
\begin{array}{l}
(m-1) m(m+1) x_{0}+(n-1) n(n+1) y_{0} \\
=(3 k+1)^{2} .
\end{array}
$$

Since $(m-1) m(m+1)$ and $(n-1) n(n+1)$ are both multiples of 3, the left side of the above equation can be divided by 3. However, the right side leaves a remainder of 1 when divided by 3, which is a contradiction.
Therefore, the line $AB$ does not contain any integer points.

Answer: C


How many integers $x$ satisfy the inequality $x^{4}-2020 x^{2}+2019<0$ ?

Solution:
## First Solution.

The given inequality can be written in factored form:

$$
\begin{gathered}
\left(x^{2}-2019\right)\left(x^{2}-1\right)<0 \\
(x-\sqrt{2019})(x+\sqrt{2019})(x-1)(x+1)<0 .
\end{gathered}
$$

Using a sign chart or graphically

![](https://cdn.mathpix.com/cropped/2024_05_30_d3836afd55956f7fff9cg-06.jpg?height=211&width=962&top_left_y=2279&top_left_x=473)
we arrive at the solution of the inequality in the set $\mathbb{R}$

$$
x \in\langle-\sqrt{2019},-1\rangle \cup\langle 1, \sqrt{2019}\rangle
$$

The desired numbers are

$$
x \in \mathbb{Z} \cap(\langle-\sqrt{2019},-1\rangle \cup\langle 1, \sqrt{2019}\rangle)
$$

or

$$
x \in\{-44,-43, \ldots,-3,-2,2,3, \ldots, 44\}
$$

1 point

There are $43 \cdot 2=86$ such integers.

Answer: 86


NÃÂµÃÂ·ÃÂ½ayka and Ponchik have the same amounts of money, composed of coins worth $1, 3, 5$, and 7 ferthings.

NÃÂµÃÂ·ÃÂ½ayka has as many 1-ferthing coins as Ponchik has 3-ferthing coins;

3-ferthing coins - as many as Ponchik has 5-ferthing coins; 5-ferthing coins - as many as Ponchik has 7-ferthing coins; and 7-ferthing coins - as many as Ponchik has 1-ferthing coins;

Determine how many 7-ferthing coins NÃÂµÃÂ·ÃÂ½ayka has, given that each has 20 coins.

Solution:
Answer: 5 coins.

Solution: Let $x, y, z, t$ be the number of 1, 3, 5, and 7-ferting coins that Nезнайка has. It is known that $x+y+z+t=20$ and $x+$ $3 y+5 z+7 t=3 x+5 y+7 z+t$. From the last equation, it follows that $6 t=2(x+y+z)$. Substituting $x+y+z=20-t$, we get the equation $6 t=2(20-t)$, the solution of which is $t=5$.

Answer: 5


Rabbits sawed several logs. They made 10 cuts and got 16 chunks. How many logs did they saw?

Solution:
Recall problem 4.

## Solution

From each log, you get one more chunk than the number of cuts made. Since there are 6 more chunks, it means there were 6 logs.

## Answer

6 logs.

Answer: 6


If $f(x) (x \in \mathbf{R})$ is an even function with a period of 2. And when $x \in [0,1]$, $f(x) = x^{\frac{1}{110}}$, then the arrangement from smallest to largest of $f\left(\frac{98}{19}\right) \cdot f\left(\frac{101}{17}\right) \cdot f\left(\frac{104}{15}\right)$ is $\qquad$ .

Solution:
Given that for $x \in \mathbf{R}, k \in \mathbf{Z}$, we have $f(2 k+x)=f(x)=f(-x)$,
thus $f\left(\frac{98}{19}\right)=f\left(6-\frac{16}{19}\right)=f\left(-\frac{16}{19}\right)=f\left(\frac{16}{19}\right), f\left(\frac{101}{17}\right)=f\left(6-\frac{1}{17}\right)=f\left(-\frac{1}{17}\right)=$ $f\left(\frac{1}{17}\right), f\left(\frac{104}{15}\right)=f\left(6+\frac{14}{15}\right)=f\left(\frac{14}{15}\right)$. Since $x \in[0,1]$, $f(x)=x$ is an increasing function.
Therefore, $f\left(\frac{1}{17}\right)<f\left(\frac{16}{19}\right)<f\left(\frac{14}{15}\right)$. That is, $f\left(\frac{101}{17}\right)<f\left(\frac{98}{19}\right)<f\left(\frac{104}{15}\right)$.
Note: The key to this problem is to use the periodicity of the function to transform the three function values to the same monotonic interval $[0,1]$ for comparison, which is a common method for comparing sizes using the monotonicity of functions.

This problem can also be solved using the graphical method as follows: By the property of even functions, the graph on $[-1,0]$ is symmetric to the graph on $[0,1]$ about the $y$-axis. Since the function has a period of 2, the graph on $[-1,0]$ is the same as the graph on $[1,2]$. Hence, we get Figure 1-2-4. Let the three known function values be denoted as $A$, $B$, and $C$. Then these three points are distributed on either side of the value 6. It is easy to see that (point is the lowest, $A$ point is the second, and $B$ point is the highest.

Answer: f(\frac{101}{17})<f(\frac{98}{19})<f(\frac{104}{15})


Solution:
1. Calculate $15 \%$ of $\frac{1+\frac{11}{24}}{\frac{5}{8}-1.5}$.

## Solution.

Adding fractions in the numerator:

$1+\frac{11}{24}=\frac{24+11}{24}=\frac{35}{24}$

Subtracting fractions in the denominator:

1 POINT

$\frac{5}{8}-1.5=\frac{5}{8}-\frac{3}{2}=\frac{5-12}{8}=-\frac{7}{8}$

Calculating the double fraction:

Simplifying in the double fraction:

$\frac{\frac{35}{24}}{-\frac{7}{8}}=-\frac{35 \cdot 8}{24 \cdot 7}=-\frac{5}{3}$

Converting the percentage to a fraction:

1 POINT

Calculating the desired percentage:

1 POINT

$\frac{15}{100} \cdot\left(-\frac{5}{3}\right)=-\frac{1}{4}$

Answer: -\frac{1}{4}


All the positive integers which are co-prime to 2012 are grouped in an increasing order in such a way that the $n^{\text {th }}$ group has $2 n-1$ numbers. So, the first three groups in this grouping are (1), $(3,5,7),(9,11,13,15,17)$. It is known that 2013 belongs to the $k^{\text {th }}$ group. Find the value of $k$.
(Note: Two integers are said to be co-prime if their greatest common divisor is 1.)

Solution:
16. Answer: 32

Solution. Note that $2012=2^{2} \times 503$, and that 503 is a prime number. There are 1006 multiples of 2 less than or equal to 2012 ; there are 4 multiples of 503 less than or equal to 2012; there are 2 multiples of 1006 less than or equal to 2012. By the Principle of Inclusion and Exclusion, there are $1006+4-2=1008$ positive integers not more than 2012 which are not co-prime to 2012 . Hence there are $2012-1008=1004$ positive integers less than 2012 which are co-prime with 2012 . Thus, 2013 is the $1005^{\text {th }}$ number co-prime with 2012. Note also that the sum of the first $n$ odd numbers equals $n_{2}^{2}$, and that $31^{2}<1005<32^{2}$, the number 2013 must be in the 326 th group. Hence $k=32$.

Answer: 32


Let $S$ be a set of $2020$ distinct points in the plane. Let
\[M=\{P:P\text{ is the midpoint of }XY\text{ for some distinct points }X,Y\text{ in }S\}.\]

Find the least possible value of the number of points in $M$.

Solution:
1. **Constructing the Set \( S \)**:
   Let \( S \) be a set of 2020 distinct points in the plane. For simplicity, consider the points to be on the x-axis:
   \[
   S = \{(1,0), (2,0), \ldots, (2020,0)\}
   \]

2. **Defining the Set \( M \)**:
   The set \( M \) consists of midpoints of all pairs of distinct points in \( S \). For any two points \( (x_i, 0) \) and \( (x_j, 0) \) in \( S \), the midpoint \( M_{i,j} \) is given by:
   \[
   M_{i,j} = \left( \frac{x_i + x_j}{2}, 0 \right)
   \]

3. **Counting the Elements in \( M \)**:
   We need to find the number of distinct midpoints. Consider the midpoints formed by the points in \( S \):
   \[
   M = \left\{ \left( \frac{x_i + x_j}{2}, 0 \right) : 1 \leq i < j \leq 2020 \right\}
   \]

4. **Range of Midpoints**:
   The smallest possible midpoint is:
   \[
   \left( \frac{1 + 2}{2}, 0 \right) = \left( \frac{3}{2}, 0 \right)
   \]
   The largest possible midpoint is:
   \[
   \left( \frac{2019 + 2020}{2}, 0 \right) = \left( \frac{4039}{2}, 0 \right)
   \]

5. **Distinct Midpoints**:
   The midpoints are of the form \( \left( \frac{k}{2}, 0 \right) \) where \( k \) ranges from 3 to 4039. The number of such midpoints is:
   \[
   4039 - 3 + 1 = 4037
   \]

6. **Proving the Bound**:
   To show that 4037 is the least possible value, consider the ordering of points \( A_1, A_2, \ldots, A_{2020} \) such that \( x_1 \leq x_2 \leq \ldots \leq x_{2020} \). The midpoints \( M_{i,j} \) for \( 1 \leq i < j \leq 2020 \) are distinct and cover all values from \( \frac{3}{2} \) to \( \frac{4039}{2} \).

   Therefore, the least possible number of distinct midpoints is indeed 4037.

\(\blacksquare\)

The final answer is \( \boxed{ 4037 } \)

Answer: 4037


Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$
f(x+y f(x+y))=y^{2}+f(x) f(y)
$$

for all $x, y \in \mathbb{R}$.

Solution:
Note that the function $f(x)=0$ for all $x$ does not satisfy the condition. Therefore, there is an $a$ such that $f(a) \neq 0$. Substitute $x=a$ and $y=0$, which gives $f(a)=f(a) f(0)$, so $f(0)=1$. Now substitute $x=1$ and $y=-1$, which gives $f(1-f(0))=1+f(1) f(-1)$. Since $f(0)=1$, this becomes $1=1+f(1) f(-1)$, so $f(1)=0$ or $f(-1)=0$. We will consider these two cases separately.

