Given a trapezoid $ABCD$ such that $\overline{AB}=\overline{AC}=\overline{BD}$. Let $M$ be the midpoint of side $CD$. Determine the angles of the trapezoid if $\measuredangle MBC=\measuredangle CAB$.

Solution:
Solution. From the condition, it follows that the given trapezoid is isosceles. Let $K$ be the midpoint of $AD$, and let $\measuredangle CAB = \measuredangle MBC = \varphi$. Then

$$
\measuredangle MKA = 180^{\circ} - \measuredangle KAC = 180^{\circ} - \measuredangle MBA.
$$

Thus, quadrilateral $ABMK$ is cyclic.

Furthermore, from the condition, it follows that triangle $\triangle ABD$ is isosceles, from which it follows that $\measuredangle AKB = 90^{\circ}$. Now, due to the cyclicity of $ABMK$, it follows that $\measuredangle AMB = \measuredangle AKB = 90^{\circ}$, i.e., triangle $\triangle AMB$ is an isosceles right triangle. Let $M_1$ be the foot of the altitude dropped from $M$. Then

$$
\overline{MM_1} = \overline{AM_1} = \frac{\overline{AB}}{2} = \frac{\overline{AC}}{2},
$$

so we get that $\varphi = 30^{\circ}$. Now easily

$$
\measuredangle ABC = 30^{\circ} + 45^{\circ} = 75^{\circ} \text{ and } \measuredangle ADC = 105^{\circ}.
$$

Answer: \measuredangleABC=75


The sofa cost 62500 rubles. Once a month, its price changed by $20 \%$ either increasing or decreasing. It is known that over the course of six months, the price increased three times and decreased three times (the order in which this happened is unknown). Can the final price of the sofa after six months be determined uniquely? If so, what did it become?

Solution:
Answer: Yes; the sofa will cost 55296 rubles.

Solution. An increase in price by $20 \%$ means the current price is multiplied by $6 / 5$, and a decrease in price by $20 \%$ means the current price is multiplied by $4 / 5$. Therefore, regardless of the order in which the price increased or decreased, the price of the sofa after six months will be $62500 \cdot(6 / 5)^{3} \cdot(4 / 5)^{3}=$ 55296 rubles.

Evaluation. If only some specific cases are considered and the correct numerical answer is obtained, 4 points. For a complete solution, 12 points.

Answer: 55296


Given that $\underbrace{\overline{abcabc\cdots abc}}_{2019 \uparrow abc}$ is a multiple of 91, then the sum of the minimum and maximum values of the three-digit number $\overline{abc}$ is

Answer: 1092


In an isosceles trapezoid, one base is equal to $40 \mathrm{~cm}$, and the other is 24 cm. The diagonals of this trapezoid are perpendicular to each other. Find its area.

Solution:
## Solution.

The area of an isosceles trapezoid, whose diagonals are perpendicular to each other, is equal to the square of its height: $S=h^{2}$. On the other hand, $S=\frac{a+b}{2} h$, from which $h=\frac{a+b}{2}=\frac{40+24}{2}=32$ (cm). And the area

$S=32^{2}=1024\left(\mathrm{~cm}^{2}\right)$.

Answer: $1024 \mathrm{~cm}^{2}$.

Answer: 1024\mathrm{~}^{2}


If $-3 \le x < \frac{3}{2}$ and $x \ne 1$, define $C(x) = \frac{x^3}{1 - x}$. The real root of the cubic $2x^3 + 3x - 7$ is of the form $p C^{-1}(q)$, where $p$ and $q$ are rational numbers. What is the ordered pair $(p, q)$?

Solution:
1. **Understanding the function \( C(x) \):**
   The function \( C(x) = \frac{x^3}{1 - x} \) is defined for \( x \in [-3, 1) \cup (1, \frac{3}{2}) \). This function is bijective on its domain, meaning it has an inverse function \( C^{-1}(y) \).

2. **Identifying the root \( r \):**
   We are given that the real root of the cubic equation \( 2x^3 + 3x - 7 = 0 \) is of the form \( r = p C^{-1}(q) \), where \( p \) and \( q \) are rational numbers. This implies \( \frac{r}{p} = C^{-1}(q) \).

3. **Expressing \( q \) in terms of \( r \) and \( p \):**
   Since \( \frac{r}{p} = C^{-1}(q) \), we have \( q = C\left(\frac{r}{p}\right) \). Substituting \( \frac{r}{p} \) into \( C(x) \), we get:
   \[
   q = C\left(\frac{r}{p}\right) = \frac{\left(\frac{r}{p}\right)^3}{1 - \frac{r}{p}} = \frac{\frac{r^3}{p^3}}{1 - \frac{r}{p}} = \frac{r^3}{p^3 - p^2 r}
   \]

4. **Relating \( r \) and \( q \):**
   From the above expression, we have:
   \[
   q = \frac{r^3}{p^3 - p^2 r}
   \]
   Rearranging this, we get:
   \[
   r^3 = q (p^3 - p^2 r)
   \]
   Simplifying further:
   \[
   r^3 = qp^3 - qp^2 r
   \]
   \[
   r^3 + qp^2 r = qp^3
   \]
   \[
   r (r^2 + qp^2) = qp^3
   \]
   \[
   r = \frac{qp^3}{r^2 + qp^2}
   \]

5. **Finding rational \( p \) and \( q \):**
   Since \( r \) is a root of \( 2x^3 + 3x - 7 = 0 \), we substitute \( r \) into the equation:
   \[
   2r^3 + 3r - 7 = 0
   \]
   Using the expression for \( r \) from above:
   \[
   2 \left(\frac{qp^3}{r^2 + qp^2}\right)^3 + 3 \left(\frac{qp^3}{r^2 + qp^2}\right) - 7 = 0
   \]
   This equation must hold true for rational \( p \) and \( q \).

6. **Solving for \( p \) and \( q \):**
   To find \( p \) and \( q \), we need to solve the system:
   \[
   2p^2q - 3 = 0
   \]
   \[
   2p^3q - 7 = 0
   \]
   Solving the first equation for \( q \):
   \[
   q = \frac{3}{2p^2}
   \]
   Substituting into the second equation:
   \[
   2p^3 \left(\frac{3}{2p^2}\right) - 7 = 0
   \]
   \[
   3p - 7 = 0
   \]
   \[
   p = \frac{7}{3}
   \]
   Substituting \( p = \frac{7}{3} \) back into the expression for \( q \):
   \[
   q = \frac{3}{2 \left(\frac{7}{3}\right)^2} = \frac{3}{2 \cdot \frac{49}{9}} = \frac{3 \cdot 9}{2 \cdot 49} = \frac{27}{98}
   \]

The final answer is \( \boxed{ (p, q) = \left(\frac{7}{3}, \frac{27}{98}\right) } \)

Answer: (p, q) = \left(\frac{7}{3}, \frac{27}{98}\right)


For a positive integer $n$, let $d_n$ be the units digit of $1 + 2 + \dots + n$. Find the remainder when
\[\sum_{n=1}^{2017} d_n\]is divided by $1000$.

Solution:
We see that $d_n$ appears in cycles of $20$ and the cycles are \[1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,\] adding a total of $70$ each cycle.
Since $\left\lfloor\frac{2017}{20}\right\rfloor=100$, we know that by $2017$, there have been $100$ cycles and $7000$ has been added. This can be discarded as we're just looking for the last three digits.
Adding up the first $17$ of the cycle of $20$, we can see that the answer is $\boxed{069}$.

~ Maths_Is_Hard

Answer: 69


Find the maximum value of
$$x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}$$

Solution:
By Cauchy-Schwarz inequality, we have
$$\left|x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right|^{2} \leqslant\left(x^{2}+y^{2}\right)\left(2-x^{2}-y^{2}\right) .$$

By the arithmetic mean inequality, we have
$$\left|x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right| \leqslant \frac{x^{2}+y^{2}+2-x^{2}-y^{2}}{2}=1 .$$

If $x=\frac{1}{2}, y=\frac{\sqrt{3}}{2}$, then
$$x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=1$$

Thus, the maximum value sought is 1.

Answer: 1


Let $ABCDEF$ be a regular hexagon. Points $P$ and $Q$ on tangents to its circumcircle at $A$ and $D$ respectively are such that $PQ$ touches the minor arc $EF$ of this circle. Find the angle between $PB$ and $QC$.

Solution:
1. **Define the problem and setup:**
   Let $ABCDEF$ be a regular hexagon inscribed in a circle with center $O$. Points $P$ and $Q$ are on the tangents to the circumcircle at $A$ and $D$ respectively, such that the line $PQ$ touches the minor arc $EF$ of the circle. We need to find the angle between the lines $PB$ and $QC$.

2. **Consider the duality transform:**
   We will use the duality transform with respect to the circumcircle. Let $X$ be any point on the minor arc $EF$. Let $XA$ meet the tangent to the circle at $B$ at point $R$, and $XD$ meet the tangent at $C$ at point $S$. We need to prove that $\angle ROS = 150^\circ$.

3. **Analyze the angles:**
   We start by analyzing the angles formed by the tangents and the lines from $X$:
   - $\angle (DC; RB) = \angle (CS; BA) = \angle (SD; AR) = \frac{\pi}{2}$.
   This means that there is a spiral similarity that maps $S$, $D$, $C$ onto $A$, $R$, $B$ respectively.

4. **Use properties of spiral similarity:**
   Since there is a spiral similarity, we have:
   \[
   |CS| \cdot |BR| = |AB| \cdot |CD| = |OC| \cdot |OB|
   \]
   This implies that $\triangle OCS \sim \triangle RBO$.

5. **Calculate the angles:**
   From the similarity, we have:
   \[
   \angle COS + \angle ROB = 90^\circ
   \]
   Since $O$ is the center of the circle, $\angle BOC = 60^\circ$ (as $B$ and $C$ are adjacent vertices of the regular hexagon).

6. **Combine the angles:**
   Therefore, we have:
   \[
   \angle ROS = 90^\circ + \angle BOC = 90^\circ + 60^\circ = 150^\circ
   \]

7. **Relate to the original problem:**
   The angle between $PB$ and $QC$ is the same as $\angle ROS$ because of the properties of the tangents and the duality transform.

The final answer is $\boxed{30^\circ}$.

Answer: 30^\circ


There are three two-digit numbers $A$, $B$, and $C$. $A$ is a perfect square, and each of its digits is also a perfect square; $B$ is a prime number, and each of its digits is also a prime number, and the sum of its digits is also a prime number; $C$ is a composite number, and each of its digits is also a composite number, the difference between the two digits is also a composite number, and $C$ is between $A$ and $B$. What is the sum of $A$, $B$, and $C$?

Solution:
【Solution】Solution: According to the analysis, first determine $A$,
$\because$ The one-digit numbers that are perfect squares are only $1,4,9$, and among them, the two-digit numbers that can form a perfect square are only $49, \therefore A=49$;
$\because$ The prime number $B$ has the sum of its two digits as a prime number and each digit is a prime number,
$\therefore$ The tens digit of $B$ can only be 2, and the units digit can only be 3, so $B=23$;
$\because$ The composite number $C$ has the difference between its two digits as a composite number and each digit is a composite number, then these digits can only be: $4,6,8,9$,
$C$ is between $A$ and $B$, i.e.,
$\therefore C=48$, hence $A+B+C=49+23+48=120$,
Therefore, the answer is: 120.

Answer: 120


In an $11 \times 11$ square, the central cell is painted black. Maxim found a rectangular grid of the largest area that is entirely within the square and does not contain the black cell. How many cells does it have?

Solution:
Answer: 55

Solution. If both sides of the rectangle are not less than 6, then it must contain the central cell, since the distance from the painted cell to the sides is 5 cells. If one of its sides is not more than 5, then the other is definitely not more than 11, and therefore its area is not more than 55.

Answer: 55


From identical isosceles triangles, where the angle opposite the base is $45^{\circ}$ and the lateral side is 1, a figure was formed as shown in the diagram. Find the distance between points $A$ and $B$.

Solution:
Answer: 2.

Solution: Let's denote the points $K, L, M$, as shown in the figure. We will construct an isosceles triangle $A K C$ equal to the original one. Connect vertex $C$ to other points as shown in the figure.

In the original triangles, the angle at the vertex is $45^{\circ}$. Therefore, the other two angles are each $62.5^{\circ}$. Then the angle $\angle C K L = 360^{\circ} - 62.5^{\circ} - 62.5^{\circ} - 45^{\circ} - 45^{\circ} - 62.5^{\circ} = 62.5^{\circ}$. Therefore, triangles $A K C$ and $K L C$ are equal by two sides and the angle between them. Similarly, it can be proven that triangles $L M C$ and $M B C$ are equal to the original ones. Then the angle $\angle A C B$ is $4 \times 45^{\circ} = 180^{\circ}$, which means points $A, C, B$ lie on the same line. Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_c10d52ec2bf2d695bd7bg-2.jpg?height=331&width=511&top_left_y=977&top_left_x=1321)
$A B = A C + C B = 2$.

## Criteria:

![](https://cdn.mathpix.com/cropped/2024_05_06_c10d52ec2bf2d695bd7bg-2.jpg?height=377&width=552&top_left_y=1419&top_left_x=1343)

A suitable construction is made, but it is not justified that points $A, C, B$ lie on the same line (or point $C$ is the midpoint of segment $A B$, but it is not justified that the resulting triangles are equal to the original ones) - 1 point.

The solution refers to the fact that a regular octagon has a center and other obvious properties of the octagon - no more than 1 point should be deducted.

Answer: 2


Let $n$ positive numbers $x_{1}, x_{2}, \cdots, x_{n}$ sum up to 1. Let $S$ be the largest of the following numbers:
$$
\frac{x_{1}}{1+x_{1}}, \frac{x_{2}}{1+x_{1}+x_{2}}, \cdots, \frac{x_{n}}{1+x_{1}+x_{2}+\cdots+x_{n}} .
$$

Find the possible minimum value of $S$. For what values of $x_{1}, x_{2}, \cdots, x_{n}$ is this minimum value achieved?

