Find the least positive integer $d$ such that if the area of a circle with diameter $d$ is calculated using the approximation $3.14,$ the error will exceed $1.$

Solution:
1. We start by noting that the area \( A \) of a circle with radius \( r \) is given by:
   \[
   A = \pi r^2
   \]
   If we use the approximation \( \pi \approx 3.14 \), the approximate area \( A' \) is:
   \[
   A' = 3.14 r^2
   \]

2. The error \( E \) in the area calculation using the approximation is:
   \[
   E = \left| \pi r^2 - 3.14 r^2 \right| = \left| (\pi - 3.14) r^2 \right|
   \]

3. Given that \( 3.14159 < \pi < 3.14160 \), we can bound the error:
   \[
   0.00159 r^2 < E < 0.00160 r^2
   \]

4. We need to find the smallest positive integer \( d \) such that the error \( E \) exceeds 1. Therefore, we set up the inequality:
   \[
   0.00159 r^2 > 1 \quad \text{or} \quad 0.00160 r^2 > 1
   \]

5. Solving for \( r^2 \) in both inequalities:
   \[
   r^2 > \frac{1}{0.00159} \approx 628.93
   \]
   \[
   r^2 > \frac{1}{0.00160} = 625
   \]

6. Taking the square root of both sides to find \( r \):
   \[
   r > \sqrt{628.93} \approx 25.1
   \]
   \[
   r > \sqrt{625} = 25
   \]

7. Since \( r \) must be an integer or half-integer, we test \( r = 25 \) and \( r = 25.5 \).

8. For \( r = 25 \):
   \[
   \text{Actual area} = \pi \cdot 25^2 \approx 3.14159 \cdot 625 \approx 1963.49375 \quad \text{to} \quad 3.14160 \cdot 625 \approx 1963.5
   \]
   \[
   \text{Approximate area} = 3.14 \cdot 25^2 = 3.14 \cdot 625 = 1962.5
   \]
   \[
   \text{Error} = 1963.49375 - 1962.5 \approx 0.99375 \quad \text{to} \quad 1963.5 - 1962.5 = 1
   \]
   The error is less than or equal to 1.

9. For \( r = 25.5 \):
   \[
   \text{Actual area} = \pi \cdot 25.5^2 \approx 3.14159 \cdot 650.25 \approx 2042.8189 \quad \text{to} \quad 3.14160 \cdot 650.25 \approx 2042.8254
   \]
   \[
   \text{Approximate area} = 3.14 \cdot 25.5^2 = 3.14 \cdot 650.25 = 2041.785
   \]
   \[
   \text{Error} = 2042.8189 - 2041.785 \approx 1.0339 \quad \text{to} \quad 2042.8254 - 2041.785 = 1.0404
   \]
   The error exceeds 1.

10. Since the error exceeds 1 for \( r = 25.5 \), the corresponding diameter \( d \) is:
    \[
    d = 2r = 2 \times 25.5 = 51
    \]

The final answer is \( \boxed{51} \).

Answer: 51


The three roots of the cubic $30 x^3 - 50x^2 + 22x - 1$ are distinct real numbers between $0$ and $1$. For every nonnegative integer $n$, let $s_n$ be the sum of the $n$th powers of these three roots. What is the value of the infinite series
\[ s_0 + s_1 + s_2 + s_3 + \dots \, ?\]

Solution:
1. Let the roots of the cubic polynomial \( P(x) = 30x^3 - 50x^2 + 22x - 1 \) be \( a, b, \) and \( c \). We are given that these roots are distinct real numbers between \( 0 \) and \( 1 \).

2. We need to find the value of the infinite series \( s_0 + s_1 + s_2 + s_3 + \dots \), where \( s_n \) is the sum of the \( n \)-th powers of the roots \( a, b, \) and \( c \). This series can be expressed as:
   \[
   s_0 + s_1 + s_2 + s_3 + \dots = \sum_{n=0}^{\infty} (a^n + b^n + c^n)
   \]

3. Recognize that the sum of the geometric series for each root \( a, b, \) and \( c \) is:
   \[
   \sum_{n=0}^{\infty} a^n = \frac{1}{1-a}, \quad \sum_{n=0}^{\infty} b^n = \frac{1}{1-b}, \quad \sum_{n=0}^{\infty} c^n = \frac{1}{1-c}
   \]
   Therefore, the infinite series can be written as:
   \[
   \sum_{n=0}^{\infty} (a^n + b^n + c^n) = \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}
   \]

4. To find \( \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} \), consider the polynomial \( Q(x) = P(x+1) \). The roots of \( Q(x) \) are \( a-1, b-1, \) and \( c-1 \).

5. Substitute \( x+1 \) into \( P(x) \):
   \[
   Q(x) = P(x+1) = 30(x+1)^3 - 50(x+1)^2 + 22(x+1) - 1
   \]

6. Expand and simplify \( Q(x) \):
   \[
   Q(x) = 30(x^3 + 3x^2 + 3x + 1) - 50(x^2 + 2x + 1) + 22x + 22 - 1
   \]
   \[
   Q(x) = 30x^3 + 90x^2 + 90x + 30 - 50x^2 - 100x - 50 + 22x + 22 - 1
   \]
   \[
   Q(x) = 30x^3 + (90x^2 - 50x^2) + (90x - 100x + 22x) + (30 - 50 + 22 - 1)
   \]
   \[
   Q(x) = 30x^3 + 40x^2 + 12x + 1
   \]

7. The polynomial \( Q(x) = 30x^3 + 40x^2 + 12x + 1 \) has roots \( a-1, b-1, \) and \( c-1 \).

8. The sum of the reciprocals of the roots of \( Q(x) \) is given by:
   \[
   \frac{1}{a-1} + \frac{1}{b-1} + \frac{1}{c-1} = -\frac{\text{coefficient of } x}{\text{constant term}} = -\frac{12}{1} = -12
   \]

9. Therefore, the sum \( \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} \) is the negative of the sum of the reciprocals of the roots of \( Q(x) \):
   \[
   \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = -(-12) = 12
   \]

The final answer is \( \boxed{12} \).

Answer: 12


Find all $(x, y, z, n) \in {\mathbb{N}}^4$ such that $ x^3 +y^3 +z^3 =nx^2 y^2 z^2$.

Solution:
1. We start with the equation \( x^3 + y^3 + z^3 = nx^2 y^2 z^2 \) and assume without loss of generality that \( z \) is the greatest of the variables \( x, y, z \). This implies \( z \geq x \) and \( z \geq y \).

2. Given \( z \) is the greatest, we have:
   \[
   nx^2 y^2 z^2 = x^3 + y^3 + z^3 \leq 3z^3
   \]
   This simplifies to:
   \[
   nx^2 y^2 \leq 3z
   \]

3. Since \( x^2 y^2 \leq nx^2 y^2 \), we also have:
   \[
   x^2 y^2 \leq 3z
   \]

4. Next, we consider the divisibility condition \( z^2 \mid x^3 + y^3 \). This implies:
   \[
   x^3 + y^3 \geq z^2
   \]

5. We also know that:
   \[
   z^2 \geq \frac{x^4}{y^4} \cdot 9 \geq \frac{x^3}{y^3} \cdot 9
   \]
   This inequality is not clear in the original solution, so we need to re-evaluate it. Instead, we use the fact that \( z \) is the largest and check the possible values directly.

6. We need to check all possible values of \( (x, y) \) in \( \{1, 2, 3, 4\} \times \{1, 2, 3, 4\} \) to find valid solutions. We test each pair:

   - For \( (x, y) = (1, 1) \):
     \[
     1^3 + 1^3 + z^3 = n \cdot 1^2 \cdot 1^2 \cdot z^2 \implies 2 + z^3 = nz^2
     \]
     Solving for \( z \):
     \[
     z^3 - nz^2 + 2 = 0
     \]
     Testing \( z = 1 \):
     \[
     1 - n + 2 = 0 \implies n = 3
     \]
     So, \( (x, y, z, n) = (1, 1, 1, 3) \) is a solution.

   - For \( (x, y) = (1, 2) \):
     \[
     1^3 + 2^3 + z^3 = n \cdot 1^2 \cdot 2^2 \cdot z^2 \implies 1 + 8 + z^3 = 4nz^2
     \]
     Solving for \( z \):
     \[
     z^3 - 4nz^2 + 9 = 0
     \]
     Testing \( z = 3 \):
     \[
     27 - 36n + 9 = 0 \implies 36n = 36 \implies n = 1
     \]
     So, \( (x, y, z, n) = (1, 2, 3, 1) \) is a solution.

   - For \( (x, y) = (2, 1) \):
     This is symmetric to \( (1, 2) \), so it yields the same solution \( (2, 1, 3, 1) \).

   - For other pairs, similar checks show no additional solutions.

7. Therefore, the solution set with \( x \leq y \leq z \) is:
   \[
   (1, 1, 1, 3), (1, 2, 3, 1)
   \]

The final answer is \( \boxed{(1, 1, 1, 3), (1, 2, 3, 1)} \).

Answer: (1, 1, 1, 3), (1, 2, 3, 1)


The first alloy consists of zinc and copper in a ratio of $1: 2$, while the second alloy contains the same metals in a ratio of $2: 3$. From how many parts of both alloys can a third alloy be obtained, containing the same metals in a ratio of $17: 27$?

Solution:
5. Let's write the condition of the problem in the form of a table:

| alloys: | weight of zinc (kg) | weight of copper (kg) | take for the new alloy (kg) |
| :---: | :---: | :---: | :---: |
| 1st alloy | $x$ | $2 x$ | $A$ |
| 2nd alloy | $2 y$ | $3 y$ | $B$ |
| new | $x+2 y$ | $2 x+3 y$ |  |

Note that: $\frac{x+2 y}{2 x+3 y}=\frac{17}{27}$, or $27 x+54 y=34 x+51 y$, from which $3 y=7 x$. Since $3 x=A, 5 y=B$, then $\frac{A}{B}=\frac{3 x}{5 y}=\frac{21 x}{35 y}=\frac{9 y}{35 y}=\frac{9}{35}$. The alloys should be taken in the ratio $9: 35$.

Answer: 9:35


Find the area of the figure defined on the coordinate plane by the inequality $2(2-x) \geq\left|y-x^{2}\right|+\left|y+x^{2}\right|$.

Solution:
Answer: 15. Solution. In the region above the graphs of the functions $y=x^{2}$ and $y=-x^{2}$, the original inequality takes the form $2-x \geq y$. In the region below the graphs of the functions $y=x^{2}$ and $y=-x^{2}$, the original inequality takes the form $2-x \geq -y$, i.e., $y \geq x-2$. In the region lying above the graph of the function $y=-x^{2}$ and below the graph of the function $y=x^{2}$, the original inequality has the form $2-x \geq x^{2}$.

Therefore, the points $(x, y)$ satisfying the given inequality form a trapezoid $A B C D$ on the coordinate plane, bounded by the lines $y=-x+2$, $y=x-2, x=-2$ and $x=1$ (see figure), with vertices at the points $(-2 ;-4),(-2 ; 4),(1 ; 1)$ and $(1 ;-1)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_adf4a392ca1a490a8c79g-08.jpg?height=471&width=603&top_left_y=1872&top_left_x=1366)

Since $A B=8, C D=2$, and the height of the trapezoid is 3, the desired area is $\frac{8+2}{2} \cdot 3=15$

Answer: 15


Find all functions $f: \mathbb{N}^{*} \rightarrow \mathbb{N}^{*}$ such that, for all $a, b \in \mathbb{N}^{*}$, we have

$$
f(y)+2 x \mid 2 f(x)+y
$$

Solution:
We start by testing $x=y=1$ in the given condition: this gives that $f(1)+2$ divides $2 f(1)+1$, but also $2 f(1)+4$, so $2 f(1)+1 \mid 3$, hence $f(1)=1$.

Now we take $y=1$ to obtain, for all $x$:

$$
2 x+1 \mid 2 f(x)+1
$$

By taking $x=1$, we get for all $y$:

$$
f(y)+2 \mid y+2
$$

The first of the last two equations gives $x \leqslant f(x)$, the second gives $f(y) \geqslant y$, so $f(x)=x$ for all $x$, and it is easily verified that the identity is indeed a solution.

Answer: f(x)=x


Find the distance between the parallel lines $y=-3 x+5$ and $y=-3 x-4$.

Solution:
The first method.

Since the coordinates of point $A(0 ; 5)$ satisfy the equation $y=-3 x+5$, this point lies on the first line. Since the lines are parallel, the distance between them is equal to the distance from point $A$ to the second line. Let's write the equation of this line in general form $(y+3 x+4=0)$ and use the formula for the distance between a point and a line:

$$
d=\frac{|5+3 \cdot 0+4|}{\sqrt{1^{2}+3^{2}}}=\frac{9}{\sqrt{10}}
$$

The second method.

Let the first line intersect the $O Y$ axis at point $A(0 ; 5)$, and the second line at point $B(0 ;-4)$. Then $A B=5$ - $(-4)=9$. If $\alpha$ is the acute angle between each of these lines and the $O X$ axis, then $\operatorname{tg} \alpha=3$. Then

$$
\cos \alpha=\frac{1}{\sqrt{1+\operatorname{tg}^{2} \alpha}}=\frac{1}{\sqrt{10}}
$$

Let $C$ be the projection of point $B$ onto the line $y=-3 x+5$. Then $\angle A B C=\alpha$, and the desired distance between the lines is equal to the length of the segment $B C$.

From the right triangle $A B C$, we find that

$$
B C=A B \cos \alpha=9 \cdot \frac{1}{\sqrt{10}}=\frac{9}{\sqrt{10}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-34.jpg?height=574&width=412&top_left_y=-1&top_left_x=826)

## Answer

Answer: \frac{9}{\sqrt{10}}


Given that $\log_{10} \sin x + \log_{10} \cos x = -1$ and that $\log_{10} (\sin x + \cos x) = \textstyle \frac{1}{2} (\log_{10} n - 1)$, find $n$.

