As shown in Figure 2, in the Cartesian coordinate system, a line segment $P Q$ of length 6 has one endpoint $P$ sliding on the ray $y=0(x \leqslant 0)$, and the other endpoint $Q$ sliding on the ray $x=0(y \leqslant 0)$. Point $M$ is on line segment $P Q$, and $\frac{P M}{M Q}=\frac{1}{2}$.
(1) Find the equation of the trajectory of point $M$;
(2) If the trajectory of point $M$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively, find the maximum area of quadrilateral $O A M B$ (where $O$ is the origin).

Solution:
(1) Let $P\left(x_{1}, 0\right)$, $Q\left(0, y_{1}\right)$, and $M(x, y)$, where $x \leqslant 0$ and $y_{1} \leqslant 0$.
From the conditions, we have
$$
x_{1}^{2}+y_{1}^{2}=36 \text{. }
$$

Also, given $\lambda=\frac{P M}{M Q}=\frac{1}{2}$, we get
$$
x=\frac{x_{1}}{1+\frac{1}{2}}, y=\frac{\frac{1}{2} y_{1}}{1+\frac{1}{2}} \text{. }
$$

Substituting $x_{1}=\frac{3}{2} x$ and $y_{1}=3 y$ into equation (1), we obtain
$$
\frac{x^{2}}{16}+\frac{y^{2}}{4}=1(x, y \leqslant 0) \text{. }
$$

Therefore, the equation of the trajectory of point $M$ is
$$
\frac{x^{2}}{16}+\frac{y^{2}}{4}=1(x \leqslant 0, y \leqslant 0) \text{. }
$$
(2) As shown in Figure 5, let point
$M(4 \cos \alpha, 2 \sin \alpha)$, where $0 \leqslant \alpha < 2 \pi$.
From the conditions $\left\{\begin{array}{l}4 \cos \alpha<0, \\ 2 \sin \alpha<0\end{array}\right.$, we get
$$
\pi<\alpha<\frac{3 \pi}{2} \text{. }
$$

Thus, $S_{\text {quadrilateral, MMB }}=S_{\triangle M M}+S_{\triangle O M M}$.
And $S_{\triangle M M}=\frac{1}{2}|O A| \cdot\left|y_{M}\right|=\frac{1}{2} \times 4 \times|2 \sin \alpha|$ $=-4 \sin \alpha$,
$S_{\text {ΔOEM }}=\frac{1}{2}|O B| \cdot\left|x_{M}\right|=\frac{1}{2} \times 2 \times|4 \cos \alpha|$
$$
=-4 \cos \alpha \text{, }
$$

Therefore, $S_{\text {quadrilateral } 04 m B}=-4(\sin \alpha+\cos \alpha)$
$$
=-4 \sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right) \leqslant 4 \sqrt{2} \text{. }
$$

The maximum area of quadrilateral $O A M B$ is $4 \sqrt{2}$ when and only when $\alpha=\frac{5 \pi}{4} \in\left(\pi, \frac{3 \pi}{2}\right)$.

Answer: 4 \sqrt{2}


For a set $S$, let $|S|$ denote the number of elements in $S$, and let $n(S)$ denote the number of subsets of $S$ including the empty set and $S$ itself. If three sets $A, B, C$ satisfy $n(A)+n(B)+n(C)=n(A \cup B \cup C), |A|=|B|=100$, find the minimum possible value of $|A \cap B \cap C|$.

Solution:
8. Since $n(x)=2^{|x|}$, combining with the given conditions we have: $2^{|A|}+2^{|B|}+2^{|C|}=2^{|A \cup B \cup C|}$, and $|A|=|B|=100$. Therefore,
(1) If $|A \cap B|=100$, and $|A|=|B|=100,|C|=101$, and $|A \cup B \cup C|=102$, then $C$ has exactly two elements not in $A \cap B$, thus $|A \cap B \cap C|=99$;
(2) If $|A \cap B|=99$, then $|A \cup B|=101$, and by $|C|=101,|A \cup B \cup C|=102$, then $C$ has exactly one element not in $A \cup B$. To minimize $|A \cap B \cap C|$, then $|C \cap[A \backslash(A \cap B)]| \geqslant 1$, and $|C \cap[B \backslash(A \cap B)]| \geqslant 1$, at this time $|A \cap B \cap C| \geqslant 98$;
(3) If $|A \cap B|=98$, to minimize $|A \cap B \cap C|$, it should be that $|C \cap(A \backslash(A \cap B))|=2, \mid C \cap (B \backslash(A \cap B)) \mid=2$, and since $|A \cup B \cup C|=102$, and $|C|=101$, thus $|A \cap B \cap C| \geqslant 97$;
(4) If $|A \cap B| \leqslant 97$, then $|A \backslash(A \cap B)| \geqslant 3,|B \backslash(A \cap B)| \geqslant 3$, thus $|A \cup B| \geqslant 100+3=103$, which is a contradiction.
Combining (1)(2)(3)(4), we know that the minimum possible value of $|A \cap B \cap C|$ is 97.

Answer: 97


Solution:
5.B.

According to the problem, we have
$$
M \geqslant x_{1}+x_{2}, M \geqslant x_{2}+x_{3}, M \geqslant x_{4}+x_{5} .
$$

Adding the three inequalities, we get $3 M \geqslant 100+x_{2} \geqslant 100$, which means $M \geqslant \frac{100}{3}$. When $x_{2}=x_{4}=0, x_{1}=x_{3}=x_{5}=\frac{100}{3}$, we have $M=\frac{100}{3}$. Therefore, the minimum value of $M$ is $m=\frac{100}{3}$.

Answer: B


The number of sets of two or more consecutive positive integers whose sum is 100 is
(A) 1.
(B) 2.
(C) 3.
(D) 4.
(E) 5.

Solution:
[Solution] In a set of consecutive integers, the number of elements can be odd or even. If it is odd, let it be $2n+1$, and their average $x$ is the central one.
Then $(2n+1)x=100$, so $x=\frac{100}{2n+1}$.
Thus $(2n+1)$ can only be 5 or 25.
If $2n+1=5$, then $x=20, n=2$, so these integers are
$$
18,19,20,21,22 \text{. }
$$

If $(2n+1)=25, x=4$ and $n=12$, this is impossible because the integers must be positive.

If the number of consecutive integers is even $2n$, let their average (between the middle pair of numbers) be represented by $x$, then $2nx=100$, i.e., $x=\frac{50}{n}$.
For $n=4$, i.e., $x=12\frac{1}{2}$, the middle pair of numbers are
$$
x-\frac{1}{2}=12, \quad x+\frac{1}{2}=13,
$$

so $9,10,11,12,13,14,15,16$ are these $2n=8$ integers.
For other integers $n$, there is no middle pair $\left(x-\frac{1}{2}\right)$ and $\left(x+\frac{1}{2}\right)$ that are positive integers.
In summary, the sets of positive integers whose sum is 100 are only the two mentioned above.
Therefore, the answer is $(B)$.

Answer: B


Solution: (1992·1996, 4·1992+8), (1992·1988, -4·1992+8) Solution a n satisfies a standard linear recurrence relation with general solution a n = n 2 + An + k. But a n = 0, so k = 0. Hence a n = n 2 + An. If you are not familiar with the general solution, then you can guess this solution and prove it by induction. Similarly, b n = Bn + 8. Hence a n 2 +b n 2 = n 4 + 2An 3 + (A 2 +B 2 )n 2 + 16Bn + 64. If this is a square, then looking at the constant and n 3 terms, it must be (n 2 + An + 8). Comparing the other terms, A = B = ±4. Thanks to Juan Ignacio Restrepo 7th Ibero 1992 © John Scholes jscholes@kalva.demon.co.uk 29 March 2004 Last corrected/updated 29 Mar 04

Answer: (1992\cdot1996,4\cdot1992+8),(1992\cdot1988,-4\cdot1992+8)


Point $A,B$ are marked on the right branch of the hyperbola $y=\frac{1}{x},x>0$. The straight line $l$ passing through the origin $O$ is perpendicular to $AB$ and meets $AB$ and given branch of the hyperbola at points $D$ and $C$ respectively. The circle through $A,B,C$ meets $l$ at $F$. Find $OD:CF$

Solution:
1. **Identify the given elements and their relationships:**
   - The hyperbola is given by \( y = \frac{1}{x} \) for \( x > 0 \).
   - Points \( A \) and \( B \) lie on this hyperbola.
   - Line \( l \) passes through the origin \( O \) and is perpendicular to \( AB \).
   - Line \( l \) intersects the hyperbola at points \( C \) and \( M \).
   - The circle through points \( A \), \( B \), and \( C \) intersects line \( l \) at point \( F \).

2. **Determine the coordinates of points \( A \) and \( B \):**
   - Let \( A = (a, \frac{1}{a}) \) and \( B = (b, \frac{1}{b}) \) where \( a, b > 0 \).

3. **Find the equation of line \( AB \):**
   - The slope of \( AB \) is given by:
     \[
     \text{slope of } AB = \frac{\frac{1}{b} - \frac{1}{a}}{b - a} = \frac{a - b}{ab(a - b)} = \frac{1}{ab}
     \]
   - Therefore, the equation of line \( AB \) is:
     \[
     y - \frac{1}{a} = \frac{1}{ab}(x - a) \implies y = \frac{1}{ab}x + \left(\frac{1}{a} - \frac{1}{ab}a\right) = \frac{1}{ab}x
     \]

4. **Find the equation of line \( l \):**
   - Since \( l \) is perpendicular to \( AB \) and passes through the origin, its slope is the negative reciprocal of the slope of \( AB \):
     \[
     \text{slope of } l = -ab
     \]
   - Therefore, the equation of line \( l \) is:
     \[
     y = -abx
     \]

5. **Find the coordinates of point \( C \):**
   - Point \( C \) is the intersection of line \( l \) and the hyperbola:
     \[
     y = -abx \quad \text{and} \quad y = \frac{1}{x}
     \]
     \[
     -abx = \frac{1}{x} \implies -abx^2 = 1 \implies x^2 = -\frac{1}{ab} \implies x = \frac{1}{\sqrt{ab}}
     \]
     \[
     y = -ab \left(\frac{1}{\sqrt{ab}}\right) = -\sqrt{ab}
     \]
     Thus, \( C = \left(\frac{1}{\sqrt{ab}}, -\sqrt{ab}\right) \).

6. **Determine the coordinates of point \( M \):**
   - Since \( M \) is another intersection of line \( l \) with the hyperbola, and considering the symmetry, \( M \) will have the same coordinates as \( C \) but with opposite signs:
     \[
     M = \left(-\frac{1}{\sqrt{ab}}, \sqrt{ab}\right)
     \]

7. **Find the orthocenter \( C \) of \( \Delta ABM \):**
   - The orthocenter \( C \) of \( \Delta ABM \) is already given as \( C = \left(\frac{1}{\sqrt{ab}}, -\sqrt{ab}\right) \).

8. **Determine the midpoint \( O \) of \( CM \):**
   - The midpoint \( O \) of \( CM \) is:
     \[
     O = \left(\frac{\frac{1}{\sqrt{ab}} + \left(-\frac{1}{\sqrt{ab}}\right)}{2}, \frac{-\sqrt{ab} + \sqrt{ab}}{2}\right) = (0, 0)
     \]

9. **Find the reflection \( F \) of \( M \) in \( AB \):**
   - Since \( F \) is the reflection of \( M \) in \( AB \), and \( AB \) is the perpendicular bisector of \( CM \), \( F \) will be at the same distance from \( C \) as \( M \) but on the opposite side of \( AB \).

10. **Determine the ratio \( OD:CF \):**
    - Since \( O \) is the midpoint of \( CM \) and \( D \) is the midpoint of \( CE \), where \( E \) is the reflection of \( C \) in \( AB \):
      \[
      ME = 2OD
      \]
    - Therefore, the ratio \( OD:CF \) is:
      \[
      OD:CF = 1:2
      \]

The final answer is \( \boxed{ 1:2 } \)

Answer: 1:2


Let $x, y$ and $z$ be real numbers that satisfy $x+\frac{1}{y}=4, y+\frac{1}{z}=1$ and $z+\frac{1}{x}=\frac{7}{3}$. Find the value of $x y z$.

Solution:
Method 1
From (1), $x=4-\frac{1}{y}=\frac{4 y-1}{y}$
$$
\Rightarrow \frac{1}{x}=\frac{y}{4 y-1}
$$

Sub. (4) into (3): $z+\frac{y}{4 y-1}=\frac{7}{3}$
$$
z=\frac{7}{3}-\frac{y}{4 y-1}
$$

From (2): $\frac{1}{z}=1-y$ $z=\frac{1}{1-y} \ldots \ldots$ (6)
(5) $=$ (6): $\frac{1}{1-y}=\frac{7}{3}-\frac{y}{4 y-1}$
$\frac{1}{1-y}=\frac{28 y-7-3 y}{3(4 y-1)}$
$3(4 y-1)=(1-y)(25 y-7)$
$12 y-3=-25 y^{2}-7+32 y$
$25 y^{2}-20 y+4=0$
$(5 y-2)^{2}=0$
$$
y=\frac{2}{5}
$$

Sub. $y=\frac{2}{5}$ into (6): $z=\frac{1}{1-\frac{2}{5}}=\frac{5}{3}$
Sub. $y=\frac{2}{5}$ into (1): $x+\frac{5}{2}=4 \Rightarrow x=\frac{3}{2}$
Method 2
$\left\{\begin{array}{l}x+\frac{1}{y}=4 \cdots \cdots(1) \\ y+\frac{1}{z}=1 \cdots \cdots(2) \\ z+\frac{1}{x}=\frac{7}{3} \cdots \cdots(3)\end{array}\right.$
(1) $\times(2): \quad x y+1+\frac{x}{z}+\frac{1}{y z}=4$
$x\left(y+\frac{1}{z}\right)+\frac{1}{y z}=3$
Sub. (2) into the eqt.: $x+\frac{x}{x y z}=3$
Let $a=x y z$, then $x+\frac{x}{a}=3 \cdots \cdots(4)$
(2) $\times(3): y\left(\frac{7}{3}\right)+\frac{y}{a}=\frac{4}{3} \Rightarrow y\left(\frac{7}{3}+\frac{1}{a}\right)=\frac{4}{3} \ldots \ldots$.
(1) $\times(3): z(4)+\frac{z}{a}=\frac{25}{3} \Rightarrow z\left(4+\frac{1}{a}\right)=\frac{25}{3} \cdots \cdots$
(4) $\times(5) \times(6): a\left(1+\frac{1}{a}\right)\left(\frac{7}{3}+\frac{1}{a}\right)\left(4+\frac{1}{a}\right)=\frac{100}{3}$
$\frac{(a+1)(7 a+3)(4 a+1)}{3 a^{2}}=\frac{100}{3}$
which reduces to $28 a^{3}-53 a^{2}+22 a+3=0$
$\Rightarrow(a-1)^{2}(28 a+3)=0$
$\therefore a=1$ $x y z=\frac{2}{5} \times \frac{5}{3} \times \frac{3}{2}=1$
Method $3(1) \times(2) \times(3)-(1)-(2)-(3)$ :
$x y z+x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x y z}-\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{28}{3}-\frac{22}{3} \Rightarrow x y z+\frac{1}{x y z}=2$
$x y z=1$

Answer: 1


Given $\min _{x \in R} \frac{a x^{2}+b}{\sqrt{x^{2}+1}}=3$.
(1) Find the range of $b$;
(2) For a given $b$, find $a$.