Solution:
【Solution】Let $y_{0}=1, y_{k}=1+x_{1}+x_{2}+\cdots+x_{k}(1 \leqslant k \leqslant n)$.
Then
$$
x_{k}=y_{k}-y_{k-1} \text {. }
$$

From the given condition,
$$
\frac{x_{k}}{1+x_{1}+x_{2}+\cdots+x_{k}} \leqslant S,
$$

which is $\frac{y_{k}-y_{k-1}}{y_{k}} \leqslant S$,
thus we have $\quad 1-S \leqslant \frac{y_{k-1}}{y_{k}} \quad(k=1,2, \cdots, n)$
From the given, we know $S \leqslant 1$ and $y_{k}>0(k=0,1,2, \cdots, n)$, so we can multiply the $n$ inequalities expressed by (1), thus obtaining
$$
(1-S)^{n} \leqslant \frac{y_{0}}{y_{n}}
$$

Noting that $y_{0}=1, y_{n}=2$, we have
$$
\begin{array}{l}
(1-S)^{n} \leqslant \frac{1}{2}, \\
S \geqslant 1-2^{-\frac{1}{n}} .
\end{array}
$$

On the other hand, if we take
$$
\frac{y_{k-1}}{y_{k}}=1-S=2^{-\frac{1}{n}} \quad(k=1,2, \cdots, n),
$$

that is, we take
$$
y_{0}=1, y_{1}=2^{\frac{1}{n}}, y_{2}=2^{\frac{2}{n}}, y_{3}=2^{\frac{3}{n}}, \cdots, y_{n}=2 \text {, }
$$

then
$$
S=1-2^{-\frac{1}{n}} \text {. }
$$

In this case,
$$
x_{k}=y_{k}-y_{k-1}=2^{\frac{k}{n}}-2^{\frac{k-1}{n}} \quad(k=1,2, \cdots, n) .
$$

This means that when $x_{k}=2^{\frac{k}{n}}-2^{\frac{k-1}{n}}(k=1,2, \cdots, n)$, $S$ can achieve the minimum value $1-2^{-\frac{1}{n}}$.

Answer: 1-2^{-\frac{1}{n}}


The figure below shows the setting of a "gesture password" on a mobile phone. In a $2 \times 2$ grid, there are 9 grid points. A "gesture password" starts from one of the grid points and connects several grid points with line segments. Each line segment must not have any unused grid points in between, and the two endpoints of the line segment cannot both be already used grid points. If a person's gesture password starts from the center grid point and only uses three grid points, how many connection methods are there?

Solution:
【Analysis】As shown in the right figure, 9 grid points are labeled with letters respectively. Starting from $O$, a line segment can first connect to any of the other 8 grid points, and then connect to a third point. By reasoning in two categories and using the principle of counting by classification, we can reach the conclusion.

【Solution】Solution: As shown in the right figure, 9 grid points are labeled with letters respectively. Starting from $O$, a line segment can first connect to any of the other 8 grid points, and then connect to a third point.
(1) When the second grid point is $A$, according to the problem, the third grid point to connect can be $B, D, E, F, H$, so there are 5 ways to connect; similarly, when the second grid point is $C, E, G$, there are 5 ways to connect each;
(2) When the second grid point is $B$, the third grid point to connect can be any of the other 7 grid points except $O$ and $B$, so there are 7 ways to connect; similarly, when the second grid point is $D, F, H$, there are 7 ways to connect each, thus the total number of connection methods that meet the conditions is $4 \times(5+7)=48$. Therefore, the answer is: 48.

Answer: 48


Determine the number of all numbers between 10000 and 99999 that, like 35453, read the same sequence of digits forwards as backwards.

Solution:
The sought number is equal to the number of all three-digit numbers, since from each three-digit number exactly one such number results if the first two digits of the number are written in reverse order behind the number, and conversely, each such five-digit number is obtained from a three-digit number.

Since there are 900 three-digit numbers, the sought number is 900.

Answer: 900


Calculate the lengths of the arcs of the curves given by the parametric equations.

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=4(2 \cos t-\cos 2 t) \\
y=4(2 \sin t-\sin 2 t)
\end{array}\right. \\
& 0 \leq t \leq \pi
\end{aligned}
$$

Solution:
## Solution

The length of the arc of a curve defined by parametric equations is determined by the formula

$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$

Let's find the derivatives with respect to $t$ for the given curve:

$$
\begin{aligned}
& x=4(2 \cos t-\cos 2 t) ; x^{\prime}(t)=4(-2 \sin t+2 \sin 2 t)=8(\sin 2 t-\sin t) \\
& y=4(2 \sin t-\sin 2 t) ; y^{\prime}(t)=4(2 \cos t-2 \cos 2 t)=8(\cos t-\cos 2 t)
\end{aligned}
$$

Then, using the above formula, we get:

$$
\begin{aligned}
L & =\int_{0}^{\pi} \sqrt{(8(\sin 2 t-\sin t))^{2}+(8(\cos t-\cos 2 t))^{2}} d t= \\
& =\int_{0}^{\pi} 8 \sqrt{\sin ^{2} 2 t-2 \sin 2 t \sin t+\sin ^{2} t+\cos ^{2} t-2 \cos 2 t \cos t+\cos ^{2} 2 t} d t= \\
& =\int_{0}^{\pi} 8 \sqrt{1-2 \sin 2 t \sin t+1-2 \cos 2 t \cos t} d t= \\
& =\int_{0}^{\pi} 8 \sqrt{2} \sqrt{1-\sin 2 t \sin t-\cos 2 t \cos t} d t= \\
& =\int_{0}^{\pi} 8 \sqrt{2} \sqrt{1-\sin t \cos t \sin t-\left(1-\sin ^{2} t\right) \cos t} d t= \\
& =\int_{0}^{\pi} 8 \sqrt{2} \sqrt{1-\sin { }^{2} t \cos t-\cos t+\sin ^{2} t \cos t} d t= \\
& =\int_{0}^{\pi} 8 \sqrt{2} \sqrt{1-\cos t} d t=\int_{0}^{\pi} 8 \sqrt{2} \sqrt{2} \sin ^{2} \frac{t}{2} d t= \\
& =\int_{0}^{\pi} 16 \sin \frac{t}{2} d t=\int_{0}^{\pi} 16 \cdot 2 \sin \frac{t}{2} d\left(\frac{t}{2}\right)=\left.32\left(-\cos \frac{t}{2}\right)\right|_{0} ^{\pi}=-32\left(\cos \frac{\pi}{2}-\cos 0\right)=-32(0-1)=32
\end{aligned}
$$

Source — "http://pluspi.org/wiki/index.php/%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%BD%D0%B8%D0%BA_%D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2%D0%B0_%D0%98%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_18-26"

Categories: Kuznetsov's Problem Book Integrals Problem 18 | Integrals

- Last modified: 13:58, 22 May 2010.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 18-27

## Material from Pluspi

Answer: 32


A scientific expedition team went to the upper reaches of a certain river to investigate an ecological area. After setting out, they advanced at a speed of $17 \mathrm{~km}$ per day, traveling upstream along the riverbank for several days before reaching their destination. They then spent several days investigating the ecological area. After completing their mission, they returned at a speed of 25 $\mathrm{km}$ per day. On the 60th day after departure, the expedition team traveled $24 \mathrm{~km}$ and returned to the starting point. How many days did the scientific expedition team spend investigating the ecological area?

Solution:
Let the expedition team took $x$ days to reach the ecological area, $y$ days to return, and spent $z$ days on the investigation. Then,
$$
x+y+z=60,
$$

and $17 x-25 y=-1$ or $25 y-17 x=1$.
(Here $x, y$ are positive integers)
First, find a special solution $\left(x_{0}, y_{0}\right)$ of equation (1) (here $x_{0}, y_{0}$ can be negative integers). Using the Euclidean algorithm, we have
$$
25=1 \times 17+8,17=2 \times 8+1 \text {. }
$$

Thus, $1=17-2 \times 8=17-2 \times(25-17)$
$$
=3 \times 17-2 \times 25 \text {. }
$$

Comparing with the left side of equation (1), we get
$$
x_{0}=-3, y_{0}=-2 \text {. }
$$

Next, find the solution of equation (1) that meets the problem's requirements.
It is easy to see that all integer solutions of equation (1) can be expressed as
$$
x=-3+25 t, y=-2+17 t \text {. }
$$

Then $x+y=42 t-5, t$ is an integer.
According to the problem, $0<x+y<60$, so only when $t=1$ does it meet the requirements, at this time $x+y=42-5=37$.
Thus, $z=60-(x+y)=23$.

Answer: 23


Let $P$ be any point on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ other than the endpoints of the major axis, $F_{1}$ and $F_{2}$ be its left and right foci respectively, and $O$ be the center. Then $\left|P F_{1}\right| \cdot\left|P F_{2}\right|+|O P|^{2}=$ $\qquad$ .

Solution:
11. Fill in 25. In $\triangle P F_{1} F_{2}$, by the median length formula, we get
$$
\left|P F_{1}\right|^{2}+\left|P F_{2}\right|^{2}=2|O P|^{2}+2\left|O F_{2}\right|^{2} \text {, }
$$

i.e., $\left(\left|P F_{1}\right|+\left|P F_{2}\right|\right)^{2}=2|O P|^{2}+2\left|O F_{2}\right|^{2}+2\left|P F_{1}\right| \cdot\left|P F_{2}\right|$, by the definition of an ellipse, we get
$$
\left|P F_{1}\right| \cdot\left|P F_{2}\right|+|O P|^{2}=2 a^{2}-\left|O F_{2}\right|^{2}=2 a^{2}-c^{2}=a^{2}+b^{2}=16+9=25 .
$$

Answer: 25


Solution:
## Solution.

By applying Gauss's trick, we calculate the sum of 100 consecutive numbers from 456 to 555:

$$
\begin{aligned}
A & =456+457+458+\ldots+554+555 \\
& =(456+555)+(457+554)+\ldots+(505+506)=50 \cdot 1011=50550
\end{aligned}
$$

By reducing to powers with base 5, we calculate the value of expression $B$.

$B=0.04^{3} \cdot \frac{1}{25^{2}} \cdot 125^{4}=\left(5^{-2}\right)^{3} \cdot \frac{1}{\left(5^{2}\right)^{2}} \cdot\left(5^{3}\right)^{4}=5^{-6} \cdot 5^{-4} \cdot 5^{12}=5^{2}=25$

Finally, the desired value is: $\frac{A}{B}=\frac{50550}{25}=2022$.

1 point

Note: The value of expression $A$ can be calculated by the student using the formula for the sum of the first $n$ natural numbers and can earn full points without proving that formula:

$$
\begin{aligned}
A & =(1+2+\ldots+555)-(1+2+\ldots+455) \\
& =\frac{555 \cdot 556}{2}-\frac{455 \cdot 456}{2}=154290-103740=50550 .
\end{aligned}
$$

Answer: 2022


Solution:
Solution. If we perform the variable substitution $x=t+13$, then in all the expressions under the square roots, 1 will be factored out:

$$
\sqrt{\frac{t}{11}+1}+\sqrt{\frac{t}{10}+1}+\sqrt{\frac{t}{9}+1}=\sqrt{\frac{t}{2}+1}+\sqrt{\frac{t}{3}+1}+\sqrt{\frac{t}{4}+1}
$$

Now it is clear that for $t>0$ the right-hand side of the equation is greater, and for $-2 \leqslant t \leqslant 0$ the left-hand side is greater. Therefore, the only solution is $t=0$, which implies $x=13$.

Estimation. For a correct solution, 13 points. If the answer is guessed but the uniqueness of the solution is not proven, 3 points.

Answer: 13


Consider the sequence $a_{1}, a_{2}, a_{3}, \ldots$ defined by $a_{1}=9$ and

$$
a_{n+1}=\frac{(n+5) a_{n}+22}{n+3}
$$

for $n \geqslant 1$.

Find all natural numbers $n$ for which $a_{n}$ is a perfect square of an integer.

Solution:
Define $b_{n}=a_{n}+11$. Then

$$
22=(n+3) a_{n+1}-(n+5) a_{n}=(n+3) b_{n+1}-11 n-33-(n+5) b_{n}+11 n+55
$$

giving $(n+3) b_{n+1}=(n+5) b_{n}$. Then

$b_{n+1}=\frac{n+5}{n+3} b_{n}=\frac{(n+5)(n+4)}{(n+3)(n+2)} b_{n-1}=\frac{(n+5)(n+4)}{(n+2)(n+1)} b_{n-2}=\cdots=\frac{(n+5)(n+4)}{5 \cdot 4} b_{1}=(n+5)(n+4)$.

Therefore $b_{n}=(n+4)(n+3)=n^{2}+7 n+12$ and $a_{n}=n^{2}+7 n+1$.

Since $(n+1)^{2}=n^{2}+2 n+1<a_{n}<n^{2}+8 n+16=(n+4)^{2}$, if $a_{n}$ is a perfect square, then $a_{n}=(n+2)^{2}$ or $a_{n}=(n+3)^{2}$.

If $a_{n}=(n+2)^{2}$, then $n^{2}+4 n+4=n^{2}+7 n+1$ giving $n=1$. If $a_{n}=(n+3)^{2}$, then $n^{2}+6 n+9=$ $n^{2}+7 n+1$ giving $n=8$.

Comment. We provide some other methods to find $a_{n}$.

Method 1: Define $b_{n}=\frac{a_{n}+11}{n+3}$. Then $b_{1}=5$ and $a_{n}=(n+3) b_{n}-11$. So

$$
a_{n+1}=(n+4) b_{n+1}-11=\frac{(n+5) a_{n}+22}{n+3}=a_{n}+\frac{2\left(a_{n}+11\right)}{n+3}=(n+3) b_{n}-11+2 b_{n}
$$

giving $(n+4) b_{n+1}=(n+5) b_{n}$. Then

$$
b_{n+1}=\frac{n+5}{n+4} b_{n}=\frac{n+5}{n+3} b_{n-1}=\cdots=\frac{n+5}{5} b_{1}=n+5
$$

Then $b_{n}=n+4$, so $a_{n}=(n+3)(n+4)-11=n^{2}+7 n+1$.

