On a horizontal plane, three points are 100 meters, 200 meters, and 300 meters away from the base of an antenna. The sum of the angles of elevation from these three points to the antenna is $90^{\circ}$. What is the height of the antenna?

Solution:
[Solution] Let the height of the antenna be $x$ meters. The angles of elevation from points 100 meters, 200 meters, and 300 meters away from the base of the antenna are $\alpha, \beta, \gamma$ respectively. Then
$$\operatorname{tg} \alpha=\frac{x}{100}, \operatorname{tg} \beta=\frac{x}{200}, \operatorname{tg} \gamma=\frac{x}{300} .$$

Since $\alpha+\beta+\gamma=90^{\circ}$, we have
$$\begin{aligned}
\operatorname{tg} \gamma & =\frac{x}{300}=\operatorname{tg}\left[90^{\circ}-(\alpha+\beta)\right] \\
& =\operatorname{ctg}(\alpha+\beta)=\frac{1}{\operatorname{tg}(\alpha+\beta)} \\
& =\frac{1-\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}{\operatorname{tg} \alpha+\operatorname{tg} \beta}=\frac{1-\frac{x}{100} \cdot \frac{x}{200}}{\frac{x}{100}+\frac{x}{200}} \\
& =\frac{100 \times 200-x^{2}}{300 x} .
\end{aligned}$$

Therefore, $2 x^{2}=100 \times 200$, and since $x>0$, we have $x=100$ meters.

Answer: 100


In each square of a $4$ by $4$ grid, you put either a $+1$ or a $-1$. If any 2 rows and 2 columns are deleted, the sum of the remaining 4 numbers is nonnegative. What is the minimum number of $+1$'s needed to be placed to be able to satisfy the conditions?

Solution:
To solve this problem, we need to determine the minimum number of $+1$'s that must be placed in a $4 \times 4$ grid such that if any 2 rows and 2 columns are deleted, the sum of the remaining 4 numbers is nonnegative.

1. **Understanding the Condition**:
   The condition states that for any 2 rows and 2 columns deleted, the sum of the remaining 4 numbers must be nonnegative. This implies that in any $2 \times 2$ subgrid formed by the intersection of any 2 rows and 2 columns, the sum of the numbers must be nonnegative.

2. **Maximum Number of $-1$'s**:
   Let's denote the number of $-1$'s in the grid by $k$. We need to find the maximum value of $k$ such that the condition is still satisfied.

3. **Constructive Example**:
   Consider the following grid configuration:
   \[
   \begin{array}{|c|c|c|c|}
   \hline
   -1 & -1 & +1 & +1 \\
   \hline
   -1 & -1 & +1 & +1 \\
   \hline
   +1 & +1 & +1 & +1 \\
   \hline
   +1 & +1 & +1 & +1 \\
   \hline
   \end{array}
   \]
   In this configuration, there are 6 $-1$'s and 10 $+1$'s. Let's verify if this configuration satisfies the condition.

4. **Verification**:
   - If we delete any 2 rows and 2 columns, we are left with a $2 \times 2$ subgrid.
   - For example, if we delete the first two rows and the first two columns, the remaining subgrid is:
     \[
     \begin{array}{|c|c|}
     \hline
     +1 & +1 \\
     \hline
     +1 & +1 \\
     \hline
     \end{array}
     \]
     The sum of the remaining numbers is $+1 + +1 + +1 + +1 = 4$, which is nonnegative.
   - Similarly, for any other combination of 2 rows and 2 columns deleted, the sum of the remaining numbers will be nonnegative.

5. **Maximality Argument**:
   - Suppose we have more than 6 $-1$'s. Then, there must be at least one $2 \times 2$ subgrid with more than 2 $-1$'s.
   - If a $2 \times 2$ subgrid has 3 or 4 $-1$'s, the sum of the numbers in that subgrid will be negative, violating the condition.
   - Therefore, the maximum number of $-1$'s that can be placed in the grid while satisfying the condition is 6.

6. **Minimum Number of $+1$'s**:
   - Since the grid has 16 cells in total, and we can have at most 6 $-1$'s, the minimum number of $+1$'s needed is:
     \[
     16 - 6 = 10
     \]

The final answer is $\boxed{10}$.

Answer: 10


Diagonals $AC$ and $BD$ of an isosceles trapezoid $ABCD$ intersect at point $O$. It is known that $AD : BC=3: 2$. Circle $\omega$ with center $O$, passing through vertices $A$ and $D$, intersects the extension of base $BC$ beyond point $B$ at point $K$. It turns out that $BK=BO$. Find the ratio of the base $AD$ to the radius of circle $\omega$.

Solution:
Answer: $1 / 4$.

Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_fc73715326e960a104bag-3.jpg?height=707&width=1022&top_left_y=1186&top_left_x=497)

Let $B C=2 a, O B=b$, then $A D=3 a, O A=3 b / 2$. Drop a perpendicular $O M$ from point $O$ to line $B C$ (from the isosceles property of the trapezoid, it follows that $M$ is the midpoint of $B C$). Next, we will find $O M^{2}$ in two ways. From triangle $O B M$ by the Pythagorean theorem, $O M^{2}=O B^{2}-B M^{2}=b^{2}-a^{2}$. From triangle $O K M$ by the Pythagorean theorem, $O M^{2}=O K^{2}-K M^{2}=\frac{9 b^{2}}{4}-(b+a)^{2}$. Equating the right sides, we have: $b^{2}-a^{2}=\frac{9 b^{2}}{4}-(b+a)^{2}, b^{2}=8 a b \cdot \frac{A D}{A O}=\frac{2 a}{b}=\frac{1}{4}$.

Answer: \frac{1}{4}


Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a+b+c=1, \\
\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}=1 .
\end{array}
$$

Then $a b c=$ . $\qquad$

Solution:
$$
\begin{array}{l}
\frac{1}{1-2 c}+\frac{1}{1-2 a}+\frac{1}{1-2 b}=1 \\
\Rightarrow(1-2 a)(1-2 b)+(1-2 b)(1-2 c)+(1-2 a)(1-2 c) \\
\quad=(1-2 a)(1-2 b)(1-2 c) \\
\Rightarrow 2-2(a+b+c)=8 a b c \\
\Rightarrow a b c=0 .
\end{array}
$$

Answer: 0


Let $T$ be the answer to question $18$. Rectangle $ZOMR$ has $ZO = 2T$ and $ZR = T$. Point $B$ lies on segment $ZO$, $O'$ lies on segment $OM$, and $E$ lies on segment $RM$ such that $BR = BE = EO'$, and $\angle BEO' = 90^o$. Compute $2(ZO + O'M + ER)$.

PS. You had better calculate it in terms of $T$.

Solution:
1. **Define the coordinates of the vertices of the rectangle:**
   - Let \( Z(0,0) \), \( O(2T,0) \), \( M(2T,T) \), and \( R(0,T) \).

2. **Place point \( B \) on segment \( ZO \):**
   - Let \( B(\lambda, 0) \).

3. **Place point \( E \) on segment \( RM \):**
   - Let \( E(2\lambda, T) \).

4. **Determine the coordinates of point \( O' \) on segment \( OM \):**
   - Let \( O'(2T, y) \).

5. **Use the condition \( BR = BE = EO' \) and \( \angle BEO' = 90^\circ \):**
   - The distance \( BR \) is given by:
     \[
     BR = \sqrt{(\lambda - 0)^2 + (0 - T)^2} = \sqrt{\lambda^2 + T^2}
     \]
   - The distance \( BE \) is given by:
     \[
     BE = \sqrt{(2\lambda - \lambda)^2 + (T - 0)^2} = \sqrt{\lambda^2 + T^2}
     \]
   - The distance \( EO' \) is given by:
     \[
     EO' = \sqrt{(2T - 2\lambda)^2 + (y - T)^2}
     \]
   - Since \( BR = BE = EO' \), we have:
     \[
     \sqrt{\lambda^2 + T^2} = \sqrt{(2T - 2\lambda)^2 + (y - T)^2}
     \]

6. **Use the condition \( \angle BEO' = 90^\circ \):**
   - This implies that \( BE \) and \( EO' \) are perpendicular. Therefore, the slopes of \( BE \) and \( EO' \) must be negative reciprocals:
     \[
     \text{slope of } BE = \frac{T - 0}{2\lambda - \lambda} = \frac{T}{\lambda}
     \]
     \[
     \text{slope of } EO' = \frac{y - T}{2T - 2\lambda}
     \]
     \[
     \frac{T}{\lambda} \cdot \frac{y - T}{2T - 2\lambda} = -1
     \]
     \[
     T(y - T) = -\lambda(2T - 2\lambda)
     \]
     \[
     Ty - T^2 = -2\lambda T + 2\lambda^2
     \]
     \[
     Ty + 2\lambda T = T^2 + 2\lambda^2
     \]
     \[
     y + 2\lambda = T + \frac{2\lambda^2}{T}
     \]
     \[
     y = T + \frac{2\lambda^2}{T} - 2\lambda
     \]

7. **Solve for \( \lambda \) using the equality of distances:**
   - From \( \sqrt{\lambda^2 + T^2} = \sqrt{(2T - 2\lambda)^2 + (y - T)^2} \):
     \[
     \lambda^2 + T^2 = (2T - 2\lambda)^2 + (y - T)^2
     \]
     \[
     \lambda^2 + T^2 = 4T^2 - 8T\lambda + 4\lambda^2 + (T + \frac{2\lambda^2}{T} - 2\lambda - T)^2
     \]
     \[
     \lambda^2 + T^2 = 4T^2 - 8T\lambda + 4\lambda^2 + (\frac{2\lambda^2}{T} - 2\lambda)^2
     \]
     \[
     \lambda^2 + T^2 = 4T^2 - 8T\lambda + 4\lambda^2 + \frac{4\lambda^4}{T^2} - 8\lambda^3 + 4\lambda^2
     \]
     \[
     \lambda^2 + T^2 = 4T^2 - 8T\lambda + 8\lambda^2 + \frac{4\lambda^4}{T^2} - 8\lambda^3
     \]
     - Solving this equation, we find \( \lambda = \frac{T}{2} \).

8. **Determine the coordinates of \( O' \):**
   - Substituting \( \lambda = \frac{T}{2} \) into the equation for \( y \):
     \[
     y = T + \frac{2(\frac{T}{2})^2}{T} - 2(\frac{T}{2})
     \]
     \[
     y = T + \frac{2(\frac{T^2}{4})}{T} - T
     \]
     \[
     y = T + \frac{T}{2} - T
     \]
     \[
     y = \frac{T}{2}
     \]
     - Therefore, \( O' = (2T, \frac{T}{2}) \).

9. **Calculate the lengths \( ZO \), \( O'M \), and \( ER \):**
   - \( ZO = 2T \)
   - \( O'M = \sqrt{(2T - 2T)^2 + (T - \frac{T}{2})^2} = \frac{T}{2} \)
   - \( ER = \sqrt{(2\lambda - 0)^2 + (T - T)^2} = \sqrt{(T)^2} = T \)

10. **Compute \( 2(ZO + O'M + ER) \):**
    \[
    2(ZO + O'M + ER) = 2(2T + \frac{T}{2} + T) = 2(3.5T) = 7T
    \]

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

Answer: 7T


Solution:
Given $a+1=\sqrt{17}$, squaring both sides yields $a^{2}+2 a+1=17$.
Then
$$
\begin{array}{l}
\left(a^{5}+2 a^{4}-17 a^{3}-a^{2}+18 a-17\right)^{1993} \\
=\left[a^{5}+2 a^{4}-\left(a^{2}+2 a+1\right) a^{3}-a^{2}+\left(a^{2}\right.\right. \\
\left.+2 a+1+1) a-\left(a^{2}+2 a+1\right)\right]^{1993} \\
=(-1)^{1993}=-1 \text {. } \\
\end{array}
$$

Note: Here, we first constructed the expression $a^{2}+2 a+1$ with a value of 17, and then substituted $a^{2}+2 a+1$ for 17 in the expression to be evaluated.

Answer: -1


Let $n$ be a strictly positive integer. Domitille has a rectangular grid divided into unit squares. Inside each unit square is written a strictly positive integer. She can perform the following operations as many times as she wishes:

- Choose a row and multiply each number in the row by $n$.
- Choose a column and subtract $n$ from each integer in the column.

Determine all values of $n$ for which the following property is satisfied:

Regardless of the dimensions of the rectangle and the integers written in the cells, Domitille can end up with a rectangle containing only 0s after a finite number of operations.

Solution:
Solution to Exercise 7 For $\mathrm{n}=1$, if we start from a configuration with one column of two rows, with different integers on these two rows, the difference between the two integers remains invariant when one of the two operations is performed. Consequently, it cannot be canceled. In particular, Domitille cannot arrive at a rectangle containing only 0s in a finite number of steps.

We now assume that $n>2$. Consider the rectangle with one column of two rows with values 1 and 0. By looking at what happens modulo $n-1$, we see that multiplication by $n$ has no effect, and the second operation preserves the difference between the two values. Thus, the difference between the two values modulo $n-1$ is an invariant, initially non-zero. Consequently, it is impossible to make the two numbers equal $\bmod n-1$, and in particular to make them both zero.

Now let's show that Domitille can always reach 0 in the case $n=2$.

First, reduce the problem. If Domitille can handle the case where there is only one column, it is sufficient for her to deal with the columns one by one. Once a column is brought to 0, the only operation that its cells will undergo when processing the other columns will be the multiplication of a row by $\mathrm{n}$. But since the elements of the column are zero, this actually has no effect.

We can therefore assume that the rectangle is in fact composed of only one column. Then, if we can make two cells of the column equal, we can merge them, that is, each time we multiply the row of the first by $\mathrm{n}$, we do the same for the second. In this case, the two cells will always contain the same value. We can therefore restrict ourselves to the case where the column contains 2 cells with the integers $a$ and $b$. By multiplying each of the rows by 2, we can actually assume that $a$ and $b$ are even. We then subtract 2 from the smallest until it is 2. This is possible because neither $a$ nor $b$ is zero. Then we apply the "double then subtract" operation: we double the one that is 2, then subtract 2 from the column, until both numbers are 2. We have thus reduced to a configuration where all the numbers in the column are 2. We then subtract 2 from the column, and we have thus nullified the column, which concludes. Note also that the integers in all the other columns remain non-zero during all these manipulations.

Comments from the graders: Very few students tackled this problem, but those who did had excellent writing. Too bad some forgot the case $n=1$. One student attempted a proof by the principle of the minimum, which unfortunately did not lead to a conclusion.

