Find all pairs of real numbers $x$ and $y$ that solve the system of equations

$$
\begin{gathered}
\frac{y^{3}+15 x^{2}}{y^{4}-x^{3}}=\frac{y^{2}+15 x}{y^{3}-x^{2}} \\
\frac{1500 y^{3}+4 x^{2}}{9 y^{4}-4}=\frac{1500 y^{2}+4 x}{9 y^{3}-4}
\end{gathered}
$$

Solution:
B3. In both equations, we eliminate the fractions and simplify to get

$$
\begin{aligned}
-x^{2} y^{3}+15 x^{2} y^{3} & =-x^{3} y^{2}+15 x y^{4} \\
-6000 y^{3}+36 x^{2} y^{3}-16 x^{2} & =-6000 y^{2}+36 x y^{4}-16 x
\end{aligned}
$$

We rewrite the first equation as \( x^{2} y^{2}(x-y) = -15 x y^{3}(x-y) \) and then as

$$
x y^{2}(x-y)(x+15 y)=0
$$

Thus, the numbers \( x \) and \( y \) satisfy one of the following four possibilities:

1. possibility: \( x=0 \). When we substitute this into the second of the above two equations, we get \( y^{3} = y^{2} \) or \( y^{2}(y-1) = 0 \). Therefore, \( y=0 \) or \( y=1 \). The first possibility is invalid because at \( x=y=0 \), some fractions in the original equations are not well-defined. Thus, only the possibility \( y=1 \) remains.
2. possibility: \( y=0 \). From the second of the above equations, we then get \( x^{2} = x \), so \( x=0 \) or \( x=1 \). The possibility \( x=0 \) is invalid for the same reason as in the previous case, so we get the solution \( x=1 \).
3. possibility: \( y=x \). Substituting this into the second equation, we get \( -6000 x^{3} - 16 x^{2} = -6000 x^{2} - 16 x \), which can be rewritten as \( 6000 x^{2}(1-x) = 16 x(x-1) \), and then as \( 16 x(x-1)(375 x + 1) = 0 \). The possibility \( x=0 \) is invalid because in this case \( y=0 \) and some fractions in the original equations are not well-defined. For the same reason, the possibility \( x=1 \) is also invalid. Therefore, \( 375 x + 1 = 0 \) or \( x = y = -\frac{1}{375} \).
4. possibility: \( x = -15 y \). Substituting this into the second equation and simplifying, we get

$$
8640 y^{5} - 6000 y^{3} - 2400 y^{2} - 240 y = 0
$$

As in the previous case, the possibility \( y=0 \) is invalid because then \( x=0 \). Therefore, we can divide the last equation by \( 240 y \) to get

$$
36 y^{4} - 25 y^{2} + 10 y - 1 = 0
$$

We observe that the last three terms form a perfect square, so we rewrite the equation as \( 36 y^{4} - (5 y - 1)^{2} = 0 \), and then factor the left side using the difference of squares to get

$$
\left(6 y^{2} - 5 y + 1\right)\left(6 y^{2} + 5 y - 1\right) = 0
$$

We factor both terms on the left side of the equation to get

$$
(3 y - 1)(2 y - 1)(6 y - 1)(y + 1) = 0
$$

The solutions are thus \( y = \frac{1}{3} \) and \( x = -5 \), \( y = \frac{1}{2} \) and \( x = -\frac{15}{2} \), \( y = \frac{1}{6} \) and \( x = -\frac{5}{2} \), and \( y = -1 \) and \( x = 15 \). In all cases, all fractions in the original equations are well-defined.

Therefore, all pairs \((x, y)\) that solve the problem are \((0,1)\), \((1,0)\), \(\left(-\frac{1}{375}, -\frac{1}{375}\right)\), \(\left(-5, \frac{1}{3}\right)\), \(\left(-\frac{15}{2}, \frac{1}{2}\right)\), \(\left(-\frac{5}{2}, \frac{1}{6}\right)\), and \((15, -1)\).

A simplified first or second equation, for example: \( x^{3} y^{2} + 14 x^{2} y^{3} - 15 x y^{4} = 0 \), \( 9 x^{2} y^{3} - 9 x y^{4} + 4 x - 4 x^{2} + 1500 y^{2} - 1500 y^{3} = 0 \)

The factored first equation, for example: \( x y^{2}(x-y)(x+15 y) = 0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ld

Answer: (0,1),(1,0),(-\frac{1}{375},-\frac{1}{375}),(-5,\frac{1}{3}),(-\frac{15}{2},\frac{1}{2}),(-\frac{5}{2},\frac{1}{6}),(15,-1)


Determine the smallest integer strictly greater than 100 that is a divisor of $100!$.

Solution: Answer: 102. Since $102=2 \times 51$, it divides 100!. However, 101 is a prime number and does not divide any of the factors of $1 \times 2 \times \cdots \times 100$, so it does not divide 100!.

Answer: 102


Determine all pairs $(x, y)$ of (real) numbers with $0 < x < 1$ and $0 <y < 1$ for which $x + 3y$ and $3x + y$ are both integer. An example is $(x,y) =( \frac{1}{8}, \frac{7}{8}) $, because $  x+3y =\frac18 +\frac{21}{8} =\frac{22}{8} =  \frac{11}{4}$ and $ 3x+y = \frac38 + \frac78 =\frac{10}{8} = \frac{5}{4}$. (Note: The example pair doesn't meet the problem's criteria as both expressions must be integers.)

Determine the integer $m > 2$ for which there are exactly $119$ pairs $(x,y)$ with $0 < x < 1$ and $0 < y < 1$ such that $x + my$ and  $mx + y$ are integers.

Remark: if $u \ne v,$ the pairs $(u, v)$ and $(v, u)$ are different.

Solution:
### Part (a)
1. We need to find all pairs \((x, y)\) of real numbers such that \(0 < x < 1\) and \(0 < y < 1\) for which \(x + 3y\) and \(3x + y\) are both integers.
2. Let \(x + 3y = a\) and \(3x + y = b\) where \(a\) and \(b\) are integers.
3. We can solve these equations for \(x\) and \(y\):
   \[
   x + 3y = a \quad \text{(1)}
   \]
   \[
   3x + y = b \quad \text{(2)}
   \]
4. Multiply equation (1) by 3:
   \[
   3x + 9y = 3a \quad \text{(3)}
   \]
5. Subtract equation (2) from equation (3):
   \[
   3x + 9y - (3x + y) = 3a - b
   \]
   \[
   8y = 3a - b
   \]
   \[
   y = \frac{3a - b}{8} \quad \text{(4)}
   \]
6. Substitute equation (4) into equation (1):
   \[
   x + 3\left(\frac{3a - b}{8}\right) = a
   \]
   \[
   x + \frac{9a - 3b}{8} = a
   \]
   \[
   x = a - \frac{9a - 3b}{8}
   \]
   \[
   x = \frac{8a - (9a - 3b)}{8}
   \]
   \[
   x = \frac{8a - 9a + 3b}{8}
   \]
   \[
   x = \frac{3b - a}{8} \quad \text{(5)}
   \]
7. For \(x\) and \(y\) to be in the interval \((0, 1)\), we need:
   \[
   0 < \frac{3b - a}{8} < 1 \quad \text{and} \quad 0 < \frac{3a - b}{8} < 1
   \]
8. This implies:
   \[
   0 < 3b - a < 8 \quad \text{and} \quad 0 < 3a - b < 8
   \]
9. Solving these inequalities:
   \[
   0 < 3b - a < 8 \implies a < 3b < a + 8
   \]
   \[
   0 < 3a - b < 8 \implies b < 3a < b + 8
   \]
10. These inequalities must be satisfied simultaneously for \(a\) and \(b\) to be integers.

### Part (b)
1. We are given the system of equations:
   \[
   x + my = a \quad \text{(1)}
   \]
   \[
   mx + y = b \quad \text{(2)}
   \]
   where \(0 < x, y < 1\) and \(a, b \in \mathbb{Z}\).
2. Solving for \(x\) and \(y\):
   \[
   x = \frac{ma - b}{m^2 - 1} \quad \text{(3)}
   \]
   \[
   y = \frac{mb - a}{m^2 - 1} \quad \text{(4)}
   \]
3. For \(x\) and \(y\) to be in the interval \((0, 1)\), we need:
   \[
   0 < \frac{ma - b}{m^2 - 1} < 1 \quad \text{and} \quad 0 < \frac{mb - a}{m^2 - 1} < 1
   \]
4. This implies:
   \[
   0 < ma - b < m^2 - 1 \quad \text{and} \quad 0 < mb - a < m^2 - 1
   \]
5. Solving these inequalities:
   \[
   0 < ma - b < m^2 - 1 \implies b < ma < b + m^2 - 1
   \]
   \[
   0 < mb - a < m^2 - 1 \implies a < mb < a + m^2 - 1
   \]
6. These inequalities must be satisfied simultaneously for \(a\) and \(b\) to be integers.
7. The number of valid pairs \((a, b)\) is given by:
   \[
   S_m = m(m - 1) + m - 2 = m^2 - 2
   \]
8. Given \(S_m = 119\):
   \[
   m^2 - 2 = 119
   \]
   \[
   m^2 = 121
   \]
   \[
   m = 11
   \]

The final answer is \( \boxed{ m = 11 } \)

Answer: m = 11


Anna and Baptiste are playing the following game. At the beginning of the game, 2022 white squares, numbered from 1 to 2022, are placed in front of them. Then, taking turns, starting with Anna, they choose a white square and color it in the color of their choice: either red or blue. The game ends after 2022 turns, that is, when the last white square is colored.

Anna's score is then equal to the number of integers $a \leqslant 2019$ such that the two squares numbered $a$ and $a+3$ are the same color. Anna wants to have the highest score possible; Baptiste wants Anna's score to be as low as possible. What is the highest score Anna can ensure, regardless of the moves Baptiste chooses to play?

Solution:
Solution to Exercise 2 For simplicity, we will say that two integers $a$ and $b$ are friends if $a=b \pm 3$. We will say that $a$ is pretty if $a=b-3$ and if $a$ and $b$ are colored the same color, and that $a$ is ugly if $a=b-3$ and if $a$ and $b$ are colored different colors: Anna therefore wants to maximize the number of pretty integers at the end of the game. If $a$ or $b$ is not yet colored, $a$ is neither pretty nor ugly.

Furthermore, for each integer $k=0,1$ and 2, we will denote by $\mathcal{E}_{k}$ the set of integers $n$ such that $1 \leqslant n \leqslant 2022$ and $n \equiv k(\bmod 3)$. We say that a set $\mathcal{E}_{k}$ is started as soon as at least one cell whose number is in $\mathcal{E}_{k}$ has been colored.

We will first demonstrate that Anna can ensure a score of at least 1008 points. To do this, the strategy Anna adopts is as follows.

1. If there exists an integer $a$ that is not yet colored, and a friend, say $b$, is colored, then Anna chooses such an integer $a$ and colors it the same color as $b$; if several integers $a$ and/or $b$ were available, any of these integers $a$ and $b$ will do.
2. If, on the contrary, no such integer $a$ exists, Anna colors a random white cell.

In case 1, the integer $\min \{a, b\}$ becomes pretty, which increases Anna's score by at least 1. In case 2, if a set $\mathcal{E}_{k}$ was already started when Anna is about to play, it means that all its elements were already colored; consequently, with her move, Anna starts a new set $\mathcal{E}_{k}$.

In particular, case 2 occurs at most 3 times during the game, and since Anna plays 1011 moves, case 1 occurs at least 1008 times during the game. As expected, Anna thus ensures a final score of at least 1008 points.

Conversely, we will now demonstrate that Baptiste can prevent Anna from scoring more than 1008 points, regardless of the moves she decides to play. To do this, the strategy Baptiste adopts is as follows.

1. If there exists an integer $a$ that is not yet colored, and a friend, say $b$, is colored, then Baptiste chooses such an integer $a$ and colors it the opposite color to that of $b$; if several integers $a$ and/or $b$ were available, any of these integers $a$ and $b$ will do.
2. If, on the contrary, no such integer $a$ exists, Baptiste colors a random white cell. In case 1, the integer $\min \{a, b\}$ becomes ugly. In case 2, any set $\mathcal{E}_{k}$ already started when Baptiste is about to play was already colored; if $n$ sets were started at that moment, it means that $2022 n / 3=674 n$ moves had already been played, and thus it was actually Anna's turn to play.

Consequently, case 2 never occurs, and since Baptiste plays 1011 moves, there are at least 1011 ugly integers at the end of the game. Furthermore, the three integers 2020, 2021, and 2022 will be neither pretty nor ugly. Among our 2022 integers, there are therefore at least $1011+3$ that will never be pretty, and Anna's score will thus never exceed $2022-(1011+3)=1008$.

Graders' Comments This problem was solved unevenly, with students falling into two main categories: on the one hand, those, quite numerous, who managed to solve the problem; on the other hand, those, equally numerous, who, although they had a reasonable intuition of what could happen, failed to correctly define strategies for Anna and Baptiste.

Students in this second category simply described how, move by move, Anna and Baptiste were supposed to play if they wanted to play optimally. However, this could in no way constitute a proof: in general, when giving a strategy for a player, it is necessary to indicate what moves this player should make even if their opponent plays in an apparently random manner. Incidentally, doing so generally leads to much simpler proofs.

Moreover, the description these students gave was almost systematically incorrect, as Baptiste is not required to directly counter Anna to prevent her from scoring more than 1008 points. For example, if Anna colors the integer 1 red, a relevant strategy for Baptiste could be to color the integer 7 blue.

It is therefore unfortunate that many students end up with very few points, even though they thought they had solved the problem. For these students, it will be very important to remember the proper way to define a strategy for a player, as described above.

Answer: 1008


Find the largest constant $\mathrm{C}$ such that for all $x, y, z \geq 0$, $(y z+z x+x y)^{2}(x+y+z) \geq C x y z\left(x^{2}+y^{2}+z^{2}\right)$.

Solution:
By setting $y=z=1, \frac{(y z+z x+x y)^{2}(x+y+z)}{x y z\left(x^{2}+y^{2}+z^{2}\right)}=\frac{(2 x+1)^{2}(x+2)}{x\left(x^{2}+2\right)}$. This quantity tends to 4 as $x$ tends to $+\infty$, so we will try to prove the inequality with $C=4$.

By expanding, we need to show that $\left(x^{3} y^{2}+x^{2} y^{3}+y^{3} z^{2}+y^{2} z^{3}+z^{3} x^{2}+z^{2} x^{3}\right)+5\left(x^{2} y^{2} z+y^{2} z^{2} x+z^{2} x^{2} y\right) \geq 2\left(x^{3} y z+y^{3} z x+z^{3} x y\right)$.