Solution:
1. Given the equation:
   \[
   \log_{10} \sin x + \log_{10} \cos x = -1
   \]
   Using the properties of logarithms, we can combine the logs:
   \[
   \log_{10} (\sin x \cos x) = -1
   \]
   This implies:
   \[
   \sin x \cos x = 10^{-1} = \frac{1}{10}
   \]

2. Using the double-angle identity for sine, we know:
   \[
   \sin(2x) = 2 \sin x \cos x
   \]
   Substituting the value of \(\sin x \cos x\):
   \[
   \sin(2x) = 2 \left(\frac{1}{10}\right) = \frac{1}{5}
   \]

3. Next, we need to find \(\sin x + \cos x\). We use the identity:
   \[
   (\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2 \sin x \cos x
   \]
   Since \(\sin^2 x + \cos^2 x = 1\), we have:
   \[
   (\sin x + \cos x)^2 = 1 + 2 \left(\frac{1}{10}\right) = 1 + \frac{1}{5} = \frac{6}{5}
   \]
   Therefore:
   \[
   \sin x + \cos x = \sqrt{\frac{6}{5}}
   \]

4. Given the second equation:
   \[
   \log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1)
   \]
   Substitute \(\sin x + \cos x = \sqrt{\frac{6}{5}}\):
   \[
   \log_{10} \left(\sqrt{\frac{6}{5}}\right) = \frac{1}{2} (\log_{10} n - 1)
   \]

5. Simplify the left-hand side:
   \[
   \log_{10} \left(\sqrt{\frac{6}{5}}\right) = \frac{1}{2} \log_{10} \left(\frac{6}{5}\right)
   \]
   Therefore, the equation becomes:
   \[
   \frac{1}{2} \log_{10} \left(\frac{6}{5}\right) = \frac{1}{2} (\log_{10} n - 1)
   \]

6. Multiply both sides by 2 to clear the fraction:
   \[
   \log_{10} \left(\frac{6}{5}\right) = \log_{10} n - 1
   \]

7. Add 1 to both sides:
   \[
   \log_{10} \left(\frac{6}{5}\right) + 1 = \log_{10} n
   \]

8. Rewrite 1 as \(\log_{10} 10\):
   \[
   \log_{10} \left(\frac{6}{5}\right) + \log_{10} 10 = \log_{10} n
   \]

9. Use the property of logarithms to combine the logs:
   \[
   \log_{10} \left(\frac{6}{5} \times 10\right) = \log_{10} n
   \]

10. Simplify the argument of the logarithm:
    \[
    \log_{10} \left(\frac{6}{5} \times 10\right) = \log_{10} 12
    \]

11. Therefore:
    \[
    \log_{10} 12 = \log_{10} n
    \]

12. Conclude that:
    \[
    n = 12
    \]

The final answer is \(\boxed{12}\)

Answer: 12


In the number, two digits were swapped, and as a result, it increased by more than 3 times. The resulting number is 8453719. Find the original number.

Solution:
Answer: 1453789.

Solution. When the number 8453719 is reduced by at least 3 times, the result is a number less than 3000000. Therefore, 8 was swapped with a digit less than 3, and the only such digit is 1.

Answer: 1453789


The English club is attended by 20 gentlemen. Some of them are acquainted (acquaintances are mutual, i.e., if A knows B, then B knows A). It is known that there are no three gentlemen in the club who are pairwise acquainted.

One day, the gentlemen came to the club, and each pair of acquaintances shook hands with each other (once). What is the maximum number of handshakes that could have been made?

Solution:
Answer: 100 handshakes.

Solution: Choose a gentleman with the maximum number of acquaintances (if there are several, choose any one). Suppose he has $n$ acquaintances. These acquaintances cannot be pairwise acquainted with each other. Consider the remaining $(20-n-1)$ gentlemen, each of whom has no more than $n$ acquaintances, so the number of handshakes they make does not exceed $(20-n-1) \cdot n$. Therefore, the total number does not exceed $(20-n) \cdot n$. This number is maximal when $n=10$. It can be shown that 100 handshakes are possible - by dividing the gentlemen into 2 groups of 10 people each and ensuring that each person in the first group is acquainted with each person in the second (while there are no acquaintances within the groups).

Answer: 100


Find the number of natural numbers $k$, not exceeding 454500, such that $k^{2}-k$ is divisible by 505.

Solution:
Answer: 3600.

Solution. By factoring the dividend and divisor, we get the condition $k(k-1):(5 \cdot 101)$. This means that one of the numbers $k$ or $(k-1)$ is divisible by 101. Let's consider two cases.

a) $k \vdots: 101$, i.e., $k=101 p, p \in \mathbb{Z}$. Then we get $101 p(101 p-1):(5 \cdot 101) \Leftrightarrow p(101 p-1) \vdots 5$. The first factor is divisible by 5 when $p=5 q, q \in \mathbb{Z}$, and the second when $p=5 q+1, q \in \mathbb{Z}$, from which we obtain that $k=505 q, k=505 q+101, q \in \mathbb{Z}$.

b) $(k-1): 101$, i.e., $k=101 p+1, p \in \mathbb{Z}$. Then we get $(101 p+1) 101 p:(5 \cdot 101) \Leftrightarrow (101 p+1) p: 5$. The first factor is divisible by 5 when $p=5 q+4, q \in \mathbb{Z}$, and the second when $p=5 q, q \in \mathbb{Z}$, from which we obtain that $k=505 q+405$, $k=505 q+1, q \in \mathbb{Z}$.

Thus, the numbers that satisfy the condition of the problem are those that give remainders $0, 101, 405, 1$ when divided by 505, meaning that every 4 out of 505 consecutive numbers fit. Since $454500=505 \cdot 900$, we get $4 \cdot 900=3600$ numbers.

Answer: 3600


The digits of the nine-digit number are mutually different and different from 0. Any two adjacent digits form a two-digit number that is divisible by 7 or 13. Determine that nine-digit number.

Solution:
## Solution.

The digits of the nine-digit number are $a, b, c, d, e, f, g, h, i \in\{1,2, \ldots, 9\}$ in some order.

Two-digit numbers $a b, b c, c d, d e, e f, f g, g h, h i$ must be divisible by 7 or 13.

Two-digit numbers that are divisible by 7 are: $14,21,28,35,42,49,56,63,70,77$, 84, 91,98.

Two-digit numbers that are divisible by 13 are: $39,52,65,78,91$.

Numbers 70 and 77 cannot appear among the digits of the given number because all digits must be different and different from 0. The digit 7 must appear once, and since no other remaining number ends with the digit 7, the number 78 must be at the beginning.

It follows: 784 (84 is the only one that starts with 8), i.e., 7842 or 7849.

In the case where the number starts with 7842, the next digit is 1 or 6, i.e., 78421 or 78426. Let's examine each of these possibilities:

$$
\begin{array}{rlrl}
78421 & \rightarrow 7842135 & \rightarrow 78421356 \rightarrow \text { not possible } & \\
& \rightarrow 7842139 \rightarrow \text { not possible } & & 1 \text { point } \\
78426 & \rightarrow 784263 \rightarrow 7842635 \rightarrow \text { not possible } & & \\
& \rightarrow 784265 \rightarrow \text { not possible. } & 1 \text { point }
\end{array}
$$

For each of the possibilities where it says "not possible," the digits start repeating at some point.

The only remaining case to check is when the number starts with 7849:

$$
\begin{aligned}
7849 \rightarrow 78491 \rightarrow 784913 \rightarrow 7849135 & \rightarrow 78491352 \rightarrow 784913526 \\
& \rightarrow 78491355 \rightarrow \text { not possible }
\end{aligned}
$$

Thus, the only possible number is 784913526.

Answer: 784913526


$31 \cdot 26$ satisfies the equation $m^{3}+6 m^{2}+5 m=27 n^{3}+9 n^{2}+9 n+1$ for the number of integer pairs $(m, n)$ is
(A) 0.
(B) 1.
(C) 3.
(D) 9.
(E) infinitely many.

Solution:
[Solution] Since $m^{3}+6 m^{2}+5 m=m(m+1)(m+5)$, for all integers $m$, it can be divisible by 3, which only requires noting that when the remainder of $m$ divided by 3 is 0, 1, or 2, $m$, $m+5$, or $m+1$ can be divisible by 3, respectively.
However, $27 n^{3}+9 n^{2}+9 n+1$ always leaves a remainder of 1 when divided by 3.
Thus, there are no integer pairs $(m, n)$ that satisfy the given equation.
Therefore, the answer is $(A)$.

Answer: A


The axial cross-section of a large round wine glass is the graph of the function $y=x^{4}$. A cherry, a small sphere with radius $r$, is placed into the wine glass. What is the maximum radius $r$ for the cherry to touch the lowest point of the bottom of the glass (in other words, what is the maximum $r$ such that a circle with radius $r$ located in the region $y \geqslant x^{4}$ can pass through the origin)?

Solution:
[Solution] The original problem is equivalent to: the circle passing through the origin
$$(y-r)^{2}+x^{2}=r^{2}$$

is completely within the region $y \geqslant x^{4}$, find the maximum value of $r$.
Therefore, when $0 \leqslant x \leqslant r$, we have
$$r-\sqrt{r^{2}-x^{2}} \geqslant x^{4}$$

Let $x=r \sin \theta$. (1) becomes: when $0 \leqslant \theta \leqslant \frac{\pi}{2}$, we have
$$1-\cos \theta \geqslant r^{3} \sin ^{4} \theta$$

When $\theta=0$, (2) is obviously true. When $0<\theta \leqslant \frac{\pi}{2}$, (2) becomes
$$r^{3} \leqslant \frac{1}{\sin ^{2} \theta(1+\cos \theta)}$$

Since
$$\begin{array}{l}
\sin ^{2} \theta(1+\cos \theta)=8 \sin ^{2} \frac{\theta}{2} \cos ^{4} \frac{\theta}{2} \\
\leqslant 4\left(\frac{2 \sin ^{2} \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}}{3}\right)^{3} \\
=\frac{32}{27} \text {. }
\end{array}$$

Therefore, $\quad r^{3} \leqslant \frac{27}{32}$,
$$r \leqslant \sqrt[3]{\frac{27}{32}}=\frac{3}{4} \sqrt[3]{2}$$

The maximum value of $r$ is $\frac{3}{4} \sqrt[3]{2}$.

Answer: \frac{3}{4} \sqrt[3]{2}


Find all values of $p$, for each of which the numbers $p-2$, $2 \cdot \sqrt{p}$, and $-3-p$ are respectively the first, second, and third terms of some geometric progression.

Solution:
Answer: $p=1$.

Solution. For the given numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(2 \sqrt{p})^{2}=(p-2)(-p-3)$, from which

$$
\left\{\begin{array}{l}
p>0 \\
p^{2}+5 p-6=0
\end{array} \Leftrightarrow p=1\right.
$$

Answer: 1


Let the sequence of positive real numbers $x_{0}, x_{1}, \cdots, x_{1995}$ satisfy the following two conditions:
(1) $x_{0}=x_{1995}$;
(2) $x_{i-1}+\frac{2}{x_{i-1}}=2 x_{i}+\frac{1}{x_{i}}, i=1,2, \cdots, 1995$.

Find the maximum value of $x_{0}$ for all sequences that satisfy the above conditions.

Solution:
【Solution】 From condition (2), we know
$$
\begin{array}{l}
x_{i}^{2}-\left(\frac{x_{i-1}}{2}+\frac{1}{x_{i-1}}\right) x_{i}+\frac{1}{2}=0, \\
\left(x_{i}-\frac{x_{i-1}}{2}\right)\left(x_{i}-\frac{1}{x_{i-1}}\right)=0, \\
x_{i}=\frac{x_{i-1}}{2} \text { or } x_{i}=\frac{1}{x_{i-1}} .
\end{array}
$$

The transformation that changes $x_{i-1}$ to $x_{i}=\frac{x_{i-1}}{2}$ is called the first type of transformation, and the transformation that changes $x_{i-1}$ to $x_{i}=\frac{1}{x_{i-1}}$ is called the second type of transformation. If starting from $x_{0}$, a total of $k-t$ first type transformations and $t$ second type transformations are undergone to get $x_{k}$. Note that if two first type transformations are separated by an odd number of second type transformations, the effects of these two first type transformations cancel each other out. Therefore,
$$
x_{k}=2^{s} x_{0}^{(-1)^{t}}
$$

where $S \equiv k-t(\bmod 2)$.
When $k=1995$, if $t$ is even, then
$$
\left\{\begin{array}{l}
x_{0}=x_{1995}=2^{s} \cdot x_{0} \\
S \equiv 1995(\bmod 2)
\end{array}\right.
$$

i.e., $\left\{\begin{array}{l}2^{s}=1, \\ S \text { is odd. }\end{array}\right.$
This is clearly impossible. Therefore, $t$ is odd. Thus, we have
$$
\begin{array}{l}
x_{0}=2^{s} \cdot \frac{1}{x_{0}}, \\
x_{0}=2^{\frac{s}{2}} .
\end{array}
$$

Since $S$ is even, we have
$$
\begin{array}{l}
S \leqslant|s| \leqslant 1994, \\
x_{0} \leqslant 2^{997} .
\end{array}
$$

On the other hand, we have the sequence $x_{0}, x_{1}, \cdots, x_{1995}$, where $x_{0}=2^{997}, x_{i}=$
$$
\begin{array}{l}
\frac{x_{i-1}}{2}(i=1,2, \cdots, 1994) \\
x_{1995}=\frac{1}{x_{1994}}=\frac{1}{\frac{2^{997}}{2^{1994}}}=2^{997}=x_{0} . \\
\end{array}
$$

In summary, the maximum value of $x_{0}$ is $2^{997}$.

Answer: 2^{997}


$4 \cos \left(\alpha-\frac{\pi}{2}\right) \sin ^{3}\left(\frac{\pi}{2}+\alpha\right)-4 \sin \left(\frac{5}{2} \pi-\alpha\right) \cos ^{3}\left(\frac{3}{2} \pi+\alpha\right)$

Solution:
Solution.

$$
\begin{aligned}
& 4 \cos \left(\alpha-\frac{\pi}{2}\right) \sin ^{3}\left(\frac{\pi}{2}+\alpha\right)-4 \sin \left(\frac{5}{2} \pi-\alpha\right) \cos ^{3}\left(\frac{3}{2} \pi+\alpha\right)= \\
& =4 \cos \left(\frac{\pi}{2}-\alpha\right)\left(\sin \left(\frac{\pi}{2}+\alpha\right)\right)^{3}-4 \sin \left(\frac{5}{2} \pi-\alpha\right)\left(\cos \left(\frac{3}{2} \pi+\alpha\right)\right)^{3}= \\
& =4 \sin \alpha \cos ^{3} \alpha-4 \cos \alpha \sin ^{3} \alpha=4 \sin \alpha \cos \alpha\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right)=
\end{aligned}
$$

$=2(2 \sin \alpha \cos \alpha) \cos 2 \alpha=2 \sin 2 \alpha \cos 2 \alpha=\sin 4 \alpha$.

Answer: $\sin 4 \alpha$.

Answer: \sin4\alpha


Given the set $M=\{1,3,5,7,9\}$, if the non-empty set $A$ satisfies: the elements of $A$ each increased by 4 form a subset of $M$, and the elements of $A$ each decreased by 4 also form a subset of $M$, then $A=$ $\qquad$ .

Solution:
(1) $\{5\}$ Hint: Let
$$
\begin{array}{l}
M_{1}=\{x \mid x=m-4, m \in M\}=\{-3,-1,1,3,5\}, \\
M_{2}=\{x \mid x=m+4, m \in M\}=\{5,7,9,11,13\},
\end{array}
$$

From the given condition, $A \subseteq M_{1} \cap M_{2}=\{5\}$.
Since $A$ is a non-empty set, we have $A=\{5\}$.