Solution:
1. Solution 1 (1) Let $f(x)=\frac{a x^{2}+b}{\sqrt{x^{2}+1}}$.
Then $f(0)=b \geqslant 3$. Furthermore, $a>0$.
(i) When $b-2 a \geqslant 0$,
$$
\begin{array}{l}
f(x)=\frac{a x^{2}+b}{\sqrt{x^{2}+1}} \\
=a \sqrt{x^{2}+1}+\frac{b-a}{\sqrt{x^{2}+1}} \\
\geqslant 2 \sqrt{a(b-a)}=3 .
\end{array}
$$

The equality holds if and only if $a \sqrt{x^{2}+1}=\frac{b-a}{\sqrt{x^{2}+1}}$, i.e., $x=$ $\pm \sqrt{\frac{b-2 a}{a}}$.
At this point, $a=\frac{b-\sqrt{b^{2}-9}}{2}$.
In particular, when $b=3$, $a=\frac{3}{2}$.
(ii) When $b-2 a<0$,
$$
a>\frac{3}{2}.
$$
In summary, the range of $b$ is $[3,+\infty)$,
(2) When $b=3$, $a \geqslant \frac{3}{2}$;

When $b>3$, $a=\frac{b-\sqrt{b^{2}-9}}{2}$.
Solution 2 (1) Let $f(x)=\frac{a x^{2}+b}{\sqrt{x^{2}+1}}$.
Since $\min _{x \in \mathbb{R}} \frac{a x^{2}+b}{\sqrt{x^{2}+1}}=3$, and $f(0)=b$, thus, $b \geqslant 3$.
It is easy to see that $a>0, f^{\prime}(x)=\frac{a x\left(x^{2}-\frac{b-2 a}{a}\right)}{\left(x^{2}+1\right)^{\frac{3}{2}}}$.
(i) When $b-2 a \leqslant 0$, let $f^{\prime}(x)=0$.

Solving gives $x_{0}=0$.
When $x<0$, $f^{\prime}(x)<0$; when $x>0$, $f^{\prime}(x)>0$.
Thus, $f(0)=b$ is the minimum value.
Therefore, $b=3$.
(ii) When $b-2 a>0$, let $f^{\prime}(x)=0$.

Solving gives $x_{0}=0, x_{1,2}= \pm \sqrt{\frac{b-2 a}{a}}$.
At this point, it is easy to see that
$f(0)=b$ is not the minimum value
$\Rightarrow b>3$, and $f\left(x_{1,2}\right)$ is the minimum value
$$
\begin{array}{l}
\Rightarrow f\left(x_{1,2}\right)=\frac{a \cdot \frac{b-2 a}{a}+b}{\sqrt{\frac{b-2 a}{a}+1}}=3 \\
\Rightarrow 2 \sqrt{a(b-a)}=3 \\
\Rightarrow a^{2}-a b+\frac{9}{4}=0 \\
\Rightarrow a=\frac{b-\sqrt{b^{2}-9}}{2} .
\end{array}
$$

In summary, the range of $b$ is $[3,+\infty)$.
(2) When $b=3$, $a \geqslant \frac{3}{2}$;

When $b>3$, $a=\frac{b-\sqrt{b^{2}-9}}{2}$.

Answer: a=\frac{b-\sqrt{b^{2}-9}}{2}


Given that $A$ is a positive integer and satisfies the following conditions: $\frac{5}{9}<\frac{9}{A}<1$. How many different values can $A$ take?

Solution:
【Analysis and Solution】
Since $\frac{5}{9}\frac{A}{9}>1$, that is, $1<\frac{A}{9}<\frac{9}{5}$;
thus $9<A<\frac{81}{5}$, which means $9<A<16 \frac{1}{5}$;
since $A$ is a positive integer;
therefore $A=10, 11, 12, 13, 14, 15, 16$; $A$ has $16-10+1=7$ different values.

Answer: 7


In the same plane, line $l_{2}$ intersects with line $l_{1}$, and line $l_{3}$ is parallel to line $l_{1}$. The number of points equidistant from these three lines is
(A) 0.
(B) 1.
(C) 2.
(D) 4.
(E) 8.

Solution:
[Solution] Let $l_{4}$ be the line parallel to and equidistant from $l_{1}$ and $l_{3}$, and let the distance be $d$.
Let $l_{5}$ and $l_{6}$ be the two lines parallel to and at a distance $d$ from $l_{2}$.
Then the intersection points $P, Q$ of $l_{4}, l_{5}, l_{6}$ are the solutions.
Therefore, the answer is $(C)$.

Answer: C


A three-digit number has its middle digit three times smaller than the sum of the other two, and the sum of the last two digits is half of the first digit. If the digits in the tens and units places are swapped, the resulting number is 18 less than the given number. What is that number?

Solution:
Solution. Let $x, y$ and $z$ denote the units, tens, and hundreds digits, respectively, of the desired three-digit number. According to the conditions of the problem,

$$
\left\{\begin{array}{l}
3 y=x+z \\
x=2(y+z) \\
100 x+10 y+z=100 x+10 z+y+18
\end{array}\right.
$$

From which we get $x=8, y=3$ and $z=1$. Therefore, the desired number is 831.

Answer: 831


A die is a cube with its faces numbered 1 through 6. One red die and one blue die are rolled. The sum of the numbers on the top face of each die is determined. What is the probability that this sum is a perfect square?

Solution:
4. When the red die is rolled, there are 6 equally likely outcomes. Similarly, when the blue die is rolled, there are 6 equally likely outcomes.
Therefore, when the two dice are rolled, there are $6 \times 6=36$ equally likely outcomes for the combination of the numbers on the top face of each. (These outcomes are Red 1 and Blue 1, Red 1 and Blue 2, Red 1 and Blue 3, ..., Red 6 and Blue 6.)
The chart below shows these possibilities along with the sum of the numbers in each case:
Blue Die
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline & & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline & 1 & 2 & 3 & 4 & 5 & 6 & \\
\hline & 2 & 3 & 4 & 5 & 6 & 7 & \\
\hline Red & 3 & 4 & 5 & 6 & 7 & 8 & \\
\hline ie & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\
\hline & 0 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
\end{tabular}

Since the only perfect squares between 2 and 12 are 4 (which equals $2^{2}$ ) and 9 (which equals $3^{2}$ ), then 7 of the 36 possible outcomes are perfect squares.
Since each entry in the table is equally likely, then the probability that the sum is a perfect square is $\frac{7}{36}$.
ANSWER: $\frac{7}{36}$

Answer: \frac{7}{36}


Calculate: $a \cdot a^{2} \cdot a^{3} \cdot \cdots \cdot a^{100}=$

Solution:
3. $a^{5050}=$

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

However, since the provided text is already in a mathematical form which is universal and does not require translation, the output remains the same:

3. $a^{5050}=$

Answer: a^{5050}


Two students agreed to meet at a certain place between 12:00 and 13:00. The one who arrives first waits for the other for $\alpha$ minutes ( $\alpha<60$ ), after which he leaves. What is the probability of their meeting (event $A$), if each of them can arrive at any time during the specified hour, and the times of their arrivals are independent? Consider the special case $\alpha=10$ minutes.

Solution:
Solution. Let the arrival time (counting from 12 o'clock) of one of the students be denoted by $x$, and the other student's arrival time by $y$. For the meeting to occur, it is necessary and sufficient that the inequality $|x-y| \leqslant \alpha$ holds, which is equivalent to the following: $-\alpha \leqslant x-y \leqslant \alpha$, or the system:

$$
\left\{\begin{array}{l}
x-y-\alpha \leqslant 0 \\
x-y+\alpha \geqslant 0
\end{array}\right.
$$

Let's represent $x$ and $y$ as Cartesian coordinates on a plane, choosing 1 minute as the unit of scale (Fig. 6.2). All possible outcomes will be represented by points within a square with a side length of 60, and the outcomes favorable for the meeting will be represented by points in the shaded area, satisfying the conditions.

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-133.jpg?height=389&width=421&top_left_y=524&top_left_x=2056)

Fig. 6.2 According to the formula in point $5^{\circ}$, we get:

$$
p(A)=\frac{60^{2}-(60-\alpha)^{2}}{60^{2}}=1-\left(1-\frac{\alpha}{60}\right)^{2}
$$

In particular, if $\alpha=10$ minutes, then $p(A)=\frac{11}{36}$.

Answer: \frac{11}{36}


A sequence $u_{1}, u_{2}, \ldots, u_{n}$ composed of the integers $1,2, \ldots, n$ is quasi-increasing if, for every index $k, u_{k} \leq u_{k+1}+2$. For example, the sequence $1,6,4,2,5,3$ is quasi-increasing but the sequence $1,4,6,2,5,3$ is not. Determine the number of quasi-increasing sequences of $n$ terms.

Solution:
We notice that a quasi-increasing sequence of $n+1$ integers can be obtained from a quasi-increasing sequence of $n$ integers by adding $n+1$ either before $n$, or before $n+1$, or in the last position (one needs to think a bit to be convinced of this point).

Thus, the number $s_{n}$ of quasi-increasing sequences of $n$ integers satisfies $s_{n+1}=3 \times s_{n}$ for all $n \geq 2$. Since $s_{2}=2$, we deduce by induction that $s_{n}=2 \times 3^{n-2}$ for $n \geq 2$ (and $s_{1}=1$).

## Binomial Coefficients

## Definition

## Definition

The binomial coefficient $\binom{n}{p}$ is the number of subsets with $p$ elements of a set with $n$ elements.

## Remark

If $p>n,\binom{n}{p}=0$ because there are no subsets with $p$ elements in a set with $n$ elements.

Answer: s_{n}=2\times3^{n-2}


Among the 2019 natural numbers from 1 to 2019, how many numbers, when added to the four-digit number 8866, will result in at least one carry?

Solution:
Reference answer: 1956
Exam point: Mathematical principle —— Addition and Multiplication Principle

Answer: 1956


In the quadrilateral $ABCD$ given in the figure, we have $AB=5, BC=17$, $CD=5$ and $DA=9$. Determine $DB$, knowing that its measure is an integer.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-034.jpg?height=214&width=440&top_left_y=2480&top_left_x=1348)

Solution:
Remember that any side of a triangle is greater than the difference and less than the sum of the other two sides. In triangle $A D B$, we have $A D - A B < B D < A D + A B$ and in triangle $C B D$, we have $B C - C D < B D < B C + C D$.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-118.jpg?height=248&width=508&top_left_y=821&top_left_x=1325)

Substituting the known values, we get $9 - 5 < B D < 5 + 9$ and $17 - 5 < B D < 17 + 5$, that is,

$$
4 < B D < 14 \text{ and } \quad 12 < B D < 22
$$

From these two inequalities, we conclude that $12 < B D < 14$ and, since $B D$ is an integer, we can only have $B D = 13$.

Answer: 13


A two-digit integer $\underline{a}\,\, \underline{b}$ is multiplied by $9$. The resulting three-digit integer is of the form $\underline{a} \,\,\underline{c} \,\,\underline{b}$ for some digit $c$. Evaluate the sum of all possible $\underline{a} \,\, \underline{b}$.

Solution:
1. Let the two-digit integer be represented as \( \underline{a}\, \underline{b} \), where \( a \) and \( b \) are the tens and units digits, respectively. This can be expressed as \( 10a + b \).

2. When this number is multiplied by 9, the resulting number is \( 9(10a + b) \).

3. According to the problem, the resulting three-digit number is of the form \( \underline{a}\, \underline{c}\, \underline{b} \), which can be expressed as \( 100a + 10c + b \).

4. Equating the two expressions, we get:
   \[
   9(10a + b) = 100a + 10c + b
   \]

5. Expanding and simplifying the equation:
   \[
   90a + 9b = 100a + 10c + b
   \]
   \[
   90a + 9b - 100a - b = 10c
   \]
   \[
   -10a + 8b = 10c
   \]
   \[
   10c = 8b - 10a
   \]
   \[
   c = \frac{8b - 10a}{10}
   \]
   \[
   c = \frac{4b - 5a}{5}
   \]

6. For \( c \) to be a digit (i.e., an integer between 0 and 9), \( 4b - 5a \) must be divisible by 5. Let \( 4b - 5a = 5k \) for some integer \( k \).

7. Solving for \( b \):
   \[
   4b = 5a + 5k
   \]
   \[
   b = \frac{5a + 5k}{4}
   \]
   Since \( b \) must be a digit (0 to 9), \( 5a + 5k \) must be divisible by 4.

8. Let’s test possible values of \( a \) (since \( a \) is a digit from 1 to 9):
   - For \( a = 1 \):
     \[
     5(1) + 5k = 5 + 5k
     \]
     \[
     5 + 5k \equiv 0 \pmod{4}
     \]
     \[
     5 \equiv -3 \pmod{4}
     \]
     \[
     -3 + 5k \equiv 0 \pmod{4}
     \]
     \[
     5k \equiv 3 \pmod{4}
     \]
     \[
     k \equiv 3 \pmod{4}
     \]
     The smallest \( k \) that satisfies this is \( k = 3 \):
     \[
     b = \frac{5(1) + 5(3)}{4} = \frac{5 + 15}{4} = 5
     \]
     So, \( a = 1 \), \( b = 5 \), and \( c = \frac{4(5) - 5(1)}{5} = 3 \).

   - For \( a = 2 \):
     \[
     5(2) + 5k = 10 + 5k
     \]
     \[
     10 + 5k \equiv 0 \pmod{4}
     \]
     \[
     10 \equiv 2 \pmod{4}
     \]
     \[
     2 + 5k \equiv 0 \pmod{4}
     \]
     \[
     5k \equiv -2 \pmod{4}
     \]
     \[
     k \equiv 2 \pmod{4}
     \]
     The smallest \( k \) that satisfies this is \( k = 2 \):
     \[
     b = \frac{5(2) + 5(2)}{4} = \frac{10 + 10}{4} = 5
     \]
     So, \( a = 2 \), \( b = 5 \), and \( c = \frac{4(5) - 5(2)}{5} = 2 \).

   - For \( a = 3 \):
     \[
     5(3) + 5k = 15 + 5k
     \]
     \[
     15 + 5k \equiv 0 \pmod{4}
     \]
     \[
     15 \equiv -1 \pmod{4}
     \]
     \[
     -1 + 5k \equiv 0 \pmod{4}
     \]
     \[
     5k \equiv 1 \pmod{4}
     \]
     \[
     k \equiv 1 \pmod{4}
     \]
     The smallest \( k \) that satisfies this is \( k = 1 \):
     \[
     b = \frac{5(3) + 5(1)}{4} = \frac{15 + 5}{4} = 5
     \]
     So, \( a = 3 \), \( b = 5 \), and \( c = \frac{4(5) - 5(3)}{5} = 1 \).