First, assume that $f(1)=0$. Substitute $x=t$ and $y=1-t$, and then $x=1-t$ and $y=t$, which gives the two equations

$$
\begin{aligned}
f(t) & =(1-t)^{2}+f(t) f(1-t) \\
f(1-t) & =t^{2}+f(t) f(1-t)
\end{aligned}
$$

Subtracting these equations gives $f(t)-f(1-t)=(1-t)^{2}-t^{2}=1-2 t$, so $f(1-t)=f(t)+2 t-1$. Substituting this into the first of the above two equations gives

$$
f(t)=(1-t)^{2}+f(t)^{2}+(2 t-1) f(t)
$$

which can be rewritten as

$$
f(t)^{2}+(2 t-2) f(t)+(1-t)^{2}=0
$$

or $(f(t)-(1-t))^{2}=0$. We conclude that $f(t)=1-t$. Checking this function in the original functional equation gives on the left $1-(x+y(1-x-y))=1-\left(x+y-x y-y^{2}\right)=1-x-y+x y+y^{2}$ and on the right $y^{2}+(1-x)(1-y)=y^{2}+1-x-y+x y$ and that is the same, so the function satisfies the equation.

Now assume $f(-1)=0$. Substitute $x=t$ and $y=-1-t$, and then $x=-1-t$ and $y=t$, which gives the two equations

$$
\begin{aligned}
f(t) & =(-1-t)^{2}+f(t) f(-1-t), \\
f(-1-t) & =t^{2}+f(t) f(-1-t)
\end{aligned}
$$

Subtracting these equations gives $f(t)-f(-1-t)=(-1-t)^{2}-t^{2}=1+2 t$, so $f(-1-t)=f(t)-2 t-1$. Substituting this into the first of the above two equations gives

$$
f(t)=(-1-t)^{2}+f(t)^{2}+(-2 t-1) f(t)
$$

which can be rewritten as

$$
f(t)^{2}-(2 t+2) f(t)+(t+1)^{2}=0
$$

or $(f(t)-(t+1))^{2}=0$. We conclude that $f(t)=t+1$. Checking this function in the original functional equation gives on the left $x+y(x+y+1)+1=x+x y+y^{2}+y+1$ and on the right $y^{2}+(x+1)(y+1)=y^{2}+x y+x+y+1$ and that is the same, so the function satisfies the equation.

We conclude that there are exactly two solutions: $f(x)=1-x$ for all $x$ and $f(x)=x+1$ for all $x$.

Answer: f(x)=1-x \text{ and } f(x)=x+1


Let $ n \geq 3$ and consider a set $ E$ of $ 2n - 1$ distinct points on a circle. Suppose that exactly $ k$ of these points are to be colored black. Such a coloring is [b]good[/b] if there is at least one pair of black points such that the interior of one of the arcs between them contains exactly $ n$ points from $ E$. Find the smallest value of $ k$ so that every such coloring of $ k$ points of $ E$ is good.

Solution:
To solve this problem, we need to determine the smallest value of \( k \) such that any coloring of \( k \) points out of \( 2n-1 \) points on a circle is "good." A coloring is "good" if there is at least one pair of black points such that the interior of one of the arcs between them contains exactly \( n \) points from \( E \).

1. **Graph Representation**:
   - Consider the \( 2n-1 \) points on the circle as vertices of a graph.
   - Connect two vertices if there are exactly \( n \) vertices between them on the circle (in either of the two arcs).

2. **Degree of Each Vertex**:
   - Each vertex will have a degree of 2 because each vertex will be connected to exactly two other vertices (one in each direction around the circle).

3. **Structure of the Graph**:
   - The graph is composed of disjoint cycles. Each cycle is formed by vertices that are \( n \) points apart.

4. **Case Analysis**:
   - **Case 1: \( 2n-1 \) is not divisible by 3**:
     - In this case, the graph consists of a single cycle of length \( 2n-1 \).
     - The maximum size of an independent set in a cycle of length \( 2n-1 \) is \( \left\lfloor \frac{2n-1}{2} \right\rfloor = n-1 \).
     - Therefore, if we color \( n \) vertices, there must be at least one pair of consecutive black vertices, ensuring that the coloring is "good."

   - **Case 2: \( 2n-1 \) is divisible by 3**:
     - In this case, the graph is the union of 3 disjoint cycles, each of length \( \frac{2n-1}{3} \).
     - The maximum size of an independent set in each cycle of length \( \frac{2n-1}{3} \) is \( \left\lfloor \frac{\frac{2n-1}{3}}{2} \right\rfloor = \frac{2n-1}{6} \).
     - Therefore, the total maximum size of an independent set in the entire graph is \( 3 \times \frac{2n-1}{6} = \frac{2n-1}{2} = n-1 \).
     - Thus, if we color \( n \) vertices, there must be at least one pair of consecutive black vertices in at least one of the cycles, ensuring that the coloring is "good."

5. **Conclusion**:
   - In both cases, the smallest value of \( k \) such that every coloring of \( k \) points is "good" is \( k = n \).

The final answer is \( \boxed{n} \).

Answer: n


As shown in the figure, the side length of the shaded small square is 2, and the side length of the largest outer square is 6. Then the area of square $ABCD$ is $\qquad$

Solution:
【Analysis】As shown in the figure: Add auxiliary lines. Since the side length of the shaded small square is 2, and the side length of the largest outer square is 6, the large square is divided into 9 small squares. The area of each triangle at the corners of the large square is equivalent to the area of a small square with a side length of 2. Therefore, the area of square $A B C D$ is equivalent to the area of 5 shaded small squares. Then, the area formula for a square can be used to solve the problem.
【Solution】Solution: $2 \times 2 \times 5=20$

Answer: The area of square $A B C D$ is 20.
Therefore, the answer is: 20.

Answer: 20


Find the derivative of the function $y=\arcsin \ln x$.

Solution:
Solution. We again apply the formula $(\arcsin u)^{\prime}=\frac{u^{\prime}}{\sqrt{1-u^{2}}}$ :

$$
y^{\prime}=(\arcsin \ln x)^{\prime}=\frac{(\ln x)^{\prime}}{\sqrt{1-(\ln x)^{2}}}=\frac{1}{x \sqrt{1-\ln ^{2} x}} .
$$

Answer: \frac{1}{x\sqrt{1-\ln^{2}x}}


If the surface area of a cube $X$ is equal to that of a regular tetrahedron $Y$, then the ratio of their volumes $\frac{V_{X}}{V_{Y}}=$ $\qquad$ .

Solution:
4. $\sqrt[4]{3}$.

Let the surface area be 12.
Then the area of each face of the cube is 2, and its side length is $\sqrt{2}$. Thus, $V_{x}=2^{\frac{3}{2}}$.

Given that the area of each face of the regular tetrahedron is 3, let its side length be a. Then
$$
\frac{\sqrt{3}}{4} a^{2}=3 \Rightarrow a=2 \times 3^{\frac{1}{4}} .
$$

Thus, $V_{Y}=2^{\frac{3}{2}} \times 3^{-\frac{1}{4}}$.
Therefore, $\frac{V_{X}}{V_{Y}}=3^{\frac{1}{4}}=\sqrt[4]{3}$.

Answer: \sqrt[4]{3}


Given that $a$ and $b$ are the roots of the equation
$$
\log _{3 x} 3+\log _{27} 3 x=-\frac{4}{3}
$$

then $a+b=$ $\qquad$

Solution:
9. $\frac{10}{81}$.

The original equation is transformed into
$$
\begin{array}{l}
\frac{1}{1+\log _{3} x}+\frac{1+\log _{3} x}{3}=-\frac{4}{3} . \\
\text { Let } 1+\log _{3} x=t . \\
\text { Then } \frac{1}{t}+\frac{t}{3}=-\frac{4}{3} \Rightarrow t=-1 \text { or }-3 \\
\Rightarrow 1+\log _{3} x=-1 \text { or }-3 .
\end{array}
$$

Thus, the two roots of the equation are $\frac{1}{9}$ and $\frac{1}{81}$. Therefore, $a+b=\frac{10}{81}$.

Answer: \frac{10}{81}


From two pieces of alloy of the same mass but with different percentage content of copper, pieces of equal mass were cut off. Each of the cut pieces was melted with the remainder of the other piece, after which the percentage content of copper in both pieces became the same. How many times smaller is the cut piece than the whole?

Solution:
Solution. Let $y$ be the mass of each alloy piece, $x$ be the mass of the cut-off piece, $p$ and $q$ be the concentration of copper in the first and second pieces of the alloy. Then $p x+q(y-x)$ and $q x+p(y-x)$ are the amounts of copper in the new alloys, respectively. According to the problem,

$$
\frac{p x+q(y-x)}{y}=\frac{q x+p(y-x)}{y} \Rightarrow(p-q) x=(y-x)(p-q) \Rightarrow x=y-x \Rightarrow y=2 x
$$

Answer: 2 times.

Answer: 2


How many positive, relatively prime number pairs (unordered) are there, where the sum of the numbers is $285$?

Solution:
We are looking for the positive integer solutions of the equation $285=a+(285-a)$, for which $a$ and $285-a$ are coprime. This precisely means that $a$ and 285 are coprime. Indeed, if an integer $d$ is a divisor of both $a$ and 285, then it is also a divisor of $(285-a)$ (and $a$); conversely, if an integer $d$ is a divisor of both $a$ and $(285-a)$, then it is also a divisor of ( $a$ and) $a+(285-a)=285$. Therefore, the greatest common divisor of $a$ and 285 is equal to the greatest common divisor of $a$ and $(285-a)$.

Thus, we need to find the numbers less than 285 that are coprime with 285.