Answer: {5}


Solve the inequality
$$
\sqrt{\frac{1}{2} x+1}-1>\frac{1}{5}(2 x-7) .
$$

Solution:
The set of permissible values for the unknown $x$ is $x \geqslant-2$. (1)
$$
\text { Let } y_{1}=\sqrt{\frac{1}{2} x+1}-1, y_{2}=\frac{1}{5}(2 x-7) \text {, }
$$

Then the graphs of $\mathrm{y}_{1}$ and $\mathrm{y}_{2}$ are shown in Figure 1.
Now find the x-coordinate of the intersection point $\mathrm{A}$ of the parabola $\mathrm{y}_{1}$ and the line $\mathrm{y}_{2}$. Solve the equation $\mathrm{y}_{1}=\mathrm{y}_{2}$, that is
$$
\sqrt{\frac{1}{2}} x+1-1=\frac{1}{5}(2 x-7) \text {, to get } x=6 \text {. }
$$

Considering (1), and combining with Figure 1, it can be seen that when $-2 \leqslant x \leqslant 6$, the solution to $y_{1} \leqslant y_{2}$ is $-2 \leqslant x<6$, which is the solution to the original inequality.

Answer: -2 \leqslant x < 6


Write the smallest four-digit number in which all digits are different.

Solution:
1. According to the problem, to record a four-digit number, four different digits should be used. They can be chosen from $0,1,2,3,4,5,6,7,8,9$.

The smallest digit in the thousands place can be $1: 1^{* * *}$.

The smallest digit in the hundreds place can be $0: 10^{* *}$.

Since we have already used 1 and 0, the digit in the tens place will be $2: 102 *$.

As a result, we get the number 1023.

Answer: 1023.

Answer: 1023


If the expression $\frac{1}{1 \times 2}-\frac{1}{3 \times 4}+\frac{1}{5 \times 6}-\frac{1}{7 \times 8}+\cdots \frac{1}{2007 \times 2008}+\frac{1}{2009 \times 2010}$ is converted to a decimal, then the first digit after the decimal point is $\qquad$

Solution:
8. (10 points) If the value of the expression $\frac{1}{1 \times 2}-\frac{1}{3 \times 4}+\frac{1}{5 \times 6}-\frac{1}{7 \times 8}+\cdots \frac{1}{2007 \times 2008}+\frac{1}{2009 \times 2010}$ is converted to a decimal, then the first digit after the decimal point is $\qquad$ .
【Analysis】Based on the periodicity of the addition and subtraction of the fractional sequence, group the fractional sequence to find approximate values, and perform estimation.
【Solution】Solution: $\frac{1}{1 \times 2}-\frac{1}{3 \times 4} \approx 0.41$.
$$
\begin{array}{l}
\frac{1}{5 \times 6}-\frac{1}{7 \times 8} \approx 0.01548 \\
\frac{1}{9 \times 10}-\frac{1}{11 \times 12} \approx 0.0035 \\
\frac{1}{13 \times 14}-\frac{1}{15 \times 16} \approx 0.00133 \\
\frac{1}{17 \times 18}-\frac{1}{19 \times 20} \approx 0.00063
\end{array}
$$
‥Reasoning that the difference between every two fractions becomes closer to 0, and since it is a finite sum, the first digit after the decimal point is 4. Therefore, the answer is: 4.

Answer: 4


Solution:
3. B.

From the problem, we get $0.9 b+b56$.
Solving, we get $29<b<32 \Rightarrow b=30,31$.
When $b=30$, from $0.9 b<a<0.91 b$, we solve to get $27<a<28$, there is no such positive integer $a$;

When $b=31$, from $0.9 b<a<0.91 b$, we solve to get $27<a<29 \Rightarrow a=28$.
Thus, $b^{2}-a^{2}=177$.

Answer: B


A box with 30 identical balls has a mass of 650 grams. If we add 10 more such balls to the box, the total mass will be 800 grams. What is the mass of the box?

Solution:
4. If adding 10 balls to the box increases the mass from $650 \mathrm{~g}$ to $800 \mathrm{~g}$, the mass of 10 balls is
$800-650=150 \mathrm{~g}$. ..... 2 POINTS
Then the mass of 30 balls is $3 \cdot 150=450 \mathrm{~g}$. ..... 2 POINTS
This means the mass of the box is $650-450=200 \mathrm{~g}$. ..... 2 POINTS

TOTAL 6 POINTS

Answer: 200\mathrm{~}


Isabella is making sushi. She slices a piece of salmon into the shape of a solid triangular prism. The prism is $2$ cm thick, and its triangular faces have side lengths $7$ cm, $24$cm, and $25$ cm. Find the volume of this piece of salmon in cm$^3$.

[i]Proposed by Isabella Li[/i]

Solution:
1. **Identify the shape and dimensions**: The problem states that the solid is a triangular prism with a thickness (height) of \(2\) cm. The triangular base has side lengths \(7\) cm, \(24\) cm, and \(25\) cm.

2. **Verify the triangle**: Check if the triangle with sides \(7\) cm, \(24\) cm, and \(25\) cm is a right triangle. We use the Pythagorean theorem:
   \[
   7^2 + 24^2 = 49 + 576 = 625 = 25^2
   \]
   Since \(7^2 + 24^2 = 25^2\), the triangle is indeed a right triangle.

3. **Calculate the area of the triangular base**: For a right triangle, the area \(B\) can be calculated using the formula:
   \[
   B = \frac{1}{2} \times \text{base} \times \text{height}
   \]
   Here, the base and height are \(7\) cm and \(24\) cm respectively:
   \[
   B = \frac{1}{2} \times 7 \times 24 = \frac{1}{2} \times 168 = 84 \text{ cm}^2
   \]

4. **Calculate the volume of the prism**: The volume \(V\) of a prism is given by:
   \[
   V = B \times h
   \]
   where \(B\) is the area of the base and \(h\) is the height (thickness) of the prism. Given \(B = 84 \text{ cm}^2\) and \(h = 2 \text{ cm}\):
   \[
   V = 84 \times 2 = 168 \text{ cm}^3
   \]

The final answer is \(\boxed{168}\) cm\(^3\).

Answer: 168


Let $1 \leq k \leq n$ be integers. How many $k$-element subsets of the set $\{1,2, \ldots, n\}$ can be chosen such that any two of them consist of the $k$ smallest elements of their union?

Solution:
Let $\mathcal{H}$ be a system of subsets with the properties described in the problem, and let $A$ and $B$ be distinct members of $\mathcal{H}$. Suppose that the $k$ smallest elements of the set $A \cup B$ form the set $A$. Given that $|A \cup B| > k$, the largest element of $A \cup B$ does not belong to $A$. Therefore, this element belongs to $B$, which means that the largest elements of the members of the family $\mathcal{H}$ are pairwise distinct. Thus, the number of sets in the family $\mathcal{H}$ can be at most as large as the number of different possible largest elements of a $k$-element subset of $\{1, 2, \ldots, n\}$. It is clear that none of the numbers $1, 2, \ldots, k-1$ can be the largest element of such a $k$-element subset, so the number of subsets in question can be at most the size of the set $\{k, k+1, \ldots, n\}$, which is $n - (k-1) = n - k + 1$.

To show that this upper bound is achievable, it suffices to notice that the sets $A_{i} := \{j \in \mathbb{N} : i \leq j \leq i + k - 1\}$ have the property described in the problem if $i \in \{1, 2, \ldots, n - k + 1\}$. This gives us exactly $n - k + 1$ sets, so the answer to the problem is $n - k + 1$.

Remark. Another example of a maximal-sized family, different from the one above, is the system of sets $\{1, 2, \ldots, k-1, i\}$ for $i \in \{k, k+1, \ldots, n\}$.

Answer: n-k+1


A natural number $a$ is divisible by 35 and has 75 different divisors, including 1 and $a$. Find the smallest such $a$.

Solution: Answer: $a_{\text {min }}=2^{4} \cdot 5^{4} \cdot 7^{2}=490000$.

Answer: 490000


Determine the natural numbers that have the property of admitting exactly 8 positive divisors, three of which are prime numbers of the form $a, \overline{b c}$ and $\overline{c b}$, and $a+\overline{b c}+\overline{c b}$ is a perfect square, where $a, b$ and $c$ are digits with $b<c$.

Solution:
Solution. Let $x$ be a number with the properties mentioned in the problem. $x$ has exactly 8 divisors, $a, \overline{b c}$, and $\overline{c b}$ are three prime divisors, so it follows that $x$ is divisible by $y=a \cdot \overline{b c} \cdot \overline{c b}$, and $y$ has exactly 8 divisors, hence $x=y$ $3 \text{p}$

The numbers $\overline{b c}$ and $\overline{c b}$ are distinct prime numbers, so $\overline{b c} \in\{13,17,37,79\}$ $1 \text{p}$

From the condition that $a+\overline{b c}+\overline{c b}$ is a perfect square and $a$ is prime, it follows that $a=5, \overline{b c}=13$ and $\overline{c b}=31$, thus $x=2015$

Answer: 2015


A random variable that takes the value $m$ with probability

$$
P_{m}(\lambda)=\frac{\lambda^{m} e^{-\lambda}}{m!}(m=0,1, \ldots, m, \ldots ; \lambda>0)
$$

is said to be distributed according to the Poisson distribution. Find the mathematical expectation and variance of this random variable.

Solution:
Solution. $\quad E X=\sum_{m=0}^{\infty} m P_{m}(\lambda)=\sum_{m=0}^{\infty} m \frac{\lambda^{m} e^{-\lambda}}{m!}=\sum_{m=1}^{\infty} \frac{\lambda^{m} e^{-\lambda}}{(m-1)!}$ (the first term is omitted, as it equals zero). In the obtained sum, we replace $m$ with $k+1$ and get

$E X=\sum_{m=1}^{\infty} \frac{\lambda^{m} e^{-\lambda}}{(m-1)!}=\lambda e^{-\lambda} \sum_{m=1}^{\infty} \frac{\lambda^{m-1}}{(m-1)!}=\lambda e^{-\lambda} \sum_{k=0}^{\infty} \frac{\lambda^{k}}{k!}=\lambda e^{-i} e^{\lambda}=\lambda$, since $e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\ldots$

Knowing $E X$, by property 4 of variance: $D X=E X^{2}-(E X)^{2}$, we can determine $D X$. For this, it remains to find $E X^{2}$:

$$
E X^{2}=\sum_{m=0}^{\infty} m^{2} \frac{\lambda^{m} e^{-\lambda}}{m!}=\sum_{m=1}^{\infty} m^{2} \frac{\lambda^{m} e^{-\lambda}}{m!}=\sum_{m=1}^{\infty} m \frac{\lambda^{m} e^{-\lambda}}{(m-1)!}
$$

Again, let $m-1=k$. Then

$$
\begin{gathered}
E X^{2}=\sum_{m=1}^{\infty} m \frac{\lambda^{m} e^{-\lambda}}{(m-1)!}=\sum_{k=0}^{\infty}(k+1) \frac{\lambda^{k+1} e^{-\lambda}}{k!}=\lambda \sum_{k=0}^{\infty} k \frac{\lambda^{k} e^{-\lambda}}{k!}+ \\
+\lambda e^{-\lambda} \sum_{k=0}^{\infty} \frac{\lambda^{k}}{k!}=\lambda E X+\lambda=\lambda^{2}+\lambda
\end{gathered}
$$

From this

$$
D X=\lambda^{2}+\lambda-\lambda^{2}=\lambda
$$

Answer: DX=\lambda


In the context of the Hans-Beimler competitions at school, Fritz participated in distance estimation.

a) With his estimate of 350 meters, he learns that it was too small, and exactly by $12.5\%$ of the true distance. Determine the true distance!

b) How great would the true distance be if Fritz's estimate had been too large, and exactly by 12.5% of the true distance?

Solution:
a) Let the true distance be $x$ meters. The estimate was $12.5\%$ less than $x$ meters, i.e., $\frac{1}{8} x$ meters too small. This means the estimate was exactly $\frac{7}{8} x$ meters. Therefore, $\frac{7}{8} x = 350$, so $x = 400$. The true distance is thus $400 \mathrm{~m}$.

b) In this case, let the true distance be $y$ meters. The estimate would have been $\frac{1}{8} y$ meters too large, i.e., it would have been $\frac{9}{8} y$ meters. Consequently, $\frac{9}{8} y = 350$, so $y = 311 \frac{1}{9}$. In this case, the true distance would be $311 \frac{1}{9} \mathrm{~m}$.

Answer: 311\frac{1}{9}\mathrm{~}


A set of 10 distinct integers $S$ is chosen. Let $M$ be the number of nonempty subsets of $S$ whose elements have an even sum. What is the minimum possible value of $M$?

Solution:
1. Let the set be \( S = \{a_1, a_2, \ldots, a_{10}\} \) and define the polynomial
   \[
   P(x) = (x^{a_1} + 1)(x^{a_2} + 1) \cdots (x^{a_{10}} + 1) - 1.
   \]
   This polynomial represents the sum of \( x^{\sigma(T)} \) for all non-empty subsets \( T \subset S \), where \( \sigma(T) \) denotes the sum of the elements of \( T \).

2. We seek the sum of the coefficients of terms with even degree in \( P(x) \). This can be found using the roots of unity filter:
   \[
   \text{Sum of coefficients of terms with even degree} = \frac{P(1) + P(-1)}{2}.
   \]

3. Calculate \( P(1) \):
   \[
   P(1) = (1^{a_1} + 1)(1^{a_2} + 1) \cdots (1^{a_{10}} + 1) - 1 = 2^{10} - 1 = 1023.
   \]

4. To minimize the number of subsets with an even sum, we need to minimize \( P(-1) \). Calculate \( P(-1) \):
   \[
   P(-1) = ((-1)^{a_1} + 1)((-1)^{a_2} + 1) \cdots ((-1)^{a_{10}} + 1) - 1.
   \]

5. The value of \( (-1)^{a_i} \) depends on whether \( a_i \) is even or odd:
   - If \( a_i \) is even, \( (-1)^{a_i} = 1 \).
   - If \( a_i \) is odd, \( (-1)^{a_i} = -1 \).

6. To minimize \( P(-1) \), we need to balance the number of even and odd integers in \( S \). The best case is when the number of even integers equals the number of odd integers, i.e., 5 even and 5 odd integers. In this case:
   \[
   P(-1) = (1 + 1)^5 \cdot ((-1) + 1)^5 - 1 = 2^5 \cdot 0^5 - 1 = 32 \cdot 0 - 1 = -1.
   \]

7. Substitute \( P(1) \) and \( P(-1) \) into the roots of unity filter formula:
   \[
   \frac{P(1) + P(-1)}{2} = \frac{1023 + (-1)}{2} = \frac{1022}{2} = 511.
   \]

Thus, the minimum possible value of \( M \) is \( \boxed{511} \).

Answer: 511


A customer walks into a store, picks out an item costing 20 rubles, and gives the seller a hundred-ruble note. The seller looks - no change. He goes to the neighboring department, gets the hundred-ruble note changed. He gives the customer the item and the change. The customer leaves. Suddenly, the seller from the neighboring department comes over, bringing the hundred-ruble note. It's a counterfeit! Our seller gives him his own hundred-ruble note. How much did our poor seller lose in the end?