We then draw the triangle:

![](https://cdn.mathpix.com/cropped/2024_05_10_da6e2ad19b1b4e3d33dcg-205.jpg?height=621&width=632&top_left_y=611&top_left_x=752)

And we see that it suffices to apply the AM-GM inequality three times with the extreme black points $\left(x^{2} z^{3}+z^{3} y^{2} \geq 2 z^{3} x y\right.$, etc.) to conclude.

Answer: 4


Evaluate $ \int_{-1}^{a^2} \frac{1}{x^2 + a^2}\ dx\ (a > 0).$

You may not use $ \tan^{-1} x$ or Complex Integral here.

Solution:
To evaluate the integral \( \int_{-1}^{a^2} \frac{1}{x^2 + a^2} \, dx \) without using \( \tan^{-1} x \) or complex integrals, we can use a substitution method.

1. **Substitution**:
   Let \( x = a \tan(\theta) \). Then, \( dx = a \sec^2(\theta) \, d\theta \).

2. **Change of Limits**:
   When \( x = -1 \):
   \[
   -1 = a \tan(\theta) \implies \tan(\theta) = -\frac{1}{a} \implies \theta = \tan^{-1}\left(-\frac{1}{a}\right)
   \]
   When \( x = a^2 \):
   \[
   a^2 = a \tan(\theta) \implies \tan(\theta) = a \implies \theta = \tan^{-1}(a)
   \]

3. **Transform the Integral**:
   Substitute \( x = a \tan(\theta) \) into the integral:
   \[
   \int_{-1}^{a^2} \frac{1}{x^2 + a^2} \, dx = \int_{\tan^{-1}(-\frac{1}{a})}^{\tan^{-1}(a)} \frac{1}{a^2 \tan^2(\theta) + a^2} \cdot a \sec^2(\theta) \, d\theta
   \]
   Simplify the integrand:
   \[
   \int_{\tan^{-1}(-\frac{1}{a})}^{\tan^{-1}(a)} \frac{a \sec^2(\theta)}{a^2 (\tan^2(\theta) + 1)} \, d\theta = \int_{\tan^{-1}(-\frac{1}{a})}^{\tan^{-1}(a)} \frac{a \sec^2(\theta)}{a^2 \sec^2(\theta)} \, d\theta = \int_{\tan^{-1}(-\frac{1}{a})}^{\tan^{-1}(a)} \frac{1}{a} \, d\theta
   \]
   \[
   = \frac{1}{a} \int_{\tan^{-1}(-\frac{1}{a})}^{\tan^{-1}(a)} d\theta
   \]

4. **Evaluate the Integral**:
   \[
   \frac{1}{a} \left[ \theta \right]_{\tan^{-1}(-\frac{1}{a})}^{\tan^{-1}(a)} = \frac{1}{a} \left( \tan^{-1}(a) - \tan^{-1}(-\frac{1}{a}) \right)
   \]

5. **Simplify the Result**:
   Recall that \( \tan^{-1}(-x) = -\tan^{-1}(x) \):
   \[
   \frac{1}{a} \left( \tan^{-1}(a) + \tan^{-1}(\frac{1}{a}) \right)
   \]
   Using the identity \( \tan^{-1}(x) + \tan^{-1}(\frac{1}{x}) = \frac{\pi}{2} \) for \( x > 0 \):
   \[
   \frac{1}{a} \cdot \frac{\pi}{2} = \frac{\pi}{2a}
   \]

The final answer is \(\boxed{\frac{\pi}{2a}}\).

Answer: \frac{\pi}{2a}


Given a sequence $\left\{a_{n}\right\}$ with all terms being positive, the sum of the first $n$ terms is $S_{n}$, and $a_{n}+1=2 \sqrt{S_{n}}\left(n \in \mathbf{Z}_{+}\right)$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=\frac{a_{n+2}}{a_{n} a_{n+1} 2^{n}}$, the sum of the first $n$ terms of the sequence $b_{n}$ is $T_{n}$, prove: $\frac{5}{6} \leqslant T_{n}<1$.

Solution:
14. (1) From $a_{n}+1=2 \sqrt{S_{n}}$, we know that when $n=1$,
$$
a_{1}+1=2 \sqrt{S_{1}}=2 \sqrt{a_{1}} \Rightarrow S_{1}=1 \text {; }
$$

When $n \geqslant 2$,
$$
\begin{array}{l}
a_{n}=S_{n}-S_{n-1}=2 \sqrt{S_{n}}-1 \\
\Rightarrow\left(\sqrt{S_{n}}-1\right)^{2}=S_{n-1} \\
\Rightarrow \sqrt{S_{n}}=\sqrt{S_{n-1}}+1\left(a_{n}>0\right) .
\end{array}
$$

Thus, $\left\{\sqrt{S_{n}}\right\}$ is an arithmetic sequence with the first term 1 and common difference 1.
Therefore, $\sqrt{S_{n}}=n, a_{n}=2 \sqrt{S_{n}}-1=2 n-1$.
(2) From (1), we know $a_{n}=2 n-1$.
$$
\begin{array}{l}
\text { Then } b_{n}=\frac{2 n+3}{(2 n-1)(2 n+1) 2^{n}} \\
=\left(\frac{2}{2 n-1}-\frac{1}{2 n+1}\right) \frac{1}{2^{n}} \\
=\frac{1}{2^{n-1}(2 n-1)}-\frac{1}{2^{n}(2 n+1)} .
\end{array}
$$

Thus, $T_{n}=b_{1}+b_{2}+\cdots+b_{n}$
$$
\begin{aligned}
= & 1-\frac{1}{6}+\frac{1}{6}-\frac{1}{20}+\cdots+\frac{1}{2^{n-1}(2 n-1)}- \\
& \frac{1}{2^{n}(2 n+1)} \\
= & 1-\frac{1}{2^{n}(2 n+1)} .
\end{aligned}
$$

Since $\frac{1}{2^{n}(2 n+1)}>0$, it is clear that $T_{n}<1$.
Also, $T_{n}=1-\frac{1}{2^{n}(2 n+1)}$ is monotonically increasing, thus,
$$
T_{n} \geqslant T_{1}=\frac{5}{6} \text {. }
$$

In summary, $\frac{5}{6} \leqslant T_{n}<1$.

Answer: \frac{5}{6}\leqslantT_{n}<1


Through the lateral edge $PC$ of a regular triangular pyramid $ABCP$, a plane parallel to the side $AB$ of the base is drawn. The lateral edge $PA$ forms an angle $\arcsin \frac{\sqrt{2}}{3}$ with this plane. Find the angle between the lateral edge and the base plane of the pyramid.

Solution:
Let $a$ and $b$ be the side of the base and the lateral edge of the given pyramid, and $\alpha$ be the required angle. Let $E$ be the orthogonal projection of point $A$ onto the given plane. According to the problem, $\sin \angle A P E = \frac{\sqrt{2}}{3}$, therefore

$$
A E = A P \sin \angle A P E = \frac{b \sqrt{2}}{3}
$$

Since the line $A B$ is parallel to the given plane, all its points are equidistant from this plane. Consider the section of the pyramid by the plane passing through points $C, P$, and the midpoint $K$ of edge $A B$. In this plane, the center $M$ of the base of the pyramid is located. The height $K D$ of triangle $C P K$ is perpendicular to the plane passing through line $C P$ parallel to $A B$, since $K D \perp P C$ and $K D \perp A B$ (line $A B$ is perpendicular to the plane $C P K$ containing $K D$). Therefore, $K D = A E = \frac{b \sqrt{2}}{3}$. Consider triangle $C P K$. In it,

$$
C K \cdot P M = C P \cdot K D, \quad \frac{a \sqrt{3}}{2} \cdot \sqrt{b^{2} - \left(\frac{a \sqrt{3}}{3}\right)^{2}} = b \cdot \frac{b \sqrt{2}}{3}
$$

After squaring and obvious simplifications, we get the equation

$$
8 b^4 - 27 a^2 b^2 + 9 a^4 = 0
$$

from which we find that $b = a \sqrt{3}$ or $b = \frac{a \sqrt{6}}{4}$. If $b = a \sqrt{3}$, then

$$
\cos \alpha = \cos \angle P A M = \frac{A M}{A P} = \frac{\frac{a \sqrt{3}}{3}}{b} = \frac{1}{3}, \quad \sin \alpha = \frac{2 \sqrt{2}}{3}
$$

If $b = \frac{a \sqrt{6}}{4}$, then

$$
\cos \alpha = \cos \angle P A M = \frac{A M}{A P} = \frac{\frac{a \sqrt{3}}{3}}{b} = \frac{2 \sqrt{2}}{3}, \quad \sin \alpha = \frac{1}{3}
$$

## Answer

$\arcsin \frac{2 \sqrt{2}}{3}$ or $\arcsin \frac{1}{3}$

Answer: \arcsin\frac{2\sqrt{2}}{3}


Determine the real value of $t$ that minimizes the expression
\[
  \sqrt{t^2 + (t^2 - 1)^2} + \sqrt{(t-14)^2 + (t^2 - 46)^2}.
\]

Solution:
1. We start by interpreting the given expression as the sum of the distances from the point \((t, t^2)\), which lies on the graph of \(y = x^2\), to the points \(A = (0, 1)\) and \(B = (14, 46)\). The expression to minimize is:
   \[
   \sqrt{t^2 + (t^2 - 1)^2} + \sqrt{(t-14)^2 + (t^2 - 46)^2}.
   \]

2. By the Triangle Inequality, the sum of the distances from any point to \(A\) and \(B\) is minimized when that point lies on the line segment \(\overline{AB}\). Therefore, we need to find the intersection points of the parabola \(y = x^2\) and the line passing through \(A\) and \(B\).

3. The equation of the line passing through \(A\) and \(B\) can be found using the slope formula:
   \[
   \text{slope} = \frac{46 - 1}{14 - 0} = \frac{45}{14}.
   \]
   Thus, the equation of the line is:
   \[
   y = \frac{45}{14}x + 1.
   \]

4. To find the intersection points of the parabola \(y = x^2\) and the line \(y = \frac{45}{14}x + 1\), we set the equations equal to each other:
   \[
   x^2 = \frac{45}{14}x + 1.
   \]

5. Rearrange the equation to form a standard quadratic equation:
   \[
   14x^2 - 45x - 14 = 0.
   \]

6. Factor the quadratic equation:
   \[
   14x^2 - 45x - 14 = (7x + 2)(2x - 7) = 0.
   \]

7. Solve for \(x\):
   \[
   7x + 2 = 0 \implies x = -\frac{2}{7},
   \]
   \[
   2x - 7 = 0 \implies x = \frac{7}{2}.
   \]

8. We need to check which of these points lies on the segment \(\overline{AB}\). The point \((x, x^2)\) must lie between \(A\) and \(B\). Since \(x = -\frac{2}{7}\) is not within the interval \([0, 14]\), it is not on the segment \(\overline{AB}\). However, \(x = \frac{7}{2}\) is within this interval.

9. Therefore, the value of \(t\) that minimizes the given expression is \(t = \frac{7}{2}\).

The final answer is \(\boxed{\frac{7}{2}}\).

Answer: \frac{7}{2}


The probability that Mark is paired up with his best friend, Mike, in a class of $16$ students randomly paired into groups of $2$ is:

(A) $\frac{1}{15}$

(B) $\frac{1}{8}$

(C) $\frac{1}{7}$

(D) $\frac{2}{15}$

(E) $\frac{1}{5}$

Solution:
1. First, we need to determine the total number of ways to pair up Mark with any of the other students in the class. Since there are 16 students in total, and Mark is one of them, there are 15 other students who can be paired with Mark.

2. The probability of Mark being paired with any specific student (including Mike) is calculated by considering that each student has an equal chance of being paired with Mark. Therefore, the probability of Mark being paired with Mike is:
   \[
   \frac{1}{15}
   \]

The final answer is \(\boxed{\frac{1}{15}}\)

Answer: \frac{1}{15}


Find such a number that its fractional part, its integer part, and the number itself form a geometric progression ${ }^{\star}$.

Solution:
289. Given the conditions $x(x-[x])=[x]^2$, hence $x=$ $=[x] \frac{(1+\sqrt{5})}{2}, |x| \geqslant |[x]|$ and $x \geqslant 0$.

Further, $[x]+1 \geqslant [x] \frac{1+\sqrt{5}}{2}$ and $[x] \leqslant \frac{2}{\sqrt{5}-1}<2$. Therefore, $[x]=0$ or 1, and $x=0$ or $\frac{1+\sqrt{5}}{2}$.

[D. Silverman, M. M., 43, 56 (January 1970).]

Answer: \frac{1+\sqrt{5}}{2}


The range of the function $f(x)=2 \cos x+\sin 2 x(x \in \mathbf{R})$ is $\qquad$ .

Solution:
6. $\left[-\frac{3 \sqrt{3}}{2}, \frac{3 \sqrt{3}}{2}\right]$.

Notice,
$$
\begin{array}{l}
f^{2}(x)=(2 \cos x+\sin 2 x)^{2} \\
=4 \cos ^{2} x(1+\sin x)^{2} \\
=\frac{4}{3}(3-3 \sin x)(1+\sin x)^{3} \\
\leqslant \frac{4}{3}\left[\frac{(3-3 \sin x)+3(1+\sin x)}{4}\right]^{4}=\frac{27}{4},
\end{array}
$$

Equality holds if and only if $3-3 \sin x=1+\sin x$, i.e., $\sin x=\frac{1}{2}$.

Thus, when $\sin x=\frac{1}{2}, \cos x=\frac{\sqrt{3}}{2}, f(x)$ achieves its maximum value $\frac{3 \sqrt{3}}{2}$;

When $\sin x=\frac{1}{2}, \cos x=-\frac{\sqrt{3}}{2}, f(x)$ achieves its minimum value $-\frac{3 \sqrt{3}}{2}$.
Therefore, the range of the function $f(x)$ is $\left[-\frac{3 \sqrt{3}}{2}, \frac{3 \sqrt{3}}{2}\right]$.

Answer: [-\frac{3\sqrt{3}}{2},\frac{3\sqrt{3}}{2}]


Let $A_{n}$ be the sum of the areas of the small triangles removed at the $n$-th step (for example, $A_{1}$ is the area of the middle small triangle removed at the first step, and $A_{2}$ is the sum of the areas of the three small triangles removed at the second step), then the sum of the areas of all the small triangles removed in the first $n$ steps is

Solution:
Answer $\frac{\sqrt{3}}{4}\left(1-\left(\frac{3}{4}\right)^{n}\right)$.
Analysis The original equilateral triangle area is $\frac{\sqrt{3}}{4}$, and in the $k$-th step, a total of $3^{k-1}$ small triangles are removed, $A_{k}=\frac{\sqrt{3}}{16}\left(\frac{3}{4}\right)^{k-1}$, therefore, the sum of a geometric series formula can be used, to get the answer as:
$$
\sum_{k=1}^{n} A_{k}=\frac{\sqrt{3}}{16} \sum_{k=1}^{n}\left(\frac{3}{4}\right)^{k-1}=\frac{\sqrt{3}}{16} \frac{1-\left(\frac{3}{4}\right)^{n}}{1-\frac{3}{4}}=\frac{\sqrt{3}}{4}\left(1-\left(\frac{3}{4}\right)^{n}\right)
$$

Answer: \frac{\sqrt{3}}{4}(1-(\frac{3}{4})^{n})


78 students stand in a row. Starting from the left, give a badminton shuttlecock to the first person, and then give a badminton shuttlecock every 2 people. Starting from the right, give a table tennis ball to the first person, and then give a table tennis ball every 3 people. The number of students who have both a badminton shuttlecock and a table tennis ball is $\qquad$ people.