Method 2: We have

$$
(n+3) a_{n+1}-(n+5) a_{n}=22
$$

and therefore

$$
\frac{a_{n+1}}{(n+5)(n+4)}-\frac{a_{n}}{(n+4)(n+3)}=\frac{22}{(n+3)(n+4)(n+5)}=11\left[\frac{1}{n+3}-\frac{2}{n+4}+\frac{1}{n+5}\right]
$$

Now define $b_{n}=\frac{a_{n}}{(n+4)(n+3)}$ to get

$$
b_{n+1}=b_{n}+11\left[\frac{1}{n+3}-\frac{2}{n+4}+\frac{1}{n+5}\right]
$$

which telescopically gives

$$
\begin{aligned}
b_{n+1} & =b_{1}+11\left[\left(\frac{1}{4}-\frac{2}{5}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{2}{6}+\frac{1}{7}\right)+\cdots+\left(\frac{1}{n+3}-\frac{2}{n+4}+\frac{1}{n+5}\right)\right] \\
& =b_{1}+11\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{n+4}+\frac{1}{n+5}\right)=\frac{9}{20}+\frac{11}{20}-\frac{11}{(n+4)(n+5)}
\end{aligned}
$$

We get $b_{n+1}=(n+4)(n+5)-11$ from which it follows that $b_{n}=(n+3)(n+4)-11=n^{2}+7 n+1$.

Answer: n=1 \text{ or } n=8


Last year, 9900 umbrellas were reported to the lost and found department. Some people lost exactly one umbrella, but there were also those who lost more than one umbrella. Specifically, $4 \%$ of them lost exactly two umbrellas, $2.5 \%$ lost three umbrellas, $0.5 \%$ lost eight umbrellas, and the rest lost one umbrella. How many people lost at least one umbrella?

Solution:
Solution. If we denote the number of people who lost umbrellas by $x$, then from the condition of the problem we have that $0.04 x$ lost two umbrellas each, $0.025 x$ lost three umbrellas each, $0.005 x$ lost eight umbrellas each, and

$$
(1-(0.04+0.025+0.005) x=0.93 x
$$

lost one umbrella each. Therefore, the number of lost umbrellas is

$$
2 \cdot 0.04 x+3 \cdot 0.025 x+8 \cdot 0.005 x+1 \cdot 0.93 x=1.125 x
$$

so $1.125 x=9900$, or $x=\frac{9900}{1.125}=8800$. Thus, last year 8800 people lost at least one umbrella.

Answer: 8800


Among the fractions with positive denominators less than 1001, which one has the smallest difference from $\frac{123}{1001}$?

Solution:
Solution. Let $p / q$ be the fraction with a positive integer denominator less than 1001, whose difference from 123/1001 is the smallest. Since 123 and 1001 are relatively prime, the latter fraction cannot be simplified, and thus

$$
\left|\frac{123}{1001}-\frac{p}{q}\right|=\left|\frac{123 q-1001 p}{1001 q}\right|=\delta
$$

cannot be zero. The numerator of the latter fraction is an integer, and thus, being non-zero, it is at least 1.

We distinguish two cases:

1. Case. $|123 q-1001 p| \geq 2$. In this case, due to the condition $1 \leq q \leq 1000$, the difference $\delta$ is at least $2 /(1001 q) \geq 1 /(1001 \cdot 500)$.
2. Case. $|123 q-1001 p|=1$. This can happen in two ways:
a) $123 q-1001 p=1$. Here $q$ and $p$ are positive integers and $q \leq 1000$, so $1001 p=123 q-1 \leq 122999$, and thus $p \leq 122$. We also see that $p$ is not divisible by three, because otherwise the left side would be divisible by three, but the right side would be 1. Thus, there are 82 possibilities for $p$, and by trying each of them, we find that only $p=94$ gives an integer for $q$: $q=765$. In this case, the fraction $p / q$ can only be $94 / 765$, and its difference from 123/1001 is $\delta=1 /(1001 \cdot 765)$.
b) $123 q-1001 p=-1$. This time, due to $q \leq 1000$, $1001 p=123 q+1 \leq 123001$, and thus $p$ can again be at most 122, and it is not divisible by three. This again gives 82 possibilities, and by trying each of them, we find that only $p=29$ gives an integer for $q$: $q=236$. The fraction $p / q=29 / 236$ has a difference from 123/1001 of $1 /(1001 \cdot 236)$.

Having examined all possibilities, the smallest difference occurs in case 2a), i.e., when $p / q=94 / 765$, so among the fractions with a positive integer denominator less than 1001, this fraction has the smallest difference from 123/1001.

Remark. The Diophantine equation $123 q-1001 p=1$ or -1 can also be solved in general, which would clearly lead to the solution of the problem. However, this also requires finding a specific solution, which can be done as described above, or by expanding the fraction 1001/123 into a continued fraction:

$$
\frac{1001}{123}=8+\frac{1}{7+\frac{1}{4+\frac{1}{4}}}
$$

The solution can be obtained by omitting the last fraction: choosing the denominator of $8+\frac{1}{7+\frac{1}{4}}=236 / 29$ as $p_{0}$ and its numerator as $q_{0}$, the general solution of the two equations is given by $p=123 r+p_{0}$ and $q=1001 r+q_{0}$, or $p=123 r-p_{0}$ and $q=1001 r-q_{0}$.

Answer: \frac{94}{765}


Find all positive integers $n$, such that there exist $a, b \in \mathbf{N}^{*}$, satisfying: $[a, b]=n!$, $(a, b)=1998 ;$

Under the condition that (1) holds, to ensure that the number of pairs of positive integers $(a, b)$ with $a \leqslant b$ does not exceed 1998, what condition should $n$ satisfy?

Solution:
4. (1) From $(a, b)=2 \times 3^{3} \times 37$, we know $37 \mid[a, b]$, which means $37 \mid n$, so $n \geqslant 37$. When $n \geqslant 37$, $1998 \mid n!$, at this time, taking $a=1998, b=n!$ will suffice. Therefore, the positive integers $n$ that satisfy the condition are all positive integers not less than 37.
(2) Let $a=1998 x, b=1998 y$, then $(x, y)=1$. When $n \geqslant 37$, we have $x y=2^{33} \times 3^{14} \times 5^{8} \times 7^{5} \times 11^{3} \times 13^{2} \times 17^{2} \times 19 \times 23 \times 29 \times 31 \times\left(\frac{n!}{37!}\right)$.

Thus, when $37 \leqslant n \leqslant 40$, the number of pairs $(x, y)$ is $\frac{1}{2} \times 2^{11}=1024$ pairs, and when $n \geqslant 41$, there are at least $\frac{1}{2} \times 2^{12}=2048$ pairs, hence $37 \leqslant n \leqslant 40$.

Answer: 37 \leqslant n \leqslant 40


Plot the graph of the function $y=\frac{1}{2} x^{2}+2 x-1$.

Solution:
Solution. Rewrite the given function as $y=\frac{1}{2}(x+2)^{2}-3$.
![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-056.jpg?height=990&width=406&top_left_y=522&top_left_x=90)

Fig. 5.12

1) $\Gamma_{1}: y=x^{2}$ - parabola (Fig. 5.12, a) with vertex at point $O(0,0)$.
2) $\Gamma_{2}: \quad y=(x+2)^{2}-\quad$ parabola (Fig. $5.12, b$ ) with vertex at point $A(-2,0)$.
3) $\Gamma_{3}: y=\frac{1}{2}(x+2)^{2}$. Obtained from $\Gamma_{2}$ by compressing it by a factor of 2 towards $O x$ (Fig. 5.12, c).
4) $\Gamma_{4}: y=\frac{1}{2}(x+2)^{2}-3 . \Gamma_{3}$ is lowered by 3 units (Fig. $5.12, d$ ).

Answer: \frac{1}{2}(x+2)^{2}-3


In the plane, let two non-intersecting circles \(k_{1}\) and \(k_{2}\) and a point \(A\) outside both circles be given.

We are looking for an equilateral triangle \(\triangle A B C\) such that \(B\) lies on \(k_{1}\) and \(C\) lies on \(k_{2}\).

a) Justify and describe a construction of such triangles.

b) Determine the largest number that can occur as the number of such triangles \(\triangle A B C\) in cases where there are not infinitely many such triangles.

Solution:
Assume there exists a \(B \in k_{1}\) and a \(C \in k_{2}\) with \(A B=B C=C A\), then I obtain the point \(C\), if \(B\) is already fixed, by rotating the segment \(A B\) around \(A\) by \(60^{\circ}\).

If I choose an arbitrary point \(B\) on \(k_{1}\) and rotate \(A B\) around \(A\) by \(60^{\circ}\), it is likely that the endpoint of the segment \(A B\) after the rotation lies on the circle \(k_{2}\). If I rotate the entire circle \(k_{1}\) around \(A\) by \(60^{\circ}\), and if there is such a triangle, then the rotated circle \(k_{1}\) intersects the circle \(k_{2}\) at a point \(C^{\prime}\) that satisfies the conditions of the problem; for the original point \(B\) of the rotation around \(A\) by \(60^{\circ}\) from \(C^{\prime}\) lies on \(k_{1}\), and it is certainly true that

\[
\angle C^{\prime} A B^{\prime}=60^{\circ} \quad \text{and} \quad C^{\prime} A=B^{\prime} A
\]

Thus, \(\triangle A B^{\prime} C^{\prime}\) is the desired triangle.

Construction: The circle \(k_{1}\) is rotated around \(A\) once in the mathematically positive direction and once in the mathematically negative direction by \(60^{\circ}\). This results in the circles \(k_{1}^{\prime}\) and \(k_{1}^{\prime \prime}\).

If points \(C\) exist with \(C \in k_{2} \cap k_{1}^{\prime}\) or \(C \in k_{2} \cap k_{1}^{\prime \prime}\), then these points are solutions to the problem. The corresponding points \(B\) are obtained by constructing the equilateral triangle over \(A C\) and the condition \(B \in k_{1}\).

Determination: The maximum number of solutions can be four equilateral triangles, and this occurs precisely when \(k_{1}^{\prime} \cap k_{2}\) and \(k_{1}^{\prime \prime} \cap k_{2}\) each contain exactly two points.

\(k_{1}^{\prime}\) and \(k_{1}^{\prime \prime}\) can each intersect \(k_{2}\) in at most two points, as it was assumed that \(k_{2}\) is not identical to \(k_{1}^{\prime}\) or \(k_{1}^{\prime \prime}\).

The case where four equilateral triangles occur is possible, as shown below (figure).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1948.jpg?height=848&width=580&top_left_y=1164&top_left_x=744)

I choose the point \(A\) on the line segment connecting \(M_{1}\) and \(M_{2}\) and will prove that \(a_{1}\) and \(a_{2}\) can be chosen such that \(M_{2} M_{1}^{\prime}0\) is found, for which \(\left(M_{1} M_{2}\right)^{2}60^{\circ}\) and \(\angle B M_{1}^{\prime} M_{1}>60^{\circ}\) also hold (by the theorem that the larger side in a triangle has the larger angle opposite it).

Thus, the sum of the angles in this triangle would be greater than \(180^{\circ}\), leading to a contradiction.

Therefore, four is the maximum number of solutions unless there are infinitely many solutions.

Adapted from \([3]\)

\subsection*{9.15 XIII. Olympiad 1973}

\subsection*{9.15.1 First Round 1973, Class 12}

Answer: 4


In the list of positive integers $1,3,9,4,11$ the positive differences between each pair of adjacent integers in the list are $3-1=2,9-3=6,9-4=5$, and $11-4=7$. In this example, the smallest positive difference between any two adjacent integers in the list is 2.
(a) Arrange the integers $1,2,3,4,5$ so that the smallest positive difference between any two adjacent integers is 2.
(b) Suppose that the twenty integers $1,2,3, \ldots, 18,19,20$ are arranged so that the smallest positive difference between any two adjacent integers is $N$.
(i) Explain why $N$ cannot be 11 or larger.
(ii) Find an arrangement with $N=10$. (Parts (i) and (ii) together prove that the maximum possible value of $N$ is 10.)
(c) Suppose that the twenty-seven integers $1,2,3, \ldots, 25,26,27$ are arranged so that the smallest positive difference between any two adjacent integers is $N$. What is the maximum possible value of $N$? Prove that your answer is correct.

Solution:
2. (a) One way of arranging these integers is $1,3,5,2,4$.
The positive differences between the pairs of adjacent integers are $2,2,3,2$.
There are many other ways of arranging these integers so that the desired property is true.
(We note that this property is equivalent to arranging the five integers so that no two consecutive integers are adjacent.)
(b) (i) In any arrangement, the integer 10 must be placed next to at least one other integer. From the list $1,2,3, \ldots, 20$, the integers "furthest away" from 10 are 1 and 20 .
The positive differences between 10 and these integers are 9 and 10 .
In other words, the positive difference between 10 and every other integer in the list is less than or equal to 10 .
Therefore, in any arrangement, there must be a positive difference that is at most 10 , and so $N$ (the smallest of these differences) is less than or equal to 10 , which means that $N$ cannot be 11 or larger.
(ii) Consider the following arrangement:
$$
10,20,9,19,8,18,7,17,6,16,5,15,4,14,3,13,2,12,1,11
$$

The positive differences between the pairs of adjacent integers are
$$
10,11,10,11,10,11,10,11,10,11,10,11,10,11,10,11,10,11,10
$$

The smallest of these positive differences is 10 .
(c) Consider the integer 14, which is the middle integer in the list $1,2,3, \ldots, 25,26,27$.
The maximum possible positive difference between 14 and another number from this list is 13 .
Therefore, in any arrangement, there must be a positive difference that is no larger than 13 . This tells us that $N$, the smallest of the positive differences, is less than or equal to 13 . To show that the maximum possible value of $N$ is 13 , we need to show that there actually exists an arrangement with $N=13$, since it is possible that, for some other reason, we could not make a list with $N=13$.
Here is such an arrangement:
$$
14,27,13,26,12,25,11,24,10,23,9,22,8,21,7,20,6,19,5,18,4,17,3,16,2,15,1
$$

The positive differences between the pairs of adjacent integers are
$$
13,14,13,14,13,14,13,14,13,14,13,14,13,14,13,14,13,14,13,14,13,14,13,14,13,14
$$

Therefore, the maximum possible value of $N$ is 13 .

Answer: 13


Find all such triples of prime numbers $p, q, r$, such that the fourth power of each of them, decreased by 1, is divisible by the product of the other two.

Solution:
Clearly, any two numbers of the triplet are distinct (if $p=q$, then $p^{4}-1$ is not divisible by $q$). Let $p$ be the smallest number of the triplet. The number

$p^{4}-1=(p-1)(p+1)\left(p^{2}+1\right)$ is divisible by $q r$. But $p-1$ is less than any of the prime numbers $q$ and $r$, and thus is coprime with them. The number $p^{2}+1$ cannot be divisible by both numbers $q$ and $r$, since $p^{2}+1<p$, this is only possible if $q=p+1$.