Answer: 2


The calculation result of the expression $101 \times 2012 \times 121 \div 1111 \div 503$ is

Solution:
【Solution】Solve: $101 \times 2012 \times 121 \div 1111 \div 503$
$$
\begin{array}{l}
=101 \times 2012 \times 121 \div(11 \times 101) \div 503 \\
=101 \times 2012 \times 121 \div 11 \div 101 \div 503 \\
=(101 \div 101) \times(2012 \div 503) \times(121 \div 11) \\
=1 \times 4 \times 11 \\
=44 .
\end{array}
$$

Therefore, the answer is: 44.

Answer: 44


On a bus rode Andrey

To the circle and back home,

Paying 115 rubles,

He bought himself a pass.

In January he didn't get it,

And therefore for several days

He bought a ticket from the driver

For 15 rubles for himself.

And on another day the conductor

Took only 11 rubles from him.

Returning from his circle

Every time he walked home our Andrey.

For January, how much money was spent,

The thrifty Andrey calculated:

To his surprise, he got

Exactly 115 rubles!

Calculate now as quickly as you can,

How many times the circle was in January

Solution:
The amount of rubles spent by Andrey on the days when he bought a ticket from the driver is divisible by 5; the total amount of rubles spent by him in January is also divisible by 5. Therefore, the total amount of money spent on other days is also divisible by 5. Hence, the number of days when Andrey bought a ticket from the conductor is divisible by 5. The numbers 0 and 10 are not suitable; numbers greater than 10 are even less suitable, so the only possible option is 5 days. Then the number of other days is $(115-11 \cdot 5): 15=4$, and the circle was 9 times.

## Answer

9 times.

Answer: 9


Suppose $f(x)$ is a rational function such that $3 f\left(\frac{1}{x}\right)+\frac{2 f(x)}{x}=x^{2}$ for $x \neq 0$. Find $f(-2)$.

Solution:
Solution: Let $x=\frac{-1}{2}$. Then
$$
\begin{aligned}
3 f(-2)+\frac{2 f\left(\frac{-1}{2}\right)}{\frac{-1}{2}} & =\frac{1}{4} \\
\Rightarrow 3 f(-2)-4 f\left(\frac{-1}{2}\right) & =\frac{1}{4}
\end{aligned}
$$

Let $x=-2$. Then
$$
\begin{array}{l}
3 f\left(\frac{-1}{2}\right)+\frac{2 f(-2)}{-2}=4 \\
\Rightarrow 3 f\left(\frac{-1}{2}\right)-f(-2)=4
\end{array}
$$

Solving this system of equations $\{(1),(2)\}$ for $f(-2)$ yields $f(-2)=\frac{67}{20}$.

Answer: \frac{67}{20}


Let $g(x) = x^5 + x^4 + x^3 + x^2 + x + 1.$ What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$?

If $k$ is a positive number and $f$ is a function such that, for every positive number $x, f(x^2 + 1 )^{\sqrt{x}} = k.$ Find the value of

\[ f\left( \frac{9 +y^2}{y^2}\right)^{\sqrt{ \frac{12}{y} }} \] for every positive number $y.$

The function $f$ satisfies the functional equation $f(x) + f(y) = f(x+y) - x \cdot y - 1$ for every pair $x,y$ of real numbers. If $f(1) = 1,$ then find the number of integers $n,$ for which $f(n) = n.$

Solution:
### Part i)
1. Given the polynomial \( g(x) = x^5 + x^4 + x^3 + x^2 + x + 1 \).
2. We need to find the remainder when \( g(x^{12}) \) is divided by \( g(x) \).
3. Notice that \( g(x) \) can be rewritten using the formula for the sum of a geometric series:
   \[
   g(x) = \frac{x^6 - 1}{x - 1}
   \]
   because \( x^6 - 1 = (x - 1)(x^5 + x^4 + x^3 + x^2 + x + 1) \).
4. Therefore, \( g(x) \) has roots at the 6th roots of unity, excluding 1.
5. Since \( g(x) \) has roots at \( \omega, \omega^2, \omega^3, \omega^4, \omega^5 \) where \( \omega = e^{2\pi i / 6} \), we need to consider \( g(x^{12}) \) at these roots.
6. Notice that \( x^{12} = (x^6)^2 \), and since \( \omega^6 = 1 \), we have \( \omega^{12} = 1 \).
7. Therefore, \( g(x^{12}) \) evaluated at any root of \( g(x) \) will be \( g(1) = 1 + 1 + 1 + 1 + 1 + 1 = 6 \).
8. Since \( g(x) \) divides \( g(x^{12}) - 6 \), the remainder when \( g(x^{12}) \) is divided by \( g(x) \) is 6.

\[
\boxed{6}
\]

### Part ii)
1. Given \( f(x^2 + 1)^{\sqrt{x}} = k \) for every positive number \( x \).
2. We need to find the value of \( f\left( \frac{9 + y^2}{y^2} \right)^{\sqrt{\frac{12}{y}}} \) for every positive number \( y \).
3. Let \( x = \frac{3}{y} \). Then \( x^2 = \left( \frac{3}{y} \right)^2 = \frac{9}{y^2} \).
4. Therefore, \( x^2 + 1 = \frac{9}{y^2} + 1 = \frac{9 + y^2}{y^2} \).
5. Substituting \( x = \frac{3}{y} \) into the given functional equation, we get:
   \[
   f\left( \frac{9 + y^2}{y^2} \right)^{\sqrt{\frac{3}{y}}} = k
   \]
6. We need to find \( f\left( \frac{9 + y^2}{y^2} \right)^{\sqrt{\frac{12}{y}}} \).
7. Notice that \( \sqrt{\frac{12}{y}} = 2 \cdot \sqrt{\frac{3}{y}} \).
8. Therefore, we have:
   \[
   f\left( \frac{9 + y^2}{y^2} \right)^{\sqrt{\frac{12}{y}}} = \left( f\left( \frac{9 + y^2}{y^2} \right)^{\sqrt{\frac{3}{y}}} \right)^2 = k^2
   \]

\[
\boxed{k^2}
\]

### Part iii)
1. Given the functional equation \( f(x) + f(y) = f(x+y) - xy - 1 \) for every pair \( x, y \) of real numbers.
2. We know \( f(1) = 1 \).
3. Set \( x = 0 \) in the functional equation:
   \[
   f(0) + f(y) = f(y) - 1 \implies f(0) = -1
   \]
4. Set \( y = 1 \) in the functional equation:
   \[
   f(x) + f(1) = f(x+1) - x - 1 \implies f(x) + 1 = f(x+1) - x - 1 \implies f(x+1) = f(x) + x + 2
   \]
5. Using the recurrence relation \( f(x+1) = f(x) + x + 2 \), we can find \( f(n) \) for any integer \( n \).
6. Calculate \( f(n) \) for positive integers:
   \[
   f(2) = f(1) + 1 + 2 = 1 + 3 = 4
   \]
   \[
   f(3) = f(2) + 2 + 2 = 4 + 4 = 8
   \]
   \[
   f(4) = f(3) + 3 + 2 = 8 + 5 = 13
   \]
   \[
   \vdots
   \]
   \[
   f(n) = \frac{(n+1)(n+2)}{2} - 2
   \]
7. We need to find \( n \) such that \( f(n) = n \):
   \[
   \frac{(n+1)(n+2)}{2} - 2 = n
   \]
   \[
   (n+1)(n+2) = 2n + 4
   \]
   \[
   n^2 + 3n + 2 = 2n + 4
   \]
   \[
   n^2 + n - 2 = 0
   \]
   \[
   (n+2)(n-1) = 0
   \]
   \[
   n = -2 \text{ or } n = 1
   \]
8. Since \( n \) must be an integer, the only solutions are \( n = -2 \) and \( n = 1 \).

\[
\boxed{2}
\]

Answer: 2


Find all real numbers $x$ for which

$$
\left(x^{2}-7 x+11\right)^{x^{2}+5 x-6}=1
$$

Solution:
IV/2. Let's recall that $a^{b}=1$ if $a=1$ or $a=-1$ and $b$ is an even number, or if $b=0$ and $a \neq 0$.

First, let's assume that the base $x^{2}-7 x+11$ is equal to 1. The equation $x^{2}-7 x+11=1$ can be rewritten as $x^{2}-7 x+10=0$, and the solutions are $x_{1}=2$ and $x_{2}=5$.

Next, we find the values of $x$ for which the base $x^{2}-7 x+11$ is equal to -1: the equation $x^{2}-7 x+11=$ -1 can be rewritten as $x^{2}-7 x+12=0$, which has solutions $x_{3}=3$ and $x_{4}=4$. The exponent $x^{2}+5 x-6$ is even for both of these values: for $x_{3}$ it is $9+15-6=18$, and for $x_{4}$ it is $16+20-6=30$. This means that for $x=3$ and $x=4$, $\left(x^{2}-7 x+11\right)^{x^{2}+5 x-6}=1$.

Finally, we find the values for which the exponent $x^{2}+5 x-6$ is equal to 0. The equation $x^{2}+$ $5 x-6=0$ has solutions $x_{5}=1$ and $x_{6}=-6$. Both solutions are valid, as the base is not equal to 0 in these cases. The expression $\left(x^{2}-7 x+11\right)^{x^{2}+5 x-6}$ has a value of 1 if $x$ is equal to $1, 2, 3, 4, 5$ or -6.

Solutions $x_{1}=2$ and $x_{2}=5$, when the base $x^{2}-7 x+11$ is equal to $1 \ldots \ldots \ldots \ldots \ldots \ldots \ldots 2$ points

Solutions $x_{3}=3$ and $x_{4}=4$, when the base is equal to -1 and the exponent is an even number . . 2 points

Solutions $x_{5}=1$ and $x_{6}=-6$, when the exponent is equal to $0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$.

(If the contestant correctly determines $x_{5}$ and $x_{6}$ but does not verify that the base is non-zero, deduct 1 point.)

Answer: 1,2,3,4,5,-6


In the acute $\triangle A B C$,
$$
\sin (A+B)=\frac{3}{5}, \sin (A-B)=\frac{1}{5}, A B=3
$$

then the area of $\triangle A B C$ is $\qquad$

Solution:
9. $\frac{3(\sqrt{6}+2)}{2}$.

From the given information,
$$
\begin{array}{l}
\left\{\begin{array}{l}
\sin A \cdot \cos B+\cos A \cdot \sin B=\frac{3}{5} \\
\sin A \cdot \cos B-\cos A \cdot \sin B=\frac{1}{5}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
\sin A \cdot \cos B=\frac{2}{5}, \\
\cos A \cdot \sin B=\frac{1}{5}
\end{array}\right. \\
\Rightarrow \tan A=2 \tan B. \\
\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}=\frac{3 \tan B}{1-2 \tan ^{2} B}, \\
\end{array}
$$

Also, $\tan (A+B)=\frac{\sin (A+B)}{\cos (A+B)}=-\frac{3}{4}$, and

thus $\tan B=\frac{\sqrt{6}+2}{2}$.
Let the height from vertex $A$ to side $B$ be $h$. Then
$$
\begin{array}{l}
3=A B=\frac{h}{\tan A}+\frac{h}{\tan B}=\frac{3 h}{2 \tan B} \\
\Rightarrow h=2 \tan B .
\end{array}
$$

Therefore, $S_{\triangle A B C}=\frac{1}{2} A B \cdot h=\frac{1}{2} \times 3 \times 2 \tan B$
$$
=3 \tan B=\frac{3(2+\sqrt{6})}{2} \text {. }
$$

Answer: \frac{3(\sqrt{6}+2)}{2}


On the blackboard, there are $n$ (where $n$ is an odd number) expressions: $* x^{2}+* x+*=0$. Two players, A and B, take turns to play the following game: each person is only allowed to replace one of the * symbols with a non-zero real number at a time. After a total of $3 n$ operations, $n$ quadratic equations are obtained. A tries to maximize the number of equations without real roots, while B tries to minimize the number of equations without real roots. Try to determine: Under the condition that A does not depend on how B operates, what is the maximum number of equations without real roots that A can achieve?

Solution:
Jia each time changes the coefficient of $x$ in an untouched equation to 1, unless Yi changes the coefficient of $x^{2}$ or the constant term in an equation Jia has already changed to any real number $a \neq 0$. In this case, Jia immediately changes the constant term or the coefficient of $x^{2}$ in this equation to $\frac{1}{a}$, so the discriminant of this equation becomes $1-4 a \cdot \frac{1}{a} = 2 \sqrt{|c|}$. Thus, the discriminant of the resulting equation is greater than zero, meaning the equation has real roots. This way, Yi can change the coefficient of $x^{2}$ in at least $\frac{n-1}{2}$ untouched equations to 1, and ensure that these $\frac{n-1}{2}$ equations have real roots, thus leaving at most $n-\frac{n-1}{2}=\frac{n+1}{2}$ equations without real roots.
Therefore, the maximum value sought is $\frac{n+1}{2}$.

Answer: \frac{n+1}{2}


Given a rectangle ABCD where $\overline{\mathrm{AB}}=2 \overline{\mathrm{BC}}$. On the side $\mathrm{AB}$, a point $\mathrm{K}$ is chosen such that $\angle \mathrm{AKD}=\angle \mathrm{DKC}$. Find this angle.

Solution:
Solution. Let's draw the height DH of triangle KCD (Fig.1). Triangles KDA and KDH are congruent (right-angled with

![](https://cdn.mathpix.com/cropped/2024_06_05_fb56a3883e438b3946fbg-1.jpg?height=272&width=449&top_left_y=1479&top_left_x=1030)

Fig. 1 common side $\mathrm{KD}$ and $\angle \mathrm{AKD}=\angle \mathrm{DKC}$ ), from which $\overline{\mathrm{DH}}=\overline{\mathrm{DA}}=\frac{\overline{\mathrm{DC}}}{2}$. Since triangle CDH is right-angled, with the leg $\overline{\mathrm{DH}}$ half the hypotenuse $\overline{\mathrm{DC}}$, we get that $\angle \mathrm{HCD}=30^{\circ}$. Then $\angle \mathrm{CKB}=\angle \mathrm{HCD}=30^{\circ}$, and therefore $\angle \mathrm{CKA}=150^{\circ}$. Thus $\angle \mathrm{AKD}=\angle \mathrm{DKC}=75^{\circ}$.

Answer: 75


$\overline{a b}$ and $\overline{b a}$ represent two-digit integers, then the number of $\overline{a b}$ that satisfy the equation $(a+b) x^{2}-\overline{a b} x+\overline{b a}=0$ with both roots being integers is ( ).
(A)0
(B) 2
(C) 4
(D) 6

Solution:
6. B.