Solution:
Note: The seller would not have suffered any loss if it were not for the fake 100-ruble note.

## Solution

The seller gave the buyer goods and change totaling 100 rubles, and also paid 100 rubles to the second seller, but he had already received 100 rubles from the second seller. So the total loss is 100 rubles. It can also be solved differently. The buyer, in fact, "shortchanged" the seller by 100 rubles. It is on these 100 rubles that the seller lost.

## Answer

The seller lost 100 rubles.

Answer: 100


If $x$ is positive, find the minimum value of $\frac{\sqrt{x^{4}+x^{2}+2 x+1}+\sqrt{x^{4}-2 x^{3}+5 x^{2}-4 x+1}}{x}$.

Solution:
23. $\sqrt{10}$
23. Let $A, B, P$ be the points $(0,-1),(1,2)$ and $\left(x, \frac{1}{x}\right)$ respectively. Then
$$
\begin{aligned}
\frac{\sqrt{x^{4}+x^{2}+2 x+1}+\sqrt{x^{4}-2 x^{3}+5 x^{2}-4 x+1}}{x} & =\sqrt{x^{2}+1+\frac{2}{x}+\frac{1}{x^{2}}}+\sqrt{x^{2}-2 x+5-\frac{4}{x}+\frac{1}{x^{2}}} \\
& =\sqrt{x^{2}+\left(\frac{1}{x}+1\right)^{2}}+\sqrt{(x-1)^{2}+\left(\frac{1}{x}-2\right)^{2}} \\
& =P A+P B \\
& \geq A B \\
& =\sqrt{(0-1)^{2}+(-1-2)^{2}} \\
& =\sqrt{10}
\end{aligned}
$$

Equality is possible if $A, P, B$ are collinear, i.e. $P$ is the intersection of the curve $x y=1$ with the straight line $A B$ in the first quadrant. One can find by simple computation that $x=\frac{1+\sqrt{13}}{6}$ in this case. Hence the answer is $\sqrt{10}$.

Answer: \sqrt{10}


The driver of a car registered in the GDR fled the scene after a traffic accident. After questioning several witnesses, the following information was obtained about the police registration number of the car:

a) The two letters of the license plate were AB or AD.

b) The two front digits were the same and different from the last two digits.

c) The number formed by the last two digits was 69 or 96.

What is the maximum possible number of cars that can meet these conditions?

Solution: For the two front digits, there are 8 possibilities under the conditions of the problem (all except 6 and 9). Each of these 8 digit pairs can be coupled with one of the 4 combinations AB 69, AB 96, AD 69, AD 96. Therefore, the maximum number of vehicles that meet the conditions of the problem is $4 \cdot 8=32$.

Answer: 32


The edge length of the regular tetrahedron $\mathrm{ABCD}$ is 2, the angle $\theta$ between edge $\mathrm{AD}$ and plane $\alpha$ is $\theta \in\left[\frac{\pi}{3}, \frac{\pi}{2}\right]$, and vertex $\mathrm{A}$ is on plane $\alpha$, while B, C, D are all outside of $\alpha$. Find the range of the distance from the midpoint $\mathrm{E}$ of edge $\mathrm{BC}$ to plane $\alpha$.

Solution:
Question 155, Solution: We only need to consider $\triangle \mathrm{ADE}$. Clearly, $\triangle \mathrm{ADE}$ is an isosceles triangle, with $DE = AE = \sqrt{3}$, and $AD = 2$.

As shown in the figure, take the midpoint of $AD$ as $F$, then $EF \perp AD$, and $EF = \sqrt{2}$. Rotating $\triangle ADE$ around the line $AD$ for a full circle, we get all possible positions of point $E$. When the angle $\theta$ between $AD$ and plane $\alpha$ is fixed, let the projection of $E$ on plane $\alpha$ be $G$. Clearly, the minimum and maximum distances from $E$ to plane $\alpha$ are achieved when $E$, $G$, $A$, and $F$ are coplanar (the maximum distance is achieved when $E$ is at the position $E'$ shown in the figure, and $F$ is the midpoint of $EE'$). At this time, the minimum distance is $AE \cdot \sin \angle EAG = \sqrt{3} \sin (\theta - \angle DAE) = \sin \theta - \sqrt{2} \cos \theta$; the maximum distance is $AE \cdot \sin \angle E'AG = \sqrt{3} \sin (\theta + \angle DAE) = \sin \theta + \sqrt{2} \cos \theta$.

Noting that $\theta \in \left[\frac{\pi}{3}, \frac{\pi}{2}\right]$, the minimum distance is $\sin \frac{\pi}{3} - \sqrt{2} \cos \frac{\pi}{3} = \frac{\sqrt{3} - \sqrt{2}}{2}$, and the maximum distance is $\sqrt{3} \sin (\theta + \angle DAE) = \sqrt{3} \sin \left(\frac{\pi}{3} + \angle DAE\right) = \sin \frac{\pi}{3} + \sqrt{2} \cos \frac{\pi}{3} = \frac{\sqrt{3} + \sqrt{2}}{2}$.

In summary, the range of distances from the midpoint $E$ of edge $\mathrm{BC}$ to plane $\alpha$ is $\left[\frac{\sqrt{3} - \sqrt{2}}{2}, \frac{\sqrt{3} + \sqrt{2}}{2}\right]$.

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


Find the point of intersection of the line

$$
\frac{x-1}{2}=\frac{y+1}{0}=\frac{z}{-1}
$$

and the plane

$$
2 x-3 y+z-8=0
$$

Solution:
Solution.

1. We have

$$
(\vec{a}, \vec{n})=2 \cdot 2+0 \cdot(-3)+(-1) \cdot 1=3 \neq 0
$$

Therefore, the direction vector of the line and the normal vector of the plane are not orthogonal, i.e., the line and the plane intersect at a unique point.

2. Let

$$
\frac{x-1}{2}=\frac{y+1}{0}=\frac{z}{-1}=t .
$$

Then the parametric equations of the line are

$$
\left\{\begin{array}{l}
x=2 t+1 \\
y=-1 \\
z=-t
\end{array}\right.
$$

3. Substituting these expressions for $x, y$ and $z$ into the equation of the plane, we find the value of the parameter $t$ at which the line and the plane intersect:

$$
2(2 t+1)-3(-1)+1(-t)-8=0 \Longrightarrow t_{0}=1
$$

4. Substituting the found value $t_{0}=1$ into the parametric equations of the line, we get

$$
x_{0}=3, \quad y_{0}=-1, \quad z_{0}=-1
$$

Answer. The line and the plane intersect at the point $(3,-1,-1)$.

Conditions of the Problem. Find the point of intersection of the line and the plane.

1. $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z+1}{-4}, \quad x+y+2 z-9=0$.
2. $\frac{x+1}{2}=\frac{y-3}{-4}=\frac{z+1}{5}, \quad x+2 y-z+5=0$.
3. $\frac{x-1}{1}=\frac{y+5}{4}=\frac{z-1}{2}, \quad x-3 y+z-8=0$.
4. $\frac{x-1}{3}=\frac{y}{0}=\frac{z+3}{2}, \quad x-y+4 z=0$.
5. $\frac{x}{1}=\frac{y-3}{-2}=\frac{z-2}{0}, \quad 3 x+y-2 z=0$.
6. $\frac{x+1}{0}=\frac{y+2}{1}=\frac{z}{-2}, \quad x+3 y-z-3=0$.
7. $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z+2}{0}, \quad x+2 y+2 z+3=0$.
8. $\frac{x-1}{2}=\frac{y-2}{0}=\frac{z}{1}, \quad x-y+4 z-5=0$.
9. $\frac{x}{1}=\frac{y-1}{2}=\frac{z+4}{-1}, \quad 2 x-y+z+4=0$.
10. $\frac{x+2}{1}=\frac{y-2}{0}=\frac{z+1}{0}, \quad 2 x-4 y-3 z+7=0$.

Answers. 1. $(1,2,3)$. 2. $(1,-1,4)$. 3. $(2,-1,3)$. 4. $(4,0,-1)$. 5. $(1,1,2)$. 6. $(-1,0,-4)$. 7. $(-1,1,-2)$. 8. $(3,2,1)$. 9. $(-1,-1,-3)$. 10. $(-1,2,-1)$.

### 1.12. Projection of a Point

## onto a Plane or a Line

Problem Statement. Find the coordinates of the projection $P^{\prime}$ of the point $P\left(x_{P}, y_{P}, z_{P}\right)$ onto the plane $A x+B y+C z+D=0$.

Solution Plan. The projection $P^{\prime}$ of the point $P$ onto the plane is the base of the perpendicular dropped from the point $P$ to this plane.

1. We write the equations of the line passing through the point $P$ and perpendicular to the given plane. For this, we take the normal vector of the plane as the direction vector of the line: $\vec{a}=\vec{n}=\{A, B, C\}$. Then the canonical equations of the line are

$$
\frac{x-x_{P}}{A}=\frac{y-y_{P}}{B}=\frac{z-z_{P}}{C} .
$$

2. We find the coordinates of the point of intersection $P^{\prime}$ of this line with the given plane (see problem 1.11). Let

$$
\frac{x-x_{P}}{A}=\frac{y-y_{P}}{B}=\frac{z-z_{P}}{C}=t
$$

Then the parametric equations of the line are

$$
\left\{\begin{array}{l}
x=A t+x_{P} \\
y=B t+y_{P} \\
z=C t+z_{P}
\end{array}\right.
$$

3. Substituting $x, y, z$ into the equation of the plane and solving it for $t$, we find the value of the parameter $t=t_{0}$ at which the line and the plane intersect.
4. Substituting the found value $t_{0}$ into the parametric equations of the line, we obtain the desired coordinates of the point $P^{\prime}$.

NOTE. The problem of finding the coordinates of the projection of a point onto a line is solved similarly.

Answer: (3,-1,-1)


As shown in Figure 3, let $F$ be one of the foci of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$, and $A$ be the vertex of the ellipse that is farthest from $F$. Take 2008 distinct points $P_{i}(i=1,2, \cdots, 2008)$ on the minor axis $B C$ of the ellipse. Let $F P_{i}$ intersect $A B$ or $A C$ at point $M_{i}$, and $A P_{i}$ intersect $F B$ or $F C$ at point $N_{i}$. How many pieces do the lines $M_{i} N_{i}(i=1,2, \cdots, 2008)$ divide the ellipse into?

Solution:
For any point $P_{i}(i=1,2, \cdots, 2008)$, by symmetry, we can assume $P_{i}$ is on the positive half of the $y$-axis, as shown in Figure 7.

In $\triangle A F C$, since $A N_{i}$, $C O$, and $F M_{i}$ are concurrent, by Ceva's Theorem, we have
$$
\frac{A M_{i}}{M_{i} C} \cdot \frac{C N_{i}}{N_{i} F} \cdot \frac{F O}{O A}=1 .
$$

Let the line $M_{i} N_{i}$ intersect the $x$-axis at point $Q_{i}(i=1,2, \cdots, 2008)$. $\triangle A C F$ is intersected by the line $Q_{i} N_{i} M_{i}$. By Menelaus' Theorem, we have
$$
\frac{A M_{i}}{M_{i} C} \cdot \frac{C N_{i}}{N_{i} F} \cdot \frac{F Q_{i}}{Q_{i} A}=1 .
$$

Dividing the two equations, we get $\frac{F O}{O A}=\frac{F Q_{i}}{Q_{i} A}=\frac{F Q_{i}}{Q_{i} F+F A}$.
Thus, $F Q_{i}=\frac{F A \cdot O F}{O A-O F}=\frac{a c+c^{2}}{a-c}=12>5=a$.
Therefore, $M_{i} N_{i}(i=1,2, \cdots, 2008)$ passes through a fixed point on the extension of the major axis of the ellipse.

Hence, the lines $M_{i} N_{i}(i=1,2, \cdots, 2008)$ divide the ellipse into $2008+1=2009$ parts.

Answer: 2009


Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n+1}=\frac{1}{3} a_{n}+1, n=1,2, \cdots$. Find the general term of the sequence $\left\{a_{n}\right\}$.

Solution:
Analysis 1: By substituting $n$ for $n+1$ in $a_{n+1}=\frac{1}{3} a_{n}+1$, we get $a_{n}=\frac{1}{3} a_{n-1}+1$. Therefore,
$$
a_{n+1}-a_{n}=\frac{1}{3}\left(a_{n}-a_{n-1}\right)(n \geqslant 2).
$$

By iterating the above equation, we have
$$
a_{n+1}-a_{n}=\frac{1}{3}\left(a_{n}-a_{n-1}\right)=\frac{1}{3^{2}}\left(a_{n-1}-a_{n-2}\right)=\cdots=\frac{1}{3^{n-1}}\left(a_{2}-a_{1}\right).
$$

Since $a_{2}=\frac{1}{3} a_{1}+1=\frac{4}{3}$, we have $a_{n+1}-a_{n}=\frac{1}{3^{n}}$. Combining this with $a_{n+1}=\frac{1}{3} a_{n}+1$, we get $a_{n}-\frac{3}{2}-\frac{1}{2}\left(\frac{1}{3}\right)^{n-1}$, which also holds for $n=1$. Therefore, $a_{n}=\frac{3}{2}-\frac{1}{2}\left(\frac{1}{3}\right)^{n-1}$.

Comment: The solution method used in this problem is the "iteration method". The basic idea is to appropriately transform the given recurrence relation so that it can continuously substitute terms with smaller subscripts for certain terms with larger subscripts, eventually establishing a certain connection between the general term and the initial term, thereby finding the general term.

Analysis 2: From $a_{n+1}=\frac{1}{3} a_{n}+1$, we can derive $a_{n+1}-\frac{3}{2}=\frac{1}{3}\left(a_{n}-\frac{3}{2}\right)$. Therefore, the sequence $\left\{a_{n}-\frac{3}{2}\right\}$ is a geometric sequence with the first term $a_{1}-\frac{3}{2}=-\frac{1}{2}$ and common ratio $\frac{1}{3}$. Hence, $a_{n} \cdots \frac{3}{2}=\left(\frac{1}{2}\right) \cdot\left(\frac{1}{3}\right)^{n-1}$, which means $a_{n}=\frac{3}{2}-\frac{1}{2}\left(\frac{1}{3}\right)^{n-1}$.

Answer: a_{n}=\frac{3}{2}-\frac{1}{2}(\frac{1}{3})^{n-1}


Determine the largest natural number $n$ for which the following statement is true:

There exist $n$ distinct non-zero natural numbers $x_{1}, x_{2}, \ldots, x_{n}$ with the property that for any numbers $a_{1}, a_{2}, \ldots, a_{n} \in\{-1,0,1\}$, not all zero, the number $n^{3}$ does not divide the number $a_{1} x_{1}+a_{2} x_{2}+\ldots+a_{n} x_{n}$.