Answer: 6


Anya wrote down 100 numbers in her notebook. Then Sonya wrote down in her notebook all the pairwise products of the numbers written by Anya. Artem noticed that there were exactly 2000 negative numbers in Sonya's notebook. How many zeros did Anya initially write down in her notebook?

Solution:
Answer: 10 zeros.

Solution: Let Anya write down $n$ positive numbers, $m$ negative numbers, and $100-n-m$ zeros in her notebook. Then, by the condition, $n m=2000$, since a negative number can only be obtained by multiplying numbers of different signs.

Let's list all the divisors of the number $2000=2^{4} * 5^{3}$:

|  | $2^{0}$ | $2^{1}$ | $2^{2}$ | $2^{3}$ | $2^{4}$ |
| :--- | :--- | :--- | :--- | :--- | :--- |
| $5^{0}$ | 1 | 2 | 4 | 8 | 16 |
| $5^{1}$ | 5 | 10 | 20 | 40 | 80 |
| $5^{2}$ | 25 | 50 | 100 | 200 | 400 |
| $5^{3}$ | 125 | 250 | 500 | 1000 | 2000 |

They can be paired such that the product of each pair is 2000. Let's write them in pairs and list their sums:

| 1 | 2 | 4 | 5 | 8 | 10 | 16 | 20 | 25 | 40 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 2000 | 1000 | 500 | 400 | 250 | 200 | 125 | 100 | 80 | 50 |

| 2001 | 1002 | 504 | 405 | 258 | 210 | 141 | 120 | 105 | 90 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |

Only in one of these cases is the sum less than 100, so this is the case that is realized. Thus, there are 90 non-zero numbers and 10 zeros.

Criteria: Only the answer - 0 points.

Only the answer with verification - 1 point.

Not all cases of divisors are considered - no more than 3 points.

No explanation of why there are no other divisors - deduct 1 point.

Answer: 10


Find the smallest positive integer $n$ for which there are exactly 2323 positive integers less than or equal to $n$ that are divisible by 2 or 23, but not both.

Solution:
17. 4644
Solution. The number of positive integers from 1 to $n$ that are divisible by 2 or 23 , but not both, is
$$
f(n)=\left\lfloor\frac{n}{2}\right\rfloor+\left\lfloor\frac{n}{23}\right\rfloor-2\left\lfloor\frac{n}{46}\right\rfloor .
$$

We need to find $f(n)=2323$, which can be done by some trial and error.
Note that if $n$ is a multiple of 2 and 23 , then the floor divisions are non-rounded exact divisions, in which case,
$$
\frac{n}{2}+\frac{n}{23}-2 \frac{n}{46}=\frac{n}{2}
$$

So, for example, $f(4646)=2323$. We can then just tick downwards.
Since 4646 and 4645 do not satisfy our criteria, we know that $f(4646)=f(4645)=f(4644)$, i.e. we can remove 4646 and 4645 and the count does not go down (we weren't counting them). However, 4644 does satisfy our criteria, so $f(4643)=2322$.
Note that by definition, this function is non-decreasing. Thus, we conclude that $n=4644$ is the first $n$ that satisfies the given conditions.

Answer: 4644


Let $ n$ be a given positive integer. Find the smallest positive integer $ u_n$ such that for any positive integer $ d$, in any $ u_n$ consecutive odd positive integers, the number of them that can be divided by $ d$ is not smaller than the number of odd integers among $ 1, 3, 5, \ldots, 2n - 1$ that can be divided by $ d$.

Solution:
1. **Restate the problem in a more structured way:**

   Let \( n \) be a positive integer. Define the set \( N_{m,k} \), where \( m \) is an odd positive integer, to be the set of the \( k \) smallest odd integers that are at least \( m \). For an arbitrary positive integer \( d \), denote by \( N_{m,k}(d) \) the number of integers in the set \( N_{m,k} \) that are divisible by \( d \). Find the smallest \( u_n \in \mathbb{N} \) (in terms of \( n \)) such that \( N_{m,u_n}(d) \ge N_{1,n}(d) \) for every positive integer \( d \) and odd positive integer \( m \).

2. **Identify the pattern and conjecture:**

   The first few values for \( u_n \) are \( 1, 3, 5, 7, 9 \), so we conjecture \( u_n = 2n - 1 \).

3. **Verify the conjecture:**

   - Clearly, \( N_{1,n}(2n-1) = 1 \) because of the presence of \( 2n-1 \) in the set \( N_{1,n} \).
   - The next odd multiple is \( 3(2n-1) = 2(3n-1) - 1 \), which means that \( N_{2n+1,2n-2}(2n-1) = 0 \).
   - Therefore, \( u_n \ge 2n-1 \).

4. **Show that \( u_n = 2n-1 \) works:**

   - Note that \( N_{1,n}(d) = 0 \) for \( d \not\in \{1, 3, 5, \ldots, 2n-1\} \), so we only need to worry about \( d \) in this set.
   - Let \( m \) be an arbitrary odd positive integer. Then the set \( N_{m,2n-1} \) consists of the integers \( m, m+2, m+4, \ldots, m+2(2n-1)-2 \).

5. **Prove that one of the first \( d \) of these integers is divisible by \( d \):**

   - Suppose none of the first \( d \) integers are multiples of \( d \). Then none are equal to \( 0 \pmod{d} \), and by the pigeonhole principle, there are two equal \( \pmod{d} \), say \( m+a \) and \( m+b \) where \( 0 \le a < b \le 2d-2 \).
   - This means \( d \mid (b-a) \). Since \( 2d-2 \ge b-a > 0 \), the only multiple of \( d \) that \( b-a \) can possibly be is \( d \) itself.
   - Now \( a \) and \( b \) are both even, so \( b-a \) is even, meaning \( d \) is even, which is a contradiction since we assumed \( d \in \{1, 3, 5, \ldots, 2n-1\} \).

6. **Conclude that one of the first \( d \) of these integers is divisible by \( d \):**

   - Let \( m+a \) be the smallest integer in the set \( N_{m,2n-1} \) such that \( d \mid (m+a) \). We have just proven \( a \le 2d-2 \).

7. **Determine the number of integers in \( N_{m,2n-1} \) divisible by \( d \):**

   - Let \( h \) be the largest integer such that \( m+a+2hd \le m+4n-4 \). Then the integers in \( N_{m,2n-1} \) which are divisible by \( d \) are \( m+a, m+a+2d, m+a+4d, \ldots, m+a+2hd \).
   - It follows that \( N_{m,2n-1}(d) = h+1 \).

8. **Compare with \( N_{1,n}(d) \):**

   - Let \( b \) be the largest odd integer such that \( bd \le 2n-1 \). If \( b = 2c-1 \), then it follows that \( N_{1,n}(d) = c \ge 1 \).
   - Now \( (2c-1)d \le 2n-1 \) so \( 2(2c-1)d \le 4n-2 \).
   - Since \( h \) is the largest integer such that \( m+a+2hd \le m+4n-4 \), we have \( a+2(h+1)d \ge 4n-4 \).
   - Now \( 2d-2 \ge a \) meaning \( 2d-2+2(h+1)d \ge a+2(h+1)d \ge 4n-4 \ge 2(2c-1)d-2 \).
   - Then it follows that \( h+1 \ge 2c-2 \ge c \) for \( c > 1 \). When \( c = 1 \), we have \( d = 2n-1 \) in which case \( h = 0 \) so \( h+1 = 1 \ge 1 = c \).

9. **Conclude that \( N_{m,2n-1}(d) \ge N_{1,n}(d) \):**

   - Therefore, \( u_n = 2n-1 \) works.

The final answer is \( \boxed{ u_n = 2n-1 } \).

Answer: u_n = 2n-1


Let $2 n$ real numbers $a_{1}, a_{2}, \cdots, a_{2 n}$ satisfy $\sum_{i=1}^{2 n-1}\left(a_{i+1}-a_{i}\right)^{2}=1$, find
$$\left(a_{n+1}+a_{n+2}+\cdots+a_{2 n}\right)-\left(a_{1}+a_{2}+\cdots+a_{n}\right)$$

the maximum value.

Solution:
When $n=1$, $\left(a_{2}-a_{1}\right)^{2}=1$, then $a_{2}-a_{1}= \pm 1$, the maximum value is 1.
When $n \geqslant 2$, let $x_{1}=a_{1}, x_{i+1}=a_{i+1}-a_{i}, i=1,2, \cdots, 2 n-1$. Then $\sum_{i=2}^{2 n} x_{i}^{2}=$ 1, and $a_{k}=x_{1}+x_{2}+\cdots+x_{k}, k=1,2, \cdots, 2 n$.

By the Cauchy-Schwarz inequality, we get
$$\begin{aligned}
& a_{n+1}+a_{n+2}+\cdots+a_{2 n}-\left(a_{1}+a_{2}+\cdots+a_{n}\right) \\
= & x_{2}+2 x_{3}+\cdots+(n-1) x_{n}+n x_{n+1}+(n-1) x_{n+2}+\cdots+x_{2 n} \\
\leqslant & {\left[1+2^{2}+\cdots+(n-1)^{2}+n^{2}+(n-1)^{2}+\cdots+1\right]^{\frac{1}{2}} } \\
& \cdot\left(x_{2}^{2}+x_{3}^{2}+\cdots+x_{2 n}^{2}\right)^{\frac{1}{2}} \\
= & {\left[n^{2}+2 \cdot \frac{1}{6}(n-1) n(2(n-1)+1)\right]^{\frac{1}{2}}=\sqrt{\frac{n\left(2 n^{2}+1\right)}{3}} . }
\end{aligned}$$

When $a_{k}=\frac{\sqrt{3} k(k-1)}{2 \sqrt{n\left(2 n^{2}+1\right)}}, k=1,2, \cdots, n+1$,
$$a_{n+k}=\frac{\sqrt{3}\left[2 n^{2}-(n-k)(n-k+1)\right]}{2 \sqrt{n\left(2 n^{2}+1\right)}}, k=1,2, \cdots, n-1$$

the equality holds, and the maximum value is $\sqrt{\frac{n\left(2 n^{2}+1\right)}{3}}$.

Answer: \sqrt{\frac{n\left(2 n^{2}+1\right)}{3}}


Find the positive integer solutions for $x$ if $x^2 - 5x + 6 = 0$.

Solution:
$\triangle$ Obviously, this is possible only with "skipping" over the nine.

Let the smaller of the sought numbers be denoted by

$$
x=\overline{a b c \ldots k l 99 \ldots 9}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_475fdfbb473422773b9eg-223.jpg?height=48&width=666&top_left_y=1004&top_left_x=341)

$$
x+1=\overline{a b c \ldots k(l+1) 00 \ldots 0}
$$

( $i$ zeros)

By the condition, each of the sums of the digits

$$
a+b+c+\ldots+k+l+9 n, \quad a+b+c+\ldots+k+(l+1)
$$

is divisible by 7. Therefore, the difference between these sums is also divisible by 7:

$$
(9 n-1): 7
$$

The smallest value of $n$ that satisfies the last relation is $n=4$. Then $9 n=36$.

For the number $\overline{a b c . . . k l}$, we need to take a single-digit number, the sum of which with 36 is divisible by 7. Clearly, this single-digit number is 6. Therefore,

$$
x=69999, \quad x+1=70000
$$

Answer: $69999,70000$.

Answer: 69999,70000


In Rt $\triangle A B C$, the altitude $C D$ on the hypotenuse $A B$ is $3$, extend $D C$ to point $P$, such that $C P=2$, connect $A P$, draw $B F \perp A P$, intersecting $C D$ and $A P$ at points $E$ and $F$ respectively. Then the length of segment $D E$ is $\qquad$

Solution:
8. $\frac{9}{5}$.
As shown in Figure 5, let $A D=a$.
In the right triangle $\triangle A B C$, given $C D=3, C D \perp A B$, we know $B D=\frac{9}{a}$.
Notice that, in the right triangles $\triangle A P D \sim \triangle E B D$,

then $\frac{D E}{B D}=\frac{A D}{P D} \Rightarrow D E$ $=\frac{9}{a} \times \frac{a}{5}=\frac{9}{5}$.

Answer: \frac{9}{5}


In the star pentagon $ABCDE$ (each side line has 2 vertices on one side from the remaining vertices, and 1 vertex on the other side). What is the sum of the angles $ABC, BCD, CDE, DEA, EAB$?

Solution:
I. solution. We use the notation of the figure (the intersection point of lines $BC$ and $DE$ is $J$, and so on).

![](https://cdn.mathpix.com/cropped/2024_05_02_ed8321d1a3fabb16dd0bg-1.jpg?height=781&width=761&top_left_y=206&top_left_x=671)

We apply the well-known theorem of the exterior angle of a triangle first to $\triangle BCN$, then to $\triangle NDL$:

$$
\begin{gathered}
\beta+\gamma=\angle ABC + \angle BCD = \angle NBC + \angle BCN = \angle BND = \angle LND, \\
\angle LND + \angle NDL = \angle NLE, \quad \text{where} \quad \angle NDL = \angle CDE = \delta,
\end{gathered}
$$

thus

$$
\beta+\gamma+\delta=\angle NLE = \angle ALE
$$

Adding $\varepsilon = \angle DEA = \angle LEA$ and $\alpha = \angle EAB = \angle EAL$ to this, we get the sum of the angles of $\triangle AEL$, so $\alpha+\beta+\gamma+\delta+\varepsilon=180^{\circ}$.

Eltető Eszter (Budapest, I. István g. I. o. t.)

II. solution. The sum of the angles in question is obtained by subtracting the sum of the interior angles at the vertices $J, K, L, M, N$ of the triangles $BJL, CKM, DNL, EMJ$, and $ANK$ from the sum of the interior angles of these triangles, which is $5 \cdot 180^{\circ}$. Among these 10 angles, 2 are vertical angles and thus equal, and each is an exterior angle of the common pentagon $JKLMN$; hence, the above subtraction is equal to twice the sum of these exterior angles.