   - For \( a = 4 \):
     \[
     5(4) + 5k = 20 + 5k
     \]
     \[
     20 + 5k \equiv 0 \pmod{4}
     \]
     \[
     20 \equiv 0 \pmod{4}
     \]
     \[
     5k \equiv 0 \pmod{4}
     \]
     \[
     k \equiv 0 \pmod{4}
     \]
     The smallest \( k \) that satisfies this is \( k = 0 \):
     \[
     b = \frac{5(4) + 5(0)}{4} = \frac{20 + 0}{4} = 5
     \]
     So, \( a = 4 \), \( b = 5 \), and \( c = \frac{4(5) - 5(4)}{5} = 0 \).

9. The possible values of \( \underline{a}\, \underline{b} \) are 15, 25, 35, and 45.

10. Summing these values:
    \[
    15 + 25 + 35 + 45 = 120
    \]

The final answer is \(\boxed{120}\).

Answer: 120


Solve the following equation, where $n \geq 2$ is a given natural number and $x$ is the unknown.

$$
\sum_{i=0}^{n-2} \frac{1}{(x+i)(x+i+1)}=x(x+1)(x+2) \cdot \ldots \cdot(x+n)+\frac{n-1}{x(x+n-1)}
$$

Solution:
The equation only makes sense with the assumption that the factors in the denominators on both sides are all different from 0. According to this, \( x \neq -i \) and \( x \neq -(i+1) \) for each \( i=0,1, \ldots, n-2 \), that is, \( x \neq 0, -1, -2, \ldots, -(n-1) \).

Each term on the left side can be written as the difference of two fractions:

\[
\frac{1}{(x+i)(x+i+1)}=\frac{(x+i+1)-(x+i)}{(x+i)(x+i+1)}=\frac{1}{x+i}-\frac{1}{x+i+1}
\]

After this decomposition, the subtrahend of each term cancels out with the minuend of the next term, resulting in 0, so only the first term of the difference with \( i=0 \) and the second term of the difference with \( i=(n-2) \) remain. These two terms combined are equal to the second term on the right side:

\[
\frac{1}{x}-\frac{1}{x+(n-1)}=\frac{n-1}{x(x+n-1)}
\]

Therefore, (1) can be rewritten as:

\[
0=x(x+1)(x+2) \cdot \ldots \cdot(x+n-1)(x+n)
\]

The first \( n \) factors being 0 cannot be considered according to the assumption, so the only solution to the equation is from \( x+n=0 \):

\[
x=-n
\]

This indeed makes both sides of (1) equal.

Margit Forró (Komárom, Czechoslovakia, General Secondary School, 3rd grade)

Answer: -n


In the $x O y$ rectangular coordinate plane, draw different lines passing through the point $(3,4)$ and the trisection points of the line segment with endpoints $(-4,5),(5,-1)$, one of the line equations is
(A) $3 x-2 y-1=0$.
(B) $4 x-5 y+8=0$.
(C) $5 x+2 y-23=0$.
(D) $x+7 y-31=0$.
(E) $x-4 y+13=0$.

Solution:
[Solution] According to the section formula, the trisection points of the line segment with endpoints $(-4,5),(5,-1)$ are: $P(-1,3), Q(2,1)$.
The equations of the lines passing through the points $(3,4)$ and $P, Q$ are respectively
$$
x-4 y+13=0 \text { and } \quad 3 x-y-5=0 .
$$

Therefore, the answer is $(E)$.

Answer: E


In a regular 2017-gon, all diagonals are drawn. Petya randomly selects some $\mathrm{N}$ diagonals. What is the smallest $N$ such that among the selected diagonals, there are guaranteed to be two of the same length?

Solution:
Answer: 1008.

Solution: Let's choose an arbitrary vertex and consider all the diagonals emanating from it. There are 2014 of them, and by length, they are divided into 1007 pairs. Clearly, by rotating the polygon, any of its diagonals can be aligned with one of these. Therefore, there are only 1007 different sizes of diagonals. Thus, by choosing 1008, Petya is guaranteed to get at least two of the same.

Answer: 1008


Positive real numbers $x, y$ satisfy $x y=1$. Then, $\frac{1}{x^{4}}+\frac{1}{4 y^{4}}$ has the minimum value of ( ).
(A) $\frac{1}{2}$
(B) $\frac{5}{8}$
(C) 1
(D) $\sqrt{2}$

Solution:
$-1 .(\mathrm{C})$.
Since $x=\frac{1}{y}$, we have,
$$
\frac{1}{x^{4}}+\frac{1}{4 y^{4}}=\frac{1}{x^{4}}+\frac{x^{4}}{4}=\left(\frac{1}{x^{2}}-\frac{x^{2}}{2}\right)^{2}+1 \text {. }
$$

When $\frac{1}{x^{2}}=\frac{x^{2}}{2}$, i.e., $x=\sqrt[4]{2}$, $\frac{1}{x^{4}}+\frac{1}{4 y^{4}}$ is minimized, and the minimum
value is 1.

Answer: C


The mole started to dig a new tunnel. First, the tunnel led 5 meters north, then $23 \mathrm{dm}$ west, $150 \mathrm{~cm}$ south, $37 \mathrm{dm}$ west, $620 \mathrm{~cm}$ south, $53 \mathrm{~cm}$ east, and $27 \mathrm{dm}$ north. How many centimeters does he still need to dig to get back to the start of the tunnel?

Solution:
After converting all data to centimeters, we can draw the mole's tunnel. The mole started digging the tunnel at point $S$ and ended at point $K$ (using the usual orientation of cardinal directions).

![](https://cdn.mathpix.com/cropped/2024_04_17_6e39d21eb867b280d8eag-1.jpg?height=711&width=634&top_left_y=1543&top_left_x=745)

The task is to determine the length of the dashed line. In the "west-east" direction, the mole moved from point $S$ by $230+370-53=547 \mathrm{~cm}$ to the west. In the "north-south" direction, the mole moved from point $S$ by $500-150-620+270=0 \mathrm{~cm}$ to the north.

The mole is $547 \mathrm{~cm}$ away from the starting point.

Answer: 547


Set $M$ consists of more than two consecutive natural numbers, and the sum of its elements is 1996. Such a set $M$ ( ).
(A) does not exist
(B) exists uniquely
(C) exists in two forms
(D) exists in more than three forms

Solution:
$-、 1$. B.
Let $M=\{a, a+1, \cdots, a+n-1\}$. From $\frac{(a+a+n-1) n}{2}=1996$, we have $n(2 a+n-1)=8 \times 499$.

Since $n$ and $2 a+n-1$ have different parities, and $n<2 a+n-1$, and 499 is a prime number, we can only have $n=8$, then $2 a+n-1=2 a+7=499$, solving for $a$ gives $a=246$.

Answer: B


Given a complete bipartite graph on $n,n$ vertices (call this $K_{n,n}$), we colour all its edges with $2$ colours, red and blue. What is the least value of $n$ such that for any colouring of the edges of the graph, there will exist at least one monochromatic $4$ cycle?

Solution:
1. **Step 1: Verify the case for \( n = 4 \)**

   We need to show that there exists a coloring of the edges of \( K_{4,4} \) that avoids any monochromatic \( C_4 \) (4-cycle). Consider the following coloring:

   - Vertices on the left side: \( \{0, 1, 2, 3\} \)
   - Vertices on the right side: \( \{0, 1, 2, 3\} \)

   Connect vertex \( i \) on the left to:
   - Vertices \( i \) and \( i+1 \) (mod 4) on the right with blue edges.
   - Vertices \( i+2 \) and \( i+3 \) (mod 4) on the right with red edges.

   This coloring avoids any monochromatic \( C_4 \). Therefore, \( n = 4 \) does not guarantee a monochromatic \( C_4 \).

2. **Step 2: Prove the case for \( n = 5 \)**

   We need to show that for any coloring of the edges of \( K_{5,5} \), there will always be a monochromatic \( C_4 \). Enumerate the vertices of one partite set as \( \{1, 2, 3, 4, 5\} \). Let \( r_i \) be the number of red edges from vertex \( i \) and \( b_i \) be the number of blue edges from vertex \( i \). Note that \( r_i + b_i = 5 \) for each \( i \).

3. **Step 3: Count the number of monochromatic \( K_{2,2} \)**

   A \( 4 \)-cycle is just a \( K_{2,2} \). If two vertices are connected with red (or blue) edges to two other vertices, we have a monochromatic \( K_{2,2} \). The number of such triples \( (u, v, w) \) where \( w \) is from the chosen partite set and \( u, v \) are from the other partite set such that both edges \( uw \) and \( vw \) are red is at most \( \binom{5}{2} = 10 \).

4. **Step 4: Use combinatorial arguments**

   The number of such triples is exactly \( \sum_{i} \binom{r_i}{2} \). Therefore, we have:
   \[
   \binom{5}{2} \geq \sum_{i} \binom{r_i}{2}
   \]
   Similarly, for blue edges:
   \[
   \binom{5}{2} \geq \sum_{i} \binom{b_i}{2}
   \]

5. **Step 5: Apply convexity**

   Since \( r_i + b_i = 5 \), by convexity, we have:
   \[
   \binom{r_i}{2} + \binom{b_i}{2} \geq 4
   \]
   with equality only when \( r_i \in \{2, 3\} \).

6. **Step 6: Sum the inequalities**

   Adding the two bounds obtained earlier:
   \[
   2 \binom{5}{2} = 5 \cdot 4 = 20
   \]
   This implies:
   \[
   \binom{5}{2} = \sum_{i} \binom{r_i}{2}
   \]
   and each \( r_i \in \{2, 3\} \).

7. **Step 7: Check for contradiction**

   If \( r_i \in \{2, 3\} \), then \( \binom{r_i}{2} \) is odd. Summing five odd numbers should yield an odd number, but \( 10 = \binom{5}{2} \) is even. This contradiction implies that no matter how we color the edges of \( K_{5,5} \), some monochromatic \( C_4 \) will be induced.

\(\blacksquare\)

The final answer is \( \boxed{ n = 5 } \)

Answer: n = 5


What is the most frequent digit in the resulting list when we write the numbers $1,2,3, \ldots, 100000$ on a sheet of paper, then replace each number by the sum of its digits, and so on, until each number consists of only one digit?

Solution: Here, it is noted that for any integer $n$, the sum of the digits of $n$ is congruent to $n$ modulo 9. The resulting sequence is therefore periodic: $1,2,3,4,5,6,7,8,9,1,2,3,4, \ldots$ and it ends with a 1, so it is the one that appears most often.

Answer: 1


Given a right-angled triangle with perimeter $18$. The sum of the squares
of the three side lengths is $128$. What is the area of the triangle?

Solution:
1. Let the side lengths of the right-angled triangle be \(a\), \(b\), and \(c\), where \(c\) is the hypotenuse. We are given the following conditions:
   \[
   a + b + c = 18
   \]
   \[
   a^2 + b^2 + c^2 = 128
   \]
   \[
   a^2 + b^2 = c^2
   \]

2. From the Pythagorean theorem, we know:
   \[
   a^2 + b^2 = c^2
   \]
   Substituting this into the second equation:
   \[
   c^2 + c^2 = 128 \implies 2c^2 = 128 \implies c^2 = 64 \implies c = \sqrt{64} = 8
   \]

3. Now, substitute \(c = 8\) into the perimeter equation:
   \[
   a + b + 8 = 18 \implies a + b = 10
   \]

4. We also know from the Pythagorean theorem that:
   \[
   a^2 + b^2 = 64
   \]

5. To find \(ab\), we use the identity \((a + b)^2 = a^2 + 2ab + b^2\):
   \[
   (a + b)^2 = 10^2 = 100
   \]
   \[
   a^2 + b^2 + 2ab = 100
   \]
   Substituting \(a^2 + b^2 = 64\) into the equation:
   \[
   64 + 2ab = 100 \implies 2ab = 36 \implies ab = 18
   \]

6. The area of the right-angled triangle is given by:
   \[
   \text{Area} = \frac{1}{2} \times a \times b = \frac{1}{2} \times 18 = 9
   \]

The final answer is \(\boxed{9}\).

Answer: 9


Given a regular tetrahedron $P-ABC$ with the center of the base being $O$, a moving plane through $O$ intersects the lateral edges or their extensions of the tetrahedron at points $S, R, Q$. If
$$
\begin{array}{l}
\overrightarrow{P S}=m \overrightarrow{P A}, \overrightarrow{P R}=n \overrightarrow{P B}, \overrightarrow{P Q}=k \overrightarrow{P C}, \\
\text { then } \frac{m n k}{m n+n k+k m}=
\end{array}
$$

Solution:
4. $\frac{1}{3}$.

Since points $S, R, Q, O$ are coplanar, we have
$$
\overrightarrow{P O}=x \overrightarrow{P S}+y \overrightarrow{P R}+z \overrightarrow{P Q},
$$

and $x+y+z=1$.
Thus, $\overrightarrow{P O}=x m \overrightarrow{P A}+y n \overrightarrow{P B}+z k \overrightarrow{P C}$.
From the fact that the center of the base of the regular tetrahedron $P-A B C$ is $O$, we get
$$
\overrightarrow{P O}=\frac{1}{3}(\overrightarrow{P A}+\overrightarrow{P B}+\overrightarrow{P C}).
$$

Since $\overrightarrow{P A}, \overrightarrow{P B}, \overrightarrow{P C}$ are not coplanar, we have
$$
m x=n y=k z=\frac{1}{3}.
$$
$$
\begin{array}{l}
\text { Therefore, } \frac{m n k}{m n+n k+k m}=\frac{1}{\frac{1}{m}+\frac{1}{n}+\frac{1}{k}} \\
=\frac{1}{3 x+3 y+3 z}=\frac{1}{3}.
\end{array}
$$

Answer: \frac{1}{3}


Determine the probability that the product of the $n$ numbers selected after $n$ choices $(n>1)$ is divisible by 10, assuming a random number selector can only choose from the numbers $1,2, \cdots, 9$, and makes these choices with equal probability.

Solution:
[Solution] To ensure that the product of the $n$ selected numbers is divisible by 10, at least one of the selections must be 5, and at least one of the selections must be an even number $2, 4, 6, 8$.
Let event $A$ represent the event that 5 is never chosen, and event $B$ represent the event that no even number is ever chosen. Let $p(E)$ denote the probability of event $E$. The probability we seek is $1-P(A \cup B)$, thus we have
$$
\begin{aligned}
& 1-p(A \cup B) \\
= & 1-p(A)-p(B)+p(A \cap B) \\
= & 1-\left(\frac{8}{9}\right)^{n}-\left(\frac{5}{9}\right)^{n}+\left(\frac{4}{9}\right)^{n} \\
= & 1-\frac{8^{n}+5^{n}-4^{n}}{9^{n}} .
\end{aligned}
$$

Therefore, the probability that the product of the $n$ selected numbers is divisible by 10 is $1-\frac{8^{n}+5^{n}-4^{n}}{9^{n}}$.