It is known that if the prime factorization of $n$ is

$$
n=p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}
$$

then the number of integers coprime with $n$ is given by the formula

$$
\varphi(n)=n \cdot\left(1-\frac{1}{p_{1}}\right)\left(1-\frac{1}{p_{2}}\right) \ldots\left(1-\frac{1}{p_{k}}\right)
$$

In our case, $n=285=3 \cdot 5 \cdot 19$, and thus

$$
\varphi(285)=285\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{19}\right)=285 \cdot \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{18}{19}=144
$$

is the number of coprime numbers. As we have seen, these can be grouped into pairs such that the sum of the pairs is 285. Thus, there are a total of 72 suitable pairs of numbers.

Springer Éva (Budapest, Szent István Gymn., 9th grade)

Remark. The problem can also be solved without knowledge of the $\varphi$ function by counting how many numbers up to 285 are divisible by 3, 5, 19, $3 \cdot 5=15$, $3 \cdot 19=57$, $5 \cdot 19=95$, and $3 \cdot 5 \cdot 19=285$:

3 is a divisor of 95 numbers, 5 is a divisor of 57 numbers, 19 is a divisor of 15 numbers, from the sum of these, we need to subtract the number of numbers divisible by 15, 57, 95, and $3 \cdot 5 \cdot 19=285$, i.e., $95+57+15-18-5-3+1=141$, because no coprime pairs can be formed from these. The remainder $285-141=144$ can indeed form such pairs.

Bohák András (Budapest, Evangelical Gymn., 10th grade)

Answer: 72


Given an integer $n \geqslant 2$. Find the smallest positive real number $c$, such that for any complex numbers $z_{1}, z_{2}, \cdots, z_{n}$, we have
$$
\left|\sum_{i=1}^{n} z_{i}\right|+c \sum_{1 \leqslant i<j \leqslant n}\left|z_{i}-z_{j}\right| \geqslant \sum_{i=1}^{n}\left|z_{i}\right| .
$$
(Supplied by Zhang Duanyang)

Solution:
3. When $n=2 m$, take
$$
\begin{array}{l}
z_{1}=z_{2}=\cdots=z_{m}=1, \\
z_{m+1}=z_{m+2}=\cdots=z_{2 m}=-1 ;
\end{array}
$$

When $n=2 m+1$, take
$$
\begin{array}{l}
z_{1}=z_{2}=\cdots=z_{m}=1, \\
z_{m+1}=z_{m+2}=\cdots=z_{2 m+1}=-\frac{m}{m+1} .
\end{array}
$$

In both cases, we have $c \geqslant \frac{2}{n}$.
We now prove that for any complex numbers $z_{1}, z_{2}, \cdots, z_{n}$, we have
$$
\left|\sum_{i=1}^{n} z_{i}\right|+\frac{2}{n} \sum_{1 \leqslant i<j \leqslant n}\left|z_{i}-z_{j}\right| \geqslant \sum_{i=1}^{n}\left|z_{i}\right| .
$$

In fact, for $1 \leqslant k \leqslant n$, we have
$$
\begin{array}{l}
\left|\sum_{i=1}^{n} z_{i}\right|+\sum_{j=1}^{n}\left|z_{k}-z_{j}\right| \\
\geqslant\left|\sum_{i=1}^{n} z_{i}+\sum_{j=1}^{n}\left(z_{k}-z_{j}\right)\right|=n\left|z_{k}\right| .
\end{array}
$$

Summing over $k$ from 1 to $n$ yields
$$
n\left|\sum_{i=1}^{n} z_{i}\right|+2 \sum_{1 \leqslant k<j \leqslant n}\left|z_{k}-z_{j}\right| \geqslant n \sum_{k=1}^{n}\left|z_{k}\right| \text {. }
$$

In conclusion, the smallest positive real number is $\frac{2}{n}$.

Answer: \frac{2}{n}


At the edge of a circular lake, there are stones numbered from 1 to 10, in a clockwise direction. Frog starts from stone 1 and jumps only on these 10 stones in a clockwise direction.

a) If Frog jumps 2 stones at a time, that is, from stone 1 to stone 3, from stone 3 to stone 5, and so on, on which stone will Frog be after 100 jumps?

b) If on the first jump, Frog goes to stone 2, on the second jump to stone 4, on the third jump to stone 7, that is, in each jump he jumps one more stone than in the previous jump. On which stone will Frog be after 100 jumps?

Solution:
Solution

a) After 5 jumps, Frog returns to stone 1 and starts the same sequence. Since 100 is a multiple of 5, on the $100^{\text{th}}$ jump, he goes to stone 1.

b) On the $1^{\text{st}}$ jump, he moves 1 stone; on the $2^{\text{nd}}$, 2 stones; on the $3^{\text{rd}}$, 3 stones, and so on until the last jump when he moves 100 stones. The total number of movements is $\frac{(1+100) \cdot 100}{2}=$ 5,050. Since every 10 movements, he returns to stone 1 and 5,050 is a multiple of 10, after 100 jumps, Frog returns to stone 1.

Answer: 1


When mailing a letter locally, a postage fee of 0.80 yuan is charged for each letter weighing no more than $20 \mathrm{~g}$, 1.60 yuan for letters weighing more than $20 \mathrm{~g}$ but no more than $40 \mathrm{~g}$, and so on, with an additional 0.80 yuan required for each additional $20 \mathrm{~g}$ (for letters weighing up to $100 \mathrm{~g}$). If a person is mailing a letter weighing $72.5 \mathrm{~g}$, then the postage fee he should pay is ( ) yuan.
(A) 2.40
(B) 2.80
(C) 3.00
(D) 3.20

Solution:
2. D.

Since $20 \times 3<72.5<20 \times 4$, according to the problem, the postage to be paid is $0.80 \times 4=3.20$ (yuan).

Answer: D


Find the number of pairs of positive integers $(x, y)$ which satisfy the equation
$$
20 x+6 y=2006 .
$$

Solution:
14. Ans: 34
$$
20 x+6 y=2006 \Longleftrightarrow 10 x+3 y=1003 .
$$

The solutions are
$$
(1,331),(4,321),(7,311), \cdots,(94,21),(97,11),(100,1) \text {. }
$$

Thus there are 34 solutions.

Answer: 34


Determine the number of all pairs \((x ; y)\) that satisfy the following conditions (1) to (4)!

(1) \(x\) and \(y\) are three-digit natural numbers.

(2) The three digits of \(x\) are all different from each other.

(3) The three digits of \(x\) are also the three digits of \(y\), but in a different order.

(4) It holds that \(x-y=45\).

Solution:
}

The conditions (1) to (4) are satisfied if and only if the digits of \(x\) are three natural numbers \(a, b, c\) for which the following holds:

\[
1 \leq a \leq 9, \quad 0 \leq b \leq 9, \quad 0 \leq c \leq 9, \quad a \neq b, \quad a \neq c, \quad b \neq c
\]

and condition (4) is satisfied by the number \(x=100 a+10 b+c\) and the number \(y\) for which exactly one of the following cases I to V applies:

(I) It holds that \(y=100 a+10 c+b\)

(II) It holds that \(y=100 b+10 a+c\) and in addition to (5) also \(b \neq 0\).

(III) It holds that \(y=100 b+10 c+a\) and in addition to (5) also (6).

(IV) It holds that \(y=100 c+10 a+b\) and in addition to (5) also \(c \neq 0\).

(V) It holds that \(y=100 c+10 b+a\) and in addition to (5) also (7).

In case I, (4) is equivalent to \(x-y=9(b-c)\) and thus to \(b-c=5\), which under the conditions (5) is satisfied exactly by

\[
(b ; c)=(5 ; 0),(6 ; 1),(7,2),(8 ; 3),(9 ; 4)
\]

together with those numbers \(a=1, \ldots, 9\) that also satisfy \(a \neq b\) and \(a \neq c\). For each of the 9 numbers \(a=1, \ldots, 9\), exactly 4 pairs \((b ; c)\) remain.

In case II, (4) is not satisfiable because 45 is not a multiple of 90.

In case III, (4) is equivalent to \(x-y=9(11 a-10 b-c)\) and thus to \(11 a-10 b-c=5\) and this to \(11 a-5=10 b+c\).

Due to (5), (6), the values \(a \geq 5\) are excluded; for these values, calculating \(11 a-5\) would result in a two-digit number with the tens digit \(a\), leading to the contradiction \(a=b\). Furthermore, \(a=1\) is excluded, as it would result in a one-digit number.

The remaining values \(a=2,3,4\) satisfy \(11 a-5=17,28,39\) and each together with

\[
(b ; c)=(1 ; 7),(2 ; 8),(3 ; 9)
\]

all conditions (5), (6). Thus, (1) to (4) are satisfied in case III by exactly 3 pairs \((x ; y)\).

In case IV, (4) is equivalent to \(x-y=9(10 a+b-11 c)\) and thus to \(10 a+b-11 c=5\) and this to \(11 c+5=10 a+b\).

Due to (5), (7), the values \(c \leq 4\) and the value \(c=9\) are excluded; \(c=0\) does not satisfy (7), and for \(1 \leq c \leq 4\) or \(c=9\), \(11 c+5\) would result in a two-digit number with the tens digit \(a=c\) or a three-digit number.

The remaining values \(c=5,6,7,8\) satisfy \(11 c+5=60,71,82,93\) and each together with

\[
(a ; b)=(6 ; 0),(7 ; 1),(8 ; 2),(9 ; 3)
\]

all conditions (5), (7). Thus, (1) to (4) are satisfied in case IV by exactly 4 pairs \((x ; y)\).

In case V, (4) is not satisfiable because 45 is not a multiple of 99.

The total number of pairs \((x ; y)\) that satisfy (1) to (4) is thus \(36+3+4=43\).