The sum of the interior and exterior angles at each vertex of a common pentagon is $180^{\circ}$, so the sum of the exterior angles is obtained by subtracting the sum of the interior angles from $5 \cdot 180^{\circ}$. The sum of the interior angles, as is well known, is $3 \cdot 180^{\circ}$, so the sum of the exterior angles is $2 \cdot 180^{\circ}$, and the above subtraction is $4 \cdot 180^{\circ}$. Therefore, the desired sum, the remainder of the subtraction, is $1 \cdot 180^{\circ} = 180^{\circ}$.

Bárász Péter (Budapest, Fazekas M. gyak. g. I. o. t.)

III. solution. Let's walk around the pentagon in the order of the labeling, starting from a point $K$ on the $EA$ side. In this way, we need to turn at the vertices $A, B, C, D, E$ by the supplementary angles of $\alpha, \beta, \gamma, \delta$, and $\varepsilon$, respectively, and we complete two full turns. Thus,

$$
\begin{gathered}
180^{\circ}-\alpha+180^{\circ}-\beta+180^{\circ}-\gamma+180^{\circ}-\delta+180^{\circ}-\varepsilon=720^{\circ} \\
\alpha+\beta+\gamma+\delta+\varepsilon=180^{\circ}
\end{gathered}
$$

Herneczki István (Sopron, Széchenyi I. g. I. o. t.)

IV. solution. The sum $\alpha+\beta+\gamma+\delta+\varepsilon$ in question can be considered as the change in direction of the segment $AE$ during the following sequence of movements. The segment $AE$

1. a) rotates around $A$ (by an angle less than $180^{\circ}$) to the ray $AB$, its new position is $A_1E_1 = AE_1$;
2. b) (if necessary,) slides (along the line $AB$) so that $E_1$ reaches $B$, its new position is $A_1'E_1' = A_1'B$;
3. a) rotates around $B$ (i.e., $E_1'$) to the ray $BC$: $A_2E_2 = A_2B$;
4. b) (if necessary,) slides to the position $A_2'E_2' = CE_2'$;
5. a) rotates to the position $A_3E_3 = CE_3$;
6. b) (if necessary,) slides to the position $A_3'E_3' = A_3'D$;
7. a) rotates to the position $A_4E_4 = A_4D$;
8. b) (if necessary,) slides to the position $A_4'E_4' = EE_4'$;
9. rotates to $EA$.

The sequence of movements has swapped the positions of $A$ and $E$, and the direction of $AE$ has rotated by $180^{\circ}$; each rotation occurred in the same direction (in the positive direction in the figure), so the value of the sum in question is $180^{\circ}$.

Answer: 180


Dasha added 158 numbers and got 1580. Then Seryozha tripled the largest of these numbers and decreased another number by 20. The resulting sum did not change. Find the smallest of the original numbers.

Solution:
# Solution:

Let x be the largest of the original numbers, and y be the number that Seryozha decreased. Then: $x+y=3 x+y-20$, i.e., $x=10$.

Since the arithmetic mean of the original numbers is 10, and the largest of these numbers is also 10, each of the given numbers is 10.

Answer: 10.

## Criteria:

| 20 points | Any complete and correct solution. |
| :--- | :--- |
| 15 points | A generally correct solution with minor flaws. |
| 5 points | The largest number is found. |
| 0 points. | The solution does not meet any of the criteria listed above or only the answer is provided. |

Answer: 10


$p, q, r$ are pairwise distinct real numbers, satisfying the system of equations
$$\left\{\begin{array}{l}
q=p(4-p) \\
r=q(4-q) \\
p=r(4-r)
\end{array}\right.$$

Find all possible values of $p+q+r$.

Solution:
[Solution] From the given,
$$p=r(4-r)=-(r-2)^{2}+4 \leqslant 4,$$

Similarly, we can get
$$q \leqslant 4, r \leqslant 4$$

Next, we prove that $p \geqslant 0$ using proof by contradiction.
If $p<0$, then from $4-p \geqslant 0$ we get
$$q \leqslant 0.$$

Again, from $4-q \geqslant 0$ we get
$$r \leqslant 0$$

Thus, we have
$$p+q+r<0$$

Adding the three equations in the problem, we get
$$p^{2}+q^{2}+r^{2}=3(p+q+r)$$

From the above two equations, we get
$$p^{2}+q^{2}+r^{2}<0$$

This is a contradiction.
Therefore, $4 \geqslant p \geqslant 0$.
Similarly, we can get
$$4 \geqslant q \geqslant 0, 4 \geqslant r \geqslant 0$$

Let
$$p=4 \sin ^{2} \theta, \theta \in\left[0, \frac{\pi}{2}\right]$$

Substituting the above into the first equation of the original system, we get
$$q=4 \sin ^{2} 2 \theta$$

Substituting the above into the second equation of the original system, we get
$$r=4 \sin ^{2} 4 \theta$$

Substituting the above into the third equation of the original system, we get
$$p=4 \sin ^{2} 8 \theta$$

Thus, we have
$$\sin 8 \theta= \pm \sin \theta$$

Since $0 \leqslant 8 \theta \leqslant 4 \pi$, there are 8 possible cases:
$$\begin{array}{l}
8 \theta=\theta, 8 \theta=\pi-\theta, 8 \theta=\pi+\theta, 8 \theta=2 \pi-\theta \\
8 \theta=2 \pi+\theta, 8 \theta=3 \pi-\theta, 8 \theta=3 \pi+\theta, 8 \theta=4 \pi-\theta . \\
\text { That is } \quad \theta \in\left\{0, \frac{\pi}{9}, \frac{\pi}{7}, \frac{2 \pi}{9}, \frac{2 \pi}{7}, \frac{\pi}{3}, \frac{3 \pi}{7}, \frac{4 \pi}{9}\right\}
\end{array}$$

When $\theta=0$ or $\frac{\pi}{3}$, $p=q=r=0$ or 3, which are discarded.
When $\theta=\frac{\pi}{9}, \frac{2 \pi}{9}, \frac{4 \pi}{9}$,
$$\begin{aligned}
p+q+r & =4\left(\sin ^{2} \frac{\pi}{9}+\sin ^{2} \frac{2 \pi}{9}+\sin ^{2} \frac{4 \pi}{9}\right) \\
& =6-2\left(\cos \frac{2 \pi}{9}+\cos \frac{4 \pi}{9}+\cos \frac{8 \pi}{9}\right) \\
& =6-2\left(2 \cos \frac{\pi}{3} \cdot \cos \frac{\pi}{9}-\cos \frac{\pi}{9}\right) \\
& =6,
\end{aligned}$$

When $\theta=\frac{\pi}{7}, \frac{2 \pi}{7}, \frac{3 \pi}{7}$,
$$\begin{aligned}
p+q+r & =4\left(\sin ^{2} \frac{\pi}{7}+\sin ^{2} \frac{2 \pi}{7}+\sin ^{2} \frac{4 \pi}{7}\right) \\
& =6-2\left(\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{8 \pi}{7}\right) \\
& =6-2\left(2 \cos \frac{3 \pi}{7} \cdot \cos \frac{\pi}{7}-\cos \frac{\pi}{7}\right) \\
& =6-2 \cos \frac{\pi}{7}\left(2 \cos \frac{3 \pi}{7}-1\right) \\
& =6-\frac{\sin \frac{2 \pi}{7}\left(2 \cos \frac{3 \pi}{7}-1\right)}{\sin \frac{\pi}{7}} \\
& =6-\frac{1}{\sin \frac{\pi}{7}} \cdot\left(\sin \frac{5 \pi}{7}-\sin \frac{\pi}{7}-\sin \frac{2 \pi}{7}\right) \\
& =6+1=7
\end{aligned}$$

Therefore, the only possible values of $p+q+r$ are 6 and 7.

Answer: 6 \text{ and } 7


In triangle $A B C$, point $D$ is on side $\overline{A B}$ such that $|A D|:|D B|=3: 4$. On segment $\overline{C D}$, point $E$ is such that $|C E|:|E D|=3: 4$. A line parallel to $A E$ passes through point $D$ and intersects $\overline{B C}$ at point $G$. Determine the ratio $|C G|:|G B|$.

Solution:
## Solution.

Let's draw the diagram.

![](https://cdn.mathpix.com/cropped/2024_05_30_a25c851e6461b5bf7f1cg-08.jpg?height=660&width=1011&top_left_y=1229&top_left_x=451)

Let the line $A E$ intersect the side $\overline{B C}$ at $F$ and let $DG$ be parallel to $EF$. By Thales' theorem, we have

$$
\frac{|C F|}{|F G|}=\frac{3}{4}=\frac{9}{12}
$$

It also holds that: $\frac{|F G|}{|G B|}=\frac{|A D|}{|D B|}=\frac{3}{4}=\frac{12}{16}$. Therefore,

$$
\begin{aligned}
& |C F|:|F G|:|G B|=9: 12: 16 \\
& |C F|=9 k,|F G|=12 k,|G B|=16 k \\
\Rightarrow & \frac{|C G|}{|G B|}=\frac{|C F|+|F G|}{|G B|}=\frac{9 k+12 k}{16 k}=\frac{21}{16}
\end{aligned}
$$

$$
\begin{aligned}
& 1 \text { point } \\
& 1 \text { point }
\end{aligned}
$$

1 point

Answer: \frac{21}{16}


The function $f(x)(x \in \mathrm{R})$ is not always greater than 0, and satisfies $f(a+b)=f(a)+f(b)$, then $f(x)$ is
(A) even function
(B) odd function
(C) both an odd function and an even function
(D) neither an odd function nor an even function

Solution:
B

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

Answer: B


Let $S$ be the area of a triangle inscribed in a circle of radius 1. Then the minimum value of $4 S+\frac{9}{S}$ is $\qquad$ .

Solution:
$10.7 \sqrt{3}$.
Let $y=4 S+\frac{9}{S}$, it is easy to know that it is a monotonically decreasing function on $\left(0, \frac{3}{2}\right]$. Also, because among the triangles inscribed in a unit circle, the equilateral triangle has the maximum area, which is equal to $\frac{3 \sqrt{3}}{4}$, hence $0<S \leqslant \frac{3 \sqrt{3}}{4}<\frac{3}{2}$.
Therefore, $y_{\min }=4 \times \frac{3 \sqrt{3}}{4}+\frac{9}{\frac{3 \sqrt{3}}{4}}=7 \sqrt{3}$.

Answer: 7 \sqrt{3}


In how many ways can we fill the cells of a $m\times n$ board with $+1$ and $-1$ such that the product of numbers on each line and on each column are all equal to $-1$?

Solution:
1. **Consider the product of all \( mn \) numbers on the board:**
   - Each cell in the \( m \times n \) board can be either \( +1 \) or \( -1 \).
   - The product of all numbers in the board is the product of the products of each row and each column.

2. **Analyze the product of each row and column:**
   - For the product of numbers in each row to be \( -1 \), each row must contain an odd number of \( -1 \)'s.
   - Similarly, for the product of numbers in each column to be \( -1 \), each column must contain an odd number of \( -1 \)'s.

3. **Consider the overall product of the board:**
   - The product of all numbers in the board is the product of the products of each row.
   - Since each row product is \( -1 \), the product of all rows is \( (-1)^m \).
   - Similarly, the product of all columns is \( (-1)^n \).

4. **Determine the feasibility based on \( m \) and \( n \):**
   - For the product of all numbers in the board to be consistent, \( (-1)^m \) must equal \( (-1)^n \).
   - This implies that \( m \) and \( n \) must have the same parity (both even or both odd).

5. **Case when \( m \not\equiv n \pmod{2} \):**
   - If \( m \) and \( n \) have different parities, it is impossible to fill the board such that the product of each row and each column is \( -1 \).

6. **Case when \( m \equiv n \pmod{2} \):**
   - If \( m \) and \( n \) have the same parity, we can construct a valid filling.
   - Consider a bijection between the fillings of the \( m \times n \) board and the \( (m-1) \times (n-1) \) board.
   - Fix the first \( m-1 \) rows and \( n-1 \) columns. The last row and column are determined by the requirement that the product of each row and column is \( -1 \).

7. **Count the number of valid fillings:**
   - There are \( 2^{(m-1)(n-1)} \) ways to fill the \( (m-1) \times (n-1) \) board with \( +1 \) and \( -1 \).
   - Each filling of the \( (m-1) \times (n-1) \) board uniquely determines a filling of the \( m \times n \) board.

Conclusion:
\[
\boxed{2^{(m-1)(n-1)}}
\]

Answer: 2^{(m-1)(n-1)}


Given the curve $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a \geqslant b>0)$ passes through the point $(\sqrt{3}, 1)$, then the area of the figure formed by all points on these curves satisfying $y \geqslant 1$ is $\qquad$

Solution:
8. $\frac{4 \pi}{3}-\sqrt{3}$.

Solution: From the problem, we have $\frac{3}{a^{2}}+\frac{1}{b^{2}}=1, b^{2}=\frac{a^{2}}{a^{2}-3}$,
From $a^{2} \geqslant b^{2}$, we get $a^{2} \geqslant 4$.
$$
\because \frac{x^{2}}{a^{2}}+\frac{\left(a^{2}-3\right) y^{2}}{a^{2}}=1, \therefore a^{2}\left(y^{2}-1\right)=3 y^{2}-
$$
$x^{2} \geqslant 4\left(y^{2}-1\right)(\because y \geqslant 1)$, thus, $x^{2}+y^{2} \leqslant 4$. The area of the figure that satisfies $x^{2}+y^{2} \leqslant 4, y \geqslant 1$ is $\frac{1}{3} \cdot \pi \cdot 2^{2}-\frac{1}{2} \times$
$$
2^{2} \sin 120^{\circ}=\frac{4 \pi}{3}-\sqrt{3}
$$

Answer: \frac{4\pi}{3}-\sqrt{3}


In the figure, $A B C D$ is a parallelogram, $E$ is on side $A B$, $F$ is on side $D C$, $G$ is the intersection of $A F$ and $D E$, and $H$ is the intersection of $C E$ and $B F$. It is known that the area of parallelogram $A B C D$ is $1$, $\frac{\mathrm{AE}}{\mathrm{EB}}=\frac{1}{4}$, and the area of triangle $B H C$ is $\frac{1}{8}$. Find the area of triangle $A D G$.

Solution:
【Analysis】Let out the base and height of the parallelogram, determine the position of point $F$, and then express the lengths of $C F$, $D F$, $B E$, and $A E$ using the base of the parallelogram. Then, use the relationship between the base and height of the parallelogram and the base and height of triangle $A D G$ to solve the problem.