Answer: 1-\frac{8^{n}+5^{n}-4^{n}}{9^{n}}


Solution: $\begin{array}{l}4 \cdot \frac{x+2013}{2013 x}=\frac{3}{2} \\ \frac{2013 x}{x+2013}=\frac{8}{3} \\ 1872+48 \times\left(\frac{2013 x}{x+2013}\right)=1872+48 \times \frac{8}{3}=1872+128=2000\end{array}$

Answer: 2000


A green chameleon always tells the truth, while a brown chameleon lies and immediately turns green after lying. In a company of 2019 chameleons (green and brown), each in turn answered the question of how many of them are currently green. The answers were the numbers $1,2,3, \ldots, 2019$ (in some order, not necessarily in the order listed above). What is the maximum number of green chameleons that could have been there initially? (R. Zhenedarov, O. Dmitriev)

Solution:
Answer: 1010.

Solution. Consider two chameleons who spoke in a row. One of them was brown at the moment of speaking; indeed, if both were green, the number of green chameleons would not have changed after the first one spoke, and the second one would have named the same number as the first. We can divide all the chameleons into 1009 pairs who spoke in a row, and one more chameleon; since in each pair there was a brown one, the initial number of green chameleons was no more than $2019-1009=1010$.

It remains to show that there could have been 1010 green chameleons. Let's number the chameleons in the order of their statements. Suppose all the odd-numbered chameleons are green, and all the even-numbered ones are brown. Then the first one will say 1010, the second one will say 1 and turn green. Then the third one will say 1011, and the fourth one will say 2 and turn green, and so on. As a result, the odd-numbered chameleons will say all the numbers from 1010 to 2019, and the even-numbered ones will say all the numbers from 1 to 1009.

Remark. From the first paragraph of the solution, it can be deduced that in any optimal example, the chameleons with even numbers must be exactly brown.

Comment. Only proving that the number of green chameleons was no more than $1010-4$ points.

Only providing an example showing that there could have been exactly 1010 green chameleons initially - 3 points.

Answer: 1010


For how many natural numbers $n$ is the value of the fraction

$$
\frac{n+2022}{n-3}
$$

an integer?

Solution:
## Solution.

Let's rewrite the expression in a different way:

$$
\frac{n+2022}{n-3}=\frac{2025}{n-3}+1
$$

If $n-3$ is positive, it must be a divisor of the number 2025. The number 2025 has 15 positive divisors:

1 point

$$
1,3,5,9,15,25,27,45,75,81,135,225,405,675,2025
$$

1 point

If the expression $n-3$ is negative, it can take the values -1 or -2 (since $n$ is a natural number). 1 point

Since 2 is not a divisor of 2025, the only possibility is $n-3=-1$ (i.e., $n=2$).

1 point

We conclude that there are a total of 16 natural numbers $n$ with the property described in the problem.

Note: The third point in the scoring scheme is given for the argument that 2025 has exactly 15 positive divisors. One alternative way to prove this claim is by using the factorization $2025=3^{4} 5^{2}$ and the formula for the number of positive divisors $\tau(m)$ of a number $m$ (where $m=p_{1}^{a_{1}} \cdots p_{k}^{a_{k}}$ is its prime factorization):

$$
\tau(m)=\left(a_{1}+1\right) \cdot \ldots \cdot\left(a_{k}+1\right)
$$

Answer: 16


There is a plate of fruit, when counted 3 at a time, 2 are left; when counted 4 at a time, 3 are left; when counted 5 at a time, 4 are left; when counted 6 at a time, 5 are left. The plate has at least $\qquad$ fruits.

Solution:
【Answer】Solution: Since $3=1 \times 3, 4=2 \times 2, 5=1 \times 5, 6=2 \times 3$, there are at least:
$$
\begin{array}{l}
3 \times 2 \times 2 \times 5-1 \\
=60-1 \\
=59 \text{ (pieces). }
\end{array}
$$

Answer: There are at least 59 pieces of fruit in this plate.
Therefore, the answer is: 59.

Answer: 59


Solution:
[Solution] By the AM-GM inequality, we have
$$2 \sin ^{2} x_{i}+\frac{1}{\sin ^{2} x_{i}}=2 \sin ^{2} x_{i}+\frac{1}{2 \sin ^{2} x_{i}}+\frac{1}{2 \sin ^{2} x_{i}} \geqslant 3 \sqrt[3]{\frac{1}{2 \sin ^{2} x_{i}}}$$

Therefore, $\prod_{i=1}^{4}\left(2 \sin ^{2} x_{i}+\frac{1}{\sin ^{2} x_{i}}\right) \geqslant 81\left(4 \sin x_{1} \sin x_{2} \sin x_{3} \sin x_{4}\right)^{-\frac{2}{3}}$.
By the concavity of the function $f(x)=\ln \sin x$ in $(0, \pi)$, we have
$$\sin x_{1} \sin x_{2} \sin x_{3} \sin x_{4} \leqslant \sin ^{4} \frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}=\frac{1}{4},$$

Thus, $\prod_{i=1}^{4}\left(2 \sin ^{2} x_{i}+\frac{1}{\sin ^{2} x_{i}}\right) \geqslant 81$.
When $x_{1}=x_{2}=x_{3}=x_{4}=\frac{\pi}{4}$, $\prod_{i=1}^{4}\left(2 \sin ^{2} x_{i}+\frac{1}{\sin ^{2} x_{i}}\right)=81$, hence 81 is the minimum value sought.

Answer: 81


A circle is inscribed in a circular sector with a central angle of $120^{\circ}$. Find the radius of the inscribed circle if the radius of the given circle is $R$.

Solution:
Solution.

$\Delta O_{2} A O_{1}$ is a right triangle, $\angle O_{1} O_{2} A=60^{\circ}, \sin 60^{\circ}=\frac{O_{1} A}{O_{2} O_{1}}=\frac{r}{R-r} ;$ $\frac{\sqrt{3}}{2}=\frac{r}{R-r}$ (Fig. 10.69). From this, $r=\frac{\sqrt{3} R}{2+\sqrt{3}}=\sqrt{3} R(2-\sqrt{3})$.

Answer: $\sqrt{3} R(2-\sqrt{3})$.

Answer: \sqrt{3}R(2-\sqrt{3})


Given the sequence $a_{n}=\sqrt{4+\frac{1}{n^{2}}}+\sqrt{4+\frac{2}{n^{2}}}+\cdots+\sqrt{4+\frac{n}{n^{2}}}-2 n, n$ is a positive integer, then the value of $\lim _{n \rightarrow \infty} a_{n}$ is $\qquad$

Solution:
9. $\frac{1}{8}$

Since $\sqrt{4+\frac{\bar{k}}{n^{2}}}\sqrt{2^{2}+\frac{k}{n^{2}}+\left(\frac{k-1}{4 n^{2}}\right)^{2}}=2+$ $\frac{k-1}{4 n^{2}}, 1 \leqslant k \leqslant n$,
thus $a_{n}>\sum_{k=1}^{n}\left(2+\frac{h-1}{4 n^{2}}\right)-2 n=\frac{n-1}{8 n}$.
Therefore, $\frac{n-1}{8 n}<a_{n}<\frac{n+1}{8 n}$, and $\lim _{n \rightarrow \infty} \frac{n-1}{8 n}=\lim _{n \rightarrow \infty} \frac{n+1}{8 n}=\frac{1}{8}$, hence $\lim _{n \rightarrow \infty} a_{n}=\frac{1}{8}$.

Answer: \frac{1}{8}


Indicate the integer closest to the larger root of the equation

$$
\operatorname{arctg}\left(\left(\frac{3 x}{22}-\frac{11}{6 x}\right)^{2}\right)-\operatorname{arctg}\left(\left(\frac{3 x}{22}+\frac{11}{6 x}\right)^{2}\right)=-\frac{\pi}{4}
$$

Solution:
Solution. The equation is equivalent to $\frac{3 x}{22}=\frac{11}{6 x} \Longleftrightarrow |x|=\frac{11}{3}$.

Answer: 4.

Answer: 4


Calculate the triple integral

$$
\iiint_{\Omega} \frac{x^{2}}{x^{2}+y^{2}} d x d y d z
$$

where the region $\Omega$ is bounded by the surfaces

$$
z=\frac{9}{2} \sqrt{x^{2}+y^{2}}, \quad z=\frac{11}{2}-x^{2}-y^{2}
$$

Solution:
Solution.

1. Since $\Omega$ is a body of revolution around the $O Z$ axis, it is convenient to switch to cylindrical coordinates

$$
\left\{\begin{array}{l}
x=\varrho \cos \varphi \\
y=\varrho \sin \varphi \\
z=z
\end{array}\right.
$$

In this case, $(\varrho, \varphi, z) \in \Omega^{\prime}$, and the desired integral is defined by the formula

$$
\iiint_{\Omega} \frac{x^{2}}{x^{2}+y^{2}} d x d y d z=\iiint_{\Omega^{\prime}} \cos ^{2} \varphi \varrho d \varrho d \varphi d z
$$

2. Let's define the region $\Omega^{\prime}$ using inequalities. For this, we first replace $x$ with $\varrho \cos \varphi$ and $y$ with $\varrho \sin \varphi$ in the equations of the surfaces. Then $\Omega^{\prime}$ is defined by the inequalities $9 \varrho / 2 \leq z \leq 11 / 2-\varrho^{2}$ or $11 / 2-\varrho^{2} \leq z \leq 9 \varrho / 2$. To choose the correct inequalities, we solve the equation

$$
\frac{9}{2} \varrho=\frac{11}{2}-\varrho^{2}
$$

This equation has a unique positive solution $\varrho=1$. Therefore, $0 \leq \varrho \leq 1$. For $0 \leq \varrho \leq 1$

$$
\frac{9}{2} \varrho \leq \frac{11}{2}-\varrho^{2}
$$

Thus, the region $\Omega^{\prime}$ is defined by the system of inequalities:

$$
\left\{\begin{array}{l}
0 \leq \varrho \leq 1 \\
9 / 2 \varrho \leq z \leq 11 / 2-\varrho^{2} \\
0 \leq \varphi \leq 2 \pi
\end{array}\right.
$$

3. We transition from the triple integral to a repeated integral:

$$
\begin{aligned}
& \iiint_{\Omega} \frac{x^{2}}{x^{2}+y^{2}} d x d y d z=\iiint_{\Omega^{\prime}} \frac{\varrho^{2} \cos ^{2} \varphi}{\varrho^{2} \cos ^{2} \varphi+\varrho^{2} \sin ^{2} \varphi} \varrho d \varrho d \varphi d z= \\
&=\int_{0}^{2 \pi} \cos ^{2} \varphi d \varphi \int_{0}^{1} \varrho d \varrho \int_{9 \varrho / 2}^{11 / 2-\varrho^{2}} d z
\end{aligned}
$$

Integrating sequentially, we get

$$
\int_{0}^{2 \pi} \cos ^{2} \varphi d \varphi \int_{0}^{1} \varrho d \varrho \int_{9 \varrho / 2}^{11 / 2-\varrho^{2}} d z=\pi \int_{0}^{1}\left(\frac{11}{2}-\varrho^{2}-\frac{9}{2} \varrho\right) \varrho d \varrho=\pi
$$

Answer. $\iiint_{\Omega} \frac{x^{2}}{x^{2}+y^{2}} d x d y d z=\pi$.

Conditions of the Problems. Compute the triple integral over the region $\Omega$ bounded by the given surfaces.

1. $\iiint_{\Omega} \frac{x}{x^{2}+y^{2}} d x d y d z, \quad x^{2}+y^{2}+4 x=0, z=8-y^{2}, z=0$.
2. $\iiint_{\Omega} \frac{y}{x^{2}+y^{2}} d x d y d z, \quad x^{2}+y^{2}-4 y=0, z=6-x^{2}, z=0$.
3. $\iiint \frac{y}{\sqrt{x^{2}+y^{2}}} d x d y d z, \quad x^{2}+y^{2}-4 y=0, z=4-x^{2}, z=0$.
4. $\iint_{\Omega}^{\Omega} \int \frac{x}{\sqrt{x^{2}+y^{2}}} d x d y d z, \quad x^{2}+y^{2}-4 x=0, z=10-y^{2}, z=0$.
5. $\iiint_{\Omega} \frac{x+y}{x^{2}+y^{2}} d x d y d z, \quad x^{2}+y^{2}-4 x=0, z=12-y^{2}, z=0$.
6. $\iiint_{\Omega} \frac{x^{2}}{x^{2}+y^{2}} d x d y d z, \quad z=\sqrt{36-x^{2}-y^{2}}, \quad z=\frac{x^{2}+y^{2}}{9}$.
7. $\iiint_{\Omega} \frac{y^{2}}{x^{2}+y^{2}} d x d y d z, \quad z=\sqrt{1-x^{2}-y^{2}}, \quad z=\frac{2\left(x^{2}+y^{2}\right)}{3}$.
8. $\iiint_{\Omega} \frac{y}{\sqrt{x^{2}+y^{2}}} d x d y d z, \quad z=\sqrt{64-x^{2}-y^{2}}, \quad z=\frac{x^{2}+y^{2}}{12}$.
9. $\iint_{\Omega} \frac{x}{\sqrt{x^{2}+y^{2}}} d x d y d z, \quad z=\sqrt{144-x^{2}-y^{2}}, \quad z=\frac{x^{2}+y^{2}}{18}$.
10. $\iint_{\Omega} \frac{2 x+3 y}{\sqrt{x^{2}+y^{2}}} d x d y d z, \quad z=\sqrt{\frac{4}{9}-x^{2}-y^{2}}, \quad z=x^{2}+y^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_5878ce417e2ca061000bg-322.jpg?height=92&width=1061&top_left_y=1147&top_left_x=118)
$\begin{array}{lllll}\text { 6. } 42 \frac{3}{4} \pi . & \text { 7. } \frac{19}{96} \pi . & \text { 8. } 0 . & \text { 9. } 0 . & 10.0 .\end{array}$

### 12.11. Triple Integral

in spherical coordinates

PROBLEM STATEMENT. Compute the triple integral

$$
\iint_{\Omega} f(x, y, z) d x d y d z
$$

where the region $\Omega$ is bounded by the surfaces

$$
x^{2}+y^{2}+z^{2}=R^{2}, \quad z= \pm \sqrt{\frac{x^{2}+y^{2}}{a^{2}}}
$$

PLAN OF SOLUTION.

1. Since $\Omega$ is bounded by a sphere and a circular cone, it is convenient to switch to spherical coordinates

$$
\left\{\begin{array}{l}
x=\varrho \cos \varphi \sin \theta \\
y=\varrho \sin \varphi \sin \theta \\
z=\varrho \cos \theta
\end{array}\right.
$$

The possible ranges for the spherical coordinates are

$$
0 \leq \varrho<\infty, \quad 0 \leq \theta \leq \pi, \quad 0 \leq \varphi \leq 2 \pi
$$

In this case, $(\varrho, \theta, \varphi) \in \Omega^{\prime}$, and the desired integral is defined by the formula

$$
\begin{aligned}
& \iiint_{\Omega} f(x, y, z) d x d y d z= \\
& \quad=\iiint_{\Omega^{\prime}} f(\varrho \cos \varphi \sin \theta, \varrho \sin \varphi \sin \theta, \varrho \cos \theta) \varrho^{2} \sin \theta d \varrho d \theta d \varphi
\end{aligned}
$$

2. Replace $x$ with $\varrho \cos \varphi \sin \theta$, $y$ with $\varrho \sin \varphi \sin \theta$, and $z$ with $\varrho \cos \theta$ in the equations of the surfaces. We get

$$
\varrho=R, \quad \operatorname{tg} \theta= \pm|a|
$$

3. Define the region $\Omega^{\prime}$ using a system of inequalities:

$$
\left\{\begin{array}{l}
0 \leq \varrho \leq R \\
\theta_{1} \leq \theta \leq \theta_{2} \\
0 \leq \varphi \leq 2 \pi
\end{array}\right.
$$

where the boundaries of $\theta$ are found by solving the equation $\operatorname{

Answer: \pi


$\odot O_{1}$ and $\odot O_{2}$ are externally tangent to each other, their radii are 7 and 14 respectively, the circle $\odot \mathrm{O}_{3}$ that contains these two circles and is tangent to them has its center on their line of centers, the radius of the fourth circle that is tangent to all three circles is $\qquad$.