Adapted from \([5]\)

Answer: 43


Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.

$$
\rho=5 \varphi, 0 \leq \varphi \leq \frac{12}{5}
$$

Solution:
## Solution

The length of the arc of a curve given by an equation in polar coordinates is determined by the formula

$$
L=\int_{\varphi_{0}}^{\varphi_{1}} \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d \varphi
$$

For the curve given by the equation $\rho=5 \varphi$, we find: $\rho^{\prime}=5$

We get:

$$
\begin{aligned}
L & =\int_{0}^{12 / 5} \sqrt{(5 \varphi)^{2}+5^{2}} d \varphi= \\
& =5 \int_{0}^{12 / 5} \sqrt{\varphi^{2}+1} d \varphi={ }^{(1)} \\
& =5\left(\left.\frac{\varphi}{2} \sqrt{\varphi^{2}+1}\right|_{0} ^{12 / 5}+\left.\frac{1}{2} \ln \left|\varphi+\sqrt{\varphi^{2}+1}\right|\right|_{0} ^{12 / 5}\right)= \\
& =5\left(\frac{1}{2} \cdot \frac{12}{5} \sqrt{\frac{144}{25}+1}-0\right)+5\left(\frac{1}{2} \ln \left|\frac{12}{5}+\sqrt{\frac{144}{25}+1}\right|-\ln |0+\sqrt{0+1}|\right)= \\
& =6 \cdot \sqrt{\frac{169}{25}}+5\left(\frac{1}{2} \ln \left|\frac{12}{5}+\sqrt{\frac{169}{25}}\right|-0\right)= \\
& =6 \cdot \frac{13}{5}+\frac{5}{2} \ln \left|\frac{12}{5}+\frac{13}{5}\right|=\frac{78}{5}+\frac{5}{2} \ln 5
\end{aligned}
$$

In (1) we used the formula: $\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \ln \left|x+\sqrt{x^{2}+a^{2}}\right|$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{\%} \% \mathrm{D} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 1 \% 82$ $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+19-25$ "

Categories: Kuznetsov's Problem Book Integrals Problem 19 | Integrals

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Answer: \frac{78}{5}+\frac{5}{2}\ln5


In a table tennis tournament, sixth-graders and seventh-graders participated, with the number of sixth-graders being twice the number of seventh-graders. The tournament was a round-robin. The number of matches won by seventh-graders was $40 \%$ more than the number of matches won by sixth-graders. How many participants were there in the tournament?

Solution:
82.16. Let $x$ be the number of seventh graders. Then the number of points $N_{7}$, scored by the seventh graders, is no more than $x \cdot 2 x + (x^2 - x) / 2$, and also $N_{6} \geqslant 2 x(2 x-1) / 2$. We get

$$
2 x^{2} + (x^{2} - x) / 2 \geqslant \frac{7}{5} x(2 x-1)
$$

from which it follows that $x \leqslant 3$.

By enumeration (or otherwise), we find that $x=3, N_{7}=21$, $N_{6}=15$. In total, there were nine participants.

Answer: 9


Suppose 32 objects are placed along a circle at equal distances. In how many ways can 3 objects be chosen from among them so that no two of the three chosen objects are adjacent nor diametrically opposite?

Solution:
Solution: One can choose 3 objects out of 32 objects in $\binom{32}{3}$ ways. Among these choices all would be together in 32 cases; exactly two will be together in $32 \times 28$ cases. Thus three objects can be chosen such that no two adjacent in $\binom{32}{3}-32-(32 \times 28)$ ways. Among these, furthrer, two objects will be diametrically opposite in 16 ways and the third would be on either semicircle in a non adjacent portion in $32-6=26$ ways. Thus required number is

$$
\binom{32}{3}-32-(32 \times 28)-(16 \times 26)=3616
$$

Answer: 3616


The function $f(x)=a x^{2}+b x+c(a>0)$ satisfies $f(x+2)=f(2-x)$, the correct statement is ( ).
A. $f(\pi-2)=f(\pi)$
B. $f\left(\frac{\sqrt{2}}{2}\right)f(\pi)$
C. $f(-2)<f(2)$
D. $f(0)<f(2)$

Solution: 2. C. From the problem, the axis of symmetry of $y=f(x)$ is the line $x=2$, and the parabola opens upwards.

Answer: C


Determine that two-digit number, the square of which is a four-digit number where the first and second digits, as well as the third and fourth digits, are identical.

Solution:
I. solution: Let the sought number be $10a + b$, then according to the problem,

$$
(10a + b)^2 = 1000c + 100c + 10d + d = 11(100c + d)
$$

where $a, b, c, d$ are non-negative single-digit numbers, and $a \neq 0, c \neq 0$.

Since the right-hand side is divisible by 11, the left-hand side must also be divisible by 11. Since 11 is a prime number, $(10a + b)^2$ can only be divisible by 11 if $10a + b$ itself is divisible by 11.

$10a + b$ can only be divisible by 11 if $a = b$, that is, $10a + b = 11a$, so (1) can be written as

$$
11^2 a^2 = 11(100c + d)
$$

which simplifies to

$$
11a^2 = 100c + d
$$

from which

$$
a^2 = \frac{100c + d}{11} = 9c + \frac{c + d}{11}
$$

where $\frac{c + d}{11}$ must be an integer. Since $c = d = 0$, and $c = d = 9$ are not possible (9999 is not a perfect square), we have $0 < c + d < 18$, and thus $c + d = 11$. Therefore,

$$
a^2 = 9c + 1
$$

from which

$$
c = \frac{a^2 - 1}{9} = \frac{(a + 1)(a - 1)}{9}
$$

Neither $(a + 1)$ nor $(a - 1)$ can be divisible by 3, so either $(a + 1)$ or $(a - 1)$ must be divisible by 9.

$a - 1 \neq 0$, otherwise $a = 1$ and $c = 0$, which is impossible. Since $a \leq 9$, $a - 1 \neq 9$. Therefore, it must be

$$
a + 1 = 9, \quad \text{hence} \quad a = 8
$$

Thus, the sought number is $11a = 88$. Indeed, $88^2 = 7744$.

Kálmán Györy (Ózd, József Attila g. II. o. t.)

II. solution: According to the first solution, the sought number is divisible by 11, and its square is a four-digit number, so only $33, 44, \ldots, 99$ are considered.

Among these, the odd numbers do not meet the criteria, because

$$
(10a + a)^2 = 100a^2 + 20a^2 + a^2
$$

and if $a$ were odd, then $a^2$ would have an odd number in the units place (09, 25, 49, 81), and the tens place would also be even due to $20a^2$ - an even number would stand, so the third and fourth digits could not match.

Therefore, only 44, 66, or 88 can be considered. Squaring each of them, we can see that only 88 satisfies the problem's requirements.

László Kovács (Bp., V., Eötvös g. III. o. t.)

Answer: 88


Given $A(-2,0)$, $B(0,-4)$, and $P$ as any point on the hyperbola $y=\frac{8}{x}$ $(x>0)$. A perpendicular line from $P$ to the x-axis meets at point $C$, and a perpendicular line from $P$ to the y-axis meets at point $D$. The minimum value of the area of quadrilateral $A B C D$ is $\qquad$

Solution:
5.16.

Let point \( P\left(x_{0}, y_{0}\right) \). Then \( y_{0}=\frac{8}{x_{0}}\left(x_{0}>0\right) \). Hence \( C\left(x_{0}, 0\right) \) and \( D\left(0, \frac{8}{x_{0}}\right) \).
From the given conditions,
\[
\begin{array}{l}
|C A|=x_{0}+2,|D B|=\frac{8}{x_{0}}+4 . \\
\text { Therefore, } S=\frac{1}{2}\left(x_{0}+2\right)\left(\frac{8}{x_{0}}+4\right) \\
=2\left(x_{0}+\frac{4}{x_{0}}\right)+8 \geqslant 16 .
\end{array}
\]

Thus, the minimum value of the area of quadrilateral \( A B C D \) is 16.

Answer: 16


Solution:
The options given in choices $\text{A}$, $\text{B}$ and $\text{E}$ don't change the initial value (13579) much, the option in choice $\text{C}$ decreases 13579 by a lot, and the one given in choice $\text{D}$ increases 13579 by a lot.
After ensuring that no trivial error was made, we have $\boxed{\text{D}}$ as the answer.

Answer: D


Find the trajectory of vertex $P$ in the first quadrant for a regular $\triangle ABP$ with side length $a$, where vertices $B$ and $A$ slide on the $x$-axis and $y$-axis respectively, with $A, B, P$ arranged clockwise.

Solution:
Let $P, A, B$ correspond to the complex numbers $x+yi, 2mi$, $2n$, respectively. Then the midpoint $C$ of $AB$ corresponds to the complex number $n+mi$, and $\overrightarrow{CP}=\overrightarrow{CB} \cdot \sqrt{3} \vec{i}$, i.e.,
$$
\begin{aligned}
& (x+yi)-(n+mi) \\
= & (2n-n-mi) \sqrt{3i} .
\end{aligned}
$$

Rearranging and applying the condition for equality of complex numbers, we get
$$
\left.n=\frac{1}{2}(\sqrt{3} y-x), m=\frac{1}{2} \sqrt{3 x}-y\right) \text {. }
$$

Given that $(2m)^{2}+(2n)^{2}=a^{2}$, we thus have
$$
\begin{array}{l}
x^{2}+y^{2}-\sqrt{3} xy=\frac{a^{2}}{4} . \quad\left(x \geqslant \frac{a}{2},\right. \\
\left.y \geqslant \frac{a}{2}\right)
\end{array}
$$
(Author's affiliation: Xiping High School, Henan)

Answer: x^{2}+y^{2}-\sqrt{3} xy=\frac{a^{2}}{4}


Let's determine the numbers $A, B, C, D$ such that

$$
\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C x+D}{x^{2}+x+1}=\frac{1}{(x+1)^{2} \cdot\left(x^{2}+x+1\right)}
$$

holds for all $x$ different from $-1$.

Solution:
I. Solution. If there exists the desired quadruple of numbers $A, B, C, D$, then the polynomial equality is also an identity, which we obtain from (1) by multiplying by the denominator on the right side (this denominator is indeed a common multiple of the denominators on the left side, so multiplication results in an integer expression):

$$
A(x+1)\left(x^{2}+x+1\right)+B\left(x^{2}+x+1\right)+(C x+D)(x+1)^{2}=1
$$

Rearranging by powers of $x$:

$$
(A+C) x^{3}+(2 A+B+2 C+D) x^{2}+(2 A+B+C+2 D) x+(A+B+D)=1
$$

Accordingly, the coefficients of the corresponding powers of $x$ on both sides must be equal, and the terms not containing $x$ must also be equal:

$$
\begin{aligned}
A+C & =0 \\
2 A+B+2 C+D & =0 \\
2 A+B+C+2 D & =0 \\
A+B+D & =1
\end{aligned}
$$

Subtract from (3) both twice (2) and (4):

$$
\begin{aligned}
& B+D=0 \\
& C-D=0
\end{aligned}
$$

From (5) and (6), $A=1$, so from (2) $C=-1$, from (7) $D=-1$, and finally from (6) $B=1$. Thus, the desired quadruple exists, and (1) becomes an identity:

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

Varga Katalin (Veszprém, Lovassy L. g. II. o. t.)