【Solution】Solution: Let the base of parallelogram $A B C D$ be $a$, and the height be $h$, with $a h=1$. $A E=\frac{a}{5}, B E=\frac{4 a}{5}, h=\frac{1}{a}$.
1. Calculate the position of point $F$ on $C D$:
$$
\begin{array}{l}
S_{\triangle B E H}=B E \times h \div 2-S_{\triangle B C H}, \\
=\frac{4}{5} a \times \frac{1}{2 a}-\frac{1}{8}, \\
=\frac{11}{40} ;
\end{array}
$$
$h_{1}=2 \times S_{\triangle B E H} \div B E$ ( $h_{1}$ is the height from $B E$ in $\triangle B E H$ ),
$$
\begin{array}{l}
=2 \times \frac{11}{40} \div \frac{4}{5} a, \\
=\frac{55}{80 a} ;
\end{array}
$$
$S_{\triangle C F H}=C F \times\left(h-h_{1}\right) \div 2$,
$$
=C F \times h \div 2-S_{\triangle B C H} \text {, }
$$

So $C F \times\left(\frac{1}{a}-\frac{55}{80 a}\right) \div 2=C F \times \frac{1}{a} \div 2-\frac{1}{8}$,
$$
\begin{array}{c}
C F \times \frac{25}{160 a}=C F \times \frac{80}{160 a}-\frac{20}{160}, \\
C F \times \frac{55}{160 a}=\frac{20}{160}, \\
C F=\frac{4 a}{11} ; \\
D F=D C-C F=\frac{7 a}{11} ;
\end{array}
$$
2. Calculate the area of $\triangle A D G$:
$$
\begin{array}{l}
S_{\triangle A D G}=S_{\triangle A D E}-S_{\triangle A E G}, \\
=A E \times h \div 2-A E \times h_{2} \div 2,\left(h_{2} \text { is the height from } A E \text { in } \triangle A E G\right) \\
=\frac{a}{5} \times \frac{1}{a} \div 2-\frac{a}{5} \times h_{2} \div 2, \\
=\frac{1}{10}-\frac{a}{10} \times h_{2}, \text { (1) } \\
S_{\triangle A D G}=S_{\triangle A D F}-S_{\triangle D F G}, \\
=D F \times h \div 2-D F \times\left(h-h_{2}\right) \div 2, \\
=\left(D F \times h_{2}\right) \div 2, \\
=\frac{7 a}{11} \times h_{2} \div 2, \\
=\frac{7 a}{22} \times h_{2},
\end{array}
$$
Substitute (2) into (1):
$$
\begin{array}{c}
\frac{7 a}{22} \times h_{2}=\frac{1}{10}-\frac{a}{10} \times h_{2}, \\
\frac{70 a}{220} \times h_{2}=\frac{22}{220}-\frac{22 a}{220} \times h_{2}, \\
h_{2}=\frac{22}{92 a}, \\
S_{\triangle A D G}=\frac{7 a}{22} \times h_{2}, \\
=\frac{7 a}{22} \times \frac{22}{92 a}, \\
=\frac{7}{92} ;
\end{array}
$$

Answer: The area of $\triangle A D G$ is $\frac{7}{92}$.

Answer: \frac{7}{92}


In rectangle $A B C D$, $A B=8, B C=4$. If rectangle $A B C D$ is folded along $A C$, then the area of the overlapping part $\triangle A F C$ is ( ).
(A) 12
(B) 10
(C) 8
(D) 6

Solution:
5.B.

Since $\triangle C D^{\prime} A \cong \triangle A B C$, therefore,
$S_{\triangle A D F}=S_{\triangle C B F}$.
And $\triangle A D^{\prime} F \backsim \triangle C B F$, so,
$\triangle A D^{\prime} F \cong \triangle C B F$.
Let $B F=x$, then $\sqrt{x^{2}+4^{2}}=8-x$.
Solving for $x$ gives $x=3$. Hence $S_{\triangle A F C}=10$.

Answer: B


Solution:
Let $x=123456$.

Thus, $x-1=123455$ and $x+1=123457$.

Therefore,

$$
123456^{2}-123455 \times 123457=x^{2}-(x-1)(x+1)=x^{2}-\left(x^{2}-1\right)=1
$$

ANSWER: 1

Answer: 1


In a row without spaces, all natural numbers are written in ascending order: $1234567891011...$. What digit stands at the 2017th place in the resulting long number?

Solution:
Answer: 7.

Solution: The first 9 digits are contained in single-digit numbers, the next 180 - in two-digit numbers. $2017-180-9=1828$. Next, $1828: 3=609 \frac{1}{3}$. This means that the 2017-th digit is the first

![](https://cdn.mathpix.com/cropped/2024_05_06_e4c8e28067279d2d2f8bg-1.jpg?height=51&width=1539&top_left_y=631&top_left_x=110)

Answer: 7


Point $D$ lies on side $A C$ of triangle $A B C$. The circle with diameter $B D$ intersects sides $A B$ and $B C$ at points $P$ and $T$ respectively. Points $M$ and $N$ are the midpoints of segments $A D$ and $C D$ respectively. It is known that $P M \| T N$.

a) Find the angle $A B C$.

b) Suppose additionally that $M P=\frac{1}{2}, N T=2, B D=\sqrt{3}$. Find the area of triangle $A B C$.

Solution:
Answer: (a) $90^{\circ};$ (b) $\frac{5 \sqrt{13}}{3 \sqrt{2}}$.

Solution. (a) Points $P$ and $T$ lie on the circle with diameter $BD$, so $\angle BPD = \angle BTD = 90^{\circ}$. Therefore, triangles $ADP$ and $DCT$ are right triangles; $PM$ and $TN$ are their medians. Since the median of a right triangle to the hypotenuse is half the hypotenuse, $TN = CN = DN$, $PM = AM = DM$. Let $\angle TCD = \gamma$. Since triangle $CTN$ is isosceles and $\angle CTN = \gamma$, $\angle TND = 2\gamma$ (as the external angle of $\triangle CTN$). Angles $\angle PMA$ and $\angle TND$ are equal due to the parallelism of lines $PM$ and $TN$. Since triangle $AMP$ is also isosceles, $\angle PAM = 90^{\circ} - \frac{1}{2} \angle PMA = 90^{\circ} - \frac{1}{2} \angle TND = 90^{\circ} - \gamma$. Therefore, the sum of angles $A$ and $C$ in triangle $ABC$ is $90^{\circ}$, and its third angle $\angle ABC$ is also $90^{\circ}$.

(b) As stated above, $CD = 2NT = 4$, $AD = 2MP = 1$. Let $\angle ADB = \psi$. Then $\angle BDC = 180^{\circ} - \psi$. By the cosine theorem for triangles $ABD$ and $ACD$, we get $AB^2 = 1 + 3 - 2\sqrt{3} \cos \psi$, $BC^2 = 16 + 3 - 8\sqrt{3} \cos (180^{\circ} - \psi)$. By the Pythagorean theorem, $AB^2 + BC^2 = AC^2 = 25$, from which it follows that $4 - 2\sqrt{3} \cos \psi + 19 + 8\sqrt{3} \cos \psi = 25$, $\cos \psi = \frac{1}{3\sqrt{3}}$. Next, we find: $\sin \psi = \sqrt{1 - \cos^2 \psi} = \frac{\sqrt{26}}{3\sqrt{3}}$, $S_{\triangle ABC} = S_{\triangle ABD} + S_{\triangle BCD} = \frac{1}{2} DA \cdot DB \sin \psi + \frac{1}{2} DC \cdot DB \sin (180^{\circ} - \psi) = \frac{1}{2} AC \cdot BD \sin \psi = \frac{1}{2} \cdot 5 \cdot \sqrt{3} \cdot \frac{\sqrt{26}}{3\sqrt{3}} = \frac{5\sqrt{13}}{3\sqrt{2}}$.

Answer: \frac{5\sqrt{13}}{3\sqrt{2}}


How many non-negative integer solutions $x$ does the equation

$$\left\lfloor \frac{x}{a}\right\rfloor = \left\lfloor \frac{x}{a+1}\right\rfloor$$

have?

$\lfloor ~ \rfloor$ ---> [url=http://en.wikipedia.org/wiki/Floor_function]Floor Function[/url].

Solution:
1. **Define the problem in terms of floor functions:**
   We need to find the number of non-negative integer solutions \( x \) such that:
   \[
   \left\lfloor \frac{x}{a} \right\rfloor = \left\lfloor \frac{x}{a+1} \right\rfloor
   \]
   where \( a \) is a positive integer.

2. **Express \( x \) in terms of \( q_1 \) and \( q_2 \):**
   For any non-negative integer \( x \), there exist non-negative integers \( q_1 \) and \( q_2 \) such that:
   \[
   q_1 a \leq x \leq (q_1 + 1)a - 1
   \]
   and
   \[
   q_2 (a+1) \leq x \leq (q_2 + 1)(a+1) - 1
   \]

3. **Set \( q_1 = q_2 \):**
   We need \( q_1 = q_2 \) for the floor functions to be equal. Therefore, we have:
   \[
   q_1 (a+1) \leq x \leq (q_1 + 1)a - 1
   \]

4. **Determine the range of \( x \):**
   The interval \( q_1 (a+1) \leq x \leq (q_1 + 1)a - 1 \) must be valid. This interval has:
   \[
   (q_1 + 1)a - 1 - q_1 (a+1) + 1 = (q_1 + 1)a - 1 - q_1 a - q_1 + 1 = a - q_1
   \]
   elements.

5. **Sum the number of solutions for each \( q_1 \):**
   We need to sum the number of solutions for each \( q_1 \) from \( 0 \) to \( a \):
   \[
   \sum_{q_1 = 0}^{a} (a - q_1)
   \]

6. **Calculate the sum:**
   The sum of the first \( a+1 \) terms of the sequence \( a, a-1, a-2, \ldots, 0 \) is:
   \[
   \sum_{q_1 = 0}^{a} (a - q_1) = a + (a-1) + (a-2) + \cdots + 1 + 0
   \]
   This is an arithmetic series with the first term \( a \) and the last term \( 0 \), and there are \( a+1 \) terms. The sum of an arithmetic series is given by:
   \[
   \frac{n}{2} (a + l)
   \]
   where \( n \) is the number of terms, \( a \) is the first term, and \( l \) is the last term. Here, \( n = a+1 \), \( a = a \), and \( l = 0 \):
   \[
   \sum_{q_1 = 0}^{a} (a - q_1) = \frac{(a+1)}{2} (a + 0) = \frac{a(a+1)}{2}
   \]

The final answer is \(\boxed{\frac{a(a+1)}{2}}\)

Answer: \frac{a(a+1)}{2}


Let $f(x)=x^{2}-2 a x-a^{2}-\frac{3}{4}$. If for any $x \in[0,1]$, we have $|f(x)| \leqslant 1$, then the range of real number $a$ is $\qquad$ .

Solution: 10. $-\frac{1}{2} \leqslant a \leqslant \frac{\sqrt{2}}{4}$

Answer: -\frac{1}{2} \leqslant a \leqslant \frac{\sqrt{2}}{4}


Determine the set of all positive real numbers \(r\) for which the following statement is true:

For every positive real number \(a\), the function \(f\) defined for all real \(x\) by \(f(x)=4-x^{2}-a x^{3}\) has a root between the numbers 2 - ar and 2.

Solution:
}
Given \(f(0)=4>0\) and since \(a>0, f(2)=-8 a<0\), we need to consider \(f(x)=-2 x-3 a x^{2}<0\).

In the interval \(2-a r<2\), it holds. Now,

\[
\begin{aligned}
f(2-a r) & =4-(2-a r)^{2}-a(2-a r)^{3} \\
& =a\left[a^{3} r^{3}+a r(12-r)-6 a^{2} r^{2}-4(2-r)\right]
\end{aligned}
\]
is greater than zero for all positive \(a\) if and only if for all positive \(a\),

\[
g(a)=a^{3} r^{3}+a r(12-r)-6 a^{2} r^{2}-4(2-r)>0
\]

holds. If \(r>0\) is a real number that satisfies the conditions of the problem, then:

\[
\lim _{a \rightarrow 0} g(a)=4 r-8 \geq 0
\]

This means \(r \geq 2\).

Conversely, any \(r \geq 2\) has the required property. For \(r=2\), we have:

\[
\begin{array}{r}
\left(a-\frac{3}{2}\right)^{2}+\frac{1}{4}>0 \\
a^{2}-3 a+\frac{5}{2}>0 \quad \text { so } \\
8 a^{3}+20 a-24 a^{2}=f(2-2 a)>0
\end{array}
\]

for all positive real \(a\).

If \(r>2\), then due to the monotonic decrease of \(f\) and the inequality \(f(2-2 a)>0\) just proven, it follows from \(0<2-a r<2\) that \(f(2-a r)>0\). Therefore, exactly all \(r \geq 2\) have the required property.

Adapted from [3]

Answer: r\geq2


In a meeting between four UN countries, let's say $A, B, C$, and $D$, country $A$ has twice as many representatives as $B$, three times as many as $C$, and four times as many as $D$. It is intended to distribute the representatives at tables with the same number of people at each one. There is only one condition: in each table, any of the countries must be in a numerical minority compared to the other three combined. How many representatives should there be at each table, at a minimum?

Solution:
Solution to Problem 3.6 The answer is 25. Let's see the proof. Let $a, b, c$ and $d$ be the number of representatives from each country. Since $a$ must be a multiple of 3 and 4, it must also be a multiple of 12. Therefore, there exists a number $k$ such that $a=12k$, then $b=6k, c=4k$ and $d=3k$. The total number of representatives is then $25k$.

If we call $M$ the number of tables, and $P$ the number of people at each table, we have $MP=25k$.

Let $a_{i}$ be the number of representatives from country $A$ at table number $i$. The condition imposed tells us that $a_{i}<\frac{P}{2}$, or $2a_{i}<P$. Since $a_{i}$ is an integer, this implies that $2a_{i} \leq P-1$. Adding all the $a_{i}$, we get:

$$
24k=2a=2a_{1}+\cdots+2a_{M} \leq M(P-1)=25k-M
$$

From this, we deduce $M \leq k$. Therefore, since $MP=25k$, we finally deduce that $P \geq 25$.

It only remains to show that, indeed, a configuration can be achieved where there are 25 people at each table. This can be achieved, for example, with a single table where there are 12 representatives from country $A$, 6 from $B$, 4 from $C$, and 3 from $D$.

Answer: 25


In triangle $ABC$, points $M$ and $N$ are chosen on sides $AB$ and $BC$ respectively, such that $AM=2MB$ and $BN=NC$. Segments $AN$ and $CM$ intersect at point $P$. Find the area of quadrilateral $MBNP$, given that the area of triangle $ABC$ is 30.

Solution:
# 6. Problem 6

In triangle $A B C$, points $M$ and $N$ are chosen on sides $A B$ and $B C$ respectively such that $A M=2 M B$ and $B N=N C$. Segments $A N$ and $C M$ intersect at point $P$. Find the area of quadrilateral $M B N P$, given that the area of triangle $A B C$ is 30.

## Answer: 7

#

Answer: 7


Determine all solutions of the diophantine equation $a^2 = b \cdot (b + 7)$ in integers $a\ge 0$ and $b \ge 0$.