Therefore, $p=2, q=3$. $r$ is a prime divisor of the number $p^{4}-1=15$, different from $q=3$, so $r=$ 5.

It remains to check that the triplet $\{2,3,5\}$ satisfies the conditions of the problem.

## Answer

$\{2,3,5\}$.

Answer: {2,3,5}


There are 9 cards with numbers $1,2,3,4,5,6,7,8$ and 9. What is the maximum number of these cards that can be laid out in some order in a row so that on any two adjacent cards, one of the numbers is divisible by the other?

Solution:
2. Answer: 8.

Note that it is impossible to arrange all 9 cards in a row as required. This follows from the fact that each of the cards with numbers 5 and 7 can only have one neighbor, the card with the number 1. Therefore, both cards 5 and 7 must be at the ends, and the card with the number 1 must be adjacent to each of them, which is impossible. It is possible to select 8 cards and arrange them in a row according to the requirements of the problem, for example: $9,3,6,2,4,8,1,5$.

Answer: 8


Find all positive integer pairs $(a, b)$ that satisfy $(a, b)=10,[a, b]=100$.

Solution:
13. $(a / 10, b / 10)=1,[a / 10, b / 10]=10 . a / 10, b / 10$ can only take the values $1,2,5,10$, and satisfy the above conditions. Therefore, we get the following table:
\begin{tabular}{|c|cccc|}
\hline$a / 10$ & 1 & 10 & 2 & 5 \\
\hline$b / 10$ & 10 & 1 & 5 & 2 \\
\hline
\end{tabular}

Answer: (a, b) = (10, 100), (100, 10), (20, 50), (50, 20)


Determine an expression for $s_{n}$ in terms of $s_{0}, r$, and $n$.

Solution:
Let's look at the first terms: $s_{0}, s_{0}+r,\left(s_{0}+r\right)+r=s_{0}+2 r,\left(s_{0}+2 r\right)+r=$ $s_{0}+3 r$ and so on.

Let's prove the conjectured formula (i.e., $s_{n}=s_{0}+n r$) by induction.

$\triangleright$ The formula is verified for $n=0$.

$\triangleright$ Let $n$ be an integer such that $s_{n}=s_{0}+n r$. Then,

$$
s_{n+1}=s_{n}+r=\left(s_{0}+n r\right)+r=s_{0}+(n+1) r
$$

hence the result by induction.

Answer: s_{n}=s_{0}+nr


Given the line $l: x=my+q, m \neq 0$ intersects the ellipse $\Gamma: 3 x^{2}+4 y^{2}=12$ at two distinct points $\mathrm{A}$ and $\mathrm{B}$. Let $\mathrm{A}$'s reflection about the major axis of the ellipse be $A_{1}$, and $F$ be the right focus of the ellipse. Find the necessary and sufficient condition for $A_{1}, F, B$ to be collinear.

Solution: 9. Analysis: Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then $A_{1}\left(x_{1},-y_{1}\right)$, then $A_{1} 、 F 、 B$ being collinear is equivalent to $\operatorname{mom}_{A_{1} F}^{\operatorname{mom}}=\lambda A_{1} B$, i.e., $\frac{y_{1}}{y_{1}+y_{2}}=\frac{1-x_{1}}{x_{2}-x_{1}}$, simplifying to: $2 m y_{1} y_{2}=(1-q)\left(y_{1}+y_{2}\right)(*)$, combining the equations of line $l$ and ellipse $\Gamma$ yields: $\left(3 m^{2}+4\right) y^{2}+6 m q y+3 q^{2}-12=0$, by Vieta's formulas we have: $y_{1}+y_{2}=-\frac{6 m q}{3 m^{2}+4}, y_{1} y_{2}=\frac{3 q^{2}-12}{3 m^{2}+4}$, substituting into $(*)$ gives: $q=4$. Therefore, the necessary and sufficient condition is $q=4\left(m^{2}>4\right)$.

Answer: 4


Determine all pairs $\{a, b\}$ of distinct real numbers such that the equations

$$
x^{2} + a x + b = 0 \quad \text{and} \quad x^{2} + b x + a = 0
$$

have at least one common solution in the set of real numbers.

Solution:
## First Solution.

Let the common solution be $\mathrm{s} x_{0}$. By equating the equations, we get:

$$
\begin{aligned}
x_{0}^{2}+a x_{0}+b & =x_{0}^{2}+b x_{0}+a \\
x_{0} a-x_{0} b+b-a & =0 \\
x_{0}(a-b)-(a-b) & =0 \\
\left(x_{0}-1\right)(a-b) & =0
\end{aligned}
$$

From the condition $a \neq b$, we get $x_{0}=1$.

Substituting into the equations, we obtain the necessary condition for $a$ and $b$:

$$
1+a+b=0
$$

Thus, the number 1 is a solution to the quadratic equations if and only if the above relation holds. 1 point

Among such pairs, we need to exclude those for which $a=b=-1-a$, i.e., $a=-\frac{1}{2}$.

Therefore, the solutions are all pairs $\{a, b\}$ such that $a+b=-1$, except when $a=b=-\frac{1}{2}$.

Answer: -1


The coefficient of $x^{4}$ in the expansion of $(\sqrt{x}+1)^{4}(x-1)^{5}$ is $\qquad$

Solution:
2, 1.45.
Notice that, in the expansion of $(\sqrt{x}+1)^{4}$, the coefficients of $x^{2}$ and $x$ are $\mathrm{C}_{4}^{4}$ and $\mathrm{C}_{4}^{2}$, respectively, and the constant term is $\mathrm{C}_{4}^{0}$.

Therefore, in the expansion of $(\sqrt{x}+1)^{4}(x-1)^{5}$, the coefficient of $x^{4}$ is
$$
\mathrm{C}_{4}^{4} \mathrm{C}_{5}^{2}(-1)^{3}+\mathrm{C}_{4}^{2} \mathrm{C}_{5}^{3}(-1)^{2}+\mathrm{C}_{4}^{0} \mathrm{C}_{5}^{4}(-1)^{1}=45
$$

Answer: 45


Find all positive integers $n$ such that there exists a sequence $\left\{x_{n}\right\}$ where $1,2, \cdots, n$ each appear once and for $k=1,2, \cdots, n$ we have $k \mid x_{1}+x_{2}+\cdots+x_{n}$.

Solution:
Analysis: This problem should make full use of special elements.
Solution: When $k=n$, $n \left\lvert\, \frac{n(n+1)}{2}\right.$, so $n+1$ is even, which means $n$ is odd. Let $n=2 p+1, p \in \mathbf{N}^{+}$. When $k=n-1=2 p$, $2 p \left\lvert\,\left(\frac{n(n+1)}{2}-x_{n}\right)\right.$, i.e., $2 p \mid\left((2 p+1)(p+1)-x_{n}\right)$
Thus, $2 p \mid(p+1)-x_{n}$
Also, $2 p \geqslant p+1>p+1-x_{n} \geqslant p+1-(2 p+1)=-p>-2 p$.
Therefore, $p+1-x_{n}=0$.
Thus, $x_{n}=p+1$.
Similarly, when $k=2 p-1$, $(2 p-1) \mid\left(2 p^{2}+2 p-x_{n-1}\right)$,
Thus, $(2 p-1) \mid(p+1)-x_{n-1}$
Also, $p+1>p+1-x_{n-1}>p+1-(2 p+1)=-p$.
Since $x_{n-1} \neq x_{n}=p+1$, we have $2 p-1<p+1 \Rightarrow p<2$. Also, $p \in \mathbf{N}^{+}$. Therefore, $p=1$. At this point, $x_{1}=1, x_{2}=3, x_{3}=2$ satisfy the conditions. Clearly, when $n=1$, $x_{1}=1$ satisfies the conditions.
In summary: $n=1, n=3$ are the positive integers that satisfy the conditions.

Answer: n=1,n=3


Given a rectangular piece of paper $ABCD$ with a perimeter of 10 (as shown in the figure), now cut it twice along directions parallel to the length and width, respectively, to form 9 rectangles of unequal sizes. The total perimeter of these 9 rectangles is $\qquad$.

Solution:
【Analysis and Solution】
Geometry, cleverly finding the perimeter.
Let the length of rectangle $A B C D$ be $a$ and the width be $b$;
then $a+b=10 \div 2=5$;
The total perimeter of these 9 small rectangles = the perimeter of the original rectangle + the segments from the cuts $\times 2$;
The total perimeter of these 9 small rectangles is $10+5 \times 2 \times 2=30$.

Answer: 30


The sum of the ages of three people, A, B, and C, represented by $x, y, z$ is 120, and $x, y, z \in (20,60)$. Then the number of ordered triples $(x, y, z)$ is $\qquad$

Solution:
8. 1141 .

Notice a basic conclusion:
The number of positive integer solutions $(a, b, c)$ to the indeterminate equation $a+b+c=n$ is $\mathrm{C}_{n-1}^{2}$.
Thus, the number of solutions to the indeterminate equation
$$
(x-20)+(y-20)+(z-20)=60
$$

satisfying $x, y, z>20$ is $\mathrm{C}_{59}^{2}$.
Among these $\mathrm{C}_{59}^{2}$ solutions, the number of solutions where $x \geqslant 60$ and do not meet the requirements is
$$
(x-59)+(y-20)+(z-20)=21
$$

satisfying $x \geqslant 60, y, z>20$, the number of such solutions is $\mathrm{C}_{20}^{2}$.
Similarly, the number of solutions where $y \geqslant 60$ is $\mathrm{C}_{20}^{2}$, and the number of solutions where $z \geqslant 60$ is $\mathrm{C}_{20}^{2}$.

Since no two individuals can simultaneously reach 60 years old, the number of ordered triples $(x, y, z)$ is $\mathrm{C}_{59}^{2}-3 \mathrm{C}_{20}^{2}=1141$.

Answer: 1141


Person A and Person B start riding bicycles from place $A$ to place $B$ at the same time. Person A's speed is 1.2 times that of Person B. After riding 5 kilometers, Person B's bicycle breaks down, and the time spent fixing it is equivalent to riding $\frac{1}{6}$ of the total distance. After fixing the bicycle, Person B's speed increases by 60%, and as a result, both Person A and Person B arrive at place $B$ at the same time. The distance between $A$ and $B$ is $\qquad$ kilometers.

Solution:
【Analysis】According to the problem, the speed ratio of A and B is $1.2: 1=6: 5$, so the time ratio is 5: 6. Let's assume A takes $5 t$, then B's original time is $6 t$. The delay time due to the fault for B is $\frac{1}{6} \times 6 t=t$, and the total time taken for the entire journey is $5 t$. Therefore, after the fault is resolved, B's increased speed saves $2 t$ of time. The speed ratio after the increase is $1.6: 1=8: 5$, so the time ratio is $5: 8$, saving three parts of time, so each part is $\frac{2}{3} t$. Therefore, the original planned time for this section is $\frac{2}{3} t \times 8=\frac{16}{3} t$, so the original planned time for the initial 5 kilometers is $6 t-\frac{16}{3} t=\frac{2}{3} t$. Therefore, the distance between A and B is $5 \times\left(6 t \div \frac{2}{3} t\right)$, and then we can calculate it.

【Solution】Solution: The speed ratio of A and B is $1.2: 1=6: 5$, so the time ratio is 5: 6; assume A takes $5 t$, then B's original time is $6 t$;

The delay time due to the fault for B is $\frac{1}{6} \times 6 t=t$, and the total time taken for the entire journey is $5 t$, so after the fault is resolved, B's increased speed saves $2 t$ of time.

The speed ratio after the increase is $1.6: 1=8: 5$, so the time ratio is $5: 8$, saving three parts of time, so each part is $\frac{2}{3} t$,
Therefore, the original planned time for this section is $\frac{2}{3} t \times 8=\frac{16}{3} t$, so the original planned time for the initial 5 kilometers is $6 t-\frac{16}{3} t=\frac{2}{3} t$, so the distance between A and B is:
$5 \times\left(6 t \div \frac{2}{3} t\right)$,
$=5 \times 9$,
$=45$ (kilometers);
Therefore, the answer is: 45.

Answer: 45


Given real numbers $x, y$ satisfy
$$
x+y=3, \frac{1}{x+y^{2}}+\frac{1}{x^{2}+y}=\frac{1}{2} \text {. }
$$

Find the value of $x^{5}+y^{5}$.

Solution:
【Analysis】Since the target algebraic expression is a symmetric algebraic expression, it can be expressed using $x+y$ and $xy$. For this purpose, we first find the value of $xy$.
Solution From the given equation, we have
$$
\begin{aligned}
2 x^{2} & +2 y+2 x+2 y^{2}=x^{3}+x y+x^{2} y^{2}+y^{3} \\
\Rightarrow & 2(x+y)^{2}-4 x y+2(x+y) \\
& =(x+y)\left((x+y)^{2}-3 x y\right)+x y+x^{2} y^{2} .
\end{aligned}
$$

Substituting $x+y=3$ into the above equation, we get
$$
\begin{array}{l}
(x y)^{2}-4 x y+3=0 \\
\Rightarrow x y=1 \text { or } x y=3 \text { (discard). } \\
\text { Also, } x^{5}+y^{5}=\left(x^{2}+y^{2}\right)\left(x^{3}+y^{3}\right)-x^{2} y^{2}(x+y), \\
x^{3}+y^{3}=(x+y)\left((x+y)^{2}-3 x y\right)=18, \\
x^{2}+y^{2}=(x+y)^{2}-2 x y=7,
\end{array}
$$

Therefore, $x^{5}+y^{5}=123$.

Answer: 123


Triangle $A B C$ is an isosceles right triangle: the angle at $C$ is right and $|A C|=|B C|=12$. Point $M$ is the midpoint of side $A B$. Point $D$ lies on side $A C$. Finally, point $E$ is the intersection of line segments $C M$ and $B D$, see the figure below.

If $|C D|=3$, what is the area of quadrilateral $A M E D$?

![](https://cdn.mathpix.com/cropped/2024_04_17_01cb006263f00c7722c1g-3.jpg?height=371&width=371&top_left_y=885&top_left_x=1522)

Solution:
B3. $\frac{162}{5}$ The area of $\triangle A M C$ is half the area of $\triangle A B C$. The area of $\triangle A B C$ is $\frac{1}{2} \cdot|A C| \cdot|B C|=\frac{1}{2} \cdot 12 \cdot 12=72$. Therefore, the area of $\triangle A M C$ is 36. To determine the area of quadrilateral $A M E D$, we now only need to determine the area of $\triangle C D E$.