Solution:
6. 6 .

It is known that $\odot O_{4}$ is externally tangent to $\odot O_{1}$ and $\odot O_{2}$, and internally tangent to $\odot O_{3}$. Let the radius of $\odot O_{4}$ be $x$, and connect $O_{1} O_{4}$, $O_{2} O_{4}$, and $O_{3} O_{4}$. Then, $O_{1} O_{3}=14$, $O_{3} O_{2}=7$, $O_{1} O_{4}=7+x$, $O_{3} O_{4}=21-x$, and $O_{2} O_{4}=14+x$. In $\triangle O_{1} O_{3} O_{4}$ and $\triangle O_{1} O_{2} O_{4}$, use the cosine rule: $\cos \angle O_{2} O_{1} O_{4}=\frac{(7+x)^{2}+14^{2}-(21-x)^{2}}{2 \cdot 14 \cdot(7+x)}$, $\cos \angle O_{2} O_{1} O_{4}=\frac{(7+x)^{2}+21^{2}-(14+x)^{2}}{2 \cdot 21 \cdot(7+x)}$. Therefore, $\frac{(7+x)^{2}+14^{2}-(21-x)^{2}}{2 \cdot 14 \cdot(7+x)}=$ $\frac{(7+x)^{2}+21^{2}-(14+x)^{2}}{2 \cdot 21 \cdot(7+x)}$, which simplifies to $196 x=1176 \Rightarrow x=6$.

Answer: 6


There are 100 students who want to sign up for the class Introduction to Acting. There are three class sections for Introduction to Acting, each of which will fit exactly 20 students. The 100 students, including Alex and Zhu, are put in a lottery, and 60 of them are randomly selected to fill up the classes. What is the probability that Alex and Zhu end up getting into the same section for the class?

Solution: Answer: $19 / 165$ There is a $\frac{60}{100}=\frac{3}{5}$ chance that Alex is in the class. If Alex is in the class, the probability that Zhu is in his section is $\frac{19}{99}$. So the answer is $\frac{3}{5} \cdot \frac{19}{99}=\frac{19}{165}$.

Answer: \frac{19}{165}


Claudia tells her friend Sabine that she has drawn a triangle $ABC$ in which the altitude from $A$ to $BC$ passes exactly through the intersection of the perpendicular bisector of $AB$ and the angle bisector of $\angle ABC$.

Sabine claims that from this information alone, one can determine the size of the angle $\angle ABC$ without seeing the drawing.

Investigate whether Sabine's claim is correct! If this is the case, determine the size of $\angle ABC$!

Solution:
Let $S$ be the mentioned intersection point, $E$ the midpoint of side $AB$, and $F$ the foot of the perpendicular from $B$ to $BC$.

Since the line through $B$ and $S$ bisects the angle $\angle ABC$, it follows that $\angle FBS = \angle SBA$. (1)

Since $S$ also lies on the perpendicular bisector of $AB$, $A$, $B$, and $S$ are the vertices of an isosceles triangle with $AS = BS$, and therefore $\angle SAB = \angle SBA$. (2)

From (1) and (2), it follows that $\angle FBS = \angle SAB = \angle SBA$. (3)

Furthermore, $A$, $B$, and $F$ are the vertices of a right triangle. The sum of the sizes of the angles mentioned in (3) is thus $90^\circ$ according to the angle sum theorem. Each of them therefore has a size of $30^\circ$. Sabine's claim is thus correct. The size of the angle $\angle ABC$ is $60^\circ$.

Solutions of the III. Round 1974 taken from [5]

### 4.17 XV. Olympiad 1975

### 4.17.1 I. Round 1975, Class 7

Answer: 60


$$
\begin{array}{l}
\frac{1}{2}(x+y)(y+z)(z+x)+(x+y+z)^{3} \\
=1-x y z
\end{array}
$$

The number of integer solutions to the equation is ( ).
(A) 2
(B) 4
(C) 5
(D) 6

Solution:
5.D.

Let $x+y=u, y+z=v, z+x=w$.
Then the equation transforms to
$$
\begin{array}{l}
4 \text { uvw }+(u+v+w)^{3} \\
=8-(u+v-w)(u-v+w) . \\
\quad(-u+v+w) .
\end{array}
$$

Simplifying, we get
$$
4\left(u^{2} v+v^{2} w+w^{2} u+u v^{2}+v w^{2}+u u^{2}\right)+8 u w=8 \text {, }
$$

which is
$$
u^{2} v+v^{2} w+w^{2} u+w w^{2}+w w^{2}+w u^{2}+2 w w=2 \text {. }
$$

Factoring the left side, we get
$$
\begin{array}{l}
(u+v)(v+w)(w+u)=2 . \\
\text { Thus, }(u+v, v+w, w+u) \\
=(1,1,2),(-1,-1,2),(1,-2,-1)
\end{array}
$$

and symmetric cases.
Solving each case, we get
$$
(u, v, w)=(1,0,1),(1,-2,1),(1,0,-2) \text {. }
$$

Thus, $(x, y, z)=(1,0,0),(2,-1,-1)$.
In summary, considering symmetry, the original equation has 6 integer solutions.

Answer: D


Each interior angle of an $N$-sided regular polygon is $140^{\circ}$. Find $N$.

Solution:
Each exterior angle $=40^{\circ}$ (adj. $\angle$ s on st. line $)$
$\frac{360^{\circ}}{N}=40^{\circ} \quad$ (sum of ext. $\angle$ s of polygon)
$$
\Rightarrow N=9
$$

Answer: 9


The following infinite sequence composed of 1s and 2s has an interesting characteristic: starting from the first term, group the terms with the same number, then write down the number of terms in each group in order, and the infinite sequence formed by these numbers is exactly the sequence itself. This sequence is called the Kolakoski sequence. According to this characteristic, continue to write the next 8 terms of this sequence (from the 14th term to the 21st term), if it is known that the sum of the first 50 terms of this sequence is 75, and the 50th term is 2, then the 73rd term, 74th term, 75th term, and 76th term can be known as $\qquad$

Solution: $\begin{array}{l}1,2,1,1,2,2,1,2 \\ 1,2,2,1\end{array}$

Answer: 1,2,2,1


Given $a, b \in R$. Try to find $s=\max \{|a+b|,|a-b|, |1-a|, |1-b|\}$, the minimum value.

Solution:
If $a b>0$, then
$$
|a-b| \leqslant|a|+|b|=|a+b| \text {. }
$$

Thus, $s=\max \{|a+b|,|1-a|,|1-b|\}$
$$
\begin{array}{l}
>\frac{1}{3}[|a+b|+|1-a|+|1-b|] \\
>\frac{1}{3}|(a+b)+(1-a)+(1-b)| \\
=\frac{2}{3},
\end{array}
$$

Equality is achieved when $a=b=\frac{1}{3}$.
If $a b<0$, then $|a-b|>|a|+|b|$. Thus,
$$
S \geqslant \max \{|1-a|,|1-b|\}>1>\frac{2}{3} .
$$

In summary, the minimum value of $S$ is $\frac{2}{3}$.

Answer: \frac{2}{3}


If the two quadratic equations $x^{2}+x+m=0$ and $m x^{2}+x+1=0$ each have two distinct real roots, but one of them is a common real root $\alpha$, then the range of the real root $\alpha$ is $\qquad$ .

Solution:
6. $\alpha=1$.

From the problem, we know $m \neq 0$, and $\Delta=1-4 m>0$, i.e., $m<\frac{1}{4}$ and $m \neq 0$; simultaneously, $\alpha^{2}+\alpha+m=0$, and $m \alpha^{2}+\alpha+1=0$.
Subtracting these equations gives $(1-m) \alpha^{2}+(m-1)=0$.

Since $m \neq 1$, we have $\alpha^{2}=1$. Solving this, we get $\alpha= \pm 1$.
If $\alpha=-1$, then $m=0$, which contradicts the problem statement.
If $\alpha=1$, then $m=-2$.

Answer: \alpha=1


In the field of real numbers, solve the system of equations

$$
\begin{aligned}
& |1-x|=y+1 \\
& |1+y|=z-2 \\
& |2-z|=x-x^{2}
\end{aligned}
$$

Solution:
1. The right side of the first equation is a non-negative number, which is reflected in the second equation, where we can remove the absolute value. Also, the right side of the second equation is a non-negative number, which, using the equality $|z-2|=|2-z|$, is reflected in the third equation, where we can remove the absolute value. The given system then has the form

$$
\begin{aligned}
|1-x| & =y+1 \\
1+y & =z-2 \\
z-2 & =x-x^{2}
\end{aligned}
$$

and from this, by simple comparison, we obtain the equation

$$
|1-x|=x-x^{2} .
$$

For $x<1$, we get the equation $1-x=x-x^{2}$, or $(1-x)^{2}=0$, whose solution $x=1$ does not satisfy the assumption $x<1$.

For $x \geq 1$, the equation $x^{2}=1$ results, from which the two solutions $x=-1$ and $x=1$ are obtained, and only $x=1$ satisfies the assumption $x \geq 1$.

From the given system, we then simply calculate the values $y=-1$ and $z=2$. The system thus has the unique solution $(x ; y ; z)=(1 ;-1 ; 2)$.

For a systematic and complete solution, award 6 points. For guessing the solution $(1 ;-1 ; 2)$, award one point.

Answer: (1;-1;2)


$\lim _{x \rightarrow 4} \frac{x+\sqrt{x}-6}{x-5 \sqrt{x}+6}$.

Solution:
12.35 The numerator and denominator of the given fraction are quadratic functions of $\sqrt{x}$. We have

$$
\lim _{x \rightarrow 4} \frac{(\sqrt{x}+3)(\sqrt{x}-2)}{(\sqrt{x}-3)(\sqrt{x}-2)}=\lim _{x \rightarrow 4} \frac{\sqrt{x}+3}{\sqrt{x}-3}=\frac{5}{-1}=-5
$$

Answer: -5.

Answer: -5


On a river, two identical sightseeing boats departed from one pier in opposite directions at 13:00. At the same time, a raft also set off from the pier. After an hour, one of the boats turned around and headed in the opposite direction. At 15:00, the second boat did the same. What is the speed of the current if, at the moment the boats met, the raft had drifted 7.5 km from the pier?

Solution:
Answer: 2.5 km/h.

Solution. Consider the reference frame associated with the river. In this frame, the boats move with equal speeds: initially, they move away from each other in opposite directions for 1 hour, then move in the same direction for another 1 hour (during which the distance between them does not change), and then approach each other until they meet. Since the distance between the boats did not change during the second time interval, they will approach each other for as long as they moved away, which is 1 hour. Thus, from the moment of departure until the meeting, 3 hours have passed. During this time, the log (moving with the current) traveled 7.5 km. Therefore, its speed is 2.5 km/h.

Answer: 2.5


Find a three-digit number that is equal to the cube of the sum of its digits.

Solution:
15. Since $4^{3}=64$ and $10^{3}=1000$, the sum of the digits can be $5,6,7,8$ or 9. From these five numbers, we will select by trial those that satisfy the condition of the problem:

$$
5^{3}=125,6^{3}=216,7^{3}=343,8^{3}=512,9^{3}=729
$$

The condition is satisfied by $512=(5+1+2)^{3}$.

Answer: 512


Two vectors $\overrightarrow{O A}$ and $\overrightarrow{O B}$ on a plane satisfy $|\overrightarrow{O A}|=a,|\overrightarrow{O B}|=b$, and $a^{2}+b^{2}=4, \overrightarrow{O A} \cdot \overrightarrow{O B}=0$. If the vector $\overrightarrow{O C}=\lambda \overrightarrow{O A}+\mu \overrightarrow{O B}(\lambda, \mu \in \mathbf{R})$, and
$$
\left(\lambda-\frac{1}{2}\right)^{2} a^{2}+\left(\mu-\frac{1}{2}\right)^{2} b^{2}=1,
$$

then the maximum value of $|\overrightarrow{O C}|$ is _________.

Solution:
4. 2 .

Since $|\overrightarrow{O A}|=a,|\overrightarrow{O B}|=b$, and $a^{2}+b^{2}=4, O A \perp O B$, therefore, points $O, A, B$ lie on a circle with the midpoint $M$ of $A B$ as the center and 1 as the radius.
Also, $\overrightarrow{O M}=\frac{1}{2}(\overrightarrow{O A}+\overrightarrow{O B}), \overrightarrow{O C}=\lambda \overrightarrow{O A}+\mu \overrightarrow{O B}$, then
$$
\overrightarrow{M C}=\overrightarrow{O C}-\overrightarrow{O M}=\left(\lambda-\frac{1}{2}\right) \overrightarrow{O A}+\left(\mu-\frac{1}{2}\right) \overrightarrow{O B}
$$

Thus,
$$
\begin{aligned}
|\overrightarrow{M C}|^{2} & =\left(\lambda-\frac{1}{2}\right)^{2} \overrightarrow{O A}^{2}+2\left(\lambda-\frac{1}{2}\right)\left(\mu-\frac{1}{2}\right) \cdot \overrightarrow{O A} \cdot \overrightarrow{O B}+\left(\mu-\frac{1}{2}\right)^{2} \overrightarrow{O B}^{2} \\
& =\left(\lambda-\frac{1}{2}\right)^{2} a^{2}+\left(\mu-\frac{1}{2}\right)^{2} b^{2}=1
\end{aligned}
$$

Therefore, point $C$ also lies on the circle with $M$ as the center and 1 as the radius. Hence, points $O, A, B, C$ are concyclic, with $M$ as the center. When $O, M, C$ are collinear, i.e., $O C$ is a diameter of $\odot M$, $|O C|$ reaches its maximum value of 2.

Answer: 2


Let $S$ be the set of two-digit numbers that do not contain the digit 0. Two numbers in $S$ are called friends if their largest digits are equal, and if the difference between their smallest digits is equal to 1. For example, 68 and 85 are friends, 78 and 88 are friends, but 58 and 75 are not friends.
Determine the largest integer $m$ such that there exists a subset $T$ of $S$ with $m$ elements, such that any two elements of $T$ are not friends.

Solution:
Answer: 45. We can take for $T$ the set of numbers whose smallest digit is odd.
Conversely, if $x=\overline{a b}$ with $1 \leqslant b<a \leqslant 9$ then $x$ and $x+1$ are friends. If $x=\overline{a b}$ with $2 \leqslant a<b \leqslant 9$ and $a$ even, then $x$ and $x+10$ are friends. We have thus found 36 disjoint pairs of friends. Consequently, among the 72 numbers $\geqslant 21, T$ can contain at most 36 numbers, so $|T| \leqslant 9+36=45$.