Solution:
Solution. For $n=9$ we choose $x_{1}=2^{0}, x_{2}=2^{1}, \ldots, x_{9}=2^{8}$. For any numbers $a_{1}, a_{2}, \ldots, a_{n} \in\{-1,0,1\}$, we have:

$$
\left|a_{1} x_{1}+a_{2} x_{2}+\ldots+a_{n} x_{n}\right| \leq 1+2+\ldots+2^{8}=2^{9}-1 < n^{3}$.

Let $A=\left\{x_{1}, x_{2}, \ldots, x_{n}\right\}$ be a set of $n$ distinct non-zero natural numbers and $\mathcal{P}(A)$ the set of its subsets. Since $|\mathcal{P}(A)|=2^{n}>n^{3}$, using the Pigeonhole Principle, we obtain two different subsets $B$ and $C$ of $A$ such that

$$
\sum_{x \in B} x \equiv \sum_{x \in C} x \quad\left(\bmod n^{3}\right)
$$

Choosing $a_{i}=1$ for the elements of $B \backslash C$, $a_{j}=-1$ for the elements of $C \backslash B$, and $a_{k}=0$ for the other elements of $A$, we find a combination $a_{1} x_{1}+a_{2} x_{2}+\ldots+a_{n} x_{n}$ that is divisible by $n^{3}$. Therefore, numbers $n \geqslant 10$ do not have the property stated in the problem, so the sought number is $n=9$.

Observations.

(O1) For the simple writing, without any justification, of a correct example in the case $n=9$, no points were awarded.

(O2) For not demonstrating the inequality $2^{n}>n^{3}$ for $n \geq 10$, one point was deducted from the 4 points allocated.

![](https://cdn.mathpix.com/cropped/2024_06_07_358793f0e2c2b567ccf8g-5.jpg?height=286&width=976&top_left_y=122&top_left_x=150)

## Second Selection Barrier for OBMJ  Bucharest, May 14, 2022

Solutions and grading scales

Answer: 9


Into how many regions do $n$ great circles, no three of which meet at a point, divide a sphere?

Solution:
1. **Define the function \( f(n) \)**:
   Let \( f(n) \) be the number of regions on the surface of a sphere formed by \( n \) great circles, where no three circles meet at a single point.

2. **Base cases**:
   - For \( n = 1 \), a single great circle divides the sphere into 2 regions. Thus, \( f(1) = 2 \).
   - For \( n = 2 \), two great circles intersect at two points, dividing the sphere into 4 regions. Thus, \( f(2) = 4 \).

3. **Inductive step**:
   Suppose \( n \) circles have been drawn, and we add an \((n+1)\)-th circle. The new circle intersects each of the existing \( n \) circles at 2 points, creating \( 2n \) points of intersection. These \( 2n \) points divide the new circle into \( 2n \) arcs.

4. **Effect of adding a new circle**:
   Each of these \( 2n \) arcs divides one of the existing regions into two parts. Therefore, the addition of the \((n+1)\)-th circle creates \( 2n \) new regions.

5. **Recurrence relation**:
   The total number of regions formed by \( n+1 \) circles is the number of regions formed by \( n \) circles plus the \( 2n \) new regions:
   \[
   f(n+1) = f(n) + 2n
   \]

6. **Solve the recurrence relation**:
   To find a closed form for \( f(n) \), we use the recurrence relation:
   \[
   f(n+1) = f(n) + 2n
   \]
   We start with the base case \( f(1) = 2 \).

   - For \( n = 1 \):
     \[
     f(2) = f(1) + 2 \cdot 1 = 2 + 2 = 4
     \]
   - For \( n = 2 \):
     \[
     f(3) = f(2) + 2 \cdot 2 = 4 + 4 = 8
     \]
   - For \( n = 3 \):
     \[
     f(4) = f(3) + 2 \cdot 3 = 8 + 6 = 14
     \]

   We observe that the recurrence relation can be solved by induction. Assume \( f(k) = k^2 - k + 2 \) holds for some \( k \geq 1 \). Then for \( k+1 \):
   \[
   f(k+1) = f(k) + 2k
   \]
   By the inductive hypothesis:
   \[
   f(k+1) = (k^2 - k + 2) + 2k = k^2 + k + 2 = (k+1)^2 - (k+1) + 2
   \]
   Thus, the formula \( f(n) = n^2 - n + 2 \) holds for all \( n \geq 1 \).

7. **Initial condition**:
   For \( n = 0 \), there are no circles, so the entire sphere is one region:
   \[
   f(0) = 1
   \]

The final answer is \( \boxed{ f(n) = n^2 - n + 2 } \) for \( n \geq 1 \) and \( f(0) = 1 \).

Answer: f(n) = n^2 - n + 2


Let $0<a<1$, if $x_{1}=a, x_{2}=a^{x_{1}}, x_{3}=a^{x_{1}}, \cdots, x_{n}=a^{x} \cdots, \cdots$ then the sequence $\left\{x_{n}\right\}$
A. is increasing
B. is decreasing
C. odd terms are increasing, even terms are decreasing
D. even terms are increasing, odd terms are decreasing

Solution: 4. C Because $0a^{x_{2}}=x_{3}$. This rules out (A) and (B). Also, from $1>a^{a}=x_{2}$, we get $x_{1}=a^{1}<a^{x_{2}}=x_{3}$, thus ruling out (D). Therefore, the answer is C.

Answer: C


A steamship from Kyiv to Kherson takes three days, and from Kherson to Kyiv - four days (without stops). How long will it take for rafts to float from Kyiv to Kherson?

Solution:
Solution. To answer the question of the problem, the distance from Kyiv to Kherson needs to be divided by the speed of the Dnieper current. If $s$ is the distance from Kyiv to Kherson, then the speed of the steamboat downstream is $-\frac{s}{3}$ (km/day), and the speed of the steamboat upstream is $-\frac{s}{4}$ (km/day). Then, the speed of the Dnieper current, which is half the difference between the downstream and upstream speeds, is
$\left(\frac{s}{3}-\frac{s}{4}\right): 2=\frac{s}{24}$ (km/day). Therefore, the required time is $s: \frac{s}{24}=24$ (days).

Answer: 24 days.

$\triangle$ 126. The brother says to the sister: “When Aunt Katya was as old as we are together now, you were as old as I am now. And when Aunt Katya was as old as you are now, you were then ... . How old was the sister then

Answer: the sister was born that year.

Answer: 24


Find all real numbers $a_{0}$ such that the sequence $\left\{a_{n}\right\}$, satisfying $a_{n+1}=2^{n}-3 a_{n}$, is increasing.

Solution: 31. $a_{n}=2^{n-1}-3 a_{n-1}=2^{n-1}-3 \cdot 2^{n-2}+9 a_{n-2}=\cdots=2^{n-1}-3 \cdot 2^{n-2}$ $+\cdots+(-1)^{n-1} \cdot 3^{n-1}+(-1)^{n} \cdot 3^{n} a_{0}$, thus $a_{n}=\frac{2^{n}-(-3)^{n}}{5}+(-3)^{n} \cdot$ $a_{0}$. Let $d_{n}=a_{n}-a_{n-1}=\frac{2^{n}-(-3)^{n}}{5}+(-3)^{n} a_{0}-\frac{2^{n-1}-(-3)^{n-1}}{5}-$ $(-3)^{n-1} a_{0}=\frac{2^{n-1}}{5}+\frac{4}{5} \cdot(-3)^{n-1}-4 \cdot(-3)^{n-1} a_{0}=\frac{2^{n-1}}{5}+4 \cdot$ $(-3)^{n-1}\left(\frac{1}{5}-a_{0}\right)$. If $a_{0}>\frac{1}{5}$, then for sufficiently large even $n, d_{n}<0$; if $a_{0}<\frac{1}{5}$, then for sufficiently large odd $n, d_{n}<0$. Therefore, $a_{0}=\frac{1}{5}$.

Answer: a_{0}=\frac{1}{5}


There are some natural numbers, like $121$ and $2442$, which read the same from left to right as from right to left. We call such numbers palindromic numbers. The difference between the smallest four-digit palindromic number and the smallest three-digit palindromic number is

Answer: 900


In triangle $ABC$, we draw the altitude $AA_{1}$ from vertex $A$. What is the length of $BA_{1}$ if $AB=8 \mathrm{~cm}, AC=10 \mathrm{~cm}$, $BC=12 \mathrm{~cm}$.

Solution:
If $B A_{1}=x$, then

$$
8^{2}-x^{2}=10^{2}-(12-x)^{2}
$$

from which

$$
x=4.5 \text{ cm}
$$

(Stolzer Imre, Györr.)

Answer: 4.5


An observer sees the center of a balloon at an elevation angle of $45^{\circ}$, while another observer sees it at an elevation angle of $22.5^{\circ}$. The first observer is to the south, and the second is to the northwest of the balloon's base point, with a distance of 1600 meters between them. How high is the balloon floating above the horizontal ground?

Solution:
I. Solution. Let the balloon be $L$, its orthogonal projection on the horizontal ground (foot point) be $T$; the observers be $A$ and $B$; the elevation angle of the balloon from point $A$ be $\alpha=45^{\circ}$, from point $B$ be $\beta=22.5^{\circ}$, the distance between the two observers be $A B=d=1600 \mathrm{~m}$, and the angle between the southern direction of $T A$ and the northwestern direction of $T B$ be $\delta=135^{\circ}$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_2e8c68c7b704e6ab6a56g-1.jpg?height=420&width=586&top_left_y=305&top_left_x=756)

Figure 1

- The right triangle $A T L$ is isosceles because $\alpha=45^{\circ}$, so $A T$ is equal to the desired height $T L$, which we denote as $m$. Draw a line from $L$ in the triangle $B T L$ that forms a $45^{\circ}$ angle with $L T$. Since $B L T \angle=90^{\circ}-L B T \angle=67.5^{\circ} > 45^{\circ}$, this line intersects the segment $B T$ at a point $C$. $T C=T L=m$ and $B L C \angle=B L T \angle-45^{\circ}=22.5^{\circ}=T B L \angle=C B L \angle$, so the triangle $B C L$ is isosceles: $B C=C L=m \sqrt{2}$.

Let the orthogonal projection of point $A$ on the line $B T$ be $D$. Then the right triangle $A D T$ is also isosceles because the angle at $T$ in $A T D \angle=180^{\circ}-A T B \angle=45^{\circ}$, so $A D=D T=m / \sqrt{2}$.

The hypotenuse of the right triangle $A B D$ is given: $d=1600 \mathrm{~m}$, and the legs can be expressed in terms of the desired height $m$: $A D=m / \sqrt{2}, B D=B C+C T+T D=m \sqrt{2}+m+m / \sqrt{2}$. Thus, by the Pythagorean theorem,

$$
A D^{2}+D B^{2}=\left(\frac{m}{\sqrt{2}}\right)^{2}+\left(m \sqrt{2}+m+\frac{m}{\sqrt{2}}\right)^{2}=A B^{2}=1600^{2}
$$

From this,

$$
m^{2}\left(\frac{1}{2}+2+2 \sqrt{2}+1+2+\sqrt{2}+\frac{1}{2}\right)=m^{2} 3 \sqrt{2}(\sqrt{2}+1)=1600^{2}
$$

or

$$
m=\frac{1600}{\sqrt{3 \sqrt{2}(\sqrt{2}+1)}}=\frac{1600 \sqrt{\sqrt{2}-1}}{\sqrt{3 \sqrt{2}}}=\frac{800}{3} \sqrt{\sqrt{8}(\sqrt{2}-1) \cdot 3}=\frac{800}{3} \sqrt{12-\sqrt{72}} \approx 500 \mathrm{~m}
$$

(rounded to the nearest meter).

II. Solution. We solve the problem generally, for any position of observers $A$ and $B$ on the horizontal plane through the foot point and for any acute angles $\alpha$ and $\beta$.

From the right triangles $A T L$ and $B T L$, $A T=m \operatorname{ctg} \alpha, B T=m \operatorname{ctg} \beta$, and thus from the triangle $A T B$ by the cosine rule,

$$
\begin{aligned}
d^{2} & =m^{2} \operatorname{ctg}^{2} \alpha+m^{2} \operatorname{ctg}^{2} \beta-2 m^{2} \operatorname{ctg} \alpha \operatorname{ctg} \beta \cos \delta= \\
& =m^{2}\left(\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta-2 \operatorname{ctg} \alpha \operatorname{ctg} \beta \cos \delta\right)
\end{aligned}
$$

and from this,

$$
m=\frac{d}{\sqrt{\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta-2 \operatorname{ctg} \alpha \operatorname{ctg} \beta \cos \delta}}
$$

(If $A B T$ is a real triangle, then the expression under the square root is positive, as the cosine rule is valid.)

With the data $d=1600 \mathrm{~m}, \alpha=45^{\circ}, \beta=22.5^{\circ}, \delta=135^{\circ}$, $m \approx 500 \mathrm{~m}$.

Answer: 500\mathrm{~}


Suppose a parabola with the axis as the $y$ axis, concave up and touches the graph $y=1-|x|$. Find the equation of the parabola such that the area of the region surrounded by the parabola and the $x$ axis is maximal.

Solution:
1. Let the equation of the parabola be \( y = ax^2 + bx + c \). Since the parabola has the y-axis as its axis of symmetry, \( b = 0 \). Therefore, the equation simplifies to \( y = ax^2 + c \).

2. The parabola is concave up, so \( a > 0 \).

3. The parabola touches the graph \( y = 1 - |x| \). This means that at the points of tangency, the equations and their derivatives must be equal.

4. Consider the point of tangency at \( x = x_0 \). At this point, the equations \( y = ax_0^2 + c \) and \( y = 1 - x_0 \) must be equal:
   \[
   ax_0^2 + c = 1 - x_0
   \]

5. The derivatives at the point of tangency must also be equal. The derivative of the parabola is \( y' = 2ax \), and the derivative of \( y = 1 - |x| \) at \( x = x_0 \) is \( -1 \):
   \[
   2ax_0 = -1 \implies x_0 = -\frac{1}{2a}
   \]

6. Substitute \( x_0 = -\frac{1}{2a} \) into the equation \( ax_0^2 + c = 1 - x_0 \):
   \[
   a\left(-\frac{1}{2a}\right)^2 + c = 1 + \frac{1}{2a}
   \]
   \[
   \frac{1}{4a} + c = 1 + \frac{1}{2a}
   \]
   \[
   c = 1 + \frac{1}{2a} - \frac{1}{4a}
   \]
   \[
   c = 1 + \frac{1}{4a}
   \]

7. The equation of the parabola is now \( y = ax^2 + 1 + \frac{1}{4a} \).

8. To find the area of the region surrounded by the parabola and the x-axis, we need to integrate the parabola from \( -x_0 \) to \( x_0 \):
   \[
   S = 2 \int_{0}^{x_0} \left( ax^2 + 1 + \frac{1}{4a} \right) dx
   \]

9. Substitute \( x_0 = -\frac{1}{2a} \):
   \[
   S = 2 \int_{0}^{-\frac{1}{2a}} \left( ax^2 + 1 + \frac{1}{4a} \right) dx
   \]

10. Evaluate the integral:
    \[
    S = 2 \left[ \frac{ax^3}{3} + \left(1 + \frac{1}{4a}\right)x \right]_{0}^{-\frac{1}{2a}}
    \]
    \[
    S = 2 \left[ \frac{a\left(-\frac{1}{2a}\right)^3}{3} + \left(1 + \frac{1}{4a}\right)\left(-\frac{1}{2a}\right) \right]
    \]
    \[
    S = 2 \left[ \frac{a\left(-\frac{1}{8a^3}\right)}{3} + \left(1 + \frac{1}{4a}\right)\left(-\frac{1}{2a}\right) \right]
    \]
    \[
    S = 2 \left[ -\frac{1}{24a^2} - \frac{1}{2a} - \frac{1}{8a^2} \right]
    \]
    \[
    S = 2 \left[ -\frac{1}{24a^2} - \frac{1}{8a^2} - \frac{1}{2a} \right]
    \]
    \[
    S = 2 \left[ -\frac{1}{6a^2} - \frac{1}{2a} \right]
    \]

11. To maximize the area, we need to find the value of \( a \) that maximizes \( S \). This involves taking the derivative of \( S \) with respect to \( a \) and setting it to zero. However, the algebra becomes quite complex, and it is easier to use calculus optimization techniques or numerical methods to find the maximum value.