Let the two integer roots be $x_{1}, x_{2}$, then
$$
x_{1}+x_{2}=\frac{10 a+b}{a+b}, x_{1} x_{2}=\frac{10 b+a}{a+b} \text {. }
$$

We have $x_{1}+x_{2}+x_{1} x_{2}=11$.
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{2}+1\right)=12=( \pm 1)( \pm 12) \\
=( \pm 2)( \pm 6)=( \pm 3)( \pm 4) .
\end{array}
$$

When $\left\{\begin{array}{l}x_{1}=2, \\ x_{2}=3\end{array}\right.$, substituting into the above equation gives $5 a=4 b$. Hence $\overline{a b}=45$.
When $\left\{\begin{array}{l}x_{1}=1 \\ x_{2}=5\end{array}\right.$, substituting into the above equation gives $4 a=5 b$. Hence $\overline{a b}=54$.

Answer: B


In $\triangle ABC$, if
\[ \tan A \cdot \tan B=\tan A \cdot \tan C+\tan C \cdot \tan B \text {, } \]
then \[ \frac{a^{2}+b^{2}}{c^{2}}= \]

Solution:
$-1.3$.
Given the equation can be transformed as
$$
\begin{array}{l}
\frac{\sin A \cdot \sin B}{\cos A \cdot \cos B}=\frac{\sin A \cdot \sin C}{\cos A \cdot \cos C}+\frac{\sin B \cdot \sin C}{\cos B \cdot \cos C} \\
\Rightarrow \sin A \cdot \sin B \cdot \cos C \\
\quad=\sin C \cdot(\sin A \cdot \cos B+\cos A \cdot \sin B) \\
\quad=\sin ^{2} C \\
\Rightarrow a b \frac{a^{2}+b^{2}-c^{2}}{2 a b}=c^{2} \\
\Rightarrow a^{2}+b^{2}=3 c^{2} .
\end{array}
$$

Therefore, the answer is 3.

Answer: 3


Determine the complex number $z$ such that

$$
|z+2|=|1-\bar{z}| \quad \text{and} \quad \operatorname{Re}\left(\frac{z}{2+3 i}\right)=\frac{1}{13}
$$

Solution:
## Solution.

From the first equation $|z+2|=|1-\bar{z}|$ we get $(x+2)^{2}+y^{2}=(1-x)^{2}+y^{2}$, which simplifies to $x=-\frac{1}{2}$

From the second equation, we have

$$
\begin{array}{cc}
\operatorname{Re} \frac{z}{2+3 i}=\operatorname{Re}\left(\frac{x+y i}{2+3 i} \cdot \frac{2-3 i}{2-3 i}\right)=\operatorname{Re}\left(\frac{2 x+3 y}{13}-\frac{3 x-2 y}{13} i\right)=\frac{2 x+3 y}{13}, & 1 \text{ point} \\
\frac{2 x+3 y}{13}=\frac{1}{13} \quad \Rightarrow \quad 2 x+3 y=1 . & 1 \text{ point }
\end{array}
$$

From $2 x+3 y=1$ and $x=-\frac{1}{2}$, it follows that $y=\frac{2}{3}$.

1 point

Therefore, $z=-\frac{1}{2}+\frac{2}{3} i$.

1 point

Answer: -\frac{1}{2}+\frac{2}{3}i


The price of a ticket to the stadium was 25 rubles. After the ticket prices were reduced, the number of spectators at the stadium increased by $50 \%$, and the revenue from ticket sales increased by $14 \%$. What is the new price of a ticket to the stadium after the price reduction?

Solution:
Answer: 19 rubles.

Let's denote: $a$ - the number of spectators coming to the stadium; $x$ - the ratio of the new ticket price to the price of 25 rubles. The new revenue is: on the one hand $(25 \cdot x) \cdot a \cdot 1.5$, on the other hand $25 \cdot 1.14 \cdot a$. From the equality $(25 \cdot x) \cdot a \cdot 1.5 = 25 \cdot 1.14 \cdot a$ we get $x=0.76$ and $25 \cdot 0.76=19$.

Answer: 19


There are seven cards in a hat, and on the card $k$ there is a number $2^{k-1}$, $k=1,2,...,7$. Solarin picks the cards up at random from the hat, one card at a time, until the sum of the numbers on cards in his hand exceeds $124$. What is the most probable sum he can get?

Solution:
1. **Identify the numbers on the cards:**
   The numbers on the cards are given by \(2^{k-1}\) for \(k = 1, 2, \ldots, 7\). Therefore, the numbers on the cards are:
   \[
   2^0, 2^1, 2^2, 2^3, 2^4, 2^5, 2^6 = 1, 2, 4, 8, 16, 32, 64
   \]

2. **Determine the sum of all cards:**
   The sum of all the numbers on the cards is:
   \[
   1 + 2 + 4 + 8 + 16 + 32 + 64 = 127
   \]

3. **Identify the stopping condition:**
   Solarin stops picking cards when the sum of the numbers on the cards in his hand exceeds 124. Therefore, the possible sums he can get are 125, 126, and 127.

4. **Analyze the possible sums:**
   - **Sum = 125:**
     To get a sum of 125, the sum of the numbers on the cards picked must be 125. This means the last card picked must be 2 (since 127 - 2 = 125).
   - **Sum = 126:**
     To get a sum of 126, the sum of the numbers on the cards picked must be 126. This means the last card picked must be 1 (since 127 - 1 = 126).
   - **Sum = 127:**
     To get a sum of 127, the sum of the numbers on the cards picked must be 127. This means the last card picked can be any of the remaining cards (3, 4, 5, 6, 7).

5. **Calculate the probabilities:**
   - **Probability of stopping at 125:**
     The probability of stopping at 125 is the probability of picking the card with the number 2 as the last card. There are 7 cards, so the probability is:
     \[
     \frac{1}{7}
     \]
   - **Probability of stopping at 126:**
     The probability of stopping at 126 is the probability of picking the card with the number 1 as the last card. There are 7 cards, so the probability is:
     \[
     \frac{1}{7}
     \]
   - **Probability of stopping at 127:**
     The probability of stopping at 127 is the probability of picking any of the remaining cards (3, 4, 5, 6, 7) as the last card. There are 5 such cards, so the probability is:
     \[
     \frac{5}{7}
     \]

6. **Determine the most probable sum:**
   Since the probability of stopping at 127 is the highest (5/7), the most probable sum Solarin can get is 127.

The final answer is \(\boxed{127}\).

Answer: 127


All natural numbers from 1 to some number were written on a board. After one of the numbers was erased, the arithmetic mean of the remaining numbers on the board is $\frac{673}{18}$. Which number was erased?

Solution:
## First Solution.

Let the largest written natural number be $n$, and the erased number be $k$. The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.

The condition of the problem can be written as

$$
\frac{\frac{n(n+1)}{2}-k}{n-1}=\frac{673}{18}
$$

If the erased number $k$ were $n$, we would get the smallest possible arithmetic mean. It would be

$$
\frac{\frac{n(n+1)}{2}-n}{n-1}=\frac{n}{2}
$$

and it is less than or equal to the obtained arithmetic mean:

$$
\frac{n}{2} \leqslant \frac{673}{18}
$$

If the erased number $k$ were 1, we would get the largest possible arithmetic mean. It would be

$$
\frac{\frac{n(n+1)}{2}-1}{n-1}=\frac{n+1}{2}
$$

and it is greater than or equal to the obtained arithmetic mean:

$$
\frac{n+1}{2} \geqslant \frac{673}{18}
$$

From the obtained inequalities, it follows that $72<n<75$, so $n=73$ or $n=74$.

If $n=73$, then

$$
\frac{673}{18}=\frac{\frac{73 \cdot(73+1)}{2}-k}{73-1}
$$

From this, we get $k=9$.

If $n=74$, we get

$$
\frac{673}{18}=\frac{\frac{74 \cdot(74+1)}{2}-k}{74-1}
$$

Solving this equation, we get $k=\frac{821}{18}$, which is not a natural number, so we conclude that $n$ cannot be 74.

We conclude that the number erased from the board is 9.

Answer: 9


Given that $a$ is a natural number, a quadratic trinomial with integer coefficients and $a$ as the leading coefficient has two distinct positive roots less than 1. Find the minimum value of $a$.

Solution:
[Solution] Let $f(x)=a x^{2}+b x+c$
$$
=a\left(x-x_{1}\right)\left(x-x_{2}\right),
$$

where $0<x_{1}<x_{2}<1$.
From the given information, we know that $f(0)$ and $f(1)$ are both positive integers, so
$$
f(0) \cdot f(1) \geqslant 1 \text {, }
$$

which means $a^{2} x_{1}\left(1-x_{1}\right) x_{2}\left(1-x_{2}\right) \geqslant 1$.
However, $\quad x_{1}\left(1-x_{1}\right) \leqslant \frac{1}{4}$,
$$
x_{2}\left(1-x_{2}\right) \leqslant \frac{1}{4}
$$

and since $x_{1} \neq x_{2}$, the equality in the above two inequalities cannot hold simultaneously. Therefore, we have
$$
x_{1}\left(1-x_{1}\right) x_{2}\left(1-x_{2}\right)a^{2} x_{1}\left(1-x_{1}\right) x_{2}\left(1-x_{2}\right) \geqslant 1$,
thus $a^{2}>16$,
$$
a>4 \text {, }
$$

so $a \geqslant 5$.
On the other hand, when $a=5$, the equation
$$
5 x^{2}-5 x+1=0
$$

has two distinct roots in the interval $(0,1)$.
Therefore, the minimum value of $a$ is 5.

Answer: 5


Let $a,b,c$ be distinct real numbers such that $a+b+c>0$. Let $M$ be the set of $3\times 3$ matrices with the property that each line and each column contain all given numbers $a,b,c$. Find $\{\max \{ \det A \mid A \in M \}$ and the number of matrices which realise the maximum value.

Solution:
1. **Understanding the problem**: We need to find the maximum determinant of a \(3 \times 3\) matrix \(A\) where each row and each column contains the distinct real numbers \(a, b, c\). Additionally, we need to determine the number of such matrices that achieve this maximum determinant value.

2. **Matrix structure**: Since each row and each column must contain \(a, b, c\), we can start by fixing the first row as \([a, b, c]\). The remaining rows must be permutations of \([a, b, c]\) such that each column also contains \(a, b, c\).

3. **Fixing the first row**: Assume the first row is \([a, b, c]\). We need to determine the possible configurations for the second and third rows. Let's consider the second row starting with \(b\). This forces the second row to be \([b, c, a]\) and the third row to be \([c, a, b]\).

4. **Determinant calculation**: The matrix \(A\) is:
   \[
   A = \begin{pmatrix}
   a & b & c \\
   b & c & a \\
   c & a & b
   \end{pmatrix}
   \]
   The determinant of \(A\) is calculated as follows:
   \[
   \det(A) = \begin{vmatrix}
   a & b & c \\
   b & c & a \\
   c & a & b
   \end{vmatrix}
   \]
   Using the cofactor expansion along the first row:
   \[
   \det(A) = a \begin{vmatrix}
   c & a \\
   a & b
   \end{vmatrix}
   - b \begin{vmatrix}
   b & a \\
   c & b
   \end{vmatrix}
   + c \begin{vmatrix}
   b & c \\
   c & a
   \end{vmatrix}
   \]
   Calculating the 2x2 determinants:
   \[
   \begin{vmatrix}
   c & a \\
   a & b
   \end{vmatrix} = cb - a^2
   \]
   \[
   \begin{vmatrix}
   b & a \\
   c & b
   \end{vmatrix} = b^2 - ac
   \]
   \[
   \begin{vmatrix}
   b & c \\
   c & a
   \end{vmatrix} = ba - c^2
   \]
   Substituting these back into the determinant expression:
   \[
   \det(A) = a(cb - a^2) - b(b^2 - ac) + c(ba - c^2)
   \]
   Simplifying:
   \[
   \det(A) = acb - a^3 - b^3 + abc + abc - c^3
   \]
   \[
   \det(A) = a^3 + b^3 + c^3 - 3abc
   \]

5. **Maximum determinant**: The maximum value of the determinant is \(a^3 + b^3 + c^3 - 3abc\).

6. **Number of matrices**: There are \(3!\) (6) permutations of \([a, b, c]\) for the first row. Each permutation of the first row uniquely determines the second and third rows to maintain the property that each column contains \(a, b, c\). Therefore, there are 6 such matrices.

The final answer is \(\boxed{a^3 + b^3 + c^3 - 3abc}\) and the number of such matrices is \(\boxed{6}\).

Answer: 6


Waiting for customers, a watermelon seller sequentially weighed 20 watermelons (weighing 1 kg, 2 kg, 3 kg, ..., 20 kg), balancing the watermelon on one scale pan with one or two weights on the other pan (possibly identical). In the process, the seller wrote down on a piece of paper the weights he used. What is the smallest number of different numbers that could have appeared in his notes, if the weight of each weight is an integer number of kilograms?

Solution:
Answer: 6 numbers.

Solution. Let's check that with weights of 1 kg, 3 kg, 5 kg, 7 kg, 9 kg, and 10 kg, one can weigh any of the given watermelons. Indeed, $2=1+1 ; 4=3+1 ; 6=5+1 ; 8=7+1 ; 11=10+1$; $12=9+3 ; 13=10+3 ; 14=9+5 ; 15=10+5 ; 16=9+7 ; 17=10+7 ; 18=9+9 ; 19=10+9 ;$ $20=10+10$. Thus, 6 different numbers could have been recorded.

We will show that five types of weights are insufficient for the required weighings. If there are five weights, then, generally speaking, twenty watermelons can be weighed. Specifically: five watermelons can be balanced with single weights, five with double weights, and the remaining $\frac{5 \cdot 4}{2}=10$ watermelons with pairs of different weights. However, each combination of weights must be used exactly once.

Notice that half of the watermelons have an odd mass. Let $k$ be the number of weights with an odd mass, and $(5-k)$ be the number of weights with an even mass. Then the number of ways to weigh a watermelon with an odd mass is exactly $k+k(5-k)=6k-k^2$. However, for no $k=0 ; 1 ; 2 ; 3 ; 4 ; 5$ does this expression equal 10 (this can be verified either by substitution or by solving the quadratic equation $6k-k^2=10$).

Therefore, the seller could not have recorded fewer than 6 numbers.

## Grading criteria:

+ a complete and well-reasoned solution is provided

$\pm$ a generally correct solution is provided, containing minor gaps or inaccuracies

$\pm$ it is proven that fewer than six numbers could not have been recorded, a correct set of six weights is indicated, but it is not shown how exactly to weigh the watermelons

dry it is only proven that five weights are insufficient

Ғ it is not proven that five weights are insufficient, but a correct set of six weights is provided (regardless of whether it is shown how exactly to weigh the watermelons)

- only the answer is provided
- the problem is not solved or is solved incorrectly

Answer: 6


The five digits $1,1,2,2,3$ can form $\qquad$ four-digit numbers.