Solution:
1. We start with the given diophantine equation:
   \[
   a^2 = b \cdot (b + 7)
   \]
   We can rewrite this equation as a quadratic in \( b \):
   \[
   b^2 + 7b - a^2 = 0
   \]

2. To solve for \( b \), we use the quadratic formula:
   \[
   b = \frac{-7 \pm \sqrt{49 + 4a^2}}{2}
   \]
   For \( b \) to be an integer, the discriminant \( 49 + 4a^2 \) must be a perfect square. Let \( m \) be an integer such that:
   \[
   49 + 4a^2 = m^2
   \]

3. Rearrange the equation to:
   \[
   m^2 - 4a^2 = 49
   \]
   This is a difference of squares, which can be factored as:
   \[
   (m - 2a)(m + 2a) = 49
   \]

4. We now consider the factor pairs of 49:
   \[
   (1, 49), (-1, -49), (7, 7), (-7, -7)
   \]
   Since \( a \geq 0 \) and \( b \geq 0 \), we only consider the positive factor pairs:
   \[
   (1, 49) \quad \text{and} \quad (7, 7)
   \]

5. For the factor pair \( (1, 49) \):
   \[
   m - 2a = 1 \quad \text{and} \quad m + 2a = 49
   \]
   Adding these equations:
   \[
   2m = 50 \implies m = 25
   \]
   Subtracting these equations:
   \[
   4a = 48 \implies a = 12
   \]

6. For the factor pair \( (7, 7) \):
   \[
   m - 2a = 7 \quad \text{and} \quad m + 2a = 7
   \]
   Adding these equations:
   \[
   2m = 14 \implies m = 7
   \]
   Subtracting these equations:
   \[
   4a = 0 \implies a = 0
   \]

7. We now check the corresponding values of \( b \):
   - For \( a = 12 \):
     \[
     b = \frac{-7 + \sqrt{49 + 4 \cdot 12^2}}{2} = \frac{-7 + \sqrt{625}}{2} = \frac{-7 + 25}{2} = 9
     \]
   - For \( a = 0 \):
     \[
     b = \frac{-7 + \sqrt{49}}{2} = \frac{-7 + 7}{2} = 0
     \]

8. Therefore, the solutions in non-negative integers are:
   \[
   (a, b) = (12, 9) \quad \text{and} \quad (0, 0)
   \]

The final answer is \(\boxed{(12, 9) \text{ and } (0, 0)}\)

Answer: (12, 9) \text{ and } (0, 0)


Condition A: $\sqrt{1+\sin \theta}=a$. Condition B: $\sin \frac{\theta}{2}+\cos \frac{\theta}{2}=a$.
(A) A is a sufficient and necessary condition for B
(B) A is a necessary condition for B
(C) A is a sufficient condition for B
(D) A is neither a necessary condition nor a sufficient condition for B

Solution:
(D)
2.【Analysis and Solution】In condition A, it is easy to know that $a \geqslant 0$, while in condition B, $a$ may take negative values. Therefore, A does not imply B, and B does not imply A, that is, A is neither a necessary condition nor a sufficient condition for B.

Answer: D


In Pascal's Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below.
\[\begin{array}{c@{\hspace{8em}} c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{4pt}}c@{\hspace{2pt}} c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{3pt}}c@{\hspace{6pt}} c@{\hspace{6pt}}c@{\hspace{6pt}}c} \vspace{4pt} \text{Row 0: } &    &    &     &     &    &    & 1 &     &     &    &    &    &  \\\vspace{4pt} \text{Row 1: } &    &    &     &     &    & 1 &    & 1  &     &    &    &    &  \\\vspace{4pt} \text{Row 2: } &    &    &     &     & 1 &    & 2 &     & 1  &    &    &    &  \\\vspace{4pt} \text{Row 3: } &    &    &     &  1 &    & 3 &    & 3  &     & 1 &    &    &  \\\vspace{4pt} \text{Row 4: } &    &    & 1  &     & 4 &    & 6 &     & 4  &    & 1 &    &  \\\vspace{4pt} \text{Row 5: } &    & 1 &     & 5  &    &10&    &10 &     & 5 &    & 1 &  \\\vspace{4pt} \text{Row 6: } & 1 &    & 6  &     &15&    &20&     &15 &    & 6 &    & 1 \end{array}\]
In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio $3 : 4 : 5$?

Solution:
Call the row $x=t+k$, and the position of the terms $t-1, t, t+1$. Call the middle term in the ratio $N = \dbinom{t+k}{t} = \frac{(t+k)!}{k!t!}$. The first term is $N \frac{t}{k+1}$, and the final term is $N \frac{k}{t+1}$. Because we have the ratio $3:4:5$,
$\frac{t}{k+1} = \frac{3}{4}$ and $\frac{k}{t+1} = \frac{5}{4}$.
$4t = 3k+3$ and $4k= 5t+5$
$4t-3k=3$
$5t-4k=-5$
Solve the equations to get $t= 27, k=35$ and $x = t+k = \boxed{062}$.
-Solution and LaTeX by jackshi2006, variables and algebra simplified by oinava

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Answer: 62


There are 10 identical urns, 9 of which contain 2 black and 2 white balls, and one contains 5 white and 1 black ball. A ball is drawn from a randomly selected urn. The drawn ball turned out to be white. What is the probability that this ball was drawn from the urn containing 5 white balls?

Solution:
S o l u t i o n. There are 2 groups of urns with different compositions of balls; 9 of the existing urns belong to the first group, one urn - to the second group. The experiment consists of drawing a ball from a randomly selected urn. Let's consider the hypotheses:

$B_{1}$ - an urn from the first group is selected;

$B_{2}$ - an urn from the second group is selected;

and the event $A$ - drawing a white ball.

The probabilities of the hypotheses $B_{1}$ and $B_{2}$ are:

$P\left(B_{1}\right)=9 / 10=0.9, P\left(B_{2}\right)=1 / 10=0.1$. Hypotheses $B_{1}$ and $B_{2}$ form a complete group of mutually exclusive events, the sum of the probabilities of the hypotheses is $P\left(B_{1}\right)+P\left(B_{2}\right)=0.9+0.1=1$.

Event $A$ can occur only either together with event $B_{1}$ or with event $B_{2}$. Before the experiment, the conditional probability of event $A$, calculated on the condition that it occurred together with event $B_{1}$, is $P_{B_{1}}(A)=2 / 4=1 / 2$ (since in each urn of the first group, among the four balls, there are 2 white balls). Before the experiment, the conditional probability of event $A$, calculated on the condition that it occurred together with event $B_{2}$, is $P_{B_{2}}(A)=5 / 6$ (since in the urn of the second group, among the six balls, there are 5 white ones).

Suppose event $A$ occurred: a white ball was drawn from some urn. Let's reassess the probability of the second hypothesis. Using formula (17), we get

$$
P_{A}\left(B_{2}\right)=\frac{P\left(B_{2}\right) P_{B_{2}}(A)}{P\left(B_{1}\right) P_{B_{1}}(A)+P\left(B_{2}\right) P_{B_{2}}(A)}=\frac{0.1 \cdot \frac{5}{6}}{0.1 \cdot \frac{5}{6}+0.9 \cdot \frac{1}{2}}=0.15625
$$

Answer: 0.15625


Suppose that $n$ is a positive integer. Determine all the possible values of the first digit after the decimal point in the decimal expression of the number $\sqrt{n^3+2n^2+n}$

Solution:
To determine the first digit after the decimal point in the decimal expression of the number \(\sqrt{n^3 + 2n^2 + n}\), we can follow these steps:

1. **Simplify the expression inside the square root:**
   \[
   \sqrt{n^3 + 2n^2 + n} = \sqrt{n(n^2 + 2n + 1)} = \sqrt{n(n+1)^2} = (n+1)\sqrt{n}
   \]

2. **Express the first digit after the decimal point:**
   The first digit after the decimal point of \((n+1)\sqrt{n}\) can be found by considering \(10(n+1)\sqrt{n}\) and taking the floor function:
   \[
   N = \lfloor 10(n+1)\sqrt{n} \rfloor
   \]
   Let \(N = 10k + r\), where \(k\) is an integer and \(0 \leq r < 10\).

3. **Set up the inequality:**
   \[
   10k + r \leq 10(n+1)\sqrt{n} < 10k + r + 1
   \]
   Dividing by 10:
   \[
   k + \frac{r}{10} \leq (n+1)\sqrt{n} < k + \frac{r+1}{10}
   \]

4. **Square the inequality:**
   \[
   \left(k + \frac{r}{10}\right)^2 \leq n(n+1)^2 < \left(k + \frac{r+1}{10}\right)^2
   \]
   Expanding both sides:
   \[
   k^2 + \frac{2kr}{10} + \frac{r^2}{100} \leq n(n+1)^2 < k^2 + \frac{2k(r+1)}{10} + \frac{(r+1)^2}{100}
   \]
   Simplifying the right-hand side:
   \[
   k^2 + \frac{2kr}{10} + \frac{r^2}{100} \leq n(n+1)^2 < k^2 + \frac{2kr}{10} + \frac{2k}{10} + \frac{r^2}{100} + \frac{2r}{100} + \frac{1}{100}
   \]
   \[
   k^2 + \frac{2kr}{10} + \frac{r^2}{100} \leq n(n+1)^2 < k^2 + \frac{2kr}{10} + \frac{2k}{10} + \frac{r^2}{100} + \frac{2r}{100} + \frac{1}{100}
   \]

5. **Analyze the inequality:**
   From the inequality, we can see that the value of \(r\) (the first digit after the decimal point) depends on the value of \(n\) and \(k\). By trial and error, we can find that all values of \(r\) from 0 to 9 are achievable by suitable choices of \(n\) and \(k\).

Conclusion:
The first digit after the decimal point in the decimal expression of \(\sqrt{n^3 + 2n^2 + n}\) can be any digit from 0 to 9.

The final answer is \(\boxed{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}\).

Answer: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9


A rhombus $\mathcal R$ has short diagonal of length $1$ and long diagonal of length $2023$. Let $\mathcal R'$ be the rotation of $\mathcal R$ by $90^\circ$ about its center. If $\mathcal U$ is the set of all points contained in either $\mathcal R$ or $\mathcal R'$ (or both; this is known as the union of $\mathcal R$ and $\mathcal R'$) and $\mathcal I$ is the set of all points contained in both $\mathcal R$ and $\mathcal R'$ (this is known as the intersection of $\mathcal R$ and $\mathcal R'$, compute the ratio of the area of $\mathcal I$ to the area of $\mathcal U$.

Solution:
1. **Substitution and Scaling**:
   - Let \( n = 2023 \) for simplicity.
   - Scale everything up by 2, noting that this will not affect the ratio of \(\mathcal{I}\) to \(\mathcal{U}\).

2. **Coordinate Setup**:
   - Define the vertices of \(\mathcal{R}\) as \((-1, 0)\), \((0, n)\), \((1, 0)\), and \((0, -n)\).
   - Define the vertices of \(\mathcal{R}'\) as \((0, 1)\), \((n, 0)\), \((0, -1)\), and \((0, -n)\).

3. **Intersection in the First Quadrant**:
   - By symmetry, the intersection point of the two sides of \(\mathcal{R}\) and \(\mathcal{R}'\) in the first quadrant will lie on the line \( y = x \).
   - The equation of one of the sides of \(\mathcal{R}\) is \( y = -\frac{1}{n}(x + 1) \).
   - Solving for the intersection point:
     \[
     y = x = -\frac{1}{n}(x + 1)
     \]
     \[
     nx = -x - 1
     \]
     \[
     (n + 1)x = -1
     \]
     \[
     x = -\frac{1}{n + 1}
     \]
     \[
     y = -\frac{1}{n + 1}
     \]
     Thus, the intersection point is \(\left(\frac{n}{n+1}, \frac{n}{n+1}\right)\).

4. **Area Calculation Using Shoelace Formula**:
   - The vertices of the intersection quadrilateral in the first quadrant are \((0, 0)\), \((0, 1)\), \(\left(\frac{n}{n+1}, \frac{n}{n+1}\right)\), and \((1, 0)\).
   - Using the Shoelace Formula to find the area:
     \[
     \text{Area} = \frac{1}{2} \left| 0 \cdot 1 + 0 \cdot \frac{n}{n+1} + \frac{n}{n+1} \cdot 0 + 1 \cdot 0 - (0 \cdot 0 + 1 \cdot \frac{n}{n+1} + \frac{n}{n+1} \cdot 1 + 0 \cdot 0) \right|
     \]
     \[
     = \frac{1}{2} \left| 0 + 0 + 0 + 0 - (0 + \frac{n}{n+1} + \frac{n}{n+1} + 0) \right|
     \]
     \[
     = \frac{1}{2} \left| -\frac{2n}{n+1} \right|
     \]
     \[
     = \frac{n}{n+1}
     \]
   - The total area of \(\mathcal{I}\) is \( 4 \cdot \frac{n}{n+1} \).

5. **Ratio Calculation**:
   - The area of \(\mathcal{U}\) is the area of \(\mathcal{R}\) plus the area of \(\mathcal{R}'\) minus the area of \(\mathcal{I}\):
     \[
     [\mathcal{U}] = 2 \cdot \left(\frac{1}{2} \cdot 2n \cdot 2\right) - 4 \cdot \frac{n}{n+1}
     \]
     \[
     = 2 \cdot 2n - 4 \cdot \frac{n}{n+1}
     \]
     \[
     = 4n - \frac{4n}{n+1}
     \]
   - The desired ratio is:
     \[
     \frac{[\mathcal{I}]}{[\mathcal{U}]} = \frac{4 \cdot \frac{n}{n+1}}{4n - \frac{4n}{n+1}}
     \]
     \[
     = \frac{\frac{4n}{n+1}}{4n - \frac{4n}{n+1}}
     \]
     \[
     = \frac{\frac{4n}{n+1}}{4n \left(1 - \frac{1}{n+1}\right)}
     \]
     \[
     = \frac{\frac{4n}{n+1}}{4n \cdot \frac{n}{n+1}}
     \]
     \[
     = \frac{4n}{4n \cdot \frac{n}{n+1}}
     \]
     \[
     = \frac{1}{n}
     \]

The final answer is \(\boxed{\frac{1}{2023}}\)

Answer: \frac{1}{2023}


In a class, each student has either 5 or 7 friends (friendship is mutual), and any two friends have a different number of friends in the class. What is the smallest number of students, greater than 0, that can be in the class?

Answer: 12

Solution:
Solution:

Let's look at some person. Suppose he has five friends. Then each of these five people has seven friends. Similarly, there are at least seven more people with five friends each. In total, there are 12 people.

It is quite easy to construct an example: two groups of 5 and 7 people, people from different groups are friends with each other, but not within the same group.

#

Answer: 12


Does there exist a six-digit natural number that, when multiplied by 9, is written with the same digits but in reverse order?