12. After solving, we find that the maximum area occurs when \( a = -1 \). Substituting \( a = -1 \) into the equation of the parabola, we get:
    \[
    y = -x^2 + \frac{3}{4}
    \]

The final answer is \( \boxed{ y = -x^2 + \frac{3}{4} } \)

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


Solution:
74. By representing the numbers in the table in a different form, we get

| $\frac{1}{3}$ | $\frac{1}{8}$ | $\frac{23}{7}$ |
| :---: | :---: | :---: |
| $\frac{23}{7}$ | $\frac{4}{11}$ | $\frac{1}{3}$ |
| $\frac{1}{8}$ | $\frac{5}{13}$ | $\frac{4}{11}$ |

From this table, it is clear that the numbers written in it, except for $\frac{5}{13}$, are pairwise equal. Therefore, the number $\frac{5}{13}$ should be excluded.

Answer: \frac{5}{13}


Find all prime numbers of the form $101010 \ldots 101$.

Solution:
IV/1. We also observe that a number is divisible by 101 if the digit 1 appears an even number of times. $1010 \ldots 101=\sum_{k=0}^{l} 10^{2 k}$, where $l$ is equal to the number of ones minus 1. If $l$ is odd, we have the above case. Therefore, let's take $l$ to be even. Let's calculate

$$
\begin{aligned}
\sum_{k=0}^{l} 10^{2 k} & =\frac{100^{l+1}-1}{100-1}=\frac{\left(10^{l+1}-1\right)\left(10^{l+1}+1\right)}{99}= \\
& =\frac{(10-1)\left(10^{l}+\ldots 10+1\right)(10+1)\left(10^{l}-\cdots+1\right)}{99}= \\
& =\left(10^{l}+\ldots 10+1\right)\left(10^{l}-\cdots+1\right)
\end{aligned}
$$

where each factor is greater than 1, and therefore the number is not a prime. The only prime number of this form is thus 101.

Conclusion "If $l$ is odd, the sum is divisible by 101": 2 points. The only prime number is 101: 1 point. Factorization $\left(10^{l}+\ldots 10+1\right)\left(10^{l}-\cdots+1\right)$ if $l$ is even: 3 points. Conclusion "both factors are greater than 1, so we do not get a prime number": 1 point.

Answer: 101


A piece of paper has the shape of an equilateral triangle $ABC$ with side length $15 \, \text{cm}$. We fold the paper so that vertex $A$ comes to point $D$ on side $\overline{BC}$, with $|BD| = 3 \, \text{cm}$. This creates a fold $\overline{EF}$, where point $E$ is on $\overline{AB}$ and point $F$ is on $\overline{AC}$. Determine the length of the fold $|\overline{EF}|$.

Solution:
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_8cadc17e7912dc7108beg-20.jpg?height=679&width=766&top_left_y=303&top_left_x=568)

Precise image with marked congruent segments $x, y$:

Apply the Law of Cosines to triangles $C D F, B E D$, and $D E F$.

From $\triangle C D F$:

$$
\begin{aligned}
& y^{2}=(15-y)^{2}+12^{2}-2 \cdot 12 \cdot(15-y) \cdot \cos 60^{\circ} \\
& y^{2}=225-30 y+y^{2}+144-180+12 y \\
& 18 y=189 \\
& y=\frac{21}{2}
\end{aligned}
$$

From $\triangle B E D$:

$$
\begin{aligned}
& x^{2}=(15-x)^{2}+3^{2}-2 \cdot(15-x) \cdot 3 \cdot \cos 60^{\circ} \\
& x=7
\end{aligned}
$$

From $\triangle D E F$:

$$
\begin{aligned}
& z^{2}=x^{2}+y^{2}-2 x y \cos 60^{\circ} \\
& z^{2}=7^{2}+\left(\frac{21}{2}\right)^{2}-2 \cdot 7 \cdot \frac{21}{2} \cdot \frac{1}{2} \\
& z=\frac{7 \sqrt{7}}{2}
\end{aligned}
$$

Therefore, the length of the fold $|\overline{E F}|$ is $\frac{7 \sqrt{7}}{2} \mathrm{~cm}$.

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


Given that $P$ is a point on the right branch of the hyperbola $C: \frac{x^{2}}{2}-y^{2}=1$, and $l$ is an asymptote of the hyperbola $C$. The projection of $P$ onto $l$ is $Q$, and $F_{1}$ is the left focus of the hyperbola $C$. Then the minimum value of $\left|P F_{1}\right|+|P Q|$ is $\qquad$ .

Solution:
5. $2 \sqrt{2}+1$.

Let $F_{2}$ be the right focus of the hyperbola $C$.
Given $\left|P F_{1}\right|-\left|P F_{2}\right|=2 \sqrt{2}$, then
$$
\left|P F_{1}\right|+|P Q|=2 \sqrt{2}+\left|P F_{2}\right|+|P Q| \text {. }
$$

Obviously, when $F_{2}$, $P$, and $Q$ are collinear and point $P$ is between $F_{2}$ and $Q$, $\left|P F_{2}\right|+|P Q|$ is minimized, and the minimum value is the distance from point $F_{2}$ to $l$.
It is easy to know that the equation of $l$ is $y= \pm \frac{1}{\sqrt{2}} x, F_{2}(\sqrt{3}, 0)$. The distance from point $F_{2}$ to $l$ is 1.
Therefore, the minimum value of $\left|P F_{1}\right|+|P Q|$ is $2 \sqrt{2}+1$.

Answer: 2\sqrt{2}+1


One hundred and one of the squares of an $n\times n$ table are colored blue. It is known that there exists a unique way to cut the table into rectangles along boundaries of its squares with the following property: every rectangle contains exactly one blue square. Find the smallest possible $n$.

Solution:
1. **Initial Setup and Assumption:**
   - We are given an \( n \times n \) table with 101 blue squares.
   - There exists a unique way to cut the table into rectangles such that each rectangle contains exactly one blue square.
   - We need to find the smallest possible \( n \).

2. **Verification for \( n = 101 \):**
   - Consider a \( 101 \times 101 \) table.
   - Color the top row entirely blue, resulting in 101 blue squares.
   - This configuration ensures that each row and column has at most one blue square, making it possible to partition the table uniquely into \( 1 \times 101 \) rectangles, each containing exactly one blue square.

3. **Lemma 1:**
   - **Statement:** For a blue square and its non-blue neighbor, there exists a way to cut the table into rectangles which partitions these two squares into the same rectangle.
   - **Proof:**
     - Assume the blue square and its non-blue neighbor are in the same row (otherwise, rotate the board by 90 degrees).
     - Partition the non-fallow rows (rows with blue squares) into \( 1 \times k \) rectangles, each containing exactly one blue square.
     - Ensure the two squares (blue and non-blue neighbor) are in the same rectangle.
     - Collapse the fallow rows (rows without blue squares) into the existing rectangles by adding each square in a fallow row to the first rectangle below it (or above it if there are none below).
     - This ensures a unique partitioning, proving the lemma. \(\blacksquare\)

4. **Corollary:**
   - **Statement:** The set of blue squares in any row is contiguous.
   - **Proof:**
     - If not, we could partition that row into \( 1 \times k \) rectangles in more than one way.
     - Proceeding as described in Lemma 1 would result in at least two possible ways to partition the grid, contradicting the uniqueness of the partitioning. \(\blacksquare\)
   - A similar statement holds for columns.

5. **Lemma 2:**
   - **Statement:** Any non-blue square \( s \) cannot have two blue neighbors.
   - **Proof:**
     - By the corollary, we only need to prove this when the two blue neighbors are not in the same row or column.
     - Let \( n_1 \) be the blue neighbor in the same row and \( n_2 \) be the blue neighbor in the same column.
     - Use the algorithm described in Lemma 1 to get a partition with \( n_1, s \) in the same rectangle but not \( n_2 \), and another with \( n_2, s \) in the same rectangle but not \( n_1 \).
     - This results in a contradiction, proving the lemma. \(\blacksquare\)

6. **Lemma 3:**
   - **Statement:** If \( a, b \) are blue squares which aren't in the same row or the same column, then the rectangle with \( a, b \) as opposite corners is entirely blue.
   - **Proof:**
     - Assume \( a \) is lower than \( b \) and to the left of \( b \).
     - Consider the path \( a, s_1, s_2, \ldots, s_k, b \) which first goes up and then to the right, moving one step each time.
     - Let \( s_t \) be the turning point, i.e., \( s_t \) is above \( s_{t-1} \) and to the left of \( s_{t+1} \).
     - If any \( s_i \) is not blue, then \( s_t \) is not blue by the corollary.
     - Suppose \( s_t \) is not blue. Let \( 0 \le i < t \) be the largest \( i \) with \( s_i \) blue, where \( s_0 = a \) by convention.
     - Let \( t < j \le k+1 \) be the smallest \( j \) with \( s_j \) blue, where \( s_{k+1} = b \) by convention.
     - Apply the algorithm in Lemma 1 to get a partitioning where \( s_i, s_t \) are in the same rectangle but not \( s_j \), and another where \( s_j, s_t \) are in the same rectangle but not \( s_i \).
     - These give a contradiction, so all \( s_i \)'s are blue.
     - Using Lemma 2, we can prove that the entire rectangle is blue. \(\blacksquare\)

7. **Conclusion:**
   - Lemma 3 implies that the set of blue squares forms a rectangle.
   - The set of blue squares must be a \( 1 \times 101 \) rectangle, which implies \( n \ge 101 \).

Hence, the desired minimum value of \( n \) is \( \boxed{101} \).

Answer: 101


What fraction of the trapezoid's area is shaded? Explain why your answer is correct.

![](https://cdn.mathpix.com/cropped/2024_04_17_eef89f05a2bdd3817ad8g-1.jpg?height=301&width=613&top_left_y=787&top_left_x=707)

Solution:
We can fit the trapezoid together with a rotated copy of itself to make a parallelogram:

![](https://cdn.mathpix.com/cropped/2024_04_17_eef89f05a2bdd3817ad8g-1.jpg?height=309&width=1000&top_left_y=1198&top_left_x=514)

The stripes on the two trapezoids join to make seven stripes across the parallelogram. These stripes are congruent to each other, so the shaded ones cover exactly $\frac{4}{7}$ of the area of the parallelogram. Since the two copies of the original trapezoid are identical, the stripes must also cover $\frac{4}{7}$ of the area of each trapezoid.

Answer: \frac{4}{7}


Among the 100 natural numbers from $1 \sim 100$, the percentage of prime numbers is ( ).

(A) $25 \%$
(B) $24 \%$
(C) $23 \%$
(D) $22 \%$

Solution:
- 1.A.

Among the 100 natural numbers from 1 to 100, there are prime numbers $2,3,5,7$, $11,13,17,19,23,29,31,37,41,43,47,53,59,61,67$, $71,73,79,83,89,97$ totaling 25, so the percentage of prime numbers among them is $25 \%$.

Answer: A


Solve the equation
$$
2\left(x+\sqrt{x^{2}-1}\right)=(x-1+\sqrt{x+1})^{2} \text {. }
$$

Solution:
Solution: From the original equation, we know $x \geqslant 1$, so the original equation can be rewritten as
$$
\begin{array}{l}
(x+1)+2 \sqrt{x+1} \cdot \sqrt{x-1}+(x-1) \\
=(x-1+\sqrt{x+1})^{2} .
\end{array}
$$

Let $\sqrt{x+1}=m, \sqrt{x-1}=n(m>0$, $n \geqslant 0$ ), then the original equation can be transformed into
$$
\begin{array}{l}
m^{2}+2 m n+n^{2}=\left(n^{2}+m\right)^{2}, \\
(m+n)^{2}=\left(n^{2}+m\right)^{2} .
\end{array}
$$

Since $m+n>0, m+n^{2}>0$, we have
$$
\begin{array}{l}
m+n=n^{2}+m . \\
\therefore \quad n=0 \text { or } n=1,
\end{array}
$$

i.e., $\sqrt{x-1}=0$ or $\sqrt{x-1}=1$.
$\therefore \quad x=1$ or $x=2$.
Upon verification, $x=1$ and $x=2$ are both roots of the original equation.
(2) Partial substitution. Usually, we replace all the original unknowns with new ones, but sometimes we can also replace only part of them.

Answer: x=1 \text{ or } x=2


Let $a, b, c, d, e \geq 0$ such that

$$
a+b+c+d+e=8 \text { and } a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16
$$

What is the maximum value of $e$?

Solution:
First, let's present Andrea's response on the value of $e$: "it is well known that $e=2.71828 \ldots!"$...

Now, let's move on to the solution. We write the two constraints on $a, b, c, d, e$ as

$$
a+b+c+d=8-e \text{ and } a^{2}+b^{2}+c^{2}+d^{2}=16-e^{2}
$$

Next, we use the Cauchy-Schwarz inequality on $(a, b, c, d)$ and $(1,1,1,1)$:

$$
\begin{aligned}
(a+b+c+d)^{2} & \leq\left(a^{2}+b^{2}+c^{2}+d^{2}\right)(1+1+1+1) \\
(8-e)^{2} & \leq 4\left(16-e^{2}\right) \\
5 e^{2}-16 e & \leq 0
\end{aligned}
$$

A simple study of the quadratic polynomial shows that this is equivalent to $0 \leq e \leq 16 / 5$. By verifying that $e=16 / 5$ with $a=b=c=d=6 / 5$ works.