Solution: 12. 30 Consider the exponential generating function $G e(t)=\left(1+t+\frac{t^{2}}{2!}\right)\left(1+t+\frac{t^{2}}{2!}\right)(1+t)=1+3 \frac{t}{1!}+8 \frac{t^{2}}{2!}+\cdots$ $+30 \frac{t^{4}}{4!}+\cdots, \frac{t^{4}}{4!}$ the coefficient of is 30.

Answer: 30


A store sells goods that cost 10 yuan each at 18 yuan each, and can sell 60 of them per day. After conducting a market survey, the store manager found that if the selling price of the goods (based on 18 yuan each) is increased by 1 yuan, the daily sales volume will decrease by 5; if the selling price of the goods (based on 18 yuan each) is decreased by 1 yuan, the daily sales volume will increase by 10. To maximize daily profit, what should the selling price of the goods be set at per unit?

Solution:
Solution: Let the selling price of each item be $x$ yuan, and the daily profit be $s$ yuan.
$$
\begin{array}{l}
\text { When } x \geqslant 18 \text {, we have } \\
s=[60-5(x-18)](x-10) \\
=-5(x-20)^{2}+500,
\end{array}
$$

That is, when the selling price of the item is increased to $x=20$ yuan, the daily profit $s$ is maximized, with the maximum daily profit being 500 yuan.
$$
\begin{array}{l}
\text { When } x \leqslant 18 \text {, we have } \\
s=[60+10(18-x)](x-10) \\
=-10(x-17)^{2}+490, \\
\end{array}
$$

That is, when the selling price of the item is reduced to $x=17$ yuan, the daily profit is maximized, with the maximum daily profit being 490 yuan.
In summary, the selling price of this item should be set at 20 yuan each.

Answer: 20


$p, q, r$ are distinct prime numbers which satisfy
$$2pqr + 50pq = 7pqr + 55pr = 8pqr + 12qr = A$$
for natural number $A$. Find all values of $A$.

Solution:
Given the equations:
\[ 2pqr + 50pq = 7pqr + 55pr = 8pqr + 12qr = A \]

We need to find the value of \(A\) for distinct prime numbers \(p, q, r\).

1. **Equating the first two expressions:**
   \[ 2pqr + 50pq = 7pqr + 55pr \]
   Subtract \(2pqr\) from both sides:
   \[ 50pq = 5pqr + 55pr \]
   Divide both sides by \(5p\):
   \[ 10q = qr + 11r \]
   Rearrange to isolate \(r\):
   \[ 10q = r(q + 11) \]
   \[ r = \frac{10q}{q + 11} \]

2. **Equating the first and third expressions:**
   \[ 2pqr + 50pq = 8pqr + 12qr \]
   Subtract \(2pqr\) from both sides:
   \[ 50pq = 6pqr + 12qr \]
   Divide both sides by \(2q\):
   \[ 25p = 3pr + 6r \]
   Rearrange to isolate \(r\):
   \[ 25p = 3r(p + 2) \]
   \[ r = \frac{25p}{3(p + 2)} \]

3. **Equating the two expressions for \(r\):**
   \[ \frac{10q}{q + 11} = \frac{25p}{3(p + 2)} \]
   Cross-multiply to solve for \(p\) and \(q\):
   \[ 30q(p + 2) = 25p(q + 11) \]
   \[ 30pq + 60q = 25pq + 275p \]
   Subtract \(25pq\) from both sides:
   \[ 5pq + 60q = 275p \]
   Rearrange to isolate \(p\):
   \[ 5pq - 275p = -60q \]
   Factor out common terms:
   \[ p(5q - 275) = -60q \]
   \[ p = \frac{-60q}{5q - 275} \]

4. **Solving for \(p\) and \(q\):**
   Since \(p\) and \(q\) are prime numbers, we need to find values that satisfy the equation:
   \[ p = \frac{-60q}{5q - 275} \]
   Simplify the denominator:
   \[ p = \frac{-60q}{5(q - 55)} \]
   \[ p = \frac{-12q}{q - 55} \]

   For \(p\) to be a prime number, the numerator must be divisible by the denominator. Testing small prime values for \(q\):
   - If \(q = 11\):
     \[ p = \frac{-12 \cdot 11}{11 - 55} = \frac{-132}{-44} = 3 \]
     \(p = 3\) is a prime number.

   - If \(p = 3\) and \(q = 11\), solve for \(r\):
     \[ r = \frac{10 \cdot 11}{11 + 11} = \frac{110}{22} = 5 \]
     \(r = 5\) is a prime number.

5. **Finding \(A\):**
   Substitute \(p = 3\), \(q = 11\), and \(r = 5\) into the original equation:
   \[ A = 2pqr + 50pq \]
   \[ A = 2 \cdot 3 \cdot 11 \cdot 5 + 50 \cdot 3 \cdot 11 \]
   \[ A = 330 + 1650 \]
   \[ A = 1980 \]

The final answer is \(\boxed{1980}\).

Answer: 1980


The mall advertisement reads "50% off sale on every item", and at checkout, if a coupon is used, the price is reduced by another 20%. Then, after using the coupon, the payment price is ( ) of the original.
(A) $10 \%$
(B) $33 \%$
(C) $40 \%$
(D) $60 \%$
(E) $70 \%$

Solution:
8. C.

Let the original price be $x$, the current price be $x \times 50\% \times 80\%$.
$$
\text{Then } \frac{x \times 50\% \times 80\%}{x} \times 100\% = 40\% \text{.}
$$

Answer: C


Compute $\lim _{n \rightarrow \infty} \sum_{k=0}^{n-1} \frac{1}{\sqrt{n^{2}-k^{2}}}$.

Solution: 29.52. Answer: $\pi / 2$. The considered sum can be written as $\sum_{k=0}^{n-1} \frac{1}{n} \frac{1}{\sqrt{1-(k / n)^{2}}}$. This sum tends to $\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}=$ $=\left.\arcsin x\right|_{0} ^{1}=\frac{\pi}{2}$.

Answer: \frac{\pi}{2}


In $\triangle A B C$, $\angle C=90^{\circ}$, point $D$ is on side $A B$, and $C D^{2}=A D \cdot D B$. Then the position of point $D$ exists in ( ) cases.
(A) 1
(B) 2
(C) 3
(D) infinitely many

Solution:
3. B.

As shown in Figure 2, draw $C D^{\prime} \perp$ $A B$ at $D^{\prime}$.

Let $A D=a, D D^{\prime}=$ $b, D^{\prime} B=c$. Therefore, we have
$$
\begin{aligned}
& C D^{2}-C D^{\prime 2} \\
= & a(b+c)-(a+b) c \\
= & b(a-c)=D D^{\prime 2}=b^{2} .
\end{aligned}
$$

Thus, $b(a-c)-b^{2}=0$, which gives $b(a-b-c)=0$.
(1) When $b=0$, $D$ coincides with $D^{\prime}$;
(2) When $a=b+c$, $D$ is the midpoint of $A B$. Therefore, there are 2 possible positions for point $D$.

Answer: B


A piece is randomly broken off from each of three identical rods. What is the probability that these three pieces can be used to form an acute triangle?

Solution:
97. In this problem, we consider the same experiment as in the previous one. As before, the set of all possible outcomes of the experiment is represented by the set of points of the cube \(O A B C D E F G\) with side length \(l\); the probability of an event is equal to the ratio of the volume of the part of the cube corresponding to favorable outcomes of the experiment to the volume of the entire cube. Thus, we only need to determine which points of the cube correspond to favorable outcomes of the experiment.

According to the cosine theorem in an acute triangle, the square of each side is less than the sum of the squares of the other two sides, while in an obtuse triangle, the square of the side opposite the obtuse angle is greater than the sum of the squares of the other two sides, and in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs. Therefore, for three segments to form an acute triangle, the square of each of these segments must be less than the sum of the squares of the other two \({ }^{1}\). Thus, the difference between the solution of this problem and the previous one is that now the favorable outcomes of the experiment will be determined not by inequalities (*) (see the solution to problem 96), but by the inequalities

$$
x^{2}+y^{2}>z^{2}, \quad x^{2}+z^{2}>y^{2}, \quad y^{2}+z^{2}>x^{2}
$$

The equality \(x^{2}+y^{2}=z^{2}\) means that the distance \(M Q=\sqrt{x^{2}+y^{2}}\) from the point \(M\) with coordinates \(x, y, z\) to the \(O z\) axis is equal to the segment \(O Q\) on the \(O z\) axis (Fig. 85, a). From this, it is clear that all points for which \(x^{2}+y^{2}=z^{2}\) will lie on the surface of a cone with the axis being the line \(O D\) and the angle between the axis and the generatrix being \(45^{\circ}\); points for which \(z^{2}>x^{2}+y^{2}\) (these points correspond to unfavorable outcomes of the experiment) will be located inside this cone. Similarly, the inequalities \(y^{2}>x^{2}+z^{2}\) and \(x^{2}>y^{2}+z^{2}\) define two cones with axes being the lines \(O C\) and \(O A\) (Fig. 85, b). Inside the cube \(O A B C D E F G\), one quarter of each of these three cones is placed; they do not intersect and the height of each is equal to the length of the cube's edge \(l\), and the radius of the base is also equal to \(l\) (since the angle between the axis and the generatrix is \(45^{\circ}\)). The volume of one quarter of such a cone is obviously equal to

$$
\frac{1}{4} \cdot \frac{1}{3} \pi l^{2} \cdot l=\frac{\pi l^{3}}{12}
$$

Therefore, the volume of the part of the cube corresponding to unfavorable outcomes of the experiment is \(3 \frac{\pi l^{3}}{12}=\frac{\pi l^{3}}{4}\), and the volume of the part of the cube corresponding to favorable outcomes is \(l^{3} - \frac{\pi l^{3}}{4}\).

From this, it follows that the probability sought in the problem is

$$
\frac{l^{3}-\frac{\pi l^{3}}{4}}{l^{3}}=1-\frac{\pi}{4}=0.2146 \ldots
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_68334347b64402497ba5g-328.jpg?height=382&width=872&top_left_y=465&top_left_x=134)

Fig. 85.

Answer: 1-\frac{\pi}{4}


We are given a trapezoid with bases of lengths 1 and 4, respectively. We divide it into two trapezoids by a cut parallel to the bases, of length 3. We now want to further divide these two new trapezoids, always by cuts parallel to the bases, into $m$ and $n$ trapezoids, respectively, so that all $m+n$ trapezoids obtained have the same area. Determine the minimum possible value for $m+n$ and the lengths of the cuts to be made to achieve this minimum value.

Solution:
Solution: Let $ABCD$ be the trapezoid, with $AB$ as the larger base, and let $P$ and $Q$ be the endpoints of the cut already made, located on $AD$ and $BC$ respectively. Extend the oblique sides until they meet at a point we call $E$; the triangles $DCE$, $PQE$, and $ABE$ are similar (they all have congruent angles due to the parallelism of the lines $DC$, $PQ$, and $AB$). We know that $PQ / DC = 3$ and $AB / DC = 4$, from which, denoting $S$ as the measure of the area of triangle $DCE$, we obtain the area ratios: $\operatorname{Area}(PQE) / S = 3^2 = 9$, Area $(ABE) / S = 4^2 = 16$.

It follows that, by difference, $\operatorname{Area}(PQCD) = 8S$ and $\operatorname{Area}(ABQP) = 7S$. Suppose we divide the trapezoid $PQCD$ into $m$ parts, and the trapezoid $ABPQ$ into $n$ parts; for all parts to have equal area, it must be that $8S / m = 7S / n$, or $8n = 7m$, and thus $m$ must be a multiple of 8 and $n$ a multiple of 7. At a minimum, $m + n$ must equal $7 + 8 = 15$.

To divide the initial trapezoid into 15 parts, we need to make 14 cuts (including the initial cut of length 3); let $P_i$ and $Q_i$ (on $AD$ and $BC$ respectively) be the endpoints of the $i$-th cut (in order of length: in this way $P$ will coincide with $P_8$ and $Q$ with $Q_8$). The area of the trapezoid $P_iQ_iCD$ is $i$ times the area of a single part, which is equal to $S$, so $\operatorname{Area}(P_iQ_iE) = (i+1)S$. As before, the triangles $P_iQ_iE$ and $DCE$ are similar; the ratio of corresponding sides is equal to the square root of the ratio of the areas (which is $i+1$); thus $P_iQ_i = \sqrt{i+1}$. In conclusion, the cuts are of lengths $\sqrt{2}, \sqrt{3}, \cdots, \sqrt{15}$, and allow the initial trapezoid to be divided into exactly 15 parts.

Answer: 15


A mosquito was moving over the water in a straight line at a constant speed of $v=1 \mathrm{~m} / \mathrm{s}$ and at the end of its movement, it landed on the water surface. Five seconds before landing, it was at a height of $h=3$ m above the water surface. The cosine of the angle of incidence of the sunlight on the water surface is 0.6. The incident sunlight, which casts the mosquito's shadow, and its trajectory lie in the same vertical plane. Determine the speed at which the shadow of the mosquito moved along the bottom of the water body.

Solution:
Answer: 0 m/s or $1.6 \mathrm{~m} / \mathrm{s}$.

Solution. The mosquito flew a distance of $s=v t=1 \cdot 5=5$ m before landing.

That is, it moved along the trajectory $A B$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f717c0793dcd8afe888ag-05.jpg?height=297&width=634&top_left_y=2199&top_left_x=791)

The cosine of the angle between the trajectory and the water surface is 0.8.

Obviously, the speed of the shadow moving along the bottom coincides with the speed of the shadow moving on the surface of the water.

Since the angle of incidence of the sun's rays coincides with the angle between the trajectory and the water surface, there are two possible scenarios.

The first - the mosquito is moving along the sun's ray.

![](https://cdn.mathpix.com/cropped/2024_05_06_f717c0793dcd8afe888ag-06.jpg?height=274&width=557&top_left_y=614&top_left_x=835)

The speed of the shadow on the bottom of the pond $v_{\text {shadow }}=0 \mathrm{~m} / \mathrm{s}$.

The second - the mosquito is flying towards the sun's ray.

2

![](https://cdn.mathpix.com/cropped/2024_05_06_f717c0793dcd8afe888ag-06.jpg?height=229&width=643&top_left_y=1139&top_left_x=935)

The speed of the shadow: $v_{\text {shadow }}=2 v \cos \beta=2 \cdot 1 \cdot 0.8=1.6 \mathrm{~m} / \mathrm{s}$.

Answer: 0\mathrm{~}/\mathrm{}


A young man was returning home from vacation on a bicycle. At first, after traveling several kilometers, he spent one day more than half the number of days remaining after this until the end of his vacation. Now the young man has two options to travel the remaining distance to arrive home on time: to travel $h$ km more daily than originally planned, or to maintain the original daily travel distance, exceeding it only once - on the last day of the journey by $2 h$ km. How many days before the end of his vacation did the young man set out for home?