Solution: Answer: it exists. Solution. Let $\overline{a b c d e f}$ be the desired number, i.e., $\overline{a b c d e f} \cdot 9=\overline{f e d c b a}$. Then it is obvious that $a=1, b=0$ (otherwise, multiplying by 9 would result in a seven-digit number). Therefore, $f=9$, and the second-to-last digit $e=8$ (which follows from column multiplication). Then the third digit $c$ can be 8 or 9. But if $c=8$, then $d=1$, since the sum of the digits is divisible by 9, but the number 108189 does not fit upon verification. If, however, $c=9$, then $d=9$ and the number 109989 is the only one that satisfies the condition, and it fits upon verification.

Answer: 109989


Solution:
2 -

![](https://cdn.mathpix.com/cropped/2024_05_01_ed653f4a7a838e105fd2g-03.jpg?height=279&width=577&top_left_y=1848&top_left_x=180)

From the statement, we have: \( AC = 50 \), \( BD = 45 \), and \( AD = 80 \). From the figure, it follows that \( BC = AC - AB \), so \( BC = 50 - AB \).

Therefore, it is enough to calculate \( AB \). Note from the figure that \( AB = AD - BD \), and thus, \( AB = 80 - 45 = 35 \).

Finally, \( BC = 50 - 35 = 15 \text{ km} \).

Solution 3 - From the figure, we have \( 45 - BC = 80 - 50 \). Therefore, \( BC = 15 \text{ km} \).

Answer: 15


Solution: 13. Let $x=\cos ^{2} 10^{\circ}+\cos ^{2} 50^{\circ}-\sin 40^{\circ} \sin 80^{\circ}, y=\sin ^{2} 10^{\circ}+\sin ^{2} 50^{\circ}+\cos 40^{\circ} \cos 80^{\circ}$, then $x+y=2-\cos 120^{\circ}=\frac{3}{2}, x-y=\cos 20^{\circ}+\cos 100^{\circ}-\cos 40^{\circ}=2 \cos 60^{\circ} \cos 40^{\circ}-\cos 40^{\circ}=$ 0. Therefore, $x=y=\frac{3}{4}$.

Answer: \frac{3}{4}


Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n+1}=a_{n}+\frac{1}{2 a_{n}}$, then $\lim _{n \rightarrow \infty}\left(a_{n}-\sqrt{n}\right)=$

Solution:
6. 0 .

From $a_{n+1}=a_{n}+\frac{1}{2 a_{n}}$, we get
$$
a_{n+1}^{2}=a_{n}^{2}+\frac{1}{4 a_{n}^{2}}+1 .
$$

By mathematical induction, we can prove $n \leqslant a_{n}^{2}<n+\sqrt[3]{n}$, thus
$$
0 \leqslant a_{n}-\sqrt{n}<\sqrt{n+\sqrt[3]{n}}-\sqrt{n}=\frac{\sqrt[3]{n}}{\sqrt{n+\sqrt[3]{n}}+\sqrt{n}}<\frac{\sqrt[3]{n}}{\sqrt{n}}=\frac{1}{\sqrt[6]{n}},
$$

and since $\lim _{n \rightarrow \infty} \frac{1}{\sqrt[6]{n}}=0$, it follows that $\lim _{n \rightarrow \infty}\left(a_{n}-\sqrt{n}\right)=0$.

Answer: 0


Points $O$ and $I$ are the centers of the circumcircle and incircle of triangle $ABC$, and $M$ is the midpoint of the arc $AC$ of the circumcircle (not containing $B$). It is known that $AB=15, BC=7$, and $MI=MO$. Find $AC$.

Solution:
Answer: $A C=13$.

Solution. (Fig. 3). First, we prove that $M I=M A$ (trident lemma).

Indeed, the external angle $A I M$ of triangle $A I B$ is equal to the sum of angles $B A I$ and $A B I$, and since $A I$ and $B I$ are angle bisectors, $\angle A I M=\frac{1}{2} \angle A+\frac{1}{2} \angle B$. Angle $I A M$ is equal to the sum of angles $I A C$ and $C A M$. But $\angle I A C=\frac{1}{2} \angle A$, and $\angle C A M=\angle C B M=\frac{1}{2} \angle B$ as inscribed angles.

From this, it follows that $\angle I A M=\frac{1}{2} \angle A+\frac{1}{2} \angle B=\angle A I M$, and therefore, triangle $A M I$ is isosceles, $M I=M A$. By the condition $M O=$ $M I$, so by the trident lemma $A O=M O=M I=M A$. This means that triangle $A O M$ is equilateral and $\angle A O M=60^{\circ}$. Since the central angle $A O M$ is twice the inscribed angle $A B M$, we have $\frac{1}{2} \angle B=30^{\circ}$, that is, $\angle B=60^{\circ}$.

By the cosine theorem, $A C^{2}=A B^{2}+B C^{2}-2 \cdot A B \cdot B C \cos 60^{\circ}=$ $13^{2}$.

Criteria. Answer without explanation - 0 points. Proved that angle $B$ is $60^{\circ}-5$ points. Points are not deducted for the absence of proof of the trident lemma. Full solution - 7 points.

![](https://cdn.mathpix.com/cropped/2024_05_06_7789e0097aa3cdbc6e57g-2.jpg?height=540&width=467&top_left_y=1779&top_left_x=1485)

Fig. 3

Answer: 13


Determine all integers $x$ for which the number $9 x^{2}-40 x+39$ is a power of a prime. (A natural number $m$ is a power of a prime if $m=p^{a}$ for some prime $p$ and non-negative integer $a$.)

Solution:
III/1. Let $9 x^{2}-40 x+39=p^{n}$ for some prime $p$ and non-negative integer $n$. From

$$
p^{n}=9 x^{2}-40 x+39=(9 x-13)(x-3)
$$

it follows that $9 x-13=p^{k}$ and $x-3=p^{l}$ or $9 x-13=-p^{k}$ and $x-3=-p^{l}$ for some numbers $k$ and $l$, where $0 \leq l<k$ and $n=k+l$.

First, let's solve the system of equations $9 x-13=p^{k}$ and $x-3=p^{l}$. We have $9\left(p^{l}+3\right)-13=p^{k}$, or $14=p^{k}-9 p^{l}=p^{l}\left(p^{k-l}-9\right)$. If $l=0$, we get $p^{k}=23$, so $p=23, k=1$ and $x=4$. Otherwise, $l \geq 1$ and $p^{l} \mid 14$, so $p^{l}=2$ and $p^{k-l}-9=7$ or $p^{l}=7$ and $p^{k-l}-9=2$. In the first case, we get $p=2, l=1, k=5$ and $x=5$, and in the second case, there is no solution.

Next, we need to consider the system of equations $9 x-13=-p^{k}$ and $x-3=-p^{l}$. Then $14=$ $p^{l}\left(9-p^{k-l}\right)$. The only possibilities are $p=2, l=1$ or $p=7, l=1$ (in this case $l=0$ is not possible). The corresponding numbers are $x=1$ and $x=-4$.

The numbers that satisfy the problem are $x=-4, x=1, x=4$ and $x=5$.

![](https://cdn.mathpix.com/cropped/2024_06_07_c45646d9586b6df064a6g-11.jpg?height=49&width=1627&top_left_y=1363&top_left_x=220)

Writing down the third solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 point

Writing down the fourth solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_c45646d9586b6df064a6g-11.jpg?height=57&width=1633&top_left_y=1511&top_left_x=220)

Conclusion $9 x-13=p^{k}$ and $x-3=p^{l}$ or $9 x-13=-p^{k}$ and $x-3=-p^{l} \ldots \ldots \ldots \ldots$.

![](https://cdn.mathpix.com/cropped/2024_06_07_c45646d9586b6df064a6g-11.jpg?height=52&width=1627&top_left_y=1613&top_left_x=223)

Complete handling of the case $9 x-13=-p^{k}$ and $x-3=-p^{l} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ point

If a contestant's approach does not allow obtaining all solutions to the problem, the contestant can earn 1 point if their method can lead to three solutions of the problem.

Answer: -4,1,4,5


Natural numbers $a$ and $b$ are such that $2a + 3b = \operatorname{LCM}(a, b)$. What values can the number $\frac{\operatorname{LCM}(a, b)}{a}$ take? List all possible options in ascending or descending order, separated by commas. If there are no solutions, write the number 0.

Solution:
Answer: 0 (no solutions)

## Solution:

Let $d=\operatorname{GCD}(a, b)$. Then $a=x d, b=y d, \operatorname{LCM}(a, b)=x y d$, where $x$ and $y$ are coprime. We get $2 x d+3 y d=x y d$. Dividing by $d$ yields $2 x+3 y=x y$.

The right side of this equation is divisible by $x$, so $2 x+3 y$ is divisible by $x$, which means $3 y$ is divisible by $x$. Since $x$ and $y$ are coprime, 3 must be divisible by $x$. Therefore, $x=3$ or $x=1$. Similarly, 2 must be divisible by $y$.

If $x=3$, the equation becomes $6+3 y=3 y$, which is false. If $x=1$, we get $2+3 y=y$, from which $y$ is negative, which is also impossible.

Therefore, there are no solutions.

Answer: 0


If positive integers $x, y$ satisfy $x^{3}+3 x^{2} y+8 x y^{2}+6 y^{3}=87$, then $x+2 y=$

Solution:
Answer: 5.
Solution: Since when $y \geqslant 3$, $6 y^{3}>87$,
at this time
$$
x^{3}+3 x^{2} y+8 x y^{2}+6 y^{3}>87 \text {, }
$$

thus
$y$ can only be 1 or 2.
(1) When $y=1$, substituting gives
$$
\begin{array}{l}
x^{3}+3 x^{2}+8 x+6=87, \\
x^{2}(x+3)+8 x=81,
\end{array}
$$

rearranging, we get

since
$x^{2}(x+3), 8 x$ are both even numbers,

thus
$$
x^{2}(x+3)+8 x \neq 81 \text {. }
$$
(2) When $y=2$, substituting gives

that is
$$
\begin{array}{l}
x^{3}+6 x^{2}+32 x+48=87, \\
x^{3}+6 x^{2}+32 x=39 .
\end{array}
$$

since
$x$ may take the value of 1.
When $x=1$, the equation holds,

thus
$$
x=1, \quad y=2, \quad x+2 y=5 \text {. }
$$

Answer: 5


In a regular triangular pyramid $SABC$ ($S$ - vertex) points $D$ and $E$ are the midpoints of edges $AC$ and $BC$ respectively. Through point $E$, a plane $\beta$ is drawn, intersecting edges $AB$ and $SB$ and being equidistant from points $D$ and $B$ by a distance of $\frac{1}{2}$. Find the lengths of the segments into which the plane divides edge $SB$, if $BC=4$ and $SC=3$.

Solution:
Since the plane $\beta$ intersects the edge $A B$, points $A$ and $B$ are located on opposite sides of it. Similarly for points $B$ and $C$. Therefore, points $A$ and $C$ (and thus all points on the edge $A C$) lie on one side of $\beta$, while point $B$ lies on the other. Thus, the segment $B D$ intersects the plane $\beta$ at some point $M$ (Fig.1). Let $B 1$ and $D 1$ be the orthogonal projections of points $B$ and $D$ onto the plane $\beta$ (Fig.2), and let the plane $\beta$ intersect the edge

$A B$ at point $N$. Then point $M$ belongs to the projection $B 1 D 1$ of the segment $B D$ onto the plane $\beta$. From the equality of right triangles $B M B 1$ and $D M D 1$ (by the leg and the adjacent acute angle), it follows that $M$ is the midpoint of $B D$, and thus the line $E M$ intersects the edge $A B$ at point $N$ and $E M \| A C$, so $N$ is the midpoint of the edge $A B$. Let the plane $\beta$ intersect the edge $S B$ at point $K$. Then $K M$ is the median and altitude of the isosceles triangle $E K N$, and since $N B 1 = E B 1$ (as projections of equal obliques $B N$ and $B E$ onto the plane $\beta$), point $B 1$ lies on $K M$. Let $S H$

- be the height of the pyramid. Denote $\angle S B H = \alpha, \angle K M B = \gamma$. From right triangles $S B H$ and $B B 1 M$, we find that

$$
\begin{gathered}
\cos \alpha = \frac{B H}{B S} = \frac{\frac{3}{B D}}{B S} = \frac{4 \sqrt{3}}{9}, \sin \gamma = \frac{B B_{1}}{B M} = \frac{B B_{1}}{\frac{1}{2} B D} = \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}} = \frac{1}{2 \sqrt{3}} \\
\sin \alpha = \sqrt{1 - \left(\frac{4 \sqrt{3}}{9}\right)^{2}} = \sqrt{1 - \frac{43}{B 1}} = \frac{\sqrt{33}}{9}, \cos \gamma = \sqrt{1 - \left(\frac{1}{2 \sqrt{3}}\right)^{2}} = \sqrt{1 - \frac{1}{12}} = \frac{\sqrt{11}}{2 \sqrt{3}} \\
\sin (\alpha + \gamma) = \sin \alpha \cos \gamma + \cos \alpha \sin \gamma = \frac{\sqrt{33}}{9} \cdot \frac{\sqrt{11}}{2 \sqrt{3}} + \frac{4 \sqrt{3}}{9} \cdot \frac{1}{2 \sqrt{3}} = \frac{5}{6}
\end{gathered}
$$

From triangle $B K M$ by the sine theorem, we find that

$$
B K = \frac{B M \sin \gamma}{\sin (\alpha + \gamma)} = \frac{\sqrt{3} \cdot \frac{1}{2 \sqrt{3}}}{\frac{1}{2}} = \frac{3}{5}.
$$

Therefore,

$$
S K = S B - B K = 3 - \frac{3}{5} = \frac{12}{5}
$$

## Answer

$\frac{12}{5}, \frac{3}{5}$.

Answer: \frac{12}{5},\frac{3}{5}


Given $2014=(a^{2}+b^{2}) \times(c^{3}-d^{3})$, where $\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}$ are four positive integers, please write a multiplication equation that satisfies the condition: $\qquad$

Solution:
【Answer】The answer is not unique
Analysis: $2014=1 \times 2014=2 \times 1007=19 \times 106=38 \times 53$
One of the solutions is $2014=\left(5^{2}+9^{2}\right) \times\left(3^{3}-2^{3}\right)$

Answer: 2014=(5^{2}+9^{2})\times(3^{3}-2^{3})


Ten people exchanged photographs. How many photographs were needed?

Solution: 21. The choice of one person can be made in ten ways, after which this person can give a photograph to nine remaining people. In total, $10 \cdot 9=90$ photographs are required.

Answer: 90


Petra and Holger stick 9 pennants together. Ines brings 4 pennants.

How many pennants can they attach to the pennant chain?

Solution:
$9+4=13$; You can attach 13 flags to the chain.