Remark. We can also obtain a system of equations for the numbers $A, B, C, D$ from $\left(1^{\prime}\right)$ by substituting permissible (i.e., not equal to -1) numbers for $x$. It is easy to compute if we take small absolute value integers, let $x=0,1,2$ and -2 (four equations are needed):

$$
\begin{array}{r}
A+B+D=1 \\
6 A+3 B+4 C+4 D=1 \\
21 A+7 B+18 C+9 D=1 \\
-3 A+3 B-2 C+D=1
\end{array}
$$

Subtract from (8) four times (5), from (9) nine times (5), and from (10) (5):

$$
\begin{array}{llr}
& 2 A-B+4 C=-3 \\
12 A-2 B+18 C=-8, & \text { i.e., } & 6 A-B+9 C=-4 \\
-4 A+2 B-2 C=0, & \text { i.e., } & -2 A+B-C=0
\end{array}
$$

Add (13) to (11) and (12): $3 C=-3$, so $C=-1$, and $4 A+8 C=4 A-8=-4, A=1$, so from (13) $B=C+2 A=1$ and from (5) $D=-1$.

Kövesi Gusztáv (Budapest, I. István g. II. o. t.) Faragó István (Budapest, Könyves Kálmán g. II. o. t.)

II. Solution. If (1) holds, then

$$
\frac{A}{x+1}+\frac{C x+D}{x^{2}+x+1}=\frac{1}{(x+1)^{2}\left(x^{2}+x+1\right)}-\frac{B}{(x+1)^{2}}=\frac{1-B\left(x^{2}+x+1\right)}{(x+1)^{2}\left(x^{2}+x+1\right)}
$$

For this to be true, the right side must become a polynomial when multiplied by $(x+1)\left(x^{2}+x+1\right)$, or it must be simplifiable by $x+1$. The numerator can be factored into $x+1$ and another (first-degree) polynomial if its value is 0 when $x=-1$, since then the $x+1$ factor will be 0. Thus, $1-B \cdot 1=0, B=1$. Indeed, with this, the right side of (14) becomes

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

Similarly, moving the first term of (14) to the right side:

$$
\frac{C x+D}{x^{2}+x+1}=\frac{-x}{(x+1)\left(x^{2}+x+1\right)}-\frac{A}{x+1}=\frac{-x-A\left(x^{2}+x+1\right)}{(x+1)\left(x^{2}+x+1\right)}
$$

the numerator on the right side must also give 0 when $x=-1$, from which $A=1$. With this, (15) becomes

$$
\frac{C x+D}{x^{2}+x+1}=\frac{-\left(x^{2}+2 x+1\right)}{(x+1)\left(x^{2}+x+1\right)}=\frac{-x-1}{x^{2}+x+1}
$$

which holds for all permissible $x$ if $C=-1, D=-1$.

Remark. This solution also sheds light on the reason why it is possible to find suitable values for $A, B, C, D$. (The systems of equations obtained above and any other similar systems of equations could have been contradictory.)

Answer: A=1,B=1,C=-1,D=-1


From the ten digits $0,1,2,3,4,5,6,7,8,9$, select six to fill in the blanks below to make the equation true, with one digit per blank, and all digits in the blanks must be different. $\square+\square \square=\square \square \square$ Then the largest three-digit number in the equation is $\qquad$ .

Solution: 【Answer】Solution: $97+8=105$ or $98+7=105$; therefore, the answer is: 105.

Answer: 105


Given a continuous function $ f(x)$ such that $ \int_0^{2\pi} f(x)\ dx = 0$.

Let $ S(x) = A_0 + A_1\cos x + B_1\sin x$, find constant numbers $ A_0,\ A_1$ and $ B_1$ for which $ \int_0^{2\pi} \{f(x) - S(x)\}^2\ dx$ is minimized.

Solution:
To find the constants \( A_0, A_1, \) and \( B_1 \) that minimize the integral \( \int_0^{2\pi} \{f(x) - S(x)\}^2 \, dx \), where \( S(x) = A_0 + A_1 \cos x + B_1 \sin x \), we need to minimize the function \( F(a, b, c) = \int_0^{2\pi} [f(x) - a - b \cos x - c \sin x]^2 \, dx \).

1. **Set up the function to minimize:**
   \[
   F(a, b, c) = \int_0^{2\pi} [f(x) - a - b \cos x - c \sin x]^2 \, dx
   \]

2. **Compute the partial derivatives and set them to zero:**
   \[
   \frac{\partial F}{\partial a} = 0, \quad \frac{\partial F}{\partial b} = 0, \quad \frac{\partial F}{\partial c} = 0
   \]

3. **Calculate \(\frac{\partial F}{\partial a}\):**
   \[
   \frac{\partial F}{\partial a} = \frac{\partial}{\partial a} \int_0^{2\pi} [f(x) - a - b \cos x - c \sin x]^2 \, dx
   \]
   Using the chain rule:
   \[
   \int_0^{2\pi} 2 [f(x) - a - b \cos x - c \sin x] (-1) \, dx = 0
   \]
   Simplifying:
   \[
   -2 \int_0^{2\pi} [f(x) - a - b \cos x - c \sin x] \, dx = 0
   \]
   Since \(\int_0^{2\pi} f(x) \, dx = 0\):
   \[
   -2 \left( \int_0^{2\pi} f(x) \, dx - a \int_0^{2\pi} 1 \, dx - b \int_0^{2\pi} \cos x \, dx - c \int_0^{2\pi} \sin x \, dx \right) = 0
   \]
   \[
   -2 \left( 0 - a \cdot 2\pi - b \cdot 0 - c \cdot 0 \right) = 0
   \]
   \[
   a = 0
   \]

4. **Calculate \(\frac{\partial F}{\partial b}\):**
   \[
   \frac{\partial F}{\partial b} = \frac{\partial}{\partial b} \int_0^{2\pi} [f(x) - a - b \cos x - c \sin x]^2 \, dx
   \]
   Using the chain rule:
   \[
   \int_0^{2\pi} 2 [f(x) - a - b \cos x - c \sin x] (-\cos x) \, dx = 0
   \]
   Simplifying:
   \[
   -2 \int_0^{2\pi} [f(x) - a - b \cos x - c \sin x] \cos x \, dx = 0
   \]
   Since \(a = 0\):
   \[
   -2 \left( \int_0^{2\pi} f(x) \cos x \, dx - b \int_0^{2\pi} \cos^2 x \, dx - c \int_0^{2\pi} \cos x \sin x \, dx \right) = 0
   \]
   \[
   -2 \left( \int_0^{2\pi} f(x) \cos x \, dx - b \cdot \pi - c \cdot 0 \right) = 0
   \]
   \[
   b = \frac{1}{\pi} \int_0^{2\pi} f(x) \cos x \, dx
   \]

5. **Calculate \(\frac{\partial F}{\partial c}\):**
   \[
   \frac{\partial F}{\partial c} = \frac{\partial}{\partial c} \int_0^{2\pi} [f(x) - a - b \cos x - c \sin x]^2 \, dx
   \]
   Using the chain rule:
   \[
   \int_0^{2\pi} 2 [f(x) - a - b \cos x - c \sin x] (-\sin x) \, dx = 0
   \]
   Simplifying:
   \[
   -2 \int_0^{2\pi} [f(x) - a - b \cos x - c \sin x] \sin x \, dx = 0
   \]
   Since \(a = 0\):
   \[
   -2 \left( \int_0^{2\pi} f(x) \sin x \, dx - b \int_0^{2\pi} \cos x \sin x \, dx - c \int_0^{2\pi} \sin^2 x \, dx \right) = 0
   \]
   \[
   -2 \left( \int_0^{2\pi} f(x) \sin x \, dx - b \cdot 0 - c \cdot \pi \right) = 0
   \]
   \[
   c = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin x \, dx
   \]

Thus, the constants \( A_0, A_1, \) and \( B_1 \) that minimize the integral are:
\[
A_0 = 0, \quad A_1 = \frac{1}{\pi} \int_0^{2\pi} f(x) \cos x \, dx, \quad B_1 = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin x \, dx
\]

The final answer is \( \boxed{ A_0 = 0, A_1 = \frac{1}{\pi} \int_0^{2\pi} f(x) \cos x \, dx, B_1 = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin x \, dx } \)

Answer: A_0 = 0, A_1 = \frac{1}{\pi} \int_0^{2\pi} f(x) \cos x \, dx, B_1 = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin x \, dx


Find all pairs $(p, q)$ of real numbers such that the polynomial $x^{2}+p x+q$ is a divisor of the polynomial $x^{4}+p x^{2}+q$.

Solution:
SOLUTION. By dividing the polynomial $x^{4}+p x^{2}+q$ by the polynomial $x^{2}+p x+q$, we find that

$x^{4}+p x^{2}+q=\left(x^{2}+p x+q\right)\left(x^{2}-p x+p^{2}+p-q\right)+\left(2 p q-p^{3}-p^{2}\right) x+q-p^{2} q-p q+q^{2}$.

The polynomial $x^{2}+p x+q$ is a divisor of the polynomial $x^{4}+p x^{2}+q$ if and only if the remainder $\left(2 p q-p^{3}-p^{2}\right) x+q-p^{2} q-p q+q^{2}$ is the zero polynomial, that is, if and only if

$$
2 p q-p^{3}-p^{2}=p\left(2 q-p^{2}-p\right)=0
$$

$$
q-p^{2} q-p q+q^{2}=q\left(1-p^{2}-p+q\right)=0 .
$$

If $p=0$, then $q=0$ or $q=-1$.