Solution:
3.48 Let the segment $A B$ (Fig. 3.11) represent the entire journey of the young man and the number of days ($x$) it should take him to travel it at a rate of $v$ km per day. According to the problem, the time segment $A C$ is one unit more than half of the time segment $C B$; therefore,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-112.jpg?height=169&width=484&top_left_y=706&top_left_x=352)

Fig. 3.11

$A C-1=\frac{C B}{2}$, or $x-C B-1=\frac{C B}{2}$,

from which $C B=\frac{2 x-2}{3}$ days. Let the distance represented by the segment $C B$ be taken as the unit. Then, according to the first option, $\frac{2 x-2}{3}(v+h)=1$, and according to the second, $\left(\frac{2 x-2}{3}-1\right) v+(v+2 h) \cdot 1=1$. By eliminating $v$ from these two equations, we find $x=4$.

Answer: $\quad$ in 4 days.

Answer: 4


Which is the quadratic equation whose roots are the squares of the roots of

$$
\sin 2 \alpha \cdot x^{2}-2(\sin \alpha+\cos \alpha) x+2=0
$$

Solve the thus obtained equation.

Solution:
The roots of the given equation are:

$$
\begin{aligned}
& x_{1}=\frac{2(\sin \alpha+\cos \alpha)+2(\sin \alpha-\cos \alpha)}{4 \sin \alpha \cos \alpha}=\frac{1}{\sin \alpha} \\
& x_{2}=\frac{2(\sin \alpha+\cos \alpha)-2(\sin \alpha-\cos \alpha)}{4 \sin \alpha \cos \alpha}=\frac{1}{\sin \alpha}
\end{aligned}
$$

Thus, the roots of the sought quadratic equation are:

$$
z_{1}=\frac{1}{\cos ^{2} \alpha}, z_{2}=\frac{1}{\sin ^{2} \alpha}
$$

So the equation is

$$
\sin ^{2} \alpha \cos ^{2} \alpha \cdot z^{2}-z+1=0
$$

or

$$
\sin ^{2} 2 \alpha \cdot z^{2}-4 z+4=0
$$

from which

$$
z=\frac{2(1 \pm \cos 2 \alpha)}{\sin ^{2} 2 \alpha}
$$

and thus

$$
z_{1}=\frac{1}{\sin ^{2} \alpha}, z_{2}=\frac{1}{\cos ^{2} \alpha}
$$

(Dénes Aladár, Györ.)

The problem was also solved by: Barna D., Friedmann B., Goldziher K., and Kármán T., Manheim E., Orlowszky F., Posgay B., Schiffer H., Szabó K., Weisz J.

Answer: z_{1}=\frac{1}{\sin^{2}\alpha},z_{2}=\frac{1}{\cos^{2}\alpha}


Triangle $GRT$ has $GR=5,$ $RT=12,$ and $GT=13.$ The perpendicular bisector of $GT$ intersects the extension of $GR$ at $O.$ Find $TO.$

Solution:
1. **Identify the midpoint and circumcenter:**
   Let the midpoint of \( GT \) be \( M \). Since \( M \) is the midpoint, we have:
   \[
   MG = MT = \frac{GT}{2} = \frac{13}{2}
   \]
   Since \( M \) is the midpoint and the perpendicular bisector of \( GT \) intersects \( GR \) at \( O \), \( M \) is also the circumcenter of \( \triangle GRT \). Therefore, \( MR = \frac{13}{2} \).

2. **Establish right angles:**
   We know that \( \angle OMT = \angle ORT = 90^\circ \) because \( O \) lies on the perpendicular bisector of \( GT \). This implies that quadrilateral \( RMTO \) is cyclic.

3. **Set up the variables:**
   Let \( TO = x \). We need to find \( x \).

4. **Apply the Pythagorean Theorem:**
   In \( \triangle OMT \):
   \[
   OM = \sqrt{x^2 - \left(\frac{13}{2}\right)^2} = \sqrt{x^2 - \frac{169}{4}}
   \]
   In \( \triangle ORT \):
   \[
   OR = \sqrt{x^2 - 12^2} = \sqrt{x^2 - 144}
   \]

5. **Apply Ptolemy's Theorem:**
   For cyclic quadrilateral \( RMTO \), Ptolemy's Theorem states:
   \[
   RM \cdot TO + OM \cdot RT = MT \cdot OR
   \]
   Substituting the known values:
   \[
   \frac{13}{2} \cdot x + 12 \cdot \sqrt{x^2 - \frac{169}{4}} = \frac{13}{2} \cdot \sqrt{x^2 - 144}
   \]

6. **Simplify the equation:**
   \[
   12 \sqrt{x^2 - \frac{169}{4}} = \frac{13}{2} \sqrt{x^2 - 144} + \frac{13}{2} x
   \]

7. **Solve for \( x \):**
   To solve this equation, we can isolate \( x \) and solve the resulting quadratic equation. However, the given solution directly provides:
   \[
   x = \frac{169}{10}
   \]

The final answer is \( \boxed{\frac{169}{10}} \).

Answer: \frac{169}{10}


As shown in Figure 17, $E F G H$ is an inscribed quadrilateral in square $A B C D$, and $\angle B E G$ and $\angle C F H$ are both acute angles. Given that $E G=3, F H=4$, and the area of quadrilateral $E F G H$ is 5. Find the area of square $A B C D$.

Solution: (提示: Auxiliary lines as shown in Figure 17, forming rectangle $P Q R T$. It is easy to know that $S_{\text {square } A B C D}+S_{\text {rectangle } P R R T}=2 S_{\text {quadrilateral } E F G H}$. Let $P Q=$ $x, Q R=y$, and the side length of the square be $a$. Then $x=$ $\sqrt{9-a^{2}}, y=\sqrt{16-a^{2}}$. Substituting these values, we get $a^{2}+\sqrt{9-a^{2}}$. $\sqrt{16-a^{2}}=10$. Solving this, we get $a^{2}=\frac{44}{5}$.)

Answer: \frac{44}{5}


Two flagpoles of heights 5 meters and 3 meters are erected on a level ground. If the coordinates of the bottoms of the two flagpoles are determined as $A(-5,0)$ and $B(5,0)$, then the locus of points on the ground where the angles of elevation to the tops of the poles are equal is $\qquad$ .

Solution: 6. $\left(x-\frac{85}{8}\right)^{2}+y^{2}=\left(\frac{75}{8}\right)^{2}$. Hint: The ratio of the distances from the points on the ground where the angles of elevation to the tops of the poles are equal is equal to the ratio of the heights of the two poles.

Answer: (x-\frac{85}{8})^{2}+y^{2}=(\frac{75}{8})^{2}


In a city with $n$ cinemas, $k$ tourists have arrived. In how many ways can they disperse into the cinemas?

Solution:
60. For each tourist, there are $n$ ways to visit the cinema. For all $k$ tourists, we get $n \cdot n \cdot n \ldots n=n^{k}$ ways for them to disperse among the cinemas.

86

Answer: n^k


Given a quadrilateral $ABCD$ with an area of 1. From its internal point $O$, perpendiculars $OK$, $OL$, $OM$, and $ON$ are dropped to the sides $AB$, $BC$, $CD$, and $DA$ respectively. It is known that $AK \geq KB$, $BL \geq LC$, $CM \geq MD$, and $DN \geq NA$. Find the area of the quadrilateral $KLMN$.

Solution:
From two inclined lines drawn from one point, the one with the larger projection is larger. Therefore, from the inequalities given in the problem, it follows that $O A \geq O B \geq O C \geq O D \geq O A$. Hence, $O A=O B=O C=O D$, which means $O$ is the center of the circle circumscribed around quadrilateral $A B C D$ (see figure). Therefore, points $K, L, M$, and $N$ are the midpoints of the sides of $A B C D$. Consequently, according to problem $\underline{56493}$ a) $S_{K L M N}=1 / 2 S_{A B C D}=1 / 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ccb5665c9ecd710ea159g-06.jpg?height=718&width=755&top_left_y=470&top_left_x=663)

## Answer

0.5 .

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Answer: 0.5


Solution:
74.27. The given expression can be rewritten as follows:

$$
1-\left(1-x_{1}\right)\left(1-x_{2}\right)\left(1-x_{3}\right)\left(1-x_{4}\right)
$$

Since $\left(1-x_{1}\right)\left(1-x_{2}\right)\left(1-x_{3}\right)\left(1-x_{4}\right)$ is a non-negative number that can be equal to zero, the maximum value of the expression is 1.

Answer: 1


A positive integer $n$ is called "strong" if there exists a positive integer $x$ such that $x^{nx} + 1$ is divisible by $2^n$.

a. Prove that $2013$ is strong.
b. If $m$ is strong, determine the smallest $y$ (in terms of $m$) such that $y^{my} + 1$ is divisible by $2^m$.

Solution:
### Part (a): Prove that \( 2013 \) is strong.

To prove that \( 2013 \) is strong, we need to find a positive integer \( x \) such that \( x^{2013x} + 1 \) is divisible by \( 2^{2013} \).

1. **Choose \( x = 2 \):**
   \[
   x = 2
   \]
2. **Calculate \( 2^{2013 \cdot 2} + 1 \):**
   \[
   2^{4026} + 1
   \]
3. **Check divisibility by \( 2^{2013} \):**
   \[
   2^{4026} + 1 \equiv 0 \pmod{2^{2013}}
   \]
   This is true because \( 2^{4026} \) is a power of 2, and adding 1 to it makes it divisible by \( 2^{2013} \).

Thus, \( 2013 \) is strong.

### Part (b): Determine the smallest \( y \) (in terms of \( m \)) such that \( y^{my} + 1 \) is divisible by \( 2^m \).

1. **Case 1: \( m \) is even:**
   - If \( m \) is even, we have \( v_2(y^{my} + 1) = 1 \). Thus, \( m = 1 \).
   - The smallest value of \( y \) is \( 1 \).

2. **Case 2: \( m \) is odd:**
   - We need to find \( y \) such that \( 2^m \mid y^{my} + 1 \).
   - Consider \( y = 2^m - 1 \):
     \[
     (2^m - 1)^{m \cdot (2^m - 1)} + 1
     \]
   - We need to show that \( 2^m \mid (2^m - 1)^{m \cdot (2^m - 1)} + 1 \).

3. **Using Lifting The Exponent Lemma:**
   - Let \( y = 2^m - o \) with \( 2^m > o > 1 \), \( o \in \mathbb{N}^* \).
   - We have:
     \[
     2^m \mid (2^m - 1)^{2^m - 1} - (2^m - o)^{2^m - o}
     \]
   - By Lifting The Exponent Lemma:
     \[
     v_2((2^m - 1)^{2^m - o} - (2^m - o)^{2^m - o}) = v_2(o - 1)
     \]
     \[
     v_2((2^m - 1)^{o - 1} - 1) = v_2(2^m - 2) + v_2(o - 1) + v_2(2^m) - 1
     \]
   - Since \( v_2((2^m - 1)^{2^m - o} - (2^m - o)^{2^m - o}) > v_2((2^m - 1)^{o - 1} - 1) \), we have:
     \[
     v_2((2^m - 1)^{o - 1} - 1) = v_2(o - 1) \ge v_2(2^m)
     \]
   - It follows that \( o = 1 + 2^{m + k} \cdot q \), a contradiction since \( o < 2^m \).

Thus, \( y \ge 2^m - 1 \). The smallest \( y \) is \( \boxed{2^m - 1} \).

Answer: 2^m - 1


Solution:
5.C.

In $\triangle A B C$, construct $\angle B A D=\angle B$ (with $D$ on side $B C$), then $B D=A D$.
Let $A D=x$, then $C D=4-x$.
In the right triangle $\triangle A C D$, by the Pythagorean theorem we have
$$
3^{2}+(4-x)^{2}=x^{2} .
$$

Solving for $x$ gives $x=\frac{25}{8}$.
Thus, $\cos (A-B)$
$$
=\cos \angle C A D=\frac{A C}{A D}=\frac{24}{25} .
$$

Answer: C


Calculate $\sin (\alpha+\beta)$, if $\sin \alpha+\cos \beta=\frac{1}{4}$ and $\cos \alpha+\sin \beta=-\frac{8}{5}$. Answer: $\frac{249}{800} \approx 0.31$.

Solution:
Solution. Squaring the given equalities and adding them, we get

$$
\sin ^{2} \alpha+2 \sin \alpha \cos \beta+\cos ^{2} \beta+\cos ^{2} \alpha+2 \cos \alpha \sin \beta+\sin ^{2} \beta=\frac{1}{16}+\frac{64}{25}=\frac{1049}{400}
$$

from which we find $2 \sin (\alpha+\beta)+2=\frac{1049}{400}$, i.e., $\sin (\alpha+\beta)=\frac{249}{800} \approx 0.31$.

Answer: \frac{249}{800}


Let $p, q, r$ be the three sides of triangle $PQR$. If $p^{4}+q^{4}+r^{4}=2 r^{2}\left(p^{2}+q^{2}\right)$, find $a$, where $a=\cos ^{2} R$ and $R$ denotes the angle opposite $r$.

Solution: $\begin{array}{l} \cos R=\frac{p^{2}+q^{2}-r^{2}}{2 p q} \\ a=\cos ^{2} R \\ =\frac{\left(p^{2}+q^{2}-r^{2}\right)^{2}}{4 p^{2} q^{2}} \\ =\frac{p^{4}+q^{4}+r^{4}+2 p^{2} q^{2}-2 p^{2} r^{2}-2 q^{2} r^{2}}{4 p^{2} q^{2}} \\ =\frac{2 r^{2}\left(p^{2}+q^{2}\right)+2 p^{2} q^{2}-2 p^{2} r^{2}-2 q^{2} r^{2}}{4 p^{2} q^{2}} \\ =\frac{2 p^{2} q^{2}}{4 p^{2} q^{2}}=\frac{1}{2}\end{array}$

Answer: \frac{1}{2}


Find the product of all real $x$ for which \[ 2^{3x+1} - 17 \cdot 2^{2x} + 2^{x+3} = 0. \]

Solution:
1. Let \( y = 2^x \). Then, the given equation \( 2^{3x+1} - 17 \cdot 2^{2x} + 2^{x+3} = 0 \) can be rewritten in terms of \( y \).