### 10.11.2 2nd Round 1975, Class 1

Answer: 13


Find the derivative.

$$
y=x(\arcsin x)^{2}+2 \sqrt{1-x^{2}} \arcsin x-2 x
$$

Solution:
## Solution

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

$=(\arcsin x)^{2}+x \cdot 2 \cdot \arcsin x \cdot \frac{1}{\sqrt{1-x^{2}}}+\frac{2}{2 \sqrt{1-x^{2}}} \cdot(-2 x) \cdot \arcsin x+2 \sqrt{1-x^{2}} \cdot \frac{1}{\sqrt{1-x^{2}}}-2=$ $=(\arcsin x)^{2}+\frac{2 x \cdot \arcsin x}{\sqrt{1-x^{2}}}-\frac{2 x \cdot \arcsin x}{\sqrt{1-x^{2}}}+2-2=(\arcsin x)^{2}$

## Problem Kuznetsov Differentiation 14-24

Answer: (\arcsinx)^{2}


For arbitrary $n \geqslant 3$ points $A_{1}$, $A_{2}, \ldots, A_{n}$ on a plane, no three of which lie on the same line, let $\alpha$ denote the smallest of the angles $\angle A_{i} A_{j} A_{k}$ formed by triples $A_{i}, A_{j}, A_{k}$ of distinct points. For each value of $n$, find the greatest value of $\alpha$. Determine for which arrangements of points this value is achieved.

Solution:
14.9. We will prove that the greatest value of $\alpha$ is $180^{\circ} / n$. Indeed, let some arrangement of points on a plane correspond to the value $\alpha$. Consider a line, say $A_{1}^{\prime} A_{2}^{\prime}$, such that all points are located in one half-plane relative to this line, and choose a point $A_{3}^{\prime}$ for which the angle $\angle A_{1}^{\prime} A_{2}^{\prime} A_{3}^{\prime}$ is maximal (Fig. 98). Then all other points lie inside this angle and

$$
\angle A_{1}^{\prime} A_{2}^{\prime} A_{3}^{\prime} \geqslant \alpha(n-2)
$$

since each of the $n-2$ angles between adjacent rays $A_{2}^{\prime} A_{i}$ ( $A_{i} \neq A_{2}^{\prime}$ ) is not less than $\alpha$. Next, choose a point $A_{4}^{\prime}$ for which the angle $\angle A_{2}^{\prime} A_{3}^{\prime} A_{4}^{\prime}$ is maximal. Then all points

$$
A_{i} \in\left\{A_{2}^{\prime} ; A_{3}^{\prime} ; A_{3}^{\prime}\right\}
$$

lie inside this angle and

$$
\angle A_{2}^{\prime} A_{3}^{\prime} A_{1}^{\prime} \geqslant \alpha(n-2)
$$

If $A_{4}^{\prime} \neq A_{1}^{\prime}$ and $A_{4}^{\prime} \neq A_{2}^{\prime}$, then we

![](https://cdn.mathpix.com/cropped/2024_05_21_03a501298610b72904feg-250.jpg?height=513&width=511&top_left_y=249&top_left_x=545)

Fig. 98 take analogously the point $A_{5}^{\prime}$ and so on. Since the number of points is $n$, one of the points in the sequence $A_{1}^{\prime}, A_{2}^{\prime}, A_{3}^{\prime}, \ldots$ will necessarily repeat for the first time. Let this happen after choosing the point $A_{k}^{\prime}$, i.e., the angle $\angle A_{k-1}^{\prime} A_{k}^{\prime} A_{j}$ is maximal for some point

$$
A_{j}=A_{i}^{\prime} \in\left\{A_{1}^{\prime} ; \ldots ; A_{k-2}^{\prime}\right\}
$$

Then, if $i \neq 1$, the point $A_{1}^{\prime}$ lies inside the angle $\angle A_{k-1}^{\prime} A_{k}^{\prime} A_{i}^{\prime}$, and hence inside the convex polygon $A_{i}^{\prime} A_{i+1}^{\prime} \ldots A_{k-1}^{\prime} A_{k}^{\prime}$, which contradicts its choice. Therefore, $i=1$ and the convex $k$-gon $A_{1}^{\prime} A_{2}^{\prime} \ldots A_{k}^{\prime}$ has the sum of angles

$$
180^{\circ}(k-2) \geqslant k \cdot \alpha(n-2)
$$

from which

$$
\alpha \leqslant \frac{180^{\circ}(k-2)}{(n-2) k}=\frac{180^{\circ}}{n-2}\left(1-\frac{2}{k}\right) \leqslant \frac{180^{\circ}}{n-2}\left(1-\frac{2}{n}\right)=\frac{180^{\circ}}{n} .
$$

Equality $\alpha=180^{\circ} / n$ is possible only if $k=n$,

$$
\angle A_{1}^{\prime} A_{2}^{\prime} A_{3}^{\prime}=\angle A_{2}^{\prime} A_{3}^{\prime} A_{4}^{\prime}=\ldots=\angle A_{k}^{\prime} A_{1}^{\prime} A_{2}^{\prime}=\alpha(n-2)
$$

and all diagonals emanating from any angle of the $n$-gon $A_{1}^{\prime} \ldots A_{n}^{\prime}$ divide it into equal angles $\alpha$. Such an $n$-gon can only be regular, since for any value $l=1, \ldots, n$ we have

$$
\angle A_{l}^{\prime} A_{l+2}^{\prime} A_{l+1}^{\prime}=\angle A_{l+2}^{\prime} A_{l}^{\prime} A_{l+1}^{\prime} \text {, i.e., } \quad A_{l}^{\prime} A_{l+1}^{\prime}=A_{l+1}^{\prime} A_{l+2}^{\prime}
$$

(we assume $A_{n+1}^{\prime}=A_{1}^{\prime}, A_{n+2}^{\prime}=A_{2}^{\prime}$ ) (Fig. 98), hence, the $n$-gon
must have not only equal angles but also equal sides. Finally, the vertices of a regular $n$-gon indeed satisfy the condition $\alpha=180^{\circ} / n$. This becomes clear if we circumscribe a circle around the $n$-gon and note that all angles between adjacent diagonals emanating from any vertex subtend an arc of $360^{\circ} / n$, and thus are equal to $180^{\circ} / n$.

![](https://cdn.mathpix.com/cropped/2024_05_21_03a501298610b72904feg-251.jpg?height=453&width=423&top_left_y=342&top_left_x=139)

Fig. 99

Answer: \alpha=\frac{180}{n}


Solution:
[Solution] $1-x+x^{2}-x^{3}+\cdots+x^{16}-x^{17}$
$$\begin{array}{l}
=\frac{1-x^{18}}{1+x} \\
\quad=\frac{1-(y-1)^{18}}{y}
\end{array}$$

The coefficient of $y^{3}$ in the numerator of the above expression is $C_{18}^{3}$, so
$$a_{2}=C_{18}^{3}=816$$

Answer: 816


Calculate the area of the figure bounded by one arch of the cycloid $x=2(t-\sin t), y=2(1-\cos t)$.

Solution:
Solution. One arch of the cycloid is obtained for \(0 \leqslant t \leqslant 2 \pi\) (Fig. 2.6). The required area is conveniently calculated using the first formula from (2.22).

\[
\begin{aligned}
S= & \left|-\int_{0}^{2 \pi} y(t) x^{\prime}(t) d t\right|=\left|-\int_{0}^{2 \pi} 2(1-\cos t) 2(1-\cos t) d t\right|= \\
= & \left|4 \int_{0}^{2 \pi}(1-\cos t)^{2} d t\right|=\left|4 \int_{0}^{2 \pi}\left(1-2 \cos t+\cos ^{2} t\right) d t\right|= \\
& =\left|4 \int_{0}^{2 \pi}(1-2 \cos t+0.5(1+\cos 2 t)) d t\right|= \\
& =\left.4|(t-2 \sin t+0.5 t+0.25 \sin 2 t)|\right|_{0} ^{2 \pi}=12 \pi
\end{aligned}
\]

![](https://cdn.mathpix.com/cropped/2024_05_22_0f84e17422d7c0df8d68g-082.jpg?height=290&width=488&top_left_y=133&top_left_x=129)

Fig. 2.7.

![](https://cdn.mathpix.com/cropped/2024_05_22_0f84e17422d7c0df8d68g-082.jpg?height=292&width=429&top_left_y=130&top_left_x=641)

Fig. 2.8.

Answer: 12\pi


In the unit circle, chords $P Q$ and $M N$ are parallel to the unit radius $O R$ through the center $O$. Chords $M P, P Q$, and $N R$ are all $s$ units long, and chord $M N$ is $d$ units long. Of the following three equations
$\mathrm{I} . d-s=1$;
II . $d s=1$;
III . $d^{2}-s^{2}=\sqrt{5}$.

which must be correct?
(A) Only I.
(B) Only II.
(C) Only III.
(D) Only I and II.
(E) I, II, and III.

Solution:
[Solution 1] As shown in the figure, the length of chord $MN$ is $d$, and each chord of length $s$ subtends an arc of $\frac{180^{\circ}}{5}=36^{\circ}$. Connect the radius $OP$, intersecting $MN$ at $T$. Draw perpendiculars from $O$ and $Q$ to $MN$, denoted as $OV$ and $QU$ respectively.
From $\triangle OVN \sim \triangle QUN$, we have
$\frac{UN}{QN}=\frac{VN}{1}$, or $\frac{d}{2}-\frac{s}{2}=s \cdot \frac{d}{2}$.
Thus, $d-s=ds$.
Since $\angle MPT = \angle MTP = 72^{\circ}$, it follows that $MT = s$.
Similarly, $\angle OTN = \angle ORN$, so $ORNT$ is a parallelogram with opposite sides $TN = OR = 1$. Therefore,
$$
MN - MT = d - s = 1.
$$

From $(d-s)^2 + 4ds = 1 + 4 = 5$ or $d^2 + 2ds + s^2 = 5$,
we get $d + s = \sqrt{5}$.
Thus, $d^2 - s^2 = (d+s)(d-s) = \sqrt{5} \cdot 1 = \sqrt{5}$.
In summary, all three equations I, II, and III are correct.
Therefore, the answer is $(E)$.

[Solution 2] Given $MT = s$ and $MN = d$, and since $OTNR$ is a parallelogram, its side $TN = MN - MT = d - s = 1$.

Let $P'$ (not shown in the figure) be the other endpoint of the diameter through $P$. Then the chord $PP'$ intersects $MN$ at $T$. By the intersecting chords theorem, we have
$PT \cdot TP' = MT \cdot TN$, or $(1-s)(1+s) = s \cdot 1$.
This quadratic equation has the positive root $s = \frac{1}{2}(\sqrt{5} - 1)$. This can be used to prove that equations II and III are correct. Therefore, all three equations I, II, and III are correct.
Hence, the answer is $(E)$.

Answer: E


Solution:
Let $P=\left(a_{1}-a_{2}^{2}\right)\left(a_{2}-a_{3}^{2}\right) \cdots\left(a_{2015}-a_{2016}^{2}\right)\left(a_{2016}-a_{1}^{2}\right)$,
Since $a_{i}>\frac{11}{9} a_{i+1}^{2} \Rightarrow a_{i}-a_{i+1}^{2}>\frac{11}{9} a_{i+1}^{2}-a_{i+1}^{2}=\frac{2}{9} a_{i+1}^{2} \geqslant 0$,
If $a_{2016}-a_{1}^{2} \leqslant 0$, then $P \leqslant 0$; If $a_{2016}-a_{1}^{2}>0$, let $a_{2017}=a_{1}$,
then $P^{\frac{1}{2016}} \leqslant \frac{1}{2016} \sum_{i=1}^{2016}\left(a_{i}-a_{i+1}^{2}\right)=\frac{1}{2016}\left(\sum_{i=1}^{2016} a_{i}-\sum_{i=1}^{2016} a_{i+1}^{2}\right)$
$=\frac{1}{2016}\left(\sum_{i=1}^{2016} a_{i}-\sum_{i=1}^{2016} a_{i}^{2}\right)=\frac{1}{2016} \sum_{i=1}^{2016}\left(a_{i}-a_{i}^{2}\right)=\frac{1}{2016} \sum_{i=1}^{2016}\left[-\left(a_{i}-\frac{1}{2}\right)^{2}+\frac{1}{4}\right]$
$\leqslant \frac{1}{2016} \sum_{i=1}^{2016} \frac{1}{4}=\frac{1}{4}$, equality holds when $a_{i}=\frac{1}{2}, i=1,2, \cdots, 2016$.
At this point, $9 a_{i}=\frac{9}{2}>\frac{11}{4}=11 a_{i+1}^{2}$ is clearly satisfied, so the maximum value of $P^{\frac{1}{2016}}$ is $\frac{1}{4}$.
In summary, the maximum value of $\left(a_{1}-a_{2}^{2}\right)\left(a_{2}-a_{3}^{2}\right) \cdots\left(a_{2015}-a_{2016}^{2}\right)\left(a_{2016}-a_{1}^{2}\right)$ is $\frac{1}{4^{2016}}$.

Answer: \frac{1}{4^{2016}}


Let $A=\left(a_{1}, a_{2}, \ldots, a_{2001}\right)$ be a sequence of positive integers. Let $m$ be the number of 3-element subsequences $\left(a_{i}, a_{j}, a_{k}\right)$ with $1 \leq i< j<k \leq 2001$ such that $a_{j}=a_{i}+1$ and $a_{k}=a_{j}+1$. Considering all such sequences $A$, find the greatest value of $m$.

Solution:
7. It is evident that arranging of $A$ in increasing order does not diminish $m$. Thus we can assume that $A$ is nondecreasing. Assume w.l.o.g. that $a_{1}=1$, and let $b_{i}$ be the number of elements of $A$ that are equal to $i$ $\left(1 \leq i \leq n=a_{2001}\right)$. Then we have $b_{1}+b_{2}+\cdots+b_{n}=2001$ and
$$ m=b_{1} b_{2} b_{3}+b_{2} b_{3} b_{4}+\cdots+b_{n-2} b_{n-1} b_{n} $$
Now if $b_{i}, b_{j}(i<j)$ are two largest $b$'s, we deduce from (1) and the AMGM inequality that $m \leq b_{i} b_{j}\left(b_{1}+\cdots+b_{i-1}+b_{i+1}+\cdots+b_{j-1}+b_{j+1}+b_{n}\right) \leq$ $\left(\frac{2001}{3}\right)^{3}=667^{3} \quad\left(b_{1} b_{2} b_{3} \leq b_{1} b_{i} b_{j}\right.$, etc.). The value $667^{3}$ is attained for $b_{1}=b_{2}=b_{3}=667$ (i.e., $a_{1}=\cdots=a_{667}=1, a_{668}=\cdots=a_{1334}=2$, $a_{1335}=\cdots=a_{2001}=3$ ). Hence the maximum of $m$ is $667^{3}$.

Answer: 667^3