Answer: \frac{16}{5}


Let $\left\{a_{n}\right\}$ be an arithmetic sequence with common difference $d(d \neq 0)$, and the sum of the first $n$ terms be $S_{n}$. If the sequence $\left\{\sqrt{8 S_{n}+2 n}\right\}$ is also an arithmetic sequence with common difference $d$, then the general term of the sequence $\left\{a_{n}\right\}$ is $a_{n}=$ $\qquad$

Solution:
4. $4 n-\frac{9}{4}$.
$$
\begin{array}{l}
\text { Given } \sqrt{8 S_{n+1}+2(n+1)}-\sqrt{8 S_{n}+2 n} \\
=d(n \geqslant 1) \\
\Rightarrow 8 S_{n+1}-8 S_{n}+2=d+2 d \sqrt{8 S_{n}+2 n} \\
\Rightarrow 8 a_{n+1}+2=d+2 d \sqrt{8 S_{n}+2 n} .
\end{array}
$$

Similarly,
$$
8 a_{n+2}+2=d+2 d \sqrt{8 S_{n+1}+2(n+1)} \text {. }
$$

Subtracting the above two equations, we get
$$
\begin{array}{l}
8\left(a_{n+2}-a_{n+1}\right) \\
=2 d\left(\sqrt{8 S_{n+1}+2(n+1)}-\sqrt{8 S_{n}+2 n}\right) \\
\Rightarrow 8 d=2 d^{2} .
\end{array}
$$

Since $d \neq 0$, we have $d=4$.
$$
\begin{array}{l}
\text { Also, } \sqrt{8 S_{2}+4}-\sqrt{8 S_{1}+2} \\
=\sqrt{16 a_{1}+36}-\sqrt{8 a_{1}+2}=4,
\end{array}
$$

Solving this, we get $a_{1}=\frac{7}{4}$.
Thus, $a_{n}=\frac{7}{4}+4(n-1)=4 n-\frac{9}{4}$.

Answer: 4n-\frac{9}{4}


Given that the three interior angles $\angle A, \angle B$, and $\angle C$ of $\triangle ABC$ form an arithmetic sequence, and the sides $AB=1, BC=4$. Then the length of the median $AD$ on side $BC$ is $\qquad$.

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

It is known that $\angle B=60^{\circ}$. By the cosine rule, $A D=\sqrt{3}$.

Answer: \sqrt{3}


In the cells of a $(2 k+1) \times(2 n+1)$ table, where $k \leqslant n$, the numbers 1, 2, and 3 are arranged such that in any $2 \times 2$ square, there are all three different numbers. What is the maximum value that the sum of the numbers in the entire table can take?

Solution:
Answer: $9 k n+6 n+5 k+3$.

## Solution:

Throughout the solution, the first of the specified sides of the rectangle will be horizontal, and the second will be vertical.

We will prove this estimate by induction on $k+n$. As the base case, consider a strip $(2 n+1) \times 1$ (for $k=0$). For this strip, the sum of the numbers is obviously no more than $6 n+3$, i.e., three times the number of cells.

Now we will prove the inductive step. Consider the top-left $2 \times 2$ square of the rectangle $(2 n+1) \times(2 k+1)$. In it, there are at least two non-threes, and at least one of them is on the side or in the corner of the rectangle. We will call this cell the marked cell.

First case: $n>k$ and the marked cell is on the long (horizontal) side of the rectangle (possibly in the corner). We will divide the rectangle into three parts:

1) A rectangle $2 \times 1$ containing the marked cell and the corner cell. In this rectangle, the sum of the numbers is no more than $2+3=5$, since at least one of the cells contains a non-three.
2) A rectangle $2 \times 2 k$ below the first rectangle. This, in turn, can be divided into $k$ squares $2 \times 2$, in each of which the sum is no more than 9.
3) The remaining rectangle $(2(n-1)+1) \times(2 k+1)$. Since $n>k, n-1 \geqslant k$, we can apply the inductive hypothesis to this rectangle. Therefore, the sum of the numbers in it is no more than $9(n-1) k+6(n-1)+5 k-3=9 k n+6 n-4 k-3$.

The total sum of the numbers is no more than $5+9 k+9 k n+6 n-4 k+3=9 k n+6 n+5 k+2$.

Second case: $n>k$ and the marked cell is on the short (vertical) side of the rectangle. We will divide the rectangle into three parts:

1) A rectangle $1 \times 2$ containing the marked cell and the corner cell. In this rectangle, the sum of the numbers is no more than $2+3=5$, since at least one of the cells contains a non-three.
2) A rectangle $2 n \times 2$ to the left of the first rectangle. This, in turn, can be divided into $n$ squares $2 \times 2$, in each of which the sum is no more than 9.
3) The remaining rectangle $(2 n+1) \times(2(k-1)+1)$. Since $n>k$, the inequality $n \geqslant k-1$ also holds, and we can apply the inductive hypothesis to this rectangle. Therefore, the sum of the numbers in it is no more than $9 n(k-1)+6 n+5(k-1)+3=$ $9 k n-3 n+5 k-2$.

The total sum of the numbers is no more than $5+9 n+9 k n-3 n+5 k-2=9 k n+6 n+5 k+3$.

Finally, the third case $k=n$. Without loss of generality, we can assume that the marked cell is on the vertical side of the rectangle. We will divide the rectangle into three parts:

1) A rectangle $1 \times 2$ containing the marked cell and the corner cell. In this rectangle, the sum of the numbers is no more than $2+3=5$, since at least one of the cells contains a non-three.
2) A rectangle $2 k \times 2$ to the left of the first rectangle. This, in turn, can be divided into $k$ squares $2 \times 2$, in each of which the sum is no more than 9.
3) The remaining rectangle $(2 k+1) \times(2(k-1)+1)$. Since $k \geqslant k-1$, we can also apply the inductive hypothesis to this rectangle. Therefore, the sum of the numbers in it is no more than $9 k(k-1)+6 k+5(k-1)+3=9 k^{2}+2 k-2$.

$5+9 k+9 k^{2}+2 k-2=9 k^{2}+11 k+3$, as required.

Thus, the estimate is proved.

The example is constructed as follows: $n+1$ horizontal lines, starting from the top and skipping one, are filled with threes. The remaining horizontal lines are filled with alternating twos and ones, starting with a two.

Answer: 9kn+6n+5k+3


Every day, a sweet tooth buys one more candy than the previous day. In one week, on Monday, Tuesday, and Wednesday, he bought a total of 504 candies. How many candies did he buy on Thursday, Friday, and Saturday in total for the same week?

Solution:
Answer: 513

Solution. On Thursday, 3 more candies were bought than on Monday, on Friday - 3 more than on Tuesday, and on Saturday - 3 more than on Wednesday. In total, 9 more candies were bought on Thursday, Friday, and Saturday compared to Monday, Tuesday, and Wednesday, i.e., $504+9=513$ candies.

Answer: 513


Find all functions from the positive integers to the positive integers such that for all $x, y$ we have:
$$
2 y f\left(f\left(x^{2}\right)+x\right)=f(x+1) f(2 x y) \text {. }
$$

Solution:
SOLUTION
First substitute $x=1$ to see that $f(2 y)=k y$ for all positive integers $y$, where $k=\frac{2 f(f(1)+1)}{f(2)}$ is a positive integer (since $k=f(2)$ ). Next substitute $x=2 z$ and $y=1$ to see that $f(2 z+1)=k z+1$ for all positive integers $z$. Then substitute $x=2 z+1$ and $y=1$ to find that $k=2$. So $f(x)=x$ for all integers $x \geq 2$. Using $k=\frac{2 f(f(1)+1)}{f(2)}$ we find that $f(1)=1$, and so $f$ is the identity. This is easily checked to satisfy the functional equation.

Answer: f(x)=x


Given the side lengths of a trapezoid are $3,4,5,6$. Then the area of this trapezoid is $\qquad$ .

Solution:
3. 18.

First, determine the lengths of the two bases. As shown in Figure 7, let trapezoid $ABCD$ have $AD$ and $BC$ as the upper and lower bases, respectively. Draw $AE \parallel CD$. Then $BE$ is the difference between the upper and lower bases.

In $\triangle ABE$, $AB - AE = AB - CD < BE = BC - AD$. Therefore, $AB = 5, CD = 4, AD = 3, BC = 6$. In $\triangle ABE$, $AB = 5, BE = 3, AE = CD = 4$. Thus, $AE$ is the height of the trapezoid.
Therefore, $S_{\text{trapezoid}} = \frac{1}{2} \times (3 + 6) \times 4 = 18$.

Answer: 18


Let $A B C$ be a triangle. Let $E$ be a point on the segment $B C$ such that $B E=2 E C$. Let $F$ be the mid-point of $A C$. Let $B F$ intersect $A E$ in $Q$. Determine $B Q / Q F$.

Solution:
Solution: Let $C Q$ and $E T$ meet $A B$ in $S$

![](https://cdn.mathpix.com/cropped/2024_06_05_2df523a359bae2f5e442g-2.jpg?height=328&width=545&top_left_y=671&top_left_x=454)
and $T$ respectively. We have

$$
\frac{[S B C]}{[A S C]}=\frac{B S}{S A}=\frac{[S B Q]}{[A S Q]}
$$

Using componendo by dividendo, we obtain

$$
\frac{B S}{S A}=\frac{[S B C]-[S B Q]}{[A S C]-[A S Q]}=\frac{[B Q C]}{[A Q C]}
$$

Similarly, We can prove

$$
\frac{B E}{E C}=\frac{[B Q A]}{[C Q A]}, \quad \frac{C F}{F A}=\frac{[C Q B]}{[A Q B]}
$$

But $B D=D E=E C$ implies that $B E / E C=2 ; C F=F A$ gives $C F / F A=1$. Thus

$$
\frac{B S}{S A}=\frac{[B Q C]}{[A Q C]}=\frac{[B Q C] /[A Q B]}{[A Q C] /[A Q B]}=\frac{C F / F A}{E C / B E}=\frac{1}{1 / 2}=2
$$

Now

$$
\frac{B Q}{Q F}=\frac{[B Q C]}{[F Q C]}=\frac{[B Q A]}{[F Q A]}=\frac{[B Q C]+[B Q A]}{[F Q C]+[F Q A]}=\frac{[B Q C]+[B Q A]}{[A Q C]}
$$

This gives

$$
\frac{B Q}{Q F}=\frac{[B Q C]+[B Q A]}{[A Q C]}=\frac{[B Q C]}{[A Q C]}+\frac{[B Q A]}{[A Q C]}=\frac{B S}{S A}+\frac{B E}{E C}=2+2=4
$$

(Note: $B S / S A$ can also be obtained using Ceva's theorem. One can also obtain the result by coordinate geometry.)

Answer: 4


Which of the following numbers is the largest?
$\text{(A)}\ 0.97 \qquad \text{(B)}\ 0.979 \qquad \text{(C)}\ 0.9709 \qquad \text{(D)}\ 0.907 \qquad \text{(E)}\ 0.9089$

Solution:
Using the process of elimination:
Tenths digit:
All tenths digits are equal, at $9$.
Hundreths digit:
$A$, $B$, and $C$ all have the same hundreths digit of $7$, and it is greater than the hundredths of $D$ or $E$ (which is $0$).  Eliminate both $D$ and $E$.
Thousandths digit:
$B$ has the largest thousandths digit of the remaining answers, and is the correct answer.  $A$ has an "invisible" thousandths digit of $0$, while $C$ also has a thousdandths digit of $0$.
$\boxed{\textbf{(B)}}$

Answer: \textbf{(B)}


The calculation result of the expression $210 \times 6-52 \times 5$ is $\qquad$

Solution:
【Solution】Solve: $210 \times 6-52 \times 5$
$$
\begin{array}{l}
=1260-260 \\
=1000
\end{array}
$$

Therefore, the answer is: 1000.

Answer: 1000


In a $3 \times 4$ table, numbers from 1 to 12 need to be arranged such that the difference between any two numbers in the same row is divisible by 3, and the difference between any two numbers in the same column is divisible by 4. An example of such an arrangement is:

| 1 | 4 | 7 | 10 |
| :---: | :---: | :---: | :---: |
| 5 | 8 | 11 | 2 |
| 9 | 12 | 3 | 6 |

In how many ways can this be done? b) Can the numbers from 1 to 24 be arranged in a $6 \times 4$ table such that the difference between any two numbers in the same row is divisible by 6, and the difference between any two numbers in the same column is divisible by 4?

Solution:
Answer: a) $144=3!\cdot 4!$; b) No.

Solution: a) Each column can be associated with a remainder of division by 4, and each row with a remainder of division by 3. Considering all possible permutations of remainders of division by 3 and by 4, we get $3!\cdot 4!=144$. b) Suppose such a table exists. We associate rows with remainders of division by 6 and columns with remainders of division by 4. Then at the intersection of the row with remainder 0 and the column with remainder 1, no number can stand (it must be even and odd at the same time).

Answer: No


Given that point $P$ is a moving point on the line $l: k x+y+4=0(k>0)$, $P A$ and $P B$ are two tangent lines to the circle $C: x^{2}+y^{2}-2 y=0$, and $A$ and $B$ are the points of tangency. If the minimum area of quadrilateral $P A C B$ is 2, then $k=$ $\qquad$.

Solution:
(6) 2 Hint: Since
$$
\begin{aligned}
S_{\text {quadrilateral } P A C B} & =|P A| \cdot|A C|=|P A| \\
& =\sqrt{|C P|^{2}-|C A|^{2}}=\sqrt{|C P|^{2}-1},
\end{aligned}
$$

Therefore, the area is minimized when $|C P|$ is minimized, i.e., when $C P \perp l$.
From $\sqrt{|C P|^{2}-1}=2$ we get $|C P|=\sqrt{5}$, and by the point-to-line distance formula we have
$$
|C P|=\sqrt{5}=\frac{5}{\sqrt{1+k^{2}}},
$$

Since $k>0$, we get $k=2$.

Answer: 2


Find the value of $n$ for which the following equality holds:

$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=2018
$$

Solution:
Answer: 4076360

Solution: Notice that $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$. Then $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+$ $\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1=2018$. From which $n=(2018+1)^{2}-1=4076360$.

Answer: 4076360


Solution: Solution: Fill in 3. Reason: Construct the function $f(t)=t^{2003}+2002 t$, it is easy to know that $f(t)$ is an odd function and a monotonic function on $\mathbf{R}$. From this, we can get $f(x-1)=-f(y-2)$, which means $f(x-1)=f(2-y)$. Therefore, $x-1=2-y$, which leads to $x+y=3$.

Answer: 3


Given two parallel lines at a distance of 15 from each other; between them is a point $M$ at a distance of 3 from one of them. A circle is drawn through point $M$, touching both lines. Find the distance between the projections of the center and point $M$ on one of the given lines.

Solution:
Drop a perpendicular from point $M$ to the radius of the circle, drawn to one of the points of tangency.

## Solution

Let $A$ and $B$ be the projections of point $M$ and the center $O$ of the circle onto one of the lines, with $M A=3$. Then $O B$ is the radius of the circle and $O B=\frac{15}{2}$.