Let $F$ be the point obtained by reflecting $D$ over $C E$. Due to symmetry, we see that $\triangle C E F$ has the same area as $\triangle C D E$. Moreover, we also know that $|C F|=3$ and $|B C|=12$. Triangles $B C E$ and $C E F$

![](https://cdn.mathpix.com/cropped/2024_04_17_fc21619166b069277c26g-5.jpg?height=382&width=371&top_left_y=203&top_left_x=1528)
have the same height, but the base of $\triangle B C E$ is four times as large. Therefore, the area of $\triangle B C E$ is four times the area of $\triangle C E F$ (or $\triangle C D E$).

If we add triangle $C D E$ to this, we see that $\triangle B C D$ has five times the area of $\triangle C D E$. The area of $\triangle B C D$ is $\frac{1}{2} \cdot|C D| \cdot|B C|=\frac{1}{2} \cdot 3 \cdot 12=18$. Altogether, we find that $\triangle C D E$ has an area of $\frac{18}{5}$, and the area of quadrilateral $A M E D$ is therefore $36-\frac{18}{5}=\frac{162}{5}$.

Answer: \frac{162}{5}


For the 2002 tax declaration, those whose annual gross income exceeded 1,050,000 forints had to pay, in addition to 40% of the excess, 267,000 forints as tax. What was the monthly gross income if the income tax was 30% of the income?

Solution:
Let $x$ be the annual gross income. Then the tax to be paid is:

$$
(x-1050000) \cdot 0.4+267000
$$

If this is equal to 30% of the gross income, then

$$
(x-1050000) \cdot 0.4+267000=0.3 \cdot x
$$

From this, we get that $x=1530000$ is the annual income. The monthly income is therefore $127500 \mathrm{Ft}$.

Answer: 127500\mathrm{Ft}


Count how many different $p$-arithmetic fractional-linear functions there are.

The function $f(f(x))$ is denoted by $f_{2}(x)$. Similarly, we introduce fractional-linear functions $f_{\mathrm{B}}(x), f_{1}(x), \ldots, f_{2}(x)$, assuming

$$
\begin{aligned}
f_{3}(x) & =f\left(f_{2}(x)\right), \\
f_{4}(x) & =f\left(f_{\mathrm{B}}(x)\right), \\
\cdots & \cdot \\
f_{n}(x) & =f\left(f_{n-1}(x)\right) .
\end{aligned}
$$

For example,

$$
\begin{aligned}
& f(x)=\frac{x+1}{x} \\
& f_{2}(x)=\frac{\frac{x+1}{x}+1}{\frac{x+1}{x}}=\frac{2 x+1}{x+1} \\
& f_{3}(x)=\frac{\frac{2 x+1}{x+1}+1}{\frac{2 x+1}{x+1}}=\frac{3 x+1}{2 x+1}
\end{aligned}
$$

The action of all functions $f_{n}(x)$ can be traced on a diagram constructed for the function $f(x)$. The action of the function $f(x)$ is such that each point of its diagram moves to the next one, i.e., makes one step in the direction of the arrows. When applying the function $f_{2}(x)$, each point on the diagram of $f(x)$ already makes two steps, landing in the second point. Generally, when applying $f_{n}(x)$, each point on the diagram of $f(x)$ makes $n$ steps in the direction of the arrows, jumping over $n-1$ points to the $n$-th. This allows constructing the diagrams of the functions $f_{2}(x), f_{3}(x), \ldots$ from the diagram of the function $f(x)$ without calculations.

Solution:
146. Let's fix some triple of different points, for example $0,1, \infty$. Each fractional-linear function maps this triple of points to some other triple of different points $y_{1}, y_{2}, y_{3}$. Conversely, for each triple of different points $y_{1}, y_{2}, y_{3}$, one can find and uniquely determine a fractional-linear function that maps this triple to $0,1, \infty$. Therefore, the number of different $p$-arithmetic fractional-linear functions is the same as the number of different triples of points in our $p$-arithmetic, extended by the symbol $\infty$ (triples differing in order should be considered different). For $y_{1}$, we can choose any of the $p+1$ points $0,1, \ldots, p-1, \infty$, for $y_{2}$ - any of the $p$ remaining points, and finally, for $y_{3}$ we can take any of the $p-1$ points not yet chosen. Thus, there will be $(p+1) p(p-1)$ combinations, i.e., different triples of points. The same number will be for $p$-arithmetic

![](https://cdn.mathpix.com/cropped/2024_05_21_a55bead9955e32b1e916g-254.jpg?height=265&width=362&top_left_y=1288&top_left_x=682)

Pic. 152. linear fractional functions.

Answer: (p+1)p(p-1)


The coefficient of $a b c^{2}$ in the expansion of $(a+2 b-3 c)^{4}$ is ( ).
(A) 208
(B) 216
(C) 217
(D) 218

Solution:
3. B.

According to the problem, the coefficient of $a b c^{2}$ is $C_{4}^{2}(-3)^{2} C_{2}^{1} \times 2 \times 1=216$

Answer: B


Let planar vectors $\boldsymbol{a}, \boldsymbol{b}$ satisfy $|\boldsymbol{a}+\boldsymbol{b}|=3$. Then the maximum value of $\boldsymbol{a} \cdot \boldsymbol{b}$ is $\qquad$.

Solution:
2. $\frac{9}{4}$.

Notice that,
$$
\boldsymbol{a} \cdot \boldsymbol{b}=\frac{1}{4}\left((\boldsymbol{a}+\boldsymbol{b})^{2}-(\boldsymbol{a}-\boldsymbol{b})^{2}\right) \leqslant \frac{9}{4},
$$

the equality holds if and only if $\boldsymbol{a}=\boldsymbol{b}$. Therefore, the maximum value of $\boldsymbol{a} \cdot \boldsymbol{b}$ is $\frac{9}{4}$.

Answer: \frac{9}{4}


Let's take a positive integer $n$, sum its digits, and then add the digits of this sum again to get an integer $S$. What is the smallest $n$ that allows us to obtain $S \geq 10?$

Solution: 11. The answer is 199. Indeed, if $n$ had 1 or 2 digits, $n \leq 99$, and then the sum of its digits would be $\leq 18$ and the sum of the digits of the number equal to the sum of its digits would be $\leq 9$, so $n$ has at least three digits. Note that for $n \leq 198$, the sum of the digits of the number is still $\leq 18$. However, if $n=199$, the sum of its digits is 19 and the sum of the digits of 19 is 10, as required.

Answer: 199


The solution to the equation $\arctan 2^{x}-\arctan 2^{-x}=\frac{\pi}{6}$ is . $\qquad$

Solution:
3. $\log _{2} \sqrt{3}$.

Taking the tangent of both sides of the known equation, we get
$$
\begin{array}{l}
\frac{2^{x}-2^{-x}}{1+2^{x} \times 2^{-x}}=\frac{\sqrt{3}}{3} \Rightarrow 2^{x}-2^{-x}=\frac{2 \sqrt{3}}{3} \\
\Rightarrow\left(2^{x}\right)^{2}-\frac{2 \sqrt{3}}{3} \times 2^{x}-1=0 .
\end{array}
$$

Since $2^{x}>0$, we have
$$
2^{x}=\sqrt{3} \Rightarrow x=\log _{2} \sqrt{3} .
$$

Answer: \log_{2}\sqrt{3}


Find the pattern and fill in the number:
$$
100,81,64,49 \text {, }
$$
$, 25,16,9,4,1$ .

Answer: 36


Let $F=\max _{1 \leqslant x \leqslant 3}\left|x^{3}-a x^{2}-b x-c\right|$. Find the minimum value of $F$ when $a, b, c$ take all real numbers. (2001, IMO China National Training Team Selection Exam)

Solution:
Let $f(x)=(x-2)^{3}-\frac{3}{4}(x-2)$, $x \in[1,3]$. Set $x-2=\cos \theta, \theta \in[0, \pi]$, then $|f(x)|=\left|\cos ^{3} \theta-\frac{3}{4} \cos \theta\right|=\frac{1}{4}|\cos 3 \theta| \leqslant \frac{1}{4}$. When $\theta=0, \frac{\pi}{3}, \frac{2 \pi}{3}, \pi$, i.e., $x=3, \frac{5}{2}, \frac{3}{2}, 1$, $|f(x)|=\frac{1}{4}$.
Also, $f(x)=(x-2)^{3}-\frac{3}{4}(x-2)$
$$
=x^{3}-6 x^{2}+\frac{45}{4} x-\frac{13}{2} \text {, }
$$

When $a=6, b=-\frac{45}{4}, c=\frac{13}{2}$, $f(x)_{\text {max }}=\frac{1}{4}$.
Now we prove: $F \geqslant \frac{1}{4}$.
Let $f^{\prime}(x)=x^{3}-a x^{2}-b x-c$. On the interval $[1,3]$, take values symmetrically about $x=2$, i.e., $x=1, \frac{3}{2}, \frac{5}{2}, 3$.
Since $f^{\prime}(1)=-a-b-c+1$,
$$
\begin{array}{l}
f^{\prime}\left(\frac{3}{2}\right)=-\frac{9}{4} a-\frac{3}{2} b-c+\frac{27}{8}, \\
f^{\prime}\left(\frac{5}{2}\right)=-\frac{25}{4} a-\frac{5}{2} b-c+\frac{125}{8}, \\
f^{\prime}(3)=-9 a-3 b-c+27,
\end{array}
$$

Thus, we have
$$
-\frac{2}{3} f^{\prime}(1)+\frac{4}{3} f^{\prime}\left(\frac{3}{2}\right)-\frac{4}{3} f^{\prime}\left(\frac{5}{2}\right)+\frac{2}{3} f^{\prime}(3)=1 .
$$

By $F \geqslant \max || f^{\prime}(1)|,| f^{\prime}\left(\frac{3}{2}\right)|,| f^{\prime}\left(\frac{5}{2}\right)|,| f^{\prime}(3)||$, we get $4 F \geqslant \frac{2}{3}\left|f^{\prime}(1)\right|+\frac{4}{3}\left|f^{\prime}\left(\frac{3}{2}\right)\right|+$
$$
\frac{4}{3}\left|f^{\prime}\left(\frac{5}{2}\right)\right|+\frac{2}{3}\left|f^{\prime}(3)\right| \geqslant 1 .
$$

Therefore, $F \geqslant \frac{1}{4}$.
Equality holds if and only if $-f^{\prime}(1)=f^{\prime}\left(\frac{3}{2}\right)=-f^{\prime}\left(\frac{5}{2}\right)$ $=f^{\prime}(3)=\frac{1}{4}$.
Hence, when $a=6, b=-\frac{45}{4}, c=\frac{13}{2}$, $F=\frac{1}{4}$.
Thus, $F_{\min }=\frac{1}{4}$.
Finally, it is noted that solving the extremum problem of multivariate functions often requires the comprehensive application of various methods. Due to the limitation of space, we will not provide more examples here.

Answer: \frac{1}{4}


The teacher gave Vasya and Petya two identical cardboard $n$-gons. Vasya cut his polygon into 33-gons along non-intersecting diagonals, while Petya cut his polygon into 67-gons along non-intersecting diagonals. Find the smallest possible value of $n$.

Solution:
Answer: 2017.

The sum of the angles of an $n$-sided polygon is $(n-2) \cdot 180^{\circ}$. If it is cut into $k$ 33-sided polygons, then $(n-2) \cdot 180^{\circ}=k \cdot (33-2) \cdot 180^{\circ}$, hence $n-2 \vdots 31$. Similarly, from the second condition, it follows that $n-2 \vdots 65$. Since 31 and 65 are coprime, the smallest possible value of $n=31 \cdot 65+2=$ 2017.

$\pm$ Correct solution with a minor error leading to an incorrect answer. 5 points

- The solution includes the idea of calculating the sum of angles, but lacks conclusions about divisibility. 2 points
- The correct answer is given, but not justified. In particular, a certain pair of cuts for which "everything fits" is provided. 1 point

Answer: 2017


Let $A B C$ be a triangle such that $A B=2, C A=3$, and $B C=4$. A semicircle with its diameter on $\overline{B C}$ is tangent to $\overline{A B}$ and $\overline{A C}$. Compute the area of the semicircle.

Solution:
Answer: $\frac{27 \pi}{40}$
Solution: Let $O, D$, and $E$ be the midpoint of the diameter and the points of tangency with $\overline{A B}$ and $\overline{A C}$ respectively. Then $[A B C]=[A O B]+[A O C]=\frac{1}{2}(A B+$ $A C) r$, where $r$ is the radius of the semicircle. Now by Heron's formula, $[A B C]=$ $\sqrt{\frac{9}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} \cdot \frac{5}{2}}=\frac{3 \sqrt{15}}{4}$. We solve for $r=\frac{3 \sqrt{15}}{10}$ and compute $\frac{1}{2} \pi r^{2}=\frac{27 \pi}{40}$.

Answer: \frac{27\pi}{40}


Tom has twelve slips of paper which he wants to put into five cups labeled $A$, $B$, $C$, $D$, $E$. He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from $A$ to $E$. The numbers on the papers are $2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4,$ and $4.5$. If a slip with $2$ goes into cup $E$ and a slip with $3$ goes into cup $B$, then the slip with $3.5$ must go into what cup?
$\textbf{(A) } A \qquad \textbf{(B) } B \qquad \textbf{(C) } C \qquad \textbf{(D) } D \qquad \textbf{(E) } E$

Solution:
The numbers have a sum of $6+5+12+4+8=35$, which averages to $7$, which means $A, B, C, D,$ and
$E$ will have the values $5, 6, 7, 8$ and $9$, respectively. Now, it's the process of elimination: Cup $A$
will have a sum of $5$, so putting a $3.5$ slip in the cup will leave $5-3.5=1.5$; however,
all of our slips are bigger than $1.5$, so this is impossible. Cup $B$ has a sum of $6$, but we are told that it already has a $3$ slip, leaving $6-3=3$, which is too small for the
$3.5$ slip. Cup $C$ is a little bit trickier but still manageable. It must have a value of $7$, so adding the $3.5$ slip leaves room for $7-3.5=3.5$. This looks good at first as we have slips smaller than that, but upon closer inspection, we see that no slip fits
exactly. And the smallest sum of two slips is $2+2=4$, which is too big, so this case is also impossible. Cup $E$ has a sum of $9$, but we are told it already has a $2$ slip. So, we
are left with $9-2=7$, which is identical to Cup C and thus also impossible.
With all other choices removed, we are left with the answer: Cup $\boxed{\textbf{(D)}~D}$.