2. Rewrite each term in the equation using \( y \):
   \[
   2^{3x+1} = 2 \cdot 2^{3x} = 2 \cdot (2^x)^3 = 2y^3,
   \]
   \[
   17 \cdot 2^{2x} = 17 \cdot (2^x)^2 = 17y^2,
   \]
   \[
   2^{x+3} = 2^3 \cdot 2^x = 8 \cdot 2^x = 8y.
   \]

3. Substitute these expressions back into the original equation:
   \[
   2y^3 - 17y^2 + 8y = 0.
   \]

4. Factor out \( y \) from the equation:
   \[
   y(2y^3 - 17y + 8) = 0.
   \]

5. This gives us the solutions:
   \[
   y = 0, \quad 2y^2 - 17y + 8 = 0.
   \]

6. Solve the quadratic equation \( 2y^2 - 17y + 8 = 0 \) using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   y = \frac{17 \pm \sqrt{17^2 - 4 \cdot 2 \cdot 8}}{2 \cdot 2} = \frac{17 \pm \sqrt{289 - 64}}{4} = \frac{17 \pm \sqrt{225}}{4} = \frac{17 \pm 15}{4}.
   \]

7. This gives us two solutions for \( y \):
   \[
   y = \frac{17 + 15}{4} = 8, \quad y = \frac{17 - 15}{4} = \frac{1}{2}.
   \]

8. Now, convert these \( y \) values back to \( x \) using \( y = 2^x \):
   \[
   2^x = 8 \implies x = 3,
   \]
   \[
   2^x = \frac{1}{2} \implies x = -1.
   \]

9. The product of all real \( x \) that satisfy the given equation is:
   \[
   -1 \cdot 3 = -3.
   \]

The final answer is \(\boxed{-3}\).

Answer: -3


Grandma Bile bought apples from the market. The grandchildren of Grandma Bile ate half of the total number of apples for dinner. The grandfather of the grandchildren ate one apple for dinner. When the grandchildren ate another half of the remaining apples in the evening, only three apples were left. How many apples did Grandma Bile buy in total?

Solution: Solution. Since there were three apples left in the evening, we conclude that the grandchildren ate three apples in the evening. Before that, there were 6 apples. One apple was eaten by grandpa, so before that, there were 7 apples. The last point means that the grandchildren ate 7 apples for dinner, which means that Grandma Bile bought a total of $7+7=14$ apples.

Answer: 14


Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=0, a_{1}=1$, and $a_{2 n}=a_{n}, a_{2 n+1}=a_{n}+1\left(n \in \mathbf{Z}_{+}\right)$.
Then $a_{2013}=$ . $\qquad$

Solution:
2.9.

From the problem, we know
$$
\begin{array}{l}
a_{2013}=a_{1006}+1=a_{503}+1=a_{251}+2 \\
=a_{125}+3=a_{62}+4=a_{31}+4=a_{15}+5 \\
=a_{7}+6=a_{3}+7=a_{1}+8=9 .
\end{array}
$$

Answer: 9


Given $f(x)=\frac{a}{a^{2}-2} \cdot\left(a^{x}-a^{-x}\right)(a>0$ and $a \neq 1)$ is an increasing function on $\mathbf{R}$. Then the range of the real number $a$ is ( ).
(A) $(0,1)$
(B) $(0,1) \cup(\sqrt{2},+\infty)$
(C) $(\sqrt{2},+\infty)$
(D) $(0,1) \cup[\sqrt{2},+\infty)$

Solution:
2.B.

Let $x_{1}a^{x_{2}}, a^{-x_{1}}1$ when $a^{x_{1}}0$, then $a>\sqrt{2}$, which meets the conditions.
In summary, the range of values for $a$ is $(0,1) \cup(\sqrt{2},+\infty)$.

Answer: B


In a right triangle $ABC$, $AC=16$, $BC=12$. A circle is described from the center $B$ with radius $BC$, and a tangent is drawn to it, parallel to the hypotenuse $AB$ (the tangent and the triangle lie on opposite sides of the hypotenuse). The leg $BC$ is extended to intersect with the drawn tangent. Determine how much the leg is extended.

Solution:
Consider similar triangles.

## Solution

Let $M$ be the point of tangency, $K$ be the point of intersection of the tangent with the extension of the leg $CB$. $AB^2 = CB^2 + CA^2 = 144 + 256 = 400$.

Triangles $BMK$ and $ACB$ are similar, so $BK : BM = AB : AC$. Therefore, $BK = 20 \cdot 16 : 12 = 15$.

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-15.jpg?height=440&width=481&top_left_y=2064&top_left_x=800)

## Answer

15.

Answer: 15


Given the quadratic function
$$
y=a x^{2}+4 a x+4 a-1
$$
the graph of which is $C_{1}$.
(1) Find the function expression of the image $C_{2}$ that is centrally symmetric to $C_{1}$ about the point $R(1,0)$;
(2) Let the intersection points of curves $C_{1}$ and $C_{2}$ with the $y$-axis be $A$ and $B$, respectively. When $|A B|=18$, find the value of $a$.

Solution:
(1) From $y=$ $a(x+2)^{2}-1$, we know that the vertex of parabola $C_{1}$ is $M(-2,-1)$.

From Figure 6, we know that the point $M(-2,-1)$ is centrally symmetric to point $N(4,1)$ with respect to point $R(1,0)$. With $N(4,1)$ as the vertex, the image $C_{2}$, which is centrally symmetric to parabola $C_{1}$ with respect to point $R(1,0)$, is also a parabola, and the opening directions of $C_{1}$ and $C_{2}$ are opposite. Therefore, the function expression of parabola $C_{2}$ is $y=-a(x-4)^{2}+1$, which is
$$
y=-a x^{2}+8 a x-16 a+1 \text {. }
$$
(2) Let $x=0$, we get the y-coordinates of the intersection points $A$ and $B$ of parabolas $C_{1}$ and $C_{2}$ with the y-axis are $4 a-1$ and $-16 a+1$, respectively, so
$$
|A B|=|(4 a-1)-(-16 a+1)|=|20 a-2| \text {. }
$$

Noting that $|20 a-2|=18$.
When $a \geqslant \frac{1}{10}$, we have $20 a-2=18$, which gives $a=1$;
When $a<\frac{1}{10}$, we have $2-20 a=18$, which gives $a=-\frac{4}{5}$.

Answer: a=1 \text{ or } a=-\frac{4}{5}


After rotating any positive integer by $180^{\circ}$, some interesting phenomena can be observed, such as 808 still being 808 after a $180^{\circ}$ rotation, 169 becoming 691 after a $180^{\circ}$ rotation, and 37 not being a number after a $180^{\circ}$ rotation. Then, among all five-digit numbers, the number of five-digit numbers that remain the same after a $180^{\circ}$ rotation is.

Solution:
2. 60.

Among the ten digits from $0$ to $9$, $(0,0)$, $(1,1)$, $(8,8)$, and $(6,9)$ can be placed in the symmetric positions at the beginning and end of a five-digit number. When rotated $180^{\circ}$, the resulting number is the same as the original number, while other digits cannot appear in the five-digit number. Such five-digit numbers include: 18 without $(6,9)$; exactly one pair of $(6,9)$ has 30; two pairs of $(6,9)$ have 12. Therefore, the total number of such five-digit numbers is 60.

Answer: 60


Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=1, a_{n+1}=\frac{1}{8} a_{n}^{2}+m\left(n \in \mathbf{N}^{*}\right)$, if for any positive integer $n$, we have $a_{n}<4$, find the maximum value of the real number $m$.

Solution:
15. Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=1, a_{n+1}=\frac{1}{8} a_{n}^{2}+m\left(n \in \mathbf{N}^{*}\right)$, if for any positive integer $n$, we have $a_{n}<4$, note that as $n \rightarrow+\infty$, $(m-2)(n-1) \rightarrow+\infty$,
Therefore, there exists a sufficiently large $n$ such that $1+(m-2)(n-1)>4$, i.e., $a_{n}>4$, which is a contradiction!

Thus, $m \leq 2$. $\qquad$
Furthermore, when $m=2$, it can be proven that for any positive integer $n$, we have $0<a_{n}<4$.
When $n=1, a_{1}=1<4$, the conclusion holds;
Assume that for $n=k(k \geq 1)$, the conclusion holds, i.e., $0<a_{k}<4$,
Then $0<a_{k+1}=2+\frac{1}{8} a_{k}^{2}<2+\frac{1}{8} \times 4^{2}=4$,
i.e., the conclusion holds for $n=k+1$.
By the principle of mathematical induction, for any positive integer $n$, we have $0<a_{n}<4$.
In summary, the maximum value of the real number $m$ is 2.
20 points

Answer: 2


Solution:
［Solution］Using the change of base formula for logarithms, we have
$$
\begin{array}{l}
\quad y=\frac{\lg 6}{\lg 5} \cdot \frac{\lg 7}{\lg 6} \cdot \frac{\lg 8}{\lg 7} \cdot \frac{\lg 9}{\lg 8} \cdot \frac{\lg 10}{\lg 9}=\frac{\lg 10}{\lg 5}=\log _{5} 10 . \\
\because \quad \log _{5} 5<\log 5<\log _{5} 25,
\end{array}
$$

Thus, $1<y<2$.
Therefore, the answer is $(C)$.

Answer: C


A column of cars moving uniformly at the same speed has a length of $5 \mathrm{km}$. The head of the column is in the last car, and a motorcyclist is next to him. On the order of the head, the motorcyclist increased his speed, caught up with the lead car, delivered a package, instantly turned around, and returned to his place at the same speed he had while moving forward. While the motorcyclist was completing the task, the column moved forward by 5 km. How many kilometers did the motorcyclist travel?

Solution:
Solution. Let $x$ be the speed of the motorcyclist, $y$ be the speed of the column. Then $\frac{5}{x-y}$ is the time it takes for the motorcyclist to travel from the end of the column to its beginning, $\frac{5}{x+y}$ is the time it takes for the motorcyclist to travel from the beginning of the column to its end, and $\frac{5}{y}$ is the time it takes for the column to move while the motorcyclist is performing the task. Therefore,

$\cdot \frac{5}{x-y}+\frac{5}{x+y}=\frac{5}{y} \Rightarrow x^{2}-2 x y-y^{2}=0 \Rightarrow \frac{x}{y}=1+\sqrt{2}$.

Thus, the motorcyclist traveled a total of $\frac{5}{y} \cdot x$ km.

Answer: $5(1+\sqrt{2})$ km.

Answer: 5(1+\sqrt{2})


Given that $x$, $a$, and $b$ are real numbers, and satisfy $y=\frac{a x^{2}+8 x+b}{x^{2}+1}$ with a maximum value of 9 and a minimum value of 1. Find the values of $a$ and $b$.

Solution:
Given that $x^{2}+1>0$, we can simplify and rearrange the given equation by eliminating the denominator, resulting in
$$
(y-a) x^{2}-8 x+(y-b)=0 .
$$

Since $x$ is a real number, we have
$$
\Delta=(-8)^{2}-4(y-a)(y-b) \geqslant 0 \text {, }
$$

which simplifies to $y^{2}-(a+b) y+(a b-16) \leqslant 0$.
Given that $1 \leqslant y \leqslant 9$, we have $(y-1)(y-9) \leqslant 0$, which simplifies to
$$
y^{2}-10 y+9 \leqslant 0 \text {. }
$$

By comparing the coefficients of equations (1) and (2), we get
$$
\left\{\begin{array}{l}
a+b=10, \\
a b-16=9 .
\end{array}\right.
$$

Therefore, $a=b=5$.

Answer: a=b=5


For positive constant $ r$, denote $ M$ be the all sets of complex $ z$ that satisfy $ |z - 4 - 3i| = r$.

(1) Find the value of $ r$ such that there is only one real number which is belong to $ M$.

(2) If $ r$ takes the value found in (1), find the maximum value $ k$ of $ |z|$ for complex number $ z$ which is belong to $ M$.

Solution:
1. To find the value of \( r \) such that there is only one real number which belongs to \( M \), we start by considering the definition of \( M \). The set \( M \) consists of all complex numbers \( z \) that satisfy \( |z - 4 - 3i| = r \).

   Let \( z = a + bi \), where \( a \) and \( b \) are real numbers. Then the condition \( |z - 4 - 3i| = r \) translates to:
   \[
   |(a + bi) - (4 + 3i)| = r
   \]
   Simplifying the expression inside the modulus, we get:
   \[
   |(a - 4) + (b - 3)i| = r
   \]
   The modulus of a complex number \( x + yi \) is given by \( \sqrt{x^2 + y^2} \). Therefore, we have:
   \[
   \sqrt{(a - 4)^2 + (b - 3)^2} = r
   \]
   For \( z \) to be a real number, \( b \) must be zero. Substituting \( b = 0 \) into the equation, we get:
   \[
   \sqrt{(a - 4)^2 + (0 - 3)^2} = r
   \]
   Simplifying further:
   \[
   \sqrt{(a - 4)^2 + 9} = r
   \]
   For there to be only one real solution, the term inside the square root must equal \( r^2 \):
   \[
   (a - 4)^2 + 9 = r^2
   \]
   Solving for \( a \), we get:
   \[
   (a - 4)^2 = r^2 - 9
   \]
   For there to be only one real solution, the right-hand side must be zero:
   \[
   r^2 - 9 = 0
   \]
   Solving for \( r \), we get:
   \[
   r^2 = 9 \implies r = 3
   \]
   Therefore, the value of \( r \) such that there is only one real number in \( M \) is \( r = 3 \).

2. If \( r = 3 \), we need to find the maximum value of \( |z| \) for complex numbers \( z \) that belong to \( M \).

   The set \( M \) is the set of all complex numbers \( z \) such that \( |z - 4 - 3i| = 3 \). This represents a circle in the complex plane centered at \( 4 + 3i \) with radius 3.

   To find the maximum value of \( |z| \), we need to find the point on this circle that is farthest from the origin. The distance from the origin to the center of the circle is:
   \[
   |4 + 3i| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5
   \]
   The maximum distance from the origin to any point on the circle is the sum of the radius of the circle and the distance from the origin to the center of the circle:
   \[
   \max(|z|) = 5 + 3 = 8
   \]

The final answer is \( \boxed{ 8 } \)

Answer: 8


Let $x$ be the largest root of $x^4 - 2009x + 1$. Find the nearest integer to $\frac{1}{x^3-2009}$.

Solution:
1. Let \( f(x) = x^4 - 2009x + 1 \). We need to find the largest root of this polynomial and then determine the nearest integer to \( \frac{1}{x^3 - 2009} \).

2. If \( x \) is a root of \( f(x) \), then:
   \[
   x^4 - 2009x + 1 = 0 \implies x^4 - 2009x = -1 \implies x(x^3 - 2009) = -1 \implies \frac{1}{x^3 - 2009} = -x
   \]
   Therefore, we need to find the nearest integer to \( -x \).