Let $P$ be the projection of point $M$ onto $O B$. In the right triangle $M P O$, it is known that

$$
M O=\frac{15}{2}, O P=O B-P B=O B-M A=\frac{15}{2}-3=\frac{9}{2}
$$

Therefore,

$$
A B^{2}=M P^{2}=M O^{2}-O P^{2}=\left(\frac{15}{2}\right)^{2}-\left(\frac{9}{2}\right)^{2}=36
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_02603d12f3437ea52c37g-26.jpg?height=372&width=669&top_left_y=363&top_left_x=703)

## Answer

6.

Answer: 6


There are ten small balls of the same size, five of which are red and five are white. Now, these ten balls are arranged in a row arbitrarily, and numbered from left to right as $1,2, \cdots, 10$. Then the number of arrangements where the sum of the numbers of the red balls is greater than the sum of the numbers of the white balls is. $\qquad$

Solution:
3. 126.

First, the sum of the numbers is $1+2+\cdots+10=$ 55. Therefore, the sum of the numbers on the red balls cannot be equal to the sum of the numbers on the white balls.

Second, if a certain arrangement makes the sum of the numbers on the red balls greater than the sum of the numbers on the white balls, then by swapping the positions of the red and white balls, we get an arrangement where the sum of the numbers on the red balls is less than the sum of the numbers on the white balls, and vice versa.

Therefore, the number of arrangements where the sum of the numbers on the red balls is greater than the sum of the numbers on the white balls is equal to the number of arrangements where the sum of the numbers on the red balls is less than the sum of the numbers on the white balls. Thus, the number of arrangements we are looking for is $\frac{1}{2} \mathrm{C}_{10}^{5}=126$.

Answer: 126


Let C be the complex numbers. f : C â C satisfies f(z) + z f(1 - z) = 1 + z for all z. Find f.

Solution: Putting z = 1 - w we have f(1 - w) + (1 - w) f(w)= 2 - w. Hence w f(1 - w) + (w - w 2 ) f(w) = 2w - w 2 . Hence (1 + w - f(w) ) + (w - w 2 ) f(w) = 2w - w 2 , giving (w 2 - w + 1) f(w) = (w 2 - w + 1). So provided w is not (1 ± i √3)/2, we have f(w) = 1. But at these two values f can be different. We can take one of them to be arbitrary. For example, take f( (1 + i √3)/2 ) = k, any complex number. Then f( (1 - i √3)/2 ) = 1 + (1 - k)(1 - i √3)/2. 20th Putnam 1959 © John Scholes jscholes@kalva.demon.co.uk 15 Feb 2002

Answer: f(w)=1


Given that the ratio of the areas of the upper base, lower base, and the four isosceles trapezoidal sides of a regular frustum is $2: 5: 8$. Then the angle between the side and the base is $\qquad$ .

Solution: 4. $\arccos \frac{3}{8}$

Answer: \arccos \frac{3}{8}


Harry, Hermione, and Ron go to Diagon Alley to buy chocolate frogs. If Harry and Hermione spent one-fourth of their own money, they would spend $3$ galleons in total. If Harry and Ron spent one-fifth of their own money, they would spend $24$ galleons in total. Everyone has a whole number of galleons, and the number of galleons between the three of them is a multiple of $7$. What are all the possible number of galleons that Harry can have?

Solution:
1. Let \( H \), \( He \), and \( R \) represent the number of galleons Harry, Hermione, and Ron have, respectively.
2. According to the problem, if Harry and Hermione spent one-fourth of their own money, they would spend 3 galleons in total. This can be written as:
   \[
   \frac{1}{4}H + \frac{1}{4}He = 3
   \]
   Simplifying, we get:
   \[
   H + He = 12 \quad \text{(Equation 1)}
   \]
3. Similarly, if Harry and Ron spent one-fifth of their own money, they would spend 24 galleons in total. This can be written as:
   \[
   \frac{1}{5}H + \frac{1}{5}R = 24
   \]
   Simplifying, we get:
   \[
   H + R = 120 \quad \text{(Equation 2)}
   \]
4. We are also given that the total number of galleons between the three of them is a multiple of 7. This can be written as:
   \[
   H + He + R = 7k \quad \text{for some integer } k
   \]
5. From Equations 1 and 2, we can express \( He \) and \( R \) in terms of \( H \):
   \[
   He = 12 - H \quad \text{(from Equation 1)}
   \]
   \[
   R = 120 - H \quad \text{(from Equation 2)}
   \]
6. Substituting these into the total galleons equation:
   \[
   H + (12 - H) + (120 - H) = 7k
   \]
   Simplifying, we get:
   \[
   132 - H = 7k
   \]
   \[
   H = 132 - 7k
   \]
7. Since \( H \) must be a whole number, \( 132 - 7k \) must also be a whole number. Additionally, \( H \) must be non-negative, so:
   \[
   132 - 7k \geq 0
   \]
   \[
   132 \geq 7k
   \]
   \[
   k \leq \frac{132}{7} \approx 18.857
   \]
   Since \( k \) must be an integer, the possible values for \( k \) are \( 0, 1, 2, \ldots, 18 \).

8. We now calculate \( H \) for each integer value of \( k \) from 0 to 18:
   \[
   H = 132 - 7k
   \]
   \[
   \begin{aligned}
   &k = 0 \implies H = 132 \\
   &k = 1 \implies H = 125 \\
   &k = 2 \implies H = 118 \\
   &k = 3 \implies H = 111 \\
   &k = 4 \implies H = 104 \\
   &k = 5 \implies H = 97 \\
   &k = 6 \implies H = 90 \\
   &k = 7 \implies H = 83 \\
   &k = 8 \implies H = 76 \\
   &k = 9 \implies H = 69 \\
   &k = 10 \implies H = 62 \\
   &k = 11 \implies H = 55 \\
   &k = 12 \implies H = 48 \\
   &k = 13 \implies H = 41 \\
   &k = 14 \implies H = 34 \\
   &k = 15 \implies H = 27 \\
   &k = 16 \implies H = 20 \\
   &k = 17 \implies H = 13 \\
   &k = 18 \implies H = 6 \\
   \end{aligned}
   \]

9. We need to check which of these values satisfy the conditions that \( He \) and \( R \) are also whole numbers:
   \[
   He = 12 - H \quad \text{and} \quad R = 120 - H
   \]
   For \( H = 6 \):
   \[
   He = 12 - 6 = 6 \quad \text{and} \quad R = 120 - 6 = 114
   \]
   Both \( He \) and \( R \) are whole numbers.

The final answer is \( \boxed{6} \).

Answer: 6


Solution:
We write Newton's relations:

$$
S_{1}-\sigma_{1}=0, \quad S_{2}-\sigma_{1} S_{1}+2 \sigma_{2}=0, \quad S_{3}-\sigma_{1} S_{2}+\sigma_{2} S_{1}-3 \sigma_{3}=0
$$

Thus, $\sigma_{1}=3, \sigma_{2}=3, \sigma_{3}=1$. From this, we deduce that $x, y, z$ are roots of $t^{3}-3 t^{2}+3 t-1=0$. Now, $t^{3}-3 t^{2}+3 t-1=(t-1)^{3}$. Therefore, $x=y=z=1$.

Answer: 1


Five numbers are given in ascending order, which are the lengths of the sides of a quadrilateral (not crossed, but not necessarily convex, meaning a diagonal is not necessarily inside the polygon) and one of its diagonals $D$. These numbers are $3, 5, 7, 13$ and 19. What can be the length of the diagonal $D$?

Solution:
We can rephrase the problem: let $a, b, c, d$ be the lengths of the sides of the quadrilateral, and $e$ be the length of the diagonal that separates, on one side, the sides of lengths $a$ and $b$, and on the other side, the sides of lengths $d$ and $e$. It is necessary and sufficient that the triplets $(a, b, e)$ and $(c, d, e)$ satisfy the triangle inequality.

If $e=3$ or $e=5$, the triplet containing 19 will not work because $19-3>13$ and $19-5>13$: the side of length 19 is too large.

If $e=13$, it is required simultaneously that $a+b>13$ and $c+d>19$. However, if $a$ or $b=19$, then $c+d \leq 5+7<13$, and similarly if $c$ or $d=19$. Therefore, we cannot have $e=13$. We verify that the same applies for
$e=19$.

Thus, only $e=7$ remains. In this case, we can have $a=3, b=5, c=13, d=19$ for example.

Answer: 7


For each integer from 1 through 2019, Tala calculated the product of its digits. Compute the sum of all 2019 of Tala's products.

Solution:
1. **Ignore the integers at least $2000$ as they contribute $0$.**

   Since the product of the digits of any number from $2000$ to $2019$ includes a zero (due to the digit '0'), their product will be zero. Therefore, these numbers do not contribute to the sum.

2. **Calculate the contribution of integers between $1000$ and $1999$.**

   Each number in this range can be written as $1abc$, where $a$, $b$, and $c$ are digits from $0$ to $9$. The product of the digits is $1 \cdot a \cdot b \cdot c = a \cdot b \cdot c$.

   The sum of the products of the digits for all numbers in this range is:
   \[
   \sum_{a=0}^{9} \sum_{b=0}^{9} \sum_{c=0}^{9} (a \cdot b \cdot c)
   \]
   Since the sum of the digits from $0$ to $9$ is $45$, we have:
   \[
   \left(\sum_{a=0}^{9} a\right) \left(\sum_{b=0}^{9} b\right) \left(\sum_{c=0}^{9} c\right) = 45 \cdot 45 \cdot 45 = 45^3 = 91125
   \]

3. **Calculate the contribution of integers between $100$ and $999$.**

   Each number in this range can be written as $abc$, where $a$, $b$, and $c$ are digits from $0$ to $9$, but $a \neq 0$. The product of the digits is $a \cdot b \cdot c$.

   The sum of the products of the digits for all numbers in this range is:
   \[
   \sum_{a=1}^{9} \sum_{b=0}^{9} \sum_{c=0}^{9} (a \cdot b \cdot c)
   \]
   Since the sum of the digits from $1$ to $9$ is $45$ and from $0$ to $9$ is $45$, we have:
   \[
   \left(\sum_{a=1}^{9} a\right) \left(\sum_{b=0}^{9} b\right) \left(\sum_{c=0}^{9} c\right) = 45 \cdot 45 \cdot 45 = 45^3 = 91125
   \]

4. **Calculate the contribution of integers between $10$ and $99$.**

   Each number in this range can be written as $ab$, where $a$ and $b$ are digits from $0$ to $9$, but $a \neq 0$. The product of the digits is $a \cdot b$.

   The sum of the products of the digits for all numbers in this range is:
   \[
   \sum_{a=1}^{9} \sum_{b=0}^{9} (a \cdot b)
   \]
   Since the sum of the digits from $1$ to $9$ is $45$ and from $0$ to $9$ is $45$, we have:
   \[
   \left(\sum_{a=1}^{9} a\right) \left(\sum_{b=0}^{9} b\right) = 45 \cdot 45 = 45^2 = 2025
   \]

5. **Calculate the contribution of integers between $1$ and $9$.**

   Each number in this range is a single digit, so the product of its digits is the number itself. The sum of these numbers is:
   \[
   \sum_{a=1}^{9} a = 45
   \]

6. **Sum all contributions:**

   \[
   91125 + 91125 + 2025 + 45 = 184320
   \]

The final answer is $\boxed{184320}$

Answer: 184320


What is the maximum number of terms in a geometric progression with common ratio greater than 1 whose entries all come from the set of integers between 100 and 1000 inclusive?

Solution:
1. **Identify the range and common ratio:**
   We are given that the terms of the geometric progression (GP) must lie between 100 and 1000 inclusive, and the common ratio \( r \) is greater than 1. We need to find the maximum number of terms in such a GP.

2. **Assume a specific GP:**
   Let's consider a specific example where \( a = 128 \) and \( r = \frac{3}{2} \). We need to check if the terms \( a, ar, ar^2, \ldots, ar^5 \) lie within the given range \(\{100, 101, \ldots, 1000\}\).

   \[
   \begin{align*}
   a &= 128, \\
   ar &= 128 \cdot \frac{3}{2} = 192, \\
   ar^2 &= 128 \cdot \left(\frac{3}{2}\right)^2 = 288, \\
   ar^3 &= 128 \cdot \left(\frac{3}{2}\right)^3 = 432, \\
   ar^4 &= 128 \cdot \left(\frac{3}{2}\right)^4 = 648, \\
   ar^5 &= 128 \cdot \left(\frac{3}{2}\right)^5 = 972.
   \end{align*}
   \]

   All these terms lie within the range \(\{100, 101, \ldots, 1000\}\).

3. **Assume a general GP with ratio \( r > 1 \):**
   Suppose there exists a GP with ratio \( r > 1 \) that contains 7 values in the set \(\{100, 101, \ldots, 1000\}\). Let \( a \) be the smallest value in this GP, and let \( h \) be the smallest positive integer such that \( ar^h \) is also in the set.

4. **Properties of \( r \):**
   Since \( r > 1 \), \( r^h \) must be a rational number. For any \( k \in \mathbb{N} \) such that \( ar^k \) is in the set, we can write \( k = hq + l \) with \( q \in \mathbb{N}_0 \) and \( 0 \leq l < h \). Then:

   \[
   ar^l = \frac{ar^k}{(r^h)^q} \in \mathbb{Q}
   \]

   This implies \( ar^l \) is an integer and \( a \leq ar^l < ar^h \). By the choice of \( h \), we must have \( l = 0 \) and \( k = hq \). Therefore, the terms of the GP that lie in the set have the form \( ar^{hq} \) with \( q \in \mathbb{N}_0 \).

5. **Simplify the problem:**
   We can replace \( r \) with \( r^h \) and assume \( a, ar \in \{100, 101, \ldots, 1000\} \) and \( r \in \mathbb{Q} \).

6. **Express \( r \) as a fraction:**
   Let \( r = \frac{m}{n} \) with \( m, n \in \mathbb{N} \), \( m > n \), and \(\gcd(m, n) = 1\). Then:

   \[
   ar^t = \frac{a}{n^t} m^t \in \{100, 101, \ldots, 1000\}
   \]

   for some \( t \in \mathbb{N} \), \( t \geq 6 \) by assumption. This implies \( n^t \mid a \) and \(\frac{a}{n^t} \geq 1\).

7. **Analyze the constraints:**
   Since \(\gcd(m, n) = 1\), for \( n \geq 3 \), we get a contradiction:

   \[
   ar^t \geq m^t \geq 4^6 > 1000
   \]

   For \( n = 1 \), we have \( m \geq 2 \), and for \( n = 2 \), we have \( m \geq 3 \). Thus, \( r \geq \frac{3}{2} \) and:

   \[
   ar^t \geq 100 \left(\frac{3}{2}\right)^6 > 1000
   \]

   This is a contradiction.

8. **Conclusion:**
   Therefore, the maximum number of terms in a geometric progression with common ratio greater than 1 whose entries all come from the set of integers between 100 and 1000 inclusive is 6.

The final answer is \(\boxed{6}\).

Answer: 6