Answer: 45


A quadruple $(a,b,c,d)$ of distinct integers is said to be $balanced$ if $a+c=b+d$. Let $\mathcal{S}$ be any set of quadruples $(a,b,c,d)$ where $1 \leqslant a<b<d<c \leqslant 20$ and where the cardinality of $\mathcal{S}$ is $4411$. Find the least number of balanced quadruples in $\mathcal{S}.$

Solution:
1. **Understanding the problem**: We need to find the least number of balanced quadruples \((a, b, c, d)\) in a set \(\mathcal{S}\) where \(1 \leq a < b < d < c \leq 20\) and the cardinality of \(\mathcal{S}\) is 4411. A quadruple \((a, b, c, d)\) is balanced if \(a + c = b + d\).

2. **Total number of quadruples**: First, we calculate the total number of quadruples \((a, b, c, d)\) where \(1 \leq a < b < d < c \leq 20\). This is given by the binomial coefficient:
   \[
   \binom{20}{4} = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = 4845
   \]

3. **Number of unbalanced quadruples**: Since the cardinality of \(\mathcal{S}\) is 4411, the number of unbalanced quadruples is:
   \[
   4845 - 4411 = 434
   \]

4. **Balanced quadruples**: We need to find the least number of balanced quadruples. We start by calculating the possible values of \(a + c\). The minimum value of \(a + c\) is \(1 + 4 = 5\) (since \(a < b < d < c\)) and the maximum value is \(17 + 20 = 37\).

5. **Counting balanced quadruples for each \(a + c\)**:
   - For \(a + c = k\), we need to count the number of ways to choose \(a\) and \(c\) such that \(a < c\) and \(a + c = k\).
   - For \(k = 5\), the only possibility is \((1, 4)\).
   - For \(k = 6\), the possibilities are \((1, 5)\) and \((2, 4)\).
   - For \(k = 7\), the possibilities are \((1, 6)\), \((2, 5)\), and \((3, 4)\).
   - This pattern continues until \(k = 20\), where the possibilities are \((10, 10)\).

6. **Summing the number of balanced quadruples**:
   - For \(k = 5\) to \(k = 20\), the number of pairs \((a, c)\) is given by \(\binom{k-1}{2}\).
   - For \(k = 21\), the number of pairs \((a, c)\) is \(\binom{9}{2}\).
   - For \(k = 22\) to \(k = 37\), the number of pairs \((a, c)\) is symmetric to the pairs for \(k = 5\) to \(k = 20\).

7. **Calculating the total number of balanced quadruples**:
   \[
   2 \cdot \left(\binom{2}{2} + \binom{3}{2} + \cdots + \binom{9}{2}\right) + \binom{10}{2} = 2 \cdot \left(1 + 3 + 6 + 10 + 15 + 21 + 28 + 36\right) + 45 = 2 \cdot 120 + 45 = 285
   \]
   - For \(a + c = 21\), we have \(\binom{9}{2} = 36\).

8. **Total balanced quadruples**:
   \[
   285 + 36 = 321
   \]

9. **Least number of balanced quadruples**:
   - Since we have 434 unbalanced quadruples, the least number of balanced quadruples is:
   \[
   525 - 434 = 91
   \]

The final answer is \(\boxed{91}\)

Answer: 91


Let $n$ be a given positive integer, $6 \mid n$. In an $n \times n$ chessboard, each square is filled with a positive integer, and the numbers in the squares of the $i$-th row from left to right are $(i-1) n+1,(i-1) n+2, \cdots,(i-1) n+n$. Now, take any 2 adjacent (sharing a common edge) squares, add 1 to one of the numbers, and add 2 to the other, which is called one operation. What is the minimum number of operations required to make all the numbers on the chessboard equal?

Solution:
Let the sum of the numbers on the chessboard be denoted as \( S \).
Obviously, when all the numbers on the chessboard are equal, each number is at least \( n^2 \), so the sum of the numbers on the chessboard \( S' \geq n^2 \cdot n^2 = n^4 \). Each operation increases \( S \) by 3, and the initial sum of the numbers on the chessboard \( S_0 = 1 + 2 + 3 + \cdots + n^2 = \frac{n^2(n^2 + 1)}{2} \). Therefore, when the numbers on the chessboard are all equal, \( S \) must increase by at least \( n^4 - \frac{n^2(n^2 + 1)}{2} = \frac{n^2(n^2 - 1)}{2} \). Thus, the number of operations required is at least \( \frac{1}{3} \cdot \frac{n^2(n^2 - 1)}{2} = \frac{n^2(n^2 - 1)}{6} \).

Next, we prove that it is possible to perform \( \frac{n^2(n^2 - 1)}{6} \) operations to make all the numbers on the chessboard equal. This is equivalent to proving that it is possible to perform a certain number of operations to make all the numbers on the chessboard equal to \( n^2 \).

Divide each row of the chessboard from left to right into groups of 3 consecutive cells, resulting in \( \frac{n}{3} \) groups, which we label as the first group, the second group, ..., the \( \frac{n}{3} \)th group.

First, note that for any 3 consecutive cells, it is possible to perform 2 operations to increase each number by 2. We call these two operations a large operation \( A: (a, b, c) \rightarrow (a+2, b+1, c) \rightarrow (a+2, b+2, c+2) \). Now, perform \( \frac{n}{2} \) large operations \( A \) on each group in a row, so that the numbers in that row all increase by \( n \). Therefore, for each row, it is possible to perform a certain number of operations to make it identical to the \( n \)th row.

Next, note that for any 2 consecutive cells, it is possible to perform 2 operations to increase each number by 3. We call these two operations a large operation \( B: (a, b) \rightarrow (a+2, b+1) \rightarrow (a+3, b+3) \). Now, assume that each row of the chessboard is identical to the \( n \)th row. Consider the \( i \)th and \( (i+1) \)th rows, and for the \( j \)th group in these 2 rows, perform a large operation \( B \) on the 2 cells in the same column, so that the numbers in the \( j \)th group of these 2 rows all increase by 3. Since the numbers in the same row but different groups differ by multiples of 3, it is possible to perform similar operations on each group of these 2 rows a certain number of times to make all the groups in these 2 rows identical to the \( \frac{n}{3} \)th group, which is \( (n^2 - 2, n^2 - 1, n^2) \). This group can be transformed into \( (n^2, n^2, n^2) \) with one operation, so these 2 rows can all become \( n^2 \). Perform such operations on every 2 consecutive rows, and all the rows will become \( n^2 \).

In conclusion, the minimum number of operations required is \( \frac{n^2(n^2 - 1)}{6} \).

Answer: \frac{n^2(n^2 - 1)}{6}


The founder of a noble family received a plot of land. Each man in the family, upon dying, divided the land he inherited equally among his sons. If he had no sons, the land went to the state. No other members of the family gained or lost any land in any other way. In total, there were 200 people in the family. What is the smallest fraction of the original plot that any member of the family could have received?

Solution:
Answer: $\frac{1}{4 \cdot 3^{65}}$

Solution: Consider the person with the smallest share. Let his father have $a_{1}$ sons, his grandfather $a_{2}$, and so on up to the founder of the lineage, who had $a_{n}$. Then the share of this person is $\frac{1}{a_{1} a_{2} \ldots a_{n}}$, and we know that $a_{1}+a_{2}+\ldots+a_{n}$ does not exceed 199 (since the founder of the lineage is definitely not included in this sum).

Thus, we are looking for a set of numbers with a sum not exceeding 199 and the maximum product. We will prove that any set, except the set consisting of 39 threes and one two, does not have the maximum product, as it can be increased.

First, if the sum of the numbers in the set is less than 199, add another number to make it equal to 199. The product will not decrease.

Second, suppose there is a number $a \geqslant 4$ in our set. Replace $a$ with a pair of numbers $b$ and $c$ greater than one, such that $b+c=a$. We will prove that $b c \geqslant b+c$. Indeed, $b c-b-c=(b-1)(c-1)-1$, which is non-negative for $b$ and $c$ greater than one.

Third, if there is a number 1 in the set, then we can replace any number $a$ and 1 with $a+1$, thereby increasing the product of these numbers by 1.

Thus, any set can be transformed into a set of twos and threes without decreasing the product of its numbers.

Furthermore, if there are at least three twos, their product is 8. If we replace three twos with two threes, the sum of the numbers does not change, but the product becomes 9, which is again an increase. Thus, we can ensure that the number of twos is no more than two.

Since $199=65 \cdot 3+2 \cdot 2$, the final set will contain exactly two twos.

Therefore, any set of natural numbers with a sum not exceeding 199 can be transformed into a set of 65 threes and two twos, and the product of the numbers in the set will not decrease. This means that the final set has the largest product, equal to $3^{65} \cdot 4$, which gives us the smallest inheritance share as the reciprocal of this number.

Answer: \frac{1}{4\cdot3^{65}}


Find the number of all integer solutions of the inequality $\sqrt{3 \cos \frac{\pi x}{2}-\cos \frac{\pi x}{4}+1}-\sqrt{6} \cdot \cos \frac{\pi x}{4} \geq 0$, belonging to the interval [1991; 2013].

Solution: Answer: 9. Comments. Solution of the inequality: $\frac{8}{3}+8 k \leq x \leq \frac{16}{3}+8 k$. From the given interval, the following numbers will be included in the answer: 1995, 1996, 1997, 2003, 2004, 2005, 2011, 2012, 2013 - 9 numbers.

Answer: 9


Let $E$ be a given $n$-element set, and $A_{1}, A_{2}, \cdots, A_{k}$ be $k$ distinct non-empty subsets of $E$, satisfying: for any $1 \leqslant i<j \leqslant k$, either $A_{i} \cap A_{j}=\varnothing$ or one of $A_{i}$ and $A_{j}$ is a subset of the other. Find the maximum value of $k$.

Solution:
3. The maximum value of $k$ is $2 n-1$.

Let $E=\{1,2, \cdots, n\}$.
For example, when $1 \leqslant i \leqslant n$, let $A_{i}=\{i\}$;
when $n+1 \leqslant i \leqslant 2 n-1$, let $A_{i}=\{x \in E \mid 1 \leqslant x \leqslant i-n+1\}$.
The $2 n-1$ sets $A_{i}$ thus selected satisfy the conditions.
We prove by mathematical induction that $k \leqslant 2 n-1$.
When $n=1$, the conclusion is obvious.
Assume that when $n \leqslant m-1$, the conclusion holds.
When $n=m$, consider the set among $A_{1}, A_{2}, \cdots, A_{k}$ that is not the full set and has the most elements (denote it as $A_{1}$, and assume $A_{1}$ has $t$ elements). Then $t \leqslant m-1$.
We classify $A_{1}, A_{2}, \cdots, A_{k}$ into three categories:
(1) The full set;
(2) Not the full set, and the intersection with $A_{1}$ is not empty;
(3) The intersection with $A_{1}$ is empty.
By the selection of $A_{1}$, we know that the sets in (2) are all subsets of $A_{1}$ and still satisfy the conditions. By the induction hypothesis, the number of sets in (2) does not exceed $2 t-1$.

The sets in (3) are all subsets of $E \backslash A_{1}$. By the induction hypothesis, the number of sets does not exceed $2(m-t)-1$.
$$
\begin{array}{l}
\text { Hence } k \leqslant 1+(2 t-1)+[2(m-t)-1] \\
=2 m-1,
\end{array}
$$

That is, when $n=m$, the conclusion also holds.
Therefore, for any $n$-element set $E$, $k \leqslant 2 n-1$.

Answer: 2n-1


On the 70 km long motorway stretch between Dresden and Karl-Marx-Stadt, some vehicles travel at an average speed of 70 km/h, while others travel at an average speed of 100 km/h. How much time do the faster vehicles save?

Is the speeding worth it, considering that the frequency and severity of accidents at 100 km/h are twice as high as at 70 km/h?

Solution:
The travel time of the slower car was 60 minutes ( $70 \mathrm{~km} : 70 \mathrm{~km} / \mathrm{h} = 1 \mathrm{~h}$ ), that of the faster one 42 minutes ( $70 \mathrm{~km} : 100 \mathrm{~km} / \mathrm{h} = 0.7 \mathrm{~h}$ ).

The time savings of 18 minutes does not outweigh the doubled accident risk.

![](https://cdn.mathpix.com/cropped/2024_06_06_69be9c4355ed0dc2557cg-55.jpg?height=383&width=397&top_left_y=1142&top_left_x=361)

Answer: 18


Suppose the equation $||x-a|-b|=2008$ has 3 distinct real roots and $a \neq 0$. Find the value of $b$.

Solution:
21. Answer: 2008
The equation is equivalent to $|x-a|=b \pm 2008$.
Case 1: If $b2008$, then both $|x-a|=b-2008$ and $|x-a|=b+2008$ has 2 real roots, which gives 4 distinct real roots $a \pm(b-2008)$ and $a \pm(b+2008)$, since $a+b+2008>a+b-2008>a-b+2008>a-b-2008$.
Case 3: If $b=2008$, then $|x-a|=b-2008=0$ has only 1 real root $x=a$, and $|x-a|=b+2008=4016$ has 2 real roots $x=a \pm 4016$.
Hence $b=2008$.

Answer: 2008


A mason has bricks with dimensions $2\times5\times8$ and other bricks with dimensions $2\times3\times7$. She also has a box with dimensions $10\times11\times14$. The bricks and the box are all rectangular parallelepipeds. The mason wants to pack bricks into the box filling its entire volume and with no bricks sticking out.

Find all possible values of the total number of bricks that she can pack.

Solution:
1. **Define the problem and variables:**
   - Brick of type A: dimensions \(2 \times 5 \times 8\)
   - Brick of type B: dimensions \(2 \times 3 \times 7\)
   - Box: dimensions \(10 \times 11 \times 14\)

2. **Calculate volumes:**
   - Volume of the box:
     \[
     10 \times 11 \times 14 = 1540 \text{ unit cubes}
     \]
   - Volume of brick A:
     \[
     2 \times 5 \times 8 = 80 \text{ unit cubes}
     \]
   - Volume of brick B:
     \[
     2 \times 3 \times 7 = 42 \text{ unit cubes}
     \]

3. **Set up the equation for the total volume:**
   \[
   80x + 42y = 1540
   \]
   where \(x\) is the number of type A bricks and \(y\) is the number of type B bricks.

4. **Simplify the equation:**
   \[
   80x + 42y = 1540 \implies 40x + 21y = 770
   \]

5. **Solve for \(y\) in terms of \(x\):**
   \[
   21y = 770 - 40x \implies y = \frac{770 - 40x}{21}
   \]
   For \(y\) to be an integer, \(770 - 40x\) must be divisible by 21.