If $q=0$, then $p=0$ or $p=-1$.

If $p \neq 0$ and $q \neq 0$, then $2 q-p^{2}-p=0$ and $1-p^{2}-p+q=0$ must hold. From the second equation, we express $q=p^{2}+p-1$. Substituting into the first equation, we get $2 p^{2}+$ $+2 p-2-p^{2}-p=p^{2}+p-2=(p+2)(p-1)=0$ and thus $p=1, q=1$ or $p=-2$, $q=1$.

Therefore, five pairs $(p, q)$ satisfy the conditions, namely $(0,0),(0,-1),(-1,0),(1,1),(-2,1)$.

Answer: (0,0),(0,-1),(-1,0),(1,1),(-2,1)


The Olimpia country is formed by $n$ islands. The most populated one is called Panacenter, and every island has a different number of inhabitants. We want to build bridges between these islands, which we'll be able to travel in both directions, under the following conditions:

a) No pair of islands is joined by more than one bridge.

b) Using the bridges we can reach every island from Panacenter.

c) If we want to travel from Panacenter to every other island, in such a way that we use each bridge at most once, the number of inhabitants of the islands we visit is strictly decreasing.

Determine the number of ways we can build the bridges.

Solution:
1. **Understanding the Problem:**
   - We have \( n \) islands, each with a unique population.
   - The most populated island is called Panacenter.
   - We need to build bridges under the following conditions:
     a) No pair of islands is joined by more than one bridge.
     b) Every island must be reachable from Panacenter.
     c) When traveling from Panacenter to any other island, the number of inhabitants must strictly decrease.

2. **Graph Representation:**
   - Represent the islands as vertices in a graph.
   - Represent the bridges as edges in the graph.
   - Condition (b) implies the graph must be connected.
   - Condition (c) implies that the graph must be a tree (a connected acyclic graph) with Panacenter as the root.

3. **Tree Structure:**
   - Since the population strictly decreases as we move away from Panacenter, each island \( I_k \) (where \( k > 1 \)) can only be connected to one of the islands \( I_1, I_2, \ldots, I_{k-1} \).
   - This ensures that there are no cycles and the population condition is maintained.

4. **Counting the Ways to Build Bridges:**
   - For \( I_2 \), it can only be connected to \( I_1 \) (Panacenter).
   - For \( I_3 \), it can be connected to either \( I_1 \) or \( I_2 \).
   - For \( I_4 \), it can be connected to \( I_1, I_2, \) or \( I_3 \).
   - Generally, \( I_k \) can be connected to any of the \( k-1 \) islands with a higher population.

5. **Total Number of Ways:**
   - The number of ways to connect \( I_2 \) is 1.
   - The number of ways to connect \( I_3 \) is 2.
   - The number of ways to connect \( I_4 \) is 3.
   - And so on, up to \( I_n \), which can be connected in \( n-1 \) ways.
   - Therefore, the total number of ways to build the bridges is the product of these choices:
     \[
     (n-1)!
     \]

Conclusion:
\[
\boxed{(n-1)!}
\]

Answer: (n-1)!


Lin Lin filled a cup with pure milk, the first time he drank $\frac{1}{3}$, then added soy milk, filled the cup and stirred evenly, the second time, Lin drank $\frac{1}{3}$ again, continued to fill the cup with soy milk and stirred evenly, repeated the above process, after the fourth time, Lin Lin drank a total of $\qquad$ (expressed as a fraction) of the total amount of pure milk.

Answer: \frac{65}{81}


Solution:
Solution.

Domain of definition: $x \neq-1, x \neq-3, x \neq-5$.

From the condition we have:

$$
\begin{aligned}
& \frac{(x+1)+1}{x+1}+\frac{(x+3)+3}{x+3}+\frac{(x+5)+5}{x+5}=6 \Leftrightarrow \\
& \Leftrightarrow \frac{x+1}{x+1}+\frac{1}{x+1}+\frac{x+3}{x+3}+\frac{3}{x+3}+\frac{x+5}{x+5}+\frac{5}{x+5}=6 \Leftrightarrow \\
& \Leftrightarrow 3+\frac{1}{x+1}+\frac{3}{x+3}+\frac{5}{x+5}=6 \Leftrightarrow \frac{3 x^{3}+18 x^{2}+23 x}{(x+1)(x+3)(x+5)}=0
\end{aligned}
$$

Considering the domain of definition, we get $3 x^{3}+18 x^{2}+23 x=0$ or $x\left(3 x^{2}+18 x+23\right)=0$, hence $x_{1}=0$, or
$3 x^{2}+18 x+23=0, x_{2}=\frac{-9+\sqrt{12}}{3}, x_{3}=\frac{-9-\sqrt{12}}{3}$.

Answer: $x_{1}=0, x_{2}=\frac{-9+\sqrt{12}}{3}, x_{3}=\frac{-9-\sqrt{12}}{3}$.

Answer: x_{1}=0,x_{2}=\frac{-9+\sqrt{12}}{3},x_{3}=\frac{-9-\sqrt{12}}{3}


Let the complex numbers $z, -z, z^{2}-z+1, z^{2}+z+1$ correspond to points $A, B, C, D$ in the complex plane, respectively. Given that $|z|=2$, quadrilateral $ABCD$ is a rhombus, and such $z=a+b \mathrm{i}$ (where $\mathrm{i}$ is the imaginary unit, $a, b \in \mathbf{R}$), then $|a|+|b|=$ $\qquad$ (Fu Lexin, problem contributor)

Solution:
5. $\frac{\sqrt{7}+3}{2}$.

As shown in Figure 3, by symmetry, it is easy to see that segment $AB$ is the diameter of the circle $|z|=2$ in the complex plane. Let the diagonals $AC$ and $BD$ of the rhombus intersect at point $E$, then $AC \perp BD$, so point $E$ is also on the circle $|z|=2$. Since $E$ is the midpoint of $AC$, the complex number corresponding to point $E$ is $\frac{z+z^{2}-z+1}{2}=\frac{z^{2}+1}{2}$, hence $\left|\frac{z^{2}+1}{2}\right|=2$, which means $\left|z^{2}+1\right|=4$.

Given $|z|=2$, we know $\left|z^{2}\right|=4$, so $z^{2}$ corresponds to a point on the circle $|w+1|=4$ and $|w|=4$. As shown in Figure 4, it is easy to see that $z^{2}=4(\cos \theta \pm \mathrm{i} \sin \theta)$, where $\cos \theta=-\frac{1}{8}$. Also,
$$
\left|\cos \frac{\theta}{2}\right|=\sqrt{\frac{1+\cos \theta}{2}}=\frac{\sqrt{7}}{4}, \quad\left|\sin \frac{\theta}{2}\right|=\sqrt{1-\cos ^{2} \frac{\theta}{2}}=\frac{3}{4} .
$$

Combining $z=2\left(\cos \frac{\theta}{2}+\mathrm{i} \sin \frac{\theta}{2}\right)$, we have
$$
|a|+|b|=2\left(\left|\cos \frac{\theta}{2}\right|+\left|\sin \frac{\theta}{2}\right|\right)=\frac{\sqrt{7}+3}{2} .
$$

Answer: \frac{\sqrt{7}+3}{2}


Maggie's car can travel 32 miles on one gallon of gasoline. At the current price of $4 per gallon, how many miles can Maggie travel with $20 worth of gasoline?
(A) 64
(B) 128
(C) 160
(D) 320
(E) 640

Solution:
5. C.
$$
20 \div 4 \times 32=160
$$

Answer: C


Side $AB$ of triangle $ABC$ is a chord of a certain circle. Sides $AC$ and $BC$ lie inside the circle, the extension of side $AC$ intersects the circle at point $D$, and the extension of side $BC$ at point $E$, with $AB=AC=CD=2, CE=\sqrt{2}$. Find the radius of the circle.

Solution:
Prove that angle $A$ is a right angle.

## Solution

By the theorem of the product of chord segments $B C=A C \cdot C D / C E=2 \sqrt{2}$.

From triangle $A B C$, we find that angle $A$ is a right angle. Therefore, $D B$ is the diameter of the circle, and $B D^{2}=A B^{2}+A D^{2}=$ 20.

![](https://cdn.mathpix.com/cropped/2024_05_06_9e7ee8b57660b9721937g-29.jpg?height=503&width=486&top_left_y=1929&top_left_x=795)

## Answer

$\sqrt{5}$.

Answer: \sqrt{5}


Among the natural numbers from 1 to 300, choose three different numbers such that their sum is divisible by 3. There are ( ) ways to do this.
(A) $\mathrm{C}_{1 \infty}^{3}$
(B) $\mathrm{C}_{1 \infty}^{1} \cdot \mathrm{C}_{100}^{1} \cdot \mathrm{C}_{1 \infty}^{1}$
(C) $3 C_{1 \infty}^{3}$
(D) $3 C_{1 \infty}^{3}+\mathrm{C}_{10}^{1} \cdot \mathrm{C}_{1 \infty}^{1} \cdot \mathrm{C}_{1 \infty}^{1}$

Solution:
5.D.

Classify the natural numbers from 1 to 300 by modulo 3. Taking one number from each class or taking three numbers from the same class can satisfy the condition.

Answer: D


$O P E N$ is a square, and $T$ is a point on side $N O$, such that triangle $T O P$ has area 62 and triangle $T E N$ has area 10. What is the length of a side of the square?

Solution:
Solution:
12
$62=P O \cdot O T / 2$ and $10=E N \cdot T N / 2=P O \cdot T N / 2$, so adding gives $72=P O \cdot(O T+$ $T N) / 2=P O \cdot O N / 2=P O^{2} / 2 \Rightarrow P O=12$.