Answer: D


Calculate the definite integral:

$$
\int_{0}^{\pi / 4} \frac{4-7 \tan x}{2+3 \tan x} d x
$$

Solution:
## Solution

Let's use the substitution:

$$
t=\operatorname{tg} x
$$

From which we get:

$$
\begin{aligned}
& d x=\frac{d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\pi}{4}=1
\end{aligned}
$$

Substitute:

$$
\int_{0}^{\pi / 4} \frac{4-7 \operatorname{tg} x}{2+3 \operatorname{tg} x} d x=\int_{0}^{1} \frac{4-7 t}{2+3 t} \cdot \frac{d t}{1+t^{2}}=\int_{0}^{1} \frac{4-7 t}{(2+3 t)\left(1+t^{2}\right)} d t=
$$

Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:

$$
\begin{aligned}
& \frac{4-7 t}{(3 t+2)\left(1+t^{2}\right)}=\frac{A}{3 t+2}+\frac{B t+C}{1+t^{2}}=\frac{A\left(1+t^{2}\right)+(B t+C)(3 t+2)}{(3 t+2)\left(1+t^{2}\right)}= \\
& =\frac{A+A t^{2}+3 B t^{2}+2 B t+3 C t+2 C}{(3 t+2)\left(1+t^{2}\right)}=\frac{(A+3 B) t^{2}+(2 B+3 C) t+(A+2 C)}{(3 t+2)\left(1+t^{2}\right)} \\
& \left\{\begin{array}{l}
A+3 B=0 \\
2 B+3 C=-7 \\
A+2 C=4
\end{array}\right. \\
& \left\{\begin{array}{l}
A=-3 B \\
2 B+3 C=-7 \\
-3 B+2 C=4
\end{array}\right.
\end{aligned}
$$

Multiply the second equation by 3 and the third by 2:

$$
\left\{\begin{array}{l}
A=-3 B \\
6 B+9 C=-21 \\
-6 B+4 C=8
\end{array}\right.
$$

Add the third equation to the second:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A=-3 B \\
13 C=-13 \\
-6 B+4 C=8
\end{array}\right. \\
& \left\{\begin{array}{l}
A=-3 B \\
C=-1 \\
-6 B-4=8
\end{array}\right. \\
& \left\{\begin{array}{l}
A=6 \\
C=-1 \\
B=-2
\end{array}\right. \\
& \frac{4-7 t}{(3 t+2)\left(1+t^{2}\right)}=\frac{6}{3 t+2}-\frac{2 t+1}{1+t^{2}}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{0}^{1}\left(\frac{6}{3 t+2}-\frac{2 t+1}{1+t^{2}}\right) d t=2 \cdot \int_{0}^{1} \frac{1}{t+\frac{2}{3}} d t-\int_{0}^{1} \frac{2 t}{1+t^{2}} d t-\int_{0}^{1} \frac{1}{1+t^{2}} d t= \\
& =\left.2 \cdot \ln \left|t+\frac{2}{3}\right|\right|_{0} ^{1}-\int_{0}^{1} \frac{d\left(t^{2}+1\right)}{1+t^{2}} d t-\left.\operatorname{arctg} t\right|_{0} ^{1}= \\
& =2 \cdot \ln \left|1+\frac{2}{3}\right|-2 \cdot \ln \left|0+\frac{2}{3}\right|-\left.\ln \left(t^{2}+1\right)\right|_{0} ^{1}-\operatorname{arctg} 1+\operatorname{arctg} 0= \\
& =2 \cdot \ln \frac{5}{3}-2 \cdot \ln \frac{2}{3}-\ln \left(1^{2}+1\right)+\ln \left(0^{2}+1\right)-\frac{\pi}{4}+0= \\
& =2 \cdot \ln \frac{5}{2}-\ln 2+\ln 1-\frac{\pi}{4}=\ln \frac{25}{4}-\ln 2-\frac{\pi}{4}=\ln \frac{25}{8}-\frac{\pi}{4}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{H} 0 \% \% 98 \% \mathrm{D} 0 \% \mathrm{BD}$ $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+9-26$ "

Categories: Kuznetsov's Problem Book Integrals Problem 9 | Integrals

- Last modified: 18:04, 13 May 2009.
- Content is available under CC-BY-SA 3.0.

Answer: \ln\frac{25}{8}-\frac{\pi}{4}


Let $t$ be a real root of the quadratic equation $a x^{2}+b x+c=0(a \neq 0)$, then the relationship between the discriminant $\Delta=b^{2}-4 a c$ and the quadratic form $M=(2 a t+b)^{2}$ is ( ).
(A) $\Delta>M$
(B) $\Delta=M$.
(C) $\Delta<M$
(D) Cannot be determined

Solution: 5. B (Hint: Since $t$ is a root of the equation, we have $a t^{2}+b t+c=0$. Thus, $M=4 a^{2} t^{2}+4 a b t+b^{2}=4 a\left[a t^{2}+b t+c\right]-4 a c$ $+b^{2}=b^{2}-4 a c=\Delta.$)

Answer: B


In the Cartesian coordinate system $x O y$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{4}=1(a>0)$, respectively. The line $l$ passes through the right vertex $A$ of the hyperbola and the point $B(0,2)$. If the sum of the distances from points $F_{1}$ and $F_{2}$ to the line $l$ equals the length of the segment $B F_{2}$, then the coordinates of point $A$ are $\qquad$

Solution:
3. $(2 \sqrt{2}, 0)$.

From the given, $c=\sqrt{a^{2}+4}$. It is easy to know that $\sin \angle O A B=\frac{2}{c},\left|B F_{2}\right|=\sqrt{c^{2}+4}$.
The sum of the distances from points $F_{1}$ and $F_{2}$ to line $l$ is
$$
\begin{array}{l}
\left|A F_{1}\right| \sin \angle O A B+\left|A F_{2}\right| \sin \angle O A B \\
=\left|F_{1} F_{2}\right| \frac{2}{c}=4 \\
\Rightarrow \sqrt{c^{2}+4}=4 \Rightarrow c=2 \sqrt{3} \Rightarrow a=2 \sqrt{2} .
\end{array}
$$

Therefore, point $A(2 \sqrt{2}, 0)$.

Answer: (2\sqrt{2},0)


Determine all quadruplets $(a, b, c, k)$ of integers such that $a, b, c$ are prime numbers and $k$ is a strictly positive integer satisfying

$$
a^{2}+b^{2}+16 c^{2}=9 k^{2}+1
$$

Solution:
A square modulo 3 is 0 or 1, and we have $1 \equiv a^{2}+b^{2}+c^{2}(\bmod 3)$, so among $(a, b, c)$, two are divisible by 3, hence two are equal to three. By symmetry, since $a, b$ play the same role, we can assume $a=3$. We then have two cases:

- If $b=3$, we have $9 k^{2}=16 c^{2}+17$ so $(3 k-4 c)(3 k+4 c)=17$. Since $3 k+4 c$ is positive and strictly greater than $3 k-4 c$ and 17 is prime, we have $3 k-4 c=1,3 k+4 c=17$ so by subtracting, $8 c=16$ hence $c=2$. We get $3 k=9$ so $k=3$.

Conversely, $(3,3,2,3)$ works because $9+9+64=1+81=1+9 \cdot 3^{2}$.

- If $c=3$, we have $9 k^{2}=b^{2}+152$ so $(3 k-b)(3 k+b)=152$. Since $152=8 \cdot 19$, we have $(3 k-b, 3 k+b)=$ $(1,152),(2,76),(4,38),(8,19)$. By summing, we get $6 k=153,78,42,27$, which excludes in particular the first and last pairs. So we have $k=13$ in the second case and $b=37$ and in the third case $k=7$ and $b=17$.

Conversely $(3,17,3,7),(17,3,3,7),(3,37,3,13)$ and $(37,3,3,13)$ work because $3^{2}+$ $17^{2}+16 \cdot 9=17 \cdot 9+17^{2}=17 \cdot 26=21^{2}+1=9 \cdot 7^{2}+1$ and because $9+16 \cdot 9+37^{2}=37^{2}+1+152=$ $1+37^{2}+4+4 \cdot 37=1+39^{2}=1+9 \cdot 13^{2}$.

The solutions are therefore $(3,3,2,3),(3,17,3,7),(17,3,3,7),(3,37,3,13)$ and $(37,3,3,13)$.

## 5 Last Courses

## 1 Homotheties

This course is taken from the excellent course on transformations of the plane by Thomas Budzinski on the POFM website (https://maths-olympiques.fr/wp-content/ uploads/2017/09/geom_transfos.pdf p.4-12).

## 2 Graph Theory

This course was inspired by the delightful course on graph theory by Pierre Bornsztein (http://maths-olympiques.fr/wp-content/uploads/2017/09/graphes.pdf).

## IV. Group B

## Content of this section

1 First Part: Algebra & Geometry . . . . . . . . . . . . . . . . . . . . . . . 80

1 Angle Hunting (Domitille) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Answer: (3,3,2,3),(3,17,3,7),(17,3,3,7),(3,37,3,13),(37,3,3,13)


To ensure that at least 2 blocks have the same color, one must take out at least $\qquad$ blocks.

Solution: 【Analysis】There are $60 \div 15=4$ colors, and it is necessary to take out $4+1=5$ pieces.

Answer: 5


Determine all triples of positive integers $(a, b, n)$ that satisfy the following equation: $a! + b! = 2^n$

Solution:
1. **Initial Considerations:**
   - We need to find all triples of positive integers \((a, b, n)\) such that \(a! + b! = 2^n\).
   - Note that \(2^n\) is a power of 2, and thus it is even. Therefore, \(a! + b!\) must also be even.

2. **Case Analysis:**
   - If both \(a\) and \(b\) are greater than 3, then \(a! + b!\) is a multiple of 3 (since any factorial of a number greater than 3 includes 3 as a factor). However, \(2^n\) is never a multiple of 3 for \(n \geq 1\). Hence, at most one of \(a\) or \(b\) can be greater than 3.

3. **Assume \(a \geq b\):**
   - This assumption simplifies our analysis since we can consider the cases for \(b\) and then determine \(a\).

4. **Case \(b = 1\):**
   - The equation becomes \(a! + 1! = 2^n\), which simplifies to \(a! + 1 = 2^n\).
   - If \(a \geq 2\), then \(a! + 1\) is odd (since \(a! \geq 2! = 2\) is even), but \(2^n\) is even, leading to a contradiction.
   - Therefore, \(a = 1\), giving the solution \((1, 1, 1)\).

5. **Case \(b = 2\):**
   - The equation becomes \(a! + 2! = 2^n\), which simplifies to \(a! + 2 = 2^n\).
   - Rearranging, we get \(a! = 2^n - 2\).
   - We need to check values of \(a\) such that \(a! = 2^n - 2\) is a factorial and a power of 2 minus 2.
   - For \(a = 2\), \(2! = 2\), so \(2 = 2^n - 2\) implies \(2^n = 4\) and \(n = 2\). This gives the solution \((2, 2, 2)\).
   - For \(a = 3\), \(3! = 6\), so \(6 = 2^n - 2\) implies \(2^n = 8\) and \(n = 3\). This gives the solution \((3, 2, 3)\).

6. **Case \(b = 3\):**
   - The equation becomes \(a! + 3! = 2^n\), which simplifies to \(a! + 6 = 2^n\).
   - Rearranging, we get \(a! = 2^n - 6\).
   - We need to check values of \(a\) such that \(a! = 2^n - 6\) is a factorial and a power of 2 minus 6.
   - For \(a = 3\), \(3! = 6\), so \(6 = 2^n - 6\) implies \(2^n = 12\), which is not a power of 2. Thus, no solution for \(a = 3\).
   - For \(a = 2\), \(2! = 2\), so \(2 = 2^n - 6\) implies \(2^n = 8\) and \(n = 3\). This gives the solution \((2, 3, 3)\).

7. **Conclusion:**
   - The only valid solutions are \((1, 1, 1)\), \((2, 2, 2)\), \((3, 2, 3)\), and \((2, 3, 3)\).

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

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


Given a cyclic quadrilateral $ABCD$. The rays $AB$ and $DC$ intersect at point $K$. It turns out that points $B$, $D$, as well as the midpoints $M$ and $N$ of segments $AC$ and $KC$, lie on the same circle. What values can the angle $ADC$ take?

Solution:
$M N$ is the midline in triangle $A K C$, so $\angle B A C = \angle N M C$. Additionally, $\angle B A C = \angle B D C$ because quadrilateral $A B C D$ is cyclic.

Let points $M$ and $N$ lie on the same side of line $B D$. Then $M$ lies inside triangle $B C D$ and, therefore, inside triangle $B N D$, and thus inside its circumcircle. But then points $B, N, D$, and $M$ cannot lie on the same circle. Therefore, $N$ and $M$ lie on opposite sides of $B D$, and $\angle B D C = \angle B M N$. From the parallelism of $M N$ and $A K$, it follows that $\angle B M N = \angle A B M$, from which $\angle B A C = \angle B D C = \angle A B M$. Hence, $A M = M B$, meaning that in triangle $A B C$, the median $B M$ is equal to half the side $A C$. Therefore, $\angle A B C = 90^{\circ}$, and thus $\angle A D C = 90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6409e68af5a1ba5e989dg-34.jpg?height=494&width=564&top_left_y=-1&top_left_x=747)

## Answer

$90^{\circ}$.

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Answer: 90


Find the smallest positive integer $n$ that satisfies the following:

We can color each positive integer with one of $n$ colors such that the equation $w + 6x = 2y + 3z$ has no solutions in positive integers with all of $w, x, y$ and $z$ having the same color. (Note that $w, x, y$ and $z$ need not be distinct.)