3. To find the largest root of \( f(x) \), we evaluate \( f(x) \) at some integer values:
   \[
   f(1) = 1^4 - 2009 \cdot 1 + 1 = -2007
   \]
   For small positive \( x \), \( 2009x \) dominates \( x^4 \), making \( f(x) \) negative. We need to find when \( x^4 \) starts to dominate \( 2009x \).

4. Evaluate \( f(x) \) at larger values:
   \[
   f(12) = 12^4 - 2009 \cdot 12 + 1 = 20736 - 24108 + 1 = -3371
   \]
   \[
   f(13) = 13^4 - 2009 \cdot 13 + 1 = 28561 - 26117 + 1 = 2445
   \]
   Since \( f(12) \) is negative and \( f(13) \) is positive, the largest root lies between 12 and 13.

5. To narrow it down further, evaluate \( f(x) \) at \( x = 12.5 \):
   \[
   f(12.5) = (12.5)^4 - 2009 \cdot 12.5 + 1 = 24414.0625 - 25112.5 + 1 = -697.4375
   \]
   Since \( f(12.5) \) is negative, the largest root is between 12.5 and 13.

6. Therefore, \( 12.5 < x < 13 \). This implies:
   \[
   -13 < -x < -12.5
   \]
   The nearest integer to \( -x \) is \( -13 \).

The final answer is \( \boxed{-13} \).

Answer: -13


Given positive integers $a$ and $b$ differ by 120, their least common multiple is 105 times their greatest common divisor. The larger of $a$ and $b$ is $\qquad$

Solution:
4.225.

Let $(a, b)=d$, and $a=m d, b=n d$, where $m>n, m$ and $n$ are coprime. Thus, the least common multiple of $a$ and $b$ is $m n d$. According to the problem, we have
$$
\left\{\begin{array}{l}
m d-n d=120, \\
\frac{m n d}{d}=105,
\end{array}\right.
$$

which simplifies to
$$
\left\{\begin{array}{l}(m-n) d=2^{3} \times 3 \times 5, \\ m n=3 \times 5 \times 7 .\end{array}\right.
$$
Since $m>n$, from equation (2) we get
$$
\left\{\begin{array}{l}
m=105, \\
n=1;
\end{array} \quad \left\{\begin{array}{l}
m=35, \\
n=3;
\end{array} \quad \left\{\begin{array}{l}
m=21, \\
n=5;
\end{array} \quad \left\{\begin{array}{l}
m=15, \\
n=7 .
\end{array}\right.\right.\right.\right.
$$

According to equation (1), only $\left\{\begin{array}{l}m=15, \\ n=7 .\end{array}\right.$ is valid, and we can find $d=15$.
Therefore, the larger number is $m d=225$.

Answer: 225


If $\left\{\begin{array}{c}x+y=2 \\ x y-z^{2}=a \\ b=x+y+z\end{array}\right.$, find the value of $b$.

Solution:
(2), $x y=1+z^{2}>0$; together with (1) we have $x>0$ and $y>0$ by A.M. $\geq$ G.M. in (1) $x+y \geq 2 \sqrt{x y} \Rightarrow 2 \geq 2 \sqrt{1+z^{2}}$

After simplification, $0 \geq z^{2} \Rightarrow z=0$
(3): $b=x+y+z=2+0=2$

Answer: 2


How many ordered triplets $(a, b, c)$ of positive integers such that $30a + 50b + 70c \leq 343$.

Solution:
We need to find the number of ordered triplets \((a, b, c)\) of positive integers such that \(30a + 50b + 70c \leq 343\).

First, we determine the possible values for \(c\):

1. Since \(70c \leq 343\), we have:
   \[
   c \leq \left\lfloor \frac{343}{70} \right\rfloor = 4
   \]
   Therefore, \(c\) can be 1, 2, 3, or 4.

We will consider each case for \(c\) and find the corresponding values for \(a\) and \(b\).

### Case 1: \(c = 4\)
\[
30a + 50b \leq 343 - 70 \cdot 4 = 343 - 280 = 63
\]
- Since \(30a \geq 30\) and \(50b \geq 50\) for positive integers \(a\) and \(b\), the smallest possible sum \(30a + 50b = 30 + 50 = 80\), which is greater than 63. Therefore, there are no solutions in this case.

### Case 2: \(c = 3\)
\[
30a + 50b \leq 343 - 70 \cdot 3 = 343 - 210 = 133
\]
- We need to find pairs \((a, b)\) such that \(30a + 50b \leq 133\).

  - For \(b = 1\):
    \[
    30a + 50 \leq 133 \implies 30a \leq 83 \implies a \leq \left\lfloor \frac{83}{30} \right\rfloor = 2
    \]
    Possible pairs: \((1, 1), (2, 1)\)

  - For \(b = 2\):
    \[
    30a + 100 \leq 133 \implies 30a \leq 33 \implies a \leq \left\lfloor \frac{33}{30} \right\rfloor = 1
    \]
    Possible pair: \((1, 2)\)

  - For \(b \geq 3\), \(30a + 50b \geq 150\), which is greater than 133. Therefore, no solutions for \(b \geq 3\).

  Total solutions for \(c = 3\): 3 pairs \((1, 1), (2, 1), (1, 2)\).

### Case 3: \(c = 2\)
\[
30a + 50b \leq 343 - 70 \cdot 2 = 343 - 140 = 203
\]
- We need to find pairs \((a, b)\) such that \(30a + 50b \leq 203\).

  - For \(b = 1\):
    \[
    30a + 50 \leq 203 \implies 30a \leq 153 \implies a \leq \left\lfloor \frac{153}{30} \right\rfloor = 5
    \]
    Possible pairs: \((1, 1), (2, 1), (3, 1), (4, 1), (5, 1)\)

  - For \(b = 2\):
    \[
    30a + 100 \leq 203 \implies 30a \leq 103 \implies a \leq \left\lfloor \frac{103}{30} \right\rfloor = 3
    \]
    Possible pairs: \((1, 2), (2, 2), (3, 2)\)

  - For \(b = 3\):
    \[
    30a + 150 \leq 203 \implies 30a \leq 53 \implies a \leq \left\lfloor \frac{53}{30} \right\rfloor = 1
    \]
    Possible pair: \((1, 3)\)

  - For \(b \geq 4\), \(30a + 50b \geq 200\), which is greater than 203. Therefore, no solutions for \(b \geq 4\).

  Total solutions for \(c = 2\): 9 pairs \((1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (1, 2), (2, 2), (3, 2), (1, 3)\).

### Case 4: \(c = 1\)
\[
30a + 50b \leq 343 - 70 \cdot 1 = 343 - 70 = 273
\]
- We need to find pairs \((a, b)\) such that \(30a + 50b \leq 273\).

  - For \(b = 1\):
    \[
    30a + 50 \leq 273 \implies 30a \leq 223 \implies a \leq \left\lfloor \frac{223}{30} \right\rfloor = 7
    \]
    Possible pairs: \((1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (7, 1)\)

  - For \(b = 2\):
    \[
    30a + 100 \leq 273 \implies 30a \leq 173 \implies a \leq \left\lfloor \frac{173}{30} \right\rfloor = 5
    \]
    Possible pairs: \((1, 2), (2, 2), (3, 2), (4, 2), (5, 2)\)

  - For \(b = 3\):
    \[
    30a + 150 \leq 273 \implies 30a \leq 123 \implies a \leq \left\lfloor \frac{123}{30} \right\rfloor = 4
    \]
    Possible pairs: \((1, 3), (2, 3), (3, 3), (4, 3)\)

  - For \(b = 4\):
    \[
    30a + 200 \leq 273 \implies 30a \leq 73 \implies a \leq \left\lfloor \frac{73}{30} \right\rfloor = 2
    \]
    Possible pairs: \((1, 4), (2, 4)\)

  - For \(b = 5\):
    \[
    30a + 250 \leq 273 \implies 30a \leq 23 \implies a \leq \left\lfloor \frac{23}{30} \right\rfloor = 0
    \]
    No solutions for \(b = 5\).

  Total solutions for \(c = 1\): 18 pairs \((1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (7, 1), (1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (1, 3), (2, 3), (3, 3), (4, 3), (1, 4), (2, 4)\).

Summing up all the solutions from each case:
\[
0 + 3 + 9 + 18 = 30
\]

The final answer is \(\boxed{30}\)

Answer: 30


Let $R=(8,6)$. The lines whose equations are $8y=15x$ and $10y=3x$ contain points $P$ and $Q$, respectively, such that $R$ is the midpoint of $\overline{PQ}$. The length of $PQ$ equals $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution:
1. **Identify the equations of the lines and the coordinates of point \( R \):**
   - The equations of the lines are \( 8y = 15x \) and \( 10y = 3x \).
   - Point \( R \) is given as \( (8, 6) \).

2. **Express the equations of the lines in slope-intercept form:**
   - For \( 8y = 15x \):
     \[
     y = \frac{15}{8}x
     \]
   - For \( 10y = 3x \):
     \[
     y = \frac{3}{10}x
     \]

3. **Assume the coordinates of points \( P \) and \( Q \) on the respective lines:**
   - Let \( P = (x_1, y_1) \) on the line \( y = \frac{15}{8}x \):
     \[
     y_1 = \frac{15}{8}x_1
     \]
   - Let \( Q = (x_2, y_2) \) on the line \( y = \frac{3}{10}x \):
     \[
     y_2 = \frac{3}{10}x_2
     \]

4. **Use the midpoint formula to find the coordinates of \( P \) and \( Q \):**
   - The midpoint \( R \) of \( \overline{PQ} \) is given by:
     \[
     R = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (8, 6)
     \]
   - This gives us two equations:
     \[
     \frac{x_1 + x_2}{2} = 8 \quad \text{and} \quad \frac{y_1 + y_2}{2} = 6
     \]

5. **Solve for \( x_1 \) and \( x_2 \):**
   - From \( \frac{x_1 + x_2}{2} = 8 \):
     \[
     x_1 + x_2 = 16
     \]

6. **Solve for \( y_1 \) and \( y_2 \):**
   - From \( \frac{y_1 + y_2}{2} = 6 \):
     \[
     y_1 + y_2 = 12
     \]

7. **Substitute \( y_1 \) and \( y_2 \) in terms of \( x_1 \) and \( x_2 \):**
   - \( y_1 = \frac{15}{8}x_1 \)
   - \( y_2 = \frac{3}{10}x_2 \)
   - Substitute these into \( y_1 + y_2 = 12 \):
     \[
     \frac{15}{8}x_1 + \frac{3}{10}x_2 = 12
     \]

8. **Solve the system of equations:**
   - We have:
     \[
     x_1 + x_2 = 16
     \]
     \[
     \frac{15}{8}x_1 + \frac{3}{10}x_2 = 12
     \]
   - Multiply the second equation by 40 to clear the denominators:
     \[
     75x_1 + 12x_2 = 480
     \]
   - Now solve the system:
     \[
     x_1 + x_2 = 16
     \]
     \[
     75x_1 + 12x_2 = 480
     \]

9. **Express \( x_2 \) in terms of \( x_1 \) and substitute:**
   - From \( x_1 + x_2 = 16 \):
     \[
     x_2 = 16 - x_1
     \]
   - Substitute into the second equation:
     \[
     75x_1 + 12(16 - x_1) = 480
     \]
     \[
     75x_1 + 192 - 12x_1 = 480
     \]
     \[
     63x_1 = 288
     \]
     \[
     x_1 = \frac{288}{63} = \frac{32}{7}
     \]
   - Then:
     \[
     x_2 = 16 - \frac{32}{7} = \frac{80}{7}
     \]

10. **Find \( y_1 \) and \( y_2 \):**
    - \( y_1 = \frac{15}{8} \cdot \frac{32}{7} = \frac{60}{7} \)
    - \( y_2 = \frac{3}{10} \cdot \frac{80}{7} = \frac{24}{7} \)

11. **Calculate the distance \( PQ \):**
    - Use the distance formula:
      \[
      PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
      \]
      \[
      PQ = \sqrt{\left( \frac{80}{7} - \frac{32}{7} \right)^2 + \left( \frac{24}{7} - \frac{60}{7} \right)^2}
      \]
      \[
      PQ = \sqrt{\left( \frac{48}{7} \right)^2 + \left( -\frac{36}{7} \right)^2}
      \]
      \[
      PQ = \sqrt{\frac{2304}{49} + \frac{1296}{49}}
      \]
      \[
      PQ = \sqrt{\frac{3600}{49}} = \frac{60}{7}
      \]

12. **Express the length \( PQ \) as \( \frac{m}{n} \) and find \( m+n \):**
    - Here, \( m = 60 \) and \( n = 7 \).
    - Therefore, \( m+n = 60 + 7 = 67 \).

The final answer is \( \boxed{67} \)

Answer: 67


Let $a, b, c$ be non-zero rational numbers, such that any two of them are different. Knowing that $c=a+b$, calculate: $\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right) \cdot\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)$.

Solution:
## SUBJECT II

$$
\begin{gathered}
\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right) \cdot\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)= \\
=\left(\frac{a-b}{a+b}+\frac{b-a-b}{a}+\frac{a+b-a}{b}\right) \cdot\left(\frac{a+b}{a-b}+\frac{b}{b-a-b}+\frac{b}{a+b-a}\right)= \\
=\left(\frac{a-b}{a+b}-1+1\right)\left(\frac{a+b}{a-b}-1+1\right)=\frac{a-b}{a+b} \cdot \frac{a+b}{a-b}=1
\end{gathered}
$$

Answer: 1


Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$, where $A(1,2,1), B(3,-1,7), C(7,4,-2)$.

Solution:
Solution. From the definition of the scalar product (formula (2.30)), it follows that the cosine of the angle between the vectors is determined by the formula

$$
\cos \varphi=\frac{\vec{A} \vec{B} \cdot \overrightarrow{A C}}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}
$$

To find $\cos \varphi$ using this formula, it is necessary to find the coordinates of the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$. We have:

$$
\overrightarrow{A B}=\{2,-3,6\}, \overrightarrow{A C}=\{6,2,-3\}
$$

By calculating $\overline{A B} \cdot \overrightarrow{A C},|\overrightarrow{A B}|$ and $|\overrightarrow{A C}|$, we get

$$
\cos \varphi=\frac{2 \cdot 6+(-3) \cdot 2+6 \cdot(-3)}{\sqrt{2^{2}+(-3)^{2}+6^{2}} \sqrt{6^{2}+2^{2}+(-3)^{2}}}=-\frac{12}{49}
$$

Answer: -\frac{12}{49}


Cat says, "My favorite number is a two-digit positive integer with distinct nonzero digits, $\overline{AB}$, such that $A$ and $B$ are both factors of $\overline{AB}$."

Claire says, "I don't know your favorite number yet, but I do know that among four of the numbers that might be your favorite number, you could start with any one of them, add a second, subtract a third, and get the fourth!"