Answer: 12


Solution:
Rewrite the system of equations as \[\frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1.\]
This equation is satisfied when $t \in \{4, 16, 36, 64\}$. After clearing fractions, for each of the values $t=4,16,36,64$, we have the equation
\[x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)=F(t),\]where $F(t)=(t-1^2)(t-3^2)(t-5^2)(t-7^2)$ and $P_k(t)=F(t)/(t-k^2)$, for $k=1,3,5,7$.
Since the polynomials on each side are equal at $t=4,16,36,64$, we can express the difference of the two polynomials by a quartic polynomial that has roots at $t=4,16,36,64$, so
\begin{align} \tag{\dag}x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)-F(t) = -(t-4)(t-16)(t-36)(t-64) \end{align}
The leading coefficient of the RHS is $-1$ because the leading coefficient of the LHS is $-1$.
Plug in $t=1^2, 3^2, 5^2, 7^2$ in succession, into $(\dag)$. In each case, most terms drop, and we end up with
\begin{align*}  x^2=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}, \quad y^2=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}},\quad z^2=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}},\quad w^2=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}} \end{align*}
Adding them up we get the sum as $3^2\cdot 4=\boxed{036}$.
Postscript for the puzzled: This solution which is seemingly unnecessarily redundant in that it computes $x^2,y^2,z^2,$ and $w^2$ separately before adding them to obtain the final answer is appealing because it gives the individual values of $x^2,y^2,z^2,$ and $w^2$ which can be plugged into the given equations to check.

Answer: 36


If a non-zero complex number $x$ satisfies $x+\frac{1}{x}=1$, then $x^{2014}+\frac{1}{x^{2014}}=$ $\qquad$

Solution: $=, 7 .-1$. Therefore $x^{2014}+\frac{1}{x^{2014}}=2 \cos \frac{4 \pi}{3}=-1$.

Answer: -1


Point $M$ is the midpoint of side $B C$ of triangle $A B C$, where $A B=17$, $A C=30$, $B C=19$. A circle is constructed with side $A B$ as its diameter. An arbitrary point $X$ is chosen on this circle. What is the minimum value that the length of segment $M X$ can take?

![](https://cdn.mathpix.com/cropped/2024_05_06_790dd471148872cd0846g-03.jpg?height=414&width=553&top_left_y=1517&top_left_x=448)

Solution:
Answer: 6.5.

Solution. For the segment $M X$ to be minimal, we need the point $X$ to lie on the segment $M O$, where $O$ is the center of the circle (and also the midpoint of side $A B$), as shown in Fig. 2.

![](https://cdn.mathpix.com/cropped/2024_05_06_790dd471148872cd0846g-04.jpg?height=430&width=555&top_left_y=310&top_left_x=449)

Fig. 2: to the solution of problem 9.5

It remains to note that $O M=\frac{1}{2} A C$ (this is the midline), and $O X=\frac{1}{2} A B$ (this is the radius of the circle constructed on $A B$ as the diameter). We get

$$
M X=O M-O X=\frac{1}{2}(A C-A B)=6.5
$$

Answer: 6.5


Let real numbers $a, b, c$ satisfy $a+b+c=0, abc=1$. Then the number of positive numbers among $a, b, c$ is
(A) 3.
(B) 2.
(C) 1.
(D) 0.

Solution:
[Solution] Given $a+b+c=0$, if there is a positive number among them, then there must be a negative number (since $abc=1$, hence $a, b, c$ do not include 0).
Also, $abc=1>0$, so the number of negative numbers is 0 or 2. But from the above, the number of negative numbers cannot be 0.
Therefore, the number of positive numbers is 1.
Hence, the answer is (C).

Answer: C


A robot is located in one of the cells of an infinite grid paper, to which the following commands can be given:

- up (the robot moves to the adjacent cell above);
- down (the robot moves to the adjacent cell below);
- left (the robot moves to the adjacent cell to the left);
- right (the robot moves to the adjacent cell to the right).

For example, if the robot executes a sequence of four commands (up, right, down, left), it will obviously return to the initial position, i.e., it will end up in the same cell from which it started. How many different sequences of 8 commands exist that return the robot to the initial position?

Solution:
Solution. For brevity, let's denote the command to move left as L, right as R, up as U, and down as D. For the robot to return to its initial position, it is necessary and sufficient that the number of L commands equals the number of R commands, and the number of U commands equals the number of D commands. Let $k$ be the number of L commands in the sequence. We will count the number $N_{k}$ of the desired sequences for $k$ from 0 to 4.

- $\boldsymbol{k}=\mathbf{0}$. The sequence consists only of U and D commands. Since their numbers are equal, on 4 out of 8 positions there should be a U, and on the remaining positions, a D. The number of ways to choose 4 positions out of 8 is $C_{8}^{4}$. Therefore, $N_{0}=C_{8}^{4}=70$;
- $\boldsymbol{k}=\mathbf{1}$. The sequence consists of one L command, one R command, and three U and three D commands. The number of ways to place the two commands L and R on 8 positions is $C_{8}^{2} \cdot C_{2}^{1}$: $C_{8}^{2}$ is the number of ways to choose 2 positions out of 8, and $C_{2}^{1}=2$ is the number of ways to place the commands L and R on these two positions. On the remaining 6 positions, 3 U commands can be placed in $C_{6}^{3}$ ways. Therefore, $N_{1}=C_{8}^{2} \cdot C_{2}^{1} \cdot C_{6}^{3}=1120$;
- $\boldsymbol{k}=\mathbf{2}$. Here, there are two L commands, two R commands, and two U and two D commands. For L and R, there are $C_{8}^{4} \cdot C_{4}^{2}$ ways to place them. On the remaining 4 positions, 2 U commands can be placed in $C_{4}^{2}$ ways. Thus, $N_{2}=C_{8}^{4} \cdot C_{4}^{2} \cdot C_{4}^{2}=2520$.

By reasoning similarly, it can be shown that $N_{3}=N_{1}$ and $N_{4}=N_{0}$. Therefore, the desired number of sequences is $2 \cdot\left(N_{0}+N_{1}\right)+N_{2}=4900$.

Answer: 4900.

Answer: 4900


In a convex quadrilateral $A B C D$, side $A B$ is equal to diagonal $B D, \angle A=65^{\circ}$, $\angle B=80^{\circ}, \angle C=75^{\circ}$. What is $\angle C A D$ (in degrees $) ?$

Solution:
Answer: 15.

Solution. Since triangle $ABD$ is isosceles, then $\angle BDA = \angle BAD = 65^{\circ}$. Therefore, $\angle DBA = 180^{\circ} - 130^{\circ} = 50^{\circ}$. Hence, $\angle CBD = 80^{\circ} - 50^{\circ} = 30^{\circ}$, $\angle CDB = 180^{\circ} - 75^{\circ} - 30^{\circ} = 75^{\circ}$. This means that triangle $DBC$ is isosceles, so $BC = BD = AB$. Therefore, $\angle BCA = \angle BAC = \frac{1}{2}(180^{\circ} - 80^{\circ}) = 50^{\circ}$. Hence, $\angle CAD = 65^{\circ} - 50^{\circ} = 15^{\circ}$.

Answer: 15


Four standard six-sided dice are rolled. Find the probability that, for each pair of dice, the product of the two numbers rolled on those dice is a multiple of 4 .

Solution:
Answer: $\square$
If any two of the dice show an odd number, then this is impossible, so at most one of the dice can show an odd number. We take two cases:
Case 1: If exactly one of the dice shows an odd number, then all three other dice must show a multiple of 4 , which can only be the number 4 . The probability that this occurs is $4 \cdot \frac{1}{2} \cdot\left(\frac{1}{6}\right)^{3}=\frac{1}{108}$.
Case 2: If all of the dice show even numbers, then the condition is satisfied. The probability that this occurs is $\left(\frac{1}{2}\right)^{4}=\frac{1}{16}$.
The total probability is $\frac{1}{108}+\frac{1}{16}=\frac{31}{432}$.

Answer: \frac{31}{432}


Solution:
2. D
2. $(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}})^{2}=6$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.

2. D
2. $(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}})^{2}=6$.

Answer: D


On a cubic planet, there live cubic mice, and they live only on the faces of the cube, not on the edges or vertices. It is known that different numbers of mice live on different faces, and the number on any two adjacent faces differs by at least 2. What is the minimum number of cubic mice that can live on this planet, given that there is at least one mouse on each face?

Solution:
Solution: We will prove that no three consecutive numbers can be the number of mice on the faces. Indeed, if there were $x, x+1$, and $x+2$ mice on some three faces, then $x$ and $x+1$ would have to be on opposite faces. But then $x+2$ mice could not be anywhere.

Consider the first 8 natural numbers. Among the first three, there is at least one missing, and among the next three, there is at least one missing. Therefore, the minimum sum is $1+2+4+5+7+8=27$. It is possible to distribute 27 mice. For this, we will divide all the faces into pairs of opposite faces. On the first pair, there will be 1 and 2 mice, on the second pair - 4 and 5, on the third pair - 7 and 8. Thus, quantities differing by 1 are on opposite faces, and on adjacent faces, the difference is at least 2.

Criteria: Only the answer - 1 point.

Only an example (with or without justification) - 2 points.

Only the estimate - 5 points

Proved that no three consecutive numbers can be the number of mice on the faces - 2 points.

Answer: 27


In each cell of a $2017 \times 2017$ board, there is a chip. In one operation, you can remove a chip that has a non-zero even number of neighbors (neighbors are chips located in cells that share a side or a corner). What is the minimum number of chips that can be left on the board using such operations?

Solution:
# Answer: 2

Solution. If there are only two chips on the board, then each of them has no more than one neighbor, and according to the rules, they cannot be removed. Therefore, at least two chips will remain. We will show how to leave exactly two chips on the board. For this, we will provide an algorithm that allows clearing two adjacent rows (or two adjacent columns) on an $m \times n$ board. Let's assume we want to clear the top two rows. We will number the cells from 1 to $n$. First, we will remove the chip from the 2nd cell of the second row (this can be done because it has 8 neighbors), then we will remove the corner chip (it now has 2 neighbors), then the 3rd chip from the first row (it has 4 neighbors), then the 2nd chip from the first row (it has 2 neighbors), and the 1st chip from the second row (it also has 2 neighbors). Next, moving from left to right, we will sequentially remove all the chips from the first row (each of them will have 4 neighbors at the moment of removal, except for the last one, which will have 2 neighbors), and then similarly remove the chips from the second row. This way, we will clear the two top rows. We will apply these actions sequentially, alternating between rows and columns, until only the chips in a $3 \times 3$ square remain. Dealing with them is quite simple: we remove the central one, then the four corner ones in any order, then the remaining top one, and finally, the remaining bottom one.