Solution:
1. **Prove that \( n > 3 \):**
   - Assume for the sake of contradiction that there are at most 3 colors, say \( c_1 \), \( c_2 \), and \( c_3 \).
   - Let 1 have color \( c_1 \) without loss of generality.
   - By considering the tuple \((1,1,2,1)\), the color of 2 must be different from \( c_1 \), so let it be \( c_2 \).
   - By considering the tuple \((3,2,3,3)\), the color of 3 cannot be \( c_2 \), and by \((3,1,3,1)\), the color of 3 cannot be \( c_1 \). Therefore, the color of 3 must be \( c_3 \).
   - By considering the tuple \((6,2,6,2)\), the color of 6 cannot be \( c_2 \), and by \((3,3,6,3)\), the color of 6 cannot be \( c_3 \). Therefore, the color of 6 must be \( c_1 \).
   - By considering the tuple \((9,6,9,9)\), the color of 9 cannot be \( c_1 \), and by \((9,3,9,3)\), the color of 9 cannot be \( c_3 \). Therefore, the color of 9 must be \( c_2 \).
   - By considering the tuple \((6,4,6,6)\), the color of 4 cannot be \( c_1 \), and by \((2,2,4,2)\), the color of 4 cannot be \( c_2 \). Therefore, the color of 4 must be \( c_3 \).
   - By considering the tuple \((6,6,12,6)\), the color of 12 cannot be \( c_1 \), and by \((12,4,12,4)\), the color of 12 cannot be \( c_3 \). Therefore, the color of 12 must be \( c_2 \).
   - However, we have reached a contradiction, as \((12,2,9,2)\) is colored with only \( c_2 \). Therefore, there must be more than 3 colors.

2. **Prove that \( n = 4 \) works:**
   - Define the following four sets:
     \[
     \begin{align*}
     c_1 &= \{3^{2a}(3b+1) \mid a, b \ge 0\} \\
     c_2 &= \{3^{2a}(3b+2) \mid a, b \ge 0\} \\
     c_3 &= \{3^{2a+1}(3b+1) \mid a, b \ge 0\} \\
     c_4 &= \{3^{2a+1}(3b+2) \mid a, b \ge 0\}
     \end{align*}
     \]
   - It is obvious that each positive integer appears in exactly one of these sets because if we divide out all powers of 3 in a number, then we will get a number that is either 1 or 2 modulo 3.
   - We assert that no quadruple of positive integers \((w, x, y, z)\) satisfying \( w + 6x = 2y + 3z \) consists of four members from the same set.
   - Assume for the sake of contradiction that \((a, b, c, d)\) are positive integers satisfying \( a + 6b = 2c + 3d \) with \( a, b, c, d \) from the same set.
   - If \((a, b, c, d)\) are from the same set, then \(\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}, \frac{d}{3}\right)\) are also from the same set. Hence, we may assume without loss of generality that at least one of \((a, b, c, d)\) is not divisible by 3, or else we can perform this operation to reduce it.
   - Let \( v_3(t) \) denote the largest integer such that \( 3^{v_3(t)} \mid t \). Since at least one of these is not divisible by 3, \( v_3(a), v_3(b), v_3(c), v_3(d) \) must be even.
   - Notice that \( w + 6x \equiv w \equiv 2y + 3z \equiv -y \pmod{3} \), so \( w + y \equiv 0 \pmod{3} \). Therefore, \( w \equiv y \equiv 0 \pmod{3} \), or else they would have different residues modulo 3. Also, they must both be divisible by 9 because \( v_3(w) \) and \( v_3(y) \) are even.
   - Since \( 2y - w = 3(2x - z) \), \( 3 \mid 2x - z \implies x + z \equiv 0 \pmod{3} \). However, this implies that they are both divisible by 3, or else they would have different residues modulo 3. This is a contradiction, as we have assumed that at least one of \( w, x, y, z \) is not divisible by 3. Therefore, this coloring works.

Thus, \( n = \boxed{4} \) is our answer.

Answer: 4


As shown in the figure, $AB \perp BC, BC \perp CD, BC$ is tangent to the circle $O$ with diameter $AD$. In which of the following cases is the area of $ABCD$ an integer?
(A) $AB=3, CD=1$.
(B) $AB=5, CD=2$.
(C) $AB=7, CD=3$.
(D) $AB=9, CD=4$.
(E) $AB=11, CD=5$.

Solution:
[Solution] If $E$ is the tangency point of $BC$ and $\odot O$, connect $OE$, and draw $DF \perp AB$.
Let $AB = a, \quad CD = b$.
Then $AD = 2OE = AB + CD = a + b$.
Also, $BC^2 = DF^2 = AD^2 - AF^2 = (a + b)^2 - (a - b)^2 = 4ab$, thus $BC = 2\sqrt{ab}$.
Then $S_{ABCD} = \frac{1}{2}(AB + CD) \cdot BC$
$$
\begin{array}{l}
= \frac{1}{2}(a + b) \cdot 2\sqrt{ab} \\
= (a + b) \sqrt{ab}.
\end{array}
$$

Clearly, among the given sets of values, only when $AB = 9, CD = 4$ is $\sqrt{ab}$ an integer.
Therefore, the answer is $(D)$.

Answer: D


Calculate the volumes of bodies bounded by surfaces.

$$
z=2 x^{2}+8 y^{2}, z=4
$$

Solution:
## Solution

In the section of the given figure by the plane $z=$ const, there is an ellipse:

$$
2 x^{2}+8 y^{2}=z
$$

The area of an ellipse described by the formula: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $\pi \cdot a \cdot b$

Let's find the radii of the ellipse:

![](https://cdn.mathpix.com/cropped/2024_05_22_348f4289c7b5f46bc246g-21.jpg?height=926&width=1105&top_left_y=979&top_left_x=818)

$$
\begin{aligned}
& \frac{2 x^{2}}{z}+\frac{8 y^{2}}{z}=1 \rightarrow a=\frac{\sqrt{z}}{\sqrt{2}} ; b=\frac{\sqrt{z}}{\sqrt{8}} \\
& \Rightarrow S=\pi a b=\pi \cdot \frac{\sqrt{z}}{\sqrt{2}} \cdot \frac{\sqrt{z}}{\sqrt{8}}=\frac{\pi \cdot z}{4}
\end{aligned}
$$

$$
\begin{aligned}
& V=\int_{0}^{4} S(z) d z=\frac{\pi}{4} \int_{0}^{4} z d z=\left.\frac{\pi}{4}\left(\frac{z^{2}}{2}\right)\right|_{0} ^{4}= \\
& =\frac{\pi}{8}\left(4^{2}-0^{2}\right)=2 \pi
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�

\%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+20-12 »

Categories: Kuznetsov's Problem Book Integrals Problem 20 | Integrals | Problems for Checking

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Answer: 2\pi


Simplify each of the fractions $\frac{1}{2014}, \frac{2}{2014}, \cdots, \frac{2012}{2014}, \frac{2013}{2014}$ to their simplest form, and find the sum of all the fractions that still have 2014 as their denominator after simplification.

Solution:
【Analysis】Since $2014=2 \times 1007=2 \times 19 \times 53$, we need to find the number of multiples of $2$, $19$, $53$, $2 \times 19$, $2 \times 53$, $19 \times 53$, and $2 \times 19 \times 53$, respectively, to determine the number of denominators that are not 2014, and thus solve the problem.
【Solution】Solution: $2014=2 \times 1007=2 \times 19 \times 53$,
Multiples of 2: 1007,
Multiples of 19: 106,
Multiples of 53: 38,
Multiples of $2 \times 19$: 53,
Multiples of $2 \times 53$: 19,
Multiples of $19 \times 53$: 2,
Multiples of $2 \times 19 \times 53$: 1,
The number of denominators that are not 2014 is:
$1007+106+38-53-19-2+1=1078$ (cases),
Excluding $\frac{2014}{2014}$, the total is: $1078-1=1077$ (cases).
The number of simplest fractions with a denominator of 2014 is:
$2013-1077=936$ (cases)
The sum is: $936 \div 2=468$.

Answer: 468


Li Shuang rides a bike at a speed of 320 meters per minute from location $A$ to location $B$. On the way, due to a bicycle malfunction, he pushes the bike and walks for 5 minutes to a place 1800 meters from $B$ to repair the bike. After 15 minutes, he continues towards $B$ at 1.5 times his original riding speed, and arrives at $B$ 17 minutes later than the expected time. What is Li Shuang's walking speed in meters per minute?

Solution:
【Solution】Solve: $1800 \div 320-1800 \div(320 \times 1.5)$
$$
\begin{array}{l}
=5.625-3.75 \\
=1.875 \text { (minutes) } \\
320 \times[5-(17-15+1.875)] \div 5 \\
=320 \times[5-3.875] \div 5 \\
=320 \times 1.125 \div 5 \\
=360 \div 5 \\
=72 \text { (meters/minute) }
\end{array}
$$

Answer: Li Shuang's speed of pushing the cart while walking is 72 meters/minute.
Therefore, the answer is: 72.

Answer: 72


Let $f(x)=x^2-2x+1.$ For some constant $k, f(x+k) = x^2+2x+1$ for all real numbers $x.$ Determine the value of $k.$

Solution:
1. Given the function \( f(x) = x^2 - 2x + 1 \), we can rewrite it in a factored form:
   \[
   f(x) = (x-1)^2
   \]

2. We are given that for some constant \( k \), the function \( f(x+k) \) is equal to \( x^2 + 2x + 1 \) for all real numbers \( x \). Therefore, we need to find \( k \) such that:
   \[
   f(x+k) = x^2 + 2x + 1
   \]

3. Substitute \( x+k \) into the function \( f \):
   \[
   f(x+k) = (x+k-1)^2
   \]

4. Expand the expression:
   \[
   (x+k-1)^2 = (x+k-1)(x+k-1) = x^2 + 2x(k-1) + (k-1)^2
   \]

5. We need this to be equal to \( x^2 + 2x + 1 \):
   \[
   x^2 + 2x(k-1) + (k-1)^2 = x^2 + 2x + 1
   \]

6. By comparing the coefficients of \( x \) and the constant terms on both sides of the equation, we get:
   \[
   2x(k-1) = 2x \quad \text{and} \quad (k-1)^2 = 1
   \]

7. From the coefficient of \( x \):
   \[
   2(k-1) = 2 \implies k-1 = 1 \implies k = 2
   \]

8. From the constant term:
   \[
   (k-1)^2 = 1 \implies k-1 = \pm 1
   \]
   This gives us two possible solutions:
   \[
   k-1 = 1 \implies k = 2
   \]
   \[
   k-1 = -1 \implies k = 0
   \]

9. However, since \( k = 0 \) does not satisfy the coefficient comparison \( 2(k-1) = 2 \), the only valid solution is:
   \[
   k = 2
   \]

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

Answer: 2


A and B play a game. A randomly selects a positive integer pair $(k, n)(2 \leqslant k \leqslant 36,1 \leqslant n \leqslant$ $216)$, and B performs the following operation: divides $[0,36]$ into $k$ segments, $[0,36]=\bigcup_{i=1}^{k}\left[a_{i-1}, a_{i}\right]\left(a_{0}=0<a_{1}<\cdots<a_{k}\right.$ $\left.=36, a_{i} \in \mathbf{N}_{+}\right)$, for each integer $a \in\left[a_{i-1}, a_{i}\right]$, takes $f(a)=2\left(a-a_{i-1}\right)$ or $a_{i}-a$, obtaining
$$
F(k)=\sum_{i=1}^{k} \sum_{a=a_{i-1}}^{a_{i}} f(a) .
$$

If $F(k) \geqslant n$, then A wins; if $F(k)<n$, then B wins. Find the probability that A wins.

Solution:
Three, first find the minimum value of $F(k)$.
For any $a \in [a_{i-1}, a_{i}](i=1,2, \cdots, k)$, take $f(a)=\min \left\{2(a-a_{i-1}), a_{i}-a\right\}$.
Let $a_{i}-a_{i-1}=b_{i}(i=1,2, \cdots, k), a-a_{i-1}=b$.
Then $f(a)=\min \left\{2 b, b_{i}-b\right\}=g(b)\left(0 \leqslant b \leqslant b_{i}\right)$.
Obviously, $g(b)=\left\{\begin{array}{ll}2 b, & 0 \leqslant b \leqslant\left[\frac{b_{i}}{3}\right] ; \\ b_{i}-b, & \left[\frac{b_{i}}{3}\right]<b \leqslant b_{i} .\end{array}\right.$
Thus, $\sum_{a=a_{i-1}}^{a_{i}} f(a) \doteq \sum_{b=0}^{b_{i}} g(b)$
$=\sum_{b=0}^{\left[\frac{b_{i}}{3}\right]} 2 b+\sum_{b=\left[\frac{b_{i}}{3}\right]+1}^{b_{i}}\left(b_{i}-b\right)=h\left(b_{i}\right)$.
It is easy to see that $h(x)=\left[\frac{x^{2}}{3}\right]\left(x \in \mathbf{N}_{+}\right)$ is convex.
Also, $F(k)=\sum_{i=1}^{k} \sum_{a=a_{i-1}}^{a_{i}} f(a)=\sum_{i=1}^{k} h\left(b_{i}\right)$,

where, $\sum_{i=1}^{k} b_{i}=\sum_{i=1}^{k}\left(a_{i}-a_{i-1}\right)=a_{k}-a_{0}=36$.
By Jensen's inequality and the adjustment method, we know that when $b_{1}, b_{2}, \cdots, b_{k}$ differ by at most 1, $F(k)$ takes its minimum value.
Let $s=\left[\frac{36}{k}\right]$. Then $36=k s+r(0 \leqslant r<k)$. Hence $F(k)_{\text {min }}=(k-r) h(s)+r h(s+1)$.
Substituting $k=2,3, \cdots, 36$ into the above formula, we get
$$
\begin{array}{l}
F(2)_{\min }=216, F(3)_{\min }=144, \\
F(4)_{\min }=108, F(5)_{\min }=85, \\
F(6)_{\min }=72, F(7)_{\min }=60, F(8)_{\min }=52, \\
F(9+i)_{\min }=45-3 i(i=0,1,2), \\
F(12+i)_{\min }=36-3 i(i=0,1, \cdots, 6), \\
F(18+i)_{\min }=18-i(i=1,2, \cdots, 18) .
\end{array}
$$

Below, we calculate the probability of player A winning.
Obviously, when player A selects the number pair $(k, n)$, and $n \leqslant F(k)_{\min }$, player A wins. Therefore, the probability of player A winning is
$$
\left(\sum_{k=2}^{36} F(k)_{\min }\right) \div(216 \times 35)=\frac{241}{1512} .
$$

Answer: \frac{241}{1512}


Barbara scored 98 points on the second-to-last test of the school year, thus increasing the average of the points she had achieved so far by 1 point. On the last test, she scored 70 points and lowered the new average by 2 points. How many tests did she take in the entire school year?