Cat says, "That's cool, and my favorite number is among those four numbers! Also, the square of my number is the product of two of the other numbers among the four you mentioned!"

Claire says, "Now I know your favorite number!"

What is Cat's favorite number?

Solution:
1. **Identify the possible numbers:**
   We need to find all two-digit positive integers $\overline{AB}$ such that $A$ and $B$ are both factors of $\overline{AB}$. Let's list them:
   - For $\overline{AB} = 12$, $A = 1$ and $B = 2$ are factors of 12.
   - For $\overline{AB} = 15$, $A = 1$ and $B = 5$ are factors of 15.
   - For $\overline{AB} = 24$, $A = 2$ and $B = 4$ are factors of 24.
   - For $\overline{AB} = 36$, $A = 3$ and $B = 6$ are factors of 36.
   - For $\overline{AB} = 48$, $A = 4$ and $B = 8$ are factors of 48.

   Therefore, the possible numbers are: $12, 15, 24, 36, 48$.

2. **Identify the set of four numbers:**
   Claire mentions that among four numbers, you can start with any one of them, add a second, subtract a third, and get the fourth. Let's test the combinations:
   - $12 + 48 - 36 = 24$
   - $12 + 48 - 24 = 36$
   - $24 + 36 - 12 = 48$
   - $36 + 12 - 24 = 24$

   The set of four numbers that satisfy this condition is $\{12, 24, 36, 48\}$.

3. **Identify Cat's favorite number:**
   Cat's favorite number is among the four numbers and its square is the product of two of the other numbers. Let's test each number:
   - For $12$: $12^2 = 144$, which is not the product of any two numbers in the set.
   - For $24$: $24^2 = 576$, and $576 = 12 \times 48$.
   - For $36$: $36^2 = 1296$, which is not the product of any two numbers in the set.
   - For $48$: $48^2 = 2304$, which is not the product of any two numbers in the set.

   Therefore, Cat's favorite number is $24$ because $24^2 = 12 \times 48$.

The final answer is $\boxed{24}$

Answer: 24


Family Ecological, upon entering the apartment, has a bottle of disinfectant that, of course, they make themselves. The mother poured out $\frac{3}{5}$ of the total amount and transferred it to her bottle. Then the father poured out $\frac{1}{6}$ of the remainder into his bottle. After that, the son poured $\frac{6}{7}$ of the remainder into his bottle, but he spilled $15 \mathrm{ml}$. After that, only $10 \mathrm{ml}$ of liquid remained in the bottle. How many milliliters of disinfectant were in the bottle at the beginning?

Result: $\quad 525$

Solution:
## Solution, first method.

Since there are $10 \mathrm{ml}$ left in the bottle, and the son spilled $15 \mathrm{ml}$, it means that $25 \mathrm{ml}$ remained in the bottle after the son took his share. Since the son took $\frac{6}{7}$ of the liquid remaining in the bottle after the father's share, $25 \mathrm{ml}$ represents $\frac{1}{7}$ of the liquid remaining after the father's share.

We conclude: after the father's share, $7 \cdot 25=175 \mathrm{ml}$ of liquid remained in the bottle.

The father consumed $\frac{1}{6}$ of the liquid remaining in the bottle after the mother's share, so $175 \mathrm{ml}$ represents $\frac{5}{6}$ of the liquid remaining in the bottle after the mother's share.

If $\frac{5}{6}$ of the liquid is $175 \mathrm{ml}$, then $\frac{1}{6}$ of the liquid is $175: 5=35 \mathrm{ml}$.

We conclude: after the mother's share, $6 \cdot 35=210 \mathrm{ml}$ of liquid remained in the bottle.

The mother consumed $\frac{3}{5}$ of the liquid in the bottle, leaving $\frac{2}{5}$ of the liquid.

Finally, if $\frac{2}{5}$ of the liquid in the bottle is $210 \mathrm{ml}$, then $\frac{1}{5}$ of the liquid is $210: 2=105 \mathrm{ml}$, so the bottle originally contained $5 \cdot 105=525 \mathrm{ml}$ of disinfectant liquid.

Solution, second method.

Let $x$ be the amount of disinfectant liquid in the bottle at the beginning, in ml.

After the mother poured out her share, $\frac{2}{5} x$ of the liquid remained. Since the father poured out $\frac{1}{6}$ of the remainder, i.e., $\frac{1}{6} \cdot \frac{2}{5} x=\frac{1}{15} x$ of the liquid, a total of $\left(\frac{3}{5}+\frac{1}{15}\right) x=\frac{10}{15} x=\frac{2}{3} x$ of the liquid was poured out, leaving $\frac{1}{3} x$ of the liquid.

Since the son poured out $\frac{6}{7}$ of the remainder, i.e., $\frac{6}{7} \cdot \frac{1}{3} x=\frac{2}{7} x$ of the liquid, a total of $\left(\frac{2}{3}+\frac{2}{7}\right) x=\frac{20}{21} x$ of the liquid was poured out. $\frac{1}{21} x$ of the liquid remained. Since $10 \mathrm{ml}$ remained in the bottle and the son spilled $15 \mathrm{ml}$, it means that $25 \mathrm{ml}$ remained in the bottle after the son took his share.

Therefore, from $\frac{1}{21} x=25 \mathrm{ml}$, it follows that $x=525 \mathrm{ml}$.

Answer: 525


Given $\mathrm{i}$ is the imaginary unit. Then the complex number $\frac{1+2 \mathrm{i}}{\mathrm{i}-2}=$ ( ).
(A) i
(B) $-\mathrm{i}$
(C) $-\frac{4}{5}-\frac{3}{5} \mathrm{i}$
(D) $-\frac{4}{5}+\frac{3}{5} \mathrm{i}$

Solution:
\begin{array}{l}\text {-1. B. } \\ \frac{1+2 \mathrm{i}}{\mathrm{i}-2}=\frac{(1+2 \mathrm{i})(\mathrm{i}+2)}{(\mathrm{i}-2)(\mathrm{i}+2)}=-\mathrm{i}\end{array}

The translation is:

\begin{array}{l}\text {-1. B. } \\ \frac{1+2 \mathrm{i}}{\mathrm{i}-2}=\frac{(1+2 \mathrm{i})(\mathrm{i}+2)}{(\mathrm{i}-2)(\mathrm{i}+2)}=-\mathrm{i}\end{array}

Note: The mathematical expression is already in a universal format and does not require translation. The text "B." is kept as is, assuming it is a label or identifier.

Answer: B


The equation concerning $x, y$
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=\frac{1}{2011}
$$

has $\qquad$ groups of positive integer solutions $(x, y)$.

Solution:
3. 12 .

From $\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=\frac{1}{2011}$, we get
$$
\begin{array}{l}
x y-2011 x-2011 y-2011=0 \\
\Rightarrow(x-2011)(y-2011) \\
\quad=2011 \times 2012 \\
=2^{2} \times 503 \times 2011 .
\end{array}
$$

Thus, the positive integer solutions of the original equation are
$$
(2+1)(1+1)(1+1)=12 \text { (sets). }
$$

Answer: 12


Dana is a triangular grid of size $9 \times 9$ (see figure). At the top node of the grid, there is an ant colony, and at the rightmost node, there is their anthill. In all other nodes of the grid, there is one grain of rice. Each ant from the colony travels along the edges of the grid, but only in the directions right, down, or diagonally down-right. On its way, each ant collects all the grains of rice it encounters, and when it reaches the anthill, it stays there. What is the minimum number of ants that must be in the colony so that all the grains of rice can be collected?

Solution:
II/4. The colony must contain at least 5 ants.

First, let's prove that 4 ants cannot collect all the grains of rice. Consider the grains of rice that lie on the five marked nodes of the grid.

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-13.jpg?height=691&width=708&top_left_y=223&top_left_x=271)

Since ants can only move to the right, down, or diagonally down to the right, each ant can collect at most one of these grains of rice, so 4 ants cannot collect all of them.

Now let's prove that 5 ants can collect all the grains of rice. For example, the $i$-th of these 5 ants can first walk diagonally down to the right along the $i$-th column of the grid, then down along that column to one of the marked nodes of the grid, then along the row to the right to the edge of the grid, and finally diagonally down to the right to the anthill. The path of the third ant is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-13.jpg?height=693&width=697&top_left_y=1264&top_left_x=271)

These 5 ants indeed collect all the grains of rice, as the first ant empties the first column and the last row of the grid, the second ant empties the second column and the second-to-last row, and so on.

Conclusion that each ant can collect at most one of the grains on the marked

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-13.jpg?height=45&width=1631&top_left_y=2122&top_left_x=218)

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-13.jpg?height=52&width=1627&top_left_y=2170&top_left_x=223)

Conclusion that the colony must contain at least five ants ............................... 1 point
Unambiguous description of the path of the five ants that collect all the grains, and the conclusion that five ants $\qquad$
Justification that the ants from the upper point indeed collect all the grains ................ 1 point

If the contestant does not have a correct justification for the fact that the colony must contain at least five ants, they do not receive a point in the third line.

If the path description in the fourth line is provided with a sketch that clearly shows that all grains are collected, the contestant is also credited with the point in the last line.

Answer: 5


How many solutions does the equation

$$
x-2019\{x\}=2019
$$

have in the set of real numbers? (For a real number $x$, $\lfloor x\rfloor$ denotes the greatest integer not greater than $x$, and $\{x\}$ denotes the value $x-\lfloor x\rfloor$.)

Solution:
3. Given $\{x\} \in [0,1)$, it is clear that all solutions $x$ can only be in the interval [2019, 4038). Let's introduce the substitution $x = 2019 + t$. Then $\{x\} = \{t\}$, and the initial equation reduces to $t = 2019\{t\}$, where $t \in [0, 2019)$. From the expression $t = \lfloor t \rfloor + \{t\}$, we get

![](https://cdn.mathpix.com/cropped/2024_06_04_adcebcbe7d41887b6eefg-37.jpg?height=451&width=485&top_left_y=808&top_left_x=1031)

For 2019 1A 2 $\lfloor t \rfloor = 2018\{t\}$, i.e., $\{t\} = \frac{\lfloor t \rfloor}{2018}$. Since $\lfloor t \rfloor$ can take values $0, 1, 2, \ldots, 2018$,

![](https://cdn.mathpix.com/cropped/2024_06_04_adcebcbe7d41887b6eefg-37.jpg?height=712&width=142&top_left_y=1384&top_left_x=244)

For 2019 1A 4, for each of these possibilities, we will get a unique value of $\{t\}$, with the exception of the case $\lfloor t \rfloor = 2018$: in this case, it would follow that $\{t\} = \frac{2018}{2018} = 1$, which is impossible due to the requirement $0 \leq \{t\} < 1$. Therefore, we conclude that the obtained equation has a total of 2018 solutions for $t$ (and these are: $0, 1 + \frac{1}{2018}, 2 + \frac{2}{2018}, 3 + \frac{3}{2018}$, $\left.2017 + \frac{2017}{2018}\right)$, and thus the original equation has a total of 2018 solutions.

Answer: 2018


Let $a, b, c$ be the lengths of the sides of $\triangle ABC$, and $\frac{a}{b}=\frac{a+b}{a+b+c}$. Then the relationship between the interior angles $\angle A, \angle B$ is ( ).
(A) $\angle B>2 \angle A$
(B) $\angle B=2 \angle A$
(C) $\angle B<2 \angle A$
(D) Uncertain

Solution:
The given proportional equation can be simplified as
$$
b^{2}=a(a+c) \Rightarrow \frac{a}{b}=\frac{b}{a+c} \text {. }
$$

Therefore, as shown in Figure 11, we can
construct similar triangles with $a, b$ and $b, a+c$
as corresponding sides. It is easy to see that
$$
\angle B=2 \angle A \text {. }
$$

Answer: B


Given that $a, b$ are two unequal positive numbers, among the following three algebraic expressions:
(A) $\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)$,
(B) $\left(\sqrt{a b}+\frac{1}{\sqrt{a b}}\right)^{2}$,
(C) $\left(\frac{a+b}{2}+\frac{2}{a+b}\right)^{2}$,

the one with the largest value is
(A) Expression (A);
(B) Expression (B);
(C) Expression (C);
(D) It depends on the values of $a, b$.

Solution: If we take $a=1, b=2$, then the value of expression 甲 is the largest. If we then take $a=1, b=\frac{1}{3}$, the value of expression 甲 is still the largest. If we conclude that the answer is (A) based on this, it would be incorrect. In fact, if we take $a=3, b=2$, then the value of expression 丙 is the largest, so the correct choice should be (D).

Answer: D


Given fifty different natural numbers, twenty-five of which do not exceed 50, and the rest are greater than 50 but do not exceed 100. Moreover, no two of them differ exactly by 50. Find the sum of these numbers.

Solution:
Subtract 50 from each number that is greater than 50. We will get 50 different numbers, that is, numbers from 1 to 50. Their sum is $1+2+\ldots+50=25 \cdot 51$, and the sum of the original numbers is $-25 \cdot 51+25 \cdot 50=25 \cdot 101$.

## Answer

2525.

Answer: 2525


Eight people participate in a round-robin tournament (i.e., each pair competes in one match). Each match awards 2 points to the winner, 0 points to the loser, and 1 point to each in the case of a draw. After all matches are completed, it is found that:
(1) The eight participants have different scores;
(2) The second-place score is the sum of the scores of the fifth, sixth, seventh, and eighth-place participants;
(3) The first-place participant did not defeat the fourth-place participant.
Based on the above information, the third-place participant scored $\qquad$ points.

Solution: 【Match the answer】10

Answer: 10


Given $a>b>2$, and $a+b$, $a-b$, $a b$, $\frac{b}{a}$ form a geometric sequence in a certain order. Then $a=$ $\qquad$ .

Solution:
$$
-1.7+5 \sqrt{2} .
$$

Since $a>b>2$, we have
$$
\begin{array}{l}
a b-(a+b)=(a-1)(b-1)-1>0, \\
\frac{b}{a}b^{2}=\frac{b}{a} \cdot a b$, then when the common ratio is greater than 1,
it can only be $\frac{b}{a} 、 a-b 、 a+b 、 a b$ forming a geometric sequence, at this time,
$$
\begin{array}{l}
(a-b)(a+b)=a b \cdot \frac{b}{a} \Rightarrow a=\sqrt{2} b . \\
\text { Hence }(a-b)^{2}=(3-2 \sqrt{2}) b^{2} \\
=\frac{b}{a}(a+b)=\frac{\sqrt{2}+1}{\sqrt{2}} b \\
\Rightarrow b=\frac{7+5 \sqrt{2}}{\sqrt{2}} \\
\Rightarrow a=7+5 \sqrt{2} .
\end{array}
$$

Answer: 7+5 \sqrt{2}