6. **Find the general solution:**
   \[
   770 - 40x \equiv 0 \pmod{21}
   \]
   Simplify \(770 \pmod{21}\):
   \[
   770 \div 21 = 36 \text{ remainder } 14 \implies 770 \equiv 14 \pmod{21}
   \]
   Thus:
   \[
   14 - 40x \equiv 0 \pmod{21} \implies -40x \equiv -14 \pmod{21} \implies 40x \equiv 14 \pmod{21}
   \]
   Simplify \(40 \pmod{21}\):
   \[
   40 \equiv 19 \pmod{21} \implies 19x \equiv 14 \pmod{21}
   \]

7. **Solve the congruence \(19x \equiv 14 \pmod{21}\):**
   Find the multiplicative inverse of 19 modulo 21. Using the Extended Euclidean Algorithm:
   \[
   19 \cdot 10 \equiv 1 \pmod{21} \implies x \equiv 14 \cdot 10 \pmod{21} \implies x \equiv 140 \pmod{21} \implies x \equiv 14 \pmod{21}
   \]
   Thus:
   \[
   x = 14 + 21k \quad \text{for integer } k
   \]

8. **Substitute \(x\) back to find \(y\):**
   \[
   y = \frac{770 - 40(14 + 21k)}{21} = \frac{770 - 560 - 840k}{21} = \frac{210 - 840k}{21} = 10 - 40k
   \]

9. **Ensure non-negative solutions:**
   \[
   x \geq 0 \implies 14 + 21k \geq 0 \implies k \geq -\frac{14}{21} \implies k \geq -1
   \]
   \[
   y \geq 0 \implies 10 - 40k \geq 0 \implies k \leq \frac{10}{40} \implies k \leq \frac{1}{4}
   \]
   Since \(k\) must be an integer, the only possible value is \(k = 0\).

10. **Substitute \(k = 0\):**
    \[
    x = 14 + 21 \cdot 0 = 14
    \]
    \[
    y = 10 - 40 \cdot 0 = 10
    \]

11. **Verify the solution:**
    \[
    80 \cdot 14 + 42 \cdot 10 = 1120 + 420 = 1540
    \]
    The solution is valid.

12. **Check the configuration:**
    - The box can be filled with 14 bricks of type A and 10 bricks of type B as described in the solution.

The final answer is \(\boxed{24}\)

Answer: 24


What is the maximum number of strings connecting adjacent nodes in a volleyball net with square cells that can be cut so that the net does not fall apart into separate pieces? The size of the net is $10 \times 100$ cells.

Solution:
1.9. Answer. 1000.

Suppose that so many strings have been torn that the net has not yet split into pieces, but no more strings can be torn. This means that there are no closed loops of strings left in the net. Prove that in this case, the number of unbroken strings is one less than the total number of nodes (including nodes located on the edges and at the corners of the net).

This can be verified, for example, as follows. Fix some node $A$. From it, you can travel along the unbroken strings to any other node $B$. Correspond to each node $B$ of the net the last string on the path leading from $A$ to $B$. This path is uniquely determined because there are no closed paths of strings. Therefore, the correspondence between the remaining strings and nodes $B$ (different from $A$) is one-to-one. Thus, the number of strings left is one less than the total number of nodes in the net, which, as is easy to calculate, is 1111. The total number of strings in the new unbroken net is 2110. Therefore, to leave 1110 strings, 1000 must be torn.

An example where 1000 strings are torn and the net has not yet split into pieces is easy to provide. For instance, you can tear all the horizontal strings in all rows except the top one. Then the net will turn into a "fringe." In essence, we have proved above that you can sequentially tear any 1000 strings, ensuring only that each time a string loop is cut, and on the 1001st time, this will not be possible.

Answer: 1000


Given arithmetic sequences $\left(x_{i}\right)_{i=1}^{\infty}$ and $\left(y_{i}\right)_{i=1}^{\infty}$ have the same first term and the following property: there exists an index $k(k>1)$, for which the equalities

$$
x_{k}^{2}-y_{k}^{2}=53, \quad x_{k-1}^{2}-y_{k-1}^{2}=78, \quad x_{k+1}^{2}-y_{k+1}^{2}=27 .
$$

hold. Find all such indices $k$.

Solution:
SOLUTION. Let $c, d$ be the differences of the first and second given arithmetic sequences, respectively. Since according to the problem statement $y_{1}=x_{1}$, the terms of both sequences have the general form

$$
x_{i}=x_{1}+(i-1) c \quad \text { and } \quad y_{i}=x_{1}+(i-1) d
$$

for any index $i$. The difference $x_{i}^{2}-y_{i}^{2}$ can therefore be rewritten as

$$
\begin{aligned}
x_{i}^{2}-y_{i}^{2} & =\left(x_{1}^{2}+2 x_{1}(i-1) c+(i-1)^{2} c^{2}\right)-\left(x_{1}^{2}+2 x_{1}(i-1) d+(i-1)^{2} d^{2}\right)= \\
& =2 x_{1}(i-1)(c-d)+(i-1)^{2}\left(c^{2}-d^{2}\right) .
\end{aligned}
$$

For the index $k$, according to the problem statement, the system of equations holds:

$$
\begin{aligned}
& 53=2 x_{1}(k-1)(c-d)+(k-1)^{2}\left(c^{2}-d^{2}\right), \\
& 78=2 x_{1}(k-2)(c-d)+(k-2)^{2}\left(c^{2}-d^{2}\right), \\
& 27=2 x_{1} k(c-d)+k^{2}\left(c^{2}-d^{2}\right) .
\end{aligned}
$$

We will now add these equations or their multiples in a suitable manner. To eliminate the terms with $x_{1}$, we subtract the sum of equations (2) and (3) from twice equation (1), since the coefficient of $2 x_{1}(c-d)$ then becomes $2(k-1)-(k-2+k)=0$. Since $2 \cdot 53-(78+27)=1$ and $2(k-1)^{2}-(k-2)^{2}-k^{2}=-2$, we obtain the simple equation $1=-2\left(c^{2}-d^{2}\right)$, from which we determine $c^{2}-d^{2}=-\frac{1}{2}$. Substituting this into equations (2) and (3), they transform into the form

$$
\begin{aligned}
& 78=2 x_{1}(k-2)(c-d)-\frac{1}{2}(k-2)^{2}, \\
& 27=2 x_{1} k(c-d)-\frac{1}{2} k^{2} .
\end{aligned}
$$

To eliminate the terms with $x_{1}$ again, we subtract $(k-2)$-times equation $\left(3^{\prime}\right)$ from $k$-times equation $\left(2^{\prime}\right)$; we then solve the resulting equation with the unknown $k$:

$$
\begin{aligned}
78 k-27(k-2) & =-\frac{1}{2}(k-2)^{2} \cdot k+\frac{1}{2} k^{2} \cdot(k-2), \\
51 k+54 & =-\frac{1}{2}\left(k^{3}-4 k^{2}+4 k\right)+\frac{1}{2}\left(k^{3}-2 k^{2}\right), \\
0 & =k^{2}-53 k-54, \\
0 & =(k+1)(k-54) .
\end{aligned}
$$

Since the index $k$ is a natural number, it must be $k=54$. This solves the problem.

Let us add that the problem statement does not require us to investigate whether for the found (unique) value of the index $k$ there exist pairs of sequences satisfying the conditions of the problem. For the sake of interest, let us mention that there are actually infinitely many such pairs of sequences; it is necessary and sufficient that their common first term $x_{1}$ and differences $c, d$ satisfy the conditions $c^{2}-d^{2}=-\frac{1}{2}$ and $x_{1}(c-d)=\frac{55}{4}$. This follows easily from any of the equations (1)-(3) after substituting the values $k=54$ and $c^{2}-d^{2}=-\frac{1}{2}$; you can verify this yourself.

GUIDING AND ADDITIONAL PROBLEMS:

Answer: 54


Two identical chessboards $(8 \times 8$ squares $)$ overlap each other. Now, one of them is rotated $45^{\circ}$ around the center. If the area of each square is 1, find the total area of the intersection of all the black squares of the two chessboards.

Solution: [Solution] The intersecting figure of the two chessboards is a regular octagon, with an area of $128(\sqrt{2}-1)$. This regular octagon can be divided into 4 parts: (white, white), (white, black), (black, white), (black, black), denoted as $S_{1}, S_{2}, S_{3}, S_{4}$ respectively. Since rotating one chessboard $90^{\circ}$ around its center swaps the two colors of the squares, when both chessboards are rotated $90^{\circ}$ around their centers, the order of the 4 parts is exactly reversed, hence we get $S_{1}=S_{4}, S_{2}=S_{3}$. Furthermore, when the two chessboards are rotated in opposite directions by $45^{\circ}$ and their positions are swapped, it is equivalent to one chessboard being stationary while the other rotates $90^{\circ}$, which can prove that $S_{1}=S_{3}$, thus the areas of the 4 parts are all equal, hence we get $S_{4}=$ $32(\sqrt{2}-1)$.

Answer: 32(\sqrt{2}-1)


Call a permutation $a_1, a_2, \ldots, a_n$ of the integers $1, 2, \ldots, n$ quasi-increasing if $a_k \leq a_{k+1} + 2$ for each $1 \leq k \leq n-1$. For example, 53421 and 14253 are quasi-increasing permutations of the integers $1, 2, 3, 4, 5$, but 45123 is not. Find the number of quasi-increasing permutations of the integers $1, 2, \ldots, 7$.

Solution:
The simple recurrence can be found.
When inserting an integer $n$ into a string with $n - 1$ integers, we notice that the integer $n$ has 3 spots where it can go: before $n - 1$, before $n - 2$, and at the very end.
EXAMPLE:
Putting 4 into the string 123:
4 can go before the 2: 1423,
Before the 3: 1243,
And at the very end: 1234.
Only the addition of the next number, n, will change anything.
Thus the number of permutations with n elements is three times the number of permutations with $n-1$ elements.
Start with $n=3$ since all $6$ permutations work. And go up: $18, 54, 162, 486$.
Thus for $n=7$ there are $2*3^5=\boxed{486}$ permutations.
When you are faced with a brain-fazing equation and combinatorics is part of the problem, use recursion! This same idea appeared on another AIME with a 8-box problem.

Answer: 486


Find all ten-digit numbers such that the 1st digit of the number equals the number of zeros in its decimal representation, the 2nd digit - the number of ones, and so on up to the 10th digit, which equals the number of nines in the number's representation.

Solution:
110. Let the desired number $X=\overline{a_{0} a_{1} a_{2} \ldots a_{9}}$ (where $a_{0}, a_{1}, \ldots, a_{9}$ are the digits of the number); in this case, $a_{0}$ is the number of zeros among the digits of $X$, $a_{1}$ is the number of ones, $a_{2}$ is the number of twos, and so on. Therefore, the sum of all the digits of $X$ is

$$
a_{0}+a_{1}+a_{2}+\ldots+a_{9}=a_{0} \cdot 0+a_{1} \cdot 1+a_{2} \cdot 2+\ldots+a_{9} \cdot 9
$$

from which we obtain

$$
a_{0}=a_{2}+2 a_{3}+3 a_{4}+4 a_{5}+5 a_{6}+6 a_{7}+7 a_{3}+8 a_{9}
$$

The equality $a_{0}=0$ is excluded by the conditions of the problem (otherwise the number $X$ would not be 10-digit; however, the condition $a_{0} \neq 0$ immediately follows from (*)); the equality $a_{0}=1$ leads to impossible values $a_{0}=a_{2}=1, a_{1}=8$ (since we should have 10 digits in total), all other digits are zeros; the equality $a_{0}=2$ is compatible only with (also impossible!) values $a_{0}=a_{2}=2, a_{1}=6$, all other digits are zeros, or $a_{0}=2, a_{3}=1, a_{1}=7$, all other digits are zeros. Let now $a_{0}=i>2$; we rewrite equality (*) as follows:

$$
a_{0}=i=a_{2}+2 a_{3}+\ldots+(i-1) a_{i}+\ldots+8 a_{9}
$$

(if $i=3$, then the terms $2 a_{3}$ and $(i-1) a_{i}$ coincide; if $i=9$, then $(i-1) a_{i}$ and $8 a_{9}$ coincide). In this case, $a_{i}$, the number of digits equal to $i$ in the number $X$, is non-zero (since $a_{0}=i$); on the other hand, (**) is impossible for $a_{i}>1$, so $a_{i}=1$. Therefore, (**) can be rewritten as:

$$
1=a_{2}+2 a_{3}+\ldots+(i-2) a_{i-1}+i a_{i+1}+\ldots+8 a_{9}, \quad(* * *)
$$

from which it immediately follows that $a_{2}=1$, and all digits of the number $X$ different from $a_{0}, a_{1}, a_{2}$ and $a_{i}$ are zero. But if $a_{2}=1$, then among the digits of the number $X$ there is a two, which, obviously, can only be the digit $a_{1}$. Thus, in the decimal representation of $X$, only the digits $a_{0}(=i), a_{1}=2, a_{2}=1$ and $a_{i}=1$ are non-zero, i.e., among the digits of $X$ there are $i$ zeros, 2 ones, 1 two, and 1 digit $i$, from which (since $X$ is 10-digit) $i=10-2-1-1=6$.

Thus, $X=6210001000$ (it is easy to check that this number indeed satisfies all the conditions of the problem).

Answer: 6210001000


From point $K$ on side $A C$ of triangle $A B C$, perpendiculars $K L_{1}$ and $K M_{1}$ were dropped to sides $A B$ and $B C$ respectively. From point $L_{1}$, a perpendicular $L_{1} L_{2}$ was dropped to $B C$, and from point $M_{1}$, a perpendicular $M_{1} M_{2}$ was dropped to $A B$.

It turned out that triangles $B L_{1} M_{1}$ and $B L_{2} M_{2}$ are similar (point $L_{1}$ in the first triangle corresponds to point $M_{2}$ in the second). In addition, $B L_{2}=6$ and $L_{2} M_{1}=4$. Find $L_{1} L_{2}$.

Solution:
# Answer: 8

## Solution:

Notice that the quadrilateral $L_{1} M_{2} L_{2} M_{1}$ is cyclic, since $\angle L_{1} M_{2} M_{1}=\angle L_{1} L_{2} M_{1}=$ $90^{\circ}$. Therefore, $\angle B M_{2} L_{2}=180^{\circ}-\angle L_{1} M_{2} L_{2}=\angle L_{2} M_{1} L_{1}=\angle B M_{1} L_{1}$. Similarly, $\angle B L_{2} M_{2}=\angle B L_{1} M_{1}$, so triangles $B L_{1} M_{1}$ and $B L_{2} M_{2}$ are similar, with point $L_{1}$ in the first triangle corresponding to point $L_{2}$ in the second. However, the problem states that they are similar in another way, which means these two triangles are isosceles. From this, we get $B L_{1}=B M_{1}=B L_{2}+L_{2} M_{1}$ and, by the Pythagorean theorem, we find $L_{1} L_{2}=\sqrt{B L_{1}^{2}-B L_{2}^{2}}=\sqrt{\left(B L_{2}+L_{2} M_{1}\right)^{2}-B L_{2}^{2}}$.

Answer: 8


If the thought number is multiplied by 3, 2 is appended to the right, the resulting number is divided by 19, and 7 is added to the quotient, the result is three times the thought number. What is this number?

Solution:
56. Let $x$ be the thought-of number. Then the condition of the problem can be written as: $(3 x \cdot 10 + 2) : 9 + 7 = 3 x$.