Solution:
II/4. Namely, if Barbara had written $n$ tests and had an average of $m$ points before the last 2 tests, then

$$
\begin{aligned}
& \frac{n \cdot m+98}{n+1}=m+1 \\
\text { and } \quad & \frac{n \cdot m+98+70}{n+2}=m+1-2=m-1
\end{aligned}
$$

From the first equation, we get $n \cdot m+98=n \cdot m+n+m+1$ or $m+n=97$, and from the second, $n \cdot m+168=$ $n \cdot m-n+2 m-2$ or $2 m-n=170$. Thus, we arrive at $3 m=267$ or $m=89$ and $n=8$. This means that Barbara wrote 10 tests in total during the school year.

The equations (1) and (2) (or equivalent) based on the problem statement: 2+2 points. Solution of the system: 3 points.

Answer: 10


Integers $a, b, c, d$, and $e$ satisfy the following three properties:
(i) $2 \leq a<b<c<d<e<100$
(ii) $\operatorname{gcd}(a, e)=1$
(iii) $a, b, c, d, e$ form a geometric sequence.
What is the value of $c$ ?

Solution:
Solution: The answer is 36 .

Let $r$ be the common ratio of the geometric sequence $a, b, c, d, e$. Since $a1$. Then $a=a, b=a r, c=a r^{2}, d=a r^{3}, e=a r^{4}$. Since $a, e$ have no common factors and $a>1, r$ is not an integer. Let $x / y$ be this common ratio, where $x, y$ are positive integers and $\operatorname{gcd}(x, y)=1$. Since $r>1$ and is not an integer, $x>y>1$. Therefore, $b=a x / y, c=a x^{2} / y^{2}, d=a x^{3} / y^{3}$ and $e=a x^{4} / y^{4}$. Since $e$ is an integer and $\operatorname{gcd}(x, y)=1$, $a$ is divisible by $y^{4}$. Then $a=k y^{4}$ for some positive integer $k$. Then $a=k y^{4}, b=k x y^{3}, c=$ $k x^{2} y^{2}, d=k x^{3} y, e=k x^{4}$. Since $\operatorname{gcd}(a, e)=1, k=1$. Hence, $a=y^{4}$ and $e=x^{4}$. Since $2 \leq a<e<100$ and $3^{4}<100<4^{4}, 2 \leq y<x \leq 3$, which implies that $x=3$ and $y=2$. Then $c=k x^{2} y^{2}=1 \cdot 3^{2} \cdot 2^{2}=6^{2}=36$.

Answer: 36


Petr glued 17 dice into a snake (as shown in the picture). He always glued the sides with the same number of dots. Then he examined the snake from all sides and correctly calculated the total number of dots on its surface. What did he get? (The sum of the dots on opposite sides of a die is 7.)

![](https://cdn.mathpix.com/cropped/2024_04_17_a36e361d6d455c7cf0e3g-2.jpg?height=203&width=717&top_left_y=2023&top_left_x=698)

Solution: Given that the sum of the dots on opposite sides of a die is always 7, we have a total of $17 \cdot 2$ such pairs on the "long" side of the snake and one pair on the sides. In total, we therefore have 35 pairs. Therefore, Petr must have counted $35 \cdot 7=245$ dots.

Answer: 245


In an isosceles triangle $ABC$, the perpendicular bisector of the lateral side $BC$ intersects the base $AB$ at point $D$ such that $AC=AD$. Find the angle $ABC$.

Write the answer in degrees without the degree symbol.

Solution:
# 3. Problem 3

In an isosceles triangle $ABC$, the perpendicular bisector of the lateral side $BC$ intersects the base $AB$ at point $D$ such that $AC = AD$. Find the angle $ABC$.

Write the answer in degrees without the degree symbol.

Answer: 36

Answer: 36


Given a triangle $OAB$ with the vertices $O(0,\ 0,\ 0),\ A(1,\ 0,\ 0),\ B(1,\ 1,\ 0)$ in the $xyz$ space.
Let $V$ be the cone obtained by rotating the triangle around the $x$-axis.
Find the volume of the solid obtained by rotating the cone $V$ around the $y$-axis.

Solution:
1. **Understanding the Problem:**
   We are given a triangle \(OAB\) with vertices \(O(0, 0, 0)\), \(A(1, 0, 0)\), and \(B(1, 1, 0)\) in the \(xyz\) space. We need to find the volume of the solid obtained by rotating the cone \(V\) (formed by rotating the triangle around the \(x\)-axis) around the \(y\)-axis.

2. **Forming the Cone:**
   When the triangle \(OAB\) is rotated around the \(x\)-axis, it forms a cone with the base on the plane \(x = 1\) and the height along the \(x\)-axis from \(x = 0\) to \(x = 1\). The radius of the base of the cone is 1 (the distance from \(O\) to \(B\) in the \(y\)-direction).

3. **Volume of the Cone:**
   The volume \(V\) of a cone is given by:
   \[
   V = \frac{1}{3} \pi r^2 h
   \]
   where \(r\) is the radius of the base and \(h\) is the height. Here, \(r = 1\) and \(h = 1\), so:
   \[
   V = \frac{1}{3} \pi (1)^2 (1) = \frac{1}{3} \pi
   \]

4. **Rotating the Cone Around the \(y\)-Axis:**
   To find the volume of the solid obtained by rotating the cone around the \(y\)-axis, we use the method of washers.

5. **Setting Up the Integral:**
   Consider a cross-section of the cone at a fixed \(y = y_0\). The region \(R(y_0)\) in the \(x, z\) plane is bounded by \(x = y_0\), \(x = 1\), and the hyperbola \(x^2 = y_0^2 + z^2\). The maximum distance from a point in \(R(y_0)\) to the \(y\)-axis is \(\sqrt{2 - y_0^2}\), and the minimum distance is \(y_0\).

6. **Volume of the Washer:**
   The volume of the washer at \(y = y_0\) is given by:
   \[
   \pi \left( (\sqrt{2 - y_0^2})^2 - (y_0)^2 \right) \, dy = \pi \left( 2 - y_0^2 - y_0^2 \right) \, dy = \pi (2 - 2y_0^2) \, dy
   \]

7. **Integrating Over the Range:**
   We integrate this expression from \(y = -1\) to \(y = 1\):
   \[
   \int_{-1}^1 \pi (2 - 2y^2) \, dy
   \]

8. **Evaluating the Integral:**
   \[
   \int_{-1}^1 \pi (2 - 2y^2) \, dy = \pi \int_{-1}^1 (2 - 2y^2) \, dy
   \]
   Split the integral:
   \[
   \pi \left( \int_{-1}^1 2 \, dy - \int_{-1}^1 2y^2 \, dy \right)
   \]
   Evaluate each part:
   \[
   \int_{-1}^1 2 \, dy = 2 \left[ y \right]_{-1}^1 = 2(1 - (-1)) = 4
   \]
   \[
   \int_{-1}^1 2y^2 \, dy = 2 \left[ \frac{y^3}{3} \right]_{-1}^1 = 2 \left( \frac{1^3}{3} - \frac{(-1)^3}{3} \right) = 2 \left( \frac{1}{3} - \left( -\frac{1}{3} \right) \right) = 2 \left( \frac{2}{3} \right) = \frac{4}{3}
   \]
   Combine the results:
   \[
   \pi \left( 4 - \frac{4}{3} \right) = \pi \left( \frac{12}{3} - \frac{4}{3} \right) = \pi \left( \frac{8}{3} \right) = \frac{8\pi}{3}
   \]

The final answer is \(\boxed{\frac{8\pi}{3}}\)

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


How many six-digit numbers are there in which only the middle two digits are the same?

Solution: 38. $9 \cdot 9 \cdot 8 \cdot 1 \cdot 7 \cdot 6=27216$ numbers.

Answer: 27216


On Monday, three bananas cost as much as a lemon and an orange together. On Tuesday, the prices of all fruits were reduced by the same amount of money, two oranges cost as much as three bananas and one lemon, and the price of half a lemon was 5 denars.

What was the price of one orange on Monday?

Solution:
Solution. Let $x, y$, and $z$ be the prices of one banana, one lemon, and one orange, respectively, on Monday. Let the price reduction on Tuesday be $r$ denars. Then from the conditions of the problem we have

$$
\left\{\begin{aligned}
3 x & =y+z \\
2(z-r) & =3(x-r)+y-r \\
\frac{1}{2}(y-r) & =5
\end{aligned}\right.
$$

If we algebraically transform the second equation, we get

$2 z=3 x-2 r+y$.

If we substitute $3 x$ from the first equation, we get $2 z=y+z-2 r+y$, so $z=2 y-2 r=2(y-r)$. But $y-r=10$, from which we get $z=2 \cdot 10=20$ denars.

Therefore, the price of one orange on Monday was 20 denars.

Answer: 20


Which of the following combinations of outcomes is not possible for this team?

(A) 2 wins, 0 losses, 1 tie

(B) 1 win, 2 losses, 0 ties

(C) 0 wins, 1 loss, 2 ties

(D) 1 win, 1 loss, 1 tie

(E) 1 win, 0 losses, 2 ties

Solution:
When Team A played Team B, if Team B won, then Team B scored more goals than Team A, and if the game ended in a tie, then Team A and Team B scored the same number of goals. Therefore, if a team has 0 wins, 1 loss, and 2 ties, then it scored fewer goals than its opponent once (the 1 loss) and the same number of goals as its oppponent twice (the 2 ties).

Combining this information, we see that the team must have scored fewer goals than were scored against them.

In other words, it is not possible for a team to have 0 wins, 1 loss, and 2 ties, and to have scored more goals than were scored against them.

We can also examine choices (A), (B), (D), (E) to see that, in each case, it is possible that the team scored more goals than it allowed.

This will eliminate each of these choices, and allow us to conclude that (C) must be correct.

(A): If the team won 2-0 and 3-0 and tied 1-1, then it scored 6 goals and allowed 1 goal.

(B): If the team won 4-0 and lost 1-2 and 2-3, then it scored 7 goals and allowed 5 goals.

(D): If the team won $4-0$, lost $1-2$, and tied 1-1, then it scored 6 goals and allowed 3 goals.

(E): If the team won 2-0, and tied 1-1 and 2-2, then it scored 5 goals and allowed 3 goals.

Therefore, it is only the case of 0 wins, 1 loss, and 2 ties where it is not possible for the team to score more goals than it allows.

ANSWER: (C)

Answer: C


The adjacent terms $a_{n}, a_{n+1}$ of the sequence $\left\{a_{n}\right\}$ are the two roots of the equation $x^{2}-c_{n} x+\left(\frac{1}{3}\right)^{n}=0$, and $a_{1}=2$. Find the sum of the infinite sequence $c_{1}, c_{2}, \cdots, c_{n}, \cdots$.

Solution:
Given the problem, we have
\[
\left\{
\begin{array}{l}
c_{n}=a_{n}+a_{n+1}, \\
a_{n} a_{n+1}=\left(\frac{1}{3}\right)^{n}.
\end{array}
\right.
\]
(1)
From (2), we get \(a_{n+1} a_{n+2} - \left(\frac{1}{3}\right)^{n+1}\), so \(\frac{a_{n+2}}{a_{n}} = \frac{1}{3}\). Therefore,
\[
\begin{aligned}
\sum_{n=1}^{\infty} c_{n} & = \sum_{n=1}^{\infty} a_{n} + \sum_{n=2}^{\infty} a_{n} = \sum_{k=0}^{\infty} a_{2 k+1} + 2 \cdot \sum_{k=1}^{\infty} a_{2 k} + \sum_{k=1}^{\infty} a_{2 k+1} \\
& = 2 \cdot \sum_{k=0}^{\infty} a_{2 k+1} + 2 \cdot \sum_{k=1}^{\infty} a_{2 k} - a_{1}.
\end{aligned}
\]

Given \(a_{1} = 2\) and from (2), we get \(a_{2} = \frac{1}{6}\), so
\[
\sum_{n=1}^{\infty} c_{n} = 2 \times \frac{2}{1 - \frac{1}{3}} + 2 \times \frac{\frac{1}{6}}{1 - \frac{1}{3}} - 2 = \frac{9}{2}.
\]

Answer: \frac{9}{2}


Given that the area of the equilateral $\triangle ABC$ is $S$, $M$ and $N$ are the midpoints of sides $AB$ and $AC$ respectively, $ME \perp BC$ at $E$, and $EF \perp AC$ at $F$. Let the areas of $\triangle AMN$, $\triangle BEM$, $\triangle EFC$, and quadrilateral $MEFN$ be $S_{1}$, $S_{2}$, $S_{3}$, and $S_{4}$ respectively. Then $\frac{S_{4}}{S_{1}+S_{2}+S_{3}}=$ $\qquad$ .

Solution:
4. $\frac{11}{21}$.

As shown in Figure 6, connect $B N$, and draw the altitude $A H$. Since $\triangle A B C$ is an equilateral triangle, we know that $B N \perp A C$.

From $\triangle M B E \backsim \triangle A B H$, we get
$$
\frac{S_{2}}{S}=\frac{1}{8},
$$

which means $S_{2}=\frac{1}{8} S$.
Since $\frac{B E}{B H}=\frac{1}{2}$, we have $\frac{C E}{B C}=\frac{3}{4}$.
Also, since $\triangle E C F \backsim \triangle B C N$, we get $\frac{S_{3}}{S}=\frac{9}{32}$, which means $S_{3}=\frac{9}{32} S$.
From $\triangle A M N \backsim \triangle A B C$, we have $\frac{S_{1}}{S}=\frac{1}{4}$, which means $S_{1}=\frac{1}{4} S$.
Therefore, $S_{4}=S-\left(S_{1}+S_{2}+S_{3}\right)=\frac{11}{32} S$. Hence, $\frac{S_{4}}{S_{1}+S_{2}+S_{3}}=\frac{11}{21}$.

Answer: \frac{11}{21}


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

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

From the problem, we know that $A \subseteq M_{1} \cap M_{2}=\{5\}$.
Since $A$ is a non-empty set, therefore, $A=\{5\}$.

Answer: \{5\}