Answer: 2


Given an integer $\mathrm{n} \geq 3$, let $\mathrm{A}_{1}, \mathrm{~A}_{2}, \ldots, \mathrm{~A}_{2 \mathrm{n}}$ be pairwise distinct non-empty subsets of the set $\{1,2, \ldots, \mathrm{n}\}$, and let $A_{2 n+1}=A_{1}$. Find the maximum value of $\sum_{i=1}^{2 n} \frac{\left|A_{i} \cap A_{i+1}\right|}{\left|A_{i}\right| \cdot\left|A_{i+1}\right|}$.

Solution:
Question 226, Solution: We first prove $\frac{\left|A_{1} \cap A_{1+1}\right|}{\left|A_{i}\right| \cdot\left|A_{i+1}\right|} \leq \frac{1}{2}(i=1, 2, 3, \ldots, 2 n)$.
If max $\left\{\left|A_{i}\right|,\left|A_{i+1}\right|\right\}=1$, then $\left|A_{i} \cap A_{i+1}\right|=0$, thus $\frac{\left|A_{1} \cap A_{i+1}\right|}{\left|A_{i}\right| \cdot\left|A_{i+1}\right|}=0<\frac{1}{2}$; on the other hand, let $A_{1}=\{1\}, A_{2}=\{1,2\}, A_{3}=\{2\}, A_{4}=\{2,3\}, A_{5}=\{3\}, A_{6}=$

In summary, the maximum value of $\sum_{i=1}^{2 n} \frac{\left|A_{1} \cap A_{i+1}\right|}{\left|A_{1}\right| \cdot \left|A_{i+1}\right|}$ is $n$.

Answer: n


In $\triangle A B C$, let $\angle C=90^{\circ}, \angle A=22.5^{\circ}, A B=4$. Then the area of $\triangle A B C$ is ( ).
(A) 2
(B) 3
(C) $2 \sqrt{2}$
(D) $2 \sqrt{3}$

Solution:
3. C.

As shown in Figure 4, draw the median $C M$ on the hypotenuse $A B$. Then
$$
\begin{array}{l}
C M=B M=2, \\
\angle B M C=2 \angle A \\
=45^{\circ} .
\end{array}
$$

Draw $C H \perp A B$ at point $H$. Then $C H=\sqrt{2}$. Therefore, $S_{\triangle A B C}=\frac{1}{2} C H \cdot A B^{\prime}=2 \sqrt{2}$.

Answer: C


The greatest integer not exceeding the real number $x$ is represented by $[x]$. Then, in the Cartesian coordinate system $x O y$, the area of the figure formed by all points $(x, y)$ satisfying $[x] \cdot [y]=2013$ is $\qquad$ .

Solution:
(2) 16 Hint: Let $[x]=a,[y]=b$, that is, the figure formed by all such points $(x, y)$ is the region enclosed by the inequalities $a \leqslant x<a+1, b \leqslant y<b+1$, with an area of 1.

Since $2013=1 \times 2013=3 \times 671=11 \times 183=33 \times 61$, the points $(x, y)$ that satisfy $[x] \cdot[y]=2013$ form 16 regions each with an area of 1, thus the total area is 16.

Answer: 16


Point $K$ lies on side $B C$ of parallelogram $A B C D$, and point $M$ lies on its side $A D$. Segments $C M$ and $D K$ intersect at point $L$, and segments $A K$ and $B M$ intersect at point $N$. Find the maximum value of the ratio of the areas of quadrilaterals $K L M N$ and $A B C D$.

Solution:
4.I.4. Without loss of generality, we can assume that the area of parallelogram $ABCD$ is 1. Let's introduce the following notations: $S_{1}$ - the area of triangle $BKN$, $S_{2}$ - the area of triangle $AMN$, $S_{3}$ - the area of triangle $CKL$, $S_{4}$ - the area of triangle $DLM$, $x$ - the area of triangle $KMN$, and $y$ - the area of triangle $KLM$ (figure).

![](https://cdn.mathpix.com/cropped/2024_05_21_46613852391e49da1fa9g-076.jpg?height=360&width=814&top_left_y=400&top_left_x=632)

Triangles $KMA$ and $ABM$ have the same area, hence triangles $KMN$ and $ABN$ also have equal areas. We have, $\frac{x}{S_{1}}=\frac{AN}{NK}$ and $\frac{S_{2}}{x}=\frac{AN}{NK}$, so $\frac{x}{S_{1}}=\frac{S_{2}}{x}$, from which $x=\sqrt{S_{1} S_{2}}$. Similarly, $y=\sqrt{S_{3} S_{4}}$. Since $S_{1}+S_{2} \geqslant 2 \sqrt{S_{1} S_{2}}$, we have

$2(x+y)=1-\left(S_{1}+S_{2}+S_{3}+S_{4}\right) \leqslant 1-2\left(\sqrt{S_{1} S_{2}}+\sqrt{S_{3} S_{4}}\right)=1-2(x+y)$, therefore $x+y \leqslant \frac{1}{4}$. Equality is achieved if and only if $S_{1}=S_{2}$ and $S_{3}=S_{4}$, i.e., if and only if $KM \| AB$.

Answer: $\frac{1}{4}$.

Answer: \frac{1}{4}


Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that
$\forall x\in \mathbb{R} \ \ f(x) = \max(2xy-f(y))$ where $y\in \mathbb{R}$.

Solution:
1. Given the functional equation:
   \[
   f(x) = \max(2xy - f(y)) \quad \forall x \in \mathbb{R}, \quad \text{where} \quad y \in \mathbb{R}
   \]
   We need to find all functions \( f: \mathbb{R} \rightarrow \mathbb{R} \) that satisfy this equation.

2. For any \( x \in \mathbb{R} \), the equation implies:
   \[
   f(x) \geq 2xy - f(y) \quad \forall y \in \mathbb{R}
   \]
   This inequality must hold for all \( y \).

3. Consider the specific case where \( y = x \):
   \[
   f(x) \geq 2x^2 - f(x)
   \]
   Rearranging this inequality, we get:
   \[
   2f(x) \geq 2x^2 \implies f(x) \geq x^2
   \]

4. Define a new function \( g(x) \) such that:
   \[
   g(x) = f(x) - x^2
   \]
   Since \( f(x) \geq x^2 \), it follows that \( g(x) \geq 0 \).

5. Substitute \( f(x) = g(x) + x^2 \) back into the original functional equation:
   \[
   g(x) + x^2 = \max(2xy - (g(y) + y^2)) \quad \forall y \in \mathbb{R}
   \]
   Simplifying the right-hand side:
   \[
   g(x) + x^2 = \max(2xy - g(y) - y^2)
   \]

6. Since \( g(x) \geq 0 \), we need to show that \( g(x) = 0 \). Consider the expression inside the maximum:
   \[
   2xy - g(y) - y^2
   \]
   For \( g(x) \) to be non-negative and the maximum to be zero, we must have:
   \[
   g(x) = \max\{-g(y) - (x-y)^2 \mid y \in \mathbb{R}\}
   \]
   Since \( g(x) \geq 0 \) and the right-hand side is non-positive, the only way for the equality to hold is if:
   \[
   g(x) = 0
   \]

7. Therefore, \( g(x) = 0 \) for all \( x \in \mathbb{R} \), which implies:
   \[
   f(x) = x^2
   \]

Conclusion:
\[
\boxed{f(x) = x^2}
\]

Answer: f(x) = x^2


Given triangle $\mathrm{ABC}$, where $2 \mathrm{BC}=\mathrm{AC}$ and angle $\mathrm{C}=106^{\circ}$. On the ray $\mathrm{BC}$, segment $\mathrm{CX}=\mathrm{CB}$ is laid out. Then, from point $\mathrm{X}$, a perpendicular is drawn to the line containing the median of triangle $\mathrm{ABC}$, drawn from vertex $\mathrm{B}$, and the intersection point is $\mathrm{Y}$. What is the measure of angle $\mathrm{CXY}$? Express your answer in degrees.

Solution:
Answer: 53

Exact match of the answer -1 point

Solution by analogy with task №8.1

#

Answer: 53


Let $A=\{1,2, \cdots, 2002\}, M=\{1001,2003,3005\}$. For any non-empty subset $B$ of $A$, if the sum of any two numbers in $B$ does not belong to $M$, then $B$ is called an $M$-free set. If $A=A_{1} \cup A_{2}, A_{1} \cap A_{2}=\varnothing$, and both $A_{1}$ and $A_{2}$ are $M$-free sets, then the ordered pair $\left(A_{1}, A_{2}\right)$ is called an $M$-partition of $A$. Try to find the number of all $M$-partitions of $A$.

Solution:
The answer is $2^{501}$.
For $m, n \in A$, if $m+n=1001$ or 2003 or 3005, then $m$ and $n$ are said to be "related".

It is easy to see that the numbers related to 1 are only 1000 and 2002, and the numbers related to 1000 and 2002 are 1 and 1003, and the numbers related to 1003 are 1000 and 2002.

Therefore, for $1, 1003, 1000, 2002$, they must be divided into two groups $\{1, 1003\}, \{1000, 2002\}$. Similarly, other groups can be divided as $\{2, 1004\}, \{999, 2001\}; \{3, 1005\}, \{998, 2000\}; \cdots; \{500, 1502\}, \{501, 1503\}; \{1001\}, \{1002\}$.
Thus, the 2002 numbers in $A$ are divided into 501 pairs, totaling 1002 groups.
Since any number is related to and only to the corresponding other group, if one group of a pair is in $A_{1}$, the other group must be in $A_{2}$.

Conversely, if one group of a pair is in $A_{2}$, the other group must be in $A_{1}$, and there are no related numbers in $A_{1}$ and $A_{2}$. Therefore, the number of $M$-partitions of $A$ is $2^{501}$.

Answer: 2^{501}