From which $x = 5$.

## $\S 3$

Answer: 5


Alyosha lists numbers where the first digit is equal to the product of the other three, while Vasya lists numbers where the last digit is equal to the product of the other three. Who will list more numbers and by how many?

Solution:
Task 3. Excellent students Alyosha and Vasya write down four-digit numbers. Alyosha writes down numbers where the first digit is equal to the product of the other three, while Vasya writes down numbers where the last digit is equal to the product of the other three. Who will write down more numbers and by how many?

| Solution | Criteria |
| :--- | :--- |
| Since Vasya can write numbers that | 1. 1 point for the correct answer. |
| end in 0, while Alyosha cannot write | 2. 1 point if it is explained why Vasya will write |
| numbers that start with 0, Vasya will | more numbers. |
| write more numbers. Let's count the | 3. -2 points if the incorrect answer is given because 00 was counted twice. |
| number of four-digit numbers that end |  |
| in 0 and contain at least one more 0 in |  |
| their representation. The first digit of |  |
| such a four-digit number can be any digit |  |
| from 1 to 9. Zero must be on the second |  |
| or third place. The number of numbers of |  |
| the form 0right is 9, the number of numbers |  |
| of the form p0 is 9 (where p>0), and there is 00. |  |
| Therefore, the total number of such numbers |  |
| is $9 \times (9 + 9 + 1) = 171$. |

Answer: 171


Find the last two digits in the decimal representation of the number $1!+2!+\ldots+2001!+2002!$.

Solution:
From some point, factorials start to be divisible by 100.

## Solution

Notice that 10! is divisible by 100, and \( n! \) for \( n > 10 \) will also be divisible by 100. Therefore, the last two digits of the number \( 1! + 2! + \ldots + 2001! + 2002! \) are the same as the last two digits of the number \( 1! + 2! + \ldots + 9! \). The last two digits of this number can be found directly (the calculations are simplified by the fact that, starting from \( n = 5 \), \( n! \) is divisible by 10).

## Answer

13.

## Angles between angle bisectors ] Problem 53388 Topics: [Sum of angles in a triangle. Theorem about the exterior angle.] [ Arithmetic progression $]$

The angles at vertices \( A, B, C \) of triangle \( ABC \) form an arithmetic progression with a common difference of \( \pi / 7 \). The angle bisectors of this triangle intersect at point \( D \). Points \( A_1, B_1, C_1 \) are located on the extensions of segments \( DA, DB, DC \) beyond points \( A, B, C \) respectively, at the same distance from point \( D \). Prove that the angles \( A_1, B_1, C_1 \) also form an arithmetic progression. Find its common difference.

## Hint

\(\angle BDC = \pi / 2 + 1 / 2 \angle A\).

## Solution

Let \(\angle A = \alpha\) be the middle angle of triangle \( ABC \), \(\angle C = \alpha - \pi / 7\), \(\angle B = \alpha + \pi / 7\). \(\angle B_1DC_1 = \pi / 2 + \alpha / 2\) (see problem \(\underline{55448}\)),

\(\angle DB_1C_1 = \angle DC_1B_1 = 1 / 2 (\pi - \pi / 2 - \alpha / 2) = \pi / 4 - \alpha / 4\). Similarly, \(\angle DA_1C_1 = \angle DC_1A_1 = \pi / 4 - \alpha / 4 - \pi / 28\), \(\angle DA_1B_1 = \angle DB_1A_1 = \pi / 4 - \alpha / 4 + \pi / 28\). Therefore,

\(\angle A_1 = \angle DA_1C_1 + \angle DA_1B_1 = \pi / 2 - \alpha / 2\), \(\angle B_1 = \pi / 2 - \alpha / 2 + \pi / 28\), \(\angle C_1 = \pi / 2 - \alpha / 2 - \pi / 28\).

![](https://cdn.mathpix.com/cropped/2024_05_06_5008c33ea747074d348fg-08.jpg?height=469&width=692&top_left_y=1043&top_left_x=680)

## Answer

\(\pi / 28\).

Answer: 13


Answer: 1


Real numbers $a$, $b$, and $c$ satisfy the equation $a + b + c = 0$. If $a$, $b$, and $c$ are all distinct and non-zero, which of the following must be true?

A) $a^2 + b^2 + c^2 = 0$.

B) Each boy takes $a^2 + b^2 + c^2 + b^2 + b^2 + c^2$0$. If $a$, $b$, and $c$ are all distinct and non-zero, which of the following must be true?

A) $a^2 + b^2 + c^2 = 0$.

B) Each boy takes $a^2 + b^2 + c^2 \neq 0$.

C) $a^2 + b^2 + c^2 = ab + bc + ca$.

D) $a^2 + b^2 + c^2 > ab + bc + ca$.

E) $a^2 + b^2 + c^2 < ab + bc + ca$.

(Note: The provided options section seems to have some formatting errors or unclear entries, but I've kept it as instructed.)

Solution:
3. $\left[-\frac{3 \sqrt{2}+\sqrt{6}}{2}, \frac{3 \sqrt{2}+\sqrt{6}}{2}\right]$.
$$
f(x, y, z)=\sqrt{x^{2}+y^{2}+2 x-2 y+2} \sin (z+\theta),
$$

where, $\cos \theta=\frac{x+1}{\sqrt{(x+1)^{2}+(y-1)^{2}}}$,
$$
\sin \theta=\frac{y-1}{\sqrt{(x+1)^{2}+(y-1)^{2}}} \text {. }
$$

Since $z$ can take any value in $\mathbf{R}$, $z+\theta$ can also take any value in $\mathbf{R}$. Therefore, the range of $\sin (z+\theta)$ is $[-1,1]$.

Next, we find the range of $g(x, y)=x^{2}+y^{2}+2 x-2 y+2$.
$$
\begin{array}{l}
\text { From } x^{2}+y^{2}-x+y=1 \text { we get } \\
\left(x-\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=\left(\sqrt{\frac{3}{2}}\right)^{2} \text {. } \\
\text { Let }\left\{\begin{array}{l}
x-\frac{1}{2}=\sqrt{\frac{3}{2}} \cos \varphi, \\
y+\frac{1}{2}=\sqrt{\frac{3}{2}} \sin \varphi .
\end{array}\right. \text { Then } \\
g(x, y)=(1+x-y)+2 x-2 y+2 \\
=3(x-y+1) \\
=3\left[\sqrt{\frac{3}{2}}(\cos \varphi-\sin \varphi)+2\right] \\
=3 \sqrt{3} \sin (\varphi+\alpha)+6 \text {. } \\
\end{array}
$$

Since $\varphi+\alpha$ can take any value in $\mathbf{R}$, the range of $g(x, y)$ is $[6-3 \sqrt{3}, 6+3 \sqrt{3}]$.
Therefore, the range of $f(x, y, z)$ is
$$
\begin{array}{l}
{[-\sqrt{6+3 \sqrt{3}}, \sqrt{6+3 \sqrt{3}}]} \\
=\left[-\frac{3 \sqrt{2}+\sqrt{6}}{2}, \frac{3 \sqrt{2}+\sqrt{6}}{2}\right] .
\end{array}
$$

Answer: \left[-\frac{3 \sqrt{2}+\sqrt{6}}{2}, \frac{3 \sqrt{2}+\sqrt{6}}{2}\right]


Tom throws a football to Wes, who is a distance $l$ away. Tom can control the time of flight $t$ of the ball by choosing any speed up to $v_{\text{max}}$ and any launch angle between $0^\circ$ and $90^\circ$. Ignore air resistance and assume Tom and Wes are at the same height. Which of the following statements is [b]incorrect[/b]?

$ \textbf{(A)}$ If $v_{\text{max}} < \sqrt{gl}$, the ball cannot reach Wes at all. $ \\ $
$ \textbf{(B)}$ Assuming the ball can reach Wes, as $v_{\text{max}}$ increases with $l$ held fixed, the minimum value of $t$ decreases. $ \\ $
$ \textbf{(C)}$ Assuming the ball can reach Wes, as $v_{\text{max}}$ increases with $l$ held fixed, the maximum value of $t$ increases. $ \\ $
$ \textbf{(D)}$ Assuming the ball can reach Wes, as $l$ increases with $v_{\text{max}}$ held fixed, the minimum value of $t$ increases. $ \\ $
$ \textbf{(E)}$ Assuming the ball can reach Wes, as $l$ increases with $v_{\text{max}}$ held fixed, the maximum value of $t$ increases.

Solution:
1. **Statement (A):** If \( v_{\text{max}} < \sqrt{gl} \), the ball cannot reach Wes at all.
   - The maximum range \( R \) of a projectile launched with speed \( v \) at an angle \( \theta \) is given by:
     \[
     R = \frac{v^2 \sin(2\theta)}{g}
     \]
   - The maximum range is achieved when \( \sin(2\theta) = 1 \), i.e., \( \theta = 45^\circ \):
     \[
     R_{\text{max}} = \frac{v^2}{g}
     \]
   - For the ball to reach Wes, \( R_{\text{max}} \geq l \):
     \[
     \frac{v_{\text{max}}^2}{g} \geq l \implies v_{\text{max}} \geq \sqrt{gl}
     \]
   - Therefore, if \( v_{\text{max}} < \sqrt{gl} \), the ball cannot reach Wes. Thus, statement (A) is correct.

2. **Statement (B):** Assuming the ball can reach Wes, as \( v_{\text{max}} \) increases with \( l \) held fixed, the minimum value of \( t \) decreases.
   - The time of flight \( t \) for a projectile is given by:
     \[
     t = \frac{2v \sin(\theta)}{g}
     \]
   - To minimize \( t \), we need to minimize \( \sin(\theta) \). As \( v_{\text{max}} \) increases, Tom can choose a smaller angle \( \theta \), reducing \( \sin(\theta) \) and thus \( t \). Therefore, statement (B) is correct.

3. **Statement (C):** Assuming the ball can reach Wes, as \( v_{\text{max}} \) increases with \( l \) held fixed, the maximum value of \( t \) increases.
   - The maximum time of flight occurs when the ball is thrown at a higher angle, approaching \( 90^\circ \). As \( v_{\text{max}} \) increases, Tom can throw the ball at a higher angle, increasing the time of flight. Therefore, statement (C) is correct.

4. **Statement (D):** Assuming the ball can reach Wes, as \( l \) increases with \( v_{\text{max}} \) held fixed, the minimum value of \( t \) increases.
   - As \( l \) increases, the ball must travel a longer distance. To achieve this with a fixed \( v_{\text{max}} \), Tom must throw the ball at a higher angle, increasing the time of flight. Therefore, statement (D) is correct.

5. **Statement (E):** Assuming the ball can reach Wes, as \( l \) increases with \( v_{\text{max}} \) held fixed, the maximum value of \( t \) increases.
   - As \( l \) increases, the ball must travel a longer distance. To achieve this with a fixed \( v_{\text{max}} \), Tom must throw the ball at a higher angle, increasing the time of flight. Therefore, statement (E) is correct.

However, the solution provided states that statement (E) is incorrect because Tom needs to throw at an angle closer to 45 degrees, making his higher possible angle go down. This reasoning is flawed because the maximum time of flight is achieved at higher angles, not closer to 45 degrees.

Therefore, the correct answer is:

The final answer is \(\boxed{\textbf{(E)}}\)

Answer: \textbf{(E)}


The polar equation of curve $C$ is $\rho=1+\cos \theta$, and the polar coordinates of point $A$ are $(2,0)$. If curve $C$ rotates once around $A$ in its plane, then the area of the figure swept by it is $\qquad$

Solution:
Let point $B(\rho, \theta)$ be on the curve $C$, then $|A B|^{2}=\rho^{2}+4-4 \rho \cos \theta$
$$
\begin{array}{l}
=(1+\cos \theta)^{2}-4 \cos \theta(1+\cos \theta) \\
=-3 \cos ^{2} \theta-2 \cos \theta+5=-3\left(\cos \theta+\frac{1}{3}\right)^{2}+\frac{16}{3} .
\end{array}
$$

Thus, when $\cos \theta=-\frac{1}{3}$, $|A B|_{\text {max }}^{2}=\frac{16}{3}$.
Therefore, the area of the figure swept by the curve $C$ is $\frac{16}{3} \pi$.

Answer: \frac{16}{3}\pi


Given that when 81849, 106392 and 124374 are divided by an integer $n$, the remainders are equal. If $a$ is the maximum value of $n$, find $a$.

Solution:
$$
\begin{array}{l}
81849=p n+k \ldots \ldots .(1) \\
106392=q n+k \ldots \ldots .(2) \\
124374=r n+k \ldots \ldots .(3)
\end{array}
$$
(2) $-(1): 24543=(q-p) n$
(3) $-(2): 17982=(r-q) n$ $\qquad$
(4): $243 \times 101=(q-p) n$
(5): $243 \times 74=(r-q) n$
$a=$ maximum value of $n=243$

Answer: 243


Knut is a very trained cyclist. On a trip, he covered on average $320 \mathrm{~m}$ per minute on his bicycle.

He set off on his bike at 7:00 AM and reached his destination at 11:00 AM. From 9:00 AM to 9:20 AM, he rested, and during the rest of the time, he cycled continuously.

How long (in $\mathrm{km}$) is the total distance covered by Knut?

Solution:
The time from 7:00 AM to 11:00 AM is 4 hours, which is 240 minutes. The break time is 20 minutes, so the driving time is 220 minutes.

Since $220 \cdot 320=70400$, the total distance traveled by Knut is thus $70400 \mathrm{~m}=70.4 \mathrm{~km}$ long.

Answer: 70.4\mathrm{~}


Write a linear function that has a zero at 4, and its graph passes through the intersection of the lines with equations $x-2y-9=0$ and $2x+y-3=0$.

Solution:
B1. The intersection of the given lines is calculated by solving a system of two equations with two unknowns: $x-2 y-9=0$ and $2 x+y-3=0$. The solution to the system is the pair $x=3, y=-3$, and the intersection point is $P(3,-3)$. The graph of the desired linear function passes through this intersection point and through the point $A(4,0)$. We calculate its slope $k=\frac{0-(-3)}{4-3}=3$ and its y-intercept $n=0-3 \cdot 4=-12$, so we have the equation of the linear function: $f(x)=3 x-12$.

Correct solution of the system ....................................................................................................................... 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_8d4f73ada7862f989e67g-12.jpg?height=57&width=1639&top_left_y=1713&top_left_x=274)

![](https://cdn.mathpix.com/cropped/2024_06_07_8d4f73ada7862f989e67g-12.jpg?height=57&width=1639&top_left_y=1765&top_left_x=274)

![](https://cdn.mathpix.com/cropped/2024_06_07_8d4f73ada7862f989e67g-12.jpg?height=57&width=1637&top_left_y=1813&top_left_x=278)

![](https://cdn.mathpix.com/cropped/2024_06_07_8d4f73ada7862f989e67g-12.jpg?height=52&width=1641&top_left_y=1867&top_left_x=276)

![](https://cdn.mathpix.com/cropped/2024_06_07_8d4f73ada7862f989e67g-12.jpg?height=51&width=1639&top_left_y=1916&top_left_x=274)

Answer: f(x)=3x-12